Cover
Half Title
Title
Dedication
Contents
Preface to the Third Edition
Authors
Acknowledgments
Chapter 1 Introduction to Transport Phenomena in Materials Processing
1.1 Transport Phenomena
1.2 Examples of Transport Phenomena in Materials Processing
1.2.1 Fluid Flow
1.2.2 Heat Transfer
1.2.3 Mass Transfer
1.2.4 Multimode Transport Phenomena
1.2.5 Nonuniform Distribution of Microstructure
1.3 Constitutive Relations for Transport Phenomena
1.3.1 Shear in Fluids
1.3.2 Modes of Heat Transfer
1.3.3 Mass Transfer
1.3.4 Summary of Constitutive Equations
1.4 Finding Solutions to Models of Transport Phenomena
1.4.1 Mathematical Solutions
1.4.2 Numerical Solutions
1.4.3 Scaling Analysis
1.4.4 A Note on Selection of Solution Approach
1.5 Engineering Units
1.6 Summary
References
Chapter 2 Steady State Conduction Heat Transfer
2.1 Introduction
2.2 Thermal Resistances
2.2.1 Conduction Resistance
2.2.2 Convection Resistance
2.2.4 Interface Resistance
2.3 Resistance Networks
2.4 General Heat Conduction Equation
2.5 Heat Transfer Boundary Conditions
2.6 One-Dimensional Heat Conduction
2.7 Conduction with Heat Generation
2.8 Multidimensional Conduction
2.8.1 Scaling Analysis
2.8.2 Two-Dimensional Paths in Resistance Networks: Branches and Shape Factors
2.8.3 Exact Solutions
2.9 Summary
2.10 Homework Problems
References
Chapter 3 Transient Conduction Heat Transfer
3.1 Introduction
3.2 Scaling Analysis of General Transient Conduction
3.3 Lumped Capacitance Analysis: Convection Resistance Dominated (Bi > 1)
3.4.1 Cooling in a Slab: Early Times for Bi >> 1
3.4.2 Cooling in a Slab: Late Times for Bi >> 1
3.4.3 Heating in a Radial System
3.5 Spatial Dependence: The General Solution with a Balance of Conduction and Convection Resistances (Bi ~ 1)
3.5.1 Heating in a Slab: Early Times for Bi ~ 1
3.5.2 Heating in a Slab: Late Times for Bi ~ 1
3.6 Solidification
3.6.1 Energy Balances with Phase Change
3.6.2 Solidification of a Pure Substance
3.7 Summary
3.8 Homework Problems
References
Chapter 4 Mass Diffusion in the Solid State
4.1 Introduction
4.2 Steady State Mass Diffusion
4.3 Fick’s Second Law of Diffusion: Transient Diffusion
4.4 Infinite Diffusion Couple
4.5 Diffusion Involving Solid-Solid Phase Change
4.6 Diffusion in Substitutional Solid Solutions
4.7 Darken’s Analysis
4.8 Self-Diffusion Coefficient
4.9 Measurement of the Interdiffusion Coefficient: Boltzmann–Matano Analysis
4.10 Influence of Temperature on the Diffusion Coefficient
4.11 Summary
4.12 Homework Problems
References
Chapter 5 Fluid Statics
5.1 Introduction
5.2 Concept of Pressure
5.2.1 Pressure at a Point and in a Column
5.2.2 Atmospheric Pressure
5.3 Measurement of Pressure
5.4 Pressure in Incompressible Fluids
5.5 Buoyancy
5.6 Summary
5.7 Homework Problems
Reference
Chapter 6 Mechanical Energy Balance in Fluid Flow
6.1 Introduction
6.2 Laminar and Turbulent Flows
6.3 Bernoulli’s Equation
6.4 Friction Losses
6.5 Influence of Bends, Fittings, and Changes in Pipe Radius
6.6 Steady-State Applications of the Modified Bernoulli Equation
6.7 Concept of Hydrostatic Head
6.8 Fluid Flow in an Open Channel
6.9 Transient Applications of the Modified Bernoulli Equation
6.10 Summary
6.11 Homework Problems
References
Chapter 7 Equations of Fluid Motion
7.1 Introduction
7.2 Conservation of Mass
7.3 Momentum Balance: The Navier-Stokes Equations
7.4 Boundary Conditions for Fluid Flow
7.5 Characteristics of Pressure-Driven Flow Behavior in a Channel
7.6 Summary
7.7 Homework Problems
References
Chapter 8 Internal Flows
8.1 Introduction
8.2 Simplifications of Equations of Motion for Internal Flows
8.3 Shear-Driven Flow between Flat Parallel Plates
8.4 Pressure-Driven Flow between Flat Parallel Plates
8.5 Fluid Flow in a Vertical Cylindrical Tube
8.6 Capillary Flowmeter
8.7 Non-Newtonian Internal Flows
8.7.1 Shear-Driven Flow of a Power-Law Fluid
8.7.2 Pressure-Driven Flow of a Power-Law Fluid
8.7.3 Pressure-Driven Flow of a Bingham Plastic
8.8 Flow through Porous Media
8.8.1 Resistance to Flow
8.8.2 Effect of Porous Media Structure on Flow
8.9 Fluidized Beds
8.10 Summary
8.11 Homework Problems
References
Chapter 9 External Flows
9.1 Introduction
9.2 Fully Developed Flow Down an Inclined Plane
9.3 Flow over a Horizontal Flat Plane
9.4 Momentum Integral Solution for Boundary Layer on a Horizontal Flat Plate
9.4.1 Entry Length at Entrance to a Pipe
9.5 Turbulent Flow
9.5.1 Characteristics of Turbulent Flows
9.5.2 Transition and Turbulent Flow over a Flat Plate
9.6 Flow Past Submerged Bluff Objects
9.7 Summary
9.8 Homework Problems
References
Chapter 10 Convection Heat Transfer
10.1 Introduction
10.2 General Energy Equation with Advection and Diffusion
10.3 Advection in Rigid, Moving Media
10.4 External Forced Convection
10.4.1 Forced Convection from a Horizontal Flat Plate
10.4.2 Forced Convection Correlations in Other Geometries
10.5 Internal Forced Convection
10.6 Natural Convection Heat Transfer
10.6.1 Natural Convection from an Isothermal Vertical Flat Plate
10.6.2 Natural Convection from Other Geometries
10.7 Boiling Heat Transfer
10.8 Summary
10.9 Homework Problems
References
Chapter 11 Mass Transfer in Fluids
11.1 Introduction
11.2 Mass and Molar Fluxes in a Fluid
11.3 Equations of Diffusion with Advection in a Binary Mixture A-B
11.4 Equimolar Counterdiffusion
11.5 One-Dimensional Steady-State Diffusion of Gas A through Stationary Gas B
11.6 Sublimation of a Sphere into a Stationary Gas
11.7 Film Model
11.8 Catalytic Surface Reactions
11.9 Diffusion and Chemical Reaction in Stagnant Film
11.10 Mass Transfer at Large Fluxes and Large Concentrations
11.11 Influence of Mass Transport on Heat Transfer in Stagnant Film
11.12 Mass Transfer Coefficient for Concentration Boundary Layer on a Flat Plate
11.13 Simultaneous Heat and Mass Transfer: Evaporative Cooling
11.14 Summary
11.15 Homework Problems
Chapter 12 Radiation Heat Transfer
12.1 Introduction
12.2 Intensity and Emissive Power
12.2.1 Emissive Flux
12.4 Surface Properties
12.5 Kirchhoff’s Law and the Hohlraum
12.6 Radiation Exchange in an Enclosure: View Factors
12.7 Radiation Exchange among Blackbodies
12.8 Radiation Exchange among Diffuse-Gray Surfaces
12.9 Notes on the Electrical Analogy
12.12 Summary
12.13 Homework Problems
References
Appendix I Math Practice for Transport Phenomena Course
Appendix II Equations of Motion and Thermal Energy Balance
Appendix III Unit Conversions
Appendix IV Selected Thermophysical Properties
Index

##### Citation preview

An Introduction to

Transport Phenomena in

Materials Engineering

This book elucidates the important role of conduction, convection, and radiation heat transfer, mass transport in solids and fluids, and internal and external fluid flow in the behavior of materials processes. These phenomena are critical in materials engineering because of the connection of transport to the evolution and distribution of microstructural properties during processing. From making choices in the derivation of fundamental conservation equations, to using scaling (order-of-magnitude) analysis showing relationships among different phenomena, to giving examples of how to represent real systems by simple models, the book takes the reader through the fundamentals of transport phenomena applied to materials processing. Fully updated, this third edition of a classic textbook offers a significant shift from the previous editions in the approach to this subject, representing an evolution incorporating the original ideas and extending them to a more comprehensive approach to the topic. FEATURES • Introduces order-of-magnitude (scaling) analysis and uses it to quickly obtain approximate solutions for complicated problems throughout the book • Focuses on building models to solve practical problems • Adds new sections on non-Newtonian flows, turbulence, and measurement of heat transfer coefficients • Offers expanded sections on thermal resistance networks, transient heat transfer, two-phase diffusion mass transfer, and flow in porous media • Features more homework problems, mostly on the analysis of practical problems, and new examples from a much broader range of materials classes and processes, including metals, ceramics, polymers, and electronic materials • Includes homework problems for the review of the mathematics required for a course based on this book and connects the theory represented by mathematics with real-world problems This book is aimed at advanced engineering undergraduates and students early in their graduate studies, as well as practicing engineers interested in understanding the behavior of heat and mass transfer and fluid flow during materials processing. While it is designed primarily for materials engineering education, it is a good reference for practicing materials engineers looking for insight into phenomena controlling their processes. A solutions manual, lecture slides, and figure slides are available for qualifying adopting professors.

An Introduction to Transport Phenomena in Materials Engineering Third Edition

David R. Gaskell Matthew John M. Krane

This work is dedicated to my father, Robert Joseph Krane, PhD Eternal rest grant unto him, O Lord, and may perpetual light shine upon him.

Contents Preface to the Third Edition .................................................................................. xiii Authors ...................................................................................................................xvii Acknowledgments ...................................................................................................xix Chapter 1

Introduction to Transport Phenomena in Materials Processing ............................................................................................1 1.1 1.2

Transport Phenomena ................................................................ 1 Examples of Transport Phenomena in Materials Processing ..................................................................................3 1.2.1 Fluid Flow .................................................................... 3 1.2.2 Heat Transfer ................................................................ 4 1.2.3 Mass Transfer ............................................................... 7 1.2.4 Multimode Transport Phenomena ................................ 8 1.2.5 Nonuniform Distribution of Microstructure ................ 9 1.3 Constitutive Relations for Transport Phenomena .................... 10 1.3.1 Shear in Fluids............................................................ 11 1.3.2 Modes of Heat Transfer .............................................. 22 1.3.3 Mass Transfer ............................................................. 33 1.3.4 Summary of Constitutive Equations .......................... 39 1.4 Finding Solutions to Models of Transport Phenomena ...........40 1.4.1 Mathematical Solutions .............................................. 41 1.4.2 Numerical Solutions ................................................... 43 1.4.3 Scaling Analysis ......................................................... 45 1.4.4 A Note on Selection of Solution Approach ................ 48 1.5 Engineering Units .................................................................... 50 1.6 Summary ................................................................................. 52 References ..........................................................................................53 Chapter 2

Steady State Conduction Heat Transfer.............................................. 55 2.1 2.2

2.3 2.4 2.5 2.6 2.7

Introduction ............................................................................. 55 Thermal Resistances................................................................ 55 2.2.1 Conduction Resistance ............................................... 55 2.2.2 Convection Resistance................................................ 59 2.2.3 Radiation Resistance ..................................................60 2.2.4 Interface Resistance ...................................................60 Resistance Networks................................................................ 63 General Heat Conduction Equation ......................................... 67 Heat Transfer Boundary Conditions ........................................ 71 One-Dimensional Heat Conduction ........................................ 73 Conduction with Heat Generation ........................................... 85 vii

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2.8

Multidimensional Conduction ................................................. 91 2.8.1 Scaling Analysis ......................................................... 91 2.8.2 Two-Dimensional Paths in Resistance Networks: Branches and Shape Factors .......................................97 2.8.3 Exact Solutions ......................................................... 105 2.9 Summary ............................................................................... 109 2.10 Homework Problems ............................................................. 109 References ........................................................................................118

Chapter 3

Transient Conduction Heat Transfer................................................. 120 3.1 3.2 3.3

Introduction ........................................................................... 120 Scaling Analysis of General Transient Conduction .............. 120 Lumped Capacitance Analysis: Convection Resistance Dominated (Bi > 1) ................................................................................. 129 3.4.1 Cooling in a Slab: Early Times for Bi >> 1 .............. 130 3.4.2 Cooling in a Slab: Late Times for Bi >> 1 ............... 137 3.4.3 Heating in a Radial System ...................................... 144 3.5 Spatial Dependence: The General Solution with a Balance of Conduction and Convection Resistances (Bi ~ 1) ................ 150 3.5.1 Heating in a Slab: Early Times for Bi ~ 1 ................ 150 3.5.2 Heating in a Slab: Late Times for Bi ~ 1 .................. 153 3.6 Solidification.......................................................................... 160 3.6.1 Energy Balances with Phase Change ....................... 161 3.6.2 Solidification of a Pure Substance............................ 165 3.7 Summary ............................................................................... 174 3.8 Homework Problems ............................................................. 174 References ........................................................................................185 Chapter 4

Mass Diffusion in the Solid State .................................................... 187 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

Introduction ........................................................................... 187 Steady State Mass Diffusion ................................................. 187 Fick’s Second Law of Diffusion: Transient Diffusion ........... 188 Infinite Diffusion Couple....................................................... 195 Diffusion Involving Solid-Solid Phase Change..................... 197 Diffusion in Substitutional Solid Solutions ...........................208 Darken’s Analysis ..................................................................209 Self-Diffusion Coefficient ..................................................... 213 Measurement of the Interdiffusion Coefficient: Boltzmann–Matano Analysis ................................................ 216 4.10 Influence of Temperature on the Diffusion Coefficient......... 221

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4.11 Summary ............................................................................... 225 4.12 Homework Problems ............................................................. 227 References ........................................................................................ 231 Chapter 5

Fluid Statics ...................................................................................... 232 5.1 5.2

Introduction ........................................................................... 232 Concept of Pressure ............................................................... 232 5.2.1 Pressure at a Point and in a Column ........................ 232 5.2.2 Atmospheric Pressure............................................... 235 5.3 Measurement of Pressure ...................................................... 237 5.4 Pressure in Incompressible Fluids ......................................... 241 5.5 Buoyancy ...............................................................................244 5.6 Summary ...............................................................................246 5.7 Homework Problems ............................................................. 247 Reference .......................................................................................... 250 Chapter 6

Mechanical Energy Balance in Fluid Flow ...................................... 251 6.1 6.2 6.3 6.4 6.5

Introduction ........................................................................... 251 Laminar and Turbulent Flows ............................................... 251 Bernoulli’s Equation .............................................................. 253 Friction Losses....................................................................... 255 Influence of Bends, Fittings, and Changes in Pipe Radius ............................................................................ 261 6.6 Steady-State Applications of the Modified Bernoulli Equation ................................................................ 263 6.7 Concept of Hydrostatic Head................................................. 269 6.8 Fluid Flow in an Open Channel ............................................ 270 6.9 Transient Applications of the Modified Bernoulli Equation ................................................................ 273 6.10 Summary ............................................................................... 277 6.11 Homework Problems ............................................................. 278 References ........................................................................................ 283 Chapter 7

Equations of Fluid Motion ............................................................... 285 7.1 7.2 7.3 7.4 7.5

Introduction ........................................................................... 285 Conservation of Mass ............................................................ 285 Momentum Balance: The Navier-Stokes Equations.............. 291 Boundary Conditions for Fluid Flow .................................... 296 Characteristics of Pressure-Driven Flow Behavior in a Channel ............................................................................... 299 7.6 Summary ............................................................................... 303 7.7 Homework Problems ............................................................. 303 References ........................................................................................306

x

Chapter 8

Contents

Internal Flows...................................................................................307 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Introduction ...........................................................................307 Simplifications of Equations of Motion for Internal Flows ...307 Shear-Driven Flow between Flat Parallel Plates ...................309 Pressure-Driven Flow between Flat Parallel Plates .............. 311 Fluid Flow in a Vertical Cylindrical Tube............................. 319 Capillary Flowmeter.............................................................. 325 Non-Newtonian Internal Flows ............................................. 328 8.7.1 Shear-Driven Flow of a Power-Law Fluid ................ 328 8.7.2 Pressure-Driven Flow of a Power-Law Fluid ........... 329 8.7.3 Pressure-Driven Flow of a Bingham Plastic ............ 334 8.8 Flow through Porous Media .................................................. 336 8.8.1 Resistance to Flow.................................................... 336 8.8.2 Effect of Porous Media Structure on Flow .............. 342 8.9 Fluidized Beds ....................................................................... 349 8.10 Summary ............................................................................... 354 8.11 Homework Problems ............................................................. 354 References ........................................................................................ 359

Chapter 9

External Flows..................................................................................360 9.1 9.2 9.3 9.4

Introduction ...........................................................................360 Fully Developed Flow Down an Inclined Plane....................360 Flow over a Horizontal Flat Plane ......................................... 363 Momentum Integral Solution for Boundary Layer on a Horizontal Flat Plate.............................................................. 367 9.4.1 Entry Length at Entrance to a Pipe .......................... 373 9.5 Turbulent Flow....................................................................... 375 9.5.1 Characteristics of Turbulent Flows........................... 376 9.5.2 Transition and Turbulent Flow over a Flat Plate ...... 383 9.6 Flow Past Submerged Bluff Objects ...................................... 390 9.7 Summary ............................................................................... 398 9.8 Homework Problems ............................................................. 399 References ........................................................................................ 403

Chapter 10 Convection Heat Transfer .................................................................405 10.1 10.2 10.3 10.4

Introduction ...........................................................................405 General Energy Equation with Advection and Diffusion .....408 Advection in Rigid, Moving Media ....................................... 413 External Forced Convection .................................................. 420 10.4.1 Forced Convection from a Horizontal Flat Plate ..... 420 10.4.2 Forced Convection Correlations in Other Geometries ............................................................... 439 10.5 Internal Forced Convection ................................................... 451

Contents

xi

10.6 Natural Convection Heat Transfer .........................................460 10.6.1 Natural Convection from an Isothermal Vertical Flat Plate................................................................... 462 10.6.2 Natural Convection from Other Geometries ............ 467 10.7 Boiling Heat Transfer ............................................................469 10.8 Summary ............................................................................... 474 10.9 Homework Problems ............................................................. 475 References ........................................................................................481 Chapter 11 Mass Transfer in Fluids ....................................................................484 11.1 Introduction ...........................................................................484 11.2 Mass and Molar Fluxes in a Fluid .........................................484 11.3 Equations of Diffusion with Advection in a Binary Mixture A-B ........................................................................... 486 11.4 Equimolar Counterdiffusion.................................................. 489 11.5 One-Dimensional Steady-State Diffusion of Gas A through Stationary Gas B ...................................................... 490 11.6 Sublimation of a Sphere into a Stationary Gas...................... 496 11.7 Film Model ............................................................................ 498 11.8 Catalytic Surface Reactions ..................................................500 11.9 Diffusion and Chemical Reaction in Stagnant Film ............. 502 11.10 Mass Transfer at Large Fluxes and Large Concentrations ....506 11.11 Influence of Mass Transport on Heat Transfer in Stagnant Film ........................................................................509 11.12 Mass Transfer Coefficient for Concentration Boundary Layer on a Flat Plate .............................................................. 512 11.13 Simultaneous Heat and Mass Transfer: Evaporative Cooling.................................................................................. 517 11.14 Summary ............................................................................... 520 11.15 Homework Problems ............................................................. 520 Chapter 12 Radiation Heat Transfer ................................................................... 523 12.1 Introduction ........................................................................... 523 12.2 Intensity and Emissive Power ................................................ 524 12.2.1 Emissive Flux ........................................................... 526 12.2.2 Irradiation ................................................................. 527 12.2.3 Radiosity................................................................... 528 12.3 Blackbody Radiation ............................................................. 528 12.4 Surface Properties ................................................................. 531 12.5 Kirchhoff’s Law and the Hohlraum ...................................... 537 12.6 Radiation Exchange in an Enclosure: View Factors ..............540 12.7 Radiation Exchange among Blackbodies .............................. 551 12.8 Radiation Exchange among Diffuse-Gray Surfaces ............. 554 12.9 Notes on the Electrical Analogy ........................................... 559

xii

Contents

12.10 Radiation Shields ................................................................... 562 12.11 Reradiating Surfaces .............................................................564 12.12 Summary ............................................................................... 568 12.13 Homework Problems ............................................................. 569 References ........................................................................................ 574 Appendix I Math Practice for Transport Phenomena Course ........................ 575 Appendix II Equations of Motion and Thermal Energy Balance ................... 579 Appendix III Unit Conversions........................................................................... 581 Appendix IV Selected Thermophysical Properties .......................................... 583 Index ...................................................................................................................... 591

Preface to the Third Edition APPROACH This new edition represents a significant shift from the previous version in the approach to this subject, which should not be unexpected with the addition of a new author. The shift is not a discontinuous one but represents an evolution over the last 15 years, incorporating the original ideas and extending them to a more comprehensive approach to the topic. In his original preface, Prof. Gaskell, who passed away in 2013, posited the need for materials engineers to study transport phenomena and, given the constraints of the materials science and engineering (MSE) curriculum, to do so in one semester. Heat and mass transfer and fluid flow in materials processing control microstructural development and the associated distribution of properties in engineered products and so understanding transport phenomena is absolutely critical to the understanding and design of those processes and products. To support this activity, Prof. Gaskell wrote this text and developed his third-year course at Purdue University. Regarding the focus of his text, he wrote in his first-edition preface that a careful balance must be made between an explanation of the fundamentals that govern the dynamics of fluid flow and the transport of heat and mass, on the one hand, and, on the other, illustration of the application of the fundamentals to specific systems of interest in materials engineering. His response to this challenge for an introductory class was to focus the first two editions of the text on the development of the governing equations and the small set of exact solutions. It is my view that he correctly eschewed a more handbook-style approach that tends to center on a catalog of processes, with ad hoc descriptions of associated transport phenomena included as needed, in favor of a more strictly fundamental approach, inspired by Bird, Stewart, and Lightfoot’s Transport Phenomena (1960). The understanding of the origins and solutions of the governing equations is vital to their application to materials processing, either in closed form or by numerical analysis. However, I believe that being restricted to this approach gives necessary, but not sufficient, coverage of the subject in response to the needs of MSE students and practicing materials engineers, and that it can be improved in two ways. First, the previous approach gives little explicit motivation or demonstration for why this subject is in the MSE curriculum. Students who have used this text at Purdue and elsewhere have repeatedly made this point, so a case must be made for the allocation of scarce time in the curriculum to this subject instead of others. New examples of processing in the text and end-of-chapter problems are included to show the applicability of this topic and help to justify its place in the plan of study. As part of this effort, examples are given from across the broad range of materials, including the processing of metals, ceramics, polymers, and electronic materials. The second aspect addressed in this edition is the need for more substantial discussion of the physical basis and xiii

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Preface to the Third Edition

meaning of the derived balance equations, which are the bedrock of the previous editions, and how to apply them to practical problems. These equations should be viewed as balances of physical phenomena, and my teaching approach in many cases has been to begin with scaling analysis to determine the order of magnitude of these phenomena and their interrelationships before attempting mathematical or numerical solutions. Further exploration is by approximating complex reality to one or more simple problems, which can be solved and from which solutions we can draw conclusions about the system’s behavior. This simple approach focuses the student on fundamental physics and provides the rationale for model simplification during application to practical problems. It helps the student connect the theory represented by the mathematics to the real world. In fact, in industrial settings, the scaling of a practical problem, perhaps followed by the approximate solution of a simplified model, frequently provides sufficient information to make an engineering decision. When not sufficient, the preliminary scaling and approximate analysis are still useful to understand the problem, guide development of a numerical model, and to interpret its results. I strongly believe that including practical examples and the practice of problem-solving methods adds depth to the framework provided by Prof. Gaskell in the previous editions and brings it more in line with a more materials processing– focused teaching approach. Many former students who took the course on which this text is based have reported the utility of this approach in exploring real situations in an industrial context.

USE OF THIS TEXT This text is aimed at junior/senior-level engineering undergraduates and students early in their graduate studies, as well as practicing engineers interested in understanding the behavior of heat and mass transfer and fluid flow during materials processing. It assumes familiarity with calculus, basic differential equations, and introductory thermodynamics and mechanics. An introductory course in materials science covering phase diagrams, time-temperature-transformation diagrams, and basic microstructural forms is very useful for understanding many applications. The course is a useful prerequisite to any materials processing lecture or lab course. The text is designed primarily for materials engineering education but will also be a good reference for anyone desiring a one-semester treatment of introductory fluids and heat transfer or for the practicing materials engineer looking for insight into the transport phenomena controlling her process. In all cases, the main purpose is to understand the basic physics of transport phenomena and how to model and apply them to materials processing. Looking through many syllabi for courses at many universities that could use this text, I  discovered many different orders and choices of topics. Looking through a variety of options, certain general restrictions became apparent: • • • •

fluid mechanics should be before convection, conduction should be before convection heat transfer, mass diffusion should be before convection mass transfer, and convection heat transfer should be before boiling.

Preface to the Third Edition

xv

One difficulty with any text is convincing students it is worth their time to read it. To incentivize reading the book, I recently added daily reading quizzes to the course. These short evaluations, usually just a few multiple choice or true/false questions, are taken online and graded by the course management software used at Purdue. They have been an effective tool to encourage reading before lecture and have improved class participation and quiz scores. As a last note on teaching, I  will address the evaluation of the students in this course, for which I have tried two different methods for testing. The first is to require weekly homework assignments, three hour-long midterms, and a two-hour final exam. This practice is common and does have the advantage of giving students direct incentives to work problems, which is the best way to learn the material. However, there are some difficulties. Many students tend to rely on others for their homework and so do not pay serious attention to the class for the five weeks between exams, but the material is too difficult to cram for exams three or four times a semester. In recent years, we have still assigned homework problems, but we did not collect the work. While it seems counterintuitive, the students have proven to be more likely to do their own work because of another structural change: we replaced the few midterms with biweekly, 25-minute quizzes. Having to perform as individuals on a much more frequent basis seems to keep the students’ attention, and their overall performance has improved (with no significant change in the types or total number of test questions).

Authors David R. Gaskell (1940–2013) was Professor of Materials Engineering at Purdue University from 1982 to 2013. Dr. Gaskell was born in Glasgow, Scotland, and attended the Royal College of Science and Technology, receiving First Class Honors in metallurgy and technical chemistry for a BSc in 1962. He moved to Hamilton, Canada, to pursue graduate studies at McMaster University and then immigrated to the United States, teaching first at the University of Pennsylvania and then at Purdue. During his career, he served as a visiting professor at the NRC Atlantic Regional Laboratory, Canada, and at the G. C. Williams Cooperative Research Centre for Extraction Metallurgy in the Department of Chemical Engineering, University of Melbourne, Australia. Professor Gaskell was dedicated to teaching and was the recipient of the Reinhardt Schumann Jr. Best Undergraduate Teacher Award in Materials Engineering several times over. Matthew John M. Krane (1964– ) is Professor of Materials Engineering at Purdue University and a member of the Purdue Center for Metal Casting Research and the Purdue Heat Treatment Consortium. Using modeling and experiments, his research focuses on the connection between macroscopic transport phenomena and defect formation during materials processes, particularly the study of the solidification of metal alloys. Professor Krane has been with Purdue’s School of Materials Engineering since 1996, but his education is in mechanical engineering (Cornell, BS, 1986; Pennsylvania, MS, 1989; Purdue, PhD, 1996), with a concentration in heat transfer and fluid flow. He has been a visiting researcher at the University of Birmingham (UK), the University of Greenwich (UK), and the Université de Lorraine (Nancy, France). In addition to consulting, research programs, and undergraduate projects with the metals processing industry during his time at Purdue, he worked in industry for three years on thermal issues in the design and manufacturing of electronic packaging.

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Acknowledgments

1 1.1

Introduction to Transport Phenomena in Materials Processing

TRANSPORT PHENOMENA

In any effort, it is wise to start by clearly stating the purpose of the endeavor. Our goal in writing this textbook is to help future and practicing engineers better understand and predict the behavior of mass, momentum, heat, and species movement, especially during materials processing. The study of the movement of these quantities is the field of transport phenomena. Before progressing with this topic, we should address an important question regarding transport phenomena: why should materials engineers find time to study this topic, either during a university degree program or on their own? There are many interesting and useful subjects to learn and a very finite amount of time available, so a case must be made, in a cramped curriculum or in a busy professional life, for the allocation of scarce time to this subject instead of many other interesting alternatives. While many engineers are concerned with transport phenomena for their own sake, the materials engineer heats, cools, pours, pumps, and diffuses materials in the pursuit of beneficial materials properties, to make the material economical to produce in the first place. (An example is age-hardened aluminum alloys, which make the aerospace industry possible and would not have the required mechanical properties without heating- and cooling-induced and mass diffusion–controlled solid-solid phase transformations [1]. Other examples are described in the next section.) The transport phenomena (along with the phase equilibria in the material) are the mechanisms that produce many material properties throughout processing by the control of microstructure development. This interrelationship is illustrated by the oft-cited “MSE triangle” shown in Figure 1.1. The three vertices of this triangle are interdependent, and understanding the physical mechanisms of transport phenomena is necessary to understanding many of these relationships. The primary approach we will use here to improve our understanding and our predictions is the making of models of the transport phenomena. These models are mathematical representations of reality, not reality itself. In an excellent description of model-making methodology in materials science and engineering, Ashby [2] wrote: A model is an idealization . . . (it) unashamedly distort(s) the inessentials in order to capture the features that really matter. . . . At best, it captures the essential physics of the problem, it illuminates the principles that underline the key observations, and it predicts behavior under conditions which have not yet been studied. DOI: 10.1201/9781003104278-1

1

2

An Introduction to Transport Phenomena in Materials Engineering

FIGURE 1.1 The “MSE triangle,” a representation of the primary, interconnected aspects of materials science and engineering.

FIGURE 1.2

Balance of generic quantity transport in control volume.

In this text, we will follow Ashby’s lead with the goal of understanding and estimating the behavior of transport phenomena in materials processing. The simplest way to arrange our thinking about many models of transport phenomena is in terms of a control volume, a fixed portion of space through and in which we observe the motion of some quantity. Figure 1.2 shows a schematic of transport in a control volume, which leads to a very general balance: quantity in quantity out + quantity generated = quantity stored

(1.1)

Mass, momentum, heat, and species each have their own possible mechanisms of transport across a control volume boundary and generation (or destruction) in the volume, and these mechanisms and how to model them are the subject of much of this text. The sum of these three phenomena is not necessarily zero, but is the amount accumulated in (or depleted from) the control volume. For example, momentum can be added to a control volume through pressure gradients, buoyancy, or friction, all of which can drive a fluid’s acceleration (storing momentum) or transport through the volume. Heat might be generated by a chemical reaction and transported across the boundary by thermal radiation. Most of the models of transport here will be derived starting from the simple balance in Eq. (1.1).

Introduction to Transport Phenomena in Materials Processing

1.2

3

EXAMPLES OF TRANSPORT PHENOMENA IN MATERIALS PROCESSING

The study of transport phenomena is sometimes divided into three areas: fluid flow, heat transfer, and mass transfer. To drive home the importance of transport phenomena in materials processing, we show in this section some examples of industrial processes in which these phenomena have a significant effect on their behavior.

1.2.1 Fluid Flow The shot peening process work hardens a metal surface by high-speed bombardment of that surface by small steel shot. One method of the production of that shot is sketched in Figure 1.3. Steel is melted in one furnace and then poured into the holding furnace in the figure, which in turn is drained in a controlled manner. The jet of liquid steel pouring from the holding furnace is intercepted by a high-speed water spray. The interaction of these two jets fractures the steel stream, atomizing it into a cloud of quickly frozen metal droplets. The size distribution of steel shot is a strong function of the water and steel stream velocities and the angle at which they interact [3]. Controlling the water spray velocity is done through the design of a piping system, taking into account frictional losses in the circuit and the spray nozzle and the behavior of the pump. That flow is typically steady, but the exit jet velocity of the steel and the angle at which it enters the water spray change substantially during the draining, as they are strong functions of the height of the fluid above the nozzle. In Chapter  6, methods for characterizing the steel jet behavior are explored. The physics of the liquid metal’s atomization are beyond the scope of this text; interested readers are referred to monographs on that topic [3, 4].

FIGURE 1.3 Atomizing a steel stream draining from a holding furnace. The water jet fractures the steel into a range of small droplet sizes.

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FIGURE 1.4

An Introduction to Transport Phenomena in Materials Engineering

Filling a cavity with polymer during injection molding.

One of the cheapest and fastest methods for making plastic parts is to form a polymer into the desired shape by injection molding. This process uses pressure to force a liquid polymer into the mold cavity, where it is quickly solidified and ejected. In Figure 1.4, we see the melt filling the mold. The process finishes when the mold is filled and perhaps some polymer has pushed into an air vent (which prevents air from building up a back pressure in the last-filled region of the mold). The front is propelled through the mold by the upstream inlet pressure and is resisted by the friction at the mold wall, the latter increasing as the filled region lengthens. For a constant fill rate, the pressure must therefore increase with time; alternatively, the flow rate can be allowed to slow while maintaining a constant pressure, although this option will slow the process and increase cycle time. Because the mold is colder than the injected melt, the polymer loses heat, increasing the melt viscosity and resistance to flow, and must fill the mold before that resistance grows enough to prevent complete filling. Models of this process in simple geometries, including the dependence of resistance to flow on the fluid deformation rate, are found in Chapter 8.

1.2.2 Heat transFer One example of using thermal transport phenomena to induce desirable materials behavior is the glass tempering process, Figure  1.5 [5]. Here, air is blown on the top and bottom of the moving glass sheet starting at a temperature around 640 oC [6]. This convection heat transfer (see Section 1.3.2 and Chapter 10) extracts heat from the glass in a way similar to how one cools hot coffee by blowing on it. The glass cools quickly at the top and bottom surfaces and, inside the solid sheet, heat moves from the hotter interior toward the edges much more slowly by conduction (see Section 1.3.2 and Chapters 2 and 3). Initially, the surface contracts and becomes stronger due to the cooling, while the centerline is still relatively hot and formable. Later, the centerline region finally cools, but that volume’s shrinkage is constrained in the in-plane directions by the stronger, colder outer layers. This constraint causes the centerline to be in tension and the surfaces in compression. With a careful control of the heating temperature and the convection process, the nonuniform residual stress field and surface compressive stresses can be tailored to improve the maximum fracture strength of the glass.

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The continuous casting of steel involves the introduction of liquid metal through a water-chilled copper mold (the primary cooling region), in which a strand is solidified to a small depth from its surface by the time it exits the bottom of the mold [7]. There the steel moves down from the mold and the strand’s direction is altered from vertical to horizontal by guide rolls (Figure 1.6). In this secondary cooling section, solidification is completed by heat extraction by forced convection to the water jets impinging on the steel surface and by conduction losses to the rolls. The convection

FIGURE 1.5 Side view of tempering process, with glass moved by rollers and cooled by air jets.

FIGURE 1.6 Schematic of steel continuous casting after the mold as the solidification of the strand is completed (not to scale).

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An Introduction to Transport Phenomena in Materials Engineering

is similar to that in glass tempering, but is much more effective because of boiling of the water on the very hot surface (see Chapter 10). The liquid-vapor phase change significantly enhances the heat transfer rate from the steel, required due to the high speed of the strand (~1 m/s); gas or single-phase liquid convection would be insufficient. The heat transfer from the steel to boil the water drives the other phase change in the system, the steel solidification (see Chapter  3). The thermal energy removed in only a few seconds is supplied by the drop in steel temperature and by the liquid-solid phase change. The secondary cooling finishes solidification but leaves a steep temperature gradient through the strand’s thickness; the leveling of that gradient and/or the further cooling of the steel is due to many other downstream convection and radiation processes. A third heat transfer example is found in Figure  1.7, which shows a green, or unfired, ceramic part in a furnace. A ceramic green body is made of small particles, formed while wet into a desired shape and then dried, producing a friable (easily crumbled) solid. Firing the body sinters the structure, growing the connections among the powder particles, increasing part density, and perhaps initiating solid-solid phase changes. One way to provide the heat for firing is to expose the body to high temperature–resistant heating elements or to a gas flame. In either case, heat is transferred from the source to the part by radiation (see Section 1.3.2 and Chapter 12), which is an electromagnetic phenomenon with wavelengths between ~0.1 μm and ~100 μm (depending on the source temperature). At these high temperatures, radiation is the dominant mechanism of heat addition to the part, but conduction controls how fast

FIGURE 1.7 A ceramic part being fired in a furnace with high-temperature interior walls.

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FIGURE 1.8 Dopant diffusing into unprotected Si, with dashed lines indicating constant concentration profiles. Most diffusion is normal to surface, but some dopant travels laterally, undercutting the SiO2 layers.

the body distributes the heat internally. Ceramics tend not to conduct heat quickly, so large temperature gradients may develop. These gradients cause a spatial distribution of thermal expansion, leading to high internal stresses and possible cracking.

1.2.3 Mass transFer One of the many steps in the production of integrated circuits is the doping of exposed silicon to form n-type junctions (e.g., with phosphorous) or p-type junctions (e.g., with boron) [8]. The dopant is supplied to the surface either as a solid or a gas and moves into the silicon layer by means of mass diffusion (see Section 1.3.3 and Chapter  4), with the maximum concentration on the surface. The dopant concentration falls exponentially deeper into the substrate, which initially has no dopant. Figure 1.8 shows such a Si layer, partially masked by silicon oxide. The dopant is applied to the unmasked regions and most diffuses normal to the surface. At the edges of the masking, the dopant diffusion becomes two dimensional, as it diffuses laterally underneath the SiO2. The junction formed by the dopant addition is somewhat wider than the exposed Si region. The spatial distribution of the dopant in the silicon is a function of the solid-state diffusion process, and control of these process outcomes is required for the reliable manufacture of integrated circuits. Mass may also be transferred by convection mass transfer, where it is carried through a vessel by a moving fluid. The chemical vapor deposition (CVD) process deposits thin solid films by encouraging a chemical reaction on a substrate in contact with various reactants mixed in a carrier gas. The flowing gas carries the reactants through a reactor vessel to the substrate and transports the products (and unused reactants) away from it. Ideally, the reaction will take place uniformly over the substrate and produce a uniform coating thickness [5]. In the case shown in Figure 1.9, in which the reaction occurs very rapidly compared to the rate of mass transfer to and from the surface, the difficulty is to configure the flow field and substrate so that the desired layer uniformity is achieved. One example of CVD is the deposition of a silicon nitride layer on silicon wafers by the reaction of dichlorosilane and ammonia at the heated substrate surface,

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 1.9 wafers.

Chemical vapor deposition vessel, where a solid reaction product coats the

with a reaction temperature around 750 oC [8]. The wafers are kept at the reaction temperature while the vessel walls are heated to levels outside the range in which the reaction occurs (so the reaction is only on the wafers), but always high enough so that there is no gas condensation. The solid coating, Si3N4, can be applied to mask areas of a wafer, preventing oxidation of the underlying silicon layer during processing and acting as a resistance to diffusion from the atmosphere to the silicon. The effectiveness of the coating is partially a function of its uniformity over the wafers, which is a strong function of how the carrier gas and reactants are moved from the gas stream to the surface.

1.2.4

MultiMode transport pHenoMena

Almost every example of a process (including the ones mentioned earlier) has more than one significant phenomenon, and usually they interact as they influence the development of microstructure. A classic example of the role of multiple transport phenomena in processing, structure, and properties is the quenching of plain carbon steel, a topic usually discussed in an introductory materials engineering course [e.g., 9]. Beginning with a carbon-bearing austenite phase (a face-centered cubic crystal structure) at an elevated temperature, the rate of cooling to room temperature controls the developing microstructure. If heat is extracted slowly, two new, lower-temperature, equilibrium phases are formed as the carbon diffuses in the austenite from regions where ferrite (BCC) forms to regions where cementite (Fe3C) forms. Chemical equilibrium predicts the formation of these two phases, given enough time for the carbon to move from one phase to the other. Rapid heat extraction will produce a cooling rate fast enough that the carbon cannot diffuse quickly enough to avoid being trapped in a new body-centered tetragonal phase (similar to a BCC structure, but elongated slightly by the trapped carbon atoms). In this case without long-range diffusion of carbon, only the BCT phase (distorted BCC) forms from FCC. This nonequilibrium phase, martensite, is much stronger than the ferrite-carbide structure and much less tough. This example illustrates the mechanism by which the relative rates of two transport phenomena (diffusion of heat and carbon) contribute to form different microstructures and properties.

Introduction to Transport Phenomena in Materials Processing

1.2.5

9

nonuniForM distribution oF Microstructure

In almost all real processes, temperatures, velocities, and/or concentrations are nonuniform and so affect microstructural development differently at different locations in a part. A simple example is during the cooling of a polyethylene oxide melt, in which this biocompatible polymer begins above its melt temperature (TM = 70 oC) and is quickly quenched to below TM. In Figure 1.10, the temperature field in such a quenched sample was not uniform. The upper left corner of the image cooled quickly enough that there was no time for crystallization to begin, while the lower right corner, farther from the chill, was hotter for long enough to nucleate and grow spherulites. This disparity in cooling rates over the sample caused the nonuniform microstructure. The notion of a distribution of microstructures in a material is central to most real processes. As will be seen throughout this text, the behavior of transport phenomena is rarely uniform and so neither are the consequent microstructures and properties. Another example of how transport phenomena affect distribution of microstructure is in Figure 1.11, which shows the solidification structure of a 2 m tall stainless

FIGURE 1.10 PeO microstructure resulting from a nonuniform quench from a melt. The upper left of the microstructure cooled quickly enough that no crystallization occurred, while the slow cooling of the right side allowed time for the nucleation and growth of spherulites. Source: (Image courtesy of John Howarter.)

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FIGURE 1.11 Microstructure distribution in a section of a 2  m tall 310 steel ingot, cast in a permanent mold, with a nonuniform microstructure. Specimen on display in School of Materials Engineering, Armstrong Hall, Purdue University. Source: (Image by Mathew John M. Krane.)

steel ingot solidified in a cast iron mold. Here the effects of fluid flow are seen in the columnar grain orientation in the outer regions being slightly up from the horizontal. A transition from a columnar to an equiaxed grain structure nearer the ingot center is caused by a decreasing temperature gradient at the solidification front. Large cavities near the top are due to metal motion caused by solidification shrinkage. Many details of the transport behavior that influence these structure distributions over a wide range of length scales are beyond the scope of this text, but are ably described in [10].

1.3 CONSTITUTIVE RELATIONS FOR TRANSPORT PHENOMENA The behavior of transport phenomena is governed by balance equations of the form of Eq. (1.1), and the terms in those balances include certain constitutive relations discovered through experimentation. A constitutive relation is a model of the response of a material to some stimulus. A  familiar example from classical mechanics is Hooke’s law,

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11

where stress ( ) and strain ( ) are related by the material stiffness (E), a resistance to elastic deformation. The definition of that material property is Hooke’s law, and it is derived from measurements of stress and strain (actually, force and displacement). This model of one-dimensional elastic deformation is only an approximation of the behavior of the material under load, and the elastic constant of the bulk sample, E, frequently varies as a function of load, crystallographic texture, and microstructure. Like most material properties, the stiffness in this model is defined by the behavior of a material during a specific, standardized test. In general, constitutive relations are not derived from first principles but are models of measured material responses integrating many smaller-scale phenomena glossed over by the model. The approximation of material behavior as the integration of influences of many atoms/molecules over some small volume eliminates the need for detailed understanding of how these atoms interact; the only concern is their aggregate behavior. Treating materials in this way is known as the continuum approximation, which models materials as if they exhibit the same behavior independent of their size (even to a point in space). While this approximation breaks down in a real material if the volume considered is not much larger than the mean free path of the constituent atoms or molecules, it is reasonable to be used down to the scale of tens of nanometers in most condensed phases. (Many texts refer to averaged, “macroscopic” behavior when discussing this approximation, but they miss the fact that many microstructural phenomena may be treated as continuous materials.) In this section, several constitutive equations and their measured properties are introduced for fluid flow and heat and mass transfer. They are observed, but simplified, representations of reality and provide the foundations for the many mathematical models of the phenomena studied in this text.

1.3.1

sHear in Fluids

Isaac Newton [11] described a fluid as “a body, parts of which are started by and give way to any force and, yielding easily, is moved among itself.”1 So, a fluid will deform and “move among itself” continuously upon application of a shear force. This response is different from a solid material, which under a constant load typically will deform to a certain point and stop. The fundamental constitutive equations for fluid motion relate the applied shear to the rate of material deformation and so help to describe the behavior of fluids during processing. The first constitutive equation is a simple linear one (defining Newtonian fluids), while other more complex (nonNewtonian) fluids may be modeled by one of many nonlinear relations. A  more complete discussion of these matters can be found in many sources, e.g., [12]. 1.3.1.1 Newtonian Fluids To begin the examination of a constitutive equation for stress in a fluid, we observe the situation in Figure 1.12, in which a fluid is contained between two horizontal flat plates separated by the vertical distance H. (This configuration is known as Couette flow.) If the lower plate is held stationary and a force of constant magnitude F is applied to the upper plate in the x-direction, the latter begins to accelerate in the same direction. Movement of the upper plate sets up a shear stress between

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FIGURE 1.12 Configuration of experiment defining viscosity in Newton’s law of viscosity, showing variation of local flow velocity with position in the fluid between the two parallel plates moving at different velocities.

the plate and the fluid, which opposes plate motion, and a steady state is reached when the applied force F is balanced by the shear force, in which state the upper plate has a constant velocity V. That shear is transmitted to the lower plate held in place by a resisting force equal and opposite to the force of the upper plate. Newton posited2 that a velocity V is directly proportional to the applied force F and to the spacing H, and is inversely proportional to the upper plate surface area in contact with the fluid, A [13]: V~

FH A

or

F V ~ . A H

(1.2)

The shear is transmitted between the plates and the fluid due to a no-slip condition, where the fluid and plate have the same velocity where they are in contact. Under this condition, the fluid at the lower plate (y = 0) is stationary and at the upper (y = H) it has the velocity, V, causing the development of a velocity gradient, V/H, in the fluid in the y-direction. So each layer is dragged by friction from the layer above it and restrained by friction with the layer below. In Eq. (1.2), the ratio V/H can be expressed in differential form as du/dy, where u is the velocity of a layer of liquid in the x-direction. Also, in Eq. (1.2), F/A is the shear stress at the interface between the upper plate and the fluid. Designating the shear stress as , Eq. (1.2) can be written as ~

du , dy

(1.3)

that is, the shear stress is proportional to the velocity gradient in the fluid. The proportionality constant, μ, defined as (1.4)

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An Introduction to Transport Phenomena in Materials Engineering

Table  1.1 shows gas viscosities two orders of magnitude lower than some liquids (e.g., water, aluminum). A gas is a population of atoms or molecules moving randomly in all directions within the vessel containing it. The interactions of molecules in a gas, by which they exchange energy and momentum, are almost entirely due to their occasional collisions while other intermolecular forces are relatively unimportant. An estimation of viscosity can be made by application of kinetic theory using a rigid-sphere molecular model (e.g., in [14]), assuming that the distribution of the atom velocities is determined only by the temperature of the gas and by the masses of the atoms. This derivation gives the functional dependence of the viscosity, ~ MT d 2 ,

(1.7)

where M is the gas atomic or molecular weight, T the temperature, and d the molecular diameter. Eq. (1.7) shows that gas viscosity is independent of the pressure of the gas and is a linear function of T1/2. The theoretical prediction that the viscosity of a gas is independent of pressure might, at first sight, seem surprising. It might seem that with increasing pressure, there would be more jostling of the atoms, such that, for a given velocity gradient, there would be a greater shear stress and hence greater viscosity in the gas. The independence of viscosity on density, and hence on pressure, is because μ is proportional to both density and , the gas mean free path [14]. However, as is also inversely proportional to density, these influences on viscosity are canceled. This canceling effect can be explained as follows: With decreasing density, there are fewer atoms, which travel a proportionately greater distance between collisions (higher ). The greater distance traveled exactly compensates for the smaller number of atoms moving, and hence the viscosity is independent of density. The theoretical calculation of the viscosity of a gas was first made by Maxwell in 1860, and the subsequent experimental observation that gas viscosity is independent of pressure (at least at reasonably low pressures) was a distinct triumph for the kinetic theory. While kinetic theory predicts that ~T 1/ 2 , experiments tell a slightly different story. As examples, measurements of viscosities of He and Ne as a function of temperature reveal that He ~T 0.65 and Ne ~T 0.67 . The dependence on temperature of the viscosities of all real gases is ~T n , with n in the range 0.6–1.0. This discrepancy occurs because the atoms of a gas are not nonattracting hard spheres, as assumed in the derivation of Eq. (1.7); they are “soft” spheres surrounded by force fields that attract at large and repel at short interatomic distances. Atoms colliding in a high-temperature gas penetrate more deeply into each other’s force fields than in a low-temperature gas. The diameters of the atoms thus appear to decrease with increasing temperature and, according to Eq. (1.7), the viscosity increases.

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In contrast with gases, there is no reasonable kinetic theory for liquids, and most theories of liquids are based on models which employ parameters that lack fundamental significance and which require a priori assumptions as to the structures, interactions, or mechanisms of transport in the fluid. For Newtonian liquids in shear, the molecular orientations and positions in the very loose (if closely packed) structure may change, the aggregate effect of the breaking and reforming of molecular interactions is that viscosity is relatively unchanged [15]. Figure 1.13 shows the measured temperature dependence of the viscosities of liquid metals in the form (1.8) in which expression H* is the activation enthalpy for the thermally activated process of viscous flow. Comparison of Eq. (1.8) to Eq. (1.7) shows that although the constitutive relation for the flow of Newtonian liquids and gases is the same, the kinetic-molecular mechanisms are very different, leading to different responses to temperature. The viscosities of gases increase and liquids decrease with increasing temperature. The elementary flow process in a liquid involves either the squeezing of material units (atoms or molecules) between pairs of other material units or the creation of small holes in the liquid into which a material unit can move, and H* is the energy barrier that a flow unit must overcome to squeeze successfully between its neighbors or is the energy required to form a hole. The fraction of flow units that are

FIGURE 1.13 Inverse temperature dependence of the dynamic viscosities of several liquid metals.

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FIGURE 1.14 Variation with temperature of viscosities of several gases and liquids at atmospheric pressure.

sufficiently energetic to overcome the energy barrier to flow increases exponentially with increasing temperature and hence the viscosity of the liquid decreases exponentially with increasing temperature. The variations of the viscosities of several common fluids with temperature at atmospheric pressure are shown in Figure  1.14. The difference in temperature dependence between liquids and gases is obvious, as is the fact that the heavier, more complicated molecules tend to have higher viscosities. Of course, the constitutive relation, Eq. (1.4), defines dynamic viscosity using only one-dimensional shear and deformation, which simplifies measurement of μ. However, in modeling multidimensional flows, the constitutive equations are more complicated. For a three-dimensional flow field, with the velocity vector written in

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FIGURE 1.15 Viscous shear stress vectors on the visible sides of a control volume.

and

where the components ui are u1  =  u, u2  =  v, and u3  =  w in Cartesian coordinates (x1 = x, x2 = y, x3 = z). These constitutive equations will be used to derive momentum balances for fluid flow in Chapter 7. They have the same proportionality constant, μ, as Newton’s law of viscosity, Eq. (1.4), which can be measured in the simpler, one-dimensional flow configuration. 1.3.1.2 Non-Newtonian Fluids Newtonian fluids are typically simple enough that their structure is little altered when sheared and so the resistance to shear stress is proportional to shear strain rate and the viscosity is constant. However, fluids with more complicated structures exhibit viscosity changes with level of shear. These non-Newtonian fluids include polymer

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melts and solutions, ceramic slurries, many foods, house paint, and magma [17]. The definition of viscosity from Eq. (1.6), (1.13) still holds, but the ratio is no longer constant. Because the viscosity varies with , μapp is termed the apparent viscosity. Rheology is the study of the behavior of different complex, non-Newtonian fluids,4 some examples of which can be seen in Figure 1.16, where they are compared to the response of a Newtonian fluid. The top curve is an example of a fluid which, as the strain rate increases, the shear stress required to maintain deformation rises faster. An example of such shear thickening behavior is a concentrated suspension, a liquid with a high solids loading [17]. With no strain, there is typically liquid between the solid particles. At low , that liquid lubricates the particles so they slide past each other easily and μapp of the suspension is low. With higher strain rates, the lubrication layers break down and there are increasing incidents of solid-solid contact, rapidly increasing the frictional resistance to fluid movement. In Eq. (1.13), it is seen that, because increases faster than , the apparent viscosity increases with higher strain rate. Porcelain slurries are examples of such suspensions; other shear thickening fluids include magma and wet cement aggregates. Behavior of shear thinning fluids is also sketched in Figure 1.16. These fluids are incrementally less difficult to shear as the strain rate increases. With higher , the

FIGURE 1.16 Shear stress and shear strain rate behavior of different types of fluids: shear thickening, Newtonian, and shear thinning.

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slope of shear curve decreases and the apparent viscosity is lower. As an example, one class of shear thinning fluids is thermoplastic polymer melts, which are generally long, flexible, covalently bonded chains the backbones of which are (mostly) carbon atoms [18]. (While important materials such as polymer solutions and uncured epoxy resins exhibit shear thinning behavior, as an example of such behavior we will focus in this discussion on polymer melts.) Simple examples of the side groups from the main chain include H atoms (polyethylene), H and Cl (polyvinyl chloride), H and CH3 (polypropylene), while polymethylmethacrylate and polycarbonate have much more complicated side structures. The polymer backbone in polycarbonate and polyethylene terephthalate includes other structures that are complex enough to affect the backbone flexibility and consequent properties. At rest, the chains are not straight but have kinks, bends, and loops leading to extensive chain entanglements. These chains are long, from thousands to tens of thousands of repeating units (mers), so the molecular weight (MW) is much larger than hydrocarbons, such as octane.

than a few mers). If these side branches are linked covalently to more than one chain, the polymer is said to be “cross-linked” and, with a sufficiently dense cross-linking, will not melt or flow. With no large-scale motion in the melt, polymer chains are in a low-energy state and so are in a random orientation, not straight, and entangled. At low and , these entanglements cause resistance to deformation and so higher viscosity. The shear tends to straighten the chains, but the strain rate is low enough that there is time for them to relax back to their original state. With no stable alteration of structure, the viscosity is not a function of shear; this constant value is the zero shear viscosity, μapp = μo. At higher strain rates, there is less time to relax, so structural changes in the melt persist as the polymer chains begin to straighten in the direction of flow. With fewer entanglements, the fluid becomes progressively easier to move even though there is more sliding friction along the chains. Recalling again Eq. (1.13), we see that μapp decreases at higher strain rates because the strain rate increases faster than the shear stress. When finally the chains are mostly aligned with the flow and the only friction is from sliding past each other in parallel, no further structural changes take place. At this very high strain rate, the fluid returns to Newtonian behavior and μapp = μ∞, a constant and the lowest value of viscosity. (This state is rarely achieved uncured epoxies.) For practical processing of thermoplastic polymers, is a typical range of strain rates. These three regimes are sketched in Figure 1.17, in polymer. This curve shows the two extreme, Newtonian regimes with constant slopes (log μ) and the thinning regime in the middle. Given the nature of the structural changes described earlier, obviously the viscosity will also be a function of the structure of the individual chains. More branched polymers will have more difficulty untangling and sliding past each other and so will

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FIGURE 1.17 The dependence of apparent viscosity and shear stress on strain rate for a shear thinning polymer.

have higher viscosities. The curve in Figure 1.18 shows the dependence of the zero shear viscosity (μo) on the molecular weight of a given polymer. Experiments confirm that below a critical molecular weight (MWc), the small chain length polymer has fewer chain interactions and μo ~ MW. Above the critical value, there are enough interactions so the resistance to flow increases rapidly as μo ~ MW3.4. This measured exponent is remarkably constant across a wide range of thermoplastic melts. The critical molecular weight is a function of temperature and polymer structure, but most useful thermoplastic polymers are found to have MW >> MWc. The relation plotted in Figure 1.18 suggests that a polymer will become less viscous the more times it is recycled. The shearing during that processing tends to break polymer chains, leading to shorter average chain lengths and broader size distributions in the plastic. This structural degradation deteriorates the plastic’s mechanical properties and necessitates the mixing of virgin material (usually above 50% by volume) with “post-consumer” plastic. It is also the reason for the “recycling cascade” in which high-quality polymer is recycled into products with less stringent mechanical requirements. This progresses down the cascade until such weak items as plastic shopping bags are produced, which must be reused, buried, or burned. How to model the behavior of such fluids? What sort of constitutive equation can we propose to relate shear stress and velocity gradient (or shear strain rate)? A simple relation that is a reasonable model for many fluids of this type is (in one dimension, as in Figure 1.12): (1.14)

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FIGURE 1.18 Dependence of zero shear viscosity of polymer melts on average molecular weight.

This relation, defining a power-law fluid, models shear as a simple nonlinear function of the velocity gradient. The coefficient, k, is the consistency index and has the odd units [kg/(m s2-n)], where n is the measured, unitless flow behavior index. (Note that if n = 1, then k = μ and Eq. (1.14) reduces to Newton’s law of viscosity, Eq. (1.4).) The power-law relation can also be written as (1.15) ening (n > 1) fluids to reproduce the behavior seen in Figure  1.16. However, the consistency and flow behavior indices generally are only a good fit to measured data over a few orders of magnitude of , and Eq. (1.14) is only applicable in the middle range shown in Figure 1.17. See table 1.2 for examples. A model that is applicable over a broader range of strain rate is due to Cross [19]: (1.16)

ters: μo, μ∞, k, and n

μo and μ∞) are

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The reference temperature, T*, is usually taken to be the glass transition temperature (Tg) of the polymer. For a wide range of thermoplastic melts, Eq. (1.20) is a surprisingly good fit, if C1 = 17.44, C2 = 51.6, the temperatures are in Kelvin, and if the temperature is between Tg and Tg + 100 K.

1.3.2

Modes oF Heat transFer

The first law of thermodynamics states that energy is conserved in any process, but the second law insists that if heat energy flows spontaneously, it moves from regions of higher to lower temperature [22, 23]. The study of heat transfer is the effort to understand the rate at which that thermal energy flows and how that rate depends on the temperatures of the heat sources and sinks. The discussion of constitutive relations for heat transfer begins with the classification of that phenomenon into its three primary modes, conduction, convection, and radiation, each of which has its own mechanism for moving thermal energy.

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TABLE 1.2 Sample Viscous Properties of Non-Newtonian, Power-Law Fluids (Compiled in [17]) Material High-density polyethylene (HDPE) Polystyrene (PS) Polypropylene (PP) Low-density polyethylene (LDPE) Nylon Polymethyl methacrylate (PMMA) Polycarbonate (PC) Toothpaste Chocolate Blood

T (oC)

n (-)

k (Pa sn)

180–220 190–225 180–200 160–230 220–235 220–260 280–320 27 30 25

0.6 0.2 0.4 0.45 0.65 0.25 0.65–0.8 0.28 0.5 0.9

3.8–6.2 × 103 3.5–7.5 × 104 4.5–7.0 × 103 4.3–9.4 × 103 1.8–2.6 × 103 2.5–9.0 × 104 1.0–8.5 × 103 1.2 × 102 7.0 × 10–1 4.0 × 10–3

1.3.2.1 Conduction Heat Transfer Heat transfer in a rigid material with no bulk motion is effected by the diffusion (or conduction) of thermal energy from a hot source to a cold sink. This mode of heat transfer requires a medium through which the energy can transit. Macroscopic experiments suggest that the heat flux, q (in W/m2), the rate at which thermal energy travels per unit cross-sectional area A, is proportional to the temperature gradient down which it moves: q ~

dT . dx

(1.21)

Inserting a proportionality constant, k, we have Fourier’s law for conduction, (1.22) which is the constitutive equation for conduction heat transfer. The negative sign is T(x); heat always flows down a temperature The proportionality constant, k, in Eq. (1.22) is the definition of a material’s thermal conductivity, which has units of [W/mK]. These units are those of the ratio of heat flux and temperature gradient, (W/m2)/(K/m). Thermal conductivities of materials are found over a range of five orders of magnitude, from 3400 W/mK for pure diamond, to 400 W/mK for pure copper, to 0.04 W/mK for fiberglass insulation and 0.015 W/mK for silica aerogel (all evaluated at room temperature). Table  1.3 has more examples of k for common materials. Thermal conductivity in solids can be measured in many ways, including the apparatus shown in Figure 1.19(a). A power source on the left (e.g., an electric resistance heater) provides a heat flow, q q A in

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TABLE 1.3 Typical Thermal Conductivity Values for Selected Materials Material Aluminum (99.5%) Aluminum (99.0%) Stainless steel (316) Copper Copper Copper Low-density polyethylene High-density polyethylene ZrO2 Al2O3 SiC Yttria-stabilized zirconia thermal barrier coating

T (oC)

k (W/mK)

Ref.

20 20 20 20 1037 (solid) 1083 (liquid) 20 20 20 20 20 700–1200

218 209 16 394 244 166 0.33 0.48 2.0 30 90 1.2–1.5

[24] [24] [9] [24] [24] [24] [9] [9] [9] [9] [9] [25]

[W], through the material with cross-sectional area A (normal to heat flow direction) and length L. The temperature gradient is obtained from the temperatures measured by the thermocouples and their positions. The steady state data may be plotted as in Figure 1.19(b) and the material thermal conductivity may be calculated as (1.23) where T1 and T4 are the temperatures measured near the hot and cold ends of the sample. Fourier’s law, like all experimentally observed constitutive equations, represents the integration of many physical mechanisms occurring at smaller scales than the experiment in Figure 1.19. The thermal conductivity, a property of the material, is dependent on the microstructure and atomic bonding. In condensed phases, when linking the value of k to the material structure, it can be broken down into two parts: (1.24) where klat is the contribution of lattice vibrations and kfe of the motion of free electrons. The first component of thermal conductivity in Eq. (1.24), klat, is present to some degree in all condensed phases, as it has its origin in the vibrational energy of the atoms. This energy increases monotonically with temperature, and its transfer is modeled as elastic waves (phonons) traveling through a material. In the hot regions, the phonons pick up more energy from the vigorously vibrating lattice and lose it heating colder, less energetic areas. In perfect crystals, the phonons would travel easily through the material and it would exhibit very high thermal conductivity. However, in real materials, imperfections such as dislocations and grain and phase boundaries scatter and misdirect these elastic waves, reducing the rate of transfer from hot to cold regions. When more imperfections exist in a volume, more scattering occurs, lowering k.

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FIGURE 1.19 A method for measuring thermal conductivity. (a) Schematic of test apparatus, showing heat input and positions of thermocouples (TC). (b) Temperature readings along length of a sample in which k is only a weak function of temperature.

The second part of the bulk thermal conductivity in Eq. (1.24) is due to the transfer of thermal energy by the motion of free electrons. These electrons move throughout a body in a way to maintain charge neutrality, but the ones from hotter regions have more kinetic energy, which they impart on the lattice and other free electrons as they move into the colder areas, thereby moving heat down a temperature gradient. As noted, the total thermal conductivity is a sum of these two physical effects. In metals, the influence of phonon motion is less than that of the free electrons characteristic of metallic bonding. In pure metals, k is higher than in alloys, as impurities strain the lattice and provide possible phonon scattering sites. Evidence for the dominance of conduction by free electron transport is found by the direct proportionality of thermal (k) and electrical ( e) conductivities in metals, as predicted by the Wiedemann–Franz law, (1.25) where L is a constant (which theory predicts as 2.44 × 10–8 ΩW/K2). Ceramics have much lower thermal conductivities than metals, as free electrons are not present and heat transport is entirely by phonons. Large differences in atomic size among different elements in a ceramic, as well as large, complicated units cells, may further lower k. At high temperatures, some ceramics are semitransparent to infrared electromagnetic

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radiation, causing an increase in heat transfer that appears (in tests such as in Figure 1.19) to be an increase in thermal conductivity. Actually, the increase is due to internal thermal radiation, not conduction. The porosity often found in bulk ceramics also affects the bulk value of k, lowering it due to the reduction of cross-sectional area and the complexity of the conduction path. Polymers, like ceramics, have almost no free electrons and also much less crystallinity than metals or ceramics. The less ordered structure of polymers gives rise to elastic waves (having their origin in the vibration and rotation of the molecular chains) more disorganized than in more crystalline materials, with many more opportunities for scattering. The less crystalline the polymer, the lower its thermal conductivity. Fourier’s law, Eq. (1.22), is written as a one-dimensional relationship between heat flux and temperature gradient, defined by experiments such as Figure 1.19. However,

The experiment shown in Figure 1.19 measures only the conductivity in a principal direction, kii.

Example 1.2 Heat Loss to a Buried Pipe Consider a fluid at temperature Tf = 85 oC, running through a polyvinyl chloride (PVC) pipe buried in soil at Ts = 75 oC. Assume the inside pipe surface is at Tf and the outside is at Ts. The inner radius of the pipe is Rf = 9 cm, the outer radius is Rs = 10 cm, and the length of pipe is L = 1 m. The thermal conductivity of the PVC is k = 1 W/mK. Find: The heat loss, q, from the fluid to the soil through the pipe wall. Solution: We start with the conduction rate equation in cylindrical coordinates:

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The area normal to the radial heat flow, unlike in Cartesian coordinates, is a function

1.3.2.2 Convection Heat Transfer Another mode of heat transfer that also requires a medium through which thermal energy is transported is convection. This mechanism is based on the transport of heat by conduction and bulk motion of the medium. If a body is hotter than the fluid surrounding it, heat conducts from the surface into the nearby fluid. If the fluid moves and the hot fluid near the surface is replaced quickly with cold fluid from elsewhere, heat is moved away from the body by advection, the motion of heat caused by bulk movement of the medium. The rapid replacement of hot fluid with cold maintains a high temperature gradient at the surface, keeping the heat transfer rate high. Convection heat transfer is a combination of the advection and conduction. Experimentally, such conditions can be observed in order to discover the relationship between the heat flux at the solid-fluid interface and the fluid temperature field. Such experiments show a dependence of the heat flux on the temperature difference between the surface (Ts) and the ambient fluid (T∞): (1.28) The exponent taken from these data depends on the mechanism driving the flow. In forced convection, in which the flow is propelled by some exterior force, such as a fan, a pump, or electromagnetic effects, n ≈ 1. In other cases, density gradients due to spatial variations in temperature or composition give rise to buoyancy-induced flows, also known as natural, or free, convection. For natural convection, n is usually in the range of 1 < n < 1.3. The experimental result represented by Eq. (1.28) is usually written as the rate equation for convection, (1.29) known as Newton’s law of cooling. The heat transfer coefficient, h [W/m2K], is the proportionality constant between the heat flux and temperature difference,

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TABLE 1.4 Approximate Ranges of Heat Transfer Coefficients (W/m2K) Convection type Forced Natural

Gas

Single phase liquid

Boiling liquid

10–1000 1–100

100–30,000 10–10,000

30,000–150,000 1000–50,000

determined by experiment. Unlike thermal conductivity or viscosity, the heat transfer coefficient is not a material property, as it is a function of fluid properties, velocity, and surface shape. For forced convection (n = 1), h is generally a very weak function of temperature (mostly through temperature-dependent fluid properties), while (1.30) for natural convection. The methodology for calculating and measuring heat transfer coefficients for forced or natural convection conditions will be discussed in Chapter 10. Typical ranges of heat transfer coefficient values for different conditions are found in Table 1.4.

Example 1.3 Cooling with a Heat Sink The surface of a heat sink of area, A = 0.1 m2, must not exceed a maximum temperature of Ts,max = 100 oC while dissipating heat at 10 W. The surrounding air temperature is T∞ = 30 oC. Find: (a) Plot the heat transfer coefficient required to dissipate the power as a function of the heat sink surface temperature and (b) find the minimum value of the heat transfer coefficient allowable. Solution: (a) We can rearrange the rate equation for convection, Eq. (1.29), and have plotted the heat transfer coefficient as a function of the surface temperature in Figure 1.20:

The value of the minimum heat transfer coefficient can be found from the previous equation or the plot:

The heat transfer coefficient must be at least that value to maintain the surface temperature below 100 oC. How to control the value of h is the topic of Chapter 10.

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FIGURE 1.20 Required heat transfer coefficient as a function of heat sink surface.

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FIGURE 1.21 The distribution of emissive flux from a perfect surface as a function of temperature and wavelength.

The emitted light first seen by the human eye as the temperature increases is red, at the higher end of the visible. As T continues to rise, the distribution continues to move to lower and the visible light changes from red to yellow to white (the whole visible spectrum). This shift can be observed by watching a high temperature furnace as it heats from room temperature. Eventually we see in Figure 1.21 that the peak of the emitted spectrum at about 5800 K is in the center of the visible range; this temperature is approximately what is found on the surface of the Sun. (From the viewpoint of the evolution of the eye of a daytime creature, this fact is probably not coincidental.) If the Planck’s distribution shown in Figure 1.21 is integrated over at a given T, the total rate of emission per unit area (qe ) turns out to be proportional to the absolute temperature of the surface (Ts) to the fourth power: (1.31) where the proportionality constant is  = 5.67 × 10–8 W/m2K4, the Stefan–Boltzmann constant. This quantity is a universal constant and not a material property. (The law was deduced from experimental data by Josef Stefan in 1879 and then derived theoretically by Ludwig Boltzmann in 1884.) The Stefan–Boltzmann law, Eq. (1.31), and Planck’s spectral distribution of radiative heat flux, Figure 1.21, assume a perfect emitter, known as a blackbody. No such surface exists (although approximations will be discussed in Chapter 12), and all real surfaces emit at some level less than Eq. (1.31). The first approximation of a lessthan-perfect emitter is a gray body, emitting (1.32) which introduces the surface property of emissivity, . This property tends to be a very strong function of temperature, wavelength, angle from the surface ( ), roughness,

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TABLE 1.5 Averaged Emissivity Values for Selected Materials (Excerpted from a Compilation in [26]) Material Aluminum (highly polished) Aluminum (oxidized) Alumina Brass (oxidized) Brick, fireclay Brick, red Cast iron (machined) Cast iron (oxidized) Concrete, rough Copper (polished) Copper (oxidized) Gold (highly polished) Graphite (pressed) Ice

T (oC) 225–575 95–500 540–1000 200–600 1000 20 22 40–250 38 115 25 225–625 250–510 0

(-) 0.039–0.057 0.20–0.31 0.65–0.45 0.6 0.75 0.93 0.44 0.95 0.94 0.023 0.78 0.018–0.035 0.98 0.985

and the atomic structure of the surface material. The value (0 < < 1) used in Eq. (1.32) is found for a specific surface, integrated over T, , and . Some typical values of this averaged for common surfaces are found in Table 1.5. The previous discussion focused on the emission of thermal radiation from matter, but if all surfaces emit, where does that radiation go? The answer shows another unique feature of radiation as a heat transfer mode: all bodies in a line of sight with each other exchange thermal radiation, both emitting and absorbing heat. The incident thermal radiation increases the vibrational energy of the atoms in the receiving material, raising its temperature. A  black body (a perfect emitter) will absorb all incident radiation, but real surfaces absorb something less. The surface absorptivity ( ), like emissivity, is presented for now as an average value, so we can write an (1.33) ∞

, then (1.34)

The range of

) not absorbed is

approximation of (1.35)

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This equation can be linearized so that it can be handled like the convection rate equation: (1.36)

(1.37) This rate equation, Eq. (1.35), while useful as a starting point to understanding radiation heat transfer, is much less general than the other constitutive relations for conduction and convection heat transfer. It is strictly applicable to one small “gray” surface at Ts in a much larger environment at T∞. The equations for a more general form of radiation exchange between two bodies will be derived in Chapter 12.

Example 1.4 Heat Transfer from a Surface by Parallel Radiation and Convection A surface at temperature TS loses heat to the environment (at T∞) by both forced convection and radiation. Assume the surface emissivity is  = 0.5 and T∞ = 300 K. Find: (a) Find an expression for the fraction (R) of heat loss due to radiation. (b) Plot R = R(TS) for h = 2, 20, and 200 W/m2K, over the range 300 K < TS < 1000 K. Solution: (a) From the rate equations, Eqs. (1.29) and (1.35), the total heat flux

Rewriting

Because h rad s can see that as T 0, R 0 (only convection). If the surface and R 1 (only radiation). These trends are temperature is very large, rad seen in Figure 1.22. For this specific example, the results show that, at the lowest h value, radiation is always important and it dominates heat transfer above 500 K, while at h = 200 W/m2K

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FIGURE 1.22 Fraction of heat loss due to radiation as a function of surface temperature and heat transfer coefficient.

radiation is only a small contribution over the entire range of surface temperatures. The trend at all convection heat transfer coefficients is that radiation, however unimportant at low temperatures, becomes increasingly important at higher temperatures.

1.3.3

Mass transFer

The purpose of materials processing is not only making a shape; it is also a quest for desirable properties, and that is accomplished by forming specific microstructures. The evolution of these microstructures in many processes (e.g., solidification, heat treatment, sintering) is governed by the movement of solute atoms in a solid solution. Equilibrium phase diagrams indicate a material’s thermodynamically favored state(s) at given process temperatures, but moving atoms to effect changes from one state to another generally takes a finite amount of time, so the study of mass transfer is necessary to understand microstructural development. Experimental observations in binary and many multicomponent metal alloys show that there is a net flow of solute atoms down concentration gradients. (In this text we will focus on binary alloys.) This diffusion is a manifestation of the random motion of atoms. In the solid state at any finite temperature, the atoms vibrate about their mean positions in a crystal lattice, and every now and then, when conditions are favorable, an atom can jump from one site to a neighboring vacant site. The probability that an atom may move in a given direction is affected by many factors, such as: crystal structure and orientation; the location and distribution of vacancies, including in and near dislocations; grain boundaries; surfaces; and the presence of other types of atoms (as in multicomponent alloys). In the end, the question is: why is there a net flow of atoms down a concentration gradient?

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The explanation for the net flow of atoms from high to low concentrations begins with a thought experiment known as the “drunkard’s walk” (or “random walk” in more polite company). Imagine a tipsy sailor leaves a bar and attempts to return to his ship. Although the sailor is capable of walking, he does not have control of the direction of his steps and thus his motion is random. To simplify the analysis, consider that the random motion of the sailor is in one dimension, in which steps are made either forward to the ship or backward to the bar. The sailor begins at the origin, x = 0, and takes n steps; he travels a distance l1 on the first step, l2 on the second step, and so on. For now, assume that li is a random distance and there is equal probability that its direction is positive or negative. After n steps, the net distance that the sailor has traveled from the origin is (1.38)

square distance

li to give the mean (1.39) -

tions of li distances at steps n

After the first step

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If the n steps are made at the rate of r steps per second, the time taken to make the n steps is t n / r (i.e., n rt ); hence, xn2

rtl 2 ,

so we see that xn2

12

~ t1 2 .

(1.40)

In this example, the mean square distance from the bar, Eq. (1.40), is the average distance from the bar traveled by randomly walking sailors, if we have averaged over enough sailors, each with their own random series of steps. The analogy to atoms moving in a perfect, uniform lattice leads us to expect that for a very large number of atoms, each making a very large number of jumps, in the aggregate the local concentration of solute atoms will move a distance proportional to the square root of time. Figure  x-direction. The space on the left has a concentration of C volume, and that on the right contains particles at the concentration of CR particles per unit volume. Over time t, the average distance traveled by all randomly 12 walking particles is the square root of the mean square distance, xn2 . For each atom, either from the right or the left, there is an equal probability of jumping, so the number of particles moving across the interface plane between the left and 12 right volumes over time t is ½C x 2 . Thus the flux of particles from left to right, jL R , is jL

R

1C 2

L

x2 t

12

.

FIGURE 1.23 Two rectangular volumes containing randomly walking particles at concentration CL in the left volume and CR in the right volume.

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Similarly, the flux of particles from the volume on the right to the volume on the left is

The net (1.41)

(1.42) -

(1.43)

(1.44) which is Fick’s Law of diffusion of component i and is a rate equation for mass diffusion in a binary solution. Examples are found in Table 1.6. Like other constitutive relations in this section, Fick’s first law can also be observed experimentally, and the diffusion coefficient, Di, is the ratio of the mass flux and the concentration gradient. It is similar to the thermal conductivity, k, in Fourier’s law, Eq. (1.22), in that it is a proportionality constant between a net flux and the gradient of a transport quantity. Unlike thermal conductivity, which is usually a weak function of temperature, Di tends to be a strong function of composition. This dependence is due to the effect on jump probabilities of strain fields in the bulk crystal lattice caused by the presence of solute atoms. The form of Eqs. (1.22) and

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TABLE 1.6 Mass Diffusion Coefficients (Excerpted from a Compilation in [9]) Material system

T (K)

Di (m2/s)

Cu in Al Zn in Cu C in -Fe C in -Fe Al+3 in Al2O3 O−2 in Al2O3 CO2 in polyvinyl chloride O2 in polyethylene

700 1200 900 1250 1000 1000 350 350

5.9 × 10–14 1.6 × 10–13 3.5 × 10–9 2.8 × 10–9 3.4 × 10–28 1.1 × 10–34 1.2 × 10–11 4.7 × 10–10

(1.44) make heat conduction and binary mass diffusion look similar from a macroscopic point of view, and similar techniques can be used to derive and solve heat and mass conservation equations. However, even though the phenomenological formulation is similar (at least in binary systems), it must be remembered that the physics of transport of these two phenomena are quite different and the similarities only apply in the simplest cases. We have also treated the jump mechanisms and probabilities as homogeneous and isotropic; if they are not, for example, due to the presence of a distribution of atoms of other elements in a multicomponent solid solution, then the net effect may be an atomic flux up concentration gradients. These issues will be addressed in Chapter 4.

Example 1.5 Diffusion of Hydrogen through a Planar Iron Wall A storage tank for hydrogen has an iron wall of thickness L = 0.001 m, as shown in Figure 1.24. The stored H2 gas is on the left side of the wall where the pressure is P1 = 1 atm, while there is practically no H2 in the outside atmosphere (P2 ≈ 0 atm). At 400 oC,

Find: Solution:

where CH(P) is the hydrogen concentration in iron at equilibrium with hydrogen at pressure P.

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FIGURE 1.24

Diffusion of hydrogen through a planar iron wall.

To calculate the concentration of hydrogen in iron in contact with H2 gas, we remember from basic thermochemistry that hydrogen dissolved in iron obeys Sievert’s law, which states that, for the equilibrium reaction between hydrogen in the gas and in solution in the iron,

the equilibrium constant is

where [H]Fe is the hydrogen concentration in iron (in kg H/kg Fe), in equilibrium with hydrogen gas at pressure PH2 . At 400 oC and a hydrogen pressure of 1.013 × 105 Pa (1 atm), the solubility of hydrogen in iron is 3 ppm by weight, or [H]Fe = 3 × 10–6 kg H/kg Fe. Thus at 400 oC,

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or

This result gives us the rate of H2 loss per unit area of tank surface, which is about 31 g H2/day.

1.3.4

suMMary oF constitutive equations

TABLE 1.7 List of Constitutive Equations Described in Section 1.3 Shear (Newtonian)

Shear (power-law fluid)

Conduction heat transfer Convection heat transfer Radiation heat transfer Mass diffusion (binary alloy)

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1.4 FINDING SOLUTIONS TO MODELS OF TRANSPORT PHENOMENA In the following chapters, we will derive several transient, multidimensional partial differential equations, which are the bases of our models of the conservation of various transport quantities (e.g., thermal energy, momentum, species). These derivations typically will begin with the constitutive relations in Section  1.3 and the general balance in Eq. (1.1). As we discuss different approaches to representing these physical phenomena, we should keep in mind that the purpose served by the models is to provide a useful theoretical framework for understanding reality and for estimating behavior. The reader should remember that the models are useful only as far as they do represent the real behavior of physical phenomena and must be supported and informed by carefully designed and executed experiments. A  researcher who helped lay both the theoretical and experimental foundations of the field of natural convection heat transfer put it best: “When nature talks, you have to listen” [27, 28]. At first, especially for those whose exposure to partial differential equations is thin, the complexity of these models can be intimidating. However, there are a variety of methods for extracting useful information from them that are within the scope of an introductory text such as this one. Because we have covered little of transport phenomena so far, but all engineers have been exposed to basic mechanics early in their university experience, a problem from that field is used to illustrate the different solution techniques.

Example 1.6.a Definition of the Falling Body Problem Consider the system shown in Figure 1.25 in which a body of mass mb is falling in a gravity field, restrained only by drag proportional to the fluid density, f, and the

(1.45) or (1.46) tion, gravity, and drag, respectively. The mass begins at some point, y = 0, with zero velocity (V = 0), and it is released at t = 0. We know from experience that the mass accelerates until the drag balances the gravity force, at which point V reaches a constant terminal velocity CD, is tion and all forces are in the y direction and

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FIGURE 1.25 Free body diagram of falling mass, subject to gravity and aerodynamic drag.

Find: The velocity of the body as a function of time, V(t), and the terminal velocity, Vterm, if there is one. (The various solutions will be produced and compared in extensions of this example throughout this section.)

1.4.1 MatHeMatical solutions The first approach to solving a mathematical model is to find an exact, closed-form solution. This type of solution gives a general equation describing the behavior of the system, valid for all cases within the assumptions used to make the model. This continuous result is more easily applicable than discrete results of individual experiments or numerical simulations, which rely on some form of statistical correlation of data to generate a mathematical expression of behavior. Unfortunately, the opportunity for exact solutions of the balance equations found in the study of transport phenomena is limited. These solutions are possible only if significantly limiting assumptions are made about the geometry, the properties, and the boundary conditions. In many cases such limitations still produce a reasonable representation of reality, which can be used for a design calculation at best and, at least, to better understand the physics or to verify a numerical solution. Because a model is only a simplification of reality, it must always be remembered that even an exact mathematical solution is only an approximation. In that light, it is apparent that, in addition to exact solutions, a wide range of mathematically approximate methods can be used to solve models. Generally, these methods involve assumptions about the shape of the function sought (e.g., temperature or velocity fields). This approach also may provide useful solutions if care is taken in the choice of these shapes. The approximations do

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not necessarily make for less desirable solutions, as long as their limitations are recognized and respected.

Example 1.6.b Exact Solution to Falling Body Problem

(1.47)

(1.48) t on the right,

(1.49)

(1.50)

regime”):

= Vterm (the “late time

(1.51)

(1.52)

at t (1.53)

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(1.54) We find (1.55) and (1.56)

(1.57) -

(1.58) or (1.59)

this approximation will work best at early times.

1.4.2

nuMerical solutions

If the model is complex enough not to admit an exact solution or even a useful approximation, it may be necessary to turn to numerical simulation. The fundamental idea with almost all numerical methods is the conversion of sets of continuous differential equations to systems of discrete algebraic equations by dividing both space and time into discrete pieces. Each simulation for specific boundary and initial conditions is similar to an individual experiment, in that the results are valid for only that set of conditions and neither method gives the continuous functions of the exact or approximate solutions. The numerical solution has many distinct advantages, beginning with allowing the relaxation of many simplifying assumptions and the prediction of much more complicated behavior. Numerical simulation can be an excellent diagnostic tool for investigating details of physical phenomena that are quite frequently unavailable from experiments due to the difficulty of observation. For opaque systems, or ones in which probes might alter the process or not survive high temperature or stress,

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many experiments in materials processing are forced to rely on post mortem analysis of samples from the final product. Even when in situ observation is possible, a well-validated simulation can probe a process with much finer spatial and temporal resolution than is possible in a real system. Trends in behavior with changes in processing parameters or materials can be suggested by the results of numerical studies, allowing the engineer to better understand the underlying mechanisms that control a process. Simulations may also be used as a stand-in for some experiments, to cheaply and easily estimate trends in process behavior; they are a helpful tool to reduce, but not eliminate, the need for expensive and sometimes dangerous experiments. Numerical simulation of processes does have these many advantages, but its use also comes with several caveats. The main difficulty from the point of view of a practicing engineer is also a major advantage: the proliferation of commercial software and freeware to perform these calculations. It is quite straightforward to obtain a code, be trained in the mechanics of its use, and begin calculations. The major pitfall may appear when the engineer does not have a good grasp of both the physics included in the model and the numerical techniques employed. Without understanding the first, the code user cannot judge the results of the simulation and may believe incorrect results arising either from his error in formulating the problem or from misapplication of the code. Without the second, unknown assumptions in the numerical treatment of the governing equations and the behavior of solution algorithms can cause non-physical artifacts in the predictions. For both these reasons, the number of engineers who have reported physically unreasonable results because of a lack of knowledge of physics and/or numerical methods is legion. There is also a surprising, if human, tendency for some engineers and managers to believe detailed and well-presented numerical results, even in the face of contrary reason and experimental data. In sum, giving a powerful code to an engineer unfamiliar with numerics and/or the problem physics is like giving a teenager whiskey and car keys: you are never sure of the outcome, but it is a safe bet that it will not be good. We need to understand numerical methods (to get started, see [29–31]) as well as the dominant physical mechanisms, aided by experiments and simpler solutions (as in this text), in order to produce correct and useful results and to interpret them properly. So, numerical solutions can be tricky and seductive, but that should not discourage us from using this powerful tool with discretion and caution. A famous meteorologist and pioneer numerical analyst, L. F. Richardson, has been quoted as writing that there are “in the mathematical country, . . . precipices and pit-shafts down which it would be possible to fall, but that need not deter us from walking about” [32]. We certainly encourage the reader to delve more deeply into this field of study.

Example 1.6.c Numerical Solution to Falling Body Problem Again, we begin with the first order, nonlinear differential equation for the velocity

(1.47)

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The approach here is to discretize time, so we only solve the equation at distinct instants. These instants are separated by a period of time (the time step, t). Denoting the velocity and the time of the nth step as Vn and tn and at the previous step as Vn-1 and tn-1 = tn – t, we can rewrite the velocity derivative as a ratio of differences and

(1.60) or (1.61) Eq. (1.61) is the formulation of a time marching problem. We start at n = 1, where Vo is known from the initial condition (here V(0) = Vo = 0). Once V1 = V( t) is found, the index is incremented to n = 2 and V2 = V(2 t) is calculated as a function of V1, a process repeated at each successive step n (tn = n t). When Vn ≈ Vn-1, the simulation has reached terminal velocity and can be ended. This technique is known as Euler’s Method, which is found in any basic text on numerical methods [e.g., 30, 32].

1.4.3

scaling analysis

In addition to using exact, approximate, or numerical solutions, one method of extracting useful information from our models very quickly is known as scaling analysis, which is described and applied extensively to transport phenomena in [33, 34]. The model equations for transport phenomena derived in later chapters (based on Eq. (1.1) and constitutive equations) represent balances of mass, momentum, energy, or species, and each term in these equations represents a physical phenomenon in the balance. One purpose of a scaling analysis is to estimate the order of magnitude5 of each term in the governing equations. The relative magnitude of each term guides us to making rational, defensible simplifications to the problem by telling us which terms are small enough to be neglected in the solution process. Another result is the functional relationships among the remaining terms. These relationships are only known to an order of magnitude, but are in the proper functional form. The procedure for this analysis is roughly outlined as follows: 1. Write the full balance equations and boundary and initial conditions for whatever quantities are transported. 2. In many problems, it is possible to break up the problem into different regions in space and/or time and examine each separately using the following steps: 3. Pick reference values for each of the dependent and independent variables, the latter based on careful definitions of the spatial and temporal extent of the problem. They should be selected so that the ratios of variables and

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4.

5.

6.

7.

their reference values vary between 0 and 1 in the region of interest. Often reference values are not known at the beginning, but are found during the analysis. Approximate the order of magnitude of each term in the equations by replacing each variable with its reference value. At this point the balance equation is no longer exact, but approximate because each term is only an order-of-magnitude estimate. Select a physical phenomenon suspected to be important throughout the problem. Divide the entire equation by the term representing that effect, making all terms dimensionless and giving the dominant term a value of (1). Compare the orders of magnitude of each term in the equation to that of the dominant physical mechanism, which is now (1). If the order of magnitude of a term is much less than 1, then the term has only a small effect and it can be neglected in further analysis. If the term is (1), then it should be retained, as it is as important as the dominant effect. Finally, if a term turns out to be much larger than (1), then that term should be considered dominant and the analysis should return to step (5). Compare the remaining terms to find the functional forms of the unknown reference values.

As will be seen in the many examples in this book, to perform this analysis we need some knowledge of system behavior in order to make many of these decisions, especially in steps (2), (3), and (5). Using this procedure, we continue with the example of the falling body.

Example 1.6.d Scaling Analysis of Falling Body Problem Starting once again with the force balance for the falling body with aerodynamic drag, (1.47)

, the velocity at the reference time,

(1.62)

three terms by dividing by g, leaving the gravity term of magnitude (1):

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(1.63)

(1.64)

(1.65)

-

(1.66)

(1.67) So, what have we done here? Without solving exactly the force balance, Eq. (1.46), the scaling analysis has extracted useful information (estimates of early behavior, terminal velocity, time to terminal velocity) from the equation quickly and easily. The estimates are only good to an order of magnitude, but the functional forms are known. Considering both the uncertainty of the answers and the short time it took to obtain them, the process gives a good return on investment [33]. The problem here is simple, almost trivial, but the method is also useful for much more complicated problems in transport phenomena with no exact solution, and this example serves as a preview of things to come.

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a note on selection oF solution approacH

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Example 1.6.e Comparison of Solutions of Falling Body Problem In order to demonstrate the kinds of results generated by the different solution methods introduced earlier, we finish the example running through Section 1.5. To easily compare the various solutions to this problem, we nondimensionalize the governing (1.68) where

or

(1.69) -

(1.70)

(1.71)

(1.72) Scaling: (1.73)

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FIGURE 1.26 Comparison of the solutions to falling body problem.

The solutions all match at low , which is not surprising, as the nonlinear drag term is less important there. The exact, numerical, and scaling analyses all report the same shape of the velocity behavior over time, although the scaling analysis is off by a factor of 2 on the terminal velocity. This difference is not surprising, as the scaling is only good to an order of magnitude (in that light, a factor of 1.4 is excellent). The scaling does predict the proper dependence of the early time and terminal velocities on the problem parameters. The approximate solution approach is more problematic. It correctly estimates the terminal velocity ( >> 1) and the behavior at early times ( < 1), but the truncated power series solution deviates from the exact result at intermediate times ( ~ 1). The series used was centered on  = 0 and so works best for early times. Up to 30 terms are needed to nurse a good agreement into later times ( ~ 3 or 4), illustrating that one must be aware of the applicability and limitations of any solution method used.

1.5

ENGINEERING UNITS

Standards and units for weights and measures were originally developed for the purposes of commerce, necessitated by the desire to get what one pays for while not relying on trust in strangers to ensure fair trade. Historian Ken Adler described this necessity when he wrote that “measures are a consequence of man’s fall, a human invention for a world outside Eden, where scarcity and mistrust rule, and labor and exchange are our lot” [35]. Over the course of history, different units have been developed, and their usefulness, first to commerce and then to engineering and science, has been a function of how well and widely people accepted the standards. The importance to commerce of standard and verifiable units is the main reason that the establishment or improvement of such standards is one of the first tasks of any new

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government. As an example, one of the few powers the U.S. Constitution explicitly grants to the Congress is “To . . . fix the Standard of Weights and Measures” [36]. The earliest standardized units, linear measure (length, area, volume), weight, and time, were developed for commercial and religious reasons. Across space and time, there have been a vast number of these types of units, but most were based on arbitrary but agreed-upon measures and were local in their use, usually within the jurisdiction of a single government that could regulate their implementation. Units for mass, force, and pressure were first defined at the time of Newton’s revolution (1600s), which provided the theory to discuss these mechanical concepts. Later work in classical thermodynamics (in the 1700–1800s) defined quantitative concepts such as temperature, energy, and power, necessitating the invention of standard units for these phenomena. Up to this point in history, standard units typically were based on what was at hand or could be easily replicated. As trade, technological, and religious practices accumulated and changed over the years, the number of units grew beyond counting. The European Age of Reason and the related (but not equivalent) Scientific Revolution in the 1600s and 1700s gave rise to a tendency to uniformity in practice and regularity in organization. One consequence of this tendency was the push to reduce the confusion of measurement systems, with their bewildering array of local options. This trend was also impelled by the simultaneous centralization of governmental power in the new, large nation-states. In France, during the ancien régime, a group of scientists in the French Royal Academy of Sciences proposed a unified, decimal-based system, with units based on unchanging natural phenomena. The first two standards proposed were the length of the meter, to be defined as 1/10,000,000th of the distance between the North Pole and the equator [35], and the gram, the mass of one cubic centimeter of water at its melting point. During the French Revolution, the Academy (no longer “Royal”) refined the metric system but worked for several years to precisely measure these quantities. (This process was slowed somewhat by the imprisonment or execution of several Academy members.) The system was approved in 1795 and formally adopted in 1799 when the measurements for the meter were complete. But the people of France were very reluctant to embrace the change in their everyday lives and widely ignored it. In a recognition of this reality, Napoleon I in 1812 allowed the old systems to continue in business, while the metric system still was required for government-related work. This arrangement survived the emperor and the metric system was fully not reintroduced until 1840. Throughout the nineteenth and twentieth centuries, these standard metric units spread across most of the world. The level of precision in these standards was sufficient for most business purposes during that time, but as technology and science have advanced, more and more precision has been required. In 1960, the 11th General Conference on Weights and Measures defined the Système International d’Unités (SI), based on the metric system. The conference began a process, ending in 2019, of using various universal constants to define the units. Five of the new base units of interest here are defined as [37]: • second (time): the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom is defined as 9,192,631,770 s-1;

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• meter (length): if the speed of light in a vacuum is defined as 299,792,458 m/s, the meter is defined by that speed and the definition of the second; • kilogram (mass): using Planck’s constant (defined as 6.62607015 × 10−34 kg m2/s), the meter and second define the kilogram; • ampere (electric current): taking the value of the elementary charge to be 1.602176634 × 10−19 A s, and using definition of a second, gives the ampere; • kelvin (thermodynamic temperature): beginning with the value of the Boltzmann constant as 1.380649 × 10−23 kg m2/s2K, the kilogram, meter, and second define Kelvin. The International System has been adopted almost universally, with the United States being the only major exception, although it does use SI in some manufacturing and more widely in technical education. Some local exceptions are occasionally made in many countries, such as the United Kingdom still using miles along roads and pints in pubs. The SI units for many quantities of interest in this text, and their equivalents in the United States Customary System of units are listed in Appendix III. In this text, the unit systems of “mks” (meter, kilogram, second) or “cgs” (centimeter, gram, second) will be used primarily, with temperatures in oC or K. Other, derived units of interest in this text include the newton (N = kg m/s2) for force, the joule (J = N m = kg m2/s2) for energy, and the watt (W = J/s = kg m2/s3) for power.

1.6 SUMMARY Introduced in this chapter is the study of transport phenomena, the purpose of which is to understand and predict the behavior of mass, momentum, heat, and species movement, especially during materials processing. We show the influence of different transport phenomena, including different modes of heat transfer (conduction, convection, radiation), fluid mechanics, and mass transfer, on examples of processing of metals, ceramics, polymers, and electronic materials. Constitutive equations for dependence of transport rates on different stimuli are introduced, and these functions are discussed in terms of relationships between material properties and structure. Anticipating the use of these constitutive equations in heat, mass, and momentum balance as models of transport mechanisms, the use of exact, approximate, numerical and scaling methods to solve these models are compared and contrasted. As an example, a simple mechanics problem is solved with all these methods and the results are compared.

NOTES 1. “Fluidum est corpus omne cuius partes cedunt vi cuicunque illatæ, & cedendo facile movetur inter se.” [11] Author’s translation. 2. Newton actually worked out the ideas behind the viscosity law in flow around a cylinder, but the result is the same. 3. Other common SI units for dynamic viscosity: kg ms= Ns m 2 = Pa s=10 g cm s=10 poise. Many other units are in limited use, each based on a different technique for measuring viscosity.

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4. Rheology is sometimes described as the “study of all fluid flows,” but this definition is fairly broad and not universally accepted. It is certainly used for the study of behavior of fluids with complex structures, but fluid mechanicians who study Newtonian flows (e.g., turbulent water flow past boats or liquid metal motion in casting processes) generally do not consider themselves to be rheologists. Such is the imprecision and, one might say, fluidity of the boundaries of technical fields. 5. The order of magnitude of a quantity, x, is written as (x). As a rough guide, one can say that y is “order x,” or (y) ~ (x), if nx < y < 10nx, where n may be somewhere approximately between 0.3 and 0.5.

REFERENCES 1. Polmear, I., D. St John, J.-F. Nie, and M. Qian, Light Alloys: Metallurgy of the Light Metals, 5th ed., Butterworth-Heinemann, 2017. 2. Ashby, M., “Physical modelling of materials problems,” Materials Science and Technology, pp. 102–111, 1992. 3. Yule, A. J., and J. J. Dunkley, Atomization of Melts for Powder Production and Spray Deposition, Clarendon Press, 1994. 4. Lavernia, E. J., and Y. Wu, Spray Atomization and Deposition, John Wiley & Sons, 1996. 5. Francis, L. F., Materials Processing: A  Unified Approach to Processing of Metals, Ceramics and Polymers, Academic Press, 2016. 6. McMaster, R. A., D. M. Shetterly, and A. G. Bueno, “Annealed and tempered glass,” in Engineered Materials Handbook. Volume 4: Ceramics and Glasses, S. J. Schneider, Jr. (ed.), ASM International, 1991. 7. Thomas, B. G., “Continuous casting,” in Modeling for Casting and Solidification Processing, K.-O. Yu (ed.), CRC Press, 2002. 8. May, G. S., and S. M. Sze, Fundamentals of Semiconductor Fabrication, John Wiley & Sons, 2004. 9. Callister, W. D., Materials Science and Engineering: An Introduction, 3rd ed., John Wiley & Sons, 1994. 10. Dantzig, J., and M. Rappaz, Solidification, 2nd ed., CRC Press, 2017. 11. Newton, I., Philosophiæ Naturalis Principia Mathematica, Book II, Section V, 1687. 12. Barnes, H. A., J. F. Hutton, and K. Walters, An Introduction to Rheology, Elsevier, 1989. 13. Newton, I., Philosophiæ Naturalis Principia Mathematica, Book II, Section IX, 1687. 14. Panton, R. L., Incompressible Flow, 2nd ed., John Wiley & Sons, 1996. 15. Rosen, S. L., Fundamental Principles of Polymeric Materials, 2nd ed., John Wiley & Sons, 1993. 16. Schlichting, H., Boundary-Layer Theory, 7th ed. (trans. J. Kestin), McGraw-Hill, 1979. 17. Chhabra, R. P., and J. F. Richardson, Non-Newtonian Flow and Applied Rheology: Engineering Applications, 2nd ed., Elsevier, 2008. 18. Callister, W. D., and D. G. Rethwisch, Callister’s Materials Science and Engineering, 10th ed., John Wiley & Sons, 2020. 19. Cross, M. M., “Rheology of non-Newtonian fluids: A new flow equation for pseudoplastic systems,” Journal of Colloid Science, v. 20, pp. 417–437, 1965. 20. Carreau, P. J., “Rheological equations from molecular network theories,” Transactions of the Society of Rheology, v. 16, p. 99, 1972. 21. Williams, M. L., R. F. Landel, and J. D. Ferry, “The temperature dependence of relaxation mechanisms in amorphous polymers and other glass-forming liquids,” Journal of the American Chemical Society, v. 77, pp. 3701–3707, 1955. 22. Gaskell, D. R., Introduction to the Thermodynamics of Materials, 5th ed., Taylor and Francis, 2008.

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2

Steady State Conduction Heat Transfer

2.1 INTRODUCTION The transfer of heat by conduction was introduced in Chapter 1, both in its physical mechanisms and the experimentally observed relationship between heat flux and temperature gradient that makes up Fourier’s law: (1.28) In this chapter, we will derive a transient, three-dimensional partial differential equation for temperature with thermal storage, heat diffusion, and volumetric heat generation, based on the first law of thermodynamics, Fourier’s law, and several idealized boundary conditions. Several solution approaches for this equation in steady state are presented to approximate the thermal behavior of real systems. To begin, however, we introduce the idea of thermal resistance based on the heat transfer rate equations in Chapter 1, and construct heat flow models using thermal resistance networks.

2.2

THERMAL RESISTANCES

We begin the study of heat transfer by modeling systems with a simple approach using resistance networks, based on an analogy with Ohm’s law for electrical circuits: (2.1) where I is the current flowing through the resistors (with overall resistance R), driven by the voltage drop, V. The thermal analogue has heat flow (q) and a potential drop, the temperature difference, T. Expressions for the thermal resistances are found here from the heat flow rate equations in Chapter 1. In an electrical resistance network, the potential (voltage) is only known at the nodes of the network, and the overall resistance represents the effect of all the individual resistances between the nodes. Similarly, we only model the temperatures at the nodes in the thermal network and treat the resistances between them. In addition, all heat inputs and outputs occur at nodes (not in the middle of a resistor) and all models will be steady state.

2.2.1

conduction resistance

For one-dimensional conduction, the rate equation can be written in terms of the heat flow, (2.2) DOI: 10.1201/9781003104278-2

55

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in Cartesian coordinates. If we assume a uniform k and q, we can separate variables in Eq. (2.2), (2.3)

(2.4)

x, so we can complete the integration. (2.5)

written as (2.6)

FIGURE 2.1

Schematic of steady conduction heat transfer through a plane wall.

57

Steady State Conduction Heat Transfer

We see that we can rewrite Eq. (2.5) in a way similar to Ohm’s law, (2.7) Here the potential is

-

(2.8) The area normal to the heat flow is Ax and L = x2 – x1. We see in Eq. (2.8) that a thicker slab, a lower thermal conductivity, and/or a smaller cross-sectional area will increase the thermal resistance.

Example 2.1 Conduction through a Glass Window The temperatures of the inner and outer surfaces of a glass window in a room are Ti = 25 oC and To = 0 oC. The glass is L = 5 mm thick, has an area normal to heat flow of 1 m2, and has a mean thermal conductivity of k = 0.84 W/mK. Find: (a) the thermal resistance of the window; (b) the rate of heat loss from the room through the window; and (c) the glass thickness (L) required to decrease the heat flow to 2000 W. Solution

We can perform a similar analysis on the annulus (the space between two concentric cylinders) in Figure 2.2. The conduction resistance in cylindrical coordinates is a bit more complicated than for the planar wall, because the cross-sectional area normal to the heat flow increases with radius. The heat flow is entirely radial (neglecting end effects by assuming L>>r2

58

FIGURE 2.2

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Schematic of steady conduction heat transfer through an annulus.

The expression of Fourier’s law of conduction (the rate equation for conduction) in radial coordinates is (2.9)

(2.10)

thickness of the pipe wall, as either increasing r2 or decreasing r1 will increase the r2/r1 ratio. We could also force the heat flow through a smaller area (increasing heat flux, q”) by shortening the pipe length (L). Finally, we could decrease the thermal conductivity by changing materials.

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Example 2.2 Radial Conduction through a Glass Tube Hot water flows through a glass tube of length L = 50 cm, inner radius r1 = 3 cm, and outer radius r2 = 5 cm. The temperatures of the inner and outer surfaces are T1 = 90 °C and T2 = 85 °C, and the glass thermal conductivity is k = 0.84 W/mK. Find: (a) the thermal resistance of the tube in the radial direction; (b) the radial heat loss rate; and (c) the heat loss rate if the thickness of the annulus is doubled, while keeping r1 constant. Solution

In contrast to the plane wall in Example 2.1, doubling the tube thickness here does not double the resistance or halve the heat flow. The resistance increases by only a factor of 1.66. The area through which the heat flows increases with the radius and so the resistance of the added material is lower than the original. A slab with uniform cross-sectional area has no analogous behavior.

2.2.2

convection resistance

An expression for the thermal resistance due to convection on a surface is found from rewriting the convection rate equation (1.31)

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as (2.11) where As and Ts are the surface area and temperature, respectively. The potential for

(2.12) The resistance can be lowered by the addition of surface area or by increasing the convective heat transfer coefficient. Unlike conduction, the convection resistance always takes the same form regardless of geometry. Changes in geometry influence Rconv through the determination of the heat transfer coefficient, and methods for controlling h are presented in Chapter 10.

2.2.3

Radiation heat transfer between a surface (at Ts) and a much larger environment (at T∞) may be characterized by the rate equation presented in Chapter  1, combining Eqs. (1.37)–(1.39).

Eq. (2.14) shows that radiation resistance is a very strong function of the surface and environmental temperatures and raising either or both reduces the surface resistance. Increasing either the total surface area, As, or the emissivity, , also will increase heat flow from the surface to the environment. We should recognize that the rate equation for radiation is not based on a linear temperature difference as a potential between the two surfaces, but on a net exchange of radiation. Such a net flow of heat is governed by a different potential (radiosity, the total radiation leaving a surface), which will be explained at length in Chapter 11. To fit radiation effects into a pattern using T as the potential driving heat flow, Eq. (2.14) is a reasonable approximation.

2.2.4

interFace resistance

When two materials are put in contact, the simplest assumption is that there is no temperature drop at the interface as heat flows through it, i.e., there is perfect thermal contact at the material interface. However, there are several reasons why real

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61

surfaces do not have perfect thermal contact, including (i) different shapes that do not fit together on the macroscopic scale (nonconforming interfaces); (ii) roughness, or microscopically uneven surfaces; or (iii) both. Two macroscopically flat, conforming surfaces with microscopic roughness are shown in Figure 2.3(a). When pressed together, as in Figure 2.3(b), there are areas of contact between the two surfaces, but gaps remain. In [1], Yovanovich reviews a plethora of models of thermal resistance at such joints, for a wide variety of combinations of conditions: conforming and nonconforming interfaces, smooth and rough surfaces, and with and without filler material in the gaps. The details of these models are beyond the scope of this text, but they do elucidate significant dependencies of

FIGURE 2.3 Surface roughness effects at a conforming interface. (a) Two microscopically rough surfaces before contact. (b) Rough surface in contact and under pressure. (c) Temperature profile influenced by interface resistance, showing the T at the interface.

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the important resistances on the characteristics of the surfaces and the materials. Three general types of mechanisms affecting the interface thermal resistance are included in these models. The first type of effect is the interface morphology. If the two surfaces are not the same size and shape (nonconformity), there is an ill-fitting interface with macroscopic gaps between the surfaces, reducing the heat transfer. The resistance across the gaps is governed by conduction and radiation heat transfers which are represented by the one dimensional resistances, Eqs. (2.8) and (2.14). At the microscale, e.g. Figure  2.3(b), a one-dimensional heat flow perpendicular to the interface becomes two dimensional as the heat approaches the joint. Because the heat travels farther and the cross-sectional area decreases, Eq. (2.8) suggests a higher interface resistance, known as a constriction resistance, due to the complex local geometry. The mechanical behavior of the materials also influences the joint thermal resistance; Figure 2.3(b) has been sketched assuming some plastic deformation in both surfaces. They could both be stronger and more brittle, which would limit the solid-solid contact area almost to point contacts at the asperities, or, at the other extreme, they both could be ductile enough to flow and fill all the gaps. The deformation of the asperities and the final solid contact area are a strong function of the clamping pressure, as well as the elastic and plastic mechanical properties of both materials. More deformation forms more contact area, making the heat path more one-dimensional and lowering the constriction resistance. The third class of mechanism contributing to the overall interface resistance is heat transfer across the gaps. These gaps may be empty (if the joint was prepared under vacuum), in which case the only mode of heat flow is radiation. The resistance is then a function of the local temperatures on either side of the void. The gaps may be filled with a gas (e.g., air, or the more conductive argon or helium), which, at lower temperatures, would have conduction as the dominant heat transfer mode. Intermediate cases with transparent gases and higher temperatures are also common, with the conduction and radiation paths in parallel. To reduce the gap conduction resistance significantly, addition of thermally conductive compounds (often known as “thermal grease”) is a common practice, especially in microelectronic applications. The removal of the parallel radiation path is more than compensated by the increase of the gap material conductivity by several orders of magnitude. The use of a malleable metal foil (e.g., indium or silver) to flow under pressure into the small voids is a logical conclusion of this approach. Such techniques and associated models are reviewed in [2]. All of these physical effects may contribute to the overall interface thermal resistance. The effect of having an imperfect contact can be seen in Figure 2.3(c), which shows temperature profiles in two solids with a steep temperature drop at the interface. This decrease is confined to the small region of the joint in which the aforementioned physical mechanisms dominate local heat transfer. The overall interface

(2.15)

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TABLE 2.1 Measured Thermal Interface Resistances ( Ri ) for Static Joints Material 1

Material 2

Fabric laminated phenolic resin Fabric laminated phenolic resin Copper Tungsten Graphite Aluminum Aluminum Aluminum Brass Mild steel SS304 SS304 Al 6061-T4 Al 6061-T4

P (MPa)

T (oC)

0.003 0.005 6.7 6.7 0.1 0.1 0.1 0.1 .01 .5 .01 .05

20 20 45 130

20 30 30 70 70

Filler air air vacuum air air helium silicone oil air air air thermal grease thermal grease

Ri (Km2/W) (×10–5) Ref 33 25 8.4 8.4 27.5 10.5 5.3 20–100 500 100 4 2

[3] [3] [4] [4] [5] [5] [5] [3] [3] [3] [3] [3]

Here TA and TB are the limits of the temperatures at the interface in materials A and B as the interface is approached from either side, hint is the interface heat transfer coefficient or interface conductance, and A is the total macroscopic area of the joint, including both gaps and contact regions. The contact resistance of an interface is defined as the overall interface resistance multiplied by the surface area, (2.16) Representative values of interface contact resistance are found in Table 2.1. The configuration assumed up to now is static contact, where the surfaces are not moving relative to each other. We can also consider dynamic contact, during which the surfaces touch only for a short period of time and may be moving relative to each other while in contact; examples include forging and rolling processes. Selections of measured thermal interface resistances are found Table 2.2, for dynamic contact during materials processing.

2.3

RESISTANCE NETWORKS

The resistances to heat transport discussed in the previous section are useful models to understand the relationships between a temperature difference and a heat flow in various configurations and by different mechanisms. They are even more useful when combined into thermal resistance networks that model systems, including several resistances. These equivalent thermal circuits follow the same rules as simple electrical circuits, namely Ohm’s and Kirchhoff’s laws. Ohm’s law has been covered in Eq. (2.1). Kirchhoff’s law applied to a thermal network is conservation of heat flow at a node. It states that all heat flowing into a node must flow out again, given

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TABLE 2.2 Measured Thermal Interface Resistances ( Ri ) for Dynamic Contacts in Materials Processing Material 1

Material 2

P (MPa)

T (oC)

Filler

Ri (Km2/W) Ref (×10–5)

Forging Al6061-O H12 tool steel Al6061-O H12 tool steel Al6061-O H12 tool steel Injection molding Polypropylene metal mold Polypropylene metal mold Plasma sprayed droplet impacting surface (T is substrate temperature) Mo (drop) In625 (substrate) Mo (drop) In625 (substrate) Zirconia (drop) glass (substrate) Zirconia (drop) glass (substrate) hot stamping Usibor 1500PTM unspecified tool steel low pressure die casting A356 H13 tool steel A356 H13 tool steel A356 H13 tool steel

0 14 28

400 400 400

MoS2 lubricant MoS2 lubricant MoS2 lubricant

50 8.3 4.8

[6] [6] [6]

air air

40 6

[7] [7]

27 400 27 400

air air air air

1.9 ± 0.15 0.12 ± 0.02 2.2 ± 0.3 0.1 ± 0.03

[8] [8] [8] [8]

10

200–400

Al-Si alloy (steel coating)

55 ± 5

[9]

7 14 21

300 300 300

ceramic die coating ceramic die coating ceramic die coating

52.6 45.5 37.7

[10] [10] [10]

2 16

that the system is assumed to be at steady state (i.e., it has no capacitance). So, at a node with N paths connected to it, (2.17) Generally, nodes are placed at surfaces, with resistors for conduction across materials and resistors for convection, radiation, and contact resistance placed at material interfaces.

Example 2.3 Heat Loss through a Furnace Window A glass window is placed in a furnace wall as a port to view the load during processing (Figure 2.4). Heat is transferred into and out of the glass by convection on both sides. Conduction is the mode of heat transfer in the glass, and we assume radiation heat transfer is negligible. For now we assume the two heat transfer coefficients, hf and hp, the furnace and plant temperatures, Tf and Tp, and the window surface area, Aw, are known.

Steady State Conduction Heat Transfer

65

FIGURE 2.4 Schematic of heat loss through a furnace window.

FIGURE 2.5

Thermal resistance network for furnace window in Figure 2.4.

When designing such a window, we should estimate the penalty for having such a window, in terms of the heat loss (q) through it. We may also wish to know, for safety reasons, the outside surface temperature of the window (To). Find: (a) an expression for the total heat loss, q; and (b) the outside window temperature, To. Solution: To begin the analysis, we approximate this system by three resistances: (i) convection from furnace gas to window, (ii) conduction through the window, and (iii) convection from window to air in the plant. The corresponding thermal resistance network is shown in Figure 2.5. Note that a complete drawing for a thermal network includes labels for all resistances and for the temperature at each node. It will also show where heat flows enter and exit the network. The direction of the flow is not known for certain until values for a specific furnace are used, so a reasonable guess is made as a sign convention. A negative value of q will be calculated where the actual heat flow moves opposite to the assumed direction.

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Kirchhoff’s law applied at any node is trivial (qin = q = qout) because there is only one path and the heat flow is the same everywhere. We can also see that, if RTOT is independent of the window temperatures (the usual first approximation), then the

(2.18) using our definitions of conduction and convection resistances, Eqs. (2.8) and (2.12). How could we decrease the heat flow? Many possibilities occur to us while examining the result in Eq. (2.18). Increasing resistance and decreasing potential are the two choices. Having a smaller window (Aw ) is an obvious solution, as is choosing a window that is thicker (L ) or a material with a lower thermal conductivity (k ). Decreasing the heat transfer coefficients also may be effective, and ways to do so are found in Chapter 10. One might also suggest reducing the overall temperature difference, but this value is usually determined by the requirements of furnace operations and worker comfort and safety in the plant. (b) Once the heat loss is known, the window temperatures can be found by applying Ohm’s law to pieces of the circuit. To write an expression for To, we examine only the convection outside of the window and equate the heat

Given that Tf and Tp are fixed, and changing hf might alter furnace performance, how do we lower the outside window temperature? Changing the window area has no effect, but a thicker or lower k of the window helps. Another alteration that might be suggested is to cool actively the outside window by blowing air on it and so increasing hp. However, while To falls as hp increases, our solution for q, Eq. (2.18), shows that the heat loss will also increase. The purpose of this example is to illustrate the use of a thermal resistance network to quickly describe gross system behavior. While crudely approximate, it does show the general dependencies of quantities of interest (e.g., To, q) on parameters we can

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control (k, L, hp, Aw). Surely that is what we typically require from a first-order engineering model. Example 2.3 shows the usefulness of making simple resistance networks to predict general thermal behavior, limited to steady conditions with uniform properties and a coarse spatial resolution. These crude models based on such networks can provide estimates for the thermal resistances governing that behavior as well as the functional dependence of resistances on the material properties, geometry, and boundary conditions.

2.4

GENERAL HEAT CONDUCTION EQUATION

The previous sections develop a simple approach that estimates heat flows and temperatures in terms of thermal resistances, but only predicts the temperatures at specific locations. During the processing of many materials, more detailed information is needed. Phase transformations and microstructural development as a function of position in a material are determined by the temperature history and spatial distribution, T(x, y, z, t). These data are also useful in predicting the development and relaxation of thermal strains and stresses. The temperature field also can be used to predict the pattern of heat flow and so can inform process and product design. To find the temperature field, we apply the first law of thermodynamics to an arbitrary control volume in space and track energy flows in that volume, as suggested in Section 1.1. This general control volume is shown in Figure 2.6, and the balance of thermal energy rates is U in

U out

U gen

Ustor .

(2.19)

The first two terms, Uin U out , are the net rate of thermal energy transfer across the surface area of the volume. Another contribution to this balance, U gen , represents heat generation inside the volume due, for example, to electrical current running through a resistor (Joule heating) or chemical energy changing to thermal during a reaction.

FIGURE 2.6

Thermal energy flows in an arbitrary control volume.

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If the sum of all the heat flows across the surface and the heat generation is not zero, then thermal energy is either stored or depleted and U stor 0 . The thermal energy balance is repeated for a two-dimensional flow in and through a differential control volume in Cartesian coordinates in Figure 2.7. Given the depth into the page is defined as dz, the differential volume is d = dx dy dz. By a sign (2.20) Assuming that conduction is the only mode of heat transfer across the boundaries, then the conduction rate equation gives (2.21) The conduction out in the x direction, qx+dx, can be rewritten in terms of the entering heat flow, qx, by the use of a Taylor series in the vicinity of x: (2.22)

(2.23)

FIGURE 2.7

Differential Cartesian control volume for thermal energy balance.

Steady State Conduction Heat Transfer

69

(2.24)

(2.25)

(2.26)

(2.27)

(2.28)

(2.29)

(2.30)

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Dividing by the differential volume, dx dy dz, gives the general form of a partial differential equation for temperature, assuming that conduction is the only heat transfer mode at the boundaries:

(2.31)

-

(2.32)

(2.33)

(2.34)

-

(2.35)

(2.36)

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(2.37) Both Eqs. (2.34) and (2.37) are equivalent to a more general form of the energy equation: (2.38)

2.5

HEAT TRANSFER BOUNDARY CONDITIONS

In whatever form we find the energy conservation equation, its solution requires an initial temperature field (if transient) and several boundary conditions. A model of a system may be one-, two-, or three-dimensional and needs two constraints on temperature in each thermally active direction. In this section, the forms of three common boundary conditions are presented. The reader must remember that the conditions described here, while mathematically convenient, are idealizations. We will describe where each form is more applicable, but boundaries are seldom if ever as uniform and unvarying as these approximations. While the mathematics may be exact, reality is messier. (This problem can be lessened by the use of numerical solutions, but it is not eliminated.) However, these simple models of reality applied to the energy conservation equation are useful for finding reasonable and useful estimates of temperature histories and distributions. The boundary condition of the first kind is that of a specified temperature, To, at a specified location, xo: (2.39) xo to be uniform, T(xo) to A specified heat flux, qo kind qo at the boundary. As

(2.40)

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FIGURE 2.8 Heat transfer boundary conditions: (a) specified temperature; (b) specified flux; (c) zero flux; and (d) specified relationship between temperature and heat flux. (The gray areas are outside the domain of interest and m is the slope of the T(x) at the boundary.)

Such a condition is a good approximation if heat is generated at the surface (as in , Figure 2.8(c). (2.41) Symmetry in a temperature field at xo is an exact example of this condition, as seen by the dashed line in Figure 2.8(c). An insulated (adiabatic) boundary can be approximated as zero flux, if the thermal conductivity of the domain is much higher than the conductivity of the surroundings (kd >> ks). The heat flux is continuous across the boundary, so

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73

where “+” indicates the side inside the domain and “-” the surroundings. Rearranging, we get

) is relatively very small. specified

or (2.42) This boundary condition of the third kind is shown in Figure 2.8(d). The resistance to heat flow is inversely proportional to the heat transfer coefficient, h. This boundary condition is a more general case in that there is a finite, non-zero resistance at xo. Looking at the extremes in values of h, we reduce Eq. (2.42) to a zero-flux case, Eq. (2.41), when h 0 (resistance ∞) and to the prescribed temperature (T(xo) T∞) case, Eq. (2.39), when h ∞ (resistance 0).

2.6 ONE-DIMENSIONAL HEAT CONDUCTION Armed with a thermal energy conservation equation, Eq. (2.38), and the appropriate number of boundary conditions from the previous section, we can begin finding temperature fields and heat flows due to conduction. The first problem to be solved will be what must be the simplest problem in heat transfer, finding the steady temperature field in a one-dimensional slab with uniform properties and two fixed temperature

(2.43)

to remind us that there is no change in temperature gradient or heat flux, if properties are uniform.

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As shown in Figure 2.9(a), the boundary conditions are (2.44)

(2.45)

(2.46) The heat flux is uniform across the domain. The solution in Eq. (2.45) shows the temperature as a function of three parameters (To, TL, L). Throughout this text, the practice of nondimensionalization of equations, boundary conditions, and solutions will be used often to reduce the solution parameter space. These transformations give more general solutions, valid for all combinations of the dimensional parameters in the original problem. In this case, we can define a nondimensional temperature and length as (2.47) Solving for T and x,

which are substituted into the original governing equation and boundary conditions, Eqs. (2.43) and (2.44).

-

or

The two solutions are plotted in Figure 2.9. Given the nondimensionalization of the problem, the curve in Figure 2.9(b) is always the same, whether TL > To or TL < To. In a slightly more complicated problem, we can examine the steady temperature field inside a tube wall with a convection condition (boundary condition of the third kind) at the outside surface and a fixed temperature at the inside surface, as in Figure 2.10. This configuration is a reasonable model for a tube with a conductive liquid flowing strongly inside, so the interior convection resistance is very small. -

or (2.49)

(2.50)

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FIGURE 2.9 Temperature fields for simple steady conduction in a slab. (a) Dimensional solutions, assuming To < TL. (b) Dimensionless solution, regardless of relative values of To and TL.

The two boundary conditions are the isothermal inside wall, (2.51)

(2.52)

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FIGURE 2.10 Schematic of geometry and boundary condition for heat flow through a tube wall.

(2.53) -

(2.54)

(2.55)

because the area normal to the flux is increasing with

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r  (A  =  2πrL, where L is the length of the tube). Note that the heat flow is still uniform in r as

, R1, R2).

R2

h

T∞

0) is

h, T1, -

(2.56) -

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79

and substituting into Eqs. (2.49), (2.51), and (2.52), we have the nondimensional

(2.57)

(2.58)

(2.59) -

(2.60) This solution can be more easily plotted. Figure  2.11 shows temperature profiles with 1 = 0.2 and various Bi values. This family of curves has individual lines corresponding to specific values of Bi. The plot shows that, under all circumstances in the stated problem, the temperature is held fixed at the inner radius, (  =  1) = 1 or T(r = R1) = T1, according to the boundary condition. In low convection resistance cases (Bi ∞), the outer surface temperature approaches the environmental condition, (  = 1) 0 or (r = R2) T∞. As the convection resistance increases (Bi 0), the body becomes isothermal as the entire tube wall goes  = 1 or T(r) = T1. Another type of one-dimensional, steady problem arises if there are two or more materials along the heat flow path. We could easily estimate thermal behavior with resistance networks, but the solution to the energy conservation equation will provide more detail about the temperature field. As an example, here we find the solution to the behavior of an oven wall with a low thermal conductivity ceramic layer supported by an outer steel shell, as in Figure 2.12. If the inside temperature is at T(x = −Lc) = To, the outside is T(x = Ls) = T∞, and To > T∞, then we know two boundary conditions in x, which we might think enough to solve the one-dimensional, steady conduction equation without heat generation.

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FIGURE 2.11

Nondimensional temperature in the tube wall for 1 = 0.2 and various Bi.

FIGURE 2.12 Temperature field in a two material oven wall.

Alas, it is not. The energy equation, Eq. (2.43), assumes uniform properties in the domain, which is not true here because kc ≠ ks. The mathematically simplest solution is to write the energy equation for the two materials separately: (2.61)

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81

where TS is the steel temperature and TC is in the ceramic. With two equations, we now need four boundary conditions, so this neat division does raise an issue: we only know the temperatures at the outer boundaries, so where do we find two more boundary conditions? While we cannot write what happens at the interface (x = 0) exactly in terms of the three conditions in Section  2.5, two boundary conditions can be obtained by assuming perfect contact between the steel and ceramic. The conditions are a continuous temperature (no contact resistance) and a continuous heat flow between the two materials. These conditions at the interface are written as

s

is approximately isothermal at the outside temperature.

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FIGURE 2.13 Influence of the temperature dependence of thermal conductivity on the temperature profile in a plane wall during heat transfer by conduction.

Up to this point, the thermal conductivity has been assumed constant and uniform, whereas it may vary significantly with temperature. Also, sometimes a geometry is such that, while the cross-sectional area varies with position, it changes gently enough that the system can be treated as one-dimensional. Imagine that the solid slab shown in Figure 2.1 is of a material with a thermal conductivity that increases with increasing temperature; the temperature profile through the slab is shown by curve (a) in Figure 2.13. If the values of q and A are uniform across the slab, then the product k (dT/dx) is also uniform, and thus k decreases with decreasing temperature and dT/dx must increase. Similarly, if k decreases with increasing temperature, the temperature profile is as shown by curve (b) in Figure 2.13. Any variation of the cross-sectional area through which heat is conducted, A, with distance, x, must also be taken into account in the integration of Eq. (2.21). Both effects are accounted for by writing the conduction rate equation as (2.65)

(2.66)

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(2.67)

(2.68)

or (2.69) x gives (2.70)

(2.71) a extends from x1 = 0.1 m to x2 = 0.5 m.

m

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FIGURE 2.14 Unidirectional heat conduction in conically shaped conductor in Example 2.4.

Find: The heat flux and temperature profile for both (a) T1 = 1000 K at x1 and T 2 = 500 K at x2 and (b) T1 = 500 K at x1 and T 2 = 1000 K at x2.

T

which is shown as curve (b) in Figure 2.15. This figure also illustrates the influence of a change in cross-sectional area on the temperature gradient; for heat flow in the +x direction, A increases with x and hence dT/dx decreases with x, and for heat flow in the -x direction, A decreases and hence dT/dx increases in the heat flow direction. If the variation of A with x is significant enough that the heat flow has a radial component, the problem can no longer be considered as one of simple unidirectional flow.

Steady State Conduction Heat Transfer

FIGURE 2.15

85

Temperature profiles in the conical conductor shown in Figure 2.14.

2.7 CONDUCTION WITH HEAT GENERATION For a one-dimensional, steady heat flow in a plane slab with internal heat generation,

(2.72)

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(2.73) -

(2.74)

the higher

(2.75)

(2.76)

(2.77)

(2.78) where the negative sign indicates that the heat flow is in the -x direction. Thus, as required by energy conservation, the sum of the magnitudes of the heat fluxes given by Eqs. (2.77) and (2.78) gives Eq. (2.75).

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Example 2.5 Steady Conduction in Slab with Heat Generation and Convection Losses Consider a plane slab of thickness 2L = 0.1 m, which generates heat at a rate of 250 kW/m3 and sheds it convectively to an environment at T∞ =15 °C. h = 60 W/m2K

k = 25 W/mK.

Find: The temperature profile in the slab and Tmax. Solution: As the heat transfer conditions are identical at both faces of the slab, the rate at which heat is transferred through unit surface area of the slab, q”, is equal to the rate of heat generation in the volume, = LA: q

q A

q L.

This profile is plotted in Figure 2.16 and the maximum temperature Tmax = T(x=0)= 236 oC. The heat flow balance is seen in the first equation in this section, where the heat generation rate inside the slab is the same as the rate of conduction out. For a higher q, the temperature gradient at the surfaces (x  =  ±L) must be higher to maintain a higher conduction rate out.

FIGURE 2.16

Temperature profile in a slab with heat generation.

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The passage of electrical current through matter causes heat generation in that body, the rate of which is related to the electrical conductivity of the material. This resistance to current flow is the source of heat dissipated by a mobile phone or a high tension transmission line. Here, we will examine the steady state relationship between that heat generation and a convective heat loss to the environment, as well as the resulting temperature field in the conductive body (Figure 2.17). The governing equation for this problem is the energy equation, here written in cylindrical coordinates: (2.79) ), , s

(2.80)

FIGURE 2.17 System geometry for heat generation in a wire.

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Two boundary conditions are required to solve this equation. The first is at the centerline (r = 0), around which the system is symmetrical, so dT dr

0.

(2.81)

r 0

(2.82)

(2.83) (2.84)

and

C2.

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(2.85) dinate using

, and

where Bi = h R/k. This solution is plotted in Figure 2.18. Examining the solution and the plot, we see certain features of the curves. First, we notice that the slope is always zero at the symmetry boundary,  = 0. The slope is negative throughout, down to the edge of the body,  = 1, where all the heat generated leaves the wire. For lower convection resistance (where h and Bi are large),

FIGURE 2.18

Temperature profiles for a heated wire for a range of Bi values.

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91

the temperature of the body is lower, as the heat generated is lost more easily to the environment. In the limit of infinite heat transfer coefficient, the surface temperature becomes the same as the environment (T∞). The opposite is true as the convection resistance increases and the heat requires a larger potential, T(L) – T∞, to force the heat out of the body. In the limit of h = 0, there is no escape for the heat. Under this condition, the temperature in the body is infinite at steady state, which means there is no steady solution.

2.8 MULTIDIMENSIONAL CONDUCTION The preceding discussion has been confined to unidirectional conduction heat flow either in the x-direction through a plane wall or radially through a cylinder or annulus. These cases show that heat flows in a direction normal to the lines of constant temperature (the isotherms) in the solid. In Figure 2.19(a), the isotherms are straight lines parallel to the faces of the plane wall (T = T(x) only), and in Figure 2.19(b) the isotherms are concentric circles (T = T(r) only). Figure 2.19(c) shows the isotherms and the directions of heat flow in a square plate, the top edge of which is maintained at the constant nondimensional temperature,  = 1, and the other three edges of which are set to  = 0. The direction of conduction heat flow is such that it is always perpendicular to the isotherms, and hence the heat flow is two dimensional. Thus, the positions of the isotherms, which are determined by the temperatures of the edges, define lanes through which heat flows and there is no transfer of heat between lanes. In a two- or three-dimensional geometry, the heat flow is a vector in the direction of the temperature gradient and must be considered in terms of its components in the multiple directions. There are many approaches to solving multidimensional conduction problems, some of which we will describe briefly in this section. First, we will demonstrate scaling analysis, and then return to examples of thermal networks to which some two-dimensional geometries may be reduced to the equivalent one-dimensional resistances. Finally, some exposure to exact solutions of Eq. (2.31) is given. Of course, all of the methods represent approximations of one sort or another. The most obvious is the approximate solution of a well-posed, exact mathematical form, but even those model formulations are only crude representations of reality. The methods here produce our “five o’clock solutions,” which may be augmented by numerical solutions if need be. All in all, the purpose of this section is to demonstrate some simple, approximate solutions for temperatures and heat flows in multidimensional conduction problems. More precise and detailed answers may be obtained for more complex geometries and boundary conditions using numerical methods.

2.8.1

scaling analysis

As a first example of scaling analysis in multidimensional systems, let us consider a case of a square body (H × H), with convection cooling on the top (h, T∞) and the other three sides held at To (> T∞), as in Figure 2.20. A general scaling procedure is outlined in Section 1.4.3.

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FIGURE 2.19 Isotherms and heat flow lines in (a) a plane wall, (b) a cylinder, and (c) a simple system undergoing heat flow in two directions.

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FIGURE 2.20 Square body cooled on the top by convection.

Starting with the steady, two-dimensional steady energy equation with no heat

(2.86) (2.87) (H), so (2.88)

where, but it is not particularly helpful. More promising is an examination of the top convection boundary condition

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If we represent the surface temperature (Ts) with Tmin, scaling this boundary condition gives (2.89)

(2.90) where Eq. (2.90) is an estimate of the minimum temperature of the body based on the Biot number (BiH). If the heat transfer coefficient is very large (no convection resistance, as BiH ∞), then Tmin T∞. When the top is insulated (h and BiH 0), then the entire body is at To. The estimate for temperatures in Eq. (2.90) is used to find the heat loss rate sup-

(2.91)

Using

(2.92) Again we look at two extreme cases, the first of which gives no heat flow (q 0) when h 0. When there is no convection resistance on the top (h ∞), the heat

(2.93)

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95

In this case, the top temperature is T∞, so this equation shows that all the resistance is due to conduction in the body. Another example of the use of scaling analysis is to obtain estimates of heat flow during multidimensional, steady conduction in a rectangular body with uniform heat generation and a uniform, fixed temperature, To, around its perimeter. The three configurations used with different aspect ratios (H/L) are shown in Figure 2.21. In this example, we are interested in an estimate of the maximum temperature rise in the body. The temperature reference is chosen as the maximum temperature difference in each body, (2.94)

FIGURE 2.21 Two-dimensional bodies at different aspect ratios with uniform heat generation and isothermal boundaries. (a) H/L = 1, (b) H/L < 1, and (c) H/L 0) and external convection (x = 0). (We  continue to use the sign convention that heat flow is positive in the positive 120 120

DOI: 10.1201/9781003104278-3

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121

FIGURE 3.1 Schematic of temperature profiles in a conducting solid heated at x  =  0 by convection to the environment at T∞ (t1 < t2 < t3 < t4).

coordinate direction, and the sign of the heat flow discovered by the analysis shows its real direction.) The T(x) curves are plotted at increasing times, as the body heats. As heat enters the body, the temperature at x = 0, To(t), rises and the interior is heated. The effect of the higher surface temperature reaches further into the body as time progresses and eventually impacts the far wall (x = L). With an insulated (zero temperature gradient) condition there, the temperature at the far wall, TL(t), increases. The heating continues until the entire body is at the ambient temperature, T∞. To make a first quantitative estimate of the thermal behavior inside this body, we select

(3.2)

(3.3)

(3.4)

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For the case with the dominant thermal resistance due to convection, BiL 0 and To TL, so there is only a very small temperature drop across the body. If BiL ∞ because the convection resistance is negligible in the heat flow path, then To T∞. If BiL ~ (1), the surface temperature is in between TL and T∞. The next three sections will explore these three Biot number regimes.

3.3

LUMPED CAPACITANCE ANALYSIS: CONVECTION RESISTANCE DOMINATED (BI > 1

Because the heat takes a finite amount of time to move through the slab, there is a period of time after the boundary condition change before it is felt at the far boundary of a finite slab ( < L, as in Figure 3.1 at t = t1). At these early times, the heat transfer behaves as if the slab is semi-infinite,4 as the far wall (at x = L) has no influence on the evolving temperature field. In this case, with one-dimensional, transient conduction and no heat generation, the thermal energy conservation is expressed by

x conduction

y conduction

z conduction

heat sensiblee heat generation storage

(2.31)

or with constant and uniform properties, (3.16) remembering that the thermal diffusivity is  = k/ c. The boundary conditions are (3.17) and the initial condition is

Scaling the system, Eqs. (3.16)–(3.17), using Tref ~ (Ti – To) and xref ~ , we obtain

or (3.18) a noticeable change in temperature. This penetration depth of the boundary condition is seen in Figure 3.6 (a). This expression for the growing size of such a region,

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131

proportional to the square root of a diffusivity multiplied by time, is ubiquitous in diffusion-controlled problems (as we will see in Chapters 4, 9, and 10). Using the penetration depth, we can scale the conduction rate equation at x = 0 to get an estimate of the behavior of the heat flux into the body as sketched in Figure 3.6 (b). q

k

T To T To dT ~ k i ~k i dx t

(3.19)

To obtain a solution for the temperature and heat flux throughout the body, we introduce an approximate technique, integral analysis, which generally agrees with the exact solution to within 10–20% (and frequently better than that). This level of

FIGURE 3.6 (a) Temperature response and (b) penetration depth and surface heat flux behavior of semi-infinite body with a sudden change in surface temperature.

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agreement is reasonable given uncertainty in the thermophysical properties and various simplifying assumptions about geometry and thermal conditions invariably made when reducing a real engineering problem to an inexact model. The integral solution consists of two steps to produce the transient profile of the dependent variable (here temperature). Step 1: The first task is to assume a profile for the dependent variable over the region in which it varies. Frequently, it is best to assume a polynomial form. (Other possible forms include exponential or trigonometric functions.) The order of the polynomial depends on the boundary conditions of the problem. If there is reason to believe from those conditions that a point of inflection exists in the profile, then a cubic function is best. However, in many cases examined in this text it is enough to assume a second order (quadratic) profile, by which three unknown coefficients to be found by application of boundary conditions. Step 2: The coefficients and/or the independent variable in the assumed profile will usually contain a quantity or quantities for which the time dependence is not known. In order to determine these functions and complete the solution to the problem, one integrates the governing conservation equation over the entire region of the assumed profile. The procedure will give rise to a differential equation for the unknown, transient quantity, the solution to which will produce the time dependence of those functions, completing the solution to the conservation equation. The integral method is applied here to the semi-infinite conduction problem introduced in this section. First, we will partially normalize temperature and position in the energy equation and boundary conditions thus: (3.20) where  = x/ (t (t), so the region of interest at any given time is x = 0 to (or  = 0 to 1). Note that the energy equation (3.20) is not entirely dimensionless, as we have no convenient reference scale for time. We assume a parabolic profile for the normalized temperature, : (3.21) This function has three unknown coefficients (C1, C2, and C3), but we have only two conditions at the boundaries of the domain. A physical argument can be made for adding a third condition, as the heat flux approaches zero at the edge of the penetration depth:

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Using these three boundary conditions, we can find the three coefficients and the temperature profile: ( = 0) = C 1(0 ) 2 + C 2(0) + C 3 = 1 ( = 1) = C 1(1) 2 + C 2(1) + C 3 = 0 ( = 1) = 2 C 1(1) + C 2 = 0 and 2

=

(3.22)

2 +1.

It may appear that we have the temperature profile, but we must remember the thermal penetration depth, , and its unknown dependence on time lurking in  = f[ (t)]. To obtain that function, we must turn to the energy integral equation, 1 2

0

2 2

1

d =

d ,

t

0

(3.23)

which is the energy conservation equation, Eq. (3.16), integrated over the region that is affected by the change in the wall temperature (x = 0 to x =  or  = 0 to  = 1). Note again that the size of this region, (t), has an unknown time dependence. Integrating the left side of the equation, we get 1

=

2 1

t

0

0

d ,

(3.24)

which is the balance of the heat flux in and out of the domain (on the left) and the temperature rise (a measure of energy storage) averaged over the domain (on the right). We can use the temperature profile ( ) from Eq. (3.22) to evaluate the three terms in this equation. The chain rule is used to find an expression for the cooling rate in terms of and the unknown functions.

t

=

t

2

= 2

= (2

2)

0

t

=

1

2

t

= 2 and

0

2

2

= 0. 1

Plugging these results into Eq. (3.24), and integrating the transient term, we get 2 2

=

1 d . 3 dt

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Solving using separation of variables and the penetration depth initial condition, (t = 0) = 0, (3.25) Now that we know the behavior of the penetration depth, we can write the temperature as a function of x and t: (3.26) We can also find the heat flux at the surface as a function of time: (3.27) Finally, an exact solution can be found by reducing the partial differential equation to an ordinary one and solving it for temperature (for details see, e.g., [4, 5]). Various procedures for this reduction of order are described in [6]. The solution to Eqs. (3.16) and (3.17) can be found as (3.28) which defines the error function in terms of an independent variable,

(The error function in Eq. (3.28) cannot be solved analytically, so its numerically calculated and usually presented in tabular form or as a math function in spreadsheets and other calculation tools.) We can also redefine the nondimensional temperature as in Eq. (3.20), (3.29) which is the definition of the complimentary error function. Using this exact solution to the energy equation, we can find the transient heat flux behavior at the surface (x = 0):

(3.30)

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FIGURE 3.7 Comparison of exact temperature solution with the integral analysis using second- and third-order polynomial profiles.

which matches the scaling analysis to a factor of 1 and the integral analysis much closer. Comparing the exact expression for the heat flux to those of the scaling and integral scaling solutions, Eqs. (3.19) and (3.27), we see that they all have the same functional dependence on material properties and thermal conditions. The coefficients for the integral and exact methods are only 2% different from each other, and the solutions are the same order of magnitude as the scaling analysis. A comparison of the nondimensional temperature profiles is shown in Figure  3.7 (remembering that E  =  √3). Given the assumptions generally made to reduce a real system to this simple configuration, as well as uncertainties in the thermophysical properties, the agreement found is acceptable. The solution from an integral analysis using a third-order polynomial is also included [5], showing that higher-order profiles are not always better. In fact, while a third-order polynomial is frequently used, it should not be expected to be as good as the quadratic because the actual temperature profile does not have a point of inflection.

Example 3.3 Deposited Metal Droplet Heating a Substrate A droplet of liquid aluminum is deposited with negligible superheat onto a steel substrate, initially at Ti = 25 oC, where the droplet freezes at a constant temperature, TAl = 660 oC. The droplet takes 10 ms to freeze. For the substrate: k = 20 W/mK,  = 8240 kg/m3, c = 470 J/kgK,  = 5.16 × 10–6 m2/s. We can assume that the heat transfer coefficient between substrate and droplet is high enough that Bi >> 1, so To = T∞ = TAl.

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Find: (a) Calculate the thermal penetration depth at the end of freezing, f. (b) Plot T(t) at half the final penetration depth: xb =  f/2. Solution: (a) Neglecting the lateral conduction in the substrate, this problem can be estimated as one-dimensional planar transient conduction with a constant temperature at x = 0, as shown in Figure 3.8 (a). Therefore, from Eq. (3.25), the penetration depth at t = 10 ms is

(b) We seek the thermal history at xb =  f/2 = 0.38 mm. The thermal effect reaches that point at

FIGURE 3.8 (a) Schematic of substrate response to a droplet freezing on its surface. (b) Temperature history during droplet freezing for substrate depth xb = 0.38 mm.

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Transient Conduction Heat Transfer

For time less than tb, the temperature at xb is still at the initial temperature, Ti = 25 oC. After that time, the temperature rises according to Eq. (3.22), t

Tb t

Ti

2 b

TAl Ti

2

b

xb2 12 t

+1

2xb 12 t

+1

The temperature at xb, Tb(t) is plotted in Figure 3.8 (b).

3.4.2

cooling in a slab: late tiMes For Bi >> 1

All bodies can be treated with the previous semi-infinite solution, as long as the penetration depth, (t), has not reached the far end of the body ( < L). At a critical time, tcrit, the thermal effect reaches the far wall, (tcrit) = L, so the solution in Eq. (3.25) for the penetration depth gives L2 12 .

tcrit

(3.31)

For times greater than tcrit, the problem becomes one of conduction in a finite slab. The temperature field behavior is now different from the semi-infinite solution in two ways: (i) the penetration depth is a constant, L, and (ii) the (as yet unknown) temperature at x = L increases in time. The complete system is 2

T x2

1 T t

T x 0, t

0

T x

To

0.

(3.32)

x L

To start, we scale the energy equation (3.32) to get estimates for the unknowns in this situation: the rising temperature at the far end, TL(t), and the falling heat flux at x = 0. To do so, we pick a length scale, xref ~ L, and the time reference is the elapsed time since the penetration depth reached the far wall, tref ~ t – tcrit. There are two temperature references used here. Conduction is controlled by the temperature difference across length L, so Tref,C ~ (TL – To), while the energy storage term is proportional to the temperature change at the far end during tref, so Tref,S ~ (Ti – TL). Scaling the thermal energy balance, Eq. (3.32), Tref ,C 2 ref

x

~

Tref , S tref

or TL

To 2

L

~

Ti TL t tcrit

.

(3.33)

Defining a nondimensional time as  =  (t – tcrit)/L2, we can rearrange Eq. (3.33) for an estimate of the far wall temperature behavior after tcrit: TL ~

To Ti . 1

(3.34)

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An Introduction to Transport Phenomena in Materials Engineering

At the beginning of the finite slab regime,  = 0, this relation shows the far wall temperature is at the initial value (TL ~ Ti), and as time goes to infinity, the TL To. The heat flux at the surface can be estimated by scaling the conduction rate equation at x T (3.35) the nondimensional version of which is (3.36) where S indicates the scaling result. An integral analysis can be performed on this conduction problem, although the unknown quantity is now the far boundary temperature value, TL(t .

(3.37)

The analysis for the thermal response for t ≥ tcrit starts by assuming a quadratic nondimensional temperature profile, (3.39) To find the three coefficients, we need one more boundary condition. Here we use the unknown nondimensional temperature at x = L (  = 1), L( ). Applying the three boundary conditions, Eq. (3.39) can be written as (3.40) So, we have the temperature profile as a function of position, but the time dependence in L is still unknown. We integrate the energy equation, Eq. (3.37), over the domain of interest (the entire slab, 0 ≤ ≤ 1) and use the assumed profile to find a differential equation for L( ):

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Transient Conduction Heat Transfer 1 1

L

0

0

d ,

L

where we have used the chain rule in the storage term. Evaluating the derivatives, 1

2

L

2

and

2

2 L

we get 0

L

3

d

1

1 3

2

L

L

2

2

0

L

2 1

1

L

2 1

3 0

3.

L

This differential equation is solved using separation of variables and a substitution u = 1 – L. u u

L

1 ln u

3

L

ln C1 u

3

or

1

L

ln

C1 exp

u C1

3

3

Applying the initial condition of TL t tcrit

Ti

or

0

L

0,

we find the unknown constant, C1 = 1. The far wall temperature is then L

1 exp

3

,

(3.41)

and the integral solution (I) for the temperature profile can be written as I

exp

3

2

2

1.

(3.42)

We note that at t = tcrit, Eq. (3.42) reduces to the same as the solution to the semi-infinite model, Eq. (3.22), because  = L at that time. Also, at large times, we see that lim

I

1,

exp

12 2 1

1

1,

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which means the entire body is at the surface temperature. The heat flux at the surface during the finite slab regime can be found using the temperature profile, Eq. (3.42):

(3.43) The nondimensional form of the integral solution for heat flux at the surface is (3.44) The exact solution (E) to this problem for both the early and late (semi-infinite and finite) regimes is given by

(3.45)

(3.46) and (3.47) Figure 3.9 is a comparison of temperature profiles from the exact and integral solutions for conduction in a body with a constant and uniform temperature at x = 0 (Bi >>1). The two solutions are plotted at three different times: (i) in the semi-infinite regime (t < tcrit, < L); (ii) at the critical time between the two regimes (t = tcrit,  = L); and (iii) in the finite slab regime (t > tcrit, L > 0). The integral solutions are plotted from Eq. (3.22) for the first two curves and Eq. (3.42) for the finite slab case; the exact solution (calculated to 10 terms in the series) is plotted from Eq.

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Transient Conduction Heat Transfer

FIGURE 3.9 Comparison of integral and exact solutions to the temperature in a cooled (or heated) slab for a constant surface temperature (Bi >> 1). Solutions are shown before, at, and after penetration of the thermal effect reaches the far wall.

(3.45) for all three curves. At each time, the predictions of temperature profiles are within a few percent of each other. There is less agreement in Figure 3.10 (a) of the heat extraction rate and the far wall temperature of the finite slab regime, when Q has slowed considerably and most of the heat has already been removed from the solid. We see that the scaling analysis does not match the other solutions quantitatively, but correctly estimates the overall behavior. Finally, the estimation of L( ) behavior by the scaling analysis is compared in Figure 3.10 (b) to the more detailed solutions. The scaling analysis does not have a close agreement to those curves, but once again does have the correct order of magnitude and the general trend. As the scaling result is roughly within a factor of two with the other estimates and shows the correct curve shape, with the proper values at the extremes (  = 0 and ∞), it fulfills its purpose, especially considering the small investment required to obtain these results.

Example 3.4 Cooling of Injection Molded Polymer To form a thin sheet of polystyrene (thickness = 2L = 1 mm), the liquid polymer is injected at Tinj = 270 oC (above its melting temperature) into a mold maintained by active cooling at Tmold = 30 oC, see Figure 3.11 (a). The heat transfer is symmetric around the centerline, so dT/dx = 0 there; our analysis will be mapped onto the top half of the sheet, where x = 0 is the top surface and x = L is the centerline. The sheet may be ejected from the mold when the polymer has cooled from its initial temperature to 10 oC below its glass transition temperature, Tg = 100 oC. k

0.33W/mK

1000 kg/m 3 c

1100 J/kgK

k/ c

3 x 10 8 m 2 /s

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FIGURE 3.10 (a) Nondimensional heat flow at surface of body cooled by a fixed temperature. The semi-infinite solution is shown for < 0 and the finite slab for > 0. (b) Nondimensional far wall temperature in the finite slab regime, comparing integral, exact, and scaling analyses.

Find: (a) Calculate the ejection time (te) required for the entire sheet to reach Te = Tg – 10 oC = 90 oC. (b) Plot the temperature profiles between the surface (x = 0) and the centerline (x = L) at t = tcrit and t = te. Solution: (a) Because the temperature at the centerline must have changed before te, we know the solution lies in the finite body regime and will use that solution. The centerline temperature is from Eq. (3.41), which can be rearranged to find the elapsed time,

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Transient Conduction Heat Transfer

FIGURE 3.11 (a) Schematic of injected polymer cooling in mold. (b) Temperature profiles when the thermal effect reaches the centerline and when the polymer is entirely below Tf.

1 ln 1 3

or

L

t

tcrit 1 4 ln 1

TL Tmold

Tinj

.

Tinj

The entire sheet is below the desired final temperature when TL = Te, so the final time is te

tcrit 1 4 ln 1 0.0005m

Te Tinj Tmold

Tinj

2

12 3 10 8 m 2 /s

1 4 ln 1

L2 12

1 4 lnn 1

90 o C 270 o C 30 o C 270 o C

Te Tinj Tmold

Tinj

4.5 s.

(b) Using Eq. (3.42) to calculate the temperature profiles in the polymer at t = tcrit = L2/12  = 0.66 s and t = te = 4.5 s, we plot T(x) at these times in Figure 3.11 (b).

144

3.4.3

An Introduction to Transport Phenomena in Materials Engineering

Heating in a radial systeM

For another example of one-dimensional transient conduction, let us examine the axisymmetric (∂/∂  = 0) heating of a cylindrical body from the outside, ignoring the end effects (∂/∂z = 0). The energy equation is

(3.48) The initial condition is a uniform temperature, (3.49) and at the outer radius the temperature is changed at t = 0, (3.50) This system is shown in Figure 3.12. Like the previous slab example, this problem can be solved in two steps. The first is the semi-infinite regime in which the thermal effect of the change at the outer boundary is not yet felt at the center of the cylinder, Figure 3.12 (a). The thermal penetration depth from r = R is (t). At the edge of the penetration depth, the temperature has not yet changed and there is no temperature gradient, so (3.51) To find the temperature response in the semi-infinite regime, we start by transforming the governing equation (3.48) and boundary conditions, Eqs. (3.50)–(3.51): (3.52) and (3.53) where (3.54) and (3.55)

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Transient Conduction Heat Transfer

FIGURE 3.12 Schematic of heated cylinder with temperature profiles for (a) semi-infinite and (b) finite body regimes.

Using the boundary conditions in Eq. (3.53) to find the coefficients in a quadratic polynomial, we have the form of the temperature field: 2

2

(3.56)

1.

Rearranging Eq. (3.52), we have

2

R

R

t

,

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An Introduction to Transport Phenomena in Materials Engineering

which is then integrated over the affected area, R – ≤ r ≤ R or 0 ≤ ≤ 1. 1 2

0

R

1

d

0

R

t

(3.57)

d

Eq. (3.56) is used in this energy integral equation, and the transient term is changed by the chain rule; thus, d . dt

t

Integrating Eq. (3.57) and using Eq. (3.56) produces a differential equation for the penetration depth, 1 2R 6

2 R 2

d , dt

which can be solved by separation of variables: 2

t

1

12

1 . 3R

(3.58)

(There is an explicit form, (t), to this equation, but it is unnecessarily complicated; for our purposes this implicit form, t( ), will suffice.) With Eqs. (3.56) and (3.58), the temperature behavior is known until the thermal effect reaches the centerline ( =R), at which time we can write from Eq. (3.58), R 2 18 .

tcrit

(3.59)

The time for the penetration depth to reach the center is less than that to travel the same distance in a slab, see Eq. (3.31). In the radial case, the incremental heated volumes are smaller near the center, so less thermal energy is needed to heat the material closer to the cylinder axis. At times after tcrit, the domain affected by the outer boundary condition reaches the center and we are in a finite body regime,Figure 3.12 (b). Nondimensionalizing the energy equation, Eq. (3.48), with R r R

1

r R

t tcrit

and

R2

,

(3.60)

we have 2

1

2

1

(3.61)

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Transient Conduction Heat Transfer

0

1,

0 , and

1

o

.

(3.62)

1

Eq. (3.62) shows that there is no radial temperature gradient (no heat flux) at r = 0 (  = 1) and that the unknown, transient temperature at the centerline, (r=0) =  o( ), is used as a boundary condition there. The assumed temperature profile in the body is now ,

2

1

2

o

1.

(3.63)

Integrating Eq. (3.61) over the radius of the body from  = 0 to  = 1 (r = R to r = 0) and applying the chain rule to the time derivative thus o o

produces a differential equation for

( ):

o

5 1 o 3 conduction

1 o 4 storage

or o

20 3

o

20 . 3

(3.64)

20 3 ,

(3.65)

The solution is 1 exp

0

where the initial condition is 0

0

0.

This expression for the centerline temperature allows us to rewrite Eq. (3.63) for the transient temperature field in the finite body regime as ,

exp

20 3

2

2

1.

(3.66)

This result is similar to Eq. (3.42) for the temperature in a finite slab, except the radial geometry has less volume per unit surface area, causing a faster centerline temperature rise than the slab (where R L for comparison). Figure 3.13 shows temperature fields during the entire process as a function of radius.

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FIGURE 3.13 cylinder.

Temperature profiles at different times through the heating process in a

Example 3.5 Curing Bakelite in a Metal Sample Mounting Press 5 A common practice to prepare metal samples for microscopic examination is to mount them in Bakelite (a thermosetting polymer resin) to make for easier polishing and etching. In a lab with a high rate of sample preparation, the cure time for the Bakelite under temperature and pressure in the mounting press can be a bottleneck. It has been suggested that the curing time be shortened by the inclusion of 0.05–0.10 volume fraction of a powder with a thermal diffusivity much higher than Bakelite. If the thermal diffusivity of both materials is known, we can make a crude first approximation that an effective value of for the mixture is a volume-weighted average of the filler ( f) and Bakelite ( B), mix

gf

f

gB

B

B

gf

f

B

,

where gf is the volume fraction of filler. The Bakelite diffusivity is B = 2 × 10−7 m2/s and f = 10−5 m2/s. The goal is to predict the time for the entire cylindrical mount to rise from the initial temperature, Ti = 25 oC, to the curing temperature, Tcure = 160 oC. The heat is supplied only at the outer radius of the cylinder and the top and bottom are treated as insulated (a dubious but simplifying approximation). Because the mount is cured under pressure and Bakelite can flow under mounting conditions, the contact resistance between the Bakelite and the press is assumed here to be negligible, and the boundary condition at R = 2 cm is treated as T r

R, t

0

TR 180 o C.

The thermal influence of the metal sample in the mount is neglected.

149

Transient Conduction Heat Transfer

Find: Compare the cure times (tcure) for the Bakelite to reach Tcure at the centerline (r = 0) for cases with no filler and g f = 0.05 and 0.10. Plot T(t) for these three cases. Solution: The first step is to use the solution for the centerline temperature after tcrit found in Eq. (3.65); that is, we assume that t > tcrit, so we use the finite body regime. Rearranging that equation and substituting the definitions of the nondimensional variables and the critical time, Eqs. (3.59) and (3.60). t tcrit

3 ln 1 20 t

o

mix

R2

R2 27 ln 1 1 18 mix 10

3 ln 1 20

T Ti TR Ti

T Ti TR Ti

To calculate the cure time, tcure, we substitute the mixture thermal diffusivity and the cure temperature into this expression. For the case with no filler, we calculate: tcure

0.02 m

2

18 2 10 7 m 2 /s

1

27 ln 1 10

160 o C 25 o C 180 o C 25 C

726 s;

the results for all cases are found in Table 3.1. The temperature histories for the three cases are plotted in Figure 3.14. The filler decreases cure time by 71–83%, a significant improvement. If the filler does not degrade mechanical properties of the mount, then this preliminary estimate suggests that further exploration through modeling and experimentation is warranted. To improve the estimates made by this simple model, several simplifications may be relaxed, including: the conduction in the axial direction of the cylinder; using a better model for the mixture thermal diffusivity based on filler size and shape; checking the contact resistance between mount and press; and accounting for the heat release upon curing. A merry band of bandits we: A merrier band you’ll never see. What is it that makes our spirits light? What else but Bakelite? When others seek for treasures old, for tarnished silver, hefty gold, What glimmers in our torches’ light? What else but Bakelite? From Devil King Kun by H. Albertus Boli [7] TABLE 3.1 Estimated Bakelite Mount Cure Times for a Range of Volume Fraction of Filler (gf ) gf 0.0 0.05 0.10

mix

(m2/s)

2.0 × 10 6.9 × 10–7 1.2 × 10–6 –7

tcrit (s)

tcure (s)

111 32 19

726 210 123

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 3.14 Centerline (r=0) temperature response as a function of filler volume fraction in a Bakelite mount with a press temperature of 180 oC.

3.5

SPATIAL DEPENDENCE: THE GENERAL SOLUTION WITH A BALANCE OF CONDUCTION AND CONVECTION RESISTANCES (BI ~ 1)

In the previous two sections, we have found solutions for transient conduction in bodies with very small and very large Biot numbers. These two extremes allowed simplifications to the transient heat transfer model, but how do we handle the case with Bi ~ 1, in which the thermal resistances inside and outside a body are the same order of magnitude? The schematic in Figure 3.1 shows this case. With the resistance to heat flow important both inside and outside the body, there are important temperature differences across the body (To – TL) and from the surface to the environment (T∞ – To), and both the boundary temperatures TL and To change throughout the process. As in Section 3.4, there is an initial period in which the effect of surface convection has not reached the far wall and the conduction appears to be acting in a semi-infinite body. This regime will be dealt with first, followed by a finite body case. In this section, we should notice a certain familiarity in the form of the solutions; at any stage, taking the limit of BiL ∞ will produce the result in the analogous step in Section 3.4, which begins with the assumption of an infinite Biot number.

3.5.1

Heating in a Slab: early timeS for Bi ~ 1

The semi-infinite regime is modeled by the energy equation (3.16), which is a balance of conduction and storage of heat. The initial condition is T x, t

0

Ti ,

(3.67)

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Transient Conduction Heat Transfer

and the boundary conditions at the edge of the penetration depth are t ,t

T x

T x

and

Ti

0.

(3.68)

x

The convection heat loss at the surface is k

T x

To ,

h T

(3.69)

x 0

where To t T x 0, t . Scaling the energy equation in the heated region 0 Tref ~ (To – Ti). As in Section 3.4.1, we have

uses xref ~ (t) and

x

(3.70)

t.

~

Using this expression and Tref , we then scale the convection boundary condition, Eq. (3.69), for the surface temperature k

To Ti

~h T

(3.71)

To

or To Ti Bi ~ ~ 1 T Ti 1 Bi

o

1 1 ~ 1 Bi 1 k h

t

,

(3.72)

where the Biot number is defined as Bi h k . We see that o(t = 0) = 0 (To = Ti). The heat flux at the surface is then found from Fourier’s law applied at x = 0: qo

k

T x

~k

To Ti

o

~k

T

Ti

~

h T

Ti

1 Bi

x 0

~

k T

Ti

Bi (3.73) 1 Bi

or QSI ~

qo L L Bi ~ . k To Ti 1 Bi

(3.74)

An approximate solution to this system can be found using the integral method introduced in Section 3.4.1. First, the energy equation (3.16) and boundary conditions, Eqs. (3.68) and (3.69), are transformed to 2 2

2

t

(3.75)

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and 1

0,

0,

and

Bi

1,

0

1

(3.76)

0

where  = (T – Ti)/(T∞ – Ti) and  = x/ . Applying these boundary conditions to find the coefficients in a second-order polynomial for produces Bi 2 Bi

T Ti T Ti

2

2

2 Bi

1,

(3.77)

the nondimensional transient temperature field in the semi-infinite regime. The integral of Eq. (3.75) over the penetration depth, 1 2

2

1

d

0

t

0

(3.78)

d

is evaluated, as was the energy integral equation (3.23) in Section 3.4.1 to find the penetration depth, 12 t.

(3.79)

Using the solution for temperature in Eq. (3.77), we can find the surface temperature as it rises during the semi-infinite regime. Defining BiL = hL/k, the nondimensional surface temperature is To Ti 1 T Ti

2 2 Bi

2

1

2 BiL

L

.

(3.80)

Applying the same definition of the part Biot number, BiL, to the temperature profile, Eq. (3.77), Bi L

L

2 Bi L

2

L

2

2 L Bi L

1.

(3.81)

Figure 3.15 (a) shows the progression of the temperature field for two different values of BiL. The shape of the temperature curves is the same as for the BiL >> 1 case seen in Figure 3.6 (a), except that instead of being instantly pegged to o = 1 (To = T∞), the surface temperature rises from zero gradually, increasing the surface temperature gradient and heat flux. At low BiL, the heat added is absorbed at almost the same rate throughout the body, so the surface temperature rise is much slower. As BiL increases, the rising conduction resistance slows the heat spreading into the body and keeps it near the surface, increasing the rate at which o rises. The surface heat flux can be found using Eqs. (3.77) and (3.79), q0

k

T x

k T 0

k T

Ti 0

Ti

2Bi , 2 Bi

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Transient Conduction Heat Transfer

FIGURE 3.15 Temperature during heating of a semi-infinite body. (a) Progression of profiles for two values of BiL. ( /L= 1 occurs at tcrit.) (b) Surface response over time for range of BiL.

which can be written in nondimensional form as QSI

3.5.2

2Bi 12 t 2 Bi

q0 L k T Ti

L

L

2Bi . 2 Bi

(3.82)

Heating in a slab: late tiMes For Bi ~ 1

Once the thermal penetration depth, , has reached the far wall (x=L) at time t = tcrit, the solution regime shifts to that of a finite body; the critical time is found from Eq. (3.79): tcrit

L2

12

.

(3.83)

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An Introduction to Transport Phenomena in Materials Engineering

The only difference between this solution and the Bi 1 case in Section 3.4.2 is the convection condition at the surface, Eq. (3.69). The far wall is still insulated, and the temperature there is an unknown function we use as a boundary condition. TL t

L, t

T x

or

1,

L

(3.84)

We normalize the energy equation and boundary conditions using   =  x/L and =(t-tcrit) /L2: 2

(3.85)

2

Bi L

1

,

0

0,

and

1

L

.

(3.86)

1

0

After tcrit, the maximum rate of temperature rise in the body always takes place at the far wall, so we scale the storage term as Tref , S

T t

TL Ti . t tcrit

t

(3.87)

The temperature reference in the conduction term is a spatial difference, not a temporal one. The temperature difference across the body (xref ~ L) is Tref,C ~ (To ~ TL), so the energy conservation equation is scaled as TL

Ti ~ 2 To TL t tcrit L storage ~ conduction

or TL

Ti

~ .

To TL Rearranging this result, TL ~ To find

Ti 1

To

and

L

~

o

1

.

(3.88)

, we scale the convection boundary condition:

o

h T hL k

T

Bi L 1

To Ti o

~ k

Ti To ~

L

TL

To L TL

~

o

.

To (3.89)

155

Transient Conduction Heat Transfer

Combining Eqs. (3.88) and (3.89) from the energy equation and the boundary condition gives

L

~

Bi L 1 Bi L Bi L

and

o

~

Bi L Bi L ~ 1 Bi L Bi L

L

Bi L . (3.90) 1 Bi L Bi L

These estimates confirm that, at the transition from semi-infinite to finite body solutions (t = tcrit), L

0

0

and

0

Bi L , 1 Bi L

0

(3.91)

consistent with Section 3.5.1. As ∞, both temperatures go to 1, with L < o always. These estimates of the surface and far wall temperatures are shown in Figure 3.16 for two values of BiL. As expected, the two temperatures for the low BiL case are within 10% of the overall temperature difference throughout the entire heating process. The high BiL case has a surface within 10% of the overall difference with a rate of far wall temperature rise, which is high at first and decreases with time. The medium value (BiL = 1) has a behavior similar to the high case, except the surface temperature begins the finite body regime at a lower value. To find better estimates of the temperature behavior, the integral method is again employed. A second-order polynomial of ( ) is used and, using the boundary conditions in Eq. (3.86), is found to be ,

FIGURE 3.16 regime.

1

L

Bi L 1 Bi L

2

2

2 Bi L

1,

(3.92)

Surface and far wall temperature histories from scaling in the finite body

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An Introduction to Transport Phenomena in Materials Engineering

where L( ) is still unknown. Integrating over the domain (0 ≤ x ≤ L or 0 ≤ ≤ 1) and using the guessed temperature field, Eq. (3.92), 1

2 2

0

1

d

d

0

gives 3Bi L 3 Bi L

L

3Bi L 3 Bi L

L

3Bi L 3 Bi L

1 exp

L

.

(3.93)

The complete expression for the transient temperature field in the finite body regime is ,

3Bi L 3 Bi L

exp

Bi L 2 Bi L

2

2 Bi L

2

1,

(3.94)

and the surface is 3Bi L 3 Bi L

1 exp

o

2 . 2 Bi L

(3.95)

The temperature profiles at different times and BiL and the histories of o( ) and L( ) are plotted in Figure 3.17. With lower relative convection resistance (BiL = 10 >> 1), the temperatures in the bodies rise faster and with steeper gradients near the surface early in the finite body regime; the results are close to the constant surface temperature case. The low Biot number results (BiL = 0.1 tcrit.

Example 3.6 Temperature Gradients during Tempering of a Glass Sheet Cooling ceramics must be done with care to prevent cracking, as they tend to have low thermal diffusivities and so can be subject to high temperature gradients during the process. These gradients cause high gradients in thermally induced strain, which may lead to brittle fracture. The models developed in this section can be used to generate the transient temperatures and temperature gradients used in thermal stress modeling to predict the likelihood of part distortion and cracking. The mechanical portion of this problem is beyond the scope of this text, but does require the information provided by these heat transfer calculations. A 2-cm thick soda-lime glass sheet is air-cooled on one side from Ti = 600 oC. External conditions are set so that the heat transfer coefficient can be either h1 = 40 W/m2K or h2 = 250 W/m2K. The ambient temperature is T∞ = 20 oC, and the glass properties are:  = 2500 kg/m3, c = 500 J/kgK, k = 1 W/mK, and  = k/ c = 8 × 10–7 m2/s. Find: (a) Calculate the Biot numbers (BiL) for the plate. (b) Plot surface temperature, T(x=0, t), the temperature in the sheet, T(x=L/4,t), and the difference between them, all as functions of time up to t = 300 s. Solution: (a) The BiL for these two cases are: BiL1

h1 L k

40 W m 2 K 0.02 m

BiL 2

h2 L k

250 W m 2 K 0.05 m

1W mK 1W mK

0.8 and 5

Both cases have BiL ~ (1), so the solutions from Section 3.5 apply.

Transient Conduction Heat Transfer

159

FIGURE 3.19 (a) T(t) for the surface (x = 0) and a point inside the glass sheet (x = L/4 = 0.5 cm) for two BiL. (b) Transient difference between temperatures at x = 0 and x = 0.5 cm.

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and so there is a large heat loss rate. The difference reaches a maximum when reaches x = L/4. With lower surface temperatures in the finite body regime, the heat loss slows and the interior (now of fixed extent) begins to catch up to the temperature of the surface. The magnitude of this temperature difference plotted in Figure 3.19 (b) is seen to increase quickly, then with a gradual decline in the finite body regime. This behavior suggests that the largest differential strains due to thermal contraction will occur early in the process. The plot also shows that increasing the heat transfer coefficient by about a factor of six affected the maximum T by a lower factor.

3.6

SOLIDIFICATION

Up to this point in our discussion of conduction heat transfer, we have assumed that any change in enthalpy of a system causes only a change in temperature, a phenomenon known as sensible heating or cooling. However, enthalpy changes can cause changes of phase as well, which require the supply or removal of latent heat, so-called because it is “hidden” and not apparent until there is a phase change. In this section we will examine the thermal behavior of materials with solid-liquid phase changes. If we extract heat at a constant rate from a pure metal at an initial temperature (Ti) above its melting point (Tm), the material begins cooling at a rate dependent on its specific heat, dH

d

cT

or

dh

cdT,

(3.97)

where the latter expression assumes constant properties. This sensible cooling of a liquid is seen at early times in Figure 3.20, and it continues until the melting temperature is reached. At that point, solid and liquid exist in equilibrium and the thermal behavior changes. In a pure substance, the solid-liquid equilibrium is at a single temperature, and the metal remains constant at Tm while the enthalpy required to effect the phase change, the latent heat Lf, is removed. This thermal arrest begins when T reaches Tm in Figure 3.20 and liquid begins to freeze and ends when the last of the liquid has solidified. Defining the mass fraction of liquid as fL

mL , mTOT

we say that the enthalpy removed during the isothermal solid-liquid phase change is dH

d mL f fL

or

dh

L f dfL ,

(3.98)

assuming constant properties. The phase change starts at fL  =  1 (all liquid). Once fL = 0 (all solid), there is no more phase transformation possible with further enthalpy removal, and sensible cooling resumes according to Eq. (3.97) with solid properties. The behavior of the freezing binary alloy has some differences compared to the pure metal, as seen in Figure 3.21. With an alloy composition of Co, the phase diagram predicts a single phase, sensible cooling until T = TLiq. At that point, the

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FIGURE 3.20 Temperature history of pure metal with a constant heat extraction rate.

nature of a binary alloy is that the liquid freezes to solid over a temperature range (TLiq – TEut). In that range, the two phases (L + ) are at equilibrium and the enthalpy removal causes simultaneous phase change and temperature drop. When the temperature reaches TEut, a second solid phase ( ) begins to freeze along with the first ( ) in a eutectic reaction. With three phases in equilibrium (L + ), there must be an isothermal change during further enthalpy removal until the last liquid disappears. When that condition occurs, the completely solid metal (in the + phases) continues to cool sensibly. The previous descriptions of the thermal behavior of metals during solidification are based on equilibrium thermodynamics and do not include non-equilibrium effects. For a much more thorough description of the physics of solidification and the relationship of heat transfer to microstructural development, see the excellent text by Dantzig and Rappaz [8]. In the following sections, we will develop one-dimensional models of the transient conduction heat transfer with phase change to illustrate the behavior of simple systems. (Here we find integral solutions to these problems; a thorough presentation of the exact solutions is found in Dantzig and Rappaz [8].) These solutions are useful for estimates of the solidification time and the mold and metal temperature histories during more complex configurations, and can provide results related to microstructural development.

3.6.1

energy balances witH pHase cHange

Earlier we assumed the mass from which we remove heat behaved in a uniform manner as it cooled and froze, like lumped capacitor. However, most practical systems exhibit temperature gradients during such a process, so here we derive a thermal energy balance from a control volume analysis, as in Section 2.4. Restricting the analysis to one dimension (x), we repeat Figure 2.6 and write a generic energy balance with no thermal energy generation: Ustor

U in U out .

(3.99)

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FIGURE 3.21 Response of a binary eutectic alloy to a constant heat extraction rate (a) phase diagram (b) temperature history.

The storage term, Ustor , is merely the time rate of change of enthalpy (∂H/∂t). For a single phase, a change in enthalpy is from Eq. (3.97), but that expression only accounts for the sensible heating. If two phases are possible, we can write the specific enthalpy (h=H/m) as a mixture of solid and liquid enthalpies weighted by mass fraction, h where fi

mi

ms

mL and fs

fs hs fs

fL hL ,

1. The solid enthalpy is hs

cT,

(3.100)

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Transient Conduction Heat Transfer

and the liquid enthalpy is hL

Lf ,

cT

assuming that cs cL c, usually a reasonable approximation near the melting point. The total enthalpy of the control volume is then H

fs cT

h

fL cT

Lf

or, where the volume is that of the infinitesimal control volume in Figure 2.7, H

dxdydz cT fL L f sensible laten heat heat

(3.101)

With that definition, we can write H t

Ustor

dxdydz c

T t

fL . t

Lf

(3.102)

In Eq. (3.102), an addition of enthalpy to the control volume (∂H/∂t > 0) results in a temperature rise (∂T/∂t > 0) or melting (∂fL/∂t > 0), or both. An expression for the net flow of heat in and out of the control volume (in one dimension) is the same as in Eq. (2.25), 2

dxdydz k

U in U out

T , x2

(3.103)

which gives us, with Eqs. (3.99) and (3.102), c

T t

2

k

T x2

fL . t

Lf

sensible conduction in heat x-direction

(3.104)

latent heat

The latent heat term is only active when and where phase change is occurring, as determined by the phase diagram, e.g., Figure 3.21 (a). When freezing isothermally, as with a pure metal (Figure 3.20) or at the eutectic in a binary alloy, ∂T/∂t = 0 and only freezing occurs. In single-phase regions, only sensible changes in heat are possible and ∂fL/∂t = 0. In two phase areas of alloy phase diagrams, all terms in Eq. (3.104) are active. Focusing on the solid-liquid interface during freezing, we can draw an infinitesimally thin control volume around that interface. In Figure 3.22, the heat flow from the solid, qs, enters the interface and qL, the heat flow into the liquid, leaves the interface, and latent heat is released as the interface position, x M t moves to the right. The heat flows are only by conduction, so qs

ks A

dTs dx

and M t

qL

kL A

dTL dx

. M t

(3.105)

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To find the latent heat release rate, we start by stating that the total latent heat released to form the solid for x = 0 to M is the decrease in total enthalpy to effect the phase change: Hf

HL

Hs

mL f

s

Lf

s

L f AM .

The latent heat release rate is then d

s

L f AM dt

s

Lf A

dM dt

s

L f AVint ,

(3.106)

when Vint is the interface velocity. The heat balance at the interface is: dTs dT dM kL L . sLf dx M dx M dt conduction conduction latent heat frrom solid to liquid release ks

(3.107)

This description of thermal behavior at the interface, known as the Stefan condition, shows that the velocity of the interface (the solidification rate) is controlled by the net heat flow in and out of the solid-liquid boundary. Eq. (3.107) also shows why Figure 3.22 is drawn with different temperature slopes in the solid and the liquid. If ks kL , then the temperature gradients will be different even without a phase change, but the added heat release (or absorption during melting) also contributes to the change in slope at the interface.

FIGURE 3.22

Heat balance in a control volume at a moving solid-liquid interface.

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Transient Conduction Heat Transfer

3.6.2

solidiFication oF a pure substance

Figure 3.23 shows a simple one-dimensional, planar configuration that is our basis for the approximate treatment of a metal freezing in a mold. We have assumed no superheat (the liquid metal enters the mold at the melting temperature, TM, the metal is pure, and the semi-infinite mold is initially at Ti. When the liquid is poured, the interface temperature between metal and mold rapidly goes to To, an as-yet unknown value. In designing such a process, we seek knowledge of the total solidification time (it is good to know when your process is finished) and the interface temperature (so we know if we might damage the mold). In the general case, the heat transferred to the mold comes from latent heat released when the liquid freezes and the sensible heat taken to cool the solid. The heat transfer down the temperature gradient in Figure 3.23 can be considered using a crude resistance network, Figure 3.24, which has heat flowing through condition resistances in the solid metal and the mold. These quantities can be approximated as Rmold

km A

and

Rsolid

M . ks A

(3.108)

The ratio of these quantities, Rsolid Rmold

M km , ks

(3.109)

can give us insight into the thermal behavior of the process and sometimes lead to simplified solutions. From Ohm’s law, Eq. (2.7), and Eq. (3.109), we can estimate the interface temperature as a function on the ratio of resistances: Rsolid Rmold

FIGURE 3.23

Ts qm qs Tm

TM To To Ti

Temperature field in the metal and mold during the freezing of a pure metal.

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FIGURE 3.24 Figure 3.23 .

Resistance network crudely representing main thermal resistances in

or To

TM

M km Ti ks

M km ks

1 .

(3.110)

These are crude estimates of the resistances and the interface temperature because the problem is transient, with unaccounted-for sensible heat changes in the solid metal, and the temperature field is not actually linear. However, the model will suffice as an initial estimate of system behavior. In the following two subsections we will examine two cases. The first is solidification in a thermally resistant mold, 1,

Rsolid Rmold

(3.111)

and the second is when both resistances play a role, 1.

Rsolid Rmold

(3.112)

The case with all significant resistance in the solid metal, Rsolid >> Rmold, is left as an exercise for the reader. 3.6.2.1 Thermally Resistant Mold In the case in which the mold is significantly more thermally resistant to conduction than the metal, Eq. (3.111), we can use Eq. (3.110) to estimate the interface temperature. To

TM

M km M km

ks

ks Ti 1

~

TM

Rsolid Rmold Ti

Rsolid Rmold

1

TM

Because To ~ TM, almost all the temperature difference between the liquid and the unheated mold is found in the mold and we can redrawFigure 3.23 with no temperature change in the solid. The first step in analyzing this special case is to look at it from the point of view of the mold, which is initially at Ti and, when the liquid enters at t = 0, has its surface at x ≥ 0 suddenly raised to TM. This case is found earlier in the chapter, where the penetration depth, temperature profile, and heat flux are 12 t ,

(3.25)

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Transient Conduction Heat Transfer

FIGURE 3.25 mold.

Temperature field during pure metal solidification with a thermally resistant

T Ti Tm Ti

2

2

(3.22)

1,

where  = −x/ , and q0

qo x

0

k

dTm dx

k TM

Ti

k TM

0

Ti

3 t

0

.

(3.27)

Thus the mold is heated during this process, but what is the source of that heat? ExaminingFigure 3.26 (a), we see the heat flux out of the solid, qo , is balanced by the latent heat release at the solid-liquid interface. k c

q0

m

3t

TM

Ti

s

Lf

dM dt

qLHR

(3.113)

This equality shows that the rate of growth of solid (dM/dt) is entirely controlled by how fast the mold can absorb heat (qo ). Solving this differential equation for the position of the interface, with the initial condition M(t=0) = 0, gives M t

2 TM

Ti

k c

sLf

m

t 3.

(3.114)

Eq. (3.113) can also be rearranged to find the interface velocity, Vint

dM dt

TM

Ti sLf

k c

m

3t .

(3.115)

Sketching M(t) and Vint(t) in Figure 3.26 (b), we see the solidification front forms instantly and grows rapidly, although it slows with time. The deceleration is caused

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FIGURE 3.26 Freezing of a pure metal in a thermally resistant mold. (a) Heat flow in and out of solid metal. (b) Sketch of solid-liquid interface position and velocity.

by the heating of the mold as the penetration depth, , lengthens and the temperature gradient at the wall decreases. If a slab has a thickness of 2Mf and is cooled by the same type of mold on both sides, then the solid fronts meet at the centerline at the final solidification time, tf, when M(t=tf)=Mf. From Eq. (3.114), the solidification time is tf

3 4

s

Lf M f

TM

Ti

2

1 k c

.

(3.116)

m

So we have an estimate for the solidification time, and how we might alter it if need be. The properties of the metal ( s, Lf, TM) and its geometry (Mf) are fixed by material selection and part design, but to slow the process, we could change the mold material to decrease its heat diffusivity, (k c)m, or preheat the mold before pouring to decrease (TM – Ti).

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Transient Conduction Heat Transfer

At the beginning of the analysis, we assumed that all significant thermal resistance was in the mold, Rsolid/Rmold > 1. (a) Scale time, temperature, and distance in this domain by selecting appropriate reference quantities. Are all of the reference values known quantities? (b) Using the results of part (a), estimate the time required for ~ L/2, where is the distance into the part that is affected by the boundary conditions (i.e., T > Ti). (c) If the slab is replaced by one with twice the thermal conductivity, what happens to the time found in part (b)? What happens if the part is halved in thickness? The copper wall of an electroslag remelting furnace is initially at T W, and the outer surface of the wall (x  =  L) is maintained at T W by the swiftly moving water in a copper cooling jacket. The temperature on the inside

Transient Conduction Heat Transfer

177

surface of the copper wall is increased at a finite rate by the relatively sudden contact with a very hot slag. Approximate the thermal contact with the slag as a constant and uniform heat flux into the wall from the furnace, q f . (a) Sketch and label the system. Using scaling analysis, find an expression for the steady state value of (Tf − T W ), where Tf is the furnace side temperature of the copper (at x = 0). (b) Scale the transient energy equation and the boundary condition on the inside wall (x = 0) to estimate (Tf (t) – T W ) before heat penetrates to far wall (t = 0). (c) Using results from parts (a) and (b), estimate the time the temperature field will take to reach steady state; comment on the dependence of that time on q f . 3.9 A square part (length  =  2L) is heated uniformly through its volume by passing a current through it, while the sides of the part are maintained at a constant and uniform temperature, Ts. We would like to quickly estimate the time dependence of the maximum temperature difference, (Tmax – Ts), at short and long times, assuming the initial condition is that the part is Ts at t = 0. A quick way to get that estimate is by a scaling analysis. (a) Sketch and label the system. Write the appropriate form of energy equation for this problem. Clearly state your assumptions and identify the sources and sinks of thermal energy. (b) Pick reference values for the temperature difference ( Tref ) and lengths (xref, yref ). (Use t, the time elapsed since heating began, for time). Which reference values are unknown? (c) Scale the 2D energy equation (it should have four terms at this point) and identify the physical phenomena represented by each of the terms. Combine any terms that are always the same order of magnitude. (d) Which of the three remaining physical mechanisms is important throughout the entire process? Divide all terms by that term representing that phenomenon, so it is (1). (e) Which term can be neglected when it reaches steady state at very large times (t ∞)? Using the remaining two terms, find the behavior of (Tmax – Ts) at large times, i.e., find its dependence on t. (f) How large must time be to neglect the term in part (e)? (g) Which term can be neglected at very small times? How does (Tmax – Ts) behave at small times? How small must time be to neglect the term? (h) Sketch (Tmax – Ts) vs. t showing behavior at small and large times. 3.10 This problem examines the question of whether, when you step on hot sand on a beach, does the sand cool down faster or slower than your foot heats up? (Think of your own experience with this problem.) The foot is initially at Tfi and the sand is at Tsi. The foot makes contact with the sand at t = 0, at which time the interface between the two (almost) immediately takes on the constant temperature To. As heat is transformed from sand to foot, we can define thermal penetration depths, s and f. (a) Write the transient energy equation and boundary conditions at x ∞ and x −∞. (b) Scale the one-dimensional transient energy equations separately in both the sand and foot. Find an estimate for the ratio of the penetration depths into the sand and the foot ( s/ f ). (c) Write the expression for heat flux continuity at the sand–foot interface. Scale this relationship using these reference values: Ts = Tsi – To and Tf = To – Tfi and the result from part (b). Find an estimate for the interface temperature, To, in terms of properties and the

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initial temperatures. (d) Sketch the temperature distribution at three times, t > 0, on the Figure 3.28. (k c)s = 3.2 × 105 WJ/m4K2 (k c)f = 2.5 × 106 WJ/m4K2 −7 2 −7 m 2/s m /s s = 2.3 × 10 f = 1.4 × 10 3.11 You are designing a sintering furnace including a small glass viewing port in the wall. You want to know how long (tcrit) after you turn on the furnace the heat propagates through the glass window and begins heating the outside of the glass. A one-dimensional semi-infinite analysis seems appropriate, and you assume that there is a constant heat flux at x = 0. Boundary conditions for the transient energy equation with no heat generation are T ( x,t

0) Ti

qw

T x

k

T (x

, t ) Ti

x 0

T x

0. x

(a) Sketch and label the system. (b) Normalize the energy equation using the following nondimensional values: T Ti (qw / k )

x

.

Keep in mind   =  (t), so the temperature reference is also not constant. Because of this time dependence, when normalized, the transient term looks like: [

] t

t

t

.

C1 2 C2 C3 , apply the normal(c) Using a second-order polynomial ized boundary conditions to obtain an expression for  = f( ). (d) Write the integral form of the normalized energy equation over  = 0 to 1. With integration, show that the energy equation reduces to: 3

d . dt

(e) Solve the differential equation for (t), and write an expression for tcrit. 3.12 An infinite area, initially at Ti is suddenly heated at t > 0 from a hole of radius R at TR > Ti (Figure 3.29). There is no heat generation, and heat flow is only radial. (a) Write the governing energy equation and boundary and initial conditions for this problem. (b) Nondimensionalize T and r with Ti

T r,t TR Ti

r R

and

,

where (t) is the thermal penetration depth. (c) Find the constants in an assumed temperature profile: C1

2

C2

C3 .

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Transient Conduction Heat Transfer

FIGURE 3.29 Schematic of geometry for radial conduction in Problem 3.12.

(d) By integrating the energy equation over the penetration depth, show that 2

t

12

1

R

,

and plot R vs. 12 t R 2 . How and why does this (t) behave differently from the slab case? 3.13 The surface temperature of an effectively semi-infinite slab of steel, initially at uniform temperature (Ti = 25 oC), is raised instantaneously to 50 oC. Find: (a) the distance from the surface at which the local temperature is 30 oC after 5 minutes; (b) heat flux into the slab at 5 minutes; and (c) total thermal energy transferred to a unit area of the slab surface from t = 0 to 5 minutes. 3.14 Liquid steel at Ts = 1650 oC is rapidly poured to a depth of d = 2.5 m in a ladle of diameter D = 3 m, which is preheated to Ti = 700 oC. The ladle lining is L = 15 cm thick. Assume that the radius of curvature in the ladle is large enough that it can be treated as a slab and that the Bi = hL//klining >> 1. (a) Calculate how much heat is transferred from the steel to the ladle by convection during the first minute. (b) Is the semi-infinite approximation still valid at the end of that minute? Ladle lining: k = 1 W/mK  = 3 × 10 –7 m2/s 3 Liquid steel:  = 7050 kg/m c = 750 J/kg K 3.15 To transform austenite near a cooled surface to martensite, this layer must be quenched from the soaking temperature of Ti = 900 oC to 400 oC in less than 2 s. (a) If we assume that the surface temperature is suddenly changed from 900 oC to Ts = 100 oC, what is the thickness of the martensite layer ( m)? (b) Find the minimum heat transfer coefficient required so that the assumption in part (a) is valid, using the layer thickness as a length scale. The steel thermal diffusivity is  = 1.2 × 10 –5 m2/s.

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3.16 The inside surface (x = 0) of an alumina furnace lining (thickness = L = 10 cm) is suddenly raised from Ti = 30 oC to To = 900 oC ( alumina = 2 × 10 −5 m2/s). (a) Sketch the geometry of interest and identify all relevant parameters. (b) What is the value of tcrit? (c) Plot T(x = L, t) from t = 0 until T(x = L, t) = To + 0.95(Ts – To), that is, L = 0.95. At what (dimensional) time does that occur? (d) For the same time period and on the same plot as in part (c), draw the curves for T(x = 0, t), T(x = L/3, t), and T(x = 2L/3, t). (e) How would the times found in parts (b) and (c) change if L was doubled? 3.17 CMSX-4, a nickel superalloy used to cast single crystal turbine blades, is poured at temperature TM into a ceramic mold of thickness L and initially at temperature Ti. Treat the mold as a slab. (a) How long before the outer surface of the mold (x = L) begins to change temperature? Assume the inside mold temperature is the melting temperature of the metal (TM ) throughout the process. (b) Once the outer surface begins to heat up, strictly speaking the finite body solutions found in this chapter do not apply because they assume an insulated boundary. Actually, heat is lost to the cold furnace by radiation (this alloy is cast in a vacuum). However, we might be able to use it for a time when the heat lost by radiation is small compared to the heat being added to the mold. Plot the dimensional radiation flux from the outer wall to the environment (x = L) and the dimensional conduction flux into the inner wall (x = 0) as functions of time on one plot and their ratio (R). When does the radiative heat loss become a significant fraction of the heat input into the ceramic (R > 0.1)? (The point here is to understand when the insulated boundary approximation is no longer reasonable.) kmold = 0.38 W/mK cmold = 750 J/kgK L = 0.02 m TM = 1700 K T∞ = 350 K

= 1500 kg/m3  = 0.5

mold

3.18 A PETE beverage bottle is made by extrusion blow molding through a ringshaped die. In Figure 3.30, a parison of PETE is extruded as a hollow tube into the center of the open mold. The mold is then closed and air is blown into the parison, coating the inside of the close mold with PETE. The bottle wall has a final thickness of 1 mm. The initial temperature of the polymer is Ti = 200 oC. The mold surface (x = 0) is maintained at To = 35 oC, and there is almost no resistance between the hot polymer and the cold mold (Bi >> 1). (a) Sketch and label the temperatures and distances during the development of T(x,t), beginning with the polymer–mold contact through the time the inside of the layer reaches 70 oC. Include at least one curve each for the semi-infinite and finite body regimes, and one for tcrit. (b) At what time (tcrit) does the cooling effect of the mold reach the inside of the PETE layer? (c) At what time (tfinal) is all of the polymer below 70 oC? (At this temperature, the polymer is strong enough to be ejected from the mold.)  = 950 kg/m3

k = 0.2 W/mK

c = 1000 J/kgK

3.19 A thick (semi-infinite) oak wall initially at the uniform temperature 25 oC is suddenly exposed to gaseous reaction products at 800 oC. Assume that

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Transient Conduction Heat Transfer

FIGURE 3.30 Schematic of the blow-molding process in Problem 3.18.

3.20

3.21

3.22

3.23

Bi ~ (1) and that the heat transfer coefficient is 20 W/m2K. Find: (a) the time to raise the surface temperature to its ignition temperature of 400 oC; (b) the thermal penetration depth at the time in part (a). A thick (semi-infinite) slab of alumina at Ti = 600 oC is exposed to an environment at 200 oC, and the heat transfer coefficient is 100 W/m2K. Find, after 10 minutes: (a) surface temperature of the slab; (b) the temperature 1 cm below the surface. Rework Example 3.3 with a heat transfer coefficient h = 5,000 W/mK. (a) How long to get s  =  0.95? 0.99? (b) Plot the time the droplet takes to change the surface temperature to s = 0.95 and 0.99 as a function of heat transfer coefficient (100 W/m2K < h < 10,000 W/m2K). (c) Is the approximation of assuming that the surface temperature changes to s = 1 “instantaneously” reasonable? Rework Example 3.3, assuming there is a drop-substrate interface resistance (1–10 Km2/W). Assume Bi ~ (1). (a) Plot the surface temperature as a function of time for Ri = 1 and 10 Km2/W. (b) Is the approximation (used in Example 3.3) of assuming that the surface temperature changes to s = 1 “instantaneously” reasonable? Is it feasible to cast iron with no superheat (at its melting temperature, TM,Fe = 1535 oC) in a thick-walled aluminum mold? Liquid Fe:  = 7200 kg/m3 k = 63 W/mK Solid Al:   =  2700 kg/m3 k = 236 W/mK

c = 790 J/kg K c  =  900 J/kg K

Lf = 2.6 × 105 J/kg TM,Al  =  660 oC

3.24 Liquid steel at its freezing temperature of 1480 oC is cast as a slab of thickness 0.1 m. Find the times required for completion of the freezing process with (a) a thick sand mold at 25 oC and (b) a water-cooled copper mold at 25 oC. (For part (b), use the solution found in Problem 3.25.) The thermophysical properties are: Liquid steel:   =  7050 kg/m3 c  =  750 J/kg K Lf  =  2.6 × 105 J/kg −5 2 m /s s = 1.1 × 10 Sand mold: k = 0.3 W/m K m = 1.1 × 10 −5 m2/s

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3.25 At t  =  0, a pure liquid metal at its melting temperature (TM ) is poured into a mold that is maintained at some lower temperature, To, as shown in Figure 3.31. Here, the mold resistance is negligible, and conduction out of the metal is controlled by the thermal resistance in the solidified metal. The moving solid-liquid interface is x = M(t). (a) Assuming a linear function of temperature in the solid, find the coefficients (C1, C2) in the temperature profile: T To Tm To

C1

C2

x M (t ).

(b) Sketch and label the normalized temperature profile in the solid and liquid metal. (c) Find the unknown function for the interface position, M(t), using the Stefan condition. (d) Sketch M vs. t and the latent heat release rate vs. t. (d) Use your result from part (b) to estimate the solidification time for the aluminum and copper slabs. Al:

TM,Al = 660 oC cAl = 1050 J/kgK Cu: TM,Cu = 1083 oC cCu = 440 J/kgK

Lf,Al = 400 kJ/kg Lf,Cu = 205 kJ/kg

= 9.2 × 10 –5

m2/s

= 1.4 × 10 –4

m2/s

Al

Cu

3.26 Consider solidification in a flat ceramic shell mold with a mold thickness D = 10 mm. There is a heat loss to the outside surface to the surroundings at T∞ = 30 oC with a constant heat transfer coefficient, h = 100 W/m2K. Except at very early times, the temperature field in the mold is at steady state. Assume that the total thermal resistance of the outside convection and the mold wall conduction can be written as RTOT

D 1 1 , UA km A hA

FIGURE 3.31 Schematic of geometry of solidifying pure metal with dominant metal resistance from Problem 3.25.

Transient Conduction Heat Transfer

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FIGURE 3.32 Schematic of the solidification fronts in a square part in Problem 3.27.

TABLE 3.2 Property Values for Problem 3.28 k (W/mK) Al Cu Fe Mullite Sand

210 170 40 3.3 1.0

(kg/m3) 2500 8000 7700 2800 1600

c (J/kgK)

Lf (kJ/kg)

1200 490 790 200 1200

400 210 280 -

Transient Conduction Heat Transfer

FIGURE 3.33 3.28.

185

Schematic of casting of pure metal in thermally resistive mold from Problem

Comment on the applicability of the assumption at t = t f. Is the approximation better or worse at earlier times? (e) We want to cast a Cu cylinder 3 cm in diameter in a 20 cm diameter mullite mold. Assuming no liquid superheat and an initial mold temperature of 300 oC, what is tf ? What two practical changes can be made to speed the solidification?

NOTES 1. It is very important to note that while we derive the solutions and work the examples in this chapter, they are described as either “heating” or “cooling,” but the analyses are independent of the direction of heat flow and may be used in either case. 2. Quite frequently, Bi makes an appearance in the discussion of conduction only when examining the validity of this lumped capacitance model. The present and the last chapter show that this parameter is more generally useful. 3. Contributed by Bailey McConnell and Jessica Scharrer, from their industry-sponsored undergraduate capstone project at Purdue University. 4. A semi-infinite body has a beginning (x = 0) but no end as x increases. 5. Contributed by Brandon Wells and Andrea Martin Tovar, from their industry-sponsored undergraduate capstone project at Purdue University.

REFERENCES 1. Hatch, J. E. (ed.), Aluminum: Properties and Physical Metallurgy, American Society for Metals, 1984. 2. Vander Voort, G. F. (ed.), Atlas of Time-Temperature Diagrams for Nonferrous Alloys, ASM International, 1991. 3. Fezi, K., and M. J. M. Krane, “Influence of a wiper on transport phenomena in direct chill casting of aluminium alloy 7050,” International Journal of Cast Metal Research, v. 30, pp. 191–200, 2017.

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4

Mass Diffusion in the Solid State

4.1 INTRODUCTION The physical and mechanical properties of all solid materials depend on their microstructure, and development of desirable engineering properties requires the ability to manipulate, for example, crystal structure, phase morphology, and grain size. Although the relative stabilities of phases are determined by thermodynamic considerations, the rates at which phase changes can occur are determined by the rates at which atoms, molecules, or ions can be transported in the solid state. An understanding of the kinetics of phase change in the solid state thus requires an appreciation of solid state diffusion. In this chapter, we will consider steady and transient diffusion behavior. The constitutive relation for binary diffusion, Fick’s first law (Section 1.3.3), is used to derive a solute conservation equation. Several transient, one-dimensional solutions are developed and applied to practical problems. These solutions at first assume uniform diffusion coefficients, but the later sections of the chapter discuss the limitations of that assumption. The chapter concludes with the measurement of composition-dependent diffusion coefficients and their dependence on temperature.

4.2 STEADY STATE MASS DIFFUSION In Section  1.3.3, we introduced Fick’s first law, a constitutive equation for binary mass diffusion in one dimension, relating the concentration gradient and flux of solute A, thus defining the mass diffusivity. jA

DA

CA x

(1.46)

Here, DA is the diffusivity of solute A, and jA is the net flux of A in the binary solution. The concentration of component A, CA, can be written with a variety of units, usually in terms of a quantity of A (e.g., kg, mole) per unit mass, moles, or volume of the solution. The flux, jA, is the rate at which that quantity of A passes through an area; its units are [(quantity of A)/m2s] and the units of mass diffusivity are then [m2/s].

Example 4.1 One-Dimensional Radial Carbon Diffusion through Wall of an Iron Tube The inner and outer surfaces of an iron tube of length L are in contact with carburizing gases (mixtures of CO and CO2) with different carburizing potentials. Local DOI: 10.1201/9781003104278-4

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thermodynamic equilibrium is established on each side of the wall between gas and surface, and thus the carbon contents of the two surfaces are fixed by the activity of carbon in the gases. Because the concentrations of carbon at the two surfaces are different and are maintained at constant values by the gases, there is a steady flux of carbon by diffusion through the wall of the cylinder given by jC

DC

dCC , dr

(4.1)

where the units of CC are [kg C/m3] and of the flux, jC, are [kg C/m2s]. Because the diffusion is steady state, the mass flow rate of carbon, JC  =  jC A, is constant throughout the annulus. Writing Eq. (4.1) in terms of carbon flow rate instead of flux, JC

2 rL DC

dCC , dr

(4.2)

we can write dr r

2 LDC dCC JC

and

ln r2 r1

2 LDC CC1 CC 2 . JC

Rearranging this result gives us: JC

2 LDC ln r2 r1

CC1 CC 2 ,

which confirms that JC is not a function of radius in the tube. A uniform JC in Eq. (4.2) should not be confused with a uniform concentration gradient. If JC is uniform and the area is proportional to radius, A = 2πrL, then the concentration gradient must go like 1/r, higher closer to r = 0 and shallowing as the area increases with r. Another cause of a nonuniform concentration gradient in steady-state mass diffusion occurs independently of geometry. If the mass diffusivity is a strong function of composition, D(C), for instance, if D is higher at lower C, then we might have a composition profile, as in Figure 4.1, where the gradient changes while the mass flow is uniform. This effect of a nonuniform D on the C(x) profile when j is uniform is shown in Figure 4.1.

4.3 FICK’S SECOND LAW OF DIFFUSION: TRANSIENT DIFFUSION In many processes, the change of microstructure is controlled by the transient diffusion of a solute. A  model for this diffusion can be developed by considering a mass balance on the one-dimensional control volume of unit cross-sectional area and length dx shown in Figure 4.2. (Although one-dimensional, this analysis is done in

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FIGURE 4.1 Sketch of composition profiles when the mass flux is uniform, one with a constant diffusion coefficient (D) and another that depends on composition (C).

FIGURE 4.2

Mass balance in a control volume of unit cross-sectional area and length x.

the same spirit as Figure 2.7.) For diffusion of a solute (i) in the x-direction, the mass balance is Ci d t solute storage

dA jx jx dx . solute flux in - out

(4.3)

Here the differential volume is d = dx dy dz and the cross-sectional area normal to the direction of the mass flux is dA = dy dz. Representing jx+ x by a Taylor series centered at x, jx

dx

jx

jx dx x

1 2 jx 2 dx 2 x2

,

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truncating the series at two terms, and substituting into Eq. (4.3) gives dx

Ci t

jx

jx dx . x

jx jx x diffusion of solute

Ci t solute storage

(4.4)

For a steady case (∂C/∂t = 0), Eq. (4.4) shows that the mass flux is uniform in x. (This result is different from Example 4.1, in which the area changes in the direction of mass flow.) Substitution of Fick’s first law, Eq. (4.1), into Eq. (4.4) gives Ci t

x

Di

Ci . x

(4.5)

In Eqs. (4.4) and (4.5), we see different versions of Fick’s second law of diffusion, in which the rate of local accumulation or depletion of species i is equal to the rate of change of solute flux due to diffusion at that location. As noted in Section 1.3.3, the diffusion coefficient is certainly a function of many microstructural features. We begin with some simple solutions by approximating D as constant and uniform, perhaps as an average of the diffusion coefficient over the composition range in the problem. With this simplification, we can readily obtain estimates for diffusion behavior in a variety of processes, including carburization, decarburizing, and nitriding in metal processing, and dopant drive-in in integrated circuit manufacturing. These and other processes can be modeled in semi-infinite or finite domains. Here we will show solutions to one-dimensional problems that begin with a uniform initial composition, Ci, and have a surface changed suddenly to a different fixed value, Co. For a semi-infinite domain, the initial and boundary conditions are C x 0, t 0

Co

,t

C x

Ci

and

C x,t 0

Ci

(4.6)

and are C x 0, t 0

Co

C x

0

and

C x, t 0

Ci

(4.7)

x L

for a finite body of thickness L. With these boundary and initial conditions and the solute conservation equation, Eq. (4.5), scaling, integral, and exact solutions can be found. A plot of composition fields in both the semi-infinite and finite regimes is shown in Figure 4.3. As we can readily see from the governing equation, the boundary and initial conditions, and Figure  4.3, the mathematics of binary diffusion with a uniform mass diffusivity, D, are the same as those of transient conduction in Section 3.4 and so the

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Mass Diffusion in the Solid State

FIGURE 4.3 Composition fields in the semi-infinite (t < tcrit) and finite body (t > tcrit) regimes.

solutions have the same form. For the entire process before and after tcrit, the exact solution for nondimensional composition ( ) can be written as C Ci C o Ci 1

2

2 n 0

1

n

2n 1

exp

1 2

n

2

1 12

2

2n 1 2

cos

, (4.8)

1

and the composition history at x = L (  = 1) as C x L

L

Ci

C o Ci

1

1

2 n 0

n

1 n 2

exp

n

1 2

2

1 12

2

,

(4.9)

where  = x/L,  = (t – tcrit)D/L2, and tcrit = L2/12D. The nondimensional solute flux (F) at the surface (x = 0) is F

jo L D C o Ci 2

exp n 0

L C o Ci 1 2

C x

2 2

x 0

1 . 12

0

(4.10)

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The approximate solutions from the integral method and scaling analysis are broken into the two regimes. The first case is the semi-infinite case (t < tcrit), in which the change in the composition at the surface (x = 0) has yet reached the far wall. The integral solution for this regime is C Ci C o Ci

2

2

1,

(4.11)

where x

and

t

12 D t .

t

(4.12)

The scaling analysis estimates a similar penetration depth as t

(4.13)

D t.

The solute flux at the surface estimated by the integral method is F

jo L D C o Ci

C x

L C o Ci

2L

L x 0

L

(4.14)

3Dt

0

and from scaling analysis F

jo L D C o Ci

L C o Ci

C x

L C o Ci

x 0

C o Ci

~

L Dt

.

(4.15)

When tcrit L , the composition at the impermeable far wall (x = L) begins to feel the effect of the change in the boundary condition at x = 0 and begins to rise. For t > tcrit, the composition field is in the finite body regime with the boundary conditions from Eq. (4.7). There the integral method gives the composition field as C Ci C o Ci

exp

2

3

1.

2

(4.16)

The far wall composition from the integral and scaling analyses are, respectively, L

1 exp

3

and

1

L

.

(4.17)

During this regime, the solute flux at the surface is F

jo L D C o Ci

L C o Ci

C x

2 1 x 0

L

2 exp

3

(4.18)

0

from the integral method and, from the scaling analysis, F 1 1

.

(4.19)

Mass Diffusion in the Solid State

193

Figure 4.4 contains plots of these nondimensional solutions. In all cases, the exact and integral solutions agree very well (within a few percent), while the scaling results have similar behavior and are within a factor of 2–3 of the other solutions.

Example 4.2 Carburization of Steel During the carburization process, carbon is diffused from a surface exposed to a carbon-rich atmosphere into the subsurface region of a steel part. The carbon profile gradually decreases from its maximum at the surface to the original level at the edge of a penetration depth. The increased carbon content lowers the martensite start

FIGURE 4.4 Plots in the finite body regime (t > tcrit) of nondimensional (a) composition profiles, and (b) far wall composition histories.

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temperature so that, when quenched, the phase transformation from austenite begins first just below the surface. Because of the higher carbon content, the surface transforms later and at a lower temperature, and a residual compressive stress is formed there, making the surface tougher and more wear resistant. The increased carbon content also makes the martensite stronger. A good introduction to all aspects of carburization is found in [1], but here we focus on the development of the carbon profile by diffusion. Consider the unidirectional carburization of a plain carbon steel at 950 °C, in which the surface of the steel is initially low (CCi = 0.2 wt%) before being brought into contact with a carburizing gas at time t = 0. This atmosphere instantaneously supplies carbon to raise its concentration in the iron at the surface to CCo = 1 wt%. The diffusion coefficient is approximated as uniform at the value of DC = 1.09 × 10–11 m2/s. Find: (a) The value of CC CCi CCo CCi at x = 0.5 mm after 1 h of carburization. (b) The time at which  = 0.5 at x = 0.8 mm. (c) Penetration depth, (t), and surface mass flux, jo(t), histories over 24 hours. Solution: The concentration profiles obtained from the integral solution, Eqs. (4.11), at t = 1, 6, 12, and 24 hours are shown in Figure 4.5. We see from the penetration depths in that plot that (24 hrs) ≈ 3.1 mm, which is much less than L, so the composition field is always in the semi-infinite regime. (a) With x = 5 × 10 –4 m and t = 3600 s, x 2 3Dt

5 10 4 m 2 3 1.09 10

11

m 2 s 3600 s

0.729

FIGURE 4.5 Carbon composition profiles diffusing into a semi-infinite iron slab in Example 4.2, based on the integral solution, Eq. (4.11).

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Mass Diffusion in the Solid State

FIGURE 4.6 Carburization penetration depth and surface mass flux as a function of time from integral solutions in Example 4.2.

and CC

CCi

2

CCo CCi

2

1

0.729

2

2 0.729

1

0.074,

which is point A in Figure 4.5. (b) From Eq. (4.11), 2

4 4 1

1

2

1

0.5

0.293

for CC CCi CCo CCi = 0.5. The time at which the composition at x = 0.8 mm is one half the value at the boundary is t

x

2

1 12D

0.0008 m 0.293

2

1 12 1.09 10

11

m 2 /s

57,0000 s

15.8 hr.

(c) The penetration depth and surface mass flux are found over 24  hours (86,400 s) in Figure 4.6, using Eqs. (4.12) and (4.14). This plot shows very diminished incremental returns after four or five hours; the duration of the process required to make significant changes increases dramatically after that time.

4.4

INFINITE DIFFUSION COUPLE

When two metal slabs with different compositions are placed in intimate contact with each other, with no resistance to solute flux, they form an infinite diffusion couple. (In practical terms, diffusion couple is apparently infinite in that, within the period of time when diffusion occurs, no changes in concentration are seen near the two actual

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ends of the couple. This configuration can be used to measure diffusion coefficients and also to model processes such as brazing and diffusion bonding. In this case, the initial conditions are C 1Ai C Ai2

C

for x 0 for x 0

t 0,

(4.20)

where C 1Ai and C Ai2 are the initial values of solute A concentration in regions 1 and 2. The boundary conditions are C

C 1Ai

for x for x

C Ai2

(t ) (t )

0 0

C x

and

for x for x

(t ) , (4.21) (t )

where (t) is the penetration depth. Nondimensionalizing the distance from the interface (x = 0) as  = x/ , these boundary conditions can be rewritten as C C 1Ai C Ai2 C 1Ai

0 1

for for

1

and

1

0

for

0

for

1 1

. (4.22)

An integral solution can be found in the range in which the composition has changed (x = − to x =  or  = −1 to  = 1), assuming that D is uniform over that region. At the edges of both species penetration depths (x = ± or  = ±1), the composition gradients are zero, so we know that there is a point of inflection in the profile and a third-order polynomial may be a good approximation. With the previous boundary conditions, we have 1 2 3 4

3

.

(4.23)

Integrating the solute conservation equation produces a differential equation for the penetration depth, the solution of which is 12Dt.

(4.24)

The symmetry of the problem gives   =  0  =  ½ at the interface (   =  0), or C A x 0 C Ao (C 1Ai C A2i ) / 2 . The flux of solute across the plane of the interface at x = 0 is obtained from Fick’s first law: jo

DA

CA x

DA x 0

3 2 C Ai C 1Ai 8

C Ai2 C 1Ai d d

DA 0

C Ai2 C 1Ai 3 12DA t 4

DA . 3t

(4.25)

The exact solution for this case is 1 x 1 erf 2 2 DA t

and

jo

C Ai2 C 1Ai 4

DA . t

(4.26)

Mass Diffusion in the Solid State

197

FIGURE 4.7 An infinite diffusion couple with uniform diffusion coefficient. (a) Normalized composition profiles (exact and integral solutions). (b) Dimensional composition profiles for carbon diffusing in an infinite couple of steel at 1000 oC.

The composition profiles in Eqs. (4.23) and (4.26) are plotted in Figure 4.7.

4.5 DIFFUSION INVOLVING SOLID-SOLID PHASE CHANGE Often we effect a change in microstructure in a solid solution by heating the material to a temperature at which there is appreciable diffusion. If the microstructure is not then at equilibrium (a likely situation), phases may grow or dissolve, which requires mass to diffuse toward, through, and away from phase boundaries. Diffusion-controlled phase change does not occur instantaneously, and the analysis presented here

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is a first step in understanding how the movement of a two-phase interface might develop. Before we look at a simple case, we need to understand the solute balance at a moving interface between two different phases. We can start with a phase diagram with a two-phase ( - ) region and draw an isothermal tie line between the two single-phase regions on its boundaries, Figure 4.8 (a). Note that no single phase has a composition anywhere on the tie line, except at the end points (C A * and C A * ); effectively, a two-phase region is a “hole in the diagram” where no single phase exists. The relative amounts of the phases are determined then by the mixture composition of the two-phase region. An interface between - , shown in Figure  4.8 (b), will have a discontinuity in composition where it jumps between the two compositions C A * , C A * at the ends of the tie line. Throughout this treatment we will assume the interface is at equilibrium, so C A xi C A * and C A xi C A * ; here we see that the two solid phases are also at equilibrium internally with no composition gradients.

FIGURE 4.8 (a) Phase diagram around a two-phase region. The tie line across the region shows equilibrium compositions for both phases. (b) The two phases at equilibrium internally and with each other for a mixture composition in the - region.

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Mass Diffusion in the Solid State

FIGURE 4.9 Schematic of interface between two phases showing movement over dx during time increment dt. The shaded area represents the solute that must be added to the interface to maintain chemical equilibrium at the interface as it moves.

In Figure 4.9, we have relaxed the internal equilibrium assumption and now have composition gradients in each phase. We know from Fick’s first law that these gradients indicate that there is a flux of solute in each phase. Let us look at what happens in Figure 4.9 after a small increment in time, dt. During dt, the interface is observed to move to the right by dx and to sweep through a volume Adx. This differential volume at time t is the phase equilibrium composition (C A * ). At t + dt, after the interface moves, there has been a phase transformation ( ) and the swept-out region has changed its composition to the phase equilibrium composition (C A * ). This change in composition represents a storage of solute A at the interface, indicating that the net solute transfer into and out of the - boundary must be non-zero for it to move. How is this “stored” mass provided to the interface? The net solute flow in and out of the interface is Jnet = Jin – Jout, where the rate equation for mass diffusion (Fick’s first law) gives J in

JA

DA A

CA x

and J out

JA

x xi

DA A

CA x

. x xi

The net solute flow at the interface is not zero if there is phase change (the interface moves), and so it will be equal to the change in composition needed to supply that phase change. The mass balance over that small time, dt, is J in

J out d t

DA A

CA x

DA A xi

CA x

dt xi

C A * C A * Adx

(4.27)

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or CA CA dx DA CA * CA * , x x x x dt i i compposition change flux of A flux of A into interface from interface from phase innto phase interface velocity DA

(4.28)

where dx/dt is the interface velocity, Vint. So, for the interface to move dx to the right in time dt, the flux into the interface, j , must exceed the flux out, j , by the amount required for phase change, C A * C A * dx dt, swept out on the C-x plot, Figure 4.9. Thus the - interface velocity (proportional to the rate of phase change) is controlled by the net diffusive flux at that interface. Eq. (4.28) is used to relate the interface motion to the solute gradients in the two phases. At chemical equilibrium, as in Figure 4.8 (b), there are no concentration gradients at the interface, and so, by Eq. (4.28), no phase boundary movement. We begin with a solution of an isothermal diffusion-controlled phase change in a semi-infinite domain initially at C Ai2 and the crystal structure in what we will call phase 2. At t = 0, the surface condition is changed to a fixed composition, C A x 0 C 1Ao. Under any circumstances such a change at the boundary will alter the composition fields in the domain, but here we also will examine the effect of a phase change between C 1Ao and C Ai2 , as shown in Figure 4.10. The change in the boundary condition to CA(x=0) = C 1Ao causes the nucleation and growth of phase 1 there. This phase will grow at the expense of phase 2, fed by the C Ai2 C 1Ao composition difference and with a composition discontinuity at the phase boundary. To this last point, we examine the phase diagram in Figure 4.10 at temperature T, and we see that the composition range along the tie line between C A2* andC 1* A is a two-phase region. Therefore, no point in Figure 4.11, a plot of composition profiles in both phases (t > 0), should have a value in that range. The minimum composition in the single-phase region for phase 1 is C 1* A , and the maximum in phase 2 is C A2* . Assuming thermodynamic equilibrium holds at the interface during the phase change, the compositions at the interface are C 1A x L C 1A* and C A2 x L C A2*. These values are the end points of the tie line across the 1–2 region of the phase diagram.

FIGURE 4.10 Figure 4.11.

Binary phase diagram for isothermal, diffusion-controlled phase change in

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Mass Diffusion in the Solid State

FIGURE 4.11 Schematic of diffusion-controlled growth of phase 1 into phase 2, initially 2 at C Ai , for the case with a surface maintained at C 1Ao for t > 0. The compositions here refer to the phase diagram in Figure 4.10.

To find the transient composition fields, C 1A ( x,t) and C A2 ( x,t), and the position and velocity of the interface, L(t) and Vint(t), we begin with Fick’s second law applied to phase 2: 2

DA2

C A2 x2

C A2 , t

(4.29)

with the initial and boundary conditions C A2 x, t 0

C A2i C A2 x L, t

C A2* C A2 x

,t

C Ai2 .

(4.30)

Here we begin to solve for this system’s behavior using the integral method. The first two steps are as we have done before, first assuming a composition profile shape and then finding a differential equation for an unknown function using that profile in the integrated conservation equation. However, in this case we also will need another equation, as we will have two unknown transient functions ( and L), so we introduce a third step in which we apply the two-phase interface condition in Eq. (4.28). We assume a nondimensional quadratic profile for solute in phase 2, 2 A

C C A2* C Ai2 C A2*

B1

2

B2

B3 .

(4.31)

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Here the spatial coordinate, , is defined in terms of y = x – L(t), which fixes the frame of reference to the interface, thus x L t

y (t )

t

.

(4.32)

With this transformation,  = 0 (y = 0 or x = L) is always at the phase boundary, and  = 1 (y =  or x = L+ ) is at the far edge of the penetration depth into phase 2. Using these variables, the two boundary conditions in Eq. (4.30) can be rewritten as 2 A

0

2 A

0 and

1

1.

(4.33)

Because there is no solute flux outside the penetration depth, C A2 y

2 A

0.

(4.34)

1

y

Applying those three conditions to Eq. (4.31), we get the approximate composition profile and its derivative: 2 A

2

2

and

d d

2 A

2 1

.

(4.35)

Substituting the previous nondimensional variables into the solute conservation Eq. (4.29), we get DA2

2

2

2 A 2

2 A

(4.36)

t

Integrating Eq. (4.36) over the penetration depth (y = 0 to y =  or  = 0 to we have 1

DA2

2

2

2 A 2

1

d

0

0

2 A

t

=1),

d .

(4.37)

The left-hand side of the integrated Eq. (4.37), representing the diffusion in the penetration region in phase 2, is LHS

DA2

1

2 0

2

2 A 2

d

DA2

2 A

2 A

DA2

2

2 1

0 2

2 DA2 2

.

0

After integration, we have the difference of the first derivatives of 2A at the boundaries of the penetration region. The derivatives are proportional to the solute fluxes in and out of the region, so the difference is the net solute flux into the region, which is stored at x = L during the phase change.

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Mass Diffusion in the Solid State

The right-hand side of the integrated Eq. (4.37) is the storage (or depletion) of solute in phase 2. It cannot be integrated directly, so d 2A dt must be changed to a function of what we can integrate ( ) and what we are looking for (L, ). To do so, we must remember that 2A is a function of time not only through (t), but L(t) also, 2 A

and both L and of 2A ,

f

L t ,

f

,

t

are hidden in the spatial variable . Writing the total derivative

d

2 A

2 A

L

2 A

dL

d ,

an expression can be found for the time derivative of 2 A

2 A

t

L

2 A

L t

2 A

t

L

2 A

:

L t

t

.

(4.38)

This expression shows the separate effects of the dependencies of L and on time on the rate of change of the composition, 2A . While the time derivatives of L and are unknown as yet, the other factors in Eq. (4.38) can be found from the definition of and the assumed 2A profile: 2 A

1

2 1

x

.

2

L

Using these functions, we get 2 A

2 1

t

1 dL dt

d t

,

and so the integrated storage term on the right-hand side of Eq. (4.37) becomes RHS

1 0

2 A

t

1

d

0

2 1

1 dL dt

2

2

1d dt

d

1 dL 1 d . dt 3 dt Equating the left- and right-hand sides of Eq. (4.37), 2 DA

1 d dL . 3 dt dt net flux growth rate of at interface of in 2 phase change

(4.39)

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So we see that the net flux in and out of the phase boundary feeds both the composition changes inside the penetration depth in phase 2 as well as the solute required to effect a phase change from 2 to 1. To simplify the solution to this problem, we will assume that 1 d 3 dt

dL . dt

Thus, we may neglect the last term in Eq. (4.39). (We may evaluate applicability of this convenient assumption after we find L(t).) This assumption allows us to solve Eq. (4.39) without the last term, with the initial condition (t=0) = 0. (t )

12 D2 t

(4.40)

Turning to the growing phase 1, we write the nondimensional composition profile there as 1 A

C 1A C 1Ao C 1A* C 1Ao

B1

x . L (t )

B2 and

(4.41)

Although we expect 1A may have some curvature, it is assumed linear in Eq. (4.41) because there are only two known boundary conditions for it, C 1Ao and C 1A x L

C 1A x 0

C 1A*

or 1 A

0

0 and

1 A

1

1.

With these conditions, we can find the two constants in Eq. (4.41): 1 A

.

(4.42)

While we now have the shapes of the composition profiles in both phases and an expression for the penetration depth in phase 2, the solution is not complete. The goal of this exercise is to find how phase 1 grows, so we still seek the layer thickness, L(t). The two-phase interface condition, Eq. (4.28), derived from the balance of solute flowing down solute gradients on either side of the interface and the solute “stored” there when the composition changes from the equilibrium value of one phase to that in the other, is used here as another equation in terms of L(t). D1A

C 1A x

DA2 x L

C A2 y

C 1A* C A2* y 0

dL dt

The gradients of composition in the two phases at the interface are C 1A x

x L

C 1A* C 1Ao L

1 A 1

C 1A* C 1Ao L

(4.43)

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Mass Diffusion in the Solid State

and C A2 y

C Ai2 C A2*

2 A

2

y 0

C Ai2 C A2*

.

0

So, D1A

C 1A* C 1Ao L

2DA2

C Ai2 C A2*

C 1A* C A2*

dL . dt

(4.44)

Rearranging Eq. (4.44) and using Eq. (4.40) for (t), the interface condition can be rewritten as a differential equation for L(t): D1A C 1A* C 1Ao L C 1A* C A2*

dL dt

C 1Ai C A2* C 1A* C A2*

2 A

3t

.

(4.45)

If we assume a solution of the form L (t )

DA2 t and

2

DA2 . t

dL dt

(4.46)

substitute it into Eq. (4.45), and rearrange the result into a quadratic in terms of the unknown parameter, , then: 2

C Ai2 C A2* 2* 1 3 CA CA *

1 D1A C 1A* C 1Ao 2 DA2 C A2* C 1A*

1

0.

In Eq. (4.46), we have assumed that L(t = 0) = 0. Solving for C Ai2 C A2* 2* 1* 2 3 CA CA 1

1

1 6

D1A DA2

gives

C 1A* C 1Ao C Ai2 C A2*

C A2* C 1A* C Ai2 i C A2*

.

(4.47)

Eqs. (4.46) and (4.47) describe the diffusion-controlled behavior of the two-phase interface. An exact solution is available for this problem in the form of the composition profiles in the two phases: 1 A

C 1A C 1Ao C 1A* C 1Ao

2 A

C A2 C Ai2 C A2* C A2 i

erf x 2 D1A t

(4.48)

DA2 D1A

erf

and erfc x 2 DA2 t erfc

.

(4.49)

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Using these exact profiles and the interface condition, Eq. (4.43), gives the exact expression for : C A2* C Ai2 C A2* C 1A* exp C 1A* C 1Ao C A2* C 1A*

1 2

erfc 1

2 A 1 A

2 A 1 A

D D exp D D

2

erf

DA2 D1A

.

(4.50)

The root of Eq. (4.50) provides the value of for the composition profiles, Eqs. (4.48) and (4.49), and the interface position, Eq. (4.46).

Example 4.3 Diffusion-Controlled Phase Change of Austenite to Ferrite during Decarburization Consider the decarburization of a plain carbon steel, with an initial carbon content of CCi  = 3 at% at 750 °C by a gas that maintains a carbon content of CCo  = 0.05 wt% at the surface (x = 0). From Figure 4.12, the carbon contents of the ferrite ( ) and austenite ( ) phases at the phase boundary are CC * = 0.09 and CC * =2.7 wt%. In dilute solutions, as concentration in atom percent is proportional to concentration in (g mol/m3) and the solutions contain only expressions for differences in concentration, the concentrations can be written as CCi 3.0 at% CC * 2.7 at% CC * 0.09 at% CCo 0.05 at%. At 750 °C, DC = 1.1 × 10–12 m2/s and DC = 5 × 10–11 m2/s.

FIGURE 4.12 Iron-carbon phase diagram near the eutectoid reaction (not to scale).

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Mass Diffusion in the Solid State

Find: (a) Plot the behavior of the interface over time, L(t), and (b) the time required for the ferrite layer to reach x = 0.1 mm and the composition profiles at that time. Solution: (a) To plot the position of the - interface over time, we use Eq. (4.46) and calculate from Eq. (4.47) for the approximate solution and from the root of Eq. (4.50) for the exact solution. The calculated values are for approx = 0.558 and exact = 0.537, a 3.9% difference. A plot of the interface position, L, in Figure 4.13, shows that the approximate solution grows slightly faster due to the neglect of the effect of the moving phase boundary in Eq. (4.39). To find the time required to grow a ferrite layer 0.1 mm thick, Eq. (4.46) is used: L 2

t

2

10 4 m 2 0.558

1 DC

2

1 12

1.1 10

m 2 /s

7300 0 s.

The approximate profiles in the and phases when L = 0.1 mm are found from Eqs. (4.35) and (4.42): CC

x L

CC * CCo

CCo

0.05 at%

0.05at%

x 0.0001m

0.09 at% 0.05at%

4.0 10 6 at% m x

and

CC

CC

*

CCi CC

2.7 at% 2

*

2

x L

x L

12 DC t

12 D t

2

3.0 at% 2.7 at% x 0.0001m

12 1.1 10

2.7 at% 2030

12

m 2 /s 7300 s

at% x 0.0001m m

x 0.0001m 12 1.1 10 3.42 106

12

2

m 2 /s 7300 s

at% 2 x 0.0001m . m2

These compositions are plotted in Figure 4.14, where they are compared to the exact solutions. It is interesting to note that, in this example, the carbon profile in ferrite is (almost) linear in the exact solution, as we assumed for convenience in the approximate. This linear shape is because DC 50 DC and so the response of the composition field to the changing interface position is much faster in than in .

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FIGURE 4.13 Growth of atmosphere.

-ferrite into austenite ( ) when exposed to a decarburizing

FIGURE 4.14 Composition profiles in ferrite ( ) and austenite ( ) phases at the time (t = 7300 s) at which the phase boundary reaches L = 0.1 mm.

4.6 DIFFUSION IN SUBSTITUTIONAL SOLID SOLUTIONS In the preceding discussion, the diffusion of interstitial solutes, such as carbon and hydrogen in iron, was considered, in which the diffusing atoms jump from one interstitial site to the next in the crystal lattice of the solvent. The solute atoms are like small animals wandering through a dense forest, in which the trees are unaware of the motion of the animals, and thus in considering the diffusion of carbon in an infinite diffusion couple comprised of steels of differing carbon contents, no consideration was given to the possible movements of the individual iron atoms. Also, because the primary locations for the small solute atoms are on the interstitial sites in the iron lattice and those sites are almost always available, the probability is roughly the same for a solute atom to jump to any of its neighboring sites. Also, the availability of sites

Mass Diffusion in the Solid State

209

into which to jump is significantly higher than in substitutional alloys, so the speed of diffusion is much higher in interstitial solutions. Consider now a diffusion couple comprising metals of the same crystal structure and similar atomic radii, such as silver and gold. At what rates do the silver and gold diffuse into each other, and how is the diffusion process described? The answer to the first question was provided by the experiments of Smigelskas and Kirkendall [2]. (The second question will be addressed in the next section.) Their experiments begin with a rectangular bar of brass (Cu-30wt%Zn), shown in Figure 4.15, wound with fine molybdenum wire and then electroplated with a layer of pure copper. Molybdenum is insoluble in both copper and brass, and the wire served as a marker of the position of the interface. During annealing it was observed that the distance, d, decreased monotonically with time, which indicated that the flux of zinc atoms from the brass outward past the markers is greater than the flux of copper atoms inward past the markers. The phenomenon is known as the Kirkendall effect.

4.7

DARKEN’S ANALYSIS

In the previous section, an experiment is described in which two metals (pure copper and Cu-30% Zn brass) form a diffusion couple, and the flux of Zn into the Cu is faster than the flux of copper into the brass, Figure 4.15. We can measure rates of mass flow in the couple from this experiment, and the theory suggested by the results was provided by later Darken [3]. To understand the motion of the interface in the Kirkendall effect, consider a train traveling due north at 60 mph in which passenger A is walking toward the front of the train at 4 mph and passenger B is walking toward the rear of the train at 5 mph. Are the passengers traveling in northerly or southerly directions? The answer depends on the frame of reference in which the passengers are observed. A stationary observer standing beside the track (the world frame of reference) sees the train traveling in a northerly direction at 60 mph, passenger A traveling in a northerly direction at 64 mph, and passenger B traveling in a northerly direction at 55 mph. However, a

FIGURE 4.15

Diffusion couple used by Smigelskas and Kirkendall [2].

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passenger seated on the train (the train frame of reference) sees passenger A traveling in a northerly direction at 4 mph and passenger B traveling in a southerly direction at 5 mph. In this analogy, the seated passengers are being transported by the bulk motion of the train and the walking passengers are being transported by both bulk flow and “diffusion” relative to it. In Kirkendall’s experiment involving an (effectively) infinite diffusion couple of metals A and B, in which the original interface is identified by inert markers, an observer seated on a marker (the marker frame of reference, like the train) sees only the movement of A and B across the marker plane, caused by diffusion down a concentration gradient given by jA

CA and jB x

DA

DB

CB . x

However, an observer near an end of the diffusion couple (in the world frame of reference), in a region where no concentration gradients occur, sees the movement of A and B at the position of the marker as being the sum of that due to diffusion across the plane of the marker and the bulk motion of the marker. Thus, if the marker is moving with a velocity, Vm, relative to the observer near one end of the diffusion couple and the concentration of A at the position of the marker is CA, this observer in the lab frame sees the flux of A as NA

DA

j A C AVm

CA x

C AVm .

(4.51)

The first term on the right-hand side of Eq. (4.51) is the rate of transport of A  across the plane of the marker by diffusion, and the second term is the rate of movement of the marker plane relative to the observer in the lab frame. For a system in which both diffusion and bulk movement are occurring, Fick’s second law, Eq. (4.4), becomes CA = t

NA x

and

CB t

NB . x

As the total number of moles per unit volume is C give C t

CA t

CB t

x

DA

CA x

DB

(4.52)

C A C B , Eqs. (4.51) and (4.52), CB x

CVm

(4.53)

If the molar volume of alloys in the system A–B is independent of composition (i.e., is constant and uniform), then ∂C/∂t = 0, in which case integration of Eq. (4.53) becomes DA

CA x

DB

CB x

CVm

I,

(4.54)

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Mass Diffusion in the Solid State

in which I is an integration constant. Eq. (4.54) is valid along the entire length of the diffusion couple, and Vm is the velocity of a marker placed at any position in the couple relative to the observer in the lab frame. By the definition of an infinite diffusion couple, no concentration gradients occur near the ends, in which case Eq. (4.54) gives I = -CVm. But with no concentration gradients, and hence no diffusion occurring near the ends, Vm = 0 and hence I = 0. Thus, Eq. (4.54) becomes CA 1 DA C x

Vm

CB x

DB m

.

(4.55)

m

Eqs. (4.55) and (4.51) are then substituted back into Eq. (4.53) for A to obtain CA t

NA x x

x CA x

DA

DA

CA x

C A Vm

CA CA DA x C

CA CB DB . x C

(4.56)

Because C is uniform, ∂CA/∂x = – ∂CB/∂x, and hence Eq. (4.54) becomes CA t

CA x

CB C

C B DA x

DA

CA x

CA CA DA C x

CA CA DB x C

C A DB C A C x

or, as CA/C = XA and CB/C = XB, the mole fractions of A and B, respectively, CA t

x

X B DA

X A DB

CA . x

(4.57)

Eq. (4.57) is identical with Fick’s second law of diffusion, Eq. (4.5), if D

X B DA

X A DB .

(4.58)

Here D is the interdiffusion coefficient for the system A–B, and DA and DB are the chemical diffusion coefficients for A and B. Also, rewriting Eq. (4.55) as Vm

CA 1 DA C x

DB

CA x

DA

DB

XA x

(4.59)

gives the velocity of the marker plane at which the gradient of the mole fraction of A is ∂XA/∂x. The marker planes in the diffusion couple are stationary only when DA = DB, which corresponds to the case in which the flux of A and B are equal but in opposite directions. Eq. (4.58) shows that the interdiffusion coefficient, D, in Fick’s law for the system A–B is not uniform because it is a function of local mole fraction. The question now

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arises: Are the chemical diffusion coefficients DA and DB uniform? Atoms move as a result of the application of a gradient in a potential field. In a system A–B with concentration gradients, the potentials of interest are the chemical potentials of A and B. The chemical potential of a gram mole of the species i in an isothermal, isobaric system is defined as i

i

RT ln ai ,

(4.60)

in which ai is the thermodynamic activity of i relative to some standard state. This state is usually pure i, in which state the chemical potential is io and the activity of i is, by definition, unity. Alternatively, the chemical potential per atom of i is written as i

=

i

o i

N0

N0

kT ln ai ,

in which N0 is Avogadro’s number. The existence of concentration gradients gives rise to gradients in the chemical potentials of the species in the system, which causes atomic motion in the direction of the negative gradient. This atomic movement down the gradient decreases the chemical potential of A. In this field, the atoms acquire a mean velocity, V , and the mobility, B, of the atom is defined as: BA

V F

o

V

i

x .

The flux of A atoms (atoms/m2s) with this mobility is thus jA

N 0 C AV

C A BA

d i . dx

(4.61)

The thermodynamic activity of A, defined in Eq. (4.60), is the product of an activity coefficient A and the mole fraction of A, XA = CA/C. Thus, Eq. (4.60) gives d A dx

RT

d ln X A dx

d ln dx

A

,

(4.62)

substitution of which into Eq. (4.61) gives the flux of A in (g mol/m2s), jA: jA

jA N0

C A BA kT

d ln X A dx

dC A d ln A BA kT 1 dx d ln C A

d ln dx

A

C A BA kT

1 dC A C A dx

d ln A dC A BA kT 1 , d ln X A dx

d ln dx

A

(4.63)

which, on comparison with Fick’s first law, gives the chemical diffusion coefficient for A as DA BA kT 1

d ln A . d ln X A

(4.64)

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Mass Diffusion in the Solid State

If the system A–B obeys Raoult’s law ( A = 1) or Henry’s law ( A = constant) over some range of composition, Eq. (4.64) becomes DA

(4.65)

BA kT,

the Nernst–Einstein equation. The thermodynamic solution behavior of a solute in a binary system approaches Henry’s law, and the behavior of the solvent approaches Raoult’s law as the concentration of the solute approaches zero. However, in many solutions it can be considered that Henry’s law is obeyed by the solute over some finite range of dilute solution, in which case Raoult’s law is obeyed by the solvent in the same range of composition. Within these ranges of composition, the chemical diffusion coefficients of the species are uniform, but in concentrated solutions they vary with composition in accordance with the nonideal thermodynamic solution properties of the system.

4.8 SELF-DIFFUSION COEFFICIENT If a thin film of the radioisotope A* of the metal A, containing moles of A* per unit cross-sectional area, is plated onto one end of a semi-infinite rod of the stable isotope A, and the rod is annealed, the radioisotope will diffuse into the rod. The concentration profile that develops is given by C A*

D*A t

exp

x2 , 4 D*A t

(4.66)

which defines the self-diffusion coefficient of A, D*A , assuming the two isotopes diffuse at the same rate. Eq. (4.66) is shown for several values of (D*A t) in Figure 4.16. With increasing time, the concentration profiles flatten as the A* spreads through the rod. From Eq. (4.66) the concentration of C A* at x = 0 is proportional to 1 t , and because 0

exp

2

2

d

then 0

C A*

dx

1;

hence the area under each of the curves in Figure 4.16 is unity. A plot of the logarithm of the experimentally measured C A* against x2 gives a straight line, with a slope of −1/ (4D*A t), which allows determination of the self-diffusion coefficient, D*A . In the rod, because A and A* are chemically identical, there are no chemical concentration gradients and hence no gradients in chemical potential. Thus, in the absence of a potential for diffusion, the concentration profiles shown in Figure  4.16 are developed by a random walking process of A atoms and the self-diffusion coefficient is a quantitative measure of this process. The probability

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FIGURE 4.16 Concentration profiles of a radioisotope A* diffusing from an initial plane source into a semi-infinite rod of A.

that an A* atom has randomly walked to a region between x and x + dx is p(x) dx is the number of A* atoms between x and x + dx divided by the total number of A* atoms in that region, p x dx

C A*

1

dx

DA* t

exp

x2 4 DA* t

dx .

The mean square distance, x 2 , over which an atom of A* has diffused in time t is thus x2

0

x 2 p x dx 1 DA* t

1 DA* t DA* t

2

3/ 2

0

x 2 exp

erf

x2 4D*A t

x 2 DA* t

dx

2DA* tx exp

x2 4DA* t

, 0

* A

2D t which is the same as the results in Eqs. (1.42) and (1.44). Is the measured self-diffusion coefficient of the radioactive atoms A*, D*A , equal to the chemical diffusion coefficient of A in the A–B alloy? Consider an alloy diffusion couple in which CB has the same value in both halves of the couple and the initial value of CA in one half equals the initial value of the sum of CA + C A* in the other half. From Eq. (4.63), the self-diffusion coefficient is D*A

B*A kT 1

d ln A* . d ln X *A

(4.67)

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Mass Diffusion in the Solid State

However, as the stable and radioactive isotopes are chemically identical, A depends only on the chemical composition of the alloy (i.e., on the sum of CA and C A* ) and is independent of the ratio C A* /CA, and thus, for a constant CA + C A* , the term d ln A* d ln X *A . in Eq. (4.67) is zero. Furthermore, the mobilities of the two isotopes of A are the same, and thus D*A

BA* kT

BA kT,

which from Eq. (4.63) leads to d ln A . d ln X A

DA* 1

DA

(4.68)

Thus, the chemical diffusion coefficient of A, DA, in a system containing concentration gradients is only equal to the self-diffusion coefficient of A, D*A, if the system obeys Raoultian (and hence Henrian) ideal behavior. As the variations of DA and DB with composition are caused by nonideal thermodynamic behavior in the system, the interdiffusion coefficient of the system A–B can be related to the self-diffusion coefficients by means of the Gibbs–Duhem equation for the binary system: XAd

A

XBd

0

B

or X A RT d ln X A

X A RT d ln

X B RT d ln X B

A

X B RT d ln

0. (4.69)

B

However, as XA XA dX A XA

1 dX A

XB

XB dX B XB

0

dX B

0 X A d ln X A

X B d ln X B

0,

the sum of the first and third terms in Eq. (4.69) is zero, and hence with some rearrangement, d ln

d ln

A

d ln X A

B

d ln X A

.

(4.70)

Thus, combination of Eqs. (4.58), (4.68), and (4.69) gives D

X B DA

X A DB

* A

* B

XB D

d ln

X B DA* 1 1

d ln

A

d ln X A A

d ln X A

.

X A DB* 1

d ln

B

d ln X B

(4.71)

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Example 4.4 Doping Silicon with Phosphorus by Diffusion To produce a semiconductor by doping silicon, a thin layer of phosphorus is placed on the surface of the silicon by vapor deposition and allowed to diffuse into the silicon. A film of thickness L = 1 μm is deposited and diffused into the silicon at 1000 °C for 5 h. At that temperature, DPSi = 1.2 10 –17 m 2 / s

P

= 2000 kg P / m 3

0.03097 kg P / g mol P

MWP

Si

= 2300 kg Si / m 3

MWSi = 0.02809 kg Si / g mol Si

Find: The depth at which the mole fraction of phosphorus has a value of 10 –3. Solution: The concentration profile developed from a plane source is given by CP

DP t

exp

x2 4DP t

(4.72)

For a 1 μm vapor-deposited layer of phosphorus, the quantity per unit area of surface, is L MWP P

2000 kg P/m 3 10 6 m

6.46 10 2 g moll P m 2 .

0.03097 kg P/g mol P

A mole fraction of 10–3 of P in Si corresponds to a concentration of CP

10

3

gmol of P gmol of Si

0.02809 kg Si gmol Si 2300 kg Si m 3

81.9

gmol of P m3

Therefore, in Eq. (4.72), with t = 5 h = 18,000 s, 81.9

gmol of P m3

6.46 10 2 g mol P m 2 1.2 x 10

17

m 2 /s 18, 000 s

exp

x2 4 1.2 x 10

17

m 2 /s 18, 000 s

,

which gives the penetration depth as x = 2.44 × 10–6 m = 2.44 μm.

4.9 MEASUREMENT OF THE INTERDIFFUSION COEFFICIENT: BOLTZMANN–MATANO ANALYSIS To measure the interdiffusion coefficient of a solute in an alloy, we might construct an infinite diffusion couple (as in Section 4.4) and measure composition profiles at several times after the start of diffusion at particular temperatures. By curve fitting these profiles to the mathematical solutions found in Eq. (4.23) or (4.26), we can obtain values for D, if it is not a strong function of composition. When the interdiffusion coefficient does vary with composition (a common case for substitutional alloys), the Boltzmann–Matano is used to obtain D from an experimentally measured

217

Mass Diffusion in the Solid State

concentration profile in an infinite diffusion couple. We begin the analysis by defining the function as t.

x

(4.73)

The use of the chain rule produces these expressions: CA t

dC A d

1 x dC A 2 t 3/ 2 d

t

and CA x

dC A d x

1 dC A . d t 1/ 2

Substitution of the latter into Fick’s second law, Eq. (4.5), gives dC A d

1 2

d d

D

dC A , d

which, being an ordinary differential equation, can be written as 1 2

dC A

d D

dC A . d

(4.74)

The infinite binary diffusion couple is made from pure A and pure B and has the initial conditions C Ao for x

CA

0

for x

0 at t 0 at t

0

.

0

The integration of Eq. (4.74) from CA = CA to CA = 0 gives 1 2

CA 0

dC A

D

CA

dC A d

0

or, as the concentration profile is always measured after diffusion has occurred for some time, t, 1 2

CA 0

x dC A

Dt

dC A dx

CA

.

(4.75)

0

At a position in the couple where CA = 0, and there is no flux (dCA/dx = 0), 1 2

CA 0

x dC A

Dt

CA dx

CA

CA dx

. 0

(4.76)

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Similarly, at a position in the couple where CA = CAo, dCA/dx is also zero, in which case Eq. (4.75) gives C Ao 0

0,

x dC A

defining the plane in the diffusion couple at which x = 0. This plane is known as the Matano interface. The plane at which x = 0 is shown in Figure 4.17 as that which makes the light gray regions of equal area. Eq. (4.76) then gives D=

1 dx 2t dC A

CA CA

0

x dC A

(4.77)

as the value of the interdiffusion coefficient for the composition CA. In Figure 4.17 the value of 0.2 C Ao 0

x dC A

is the dark gray area under the concentration profile between CA = 0.2CAo and CA = 0 and is the slope of the tangent to the concentration profile at CA = 0.2CAo. The measured self-diffusion coefficients of nickel and gold in the system Ni-Au at 900 °C are shown in Figure 4.18, and the measured interdiffusion coefficients are

FIGURE 4.17 Concentration profiles in an infinite diffusion couple used in the Boltzmann– Matano analysis for determination of the interdiffusion coefficient.

219

Mass Diffusion in the Solid State

FIGURE 4.18 Variation with mole fraction Ni of self-diffusion coefficients of Ni and Au in the Ni-Au system at 900 oC.

shown in comparison with values calculated using Eq. (10.34) in Figure  4.19 [4]. The thermodynamic activities of nickel and gold and the thermodynamic factor in Eq. (10.34), 1

d ln

Ni

d ln X Ni

,

are shown in Figure 4.20. The system Au-Ni has a miscibility gap in the solid state that begins at XNi = 0.8 and T = 820°C, and the minimum in the thermodynamic factor at this composition at 900 °C reflects the tendency of the system toward immiscibility. The minimum in the interdiffusion coefficient in Figure 4.20 is caused by the thermodynamic factor and not by any peculiarity in the composition dependence of the term, * X Ni DAu

* X Au DNi ,

in Eq. (4.71). The very low interdiffusion coefficients in the composition range over the peak of the immiscibility gap are thus caused by a low thermodynamic driving force in this range and not by any anomaly in the atomic mobilities. The variations of D with composition in four other binary metal systems are shown in Figure 4.21.

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FIGURE 4.19 Comparison of the measured and calculated interdiffusion coefficients in the Ni-Au system at 900 oC.

FIGURE 4.20 Variation of the thermodynamic activities and the thermodynamic factor in the Ni-Au at 900 oC.

Mass Diffusion in the Solid State

221

FIGURE 4.21 Composition dependencies of the interdiffusion coefficients in four binary systems.

4.10

INFLUENCE OF TEMPERATURE ON THE DIFFUSION COEFFICIENT

The examination of temperature dependence of D begins with interstitial diffusion through a face-centered cubic (FCC) iron unit cell. Figure 4.23 (a) shows a different perspective of the unit cell of FCC iron shown in Figure 4.22 and identifies three interstitial sites (1, 2, and 3) along the x-axis available for occupancy by a small impurity atom such as carbon. At 950 °C the diameter of the Fe atom in the structure is 2.58 Å, and thus the diameter of the interstitial site is 1.07 Å. The equilibrium distance between atoms a and b in Figure 4.23 (a) is 1.04 Å. The carbon atom has a diameter of 1.54 Å, so the presence of a carbon atom on an interstitial site causes local distortion of the iron lattice, which is why the solubility

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FIGURE 4.22 Face-centered cubic crystal structure of Fe showing locations of the octahedral interstitial sites.

of carbon in iron is relatively small. An interstitial atom on an interstitial site such as site 2 vibrates in all directions about the center of the site with a frequency of vibration, v, which, typically, is on the order of 1013Hz. For most of the time the vibrating atom is contained in an “energy well” that arises from the presence of the six neighboring Fe atoms at locations a to f (i.e., if the interstitial moves in any direction from the center of the well it experiences a repulsive force exerted by the neighboring Fe atoms). The variation of the energy of an interstitial atom with position along the x-axis, Figure 4.23 (b), shows that if an interstitial atom is to squeeze successfully between the atoms a and b and jump to the interstitial site 1 as a result of its vibration, it must have at least the energy Ea, the activation energy required for the jump. With that qualitative description of how a jump might occur, the question now arises: at what rate do interstitial atoms jump successfully from one interstitial site to another? In a large population of n atoms, the number of atoms that have energies greater or equal to Ei is ni n exp( Ei kT ). Thus, the fraction of interstitial atoms in a solution in a host lattice that have an energy greater than or equal to the required activation energy for a successful jump from one site to another is exp( Ea kT ), which also represents the probability that any one atom is sufficiently energetic for a successful jump. In Figure 4.22, the interstitial atom at the center of the unit cell can jump to any one of 12 neighboring interstitial sites, and thus the frequency, , with which an interstitial atom at a site such as 2 in Figure 4.23 (a) jumps to site 1 is: (frequency of vibration) x (probability the atom is sufficiently energetic to make a successful jump) x (probability that jump is in direction 1 2),

223

Mass Diffusion in the Solid State

FIGURE 4.23 (a) Interstitial sites in the face-centered cubic Fe lattice. Atoms on cell faces denoted by letters a–f; some interstitial sites denoted by numbers 1–3. (b) Variation of energy with position between interstitial lattice sites.

that is, v exp

Ea / kT 12.

(4.78)

The activation energy for diffusion of carbon in FCC iron is 2.223 × 10–19 J/atom or 133,900 J/g mol and thus at 950°C, 1013 s exp

2.223 10 19 J Catoms 1.381 10 23 J Catoms K 1223K

(i.e., a carbon atom makes 1.59 × 106 jumps/second).

1 12

1.59 106 s 1 , (4.79)

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If the number of interstitials per unit area on the plane R in Figure 4.23 (b) is NR, the flux of interstitials from the plane R to the plane L in the -x-direction is jR

NR ,

L

and, if the number of interstitials per unit area on the plane L in Figure 4.23 (b) is NL, the flux of interstitials from the plane L to the plane R in the x-direction is jL

NL ,

R

The net flux of interstitials in the x-direction is thus j

jL

jR

R

NL

L

NR ,

(4.80)

with the planes L and R being separated by the jump distance , the concentration of interstitials on a plane, CI, is CI

,

N

(4.81)

substitution of which into Eq. (4.80) gives j

CL

2

CR

CR CL

.

(4.82)

In Eq. (4.82), the term in parentheses is the concentration gradient of the interstitials in the x direction, and hence Eq. (4.82) can be written as 2

j

C . x

Comparing Eq. (4.83) to Fick’s first law, j D

(4.83) D C x , shows that

2

or, from Eq. (4.78), D

v

2

exp

Ea / kT 12.

(4.84)

The preexponential terms in Eq. (4.84) can be lumped together as D0 to give the exponential dependence of the diffusion coefficient on temperature as D

D0 exp

Ea / kT .

(4.85)

In Eq. (4.85), Ea is the activation energy for diffusion per atom, and multiplying this quantity by Avogadro’s number, N0, gives the activation energy for diffusion per gram mole of diffusing species, in which case the diffusion coefficient is given by D = D0 exp

Ea / RT ,

(4.86)

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Mass Diffusion in the Solid State

FIGURE 4.24 Variation with temperature of the diffusion coefficient DC in body-centered cubic iron ( ferrite).

Fe

for carbon

in which R N 0 k is the universal gas constant. Whether or not the activation energy is in units of energy per atom or energy per gram mole can be determined by observing whether the expression contains Boltzmann’s constant, k , or the gas constant, R. The variation of the logarithm of the diffusion coefficient of carbon in bodycentered cubic iron with temperature over the range −38 °C to 800 °C is shown in Figure 4.24. The variation is linear with the equation log10 D

Ea R

1 + log10 Do T

4400 K T

5.70.

From the slope of the line the activation energy for diffusion is 4400 K 2.303 8.3144 J gmol K

4.11

84, 300 J / g mol.

SUMMARY

Diffusion in the solid state occurs by the random motions of atoms, and the mean square distance through which an atom moves by random motion is proportional to the time for which the diffusion has occurred. In binary alloys diffusion occurs down concentration gradients, and Fick’s first law states that the diffusion flux is proportional to the concentration gradient, with the proportionality constant being the diffusion coefficient of the diffusing species. The equation for solute conservation (Fick’s second law) states that the rate of change of concentration of the diffusing species at

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any location is equal to the gradient of the diffusion flux (which itself is proportional to the second derivative of concentration with respect to distance). Thus, although the phenomenon of mass transport by diffusion is intrinsically different from the phenomenon of heat transport by conduction, the similarity of the laws governing the phenomena makes the mathematical treatment of the diffusion of interstitials in a solid very much like the mathematical treatment of heat conduction. Scaling and other approximate analyses are applied to single-phase, diffusion-controlled processes, producing solutions to problems in carburization, doping of silicon, nitriding, and diffusion bonding. A control volume analysis applied to the interface of two solid phases gives the solute conservation at that boundary, where the net flux across the phase boundary may cause growth and dissolution of the phases and the motion of that interface. Cases involving alloy homogenization and precipitate growth are studied using the interface solute balance and Fick’s second law. Observation of diffusion in substitutional solid solution shows that the diffusion coefficient, defined by Fick’s first law, can vary significantly with composition. This diffusion, or interdiffusion, coefficient, is thus only a measure of the rate at which concentration gradients in the alloy are eliminated by diffusion processes, and it is not an intrinsic property. It is instead a function of the chemical diffusion coefficients of the individual component species and of local composition. Fick’s laws hold only for thermodynamically ideal solutions in which the thermodynamic activity of the solute is proportional to its concentration (i.e., the solute obeys Henry’s law and the solvent obeys Raoult’s law). In general, mass transport by diffusion occurs down gradients of chemical potential (thermodynamic activity gradients), with the consequence that the chemical diffusion coefficients of the diffusing species are only independent of composition if the system is thermodynamically ideal. In such systems, the chemical diffusion coefficient of a species is proportional to its mobility, where the mobility is defined as the velocity of the diffusing atom per unit force acting on it, and the force arises from the gradient in chemical potential. Darken’s analysis models these phenomena. The chemical diffusion coefficient of a diffusing species in an alloy is also only equal to its self-diffusion coefficient (determined by measuring isotope diffusion in a one-component system) if the system is thermodynamically ideal. Kirkendall’s experiments show that, in systems in which the chemical diffusivities of the components have different values, the diffusion process causes movement of the positions of lattice planes relative to a fixed reference and hence mass transport in the system is the sum of motion by diffusion down a concentration gradient and motion of the lattice planes. Composition-dependent interdiffusion coefficients can be found directly by measuring composition profiles some time after transport begins in an infinite diffusion couple and applying the Boltzmann–Mantano analysis. Diffusion is a thermally activated process, and hence the diffusion coefficient increases exponentially with increasing temperature, with the rate of increase being determined by the activation energy required for an atom to jump from one site to another. This phenomenon shows again that there is not a physical analogy between mass diffusion and heat conduction. Conduction is not a thermally activated process, and the temperature dependence of the thermal diffusivity is determined by completely different phenomena.

Mass Diffusion in the Solid State

4.12

227

HOMEWORK PROBLEMS

(Always begin with a clear sketch of the physical problem. Do not plug in values for variables until absolutely necessary.) 4.1

4.2

4.3

4.4

A carbon steel containing 0.1 wt% C is carburized at 950 °C by a gas in which the carbon activity is equivalent to 1 wt% C in iron. (a) Calculate the time required to obtain a carbon content of 0.5 wt% at a depth of penetration of 0.05 cm. (b) What carburizing temperature is required to get a concentration of 0.5 wt% at a depth of penetration of 0.1 cm in the same time that 0.5 wt% was attained at 0.05 cm at 950 °C? Assume that the carbon activity in the carburizing gas at the higher temperature still corresponds to 1 wt% C in iron. The diffusion coefficient for carbon in austenitic iron is DC 7 10 6 m 2 s exp 133, 900 J/gmol RT . Hydrogen is stored at 400 °C in a long cylindrical steel vessel of inner diameter 0.1  m. Calculate the wall thickness of the vessel at which the hydrogen leaks from the vessel by diffusion through the wall at the rate less than 0.1 cm3(STP)/s per meter length of the vessel. The pressure of hydrogen inside the vessel is maintained at 101.3 kPa, and the hydrogen pressure outside the vessel is negligibly small. At 400 °C, DCFe 10 8 m 2 s , and,

7.29 10 5 P kg s2 m 5 . from Sievert’s law, CHFe kg m 3 Hydrogen in solution in a thick slab of solid nickel is removed by subjecting one face of the slab to a vacuum at 600 °C. (a) If the initial concentration of H in the nickel is 8 gmol/m3, after what time is the local concentration of H at a depth of 1 mm below the surface equal to C HNi = 4 gmol/m3? (b) At this diffusion time, what is the flux of hydrogen from the surface? (c) How much hydrogen has been removed from the slab per unit area by the time found in part (a)? For hydrogen in nickel, 41, 200 J/gmol RT . D 7.77 10 7 m 2 s exp In March 1944, an organization of mostly British prisoners of war (POWs) held in a camp in Silesia were planning a mass escape. One difficulty was fabricating steel tools that could hold an edge, and one POW with metallurgy knowledge came up with a solution: Travis made the wire-clippers .  .  . out of tie bars which he ripped off the huts. He riveted them together like scissors and filed cutting notches in them. It was very soft steel, so he hardened the metal himself. In a home-made forge he heated the clippers till they were red hot, poured a few grains of sugar on the metal around the cutting notches, and heated it up again so the carbon in the sugar was baked into the metal. Then he plunged it into cold water, and the steel came out hard enough to cut wire [5]. This process was a crude approximation of pack carburization, during which solid carbon is packed on a surface and, when heated, it diffuses into the steel.

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229

Mass Diffusion in the Solid State

FIGURE 4.25

Infinite diffusion couple for Problems 4.6 and 4.7.

edge of the appropriate penetration depth ( and these variables: B1

2

B2

B3

F1

2

F2

F3

A

or

) and one at the interface

B

.

Assume that the composition at the interface, CInt, is a constant throughout the entire process. (c) Show that integrating the conservation equations separately in each alloy over the two penetration depths ( A and B), using the assumed profiles in part (a), gives: A

12DA t

and

B

12DB t .

Use the flux continuity condition at x = 0 to obtain an expression for the nondimensional interface composition, Int, in terms of only the diffusion coefficients. (d) Based on your results, sketch C(x) at a time t > 0 for these three conditions: (i) DA /DB > 1 4.8

(ii) DA /DB < 1

(iii) DA /DB = 1.

An iron film (with no nitrogen in it) is vapor-deposited on a silicon wafer with a thickness of d = 10 μm. The film is then exposed at time t = 0 to a nitriding atmosphere, which fixes the nitrogen concentration in the iron on the exposed surface (x = 0) at C NFe  = Cs. You would like to know the changing nitrogen field in the iron film as a function of position and time. You may assume that: (i) Cs is much higher than the nitrogen solubility in the silicon, so C NSi ≈ 0, and there is no nitrogen flux of nitrogen into the silicon; and (ii) Cs is sufficiently low that there are no phase transformations in

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FIGURE 4.26

Schematic for Problem 4.8.

4.9

the iron and no significant volume change. The diffusion coefficient for nitrogen in iron is DNFe  = 10 −12 m2/s. (a) Sketch the concentration profile of nitrogen at the start of the process (t = 0), and for several times leading to steady state. (b) Write the solution of the diffusion equation that is valid for a short time after the nitriding atmosphere is introduced. What is the thickness of the sublayer in which nitrogen concentration is at least half of the surface concentration at 1 s and at 4 s? (c) At what time, tcrit, does your solution in part (b) cease to be valid? (d) Write the solution of the diffusion equation that is valid after tcrit. At what time, t, does the nitrogen concentration on the iron side of the iron-silicon interface reach 0.9Cs? (contributed by Adam Powell, Worcester Polytechnic Institute) During the solidification of a binary alloy, the two components freeze at different rates, leaving a composition profile that is solute-poor in the center of a dendrite and solute-rich at the position where a dendrite impinges on the one growing next to it (the last region to freeze). This effect, known as coring, may set up a repetitive composition field, as shown in the figure. Because coring causes nonuniform properties, the alloy is homogenized by heat treating at a high enough temperature to allow diffusion to occur. (a) In Figure 4.27, sketch the composition profile, C(x), between x = 0 and x = L after the alloy has been held at an elevated temperature for a period of time such that the composition at x = 0 has fallen 50% of the difference between Cave and . (Note that Cmax is C(x=0) and = Cmax(t  =  0).) What is the average composition, Cave, at that time? (b) Using these normalizations: C Cave * Cmax Cave

x L

tD L2

* Cmax

Cave

* Cmin

2

,

find the composition profile, ( , ) = B1 cos(π ), as a function of the normalized composition at x = 0, o ( ): o

0,

C x C

0, t * max

Cave Cave

.

Mass Diffusion in the Solid State

231

FIGURE 4.27 Schematic of composition variation after solidification at the dendrite scale for homogenization in Problem 4.9.

(c) Using the normalized mass conservation equation, find an expression for the normalized composition at x = 0, o ( ). (Hint: integrate between  = 0 and  = 1/2.) (c) For D = 10–13 m2/s and L = 100 μm, how long will it take to get o = 0.5? o = 0.01? 4.10 In a Mg-9wt%Al alloy, blocky structures of phase (nominally Mg4Al3) grow on grain boundaries near the end of solidification. One route to hardening this alloy is to dissolve this phase and spread aluminum through the matrix (homogenization) and then precipitate many, smaller particles throughout the metal (aging). To model the aging process, we begin with present. At t = 0, begins to grow at Al,i = 42 Al,i = 9 wt% Al and no wt% Al because the homogenized metal has been quenched to 150 oC, at which temperature the metal is in the + region on the phase diagram, with * Al = 5 wt% Al. (a) Sketch the system at t = 0 and two later times, showing the position of the - interface and CAl(x) in both phases. (b) Simplify the solution given in Section 4.5, given that C Al x ≈0. (c) How thick will the precipitate be after 48 h (which time gives the maximum hardness for this m2 155, 00 J mole exp [6]. heat treat)? DAl 3.9 10 5 RT s 4.11 Show that the expression in (a) Eq. (4.23) and (b) Eq. (4.24) are correct.

REFERENCES 1. Dossett, J., and G. E. Totten (eds.), ASM Handbook: Volume 4A: Steel Heat Treating Fundamentals and Processes, ASM International, 2013. 2. Smigelskas, D., and E. O. Kirkendall, “Zinc diffusion in alpha brass,” Transactions of the AIME, v. 171, pp. 130–142, 1947. 3. Darken, L. S., “Diffusion, mobility, and their interrelation through free energy in binary metallic systems,” Transactions of the AIME, v. 175, pp. 184, 1948. 4. Reynolds, J. E., B. L. Averbach, and M. Cohen, “Self-diffusion and interdiffusion in gold-nickel alloys,” Acta Metallurgica, v. 5, pp. 29–40, 1957. 5. Brickhill, P., The Great Escape, Faber & Faber, 1951. 6. Brennan, S., A. P. Warren, K. R. Coffey, N. Kulkarni, P. Todd, M. Kilmov, and Y. Sohn, “Aluminum impurity diffusion in magnesium,” Journal of Phase Equilibria and Diffusion, v. 33, pp. 121–125, 2012.

5

Fluid Statics

5.1 INTRODUCTION The study of fluid mechanics begins with the examination of fluids that are not moving, or at least in which no part of the fluid is moving relative to any other part. Without such relative motion, there are no velocity gradients and so no shear stresses. Without shear, the constitutive equations for a Newtonian fluid, Eqs. (1.10) and (1.11) reduce to only the normal force per unit area, or pressure, throughout the fluid. In this chapter, we look at the pressure distribution in a static fluid and some applications.

5.2 CONCEPT OF PRESSURE 5.2.1 pressure at a point and in a coluMn The force exerted by a gas on the walls of a vessel is caused by collisions of the individual atoms or molecules of the gas with the wall. An individual atom of mass m approaches the wall from some direction with the velocity V , and, after an elastic collision with the wall, rebounds in another direction at the same velocity magnitude. As momentum, mV , is a vector quantity that depends on direction as well as magnitude, the change in the direction of motion of the atom causes a change in its momentum. For example, if the atom of mass m is traveling in the +x direction normal to the wall, its momentum before the collision is mu, and after the collision is −mu. The change in the momentum of the atom as a result of the collision is thus (mu) − (−mu) = 2mu. The population of atoms in the gas is continuously colliding with the walls of the containing vessel, and hence there is a continuous rate of change of momentum, which has the units momentum time

mass

velocity time

kg

m s2

N .

These are also the units of force, and thus the force exerted by the gas on the walls of the container arises from, and is equal to, the rate of change of momentum of the gas atoms or molecules caused by their collisions with the walls. The force exerted by a liquid on the walls of its container arises from a similar phenomenon, and the fluid pressure is this force per unit area. Any imaginary surface within the body of the fluid experiences the same force in the direction normal to the plane of the imaginary surface. The pressure at any point in a static fluid is the same in all directions, as can be seen by considering the triangular element of fluid with sides of lengths a, b, and c and unit depth, located within a body of fluid, as shown in Figure 5.1. Mechanical equilibrium requires that the algebraic sum of the forces acting in the x-direction be zero, Pa sin 232

a

Pb sin

b,

(5.1)

DOI: 10.1201/9781003104278-5

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Fluid Statics

FIGURE 5.1 Pressures acting on a volume element of a fluid.

and the algebraic sum of the forces acting in the y-direction be zero, Pa cos

a

Pb cos

b

w

Pc

c.

(5.2)

In Eq. (5.2), w is the body force, or the weight of the element of fluid, which arises from the influence of the gravitational field on the mass of fluid in the element. For a mass of fluid m of density in an element of volume , w = mg =

g

=

g

1 a b . 2

(5.3)

From the geometry of the triangular cross-section of the element: sin

y b

cos

c2 b

sin

y a

cos

c1 . a

Thus, from Eq. (5.1), Pa

(5.4)

Pb

and from Eq. (5.2), Pa

c1 Pb

c2

g

1 a b 2

Pc

c.

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As Pa

Pb and

Pa

g

c1

c2

c,

a b2 c

Pc .

(5.5)

As the element is made infinitesimally small, the body force approaches zero, and hence in the limit of a = b = c 0, Pa

Pc ,

Pb

Pc ,

and thus Pa

(5.6)

that is, the pressure at any point in the static fluid is the same in all directions. The only forces acting in a static fluid are normal forces (such as those giving rise to the pressures Pa, Pb, and Pc in Figure  5.1) and the body forces that arise from the influence of gravity on the mass of the fluid. (Note that the normal forces act on and perpendicular to the infinitesimal control volume surfaces, while body forces act throughout the entire volume.) This influence of gravity is such that the pressure on any plane in a static fluid normal to the gravitational field arises from the weight of fluid above the plane, and thus the pressure at a point in a static fluid in a gravitational field varies with position along the direction of the field. Consider the column of fluid of cross-sectional area A in a gravitational field above the level z shown in Figure 5.2 (a). The body force exerted by the column of fluid above the level z on the fluid below the level z equals P |z A, where P |z is the normal pressure in the fluid at the level z. In Figure 5.2 (b) the body force exerted by the column of fluid above the level z + z on the fluid below the level z + z equals P |z+ z A, where P |z+ z is the normal pressure in the fluid at the level z + z. Thus, in Figure 5.2(c), the difference between the normal force at level z and the normal force at level z + z equals the body force arising from the weight of the fluid between the two levels: Pz

z

gA z

A

Pz A

or Pz

z

Pz

g,

z Which, in the limit of z

0, becomes dP dz

g.

(5.7)

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Fluid Statics

FIGURE 5.2 Derivation of the barometric formula from consideration of the body force of a fluid in a gravitational field.

5.2.2

atMospHeric pressure

Consider atmospheric air above the surface of the earth. The standard atmospheric pressure at sea level is 1.01325 × 105 Pa (1 standard atmosphere) at a temperature of 15 °C (288 K) and the standard acceleration due to gravity is g = 9.8067 m/s2 (which is defined in terms of the pull of the earth’s gravitational field on a mass of 1 lb at sea level and 45 °N latitude). Atmospheric air obeys the ideal gas law, RT M ,

P

(5.8)

where P = pressure of the gas (Pa), = volume of the gas per unit mass (= 1/ in m3/ kg), R = universal gas constant (= 8.3144 J/(g mol K)), T = absolute temperature of the gas (K), M = molecular weight of the gas (kg/g mol). Rearrangement of Eq. (5.8) as 1

PM . RT

And substituting it into Eq. (5.7) gives dP dz

PMg RT

or dP P

Mg dz. RT

(5.9)

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Assuming that T and g are independent of z, integration of Eq. (5.9) between the limits P = Po (standard atmospheric pressure) at z = 0 (sea level) and P = P at z = z gives ln

P P0

M gz RT

or P

P0 e

M g z/RT

,

(5.10)

which is known as the barometric formula. Air is a mixture of 21 vol% O2 (molecular weight 0.032 kg/g mol) and 79 vol% N2 (molecular weight 0.028 kg/g mol), and thus the molecular weight of air is 0.21 0.032

M

0.79 0.028 kg/g mol = 0.02884 kg/g mol.

At T = 288 K, Mg RT

0.02884 kg/gmol

9.81m/s2

8.3144 J/gmol K 288 K

1.181 10 4 / m,

and hence Eq. (5.10) becomes (where z = height in meters) P

P0 e

1.181

10

4

/m z

,

(5.11)

which is shown as line A in Figure 5.3.

FIGURE 5.3 Variation of atmospheric pressure with altitude. Line A, which is given by Eq. (5.11), is drawn for a constant air temperature of 288 K. Line B, which is given by Eq. (5.14), is drawn for an air temperature that decreases at the rate of 6.5 °C per 1000 m to an altitude of 11,000 m.

237

Fluid Statics

By convention, the international standard atmosphere is defined on the basis of a uniform temperature gradient of −6.5 °C per 10,00 m of altitude from sea level to a height of 11,000 m and a constant temperature of −56.5 °C for heights greater than 11,000 m. Thus, in the range z = 0 to z = 11,000 m, 288 K 0.0065 K/m z,

T

(5.12)

which, when substituted into Eq. (5.9) gives dP P

Mg dz, 0.0065 K m z

R 288 K

(5.13)

integration of which produces P

P0

288 K 0.0065 K m z

5.245

288 K

.

(5.14)

Equation (5.14) is shown as line B in Figure 0.3. As an example, from Eq. (5.14) it is calculated that the air pressure at the summit of Mount Phillips in the Sangre de Christo Range in New Mexico (elevation 3,579 m, or 11,742 ft) is 0.64 atm, which illustrates why hikers have some difficulty breathing near the summit. At 17,000 ft (5,182 m), which is the maximum altitude at which passengers without oxygen masks can fly in an unpressurized airplane, the air pressure is 0.52 atm. For completeness, it must be pointed out that the gravitational constant itself varies with distance above sea level as g

m 9.8067 2 s

6388 km 6388 km z

2

,

where z is in kilometers. The gravitation acceleration at the summit of Mount Phillips is thus 9.7957 m/s2. This effect on pressure is only a few percent at altitude of 50 km, and so has little effect on problems in this text.

5.3

MEASUREMENT OF PRESSURE

In 1643, the Italian natural philosopher Evangelista Torricelli filled a 4-foot-long glass tube with mercury and immersed the open end of the tube in a bath of mercury as shown in Figure 5.4. He observed that the mercury did not run out of the tube and concluded that a vacuum had been created above the mercury in the tube and that the column of mercury was supported by the atmospheric pressure acting on the free surface of the mercury in the bath. At equilibrium, the normal force per unit area on the surface of the mercury (point A) exerted by the column of air above the surface is equal to the body force per unit area at point B exerted by the column of mercury above B. The body force F exerted by the column of mercury of mass m, height h, and cross-sectional area A is F

mg

A h g,

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An Introduction to Transport Phenomena in Materials Engineering

which gives rise to a static pressure at point B of P

F A

gh;

hence, measurement of the height h allows determination of the atmospheric pressure at point B. Liquid mercury, like most condensed phases, is considered to be incompressible, in which case its density is independent of position within the column. Thus, the static pressure in the mercury column decreases linearly with z from the value pgh at point B to virtually zero at the free surface at the height h. Standard atmospheric pressure at 0 °C corresponds to h = 760 mm and is referred to as “760 mm Hg” or “29.92 in Hg.” The arrangement shown in Figure 5.4 gave rise to the development of the barometer, which provides a means of measuring the atmospheric pressure.1 Many devices that measure pressure actually measure the difference between the pressures of two fluids, giving rise to the distinction between absolute pressure and gauge pressure. Inflating a tire to a measured gauge pressure (Pg) of 240 kPa (35 psi) means that the difference between the pressure inside the tire and atmospheric pressure is 240 kPa. If the atmospheric pressure outside the tire is 100 kPa (14.7 psi), the actual pressure inside the tire is 340 kPa. This value is the absolute pressure (Pabs) and is determined by the rate of change of momentum of the atoms or molecules of the gas caused by their collisions with the walls of the container. Thus, Pabs

Pg

Patm .

FIGURE 5.4 Torricelli vacuum created in an inverted flask containing a liquid.

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Fluid Statics

FIGURE 5.5 Simple U-tube manometer for measuring differences in pressure.

Figure  5.5 shows a simple U-tube manometer, which allows measurement of the difference between the pressure of a fluid in a tank, Pt, and the atmospheric pressure, Po. The U-tube is partially filled with an incompressible fluid of density (which is immiscible with the fluid in the tank), and, at equilibrium, the absolute pressure at level A is the same in both legs of the manometer. At level A, the pressure is Pt in the left leg while, in the right leg, it is that due to the body force of the column of manometric fluid (the black region in Figure 5.5) of height h plus the atmospheric pressure, Po, exerted on the surface of the manometric fluid. Thus, Pt

gh Po .

It has been assumed here that the fluid density at the bottom of the manometer is much greater than the fluid in the tank or the atmosphere, thus allowing us to ignore the hydrostatic pressure in those fluids. This assumption is valid, for example, if the fluids in the atmosphere and tank are gases and in the manometer it is a liquid. The Torricelli barometer shown in Figure 5.4 is also a manometer, in that it measures the difference between the atmospheric pressure at A and the pressure in the volume above the column of liquid in the closed end of the tube. While we have a zero absolute pressure at the top of the column, the pressure of that “vacuum” is actually the saturated vapor pressure of the liquid in the barometer, which for liquid mercury at 298 K is 0.76 Pa. This value is negligible in comparison to an atmospheric pressure of approximately 100 kPa. Figure 5.6 shows a U-tube manometer containing two immiscible liquids, 1 and 2, of densities 1 and 2, (where 1 > 2). The cross-sectional area of the tubing connecting the two reservoirs is At, and the cross-sectional areas of each reservoir is Ar. With atmospheric pressure, Po, exerted on the free surfaces of both liquids, the equilibrium

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FIGURE 5.6 U-tube manometer containing two immiscible liquids.

state is as shown in Figure 5.6(a). The pressure at level z is the same in both legs of the manometer, Po

1

g h1

Po

2

g h2

or h.

h

1 1

(5.15)

2 2

In Figure  5.6 (b) the pressure exerted on the surface of liquid 2 has been increased to Po+ P, which causes the level of the surface of liquid 2 to fall by the distance Z, the position of contact of the two liquids in the right leg of the U-tube to change by the distance z, and the level of the surface of liquid 1 to rise the distance Z. At the new state of equilibrium, the pressures in the two legs are equal at level 2 and hence P0

1

g x1

Po

P

2

g x2

or P

g

1

x1

2

x2 .

Z

z

From geometric considerations x1

h1

(5.16)

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Fluid Statics

and x2

h2

Z

z.

Substitution into Eq. (5.16) gives P

h1

g

1

g

1 1

h

Z

z Z

h

2 2

h2

2 1

Z z

2

2

z 1

.

which in view of the equality in Eq. (5.15), becomes P

g

Z

1

2

z

2

.

1

(5.17)

Conservation of total volume also give z Ar

Z At ,

substitution of which into Eq. (5.17) gives

P

g

Z

1

g Z

ZAt Ar

2

1

2

At Ar

2

2

1

1

.

(5.18)

As Z is the measured quantity, the sensitivity of the manometer is increased by decreasing the difference between 1 and 2 or the ratio At/Ar.

5.4

PRESSURE IN INCOMPRESSIBLE FLUIDS

In the case of incompressible fluids, it is often necessary to know the pressure at some depth below the free surface of the liquid, in which case it is convenient to consider that h increases in the -z direction (i.e., in the direction of the gravitational field). Thus, taking Po to be pressure at the free surface, the pressure at a depth h below the surface is obtained from integration of Eq. (5.7) as P

P0

g h,

(5.19)

that is, in a fluid of constant density, the pressure increases linearly with increasing depth below the free surface. Figure  5.7 shows a plate of length L and width W immersed vertically in a liquid of density . From Eq. (5.19), the pressure exerted by the liquid on each face of the plate varies linearly with depth below the free surface from a gauge pressure of zero at the free surface (z  =  0) to a gauge pressure of gL at the

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FIGURE 5.7 Vertical plate immersed in an incompressible fluid and the variation, with depth, of the pressures exerted by the fluid on both sides of the plate.

bottom of the plate. The total force, exerted on each side of the plate is obtained from the definition Ftotal

A

P dA,

In which A is area. For a plate of constant width W, dA = W dh, and hence L

Ftotal

0

ghW dh

gL2W 2.

The average pressure exerted on each side of the plate is thus Pave

Ftotal WL

gL 2,

which is the local gauge pressure exerted halfway down the plate.

Example 5.1 Force on a Submerged Wall The rectangular metal water tank shown in Figure 5.8 has a glass window of dimensions 2 m × 2 m located at a distance of 1 m below the waterline in the tank. Find: What force does the water,  = 1000 kg/m3, exert on the window? Solution: From Eq. (5.19), the gauge pressure at the window centroid is P P0

gzc

1000 kg/m 3 9.81m/s2 zc

9810 Pa/m zc .

Thus, the local pressure exerted at the centroid of the window (at zc = 2 m) is 19,620 Pa and the total force exerted on the window by the water is Ftotal

19, 620 Pa 4 m 2

78, 480 N.

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Fluid Statics

FIGURE 5.8 Variation with depth of the pressure exerted by an incompressible fluid on the walls of its container.

Example 5.2 Force inside Oil Drum A cylindrical tank of radius 2 m is placed horizontally and is half-filled with oil of density 888 kg/m3. Find: Calculate the force exerted by the oil on one end of the tank. Solution: The force on the submerged wall of the oil drum is F

area

P dA,

where in Figure 5.9 P

gy

and dA

2 x dy.

From the equation of a circle, x2

y2

R2

and x

R2 y2

1/ 2

.

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FIGURE 5.9 drum.

Consideration of the pressure exerted by a fluid on the end of a cylindrical

Thus, 2 R2 y2

dA

1/ 2

dy

and F

2 g y R2 y2 F

2 gR 3 3

1/ 2

dy

2 g

1 2 2 R y 3

3/ 2

2 888 kg/m 3 9.81 m/s2 3

R

2 g 0 0

2m

3

R3 3

2 gR 3 . 3

46, 450 N

In this example, Pave

2 g R3 3 R2 2

F A

4 gR , 3

which is the local pressure at the centroid of the semicircle, y

4R x 3

0.

5.5 BUOYANCY The phenomenon of buoyancy was first considered in the third century B.C. by Archimedes of Syracuse, who formulated what is known as Archimedes’ principle: a body immersed or floating in a fluid is acted upon by an upward buoyant force which is equal to the weight of the fluid displaced and which acts through the center of gravity of the displaced volume. Figure 5.10 shows a solid object of density s immersed in a fluid of density f. Consider the cylindrical volume element of cross-sectional

245

Fluid Statics

FIGURE 5.10

Illustration of Archimedes’ principle.

area dA located in the object. The net differential upward force dFup exerted on the element by the fluid is given by dFup

P2 dA P1 dA.

But this is also the body force of an identical volume element of fluid given by f

g z2 z1 dA.

Thus, integrating over the entire surface of a solid object gives the total upward force exerted on the object by the fluid as Fup where

f

g ,

is the volume of the object.

Example 5.3 Calculation of Density from Buoyancy Force Measurement In air an object weighs wa = 3 N and, when immersed in water, it weighs ww = 1.5 N. The density of water is f = 1000 kg/m3. The force balance is illustrated in Figure 5.11. Find: What is the density of the object? Solution: The decrease in the weight of the object when immersed in water equals the buoyancy force, which is the weight of displaced water, W

Wa – Ww

buoyancy = weight of displaced water

Wdw

f

g

objj

.

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FIGURE 5.11 Difference between the weights obtained when an object is (a) weighed in air and (b) weighed in water.

Therefore, buoyancy , fg

obj

but wa

s

g

s

buoyancy

obj

,

f

so, f s

wa

buoyancy

1000 kg m 3 3 N 3N

1.5 N

2000 kg / m 3 .

In this example the influence of the buoyancy force of air on the measurement of the weight of the object in air is ignored.

5.6 SUMMARY The pressure exerted by a fluid on the walls of its container arises from the collisions of the individual atoms or molecules with the walls, and the force exerted on the wall is equal to the rate of change of momentum of the atoms or molecules due to the collisions. The pressure at any point in a fluid is the same in all directions. The influence of gravity is such that the pressure at any plane in a static fluid, normal to the gravitational field, arises from the weight of the fluid above the plane, and thus the pressure at a point in a static fluid varies with position in the direction of the gravitational field. In incompressible fluids, the pressure is a linear function of position in the direction of the gravitational field, and thus in liquids, the pressure increases linearly with

Fluid Statics

247

depth below the free surface of the liquid. The average pressure exerted by a liquid on a plate that is submerged vertically in the liquid is equal to the local value exerted at the centroid of the plate. The phenomenon of buoyancy is explained by Archimedes’ principle, which states that a body immersed in or floating in a fluid is acted upon by an upward force equal to the weight of the fluid displaced by the body, which acts through the center of gravity of the displaced volume.

5.7

HOMEWORK PROBLEMS

(Always begin with a clear sketch of the physical problem. Do not plug in values for variables until absolutely necessary.) 5.1

5.2

5.3 5.4

5.5

5.6

5.7

5.8

A 2 m × 3 m rectangular floodgate is placed vertically in water with the 2-m side at the free surface of the water. Calculate the force exerted by the water on one side of the floodgate. The density of the water is 997 kg/m 3. The center of a circular floodgate of radius 0.5 m is located in a dam at a depth of 2 m beneath the free surface of the water contained by the dam. Calculate the force exerted by the water on the floodgate. The density of water is 997 kg/m3. A hemispherical bowl of radius 0.5 m is filled with water of density 997 kg/m3. Calculate the force exerted on the bowl. A submariner is escaping from a damaged submarine at a depth of 50 m beneath the surface of the ocean. To prevent experiencing the “bends,” the rate of decrease in pressure exerted on his body must not exceed 3,400 Pa/s. What is the maximum rate at which he can ascend without experiencing the bends? The density of salt water is 1,030 kg/m3. When immersed in water of density 997 kg/m3 a solid sphere is balanced with a mass of 10 g, and when immersed in an oil of density 800 kg/m3 it balances with a mass of 11 g. Calculate (a) the volume and (b) the density of the sphere. When placed in water of density 997 kg/m3, a hollow sphere of inner radius 10 cm and outer radius 11 cm floats with half of its volume below the level of the free surface of the water. (a) Calculate the density of the material from which the sphere is made. (b) When placed in an oil the sphere floats with 60% of its volume below the level of the free surface of the oil. Calculate the density of the oil. A U-tube manometer contains mercury of density 13,600 kg/m3, and oil. (a) When both ends of the tube are open to the atmosphere, h1 = 17 cm and h = 1 cm. Calculate the density of the oil. (b) The open end of the left leg is sealed and is pressurized to a gauge pressure, P, which causes the level of the free surface of the oil in the left leg to fall 0.5 cm. Calculate the value of P. The altimeter in an airplane records a pressure of 70 kPa when the absolute pressure at sea level is 101 kPa and the air temperature is 288 K. Calculate

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FIGURE 5.12 Balance for measurements of sphere density in Problem 5.5.

FIGURE 5.13 The U-tube manometer in Problem 5.7.

FIGURE 5.14 Hydraulic press in Problem 5.9.

5.9

the altitude of the airplane, assuming that the air temperature is not a function of altitude. In the hydraulic press shown, what mass can be raised by the ram when a mass of 1 kg is placed on the plunger? (See Figure 5.14.)

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Fluid Statics

5.10 Calculate the gauge pressure in the oil at point D in Figure 5.15. The manometric fluids are mercury ( Hg = 13,600 kg/m3) and water ( w = 997 kg/m3). The density of the oil is 800 kg/m3.

FIGURE 5.15 Two-fluid manometer in Problem 5.10.

FIGURE 5.16

Electroslag remelting with atomization of the liquid metal in Problem 5.11.

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5.11 A metal electrode is melted in a hot slag and liquid metal droplets fall through the slag and form a reservoir. The metal then exits this reservoir through a small nozzle at the bottom, and this metal stream is atomized into a spray of droplets. (The chamber below is in vacuum, and the spray builds up an ingot with a very refined, but somewhat porous, microstructure.) Here we will find only the pressures in the system; the metal flow rate and its dependencies will be considered inChapter 6. (a) Calculate the pressure at the free surface on the top of the slag. (b) Calculate the pressure at the interface of the slag and metal. (c) Calculate the pressure at the bottom of the liquid metal pool. Use metal = 8,900 kg/m3, slag = 2000 kg/m3, L M = 0.3 m, LS = 0.2 m, and Patm = 101 kPa. (This problem is based on information from Carter and Jones [1]).

NOTE 1. Weather reports give barometric pressure in a number, usually without units. In the USA, the number is around 29 and elsewhere around 760. These reports assume pressure is in units of the height of a column of mercury (inches in the USA and millimeters in most other countries).

REFERENCE 1. Carter, W. T., Jr., and R. M. F. Jones, “Nucleated casting for the production of large superalloy ingots,” JOM, v. 55 (4), pp. 52–57, 2005.

6

Mechanical Energy Balance in Fluid Flow

6.1 INTRODUCTION A fluid moving in a conduit has, by virtue of its mass and motion, a kinetic energy that may change along the flow direction. If the fluid flow is not horizontal, the potential energy of the fluid also varies in the flow direction, which change may work to impel or retard motion. In all cases of real fluids, the friction arising from the shear stress exerted on the fluid by the containing conduit causes a transformation of mechanical to thermal energy. All of the mechanical energy added to a flowing fluid to maintain motion is obtained from the work done on the fluid by some external agency, and the rate at which this work is provided is the power required to sustain the flow. The most common power supplies are fans and pumps, which effect an increase in the pressure of the fluid at some point in the pipeline; the flows can also be driven by potential energy. Whatever drives or restrains the flow, we are concerned with how a given fluid responds in a given geometrical configuration. Understanding that behavior begins with a brief discussion of some flow characteristics, followed by examining an energy balance over the entire flow system. We derive that mechanical energy balance and apply it to many steady and transient flows.

6.2 LAMINAR AND TURBULENT FLOWS Up to this point in the text, we have examined stationary fluid systems, but now we will discuss moving flows. The nature of these fluid flows is dependent on the magnitude of the forces driving and restraining them, which determine the velocity fields. For the flow of a fluid of given physical properties in a given geometry, a critical velocity range exists below which the flow is laminar and above which the flow is turbulent. The two regimes are illustrated in Figure 6.1, which shows a thin stream of dye being introduced into a liquid flowing in a duct. At lower flow rates, shown in Figure 6.1 (a), the dye moves with the fluid in a stable line parallel to the axis of the tube. This stable behavior indicates that, at this low velocity, flow can be regarded as being the unidirectional movement of lamellae of fluid sliding over one another, with no macroscopic mixing or intermingling of the fluid in the vertical direction. This type of flow is known as laminar flow. If the flow velocity is increased beyond the critical range, then, as in Figure  6.1 (b), the flow becomes turbulent and the emerging dye stream is rapidly broken up and mixed with the fluid in the direction perpendicular to the dominant velocity. In the critical range, a transition from laminar to turbulent flow occurs. Experiments similar to those illustrated in Figure 6.1 were published in 1883 by Osborne Reynolds [1], who established a criterion for the transition from laminar DOI: 10.1201/9781003104278-6

251

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 6.1 Schematic of dye injected into (a) laminar flow and (b) turbulent flow in a duct, as seen in a series of experiments by Reynolds [1].

to turbulent flow in terms of a dimensionless quantity, known now as the Reynolds number: Re

U Lc

.

(6.1)

The characteristic length (Lc) in this flow is the height of the duct. The magnitude of the average velocity of the fluid, U, is measured as the volume flow rate divided by the cross-sectional area of the duct, and is the fluid density and μ its dynamic viscosity defined by Newton’s law of viscosity in Section 1.3.1. For fluid flow in a smooth circular pipe of diameter D, the transition from laminar to turbulent flow begins at a Reynolds number of approximately 2100, using the pipe diameter as the characteristic length (Lc = D). From Eq. (6.1), we see that (i) increasing the diameter of the tube decreases the average velocity at which the transition occurs, or (ii) increasing the flow velocity decreases the maximum pipe diameter in which laminar flow persists. For flow of a fluid at a given velocity in a pipe of given diameter, the velocity at which the transition occurs is proportional to the viscosity of the fluid and is inversely proportional to its density.

Mechanical Energy Balance in Fluid Flow

253

For now, we leave our description of turbulent flow as being more chaotic and disordered than laminar, with significantly more mixing, and we emphasize that for any specific geometry there is a Reynolds number range in which the flow regime changes from laminar (at low Re) to turbulent (at high Re). In Chapter 9, characteristics of turbulent flow and mechanisms of transition will be further discussed.

6.3

BERNOULLI’S EQUATION

Figure  6.2 shows the flow of an incompressible fluid through a section of pipe between locations 1 and 2. We assume initially that the flow is inviscid, i.e., that viscous effects are not important to the fluid behavior. This assumption allows the simplification of characterizing the velocity in the tube as the average value across the tube cross-section, , so we can neglect local variation of fluid velocity in that area. The fluid enters at 1 with the average flow velocity 1 and leaves at 2 with 2, and the static pressures in the fluid at planes 1 and 2 are P1 and P2. The fluid entering the section at plane 1 is being pushed by the fluid behind it, and the work done on a unit mass of the fluid entering is P1/ . Similarly, the fluid leaving the section at plane 2 is pushing on the fluid in front of it (downstream of section 1–2) and the work done by a unit mass of the fluid leaving is P2/ ). The difference between the work done on the fluid at section 1 and the work done by the fluid at section 2 is the net flow work done on the fluid. The kinetic energy of the fluid entering at plane 1 is EK

1 m 12 , 2

and the kinetic energy per unit mass of fluid entering is 12 2. Similarly, the kinetic energy per unit mass of fluid leaving at plane 2 is 22 2. The fluid potential energy at plane 1 is Ep

m g z1 ,

FIGURE 6.2 Fluid flowing through a section of pipe from location 1 to 2.

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An Introduction to Transport Phenomena in Materials Engineering

and the potential energy per unit mass of fluid entering is gz1. Similarly, the potential energy per unit mass of fluid leaving at plane 2 is gz2. Conservation of mechanical energy between locations 1 and 2 requires that net flow work done on the fluid = change in kinetic energgy + change in potential energy , that is, 1 2 2 g z2 z1 0. 2 1 2 2 1 net work change in change in on fluid kinetic enerrgy potential energy

P

P

+

(6.2)

Equation (6.2) is Bernoulli’s equation, a mechanical energy balance in frictionless flow through a conduit with changes in pressure, kinetic energy, and potential energy. To apply this equation to a flow in a conduit, first one must identify the starting point upstream (i) and the end point downstream (ii) and then be able to trace a single, continuous path through the system between these two points. Rearranging Eq. (6.2), we can see that the sum of the work done on the fluid and the changes in kinetic and potential energies do not vary along that continuous path: P 2

1 2

2 2

+ g z2 =

P

+ 1

1 2

2 1

+ g z1

CB ,

(6.3)

where CB is a constant. In this development of Bernoulli’s equation, we have neglected several possible physical phenomena. The density is assumed to be not a function of pressure, which is valid for liquids or if the change in the pressure in a gas is small enough. (This approximation is reasonable in most applications in materials processing; supersonic flow in an oxygen lance during oxygen steelmaking is one exception.) A steady flow has been assumed, so no work must be done (or recovered) from an acceleration (or deceleration) of the fluid. No work or heat has been directly added to the system, except due to the imposed pressure difference between points 1 and 2, and we have neglected any friction in the conduit. Using the average velocity in the pipe, , we may obtain the volume flow rate (Q) in the pipe, Q

(6.4)

A,

where A  is the cross-sectional area perpendicular to the flow direction and Q has units of m3/s. The mass flow rate in the pipe, m, is the product of the volume flow rate and the fluid density, m

Q

A

1

A 2,

(6.5)

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Mechanical Energy Balance in Fluid Flow

where conservation of mass requires m in and out of the pipe be the same, assuming no accumulation of fluid. In any real system, the second law of thermodynamics requires that some of the mechanical energy be degraded to thermal energy due to the friction in the moving fluid. The existence of this effect means that the Bernoulli equation, Eqs. (6.2) and (6.3), is not strictly valid, i.e., CB in Eq. (6.3) is not a constant. Eq. (6.2) must be modified to account for the dissipation of energy caused by the viscous drag exerted on the flowing fluid by the tube wall. If, between planes 1 and 2, this quantity, which is termed the friction loss, is ef, the energy balance becomes net flow work done on the fluid

increase in kinetiic energy

increase in frictio i n + potential energy loss

+

0.

Furthermore, if between locations 1 and 2, work w is done on unit mass of the fluid, the mechanical energy balance becomes flow work

w

ke

pe

g z2

z1

ef

or P

P 1

= 2

2 2

2 1

2

2

w

ef .

(6.6)

Eq. (6.6) is called the modified Bernoulli equation.

6.4 FRICTION LOSSES To model the contribution of wall friction in a straight length of pipe to the friction loss term, ef, first consider a length of horizontal pipe, z z , with a uniform cross-sectional area, A A , in which the flow velocity is uniform 1 2 , and there is no added work w 0 . For this situation, Eq. (6.6) is reduced to 1

1

2

2

P1

P2

ef ,

(6.7)

Illustrating that a pressure drop in the flowing fluid is required to supply the mechanical energy necessary to overcome the viscous drag exerted on the fluid by the pipe wall. In Eq. (6.7), P1 > P2, and hence ef > 0. The decrease in the normal force exerted on the fluid over the length of pipe, L, is equal to the shear force exerted on the fluid at the pipe wall: normal force = P1

P2

R2

0

2 R L = shear force,

and Eq. (6.7) becomes ef

2

o

L . R

(6.8)

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If we define the nondimensional shear stress, or Darcy friction factor, f, as 8

f

D P , L 1 2 2

0 2

(6.9)

and substitute it into Eq. (6.8), ef can be written as 1 L f 2 D

ef

2

,

(6.10)

which is a general expression for calculating the friction losses. The friction factor for laminar flow in a tube can be found from the laminar solutions for the average velocity, , and wall shear, o, to be later developed in Section  8.4 (if the flow is horizontal). In that case, 64

f

D

64 . Re D

(6.11)

Substitution of this friction factor into Eq. (6.10) gives ef

L 32 , D2

which shows that the friction loss per unit mass of fluid per unit length of flow is proportional to the viscosity and the velocity, and is inversely proportional to the cross-sectional area of flow and the density. This solution is valid in the laminar regime, which holds for flows with ReD < 2100. For turbulent flow between ReD ≈ 5 × 103 and 105, the friction factor in a smooth-walled pipe is given by a correlation of experimental data 0.316 Re D1/ 4 .

f

(6.12)

The Darcy friction factor, f, is defined in Eq. (6.9) and is frequently used in the literature. However, the Fanning friction factor, fF, is also used and is defined as fF

0

1 2

2

1 D 2 L

P 2

.

(6.13)

The relationship between the two definitions of a friction factor is a factor of four: f = 4fF. In this text, we use by default the Darcy friction factor, f, defined in Eq. (6.9).

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Mechanical Energy Balance in Fluid Flow

Example 6.1 Calculation of Pressure Drop in Pipe Water at 300 K is forced through 300 m of smooth pipe with an inside diameter of 0.05 m at a volume flow rate of 0.0015 m3/s. Use  = 997 kg/m3 and μ = 8.57 × 10–4 kg/ms. Find: Pressure drop needed to maintain flow through pipe. Solution: First, find the average velocity from the volume flow rate, Q, in Eq. (6.4): 0.0015m 3 /s

Q D/2

2

0.05 m / 2

2

0.764 m/s,

so we can calculate the Reynolds number. 997 kg/m 3 0.764 m/s 0.05m

D

Re D

8.57 10 4 kg/ms

4.44 10

At this value of the Reynolds number, the flow is turbulent (ReD > 2100) and the friction factor can be calculated from the empirical correlation in Eq. (6.12): f

0.316 44, 400

1/ 4

0.0218.

Rearrangement of Eq. (6.9) gives an expression for the pressure drop: P

1 f 2

2

L D

1 0.0218 997 kg/m 3 0.764 m/s 2

2

300 m 0.05m

3.80 10 4 Pa

which is roughly a third of a standard atmosphere (Patm = 1.01 × 105 Pa). Example 6.1 was straightforward because enough information was available to calculate the Reynolds number directly. If a friction factor cannot be found directly from the information given in a problem statement, the choice of correct expression for it must be found by trial and error.

Example 6.2 Calculation of Flow Rate in Pipe for Given Pressure Drop Water at 300 K is forced through a smooth pipe with an inside diameter of 0.07 m by a pressure gradient of P/L = 125 Pa/m. Use  = 997 kg/m3 and μ = 8.57 × 10–4 kg/ms. Find: (a) Average velocity, (b) volume flow rate, and (c) mass flow rate of stream. Solution: (a) Because we do not know the flow rate a priori, we cannot immediately calculate the average velocity or Reynolds number. Instead, we assume a flow regime, which in turn gives us a relationship between the friction

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factor (based on pressure gradient) and Reynolds number (based on the average velocity). If our assumption is correct, then we have the flow rates; if not, we try another flow regime. Assumption 1: Laminar Flow (ReD < 2100) For laminar flow, the combination of Eqs. (6.9) and (6.11) gives D P L 1 2 2

f or Re D

1 32

D L

P D2 2

1 Re 2D

64 Re D

997 kg/m 3 0.07 m 1 125 Pa/m 2 32 8.57 100 4 kg/ms

D3

P L

2

2

3

1.82 106.

Because ReD > 2100 when we assumed the opposite, the flow is definitely turbulent and we can not use laminar flow friction factor. Assumption 2: Turbulent Flow (ReD > 5 × 103) For this flow regime, to find an expression for the Reynolds number, we combine Eqs. (6.9) and (6.12). D P L 1 2 2

f

2

D L

P D2 2

1 Re 2D

0.316 Re D1/ 4

Manipulation of this relationship by isolating the Reynolds number gives 2 0.316

Re D

D3

P L

4/7

997 kg/m 3 0.07 m 2 125 Pa/m 2 0.316 8.57 10 4 kg/ms

2

3

4/7

7.85 10 4. This Reynolds number is within the range over which the friction factor correlation is applicable (5 × 103 > ReD > 105), so the assumption is self-consistent. Using this value of ReD, we can find the average velocity in the pipe, Re D D

v

7.85 10 4 8.57 10 4 kg/ms 997 kg/m 3 0.07 m

m , 0.964 m/s

as well as the volume and mass flow rates: Q

( D / 2 )2

A m

Q

0.964 m/s

997 kg/m 3

(0.07 m / 2)2

3.71 10 3 m 3 /s

3.71 10 3 m 3 /s 3.70 kg/s.

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Mechanical Energy Balance in Fluid Flow

The foregoing discussion has been limited to consideration of fluid flow in smoothwalled tubes. Any roughness of the tube wall increases the friction factor and hence decreases the flow rate obtained for a given value of P/L. The influence of surface roughness on the flow is correlated by the relative roughness, /D, the quotient of a characteristic height of protuberances on the surface, , and the diameter of the tube, D. Typically reported values of are listed in Table  6.1. Measurements of roughness’ effect on flow were carried out originally by Nikurasde [2] and Colebrook [3], controlling roughness by attaching sand grains of a known diameter and a tight size distribution to the inside of smooth pipes and using that diameter for . How these results translate to pipes in service, with much more complicated inner surfaces, is less clear. In the absence of clear guidance from the literature, most practitioners assume represents some height of protuberances on the pipe wall, but that height does not seem to be found by standard roughness measurement techniques (e.g., a profilometer). Also, the absolute roughness values in Table 6.1 are only approximate values for new pipes. As the pipes are used, they inevitably begin to foul due to debris gradually accumulating on the walls. Depending on the fluid and the pipe materials, significant corrosion also may occur, further roughening the conduit interior. Extending our scope to more vigorous flows than Eq. (6.12) and to pipes with rough surfaces, we can see the dependence of the friction factor on ReD and relative roughness, /D, in Figure 6.3, which is known now as the Moody diagram [5]. This diagram is based on data presented by Nikurasde [2] and Colebrook [3], the latter of whom curve-fit these data to find an implicit correlation for the friction factor: 1 2 log 3.7 D

f

2

2.51 Re D

.

f

(6.14)

The data fit by this correlation are within ±3–5% of Eq. (6.14). Moody developed this diagram to simplify finding f; the plot acts as an explicit replacement for the implicit Eq. (6.14), eliminating the need for iterative root finding for the friction factor. However, using such a diagram is not possible when calculating with a computer, so several researchers have reported different correlations that are explicit in f. For example, f

1 2 log 3.7 D

6.97 Re D

0.9

2

[6]

TABLE 6.1 Approximate Absolute Roughness Values for Common New Surfaces [4] Material drawn tubing steel cast iron concrete plastic, glass

(mm) 0.00013 0.0038 0.022 0.025–0.25 0.0 (effectively smooth)

(6.15)

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FIGURE 6.3 Moody diagram for Darcy friction factor dependence on Reynolds number and relative roughness.

and f

1 1.805 log 4.267 D

1.108

5.472 Re D

0.966

2

[7].

(6.16)

Examination of these correlations indicates that, at sufficiently high Reynolds numbers, the roughness effect in the first term inside the log functions dominates the second term. In that “fully turbulent” regime, the friction factor is only a function of the surface roughness. (Of course, in a smooth pipe, /D 0, that regime never develops and the friction factor is always Reynolds number dependent.) All of the correlations are valid in the range 5 × 103 < ReD < 108 and 10–6 < /D < 10–2 and agree with each other well within the uncertainty of the original experimental data. All in all, with the scatter in the original data and the lack of clarity in the definition of , the calculation of f has significant uncertainty, estimated in various textbooks (e.g., [8, 9]) as around ±10%. The foregoing discussion is based on the assumption of a circular conduit cross-section, while other cross-section shapes (especially rectangular) are not uncommon. In these latter cases, we can use the previous friction factor correlations with some alteration and within certain limits. To approximate the response in a noncircular channel, we replace the round pipe diameter with the hydraulic diameter, DH

4 A / Pw .

(6.17)

The area (A) is of the conduit cross-section normal to the flow direction, and Pw is the wetted perimeter, the length of the conduit perimeter in contact with the moving

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TABLE 6.2 Coefficients for Laminar Friction Factors for Noncircular Conduits [9] Shape ANNULUS D1 = inner diameter D2 = outer diameter

RECTANGLE a, b = lengths of sides

DH

Parameters values

D2 – D1

D1/D2 0.0001 0.01 0.1 0.6 1.0 a/b 0 0.05 0.25 0.50 0.75 1.00

2ab a b

C Eq. (6.18) 71.8 80.1 89.4 95.6 96.0 96.0 89.9 72.9 62.2 57.9 56.9

fluid. In turbulent flows, more of the pressure drop is due to extensive mixing at different length scales through the cross-section; this mixing is absent from laminar flow, in which most of the frictional losses are near the wall. Therefore, the simple replacement of DH for D in existing circular pipe correlations works better with turbulent flows (Re DH 5 103 ) than in the laminar regime. It works best if the conduit shape is somewhat close (within a factor of four or so) to an aspect ratio of one. For laminar flow, the friction factor can be found in the functional form of Eq. (6.11) as f

C Re DH ,

(6.18)

where C values are found from the exact solutions of the velocity field in the different conduits. Many values for C are presented in [9], from which a few are shown in Table 6.2.

6.5 INFLUENCE OF BENDS, FITTINGS, AND CHANGES IN PIPE RADIUS Eq. (6.8) shows the frictional losses in a pipe flow as a function of only the shear introduced at the conduit walls. However, in typical piping systems there are other sources of flow disruptions acting as sinks of mechanical energy. The occurrence of bends, valves, and abrupt changes of radius in a pipe system increases the extent to which the energy of the fluid is dissipated and hence increases the magnitude of the friction loss. The increase in the friction loss caused by the presence of a bend in a pipe is presented in terms of the increase in the length of a straight pipe, which would be required to produce the same increase in friction loss. Thus, any bend or fitting has

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an equivalent length, Le, of straight pipe. For example, the presence of a 45° elbow in a cylindrical pipe is equivalent to increasing the length of straight pipe by 16 pipe diameters, and thus for a 45° elbow, Le D

16.

While the pressure drop over a length of pipe is proportional to 2 , experiments show the pressure drop through different devices goes like P ~ n , where 1.8 < n < 2.1 [4]. For simplicity’s sake, we will treat n = 2 and Le/D is a constant for a given flow disruption. These equivalent lengths have been measured over a range of ReD > 1000. To apply to slower laminar flows (ReD < 1000), the simple linear relation Le D

laminar

Re D 1000

Le D

(6.19)

is used. The values of Le/D for various fittings are listed in Table 6.3, and the friction loss for a pipe system containing such fittings is calculated as ef

Le 2 1 L 1 f v2 f v . 2 2 D D pipe i wall friction friction loss through fittings

(6.20)

As noted in [10], these equivalent length values have an uncertainty up to ±25%. Given that level of uncertainty, estimates made with these models are approximate; measurements of the pressure drop through a system as a function of the flow rate must be made to obtain more accurate system performance curves. However, calculations done with the method presented here are useful for a first pass design to show trends in system behavior. Sudden changes in the radius of pipe through which the fluid is flowing, such as sudden expansions or contractions, are dealt with in terms of the friction loss factor, K, using which the consequent friction loss is calculated as ef

1 K v 2. 2

(6.21)

TABLE 6.3 Values for Le/D for Various Fittings [4] Fitting 45° standard elbow 90° elbow, standard radius 90° elbow, long radius 90° square elbow 180° close return bend

Le/D

Fitting

Le/D

16 30 20 57 50

Gate valve, fully open Gate valve, ¼ closed Gate valve, ½ closed Gate valve, ¾ closed Globe valve, fully open

13–17 35–50 160–200 900–1200 340

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For a sudden contraction, K

0.5 1

As AL

(6.22)

and for a sudden expansion 1

K

As AL

2

,

(6.23)

where As and AL are the cross-sectional areas of the smaller and larger diameter pipes, respectively. Eqs. (6.22) and (6.23) apply to turbulent flow, and the velocity used in Eq. (6.21) is the average value in the smaller diameter pipe.

6.6 STEADY-STATE APPLICATIONS OF THE MODIFIED BERNOULLI EQUATION Now that we have models of how frictional losses in conduits bleed mechanical energy from a moving fluid, we can examine the steady-state behavior of such a fluid as it flows through a conduit. First, we explore an example in which fluid drains from a vessel while being replenished, so that the flow rate and the amount of liquid in the vessel is steady.

Example 6.3 Drainage of a Replenished Ladle Figure 6.4 shows fluid draining through an orifice in the base of a ladle. The fluid is being poured into the ladle at the same rate it drains, such that the depth of the fluid, ho, remains constant. The diameters of the ladle and the orifice are, respectively, D and d. Assume the liquid is an aluminum alloy with μ  =  1.3 × 10–3 kg/ms  and  = 2500 kg/m3 and the diameters are D = 2 m and d = 0.1 m. The depth of the vessel is maintained at ho = 1 m. Find: What is the mass flow rate of the aluminum into and out of the ladle? Solution: If plane 1 is located at the liquid-free surface in the vessel and plane 2 is located at the level of the exit of the orifice, the modified Bernoulli equation in Eq. (6.6) becomes P

P 2

1

2 2

2 1

2

2

g z2

z1

ef

0.

The aluminum at the top surface in the ladle (P1) is at the pressure of the surrounding air, Patm. The fluid requires some pressure to push it through the bottom opening at level 2, but free expansion of the metal jet once it exits the orifice gives P2 = Patm

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FIGURE 6.4 Fluid draining through an opening in the base of a replenished vessel.

also. From mass balance considerations, the mass flow rate through the orifice equals m through the vessel: m

1

D2 4

2

d2 4

or

1

If d > As for both ends of the pipe. Thus, from Eqs. (6.22) and (6.23), respectively, K = 0.5 at the entrance (a contraction of area) and K = 1 at the exit (an expansion): 1 1 2

1 K entry K exit v 2 2

2

0.5 0.764 m s

2

0.44 m s .

The friction loss is then ef =

1.5 m +100 m 35m 25 m 15 m 1 0.0183 2 0.1 m 6 0.764

m s

2

0.44

m s

2

10.4

m s

3 31

2

.

Substitution into Eq. (6.27) gives w

211 m / s

2

10.4 m / s

2

221 m / s

2

221 J/kg,

which, being a positive quantity, indicates that work is done on the fluid by the pump. The mass flow rate is m

Q

997 kg m 3 6 10 3 m 3 s

5.98 kg/s,

and thus the power required of the pump, W, is W

wm

221J kg 5.98 kg s

1324 W.

Of this power, 211/221, or 95.3%, is needed to pump the water to a higher elevation, increasing the potential energy of the water by the required amount at the required rate, and the remaining 4.7% is dissipated as friction loss. The contribution of the frictional losses at the entrance and exit of the pipe is negligible. Just as work must be done on a fluid (by means of a pump) to increase its potential energy, useful work can be obtained from a fluid (using a turbine) that is flowing downward under the influence of gravity. In such a case, w is a negative quantity.

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Example 6.5 Using Potential Energy of Water to Drive a Turbine Water is fed from a reservoir to a turbine through a vertical concrete pipe of diameter D = 0.5 m and average roughness  = 5 × 10–4 m at a mass flow rate of m = 1000 kg/s. The surface of the reservoir is L = 50 m above the entrance of the turbine, and the pressure at the reservoir surface and at the turbine exit are both P = 105 Pa. The density and viscosity of the water are, respectively,  = 997 kg/m3 and μ = 8.57 × 10–4 Pa∙s. Find: Power generated by the turbine. Solution: With m = 1000 kg/s and D = 0.5 m, the linear flow velocity is 1000 kg s

m

v

D/2

2

2

0.25m

5.11 m / s

997 kg m 3

and thus the Reynolds number is Re D

997 kg s 5.11m s 0.5m

vD

8.57

2.98

-4

10 kg m s

106 ,

which indicates that the flow is turbulent. The friction factor is obtained from Eq. (6.15) as

f

1 2 log 3.7 D

6.9 Re D

1 0.0005m 2 log 3.7 0.5 m

2

0.9

6.9 2.98 106

0.9

2

, 0.9

3.7 D 6.9 Re D in this case, showing which gives f = 0.0198. (Note that that we are in the fully turbulent regime where f is only a very weak function of Re D .) For the path between the reservoir surface and the turbine exit, ef

1 L 2 f v 2 D

1 50 m 0.0198 2 0.5m

5.11 m s

For a constant and uniform m, Eq. (6.5) shows that v1 g h2 h1

w ef

2

2

25.8 m s .

v2 and Eq. (6.6) is

0

or w

9.81 m s2

50 m

25.8 m s

2

465 J kg.

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Mechanical Energy Balance in Fluid Flow

The negative sign indicates that the water is transferring work to the turbine. The power extracted from the water is thus power

465 J kg 1000 kg s

wm

465 kW.

Note that only 1.3% of the potential energy is lost to frictional losses in the pipe. The friction loss in the turbine is expressed in terms of the efficiency with which the energy extracted from the water is converted to shaft energy. The turbine efficiency, , is defined as shaft energy delivered by turbine . o water energy extracted from Thus, if the turbine has an efficiency of 90%, the shaft work delivered would be 0.9 465 kW

418 kW.

6.7 CONCEPT OF HYDROSTATIC HEAD The total energy of a fluid at any point in a flow system is

1 2 mgz. mv 2 flow kinetic potential energy energy energy Pm

(6.28)

Dividing Eq. (6.28) by the mass of the fluid multiplied by gravity, mg, gives H

P 1 v2 g 2 g

(6.29)

z.

Eq. (6.29) has the units of length and defines the hydrostatic head, H, of the flow at a point of interest. The first term in Eq. (6.29) is the pressure head, the second term the velocity head, and the third term the elevation head. In considering flow between plane 1 and plane 2 in a flow system, the loss of head, – HL, is obtained from Eq. (6.29) as HL

H 2 H1

1 v22 v12 2 g

P2 P1 g

z2 z1 .

(6.30)

Eq. (6.29) shows that in ideal flow with no frictional loss or work addition, H is a constant along the flow direction, so the head loss is zero (HL = 0). However, comparison of Eq. (6.30) with Eq. (6.6), for a flow system in which w is zero between planes 1 and 2, shows that HL

ef g

;

(6.31)

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hence the loss of head between two points in a flow system is an alternative measure of the dissipation of energy by friction loss between the two points. Application of the concept of head loss is restricted to the flow of incompressible fluids in flow systems in which there is no work transfer into or out of the fluid (e.g., no pumps or turbines).

6.8 FLUID FLOW IN AN OPEN CHANNEL Consider a steady flow of fluid in the inclined open channel shown in Figure 6.6. The channel, of length L and width W, is inclined at an angle from the horizontal, and under conditions of uniform flow, the depth h of the fluid is constant along the channel. The gravitational force acting on the fluid in the channel in the direction of flow equals the frictional force exerted on the fluid by the walls of the channel: hWL g sin

o

L 2h W

1 f v 2 L 2h W , 8

(6.32)

where the wetted perimeter is the sum of the width (W) and twice the height of the liquid layer (2h). From Eq. (6.17), the hydraulic diameter of the flow, DH, is DH

4A Pw

4hW , 2h W

FIGURE 6.6 Gravity-induced flow in an open channel.

(6.33)

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Mechanical Energy Balance in Fluid Flow

substitution of which into Eq. (6.32) gives 1 L f 8 DH

hWLg sin

v 2 4 hW

or 2 DH g sin f

v

1/ 2

.

(6.34)

For fully turbulent flow, the friction factor is given by Eq. (6.15) when Re DH f

2 log

1 3.7 DH

:

2

,

(6.35)

where is the absolute roughness of the channel wall. Substitution of Eq. (6.35) into Eq. (6.34) gives v

8 DH g sin

1/ 2

log

1 3.7 DH

.

(6.36)

Example 6.6 Flow of Liquid Lead Tapped from a Furnace Liquid lead, tapped from a blast furnace, runs down an open channel (of width W = 0.3 m and at an inclination from the horizontal of angle ) into a kettle. The maximum allowable depth of the moving lead layer is h = 0.1 m. The absolute roughness of the channel wall is  = 0.0003 m, and the density and viscosity of liquid lead are  = 10,050 kg/m3 and μ = 1.82 × 10–3 Pa∙s. Find: The depth of the mass flow rate of lead down the channel, m, over the range 5° < < 25°. Solution: We will use the correlation for friction factor for turbulent flow in a circular pipe, but using the hydraulic diameter as the length scale. That quantity at the maximum depth of lead is DH

4A Pw

4hW 2h W

4 0.1m 0.3m 2 0.1m

0.3 m

0.24 m.

The aspect ratio of this layer is AR h W 0.1 m 0.3 m 1 3, which is close enough to unity to use the pipe flow correlations. The friction factor, assuming a fully turbulent flow, is from Eq. (6.15): f

1 2 log 3.7 DH

2

1 0.0003m 2 log 3.7 0.24 m

2

0.0207.

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From Eq. (6.34), we find the average velocity and mass flow rate as functions of inclination angle: v

1/ 2

2DH g sin f

2 0.24 m 9.81m/s2

1/ 2 1/ 2

s sin

0.0207

15.1

m s

sin

1/ 2

and m

vA

vhW

hW

10, 050 kg/m 3 0.1 m 4550

kg s

sin

1/ 2

2DH g sin f 0.3m

1/ 2

2 0.24 m 9.81m/s2

1/ 2

sin

0.0207

1/ 2

.

These results are plotted in Figure 6.7, which shows that the rate of lead transfer down this inclined channel more than doubles over the range of possible inclination angles. To check if the flow really is fully turbulent, we compare the two terms in the expression inside the log function in Eq. (6.15). For the flow to be in that regime,

3.7 DH

6.9 Re DH

0.9

or

1 3.7

Re DH DH

6.9

0.9

1.

FIGURE 6.7 Average velocity and mass flow rate down inclined open channel in this example as a function of inclination angle.

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Mechanical Energy Balance in Fluid Flow

FIGURE 6.8 Ratio of terms inside log function in friction factor correlation, Eq. (6.15). 0.9 3.7 DH Re DH 6.9 >> 1, the assumption that the flow is fully turbulent Because the ratio is correct over the range of possible channel inclination angles.

A plot of this expression over the range of inclination angles is shown in Figure 6.8. Because the curve is much greater than unity over the entire range, we can safely assume the flow is fully turbulent and f ≠ f (Re).

6.9 TRANSIENT APPLICATIONS OF THE MODIFIED BERNOULLI EQUATION In many cases the purpose of the analysis shown in previous sections is to estimate the rate of change of liquid levels in a system such as liquid metal filling a mold or draining from a ladle. Using the latter case as an example, Figure 6.9, we will analyze the transient liquid behavior in a ladle. For the round ladle shown, the liquid level begins at h1(t=0) = ho and falls to h1 (t=tf) = 0, where tf is the draining finish time. The outlet of the free jet existing the ladle is at z2 = h2 = 0. The modified Bernoulli equations for this configuration (assuming w = 0) is written as: P2 P1

1 2 v2 v12 2

g h2 h1 t

1 K1 v12 2

1 K 2 v22 2

0.

(6.37)

All the possible frictional losses in the ladle and the orifice discussed in earlier sections are lumped into the coefficients K1 and K2, respectively, which are assumed to be not functions of liquid velocity. Rearranging Eq. (6.37) and using h2 = 0, gives P1 P2

g h1 t

h2

1 2 v2 v12 2

1 K1 v12 2

1 K 2 v22 . 2

(6.38)

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FIGURE 6.9

A ladle emptied by draining through an orifice.

The left-hand side of Eq. (6.38) has the terms driving the flow, the pressure difference and potential energy, which provide the mechanical energy necessary to change the kinetic energy and overcome frictional losses on the right-hand side. The two velocities in Eqs. (6.37) and (6.38) are related by the volumetric flow rate, which is uniform along the path from 1 to 2 if the fluid is incompressible:

Q

v1 A1

v2

A2

or v2

v1 A1 A2 .

(6.39)

Substituting Eq. (6.39) into (6.38) and rearranging gives P1 P2

g h1 t

h2

1 2

A1 A2

2 2 1

1 v

1 K1 2

A1 A2

2

1 K2

2

K 2 v12

or 2 P1 P2

2 g h1 t

K1 1

h2

A1 A2

v12 .

The velocities at planes 1 and 2 are v1

2 P1 P2 K1 1

2 g h1 t A1 A2

2

h2

1 K2

v2 A2 A1 .

(6.40)

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Mechanical Energy Balance in Fluid Flow

So far this derivation follows the case of the replenished vessel in Section 6.6, except h1(t) is still an unknown function of time. However, we observe that the liquid velocity at 1 is the rate of change of the plane 1 height: dh1 , dt

v1

which gives us a differential equation for h1(t): 2 P1 P2

dh1 dt

2 g h1 t

K1 1

2

A1 A2

h2

C1 h1 (t )

1 K2

C2 ,

(6.41)

where C1

2g K1 1

2

A1 A2

2 P1 P2

and C2

1 K2

K1 1

A1 A2

2 g h2 1 K2

. (6.42)

Using separation of variables, the solution of Eq. (6.41) can be written as 2 C1

C1 h1

C2

1/ 2

C3 .

t

Applying the initial condition, h1(t = 0) = h0, we obtain the constant of integration, 2 C1

C3

C2

C1 ho

1/ 2

.

The complete solution for the transient liquid level, h1, is thus 1 C1

h1 t

C1 ho

C2

1/ 2

C1 t 2

2

C2 . C1

(6.43)

To find the time, tD, needed to drain the vessel to level h2, we set h1(tf) = h2 and solve Eq. (6.43) for time, giving tf

2 C1

C2

C1 ho

1/ 2

C1 hD

C2

1/ 2

or, using the definitions of the constants in Eq. (6.42), tD

2 K1 1 g g ho h2

A1 A2

2

1 K2

P1 P2

1/ 2

g hD

P1 P2 2

1/ 2

.

(6.44)

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The transient velocity at the orifice is the time derivative of Eq. (6.43) combined with the uniform volumetric flow rate, Eq. (6.40): v2 v1

A1 A2

dh1 A1 dt A2

A1 A2

C1ho C2

C1 t 2

1/ 2

(6.45)

or v2

A1 A2

2 K1 1

g ho

h2

A1 A2

2

1 K2 (6.46)

gt . 2

1/ 2

P1 P2

The results of the previous analysis can be simplified in several ways. If there is atmospheric pressure on the top surface in the vessel and at the jet exit, then P1 = P2 = Patm. If the orifice is much smaller than the area of the vessel, then A1/A2 2 >> 1 and A1 / A2 1 K 2 K 2 1 0 or v2 > v1. In this case, the flow enters from the left and splits into two different directions; mass leaving out the top of the volume (v2 > 0) slows the horizontal velocity. In the opposite case, if u1 < u2 (accelerating

FIGURE 7.2

Control volume illustrating conservation of mass in two dimensions.

289

Equations of Fluid Motion

flow), then ∂u/∂x > 0 and ∂v/∂y < 0, so v2 < v1. To accomplish this increased flow rate in the x direction, mass must be entrained into the volume through the top with a downward velocity, v2 < 0. Beginning with Eq. (7.13) for two-dimensional steady flow (w = 0), we can define the streamfunction, (x, y), in Cartesian coordinates as u

and

y

v

x

.

(7.15)

This quantity is defined thus to satisfy the continuity equation identically: u x

v y

2

x

y

y

x

x

y

y

x

x y

2

y x

0.

The total differential of (x,y) is d

x

dx

dy

y

v dx

(7.16)

u dy.

We also define a streamline as a line in a velocity field everywhere tangent to the flow, as shown in the example flow field in Figure 7.3 (a). This tangency condition means that velocity has no normal component on a streamline and so streamlines are impermeable. The slope of the streamline can be defined as dy dx

v , u

(7.17)

where the rise over run is determined by the ratio of velocity components. Eqs. (7.16) and (7.17) combined gives v dx

u dy

0,

d

which shows that the value of the streamfunction is uniform along a streamline. The fact that no fluid crosses a streamline (a line of constant ) is useful for displaying the distribution of the volume flow rate, Q =  V A, in a flow field. Any two streamlines form a stream tube, and we find that the flow rate is the same across any line drawn from one streamline to the other. This result is illustrated in Figure 7.3 (b) by calculating Q across the two lines, A and B. For line A, the only normal velocity component is u, so y2

QA

y2

u dy y1

y1

y

dy

2

1

,

For line B, with no horizontal component of velocity, x1

QB

x2

v dx x2

x2

v dx x1

x1

x

dx

2

1

.

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FIGURE 7.3 (a) Streamlines (lines of constant ) in a flow field, showing tangent velocity vectors. The surface also acts as a streamline, as no flow passes through it. (b) Flow through a stream tube between two streamlines ( 1 and 2).

FIGURE 7.4 Streamlines in a flow field moving from left to right. The flow first accelerates as streamlines converge, then decelerates as they diverge.

291

Equations of Fluid Motion

We see that QA = QB, so the volume flow rate does not change along the stream tube constrained by 1 and 2. The value of the volume flow rate, Q, is also the difference of the streamfunction values for the two streamlines. Figure 7.4 shows a family of streamlines representing a flow field moving left to right. Moving from the left, the streamlines converge then diverge. As they move closer together, the cross sectional areas of the stream tubes decrease, causing an increase in the velocities along the tubes. When they then diverge farther downstream, those areas increase and the flow decelerates.

7.3 MOMENTUM BALANCE: THE NAVIER–STOKES EQUATIONS In this section, we apply Newton’s second law of motion to a fluid control volume to derive a momentum balance equation. This balance and the continuity equation (7.13) form a basis for a description of the mechanisms controlling fluid motion. In most early physics courses, Newton’s second law is written as F

ma,

where the mass is constant and a is the acceleration vector. A more general version is that the sum of the forces causes a change in the momentum vector, , with time: d . dt

F

(7.18)

We can write the momentum field as x, y,t

mV ,

and its total derivative in two-dimensional Cartesian coordinates as d

mV t

mV

dt

x

mV

dx

y

dy.

(7.19)

The total change in momentum with time is then d dt

mV dt t dt mV t

mV dx x dt u

mV dy y dt

mV

mV

x

y

(7.20)

,

where u = dx/dt and v = dy/dt. This form of the total time derivative in a flow field is known as the substantial, or material, derivative, which is usually written for any transported quantity as

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D Dt

u

t

x

v

.

y

(7.21)

In Cartesian coordinates, the vector definition can be written in its x and y components: mu

D Dt

u

t

mu x

v

mu

mv

y

u

t

mv

v

x

mv y

ˆj (7.22)

In each direction, the first term is the local acceleration in the flow. It represents changes in the fluid velocity with time at a point; it is zero in a steady flow. The next two terms show how a fluid moving in a velocity field can accelerate or decelerate by moving from one point in that field to another with a different velocity. They represent the rate of change of mass flux through a point because there is a velocity gradient there. To examine this idea more closely, let us focus on the term u mu x iˆ in Eq. (7.22). The derivative represents the rate of change of the x direction momentum (mu) with a change in position along the x axis. How fast does that change occur? As noted earlier, the rate of position change is the velocity, u dx dt , and so the product u

mu

dx dt

x

mu

mu

x

t

Is the rate of change of mu as a result of horizontal fluid motion through a nonuniform momentum field. If there is a vertical motion, v dy dt, then the x momentum will change at the rate v mu y as the fluid moves in the vertical direction. Similar arguments are made for the time rate of change of the y momentum, mv, due to change in position. These four terms in Eq. (7.22) represent the advective acceleration, or the change in momentum with position in a flow, which occurs where there is a nonuniform velocity field, even if that flow is steady. Having found an expression in terms of the velocity field for the change of momentum in Newton’s second law, Eq. (7.22), we turn to the sum of the forces. The forces on a two-dimensional control volume (with depth dz into the page) are shown in Figure 7.5, and they fall into three types: normal, shear, and body forces. The normal forces act perpendicular to surfaces, the shear parallel to them, and the body forces throughout the volume. The net normal force in the x direction, using a truncated Taylor series, is

xx x dx

xx x

dydz

xx xx x

x

dx

xx x

dydz

x

xx

x

dx dydz. (7.23)

Similarly, the shear forces in x are

yx y dy

yx y

dxdz

yx yx y

y

dy y

yx y

dxdz

yx

y

dy dxdz. (7.24)

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Equations of Fluid Motion

FIGURE 7.5 Two-dimensional control volume showing forces on a moving fluid.

The body force in the x direction depends on what physical effects act on the entire volume (as opposed to acting on the surfaces) in the system we are modeling. In many processes, electromagnetic forces are used to either damp or impel fluid flow, and gravity is always included. Here, we include only gravity: FBx

g x dxdydz ,

mg x

(7.25)

where the gravity vector is g g x iˆ g y ˆj.. The sum of these forces in the x direction is Fx iˆ

yx xx g x dxdydz iˆ, x y normal shear gravity

(7.26)

and a similar process in the y direction will give Fy ˆj

yy xy g y dxdydz ˆj. y x normal shear gravity

(7.27)

Applying the substantial derivative of the momentum field, Eq. (7.22), at the location of the control volume subject to the forces in Eqs. (7.26) and (7.27), and using m =  dxdydz, gives us a new form of Newton’s second law in x and y: u

u

u

v

u

t x y channge in momentum

xx

x

yx

gx y sum of forces

(7.28)

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and v v v u v t x y chaange in momentum

yy

y

xy

gy . x sum of forces

(7.29)

To progress from this point, we require constitutive equations for the shear and normal stresses in two dimensions; these relations are found from those in Section 1.3.1.1 assuming ∂/∂z = 0. P

xx

u x

2

P

yy

2

v y

xy

u y

yx

v x

(7.30)

A derivation of these equations for a Newtonian fluid is found in Schlichting [1]. Using these relations between forces on the fluid and its velocity gradients, gives us the momentum balances in the x direction, u t

u

u x

v

u y

P 2 x x

u x

v x

y

y

u y

y

v y

(7.31)

gx ,

and in the y direction, v t

u

v x

v

v y

P y

v x

x

x

u y

2

(7.32)

gy .

For many applications, the temperature variation in a fluid is small enough that the dynamic viscosity, μ, may be treated as uniform, and the range of pressure is small enough that the fluid is effectively incompressible ( ≈ constant and uniform). Within those limitations, we can reduce the equations to a more simplified form: u u u u v t x y local convecctive acceleration

1 P x pressure

2

2 u u 2 x y2 viscous friction

v v v u v t x y local conveective acceleration

1 P y pressure

2

gx

(7.33)

gravity

and v

2

v x y2 viscous friction 2

gy . gravity

(7.34)

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Equations of Fluid Motion

These equations, expressions of Newton’s second law in terms of fluid velocity and pressure, are the Navier–Stokes equations. In each direction, the left-hand side is the time rate of change of momentum. Causing these local and advective accelerations are the forces on the right-hand sides of Eqs. (7.33) and (7.34): pressure, viscous friction arising from shear stress, and gravity. Extending the previous derivation to three dimensions, using the full constitutive equations in Section 1.3.1.1, gives us 2

2

2

u t

u

u x

v

u y

w

u z

1 P x

u x2

u y2

u z2

v t

v x

1 P y

2

w

v z

2

v

v y

2

u

w x

v

w y

w

w z

1 P z

v

x2

v y2

2

2

v z2

gx ,

(7.35)

gy ,

(7.36)

and w t

u

w x2

2

w y2

w z2

gz .

(7.37)

Written in cylindrical coordinates (r- -z), these equations are u u u t r

v u r

v2 r

1 ru r r r

p r v v u t r

v v r

uv r

u z

w

w

1 r2

2

2 v r2

u 2

(7.38)

2

u z2

gr ,

v z

1 p r

1 rv r r r

2

1 r2

2 u r2

v 2

2

(7.39)

v

g ,

z2

and w w u t r

v w r

w p z

w z w 1 r r r r

1 r2

2

w 2

(7.40)

2

w z2

gz .

The Navier–Stokes equations and the continuity equation govern the flow of a fluid, but a general solution for the velocity field does not exist. This system consists of four equations and four unknowns (u, v, w, P) and the momentum equations might

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be solved for the velocity components, if P(x, y, z) is known. Unfortunately, the mass conservation equation does not include pressure, complicating the solution to the system. Another difficulty arises in the advective acceleration, which includes the nonlinear terms u u x iˆ and v v y ˆj . These nonlinearities are an important reason that configurations that admit solutions are few and only for idealized conditions. However, like other efforts in this text, a combination of exact, approximate, and scaling results (the “five o’clock solutions” of Section  1.4) may be used to make estimates about the flow behavior. These efforts may provide enough information for an engineering decision or, if these results are not sufficient, they will inform modeling choices made in subsequent studies using computational fluid dynamics (CFD).

7.4

BOUNDARY CONDITIONS FOR FLUID FLOW

To attempt any exact, approximate, or numerical solutions to the Navier–Stokes equations and the continuity equation, boundary conditions for the velocity components are required. The two types of conditions considered here are based on the velocity and the shear behavior at the domain boundaries. The most common boundary condition for fluids is that of a continuous velocity between the fluid and whatever material defines its volume. While in most situations the boundary is a solid, it can be another fluid. In any case, in the limit as the boundary is approached from both sides, the velocities in the two fluids approach one another. Thus, a fluid at the macroscopic scale appears to “stick” to whatever bounds it; this behavior is known as the no-slip condition. lim ut n

0

lim ut

fluid

n

0

surroundings

,

(7.41)

where ut is the velocity tangential to the interface and n is the direction normal to it. In the special case of a solid, stationary boundary, the boundary conditions are ut fluid 0, the no-slip condition, and un

fluid

0,

(7.42)

because the boundary is impermeable and flow does not pass through it. At the fluid volume boundary, we usually find a continuous shear also. For a Newtonian fluid, this continuity can be written with the aid of Eq. (1.5) as

interface

ut n

,

(7.43)

fluid

where the interface shear is proportional to the change of the tangential velocity in the direction normal to the surface. At a solid boundary, this shear is not generally known and indeed may be one of the important outcomes of the analysis, but it can be found once the velocity field

297

Equations of Fluid Motion

is known. At a fluid-fluid boundary, we can equate the continuous Newtonian shear from both sides ut n

interface

ut n

fluid 1

.

(7.44)

fluid 2

In some rare cases, these conditions can be modified as a velocity jump at a solid wall proportional to the shear stress at that boundary: ut

fluid

ut

wall

interface

LS

ut

fluid

n

.

(7.45)

interface

The coefficient LS is the slip length, a strong function of fluid and wall properties and flow geometry. If LS ≠ 0, then there is slip at the wall and the fluid is not perfectly “sticky” there. To define a regime in which there is this velocity discontinuity, we examine the value of the Knudsen number (Kn) in a flow, Kn

/ LC ,

(7.46)

where is the mean free path of atoms or molecules in the fluid and LC is a characteristic length scale based on the flow geometry. In most engineering applications, Kn < 0.01 and the no-slip condition is a reasonable approximation. In rarified gases or in very small spaces, the Knudsen number may be larger, as the mean free path becomes a significant fraction of the length scale of the flow. If 0.01 < Kn < 0.3, the use of the slip boundary condition (7.45) is a better model than (7.41). If Kn > 0.3 or so, then the assumptions of a continuum fluid break down and we require for this regime a different model of flow altogether. Discussions about flows in these regimes (Kn > 0.01), including estimates of LC and LS, can be found in books on microfluidics [2, 3] and rarefied gases [4, 5]. In this text, all problems are in the range of Kn > 1.

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FIGURE 8.6 Nondimensionalized average velocity at blade exit and final wet tape thickness ( = u /V = Hf/H) as a function of the pressure to shear ratio, .

FIGURE 8.7 Effect of flow on fiber orientation during tape casting in a Π > 1), as in Figure 8.7,

319

Internal Flows

the velocity gradients rotate the fibers so that their long axes align with the direction of the flow. The drag on each part of a fiber in the flow direction is proportional to the local velocity squared. In a uniform velocity field, the drag is uniform and the fiber is carried with the flow without change in orientation. In a nonuniform velocity field, the drag varies along the length of the fiber, which variation imparts a moment to the solid, the net effect of which is rotation of the fiber. That moment and rotation ceases when the fiber is stably aligned with the local flow. This effect can be seen in a schematic representing an experiment with fibers in transparent corn syrup (Figure 8.7).

8.5

FLUID FLOW IN A VERTICAL CYLINDRICAL TUBE

Flow in a cylindrical tube is another common configuration, and we use the same approach as in the previous section to find a solution for flow rates and shear stress. To illustrate the effect of gravity in non-horizontal tubes, this mechanism is also included. In Figure 8.8, a vertical tube is shown, in which the sign convention is that a positive axial velocity, w > 0, is upwards against gravity in the positive z direction. The fluid velocity in contact with the inner tube wall (r = R) is zero and, by symmetry, w has its maximum value at the tube axis (i.e., at r = 0). The flow is driven by an imposed pressure gradient along the pipe and is resisted by the shear stress, rz, arising from friction at the no-slip boundary at the outer wall; gravity may assist or resist motion depending on the flow direction. We assume a steady, fully developed (∂/∂z = 0) flow, and Figure 8.8 shows the pipe in a cylindrical coordinate system (z, r, ). If the flow is also axisymmetric (∂/∂  = 0), the continuity equation (7.14) is ru

1 r

1 v r

r

w z

0

or

ru

(8.25)

C,

So the radial velocity u = 0 everywhere, given the no-slip condition at the outer surface (r = R). These results and simplifications are applied to the z direction momentum equation in cylindrical coordinates, Eq. (7.40): w t

u

w r

v w r

w 1 r r r r

p z

w z

w 1 r2

2

w 2

(8.26)

2

w z2

gz

or 0

p w r z r r r pressure friction

gz gravity

(8.27)

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 8.8 The geometry for a one-dimensional flow in a cylindrical tube.

Multiplying by and integrating over r, w r

r

1 2

dp 2 r dz

g

C1 .

(8.28)

The integration constant C1 = 0 because symmetry gives ∂w/∂r = 0 at the centerline. For a Newtonian fluid, w 1 r 2

dp r. dz

g

(8.29)

A second integration then gives 1 4

w

g

dP 2 r dz

C2 .

(8.30)

Applying the no-slip condition, w(r = R) = 0, C2

1 4

g

dP 2 R dz

(8.31)

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Internal Flows

and thus R2 4

w

g

dP dz

2

r R

1

.

(8.32)

Defining the uniform pressure gradient in terms of the pressure difference ΔP  = PU – PD over a length of pipe L, R2 4

w

P L

r R

1

g

2

.

(8.33)

(Note that our sign conventions in Figure 8.8 has the upstream pressure at the bottom and the downstream at the top.) The axial velocity thus has a parabolic shape, as shown in Figure 8.9, which also shows the linear variation of rz across the diameter of the tube. The maximum value of w, which occurs at r = 0 is R2 4

wmax

P L

(8.34)

g .

The average value of the flow velocity is defined as w

R2

A

w dA

R2 2

2

R

0

0

P L

g

R2 2

w r dr d 1 2 r 2

1 r4 4 R2

R2 8

P L

R

0

P L

g

R2 2

P L

R 0

1

r2 rd R2

g

R2 . 4

Finally, w

g ,

(8.35)

which is half the value of wmax. The volume flow rate Q is then Q

w

R2

R4 8

P L

g ;

(8.36)

this relationship is known as the Hagen–Poiseuille equation. The mass flow rate is m

w

R2

R4 8

P L

g .

(8.37)

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If we nondimensionalize the profiles of axial velocity, Eq. (8.33), and shear stress, Eq. (8.29), we obtain these forms: w gR 2

w*

1 4

1 1

r R

2

(8.38)

and T*

rz

gR

1 1 2

r , R

(8.39)

where P , gL a ratio of the applied pressure drop along the tube and the hydrostatic head. With these expressions, plotted in Figure 8.9, we can see that the shear stress and velocity profiles are only functions of the nondimensional radius, r/R, and . In this form, it is easy to see that there is no flow (w = 0) and so no shear ( rz = 0) when  = 1, or the imposed pressure drop is positive (ΔP > 0) and the same absolute value as the hydrostatic head. In that case, the pressure drop is imposed to keep the fluid from draining through the tube. If the pressure difference increases (larger ), we see in Figure 8.9 that the velocity increases as ( −1) and the shear stress opposing the flow changes with (1− ). The reverse is true as the pressure drop is imposed so that PD > PU (ΔP and < 0) and is steadily more and more negative. If | | >> 1 and so −1 ≈ , then the hydrostatic contribution is negligible, regardless of the flow direction.

Example 8.2 Introducing Chlorine Gas into a Metal Bath Zinc is removed from liquid lead by reaction with chlorine gas to form ZnCl2. Figure 8.10 shows the chlorine gas being bubbled gently through a bath of liquid lead at 400 oC. The graphite bubbling tube has an internal diameter of 1.5  mm and is immersed to a depth of 1 m in the lead. The atmospheric pressure at the liquid surface is 101.3 kPa. At 400 oC, Pb

10, 560 kg/m 3

Cl2

2.87 10 5 kg/ms.

Find: The static pressure in the tube at the level of the surface of the bath required to give a bubbling rate of 0.2 kg Cl2/hour. Solution: The first step is to write the proper relationship among the material properties, process geometry, desired flow rate, and required pressure at the top of the tube. Assuming a laminar flow, Eq. (8.37) gives that relationship, here rearranged as Ptop

Pbottom

8m L R4

g L,

Internal Flows

323

FIGURE 8.9 Radial profiles of nondimensional (a) axial velocity and (b) shear stress in a cylindrical pipe, over a range of imposed pressure drop-hydrostatic head ratios ( ).

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 8.10 Schematic of system for bubbling chlorine gas through liquid lead to capture zinc dissolved in the bath.

where the pressures are at the top and bottom of the inside of the tube. To push the chlorine out of the tube, the pressure needed at the top (Ptop) is a sum of the three terms on the right-hand side of this equation, representing pressure at the exit, the pressure drop due to friction in the tube, and the weight of the column of gas. The pressure at the bottom of the tube is the hydrostatic pressure under the weight of the lead to the depth of the tube: Pbottom Pbottom

Patm

Pb

gL

101, 000 Pa 10, 600 kg/m 3 9.81 m/s2 1 m

205, 000 Pa.

To find the value of the next two terms, the density of the gas is needed. Treating chlorine gas as an ideal gas of molecular weight (MW) 0.07091 kg/gmol, the density of the gas is

1 kg V

P MW RT

205, 000 Pa 70.91

kg kgmol

kg m 2 673K 8.314 2 s K kgmoll

2.60 kg/m 3 .

325

Internal Flows

Plugging these values into the previous expression for the top pressure gives: 8 Ptop

205,000 Pa

kg Cl 2 2.87 10 5 Pa s 1 m hr s kg 2.60 3 (0.00075m )4 3600 h m 0.2

2.596

kg m3

9.81

m s2

1m .

Note that the sign of the mass flow rate is negative, as the mean velocity is in the negative z-direction. Ptop

205, 000 Pa

4, 900 Pa

25 Pa = 210, 000 Pa

So the pressure needed at the top of the tube is a little over two standard atmospheres. Most of this pressure is required to overcome the high pressure due to the hydrostatic head in the lead at the tube exit (205 kPa). The pressure drop in the tube (4.9 kPa) and the effect of the weight of the gas (0.025 kPa) are very small in comparison due, respectively, to the low viscosity and density of the gas compared to liquids. This result should be expected because the quantity ≈ 8200 in this case, showing that the hydrostatic term in the tube is negligible compared to the pressure drop required by the lead. We have assumed that the analysis in this section applies to the conditions in this example, particularly that the flow is laminar. For laminar flow in a tube, the Reynolds number based on the diameter and the mean velocity must be below approximately 2100. Writing ReD in terms of the mass flow rate and calculating its value, Re D

wD

4m D

4 0.2 kg/hour 3600 s/hour

0.0015m 2.87 10 5 kg/ms

1640,

we see that the flow is indeed laminar.

8.6

CAPILLARY FLOWMETER

In the preceding derivations, fluid flow was considered between horizontal plates, and in a vertical cylindrical tube. In both cases, both the shear stress and velocity of the fluid are proportional to ΔP/L – g cos . In the case of horizontal flow,  = 90° and only ΔP/L appears in the expressions for and w; in the case of inclined flow, ≠ 90° and cos ≠ 0, so the expressions for and w contain both terms. The significance of the terms ΔP/L – g cos can be appreciated from an examination of the capillary flowmeter shown in Figure 8.11. This simple device is a tube, narrowed over a length L to a capillary tube of inside diameter d inclined at an angle to the vertical, with a U-tube manometer containing a fluid of density m, placed across the length of capillary. The diameter of the capillary section is much less than

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 8.11

Schematic of capillary flowmeter.

the diameter of the rest of the tube so that practically all the resistance to flow over the manometer is due to friction there. When a second fluid of density f, immiscible with the manometer fluid, flows through the capillary tube, a difference in height h occurs between the levels of the fluid in the two arms of the manometer. If Po is the static pressure in the flowing fluid at the entrance to the capillary and PL is the static pressure in the fluid at the point of exit from the capillary, the pressure at the level x at the bottom of the manometer, Px, is found from consideration of the left arm of the manometer, Px

pressure due to the weight of the column of manometer flluid of height h1

h

pressure due to the weight of the column of second fluid of height h2

PL

h1 h

m

g

h2

f

PL .

g

From consideration of the right arm of the manometer, that same pressure can be written as Px

pressure due to the weight of the column of manometer flluid of height h1 pressure due to the weight of the column of second fluid of height h h2 h1

m

g

Lcos

Po

h

Lcos

h2

f

g Po .

327

Internal Flows

Equating these expressions for pressure at the bottom of the manometer and rearranging gives PL Po

h

m

f

g L cos

f

g,

and hence the pressure difference measured at the manometer, h ( h

m

f

g

L cos

PL Po

f

m

– f) g, is

g

or h

m

f

g

P L

L

f

cos g.

(8.40)

(The pressure difference is defined as in Eq. (8.33), where PL = PD and Po = PU.) Thus, the pressure difference measured at the manometer includes the decrease in the static pressure across the capillary and the contribution due to the influence of gravity on the fluid, and consequently, use of the flowmeter does not require knowledge of the angle of inclination of the capillary.

Example 8.3 Measuring Volume Flow Rate Using a Capillary Flowmeter Consider the use of the capillary flowmeter shown in Figure 8.11 to measure the volume flow rate of water. The manometer fluid is an organic liquid, which is immiscible with water. When the water flows through the capillary (L = 0.1 m and d = 0.005 m), the height difference between the levels of the manometer fluid is h = 0.030 m. The thermophysical data are 1154 kg/m 3

m

998 kg/m 3

f

f

9.44

10 4 kg/ms.

Find: The volume flow rate of water, Q. Solution: The pressure difference measured by the height difference in the U-tube, h, is

m

f

gh

1154 998

kg m3

9.81

m s2

0.03 m

45.9

kg m s2

and, using Eq. (8.40), m

f

L

gh

45.9

kg m s2

1 0.1 m

459 kg / m 2 s2

P L

f

g cos .

From Eq. (8.36), modified to include the inclination angle (g cos for the inclined pipe rather than g for a vertical one), the volume flow rate is

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An Introduction to Transport Phenomena in Materials Engineering

P L

Q

f

g cos

R4 8 Q

0.0025 m

kg 459 2 2 ms 7.46 10

6

4

8 9.44 10 4 ms/kg

m 3 /s.

The use of this relation for flow in the capillary tube assumes laminar flow. To confirm this assumption, we find the Reynolds number based on the average flow velocity. That velocity is Q R2

w

7.46 10

6

m 3 /s

0.0025m

0.38 m/s,

2

and Red is Re d

f

998 kg/m 3 0.38 m/s 0.005m

wd

9.44 10 4 kg/ms

f

2008.

Because Red < Red,trans (≈ 2100 for flow in a tube), we see that the flow is indeed laminar.

8.7 NON-NEWTONIAN INTERNAL FLOWS So far we have concentrated our effort on Newtonian fluids with their simple constitutive relation between shear stress and velocity gradient (or strain rate). In this section, we explore the flow of non-Newtonian fluids (introduced in Section 1.3.1.2) in some of the simple, one-dimensional geometries in previous sections. The first two examples are with power-law fluids and the third with a Bingham plastic. More such solutions are found in books on rheology (e.g., [7, 8]) and in some transport phenomena texts [9, 10].

8.7.1

sHear-driven Flow oF a power-law Fluid

We start with the same conditions as the shear-driven flow of a Newtonian fluid between flat parallel plates (Section 8.3). Figure 8.1 shows the geometry and boundary conditions and the appropriate static force balance is 0.

y

(8.8)

For a power-law fluid,

y

0

y

k

u y

n

,

(8.41)

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Internal Flows

which, when integrated over the layer thickness, is C1*

y

k

u y

n

dy

1/ n

C1* k

u y

k

n

.

(8.42)

u y

C1

Integrating again gives u

C1 y C2 ,

and applying the boundary conditions of no slip on the bottom plate and u = V on the top, u

y . H

V

(8.43)

This velocity solution is plotted in Figure 8.1 and is the same as for Newtonian fluids, Eq. (8.10), as are the mass and volume flow rates, Eq. (8.14). The shear stress is different: V H

k

yx

n

.

(8.44)

For shear thinning fluids, the shear is less sensitive to V and H than in Newtonian fluids; for shear thickening, the opposite is true.

8.7.2

pressure-driven Flow oF a power-law Fluid

Once again we revisit a solution from Section 8.4, in which we use the power-law constitutive relation, Eq. (1.15), to find the flow driven by a uniform pressure gradient between two stationary, parallel flat plates. For a power-law fluid, the force balance dP dx

y

(8.15)

becomes y

k

u y

n

dP . dx

(8.45)

In the Couette flow example in Section 8.7.1, the velocity gradient is positive, so its presence under the non-integer exponent in Eq. (8.45) is not a mathematical difficulty. However, in this case, the velocity gradient is negative for y > 0. To address this issue, we rewrite Eq. (1.15) as

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An Introduction to Transport Phenomena in Materials Engineering

yx

u y

app

u k y

n 1

u y

(8.46)

so the shear stress and velocity gradient have the same sign (as expected) and the apparent viscosity is always positive. Eq. (8.45) can be written anew as

k

y

n 1

u y

u y

dP , dx

which integrates to u y

k

n 1

u y

dP y dx

C1* .

Because the velocity gradient is negative for y > 0, u y

u y

and k

u y

n 1

u y

u y

k

n

dP y dx

C1* .

(8.47)

Applying the centerline symmetry condition u y

0 y 0

to Eq. (8.47) gives C1* = 0. Rearranging Eq. (8.47) gives 1 dP k dx

u y

1 n

y

1 n

1 P k L

1 n

1

yn ,

which after integration produces u

1 P k L

1 n

n n 1

y

1 n n

C2 .

The integration constant is found by applying the no-slip condition at y = H, and the complete velocity solution is

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Internal Flows

1 P k L

u

1/ n

n n 1

H

1 n n

y H

1

1 n n

.

(8.48)

The average velocity across the layer and the mass flow rate are: 1/ n

1 P k L

u

1 n n H n 2n 1

(8.49)

and m

W

1 P k L

1/ n

1 2n 2n H n , 2n 1

(8.50)

where W is the width of the channel into the page and the cross-sectional area is (2HW). Eq. (8.50) gives us the relationship between the driving force, ΔP, and the flow rate, m: m ~ P1/ n . For a shear thinning fluid (n < 1), the exponent on ΔP is greater than one, so the mass flow rate increases faster than any change in pressure drop; that is, the fluid becomes progressively easier to push through the channel as the flow rate increases. The opposite is true with shear thickening fluids (n > 1). If we define a normalized velocity as u u

1 2n 1 n

1

y H

1 n n

,

(8.51)

we can examine the effect of shear thinning and thickening behavior on the velocity profile shape compared to the Newtonian case in Figure 8.3. Velocity profiles for a range of values for the flow behavior index of power-law fluids from n = 0.1 to n = 10 are compared in Figure 8.12. As n decreases from the Newtonian case (n = 1), the profile at the centerline is more and more blunted and the velocity gradient steepens near the walls. With the highest shear stress at the walls, μapp is lowest there, whereas the relatively low stress around y = 0 results in a high viscosity and low velocity gradient there. The opposite effect is seen in a shear thickening fluid.

Example 8.4 Finding the Pressure Drop Necessary to Extrude a Polymer through a Circular Die Most of the pressure drop in a polymer extrusion process is in the narrowest passage for a fluid, that is, in the die through which the polymer leaves the extruder. In this case, we are interested in extruding a high-density polyethylene (HDPE) fiber

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FIGURE 8.12 Velocity profiles of power-law fluids between parallel flat plate driven by uniform pressure gradient over a range of flow behavior index values (0.1 < n < 10).

FIGURE 8.13 Polymer extrusion through and exiting a circular die.

through the circular die shown in Figure 8.13. Assuming a fully developed flow in the die and treating HDPE as a power-law fluid, we would like to know the pressure drop over the die length L from point 1 to 2 necessary to maintain a volumetric flow rate of Q = 10–4 m3/s. For comparison to the viscosities of more common fluids (e.g., water), it is interesting to calculate the apparent viscosity profile in the flow. R

0.005m

0.025m

L

n

0.4

k

In this cylindrical geometry, the static force balance is P z

1 r r r

rz

or, for a power-law fluid, P L

1 rk r r

w r

n

.

20 kPa s

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Internal Flows

The solution, with no slip at r = R and symmetry at r = 0, is 1 P 2k L

n

w

1 n

1 n

R

1 n n

r R

1

1 n n

.

Find: (a) The pressure drop to maintain the specified flow rate and (b) apparent viscosity as a function of radial position. Solution: (a) The volumetric flow rate is related to the pressure drop through the average velocity across the die, so we begin by finding an expression for w. 1 R2

w

2

R

1 n

1 n n 1 P R n 2k L 1 3n

w r dr d 0 0

The volumetric flow rate is then w Ac

Q

w R

n

2

1 3n

R

1 3n n

1 P 2k L

1 n

.

Rearranging this equation for the pressure drop, P

Q

n

1 3n n

n

2k L . R1 3n

Using the data given, we can calculate the pressure drop: P

10

4

m3 s

0.0251

0.4

1 3(0.4) (0.4)

m1.2 s0.4

0.4

2 20,000 Pa s0.4 0.025m 0.005m

1.25 1.15 108

Pa s0.4 m1.2

1 3( 0.4)

3.63 MPa = 36.3 bar.

This result is useful not only for selecting an extruder that can generate that pressure but also for selecting die material and designing the fixture to withstand this internal pressure. (b) The apparent viscosity is a function of the velocity gradient in the die: w r

1 P 2k L 7.94 10

1 n

8

r

1 n

1 2 20,000 Pa s0.4 r m

1 0.4

1 . s

6

3.63 10 Pa 0.025m

1 0.4

1

r 0.4

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 8.14 Variation of apparent viscosity and velocity gradient for flow of HDPE through a circular die in Example 8.4.

The apparent viscosity for a power-law fluid is defined by Eq. (8.46):

app

w k r

n 1

20, 000Pa s

0.0914 Pa s

0.4

7.94 10 r m

r m

8

1 0.4

0.4 1

1 s

1.5

.

Figure 8.14 shows both the velocity gradient and apparent viscosity across the die. The gradient is highest at the outer wall (r = R), where the viscosity is lowest, as we expect in a shear thinning fluid. As r 0, the velocity gradient goes to zero due to the system’s symmetry, which causes the apparent viscosity there to approach infinity. As noted in Chapter 1, the flow field is predicted well at the wall, but is less good near the center line, where the viscosity really should approach the zero shear viscosity. Because the shear is so small in this region, the use of a power-law model in this case has little effect on the important result (the required pressure drop).

8.7.3

pressure-driven Flow oF a bingHaM plastic

Another type of non-Newtonian fluid requires a minimum critical shear stress ( c) below which no deformation occurs and above which it flows according to one of the many models presented in Chapter 1. A Bingham plastic is a fluid that does not deform below c and behaves like a Newtonian fluid above that critical shear stress [8]. For one-dimensional flows, yx

c

u y

for

yx

c

.

(8.52)

335

Internal Flows

and u y

0

for

yx

c

.

(8.53)

Some fluids have a nonlinear response past the critical shear stress, and the Herschel–Bulkley model may be used [8]:

yx

c

k

u y

n

for

yx

c

,

(8.54)

The approach to solving flow fields in fluids with a minimum critical stress for motion is the same as previous cases, but with a few complications. Taking pressure-driven flow between the two flat parallel plates (a distance 2H apart) as an example (Figure 8.15), the motion of a Bingham plastic is divided into two sections, one with high enough shear stress to cause deformation and one in which there is no relative motion (slug flow). Starting with the familiar static force balance for this configuration, yx

y

P , x

(8.15)

the constitutive relation in Eq. (8.52) is inserted and the procedure developed in previous sections is applied. Integrating twice and applying the no-slip condition at the walls (y = ±H) and zero shear at the centerline (due to symmetry), gives the velocity field in the sheared region as u( y)

1 2

P H2 1 L

y H

2 c

H 1

y , H

(8.55)

FIGURE 8.15 Nondimensional velocity profiles for the flow of Bingham plastic between parallel flat plates, showing regions of deformation (|y| > Hc) and slug flow (|y| < Hc).

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An Introduction to Transport Phenomena in Materials Engineering

and the velocity at the critical shear is uc

1 2

Hc H

P H2 1 L

2 c

H 1

Hc . H

(8.56)

This velocity is uniform in what is called the slug flow region around the centerline, |y| ≤ Hc, where the position of the transition between the two regimes is Hc

L

P dx

c

c

P

.

(8.57)

In the low stress, slug flow region where |y| < Hc, there is no relative motion, so the velocity has uniformly the value from Eq. (8.56). Eq. (8.58) shows the normalized velocity profiles ( ) vs. the normalized height ( ), 1 1 2

2 c

1

where 1

(8.58)

c

and the core velocity, c, is given by Eq. (8.59):

c

1 1 2

2 c

c

1

where

c

c

.

(8.59)

The normalized variables are defined by u

L , P H2

y , H

and

c

Hc H

c

H

L P

.

(8.60)

Figure 8.15 shows the steep velocity gradients near the walls, where the shear stress is high enough to initiate deformation. The flat regions near the center indicate stress lower than critical. When c is high, the pressure gradient driving the flow is weak compared to the shear required to begin deformation, so the slug flow covers most of the channel and the maximum flow velocity is low. A flow with lower values of c is easier to deform because the pressure gradient is stronger relative to the critical shear stress. In that case, more of the channel has deforming fluid and the maximum velocity increases. As c 0, the solution approaches the behavior of a Newtonian fluid derived in Section 8.4.

8.8

FLOW THROUGH POROUS MEDIA

8.8.1 resistance to Flow A porous medium is a rigid, solid structure with a complimentary, continuous system of void space through which a fluid may move. Interest in flow through such structures

337

Internal Flows

began with the study by Henry Darcy of flows in the water system in Dijon, France [11], and has found wide application in the field of groundwater flows. Applications in materials processing include infiltration of metal into ceramic preforms or polymers into mattes of carbon fiber (to make metal-matrix or carbon-fiber-reinforced composites, respectively), the movement of liquid metal through dendritic arrays during alloy solidification, and the flow of water into a mold during slip casting of a ceramic slurry. The structure of the solid in a porous medium, or packed bed, may be as regular as uniformly sized spheres or fibers, or may be made of particles of much more irregular shapes and over a distribution of sizes. The structure may also be made of such particles and then sintered to give it some rigidity. Whatever the size and shape of the solid and the voids, we assume that their length scale is much smaller than the size of the entire bed (as in Figure 8.16). Also, the porosity of the bed is continuous along its length and the structure is evenly distributed over the cross-section so that there is a spatially uniform resistance to flow. The latter feature prevents a preferential flow, or channeling, in some areas of the bed. These assumptions are made to describe the simplest of porous media. An example of a cross-section of a bed is shown in Figure 8.16, which has an area normal to the bulk flow of Abed, there is a much smaller area through which the fluid can flow (the cross-sectional area of the voids, Avoid). The void fraction of the bed, , is defined as Avoid Abed .

(8.61)

Measurement of the volume flow rate of fluid through the bed, Q, allows the calculation of two average velocities: the superficial velocity, us , calculated as us

Q Abed

and the actual average velocity of the fluid through the voids, u , Q Avoid .

u

From Eq. (8.61) the two velocities are related as Q

u

Abed

us Abed

or u

us .

(8.62)

Given the geometric complexity of a packed bed, with its many twisty passages mostly alike, we immediately see that the motion of the fluid in it will be much more complicated than the relatively simple flows discussed earlier in this chapter. The

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An Introduction to Transport Phenomena in Materials Engineering

FIGURE 8.16 Example of structure of porous medium, showing the difference in length scale of the entire packed bed (on left) and the detailed structure (on right), assuming ℓ/L 1, the permeability is very small, Darcy drag dominates the flow resistance, and the velocity profile is very close to slug flow (a uniform velocity field).

8.8.2

eFFect oF porous Media structure on Flow

While we can write force balances and their solutions for the macroscopic velocities in a porous medium, one aspect that must be addressed is the connection between the flow resistance (or permeability) as a function of the structure of the medium. The average axial velocity w along a horizontal pipe (Eq. (8.35) with no gravity term) is written as R2 P . 8 L

w

This relation can be adapted to handle arbitrarily shaped passages thus w

K1

Rh2

P L

(8.80)

Rh2

P , L

(8.81)

or ws

K1

where K1 is an empirical constant of proportionality. The hydraulic radius, Rh, of the passage is defined as the ratio of the void volume (Vvoid) and the solid surface area wetted by the fluid (S ): Rh

Vvoid . S

(8.82)

This definition approximates the effect of the porous medium geometry through use of a simple quantity into which is lumped more complex information about that geometry. With a bed of volume Vbed and porosity , the volume of the voids is Vvoid

Vbed

343

Internal Flows

and the volume of the solid is Vs = (1 – )Vbed. Thus, if S’ is the total surface area of the solid, then S0, the total surface area of the solid per unit volume of solid, is S

So

1

(8.83)

Vbed

and Rh

Vbed So 1

.

So 1

Vbed

(8.84)

Substitution of Eq. (8.84) into Eq. (8.81) gives 3

K1 P L So2 1

ws

2

.

For laminar fluid flow through the packed bed, K1 has been evaluated as 1/4.2, so that 1 4.2

ws

P L So2 1

3 2

(8.85)

.

Equation (8.85) is known as the Blake–Kozeny equation. Comparison with Darcy’s law in Eq. (8.63), shows that the Blake–Kozeny equation gives us a model for the permeability, 1 4.2 So2 1

K

3

(8.86)

.

2

The permeability is thus determined only by the porosity of the bed and geometry of the solid structure. Again, transition from laminar to turbulent flow begins at a critical value of the Reynolds number, which is defined for packed beds as follows: Re

wD

ws

2 Rh

2

ws So 1

.

(8.87)

The length scale used, D = 2 Rh from Eq. (8.84), reflects the size of the passages through which fluid flows, not the size of the bed. For packed beds, the factor of 2 in Eq. (8.87) is omitted and a new Reynolds number, Rec, is defined as Re c

ws So 1

,

(8.88)

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An Introduction to Transport Phenomena in Materials Engineering

where the passage length scale is Rh. Flows are laminar through the packed bed of Rec < 2 and the transition from laminar to fully turbulent flow occurs in the range 2 < Rec. < 1000. Turbulent flow requires the introduction of the Darcy friction factor, which is obtained from Eq. (6.9) as f

D L

4 So 1

P 1 w2 2

2

P L

ws2

.

(8.89)

For fully turbulent flow, the friction factor has a constant value of 0.292, and for Rec > 1000, ws2

0.292 So 1

P L

4

3

.

(8.90)

Eq. (8.90) is known as the Burke–Plummer equation. Note that in this regime, the drag goes like the square of the more vigorous velocity, compared to the linear drag of Darcy’s law, Eq. (8.63). The variation of ws with ΔP/L over the entire range of Reynolds numbers is given by the sum of Eqs. (8.85) and (8.90) as P L

2

4.2 So2 1

ws

3

4

Equation (8.91) is simplified by dividing both sides by bining with Eqs. (8.88) and (8.89) to get f

16.8 Re c

ws2

0.292 So 1

0.292,

3

ws2 1

.

(8.91)

So 4

3

and com-

(8.92)

which again shows that the friction factor is a function only of the Reynolds number. Thus, for a required value of ws through the packed bed, Rec is obtained from Eq. (8.88), f from Eq. (8.92), and ΔP/L from Eq. (8.91). At very low Reynolds numbers, the term 16.8/Rec in Eq. (8.92) is much larger than 0.292 and thus the Blake–Kozeny contribution to Eq. (8.91) predominates, and at very large Reynolds numbers, 16.8/Rec 1, we set the porosity at its maximum value of  = 1. The variation of with us in the range of 0.75 m/s < us < 10 m/s, is obtained from Eq. (8.106) and is plotted in Figure 8.21. As in Figure 8.20, minimum fluidization conditions in Figure 8.21 are marked as point C and the elutriation as point D.

FIGURE 8.21 Variation of fluidized bed porosity with average velocity and Red .

354

8.10

An Introduction to Transport Phenomena in Materials Engineering

SUMMARY

The behavior of simple internal fluid flows is determined by the geometry of the containing duct, the magnitude of the force causing the flow, and the properties of the fluid. At low enough flow velocities laminar flow occurs and is characterized by lamellae of fluid sliding over one another, with no macroscopic mixing of the fluid in directions normal to the direction of flow. The fluid in contact with the walls of a containing duct is generally stationary, and velocity gradients are established in the fluid in directions normal to the flow. The shear stresses, velocity distributions, and hence average linear velocities and volume and mass flow rates are determined solutions to the continuity and momentum equations. These governing equations for flow are simplified by the fully developed and other approximations and then integrated twice to yield the functions of shear stress and velocity with position in the flow. The two integration constants are obtained from boundary conditions for the particular type of flow. Applications shown here include combinations of shear-, pressure-, and gravity-driven flows in Cartesian and cylindrical conduits. These flows are found in both Newtonian and non-Newtonian fluids, and in packed and fluidized beds.

8.11

HOMEWORK PROBLEMS

(Always begin with a clear sketch of the physical problem. Do not plug in values for variables until absolutely necessary.) 8.1

8.2

8.3

8.4

A shaft of outer diameter D is rotating at revolutions per second inside an outer, cylindrical casing with a thin film of lubricant (viscosity μ and thickness ) in the annulus between the shaft and the casing. (a) Write an expression that relates the torque, T, required to overcome the friction in the fluid to D, r, μ, and . (b) Calculate the torque when D = 2.5 cm,  = 0.1 mm, r = 1000 rpm, and μ = 3 × 10–3 Pa∙s. (c) Show that Couette flow is a reasonable approximation. The shaft in Problem 8.1 is rotated in the hole, and it is found that a torque of 1 N∙m is required to sustain a speed of rotation of 60 rpm. (a) Calculate the viscosity of the lubricant in the annulus between the shaft and the block. Ignore any end effects. (b) Assume a linear velocity gradient through the lubricant and show why that approximation is reasonable. Two immiscible incompressible fluids are in horizontal layers between two parallel plates. Both plates are stationary. The two layers between the plates of length L, width W, and vertical separation 2 are subject to a uniform pressure gradient dp/dx. Fluid 1 in the lower layer is more dense and more viscous than fluid 2 in the upper layer, and the two layers have the same thickness, . (a) Sketch the velocity fields in fluids 1 and 2. (b) Simplify the Navier–Stokes equations and solve for u1(y) and u2(y). (c) Write expressions for the mass flow rate in each layer. Sketch the probable orientation patterns (textures) of long fibers in the reservoir and under the doctor blade in tape casting for cases of Π >> 1 or Π 1. (Magmas, while not engineering fluids, also have large Pr; being molten ceramic mixtures, they have low k and high μ.) On the other hand, liquid metals have a kinematic viscosity on the order of magnitude as water, but possess a much higher thermal diffusivity, so their Pr 100 polymers, magma

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An Introduction to Transport Phenomena in Materials Engineering

dimension (the direction of motion, or the axial direction) and treating heat losses in perpendicular directions as heat sinks. In order for this approximation to be valid, the heat flow in the body must be oriented so that it is mainly in the axial direction. If the heat flow in the direction of motion is much greater than the directions normal to motion, then this one-dimensional approximation is reasonable. If the moving body can be modeled as one-dimensional, then we can define a control volume over which we can perform an energy balance in order to derive a conservation equation for thermal energy in terms of temperature. Figure 10.5 shows such a control volume, defined by length dx, cross-sectional area Ac, and perimeter p, and in which thermal energy is transferred by conduction (qx) and advection in the x direction. The amount of energy that is brought into the control volume at location x VAc cT x , where ex is the specific enthalpy at x and by bulk solid motion is m ex V is the speed of the moving body. The rate at which energy is advected out of the volume at (x + dx) can be different and is written as m ex dx VAc cT x dx . Also, heat can be generated in the volume (q) and also lost to the ambient by convection from the surface. (This convection loss moves normal to the axial motion. To use a simple one-dimensional model, we treat convection as a heat sink, a “negative generation,” rather than as a boundary condition in y or z for a two-dimensional or three-dimensional model.) For the rest of this derivation, we will assume mass flow rate, m, material properties, heat transfer coefficient (h), and geometry do not change along the direction of motion (x). The first law of thermodynamics (energy is conserved) for this control volume can be written as the usual energy rate balance: Uin

Uout

U gen

U stor ,

(2.19)

where heat flows in the x direction by conduction and advection, and the process is steady state. Uin U out

q x m ex qx

dx

qx

m ex

dx

VAc cT qx

dx

x

U gen VAc cT

q Ac dx x dx

U stor

h p dx (T T ) 0.

FIGURE 10.5 Control volume (in gray) for analysis of moving rigid body.

(10.13)

415

Convection Heat Transfer

Note that only the part of the body that actually exchanges heat with the environment is counted in the perimeter, p. Writing a Taylor series expansion of the specific enthalpy at (x + dx), we obtain ex

dex dx dx

ex

dx

higher order terms.

Using this expression to find the difference between the energy advected in and out of the control volume, we get m (e x

ex )

dx

VAc c

dT dx. dx

(10.14)

Similarly, we can find the change in the diffusion heat transfer over dx: qx

dx

qx

d dT k Ac dx. dx dx

(10.15)

Putting these relations into Eq. (2.18), we get the energy conservation equation that describes the temperature along the length of the moving body. d

VAc cT dx advection

d dT kAc dx dx conductiion

hp T T

q Ac

convection heat loss generation

(10.16)

It is useful to look carefully at this energy equation to remind ourselves of the physical phenomena that govern it. Eq. (10.16) is equivalent to Eq. (10.10) for a steady, one-dimensional flow. The first term is the change in the thermal energy of a mass as it moves through space, i.e., it is the rate of change of thermal energy advected as it passes a specified point in space. The second term represents the diffusion of thermal energy along the length of the body due to the axial temperature gradient. This conduction term happens independent of the body motion and its magnitude. The third term is the convection heat loss from the outer surface to the environment and the final term is heat generated inside the body. For uniform properties and cross-sectional area, the energy balance in Eq. (10.16) reduces to dT dx advection cV

d dT dx dx conduction k

hp T T q. Ac connvection heat loss generation

(10.17)

The boundary conditions in x for this equation are found by considering the physical configuration. As the body leaves a constant temperature source (e.g., a furnace or extruder), it likely has the temperature of that device. As the body loses heat to the environment, eventually it will cool to the ambient temperature. These boundary conditions are written as

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An Introduction to Transport Phenomena in Materials Engineering

T (x

0)

and

To

T (x

)

T .

At this point, we have an equation that can be solved for temperature as a function of 11 parameters: T(x, k, Ac, , c, V, h, p, T∞, To, q). It would simplify matters to reduce the parameter space and produce a more generalized result by nondimensionalizing the equation and boundary conditions, using carefully chosen reference values. We start here by dropping the heat generation term. (This step is not necessary, but most applications do not have this effect and it does simplify the following procedure.) We pick references for the temperature difference, T(x) – T∞, and for the axial coordinate, x, thus T T To T

and

x l.

(10.18)

The characteristic length, ℓ, is initially unknown, and its form will be chosen to reduce the number of parameters. Inserting Eq. (10.18) into Eq. (10.17) and rearranging, we get kAc To l

2

d2

VAc c To l

2

d

d d

hp To

0,

(10.19)

a second-order differential equation for  = f( ). We nondimensionalize the equation by dividing by the coefficient of the first term, and setting ℓ = (kAc/hp)½ and Pe = Vℓ/ : d2 d

V

d d

2

0,

(10.20)

we have reduced the problem to one of finding temperature, , as a function of the axial coordinate, , and the Peclet number, Pe. Introducing the Peclet number as the ratio of the speed of advection, V, and an effective thermal diffusion speed, /ℓ, the energy equation and boundary conditions are then written as d2 2

d axial

conduction

d d axiial

0

Pe

0

1

(10.21)

convection

0.

loss

The general solution for Eq. (10.21) is  = exp(-B ) where B

( Pe 2) 2 1

( Pe 2).

(10.22)

We can also find approximate solutions for different physical situations, depending on which terms (i.e., which physical phenomena) dominate the energy conservation equation. There are three interesting cases that are simplifications of the general case [3]. 1 Pe 0: In a slow-moving solid (low V), or one with a large thermal diffusivity ( ), the advection term is negligible, and relatively little heat is carried

417

Convection Heat Transfer

by the motion of the body. The heat flow is then a balance of conduction along the body and the convection heat loss: d2 d

1 2

0

1

0

1

1

0,

1

(10.23)

and the solution is 1 = e- (i.e., B = 1). This result is the classic solution for a stationary fin of infinite length [4]. 2 Pe ∞: In this situation, the advection term completely overwhelms the other effects and almost all of the heat moving downstream in x is carried by bulk motion. While conduction always occurs, its effect is negligible. If the body is moving fast enough (Pe ∞), there is not time even for convective losses to occur and the conservation equation (10.21) is: d d

2

0.

(10.24)

Integrating this equation and applying the condition at  = 0, we find 2 = 1 (B = 0), implying that there is simply no time to lose any thermal energy to the environment, there is effectively no drop in temperature. This result is interesting as a limit, but is not applicable to systems that do have a measurable heat loss. 3 Pe large, but not infinite: In this case, advection is much more important than axial conduction (as shown earlier), but is still balanced by heat loss to the ambient. Most practical materials processing problems fall into this case. Pe

d d

3

3

0

3

0

1

(10.25)

The solution for this regime is 3 = e- /Pe (B = 1/Pe), which is simpler than the general case. The validity of these three approximate cases can be seen in comparison to the exact solution in Figure 10.6.

FIGURE 10.6 Coefficient (B) in exponent in solutions of general and special cases of a onedimensional advection-diffusion problem.

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An Introduction to Transport Phenomena in Materials Engineering

Example 10.2 Cooling of a Steel Sheet Exiting a Heat Treat Furnace A steel slab emerging from a soaking furnace at 1400 K cools as it moves away from the furnace. We would like to roll the slab when its temperature is no lower than 1350 K. As a first approximation, assume that all the heat loss is by radiation and that the average temperature of the slab is 1375 K for the purpose of estimating a radiation heat transfer coefficient. The material property data, processing parameters, and geometry are: k = 31 W/mK c = 1200 J/kg K  = 7800 kg/m3  = 3.3 × 10–6 m2/s To = 1400 K T∞ = 300 K V = 0.05 m/s  = 0.7 d = 0.05 m (slab thickness) w = 2 m (slab width) Find: The approximate distance, L, the roll stand should be from the furnace to roll at T(L) ≥ 1350 K. Solution: First, we estimate the “radiation heat transfer coefficient,” taking the slab surface temperature as the average of the furnace temperature and the rolling temperature (T = 1375 K), TS2 T

(0.7)(5.67 10

8

2

TS

W/m 2 K 4 ) 1375 K

T 2

. 300 K

(1.38) 2

1375 K 300 K

132 W/m 2 K Because it is a thin, wide slab (d 1, but in general also may be / T < 1.

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Convection Heat Transfer

The primary goals here are (i) to understand the physical mechanisms important to the fluid flow and heat transfer and (ii) to estimate the surface heat flux and heat transfer coefficient. To do so, we begin with a scaling analysis of the governing equations inside the boundary layers. The analysis in Section 9.4 gives results for the flow field: vref ~ U

~ x Re x 1/ 2 ,

and

x

(10.26)

based on the boundary layer equation with no pressure gradient, u x

u

v

2

u y

u

,

(10.27)

yref ~ .

(10.28)

x2

and the reference values uref ~ U

The thermal boundary layer thickness ( T) is as yet unknown, but we might find an order of magnitude estimate if we scale the energy equation in that near-wall region, 0 ≤ y ≤ T. The thickness of the thermal boundary layer will help us find an expression for the heat transfer coefficient but, as we will see later, T is not usually the same as . In a steady condition without heat generation, the energy equation is u

2

T T v x y advection

2

T

T

2

x y2 conduction

.

(10.29)

Applying reference values from Eqs. (10.26) and using T T as a reference for the temperature difference, we get U

T , x

U

T x

T

~

2 T

T

,

T x

2

.

ref

~ TS

T

T

(10.30)

Note that the y-direction reference is defined for this equation as yref ~ T, because the thermal boundary thickness is the distance over which the temperature varies. 2 The ratio of the two conduction terms on the right-hand side is T x which we assume is very small. Except near the plate’s leading edge, x = 0, this assumption is supported by experimental observation, and we conclude that there is no significant conduction in the flow direction. Using this result, we drop the x-direction conduction term in Eq. (10.29), u

T x

v

T y

2

T

y2

,

(10.31)

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which scales as T , x advection in y

U

T x T advection in x

T

~

U

T

2

.

conducttion in y

(10.32)

The relationship between the advection terms depends on the as-yet-unknown / T ratio. Because we do not know its value, we examine thermal behavior at its limits, / T >> 1 and / T > 1 We start with / T >> 1, as shown in Figure 10.10. We see that the maximum velocity in the thermal boundary layer shown, u y T , is much less than the freestream value, so U∞ is a poor velocity reference there. We scale the velocity gradient inside the thermal layer thus, U u uref ~ ~ , y T

(10.33)

where the reference velocity for each layer is chosen to be the velocity at its edge. So, uref ~ U

T

,

(10.34)

FIGURE 10.10 Temperature and velocity profiles near an isothermal flat plate where the thermal boundary layer is much smaller than the velocity layer ( 1). T

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which is much less than U∞. If we use Eq. (10.34) to rescale the continuity equation, we find a new reference for the vertical velocity also: u x

v y

uref

0 or

~

xref

vref

and

yref

T

vref ~ uref

x

~U

T

T

x

. (10.35)

So we have the proper velocity scales inside the thermal boundary layer and we re-examine the heat balance, Eq. (10.31). T , x

uref

U

T

T

~

T

T , x

T

U

T

vref

T , x

T

T

T

U

T

x

U

2

T

~

T

T

T x

T

2

T

~ T

2

conduction in y

(10.36)

The two advection terms have the same order of magnitude everywhere, so they can be combined: T

U

T x

T

~ T

2

.

(10.37)

~ conduction

This result shows the fundamental mechanisms of convection heat transfer in the thermal boundary layer. Heat diffuses normal to the surface into the layer, and then is swept away by the advecting fluid. Rearrangement of Eq. (10.37) gives us an estimate of T, T

~ x Re x 1 2 Pr

13

.

(10.38)

Comparing the two boundary layer thicknesses, we see that their ratio is

~ T

x Re x1 2 x Re x

12

Pr

13

~ Pr1 3 .

(10.39)

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We had assumed that / T >> 1, so Eq. (10.39) tells us that this assumption is valid if Pr >> 1. One use for the previous relation is the estimation of the heat transfer coefficient, which we find from scaling the surface boundary condition. q h

h TS T

kf

dT f

TS T

dy

h~

kf

~ 0

kf

dT f dy

kf TS T

0

TS T T

~

kf T

Re1x 2 Pr1 3 x

(10.40)

hx ~ Re1x 2 Pr1 3 . kf

(10.41)

~ kf

T

The local Nusselt number is then Nux ~

A sketch of h(x), Figure 10.11, shows that h ~ x-1/2 and is infinite at the leading edge (x = 0) because the boundary layer thickness is predicted to go to zero there, suggesting an infinite temperature gradient. (Of course, in reality h remains finite at the leading edge. There the boundary layer approximations fail, especially that /x and T/x > 1. T However, this analysis actually is applicable for much lower values of Pr, indeed all the way to Pr ~ (1). The solution then is used successfully for fluids such as water and gases. 10.4.1.2 Scaling in the Thermal Boundary Layer for / T 0.2 and ReD < 10,000 (slow, laminar flow over the cylinder), FD ReD

1,

(10.79)

and, for a more general result (ReD Pr > 0.4), FD ReD

1

Re D 282, 000

58

45

.

(10.80)

This general correlation matches well the experimental data in [17]. The largest deviation from data occurs in ReD ranges where the flow pattern suddenly shifts (e.g., 7 × 104 < ReD < 105), which is where the separation point changes rapidly to a location farther from the leading stagnation point. The authors suggest that, except in that range, changes in flow patterns cause only weak transitions in value or slope of Nu D Re D and the given correlation is sufficient. For liquid metals, the Reynolds and Prandtl numbers are expected to have the same exponents (as in our previous analysis for Pr qCHF , the boundary condition jumps over to region (iii) in Figure 10.28, film boiling, which is characterized by a continuous surface vapor layer, which forms at and above the critical heat flux. This layer both insulates and lubricates the surface, requiring a ΔTe much higher than that needed for nucleate boiling and disconnecting the liquid from a no-slip condition at the surface. For film boiling on a horizontal cylinder, a combination of analysis and experimentation gives an expression for the heat transfer: Nu D, FB

hFB D 0.62 kv

vg v kv

L

v

TS

h*fg D3

TSAT

14

,

(10.162)

where D is the diameter and h*fg is the latent heat of vaporization modified to include the sensible heat required to raise the liquid temperature from Tsat to TS,FB, h*fg h fg 0.4cv Te .

(10.163)

At this point, we have seen the effect of increasing the heat flux from zero, with the heat transfer to the liquid changing regime from natural convection to nucleate boiling, then at the critical heat flux a sudden change to film boiling. This description of a heat flux–controlled process with boiling is useful to familiarize us with its different regimes, but the most common application for boiling in materials processing is quenching. Here, we begin with a surface at a high temperature (let us assume high enough to be in the film boiling regime) and the excess temperature falls continuously to zero through several boiling regimes. With the very high heat transfer rates in boiling, parts are cooled very quickly during a quench. The fast cooling of an aluminum alloy or nickel superalloy after solutioning avoids unwanted secondary solid-phase precipitation, and in steel the high cooling rate enables martensite production. The progression of boiling regimes during a quench begins with film boiling when the part is put in the quenchant. As the surface temperature drops to the Leidenfrost point (or minimum heat flux), the film begins to break up and move into the transition region. There a mixture of patches of vapor and of vigorous nucleate boiling, moving around on the surface in an unpredictable way; their motion is highly unstable and transient. This behavior also explains why the heat transfer coefficient is so high during transition and why it has the highest cooling rate. (This unpredictability makes it difficult to design devices to operate in this excess temperature range.) Once the excess temperature drops to the point of the maximum heat flux, the surface reverts to nucleate boiling and then single phase natural convection, and cooling until the surface is near the bulk temperature of the liquid.

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FIGURE 10.30 The pool boiling curve for liquid water at 1 atm, where excess temperature ΔTe= TS – Tsat. The quenching path is temperature-controlled, with an isothermal jump from film transition boiling to nucleate transition boiling.

10.8 SUMMARY Convection is heat transfer between a solid and fluid with a non-zero relative velocity. The primary physical mechanisms in convection are conduction of heat into and through the fluid near the solid surface and the advection of heat toward or away from that surface. The interaction at the surface is defined by the heat transfer coefficient, which is proportional to the temperature gradient at the surface in the fluid; the Nusselt number is the nondimensional temperature gradient. A useful fluid property in convection is the Prandtl number, the ratio of momentum and thermal diffusivities. External forced convection is heat transfer at a surface with unbounded fluid flow on one side, where that flow is driven by an imposed shear or pressure gradient unrelated to the temperature field. To understand the mechanisms of convection, we modeled flow over a semi-infinite flat plate, defining a thermal boundary layer as the region in which the temperature ranges from the surface to the freestream values. Scaling analysis in that region produces functions for thermal boundary layer thickness, the heat transfer coefficients, and Nusselt number in terms of the Reynolds and Prandtl numbers and distance from the plate leading edge. The scaling analysis also gives a relationship between Pr and the ratio of the two boundary layer thicknesses. More detailed analyses of the flat plate boundary layers are performed using integral

Convection Heat Transfer

475

analysis. However, in most realistic geometries and flow configurations, no mathematical solution is available (short of numerical analysis), but we can use the flat plate solutions as a guide to how we treat these situations using experimental data. Some experimental correlations of measurements are shown, using as correlation parameters the nondimensional variable found from scaling, e.g., Nu = f(Re,Pr). Internal forced convection occurs when a fluid at some inlet temperature interacts with the walls of a conduit at a different temperature. The heat transfer coefficient is redefined as using a different temperature difference than external flows because there is no freestream value; instead we use the local mean temperature in the conduit, which changes along the flow direction. Heat transfer coefficient correlations are presented here from analysis and experiments. Natural convection is driven by forces dependent on the density field, as it is induced by the effect of spatial variations in temperature (or composition) on the local density. Scaling analysis on the coupled thermal and momentum boundary layers on a vertical flat plate in a gravity field produces a new vigor parameter similar to the Reynolds number in forced flow. The Rayleigh number, Ra, includes the effect of buoyancy due to temperature differences. The heat transfer correlations in the form of Nu = f(Ra, Pr) are given from the scaling analysis and experimentation. Finally, the topic of boiling heat transfer is presented, examining the many boiling regimes in stagnant pools. These regimes each have different vapor-liquid interface morphologies that control the heat flux–temperature different correlation. Using the boiling curve, which shows these different regimes in heat flux–temperature space, related experimental correlations are discussed. Quenching is also discussed in terms of the temperature path on the boiling curve.

10.9

HOMEWORK PROBLEMS

(Always begin with a clear sketch of the physical problem. Do not plug in values for variables until absolutely necessary.) 10.1

10.2

A cylindrical cobalt rod is cold-drawn through a die at a velocity of V = 0.5 m/s. The resulting diameter is D = 0.002 m and the heat generation due to plastic deformation raises the temperature of the rod to 430 oC as it leaves the die. The rod is cooled by an air flow at U = 1 m/s and 30 oC perpendicular to it. (a) Estimate the heat transfer coefficient on the rod. (b) If the rod temperature must be lower than 130 oC when it is drawn again, how far apart should the dies be? Air:  = 0.8 kg/m3, c = 1000 J/kg K, k = 0.03 W/m K, μ = 2.0 × 10–5 kg/ms. Cobalt alloy:  = 8900 kg/m3, c = 450 J/kg K, k = 67 W/m K. We wish to continuously heat treat a nickel superalloy rod by induction hardening as it moves down a production line. The initial condition at x = 0 is that the metal temperature is at ambient. (a) For Vrod = 0.01 m/s and Uair = 0.2 m/s (normal to rod), calculate the heat transfer coefficient and the Peclet number. (b) Solve Eq. (10.16) for the temperature along the moving rod, except with heat generation. (c) Using the solution in part (b), estimate the required uniform volumetric heat generation rate in the rod

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An Introduction to Transport Phenomena in Materials Engineering

if we want it heated from 30 oC to 900 oC over a distance L = 10 m. Air:  = 0.8 kg/m3, c = 1000 J/kg K, k = 0.03 W/m K, μ = 2.0 X 10 –5 kg/ms. Nickel alloy:  = 8500 kg/m3, c = 450 J/kg K, k = 10 W/m K. Repeat the scaling analysis of Section 10.4.1, except assume that / T ~ O(1). Show that the estimates for T(x), h(x), and Nu(x) where / T >> 1 are still valid. Repeat the scaling analysis of Section 10.4.1, except assume that / T