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F U N D A M E N TA L S O F

ENGINEERING

THERMODYNAMICS SEVENTH EDITION

MICHAEL J. MORAN The Ohio State University

HOWARD N. SHAPIRO Wayne State University

DAISIE D. BOET TNER Colonel, U.S. Army

MARGARET B. BAILEY Rochester Institute of Technology

John Wiley & Sons, Inc.

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Publisher Executive Editor Editorial Assistant Marketing Manager Content Manager Production Editor Designer Executive Media Editor Photo Department Manager Photo Editor Production Management Services

Don Fowley Linda Ratts Renata Marchione Christopher Ruel Dorothy Sinclair Sandra Dumas James O’Shea Thomas Kulesa Hilary Newman Sheena Goldstein Ingrao Associates

Cover Photo Top photo: © Andrey Prokhorov/iStockphoto, Bottom photos clockwise from top left: © Liane Cary/Age Fotostock America, Inc., © BgDigital/iStockphoto, © Philip and Karen Smith/ Getty Images, Inc., © Achim Baqué/iStockphoto, © Mike Kemp RubberBall/Age Fotostock America, Inc., © Fertnig/iStockphoto This book was typeset in 10/12 Times Ten Roman at Aptara®, Inc. and printed and bound by Courier/ Westford. The cover was printed by Courier/Westford. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. The paper in this book was manufactured by a mill whose forest management programs include sustained yield-harvesting of its timberlands. Sustained yield harvesting principles ensure that the number of trees cut each year does not exceed the amount of new growth. This book is printed on acid-free paper.

`

Copyright © 2011, 2008, 2004, 2000, 1996, 1993, 1988 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 6468600. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/ returnlabel. Outside of the United States, please contact your local representative. ISBN ISBN

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Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

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Preface Professors Moran and Shapiro are delighted to welcome two new co-authors for the seventh edition of Fundamentals of Engineering Thermodynamics. Dr. Daisie D. Boettner, PE, professor of mechanical engineering at the United States Military Academy at West Point, and Dr. Margaret B. Bailey, PE, professor of mechanical engineering at the Rochester Institute of Technology, bring outstanding experience in engineering education, research, and service to the team. Their perspectives enrich the presentation and build upon our existing strengths in exciting new ways.

A Textbook for the 21st Century In the twenty-first century, engineering thermodynamics plays a central role in developing improved ways to provide and use energy, while mitigating the serious human health and environmental consequences accompanying energy—including air and water pollution and global climate change. Applications in bioengineering, biomedical systems, and nanotechnology also continue to emerge. This book provides the tools needed by specialists working in all such fields. For non-specialists, this book provides background for making decisions about technology related to thermodynamics—on the job and as informed citizens. Engineers in the twenty-first century need a solid set of analytical and problem-solving skills as the foundation for tackling important societal issues relating to engineering thermodynamics. The seventh edition develops these skills and significantly expands our coverage of their applications to provide • current context for the study of thermodynamic principles. • relevant background to make the subject meaningful for meeting the challenges of the decades ahead. • significant material related to existing technologies in light of new challenges. In the seventh edition, we build on the core features that have made the text the global leader in engineering thermodynamics education. (The present discussion of core features centers on new aspects; see the Preface to the sixth edition for more.) We are known for our clear and concise explanations grounded in the fundamentals, pioneering pedagogy for effective learning, and relevant, up-to-date applications. Through the creativity and experience of our newly expanded author team, and based on excellent feedback from instructors and students, we continue to enhance what has become the leading text in the field.

New in the Seventh Edition In a major departure from previous editions of this book and all other texts intended for the same student population, we have introduced animations that strengthen students’ understanding of basic phenomena and applications. The seventh edition also features a crisp new interior design aimed at helping students • better understand and apply the subject matter, and • fully appreciate the relevance of the topics to engineering practice and to society. This edition also provides, inside the front cover under the heading How to Use This Book Effectively, an updated roadmap to core features of this text that make it so effective for student learning. To fully understand all of the many features we have built into the book, be sure to see this important element. In this edition, several enhancements to improve student learning have been introduced or upgraded: • New animations are offered at key subject matter locations to improve student learning. When viewing the animations, students will develop deeper understanding by visualizing key processes and phenomena. • Special text elements feature important illustrations of engineering thermodynamics applied to our environment, society, and world: • New ENERGY & ENVIRONMENT presentations explore topics related to energy resource use and environmental issues in engineering. • Updated BIOCONNECTIONS discussions tie textbook topics to contemporary applications in biomedicine and bioengineering. • Additional Horizons features have been included that link subject matter to thoughtprovoking 21st century issues and emerging technologies. Suggestions for additional reading and sources for topical content presented in these elements provided on request. • End-of-Chapter problems in each of the three modes: conceptual, skill building, and design have been extensively revised and hundreds of new problems added.

v

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Preface • New and revised class-tested material contributes to student learning and instructor effectiveness: • Significant new content explores how thermodynamics contributes to meet the challenges of the 21st century.

errata for both the text and problems.

chapter summary information, including Key Terms and Key Equations.

chapter learning outcomes.

correlation guides to ease transition between editions of this text and for switching to this edition from another book.

text Preface.

• Key aspects of fundamentals and applications within the text have been enhanced. • In response to instructor and student needs, classtested changes that contribute to a more just-intime presentation have been introduced: • TAKE NOTE... entries in the margins are expanded throughout the textbook to improve student learning. For example, see p. 8. • Boxed material allows students and instructors to explore topics in greater depth. For example, see p. 188. • New margin terms at many locations aid in navigating subject matter.

• Student Companion Site: Helps students learn the subject matter with resources including

animations—new in this edition.

answers to selected problems.

errata for both the text and problems.

chapter summary information, including Key Terms and Key Equations.

chapter learning outcomes.

chapter-by-chapter summary of Special Features as listed in the Instructor Companion Site.

text Preface.

Supplements The following supplements are available with the text: • Outstanding Instructor and Student companion web sites (visit www.wiley.com/college/moran) that greatly enhance teaching and learning: • Instructor Companion Site: Assists instructors in delivering an effective course with resources including

animations—new in this edition.

chapter-by-chapter summary of Special Features, including

the subject of each solved example,

the topics of all ENERGY & ENVIRONMENT, BIOCONNECTIONS, and Horizons features,

the themes of the accompanying DESIGN

• Interactive Thermodynamic: IT software is available as a stand-alone product or with the textbook. IT is a highly-valuable learning tool that allows students to develop engineering models, perform “what-if” analyses, and examine principles in more detail to enhance their learning. Brief tutorials of IT are included within the text and the use of IT is illustrated within selected solved examples. • WileyPLUS is an online set of instructional, practice, and course management resources, including the full text, for students and instructors. Visit www.wiley.com/college/moran or contact your local Wiley representative for information on the above-mentioned supplements.

& OPEN ENDED PROBLEMS

Ways to Meet Different Course Needs

a complete solution manual that is easy to navigate.

solutions to computer-based problems for use with both IT: Interactive Thermodynamics as well as EES: Engineering Equation Solver.

image galleries with text images available in various helpful electronic formats.

sample syllabi on semester and quarter bases.

In recognition of the evolving nature of engineering curricula, and in particular of the diverse ways engineering thermodynamics is presented, the text is structured to meet a variety of course needs. The following table illustrates several possible uses of the textbook assuming a semester basis (3 credits). Courses could be taught using this textbook to engineering students with appropriate background beginning in their second year of study.

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Preface Type of course

Intended audience

Chapter coverage

Nonmajors

• Principles. Chaps. 1–6. • Applications. Selected topics from Chaps. 8–10 (omit compressible flow in Chap. 9).

Majors

• Principles. Chaps. 1–6. • Applications. Same as above plus selected topics from Chaps. 12 and 13.

Majors

• First course. Chaps. 1–7. (Chap. 7 may be deferred to second course or omitted.) • Second course. Selected topics from Chaps. 8–14 to meet particular course needs.

Surveys

Two-course sequences

vii

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Acknowledgments We thank the many users of our previous editions, located at hundreds of universities and colleges in the United States, Canada, and world-wide, who continue to contribute to the development of our text through their comments and constructive criticism. The following colleagues have assisted in the development of this edition. We greatly appreciate their contributions: John Abbitt, University of Florida Ralph Aldredge, University of California-Davis Leticia Anaya, University of North Texas Kendrick Aung, Lamar University Cory Berkland, The University of Kansas Justin Barone, Virginia Polytechnic Institute and State University William Bathie, Iowa State University Leonard Berkowitz, California State Polytechnic University, Pomona Eugene F. Brown, Virginia Polytechnic Institute and State University David L. Ernst, Texas Tech University Sebastien Feve, Iowa State University Timothy Fox, California State UniversityNorthridge Nick Glumac, University of Illinois at UrbanaChampaign Tahereh S. Hall, Virginia Polytechnic Institute and State University Daniel W. Hoch, University of North CarolinaCharlotte Timothy J. Jacobs, Texas A&M University Fazal B. Kauser, California State Polytechnic University, Pomona MinJun Kim, Drexel University Joseph F. Kmec, Purdue University Feng C. Lai, University of Oklahoma Kevin Lyons, North Carolina State University Pedro Mago, Mississippi State University Raj M. Manglik, University of Cincinnati Thuan Nguyen, California State Polytechnic University, Pomona John Pfotenhauer, University of Wisconsin-Madison

viii

Paul Puzinauskas, University of Alabama Muhammad Mustafizur Rahman, University of South Florida Jacques C. Richard, Texas A&M University Charles Ritz, California State Polytechnic University, Pomona Francisco Ruiz, Illinois Institute of Technology Iskender Sahin, Western Michigan University Will Schreiber, University of Alabama Enrico Sciubba, University of Rome (Italy) Tien-Mo Shih, University of Maryland Larry Sobel, Raytheon Missile Systems Thomas Twardowski, Widener University V. Ismet Ugursal, Dalhousie University, Nova Scotia. Angela Violi, University of Michigan K. Max Zhang, Cornell University The views expressed in this text are those of the authors and do not necessarily reflect those of individual contributors listed, The Ohio State University, Wayne State University, Rochester Institute of Technology, the United States Military Academy, the Department of the Army, or the Department of Defense. We also acknowledge the efforts of many individuals in the John Wiley and Sons, Inc., organization who have contributed their talents and energy to this edition. We applaud their professionalism and commitment. We continue to be extremely gratified by the reception this book has enjoyed over the years. With this edition we have made the text more effective for teaching the subject of engineering thermodynamics and have greatly enhanced the relevance of the subject matter for students who will shape the 21st century. As always, we welcome your comments, criticisms, and suggestions. Michael J. Moran [email protected] Howard N. Shapiro [email protected] Daisie D. Boettner [email protected] Margaret B. Bailey [email protected]

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Contents 2.1.2 Potential Energy

1 Getting Started: Introductory

2.1.3 Units for Energy

Concepts and Definitions 3

2.1.4 Conservation of Energy in Mechanics

1.1 Using Thermodynamics 4 1.2 Deﬁning Systems 4 1.2.1 Closed Systems 1.2.2 Control Volumes

7 1.3 Describing Systems and Their Behavior 8 1.2.3 Selecting the System Boundary

1.3.1 Macroscopic and Microscopic Views

1.6

1.7

9

45

20

21 1.8 Engineering Design and Analysis 22 1.8.1 Design 23 1.8.2 Analysis 23 1.9 Methodology for Solving Thermodynamics Problems 24 Chapter Summary and Study Guide 26

2 Energy and the First Law of Thermodynamics 37 2.1 Reviewing Mechanical Concepts of Energy 38 2.1.1 Work and Kinetic Energy 38

46 Further Examples of Work 50

Quasiequilibrium Processes

9

1.7.2 Kelvin and Rankine Temperature 1.7.3 Celsius and Fahrenheit Scales

46

2.2.5 Expansion or Compression Work in

10 Measuring Mass, Length, Time, and Force 11 1.4.1 SI Units 11 1.4.2 English Engineering Units 12 Speciﬁc Volume 13 Pressure 14 1.6.1 Pressure Measurement 15 1.6.2 Buoyancy 16 1.6.3 Pressure Units 17 Temperature 18 1.7.1 Thermometers 19 Scales

Work

2.2.4 Expansion or Compression Work in Actual

1.3.4 Equilibrium

1.5

2.2.3 Modeling Expansion or Compression

Processes

1.3.3 Extensive and Intensive Properties

1.4

42 2.2 Broadening Our Understanding of Work 42 2.2.1 Sign Convention and Notation 43 2.2.2 Power 44

8

1.3.2 Property, State, and Process

41

2.1.5 Closing Comment

6 6

of Thermodynamics

40 41

2.2.6

2.2.7 Further Examples of Work in

Quasiequilibrium Processes

51

2.2.8 Generalized Forces and Displacements

52

2.3 Broadening Our Understanding of Energy 53 2.4 Energy Transfer by Heat 54 2.4.1 Sign Convention, Notation, and

54 55 2.4.3 Closing Comments 57 2.5 Energy Accounting: Energy Balance for Closed Systems 58 Heat Transfer Rate

2.4.2 Heat Transfer Modes

2.5.1 Important Aspects of the Energy Balance

60

2.5.2 Using the Energy Balance: Processes

of Closed Systems

62

2.5.3 Using the Energy Rate Balance:

Steady-State Operation

66

2.5.4 Using the Energy Rate Balance:

68 2.6 Energy Analysis of Cycles 70 2.6.1 Cycle Energy Balance 71 2.6.2 Power Cycles 71 Transient Operation

2.6.3 Refrigeration and Heat Pump Cycles

72

2.7 Energy Storage 74 2.7.1 Overview 74 2.7.2 Storage Technologies

74

Chapter Summary and Study Guide 75 ix

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Contents

3 Evaluating Properties

Evaluating Properties Using the Ideal Gas Model 127 3.12 Introducing the Ideal Gas Model 127

91

3.1 Getting Started 92 3.1.1 Phase and Pure Substance 3.1.2 Fixing the State

92

92

3.12.1 Ideal Gas Equation of State

Evaluating Properties: General Considerations 93 3.2 p–y–T Relation 93 3.2.1 p–y–T Surface 94 96

108 3.7 Evaluating Properties Using Computer Software 109 3.8 Applying the Energy Balance Using Property Tables and Software 110 3.8.1 Using Property Tables 112 3.8.2 Using Software 115 3.9 Introducing Speciﬁc Heats cy and cp 117 3.10 Evaluating Properties of Liquids and Solids 118 Values

3.10.1 Approximations for Liquids Using

3.11 Generalized Compressibility Chart 122 3.11.1 Universal Gas Constant, R 122 3.11.2 Compressibility Factor, Z 122 3.11.3 Generalized Compressibility Data,

123

3.14.1 Using Ideal Gas Tables

133

3.14.2 Using Constant Speciﬁc Heats

135

137 3.15 Polytropic Process Relations 141 Chapter Summary and Study Guide 143 3.14.3 Using Computer Software

4 Control Volume Analysis 4.1 Conservation of Mass for a Control Volume 164 4.1.1 Developing the Mass Rate

Balance

164

4.1.2 Evaluating the Mass Flow

165 4.2 Forms of the Mass Rate Balance 166 Rate

4.2.1 One-Dimensional Flow Form of the Mass Rate

Balance

166

4.2.2 Steady-State Form of the Mass Rate

Balance

167

4.2.3 Integral Form of the Mass Rate

167 4.3 Applications of the Mass Rate Balance 168 4.3.1 Steady-State Application 168 Balance

118

3.10.2 Incompressible Substance Model

Gas Tables, Constant Speciﬁc Heats, and Software 133

Using Energy 163

3.6.3 Reference States and Reference

Z Chart

130 3.13 Internal Energy, Enthalpy, and Speciﬁc Heats of Ideal Gases 130 3.13.1 Du, Dh, cy, and cp Relations 130 3.13.2 Using Speciﬁc Heat Functions 132 3.14 Applying the Energy Balance Using Ideal 3.12.3 Microscopic Interpretation

3.3 Studying Phase Change 97 3.4 Retrieving Thermodynamic Properties 100 3.5 Evaluating Pressure, Speciﬁc Volume, and Temperature 100 3.5.1 Vapor and Liquid Tables 100 3.5.2 Saturation Tables 103 3.6 Evaluating Speciﬁc Internal Energy and Enthalpy 106 3.6.1 Introducing Enthalpy 106 3.6.2 Retrieving u and h Data 107

3.11.4 Equations of State

128

3.12.2 Ideal Gas Model

3.2.2 Projections of the p–y–T Surface

Saturated Liquid Data

127

119

4.3.2 Time-Dependent (Transient)

169 4.4 Conservation of Energy for a Control Volume 172 Application

4.4.1 Developing the Energy Rate Balance for a

126

Control Volume

172

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Contents 4.4.2 Evaluating Work for a Control

Volume

5 The Second Law

173

4.4.3 One-Dimensional Flow Form of the Control

Volume Energy Rate Balance

173

4.4.4 Integral Form of the Control Volume Energy

174 4.5 Analyzing Control Volumes at Steady State 175 Rate Balance

175 176

Law

178

4.7 Turbines 180 4.7.1 Steam and Gas Turbine Modeling

182

4.7.2 Application to a Steam Turbine

182

4.8 Compressors and Pumps 184

4.8.2 Applications to an Air Compressor and a

184

4.8.3 Pumped-Hydro and Compressed-Air Energy

188 4.9 Heat Exchangers 189 Storage

4.9.1 Heat Exchanger Modeling

190

4.9.2 Applications to a Power Plant Condenser

190

194

4.10.2 Using a Throttling Calorimeter to

195 4.11 System Integration 196 4.12 Transient Analysis 199 Determine Quality

246

5.5 Applying the Second Law to Thermodynamic Cycles 248 5.6 Second Law Aspects of Power Cycles Interacting with Two Reservoirs 249 5.6.1 Limit on Thermal Efﬁciency 249

249

5.7 Second Law Aspects of Refrigeration and Heat Pump Cycles Interacting with Two Reservoirs 251 5.7.1 Limits on Coefﬁcients of Performance 251 Refrigeration and Heat Pump Cycles 252

199

4.12.2 The Energy Balance in Transient

200

4.12.3 Transient Analysis Applications

245

5.7.2 Corollaries of the Second Law for

4.12.1 The Mass Balance in Transient

Analysis

244

5.4 Interpreting the Kelvin–Planck Statement 247

Cycles

4.10.1 Throttling Device Modeling

Analysis

5.3.2 Demonstrating Irreversibility

5.6.2 Corollaries of the Second Law for Power

4.10 Throttling Devices 194 Considerations

242 5.3 Irreversible and Reversible Processes 242 5.3.1 Irreversible Processes 242 5.2.4 Second Law Summary

5.3.4 Internally Reversible Processes

184

and Computer Cooling

241

5.3.3 Reversible Processes

4.8.1 Compressor and Pump Modeling

Considerations

239

5.2.3 Entropy Statement of the Second

178

4.6.2 Application to a Steam Nozzle

239

Second Law

4.6.1 Nozzle and Diffuser Modeling

Pump System

Law

5.2.2 Kelvin–Planck Statement of the

4.6 Nozzles and Diffusers 177

Considerations

238 5.2 Statements of the Second Law 239 5.2.1 Clausius Statement of the Second

Volumes at Steady State

Considerations

238

Work

5.1.3 Aspects of the Second Law

4.5.2 Modeling Considerations for Control

Considerations

5.1 Introducing the Second Law 236 5.1.1 Motivating the Second Law 236 5.1.2 Opportunities for Developing

4.5.1 Steady-State Forms of the Mass and Energy

Rate Balances

of Thermodynamics 235

201

Chapter Summary and Study Guide 209

5.8 The Kelvin and International Temperature Scales 253 5.8.1 The Kelvin Scale 253 5.8.2 The Gas Thermometer 255 5.8.3 International Temperature Scale

256

xi

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Contents 6.7.2 Evaluating Entropy Production and

5.9 Maximum Performance Measures for Cycles Operating Between Two Reservoirs 256 5.9.1 Power Cycles 257 5.9.2 Refrigeration and Heat Pump Cycles

Transfer Balance

259

5.10 Carnot Cycle 262 5.10.1 Carnot Power Cycle

262

5.10.2 Carnot Refrigeration and Heat Pump

Cycles

264

5.10.3 Carnot Cycle Summary

264

5.11 Clausius Inequality 264

Chapter Summary and Study Guide 266

6 Using Entropy

284

6.6

285 Introducing the T dS Equations 286 Entropy Change of an Incompressible Substance 288 Entropy Change of an Ideal Gas 289 6.5.1 Using Ideal Gas Tables 289 6.5.2 Assuming Constant Speciﬁc Heats 291 6.5.3 Computer Retrieval 291 Entropy Change in Internally Reversible Processes of Closed Systems 292 6.6.1 Area Representation of Heat

Transfer

292

6.6.2 Carnot Cycle Application

Reversible Process of Water

293

6.7 Entropy Balance for Closed Systems 295 6.7.1 Interpreting the Closed System Entropy

296

at Steady State

308

6.10.2 Applications of the Rate Balances to

Control Volumes at Steady State 309

6.11 Isentropic Processes 315 6.11.1 General Considerations

316 316

318 6.12 Isentropic Efﬁciencies of Turbines, Nozzles, Compressors, and Pumps 322 6.12.1 Isentropic Turbine Efﬁciency 322 6.12.2 Isentropic Nozzle Efﬁciency 325 6.12.3 Isentropic Compressor and Pump

327 6.13 Heat Transfer and Work in Internally Reversible, Steady-State Flow Processes 329 6.13.1 Heat Transfer 329 6.13.2 Work 330 6.13.3 Work In Polytropic Processes 331 Chapter Summary and Study Guide 333 Efﬁciencies

292

6.6.3 Work and Heat Transfer in an Internally

Balance

305 6.9 Entropy Rate Balance for Control Volumes 307 6.10 Rate Balances for Control Volumes at Steady State 308 of Entropy

of Air

6.2.5 Using Graphical Entropy Data

6.5

6.8.2 Statistical Interpretation

6.11.3 Illustrations: Isentropic Processes

284 6.2.3 Liquid Data 284 6.2.4 Computer Retrieval 285

6.4

300 6.8 Directionality of Processes 302 6.8.1 Increase of Entropy Principle 302 Balance

6.11.2 Using the Ideal Gas Model

6.2.2 Saturation Data

6.3

297

6.7.4 Closed System Entropy Rate

6.10.1 One-Inlet, One-Exit Control Volumes

281

6.1 Entropy–A System Property 282 6.1.1 Deﬁning Entropy Change 282 6.1.2 Evaluating Entropy 283 6.1.3 Entropy and Probability 283 6.2 Retrieving Entropy Data 283 6.2.1 Vapor Data

297

6.7.3 Applications of the Closed System Entropy

7 Exergy Analysis

359

7.1 Introducing Exergy 360 7.2 Conceptualizing Exergy 361 7.2.1 Environment and Dead State 7.2.2 Deﬁning Exergy

362

362

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Contents

8.4 Improving Performance—Regenerative Vapor Power Cycle 453 8.4.1 Open Feedwater Heaters 453 8.4.2 Closed Feedwater Heaters 458 8.4.3 Multiple Feedwater Heaters 459 8.5 Other Vapor Power Cycle Aspects 463

7.3 Exergy of a System 362 7.3.1 Exergy Aspects 365 7.3.2 Speciﬁc Exergy 366 7.3.3 Exergy Change 368 7.4 Closed System Exergy Balance 368 7.4.1 Introducing the Closed System Exergy

Balance

369

8.5.2

373

7.4.3 Exergy Destruction and Loss

376 7.5 Exergy Rate Balance for Control Volumes at Steady State 377 7.5.1 Comparing Energy and Exergy for Control

Volumes at Steady State

380

7.5.2 Evaluating Exergy Destruction in Control

Volumes at Steady State

380

7.5.3 Exergy Accounting in Control Volumes at

385 7.6 Exergetic (Second Law) Efﬁciency 389 7.6.1 Matching End Use to Source 390 Steady State

7.6.2 Exergetic Efﬁciencies of Common

392

7.6.3 Using Exergetic Efﬁciencies

394

7.7 Thermoeconomics 395 7.7.1 Costing 395 7.7.2 Using Exergy in Design

493

Considering Internal Combustion Engines 494 9.1 Introducing Engine Terminology 494 9.2 Air-Standard Otto Cycle 497 9.3 Air-Standard Diesel Cycle 502 9.4 Air-Standard Dual Cycle 506 Considering Gas Turbine Power Plants 509 9.5 Modeling Gas Turbine Power Plants 509 9.6 Air-Standard Brayton Cycle 511 Transfers

396

512

9.6.3 Considering Gas Turbine Irreversibilities and

398 Chapter Summary and Study Guide 403

Losses

518

9.7 Regenerative Gas Turbines 521 9.8 Regenerative Gas Turbines with Reheat and Intercooling 525 9.8.1 Gas Turbines with Reheat 526 9.8.2 Compression with Intercooling 528 9.8.3 Reheat and Intercooling 532 9.8.4 Ericsson and Stirling Cycles 535 9.9 Gas Turbine–Based Combined Cycles 537

425

Introducing Power Generation 426 Considering Vapor Power Systems 430 8.1 Introducing Vapor Power Plants 430 8.2 The Rankine Cycle 433 8.2.1 Modeling the Rankine Cycle 434 8.2.2 Ideal Rankine Cycle 437

9.9.1 Combined Gas Turbine–Vapor Power Cycle

8.2.3 Effects of Boiler and Condenser Pressures on

441

8.2.4 Principal Irreversibilities and Losses

511

9.6.2 Ideal Air-Standard Brayton Cycle

System

the Rankine Cycle

9 Gas Power Systems

9.6.1 Evaluating Principal Work and Heat

7.7.3 Exergy Costing of a Cogeneration

8 Vapor Power Systems

465 8.6 Case Study: Exergy Accounting of a Vapor Power Plant 468 Chapter Summary and Study Guide 475 8.5.3 Carbon Capture and Storage

374

7.4.4 Exergy Accounting

Components

463 Cogeneration 465

8.5.1 Working Fluids

7.4.2 Closed System Exergy Rate

Balance

xiii

443

8.3 Improving Performance— Superheat, Reheat, and Supercritical 447

537

544 9.10 Integrated Gasiﬁcation Combined-Cycle Power Plants 544 9.11 Gas Turbines for Aircraft Propulsion 546 9.9.2 Cogeneration

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Contents

Considering Compressible Flow Through Nozzles and Diffusers 550 9.12 Compressible Flow Preliminaries 551 9.12.1 Momentum Equation for Steady

One-Dimensional Flow

551

608 10.7 Gas Refrigeration Systems 612 10.7.1 Brayton Refrigeration Cycle 612

9.12.3 Determining Stagnation State

555 9.13 Analyzing One-Dimensional Steady Flow in Nozzles and Diffusers 555 Properties

9.13.1 Exploring the Effects of Area Change in

555

9.13.2 Effects of Back Pressure on Mass Flow

Rate

608

Pumps

552

Subsonic and Supersonic Flows

10.6.1 Carnot Heat Pump Cycle 10.6.2 Vapor-Compression Heat

9.12.2 Velocity of Sound and Mach

Number

10.5 Absorption Refrigeration 606 10.6 Heat Pump Systems 608

558

560 9.14 Flow in Nozzles and Diffusers of Ideal Gases with Constant Speciﬁc Heats 561 9.14.1 Isentropic Flow Functions 562 9.14.2 Normal Shock Functions 565 Chapter Summary and Study Guide 569 9.13.3 Flow Across a Normal Shock

10.7.2 Additional Gas Refrigeration

Applications

616

10.7.3 Automotive Air Conditioning Using Carbon

617 Chapter Summary and Study Guide 619 Dioxide

11 Thermodynamic Relations

631

11.1 Using Equations of State 632 11.1.1 Getting Started 632

633 11.1.3 Multiconstant Equations of State 637 11.2 Important Mathematical Relations 638 11.3 Developing Property Relations 641 11.3.1 Principal Exact Differentials 642 11.1.2 Two-Constant Equations of State

11.3.2 Property Relations from Exact

10 Refrigeration and Heat Pump Systems 589

591

10.2 Analyzing Vapor-Compression Refrigeration Systems 592 10.2.1 Evaluating Principal Work and Heat

Transfers

592

10.2.2 Performance of Ideal Vapor-

Compression Systems

593

10.2.3 Performance of Actual Vapor-

596 10.2.4 The p –h Diagram 600 10.3 Selecting Refrigerants 600 10.4 Other Vapor-Compression Applications 603 10.4.1 Cold Storage 603 10.4.2 Cascade Cycles 604 Compression Systems

10.4.3 Multistage Compression with

Intercooling

605

642

11.3.3 Fundamental Thermodynamic

647 11.4 Evaluating Changes in Entropy, Internal Energy, and Enthalpy 648 11.4.1 Considering Phase Change 648 Functions

10.1 Vapor Refrigeration Systems 590 10.1.1 Carnot Refrigeration Cycle 590 10.1.2 Departures from the Carnot Cycle

Differentials

11.4.2 Considering Single-Phase

651 11.5 Other Thermodynamic Relations 656 Regions

11.5.1 Volume Expansivity, Isothermal and

Isentropic Compressibility

657

11.5.2 Relations Involving Speciﬁc Heats

658

661 11.6 Constructing Tables of Thermodynamic Properties 663 11.5.3 Joule–Thomson Coefﬁcient

11.6.1 Developing Tables by Integration Using

p–y–T and Speciﬁc Heat Data

664

11.6.2 Developing Tables by Differentiating

a Fundamental Thermodynamic Function 665

11.7 Generalized Charts for Enthalpy and Entropy 668

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Contents

11.8 p–y–T Relations for Gas Mixtures 675 11.9 Analyzing Multicomponent Systems 679 11.9.1 Partial Molal Properties 680 11.9.2 Chemical Potential 682

12.7 Psychrometric Charts 740 12.8 Analyzing Air-Conditioning Processes 741 12.8.1 Applying Mass and Energy Balances

to Air-Conditioning Systems

11.9.4 Fugacity

Composition

683

746 12.8.4 Humidiﬁcation 750 12.8.5 Evaporative Cooling 752

688

11.9.6 Chemical Potential for Ideal

689 Chapter Summary and Study Guide 690

12.8.6 Adiabatic Mixing of Two Moist Air

Solutions

755 12.9 Cooling Towers 758 Chapter Summary and Study Guide 761 Streams

12 Ideal Gas Mixture and Psychrometric Applications 705 Ideal Gas Mixtures: General Considerations 706 12.1 Describing Mixture Composition 706 12.2 Relating p, V, and T for Ideal Gas Mixtures 710 12.3 Evaluating U, H, S, and Speciﬁc Heats 711 12.3.1 Evaluating U and H 711 12.3.2 Evaluating cy and cp 712 12.3.3 Evaluating S 712 12.3.4 Working on a Mass Basis 713 12.4 Analyzing Systems Involving Mixtures 714

Combustion Fundamentals 778 13.1 Introducing Combustion 778 13.1.1 Fuels 779 13.1.2 Modeling Combustion Air 779 13.1.3 Determining Products of

Combustion Systems

786

13.2 Conservation of Energy—Reacting Systems 787 13.2.1 Evaluating Enthalpy for Reacting

787

Systems

13.2.2 Energy Balances for Reacting

Systems

789

13.2.3 Enthalpy of Combustion and Heating

721 Psychrometric Applications 727 12.5 Introducing Psychrometric Principles 727 12.5.1 Moist Air 727 12.4.2 Mixing of Ideal Gases

12.5.2 Humidity Ratio, Relative Humidity, Mixture

728

12.5.3 Modeling Moist Air in Equilibrium with

Liquid Water

782

13.1.4 Energy and Entropy Balances for Reacting

714

Enthalpy, and Mixture Entropy

13 Reacting Mixtures and Combustion 777

12.4.1 Mixture Processes at Constant

Composition

743

12.8.3 Dehumidiﬁcation

685

11.9.5 Ideal Solution

741

12.8.2 Conditioning Moist Air at Constant

11.9.3 Fundamental Thermodynamic Functions

for Multicomponent Systems

xv

730

12.5.4 Evaluating the Dew Point Temperature

797

13.3 Determining the Adiabatic Flame Temperature 800 13.3.1 Using Table Data 801 13.3.2 Using Computer Software 801 13.3.3 Closing Comments 804 13.4 Fuel Cells 804 13.4.1 Proton Exchange Membrane Fuel Cell

731

12.5.5 Evaluating Humidity Ratio Using the

737 12.6 Psychrometers: Measuring the Wet-Bulb and Dry-Bulb Temperatures 738 Adiabatic-Saturation Temperature

Values

806

808 13.5 Absolute Entropy and the Third Law of Thermodynamics 808 13.4.2 Solid Oxide Fuel Cell

13.5.1 Evaluating Entropy for Reacting

Systems

809

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Contents 13.5.2 Entropy Balances for Reacting

Systems

14.3 Calculating Equilibrium Compositions 855

810

14.3.1 Equilibrium Constant for Ideal Gas

13.5.3 Evaluating Gibbs Function for Reacting

815 Chemical Exergy 816 13.6 Conceptualizing Chemical Exergy 817 Systems

13.6.1 Working Equations for Chemical

Exergy

819

14.3.2 Illustrations of the Calculation of

Equilibrium Compositions for Reacting Ideal Gas Mixtures 858 14.3.3 Equilibrium Constant for Mixtures and

863 14.4 Further Examples of the Use of the Equilibrium Constant 865 14.4.1 Determining Equilibrium Flame

821 13.7 Standard Chemical Exergy 821 13.6.3 Closing Comments

Temperature

13.7.1 Standard Chemical Exergy of a Hydrocarbon:

C aH b

855

Solutions

819

13.6.2 Evaluating Chemical Exergy for Several

Cases

Mixtures

822

13.7.2 Standard Chemical Exergy of Other

825 13.8 Applying Total Exergy 826 13.8.1 Calculating Total Exergy 826 Substances

13.8.2 Calculating Exergetic Efﬁciencies

829 Chapter Summary and Study Guide 832 of Reacting Systems

865

14.4.2 Van’t Hoff Equation

870

14.4.3 Ionization

14.4.4 Simultaneous Reactions

871

Phase Equilibrium 874 14.5 Equilibrium between Two Phases of a Pure Substance 874 14.6 Equilibrium of Multicomponent, Multiphase Systems 876 14.6.1 Chemical Potential and Phase

Equilibrium

876

14.6.2 Gibbs Phase Rule

14 Chemical and Phase

869

879

Equilibrium 847

Appendix Tables, Figures,

Equilibrium Fundamentals 848 14.1 Introducing Equilibrium Criteria 848

and Charts 889

14.1.1 Chemical Potential and

Equilibrium

849

14.1.2 Evaluating Chemical Potentials

Chemical Equilibrium 853 14.2 Equation of Reaction Equilibrium 853 14.2.1 Introductory Case 853 14.2.2 General Case 854

850

Index to Tables in SI Units 889 Index to Tables in English Units 937 Index to Figures and Charts 985 Index 996 Answers to Selected Problems: Visit the student companion site at www.wiley.com/ college/moran.

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F U N D A M E N TA L S O F

ENGINEERING

THERMODYNAMICS SEVENTH EDITION

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Fluids such as air and water exert pressure, introduced in Sec. 1.6. © Jeffrey Warrington/Alamy

ENGINEERING CONTEXT Although aspects of thermodynamics have been studied since ancient times, the formal study of thermodynamics began in the early nineteenth century through consideration of the capacity of hot objects to produce work. Today the scope is much larger. Thermodynamics now provides essential concepts and methods for addressing critical twenty-first-century issues, such as using fossil fuels more effectively, fostering renewable energy technologies, and developing more fuel-efficient means of transportation. Also critical are the related issues of greenhouse gas emissions and air and water pollution. Thermodynamics is both a branch of science and an engineering specialty. The scientist is normally interested in gaining a fundamental understanding of the physical and chemical behavior of fixed quantities of matter at rest and uses the principles of thermodynamics to relate the properties of matter. Engineers are generally interested in studying systems and how they interact with their surroundings. To facilitate this, thermodynamics has been extended to the study of systems through which matter flows, including bioengineering and biomedical systems. The objective of this chapter is to introduce you to some of the fundamental concepts and definitions that are used in our study of engineering thermodynamics. In most instances this introduction is brief, and further elaboration is provided in subsequent chapters.

2

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1 Getting Started Introductory Concepts and Definitions

LEARNING OUTCOMES When you complete your study of this chapter, you will be able to... c

demonstrate understanding of several fundamental concepts used throughout the book . . . including closed system, control volume, boundary and surroundings, property, state, process, the distinction between extensive and intensive properties, and equilibrium.

c

apply SI and English Engineering units, including units for specific volume, pressure, and temperature.

c

work with the Kelvin, Rankine, Celsius, and Fahrenheit temperature scales.

c

apply the problem-solving methodology used in this book.

3

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Chapter 1 Getting Started

1.1 Using Thermodynamics Engineers use principles drawn from thermodynamics and other engineering sciences, including fluid mechanics and heat and mass transfer, to analyze and design things intended to meet human needs. Throughout the twentieth century, engineering applications of thermodynamics helped pave the way for significant improvements in our quality of life with advances in major areas such as surface transportation, air travel, space flight, electricity generation and transmission, building heating and cooling, and improved medical practices. The wide realm of these applications is suggested by Table 1.1. In the twenty-first century, engineers will create the technology needed to achieve a sustainable future. Thermodynamics will continue to advance human well-being by addressing looming societal challenges owing to declining supplies of energy resources: oil, natural gas, coal, and fissionable material; effects of global climate change; and burgeoning population. Life in the United States is expected to change in several important respects by midcentury. In the area of power use, for example, electricity will play an even greater role than today. Table 1.2 provides predictions of other changes experts say will be observed. If this vision of mid-century life is correct, it will be necessary to evolve quickly from our present energy posture. As was the case in the twentieth century, thermodynamics will contribute significantly to meeting the challenges of the twenty-first century, including using fossil fuels more effectively, advancing renewable energy technologies, and developing more energy-efficient transportation systems, buildings, and industrial practices. Thermodynamics also will play a role in mitigating global climate change, air pollution, and water pollution. Applications will be observed in bioengineering, biomedical systems, and the deployment of nanotechnology. This book provides the tools needed by specialists working in all such fields. For nonspecialists, the book provides background for making decisions about technology related to thermodynamics—on the job, as informed citizens, and as government leaders and policy makers.

1.2 Defining Systems

system

surroundings boundary

The key initial step in any engineering analysis is to describe precisely what is being studied. In mechanics, if the motion of a body is to be determined, normally the first step is to define a free body and identify all the forces exerted on it by other bodies. Newton’s second law of motion is then applied. In thermodynamics the term system is used to identify the subject of the analysis. Once the system is defined and the relevant interactions with other systems are identified, one or more physical laws or relations are applied. The system is whatever we want to study. It may be as simple as a free body or as complex as an entire chemical refinery. We may want to study a quantity of matter contained within a closed, rigid-walled tank, or we may want to consider something such as a pipeline through which natural gas flows. The composition of the matter inside the system may be fixed or may be changing through chemical or nuclear reactions. The shape or volume of the system being analyzed is not necessarily constant, as when a gas in a cylinder is compressed by a piston or a balloon is inflated. Everything external to the system is considered to be part of the system’s surroundings. The system is distinguished from its surroundings by a specified boundary, which may be at rest or in motion. You will see that the interactions between a system and its surroundings, which take place across the boundary, play an important part in engineering thermodynamics. Two basic kinds of systems are distinguished in this book. These are referred to, respectively, as closed systems and control volumes. A closed system refers to a fixed quantity of matter, whereas a control volume is a region of space through which mass may flow. The term control mass is sometimes used in place of closed system, and the term open system is used interchangeably with control volume. When the terms control mass and control volume are used, the system boundary is often referred to as a control surface.

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1.2 Defining Systems

5

TABLE 1.1

Selected Areas of Application of Engineering Thermodynamics Aircraft and rocket propulsion Alternative energy systems Fuel cells Geothermal systems Magnetohydrodynamic (MHD) converters Ocean thermal, wave, and tidal power generation Solar-activated heating, cooling, and power generation Thermoelectric and thermionic devices Wind turbines Automobile engines Bioengineering applications Biomedical applications Combustion systems Compressors, pumps Cooling of electronic equipment Cryogenic systems, gas separation, and liquefaction Fossil and nuclear-fueled power stations Heating, ventilating, and air-conditioning systems Absorption refrigeration and heat pumps Vapor-compression refrigeration and heat pumps Steam and gas turbines Power production Propulsion control coatings

Solar-cell arrays

Surfaces with thermal control coatings International Space Station

International Space Station

Steam generator

Combustion gas cleanup

Coal

Air Steam

Turbine Generator

Electric power

Cooling tower

Condenser

Ash Condensate Refrigerator

Stack

Cooling water

Electrical power plant

Vehicle engine Trachea Lung

Fuel in Combustor Compressor Air in

Turbine Hot gases out Heart

Turbojet engine

Biomedical applications

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Chapter 1 Getting Started TABLE 1.2

Predictions of Life in the United States in 2050 At home c Homes are constructed better to reduce heating and cooling needs. c Homes have systems for electronically monitoring and regulating energy use. c Appliances and heating and air-conditioning systems are more energy-efficient. c Use of solar energy for space and water heating is common. c More food is produced locally. Transportation c Plug-in hybrid vehicles and all-electric vehicles dominate. c Hybrid vehicles mainly use biofuels. c Use of public transportation within and between cities is common. c An expanded passenger railway system is widely used. Lifestyle c Efficient energy-use practices are utilized throughout society. c Recycling is widely practiced, including recycling of water. c Distance learning is common at most educational levels. c Telecommuting and teleconferencing are the norm. c The Internet is predominately used for consumer and business commerce. Power generation c Electricity plays a greater role throughout society. c Wind, solar, and other renewable technologies contribute a significant share of the nation’s electricity needs. c A mix of conventional fossil-fueled and nuclear power plants provide a smaller, but still significant, share of the nation’s electricity needs. c A smart and secure national power transmission grid is in place.

1.2.1 Closed Systems closed system

isolated system

Gas

Boundary

A closed system is defined when a particular quantity of matter is under study. A closed system always contains the same matter. There can be no transfer of mass across its boundary. A special type of closed system that does not interact in any way with its surroundings is called an isolated system. Figure 1.1 shows a gas in a piston–cylinder assembly. When the valves are closed, we can consider the gas to be a closed system. The boundary lies just inside the piston and cylinder walls, as shown by the dashed lines on the figure. Since the portion of the boundary between the gas and the piston moves with the piston, the system volume varies. No mass would cross this or any other part of the boundary. If combustion occurs, the composition of the system changes as the initial combustible mixture becomes products of combustion.

1.2.2 Control Volumes

Fig. 1.1 Closed system: A gas in a piston–cylinder assembly.

control volume

In subsequent sections of this book, we perform thermodynamic analyses of devices such as turbines and pumps through which mass flows. These analyses can be conducted in principle by studying a particular quantity of matter, a closed system, as it passes through the device. In most cases it is simpler to think instead in terms of a given region of space through which mass flows. With this approach, a region within a prescribed boundary is studied. The region is called a control volume. Mass may cross the boundary of a control volume. A diagram of an engine is shown in Fig. 1.2a. The dashed line defines a control volume that surrounds the engine. Observe that air, fuel, and exhaust gases cross the boundary. A schematic such as in Fig. 1.2b often suffices for engineering analysis.

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1.2 Defining Systems Air in

Drive shaft

7

Air in

Exhaust gas out Fuel in

Fuel in

Drive shaft Exhaust gas out Boundary (control surface)

Boundary (control surface)

(a)

(b)

Fig. 1.2 Example of a control volume (open system). An automobile engine.

BIOCONNECTIONS Living things and their organs can be studied as control volumes. For the pet shown in Fig. 1.3a, air, food, and drink essential to sustain life and for activity enter across the boundary, and waste products exit. A schematic such as Fig. 1.3b can suffice for biological analysis. Particular organs, such as the heart, also can be studied as control volumes. As shown in Fig. 1.4, plants can be studied from a control volume viewpoint. Intercepted solar radiation is used in the production of essential chemical substances within plants by photosynthesis. During photosynthesis, plants take in carbon dioxide from the atmosphere and discharge oxygen to the atmosphere. Plants also draw in water and nutrients through their roots.

1.2.3 Selecting the System Boundary The system boundary should be delineated carefully before proceeding with any thermodynamic analysis. However, the same physical phenomena often can be analyzed in terms of alternative choices of the system, boundary, and surroundings. The choice of a particular boundary defining a particular system depends heavily on the convenience it allows in the subsequent analysis.

CO2, other gases

Ingestion (food, drink)

Air

Air Gut

Ingestion (food, drink)

Boundary (control surface)

Lungs CO2

Boundary (control surface)

O2 CO2

Circulatory system Kidneys

Excretion (waste products)

Body tissues

O2

Photosynthesis (leaf)

Heart Excretion (undigested food)

(a)

Solar radiation

CO2, other gases

Excretion (urine) (b)

Fig. 1.3 Example of a control volume (open system) in biology.

H2O, minerals

Fig. 1.4 Example of a control volume (open system) in botany.

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Chapter 1 Getting Started

8

Air

TAKE NOTE...

Animations reinforce many of the text presentations. You can view these animations by going to the student companion site for this book. Animations are keyed to specific content by an icon in the margin. The first of these icons appears directly below. In this example, the label System_Types refers to the text content while A.1–Tabs a,b&c refers to the particular animation (A.1) and the tabs (Tabs a,b&c) of the animation recommended for viewing now to enhance your understanding.

A

System_Types A.1 – Tabs a, b, & c

Tank Air compressor

Fig. 1.5 Air compressor and storage tank.

–

+

In general, the choice of system boundary is governed by two considerations: (1) what is known about a possible system, particularly at its boundaries, and (2) the objective of the analysis. Figure 1.5 shows a sketch of an air compressor connected to a storage tank. The system boundary shown on the figure encloses the compressor, tank, and all of the piping. This boundary might be selected if the electrical power input is known, and the objective of the analysis is to determine how long the compressor must operate for the pressure in the tank to rise to a specified value. Since mass crosses the boundary, the system would be a control volume. A control volume enclosing only the compressor might be chosen if the condition of the air entering and exiting the compressor is known, and the objective is to determine the electric power input. b b b b b

1.3 Describing Systems and Their Behavior Engineers are interested in studying systems and how they interact with their surroundings. In this section, we introduce several terms and concepts used to describe systems and how they behave.

1.3.1 Macroscopic and Microscopic Views of Thermodynamics Systems can be studied from a macroscopic or a microscopic point of view. The macroscopic approach to thermodynamics is concerned with the gross or overall behavior. This is sometimes called classical thermodynamics. No model of the structure of matter at the molecular, atomic, and subatomic levels is directly used in classical thermodynamics. Although the behavior of systems is affected by molecular structure, classical thermodynamics allows important aspects of system behavior to be evaluated from observations of the overall system. The microscopic approach to thermodynamics, known as statistical thermodynamics, is concerned directly with the structure of matter. The objective of statistical thermodynamics is to characterize by statistical means the average behavior of the particles making up a system of interest and relate this information to the observed macroscopic behavior of the system. For applications involving lasers, plasmas, high-speed gas flows, chemical kinetics, very low temperatures (cryogenics), and others, the methods of statistical thermodynamics are essential. The microscopic approach is used in this text to interpret internal energy in Chap. 2 and entropy in Chap 6. Moreover, as

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1.3 Describing Systems and Their Behavior

9

noted in Chap. 3, the microscopic approach is instrumental in developing certain data, for example ideal gas specific heats. For a wide range of engineering applications, classical thermodynamics not only provides a considerably more direct approach for analysis and design but also requires far fewer mathematical complications. For these reasons the macroscopic viewpoint is the one adopted in this book. Finally, relativity effects are not significant for the systems under consideration in this book.

1.3.2 Property, State, and Process To describe a system and predict its behavior requires knowledge of its properties and how those properties are related. A property is a macroscopic characteristic of a system such as mass, volume, energy, pressure, and temperature to which a numerical value can be assigned at a given time without knowledge of the previous behavior (history) of the system. The word state refers to the condition of a system as described by its properties. Since there are normally relations among the properties of a system, the state often can be specified by providing the values of a subset of the properties. All other properties can be determined in terms of these few. When any of the properties of a system change, the state changes and the system is said to have undergone a process. A process is a transformation from one state to another. However, if a system exhibits the same values of its properties at two different times, it is in the same state at these times. A system is said to be at steady state if none of its properties change with time. Many properties are considered during the course of our study of engineering thermodynamics. Thermodynamics also deals with quantities that are not properties, such as mass flow rates and energy transfers by work and heat. Additional examples of quantities that are not properties are provided in subsequent chapters. For a way to distinguish properties from nonproperties, see the box on p. 10.

property

state

process steady state

Prop_State_Process A.2 – Tab a

A

1.3.3 Extensive and Intensive Properties Thermodynamic properties can be placed in two general classes: extensive and intensive. A property is called extensive if its value for an overall system is the sum of its values for the parts into which the system is divided. Mass, volume, energy, and several other properties introduced later are extensive. Extensive properties depend on the size or extent of a system. The extensive properties of a system can change with time, and many thermodynamic analyses consist mainly of carefully accounting for changes in extensive properties such as mass and energy as a system interacts with its surroundings. Intensive properties are not additive in the sense previously considered. Their values are independent of the size or extent of a system and may vary from place to place within the system at any moment. Thus, intensive properties may be functions of both position and time, whereas extensive properties can vary only with time. Specific volume (Sec. 1.5), pressure, and temperature are important intensive properties; several other intensive properties are introduced in subsequent chapters. to illustrate the difference between extensive and intensive properties, consider an amount of matter that is uniform in temperature, and imagine that it is composed of several parts, as illustrated in Fig. 1.6. The mass of the whole is the sum of the masses of the parts, and the overall volume is the sum of the volumes of the parts. However, the temperature of the whole is not the sum of the temperatures of the parts; it is the same for each part. Mass and volume are extensive, but temperature is intensive. b b b b b

extensive property

intensive property

Ext_Int_Properties A.3 – Tab a

A

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Chapter 1 Getting Started

Fig. 1.6 Figure used to discuss the extensive and intensive property concepts.

(a)

(b)

Distinguishing Properties from Nonproperties At a given state each property has a definite value that can be assigned without knowledge of how the system arrived at that state. Therefore, the change in value of a property as the system is altered from one state to another is determined solely by the two end states and is independent of the particular way the change of state occurred. That is, the change is independent of the details of the process. Conversely, if the value of a quantity is independent of the process between two states, then that quantity is the change in a property. This provides a test for determining whether a quantity is a property: A quantity is a property if, and only if, its change in value between two states is independent of the process. It follows that if the value of a particular quantity depends on the details of the process, and not solely on the end states, that quantity cannot be a property.

1.3.4 Equilibrium equilibrium

equilibrium state

Classical thermodynamics places primary emphasis on equilibrium states and changes from one equilibrium state to another. Thus, the concept of equilibrium is fundamental. In mechanics, equilibrium means a condition of balance maintained by an equality of opposing forces. In thermodynamics, the concept is more far-reaching, including not only a balance of forces but also a balance of other influences. Each kind of influence refers to a particular aspect of thermodynamic, or complete, equilibrium. Accordingly, several types of equilibrium must exist individually to fulfill the condition of complete equilibrium; among these are mechanical, thermal, phase, and chemical equilibrium. Criteria for these four types of equilibrium are considered in subsequent discussions. For the present, we may think of testing to see if a system is in thermodynamic equilibrium by the following procedure: Isolate the system from its surroundings and watch for changes in its observable properties. If there are no changes, we conclude that the system was in equilibrium at the moment it was isolated. The system can be said to be at an equilibrium state. When a system is isolated, it does not interact with its surroundings; however, its state can change as a consequence of spontaneous events occurring internally as its intensive properties, such as temperature and pressure, tend toward uniform values. When all such changes cease, the system is in equilibrium. At equilibrium, temperature is uniform throughout the system. Also, pressure can be regarded as uniform throughout as long as the effect of gravity is not significant; otherwise a pressure variation can exist, as in a vertical column of liquid. There is no requirement that a system undergoing a process be in equilibrium during the process. Some or all of the intervening states may be nonequilibrium states. For many such processes we are limited to knowing the state before the process occurs and the state after the process is completed.

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1.4 Measuring Mass, Length, Time, and Force

1.4 Measuring Mass, Length, Time, and Force When engineering calculations are performed, it is necessary to be concerned with the units of the physical quantities involved. A unit is any specified amount of a quantity by comparison with which any other quantity of the same kind is measured. For example, meters, centimeters, kilometers, feet, inches, and miles are all units of length. Seconds, minutes, and hours are alternative time units. Because physical quantities are related by definitions and laws, a relatively small number of physical quantities suffice to conceive of and measure all others. These are called primary dimensions. The others are measured in terms of the primary dimensions and are called secondary. For example, if length and time were regarded as primary, velocity and area would be secondary. A set of primary dimensions that suffice for applications in mechanics are mass, length, and time. Additional primary dimensions are required when additional physical phenomena come under consideration. Temperature is included for thermodynamics, and electric current is introduced for applications involving electricity. Once a set of primary dimensions is adopted, a base unit for each primary dimension is specified. Units for all other quantities are then derived in terms of the base units. Let us illustrate these ideas by considering briefly two systems of units: SI units and English Engineering units.

base unit

1.4.1 SI Units In the present discussion we consider the system of units called SI that takes mass, length, and time as primary dimensions and regards force as secondary. SI is the abbreviation for Système International d’Unités (International System of Units), which is the legally accepted system in most countries. The conventions of the SI are published and controlled by an international treaty organization. The SI base units for mass, length, and time are listed in Table 1.3 and discussed in the following paragraphs. The SI base unit for temperature is the kelvin, K. The SI base unit of mass is the kilogram, kg. It is equal to the mass of a particular cylinder of platinum–iridium alloy kept by the International Bureau of Weights and Measures near Paris. The mass standard for the United States is maintained by the National Institute of Standards and Technology. The kilogram is the only base unit still defined relative to a fabricated object. The SI base unit of length is the meter (metre), m, defined as the length of the path traveled by light in a vacuum during a specified time interval. The base unit of time is the second, s. The second is defined as the duration of 9,192,631,770 cycles of the radiation associated with a specified transition of the cesium atom.

TABLE 1.3

Units for Mass, Length, Time, and Force SI Quantity

Unit

mass length time force

kilogram meter second newton (5 1 kg · m/s2)

English Symbol

kg m s N

Unit

pound mass foot second pound force (5 32.1740 lb · ft/s2)

Symbol

lb ft s lbf

SI base units

11

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Chapter 1 Getting Started The SI unit of force, called the newton, is a secondary unit, defined in terms of the base units for mass, length, and time. Newton’s second law of motion states that the net force acting on a body is proportional to the product of the mass and the acceleration, written F r ma. The newton is defined so that the proportionality constant in the expression is equal to unity. That is, Newton’s second law is expressed as the equality F 5 ma

(1.1)

The newton, N, is the force required to accelerate a mass of 1 kilogram at the rate of 1 meter per second per second. With Eq. 1.1 1 N 5 11 kg211 m/ s22 5 1 kg ? m/ s2

TAKE NOTE...

Observe that in the calculation of force in newtons, the unit conversion factor is set off by a pair of vertical lines. This device is used throughout the text to identify unit conversions.

(1.2)

to illustrate the use of the SI units introduced thus far, let us determine the weight in newtons of an object whose mass is 1000 kg, at a place on the earth’s surface where the acceleration due to gravity equals a standard value defined as 9.80665 m/s2. Recalling that the weight of an object refers to the force of gravity, and is calculated using the mass of the object, m, and the local acceleration of gravity, g, with Eq. 1.1 we get F 5 mg 5 11000 kg219.80665 m / s22 5 9806.65 kg ? m / s2 This force can be expressed in terms of the newton by using Eq. 1.2 as a unit conversion factor. That is, F 5 a9806.65

kg ? m s2

b`

1N ` 5 9806.65 N 1 kg ? m / s2

b b b b b

Since weight is calculated in terms of the mass and the local acceleration due to gravity, the weight of an object can vary because of the variation of the acceleration of gravity with location, but its mass remains constant.

TABLE 1.4

SI Unit Prefixes Factor

Prefix

Symbol

1012 109 106 103 102 1022 1023 1026 1029 10212

tera giga mega kilo hecto centi milli micro nano pico

T G M k h c m m n p

if the object considered previously were on the surface of a planet at a point where the acceleration of gravity is one-tenth of the value used in the above calculation, the mass would remain the same but the weight would be onetenth of the calculated value. b b b b b SI units for other physical quantities are also derived in terms of the SI base units. Some of the derived units occur so frequently that they are given special names and symbols, such as the newton. SI units for quantities pertinent to thermodynamics are given as they are introduced in the text. Since it is frequently necessary to work with extremely large or small values when using the SI unit system, a set of standard prefixes is provided in Table 1.4 to simplify matters. For example, km denotes kilometer, that is, 103 m.

1.4.2 English Engineering Units

English base units

Although SI units are the worldwide standard, at the present time many segments of the engineering community in the United States regularly use other units. A large portion of America’s stock of tools and industrial machines and much valuable engineering data utilize units other than SI units. For many years to come, engineers in the United States will have to be conversant with a variety of units. In this section we consider a system of units that is commonly used in the United States, called the English Engineering system. The English base units for mass, length, and time are listed in Table 1.3 and discussed in the following paragraphs. English units for other quantities pertinent to thermodynamics are given as they are introduced in the text.

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1.5 Specific Volume

13

The base unit for length is the foot, ft, defined in terms of the meter as 1 ft 5 0.3048 m

(1.3)

The inch, in., is defined in terms of the foot 12 in. 5 1 ft One inch equals 2.54 cm. Although units such as the minute and the hour are often used in engineering, it is convenient to select the second as the English Engineering base unit for time. The English Engineering base unit of mass is the pound mass, lb, defined in terms of the kilogram as 1 lb 5 0.45359237 kg

(1.4)

The symbol lbm also may be used to denote the pound mass. Once base units have been specified for mass, length, and time in the English Engineering system of units, a force unit can be defined, as for the newton, using Newton’s second law written as Eq. 1.1. From this viewpoint, the English unit of force, the pound force, lbf, is the force required to accelerate one pound mass at 32.1740 ft/s2, which is the standard acceleration of gravity. Substituting values into Eq. 1.1 1 lbf 5 11 lb2132.1740 ft/ s22 5 32.1740 lb ? ft/ s2

(1.5)

With this approach force is regarded as secondary. The pound force, lbf, is not equal to the pound mass, lb, introduced previously. Force and mass are fundamentally different, as are their units. The double use of the word “pound” can be confusing, however, and care must be taken to avoid error. to show the use of these units in a single calculation, let us determine the weight of an object whose mass is 1000 lb at a location where the local acceleration of gravity is 32.0 ft/s2. By inserting values into Eq. 1.1 and using Eq. 1.5 as a unit conversion factor, we get ft 1 lbf F 5 mg 5 11000 lb2a32.0 2 b ` ` 5 994.59 lbf s 32.1740 lb ? ft/ s2 This calculation illustrates that the pound force is a unit of force distinct from the pound mass, a unit of mass. b b b b b

1.5 Specific Volume Three measurable intensive properties that are particularly important in engineering thermodynamics are specific volume, pressure, and temperature. Specific volume is considered in this section. Pressure and temperature are considered in Secs. 1.6 and 1.7, respectively. From the macroscopic perspective, the description of matter is simplified by considering it to be distributed continuously throughout a region. The correctness of this idealization, known as the continuum hypothesis, is inferred from the fact that for an extremely large class of phenomena of engineering interest the resulting description of the behavior of matter is in agreement with measured data. When substances can be treated as continua, it is possible to speak of their intensive thermodynamic properties “at a point.” Thus, at any instant the density r at a point is defined as m r 5 lim a b VSV¿ V

(1.6)

where V 9 is the smallest volume for which a definite value of the ratio exists. The volume V 9 contains enough particles for statistical averages to be significant. It is the smallest

Ext_Int_Properties A.3 – Tabs b & c

A

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Chapter 1 Getting Started volume for which the matter can be considered a continuum and is normally small enough that it can be considered a “point.” With density defined by Eq. 1.6, density can be described mathematically as a continuous function of position and time. The density, or local mass per unit volume, is an intensive property that may vary from point to point within a system. Thus, the mass associated with a particular volume V is determined in principle by integration m5

specific volume

molar basis

# r dV V

(1.7)

and not simply as the product of density and volume. The specific volume y is defined as the reciprocal of the density, y 5 1yr. It is the volume per unit mass. Like density, specific volume is an intensive property and may vary from point to point. SI units for density and specific volume are kg/m3 and m3/kg, respectively. However, they are also often expressed, respectively, as g/cm3 and cm3/g. English units used for density and specific volume in this text are lb/ft3 and ft3/lb, respectively. In certain applications it is convenient to express properties such as specific volume on a molar basis rather than on a mass basis. A mole is an amount of a given substance numerically equal to its molecular weight. In this book we express the amount of substance on a molar basis in terms of the kilomole (kmol) or the pound mole (lbmol), as appropriate. In each case we use n5

m M

(1.8)

The number of kilomoles of a substance, n, is obtained by dividing the mass, m, in kilograms by the molecular weight, M, in kg/kmol. Similarly, the number of pound moles, n, is obtained by dividing the mass, m, in pound mass by the molecular weight, M, in lb/lbmol. When m is in grams, Eq. 1.8 gives n in gram moles, or mol for short. Recall from chemistry that the number of molecules in a gram mole, called Avogadro’s number, is 6.022 3 1023. Appendix Tables A-1 and A-1E provide molecular weights for several substances. To signal that a property is on a molar basis, a bar is used over its symbol. Thus, y signifies the volume per kmol or lbmol, as appropriate. In this text, the units used for y are m3/kmol and ft3/lbmol. With Eq. 1.8, the relationship between y and y is y 5 My

(1.9)

where M is the molecular weight in kg/kmol or lb/lbmol, as appropriate.

1.6 Pressure

pressure

Next, we introduce the concept of pressure from the continuum viewpoint. Let us begin by considering a small area A passing through a point in a fluid at rest. The fluid on one side of the area exerts a compressive force on it that is normal to the area, Fnormal. An equal but oppositely directed force is exerted on the area by the fluid on the other side. For a fluid at rest, no other forces than these act on the area. The pressure p at the specified point is defined as the limit p 5 lim a ASA¿

A

Ext_Int_Properties A.3 – Tab d

Fnormal b A

(1.10)

where A9 is the area at the “point” in the same limiting sense as used in the definition of density.

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1.6 Pressure

15

Big Hopes For Nanotechnology Nanoscience is the study of molecules and molecular structures, called nanostructures, having one or more dimensions less than about 100 nanometers. One nanometer is one billionth of a meter: 1 nm 5 1029 m. To grasp this level of smallness, a stack of 10 hydrogen atoms would have a height of 1 nm, while a human hair has a diameter about 50,000 nm. Nanotechnology is the engineering of nanostructures into useful products. At the nanotechnology scale, behavior may differ from our macroscopic expectations. For example, the averaging used to assign property values at a point in the

continuum model may no longer apply owing to the interactions among the atoms under consideration. Also at these scales, the nature of physical phenomena such as current flow may depend explicitly on the physical size of devices. After many years of fruitful research, nanotechnology is now poised to provide new products with a broad range of uses, including implantable chemotherapy devices, biosensors for glucose detection in diabetics, novel electronic devices, new energy conversion technologies, and ‘smart materials’, as for example fabrics that allow water vapor to escape while keeping liquid water out.

If the area A9 was given new orientations by rotating it around the given point, and the pressure determined for each new orientation, it would be found that the pressure at the point is the same in all directions as long as the fluid is at rest. This is a consequence of the equilibrium of forces acting on an element of volume surrounding the point. However, the pressure can vary from point to point within a fluid at rest; examples are the variation of atmospheric pressure with elevation and the pressure variation with depth in oceans, lakes, and other bodies of water. Consider next a fluid in motion. In this case the force exerted on an area passing through a point in the fluid may be resolved into three mutually perpendicular components: one normal to the area and two in the plane of the area. When expressed on a unit area basis, the component normal to the area is called the normal stress, and the two components in the plane of the area are termed shear stresses. The magnitudes of the stresses generally vary with the orientation of the area. The state of stress in a fluid in motion is a topic that is normally treated thoroughly in fluid mechanics. The deviation of a normal stress from the pressure, the normal stress that would exist were the fluid at rest, is typically very small. In this book we assume that the normal stress at a point is equal to the pressure at that point. This assumption yields results of acceptable accuracy for the applications considered. Also, the term pressure, unless stated otherwise, refers to absolute pressure: pressure with respect to the zero pressure of a complete vacuum.

absolute pressure patm Gas at pressure p

L

Tank a

b

Manometer liquid

1.6.1 Pressure Measurement

Fig. 1.7 Manometer.

Manometers and barometers measure pressure in terms of the length of a column of liquid such as mercury, water, or oil. The manometer shown in Fig. 1.7 has one end open to the atmosphere and the other attached to a tank containing a gas at a uniform pressure. Since pressures at equal elevations in a continuous mass of a liquid or gas at rest are equal, the pressures at points a and b of Fig. 1.7 are equal. Applying an elementary force balance, the gas pressure is p 5 patm 1 rgL

Mercury vapor, pvapor

L

patm

(1.11)

where patm is the local atmospheric pressure, r is the density of the manometer liquid, g is the acceleration of gravity, and L is the difference in the liquid levels. The barometer shown in Fig. 1.8 is formed by a closed tube filled with liquid mercury and a small amount of mercury vapor inverted in an open container of liquid mercury. Since the pressures at points a and b are equal, a force balance gives the

b Mercury, ρm

Fig. 1.8 Barometer.

a

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Chapter 1 Getting Started

Elliptical metal Bourdon tube

Pointer

atmospheric pressure as patm 5 pvapor 1 rmgL

Pinion gear Support Linkage

where rm is the density of liquid mercury. Because the pressure of the mercury vapor is much less than that of the atmosphere, Eq. 1.12 can be approximated closely as patm 5 rmgL. For short columns of liquid, r and g in Eqs. 1.11 and 1.12 may be taken as constant. Pressures measured with manometers and barometers are frequently expressed in terms of the length L in millimeters of mercury (mmHg), inches of mercury (inHg), inches of water (inH2O), and so on. a barometer reads 750 mmHg. If rm 5 13.59 g/cm3 and g 5 9.81 m/s , the atmospheric pressure, in N/m2, is calculated as follows: 2

Gas at pressure p

Fig. 1.9 Pressure measurement by a Bourdon tube gage.

(1.12)

patm 5 rmgL 5 c a13.59

1 kg 102 cm 3 m 1m 1N `d ` b ` `` ` d c9.81 2 d c 1750 mmHg2 ` 3 ` 3 3 s 10 mm 1 kg ? m/ s2 cm 10 g 1 m

5 105 Nym2

g

b b b b b

A Bourdon tube gage is shown in Fig. 1.9. The figure shows a curved tube having an elliptical cross section with one end attached to the pressure to be measured and the other end connected to a pointer by a mechanism. When fluid under pressure fills the tube, the elliptical section tends to become circular, and the tube straightens. This motion is transmitted by the mechanism to the pointer. By calibrating the deflection of the pointer for known pressures, a graduated scale can be determined from which any applied pressure can be read in suitable units. Because of its construction, the Bourdon tube measures the pressure relative to the pressure of the surroundings existing at the instrument. Accordingly, the dial reads zero when the inside and outside of the tube are at the same pressure. Pressure can be measured by other means as well. An important class of sensors utilize the piezoelectric effect: A charge is generated within certain solid materials when they are deformed. This mechanical input/electrical output provides the basis for pressure measurement as well as displacement and force measurements. Another important type of sensor employs a diaphragm that deflects when a force is applied, altering an inductance, resistance, or capacitance. Fig. 1.10 Pressure sensor with automatic data Figure 1.10 shows a piezoelectric pressure sensor together with an acquisition. automatic data acquisition system.

1.6.2 Buoyancy buoyant force

When a body is completely, or partially, submerged in a liquid, the resultant pressure force acting on the body is called the buoyant force. Since pressure increases with depth from the liquid surface, pressure forces acting from below are greater than pressure forces acting from above; thus the buoyant force acts vertically upward. The buoyant force has a magnitude equal to the weight of the displaced liquid (Archimedes’ principle). applying Eq. 1.11 to the submerged rectangular block shown in Fig. 1.11, the magnitude of the net force of pressure acting upward, the buoyant

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1.6 Pressure

17

patm

force, is F 5 A1p2 2 p12 5 A1patm 1 rgL22 2 A1patm 1 rgL12 5 rgA1L2 2 L12 5 rgV

Liquid with density ρ

where V is the volume of the block and r is the density of the surrounding liquid. Thus, the magnitude of the buoyant force acting on the block is equal to the weight of the displaced liquid. b b b b b

p1A

L1 L2

Block

1.6.3 Pressure Units

Area = A p2A

The SI unit of pressure and stress is the pascal. 1 pascal 5 1 N/ m2

Fig. 1.11 Evaluation of buoyant force for a submerged body.

However, multiples of the pascal: the kPa, the bar, and the MPa are frequently used. 1 kPa 5 103 N/ m2 1 bar 5 105 N/ m2 1 MPa 5 106 N/ m2 Commonly used English units for pressure and stress are pounds force per square foot, lbf/ft2, and pounds force per square inch, lbf/in.2 Although atmospheric pressure varies with location on the earth, a standard reference value can be defined and used to express other pressures. 1.01325 3 105 N/ m2 1 standard atmosphere 1atm2 5 • 14.696 lbf/ in.2 760 mmHg 5 29.92 inHg

(1.13)

Since 1 bar (105 N/m2) closely equals one standard atmosphere, it is a convenient pressure unit despite not being a standard SI unit. When working in SI, the bar, MPa, and kPa are all used in this text. Although absolute pressures must be used in thermodynamic relations, pressuremeasuring devices often indicate the difference between the absolute pressure of a system and the absolute pressure of the atmosphere existing outside the measuring device. The magnitude of the difference is called a gage pressure or a vacuum pressure. The term gage pressure is applied when the pressure of the system is greater than the local atmospheric pressure, patm. p1gage2 5 p1absolute2 2 patm1absolute2

(1.14)

When the local atmospheric pressure is greater than the pressure of the system, the term vacuum pressure is used. p1vacuum2 5 patm1absolute2 2 p1absolute2

gage pressure vacuum pressure

(1.15)

Engineers in the United States frequently use the letters a and g to distinguish between absolute and gage pressures. For example, the absolute and gage pressures in pounds force per square inch are written as psia and psig, respectively. The relationship among the various ways of expressing pressure measurements is shown in Fig. 1.12.

TAKE NOTE...

In this book, the term pressure refers to absolute pressure unless indicated otherwise.

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p (gage)

Absolute pressure that is greater than the local atmospheric pressure

Atmospheric pressure

p (vacuum)

p (absolute)

Absolute pressure that is less than the local atmospheric pressure

patm (absolute)

p (absolute)

Zero pressure

Zero pressure

Fig. 1.12 Relationships among the absolute, atmospheric, gage, and vacuum pressures.

BIOCONNECTIONS One in three Americans is said to have high blood pressure. Since this can lead to heart disease, strokes, and other serious medical complications, medical practitioners recommend regular blood pressure checks for everyone. Blood pressure measurement aims to determine the maximum pressure (systolic pressure) in an artery when the heart is pumping blood and the minimum pressure (diastolic pressure) when the heart is resting, each pressure expressed in millimeters of mercury, mmHg. The systolic and diastolic pressures of healthy persons should be less than about 120 mmHg and 80 mmHg, respectively. The standard blood pressure measurement apparatus in use for decades involving an inflatable cuff, mercury manometer, and stethoscope is gradually being replaced because of concerns over mercury toxicity and in response to special requirements, including monitoring during clinical exercise and during anesthesia. Also, for home use and self-monitoring, many patients prefer easy-to-use automated devices that provide digital displays of blood pressure data. This has prompted biomedical engineers to rethink blood pressure measurement and develop new mercury-free and stethoscope-free approaches. One of these uses a highly-sensitive pressure transducer to detect pressure oscillations within an inflated cuff placed around the patient’s arm. The monitor’s software uses these data to calculate the systolic and diastolic pressures, which are displayed digitally.

1.7 Temperature In this section the intensive property temperature is considered along with means for measuring it. A concept of temperature, like our concept of force, originates with our sense perceptions. Temperature is rooted in the notion of the “hotness” or “coldness” of objects. We use our sense of touch to distinguish hot objects from cold objects and to arrange objects in their order of “hotness,” deciding that 1 is hotter than 2, 2 hotter

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1.7 Temperature than 3, and so on. But however sensitive human touch may be, we are unable to gauge this quality precisely. A definition of temperature in terms of concepts that are independently defined or accepted as primitive is difficult to give. However, it is possible to arrive at an objective understanding of equality of temperature by using the fact that when the temperature of an object changes, other properties also change. To illustrate this, consider two copper blocks, and suppose that our senses tell us that one is warmer than the other. If the blocks were brought into contact and isolated from their surroundings, they would interact in a way that can be described as a thermal (heat) interaction. During this interaction, it would be observed that the volume of the warmer block decreases somewhat with time, while the volume of the colder block increases with time. Eventually, no further changes in volume would be observed, and the blocks would feel equally warm. Similarly, we would be able to observe that the electrical resistance of the warmer block decreases with time, and that of the colder block increases with time; eventually the electrical resistances would become constant also. When all changes in such observable properties cease, the interaction is at an end. The two blocks are then in thermal equilibrium. Considerations such as these lead us to infer that the blocks have a physical property that determines whether they will be in thermal equilibrium. This property is called temperature, and we postulate that when the two blocks are in thermal equilibrium, their temperatures are equal. It is a matter of experience that when two objects are in thermal equilibrium with a third object, they are in thermal equilibrium with one another. This statement, which is sometimes called the zeroth law of thermodynamics, is tacitly assumed in every measurement of temperature. Thus, if we want to know if two objects are at the same temperature, it is not necessary to bring them into contact and see whether their observable properties change with time, as described previously. It is necessary only to see if they are individually in thermal equilibrium with a third object. The third object is usually a thermometer.

Ext_Int_Properties A.3 – Tab e

A

thermal (heat) interaction

thermal equilibrium

temperature

zeroth law of thermodynamics

1.7.1 Thermometers Any object with at least one measurable property that changes as its temperature changes can be used as a thermometer. Such a property is called a thermometric property. The particular substance that exhibits changes in the thermometric property is known as a thermometric substance. A familiar device for temperature measurement is the liquid-in-glass thermometer pictured in Fig. 1.13a, which consists of a glass capillary tube connected to a bulb filled with a liquid such as alcohol and sealed at the other end. The space above the liquid is occupied by the vapor of the liquid or an inert gas. As temperature increases, the liquid expands in volume and rises in the capillary. The length L of the liquid in the capillary depends on the temperature. Accordingly, the liquid is the thermometric substance and L is the thermometric property. Although this type of thermometer is commonly used for ordinary temperature measurements, it is not well suited for applications where extreme accuracy is required. More accurate sensors known as thermocouples are based on the principle that when two dissimilar metals are joined, an electromotive force (emf) that is primarily a function of temperature will exist in a circuit. In certain thermocouples, one thermocouple wire is platinum of a specified purity and the other is an alloy of platinum and rhodium. Thermocouples also utilize copper and constantan (an alloy of copper and nickel), iron and constantan, as well as several other pairs of materials. Electricalresistance sensors are another important class of temperature measurement devices. These sensors are based on the fact that the electrical resistance of various materials changes in a predictable manner with temperature. The materials used for this purpose are normally conductors (such as platinum, nickel, or copper) or semiconductors.

19

thermometric property

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L

Liquid (a)

(b)

(c)

Fig. 1.13 Thermometers. (a) Liquid-in-glass. (b) Electrical-resistance (c) Infrared-sensing ear thermometer. Devices using conductors are known as resistance temperature detectors. Semiconductor types are called thermistors. A battery-powered electrical-resistance thermometer commonly used today is shown in Fig. 1.13b. A variety of instruments measure temperature by sensing radiation, such as the ear thermometer shown in Fig. 1.13c. They are known by terms such as radiation thermometers and optical pyrometers. This type of thermometer differs from those previously considered because it is not required to come in contact with the object whose temperature is to be determined, an advantage when dealing with moving objects or objects at extremely high temperatures.

ENERGY & ENVIRONMENT The mercury-in-glass fever thermometers, once found in nearly every medicine cabinet, are a thing of the past. The American Academy of Pediatrics has designated mercury as too toxic to be present in the home. Families are turning to safer alternatives and disposing of mercury thermometers. Proper disposal is an issue, experts say. The safe disposal of millions of obsolete mercury-filled thermometers has emerged in its own right as an environmental issue. For proper disposal, thermometers must be taken to hazardouswaste collection stations rather than simply thrown in the trash where they can be easily broken, releasing mercury. Loose fragments of broken thermometers and anything that contacted mercury should be transported in closed containers to appropriate disposal sites. The present generation of liquid-in-glass fever thermometers for home use contains patented liquid mixtures that are nontoxic, safe alternatives to mercury. Other types of thermometers also are used in the home, including battery-powered electrical-resistance thermometers.

1.7.2 Kelvin and Rankine Temperature Scales Empirical means of measuring temperature such as considered in Sec. 1.7.1 have inherent limitations. the tendency of the liquid in a liquid-in-glass thermometer to freeze at low temperatures imposes a lower limit on the range of temperatures that can be

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1.7 Temperature measured. At high temperatures liquids vaporize, and therefore these temperatures also cannot be determined by a liquid-in-glass thermometer. Accordingly, several different thermometers might be required to cover a wide temperature interval. b b b b b In view of the limitations of empirical means for measuring temperature, it is desirable to have a procedure for assigning temperature values that does not depend on the properties of any particular substance or class of substances. Such a scale is called a thermodynamic temperature scale. The Kelvin scale is an absolute thermodynamic temperature scale that provides a continuous definition of temperature, valid over all ranges of temperature. The unit of temperature on the Kelvin scale is the kelvin (K). The kelvin is the SI base unit for temperature. To develop the Kelvin scale, it is necessary to use the conservation of energy principle and the second law of thermodynamics; therefore, further discussion is deferred to Sec. 5.8 after these principles have been introduced. However, we note here that the Kelvin scale has a zero of 0 K, and lower temperatures than this are not defined. By definition, the Rankine scale, the unit of which is the degree rankine (°R), is proportional to the Kelvin temperature according to T18R2 5 1.8T1K2

Kelvin scale

Rankine scale

(1.16)

As evidenced by Eq. 1.16, the Rankine scale is also an absolute thermodynamic scale with an absolute zero that coincides with the absolute zero of the Kelvin scale. In thermodynamic relationships, temperature is always in terms of the Kelvin or Rankine scale unless specifically stated otherwise. Still, the Celsius and Fahrenheit scales considered next are commonly encountered.

1.7.3 Celsius and Fahrenheit Scales The relationship of the Kelvin, Rankine, Celsius, and Fahrenheit scales is shown in Fig. 1.14 together with values for temperature at three fixed points: the triple point, ice point, and steam point. By international agreement, temperature scales are defined by the numerical value assigned to the easily reproducible triple point of water: the state of equilibrium between

°F 212

671.67

°R

100.0

°C

373.15

K

–459.67

0.00

Rankine

Fahrenheit

32.0

32.02

491.69 491.67

0.01 0.00 –273.15

Celsius 0.00

Kelvin

Ice point

273.15

Triple point of water

273.16

Steam point

Absolute zero

Fig. 1.14 Comparison of temperature scales.

triple point

21

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Chapter 1 Getting Started

Celsius scale

steam, ice, and liquid water (Sec. 3.2). As a matter of convenience, the temperature at this standard fixed point is defined as 273.16 kelvins, abbreviated as 273.16 K. This makes the temperature interval from the ice point 1 (273.15 K) to the steam point 2 equal to 100 K and thus in agreement over the interval with the Celsius scale, which assigns 100 Celsius degrees to it. The Celsius temperature scale uses the unit degree Celsius (°C), which has the same magnitude as the kelvin. Thus, temperature differences are identical on both scales. However, the zero point on the Celsius scale is shifted to 273.15 K, as shown by the following relationship between the Celsius temperature and the Kelvin temperature T18C2 5 T1K2 2 273.15

Fahrenheit scale

(1.17)

From this it can be concluded that on the Celsius scale the triple point of water is 0.01°C and that 0 K corresponds to −273.15°C. These values are shown on Fig. 1.14. A degree of the same size as that on the Rankine scale is used in the Fahrenheit scale, but the zero point is shifted according to the relation T18F2 5 T18R2 2 459.67

(1.18)

TAKE NOTE...

When making engineering calculations, it’s usually okay to round off the last numbers in Eqs. 1.17 and 1.18 to 273 and 460, respectively. This is frequently done in this book.

Substituting Eqs. 1.17 and 1.18 into Eq. 1.16, we get T18F2 5 1.8T18C2 1 32

(1.19)

This equation shows that the Fahrenheit temperature of the ice point (0°C) is 32°F and of the steam point (100°C) is 212°F. The 100 Celsius or Kelvin degrees between the ice point and steam point correspond to 180 Fahrenheit or Rankine degrees, as shown in Fig. 1.14.

BIOCONNECTIONS Cryobiology, the science of life at low temperatures, comprises the study of biological materials and systems (proteins, cells, tissues, and organs) at temperatures ranging from the cryogenic (below about 120 K) to the hypothermic (low body temperature). Applications include freeze-drying pharmaceuticals, cryosurgery for removing unhealthy tissue, study of cold-adaptation of animals and plants, and long-term storage of cells and tissues (called cryopreservation). Cryobiology has challenging engineering aspects owing to the need for refrigerators capable of achieving the low temperatures required by researchers. Freezers to support research requiring cryogenic temperatures in the low-gravity environment of the International Space Station, shown in Table 1.1, are illustrative. Such freezers must be extremely compact and miserly in power use. Further, they must pose no hazards. On-board research requiring a freezer might include the growth of near-perfect protein crystals, important for understanding the structure and function of proteins and ultimately in the design of new drugs.

1.8 Engineering Design and Analysis The word engineer traces its roots to the Latin ingeniare, relating to invention. Today invention remains a key engineering function having many aspects ranging from developing new devices to addressing complex social issues using technology. In pursuit of many such activities, engineers are called upon to design and analyze things intended to meet human needs. Design and analysis are considered in this section. 1

The state of equilibrium between ice and air-saturated water at a pressure of 1 atm. The state of equilibrium between steam and liquid water at a pressure of 1 atm.

2

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1.8 Engineering Design and Analysis

1.8.1 Design Engineering design is a decision-making process in which principles drawn from engineering and other fields such as economics and statistics are applied, usually iteratively, to devise a system, system component, or process. Fundamental elements of design include the establishment of objectives, synthesis, analysis, construction, testing, and evaluation. Designs typically are subject to a variety of constraints related to economics, safety, environmental impact, and so on. Design projects usually originate from the recognition of a need or an opportunity that is only partially understood initially. Thus, before seeking solutions it is important to define the design objectives. Early steps in engineering design include pinning down quantitative performance specifications and identifying alternative workable designs that meet the specifications. Among the workable designs are generally one or more that are “best” according to some criteria: lowest cost, highest efficiency, smallest size, lightest weight, etc. Other important factors in the selection of a final design include reliability, manufacturability, maintainability, and marketplace considerations. Accordingly, a compromise must be sought among competing criteria, and there may be alternative design solutions that are feasible.3

design constraints

1.8.2 Analysis Design requires synthesis: selecting and putting together components to form a coordinated whole. However, as each individual component can vary in size, performance, cost, and so on, it is generally necessary to subject each to considerable study or analysis before a final selection can be made. a proposed design for a fire-protection system might entail an overhead piping network together with numerous sprinkler heads. Once an overall configuration has been determined, detailed engineering analysis is necessary to specify the number and type of the spray heads, the piping material, and the pipe diameters of the various branches of the network. The analysis also must aim to ensure all components form a smoothly working whole while meeting relevant cost constraints and applicable codes and standards. b b b b b Engineers frequently do analysis, whether explicitly as part of a design process or for some other purpose. Analyses involving systems of the kind considered in this book use, directly or indirectly, one or more of three basic laws. These laws, which are independent of the particular substance or substances under consideration, are 1. the conservation of mass principle 2. the conservation of energy principle 3. the second law of thermodynamics In addition, relationships among the properties of the particular substance or substances considered are usually necessary (Chaps. 3, 6, 11–14). Newton’s second law of motion (Chaps. 1, 2, 9), relations such as Fourier’s conduction model (Chap. 2), and principles of engineering economics (Chap. 7) also may play a part. The first steps in a thermodynamic analysis are definition of the system and identification of the relevant interactions with the surroundings. Attention then turns to the pertinent physical laws and relationships that allow the behavior of the system to be described in terms of an engineering model. The objective in modeling is to obtain a simplified representation of system behavior that is sufficiently faithful for the purpose 3

For further discussion, see A. Bejan, G. Tsatsaronis, and M. J. Moran, Thermal Design and Optimization, John Wiley & Sons, New York, 1996, Chap. 1.

engineering model

23

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Chapter 1 Getting Started of the analysis, even if many aspects exhibited by the actual system are ignored. For example, idealizations often used in mechanics to simplify an analysis and arrive at a manageable model include the assumptions of point masses, frictionless pulleys, and rigid beams. Satisfactory modeling takes experience and is a part of the art of engineering. Engineering analysis is most effective when it is done systematically. This is considered next.

1.9 Methodology for Solving Thermodynamics Problems A major goal of this textbook is to help you learn how to solve engineering problems that involve thermodynamic principles. To this end, numerous solved examples and endof-chapter problems are provided. It is extremely important for you to study the examples and solve problems, for mastery of the fundamentals comes only through practice. To maximize the results of your efforts, it is necessary to develop a systematic approach. You must think carefully about your solutions and avoid the temptation of starting problems in the middle by selecting some seemingly appropriate equation, substituting in numbers, and quickly “punching up” a result on your calculator. Such a haphazard problemsolving approach can lead to difficulties as problems become more complicated. Accordingly, it is strongly recommended that problem solutions be organized using the five steps in the box below, which are employed in the solved examples of this text. cccc

➊ Known: State briefly in your own words what is known. This requires that you read the problem carefully and think about it. ➋ Find: State concisely in your own words what is to be determined. ➌ Schematic and Given Data: Draw a sketch of the system to be considered. Decide whether a closed system or control volume is appropriate for the analysis, and then carefully identify the boundary. Label the diagram with relevant information from the problem statement. Record all property values you are given or anticipate may be required for subsequent calculations. Sketch appropriate property diagrams (see Sec. 3.2), locating key state points and indicating, if possible, the processes executed by the system. The importance of good sketches of the system and property diagrams cannot be overemphasized. They are often instrumental in enabling you to think clearly about the problem. ➍ Engineering Model: To form a record of how you model the problem, list all simplifying assumptions and idealizations made to reduce it to one that is manageable. Sometimes this information also can be noted on the sketches of the previous step. The development of an appropriate model is a key aspect of successful problem solving. ➎ Analysis: Using your assumptions and idealizations, reduce the appropriate governing equations and relationships to forms that will produce the desired results. It is advisable to work with equations as long as possible before substituting numerical data. When the equations are reduced to final forms, consider them to determine what additional data may be required. Identify the tables, charts, or property equations that provide the required values. Additional property diagram sketches may be helpful at this point to clarify states and processes. When all equations and data are in hand, substitute numerical values into the equations. Carefully check that a consistent and appropriate set of units is being employed. Then perform the needed calculations. Finally, consider whether the magnitudes of the numerical values are reasonable and the algebraic signs associated with the numerical values are correct.

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1.9 Methodology for Solving Thermodynamics Problems

25

The problem solution format used in this text is intended to guide your thinking, not substitute for it. Accordingly, you are cautioned to avoid the rote application of these five steps, for this alone would provide few benefits. Indeed, as a particular solution evolves you may have to return to an earlier step and revise it in light of a better understanding of the problem. For example, it might be necessary to add or delete an assumption, revise a sketch, determine additional property data, and so on. The solved examples provided in the book are frequently annotated with various comments intended to assist learning, including commenting on what was learned, identifying key aspects of the solution, and discussing how better results might be obtained by relaxing certain assumptions. In some of the earlier examples and end-of-chapter problems, the solution format may seem unnecessary or unwieldy. However, as the problems become more complicated you will see that it reduces errors, saves time, and provides a deeper understanding of the problem at hand. The example to follow illustrates the use of this solution methodology together with important system concepts introduced previously, including identification of interactions occurring at the boundary.

cccc

EXAMPLE 1.1 c

Using the Solution Methodology and System Concepts A wind turbine–electric generator is mounted atop a tower. As wind blows steadily across the turbine blades, electricity is generated. The electrical output of the generator is fed to a storage battery. (a) Considering only the wind turbine–electric generator as the system, identify locations on the system boundary

where the system interacts with the surroundings. Describe changes occurring within the system with time. (b) Repeat for a system that includes only the storage battery. SOLUTION Known: A wind turbine–electric generator provides electricity to a storage battery. Find: For a system consisting of (a) the wind turbine–electric generator, (b) the storage battery, identify locations

where the system interacts with its surroundings, and describe changes occurring within the system with time. Schematic and Given Data: Engineering Model:

Part (a)

1. In part (a), the system is the control volume shown by

the dashed line on the figure. Air flow

Turbine–generator

2. In part (b), the system is the closed system shown by the

dashed line on the figure. 3. The wind is steady.

Electric current flow

Part (b) Storage battery

Thermal interaction

Fig. E1.1 Analysis: (a) In this case, the wind turbine is studied as a control volume with air flowing across the boundary. Another

principal interaction between the system and surroundings is the electric current passing through the wires. From the macroscopic perspective, such an interaction is not considered a mass transfer, however. With a

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Chapter 1 Getting Started steady wind, the turbine–generator is likely to reach steady-state operation, where the rotational speed of the blades is constant and a steady electric current is generated. ➊ (b) In this case, the battery is studied as a closed system. The principal interaction between the system and its surroundings is the electric current passing into the battery through the wires. As noted in part (a), this interaction is not considered a mass transfer. As the battery is charged and chemical reactions occur within it, the temperature of the battery surface may become somewhat elevated and a thermal interaction might occur between the battery and its surroundings. This interaction is likely to be of secondary importance. Also, as the battery is charged, the state within changes with time. The battery is not at steady state. ➊ Using terms familiar from a previous physics course, the system of part (a) involves the conversion of kinetic energy to electricity, whereas the system of part (b) involves energy storage within the battery.

✓ Skills Developed Ability to… ❑ apply the problem-solving

methodology used in this book. ❑ define a control volume and identify interactions on its boundary. ❑ define a closed system and identify interactions on its boundary. ❑ distinguish steady-state operation from nonsteady operation.

May an overall system consisting of the turbine-generator and battery be considered as operating at steady state? Explain. Ans. No. A system is at steady state only if none of its properties change with time.

c CHAPTER SUMMARY AND STUDY GUIDE In this chapter, we have introduced some of the fundamental concepts and definitions used in the study of thermodynamics. The principles of thermodynamics are applied by engineers to analyze and design a wide variety of devices intended to meet human needs. An important aspect of thermodynamic analysis is to identify systems and to describe system behavior in terms of properties and processes. Three important properties discussed in this chapter are specific volume, pressure, and temperature. In thermodynamics, we consider systems at equilibrium states and systems undergoing processes (changes of state). We study processes during which the intervening states are not equilibrium states and processes during which the departure from equilibrium is negligible. In this chapter, we have introduced SI and English Engineering units for mass, length, time, force, and temperature. You will need to be familiar with both sets of units as you use this book. For Conversion Factors, see inside the front cover of the book. Chapter 1 concludes with discussions of how thermodynamics is used in engineering design and how to solve thermodynamics problems systematically.

This book has several features that facilitate study and contribute to understanding. For an overview, see How To Use This Book Effectively inside the front cover of the book. The following checklist provides a study guide for this chapter. When your study of the text and the end-of-chapter exercises has been completed you should be able to c write out the meanings of the terms listed in the margin

c

c c c

throughout the chapter and understand each of the related concepts. The subset of key concepts listed below is particularly important in subsequent chapters. use SI and English units for mass, length, time, force, and temperature and apply appropriately Newton’s second law and Eqs. 1.16–1.19. work on a molar basis using Eq. 1.8. identify an appropriate system boundary and describe the interactions between the system and its surroundings. apply the methodology for problem solving discussed in Sec. 1.9.

c KEY ENGINEERING CONCEPTS system, p. 4 surroundings, p. 4 boundary, p. 4 closed system, p. 6 control volume, p. 6 property, p. 9

state, p. 9 process, p. 9 extensive property, p. 9 intensive property, p. 9 equilibrium, p. 10 specific volume, p. 14

pressure, p. 14 temperature, p. 19 Kelvin scale, p. 21 Rankine scale, p. 21

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c KEY EQUATIONS n 5 m/M

(1.8) p. 14

Relation between amounts of matter on a mass basis, m, and on a molar basis, n.

T(8R) 5 1.8T(K)

(1.16) p. 21

Relation between the Rankine and Kelvin temperatures.

T(8C) 5 T(K) 2 273.15

(1.17) p. 22

Relation between the Celsius and Kelvin temperatures.

T(8F) 5 T(8R) 2 459.67

(1.18) p. 22

Relation between the Fahrenheit and Rankine temperatures.

T(8F) 5 1.8T(8C) 1 32

(1.19) p. 22

Relation between the Fahrenheit and Celsius temperatures.

c EXERCISES: THINGS ENGINEERS THINK ABOUT 1. In 1998, owing to a mix-up over units, the NASA Mars Climate Orbiter veered off-course and was lost. What was the mix-up? 2. Operating rooms in hospitals typically have a positive pressure relative to adjacent spaces. What does this mean and why is it done? 3. The driver’s compartment of race cars can reach 60°C during a race. Why? 4. You may have used the mass unit slug in previous engineering or physics courses. What is the relation between the slug and pound mass? Is the slug a convenient mass unit? 5. Based on the macroscopic view, a quantity of air at 100 kPa, 20°C is in equilibrium. Yet the atoms and molecules of the air are in constant motion. How do you reconcile this apparent contradiction? 6. Laura takes an elevator from the tenth floor of her office building to the lobby. Should she expect the air pressure on the two levels to differ much? 7. How do dermatologists remove pre-cancerous skin blemishes cryosurgically?

8. When one walks barefoot from a carpet onto a ceramic tile floor, the tiles feel colder than the carpet even though each surface is at the same temperature. Explain. 9. Air at 1 atm, 70°F in a closed tank adheres to the continuum hypothesis. Yet when sufficient air has been drawn from the tank, the hypothesis no longer applies to the remaining air. Why? 10. Are the systolic and diastolic pressures reported in blood pressure measurements absolute, gage, or vacuum pressures? 11. When the instrument panel of a car provides the outside air temperature, where is the temperature sensor located? 12. How does a pressure measurement of 14.7 psig differ from a pressure measurement of 14.7 psia? 13. What is a nanotube? 14. If a system is at steady state, does this mean intensive properties are uniform with position throughout the system or constant with time? Both uniform with position and constant with time? Explain.

c PROBLEMS: DEVELOPING ENGINEERING SKILLS Exploring System Concepts 1.1 Using the Internet, obtain information about the operation of an application listed or shown in Table 1.1. Obtain sufficient information to provide a full description of the application, together with relevant thermodynamic aspects. Present your findings in a memorandum. 1.2 As illustrated in Fig. P1.2, water circulates through a piping system, servicing various household needs. Considering the water heater as a system, identify locations on the system boundary where the system interacts with its surroundings and describe significant occurrences within the system. Repeat for the dishwasher and for the shower. Present your findings in a memorandum.

1.3 Reef aquariums such as shown in Fig. P1.3 are popular attractions. Such facilities employ a variety of devices, including heaters, pumps, filters, and controllers, to create a healthy environment for the living things residing in the aquarium, which typically include species of fish, together with corals, clams, and anemone. Considering a reef aquarium as a system, identify locations on the system boundary where the system interacts with its surroundings. Using the Internet, describe significant occurrences within the system, and comment on measures for the health and safety of the aquatic life. Present your findings in a memorandum.

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Chapter 1 Getting Started Working with Force and Mass 1.6 If Superman has a mass of 100 kg on his birth planet Krypton, where the acceleration of gravity is 25 m/s2, determine (a) his weight on Krypton, in N, and (b) his mass, in kg, and weight, in N, on Earth where g 5 9.81 m/s2. Hot

Shower + –

Dishwasher

1.8 A gas occupying a volume of 25 ft3 weighs 3.5 lbf on the moon, where the acceleration of gravity is 5.47 ft/s2. Determine its weight, in lbf, and density, in lb/ft3, on Mars, where g 5 12.86 ft/s2.

Cold Water meter

Hot water heater + Electric meter – Drain lines

Fig. P1.2

1.7 A person whose mass is 150 lb weighs 144.4 lbf. Determine (a) the local acceleration of gravity, in ft/s2, and (b) the person’s mass, in lb and weight, in lbf, if g 5 32.174 ft/s2.

1.9 Atomic and molecular weights of some common substances are listed in Appendix Tables A-1 and A-1E. Using data from the appropriate table, determine (a) the mass, in kg, of 20 kmol of each of the following: air, C, H2O, CO2. (b) the number of lbmol in 50 lb of each of the following: H2, N2, NH3, C3H8. 1.10 In severe head-on automobile accidents, a deceleration of 60 g’s or more (1 g 5 32.2 ft/s2) often results in a fatality. What force, in lbf, acts on a child whose mass is 50 lb, when subjected to a deceleration of 60 g’s? 1.11 At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates such that for each 4.7 lbf applied, the spring elongates one inch. If local acceleration of gravity is 32.2 ft/s2, what distance, in inches, did the spring elongate? 1.12 A spring compresses in length by 0.14 in. for every 1 lbf of applied force. Determine the mass of an object, in pounds mass, that causes a spring deflection of 1.8 in. The local acceleration of gravity 5 31 ft/s2.

Fig. P1.3 Working with Units 1.4 Perform the following unit conversions: (a) 1 L to in.3 (b) 650 J to Btu (c) 0.135 kW to ft ? lbf/s (d) 378 g/s to lb/min (e) 304 kPa to lbf/in.2 (f) 55 m3/h to ft3/s (g) 50 km/h to ft/s (h) 8896 N to ton (52000 lbf) 1.5 Perform the following unit conversions: 3

(a) 122 in. to L (b) 778.17 ft ? lbf to kJ (c) 100 hp to kW (d) 1000 lb/h to kg/s (e) 29.392 lbf/in.2 to bar (f) 2500 ft3/min to m3/s (g) 75 mile/h to km/h (h) 1 ton (52000 lbf) to N

1.13 At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g 5 32.05 ft/s2, what is her weight, in lbf, and mass, in lb? 1.14 Estimate the magnitude of the force, in lbf, exerted on a 12-lb goose in a collision of duration 10−3 s with an airplane taking off at 150 miles/h. 1.15 Determine the upward applied force, in lbf, required to accelerate a 4.5-lb model rocket vertically upward, as shown in Fig. P1.15, with an acceleration of 3 g’s. The only other significant force acting on the rocket is gravity, and 1 g 5 32.2 ft/s2. 1.16 An object is subjected to an applied upward force of 10 lbf. The only other force acting on the object is the force of gravity. The acceleration of gravity is 32.2 ft/s2. If the object has a mass of 50 lb, determine the net acceleration of the object, in ft/s2. Is the net acceleration upward or downward? 1.17 An astronaut weighs 700 N on Earth where g 5 9.81 m/s2. What is the astronaut’s weight, in N, on an orbiting space station where the acceleration of gravity is 6 m/s2? Express each weight in lbf.

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m = 4.5 lb a = 3g

29

1.24 The pressure of the gas contained in the piston–cylinder assembly of Fig. 1.1 varies with its volume according to p 5 A 1 (B/V), where A, B are constants. If pressure is in lbf/ft2 and volume is in ft3, what are the units of A and B? 1.25 As shown in Figure P1.25, a gas is contained in a piston– cylinder assembly. The piston mass and cross-sectional area are denoted m and A, respectively. The only force acting on the top of the piston is due to atmospheric pressure, patm. Assuming the piston moves smoothly in the cylinder and the local acceleration of gravity g is constant, show that the pressure of the gas acting on the bottom of the piston remains constant as gas volume varies. What would cause the gas volume to vary? Piston m, A

patm

Gas pgas

Fig. P1.25

Fig. P1.15 1.18 Using local acceleration of gravity data from the Internet, determine the weight, in N, of a person whose mass is 80 kg living in: (a) Mexico City, Mexico (b) Cape Town, South Africa (c) Tokyo, Japan (d) Chicago, IL (e) Copenhagen, Denmark

1.26 As shown in Fig. P1.26, a vertical piston–cylinder assembly containing a gas is placed on a hot plate. The piston initially rests on the stops. With the onset of heating, the gas pressure increases. At what pressure, in bar, does the piston start rising? The piston moves smoothly in the cylinder and g 5 9.81 m/s2. patm = 1 bar

1.19 A town has a 1-million-gallon storage capacity water tower. If the density of water is 62.4 lb/ft3 and local acceleration of gravity is 32.1 ft/s2, what is the force, in lbf, the structural base must provide to support the water in the tower?

Stops

1.21 A spherical balloon holding 35 lb of air has a diameter of 10 ft. For the air, determine (a) the specific volume, in ft3/lb and ft3/lbmol, and (b) the weight, in lbf. Let g 5 31.0 ft/s2. 1.22 A closed vessel having a volume of 1 liter holds 2.5 3 1022 molecules of ammonia vapor. For the ammonia, determine (a) the amount present, in kg and kmol, and (b) the specific volume, in m3/kg and m3/kmol. 1.23 The specific volume of water vapor at 0.3 MPa, 160°C is 0.651 m3/kg. If the water vapor occupies a volume of 2 m3, determine (a) the amount present, in kg and kmol, and (b) the number of molecules.

m = 50 kg A = 0.01 m2

Gas

Using Specific Volume, Volume, and Pressure 1.20 A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.

Piston

+

–

Hot plate

Fig. P1.26 1.27 Three kg of gas in a piston–cylinder assembly undergo a process during which the relationship between pressure and specific volume is py0.5 5 constant. The process begins with p1 5 250 kPa and V1 5 1.5 m3 and ends with p2 5 100 kPa. Determine the final specific volume, in m3/kg. Plot the process on a graph of pressure versus specific volume. 1.28 A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 constant. The process begins with p1 5 20 lbf/in.2, V1 5 10 ft3 and ends with p2 5 100 lbf/in.2, V2 5 2.9 ft3. Determine (a) the value of n and (b) the specific volume at

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Chapter 1 Getting Started

states 1 and 2, each in ft3/lb. (c) Sketch the process on pressurevolume coordinates.

mercury is 13.59 g/cm3, and g 5 9.81 m/s2, determine the pressure of the natural gas, in kPa.

1.29 A system consists of carbon monoxide (CO) in a piston– cylinder assembly, initially at p1 5 200 lbf/in.2, and occupying a volume of 2.0 m3. The carbon monoxide expands to p2 5 40 lbf/in.2 and a final volume of 3.5 m3. During the process, the relationship between pressure and volume is linear. Determine the volume, in ft3, at an intermediate state where the pressure is 150 lbf/in.2, and sketch the process on a graph of pressure versus volume. 1.30 Figure P1.30 shows a gas contained in a vertical piston– cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g 5 9.81 m/s2. F A = 0.8 cm2

Natural gas

Instrument room

Fig. P1.33 1.34 As shown in Figure P1.34, the exit of a gas compressor empties into a receiver tank, maintaining the tank contents at a pressure of 200 kPa. If the local atmospheric pressure is 1 bar, what is the reading of the Bourdon gage mounted on the tank wall in kPa? Is this a vacuum pressure or a gage pressure? Explain.

Shaft

patm = 1 bar

patm = 1 bar

Piston

D = 10 cm

Inlet

Gas compressor

Exit

Receiver tank at 200 kPa

Gas at p = 3 bar

Fig. P1.34 Fig. P1.30 1.31 A gas contained within a piston–cylinder assembly undergoes three processes in series:

1.35 The barometer shown in Fig. P1.35 contains mercury (r 5 13.59 g/cm3). If the local atmospheric pressure is 100 kPa and g 5 9.81 m/s2, determine the height of the mercury column, L, in mmHg and inHg.

Process 1–2: Compression with pV 5 constant from p1 5 1 bar, V1 5 1.0 m3 to V2 5 0.2 m3

Mercury vapor

Process 2–3: Constant-pressure expansion to V3 5 1.0 m3 Process 3–1: Constant volume Sketch the processes in series on a p–V diagram labeled with pressure and volume values at each numbered state. 1.32 Referring to Fig. 1.7, (a) if the pressure in the tank is 1.5 bar and atmospheric pressure is 1 bar, determine L, in m, for water with a density of 997 kg/m3 as the manometer liquid. Let g 5 9.81 m/s2. (b) determine L, in cm, if the manometer liquid is mercury with a density of 13.59 g/cm3 and the gas pressure is 1.3 bar. A barometer indicates the local atmospheric pressure is 750 mmHg. Let g 5 9.81 m/s2. 1.33 Figure P1.33 shows a storage tank holding natural gas. In an adjacent instrument room, a U-tube mercury manometer in communication with the storage tank reads L 5 1.0 m. If the atmospheric pressure is 101 kPa, the density of the

L patm = 100 kPa

Liquid mercury, ρ m = 13.59 g/cm3

Fig. P1.35 1.36 Water flows through a Venturi meter, as shown in Fig. P1.36. The pressure of the water in the pipe supports columns of water that differ in height by 10 in. Determine the difference in pressure between points a and b, in lbf/in.2 Does the

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patm = 1 atm, g = 32.2 ft/s2

pressure increase or decrease in the direction of flow? The atmospheric pressure is 14.7 lbf/in.2, the specific volume of water is 0.01604 ft3/lb, and the acceleration of gravity is g 5 32.0 ft/s2.

1000 ft patm = 14.7 lbf/in.2 g = 32.0 ft/s2 L = 10 in.

Water v = 0.01604 ft3/lb

b

a

Fig. P1.36

1.37 Figure P1.37 shows a tank within a tank, each containing air. The absolute pressure in tank A is 267.7 kPa. Pressure gage A is located inside tank B and reads 140 kPa. The Utube manometer connected to tank B contains mercury. Using data on the diagram, determine the absolute pressure inside tank B, in kPa, and the column length L, in cm. The atmospheric pressure surrounding tank B is 101 kPa. The acceleration of gravity is g 5 9.81 m/s2.

Tank B

patm = 101 kPa

Fig. P1.38 1.40 A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the gas pressure at the inlet is 5.5 psig, what is the absolute pressure, in psia, of the gas at the exit? Atmospheric pressure is 14.5 lbf/in.2 1.41 As shown in Fig. P1.41, air is contained in a vertical pistoncylinder assembly fitted with an electrical resistor.The atmosphere exerts a pressure of 14.7 lbf/in.2 on the top of the piston, which has a mass of 100 lb and face area of 1 ft2. As electric current passes through the resistor, the volume of the air increases while the piston moves smoothly in the cylinder. The local acceleration of gravity is g 5 32.0 ft/s2. Determine the pressure of the air in the piston-cylinder assembly, in lbf/in.2 and psig. Piston

L Tank A, pA = 267.7 kPa

patm = 14.7 lbf/in.2 mpiston = 100 lb Apiston = 1 ft2

Gage A + Air

pgage, A = 140 kPa

Mercury ( ρ = 13.59 g/cm3) g = 9.81 m/s2

–

Fig. P1.37 Fig. P1.41 1.38 As shown in Fig. P1.38, an underwater exploration vehicle submerges to a depth of 1000 ft. If the atmospheric pressure at the surface is 1 atm, the water density is 62.4 lb/ft3, and g 5 32.2 ft/s2, determine the pressure on the vehicle, in atm. 1.39 A vacuum gage indicates that the pressure of carbon dioxide in a closed tank is −10 kPa. A mercury barometer gives the local atmospheric pressure as 750 mmHg. Determine the absolute pressure of the carbon dioxide, in kPa. The density of mercury is 13.59 g/cm3 and g 5 9.81 m/s2.

1.42 Warm air is contained in a piston-cylinder assembly oriented horizontally as shown in Fig. P1.42. The air cools slowly from an initial volume of 0.003 m3 to a final volume of 0.002 m3. During the process, the spring exerts a force that varies linearly from an initial value of 900 N to a final value of zero. The atmospheric pressure is 100 kPa, and the area of the piston face is 0.018 m2. Friction between the piston and the cylinder wall can be neglected. For the air in the piston-cylinder assembly, determine the initial and final pressures, each in kPa and atm.

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Chapter 1 Getting Started A = 0.018 m2

patm = 100 kPa Air Spring force varies linearly from 900 N when V1 = 0.003 m3 to zero when V2 = 0.002 m3

1.46 As shown in Figure P1.46, an inclined manometer is used to measure the pressure of the gas within the reservoir. (a) Using data on the figure, determine the gas pressure, in lbf/in.2 (b) Express the pressure as a gage or a vacuum pressure, as appropriate, in lbf/in.2 (c) What advantage does an inclined manometer have over the U-tube manometer shown in Figure 1.7?

patm = 14.7 lbf/in.2

Fig. P1.42 Gas reservoir

1.43 The pressure from water mains located at street level may be insufficient for delivering water to the upper floors of tall buildings. In such a case, water may be pumped up to a tank that feeds water to the building by gravity. For an open storage tank atop a 300-ft-tall building, determine the pressure, in lbf/in.2, at the bottom of the tank when filled to a depth of 20 ft. The density of water is 62.2 lb/ft3, g 5 32.0 ft/s2, and the local atmospheric pressure is 14.7 lbf/in.2 1.44 Figure P1.44 shows a tank used to collect rainwater having a diameter of 4 m. As shown in the figure, the depth of the tank varies linearly from 3.5 m at its center to 3 m along the perimeter. The local atmospheric pressure is 1 bar, the acceleration of gravity is 9.8 m/s2, and the density of the water is 987.1 kg/m3. When the tank is filled with water, determine (a) the pressure, in kPa, at the bottom center of the tank. (b) the total force, in kN, acting on the bottom of the tank.

g = 32.2 ft/s2

b

a

6 in. 10 in.

30° Mercury ( ρ = 845 lb/ft3)

Fig. P1.46

1.47 Figure P1.47 shows a spherical buoy, having a diameter of 1.5 m and weighing 8500 N, anchored to the floor of a lake by a cable. Determine the force exerted by the cable, in N. The density of the lake water is 103 kg/m3 and g 5 9.81 m/s2.

patm = 1 bar Water ρ = 103 kg/m3

4m

3m

Tank

Buoy

3.5 m

D = 1.5 m Weight = 8500 N

Fig. P1.44 1.45 If the water pressure at the base of the water tower shown in Fig. P1.45 is 4.15 bar, determine the pressure of the air trapped above the water level, in bar. The density of the water is 103 kg/m3 and g 5 9.81 m/s2.

Cable

Air

Fig. P1.47 L = 30 m Water p = 4.15 bar

Fig. P1.45

1.48 Because of a break in a buried oil storage tank, groundwater has leaked into the tank to the depth shown in Fig. P1.48. Determine the pressure at the oil–water interface and at the bottom of the tank, each in lbf/in.2 (gage). The densities of the water and oil are, respectively, 62 and 55, each in lb/ft3. Let g 5 32.2 ft/s2.

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Problems: Developing Engineering Skills patm

33

1.55 Figure P1.55 shows a system consisting of a cylindrical copper rod insulated on its lateral surface while its ends are in contact with hot and cold walls at temperatures 1000°R and 500°R, respectively.

Standpipe open to atmosphere 17 ft Oil

(a) Sketch the variation of temperature with position through the rod, x. (b) Is the rod in equilibrium? Explain.

3 ft

Water

Insulation

Fig. P1.48 1000°R

1.49 Figure P1.49 shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the reading of the pressure gage, in lbf/in.2 (gage). The densities of the oil and mercury are 55 and 845, respectively, each in lb/ft3. Let g 5 32.2 ft/s2.

Rod

500°R

Fig. P1.55 1.56 What is (a) the lowest naturally occurring temperature recorded on Earth, (b) the lowest temperature recorded in a laboratory on Earth, (c) the lowest temperature recorded in the Earth’s solar system, and (d) the temperature of deep space, each in K?

Pressure gage

Air L1 = 3 ft L2 = 0.5 ft L3 = 0.75 ft Oil (ρ = 55 lb/ft3)

x

L1 patm

L2

L3

1.57 What is the maximum increase and maximum decrease in body temperature, each in °C, from a normal body temperature of 37°C that humans can experience before serious medical complications result? 1.58 For liquid-in-glass thermometers, the thermometric property is the change in length of the thermometer liquid with temperature. However, other effects are present that can affect the temperature reading of such thermometers. What are some of these?

Reviewing Concepts 1.59 Answer the following true or false. Explain.

Fig. P1.49

Mercury ( ρ = 845 lb/ft3) g = 32.2 ft/s2

Exploring Temperature 1.50 The 30-year average temperature in Toronto, Canada, during summer is 19.5°C and during winter is −4.9°C. What are the equivalent average summer and winter temperatures in °F and °R? 1.51 Convert the following temperatures from °F to °C: (a) 86°F, (b) −22°F, (c) 50°F, (d) −40°F, (e) 32°F, (f) −459.67°F. Convert each temperature to K. 1.52 Natural gas is burned with air to produce gaseous products at 1985°C. Express this temperature in K, °R, and °F. 1.53 The temperature of a child ill with a fever is measured as 40°C. The child’s normal temperature is 37°C. Express both temperatures in °F. 1.54 Does the Rankine degree represent a larger or smaller temperature unit than the Kelvin degree? Explain.

(a) A closed system always contains the same matter; there is no transfer of matter across its boundary. (b) The volume of a closed system can change. (c) One nanosecond equals 109 seconds. (d) When a closed system undergoes a process between two specified states, the change in temperature between the end states is independent of details of the process. (e) Body organs, such as the human heart, whose shapes change as they perform their normal functions can be studied as control volumes. (f) This book takes a microscopic approach to thermodynamics. 1.60 Answer the following true or false. Explain. (a) 1 N equals 1 kg ? m/s2 but 1 lbf does not equal 1 lb ? ft/s2. (b) Specific volume, the volume per unit of mass, is an intensive property while volume and mass are extensive properties. (c) The kilogram for mass and the meter for length are examples of SI base units defined relative to fabricated objects. (d) If the value of any property of a system changes with time, that system cannot be at steady state.

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Chapter 1 Getting Started

(e) The composition of a closed system cannot change. (f) According to Archimedes’ principle, the magnitude of the buoyant force acting on a submerged body is equal to the weight of the body. 1.61 Answer the following true or false. Explain. (a) A refrigerant at an absolute pressure of 0.8 atm is at a gage pressure of 0.2 atm. (b) Temperature is the property that is the same for each of two systems when they are in thermal equilibrium.

(c) The Rankine degree is a smaller temperature unit than the Kelvin degree. (d) A vessel holding 0.5 kmol of oxygen (O2) contains 16 lb of O2. (e) A control volume is a special type of closed system that does not interact in any way with its surroundings. (f) The pressure unit psia indicates an absolute pressure expressed in pounds force per square inch.

c DESIGN & OPEN ENDED PROBLEMS: EXPLORING ENGINEERING PRACTICE 1.1D In the United States today, nearly all of our electricity is produced by fossil-fuel power plants burning coal or natural gas, nuclear power plants, and hydroelectric power plants. Using the Internet, determine the percent contributions of these types of electricity generation to the U.S. total. For each of the four types, determine at least three significant environmental considerations associated with it and how such environmental aspects affect the respective plant design, operation, and cost. Write a report with at least three references.

1.6D Magnetic resonance imaging (MRI) employs a strong magnetic field to produce detailed pictures of internal organs and tissues. As shown in Fig. P1.6D, the patient reclines on a table that slides into the cylindrical opening where the field is created. Considering a MRI scanner as a system, identify locations on the system boundary where the system interacts with its surroundings. Also describe significant occurrences within the system and the measures taken for patient comfort and safety. Write a report including at least three references.

1.2D Mercury is recognized as a significant biohazard. This has led to the elimination of mercury-in-glass thermometers (see Energy & Environment in Sec.1.7.1) and increasing regulation of coal-fired power plants, emissions from which are the major source of U.S. soil and water mercury contamination. Investigate medical complications of mercury exposure and their economic impact. Report your findings in a well-documented PowerPoint presentation. 1.3D Ecological footprints measure humankind’s demands on nature. Using the Internet, estimate the amount of land and water needed annually to support your consumption of goods and services and to absorb your wastes. Prepare a memorandum reporting your estimates and listing at least three things you can do to reduce your footprint. 1.4D One type of prosthetic limb relies on suction to attach to an amputee’s residual limb. The engineer must consider the required difference between atmospheric pressure and the pressure inside the socket of the prosthetic limb to develop suction sufficient to maintain attachment. What other considerations are important as engineers design this type of prosthetic device? Write a report of your findings including at least three references. 1.5D Design a low-cost, compact, light-weight, hand-held, human-powered air pump capable of directing a stream of air for cleaning computer keyboards, circuit boards, and hard-to-reach locations in electronic devices. The pump cannot use electricity, including batteries, nor employ any chemical propellants. All materials must be recyclable. Owing to existing patent protections, the pump must be a distinct alternative to the familiar tube and plunger bicycle pump and to existing products aimed at accomplishing the specified computer and electronic cleaning tasks.

MRI scanner

Fig. P1.6D 1.7D A major barrier to wider deployment of solar power systems by homeowners and small businesses is the initial cost to purchase and install rooftop components. Today some U.S. municipalities and utilities are developing plans to assist property owners to acquire such components through loans and leasing arrangements. Investigate and critically evaluate these and other options for fostering deployment of solar power systems discovered through design-group brainstorming and use of the Internet. Report your findings in a poster presentation. 1.8D The sphygmomanometer commonly used to measure blood pressure is shown in Fig. P1.8D. During testing, the cuff is placed around the patient’s arm and fully inflated by repeated squeezing of the inflation bulb. Then, as the cuff pressure is gradually reduced, arterial sounds known as Korotkoff sounds are monitored with a stethoscope. Using these sounds as cues, the systolic and diastolic pressures can be identified. These pressures are reported in terms of the mercury column length, in mmHg. Investigate the physical basis for the Korotkoff sounds, their role in identifying the systolic and diastolic pressures, and why these pressures are significant in medical practice. Write a report including at least three references.

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Design & Open Ended Problems: Exploring Engineering Practice patm

Cuff pressure, pc

Mercury column L pc

Inflation bulb Air

Pressure release valve

One-way valve

Fig. P1.8D

1.9D Over the globe, unsafe levels of arsenic, which is a tasteless, odorless, and colorless poison, are present in underground wells providing drinking water to millions of people. The object of this project is to identify low-cost, easy-to-deploy, and easyto-use treatment technologies for removing arsenic from drinking water. Technologies considered should include, but not be limited to, applications of smart materials and other nanotechnology approaches. Write a report critically evaluating existing and proposed technologies. List at least three references. 1.10D Conduct a term-length design project in the realm of bioengineering done on either an independent or a small-

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group basis. The project might involve a device or technique for minimally invasive surgery, an implantable drug-delivery device, a biosensor, artificial blood, or something of special interest to you or your design group. Take several days to research your project idea and then prepare a brief written proposal, including several references, that provides a general statement of the core concept plus a list of objectives. During the project, observe good design practices such as discussed in Sec. 1.3 of Thermal Design and Optimization, John Wiley & Sons Inc., New York, 1996, by A. Bejan, G. Tsatsaronis, and M. J. Moran. Provide a well-documented final report, including several references. 1.11D Conduct a term-length design project involving the International Space Station pictured in Table 1.1 done on either an independent or a small-group basis. The project might involve an experiment that is best conducted in a lowgravity environment, a device for the comfort or use of the astronauts, or something of special interest to you or your design group. Take several days to research your project idea and then prepare a brief written proposal, including several references, that provides a general statement of the core concept plus a list of objectives. During the project, observe good design practices such as discussed in Sec. 1.3 of Thermal Design and Optimization, John Wiley & Sons Inc., New York, 1996, by A. Bejan, G. Tsatsaronis, and M. J. Moran. Provide a well-documented final report, including several references.

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In Sec. 2.1, the relationship between kinetic and gravitational potential energies is considered. © Liane Cary/Age Fotostock America, Inc.

ENGINEERING CONTEXT Energy is a fundamental concept of thermodynamics and one of the most significant aspects of engineering analysis. In this chapter we discuss energy and develop equations for applying the principle of conservation of energy. The current presentation is limited to closed systems. In Chap. 4 the discussion is extended to control volumes. Energy is a familiar notion, and you already know a great deal about it. In the present chapter several important aspects of the energy concept are developed. Some of these you have encountered before. A basic idea is that energy can be stored within systems in various forms. Energy also can be converted from one form to another and transferred between systems. For closed systems, energy can be transferred by work and heat transfer. The total amount of energy is conserved in all conversions and transfers. The objective of this chapter is to organize these ideas about energy into forms suitable for engineering analysis. The presentation begins with a review of energy concepts from mechanics. The thermodynamic concept of energy is then introduced as an extension of the concept of energy in mechanics.

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2 Energy and the First Law of Thermodynamics LEARNING OUTCOMES When you complete your study of this chapter, you will be able to… c

demonstrate understanding of key concepts related to energy and the first law of thermodynamics . . . including internal, kinetic, and potential energy, work and power, heat transfer and heat transfer modes, heat transfer rate, power cycle, refrigeration cycle, and heat pump cycle.

c

apply closed system energy balances, appropriately modeling the case at hand, and correctly observing sign conventions for work and heat transfer.

c

conduct energy analyses of systems undergoing thermodynamic cycles, evaluating as appropriate thermal efficiencies of power cycles and coefficients of performance of refrigeration and heat pump cycles.

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Chapter 2 Energy and the First Law of Thermodynamics

2.1 Reviewing Mechanical Concepts of Energy Building on the contributions of Galileo and others, Newton formulated a general description of the motions of objects under the influence of applied forces. Newton’s laws of motion, which provide the basis for classical mechanics, led to the concepts of work, kinetic energy, and potential energy, and these led eventually to a broadened concept of energy. The present discussion begins with an application of Newton’s second law of motion.

2.1.1 Work and Kinetic Energy TAKE NOTE...

Boldface symbols denote vectors. Vector magnitudes are shown in lightface type.

The curved line in Fig. 2.1 represents the path of a body of mass m (a closed system) moving relative to the x–y coordinate frame shown. The velocity of the center of mass of the body is denoted by V. The body is acted on by a resultant force F, which may vary in magnitude from location to location along the path. The resultant force is resolved into a component Fs along the path and a component Fn normal to the path. The effect of the component Fs is to change the magnitude of the velocity, whereas the effect of the component Fn is to change the direction of the velocity. As shown in Fig. 2.1, s is the instantaneous position of the body measured along the path from some fixed point denoted by 0. Since the magnitude of F can vary from location to location along the path, the magnitudes of Fs and Fn are, in general, functions of s. Let us consider the body as it moves from s 5 s1, where the magnitude of its velocity is V1, to s 5 s2, where its velocity is V2. Assume for the present discussion that the only interaction between the body and its surroundings involves the force F. By Newton’s second law of motion, the magnitude of the component Fs is related to the change in the magnitude of V by dV dt

(2.1)

dV ds dV 5 mV ds dt ds

(2.2)

Fs 5 m Using the chain rule, this can be written as Fs 5 m

where V 5 ds/dt. Rearranging Eq. 2.2 and integrating from s1 to s2 gives

#

V2

mV dV 5

V1

y

# F ds s1

Fs ds

s2

s

(2.3)

Path

V F

s

0

Body Fn

x

Fig. 2.1 Forces acting on a moving system.

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2.1 Reviewing Mechanical Concepts of Energy

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The integral on the left of Eq. 2.3 is evaluated as follows

#

V2

V1

V2 1 1 mV dV 5 mV2 d 5 m1V22 2 V212 2 2 V1

(2.4)

The quantity 12 mV2 is the kinetic energy, KE, of the body. Kinetic energy is a scalar quantity. The change in kinetic energy, ΔKE, of the body is 1 ¢KE 5 KE2 2 KE1 5 m1V22 2 V212 2

(2.5)

The integral on the right of Eq. 2.3 is the work of the force Fs as the body moves from s1 to s2 along the path. Work is also a scalar quantity. With Eq. 2.4, Eq. 2.3 becomes 1 m1V22 2 V212 5 2

#

s2

F ? ds

(2.6)

s1

where the expression for work has been written in terms of the scalar product (dot product) of the force vector F and the displacement vector ds. Equation 2.6 states that the work of the resultant force on the body equals the change in its kinetic energy. When the body is accelerated by the resultant force, the work done on the body can be considered a transfer of energy to the body, where it is stored as kinetic energy. Kinetic energy can be assigned a value knowing only the mass of the body and the magnitude of its instantaneous velocity relative to a specified coordinate frame, without regard for how this velocity was attained. Hence, kinetic energy is a property of the body. Since kinetic energy is associated with the body as a whole, it is an extensive property.

ENERGY & ENVIRONMENT Did you ever wonder what happens to the kinetic energy when you step on the brakes of your moving car? Automotive engineers have, and the result is the hybrid vehicle combining regenerative braking, batteries, an electric motor, and a conventional engine. When brakes are applied in a hybrid vehicle, some of the vehicle’s kinetic energy is harvested and stored on board electrically for use when needed. Through regenerative braking and other innovative features, hybrids get much better mileage than comparably sized conventional vehicles. Hybrid vehicle technology is quickly evolving. Today’s hybrids use electricity to supplement conventional engine power, while future plug-in hybrids will use the power of a smaller engine to supplement electricity. The hybrids we now see on the road have enough battery power on board for acceleration to about 20 miles per hour and after that assist the engine when necessary. This improves fuel mileage, but the batteries are recharged by the engine—they are never plugged in. Plug-in hybrids achieve even better fuel economy. Instead of relying on the engine to recharge batteries, most recharging will be received from an electrical outlet while the car is idle—overnight, for example. This will allow cars to get the energy they need mainly from the electrical grid, not the fuel pump. Widespread deployment of plug-ins awaits development of a new generation of batteries and ultra-capacitors (see Sec. 2.7). Better fuel economy not only allows our society to be less reliant on oil to meet transportation needs but also reduces release of CO2 into the atmosphere from vehicles. Each gallon of gasoline burned by a car’s engine produces about 9 kg (20 lb) of CO2. A conventional vehicle produces several tons of CO2 annually; fuel-thrifty hybrids produce much less. Still, since hybrids use electricity from the grid, we will have to make a greater effort to reduce power plant emissions by including more wind power, solar power, and other renewables in the national mix.

kinetic energy

TAKE NOTE...

The symbol D always means “final value minus initial value.”

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Chapter 2 Energy and the First Law of Thermodynamics

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2.1.2

R z2

z

mg

z1

Earth's surface

Fig. 2.2 Illustration used to introduce the potential energy concept.

Potential Energy

Equation 2.6 is a principal result of the previous section. Derived from Newton’s second law, the equation gives a relationship between two defined concepts: kinetic energy and work. In this section it is used as a point of departure to extend the concept of energy. To begin, refer to Fig. 2.2, which shows a body of mass m that moves vertically from an elevation z1 to an elevation z2 relative to the surface of the earth. Two forces are shown acting on the system: a downward force due to gravity with magnitude mg and a vertical force with magnitude R representing the resultant of all other forces acting on the system. The work of each force acting on the body shown in Fig. 2.2 can be determined by using the definition previously given. The total work is the algebraic sum of these individual values. In accordance with Eq. 2.6, the total work equals the change in kinetic energy. That is 1 m1V22 2 V212 5 2

#

z2

R dz 2

z1

#

z2

mg dz

(2.7)

z1

A minus sign is introduced before the second term on the right because the gravitational force is directed downward and z is taken as positive upward. The first integral on the right of Eq. 2.7 represents the work done by the force R on the body as it moves vertically from z1 to z2. The second integral can be evaluated as follows: z2

# mg dz 5 mg1z z1

2

2 z1 2

(2.8)

where the acceleration of gravity has been assumed to be constant with elevation. By incorporating Eq. 2.8 into Eq. 2.7 and rearranging 1 m1V22 2 V212 1 mg1z2 2 z12 5 2 gravitational potential energy

z2

# R dz

(2.9)

z1

The quantity mgz is the gravitational potential energy, PE. The change in gravitational potential energy, ΔPE, is ¢PE 5 PE2 2 PE1 5 mg1z2 2 z12

(2.10)

Potential energy is associated with the force of gravity and is therefore an attribute of a system consisting of the body and the earth together. However, evaluating the force of gravity as mg enables the gravitational potential energy to be determined for a specified value of g knowing only the mass of the body and its elevation. With this view, potential energy is regarded as an extensive property of the body. Throughout this book it is assumed that elevation differences are small enough that the gravitational force can be considered constant. The concept of gravitational potential energy can be formulated to account for the variation of the gravitational force with elevation, however. To assign a value to the kinetic energy or the potential energy of a system, it is necessary to assume a datum and specify a value for the quantity at the datum. Values of kinetic and potential energy are then determined relative to this arbitrary choice of datum and reference value. However, since only changes in kinetic and potential energy between two states are required, these arbitrary reference specifications cancel.

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2.1 Reviewing Mechanical Concepts of Energy

2.1.3 Units for Energy Work has units of force times distance. The units of kinetic energy and potential energy are the same as for work. In SI, the energy unit is the newton-meter, N ? m, called the joule, J. In this book it is convenient to use the kilojoule, kJ. Commonly used English units for work, kinetic energy, and potential energy are the foot-pound force, ft ? lbf, and the British thermal unit, Btu. When a system undergoes a process where there are changes in kinetic and potential energy, special care is required to obtain a consistent set of units. to illustrate the proper use of units in the calculation of such terms, consider a system having a mass of 1 kg whose velocity increases from 15 m/s to 30 m/s while its elevation decreases by 10 m at a location where g 5 9.7 m/s2. Then 1 m1V22 2 V212 2 1 m 2 m 2 1N 1 kJ ` 5 11 kg2 c a30 b 2 a15 b d ` ` ` 3 2 s s 2 1 kg ? m/ s 10 N ? m 5 0.34 kJ ¢PE 5 mg1z2 2 z12

¢KE 5

5 11 kg2a9.7

m 1N 1 kJ ` b1210 m2 ` ` ` s2 1 kg ? m/ s2 103 N ? m

5 20.10 kJ For a system having a mass of 1 lb whose velocity increases from 50 ft/s to 100 ft/s while its elevation decreases by 40 ft at a location where g 5 32.0 ft/s2, we have 1 ft 2 ft 2 1 lbf 1 Btu ¢KE 5 11 lb2 c a100 b 2 a50 b d ` ` ` ` 2 s s 2 778 ft ? lbf 32.2 lb ? ft/ s 5 0.15 Btu ft 1 lbf 1 Btu ¢PE 5 11 lb2 a32.0 2 b 1240 ft2 ` ` ` ` 2 s 32.2 lb ? ft/ s 778 ft ? lbf 5 20.05 Btu

b b b b b

2.1.4 Conservation of Energy in Mechanics Equation 2.9 states that the total work of all forces acting on the body from the surroundings, with the exception of the gravitational force, equals the sum of the changes in the kinetic and potential energies of the body. When the resultant force causes the elevation to be increased, the body to be accelerated, or both, the work done by the force can be considered a transfer of energy to the body, where it is stored as gravitational potential energy and/or kinetic energy. The notion that energy is conserved underlies this interpretation. The interpretation of Eq. 2.9 as an expression of the conservation of energy principle can be reinforced by considering the special case of a body on which the only force acting is that due to gravity, for then the right side of the equation vanishes and the equation reduces to 1 m1V22 2 V212 1 mg1z2 2 z12 5 0 2 (2.11)

or 1 1 mV22 1 mgz2 5 mV21 1 mgz1 2 2

z

mg

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Chapter 2 Energy and the First Law of Thermodynamics Under these conditions, the sum of the kinetic and gravitational potential energies remains constant. Equation 2.11 also illustrates that energy can be converted from one form to another: For an object falling under the influence of gravity only, the potential energy would decrease as the kinetic energy increases by an equal amount.

2.1.5 Closing Comment The presentation thus far has centered on systems for which applied forces affect only their overall velocity and position. However, systems of engineering interest normally interact with their surroundings in more complicated ways, with changes in other properties as well. To analyze such systems, the concepts of kinetic and potential energy alone do not suffice, nor does the rudimentary conservation of energy principle introduced in this section. In thermodynamics the concept of energy is broadened to account for other observed changes, and the principle of conservation of energy is extended to include a wide variety of ways in which systems interact with their surroundings. The basis for such generalizations is experimental evidence. These extensions of the concept of energy are developed in the remainder of the chapter, beginning in the next section with a fuller discussion of work.

2.2 Broadening Our Understanding of Work The work W done by, or on, a system evaluated in terms of macroscopically observable forces and displacements is W5

#

s2

F ? ds

(2.12)

s1

thermodynamic definition of work

This relationship is important in thermodynamics, and is used later in the present section to evaluate the work done in the compression or expansion of gas (or liquid), the extension of a solid bar, and the stretching of a liquid film. However, thermodynamics also deals with phenomena not included within the scope of mechanics, so it is necessary to adopt a broader interpretation of work, as follows. A particular interaction is categorized as a work interaction if it satisfies the following criterion, which can be considered the thermodynamic definition of work: Work is done by a system on its surroundings if the sole effect on everything external to the system could have been the raising of a weight. Notice that the raising of a weight is, in effect, a force acting through a distance, so the concept of work in thermodynamics is a natural extension of the concept of work in mechanics. However, the test of whether a work interaction has taken place is not that the elevation of a weight has actually taken place, or that a force has actually acted through a distance, but that the sole effect could have been an increase in the elevation of a weight. consider Fig. 2.3 showing two systems labeled A and B. In system A, a gas is stirred by a paddle wheel: the paddle wheel does work on the gas. In principle, the work could be evaluated in terms of the forces and motions at the boundary between the paddle wheel and the gas. Such an evaluation of work is consistent with Eq. 2.12, where work is the product of force and displacement. By contrast, consider system B, which includes only the battery. At the boundary of system B, forces and motions are not evident. Rather, there is an electric current i driven by an electrical potential difference existing across the terminals a and b. That this type of interaction at the boundary can be classified as work follows from the thermodynamic

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2.2 Broadening Our Understanding of Work

Paddle wheel

43

System A i

Gas

System B a

b

Battery

Fig. 2.3 Two examples of work. definition of work given previously: We can imagine the current is supplied to a hypothetical electric motor that lifts a weight in the surroundings. b b b b b Work is a means for transferring energy. Accordingly, the term work does not refer to what is being transferred between systems or to what is stored within systems. Energy is transferred and stored when work is done.

2.2.1 Sign Convention and Notation Engineering thermodynamics is frequently concerned with devices such as internal combustion engines and turbines whose purpose is to do work. Hence, in contrast to the approach generally taken in mechanics, it is often convenient to consider such work as positive. That is, W . 0: work done by the system W , 0: work done on the system This sign convention is used throughout the book. In certain instances, however, it is convenient to regard the work done on the system to be positive, as has been done in the discussion of Sec. 2.1. To reduce the possibility of misunderstanding in any such case, the direction of energy transfer is shown by an arrow on a sketch of the system, and work is regarded as positive in the direction of the arrow. To evaluate the integral in Eq. 2.12, it is necessary to know how the force varies with the displacement. This brings out an important idea about work: The value of W depends on the details of the interactions taking place between the system and surroundings during a process and not just the initial and final states of the system. It follows that work is not a property of the system or the surroundings. In addition, the limits on the integral of Eq. 2.12 mean “from state 1 to state 2” and cannot be interpreted as the values of work at these states. The notion of work at a state has no meaning, so the value of this integral should never be indicated as W2 − W1.

sign convention for work

work is not a property

Nanoscale Machines on the Move Engineers working in the field of nanotechnology, the engineering of molecular-sized devices, look forward to the time when practical nanoscale machines can be fabricated that are capable of movement, sensing and responding to stimuli such as light and sound, delivering medication within the body, performing computations, and numerous other functions that promote human well being. For inspiration, engineers study biological nanoscale machines in living things that perform functions

such as creating and repairing cells, circulating oxygen, and digesting food. These studies have yielded positive results. Molecules mimicking the function of mechanical devices have been fabricated, including gears, rotors, ratchets, brakes, switches, and abacus-like structures. A particular success is the development of molecular motors that convert light to rotary or linear motion. Although devices produced thus far are rudimentary, they do demonstrate the feasibility of constructing nanomachines, researchers say.

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Chapter 2 Energy and the First Law of Thermodynamics The differential of work, dW, is said to be inexact because, in general, the following integral cannot be evaluated without specifying the details of the process 2

# dW 5 W 1

On the other hand, the differential of a property is said to be exact because the change in a property between two particular states depends in no way on the details of the process linking the two states. For example, the change in volume between two states can be determined by integrating the differential dV, without regard for the details of the process, as follows

#

V2

dV 5 V2 2 V1

V1

where V1 is the volume at state 1 and V2 is the volume at state 2. The differential of every property is exact. Exact differentials are written, as above, using the symbol d. To stress the difference between exact and inexact differentials, the differential of work is written as dW. The symbol d is also used to identify other inexact differentials encountered later.

2.2.2 Power power

Many thermodynamic analyses are concerned with the time rate at which energy transfer occurs. The rate of energy transfer by work is called power and is denoted ? by W. When a work interaction involves a macroscopically observable force, the rate of energy transfer by work is equal to the product of the force and the velocity at the point of application of the force ?

W5F?V

(2.13)

?

A dot appearing over a symbol, as in W, is used throughout this book to indicate a time rate. In principle, Eq. 2.13 can be integrated from time t1 to time ts to get the total work done during the time interval W5

#

t2

?

W dt 5

t1

#

t2

F ? V dt

(2.14)

t1

?

units for power

The same sign convention applies for W as for W. Since power is a time rate of doing work, it can be expressed in terms of any units for energy and time. In SI, the unit for power is J/s, called the watt. In this book the kilowatt, kW, is generally used. Commonly used English units for power are ft ? lbf/s, Btu/h, and horsepower, hp.

to illustrate the use of Eq. 2.13, let us evaluate the power required for a bicyclist traveling at 20 miles per hour to overcome the drag force imposed by the surrounding air. This aerodynamic drag force is given by Fd 5 12 CdArV 2 where Cd is a constant called the drag coefficient, A is the frontal area of the bicycle and rider, and r is the air density. By Eq. 2.13 the required power is Fd ? V or ?

W 5 112 CdArV22V 5 12Cd ArV3

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2.2 Broadening Our Understanding of Work Using typical values: Cd 5 0.88, A 5 3.9 ft2, and r 5 0.075 lb/ft3, together with V 5 20 mi/h 5 29.33 ft/s, and also converting units to horsepower, the power required is ?

1 hp 1 lb ft 3 1 lbf 10.88213.9 ft22a0.075 3 ba29.33 b ` ` ` ` 2 s 2 ft 32.2 lb ? ft / s 550 ft ? lbf/ s 5 0.183 hp b b b b b

W5

Drag can be reduced by streamlining the shape of a moving object and using the strategy known as drafting (see box).

Drafting Drafting occurs when two or more moving vehicles or individuals align closely to reduce the overall effect of drag. Drafting is seen in competitive events such as auto racing, bicycle racing, speed-skating, and running. Studies show that air flow over a single vehicle or individual in motion is characterized by a high-pressure region in front and a low-pressure region behind. The difference between these pressures creates a force, called drag, impeding motion. During drafting, as seen in the sketch below, a second vehicle or individual is closely aligned with another, and air flows over the pair nearly as if they were a single entity, thereby altering the pressure between them and reducing the drag each experiences. While race-car drivers use drafting to increase speed, non–motor sport competitors usually aim to reduce demands on their bodies while maintaining the same speed.

2.2.3

Modeling Expansion or Compression Work

There are many ways in which work can be done by or on a system. The remainder of this section is devoted to considering several examples, beginning with the important case of the work done when the volume of a quantity of a gas (or liquid) changes by expansion or compression. Let us evaluate the work done by the closed system shown in Fig. 2.4 consisting of a gas (or liquid) contained in a piston–cylinder assembly as the gas expands. During the process the gas pressure exerts a normal force on the piston. Let p denote the pressure acting at the interface between the gas and the piston. The force exerted System boundary Area = A

Average pressure at the piston face = p

F = pA Gas or liquid x

x1

x2

Fig. 2.4 Expansion or compression of a gas or liquid.

45

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Chapter 2 Energy and the First Law of Thermodynamics by the gas on the piston is simply the product pA, where A is the area of the piston face. The work done by the system as the piston is displaced a distance dx is dW 5 pA dx

(2.15)

The product A dx in Eq. 2.15 equals the change in volume of the system, dV. Thus, the work expression can be written as dW 5 p dV

(2.16)

Since dV is positive when volume increases, the work at the moving boundary is positive when the gas expands. For a compression, dV is negative, and so is work found from Eq. 2.16. These signs are in agreement with the previously stated sign convention for work. For a change in volume from V1 to V2, the work is obtained by integrating Eq. 2.16 W5

#

V2

p dV

(2.17)

V1

Although Eq. 2.17 is derived for the case of a gas (or liquid) in a piston–cylinder assembly, it is applicable to systems of any shape provided the pressure is uniform with position over the moving boundary.

2.2.4 Expansion or Compression Work in Actual Processes There is no requirement that a system undergoing a process be in equilibrium during the process. Some or all of the intervening states may be nonequilibrium states. For many such processes we are limited to knowing the state before the process occurs and the state after the process is completed. Typically, at a nonequilibrium state intensive properties vary with position at a given time. Also, at a specified position intensive properties may vary with time, sometimes chaotically. In certain cases, spatial and temporal variations in properties such as temperature, pressure, and velocity can be measured, or obtained by solving appropriate governing equations, which are generally differential equations. To perform the integral of Eq. 2.17 requires a relationship between p the gas pressure at the moving boundary and the system volume. However, due to nonequilibrium effects during an actual expansion or comMeasured data Curve fit pression process, this relationship may be difficult, or even impossible, to obtain. In the cylinder of an automobile engine, for example, combustion and other nonequilibrium effects give rise to nonuniformities throughout the cylinder. Accordingly, if a pressure transducer were mounted on the cylinder head, the recorded output might provide only an approximation for the pressure at the piston face required by Eq. 2.17. Moreover, even when the measured pressure is essentially equal to that at the piston face, scatter might exist in the pressure–volume data, as illustrated in Fig. 2.5. Still, performing the integral of Eq. 2.17 based on V a curve fitted to the data could give a plausible estimate of the work. Fig. 2.5 Pressure at the piston face versus We will see later that in some cases where lack of the required pressure– cylinder volume. volume relationship keeps us from evaluating the work from Eq. 2.17, the work can be determined alternatively from an energy balance (Sec. 2.5).

2.2.5 Expansion or Compression Work in Quasiequilibrium Processes quasiequilibrium process

Processes are sometime modeled as an idealized type of process called a quasiequilibrium (or quasistatic) process. A quasiequilibrium process is one in which the departure

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2.2 Broadening Our Understanding of Work from thermodynamic equilibrium is at most infinitesimal. All states through which the system passes in a quasiequilibrium process may be considered equilibrium states. Because nonequilibrium effects are inevitably present during actual processes, systems of engineering interest can at best approach, but never realize, a quasiequilibrium process. Still the quasiequilibrium process plays a role in our study of engineering thermodynamics. For details, see the box. To consider how a gas (or liquid) might be expanded or compressed in a quasiequilibrium fashion, refer to Fig. 2.6, which shows a system consisting of a gas initially at an equilibrium state. As shown in the figure, the gas pressure is maintained uniform throughout by a number of small masses resting on the freely moving piston. Imagine that one of the masses is removed, allowing the piston to move upward as the gas expands slightly. During such an expansion the state of the gas would depart only slightly from equilibrium. The system would eventually come to a new equilibrium state, where the pressure and all other intensive properties would again be uniform in value. Moreover, were the mass replaced, the gas would be restored to its initial state, while again the departure from equilibrium would be slight. If several of the masses were removed one after another, the gas would pass through a sequence of equilibrium states without ever being far from equilibrium. In the limit as the increments of mass are made vanishingly small, the gas would undergo a quasiequilibrium expansion process. A quasiequilibrium compression can be visualized with similar considerations.

Using the Quasiequilibrium Process Concept Our interest in the quasiequilibrium process concept stems mainly from two considerations: c

Simple thermodynamic models giving at least qualitative information about the behavior of actual systems of interest often can be developed using the quasiequilibrium process concept. This is akin to the use of idealizations such as the point mass or the frictionless pulley in mechanics for the purpose of simplifying an analysis.

c

The quasiequilibrium process concept is instrumental in deducing relationships that exist among the properties of systems at equilibrium (Chaps. 3, 6, and 11).

Equation 2.17 can be applied to evaluate the work in quasiequilibrium expansion or compression processes. For such idealized processes the pressure p in the equation is the pressure of the entire quantity of gas (or liquid) undergoing the process, and not just the pressure at the moving boundary. The relationship between the pressure and volume may be graphical or analytical. Let us first consider a graphical relationship. A graphical relationship is shown in the pressure–volume diagram ( p–V diagram) of Fig. 2.7. Initially, the piston face is at position x1, and the gas pressure is p1; at the conclusion of a quasiequilibrium expansion process the piston face is at position x2, and the pressure is reduced to p2. At each intervening piston position, the uniform pressure throughout the gas is shown as a point on the diagram. The curve, or path, connecting states 1 and 2 on the diagram represents the equilibrium states through which the system has passed during the process. The work done by the gas on the piston during the expansion is given by ep dV , which can be interpreted as the area under the curve of pressure versus volume. Thus, the shaded area on Fig. 2.7 is equal to the work for the process. Had the gas been compressed from 2 to 1 along the same path on the p–V diagram, the magnitude of the work would be the same, but the sign would be negative, indicating that for the compression the energy transfer was from the piston to the gas. The area interpretation of work in a quasiequilibrium expansion or compression process allows a simple demonstration of the idea that work depends on the process.

Incremental masses removed during an expansion of the gas or liquid

Gas or liquid Boundary

Fig. 2.6 Illustration of a quasiequilibrium expansion or compression.

47

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Chapter 2 Energy and the First Law of Thermodynamics

48

p1

1 Path

Pressure

δ W = p dV p 1 p2

2 Area = 2 ∫1 p dV V1

dV

V2

Volume

A

Gas or liquid

x

2 Area = work for process A x1

x2

Fig. 2.7 Work of a quasiequilibrium expansion or compression process.

A

Comp_Work A.4 – All Tabs Exp_Work A.5 – All Tabs

polytropic process

cccc

B

V

Fig. 2.8 Illustration that work depends on the process.

This can be brought out by referring to Fig. 2.8. Suppose the gas in a piston–cylinder assembly goes from an initial equilibrium state 1 to a final equilibrium state 2 along two different paths, labeled A and B on Fig. 2.8. Since the area beneath each path represents the work for that process, the work depends on the details of the process as defined by the particular curve and not just on the end states. Using the test for a property given in Sec. 1.3.3, we can conclude again (Sec. 2.2.1) that work is not a property. The value of work depends on the nature of the process between the end states. The relation between pressure and volume, or pressure and specific volume, also can be described analytically. A quasiequilibrium process described by pV n 5 constant, or pyn 5 constant, where n is a constant, is called a polytropic process. Additional analytical forms for the pressure–volume relationship also may be considered. The example to follow illustrates the application of Eq. 2.17 when the relationship between pressure and volume during an expansion is described analytically as pV n 5 constant.

EXAMPLE 2.1 c

Evaluating Expansion Work A gas in a piston–cylinder assembly undergoes an expansion process for which the relationship between pressure and volume is given by pV n 5 constant The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. Determine the work for the process, in kJ, if (a) n 5 1.5, (b) n 5 1.0, and (c) n 5 0. SOLUTION Known: A gas in a piston–cylinder assembly undergoes an expansion for which pV n 5 constant. Find: Evaluate the work if (a) n 5 1.5, (b) n 5 1.0, (c) n 5 0.

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2.2 Broadening Our Understanding of Work

49

Schematic and Given Data: The given p–V relationship and the given data for pressure and volume can be used

to construct the accompanying pressure–volume diagram of the process. Engineering Model: 3.0

1

1. The gas is a closed system.

2c p1 = 3.0 bar V1 = 0.1 m3

p (bar)

Gas

➊

2.0 pVn

= constant

2b 1.0

Area = work for part a

0.1 V

V2 = 0.2 m3

2. The moving boundary is the

only work mode. 3. The expansion is a polytropic

process.

2a

➋

0.2

Fig. E2.1

(m3)

Analysis: The required values for the work are obtained by integration of Eq. 2.17 using the given pressure–volume

relation. (a) Introducing the relationship p 5 constant/V n into Eq. 2.17 and performing the integration

W5

#

V2

p dV 5

V1

#

V2

V1

constant dV Vn

1constant2V12n 2 1constant2V12n 2 1 5 12n The constant in this expression can be evaluated at either end state: constant 5 p1V n1 5 p2V n2. The work expression then becomes W5

1p2V 2n2V12n 2 1p1V 1n2V12n p2V2 2 p1V1 2 1 5 12n 12n

(a)

This expression is valid for all values of n except n 5 1.0. The case n 5 1.0 is taken up in part (b). To evaluate W, the pressure at state 2 is required. This can be found by using p1V n1 5 p2V n2, which on rearrangement yields p2 5 p1a

V1 n 0.1 1.5 b 5 13 bar2a b 5 1.06 bar V2 0.2

Accordingly, ➌

11.06 bar210.2 m32 2 13210.12 105 N/ m2 1 kJ ` ` 3 ` b` 1 bar 1 2 1.5 10 N ? m 5 117.6 kJ

W5a

(b) For n 5 1.0, the pressure–volume relationship is pV 5 constant or p 5 constant/V. The work is

W 5 constant

#

V2

V1

V2 V2 dV 5 1constant2 ln 5 1p1V12 ln V V1 V1

(b)

Substituting values W 5 13 bar210.1 m32 `

105 N/ m2 1 kJ 0.2 ` ` 3 ` ln a b 5 120.79 kJ 1 bar 0.1 10 N ? m

(c) For n 5 0, the pressure–volume relation reduces to p 5 constant, and the integral becomes W 5 p(V2 2 V1), which is a special case of the expression found in part (a). Substituting values and converting units as above, W 5 130 kJ. ➍

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Chapter 2 Energy and the First Law of Thermodynamics ➊ In each case, the work for the process can be interpreted as the area under the curve representing the process on the accompanying p–V diagram. Note that the relative areas are in agreement with the numerical results. ➋ The assumption of a polytropic process is significant. If the given pressure– volume relationship were obtained as a fit to experimental pressure–volume data, the value of e p dV would provide a plausible estimate of the work only when the measured pressure is essentially equal to that exerted at the piston face. ➌ Observe the use of unit conversion factors here and in part (b).

✓ Skills Developed Ability to… ❑ apply the problem-solving

methodology. ❑ define a closed system and

identify interactions on its boundary. ❑ evaluate work using Eq. 2.17. ❑ apply the pressure–volume relation pV n 5 constant.

➍ In each of the cases considered, it is not necessary to identify the gas (or liquid) contained within the piston–cylinder assembly. The calculated values for W are determined by the process path and the end states. However, if it is desired to evaluate a property such as temperature, both the nature and amount of the substance must be provided because appropriate relations among the properties of the particular substance would then be required. Evaluate the work, in kJ, for a two-step process consisting of an expansion with n 5 1.0 from p1 5 3 bar, V1 5 0.1 m3 to V 5 0.15 m3, followed by an expansion with n 5 0 from V 5 0.15 m3 to V2 5 0.2 m3. Ans. 22.16 kJ.

2.2.6 Further Examples of Work To broaden our understanding of the work concept, we now briefly consider several other examples of work.

Extension of a Solid Bar Area = A

F

x x1

x2

Fig. 2.9 Elongation of a solid bar.

Consider a system consisting of a solid bar under tension, as shown in Fig. 2.9. The bar is fixed at x 5 0, and a force F is applied at the other end. Let the force be represented as F 5 sA, where A is the cross-sectional area of the bar and s the normal stress acting at the end of the bar. The work done as the end of the bar moves a distance dx is given by dW 5 2sA dx. The minus sign is required because work is done on the bar when dx is positive. The work for a change in length from x1 to x2 is found by integration W52

#

x2

sA dx

(2.18)

x1

Equation 2.18 for a solid is the counterpart of Eq. 2.17 for a gas undergoing an expansion or compression.

Rigid wire frame

Stretching of a Liquid Film

Surface of film

Movable wire F

l dx x

Fig. 2.10 Stretching of a liquid film.

Figure 2.10 shows a system consisting of a liquid film suspended on a wire frame. The two surfaces of the film support the thin liquid layer inside by the effect of surface tension, owing to microscopic forces between molecules near the liquid–air interfaces. These forces give rise to a macroscopically measurable force perpendicular to any line in the surface. The force per unit length across such a line is the surface tension. Denoting the surface tension acting at the movable wire by t, the force F indicated on the figure can be expressed as F 5 2lt, where the factor 2 is introduced because two film surfaces act at the wire. If the movable wire is displaced by dx, the work is given by dW 5 22lt dx. The minus sign is required because work is done on the system when dx is positive. Corresponding to a displacement dx is a change in the total area of the surfaces in contact with

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2.2 Broadening Our Understanding of Work

51

the wire of dA 5 2l dx, so the expression for work can be written alternatively as dW 5 2t dA. The work for an increase in surface area from A1 to A2 is found by integrating this expression W52

#

A2

t dA

(2.19)

A1

Power Transmitted by a Shaft A rotating shaft is a commonly encountered machine element. Consider a shaft + rotating with angular velocity v and exerting a torque t on its surroundings. Let the torque be expressed in terms of a tangential force Ft and radius R: t 5 Ft R. – The velocity at the point of application of the force is V 5 Rv, where v is in radians per unit time. Using these relations with Eq. 2.13, we obtain an expression for the power transmitted from the shaft to the surroundings # W 5 FtV 5 1t/ R21Rv2 5 tv (2.20)

W˙ shaft Motor ,ω

A related case involving a gas stirred by a paddle wheel is considered in the discussion of Fig. 2.3.

Electric Power Shown in Fig. 2.11 is a system consisting of an electrolytic cell. The cell is connected to an external circuit through which an electric current, i, is flowing. The current is driven by the electrical potential difference e existing across the terminals labeled a and b. That this type of interaction can be classed as work is considered in the discussion of Fig. 2.3. The rate of energy transfer by work, or the power, is # W 5 2ei (2.21)

–

i Ᏹ

+ a

b

System boundary Electrolytic cell

Since the current i equals dZ/dt, the work can be expressed in differential form as dW 5 2e dZ

(2.22)

where dZ is the amount of electrical charge that flows into the system. The minus signs appearing in Eqs. 2.21 and 2.22 are required to be in accord with our previously stated sign convention for work.

Work Due to Polarization or Magnetization Let us next refer briefly to the types of work that can be done on systems residing in electric or magnetic fields, known as the work of polarization and magnetization, respectively. From the microscopic viewpoint, electrical dipoles within dielectrics resist turning, so work is done when they are aligned by an electric field. Similarly, magnetic dipoles resist turning, so work is done on certain other materials when their magnetization is changed. Polarization and magnetization give rise to macroscopically detectable changes in the total dipole moment as the particles making up the material are given new alignments. In these cases the work is associated with forces imposed on the overall system by fields in the surroundings. Forces acting on the material in the system interior are called body forces. For such forces the appropriate displacement in evaluating work is the displacement of the matter on which the body force acts.

2.2.7 Further Examples of Work in Quasiequilibrium Processes Systems other than a gas or liquid in a piston–cylinder assembly also can be envisioned as undergoing processes in a quasiequilibrium fashion. To apply the quasiequilibrium process concept in any such case, it is necessary to conceive of an ideal situation in

Fig. 2.11 Electrolytic cell used to discuss electric power.

TAKE NOTE...

When power is evaluated in terms of the watt, and the unit of current is the ampere (an SI base unit), the unit of electric potential is the volt, defined as 1 watt per ampere.

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Chapter 2 Energy and the First Law of Thermodynamics

TAKE NOTE...

Some readers may elect to defer Secs. 2.2.7 and 2.2.8 and proceed directly to Sec. 2.3 where the energy concept is broadened.

which the external forces acting on the system can be varied so slightly that the resulting imbalance is infinitesimal. As a consequence, the system undergoes a process without ever departing significantly from thermodynamic equilibrium. The extension of a solid bar and the stretching of a liquid surface can readily be envisioned to occur in a quasiequilibrium manner by direct analogy to the piston– cylinder case. For the bar in Fig. 2.9 the external force can be applied in such a way that it differs only slightly from the opposing force within. The normal stress is then essentially uniform throughout and can be determined as a function of the instantaneous length: s 5 s(x). Similarly, for the liquid film shown in Fig. 2.10 the external force can be applied to the movable wire in such a way that the force differs only slightly from the opposing force within the film. During such a process, the surface tension is essentially uniform throughout the film and is functionally related to the instantaneous area: t 5 t(A). In each of these cases, once the required functional relationship is known, the work can be evaluated using Eq. 2.18 or 2.19, respectively, in terms of properties of the system as a whole as it passes through equilibrium states. Other systems also can be imagined as undergoing quasiequilibrium processes. For example, it is possible to envision an electrolytic cell being charged or discharged in a quasiequilibrium manner by adjusting the potential difference across the terminals to be slightly greater, or slightly less, than an ideal potential called the cell electromotive force (emf). The energy transfer by work for passage of a differential quantity of charge to the cell, dZ, is given by the relation dW 5 2e dZ

(2.23)

In this equation e denotes the cell emf, an intensive property of the cell, and not just the potential difference across the terminals as in Eq. 2.22. Consider next a dielectric material residing in a uniform electric field. The energy transferred by work from the field when the polarization is increased slightly is dW 5 2E ? d1VP2

(2.24)

where the vector E is the electric field strength within the system, the vector P is the electric dipole moment per unit volume, and V is the volume of the system. A similar equation for energy transfer by work from a uniform magnetic field when the magnetization is increased slightly is dW 5 2m0H ? d1VM2

(2.25)

where the vector H is the magnetic field strength within the system, the vector M is the magnetic dipole moment per unit volume, and m0 is a constant, the permeability of free space. The minus signs appearing in the last three equations are in accord with our previously stated sign convention for work: W takes on a negative value when the energy transfer is into the system.

2.2.8 Generalized Forces and Displacements The similarity between the expressions for work in the quasiequilibrium processes considered thus far should be noted. In each case, the work expression is written in the form of an intensive property and the differential of an extensive property. This is brought out by the following expression, which allows for one or more of these work modes to be involved in a process dW 5 p dV 2 sd1A x2 2 t dA 2 e dZ 2 E ? d1V P2 2 m0H ? d1V M2 1 p (2.26) where the last three dots represent other products of an intensive property and the differential of a related extensive property that account for work. Because of the notion of work being a product of force and displacement, the intensive property in these relations is sometimes referred to as a “generalized” force and the extensive

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2.3 Broadening Our Understanding of Energy

53

F

property as a “generalized” displacement, even though the quantities making up the work expressions may not bring to mind actual forces and displacements. Owing to the underlying quasiequilibrium restriction, Eq. 2.26 does not represent every type of work of practical interest. An example is provided by a paddle wheel that stirs a gas or liquid taken as the system. Whenever any shearing action takes place, the system necessarily passes through nonequilibrium states. To appreciate more fully the implications of the quasiequilibrium process concept requires consideration of the second law of thermodynamics, so this concept is discussed again in Chap. 5 after the second law has been introduced.

i

2.3 Broadening Our Understanding of Energy The objective in this section is to use our deeper understanding of work developed in Sec. 2.2 to broaden our understanding of the energy of a system. In particular, we consider the total energy of a system, which includes kinetic energy, gravitational potential energy, and other forms of energy. The examples to follow illustrate some of these forms of energy. Many other examples could be provided that enlarge on the same idea. When work is done to compress a spring, energy is stored within the spring. When a battery is charged, the energy stored within it is increased. And when a gas (or liquid) initially at an equilibrium state in a closed, insulated vessel is stirred vigorously and allowed to come to a final equilibrium state, the energy of the gas is increased in the process. In keeping with the discussion of work in Sec. 2.2, we can also think of other ways in which work done on systems increases energy stored within those systems— work related to magnetization, for example. In each of these examples the change in system energy cannot be attributed to changes in the system’s overall kinetic or gravitational potential energy as given by Eqs. 2.5 and 2.10, respectively. The change in energy can be accounted for in terms of internal energy, as considered next. In engineering thermodynamics the change in the total energy of a system is considered to be made up of three macroscopic contributions. One is the change in kinetic energy, associated with the motion of the system as a whole relative to an external coordinate frame. Another is the change in gravitational potential energy, associated with the position of the system as a whole in the earth’s gravitational field. All other energy changes are lumped together in the internal energy of the system. Like kinetic energy and gravitational potential energy, internal energy is an extensive property of the system, as is the total energy. Internal energy is represented by the symbol U, and the change in internal energy in a process is U2 2 U1. The specific internal energy is symbolized by u or u, respectively, depending on whether it is expressed on a unit mass or per mole basis. The change in the total energy of a system is E2 2 E1 5 1U2 2 U12 1 1KE2 2 KE12 1 1PE2 2 PE12

Battery

Paddle wheel

Gas

internal energy

(2.27a) Total_Energy A.6 – Tab a

or ¢E 5 ¢U 1 ¢KE 1 ¢PE

(2.27b)

All quantities in Eq. 2.27 are expressed in terms of the energy units previously introduced. The identification of internal energy as a macroscopic form of energy is a significant step in the present development, for it sets the concept of energy in thermodynamics apart from that of mechanics. In Chap. 3 we will learn how to evaluate changes

A

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microscopic interpretation of internal energy for a gas

in internal energy for practically important cases involving gases, liquids, and solids by using empirical data. To further our understanding of internal energy, consider a system we will often encounter in subsequent sections of the book, a system consisting of a gas contained in a tank. Let us develop a microscopic interpretation of internal energy by thinking of the energy attributed to the motions and configurations of the individual molecules, atoms, and subatomic particles making up the matter in the system. Gas molecules move about, encountering other molecules or the walls of the container. Part of the internal energy of the gas is the translational kinetic energy of the molecules. Other contributions to the internal energy include the kinetic energy due to rotation of the molecules relative to their centers of mass and the kinetic energy associated with vibrational motions within the molecules. In addition, energy is stored in the chemical bonds between the atoms that make up the molecules. Energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus. In dense gases, liquids, and solids, intermolecular forces play an important role in affecting the internal energy.

2.4 Energy Transfer by Heat Thus far, we have considered quantitatively only those interactions between a system and its surroundings that can be classed as work. However, closed systems also can interact with their surroundings in a way that cannot be categorized as work.

Gas

Hot plate

energy transfer by heat

when a gas in a rigid container interacts with a hot plate, the energy of the gas is increased even though no work is done. b b b b b This type of interaction is called an energy transfer by heat. On the basis of experiment, beginning with the work of Joule in the early part of the nineteenth century, we know that energy transfers by heat are induced only as a result of a temperature difference between the system and its surroundings and occur only in the direction of decreasing temperature. Because the underlying concept is so important in thermodynamics, this section is devoted to a further consideration of energy transfer by heat.

2.4.1 Sign Convention, Notation, and Heat Transfer Rate The symbol Q denotes an amount of energy transferred across the boundary of a system in a heat interaction with the system’s surroundings. Heat transfer into a system is taken to be positive, and heat transfer from a system is taken as negative. sign convention for heat transfer

A

HT_Modes A.7 – Tab a

Q . 0: heat transfer to the system Q , 0: heat transfer from the system This sign convention is used throughout the book. However, as was indicated for work, it is sometimes convenient to show the direction of energy transfer by an arrow on a sketch of the system. Then the heat transfer is regarded as positive in the direction of the arrow. The sign convention for heat transfer is just the reverse of the one adopted for work, where a positive value for W signifies an energy transfer from the system to the surroundings. These signs for heat and work are a legacy from engineers and scientists who were concerned mainly with steam engines and other devices that develop a work output from an energy input by heat transfer. For such applications, it was convenient to regard both the work developed and the energy input by heat transfer as positive quantities.

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2.4 Energy Transfer by Heat The value of a heat transfer depends on the details of a process and not just the end states. Thus, like work, heat is not a property, and its differential is written as dQ. The amount of energy transfer by heat for a process is given by the integral

heat is not a property

2

Q5

# dQ

(2.28)

1

where the limits mean “from state 1 to state 2” and do not refer to the values of heat at those states. As for work, the notion of “heat” at a state has no meaning, and the integral should never be evaluated as Q2 2 Q1. ? The net rate of heat transfer is denoted by Q. In principle, the amount of energy transfer by heat during a period of time can be found by integrating from time t1 to time t2 t2

Q5

# Q dt ?

rate of heat transfer

(2.29)

t1

To perform the integration, it is necessary to know how the rate of heat transfer varies with time. In some cases it is convenient to use the heat flux, q? , which is the heat transfer ? rate per unit of system surface area. The net rate of heat transfer, Q, is related to the ? heat flux q by the integral ?

Q5

# q dA ?

(2.30)

A

where A represents the area on the boundary of the system where heat transfer occurs. ? The units for heat transfer Q and heat transfer rate Q are the same as those introduced ? previously for W and W, respectively. The units for the heat flux are those of the heat transfer rate per unit area: kW/m2 or Btu/h ? ft2. The word adiabatic means without heat transfer. Thus, if a system undergoes a process involving no heat transfer with its surroundings, that process is called an adiabatic process.

BIOCONNECTIONS Medical researchers have found that by gradually increasing the temperature of cancerous tissue to 41–458C the effectiveness of chemotherapy and radiation therapy is enhanced for some patients. Different approaches can be used, including raising the temperature of the entire body with heating devices and, more selectively, by beaming microwaves or ultrasound onto the tumor or affected organ. Speculation about why a temperature increase may be beneficial varies. Some say it helps chemotherapy penetrate certain tumors more readily by dilating blood vessels. Others think it helps radiation therapy by increasing the amount of oxygen in tumor cells, making them more receptive to radiation. Researchers report that further study is needed before the efficacy of this approach is established and the mechanisms whereby it achieves positive results are known.

2.4.2 Heat Transfer Modes Methods based on experiment are available for evaluating energy transfer by heat. These methods recognize two basic transfer mechanisms: conduction and thermal radiation. In addition, empirical relationships are available for evaluating energy transfer involving a combined mode called convection. A brief description of each of these is given next. A detailed consideration is left to a course in engineering heat transfer, where these topics are studied in depth.

adiabatic

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56

Conduction

. Qx

T1

L

T2 Area, A x

Fig. 2.12 Illustration of Fourier’s conduction law.

Fourier’s law

Energy transfer by conduction can take place in solids, liquids, and gases. Conduction can be thought of as the transfer of energy from the more energetic particles of a substance to adjacent particles that are less energetic due to interactions between particles. The time rate of energy transfer by conduction is quantified macroscopically by Fourier’s law. As an elementary application, consider Fig. 2.12 showing a plane wall of thickness L at steady state, where the temperature T(x) varies linearly with position x. By Fourier’s law, the rate of heat transfer across any plane normal to the x ? direction, Qx, is proportional to the wall area, A, and the temperature gradient in the x direction, dT/dx: ?

Qx 5 2kA

dT dx

(2.31)

where the proportionality constant k is a property called the thermal conductivity. The minus sign is a consequence of energy transfer in the direction of decreasing temperature. in the case of Fig. 2.12 the temperature varies linearly; thus, the temperature gradient is T2 2 T1 dT 5 1, 02 dx L and the rate of heat transfer in the x direction is then ?

Qx 5 2kA c

A

HT_Modes A.7 – Tab b

T2 2 T1 d L

b b b b b

Values of thermal conductivity are given in Table A-19 for common materials. Substances with large values of thermal conductivity such as copper are good conductors, and those with small conductivities (cork and polystyrene foam) are good insulators.

Radiation Thermal radiation is emitted by matter as a result of changes in the electronic configurations of the atoms or molecules within it. The energy is transported by electromagnetic waves (or photons). Unlike conduction, thermal radiation requires no intervening medium to propagate and can even take place in a vacuum. Solid surfaces, gases, and liquids all emit, absorb, and transmit thermal radiation to varying degrees. The rate ? at which energy is emitted, Qe, from a surface of area A is quantified macroscopically by a modified form of the Stefan–Boltzmann law ?

Qe 5 esAT b4

Stefan–Boltzmann law

(2.32)

which shows that thermal radiation is associated with the fourth power of the absolute temperature of the surface, Tb. The emissivity, e, is a property of the surface that indicates how effectively the surface radiates 10 # e # 1.02, and s is the Stefan– Boltzmann constant: s 5 5.67 3 1028 W/ m2 ? K4 5 0.1714 3 1028 Btu/ h ? ft2 ? 8R4

A

HT_Modes A.7 – Tab d

In general, the net rate of energy transfer by thermal radiation between two surfaces involves relationships among the properties of the surfaces, their orientations with respect to each other, the extent to which the intervening medium scatters, emits, and absorbs thermal radiation, and other factors. A special case that occurs frequently is radiation exchange between a surface at temperature Tb and a much larger surrounding surface at Ts, as shown in Fig. 2.13. The net rate of radiant exchange between the smaller surface, whose area is A and emissivity is e, and the larger surroundings is ?

Qe 5 esA3T b4 2 T s44

(2.33)

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2.4 Energy Transfer by Heat

Cooling air flow Tf < Tb

57

. Qc

Surrounding surface at Ts

Tb

. Qe

A Surface of emissivity ε, area A, and temperature Tb

Wire leads Transistor Circuit board

Fig. 2.14 Illustration of Newton’s law of cooling.

Fig. 2.13 Net radiation exchange.

Convection Energy transfer between a solid surface at a temperature Tb and an adjacent gas or liquid at another temperature Tf plays a prominent role in the performance of many devices of practical interest. This is commonly referred to as convection. As an illustration, consider Fig. 2.14, where Tb . Tf . In this case, energy is transferred in the direction indicated by the arrow due to the combined effects of conduction within the air and the bulk motion of the air. The rate of energy transfer from the surface to the air can be quantified by the following empirical expression: ?

Qc 5 hA1Tb 2 Tf2

(2.34)

known as Newton’s law of cooling. In Eq. 2.34, A is the surface area and the proportionality factor h is called the heat transfer coefficient. In subsequent applications of Eq. 2.34, a minus sign may be introduced on the right side to conform to the sign convention for heat transfer introduced in Sec. 2.4.1. The heat transfer coefficient is not a thermodynamic property. It is an empirical parameter that incorporates into the heat transfer relationship the nature of the flow pattern near the surface, the fluid properties, and the geometry. When fans or pumps cause the fluid to move, the value of the heat transfer coefficient is generally greater than when relatively slow buoyancy-induced motions occur. These two general categories are called forced and free (or natural) convection, respectively. Table 2.1 provides typical values of the convection heat transfer coefficient for forced and free convection.

2.4.3 Closing Comments The first step in a thermodynamic analysis is to define the system. It is only after the system boundary has been specified that possible heat interactions with the surroundings are considered, for these are always evaluated at the system boundary.

TABLE 2.1

Typical Values of the Convection Heat Transfer Coefficient Applications

Free convection Gases Liquids Forced convection Gases Liquids

h (W/m2 ? K)

2–25 50–1000 25–250 50–20,000

h (Btu/h ? ft2 ? 8R)

0.35–4.4 8.8–180 4.4–44 8.8–3500

Newton’s law of cooling

HT_Modes A.7 – Tab c

A

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Chapter 2 Energy and the First Law of Thermodynamics In ordinary conversation, the term heat is often used when the word energy would be more correct thermodynamically. For example, one might hear, “Please close the door or ‘heat’ will be lost.” In thermodynamics, heat refers only to a particular means whereby energy is transferred. It does not refer to what is being transferred between systems or to what is stored within systems. Energy is transferred and stored, not heat. Sometimes the heat transfer of energy to, or from, a system can be neglected. This might occur for several reasons related to the mechanisms for heat transfer discussed above. One might be that the materials surrounding the system are good insulators, or heat transfer might not be significant because there is a small temperature difference between the system and its surroundings. A third reason is that there might not be enough surface area to allow significant heat transfer to occur. When heat transfer is neglected, it is because one or more of these considerations apply. In the discussions to follow, the value of Q is provided or it is an unknown in the analysis. When Q is provided, it can be assumed that the value has been determined by the methods introduced above. When Q is the unknown, its value is usually found by using the energy balance, discussed next.

2.5 Energy Accounting: Energy Balance for Closed Systems

first law of thermodynamics

As our previous discussions indicate, the only ways the energy of a closed system can be changed are through transfer of energy by work or by heat. Further, based on the experiments of Joule and others, a fundamental aspect of the energy concept is that energy is conserved; we call this the first law of thermodynamics. For further discussion of the first law, see the box.

Joule’s Experiments and the First Law In classic experiments conducted in the early part of the nineteenth century, Joule studied processes by which a closed system can be taken from one equilibrium state to another. In particular, he considered processes that involve work interactions but no heat interactions between the system and its surroundings. Any such process is an adiabatic process, in keeping with the discussion of Sec. 2.4.1. Based on his experiments Joule deduced that the value of the net work is the same for all adiabatic processes between two equilibrium states. In other words, the value of the net work done by or on a closed system undergoing an adiabatic process between two given states depends solely on the end states and not on the details of the adiabatic process. If the net work is the same for all adiabatic processes of a closed system between a given pair of end states, it follows from the definition of property (Sec. 1.3) that the net work for such a process is the change in some property of the system. This property is called energy. Following Joule’s reasoning, the change in energy between the two states is defined by E2 2 E1 5 2Wad

(a)

where the symbol E denotes the energy of a system and Wad represents the net work for any adiabatic process between the two states. The minus sign before the work term is in

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59

accord with the previously stated sign convention for work. Finally, note that since any arbitrary value E1 can be assigned to the energy of a system at a given state 1, no particular significance can be attached to the value of the energy at state 1 or at any other state. Only changes in the energy of a system have significance. The foregoing discussion is based on experimental evidence beginning with the experiments of Joule. Because of inevitable experimental uncertainties, it is not possible to prove by measurements that the net work is exactly the same for all adiabatic processes between the same end states. However, the preponderance of experimental findings supports this conclusion, so it is adopted as a fundamental principle that the work actually is the same. This principle is an alternative formulation of the first law, and has been used by subsequent scientists and engineers as a springboard for developing the conservation of energy concept and the energy balance as we know them today.

Summarizing Energy Concepts All energy aspects introduced in this book thus far are summarized in words as follows: net amount of energy net amount of energy change in the amount transferred in across transferred out across of energy contained ≥ within a system ¥ 5 ≥the system boundary by¥ 2 ≥the system boundary ¥ heat transfer during by work during the during some time the time interval time interval interval This word statement is just an accounting balance for energy, an energy balance. It requires that in any process of a closed system the energy of the system increases or decreases by an amount equal to the net amount of energy transferred across its boundary. The phrase net amount used in the word statement of the energy balance must be carefully interpreted, for there may be heat or work transfers of energy at many different places on the boundary of a system. At some locations the energy transfers may be into the system, whereas at others they are out of the system. The two terms on the right side account for the net results of all the energy transfers by heat and work, respectively, taking place during the time interval under consideration. The energy balance can be expressed in symbols as E2 2 E1 5 Q 2 W

(2.35a) energy balance

Introducing Eq. 2.27 an alternative form is ¢KE 1 ¢PE 1 ¢U 5 Q 2 W

(2.35b)

which shows that an energy transfer across the system boundary results in a change in one or more of the macroscopic energy forms: kinetic energy, gravitational potential energy, and internal energy. All previous references to energy as a conserved quantity are included as special cases of Eqs. 2.35. Note that the algebraic signs before the heat and work terms of Eqs. 2.35 are different. This follows from the sign conventions previously adopted. A minus sign appears before W because energy transfer by work from the system to the surroundings is taken to be positive. A plus sign appears before Q because it is regarded to be positive when the heat transfer of energy is into the system from the surroundings.

Energy_Bal_Closed _Sys A.8 – All Tabs

A

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Chapter 2 Energy and the First Law of Thermodynamics

BIOCONNECTIONS The energy required by animals to sustain life is derived from oxidation of ingested food. We often speak of food being burned in the body. This is an appropriate expression because experiments show that when food is burned with oxygen, approximately the same energy is released as when the food is oxidized in the body. Such an experimental device is the well-insulated, constant-volume calorimeter shown in Fig. 2.15. Thermometer

Access port – +

O2 Igniter Stirrer

Sample Water bath

Fig. 2.15 Constant-volume calorimeter.

Insulation

A carefully weighed food sample is placed in the chamber of the calorimeter together with oxygen (O2). The entire chamber is submerged in the calorimeter’s water bath. The chamber contents are then electrically ignited, fully oxidizing the food sample. The energy released during the reaction within the chamber results in an increase in calorimeter temperature. Using the measured temperature rise, the energy released can be calculated from an energy balance for the calorimeter as the system. This is reported as the calorie value of the food sample, usually in terms of kilocalorie (kcal), which is the “Calorie” seen on food labels.

2.5.1 Important Aspects of the Energy Balance Various special forms of the energy balance can be written. For example, the energy balance in differential form is dE 5 dQ 2 dW

(2.36)

where dE is the differential of energy, a property. Since Q and W are not properties, their differentials are written as dQ and dW, respectively. The instantaneous time rate form of the energy balance is time rate form of the energy balance

dE ? ? 5Q2W dt The rate form of the energy balance expressed in words is time rate of change net rate at which net rate at which of the energy energy is being energy is being contained within transferred in 5 2 ≥ ¥ ≥ ¥ ≥ transferred out ¥ the system at by heat transfer by work at time t at time t time t

(2.37)

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2.5 Energy Accounting: Energy Balance for Closed Systems Since the time rate of change of energy is given by dE d KE d PE dU 5 1 1 dt dt dt dt Equation 2.37 can be expressed alternatively as # # d KE d PE dU 1 1 5Q2W dt dt dt

(2.38)

Equations 2.35 through 2.38 provide alternative forms of the energy balance that are convenient starting points when applying the principle of conservation of energy to closed systems. In Chap. 4 the conservation of energy principle is expressed in forms suitable for the analysis of control volumes. When applying the energy balance in any of its forms, it is important to be careful about signs and units and to distinguish carefully between rates and amounts. In addition, it is important to recognize that the location of the system boundary can be relevant in determining whether a particular energy transfer is regarded as heat or work. consider Fig. 2.16, in which three alternative systems are shown that include a quantity of a gas (or liquid) in a rigid, well-insulated container. In Fig. 2.16a, the gas itself is the system. As current flows through the copper plate, there is an energy transfer from the copper plate to the gas. Since this energy transfer occurs as a result of the temperature difference between the plate and the gas, it is classified as a heat transfer. Next, refer to Fig. 2.16b, where the boundary is drawn to include the copper plate. It follows from the thermodynamic definition of work that the energy transfer that occurs as current crosses the boundary of this system must be regarded as work. Finally, in Fig. 2.16c, the boundary is located so that no energy is transferred across it by heat or work. b b b b b

Copper plate Gas or liquid

Rotating shaft

+

Gas W or liquid

Q W=0

– Electric generator

System boundary

System boundary

Insulation

+ –

Q=0

Mass (b)

(a)

+

Gas or liquid

System boundary

–

Q = 0, W = 0 (c)

Fig. 2.16 Alternative choices for system boundaries.

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62

Closing Comments Thus far, we have been careful to emphasize that the quantities symbolized by W and Q in the foregoing equations account for transfers of energy and not transfers of work and heat, respectively. The terms work and heat denote different means whereby energy is transferred and not what is transferred. However, to achieve economy of expression in subsequent discussions, W and Q are often referred to simply as work and heat transfer, respectively. This less formal manner of speaking is commonly used in engineering practice. The five solved examples provided in Secs. 2.5.2–2.5.4 bring out important ideas about energy and the energy balance. They should be studied carefully, and similar approaches should be used when solving the end-of-chapter problems. In this text, most applications of the energy balance will not involve significant kinetic or potential energy changes. Thus, to expedite the solutions of many subsequent examples and end-of-chapter problems, we indicate in the problem statement that such changes can be neglected. If this is not made explicit in a problem statement, you should decide on the basis of the problem at hand how best to handle the kinetic and potential energy terms of the energy balance.

2.5.2 Using the Energy Balance: Processes of Closed Systems The next two examples illustrate the use of the energy balance for processes of closed systems. In these examples, internal energy data are provided. In Chap. 3, we learn how to obtain internal energy and other thermodynamic property data using tables, graphs, equations, and computer software. cccc

EXAMPLE 2.2 c

Cooling a Gas in a Piston–Cylinder Four-tenths kilogram of a certain gas is contained within a piston–cylinder assembly. The gas undergoes a process for which the pressure–volume relationship is pV1.5 5 constant The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. The change in specific internal energy of the gas in the process is u2 2 u1 5 255 kJ/kg. There are no significant changes in kinetic or potential energy. Determine the net heat transfer for the process, in kJ. SOLUTION Known: A gas within a piston–cylinder assembly undergoes an expansion process for which the pressure–volume

relation and the change in specific internal energy are specified. Find: Determine the net heat transfer for the process. Schematic and Given Data: p

1

u2 – u1 = –55 kJ/kg

Engineering Model: 1. The gas is a closed system. 2. The process is described by

pV1.5 = constant

Gas

➊

pV1.5 = constant

2 Area = work

V

pV1.5 5 constant. 3. There is no change in the

kinetic or potential energy of the system. Fig. E2.2

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Analysis: An energy balance for the closed system takes the form

¢KE 0 1 ¢PE 0 1 ¢U 5 Q 2 W where the kinetic and potential energy terms drop out by assumption 3. Then, writing DU in terms of specific internal energies, the energy balance becomes m1u2 2 u12 5 Q 2 W where m is the system mass. Solving for Q Q 5 m1u2 2 u12 1 W The value of the work for this process is determined in the solution to part (a) of Example 2.1: W 5 117.6 kJ. The change in internal energy is obtained using given data as m1u2 2 u12 5 0.4 kg a255

kJ b 5 222 kJ kg

Substituting values ➋

Q 5 222 1 17.6 5 24.4 kJ

➊ The given relationship between pressure and volume allows the process to be represented by the path shown on the accompanying diagram. The area under the curve represents the work. Since they are not properties, the values of the work and heat transfer depend on the details of the process and cannot be determined from the end states only. ➋ The minus sign for the value of Q means that a net amount of energy has been transferred from the system to its surroundings by heat transfer.

✓ Skills Developed Ability to… ❑ define a closed system and

identify interactions on its boundary. ❑ apply the closed-system energy balance.

If the gas undergoes a process for which pV 5 constant and Du 5 0, determine the heat transfer, in kJ, keeping the initial pressure and given volumes fixed. Ans. 20.79 kJ.

In the next example, we follow up the discussion of Fig. 2.16 by considering two alternative systems. This example highlights the need to account correctly for the heat and work interactions occurring on the boundary as well as the energy change.

cccc

EXAMPLE 2.3 c

Considering Alternative Systems Air is contained in a vertical piston–cylinder assembly fitted with an electrical resistor. The atmosphere exerts a pressure of 14.7 lbf/in.2 on the top of the piston, which has a mass of 100 lb and a face area of 1 ft2. Electric current passes through the resistor, and the volume of the air slowly increases by 1.6 ft3 while its pressure remains constant. The mass of the air is 0.6 lb, and its specific internal energy increases by 18 Btu/lb. The air and piston are at rest initially and finally. The piston–cylinder material is a ceramic composite and thus a good insulator. Friction between the piston and cylinder wall can be ignored, and the local acceleration of gravity is g 5 32.0 ft/s2. Determine the heat transfer from the resistor to the air, in Btu, for a system consisting of (a) the air alone, (b) the air and the piston. SOLUTION Known: Data are provided for air contained in a vertical piston–cylinder fitted with an electrical resistor. Find: Considering each of two alternative systems, determine the heat transfer from the resistor to the air.

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Chapter 2 Energy and the First Law of Thermodynamics Schematic and Given Data: Piston

Piston

patm = 14.7 lbf/in.2

System boundary for part (a)

System boundary for part (b)

mpiston = 100 lb Apiston = 1 ft2 +

+

Air

Air –

– mair = 0.6 lb V2 – V1 = 1.6 ft3 Δuair = 18 Btu/lb

(a)

(b)

Fig. E2.3

Engineering Model: 1. Two closed systems are under consideration, as shown in the schematic. 2. The only significant heat transfer is from the resistor to the air, during which the air expands slowly and

its pressure remains constant. 3. There is no net change in kinetic energy; the change in potential energy of the air is negligible; and since the

piston material is a good insulator, the internal energy of the piston is not affected by the heat transfer. 4. Friction between the piston and cylinder wall is negligible. 5. The acceleration of gravity is constant; g 5 32.0 ft/s2. Analysis: (a) Taking the air as the system, the energy balance, Eq. 2.35, reduces with assumption 3 to

1¢KE 0 1 ¢PE 0 1 ¢U2air 5 Q 2 W Or, solving for Q Q 5 W 1 ¢Uair For this system, work is done by the force of the pressure p acting on the bottom of the piston as the air expands. With Eq. 2.17 and the assumption of constant pressure W5

#

V2

p dV 5 p1V2 2 V12

V1

To determine the pressure p, we use a force balance on the slowly moving, frictionless piston. The upward force exerted by the air on the bottom of the piston equals the weight of the piston plus the downward force of the atmosphere acting on the top of the piston. In symbols pApiston 5 mpiston g 1 patmApiston Solving for p and inserting values p5 5

mpiston g Apiston

1 patm

1100 lb2132.0 ft / s22 2

1 ft

`

1 lbf 1 ft2 lbf lbf ` ` ` 1 14.7 2 5 15.4 2 2 32.2 lb ? ft / s 144 in.2 in. in.

Thus, the work is W 5 p1V2 2 V12 5 a15.4

2 lbf 1 Btu 3 144 in. b11.6 ft 2 ` ` ` ` 5 4.56 Btu 2 2 778 ft ? lbf in. 1 ft

With DUair 5 mair(Duair), the heat transfer is Q 5 W 1 mair1¢uair2 5 4.56 Btu 1 10.6 lb2a18

Btu b 5 15.36 Btu lb

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(b) Consider next a system consisting of the air and the piston. The energy change of the overall system is the sum of the energy changes of the air and the piston. Thus, the energy balance, Eq. 2.35, reads 1¢KE 0 1 ¢PE 0 1 ¢U2air 1 1¢KE 0 1 ¢PE 1 ¢U 02piston 5 Q 2 W where the indicated terms drop out by assumption 3. Solving for Q Q 5 W 1 1¢PE2piston 1 1¢U2air For this system, work is done at the top of the piston as it pushes aside the surrounding atmosphere. Applying Eq. 2.17 W5

#

V2

p dV 5 patm1V2 2 V12

V1

5 a14.7

lbf 144 in.2 1 Btu b11.6 ft32 ` ` ` ` 5 4.35 Btu 2 778 ft ? lbf 1 ft2 in.

The elevation change, Dz, required to evaluate the potential energy change of the piston can be found from the volume change of the air and the area of the piston face as ¢z 5

V2 2 V1 1.6 ft3 5 5 1.6 ft Apiston 1 ft2

Thus, the potential energy change of the piston is 1¢PE2piston 5 mpiston g¢z 5 1100 lb2a32.0

ft 1 lbf 1 Btu b11.6 ft2 ` ` ` ` 5 0.2 Btu 2 2 s 32.2 lb ? ft / s 778 ft ? lbf

Finally, Q 5 W 1 1¢PE2piston 1 mair ¢uair 5 4.35 Btu 1 0.2 Btu 1 10.6 lb2a18

Btu b 5 15.35 Btu lb

➊ ➋ To within round-off, this answer agrees with the result of part (a). ➊ Although the value of Q is the same for each system, observe that the values for W differ. Also, observe that the energy changes differ, depending on whether the air alone or the air and the piston is the system. ➋ For the system of part (b), the following energy balance sheet gives a full accounting of the heat transfer of energy to the system: Energy In by Heat Transfer

15.35 Btu Disposition of the Energy In

• Energy stored Internal energy of the air Potential energy of the piston • Energy out by work

10.8 0.2 4.35 15.35

Btu Btu Btu Btu

(70.4%) ( 1.3%) (28.3%) (100%)

What is the change in potential energy of the air, in Btu?

Ans. 1 Supersonic

M 1 on lower branches s

Fig. 9.34 Intersection of Fanno and Rayleigh lines as a solution to the normal shock equations.

When combined with property relations for the particular fluid under consideration, Eqs. 9.46, 9.47, and 9.48 allow the downstream conditions to be determined for specified upstream conditions. Equation 9.49, which corresponds to Eq. 6.39, leads to the important conclusion that the downstream state must have greater specific entropy than the upstream state, or sy . sx.

FANNO AND RAYLEIGH LINES. The mass and energy equations, Eqs. 9.46 and 9.47, can be combined with property relations for the particular fluid to give an equation that when plotted on an h–s diagram is called a Fanno line. Similarly, the mass and momentum equations, Eqs. 9.46 and 9.48, can be combined to give an equation that when plotted on an h–s diagram is called a Rayleigh line. Fanno and Rayleigh lines are sketched on h–s coordinates in Fig. 9.34. It can be shown that the point of maximum entropy on each line, points a and b, corresponds to M 5 1. It also can be shown that the upper and lower branches of each line correspond, respectively, to subsonic and supersonic velocities. The downstream state y must satisfy the mass, energy, and momentum equations simultaneously, so state y is fixed by the intersection of the Fanno and Rayleigh lines passing through state x. Since sy . sx, it can be concluded that the flow across the shock can only pass from x to y. Accordingly, the velocity changes from supersonic before the shock (Mx . 1) to subsonic after the shock (My , 1). This conclusion is consistent with the discussion of cases e, f, and g in Fig. 9.32. A significant increase in pressure across the shock accompanies the decrease in velocity. Figure 9.34 also locates the stagnation states corresponding to the states upstream and downstream of the shock. The stagnation enthalpy does not change across the shock, but there is a marked decrease in stagnation pressure associated with the irreversible process occurring in the normal shock region.

9.14

Flow in Nozzles and Diffusers of Ideal Gases with Constant Specific Heats

The discussion of flow in nozzles and diffusers presented in Sec. 9.13 requires no assumption regarding the equation of state, and therefore the results obtained hold generally. Attention is now restricted to ideal gases with constant specific heats. This case is appropriate for many practical problems involving flow through nozzles and diffusers. The assumption of constant specific heats also allows the derivation of relatively simple closed-form equations.

Fanno line Rayleigh line

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9.14.1 Isentropic Flow Functions Let us begin by developing equations relating a state in a compressible flow to the corresponding stagnation state. For the case of an ideal gas with constant cp, Eq. 9.39 becomes To 5 T 1

V2 2cp

where To is the stagnation temperature. Using Eq. 3.47a, cp 5 kR/(k 2 1), together with Eqs. 9.37 and 9.38, the relation between the temperature T and the Mach number M of the flowing gas and the corresponding stagnation temperature To is To k21 2 511 M T 2

(9.50)

With Eq. 6.43, a relationship between the temperature T and pressure p of the flowing gas and the corresponding stagnation temperature To and the stagnation pressure po is po To k/1k212 5a b p T Introducing Eq. 9.50 into this expression gives po k 2 1 2 k/1k212 5 a1 1 M b p 2

(9.51)

Although sonic conditions may not actually be attained in a particular flow, it is convenient to have an expression relating the area A at a given section to the area A* that would be required for sonic flow (M 5 1) at the same mass flow rate and stagnation state. These areas are related through rAV 5 r*A*V* where r* and V* are the density and velocity, respectively, when M 5 1. Introducing the ideal gas equation of state, together with Eqs. 9.37 and 9.38, and solving for A/A* A 1 p* T 1y2 1 p*/ po T / To 1y2 5 a b 5 a b A* M p T* M p/ po T */ To where T* and p* are the temperature and pressure, respectively, when M 5 1. Then with Eqs. 9.50 and 9.51 A 1 2 k 2 1 2 1k112/21k212 5 ca b a1 1 M bd A* M k11 2

3.0 2.5 2.0 A 1.5 A* 1.0 0.5 0

0

0.5

1.0

1.5 M

2.0

2.5

Fig. 9.35 Variation of A/A* with Mach number in isentropic flow for k 5 1.4.

3.0

(9.52)

The variation of A/A* with M is given in Fig. 9.35 for k 5 1.4. The figure shows that a unique value of A/A* corresponds to any choice of M. However, for a given value of A/A* other than unity, there are two possible values for the Mach number, one subsonic and one supersonic. This is consistent with the discussion of Fig. 9.30, where it was found that a converging–diverging passage with a section of minimum area is required to accelerate a flow from subsonic to supersonic velocity. Equations 9.50, 9.51, and 9.52 allow the ratios T/To, p/po, and A/A* to be computed and tabulated with the Mach number as the single independent variable for a specified value of k. Table 9.2 provides a tabulation of this kind for k 5 1.4. Such a table facilitates the analysis of flow through nozzles and diffusers. Equations 9.50, 9.51, and 9.52 also can be readily evaluated using calculators and computer software such as Interactive Thermodynamics: IT.

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TABLE 9.2

Isentropic Flow Functions for an Ideal Gas with k 5 1.4 M

T/To

p/po

A/A*

0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40

1.000 00 0.998 00 0.992 06 0.982 32 0.968 99 0.952 38 0.932 84 0.910 75 0.886 52 0.860 58 0.833 33 0.805 15 0.776 40 0.747 38 0.718 39 0.689 65 0.661 38 0.633 72 0.606 80 0.580 72 0.555 56 0.531 35 0.508 13 0.485 91 0.464 68

1.000 00 0.993 03 0.972 50 0.939 47 0.895 62 0.843 02 0.784 00 0.720 92 0.656 02 0.591 26 0.528 28 0.468 35 0.412 38 0.360 92 0.314 24 0.272 40 0.235 27 0.202 59 0.174 04 0.149 24 0.127 80 0.109 35 0.093 52 0.079 97 0.068 40

` 5.8218 2.9635 2.0351 1.5901 1.3398 1.1882 1.094 37 1.038 23 1.008 86 1.000 00 1.007 93 1.030 44 1.066 31 1.1149 1.1762 1.2502 1.3376 1.4390 1.5552 1.6875 1.8369 2.0050 2.1931 2.4031

In Example 9.14, we consider the effect of back pressure on flow in a converging nozzle. The first step of the analysis is to check whether the flow is choked. cccc

EXAMPLE 9.14 c

Determining the Effect of Back Pressure: Converging Nozzle A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1.0 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k 5 1.4, determine the mass flow rate, in kg/s, and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa. SOLUTION Known: Air flows isentropically from specified stagnation conditions through a converging nozzle with a known exit area. Find: For back pressures of 500 and 784 kPa, determine the mass flow rate, in kg/s, and the exit Mach number. Schematic and Given Data: T 1

1

po

T

po

1

To

2 2

V1 ≈ 0 A2 = 0.001 m2

T1 = To = 360 K p1 = po = 1.0 MPa

2

To pB = p2 = 784 kPa

p2 = p* = 528 kPa pB = 500 kPa

s

p* = 528 kPa

s

Fig. E9.14

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Engineering Model: 1. The control volume shown in the accompanying sketch operates at steady state. 2. The air is modeled as an ideal gas with k 5 1.4. 3. Flow through the nozzle is isentropic. Analysis: The first step is to check whether the flow is choked. With k 5 1.4 and M 5 1.0, Eq. 9.51 gives ➊ p*/po 5 0.528. Since po 5 1.0 MPa, the critical pressure is p* 5 528 kPa. Thus, for back pressures of 528 kPa or less, the Mach number is unity at the exit and the nozzle is choked. (a) From the above discussion, it follows that for a back pressure of 500 kPa, the nozzle is choked. At the exit,

M2 5 1.0 and the exit pressure equals the critical pressure, p2 5 528 kPa. The mass flow rate is the maximum value that can be attained for the given stagnation properties. With the ideal gas equation of state, the mass flow rate is p2 # m 5 r2A2V2 5 A2 V2 RT2 The exit area A2 required by this expression is specified as 1023 m2. Since M 5 1 at the exit, the exit temperature T2 can be found from Eq. 9.50, which on rearrangement gives T2 5

To 360 K 5 5 300 K k21 2 1.4 2 1 2 M 11 b 112 11a 2 2

Then, with Eq. 9.37, the exit velocity V2 is V2 5 2kRT2 5

B

1.4 a

1 kg ? m/ s2 8314 N ? m b1300 K2 ` ` 5 347.2 m/ s 28.97 kg ? K 1N

Finally 1528 3 103 N/ m2211023 m221347.2 m/ s2 # 5 2.13 Kg/ s m5 8314 N ? m a b1300 K2 28.97 kg ? K (b) Since the back pressure of 784 kPa is greater than the critical pressure determined above, the flow throughout the nozzle is subsonic and the exit pressure equals the back pressure, p2 5 784 kPa. The exit Mach number can be found by solving Eq. 9.51 to obtain

M2 5 e

1y2 po 1k212yk 2 ca b 2 1d f k 2 1 p2

Inserting values 1/ 2 2 1 3 106 0.286 M2 5 e ca b 2 1 d f 5 0.6 1.4 2 1 7.84 3 105

With the exit Mach number known, the exit temperature T2 can be found from Eq. 9.50 as 336 K. The exit velocity is then V2 5 M2c2 5 M2 2kRT2 5 0.6

B

1.4a

8314 b13362 28.97

5 220.5 m/ s The mass flow rate is 1784 3 10321102321220.52 p2 # m 5 r2A2V2 5 A2 V2 5 RT2 18314/ 28.97213362 5 1.79 kg/ s

✓ Skills Developed Ability to… ❑ apply the ideal gas model

with constant k in the analysis of isentropic flow through a converging nozzle. ❑ understand when choked flow occurs in a converging nozzle for different back pressures. ❑ determine conditions at the throat and the mass flow rate for different back pressures and a fixed stagnation state.

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➊ The use of Table 9.2 reduces some of the computation required in the solution. It is left as an exercise to develop a solution using this table. Also, observe that the first step of the analysis is to check whether the flow is choked.

Using the isentropic flow functions in Table 9.2, determine the exit temperature and Mach number for a back pressure of 843 kPa. Ans. 342.9 K, 0.5.

9.14.2 Normal Shock Functions Next, let us develop closed-form equations for normal shocks for the case of an ideal gas with constant specific heats. For this case, it follows from the energy equation, Eq. 9.47b, that there is no change in stagnation temperature across the shock, Tox 5 Toy. Then, with Eq. 9.50, the following expression for the ratio of temperatures across the shock is obtained k21 2 Mx 2 5 Tx k21 2 11 My 2 11

Ty

(9.53)

Rearranging Eq. 9.48 px 1 rxV2x 5 py 1 ryV2y Introducing the ideal gas equation of state, together with Eqs. 9.37 and 9.38, the ratio of the pressure downstream of the shock to the pressure upstream is py 1 1 kM 2x 5 px 1 1 kM 2y

(9.54)

Similarly, Eq. 9.46 becomes py Ty Mx 5 px B Tx My The following equation relating the Mach numbers Mx and My across the shock can be obtained when Eqs. 9.53 and 9.54 are introduced in this expression

M2y 5

M2x 1

2 k21

2k M2x 2 1 k21

(9.55)

The ratio of stagnation pressures across a shock poy/pox is often useful. It is left as an exercise to show that

poy pox

k 2 1 2 1k112/21k212 My Mx 2 5 ± ≤ My k21 2 Mx 11 2 11

(9.56)

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Normal Shock Functions for an Ideal Gas with k 5 1.4 Mx

My

py/px

Ty/Tx

poy/pox

1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.10 2.20 2.30 2.40 2.50 2.60 2.70 2.80 2.90 3.00 4.00 5.00 10.00 `

1.000 00 0.911 77 0.842 17 0.785 96 0.739 71 0.701 09 0.668 44 0.640 55 0.616 50 0.595 62 0.577 35 0.561 28 0.547 06 0.534 41 0.523 12 0.512 99 0.503 87 0.495 63 0.488 17 0.481 38 0.475 19 0.434 96 0.415 23 0.387 57 0.377 96

1.0000 1.2450 1.5133 1.8050 2.1200 2.4583 2.8201 3.2050 3.6133 4.0450 4.5000 4.9784 5.4800 6.0050 6.5533 7.1250 7.7200 8.3383 8.9800 9.6450 10.333 18.500 29.000 116.50 `

1.0000 1.0649 1.1280 1.1909 1.2547 1.3202 1.3880 1.4583 1.5316 1.6079 1.6875 1.7704 1.8569 1.9468 2.0403 2.1375 2.2383 2.3429 2.4512 2.5632 2.6790 4.0469 5.8000 20.388 `

1.000 00 0.998 92 0.992 80 0.979 35 0.958 19 0.929 78 0.895 20 0.855 73 0.812 68 0.767 35 0.720 88 0.674 22 0.628 12 0.583 31 0.540 15 0.499 02 0.460 12 0.423 59 0.389 46 0.357 73 0.328 34 0.138 76 0.061 72 0.003 04 0.0

Since there is no area change across a shock, Eqs. 9.52 and 9.56 combine to give A*x A*y

5

poy pox

(9.57)

For specified values of Mx and specific heat ratio k, the Mach number downstream of a shock can be found from Eq. 9.55. Then, with Mx, My, and k known, the ratios Ty/Tx, py/px, and poy/pox can be determined from Eqs. 9.53, 9.54, and 9.56. Accordingly, tables can be set up giving My, Ty/Tx, py/px, and poy/pox versus the Mach number Mx as the single independent variable for a specified value of k. Table 9.3 is a tabulation of this kind for k 5 1.4. In the next example, we consider the effect of back pressure on flow in a converging– diverging nozzle. Key elements of the analysis include determining whether the flow is choked and if a normal shock exists.

cccc

EXAMPLE 9.15 c

Determining the Effect of Back Pressure: Converging–Diverging Nozzle A converging–diverging nozzle operating at steady state has a throat area of 1.0 in.2 and an exit area of 2.4 in.2 Air enters the nozzle with a negligible velocity at a pressure of 100 lbf/in.2 and a temperature of 500°R. For air as an ideal gas with k 5 1.4, determine the mass flow rate, in lb/s, the exit pressure, in lbf/in.2, and exit Mach ➊ number for each of the five following cases. (a) Isentropic flow with M 5 0.7 at the throat. (b) Isentropic flow with M 5 1 at the throat and the diverging portion acting as a diffuser. (c) Isentropic flow with M 5 1 at the throat and the diverging portion acting as a nozzle. (d) Isentropic flow through the nozzle with a normal

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shock standing at the exit. (e) A normal shock stands in the diverging section at a location where the area is 2.0 in.2 Elsewhere in the nozzle, the flow is isentropic. SOLUTION Known: Air flows from specified stagnation conditions through a converging–diverging nozzle having a known

throat and exit area. Find: The mass flow rate, exit pressure, and exit Mach number are to be determined for each of five cases. Schematic and Given Data: Engineering Model: A2 = 2.4 in.2

V1 ≈ 0 lbf/in.2

p1 = po = 100 T1 = To = 500°R

T

1

At = 1.0 in.2

T

2 p = 95.9 lbf/in.2 2

2

pt Mt = 0.7

throughout, except for case (e), where a shock stands in the diverging section.

p2 = 95.3 lbf/in.2 pt = p*

s

po = 100 lbf/in.2 To = 500°R

s

Case (b)

T

Stagnation state Stagnation state associated with associated with state x state y 1

pox

poy

p2

2

y Sonic state associated with state y

pt = p* Mt = 1 x p2 = 6.84 lbf/in.2 2 Cases (c) and (d)

3. Flow through the nozzle is isentropic

Mt = 1

T 1

2. The air is modeled as an ideal gas with

po = 100 lbf/in.2 To = 500°R

1

To = 500°R

Case (a)

nying sketch operates at steady state. The T–s diagrams provided locate states within the nozzle. k 5 1.4.

po = 100 lbf/in.2

1

2

1. The control volume shown in the accompa-

Normal shock

Sonic state associated with state x s

s

Case (e)

Fig. E9.15

Analysis: (a) The accompanying T–s diagram shows the states visited by the gas in this case. The following are known:

the Mach number at the throat, Mt 5 0.7, the throat area, At 5 1.0 in.2, and the exit area, A2 5 2.4 in.2 The exit Mach number M2, exit temperature T2, and exit pressure p2 can be determined using the identity A2 A 2 At 5 A* At A* With Mt 5 0.7, Table 9.2 gives At /A* 5 1.09437. Thus A2 2.4 in.2 5a b 11.094372 5 2.6265 A* 1.0 in.2

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The flow throughout the nozzle, including the exit, is subsonic. Accordingly, with this value for A2/A*, Table 9.2 gives M2 < 0.24. For M2 5 0.24, T2/To 5 0.988, and p2/po 5 0.959. Since the stagnation temperature and pressure are 500°R and 100 lbf/in.2, respectively, it follows that T2 5 494°R and p2 5 95.9 lbf/in.2 The velocity at the exit is V2 5 M2c2 5 M2 2kRT2 5 0.24

B

1.4 a

1545 ft ? lbf 32.2 lb ? ft / s2 b 14948R2 ` ` 28.97 lb ? 8R 1 lbf

5 262 ft/ s The mass flow rate is p2 # m 5 r2 A2V2 5 A2 V2 RT2 5

195.9 lbf/ in.2212.4 in.221262 ft/ s2 1545 ft ? lbf a b 14948R2 28.97 lb ? 8R

5 2.29 lb/ s

(b) The accompanying T–s diagram shows the states visited by the gas in this case. Since M 5 1 at the throat,

we have At 5 A*, and thus A2/A* 5 2.4. Table 9.2 gives two Mach numbers for this ratio: M < 0.26 and M < 2.4. The diverging portion acts as a diffuser in the present part of the example; accordingly, the subsonic value is appropriate. The supersonic value is appropriate in part (c). Thus, from Table 9.2 we have at M2 5 0.26, T2/To 5 0.986, and p2/po 5 0.953. Since To 5 500°R and po 5 100 lbf/in.2, it follows that T2 5 493°R and p2 5 95.3 lbf/in.2 The velocity at the exit is V2 5 M2c2 5 M2 1kRT2 5 0.26

B

11.42 a

1545 b 14932Z32.2Z 5 283 ft/ s 28.97

The mass flow rate is 195.3212.4212832 p2 # 5 2.46 lb/ s m5 A2 V2 5 RT2 1545 a b14932 28.97 This is the maximum mass flow rate for the specified geometry and stagnation conditions: the flow is choked. (c) The accompanying T–s diagram shows the states visited by the gas in this case. As discussed in part (b), the

exit Mach number in the present part of the example is M2 5 2.4. Using this, Table 9.2 gives p2/p0 5 0.0684. With po 5 100 lbf/in.2, the pressure at the exit is p2 5 6.84 lbf/in.2 Since the nozzle is choked, the mass flow rate is the same as found in part (b). (d) Since a normal shock stands at the exit and the flow upstream of the shock is isentropic, the Mach number Mx

and the pressure px correspond to the values found in part (c), Mx 5 2.4, px 5 6.84 lbf/in.2 Then, from Table 9.3, My < 0.52 and py/px 5 6.5533. The pressure downstream of the shock is thus 44.82 lbf/in.2 This is the exit pressure. The mass flow is the same as found in part (b). (e) The accompanying T–s diagram shows the states visited by the gas. It is known that a shock stands in the

diverging portion where the area is Ax 5 2.0 in.2 Since a shock occurs, the flow is sonic at the throat, so Ax* 5 At 5 1.0 in.2 The Mach number Mx can then be found from Table 9.2, by using Ax / A*x 5 2, as Mx 5 2.2. The Mach number at the exit can be determined using the identity A2 A2 A*x 5 a ba b A*y A*x A*y

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Introducing Eq. 9.57 to replace A*x /A*y , this becomes A2 A2 poy 5 a ba b A*y A*x pox where pox and poy are the stagnation pressures before and after the shock, respectively. With Mx 5 2.2, the ratio of stagnation pressures is obtained from Table 9.3 as poy /pox 5 0.62812. Thus 2

A2 2.4 in. 5a b 10.628122 5 1.51 A*y 1.0 in.2 Using this ratio and noting that the flow is subsonic after the shock, Table 9.2 gives M2 < 0.43, for which p2 /poy 5 0.88. The pressure at the exit can be determined using the identity p2 5 a

poy p2 lbf ba b pox 5 10.88210.6282 a100 2 b 5 55.3 lbf / in.2 poy pox in.

Since the flow is choked, the mass flow rate is the same as that found in part (b). ➊ With reference to cases labeled on Fig. 9.32, part (a) of the present example corresponds to case c on the figure, part (b) corresponds to case d, part (c) corresponds to case i, part (d) corresponds to case g, and part (e) corresponds to case f.

✓ Skills Developed Ability to… ❑ analyze isentropic flow

through a converging– diverging nozzle for an ideal gas with constant k. ❑ understand the occurrence of choked flow and normal shocks in a converging– diverging nozzle for different back pressures. ❑ analyze the flow through a converging–diverging nozzle when normal shocks are present for an ideal gas with constant k.

What is the stagnation temperature, in 8R, corresponding to the exit state for case (e)? Ans. 500°R.

c CHAPTER SUMMARY AND STUDY GUIDE In this chapter, we have studied the thermodynamic modeling of internal combustion engines, gas turbine power plants, and compressible flow in nozzles and diffusers. The modeling of cycles is based on the use of air-standard analysis, where the working fluid is considered to be air as an ideal gas. The processes in internal combustion engines are described in terms of three air-standard cycles: the Otto, Diesel, and dual cycles, which differ from each other only in the way the heat addition process is modeled. For these cycles, we have evaluated the principal work and heat transfers along with two important performance parameters: the mean effective pressure and the thermal efficiency. The effect of varying compression ratio on cycle performance is also investigated. The performance of simple gas turbine power plants is described in terms of the air-standard Brayton cycle. For this cycle, we evaluate the principal work and heat transfers along with two important performance parameters: the back work ratio and the thermal efficiency. We also consider the effects on performance of irreversibilities and of varying compressor pressure ratio. Three modifications of the simple cycle to improve performance are introduced: regeneration, reheat, and compression with intercooling. Applications related to gas turbines are also considered, including combined gas turbine–vapor power cycles, integrated gasification combined-cycle (IGCC) power plants, and

gas turbines for aircraft propulsion. In addition, the Ericsson and Stirling cycles are introduced. The chapter concludes with the study of compressible flow through nozzles and diffusers. We begin by introducing the momentum equation for steady, one-dimensional flow, the velocity of sound, and the stagnation state. We then consider the effects of area change and back pressure on performance in both subsonic and supersonic flows. Choked flow and the presence of normal shocks in such flows are investigated. Tables are introduced to facilitate analysis for the case of ideal gases with constant specific heat ratio, k 5 1.4. The following list provides a study guide for this chapter. When your study of the text and end-of-chapter exercises has been completed, you should be able to c write out the meanings of the terms listed in the margin

throughout the chapter and understand each of the related concepts. The subset of key concepts listed on p. 570 is particularly important. c sketch p–y and T–s diagrams of the Otto, Diesel, and dual cycles. Apply the closed system energy balance and the second law along with property data to determine the performance of these cycles, including mean effective pressure, thermal efficiency, and the effects of varying compression ratio.

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c sketch schematic diagrams and accompanying T–s diagrams of

the Brayton cycle and modifications involving regeneration, reheat, and compression with intercooling. In each case, be able to apply mass and energy balances, the second law, and property data to determine gas turbine power cycle performance, including thermal efficiency, back work ratio, net power output, and the effects of varying compressor pressure ratio. c analyze the performance of gas turbine–related applications involving combined gas turbine–vapor power plants, IGCC

power plants, and aircraft propulsion. You also should be able to apply the principles of this chapter to Ericsson and Stirling cycles. c discuss for nozzles and diffusers the effects of area change in subsonic and supersonic flows, the effects of back pressure on mass flow rate, and the appearance and consequences of choking and normal shocks. c analyze the flow in nozzles and diffusers of ideal gases with constant specific heats, as in Examples 9.14 and 9.15.

c KEY ENGINEERING CONCEPTS mean effective pressure, p. 495 air-standard analysis, p. 495 Otto cycle, p. 497 Diesel cycle, p. 502 dual cycle, p. 506 Brayton cycle, p. 511 regenerator, p. 521

regenerator effectiveness, p. 523 reheat, p. 526 intercooler, p. 528 combined cycle, pp. 537, 544 turbojet engine, p. 546 compressible flow, p. 550 momentum equation, p. 552

velocity of sound, p. 553 Mach number, p. 554 subsonic and supersonic flow, p. 554 stagnation state, p. 555 choked flow, p. 559 normal shock, p. 559

c KEY EQUATIONS mep 5

net work for one cycle displacement volume

(9.1) p. 495

Mean effective pressure for reciprocating piston engines

Otto Cycle h5

1u3 2 u22 2 1u4 2 u12 u4 2 u1 512 u3 2 u2 u3 2 u2 1 h 5 1 2 k21 r

(9.3) p. 498

Thermal efficiency (Figure 9.3)

(9.8) p. 499

Thermal efficiency (cold-air standard basis)

Diesel Cycle h5

Wcycle / m

Q41/ m u4 2 u1 512 Q23 / m Q23 / m h3 2 h2 k rc 2 1 1 h 5 1 2 k21 c d k1rc 2 12 r 512

(9.11) p. 503

Thermal efficiency (Figure 9.5)

(9.13) p. 503

Thermal efficiency (cold-air standard basis)

Brayton Cycle # # # # 1h3 2 h42 2 1h2 2 h12 Wt / m 2 Wc / m # # 5 h5 h3 2 h2 Qin/ m # # Wc / m h2 2 h1 bwr 5 # # 5 h3 2 h4 Wt / m 1 h512 1p2 / p121k212/k hx 2 h2 hreg 5 h4 2 h2

(9.19) p. 512

Thermal efficiency (Figure 9.9)

(9.20) p. 512

Back work ratio (Figure 9.9)

(9.25) p. 516

Thermal efficiency (cold-air standard basis)

(9.27) p. 523

Regenerator effectiveness for the regenerative gas turbine cycle (Figure 9.14)

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Compressible Flow in Nozzles and Diffusers # F 5 m 1V2 2 V12

(9.31) p. 552

Momentum equation for steady-state one-dimensional flow

c 5 2kRT M 5 V/ c ho 5 h 1 V 2/ 2 To k21 2 511 M T 2

(9.37) p. 554 (9.38) p. 554 (9.39) p. 555

Ideal gas velocity of sound

(9.50) p. 562

Isentropic flow function relating temperature and stagnation temperature (constant k)

po To k/1k212 k 2 1 2 k/1k212 5a b 5 a1 1 M b p T 2

(9.51) p. 562

Isentropic flow function relating pressure and stagnation pressure (constant k)

Mach number Stagnation enthalpy

c EXERCISES: THINGS ENGINEERS THINK ABOUT 1. Diesel engines are said to produce higher torque than gasoline engines. What does that mean? 2. Formula One race cars have 2.4 liter engines. What does that signify? How is your car’s engine sized in liters? 3. The ideal Brayton and Rankine cycles are composed of the same four processes, yet look different when represented on a T–s diagram. Explain. 4. The term regeneration is used to describe the use of regenerative feedwater heaters in vapor power plants and regenerative heat exchangers in gas turbines. In what ways are the purposes of these devices similar? How do they differ? 5. You jump off a raft into the water in the middle of a lake. What direction does the raft move? Explain. 6. What is the purpose of a rear diffuser on a race car? 7. What is the meaning of the octane rating that you see posted on gas pumps? Why is it important to consumers? 8. Why aren’t jet engines of airliners fitted with screens to avoid birds being pulled into the intake? 9. When did the main power plant providing electricity to your residence begin generating power? How long is it expected to continue operating?

11. A nine-year-old camper is suddenly awakened by a metallic click coming from the direction of a railroad track passing close to her camping area; soon afterward, she hears the deep growling of a diesel locomotive pulling an approaching train. How would you interpret these different sounds to her? 12. Automakers have developed prototype gas turbine– powered vehicles, but the vehicles have not been generally marketed to consumers. Why? 13. In making a quick stop at a friend’s home, is it better to let your car’s engine idle or turn it off and restart when you leave? 14. How do today’s more effective diesel engine exhaust treatment systems work? 15. What is the range of fuel efficiencies, in miles per gallon, you get with you car? At what speeds, in miles per hour, is the peak achieved? 16. Where is Marcellus shale and why is it significant? 17. Does your state regulate the practice of venting highpressure natural gas to clean debris from pipelines leading to power plant gas turbines? What hazards are associated with this practice?

10. What is the purpose of the gas turbine–powered auxiliary power units commonly seen at airports near commercial aircraft?

c PROBLEMS: DEVELOPING ENGINEERING SKILLS Otto, Diesel, and Dual Cycles 9.1 An air-standard Otto cycle has a compression ratio of 9. At the beginning of compression, p1 5 100 kPa and T1 5 300 K. The heat addition per unit mass of air is 1350 kJ/kg. Determine (a) the net work, in kJ per kg of air. (b) the thermal efficiency of the cycle. (c) the mean effective pressure, in kPa.

(d) the maximum temperature in the cycle, in K. (e) To investigate the effects of varying compression ratio, plot each of the quantities calculated in parts (a) through (d) for compression ratios ranging from 1 to 12. 9.2 Solve Problem 9.1 on a cold air-standard basis with specific heats evaluated at 300 K. 9.3 At the beginning of the compression process of an airstandard Otto cycle, p1 5 1 bar, T1 5 290 K, V1 5 400 cm3.

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The maximum temperature in the cycle is 2200 K and the compression ratio is 8. Determine (a) (b) (c) (d) (e) the (f)

the heat addition, in kJ. the net work, in kJ. the thermal efficiency. the mean effective pressure, in bar. Develop a full accounting of the exergy transferred to air during the heat addition, in kJ. Devise and evaluate an exergetic efficiency for the cycle.

Let T0 5 290 K, p0 5 1 bar. 9.4 Plot each of the quantities specified in parts (a) through (d) of Problem 9.3 versus the compression ratio ranging from 2 to 12.

for maximum cycle temperatures ranging from 2000 to 50008R and compression ratios of 6, 8, and 10. 9.10 Solve Problem 9.9 on a cold air-standard basis using k 5 1.4. 9.11 Consider an air-standard Otto cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.3. The mass of air is 0.002 kg. Determine (a) (b) (c) (d)

9.5 Solve Problem 9.3 on a cold air-standard basis with specific heats evaluated at 300 K. 9.6 A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.7 lbf/in.2, a temperature of 808F, and a volume of 0.0196 ft3 at the beginning of compression. The compression ratio is 10, and maximum pressure in the cycle is 1080 lbf/in.2 Determine, using a cold air-standard analysis with k 5 1.4, the power developed by the engine, in horsepower, and the mean effective pressure, in lbf/in.2

Fig. P9.6 9.7 An air-standard Otto cycle has a compression ratio of 8 and the temperature and pressure at the beginning of the compression process are 5208R and 14.2 lbf/in.2, respectively. The mass of air is 0.0015 lb. The heat addition is 0.9 Btu. Determine (a) the maximum temperature, in 8R. (b) the maximum pressure, in lbf/in.2 (c) the thermal efficiency. (d) To investigate the effects of varying compression ratio, plot each of the quantities calculated in parts (a) through (c) for compression ratios ranging from 2 to 12. 9.8 Solve Problem 9.7 on a cold air-standard basis with specific heats evaluated at 5208R. 9.9 At the beginning of the compression process in an airstandard Otto cycle, p1 5 14.7 lbf/in.2 and T1 5 5308R. Plot the thermal efficiency and mean effective pressure, in lbf/in.2,

the the the the

heat addition and the heat rejection, each in kJ. net work, in kJ. thermal efficiency. mean effective pressure, in kPa.

State

T (K)

p (kPa)

u (kJ/kg)

1 2 3 4

305 367.4 960 458.7

85 767.9 2006 127.8

217.67 486.77 725.02 329.01

9.12 Consider a cold air-standard Otto cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.3. The heat rejection from the cycle is 86 Btu per lb of air. Assuming cy 5 0.172 Btu/lb ? 8R, determine (a) (b) (c) (d)

the the the the

compression ratio. net work per unit mass of air, in Btu/lb. thermal efficiency. mean effective pressure, in lbf/in.2 State

T (8R)

p (lbf/in.2)

1 2 3 4

500 1204.1 2408.2 1000

47.50 1030 2060 95

9.13 Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having n 5 1.3. The compression ratio is 9 for the modified cycle. At the beginning of compression, p1 5 1 bar and T1 5 300 K and V1 5 2270 cm3. The maximum temperature during the cycle is 2000 K. Determine (a) the heat transfer and work in kJ, for each process in the modified cycle. (b) the thermal efficiency. (c) the mean effective pressure, in bar. 9.14 A four-cylinder, four-stroke internal combustion engine has a bore of 2.55 in. and a stroke of 2.10 in. The clearance volume is 12% of the cylinder volume at bottom dead center and the crankshaft rotates at 3600 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.6 lbf/in.2 and a temperature of 1008F at the beginning of compression. The maximum temperature in the cycle is 52008R. Based on this model, calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower. 9.15 At the beginning of the compression process in an airstandard Otto cycle, p1 5 1 bar and T1 5 300 K. The maximum

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Problems: Developing Engineering Skills cycle temperature is 2000 K. Plot the net work per unit of mass, in kJ/kg, the thermal efficiency, and the mean effective pressure, in bar, versus the compression ratio ranging from 2 to 14. 9.16 Investigate the effect of maximum cycle temperature on the net work per unit mass of air for air-standard Otto cycles with compression ratios of 5, 8, and 11. At the beginning of the compression process, p1 5 1 bar and T1 5 295 K. Let the maximum temperature in each case vary from 1000 to 2200 K. 9.17 The pressure-specific volume diagram of the air-standard Lenoir cycle is shown in Fig. P9.17. The cycle consists of constant volume heat addition, isentropic expansion, and constant pressure compression. For the cycle, p1 5 14.7 lbf/in.2 and T1 5 5408R. The mass of air is 4.24 3 1023 lb, and the maximum cycle temperature is 16008R. Assuming cy 5 0.171 Btu/lb ? 8R, determine for the cycle

573

9.19 On a cold air-standard basis, derive an expression for the thermal efficiency of the Atkinson cycle (see Fig. P9.18) in terms of the volume ratio during the isentropic compression, the pressure ratio for the constant volume process, and the specific heat ratio. Compare the thermal efficiencies of the cold air-standard Atkinson and Otto cycles, each having the same compression ratio and maximum temperature. Discuss. 9.20 The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are 95 kPa and 300 K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2150 K. Determine (a) (b) (c) (d)

the the the the

compression ratio. cutoff ratio. thermal efficiency of the cycle. mean effective pressure, in kPa.

9.21 Solve Problem 9.20 on a cold air-standard basis with specific heats evaluated at 300 K.

(a) the net work, in Btu. (b) the thermal efficiency.

9.22 Consider an air-standard Diesel cycle. At the beginning of compression, p1 5 14.0 lbf/in.2 and T1 5 5208R. The mass of air is 0.145 lb and the compression ratio is 17. The maximum temperature in the cycle is 40008R. Determine

p 2 1600°R

(a) the heat addition, in Btu. (b) the thermal efficiency. (c) the cutoff ratio.

s=c

9.23 Solve Problem 9.22 on a cold air-standard basis with specific heats evaluated at 5208R. 1

9.24 Consider an air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5. Determine

3

540°R

v

Fig. P9.17

9.18 The pressure-specific volume diagram of the air-standard Atkinson cycle is shown in Fig. P9.18. The cycle consists of isentropic compression, constant volume heat addition, isentropic expansion, and constant pressure compression. For a particular Atkinson cycle, the compression ratio during isentropic compression is 8.5. At the beginning of this compression process, p1 5 100 kPa and T1 5 300 K. The constant volume heat addition per unit mass of air is 1400 kJ/kg. (a) Sketch the cycle on T–s coordinates. Determine (b) the net work, in kJ per kg of air, (c) the thermal efficiency of the cycle, and (d) the mean effective pressure, in kPa. p 3

2

the the the the

cutoff ratio. heat addition per unit mass, in kJ/kg. net work per unit mass, in kJ/kg. thermal efficiency.

State

T (K)

p (kPa)

u (kJ/kg)

h (kJ/kg)

1 2 3 4

380 1096.6 1864.2 875.2

100 5197.6 5197.6 230.1

271.69 842.40 1548.47 654.02

380.77 1157.18 2082.96 905.26

9.25 Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5. For k 5 1.4, cy 5 0.718 kJ/(kg ? K), and cp 5 1.005 kJ/(kg ? K), determine (a) the heat transfer per unit mass and work per unit mass for each process, in kJ/kg, and the cycle thermal efficiency. (b) the exergy transfers accompanying heat and work for each process, in kJ/kg. Devise and evaluate an exergetic efficiency for the cycle. Let T0 5 300 K and p0 5 100 kPa.

s=c

s=c 1

(a) (b) (c) (d)

4 v

Fig. P9.18

State

T (K)

p (kPa)

y (m3/kg)

1 2 3 4

340 1030.7 2061.4 897.3

100 4850.3 4850.3 263.9

0.9758 0.06098 0.1220 0.9758

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9.26 Consider an air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5. Determine (a) (b) (c) (d)

the the the the

9.32 The thermal efficiency, h, of a cold air-standard diesel cycle can be expressed by Eq. 9.13:

cutoff ratio. heat addition per unit mass, in Btu/lb. net work per unit mass, in Btu/lb. thermal efficiency.

State

T (8R)

1 2 3 4

520 1502.5 3000 1527.1

2

p (lbf/in. )

h512

u (Btu/lb)

h (Btu/lb)

88.62 266.84 585.04 271.66

124.27 369.84 790.68 376.36

14.2 657.8 657.8 41.8

9.27 Consider a cold air-standard Diesel cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.5. For k 5 1.4, cy 5 0.172 Btu/(lb ? 8R), and cp 5 0.240 Btu/(lb ? 8R), determine (a) heat transfer per unit mass and work per unit mass for each process, in Btu/lb, and the cycle thermal efficiency. (b) exergy transfers accompanying heat and work for each process, in Btu/lb. Devise and evaluate an exergetic efficiency for the cycle. Let T0 5 5408R and p0 5 14.7 lbf/in.2 State

T (8R)

p (lbf/in.2)

y (ft3/lb)

1 2 3 4

540 1637 3274 1425.1

14.7 713.0 713.0 38.8

13.60 0.85 1.70 13.60

9.28 The displacement volume of an internal combustion engine is 5.6 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.4. The state of the air at the beginning of compression is fixed by p1 5 95 kPa, T1 5 278C, and V1 5 6.0 liters. Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1500 times per min. 9.29 The state at the beginning of compression of an air-standard Diesel cycle is fixed by p1 5 100 kPa and T1 5 310 K. The compression ratio is 15. For cutoff ratios ranging from 1.5 to 2.5, plot (a) (b) (c) (d)

the the the the

(a) the mass of air, in kg. (b) the heat addition and heat rejection per cycle, each in kJ. (c) the net work, in kJ, and the thermal efficiency.

maximum temperature, in K. pressure at the end of the expansion, in kPa. net work per unit mass of air, in kJ/kg. thermal efficiency.

9.30 An air-standard Diesel cycle has a maximum temperature of 1800 K. At the beginning of compression, p1 5 95 kPa and T1 5 300 K. The mass of air is 12 g. For compression ratios ranging from 15 to 25, plot (a) the net work of the cycle, in kJ. (b) the thermal efficiency. (c) the mean effective pressure, in kPa. 9.31 At the beginning of compression in an air-standard Diesel cycle, p1 5 170 kPa, V1 5 0.016 m3, and T1 5 315 K. The compression ratio is 15 and the maximum cycle temperature is 1400 K. Determine

1 r

k21

c

r ck 2 1 d k1rc 2 12

where r is compression ratio and rc is cutoff ratio. Derive this expression. 9.33 At the beginning of the compression process in an airstandard Diesel cycle, p1 5 1 bar and T1 5 300 K. For maximum cycle temperatures of 1200, 1500, 1800, and 2100 K, plot the heat addition per unit of mass, in kJ/kg, the net work per unit of mass, in kJ/kg, the mean effective pressure, in bar, and the thermal efficiency, each versus compression ratio ranging from 5 to 20. 9.34 An air-standard dual cycle has a compression ratio of 9. At the beginning of compression, p1 5 100 kPa, T1 5 300 K, and V1 5 14 L. The heat addition is 22.7 kJ, with one half added at constant volume and one half added at constant pressure. Determine (a) the temperatures at the end of each heat addition process, in K. (b) the net work of the cycle per unit mass of air, in kJ/kg. (c) the thermal efficiency. (d) the mean effective pressure, in kPa. 9.35 For the cycle in Problem 9.34, plot each of the quantities calculated in parts (a) through (d) versus the ratio of constant-volume heat addition to total heat addition varying from 0 to 1. Discuss. 9.36 Solve Problem 9.34 on a cold air-standard basis with specific heats evaluated at 300 K. 9.37 The thermal efficiency, h, of a cold air-standard dual cycle can be expressed as h512

1

c

rpr kc 2 1

r k21 1rp 2 12 1 krp1rc 2 12

d

where r is compression ratio, rc is cutoff ratio, and rp is the pressure ratio for the constant volume heat addition. Derive this expression. 9.38 Consider an air-standard dual cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.7. If the mass of air is 0.05 kg, determine (a) (b) (c) (d) (e)

the the the the the

cutoff ratio. heat addition to the cycle, in kJ. heat rejection from the cycle, in kJ. net work, in kJ. thermal efficiency.

State

T (K)

p (kPa)

u (kJ/kg)

h (kJ/kg)

1 2 3 4 5

300 862.4 1800 1980 840.3

95 4372.8 9126.9 9126.9 265.7

214.07 643.35 1487.2 1659.5 625.19

300.19 890.89 2003.3 2227.1 866.41

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9.39 The pressure and temperature at the beginning of compression in an air-standard dual cycle are 14.0 lbf/in.2 and 5208R, respectively. The compression ratio is 15 and the heat addition per unit mass of air is 800 Btu/lb. At the end of the constant volume heat addition process, the pressure is 1200 lbf/in.2 Determine

(a) determine the net power developed per unit mass flowing, in kJ/kg, and the thermal efficiency for a compressor pressure ratio of 8. (b) plot the net power developed per unit mass flowing, in kJ/kg, and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50.

(a) the net work of the cycle per unit mass of air, in Btu/lb. (b) the heat rejection for the cycle per unit mass of air, in Btu/lb. (c) the thermal efficiency. (d) the cutoff ratio. (e) To investigate the effects of varying compression ratio, plot each of the quantities calculated in parts (a) through (d) for compression ratios ranging from 10 to 28.

9.44 An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 300 K and 100 kPa and a fixed turbine inlet temperature of 1700 K. For the cycle,

9.40 An air-standard dual cycle has a compression ratio of 16. At the beginning of compression, p1 5 14.5 lbf/in.2, V1 5 0.5 ft3, and T1 5 508F. The pressure doubles during the constant volume heat addition process. For a maximum cycle temperature of 30008R, determine (a) the heat addition to the cycle, in Btu. (b) the net work of the cycle, in Btu. (c) the thermal efficiency. (d) the mean effective pressure, in lbf/in.2 (e) To investigate the effects of varying maximum cycle temperature, plot each of the quantities calculated in parts (a) through (d) for maximum cycle temperatures ranging from 3000 to 40008R. 9.41 At the beginning of the compression process in an airstandard dual cycle, p1 5 1 bar and T1 5 300 K. The total heat addition is 1000 kJ/kg. Plot the net work per unit of mass, in kJ/kg, the mean effective pressure, in bar, and the thermal efficiency versus compression ratio for different fractions of constant volume and constant pressure heat addition. Consider compression ratio ranging from 10 to 20.

Brayton Cycle 9.42 An ideal air-standard Brayton cycle operating at steady state produces 10 MW of power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.9. Sketch the T–s diagram for the cycle and determine (a) the (b) the passing (c) the

mass flow rate of air, in kg/s. rate of heat transfer, in kW, to the working fluid through the heat exchanger. thermal efficiency.

State

p (kPa)

T (K)

h (kJ/kg)

1 2 3 4

100 1200 1200 100

300 603.5 1450 780.7

300.19 610.65 1575.57 800.78

9.43 An ideal cold air-standard Brayton cycle operates at steady state with compressor inlet conditions of 300 K and 100 kPa, fixed turbine inlet temperature of 1700 K, and k 5 1.4. For the cycle,

(a) determine the net power developed per unit mass flowing, in kJ/kg, and the thermal efficiency for a compressor pressure ratio of 8. (b) plot the net power developed per unit mass flowing, in kJ/kg, and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50. 9.45 Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. 9.46 For the Brayton cycle of Problem 9.45, investigate the effects of varying compressor pressure ratio and turbine inlet temperature. Plot the same quantities calculated in Problem 9.45 for (a) a compressor pressure ratio of 10 and turbine inlet temperatures ranging from 1000 to 1600 K. (b) a turbine inlet temperature of 1400 K and compressor pressure ratios ranging from 2 to 20. Discuss. 9.47 The rate of heat addition to an ideal air-standard Brayton cycle is 3.4 3 109 Btu/h. The pressure ratio for the cycle is 14 and the minimum and maximum temperatures are 5208R and 30008R, respectively. Determine (a) the thermal efficiency of the cycle. (b) the mass flow rate of air, in lb/h. (c) the net power developed by the cycle, in Btu/h. 9.48 Solve Problem 9.47 on a cold air-standard basis with specific heats evaluated at 5208R. 9.49 On the basis of a cold air-standard analysis, show that the back work ratio of an ideal air-standard Brayton cycle equals the ratio of absolute temperatures at the compressor inlet and the turbine outlet. 9.50 The compressor inlet temperature of an ideal air-standard Brayton cycle is 5208R and the maximum allowable turbine inlet temperature is 26008R. Plot the net work developed per unit mass of air flow, in Btu/lb, and the thermal efficiency versus compressor pressure ratio for pressure ratios ranging from 12 to 24. Using your plots, estimate the pressure ratio for maximum net work and the corresponding value of thermal efficiency. Compare the results to those obtained in analyzing the cycle on a cold air-standard basis. 9.51 The compressor inlet temperature for an ideal Brayton cycle is T1 and the turbine inlet temperature is T3. Using a

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cold air-standard analysis, show that the temperature T2 at the compressor exit that maximizes the net work developed per unit mass of air flow is T2 5 (T1T3)1/2.

and the temperature is 1400 K. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis of an air-standard analysis,

9.52 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate

(a) develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in kW. (b) devise and evaluate an exergetic efficiency for the gas turbine cycle.

(a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.

9.58 Air enters the compressor of a simple gas turbine at 14.5 lbf/in.2, 808F, and exits at 87 lbf/in.2, 5148F. The air enters the turbine at 15408F, 87 lbf/in.2 and expands to 9178F, 14.5 lbf/in.2 The compressor and turbine operate adiabatically, and kinetic and potential energy effects are negligible. On the basis of an air-standard analysis,

Plot the quantities calculated in parts (a) through (d) versus isentropic efficiency for equal compressor and turbine isentropic efficiencies ranging from 70 to 100%. Discuss. 9.53 The cycle of Problem 9.42 is modified to include the effects of irreversibilities in the adiabatic expansion and compression processes. If the states at the compressor and turbine inlets remain unchanged, the cycle produces 10 MW of power, and the compressor and turbine isentropic efficiencies are both 80%, determine (a) the pressure, in kPa, temperature, in K, and specific enthalpy, in kJ/kg, at each principal state of the cycle and sketch the T–s diagram. (b) the mass flow rate of air, in kg/s. (c) the rate of heat transfer, in kW, to the working fluid passing through the heat exchanger. (d) the thermal efficiency. 9.54 Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 2100 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. Determine (a) the net power developed, in MW. (b) the rate of heat addition in the combustor, in MW. (c) the thermal efficiency of the cycle. 9.55 Air enters the compressor of a simple gas turbine at p1 5 14 lbf/in.2, T1 5 5208R. The isentropic efficiencies of the compressor and turbine are 83 and 87%, respectively. The compressor pressure ratio is 14 and the temperature at the turbine inlet is 25008R. The net power developed is 5 3 106 Btu/h. On the basis of an air-standard analysis, calculate (a) the volumetric flow rate of the air entering the compressor, in ft3/min. (b) the temperatures at the compressor and turbine exits, each in 8R. (c) the thermal efficiency of the cycle. 9.56 Solve Problem 9.55 on a cold air-standard basis with specific heats evaluated at 5208R. 9.57 Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 kPa,

Let T0 5 300 K, p0 5 100 kPa.

(a) develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in Btu/lb. (b) devise and evaluate an exergetic efficiency for the gas turbine cycle. Let T0 5 808F, p0 5 14.5 lbf/in.2

Regeneration, Reheat, and Compression with Intercooling 9.59 An ideal air-standard regenerative Brayton cycle produces 10 MW of power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.14. Sketch the T–s diagram and determine (a) the (b) the passing (c) the

mass flow rate of air, in kg/s. rate of heat transfer, in kW, to the working fluid through the combustor. thermal efficiency. State

p (kPa)

T (K)

h (kJ/kg)

1 2 x 3 4 y

100 1200 1200 1200 100 100

300 603.5 780.7 1450 780.7 603.5

300.19 610.65 800.78 1575.57 800.78 610.65

9.60 The cycle of Problem 9.59 is modified to include the effects of irreversibilities in the adiabatic expansion and compression process. The regenerator effectiveness is 100%. If the states at the compressor and turbine inlets remain unchanged, the cycle produces 10 MW of power, and the compressor and turbine isentropic efficiencies are both 80%, determine (a) the pressure, in kPa, temperature, in K, and enthalpy, in kJ/kg, at each principal state of the cycle and sketch the T–s diagram. (b) the mass flow rate of air, in kg/s. (c) the rate of heat transfer, in kW, to the working fluid passing through the combustor. (d) the thermal efficiency. 9.61 The cycle of Problem 9.60 is modified to include a regenerator with an effectiveness of 70%. Determine (a) the specific enthalpy, in kJ/kg, and the temperature, in K, for each stream exiting the regenerator and sketch the T–s diagram.

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Problems: Developing Engineering Skills (b) the (c) the passing (d) the

mass flow rate of air, in kg/s. rate of heat transfer, in kW, to the working fluid through the combustor. thermal efficiency.

9.62 Air enters the compressor of a cold air-standard Brayton cycle with regeneration at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rate of exergy destruction in the regenerator, in kW, for T0 5 300 K. 9.63 Air enters the compressor of a regenerative air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 2100 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. For a regenerator effectiveness of 85%, determine (a) the net power developed, in MW. (b) the rate of heat addition in the combustor, in MW. (c) the thermal efficiency of the cycle.

577

(a) the thermal efficiency. (b) the percent decrease in heat addition to the air. 9.65 On the basis of a cold air-standard analysis, show that the thermal efficiency of an ideal regenerative gas turbine can be expressed as h512a

T1 b 1r21k212/k T3

where r is the compressor pressure ratio, and T1 and T3 denote the temperatures at the compressor and turbine inlets, respectively. 9.66 An air-standard Brayton cycle has a compressor pressure ratio of 10. Air enters the compressor at p1 5 14.7 lbf/in.2, T1 5 708F with a mass flow rate of 90,000 lb/h. The turbine inlet temperature is 22008R. Calculate the thermal efficiency and the net power developed, in horsepower, if (a) the turbine and compressor isentropic efficiencies are each 100%. (b) the turbine and compressor isentropic efficiencies are 88 and 84%, respectively. (c) the turbine and compressor isentropic efficiencies are 88 and 84%, respectively, and a regenerator with an effectiveness of 80% is incorporated.

Plot the quantities calculated in parts (a) through (c) for regenerator effectiveness values ranging from 0 to 100%. Discuss.

9.67 Fig. P9.67 illustrates a gas turbine power plant that uses solar energy as the source of heat addition (see U.S. Patent 4,262,484). Operating data are given on the figure. Modeling the cycle as a Brayton cycle, and assuming no pressure drops in the heat exchanger or interconnecting piping, determine

9.64 Reconsider Problem 9.55, but include a regenerator in the cycle. For regenerator effectiveness values ranging from 0 to 100%, plot

(a) the thermal efficiency. (b) the air mass flow rate, in kg/s, for a net power output of 500 kW.

Solar energy 4 T4 = 1140 K Pressurized solar storage medium

Compressor Air T1 = 310 K p1 = 1 bar 1

Turbine ηt = 0.8 2 T2 = 520 K p2 = 5 bar

3

5 T3 = 760 K

Heat exchanger Exhaust 1 bar

Fig. P9.67

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9.68 Air enters the compressor of a regenerative gas turbine with a volumetric flow rate of 3.2 3 105 ft3/min at 14.5 lbf/in.2, 778F, and is compressed to 60 lbf/in.2 The air then passes through the regenerator and exits at 11208R. The temperature at the turbine inlet is 17008R. The compressor and turbine each has an isentropic efficiency of 84%. Using an air-standard analysis, calculate (a) the thermal efficiency of the cycle. (b) the regenerator effectiveness. (c) the net power output, in Btu/h. 9.69 Air enters the turbine of a gas turbine at 1200 kPa, 1200 K, and expands to 100 kPa in two stages. Between the stages, the air is reheated at a constant pressure of 350 kPa to 1200 K. The expansion through each turbine stage is isentropic. Determine, in kJ per kg of air flowing, (a) the work developed by each stage. (b) the heat transfer for the reheat process. (c) the increase in net work as compared to a single stage of expansion with no reheat. 9.70 Reconsider Problem 9.69 and include in the analysis that each turbine stage might have an isentropic efficiency less than 100%. Plot each of the quantities calculated in parts (a) through (c) of Problem 9.69 for values of the interstage pressure ranging from 100 to 1200 kPa and for isentropic efficiencies of 100%, 80%, and 60%. 9.71 Consider a two-stage turbine operating at steady state with reheat at constant pressure between the stages. Show that the maximum work is developed when the pressure ratio is the same across each stage. Use a cold air-standard analysis, assuming the inlet state and the exit pressure are specified, each expansion process is isentropic, and the temperature at the inlet to each turbine stage is the same. Kinetic and potential energy effects can be ignored. 9.72 Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the inlet temperature for each turbine stage is 1400 K. The pressure ratios across each turbine stage are equal. The turbine stages and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and each turbine stage as well as the regenerator, in kW, for T0 5 300 K. 9.73 Air enters a two-stage compressor operating at steady state at 5208R, 14 lbf/in.2 The overall pressure ratio across the stages is 12 and each stage operates isentropically. Intercooling occurs at constant pressure at the value that minimizes compressor work input as determined in Example 9.10, with air exiting the intercooler at 5208R. Assuming ideal gas behavior, with k 5 1.4, determine the work per unit mass of air flowing for the two-stage compressor. Kinetic and potential energy effects can be ignored.

9.74 A two-stage air compressor operates at steady state, compressing 10 m3/min of air from 100 kPa, 300 K, to 1200 kPa. An intercooler between the two stages cools the air to 300 K at a constant pressure of 350 kPa. The compression processes are isentropic. Calculate the power required to run the compressor, in kW, and compare the result to the power required for isentropic compression from the same inlet state to the same final pressure. 9.75 Reconsider Problem 9.74 and include in the analysis that each compressor stage might have an isentropic efficiency less than 100%. Plot, in kW, (a) the power input to each stage, (b) the heat transfer rate for the intercooler, and (c) the decrease in power input as compared to a single stage of compression with no intercooling, for values of interstage pressure ranging from 100 to 1200 kPa and for isentropic efficiencies of 100%, 80%, and 60%. 9.76 Air enters a compressor operating at steady state at 14 lbf/in.2, 608F, with a volumetric flow rate of 6000 ft3/min. The compression occurs in two stages, with each stage being a polytropic process with n 5 1.27. The air is cooled to 808F between the stages by an intercooler operating at 45 lbf/in.2 Air exits the compressor at 150 lbf/in.2 Determine, in Btu per min (a) the power and heat transfer rate for each compressor stage. (b) the heat transfer rate for the intercooler. 9.77 Air enters the first compressor stage of a cold air-standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T0 5 300 K. 9.78 Referring to Example 9.10, show that if Td . T1 the pressure ratios across the two compressor stages are related by pi p2 Td k/1k212 5 a ba b p1 pi T1 9.79 Rework Example 9.10 for the case of a three-stage compressor with intercooling between stages. 9.80 An air-standard regenerative Brayton cycle operating at steady state with intercooling and reheat produces 10 MW of power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.19. Sketch the T–s diagram for the cycle and determine (a) the (b) the passing (c) the

mass flow rate of air, in kg/s. rate of heat transfer, in kW, to the working fluid through each combustor. thermal efficiency.

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p (kPa)

T (K)

h (kJ/kg)

1 2 3 4 5 6 7 8 9 10

100 300 300 1200 1200 1200 300 300 100 100

300 410.1 300 444.8 1111.0 1450 1034.3 1450 1111.0 444.8

300.19 411.22 300.19 446.50 1173.84 1575.57 1085.31 1575.57 1173.84 446.50

579

The turbine expansion also occurs in two stages, with reheat to 20008R between the stages at 50 lbf/in.2 The isentropic efficiencies of the turbine and compressor stages are 88 and 84%, respectively.

Other Gas Power System Applications

9.81 Air enters the compressor of a cold air-standard Brayton cycle with regeneration, intercooling, and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The intercooler and reheater both operate at the same pressure. The temperature at the inlet to the second compressor stage is 300 K, and the inlet temperature for each turbine stage is 1400 K. The compressor and turbine stages each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine stages as well as the regenerator, in kW, for T0 5 300 K. 9.82 An air-standard Brayton cycle produces 10 MW of power. The compressor and turbine isentropic efficiencies are both 80%. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.9. (a) Fill in the missing data in the table and sketch the T–s diagram for the cycle. (b) Determine the mass flow rate of air, in kg/s. (c) Perform a full accounting for the net rate of exergy increase as the air passes through the combustor. Let T0 5 300 K, p0 5 100 kPa. State

p (kPa)

T (K)

h (kJ/kg)

s8 [kJ/(kg ? K)]

pr

1 2 3 4

100 1200 1200 100

300

300.19

1.70203

1.3860

1450

1575.57

3.40417

522

9.83 For each of the following modifications of the cycle of part (c) of Problem 9.66, determine the thermal efficiency and net power developed, in horsepower. (a) Introduce a two-stage turbine expansion with reheat between the stages at a constant pressure of 50 lbf/in.2 Each turbine stage has an isentropic efficiency of 88% and the temperature of the air entering the second stage is 20008R. (b) Introduce two-stage compression, with intercooling between the stages at a pressure of 50 lbf/in.2 Each compressor stage has an isentropic efficiency of 84% and the temperature of the air entering the second stage is 708F. (c) Introduce both compression with intercooling and reheat between turbine stages. Compression occurs in two stages, with intercooling to 708F between the stages at 50 lbf/in.2

9.84 Air at 26 kPa, 230 K, and 220 m/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressure ratio is 11, the turbine inlet temperature is 1400 K, and air exits the nozzle at 26 kPa. The diffuser and nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of 85% and 90%, respectively, and there is no pressure drop for flow through the combustor. Kinetic energy is negligible everywhere except at the diffuser inlet and the nozzle exit. On the basis of air-standard analysis, determine (a) the pressures, in kPa, and temperatures, in K, at each principal state. (b) the rate of heat addition to the air passing through the combustor, in kJ/s. (c) the velocity at the nozzle exit, in m/s. 9.85 For the turbojet in Problem 9.84, plot the velocity at the nozzle exit, in m/s, the pressure at the turbine exit, in kPa, and the rate of heat input to the combustor, in kW, each as a function of compressor pressure ratio in the range of 6 to 14. Repeat for turbine inlet temperatures of 1200 K and 1000 K. 9.86 Air enters the diffuser of a turbojet engine with a mass flow rate of 85 lb/s at 9 lbf/in.2, 4208R, and a velocity of 750 ft/s. The pressure ratio for the compressor is 12, and its isentropic efficiency is 88%. Air enters the turbine at 24008R with the same pressure as at the exit of the compressor. Air exits the nozzle at 9 lbf/in.2 The diffuser operates isentropically and the nozzle and turbine have isentropic efficiencies of 92% and 90%, respectively. On the basis of an air-standard analysis, calculate (a) (b) (c) (d)

the the the the

rate of heat addition, in Btu/h. pressure at the turbine exit, in lbf/in.2 compressor power input, in Btu/h. velocity at the nozzle exit, in ft/s.

Neglect kinetic energy except at the diffuser inlet and the nozzle exit. 9.87 Consider the addition of an afterburner to the turbojet in Problem 9.84 that raises the temperature at the inlet of the nozzle to 1300 K. Determine the velocity at the nozzle exit, in m/s. 9.88 Consider the addition of an afterburner to the turbojet in Problem 9.86 that raises the temperature at the inlet of the nozzle to 22008R. Determine the velocity at the nozzle exit, in ft/s. 9.89 Air enters the diffuser of a ramjet engine at 6 lbf/in.2, 4208R, with a velocity of 1600 ft/s, and decelerates essentially to zero velocity. After combustion, the gases reach a temperature of 20008R before being discharged through the nozzle at 6 lbf/in.2 On the basis of an air-standard analysis, determine (a) the pressure at the diffuser exit, in lbf/in.2 (b) the velocity at the nozzle exit, in ft/s. Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

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9.90 Air enters the diffuser of a ramjet engine at 40 kPa, 240 K, with a velocity of 2500 km/h and decelerates to negligible velocity. On the basis of an air-standard analysis, the heat addition is 1080 kJ per kg of air passing through the engine. Air exits the nozzle at 40 kPa. Determine

· Qin,1 p2 = 800 lbf/in.2 2

(a) the pressure at the diffuser exit, in kPa. (b) the velocity at the nozzle exit, in m/s.

Helium cycle

Neglect kinetic energy except at the diffuser inlet and the nozzle exit. 9.91 A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as the compressor. Air enters the diffuser with a volumetric flow rate of 83.7 m3/s at 40 kPa, 240 K, and a velocity of 180 m/s, and decelerates essentially to zero velocity. The compressor pressure ratio is 10 and the compressor has an isentropic efficiency of 85%. The turbine inlet temperature is 1140 K, and its isentropic efficiency is 85%. The turbine exit pressure is 50 kPa. Flow through the diffuser and nozzle is isentropic. Using an air-standard analysis, determine

4 p4 = p1

Helium 1 p1 = 200 lbf/in.2 T1 = 180°F m· 1 = 8×105 lb/h

5

Neglect kinetic energy except at the diffuser inlet and the nozzle exit. 9.93 Helium is used in a combined cycle power plant as the working fluid in a simple closed gas turbine serving as the topping cycle for a vapor power cycle. A nuclear reactor is the source of energy input to the helium. Figure P9.93 provides steady-state operating data. Helium enters the compressor of the gas turbine at 200 lbf/in.2, 1808F with a mass flow rate of 8 3 105 lb/h and is compressed to 800 lbf/in.2 The isentropic efficiency of the compressor is 80%. The helium then passes through the reactor with a negligible decrease in pressure, exiting at 14008F. Next, the helium expands through the turbine, which has an isentropic efficiency of 80%, to a pressure of 200 lbf/in.2 The helium then passes through the interconnecting heat exchanger. A separate stream of liquid water enters the heat exchanger and exits as saturated vapor at 1200 lbf/in.2 The vapor is superheated before entering the turbine at 8008F, 1200 lbf/in.2 The steam expands through the turbine to 1 lbf/in.2 and a

Saturated vapor p5 = 1200 lbf/in.2

Heat exchanger 9

· Qin,2

Superheater

p6 = p5 6 T = 800°F 6

Pump ηp = 100% Steam cycle

Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

(a) the power delivered to the propeller, in hp. (b) the velocity at the nozzle exit, in ft/s.

Turbine ηt = 80%

Compressor ηc = 80%

(a) the power delivered to the propeller, in MW. (b) the velocity at the nozzle exit, in m/s.

9.92 A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as the compressor. Air enters the diffuser at 12 lbf/ in.2, 4608R, with a volumetric flow rate of 30,000 ft3/min and a velocity of 520 ft/s. In the diffuser, the air decelerates isentropically to negligible velocity. The compressor pressure ratio is 9, and the turbine inlet temperature is 21008R. The turbine exit pressure is 25 lbf/in.2, and the air expands to 12 lbf/in.2 through a nozzle. The compressor and turbine each has an isentropic efficiency of 87%, and the nozzle has an isentropic efficiency of 95%. Using an air-standard analysis, determine

p 3 = p2 T3 = 1400°F

3

Turbine

8 Condenser

7 p7 = 1 lbf/in.2 x7 = 0.90

Saturated liquid p8 = p7

Cooling water T10 = 60°F

10

11 T11 = 90°F

Fig. P9.93

quality of 0.9. Saturated liquid exits the condenser at 1 lbf/in.2 Cooling water passing through the condenser experiences a temperature rise from 60 to 908F. The isentropic pump efficiency is 100%. Stray heat transfer and kinetic and potential energy effects can be ignored. Determine (a) the mass flow rates of the steam and the cooling water, each in lb/h. (b) the net power developed by the gas turbine and vapor cycles, each in Btu/h. (c) the thermal efficiency of the combined cycle. 9.94 A combined gas turbine-vapor power plant operates as shown in Fig. P9.94. Pressure and temperature data are given at principal states, and the net power developed by the gas turbine is 147 MW. Using air-standard analysis for the gas turbine, determine (a) the net power, in MW, developed by the power plant. (b) the overall thermal efficiency of the plant. Develop a full accounting of the net rate of exergy increase of the air passing through the gas turbine combustor. Let T0 5 300 K, p0 5 1 bar.

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T2 = 690 K p5 = 13.6 bar 2

T3 = 1580 K 3 p3 = 13 bar

Combustor

Gas turbine Compressor

Turbine

(a) the mass flow rates of air and steam, each in kg/s. (b) the thermal efficiency of the combined cycle. (c) a full accounting of the net exergy increase of the air passing through the combustor of the gas turbine, # mair3ef3 2 ef24, in MW. Discuss.

4 T4 = 900 K p4 = 1 bar

Air inlet

5

Let T0 5 300 K, p0 5 100 kPa.

T7 = 520°C 7 p7 = 100 bar Heat exchanger Turbine

p6 = p7

6

1200 kPa. The isentropic efficiency of the compressor is 84%. The conditions at the inlet to the turbine are 1200 kPa and 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a pressure of 100 kPa. The air then passes through the interconnecting heat exchanger, and is finally discharged at 480 K. Steam enters the turbine of the vapor power cycle at 8 MPa, 4008C, and expands to the condenser pressure of 8 kPa. Water enters the pump as saturated liquid at 8 kPa. The turbine and pump have isentropic efficiencies of 90 and 80%, respectively. Determine

147 MW

T1 = 300 K 1 p1 = 1 bar

Exhaust T5 = 400 K p5 = 1 bar

· Wgas =

581

Steam cycle

8

ηt = 85%

Condenser

Pump

· Wvap

· Qout

9 ηp = 80%

p9 = p8 = 0.08 bar

Fig. P9.94 9.95 Air enters the compressor of a combined gas turbine–vapor power plant (Fig. 9.22) at 1 bar, 258C. The isentropic compressor efficiency is 85% and the compressor pressure ratio is 14. The air passing through the combustor receives energy by heat transfer at a rate of 50 MW with no significant decrease in pressure. At the inlet to the turbine the air is at 12508C. The air expands through the turbine, which has an isentropic efficiency of 87%, to a pressure of 1 bar. Then, the air passes through the interconnecting heat exchanger and is finally discharged at 2008C, 1 bar. Steam enters the turbine of the vapor cycle at 12.5 MPa, 5008C, and expands to a condenser pressure of 0.1 bar. Water enters the pump as a saturated liquid at 0.1 bar. The turbine and pump have isentropic efficiencies of 90 and 100%, respectively. Cooling water enters the condenser at 208C and exits at 358C. Determine (a) the mass flow rates of the air, steam, and cooling water, each in kg/s. (b) the net power developed by the gas turbine cycle and the vapor cycle, respectively, each in MW. (c) the thermal efficiency of the combined cycle. (d) the net rate at which exergy is carried out with the # exhaust air, mair3ef5 2 ef14, in MW. (e) the net rate at which exergy is carried out with the cooling water, in MW. Let T0 5 208C, p0 5 1 bar. 9.96 A combined gas turbine–vapor power plant (Fig. 9.22) has a net power output of 100 MW. Air enters the compressor of the gas turbine at 100 kPa, 300 K, and is compressed to

9.97 A simple gas turbine is the topping cycle for a simple vapor power cycle (Fig. 9.22). Air enters the compressor of the gas turbine at 608F, 14.7 lbf/in.2, with a volumetric flow rate of 40,000 ft3/min. The compressor pressure ratio is 12 and the turbine inlet temperature is 26008R. The compressor and turbine each have isentropic efficiencies of 88%. The air leaves the interconnecting heat exchanger at 8408R, 14.7 lbf/ in.2 Steam enters the turbine of the vapor cycle at 1000 lbf/ in.2, 9008F, and expands to the condenser pressure of 1 lbf/in.2 Water enters the pump as saturated liquid at 1 lbf/in.2 The turbine and pump efficiencies are 90 and 70%, respectively. Cooling water passing through the condenser experiences a temperature rise from 60 to 808F with a negligible change in pressure. Determine (a) the mass flow rates of the air, steam, and cooling water, each in lb/h. (b) the net power developed by the gas turbine cycle and the vapor cycle, respectively, each in Btu/h. (c) the thermal efficiency of the combined cycle. (d) a full accounting of the net exergy increase of the air # passing through the combustor of the gas turbine, mair 3ef3 2 ef24, in Btu/h. Discuss. Let T0 5 5208R, p0 5 14.7 lbf/in.2 9.98 Air enters the compressor of an Ericsson cycle at 300 K, 1 bar, with a mass flow rate of 5 kg/s. The pressure and temperature at the inlet to the turbine are 10 bar and 1400 K, respectively. Determine (a) the net power developed, in kW. (b) the thermal efficiency. (c) the back work ratio. 9.99 For the cycle in Problem 9.98, plot the net power developed, in kW, for compressor pressure ratios ranging from 2 to 15. Repeat for turbine inlet temperatures of 1200 K and 1000 K. 9.100 Air is the working fluid in an Ericsson cycle. Expansion through the turbine takes place at a constant temperature of 22508R. Heat transfer from the compressor occurs at 5608R. The compressor pressure ratio is 12. Determine (a) the net work, in Btu per lb of air flowing. (b) the thermal efficiency.

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9.101 Nitrogen (N2) is the working fluid of a Stirling cycle with a compression ratio of nine. At the beginning of the isothermal compression, the temperature, pressure, and volume are 310 K, 1 bar, and 0.008 m3, respectively. The temperature during the isothermal expansion is 1000 K. Determine (a) the net work, in kJ. (b) the thermal efficiency. (c) the mean effective pressure, in bar. 9.102 Helium is the working fluid in a Stirling cycle. In the isothermal compression, the helium is compressed from 15 lbf/in.2, 1008F, to 150 lbf/in.2 The isothermal expansion occurs at 15008F. Determine (a) the work and heat transfer, in Btu per lb of helium, for each process in the cycle. (b) the thermal efficiency.

Compressible Flow 9.103 Calculate the thrust developed by the turbojet engine in Problem 9.84, in kN. 9.104 Calculate the thrust developed by the turbojet engine in Problem 9.86, in lbf. 9.105 Calculate the thrust developed by the turbojet engine with afterburner in Problem 9.87, in kN. 9.106 Referring to the turbojet in Problem 9.86 and the modified turbojet in Problem 9.88, calculate the thrust developed by each engine, in lbf. Discuss. 9.107 Air enters the diffuser of a turbojet engine at 18 kPa, 216 K, with a volumetric flow rate of 230 m3/s and a velocity of 265 m/s. The compressor pressure ratio is 15, and its isentropic efficiency is 87%. Air enters the turbine at 1360 K and the same pressure as at the exit of the compressor. The turbine isentropic efficiency is 89%, and the nozzle isentropic efficiency is 97%. The pressure at the nozzle exit is 18 kPa. On the basis of an air-standard analysis, calculate the thrust, in kN. 9.108 Calculate the ratio of the thrust developed to the mass flow rate of air, in N per kg/s, for the ramjet engine in Problem 9.90. 9.109 Air flows at steady state through a horizontal, wellinsulated, constant-area duct of diameter 0.25 m. At the inlet, p1 5 2.4 bar, T1 5 430 K. The temperature of the air leaving the duct is 370 K. The mass flow rate is 600 kg/min. Determine the magnitude, in N, of the net horizontal force exerted by the duct wall on the air. In which direction does the force act? 9.110 Liquid water at 708F flows at steady state through a 2-in.-diameter horizontal pipe. The mass flow rate is 25 lb/s. The pressure decreases by 2 lbf/in.2 from inlet to exit of the pipe. Determine the magnitude, in lbf, and direction of the horizontal force required to hold the pipe in place. 9.111 Air enters a horizontal, well-insulated nozzle operating at steady state at 12 bar, 500K, with a velocity of 50 m/s and exits at 7 bar, 440 K. The mass flow rate is 1 kg/s. Determine the net force, in N, exerted by the air on the duct in the direction of flow.

9.112 Using the ideal gas model, determine the sonic velocity of (a) air at 608F. (b) oxygen (O2) at 9008R. (c) argon at 5408R. 9.113 A flash of lightning is sighted and 3 seconds later thunder is heard. Approximately how far away was the lightning strike? 9.114 Using data from Table A-4, estimate the sonic velocity, in m/s, of steam of 60 bar, 3608C. Compare the result with the value predicted by the ideal gas model. 9.115 Plot the Mach number of carbon dioxide at 1 bar, 460 m/s, as a function of temperature in the range 250 to 1000 K. 9.116 An ideal gas flows through a duct. At a particular location, the temperature, pressure, and velocity are known. Determine the Mach number, stagnation temperature, in 8R, and the stagnation pressure, in lbf/in.2, for (a) air at 3108F, 100 lbf/in.2, and a velocity of 1400 ft/s. (b) helium at 5208R, 20 lbf/in.2, and a velocity of 900 ft/s. (c) nitrogen at 6008R, 50 lbf/in.2, and a velocity of 500 ft/s. 9.117 For Problem 9.111, determine the values of the Mach number, the stagnation temperature, in K, and the stagnation pressure, in bar, at the inlet and exit of the duct, respectively. 9.118 Using the Mollier diagram, Fig. A-8E, determine for water vapor at 500 lbf/in.2, 6008F, and 1000 ft/s (a) the stagnation enthalpy, in Btu/lb. (b) the stagnation temperature, in 8F. (c) the stagnation pressure, in lbf/in.2 9.119 Steam flows through a passageway, and at a particular location the pressure is 3 bar, the temperature is 281.48C, and the velocity is 688.8 m/s. Determine the corresponding specific stagnation enthalpy, in kJ/kg, and stagnation temperature, in 8C, if the stagnation pressure is 7 bar. 9.120 For the isentropic flow of an ideal gas with constant specific heat ratio k, the ratio of the temperature T* to the stagnation temperature To is T*/To 5 2/(k 1 1). Develop this relationship. 9.121 A gas expands isentropically through a converging nozzle from a large tank at 8 bar, 500 K. Assuming ideal gas behavior, determine the critical pressure p*, in bar, and the corresponding temperature, in K, if the gas is (a) air. (b) carbon dioxide (CO2). (c) water vapor. 9.122 Carbon dioxide is contained in a large tank, initially at 100 lbf/in.2, 8008R. The gas discharges through a converging nozzle to the surroundings, which are at 14.7 lbf/in.2, and the pressure in the tank drops. Estimate the pressure in the tank, in lbf/in.2, when the flow first ceases to be choked. 9.123 Steam expands isentropically through a converging nozzle operating at steady state from a large tank at 1.83 bar, 2808C. The mass flow rate is 2 kg/s, the flow is choked, and the exit plane pressure is 1 bar. Determine the diameter of the nozzle, in cm, at locations where the pressure is 1.5 bar, and 1 bar, respectively.

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Problems: Developing Engineering Skills 9.124 An ideal gas mixture with k 5 1.31 and a molecular weight of 23 is supplied to a converging nozzle at po 5 5 bar, To 5 700 K, which discharges into a region where the pressure is 1 bar. The exit area is 30 cm2. For steady isentropic flow through the nozzle, determine (a) the exit temperature of the gas, in K. (b) the exit velocity of the gas, in m/s. (c) the mass flow rate, in kg/s. 9.125 An ideal gas expands isentropically through a converging nozzle from a large tank at 120 lbf/in.2, 6008R, and discharges into a region at 60 lbf/in.2 Determine the mass flow rate, in lb/s, for an exit flow area of 1 in.2, if the gas is (a) air, with k 5 1.4. (b) carbon dioxide, with k 5 1.26. (c) argon, with k 5 1.667. 9.126 Air at po 5 1.4 bar, To 5 280 K expands isentropically through a converging nozzle and discharges to the atmosphere at 1 bar. The exit plane area is 0.0013 m2. (a) Determine the mass flow rate, in kg/s. (b) If the supply region pressure, po, were increased to 2 bar, what would be the mass flow rate, in kg/s? 9.127 Air enters a nozzle operating at steady state at 45 lbf/in.2, 8008R, with a velocity of 480 ft/s, and expands isentropically to an exit velocity of 1500 ft/s. Determine (a) the exit pressure, in lbf/in.2 (b) the ratio of the exit area to the inlet area. (c) whether the nozzle is diverging only, converging only, or converging–diverging in cross section. 9.128 Air as an ideal gas with k 5 1.4 enters a converging– diverging nozzle operating at steady state and expands isentropically as shown in Fig. P9.128. Using data from the figure and from Table 9.2 as needed, determine (a) the stagnation pressure, in lbf/in.2, and the stagnation temperature, in 8R. (b) the throat area, in in.2 (c) the exit area, in in.2 Throat M1 = 0.2 T1 = 595°R p1 = 77.8 lbf/in.2 m· = 1 lb/s

M2 = 1.5

1

2

Fig. P9.128 9.129 Air as an ideal gas with k 5 1.4 enters a diffuser operating at steady state at 4 bar, 290 K, with a velocity of 512 m/s. Assuming isentropic flow, plot the velocity, in m/s, the Mach number, and the area ratio A/A* for locations in the flow corresponding to pressures ranging from 4 to 14 bar. 9.130 A converging–diverging nozzle operating at steady state has a throat area of 3 cm2 and an exit area of 6 cm2. Air as an ideal gas with k 5 1.4 enters the nozzle at 8 bar, 400 K, and a Mach number of 0.2, and flows isentropically throughout. If

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the nozzle is choked, and the diverging portion acts as a supersonic nozzle, determine the mass flow rate, in kg/s, and the Mach number, pressure, in bar, and temperature, in K, at the exit. Repeat if the diverging portion acts as a supersonic diffuser. 9.131 A converging–diverging nozzle operates at steady state. Air as an ideal gas with k 5 1.4 enters the nozzle at 500 K, 6 bar, and a Mach number of 0.3. The air flows isentropically to the exit plane, where a normal shock stands. The temperature just upstream of the shock is 380.416 K. Determine the back pressure, in bar. 9.132 A converging–diverging nozzle operates at steady state. Air as an ideal gas with k 5 1.4 enters the nozzle at 500 K, 6 bar, and a Mach number of 0.3. A normal shock stands in the diverging section at a location where the Mach number 1.40. The cross-sectional areas of the throat and the exit plane are 4 cm2 and 6 cm2, respectively. The flow is isentropic, except where the shock stands. Determine the exit pressure, in bar, and the mass flow rate, in kg/s. 9.133 Air as an ideal gas with k 5 1.4 enters a convergingdiverging duct with a Mach number of 2. At the inlet, the pressure is 26 lbf/in.2 and the temperature is 4458R. A normal shock stands at a location in the converging section of the duct, with Mx 5 1.5. At the exit of the duct, the pressure is 150 lbf/in.2 The flow is isentropic everywhere except in the immediate vicinity of the shock. Determine temperature, in 8R, and the Mach number at the exit. 9.134 Air as an ideal gas with k 5 1.4 undergoes a normal shock. The upstream conditions are px 5 0.5 bar, Tx 5 280 K, and Mx 5 1.8. Determine (a) the pressure py, in bar. (b) the stagnation pressure pox, in bar. (c) the stagnation temperature Tox, in K. (d) the change in specific entropy across the shock, in kJ/kg ? K. (e) Plot the quantities of parts (a)–(d) versus Mx ranging from 1.0 to 2.0. All other upstream conditions remain the same. 9.135 A converging–diverging nozzle operates at steady state. Air as an ideal gas with k 5 1.4 flows through the nozzle, discharging to the atmosphere at 14.7 lbf/in.2 and 5208R. A normal shock stands at the exit plane with Mx 5 1.5. The exit plane area is 1.8 in.2 Upstream of the shock, the flow is isentropic. Determine (a) the stagnation pressure pox, in lbf/in.2 (b) the stagnation temperature Tox, in 8R. (c) the mass flow rate, in lb/s. 9.136 A converging–diverging nozzle operates at steady state. Air as an ideal gas with k 5 1.4 flows through the nozzle, discharging to the atmosphere at 14.7 lbf/in.2 and 5108R. A normal shock stands at the exit plane with px 5 9.714 lbf/in.2 The exit plane area is 2 in.2 Upstream of the shock, the flow is isentropic. Determine (a) the throat area, in in.2 (b) the entropy produced in the nozzle, in Btu/8R per lb of air flowing.

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9.137 Air at 3.4 bar, 530 K, and a Mach number of 0.4 enters a converging–diverging nozzle operating at steady state. A normal shock stands in the diverging section at a location where the Mach number is Mx 5 1.8. The flow is isentropic, except where the shock stands. If the air behaves as an ideal gas with k 5 1.4, determine (a) (b) (c) (d) (e) (f)

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the the the the the the

stagnation temperature Tox, in K. stagnation pressure pox, in bar. pressure px, in bar. pressure py, in bar. stagnation pressure poy, in bar. stagnation temperature Toy, in K.

If the throat area is 7.6 3 1024 m2, and the exit plane pressure is 2.4 bar, determine the mass flow rate, in kg/s, and the exit area, in m2.

stands at the inlet to the channel. Downstream of the shock the flow is isentropic; the Mach number is unity at the throat; and the air exits at 20 lbf/in.2, 7008R, with negligible velocity. If the mass flow rate is 45 lb/s, determine the inlet and throat areas, in ft2. 9.139 Derive the following expressions: (a) Eq. 9.55, (b) Eq. 9.56, (c) Eq. 9.57. 9.140 Using Interactive Thermodynamics: IT, generate tables of the same isentropic flow functions as in Table 9.2 for specific heat ratios of 1.2, 1.3, 1.4, and 1.67 and Mach numbers ranging from 0 to 5. 9.141 Using Interactive Thermodynamics: IT, generate tables of the same normal shock functions as in Table 9.3 for specific heat ratios of 1.2, 1.3, 1.4, and 1.67 and Mach numbers ranging from 1 to 5.

9.138 Air as an ideal gas with k 5 1.4 enters a converging– diverging channel at a Mach number of 1.6. A normal shock

c DESIGN & OPEN-ENDED PROBLEMS: EXPLORING ENGINEERING PRACTICE 9.1D Congress has mandated the average fuel economy for passenger cars sold in the United States to be 35 miles per gallon beginning in 2020. In Europe, the goal is 47 miles per gallon by 2012. In each case, identify the major factors spurring legislative action, including, as appropriate, technical, economic, societal, and political factors. Analyze the disparity between these goals. Comment on their likely effectiveness in achieving the respective legislative aims. Report your findings in a PowerPoint presentation. 9.2D Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact. Summarize your findings in a report with at least three references. 9.3D The Annual Energy Outlook with Projections report released by the U.S. Energy Information Administration projects annual consumption estimates for various fuel types through the next 25 years. According to the report, biofuels will play an increasing role in the liquid fuel supply over that time period. Based on technologies commercially available or reasonably expected to become available in the next decade, identify the most viable options for producing biofuels. Compare several options based on energy return on energy invested (EROEI), water and land requirements, and effects on global climate change. Draw conclusions based on your study, and present your findings in a report with at least three references. 9.4D Investigate the following technologies: plug-in hybrid vehicles, all-electric vehicles, hydrogen fuel cell vehicles, diesel-powered vehicles, natural gas–fueled vehicles, and ethanol-fueled vehicles, and make recommendations on which of these technologies should receive federal research,

development, and deployment support over the next decade. Base your recommendation on the result of a decision matrix method such as the Pugh method to compare the various technologies. Clearly identify and justify the criteria used for the comparison and the logic behind the scoring process. Prepare a 15-minute briefing and an executive summary suitable for a conference with your local congressperson. 9.5D Figure P9.5D shows a wheeled platform propelled by thrust generated using an onboard water tank discharging water through an elbow to which a nozzle is attached. Design and construct such an apparatus using easily obtained materials like a skateboard and one-gallon milk jug. Investigate the effects of elbow angle and nozzle exit area on volumetric flow and thrust. Prepare a report including results and conclusions together with an explanation of the measurement techniques and experimental procedures.

Opening for atmosphere exposure

Elbow (tubing) Water tank stand

Fig. P9.5D

Nozzle

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Design & Open-Ended Problems: Exploring Engineering Practice 9.6D Owing to its very low temperature relative to seawater, liquid natural gas (LNG) arriving at U.S. ports by ship has considerable thermomechanical exergy. Yet when LNG is regasified in heat exchangers where seawater is the other stream, that exergy is largely destroyed. Conduct a search of the patent literature for methods to recover a substantial portion of LNG exergy during regasification. Consider patents both granted and pending. Critically evaluate the technical merit and economic feasibility of two different methods found in your search. Report your conclusions in an executive summary and PowerPoint presentation.

company at a cost of $0.06 per kW ? h for a plant expansion or to purchase and operate a diesel engine generator set that runs on natural gas. Assume natural gas is priced at 6 dollars per thousand cubic feet, each cubic foot of gas has a heating value of 1000 Btu, and the plant operates 7800 h per year. Specify diesel-generator equipment that would supply the required power and perform an economic analysis to determine which alternative is best. Present your recommendation in a memorandum with supporting analysis and calculations, including at least three references. 9.9D A financial services company has a computer server facility that requires very reliable electric power. The power demand is 3000 kW. The company has hired you as a consultant to study the feasibility of using 250-kW microturbines for this application. Write a report discussing the pros and cons of such an arrangement compared to purchasing power from the local utility.

9.7D Develop the preliminary specifications for a 160-MW closed-cycle gas turbine power plant. Consider carbon dioxide and helium as possible gas turbine working fluids that circulate through a nuclear power unit where they absorb energy by heat transfer. Sketch the schematic of your proposed cycle. For each of the two working fluids, determine key operating pressures and temperatures of the gas turbine cycle and estimate the expected performance of the cycle. Write a report that includes your analysis and design, and recommend a working fluid. Include at least three references.

9.10D Figure P9.10D shows a combined cycle formed by a topping gas turbine and an organic Rankine bottoming cycle. Steady-state operating data are labeled on the figure. Owing to internal irreversibilities, the generator electricity output is 95% of the input shaft power. The regenerator preheats air entering the combustor. In the evaporator, hot exhaust gas from the regenerator vaporizes the bottoming

9.8D As an engineer you must recommend whether or not to purchase 2 MW of electric power from the local utility

t = 85%

c = 85%

· WGTC

gen = 95%

Gas turbine cycle Compressor

Turbine

Generator

· We = 100 kW

· Qin 1

2

T1 = 300 K p1 = 100 kPa

4

T4 = 1200 K

reg = 80% 3

Combustor Regenerator

5 8 p8, T8

6

t = 85%

Evaporator

· WORC

gen = 95%

Turbine

Generator

· We

Organic Rankine cycle 9

7

p9

11 T7 = T8 + 20 K p7 = p1

Condenser 10 Pump

· Qout

Saturated liquid p = 85%

˙p W

585

Fig. P9.10D

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Chapter 9 Gas Power Systems

cycle working fluid. For each of three such working fluids—propane, Refrigerant 22, and Refrigerant 134a— specify appropriate ranges for p8, turbine inlet pressure, and T8, turbine inlet temperature; also determine turbine exit pressure p9. For each working fluid, investigate the influence on net combined-cycle electricity production and on combined-cycle thermal efficiency of varying p8, T8, and compressor pressure ratio. Identify the bottoming working fluid and operating conditions with greatest net combinedcycle electricity production. Repeat for greatest combinedcycle thermal efficiency. Apply engineering modeling compatible with that used in the text for Rankine cycles and air-standard analysis of gas turbines. Present your analyses, results, and recommendations in a technical

article adhering to ASME standards with at least three references. 9.11D Micro-CHP (combined heat and power) units capable of producing up to 1.8 kW of electric power are now commercially available for use in the home. Such units contribute to domestic space or water heating needs while providing electricity as a by-product. They operate on an internal combustion engine fueled by natural gas. By hybridizing a micro-CHP unit with a gas furnace, all domestic heating needs can be met while generating a substantial portion of the annual electric power requirement. Evaluate this hybrid form for application to a typical single-family dwelling in your locale having natural gas service. Consider on-site battery

Engine exhaust e

Low-temperature superheater, High-temperature vapor generator

8 4

Four-cylinder engine

High-temperature superheater

Pump

Fuel Exhaust gas

Air

d Coolant c

a

5

1 Pump High-temperature condenser, Low-temperature vapor generator

2

Low-temperature expander

3

6 High-temperature expander 7 b

1-2-3-4-1: High-temperature cycle 5-6-7-8-5: Low-temperature cycle a-b-c: Engine coolant d-e: Engine exhaust gas

Pump

Low-temperature vapor generator

Radiator, Low-temperature condenser

Fig. P9.12D

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Design & Open-Ended Problems: Exploring Engineering Practice storage of excess electricity generated and the possibility of net metering. Specify equipment and determine costs, including initial cost and installation cost. Estimate the annual cost for heating and power using the hybrid and compare it to the annual cost for heating and power with a stand-alone gas furnace and grid power. Using your findings, recommend the better approach for the dwelling. 9.12D Figure P9.12D provides the schematic of an internal combustion automobile engine fitted with two Rankine vapor power bottoming cycles: a high-temperature cycle 1-2-3-4-1 and a low-temperature cycle 5-6-7-8-5. These cycles develop additional power using waste heat derived

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from exhaust gas and engine coolant. Using operating data for a commercially available car having a conventional four-cylinder internal combustion engine with a size of 2.5 liters or less, specify cycle working fluids and state data at key points sufficient to produce at least 15 hp more power. Apply engineering modeling compatible with that used in the text for Rankine cycles and air-standard analysis of internal combustion engines. Write a final report justifying your specifications together with supporting calculations. Provide a critique of the use of such bottoming cycles on car engines and a recommendation about whether such technology should be actively pursued by automakers.

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Refrigeration systems commonly used for preserving food are introduced in Sec. 10.1. © Mike Kemp RubberBall/Age Fotostock America, Inc.

ENGINEERING CONTEXT Refrigeration systems for food preservation and air conditioning play prominent roles in our everyday lives. Heat pumps also are used for heating buildings and for producing industrial process heat. There are many other examples of commercial and industrial uses of refrigeration, including air separation to obtain liquid oxygen and liquid nitrogen, liquefaction of natural gas, and production of ice. To achieve refrigeration by most conventional means requires an electric power input. Heat pumps also require power to operate. Referring again to Table 8.1, we see that in the United States electricity is obtained today primarily from coal, natural gas, and nuclear, all of which are nonrenewable. These nonrenewables have significant adverse effects on human health and the environment associated with their use. Depending on the type of resource, such effects are related to extraction from the earth, processing and distribution, emissions during power production, and waste products. Ineffective refrigeration and heat pump systems, excessive building cooling and heating, and other wasteful practices and lifestyle choices not only misuse increasingly scarce nonrenewable resources but also endanger our health and burden the environment. Accordingly, refrigeration and heat pump systems is an area of application where more effective systems and practices can significantly improve our national energy posture. The objective of this chapter is to describe some of the common types of refrigeration and heat pump systems presently in use and to illustrate how such systems can be modeled thermodynamically. The three principal types described are the vapor-compression, absorption, and reversed Brayton cycles. As for the power systems studied in Chaps. 8 and 9, both vapor and gas systems are considered. In vapor systems, the refrigerant is alternately vaporized and condensed. In gas refrigeration systems, the refrigerant remains a gas.

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10 Refrigeration and Heat Pump Systems LEARNING OUTCOMES When you complete your study of this chapter, you will be able to... c

Demonstrate understanding of basic vapor-compression refrigeration and heat pump systems.

c

Develop and analyze thermodynamic models of vapor-compression systems and their modifications, including c

sketching schematic and accompanying T–s diagrams.

c

evaluating property data at principal states of the systems.

c

applying mass, energy, entropy, and exergy balances for the basic processes.

c

determining refrigeration and heat pump system performance, coefficient of performance and capacity.

c

Explain the effects on vapor-compression system performance of varying key parameters.

c

Demonstrate understanding of the operating principles of absorption and gas refrigeration systems, and perform thermodynamic analysis of gas systems.

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Chapter 10 Refrigeration and Heat Pump Systems Refrig_Cycle A.10 – All Tabs

10.1

Vapor Refrigeration Systems

The purpose of a refrigeration system is to maintain a cold region at a temperature below the temperature of its surroundings. This is commonly achieved using the vapor refrigeration systems that are the subject of the present section.

10.1.1

TAKE NOTE...

See Sec. 6.13.1 for the area interpretation of heat transfer on a T–s diagram for the case of internally reversible flow though a control volume at steady state.

Carnot Refrigeration Cycle

To introduce some important aspects of vapor refrigeration, let us begin by considering a Carnot vapor refrigeration cycle. This cycle is obtained by reversing the Carnot vapor power cycle introduced in Sec. 5.10. Figure 10.1 shows the schematic and accompanying T–s diagram of a Carnot refrigeration cycle operating between a region at temperature TC and another region at a higher temperature TH. The cycle is executed by a refrigerant circulating steadily through a series of components. All processes are internally reversible. Also, since heat transfers between the refrigerant and each region occur with no temperature differences, there are no external irreversibilities. The energy transfers shown on the diagram are positive in the directions indicated by the arrows. Let us follow the refrigerant as it passes steadily through each of the components in the cycle, beginning at the inlet to the evaporator. The refrigerant enters the evaporator as a two-phase liquid–vapor mixture at state 4. In the evaporator some of the refrigerant changes phase from liquid to vapor as a result of heat transfer from the region at temperature TC to the refrigerant. The temperature and pressure of the refrigerant remain constant during the process from state 4 to state 1. The refrigerant is then compressed adiabatically from state 1, where it is a two-phase liquid–vapor mixture, to state 2, where it is a saturated vapor. During this process, the temperature of the refrigerant increases from TC to TH, and the pressure also increases. The refrigerant passes from the compressor into the condenser, where it changes phase from saturated vapor to saturated liquid as a result of heat transfer to the region at temperature TH. The temperature and pressure remain constant in the process from state 2 to state 3. The refrigerant returns to the state at the inlet of the evaporator by expanding adiabatically through a turbine. In this process, from state 3 to state 4, the temperature decreases from TH to TC, and there is a decrease in pressure.

· Qout

Warm region at TH

2 3

Condenser T

· Wt

· Wc

Turbine

TH

3

2

4

1

b

a

Compressor TC

Evaporator

4 Cold region at TC

1 · Qin

Fig. 10.1 Carnot vapor refrigeration cycle.

s

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10.1 Vapor Refrigeration Systems

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Since the Carnot vapor refrigeration cycle is made up of internally reversible processes, areas on the T–s diagram can be interpreted as heat transfers. Area 1–a–b–4–1 is the heat added to the refrigerant from the cold region per unit mass of refrigerant flowing. Area 2–a–b–3–2 is the heat rejected from the refrigerant to the warm region per unit mass of refrigerant flowing. The enclosed area 1–2–3–4–1 is the net heat transfer from the refrigerant. The net heat transfer from the refrigerant equals the net work done on the refrigerant. The net work is the difference between the compressor work input and the turbine work output. The coefficient of performance b of any refrigeration cycle is the ratio of the refrigeration effect to the net work input required to achieve that effect. For the Carnot vapor refrigeration cycle shown in Fig. 10.1, the coefficient of performance is # # TC1sa 2 sb2 Qin / m area 1–a–b–4–1 # # 5 bmax 5 # # 5 area 1–2–3–4–1 1TH 2 TC21sa 2 sb2 Wc / m 2 Wt / m which reduces to bmax 5

TC TH 2 TC

(10.1)

This equation, which corresponds to Eq. 5.10, represents the maximum theoretical coefficient of performance of any refrigeration cycle operating between regions at TC and TH.

10.1.2

Departures from the Carnot Cycle

Actual vapor refrigeration systems depart significantly from the Carnot cycle considered above and have coefficients of performance lower than would be calculated from Eq. 10.1. Three ways actual systems depart from the Carnot cycle are considered next. T

c One of the most significant departures is related to the heat trans-

3′

2′

Condenser temperature, TH′

fers between the refrigerant and the two regions. In actual systems, TH′ Temperature of warm these heat transfers are not accomplished reversibly as presumed region, TH above. In particular, to achieve a rate of heat transfer sufficient to Temperature of cold maintain the temperature of the cold region at TC with a practiregion, TC cal-sized evaporator requires the temperature of the refrigerant in TC′ 4′ 1′ Evaporator the evaporator, T9C, to be several degrees below TC. This is illustemperature, TC′ trated by the placement of the temperature T9C on the T–s diagram s b a of Fig. 10.2. Similarly, to obtain a sufficient heat transfer rate from the refrigerant to the warm region requires that the refrigerant Fig. 10.2 Comparison of the condenser and temperature in the condenser, T9H, be several degrees above TH. evaporator temperatures with those of the warm This is illustrated by the placement of the temperature T9H on the and cold regions. T–s diagram of Fig. 10.2. Maintaining the refrigerant temperatures in the heat exchangers at T9C and T9H rather than at TC and TH, respectively, has the effect of reducing the coefficient of performance. This can be seen by expressing the coefficient of performance of the refrigeration cycle designated by 19–29–39–49–19 on Fig. 10.2 as b¿ 5

T¿C area 1¿–a–b–4¿–1 5 area 1¿–2¿–3¿–4¿–1¿ T¿H 2 T¿C

(10.2)

Comparing the areas underlying the expressions for bmax and b9 given above, we conclude that the value of b9 is less than bmax. This conclusion about the effect of refrigerant temperature on the coefficient of performance also applies to other refrigeration cycles considered in the chapter.

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Chapter 10 Refrigeration and Heat Pump Systems · Qout

c Even when the temperature differences between the refrigerant and

warm and cold regions are taken into consideration, there are other features that make the vapor refrigeration cycle of Fig. 10.2 impractical 3 2 as a prototype. Referring again to the figure, note that the compression process from state 19 to state 29 occurs with the refrigerant as a two-phase liquid–vapor mixture. This is commonly referred to as wet Condenser compression. Wet compression is normally avoided because the pres· ence of liquid droplets in the flowing liquid–vapor mixture can damW Expansion c Compressor age the compressor. In actual systems, the compressor handles vapor valve only. This is known as dry compression. Evaporator c Another feature that makes the cycle of Fig. 10.2 impractical is the expansion process from the saturated liquid state 39 to the low-quality, twophase liquid–vapor mixture state 49. This expansion typically produces a 4 1 Saturated or relatively small amount of work compared to the work input in the comsuperheated vapor pression process. The work developed by an actual turbine would be · smaller yet because turbines operating under these conditions have low Qin isentropic efficiencies. Accordingly, the work output of the turbine is norFig. 10.3 Components of a vapormally sacrificed by substituting a simple throttling valve for the expansion compression refrigeration system. turbine, with consequent savings in initial and maintenance costs. The components of the resulting cycle are illustrated in Fig. 10.3, where dry compression is presumed. This cycle, known as the vapor-compression refrigeration cycle, is the subject of the section to follow.

10.2 vapor-compression refrigeration

Analyzing Vapor-Compression Refrigeration Systems

Vapor-compression refrigeration systems are the most common refrigeration systems in

use today. The object of this section is to introduce some important features of systems of this type and to illustrate how they are modeled thermodynamically.

10.2.1

Evaluating Principal Work and Heat Transfers

Let us consider the steady-state operation of the vapor-compression system illustrated in Fig. 10.3. Shown on the figure are the principal work and heat transfers, which are positive in the directions of the arrows. Kinetic and potential energy changes are neglected in the following analyses of the components. We begin with the evaporator, where the desired refrigeration effect is achieved. c As the refrigerant passes through the evaporator, heat transfer from the refrigerated

refrigeration capacity ton of refrigeration

space results in the vaporization of the refrigerant. For a control volume enclosing the refrigerant side of the evaporator, the mass and energy rate balances reduce to give the rate of heat transfer per unit mass of refrigerant flowing as # Qin (10.3) # 5 h1 2 h4 m # # where m is the mass flow rate of the refrigerant. The heat transfer rate Qin is referred to as the refrigeration capacity. In the SI unit system, the capacity is normally expressed in kW. In the English unit system, the refrigeration capacity may be expressed in Btu/h. Another commonly used unit for the refrigeration capacity is the ton of refrigeration, which is equal to 200 Btu/min or about 211 kJ/min. c The refrigerant leaving the evaporator is compressed to a relatively high pressure and temperature by the compressor. Assuming no heat transfer to or from the

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compressor, the mass and energy rate balances for a control volume enclosing the compressor give # Wc (10.4) # 5 h2 2 h1 m # # where Wc / m is the rate of power input per unit mass of refrigerant flowing. c Next, the refrigerant passes through the condenser, where the refrigerant condenses and there is heat transfer from the refrigerant to the cooler surroundings. For a control volume enclosing the refrigerant side of the condenser, the rate of heat transfer from the refrigerant per unit mass of refrigerant flowing is # Qout (10.5) # 5 h2 2 h3 m c Finally, the refrigerant at state 3 enters the expansion valve and expands to the

evaporator pressure. This process is usually modeled as a throttling process for which h4 5 h3

(10.6)

The refrigerant pressure decreases in the irreversible adiabatic expansion, and there is an accompanying increase in specific entropy. The refrigerant exits the valve at state 4 as a two-phase liquid–vapor mixture. In the vapor-compression system, the net power input is equal to the compressor power, since the expansion valve involves no power input or output. Using the quantities and expressions introduced above, the coefficient of performance of the vaporcompression refrigeration system of Fig. 10.3 is # # Qin / m h1 2 h4 b5 # # 5 h2 2 h1 Wc / m

(10.7)

Provided states 1 through 4 are fixed, Eqs. 10.3 through 10.7 can be used to evaluate the principal work and heat transfers and the coefficient of performance of the vapor-compression system shown in Fig. 10.3. Since these equations have been developed by reducing mass and energy rate balances, they apply equally for actual performance when irreversibilities are present in the evaporator, compressor, and condenser and for idealized performance in the absence of such effects. Although irreversibilities in the evaporator, compressor, and condenser can have a pronounced effect on overall performance, it is instructive to consider an idealized cycle in which they are assumed absent. Such a cycle establishes an upper limit on the performance of the vapor-compression refrigeration cycle. It is considered next. T 2s

10.2.2

Performance of Ideal Vapor-Compression Systems

If irreversibilities within the evaporator and condenser are ignored, there are no frictional pressure drops, and the refrigerant flows at constant pressure through the two heat exchangers. If compression occurs without irreversibilities, and stray heat transfer to the surroundings is also ignored, the compression process is isentropic. With these considerations, the vapor-compression refrigeration cycle labeled 1–2s–3–4–1 on the T–s diagram of Fig. 10.4 results. The cycle consists of the following series of processes: Process 1–2s: Isentropic compression of the refrigerant from state 1 to the condenser pressure at state 2s.

3

4

1

s

Fig. 10.4 T–s diagram of an ideal vaporcompression cycle.

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A

VCRC A.30 – Tab a

Process 2s–3: Heat transfer from the refrigerant as it flows at constant pressure through the condenser. The refrigerant exits as a liquid at state 3. Process 3–4: Throttling process from state 3 to a two-phase liquid–vapor mixture at 4. Process 4–1: Heat transfer to the refrigerant as it flows at constant pressure through the evaporator to complete the cycle. All processes of the cycle shown in Fig. 10.4 are internally reversible except for the throttling process. Despite the inclusion of this irreversible process, the cycle is commonly referred to as the ideal vapor-compression cycle. The following example illustrates the application of the first and second laws of thermodynamics along with property data to analyze an ideal vapor-compression cycle.

ideal vaporcompression cycle

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EXAMPLE 10.1 c

Analyzing an Ideal Vapor-Compression Refrigeration Cycle Refrigerant 134a is the working fluid in an ideal vapor-compression refrigeration cycle that communicates thermally with a cold region at 08C and a warm region at 268C. Saturated vapor enters the compressor at 08C and saturated liquid leaves the condenser at 268C. The mass flow rate of the refrigerant is 0.08 kg/s. Determine (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance, and (d) the coefficient of performance of a Carnot refrigeration cycle operating between warm and cold regions at 26 and 08C, respectively. SOLUTION Known: An ideal vapor-compression refrigeration cycle operates with Refrigerant 134a. The states of the refrigerant entering the compressor and leaving the condenser are specified, and the mass flow rate is given. Find: Determine the compressor power, in kW, the refrigeration capacity, in tons, coefficient of performance, and

the coefficient of performance of a Carnot vapor refrigeration cycle operating between warm and cold regions at the specified temperatures. Schematic and Given Data: Warm region TH = 26°C = 299 K · Qout

3

Condenser

2s

T Expansion valve

2s

· Wc

Compressor

26°C

0°C 4

Evaporator

3

Temperature of warm region

a

4

1

Temperature of cold region

1 s

· Qin Cold region TC = 0°C = 273 K

Fig. E10.1

Engineering Model: 1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are

indicated by dashed lines on the accompanying sketch.

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2. Except for the expansion through the valve, which is a throttling process, all processes of the refrigerant

are internally reversible. 3. The compressor and expansion valve operate adiabatically. 4. Kinetic and potential energy effects are negligible. 5. Saturated vapor enters the compressor, and saturated liquid leaves the condenser. Analysis: Let us begin by fixing each of the principal states located on the accompanying schematic and T–s diagrams. At the inlet to the compressor, the refrigerant is a saturated vapor at 08C, so from Table A-10, h1 5 247.23 kJ/kg and s1 5 0.9190 kJ/kg ? K. The pressure at state 2s is the saturation pressure corresponding to 268C, or p2 5 6.853 bar. State 2s is fixed by p2 and the fact that the specific entropy is constant for the adiabatic, internally reversible compres➊ sion process. The refrigerant at state 2s is a superheated vapor with h2s 5 264.7 kJ/kg. State 3 is saturated liquid at 268C, so h3 5 85.75 kJ/kg. The expansion through the valve is a throttling process (assumption 2), so h4 5 h3. (a) The compressor work input is

# # Wc 5 m1h2s 2 h12 # where m is the mass flow rate of refrigerant. Inserting values # 1 kW Wc 5 10.08 kg/ s21264.7 2 247.232 kJ/ kg ` ` 1 kJ/ s 5 1.4 kW (b) The refrigeration capacity is the heat transfer rate to the refrigerant passing through the evaporator. This is given by # # Qin 5 m1h1 2 h42 1 ton 5 10.08 kg/ s2Z60 s / minZ1247.23 2 85.752 kJ/ kg ` ` 211 kJ/ min 5 3.67 ton (c) The coefficient of performance b is

# Qin h1 2 h4 247.23 2 85.75 b5 # 5 5 5 9.24 h 2 h 264.7 2 247.23 Wc 2s 1 (d) For a Carnot vapor refrigeration cycle operating at TH 5 299 K and

TC 5 273 K, the coefficient of performance determined from Eq. 10.1 is ➋

bmax 5

TC 5 10.5 TH 2 TC

➊ The value for h2s can be obtained by double interpolation in Table A-12 or by using Interactive Thermodynamics: IT. ➋ As expected, the ideal vapor-compression cycle has a lower coefficient of performance than a Carnot cycle operating between the temperatures of the warm and cold regions. The smaller value can be attributed to the effects of the external irreversibility associated with desuperheating the refrigerant in the condenser (Process 2s–a on the T–s diagram) and the internal irreversibility of the throttling process.

Keeping all other given data the same, determine the mass flow rate of refrigerant, in kg/s, for a 10-ton refrigeration capacity. Ans. 0.218 kg/s.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of

the ideal vapor compression refrigeration cycle. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate refrigeration capacity and coefficient of performance. ❑ compare with the corresponding Carnot refrigeration cycle.

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T

2s

2

3

Temperature of warm region, TH

4

1

Fig. 10.5 T–s diagram of an actual vapor-compression cycle.

10.2.3

A

cccc

Temperature of cold region, TC

s

Performance of Actual Vapor-Compression Systems

Figure 10.5 illustrates several features exhibited by actual vapor-compression systems. As shown in the figure, the heat transfers between the refrigerant and the warm and cold regions are not accomplished reversibly: the refrigerant temperature in the evaporator is less than the cold region temperature, TC, and the refrigerant temperature in the condenser is greater than the warm region temperature, TH. Such irreversible heat transfers have a significant effect on performance. In particular, the coefficient of performance decreases as the average temperature of the refrigerant in the evaporator decreases and as the average temperature of the refrigerant in the condenser increases. Example 10.2 provides an illustration.

VCRC A.30 – Tab b

EXAMPLE 10.2 c

Considering the Effect of Irreversible Heat Transfer on Performance Modify Example 10.1 to allow for temperature differences between the refrigerant and the warm and cold regions as follows. Saturated vapor enters the compressor at 210°C. Saturated liquid leaves the condenser at a pressure of 9 bar. Determine for the modified vapor-compression refrigeration cycle (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance. Compare results with those of Example 10.1. SOLUTION Known: An ideal vapor-compression refrigeration cycle operates with Refrigerant 134a as the working fluid. The evaporator temperature and condenser pressure are specified, and the mass flow rate is given. Find: Determine the compressor power, in kW, the refrigeration capacity, in tons, and the coefficient of perfor-

mance. Compare results with those of Example 10.1. Schematic and Given Data: T

Engineering Model:

2s

1. Each component of the cycle is analyzed as a control volume at 3

steady state. The control volumes are indicated by dashed lines on the sketch accompanying Example 10.1.

9 bar 26°C

2. Except for the process through the expansion valve, which is a throt-

tling process, all processes of the refrigerant are internally reversible. 0°C –10°C

4

3. The compressor and expansion valve operate adiabatically.

1

4. Kinetic and potential energy effects are negligible. s

Fig. E10.2

5. Saturated vapor enters the compressor, and saturated liquid exits

the condenser.

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Analysis: Let us begin by fixing each of the principal states located on the accompanying T–s diagram. Starting at

the inlet to the compressor, the refrigerant is a saturated vapor at 2108C, so from Table A-10, h1 5 241.35 kJ/kg and s1 5 0.9253 kJ/kg ? K. The superheated vapor at state 2s is fixed by p2 5 9 bar and the fact that the specific entropy is constant for the adiabatic, internally reversible compression process. Interpolating in Table A-12 gives h2s 5 272.39 kJ/kg. State 3 is a saturated liquid at 9 bar, so h3 5 99.56 kJ/kg. The expansion through the valve is a throttling process; thus, h4 5 h3. (a) The compressor power input is

# # Wc 5 m1h2s 2 h12 # where m is the mass flow rate of refrigerant. Inserting values # 1 kW Wc 5 10.08 kg/ s21272.39 2 241.352 kJ/ kg ` ` 1 kJ/ s 5 2.48 kW (b) The refrigeration capacity is

# # Qin 5 m1h1 2 h42 5 10.08 kg/ s2Z60 s/ minZ1241.35 2 99.562 kJ/ kg `

1 ton ` 211 kJ/ min

5 3.23 ton (c) The coefficient of performance b is

# Qin h1 2 h4 241.35 2 99.56 b5 # 5 5 5 4.57 h 2 h 272.39 2 241.35 Wc 2s 1 Comparing the results of the present example with those of Example 10.1, we see that the power input required by the compressor is greater in the present case. Furthermore, the refrigeration capacity and coefficient of performance are smaller in this example than in Example 10.1. This illustrates the considerable influence on performance of irreversible heat transfer between the refrigerant and the cold and warm regions.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of

the ideal vapor compression refrigeration cycle. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate compressor power, refrigeration capacity, and coefficient of performance.

Determine the rate of heat transfer from the refrigerant passing through the condenser to the surroundings, in kW. Ans. 13.83 kW.

Referring again to Fig. 10.5, we can identify another key feature of actual vaporcompression system performance. This is the effect of irreversibilities during compression, suggested by the use of a dashed line for the compression process from state 1 to state 2. The dashed line is drawn to show the increase in specific entropy that accompanies an adiabatic irreversible compression. Comparing cycle 1–2–3–4–1 with cycle 1–2s–3–4–1, the refrigeration capacity would be the same for each, but the work input would be greater in the case of irreversible compression than in the ideal cycle. Accordingly, the coefficient of performance of cycle 1–2–3–4–1 is less than that of cycle 1–2s–3–4–1. The effect of irreversible compression can be accounted for by using the isentropic compressor efficiency, which for states designated as in Fig. 10.5 is given by # # 1Wc/ m2s h2s 2 h1 hc 5 # # 5 h2 2 h1 1Wc/ m2

TAKE NOTE...

The isentropic compressor efficiency is introduced in Sec. 6.12.3. See Eq. 6.48.

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VCRC A.30 – Tabs c & d

Additional departures from ideality stem from frictional effects that result in pressure drops as the refrigerant flows through the evaporator, condenser, and piping connecting the various components. These pressure drops are not shown on the T–s diagram of Fig. 10.5 and are ignored in subsequent discussions for simplicity. Finally, two additional features exhibited by actual vapor-compression systems are shown in Fig. 10.5. One is the superheated vapor condition at the evaporator exit (state 1), which differs from the saturated vapor condition shown in Fig. 10.4. Another is the subcooling of the condenser exit state (state 3), which differs from the saturated liquid condition shown in Fig. 10.4. Example 10.3 illustrates the effects of irreversible compression and condenser exit subcooling on the performance of the vapor-compression refrigeration system.

EXAMPLE 10.3 c

Analyzing an Actual Vapor-Compression Refrigeration Cycle Reconsider the vapor-compression refrigeration cycle of Example 10.2, but include in the analysis that the compressor has an isentropic efficiency of 80%. Also, let the temperature of the liquid leaving the condenser be 308C. Determine for the modified cycle (a) the compressor power, in kW, (b) the refrigeration capacity, in tons, (c) the coefficient of performance, and (d) the rates of exergy destruction within the compressor and expansion valve, in kW, for T0 5 299 K (268C). SOLUTION Known: A vapor-compression refrigeration cycle has an isentropic compressor efficiency of 80%. Find: Determine the compressor power, in kW, the refrigeration capacity, in tons, the coefficient of performance,

and the rates of exergy destruction within the compressor and expansion valve, in kW. Schematic and Given Data: Engineering Model:

T 2

p2 = 9 bar

1. Each component of the cycle is analyzed as a control

volume at steady state.

2s

2. There are no pressure drops through the evaporator and

condenser. 30°C

3

–10°C

T0 = 26°C = 299 K

4

1

efficiency of 80%. The expansion through the valve is a throttling process. 4. Kinetic and potential energy effects are negligible.

s

Fig. E10.3

3. The compressor operates adiabatically with an isentropic

5. Saturated vapor at 2108C enters the compressor, and

liquid at 308C leaves the condenser. 6. The environment temperature for calculating exergy is

T0 5 299 K (268C). Analysis: Let us begin by fixing the principal states. State 1 is the same as in Example 10.2, so h1 5 241.35 kJ/kg

and s1 5 0.9253 kJ/kg ? K. Owing to the presence of irreversibilities during the adiabatic compression process, there is an increase in specific entropy from compressor inlet to exit. The state at the compressor exit, state 2, can be fixed using the isentropic compressor efficiency # # 1h2s 2 h12 1Wc / m2s hc 5 # # 5 1h2 2 h12 Wc / m

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where h2s is the specific enthalpy at state 2s, as indicated on the accompanying T–s diagram. From the solution to Example 10.2, h2s 5 272.39 kJ/kg. Solving for h2 and inserting known values h2 5

1272.39 2 241.352 h2s 2 h1 1 h1 5 1 241.35 5 280.15 kJ/ kg hc 10.802

State 2 is fixed by the value of specific enthalpy h2 and the pressure, p2 5 9 bar. Interpolating in Table A-12, the specific entropy is s2 5 0.9497 kJ/kg ? K. The state at the condenser exit, state 3, is in the liquid region. The specific enthalpy is approximated using Eq. 3.14, together with saturated liquid data at 308C, as follows: h3 < hf 5 91.49 kJ/ kg. Similarly, with Eq. 6.5, s3 < sf 5 0.3396 kJ/ kg ? K. The expansion through the valve is a throttling process; thus, h4 5 h3. The quality and specific entropy at state 4 are, respectively x4 5

h4 2 hf4 91.49 2 36.97 5 5 0.2667 hg4 2 hf4 204.39

and s4 5 sf4 1 x41sg4 2 sf42 5 0.1486 1 10.2667210.9253 2 0.14862 5 0.3557 kJ/ kg ? K (a) The compressor power is

# # Wc 5 m1h2 2 h12 5 10.08 kg/ s21280.15 2 241.352 kJ/ kg `

1 kW ` 5 3.1 kW 1 kJ/ s

(b) The refrigeration capacity is

# # Qin 5 m1h1 2 h42 5 10.08 kg/ s2Z60 s/ minZ1241.35 2 91.492 kJ/ kg ` 5 3.41 ton

1 ton ` 211 kJ/ min

(c) The coefficient of performance is

➊

b5

1h1 2 h42 1241.35 2 91.492 5 5 3.86 1h2 2 h12 1280.15 2 241.352

(d) The rates of exergy destruction in the compressor and expansion valve can be # #

found by reducing the exergy rate balance or using the relationship Ed 5 T0scv , # where scv is the rate of entropy production from an entropy rate balance. With either approach, the rates or exergy destruction for the compressor and valve are, respectively # # # # 1Ed2c 5 mT01s2 2 s12 and 1Ed2valve 5 mT01s4 2 s32 Substituting values ➋

# kg kJ 1 kW 1Ed2c 5 a0.08 b1299 K210.9497 2 0.92532 ` ` 5 0.58 kW s kg ? K 1 kJ/ s

and # 1Ed2valve 5 10.0821299210.3557 2 0.33962 5 0.39 kW ➊ While the refrigeration capacity is greater than in Example 10.2, irreversibilities in the compressor result in an increase in compressor power compared to isentropic compression. The overall effect is a lower coefficient of performance than in Example 10.2.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of

the vapor compression refrigeration cycle with irreversibilities in the compressor and subcooled liquid exiting the condenser. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate compressor power, refrigeration capacity, and coefficient of performance. ❑ calculate exergy destruction in the compressor and expansion valve.

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➋ The exergy destruction rates calculated in part (d) measure the effect of irreversibilities as the refrigerant flows through the compressor and valve. The percentages of the power input (exergy input) to the compressor destroyed in the compressor and valve are 18.7 and 12.6%, respectively.

What would be the coefficient of performance if the isentropic compressor efficiency were 100%? Ans. 4.83.

10.2.4 The p–h Diagram p–h diagram

A thermodynamic property diagram widely used in the refrigeration field is the pressure– enthalpy or p–h diagram. Figure 10.6 shows the main features of such a property diagram. The principal states of the vapor-compression cycles of Fig. 10.5 are located on this p–h diagram. It is left as an exercise to sketch the cycles of Examples 10.1, 10.2, and 10.3 on p–h diagrams. Property tables and p–h diagrams for many refrigerants are given in handbooks dealing with refrigeration.

Selecting Refrigerants

10.3

Co

ns

ta

nt

s

Refrigerant selection for a wide range of refrigeration and air-conditioning applications is generally based on three factors: performance, safety, and environmental impact. The term performance refers to providing the required cooling or heating capacity reliably and cost effectively. Safety refers to avoiding hazards such as toxicity and flammability. Finally, environmental impact prip marily refers to using refrigerants that do not harm the stratospheric ozone layer or contribute significantly to global climate change. We Condenser begin by considering some performance aspects. pressure 3 2s 2 The temperatures of the refrigerant in the evaporator and condenser of vapor-compression cycles are governed by the temperatures Evaporator of the cold and warm regions, respectively, with which the system pressure interacts thermally. This, in turn, determines the operating pressures 4 1 Constant T in the evaporator and condenser. Consequently, the selection of a refrigerant is based partly on the suitability of its pressure–temperah ture relationship in the range of the particular application. It is generally desirable to avoid excessively low pressures in the evaporator Fig. 10.6 Principal features of the pressure– enthalpy diagram for a typical refrigerant, with and excessively high pressures in the condenser. Other considerations in refrigerant selection include chemical stability, corrosiveness, and vapor-compression cycles superimposed. cost. The type of compressor also affects the choice of refrigerant. Centrifugal compressors are best suited for low evaporator pressures and refrigerants with large specific volumes at low pressure. Reciprocating compressors perform better over large pressure ranges and are better able to handle low specific volume refrigerants.

Refrigerant Types and Characteristics Prior to the 1930s, accidents were prevalent among those who worked closely with refrigerants due to the toxicity and flammability of most refrigerants at the time. Because of such hazards, two classes of synthetic refrigerants were developed, each containing chlorine and possessing highly stable molecular structures: CFCs (chlorofluorocarbons) and HCFCs (hydrochlorofluorocarbons). These refrigerants were widely known as “freons,” the common trade name. In the early 1930s, CFC production began with R-11, R-12, R-113, and R-114. In 1936, the first HCFC refrigerant, R-22, was introduced. Over the next several decades,

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10.3 Selecting Refrigerants nearly all of the synthetic refrigerants used in the U.S. were either CFCs or HCFCs, with R-12 being most commonly used. To keep order with so many new refrigerants having complicated names, the “R” numbering system was established in 1956 by DuPont and persists today as the industry standard. Table 10.1 lists information, including refrigerant number, chemical composition, and global warming potential for selected refrigerants.

Environmental Considerations After decades of use, compelling scientific data indicating that release of chlorinecontaining refrigerants into the atmosphere is harmful became widely recognized. Concerns focused on released refrigerants depleting the stratospheric ozone layer and contributing to global climate change. Because of the molecular stability of the CFC and HCFC molecules, their adverse effects are long-lasting. In 1987, an international agreement was adopted to ban production of certain chlorinecontaining refrigerants. In response, a new class of chlorine-free refrigerants was developed: the HFCs (hydrofluorocarbons). One of these, R-134a, has been used for over 20 years as the primary replacement for R-12. Although R-134a and other HFC refrigerants do not contribute to atmospheric ozone depletion, they do contribute to global climate change. Owing to a relatively high Global Warming Potential of about 1430 for R-134a, we may soon see reductions in its use in the United States despite widespread deployment in refrigeration and air-conditioning systems, including automotive air conditioning. Carbon dioxide (R-744) and R-1234yf are potential replacements for R-134a in automotive systems. See Sec. 10.7.3 for discussion of CO2-charged automotive air-conditioning systems. Another refrigerant that has been used extensively in air-conditioning and refrigeration systems for decades, R-22, is being phased out under a 1995 amendment to the international agreement on refrigerants because of its chlorine content. Effective in 2010, R-22 cannot be installed in new systems. However, recovered and recycled R-22 can be used to service existing systems until supplies are no longer available. As R-22 is phased out, replacement refrigerants are being introduced, including R-410A and R-407C, both HFC blends.

Natural Refrigerants Nonsynthetic, naturally occurring substances also can be used as refrigerants. Called natural refrigerants, they include carbon dioxide, ammonia, and hydrocarbons. As TABLE 10.1

Refrigerant Data Including Global Warming Potential (GWP) Refrigerant Number

Type

Chemical Formula

R-12 R-11 R-114 R-113 R-22 R-134a R-1234yf R-410A

CFC CFC CFC CFC HCFC HFC HFC HFC blend

R-407C

HFC blend

R-744 (carbon dioxide) R-717 (ammonia) R-290 (propane) R-50 (methane) R-600 (butane)

Natural Natural Natural Natural Natural

CCl2F2 CCl3F CClF2CClF2 CCl2FCCIF2 CHClF2 CH2FCF3 CF3CF5CH2 R-32, R-125 (50/50 Weight %) R-32, R-125, R-134a (23/25/52 Weight %) CO2 NH3 C3H8 CH4 C4H10

Approx. GWPa

10900 4750 10000 6130 1810 1430 4 1725 1526 1 0 10 25 10

a The Global Warming Potential (GWP) depends on the time period over which the potential influence on global warming is estimated. The values listed are based on a 100 year time period, which is an interval favored by some regulators.

601

TAKE NOTE...

Global warming refers to an increase in global average temperature due to a combination of natural phenomena and human industrial, agricultural, and lifestyle activities. The Global Warming Potential (GWP) is a simplified index that aims to estimate the potential future influence on global warming of different gases when released to the atmosphere. The GWP of a gas refers to how much that gas contributes to global warming in comparison to the same amount of carbon dioxide. The GWP of carbon dioxide is taken to be 1.

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Chapter 10 Refrigeration and Heat Pump Systems indicated by Table 10.1, natural refrigerants typically have low Global Warming Potentials. Ammonia (R-717), which was widely used in the early development of vaporcompression refrigeration, continues to serve today as a refrigerant for large systems used by the food industry and in other industrial applications. In the past two decades, ammonia has been increasingly used because of the R-12 phase out and is receiving even greater interest today due to the R-22 phase out. Ammonia is also used in the absorption systems discussed in Sec. 10.5. Hydrocarbons, such as propane (R-290), are used worldwide in various refrigeration and air-conditioning applications including commercial and household appliances. In the United States, safety concerns limit propane use to niche markets like industrial process refrigeration. Other hydrocarbons—methane (R-50) and butane (R-600)—are also under consideration for use as refrigerants.

Refrigeration with No Refrigerant Needed Alternative cooling technologies aim to achieve a refrigerating effect without use of refrigerants, thereby avoiding adverse effects associated with release of refrigerants to the atmosphere. One such technology is thermoelectric cooling. See the box.

New Materials May Improve Thermoelectric Cooling You can buy a thermoelectric cooler powered from the cigarette lighter outlet of your car. The same technology is used in space flight applications and in power amplifiers and microprocessors. Figure 10.7 shows a thermoelectric cooler separating a cold region at temperature TC and a warm region at temperature TH. The cooler is formed from two n-type and two p-type semiconductors with low thermal conductivity, five metallic interconnects with high electrical conductivity and high thermal conductivity, two electrically insulating ceramic substrates, and a power source. When power is provided by the source, current flows through the resulting electric circuit, giving a refrigeration effect: a heat transfer of energy from the cold region. This is known as the Peltier effect. The p-type semiconductor material in the right leg of the cooler shown in Fig. 10.7 has electron vacancies, called holes. Electrons move through this material by filling individual holes, slowing electron motion. In the adjacent n-type semiconductor, no holes exist in its

+

–

i

i Warm Region at TH · Qout

Negatively charged electron

e–

e– n

· Qout

e–

e– n

p

p

+ e– Metallic interconnect

Ceramic substrate

+ e–

e–

· Qin

e– · Qin

Cold Region at TC

Fig. 10.7 Schematic of a thermoelectric cooler.

Positively charged hole

Ceramic substrate

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10.4 Other Vapor-Compression Applications

material structure, so electrons move freely and more rapidly through that material. When power is provided by the power source, positively charged holes move in the direction of current while negatively charged electrons move opposite to the current; each transfers energy from the cold region to the warm region. The process of Peltier refrigeration may be understood by following the journey of an electron as it travels from the negative terminal of the power source to the positive terminal. On flowing through the metallic interconnect and into the p-type material, the electron slows and loses energy, causing the surrounding material to warm. At the other end of the p-type material, the electron accelerates as it enters the metallic interconnect and then the n-type material. The accelerating electron acquires energy from the surrounding material and causes the end of the p-type leg to cool. While the electron traverses the p-type material from the hot end to the cold end, holes are moving from the cold end to the hot end, transferring energy away from the cold end. While traversing the n-type material from the cold end to the hot end, the electron also transfers energy away from the cold end to the hot end. When it reaches the warm end of the n-type leg, the electron flows through the metallic interconnect and enters the next p-type material, where it slows and again loses energy. This scenario repeats itself at each pair of p-type and n-type legs, resulting in more removal of energy from the cold end and its deposit at the hot end. Thus, the overall effect of the thermoelectric cooler is heat transfer from the cold region to the warm region. These simple coolers have no moving parts at a macroscopic level and are compact. They are reliable and quiet. They also use no refrigerants that harm the ozone layer or contribute to global climate change. Despite such advantages, thermoelectric coolers have found only specialized application because of low coefficients of performance compared to vapor-compression systems. However new materials and production methods may make this type of cooler more effective, material scientists report. As shown in Fig. 10.7, at the core of a thermoelectric cooler are two dissimilar materials, in this case n-type and p-type semiconductors. To be effective for thermoelectric cooling, the materials must have low thermal conductivity and high electrical conductivity, a rare combination in nature. However, new materials with novel microscopic structures at the nanometer level may lead to improved cooler performance. With nanotechnology and other advanced techniques, material scientists are striving to find materials with the favorable characteristics needed to improve the performance of thermoelectric cooling devices.

10.4

Other Vapor-Compression Applications

The basic vapor-compression refrigeration cycle can be adapted for special applications. Three are presented in this section. The first is cold storage, which is a thermal energy storage approach that involves chilling water or making ice. The second is a combined-cycle arrangement where refrigeration at relatively low temperature is achieved through a series of vapor-compression systems, with each normally employing a different refrigerant. In the third, compression work is reduced through multistage compression with intercooling between stages. The second and third applications considered are analogous to power cycle applications considered in Chaps. 8 and 9.

10.4.1 Cold Storage Chilling water or making ice during off-peak periods, usually overnight or over weekends, and storing chilled water/ice in tanks until needed for cooling is known as cold storage. Cold storage is an aspect of thermal energy storage considered in the box on p. 111. Applications of cold storage include cooling of office and commercial buildings, medical centers, college campus buildings, and shopping malls.

603

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Chapter 10 Refrigeration and Heat Pump Systems Return refrigerant

Refrigeration unit

Ice maker Cold refrigerant Ice storage + –

Chilled coolant

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Pump

Coolant loop

Cool, less humid air to occupied spaces

Warm, humid air from occupied spaces

Condensate

Fig. 10.8 Cold storage applied to comfort cooling.

Figure 10.8 illustrates a cold storage system intended for the comfort cooling of an occupied space. It consists of a vapor-compression refrigeration unit, ice maker and ice storage tank, and coolant loop. Running at night, when less power is required for its operation due to cooler ambient temperatures and when electricity rates are lowest, the refrigeration unit freezes water. The ice produced is stored in the accompanying tank. When cooling is required by building occupants during the day, the temperature of circulating building air is reduced as it passes over coils carrying chilled coolant flowing from the ice storage tank. Depending on local climate, some moisture also may be removed or added (see Secs. 12.8.3 and 12.8.4). Cool storage can provide all cooling required by the occupants or work in tandem with vaporcompression or other comfort cooling systems to meet needs.

10.4.2 Cascade Cycles Combined cycle arrangements for refrigeration are called cascade cycles. In Fig. 10.9 a cascade cycle is shown in which two vapor-compression refrigeration cycles, labeled A and B, are arranged in series with a counterflow heat exchanger linking them. In the intermediate heat exchanger, the energy rejected during condensation of the refrigerant in the lower-temperature cycle A is used to evaporate the refrigerant in the higher-temperature cycle B. The desired refrigeration effect occurs in the low-temperature evaporator, and heat rejection from the overall cycle occurs in the high-temperature condenser. The coefficient of performance is the ratio of the refrigeration effect to the total work input # Qin # b5 # WcA 1 WcB

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10.4 Other Vapor-Compression Applications · Qout

The mass flow rates in cycles A and B are normally different. However, the mass flow rates are related by mass and energy rate balances on the interconnecting counterflow heat exchanger serving as the condenser for cycle A and the evaporator for cycle B. Although only two cycles are shown in Fig. 10.9, cascade cycles may employ three or more individual cycles. A significant feature of the cascade system illustrated in Fig. 10.9 is that the refrigerants in the two or more stages can be selected to have advantageous evaporator and condenser pressures in the two or more temperature ranges. In a double cascade system, a refrigerant would be selected for cycle A that has a saturation pressure–temperature relationship that allows refrigeration at a relatively low temperature without excessively low evaporator pressures. The refrigerant for cycle B would have saturation characteristics that permit condensation at the required temperature without excessively high condenser pressures.

10.4.3

High-temperature condenser

7

6

Expansion valve Compressor

Intermediate heat exchanger

8

3

5

2 Cycle A Compressor

The advantages of multistage compression with intercooling between Expansion stages have been cited in Sec. 9.8, dealing with gas power systems. Intervalve cooling is achieved in gas power systems by heat transfer to the lower4 1 Low-temperature temperature surroundings. In refrigeration systems, the refrigerant temevaporator perature is below that of the surroundings for much of the cycle, so other means must be employed to accomplish intercooling and achieve the attendant savings in the required compressor work input. An · arrangement for two-stage compression using the refrigerant itself for Qin intercooling is shown in Fig. 10.10. The principal states of the refrigerant Fig. 10.9 Example of a cascade vaporfor an ideal cycle are shown on the accompanying T–s diagram. compression refrigeration cycle.

· Qout (1) Condenser 4 Expansion valve Compressor

· Wc2

(1) 6

3

(1)

9 (x)

Flash chamber

a

T Direct contact heat exchanger 4 2

7

(1 – x)

(1 – x) Compressor

5 · Wc1

2 7

3 6

9

Expansion valve 8

1 8

Evaporator

· WcB

Cycle B

Multistage Compression with Intercooling

5

605

1

(1 – x) s · Qin

Fig. 10.10 Refrigeration cycle with two stages of compression and flash intercooling.

· WcA

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Chapter 10 Refrigeration and Heat Pump Systems

flash chamber

Intercooling is accomplished in this cycle by means of a direct contact heat exchanger. Relatively low-temperature saturated vapor enters the heat exchanger at state 9, where it mixes with higher-temperature refrigerant leaving the first compression stage at state 2. A single mixed stream exits the heat exchanger at an intermediate temperature at state 3 and is compressed in the second compressor stage to the condenser pressure at state 4. Less work is required per unit of mass flow for compression from 1 to 2 followed by compression from 3 to 4 than for a single stage of compression 1–2–a. Since the refrigerant temperature entering the condenser at state 4 is lower than for a single stage of compression in which the refrigerant would enter the condenser at state a, the external irreversibility associated with heat transfer in the condenser is also reduced. A central role is played in the cycle of Fig. 10.10 by a liquid–vapor separator, called a flash chamber. Refrigerant exiting the condenser at state 5 expands through a valve and enters the flash chamber at state 6 as a two-phase liquid–vapor mixture with quality x. In the flash chamber, the liquid and vapor components separate into two streams. Saturated vapor exiting the flash chamber enters the heat exchanger at state 9, where intercooling is achieved as discussed above. Saturated liquid exiting the flash chamber at state 7 expands through a second valve into the evaporator. On the basis of a unit of mass flowing through the condenser, the fraction of the vapor formed in the flash chamber equals the quality x of the refrigerant at state 6. The fraction of the liquid formed is then (1 2 x). The fractions of the total flow at various locations are shown in parentheses on Fig. 10.10.

10.5 absorption refrigeration

Absorption Refrigeration

Absorption refrigeration cycles are the subject of this section. These cycles have some features in common with the vapor-compression cycles considered previously but differ in two important respects: c One is the nature of the compression process. Instead of compressing a vapor

between the evaporator and the condenser, the refrigerant of an absorption system is absorbed by a secondary substance, called an absorbent, to form a liquid solution. The liquid solution is then pumped to the higher pressure. Because the average specific volume of the liquid solution is much less than that of the refrigerant vapor, significantly less work is required (see the discussion of Eq. 6.51b in Sec. 6.13.2). Accordingly, absorption refrigeration systems have the advantage of relatively small work input compared to vapor-compression systems. c The other main difference between absorption and vapor-compression systems is that some means must be introduced in absorption systems to retrieve the refrigerant vapor from the liquid solution before the refrigerant enters the condenser. This involves heat transfer from a relatively high-temperature source. Steam or waste heat that otherwise would be discharged to the surroundings without use is particularly economical for this purpose. Natural gas or some other fuel can be burned to provide the heat source, and there have been practical applications of absorption refrigeration using alternative energy sources such as solar and geothermal energy. The principal components of an absorption refrigeration system are shown schematically in Fig. 10.11. In this case, ammonia is the refrigerant and water is the absorbent. Ammonia circulates through the condenser, expansion valve, and evaporator as in a vapor-compression system. However, the compressor is replaced by the absorber, pump, generator, and valve shown on the right side of the diagram. c In the absorber, ammonia vapor coming from the evaporator at state 1 is absorbed

by liquid water. The formation of this liquid solution is exothermic. Since the amount of ammonia that can be dissolved in water increases as the solution temperature

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10.5 Absorption Refrigeration · QG

· Qout

607

Hightemperature source

Generator Condenser 2

Weak solution

b

Expansion valve

c Valve

Strong solution

3

Pump

Absorber

· Wp

4 Evaporator 1 · Qin Refrigerated region

a Cooling water

Fig. 10.11 Simple ammonia–water absorption refrigeration system. decreases, cooling water is circulated around the absorber to remove the energy released as ammonia goes into solution and maintain the temperature in the absorber as low as possible. The strong ammonia– water solution leaves the absorber at point a and enters the pump, where its pressure is increased to that of the generator. c In the generator, heat transfer from a high-temperature source drives ammonia vapor out of the solution (an endothermic process), leaving a weak ammonia–water solution in the generator. The vapor liberated passes to the condenser at state 2, and the remaining weak solution at c flows back to the absorber through a valve. The only work input is the power required to operate the pump, and this is small in comparison to the work that would be required to compress refrigerant vapor between the same pressure levels. However, costs associated with the heat source and extra equipment not required by vapor-compressor systems can cancel the advantage of a smaller work input.

· Qout

2 Condenser

3

Rectifier · QG

Generator

Expansion

Ammonia–water systems normally employ several modifications of valve the simple absorption cycle considered above. Two common modificaHeat tions are illustrated in Fig. 10.12. In this cycle, a heat exchanger is included exchanger between the generator and the absorber that allows the strong water– ammonia solution entering the generator to be preheated by the weak 4 solution returning from the generator # to the absorber, thereby reducing the heat transfer to the generator, QG. The other modification shown on Valve the figure is the rectifier placed between the generator and the condenser. The function of the rectifier is to remove any traces of water from the refrigerant before it enters the condenser. This eliminates the possibility 1 Evaporator Absorber of ice formation in the expansion valve and the evaporator. Another type of absorption system uses lithium bromide as the absor· · bent and water as the refrigerant. The basic principle of operation is the Qcw Qin same as for ammonia–water systems. To achieve refrigeration at lower temperatures than are possible with water as the refrigerant, a lithium Fig. 10.12 Modified ammonia–water bromide–water absorption system may be combined with another cycle absorption system.

Pump · Wp

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Chapter 10 Refrigeration and Heat Pump Systems using a refrigerant with good low-temperature characteristics, such as ammonia, to form a cascade refrigeration system.

10.6

A

Heat_Pump_Cycle A.11 – All Tabs

Heat Pump Systems

The objective of a heat pump is to maintain the temperature within a dwelling or other building above the temperature of the surroundings or to provide a heat transfer for certain industrial processes that occur at elevated temperatures. Heat pump systems have many features in common with the refrigeration systems considered thus far and may be of the vapor-compression or absorption type. Vapor-compression heat pumps are well suited for space heating applications and are commonly used for this purpose. Absorption heat pumps have been developed for industrial applications and are also increasingly being used for space heating. To introduce some aspects of heat pump operation, let us begin by considering the Carnot heat pump cycle.

10.6.1

Carnot Heat Pump Cycle

By simply changing our viewpoint, we can regard the cycle shown in Fig. 10.1 as a heat pump. The objective of the cycle now, however, is to deliver the heat transfer # Qout to the warm region, which is the space to be heated. At steady state, the rate at which energy is supplied to the warm region by heat# transfer is the sum of the energy supplied to the# working fluid from the cold region, Qin, and the net rate of work input to the cycle, Wnet. That is # # # Qout 5 Qin 1 Wnet (10.8) The coefficient of performance of any heat pump cycle is defined as the ratio of the heating effect to the net work required to achieve that effect. For the Carnot heat pump cycle of Fig. 10.1 # # Qout / m area 2–a–b–3–2 # # 5 gmax 5 # # area 1–2–3–4–1 Wc / m 2 Wt / m which reduces to gmax 5

TH1sa 2 sb2 TH 5 TH 2 TC 1TH 2 TC21sa 2 sb2

(10.9)

This equation, which corresponds to Eq. 5.11, represents the maximum theoretical coefficient of performance for any heat pump cycle operating between two regions at temperatures TC and TH. Actual heat pump systems have coefficients of performance that are lower than calculated from Eq. 10.9. A study of Eq. 10.9 shows that as the temperature TC of the cold region decreases, the coefficient of performance of the Carnot heat pump decreases. This trait is also exhibited by actual heat pump systems and suggests why heat pumps in which the role of the cold region is played by the local atmosphere (air-source heat pumps) normally require backup systems to provide heating on days when the ambient temperature becomes very low. If sources such as well water or the ground itself are used, relatively high coefficients of performance can be achieved despite low ambient air temperatures, and backup systems may not be required.

10.6.2

Vapor-Compression Heat Pumps

Actual heat pump systems depart significantly from the Carnot cycle model. Most systems in common use today are of the vapor-compression type. The method of analysis of vapor-compression heat pumps is the same as that of vapor-compression

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10.6 Heat Pump Systems

Inside air Condenser · Qout

Outside air 3

Evaporator

4 Expansion valve

· Wc

· Qin

Compressor

2

1

Fig. 10.13 Air-source vapor-compression heat pump system.

refrigeration cycles considered previously. Also, the previous discussions concerning the departure of actual systems from ideality apply for vapor-compression heat pump systems as for vapor-compression refrigeration cycles. As illustrated by Fig. 10.13, a typical vapor-compression heat pump for space heating has the same basic components as the vapor-compression refrigeration system: compressor, condenser, expansion valve, and evaporator. The objective of the system# is # different, however. In a heat pump system, Qin comes from the surroundings, and Qout is directed to the dwelling as the desired effect. A net work input is required to accomplish this effect. The coefficient of performance of a simple vapor-compression heat pump with states as designated on Fig. 10.13 is # # h2 2 h3 Qout / m g5 # # 5 h2 2 h1 Wc / m

vapor-compression heat pump

(10.10)

The value of g can never be less than unity. Many possible sources are available for heat transfer to the refrigerant passing through the evaporator, including outside air; the ground; and lake, river, or well water. Liquid circulated through a solar collector and stored in an insulated tank also can be used as a source for a heat pump. Industrial heat pumps employ waste heat or warm liquid or gas streams as the low-temperature source and are capable of achieving relatively high condenser temperatures. In the most common type of vapor-compression heat pump for space heating, the evaporator communicates thermally with the outside air. Such air-source heat pumps also can be used to provide cooling in the summer with the use of a reversing valve, as illustrated in Fig. 10.14. The solid lines show the flow path of the refrigerant in the heating mode, as described previously. To use the same components as an air conditioner, the valve is actuated, and the refrigerant follows the path indicated by the dashed line. In the cooling mode, the outside heat exchanger becomes the condenser, and the inside heat exchanger becomes the evaporator. Although heat pumps can be more costly to install and operate than other direct heating systems, they can be competitive when the potential for dual use is considered. Example 10.4 illustrates use of the first and second laws of thermodynamics together with property data to analyze the performance of an actual heat pump cycle, including cost of operation.

air-source heat pump

609

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Chapter 10 Refrigeration and Heat Pump Systems

Expansion valve

Inside heat exchanger

Outside heat exchanger

Reversing valve

Heating mode Cooling mode

Fig. 10.14 Example of an air-to-air reversing

Compressor

· Wc

heat pump.

cccc

EXAMPLE 10.4 c

Analyzing an Actual Vapor-Compression Heat Pump Cycle Refrigerant 134a is the working fluid in an electric-powered, air-source heat pump that maintains the inside temperature of a building at 228C for a week when the average outside temperature is 58C. Saturated vapor enters the compressor at 288C and exits at 508C, 10 bar. Saturated liquid exits the condenser at 10 bar. The refrigerant mass flow rate is 0.2 kg/s for steady-state operation. Determine (a) the compressor power, in kW, (b) the isentropic compressor efficiency, (c) the heat transfer rate provided to the building, in kW, (d) the coefficient of performance, and (e) the total cost of electricity, in $, for 80 hours of operation during that week, evaluating electricity at 15 cents per kW ? h. SOLUTION Known: A heat pump cycle operates with Refrigerant 134a. The states of the refrigerant entering and exiting the compressor and leaving the condenser are specified. The refrigerant mass flow rate and interior and exterior temperatures are given. Find: Determine the compressor power, the isentropic compressor efficiency, the heat transfer rate to the build-

ing, the coefficient of performance, and the cost to operate the electric heat pump for 80 hours of operation. Schematic and Given Data: Building inside Tinside = 22°C · Qout

Condenser 3

2

Expansion valve

T · Wc

Compressor

4

2

50°C

2s 3

1

–8°C

P2 = 10 bar

Tinside = 22°C Toutside = 5°C 4

1

Evaporator

· Qin Toutside = 5°C

Outside

s

Fig. E10.4

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10.6 Heat Pump Systems

611

Engineering Model: 1. Each component of the cycle is analyzed as a control volume at steady state. 2. There are no pressure drops through the evaporator and condenser. 3. The compressor operates adiabatically. The expansion through the valve is a throttling process. 4. Kinetic and potential energy effects are negligible. 5. Saturated vapor enters the compressor and saturated liquid exits the condenser. 6. For costing purposes, conditions provided are representative of the entire week of operation and the

value of electricity is 15 cents per kW ? h. Analysis: Let us begin by fixing the principal states located on the accompanying schematic and T–s diagram.

State 1 is saturated vapor at 288C; thus h1 and s1 are obtained directly from Table A-10. State 2 is superheated vapor; knowing T2 and p2, h2 is obtained from Table A-12. State 3 is saturated liquid at 10 bar and h3 is obtained from Table A-11. Finally, expansion through the valve is a throttling process; therefore, h4 5 h3. A summary of property values at these states is provided in the following table: State

T (°C)

p (bar)

h (kJ/kg)

s (kJ/kg ? K)

1 2 3 4

28 50 — —

2.1704 10 10 2.1704

242.54 280.19 105.29 105.29

0.9239 — — —

(a) The compressor power is

# kg kJ 1 kW # Wc 5 m1h2 2 h12 5 0.2 1280.19 2 242.542 ` ` 5 7.53 kW s kg 1 kJ/ s (b) The isentropic compressor efficiency is

# # 1h2s 2 h12 1Wc / m2s hc 5 # # 5 1h2 2 h12 1Wc / m2 where h2s is the specific entropy at state 2s, as indicated on the accompanying T–s diagram. State 2s is fixed using p2 and s2s 5 s1. Interpolating in Table A-12, h2s 5 274.18 kJ/kg. Solving for compressor efficiency 1h2s 2 h12 1274.18 2 242.542 hc 5 5 5 0.84 184%2 1h2 2 h12 1280.19 2 242.542 (c) The heat transfer rate provided to the building is

# kg kJ 1 kW # Qout 5 m1h2 2 h32 5 a0.2 b1280.19 2 105.292 ` ` 5 34.98 kW s kg 1 kJ/ s (d) The heat pump coefficient of performance is

g5

# Qout 34.98 kW # 5 5 4.65 7.53 kW Wc

(e) Using the result from part (a) together with the given cost and use data

3electricity cost for 80 hours of operation4 5 17.53 kW2180 h2a0.15

$ b 5 $90.36 kW ? h

If the cost of electricity is 10 cents per kW ? h, which is the U. S. average for the period under consideration, evaluate the cost to operate the heat pump, in $, keeping all other data the same. Ans. $60.24.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of

the vapor-compression heat pump cycle with irreversibilities in the compressor. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate the compressor power, heat transfer rate delivered, and coefficient of performance. ❑ calculate isentropic compressor efficiency. ❑ conduct an elementary economic evaluation.

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Chapter 10 Refrigeration and Heat Pump Systems

Gas Refrigeration Systems

10.7 gas refrigeration systems

All refrigeration systems considered thus far involve changes in phase. Let us now turn to gas refrigeration systems in which the working fluid remains a gas throughout. Gas refrigeration systems have a number of important applications. They are used to achieve very low temperatures for the liquefaction of air and other gases and for other specialized applications such as aircraft cabin cooling. The Brayton refrigeration cycle illustrates an important type of gas refrigeration system.

10.7.1 Brayton Refrigeration Cycle Brayton refrigeration cycle

The Brayton refrigeration cycle is the reverse of the closed Brayton power cycle introduced in Sec. 9.6. A schematic of the reversed Brayton cycle is provided in Fig. 10.15a. The refrigerant gas, which may be air, enters the compressor at state 1, where the temperature is somewhat below the temperature of the cold region, TC, and is compressed to state 2. The gas is then cooled to state 3, where the gas temperature approaches the temperature of the warm region, TH. Next, the gas is expanded to state 4, where the temperature, T4, is well below that of the cold region. Refrigeration is achieved through heat transfer from the cold region to the gas as it passes from state 4 to state 1, completing the cycle. The T–s diagram in Fig. 10.15b shows an ideal Brayton refrigeration cycle, denoted by 1–2s–3–4s–1, in which all processes are assumed to be internally reversible and the processes in the turbine and compressor are adiabatic. Also shown is the cycle 1–2–3–4–1, which suggests the effects of irreversibilities during adiabatic compression and expansion. Frictional pressure drops have been ignored.

CYCLE ANALYSIS. The method of analysis of the Brayton refrigeration cycle is similar to that of the Brayton power cycle. Thus, at steady state the work of the compressor and the turbine per unit of mass flow are, respectively # # Wc Wt # 5 h2 2 h1 and # 5 h3 2 h4 m m Warm region at TH · Qout

2

T

2s Constant pressure

Heat exchanger 2

3

Turbine

Compressor

3 TH TC 1

Constant pressure

4s 4

Heat exchanger 4

· Wcycle = · · Wc – Wt

1 s · Qin Cold region at TC (a)

Fig. 10.15 Brayton refrigeration cycle.

(b)

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10.7 Gas Refrigeration Systems

613

In obtaining these expressions, heat transfer with the surroundings and changes in kinetic and potential energy have been ignored. The magnitude of the work developed by the turbine of a Brayton refrigeration cycle is typically significant relative to the compressor work input. Heat transfer from the cold region to the refrigerant gas circulating through the low-pressure heat exchanger, the refrigeration effect, is # Qin # 5 h1 2 h4 m The coefficient of performance is the ratio of the refrigeration effect to the net work input: # # 1h1 2 h42 Qin / m # # 5 b5 # # 1h2 2 h12 2 1h3 2 h42 Wc / m 2 Wt / m

(10.11)

In the next example, we illustrate the analysis of an ideal Brayton refrigeration cycle.

cccc

EXAMPLE 10.5 c

Analyzing an Ideal Brayton Refrigeration Cycle Air enters the compressor of an ideal Brayton refrigeration cycle at 1 atm, 480°R, with a volumetric flow rate of 50 ft3/s. If the compressor pressure ratio is 3 and the turbine inlet temperature is 540°R, determine (a) the net power input, in Btu/min, (b) the refrigeration capacity, in Btu/min, (c) the coefficient of performance. SOLUTION Known: An ideal Brayton refrigeration cycle operates with air. Compressor inlet conditions, the turbine inlet temperature, and the compressor pressure ratio are given. Find: Determine the net power input, in Btu/min, the refrigeration capacity, in Btu/min, and the coefficient of

performance. Schematic and Given Data: Engineering Model:

· Qout T3 = 540°R

1. Each component of the

Heat exchanger

T

cycle is analyzed as a control volume at steady state. The control volumes are indicated by dashed lines on the accompanying sketch.

2s

2s

3

Turbine

Compressor

p = 3 atm

· Wcycle

2. The turbine and compressor

T3 = 540°R

processes are isentropic. 3 p = 1 atm

Heat exchanger 4s · Qin

1 (AV)1 = 50 ft3/s T1 = 480°R p1 = 1 atm

3. There are no pressure drops

1

through the heat exchangers.

T1 = 480°R

4. Kinetic and potential energy

effects are negligible. 4s

5. The working fluid is air s

Fig. E10.5

modeled as an ideal gas.

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Analysis: The analysis begins by determining the specific enthalpy at each numbered state of the cycle. At state

1, the temperature is 4808R. From Table A-22E, h1 5 114.69 Btu/lb, pr1 5 0.9182. Since the compressor process is isentropic, h2s can be determined by first evaluating pr at state 2s. That is pr2 5

p2 p 5 13210.91822 5 2.755 p1 r1

Then, interpolating in Table A-22E, we get h2s 5 157.1 Btu/lb. The temperature at state 3 is given as T3 5 5408R. From Table A-22E, h3 5 129.06 Btu/lb, pr3 5 1.3860. The specific enthalpy at state 4s is found by using the isentropic relation pr4 5 pr3

p4 5 11.3860211/ 32 5 0.462 p3

Interpolating in Table A-22E, we obtain h4s 5 94.1 Btu/lb. (a) The net power input is

# # Wcycle 5 m31h2s 2 h12 2 1h3 2 h4s24 # This requires the mass flow rate m, which can be determined from the volumetric flow rate and the specific volume at the compressor inlet: 1AV21 # m5 y1 Since y1 5 1R / M2T1 / p1 1AV21 p1 # m5 1R/ M2T1 5

150 ft3/ s2Z60 s/ minZ114.7 lbf/ in.22Z144 in.2/ ft2Z a

1545 ft ? lbf b14808R2 28.97 lb ? 8R

5 248 lb/ min Finally # Wcycle 5 1248 lb/ min231157.1 2 114.692 2 1129.06 2 94.124 Btu/ lb 5 1848 Btu/ min (b) The refrigeration capacity is

# # Qin 5 m1h1 2 h4s2 5 1248 lb/ min21114.69 2 94.12 Btu/ lb 5 5106 Btu / min (c) The coefficient of performance is

# Qin 5106 b5 # 5 5 2.76 1848 Wcycle

➊

➊ Irreversibilities within the compressor and turbine serve to decrease the coefficient of performance significantly from that of the corresponding ideal cycle because the compressor work requirement is increased and the turbine work output is decreased. This is illustrated in Example 10.6.

Determine the refrigeration capacity in tons of refrigeration.

Ans. 25.53 ton.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of the

ideal Brayton refrigeration cycle. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate net power input, refrigeration capacity, and coefficient of performance.

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615

Example 10.6 illustrates the effects of irreversible compression and turbine expansion Building on Example 10.5, on the performance of Brayton cycle refrigeration. For this, we apply the isentropic compressor and turbine efficiencies introduced in Sec. 6.12.

cccc

EXAMPLE 10.6 c

Evaluating Performance of a Brayton Refrigeration Cycle with Irreversibilities Reconsider Example 10.5, but include in the analysis that the compressor and turbine each have an isentropic efficiency of 80%. Determine for the modified cycle (a) the net power input, in Btu/min, (b) the refrigeration capacity, in Btu/min, (c) the coefficient of performance, and discuss its value. SOLUTION Known: A Brayton refrigeration cycle operates with air. Compressor inlet conditions, the turbine inlet temperature, and the compressor pressure ratio are given. The compressor and turbine each have an isentropic efficiency of 80%. Find: Determine the net power input and the refrigeration capacity, each in Btu/min. Also, determine the coef-

ficient of performance and discuss its value. Schematic and Given Data: Engineering Model:

2

T 2s

1. Each component of the cycle is analyzed as a control

volume at steady state. 2. The compressor and turbine are adiabatic.

p = 3 atm

3. There are no pressure drops through the heat exchangers. 3

4. Kinetic and potential energy effects are negligible.

T3 = 540°R 1

5. The working fluid is air modeled as an ideal gas.

T1 = 480°R

p = 1 atm

4 4s s

Fig. E10.6

Analysis: (a) The power input to the compressor is evaluated using the isentropic compressor efficiency, hc. That is

# # # 1Wc / m2s Wc # 5 hc m # # The value of the work per unit mass for the isentropic compression, 1Wc / m2s, is determined with data from the solution in Example 10.5 as 42.41 Btu/lb. The actual power required is then # # # m1Wc / m2s 1248 lb/ min2142.41 Btu/ lb2 # Wc 5 5 hc 10.82 5 13,147 Btu / min # # # # The turbine power output is determined using the # turbine isentropic efficiency ht. Thus, Wt / m 5 h t1Wt / m2s. # Using data from the solution to Example 10.5 gives 1Wt / m2s 5 34.96 Btu/ lb. The actual turbine work is then # # # # Wt 5 mht1Wt / m2s 5 1248 lb / min210.82134.96 Btu / lb2 5 6936 Btu/ min

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The net power input to the cycle is # Wcycle 5 13,147 2 6936 5 6211 Btu/ min (b) The specific enthalpy at the# turbine exit, h4, is required to evaluate # the refrigeration capacity. This enthalpy

# # can be determined by solving Wt 5 m1h3 2 h42 to obtain h4 5 h3 2 Wt / m. Inserting known values h4 5 129.06 2 a

6936 b 5 101.1 Btu/ lb 248

The refrigeration capacity is then # # Qin 5 m1h1 2 h42 5 124821114.69 2 101.12 5 3370 Btu / min (c) The coefficient of performance is

# Qin 3370 b5 # 5 5 0.543 6211 Wcycle The value of the coefficient of performance in this case is less than unity. This means that the refrigeration effect is smaller than the net work required to achieve it. Additionally, note that irreversibilities in the compressor and turbine have a significant effect on the performance of gas refrigeration systems. This is brought out by comparing the results of the present example with those of Example 10.5. Irreversibilities result in an increase in the work of compression and a reduction in the work output of the turbine. The refrigeration capacity is also reduced. The overall effect is that the coefficient of performance is decreased significantly.

✓ Skills Developed Ability to… ❑ sketch the T–s diagram of

the Brayton refrigeration cycle with irreversibilities in the turbine and compressor. ❑ fix each of the principal states and retrieve necessary property data. ❑ calculate net power input, refrigeration capacity, and coefficient of performance.

Determine the coefficient of performance for a Carnot refrigeration cycle operating between reservoirs at 4808R and 5408R. Ans. 8.

10.7.2

Additional Gas Refrigeration Applications

To obtain even moderate refrigeration capacities with the Brayton refrigeration cycle, equipment capable of achieving relatively high pressures and volumetric flow rates is needed. For most applications involving air conditioning and for ordinary refrigeration processes, vapor-compression systems can be built more cheaply and can operate with higher coefficients of performance than gas refrigeration systems. With suitable modifications, however, gas refrigeration systems can be used to achieve temperatures of about 21508C (22408F), which are well below the temperatures normally obtained with vapor systems. Figure 10.16 shows the schematic and T–s diagram of an ideal Brayton cycle modified by introduction of a heat exchanger. The heat exchanger allows the air exiting the compressor at state 2 to cool below the warm region temperature TH, giving a low turbine inlet temperature, T3. Without the heat exchanger, air could be cooled only close to TH, as represented on the figure by state a. In the subsequent expansion through the turbine, the air achieves a much lower temperature at state 4 than would have been possible without the heat exchanger. Accordingly, the refrigeration effect, achieved from state 4 to state b, occurs at a correspondingly lower average temperature. An example of the application of gas refrigeration to cabin cooling in an aircraft is illustrated in Fig. 10.17. As shown in the figure, a small amount of high-pressure air is extracted from the main jet engine compressor and cooled by heat transfer to the ambient. The high-pressure air is then expanded through an auxiliary turbine to the pressure maintained in the cabin. The air temperature is reduced in the expansion and thus is able to fulfill its cabin cooling function. As an additional benefit, the turbine expansion can provide some of the auxiliary power needs of the aircraft.

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10.7 Gas Refrigeration Systems · Qin from refrigerated region

T 2

Heat exchanger

b

· Qout

a

1

a 4

3 · Qout to warm region

3

1

2

· Wcycle Turbine

Temperature of warm region, TH

b · Qin 4

Compressor

Fig. 10.16 Brayton refrigeration cycle modified with a heat exchanger. Size and weight are important considerations in the selection of equipment for use in aircraft. Open-cycle systems, like the example given here, utilize compact highspeed rotary turbines and compressors. Furthermore, since the air for cooling comes directly from the surroundings, there are fewer heat exchangers than would be needed if a separate refrigerant were circulated in a closed vapor-compression cycle.

10.7.3

Automotive Air Conditioning Using Carbon Dioxide

Owing primarily to environmental concerns, automotive air-conditioning systems using CO2 are currently under active consideration. Carbon dioxide causes no harm to the ozone layer, and its Global Warming Potential of 1 is small compared to that of R-134a, commonly used in automotive air-conditioning systems. Carbon dioxide is nontoxic and nonflammable. As it is abundant in the atmosphere and the exhaust gas of coal-burning power and industrial plants, CO2 is a relatively inexpensive choice as a refrigerant. Still, automakers considering a shift to CO2 away from R-134a must Ambient air in

To combustor

Main jet engine compressor

Heat transfer to ambient Air extracted for cabin cooling

Auxiliary turbine

Auxiliary power

Cool air to cabin

Fig. 10.17 An application of gas refrigeration to aircraft cabin cooling.

s

617

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Chapter 10 Refrigeration and Heat Pump Systems · Qout

Gas cooler 2

3

Compressor

· Wc 2

T

1

pc

3

Heat exchanger Tc (31°C, 88°F)

Ambient temperature, TH

4 1

5

4

6

Temperature of passenger cabin, TC

Expansion valve 6

5 Evaporator

s · Qin

Fig. 10.18 Carbon dioxide automotive air-conditioning system. weigh system performance, equipment costs, and other key issues before embracing such a change in longstanding practice. Figure 10.18 shows the schematic of a CO2-charged automotive air-conditioning system with an accompanying T–s diagram labeled with the critical temperature Tc and critical pressure pc of CO2: 31°C (88°F) and 72.9 atm, respectively. The system combines aspects of gas refrigeration with aspects of vapor-compression refrigeration. Let us follow the CO2 as it passes steadily through each of the components, beginning with the inlet to the compressor. Carbon dioxide enters the compressor as superheated vapor at state 1 and is compressed to a much higher temperature and pressure at state 2. The CO2 passes from the compressor into the gas cooler, where it cools at constant pressure to state 3 as a result of heat transfer to the ambient. The temperature at state 3 approaches that of the ambient, denoted on the figure by TH. CO2 is further cooled in the interconnecting heat exchanger at constant pressure to state 4, where the temperature is below that of ambient. Cooling is provided by low-temperature CO2 in the other stream of the heat exchanger. During this portion of the refrigeration cycle, the processes are like those seen in gas refrigeration. This similarity ends abruptly as the CO2 next expands through the valve to state 5 in the liquid–vapor region and then enters the evaporator, where it is vaporized to state 6 by heat transfer from the passenger cabin at temperature TC, thereby cooling the passenger cabin. These processes are like those seen in vaporcompression refrigeration systems. Finally, at state 6 the CO2 enters the heat exchanger, exiting at state 1. The heat exchanger increases the cycle’s performance in two ways: by delivering lower-quality two-phase mixture at state 5, increasing the refrigerating effect through the evaporator, and by producing higher-temperature superheated vapor at state 1, reducing the compressor power required.

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c CHAPTER SUMMARY AND STUDY GUIDE In this chapter we have considered refrigeration and heat pump systems, including vapor systems where the refrigerant is alternately vaporized and condensed, and gas systems where the refrigerant remains a gas. The three principal types of systems discussed are the vapor-compression, absorption, and reversed Brayton cycles. The performance of simple vapor refrigeration systems is described in terms of the vapor-compression cycle. For this cycle, we have evaluated the principal work and heat transfers along with two important performance parameters: the coefficient of performance and the refrigeration capacity. We have considered the effect on performance of irreversibilities during the compression process and in the expansion across the valve, as well as the effect of irreversible heat transfer between the refrigerant and the warm and cold regions. Variations of the basic vaporcompression refrigeration cycle also have been considered, including cold storage, cascade cycles, and multistage compression with intercooling. A discussion of vapor-compression heat pump systems is also provided. Qualitative discussions are presented of refrigerant properties and of considerations in selecting refrigerants. Absorption refrigeration and heat pump systems are also discussed qualitatively. The chapter concludes with a study of gas refrigeration systems.

The following list provides a study guide for this chapter. When your study of the text and end-of-chapter exercises has been completed, you should be able to c write out the meanings of the terms listed in the margin

c

c

c

c

throughout the chapter and understand each of the related concepts. The subset of key concepts listed below is particularly important. sketch the T–s diagrams of vapor-compression refrigeration and heat pump cycles and of Brayton refrigeration cycles, correctly showing the relationship of the refrigerant temperature to the temperatures of the warm and cold regions. apply the first and second laws along with property data to determine the performance of vapor-compression refrigeration and heat pump cycles and of Brayton refrigeration cycles, including evaluation of the power required, the coefficient of performance, and the capacity. sketch schematic diagrams of vapor-compression cycle modifications, including cascade cycles and multistage compression with intercooling between the stages. In each case be able to apply mass and energy balances, the second law, and property data to determine performance. explain the operation of absorption refrigeration systems.

c KEY ENGINEERING CONCEPTS vapor-compression refrigeration, p. 592 ton of refrigeration, p. 592 refrigeration capacity, p. 592 absorption refrigeration, p. 606

vapor-compression heat pump, p. 609 Brayton refrigeration cycle, p. 612

c KEY EQUATIONS TC TH 2 TC

(10.1) p. 591

Coefficient of performance of the Carnot refrigeration cycle (Fig. 10.1)

(10.7) p. 593

Coefficient of performance of the vapor-compression refrigeration cycle (Fig. 10.3)

(10.9) p. 608

Coefficient of performance of the Carnot heat pump cycle (Fig. 10.1)

# # h2 2 h3 Qout/ m g5 # # 5 h2 2 h1 Wc / m

(10.10) p. 609

Coefficient of performance of the vapor-compression heat pump cycle (Fig. 10.13)

# # 1h1 2 h42 Qin/ m # # # 5 b5 # 2 h 1h Wc/ m 2 Wt/ m 2 12 2 1h3 2 h42

(10.11) p. 613

Coefficient of performance of the Brayton refrigeration cycle (Fig. 10.15)

bmax 5

# # Qin / m h1 2 h4 b5 # # 5 h Wc / m 2 2 h1 gmax 5

TH TH 2 TC

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Chapter 10 Refrigeration and Heat Pump Systems

c EXERCISES: THINGS ENGINEERS THINK ABOUT 1. What are the temperatures inside the fresh food and freezer compartments of your refrigerator? Do you know what values are recommended for these temperatures? 2. You have a refrigerator located in your garage. Does it perform differently in the summer than in the winter? Explain.

8. What are n-type and p-type semiconductors? 9. What qualifies a refrigerator to be an Energy Star ® appliance? 10. You see an advertisement claiming that heat pumps are particularly effective in Atlanta, Georgia. Why might that be true?

3. Abbe installs a dehumidifier to dry the walls of a small, closed basement room. When she enters the room later, it feels warm. Why?

11. If your car’s air conditioner discharges only warm air while operating, what might be wrong with it?

4. Why does the indoor unit of a central air conditioning system have a drain hose?

12. Large office buildings often use air conditioning to cool interior areas even in winter in cold climates. Why?

5. Your air conditioner has a label that lists an EER of 10 Btu/h per watt. What does that mean? 6. Can the coefficient of performance of a heat pump have a value less than one? 7. You see an advertisement for a natural gas–fired absorption refrigeration system. How can burning natural gas play a role in achieving cooling?

13. In what North American locations are heat pumps not a good choice for heating dwellings? Explain. 14. If the heat exchanger is omitted from the system of Fig. 10.16, what is the effect on the coefficient of performance?

c PROBLEMS: DEVELOPING ENGINEERING SKILLS Vapor Refrigeration Systems 10.1 Refrigerant 22 is the working fluid in a Carnot refrigeration cycle operating at steady state. The refrigerant enters the condenser as saturated vapor at 32°C and exits as saturated liquid. The evaporator operates at 0°C. What is the coefficient of performance of the cycle? Determine, in kJ per kg of refrigerant flowing (a) the work input to the compressor. (b) the work developed by the turbine. (c) the heat transfer to the refrigerant passing through the evaporator. 10.2 Refrigerant 22 is the working fluid in a Carnot vapor refrigeration cycle for which the evaporator temperature is 230°C. Saturated vapor enters the condenser at 36°C, and saturated liquid exits at the same temperature. The mass flow rate of refrigerant is 10 kg/min. Determine (a) the rate of heat transfer to the refrigerant passing through the evaporator, in kW. (b) the net power input to the cycle, in kW. (c) the coefficient of performance. (d) the refrigeration capacity, in tons. 10.3 A Carnot vapor refrigeration cycle operates between thermal reservoirs at 40°F and 90°F. For (a) Refrigerant 134a, (b) propane, (c) water, (d) Refrigerant 22, and (e) ammonia as the working fluid, determine the operating pressures in the condenser and evaporator, in lbf/in.2, and the coefficient of performance.

10.4 Consider a Carnot vapor refrigeration cycle with Refrigerant 134a as the working fluid. The cycle maintains a cold region at 40°F when the ambient temperature is 90°F. Data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 10.1. Sketch the T–s diagram for the cycle and determine the (a) temperatures in the evaporator and condenser, each in °R. (b) compressor and turbine work, each in Btu per lb of refrigerant flowing. (c) coefficient of performance. (d) coefficient of performance for a Carnot cycle operating at the reservoir temperatures. Compare the coefficients of performance determined in (c) and (d), and comment. State

p (lbf/in.2)

h (Btu/lb)

s (Btu/lb ? 8R)

1 2 3 4

40 140 140 40

104.12 114.95 44.43 42.57

0.2161 0.2161 0.0902 0.0902

10.5 For the cycle in Problem 10.4, determine (a) the rates of heat transfer, in Btu per lb of refrigerant flowing, for the refrigerant flowing through the evaporator and condenser, respectively. (b) the rates and directions of exergy transfer accompanying each of these heat transfers, in Btu per lb of refrigerant flowing. Let T0 5 90°F.

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Problems: Developing Engineering Skills 10.6 An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 2 bar, and saturated liquid exits the condenser at 8 bar. The mass flow rate of refrigerant is 7 kg/min. Determine (a) the compressor power, in kW. (b) the refrigerating capacity, in tons. (c) the coefficient of performance. 10.7 Plot each of the quantities in Problem 10.6 versus evaporator temperature for evaporator pressures ranging from 0.6 to 4 bar, while the condenser pressure remains fixed at 8 bar. 10.8 Refrigerant 134a is the working fluid in an ideal vaporcompression refrigeration cycle operating at steady state. Refrigerant enters the compressor at 1.4 bar, 2128C, and the condenser pressure is 9 bar. Liquid exits the condenser at 328C. The mass flow rate of refrigerant is 7 kg/min. Determine (a) the compressor power, in kW. (b) the refrigeration capacity, in tons. (c) the coefficient of performance.

Hot region · Qout

Condenser 2

Expansion valve

· Wc

Compressor

4

lb m· 1 = 30.59 ––– min 1 Evaporator

Cold region p (lbf/in.2)

1 2 3 4

10 180 180 10

Fig. P10.9

T (°F) 0 --Sat. Sat.

(a) the compressor power, in horsepower. (b) the rate of heat transfer, from the working fluid passing through the condenser, in Btu/min. (c) the coefficient of performance. 10.10 Refrigerant 22 enters the compressor of an ideal vaporcompression refrigeration system as saturated vapor at 2408C with a volumetric flow rate of 15 m3/min. The refrigerant leaves the condenser at 198C, 9 bar. Determine (a) the compressor power, in kW. (b) the refrigerating capacity, in tons. (c) the coefficient of performance. 10.11 An ideal vapor-compression refrigeration cycle, with ammonia as the working fluid, has an evaporator temperature of 2208C and a condenser pressure of 12 bar. Saturated vapor enters the compressor, and saturated liquid exits the condenser. The mass flow rate of the refrigerant is 3 kg/min. Determine

10.12 Refrigerant 134a enters the compressor of an ideal vapor-compression refrigeration cycle as saturated vapor at 2108F. The condenser pressure is 160 lbf/in.2 The mass flow rate of refrigerant is 6 lb/min. Plot the coefficient of performance and the refrigerating capacity, in tons, versus the condenser exit temperature ranging from the saturation temperature at 160 lbf/in.2 to 908F. 10.13 To determine the effect of changing the evaporator temperature on the performance of an ideal vaporcompression refrigeration cycle, plot the coefficient of performance and the refrigerating capacity, in tons, for the cycle in Problem 10.11 for saturated vapor entering the compressor at temperatures ranging from 240 to 2108C. All other conditions are the same as in Problem 10.11. 10.14 To determine the effect of changing condenser pressure on the performance of an ideal vapor-compression refrigeration cycle, plot the coefficient of performance and the refrigerating capacity, in tons, for the cycle in Problem 10.11 for condenser pressures ranging from 8 to 16 bar. All other conditions are the same as in Problem 10.11. 10.15 A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 2 bar, and saturated liquid exits the condenser at 8 bar. The isentropic compressor efficiency is 80%. The mass flow rate of refrigerant is 7 kg/min. Determine

· Qin

State

is 30.59 lb/min. Sketch the T–s diagram for the cycle and determine

(a) the coefficient of performance. (b) the refrigerating capacity, in tons.

10.9 Figure P10.9 provides steady-state operating data for an ideal vapor-compression refrigeration cycle with Refrigerant 134a as the working fluid. The mass flow rate of refrigerant

3

621

h s (Btu/lb) (Btu/lb·°R) 102.94 131.04 50.64 50.64

0.2391 0.2391 0.1009 ---

(a) the compressor power, in kW. (b) the refrigeration capacity, in tons. (c) the coefficient of performance. 10.16 Modify the cycle in Problem 10.9 to have an isentropic compressor efficiency of 83% and let the temperature of the liquid leaving the condenser be 1008F. Determine, for the modified cycle,

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(a) the compressor power, in horsepower. (b) the rate of heat transfer from the working fluid passing through the condenser, in Btu/min. (c) the coefficient of performance. (d) the rates of exergy destruction in the compressor and expansion valve, each in Btu/min, for T0 5 908F. 10.17 Data for steady-state operation of a vapor-compression refrigeration cycle with Refrigerant 134a as the working fluid are given in the table below. The states are numbered as in Fig. 10.3. The refrigeration capacity is 4.6 tons. Ignoring heat transfer between the compressor and its surroundings, sketch the T–s diagram of the cycle and determine (a) the mass flow rate of the refrigerant, in kg/min. (b) the isentropic compressor efficiency. (c) the coefficient of performance. (d) the rates of exergy destruction in the compressor and expansion valve, each in kW. (e) the net changes in flow exergy rate of the refrigerant passing through the evaporator and condenser, respectively, each in kW. Let T0 5 21°C, p0 5 1 bar. State

p (bar)

T (°C)

h (kJ/kg)

s (kJ/kg ? K)

1 2 3 4

1.4 7 7 1.4

210 58.5 24 218.8

243.40 295.13 82.90 82.90

0.9606 1.0135 0.3113 0.33011

10.18 A vapor-compression refrigeration system, using ammonia as the working fluid, has evaporator and condenser pressures of 30 and 200 lbf/in.2, respectively. The refrigerant passes through each heat exchanger with a negligible pressure drop. At the inlet and exit of the compressor, the temperatures are 108F and 3008F, respectively. The heat transfer rate from the working fluid passing through the condenser is 50,000 Btu/h, and liquid exits at 200 lbf/in.2, 908F. If the compressor operates adiabatically, determine (a) the compressor power input, in hp. (b) the coefficient of performance. 10.19 If the minimum and maximum allowed refrigerant pressures are 1 and 10 bar, respectively, which of the following can be used as the working fluid in a vapor-compression refrigeration system that maintains a cold region at 08C, while discharging energy by heat transfer to the surrounding air at 308C: Refrigerant 22, Refrigerant 134a, ammonia, propane? 10.20 Consider the following vapor-compression refrigeration cycle used to maintain a cold region at temperature TC when the ambient temperature is 80°F: Saturated vapor enters the compressor at 15°F below TC, and the compressor operates adiabatically with an isentropic efficiency of 80%. Saturated liquid exits the condenser at 958F. There are no pressure drops through the evaporator or condenser, and the refrigerating capacity is 1 ton. Plot refrigerant mass flow rate, in lb/min, coefficient of performance, and refrigerating efficiency, versus TC ranging from 408F to 2258F if the refrigerant is

(a) Refrigerant 134a. (b) propane. (c) Refrigerant 22. (d) ammonia. The refrigerating efficiency is defined as the ratio of the cycle coefficient of performance to the coefficient of performance of a Carnot refrigeration cycle operating between thermal reservoirs at the ambient temperature and the temperature of the cold region. 10.21 In a vapor-compression refrigeration cycle, ammonia exits the evaporator as saturated vapor at 2228C. The refrigerant enters the condenser at 16 bar and 1608C, and saturated liquid exits at 16 bar. There is no significant heat transfer between the compressor and its surroundings, and the refrigerant passes through the evaporator with a negligible change in pressure. If the refrigerating capacity is 150 kW, determine (a) the (b) the (c) the (d) the

mass flow rate of refrigerant, in kg/s. power input to the compressor, in kW. coefficient of performance. isentropic compressor efficiency.

10.22 A vapor-compression refrigeration system with a capacity of 10 tons has superheated Refrigerant 134a vapor entering the compressor at 158C, 4 bar, and exiting at 12 bar. The compression process can be modeled by py1.01 5 constant. At the condenser exit, the pressure is 11.6 bar, and the temperature is 448C. The condenser is water-cooled, with water entering at 208C and leaving at 308C with a negligible change in pressure. Heat transfer from the outside of the condenser can be neglected. Determine (a) the mass flow rate of the refrigerant, in kg/s. (b) the power input and the heat transfer rate for the compressor, each in kW. (c) the coefficient of performance. (d) the mass flow rate of the cooling water, in kg/s. (e) the rates of exergy destruction in the condenser and expansion valve, each expressed as a percentage of the power input. Let T0 5 208C. 10.23 Data for steady-state operation of a vapor-compression refrigeration cycle with propane as the working fluid are given in the table below. The states are numbered as in Fig. 10.3. The mass flow rate of refrigerant is 8.42 lb/min. Heat transfer from the compressor to its surroundings occurs at a rate of 3.5 Btu per lb of refrigerant passing through the compressor. The condenser is water-cooled, with water entering at 658F and leaving at 808F with negligible change in pressure. Sketch the T–s diagram of the cycle and determine (a) the refrigeration capacity, in tons. (b) the compressor power, in horsepower. (c) the mass flow rate of the condenser cooling water, in lb/min. (d) the coefficient of performance. State

p (lbf/in.2)

T (8F)

h (Btu/lb)

1 2 3 4

38.4 180 180 38.4

0 120 85 0

193.2 229.8 74.41 74.41

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Problems: Developing Engineering Skills 10.24 A window-mounted air conditioner supplies 19 m3/min of air at 158C, 1 bar to a room. Air returns from the room to the evaporator of the unit at 228C. The air conditioner operates at steady state on a vapor-compression refrigeration cycle with Refrigerant-22 entering the compressor at 4 bar, 108C. Saturated liquid refrigerant at 9 bar leaves the condenser. The compressor has an isentropic efficiency of 70%, and refrigerant exits the compressor at 9 bar. Determine the compressor power, in kW, the refrigeration capacity, in tons, and the coefficient of performance.

623

Surroundings TH = 550°R (90°F) · Qout TH

Saturated liquid

3

Condenser

Expansion valve

4 p4 = p1

2

p2 = p3 = 160 lbf/in.2

· Wc

Compressor

Evaporator · Qin

ηc = 80%

1 Saturated vapor T1 = 40°F

TC Cool region TC = 520°R (60°F)

Fig. P10.24 10.25 A vapor-compression refrigeration system for a household refrigerator has a refrigerating capacity of 1000 Btu/h. Refrigerant enters the evaporator at 2108F and exits at 08F. The isentropic compressor efficiency is 80%. The refrigerant condenses at 958F and exits the condenser subcooled at 908F. There are no significant pressure drops in the flows through the evaporator and condenser. Determine the evaporator and condenser pressures, each in lbf/in.2, the mass flow rate of refrigerant, in lb/min, the compressor power input, in horsepower, and the coefficient of performance for (a) Refrigerant 134a and (b) propane as the working fluid. 10.26 A vapor-compression air conditioning system operates at steady state as shown in Fig. P10.26. The system maintains a cool region at 608F and discharges energy by heat transfer to the surroundings at 908F. Refrigerant 134a enters the compressor as a saturated vapor at 408F and is compressed adiabatically to 160 lbf/in.2 The isentropic compressor efficiency is 80%. Refrigerant exits the condenser as a saturated liquid at 160 lbf/in.2 The mass flow rate of the refrigerant is 0.15 lb/s. Kinetic and potential energy changes are negligible as are changes in pressure for flow through the evaporator and condenser. Determine (a) the power required by the compressor, in Btu/s. (b) the coefficient of performance. (c) the rates of exergy destruction in the compressor and expansion valve, each in Btu/s. (d) the rates of exergy destruction and exergy transfer accompanying heat transfer, each in Btu/s, for a control volume comprising the evaporator and a portion of the cool region such that heat transfer takes place at TC 5 5208R (608F). (e) the rates of exergy destruction and exergy transfer accompanying heat transfer, each in Btu/s, for a control volume enclosing the condenser and a portion of the surroundings such that heat transfer takes place at TH 5 5508R (908F). Let T0 5 5508R.

Fig. P10.26 10.27 A vapor-compression refrigeration cycle with Refrigerant 134a as the working fluid operates with an evaporator temperature of 508F and a condenser pressure of 180 lbf/in.2 Saturated vapor enters the compressor. Refrigerant enters the condenser at 1408F and exits as saturated liquid. The cycle has a refrigeration capacity of 5 tons. Determine (a) the (b) the (c) the (d) the

refrigerant mass flow rate, in lb/min. compressor isentropic efficiency. compressor power, in horsepower. coefficient of performance.

Plot each of the quantities calculated in parts (b) through (d) for compressor exit temperatures varying from 1308F to 1408F.

Cascade and Multistage Systems 10.28 A vapor-compression refrigeration system operates with the cascade arrangement of Fig. 10.9. Refrigerant 22 is the working fluid in the high-temperature cycle and Refrigerant 134a is used in the low-temperature cycle. For the Refrigerant 134a cycle, the working fluid enters the compressor as saturated vapor at 2308F and is compressed isentropically to 50 lbf/in.2 Saturated liquid leaves the intermediate heat exchanger at 50 lbf/in.2 and enters the expansion valve. For the Refrigerant 22 cycle, the working fluid enters the compressor as saturated vapor at a temperature 58F below that of the condensing temperature of the Refrigerant 134a in the intermediate heat exchanger. The Refrigerant 22 is compressed isentropically to 250 lbf/ in.2 Saturated liquid then enters the expansion valve at 250 lbf/in.2 The refrigerating capacity of the cascade system is 20 tons. Determine (a) the power input to each compressor, in Btu/min. (b) the overall coefficient of performance of the cascade cycle. (c) the rate of exergy destruction in the intermediate heat exchanger, in Btu/min. Let T0 5 808F, p0 5 14.7 lbf/in.2

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10.29 A vapor-compression refrigeration system uses the arrangement shown in Fig. 10.10 for two-stage compression with intercooling between the stages. Refrigerant 134a is the working fluid. Saturated vapor at 2308C enters the first compressor stage. The flash chamber and direct contact heat exchanger operate at 4 bar, and the condenser pressure is 12 bar. Saturated liquid streams at 12 and 4 bar enter the high- and low-pressure expansion valves, respectively. If each compressor operates isentropically and the refrigerating capacity of the system is 10 tons, determine

· Qout

Condenser 4 5

Expansion valve

3 6

(a) the power input to each compressor, in kW. (b) the coefficient of performance. 10.30 Figure P10.30 shows a two-stage vapor-compression refrigeration system with ammonia as the working fluid. The system uses a direct contact heat exchanger to achieve intercooling. The evaporator has a refrigerating capacity of 30 tons and produces 2208F saturated vapor at its exit. In the first compressor stage, the refrigerant is compressed adiabatically to 80 lbf/in.2, which is the pressure in the direct contact heat exchanger. Saturated vapor at 80 lbf/in.2 enters the second compressor stage and is compressed adiabatically to 250 lbf/in.2 Each compressor stage has an isentropic efficiency of 85%. There are no significant pressure drops as the refrigerant passes through the heat exchangers. Saturated liquid enters each expansion valve. Determine # # (a) the ratio of mass flow rates, m3/ m1. (b) the power input to each compressor stage, in horsepower. (c) the coefficient of performance. (d) Plot each of the quantities calculated in parts (a)–(c) versus the direct-contact heat exchanger pressure ranging from 20 to 200 lbf/in.2 Discuss. · Qout

4 Condenser

5 Expansion valve

Direct contact heat exchanger

Comp 2

· Wc2

6 3 7

2

Expansion valve

Comp 1 Evaporator 1

8

· Wc1

· Qin = 30 tons

Fig. P10.30 10.31 Figure P10.31 shows a two-stage, vapor-compression refrigeration system with two evaporators and a direct contact heat exchanger. Saturated vapor ammonia from evaporator

· Wc,2

Compressor 2

8

Evaporator 2

· Qin,2 = 10 tons Expansion valve

Direct contact heat exchanger

2 · Wc,1

Compressor 1

7 Evaporator 1

1

· Qin,1 = 5 tons

Fig. P10.31 1 enters compressor 1 at 18 lbf/in.2 and exits at 70 lbf/in.2 Evaporator 2 operates at 70 lbf/in.2, with saturated vapor exiting at state 8. The condenser pressure is 200 lbf/in.2, and saturated liquid refrigerant exits the condenser. Each compressor stage has an isentropic efficiency of 80%. The refrigeration capacity of each evaporator is shown on the figure. Sketch the T–s diagram of the cycle and determine (a) the temperatures, in 8F, of the refrigerant in each evaporator. (b) the power input to each compressor stage, in horsepower. (c) the overall coefficient of performance. 10.32 Figure P10.32 shows the schematic diagram of a vaporcompression refrigeration system with two evaporators using Refrigerant 134a as the working fluid. This arrangement is used to achieve refrigeration at two different temperatures with a single compressor and a single condenser. The lowtemperature evaporator operates at 2188C with saturated vapor at its exit and has a refrigerating capacity of 3 tons. The higher-temperature evaporator produces saturated vapor at 3.2 bar at its exit and has a refrigerating capacity of 2 tons. Compression is isentropic to the condenser pressure of 10 bar. There are no significant pressure drops in the flows through the condenser and the two evaporators, and the refrigerant leaves the condenser as saturated liquid at 10 bar. Calculate (a) the mass flow rate of refrigerant through each evaporator, in kg/min. (b) the compressor power input, in kW. (c) the rate of heat transfer from the refrigerant passing through the condenser, in kW.

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Problems: Developing Engineering Skills · Qout 3

exchanger, entering the expansion valve at 18 bar. If the mass flow rate of refrigerant is 12 kg/min, determine 2

Condenser

· Wc

Compressor

4

1 7

Vapor-Compression Heat Pump Systems

8

10.34 Figure P10.34 gives data for an ideal vapor-compression heat pump cycle operating at steady state with Refrigerant 134a as the working fluid. The heat pump provides heating at a rate of 15 kW to maintain the interior of a building at 208C when the outside temperature is 58C. Sketch the T–s diagram for the cycle and determine the

Expansion valve

· Qin,2 = 2 tons Expansion valve 5

(a) temperatures at the principal states of the cycle, each in 8C. (b) the power input to the compressor, in kW. (c) the coefficient of performance. (d) the coefficient of performance for a Carnot heat pump cycle operating between reservoirs at the building interior and outside temperatures, respectively.

6

Evaporator 1

(a) the refrigeration capacity, in tons of refrigeration. (b) the compressor power input, in kW. (c) the coefficient of performance. Discuss possible advantages and disadvantages of this arrangement.

Expansion valve Evaporator 2

625

· Qin,1 = 3 tons

Fig. P10.32

Compare the coefficients of performance determined in (c) and (d). Discuss

10.33 An ideal vapor-compression refrigeration cycle is modified to include a counterflow heat exchanger, as shown in Fig. P10.33. Ammonia leaves the evaporator as saturated vapor at 1.0 bar and is heated at constant pressure to 58C before entering the compressor. Following isentropic compression to 18 bar, the refrigerant passes through the condenser, exiting at 408C, 18 bar. The liquid then passes through the heat

Building interior TH = 20°C = 293 K · Qout = 15 kW Saturated liquid Condenser 3

2

· Qout Expansion valve

2

Condenser

· Wc

Compressor

Compressor

4

1 Saturated vapor

Evaporator 3

· Wc

1

TC = 5°C = 278 K Outside

· Qin

Heat exchanger

4

Expansion valve 5

p (bar)

h (kJ/kg)

1 2 3 4

2.4 8 8 2.4

244.09 268.97 93.42 93.42

Fig. P10.34 Evaporator · Qin

Fig. P10.33

State

6

10.35 Refrigerant 134a is the working fluid in a vapor-compression heat pump system with a heating capacity of 60,000 Btu/h. The condenser operates at 200 lbf/in.2, and the evaporator temperature is 08F. The refrigerant is a saturated vapor at the evaporator exit and a liquid at 1108F at the condenser exit. Pressure drops in the

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flows through the evaporator and condenser are negligible. The compression process is adiabatic, and the temperature at the compressor exit is 1808F. Determine (a) the (b) the (c) the (d) the

mass flow rate of refrigerant, in lb/min. compressor power input, in horsepower. isentropic compressor efficiency. coefficient of performance.

10.36 Refrigerant 134a is the working fluid in a vaporcompression heat pump that provides 35 kW to heat a dwelling on a day when the outside temperature is below freezing. Saturated vapor enters the compressor at 1.6 bar, and saturated liquid exits the condenser, which operates at 8 bar. Determine, for isentropic compression (a) the refrigerant mass flow rate, in kg/s. (b) the compressor power, in kW. (c) the coefficient of performance.

compression heat pump be used to develop the process heating using a wastewater stream at 1258F as the lowertemperature source. Figure P10.39 provides data for this cycle operating at steady state. The compressor isentropic efficiency is 80%. Sketch the T–s diagram for the cycle and determine the (a) specific enthalpy at the compressor exit, in Btu/lb. (b) temperatures at each of the principal states, in 8F. (c) mass flow rate of the refrigerant, in lb/h. (d) compressor power, in Btu/h. (e) coefficient of performance and compare with the coefficient of performance for a Carnot heat pump cycle operating between reservoirs at the process temperature and the wastewater temperature, respectively. Process TH = 170°F

Recalculate the quantities in parts (b) and (c) for an isentropic compressor efficiency of 75%. 10.37 An office building requires a heat transfer rate of 20 kW to maintain the inside temperature at 218C when the outside temperature is 08C. A vapor-compression heat pump with Refrigerant 134a as the working fluid is to be used to provide the necessary heating. The compressor operates adiabatically with an isentropic efficiency of 82%. Specify appropriate evaporator and condenser pressures of a cycle for this purpose assuming DTcond 5 DTevap 5 108C, as shown in Figure P10.37. The states are numbered as in Fig. 10.13. The refrigerant exits the evaporator as saturated vapor and exits the condenser as saturated liquid at the respective pressures. Determine the (a) mass flow rate of refrigerant, in kg/s. (b) compressor power, in kW. (c) coefficient of performance and compare with the coefficient of performance for a Carnot heat pump cycle operating between reservoirs at the inside and outside temperatures, respectively.

· Qout = 3 ⫻ 106 Btu/h Condenser 3

2

Expansion valve

4

1 Evaporator

TC = 125°F Wastewater

State T 1 2 3 4

2 ⌬Tcond 21°C

2s

· Wc

Compressor c = 80%

· Qin

p h (lbf/in.2) (Btu/lb) 180 400 400 180

116.74 ? 76.11 76.11

3

Fig. P10.39

0°C 4

1 ⌬Tevap

s

Fig. P10.37 10.38 Repeat the calculations of Problem 10.37 for Refrigerant 22 as the working fluid. Compare the results with those of Problem 10.37 and discuss. 10.39 A process requires a heat transfer rate of 3 3 106 Btu/h at 1708F. It is proposed that a Refrigerant 134a vapor-

10.40 A vapor-compression heat pump with a heating capacity of 500 kJ/min is driven by a power cycle with a thermal efficiency of 25%. For the heat pump, Refrigerant 134a is compressed from saturated vapor at 2108C to the condenser pressure of 10 bar. The isentropic compressor efficiency is 80%. Liquid enters the expansion valve at 9.6 bar, 348C. For the power cycle, 80% of the heat rejected is transferred to the heated space. (a) Determine the power input to the heat pump compressor, in kW. (b) Evaluate the ratio of the total rate that heat is delivered to the heated space to the rate of heat input to the power cycle. Discuss.

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10.41 Refrigerant 134a enters the compressor of a vaporcompression heat pump at 15 lbf/in.2, 08F and is compressed adiabatically to 160 lbf/in.2, 1608F. Liquid enters the expansion valve at 160 lbf/in.2, 958F. At the valve exit, the pressure is 15 lbf/in.2

10.44 Air enters the compressor of a Brayton refrigeration cycle at 100 kPa, 270 K. The compressor pressure ratio is 3, and the temperature at the turbine inlet is 315 K. The compressor and turbine have isentropic efficiencies of 82% and 85%, respectively. Determine the

(a) Determine the isentropic compressor efficiency. (b) Determine the coefficient of performance. (c) Perform a full exergy accounting of the compressor power input, in Btu per lb of refrigerant flowing. Discuss.

(a) net work input, per unit mass of air flow, in kJ/kg. (b) exergy accounting of the net power input, in kJ per kg of air flowing. Discuss.

Let T0 5 4808R. 10.42 A geothermal heat pump operating at steady state with Refrigerant-22 as the working fluid is shown schematically in Fig. P10.42. The heat pump uses 558F water from wells as the thermal source. Operating data are shown on the figure for a day in which the outside air temperature is 208F. Assume adiabatic operation of the compressor. For the heat pump, determine (a) the volumetric flow rate of heated air to the house, in ft3/min. (b) the isentropic compressor efficiency. (c) the compressor power, in horsepower. (d) the coefficient of performance. (e) the volumetric flow rate of water from the geothermal wells, in gal/min. For T0 5 208F, perform a full exergy accounting of the compressor power input, and devise and evaluate a second law efficiency for the heat pump system. Return air from house 70°F

lbf/in.2

p3 = 180 T3 = 75 °F

· Qout = 4.2 tons

2

3 Condenser Expansion valve

p4 = 70 lbf/in.2

Heated air to house 110°F

4

p2 = 180 lbf/in.2 T2 = 140°F · Wc

Condenser Evaporator 1

Water discharged at 45°F

lbf/in.2

p1 = 70 T1 = 40°F

Water from geothermal wells enters at 55°F

Fig. P10.42 Gas Refrigeration Systems 10.43 Air enters the compressor of an ideal Brayton refrigeration cycle at 100 kPa, 300 K. The compressor pressure ratio is 3.75, and the temperature at the turbine inlet is 350 K. Determine the (a) net work input, per unit mass of air flow, in kJ/kg. (b) refrigeration capacity, per unit mass of air flow, in kJ/kg. (c) coefficient of performance. (d) coefficient of performance of a Carnot refrigeration cycle operating between thermal reservoirs at TC 5 300 K and TH 5 350 K, respectively.

Let T0 5 315 K. 10.45 Plot the quantities calculated in parts (a) through (c) of Problem 10.43 versus the compressor pressure ratio ranging from 3 to 6. Repeat for compressor and turbine isentropic efficiencies of 90%, 85%, and 80%. 10.46 An ideal Brayton refrigeration cycle has a compressor pressure ratio of 6. At the compressor inlet, the pressure and temperature of the entering air are 20 lbf/in.2 and 4608R. The temperature at the inlet of the turbine is 7008R. For a refrigerating capacity of 15 tons, determine (a) the mass flow rate, in lb/min. (b) the net power input, in Btu/min. (c) the coefficient of performance. 10.47 Reconsider Problem 10.46, but include in the analysis that the compressor and turbine have isentropic efficiencies of 78% and 92%, respectively. 10.48 The table below provides steady-state operating data for an ideal Brayton refrigeration cycle with air as the working fluid. The principal states are numbered as in Fig. 10.15. The volumetric flow rate at the turbine inlet is 0.4 m3/s. Sketch the T–s diagram for the cycle and determine the (a) specific enthalpy, in kJ/kg, at the turbine exit. (b) mass flow rate, in kg/s. (c) net power input, in kW. (d) refrigeration capacity, in kW. (e) coefficient of performance. State

p (kPa)

T (K)

h (kJ/kg)

pr

1 2 3 4

140 420 420 140

270 — 320 —

270.11 370.10 320.29 ?

0.9590 2.877 1.7375 —

10.49 Air enters the compressor of a Brayton refrigeration cycle at 100 kPa, 260 K, and is compressed adiabatically to 300 kPa. Air enters the turbine at 300 kPa, 300 K, and expands adiabatically to 100 kPa. For the cycle (a) determine the net work per unit mass of air flow, in kJ/kg, and the coefficient of performance if the compressor and turbine isentropic efficiencies are both 100%. (b) plot the net work per unit mass of air flow, in kJ/kg, and the coefficient of performance for equal compressor and turbine isentropic efficiencies ranging from 80 to 100%. 10.50 The Brayton refrigeration cycle of Problem 10.43 is modified by the introduction of a regenerative heat exchanger. In the modified cycle, compressed air enters the regenerative heat exchanger at 350 K and is cooled

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to 320 K before entering the turbine. Determine, for the modified cycle, (a) the lowest temperature, in K. (b) the net work input per unit mass of air flow, in kJ/kg. (c) the refrigeration capacity, per unit mass of air flow, in kJ/kg. (d) the coefficient of performance. 10.51 Reconsider Problem 10.50, but include in the analysis that the compressor and turbine have isentropic efficiencies of 85 and 88% respectively. Answer the same questions as in Problem 10.50. 10.52 Plot the quantities calculated in parts (a) through (d) of Problem 10.50 versus the compressor pressure ratio ranging from 4 to 7. Repeat for equal compressor and turbine isentropic efficiencies of 95%, 90%, and 80%. 10.53 Consider a Brayton refrigeration cycle with a regenerative heat exchanger. Air enters the compressor at 4808R, 15 lbf/in.2 and is compressed isentropically to 40 lbf/in.2 Compressed air enters the regenerative heat exchanger at 5408R and is cooled to 4808R before entering the turbine. The expansion through the turbine is isentropic. If the refrigeration capacity is 15 tons, calculate (a) the volumetric flow rate at the compressor inlet, in ft3/min. (b) the coefficient of performance.

(a) the power developed by the turbine, in horsepower. (b) the isentropic turbine efficiency. (c) the rate of heat transfer from the air to the ambient, in Btu/min. 10.57 Air within a piston–cylinder assembly undergoes a Stirling refrigeration cycle, which is the reverse of the Stirling power cycle introduced in Sec. 9.8.4. At the beginning of the isothermal compression, the pressure and temperature are 100 kPa and 350 K, respectively.The compression ratio is 7, and the temperature during the isothermal expansion is 150 K. Determine the (a) heat transfer for the isothermal compression, in kJ per kg of air. (b) net work for the cycle, in kJ per kg of air. (c) coefficient of performance. 10.58 Air undergoes an Ericsson refrigeration cycle, which is the reverse of the Ericsson power cycle introduced in Sec. 9.8.4. Figure P10.58 provides data for the cycle operating at steady state. Sketch the p–y diagram for the cycle and determine the (a) heat transfer for the isothermal expansion, per unit mass of air flow, in kJ/kg. (b) net work, per unit mass of air flow, in kJ/kg. (c) coefficient of performance.

10.54 Reconsider Problem 10.53, but include in the analysis that the compressor and turbine each have isentropic efficiencies of 88%. Answer the same questions for the modified cycle as in Problem 10.53. 10.55 Air at 2 bar, 380 K is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchanger where it is cooled at constant pressure to 320 K through heat transfer with the ambient. It then expands adiabatically to 0.95 bar through a turbine and is discharged into the cabin. The turbine has an isentropic efficiency of 75%. If the mass flow rate of the air is 1.0 kg/s, determine

2

3 · Wt

Turbine

4

Ideal regenerator

· Qout · Wc

Compressor

p2 p1 = 3.5

· Qin 1

Air

(a) the power developed by the turbine, in kW. (b) the rate of heat transfer from the air to the ambient, in kW. 10.56 Air at 32 lbf/in.2, 6808R is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchanger where it is cooled at constant pressure to 600°R through heat transfer with the ambient. It then expands adiabatically to 14 lbf/in.2 through a turbine and is discharged into the cabin at 5008R with a mass flow rate of 200 lb/min. Determine

State

p (kPa)

T (K)

1 2 3 4

100 350 350 100

310 310 270 270

Fig. P10.58

c DESIGN & OPEN-ENDED PROBLEMS: EXPLORING ENGINEERING PRACTICE 10.1D Children may wonder how a household refrigerator works to keep food cold in a warm kitchen. Prepare a 20minute presentation suitable for an elementary school science class to explain the principles of operation of a refrigerator. Include instructional aids to enhance your presentation.

two other students living in the same residence as you do. Survey the other students to determine their needs in order to size the unit. Critically evaluate competing brands. What type and number of thermoelectric modules are used in the unit selected, and what is its power requirement? Summarize your findings in a memorandum.

10.2D The object of this project is to select a compact thermoelectric refrigerator to be shared by you and at least

10.3D In cases involving cardiac arrest, stroke, heart attack, and hyperthermia, hospital medical staff must move quickly to

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Design & Open-Ended Problems: Exploring Engineering Practice reduce the patient’s body temperature by several degrees. A system for this purpose featuring a disposable plastic body suit is described in BIOCONNECTIONS in Sec. 4.9.1. Conduct a search of the patent literature for alternative ways to achieve cooling of medically distressed individuals. Consider patents both granted and pending. Critically evaluate two different methods found in your search relative to each other and the body suit approach. Write a report including at least three references. 10.4D Identify and visit a local facility that uses cold thermal storage. Conduct a forensic study to determine if the cold storage system is well-suited for the given application today. Consider costs, effectiveness in providing the desired cooling, contribution to global climate change, and other pertinent issues. If the cold storage system is well suited for the application, document that. If the cold storage system is not well suited, recommend system upgrades or an alternative approach for obtaining the desired cooling. Prepare a PowerPoint presentation of your findings. 10.5D A vertical, closed-loop geothermal heat pump is under consideration for a new 50,000-ft2 school building. The design capacity is 100 tons for both heating and cooling. The local water table is 150 ft, and the ground water temperature is 558F. Specify a ground source heat pump as well as the number and depth of wells for this application and develop a layout of vertical wells and piping required by the system. 10.6D Investigate the economic feasibility of using a waste heat-recovery heat pump for domestic water heating that employs ventilation air being discharged from a dwelling as the source. Assume typical hot water use of a family of four living in a 2200-ft2 single-family dwelling in your locale. Write a report of your findings. 10.7D Food poisoning is on the rise and can be fatal. Many of those affected have eaten recently at a restaurant, café, or fast-food outlet serving food that has not been cooled properly by the food supplier or restaurant food-handlers. To be safe, foods should not be allowed to remain in the temperature range where bacteria most quickly multiply. Standard refrigerators typically do not have the ability to provide the rapid cooling needed to ensure dangerous levels of bacteria are not attained. A food processing company supplying a wide range of fish products to restaurants has requested your project group to provide advice on how to achieve best cooling practices in its factory. In particular, you are asked to consider applicable health regulations, suitable equipment, typical operating costs, and other pertinent issues. A written report providing your recommendations is required, including an annotated list of food-cooling Dos and Don’ts for restaurants supplied by the company with fish. 10.8D According to researchers, advances in nanomaterial fabrication are leading to development of tiny thermoelectric modules that could be used in various applications, including integrating nanoscale cooling devices within the uniforms of firefighters, emergency workers, and military personnel; embedding thermoelectric modules in facades of a building; and using thermoelectric modules to recover waste heat in automobiles. Research two applications for this technology proposed within the past five years. Investigate the technical

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readiness and economic feasibility for each concept. Report your findings in an executive summary and a PowerPoint presentation with at least three references. 10.9D In induced hypothermia, the temperature of a particular organ, such as the heart, is lowered to reduce the metabolic rate during surgery. Cooling is achieved by circulating blood through a heat exchanger outside the body. When the cooled blood is reintroduced through blood vessels in the organ, it cools the organ to the desired temperature. Develop the preliminary design of a vapor-compression refrigeration system to cool blood during heart surgery. Determine the necessary temperature requirements and specify a refrigerant, operating pressures and temperatures for the working fluid, and the refrigeration capacity. 10.10D A vapor-compression refrigeration system operating continuously is being considered to provide a minimum of 80 tons of refrigeration for an industrial refrigerator maintaining a space at 28C. The surroundings to which the system rejects energy by heat transfer reach a maximum temperature of 408C. For effective heat transfer, the system requires a temperature difference of at least 208C between the condensing refrigerant and surroundings and between the vaporizing refrigerant and refrigerated space. The project manager wishes to install a system that minimizes the annual cost for electricity (monthly electricity cost is fixed at 5.692 cents for the first 250 kW ? h and 6.006 cents for any usage above 250 kW ? h). You are asked to evaluate two alternative designs: a standard vaporcompression refrigeration cycle and a vapor-compression refrigeration cycle that employs a power-recovery turbine in lieu of an expansion valve. For each alternative, consider three refrigerants: ammonia, Refrigerant 22, and Refrigerant 134a. Based on electricity cost, recommend the better choice between the two alternatives and a suitable refrigerant. Other than electricity cost, what additional factors should the manager consider in making a final selection? Prepare a written report including results, conclusions, and recommendations. 10.11D High-performance aircraft increasingly feature electronics that assist flight crews in performing their duties and reducing their fatigue. While these electronic devices improve aircraft performance, they also add greatly to the thermal load that must be managed within the aircraft. Cooling technologies currently used on aircraft are approaching their limits and other means are being considered, including vapor-compression refrigeration systems. However, unlike cooling systems used on earth, systems employed on aircraft must meet rapidly changing conditions. For instance, as onboard electronic devices switch on and off, the energy they emit by heat transfer alters the thermal load; additionally, the temperature of the air outside the aircraft into which such waste heat is discarded changes with altitude and flight speed. Accordingly, for vapor-compression systems to be practical for aircraft use, engineers must determine if the systems can quickly adapt to rapidly changing thermal loads and temperatures. The object of this project is to develop the preliminary design of a bench-top laboratory set-up with which to evaluate the performance of a vapor-compression refrigeration system subject to broadly variable thermal inputs and changing ambient conditions. Document your design in a report having at least three references.

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Systems often involve mixtures of several different entities; mixture rules for gases are provided in Sec. 11.8. Image Source/Getty Images, Inc. ENGINEERING CONTEXT As seen in previous chapters, application of thermodynamic principles to engineering systems requires data for specific internal energy, enthalpy, entropy, and other properties. The objective of this chapter is to introduce thermodynamic relations that allow u, h, s, and other thermodynamic properties of simple compressible systems to be evaluated from data that are more readily measured. Primary emphasis is on systems involving a single chemical species such as water or a mixture such as air. An introduction to general property relations for mixtures and solutions is also included. Means are available for determining pressure, temperature, volume, and mass experimentally. In addition, the relationships between the specific heats cy and cp and temperature at relatively low pressure are accessible experimentally. Values for certain other thermodynamic properties also can be measured without great difficulty. However, specific internal energy, enthalpy, and entropy are among those properties that are not easily obtained experimentally, so we resort to computational procedures to determine values for these.

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11 Thermodynamic Relations LEARNING OUTCOMES When you complete your study of this chapter, you will be able to... c

calculate p–y–T data using equations of state involving two or more constants.

c

demonstrate understanding of exact differentials involving properties and utilize the property relations developed from the exact differentials summarized in Table 11.1.

c

evaluate Du, Dh, and Ds, using the Clapeyron equation when considering phase change, and using equations of state and specific heat relations when considering single phases.

c

demonstrate understanding of how tables of thermodynamic properties are constructed.

c

evaluate Dh and Ds using generalized enthalpy and entropy departure charts.

c

utilize mixture rules, such as Kay’s Rule, to relate pressure, volume, and temperature of mixtures.

c

apply thermodynamic relations to multicomponent systems.

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Chapter 11 Thermodynamic Relations

11.1

equations of state TAKE NOTE...

Using the generalized compressibility chart, virial equations of state and the ideal gas model are introduced in Chap. 3. See Secs. 3.11 and 3.12.

Using Equations of State

An essential ingredient for the calculation of properties such as the specific internal energy, enthalpy, and entropy of a substance is an accurate representation of the relationship among pressure, specific volume, and temperature. The p–y–T relationship can be expressed alternatively: There are tabular representations, as exemplified by the steam tables. The relationship also can be expressed graphically, as in the p–y–T surface and compressibility factor charts. Analytical formulations, called equations of state, constitute a third general way of expressing the p–y–T relationship. Computer software such as Interactive Thermodynamics: IT also can be used to retrieve p–y–T data. The virial equation and the ideal gas equation are examples of analytical equations of state introduced in previous sections of the book. Analytical formulations of the p–y–T relationship are particularly convenient for performing the mathematical operations required to calculate u, h, s, and other thermodynamic properties. The object of the present section is to expand on the discussion of p–y–T relations for simple compressible substances presented in Chap. 3 by introducing some commonly used equations of state.

11.1.1

Getting Started

Recall from Sec. 3.11 that the virial equation of state can be derived from the principles of statistical mechanics to relate the p–y–T behavior of a gas to the forces between molecules. In one form, the compressibility factor Z is expanded in inverse powers of specific volume as

virial equation

Z511

D1T2 . . . C1T2 B1T2 1 1 1 2 y y y3

(11.1)

The coefficients B, C, D, etc. are called, respectively, the second, third, fourth, etc. virial coefficients. Each virial coefficient is a function of temperature alone. In principle, the virial coefficients are calculable if a suitable model for describing the forces of interaction between the molecules of the gas under consideration is known. Future advances in refining the theory of molecular interactions may allow the virial coefficients to be predicted with considerable accuracy from the fundamental properties of the molecules involved. However, at present, just the first few coefficients can be calculated and only for gases consisting of relatively simple molecules. Equation 11.1 also can be used in an empirical fashion in which the coefficients become parameters whose magnitudes are determined by fitting p–y–T data in particular realms of interest. Only a few coefficients can be found this way, and the result is a truncated equation valid only for certain states. In the limiting case where the gas molecules are assumed not to interact in any way, the second, third, and higher terms of Eq. 11.1 vanish and the equation reduces to Z 5 1. Since Z 5 py / RT, this gives the ideal gas equation of state py 5 R / T. The ideal gas equation of state provides an acceptable approximation at many states, including but not limited to states where the pressure is low relative to the critical pressure and/or the temperature is high relative to the critical temperature of the substance under consideration. At many other states, however, the ideal gas equation of state provides a poor approximation. Over 100 equations of state have been developed in an attempt to improve on the ideal gas equation of state and yet avoid the complexities inherent in a full virial series. In general, these equations exhibit little in the way of fundamental physical significance and are mainly empirical in character. Most are developed

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11.1 Using Equations of State

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for gases, but some describe the p–y–T behavior of the liquid phase, at least qualitatively. Every equation of state is restricted to particular states. This realm of applicability is often indicated by giving an interval of pressure, or density, where the equation can be expected to represent the p–y–T behavior faithfully. When it is not stated, the realm of applicability of a given equation can be approximated by expressing the equation in terms of the compressibility factor Z and the reduced properties pR, TR, y9R and superimposing the result on a generalized compressibility chart or comparing with tabulated compressibility data obtained from the literature.

11.1.2

Two-Constant Equations of State

Equations of state can be classified by the number of adjustable constants they include. Let us consider some of the more commonly used equations of state in order of increasing complexity, beginning with two-constant equations of state.

van der Waals Equation An improvement over the ideal gas equation of state based on elementary molecular arguments was suggested in 1873 by van der Waals, who noted that gas molecules actually occupy more than the negligibly small volume presumed by the ideal gas model and also exert long-range attractive forces on one another. Thus, not all of the volume of a container would be available to the gas molecules, and the force they exert on the container wall would be reduced because of the attractive forces that exist between molecules. Based on these elementary molecular arguments, the van der Waals equation of state is p5

RT a 2 2 y2b y

(11.2)

van der Waals equation

The constant b is intended to account for the finite volume occupied by the molecules, the term a/ y 2 accounts for the forces of attraction between molecules, and R is the universal gas constant. Note than when a and b are set to zero, the ideal gas equation of state results. The van der Waals equation gives pressure as a function of temperature and specific volume and thus is explicit in pressure. Since the equation can be solved for temperature as a function of pressure and specific volume, it is also explicit in temperature. However, the equation is cubic in specific volume, so it cannot generally be solved for specific volume in terms of temperature and pressure. The p van der Waals equation is not explicit in specific volume. T > Tc

EVALUATING a AND b. The van der Waals equation is a two-constant equation of state. For a specified substance, values for the constants a and b can be found by fitting the equation to p–y–T data. With this approach several sets of constants might be required to cover all states of interest. Alternatively, a single set of constants for the van der Waals equation can be determined by noting that the critical isotherm passes through a point of inflection at the critical point, and the slope is zero there. Expressed mathematically, these conditions are, respectively a

0 2p 0p b 5 0, a b 5 0 1critical point2 2 0y y 0 T T

T = Tc T < Tc

∂p _( (––– ∂v T = 0 Critical point

(11.3)

Although less overall accuracy normally results when the constants a and b are determined using critical point behavior than when they are determined

v

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Chapter 11 Thermodynamic Relations by fitting p–y–T data in a particular region of interest, the advantage of this approach is that the van der Waals constants can be expressed in terms of the critical pressure pc and critical temperature Tc, as demonstrated next. For the van der Waals equation at the critical point pc 5

RTc a 2 2 yc 2 b yc

Applying Eqs. 11.3 with the van der Waals equation gives a

0 2p 2

2RTc

b 5

3

2

6a 50 y 4c

0y T 1yc 2 b2 0p RTc 2a a b 52 1 3 50 2 0y T 1yc 2 b2 yc

Solving the foregoing three equations for a, b, and yc in terms of the critical pressure and critical temperature 27 R 2T 2c 64 pc RTc b5 8pc 3 RTc yc 5 8 pc a5

(11.4a) (11.4b) (11.4c)

Values of the van der Waals constants a and b determined from Eqs. 11.4a and 11.4b for several common substances are given in Table A-24 for pressure in bar, specific volume in m3/kmol, and temperature in K. Values of a and b for the same substances are given in Table A-24E for pressure in atm, specific volume in ft3/lbmol, and temperature in 8R.

GENERALIZED FORM. Introducing the compressibility factor Z 5 py/ RT, the reduced temperature TR 5 T/Tc, the pseudoreduced specific volume y R¿ 5 pcy / RTc, and the foregoing expressions for a and b, the van der Waals equation can be written in terms of Z, y9R, and TR as Z5

yR¿ 27/ 64 2 yR¿ 2 1/ 8 TRyR¿

(11.5)

or alternatively in terms of Z, TR, and pR as Z3 2 a

pR 27pR 27p 2R 1 1b Z 2 1 a b Z 2 50 8TR 64T 2R 512T 3R

(11.6)

The details of these developments are left as exercises. Equation 11.5 can be evaluated for specified values of y9R and TR and the resultant Z values located on a generalized compressibility chart to show approximately where the equation performs satisfactorily. A similar approach can be taken with Eq. 11.6. The compressibility factor at the critical point yielded by the van der Waals equation is determined from Eq. 11.4c as Zc 5

pcyc RTc

5 0.375

Actually, Zc varies from about 0.23 to 0.33 for most substances (see Tables A-1). Accordingly, with the set of constants given by Eqs. 11.4, the van der Waals equation is inaccurate in the vicinity of the critical point. Further study would show inaccuracy in other regions as well, so this equation is not suitable for many thermodynamic evaluations. The van der Waals equation is of interest to us primarily because it is

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11.1 Using Equations of State the simplest model that accounts for the departure of actual gas behavior from the ideal gas equation of state.

Redlich–Kwong Equation Three other two-constant equations of state that have been widely used are the Berthelot, Dieterici, and Redlich–Kwong equations. The Redlich–Kwong equation, considered by many to be the best of the two-constant equations of state, is p5

RT a 2 y2b y1y 1 b2T 1y2

(11.7)

This equation, proposed in 1949, is mainly empirical in nature, with no rigorous justification in terms of molecular arguments. The Redlich–Kwong equation is explicit in pressure but not in specific volume or temperature. Like the van der Waals equation, the Redlich–Kwong equation is cubic in specific volume. Although the Redlich–Kwong equation is somewhat more difficult to manipulate mathematically than the van der Waals equation, it is more accurate, particularly at higher pressures. The two-constant Redlich–Kwong equation performs better than some equations of state having several adjustable constants; still, two-constant equations of state tend to be limited in accuracy as pressure (or density) increases. Increased accuracy at such states normally requires equations with a greater number of adjustable constants. Modified forms of the Redlich–Kwong equation have been proposed to achieve improved accuracy.

EVALUATING a AND b. As for the van der Waals equation, the constants a and b in Eq. 11.7 can be determined for a specified substance by fitting the equation to p–y–T data, with several sets of constants required to represent accurately all states of interest. Alternatively, a single set of constants in terms of the critical pressure and critical temperature can be evaluated using Eqs. 11.3, as for the van der Waals equation. The result is a 5 a¿

R 2Tc5y2 pc

and

b 5 b¿

RTc pc

(11.8)

where a9 5 0.42748 and b9 5 0.08664. Evaluation of these constants is left as an exercise. Values of the Redlich–Kwong constants a and b determined from Eqs. 11.8 for several common substances are given in Table A-24 for pressure in bar, specific volume in m3/kmol, and temperature in K. Values of a and b for the same substances are given in Table A-24E for pressure in atm, specific volume in ft3/lbmol, and temperature in 8R.

GENERALIZED FORM. Introducing the compressibility factor Z, the reduced temperature TR, the pseudoreduced specific volume y9R, and the foregoing expressions for a and b, the Redlich–Kwong equation can be written as Z5

y¿R a¿ 2 y¿R 2 b¿ 1y¿R 1 b¿2T 3R/2

(11.9)

Equation 11.9 can be evaluated at specified values of y9R and TR and the resultant Z values located on a generalized compressibility chart to show the regions where the equation performs satisfactorily. With the constants given by Eqs. 11.8, the compressibility factor at the critical point yielded by the Redlich–Kwong equation is Zc 5 0.333, which is at the high end of the range of values for most substances, indicating that inaccuracy in the vicinity of the critical point should be expected. In Example 11.1, the pressure of a gas is determined using three equations of state and the generalized compressibility chart. The results are compared.

Redlich–Kwong equation

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Chapter 11 Thermodynamic Relations

EXAMPLE 11.1 c

Comparing Equations of State A cylindrical tank containing 4.0 kg of carbon monoxide gas at 2508C has an inner diameter of 0.2 m and a length of 1 m. Determine the pressure, in bar, exerted by the gas using (a) the generalized compressibility chart, (b) the ideal gas equation of state, (c) the van der Waals equation of state, (d) the Redlich–Kwong equation of state. Compare the results obtained. SOLUTION Known: A cylindrical tank of known dimensions contains 4.0 kg of CO gas at 2508C. Find: Determine the pressure exerted by the gas using four alternative methods. Schematic and Given Data: Engineering Model:

D = 0.2 m

1. As shown in the accompanying figure, the closed system is taken as the gas. 2. The system is at equilibrium.

L=1m

4 kg CO gas at –50°C

Fig. E11.1 Analysis: The molar specific volume of the gas is required in each part of the solution. Let us begin by evaluat-

ing it. The volume occupied by the gas is V5a

p10.2 m2211.0 m2 pD2 bL 5 5 0.0314 m3 4 4

The molar specific volume is then kg V 0.0314 m3 m3 y 5 My 5 M a b 5 a28 ba b 5 0.2198 m kmol 4.0 kg kmol (a) From Table A-1 for CO, Tc 5 133 K, pc 5 35 bar. Thus, the reduced temperature TR and pseudoreduced

specific volume y9R are, respectively 223 K 5 1.68 133 K 10.2198 m3/ kmol2135 3 105 N/ m22 ypc y¿R 5 5 5 0.696 18314 N ? m/ kmol ? K21133 K2 RTc

TR 5

Turning to Fig. A-2, Z < 0.9. Solving Z 5 py / RT for pressure and inserting known values p5

0.918314 N ? m/ kmol ? K21223 K2 1 bar Z RT 5 ` 5 ` 5 75.9 bar y 10.2198 m3/ kmol2 10 N/ m2

(b) The ideal gas equation of state gives

p5

18314 N ? m/ kmol ? K21223 K2 1 bar RT 5 ` 5 ` 5 84.4 bar y 10.2198 m3/ kmol2 10 N/ m2

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11.1 Using Equations of State

637

(c) For carbon monoxide, the van der Waals constants a and b given by Eqs. 11.4 can be read directly from Table A-24. Thus

a 5 1.474 bar a

m3 2 b and kmol

b 5 0.0395

m3 kmol

Substituting into Eq. 11.2 p5 5

RT a 2 2 y2b y 1.474 bar1m3/ kmol22 1 bar ` 2 10.2198 m3/ kmol22 10.2198 2 0.039521m3/ kmol2 105 N/ m2

18314 N ? m/ kmol ? K21223 K2

`

5 72.3 bar Alternatively, the values for y9R and TR obtained in the solution of part (a) can be substituted into Eq. 11.5, giving Z 5 0.86. Then, with p 5 ZRT/ y, p 5 72.5 bar. The slight difference is attributed to roundoff. (d) For carbon monoxide, the Redlich–Kwong constants given by Eqs. 11.8 can be read directly from Table

A-24. Thus a5

17.22 bar1m621K21/2 1kmol22

and

b 5 0.02737 m3/ kmol

Substituting into Eq. 11.7 p5 5

RT a 2 y2b y1y 1 b2T 1/2 18314 N ? m/ kmol ? K21223 K2

10.2198 2 0.027372 m3/ kmol 5 75.1 bar

✓ Skills Developed `

1 bar 17.22 bar ` 2 5 2 10 N/ m 10.2198210.247172122321/2

Alternatively, the values for y9R and TR obtained in the solution of part (a) can be substituted into Eq. 11.9, giving Z 5 0.89. Then, with p 5 ZRT/ y, p 5 75.1 bar. In comparison to the value of part (a), the ideal gas equation of state predicts a pressure that is 11% higher and the van der Waals equation gives a value that is 5% lower. The Redlich–Kwong value is about 1% less than the value obtained using the compressibility chart.

Ability to… ❑ determine pressure using

the compressibility chart, ideal gas model, and the van der Waals and Redlich– Kwong equations of state. ❑ perform unit conversions correctly.

Using the given temperature and the pressure value determined in part (a), check the value of Z using Fig. A-2. Ans. Z < 0.9.

11.1.3

Multiconstant Equations of State

To fit the p–y–T data of gases over a wide range of states, Beattie and Bridgeman proposed in 1928 a pressure-explicit equation involving five constants in addition to the gas constant. The Beattie–Bridgeman equation can be expressed in a truncated virial form as p5

b g RT d 1 21 31 4 y y y y

(11.10)

where b 5 BRT 2 A 2 cR/ T 2 g 5 2BbRT 1 Aa 2 BcR / T 2 d 5 BbcR / T 2

(11.11)

Beattie–Bridgeman equation

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Chapter 11 Thermodynamic Relations The five constants a, b, c, A, and B appearing in these equations are determined by curve fitting to experimental data. Benedict, Webb, and Rubin extended the Beattie–Bridgeman equation of state to cover a broader range of states. The resulting equation, involving eight constants in addition to the gas constant, has been particularly successful in predicting the p–y–T behavior of light hydrocarbons. The Benedict–Webb–Rubin equation is

Benedict–Webb–Rubin equation

p5

1bRT 2 a2 aa g g RT C 1 c 1 aBRT 2 A 2 2 b 2 1 1 6 1 3 2 a1 1 2 b exp a2 2 b 3 y y yT y y y T y (11.12)

Values of the constants appearing in Eq. 11.12 for five common substances are given in Table A-24 for pressure in bar, specific volume in m3/kmol, and temperature in K. Values of the constants for the same substances are given in Table A-24E for pressure in atm, specific volume in ft3/lbmol, and temperature in 8R. Because Eq. 11.12 has been so successful, its realm of applicability has been extended by introducing additional constants. Equations 11.10 and 11.12 are merely representative of multiconstant equations of state. Many other multiconstant equations have been proposed. With high-speed computers, equations having 50 or more constants have been developed for representing the p–y–T behavior of different substances.

11.2

TAKE NOTE...

The state principle for simple systems is introduced in Sec. 3.1.

exact differential

Important Mathematical Relations

Values of two independent intensive properties are sufficient to fix the intensive state of a simple compressible system of specified mass and composition—for instance, temperature and specific volume (see Sec. 3.1). All other intensive properties can be determined as functions of the two independent properties: p 5 p(T, y), u 5 u(T, y), h 5 h(T, y), and so on. These are all functions of two independent variables of the form z 5 z (x, y), with x and y being the independent variables. It might also be recalled that the differential of every property is exact (Sec. 2.2.1). The differentials of nonproperties such as work and heat are inexact. Let us review briefly some concepts from calculus about functions of two independent variables and their differentials. The exact differential of a function z, continuous in the variables x and y, is dz 5 a

0z 0z b dx 1 a b dy 0x y 0y x

(11.13a)

This can be expressed alternatively as dz 5 M dx 1 N dy

(11.13b)

where M 5 10z / 0x2y and N 5 10z / 0y2x . The coefficient M is the partial derivative of z with respect to x (the variable y being held constant). Similarly, N is the partial derivative of z with respect to y (the variable x being held constant). If the coefficients M and N have continuous first partial derivatives, the order in which a second partial derivative of the function z is taken is immaterial. That is 0 0z 0 0z ca b d 5 ca b d 0y 0x y x 0x 0y x y

(11.14a)

0N 0M b 5a b 0y x 0x y

(11.14b)

or a test for exactness

which can be called the test for exactness, as discussed next.

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11.2 Important Mathematical Relations In words, Eqs. 11.14 indicate that the mixed second partial derivatives of the function z are equal. The relationship in Eqs. 11.14 is both a necessary and sufficient condition for the exactness of a differential expression, and it may therefore be used as a test for exactness. When an expression such as M dx 1 N dy does not meet this test, no function z exists whose differential is equal to this expression. In thermodynamics, Eq. 11.14 is not generally used to test exactness but rather to develop additional property relations. This is illustrated in Sec. 11.3 to follow. Two other relations among partial derivatives are listed next for which applications are found in subsequent sections of this chapter. These are 0y 0x b a b 51 0y z 0x z

(11.15)

0y 0z 0x b a b a b 5 21 0z x 0x y 0y z

(11.16)

a and a

consider the three quantities x, y, and z, any two of which may be selected as the independent variables. Thus, we can write x 5 x( y, z) and y 5 y(x, z). The differentials of these functions are, respectively dx 5 a

0x 0x b dy 1 a b dz 0y z 0z y

and

dy 5 a

0y 0y b dx 1 a b dz 0x z 0z x

Eliminating dy between these two equations results in c1 2 a

0y 0y 0x 0x 0x b a b d dx 5 c a b a b 1 a b d dz 0y z 0x z 0y z 0z x 0z y

(11.17)

Since x and z can be varied independently, let us hold z constant and vary x. That is, let dz 5 0 and dx ? 0. It then follows from Eq. 11.17 that the coefficient of dx must vanish, so Eq. 11.15 must be satisfied. Similarly, when dx 5 0 and dz ? 0, the coefficient of dz in Eq. 11.17 must vanish. Introducing Eq. 11.15 into the resulting expression and rearranging gives Eq. 11.16. The details are left as an exercise. b b b b b

APPLICATION. An equation of state p 5 p(T, y) provides a specific example of a function of two independent variables. The partial derivatives 10p/ 0T2y and 10p/ 0y2T of p(T, y) are important for subsequent discussions. The quantity 10p/ 0T2y is the partial derivative of p with respect to T (the variable y being held constant). This partial derivative represents the slope at a point on a line of constant specific volume (isometric) projected onto the p–T plane. Similarly, the partial derivative 10p/ 0y2T is the partial derivative of p with respect to y (the variable T being held constant). This partial derivative represents the slope at a point on a line of constant temperature (isotherm) projected on the p–y plane. The partial derivatives 10p/ 0T2y and 10p/ 0y2T are themselves intensive properties because they have unique values at each state. The p–y–T surfaces given in Figs. 3.1 and 3.2 are graphical representations of functions of the form p 5 p(y, T). Figure 11.1 shows the liquid, vapor, and two-phase regions of a p–y–T surface projected onto the p–y and p–T planes. Referring first to Fig. 11.1a, note that several isotherms are sketched. In the single-phase regions, the partial derivative 10p/ 0y2T giving the slope is negative at each state along an isotherm except at the critical point, where the partial derivative vanishes. Since the isotherms are horizontal in the two-phase liquid–vapor region, the partial derivative 10p/ 0y2T vanishes there as well. For these states, pressure is independent of specific volume and is a function of temperature only: p 5 psat(T ). Figure 11.1b shows the liquid and vapor regions with several isometrics (constant specific volume lines) superimposed. In the single-phase regions, the isometrics are

639

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Chapter 11 Thermodynamic Relations p

p

v < vc

T = Tc T < Tc

∂p ––– < 0 ∂v T

( (

v = vc

∂p (––– ( ∂T v > 0

T > Tc ∂p (––– ( ∂v T < 0

Critical point

∂p (––– ( ∂v T = 0 Critical point (∂p/∂v)T = 0 Liquid-vapor

Isome tric

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∂p (––– ( ∂T v > 0 v > vc ric

et Isom

Iso

the

rm

Liquid dp (––– ( dT sat

Iso

the

rm

Triple point (a)

v

Vapor

Locus of saturation states T (b)

Fig. 11.1 Diagrams used to discuss (≠p/≠y)T and (≠p/≠T )y. (a) p–y diagram. (b) Phase diagram. nearly straight or are slightly curved and the partial derivative 10p/ 0T2y is positive at each state along the curves. For the two-phase liquid–vapor states corresponding to a specified value of temperature, the pressure is independent of specific volume and is determined by the temperature only. Hence, the slopes of isometrics passing through the two-phase states corresponding to a specified temperature are all equal, being given by the slope of the saturation curve at that temperature, denoted simply as (dp/dT)sat. For these two-phase states, 10p/ 0T2y 5 1dp/ dT2sat. In this section, important aspects of functions of two variables have been introduced. The following example illustrates some of these ideas using the van der Waals equation of state.

cccc

EXAMPLE 11.2 c

Applying Mathematical Relations For the van der Waals equation of state, (a) determine an expression for the exact differential dp, (b) show that the mixed second partial derivatives of the result obtained in part (a) are equal, and (c) develop an expression for the partial derivative 10y/ 0T2p. SOLUTION Known: The equation of state is the van der Waals equation. Find: Determine the differential dp, show that the mixed second partial derivatives of dp are equal, and develop

an expression for 10y/ 0T2p.

Analysis: (a) By definition, the differential of a function p 5 p(T, y) is

dp 5 a

0p 0p b dT 1 a b dy 0y T 0T y

The partial derivatives appearing in this expression obtained from the van der Waals equation expressed as p 5 RT/(y 2 b) 2 a/y2 are M5a

0p 0p R RT 2a b 5 , N 5 a b 5 2 1 3 2 0T y y 2 b 0y T 1y 2 b2 y

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11.3 Developing Property Relations

641

Accordingly, the differential takes the form dp 5 a

R 2RT 2a b dT 1 c 1 3 d dy 2 y2b 1y 2 b2 y

(b) Calculating the mixed second partial derivatives

0M b 5 0y T 0N a b 5 0T y

a

0p 0 R ca b d 5 2 0y 0T y T 1y 2 b22 0p 0 R ca b d 5 2 0T 0y T y 1y 2 b22

Thus, the mixed second partial derivatives are equal, as expected. (c) An expression for 10y/ 0T2p can be derived using Eqs. 11.15 and 11.16. Thus, with x 5 p, y 5 y, and z 5

T, Eq. 11.16 gives a

0p 0y 0T b a b a b 5 21 0T p 0y T 0p y

or a

0y 1 b 52 0T p 10p/ 0y2T 10T/ 0p2y

Then, with x 5 T, y 5 p, and z 5 y, Eq. 11.15 gives a

0T 1 b 5 0p y 10p/ 0T2y

a

10p/ 0T2y 0y b 5 0T p 10p/ 0y2T

Combining these results

The numerator and denominator of this expression have been evaluated in part (a), so ➊

a

R/ 1y 2 b2 0y b 52 0T p 32RT/ 1y 2 b22 1 2a/ y34

which is the desired result. ➊ Since the van der Waals equation is cubic in specific volume, it can be solved for y (T, p) at only certain states. Part (c) shows how the partial derivative 10y/ 0T2p can be evaluated at states where it exists.

Using the results obtained, develop an expression for 0y a b of an ideal gas. Ans. y/T. 0T p

11.3

Developing Property Relations

In this section, several important property relations are developed, including the expressions known as the Maxwell relations. The concept of a fundamental thermodynamic function is also introduced. These results, which are important for subsequent discussions, are obtained for simple compressible systems of fixed chemical composition using the concept of an exact differential.

✓ Skills Developed Ability to… ❑ use Eqs. 11.15 and 11.16

together with the van der Waals equation of state to develop a thermodynamic property relation.

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Chapter 11 Thermodynamic Relations

11.3.1

Principal Exact Differentials

The principal results of this section are obtained using Eqs. 11.18, 11.19, 11.22, and 11.23. The first two of these equations are derived in Sec. 6.3, where they are referred to as the T ds equations. For present purposes, it is convenient to express them as du 5 T ds 2 p dy dh 5 T ds 1 y dp

(11.18) (11.19)

The other two equations used to obtain the results of this section involve, respectively, the specific Helmholtz function c defined by c 5 u 2 Ts

Helmholtz function

(11.20)

and the specific Gibbs function g defined by g 5 h 2 Ts

Gibbs function

(11.21)

The Helmholtz and Gibbs functions are properties because each is defined in terms of properties. From inspection of Eqs. 11.20 and 11.21, the units of c and g are the same as those of u and h. These two new properties are introduced solely because they contribute to the present discussion, and no physical significance need be attached to them at this point. Forming the differential dc dc 5 du 2 d1Ts2 5 du 2 T ds 2 s dT Substituting Eq. 11.18 into this gives dc 5 2p dy 2 s dT

(11.22)

Similarly, forming the differential dg dg 5 dh 2 d1Ts2 5 dh 2 T ds 2 s dT Substituting Eq. 11.19 into this gives dg 5 y dp 2 s dT

(11.23)

11.3.2 Property Relations from Exact Differentials The four differential equations introduced above, Eqs. 11.18, 11.19, 11.22, and 11.23, provide the basis for several important property relations. Since only properties are involved, each is an exact differential exhibiting the general form dz 5 M dx 5 N dy considered in Sec. 11.2. Underlying these exact differentials are, respectively, functions of the form u(s, y), h(s, p), c(y, T), and g(T, p). Let us consider these functions in the order given. The differential of the function u 5 u(s, y) is du 5 a

0u 0u b ds 1 a b dy 0y s 0s y

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11.3 Developing Property Relations By comparison with Eq. 11.18, we conclude that 0u b 0s y 0u 2p 5 a b 0y s T5a

(11.24) (11.25)

The differential of the function h 5 h(s, p) is dh 5 a

0h 0h b ds 1 a b dp 0s p 0p s

By comparison with Eq. 11.19, we conclude that 0h b 0s p 0h y5a b 0p s

T5a

(11.26) (11.27)

Similarly, the coefficients 2p and 2s of Eq. 11.22 are partial derivatives of c(y, T) 0c b 0y T 0c 2s 5 a b 0T y

2p 5 a

(11.28) (11.29)

and the coefficients y and 2s of Eq. 11.23 are partial derivatives of g(T, p) 0g b 0p T 0g 2s 5 a b 0T p y5a

(11.30) (11.31)

As each of the four differentials introduced above is exact, the second mixed partial derivatives are equal. Thus, in Eq. 11.18, T plays the role of M in Eq. 11.14b and 2p plays the role of N in Eq. 11.14b, so a

0p 0T b 5 2a b 0y s 0s y

(11.32)

In Eq. 11.19, T and y play the roles of M and N in Eq. 11.14b, respectively. Thus a

0T 0y b 5a b 0s p 0p s

(11.33)

Similarly, from Eqs. 11.22 and 11.23 follow 0p 0s b 5a b 0y T 0T y 0s 0y a b 5 2a b 0T p 0p T

a

(11.34) (11.35)

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Chapter 11 Thermodynamic Relations TABLE 11.1

Summary of Property Relations from Exact Differentials Basic relations: from u 5 u(s, y)

from h 5 h(s, p) T5a

0u b 0s y

(11.24)

T5a

0h b 0s p

(11.26)

2p 5 a

0u b 0y s

(11.25)

y5a

0h b 0p s

(11.27)

from c 5 c(y, T )

from g 5 g(T, p) 2p 5 a 2s 5 a

0c 0y 0c 0T

b

y5a

(11.28) T

b

2s 5 a

(11.29)

y

0g 0p 0g 0T

b

(11.30) T

b

(11.31)

p

Maxwell relations: a

0p 0T b 5 2a b 0s y 0y s

(11.32)

a

0p 0s b 5a b 0T y 0y T

(11.34)

a

0T 0y b 5a b 0p s 0s p

(11.33)

a

0y 0s b 5 2a b 0T p 0p T

(11.35)

(11.36)

Additional relations:

Maxwell relations

a

0u 0h b 5a b 0s y 0s p

a

0c 0u b 5a b 0y T 0y s

a

0g 0h b 5a b 0p s 0p T

a

0c 0T

b 5a y

0g 0T

b p

Equations 11.32 through 11.35 are known as the Maxwell relations. Since each of the properties T, p, y, s appears on the left side of two of the eight equations, Eqs. 11.24 through 11.31, four additional property relations can be obtained by equating such expressions. They are

a

0c 0u 0h 0u b 5 a b , a b 5 a b 0y T 0s y 0s p 0y s

a

0c 0g 0g 0h b 5 a b , a b 5 a b 0T y 0p s 0p T 0T p

(11.36)

Equations 11.24 through 11.36, which are listed in Table 11.1 for ease of reference, are 16 property relations obtained from Eqs. 11.18, 11.19, 11.22, and 11.23, using the concept of an exact differential. Since Eqs. 11.19, 11.22, and 11.23 can themselves be derived from Eq. 11.18, the important role of the first T dS equation in developing property relations is apparent. The utility of these 16 property relations is demonstrated in subsequent sections of this chapter. However, to give a specific illustration at this point, suppose the partial derivative 10s/ 0y2T involving entropy is required for a certain purpose. The Maxwell relation Eq. 11.34 would allow the derivative to be determined by evaluating the partial derivative 10p/ 0T2y, which can be obtained using p–y–T data only. Further elaboration is provided in Example 11.3.

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11.3 Developing Property Relations cccc

645

EXAMPLE 11.3 c

Applying the Maxwell Relations Evaluate the partial derivative 10s/ 0y2T for water vapor at a state fixed by a temperature of 2408C and a specific volume of 0.4646 m3/kg. (a) Use the Redlich–Kwong equation of state and an appropriate Maxwell relation. (b) Check the value obtained using steam table data. SOLUTION Known: The system consists of a fixed amount of water vapor at 2408C and 0.4646 m3/kg. Find: Determine the partial derivative 10s/ 0y2T employing the Redlich–Kwong equation of state, together with

a Maxwell relation. Check the value obtained using steam table data. Engineering Model: 1. The system consists of a fixed amount of water at a known equilibrium state. 2. Accurate values for 10s/ 0T2y in the neighborhood of the given state can be determined from the Redlich–

Kwong equation of state. Analysis: (a) The Maxwell relation given by Eq. 11.34 allows 10s/ 0y2T to be determined from the p–y–T relationship. That is

a

0p 0s b 5a b 0y T 0T y

The partial derivative 10p/ 0T2y obtained from the Redlich–Kwong equation, Eq. 11.7, is a

0p R a 1 b 5 0T y y 2 b 2y1y 1 b2T 3/2

At the specified state, the temperature is 513 K and the specific volume on a molar basis is y 5 0.4646

m3 18.02 kg m3 a b 5 8.372 kg kmol kmol

From Table A-24 a 5 142.59 bar a

m3 2 m3 b 1K21/2, b 5 0.0211 kmol kmol

Substituting values into the expression for 10p/ 0T2y N?m m3 2 b 142.59 bar a b 1K21/2 0p kmol ? K 105 N/ m2 kmol a b 5 1 ` ` 3 2 3 1 bar 0T y m m m 3/ 2 18.372 2 0.02112 2a8.372 b a8.3931 b1513 K2 kmol kmol kmol N?m 1 kJ b ` 3 ` 5 a1004.3 3 m ? K 10 N ? m kJ 5 1.0043 3 m ?K a8314

Accordingly a

0s kJ b 5 1.0043 3 0y T m ?K

(b) A value for 10s/ 0y2T can be estimated using a graphical approach with steam table data, as follows: At 2408C, Table A-4 provides the values for specific entropy s and specific volume y tabulated below

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Chapter 11 Thermodynamic Relations

646

T 5 2408C p (bar)

s (kJ/kg ? K)

y (m3/kg)

1.0 1.5 3.0 5.0 7.0 10.0

7.9949 7.8052 7.4774 7.2307 7.0641 6.8817

2.359 1.570 0.781 0.4646 0.3292 0.2275

With the values for s and y listed in the table, the plot in Fig. E11.3a giving s versus y can be prepared. Note that a line representing the tangent to the curve at the given state is shown on the plot. The pressure at this state is 5 bar. The slope of the tangent is 10s/ 0y2T < 1.0 kJ/ m3 ? K. Thus, the value of 10s/ 0y2T obtained using the Redlich–Kwong equation agrees closely with the result determined graphically using steam table data.

Specific entropy, kJ/kg·K

T = 240°C

1.5 bar

7.5 3 bar 5 bar 7 bar 7.0

0.5

1.0 Specific volume, m3/kg

1.5

Fig. E11.3a

Alternative Solution: Alternatively, the partial derivative 10s/ 0y2T can be estimated using numerical methods and computer-generated data. The following IT code illustrates one way the partial derivative, denoted dsdv, can be estimated:

v 5 0.4646 // m3/kg T 5 240 // 8C v2 5 v 1 dv v1 5 v 2 dv dv 5 0.2 v2 5 v_PT (“Water/Steam”, p2, T) v1 5 v_PT (“Water/Steam”, p1, T) s2 5 s_PT (“Water/Steam”, p2, T) s1 5 s_PT (“Water/Steam”, p1, T) dsdv 5 (s2 2 s1)/(v2 2 v1) Using the Explore button, sweep dv from 0.0001 to 0.2 in steps of 0.001. Then, using the Graph button, the following graph can be constructed: 1.12 1.10 T = 240°C

dsdv

1.08 1.06 1.04 1.02 1.00 0.00

0.05

0.10 dv

0.15

0.20

Fig. E11.3b

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11.3 Developing Property Relations From the computer data, the y-intercept of the graph is ➊

a

647

✓ Skills Developed

0s ¢s kJ b 5 lim a b < 1.033 3 0y T ¢yS0 ¢y T m ?K

This answer is an estimate because it relies on a numerical approximation of the partial derivative based on the equation of state that underlies the steam tables. The values obtained using the Redlich–Kwong equation of state and the graphical method using steam table data agree with this result. ➊ It is left as an exercise to show that, in accordance with Eq. 11.34, the value of 10p/ 0T2y estimated by a procedure like the one used for 10s/ 0y2T agrees with the value given here.

Ability to… ❑ apply a Maxwell relation to

evaluate a thermodynamic quantity. ❑ apply the Redlich–Kwong equation. ❑ perform a comparison with data from the steam table using graphical and computer-based methods.

For steam at T 5 2408C, y 5 0.4646 m3/kg, p 5 5 bar, calculate the value of the compressibility factor Z. Ans. 0.981.

11.3.3 Fundamental Thermodynamic Functions A fundamental thermodynamic function provides a complete description of the thermodynamic state. In the case of a pure substance with two independent properties, the fundamental thermodynamic function can take one of the following four forms: u 5 u1s, y2 h 5 h1s, p2 c 5 c1T, y2 g 5 g1T, p2

(11.37)

Of the four fundamental functions listed in Eqs. 11.37, the Helmholtz function c and the Gibbs function g have the greatest importance for subsequent discussions (see Sec. 11.6.2). Accordingly, let us discuss the fundamental function concept with reference to c and g. In principle, all properties of interest can be determined from a fundamental thermodynamic function by differentiation and combination. consider a fundamental function of the form c(T, y). The properties y and T, being the independent variables, are specified to fix the state. The pressure p at this state can be determined from Eq. 11.28 by differentiation of c(T, y). Similarly, the specific entropy s at the state can be found from Eq. 11.29 by differentiation. By definition, c 5 u 2 Ts, so the specific internal energy is obtained as u 5 c 1 Ts With u, p, and y known, the specific enthalpy can be found from the definition h 5 u 1 py. Similarly, the specific Gibbs function is found from the definition, g 5 h 2 Ts. The specific heat cy can be determined by further differentiation, cy 5 10u/ 0T2y. Other properties can be calculated with similar operations. b b b b b consider a fundamental function of the form g(T, p). The properties T and p are specified to fix the state. The specific volume and specific entropy at this state can be determined by differentiation from Eqs. 11.30 and 11.31, respectively. By definition, g 5 h 2 Ts, so the specific enthalpy is obtained as h 5 g 1 Ts

fundamental thermodynamic function

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Chapter 11 Thermodynamic Relations With h, p, and y known, the specific internal energy can be found from u 5 h 2 py. The specific heat cp can be determined by further differentiation, cp 5 10h/ 0T2p. Other properties can be calculated with similar operations. b b b b b Like considerations apply for functions of the form u(s, y) and h(s, p), as can readily be verified. Note that a Mollier diagram provides a graphical representation of the fundamental function h(s, p).

11.4

Evaluating Changes in Entropy, Internal Energy, and Enthalpy

With the introduction of the Maxwell relations, we are in a position to develop thermodynamic relations that allow changes in entropy, internal energy, and enthalpy to be evaluated from measured property data. The presentation begins by considering relations applicable to phase changes and then turns to relations for use in single-phase regions.

11.4.1 Considering Phase Change The objective of this section is to develop relations for evaluating the changes in specific entropy, internal energy, and enthalpy accompanying a change of phase at fixed temperature and pressure. A principal role is played by the Clapeyron equation, which allows the change in enthalpy during vaporization, sublimation, or melting at a constant temperature to be evaluated from pressure-specific volume–temperature data pertaining to the phase change. Thus, the present discussion provides important examples of how p–y–T measurements can lead to the determination of other property changes, namely Ds, Du, and Dh for a change of phase. Consider a change in phase from saturated liquid to saturated vapor at fixed temperature. For an isothermal phase change, pressure also remains constant, so Eq. 11.19 reduces to dh 5 T ds Integration of this expression gives sg 2 sf 5

hg 2 hf T

(11.38)

Hence, the change in specific entropy accompanying a phase change from saturated liquid to saturated vapor at temperature T can be determined from the temperature and the change in specific enthalpy. The change in specific internal energy during the phase change can be determined using the definition h 5 u 1 py. ug 2 uf 5 hg 2 hf 2 p1yg 2 yf2

(11.39)

Thus, the change in specific internal energy accompanying a phase change at temperature T can be determined from the temperature and the changes in specific volume and enthalpy.

CLAPEYRON EQUATION. The change in specific enthalpy required by Eqs. 11.38 and 11.39 can be obtained using the Clapeyron equation. To derive the Clapeyron equation, begin with the Maxwell relation a

0p 0s b 5a b 0y T 0T y

(11.34)

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11.4 Evaluating Changes in Entropy, Internal Energy, and Enthalpy During a phase change at fixed temperature, the pressure is independent of specific volume and is determined by temperature alone. Thus, the quantity 10p/ 0T2y is determined by the temperature and can be represented as a

0p dp b 5a b 0T y dT sat

where “sat” indicates that the derivative is the slope of the saturation pressure– temperature curve at the point determined by the temperature held constant during the phase change (Sec. 11.2). Combining the last two equations gives a

dp 0s b 5a b 0y T dT sat

Since the right side of this equation is fixed when the temperature is specified, the equation can be integrated to give sg 2 s f 5 a

dp b 1yg 2 yf2 dT sat

Introducing Eq. 11.38 into this expression results in the Clapeyron equation a

hg 2 hf dp b 5 dT sat T1yg 2 yf2

Clapeyron equation

(11.40)

Equation 11.40 allows (hg 2 hf) to be evaluated using only p–y–T data pertaining to the phase change. In instances when the enthalpy change is also measured, the Clapeyron equation can be used to check the consistency of the data. Once the specific enthalpy change is determined, the corresponding changes in specific entropy and specific internal energy can be found from Eqs. 11.38 and 11.39, respectively. Equations 11.38, 11.39, and 11.40 also can be written for sublimation or melting occurring at constant temperature and pressure. In particular, the Clapeyron equation would take the form a

dp h– 2 h¿ b 5 dT sat T1y– 2 y¿2

(11.41)

where 0 and 9 denote the respective phases, and (dp/dT)sat is the slope of the relevant saturation pressure–temperature curve. The Clapeyron equation shows that the slope of a saturation line on a phase diagram depends on the signs of the specific volume and enthalpy changes accompanying the phase change. In most cases, when a phase change takes place with an increase in specific enthalpy, the specific volume also increases, and (dp/dT)sat is positive. However, in the case of the melting of ice and a few other substances, the specific volume decreases on melting. The slope of the saturated solid–liquid curve for these few substances is negative, as pointed out in Sec. 3.2.2 in the discussion of phase diagrams. An approximate form of Eq. 11.40 can be derived when the following two idealizations are justified: (1) yf is negligible in comparison to yg, and (2) the pressure is low enough that yg can be evaluated from the ideal gas equation of state as yg 5 RT/p. With these, Eq. 11.40 becomes a

hg 2 hf dp b 5 dT sat RT 2 / p

which can be rearranged to read a

hg 2 hf d ln p b 5 dT sat RT 2

(11.42)

Equation 11.42 is called the Clausius–Clapeyron equation. A similar expression applies for the case of sublimation.

Clausius–Clapeyron equation

649

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Chapter 11 Thermodynamic Relations

650

The use of the Clapeyron equation in any of the foregoing forms requires an accurate representation for the relevant saturation pressure–temperature curve. This must not only depict the pressure–temperature variation accurately but also enable accurate values of the derivative (dp/dT)sat to be determined. Analytical representations in the form of equations are commonly used. Different equations for different portions of the pressure– temperature curves may be required. These equations can involve several constants. One form that is used for the vapor-pressure curves is the four-constant equation ln psat 5 A 1

B 1 C ln T 1 DT T

in which the constants A, B, C, D are determined empirically. The use of the Clapeyron equation for evaluating changes in specific entropy, internal energy, and enthalpy accompanying a phase change at fixed T and p is illustrated in the next example.

EXAMPLE 11.4 c

Applying the Clapeyron Equation Using p–y–T data for saturated water, calculate at 1008C (a) hg 2 hf, (b) ug 2 uf, (c) sg 2 sf. Compare with the respective steam table value. SOLUTION Known: The system consists of a unit mass of saturated water at 1008C. Find: Using saturation data, determine at 1008C the change on vaporization of the specific enthalpy, specific

internal energy, and specific entropy, and compare with the respective steam table value. Analysis: For comparison, Table A-2 gives at 1008C, hg 2 hf 5 2257.0 kJ/kg, ug 2 uf 5 2087.6 kJ/kg, sg 2 sf 5 6.048 kJ/kg ? K. (a) The value of hg 2 hf can be determined from the Clapeyron equation, Eq. 11.40, expressed as

hg 2 hf 5 T1yg 2 yf2a

dp b dT sat

This equation requires a value for the slope (dp/dT)sat of the saturation pressure–temperature curve at the specified temperature. The required value for (dp/dT)sat at 1008C can be estimated graphically as follows. Using saturation pressure– temperature data from the steam tables, the accompanying plot can be prepared. Note that a line drawn tangent to the curve at 1008C is shown on the plot. The slope of this tangent line is about 3570 N/m2 ? K. Accordingly, at 1008C a

dp N b < 3570 2 dT sat m ?K

4 Saturation pressure (bar)

cccc

3

2

1

0 40

60

80 100 Temperature (°C)

120

140

Fig. E11.4

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11.4 Evaluating Changes in Entropy, Internal Energy, and Enthalpy

651

Inserting property data from Table A-2 into the equation for hg 2 hf gives hg 2 hf 5 1373.15 K211.673 2 1.0435 3 10232a

m3 N 1 kJ b a3570 2 b` 3 ` kg m ? K 10 N ? m

5 2227 kJ/ kg This value is about 1% less than the value read from the steam tables. ➊ Alternatively, the derivative (dp/dT)sat can be estimated using numerical methods and computer-generated data. The following IT code illustrates one way the derivative, denoted dpdT, can be estimated: T 5 100 // 8C dT 5 0.001 T1 5 T 2 dT T2 5 T 1 dT p1 5 Psat (“Water/Steam”, T1) // bar p2 5 Psat (“Water/Steam”, T2) // bar dpdT 5 ((p2 2 p1) / (T2 2 T1)) * 100000 Using the Explore button, sweep dT from 0.001 to 0.01 in steps of 0.001. Then, reading the limiting value from the computer data a

dp N b < 3616 2 dT sat m ?K

When this value is used in the above expression for hg 2 hf, the result is hg 2 hf 5 2256 kJ/kg, which agrees closely with the value read from the steam tables. (b) With Eq. 11.39

ug 2 uf 5 hg 2 hf 2 psat1yg 2 yf2 Inserting the IT result for (hg 2 hf) from part (a) together with saturation data at 1008C from Table A-2 kJ N m3 1 kJ 2 a1.014 3 105 2 b a1.672 b ` 3 ` kg kg 10 N ? m m kJ 5 2086.5 kg

ug 2 uf 5 2256

which also agrees closely with the value from the steam tables. (c) With Eq. 11.38 and the IT result for (hg 2 hf) from part (a)

sg 2 s f 5

hg 2 hf T

2256 kJ/ kg kJ 5 5 6.046 kg ? K 373.15 K

which again agrees closely with the steam table value. ➊ Also, (dp/dT )sat might be obtained by differentiating an analytical expression for the vapor pressure curve, as discussed on page 650.

Use the IT result (dp/dT )sat 5 3616 N/m2 ? K to extrapolate the saturation pressure, in bar, at 1058C. Ans. 1.195 bar.

11.4.2 Considering Single-Phase Regions The objective of the present section is to derive expressions for evaluating D s, Du, and Dh between states in single-phase regions. These expressions require both p–y–T data and appropriate specific heat data. Since single-phase regions are under present

✓ Skills Developed Ability to… ❑ use the Clapeyron Equation

with p–y–T data for saturated water to evaluate, ufg, hfg, and sfg. ❑ use graphical and computerbased methods to evaluate thermodynamic property data and relations.

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Chapter 11 Thermodynamic Relations consideration, any two of the properties pressure, specific volume, and temperature can be regarded as the independent properties that fix the state. Two convenient choices are T, y and T, p.

T AND y AS INDEPENDENT PROPERTIES. With temperature and specific volume as the independent properties that fix the state, the specific entropy can be regarded as a function of the form s 5 s(T, y). The differential of this function is ds 5 a

0s 0s b dT 1 a b dy 0T y 0y T

The partial derivative 10s/ 0y2T appearing in this expression can be replaced using the Maxwell relation, Eq. 11.34, giving ds 5 a

0p 0s b dT 1 a b dy 0T y 0T y

(11.43)

The specific internal energy also can be regarded as a function of T and y: u 5 u(T, y). The differential of this function is du 5 a

0u 0u b dT 1 a b dy 0T y 0y T

With cy 5 10u/ 0T2y du 5 cy dT 1 a

0u b dy 0y T

(11.44)

Substituting Eqs. 11.43 and 11.44 into du 5 T ds 2 p dy and collecting terms results in ca

0p 0u 0s b 1 p 2 T a b d dy 5 c T a b 2 cy d dT 0y T 0T y 0T y

(11.45)

Since specific volume and temperature can be varied independently, let us hold specific volume constant and vary temperature. That is, let dy 5 0 and dT fi 0. It then follows from Eq. 11.45 that a

cy 0s b 5 0T y T

(11.46)

Similarly, suppose that dT 5 0 and dy ? 0. It then follows that a

0p 0u b 5 Ta b 2 p 0y T 0T y

(11.47)

Equations 11.46 and 11.47 are additional examples of useful thermodynamic property relations. TAKE NOTE...

Here we demonstrate that the specific internal energy of a gas whose equation of state is py 5 RT depends on temperature alone, thereby confirming a claim made in Sec. 3.12.2.

Equation 11.47, which expresses the dependence of the specific internal energy on specific volume at fixed temperature, allows us to demonstrate that the internal energy of a gas whose equation of state is py 5 RT depends on temperature alone, a result first discussed in Sec. 3.12.2. Equation 11.47 requires the partial derivative 10p/ 0T2y. If p 5 RT/y, the derivative is 10p/ 0T2y 5 R/ y. Introducing this, Eq. 11.47 gives a

0p 0u R b 5 Ta b 2 p 5 Ta b 2 p 5 p 2 p 5 0 y 0y T 0T y

This demonstrates that when py 5 RT, the specific internal energy is independent of specific volume and depends on temperature alone. b b b b b

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11.4 Evaluating Changes in Entropy, Internal Energy, and Enthalpy

653

Continuing the discussion, when Eq. 11.46 is inserted in Eq. 11.43, the following expression results. ds 5

0p cy dT 1 a b dy T 0T y

(11.48)

Inserting Eq. 11.47 into Eq. 11.44 gives du 5 cy dT 1 c T a

0p b 2 p d dy 0T y

(11.49)

Observe that the right sides of Eqs. 11.48 and 11.49 are expressed solely in terms of p, y, T, and cy. Changes in specific entropy and internal energy between two states are determined by integration of Eqs. 11.48 and 11.49, respectively. s2 2 s 1 5

#

2

1

cy dT 1 T

2

u2 2 u1 5

2

# a 0T b 1 2

0p

dy

# c dT 1 # c T a 0T b 0p

y

1

(11.50)

y

2 p d dy

(11.51)

y

1

To integrate the first term on the right of each of these expressions, the variation of cy with temperature at one fixed specific volume (isometric) is required. Integration of the second term requires knowledge of the p–y–T relation at the states of interest. An equation of state explicit in pressure would be particularly convenient for evaluating the integrals involving 10p/ 0T2y . The accuracy of the resulting specific entropy and internal energy changes would depend on the accuracy of this derivative. In cases where the integrands of Eqs. 11.50 and 11.51 are too complicated to be integrated in closed form they may be evaluated numerically. Whether closed-form or numerical integration is used, attention must be given to the path of integration. let us consider the evaluation of Eq. 11.51. Referring to Fig. 11.2, if the specific heat cy is known as a function of temperature along the isometric (constant specific volume) passing through the states x and y, one possible path of integration for determining the change in specific internal energy between states 1 and 2 is 1–x–y–2. The integration would be performed in three steps. Since temperature is constant from state 1 to state x, the first integral of Eq. 11.51 vanishes, so

#

ux 2 u1 5

yx

cTa

y1

0p b 2 p d dy 0T y

From state x to y, the specific volume is constant and cy is known as a function of temperature only, so uy 2 ux 5

#

Ty

p

cy dT

Tx

where Tx 5 T1 and Ty 5 T2. From state y to state 2, the temperature is constant once again, and u2 2 uy 5

#

2

y2

yy 5yx

cTa

0p b 2 p d dy 0T y

When these are added, the result is the change in specific internal energy between states 1 and 2. b b b b b

1

v x = vy y T2 = Ty

x cv = cv(T, vx)

T1 = Tx

T AND p AS INDEPENDENT PROPERTIES. In this section a presentation parallel to that considered above is provided for the Fig. 11.2 Integration path between two vapor choice of temperature and pressure as the independent properties. states.

v

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Chapter 11 Thermodynamic Relations With this choice for the independent properties, the specific entropy can be regarded as a function of the form s 5 s(T, p). The differential of this function is ds 5 a

0s 0s b dT 1 a b dp 0T p 0p T

The partial derivative 10s/ 0p2T appearing in this expression can be replaced using the Maxwell relation, Eq. 11.35, giving ds 5 a

0s 0y b dT 2 a b dp 0T p 0T p

(11.52)

The specific enthalpy also can be regarded as a function of T and p: h 5 h(T, p). The differential of this function is dh 5 a

0h 0h b dT 1 a b dp 0T p 0p T

With cp 5 10h/ 0T2p dh 5 cp dT 1 a

0h b dp 0p T

(11.53)

Substituting Eqs. 11.52 and 11.53 into dh 5 T ds 1 y dp and collecting terms results in ca

0h 0y 0s b 1 T a b 2 y d dp 5 c T a b 2 cp d dT 0p T 0T p 0T p

(11.54)

Since pressure and temperature can be varied independently, let us hold pressure constant and vary temperature. That is, let dp 5 0 and dT ? 0. It then follows from Eq. 11.54 that a

cp 0s b 5 0T p T

(11.55)

Similarly, when dT 5 0 and dp ? 0, Eq. 11.54 gives a

0h 0y b 5 y 2 Ta b 0p T 0T p

(11.56)

Equations 11.55 and 11.56, like Eqs. 11.46 and 11.47, are useful thermodynamic property relations. When Eq. 11.55 is inserted in Eq. 11.52, the following equation results: ds 5

cp T

dT 2 a

0y b dp 0T p

(11.57)

Introducing Eq. 11.56 into Eq. 11.53 gives dh 5 cp dT 1 c y 2 T a

0y b d dp 0T p

(11.58)

Observe that the right sides of Eqs. 11.57 and 11.58 are expressed solely in terms of p, y, T, and cp. Changes in specific entropy and enthalpy between two states are found by integrating Eqs. 11.57 and 11.58, respectively s2 2 s 1 5

#

1

2

cp T

2

dT 2

# a 0T b 1

0y

p

dp

(11.59)

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11.4 Evaluating Changes in Entropy, Internal Energy, and Enthalpy 2

h2 2 h1 5

#c

dT 1

p

1

#

2

cy 2 Ta

1

0y b d dp 0T p

655

(11.60)

To integrate the first term on the right of each of these expressions, the variation of cp with temperature at one fixed pressure (isobar) is required. Integration of the second term requires knowledge of the p–y–T behavior at the states of interest. An equation of state explicit in y would be particularly convenient for evaluating the integrals involving 10y/ 0T2p. The accuracy of the resulting specific entropy and enthalpy changes would depend on the accuracy of this derivative. Changes in specific enthalpy and internal energy are related through h 5 u 1 py by h2 2 h1 5 1u2 2 u12 1 1p2y2 2 p1y12

(11.61)

Hence, only one of Dh and Du need be found by integration. Then, the other can be evaluated from Eq. 11.61. Which of the two property changes is found by integration depends on the information available. Dh would be found using Eq. 11.60 when an equation of state explicit in y and cp as a function of temperature at some fixed pressure are known. Du would be found from Eq. 11.51 when an equation of state explicit in p and cy as a function of temperature at some specific volume are known. Such issues are considered in Example 11.5. cccc

EXAMPLE 11.5 c

Evaluating Ds, Du, and Dh of a Gas Using the Redlich–Kwong equation of state, develop expressions for the changes in specific entropy, internal energy, and enthalpy of a gas between two states where the temperature is the same, T1 5 T2, and the pressures are p1 and p2, respectively. SOLUTION Known: Two states of a unit mass of a gas as the system are fixed by p1 and T1 at state 1 and p2, T2 (5 T1) at

state 2. Find: Determine the changes in specific entropy, internal energy, and enthalpy between these two states. Schematic and Given Data: Engineering Model: The Redlich–Kwong equation of state represents the p–y–T behavior at these states and yields accurate values for 10p/ 0T2y.

p

p2 p1

2 1 Iso

the

v2

v1 v

rm

Fig. E11.5

Analysis: The Redlich–Kwong equation of state is explicit in pressure, so Eqs. 11.50 and 11.51 are selected for

determining s2 2 s1 and u2 2 u1. Since T1 5 T2, an isothermal path of integration between the two states is convenient. Thus, these equations reduce to give s2 2 s1 5

#

2

a

1

u2 2 u1 5

#

1

2

cT a

0p b dy 0T y

0p b 2 p d dy 0T y

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Chapter 11 Thermodynamic Relations

The limits for each of the foregoing integrals are the specific volumes y1 and y2 at the two states under consideration. Using p1, p2, and the known temperature, these specific volumes would be determined from the Redlich–Kwong equation of state. Since this equation is not explicit in specific volume, the use of an equation solver such as Interactive Thermodynamics: IT is recommended. The above integrals involve the partial derivative 10p/ 0T2y, which can be determined from the Redlich–Kwong equation of state as a

0p R a 1 b 5 0T y y 2 b 2y1y 1 b2T 3 /2

Inserting this into the expression for (s2 2 s1) gives s2 2 s1 5

#

y2

#

y2

c

R a 1 d dy y2b 2y1y 1 b2T 3/2

c

R a 1 1 1 a 2 b d dy 3/ 2 y y2b y1b 2bT

y1

5

y1

5 R ln a

y2 2 b y2 y2 1 b a b1 c ln a b 2 ln a bd 3/ 2 y y1 2 b y1 1 b 2bT 1

5 R ln a

y21y1 1 b2 y2 2 b a b1 ln c d 3/ 2 y1 2 b y11y2 1 b2 2bT

With the Redlich–Kwong equation, the integrand of the expression for (u2 2 u1) becomes cTa

0p R a RT a 1 d 2 c 2 d b 2 pd 5 T c y2b y2b 0T y 2y1y 1 b2T 3 /2 y1y 1 b2T1/2 3a 5 2y1y 1 b2T1/2

Accordingly u2 2 u1 5

#

y2

y1

5

3a 3a dy 5 1/ 2 2y1y 1 b2T 2bT 1/2

#

y2

y1

1 1 a 2 b dy y y1b

y21y1 1 b2 y2 y2 1 b 3a 3a c ln 2 ln a bd 5 ln c d 1/ 2 1/ 2 y y1 1 b y11y2 1 b2 2bT 2bT 1

Finally, (h2 2 h1) would be determined using Eq. 11.61 together with the known values of (u2 2 u1), p1, y1, p2, and y2.

✓ Skills Developed Ability to… ❑ perform differentiations and

integrations required to evaluate Du and Ds using the two-constant Redlich– Kwong equation of state.

Using results obtained, develop expressions for Du and Ds of an ideal gas. Ans. Du 5 0, Ds 5 R ln (y2/y1 ).

11.5

Other Thermodynamic Relations

The presentation to this point has been directed mainly at developing thermodynamic relations that allow changes in u, h, and s to be evaluated from measured property data. The objective of the present section is to introduce several other thermodynamic relations that are useful for thermodynamic analysis. Each of the properties considered has a common attribute: it is defined in terms of a partial derivative of some other property. The specific heats cy and cp are examples of this type of property.

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11.5 Other Thermodynamic Relations

657

11.5.1 Volume Expansivity, Isothermal and Isentropic Compressibility In single-phase regions, pressure and temperature are independent, and we can think of the specific volume as being a function of these two, y 5 y(T, p). The differential of such a function is dy 5 a

0y 0y b dT 1 a b dp 0T p 0p T

Two thermodynamic properties related to the partial derivatives appearing in this differential are the volume expansivity b, also called the coefficient of volume expansion b5

1 0y a b y 0T p

volume expansivity

(11.62)

and the isothermal compressibility k

isothermal compressibility

1 0y k52 a b y 0p T

(11.63)

By inspection, the unit for b is seen to be the reciprocal of that for temperature and the unit for k is the reciprocal of that for pressure. The volume expansivity is an indication of the change in volume that occurs when temperature changes while pressure remains constant. The isothermal compressibility is an indication of the change in volume that takes place when pressure changes while temperature remains constant. The value of k is positive for all substances in all phases. The volume expansivity and isothermal compressibility are thermodynamic properties, and like specific volume are functions of T and p. Values for b and k are provided in compilations of engineering data. Table 11.2 gives values of these properties for liquid water at a pressure of 1 atm versus temperature. For a pressure of 1 atm, water has a state of maximum density at about 48C. At this state, the value of b is zero. The isentropic compressibility a is an indication of the change in volume that occurs when pressure changes while entropy remains constant: 1 0y a52 a b y 0p s

(11.64)

The unit for a is the reciprocal of that for pressure.

TABLE 11.2

Volume Expansivity B and Isothermal Compressibility K of Liquid Water at 1 atm Versus Temperature T (8C)

Density (kg/m3)

b 3 106 (K)21

k 3 106 (bar)21

0 10 20 30 40 50

999.84 999.70 998.21 995.65 992.22 988.04

268.14 87.90 206.6 303.1 385.4 457.8

50.89 47.81 45.90 44.77 44.24 44.18

isentropic compressibility

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Chapter 11 Thermodynamic Relations

velocity of sound

The isentropic compressibility is related to the speed at which sound travels in a substance, and such speed measurements can be used to determine a. In Sec. 9.12.2, the velocity of sound, or sonic velocity, is introduced as c5

B

2y2 a

0p b 0y s

(9.36b)

The relationship of the isentropic compressibility and the velocity of sound can be obtained using the relation between partial derivatives expressed by Eq. 11.15. Identifying p with x, y with y, and s with z, we have TAKE NOTE...

Through the Mach number, the sonic velocity c plays an important role in analyzing flow in nozzles and diffusers. See Sec. 9.13.

a

0p 1 b 5 0y s 10y/ 0p2s

With this, the previous two equations can be combined to give c 5 1y/ a

(11.65)

The details are left as an exercise.

BIOCONNECTIONS The propagation of elastic waves, such as sound waves, has important implications related to injury in living things. During impact such as a collision in a sporting event (see accompanying figure), elastic waves are created that cause some bodily material to move relative to the rest of the body. The waves can propagate at supersonic, transonic, or subsonic speeds depending on the nature of the impact, and the resulting trauma can cause serious damage. The waves may be focused into a small area, causing localized damage, or they may be reflected at the boundary of organs and cause more widespread damage. An example of the focusing of waves occurs in some head injuries. An impact to the skull causes flexural and compression waves to move along the curved surface and arrive at the far side of the skull simultaneously. Waves also propagate through the softer brain tissue. Consequently, concussions, skull fractures, and other injuries can appear at locations away from the site of the original impact. Central to an understanding of traumatic injury is data on speed of sound and other elastic characteristics of organs and tissues. For humans the speed of sound varies widely, from approximately 30–45 m/s in spongy lung tissue to about 1600 m/s in muscle and 3500 m/s in bone. Because the speed of sound in the lungs is relatively low, impacts such as in automobile collisions or even air-bag deployment can set up waves that propagate supersonically. Medical personnel responding to traumas are trained to check for lung injuries. The study of wave phenomena in the body constitutes an important area in the field of biomechanics.

11.5.2 Relations Involving Specific Heats In this section, general relations are obtained for the difference between specific heats (cp 2 cy) and the ratio of specific heats cp/cy.

EVALUATING (cP 2 cY). An expression for the difference between cp and cy can be obtained by equating the two differentials for entropy given by Eqs. 11.48 and 11.57 and rearranging to obtain 1cp 2 cy2 dT 5 T a

0p 0y b dy 1 T a b dp 0T p 0T y

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11.5 Other Thermodynamic Relations Considering the equation of state p 5 p(T, y), the differential dp can be expressed as dp 5

0p 0p b dT 1 b dy 0T y 0y T

Eliminating dp between the last two equations and collecting terms gives c 1cp 2 cy2 2 T a

0p 0p 0p 0y 0y b a b d dT 5 T c a b a b 1 a b d dy 0T p 0T y 0T p 0y T 0T y

Since temperature and specific volume can be varied independently, the coefficients of the differentials in this expression must vanish, so 0p 0y ba b 0T p 0T y

(11.66)

0p 0p 0y b 5 2a b a b 0T p 0y T 0T y

(11.67)

cp 2 cy 5 T a a

Introducing Eq. 11.67 into Eq. 11.66 gives cp 2 cy 5 2T a

0y 2 0p b a b 0T p 0y T

(11.68)

This equation allows cy to be calculated from observed values of cp knowing only p–y–T data, or cp to be calculated from observed values of cy. for the special case of an ideal gas, Eq. 11.68 reduces to Eq. 3.44: cp1T2 5 cy1T2 1 R, as can readily be shown. b b b b b The right side of Eq. 11.68 can be expressed in terms of the volume expansivity b and the isothermal compressibility k. Introducing Eqs. 11.62 and 11.63, we get

cp 2 cy 5 y

Tb2 k

(11.69)

In developing this result, the relationship between partial derivatives expressed by Eq. 11.15 has been used. Several important conclusions about the specific heats cp and cy can be drawn from Eq. 11.69. since the factor b2 cannot be negative and k is positive for all substances in all phases, the value of cp is always greater than, or equal to, cy. The specific heats are equal when b 5 0, as occurs in the case of water at 1 atmosphere and 48C, where water is at its state of maximum density. The two specific heats also become equal as the temperature approaches absolute zero. For some liquids and solids at certain states, cp and cy differ only slightly. For this reason, tables often give the specific heat of a liquid or solid without specifying whether it is cp or cy. The data reported are normally cp values, since these are more easily determined for liquids and solids. b b b b b

EVALUATING cp/cY. Next, let us obtain expressions for the ratio of specific heats, k. Employing Eq. 11.16, we can rewrite Eqs. 11.46 and 11.55, respectively, as cy 0s 21 5a b 5 T 0T y 10y/ 0s2T10T/ 0y2s cp 0s 21 5a b 5 T 0T p 10p/ 0s2T10T/ 0p2s

659

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Chapter 11 Thermodynamic Relations Forming the ratio of these equations gives cp 10y/ 0s2T 10T/ 0y2s 5 cy 10p/ 0s2T 10T/ 0p2s

(11.70)

Since 10s/ 0p2T 5 1/ 10p/ 0s2T and 10p/ 0T2s 5 1/ 10T/ 0p2s, Eq. 11.70 can be expressed as cp 0p 0y 0s 0T 5 ca b a b d ca b a b d cy 0s T 0p T 0T s 0y s

(11.71)

Finally, the chain rule from calculus allows us to write 10y/ 0p2T 5 10y/ 0s2T 10s/ 0p2T and 10p/ 0y2s 5 10p/ 0T2s10T/ 0y2s, so Eq. 11.71 becomes k5

cp 0p 0y 5a ba b cy 0p T 0y s

(11.72)

This can be expressed alternatively in terms of the isothermal and isentropic compressibilities as k5

k a

(11.73)

Solving Eq. 11.72 for 10p/ 0y2s and substituting the resulting expression into Eq. 9.36b gives the following relationship involving the velocity of sound c and the specific heat ratio k c 5 22ky210p/ 0y2T

(11.74)

Equation 11.74 can be used to determine c knowing the specific heat ratio and p–y–T data, or to evaluate k knowing c and 10p/ 0y2T. in the special case of an ideal gas, Eq. 11.74 reduces to give Eq. 9.37 (Sec. 9.12.2): c 5 1kRT 1ideal gas2 as can easily be verified. b

(9.37)

b b b b

In the next example we illustrate the use of specific heat relations introduced above.

cccc

EXAMPLE 11.6 c

Using Specific Heat Relations For liquid water at 1 atm and 208C, estimate (a) the percent error in cy that would result if it were assumed that cp 5 cy, (b) the velocity of sound, in m/s. SOLUTION Known: The system consists of a fixed amount of liquid water at 1 atm and 208C. Find: Estimate the percent error in cy that would result if cy were approximated by cp, and the velocity of sound,

in m/s.

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11.5 Other Thermodynamic Relations

661

Analysis: (a) Equation 11.69 gives the difference between cp and cy. Table 11.2 provides the required values for the

volume expansivity b, the isothermal compressibility k, and the specific volume. Thus cp 2 cy 5 y

Tb2 k

1 206.6 3 1026 2 bar b1293 K2a ba b 3 K 998.21 kg/ m 45.90 3 1026 bar ? m3 105 N/ m2 1 kJ 5 a272.96 3 1026 b` ` ` 3 ` kg ? K 1 bar 10 N ? m

5a

5 0.027

kJ kg ? K

➊ Interpolating in Table A-19 at 208C gives cp 5 4.188 kJ/kg ? K Thus, the value of cy is cy 5 4.188 2 0.027 5 4.161 kJ/ kg ? K Using these values, the percent error in approximating cy by cp is ➋

a

cp 2 cy 0.027 b11002 5 a b11002 5 0.6% cy 4.161

(b) The velocity of sound at this state can be determined using Eq. 11.65. The required value for the isentropic compressibility a is calculable in terms of the specific heat ratio k and the isothermal compressibility k. With Eq. 11.73, a 5 k/k. Inserting this into Eq. 11.65 results in the following expression for the velocity of sound

c5

ky Bk

The values of y and k required by this expression are the same as used in part (a). Also, with the values of cp and cy from part (a), the specific heat ratio is k 5 1.006. Accordingly ❸

c5

2 105 N/ m2 1 kg ? m/ s ` ` ` 5 1482 m/ s B 1998.21 kg/ m32145.902 1 bar 1N

11.006211062 bar

`

➊ Consistent with the discussion of Sec. 3.10.1, we take cp at 1 atm and 208C as the saturated liquid value at 208C.

✓ Skills Developed

➋ The result of part (a) shows that for liquid water at the given state, cp and cy are closely equal.

Ability to… ❑ apply specific heat relations

❸ For comparison, the velocity of sound in air at 1 atm, 208C is about 343 m/s, which can be checked using Eq. 9.37.

❑ evaluate velocity of sound

to liquid water. for liquid water.

A submarine moves at a speed of 20 knots (1 knot 5 1.852 km/h). Using the sonic velocity calculated in part (b), estimate the Mach number of the vessel relative to the water. Ans. 0.0069.

11.5.3 Joule–Thomson Coefficient The value of the specific heat cp can be determined from p–y–T data and the Joule– Thomson coefficient. The Joule–Thomson coefficient mJ is defined as mJ 5 a

0T b 0p h

(11.75)

Joule–Thomson coefficient

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Chapter 11 Thermodynamic Relations Like other partial differential coefficients introduced in this section, the Joule– Thomson coefficient is defined in terms of thermodynamic properties only and thus is itself a property. The units of mJ are those of temperature divided by pressure. A relationship between the specific heat cp and the Joule–Thomson coefficient mJ can be established by using Eq. 11.16 to write a

0p 0T 0h b a b a b 5 21 0p h 0h T 0T p

The first factor in this expression is the Joule–Thomson coefficient and the third is cp. Thus cp 5

21 mJ10p/ 0h2T

With 10h/ 0p2T 5 1/ 10p/ 0h2T from Eq. 11.15, this can be written as cp 5 2

1 0h a b mJ 0p T

(11.76)

The partial derivative 10h/ 0p2T, called the constant-temperature coefficient, can be eliminated from Eq. 11.76 by use of Eq. 11.56. The following expression results:

cp 5

1 0y c T a b 2 yd mJ 0T p

(11.77)

Equation 11.77 allows the value of cp at a state to be determined using p–y–T data and the value of the Joule–Thomson coefficient at that state. Let us consider next how the Joule–Thomson coefficient can be found experimentally.

EXPERIMENTAL EVALUATION. The Joule–Thomson coefficient can be evaluated experimentally using an apparatus like that pictured in Fig. 11.3. Consider first Fig. 11.3a, which shows a porous plug through which a gas (or liquid) may pass. During operation at steady state, the gas enters the apparatus at a specified temperature T1 and pressure p1 and expands through the plug to a lower pressure p2, which is controlled by an outlet valve. The temperature T2 at the exit is measured. The apparatus is designed T

Inversion state Inlet T1, p1

Inversion state Porous plug

Inversion state Critical point

T2, p2

Inlet state (T1, p1) ∂T –– h ∂p

( )

Vapor Liquid

Valve (a)

Triple point

Solid p (b)

Fig. 11.3 Joule–Thomson expansion. (a) Apparatus. (b) Isenthalpics on a T–p diagram.

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11.6 Constructing Tables of Thermodynamic Properties so that the gas undergoes a throttling process (Sec. 4.10) as it expands from 1 to 2. Accordingly, the exit state fixed by p2 and T2 has the same value for the specific enthalpy as at the inlet, h2 5 h1. By progressively lowering the outlet pressure, a finite sequence of such exit states can be visited, as indicated on Fig. 11.3b. A curve may be drawn through the set of data points. Such a curve is called an isenthalpic (constant enthalpy) curve. An isenthalpic curve is the locus of all points representing equilibrium states of the same specific enthalpy. The slope of an isenthalpic curve at any state is the Joule–Thomson coefficient at that state. The slope may be positive, negative, or zero in value. States where the coefficient has a zero value are called inversion states. Notice that not all lines of constant h have an inversion state. The uppermost curve of Fig. 11.3b, for example, always has a negative slope. Throttling a gas from an initial state on this curve would result in an increase in temperature. However, for isenthalpic curves having an inversion state, the temperature at the exit of the apparatus may be greater than, equal to, or less than the initial temperature, depending on the exit pressure specified. For states to the right of an inversion state, the value of the Joule–Thomson coefficient is negative. For these states, the temperature increases as the pressure at the exit of the apparatus decreases. At states to the left of an inversion state, the value of the Joule–Thomson coefficient is positive. For these states, the temperature decreases as the pressure at the exit of the device decreases. This can be used to advantage in systems designed to liquefy gases.

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inversion states

Small Power Plants Pack Punch An innovation in power systems moving from concept to reality promises to help keep computer networks humming, hospital operating rooms lit, and shopping centers thriving. Called distributed generation systems, compact power plants provide electricity for small loads or are linked for larger applications. With distributed generation, consumers hope to avoid unpredictable price swings and brownouts. Distributed generation includes a broad range of technologies that provide relatively small levels of power at sites close to users, including but not limited to internal combustion engines, microturbines, fuel cells, and photovoltaic systems.

11.6

Although the cost per kilowatt-hour may be higher with distributed generation, some customers are willing to pay more to gain control over their electric supply. Computer networks and hospitals need high reliability, since even short disruptions can be disastrous. Businesses such as shopping centers also must avoid costly service interruptions. With distributed generation, the needed reliability is provided by modular units that can be combined with power management and energy storage systems to ensure quality power is available when needed.

Constructing Tables of Thermodynamic Properties

The objective of this section is to utilize the thermodynamic relations introduced thus far to describe how tables of thermodynamic properties can be constructed. The characteristics of the tables under consideration are embodied in the tables for water and the refrigerants presented in the Appendix. The methods introduced in this section are extended in Chap. 13 for the analysis of reactive systems, such as gas turbine and vapor power systems involving combustion. The methods of this section also provide the basis for computer retrieval of thermodynamic property data. Two different approaches for constructing property tables are considered: c The presentation of Sec. 11.6.1 employs the methods introduced in Sec. 11.4 for

assigning specific enthalpy, specific internal energy, and specific entropy to states of pure, simple compressible substances using p–y–T data, together with a limited amount of specific heat data. The principal mathematical operation of this approach is integration.

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Chapter 11 Thermodynamic Relations

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c The approach of Sec. 11.6.2 utilizes the fundamental thermodynamic function con-

cept introduced in Sec. 11.3.3. Once such a function has been constructed, the principal mathematical operation required to determine all other properties is differentiation.

11.6.1 Developing Tables by Integration Using p–y–T and Specific Heat Data In principle, all properties of present interest can be determined using cp 5 cp01T2 p 5 p1y, T2, y 5 y1p, T2

T 8

7

1 Arbitrary datum state h1 = s1 = 0

2

(11.78)

In Eqs. 11.78, cp0(T ) is the specific heat cp for the substance under consideration extrapolated to zero pressure. This function might be determined from data obtained calorimetrically or from spectroscopic data, using equations supplied by statistical mechanics. Specific heat expressions for several gases are given in Tables A-21. The expressions p(y, T) and y(p, T) represent functions that describe the saturation pressure–temperature curves, as well as the p–y–T relations for the single-phase regions. These functions may be tabular, graphical, or analytical in character. Isobar– Whatever their forms, however, the functions must not only represent reduced pressure pR low enough the p–y–T data accurately but also yield accurate values for derivafor the ideal gas tives such as 10y/ 0T2p and 1dp/ dT2sat. model to be Figure 11.4 shows eight states of a substance. Let us consider how appropriate values can be assigned to specific enthalpy and specific entropy at 6 5 these states. The same procedures can be used to assign property values at other states of interest. Note that when h has been assigned to a state, the specific internal energy at that state can be found from 3 4 u 5 h 2 py. c Let the state denoted by 1 on Fig. 11.4 be selected as the datum

state for enthalpy and entropy. Any value can be assigned to h and s at this state, but a value of zero would be usual. It should be noted that the use of an arbitrary datum state and arbitrary referFig. 11.4 T–y diagram used to discuss how h ence values for specific enthalpy and specific entropy suffices only and s can be assigned to liquid and vapor for evaluations involving differences in property values between states. states of the same composition, for then datums cancel. c Once a value is assigned to enthalpy at state 1, the enthalpy at the saturated vapor state, state 2, can be determined using the Clapeyron equation, Eq. 11.40 v

h2 2 h1 5 T11y2 2 y12a

dp b dT sat

where the derivative (dp/dT)sat and the specific volumes y1 and y2 are obtained from appropriate representations of the p–y–T data for the substance under consideration. The specific entropy at state 2 is found using Eq. 11.38 in the form s2 2 s 1 5

h2 2 h1 T1

c Proceeding at constant temperature from state 2 to state 3, the entropy and enthalpy

are found by means of Eqs. 11.59 and 11.60, respectively. Since temperature is fixed, these equations reduce to give s3 2 s2 5 2

#

p3

p2

a

0y b dp 0T p

and

h3 2 h2 5

#

p3

p2

With the same procedure, s4 and h4 can be determined.

cy 2 T a

0y b d dp 0T p

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11.6 Constructing Tables of Thermodynamic Properties c The isobar (constant-pressure line) passing through state 4 is assumed to be at a

low enough pressure for the ideal gas model to be appropriate. Accordingly, to evaluate s and h at states such as 5 on this isobar, the only required information would be cp0(T) and the temperatures at these states. Thus, since pressure is fixed, Eqs. 11.59 and 11.60 give, respectively s5 2 s4 5

#

T5

T4

cp0

dT T

and

h5 2 h4 5

#

T5

cp0 dT

T4

c Specific entropy and enthalpy values at states 6 and 7 are found from those at state

5 by the same procedure used in assigning values at states 3 and 4 from those at state 2. Finally, s8 and h8 are obtained from the values at state 7 using the Clapeyron equation.

11.6.2 Developing Tables by Differentiating a Fundamental Thermodynamic Function Property tables also can be developed using a fundamental thermodynamic function. It is convenient for this purpose to select the independent variables of the fundamental function from among pressure, specific volume (density), and temperature. This indicates the use of the Helmholtz function c(T, y) or the Gibbs function g(T, p). The properties of water tabulated in Tables A-2 through A-6 have been calculated using the Helmholtz function. Fundamental thermodynamic functions also have been employed successfully to evaluate the properties of other substances in the Appendix tables. The development of a fundamental thermodynamic function requires considerable mathematical manipulation and numerical evaluation. Prior to the advent of highspeed computers, the evaluation of properties by this means was not feasible, and the approach described in Sec. 11.6.1 was used exclusively. The fundamental function approach involves three steps: 1. The first step is the selection of a functional form in terms of the appropriate pair of independent properties and a set of adjustable coefficients, which may number 50 or more. The functional form is specified on the basis of both theoretical and practical considerations. 2. Next, the coefficients in the fundamental function are determined by requiring that a set of carefully selected property values and/or observed conditions be satisfied in a least-squares sense. This generally involves the use of property data requiring the assumed functional form to be differentiated one or more times, such as p–y–T and specific heat data. 3. When all coefficients have been evaluated, the function is carefully tested for accuracy by using it to evaluate properties for which accepted values are known. These may include properties requiring differentiation of the fundamental function two or more times. For example, velocity of sound and Joule–Thomson data might be used. This procedure for developing a fundamental thermodynamic function is not routine and can be accomplished only with a computer. However, once a suitable fundamental function is established, extreme accuracy in and consistency among the thermodynamic properties is possible. The form of the Helmholtz function used in constructing the steam tables from which Tables A-2 through A-6 have been extracted is c1r, T2 5 c01T2 1 RT 3ln r 1 rQ1r,t24

(11.79)

where c0 and Q are given as the sums listed in Table 11.3. The independent variables are density and temperature. The variable t denotes 1000/T. Values for pressure, specific

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Chapter 11 Thermodynamic Relations TABLE 11.3

Fundamental Equation Used to Construct the Steam Tablesa,b c 5 c 01T2 1 RT 3In r 1 rQ1r, t24

(1)

where 6

c 0 5 a C i / t i 21 1 C 7 In T 1 C 8 In T / t

(2)

i5 1

and 7

8

10

Q 5 1t 2 t c2 a 1t 2 t aj2 j 2 2 c a A ij 1r 2 r aj2i 2 1 1 e 2Er a A ij r i 2 9 d j51

i5 1

(3)

i5 9

In (1), (2), and (3), T denotes temperature on the Kelvin scale, t denotes 1000/T, r denotes density in g/cm3, R 5 4.6151 bar ? cm3/g ? K or 0.46151 J/g ? K, tc 5 1000/Tc 5 1.544912, E 5 4.8, and t aj 5 t c1j 5 12

raj 5 0.6341j 5 12

5 2.51j . 12

5 1.01j . 12

The coefficients for c 0 in J/g are given as follows: C1 5 1857.065

C4 5

36.6649

C2 5 3229.12

C5 5 220.5516

C3 5 2419.465

C6 5

C7 5

46.0

C8 5 21011.249

4.85233

Values for the coefficients Aij are listed in the original source.a a

J. H. Keenan, F. G. Keyes, P. G. Hill, and J. G. Moore, Steam Tables, Wiley, New York, 1969. Also see L. Haar, J. S. Gallagher, and G. S. Kell, NBS/NRC Steam Tables, Hemisphere, Washington, D.C., 1984. The properties of water are determined in this reference using a different functional form for the Helmholtz function than given by Eqs. (1)–(3). b

internal energy, and specific entropy can be determined by differentiation of Eq. 11.79. Values for the specific enthalpy and Gibbs function are found from h 5 u 1 py and g 5 c 1 py, respectively. The specific heat cy is evaluated by further differentiation, cy 5 10u/ 0T2y. With similar operations, other properties can be evaluated. Property values for water calculated from Eq. 11.79 are in excellent agreement with experimental data over a wide range of conditions.

ENERGY & ENVIRONMENT Due to the phase-out of chlorine-containing CFC refrigerants because of ozone layer and global climate change concerns, new substances and mixtures with no chlorine have been developed in recent years as possible alternatives (see Sec. 10.3). This has led to significant research efforts to provide the necessary thermodynamic property data for analysis and design. The National Institute of Standards and Technology (NIST) has led governmental efforts to provide accurate data. Specifically, data have been developed for high-accuracy p–y–T equations of state from which fundamental thermodynamic functions can be obtained. The equations are carefully validated using data for velocity of sound, Joule–Thomson coefficient, saturation pressure-temperature relations, and specific heats. Such data were used to calculate the property values in Tables A-7 to A-18 in the Appendix. NIST has developed REFPROP, a computer data base that is the current standard for refrigerant and refrigerant mixture properties.

Example 11.7 illustrates the use of a fundamental function to determine thermodynamic properties for computer evaluation and to develop tables.

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11.6 Constructing Tables of Thermodynamic Properties cccc

667

EXAMPLE 11.7 c

Determining Properties Using a Fundamental Function The following expression for the Helmholtz function has been used to determine the properties of water: c1r, T 2 5 c01T2 1 RT 3ln r 1 rQ1r, t24 where r denotes density and t denotes 1000/T. The functions c0 and Q are sums involving the indicated independent variables and a number of adjustable constants (see Table 11.3). Obtain expressions for (a) pressure, (b) specific entropy, and (c) specific internal energy resulting from this fundamental thermodynamic function. SOLUTION Known: An expression for the Helmholtz function c is given. Find: Determine the expressions for pressure, specific entropy, and specific internal energy that result from this

fundamental thermodynamic function. Analysis: The expressions developed below for p, s, and u require only the functions c0(T ) an Q(r, t). Once these

functions are determined, p, s, and u can each be determined as a function of density and temperature using elementary mathematical operations. (a) When expressed in terms of density instead of specific volume, Eq. 11.28 becomes

p 5 r2 a

0c b 0r T

as can easily be verified. When T is held constant t is also constant. Accordingly, the following is obtained on differentiation of the given function: a

0c 0Q 1 b 5 RT c 1 Q1r, t2 1 ra b d r 0r T 0r t

Combining these equations gives an expression for pressure p 5 rRT c 1 1 rQ 1 r2 a

0Q b d 0r t

(a)

(b) From Eq. 11.29

s 5 2a

0c b 0T r

Differentiation of the given expression for c yields a

0c dc0 0Q dt b 5 1 c R1ln r 1 rQ2 1 RTr a b d 0T r dT 0t r dT dc0 0Q 1000 5 1 c R1ln r 1 rQ2 1 RTr a b a2 2 b d dT 0t r T dc0 0Q 5 1 R c ln r 1 rQ 2 rta b d dT 0t r

Combining results gives s52

dc0 0Q 2 R c ln r 1 rQ 2 rta b d dT 0t r

(b)

(c) By definition, c 5 u 2 Ts. Thus, u 5 c 1 Ts. Introducing the given expression for c together with the expression for s from part (b) results in

u 5 3c0 1 RT 1ln r 1 rQ24 1 T e2 5 c0 2 T

dc0 0Q 1 RTrt a b dT 0t r

dc0 0Q 2 R c ln r 1 rQ 2 rta b d f dT 0t r

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Chapter 11 Thermodynamic Relations

This can be written more compactly by noting that T

dc0 dc0 dt dc0 dc0 1000 5T 5T a2 2 b 5 2t dT dt dT dt dt T

✓ Skills Developed

Thus,

Ability to… ❑ derive expressions for pres-

d1c0t2 dc0 dc0 c0 2 T 5 c0 1 t 5 dT dt dt

sure, specific entropy, and specific internal energy based on a fundamental thermodynamic function.

Finally, the expression for u becomes u5

d1c0t2 0Q 1 RTrt a b 0t r dt

(c)

Using results obtained, how can an expression be developed for h? Ans. h 5 u 1 p/r. Substitute Eq. (c) for u and Eq. (a) for p and collect terms.

11.7 TAKE NOTE...

Generalized compressibility charts are provided in Appendix figures A-1, A-2, and A-3. See Example 3.7 for an application.

Generalized Charts for Enthalpy and Entropy

Generalized charts giving the compressibility factor Z in terms of the reduced properties pR, TR, and y9R are introduced in Sec. 3.11. With such charts, estimates of p–y–T data can be obtained rapidly knowing only the critical pressure and critical temperature for the substance of interest. The objective of the present section is to introduce generalized charts that allow changes in enthalpy and entropy to be estimated.

Generalized Enthalpy Departure Chart The change in specific enthalpy of a gas (or liquid) between two states fixed by temperature and pressure can be evaluated using the identity h1T2, p22 2 h1T1, p12 5 3h*1T22 2 h*1T124 1 53h1T2, p22 2 h*1T224 2 3h1T1, p12 2 h*1T1246

(11.80)

The term [h(T, p) 2 h*(T)] denotes the specific enthalpy of the substance relative to that of its ideal gas model when both are at the same temperature. The superscript * is used in this section to identify ideal gas property values. Thus, Eq. 11.80 indicates that the change in specific enthalpy between the two states equals the enthalpy change determined using the ideal gas model plus a correction that accounts for the departure from ideal gas behavior. The correction is shown underlined in Eq. 11.80. The ideal gas term can be evaluated using methods introduced in Chap. 3. Next, we show how the correction term is evaluated in terms of the enthalpy departure.

DEVELOPING THE ENTHALPY DEPARTURE. The variation of enthalpy with pressure at fixed temperature is given by Eq. 11.56 as a

0y 0h b 5 y 2 Ta b 0p T 0T p

Integrating from pressure p9 to pressure p at fixed temperature T p

h1T, p2 2 h1T, p¿2 5

# cy 2 T a 0T b d dp p¿

0y

p

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11.7 Generalized Charts for Enthalpy and Entropy This equation is not altered fundamentally by adding and subtracting h*(T) on the left side. That is 3h1T, p2 2 h*1T24 2 3h1T, p¿2 2 h*1T24 5

p

# c y 2 T a 0T b d dp 0y

(11.81)

p

p¿

As pressure tends to zero at fixed temperature, the enthalpy of the substance approaches that of its ideal gas model. Accordingly, as p9 tends to zero lim 3h1T, p¿2 2 h*1T24 5 0

p¿S0

In this limit, the following expression is obtained from Eq. 11.81 for the specific enthalpy of a substance relative to that of its ideal gas model when both are at the same temperature: h1T, p2 2 h*1T2 5

p

# c y 2 T a 0T b d dp 0y

(11.82)

p

0

This also can be thought of as the change in enthalpy as the pressure is increased from zero to the given pressure while temperature is held constant. Using p–y–T data only, Eq. 11.82 can be evaluated at states 1 and 2 and thus the correction term of Eq. 11.80 evaluated. Let us consider next how this procedure can be conducted in terms of compressibility factor data and the reduced properties TR and pR. The integral of Eq. 11.82 can be expressed in terms of the compressibility factor Z and the reduced properties TR and pR as follows. Solving Z 5 py/RT gives y5

ZRT p

On differentiation a

0y RZ RT 0Z b 5 1 a b p p 0T p 0T p

With the previous two expressions, the integrand of Eq. 11.82 becomes y 2 Ta

RT 0Z 0y ZRT RZ RT 2 0Z 1 a b d 52 b 5 2Tc a b p p 0T p p 0T p p 0T p

(11.83)

Equation 11.83 can be written in terms of reduced properties as y 2 Ta

RTc T 2R 0Z 0y b 52 ? a b pc pR 0TR pR 0T p

Introducing this into Eq. 11.82 gives on rearrangement h*1T 2 2 h1T, p2 RTc

5 T 2R

#

pR

a

0

0Z dpR b 0TR pR pR enthalpy departure

Or, on a per mole basis, the enthalpy departure is h*1T 2 2 h1T, p2 5 T R2 RTc

#

0

pR

a

dpR 0Z b 0TR pR pR

(11.84)

The right side of Eq. 11.84 depends only on the reduced temperature TR and reduced pressure pR. Accordingly, the quantity 1h* 2 h2/ RTc, the enthalpy departure, is a function only of these two reduced properties. Using a generalized equation of state giving Z as a function of TR and pR, the enthalpy departure can readily be evaluated with a computer. Tabular representations are also found in the literature. Alternatively, the graphical representation provided in Fig. A-4 can be employed.

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Chapter 11 Thermodynamic Relations

EVALUATING ENTHALPY CHANGE. The change in specific enthalpy between two states can be evaluated by expressing Eq. 11.80 in terms of the enthalpy departure as h2 2 h1 5 h2* 2 h1* 2 RTc c a

h* 2 h h* 2 h b 2a b d RTc 2 RTc 1

(11.85)

The first underlined term in Eq. 11.85 represents the change in specific enthalpy between the two states assuming ideal gas behavior. The second underlined term is the correction that must be applied to the ideal gas value for the enthalpy change to obtain the actual value for the enthalpy change. Referring to the engineering literature, the quantity 1h* 2 h2/ RTc at states 1 and 2 can be calculated with an equation giving Z(TR, pR) or obtained from tables. This quantity also can be evaluated at state 1 from the generalized enthalpy departure chart, Fig. A-4, using the reduced temperature TR1 and reduced pressure pR1 corresponding to the temperature T1 and pressure p1 at the initial state, respectively. Similarly, 1h* 2 h2/ RTc at state 2 can be evaluated from Fig. A-4 using TR2 and pR2. The use of Eq. 11.85 is illustrated in the next example. cccc

EXAMPLE 11.8 c

Using the Generalized Enthalpy Departure Chart Nitrogen enters a turbine operating at steady state at 100 bar and 300 K and exits at 40 bar and 245 K. Using the enthalpy departure chart, determine the work developed, in kJ per kg of nitrogen flowing, if heat transfer with the surroundings can be ignored. Changes in kinetic and potential energy from inlet to exit also can be neglected. SOLUTION Known: A turbine operating at steady state has nitrogen entering at 100 bar and 300 K and exiting at 40 bar and 245 K. Find: Using the enthalpy departure chart, determine the work developed. Schematic and Given Data: T

Engineering Model:

1

· Wcv ––– m·

1. The control volume shown on the accom-

panying figure operates at steady state. 2. There is no significant heat transfer between

1 N2 p1 = 100 bar T1 = 300 K

2

the control volume and its surroundings. 3. Changes in kinetic and potential energy

p2 = 40 bar T2 = 245 K

between inlet and exit can be neglected.

2 2s s

4. Equilibrium property relations apply at the

inlet and exit.

Fig. E11.8 Analysis: The mass and energy rate balances reduce at steady state to give

# # Wcv Qcv V21 2 V22 0 5 # 2 # 1 c h1 2 h2 1 1 g1z1 2 z22 d 2 m m # where m is the mass flow rate. Dropping the heat transfer term by assumption 2 and the kinetic and potential energy terms by assumption 3 gives on rearrangement # Wcv # 5 h1 2 h2 m

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11.7 Generalized Charts for Enthalpy and Entropy

671

The term h1 2 h2 can be evaluated as follows: h1 2 h2 5

1 h* 2 h h* 2 h e h*1 2 h2* 2 RTc c a b 2a b df RTc 1 RTc 2 M

In this expression, M is the molecular weight of nitrogen and the other terms have the same significance as in Eq. 11.85. With specific enthalpy values from Table A-23 at T1 5 300 K and T2 5 245 K, respectively h*1 2 h2* 5 8723 2 7121 5 1602 kJ/ kmol The terms 1h* 2 h2/ RTc at states 1 and 2 required by the above expression for h1 2 h2 can be determined from Fig. A-4. First, the reduced temperature and reduced pressure at the inlet and exit must be determined. From Tables A-1, Tc 5 126 K, pc 5 33.9 bar. Thus, at the inlet TR1 5

300 100 5 2.38, pR1 5 5 2.95 126 33.9

TR2 5

245 40 5 1.94, pR2 5 5 1.18 126 33.9

At the exit

By inspection of Fig. A-4 ➊

a

h* 2 h h* 2 h b < 0.5, a b < 0.31 RTc 1 RTc 2

Substituting values # Wcv 1 kJ kJ c 1602 2 a8.314 b 1126 K210.5 2 0.312 d 5 50.1 kJ/ kg # 5 kg kmol kmol ?K m 28 kmol ➊ Due to inaccuracy in reading values from a graph such as Fig. A-4, we cannot expect extreme accuracy in the final calculated result.

Determine the work developed, in kJ per kg of nitrogen flowing, assuming the ideal gas model. Ans. 57.2 kJ/kg.

Generalized Entropy Departure Chart A generalized chart that allows changes in specific entropy to be evaluated can be developed in a similar manner to the generalized enthalpy departure chart introduced earlier in this section. The difference in specific entropy between states 1 and 2 of a gas (or liquid) can be expressed as the identity s1T2, p22 2 s1T1, p12 5 s*1T2, p22 2 s*1T1, p12 1 53s1T2, p22 2 s*1T2, p224 2 3s1T1, p12 2 s*1T1, p1246 (11.86) where 3s1T, p2 2 s*1T, p24 denotes the specific entropy of the substance relative to that of its ideal gas model when both are at the same temperature and pressure. Equation 11.86 indicates that the change in specific entropy between the two states equals the entropy change determined using the ideal gas model plus a correction (shown underlined) that accounts for the departure from ideal gas behavior. The ideal gas term can be evaluated using methods introduced in Sec. 6.5. Let us consider next how the correction term is evaluated in terms of the entropy departure.

✓ Skills Developed Ability to… ❑ use data from the generalized

enthalpy departure chart to calculate the change in enthalpy of nitrogen.

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Chapter 11 Thermodynamic Relations

DEVELOPING THE ENTROPY DEPARTURE. The following Maxwell relation gives the variation of entropy with pressure at fixed temperature: a

0s 0y b 5 2a b 0p T 0T p

(11.35)

Integrating from pressure p9 to pressure p at fixed temperature T gives p

s1T, p2 2 s1T, p¿2 5 2

# a 0T b 0y

p¿

dp

(11.87)

p

For an ideal gas, y 5 RT/p, so 10y/ 0T2p 5 R/ p. Using this in Eq. 11.87, the change in specific entropy assuming ideal gas behavior is s*1T, p2 2 s*1T, p¿2 5 2

#

p

p¿

R dp p

(11.88)

Subtracting Eq. 11.88 from Eq. 11.87 gives p

R

# c p 2 a 0T b d dp

3s1T, p2 2 s*1T, p24 2 3s1T, p¿2 2 s*1T, p¿24 5

0y

(11.89)

p

p¿

Since the properties of a substance tend to merge into those of its ideal gas model as pressure tends to zero at fixed temperature, we have lim 3s1T, p¿2 2 s*1T, p¿24 5 0

p¿S0

Thus, in the limit as p9 tends to zero, Eq. 11.89 becomes s1T, p2 2 s*1T, p2 5

p

R

# c p 2 a 0T b d dp 0y

entropy departure

(11.90)

p

0

Using p–y–T data only, Eq. 11.90 can be evaluated at states 1 and 2 and thus the correction term of Eq. 11.86 evaluated. Equation 11.90 can be expressed in terms of the compressibility factor Z and the reduced properties TR and pR. The result, on a per mole basis, is the entropy departure h*1T2 2 h1T, p2 s*1T, p2 2 s1T, p2 5 1 R RTRTc

#

pR

1Z 2 12

0

dpR pR

(11.91)

The right side of Eq. 11.91 depends only on the reduced temperature TR and reduced pressure pR. Accordingly, the quantity 1s* 2 s2/ R, the entropy departure, is a function only of these two reduced properties. As for the enthalpy departure, the entropy departure can be evaluated with a computer using a generalized equation of state giving Z as a function of TR and pR. Alternatively, tabular data from the literature or the graphical representation provided in Fig. A-5 can be employed.

EVALUATING ENTROPY CHANGE. The change in specific entropy between two states can be evaluated by expressing Eq. 11.86 in terms of the entropy departure as s2 2 s1 5 s *2 2 s1* 2 R c a

s* 2 s s* 2 s b 2a b d R R 2 1

(11.92)

The first underlined term in Eq. 11.92 represents the change in specific entropy between the two states assuming ideal gas behavior. The second underlined term is the correction that must be applied to the ideal gas value for entropy change to obtain the actual value for the entropy change. The quantity 1s* 2 s21/ R appearing in Eq. 11.92 can be evaluated from the generalized entropy departure chart, Fig. A-5, using the reduced temperature TR1 and reduced pressure pR1 corresponding to the temperature T1 and pressure p1 at the initial state, respectively. Similarly, 1s* 2 s22/ R

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11.7 Generalized Charts for Enthalpy and Entropy

673

can be evaluated from Fig. A-5 using TR2 and pR2. The use of Eq. 11.92 is illustrated in the next example.

cccc

EXAMPLE 11.9 c

Using the Generalized Entropy Departure Chart For the case of Example 11.8, determine (a) the rate of entropy production, in kJ/kg ? K, and (b) the isentropic turbine efficiency. SOLUTION Known: A turbine operating at steady state has nitrogen entering at 100 bar and 300 K and exiting at 40 bar and 245 K. Find: Determine the rate of entropy production, in kJ/kg ? K, and the isentropic turbine efficiency. Schematic and Given Data: See Fig. E11.8. Engineering Model: See Example 11.8. Analysis: (a) At steady state, the control volume form of the entropy rate equation reduces to give

# scv # 5 s2 2 s1 m The change in specific entropy required by this expression can be written as s2 2 s 1 5

1 s* 2 s s* 2 s e s2* 2 s1* 2 R c a b 2a b df M R R 2 1

where M is the molecular weight of nitrogen and the other terms have the same significance as in Eq. 11.92. The change in specific entropy s2* 2 s 1* can be evaluated using s2* 2 s1* 5 s81T22 2 s 81T12 2 R ln

p2 p1

With values from Table A-23 s2* 2 s*1 5 185.775 2 191.682 2 8.314 ln

40 kJ 5 1.711 100 kmol ? K

The terms 1s* 2 s2/ R at the inlet and exit can be determined from Fig. A-5. Using the reduced temperature and reduced pressure values calculated in the solution to Example 11.8, inspection of Fig. A-5 gives a

s* 2 s s* 2 s b < 0.21, a b < 0.14 R R 1 2

Substituting values # scv 1 kJ kJ c 1.711 2 8.314 10.14 2 0.212 d # 5 128 kg/ kmol2 kmol ? K kmol ? K m kJ 5 0.082 kg ? K (b) The isentropic turbine efficiency is defined in Sec. 6.12 as

# # 1Wcv/ m2 ht 5 # # 1Wcv/ m2s

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where the denominator is the work that would be developed by the turbine if the nitrogen expanded isentropically from the specified inlet state to the specified exit pressure. Thus, it is necessary to fix the state, call it 2s, at the turbine exit for an expansion in which there is no change in specific entropy from inlet to exit. With 1s2s 2 s12 5 0 and procedures similar to those used in part (a) s* 2 s s* 2 s b 2a b d R R 2s 1 p2 s* 2 s s* 2 s 0 5 c s 81T2s2 2 s 81T12 2 R ln a b d 2 R c a b 2a b d p1 R R 2s 1

0 5 s *2s 2 s1* 2 R c a

Using values from part (a), the last equation becomes 0 5 s 81T2s2 2 191.682 2 8.314 ln

40 s* 2 s 2 Ra b 1 1.746 100 R 2s

or s 81T2s2 2 R a

s* 2 s b 5 182.3 R 2s

The temperature T2s can be determined in an iterative procedure using s 8 data from Table A-23 and 1s* 2 s2/ R from Fig. A-5 as follows: First, a value for the temperature T2s is assumed. The corresponding value of s 8 can then be obtained from Table A-23. The reduced temperature (TR)2s 5 T2s/Tc, together with pR2 5 1.18, allows a value for 1s* 2 s2/ R to be obtained from Fig. A-5. The procedure continues until agreement with the value on the right side of the above equation is obtained. Using this procedure, T2s is found to be closely 228 K. With the temperature T2s known, the work that would be developed by the turbine if the nitrogen expanded isentropically from the specified inlet state to the specified exit pressure can be evaluated from # Wcv a # b 5 h1 2 h2s m s 5

1 h* 2 h h* 2 h *2 2 RTc c a e 1h1* 2 h2s b 2a b df M RTc 1 RTc 2s

* 5 6654 kJ/ kmol. From Fig. A-4 at pR2 5 1.18 and (TR)2s 5 From Table A-23, h2s 228/126 5 1.81 a

h* 2 h b < 0.36 RTc 2s

# # Values for the other terms in the expression for 1Wcv / m2s are obtained in the solution to Example 11.8. Finally # Wcv 1 a # b 5 38723 2 6654 2 18.31421126210.5 2 0.3624 5 68.66 kJ/ kg 28 m s With the work value from Example 11.8, the turbine efficiency is # # 1Wcv ym2 50.1 5 0.73173%2 ➊ ht 5 # # 5 68.66 1Wcv ym2s ➊ We cannot expect extreme accuracy when reading data from a generalized chart such as Fig. A-5, which affects the final calculated result.

Determine the rate of entropy production, in kJ/K per kg of nitrogen flowing, assuming the ideal gas model. Ans. 0.061 kJ/kg ? K.

✓ Skills Developed Ability to… ❑ use data from the general-

ized entropy departure chart to calculate the entropy production. ❑ use data from the generalized enthalpy and entropy departure charts to calculate isentropic turbine efficiency. ❑ use an iterative procedure to calculate the temperature at the end of an isentropic process using data from the generalized entropy departure chart.

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11.8 p–y–T Relations for Gas Mixtures

11.8

675

p–y–T Relations for Gas Mixtures

Many systems of interest involve mixtures of two or more components. The principles of thermodynamics introduced thus far are applicable to systems involving mixtures, but to apply such principles requires that mixture properties be evaluated. Since an unlimited variety of mixtures can be formed from a given set of pure components by varying the relative amounts present, the properties of mixtures are available in tabular, graphical, or equation forms only in particular cases such as air. Generally, special means are required for determining mixture properties. In this section, methods for evaluating the p–y–T relations for pure components introduced in previous sections of the book are adapted to obtain plausible estimates for gas mixtures. In Sec. 11.9 some general aspects of property evaluation for multicomponent systems are introduced. To evaluate the properties of a mixture requires knowledge of the composition. The composition can be described by giving the number of moles (kmol or lbmol) of each component present. The total number of moles, n, is the sum of the number of moles of each of the components

TAKE NOTE...

The special case of ideal gas mixtures is considered in Secs. 12.1–12.4, with applications to psychrometrics in the second part of Chap. 12 and reacting mixtures in Chaps. 13 and 14.

j

n 5 n1 1 n2 1 . . . 1 nj 5 a ni

(11.93)

i51

The relative amounts of the components present can be described in terms of mole fractions. The mole fraction yi of component i is defined as ni n

yi 5

(11.94)

Dividing each term of Eq. 11.93 by the total number of moles and using Eq. 11.94 j

1 5 a yi

(11.95)

i51

That is, the sum of the mole fractions of all components present is equal to unity. Most techniques for estimating mixture properties are empirical in character and are not derived from fundamental principles. The realm of validity of any particular technique can be established only by comparing predicted property values with empirical data. The brief discussion to follow is intended only to show how certain of the procedures for evaluating the p–y–T relations of pure components introduced previously can be extended to gas mixtures.

MIXTURE EQUATION OF STATE. One way the p–y–T relation for a gas mixture can be estimated is by applying to the overall mixture an equation of state such as introduced in Sec. 11.1. The constants appearing in the equation selected would be mixture values determined with empirical combining rules developed for the equation. For example, mixture values of the constants a and b for use in the van der Waals and Redlich–Kwong equations would be obtained using relations of the form j

2

j

a 5 a a yia1i /2 b , b 5 a a yibi b i51

i51

(11.96)

where ai and bi are the values of the constants for component i and yi is the mole fraction. Combination rules for obtaining mixture values for the constants in other equations of state also have been suggested.

KAY’S RULE. The principle of corresponding states method for single components introduced in Sec. 3.11.3 can be extended to mixtures by regarding the mixture as if it were a single pure component having critical properties calculated by one of several mixture rules. Perhaps the simplest of these, requiring only the determination of a mole fraction averaged critical temperature Tc and critical pressure pc, is Kay’s rule

Kay’s rule

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Chapter 11 Thermodynamic Relations j

j

Tc 5 a yiTc,i, pc 5 a yi pc,i i51

(11.97)

i51

where Tc,i, pc,i, and yi are the critical temperature, critical pressure, and mole fraction of component i, respectively. Using Tc and pc, the mixture compressibility factor Z is obtained as for a single pure component. The unknown quantity from among the pressure p, volume V, temperature T, and total number of moles n of the gas mixture can then be obtained by solving Z5

pV nRT

(11.98)

Mixture values for Tc and pc also can be used to enter the generalized enthalpy departure and entropy departure charts introduced in Sec. 11.7.

additive pressure rule

ADDITIVE PRESSURE RULE. Additional means for estimating p–y–T relations for mixtures are provided by empirical mixture rules, of which several are found in the engineering literature. Among these are the additive pressure and additive volume rules. According to the additive pressure rule, the pressure of a gas mixture occupying volume V at temperature T is expressible as a sum of pressures exerted by the individual components: p 5 p1 1 p2 1 p3 1 . . .4 T,V

(11.99a)

where the pressures p1, p2, etc. are evaluated by considering the respective components to be at the volume and temperature of the mixture. These pressures would be determined using tabular or graphical p–y–T data or a suitable equation of state. An alternative expression of the additive pressure rule in terms of compressibility factors can be obtained. Since component i is considered to be at the volume and temperature of the mixture, the compressibility factor Zi for this component is Zi 5 piV/ niRT, so the pressure pi is pi 5

ZiniRT V

p5

ZnRT V

Similarly, for the mixture

Substituting these expressions into Eq. 11.99a and reducing gives the following relationship between the compressibility factors for the mixture Z and the mixture components Zi j

Z 5 a yiZi4T,V

(11.99b)

i51

The compressibility factors Zi are determined assuming that component i occupies the entire volume of the mixture at the temperature T. additive volume rule

ADDITIVE VOLUME RULE. The underlying assumption of the

additive volume rule is that the volume V of a gas mixture at temperature T and pressure p is express-

ible as the sum of volumes occupied by the individual components: V 5 V1 1 V2 1 V3 1 . . .4 p,T

(11.100a)

where the volumes V1, V2, etc. are evaluated by considering the respective components to be at the pressure and temperature of the mixture. These volumes would be determined from tabular or graphical p–y–T data or a suitable equation of state.

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11.8 p–y–T Relations for Gas Mixtures

677

An alternative expression of the additive volume rule in terms of compressibility factors can be obtained. Since component i is considered to be at the pressure and temperature of the mixture, the compressibility factor Zi for this component is Zi 5 pVi/ niRT, so the volume Vi is Vi 5

ZiniRT p

V5

ZnRT p

Similarly, for the mixture

Substituting these expressions into Eq. 11.100a and reducing gives j

Z 5 a yiZi4p,T

(11.100b)

i51

The compressibility factors Zi are determined assuming that component i exists at the temperature T and pressure p of the mixture. The next example illustrates alternative means for estimating the pressure of a gas mixture.

cccc

EXAMPLE 11.10 c

Estimating Mixture Pressure by Alternative Means A mixture consisting of 0.18 kmol of methane (CH4) and 0.274 kmol of butane (C4H10) occupies a volume of 0.241 m3 at a temperature of 2388C. The experimental value for the pressure is 68.9 bar. Calculate the pressure, in bar, exerted by the mixture by using (a) the ideal gas equation of state, (b) Kay’s rule together with the generalized compressibility chart, (c) the van der Waals equation, and (d) the rule of additive pressures employing the generalized compressibility chart. Compare the calculated values with the known experimental value. SOLUTION Known: A mixture of two specified hydrocarbons with known molar amounts occupies a known volume at a specified temperature. Find: Determine the pressure, in bar, using four alternative methods, and compare the results with the experi-

mental value. Schematic and Given Data: Engineering Model: As shown in the accompanying figure, the system is the T = 238°C

mixture. p=?

0.18 kmol CH4 0.274 kmol C4H10 V = 0.241 m3

Fig. E11.10 Analysis: The total number of moles of mixture n is

n 5 0.18 1 0.274 5 0.454 kmol

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Thus, the mole fractions of the methane and butane are, respectively y1 5 0.396 and y2 5 0.604 The specific volume of the mixture on a molar basis is y5

0.241 m3 m3 5 0.531 10.18 1 0.2742 kmol kmol

(a) Substituting values into the ideal gas equation of state

18314 N ? m/ kmol ? K21511 K2 1 bar RT ` 5 5 ` y 10.531 m3/ kmol2 10 N/ m2 5 80.01 bar

p5

(b) To apply Kay’s rule, the critical temperature and pressure for each component are required. From Table A-1, for methane

Tc1 5 191 K, pc1 5 46.4 bar and for butane Tc2 5 425 K, pc2 5 38.0 bar Thus, with Eqs. 11.97 Tc 5 y1Tc1 1 y2Tc2 5 10.396211912 1 10.604214252 5 332.3 K pc 5 y1pc1 1 y2 pc2 5 10.3962146.42 1 10.6042138.02 5 41.33 bar Treating the mixture as a pure component having the above values for the critical temperature and pressure, the following reduced properties are determined for the mixture: T 511 5 5 1.54 Tc 332.3 10.5312141.332Z105Z ypc y¿R 5 5 RTc 1831421332.32 5 0.794 TR 5

Turning to Fig. A-2, Z < 0.88. The mixture pressure is then found from 18314215112 ZnRT RT 5Z 5 0.88 V y 10.5312 Z105Z 5 70.4 bar

p5

(c) Mixture values for the van der Waals constants can be obtained using Eqs. 11.96. This requires values of the van der Waals constants for each of the two mixture components. Table A-24 gives the following values for methane:

a1 5 2.293 bar a

m3 2 m3 b , b1 5 0.0428 kmol kmol

Similarly, from Table A-24 for butane a2 5 13.86 bar a

m3 2 m3 b , b2 5 0.1162 kmol kmol

Then, the first of Eqs. 11.96 gives a mixture value for the constant a as a 5 1y1a11/2 1 y2a12/222 5 30.39612.29321/2 1 0.604113.8621/242 m3 2 5 8.113 bar a b kmol Substituting into the second of Eqs. 11.96 gives a mixture value for the constant b b 5 y1b1 1 y2b2 5 10.396210.04282 1 10.604210.11622 m3 5 0.087 kmol

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679

Inserting the mixture values for a and b into the van der Waals equation together with known data p5 5

RT a 2 2 y2b y 18314 N ? m/ kmol ? K21511 K2

10.531 2 0.08721m3/ kmol2 5 66.91 bar

`

8.113 bar 1m3/ kmol22 1 bar ` 2 10.531 m3/ kmol22 105 N/ m2

(d) To apply the additive pressure rule with the generalized compressibility chart requires that the compress-

ibility factor for each component be determined assuming that the component occupies the entire volume at the mixture temperature. With this assumption, the following reduced properties are obtained for methane T 511 5 5 2.69 Tc1 191 10.241 m3/ 0.18 kmol2146.4 bar2 105 N/ m2 y1pc1 y¿R1 5 5 ` ` 5 3.91 RTc1 18314 N ? m/ kmol ? K21191 K2 1 bar

TR1 5

With these reduced properties, Fig. A-2 gives Z1 < 1.0. Similarly, for butane T 511 5 5 1.2 Tc2 425 10.8821382Z105Z y2 pc2 5 5 0.95 y¿R2 5 RTc2 18314214252

TR2 5

From Fig. A-2, Z2 < 0.8. The compressibility factor for the mixture determined from Eq. 11.99b is Z 5 y1 Z1 1 y2 Z2 5 10.396211.02 1 10.604210.82 5 0.88. Accordingly, the same value for pressure as determined in part (b) using Kay’s rule results: p 5 70.4 bar. In this particular example, the ideal gas equation of state gives a value for pressure that exceeds the experimental value by nearly 16%. Kay’s rule and the rule of additive pressures give pressure values about 3% greater than the experimental value. The van der Waals equation with mixture values for the constants gives a pressure value about 3% less than the experimental value.

✓ Skills Developed Ability to… ❑ calculate the pressure of

a gas mixture using four alternative methods.

Convert the mixture analysis from a molar basis to a mass fraction basis. Ans. Methane: 0.153, Butane: 0.847.

11.9

Analyzing Multicomponent Systems

In the preceding section we considered means for evaluating the p–y–T relation of gas mixtures by extending methods developed for pure components. The current section is devoted to the development of some general aspects of the properties of systems with two or more components. Primary emphasis is on the case of gas mixtures, but the methods developed also apply to solutions. When liquids and solids are under consideration, the term solution is sometimes used in place of mixture. The present discussion is limited to nonreacting mixtures or solutions in a single phase. The effects of chemical reactions and equilibrium between different phases are taken up in Chaps. 13 and 14. To describe multicomponent systems, composition must be included in our thermodynamic relations. This leads to the definition and development of several new concepts, including the partial molal property, the chemical potential, and the fugacity.

TAKE NOTE...

Section 11.9 may be deferred until Secs. 12.1–12.4 have been studied.

solution

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Chapter 11 Thermodynamic Relations

11.9.1

Partial Molal Properties

In the present discussion we introduce the concept of a partial molal property and illustrate its use. This concept plays an important role in subsequent discussions of multicomponent systems.

DEFINING PARTIAL MOLAL PROPERTIES. Any extensive thermodynamic property X of a single-phase, single-component system is a function of two independent intensive properties and the size of the system. Selecting temperature and pressure as the independent properties and the number of moles n as the measure of size, we have X 5 X(T, p, n). For a single-phase, multicomponent system, the extensive property X must then be a function of temperature, pressure, and the number of moles of each component present, X 5 X(T, p, n1, n2, . . . , nj). If each mole number is increased by a factor a, the size of the system increases by the same factor, and so does the value of the extensive property X. That is aX1T, p, n1, n2, . . . , nj2 5 X1T, p, an1, an2, . . . , anj2 Differentiating with respect to a while holding temperature, pressure, and the mole numbers fixed and using the chain rule on the right side gives X5

0X 0X 0X n1 1 n2 1 . . . 1 nj 01an12 01an22 01anj2

This equation holds for all values of a. In particular, it holds for a 5 1. Setting a 5 1 j 0X X 5 a ni b 0n i T, p, nl i51

partial molal property

(11.101)

where the subscript nl denotes that all n’s except ni are held fixed during differentiation. The partial molal property Xi is by definition Xi 5

0X b 0ni T, p, nl

(11.102)

The partial molal property Xi is a property of the mixture and not simply a property of component i, for Xi depends in general on temperature, pressure, and mixture composition: Xi1T, p, n1, n2, . . . , nj2. Partial molal properties are intensive properties of the mixture. Introducing Eq. 11.102, Eq. 11.101 becomes j

X 5 a ni Xi

(11.103)

i51

This equation shows that the extensive property X can be expressed as a weighted sum of the partial molal properties Xi. Selecting the extensive property X in Eq. 11.103 to be volume, internal energy, enthalpy, and entropy, respectively, gives j

j

j

V 5 a ni Vi, U 5 a ni Ui, H 5 a ni Hi, i51

i51

i51

j

S 5 a ni Si

(11.104)

i51

where Vi, Ui, Hi, Si denote the partial molal volume, internal energy, enthalpy, and entropy. Similar expressions can be written for the Gibbs function G and the Helmholtz function C. Moreover, the relations between these extensive properties: H 5 U 1 pV, G 5 H 2 TS, C = U 2 TS can be differentiated with respect to ni while holding temperature, pressure, and the remaining n’s constant to produce corresponding relations

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11.9 Analyzing Multicomponent Systems among partial molal properties: Hi 5 Ui 1 pVi, Gi 5 Hi 2 T Si, °i 5 Ui 2 T Si , where Gi and °i are the partial molal Gibbs function and Helmholtz function, respectively. Several additional relations involving partial molal properties are developed later in this section.

EVALUATING PARTIAL MOLAL PROPERTIES. Partial molal properties can be evaluated by several methods, including the following: c If the property X can be measured, Xi can be found by extrapolating a plot giving

1¢X /¢ni2T, p, nl versus Dni. That is Xi 5 a

0X ¢X b 5 lim a b 0ni T, p, nl ¢ni S0 ¢ni T, p, nl

c If an expression for X in terms of its independent variables is known, Xi can be

evaluated by differentiation. The derivative can be determined analytically if the function is expressed analytically or found numerically if the function is in tabular form. c When suitable data are available, a simple graphical procedure known as the method of intercepts can be used to evaluate partial molal properties. In principle, the method can be applied for any extensive property. To introduce this method, let us consider the volume of a system consisting of two components, A and B. For this system, Eq. 11.103 takes the form V 5 nAVA 1 nBVB where VA and VB are the partial molal volumes of A and B, respectively. Dividing by the number of moles of mixture n V 5 yAVA 1 yBVB n where yA and yB denote the mole fractions of A and B, respectively. Since yA 1 yB 5 1, this becomes V 5 11 2 yB2VA 1 yBVB 5 VA 1 yB1VB 2 VA2 n This equation provides the basis for the method of intercepts. For example, refer to Fig. 11.5, in which V/n is plotted as a function of yB at constant T and p. At a specified value for yB, a tangent to the curve is shown on the figure. When extrapolated, the tangent line intersects the axis on the left at VA and the axis on the right at VB. These values for the partial molal volumes correspond to the particular specifications for T, p, and yB. At fixed temperature and pressure, VA and VB vary with yB and are not equal to the molar specific volumes of pure A and pure B, denoted on the figure as yA and yB , respectively. The values of yA and yB are fixed by temperature and pressure only. V T and p constant –– n V as a function of y –– B n

vB(T, p) VB(T, p, yB)

vA(T, p)

Tangent line

VA(T, p, yB)

0 (pure A)

yB Mole fraction of B

Fig. 11.5 Illustration of the evaluation of 1.0 (pure B)

partial molal volumes by the method of intercepts.

method of intercepts

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EXTENSIVE PROPERTY CHANGES ON MIXING. Let us conclude the present discussion by evaluating the change in volume on mixing of pure components at the same temperature and pressure, a result for which an application is given in the discussion of Eq. 11.135. The total volume of the pure components before mixing is j

Vcomponents 5 a ni yi i51

where yi is the molar specific volume of pure component i. The volume of the mixture is j

Vmixture 5 a ni Vi i51

where Vi is the partial molal volume of component i in the mixture. The volume change on mixing is j

j

¢Vmixing 5 Vmixture 2 Vcomponents 5 a ni Vi 2 a ni yi i51

i51

or j

¢Vmixing 5 a ni1Vi 2 yi2

(11.105)

i51

Similar results can be obtained for other extensive properties, for example, j

¢Umixing 5 a ni1Ui 2 ui2 i51 j

¢Hmixing 5 a ni 1Hi 2 hi2 i51 j

(11.106)

¢Smixing 5 a ni 1Si 2 si2 i51

In Eqs. 11.106, ui, hi, and si denote the molar internal energy, enthalpy, and entropy of pure component i, respectively. The symbols Ui, Hi, and Si denote the respective partial molal properties.

11.9.2 Chemical Potential

chemical potential

Of the partial molal properties, the partial molal Gibbs function is particularly useful in describing the behavior of mixtures and solutions. This quantity plays a central role in the criteria for both chemical and phase equilibrium (Chap. 14). Because of its importance in the study of multicomponent systems, the partial molal Gibbs function of component i is given a special name and symbol. It is called the chemical potential of component i and symbolized by mi mi 5 Gi 5

0G b 0ni T, p, nl

(11.107)

Like temperature and pressure, the chemical potential mi is an intensive property. Applying Eq. 11.103 together with Eq. 11.107, the following expression can be written: j

G 5 a ni mi i51

(11.108)

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11.9 Analyzing Multicomponent Systems Expressions for the internal energy, enthalpy, and Helmholtz function can be obtained from Eq. 11.108, using the definitions H 5 U 1 pV, G 5 H 2 TS, and C 5 U 2 TS. They are j

U 5 TS 2 pV 1 a ni mi i51

j

H 5 TS 1 a ni mi

(11.109)

i51

j

° 5 2pV 1 a ni mi i51

Other useful relations can be obtained as well. Forming the differential of G(T, p, n1, n2, . . . , nj) dG 5

j 0G 0G 0G b dp 1 b dT 1 a a b dni 0p T, n 0T p, n i51 0ni T, p, nl

(11.110)

The subscripts n in the first two terms indicate that all n’s are held fixed during differentiation. Since this implies fixed composition, it follows from Eqs. 11.30 and 11.31 (Sec. 11.3.2) that V5a

0G 0G b and 2S 5 a b 0p T, n 0T p, n

(11.111)

With Eqs. 11.107 and 11.111, Eq. 11.110 becomes j

dG 5 V dp 2 S dT 1 a mi dni

(11.112)

i51

which for a multicomponent system is the counterpart of Eq. 11.23. Another expression for dG is obtained by forming the differential of Eq. 11.108. That is j

j

dG 5 a ni dmi 1 a mi dni i51

i51

Gibbs–Duhem equation

Combining this equation with Eq. 11.112 gives the Gibbs–Duhem equation j

a ni dmi 5 V dp 2 S dT

(11.113)

i51

11.9.3 Fundamental Thermodynamic Functions for Multicomponent Systems A fundamental thermodynamic function provides a complete description of the thermodynamic state of a system. In principle, all properties of interest can be determined from such a function by differentiation and/or combination. Reviewing the developments of Sec. 11.9.2, we see that a function G(T, p, n1, n2, . . . , nj) is a fundamental thermodynamic function for a multicomponent system. Functions of the form U(S, V, n1, n2, . . . , nj), H(S, p, n1, n2, . . . , nj), and C(T, V, n1, n2, . . . , nj) also can serve as fundamental thermodynamic functions for multicomponent systems. To demonstrate this, first form the differential of each of Eqs. 11.109 and use the Gibbs–Duhem equation, Eq. 11.113, to reduce the resultant expressions to obtain j

dU 5 T dS 2 p dV 1 a mi dni

(11.114a)

i51 j

dH 5 T dS 1 V dp 1 a mi dni

(11.114b)

i51

j

d° 5 2p dV 2 S dT 1 a mi dni i51

(11.114c)

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Chapter 11 Thermodynamic Relations For multicomponent systems, these are the counterparts of Eqs. 11.18, 11.19, and 11.22, respectively. The differential of U(S, V, n1, n2, . . . , nj) is dU 5

j 0U 0U 0U b dS 1 b dV 1 a a b dni 0S V, n 0V S, n i51 0ni S,V, nl

Comparing this expression term by term with Eq. 11.114a, we have 0U 0U 0U b , 2p 5 b , mi 5 b 0S V, n 0V S, n 0ni S,V, nl

T5

(11.115a)

That is, the temperature, pressure, and chemical potentials can be obtained by differentiation of U(S, V, n1, n2, . . . , nj). The first two of Eqs. 11.115a are the counterparts of Eqs. 11.24 and 11.25. A similar procedure using a function of the form H(S, p, n1, n2, . . . , nj) together with Eq. 11.114b gives T5

0H 0H 0H b , V 5 b , mi 5 b 0S p, n 0p S, n 0ni S, p, nl

(11.115b)

where the first two of these are the counterparts of Eqs. 11.26 and 11.27. Finally, with C(S, V, n1, n2, . . . , nj) and Eq. 11.114c 2p 5

0° 0° 0° b , 2S 5 b , mi 5 b 0V T, n 0T V, n 0ni T,V, nl

(11.115c)

The first two of these are the counterparts of Eqs. 11.28 and 11.29. With each choice of fundamental function, the remaining extensive properties can be found by combination using the definitions H 5 U 1 pV, G 5 H 2 TS, C = U 2 TS. The foregoing discussion of fundamental thermodynamic functions has led to several property relations for multicomponent systems that correspond to relations obtained previously. In addition, counterparts of the Maxwell relations can be obtained by equating mixed second partial derivatives. For example, the first two terms on the right of Eq. 11.112 give 0V 0S b 52 b 0T p, n 0p T, n

(11.116)

which corresponds to Eq. 11.35. Numerous relationships involving chemical potentials can be derived similarly by equating mixed second partial derivatives. An important example from Eq. 11.112 is 0mi 0V b 5 b 0p T, n 0ni T, p, nl Recognizing the right side of this equation as the partial molal volume, we have 0mi b 5 Vi 0p T, n

(11.117)

This relationship is applied in the development of Eqs. 11.126. The present discussion concludes by listing four different expressions derived above for the chemical potential in terms of other properties. In the order obtained, they are mi 5

0G 0U 0H 0° b 5 b 5 b 5 b 0ni T, p, nl 0ni S, V, nl 0ni S, p, nl 0ni T, V, nl

(11.118)

Only the first of these partial derivatives is a partial molal property, however, for the term partial molal applies only to partial derivatives where the independent

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11.9 Analyzing Multicomponent Systems variables are temperature, pressure, and the number of moles of each component present.

11.9.4 Fugacity The chemical potential plays an important role in describing multicomponent systems. In some instances, however, it is more convenient to work in terms of a related property, the fugacity. The fugacity is introduced in the present discussion.

Single-Component Systems Let us begin by taking up the case of a system consisting of a single component. For this case, Eq. 11.108 reduces to give G 5 nm or m 5

G 5g n

That is, for a pure component the chemical potential equals the Gibbs function per mole. With this equation, Eq. 11.30 written on a per mole basis becomes 0m b 5y 0p T

(11.119)

For the special case of an ideal gas, y 5 RT/ p, and Eq. 11.119 assumes the form 0m* RT b 5 p 0p T where the asterisk denotes ideal gas. Integrating at constant temperature m* 5 RT ln p 1 C 1T2

(11.120)

where C(T ) is a function of integration. Since the pressure p can take on values from zero to plus infinity, the ln p term of this expression, and thus the chemical potential, has an inconvenient range of values from minus infinity to plus infinity. Equation 11.120 also shows that the chemical potential can be determined only to within an arbitrary constant.

INTRODUCING FUGACITY. Because of the above considerations, it is advantageous for many types of thermodynamic analyses to use fugacity in place of the chemical potential, for it is a well-behaved function that can be more conveniently evaluated. We introduce the fugacity f by the expression m 5 RT ln f 1 C1T2

(11.121)

Comparing Eq. 11.121 with Eq. 11.120, the fugacity is seen to play the same role in the general case as pressure plays in the ideal gas case. Fugacity has the same units as pressure. Substituting Eq. 11.121 into Eq. 11.119 gives RT a

0 ln f b 5y 0p T

(11.122)

Integration of Eq. 11.122 while holding temperature constant can determine the fugacity only to within a constant term. However, since ideal gas behavior is approached as pressure tends to zero, the constant term can be fixed by requiring that the fugacity of a pure component equals the pressure in the limit of zero pressure. That is f 51 p pS 0 lim

(11.123)

Equations 11.122 and 11.123 then completely determine the fugacity function.

fugacity

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Chapter 11 Thermodynamic Relations

EVALUATING FUGACITY. Let us consider next how the fugacity can be evaluated. With Z 5 py/ RT , Eq. 11.122 becomes RT a

0 ln f RT Z b 5 p 0p T

or a

0 ln f Z b 5 p 0p T

Subtracting 1/p from both sides and integrating from pressure p9 to pressure p at fixed temperature T p

# 1Z 2 12d ln p

3ln f 2 ln p4pp¿ 5

p¿

or cln

f p d 5 p p¿

p

# 1Z 2 12d ln p p¿

Taking the limit as p9 tends to zero and applying Eq. 11.123 results in p f ln 5 1Z 2 12d ln p p 0 When expressed in terms of the reduced pressure, pR 5 p/pc, the above equation is

#

ln

f 5 p

#

pR

1Z 2 12d ln pR

(11.124)

0

Since the compressibility factor Z depends on the reduced temperature TR and reduced pressure pR, it follows that the right side of Eq. 11.124 depends on these properties only. Accordingly, the quantity ln f/p is a function only of these two reduced properties. Using a generalized equation of state giving Z as a function of TR and pR, ln f /p can readily be evaluated with a computer. Tabular representations are also found in the literature. Alternatively, the graphical representation presented in Fig. A-6 can be employed. to illustrate the use of Fig. A-6, consider two states of water vapor at the same temperature, 4008C. At state 1 the pressure is 200 bar, and at state 2 the pressure is 240 bar. The change in the chemical potential between these states can be determined using Eq. 11.121 as m2 2 m1 5 RT ln

f2 f2 p2 p1 5 RT ln a b p2 p1 f1 f1

Using the critical temperature and pressure of water from Table A-1, at state 1 pR1 5 0.91, TR1 5 1.04, and at state 2 pR2 5 1.09, TR2 5 1.04. By inspection of Fig. A-6, f1/p1 5 0.755 and f2/p2 5 0.7. Inserting values in the above equation m2 2 m1 5 18.31421673.152 ln c 10.72a

240 1 ba b d 5 597 kJ/ kmol 200 0.755

For a pure component, the chemical potential equals the Gibbs function per mole, g 5 h 2 T s. Since the temperature is the same at states 1 and 2, the change in the chemical potential can be expressed as m2 2 m1 5 h2 2 h1 2 T1s2 2 s12. Using steam table data, the value obtained with this expression is 597 kJ/kmol, which agrees with the value determined from the generalized fugacity coefficient chart. b b b b b

Multicomponent Systems The fugacity of a component i in a mixture can be defined by a procedure that parallels the definition for a pure component. For a pure component, the development

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11.9 Analyzing Multicomponent Systems begins with Eq. 11.119, and the fugacity is introduced by Eq. 11.121. These are then used to write the pair of equations, Eqs. 11.122 and 11.123, from which the fugacity can be evaluated. For a mixture, the development begins with Eq. 11.117, the counterpart of Eq. 11.119, and the fugacity fi of component i is introduced by mi 5 RT ln fi 1 Ci 1T2

(11.125)

which parallels Eq. 11.121. The pair of equations that allow the fugacity of a mixture component, f i , to be evaluated are

RT a

0 ln fi b 5 Vi 0p T, n

lim a

pS 0

fi b 51 yi p

(11.126a) (11.126b)

The symbol fi denotes the fugacity of component i in the mixture and should be carefully distinguished in the presentation to follow from fi, which denotes the fugacity of pure i.

DISCUSSION. Referring to Eq. 11.126b, note that in the ideal gas limit the fugacity f i is not required to equal the pressure p as for the case of a pure component, but to equal the quantity yi p. To see that this is the appropriate limiting quantity, consider a system consisting of a mixture of gases occupying a volume V at pressure p and temperature T. If the overall mixture behaves as an ideal gas, we can write p5

nRT V

(11.127)

where n is the total number of moles of mixture. Recalling from Sec. 3.12.3 that an ideal gas can be regarded as composed of molecules that exert negligible forces on one another and whose volume is negligible relative to the total volume, we can think of each component i as behaving as if it were an ideal gas alone at the temperature T and volume V. Thus, the pressure exerted by component i would not be the mixture pressure p but the pressure pi given by pi 5

ni RT V

(11.128)

where ni is the number of moles of component i. Dividing Eq. 11.128 by Eq. 11.127 pi ni RT/ V ni 5 5 5 yi p n nRT/ V On rearrangement pi 5 yi p

(11.129)

Accordingly, the quantity yi p appearing in Eq. 11.126b corresponds to the pressure pi. Summing both sides of Eq. 11.129, we obtain j

j

j

a pi 5 a yi p 5 p a yi

i51

i51

i51

Or, since the sum of the mole fractions equals unity j

p 5 a pi

(11.130)

i51

In words, Eq. 11.130 states that the sum of the pressures pi equals the mixture pressure. This gives rise to the designation partial pressure for pi. With this background, we now see that Eq. 11.126b requires the fugacity of component i to approach the partial pressure of component i as pressure p tends to zero. Comparing Eqs. 11.130

fugacity of a mixture component

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Chapter 11 Thermodynamic Relations and 11.99a, we also see that the additive pressure rule is exact for ideal gas mixtures. This special case is considered further in Sec. 12.2 under the heading Dalton model.

EVALUATING FUGACITY IN A MIXTURE. Let us consider next how the fugacity of component i in a mixture can be expressed in terms of quantities that can be evaluated. For a pure component i, Eq. 11.122 gives RT a

0 ln fi b 5 yi 0p T

(11.131)

where yi is the molar specific volume of pure i. Subtracting Eq. 11.131 from Eq. 11.126a RT c

0 ln 1 fi / fi2 0p

d

T, n

5 Vi 2 yi

(11.132)

Integrating from pressure p9 to pressure p at fixed temperature and mixture composition fi p RT c lna b d 5 fi p¿

p

# 1V 2 y 2 dp i

i

p¿

In the limit as p9 tends to zero, this becomes fi fi RT c ln a b 2 lim ln a b d 5 fi p¿S0 fi

p

# 1V 2 y 2 dp i

i

0

Since fi S p¿ and fi S yi p¿ as pressure p9 tends to zero yi p¿ fi lim ln a b S ln a b 5 ln yi p¿S0 fi p¿ Therefore, we can write fi RT cln a b 2 ln yi d 5 fi

p

# 1V 2 y 2 dp i

i

0

or RT ln a

fi b5 yi fi

p

# 1V 2 y 2 dp i

i

(11.133)

0

in which fi is the fugacity of component i at pressure p in a mixture of given composition at a given temperature, and fi is the fugacity of pure i at the same temperature and pressure. Equation 11.133 expresses the relation between fi and fi in terms of the difference between Vi and yi , a measurable quantity.

11.9.5 Ideal Solution ideal solution

The task of evaluating the fugacities of the components in a mixture is considerably simplified when the mixture can be modeled as an ideal solution. An ideal solution is a mixture for which f i 5 yi fi 1ideal solution2

Lewis–Randall rule

(11.134)

Equation 11.134, known as the Lewis–Randall rule, states that the fugacity of each component in an ideal solution is equal to the product of its mole fraction and the fugacity of the pure component at the same temperature, pressure, and state of aggregation (gas, liquid, or solid) as the mixture. Many gaseous mixtures at low to moderate pressures are adequately modeled by the Lewis–Randall rule. The ideal gas mixtures

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11.9 Analyzing Multicomponent Systems considered in Chap. 12 are an important special class of such mixtures. Some liquid solutions also can be modeled with the Lewis–Randall rule. As consequences of the definition of an ideal solution, the following characteristics are exhibited: c Introducing Eq. 11.134 into Eq. 11.132, the left side vanishes, giving Vi 2 yi 5 0, or

Vi 5 yi

(11.135)

Thus, the partial molal volume of each component in an ideal solution is equal to the molar specific volume of the corresponding pure component at the same temperature and pressure. When Eq. 11.135 is introduced in Eq. 11.105, it can be concluded that there is no volume change on mixing pure components to form an ideal solution. With Eq. 11.135, the volume of an ideal solution is j

j

j

V 5 a ni Vi 5 a ni yi 5 a Vi 1ideal solution2 i51

i51

(11.136)

i51

where Vi is the volume that pure component i would occupy when at the temperature and pressure of the mixture. Comparing Eqs. 11.136 and 11.100a, the additive volume rule is seen to be exact for ideal solutions. c It also can be shown that the partial molal internal energy of each component in an ideal solution is equal to the molar internal energy of the corresponding pure component at the same temperature and pressure. A similar result applies for enthalpy. In symbols Ui 5 ui, Hi 5 hi

(11.137)

With these expressions, it can be concluded from Eqs. 11.106 that there is no change in internal energy or enthalpy on mixing pure components to form an ideal solution. With Eqs. 11.137, the internal energy and enthalpy of an ideal solution are j

j

U 5 a niui and H 5 a ni hi 1ideal solution2 i51

(11.138)

i51

where ui and hi denote, respectively, the molar internal energy and enthalpy of pure component i at the temperature and pressure of the mixture. Although there is no change in V, U, or H on mixing pure components to form an ideal solution, we expect an entropy increase to result from the adiabatic mixing of different pure components because such a process is irreversible: The separation of the mixture into the pure components would never occur spontaneously. The entropy change on adiabatic mixing is considered further for the special case of ideal gas mixtures in Sec. 12.4.2. The Lewis–Randall rule requires that the fugacity of mixture component i be evaluated in terms of the fugacity of pure component i at the same temperature and pressure as the mixture and in the same state of aggregation. For example, if the mixture were a gas at T, p, then fi would be determined for pure i at T, p and as a gas. However, at certain temperatures and pressures of interest a component of a gaseous mixture may, as a pure substance, be a liquid or solid. An example is an air–water vapor mixture at 208C (688F) and 1 atm. At this temperature and pressure, water exists not as a vapor but as a liquid. Although not considered here, means have been developed that allow the ideal solution model to be useful in such cases.

11.9.6

Chemical Potential for Ideal Solutions

The discussion of multicomponent systems concludes with the introduction of expressions for evaluating the chemical potential for ideal solutions used in Sec. 14.3.3.

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Chapter 11 Thermodynamic Relations Consider a reference state where component i of a multicomponent system is pure at the temperature T of the system and a reference-state pressure pref. The difference in the chemical potential of i between a specified state of the multicomponent system and the reference state is obtained with Eq. 11.125 as mi 2 m8i 5 RT ln

activity

fi f 8i

(11.139)

where the superscript 8 denotes property values at the reference state. The fugacity ratio appearing in the logarithmic term is known as the activity, ai, of component i in the mixture. That is ai 5

fi f 8i

(11.140)

For subsequent applications, it suffices to consider the case of gaseous mixtures. For gaseous mixtures, pref is specified as 1 atm, so m8i and f 8i in Eq. 11.140 are, respectively, the chemical potential and fugacity of pure i at temperature T and 1 atm. Since the chemical potential of a pure component equals the Gibbs function per mole, Eq. 11.139 can be written as mi 5 g i8 1 RT ln ai

(11.141)

where g8i is the Gibbs function per mole of pure component i evaluated at temperature T and 1 atm: g8i 5 gi (T, 1 atm). For an ideal solution, the Lewis–Randall rule applies and the activity is ai 5

yi fi f 8i

(11.142)

where fi is the fugacity of pure component i at temperature T and pressure p. Introducing Eq. 11.142 into Eq. 11.141 mi 5 g8i 1 RT ln

yi fi f 8i

or fi pref yi p mi 5 g8i 1 RT ln c a b a b d 1ideal solution2 p f 8i pref

(11.143)

In principle, the ratios of fugacity to pressure shown underlined in this equation can be evaluated from Eq. 11.124 or the generalized fugacity chart, Fig. A-6, developed from it. If component i behaves as an ideal gas at both T, p and T, pref, we have fi / p 5 f 8i / pref 5 1; Eq. 11.143 then reduces to mi 5 g8i 1 RT ln

yi p 1ideal gas2 pref

(11.144)

c CHAPTER SUMMARY AND STUDY GUIDE In this chapter, we introduce thermodynamic relations that allow u, h, and s as well as other properties of simple compressible systems to be evaluated using property data that are more readily measured. The emphasis is on systems involving a single chemical species such as water or a mixture such as air. An intro-

duction to general property relations for mixtures and solutions is also included. Equations of state relating p, y, and T are considered, including the virial equation and examples of two-constant and multiconstant equations. Several important property relations based

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Key Equations on the mathematical characteristics of exact differentials are developed, including the Maxwell relations. The concept of a fundamental thermodynamic function is discussed. Means for evaluating changes in specific internal energy, enthalpy, and entropy are developed and applied to phase change and to single-phase processes. Property relations are introduced involving the volume expansivity, isothermal and isentropic compressibilities, velocity of sound, specific heats and specific heat ratio, and the Joule–Thomson coefficient. Additionally, we describe how tables of thermodynamic properties are constructed using the property relations and methods developed in this chapter. Such procedures also provide the basis for data retrieval by computer software. Also described are means for using the generalized enthalpy and entropy departure charts and the generalized fugacity coefficient chart to evaluate enthalpy, entropy, and fugacity, respectively. We also consider p–y–T relations for gas mixtures of known composition, including Kay’s rule. The chapter concludes with a discussion of property relations for multicomponent systems, including partial molal properties, chemical potential, fugacity, and activity. Ideal solutions and the Lewis–Randall rule are introduced as a part of that presentation. The following checklist provides a study guide for this chapter. When your study of the text and end-of-chapter exercises has been completed you should be able to write out the meanings of the terms listed in the margins throughout the chapter and understand each of the related concepts. The subset of key

concepts listed below is particularly important. Additionally, for systems involving a single species you should be able to c calculate p–y–T data using equations of state such as the

Redlich–Kwong and Benedict–Webb–Rubin equations. c use the 16 property relations summarized in Table 11.1 and

explain how the relations are obtained. c evaluate Ds, Du, and Dh, using the Clapeyron equation when

considering phase change, and using equations of state and specific heat relations when considering single phases. c use the property relations introduced in Sec. 11.5, such as those involving the specific heats, the volume expansivity, and the Joule–Thomson coefficient. c explain how tables of thermodynamic properties, such as Tables A-2 through A-18, are constructed. c use the generalized enthalpy and entropy departure charts, Figs. A-4 and A-5, to evaluate Dh and Ds. For a gas mixture of known composition, you should be able to c apply the methods introduced in Sec. 11.8 for relating pressure,

specific volume, and temperature—Kay’s rule, for example. For multicomponent systems, you should be able to c evaluate extensive properties in terms of the respective par-

tial molal properties. c evaluate partial molal volumes using the method of intercepts. c evaluate fugacity using data from the generalized fugacity

coefficient chart, Fig. A-6. c apply the ideal solution model.

c KEY ENGINEERING CONCEPTS equation of state, p. 632 exact differential, p. 638 test for exactness, p. 638 Helmholtz function, p. 642 Gibbs function, p. 642 Maxwell relations, p. 644

fundamental thermodynamic function, p. 647 Clapeyron equation, p. 649 Joule-Thomson coefficient, p. 661 enthalpy departure, p. 669 entropy departure, p. 672

Kay’s rule, p. 675 method of intercepts, p. 681 chemical potential, p. 682 fugacity, p. 685 Lewis–Randall rule, p. 688

c KEY EQUATIONS

Equations of State Z511

B1T 2 C1T2 D1T 2 1 1 1... 2 y y y3

(11.1) p. 632

Virial equation of state

RT a 2 2 y2b y

(11.2) p. 633

van der Waals equation of state

RT a 2 y2b y1y 1 b2T 1/2

(11.7) p. 635

Redlich–Kwong equation of state

p5

p5

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Chapter 11 Thermodynamic Relations

Mathematical Relations for Properties 0 0z 0 0z ca b d 5 ca b d 0y 0x y x 0x 0y x y

(11.14a) p. 638

0M 0N a b 5a b 0y x 0x y

(11.14b) p. 638

a

Test for exactness

0y 0x b a b 5 1 0y z 0x z

(11.15) p. 639

0y 0z 0x a b a b a b 5 21 0z x 0x y 0y z

(11.16) p. 639

Table 11.1

(11.24– 11.36) p. 644

Important relations among partial derivatives of properties

Summary of property relations from exact differentials

Expressions for Du, Dh, and Ds a

hg 2 hf dp b 5 dT sat T1yg 2 yf2

#

s2 2 s1 5

2

1

2

# c dT 1 #

u2 2 u1 5

1

2

cp

#T

# a 0T b dy

# a 0T b dp

2

0y

1

s2 2 s1 5 s*2 2 s *1 2 R c a

Expressions for changes in s and u with T and y as independent variables

(11.59) p. 654

0y

Expressions for changes in s and h with T and p as independent variables

(11.60) p. 655

p

1

h2 2 h1 5 h*2 2 h*1 2 RTc c a

(11.51) p. 653

p

# c dT 1 # c y 2 T a 0T b d dp p

Clapeyron equation

(11.50) p. 653

y

0p c T a b 2 p d dy 0T y

1

2

0p

1

2

dT 2

1

h2 2 h1 5

2

y

1

s2 2 s1 5

2

cy dT 1 T

(11.40) p. 649

h* 2 h h* 2 h b 2a b d RTc 2 RTc 1

s* 2 s s* 2 s b 2a b d R R 2 1

(11.85) p. 670

(11.92) p. 672

Evaluating enthalpy and entropy changes in terms of generalized enthalpy and entropy departures and data from Figs. A-4 and A-5, respectively

Additional Thermodynamic Relations

c5

B

c 5 u 2 Ts

(11.20) p. 642

Helmholtz function

g 5 h 2 Ts

(11.21) p. 642

Gibbs function

(9.36b) p. 658 (11.74) p. 660

Expressions for velocity of sound

(11.75) p. 661

Joule–Thomson coefficient

2y2 a

0p 0p b 5 2ky2 a b 0y s B 0y T

mJ 5 a

0T b 0p h

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Exercises: Things Engineers Think About

693

Properties of Multicomponent Mixtures j

j

Tc 5 a yiTc, i, pc 5 a yi pc, i i51

(11.97) p. 676

Kay’s rule for critical temperature and pressure of mixtures

(11.102) p. 680

— Partial molal property Xi and its relation to extensive property X

(11.103) p. 680

X as a weighted sum of partial molal properties

(11.107) p. 682

Chemical potential of species i in a mixture

i51

Xi 5

0X b 0ni T, p, nl j

X 5 a ni Xi i51

mi 5 Gi 5

0ln f b 5y 0p T

(11.122) p. 685

f lim 5 1 pS0 p

(11.123) p. 685

0 ln fi b 5 Vi 0p T, n

(11.126a) p. 687

RT a

RT a

0G b 0ni T, p, nl

Expressions for evaluating fugacity of a single-component system

Expressions for evaluating fugacity of mixture component i

fi lim a b 5 1 pS 0 yi p

(11.126b) p. 687

fi 5 yi fi

(11.134) p. 688

Lewis–Randall rule for ideal solutions

(11.144) p. 690

Chemical potential of component i in an ideal gas mixture

mi 5 gi8 1 RT ln

yi p pref

c EXERCISES: THINGS ENGINEERS THINK ABOUT 1. What is an advantage of using the Redlich–Kwong equation of state in the generalized form given by Eq. 11.9 instead of Eq. 11.7? A disadvantage?

9. Can you devise a way to determine the specific heat cp of a gas by direct measurement? Indirectly, using other measured data?

2. To determine the specific volume of superheated water vapor at a known pressure and temperature, when would you use each of the following: the steam tables, the generalized compressibility chart, an equation of state, the ideal gas model?

10. For an ideal gas, what is the value of the Joule–Thomson coefficient?

3. If the function p 5 p(T, y) is an equation of state, is 10p/ 0T2y a property? What are the independent variables of 10p/ 0T2y? 4. In the expression 10u/ 0T2y, what does the subscript y signify? 5. Explain how a Mollier diagram provides a graphical representation of the fundamental function h(s, p). 6. How is the Clapeyron equation used? 7. For a gas whose equation of state is py 5 RT, are the specific heats cp and cy necessarily functions of temperature alone? 8. Referring to the phase diagram for water (Fig. 3.5), explain why ice melts under the blade of an ice skate.

11. At what states is the entropy departure negligible? The fugacity coefficient, f/p, closely equal to unity? 12. In Eq. 11.107, what do the subscripts T, p, and nl signify? What does i denote? 13. How does Eq. 11.108 reduce for a system consisting of a pure substance? Repeat for an ideal gas mixture. 14. If two different liquids of known volumes are mixed, is the final volume necessarily equal to the sum of the original volumes? 15. For a binary solution at temperature T and pressure p, how would you determine the specific heat cp? Repeat for an ideal solution and for an ideal gas mixture.

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c PROBLEMS: DEVELOPING ENGINEERING SKILLS Using Equations of State 11.1 Owing to safety requirements, the pressure within a 19.3 ft3 cylinder should not exceed 52 atm. Check the pressure within the cylinder if filled with 100 lb of CO2 maintained at 2128F using the (a) van der Waals equation. (b) compressibility chart. (c) ideal gas equation of state. 11.2 Ten pounds mass of propane have a volume of 2 ft3 and a pressure of 600 lbf/in.2 Determine the temperature, in 8R, using the (a) van der Waals equation. (b) compressibility chart. (c) ideal gas equation of state. (d) propane tables. 11.3 The pressure within a 23.3-m3 tank should not exceed 105 bar. Check the pressure within the tank if filled with 1000 kg of water vapor maintained at 3608C using the (a) ideal gas equation of state. (b) van der Waals equation. (c) Redlich–Kwong equation. (d) compressibility chart. (e) steam tables. 11.4 Estimate the pressure of water vapor at a temperature of 5008C and a density of 24 kg/m3 using the (a) steam tables. (b) compressibility chart. (c) Redlich–Kwong equation. (d) van der Waals equation. (e) ideal gas equation of state. 11.5 Methane gas flows through a pipeline with a volumetric flow rate of 11 ft3/s at a pressure of 183 atm and a temperature of 568F. Determine the mass flow rate, in lb/s, using the (a) ideal gas equation of state. (b) van der Waals equation. (c) compressibility chart. 11.6 Determine the specific volume of water vapor at 20 MPa and 4008C, in m3/kg, using the (a) steam tables. (b) compressibility chart. (c) Redlich–Kwong equation. (d) van der Waals equation. (e) ideal gas equation of state. 11.7 A vessel whose volume is 1 m3 contains 4 kmol of methane at 1008C. Owing to safety requirements, the pressure of the methane should not exceed 12 MPa. Check the pressure using the (a) ideal gas equation of state. (b) Redlich–Kwong equation. (c) Benedict–Webb–Rubin equation.

11.8 Methane gas at 100 atm and 2188C is stored in a 10-m3 tank. Determine the mass of methane contained in the tank, in kg, using the (a) ideal gas equation of state. (b) van der Waals equation. (c) Benedict–Webb–Rubin equation. 11.9 Using the Benedict–Webb–Rubin equation of state, determine the volume, in m3, occupied by 165 kg of methane at a pressure of 200 atm and temperature of 400 K. Compare with the results obtained using the ideal gas equation of state and the generalized compressibility chart. 11.10 A rigid tank contains 1 kg of oxygen (O2) at p1 5 40 bar, T1 5 180 K. The gas is cooled until the temperature drops to 150 K. Determine the volume of the tank, in m3, and the final pressure, in bar, using the (a) ideal gas equation of state. (b) Redlich–Kwong equation. (c) compressibility chart. 11.11 One pound mass of air initially occupying a volume of 0.4 ft3 at a pressure of 1000 lbf/in.2 expands in a piston–cylinder assembly isothermally and without irreversibilities until the volume is 2 ft3. Using the Redlich–Kwong equation of state, determine the (a) temperature, in 8R. (b) final pressure, in lbf/in.2 (c) work developed in the process, in Btu. 11.12 Water vapor initially at 2408C, 1 MPa expands in a piston–cylinder assembly isothermally and without internal irreversibilities to a final pressure of 0.1 MPa. Evaluate the work done, in kJ/kg. Use a truncated virial equation of state with the form B C 1 2 y y where B and C are evaluated from steam table data at 2408C and pressures ranging from 0 to 1 MPa. Z511

11.13 Referring to the virial series, Eqs. 3.30 and 3.31, show that Bˆ 5 B/ RT, Cˆ 5 1C 2 B22/ R 2 T 2. 11.14 Express Eq. 11.5, the van der Waals equation, in terms of the compressibility factor Z (a) as a virial series in y¿R. [Hint: Expand the 1y¿R 2 1/ 8221 term of Eq. 11.5 in a series.] (b) as a virial series in pR. (c) Dropping terms involving ( pR)2 and higher in the virial series of part (b), obtain the following approximate form: 1 27/ 64 pR Z511a 2 b 8 TR TR (d) Compare the compressibility factors determined from the equation of part (c) with tabulated compressibility factors from the literature for 0 , pR , 0.6 and each of TR 5 1.0, 1.2, 1.4, 1.6, 1.8, 2.0. Comment on the realm of validity of the approximate form.

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Problems: Developing Engineering Skills 1 1 b 1 b2 2 b3 py a 2 5 RT1y 1 b2 RT 11 1 b23

11.15 The Berthelot equation of state has the form p5

a RT 2 y2b T y2

(a) Using Eqs. 11.3, show that

a5

27 R 2 Tc3 1 RTc , b 5 64 pc 8 pc

(b) Express the equation in terms of the compressibility factor Z, the reduced temperature TR, and the pseudoreduced specific volume, y¿R. 11.16 The Beattie–Bridgeman equation of state can be expressed as p5

RT 11 2 e21y 1 B2 y2

2

A y2

where b 5 b/ 4y, a 5 a0 exp (a1T 1 a2T 2), and b 5 b0 1 b1T 1 b2T 2. For Refrigerants 12 and 13, the required coefficients for T in K, a in J ? L/(mol)2, and b in L/mol are given in Table P11.19. Specify which of the two refrigerants would allow the smaller amount of mass to be stored in a 10-m3 vessel at 0.2 MPa, 808C.

Using Relations from Exact Differentials 11.20 The differential of pressure obtained from a certain equation of state is given by one of the following expressions. Determine the equation of state. dp 5

a b A 5 A0 a1 2 b , B 5 B0 a1 2 b y y c e5 yT 3

dQ int 5 dU 1 p dV rev

Using this expression together with the test for exactness, demonstrate that Qint is not a property. rev 11.22 Show that Eq. 11.16 is satisfied by an equation of state with the form p 5 [RT/(y 2 b)] 1 a. 11.23 For the functions x 5 x(y, w), y 5 y(z, w), z 5 z(x, w), demonstrate that

RT 2a b exp a b y2b RTy

0x 0y 0z b b b 51 0y w 0z w 0x w

(a) Using Eqs. 11.3, show that pce

2

, b 5

RTc

11.24 Using Eq. 11.35, check the consistency of

pce2

(b) Show that the equation of state can be expressed in terms of compressibility chart variables as Z5a

y¿R y¿R 2 1/ e

1y 2 b22

rev

11.17 The Dieterici equation of state is

4R2T 2c

dy 1

11.21 Introducing dQ int 5 T dS into Eq. 6.8 gives

and A0, B0, a, b, and c are constants. Express this equation of state in terms of the reduced pressure, pR, reduced temperature, TR, pseudoreduced specific volume, y¿R, and appropriate dimensionless constants.

a5

21y 2 b2

dT RT RT 2 RT R dp 5 2 dy 1 dT y2b 1y 2 b22

where

p5a

695

b exp a 2

(a) the steam tables at 2 MPa, 4008C. (b) the Refrigerant 134a tables at 2 bar, 508C. 11.25 Using Eq. 11.35, check the consistency of

24 b TRy¿Re2

(c) Convert the result of part (b) to a virial series in y¿R. (Hint: Expand the 1y¿R 2 1/ e2221 term in a series. Also expand the exponential term in a series.) 11.18 The Peng–Robinson equation of state has the form

(a) the steam tables at 100 lbf/in.2, 6008F. (b) the Refrigerant 134a tables at 40 lbf/in.2, 1008F. 11.26 At a pressure of 1 atm, liquid water has a state of maximum density at about 48C. What can be concluded about 10s/ 0p2T at (a) 38C? (b) 48C? (c) 58C?

RT a p5 2 2 y2b y 2 c2 Using Eqs. 11.3, evaluate the constants a, b, c in terms of the critical pressure pc, critical temperature Tc, and critical compressibility factor Zc.

11.27 A gas enters a compressor operating at steady state and is compressed isentropically. Does the specific enthalpy increase or decrease as the gas passes from the inlet to the exit?

11.19 The p–y–T relation for chlorofluorinated hydrocarbons can be described by the Carnahan–Starling–DeSantis equation of state

11.28 Show that T, p, h, c, and g can each be determined from a fundamental thermodynamic function of the form u 5 u(s, y).

R-12 R-13

a0 3 1023

a1 3 103

a2 3 106

b0

b1 3 104

b2 3 108

3.52412 2.29813

22.77230 23.41828

20.67318 21.52430

0.15376 0.12814

21.84195 21.84474

25.03644 210.7951

Table P11.19

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Chapter 11 Thermodynamic Relations

11.29 Evaluate p, s, u, h, cy, and cp for a substance for which the Helmholtz function has the form c 5 2RT ln

y T T T 2 cT ¿ c 1 2 1 ln d y¿ T¿ T¿ T¿

where y9 and T9 denote specific volume and temperature, respectively, at a reference state, and c is a constant. 11.30 The Mollier diagram provides a graphical representation of the fundamental thermodynamic function h 5 h(s, p). Show that at any state fixed by s and p the properties T, y, u, c, and g can be evaluated using data obtained from the diagram. 11.31 Derive the relation cp 5 2T10 2gy0T 22p.

Evaluating Ds, Du, and Dh 11.32 Using p–y–T data for saturated ammonia from Table A-13E, calculate at 208F (a) hg 2 hf. (b) ug 2 uf. (c) sg 2 sf. Compare with values obtained from Table A-13E. 11.33 Using p–y–T data for saturated water from the steam tables, calculate at 508C (a) hg 2 hf. (b) ug 2 uf. (c) sg 2 sf. Compare with values obtained from the steam tables. 11.34 Using hfg, yfg, and psat at 108F from the Refrigerant 134a tables, estimate the saturation pressure at 208F. Comment on the accuracy of your estimate. 11.35 Using hfg, yfg, and psat at 268C from the ammonia tables, estimate the saturation pressure at 308C. Comment on the accuracy of your estimate. 11.36 Using triple-point data for water from Table A-6E, estimate the saturation pressure at 2408F. Compare with the value listed in Table A-6E. 11.37 At 08C, the specific volumes of saturated solid water (ice) and saturated liquid water are, respectively, yi 5 1.0911 3 1023 m3/kg and yf 5 1.0002 3 1023 m3/kg, and the change in specific enthalpy on melting is hif 5 333.4 kJ/kg. Calculate the melting temperature of ice at (a) 250 bar, (b) 500 bar. Locate your answers on a sketch of the p–T diagram for water. 11.38 The line representing the two-phase solid–liquid region on the phase diagram slopes to the left for substances that expand on freezing and to the right for substances that contract on freezing (Sec. 3.2.2). Verify this for the cases of lead that contracts on freezing and bismuth that expands on freezing. 11.39 Consider a four-legged chair at rest on an ice rink. The total mass of the chair and a person sitting on it is 80 kg. If the ice temperature is 228C, determine the minimum total area, in cm2, the tips of the chair legs can have before the ice in contact with the legs would melt. Use data from Problem 11.37 and let the local acceleration of gravity be 9.8 m/s2.

11.40 Over a certain temperature interval, the saturation pressure–temperature curve of a substance is represented by an equation of the form ln psat 5 A 2 B/T, where A and B are empirically determined constants. (a) Obtain expressions for hg 2 hf and sg 2 sf in terms of p–y–T data and the constant B. (b) Using the results of part (a), calculate hg 2 hf and sg 2 sf for water vapor at 258C and compare with steam table data. 11.41 Using data for water from Table A-2, determine the constants A and B to give the best fit in a least-squares sense to the saturation pressure in the interval from 20 to 308C by the equation ln psat 5 A 2 B/T. Using this equation, determine dpsat/dT at 258C. Calculate hg 2 hf at 258C and compare with the steam table value. 11.42 Over limited intervals of temperature, the saturation pressure–temperature curve for two-phase liquid–vapor states can be represented by an equation of the form ln psat 5 A 2 B/T, where A and B are constants. Derive the following expression relating any three states on such a portion of the curve: psat, 3 psat, 2 t 5a b psat, 1 psat,1 where t = T2(T3 2 T1)/T3(T2 2 T1). 11.43 Use the result of Problem 11.42 to determine (a) the saturation pressure at 308C using saturation pressure– temperature data at 20 and 408C from Table A-2. Compare with the table value for saturation pressure at 308C. (b) the saturation temperature at 0.006 MPa using saturation pressure–temperature data at 20 to 408C from Table A-2. Compare with the saturation temperature at 0.006 MPa given in Table A-3. 11.44 Complete the following exercises dealing with slopes: (a) At the triple point of water, evaluate the ratio of the slope of the vaporization line to the slope of the sublimation line. Use steam table data to obtain a numerical value for the ratio. (b) Consider the superheated vapor region of a temperature– entropy diagram. Show that the slope of a constant specific volume line is greater than the slope of a constant pressure line through the same state. (c) An enthalpy–entropy diagram (Mollier diagram) is often used in analyzing steam turbines. Obtain an expression for the slope of a constant-pressure line on such a diagram in terms of p–y–T data only. (d) A pressure–enthalpy diagram is often used in the refrigeration industry. Obtain an expression for the slope of an isentropic line on such a diagram in terms of p–y–T data only. 11.45 Using only p–y–T data from the ammonia tables, evaluate the changes in specific enthalpy and entropy for a process from 70 lbf/in.2, 408F to 14 lbf/in.2, 408F. Compare with the table values. 11.46 One kmol of argon at 300 K is initially confined to one side of a rigid, insulated container divided into equal volumes of 0.2 m3 by a partition. The other side is initially evacuated. The partition is removed and the argon expands to fill the

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Problems: Developing Engineering Skills entire container. Using the van der Waals equation of state, determine the final temperature of the argon, in K. Repeat using the ideal gas equation of state. 11.47 Obtain the relationship between cp and cy for a gas that obeys the equation of state p(y 2 b) 5 RT. 11.48 The p–y–T relation for a certain gas is represented closely by y = RT/p 1 B 2 A/RT, where R is the gas constant and A and B are constants. Determine expressions for the changes in specific enthalpy, internal energy, and entropy, [h(p2, T) 2 h(p1, T)], [u(p2, T) 2 u(p1, T)], and [s(p2, T) 2 s(p1, T)], respectively. 11.49 Develop expressions for the specific enthalpy, internal energy, and entropy changes [h(y2, T) 2 h(y1, T)], [u(y2, T) 2 u(y1, T)], [s(y2, T) 2 s(y1, T)], using the (a) van der Waals equation of state. (b) Redlich–Kwong equation of state. 11.50 At certain states, the p–y–T data of a gas can be expressed as Z 5 1 2 Ap/T 4, where Z is the compressibility factor and A is a constant. (a) Obtain an expression for ( 0p/ 0T ) y in terms of p, T, A, and the gas constant R. (b) Obtain an expression for the change in specific entropy, [s( p2, T) 2 s ( p1, T)]. (c) Obtain an expression for the change in specific enthalpy, [h(p2, T) 2 h ( p1, T)]. 11.51 For a gas whose p–y–T behavior is described by Z 5 1 1 Bp/RT, where B is a function of temperature, derive expressions for the specific enthalpy, internal energy, and entropy changes, [h( p2, T) 2 h( p1, T)], [u( p2, T) 2 u( p1, T)], and [s(p2, T) 2 s(p1, T)]. 11.52 For a gas whose p–y–T behavior is described by Z 5 1 1 B/y 1 C/y2, where B and C are functions of temperature, derive an expression for the specific entropy change, [s(y2, T) 2 s(y1, T)].

697

partial derivatives of Z. For gas states with pR , 3, TR , 2, determine the sign of k. Discuss. 11.57 Show that the isothermal compressibility k is always greater than or equal to the isentropic compressibility a. 11.58 Prove that 10b/ 0p2T 5 210k/ 0T2p. 11.59 For aluminum at 08C, r 5 2700 kg/m3, b 5 71.4 3 1028 (K)21, k 5 1.34 3 10213 m2/N, and cp 5 0.9211 kJ/kg ? K. Determine the percent error in cy that would result if it were assumed that cp 5 cy. 11.60 Estimate the temperature rise, in 8C, of mercury, initially at 08C and 1 bar if its pressure were raised to 1000 bar isentropically. For mercury at 08C, cp 5 28.0 kJ/kmol ? K, y 5 0.0147 m3/ kmol, and b 5 17.8 3 1025 (K)21. 11.61 At certain states, the p–y–T data for a particular gas can be represented as Z 5 1 2 Ap/T 4, where Z is the compressibility factor and A is a constant. Obtain an expression for the specific heat cp in terms of the gas constant R, specific heat ratio k, and Z. Verify that your expression reduces to Eq. 3.47a when Z 5 1. 11.62 For a gas obeying the van der Waals equation of state, (a) show that 10cy / 0y2T 5 0. (b) develop an expression for cp 2 cy. (c) develop expressions for [u(T2, y2) 2 u(T1, y1)] and [s(T2, y2) 2 s(T1, y1)]. (d) complete the Du and Ds evaluations if cy 5 a 1 bT, where a and b are constants. 11.63 If the value of the specific heat cy of air is 0.1965 Btu/ lb ? 8R at T1 5 10008F, y1 5 36.8ft3/lb, determine the value of cy at T2 5 10008F, y2 5 0.0555 ft3/lb. Assume that air obeys the Berthelot equation of state p5

RT a 2 2 y2b Ty

where Using Other Thermodynamic Relations 11.53 The volume of a 1-kg copper sphere is not allowed to vary by more than 0.1%. If the pressure exerted on the sphere is increased from 10 bar while the temperature remains constant at 300 K, determine the maximum allowed pressure, in bar. Average values of r, b, and k are 8888 kg/m3, 49.2 3 1026 (K)21, and 0.776 3 10211 m2/N respectively. 11.54 The volume of a 1-lb copper sphere is not allowed to vary by more than 0.1%. If the pressure exerted on the sphere is increased from 1 atm while the temperature remains constant at 808F, determine the maximum allowed pressure, in atm. Average values of r, b, and k are 555 lb/ft3, 2.75 3 1025 (8R)21, and 3.72 3 10210 ft2/lbf, respectively. 11.55 Develop expressions for the volume expansivity b and the isothermal compressibility k for (a) an ideal gas. (b) a gas whose equation of state is p(y 2 b) 5 RT. (c) a gas obeying the van der Waals equation. 11.56 Derive expressions for the volume expansivity b and the isothermal compressibility k in terms of T, p, Z, and the first

a5

1 RTc 27 R2T 3c , b 5 64 pc 8 pc

11.64 Show that the specific heat ratio k can be expressed as k 5 cpk/(cpk 2 Tyb2). Using this expression together with data from the steam tables, evaluate k for water vapor at 200 lbf/in.2, 5008F. 11.65 For liquid water at 408C, 1 atm estimate (a) cy, in kJ/kg ? K. (b) the velocity of sound, in m/s. Use Data from Table 11.2, as required. 11.66 Using steam table data, estimate the velocity of sound in liquid water at (a) 208C, 50 bar, (b) 508F, 1500 lbf/in.2 11.67 At a certain location in a wind tunnel, a stream of air is at 5008F, 1 atm and has a velocity of 2115 ft/s. Determine the Mach number at this location. 11.68 For a gas obeying the equation of state p(y 2 b) 5 RT, where b is a positive constant, can the temperature be reduced in a Joule–Thomson expansion?

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11.69 A gas is described by y 5 RT/p 2 A/T 1 B, where A and B are constants. For the gas (a) obtain an expression for the temperatures at the Joule– Thomson inversion states. (b) obtain an expression for cp 2 cy. 11.70 Determine the maximum Joule–Thomson inversion temperature in terms of the critical temperature Tc predicted by the (a) van der Waals equation. (b) Redlich–Kwong equation. (c) Dieterici equation given in Problem 11.17. 11.71 Derive an equation for the Joule–Thomson coefficient as a function of T and y for a gas that obeys the van der Waals equation of state and whose specific heat cy is given by cy 5 A 1 BT 1 CT 2, where A, B, C are constants. Evaluate the temperatures at the inversion states in terms of R, y, and the van der Waals constants a and b. 11.72 Show that Eq. 11.77 can be written as mJ 5

T 2 01y/ T2 b a cp 0T p

(a) Using this result, obtain an expression for the Joule– Thomson coefficient for a gas obeying the equation of state Ap RT 2 2 y5 p T where A is a constant. (b) Using the result of part (a), determine cp, in kJ/kg ? K, for CO2 at 400 K, 1 atm, where mJ 5 0.57 K/atm. For CO2, A 5 2.78 3 1023 m5 ? K2/kg ? N.

Developing Property Data 11.73 If the specific heat cy of a gas obeying the van der Waals equation is given at a particular pressure, p9, by cy 5 A 1 BT, where A and B are constants, develop an expression for the change in specific entropy between any two states 1 and 2: [s(T2, p2) 2 s(T1, p1)]. 11.74 For air, write a computer program that evaluates the change in specific enthalpy from a state where the temperature is 258C and the pressure is 1 atm to a state where the temperature is T and the pressure is p. Use the van der Waals equation of state and account for the variation of the ideal gas specific heat as in Table A-21. 11.75 Using the Redlich–Kwong equation of state, determine the changes in specific enthalpy, in kJ/kmol, and entropy, in kJ/kmol ? K, for ethylene between 400 K, 1 bar and 400 K, 100 bar. 11.76 Using the Benedict–Webb–Rubin equation of state together with a specific heat relation from Table A-21, determine the change in specific enthalpy, in kJ/kmol, for methane between 300 K, 1 atm and 400 K, 200 atm. 11.77 A certain pure, simple compressible substance has the following property relations. The p–y–T relationship in the vapor phase is Bp RT y5 2 2 p T

where y is in ft3/lb, T is in 8R, p is in lbf/ft2, R 5 50 ft ? lbf/ lb ? 8R, and B 5 100 ft5 ? (8R)2/lb ? lbf. The saturation pressure, in lbf/ft2, is described by ln psat 5 12 2

2400 T

The Joule–Thomson coefficient at 10 lbf/in.2, 2008F is 0.0048R ? ft2/lbf. The ideal gas specific heat cp0 is constant over the temperature range 0 to 3008F. (a) Complete the accompanying table of property values T

p

08F 1008F

vf

0.03 0.03

vg

hf

hg

0

sf

sg

0.000

for p in lbf/in.2, y in ft3/lb, h in Btu/lb, and s in Btu/lb ? 8R. (b) Evaluate y, h, s at the state fixed by 15 lbf/in. 2, 3008F. 11.78 In Table A-2, at temperatures up to 508C, the values of uf and hf differ in most cases by 0.01 kJ/kg. Yet at each of these temperatures the product psatyf is small enough to be neglected and the table values of uf and hf should be the same. In a memorandum, provide a plausible explanation.

Using Enthalpy and Entropy Departures 11.79 Beginning with Eq. 11.90, derive Eq. 11.91. 11.80 Derive an expression giving (a) the internal energy of a substance relative to that of its ideal gas model at the same temperature: [u(T, y) 2 u*(T)]. (b) the entropy of a substance relative to that of its ideal gas model at the same temperature and specific volume: [s(T, y) 2 s*(T, y)]. 11.81 Derive expressions for the enthalpy and entropy departures using an equation of state with the form Z 5 1 1 BpR, where B is a function of the reduced temperature, TR. 11.82 The following expression for the enthalpy departure is convenient for use with equations of state that are explicit in pressure: h*1T2 2 h1T, y2 RTc

5 TR c 1 2 Z 2

1 RT

y

# c T a 0T b q

0p

2 p d dy d

y

(a) Derive this expression. (b) Using the given expression, evaluate the enthalpy departure for a gas obeying the Redlich–Kwong equation of state. (c) Using the result of part (b), determine the change in specific enthalpy, in kJ/kmol, for CO2 undergoing an isothermal process at 300 K from 50 to 20 bar. 11.83 Using the equation of state of Problem 11.14 (c), evaluate y and cp for water vapor at 5508C, 20 MPa and compare with data from Table A-4 and Fig. 3.9, respectively. Discuss. 11.84 Ethylene at 678C, 10 bar enters a compressor operating at steady state and is compressed isothermally without internal irreversibilities to 100 bar. Kinetic and potential energy changes are negligible. Evaluate in kJ per kg of ethylene flowing through the compressor (a) the work required. (b) the heat transfer.

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11.85 Methane at 278C, 10 MPa enters a turbine operating at steady state, expands adiabatically through a 5 : 1 pressure ratio, and exits at 2488C. Kinetic and potential energy effects are negligible. If cpo 5 35 kJ/ kmol ? K, determine the work developed per kg of methane flowing through the turbine. Compare with the value obtained using the ideal gas model.

Owing to safety requirements, the pressure should not exceed 180 bar. Check the pressure using

11.86 Nitrogen (N2) enters a compressor operating at steady state at 1.5 MPa, 300 K and exits at 8 MPa, 500 K. If the work input is 240 kJ per kg of nitrogen flowing, determine the heat transfer, in kJ per kg of nitrogen flowing. Ignore kinetic and potential energy effects.

Compare and discuss these results.

11.87 Oxygen (O2) enters a control volume operating at steady state with a mass flow rate of 9 kg/min at 100 bar, 287 K and is compressed adiabatically to 150 bar, 400 K. Determine the power required, in kW, and the rate of entropy production, in kW/K. Ignore kinetic and potential energy effects. 11.88 Argon gas enters a turbine operating at steady state at 100 bar, 325 K and expands adiabatically to 40 bar, 235 K with no significant changes in kinetic or potential energy. Determine (a) the work developed, in kJ per kg of argon flowing through the turbine. (b) the amount of entropy produced, in kJ/K per kg of argon flowing. 11.89 Oxygen (O2) undergoes a throttling process from 100 bar, 300 K to 20 bar. Determine the temperature after throttling, in K, and compare with the value obtained using the ideal gas model. 11.90 Water vapor enters a turbine operating at steady state at 30 MPa, 6008C and expands adiabatically to 6 MPa with no significant change in kinetic or potential energy. If the isentropic turbine efficiency is 80%, determine the work developed, in kJ per kg of steam flowing, using the generalized property charts. Compare with the result obtained using steam table data. Discuss. 11.91 Oxygen (O2) enters a nozzle operating at steady state at 60 bar, 300 K, 1 m/s and expands isentropically to 30 bar. Determine the velocity at the nozzle exit, in m/s. 11.92 A quantity of nitrogen gas in a piston–cylinder assembly undergoes a process at a constant pressure of 80 bar from 220 to 300 K. Determine the work and heat transfer for the process, each in kJ per kmol of nitrogen. 11.93 A closed, rigid, insulated vessel having a volume of 0.142 m3 contains oxygen (O2) initially at 100 bar, 78C. The oxygen is stirred by a paddle wheel until the pressure becomes 150 bar. Determine the (a) final temperature, in 8C. (b) work, in kJ. (c) amount of exergy destroyed in the process, in kJ. Let T0 5 78C.

Evaluating p–y–T for Gas Mixtures 11.94 A preliminary design calls for a 1 kmol mixture of CO2 and C2H6 (ethane) to occupy a volume of 0.15 m3 at a temperature of 400 K. The mole fraction of CO2 is 0.3.

(a) the ideal gas equation of state. (b) Kay’s rule together with the generalized compressibility chart. (c) the additive pressure rule together with the generalized compressibility chart.

11.95 A gaseous mixture with a molar composition of 60% CO and 40% H2 enters a turbine operating at steady state at 3008F, 2000 lbf/in.2 and exits at 2128F, 1 atm with a volumetric flow rate of 20,000 ft3/min. Estimate the volumetric flow rate at the turbine inlet, in ft3/min, using Kay’s rule. What value would result from using the ideal gas model? Discuss. 11.96 A 0.1-m3 cylinder contains a gaseous mixture with a molar composition of 97% CO and 3% CO2 initially at 138 bar. Due to a leak, the pressure of the mixture drops to 129 bar while the temperature remains constant at 308C. Using Kay’s rule, estimate the amount of mixture, in kmol, that leaks from the cylinder. 11.97 A gaseous mixture consisting of 0.75 kmol of hydrogen (H2) and 0.25 kmol of nitrogen (N2) occupies 0.085 m3 at 258C. Estimate the pressure, in bar, using (a) the ideal gas equation of state. (b) Kay’s rule together with the generalized compressibility chart. (c) the van der Waals equation together with mixture values for the constants a and b. (d) the rule of additive pressure together with the generalized compressibility chart. 11.98 A gaseous mixture of 0.5 lbmol of methane and 0.5 lbmol of propane occupies a volume of 7.65 ft3 at a temperature of 1948F. Estimate the pressure using the following procedures and compare each estimate with the measured value of pressure, 50 atm: (a) the ideal gas equation of state. (b) Kay’s rule together with the generalized compressibility chart. (c) the van der Waals equation together with mixture values for the constants a and b. (d) the rule of additive pressures together with the van der Waals equation. (e) the rule of additive pressures together with the generalized compressibility chart. (f) the rule of additive volumes together with the van der Waals equation. 11.99 One lbmol of a gaseous mixture occupies a volume of 1.78 ft3 at 2128F. The mixture consists of 69.5% carbon dioxide and 30.5% ethylene (C2H4) on a molar basis. Estimate the mixture pressure, in atm, using (a) the ideal gas equation of state. (b) Kay’s rule together with the generalized compressibility chart. (c) the additive pressure rule together with the generalized compressibility chart. (d) the van der Waals equation together with mixture values for the constants a and b.

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11.100 Air having an approximate molar composition of 79% N2 and 21% O2 fills a 0.36-m3 vessel. The mass of mixture is 100 kg. The measured pressure and temperature are 101 bar and 180 K, respectively. Compare the measured pressure with the pressure predicted using (a) the ideal gas equation of state. (b) Kay’s rule. (c) the additive pressure rule with the Redlich–Kwong equation. (d) the additive volume rule with the Redlich–Kwong equation. 11.101 A gaseous mixture consisting of 50% argon and 50% nitrogen (molar basis) is contained in a closed tank at 20 atm, 21408F. Estimate the specific volume, in ft3/lb, using (a) the ideal gas equation of state. (b) Kay’s rule together with the generalized compressibility chart. (c) the Redlich–Kwong equation with mixture values for a and b. (d) the additive volume rule together with the generalized compressibility chart. 11.102 Using the Carnahan–Starling–DeSantis equation of state introduced in Problem 11.19, together with the following expressions for the mixture values of a and b: a 5 y21a1 1 2y1y211 2 f1221a1a221/2 1 y22a2 b 5 y1b1 1 y2b2 where f12 is an empirical interaction parameter, determine the pressure, in kPa, at y = 0.005 m3/kg, T 5 1808C for a mixture of Refrigerants 12 and 13, in which Refrigerant 12 is 40% by mass. For a mixture of Refrigerants 12 and 13, f12 5 0.035. 11.103 A rigid vessel initially contains carbon dioxide gas at 328C and pressure p. Ethylene gas is allowed to flow into the tank until a mixture consisting of 20% carbon dioxide and 80% ethylene (molar basis) exists within the tank at a temperature of 438C and a pressure of 110 bar. Determine the pressure p, in bar, using Kay’s rule together with the generalized compressibility chart. 11.104 Two tanks having equal volumes are connected by a valve. One tank contains carbon dioxide gas at 1008F and pressure p. The other tank contains ethylene gas at 1008F and 1480 lbf/in.2 The valve is opened and the gases mix, eventually attaining equilibrium at 1008F and pressure p9 with a composition of 20% carbon dioxide and 80% ethylene (molar basis). Using Kay’s rule and the generalized compressibility chart, determine in lbf/in.2 (a) the initial pressure of the carbon dioxide, p. (b) the final pressure of the mixture, p9. Analyzing Multicomponent Systems 11.105 A binary solution at 258C consists of 59 kg of ethyl alcohol (C2H5OH) and 41 kg of water. The respective partial molal volumes are 0.0573 and 0.0172 m3/kmol. Determine the total volume, in m3. Compare with the volume calculated using the molar specific volumes of the pure components, each a liquid at 258C, in the place of the partial molal volumes.

11.106 The following data are for a binary solution of ethane (C2H6) and pentane (C5H12) at a certain temperature and pressure: mole fraction of ethane

0.2

0.3

0.4

0.5

0.6

0.7

0.8

volume (in m3) per 0.119 0.116 0.112 0.109 0.107 0.107 0.11 kmol of solution

Estimate (a) the specific volumes of pure ethane and pure pentane, each in m3/kmol. (b) the partial molal volumes of ethane and pentane for an equimolar solution, each in m3/kmol. 11.107 The following data are for a binary mixture of carbon dioxide and methane at a certain temperature and pressure: mole fraction of methane

0.000

0.204

0.406 0.606 0.847

1.000

volume (in ft3) per lbmol of mixture

1.506

3.011

3.540 3.892 4.149

4.277

Estimate (a) the specific volumes of pure carbon dioxide and pure methane, each in ft3/lbmol. (b) the partial molal volumes of carbon dioxide and methane for an equimolar mixture, each in ft3/lbmol. 11.108 Using p–y–T data from the steam tables, determine the fugacity of water as a saturated vapor at (a) 2808C, (b) 5008F. Compare with the values obtained from the generalized fugacity chart. 11.109 Determine the fugacity, in atm, for (a) butane at 555 K, 150 bar. (b) methane at 1208F, 800 lbf/in.2 (c) benzene at 8908R, 135 atm. 11.110 Using the equation of state of Problem 11.14 (c), evaluate the fugacity of ammonia at 750 K, 100 atm and compare with the value obtained from Fig. A-6. 11.111 Using tabulated compressibility factor data from the literature, evaluate f/p at TR 5 1.40 and pR 5 2.0. Compare with the value obtained from Fig. A-6. 11.112 Consider the truncated virial expansion Z 5 1 1 Bˆ 1TR2pR 1 Cˆ 1TR2p2R 1 Dˆ 1TR2p3R (a) Using tabulated compressibility factor data from the ˆ for 0 , pR , 1.0 literature, evaluate the coefficients Bˆ , Cˆ , and D and each of TR 5 1.0, 1.2, 1.4, 1.6, 1.8, 2.0. (b) Obtain an expression for ln (f/p) in terms of TR and pR. Using the coefficients of part (a), evaluate f/p at selected states and compare with tabulated values from the literature. 11.113 Derive the following approximation for the fugacity of a liquid at temperature T and pressure p: L f 1T, p2 < fsat 1T 2 exp e

yf 1T 2 RT

3p 2 psat1T 24 f

L 1T 2 is the fugacity of the saturated liquid at temperature where fsat T. For what range of pressures might the approximation L f1T, p2 < fsat 1T 2 apply?

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Problems: Developing Engineering Skills 11.114 Beginning with Eq. 11.122, (a) evaluate ln f for a gas obeying the Redlich–Kwong equation of state. (b) Using the result of part (a), evaluate the fugacity, in bar, for Refrigerant 134a at 908C, 10 bar. Compare with the fugacity value obtained from the generalized fugacity chart. 11.115 Consider a one-inlet, one-exit control volume at steady state through which the flow is internally reversible and isothermal. Show that the work per unit of mass flowing can be expressed in terms of the fugacity f as # f2 Wcv V21 2 V22 a # b int 5 2RT ln a b 1 1 g1z1 2 z22 f1 2 m rev 11.116 Methane expands isothermally and without irreversibilities through a turbine operating at steady state, entering at 60 atm, 778F and exiting at 1 atm. Using data from the generalized fugacity chart, determine the work developed, in Btu per lb of methane flowing. Ignore kinetic and potential energy effects. 11.117 Propane (C3H8) enters a turbine operating at steady state at 100 bar, 400 K and expands isothermally without irreversibilities to 10 bar. There are no significant changes in kinetic or potential energy. Using data from the generalized fugacity chart, determine the power developed, in kW, for a mass flow rate of 50 kg/min. 11.118 Ethane (C2H6) is compressed isothermally without irreversibilities at a temperature of 320 K from 5 to 40 bar. Using data from the generalized fugacity and enthalpy departure charts, determine the work of compression and the heat transfer, each in kJ per kg of ethane flowing. Assume steady-state operation and neglect kinetic and potential energy effects. 11.119 Methane enters a turbine operating at steady state at 100 bar, 275 K and expands isothermally without irreversibilities to 15 bar. There are no significant changes in kinetic or potential energy. Using data from the generalized fugacity and enthalpy departure charts, determine the power developed and heat transfer, each in kW, for a mass flow rate of 0.5 kg/s. 11.120 Methane flows isothermally and without irreversibilities through a horizontal pipe operating at steady state, entering at 50 bar, 300 K, 10 m/s and exiting at 40 bar. Using data from the generalized fugacity chart, determine the velocity at the exit, in m/s. 11.121 Determine the fugacity, in atm, for pure ethane at 310 K, 20.4 atm and as a component with a mole fraction of 0.35 in an ideal solution at the same temperature and pressure. 11.122 Denoting the solvent and solute in a dilute binary liquid solution at temperature T and pressure p by the subscripts 1 and 2, respectively, show that if the fugacity of the solute is proportional to its mole fraction in the solution: f2 5 ky2, where k is a constant (Henry’s rule), then the fugacity of the solvent is f1 5 y1 f1 , where y1 is the solvent mole fraction and f1 is the fugacity of pure 1 at T, p. 11.123 A tank contains 310 kg of a gaseous mixture of 70% ethane and 30% nitrogen (molar basis) at 311 K and 170 atm. Determine the volume of the tank, in m3, using data from the generalized compressibility chart together with (a)

Kay’s rule, (b) the ideal solution model. Compare with the measured tank volume of 1 m3. 11.124 A tank contains a mixture of 75% argon and 25% ethylene on a molar basis at 778F, 81.42 atm. For 157 lb of mixture, estimate the tank volume, in ft3, using (a) the ideal gas equation of state. (b) Kay’s rule together with data from the generalized compressibility chart. (c) the ideal solution model together with data from the generalized compressibility chart. 11.125 A tank contains a mixture of 70% ethane and 30% nitrogen (N2) on a molar basis at 400 K, 200 atm. For 2130 kg of mixture, estimate the tank volume, in m3, using (a) the ideal gas equation of state. (b) Kay’s rule together with data from the generalized compressibility chart. (c) the ideal solution model together with data from the generalized compressibility chart. 11.126 An equimolar mixture of O2 and N2 enters a compressor operating at steady state at 10 bar, 220 K with a mass flow rate of 1 kg/s. The mixture exits at 60 bar, 400 K with no significant change in kinetic or potential energy. Stray heat transfer from the compressor can be ignored. Determine for the compressor (a) the power required, in kW. (b) the rate of entropy production, in kW/K. Assume the mixture is modeled as an ideal solution. For the pure components: 10 bar, 220 K

Oxygen Nitrogen

60 bar, 400 K

h (kJ/kg)

s (kJ/kg ? K)

h (kJ/kg)

s (kJ/kg ? K)

195.6 224.1

5.521 5.826

358.2 409.8

5.601 5.911

11.127 A gaseous mixture with a molar analysis of 70% CH4 and 30% N2, enters a compressor operating at steady state at 10 bar, 250 K and a molar flow rate of 6 kmol/h. The mixture exits the compressor at 100 bar. During compression, the temperature of the mixture departs from 250 K by no more than 0.1 K. The power required by the compressor is reported to be 6 kW. Can this value be correct? Explain. Ignore kinetic and potential energy effects. Assume the mixture is modeled as an ideal solution. For the pure components at 250 K: h (kJ/kg)

Methane Nitrogen

s (kJ/kg ? K)

10 bar

100 bar

10 bar

100 bar

506.0 256.18

358.6 229.68

10.003 5.962

8.3716 5.188

11.128 The departure of a binary solution from ideal solution behavior is gauged by the activity coefficient, gi 5 ai /yi, where ai is the activity of component i and yi is its mole fraction in the solution (i 5 1, 2). Introducing Eq. 11.140, the activity coefficient can be expressed alternatively as gi 5 fi / yi f 8i . Using this expression together with the Gibbs–Duhem equation, derive the following relation among the activity coefficients and the mole fractions for a solution at temperature T and pressure p: ay1

0 ln g2 0 ln g1 b 5 ay2 b 0y1 p, T 0y2 p, T

How might this expression be used?

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Chapter 11 Thermodynamic Relations

c DESIGN & OPEN-ENDED PROBLEMS: EXPLORING ENGINEERING PRACTICE 11.1D Design a laboratory flask for containing up to 10 kmol of mercury vapor at pressures up to 3 MPa, and temperatures from 900 to 1000 K. Consider the health and safety of the technicians who would be working with such a mercury vapor-filled container. Use a p-y-T relation for mercury vapor obtained from the literature, including appropriate property software. Write a report including at least three references. 11.2D The p–h diagram (Sec. 10.2.4) used in the refrigeration engineering field has specific enthalpy and the natural logarithm of pressure as coordinates. Inspection of such a diagram suggests that in portions of the vapor region constantentropy lines are nearly linear, and thus the relation between h, ln p and s might be expressible there as h1s, p2 5 1A s 1 B2 ln p 1 1C s 1 D2 Investigate the viability of this expression for pressure ranging up to 10 bar, using data for Refrigerant 134a. Summarize your conclusions in a memorandum. 11.3D Compressed natural gas (CNG) is being used as a fuel to replace gasoline for automobile engines. Aluminum cylinders wrapped in a fibrous composite can provide lightweight, economical, and safe on-board storage. The storage vessels should hold enough CNG for 100 to 125 miles of urban travel, at storage pressures up to 3000 lbf/in.2, and with a maximum total mass of 150 lb. Adhering to applicable U.S. Department of Transportation standards, specify both the size and number of cylinders that would meet the above design constraints. 11.4D A portable refrigeration machine requiring no external power supply and using carbon dioxide at its triple point is described in U.S. Patent No. 4,096,707. Estimate the cost of the initial carbon dioxide charge required by such a machine to maintain a 6 ft by 8 ft by 15 ft cargo container at 358F for up to 24 hours, if the container is fabricated from sheet metal covered with a 1-in. layer of polystyrene. Would you recommend the use of such a refrigeration machine? Report your findings in a PowerPoint presentation. 11.5D A power plant located at a river’s mouth where freshwater river currents meet saltwater ocean tides can generate electricity by exploiting the difference in composition of the freshwater and saltwater. The technology for generating power is called reverse electrodialysis. While only small-scale demonstration power plants using reverse electrodialysis have been developed thus far, some observers have high expectations for the approach. Investigate the technical readiness and economic feasibility of this renewable power source for providing 3%, or more, of annual U.S. electricity by 2030. Present your conclusions in a report, including a discussion of potential adverse environmental effects of such power plants and at least three references. 11.6D During a phase change from liquid to vapor at fixed pressure, the temperature of a binary nonazeotropic solution such as an ammonia–water solution increases rather than remains constant as for a pure substance. This attribute is exploited in both the Kalina power cycle and in the Lorenz refrigeration cycle. Write a report assessing the status of

technologies based on these cycles. Discuss the principal advantages of using binary nonazeotropic solutions. What are some of the main design issues related to their use in power and refrigeration systems? 11.7D The following data are known for a 100-ton ammonia– water absorption system like the one shown in Fig. 10.12. The pump is to handle 570 lb of strong solution per minute. The generator conditions are 175 lbf/in.2, 2208F. The absorber is at 29 lbf/in.2 with strong solution exiting at 808F. For the evaporator, the pressure is 30 lbf/in.2 and the exit temperature is 108F. Specify the type and size, in horsepower, of the pump required. Present your findings in a memorandum. 11.8D The Servel refrigerator works on an absorption principle and requires no moving parts. An energy input by heat transfer is used to drive the cycle, and the refrigerant circulates due to its natural buoyancy. This type of refrigerator is commonly employed in mobile applications, such as recreational vehicles. Liquid propane is burned to provide the required energy input during mobile operation, and electric power is used when the vehicle is parked and can be connected to an electrical outlet. Investigate the principles of operation of commercially available Servel-type systems, and study their feasibility for solar-activated operation. Consider applications in remote locations where electricity or gas is not available. Write a report summarizing your findings. 11.9D In the experiment for the regelation of ice, a small-diameter wire weighted at each end is draped over a block of ice. The loaded wire is observed to cut slowly through the ice without leaving a trace. In one such set of experiments, a weighted 1.00-mm-diameter wire is reported to have passed through 08C ice at a rate of 54 mm/h. Perform the regelation experiment and propose a plausible explanation for this phenomenon. 11.10D Figure P11.10D shows the schematic of a hydraulic accumulator in the form of a cylindrical pressure vessel with a piston separating a hydraulic fluid from a charge of nitrogen gas. The device has been proposed as a means for storing some of the exergy of a decelerating vehicle as it comes to rest. The exergy is stored by compressing the nitrogen. When the vehicle accelerates again, the gas expands and returns some exergy to the hydraulic fluid which is in communication with the vehicle’s drive train, thereby assisting the vehicle to accelerate. In a proposal for one such device, the nitrogen operates in the range 50–150 bar and 200–350 K. Develop a thermodynamic model of the accumulator and use the model to assess its suitability for vehicle deceleration/acceleration.

Piston Hydraulic fluid

Fig. P11.10D

Nitrogen gas

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Library stacks require careful temperature and humidity control using air-conditioning processes considered in Sec. 12.8. © GreHinsdale Corbis RF/Age Fotostock America, Inc.

ENGINEERING CONTEXT Many systems of interest involve gas mixtures of two or more components. To apply the principles of thermodynamics introduced thus far to these systems requires that we evaluate properties of the mixtures. Means are available for determining the properties of mixtures from the mixture composition and the properties of the individual pure components from which the mixtures are formed. Methods for this purpose are discussed both in Chap. 11 and in the present chapter. The objective of the present chapter is to study mixtures where the overall mixture and each of its components can be modeled as ideal gases. General ideal gas mixture considerations are provided in the first part of the chapter. Understanding the behavior of ideal gas mixtures of dry air and water vapor is prerequisite to considering air-conditioning processes in the second part of the chapter, which is identified by the heading, Psychrometric Applications. In those processes, we sometimes must consider the presence of liquid water as well. We will also need to know how to handle ideal gas mixtures when we study the subjects of combustion and chemical equilibrium in Chaps. 13 and 14, respectively.

704

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12 Ideal Gas Mixture and Psychrometric Applications LEARNING OUTCOMES When you complete your study of this chapter, you will be able to... c

describe ideal gas mixture composition in terms of mass fractions or mole fractions.

c

use the Dalton model to relate pressure, volume, and temperature and to calculate changes in U, H, and S for ideal gas mixtures.

c

apply mass, energy, and entropy balances to systems involving ideal gas mixtures, including mixing processes.

c

demonstrate understanding of psychrometric terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature.

c

use the psychrometric chart to represent common air-conditioning processes and to retrieve data.

c

apply mass, energy, and entropy balances to analyze air-conditioning processes and cooling towers.

705

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Ideal Gas Mixtures: General Considerations TAKE NOTE... c In Secs. 12.1–12.3, we

introduce mixture concepts required for study of psychrometrics in the second part of this chapter and combustion in Chap. 13. c In Sec. 12.4, we extend the discussion of mixtures and provide several solved examples illustrating important types of mixture applications. For economy of effort, some readers may elect to defer Sec. 12.4 and proceed directly to content having more immediate interest for them: psychrometrics beginning in Sec. 12.5 or combustion beginning in Sec. 13.1.

12.1

Describing Mixture Composition

To specify the state of a mixture requires the composition and the values of two independent intensive properties such as temperature and pressure. The object of the present section is to consider ways for describing mixture composition. In subsequent sections, we show how mixture properties other than composition can be evaluated. Consider a closed system consisting of a gaseous mixture of two or more components. The composition of the mixture can be described by giving the mass or the number of moles of each component present. With Eq. 1.8, the mass, the number of moles, and the molecular weight of a component i are related by mi Mi

ni 5

(12.1)

where mi is the mass, ni is the number of moles, and Mi is the molecular weight of component i, respectively. When mi is expressed in terms of the kilogram, ni is in kmol. When mi is in terms of the pound mass, ni is in lbmol. However, any unit of mass can be used in this relationship. The total mass of the mixture, m, is the sum of the masses of its components j

m 5 m 1 1 m2 1 . . . 1 mj 5 a mi

(12.2)

i51

mass fractions

The relative amounts of the components present in the mixture can be specified in terms of mass fractions. The mass fraction mfi of component i is defined as mfi 5

gravimetric analysis

mi m

(12.3)

A listing of the mass fractions of the components of a mixture is sometimes referred to as a gravimetric analysis. Dividing each term of Eq. 12.2 by the total mass of mixture m and using Eq. 12.3 j

1 5 a mfi

(12.4)

i51

That is, the sum of the mass fractions of all the components in a mixture is equal to unity. The total number of moles in a mixture, n, is the sum of the number of moles of each of its components j

n 5 n1 1 n2 1 . . . 1 nj 5 a ni

(12.5)

i51

mole fractions

The relative amounts of the components present in the mixture also can be described in terms of mole fractions. The mole fraction yi of component i is defined as yi 5

ni n

(12.6)

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12.1 Describing Mixture Composition A listing of the mole fractions of the components of a mixture may be called a molar analysis. An analysis of a mixture in terms of mole fractions is also called a volumetric analysis.

707

molar analysis volumetric analysis

Dividing each term of Eq. 12.5 by the total number of moles of mixture n and using Eq. 12.6 j

1 5 a yi

(12.7)

i51

That is, the sum of the mole fractions of all the components in a mixture is equal to unity. The apparent (or average) molecular weight of the mixture, M, is defined as the ratio of the total mass of the mixture, m, to the total number of moles of mixture, n M5

m n

apparent molecular weight

(12.8)

Equation 12.8 can be expressed in a convenient alternative form. With Eq. 12.2, it becomes m1 1 m2 1 . . . 1 mj M5 n Introducing mi 5 niMi from Eq. 12.1 M5

n1M1 1 n2 M2 1 . . . 1 nj Mj n

Finally, with Eq. 12.6, the apparent molecular weight of the mixture can be calculated as a mole-fraction average of the component molecular weights j

M 5 a yi Mi

(12.9)

i51

consider the case of air. A sample of atmospheric air contains several gaseous components including water vapor and contaminants such as dust, pollen, and pollutants. The term dry air refers only to the gaseous components when all water vapor and contaminants have been removed. The molar analysis of a typical sample of dry air is given in Table 12.1. Selecting molecular weights for nitrogen, oxygen, argon, and carbon dioxide from Table A-1, and neglecting the trace substances neon, helium, etc., the apparent molecular weight of dry air obtained from Eq. 12.9 is M < 0.7808128.022 1 0.2095132.002 1 0.0093139.942 1 0.0003144.012 5 28.97 kg/ kmol 5 28.97 lb/ lbmol This value, which is the entry for air in Tables A-1, would not be altered significantly if the trace substances were also included in the calculation. b b b b b TABLE 12.1

Approximate Composition of Dry Air Component

Nitrogen Oxygen Argon Carbon dioxide Neon, helium, methane, and others

Mole Fraction (%)

78.08 20.95 0.93 0.03 0.01

dry air

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Next, we consider two examples illustrating, respectively, the conversion from an analysis in terms of mole fractions to an analysis in terms of mass fractions, and conversely.

cccc

EXAMPLE 12.1 c

Converting Mole Fractions to Mass Fractions The molar analysis of the gaseous products of combustion of a certain hydrocarbon fuel is CO2, 0.08; H2O, 0.11; O2, 0.07; N2, 0.74. (a) Determine the apparent molecular weight of the mixture. (b) Determine the composition in terms of mass fractions (gravimetric analysis). SOLUTION Known: The molar analysis of the gaseous products of combustion of a hydrocarbon fuel is given. Find: Determine (a) the apparent molecular weight of the mixture, (b) the composition in terms of mass fractions. Analysis: (a) Using Eq. 12.9 and molecular weights (rounded) from Table A-1

M 5 0.081442 1 0.111182 1 0.071322 1 0.741282 5 28.46 kg/ kmol 5 28.46 lb/ lbmol (b) Equations 12.1, 12.3, and 12.6 are the key relations required to determine the composition in terms of mass fractions.

➊

Although the actual amount of mixture is not known, the calculations can be based on any convenient amount. Let us base the solution on 1 kmol of mixture. Then, with Eq. 12.6 the amount ni of each component present in kmol is numerically equal to the mole fraction, as listed in column (ii) of the accompanying table. Column (iii) of the table gives the respective molecular weights of the components. Column (iv) of the table gives the mass mi of each component, in kg per kmol of mixture, obtained with Eq. 12.1 in the form mi 5 Mini. The values of column (iv) are obtained by multiplying each value of column (ii) by the corresponding value of column (iii). The sum of the values in column (iv) is the mass of the mixture: kg of mixture per kmol of mixture. Note that this sum is just the apparent mixture molecular weight determined in part (a). Finally, using Eq. 12.3, column (v) gives the mass fractions as a percentage. The values of column (v) are obtained by dividing the values of column (iv) by the column (iv) total and multiplying by 100. (i) Component

(ii)* ni

3

(iii) Mi

5

(iv)** mi

(v) mfi %

CO2 H2O O2 N2

0.08 0.11 0.07 0.74

3 3 3 3

44 18 32 28

5 5 5 5

3.52 1.98 2.24 20.72

12.37 6.96 7.87 72.80

28.46

100.00

1.00

*Entries in this column have units of kmol per kmol of mixture. For example, the first entry is 0.08 kmol of CO2 per kmol of mixture. **Entries in this column have units of kg per kmol of mixture. For example, the first entry is 3.52 kg of CO2 per kmol of mixture. The column sum, 28.46, has units of kg of mixture per kmol of mixture.

✓ Skills Developed Ability to… ❑ calculate the apparent

molecular weight with known mole fractions. ❑ determine the gravimetric analysis given the molar analysis.

➊ If the solution to part (b) were conducted on the basis of some other assumed amount of mixture—for example, 100 kmol or 100 lbmol—the same result for the mass fractions would be obtained, as can be verified.

Determine the mass, in kg, of CO2 in 0.5 kmol of mixture.

Ans. 1.76 kg.

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12.1 Describing Mixture Composition cccc

709

EXAMPLE 12.2 c

Converting Mass Fractions to Mole Fractions A gas mixture has the following composition in terms of mass fractions: H2, 0.10; N2, 0.60; CO2, 0.30. Determine (a) the composition in terms of mole fractions and (b) the apparent molecular weight of the mixture. SOLUTION Known: The gravimetric analysis of a gas mixture is known. Find: Determine the analysis of the mixture in terms of mole fractions (molar analysis) and the apparent molec-

ular weight of the mixture. Analysis: (a) Equations 12.1, 12.3, and 12.6 are the key relations required to determine the composition in terms of mole

fractions. Although the actual amount of mixture is not known, the calculation can be based on any convenient ➊ amount. Let us base the solution on 100 kg. Then, with Eq. 12.3, the amount mi of each component present, in kg, is equal to the mass fraction multiplied by 100 kg. The values are listed in column (ii) of the accompanying table. Column (iii) of the table gives the respective molecular weights of the components. Column (iv) of the table gives the amount ni of each component, in kmol per 100 kg of mixture, obtained using Eq. 12.1. The values of column (iv) are obtained by dividing each value of column (ii) by the corresponding value of column (iii). The sum of the values of column (iv) is the total amount of mixture, in kmol per 100 kg of mixture. Finally, using Eq. 12.6, column (v) gives the mole fractions as a percentage. The values of column (v) are obtained by dividing the values of column (iv) by the column (iv) total and multiplying by 100. ➋

(i) Component

(ii)* mi

4

(iii) Mi

5

(iv)** ni

(v) yi %

H2 N2 CO2

10 60 30

4 4 4

2 28 44

5 5 5

5.00 2.14 0.68

63.9 27.4 8.7

7.82

100.0

100

*Entries in this column have units of kg per 100 kg of mixture. For example, the first entry is 10 kg of H2 per 100 kg of mixture. **Entries in this column have units of kmol per 100 kg of mixture. For example, the first entry is 5.00 kmol of H2 per 100 kg of mixture. The column sum, 7.82, has units of kmol of mixture per 100 kg of mixture.

(b) The apparent molecular weight of the mixture can be found by using Eq. 12.9 and the calculated mole fractions. The value can be determined alternatively by using the column (iv) total giving the total amount of mixture in kmol per 100 kg of mixture. Thus, with Eq. 12.8

M5

100 kg kg m lb 5 5 12.79 5 12.79 n lbmol 7.82 kmol kmol

➊ If the solution to part (a) were conducted on the basis of some other assumed amount of mixture, the same result for the mass fractions would be obtained, as can be verified. ➋ Although H2 has the smallest mass fraction, its mole fraction is the largest. How many kmol of H2 would be present in 200 kg of mixture?

Ans. 10 kmol.

✓ Skills Developed Ability to… ❑ determine the molar analysis

given the gravimetric analysis.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

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12.2

Temperature = T Pressure = p

Gas 1 : n1 Gas 2 : n2

Volume = V

Fig. 12.1 Mixture of several gases.

Dalton model

partial pressure

The definitions given in Sec. 12.1 apply generally to mixtures. In the present section we are concerned only with ideal gas mixtures and introduce a model commonly used in conjunction with this idealization: the Dalton model. Consider a system consisting of a number of gases contained within a closed vessel of volume V as shown in Fig. 12.1. The temperature of the gas mixture is T and the pressure is p. The overall mixture is considered an ideal gas, so p, V, T, and the total number of moles of mixture n are related by the ideal gas equation of state

n moles mixture

Gas j : nj Boundary

Relating p, V, and T for Ideal Gas Mixtures1

p5n

RT V

(12.10)

With reference to this system let us consider the Dalton model. The Dalton model is consistent with the concept of an ideal gas as being made up of molecules that exert negligible forces on one another and whose volume is negligible relative to the volume occupied by the gas (Sec. 3.12.3). In the absence of significant intermolecular forces, the behavior of each component is unaffected by the presence of the other components. Moreover, if the volume occupied by the molecules is a very small fraction of the total volume, the molecules of each gas present may be regarded as free to roam throughout the full volume. In keeping with this simple picture, the Dalton model assumes that each mixture component behaves as an ideal gas as if it were alone at the temperature T and volume V of the mixture. It follows from the Dalton model that the individual components would not exert the mixture pressure p but rather a partial pressure. As shown below, the sum of the partial pressures equals the mixture pressure. The partial pressure of component i, pi, is the pressure that ni moles of component i would exert if the component were alone in the volume V at the mixture temperature T. The partial pressure can be evaluated using the ideal gas equation of state pi 5

ni RT V

(12.11)

Dividing Eq. 12.11 by Eq. 12.10 pi ni RT / V ni 5 5 5 yi p n nRT/ V Thus, the partial pressure of component i can be evaluated in terms of its mole fraction yi and the mixture pressure p pi 5 yi p

(12.12)

To show that the sum of partial pressures equals the mixture pressure, sum both sides of Eq. 12.12 to obtain j

j

j

a pi 5 a yi p 5 p a yi

i51

i51

i51

Since the sum of the mole fractions is unity (Eq. 12.7), this becomes j

p 5 a pi

(12.13)

i51

The Dalton model is a special case of the additive pressure rule for relating the pressure, specific volume, and temperature of gas mixtures introduced in Sec. 11.8. 1

The concept of an ideal gas mixture is a special case of the ideal solution concept introduced in Sec. 11.9.5.

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12.3 Evaluating U, H, S, and Specific Heats Among numerous other mixture rules found in the engineering literature is the Amagat model considered in the box that follows.

Introducing the Amagat Model The underlying assumption of the Amagat model is that each mixture component behaves as an ideal gas as if it existed separately at the pressure p and temperature T of the mixture. The volume that ni moles of component i would occupy if the component existed at p and T is called the partial volume, Vi, of component i. As shown below, the sum of the partial volumes equals the total volume. The partial volume can be evaluated using the ideal gas equation of state Vi 5

niRT p

(12.14)

Dividing Eq. 12.14 by the total volume V niRT/ p Vi ni 5 5 5 yi V n nRT/ p Thus, the partial volume of component i also can be evaluated in terms of its mole fraction yi and the total volume (12.15)

Vi 5 yiV

This relationship between volume fraction and mole fraction underlies the use of the term volumetric analysis as signifying an analysis of a mixture in terms of mole fractions. To show that the sum of partial volumes equals the total volume, sum both sides of Eq. 12.15 to obtain j

j

j

a Vi 5 a yi V 5 V a yi

i51

i51

i51

Since the sum of the mole fractions equals unity, this becomes j

V 5 a Vi

(12.16)

i51

Finally, note that the Amagat model is a special case of the additive volume model introduced in Sec. 11.8.

12.3

Evaluating U, H, S, and Specific Heats

To apply the conservation of energy principle to a system involving an ideal gas mixture requires evaluation of the internal energy, enthalpy, or specific heats of the mixture at various states. Similarly, to conduct an analysis using the second law normally requires the entropy of the mixture. The objective of the present section is to develop means to evaluate these properties for ideal gas mixtures.

12.3.1 Evaluating U and H Consider a closed system consisting of an ideal gas mixture. Extensive properties of the mixture, such as U, H, or S, can be found by adding the contribution of each component at the condition at which the component exists in the mixture. Let us apply this model to internal energy and enthalpy.

711

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Since the internal energy and enthalpy of ideal gases are functions of temperature only, the values of these properties for each component present in the mixture are determined by the mixture temperature alone. Accordingly j

U 5 U1 1 U2 1 . . . 1 Uj 5 a Ui

(12.17)

i51 j

H 5 H1 1 H 2 1 . . . 1 H j 5 a H i

(12.18)

i51

where Ui and Hi are the internal energy and enthalpy, respectively, of component i evaluated at the mixture temperature. Equations 12.17 and 12.18 can be rewritten on a molar basis as j

nu 5 n1u1 1 n2 u2 1 . . . 1 nj uj 5 a ni ui

(12.19)

i51

and j

nh 5 n1h1 1 n2 h2 1 . . . 1 nj hj 5 a ni hi

(12.20)

i51

where u and h are the specific internal energy and enthalpy of the mixture per mole of mixture, and ui and hi are the specific internal energy and enthalpy of component i per mole of i. Dividing by the total number of moles of mixture, n, gives expressions for the specific internal energy and enthalpy of the mixture per mole of mixture, respectively j

u 5 a yi ui

(12.21)

i51 j

h 5 a yi hi

(12.22)

i51

Each of the molar internal energy and enthalpy terms appearing in Eqs. 12.19 through 12.22 is evaluated at the mixture temperature only.

12.3.2 Evaluating cy and cp Differentiation of Eqs. 12.21 and 12.22 with respect to temperature results, respectively, in the following expressions for the specific heats cy and cp of the mixture on a molar basis j

cy 5 a yi cy,i

(12.23)

i51 j

cp 5 a yi cp,i

(12.24)

i51

That is, the mixture specific heats cp and cy are mole-fraction averages of the respective component specific heats. The specific heat ratio for the mixture is k 5 cp / cy.

12.3.3 Evaluating S The entropy of a mixture can be found, as for U and H, by adding the contribution of each component at the condition at which the component exists in the mixture.

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12.3 Evaluating U, H, S, and Specific Heats The entropy of an ideal gas depends on two properties, not on temperature alone as for internal energy and enthalpy. Accordingly, for the mixture j

S 5 S1 1 S2 1 . . . 1 Sj 5 a Si i51

(12.25)

where Si is the entropy of component i evaluated at the mixture temperature T and partial pressure pi (or at temperature T and total volume V). Equation 12.25 can be written on a molar basis as j

ns 5 n1s1 1 n2s2 1 . . . 1 nj sj 5 a ni si

(12.26)

i51

where s is the entropy of the mixture per mole of mixture and si is the entropy of component i per mole of i. Dividing by the total number of moles of mixture, n, gives an expression for the entropy of the mixture per mole of mixture j

s 5 a yi si

(12.27)

i51

In subsequent applications, the specific entropies si of Eqs. 12.26 and 12.27 are evaluated at the mixture temperature T and the partial pressure pi.

12.3.4 Working on a Mass Basis In cases where it is convenient to work on a mass basis, the foregoing expressions are written with the mass of the mixture, m, and the mass of component i in the mixture, mi, replacing, respectively, the number of moles of mixture, n, and the number of moles of component i, ni. Similarly, the mass fraction of component i, mfi, replaces the mole fraction, yi. All specific internal energies, enthalpies, and entropies are evaluated on a unit mass basis rather than on a per mole basis as above. To illustrate, Table 12.2 TABLE 12.2

Property Relations on a Mass Basis for Binary Ideal Gas Mixtures Notation:

m1 5 mass of gas 1, M1 5 molecular weight of gas 1 m2 5 mass of gas 2, M2 5 molecular weight of gas 2 m 5 mixture mass 5 m1 1 m2, mf1 5 (m1/m), mf2 5 (m2/m) T 5 mixture temperature, p 5 mixture pressure, V 5 mixture volume

Equation of state: p 5 m( R/M)T/V

(a)

where M 5 ( y1 M1 1 y2 M2 ) and the mole fractions y1 and y2 are given by y1 5 n1/(n1 1 n2), y2 5 n2/(n1 1 n2)

(b)

where n1 5 m1/M1 and n2 5 m2/M2. Partial pressures:

p1 5 y1 p, p2 5 y2p

(c)

Properties on a mass basis: Mixture enthalpy:

H 5 m1h1(T ) 1 m2h2(T )

Mixture internal energy:

U 5 m1u1(T ) 1 m2u2(T )

Mixture specific heats:

cp 5 (m1/m)cp1(T ) 1 (m2/m)cp2(T ) 5 (mf1 )cp1(T ) 1 (mf2 )cp2(T )

(d) (e) (f)

cy 5 (m1/m)cy1(T ) 1 (m2/m)cy2(T ) 5 (mf1 )cy1(T ) 1 (mf2 )cy2(T ) Mixture entropy:

S 5 m1s1(T, p1 ) 1 m2s2(T, p2 )

(g) (h)

713

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714

provides property relations on a mass basis for binary mixtures. These relations are applicable, in particular, to moist air, introduced in Sec. 12.5. By using the molecular weight of the mixture or of component i, as appropriate, data can be converted from a mass basis to a molar basis, or conversely, with relations of the form u 5 Mu, h 5 Mh, cp 5 Mcp, cy 5 Mcy, s 5 Ms

(12.28)

for the mixture, and ui 5 Mi ui, hi 5 Mi hi, cp,i 5 Micp,i, cy,i 5 Mi cy,i, si 5 Mi si

(12.29)

for component i.

12.4

Analyzing Systems Involving Mixtures

To perform thermodynamic analyses of systems involving nonreacting ideal gas mixtures requires no new fundamental principles. The conservation of mass and energy principles and the second law of thermodynamics are applicable in the forms previously introduced. The only new aspect is the proper evaluation of the required property data for the mixtures involved. This is illustrated in the present section, which deals with two classes of problems involving mixtures: In Sec. 12.4.1 the mixture is already formed, and we study processes in which there is no change in composition. Section 12.4.2 considers the formation of mixtures from individual components that are initially separate.

12.4.1 Mixture Processes at Constant Composition State 1

State 2

(n1, n2, …, nj) at T1, p1

(n1, n2, …, nj) at T2, p2

j _ U1 = Σ niu i(T1)

j _ U2 = Σ niu i(T2)

_ H1 = Σ nihi(T1)

j _ H2 = Σ nihi(T2)

i =1 j

i =1 j

_ S1 = Σ nis i(T1, pi1) i =1

i =1

i =1

j _ S2 = Σ nis i(T2, pi2) i =1

In the present section we are concerned with the case of ideal gas mixtures undergoing processes during which the composition remains constant. The number of moles of each component present, and thus the total number of moles of mixture, remain the same throughout the process. This case is shown schematically in Fig. 12.2, which is labeled with expressions for U, H, and S of a mixture at the initial and final states of a process undergone by the mixture. In accordance with the discussion of Sec. 12.3, the specific internal energies and enthalpies of the components are evaluated at the temperature of the mixture. The specific entropy of each component is evaluated at the mixture temperature and the partial pressure of the component in the mixture. The changes in the internal energy and enthalpy of the mixture during the process are given, respectively, by j

U2 2 U1 5 a ni3ui1T22 2 ui1T124

(12.30)

H2 2 H1 5 a ni3hi1T22 2 hi1T124

(12.31)

i51 j

Fig. 12.2 Process of an ideal gas mixture.

i51

where T1 and T2 denote the temperature at the initial and final states. Dividing by the number of moles of mixture, n, expressions for the change in internal energy and enthalpy of the mixture per mole of mixture result j

¢u 5 a yi3ui1T22 2 ui1T124

(12.32)

i51 j

¢h 5 a yi3hi1T22 2 hi1T124 i51

(12.33)

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Similarly, the change in entropy for the mixture is j

S2 2 S1 5 a ni3si1T2, pi22 2 si1T1, pi124

(12.34)

i51

where pi1 and pi2 denote, respectively, the initial and final partial pressures of component i. Dividing by the total moles of mixture, Eq. 12.34 becomes j

¢s 5 a yi3si1T2, pi22 2 si1T1, pi124

(12.35)

i51

Companion expressions for Eqs. 12.30 through 12.35 on a mass basis also can be written. This is left as an exercise. The foregoing expressions giving the changes in internal energy, enthalpy, and entropy of the mixture are written in terms of the respective property changes of the components. Accordingly, different datums might be used to assign specific enthalpy values to the various components because the datums would cancel when the component enthalpy changes are calculated. Similar remarks apply to the cases of internal energy and entropy.

Using Ideal Gas Tables For several common gases modeled as ideal gases, the quantities ui and hi appearing in the foregoing expressions can be evaluated as functions of temperature only from Tables A-22 and A-23. Table A-22 for air gives these quantities on a mass basis. Table A-23 gives them on a molar basis. The ideal gas tables also can be used to evaluate the entropy change. The change in specific entropy of component i required by Eqs. 12.34 and 12.35 can be determined with Eq. 6.20b as ¢si 5 s 8i 1T22 2 s 8i 1T12 2 R ln

pi2 pi1

Since the mixture composition remains constant, the ratio of the partial pressures in this expression is the same as the ratio of the mixture pressures, as can be shown by using Eq. 12.12 to write pi2 yi p2 p2 5 5 pi1 yi p1 p1 Accordingly, when the composition is constant, the change in the specific entropy of component i is simply ¢si 5 s 8i 1T22 2 s 8i 1T12 2 R ln

p2 p1

(12.36)

where p1 and p2 denote, respectively, the initial and final mixture pressures. The terms s8i of Eq. 12.36 can be obtained as functions of temperature for several common gases from Table A-23. Table A-22 for air gives s8 versus temperature.

Assuming Constant Specific Heats When the component specific heats cy,i and cp,i are taken as constants, the specific internal energy, enthalpy, and entropy changes of the mixture and the components of the mixture are given by ¢u 5 cy1T2 2 T12, ¢h 5 cp1T2 2 T12, p2 T2 ¢s 5 cp ln 2 R ln , p T1 1

¢ui 5 cy,i1T2 2 T12 ¢hi 5 cp,i1T2 2 T12 p2 T2 ¢si 5 cp,i ln 2 R ln p T1 1

(12.37) (12.38) (12.39)

TAKE NOTE...

When mixture composition remains constant, a ratio of partial pressures, pi2/p , i1 equals the ratio of mixture pressures, p2/p1.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications where the mixture specific heats cy and cp are evaluated from Eqs. 12.23 and 12.24, respectively, using data from Tables A-20 or the literature, as required. The expression for ¢u can be obtained formally by substituting the above expression for ¢ui into Eq. 12.32 and using Eq. 12.23 to simplify the result. Similarly, the expressions for ¢h and ¢s can be obtained by inserting ¢hi and ¢si into Eqs. 12.33 and 12.35, respectively, and using Eq. 12.24 to simplify. In the equations for entropy change, the ratio of mixture pressures replaces the ratio of partial pressures as discussed above. Similar expressions can be written for the mixture specific internal energy, enthalpy, and entropy changes on a mass basis. This is left as an exercise.

Using Computer Software The changes in internal energy, enthalpy, and entropy required in Eqs. 12.32, 12.33, and 12.35, respectively, also can be evaluated using computer software. Interactive Thermodynamics: IT provides data for a large number of gases modeled as ideal gases, and its use is illustrated in Example 12.4 below. The next example illustrates the use of ideal gas mixture relations for analyzing a compression process.

cccc

EXAMPLE 12.3 c

Analyzing an Ideal Gas Mixture Undergoing Compression A mixture of 0.3 lb of carbon dioxide and 0.2 lb of nitrogen is compressed from p1 5 1 atm, T1 5 540°R to p2 5 3 atm in a polytropic process for which n 5 1.25. Determine (a) the final temperature, in 8R, (b) the work, in Btu, (c) the heat transfer, in Btu, (d) the change in entropy of the mixture, in Btu/°R. SOLUTION Known: A mixture of 0.3 lb of CO2 and 0.2 lb of N2 is compressed in a polytropic process for which n 5 1.25.

At the initial state, p1 5 1 atm, T1 5 540°R. At the final state, p2 5 3 atm. Find: Determine the final temperature, in °R, the work, in Btu, the heat transfer, in Btu, and the entropy change

of the mixture in, Btu/°R. Schematic and Given Data: Boundary

p

Engineering Model: 2

1. As shown in the accompanying figure,

n = 1.25

the system is the mixture of CO2 and N2. The mixture composition remains constant during the compression.

0.3 lb CO2 0.2 lb N2

p1 = 1 atm, T1 = 540°R, p2 = 3 atm

Fig. E12.3

2. The Dalton model applies: Each mix-

States of the mixture 1 v

ture component behaves as if it were an ideal gas occupying the entire system volume at the mixture temperature. The overall mixture acts as an ideal gas. 3. The compression process is a poly-

tropic process for which n 5 1.25. 4. The changes in kinetic and potential

energy between the initial and final states can be ignored.

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Analysis: (a) For an ideal gas, the temperatures and pressures at the end states of a polytropic process are related by

Eq. 3.56 T2 5 T 1 a

p2 1n212/n b p1

Inserting values 3 0.2 T2 5 540a b 5 6738R 1 (b) The work for the compression process is given by

W5

#

2

p dV

1

Introducing pV n 5 constant and performing the integration W5

p2V2 2 p1V1 12n

With the ideal gas equation of state, this reduces to W5

m1R/ M21T2 2 T12 12n

The mass of the mixture is m 5 0.3 1 0.2 5 0.5 lb. The apparent molecular weight of the mixture can be calculated using M 5 m/n, where n is the total number of moles of mixture. With Eq. 12.1, the numbers of moles of CO2 and N2 are, respectively nCO2 5

0.3 5 0.0068 lbmol, 44

nN2 5

0.2 5 0.0071 lbmol 28

The total number of moles of mixture is then n 5 0.0139 lbmol. The apparent molecular weight of the mixture is M 5 0.5/0.0139 5 35.97. Calculating the work 1545 ft ? lbf b16738R 2 5408R2 35.97 lb ? 8R 1 Btu W5 ` ` 1 2 1.25 778 ft ? lbf 5 214.69 Btu 10.5 lb2a

where the minus sign indicates that work is done on the mixture, as expected. (c) With assumption 4, the closed system energy balance can be placed in the form

Q 5 ¢U 1 W where DU is the change in internal energy of the mixture. The change in internal energy of the mixture equals the sum of the internal energy changes of the components. With Eq. 12.30 ➊

¢U 5 nCO23uCO21T22 2 uCO21T124 1 nN23uN21T22 2 uN21T124

This form is convenient because Table A-23E gives internal energy values for N2 and CO2, respectively, on a molar basis. With values from this table ¢U 5 10.0068213954 2 29842 1 10.0071213340 2 26782 5 11.3 Btu Inserting values for DU and W into the expression for Q Q 5 111.3 2 14.69 5 23.39 Btu where the minus sign signifies a heat transfer from the system.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

(d) The change in entropy of the mixture equals the sum of the entropy changes of the components. With Eq. 12.34

¢S 5 nCO2 ¢sCO2 1 nN2 ¢sN2 where ¢sN2 and ¢sCO2 are evaluated using Eq. 12.36 and values of s 8 for N2 and CO2 from Table A-23E. That is 3 ¢S 5 0.0068a53.123 2 51.082 2 1.986 ln b 1 3 1 0.0071a47.313 2 45.781 2 1.986 ln b 1 5 20.0056 Btu/ 8R

➋

Entropy decreases in the process because entropy is transferred from the system accompanying heat transfer. ➊ In view of the relatively small temperature change, the changes in the internal energy and entropy of the mixture can be evaluated alternatively using the constant specific heat relations, Eqs. 12.37 and 12.39, respectively. In these equations, cy and cp are specific heats for the mixture determined using Eqs. 12.23 and 12.24 together with appropriate specific heat values for the components chosen from Table A-20E. ➋ Since the composition remains constant, the ratio of partial pressures equals the ratio of mixture pressures, so Eq. 12.36 can be used to evaluate the component specific entropy changes required here.

✓ Skills Developed Ability to… ❑ analyze a polytropic pro-

cess of a closed system for a mixture of ideal gases. ❑ apply ideal gas mixture principles. ❑ determine changes in internal energy and entropy for ideal gas mixtures using tabular data.

Recalling that polytropic processes are internally reversible, determine for the system the amount of entropy transfer accompanying heat transfer, in Btu/°R. Ans. 20.0056 Btu/°R.

The next example illustrates the application of ideal gas mixture principles for the analysis of a mixture expanding isentropically through a nozzle. The solution features the use of table data and IT as an alternative.

cccc

EXAMPLE 12.4 c

Considering an Ideal Gas Mixture Expanding Isentropically Through a Nozzle A gas mixture consisting of CO2 and O2 with mole fractions 0.8 and 0.2, respectively, expands isentropically and at steady state through a nozzle from 700 K, 5 atm, 3 m/s to an exit pressure of 1 atm. Determine (a) the temperature at the nozzle exit, in K, (b) the entropy changes of the CO2 and O2 from inlet to exit, in kJ/kmol ? K, (c) the exit velocity, in m/s. SOLUTION Known: A gas mixture consisting of CO2 and O2 in specified proportions expands isentropically through a nozzle

from specified inlet conditions to a given exit pressure. Find: Determine the temperature at the nozzle exit, in K, the entropy changes of the CO2 and O2 from inlet to

exit, in kJ/kmol ? K, and the exit velocity, in m/s.

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Schematic and Given Data: T

V1 = 3 m/s p1 = 5 atm T1 = 700 K

1

p2 = 1 atm

p1 = 5 atm T1 = 700 K

States of the mixture 1 2 2

p2 = 1 atm T2 = ? s

Fig. E12.4 Engineering Model: 1. The control volume shown by the dashed line on the accompanying figure operates at steady state. 2. The mixture composition remains constant as the mixture expands isentropically through the nozzle. 3. The Dalton model applies: The overall mixture and each mixture component act as ideal gases. The state

of each component is defined by the temperature and the partial pressure of the component. 4. The change in potential energy between inlet and exit can be ignored. Analysis: (a) The temperature at the exit can be determined using the fact that the expansion occurs isentropically: s2 2 s1 5 0.

As there is no change in the specific entropy of the mixture between inlet and exit, Eq. 12.35 can be used to write s2 2 s1 5 yO2 ¢sO2 1 yCO2 ¢sCO2 5 0

(a)

Since composition remains constant, the ratio of partial pressures equals the ratio of mixture pressures. Accordingly, the change in specific entropy of each component can be determined using Eq. 12.36. Equation (a) then becomes yO2 c s O8 21T22 2 s O8 21T12 2 R ln

p2 p2 d 1 yCO2 c s 8CO2 1T22 2 s C8 O2 1T12 2 R ln d 5 0 p1 p1

On rearrangement yO2 s O8 21T22 1 yCO2 s C8 O21T22 5 yO2 s O8 21T12 1 yCO2 s 8CO21T12 1 1yO2 1 yCO22R ln

p2 p1

The sum of mole fractions equals unity, so the coefficient of the last term on the right side is 1yO2 1 yCO22 5 1. Introducing given data, and values of s 8 for O2 and CO2 at T1 5 700 K from Table A-23 0.2s 8O21T22 1 0.8s 8CO21T22 5 0.21231.3582 1 0.81250.6632 1 8.314 ln

1 5

or 0.2s 8O2 1T22 1 0.8s C8 O2 1T22 5 233.42 kJ/ kmol ? K To determine the temperature T2 requires an iterative approach with the above equation: A final temperature T2 is assumed, and the s 8 values for O2 and CO2 are found from Table A-23. If these two values do not satisfy the equation, another temperature is assumed. The procedure continues until the desired agreement is attained. In the present case at T 5 510 K: at T 5 520 K:

0.21221.2062 1 0.81235.7002 5 232.80 0.21221.8122 1 0.81236.5752 5 233.62

Linear interpolation between these values gives T2 5 517.6 K.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Alternative Solution:

Alternatively, the following IT program can be used to evaluate T2 without resorting to iteration with table data. In the program, yO2 denotes the mole fraction of O2, p1_O2 denotes the partial pressure of O2 at state 1, s1_O2 denotes the entropy per mole of O2 at state 1, and so on. T1 5 700 // K p1 5 5 // bar p2 5 1 // bar yO2 5 0.2 yCO2 5 0.8 p1_O2 5 yO2 * p1 p1_CO2 5 yCO2 * p1 p2_O2 5 yO2 * p2 p2_CO2 5 yCO2 * p2 s1_O2 5 s_TP (“O2”,T1,p1_O2) s1_CO2 5 s_TP (“CO2”,T1,p1_CO2) s2_O2 5 s_TP (“O2”,T2,p2_O2) s2_CO2 5 s_TP (“CO2”,T2,p2_CO2) // When expressed in terms of these quantities, Eq. (a) takes the form yO2 * (s2_O2 2 s1_O2) 1 yCO2 * (s2_CO2 2 s1_CO2) 5 0 Using the Solve button, the result is T2 5 517.6 K, which agrees with the value obtained using table data. Note that IT provides the value of specific entropy for each component directly and does not return s 8 of the ideal gas tables. ➊ (b) The change in the specific entropy for each of the components can be determined using Eq. 12.36. For O2 ¢sO2 5 s O8 21T22 2 s O8 21T12 2 R ln

p2 p1

Inserting s 8 values for O2 from Table A-23 at T1 5 700 K and T2 5 517.6 K ¢sO2 5 221.667 2 231.358 2 8.314 ln10.22 5 3.69 kJ/ kmol ? K Similarly, with CO2 data from Table A-23 ¢sCO2 5 s C8 O21T22 2 s 8CO21T12 2 R ln

p2 p1

5 236.365 2 250.663 2 8.314 ln10.22 5 20.92 kJ/ kmol ? K

➋

(c) Reducing the energy rate balance for the one-inlet, one-exit control volume at steady state

0 5 h1 2 h2 1

V21 2 V22 2

where h1 and h2 are the enthalpy of the mixture, per unit mass of mixture, at the inlet and exit, respectively. Solving for V2 V2 5 2V21 1 21h1 2 h22 The term (h1 2 h2) in the expression for V2 can be evaluated as h1 2 h2 5

h1 2 h2 1 5 3yO21h1 2 h22O2 1 yCO21h1 2 h22CO24 M M

where M is the apparent molecular weight of mixture, and the molar specific enthalpies of O2 and CO2 are from Table A-23. With Eq. 12.9, the apparent molecular weight of the mixture is M 5 0.81442 1 0.21322 5 41.6 kg/ kmol Then, with enthalpy values at T1 5 700 K and T2 5 517.6 K from Table A-23 1 30.2121,184 2 15,3202 1 0.8127,125 2 18,46824 41.6 5 194.7 kJ/ kg

h1 2 h2 5

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12.4 Analyzing Systems Involving Mixtures Finally, ➌

V2 5

B

a3

2 m 2 kJ 1 kg ? m/ s 103 N ? m b 1 2a194.7 b ` ` ` ` 5 624 m/ s s kg 1N 1 kJ

➊ Parts (b) and (c) can be solved alternatively using IT. These parts also can be solved using a constant cp together with Eqs. 12.38 and 12.39. Inspection of Table A-20 shows that the specific heats of CO2 and O2 increase only slightly with temperature over the interval from 518 to 700 K, and so suitable constant values of cp for the components and the overall mixture can be readily determined. These alternative solutions are left as exercises. ➋ Each component experiences an entropy change as it passes from inlet to exit. The increase in entropy of the oxygen and the decrease in entropy of the carbon dioxide are due to entropy transfer accompanying heat transfer from the CO2 to the O2 as they expand through the nozzle. However, as indicated by Eq. (a), there is no change in the entropy of the mixture as it expands through the nozzle.

721

✓ Skills Developed Ability to… ❑ analyze the isentropic expan-

sion of an ideal gas mixture flowing through a nozzle. ❑ apply ideal gas mixture principles together with mass and energy balances to calculate the exit velocity of a nozzle. ❑ determine the exit temperature for a given inlet state and a given exit pressure using tabular data and alternatively using IT.

➌ Note the use of unit conversion factors in the calculation of V2. What would be the exit velocity, in m/s, if the isentropic nozzle efficiency were 90%? Ans. 592 m/s.

12.4.2

Mixing of Ideal Gases

Thus far, we have considered only mixtures that have already been formed. Now let us take up cases where ideal gas mixtures are formed by mixing gases that are initially separate. Such mixing is irreversible because the mixture forms spontaneously, and a work input from the surroundings would be required to separate the gases and return them to their respective initial states. In this section, the irreversibility of mixing is demonstrated through calculations of the entropy production. Three factors contribute to the production of entropy in mixing processes: 1. The gases are initially at different temperatures. 2. The gases are initially at different pressures. 3. The gases are distinguishable from one another. Entropy is produced when any of these factors is present during a mixing process. This is illustrated in the next example, where different gases, initially at different temperatures and pressures, are mixed. cccc

EXAMPLE 12.5 c

Investigating Adiabatic Mixing of Gases at Constant Total Volume Two rigid, insulated tanks are interconnected by a valve. Initially 0.79 lbmol of nitrogen at 2 atm and 4608R fills one tank. The other tank contains 0.21 lbmol of oxygen at 1 atm and 5408R. The valve is opened and the gases are allowed to mix until a final equilibrium state is attained. During this process, there are no heat or work interactions between the tank contents and the surroundings. Determine (a) the final temperature of the mixture, in 8R, (b) the final pressure of the mixture, in atm, (c) the amount of entropy produced in the mixing process, in Btu/8R. SOLUTION Known: Nitrogen and oxygen, initially separate at different temperatures and pressures, are allowed to mix without heat or work interactions with the surroundings until a final equilibrium state is attained.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Find: Determine the final temperature of the mixture, in 8R, the final pressure of the mixture, in atm, and the

amount of entropy produced in the mixing process, in Btu/8R. Schematic and Given Data: Engineering Model:

Insulation

1. The system is taken to be the nitrogen and the oxygen together. 2. When separate, each of the gases behaves as an ideal gas.

Valve

3. The final mixture acts as an ideal gas and the Dalton model applies:

Each mixture component occupies the total volume and exhibits the mixture temperature.

Initially 0.79 lbmol Initially 0.21 lbmol of O2 at 1 atm of N2 at 2 atm and 460°R and 540°R

4. No heat or work interactions occur with the surroundings, and there

are no changes in kinetic and potential energy. Fig. E12.5 Analysis: (a) The final temperature of the mixture can be determined from an energy balance. With assumption 4, the

closed system energy balance reduces to 0

0

¢U 5 Q 2 W

or

U2 2 U1 5 0

The initial internal energy of the system, U1, equals the sum of the internal energies of the two gases when separate U1 5 nN2 uN21TN22 1 nO2uO21TO22 where TN2 5 4608R is the initial temperature of the nitrogen and TO2 5 5408R is the initial temperature of the oxygen. The final internal energy of the system, U2, equals the sum of the internal energies of the two gases evaluated at the final mixture temperature T2 U2 5 nN2uN21T22 1 nO2uO21T22 Collecting the last three equations nN23uN21T22 2 uN21TN224 1 nO23uO21T22 2 uO21TO224 5 0 The temperature T2 can be determined using specific internal energy data from Table A-23E and an iterative procedure like that employed in part (a) of Example 12.4. However, since the specific heats of N2 and O2 vary little over the temperature interval from 460 to 5408R, the solution can be conducted accurately on the basis of constant specific heats. Hence, the foregoing equation becomes nN2cy, N21T2 2 TN22 1 nO2cy,O21T2 2 TO22 5 0 Solving for T2 T2 5

nN2cy, N2TN2 1 nO2cy, O2TO2 nN2cy, N2 1 nO2cy,O2

Selecting cy values for N2 and O2 from Table A-20E at the average of the initial temperatures of the gases, 5008R, and using the respective molecular weights to convert to a molar basis cy, N2 5 a28.01 cy, O2 5 a32.0

lb Btu Btu b a0.177 b 5 4.96 lbmol lb ? 8R lbmol ? 8R

lb Btu Btu b 5 4.99 b a0.156 lb ? 8R lbmol lbmol ? 8R

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Substituting values into the expression for T2 Btu Btu b14608R2 1 10.21 lbmol2a4.99 b15408R2 lbmol ? 8R lbmol ? 8R Btu Btu b 1 10.21 lbmol2a4.99 b 10.79 lbmol2a4.96 lbmol ? 8R lbmol ? 8R

10.79 lbmol2a4.96 T2 5

5 4778R (b) The final mixture pressure p2 can be determined using the ideal gas equation of state, p2 5 nRT2 / V, where n

is the total number of moles of mixture and V is the total volume occupied by the mixture. The volume V is the sum of the volumes of the two tanks, obtained with the ideal gas equation of state as follows V5

nN2RTN2 nO2RTO2 1 pN2 pO2

where pN2 5 2 atm is the initial pressure of the nitrogen and pO2 5 1 atm is the initial pressure of the oxygen. Combining results and reducing p2 5

1nN2 1 nO22T2 nN2TN2 nO2TO2 a 1 b pN2 pO2

Substituting values 11.0 lbmol214778R2 10.79 lbmol214608R2 10.21 lbmol215408R2 c 1 d 2 atm 1 atm 5 1.62 atm

p2 5

(c) Reducing the closed system form of the entropy balance

S2 2 S1 5

#

0

2

1

a

dQ b 1s T b

where the entropy transfer term drops out for the adiabatic mixing process. The initial entropy of the system, S1, is the sum of the entropies of the gases at the respective initial states S1 5 nN2sN21TN2, pN22 1 nO2sO21TO2, pO22 The final entropy of the system, S2, is the sum of the entropies of the individual components, each evaluated at the final mixture temperature and the partial pressure of the component in the mixture S2 5 nN2sN21T2, yN2 p22 1 nO2sO21T2, yO2 p22 Collecting the last three equations s 5 nN23sN21T2, yN2 p22 2 sN21TN2, pN224 1 nO23sO21T2, yO2 p22 2 sO21TO2, pO224 Evaluating the change in specific entropy of each gas in terms of a constant specific heat cp, this becomes s 5 nN2acp, N2 ln

yN2 p2 T2 2 R ln b pN2 TN2

1 nO2 acp, O2 ln

yO2 p2 T2 2 R ln b pO2 TO2

The required values for cp can be found by adding R to the cy values found previously (Eq. 3.45) cp, N2 5 6.95

Btu , lbmol ? 8R

cp, O2 5 6.98

Btu lbmol ? 8R

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Since the total number of moles of mixture n 5 0.79 1 0.21 5 1.0, the mole fractions of the two gases are yN2 5 0.79 and yO2 5 0.21. Substituting values into the expression for s gives s 5 0.79 lbmol c6.95

10.79211.62 atm2 Btu 4778R Btu ln a b 2 1.986 ln a bd lbmol ? 8R 4608R lbmol ? 8R 2 atm

10.21211.62 atm2 4778R Btu Btu ln a b 21.986 ln a bd 1 0.21 lbmol c6.98 lbmol ? 8R 5408R lbmol ? 8R 1 atm ➊ 5 1.168 Btu / 8R ➊ Entropy is produced when different gases, initially at different temperatures and pressures, are allowed to mix.

✓ Skills Developed Ability to… ❑ analyze the adiabatic

mixing of two ideal gases at constant total volume. ❑ apply energy and entropy balances to the mixing of two gases. ❑ apply ideal gas mixture principles assuming constant specific heats.

Determine the total volume of the final mixture, in ft3. Ans. 215 ft . 3

In the next example, we consider a control volume at steady state where two incoming streams form a mixture. A single stream exits.

cccc

EXAMPLE 12.6 c

Analyzing Adiabatic Mixing of Two Streams At steady state, 100 m3/min of dry air at 32°C and 1 bar is mixed adiabatically with a stream of oxygen (O2) at 127°C and 1 bar to form a mixed stream at 478C and 1 bar. Kinetic and potential energy effects can be ignored. Determine (a) the mass flow rates of the dry air and oxygen, in kg/min, (b) the mole fractions of the dry air and oxygen in the exiting mixture, and (c) the time rate of entropy production, in kJ/K ? min. SOLUTION Known: At steady state, 100 m3/min of dry air at 328C and 1 bar is mixed adiabatically with an oxygen stream

at 1278C and 1 bar to form a mixed stream at 478C and 1 bar. Find: Determine the mass flow rates of the dry air and oxygen, in kg/min, the mole fractions of the dry air and

oxygen in the exiting mixture, and the time rate of entropy production, in kJ/K ? min. Schematic and Given Data: Air T1 = 32°C p1 = 1 bar (AV)1 = 100 m3/min

1

3 Mixed stream

T2 = 127°C p2 = 1 bar Oxygen

Insulation

2

T3 = 47°C p3 = 1 bar

Fig. E12.6

Engineering Model: 1. The control volume identified by the dashed line on the accompanying figure operates at steady state. 2. No heat transfer occurs with the surroundings.

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#

3. Kinetic and potential energy effects can be ignored, and Wcv 5 0. 4. The entering gases can be regarded as ideal gases. The exiting mixture can be regarded as an ideal gas

mixture adhering to the Dalton model. 5. The dry air is treated as a pure component. Analysis: (a) The mass flow rate of the dry air entering the control volume can be determined from the given volumetric

flow rate (AV)1 1AV21 # ma1 5 ya1 where ya1 is the specific volume of the air at 1. Using the ideal gas equation of state 8314 N ? m a b 1305 K2 1R/ Ma2T1 28.97 kg ? K m3 ya1 5 5 5 0.875 p1 kg 105 N/ m2 The mass flow rate of the dry air is then kg 100 m3/ min # 5 114.29 ma1 5 3 min 0.875 m / kg The mass flow rate of the oxygen can be determined using mass and energy rate balances. At steady state, the amounts of dry air and oxygen contained within the control volume do not vary. Thus, for each component individually it is necessary for the incoming and outgoing mass flow rates to be equal. That is # # ma1 5 ma3 1dry air2 # # mo2 5 mo3 1oxygen2 Using assumptions 1–3 together with the foregoing mass flow rate relations, the energy rate balance reduces to # # # # 0 5 ma ha1T12 1 mo ho1T22 2 3ma ha1T32 1 mo ho1T324 # # where ma and mo denote the mass flow rates of the dry air and oxygen, respectively. The enthalpy of the mixture at the exit is evaluated by summing the contributions of the air and oxygen, each at the mixture temperature. # Solving for mo # # ha1T32 2 ha1T12 mo 5 ma c d ho1T22 2 ho1T32 The specific enthalpies can be obtained from Tables A-22 and A-23. Since Table A-23 gives enthalpy values on a molar basis, the molecular weight of oxygen is introduced into the denominator to convert the molar enthalpy values to a mass basis 1114.29 kg/ min21320.29 kJ/ kg 2 305.22 kJ/ kg2 # mo 5 1 a b111,711 kJ/ kmol 2 9,325 kJ/ kmol2 32 kg/ kmol 5 23.1

kg min

(b) To obtain the mole fractions of the dry air and oxygen in the exiting mixture, first convert the mass flow rates to molar flow rates using the respective molecular weights # 114.29 kg/ min ma # na 5 5 5 3.95 kmol/ min Ma 28.97 kg/ kmol # 23.1 kg/ min mo # no 5 5 5 0.72 kmol/ min Mo 32 kg/ kmol

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# # where n denotes molar flow rate. The molar flow rate of the mixture n is the sum # # # n 5 na 1 no 5 3.95 1 0.72 5 4.67 kmol/ min The mole fractions of the air and oxygen in the exiting mixture are, respectively # # na 3.95 no 0.72 ➊ ya 5 # 5 5 0.846 and yo 5 # 5 5 0.154 4.67 4.67 n n (c) For the control volume at steady state, the entropy rate balance reduces to

# # # # # 0 5 ma sa1T1,p12 1 mo so1T2, p22 2 3ma sa1T3, ya p32 1 mo so1T3, yo p324 1 s The specific entropy of each component in the exiting ideal gas mixture is evaluated at its partial pressure in # the mixture and at the mixture temperature. Solving for s # # # s 5 ma3sa1T3, ya p32 2 sa1T1, p124 1 mo3so1T3, yo p32 2 so1T2, p224 Since p1 5 p3, the specific entropy change of the dry air is sa1T3, ya p32 2 sa1T1, p12 5 s8a 1T322s8a 1T12 2

ya p3 R ln p1 Ma

5 s8a1T32 2 s8a1T12 2 R ln ya Ma The s8a terms are evaluated from Table A-22. Similarly, since p2 5 p3, the specific entropy change of the oxygen is so1T3, yo p32 2 so1T2, p22 5

1 3s 8o 1T32 2 s 8o 1T22 2 R ln yo4 Mo

The s 8o terms are evaluated from Table A-23. Note the use of the molecular weights Ma and Mo in the last two equations to obtain the respective entropy changes on a mass basis. The expression for the rate of entropy production becomes # mo R # # s 5 ma c s8a1T32 2 s8a 1T12 2 ln ya d 1 3s 8o1T32 2 s 8o1T22 2 R ln yo4 Ma Mo Substituting values kg kJ 8.314 kJ kJ # s 5 a114.29 21.71865 2a b ln 0.846 d b c 1.7669 kg ? K kg ? K 28.97 kg ? K min 1a ➋

23.1 kg/ min kJ kJ kJ 2 213.765 2 a8.314 b ln 0.154 d b c 207.112 kmol ? K kmol ? K kmol ? K 32 kg/ kmol

5 17.42

kJ K ? min

➊ This calculation is based on dry air modeled as a pure component (assumption 5). However, since O2 is a component of dry air (Table 12.1), the actual mole fraction of O2 in the exiting mixture is greater than given here. ➋ Entropy is produced when different gases, initially at different temperatures, are allowed to mix.

What are the mass fractions of air and oxygen in the exiting mixture? Ans. mfair 5 0.832, mfO2 5 0.168.

✓ Skills Developed Ability to… ❑ analyze the adiabatic mixing

of two ideal gas streams at steady state. ❑ apply ideal mixture principles together with mass, energy, and entropy rate balances.

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BIOCONNECTIONS Does spending time inside a building cause you to sneeze, cough, or develop a headache? If so, the culprit could be the ambient air. The term sick building syndrome (SBS) describes a condition where indoor air quality leads to acute health problems and comfort issues for building occupants. Effects of SBS are often linked to the amount of time an occupant spends within the space; yet while the specific cause and illness are frequently unidentifiable, symptoms typically subside after the occupant leaves the building. If the symptoms persist even after leaving the space and are diagnosed as a specific illness attributed to an airborne contaminant, the term building-related illness is a more accurate descriptor. The U.S. Environmental Protection Agency (EPA) recently conducted a study of 100 domestic office buildings, the largest study of its kind. The results agree with most previous findings that relate lower ventilation rates per person within office buildings to higher rates of SBS symptom reporting. Building codes and guidelines in the U.S. generally recommend ventilation rates within office buildings in the range of 15–20 ft3/min per occupant. Some of the spaces studied had ventilation rates below the guidelines. Careful design is needed to ensure that air distribution systems deliver acceptable ventilation for each space. Inadequate system installation and improper maintenance also can give rise to indoor air quality problems, even when appropriate standards have been applied in the design. The EPA study found this to be the case with the systems of many buildings in the study group. Also, testing and balancing of the installed systems were never conducted in several of the buildings to ensure that systems were operating according to design intent. Indoor air quality continues to be a significant concern for both building occupants and engineers who design and operate building air delivery systems.

Psychrometric Applications The remainder of this chapter is concerned with the study of systems involving mixtures of dry air and water vapor. A condensed water phase also may be present. Knowledge of the behavior of such systems is essential for the analysis and design of air-conditioning devices, cooling towers, and industrial processes requiring close control of the vapor content in air. The study of systems involving dry air and water is known as psychrometrics.

12.5

psychrometrics

Introducing Psychrometric Principles

The object of the present section is to introduce some important definitions and principles used in the study of systems involving dry air and water.

12.5.1 Moist Air The term moist air refers to a mixture of dry air and water vapor in which the dry air is treated as if it were a pure component. As can be verified by reference to appropriate property data, the overall mixture and each mixture component behave as ideal gases at the states under present consideration. Accordingly, for the applications to be considered, the ideal gas mixture concepts introduced previously apply directly. In particular, the Dalton model and the relations provided in Table 12.2 are applicable to moist air mixtures. Simply by identifying gas 1 with dry air, denoted by the subscript a, and gas 2 with water vapor, denoted by the subscript v, the table gives a useful set of moist air property relations. Referring to Fig. 12.3, let’s verify this point by obtaining a sampling of moist air relations and relating them to entries in Table 12.2.

moist air

TAKE NOTE...

Moist air is a binary mixture of dry air and water vapor, and the property relations of Table 12.2 apply.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Shown in Fig. 12.3—a special case of Fig. 12.1—is a closed system consisting of moist air occupying a volume V at mixture pressure p and mixture temperature T. The overall mixture is assumed to obey the ideal gas equation of state. Thus

Temperature = T Pressure = p

p5 na, ma: dry air nv, mv: water vapor n, m: mixture

Boundary

Volume = V

water vapor.

pa 5

pg

ma1R / Ma2T mv1R / Mv2T na RT nv RT 5 5 , pv 5 V V V V

v

Mixture temperature

Typical state of the water vapor in moist air

v

Fig. 12.4 T–y diagram for water vapor in an air–water mixture.

(12.41a)

where na and nv denote the moles of dry air and water vapor, respectively; ma, mv, Ma, and Mv are the respective masses and molecular weights. The amount of water vapor present is normally much less than the amount of dry air. Accordingly, the values of nv, mv, and pv are small relative to the corresponding values of na, ma, and pa. Forming ratios with Eqs. 12.40 and 12.41a, we get the following alternative expressions for pa and pv p pa 5 ya p, pv 5 yv p

State of the water vapor in a saturated mixture

(12.40)

where n, m, and M denote the moles, mass, and molecular weight of the mixture, respectively, and n 5 m/M. Each mixture component is considered to act as if it existed alone in the volume V at the mixture temperature T while exerting a part of the pressure. The mixture pressure is the sum of the partial pressures of the dry air and the water vapor: p 5 pa 1 pv. That is, the Dalton model applies. Using the ideal gas equation of state, the partial pressures pa and pv of the dry air and water vapor are, respectively

Fig. 12.3 Mixture of dry air and

T

m1R/ M2T nRT 5 V V

(12.41b)

where ya and yv are the mole fractions of the dry air and water vapor, respectively. These moist air expressions conform to Eqs. (c) in Table 12.2. A typical state of water vapor in moist air is shown in Fig. 12.4. At this state, fixed by the partial pressure pv and the mixture temperature T, the vapor is superheated. When the partial pressure of the water vapor corresponds to the saturation pressure of water at the mixture temperature, pg of Fig. 12.4, the mixture is said to be saturated. Saturated air is a mixture of dry air and saturated water vapor. The amount of water vapor in moist air varies from zero in dry air to a maximum, depending on the pressure and temperature, when the mixture is saturated.

saturated air

12.5.2 Humidity Ratio, Relative Humidity, Mixture Enthalpy, and Mixture Entropy

humidity ratio

A given moist air sample can be described in a number of ways. The mixture can be described in terms of the moles of dry air and water vapor present or in terms of the respective mole fractions. Alternatively, the mass of dry air and water vapor, or the respective mass fractions, can be specified. The composition also can be indicated by means of the humidity ratio v, defined as the ratio of the mass of the water vapor to the mass of dry air v5

mv ma

The humidity ratio is sometimes referred to as the specific humidity.

(12.42)

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12.5 Introducing Psychrometric Principles The humidity ratio can be expressed in terms of partial pressures and molecular weights by solving Eqs. 12.41a for ma and mv, respectively, and substituting the resulting expressions into Eq. 12.42 to obtain v5

Mv pvV/ RT Mv pv mv 5 5 ma Ma pa Ma paV/ RT

Introducing pa 5 p 2 pv and noting that the ratio of the molecular weight of water to that of dry air, Mv /Ma, is approximately 0.622, this expression can be written as v 5 0.622

pv p 2 pv

(12.43)

Moist air also can be described in terms of the relative humidity f, defined as the ratio of the mole fraction of water vapor yv in a given moist air sample to the mole fraction yv, sat in a saturated moist air sample at the same mixture temperature and pressure: f5

yv b yv,sat T, p

Since pv 5 yv p and pg 5 yv,sat p, the relative humidity can be expressed as f5

pv b pg T, p

relative humidity

(12.44)

The pressures in this expression for the relative humidity are labeled on Fig. 12.4. The humidity ratio and relative humidity can be measured. For laboratory measurements of humidity ratio, a hygrometer can be used in which a moist air sample is exposed to suitable chemicals until the moisture present is absorbed. The amount of water vapor is determined by weighing the chemicals. Continuous recording of the relative humidity can be accomplished by means of transducers consisting of resistance- or capacitancetype sensors whose electrical characteristics change with relative humidity.

Relative humidity

Temperature

Evaluating H, U, and S for Moist Air The values of H, U, and S for moist air modeled as an ideal gas mixture can be found by adding the contribution of each component at the condition at which the component exists in the mixture. For example, the enthalpy H of a given moist air sample is H 5 Ha 1 Hv 5 ma ha 1 m v hv

(12.45)

This moist air expression conforms to Eq. (d) in Table 12.2. Dividing by ma and introducing the humidity ratio gives the mixture enthalpy per unit mass of dry air mv H 5 ha 1 h 5 ha 1 vhv ma v ma

(12.46)

The enthalpies of the dry air and water vapor appearing in Eq. 12.46 are evaluated at the mixture temperature. An approach similar to that for enthalpy also applies to the evaluation of the internal energy of moist air. Reference to steam table data or a Mollier diagram for water shows that the enthalpy of superheated water vapor at low vapor pressures is very closely given by the saturated vapor value corresponding to the given temperature. Hence, the enthalpy of the water vapor hv in Eq. 12.46 can be taken as hg at the mixture temperature. That is hv < hg1T2

Sensing element

(12.47)

mixture enthalpy

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

mixture entropy

Equation 12.47 is used in the remainder of the chapter. Enthalpy data for water vapor as an ideal gas from Table A-23 are not used for hv because the enthalpy datum of the ideal gas tables differs from that of the steam tables. These different datums can lead to error when studying systems that contain both water vapor and a liquid or solid phase of water. The enthalpy of dry air, ha, can be obtained from the appropriate ideal gas table, Table A-22 or Table A-22E, however, because air is a gas at all states under present consideration and is closely modeled as an ideal gas at these states. In accord with Eq. (h) in Table 12.2, the moist air mixture entropy has two contributions: water vapor and dry air. The contribution of each component is determined at the mixture temperature and the partial pressure of the component in the mixture. Using Eq. 6.18 and referring to Fig. 12.4 for the states, the specific entropy of the water vapor is given by sv(T, pv) 5 sg(T) 2 R ln pv/pg, where sg is the specific entropy of saturated vapor at temperature T. Observe that the ratio of pressures, pv/pg, can be replaced by the relative humidity f, giving an alternative expression.

Using Computer Software Property functions for moist air are listed under the Properties menu of Interactive Thermodynamics: IT. Functions are included for humidity ratio, relative humidity, specific enthalpy and entropy as well as other psychrometric properties introduced later. The methods used for evaluating these functions correspond to the methods discussed in this chapter, and the values returned by the computer software agree closely with those obtained by hand calculations with table data. The use of IT for psychrometric evaluations is illustrated in examples later in the chapter.

BIOCONNECTIONS Medical practitioners and their patients have long noticed that influenza cases peak during winter. Speculation about the cause ranged widely, including the possibility that people spend more time indoors in winter and thus transmit the flu virus more easily, or that the peak might be related to less sunlight exposure during winter, perhaps affecting human immune responses. Since air is drier in winter, others suspected a link between relative humidity and influenza virus survival and transmission. In a 2007 study, using influenza-infected guinea pigs in climate-controlled habitats, researchers investigated the effects of variable habitat temperature and humidity on the aerosol spread of influenza virus. The study showed there were more infections when it was colder and drier, but relative humidity was a relatively weak variable in explaining findings. This prompted researchers to look for a better rationale. When data from the 2007 study were reanalyzed, a significant correlation was found between humidity ratio and influenza. Unlike relative humidity, humidity ratio measures the actual amount of moisture present in air. When humidity ratio is low, as in peak winter flu months, the virus survives longer and transmission rates increase, researchers say. These findings strongly point to the value of humidifying indoor air in winter, particularly in highrisk places such as nursing homes.

12.5.3

Modeling Moist Air in Equilibrium with Liquid Water

Thus far, our study of psychrometrics has been conducted as an application of the ideal gas mixture principles introduced in the first part of this chapter. However, many systems of interest are composed of a mixture of dry air and water vapor in contact with a liquid (or solid) water phase. To study these systems requires additional considerations.

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12.5 Introducing Psychrometric Principles Shown in Fig. 12.5 is a vessel containing liquid water, above which is a mixture of water vapor and dry air. If no interactions with the surroundings are allowed, liquid will evaporate until eventually the gas phase becomes saturated and the system attains an equilibrium state. For many engineering applications, systems consisting of moist air in equilibrium with a liquid water phase can be described simply and accurately with the following idealizations:

System boundary

Gas phase: Dry air and water vapor

c The dry air and water vapor behave as independent ideal gases. c The equilibrium between the liquid phase and the water vapor is not signifi-

cantly disturbed by the presence of the air. c The partial pressure of the water vapor equals the saturation pressure of water

corresponding to the temperature of the mixture: pv 5 pg(T ).

Fig. 12.5 System consisting of

Similar considerations apply for systems consisting of moist air in equilibrium with a solid water phase. The presence of the air actually alters the partial pressure of the vapor from the saturation pressure by a small amount whose magnitude is calculated in Sec. 14.6.

12.5.4

Liquid water

moist air in contact with liquid water.

Evaluating the Dew Point Temperature

A significant aspect of the behavior of moist air is that partial condensation of the water vapor can occur when the temperature is reduced. This type of phenomenon is commonly encountered in the condensation of vapor on windowpanes and on pipes carrying cold water. The formation of dew on grass is another familiar example. To study such condensation, consider a closed system consisting of a sample of moist air that is cooled at constant pressure, as shown in Fig. 12.6. The property diagram given on this figure locates states of the water vapor. Initially, the water vapor is superheated at state 1. In the first part of the cooling process, both the system pressure and the composition of the moist air remain constant. Accordingly, since pv 5 yvp, the partial pressure of the water vapor remains constant, and the water vapor cools at constant pv from state 1 to state d, called the dew point. The saturation temperature corresponding to pv is called the dew point temperature. This temperature is labeled on Fig. 12.6. In the next part of the cooling process, the system cools below the dew point temperature and some of the water vapor initially present condenses. At the final state, the system consists of a gas phase of dry air and water vapor in equilibrium with a liquid water phase. In accord with the discussion of Sec. 12.5.3, the vapor that remains is saturated vapor at the final temperature, state 2 of Fig. 12.6, with a partial pressure equal to the saturation pressure pg2 corresponding to this temperature. The condensate is saturated liquid at the final temperature: state 3 of Fig. 12.6. T

dew point temperature

pg1 pv1 < pg1 1

d Dew point 3 Condensate Final state 2 of the water vapor

Initial temperature Initial state of the water vapor Dew point temperature pg2 < pv1

p

p

Air and saturated vapor at final temperature

Final temperature

v

Dry air and superheated vapor at the initial temperature

Initial state

Fig. 12.6 States of water for moist air cooled at constant mixture pressure.

Condensate: saturated liquid at final temperature

Final state

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Referring again to Fig. 12.6, note that the partial pressure of the water vapor at the final state, pg2, is less than the initial value, pv1. Owing to condensation, the partial pressure decreases because the amount of water vapor present at the final state is less than at the initial state. Since the amount of dry air is unchanged, the mole fraction of water vapor in the moist air also decreases. In the next two examples, we illustrate the use of psychrometric properties introduced thus far. The examples consider, respectively, cooling moist air at constant pressure and at constant volume.

cccc

EXAMPLE 12.7 c

Cooling Moist Air at Constant Pressure A 1-lb sample of moist air initially at 708F, 14.7 lbf/in.2, and 70% relative humidity is cooled to 408F while keeping the pressure constant. Determine (a) the initial humidity ratio, (b) the dew point temperature, in 8F, and (c) the amount of water vapor that condenses, in lb. SOLUTION Known: A 1-lb sample of moist air is cooled at a constant mixture pressure of 14.7 lbf/in.2 from 70 to 408F. The

initial relative humidity is 70%. Find: Determine the initial humidity ratio, the dew point temperature, in 8F, and the amount of water vapor that condenses, in lb. Schematic and Given Data: pg1 = 0.3632 lbf/in.2 pv1 = 0.2542 lbf/in.2

T

m = 1 lb T1 = 70°F φ 1 = 70% T2 = 40°F

70°F Initial state of vapor Dewpoint temperature = 60°F pg2 = 0.1217 lbf/in.2 40°F Condensate Final state of vapor v

Fig. E12.7

Engineering Model: 1. The 1-lb sample of moist air is taken as the closed system. The system pressure remains constant at 14.7

lbf/in.2 2. The gas phase can be treated as an ideal gas mixture. The Dalton model applies: Each mixture component

acts as an ideal gas existing alone in the volume occupied by the gas phase at the mixture temperature. 3. When a liquid water phase is present, the water vapor exists as a saturated vapor at the system temperature.

The liquid present is a saturated liquid at the system temperature. Analysis: (a) The initial humidity ratio can be evaluated from Eq. 12.43. This requires the partial pressure of the water

vapor, pv1, which can be found from the given relative humidity and pg from Table A-2E at 708F as follows pv1 5 fpg 5 10.72a0.3632

lbf lbf b 5 0.2542 2 2 in. in.

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Inserting values in Eq. 12.43 v1 5 0.622 a

lb 1vapor2 0.2542 b 5 0.011 14.7 2 0.2542 lb 1dry air2

(b) The dew point temperature is the saturation temperature corresponding to the partial pressure, pv1. Interpolation

in Table A-2E gives T 5 608F. The dew point temperature is labeled on the accompanying property diagram. (c) The amount of condensate, mw, equals the difference between the initial amount of water vapor in the

sample, mv1, and the final amount of water vapor, mv2. That is mw 5 mv1 2 mv2 To evaluate mv1, note that the system initially consists of 1 lb of dry air and water vapor, so 1 lb 5 ma 1 mv1, where ma is the mass of dry air present in the sample. Since v1 5 mv1/ ma, ma 5 mv1/ v1. With this we get 1 lb 5

mv1 1 1 mv1 5 mv1a 1 1b v1 v1

Solving for mv1 mv1 5

1 lb 11/ v12 1 1

Inserting the value of v1 determined in part (a) mv1 5

1 lb 5 0.0109 lb 1vapor2 11/ 0.0112 1 1

➊ The mass of dry air present is then ma 5 1 2 0.0109 5 0.9891 lb (dry air). Next, let us evaluate mv2. With assumption 3, the partial pressure of the water vapor remaining in the system at the final state is the saturation pressure corresponding to 408F: pg 5 0.1217 lbf/in.2 Accordingly, the humidity ratio after cooling is found from Eq. 12.43 as v2 5 0.622 a

lb 1vapor2 0.1217 b 5 0.0052 14.7 2 0.1217 lb 1dry air2

The mass of the water vapor present at the final state is then mv2 5 v2ma 5 10.0052210.98912 5 0.0051 lb 1vapor2 Finally, the amount of water vapor that condenses is mw 5 mv1 2 mv2 5 0.0109 2 0.0051 5 0.0058 lb 1condensate2 ➊ The amount of water vapor present in a typical moist air mixture is considerably less than the amount of dry air present.

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ demonstrate understanding

of the dew point temperature and the formation of liquid condensate when pressure is constant. ❑ retrieve property data for water.

Determine the quality of the two-phase, liquid–vapor mixture and the relative humidity of the gas phase at the final state. Ans. 47%, 100%.

cccc

EXAMPLE 12.8 c

Cooling Moist Air at Constant Volume An air–water vapor mixture is contained in a rigid, closed vessel with a volume of 35 m3 at 1.5 bar, 1208C, and f 5 10%. The mixture is cooled at constant volume until its temperature is reduced to 228C. Determine (a) the dew point temperature corresponding to the initial state, in 8C, (b) the temperature at which condensation actually begins, in 8C, and (c) the amount of water condensed, in kg.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

SOLUTION Known: A rigid, closed tank with a volume of 35 m3 containing moist air initially at 1.5 bar, 1208C, and f 5 10%

is cooled to 228C. Find: Determine the dew point temperature at the initial state, in 8C, the temperature at which condensation actually begins, in 8C, and the amount of water condensed, in kg. Schematic and Given Data: T

Engineering Model:

pv1

V = 35 m3 Dew point temperature Initially moist air at 1.5 bar, 120°C φ = 10%

1

closed system. The system volume remains constant.

T1 Condensation begins

1′ 2

f2

1. The contents of the tank are taken as a

mixture. The Dalton model applies: Each mixture component acts as an ideal gas existing alone in the volume occupied by the gas phase at the mixture temperature.

g2

Boundary v

Fig. E12.8

2. The gas phase can be treated as an ideal gas

3. When a liquid water phase is present, the

water vapor exists as a saturated vapor at the system temperature. The liquid is a saturated liquid at the system temperature.

Analysis: (a) The dew point temperature at the initial state is the saturation temperature corresponding to the partial

pressure pv1. With the given relative humidity and the saturation pressure at 1208C from Table A-2 pv1 5 f1pg1 5 10.10211.9852 5 0.1985 bar Interpolating in Table A-2 gives the dew point temperature as 608C, which is the temperature condensation would begin if the moist air were cooled at constant pressure. (b) Whether the water exists as a vapor only, or as liquid and vapor, it occupies the full volume, which remains

constant. Accordingly, since the total mass of the water present is also constant, the water undergoes the constant specific volume process illustrated on the accompanying T–y diagram. In the process from state 1 to state 19, the water exists as a vapor only. For the process from state 19 to state 2, the water exists as a twophase liquid–vapor mixture. Note that pressure does not remain constant during the cooling process from state 1 to state 2. State 19 on the T–y diagram denotes the state where the water vapor first becomes saturated. The saturation temperature at this state is denoted as T 9. Cooling to a temperature less than T 9 results in condensation of some of the water vapor present. Since state 19 is a saturated vapor state, the temperature T 9 can be found by interpolating in Table A-2 with the specific volume of the water at this state. The specific volume of the vapor at state 19 equals the specific volume of the vapor at state 1, which can be evaluated from the ideal gas equation 1R/ Mv2T1 8314 N ? m 393 K 5a ba b pv1 18 kg ? K 0.1985 3 105 N/ m2 m3 5 9.145 kg

yv1 5

➊ Interpolation in Table A-2 with yv1 5 yg gives T 9 5 568C. This is the temperature at which condensation begins. (c) The amount of condensate equals the difference between the initial and final amounts of water vapor present. The mass of the water vapor present initially is

mv1 5

V 35 m3 5 5 3.827 kg yv1 9.145 m3/ kg

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735

The mass of water vapor present finally can be determined from the quality. At the final state, the water forms a two-phase liquid–vapor mixture having a specific volume of 9.145 m3/kg. Using this specific volume value, the quality x2 of the liquid–vapor mixture can be found as x2 5

yv2 2 y f 2 9.145 2 1.0022 3 10 23 5 5 0.178 yg2 2 y f 2 51.447 2 1.0022 3 10 23

where yf2 and yg2 are the saturated liquid and saturated vapor specific volumes at T2 5 228C, respectively. Using the quality together with the known total amount of water present, 3.827 kg, the mass of the water vapor contained in the system at the final Ski lls Dev elo ped state is

✓

mv2 5 10.178213.8272 5 0.681 kg The mass of the condensate, mw2, is then mw2 5 mv1 2 mv2 5 3.827 2 0.681 5 3.146 kg ➊ When a moist air mixture is cooled at constant mixture volume, the temperature at which condensation begins is not the dew point temperature corresponding to the initial state. In this case, condensation begins at 568C, but the dew point temperature at the initial state, determined in part (a), is 608C.

Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ demonstrate understanding

of the onset of condensation when cooling moist air at constant volume. ❑ retrieve property data for water.

Determine the humidity ratio at the initial state and the amount of dry air present, in kg. Ans. 0.0949, 40.389 kg.

No additional fundamental concepts are required for the study of closed systems involving mixtures of dry air and water vapor. Example 12.9, which builds on Example 12.8, brings out some special features of the use of conservation of mass and conservation of energy in analyzing this kind of system. Similar considerations can be used to study other closed systems involving moist air.

cccc

EXAMPLE 12.9 c

Evaluating Heat Transfer for Moist Air Cooling at Constant Volume An air–water vapor mixture is contained in a rigid, closed vessel with a volume of 35 m3 at 1.5 bar, 1208C, and f 5 10%. The mixture is cooled until its temperature is reduced to 228C. Determine the heat transfer during the process, in kJ. SOLUTION Known: A rigid, closed tank with a volume of 35 m3 containing moist air initially at 1.5 bar, 1208C, and f 5 10%

is cooled to 228C. Find: Determine the heat transfer for the process, in kJ. Schematic and Given Data: See the figure for Example 12.8. Engineering Model: 1. The contents of the tank are taken as a closed system. The system volume remains constant. 2. The gas phase can be treated as an ideal gas mixture. The Dalton model applies: Each component acts as

an ideal gas existing alone in the volume occupied by the gas phase at the mixture temperature.

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736

3. When a liquid water phase is present, the water vapor exists as a saturated vapor and the liquid is a satu-

rated liquid, each at the system temperature. 4. There is no work during the cooling process and no change in kinetic or potential energy. Analysis: Reduction of the closed system energy balance using assumption 4 results in

¢U 5 Q 2 W

0

or Q 5 U2 2 U1 where U1 5 maua1 1 mv1uv1 5 maua1 1 mv1ug1 and U2 5 maua2 1 mv2uv2 1 mw2uw2 5 maua2 1 mv2ug2 1 mw2uf2 In these equations, the subscripts a, v, and w denote, respectively, dry air, water vapor, and liquid water. The specific internal energy of the water vapor at the initial state can be approximated as the saturated vapor value at T1. At the final state, the water vapor is assumed to exist as a saturated vapor, so its specific internal energy is ug at T2. The liquid water at the final state is saturated, so its specific internal energy is uf at T2. Collecting the last three equations ➊

Q 5 ma1ua2 2 ua12 1 mv2ug2 1 mw2uf 2 2 mv1ug1

The mass of dry air, ma, can be found using the ideal gas equation of state together with the partial pressure of the dry air at the initial state obtained using pv1 5 0.1985 bar from the solution to Example 12.8 as follows: 311.5 2 0.19852 3 105 N/ m24135 m32 pa1V 5 1R/ Ma2T1 18314/ 28.97 N ? m/ kg ? K21393 K2 5 40.389 kg

ma 5

Then, evaluating internal energies of dry air and water from Tables A-22 and A-2, respectively Q 5 40.3891210.49 2 281.12 1 0.68112405.72 1 3.146192.322 2 3.82712529.32 5 22851.87 1 1638.28 1 290.44 2 9679.63 5 210,603 kJ The values for mv1, mv2, and mw2 are from the solution to Example 12.8. ➊ The first underlined term in this equation for Q is evaluated with specific internal energies from the ideal gas table for air, Table A-22. Steam table data are used to evaluate the second underlined term. The different datums for internal energy underlying these tables cancel because each of these two terms involves internal energy differences. Since the specific heat cya for dry air varies only slightly over the interval from 120 to 228C (Table A-20), the specific internal energy change of the dry air could be evaluated alternatively using a constant cya value. See the QuickQuiz that follows.

Calculate the change in internal energy of the dry air, in kJ, assuming a constant specific heat cya interpolated from Table A-20 at the average of the initial and final temperatures. Ans. 22854 kJ.

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ apply the energy balance to

the cooling of moist air at constant volume. ❑ retrieve property data for water.

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12.5 Introducing Psychrometric Principles

12.5.5 Evaluating Humidity Ratio Using the AdiabaticSaturation Temperature The humidity ratio v of an air–water vapor mixture can be determined, in principle, knowing the values of three mixture properties: the pressure p, the temperature T, and the adiabatic-saturation temperature Tas introduced in this section. The relationship among these mixture properties is obtained by applying conservation of mass and conservation of energy to an adiabatic saturator (see box). Equations 12.48 and 12.49 give the humidity ratio v in terms of the adiabaticsaturation temperature and other quantities: v5

ha1Tas2 2 ha1T2 1 v¿3hg1Tas2 2 hf 1Tas24

(12.48)

hg1T2 2 hf 1Tas2

where hf and hg denote the enthalpies of saturated liquid water and saturated water vapor, respectively, obtained from the steam tables at the indicated temperatures. The enthalpies of the dry air ha can be obtained from the ideal gas table for air. Alternatively, ha(Tas) 2 ha(T ) 5 cpa(Tas 2 T ), where cpa is an appropriate constant value for the specific heat of dry air. The humidity ratio v9 appearing in Eq. 12.48 is v¿ 5 0.622

pg1Tas2

(12.49)

p 2 pg1Tas2

where pg(Tas) is the saturation pressure at the adiabatic-saturation temperature and p is the mixture pressure.

Modeling an Adiabatic Saturator Figure 12.7 shows the schematic and process representations of an adiabatic saturator, which is a two-inlet, single-exit device through which moist air passes. The device is assumed to operate at steady state and without significant heat transfer with its surroundings. An air–water vapor mixture of unknown humidity ratio v enters the adiabatic saturator at a known pressure p and temperature T. As the mixture passes through the device, it comes into contact with a pool of water. If the entering mixture is not saturated (f , 100%), some of the water would evaporate. The energy required to evaporate the water would come from the moist air, so the mixture temperature would decrease as the air passes through the duct. For a sufficiently long duct, the mixture would be saturated as it exits (f 5 100%). Since a saturated mixture would be achieved without heat transfer with the surroundings, the temperature of the exiting mixture is the adiabatic-saturation temperature. As indicated on Fig. 12.7, a steady flow of makeup water at temperature Tas is added at the same rate at which water is evaporated. The pressure of the mixture is assumed to remain constant as it passes through the device. Equation 12.48 giving the humidity ratio v of the entering moist air in terms of p, T, and Tas can be obtained by applying conservation of mass and conservation of energy to the adiabatic saturator, as follows: At steady state, the mass flow rate of the dry air entering the device, m? a, must equal the mass flow rate of the dry air exiting. The mass flow rate of the makeup water is the difference ? , respectively. These between the exiting and entering vapor flow rates denoted by m? v and m9 v flow rates are labeled on Fig. 12.7a. At steady state, the energy rate balance reduces to # # # # # # (maha 1 mvhv) moist air 1 [(m v¿ 2 mv)hw] makeup 5 (maha 1 m v¿ hv) moist air entering

water

exiting

Several assumptions underlie this expression: Each of the two moist air streams is modeled as an ideal gas mixture of dry air and water vapor. Heat transfer with the surroundings

adiabatic-saturation temperature

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Moist air p, T, ω

State of the water vapor in the incoming moist air stream

Saturated mixture Tas, ω ′, p

m·a m· v

State of the water vapor in the exiting moist air stream

m·a m· ′v

T

pg(Tas)

Tas Insulation State of the makeup water

Makeup water — saturated liquid at Tas, mass flow rate = m· ′v – m· v

v

(a)

(b)

Fig. 12.7 Adiabatic saturator. (a) Schematic. (b) Process representation.

TAKE NOTE...

Although derived with reference to the adiabatic saturator in Fig. 12.7, the relationship provided by Eq. 12.48 applies generally to moist air mixtures and is not restricted to this type of system or even to control volumes. The relationship allows the humidity ratio v to be determined for any moist air mixture for which the pressure p, temperature T, and adiabatic-saturation temperature Tas are known.

? is assumed to be negligible. There is no work Wcv, and changes in kinetic and potential energy are ignored. Dividing by the mass flow rate of the dry air, m? a, the energy rate balance can be written on the basis of a unit mass of dry air passing through the device as

(ha 1 vhg ) moist air 1 [(v¿ 2 v)hf ] makeup 5 (ha 1 v¿hg ) moist air entering

dry-bulb temperature psychrometer

exiting

(12.50)

? ym? . ? ym? and v9 5 m9 where v 5 m v a v a For the exiting saturated mixture, the partial pressure of the water vapor is the saturation pressure corresponding to the adiabatic-saturation temperature, pg(Tas ). Accordingly, the humidity ratio v9 can be evaluated knowing Tas and the mixture pressure p, as indicated by Eq. 12.49. In writing Eq. 12.50, the specific enthalpy of the entering water vapor has been evaluated as that of saturated water vapor at the temperature of the incoming mixture, in accordance with Eq. 12.47. Since the exiting mixture is saturated, the enthalpy of the water vapor at the exit is given by the saturated vapor value at Tas. The enthalpy of the makeup water is evaluated as that of saturated liquid at Tas. When Eq. 12.50 is solved for v, Eq. 12.48 results. The details of the solution are left as an exercise.

12.6

wet-bulb temperature

water

Psychrometers: Measuring the Wet-Bulb and Dry-Bulb Temperatures

For moist air mixtures in the normal pressure and temperature ranges of psychrometrics, the readily-measured wet-bulb temperature is an important parameter. The wet-bulb temperature is read from a wet-bulb thermometer, which is an ordinary liquid-in-glass thermometer whose bulb is enclosed by a wick moistened with water. The term dry-bulb temperature refers simply to the temperature that would be measured by a thermometer placed in the mixture. Often a wet-bulb thermometer is mounted together with a dry-bulb thermometer to form an instrument called a psychrometer. The psychrometer of Fig. 12.8a is whirled in the air whose wet- and dry-bulb temperatures are to be determined. This induces air flow over the two thermometers. For the psychrometer of Fig. 12.8b, the air flow is induced by a battery-operated fan.

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12.6 Psychrometers: Measuring the Wet-Bulb and Dry-Bulb Temperatures

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Dry-bulb thermometer Wet-bulb thermometer

Bearing Handle Wet-bulb thermometer

Air in

Dry-bulb thermometer

Batteryoperated fan

Switch Air out

Wick

(a)

(b)

Fig. 12.8 Psychrometers. (a) Sling psychrometer. (b) Aspirating psychrometer. In each type of psychrometer, if the surrounding air is not saturated, water in the wick of the wet-bulb thermometer evaporates and the temperature of the remaining water falls below the dry-bulb temperature. Eventually a steady-state condition is attained by the wet-bulb thermometer. The wet- and dry-bulb temperatures are then read from the respective thermometers. The wet-bulb temperature depends on the rates of heat and mass transfer between the moistened wick and the air. Since these depend in turn on the geometry of the thermometer, air velocity, supply water temperature, and other factors, the wet-bulb temperature is not a mixture property. For moist air mixtures in the normal pressure and temperature ranges of psychrometric applications, the adiabatic saturation temperature introduced in Sec. 12.5.5 is closely approximated by the wet-bulb temperature. Accordingly, the humidity ratio for any such mixture can be calculated by using the wet-bulb temperature in Eqs. 12.48 and 12.49 in place of the adiabatic-saturation temperature. Close agreement between the adiabatic-saturation and wet-bulb temperatures is not generally found for moist air departing from normal psychrometric conditions.

BIOCONNECTIONS The National Weather Service is finding better ways to help measure our misery during cold snaps so we can avoid weather dangers. The wind chill index, for many years based on a single 1945 study, has been upgraded using new physiological data and computer modeling to better reflect the perils of cold winds and freezing temperatures. The new wind chill index is a standardized “temperature” that accounts for both the actual air temperature and the wind speed. The formula on which it is based uses measurements of skin tissue thermal resistance and computer models of the wind patterns over the human face, together with principles of heat transfer. Using the new index, an air temperature of 58F and a wind speed of 25 miles per hour correspond to a wind chill temperature of 2408F. The old index assigned a wind chill of only 2208F to the same conditions. With the new information, people are better armed to avoid exposure that can lead to such serious medical problems as frostbite. The improved measure was developed by universities, international scientific societies, and government in an effort that led to the new standard being adopted in the United States. Further upgrades are in the works to include the amount of cloud cover in the formula, since solar radiation is also an important factor in how cold it feels.

TAKE NOTE...

The humidity ratio for moist air mixtures considered in this book can be calculated by using the wet-bulb temperature in Eqs. 12.48 and 12.49 in place of the adiabatic saturation temperature.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

12.7

Graphical representations of several important properties of moist air are provided by psychrometric charts. The main features of one form of chart are shown in Fig. 12.9. Complete charts in SI and English units are given in Figs. A-9 and A-9E. These charts are constructed for a mixture pressure of 1 atm, but charts for other mixture pressures are also available. When the mixture pressure differs only slightly from 1 atm, Figs. A-9 remain sufficiently accurate for engineering analyses. In this text, such differences are ignored. Let us consider several features of the psychrometric chart: c Referring to Fig. 12.9, note that the abscissa gives the dry-bulb temperature and

the ordinate provides the humidity ratio. For charts in SI, the temperature is in 8C and v is expressed in kg, or g, of water vapor per kg of dry air. Charts in English units give temperature in 8F and v in lb, or grains, of water vapor per lb of dry air, where 1 lb 5 7000 grains. c Equation 12.43 shows that for fixed mixture pressure there is a direct correspondence between the partial pressure of the water vapor and the humidity ratio. Accordingly, the vapor pressure also can be shown on the ordinate, as illustrated on Fig. 12.9. c Curves of constant relative humidity are shown on psychrometric charts. On Fig. 12.9, curves labeled f 5 100, 50, and 10% are indicated. Since the dew point is the state where the mixture becomes saturated when cooled at constant vapor pressure, the dew point temperature corresponding to a given moist air state can be determined by following a line of constant v (constant pv) to the saturation line, f 5 100%. The dew point temperature and dry-bulb temperature are identical for states on the saturation curve. c Psychrometric charts also give values of the mixture enthalpy per unit mass of dry air in the mixture: ha 1 vhv. In Figs. A-9 and A-9E, the mixture enthalpy has units of kJ per kg of dry air and Btu per lb of dry air, respectively. The numerical values provided on these charts are determined relative to the following special reference states and reference values. In Fig. A-9, the enthalpy of the dry air ha is determined relative to a zero value at 08C, and not 0 K as in Table A-22. Accordingly, in place of Eq. 3.49 used to develop the enthalpy data of Tables A-22, the following expression is employed to evaluate the enthalpy of the dry air for use on the psychrometric chart: ha 5

#

T

cpa dT 5 cpaT18C2

(12.51)

273.15 K

Barometric pressure = 1 atm

φ

psychrometric chart

Psychrometric Charts

=1 00 %

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Scale for the mixture enthalpy per unit mass of dry air

Wet-bulb and dew point temperature scales Twb

%

Twb

φ

=

Twb

φ = 10% Dry-bulb temperature

Fig. 12.9 Psychrometric chart.

50

Volume per unit mass of dry air

ω

pv

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12.8 Analyzing Air-Conditioning Processes where cpa is a constant value for the specific heat cp of dry air and T(8C) denotes the temperature in 8C. For the chart in English units, Fig. A-9E, ha is determined relative to a datum of 08F, using ha 5 cpa T(8F), where T(8F) denotes the temperature in 8F. In the temperature ranges of Figs. A-9 and A-9E, cpa can be taken as 1.005 kJ/kg ? K and 0.24 Btu/lb ? 8R, respectively. On Figs. A-9 the enthalpy of the water vapor hv is evaluated as hg at the dry-bulb temperature of the mixture from Table A-2 or A-2E, as appropriate. c Another important parameter on psychrometer charts is the wet-bulb temperature. As illustrated by Figs. A-9, constant Twb lines run from the upper left to the lower right of the chart. The relationship between the wet-bulb temperature and other chart quantities is provided by Eq. 12.48. The wet-bulb temperature can be used in this equation in place of the adiabatic-saturation temperature for the states of moist air located on Figs. A-9. c Lines of constant wet-bulb temperature are approximately lines of constant mixture enthalpy per unit mass of dry air. This feature can be brought out by study of the energy balance for the adiabatic saturator, Eq. 12.50. Since the contribution of the energy entering the adiabatic saturator with the makeup water is normally much smaller than that of the moist air, the enthalpy of the entering moist air is very nearly equal to the enthalpy of the saturated mixture exiting. Accordingly, all states with the same value of the wet-bulb temperature (adiabatic-saturation temperature) have nearly the same value for the mixture enthalpy per unit mass of dry air. Although Figs. A-9 ignore this slight effect, some psychrometric charts are drawn to show the departure of lines of constant wet-bulb temperature from lines of constant mixture enthalpy. c As shown on Fig. 12.9, psychrometric charts also provide lines representing volume per unit mass of dry air, V/ma. Figures A-9 and A-9E give this quantity in units of m3/kg and ft3/lb, respectively. These specific volume lines can be interpreted as giving the volume of dry air or of water vapor, per unit mass of dry air, since each mixture component is considered to fill the entire volume. The psychrometric chart is easily used. a psychrometer indicates that in a classroom the dry-bulb temperature is 688F and the wet-bulb temperature is 608F. Locating the mixture state on Fig. A-9E corresponding to the intersection of these temperatures, we read v 5 0.0092 lb(vapor)/lb(dry air) and f 5 63%. b b b b b

12.8

Analyzing Air-Conditioning Processes

The purpose of the present section is to study typical air-conditioning processes using the psychrometric principles developed in this chapter. Specific illustrations are provided in the form of solved examples involving control volumes at steady state. In each example, the methodology introduced in Sec. 12.8.1 is employed to arrive at the solution. To reinforce principles developed in this chapter, the psychrometric parameters required by these examples are determined in most cases using tabular data from Appendix tables. Where a full psychrometric chart solution is not also provided, we recommend the example be solved using the chart, checking results with values from the solution presented.

12.8.1 Applying Mass and Energy Balances to Air-Conditioning Systems The object of this section is to illustrate the use of the conservation of mass and conservation of energy principles in analyzing systems involving mixtures of dry air and water vapor in which a condensed water phase may be present. The same basic

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications · Q cv

Moist air 1 m·a, m· v1

Boundary 2 Moist air m·a, m· v2

3 Liquid or vapor, m· w

Fig. 12.10 System for conditioning moist air.

solution approach that has been used in thermodynamic analyses considered thus far is applicable. The only new aspect is the use of the special vocabulary and parameters of psychrometrics. Systems that accomplish air-conditioning processes such as heating, cooling, humidification, or dehumidification are normally analyzed on a control volume basis. To consider a typical analysis, refer to Fig. 12.10, which shows a two-inlet, single-exit control volume at steady state. A moist air stream enters at 1, a moist air stream exits at 2, and a wateronly stream enters (or exits) at 3. The water-only stream may be a liquid # or a vapor. Heat transfer at the rate Qcv can occur between the control volume and its surroundings. Depending on the application, the value of # Qcv might be positive, negative, or zero.

Mass Balance At steady state, the amounts of dry air and water vapor contained within the control volume cannot vary. Thus, for each component individually it is necessary for the total incoming and outgoing mass flow rates to be equal. That is # # ma1 5 ma2 1dry air2 # # # mv1 1 mw 5 mv2 1water2 # For simplicity, the constant mass flow rate of the dry air is denoted by ma. The mass flow rates of the water vapor can be expressed conveniently in terms of humidity # # # # ratios as mv1 5 v1ma and mv2 5 v2ma. With these expressions, the mass balance for water becomes # # mw 5 ma1v2 2 v12 1water2 (12.52) When water is added at 3, v2 is greater than v1. TAKE NOTE...

As indicated by the developments of Sec. 12.8.1, several simplifying assumptions are made when analyzing air-conditioning systems considered in Examples 12.10–12.14 to follow. They include: c The control volume is at steady-state. c Moist air streams are ideal gas mixtures of dry air and water vapor adhering to the Dalton model. c Flow is one-dimensional where mass crosses the boundary of the control volume, and the effects of kinetic and potential energy at these locations are neglected. c The only work is flow work (Sec. 4.4.2) where mass crosses the boundary of the control volume.

Energy Balance

# Assuming Wcv 5 0 and ignoring all kinetic and potential energy effects, the energy rate balance reduces at steady state to # # # # # # 0 5 Qcv 1 1maha1 1 mv1hv12 1 mwhw 2 1maha2 1 mv2hv22 (12.53) In this equation, the entering and exiting moist air streams are regarded as ideal gas mixtures of dry air and water vapor. Equation 12.53 can be cast into a form that is particularly convenient for the analysis of air-conditioning systems. First, with Eq. 12.47 the enthalpies of the entering and exiting water vapor can be evaluated as the saturated vapor enthalpies corresponding to the temperatures T1 and T2, respectively, giving # # # # # # 0 5 Qcv 1 1m aha1 1 mv1hg12 1 mwhw 2 1maha2 1 mv2hg22 # # # # Then, with mv1 5 v1ma and mv2 5 v2ma, the equation can be expressed as # # # # 0 5 Qcv 1 ma1ha1 1 v1hg12 1 mwhw 2 ma1ha2 1 v2hg22 (12.54) Finally, introducing Eq. 12.52, the energy rate balance becomes # # 0 5 Qcv 1 ma 31ha1 2 ha22 1 v1hg1 1 1v2 2 v12hw 2 v2hg24

(12.55)

The first underlined term of Eq. 12.55 can be evaluated from Tables A-22 giving the ideal gas properties of air. Alternatively, since relatively small temperature differences are normally encountered in the class of systems under present consideration, this term can be evaluated as ha1 2 ha2 5 cpa (T1 2 T2), where cpa is a constant value for the specific heat of dry air. The second underlined term of Eq. 12.55 can be evaluated using steam table data together with known values for v1 and v2. As illustrated in discussions to follow, Eq. 12.55 also can be evaluated using the psychrometric chart or IT.

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12.8 Analyzing Air-Conditioning Processes

12.8.2

743

Conditioning Moist Air at Constant Composition

Building air-conditioning systems frequently heat or cool a moist air stream with no change in the amount of water vapor present. In such cases the humidity ratio v remains constant, while relative humidity and other moist air parameters vary. Example 12.10 gives an elementary illustration using the methodology of Sec. 12.8.1.

cccc

EXAMPLE 12.10 c

Heating Moist Air in a Duct Moist air enters a duct at 108C, 80% relative humidity, and a volumetric flow rate of 150 m3/min. The mixture is heated as it flows through the duct and exits at 308C. No moisture is added or removed, and the mixture pressure remains approximately constant at 1 bar. For steady-state operation, determine (a) the rate of heat transfer, in kJ/min, and (b) the relative humidity at the exit. Changes in kinetic and potential energy can be ignored. SOLUTION Known: Moist air that enters a duct at 108C and f 5 80% with a volumetric flow rate of 150 m3/min is heated

at constant pressure and exits at 308C. No moisture is added or removed. Find: Determine the rate of heat transfer, in kJ/min, and the relative humidity at the exit. Schematic and Given Data: m3 ___ (AV)1 = 150 min T1 = 10°C φ 1 = 80%

1

T

· Q cv T2 = 30°C

Boundary

2

pg(T2)

pv

T2 pg(T1) T1

Engineering Model: 1. The control volume shown in

the accompanying figure operates at steady state.

2

2. The changes in kinetic and

potential energy between inlet and # exit can be ignored and Wcv 5 0.

1

v

3. The entering and exiting

moist air streams are regarded as ideal gas mixtures adhering to the Dalton model.

Fig. E12.10a

Analysis:

#

(a) The heat transfer rate Q cv can be determined from the mass and energy rate balances. At steady state, the

amounts of dry air and water vapor contained within the control volume cannot vary. Thus, for each component individually it is necessary for the incoming and outgoing mass flow rates to be equal. That is # # ma1 5 ma2 1dry air2 # # mv1 5 mv2 1water vapor2 # # For simplicity, the constant mass flow rates of the dry air and water vapor are denoted, respectively, by m a and m v. From these considerations, it can be concluded that the humidity ratio is the same at the inlet and exit: v1 5 v2. The steady-state form of the energy rate balance reduces with assumption 2 to # #0 # # # # 0 5 Qcv 2 Wcv 1 1maha1 1 mvhv12 2 1maha2 1 mvhv22 In writing this equation, the incoming and outgoing moist air streams are regarded as ideal gas mixtures of dry air and water vapor. # Solving for Qcv # # # Qcv 5 ma1ha2 2 ha12 1 mv1hv2 2 hv12

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

# # # Noting that mv 5 vma, where v is the humidity ratio, the expression for Qcv can be written in the form # # ➊ Qcv 5 ma31ha2 2 ha12 1 v1hv2 2 hv124 (a) # To evaluate Qcv from this expression requires the specific enthalpies of the dry air and water vapor at the inlet and exit, the mass flow rate of the dry air, and the humidity ratio. The specific enthalpies of the dry air are obtained from Table A-22 at the inlet and exit temperatures T1 and T2, respectively: ha1 5 283.1 kJ/kg, ha2 5 303.2 kJ/kg. The specific enthalpies of the water vapor are found using hv < hg and data from Table A-2 at T1 and T2, respectively: hg1 5 2519.8 kJ/kg, hg2 5 2556.3 kJ/kg. The mass flow rate of the dry air can be determined from the volumetric flow rate at the inlet (AV)1 1AV21 # ma 5 ya1 In this equation, ya1 is the specific volume of the dry air evaluated at T1 and the partial pressure of the dry air pa1. Using the ideal gas equation of state ya1 5

1R/ M2T1 pa1

The partial pressure pa1 can be determined from the mixture pressure p and the partial pressure of the water vapor pv1: pa1 5 p 2 pv1. To find pv1, use the given inlet relative humidity and the saturation pressure at 108C from Table A-2 pv1 5 f1pg1 5 10.8210.01228 bar2 5 0.0098 bar Since the mixture pressure is 1 bar, it follows that pa1 5 0.9902 bar. The specific volume of the dry air is then a ya1 5

8314 N ? m b1283 K2 28.97 kg ? K

10.9902 3 105 N/ m22

5 0.82 m3/ kg

Using this value, the mass flow rate of the dry air is 150 m3/ min # ma 5 5 182.9 kg/ min 0.82 m3/ kg The humidity ratio v can be found from pv1 0.0098 b 5 0.622a b p 2 pv1 1 2 0.0098 kg 1vapor2 5 0.00616 kg 1dry air2

v 5 0.622a

Finally, substituting values into Eq. (a) we get # Qcv 5 182.931303.2 2 283.12 1 10.00616212556.3 2 2519.824 5 3717 kJ/ min (b) The states of the water vapor at the duct inlet and exit are located on the accompanying T–y diagram. Since the composition of the moist air and the mixture pressure remain constant, the partial pressure of the water vapor at the exit equals the partial pressure of the water vapor at the inlet: pv2 5 pv1 5 0.0098 bar. The relative humidity at the exit is then

➋

f2 5

pv2 0.0098 5 5 0.231123.1%2 pg2 0.04246

where pg2 is from Table A-2 at 308C. Alternative Psychrometric Chart Solution: Let us consider an alternative solution using the psychrometric chart. As shown on the sketch of the psychrometric chart, Fig. E12.10b, the state of the moist air at the inlet is defined by f1 5 80% and a dry-bulb temperature of 108C. From the solution to part (a), we know that the humidity ratio has the same value at the exit as at the inlet. Accordingly, the state of the moist air at the exit is fixed by

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12.8 Analyzing Air-Conditioning Processes

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v2 5 v1 and a dry-bulb temperature of 308C. By inspection of Fig. A-9, the relative humidity at the duct exit is about 23%, and thus in agreement with the result of part (b). The rate of heat transfer can be evaluated from the psychrometric chart using the following expression obtained by rearranging Eq. (a) of part (a) to read # # (b) Qcv 5 ma31ha 1 vhv22 2 1ha 1 vhv214 # To evaluate Qcv from this expression requires values for the mixture enthalpy per unit mass of dry air (ha 1 vhv) at the inlet and exit. These can be determined by inspection of the psychrometric chart, Fig. A-9, as (ha 1 vhv)1 5 25.7 kJ/kg(dry air), (ha 1 vhv)2 5 45.9 kJ/kg(dry air). Using the specific volume value ya1 at the inlet state read from the chart together with the given volumetric flow rate at the inlet, the mass flow rate of the dry air is found as # ma 5

kg1dry air2 150 m3/ min 5 185 3 min 0.81 m / kg1dry air2

Substituting values into the energy rate balance, Eq. (b), we get kg1dry air2 # kJ 145.9 2 25.72 Qcv 5 185 min kg1dry air2 kJ 5 3737 min which agrees closely with the result obtained in part (a), as expected.

lpy

➌

tha

r

xtu

Mi

1 10°C

n ee

φ

=

0% % 80 =

10

φ

2

ω

ω2 = ω1

30°C Dry-bulb temperature

Fig. E12.10b

# ➊ The first underlined term in this equation for Qcv is evaluated with specific enthalpies from the ideal gas table for air, Table A-22. Steam table data are used to evaluate the second underlined term. Note that the different datums for enthalpy underlying these tables cancel because each of the two terms involves enthalpy differences only. Since the specific heat cpa for dry air varies only slightly over the interval from 10 to 308C (Table A-20), the specific enthalpy change of the dry air could be evaluated alternatively with cpa 5 1.005 kJ/kg ? K. ➋ No water is added or removed as the moist air passes through the duct at constant pressure; accordingly, the humidity ratio v and the partial pressures pv and pa remain constant. However, because the saturation pressure increases as the temperature increases from inlet to exit, the relative humidity decreases: f2 , f1. ➌ The mixture pressure, 1 bar, differs slightly from the pressure used to construct the psychrometric chart, 1 atm. This difference is ignored. Using the psychrometric chart, what is the dew point temperature, in 8C, for the moist air entering? At the exit? Ans. < 78C, same.

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ apply mass and energy bal-

ances for heating at constant composition in a control volume at steady state. ❑ retrieve necessary property data.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Cooling coil m· r i 1

Heating coil

e 2

Moist air m· a, T1, ω 1, p = 1 atm

φ 2 = 100% T 2 < T1 ω 2 T2 ω3 = ω2 Initial dew point

φ1 1

φ3

2

φ=

% 100

ω

3

m· w Condensate – saturated at T2

T2

(Dehumidifier section)

(Heating section) (a)

T3 Dry-bulb temperature

T1

(b)

Fig. 12.11 Dehumidification. (a) Equipment schematic. (b) Psychrometric chart representation.

12.8.3 TAKE NOTE...

A dashed line on a property diagram signals only that a process has occurred between initial and final equilibrium states, and does not define a path for the process.

Dehumidification

When a moist air stream is cooled at constant mixture pressure to a temperature below its dew point temperature, some condensation of the water vapor initially present will occur. Figure 12.11 shows the schematic of a dehumidifier using this principle. Moist air enters at state 1 and flows across a cooling coil through which a refrigerant or chilled water circulates. Some of the water vapor initially present in the moist air condenses, and a saturated moist air mixture exits the dehumidifier section at state 2. Although water condenses at various temperatures, the condensed water is assumed to be cooled to T2 before it exits the dehumidifier. Since the moist air leaving the humidifier is saturated at a temperature lower than the temperature of the moist air entering, the moist air stream at state 2 might be uncomfortable for direct use in occupied spaces. However, by passing the stream through a following heating section, it can be brought to a condition—state 3—most occupants would regard as comfortable. Let us sketch the procedure for evaluating the rates at which condensate exits and refrigerant circulates. This requires the use of mass and energy rate balances for the dehumidifier section. They are developed next.

Mass Balance # The mass flow rate of the condensate mw can be related to the mass flow rate of the # dry air ma by applying conservation of mass separately for the dry air and water passing through the dehumidifier section. At steady state # # ma1 5 ma2 (dry air) # # # mv1 5 mw 1 mv2 (water) # The common mass flow rate of the dry air is denoted as ma. Solving for the mass flow rate of the condensate # # # mw 5 mv1 2 mv2 # # # # Introducing mv1 5 v1ma and mv2 5 v2ma, the amount of water condensed per unit mass of dry air passing through the device is # mw # 5 v1 2 v2 ma

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This expression requires the humidity ratios v1 and v2. Because no moisture is added or removed in the heating section, it can be concluded from conservation of mass that v2 5 v3, so v3 can be used in the above equation in place of v2.

Energy Balance # The mass flow rate of the refrigerant through the cooling coil mr can be related to # the mass flow rate of the dry #air ma by means of an energy rate balance applied to the dehumidifier section. With Wcv 5 0, negligible heat transfer with the surroundings, and no significant kinetic and potential energy changes, the energy rate balance reduces at steady state to # # # # # # 0 5 mr1hi 2 he2 1 1maha1 1 mv1hv12 2 mwhw 2 1maha2 1 mv2hv22 where hi and he denote the specific enthalpy values of the refrigerant entering and # # # # exiting the dehumidifier section, respectively. Introducing mv1 5 v1ma, mv2 5 v2ma, # # and mw 5 1v1 2 v22ma # # 0 5 mr1hi 2 he2 1 ma31ha1 2 ha22 1 v1hg1 2 v2hg2 2 1v1 2 v22hf24 where the specific enthalpies of the water vapor at 1 and 2 are evaluated at the saturated vapor values corresponding to T1 and T2, respectively. Since the condensate is assumed to exit as a saturated liquid at T2, hw 5 hf2. Solving for the refrigerant mass flow rate per unit mass of dry air flowing through the device # 1ha1 2 ha22 1 v1hg1 2 v2hg2 2 1v1 2 v22hf 2 mr # 5 he 2 hi ma The accompanying psychrometric chart, Fig. 12.11b, illustrates important features of the processes involved. As indicated by the chart, the moist air first cools from state 1, where the temperature is T1 and the humidity ratio is v1, to state 2, where the mixture is saturated (f2 5 100%), the temperature is T2 , T1, and the humidity ratio is v2 , v1. During the subsequent heating process, the humidity ratio remains constant, v2 5 v3, and the temperature increases to T3. Since all states visited are not equilibrium states, these processes are indicated on the psychrometric chart by dashed lines. The example that follows provides an illustration involving dehumidification where one of the objectives is the refrigerating capacity of the cooling coil.

cccc

EXAMPLE 12.11 c

Assessing Dehumidifier Performance Moist air at 308C and 50% relative humidity enters a dehumidifier operating at steady state with a volumetric flow rate of 280 m3/min. The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated at 108C. Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat transfer to the surroundings and pressure remains constant at 1.013 bar. Determine (a) the mass flow rate of the dry air, in kg/min, (b) the rate at which water is condensed, in kg per kg of dry air flowing through the control volume, and (c) the required refrigerating capacity, in tons. SOLUTION Known: Moist air enters a dehumidifier at 308C and 50% relative humidity with a volumetric flow rate of 280 m3/min.

Condensate and moist air exit in separate streams at 108C. Determine: Find the mass flow rate of the dry air, in kg/min, the rate at which water is condensed, in kg per kg of dry air, and the required refrigerating capacity, in tons.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Schematic and Given Data: Cooling coil

Engineering Model:

Heating coil

1. The control volume shown in the accompanying fig-

ure operates at steady state. Changes in kinetic and # potential energy can be neglected, and Wcv 5 0. Saturated mixture 10°C

m3/min

(AV)1 = 280 T1 = 30°C φ 1 = 50%

2. There is no significant heat transfer to the sur-

roundings. 3. The pressure remains constant throughout at

1.013 bar. 4. At location 2, the moist air is saturated. The con1

3

densate exits at location 3 as a saturated liquid at temperature T2.

2 Control volume

5. The moist air streams are regarded as ideal gas

Condensate, saturated at T2 = 10°C

mixtures adhering to the Dalton model.

Fig. E12.11a Analysis: (a) At steady state, the mass flow rates of the dry air entering and exiting are equal. The common mass flow

rate of the dry air can be determined from the volumetric flow rate at the inlet 1AV21 # ma 5 ya1 The specific volume of the dry air at inlet 1, ya1, can be evaluated using the ideal gas equation of state, so # ma 5

1AV21 1R/ Ma21T1/ pa12

The partial pressure of the dry air pa1 can be determined from pa1 5 p1 2 pv1. Using the relative humidity at the inlet f1 and the saturation pressure at 308C from Table A-2 pv1 5 f1pg1 5 10.5210.042462 5 0.02123 bar # Thus, pa1 5 1.013 2 0.02123 5 0.99177 bar. Inserting values into the expression for ma gives 1280 m3/ min210.99177 3 105 N/ m22 # ma 5 5 319.35 kg/ min 18314/ 28.97 N ? m/ kg ? K21303 K2 # # # # # # # (b) Conservation of mass for the water requires mv1 5 mv2 1 mw. With mv1 5 v1ma and mv2 5 v2ma, the rate at which water is condensed per unit mass of dry air is # mw # 5 v1 2 v2 ma The humidity ratios v1 and v2 can be evaluated using Eq. 12.43. Thus, v1 is v1 5 0.662 a

kg1vapor2 pv1 0.02123 b 5 0.622a b 5 0.0133 p1 2 pv1 0.99177 kg1dry air2

Since the moist air is saturated at 108C, pv2 equals the saturation pressure at 108C: pg 5 0.01228 bar from Table A-2. Equation 12.43 then gives v2 5 0.0076 kg(vapor)/kg(dry air). With these values for v1 and v2 # kg1condensate2 mw # 5 0.0133 2 0.0076 5 0.0057 kg1dry air2 ma

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#

(c) The rate of heat transfer Qcv between the moist air stream and the refrigerant coil can be determined

using an energy rate balance. With assumptions 1 and 2, the steady-state form of the energy rate balance reduces to # # # # # # (a) 0 5 Qcv 1 1maha1 1 mv1hv12 2 mwhw 2 1maha2 1 mv2hv22 # # # # # # With mv1 5 v1ma, mv2 5 v2ma, and mw 5 1v1 2 v22ma, this becomes # # Qcv 5 ma31ha2 2 ha12 2 v1hg1 1 v2hg2 1 1v1 2 v22hf24 (b) which agrees with Eq. 12.55. In Eq. (b), the specific enthalpies of the water vapor at 1 and 2 are evaluated at the saturated vapor values corresponding to T1 and T2, respectively, and the specific enthalpy of the exiting condensate is evaluated as hf at T2. Selecting enthalpies from Tables A-2 and A-22, as appropriate, Eq. (b) reads # Qcv 5 1319.35231283.1 2 303.22 2 0.013312556.32 1 0.007612519.82 1 0.0057142.0124 5 211,084 kJ/ min Since 1 ton of refrigeration equals a heat transfer rate of 211 kJ/min (Sec. 10.2.1), the required refrigerating capacity is 52.5 tons. Alternative Psychrometric Chart Solution: Let us consider an alternative solution using the psychrometric chart. As shown on the sketch of the psychrometric chart, Fig. E12.11b, the state of the moist air at the inlet 1 is defined by f 5 50% and a dry-bulb temperature of 308C. At 2, the moist air is saturated at 108C. Rearranging Eq. (a), we get # # (c) Qcv 5 ma31ha 1 vhv22 2 1ha 1 vhv21 1 1v1 2 v22hw4

The underlined terms and humidity ratios, v1 and v2, can be read directly from the chart. The mass flow rate of the dry air can be determined using the volumetric flow rate at the inlet and ya1 read from the chart. The specific enthalpy hw is obtained (as above) from Table A-2: hf at T2. The details are left as an exercise.

(ha + w hv)2

(ha + w hv)1

y alp nth in e , ific air ) ec st Sp moi y air of (dr /kg J k 0% 10 = f % 50 = f 1

w1 w2

2

10°C

w

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ apply mass and energy

30°C Dry-bulb temperature

Fig. E12.11b

Using the psychrometric chart, determine the wet-bulb temperature of the moist air entering the dehumidifier, in °C. Ans. ω 1

T2 > T1 ω2 > ω1

ω

T2 < T1 ω2 > ω1 ω

2

1

1

Water injected (vapor or liquid)

Dry-bulb temperature

Dry-bulb temperature

(b)

(c)

(a)

Fig. 12.12 Humidification. (a) Control volume. (b) Steam injected. (c) Liquid injected.

12.8.4

Humidification

It is often necessary to increase the moisture content of the air circulated through occupied spaces. One way to accomplish this is to inject steam. Alternatively, liquid water can be sprayed into the air. Both cases are shown schematically in Fig. 12.12a. The temperature of the moist air as it exits the humidifier depends on the condition of the water introduced. When relatively high-temperature steam is injected, both the humidity ratio and the dry-bulb temperature are increased. This is illustrated by the accompanying psychrometric chart of Fig. 12.12b. If liquid water is injected instead of steam, the moist air may exit the humidifier with a lower temperature than at the inlet. This is illustrated in Fig. 12.12c. The example to follow illustrates the case of steam injection. The case of liquid water injection is considered further in the next section.

cccc

EXAMPLE 12.12 c

Analyzing a Steam-Spray Humidifier Moist air with a temperature of 228C and a wet-bulb temperature of 98C enters a steam-spray humidifier. The mass flow rate of the dry air is 90 kg/min. Saturated water vapor at 1108C is injected into the mixture at a rate of 52 kg/h. There is no heat transfer with the surroundings, and the pressure is constant throughout at 1 bar. Using the psychrometric chart, determine at the exit (a) the humidity ratio and (b) the temperature, in 8C. SOLUTION Known: Moist air enters a humidifier at a temperature of 228C and a wet-bulb temperature of 98C. The mass flow rate of the dry air is 90 kg/min. Saturated water vapor at 1108C is injected into the mixture at a rate of 52 kg/h. Find: Using the psychrometric chart, determine at the exit the humidity ratio and the temperature, in 8C. Schematic and Given Data: 1

Engineering Model:

2 Moist air T2 = ? ω2 = ?

m· a = 90 kg/min T1 = 22°C Twb = 9°C

1. The control volume shown in the accompanying figure oper-

ates at steady state. Changes in kinetic and potential energy # can be neglected and Wcv 5 0. 2. There is no heat transfer with the surroundings.

3 Saturated water vapor at 110°C, m· = 52 kg/h

Boundary

3. The pressure remains constant throughout at 1 bar. Figure A-9

remains valid at this pressure.

st

Fig. E12.12a

4. The moist air streams are regarded as ideal gas mixtures adher-

ing to the Dalton model.

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Analysis: (a) The humidity ratio at the exit v2 can be found from mass rate balances on the dry air and water individually. Thus

# # ma1 5 ma2 1dry air2 # # # mv1 1 mst 5 mv2 1water2 # # # # # With mv1 5 v1ma, and mv2 5 v2ma, where ma is the mass flow rate of the air, the second of these becomes # mst v2 5 v1 1 # ma Using the inlet dry-bulb temperature, 228C, and the inlet wet-bulb temperature, 98C, the value of the humidity ratio v1 can be found by inspection of the psychrometric chart, Fig. A-9. The result is v1 5 0.002 kg (vapor)/kg(dry air). This value should be verified as an exercise. Inserting values into the expression for v2 v2 5 0.002 1

152 kg/ h2Z1 h/ 60 minZ 90 kg/ min

5 0.0116

kg1vapor2 kg1dry air2

(b) The temperature at the exit can be determined using an energy rate balance. With assumptions 1 and 2, the

steady-state form of the energy rate balance reduces to a special case of Eq. 12.55. Namely 0 5 ha1 2 ha2 1 v1hg1 1 1v2 2 v12hg3 2 v2hg2

(a)

In writing this, the specific enthalpies of the water vapor at 1 and 2 are evaluated as the respective saturated vapor values, and hg3 denotes the enthalpy of the saturated vapor injected into the moist air. Equation (a) can be rearranged in the following form suitable for use with the psychrometric chart. ➊

1ha 1 vhg22 5 1ha 1 vhg21 1 1v2 2 v12hg3

(b)

As shown on the sketch of the psychrometric chart, Fig. E12.12b, the first term on the right of Eq. (b) can be obtained from Fig. A-9 at the inlet state, defined by the intersection of the inlet dry-bulb temperature, 228C, and the inlet wet-bulb temperature, 98C; the value is 27.2 kJ/kg(dry air). The second term on the right can be evaluated using the known humidity ratios v1 and v2 and hg3 from Table A-2: 2691.5 kJ/kg(vapor). The value of the second term of Eq. (b) is 25.8 kJ/kg (dry air). The state at the exit is then fixed by v2 and (ha 1 vhg)2 5 53 kJ/kg(dry air), calculated from the two values just determined. Finally, the temperature at the exit can be read directly from the chart. The result is T2 < 23.58C . ir,

ta

of lpy ) air ry

is mo

a nth

(ha + w hg)2 = 53 kJ/kg(dry air)

e ic cif d g( J/k k in

e Sp

(ha + w hg)1 = 27.2 kJ/kg(dry air)

w2

2

1 Twb = 9°C T1 = 22°C

w

T2 = 23.5°C

w1

Fig. E12.12b

Alternative IT Solution:

➋ The following program allows T2 to be determined using IT, where m# a is denoted as mdota, m# st is denoted as mdotst, w1 and w2 denote v1 and v2, respectively, and so on.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications // Given data T1 5 22 // 8C Twb1 5 9 // 8C mdota 5 90 // kg/min p 5 1 // bar Tst 5 110 // 8C mdotst 5 (52 / 60) // converting kg/h to kg/min // Evaluate humidity ratios w1 5 w_TTwb (T1,Twb1,p) w2 5 w1 1 (mdotst / mdota) // Denoting the enthalpy of moist air at state 1 by // h1, etc., the energy balance, Eq. (a), becomes 0 5 h1 2 h2 1 (w2 2 w1)*hst // Evaluate enthalpies h1 5 ha_Tw(T1,w1) h2 5 ha_Tw(T2,w2) hst 5 hsat_Px(“Water/Steam”,psat,1) psat 5 Psat_T(“Water/Steam ”,Tst)

Using the Solve button, the result is T2 5 23.48C, which agrees closely with the values obtained above, as expected. ➊ A solution of Eq. (b) using data from Tables A-2 and A-22 requires an iterative (trial) procedure. The result is T2 5 248C, as can be verified. ➋ Note the use of special Moist Air functions listed in the Properties menu of IT.

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. appl y mass and energy ❑ balances for a spray humidification process in a control volume at steady state. ❑ retrieve necessary property data using the psychrometric chart. ❑ apply IT for psychrometric analysis.

Using the psychrometric chart, what is the relative humidity at the exit? Ans. ω1

Moist air m· a, T1, ω 1

1

ω1

T2 T1 Dry-bulb temperature

Soaked pad (a)

(b)

Fig. 12.13 Evaporative cooling. (a) Equipment schematic. (b) Psychrometric chart representation.

mixture enthalpy are closely lines of constant wet-bulb temperature (Sec. 12.7), it follows that evaporative cooling takes place at a nearly constant wet-bulb temperature. In the next example, we consider the analysis of an evaporative cooler.

cccc

EXAMPLE 12.13 c

Considering an Evaporative Cooler Air at 1008F and 10% relative humidity enters an evaporative cooler with a volumetric flow rate of 5000 ft3/min. Moist air exits the cooler at 708F. Water is added to the soaked pad of the cooler as a liquid at 708F and evaporates fully into the moist air. There is no heat transfer with the surroundings and the pressure is constant throughout at 1 atm. Determine (a) the mass flow rate of the water to the soaked pad, in lb/h, and (b) the relative humidity of the moist air at the exit to the evaporative cooler. SOLUTION Known: Air at 1008F and f 5 10% enters an evaporative cooler with a volumetric flow rate of 5000 ft3/min.

Moist air exits the cooler at 708F. Water is added to the soaked pad of the cooler at 708F. Find: Determine the mass flow rate of the water to the soaked pad, in lb/h, and the relative humidity of the

moist air at the exit of the cooler. Schematic and Given Data: Engineering Model:

Water at 70°F

1. The control volume shown in the accompanying figure operT2 = 70°F

T1 = 100°F φ 1 = 10% 3 ft (AV)1 = 5000 ___ min

ates at steady state. Changes in kinetic and potential energy # can be neglected and Wcv 5 0. 2. There is no heat transfer with the surroundings. 3. The water added to the soaked pad enters as a liquid and

evaporates fully into the moist air. 1 Soaked pad

Fig. E12.13

2 Boundary

4. The pressure remains constant throughout at 1 atm. 5. The moist air streams are regarded as ideal gas mixtures

adhering to the Dalton model.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Analysis: (a) Applying conservation of mass to the dry air and water individually as in previous examples gives

# # mw 5 ma1v2 2 v12 # # # where mw is the mass flow rate of the water to the soaked pad. To find mw requires v1, ma, and v2. These will now be evaluated in turn. The humidity ratio v1 can be found from Eq. 12.43, which requires pv1, the partial pressure of the moist air entering the control volume. Using the given relative humidity f1 and pg at T1 from Table A-2E, we have pv1 5 f1pg1 5 0.095 lbf/in.2 With this, v1 5 0.00405 lb(vapor)ylb(dry air). # The mass flow rate of the dry air ma can be found as in previous examples using the volumetric flow rate and specific volume of the dry air. Thus 1AV21 # ma 5 ya1 The specific volume of the dry air can be evaluated from the ideal gas equation of state. The result is ya1 5 14.2 ft3/lb (dry air). Inserting values, the mass flow rate of the dry air is # ma 5

lb1dry air2 5000 ft3/ min 5 352.1 3 min 14.2 ft / lb1dry air2

To find the humidity ratio v2, reduce the steady-state forms of the mass and energy rate balances using assumption 1 to obtain # # # # # 0 5 1maha1 1 mv1hv12 1 mwhw 2 1maha2 1 mv2hv22 With the same reasoning as in previous examples, this can be expressed as the following special form of Eq. 12.55: 0 5 1ha 1 vhg21 1 1v2 2 v12hf 2 1ha 1 vhg22

(a)

where hf denotes the specific enthalpy of the water entering the control volume at 708F. Solving for v2 ➊

v2 5

ha1 2 ha2 1 v11hg1 2 hf2 hg2 2 hf

5

cpa1T1 2 T22 1 v11hg1 2 hf2 hg2 2 hf

where cpa 5 0.24 Btu/lb ? 8R. With hf, hg1, and hg2 from Table A-2E v2 5

0.241100 2 702 1 0.0040511105 2 38.12 11092 2 38.12

5 0.0109

lb1vapor2 lb1dry air2

# # Substituting values for ma, v1, and v2 into the expression for mw lb1dry air2 60 min lb1water2 # mw 5 c 352.1 ` d 10.0109 2 0.004052 ` 1h min lb1dry air2 5 144.7

lb1water2 h

(b) The relative humidity of the moist air at the exit can be determined using Eq. 12.44. The partial pressure of the water vapor required by this expression can be found by solving Eq. 12.43 to obtain

pv2 5

v2 p v2 1 0.622

Inserting values pv2 5

10.01092114.696 lbf/ in.22 5 0.253 lbf/ in.2 10.0109 1 0.6222

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At 708F, the saturation pressure is 0.3632 lbf/in.2 Thus, the relative humidity at the exit is f2 5

0.253 5 0.697169.7%2 0.3632

✓ Skills Developed

Alternative Psychrometric Chart Solution: Since the underlined term in Eq. (a)

is much smaller than either of the moist air enthalpies, the enthalpy of the moist air remains nearly constant, and thus evaporative cooling takes place at a nearly constant wet-bulb temperature. See Fig. 12.13b and the accompanying discussion. Using this approach with the psychrometric chart, Fig. A-9E, determine humidity ratio and relative humidity at the exit, and compare with the previously determined values. The details are left as an exercise.

Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ apply mass and energy balances for an evaporative cooling process in a control volume at steady state. retr ieve property data for ❑ dry air and water.

➊ A constant value of the specific heat cpa has been used here to evaluate the term (ha1 2 ha2). As shown in previous examples, this term can be evaluated alternatively using the ideal gas table for air.

Using steam table data, what is the dew point temperature at the exit, in 8F? Ans. 59.68F.

12.8.6

Adiabatic Mixing of Two Moist Air Streams

A common process in air-conditioning systems is the mixing of moist air streams, as shown in Fig. 12.14. The objective of the thermodynamic analysis of such a process is normally to fix the flow rate and state of the exiting stream for specified flow rates and states of each of the two inlet streams. The case of adiabatic mixing is governed by Eqs. 12.56 to follow. The mass rate balances for the dry air and water vapor at steady state are, respectively, # # # ma1 1 ma2 5 ma3 1dry air2 (12.56a) # # # mv1 1 mv2 5 mv3 1water vapor2

Mixture enthalpy per unit mass of dry air (ha + w hg)1 (ha + w hg)3 (ha + w hg)2

1 m· a1, T1, ω 1

1

w1

3 m· a3 T3 ω3 2 m· a2, T2, ω 2

Insulation (a)

w3

3

w2

2

T2

T3

T1 (b)

Fig. 12.14 Adiabatic mixing of two moist air streams. (a) Equipment representation. (b) Psychrometric chart representation.

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications # # With mv 5 vma, the water vapor mass balance becomes # # # v1ma1 1 v2ma2 5 v3ma3 1water vapor2 (12.56b) # # Assuming Qcv 5 Wcv 5 0 and ignoring the effects of kinetic and potential energy, the energy rate balance reduces at steady state to # # # ma11ha1 1 v1hg12 1 ma21ha2 1 v2hg22 5 ma31ha3 1 v3hg32 (12.56c) where the enthalpies of the entering and exiting water vapor are evaluated as the saturated vapor values at the respective dry-bulb temperatures. If the inlet flow rates and states are known, Eqs. 12.56 are three equations in three # unknowns: ma3, v3, and 1ha3 1 v3hg32. The solution of these equations is illustrated by Example 12.14. Let us also consider how Eqs. 12.56 can be solved geometrically with the psychro# metric chart: Using Eq. 12.56a to eliminate ma3, the mass flow rate of dry air at 3, from Eqs. 12.56b and 12.56c, we get # 1ha3 1 v3hg32 2 1ha2 1 v2hg22 v3 2 v2 ma1 (12.57) # 5v 2v 5 1ha1 1 v1hg12 2 1ha3 1 v3hg32 ma2 1 3 From the relations of Eqs. 12.57, we conclude that on a psychrometric chart state 3 of the mixture lies on a straight line connecting states 1 and 2 of the two streams before mixing (see end-of-chapter Prob. 12.93). This is shown in Fig. 12.14b.

cccc

EXAMPLE 12.14 c

Analyzing Adiabatic Mixing of Two Moist Air Streams A stream consisting of 142 m3/min of moist air at a temperature of 58C and a humidity ratio of 0.002 kg(vapor)/ kg(dry air) is mixed adiabatically with a second stream consisting of 425 m3/min of moist air at 248C and 50% relative humidity. The pressure is constant throughout at 1 bar. Determine (a) the humidity ratio and (b) the temperature of the exiting mixed stream, in 8C. SOLUTION Known: A moist air stream at 58C, v 5 0.002 kg(vapor)/kg(dry air), and a volumetric flow rate of 142 m3/min is

mixed adiabatically with a stream consisting of 425 m3/min of moist air at 248C and f 5 50%. Find: Determine the humidity ratio and the temperature, in 8C, of the mixed stream exiting the control volume. Schematic and Given Data: Engineering Model:

1 Insulation (AV)1 = 142 m3/min T1 = 5°C kg (vapor) ω1 = 0.002 __________ kg (dry air)

1. The control volume shown in the accompanying

3

ω3 = ? T3 = ?

2 (AV)2 = 425 m3/min T2 = 24°C φ2 = 50%

figure operates at steady state. Changes in kinetic # and potential energy can be neglected and Wcv 5 0. 2. There is no heat transfer with the surroundings. 3. The pressure remains constant throughout at 1 bar. 4. The moist air streams are regarded as ideal gas

mixtures adhering to the Dalton model.

Fig. E12.14 Analysis: (a) The humidity ratio v3 can be found by means of mass rate balances for the dry air and water vapor,

respectively

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# # # ma1 1 ma2 5 ma3 1dry air2 # # # mv1 1 mv2 5 mv3 1water vapor2 # # # # # # With mv1 5 v1ma1, mv2 5 v2ma2, and mv3 5 v3ma3, the second of these balances becomes (Eq. 12.56b) # # # v1ma1 1 v2ma2 5 v3ma3 Solving v3 5

# # v1ma1 1 v2ma2 # ma3

# # # Since ma3 5 ma1 1 ma2, this can be expressed as # # v1ma1 1 v2ma2 # # ma1 1 ma2 # # # # To determine v3 requires values for v2, ma1, and ma2. The mass flow rates of the dry air, ma1 and ma2, can be found as in previous examples using the given volumetric flow rates v3 5

1AV21 1AV22 # # ma1 5 , ma2 5 ya1 ya2 The values of ya1, ya2, and v2 are readily found from the psychrometric chart, Fig. A-9. Thus, at v1 5 0.002 and T1 5 58C, ya1 5 0.79 m3/ kg1dry air2. At f2 5 50% and T2 5 248C, ya2 5 0.855 m3/ kg1dry air2 and v2 5 0.0094. The # # mass flow rates of the dry air are then ma1 5 180 kg1dry air2/ min and ma2 5 497 kg1dry air2/ min. Inserting values into the expression for v3 v3 5

kg1vapor2 10.002211802 1 10.0094214972 5 0.0074 180 1 497 kg1dry air2

(b) The temperature T3 of the exiting mixed stream can be found from an energy rate balance. Reduction of the

energy rate balance using assumptions 1 and 2 gives (Eq. 12.56c) m? a11ha 1 vhg21 1 m? a21ha 1 vhg22 5 m? a31ha 1 vhg23

(a)

Solving 1ha 1 vhg23 5

# # ma11ha 1 vhg21 1 ma21ha 1 vhg22 # # ma1 1 ma2

(b)

With 1ha 1 vhg21 5 10 kJ/ kg1dry air2 and 1ha 1 vhg22 5 47.8 kJ/ kg1dry air2 from Fig. A-9 and other known values 1ha 1 vhg23 5

1801102 1 497147.82 kJ 5 37.7 180 1 497 kg1dry air2

➊ This value for the enthalpy of the moist air at the exit, together with the previously determined value for v3, fixes the state of the exiting moist air. From inspection of Fig. A-9, T3 5 198C. Alternative Solutions:

The use of the psychrometric chart facilitates the solution for T3. Without the chart, an iterative solution of Eq. (b) using data from Tables A-2 and A-22 could be used. Alternatively, T3 can be determined using the following IT program, where f2 is denoted as phi2, the volumetric flow rates at 1 and 2 are denoted as AV1 and AV2, respectively, and so on. // Given data T1 5 5 // 8C w1 5 0.002 // kg(vapor) / kg(dry air) AV1 5 142 // m3/min T2 5 24 // 8C phi2 5 0.5 AV2 5 425 // m3/min p 5 1 // bar

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

// Mass balances for water vapor and dry air: w1 * mdota1 1 w2 * mdota2 5 w3 * mdota3 mdota1 1 mdota2 5 mdota3 // Evaluate mass flow rates of dry air mdota1 5 AV1 / va1 ➋ va1 5 va_Tw(T1, w1, p) mdota2 5 AV2 / va2 va2 5 va_Tphi(T2, phi2, p) // Determine w2 w2 5 w_Tphi(T2, phi2, p) // The energy balance, Eq. (a), reads mdota1 * h1 1 mdota2 * h2 5 mdota3 * h3 h1 5 ha_Tw(T1, w1) h2 5 ha_Tphi(T2, phi2, p) h3 5 ha_Tw(T3, w3) Using the Solve button, the result is T3 5 19.018C and v3 5 0.00745 kg(vapor)/ kg (dry air), which agree with the psychrometric chart solution. ➊ A solution using the geometric approach based on Eqs. 12.57 is left as an exercise. ➋ Note the use here of special Moist Air functions listed in the Properties menu of IT.

✓ Skills Developed Ability to… ❑ apply psychrometric termi-

nology and principles. ❑ apply mass and energy

balances for an adiabatic mixing process of two moist air streams in a control volume at steady state. ❑ retrieve property data for moist air using the psychrometric chart. ❑ apply IT for psychrometric analysis.

Using the psychrometric chart, what is the relative humidity at the exit? Ans. < 53%.

12.9

Cooling Towers

Power plants invariably discharge considerable energy to their surroundings by heat transfer (Chap. 8). Although water drawn from a nearby river or lake can be employed to carry away this energy, cooling towers provide an alternative in locations where sufficient cooling water cannot be obtained from natural sources or where concerns for the environment place a limit on the temperature at which cooling water can be returned to the surroundings. Cooling towers also are frequently employed to provide chilled water for applications other than those involving power plants. Cooling towers can operate by natural or forced convection. Also they may be counterflow, cross-flow, or a combination of these. A schematic diagram of a forcedconvection, counterflow cooling tower is shown in Fig. 12.15. The warm water to be cooled enters at 1 and is sprayed from the top of the tower. The falling water usually passes through a series of baffles intended to keep it broken up into fine drops to promote evaporation. Atmospheric air drawn in at 3 by the fan flows upward, counter to the direction of the falling water droplets. As the two streams interact, a fraction of the entering liquid water stream evaporates into the moist air, which exits at 4 with a greater humidity ratio than the incoming moist air at 3, while liquid water exits at 2 with a lower temperature than the water entering at 1. Since some of the incoming water is evaporated into the moist air stream, an equivalent amount of

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12.9 Cooling Towers

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Discharged moist air m· a, T4, ω 4 > ω 3

4 Fan

Warm water inlet T1, m· w 1

Atmospheric air m· a, T3, ω 3 3

2 Liquid

5 Makeup water

Return water m· w T2 < T1

Fig. 12.15 Schematic of a cooling tower.

makeup water is added at 5 so that the return mass flow rate of the cool water equals the mass flow rate of the warm water entering at 1. For operation at steady state, mass balances for the dry air and water and an energy balance on the overall cooling tower provide information about cooling tower performance. In applying the energy balance, heat transfer with the surroundings is usually neglected. The power input to the fan of forced-convection towers also may be negligible relative to other energy rates involved. The example to follow illustrates the analysis of a cooling tower using conservation of mass and energy together with property data for the dry air and water.

cccc

EXAMPLE 12.15 c

Determining Mass Flow Rates for a Power Plant Cooling Tower Water exiting the condenser of a power plant at 388C enters a cooling tower with a mass flow rate of 4.5 3 107 kg/h. A stream of cooled water is returned to the condenser from a cooling tower with a temperature of 308C and the same flow rate. Makeup water is added in a separate stream at 208C. Atmospheric air enters the cooling tower at 258C and 35% relative humidity. Moist air exits the tower at 358C and 90% relative humidity. Determine the mass flow rates of the dry air and the makeup water, in kg/h. The cooling tower operates at steady state. Heat transfer with the surroundings and the fan power can each be neglected, as can changes in kinetic and potential energy. The pressure remains constant throughout at 1 atm. SOLUTION Known: A liquid water stream enters a cooling tower from a condenser at 388C with a known mass flow rate. A stream of cooled water is returned to the condenser at 308C and the same flow rate. Makeup water is added at 208C. Atmospheric air enters the tower at 258C and f 5 35%. Moist air exits the tower at 358C and f 5 90%. Find: Determine the mass flow rates of the dry air and the makeup water, in kg/h.

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Schematic and Given Data: Engineering Model:

Moist air T4 = 35°C φ 4 = 90% 4

1. The control volume shown in the accompanying figure 1

3

2 5 Makeup water T5 = 20°C

Liquid water, T1 = 38°C m· 1 = 4.5 × 107 kg/h Atmospheric air T3 = 25°C, φ 3 = 35% Liquid water, T2 = 30°C m· 2 = 4.5 × 107 kg/h

operates at steady state. Heat transfer with the surroundings can be neglected, as # can changes in kinetic and potential energy; also Wcv 5 0. 2. To evaluate specific enthalpies, each liquid stream is

regarded as a saturated liquid at the corresponding specified temperature. 3. The moist air streams are regarded as ideal gas mixtures

adhering to the Dalton model. 4. The pressure is constant throughout at 1 atm.

Fig. E12.15 Analysis: The required mass flow rates can be found from mass and energy rate balances. Mass balances for the

dry air and water individually reduce at steady state to # # ma3 5 ma4 1dry air2 # # # # # m1 1 m5 1 mv3 5 m2 1 mv4 1water2 # # # The common mass flow rate of the dry air is denoted as ma. Since m1 5 m2, the second of these equations becomes # # # m5 5 mv4 2 mv3 # # # # With mv3 5 v3 ma and mv4 5 v4 ma # # m5 5 ma 1v4 2 v32 # # Accordingly, the two required mass flow rates, ma and m5, are related by this equation. Another equation relating the flow rates is provided by the energy rate balance. Reducing the energy rate balance with assumption 1 results in # # # # # # # 0 5 m1hw1 1 1maha3 1 mv3hv32 1 m5hw5 2 m2hw2 2 1maha4 1 mv4hv42 Evaluating the enthalpies of the water vapor as the saturated vapor values at the respective temperatures and the enthalpy of each liquid stream as the saturated liquid enthalpy at each respective temperature, the energy rate equation becomes # # # # # # # 0 5 m1hf1 1 1maha3 1 mv3hg32 1 m5hf5 2 m2hf 2 2 1maha4 1 mv4hg42 # # # # # # # # # Introducing m1 5 m2, m5 5 ma1v4 2 v32, mv3 5 v3ma, and mv4 5 v4 ma and solving for ma # m11hf1 2 hf22 # ma 5 (a) ha4 2 ha3 1 v4hg4 2 v3hg3 2 1v4 2 v32hf5 The humidity ratios v3 and v4 required by this expression can be determined from Eq. 12.43, using the partial pressure of the water vapor obtained with the respective relative humidity. Thus, v3 5 0.00688 kg(vapor)/kg(dry air) and v4 5 0.0327 kg(vapor)/kg(dry air). # With enthalpies from Tables A-2 and A-22, as appropriate, and the known values for v3, v4, and m1, the expres# sion for ma becomes 14.5 3 10721159.21 2 125.792 1308.2 2 298.22 1 10.0327212565.32 2 10.00688212547.22 2 10.02582183.962 5 2.03 3 107 kg/ h # Finally, inserting known values into the expression for m5 results in # m5 5 12.03 3 107210.0327 2 0.006882 5 5.24 3 105 kg/ h m? a 5

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Key Engineering Concepts Alternative Psychrometric Chart Solution: Equation (a) can be rearranged to

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✓ Skills Developed

read # m11hf1 2 hf22 # ma 5 1ha4 1 v4hg42 2 1ha3 1 v3hg32 2 1v4 2 v32hf5

Ability to… ❑ apply psychrometric termi-

nology and principles.

The specific enthalpy terms hf1, hf2, and hf5 are obtained from Table A-2, as above. The underlined terms and v3 and v4 can be obtained by inspection of a psychrometric chart from the engineering literature providing data at states 3 and 4. Figure A-9 does not suffice in this application at state 4. The details are left as an exercise.

❑ apply mass and energy

balances for a cooling tower process in a control volume at steady state. ❑ retrieve property data for dry air and water.

Using steam table data, determine the partial pressure of the water vapor in the entering moist air stream, pv3, in bar. Ans. 0.0111 bar.

c CHAPTER SUMMARY AND STUDY GUIDE In this chapter we have applied the principles of thermodynamics to systems involving ideal gas mixtures, including the special case of psychrometric applications involving air–water vapor mixtures, possibly in the presence of liquid water. Both closed system and control volume applications are presented. The first part of the chapter deals with general ideal gas mixture considerations and begins by describing mixture composition in terms of the mass fractions or mole fractions. The Dalton model, which brings in the partial pressure concept, is then introduced for the p–y–T relation of ideal gas mixtures. Means are also introduced for evaluating the enthalpy, internal energy, and entropy of a mixture by adding the contribution of each component at its condition in the mixture. Applications are considered where ideal gas mixtures undergo processes at constant composition as well as where ideal gas mixtures are formed from their component gases. In the second part of the chapter, we study psychrometrics. Special terms commonly used in psychrometrics are introduced, including moist air, humidity ratio, relative humidity, mixture enthalpy, and the dew point, dry-bulb, and wet-bulb temperatures. The psychrometric chart, which gives a graphical representation of important moist air properties, is introduced. The principles of conservation of mass and energy are formulated in terms of psychrometric quantities, and typical air-conditioning applications are considered, including dehumidification and humidification, evaporative cooling, and mix-

ing of moist air streams. A discussion of cooling towers is also provided. The following list provides a study guide for this chapter. When your study of the text and end-of-chapter exercises has been completed, you should be able to c write out the meanings of the terms listed in the margin

throughout the chapter and understand each of the related concepts. The subset of key concepts listed below is particularly important. c describe mixture composition in terms of mass fractions or mole fractions. c relate pressure, volume, and temperature of ideal gas mixtures using the Dalton model, and evaluate U, H, cy, cp, and S of ideal gas mixtures in terms of the mixture composition and the respective contribution of each component. c apply the conservation of mass and energy principles and the second law of thermodynamics to systems involving ideal gas mixtures. For psychrometric applications, you should be able to c evaluate the humidity ratio, relative humidity, mixture enthalpy,

and dew point temperature. c use the psychrometric chart. c apply the conservation of mass and energy principles and the

second law of thermodynamics to analyze air-conditioning processes and cooling towers.

c KEY ENGINEERING CONCEPTS mass fraction, p. 706 gravimetric analysis, p. 706 mole fraction, p. 706 molar (volumetric) analysis, p. 707 apparent molecular weight, p. 707 Dalton model, p. 710

partial pressure, p. 710 psychrometrics, p. 727 moist air, p. 727 humidity ratio, p. 728 relative humidity, p. 729 mixture enthalpy, p. 729

dew point temperature, p. 731 dry-bulb temperature, p. 738 wet-bulb temperature, p. 738 psychrometric chart, p. 740

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

c KEY EQUATIONS

Ideal Gas Mixtures: General Considerations mfi 5 mi / m

(12.3) p. 706

j

Analysis in terms of mass fractions

1 5 a mfi

(12.4) p. 706

yi 5 ni / n

(12.6) p. 706

i51

j

1 5 a yi

(12.7) p. 707

Analysis in terms of mole fractions

i51 j

M 5 a yiMi

(12.9) p. 707

pi 5 yi p

(12.12) p. 710

Apparent molecular weight

i51

j

p 5 a pi

(12.13) p. 710

Partial pressure of component i and relation to mixture pressure p

i51 j

u 5 a yiui

(12.21) p. 712

h 5 a yihi

(12.22) p. 712

s 5 a yi si

(12.27) p. 713

i51 j i51 j

Internal energy, enthalpy, and entropy per mole of mixture. — ui — and hi evaluated at mixture temperature T. — si evaluated at T and partial pressure pi.

i51 j

cy 5 a yi cy,i

(12.23) p. 712

i51 j

cp 5 a yi cp,i

Mixture specific heats on a molar basis (12.24) p. 712

i51

Psychrometric Applications v5

pv mv 5 0.622 ma p 2 pv f5

pv b pg T, p

H 5 ha 1 vhv ma

(12.42, 12.43) p. 729

Humidity ratio

(12.44) p. 729

Relative humidity

(12.46) p. 729

Mixture enthalpy per unit mass of dry air

c EXERCISES: THINGS ENGINEERS THINK ABOUT 1. In an equimolar mixture of O2 and N2, are the mass fractions equal? Explain.

3. Which component of the fuel–air mixture in a cylinder of an automobile engine would have the greater mass fraction?

2. If two different ideal gases mix spontaneously, is the process irreversible? Explain.

4. A rigid, insulated container is divided into two compartments by a partition, and each compartment contains air at the

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same temperature and pressure. If the partition is removed is entropy produced within the container? Explain.

11. Does operating a car’s air-conditioning system affect its fuel economy? Explain.

5. Which do you think is most closely related to human comfort, the humidity ratio or the relative humidity? Explain.

12. What is a food dehydrator and when might you use one?

6. How do you explain the different rates of evaporation from a dish of water in winter and summer? 7. Can the dry-bulb and wet-bulb temperatures be equal? Explain.

13. What is meant by a zero-energy building? 14. What is the difference between a steam sauna and a steam room?

8. How do you explain the water dripping from the tailpipe of an automobile on a cold morning?

15. Your local weather report gives the temperature, relative humidity, and dew point. When planning summertime outdoor activities, are these equally important? Explain.

9. Would you recommend an evaporative cooling system for use in Florida? In Arizona? Explain.

16. Under what conditions would frost accumulate on the interior of a car’s windshield?

10. During winter, why do eyeglasses fog up when the wearer enters a warm building?

c PROBLEMS: DEVELOPING ENGINEERING SKILLS Determining Mixture Composition 12.1 The analysis on a mass basis of an ideal gas mixture at 508F, 25 lbf/in.2 is 60% CO2, 25% SO2, and 15% N2. Determine (a) the (b) the (c) the (d) the

analysis in terms of mole fractions. apparent molecular weight of the mixture. partial pressure of each component, in lbf/in.2 volume occupied by 20 lb of the mixture, in ft3.

12.2 The molar analysis of a gas mixture at 308C, 2 bar is 40% N2, 50% CO2, and 10% CH4. Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in bar. (c) the volume occupied by 10 kg of mixture, in m3. 12.3 The analysis on a molar basis of a gas mixture at 508F, 1 atm is 20% Ar, 35% CO2, and 45% O2. Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in lbf/in.2 (c) the volume occupied by 10 lb of mixture, in ft3. 12.4 The molar analysis of a gas mixture at 258C, 0.1 MPa is 60% N2, 30% CO2, and 10% O2. Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in MPa. (c) the volume occupied by 50 kg of the mixture, in m3. 12.5 The analysis on a mass basis of an ideal gas mixture at 308F, 15 lbf/in.2 is 55% CO2, 30% CO, and 15% O2. Determine (a) the (b) the (c) the (d) the

analysis in terms of mole fractions. apparent molecular weight of the mixture. partial pressure of each component, in lbf/in.2 volume occupied by 10 lb of the mixture, in ft3.

12.7 A rigid vessel having a volume of 3 m3 initially contains a mixture at 218C, 1 bar consisting of 79% N2 and 21% O2 on a molar basis. Helium is allowed to flow into the vessel until the pressure is 2 bar. If the final temperature of the mixture within the vessel is 278C, determine the mass, in kg, of each component present. 12.8 Nitrogen (N2) at 150 kPa, 408C occupies a closed, rigid container having a volume of 1 m3. If 2 kg of oxygen (O2) is added to the container, what is the molar analysis of the resulting mixture? If the temperature remains constant, what is the pressure of the mixture, in kPa? 12.9 A flue gas in which the mole fraction of H2S is 0.002 enters a scrubber operating at steady state at 2008F, 1 atm and a volumetric flow rate of 20,000 ft3/h. If the scrubber removes 92% (molar basis) of the entering H2S, determine the rate at which H2S is removed, in lb/h. Comment on why H2S should be removed from the gas stream. 12.10 A gas mixture with a molar analysis of 20% C3H8 (propane) and 80% air enters a control volume operating at steady state at location 1 with a mass flow rate of 5 kg/min, as shown in Fig. P12.10. Air enters as a separate stream at 2 and dilutes the mixture. A single stream exits with a mole fraction of propane of 3%. Assuming air has a molar analysis of 21% O2 and 79% N2, determine (a) the molar flow rate of the entering air at 2, in kmol/min. (b) the mass flow rate of oxygen in the exiting stream, in kg/min.

1

12.6 Natural gas at 238C, 1 bar enters a furnace with the following molar analysis: 40% propane (C3H8), 40% ethane (C2H6), 20% methane (CH4). Determine

20% C3H8, 80% Air m· = 5 kg/min

(a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in bar. (c) the mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.

Air (21% O2, 79% N2)

1

3 2

Fig. P12.10

3% C3H8, 97% Air

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Considering Constant-Composition Processes 12.11 A gas mixture in a piston–cylinder assembly consists of 2 lb of N2 and 3 lb of He. Determine (a) the composition in terms of mass fractions. (b) the composition in terms of mole fractions. (c) the heat transfer, in Btu, required to increase the mixture temperature from 70 to 1508F, while keeping the pressure constant. (d) the change in entropy of the mixture for the process of part (c), in Btu/ 8R. For parts (c) and (d), use the ideal gas model with constant specific heats. 12.12 Two kg of a mixture having an analysis on a mass basis of 30% N2, 40% CO2, 30% O2 is compressed adiabatically from 1 bar, 300 K to 4 bar, 500 K. Determine (a) the work, in kJ. (b) the amount of entropy produced, in kJ/K. 12.13 As illustrated in Fig. P12.13, an ideal gas mixture in a piston–cylinder assembly has a molar analysis of 30% carbon dioxide (CO2) and 70% nitrogen (N2). The mixture is cooled at constant pressure from 425 to 325 K. Assuming constant specific heats evaluated at 375 K, determine the heat transfer and the work, each in kJ per kg of mixture.

xCO2 = 0.3 xN = 0.7 2

p = constant T1 = 425K T2 = 325K

Fig. P12.13 12.14 A mixture consisting of 0.6 lbmol of N2 and 0.4 lbmol of O2 is compressed isothermally at 10008R from 1 to 3 atm. During the process, there is energy transfer by heat from the mixture to the surroundings, which are at 408F. For the mixture, determine (a) the work, in Btu. (b) the heat transfer, in Btu. (c) the amount of entropy produced, in Btu/8R. For an enlarged system that includes the mixture and enough of its immediate surroundings that heat transfer occurs at 408F, determine the amount of entropy produced, in Btu/8R. Discuss. 12.15 A mixture consisting of 2.8 kg of N2 and 3.2 kg of O2 is compressed from 1 bar, 300 K to 2 bar, 600 K. During the process there is heat transfer from the mixture to the surroundings, which are at 278C. The work done on the mixture is claimed to be 2300 kJ. Can this value be correct? 12.16 A mixture having a molar analysis of 50% CO2, 33.3% CO, and 16.7% O2 enters a compressor operating at steady state at 378C, 1 bar, 40 m/s with a mass flow rate of 1 kg/s and exits at 2378C, 30 m/s. The rate of heat transfer from the compressor to its surroundings is 5% of the power input.

(a) Neglecting potential energy effects, determine the power input to the compressor, in kW. (b) If the compression is polytropic, evaluate the polytropic exponent n and the exit pressure, in bar. 12.17 A mixture of 5 kg of H2 and 4 kg of O2 is compressed in a piston–cylinder assembly in a polytropic process for which n 5 1.6. The temperature increases from 40 to 2508C. Using constant values for the specific heats, determine (a) the heat transfer, in kJ. (b) the entropy change, in kJ/K. 12.18 A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2 at 720 K, 0.35 MPa and a volumetric flow rate of 3.2 m3/s. The mixture exits the turbine at 380 K, 0.11 MPa. For adiabatic operation with negligible kinetic and potential energy effects, determine the power developed at steady state, in kW. 12.19 A gas mixture at 1500 K with the molar analysis 10% CO2, 20% H2O, 70% N2 enters a waste-heat boiler operating at steady state, and exits the boiler at 600 K. A separate stream of saturated liquid water enters at 25 bar and exits as saturated vapor with a negligible pressure drop. Ignoring stray heat transfer and kinetic and potential energy changes, determine the mass flow rate of the exiting saturated vapor, in kg per kmol of gas mixture. 12.20 An equimolar mixture of helium and carbon dioxide enters an insulated nozzle at 2608F, 5 atm, 100 ft/s and expands isentropically to a pressure of 3.24 atm. Determine the temperature, in 8F, and the velocity, in ft/s, at the nozzle exit. Neglect potential energy effects. 12.21 An equimolar mixture of helium (He) and carbon dioxide (CO2) enters an insulated nozzle at 2608F, 5 atm, 100 ft/s and expands isentropically to a velocity of 1110 ft/s. Determine the temperature, in 8F, and the pressure, in atm, at the nozzle exit. Neglect potential energy effects. 12.22 A gas mixture having a molar analysis of 60% O2 and 40% N2 enters an insulated compressor operating at steady state at 1 bar, 208C with a mass flow rate of 0.5 kg/s and is compressed to 5.4 bar. Kinetic and potential energy effects are negligible. For an isentropic compressor efficiency of 78%, determine (a) the temperature at the exit, in 8C. (b) the power required, in kW. (c) the rate of entropy production, in kW/K. 12.23 A mixture having a molar analysis of 60% N2, 17% CO2, and 17% H2O enters a turbine at 1000 K, 8 bar, with a mass flow rate of 2 kg/s and expands isentropically to a pressure of 1 bar. Ignoring kinetic and potential energy effects, determine for steady-state operation (a) the temperature at the exit, in K. (b) the power developed by the turbine, in kW. 12.24 A mixture having a molar analysis of 60% N2 and 40% CO2 enters an insulated compressor operating at steady state at 1 bar, 308C with a mass flow rate of 1 kg/s and is compressed to 3 bar, 1478C. Neglecting kinetic and potential energy effects, determine

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Problems: Developing Engineering Skills (a) the power required, in kW. (b) the isentropic compressor efficiency. (c) the rate of exergy destruction, in kW, for T0 5 300 K. 12.25 An equimolar mixture of N2 and CO2 enters a heat exchanger at 2408F, 500 lbf/in.2 and exits at 5008F, 500 lbf/in.2 The heat exchanger operates at steady state, and kinetic and potential energy effects are negligible. (a) Using the ideal gas mixture concepts of the present chapter, determine the rate of heat transfer to the mixture, in Btu per lbmol of mixture flowing. (b) Compare with the value of the heat transfer determined using the generalized enthalpy chart (Fig. A-4), together with Kay’s rule (see Sec. 11.8). 12.26 Natural gas having a molar analysis of 60% methane (CH4) and 40% ethane (C2H6) enters a compressor at 340 K, 6 bar and is compressed isothermally without internal irreversibilities to 20 bar. The compressor operates at steady state, and kinetic and potential energy effects are negligible. (a) Assuming ideal gas behavior, determine for the compressor the work and heat transfer, each in kJ per kmol of mixture flowing. (b) Compare with the values for work and heat transfer, respectively, determined assuming ideal solution behavior (Sec. 11.9.5). For the pure components at 340 K: h (kJ/kg)

Methane Ethane

s (kJ/kg ? K)

6 bar

20 bar

6 bar

20 bar

715.33 462.39

704.40 439.13

10.9763 7.3493

10.3275 6.9680

Forming Mixtures 12.27 One kilogram of argon at 278C, 1 bar is contained in a rigid tank connected by a valve to another rigid tank containing 0.8 kg of O2 at 1278C, 5 bar. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 878C. Determine (a) the volume of each tank, in m3. (b) the final pressure, in bar. (c) the heat transfer to or from the gases during the process, in kJ. (d) the entropy change of each gas, in kJ/K. 12.28 Using the ideal gas model with constant specific heats, determine the mixture temperature, in K, for each of two cases: (a) Initially, 0.6 kmol of O2 at 500 K is separated by a partition from 0.4 kmol of H2 at 300 K in a rigid insulated vessel. The partition is removed and the gases mix to obtain a final equilibrium state. (b) Oxygen (O2) at 500 K and a molar flow rate of 0.6 kmol/s enters an insulated control volume operating at steady state and mixes with H2 entering as a separate stream at 300 K and a molar flow rate of 0.4 kmol/s. A single mixed stream exits. Kinetic and potential energy effects can be ignored. 12.29 A system consists initially of nA moles of gas A at pressure p and temperature T and nB moles of gas B separate

765

from gas A but at the same pressure and temperature. The gases are allowed to mix with no heat or work interactions with the surroundings. The final equilibrium pressure and temperature are p and T, respectively, and the mixing occurs with no change in total volume. (a) Assuming ideal gas behavior, obtain an expression for the entropy produced in terms of R, nA, and nB. (b) Using the result of part (a), demonstrate that the entropy produced has a positive value. (c) Would entropy be produced when samples of the same gas at the same temperature and pressure mix? Explain. 12.30 Determine the amount of entropy produced, in Btu/ 8R, when 1 lb of H2 at 708F, 1 atm is allowed to mix adiabatically to a final equilibrium state with 20 lb of (a) CO2 and (b) H2, initially at the same temperature and pressure. 12.31 Two kg of N2 at 450 K, 7 bar is contained in a rigid tank connected by a valve to another rigid tank holding 1 kg of O2 at 300 K, 3 bar. The valve is opened and gases are allowed to mix, achieving an equilibrium state at 370 K. Determine (a) the volume of each tank, in m3. (b) the final pressure, in bar. (c) the heat transfer to or from the gases during the process, in kJ. (d) the entropy change of each gas, in kJ/K. 12.32 An insulated tank having a total volume of 60 ft3 is divided into two compartments. Initially one compartment having a volume of 20 ft3 contains 4 lb of carbon monoxide (CO) at 5008F and the other contains 0.8 lb of helium (He) at 608F. The gases are allowed to mix until an equilibrium state is attained. Determine (a) the final temperature, in 8F. (b) the final pressure, in lbf/in.2 (c) the exergy destruction, in Btu, for T0 5 608F. 12.33 A rigid insulated tank has two compartments. Initially one compartment is filled with 2.0 lbmol of argon at 1508F, 50 lbf/in.2 and the other is filled with 0.7 lbmol of helium at 08F, 15 lbf/in.2 The gases are allowed to mix until an equilibrium state is attained. Determine (a) the final temperature, in 8F. (b) the final pressure, in atm. (c) the amount of entropy produced, in Btu/ 8R. 12.34 A rigid insulated tank has two compartments. Initially one contains 0.5 kmol of carbon dioxide (CO2) at 278C, 2 bar and the other contains 1 kmol of oxygen (O2) at 1528C, 5 bar. The gases are allowed to mix while 500 kJ of energy are added by electrical work. Determine (a) the (b) the (c) the (d) the

final temperature, in 8C. final pressure, in bar. change in exergy, in kJ, for T0 5 208C. exergy destruction, in kJ.

12.35 Air at 408C, 1 atm and a volumetric flow rate of 50 m3/min enters an insulated control volume operating at steady state and mixes with helium entering as a separate stream at 1008C, 1 atm and a volumetric flow rate of 20 m3/min. A single mixed

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stream exits at 1 atm. Ignoring kinetic and potential energy effects, determine for the control volume (a) the temperature of the exiting mixture, in 8C. (b) the rate of entropy production, in kW/K. 12.36 Argon (Ar), at 300 K, 1 bar with a mass flow rate of 1 kg/s enters the insulated mixing chamber shown in Fig. P12.36 and mixes with carbon dioxide (CO2) entering as a separate stream at 575 K, 1 bar with a mass flow rate of 0.5 kg/s. The mixture exits at 1 bar. Assume ideal gas behavior with k 5 1.67 for Ar and k 5 1.25 for CO2. For steady-state operation, determine (a) the molar analysis of the exiting mixture. (b) the temperature of the exiting mixture, in K. (c) the rate of entropy production, in kW/K.

1

3

2

Kinetic and potential energy effects can be ignored. 12.40 Helium at 400 K, 1 bar enters an insulated mixing chamber operating at steady state, where it mixes with argon entering at 300 K, 1 bar. The mixture exits at a pressure of 1 bar. If the argon mass flow rate is x times that of helium, plot versus x (a) the exit temperature, in K. (b) the rate of exergy destruction within the chamber, in kJ per kg of helium entering. Kinetic and potential energy effects can be ignored. Let T0 5 300 K.

insulation

Argon (Ar) T1 = 300 K p1 = 1 bar m· 1 = 1 kg/s

(a) the mass and molar analyses of the mixture. (b) the temperature of the mixture at the exit of the valve, in 8F. (c) the rates of exergy destruction for the mixing chamber and the valve, each in Btu per lb of mixture, for T0 5 408F.

mixture exiting p3 = 1 bar

Carbon dioxide (CO2) T2 = 575 K p2 = 1 bar m· 2 = 0.5 kg/s

Fig. P12.36 12.37 Nitrogen (N2) at 1208F, 20 lbf/in.2 and a volumetric flow rate of 300 ft3/min enters an insulated control volume operating at steady state and mixes with oxygen (O2) entering as a separate stream at 2008F, 20 lbf/in.2 and a mass flow rate of 50 lb/min. A single mixed stream exits at 17 lbf/in.2 Kinetic and potential energy effects can be ignored. Using the ideal gas model with constant specific heats, determine for the control volume (a) the temperature of the exiting mixture, in 8F. (b) the rate of exergy destruction, in Btu/min, for T0 5 408F. 12.38 Air at 778C, 1 bar, and a molar flow rate of 0.1 kmol/s enters an insulated mixing chamber operating at steady state and mixes with water vapor entering at 2778C, 1 bar, and a molar flow rate of 0.3 kmol/s. The mixture exits at 1 bar. Kinetic and potential energy effects can be ignored. For the chamber, determine (a) the temperature of the exiting mixture, in 8C. (b) the rate of entropy production, in kW/K. 12.39 A gas mixture required in an industrial process is prepared by first allowing carbon monoxide (CO) at 808F, 18 lbf/in.2 to enter an insulated mixing chamber operating at steady state and mix with argon (Ar) entering at 3808F, 18 lbf/in.2 The mixture exits the chamber at 1408F, 16 lbf/in.2 and is then allowed to expand in a throttling process through a valve to 14.7 lbf/in.2 Determine

12.41 Hydrogen (H2) at 778C, 4 bar enters an insulated chamber at steady state, where it mixes with nitrogen (N2) entering as a separate stream at 2778C, 4 bar. The mixture exits at 3.8 bar with the molar analysis 75% H2, 25% N2. Kinetic and potential energy effects can be ignored. Determine (a) the temperature of the exiting mixture, in 8C. (b) the rate at which entropy is produced, in kJ/K per kmol of mixture exiting. 12.42 An insulated, rigid tank initially contains 1 kmol of argon (Ar) at 300 K, 1 bar. The tank is connected by a valve to a large vessel containing nitrogen (N2) at 500 K, 4 bar. A quantity of nitrogen flows into the tank, forming an argonnitrogen mixture at temperature T and pressure p. Plot T, in K, and p, in bar, versus the amount of N2 within the tank, in kmol. 12.43 A stream of oxygen (O2) at 1008F, 2 atm enters an insulated chamber at steady state with a mass flow rate of 1 lb/min and mixes with a stream of air entering separately at 2008F, 1.5 atm with a mass flow rate of 2 lb/min. The mixture exits at a pressure of 1 atm. Kinetic and potential energy effects can be ignored. On the basis of constant specific heats, determine (a) the temperature of the exiting mixture, in 8F. (b) the rate of exergy destruction, in Btu/min, for T0 5 408F. 12.44 A device is being designed to separate into components a natural gas consisting of CH4 and C2H6 in which the mole fraction of C2H6, denoted by y, may vary from 0.05 to 0.50. The device will receive natural gas at 208C, 1 atm with a volumetric flow rate of 100 m3/s. Separate streams of CH4 and C2H6 will exit, each at 208C, 1 atm. Heat transfer between the device and its surroundings occurs at 208C. Ignoring kinetic and potential energy effects, plot versus y the minimum theoretical work input required at steady state, in kW.

Exploring Psychrometric Principles 12.45 A water pipe at 58C runs above ground between two buildings. The surrounding air is at 358C. What is the maximum relative humidity the air can have before condensation occurs on the pipe?

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Fig. P12.47

12.46 The inside temperature of a wall in a dwelling is 168C. If the air in the room is at 218C, what is the maximum relative humidity the air can have before condensation occurs on the wall? 12.47 A lecture hall having a volume of 106 ft3 contains air at 808F, 1 atm, and a humidity ratio of 0.01 lb of water vapor per lb of dry air. Determine (a) the relative humidity. (b) the dew point temperature, in 8F. (c) the mass of water vapor contained in the room, in lb. 12.48 A large room contains moist air at 308C, 102 kPa. The partial pressure of water vapor is 1.5 kPa. Determine (a) the (b) the (c) the (d) the 10 kg.

relative humidity. humidity ratio, in kg(vapor) per kg(dry air). dew point temperature, in 8C. mass of dry air, in kg, if the mass of water vapor is

12.49 To what temperature, in 8C, must moist air with a humidity ratio of 5 3 1023 kg(vapor) per kg(dry air) be cooled at a constant pressure of 2 bar to become saturated moist air? 12.50 A fixed amount of moist air initially at 1 bar and a relative humidity of 60% is compressed isothermally until condensation of water begins. Determine the pressure of the mixture at the onset of condensation, in bar. Repeat if the initial relative humidity is 90%. 12.51 As shown in Fig. P12.51, moist air at 308C, 2 bar, and 50% relative humidity enters a heat exchanger operating at

steady state with a mass flow rate of 600 kg/h and is cooled at constant pressure to 208C. Ignoring kinetic and potential energy effects, determine the rate of heat transfer from the moist air stream, in kJ/h.

m· 1 = 600 kg/h T1 = 30°C 1 p1 = 2 bar f1 = 50%

2 T2 = 20°C p2 = 2 bar

Fig. P12.51 12.52 One pound of moist air initially at 808F, 1 atm, 50% relative humidity is compressed isothermally to 3 atm. If condensation occurs, determine the amount of water condensed, in lb. If there is no condensation, determine the final relative humidity. 12.53 A vessel whose volume is 0.5 m3 initially contains dry air at 0.2 MPa and 208C. Water is added to the vessel until the air is saturated at 208C. Determine the (a) mass of water added, in kg. (b) final pressure in the vessel, in bar. 12.54 Wet grain at 208C containing 40% moisture by mass enters a dryer operating at steady state. Dry air enters the dryer at 908C, 1 atm at a rate of 15 kg per kg of wet grain entering. Moist air exits the dryer at 388C, 1 atm, and 52% relative humidity. For the grain exiting the dryer, determine the percent moisture by mass.

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12.55 Figure P12.55 shows a spray dryer operating at steady state. The mixture of liquid water and suspended solid particles entering through the spray head contains 30% solid matter by mass. Dry air enters at 1778C, 1 atm, and moist air exits at 858C, 1 atm, and 21% relative humidity with a volumetric flow rate of 310 m3/min. The dried particles exit separately. Determine (a) the volumetric flow rate of the entering dry air, in m3/min. (b) the rate that dried particles exit, in kg/min.

Solid–liquid mixture 70% liquid 30% solid

Dry air, 177°C, 1 atm

Moist air 310 m3/min, 85°C, 1 atm, φ = 21%

Dried particles

Fig. P12.55 12.56 A mixture of nitrogen and water vapor at 2008F, 1 atm has the molar analysis 80% N2, 20% water vapor. If the mixture is cooled at constant pressure, determine the temperature, in 8F, at which water vapor begins to condense. 12.57 A system consisting initially of 0.5 m3 of air at 358C, 1 bar, and 70% relative humidity is cooled at constant pressure to 298C. Determine the work and heat transfer for the process, each in kJ. 12.58 Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents are cooled. Determine the heat transfer, in kJ, if the final temperature in the tank is (a) 1108C, (b) 308C. 12.59 A closed, rigid tank initially contains 0.5 m3 of moist air in equilibrium with 0.1 m3 of liquid water at 808C and 0.1 MPa. If the tank contents are heated to 2008C, determine (a) the final pressure, in MPa. (b) the heat transfer, in kJ. 12.60 A closed, rigid cylindrical tank having a height of 6 ft and a diameter of 2 ft initially contains air at 3008F, 80 lbf/in.2, and 10% relative humidity. If the tank contents are cooled to 1708F, determine (a) if condensation occurs. (b) the final pressure, in lbf/in.2 (c) the heat transfer, in Btu. (d) the change in entropy, in Btu/8R. 12.61 Gaseous combustion products with the molar analysis of 15% CO2, 25% H2O, 60% N2 enter an engine’s exhaust pipe at 11008F, 1 atm and are cooled as they pass through the pipe, to 1258F, 1 atm. Determine the heat transfer at steady state, in Btu per lb of entering mixture. 12.62 Air at 608F, 14.7 lbf/in.2, and 75% relative humidity enters an insulated compressor operating at steady state and

is compressed to 100 lbf/in.2 The isentropic compressor efficiency is hc. (a) For hc 5 0.8, determine the temperature, in 8R, of the exiting air, and the work input required and the exergy destruction, each in Btu per lb of dry air flowing. Let T0 5 5208R. (b) Plot each of the quantities determined in part (a) versus hc ranging from 0.7 to 1.0. 12.63 Air at 358C, 3 bar, 30% relative humidity, and a velocity of 50 m/s expands isentropically through a nozzle. Determine the lowest exit pressure, in bar, that can be attained without condensation. For this exit pressure, determine the exit velocity, in m/s. The nozzle operates at steady state and without significant potential energy effects. 12.64 A closed, rigid tank having a volume of 1 m3 contains a mixture of carbon dioxide (CO2) and water vapor at 758C. The respective masses are 12.3 kg of carbon dioxide and 0.05 kg of water vapor. If the tank contents are cooled to 208C, determine the heat transfer, in kJ, assuming ideal gas behavior. 12.65 Gaseous combustion products at 8008F, 1 atm and a volumetric flow rate of 5 ft3/s enter a counterflow heat exchanger operating at steady state and exit at 2008F. The molar analysis of the products is 7.1% CO2, 4.3% O2, 14.3% H2O, 74.3% N2. A separate moist-air stream enters the heat exchanger at 608F, 1 atm, and 30% relative humidity and exits at 1008F. Determine the mass flow rate of the entering moist-air stream, in lb/s. The pressure drops of the two streams can be ignored, as can stray heat transfer and kinetic and potential energy effects. 12.66 Air enters a compressor operating at steady state at 508C, 0.9 bar, 70% relative humidity with a volumetric flow rate of 0.8 m3/s. The moist air exits the compressor at 1958C, 1.5 bar. Assuming the compressor is well insulated, determine (a) the relative humidity at the exit. (b) the power input, in kW. (c) the rate of entropy production, in kW/K. 12.67 Moist air enters a control volume operating at steady state with a volumetric flow rate of 3500 ft3/min. The moist air enters at 1208F, 1.2 atm, and 75% relative humidity. Heat transfer occurs through a surface maintained at 508F. Saturated moist air and condensate exit the control volume, # each at 688F. Assuming Wcv 5 0, and kinetic and potential energy effects are negligible, determine (a) the (b) the (c) the (d) the

mass flow rate of condensate, in lb/min. rate of heat transfer, in Btu/min. rate of entropy production, in Btu/8R ? min. rate of exergy destruction, in Btu/min, for T0 5 508F.

12.68 Moist air at 158C, 1.3 atm, 63% relative humidity and a volumetric flow rate of 770 m3/h enters a control volume at steady state and flows along a surface maintained at 1878C, through which heat transfer occurs. Liquid water at 158C is injected at a rate of 7 kg/h and evaporates into the flowing # stream. For the control volume, Wcv 5 0, and kinetic and potential energy effects are negligible. If moist air exits at 458C, 1.3 atm, determine (a) the rate of heat transfer, in kW. (b) the rate of entropy production, in kW/K.

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Problems: Developing Engineering Skills 12.69 Using Eq. 12.48, determine the humidity ratio and relative humidity for each case below. (a) The dry-bulb and wet-bulb temperatures in a conference room at 1 atm are 24 and 168C, respectively. (b) The dry-bulb and wet-bulb temperatures in a factory space at 1 atm are 75 and 608F, respectively. (c) Repeat parts (a) and (b) using the psychrometric chart. (d) Repeat parts (a) and (b) using Interactive Thermodynamics: IT. 12.70 Using the psychrometric chart, Fig. A-9, determine (a) the relative humidity, the humidity ratio, and the specific enthalpy of the mixture, in kJ per kg of dry air, corresponding to dry-bulb and wet-bulb temperatures of 30 and 258C, respectively. (b) the humidity ratio, mixture specific enthalpy, and wetbulb temperature corresponding to a dry-bulb temperature of 308C and 60% relative humidity. (c) the dew point temperature corresponding to dry-bulb and wet-bulb temperatures of 30 and 208C, respectively. (d) Repeat parts (a)–(c) using Interactive Thermodynamics: IT. 12.71 Using the psychrometric chart, Fig. A-9E, determine (a) the dew point temperature corresponding to dry-bulb and wet-bulb temperatures of 80 and 708F, respectively. (b) the humidity ratio, the specific enthalpy of the mixture, in Btu per lb of dry air, and the wet-bulb temperature corresponding to a dry-bulb temperature of 808F and 70% relative humidity. (c) the relative humidity, humidity ratio, and mixture specific enthalpy corresponding to dry-bulb and wet-bulb temperatures of 80 and 658F, respectively. (d) Repeat parts (a)–(c) using Interactive Thermodynamics: IT. 12.72 A fixed amount of air initially at 528C, 1 atm, and 10% relative humidity is cooled at constant pressure to 158C. Using the psychrometric chart, determine whether condensation occurs. If so, evaluate the amount of water condensed, in kg per kg of dry air. If there is no condensation, determine the relative humidity at the final state. 12.73 A fan within an insulated duct delivers moist air at the duct exit at 358C, 50% relative humidity, and a volumetric flow rate of 0.4 m3/s. At steady state, the power input to the fan is 1.7 kW. The pressure in the duct is nearly 1 atm throughout. Using the psychrometric chart, determine the temperature, in 8C, and relative humidity at the duct inlet. 12.74 The mixture enthalpy per unit mass of dry air, in kJ/kg(a), represented on Fig. A-9 can be approximated closely from the expression H 5 1.005 T18C2 1 v32501.7 1 1.82 T18C24 ma When using Fig. A-9E, the corresponding expression, in Btu/lb(a), is H 5 0.24 T18F2 1 v31061 1 0.444 T18F24 ma Noting all significant assumptions, develop the above expressions.

769

Considering Air-Conditioning Applications 12.75 Each case listed gives the dry-bulb temperature and relative humidity of the moist-air stream entering an airconditioning system: (a) 308C, 40%, (b) 178C, 60%, (c) 258C, 70%, (d) 158C, 40%, (e) 278C, 30%. The condition of the moist-air stream exiting the system must satisfy these constraints: 22 # Tdb # 278C, 40 # f # 60%. In each case, develop a schematic of equipment and processes from Sec. 12.8 that would achieve the desired result. Sketch the processes on a psychrometric chart. 12.76 Moist air enters an air-conditioning system as shown in Fig. 12.11 at 268C, f 5 80% and a volumetric flow rate of 0.47 m3/s. At the exit of the heating section the moist air is at 268C, f 5 50%. For operation at steady state, and neglecting kinetic and potential energy effects, determine (a) the rate energy is removed by heat transfer in the dehumidifier section, in tons. (b) the rate energy is added by heat transfer in the heating section, in kW. 12.77 Air at 1 atm with dry-bulb and wet-bulb temperatures of 82 and 688F, respectively, enters a duct with a mass flow rate of 10 lb/min and is cooled at essentially constant pressure to 628F. For steady-state operation and negligible kinetic and potential energy effects, determine (a) the relative humidity at the duct inlet. (b) the rate of heat transfer, in Btu/min. (c) Check your answers using data from the psychrometric chart. (d) Check your answers using Interactive Thermodynamics: IT. 12.78 Air at 358C, 1 atm, and 50% relative humidity enters a dehumidifier operating at steady state. Saturated moist air and condensate exit in separate streams, each at 158C. Neglecting kinetic and potential energy effects, determine (a) the heat transfer from the moist air, in kJ per kg of dry air. (b) the amount of water condensed, in kg per kg of dry air. (c) Check your answers using data from the psychrometric chart. (d) Check your answers using Interactive Thermodynamics: IT. 12.79 Air at 808F, 1 atm, and 70% relative humidity enters a dehumidifier operating at steady state with a mass flow rate of 1 lb/s. Saturated moist air and condensate exit in separate streams, each at 508F. Neglecting kinetic and potential energy effects, determine (a) the rate of heat transfer from the moist air, in tons. (b) the rate water is condensed, in lb/s. (c) Check your answers using data from the psychrometric chart. (d) Check your answers using Interactive Thermodynamics: IT. 12.80 Moist air at 288C, 1 bar, and 50% relative humidity flows through a duct operating at steady state. The air is cooled at essentially constant pressure and exits at 208C. Determine the heat transfer rate, in kJ per kg of dry air flowing, and the relative humidity at the exit. 12.81 An air conditioner operating at steady state takes in moist air at 288C, 1 bar, and 70% relative humidity. The moist air first passes over a cooling coil in the dehumidifier unit and

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some water vapor is condensed. The rate of heat transfer between the moist air and the cooling coil is 11 tons. Saturated moist air and condensate streams exit the dehumidifier unit at the same temperature. The moist air then passes through a heating unit, exiting at 248C, 1 bar, and 40% relative humidity. Neglecting kinetic and potential energy effects, determine

12.84 An air-conditioning system consists of a spray section followed by a reheater. Moist air at 328C and f 5 77% enters the system and passes through a water spray, leaving the spray section cooled and saturated with water. The moist air is then heated to 258C and f 5 45% with no change in the amount of water vapor present. For operation at steady state, determine

(a) the temperature of the moist air exiting the dehumidifier unit, in 8C. (b) the volumetric flow rate of the air entering the air conditioner, in m3/min. (c) the rate water is condensed, in kg/min. (d) the rate of heat transfer to the air passing through the heating unit, in kW.

(a) the temperature of the moist air leaving the spray section, in 8C. # (b) the change m the amount of water vapor contained in the moist air passing through the system, in kg per kg of dry air.

12.82 Figure P12.82 shows a compressor followed by an aftercooler. Atmospheric air at 14.7 lbf/in.2, 908F, and 75% relative humidity enters the compressor with a volumetric flow rate of 100 ft3/min. The compressor power input is 15 hp. The moist air exiting the compressor at 100 lbf/in.2, 4008F flows through the aftercooler, where it is cooled at constant pressure, exiting saturated at 1008F. Condensate also exits the aftercooler at 1008F. For steady-state operation and negligible kinetic and potential energy effects, determine (a) the rate of heat transfer from the compressor to its surroundings, in Btu/min. (b) the mass flow rate of the condensate, in lb/min. (c) the rate of heat transfer from the moist air to the refrigerant circulating in the cooling coil, in tons of refrigeration. Refrigerant circulating in a cooling coil T2 = 400°F p3 = 100 lbf/in.2 2

Aftercooler 3 Air-water vapor mixture T3 = 100°F φ3 = 100% p3 = 100 lbf/in.2

Saturated liquid at 100°F Power in Compressor 15 hp 1

Heat transfer to the surroundings

Atmospheric air, 100 ft3/min at T1 = 90°F, φ1 = 75%, p1 = 14.7 lbf/in.2

Fig. P12.82 12.83 Outside air at 508F, 1 atm, and 40% relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at 458F and a moist air stream exits with a volumetric flow rate of 1000 ft3/min at 908F, 1 atm and a relative humidity of 40%. Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in lb/min. (b) the rate of heat transfer to the moist air, in Btu/h.

Locate the principal states on a psychrometric chart. 12.85 Moist air at 958F, 1 atm, and a relative humidity of 30% enters a steam-spray humidification device operating at steady state with a volumetric flow rate of 5700 ft3/min. Saturated water vapor at 2308F is sprayed into the moist air, which then exits the device at a relative humidity of 50%. Heat transfer between the device and its surroundings can be ignored, as can kinetic and potential energy effects. Determine (a) the temperature of the exiting moist air stream, in 8F. (b) the rate at which steam is injected, in lb/min. 12.86 For the steam-spray humidifier in Problem 12.85, determine the exergy destruction rate, in Btu/min. Let T0 5 958F. 12.87 Outside air at 508F, 1 atm, and 40% relative humidity enters an air conditioner operating at steady state with a mass flow rate of 3.3 lb/s. The air is first heated at essentially constant pressure to 908F. Liquid water at 608F is then injected, bringing the air to 708F, 1 atm. Determine (a) the rate of heat transfer to the air passing through the heating section, in Btu/s. (b) the rate water is injected, in lb/s. (c) the relative humidity at the exit of the humidification section. Kinetic and potential energy effects can be ignored. 12.88 Moist air at 278C, 1 atm, and 50% relative humidity enters an evaporative cooling unit operating at steady state consisting of a heating section followed by a soaked pad evaporative cooler operating adiabatically. The air passing through the heating section is heated to 458C. Next, the air passes through a soaked pad exiting with 50% relative humidity. Using data from the psychrometric chart, determine (a) the humidity ratio of the entering moist air mixture, in kg(vapor) per kg(dry air). (b) the rate of heat transfer to the moist air passing through the heating section, in kJ per kg of mixture. (c) the humidity ratio and temperature, in 8C, at the exit of the evaporative cooling section. 12.89 In an industrial dryer operating at steady state, atmospheric air at 808F, 1 atm, and 65% relative humidity is first heated to 2808F at constant pressure. The heated air is then allowed to pass over the materials being dried, exiting the dryer at 1508F, 1 atm, and 30% relative humidity. If moisture is to be removed from the materials at a rate of 2700 lb/h, determine (a) the mass flow rate of dry air required, in lb/h. (b) the rate of heat transfer to the air as it passes through the heating section, in Btu/h. Neglect kinetic and potential energy effects.

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Problems: Developing Engineering Skills 12.90 At steady state, moist air is to be supplied to a classroom at a specified volumetric flow rate and temperature T. Air is removed from the classroom in a separate stream at a temperature of 278C and 50% relative humidity. Moisture is added to the air in the room from the occupants at a rate of 4.5 kg/h. The moisture can be regarded as saturated vapor at 338C. Heat transfer into the occupied space from all sources is estimated to occur at a rate of 34,000 kJ/h. The pressure remains uniform at 1 atm.

771

Employ this relation to demonstrate on a psychrometric chart that state 3 of the mixture lies on a straight line connecting the initial states of the two streams before mixing. 12.94 For the adiabatic mixing process in Example 12.14, plot the exit temperature, in 8C, versus the volumetric flow rate of stream 2 ranging from 0 to 1400 m3/min. Discuss the plot as (AV)2 goes to zero and as (AV)2 becomes large.

(a) For a supply air volumetric flow rate of 40 m3/min, determine the supply air temperature T, in 8C, and the relative humidity. (b) Plot the supply air temperature, in 8C, and relative humidity, each versus the supply air volumetric flow rate ranging from 35 to 90 m3/min.

12.95 A stream consisting of 35 m3/min of moist air at 148C, 1 atm, 80% relative humidity mixes adiabatically with a stream consisting of 80 m3/min of moist air at 408C, 1 atm, 40% relative humidity, giving a single mixed stream at 1 atm. Using the psychrometric chart together with the procedure of Problem 12.93, determine the relative humidity and temperature, in 8C, of the exiting stream.

12.91 At steady state, a device for heating and humidifying air has 250 ft3/min of air at 408F, 1 atm, and 80% relative humidity entering at one location, 1000 ft3/min of air at 608F, 1 atm, and 80% relative humidity entering at another location, and liquid water injected at 558F. A single moist air stream exits at 858F, 1 atm, and 35% relative humidity. Determine

12.96 At steady state, a stream of air at 608F, 1 atm, 30% relative humidity is mixed adiabatically with a stream of air at 908F, 1 atm, 70% relative humidity. The mass flow rate of the higher-temperature stream is twice that of the other stream. A single mixed stream exits at 1 atm. Using the result of Problem 12.74, determine for the exiting stream

(a) the rate of heat transfer to the device, in Btu/min. (b) the rate at which liquid water is injected, in lb/min.

(a) the temperature, in 8F. (b) the relative humidity.

Neglect kinetic and potential energy effects.

Neglect kinetic and potential energy effects.

12.92 Air at 358C, 1 bar, and 10% relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is 50 m3/min. Liquid water at 208C enters the cooler and fully evaporates. Moist air exits the cooler at 258C, 1 bar. If there is no significant heat transfer between the device and its surroundings, determine (a) the rate at which liquid enters, in kg/min. (b) the relative humidity at the exit. (c) the rate of exergy destruction, in kJ/min, for T0 5 208C. Neglect kinetic and potential energy effects. 12.93 Using Eqs. 12.56, show that # 1ha3 1 v3hg32 2 1ha2 1 v2hg22 ma1 v3 2 v2 # 5 v 2v 5 1ha1 1 v1hg12 2 1ha3 1 v3hg32 ma2 1 3

12.97 At steady state, moist air at 428C, 1 atm, 30% relative humidity is mixed adiabatically with a second moist air stream entering at 1 atm. The mass flow rates of the two streams are the same. A single mixed stream exits at 298C, 1 atm, 40% relative humidity with a mass flow rate of 2 kg/s. Kinetic and potential energy effects are negligible. For the second entering moist air stream, determine, using data from the psychrometric chart, (a) the relative humidity. (b) the temperature, in 8C. 12.98 Figure P12.98 shows two options for conditioning atmospheric air at steady state. In each case, air enters at 158C, 1 atm, and 20% relative humidity with a volumetric flow rate of 150 m3/min and exits at 308C, 1 atm, and 40%

Saturated water vapor at 1 atm 3 Air T1 = 15°C = 288 K 1 p1 = 1 atm f1 = 20% (AV)1 = 150 m3/min

2

T2 = 30°C p2 = 1 atm f2 = 40%

Option A

Liquid water at 20°C 3 Air T1 = 15°C = 288 K 1 p1 = 1 atm f1 = 20% (AV)1 = 150 m3/min

2

+ – Electric resistor

T2 = 30°C p2 = 1 atm f2 = 40%

Option B

Fig. P12.98

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relative humidity. One method conditions the air by injecting saturated water vapor at 1 atm. The other method allows the entering air to pass through a soaked pad replenished by liquid water entering at 208C. The moist air stream is then heated by an electric resistor. For T0 5 288 K, which of the two options is preferable from the standpoint of having less exergy destruction? Discuss. 12.99 Air at 308C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 58C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in 8C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0 5 208C. Neglect kinetic and potential energy effects. 12.100 Moist air enters a dehumidifier at 808F, 1 atm, and f 5 60% and exits at 588F, 1 atm, and f 5 90% with a volumetric flow rate of 10,000 ft3/min. The stream then mixes adiabatically with a moist air stream at 958F, 1 atm, and f 5 47% having a volumetric flow rate of 2000 ft3/min. A single moist-air stream exits the mixing chamber. Stray heat transfer and kinetic and potential energy effects can be ignored. Determine at steady state

Analyzing Cooling Towers 12.103 In the condenser of a power plant, energy is discharged by heat transfer at a rate of 836 MW to cooling water that exits the condenser at 408C into a cooling tower. Cooled water at 208C is returned to the condenser. Atmospheric air enters the tower at 258C, 1 atm, 35% relative humidity. Moist air exits at 358C, 1 atm, 90% relative humidity. Makeup water is supplied at 208C. For operation at steady state, determine the mass flow rate, in kg/s, of (a) the entering atmospheric air. (b) the makeup water. Ignore kinetic and potential energy effects. 12.104 Liquid water at 1008F enters a cooling tower operating at steady state, and cooled water exits the tower at 808F. Data for the various streams entering and exiting the tower are shown in Fig. P12.104. No makeup water is provided. Determine (a) the mass flow rate of the entering atmospheric air, in lb/h. (b) the rate at which water evaporates, in lb/h. (c) the mass flow rate of the exiting liquid stream, in lb/h. T3 = 85°F p3 = 1 atm f3 = 90%

(a) the rate water is removed from the moist air passing through the dehumidifier, in lb/h. (b) the temperature, in 8F, and the relative humidity of the moist air at the mixing chamber exit. 12.101 A stream of air (stream 1) at 608F, 1 atm, 30% relative humidity is mixed adiabatically with a stream of air (stream 2) at 908F, 1 atm, 80% relative humidity. A single stream (stream 3) exits the mixing chamber at temperature T3 and 1 atm. Assume steady state and ignore kinetic and potential energy effects. Letting r denote the ratio of dry air mass flow rates # # ma1/ ma2 (a) determine T3, in 8F, for r 5 2. (b) plot T3, in 8F, versus r ranging from 0 to 10. 12.102 Figure P12.102 shows the adiabatic mixing of two moist-air streams at steady state. Kinetic and potential energy effects are negligible. Determine the rate of exergy destruction, in Btu/min, for T0 5 958F. (AV) = 2770 ft3/min T = 80°F p = 1 atm φ = 50% 1 3

p = 1 atm 2 (AV) = 3000 ft3/min T = 120°F p = 1 atm Twb = 85°F

Fig. P12.102

3

Atmospheric air T1 = 70°F p1 = 1 atm 1 f1 = 40%

2

Liquid water at T2 = 100°F m· 2 = 10,000 lb/h

4

Liquid water at T4 = 80°F

Fig. P12.104 12.105 Liquid water at 1308F and a mass flow rate of 105 lb/h enters a cooling tower operating at steady state. Liquid water exits the tower at 708F. No makeup water is provided. Atmospheric air enters at 1 atm with a dry-bulb temperature of 508F and a wet-bulb temperature of 358F. Saturated air exits at 1208F, 1 atm. Ignoring kinetic and potential energy effects, determine the mass flow rate of the cooled water stream exiting the tower, in lb/h. 12.106 Liquid water at 1008F and a volumetric flow rate of 200 gal/min enters a cooling tower operating at steady state. Atmospheric air enters at 1 atm with a dry-bulb temperature of 808F and a wet-bulb temperature of 608F. Moist air exits the cooling tower at 908F and 90% relative humidity. Makeup water is provided at 808F. Plot the mass flow rates of the dry air and makeup water, each in lb/min, versus return water temperature ranging from 80 to 1008F. Ignore kinetic and potential energy effects. 12.107 Liquid water enters a cooling tower operating at steady state at 408C with a mass flow rate of 105 kg/h. Cooled water at 258C exits the cooling tower at the same mass flow rate.

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Design & Open-Ended Problems: Exploring Engineering Practice Makeup water is supplied at 238C. Atmospheric air enters the tower at 308C, 1 bar, 35% relative humidity. A saturated moist air stream exits at 348C, 1 bar. Determine (a) the mass flow rates of the dry air and makeup water, each in kg/h. (b) the rate of exergy destruction within the cooling tower, in kW, for T0 5 238C. Ignore kinetic and potential energy effects. 12.108 Liquid water at 1108F and a volumetric flow rate of 250 ft3/min enters a cooling tower operating at steady state.

773

Cooled water exits the cooling tower at 888F. Atmospheric air enters the tower at 808F, 1 atm, 40% relative humidity, and saturated moist air at 1058F, 1 atm exits the cooling tower. Determine (a) the mass flow rates of the dry air and the cooled water, each in lb/min. (b) the rate of exergy destruction within the cooling tower, in Btu/s, for T0 5 778F. Ignore kinetic and potential energy effects.

c DESIGN & OPEN-ENDED PROBLEMS: EXPLORING ENGINEERING PRACTICE 12.1D About half the air we breathe on some airplanes is fresh air, and the rest is recirculated. Investigate typical equipment schematics for providing a blend of fresh and recirculated filtered air to the passenger cabins of commercial airplanes. What types of filters are used and how do they work? Write a report including at least three references.

discuss designs of personal protective clothing for fire fighters and other first responders. How are such clothing systems designed to provide protection against chemical or biological agents while maintaining sufficient thermal comfort to allow for strenuous physical activity? Include in your report at least three references.

12.2D Identify a campus, commercial, or other building in your locale with an air-conditioning system installed 20 or more years ago. Critically evaluate the efficacy of the system in terms of comfort level provided, operating costs, maintenance costs, global warming potential of the refrigerant used, and other pertinent issues. On this basis, recommend specific system upgrades or a full system replacement, as warranted. Present your findings in a PowerPoint presentation.

12.7D Tunnel-type spray coolers are used to cool vegetables. In one application, tomatoes on a 4-ft-wide conveyor belt pass under a 348F water spray and cool from 70 to 408F. The water is collected in a reservoir below and recirculated through the evaporator of an ammonia refrigeration unit. Recommend the length of the tunnel and the conveyor speed. Estimate the number of tomatoes that can be cooled in an hour with your design. Write a report including sample calculations and at least three references.

12.3D Study the air-conditioning system for one of the classrooms you frequent where occupant comfort is unsatisfactory and describe the system in detail, including the control strategy used. Propose modifications aimed at improving occupant satisfaction, including a new heating, ventilation, and air conditioning (HVAC) system for the room, if warranted. Compare the proposed and existing systems in terms of occupant comfort, potential impact on productivity, and energy requirements. Detail your findings in an executive summary and PowerPoint presentation. 12.4D Critically evaluate a cooling tower on your campus, or nearby, in terms of effectiveness in providing the required range of cooled water, operating costs, maintenance costs, and other relevant issues. If warranted, recommend costeffective upgrades of the existing cooling tower or alternative cooling technologies to achieve the desired level of performance, including options for minimizing water loss. Present your findings in a report including at least three references. 12.5D Using methods in the most current ASHRAE Handbook of HVAC Applications, estimate the rate of evaporation from the surface of an Olympic-size swimming pool. The swimming hall is maintained at 308C with a dew point temperature of 218C. Based on your estimate, determine the load in kW, that the added moisture places on the air-conditioning system. Write a memorandum that summarizes your analysis and include at least three references. 12.6D Write a report explaining the way the human body regulates its temperature, both under cold and hot climate conditions. Based on these thermoregulation mechanisms,

12.8D Nearly 20 years ago, eight scientists entered Biosphere 2, located in Oracle, Arizona, for a planned two-year period of isolation. The three-acre biosphere had several ecosystems, including a desert, tropical rain forest, grassland, and saltwater wetlands. It also included species of plants and microorganisms intended to sustain the ecosystems. According to plan, the scientists would produce their own food using intensive organic farming, fish farmed in ponds, and a few barnyard animals. Occupants also would breathe oxygen produced by the plants and drink water cleansed by natural processes. Sunlight and a natural gas–fueled generator were to meet all energy needs. Numerous difficulties were encountered with the ecosystems and by the scientists, including insufficient oxygen, hunger and loss of body weight, and animosities among individuals. Study the record of Biosphere 2 for lessons that would substantially assist in designing a selfsustaining enclosed biosphere for human habitation on Mars. Present your results in a report including at least three references. 12.9D In a 2007 study using influenza-infected guinea pigs in climate-controlled habitats, researchers investigated the aerosol transmission of influenza virus while varying the temperature and humidity within the habitats. The research showed there were more infections when it was colder and drier, and based on this work a significant correlation was found between humidity ratio and influenza. (See BIOCONNECTIONS on p. 730) For the aerosol transmission experiments, guinea pigs were housed within cabinets like the one shown in Figure P12.9D. Each cabinet is fitted with a dedicated compressed-air

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Chapter 12 Ideal Gas Mixture and Psychrometric Applications Display panel consists of a power switch, refrigeration switch (on/off), temperature controller, and humidity controller

Air vent Guinea pigs housed here Five adjustable shelves Access port Port for condensate drain and capture (unit located on back of cabinet)

Fig. P12.9D line and an associated compressed-air dryer that provides precise and rapid control for the cabinet’s humidity injection and dehumidification systems. A condensate recirculator collects and recycles the condensate that forms in the base of the chamber and provides continuous, clean, filtered water to the cabinet’s humidity injection system. The cabinets were placed within an isolated room with an ambient temperature of approximately 208C. Experimenters say ambient temperatures in excess of 258C can result in chamber failure. The objective of this project is to specify the heating, ventilation, and air conditioning (HVAC) system for the room housing the cabinets, assuming the room encompasses 500 ft2 and houses five environmental cabinets with up to eight guinea pigs per cabinet. Each cabinet delivers a maximum heat transfer rate of 4000 Btu/h to the room. Document your design in a report including a minimum of three references that substantiate assumptions made during the design process. 12.10D Building energy use is significant in the United States, consuming about 70% of all electricity generated. An increase in electricity use of about 50% is expected by the end of the current decade. In response to the adverse impact that buildings have on energy use and the environment, in 1998 the U.S. Green Building Council developed LEED® (Leadership in Energy and Environmental Design), a certification system aimed at improving the performance of buildings across several measures, including energy and water use, greenhouse gas emission, and indoor environmental quality. Thousands of buildings throughout the world have now gained LEED®

certification. Identify a newly constructed LEED®-certified building on your campus or in a nearby locale. Determine the level of LEED® certification the building achieves: certified, silver, gold, or platinum. Prepare a summary of the building’s design, focusing on elements incorporated to improve building energy and environmental performance and associated costs. Present your findings in a written report including at least three references. 12.11D In 2010, the U.S. Department of Energy focused its research agenda on innovative technologies to provide energyefficient cooling for buildings and reduce greenhouse gas emissions. Key agenda items include developing the following: 1. Cooling systems using refrigerants with global warming potential less than or equal to 1 2. Air conditioning systems for warm and humid climates that increase the coefficient of performance of ventilation air cooling by 50% or more, based on current technology 3. Hot-climate vapor-compression air-conditioning systems that condition recirculated air while increasing the coefficient of performance by 50% or more, based on current technology For one such project supported by the Department of Energy, prepare a report that summarizes the project goals and objectives, research plan, and expected outcomes. Also critically evaluate the feasibility of incorporating the resulting technology into existing cooling systems.

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Design & Open-Ended Problems: Exploring Engineering Practice 12.12D An air-handling system is being designed for a 40 ft 3 40 ft 3 8 ft biological research facility that houses 3000 laboratory mice. The indoor conditions must be maintained at 758F, 60% relative humidity when the outdoor air conditions are 908F, 70% relative humidity. Develop a preliminary design of an air conditioning and distribution system that satisfies National Institute of Health (NIH) standards for animal facilities. Assume a biological safety level of one (BSL-1), and that two thirds of the floor space is devoted to animal care. Since an interruption in ventilation or air conditioning could place the laboratory animals under stress and compromise the research under way in the facility, account for redundancy in your design. 12.13D Adequate levels of ventilation reduce the likelihood of sick building syndrome. (See BIOCONNECTIONS on p. 727.) The outdoor air used for ventilation must be conditioned, and this requires energy. Consider the air-handling system for the commercial building illustrated in Figure P12.13D, consisting of ducting, two dampers labeled A and B, a vapor-compression dehumidifier, and a heater. The system supplies 25 m3/s of conditioned air at 208C and a relative humidity of 55% to maintain the interior space at 258C and a relative humidity of

50%. The recirculated air has the same conditions as the air in the interior space. A minimum of 5 m3/s of outdoor air is required to provide adequate ventilation. Dampers A and B can be set to provide alternative operating modes to maintain required ventilation rates. On a given summer day when the outside air dry-bulb temperature and relative humidity are 258C and 60%, respectively, which of the following three operating modes is best from the standpoint of minimizing the total heat transfer of energy from the conditioned air to the cooling coil and to the conditioned air from the heating coil? 1. Dampers A and B closed. 2. Damper A open and damper B closed with outside air contributing one-quarter of the total supply air. 3. Dampers A and B open. One-quarter of the conditioned air comes from outside air and one-third of the recirculated air bypasses the dehumidifier via open damper B; the rest flows through damper A. Present your recommendation together with your reasoning in a PowerPoint presentation suitable for your class. Additionally, in an accompanying memorandum, provide well-documented sample calculations in support of your recommendation. Recirculated air

Damper A Cooling coil

Damper B Heating coil

Supply air, 25 m3/s at 20°C, f = 55%

Outside air at 25°C, f = 60%

Condensate

Fig. P12.13D

775

Interior air-conditioned space at 25°C, f = 50%

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Combustion fundamentals are introduced in Sec. 13.1. © Digital Vision/Age Fotostock America, Inc.

ENGINEERING CONTEXT The objective of this chapter is to study systems involving chemical reactions. Since the combustion of hydrocarbon fuels occurs in most power-producing devices (Chaps. 8 and 9), combustion is emphasized. The thermodynamic analysis of reacting systems is primarily an extension of principles introduced thus far. The concepts applied in the first part of the chapter dealing with combustion fundamentals remain the same: conservation of mass, conservation of energy, and the second law. It is necessary, though, to modify the methods used to evaluate specific enthalpy, internal energy, and entropy, by accounting for changing chemical composition. Only the manner in which these properties are evaluated represents a departure from previous practice, for once appropriate values are determined they are used as in earlier chapters in the energy and entropy balances for the system under consideration. In the second part of the chapter, the exergy concept of Chap. 7 is extended by introducing chemical exergy. The principles developed in this chapter allow the equilibrium composition of a mixture of chemical substances to be determined. This topic is studied in Chap. 14. The subject of dissociation is also deferred until then. Prediction of reaction rates is not within the scope of classical thermodynamics, so the topic of chemical kinetics, which deals with reaction rates, is not discussed in this text.

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13 Reacting Mixtures and Combustion LEARNING OUTCOMES When you complete your study of this chapter, you will be able to… c

demonstrate understanding of key concepts, including complete combustion, theoretical air, enthalpy of formation, and adiabatic flame temperature.

c

determine balanced reaction equations for combustion of hydrocarbon fuels.

c

apply mass, energy, and entropy balances to closed systems and control volumes involving chemical reactions.

c

perform exergy analyses, including chemical exergy and the evaluation of exergetic efficiencies.

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Chapter 13 Reacting Mixtures and Combustion

Combustion Fundamentals 13.1 reactants products

complete combustion

Introducing Combustion

When a chemical reaction occurs, the bonds within molecules of the reactants are broken, and atoms and electrons rearrange to form products. In combustion reactions, rapid oxidation of combustible elements of the fuel results in energy release as combustion products are formed. The three major combustible chemical elements in most common fuels are carbon, hydrogen, and sulfur. Sulfur is usually a relatively unimportant contributor to the energy released, but it can be a significant cause of pollution and corrosion problems. Combustion is complete when all the carbon present in the fuel is burned to carbon dioxide, all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other combustible elements are fully oxidized. When these conditions are not fulfilled, combustion is incomplete. In this chapter, we deal with combustion reactions expressed by chemical equations of the form reactants S products or fuel 1 oxidizer S products When dealing with chemical reactions, it is necessary to remember that mass is conserved, so the mass of the products equals the mass of the reactants. The total mass of each chemical element must be the same on both sides of the equation, even though the elements exist in different chemical compounds in the reactants and products. However, the number of moles of products may differ from the number of moles of reactants. consider the complete combustion of hydrogen with oxygen 1H2 1 12 O2 S 1H2O

stoichiometric coefficients

(13.1)

In this case, the reactants are hydrogen and oxygen. Hydrogen is the fuel and oxygen is the oxidizer. Water is the only product of the reaction. The numerical coefficients in the equation, which precede the chemical symbols to give equal amounts of each chemical element on both sides of the equation, are called stoichiometric coefficients. In words, Eq. 13.1 states 1 kmol H2 1 12 kmol O2 S 1 kmol H2O or in English units 1 lbmol H2 1 12 lbmol O2 S 1 lbmol H2O Note that the total numbers of moles on the left and right sides of Eq. 13.1 are not equal. However, because mass is conserved, the total mass of reactants must equal the total mass of products. Since 1 kmol of H2 equals 2 kg, 12 kmol of O2 equals 16 kg, and 1 kmol of H2O equals 18 kg, Eq. 13.1 can be interpreted as stating 2 kg H2 1 16 kg O2 S 18 kg H2O or in English units 2 lb H2 1 16 lb O2 S 18 lb H2O

b b b b b

In the remainder of this section, consideration is given to the makeup of the fuel, oxidizer, and combustion products typically involved in engineering combustion applications.

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13.1 Introducing Combustion

13.1.1

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Fuels

A fuel is simply a combustible substance. In this chapter emphasis is on hydrocarbon fuels, which contain hydrogen and carbon. Sulfur and other chemical substances also may be present. Hydrocarbon fuels can exist as liquids, gases, and solids. Liquid hydrocarbon fuels are commonly derived from crude oil through distillation and cracking processes. Examples are gasoline, diesel fuel, kerosene, and other types of fuel oils. Most liquid fuels are mixtures of hydrocarbons for which compositions are usually given in terms of mass fractions. For simplicity in combustion calculations, gasoline is often modeled as octane, C8H18, and diesel fuel as dodecane, C12H26. Gaseous hydrocarbon fuels are obtained from natural gas wells or are produced in certain chemical processes. Natural gas normally consists of several different hydrocarbons, with the major constituent being methane, CH4. The compositions of gaseous fuels are usually given in terms of mole fractions. Both gaseous and liquid hydrocarbon fuels can be synthesized from coal, oil shale, and tar sands. Coal is a familiar solid fuel. Its composition varies considerably with the location from which it is mined. For combustion calculations, the composition of coal is usually expressed as an ultimate analysis. The ultimate analysis gives the composition on a mass basis in terms of the relative amounts of chemical elements (carbon, sulfur, hydrogen, nitrogen, oxygen) and ash.

ultimate analysis

13.1.2 Modeling Combustion Air Oxygen is required in every combustion reaction. Pure oxygen is used only in special applications such as cutting and welding. In most combustion applications, air provides the needed oxygen. The composition of a typical sample of dry air is given in Table 12.1. For the combustion calculations of this book, however, the following model is used for simplicity: c All components of dry air other than oxygen are lumped together with nitrogen.

Accordingly, air is considered to be 21% oxygen and 79% nitrogen on a molar basis. With this idealization the molar ratio of the nitrogen to the oxygen is 0.79/0.21 5 3.76. When air supplies the oxygen in a combustion reaction, therefore, every mole of oxygen is accompanied by 3.76 moles of nitrogen. c We also assume that the nitrogen present in the combustion air does not undergo chemical reaction. That is, nitrogen is regarded as inert. The nitrogen in the products is at the same temperature as the other products, however. Accordingly, nitrogen undergoes a change of state if the products are at a temperature other than the reactant air temperature. If a high enough product temperature is attained, nitrogen can form compounds such as nitric oxide and nitrogen dioxide. Even trace amounts of oxides of nitrogen appearing in the exhaust of internal combustion engines can be a source of air pollution.

TAKE NOTE...

In this model, air is assumed to contain no water vapor. When moist air is involved in combustion, the water vapor present must be considered in writing the combustion equation.

Air–Fuel Ratio Two parameters that are frequently used to quantify the amounts of fuel and air in a particular combustion process are the air–fuel ratio and its reciprocal, the fuel–air ratio. The air–fuel ratio is simply the ratio of the amount of air in a reaction to the amount of fuel. The ratio can be written on a molar basis (moles of air divided by moles of fuel) or on a mass basis (mass of air divided by mass of fuel). Conversion between these values is accomplished using the molecular weights of the air, Mair, and fuel, Mfuel, moles of air 3 Mair mass of air 5 mass of fuel moles of fuel 3 Mfuel moles of air Mair 5 a b moles of fuel Mfuel

air–fuel ratio

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Chapter 13 Reacting Mixtures and Combustion or AF 5 AF a

Mair b Mfuel

(13.2)

where AF is the air–fuel ratio on a molar basis and AF is the ratio on a mass basis. For the combustion calculations of this book the molecular weight of air is taken as 28.97. Tables A-1 provide the molecular weights of several important hydrocarbons. Since AF is a ratio, it has the same value whether the quantities of air and fuel are expressed in SI units or English units.

Theoretical Air theoretical air

The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and sulfur present in the fuel is called the theoretical amount of air. For complete combustion with the theoretical amount of air, the products consist of carbon dioxide, water, sulfur dioxide, the nitrogen accompanying the oxygen in the air, and any nitrogen contained in the fuel. No free oxygen appears in the products. let us determine the theoretical amount of air for the complete combustion of methane. For this reaction, the products contain only carbon dioxide, water, and nitrogen. The reaction is CH4 1 a1O2 1 3.76N22 S bCO2 1 cH2O 1 dN2

(13.3)

where a, b, c, and d represent the numbers of moles of oxygen, carbon dioxide, water, and nitrogen. In writing the left side of Eq. 13.3, 3.76 moles of nitrogen are considered to accompany each mole of oxygen. Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, results in four equations among the four unknowns C: H: O: N:

b51 2c 5 4 2b 1 c 5 2a d 5 3.76a

Solving these equations, the balanced chemical equation is CH4 1 21O2 1 3.76N22 S CO2 1 2H2O 1 7.52N2

(13.4)

The coefficient 2 before the term (O2 1 3.76N2) in Eq. 13.4 is the number of moles of oxygen in the combustion air, per mole of fuel, and not the amount of air. The amount of combustion air is 2 moles of oxygen plus 2 3 3.76 moles of nitrogen, giving a total of 9.52 moles of air per mole of fuel. Thus, for the reaction given by Eq. 13.4 the air–fuel ratio on a molar basis is 9.52. To calculate the air–fuel ratio on a mass basis, use Eq. 13.2 to write AF 5 AF a

percent of theoretical air percent excess air

Mair 28.97 b 5 9.52 a b 5 17.19 Mfuel 16.04

b b b b b

Normally the amount of air supplied is either greater or less than the theoretical amount. The amount of air actually supplied is commonly expressed in terms of the percent of theoretical air. For example, 150% of theoretical air means that the air actually supplied is 1.5 times the theoretical amount of air. The amount of air supplied can be expressed alternatively as a percent excess air or a percent deficiency of air. Thus, 150% of theoretical air is equivalent to 50% excess air, and 80% of theoretical air is the same as a 20% deficiency of air.

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13.1 Introducing Combustion

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consider the complete combustion of methane with 150% theoretical air (50% excess air). The balanced chemical reaction equation is CH4 1 11.521221O2 1 3.76N22 S CO2 1 2H2O 1 O2 1 11.28N2

(13.5)

In this equation, the amount of air per mole of fuel is 1.5 times the theoretical amount determined by Eq. 13.4. Accordingly, the air–fuel ratio is 1.5 times the air–fuel ratio determined for Eq. 13.4. Since complete combustion is assumed, the products contain only carbon dioxide, water, nitrogen, and oxygen. The excess air supplied appears in the products as uncombined oxygen and a greater amount of nitrogen than in Eq. 13.4, based on the theoretical amount of air. b b b b b The equivalence ratio is the ratio of the actual fuel–air ratio to the fuel–air ratio for complete combustion with the theoretical amount of air. The reactants are said to form a lean mixture when the equivalence ratio is less than unity. When the ratio is greater than unity, the reactants are said to form a rich mixture. In Example 13.1, we use conservation of mass to obtain balanced chemical reactions. The air–fuel ratio for each of the reactions is also calculated.

cccc

equivalence ratio

EXAMPLE 13.1 c

Determining the Air–Fuel Ratio for Complete Combustion of Octane Determine the air–fuel ratio on both a molar and mass basis for the complete combustion of octane, C8H18, with (a) the theoretical amount of air, (b) 150% theoretical air (50% excess air). SOLUTION Known: Octane, C8H18, is burned completely with (a) the theoretical amount of air, (b) 150% theoretical air. Find: Determine the air–fuel ratio on a molar and a mass basis. Engineering Model: 1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen. 2. The nitrogen is inert. 3. Combustion is complete. Analysis: (a) For complete combustion of C8H18 with the theoretical amount of air, the products contain carbon dioxide,

water, and nitrogen only. That is C8H18 1 a1O2 1 3.76N22 S bCO2 1 cH2O 1 dN2 Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, gives C: H: O: N:

b58 2c 5 18 2b 1 c 5 2a d 5 3.76a

Solving these equations, a 5 12.5, b 5 8, c 5 9, d 5 47. The balanced chemical equation is C8H18 1 12.51O2 1 3.76N22 S 8CO2 1 9H2O 1 47N2 The air–fuel ratio on a molar basis is AF 5

12.5 1 12.513.762 1

5

12.514.762 kmol 1air2 5 59.5 1 kmol 1fuel2

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Chapter 13 Reacting Mixtures and Combustion

The air–fuel ratio expressed on a mass basis is

AF 5 c 59.5

kmol 1air2 kmol 1fuel2

28.97 d≥ 114.22

kg 1air2 kmol 1air2 kg 1air2 ¥ 5 15.1 kg 1fuel2 kg 1fuel2 kmol 1fuel2

(b) For 150% theoretical air, the chemical equation for complete combustion takes the form

➊

C8H18 1 1.5112.521O2 1 3.76N22 S bCO2 1 cH2O 1 dN2 1 eO2

Applying conservation of mass C: H: O: N:

b58 2c 5 18 2b 1 c 1 2e 5 11.52112.52122 d 5 11.52112.5213.762

Solving this set of equations, b 5 8, c 5 9, d 5 70.5, e 5 6.25, giving a balanced chemical equation C8H18 1 18.751O2 1 3.76N22 S 8CO2 1 9H2O 1 70.5N2 1 6.25O2 The air–fuel ratio on a molar basis is AF 5

18.7514.762 1

kmol 1air2 5 89.25 kmol 1fuel2

On a mass basis, the air–fuel ratio is 22.6 kg (air)/kg (fuel), as can be verified. ➊ When complete combustion occurs with excess air, oxygen appears in the products, in addition to carbon dioxide, water, and nitrogen.

✓ Skills Developed Ability to… ❑ balance a chemical reaction

equation for complete combustion with theoretical air and with excess air. appl y definitions of air–fuel ❑ ratio on mass and molar bases.

For the condition in part (b), determine the equivalence ratio.

Ans. 0.67.

13.1.3 Determining Products of Combustion

TAKE NOTE...

In actual combustion processes, the products of combustion and their relative amounts can be determined only through measurement.

In each of the illustrations given above, complete combustion is assumed. For a hydrocarbon fuel, this means that the only allowed products are CO2, H2O, and N2, with O2 also present when excess air is supplied. If the fuel is specified and combustion is complete, the respective amounts of the products can be determined by applying the conservation of mass principle to the chemical equation. The procedure for obtaining the balanced reaction equation of an actual reaction where combustion is incomplete is not always so straightforward. Combustion is the result of a series of very complicated and rapid chemical reactions, and the products formed depend on many factors. When fuel is burned in the cylinder of an internal combustion engine, the products of the reaction vary with the temperature and pressure in the cylinder. In combustion equipment of all kinds, the degree of mixing of the fuel and air is a controlling factor in the reactions that occur once the fuel and air mixture is ignited. Although the amount of air supplied in an actual combustion process may exceed the theoretical amount, it is not uncommon for some carbon monoxide and unburned oxygen to appear in the products. This can be due to incomplete mixing, insufficient time for complete combustion, and other factors. When the amount of air supplied is less than the theoretical amount of air, the products

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13.1 Introducing Combustion may include both CO2 and CO, and there also may be unburned fuel in the products. Unlike the complete combustion cases considered above, the products of combustion of an actual combustion process and their relative amounts can be determined only by measurement. Among several devices for measuring the composition of products of combustion are the Orsat analyzer, gas chromatograph, infrared analyzer, and flame ionization detector. Data from these devices can be used to determine the mole fractions of the gaseous products of combustion. The analyses are often reported on a “dry” basis. In a dry product analysis, the mole fractions are given for all gaseous products except the water vapor. In Examples 13.2 and 13.3, we show how analyses of the products of combustion on a dry basis can be used to determine the balanced chemical reaction equations. Since water is formed when hydrocarbon fuels are burned, the mole fraction of water vapor in the gaseous products of combustion can be significant. If the gaseous products of combustion are cooled at constant mixture pressure, the dew point temperature is reached when water vapor begins to condense. Since water deposited on duct work, mufflers, and other metal parts can cause corrosion, knowledge of the dew point temperature is important. Determination of the dew point temperature is illustrated in Example 13.2, which also features a dry product analysis.

cccc

783

dry product analysis

TAKE NOTE...

For cooling of combustion products at constant pressure, the dew point temperature marks the onset of condensation of water vapor present in the products. See Sec. 12.5.4 to review this concept.

EXAMPLE 13.2 c

Using a Dry Product Analysis for Combustion of Methane Methane, CH4, is burned with dry air. The molar analysis of the products on a dry basis is CO2, 9.7%; CO, 0.5%; O2, 2.95%; and N2, 86.85%. Determine (a) the air–fuel ratio on both a molar and a mass basis, (b) the percent theoretical air, (c) the dew point temperature of the products, in 8F, if the mixture were cooled at 1 atm. SOLUTION Known: Methane is burned with dry air. The molar analysis of the products on a dry basis is provided. Find: Determine (a) the air–fuel ratio on both a molar and a mass basis, (b) the percent theoretical air, and (c) the

dew point temperature of the products, in 8F, if cooled at 1 atm. Engineering Model: 1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert. 2. The products form an ideal gas mixture and the dew point temperature of the mixture is conceptualized as

in Sec. 12.5.4. Analysis:

➊ (a) The solution is conveniently conducted on the basis of 100 lbmol of dry products. The chemical equation then reads aCH4 1 b1O2 1 3.76N22 S 9.7CO2 1 0.5CO 1 2.95O2 1 86.85N2 1 cH2O In addition to the assumed 100 lbmol of dry products, water must be included as a product. Applying conservation of mass to carbon, hydrogen, and oxygen, respectively C: H: O:

9.7 1 0.5 5 a 2c 5 4a 19.72122 1 0.5 1 212.952 1 c 5 2b

➋ Solving this set of equations gives a 5 10.2, b 5 23.1, c 5 20.4. The balanced chemical equation is 10.2CH4 1 23.11O2 1 3.76N22 S 9.7CO2 1 0.5CO 1 2.95O2 1 86.85N2 1 20.4H2O

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On a molar basis, the air–fuel ratio is AF 5

23.114.762 10.2

5 10.78

lbmol 1air2 lbmol 1fuel2

On a mass basis AF 5 110.782 a

lb 1air2 28.97 b 5 19.47 16.04 lb 1fuel2

(b) The balanced chemical equation for the complete combustion of methane with the theoretical amount of air is

CH4 1 21O2 1 3.76N22 S CO2 1 2H2O 1 7.52N2 The theoretical air–fuel ratio on a molar basis is 1AF2theo 5

214.762 1

5 9.52

lbmol 1air2 lbmol 1fuel2

The percent theoretical air is then found from 1AF2 1AF2theo 10.78 lbmol 1air2/ lbmol 1fuel2 5 1.13 1113%2 5 9.52 lbmol 1air2/ lbmol 1fuel2

% theoretical air 5

(c) To determine the dew point temperature requires the partial pressure of the water vapor pv. The partial pressure pv

is found from pv 5 yvp, where yv is the mole fraction of the water vapor in the combustion products and p is 1 atm. Referring to the balanced chemical equation of part (a), the mole fraction of the water vapor is yv 5

20.4 5 0.169 100 1 20.4

➌ Thus, pv 5 0.169 atm 5 2.484 lbf/in.2 Interpolating in Table A-2E, T 5 134°F. ➊ The solution could be obtained on the basis of any assumed amount of dry products—for example, 1 lbmol. With some other assumed amount, the values of the coefficients of the balanced chemical equation would differ from those obtained in the solution, but the air–fuel ratio, the value for the percent of theoretical air, and the dew point temperature would be unchanged. ➋ The three unknown coefficients, a, b, and c, are evaluated here by application of conservation of mass to carbon, hydrogen, and oxygen. As a check, note that the nitrogen also balances N:

b(3.76) 5 86.85

This confirms the accuracy of both the given product analysis and the calculations conducted to determine the unknown coefficients. ➌ If the products of combustion were cooled at constant pressure below the dew point temperature of 1348F, some condensation of the water vapor would occur.

✓ Skills Developed Ability to… ❑ balance a chemical reaction

equation for incomplete combustion given the analysis of dry products of combustion. ❑ apply definitions of air–fuel ratio on mass and molar bases as well as percent theoretical air. ❑ determine the dew point temperature of combustion products.

Recalculate the dew point temperature as in part (c) if the air supply were moist air, including 3.53 kmol of additional water vapor. Ans. 139°F. In Example 13.3, a fuel mixture having a known molar analysis is burned with air, giving products with a known dry analysis.

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13.1 Introducing Combustion cccc

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EXAMPLE 13.3 c

Burning Natural Gas with Excess Air A natural gas has the following molar analysis: CH4, 80.62%; C2H6, 5.41%; C3H8, 1.87%; C4H10, 1.60%; N2, 10.50%. The gas is burned with dry air, giving products having a molar analysis on a dry basis: CO2, 7.8%; CO, 0.2%; O2, 7%; N2, 85%. (a) Determine the air–fuel ratio on a molar basis. (b) Assuming ideal gas behavior for the fuel mixture, determine the amount of products in kmol that would be formed from 100 m3 of fuel mixture at 300 K and 1 bar. (c) Determine the percent of theoretical air. SOLUTION Known: A natural gas with a specified molar analysis burns with dry air, giving products having a known molar analysis on a dry basis. Find: Determine the air–fuel ratio on a molar basis, the amount of products in kmol that would be formed from

100 m3 of natural gas at 300 K and 1 bar, and the percent of theoretical air. Engineering Model: 1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert. 2. The fuel mixture can be modeled as an ideal gas. Analysis: (a) The solution can be conducted on the basis of an assumed amount of fuel mixture or on the basis of an assumed amount of dry products. Let us illustrate the first procedure, basing the solution on 1 kmol of fuel mixture. The chemical equation then takes the form

10.8062CH4 1 0.0541C2H6 1 0.0187C3H8 1 0.0160C4H10 1 0.1050N22 1 a1O2 1 3.76N22 S b10.078CO2 1 0.002CO 1 0.07O2 1 0.85N22 1 cH2O The products consist of b kmol of dry products and c kmol of water vapor, each per kmol of fuel mixture. Applying conservation of mass to carbon b10.078 1 0.0022 5 0.8062 1 210.05412 1 310.01872 1 410.01602 Solving gives b 5 12.931. Conservation of mass for hydrogen results in 2c 5 410.80622 1 610.05412 1 810.01872 1 1010.01602 which gives c 5 1.93. The unknown coefficient a can be found from either an oxygen balance or a nitrogen balance. Applying conservation of mass to oxygen 12.9313210.0782 1 0.002 1 210.0724 1 1.93 5 2a ➊ giving a 5 2.892. The balanced chemical equation is then 10.8062CH4 1 0.0541C2H6 1 0.0187C3H8 1 0.0160C4H10 1 0.1050N22 1 2.8921O2 1 3.76N22 S 12.93110.078CO2 1 0.002CO 1 0.07O2 1 0.85N22 1 1.93H2O The air–fuel ratio on a molar basis is AF 5

12.892214.762 1

5 13.77

kmol 1air2 kmol 1fuel2

(b) By inspection of the chemical reaction equation, the total amount of products is b 1 c 5 12.931 1 1.93 5

14.861 kmol of products per kmol of fuel. The amount of fuel in kmol, nF, present in 100 m3 of fuel mixture

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at 300 K and 1 bar can be determined from the ideal gas equation of state as nF 5 5

pV RT 1105 N/ m221100 m32 5 4.01 kmol 1fuel2 18314 N ? m/ kmol ? K21300 K2

Accordingly, the amount of product mixture that would be formed from 100 m3 of fuel mixture is (14.861)(4.01) 5 59.59 kmol of product gas. (c) The balanced chemical equation for the complete combustion of the fuel mixture with the theoretical amount of air is

10.8062CH4 1 0.0541C2H6 1 0.0187C3H8 1 0.0160C4H10 1 0.1050N22 1 21O2 1 3.76N22 S 1.0345CO2 1 1.93H2O 1 7.625N2 The theoretical air–fuel ratio on a molar basis is 1AF2theo 5

214.762 1

5 9.52

kmol 1air2 kmol 1fuel2

The percent theoretical air is then % theoretical air 5

13.77 kmol 1air2/ kmol 1fuel2 5 1.45 1145%2 9.52 kmol 1air2/ kmol 1fuel2

✓ Skills Developed

➊ A check on both the accuracy of the given molar analyses and the calculations conducted to determine the unknown coefficients is obtained by applying conservation of mass to nitrogen. The amount of nitrogen in the reactants is 0.105 1 13.76212.8922 5 10.98 kmol/ kmol of fuel The amount of nitrogen in the products is (0.85)(12.931) 5 10.99 kmol/kmol of fuel. The difference can be attributed to round-off.

Ability to… ❑ balance a chemical reaction

equation for incomplete combustion of a fuel mixture given the analysis of dry products of combustion. ❑ apply the definition of air– fuel ratio on a molar basis as well as percent theoretical air.

Determine the mole fractions of the products of combustion.

Ans. yCO 5 0.0679, yCO 5 0.0017, yO 5 0.0609, yN 5 0.7396, yH O 5 2

2

2

2

0.1299.

13.1.4 Energy and Entropy Balances for Reacting Systems Thus far our study of reacting systems has involved only the conservation of mass principle. A more complete understanding of reacting systems requires application of the first and second laws of thermodynamics. For these applications, energy and entropy balances play important roles, respectively. Energy balances for reacting systems are developed and applied in Secs. 13.2 and 13.3; entropy balances for reacting systems are the subject of Sec. 13.5. To apply such balances, it is necessary to take special care in how internal energy, enthalpy, and entropy are evaluated. For the energy and entropy balances of this chapter, combustion air and (normally) products of combustion are modeled as ideal gas mixtures. Accordingly, ideal gas mixture principles introduced in the first part of Chap. 12 play a role. For ease of reference, Table 13.1 summarizes ideal gas mixture relations introduced in Chap. 12 that are used in this chapter.

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TABLE 13.1

Internal Energy, Enthalpy, and Entropy for Ideal Gas Mixtures Notation: ni 5 moles of gas i, yi 5 mole fraction of gas i T 5 mixture temperature, p 5 mixture pressure pi 5 yi p 5 partial pressure of gas i ui 5 specific internal energy of gas i, per mole of i hi 5 specific enthalpy of gas i, per mole of i si 5 specific entropy of gas i, per mole of i Mixture internal energy: j

U 5 n 1u 1 1 n 2u 2 1 . . . 1 n j u j 5 a ni u i1T2

(12.19)

i5 1

Mixture enthalpy: j

H 5 n 1h 1 1 n 2h 2 1 . . . 1 n jh j 5 a n ih i1T2

(12.20)

i5 1

Mixture entropy: j

S 5 n 1s1 1 n 2s2 1 . . . 1 n j sj 5 a n i si1T, p i2

(12.26)

i5 1

c With Eq. 6.18:

pi p 5 si 1T, p2 2 R ln yi

si1T, p i2 5 si1T, p2 2 R ln

(a)

c With Eq. 6.18 and pref 5 1 atm:

pi p ref yi p

si1T, p i2 5 si1T, p ref2 2 R ln 5 s8i 1T 2 2 R ln

p ref

(b)1

where s8i is obtained from Table A-23 and A-23E, as appropriate. 1

Equation (b) corresponds to Eq. 13.23.

13.2

Conservation of Energy— Reacting Systems

The objective of the present section is to illustrate the application of the conservation of energy principle to reacting systems. The forms of the conservation of energy principle introduced previously remain valid whether or not a chemical reaction occurs within the system. However, the methods used for evaluating the properties of reacting systems differ somewhat from the practices used to this point. TAKE NOTE...

13.2.1 Evaluating Enthalpy for Reacting Systems In most tables of thermodynamic properties used thus far, values for the specific internal energy, enthalpy, and entropy are given relative to some arbitrary datum state where the enthalpy (or alternatively the internal energy) and entropy are set to zero. This approach is satisfactory for evaluations involving differences in property values between states of the same composition, for then arbitrary datums cancel. However,

When applying energy and entropy balances to reacting systems, it is necessary to take special care in how internal energy, enthalpy, and entropy are evaluated.

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standard reference state

when a chemical reaction occurs, reactants disappear and products are formed, so differences cannot be calculated for all substances involved. For reacting systems, it is necessary to evaluate h, u, and s in such a way that there are no subsequent ambiguities or inconsistencies in evaluating properties. In this section, we will consider how this is accomplished for h and u. The case of entropy is handled differently and is taken up in Sec. 13.5. An enthalpy datum for the study of reacting systems can be established by assigning arbitrarily a value of zero to the enthalpy of the stable elements at a state called the standard reference state and defined by Tref 5 298.15 K (258C) and pref 5 1 atm. In English units the temperature at the standard reference state is closely 5378R (778F). Note that only stable elements are assigned a value of zero enthalpy at the standard state. The term stable simply means that the particular element is in a chemically stable form. For example, at the standard state the stable forms of hydrogen, oxygen, and nitrogen are H2, O2, and N2 and not the monatomic H, O, and N. No ambiguities or conflicts result with this choice of datum.

ENTHALPY OF FORMATION. Using the datum introduced above, enthalpy enthalpy of formation

values can be assigned to compounds for use in the study of reacting systems. The enthalpy of a compound at the standard state equals its enthalpy of formation, symbolized h8f . The enthalpy of formation is the energy released or absorbed when the compound is formed from its elements, the compound and elements all being at Tref and pref. The enthalpy of formation is usually determined by application of procedures from statistical thermodynamics using observed spectroscopic data. The enthalpy of formation also can be found in principle by measuring the heat transfer in a reaction in which the compound is formed from the elements. consider the simple reactor shown in Fig. 13.1, in which carbon and oxygen each enter at Tref and pref and react completely at steady state to form carbon dioxide at the same temperature and pressure. Carbon dioxide is formed from carbon and oxygen according to C 1 O2 S CO2

C Tref, pref O2 Tref, pref

(13.6)

This reaction is exothermic, so for the carbon dioxide to exit at the same temperature as the entering elements, there would be a heat transfer from the reactor to its surroundings. The rate of heat transfer and the enthalpies of the incoming and exiting streams are related by the energy rate balance # # # # 0 5 Qcv 1 mChC 1 mO2hO2 2 mCO2hCO2 # where m and h denote, respectively, mass flow # rate and specific enthalpy. In writing this equation, we have assumed no work Wcv and negligible effects of kinetic and potential energy. For enthalpies on a molar basis, the energy rate balance appears as # # # # 0 5 Qcv 1 nChC 1 nO2hO2 2 nCO2hCO2 # where n and h denote, respectively, the molar flow rate and specific enthalpy per mole. Solving for the specific enthalpy of carbon dioxide and noting from Eq. 13.6 that all molar flow rates are equal # # # # nO2 nC Qcv Qcv (13.7) hCO2 5 # 1 # hC 1 # hO 5 # 1 hC 1 hO2 nCO2 nCO2 nCO2 2 nCO2 CO2 Tref, pref

Fig. 13.1 Reactor used to discuss the enthalpy of formation concept.

Since carbon and oxygen are stable elements at the standard state, hC 5 hO2 5 0, and Eq. 13.7 becomes # Qcv hCO2 5 # (13.8) nCO2 Accordingly, the value assigned to the specific enthalpy of carbon dioxide at the standard state, the enthalpy of formation, equals the heat transfer,

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per mole of CO2, between the reactor and its surroundings. If the heat transfer could be measured accurately, it would be found to equal 2393,520 kJ per kmol of carbon dioxide formed (2169,300 Btu per lbmol of CO2 formed). b b b b b Tables A-25 and A-25E give values of the enthalpy of formation for several compounds in units of kJ/kmol and Btu/lbmol, respectively. In this text, the superscript 8 is used to denote properties at 1 atm. For the case of the enthalpy of formation, the reference temperature Tref is also intended by this symbol. The values of h8f listed in Tables A-25 and A-25E for CO2 correspond to those given in the previous example. The sign associated with the enthalpy of formation values appearing in Tables A-25 corresponds to the sign convention for heat transfer. If there is heat transfer from a reactor in which a compound is formed from its elements (an exothermic reaction as in the previous example), the enthalpy of formation has a negative sign. If a heat transfer to the reactor is required (an endothermic reaction), the enthalpy of formation is positive.

Evaluating Enthalpy The specific enthalpy of a compound at a state other than the standard state is found by adding the specific enthalpy change ¢h. between the standard state and the state of interest to the enthalpy of formation h1T, p2 5 h8f 1 3h1T, p2 2 h1Tref, pref24 5 h8f 1 ¢h

(13.9)

That is, the enthalpy of a compound is composed of h8f, associated with the formation of the compound from its elements, and ¢h, associated with a change of state at constant composition. An arbitrary choice of datum can be used to determine ¢h, since it is a difference at constant composition. Accordingly, ¢h can be evaluated from tabular sources such as the steam tables, the ideal gas tables when appropriate, and so on. Note that as a consequence of the enthalpy datum adopted for the stable elements, the specific enthalpy determined from Eq. 13.9 is often negative. Tables A-25 provide two values of the enthalpy of formation of water: h8f 1l2, h8f 1g2. The first is for liquid water and the second is for water vapor. Under equilibrium conditions, water exists only as a liquid at 258C (778F) and 1 atm. The vapor value listed is for a hypothetical ideal gas state in which water is a vapor at 258C (778F) and 1 atm. The difference between the two enthalpy of formation values is given closely by the enthalpy of vaporization hfg at Tref. That is, h8f 1g2 2 h8f 1l2 < hfg1258C2

(13.10)

Similar considerations apply to other substances for which liquid and vapor values for hof are listed in Tables A-25.

13.2.2 Energy Balances for Reacting Systems Several considerations enter when writing energy balances for systems involving combustion. Some of these apply generally, without regard for whether combustion takes place. For example, it is necessary to consider if significant work and heat transfers take place and if the respective values are known or unknown. Also, the effects of kinetic and potential energy must be assessed. Other considerations are related directly to the occurrence of combustion. For example, it is important to know the state of the fuel before combustion occurs. Whether the fuel is a liquid, a gas, or a solid is important. It is also necessary to consider whether the fuel is premixed with the combustion air or the fuel and air enter a reactor separately.

TAKE NOTE...

When applying Eq. 13.9 to water vapor, we use the vapor value of the enthalpy of formation of water, h8f 1g2, from Tables A-25 together with ¢h for water vapor from Tables A-23.

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Chapter 13 Reacting Mixtures and Combustion The state of the combustion products also must be assessed. For instance, the presence of water vapor should be noted, for some of the water present will condense if the products are cooled sufficiently. The energy balance must then be written to account for the presence of water in the products as both a liquid and a vapor. For cooling of combustion products at constant pressure, the dew point temperature method of Example 13.2 is used to determine the temperature at the onset of condensation.

Analyzing Control Volumes at Steady State To illustrate the many considerations involved when writing energy balances for reacting systems, we consider special cases of broad interest, highlighting the underlying assumptions. Let us begin by considering the steady-state reactor shown in Fig. 13.2, in which a hydrocarbon fuel CaHb burns completely with the theoretical amount of air according to b b b CaHb 1 aa 1 b1O2 1 3.76N22 S aCO2 1 H2O 1 aa 1 b 3.76N2 (13.11) 4 2 4 The fuel enters the reactor in a stream separate from the combustion air, which is regarded as an ideal gas mixture. The products of combustion also are assumed to form an ideal gas mixture. Kinetic and potential energy effects are ignored. With the foregoing idealizations, the mass and energy rate balances for the twoinlet, single-exit reactor can be used to obtain the following equation on a per mole of fuel basis: # # Wcv Qcv b b # 2 # 5 c ahCO2 1 hH2O 1 aa 1 b 3.76 hN2 d 2 4 nF nF b b 2 hF 2 c aa 1 b hO2 1 aa 1 b 3.76 hN2 d 4 4

(13.12a)

# where nF denotes the molar flow rate of the fuel. Note that each coefficient on the right side of this equation is the same as the coefficient of the corresponding substance in the reaction equation. The first underlined term on the right side of Eq. 13.12a is the enthalpy of the exiting gaseous products of combustion per mole of fuel. The second underlined term on the right side is the enthalpy of the combustion air per mole of fuel. In accord with Table 13.1, the enthalpies of the combustion products and the air have been evaluated by adding the contribution of each component present in the respective ideal gas mixtures. The symbol hF denotes the molar enthalpy of the fuel. Equation 13.12a can be expressed more concisely as # # Qcv Wcv (13.12b) # 2 # 5 hP 2 hR nF nF where hP and hR denote, respectively, the enthalpies of the products and reactants per mole of fuel.

EVALUATING ENTHALPY TERMS. Once the energy balance has been written, the next step is to evaluate the individual enthalpy terms. Since each component of CaHb at TF · Qcv Air at TA

Fig. 13.2 Reactor at steady state.

· Wcv

Combustion products at TP

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the combustion products is assumed to behave as an ideal gas, its contribution to the enthalpy of the products depends solely on the temperature of the products, TP. Accordingly, for each component of the products, Eq. 13.9 takes the form h 5 h8f 1 3h1TP2 2 h1Tref24

(13.13)

In Eq. 13.13, h8f is the enthalpy of formation from Table A-25 or A-25E, as appropriate. The second term accounts for the change in enthalpy from the temperature Tref to the temperature TP. For several common gases, this term can be evaluated from tabulated values of enthalpy versus temperature in Tables A-23 and A-23E, as appropriate. Alternatively, the term can be obtained by integration of the ideal gas specific heat cp obtained from Tables A-21 or some other source of data. A similar approach is employed to evaluate the enthalpies of the oxygen and nitrogen in the combustion air. For these 0

h 5 h8f 1 3h1TA2 2 h1Tref24

(13.14)

where TA is the temperature of the air entering the reactor. Note that the enthalpy of formation for oxygen and nitrogen is zero by definition and thus drops out of Eq. 13.14 as indicated. The evaluation of the enthalpy of the fuel is also based on Eq. 13.9. If the fuel can be modeled as an ideal gas, the fuel enthalpy is obtained using an expression of the same form as Eq. 13.13, with the temperature of the incoming fuel replacing TP . With the foregoing considerations, Eq. 13.12a takes the form # # 0 Wcv Qcv b b # 2 # 5 a1h8f 1 ¢h2CO2 1 1h8f 1 ¢h2H2O 1 aa 1 b 3.761h8f 1 ¢h2N2 2 4 nF nF 0 0 b b 21h8f 1 ¢h2F 2 aa 1 b 1h8f 1 ¢h2O2 2 aa 1 b 3.761h8f 1 ¢h2N2 (13.15a) 4 4 The terms set to zero in this expression are the enthalpies of formation of oxygen and nitrogen. Equation 13.15a can be written more concisely as # # Qcv Wcv # 2 # 5 a ne1h8f 1 ¢h2e 2 a ni 1h8f 1 ¢h2i nF nF P R

(13.15b)

where i denotes the incoming fuel and air streams and e the exiting combustion products. Although Eqs. 13.15 have been developed with reference to the reaction of Eq. 13.11, equations having the same general forms would be obtained for other combustion reactions. In Examples 13.4 and 13.5, the energy balance is applied together with tabular property data to analyze control volumes at steady state involving combustion. Example 13.4 involves a reciprocating internal combustion engine while Example 13.5 involves a simple gas turbine power plant.

cccc

TAKE NOTE...

The coefficients ni and n of e Eq. 13.15b correspond to the respective coefficients of the reaction equation giving the moles of reactants and products per mole of fuel, respectively.

EXAMPLE 13.4 c

Analyzing an Internal Combustion Engine Fueled with Liquid Octane Liquid octane enters an internal combustion engine operating at steady state with a mass flow rate of 0.004 lb/s and is mixed with the theoretical amount of air. The fuel and air enter the engine at 778F and 1 atm. The mixture burns completely and combustion products leave the engine at 11408F. The engine develops a power output of 50 horsepower. Determine the rate of heat transfer from the engine, in Btu/s, neglecting kinetic and potential energy effects.

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SOLUTION Known: Liquid octane and the theoretical amount of air enter an internal combustion engine operating at steady

state in separate streams at 778F, 1 atm. Combustion is complete and the products exit at 11408F. The power developed by the engine and fuel mass flow rate are specified. Find: Determine the rate of heat transfer from the engine, in Btu/s. Schematic and Given Data:

Air at 77°F, 1 atm

Drive shaft

Engineering Model: 50 hp Combustion products at 1140°F Fuel at 77°F, 1 atm

1. The control volume identified by a dashed line on

the accompanying figure operates at steady state. 2. Kinetic and potential energy effects can be ignored. 3. The combustion air and the products of combus-

tion each form ideal gas mixtures. 4. Each mole of oxygen in the combustion air is

accompanied by 3.76 moles of nitrogen. The nitrogen is inert and combustion is complete.

Fig. E13.4 Analysis: The balanced chemical equation for complete combustion with the theoretical amount of air is obtained

from the solution to Example 13.1 as C8H18 1 12.5O2 1 47N2 S 8CO2 1 9H2O1g2 1 47N2 The energy rate balance reduces, with assumptions 1–3, to give # # Qcv Wcv # 5 # 1 hP 2 hR nF nF # 0 Wcv 5 # 1 583h8f 1 ¢h4CO2 1 93h8f 1 ¢h4H2O1g2 1 473h8f 1 ¢h4N2 6 ➊ nF 0

0