Diffusion, segregation and solid-state phase transformations: Course reminders and solved problems 9782759827442

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Current Natural Sciences

Didier BLAVETTE and Thomas PHILIPPE

Diffusion, Segregation and Solid-State Phase Transformations Course Reminders and Solved Problems

Didier BLAVETTE is Professor of Physics at the University of Rouen-Normandy. He conducts his research in the Materials Physics Group (Groupe de physique des matériaux (GPM), UMR CNRS 6634) of which he was the director from 2003 to 2012. He taught for 30 years general physics, solid state physics, electronics and material sciences, in particular diffusion and phase transformations in master. He was the director of the CNRS research group "transdiff" on diffusive phase transformations (2004-2008) and of the CNRS school on this subject that was held in 2008. Thomas PHILIPPE is a CNRS research fellow at the Condensed Matter Laboratory of the Ecole polytechnique in Palaiseau, France. After his PhD with Didier Blavette at the University of Rouen-Normandy, he joined Peter W. Voorhees at Northwestern University (USA) for a post-doc. He is a specialist of nucleation, growth and coarsening in multicomponent systems. Cover illustration: The legend of cover image 1 is: 3D reconstruction of a boron enriched Cottrell atmosphere in ordered FeAl alloys analysed by atom probe tomography [24]. Boron atoms are in red whereas Al atoms are in green (iron is omitted for the sake of clarity). Al-rich (001) planes of the ordered structure are evidenced. A close examination shows the presence of an edge dislocation. Boron is shown to segregate along the dislocation line (courtesy E. Cadel, GPM, [24]). The legend of cover image 2 is: 3D reconstruction of a model nickel base superalloy exhibiting 7 nm in size aluminium enriched precipitates finely dispersed in the chromium enriched parent phase (courtesy A. Azzam, GPM, [44]). Printed in France

EDP Sciences – ISBN(print): 978-2-7598-2743-5 – ISBN(ebook): 978-2-7598-2744-2 DOI: 10.1051/978-2-7598-2743-5 All rights relative to translation, adaptation and reproduction by any means whatsoever are reserved, worldwide. In accordance with the terms of paragraphs 2 and 3 of Article 41 of the French Act dated March 11, 1957, “copies or reproductions reserved strictly for private use and not intended for collective use” and, on the other hand, analyses and short quotations for example or illustrative purposes, are allowed. Otherwise, “any representation or reproduction – whether in full or in part – without the consent of the author or of his successors or assigns, is unlawful” (Article 40, paragraph 1). Any representation or reproduction, by any means whatsoever, will therefore be deemed an infringement of copyright punishable under Articles 425 and following of the French Penal Code. Ó Science Press, EDP Sciences, 2022

Preface

This book is a translation of a previous version published in French recently. Some errors have been corrected. There remain of course some typing errors and less obvious errors, which we ask the reader to forgive. This book brings together problems that are closely related to real issues in materials science. The book also presents succinct course reminders. It is intended for teachers and students of materials science. It can also be very useful for engineers, PhD students and researchers. Diffusion and phase transformations in solids (precipitation of a new phase, ordering, growth of a thin film on a surface, oxidation of metals, etc.) constitute essential foundations in materials science. This body of knowledge is absolutely necessary for the design and optimization of materials and heat treatments leading to the desired properties whether they are structural materials (steels, Ni, Al, Cu, Ti based alloys…) or functional materials (semiconductors, quantum wells for optronics, and magnetic multilayers). Diffusive phase transformations are therefore essential in physical metallurgy. They are the foundations taught in particular in all the training courses of materials science in engineering schools and universities [1, 2]. After nearly 30 years of teaching in materials physics and designing exercises for tutorials and exam problems, the idea came to me to collect these exam subjects with detailed solutions. The problems proposed are often based on research problems with illustrations from electron microscopy or atom probe tomography observations, the flagship instrument designed in the Materials Physics Group in the 1990s. Links are frequently made with “everyday materials”. This book gathers a set of solved problems dealing with the thermodynamics of phase transformations, diffusion and solid-state transport phenomena in alloys and phase transformation kinetics. The book is organized into three main chapters. Each chapter begins with a course reminder that concisely gives the theoretical background necessary to solve the problems. There are of course many simplifications and omissions, which the reader will forgive. It is by no means an exhaustive course. Moreover, we have limited ourselves to binary alloys. To teach is to choose and DOI: 10.1051/978-2-7598-2743-5.c901 Ó Science Press, EDP Sciences, 2022

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Preface

therefore to exclude. Only the notions necessary to solve the problems of the book are introduced. The demonstrations of these reminders are sometimes not detailed. On the other hand, the steps of the calculations are most often given, which allow the reader to easily re-demonstrate the given expressions. The demonstration of some equations is also included in the problems. The book contains some bibliographical references of books or articles bringing a complementary addition. The book gathers about thirty problems. They are often preceded by a preamble that sets the context by introducing the new concepts used. In a way, problems constitute a complement to course reminders. They are designed to guide the reader step by step and usually lead to the study of a real industrial problem (aeronautics, nuclear power plants, microelectronics…). The detailed solution is given at the end of each problem. The problems of this book were inspired by other books, in particular those of Jean Philibert on diffusion, Yves Quéré on the physics of materials, Porter and Easteling on phase transformations, whom I wish to thank. This book was also inspired by exchanges with many colleagues, in particular Georges Martin, François Ducastelle, Alphonse Finel, Yann Lebouar, Annick Loiseau, Michel Guttmann, and many others who will forgive me for not being able to mention them all. My thanks also go to my former students and doctoral students who contributed directly or indirectly to this book. Thanks to Manon Bonvalet, Emmanuel Cadel, Fréderic Danoix, Bernard Deconihout, Fréderic De Geuser, Williams Lefebvre, Cristelle Pareige, Philippe Pareige, Thomas Philippe, Bertrand Radiguet, Xavier Sauvage and François Vurpillot for the enriching discussions we had and the vivid illustrations resulting from their research included in this book. This book owes a lot to Thomas Philippe, co-author of this book, who agreed to embark on the adventure by revisiting and enriching the problems and designing new ones. Didier Blavette

Preface by Yves Bréchet Some thoughts on the virtues of exercises and problems…

Blaise Pascal, who knew a lot about persuasion as well as about constructing science knowledge, wrote: “One is usually better persuaded by reasons one has found oneself, than by those that have come into the minds of others.”

All those who have been involved in teaching know that the aim of a lecture is to make students want to understand, but that it is through the practice of exercises, in a less theatrical relationship, that the student has really understood. That is to say that he has not limited himself to remembering, but that he knows how to reconstruct the reasoning that allows him to appropriate the acquired knowledge of the discipline, the only way to be able to go one step further himself, either in the application or in the enrichment of the knowledge. This shows the importance of “exercises”. Similarly, “problems” get a bad press because they are too often associated with the need to assess a student’s learning. This is certainly one of their functions, but in my opinion, it is the least of them: a well-designed problem should allow the student to continue learning, and he should know more when he leaves the exam than he knew when he entered it. Any teacher who sees a problem only as a control tool is condemning himself to a sterile psittacism which, in its most perverse form, ends up shaping the course itself not according to what is useful to know, but according to what is easy to control. The volume of exercises and problems proposed by Didier Blavette and Thomas Philippe places them firmly in the tradition of Pascal. The theme of the collection presented here, on phase transformations and diffusion, allows for a treatment which, without claiming to be exhaustive, has the merit of being coherent. The alternation of concise and enlightening reminders of the course, and exercises and

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problems of varying difficulty, corresponding either to “classics” or to lesser-known results, offers a progression that allows the student to “construct” his own reasons. The illustrations, often coming from the field of the Atom probe tomography where the Rouen team has taken the lion’s share, fulfil this beautiful mission that D. Blavette called “to make visible the invisible” in a distant echo of Jean Perrin who gave himself for the atomic theory “to explain the complicated visible by the simple invisible”. Finally, the structure of the book, in a beautiful classical style, declines three pillars of phase transformations: what are the driving forces that are responsible for them, what are the mechanisms of mobilities that make it operational, what are the kinetics that result from the combination of these two aspects. Does it need to be said? It is so obvious to all those who work in the “material” industry that it is fascinating that it can be so easily forgotten in high places. However, pedagogy is the art of repetition, and we have too often heard the Trissotins say with aplomb that is matched only by their ignorance, that physical metallurgy was a discipline with no future. It is always necessary to remind them of the obvious, at the risk of offending their certainties, and particularly at a time when “industrial sovereignty” has become a fashionable slogan: metallurgy is the indispensable foundation in many industries of sovereignty. This sovereignty begins with matter and materials, with manufacturing knowledge. The practical implications of the phenomena illustrated here cover immense fields of industrial application, from aircraft fuselage alloys to superalloys for turbines, from ageing materials for nuclear power plants to high-strength steels for automobiles. However, it is also in magnetic memories, in the realization of semiconductor nanostructures, in the development of hard magnets, or in the optimization of stainless steel thermal engine injectors, that we find the importance of phase transformations. Moreover, in the great movement of materials science towards the “tailor-made material”, physical metallurgy, and in particular that of phase transformations, shows an ever-innovative way, and it is by a practice of the exercise and the problem as much as by the lecture, however brilliant it may be, that we will train the engineers and the scientists who will open this way… Yves Bréchet, member of the Academy of Sciences

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

III

CHAPTER 1 Thermodynamics of Equilibria and Phase Diagrams . . . . . . . . . . . . . . . . . . 1.1 Course Reminders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Precipitation and Chemical Ordering . . . . . . . . . . . . . . . . . . . 1.1.2 Thermodynamic Functions and Potentials . . . . . . . . . . . . . . . 1.1.3 Binary Equilibria: Ideal and Regular Solid Solutions . . . . . . . . 1.1.4 The Critical Transition Theory and the Landau Theory . . . . . 1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Equilibrium Vacancy Concentration . . . . . . . . . . . . . . . . . . . . 1.2.2 Intergranular Segregation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Phase Diagrams and Regular Solution . . . . . . . . . . . . . . . . . . 1.2.4 Order–Disorder Transformations and Bragg–Williams Theory 1.2.5 Ordering in FCCs and Landau’s Theory . . . . . . . . . . . . . . . . . 1.2.6 Ordering in BCCs and Landau’s Theory . . . . . . . . . . . . . . . . .

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1 1 1 9 13 19 23 23 27 32 40 49 55

CHAPTER 2 Diffusion and Transport in Solids . . . . . . . . . . . . . . . . . . . . . . . 2.1 Course Reminders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Diffusion at the Atomic Scale . . . . . . . . . . . . . . . . 2.1.2 Fick’s Two Laws . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Hetero-Diffusion, Kirkendall Effect . . . . . . . . . . . . 2.1.4 Diffusion Short Circuits: Influence of Defects . . . . 2.1.5 Driven Diffusion in a Force Field . . . . . . . . . . . . . 2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Fundamental Mechanisms of Diffusion – Einstein’s 2.2.2 Elementary Mechanisms and Self-Diffusion . . . . . . 2.2.3 Diffusion in CuAl Alloys . . . . . . . . . . . . . . . . . . . . 2.2.4 Surface Hardening by Cementation . . . . . . . . . . . 2.2.5 Diffusion in Thin Sandwich Films . . . . . . . . . . . . . 2.2.6 Kirkendall Effect and Diffusion in CuZn . . . . . . . . 2.2.7 Short-Circuits of Diffusion: Grain Boundaries and Dislocations . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

VIII

2.2.8 Driven Diffusion in a Force Field and Spinodal Decomposition 2.2.9 A New Fick Law According to Howe . . . . . . . . . . . . . . . . . . . 2.2.10 Driven Diffusion in a Force Field: Application to Nabarro-Herring Creep . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.11 Driven Diffusion of Impurities in a Force Field, the Formation of Cottrell Atmospheres . . . . . . . . . . . . . . . . . 2.2.12 Diffusion in Semiconductors: The Generation of an Internal Nernst Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.13 Driven Diffusion in a Force Field: Oxidation of Nickel . . . . . .

. 108 . 115 . 122 . 126 . 132 . 139

CHAPTER 3 Kinetics of Formation of a New Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Course Reminder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Continuous and Discontinuous Precipitation . . . . . . . . . . . . . 3.1.2 Instability and Metastability . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Homogeneous Nucleation . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Heterogeneous Nucleation . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Nucleation of Metastable Phases . . . . . . . . . . . . . . . . . . . . . 3.1.6 Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.7 The Spinodal Decomposition . . . . . . . . . . . . . . . . . . . . . . . . 3.1.8 Coarsening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.9 Overall Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Eutectoid Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Nucleation of Precipitates in Ni-Al Alloys . . . . . . . . . . . . . . 3.2.3 Nucleation and Growth of Precipitates in Dilute Cu-Co Alloys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Heterogeneous Nucleation of Precipitates in AlCu . . . . . . . . 3.2.5 Deal and Grove’s Growth Law of a Spherical Precipitate . . . 3.2.6 Growth of an Oxide Film on Surface, Deal and Grove’s law . 3.2.7 Growth and Coarsening of Spherical Precipitates . . . . . . . . . 3.2.8 Coarsening in NiAl Superalloys . . . . . . . . . . . . . . . . . . . . . . 3.2.9 Zener’s Model for the Growth of a Precipitate with Incoherent Planar Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.10 The Spinodal Decomposition . . . . . . . . . . . . . . . . . . . . . . . 3.2.11 Zeldovich’s Kinetic Theory of Nucleation and Incubation Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.12 Capture Coefficient, Nucleation Flux and Incubation Time .

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References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

Chapter 1 Thermodynamics of Equilibria and Phase Diagrams 1.1 1.1.1

Course Reminders Precipitation and Chemical Ordering

Water and alcohol are miscible in any proportion. The system remains single-phase. On the other hand, water and oil are immiscible. The water-salt mixture remains single-phase if the solution is diluted. Above a certain salt concentration, called the solubility limit, there are two phases, liquid and solid, as crystals of salt precipitate. We then have the H2O + NaCl solution saturated in salt and NaCl crystals. It is the same in alloys, especially in the solid-state. When two chemical species A and B are mixed, there is often unmixing at low temperatures [3]. The solubility of B in A is generally limited and increases with increasing temperature T because the entropy (i.e. disorder) of the system increases. Above a certain temperature called critical temperature (TC ), there can be complete miscibility in the solid phase. On the other hand, the solubility is zero at 0 K because the entropy vanishes according to the third principle of thermodynamics. In this book, we will restrict ourselves to binary alloys which very often give rise to a phase separation [4–6]. A new phase (b) precipitates from the supersaturated a parent phase (figure 1.1). It is quite rare to have complete miscibility in the solid state. However, it is possible when elements A and B have the same crystallographic structure. This is the case of CuNi alloys. This b phase can be chemically ordered, such as the c0 phase (Ni3Al, L12 order) precipitating in nickel-based superalloys used in aircraft engines. This is illustrated in figure 1.2. Another phenomenon is highlighted, the intergranular segregation of

DOI: 10.1051/978-2-7598-2743-5.c001 © Science Press, EDP Sciences, 2022

2

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.1 – 3D reconstruction of a volume of material analysed by atom probe tomography [7] in a model superalloy based on Al supersaturated nickel. Small Al-enriched precipitates of about 7 nm in diameter are highlighted in dark contrast (F. Vurpillot, GPM).

FIG. 1.2 – Atom probe tomography images revealing the spatial distribution of Al, Cr, Mo, and B atoms in a nickel-based N18 superalloy (Ni is not shown). A small Al-rich Ni3Al ordered precipitate of about 10 nm appears to the left of a grain boundary in grain 1. The L12 chemical order in the precipitate is evidenced by the alternation of Al-rich planes with Al-poor planes. The parent phase is Al depleted and Cr enriched. A large Al-rich precipitate of the same nature exceeding the size of the analysed volume adjoins the GB in grain 2. This analysis also shows the segregation of boron (and Mo) at the GB (E. Cadel, GPM) [9].

additional elements like boron. The segregation of boron to grain boundaries (GB) strengthens GBs and thus prevents from intergranular decohesion [8]. The segregation of an additive element on crystal defects can occur even when the alloy is not supersaturated in this element. It will be studied in one of the problems of this chapter. The phenomenon of unmixing, or precipitation of a second phase in a B supersaturated AB alloy, occurs when atoms of the same nature have a greater affinity than atoms of different nature. In the simplest mean field approach, and limiting

Thermodynamics of Equilibria and Phase Diagrams

3

ourselves to interactions between first neighbour atoms, this results in energies of homo-atomic interaction AA and BB (eAA , eBB ) that are weak compared to hetero-atomic interactions eAB 1. We then define the order energy e. This is an essential interaction parameter in the approaches developed later (regular solid solutions). e can be expressed from a model of broken bonds. It is easy to see that the energy necessary to cut the two bonds AA and BB to form two bonds AB is 2e with: 1 e ¼ eAB  ðeAA þ eBB Þ 2 If e [ 0, the AA and BB pairs are favoured and there is a tendency for A and B atoms to unmix, leading to the formation of two phases, one rich in A (say the parent solid solution) and the other rich in B (precipitates). On the other hand, if e\0, the heteroatomic AB pairs are less costly in terms of energy and there is a tendency to ordering. A and B atoms then occupy preferential sites in the host structure. Note that even if we limit ourselves to this simple approach where interactions between atoms in second or third neighbours are omitted, we can have both unmixing and ordering. Thus, if e is not too positive, we can have precipitation of an ordered phase (F. Ducastelle [10]). The ordered B2 structure (stoichiometry AB) is the simplest in the centered cubic structures (CC). The two types of atoms, A and B order themselves on the centered cubic lattice by occupying respectively the centre sites and the corner sites of the BCC structure (figure 1.3).

FIG. 1.3 – Ordered structure B2 of stoichiometry AB (atoms A in light grey, B in dark grey).

1

Note that these interaction energies are potential energies and are therefore negative. Strong bonds lead to more negative interaction energies (cohesive energies of pure A and B materials are −εAA and −εBB respectively). These interaction energies correspond to the difference between the final state (crystal) and the reference state (being set to zero), for which the atoms are at an infinite distance from each other.

4

Diffusion, Segregation and Solid-State Phase Transformations

These ordered B2 phases are for example observed in FeAl (figure 1.4), NiAl or ClCs alloys. We speak then of lattice decoration and symmetry breaking. The system initially BCC in the disordered state becomes simple cubic (SC) by ordering with a two-atom pattern: B in (000) and A in (½,½,½). This symmetry breaking is reflected in diffraction (XRD or electrons), in reciprocal space, by additional superstructure spots. For BCC, the reciprocal lattice is FCC and the Miller indices must satisfy the condition h þ k þ l ¼ 2n. This condition falls when ordering develops and superlattice spots in (100), (010), (001) appear. The reciprocal lattice becomes SC (figure 1.5).

FIG. 1.4 – 3D tomographic atom probe image showing the alternation of Fe and Al rich planes along the analysis direction in ordered Fe40at.%.Al intermetallics with B2 structure (E. Cadel, GPM). Each black (Fe) or grey (Al) dot represents one atom. The distance between the Al-rich planes is equal to the lattice parameter (a = 0.29 nm).

FIG. 1.5 – Reciprocal lattice of BCC structure (black dots) and superstructure spots (Χ) related to B2 ordering.

Note that the ordered B2 structure can be described by a periodic static wave of concentration (in B atoms) along accounting for the alternation of B-rich planes with A-rich planes (BABAB…) along this direction. The atomic plane by atomic plane analysis of a FeAl alloy of B2 structure using atom probe tomography

Thermodynamics of Equilibria and Phase Diagrams

5

illustrates this stacking sequence (figure 1.6). The concept of static wave of concentration was introduced by Katchaturyan [11]. Given the symmetries of the structure, the same concentration wave is observed along the two other equivalent cubic directions and .

FIG. 1.6 – Concentration profile (in Al) from figure 1.4 highlighting the concentration wave along the direction in Fe40at.%.Al alloys of ordered structure B2 (E. Cadel, GPM). The wavelength is equal to the crystal parameter a = 0.29 nm (a = 2d002 with a the lattice parameter). Figure 1.7 shows, for simplicity, such a periodic wave C ðx Þ for a one-dimensional crystal of periodicity a. C is the atomic fraction of B atoms. In this 1D crystal, A and B atoms alternate with a period2 k ¼ 2a along x. The position of B atoms is given by x ¼ 2na with n an integer. x ¼ ð2n þ 1Þa for A atoms. Figure 1.7b shows this concentration wave as well as that of the second variant of order (dotted line). This time, B atoms are located in x ¼ ð2n þ 1Þa and A atoms in x ¼ 2na. Concentration waves are out of phase (shift of k=2 ¼ a ). This concept of static concentration waves is sometimes difficult to understand. In the formalism of crystallography, atoms are exactly located on the crystal positions contrary to the wave description. One way to see things is to consider that atoms vibrate around lattice positions. C(x) may be interpreted as proportional to the probability density to find B atoms at the position x within dx. Let us return to the real case of the three-dimensional crystal (with lattice parameter a). The (001) planes follow each other this time with a periodicity equal to a/2, so that the periodicity of the concentration wave, twice this value, is k ¼ a. The associated wave vector is therefore q ¼ ð2p=a Þð100Þ. The symmetries of the BCC lattice make the , , and directions indistinguishable. Thus, the wave along also generates the two other waves along and . 2

It is worth noting that in the 3D crystal, the period of the concentration wave according to is equal to the lattice parameter (k ¼ a) and (002) plane spacing is a/2. In 1D, the period of the static wave is k ¼ 2a because atoms are distant by a. With the same notation, there is therefore a factor 2 between 1D and 3D crystals if we keep the same notation for the crystal parameter (a).

6

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.7 – (a) 1D ordering in a stoichiometric AB one-dimensional crystal (periodicity is

a and B atoms are located at x ¼ 2na with n integer). (b) Corresponding C ðx Þ static wave of concentration (solid line). C ðx Þ denotes the atomic fraction of B atoms at the position x. The average atomic fraction of the alloy is here C0 ¼ 1=2. In a stoichiometric alloy, the atomic fraction of B atoms (or A atoms) is C0 ¼ 1=2 and the extrema are C ¼ 1 and 0 for a perfect chemical order (g ¼ 1). The dotted line shows the concentration wave corresponding to the second ordering variant that is in antiphase with the concentration wave for the first variant (coordinates of B atoms are x ¼ ð2n þ 1Þa).

Only the wave along is therefore needed to define the B2 order. When B atoms are at the corners of the B2 structure, this wave is written:
 C ðrÞ ¼ C0 1 þ geiq:r C ðrÞ is the concentration of B atoms at a given position r. r is the position vector of atoms in the BCC structure: (000) and ½(111). The wave amplitude g is the order parameter. For a perfect stoichiometric structure, C0 ¼ 1=2. For a perfect order (g ¼ 1), C ðrÞ ¼ 1 for r ¼ ð000Þ and all equivalent sites of the BCC structure. For a disordered structure (g ¼ 0), C ¼ 1=2 whatever r. As shown in figure 1.5, the wave vector q ¼ ð2p=a Þð100Þ generates all superstructure spots in the reciprocal lattice. These are derived from the fundamental BCC spots in the following way: (100) = (000) + (100), (210) = (110) + (100) etc.… We omit here the factor 2π/a that is common to all terms. There are two ordering variants depending on whether B atoms are located at the centres or at the corners of the BCC structure. When B atoms are at the centres of the BCC structure (instead of corners) and A at the corners, C ðrÞ is then in antiphase and:

Thermodynamics of Equilibria and Phase Diagrams

7


 C ðrÞ ¼ C0 1  geiq:r These two variants are equivalent and have the same free energy, thus F ðgÞ ¼ F ðgÞ. The free energy functional is even. As we will see at the end of this chapter, the phase transition can be of second order according to Landau’s theory (but not necessarily). In a second order transition, the order parameter gðT Þ decreases continuously with temperature T until it becomes zero at critical temperature TC . Moreover, there is no two-phase domain in the phase diagram. The situation is quite different in FCCs. The simplest FCC ordered structure, called L12, has the A3B stoichiometry. A atoms occupy the centres of the faces and B the corners of the FCC structure (figure 1.8). There are numerous alloys where L12 ordering is observed (Ni3Al, Cu3Al). Again, this long-range order results in periodic concentration waves (in A or B) in the three equivalent directions. For stoichiometric A3B alloys, there are alternating mixed (M) B-rich planes (50% B, 50% A) with pure (P) A planes. The plane-by-plane analysis of a Ni3Al alloy by atom probe illustrates such static waves along (figure 1.9).

FIG. 1.8 – L12 ordered structure (A3B stoichiometry, B atoms are represented in black and A atoms in grey).

There are three concentration waves along x, y, z whose related vectors are q1 ¼ ð2p=a Þð100Þ, q2 ¼ ð2p=a Þð010Þ, and q3 ¼ ð2p=a Þð001Þ. These vectors generate all the superstructure spots associated with L12 order in the reciprocal space. Such superstructure spots are observed in diffraction experiments. The concentration of B atoms C ðrÞ at a given site r is written as the sum of three components of amplitudes g1 , g2 , g3 related to the three wave vectors:
 C ðrÞ ¼ C0 1 þ g1 eiq1 :r þ g2 eiq2 :r þ g3 eiq3 :r

8

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.9 – Concentration profiles as given by atom probe analysis of an ordered Ni3Al phase along the direction. For each analysed plane, 30 atoms are detected in average. The long-range ordering in this L12 phase results in a static concentration wave. Al-rich mixed planes (M) that contain both Al and Ni alternate with almost Ni pure planes (P). Additional elements (Ti, W, Ta) are observed in Al-rich planes, suggesting that these atoms replace Al in the ordered Ni3Al structure. For a perfect stoichiometric structure, C0 ¼ 1=4. r is the position vector of atoms in the FCC structure: (000), ½(110), ½(011), ½(101). The norm of base vectors is equal to the lattice parameter (a). There are now three order parameters that define an order vector g ¼ ðg1 ; g2 ; g3 Þ. There are 4 order variants depending on whether atom B of structure A3B
is placed in (000), ½(110), ½(011), or ½(101). Writing the four concentrations C rj corresponding to these four locations (j ¼ 1; 2; 3; 4) shows that one variant is derived

Thermodynamics of Equilibria and Phase Diagrams

9

from the other by a change in signs of some components of the order vectors. In contrast to B2 ordering, the change in variants does not result in a simple change in the sign of vector η (see problem 1.2.5). Thus, unlike BCC structures, the free energy functional F ðgÞ has no parity in FCC structures. As we will see at the end of this chapter, Landau’s theory predicts that the transition is of first order. There is a discontinuity of gðT Þ at critical temperature TC . Moreover, the disordered and ordered phases can coexist in the phase diagram at the transition.

1.1.2

Thermodynamic Functions and Potentials

Consider a closed system (no matter exchange). The variation of internal energy dU with temperature is written for one mole: dU ¼ CV dT CV is the molar specific heat at constant volume (quantity of heat necessary to raise temperature by 1 °C). In solids, unlike gases, C V is very close to the specific heat  at constant pressure Cp , we remind that Cp ¼ @H @T P . The increase in Cp with temperature T can be described by the Debye model (Kittel [12]). Cp tends toward 3R for T [ hD (the Dulong and Petit law). θD is the Debye temperature of the material. Let us examine thermodynamic functions. Enthalpy is written H ¼ U þ PV , free enthalpy G ¼ H  TS and free energy F ¼ U  TS. In a closed system (without exchange of matter), these functions can be written as exact total differentials as a function of T , V and P. One can remember how to write these exact differentials using the following memo “Good Physicists Have Started Under Very Fine Teachers” (figure 1.10). In front of extensive variables S and V we find the conjugate intensive variables T and P, respectively. In the following fundamental expressions of differentials of thermodynamic functions (U , H , F, G), extensive variables are always multiplied by their conjugate variables: Thus, dU ¼ PdV þ TdS dH ¼ TdS þ VdP dG ¼ VdP  SdT dF ¼ PdV  SdT In principle H ¼ 0 for 25 °C. Note the presence of two negative terms for SdT and PdV . At constant pressure, both entropy S and enthalpy H increase with T because their derivatives are positive: @H =@T ¼ Cp and as @H ¼ T @S, thus @S=@T ¼ Cp =T . Both G and F decrease with T with a local slope (first derivative) equal to S. A disordered phase sees its free enthalpy G decreasing more rapidly with T than the ordered state because the entropy of the ordered phase So is lower than that of the

10

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.10 – Cyclic diagram “Good Physicists Have Started Under Very Fine Teachers” giving the variables on which the thermodynamic functions U , H , F, and G depend. The opposite variables in this diagram are conjugate (S and T as well as P and V ). disordered phase Sd . G ðT Þ curves related to both ordered and disordered phases intersect at TC . TC is the critical temperature. Above TC , the free enthalpy related to the ordered phase is smaller, the disordered phase is therefore the most stable (figure 1.11).

FIG. 1.11 – Free enthalpy G ðT Þ of the disordered phase decreases faster with temperature (T ) than that of the ordered phase. The intersection of the two curves gives the critical temperature of the order–disorder transition (TC ). Above TC , the ordered phase becomes disordered.

Thermodynamics of Equilibria and Phase Diagrams

11

Differentials of thermodynamic functions given above show that U is minimum when V and S are constant and H is minimum when S and P are constant. In both cases, requiring constant entropy is unpractical because it is generally not verified in phase transformation processes. We should use F or G that do not depend anymore on entropy S. They both get minimum under realistic conditions. Writing the total differentials shows that dG ¼ 0 if P and T are constant. Thus, the thermodynamic equilibrium results in a minimum of G. The free energy F is minimum at equilibrium when T and V are constant (dF ¼ 0). In materials science, F or G are commonly used. However, G is more often used in practice when it is easier to impose the pressure of the system than its volume. Using F or G, it is thus possible to theoretically predict, for instance, the concentration of vacancies in a crystal (see problem 1.2.1). Let us now consider a binary alloy composed of nA moles of A and nB moles of B. We define the related atomic fractions: nA nB XA ¼ and XB ¼ nA þ nB nA þ nB We have of course XA þ XB ¼ 1. In binary alloys, there is only one independent concentration variable, for example XB . The subscript B will sometimes be omitted to lighten the notation (X ¼ XB ). In the phenomenon of precipitation of a new phase (noted b) in a supersaturated a solid solution, both a and b phases are open systems with exchanges of matter between both phases. The chemical potential of species i (li , i ¼ A, B) is then to be considered. li is the partial free energy or the partial free enthalpy of species i. It is an intensive quantity and can be written as a partial derivative of F and G:  @F  li ¼ @ni T;V li ¼

 @G  @ni T ;P

When li is the chemical potential per atom, ni becomes the number of atoms (i = A, B). If li is expressed for one mole, ni is then the number of moles. For T , P constant and for nA moles of A and nB moles of B, free enthalpy G reads: G ¼ nA lA þ nB lB If we express G and the chemical potentials for one mole, then we have: G ¼ XA lA þ XB lB As XA þ XB ¼ 1: G ¼ lA þ XB ðlB  lA Þ:

12

Diffusion, Segregation and Solid-State Phase Transformations

The local slope of G (tangent) at a given atomic fraction (XB ) is therefore equal to lB  lA (figure 1.12). Thus chemical potential lA (resp. lB ) is given graphically by the intersection of the tangent with the ordinate at XB ¼ 0 (resp. XB ¼ 1). It should be kept in mind that chemical potentials in a binary alloy depend on the atomic fraction of B atoms.

FIG. 1.12 – Representation of free enthalpy G ðXB Þ of a solid solution AB as a function of the atomic fraction of B atoms at T and P constant. The intersection of the tangent of G ðXB Þ with vertical axes at XB ¼ 0 and XB ¼ 1 gives chemical potentials lA and lB respectively. For an infinitesimal variation of dμA, tangent point XB does not change and dμB is deduced from dμA by the inverse lever rule: XB dlB ¼ XA dlA (Gibbs–Duhem equation). The chemical potentials being partial free enthalpies, the variation dG associated with the exchange of dnA and dnB atoms is written: dG ¼ lA dnA þ lB dnB Given the previous expression of G, we end up with nA dlA þ nB dlB ¼ 0 ðGibbs – Duhem equationÞ Dividing by the number of moles ðnA þ nB Þ, we obtain: XA dlA þ XB dlB ¼ 0 This well-known lever rule can be deduced from diagram G ðXB Þ by noticing that the point of tangency XB remains immobile for infinitesimal variations of XB (figure 1.12). The thermodynamic equilibrium between both the α parent phase and β precipitates results in the equality of chemical potentials of each species (A, B) in both α

Thermodynamics of Equilibria and Phase Diagrams

13

and β phases. This can be demonstrated as follows. Consider a two phase system composed of α and β phases of volume V a and V b , respectively. At constant pressure P and temperature T , the Gibbs energy variation is given by: dG ¼ dU  TdS þ PdV The change in internal energy in the a phase is dU a ¼ TdS a  PdV a þ laA dnAa þ laB dnBa laA and laB are chemical potentials of A and B in the a phase. nAa and nBa are the number of moles of A and B, respectively, in a. S a is the entropy related to the a phase. Similarly, one can write for b: dU b ¼ TdS b  PdV b þ lbA dnAb þ lbB dnBb þ rdAb where the surface energy r is assigned to b. Ab is the area of the interface separating the two phases. Requiring dV b ¼ dV a and since mass conservation implies dnAb ¼ dnAa and dnBb ¼ dnBa , one gets for dG:     dG ¼ lbA  laA dnAb þ lbB  laB dnBb þ rdAb The thermodynamic equilibrium, for dAβ = 0, requires that partial derivatives of G with respect to the other variables are zero. Thus, dG ¼ 0, and this necessarily leads to the equality of chemical potentials of A and B atoms3.

1.1.3

Binary Equilibria: Ideal and Regular Solid Solutions

Let us consider a random solid solution composed of A and B atoms. We shall first consider that A and B components have the same structure. Thus, the AB solid solution adopts this common structure (for example FCC or BCC). The free enthalpy of a random solid solution composed of atoms A and B is written: G ¼ XA GA þ XB GB þ DGm GA and GB are the free enthalpies of the pure A and B phases. The third term is the enthalpy of mixing DGm : DGm ¼ DHm  T DSm

When considering a spherical nucleus (b phase), the pressures in the phases are different P b 6¼ P a (Gibbs–Thomson effect). Moreover, one has for a spherical nucleus    with radius  R: dAb ¼ 2dV b =R. In this case, dG is given by dG ¼ lbA  laA dnAb þ lbB  laB dnBb
  P b  P a  2r=R dV b . Equilibrium requires that partial derivatives of G with respect to all variables are zero. dG ¼ 0, and this leads to the equality of chemical potentials but this time at different pressures. This also gives the Young–Laplace relation P b  P a ¼ 2r=R. These two conditions determine the properties of the critical nucleus, see chapter 3. 3

14

Diffusion, Segregation and Solid-State Phase Transformations

DSm is the configuration entropy of the mixture of atoms A and B. If the enthalpy of mixing DHm is zero we have an ideal solution, the order energy e is zero. Everything happens as if A and B atoms were identical (mechanical mixing). This is the case for example of CuNi alloys where Cu and Ni are miscible in the solid phase in any proportion. In the simplest regular solution model, we express DHm in a mean field approach in considering only the interactions between first neighbour atoms. We write that each atom is on average surrounded by ZXA A atoms and ZXB B atoms (Z is the lattice coordination). For N atomic sites, we have NZXA XB first neighbour A–B pairs. The order energy e being the energy required to form one A–B pair from AA and BB pairs, we have: DHm ¼ NZXA XB e The entropy of mixing can be easily expressed as a function of XA and XB using the statistical expression of entropy: X DS ¼ kN pi ln pi i

N denotes the number of atomic sites and i the configurations. k is the Boltzmann constant. In a random solid solution, each atomic site is occupied by A or B with occupation frequencies pA ,pB (i ¼ A, B) equal to the atomic fractions XA and XB . The atomic fraction of vacancies is generally very small and can be neglected in the previous equation. For one mole, N is the Avogadro number (N = 6.02 × 1023) and writing the constant of perfect gases as R ¼ kN , we can express the entropy of mixing as: DSm ¼ RðXA ln XA þ XB ln XB Þ From these expressions we can derive the expression of free enthalpy G in the regular solution model: G ¼ XA GA þ XB GB þ NZXA XB e þ RT ðXA ln XA þ XB ln XB Þ Multiplying the mixing enthalpy term by XA þ XB ð¼ 1Þ, G can be expressed as the sum of two terms depending on A and B, respectively. Let us write G as a function of chemical potentials lA and lB : G ¼ XA lA þ XB lB The identification of the two previous equations leads to the expression of chemical potentials: li ¼ Gi þ NZ eð1  Xi Þ2 þ RT ln Xi ; i ¼ A; B:

Thermodynamics of Equilibria and Phase Diagrams

15

This expression is that of the chemical potential of species A, B within the regular solution approximation. For an ideal solution (e ¼ 0), we have: li ¼ Gi þ RT ln Xi . In the general case, the chemical potential can be written in an analogous way as a function of a parameter called chemical activity (ai ): li ¼ Gi þ RT ln ai For any solution, we can write the activity coefficient as: ai ¼ ci Xi where ci is the activity coefficient that depends on e. Note that this parameter depends on atomic fractions. We see that ai ¼ Xi for an ideal solution. Consequently, ci ¼ 1. It is also worth noting that DSm [ 0, as a consequence T DSm \0. On the other hand, the sign of DHm is driven by that of e. If e\0, the enthalpic and entropic terms are both negative and DGm ðXB Þ as well as G ðXB Þ have only one minimum. We have thus the formation of an ordered phase (figure 1.13). On the other hand, for e [ 0, there is competition between these two terms (DHm and T DSm ). The entropy term that results from disorder is proportional to T . Hence, for low temperatures (T \TC the critical temperature), the entropy is small and the enthalpy term dominates. As a result, DGm ðXB Þ and G ðXB Þ show two minima (figure 1.14a). The free enthalpy of the system can therefore be minimized by the precipitation of a new phase (b) enriched in B atoms within the A-rich parent phase (a) that is depleted in B atoms. This is accompanied by a gain in free enthalpy DGp called the driving force for precipitation (figure 1.15a).

FIG. 1.13 – Free enthalpy of mixing DGm as a function of XB for e\0. The tendency is chemical order.

16

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.14 – Free enthalpy of mixing DGm for e [ 0. (a) If T1 \TC , DGm shows two minima. The system is composed of two phases that have the same structure, one rich in A, the other rich in B whose compositions (Xa and Xb ) are given by the two minima (for GA ¼ GB ). (b) For higher temperatures above the critical temperature (T2 [ TC ), only one minimum remains, the system is made of a single phase.

For T [ TC (figure 1.14b), the entropy term prevails and there is only one minimum of G ðXB Þ. The system becomes a single-phase system (the solid solution). The inflection points of G ðXB Þ disappear for T ¼ TC . TC can be derived from the study of the second derivative of G ðXB Þ (no more inflexion points at T  TC ): TC ¼

Ze 2k

TC is proportional to the order energy. For T \TC , two phases (α, β) of same structure are formed, that only the chemical compositions distinguish. The equilibrium compositions of these two phases, XBa and XBb , are given by the common tangent to the free enthalpy which expresses the equality of chemical potentials of atoms A or B in both phases (figure 1.15a). When laB ¼ lbB and laA ¼ lbA , exchanging A and B atoms between the a and b phases does not change the overall Gibbs energy of the system. Equilibrium is reached and the Gibbs energy of the system, as given by the common tangent (figure 1.15a), is minimum. This argument justifies the equality of chemical potentials and the tangent rule. Note that the overall energy of the system with composition XB0 can be expressed as the weighted free enthalpy:  

 G XB0 ¼ fG XBb þ ð1  f ÞG XBa with f the molar fraction of b, as given by the inverse lever rule 
0   b a a f ¼ XB  X B = XB  XB .

Thermodynamics of Equilibria and Phase Diagrams

17

For XBa \\1, the solubility limit of B atoms in the solid solution α reads: XBa ¼ e2TC =T XBa grows rapidly with T . Lower TC, more rapid the increase in XBa with T . When pure A and B phases have different structures, both α and β phases have different structures, and they can co-exist if T \TC (figure 1.15b). It is then appropriate to consider two functions G ðXB Þ, one for each phases α and β. The common tangent rule gives the composition of both α and β phases in the equilibrium state at a given temperature. Under some approximations (XBa 1 and XBb  1), it is quite easy to demonstrate that the solubility can be written as (problem 1.2.3): XBa ¼ eDG=RT e2TC =T DG ¼ Gb  Ga is the difference of free enthalpies of α and β phases for XB ¼ 1. When α and β have the same structure, DG ¼ 0 and the previous expression is recovered. Gibbs’ phase law. We are going to calculate the number of possible equilibrium phases in an alloy containing n distinct chemical species A, B, C… Let u be the number of equilibrium phases. The number of independent variables including T and P is then equal to: N ¼ ðn  1Þu þ 2 Writing equilibrium between the u phases leads to u  1 equalities of chemical potential for the constituents. There are thus n ðu  1Þ relations resulting from equilibrium. The variance, being the number of degrees of freedom, is the difference between these last two terms, thus: v ¼ nþ2  u Generally, the pressure P is fixed, so that the new variance is written v 0 ¼ n þ 1  u. For a binary alloy (n ¼ 2), we have 3 phases at most in equilibrium (v 0 ¼ 0). This is a point (T and XB are fixed) in the phase diagram T ðXB Þ. It is for example the eutectoid point in steels. At this point the austenite is in equilibrium with ferrite a and cementite Fe3C. In the two-phase domain (u ¼ 2), the variance v 0 ¼ 1 and therefore XB and T are linked. This defines the solubility limit XBa ðT Þ. v 0 ¼ 0 at a given temperature. The latter does not depend on the nominal composition XB0 . Only the mole fraction of precipitated phase (f ) changes with XB0 . As mentioned earlier, the mole fraction f is given by the inverse lever rule: 
  f ¼ XB0  XBa = XBb  XBa where once again XBa and XBb are the equilibrium atomic fractions in B atoms in the α parent phase and β precipitates, respectively.

18

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.15 – (a) Representation of free enthalpy G ðXB Þ for a regular solution and an order energy e [ 0 (T and P are held constant). At T \TC , the curve presents two minima. There is unmixing between the two α and β phases, which have the same structure but different compositions. The atomic fractions of B atoms in the two phases (XBa , XBb ) are given by the common tangent. DGp (driving force for precipitation) is the free enthalpy gain realized for a nominal fraction XB0 after precipitation when thermodynamic equilibrium is reached. The equilibrium mole fraction of b phase is given by the lever rule, 
  f ¼ XB0  XBa = XBb  XBa . (b) The free enthalpy curves for the two α and β phases that have distinct structures. Again, the common tangent rule gives the composition of both phases at equilibrium (XBa , XBb ).

The situation is quite different in ternary alloys because there is an additional degree of freedom (v 0 ¼ 1 when T and P are imposed). In this case, the composition of α and β phases depend on the nominal composition XB0 .

Thermodynamics of Equilibria and Phase Diagrams

1.1.4

19

The Critical Transition Theory and the Landau Theory

The order parameter g of a given phase can be defined for many processes, such as chemical ordering in alloys, ferromagnetism, superconductivity or solidification. In order–disorder transitions, g ¼ 0 if the considered phase is random (disordered) and g ¼ 1 if a perfect long-range order is established. Let us take an example. For B2 ordering in a BCC alloy, g ¼ 1 if A and B atoms are located at the corners (sites 1) and at the centres (sites 2) of the BCC structure, respectively. Let us consider the probability to have B on site 2 (PB2 ). We can then define the order parameter as follows: g¼

PB2  XB 1  XB

In this definition of g, the order can no longer be perfect if XB \1=2 because sites 2 are partially occupied by atoms B. Sites 2 are also partially occupied by atoms A or by structural vacancies. According to Ehrenfest (1933), order–disorder transitions are of two types, first or second order. The transition is of first order if the partial derivative of G (or F) with respect to intensive variables (T , P, lA , lB ) is discontinuous at the transition temperature TC . In contrast, it is of second order when the first derivative is continuous and the second derivative discontinuous (e.g. @ 2 G=@T 2 ). In first-order transitions, as @G=@T ¼ S, at TC there is an abrupt change in entropy between the disordered and ordered states (DS ¼ Sd  So ). This results in an abrupt and discontinuous change in the slope of both G ðT Þ as shown in figure 1.11 and of the order parameter g (figure 1.16a). As DH ¼ T DS at constant P, the order–disorder transition is accompanied by a latent heat of transformation (L ¼ DH figure 1.16b). At TC , DG ¼ Gd  Go ¼ 0 (figure 1.11) and thus L ¼ T DS. In first-order transitions, there can be coexistence of ordered domains with the disordered phase. When XB differs from the exact stoichiometry (XBC , figure 1.16), the disordered and ordered phases have different compositions because of the discontinuity of the first derivative of G with respect to chemical potentials (lA , lB )4. There is therefore a two-phase domain in the phase diagram (figure 1.16c). In second order transitions (figure 1.17), @G=@T is continuous at TC , so DS ¼ 0. As entropy is a measure of disorder, there is no discontinuity of the order parameter g at TC (figure 1.17a). There is a continuous evolution of g close to TC unlike first order transitions (figure 1.16a). At TC we have DG ¼ DH  T DS ¼ 0. As DS ¼ 0, thus L ¼ 0. There is no enthalpy excess and thus no latent heat of transformation at the transition. On the other hand, as Cp is different in the ordered and disordered states, @H =@T is thus discontinuous at TC . Since @ 2 G=@T 2 is discontinuous at TC in 2nd order transition, @S=@T and consequently @g=@T are discontinuous at TC .

Indeed, we have dG=dlaB ¼ XBa and dG=dlbB ¼ XBb , therefore XBb 6¼ XBa .

4

20

Diffusion, Segregation and Solid-State Phase Transformations

Another feature of second order transitions is that no ordered and disordered phases can coexist because there is no discontinuity of the first derivative of the free enthalpy with respect to chemical potentials (lA , lB ). There is no two-phase domain in phase diagram. We can now state Landau’s two rules. In the framework of this theory, it is customary to consider the free energy functional F ðgÞ rather than G ðgÞ. However, this choice has no impact on what follows. The Landau theory is based on both the system symmetries related to order and the parity of F ðgÞ. We then define a temperature T0 at which the disordered state (g ¼ 0) becomes unstable when T \T0 . The second derivative of the energy functional F ðgÞ with respect to g is then negative. It becomes positive for T [ T0 (cf. problems 1.2.5 and 1.2.6). The disordered state becomes stable (F is a minimum minimorum at g ¼ 0) or metastable (F has a minimum but it is not a minimum minimorum). At T0 we then have @ 2 F=@g2 ¼ 0. This is the instability temperature of the disordered state, not to be confused with the critical temperature of the order–disorder transition TC . Landau’s first rule states that if F ðgÞ is even, the transition can be of first or second order. This is the case for the B2 ordering of BCC alloys (problem 1.2.6). We shall examine why. There are two equivalent order variants depending on whether B atoms are located at the centres or at the corners of centred cubic (CC) lattice. Let us calculate g for XB ¼ 1=2. We see that if PB2 ¼ 1 (1st order variant, B is located at sites 2) then g ¼ 1. On the other hand if PB2 ¼ 0 (B atoms are not located at sites 2 but at sites 1), then g ¼ 1. As a result, one order variant transforms into the other by simply changing the sign of the order parameter g. This result can be generalized regardless the value of g. Only a translation ½(111) distinguishes the two variants which are simply in phase opposition in the sense of static concentration waves. They therefore have the same free energy, thus F ðgÞ ¼ F ðgÞ. The functional F ðgÞ is therefore even. The interface between two different variants is called antiphase boundary. Let us now examine the situation where F ðgÞ has no parity. As demonstrated in problem 1.2.5, the transition is necessarily of first order. This is the case of L12 ordering in FCC materials. In an ordered alloy of stoichiometry A3B, there are 4 variants because the B atoms can be placed on the corners of the cube (position (000)) or on one of the three face centres of the FCC lattice. It is now necessary to consider the 3 associated translation vectors: ½(110), ½(011), ½(101). The order parameter is no longer a scalar but a vector g whose three components are associated with the three static waves along x, y, z directions of the real space. In contrast to second order transitions in BCC structures, F ðgÞ no longer has parity in FCC structures because variants are no longer deduced by a simple change in the sign of vector g. The transition is thus necessarily of first order in FCC solids and a two-phase domain is expected in the phase diagram (e.g. the Ni3Al phase co-exists with the Ni (Al) solid solution in supersaturated binary NiAl alloys) contrary to transitions of second order that may occur (but not necessarily, first Landau’s rule) in BCC alloys. For second order transitions, ordered precipitates cannot coexist with the disordered phase.

Thermodynamics of Equilibria and Phase Diagrams

21

FIG. 1.16 – First order transitions. (a) The order parameter gðT Þ shows a discontinuity at critical temperature TC . (b) The enthalpy H ðT Þ also shows a discontinuity at temperature TC resulting in a sharp increase (DH ¼ L, the latent heat of transformation). (c) The phase diagram T ðXB Þ shows two two-phase domains (α disordered + β ordered). For an atomic fraction XBC , there is a congruent transformation (direct) at TC , the disordered phase α transforms into the ordered phase b without crossing the two-phase domain a þ b.

22

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 1.17 – Second order transitions. (a) The order parameter gðT Þ decreases gradually until critical temperature TC . H ðT Þ is continuous at critical temperature TC . (c) There is no two-phase domain a þ b in the phase diagram T ðXB Þ.

Thermodynamics of Equilibria and Phase Diagrams

1.2 1.2.1

23

Problems Equilibrium Vacancy Concentration

Both interstitials and vacancies (absence of an atom on a site of the crystal) are point defects. Vacancies lead to an elastic deformation of the lattice. The creation of a vacancy requires an energy of formation and is accompanied by an entropy of vibration. To those terms is added a configuration entropy because the vacancy can be placed in any N sites of the crystal. We propose to calculate the concentration of vacancies at thermodynamic equilibrium. 1. Enthalpy and entropy of vacancy formation 1.1. Enthalpy. Consider a crystal of coordination number Z . Calculate the ðvÞ enthalpy of formation of a vacancy DHf by estimating the work necessary to bring an atom from the volume to the surface in a terrace edge situation. In this situation, we will make the approximation that on average, atoms (A) have (Z =2) bonds at the surface. Only first neighbour interactions between atoms (eAA ) are considered. The elastic relaxation term will be accounted for by introducing a “phantom” interaction energy eAv of the vacancy ðvÞ with its first neighbouring A atoms. 1.2. Entropy. When a vacancy is created by moving an atom from the volume to the surface, one creates a vibrational disorder that leads to an entropy ðvÞ term DSf . Give the expression of this entropy as a function of heat ðvÞ

exchange dQ and temperature T . Calculate DSf by writing that dQ is of the order of the thermal quantum kT (k is the Boltzmann constant). 2. Configuration entropy DSc 2.1. The classical expression of entropy5 is DS ¼ k ln X, X is the configurational number, that is the number of equivalent states from the macroscopic point of view. Calculate the configurational entropy term DSc for n vacancies distributed among N crystal sites. Then express this term as a function of the atomic fraction of vacancies Xv ¼ n=N . It will be assumed that N and n are large so that the Stirling formula can be used:

5

The Boltzmann formula makes the link between the macroscopic and microscopic worlds. A given macroscopic state, of entropy S, can be generated by a very large number of microscopic states. Boltzmann notes this number X and the thermodynamic entropy is proportional to the logarithm of X, the Boltzmann constantPk being the proportionality factor (k = 1,38 × 10–23). The statistical entropy, DSc ¼ kN i pi ln pi , can be derived from the theory of Shannon which proposes to measure the lack of information. When all probabilities pi are equal, we find the formula of Boltzmann: no event is favored, the information is minimal and thus the entropy is maximum. Thus, the thermodynamic entropy of a system can be interpreted as the measure of the lack of information that remains at the microscopic level for a given macroscopic state of this system.

Diffusion, Segregation and Solid-State Phase Transformations

24

ln u! ¼ u ln u  u when u ! þ 1: 2.2. Demonstrate the previous result DSc ðXv Þ using the expression of the statistical entropy given below: X pi ln pi : DSc ¼ kN i

What is the significance of the probabilities pi ? Deduce DSc ðXv Þ. 2.3. What is the maximum value of DSc and for which value of Xv is it obtained? 2.4. Represent graphically DSc as a function of Xv . First, look for the asymptotes of DSc when Xv ! 0 and Xv ! 1. What do you think of the evolution of the free energy at low vacancy concentration? Is purifying a crystal of its vacancies (Xv ¼ 0) only a technological challenge? 3. Free enthalpy of vacancy formation DGf 3.1. The free enthalpy required to create n vacancies can be written as: ðvÞ ðvÞ DGf ¼ nDGf  T DSc . DGf is the free enthalpy of formation of a single vacancy6. Show graphically that DGf has a minimum. To what does this minimum correspond? Show graphically that this minimum of DGf occurs for a value of Xv which depends on T . 3.2. Show without calculation how both the equilibrium concentration and the entropy term vary with temperature T . 4. Vacancy concentration at equilibrium Xve 4.1. Calculate the expression of the vacancy concentration (Xve ) as a function of T for Xve \\1. 5. Application to copper 5.1. A good order of magnitude of the free enthalpy required to create a single ðvÞ ðvÞ vacancy is DHf  1 eV, and we assume DSf  k. Calculate Xve for the following temperatures: −273 °C, 0 °C, 500 °C, 800 °C and 1083 °C (melting temperature of copper). The results will be presented in a table. ðvÞ 5.2. Show that DGf remains positive up to the highest accessible temperatures.

ðvÞ

We can equate the enthalpy DHf

6

ðvÞ DHf

ðvÞ DEf

of formation of the vacancies to the internal energy. Indeed,

¼ þ pDV , with DV the change in volume when a vacancy is introduced, that is of same order than the atomic volume. p is the pressure. We find that pDV is negligible at atmospheric pressure.

Thermodynamics of Equilibria and Phase Diagrams

25

Solution to the problem 1. Enthalpy and entropy of vacancy formation 1.1. Considering the energy difference between the final state corresponding to the crystal after introduction of the vacancy and the initial state of the crystal (see diagram below), we obtain:  eAA  ðvÞ : DHf ¼ Z eAv  2

ðvÞ

1.2. The entropy associated with the creation of a vacancy is written DSf dQ T

and in first approximation becomes

ðvÞ DSf

¼

 k.

2. Configuration entropy DSc N! 2.1. There are CnN ¼ ðN n Þ!n! ways to distribute n vacancies among the N sites

N! of the crystal, therefore the entropy reads DSc ¼ k ln ðN n Þ!n!. Using the Stirling formula and the atomic fraction of vacancies, we find that DSc ¼ kN ðXv ln Xv þ ð1  Xv Þ lnð1  Xv ÞÞ. We have also assumed that N is large and that N  n remains large enough to apply the Stirling formula. 2.2. Using the expression of the statistical entropy we obtain directly that DSc ¼ kN ðp1 ln p1 þ p2 ln p2 Þ with p1 ¼ XV the probability of having a vacancy in one of the sites of the crystal and p2 ¼ 1  p1 the probability that a site is occupied by an atom. 2.3. The maximum value DSc ¼ kN ln 2 is obtained for Xv ¼ 1=2. Note that generally Xv 1=2 so that nN . 2.4. The configuration entropy can be represented schematically as follows:

26

Diffusion, Segregation and Solid-State Phase Transformations

Xv !0

Xv !1

dDSc c Note that dDS dXv ! 1 and dXv ! 1. Note that Xv tends toward 1 has no physical meaning. The free enthalpy of the system being written as G ¼ H  T DS and because of the asymptote in Xv ¼ 0, we understand that G always decreases with Xv if T [ 0. Consequently, the minimum of G is obtained for Xv [ 0. It is only at T ¼ 0 that the minimum of G is observed at Xv ¼ 0. In other words, we cannot theoretically find a perfect crystal (without vacancies) at non-zero temperature, T = 0 being not reachable in practice.

3. Free enthalpy of vacancy formation DGf h i ðvÞ 3.1. We can write DGf ¼ N Xv DGf þ kT ðXv ln Xv þ ð1  Xv Þ lnð1  Xv ÞÞ . A graphical construction (see below) shows that for a certain value of Xv the free enthalpy is minimal. This defines the concentration of vacancies at equilibrium.

Thermodynamics of Equilibria and Phase Diagrams

27

3.2. The entropy term varies linearly with temperature, this has the effect of moving the minimum of the free enthalpy of formation of the vacancies towards greater values of Xv . The concentration of vacancies at equilibrium therefore increases with temperature. 4. Vacancy concentration at equilibrium Xve



4.1. The equilibrium concentration is determined by Xve 1Xve

ðvÞ

¼ eDGf

writing

dDGf  dXv X e ¼

0. We find

v

=kT

ðvÞ DGf

¼

ðvÞ DSf =k

ðvÞ

. Assuming that Xve \\1 we obtain: Xve ¼ eDGf ðvÞ DHf



ðvÞ T DSf ,

we find that Xve ¼ Ae

=kT

ðvÞ

DHf =kT

. By with

A¼e (a constant). The equilibrium concentration of vacancies increases with temperature. 5. Application to copper ðlÞ

5.1. For DSf  k, the prefactor of the vacancy concentration is e1  2:7 which allows us to calculate the following equilibrium concentrations: T (°C) Xve

−273 0

0 10–18

500 8 × 10–7

ðvÞ

800 5 × 10–5

ðvÞ

1083 5.2 × 10–4

ðvÞ

5.2. We note that T \T  ¼ DHf =DSf and DGf [ 0. We find T   11 600 K that is well above the melting temperature of copper (1356 K), ðvÞ DGf is therefore always positive in the considered temperature range.

1.2.2

Intergranular Segregation

After solidification, materials are most often polycrystalline, that is to say, made up of grains with different crystal orientations. Interfaces between grains, called grain boundaries (GB), extend over a small thickness of a few interatomic distances. When the disorientation between grains is low, the interfacial energy of such low-angle GBs is lower. Impurity segregation to GBs are often observed as illustrated in the figure shown below [13]. Alloying elements or impurities often lead to lattice strain in the grain and therefore to an additional elastic energy. This is the case of boron for example in steels or nickel-based alloys. The segregation of these impurities to GBs allows to relax elastic stresses that ends up to a decrease in the system enthalpy. This partition of impurities however decreases the configurational entropy of the system (lower disorder) which increases the free enthalpy G ¼ H  TS. This competition between enthalpy and entropy in the overall free enthalpy of the system (G) leads to

28

Diffusion, Segregation and Solid-State Phase Transformations

Atom probe tomography image (reconstructed volume of 60 × 60 × 150 nm3) showing the segregation of impurities (Si, P and Ni) in a Fe9at.%.Cr alloy (C. Pareige, GPM). The decoration of GBs shows three GBs meeting at a triple point. Small Si precipitates of a few nm in diameter are also observed [14].

a minimum of G that determines the equilibrium atomic fraction of impurities (XGB ) in GBs. We shall consider a polycrystalline alloy (AB) whose atomic fraction of B atoms of solutes is low so that X0 \\1. The model will then be applied to the segregation of boron in the nickel. 1. Qualitative approach. How important is the equilibrium segregation of impurities at low temperatures compared to that at high temperatures? Justify your answer in terms of configuration entropy. 2. Quantitative approach. We shall demonstrate McLean’s formula [15] which gives the evolution of the equilibrium intergranular segregation rate as a function of temperature (T ) in binary alloys. Let us consider the free enthalpy gain DG ð1Þ ¼ DE  T DS ð1Þ related to the segregation of one B atom to grain boundary (DG ð1Þ \0). DS ð1Þ is the thermal entropy of segregation (segregated B atoms modifies the vibrational states in GBs). We shall neglect DS ð1Þ in the following calculations. DE is the energy7 that is gained when moving one B atom from the dDS bulk to the grain boundary (DE\0). Show that at equilibrium DE ¼ T dn . nGB GB is the number of B atoms in the grain boundary and DS is here the configuration entropy that is not to be confused with the segregation entropy DS ð1Þ per atom. 3. Let us define the number of B atoms (nc ) in the crystal (grains). Nc and NGB are the number of atomic sites in the crystal and in GBs, respectively. We shall assume that all GB sites are favourable to segregation and their number will be

DE ¼ DH at P and V help constant.

7

Thermodynamics of Equilibria and Phase Diagrams

29

taken equal to NGB . Calculate the values of the configuration entropy in the grain (DSc ) and in GBs (DSGB ). The following expression for the statistical entropy will be used: X DS ¼ kN pi ln pi i

4.

5.

6.

7.

k is the Boltzmann constant. Let us introduce the atomic fractions of B atoms Xi ¼ ni =Ni with i ¼ c; GB. Express the associated total entropy variation as a function of the variations of S in GBs and in the crystal (DSGB and DSc ). dDSc GB Express dDS dnGB and dnGB in terms of the atomic fractions XGB and Xc respectively. Note that the number of solute B is conserved (nGB þ nc ¼ n ¼ cste). Derive the equilibrium value of XGB as a function of temperature, the absolute value of DE (jDE j) and Xc . Give an approximation for Xc \\1. Sketch the segregation rate sðT Þ ¼ XGB =Xc as a function of T . Through the study of s ðT Þ function, show qualitatively how does s ðT Þ curve vary with DE. Sketch s ðT Þ for two increasing values of jDE j (jDE2 j [ jDE1 j). To which values does s ðT Þ tend for large values and very small values of jDE j. Numerical application. Let us consider the segregation of boron (B) in nickel. The atomic fraction of boron in the alloy is Xc = 0.05at.%. and DE = −0.6 eV. Calculate the value of the segregation rate s ðT Þ for T = 200 °C, 400 °C, 600 °C, 800 °C and 1000 °C. Influence of grain size ( U). Establish a simple expression of the volume fraction fv occupied by GBs as a function of GB width e. Generally, e\\U, the grain size, even for 10 nm nanograins. For the sake of simplicity, we shall consider cuboidal grains of side U. GBs will be represented as corridors of width e (e  0.5 nm). Sketch this simplified microstructure. Express qualitatively the condition on fv that is required for Xc to be close to and approximated to the nominal atomic fraction X0 . Give the minimum grain size for which the condition Xc  X0 remains valid, within an error of 1%.

Solution to the problem 1. Configurational order increases when B atoms segregate at the grain boundaries. Thermodynamics teaches us that entropy, which expresses disorder, decreases when the temperature decreases. Consequently, the segregation of B atoms to GBs (order) will be favoured at low temperatures for which configuration order is greater. The third principle of thermodynamics also states that entropy is zero at 0 K. Segregation level will therefore be maximum at 0 K. 2. Let us move nGB B atoms in GBs. The related Gibbs free enthalpy change is given by DG ¼ nGB DG ð1Þ  T DS. DS is the additional configuration entropy related to the way one distributes B atoms in GBs. Equilibrium state is reached when dDG ¼ 0. Assuming that DS ð1Þ ¼ 0, therefore DG is minimum, consequently dn GB DG ð1Þ ¼ DE, as a result we get at equilibrium:

Diffusion, Segregation and Solid-State Phase Transformations

30

DE ¼ T

dDS dnGB

3. Let us place nGB B atoms in the available GB sites (NGB ) and nc B atoms in the Nc grain sites. Atomic fractions in GBs and in grains can be expresses as: XGB ¼ nGB nc NGB and Xc ¼ Nc . The entropy of configuration related to GBs (DSGB ) may easily be derived from the expression of the statistical entropy: DSGB ¼ kNGB ½XGB ln XGB þ ð1  XGB Þ lnð1  XGB Þ In the same way the crystal entropy reads: DSc ¼ kNc ½Xc ln Xc þ ð1  Xc Þ lnð1  Xc Þ As entropy is an extensive variable, the total entropy of configuration (GB + grains) writes as: DS ¼ DSGB þ DSc . dDSc dDS GB 4. Let us write that dn ¼ DS dnGB þ dnGB . As the number of atoms is conserved, GB dnGB ¼ dnc . We can also write that: dDSGB dDSGB dXGB 1 dDSGB XGB ¼ ¼ ¼ k ln : dnGB dXGB dnGB NGB dXGB 1  XGB In the same way, dDSc dDSc dDXc 1 dDSc Xc ¼ ¼ ¼ k ln : Nc dXc dnGB dXc dnc 1  Xc Using results of question 2, we find that:   DE ð1  XGB ÞXc ¼ ln : kT ð1  Xc ÞXGB Assuming that Xc \\1 and replacing the energy gain related to segregation ΔE by jDE j (ΔE < 0), then we get the following approximate expression: Xc : XGB ¼ Xc þ ejDE j=kT 5. The segregation rate is written s ¼ XXGBc ¼ Xc þ e1jDE j=kT . It is quite easy to see that: ds T !0 dT ! 0

T! þ 1

T !0

T! þ 1

ds and dT ! 0. Similarly, s ! 1=Xc and s ! 1 þ1Xc  1 (as X0 \\1). We can thus sketch the variation of the segregation rate sðT Þ as follows:

Thermodynamics of Equilibria and Phase Diagrams

31

We observe that sðT Þ is greater for the highest value of jDE j (segregation level more important) whatever temperature. The curve related to DE2 is therefore above that related to DE1 as shown in the figure. 6. Segregation rates sðT Þ for boron in nickel as provided by the table below shows a significant increase with decreasing temperatures (one order of magnitude from 1000 °C to 200 °C): T (°C) s

200 2000

400 1878

600 1181

800 492

1000 211

7. The schematic representation of the microstructure is given below (2D section):

From this simplified periodic structure, we can derive the volume fraction using a unit cell (a cube of side U þ e) composed of one grain with the three related

Diffusion, Segregation and Solid-State Phase Transformations

32

corridors of width e. Assuming that e\\U and keeping only 1st order terms, we reach the following approximated expressions: fv ¼ 3e U. The rule of levers applied to the 2 co-existing regions gives X0 ¼ fv XGB þ ð1  fv ÞXc . Note that X0  Xc when fv \\1. The error on this approximation will be less than 1% when the volume fraction does not exceed 0.01 (grains larger than 150 nm for e = 0.5 nm).

1.2.3

Phase Diagrams and Regular Solution

1. Let us consider phase separation in a binary alloy in which a and b have distinct compositions but the same structure (coordination number Z ). The atomic fractions of B atoms are designated by X. We will model the thermodynamics of the system by a regular solid solution. The quantities will be expressed for one mole. Give the expression for the order energy e and give its sign for a precipitation reaction. What would happen for an opposite sign of e? 2. Recall the expression of the enthalpy of mixing DHm and plot it as a function of X. Is the graph symmetric? Why is it so? 3. Recall the expression of the entropy of mixing DSm . Represent DHm ðX Þ and T DSm ðX Þ on the same sketch at a given temperature T . 4. Study the asymptotic slopes of both DHm and T DSm for X ¼ 0 and X ¼ 1. Describe the competition between these two terms for increasing values of the composition X. Explain qualitatively why there is complete miscibility between atoms A and B when T [ TC , TC being the critical temperature of the regular solution. Represent the free enthalpy of mixing DGm ðX Þ for two temperatures T1 \TC and T2 [ TC . 5. It is assumed that the free enthalpies of pure A and B elements are equal (GA ¼ GB ), as well as HA ¼ HB . Represent the free enthalpy of the alloy. Show how the limits of solubility of B (respectively A) in the A-rich parent phase a (respectively in B-rich b precipitates) can be derived from this graph. How do these limits vary with T ? What is their value when the temperature approaches 0 K? 6. Let X0 be the nominal atomic fraction of B atoms in the alloy. Give the expression of the mole fraction of b phase fm as a function of X0 and of the equilibrium atomic fractions in the a and b phases, Xa and Xb . Give the equilibrium enthalpy H0 of the system (of nominal atomic fraction X0 ) as a function of the enthalpies Ha and Hb of the phases and of fm . What is the sign of the enthalpy variation DH accompanying the precipitation of b in the supersaturated a solid solution? Represent DH on the graph of DHm ðX Þ. Is the precipitation reaction exothermic or endothermic? Justify. Give, without calculation, the sign of the entropy variation DS caused by the precipitation reaction. Does the sign of DS contradict the second principle of thermodynamics? 7. Study the extrema of G ðX Þ and calculate the shape of the graph T ¼ f ðX Þ that represents the solubility gap. Find the symmetries of the graph T ¼ f ðX Þ. What is the relationship between the solubility limits of B atoms in α and β phases, Xa and Xb ?

Thermodynamics of Equilibria and Phase Diagrams

33

8. Represent G ðX Þ when GA differs from GB with GA \GB . Does this change (i) the shape of the miscibility gap T ðX Þ, (ii) its symmetry properties and (iii) the value of solubility limits Xa and Xb ? Demonstrate these properties using thermodynamic equilibrium conditions (equality of chemical potentials of A and B atoms). For that purpose, use the form of the chemical potentials in regular solutions: li ¼ Gi þ NZ eð1  Xi Þ2 þ RT ln Xi , with i = A or B. 9. Give the equilibrium law T ¼ f ðX Þ. For what value of X ¼ XC , is the solubility Xa maximum (and Xb minimum)? By doing a Taylor expansion to the first order around XC , calculate the expression for critical temperature TC ¼ T ðXC Þ as a function of e and Z . Represent the miscibility gap T ðX Þ. 10. When T \TC , G ðX Þ have two inflection points between which the curvature of G ðX Þ is negative. This is a zone of thermodynamic instability. Very small fluctuations in concentration around the nominal composition (X0 ) leads to a decrease in the overall free enthalpy averaged over the two extreme concentrations (X , X þ ). The system then is unstable in this region of the phase diagram. Very small fluctuations in concentration will then spontaneously increase with time. This regime is known as the spinodal decomposition. Calculate the shape of the spinodal line of instability by finding the location of the inflection points of G ðX Þ. Deduce the locations of the spinodal X ðT Þ. For what temperature do the inflection points disappear? Find the expression of the critical temperature TC . 11. Is it possible to derive the solubility limits Xa ðT Þ and Xb ðT Þ from the previous expression of T ðX Þ without any approximation? Calculate them in the low solubility approximation Xa \\1. 12. The structure of the precipitating b phase is now different from that of the parent phase a. The free enthalpies of pure B in a and b phases are now noted Ga and Gb respectively. Find the simplified expression of the solubility of B atoms in a phase Xa ðT Þ. For this purpose, let assume that the solid solution a is dilute (Xa \\1) and that b is nearly pure in B (Xb  1). Write down the equality of chemical potentials in both a and b phases and show that the solubility limit takes the form Xa ¼ AeQ=RT . Give the expressions of A and Q as a function of (i) entropies Sa and Sb , (ii) enthalpies Ha and Hb of pure phases, (iii) and of e. 13. We will apply the calculations to AlSi alloys (FCC structure) in which silicon (of diamond structure) precipitates. The phase diagram shows a very large miscibility gap (Al-Si + pure Si solid solution) ending up with a eutectic plateau at 577 °C with a eutectic point located in X = 12.6at.%. silicon. The melting point of aluminium is 660 °C and that of silicon 1414 °C. The solubility of silicon in aluminium, as given by the phase diagram, is 1.25at.%. at 550 °C and 0.46at.%. at 450 °C. Represent schematically the phase diagram. Deduce the values of the two constants Q and A involved in the expression of Xa ðT Þ. Calculate the solubility of Si in Al at 200 °C. 14. Let us assume that Ha  Hb . Deduce the value of both the order energy e and the theoretical critical temperature TC . Compare TC to the melting point of aluminium (660 °C) and that of silicon (1414 °C). What is your conclusion? Has TC any physical meaning here?

34

Diffusion, Segregation and Solid-State Phase Transformations

Solution to the problem 1. The order energy is written: 1 e ¼ eAB  ðeAA þ eBB Þ 2 e\0 reflects a tendency to ordering, A atoms “prefer” to surround themselves with B. In contrast when e [ 0, the trend is to clustering of B atoms (precipitation). 2. The number of bonds is NZ =2 and the probability to have A–B bonds writes X ð1  X Þ. X is the atomic fraction of B atoms, Z the coordination number and N the Avogadro number. The enthalpy of mixing is therefore: DHm ¼ NZ eX ð1  X Þ DHm has therefore a parabolic shape and presents a symmetry around X ¼ 1=2 because replacing X by 1  X does not change DHm . Schematically, we can represent DHm as follows:

3. The entropy of mixing is written: DSm ¼ R½X ln X þ ð1  X Þ lnð1  X Þ with R the gas constant. Again this term is symmetric around X = 1/2 so that the corresponding free enthalpy is also symmetric (see following figure). We have represented the entropy of mixing in the previous graph for a given temperature, say T1. m 4. We obtain dDH dX ¼ NZ eð1  2X Þ and we can therefore deduce the slopes of the enthalpy of mixing in 0 and 1:

Thermodynamics of Equilibria and Phase Diagrams

35

dDHm X!0 ! NZ e dX and dDHm X!1 ! NZ e: dX The entropy term T DSm has the following slopes: T

dDSm X X!0 ! 1 ¼ RT ln 1X dX

and T

dDSm X!1 ! þ 1: dX

In addition, T

dDSm X!1=2 ! 0: dX

The term T DSm is represented in the previous graph for two temperatures T1 and T2 , with T2 [ T1 . It is easy to see that the entropic term dominates the enthalpic term for low atomic fractions of B. In contrast, when X increases, it is the enthalpic term that can take over at low temperatures. As a result, DGm ¼ DHm  T DSm presents 2 minima (see the following graph) below a certain critical temperature TC . For T [ TC , entropy always dominates, whatever X, and thus DGm has only one minimum. We speak then of complete miscibility between A and B chemical species.

36

Diffusion, Segregation and Solid-State Phase Transformations

5. The limits of solubility are given by the rule of the common tangent that expresses thermodynamic equilibrium (equality of chemical potentials: laA ¼ lbA and laB ¼ lbB ), see graph below (for GA ¼ GB ). Moreover, if we define Xa and Xb as the solubility limits of B atoms in both α and β phases, then 1  Xa and 1  Xb give those of A (ΔGm is symmetric). We notice that at low temperatures, the enthalpic term dominates DGm . However, the entropy term dominates DGm for X ¼ 0 and X ¼ 1 because of asymptotes related to the entropy term (previous question). Thus, the limits of solubilities tend towards 0 and 1 when temperature approaches 0 K. This agrees with the 3rd principle of thermodynamics that states that entropy is zero at 0 K. It is then impossible to mix A and B species. On the other hand, when temperature increases, solubilities tend towards X ¼ 1=2, entropy of mixing dominates and the mixture is maximum.

6. At equilibrium, X0 is the weighted sum of the atomic fractions of B in a and b and X0 Xa . By definition, fm is written: X0 ¼ ð1  fm ÞXa þ fm Xb . We deduce that fm ¼ X b Xa

is the equilibrium molar function of β phase and 1  fm that of a. The enthalpy of the alloy is therefore a weighted sum of phase enthalpies: H0 ¼ fm Hb þ ð1  fm ÞHa . As shown in the graph below, DHm ðX0 Þ [ H0 in the two-phase domain:

Consequently, the enthalpy variation associated with precipitation DH ¼ H0  DHm ðX0 Þ is negative (the system decreases its enthalpy by releasing heat: DH ¼ DQ\0). The reaction is then exothermic. Moreover, we notice that H0 ¼ Ha ¼ Hb , because GA ¼ GB and HA ¼ HB .

Thermodynamics of Equilibria and Phase Diagrams

37

Precipitation of the β phase leads to the formation of regions rich in A and others rich in B. Because the system partitions, disorder is smaller. Consequently, configuration entropy decreases when precipitation develops: DS\0. We can show that DS\0 is compatible with the second principle of thermodynamics. For that purpose, let us write the required condition for the phase transformation to occur: DG ¼ DH  T DS\0 and DH ¼ DQ\0, we then retrieve the second principle: DS [ DQ T (equality is achieved when DG ¼ 0). Consequently, the entropy variation can be negative because the system is not isolated (DQ\0). If the system is isolated (no heat exchanged with outside: DQ ¼ 0), we find DS [ 0. Entropy necessarily increases. 7. The free enthalpy of the system can be written in a general way as follows: G ¼ DHm  T DSm þ ð1  X ÞGA þ XGB When GA ¼ GB , the minima of G gives the phase solubility limits (Xa and Xb ). m Writing dDG dX ¼ 0 leads to the expression of the solubility: T ðX Þ ¼ 

Z eð1  2X Þ X k ln 1X

where k ¼ R=N is the Boltzmann constant. We easily see that T ðX Þ ¼ T ð1  X Þ, which again reflects the symmetry around X ¼ 1=2. Indeed, if GA ¼ GB , atoms A and B are not distinguishable and are interchangeable. Swapping X and 1  X does not change results (this was already visible in the enthalpies and entropies of mixing). As a result, we have Xa ¼ 1  Xb . 8. When GA \GB , G ðX Þ can be sketched as follows:

The equilibrium condition is written as the equality of chemical potentials in both α and β phases, laA ¼ lbA and laB ¼ lbB . For a regular solution, chemical potentials are given by li ¼ Gi þ NZ eð1  Xi Þ2 þ RT ln Xi, with i = A or B. It is immediate to see that free enthalpy of pure compounds (Gi ) are eliminated when writing equilibrium. The values of Gi have therefore no influence on solubility

38

Diffusion, Segregation and Solid-State Phase Transformations

limits and the shape of the phase diagram is preserved8. It is worth mentioning that when GA 6¼ GB the solubility limits (as given by the common tangent) are no more given by the minima of G ðX Þ. 9. For a regular solution, it has been shown that the solubility limits remain unchanged when GA 6¼ GB . Thus, we get the same equation of T ðX Þ as that demonstrated in question 7. Since T ðX Þ ¼ T ð1  X Þ, the solubility gap is symmetric with respect to X ¼ 1=2 and then the maximum solubility is necessarily obtained for X ¼ 1=2 and Xa ¼ 1  Xb . The critical temperature can easily be calculated as follows. Let us write X ¼ X¼1=2

1=2 þ x with x a very small quantity. As T ðX Þ ! TC , 1st order development of T ðX Þ for x ! 0 leads to the following limit: T ðx Þ ¼ 

Z eð2x Þ Z e 2x Z e ¼ ¼ : k lnð1 þ 2x Þ  k lnð1  2x Þ k 4x 2k

Thus TC ¼ Z2ke. We can therefore graphically represent the miscibility gap:

10. The locations of the spinodal decomposition are given by the inflection points of 2 the free enthalpy, i.e. when ddXG2 ¼ 0, which leads to the following condition: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  1  2kT =Z e : X¼ 2

8 This can be demonstrated graphically. The first derivative of G ¼ DHm  T DSm þ ð1  X ÞGA þ XGB with respect to X shows that the slope is increased by GB  GA when GA 6¼ GB . We have therefore dG dX ¼ GB  GA both in Xa and Xb (the solubility for GB ¼ GA ) because the slope of the common tangent was zero for GA ¼ GB . Free enthalpies are increased by ðGB  GA ÞX in Xb and Xa and therefore are on the same straight line (the common tangent of G of slope GB  GA ). The solubility limits are therefore unchanged for GA 6¼ GB .

Thermodynamics of Equilibria and Phase Diagrams

39

The spinodal limit is shown in the previous figure (dotted
line). Two solutions are  possible if the discriminant of the equation is positive T \ Z2ke . They merge in T ¼ Z2ke ¼ TC as expected. Using again X ¼ 1=2 þ x, one gets the following relation:
 T ðx Þ ¼ TC 1  4x 2 : The shape of the spinodal line is thus parabolic near the critical temperature. . This equation has no analytical solution for 11. We found that T ðX Þ ¼ 2TC 12X ln X 1X

X ðT Þ. An approximated solution can be found for X\\1: X  e2TC =T . 12. We write the equality of chemical potentials for B atoms: lbB ¼ laB . Using the expressions of chemical potentials for Xa \\1 and Xb  1, we obtain: Ga þ RT ln Xa þ NZ e ¼ Gb and therefore, Xa ¼ eðGa Gb þ NZ eÞ=RT : As G ¼ H  TS, we get Xa ¼ AeQ=RT where A ¼ eðSa Sb Þ=R and Q ¼ Ha  Hb þ NZ e. 13. Composition data lead to the following schematic Al-Si phase diagram that exhibits an eutectic reaction (liquid → α + Si) at 577 °C:

Let us write the solubility limits for two different temperatures: Xa1 ¼ AeQ=RT1 X1

R and Xa2 ¼ AeQ=RT2 . We can then deduce Q ¼ 1=T2 1=T ln Xa2  49 431 J/mole 1 a (we have used R = 8.31). As a consequence, A ¼ 17:21. The solubility limit of Si in the phase a at 200 °C is very low  0.006at.%.

Diffusion, Segregation and Solid-State Phase Transformations

40

14. The value of Q (question 12) for Ha  Hb gives e  43 meV. The theoretical critical temperature is thus TC ¼ Z2ke  2991 K for Z ¼ 12. TC is well above the melting temperature of Si. The miscibility gap is therefore only partially visible because it intercepts solidus and liquidus lines. However, TC has no physical meaning here as Al and Si have different structures. Note that Al-Si phase diagram is not symmetrical. This is because AlSi alloy does not behave as a single regular solution since Al and Si have different structures.

1.2.4

Order–Disorder Transformations and Bragg–Williams Theory

1. Generalities on first order transformations. We consider in a general way first order transitions. TC is the critical temperature. Let us define the gain of free enthalpy DG ¼ Gf  Gi ad that of enthalpy DH ¼ Hf  Hi related to ordering or disordering transformations. Subscripts i and f refer to the initial and final states, respectively, before and after transformations. 1.1. Represent H ðT Þ and indicate DH at TC . How is called jDH ðTC Þj? 1.2. What is the sign of DH for ordering at a temperature T \TC ? Is the transformation endothermic or exothermic? T is assumed hereafter to be very close to TC . 1.3. Calculate the related entropy change DS ¼ Sf  Si at TC . What is its sign for ordering at the temperature T ? Does this result contradict the second principle of thermodynamics? 1.4. Answer the same two previous questions for disordering at a temperature T þ [ TC . T þ is also assumed to be very close to TC . 1.5. Express DG for T close to TC as a function of DH , TC and T . Both DH and DS will be assumed to be equal to their values at TC since T is close to TC . 1.6. Demonstrate that DG ðT Þ remains negative for both ordering and disordering transformations. Represent DG ðT Þ for both ordered and disordered phases and show on this diagram that DG\0 for both transformations at T and T þ : 2. B2 ordered structures and CuPd alloys. We consider a disordered cubic centred (BCC) AB alloy transforming into an ordered phase of type B2 in which A and B atoms are preferentially placed in sites 1 (corners) and 2 (centres). 2.1. Represent this structure and give the expression of the related static concentration wave. Remind the meaning of each term. Is the order parameter a scalar or a vector? 2.2. According to Landau’s rules, can the order–disorder transition be of 1st or 2nd order? 2.3. Give the expression of the energy e necessary to create a pair AB as a function of pair energies (eii , i = A, B). What is the sign of ε for ordering?

Thermodynamics of Equilibria and Phase Diagrams

41

2.4. We now consider ordering in CuPd alloys (A = Cu, B = Pd). The phase diagram (figure below) shows the domain of existence of the β Cu-Pd ordered phase (B2). Cu and Pd form a random solid solution (α) outside of this order domain. Reproduce schematically the phase diagram and indicate two-phase domains.

2.5. As shown in the phase diagram, the ordered phase b can deviate significantly from the ideal 50–50 stoichiometry. We know that copper atom (rCu ¼ 1.28 Å) is smaller than palladium atom (rPd ¼ 1.37 Å). Considering this size effect, how are the deviations from the ideal stoichiometry accommodated (vacancies, anti-sites?). Give the answer for an atomic fraction of Pd X [ 0:5 and X\0:5? 2.6. Considering the phase diagram, is the transition of 1st or 2nd order? Justify your answer using Ehrenfest’s rules. 3. The Bragg–Williams mean field theory [16] allows to calculate the temperature evolution of the equilibrium order parameter gðT Þ. For the sake of simplicity, we consider a stoichiometric alloy (AB, with X ¼ 0:5). We shall calculate successively (i) the molar enthalpy H, (ii) the molar entropy S and finally (iii) the free enthalpy G ¼ H  TS as a function of the order parameter g. 3.1. We will calculate the molar enthalpy of the system H ðgÞ. To do this, express H as a function of both the numbers of pairs of first neighbour atoms nij and the interaction potentials between first neighbours (eij with i = A, B, j = A, B). Note that eij are negative. Give the expression of the number of pairs of first neighbour atoms (Np ) for one mole of atoms as a function of both the coordination number Z and the Avogadro number N . Deduce the expression for nij as a function of the probabilities of the presence of atoms A and B in sites 1 and 2 (PA1 ; PA2 ; PB1 ; PB2 ). Then express the probabilities PA2 , PB1 and PB2 as a function of PA1 . Deduce the expressions of the four probabilities PA1 ; PA2 ; PB1 ; PB2 as a function of the order parameter g as defined by g ¼

PA1 XB 1XB .

Deduce the expression of H ðgÞ as a

Diffusion, Segregation and Solid-State Phase Transformations

42

3.2. 3.3.

3.4. 3.5.

3.6.

3.7.

3.8.

3.9.

function of effective interactions eij . Show that H ðgÞ ¼ H0 þ DH ðgÞ. Give the expressions of H0 as a function of eij and DH ðgÞ as a function of e and η. What is your interpretation of H0? Does H ðgÞ have parity? What are your interpretations of H0 and DH ðgÞ? What is the sign of DH ðgÞ? Give the value of the ordering enthalpy for perfect order: H ðg ¼ 1Þ. Does it depend on homo-atomic interactions (between atoms of same nature)? Justify your answer. Represent schematically H ðgÞ. Is the ordering reaction exothermic or endothermic? We are now interested in the molar configuration entropy of mixture (S) as a function of g. As entropy is an extensive variable, S can be written as the sum of entropies related to atomic sites 1 and 2: S ¼ S1 þ S2 . It is reminded that for one
imole:i  i i Si ¼  kN 2 PA ln PA þ PB ln PB . k is the Boltzmann constant. Compare S1 to S2 . Deduce the expression of entropy S ðgÞ. Does S ðgÞ have parity? Calculate the derivative of S with respect to g. Study the limits for g ¼ 0 and g ¼ 1. Represent S ðgÞ. What is the sign of the entropy variation related to ordering? Is it expected and why? Has the free enthalpy G ðgÞ ¼ H ðgÞ  TS ðgÞ a parity? What is the consequence (Landau’s rules)? So as to study the evolution of the equilibrium order parameter g as a function of T , calculate the minimum of G ðgÞ. Show that writing thermodynamic equilibrium leads to the following master equation: f ðgÞ ¼ AðT Þg with f ðgÞ a function depending only on g and AðT Þ [ 0. Give the expression of AðT Þ and f ðgÞ. The equation f ðgÞ ¼ AðT Þg is transcendental and therefore has no analytical solution. We shall solve this problem graphically. Represent graphically both f ðgÞ and AðT Þg for decreasing temperatures from T [ TC to low temperatures T \\TC . Sketch the evolution of gðT Þ. Derive the critical temperature TC from the first derivative of f ðgÞ in g ¼ 0. Deduce the expression of AðT Þ as a function of TC . Calculate the expression of the variation of enthalpy DH ðg ¼ 1Þ as a function of TC . What is its sign? Conclusion? Determine the expression of DCp ¼ dDH =dT as a function of g and TC . Sketch DCp ¼ f ðT Þ.

4. Application of the Bragg–Williams theory to CuPd system. Despite the Bragg– Williams model does not account properly for first order transition (no discontinuity of H ðT Þ or gðT Þ at TC is predicted), we shall apply this theory to CuPd but care should be taken in the interpretation of the result. 4.1. Knowing that TC = 550 °C, calculate the related value of the order energy (e) in eV. 4.2. Give the expression for the binding energy UA as a function of eAA . For that purpose, use the previous expression of the number of pairs Np . We give UCu = 336 kJ/mole and UPd = 376 kJ/mole. Deduce the two homo-atomic interaction energies eii with i = Cu, Pd (in eV). Deduce the value of the pair interaction potential eCuPd (in eV).

Thermodynamics of Equilibria and Phase Diagrams

43

4.3. What amount of molar heat is exchanged with outside (Q ¼ DH ) for total ordering (g ¼ 1)? 4.4. Show how ordering appears in X-ray or electron diffraction. For that purpose, sketch the cross-section of reciprocal lattice (h k 0). Solution to the problem 1. Generalities on first order transformations. 1.1. H ðT Þ increases with temperature since its derivative is equal to the   . specific heat that is a positive quantity: Cp ¼ dH dT p

Hd and Ho are the enthalpies of the disordered and ordered states at TC . For first order transitions, there is discontinuity of H ðT Þ at TC . H ðT Þ increases abruptly of DH . jDH ðTC Þj is the latent heat of transformation. 1.2. DH ¼ Hf  Hi ¼ Ho  Hd \0 for ordering (see previous figure). The enthalpy of the system decreases through the release of heat (Q ¼ jDH j). Ordering is exothermic. DH ðT Þ  DH ðTC Þ for T very close to TC , and DH is negative at T ¼ T . 1.3. DG ¼ DH  T DS. At T ¼ TC , we have DG ¼ 0 (figure below), then DS ðTC Þ ¼ DH ðTC Þ=TC . As DH ðT Þ  DH ðTC Þ\0, therefore DS ðT Þ\0. As expected the entropy decreases since the system gets ordered. This does not contradict the second principle of thermodynamics that states DS  0 for isolated system because here the system exchanges heat (Q) with the outside and is not isolated. 1.4. For T ¼ T þ slightly above TC , DH ¼ Hf  Hi ¼ Hd  Ho [ 0. The system needs energy for the disordering transformation to operate. Disordering is endothermic. DS ðT þ Þ [ 0 since DH ðT þ Þ [ 0. This does not contradict the second principle of thermodynamics.

44

Diffusion, Segregation and Solid-State Phase Transformations 1.5. DG ¼ DH  T DS. Since we work near TC , DS ’ DH ðTC Þ=TC and DH ’ DH ðTC Þ. Thus, DG ¼ DH ðTC  T Þ=T . When ordering the system at T , TC  T [ 0 and DH \0 (exothermic reaction). Consequently, DG ðT Þ\0. Conversely, for disordering at T þ , TC  T \0 and DH [ 0 (endothermic reaction) and then again DG ðT þ Þ\0. G necessarily decreases for any transition to occur. As expected, the curve G ðT Þ shows that DG ðT Þ\0 for both ordering and disordering transformations (figure below). Since the slope of G ðT Þ is equal to the entropy and Sd [ So , G ðT Þ decreases more rapidly with T for the disordered state as compared to the ordered state. The critical temperature is that for which free enthalpies related to ordered and disordered states are equal. The equilibrium state at a given temperature is given by the smallest value of G ðT Þ (ordered at T \TC and disordered at T [ TC ).

2. B2 ordered structures and CuPd alloys. 2.1. The B2 structure is shown below:

Thermodynamics of Equilibria and Phase Diagrams

45

A atoms are in dark grey (site 1) and B in light grey (site 2). Ordered structure B2 can be described by a static wave of concentration along (alternated planes composed of A and B): Cn ¼ C ð1 þ geiq:rn Þ with C the atomic fraction of A in the alloy (C = 1/2 here). The order parameter g is scalar. There is only one wave vector q ¼ 2p a ð1 0 0Þ that generates all superstructure spots from fundamental ones. As a reminder, the reciprocal lattice of the real centred cubic (BCC) lattice is a face-centred cubic (FCC). r1 ¼ ð0 0 0Þ and r2 ¼ a2 ð1 1 1Þ are the coordinates of corners and centres of the B2 structure (2 atoms per unit cell). We have C1 ¼ C ð1 þ gÞ and C2 ¼ C ð1  gÞ. For a perfect order (η = 1), A atoms occupy only sites 1 and B sites 2. In the second order variant, conversely, sites 1 and sites 2 are, respectively, occupied by B and A atoms. One variant can be derived from the other, generated through the translation vector 1=2ð1 1 1Þ, and g transforms into g in the expressions of concentration waves. 2.2. The two variants, which are out of phase obviously have the same free energy, and thus F ðgÞ ¼ F ðgÞ. The functional F ðgÞ is therefore even. According to Landau’s rules, order–disorder transition can be either of second or first order. 2.3. The energy to create a heteroatomic AB pair e ¼ eAB  12 ðeAA þ eBB Þ is negative when chemical ordering prevails (A atoms “prefer” to be surrounded by B atoms and vice versa). 2.4. The phase diagram can be represented as follows:

2.5. When the system is not stochiometric with a Pd excess (X [ 0:5), it is difficult to place big Pd atoms in excess in small Cu sites. The presence of vacancies on the Cu sites then allows to accommodate the substoichiometry in Cu. In contrast, when X\0:5 small Cu atoms in excess are located in large Pd sites. One then has the formation of anti-sites that accommodate the Pd substoichiometry. 2.6. Since there is a two-phase region according to the phase diagram, where ordered and disordered phases coexist, this is a 1st order transition.

46

Diffusion, Segregation and Solid-State Phase Transformations

3. The Bragg–Williams mean field theory 3.1. The number of pairs of first neighbour atoms is: Np ¼ NZ 2 . We deduce the number of pairs nAA ¼ Np PA1 PA2 , nBB ¼ Np PB1 PB2 and nAB ¼ Np PA1 PB2 þ PA1 XB 1XB , when PA1 þ PA2 1g ¼1 2 2 . As

Np PA2 PB1 . According to the expression of g ¼

XB ¼ 1=2 we

have PA1 ¼ 1 þ2 g, and thus PB1 ¼ 1  PA1 ¼  XB ¼ 12, we 2 1 1 2 have PA ¼ PB , likewise PA ¼ PB . We can deduce the enthalpy: H ¼ nAA eAA þ nAB eAB þ nBB eBB ¼ H0 þ DH ðgÞ NZ 2 with DH ðgÞ ¼ NZ 4 g e and H0 ¼ H ðg ¼ 0Þ ¼ 8 ðeAA þ eBB þ 2eAB Þ. As interaction energies are negative (eij are potential energies), H0 is negative. Note also that DH ðgÞ ¼ DH ðgÞ, and therefore DH ðgÞ has parity. 3.2. H0 ¼ H ðg ¼ 0Þ corresponds to the enthalpy of the disordered state and since e\0, we have DH ðgÞ\0. Consequently, enthalpy decreases when order increases. 3.3. H ðg ¼ 1Þ ¼ NZ 2 eAB . This term does not depend on eAA and eBB . It is immediate to see that a perfect ordered system does not contain any AA and BB pairs. 3.4. We can represent H ðgÞ in the following way:

DH ðgÞ ¼ H ðgÞ  H ð0Þ\0. The system loses energy and releases the following transformation heat: DQ ¼ jDH j. Ordering is exothermic. 3.5. Because sites 1 and 2 are indiscernible, S1 ¼ S2 . We then find that

 1g S ¼ R 1 þ2 g ln 1 þ2 g þ 1g 2 ln 2 . Moreover, S ðgÞ ¼ S ðgÞ. 1þg R 3.6. We find dS dg ¼  2 ln 1g that tends towards 0 when g ! 0 and towards 1 when g ! 1. We see that dS dg \0. This could be anticipated as entropy is known to decrease when order increases. S ðgÞ is represented below.

Thermodynamics of Equilibria and Phase Diagrams

47

3.7. As G ðgÞ ¼ G ðgÞ, according to Landau’s rules, B2 ordering in BCC can  ¼ 0 , be either a first or second order transition. Writing equilibrium dG dg þg Ze ¼  kT g. This latter equation can be one gets the master equation ln 11g

þg Ze and A ¼  kT . written in the form f ðgÞ ¼ Ag with f ðgÞ ¼ ln 11g 3.8. f ðgÞ and Ag are plotted below in order to determine graphically the solutions of the master equation f ðgÞ ¼ AðT Þg:

AðT Þ decreases for increasing temperatures. This representation allows us to understand the behaviour of temperature dependence of the order parameter: gðT Þ. At T2  TC , A is small and the only solution of f ðgÞ ¼ Ag, is g ¼ 0. When T1 \TC one gets another non-zero value of g, which increases with decreasing values of temperature. The order parameter decreases in a continuous way with T up to TC as shown below. The function f ðgÞ presents a vertical asymptote in g ! 1, therefore at zero temperature the equilibrium value of the order parameter tends toward 1.

48

Diffusion, Segregation and Solid-State Phase Transformations

The derivative of f ðgÞ reads f 0 ðgÞ ¼ ð1gÞ2ð1 þ gÞ. We have f 0 ðgÞ ¼ 2 at TC

when g ¼ 0. As f 0 ðg ¼ 0Þ = AðT ¼ TC Þ, then TC ¼  Z2ke and A ¼ 2TC =T . 3.9. We have DH ðg ¼ 1Þ ¼  RT2 C \0 that shows again that, as expected, we find enthalpy decreases when order increases. DCp ¼ dDH dT , dg DCp ¼ RTC g dT .

4. Application of the Bragg–Williams theory to the CuPd system. C 4.1. e ¼  2kT Z so that e ¼ 17:7 meV for Z ¼ 8 (BCC coordination number). NZ 4.2. UA ¼  2 eAA so that eCuCu ¼ 0:87 eV and ePdPd ¼ 0:98 eV. Note that eAA are potential energies and are therefore negative. The value of eCuPd ¼ 0:94 eV can easily be derived from that of ε. 4.3. Q ¼ 3:42 kJ. 4.4. The diffraction pattern (for l ¼ 0) is the cross-section of the reciprocal lattice for l ¼ 0:

Thermodynamics of Equilibria and Phase Diagrams

1.2.5

49

Ordering in FCCs and Landau’s Theory

1. We consider a FCC alloy forming an ordered phase A3B of type L12. Let ri (i = 1, 2, 3, 4) be the position vector of atoms in the FCC structure and a is the lattice parameter. Represent the structure and give the expression of the concentration waves of B atoms C ðri Þ. Write down the latter for the four order variants of the L12 structure. Deduce the expression of the four related order vectors g. Is the change from one variant to another simply reduced to a change in the sign of g as in BCC? Does the free energy functional F ðgÞ have parity? According to Landau’s rules, what is the order of the transition? 2. We consider the following energy functional: F ðgÞ ¼ F0 þ a ðT  T0 Þg2 þ bg3 þ cg4 , with T0 the “instability” temperature of the disordered state. a, b and c are constants. For the sake of simplicity, g is here a scalar that is taken as the value of one of the three components of g. Considering Landau’s rules, is it a 1st or a 2nd order transition? Justify your answer qualitatively, without calculation. 3. What must be the concavity of F ðgÞ at g ¼ 0 for temperatures below and above T0 respectively? What shall be the sign of a ? 4. Study the extrema of the functional F ðgÞ. What must be the sign of c so that only the disordered state remains stable above a certain temperature T þ . Calculate temperature T þ ¼ f ðT0 Þ. Among these extrema, denoted by g (the two solutions g þ and g of a second order equation), specify without calculation which one corresponds to a minimum? Represent schematically F ðgÞ. 5. Give the value of gm related to the minimum of F ðgÞ for η ≠ 0. What is the sign of b? Justify. What is the value of the minimum state g0 at T0 ? Compare g0 and g þ þ, the value of g þ in T ¼ T þ. Compare also F ðg0 Þ and F0 . How does gm vary when T decreases? Same question for g . Is there a temperature Tc where there is coexistence of both ordered and disordered states? So as to calculate Tc , write down a first equation expressing this coexistence. Write down a second equation showing the existence of a minimum of F for a non-zero value of gc .

50

Diffusion, Segregation and Solid-State Phase Transformations

6. We shall now calculate Tc and gc as a function of constants (a, b, c) using the two latter 2nd order equations. Subtract both equations so that to express Tc as a function of T0 , gc , b and a. Compare both temperatures. Deduce the expression of gc as a function of constants and shows that it is proportional to g0 . Compare the three values gc , g0 , g þ þ . Is that expected? Give the final expression of TC as a function of constants T0 , a, b and c. To which energy
 is proportional Tc in the Bragg–Williams mean field theory? Show that F g þ [ F0 at T [ TC so that the ordered state is metastable for such temperatures. 7. Represent qualitatively the evolution of gðT Þ and below that of the specific heat Cp ðT Þ without calculation. Is disordering endothermic or exothermic? Why is it so? 8. Synthesis – Represent F ðgÞ for various decreasing temperatures from T [ T þ to T \T0 . Indicate stable, unstable and metastable states on the graph using symbols S, U, and M, respectively. Represent schematically gðT Þ and show hysteresis effects due to metastability of both order and disorder states. Describe qualitatively and succinctly the various ordering kinetic regimes for decreasing temperatures. 9. We now consider a perfectly ordered alloy. What happens when temperature increases? Describe the different kinetic modes of transformation using carefully chosen temperature ranges. What will be the influence of antiphase boundaries in the early stages of disordering? Solution to the problem 1. The concentration waves related to B atoms are written:

 C rj ¼ C0 1 þ g1 eiq1 :rj þ g2 eiq2 :rj þ g3 eiq3 :rj 2p 2p with q1 ¼ 2p a ð100Þ, and q2 ¼ a ð010Þ q3 ¼ a ð001Þ the three wave vectors related to the L12 ordered structure sketched below (B in black). C0 ¼ 1=4 (Stoichiometry of A3B ordered structure).

Thermodynamics of Equilibria and Phase Diagrams

51

r1 ¼ a ð000Þ, r2 ¼ a2 ð110Þ, r3 ¼ a2 ð011Þ, r4 ¼ a2 ð101Þ are the position vectors of atomic sites. Occupation probabilities of the four sites read: site 1: C ðr1 Þ ¼ 14 ð1 þ g1 þ g2 þ g3 Þ. site 2: C ðr2 Þ ¼ 14 ð1  g1  g2 þ g3 Þ. site 3: C ðr3 Þ ¼ 14 ð1 þ g1  g2  g3 Þ. site 4: C ðr4 Þ ¼ 14 ð1  g1 þ g2  g3 Þ. It is straightforward to see that the order parameter gi according to wave vector qi is the difference in concentration between mixed planes and pure planes of the ordered structure. Along x, y, and z axes of FCC we have, respectively: g1 ¼ C ðr1 Þ þ C ðr3 Þ  C ðr2 Þ  C ðr4 Þ, g2 ¼ C ðr1 Þ þ C ðr4 Þ  C ðr2 Þ  C ðr3 Þ, g3 ¼ C ðr1 Þ þ C
ðr2Þ  C ðr3 Þ  C ðr4 Þ. g ¼ 0 when C rj ¼ 1=4 whatever j (disorder). The first order variant is such that B is located in site 1. The concentration (occupancy probability) is thus C ðr1 Þ ¼ 1 and the other occupancy probabilities of sites 2, 3 and 4 are zero. For this variant, the order vector is: g ¼ ð111Þ. The second variant is generated in placing B on site 2. Then C ðr2 Þ ¼ 1 and the other site occupancy probabilities are zero. The order vector reads g ¼ ð1  1 1Þ. Other variants are written by circular permutation: g ¼ ð1  1  1Þ; g ¼ ð1 1  1Þ. In contrast to BCC-B2 ordering it is not possible to derive one variant from the other by simply changing the sign of vector g. Thus, the functional has no parity. According to Landau’s rules, such a functional generates a 1st order transition. 2. It is immediate to see that F ðgÞ has no parity. According to Landau’s second rule, the transition is of the first order. 3. The concavity of F ðgÞ in g ¼ 0 must be positive so as to get a minimum of F ðgÞ when temperature is above T0 (denoted as T0þ hereafter). This is characteristic of a stable or metastable state. Conversely, the second derivative should be negative so as to have a maximum of F ðgÞ when temperature is below T0 (denoted as T0 ). This maximum of F ðgÞ corresponds to an unstable state. As a 2 2 summary, @@gF2 [ 0 for T ¼ T0þ and @@gF2 \0 when T ¼ T0 .  2 2  We find that @@gF2 ¼ 2a ðT  T0 Þ þ 6bg þ 12cg2 and therefore @@gF2  ¼ 2a ðT  T0 Þ. g¼0

Previous conditions on second derivative impose that a [ 0. F ðgÞ for η close to zero and T close to T0 can be sketched as follows:

Diffusion, Segregation and Solid-State Phase Transformations

52 4. Writing

@F @g

¼ 0 to determine extrema leads to: 2agðT  T0 Þ þ 3bg2 þ 4cg3 ¼ 0.

The solution g ¼ 0 is trivial. Other solutions are given by: 4cg2 þ 3bg þ 2a ðT  T0 Þ ¼ 0. The discriminant of this 2nd degree equation reads: D ¼ 9b2  32acðT  T0 Þ. As a [ 0, it is necessary that c [ 0. Indeed, positive value of D whatever T > T0 would lead to an extremum of F(η) with η ≠ 0 whatever T > T0. This is not physical as order state should disappear at high temperatures. The other two non-trivial extrema (g 6¼ 0) are therefore in pffiffiffi D g ¼ 3b . 8c In order that only the disordered state remains for T [ T þ, D should then be negative for such temperatures. T þ is therefore such as D ¼ 0. Consequently: T þ ¼ T0 þ 9b2 =32ac. Let us return to the case for which there are 3 extrema of F for T \T þ and T > T0. As F is convex (@ 2 F=@g2 [ 0) in g ¼ 0, this trivial extremum of F(η) is a minimum. Consequently, the second extremum of F ðgÞ in g is necessarily a maximum and the third extremum in g þ is a minimum as shown below9:

The smallest value of η is g ¼ 3b 8c

pffiffiffi D

. F ðgÞ is concave (@ 2 F=@g2 \0). This pffiffiffi þ D is indeed a maximum of F ðgÞ is an unstable state. F ðgÞ in g þ ¼ 3b8c minimum. 5. The non-trivial minimum of F for η ≠ 0 is in gm ¼ g þ. As D ¼ 9b2  32acðT  T0 Þ and ac [ 0, gm increases when T decreases since D increases. As expected, order increases (entropy decreases) when decreasing T. Inversely, g decreases with decreasing T. 3b In T ¼ T þ, D ¼ 0. The only extremum of F for non-zero η is given by g þ þ ¼ 8c . As g þ [ 0, necessarily b\0 because c [ 0.
 In this figure, we have chosen F g þ [ F0 , the ordered state is therefore metastable. This is the case for temperatures above T0 , but not below the temperature TC , as shown in the next question. 9

Thermodynamics of Equilibria and Phase Diagrams

53

pffiffiffiffiffi pffiffiffiffiffiffiffi 2 Let us take T ¼ T0 . We find g0 ¼ 3b þ8c 9b ¼ 3b 9b2 ¼ 4c where we have used þ 2 2 3b [ 0 since b\0. We see that g0 ¼ 2g þ. F is a minimum since @ F=@g in g0 is equal to 9b2 =4c, and is therefore positive. It can also be seen that the other extremum is g ¼ 0 at this temperature, for which F ¼ F0 and is a maximum. As g0 is a minimum, it necessarily implies F ðg0 Þ\F0 . Indeed, we get F ðg0 Þ ¼ 4

27 b F0  256 c3 \F0 as c [ 0. According to Ehrenfest’s theory, both disordered (g ¼ 0) and ordered (g ¼ gc ) states can coexist at Tc for a 1st order transition. We shall demonstrate that it is the case for the considered functional F(η). gc is an extremum of F and thus necessarily satisfies the condition 4cg2c þ 3bgc þ 2a ðTc  T0 Þ ¼ 0. For the two states to coexist, the related free energies should be equal: F ð gc ; T

c Þ2¼ F ðg ¼ 0; Tc Þ ¼ F0. This condition leads to the second master equation 4 cgc þ bgc þ a ðTc  T0 Þ ¼ 0 for gc 6¼ 0. 6. We deduce by subtraction of the two previous master equations that bgc ¼ 2a ðTc  T0 Þ, and thus Tc ¼ T0  bgc =2a. As b\0 and a [ 0, Tc [ T0 . Finally, we can calculate the value of the ordered state gc using the previous 2 expression of gm for T ¼ Tc . We then find: gc ¼ b 2c ¼ 3 g0 . We see that both þ gc \g0 and gc [ g þ. Injecting this expression of gc in that of Tc gives the expression of the critical temperature: Tc ¼ T0 þ b2 =4ac. The latter is proportional to the order energy e ¼ eAB  12 ðeAA þ eBB Þ in the Bragg–Williams theory where eAA , eBB and eAB are the effective interaction energies of AA, BB and AB pairs. We can check that the ordered
state is metastable when T [ Tc and stable otherwise. Indeed, F g þ  F0 [ 0 implies a ðT  T0 Þ þ bg þ þ cg2þ [ 0. Substituting g þ by its expression leads to T [ Tc with Tc ¼ T0 þ b2 =4ac. 7. Both gm ðT Þ and Cp ðT Þ are represented below. As shown previously, order– disorder transition is endothermic (DCp [ 0). Energy is required to destroy the ordered state (selfish rule: DH [ 0 when the system receives energy).

54

Diffusion, Segregation and Solid-State Phase Transformations

8. Synthesis. We can now represent F ðgÞ schematically:

S, M, and U refer to stable, metastable and unstable states, respectively. Due to the presence of metastable states, gm ðT Þ shows hysteresis (see below) depending on whether temperature is decreased (ordering) or increased (the system gets disordered).

Thermodynamics of Equilibria and Phase Diagrams

55

When ordering the system from a high temperature disordered state to lower temperatures, the initial disordered state (g ¼ 0) becomes metastable (M) for T \Tc and T [ T0 (sketch of F(η) above). There is a barrier between the disordered metastable state (S) and the stable ordered state (S). The transformation kinetics proceeds by nucleation and growth of small ordered domains in the disordered parent phase. Below T0 , the state g ¼ 0 becomes unstable and spinodal ordering develops. In this spinodal regime, the amplitude of concentration waves increases gradually in the parent phase. There is a global and gradual ordering of the whole disordered phase. 9. When T increases from low temperatures and disordering arises for T [ Tc , the initial ordered state (g ¼ gm ) is metastable (M) for T \T þ. Again, there is a barrier between the metastable ordered state and the stable disordered state. We then have the nucleation and growth of disordered domains in the parent ordered phase. The transformation becomes continuous above T þ (spinodal disordering). The amplitude of the order concentration waves gradually decreases during the disordering kinetics. If the system contains antiphase boundaries (APBs), disordering can occur preferentially in these regions because APBs are more disordered regions.

1.2.6

Ordering in BCCs and Landau’s Theory

1. We consider a BCC alloy in which an AB ordered phase of type B2 forms (for example CuZn). Does the free energy functional F ðgÞ have parity and why? 2. We consider the following free energy functional: F ðgÞ ¼ F0 þ a ðT  T0 Þg2 þ bg4 . T0 is the temperature below which the disordered state becomes unstable and F ðg ¼ 0Þ ¼ F0 . According to Landau’s rules, can the transition be of first order or second order?

Diffusion, Segregation and Solid-State Phase Transformations

56

3. What is the sign of the second derivative (concavity) of F ðgÞ in g ¼ 0 for T below and above T0 . Deduce the sign of a. Sketch F ðgÞ close to g ¼ 0 for T below and above T0 . 4. Study the extrema ge of the functional F ðgÞ. Calculate F ðge Þ and the curvature of F in g ¼ ge . Considering the sign of T  T0 , what should be the sign of b? What is your interpretation of each extrema ge ? 5. Is the energy F ðgÞ maximum or minimum in ge ? Sketch F ðgÞ for T \T0 . 6. For which temperature does order vanish? Express the critical temperature of the order–disorder transition (TC ) as a function of T0 . What is the value ge for T T0 . Deduce the expression of TC as a function of constants a and b. Deduce the expression of ge as a function of T and TC . Represent schematically the evolution of jge ðT Þj. Is the evolution continuous? What do you deduce from it? What is the critical exponent? 7. How does the concavity of F ðgÞ vary for g ¼ 0 and g ¼ ge for decreasing temperatures? Deduce the evolution of F ðgÞ for decreasing temperatures from T [ T0 to T \\T0 . Can there be co-existence of ordered and disordered phases? Conclusion? Is it a 1st or 2nd order transition? 8. Is there a latent heat of transformation (L ¼ DH ) at the order–disorder transition? Calculate the expression of the specific heat Cp ðT Þ. For that purpose, express F ðT Þ at equilibrium by considering the evolution of ge ðT Þ. Show that Cp ðT Þ ¼ Cp0 ðT Þ þ DCp ðT Þ. Cp0 ðT Þ is the specific heat for g ¼ 0 and DCp ðT Þ is the specific heat excess due to ordering. Represent schematically Cp ðT Þ. Show that DCp ðT Þ at T0 can be expressed as a function of a. 9. Does the ordering of the initially disordered alloy involve a barrier? Is the regime unstable (spinodal order) or metastable (nucleation and growth of ordered domains)? 10. We now choose b\0. Obtaining solutions to our problem requires adding an additional term cg6 to the previous expression of F ðgÞ. A metastable (or stable) order is then possible above the instability temperature of the disordered state T0 . The study of the new functional F ðgÞ shows that the transition is then of 1st order. Without calculations, represent schematically F ðgÞ at T ¼ TC and for the three following temperatures: T1 [ TC , T2 \TC , T3 \T0 with T2 [ T0 . Solution to the problem 1. In centred cubic structures (BCC) the functional F ðgÞ is even because F ðgÞ ¼ F ðgÞ with g and g the order parameters associated with the two translation variants of the B2 structure (A atoms are located at either the centre or at the corners of BCC structure). 2. F ðgÞ is even, so the transition can be either 1st or 2nd order. 3 @2 F 2 3. @F and For g ¼ 0, @g ¼ 2agðT  T0 Þ þ 4bg @g2 ¼ 2a ðT  T0 Þ þ 12bg . @2 F @g2 @2 F @g2

¼ 2a ðT  T0 Þ > 0 for T [ T0 , F ðgÞ is therefore a minimum. In contrast, \0 for T \T0 since F ðgÞ is a maximum. We can then conclude that a [ 0.

Thermodynamics of Equilibria and Phase Diagrams

4. Let

us

examine

the

extrema

of

the

functional.

57

@F @g

¼0

leads

to

gð2a ðT  T0 Þ þ 4bg Þ ¼ 0. g ¼ 0 is an obvious solution and the two other soluqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tions are ge ¼  aðT2bT0 Þ. Ordering is expected at low temperatures, in particular when T \T0 . We therefore have b [ 0 for T \T0 because a [ 0. These two solutions (ge ) are related to both ordering variants. Moreover, we have: 2

2

T0 Þ F ðge Þ ¼ F0  a ðT4b and  @2 F  @g2  ¼ 4a ðT  T0 Þ. 2

ge

For T \T0 , F ðge Þ is smaller than F0 because b is positive. Moreover, the curvature of F for g ¼ ge is positive as T < T0. The two possible variants are therefore stable. Note that for T [ T0 , it would be necessary that b \ 0 to get non trivial (g 6¼ 0) extrema of F. This would imply that F ðge Þ [ F0 and the curvature of F would then be negative for g ¼ ge . It would be a maximum of F related to an unstable state. There would be no stable ordered state for b \ 0. This confirms that b [ 0 is the valid solution. 5. Let us examine the concavities of the functional for both variants. We have found 2 2 that @@gF2 ¼ 4a ðT  T0 Þ for g ¼ ge , and therefore @@gF2 [ 0 when T \T0 . As a consequence, F is minimum for ge > 0. This an ordered state, as shown below.

58

Diffusion, Segregation and Solid-State Phase Transformations

6. ge ¼ 0 for T ¼ T0 , which implies TC ¼ T0 , disorder is stable above this temperature. For T \\T0 , the order parameter tends to 1 (and 1). Then, when T ! 0, ge ! 1 and thus aT0 =2b ¼ 1. We deduce that TC ¼ 2b=a (TC ¼ T0 ). We can now represent ge ðT Þ as follows:

ge ðT Þ is therefore continuous, which implies that it is a 2nd order transition. It can be shown also that there is a vertical asymptote in T ¼ T0 and the slope is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0Þ 1=2T0 in T ¼ 0. We found that jge j ¼ aðTT . jge ðT Þj decreases with T as 2b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T0  T . The critical exponent is therefore ½. Replacing T0 by its expression pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi leads to jge j ¼ 1  T =T0 . 7. Let us remind that T0 is nothing else than the critical temperature. In the following, we shall keep the notation T0 for the critical temperature. We have 2 2 demonstrated that @@gF2 ¼ 2a ðT  T0 Þ when g ¼ 0. Thus, @@gF2 gets more and more negative when temperature decreases. The curvature of F evolves as shown below:

Thermodynamics of Equilibria and Phase Diagrams For g ¼ ge , we have shown that decreases,

@ F @g2 2

@2F @g2

59

¼ 4a ðT0  T Þ. Thus, when the temperature

increases. We can now represent F ðgÞ for various temperatures10:

As shown in this figure, there is no temperature for which an ordered state has the same free energy as the disordered state: F ðgÞ ¼ F ð0Þ is never observed. There is therefore no possible coexistence of the ordered phase with the disordered phase. We then confirm that the transition is of 2nd order, as previously discussed. The function jge ðT Þj is continuous at the transition temperature T0 ð¼ TC Þ. 8. In a 2nd order transition,   the first derivative of F ðT Þ is continuous in T0 . For @F  @F  T ¼ T0 , then @T T þ ¼ @T T  . As these partial derivatives are equal to entropy, 0

0

then SjT þ ¼ SjT  . There is no entropy variation at T0 . Let us remind that DG ¼ 0 0 DH  T DS ¼ 0 at critical temperature T0 so L ¼ T DS ¼ DH ¼ 0, there is no latent heat.

It is worth reminding that, in contrast to previous representations of F ðgÞ (questions 3, 7 and 10), where F0 is either taken, for sake of clarity, as constant or increasing with temperature, in reality F0 ðT Þ decreases when temperature increases: 10

60

Diffusion, Segregation and Solid-State Phase Transformations   By definition Cp ¼ @H @T P and we have dH ¼ VdP þ TdS, thus @H ¼ T @S at @F  and replacing @H by its expression in the constant pressure P. As S ¼  @T V @ F expression of the specific heat leads to Cp ¼ T @T 2. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ðT T0 Þ Replacing g ¼ ge ðT Þ ¼  in the expression 2b 2

of

F ð gÞ ¼

2

T0 Þ . F0 þ a ðT  T0 Þg2 þ bg4 leads to F ðT Þ ¼ F0  a ðT4b It can be concluded that for T T0 , where non-trivial minima of F ðgÞ exist, we 2 2 get for the specific heat: Cp ¼ T @@TF20 þ ab T ¼ Cp0 ðT Þ þ DCp ðT Þ where 2

2

Cp0 ðT Þ ¼ Cp ðT Þ for g ¼ 0 (F ¼ F0 at T0 ) and DCp ðT Þ ¼ ab T is the specific heat excess due to ordering (b > 0). The latter increases with the temperature as shown in the figure below. As demonstrated before T0 ¼ TC and TC ¼ 2b=a, we 2 2 then have: DCp ðT0 Þ ¼ ab T0 ¼ 2a. When T [ T0 , we have Cp ¼ T @@TF20 . The temperature dependence of Cp ðT Þ is represented below, near TC :

9. The disordered state is unstable and has a negative concavity, there is no nucleation barrier. The ordering kinetic develops in a spinodal regime. The amplitude of concentration waves gradually increases with time in the parent disordered phase. 10. As the transition becomes 1st order, we can anticipate that both ordered and disordered states can coexist at TC as shown below:

Thermodynamics of Equilibria and Phase Diagrams

61

Chapter 2 Diffusion and Transport in Solids 2.1

Course Reminders

Diffusion of atoms (A, B) in alloys can easily be exhibited by making a diffusion couple (A–B) and looking at the evolution of the concentration profiles with time. Let us take for example Cu and Ni, which are two miscible metals in any proportions. Let us bring this couple to 800 °C. The concentration profiles across the Cu–Ni interface show an interdiffusion (migration) of chemical species that increases with time and temperature (figure 2.1). As demonstrated latter, the diffusion flux of ! Ni is given by the first Fick law, ~ J ¼ D r C with D the diffusion coefficient of Ni in Cu. This equation expresses that Ni atoms migrate down the concentration gradient ! (r C ), as shown in figure 2.1. Experiments show that the diffusion rate (diffusion length per unit of time) increases rapidly with temperature. Diffusion [17–21] is thermally activated.

2.1.1

Diffusion at the Atomic Scale

The migration of atoms in a solid in the absence of a driving force (electric, chemical, gravitational…) is a stochastic phenomenon. Atoms move randomly in the three directions of space. This is what happens for instance for small particles suspended in a liquid in the absence of gravitation. This stochastic movement of particles is called the Brownian motion. The distribution of the particle positions along an arbitrary axis x follows a Gaussian distribution (figure 2.2): x 2 1 P ðx Þ ¼ pffiffiffiffiffiffiffiffiffiffi e 2r2 2pr2

P ðx Þ is here the probability density to find the particle at x within dx. The standard deviation of the distribution σ is proportional to the diffusion distance. It pffiffiffiffiffiffiffiffi can be shown that r ¼ 2Dt with D the diffusion coefficient and t the diffusion time.

DOI: 10.1051/978-2-7598-2743-5.c002 © Science Press, EDP Sciences, 2022

64

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 2.1 – (a) Cu–Ni diffusion couple. (b) Ni concentration profile across the Cu–Ni interface. Kinetic evolution at a fixed temperature T for three heat treatment times ðt ¼ 0; t1 [ 0; t2 [ t1 Þ.

FIG. 2.2 – Gaussian distribution P ðx Þ. Particles, initially located at x ¼ 0, have a 5% probability to reach a distance greater than twice the standard deviation σ of the distribution pffiffiffiffiffiffiffiffi 2Dt ) after a diffusion time t.

(r ¼

Diffusion and Transport in Solids

65

Let us now examine the process of random diffusion at the atomic scale. At T > 0 K, atoms vibrate around their position and manage after many attempts to migrate to a neighbouring atomic site in all the three directions of space. The time scale of diffusion process (0.1 ps) is given by the frequency of vibration of atoms which is of the order of 1013 Hz. In crystalline solids, we distinguish two modes of diffusion, the diffusion of interstitials (case of carbon in steels, figure 2.3(b)) and that of atoms in substitution on the sites of the lattice (case of copper in nickel in CuNi alloys). In metals, the diffusion of substitutional atoms proceeds via vacancies. A vacancy should be present in a first neighbour site for atoms to migrate (figure 2.3(a)). In semiconductors such as silicon, the diffusion mechanism is more complex (kick-out mechanism) and can be mixed (interstitial and substitutional). Besides, we speak of self-diffusion for the migration of A atoms in pure A materials and of interdiffusion when B atoms diffuse in A.

FIG. 2.3 – (a) Vacancy diffusion. An atom can only migrate to an adjacent site if it is

unoccupied (occupied by a vacancy). (b) Diffusion of atoms in interstitial position. DGm is the migration barrier.

Einstein was very quickly interested in Brownian motion and proposed an atomistic interpretation of the diffusion coefficient given below: D¼

Z 1X Ci xi2 2 i¼1

The diffusion coefficient is proportional to the mean square distance. The sum is made over the Z first neighbours of atomic sites. xi is the corresponding jump distance projected along the x axis. Ci is the jump frequency of atoms toward site i:

Diffusion, Segregation and Solid-State Phase Transformations

66

Ci ¼

1 si

si is the corresponding relaxation time. In the case of cubic solids, Ci ¼ C0 for all jumps i and: D ¼ C0 a 2 with a the lattice parameter. Let us introduce the average relaxation time s:  2 Z X x xi2 ¼ s s i¼1 i Thus:

 2 x D¼ 2s

Let us consider the mean square diffusion distance q after n jumps in a cubic crystal. The related diffusion time after n random jumps is t ¼ ns, we have: q ¼ pffiffiffiffiffiffiffiffiffi hq2 i with  2   q ¼ 3n x 2 pffiffiffiffiffiffiffiffi       Because x 2 ¼ y 2 ¼ z 2 thus, q ¼ 6Dt . p ffiffiffiffiffiffiffiffi For a 1D diffusion process, q ¼ 2Dt . For interstitial diffusion, the jump frequency C is proportional to the probability for atoms to jump through the migration barrier DGm (figure 2.3), thus: DGm

C0 ¼ te kT

t is the vibration frequency of atoms (  1013 Hz). As mentioned earlier, the diffusion of substitution atoms proceeds via a vacancy mechanism. The jump probability is therefore much lower and is decreased by a factor equal to the atomic fraction of vacancies (probability of presence of a vacancy in a neighbouring site): DGf

Xv ¼ e kT

DGf is the free enthalpy of formation of a vacancy (see problem 1.2.1). Writing DG ¼ DH  T DS in the same way for both the migration and formation terms, we obtain: DS

DH

C0 ¼ te k e kT

DS ¼ DSm þ DSf DH ¼ DHm þ DHf

Diffusion and Transport in Solids

67

DHm and DHf (expressed per atom) are of the order of 1 to a few eV. Diffusion data are however more commonly expressed for one mole, as shown thereafter. For a cubic crystal D ¼ C0 a 2 and D can eventually be put in the form: Q

D ¼ D0 eRT DS

with D0 ¼ ta 2 e R . Q and DS are now expressed for one mole and not per atom. This expression of D as a function of temperature remains valid whatever the structure of the material. Only the expression for D0 changes. The diffusion is a thermally activated phenomenon with a molar energy of activation Q ¼ N DH (N is the Avogadro number). The activation energy (DHm þ DHf ) is larger for substitutional diffusion than for interstitial diffusion (DHm ). This results in a diffusivity D that can be several orders of magnitude smaller. Activation energies and D0 can be derived from the measurement of the diffusion rate at various temperatures. These data can be deduced from an Arrhenius plot of D as represented in figure 2.4.

ln D

ln D0

Q/R

1/ T FIG. 2.4 – Arrhenius representation: ln D is plotted as a function of 1=T . The intersection of the line with ordinate gives ln D0 and its slope is proportional to the activation energy for diffusion Q.

Note that substitutional self-diffusion of A atoms is much slower than the migration of vacancies. The ratio of diffusion rates is equal to Xv . The vacancy diffusivity Dv is written: Dv ¼ DA =Xv DA is the self-diffusion coefficient of A atoms in A. The activation energy Q for self-diffusion of A atoms is approximately proportional to its melting temperature Tf (e.g. Q  18 RTf in cubic crystals). This can be justified by arguing that enthalpies DHm and DHf increase with the binding

68

Diffusion, Segregation and Solid-State Phase Transformations

energy of the crystal and thus with the melting temperature. It should be mentioned that the interdiffusion data (B diffusion in A) as given in literature are generally for dilute alloys in which B concentration is very low (infinite dilution). Things become more complicated for concentrated alloys as we shall see later.

2.1.2

Fick’s Two Laws

The expression for the diffusion flux J per unit area (atoms/m2s) in the presence of a concentration gradient ðC ðx ÞÞ can be found by writing the net flux between two regions distant of dx (figure 2.5): J ¼ J ðx Þ  J ðx þ dx Þ ¼ C0 dx ðC ðx Þ  C ðx þ dx ÞÞ C ðx Þ is the concentration per unit volume of considered species along x direction and C0 is the jump frequency. Unities of J are therefore [m−2s−1]. Writing that dx ¼ a (lattice parameter) we then get for interstitial diffusion in a cubic crystal: J ¼ D

dC dx

We introduce here the diffusion coefficient D ¼ C0 a 2 (for cubic crystals as previously introduced). Unities of D is [m2s−1]. dx being very small, dC =dx is equal to the gradient of C ðx Þ. By generalizing the approach to three dimensions, we find the first law of Fick which links the flux vector to the concentration gradient of the diffusing species: ! ~ J ¼ Dr C Atoms (solutes B or solvent atoms A) diffuse down the gradient of concentration C ðx; y; z Þ. Fick’s second law is the consequence of the conservation of matter at a given location. The temporal evolution of concentration C ðx; y; z Þ in an infinitesimal volume dxdydz is given by the difference between the inflow and the outflow in this elemental volume. This leads to the continuity equation governing the temporal evolution of the concentration in x; y; z: @C ¼ div~ J @t Thus, if D is independent of the space and therefore independent of concentrations, D can be taken out of the divergence and we end up with: @C ¼ DDC @t DC denotes here the scalar Laplacian of C . The time evolution of C , at a given location, is proportional to the local curvature of concentration profile. In a 1D diffusion problem, if the curvature @ 2 C ðx Þ=@x 2 \0 (resp. >0), then C ðx Þ decreases

Diffusion and Transport in Solids

69

FIG. 2.5 – (a) Diffusion of interstitial B atoms in the presence of a concentration gradient.

JB ) migrate (b) Related concentration profile CB ðx Þ. The first Fick law states that B atoms (~ down the concentration gradient. The process stops when the concentration CB ðx Þ becomes constant with x. The flux gets then zero (~ JB ¼ 0) and the system reaches its thermodynamic equilibrium.

(resp. increases) with time. Diffusion therefore tends to attenuate the concentration fluctuations that may exist. Their amplitude decreases exponentially with time as we shall see. Let us consider a binary alloy of nominal concentration C0 in which there are periodic fluctuations (wavelength k, figure 2.6). Concentration at a given location x can be written: C ðx; t Þ ¼ C0 þ Aðt Þ sin

2px k

At t = 0, Aðt Þ ¼ A0 . A0 is the initial amplitude of fluctuations before heat treatment (t ¼ 0) at a given temperature T.

70

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 2.6 – Sinusoidal concentration profile of wavelength k. Aðt Þ is the amplitude at a given time t during heat treatment.

It is easy to show that the solution of second Fick’s equation (problem 2.2.3) leads to an exponential temporal decay of amplitude Aðt Þ, similarly to the discharge of a capacitor (figure 2.7): A ð t Þ ¼ A0 e

tt

k

:

Equilibrium is reached for t ! 1. The characteristic time constant tk is proportional to the square of the spatial period of fluctuations (k): tk ¼

k2 4p2 D

D is the diffusion coefficient of chemical species considered at the chosen heat treatment temperature T. It can be shown that amplitude falls to 5% of its initial value A0 after t ’ 3tk . Real fluctuations in solids are obviously more complex than a simple sinusoidal. For more complex fluctuations, we can then write C ðx Þ as a Fourier series with components of increasing spatial frequency qi ¼ 2p=ki. The two previous equations show that higher rank Fourier components (high spatial frequency, small wavelength k) are the first to decrease with time. Let us take a second example, that of the diffusion of B atoms from the surface of a material (alloy AB) with an initial concentration in B atoms equal to C0 . This problem is of great importance, for instance when heat treatments are applied to materials in order to harden their surface (e.g. ball bearings). Surfaces can be hardened thanks to diffusion and the enrichment of surface regions in carbon (carburizing) or nitrogen (nitriding). Materials are placed in a controlled atmosphere that is rich in carbon or nitrogen so that the surface concentration remains constant (CS ). The solution of the second Fick equation (see problem 2.2.4) shows that concentration amplitudes decrease with x and t according to the error function pffiffiffiffiffiffi erf ðu Þ with u ¼ x=2 Dt (figure 2.8):

Diffusion and Transport in Solids

71

A t

A0

t

0

FIG. 2.7 – Exponential decay for the amplitude of sinusoidal fluctuations Aðt Þ as a function of time t.



x C ðx; t Þ ¼ CS  ðCS  C0 Þerf pffiffiffiffiffiffi 2 Dt



pffiffiffiffiffiffi We can then show that at a diffusion distance x  ¼ Dt from the surface, the concentration is close to ðCS þ C0 Þ=2. Let us remind that the error function erf ðu Þ is twice the integral between 0 and þ u of a Gauss distribution with a standard pffiffiffiffiffiffiffiffi deviation r ¼ 1=2: Z þu 2 2 erf ðu Þ ¼ pffiffiffi ev dv p 0 The latter being non-integrable, erf ðu Þ is tabulated as a function of u (table is provided in problem 2.2.4).

FIG. 2.8 – (a) Error function erf ðu Þ also called Gauss integral. (b) Concentration profile C ðx Þ related to the diffusion of B atoms for which the surface concentration is kept constant (CS ). C0 is the initial concentration in the material before diffusion.

Diffusion, Segregation and Solid-State Phase Transformations

72

2.1.3

Hetero-Diffusion, Kirkendall Effect

Let us consider an AB alloy in which the diffusion coefficients of A and B are different (DA ,DB ). Related fluxes J A and J B are then not equal in norm,11 leading to a vacancy flux ~ J v such that: ~ JB þ~ Jv ¼ 0 JA þ~ The vacancy flux can be written in the following general form ~ J v ¼ C0~ v. C0 is the concentration of atoms per unit volume and ~ v denotes the velocity vector of vacancies which is therefore written as: ! ~ v ¼ ðDA  DB Þr XA DA and DB being partial diffusion coefficients (also called intrinsic coefficients): ! ~ J A ¼ DA r CA ! ~ J B ¼ DB r CB XA is the atomic fraction of A. This “wind” of vacancies causes a displacement of the lattice planes (figure 2.9) that have a velocity equal to that of vacancies (v). This is the so-called Kirkendall effect (problem 2.2.6). The vacancy flux as well as the plane displacement follow the inverse direction of that of the migration of A atoms (resp. B) when DA > DB (resp. DB > DA). Figure 2.9 shows the mechanism when vacancies migrate from the right side to the left side (v < 0). Vacancies leave an interstitial loop and migrate towards a vacancy loop. This vacancy migration leads to both the annihilation of lattice planes on the left side and the creation of planes on the right side (figure 2.9). This induces a displacement of lattice planes from right to left, in the same direction as vacancies. It is worth mentioning that this vacancy migration, due to the different mobilities of A and B species, may also lead to the formation of voids resulting from the agglomeration of vacancies. If it is the case, previous equations cannot be directly used. Note also that vacancies may be eliminated to point defect sinks such as grain boundaries or sample surfaces. This lattice displacement will modify the previously established diffusion laws. Let us remind that Fick’s equations are given with respect to the lattice frame which 0 is now mobile. The new expression of fluxes ~ J i for each species (i ¼ A; B) in an immobile coordinate system (fixed reference frame) attached to the sample is then v): obtained by adding a transport term (Ci~ 0 ! ~ J i ¼ Di r Ci þ Ci~ v

11

For the sake of simplicity, we often make a language abuse, talking of flux whereas J is in reality a flux density per unit of area that is expressed in at/m2s.

Diffusion and Transport in Solids

73

FIG. 2.9 – Displacement of lattice planes resulting from the vacancy flux from the right to the left side. Vertical arrows show the induced displacement of dislocation lines resulting from the vacancy flux. Vacancies coming from the loop (vacancy source) on the right side migrate towards dislocation lines (vacancy sinks) on the left side. Substituting the expression of the transport velocity ~ v, we can write again the diffusion flux in the same form as Fick’s equation but with an interdiffusion coeffi~ now depending on local concentration Ci ðx; y; z Þ: cient D 0 ~ !Ci ~ J i ¼ Dr

~ ¼ X A D B þ XB D A . with D This equation is known as Darken’s equation. For a diluted alloy XB \\1 and ~ = DB . In the fixed frame of reference of the sample, we have: XA  1 we find D 0 0 ~ JB ¼ 0 JA þ~

~ now depends on concentrations, Fick’s second law takes a Note that since D more complex form. In one dimension, it is easy to show that:

74

Diffusion, Segregation and Solid-State Phase Transformations  2 2 @C ~ @ C þ DB  DA @C ¼D @t @x 2 @x C0

2.1.4

Diffusion Short Circuits: Influence of Defects

Extended defects like grain boundaries (GBs) or dislocations constitute diffusion short circuits because the species are much more mobile there. There is indeed a “free” volume in GBs and dislocations, which can be seen as vacancies. This accelerates the diffusion in the GB plane or along dislocation lines (figure 2.10). This results in a lower activation energy QD for diffusion in the defect than in the grain QV (bulk). Short circuits play an important role at low temperatures where the bulk diffusion in grains is low. There is thus a “pivot” temperature T0 below which the defect influence on diffusion is dominant. In nanomaterials where the grain size (d) is small, the relative importance of GBs compared to that of grains is larger. The role of GBs is then much more important. The pivot temperature T0 below which GBs have a major influence on diffusion increases. Let e be the thickness of GBs (say for instance e  0.5 nm). Temperature T0 for which the contributions in volume and along the grain boundaries (weighted by the surfaces seen for diffusion) is written (see problem 2.2.7): T0 ¼

QV  QD R lnðd=2eÞ

It is the same expression for dislocations. T0 increases when the dislocation density increases (e.g. hard drawn materials).

FIG. 2.10 – (a) Diffusion short circuit along a grain boundary (perpendicular to the figure). Diffusion is fastest in the GB region (0.5 nm thick). (b) Short circuit diffusion along an edge dislocation line (perpendicular to the figure). In this so-called “pipe diffusion” phenomenon, atoms migrate along a tube whose radius is of the same order as the norm of the Burgers vector of dislocation.

Diffusion and Transport in Solids

2.1.5

75

Driven Diffusion in a Force Field

The previous equations are valid only if atomic species (i) are not subject to any forces (chemical, electrical, thermal, elastic…). Let us consider first the case of chemical interactions in an AB alloy (of energy order ε ≠ 0, tendency to unmix). Chemical species will diffuse to minimize the free energy of the system. Diffusion will be driven so as to reach equilibrium for which chemical potentials of A and B atoms are both uniform everywhere. Chemical species hence migrate to decrease their chemical potential. In a first approach, it is then reasonable to write that the migration velocity of chemical species ~ v i (i ¼ A; B) is proportional to the chemical potential gradient (see problem 2.2.8): ! ~ v i ¼ Mi r li li is here the atomic chemical potential of A or B and Mi their atomic mobility, a coefficient that will be shown latter to be proportional to diffusivity. By expressing fluxes (~ J i ¼ Ci~ v i ) we end up with: ! ~ J i ¼ Mi Ci r li Let us remind that Ci is the concentration of i atoms per unit volume. The latter equation shows that A or B atoms move from high chemical potential regions to low chemical potential regions. The free enthalpy of the system then decreases towards its minimum (thermodynamic equilibrium). At equilibrium, chemical potentials of each species become equal everywhere, then Ji ¼ 0 as expected. The atomic mobility Mi of chemical species i can easily be expressed as a function of diffusion coefficients. Let us write the diffusion flux for an ideal solution for which there are no chemical interactions (the chemical energy e ¼ 0 in the mean field model of regular solutions). We must then find the classical Fick equation. By injecting the expression of chemical potential li for an ideal solution (chemical activity ai ¼ Xi ), and by identification to Fick’s first law, we find the following expression for Mi : Mi ¼ Di =kT Note that if chemical potentials are expressed for one mole then Mi ¼ Di =RT with R ¼ kN (N is the Avogadro number). Let us now express the fluxes for a non-ideal solid solution (e 6¼ 0). The chemical potentials are written: li ¼ Gi þ kT ln Xi þ ui ui accounts for the non-ideality of AB solid solution. ui = kT ln ci with ci the activity coefficient. ui is related to chemical interactions between atoms. However, additional interaction terms should be included in the ui coefficient if necessary. The electromigration of charged species in an electric field is an example (problems 2.2.12 and 2.2.13). Fluxes are then expressed as the sum of a diffusive term given by the Fick equation and a transport term under the action of the force fields:

76

Diffusion, Segregation and Solid-State Phase Transformations ! ~ J i ¼ Di r Ci þ Ci~ vi

Di ! with ~ v i ¼  kT r ui . This leads to the Nernst–Einstein equation:

C i Di ~ ! ~ Fi J i ¼ Di r Ci þ kT ~i is the force exerted on the species i: where F ! ~i ¼ r ui F Note that the transport velocity takes the simple form: ~i ~ v i ¼ Mi F As an example, let us consider the electromigration of a given chemical species of charge qi that diffuses in an electric field E. For the sake of simplicity, no chemical ~i is the interactions with other species will be considered. In the present case, F ! ~ with E ~ ¼ r V . The interaction term ~ i ¼ qi E Coulomb force that classically reads F is ui ¼ qi V with V the electric potential. The velocity of charged species is then ~ The latter equation can be written as ~ ~ with ui ¼ Mi qi the ~ v i ¼ ui E v i ¼ Mi qi E. electric mobility of the considered species i. Let us now consider only the transport term (Fick’s term is neglected), the charge flux reads ~ J q ¼ qi ~ J so that: ~ ~ J q ¼ rE Di 2 r ¼ kT qi Ci is the so-called electrical conductivity. The latter equation that gives the charge flux is an alternative expression of the Ohm law: I ¼ U =R with I the current and U the voltage applied to the resistance R. It can be easily shown that R ¼ ql=S. q ¼ 1=r is the resistivity, l the length of the conductor and S its section area. We shall come back to the general case where φi refers to any interaction (not necessarily electric). Let us examine the equilibrium state such as ~ J i ¼ 0. The diffusive and transport terms compensate each other. Thus, a non-homogeneous equilibrium concentration Ci ðx Þ can be sustained (figure 2.11) and takes the following form:

C i ðx Þ ¼ C 0 e

ui ðx Þ kT

At equilibrium, and as expected, we thus end up with the well-known Boltzmann distribution.

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77

~i acting on i atoms, can stabilize the concentration profile Ci ðx Þ FIG. 2.11 – A force field F

such that ~ J i ¼ 0: the transport term under the action of the force compensates the Fickian diffusive term.

If we now consider the chemical forces alone (non-ideal solution), we can then show that in a binary AB alloy, the expression of the solute flux B can again be expressed as Fick’s law but with a diffusion coefficient including the thermodynamics of the system. Previous equation of the total flux (Fick’s term + the transport term) leads to (problem 2.2.8): ! ~ J B ¼ DB r CB d G with DB ¼ MB XB ð1  XB Þ dX 2. 2

B

We then discover that the apparent diffusion coefficient DB may be negative if the curvature of Gibbs energy G ðXB Þ is negative. This was shown previously to occur in supersaturated solid solutions (chapter 1) that leads to a miscibility gap. Considering a regular solution, we showed that the alloy decomposes into two phases of the same structure (e.g. FeCr). The flux of B atoms then goes up the concentration gradients. The system becomes unstable, and the amplitude of concentration fluctuations increases. This is the spinodal regime already mentioned in chapter 1. For a system where the mobilities of A and B atoms are different, leading to a Kirkendall effect, we can show (problem 2.2.8) that: ! ~ J B ¼ MC0 XB ð1  XB Þr ðlB  lA Þ with an intermobility of the same form as Darken’s equation: M ¼ MA XB þ MB XA

Diffusion, Segregation and Solid-State Phase Transformations

78

and again, we can introduce an apparent diffusion coefficient including the thermodynamics of the system: DB ¼ MXB ð1  XB Þ

2.2 2.2.1

@2G @XB2

Problems Fundamental Mechanisms of Diffusion – Einstein’s Equation

We will establish the expressions of the atomic diffusion coefficients using the equation proposed by Einstein: Dx ¼

Z 1X Ci Xi2 2 i¼1

This relation gives the diffusion coefficient of a given species along a direction x. Z is the number of possible atomic jumps (i) between the atom and the first neighbouring sites. Ci is the jump frequency of atoms in the first neighbouring site i. Xi is the related jump distance projected on the considered direction x. We consider hereafter the diffusion of substitutional atoms (vacancy mechanism). 1. Diffusion in BCC and FCC lattices We consider a centred cubic crystal (BCC) of parameter a that we place in an orthonormal reference frame x; y; z. We consider the diffusion of an atom from a site of the lattice towards the first neighbouring sites. 1.1. Given the symmetries of the structure, show that Ci ¼ C0 whatever i. Deduce that Dx ¼ Dy ¼ Dz . How many jumps are to be considered, and how many of these have a non-zero projection along x? Make a diagram showing these jumps in the structure. Give the expression of Dx as a function of a and C0 . 1.2. Answer to same questions for a face-centred cubic crystal (FCC). Compare with results related to BCC. Conclusion? 2. Atomic diffusion in hexagonal structures We now consider a material that has a hexagonal structure. Lattice parameters are a in the (0001) base plane (xOy) and c along z direction (perpendicular to xOy). Let us state C1 as the jump frequency between first neighbour atoms in the base plane and C2 that for the first neighbour jumps between (0001) planes below and above the considered basal plane.

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2.1. Sketch the hexagonal structure. Represent the jumps from the origin (0,0) towards the first neighbours in the three directions of space. Examine jumps that have a non-zero projection on x or y directions. Calculate the diffusion coefficients related to x and y directions (Dx , Dy ) as a function of C1 and C2 . Show that they are identical in considering the crystal symmetries. 2.2. Calculate the diffusion coefficient related to z (Dz ) as a function of jump frequencies. 2.3. Consider an ideal hexagonal structure (compact hexagonal with pffiffiffiffiffiffiffiffi c=a ¼ 8=3). Calculate jump distance d between the basal plane and (0001) planes above and below. Show that d is equal to the first neighbour distance in the basal plane. Explain why C1 ¼ C2. Give expressions of Dx , Dy and Dz as a function of C1 and a. What is your conclusion? Solution to the problem 1. Diffusion in BCC and FCC structures 1.1. Atoms of the BCC lattice have Z = 8 first neighbours placed at a distance equal to half the diagonal of the cubic unit cell:

These 8 jumps are strictly identical in the BCC structure because x, y and z directions are indistinguishable. Consequently, jump frequencies related to each of these 8 jumps are identical: Ci ¼ C0 whatever the first neighbour jump (i) considered. The projected distance of these jumps on x axis is a=2 whatever the jump considered. All of them have a non-zero projection because none of the jumps is realized in the (200) plane perpendicular to x axis. Finally, we have: a 2 1X 1 Dx ¼ Ci Xi 2 ¼ 8C0 ¼ C0 a 2 ; 2 i 2 2

80

Diffusion, Segregation and Solid-State Phase Transformations Because of the crystal symmetries, we conclude that Dx ¼ Dy ¼ Dz . 1.2. In FCCs, atoms have 12 first neighbours (coordination number Z = 12). The jump distance is equal to half the diagonal of faces of the FCC structure (figure below).

It is easy to see that among the 12 possible jumps there are 4 jumps in the xOz plane. Their related projected distance on the x axis is equal to a=2. In the same way, there are 4 jumps in the xOy plane of projection a=2 on the x axis. In contrast the 4 jumps in plane yOz have a zero projection on x direction (see schematic of the FCC crystal). Because of FCC symmetries, diffusion coefficients are identical along x, y and z and are given by:

a 2 C0 a 2 Dx ¼ 4 þ4 þ 0 ¼ C0 a 2 ¼ Dy ¼ Dz 2 2 2 These expressions are identical to those in BCC. 2. Atomic diffusion in hexagonal structures 2.1. Let us consider the basal plane (xOy) as shown below:

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81

z y

x

O a

:z :z :z

0 c/2 c/2

Let us consider the centre site ð0; 0Þ of the (0001) basal plane and projected distances on Ox. Among the 6 possible jumps in this plane, there are 4 that have a projected distance equal to a=2, the 2 others have a projected distance equal to a. In the basal plane, the jump frequency is C1 . There are 6 possible jumps from (0,0) towards the neighbour basal planes (below and above) located at z ¼ c=2 (jump frequency C2 ). Among them, 4 have a non-zero projection (a/2) and 2 have a zero-projection distance. This leads to:



C1 a 2 C2 a 2 3C1 þ C2 2 4 4 a þ 2a 2 þ þ0 ¼ Dx ¼ 2 2 2 2 2 with Dx ¼ Dy because of symmetries (x and y are indistinguishable). However, as we shall see in the following question, diffusion along z may be different. 2.2. All the 6 neighbour jumps between basal planes have a projected distance equal toh c=2.i The diffusion coefficient along z is therefore given by:  2 Dz ¼ C22 6 2c ¼ 3C4 2 c2 2.3. Let us calculate distance d:

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82

pffiffi pffiffiffiffiffiffiffiffi  2 We have d 2 ¼ 2c þ b2 and c=a ¼ 8=3. We see that b ¼ a 3 3 as pffiffi a cos p6 ¼ 23 ¼ 2b . As a result, we have d ¼ a. Since jump distances in the basal planes and between basal planes are identical, related frequencies are also identical and C1 ¼ C2 . We therefore deduce that Dx ¼ 2C1 a 2 ¼ Dy ¼ Dz . Diffusion in ideal compact hexagonal structures is therefore isotropic as for BCC and FCC.

2.2.2

Elementary Mechanisms and Self-Diffusion

1. The frequency of vibration of atoms t in a solid can be estimated using a simple harmonic model. Let us consider an oscillating mass m attached to a spring of stiffness constant k. Sketch this model. Calculate t and show by analogy that we qffiffiffi 1 E can write for a solid: t ¼ 2pb q , b is the interatomic distance, E is the elas-

2.

3.

4.

5. 6.

tic modulus (Young’s modulus) and q the density. Calculate b for nickel (FCC structure of crystal parameter a = 0.35 nm). Calculate the frequency t for nickel with E = 200 GPa and m = 58.7 amu. Q The self-diffusion coefficient is written: D ¼ D0 eRT . We consider that the molar entropies of formation (DSf ) and migration (DSm ) of vacancies in nickel are both close to 3.35 R (R: gas constant). Calculate the pre-exponential factor D0 and compare to the tabulated value of 1.9 cm2/s reported in diffusion data for nickel. It is generally observed that the activation energy for self-diffusion Q is proportional to the melting temperature of the material (Tf ). In FCC materials, Q  18 RTf . The melting temperature of nickel (Tf ) is 1453 °C. Calculate the activation energy Q in J/mol and eV/atom and compare it to the measured value (68 kcal/mol). Compare the activation energy Q per atom (in eV) to the value of the thermal quantum kT at 1000 °C. What is your conclusion? The enthalpy of formation of a vacancy is 1 eV. Calculate the enthalpy of migration (in eV). Calculate the theoretical value of the self-diffusion coefficient (in cm2/s) at 1000 °C. What is the diffusion distance r (in μm) of nickel atoms after 30 min at 1000 °C? Calculate the jump frequency of atoms (C0 ) at 1000 °C. Compare t to C0 . Why is there such a difference between both quantities? The enthalpy of vacancy formation is 1 eV. Calculate the atomic fraction of vacancies at 1000 °C. Deduce the diffusion coefficient of vacancies and their jump frequency CV . What distance (in μm) is effectively covered by a vacancy after 30 min at 1000 °C? Compared to the distance covered by an atom.

Diffusion and Transport in Solids

83

Solution to the problem 1. Let us define the initial position of the mass in x ¼ 0.

k

m

0 x

Let us apply the fundamental principle of dynamics. We then get the following 2 differential equation: F ¼ kx ¼ m ddt 2x . Let ɷ be the angular frequency. x ¼ x0 sinðxt Þ is a solution of the dynamic equation. By injecting this solution into the previous equation, we find: k ¼ mx2 so that the angular frequency is qffiffiffi given by: x ¼ mk . Let us write that kx ¼ jF j ¼ rS with S the surface on which the force F is applied. By definition of Young’s modulus, the applied stress reads: r ¼ Ex=b where x=b is the relative deformation. The spring stiffness can thus be expressed as a function of both E and the interatomic distance b: k ¼ ES=b, qffiffiffiffiffi qffiffiffi 2 3 1 E Thus, x ¼ ES bm . Writing that S ¼ b and q ¼ m=b leads to x ¼ b q . As qffiffiffi 1 E t ¼ x=2p, the frequency reads: t ¼ 2pb q. The nickel structure is FCC and the unit cell contains 4 atoms (multiplicity = 4). pffiffi As a result, the interatomic distance is b ¼ a 2 2  0.25 nm, and the density reads q ¼ ða3m=4Þ with m the mass of Ni atoms (58.7 amu). Consequently, the mass expressed in kg is m = 58.7 × 10–3/(6.02 × 1023) so that q  9.09 × 103 kg/m3 and finally t  3 × 1012 Hz that is the good order of magnitude for the vibration frequency in solids. Q DS 2. Let us write the classical expression for the diffusion coefficient: D ¼ ta 2 e R eRT with DS ¼ DSf þ DSm . Then we get D0  3 × 10–4 m2/s, a value that is close to the tabulated value of 1.9 cm2/s in diffusion data.

Diffusion, Segregation and Solid-State Phase Transformations

84

3. Knowing that Q  18 RTf , we deduce Q  2.58 × 105 J/mol that is equivalent to 2.68 eV/atom. This value is close to that measured (2.84 × 105 J/mol, 1 cal = 4.18 J). At 1000 °C, kT  0.11 eV, which is very low compared to Q. Q Consequently, the term eRT is very small (  10–11). As Q ¼ DH ¼ DHf þ DHm and DHf  1 eV, we find DHm  1.68 eV/atom. 18Tf

4. As D ¼ D0 e T then D  7.6 × 10–15 m2/s (7.6 × 10–11 cm2/s). We then get pffiffiffiffiffiffiffiffi the following diffusion length r = 6Dt  9.1 µm. 5. As D ¼ C0 a 2 then C0  6.2 × 104 Hz. The latter value is very small compared to the vibration frequency that was calculated previously (t = 3 × 1012 Hz). Q DS This large difference is expected because C0 ¼ te R eRT expresses the probability of successful jumps (per unit of time) across the diffusion barrier Q whereas t is the number of jump attempts per second that evidently is much larger. 6. The diffusion coefficient of vacancies is given by: DV ¼ D=XV with XV the DSf

DHf

atomic fraction of vacancies that reads: XV ¼ e k e kT . As DHf  1 eV and DSf  3.35R, then XV = 3.2 × 10–3. We finally find that: DV = 2.4 × 10–12 m2/s and CV ¼ XCV0  2 × 107 Hz. The diffusion length for a vacancy is therefore close to 1.6 × 10–4 m, that is much larger than that of atoms.

2.2.3

Diffusion in CuAl Alloys

Copper is a good conductor of electricity. The addition of aluminium to copper hardens the material which allows the production of electrical cables with a higher mechanical strength. Cu and Al have the same structure (FCC). We therefore expect a good solubility of aluminium in copper. The phase diagram shows indeed the existence of a Cu (Al) solid solution with a solubility greater than 15% aluminium. When the Al concentration exceeds its solubility, many ordered phases are formed. In this problem, we consider a CuAl alloy in solid solution whose aluminium concentration C0 is lower than the solubility limit of aluminium in copper. 1. We will estimate some quantities related to the diffusion of aluminium in copper from measurements made at two temperatures. The diffusion coefficient of aluminium in copper at 500 °C is D1 = 2.6 10–17 m2/s and D2 = 1.6 10–12 m2/s at 1000 °C. 1.1. Calculate the values of the activation energy for diffusion (Q in J/mol) and of the pre-exponential diffusion coefficient (D0 ). 1.2. Remind the classical expression of D0 for a cubic crystal. Deduce the expression of the entropy variation DS as a function of D0 , the frequency of vibration of the atoms (t = 1013 Hz) and the crystalline parameter of FCC structure for copper (a = 0.35 nm). It is reminded that DS is the sum of both the migration entropy and the vacancy formation entropy. Calculate DS as a function of R (gas constant). 1.3. The entropy of vacancy formation is DSf ¼ R. Deduce the migration entropy DSm as a function of R.

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85

1.4. Calculate the value of the diffusion coefficient at 800 °C. 1.5. What is the effective diffusion distance of aluminium atoms after one hour at 800 °C? 2. The influence of quench. We shall now study the influence of quench on the diffusion coefficient D of aluminium in copper. 2.1. The energy of formation of a vacancy in copper being 1 eV, calculate the vacancy concentration at 1000 °C. 2.2. CuAl alloy is now quenched rapidly from 1000 °C to 800 °C. The quench rate is so high that vacancies that exist at 1000 °C do not have time to eliminate to defect sinks (dislocations, grain boundaries, surface). Calculate the ratio (r) between the non-equilibrium concentration of vacancies after quenching and the equilibrium concentration at 800 °C. 2.3. Given the vacancy supersaturation, express the effective diffusion coefficient D  as a function of D at 800 °C. 2.4. Considering point defect sinks, do you think that the atomic mobility remains constant with time during ageing at 800 °C? Justify and explain what happens. Represent qualitatively the evolution with time of the diffusion coefficient. 2.5. Let us consider a polycrystalline material. d is the diameter of the grains. Give an order of magnitude for the time te required to reach the asymptotic equilibrium value of D when vacancy supersaturation gets negligible. Calculate te for d = 100 μm. Experimental data related to the self-diffusion of copper are D0 = 0.78 cm2/s and Q = 50.5 kcal/mol (1 cal = 4.18 J). For this calculation, we will assume that the diffusion coefficient for vacancies is kept constant, and equal to that at 800 °C. What is the influence of the grain size on Dðt Þ? 3. Homogenization. After solidification the alloy is inhomogeneous. Concentration fluctuations are present. The concentration profile C ðx; t Þ related to Al along an arbitrary direction (Ox) is modelled by a simple sinusoidal function of periodicity k:  

2p x C ðx; t Þ ¼ C0 1 þ Aðt Þ sin k C0 is the bulk Al concentration. At t ¼ 0, the fluctuation amplitude is Aðt Þ ¼ A0 . 3.1. The alloy is submitted to a thermal treatment at T = 1000 °C in order to homogenize the material. Using Fick’s second law and examining the local curvature of C ðx Þ, explain how concentration fluctuations evolve with time. 3.2. We consider that the diffusion coefficient is independent of the local composition C ðx Þ and consequently of x. Solve the second Fick equation and show that the relative amplitude can be written as: Aðt Þ ¼ A0 et=s . Give the expression of the relaxation time s. Calculate the time (t) required to have Aðt Þ equal to 1% of its initial value A0 (at t = 0). We will take k ¼ k0 = 100 μm.

Diffusion, Segregation and Solid-State Phase Transformations

86

3.3. We now take a more realistic form of C ðx; t Þ. The latter can be expressed as a Fourier series. Fourier components have amplitudes Ai and related wavelengths given by ki ¼ k0 =i (i is an integer). Keeping in mind the previous results, give the expression of both Ai ðt Þ and related relaxation time si . Show schematically how the concentration signal C ðx; t Þ evolves with time. Solution to the problem 1.1. The expressions for the diffusion coefficients of aluminium in copper, D1 = 2.6 10−17 m2/s at T1 = 500 °C and D2 = 1.6 10−12m2/s at T2 = 1000 °C, lead to the following system of equations where D0 and Q are the two unknown quantities to be derived: D1 ¼ D0 e

Q RT

1

D2 ¼ D0 e

Q RT

2

The combination of both equations leads to:   T1 T2 D1 Q ¼ R T1 T2 ln D2  1.8 × 105 J/mol Q

and D0 ¼ D1 eRT1  4 × 10–5 m2/s.

 D0 DS 1.2. For a cubic crystal we have: D0 ¼ ta 2 e R . We deduce DS ¼ R ln ta = 3.5 R. 2 1.3. Knowing that DSf ¼ R and that DS ¼ DSf þ DSm , we find for the migration entropy of vacancies: DSm = 2.5 R. 1.4. As D0 and Q are now known, we can calculate the diffusion coefficient of Q aluminium in copper at 800 °C: Dð800  CÞ ¼ D0 eRT  6.7 × 10–14 m2/s. 1.5. The effective distance travelled by aluminium over a time t is: pffiffiffiffiffiffiffiffi X ¼ 6Dt = 38 μm for t = 1 h at 800 °C. 2. The influence of quench DSf

DHf

2.1. The concentration at equilibrium is given by XV ¼ e R e kT . DHf = 1 eV (for one atom) and ΔSf = R (for one mole, see question 1.3). The atomic fraction of vacancies at 1000 °C is then XV  3 × 10–4. 2.2. It is assumed that during quenching excess vacancies do not have time to eliminate to defect sinks. The initial atomic fraction of vacancies at t = 0 just after quenching from 1000 °C to 800 °C is therefore that of equilibrium at 1000 °C: XV ð1000  CÞ = 3 × 10–4. However, the equilibrium concentration at 800 °C is XV ð800  CÞ  5.5 × 10–5. The vacancy supersaturation is therefore very high when the heat treatment starts. The  CÞ related ratio of concentrations is r ¼ XXVVðð1000 800  CÞ  5.45.

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87

2.3. Since this is a diffusion mechanism that involves vacancies, the diffusion rate of aluminium is directly proportional to the vacancy concentration, so we have D  ¼ rD. 2.4. In proportion as excess vacancies annihilate to defect sinks, the mobility of atoms decreases and tends to its equilibrium value. This is achieved when the concentration of vacancies also reaches its equilibrium value.

2.5. The elimination of vacancies in supersaturation to grain boundaries (GBs) can be sketched as follows:

In a first approximation vacancies must diffuse over a maximum distance close to half the grain diameter (d = 100 μm). As the effective diffusion pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi distance is 6DV t one can write that d2 ¼ 6DV te . te is the time required to reach the equilibrium value of the diffusivity (D) of Al in Cu, during d2 . Let us deduce which excess vacancies eliminate to GBs. Then te ¼ 24D V Q

with DCu ¼ D0 eRT . DV from the self-diffusion coefficient of Cu: DV ¼ DXCu V –11 2 m /s and we deduce te  5.6 s. When We then find DV  7.4 × 10 the grain size increases, the diffusion of aluminium remains longer above its equilibrium value because te increases as d 2 .

Diffusion, Segregation and Solid-State Phase Transformations

88 3. Homogenization

3.1. Let us represent schematically the sinusoidal profile of aluminium concentration:

@ C According to Fick’s law, @C @t ¼ D @x 2 . The temporal variations of concentration are thus governed by the curvature of C ðx; t Þ. When it is negative, the concentration decreases. Conversely, it increases when the curvature is positive. The amplitude of concentration fluctuations will therefore decrease. For very long times, concentrations tend towards the nominal concentration C0 . 3.2. Let us replace C ðx; t Þ by its expression into Fick’s second law. We then 2 find: dAdtðt Þ ¼  4pk2D Aðt Þ. The solution of this equation is: Aðt Þ ¼ A0 et=s 2

2

with s ¼ 4pk2 D and A0 ¼ Aðt ¼ 0Þ. The time necessary to have an amplitude

A can be written as t ¼  4pk2 D lnðA=A0 Þ. For Aðt Þ ¼ 102 A0 we find t = 730 s (  12.7 min). 3.3. A more realistic composition profile is sketch below (dotted line): 2

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89

High rank Fourier components are superimposed to fluctuations that have a larger wavelength. The concentration signal tends towards a sine like profile for increasing times. The concentration is now written as a Fourier series and is the sum of Fourier components: "  # X 2p C ðx; t Þ ¼ C0 1 þ Ai ðt Þ sin x : ki i   It is easy to check that each component C0 Ai ðt Þ sin 2p ki x is a solution of the second Fick law. As a result, the sum of these Fourier terms is also a solution. It is easy to see that during heat treatment, small wavelength fluctuations (dotted oscillations in figure) attenuate first because the required diffusion distance is smaller for these fluctuations to decrease. k2

Indeed, as previously demonstrated, the relaxation time si ¼ 4p2i D is smaller for small values of ki . In other words, the smaller the wavelength ki the faster the attenuation. The composition “signal” will therefore be first “purified” of high rank Fourier components (high spatial frequency 2p=ki means small wavelengths). Low rank Fourier components will shrink for longer ageing time.

2.2.4

Surface Hardening by Cementation

Low carbon austenitic steel (FCC structure) is held at high temperatures for several hours in a CO atmosphere with constant pressure. The carbon concentration at the surface Cs is then maintained constant despite the diffusion of carbon from the sample surface to the bulk. A carbon enriched layer (a fraction of a millimetre thick) is formed and harden the surface.12 This so-called cementation process is used to harden surfaces where there is friction. This is for instance the case of rolling balls. It will be assumed that the diffusion coefficient of carbon D, does not depend on the local concentration of carbon and therefore does not depend on the depth (x) from the surface.

12

For large carbon concentrations, the precipitation of cementite (Fe3C) may take place depending on the ageing temperature. In the industrial process, the steel is sometimes quenched to transform the carbon-rich surface layer into a new phase called martensite which hardens the material in surface. This phase has the same composition as the initial FCC austenite (high temperature iron structure) but its structure is different (quadratic centered).

Diffusion, Segregation and Solid-State Phase Transformations

90

1. Solving using reduced variable 1.1. We shall calculate the solution C ðx; t Þ of the second Fick law. The latter involves partial derivatives with respect to both x and t variables. Solving this equation is not always immediate. In order to make it easier, let us reduce the number of variables from two to one. It is intuitive to consider that the layer thickness grows as the diffusion distance that is proportional pffiffiffiffiffiffi pffiffiffiffiffiffi to Dt . It is therefore clever to set a reduced variable u ¼ x=2 Dt . The concentration then depends only on the reduced variable u. We will solve the second Fick equation to calculate the shape of the concentration profile 2 2 dC C ðx Þ. First, calculate @@xC2 as a function of dduC2 , then @C @t as a function of du . Replace these expressions in the second Fick equation so as to get a second order differential equation where u is the only variable. 0 1.2. Let write C 0 ¼ dC du . Give the expression of d ln C as a function of u. Deduce 0 the expression of C ðu Þ and then that of C ðu Þ. 1.3. Show that C ðu Þ can easily be expressed as a function of the error function Ru 2 erf ðu Þ. Let us remind that erf ðu Þ ¼ p2ffiffip 0 ev dv. Determine the integration constants as a function of Cs and C0 . Note that: C ðx ¼ 0; t Þ ¼ Cs whatever t, as well as C ðx; t ¼ 0Þ ¼ C0 whatever x [ 0 and C ðx; t Þ ! C0 when x ! þ 1. 1.4. Represent schematically both erf ðu Þ, and C ðx Þ for various increasing times. Describe the diffusion kinetics and examine C ðx Þ for small depths. We remind that erf ðu Þ is the primitive of Gaussian distribution that is not analytically integrable. erf ðu Þ is tabulated below: u erf ðu Þ

0 0

0,2 0,22

0,4 0,43

0,6 0,6

0,8 0,74

1 0,84

1,2 0,91

1,4 0,95

∞ 1

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91

2. Solving using Laplace transform 2.1. Write the Laplace transform (LT) of the differential equation governing C ðx; t Þ as given by Fick’s second law. We then get a new differential equation that only depends on the variable x. Let us define Cðx; pÞ as the LT of C ðx; t Þ with p ¼ jx (ɷ is the angular frequency, j2 = −1) the Laplace variable. Laplace transform is defined as follows: Z 1 LT ðf ðt ÞÞ ¼ f ðt Þept dt: 0

In addition, it is reminded that: LT ðK Þ ¼ K =p where K is a constant, LT ð@C =@t Þ ¼ pCðx; pÞ þ H ; with H a constant,

 LT @ 2 C =@x 2 ¼ @ 2 Cðx; pÞ=@x 2

pffi   pffiffi a p LT 1  erf a=2 t ¼ e p . 2.2. Show that Cðx; pÞ ¼ Aeax þ Beax is solution of the new differential equation without second member. Deduce the expression of a as a function of p and D. Do both terms Aeax and Beax have a physical meaning? 2.3. Find a particular solution, deduce the general solution Cðx; pÞ and determine the value of the integration constants. Give the general solution C ðx; t Þ using the inverse Laplace transform (LT−1) and compare to the solution established in question 1.

3. Application to cementation A piece of steel containing 0.1% carbon is heat treated at 930 °C. The CO enriched atmosphere maintains the surface composition at 1at.% of carbon whatever the time t. The carbon diffusivity is assumed to be independent of carbon concentration: D = 1.4 10–7 cm2s−1 whatever x. 3.1. Calculate the time required to get a carbon content of 0.45at.% at a depth of 0.05 cm. 3.2. How long does it take at the same temperature to double this penetration depth? 3.3. Let us now take D ¼ 0:27e17400=T cm2s−1. What temperature increase is needed to obtain 0.45at.% C at a depth of 0.1 cm after the same time as that required to get 0.45at.% at 0.05 cm at 930 °C?

92

Diffusion, Segregation and Solid-State Phase Transformations

Solution to the problem 1. Solving using reduced variable

pffiffiffiffiffiffi dC @u p1ffiffiffiffi dC 1.1. Using the reduced variable u ¼ x=2 Dt , we get @C @x ¼ du @x ¼ 2 Dt du and:   @2C d 1 @C @u d 2 C 1 p ffiffiffiffiffi ffi ¼ : ¼ @x 2 du 2 Dt du @x du 2 4Dt dC @u x ffiffiffiffi dC p Moreover, we have @C @t ¼ du @t ¼  4t Dt du . Reinjecting these expressions in

the second Fick equation

@C @t

¼ D @@xC2 leads to: pffiffiffiffiffiffi 2 Dt d C dC ¼ x du 2 du 2

and therefore, d2C 2u dC du ¼ du 2 . dC 0 0 1.2. Writing C 0 ¼ @C @u one obtains C 0 ¼ d ln C ¼ 2udu. Integration of the latter leads to lnjC 0 j ¼ u 2 þ K , with K a constant. As a result, we can 2 write: C 0 ¼ Aeu . A is a constant. Second integration ends up to: R u v 2 C ðu Þ ¼ A 0 e dv þ B. 1.3. Expressing the latter equation as a function of the error function erf ðu Þ ¼ pffiffi R u v 2 p p2ffiffi e dv gives: C ð u Þ ¼ A 2 erf ðu Þ þ B. p 0 A and B constants can be derived from limit conditions. C ðx ¼ 0; t Þ ¼ Cs then C ðu ¼ 0Þ ¼ Cs ¼ B. In addition, we know that C ðx; t Þ ! C0 when t ! 0. Yet limt!0 C ðx; t Þ ¼ limu! þ 1 C ðu Þ. Because erf ðu Þ ! 1 when u ! pffiffi pffiffi þ 1 we can write A 2p þ B ¼ C0 . Replacing B gives A 2p ¼ C0  Cs and finally: C ðu Þ ¼ Cs  ðCs  C0 Þerf ðu Þ 1.4. We can now represent schematically erf ðu Þ and C ðx Þ for increasing times:

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C ðx Þ increases with time at a given depth x. For a given value of pffiffiffiffiffiffi u ¼ x=2 Dt , C(u) is a constant so the distance x that is required to get a pffiffiffiffiffiffi given concentration increases as the diffusion length Dt , as expected. In addition, the table shows that erf ðu Þ is proportional to u when u is small as shown above. This implies that C ðx; t Þ decreases linearly with x for small depths. 2. Solving using Laplace transform 2.1. We start again from Fick’s equation

@C @t

¼ D @@xC2 . Applying the Laplace 2

Þ with H a constant of transformation leads to pCðx; pÞ þ H ¼ D @ C@xðx;p 2 integration. 2.2. Let us consider the differential equation without second member (H ¼ 0): 2

Þ D @ C@xðx;p  pCðx; pÞ ¼ 0. This is a classical equation that is known to have 2 the following solution: 2

Cðx; pÞ ¼ Aeax þ Beax pffiffiffi Replacing the latter expression into Fick’s equation gives a ¼ Dp . Moreover Cðx; pÞ should not diverge when x ! þ 1, then necessarily A ¼ 0. 2.3. The general solution is the sum of the solution without second member (H = 0) and the particular solution Cðx; pÞ ¼ H =p. As a result, pffiffip Cðx; pÞ ¼ H =p þ Be Dx . When x ! þ 1 we get C ! C0 . Then its Laplace transform (C0 ) necessarily tends toward C0 =p and therefore H ¼ C0 . When x ¼ 0, we have C ¼ CS . Thus, when taking the Laplace transform we obtain H =p þ B ¼ 0 CS =p and therefore B ¼ CS C p . Substituting the values of the integration constants in the expression for Cðx; pÞ and applying the inverse Laplace transformation leads to:    x C ðx; t Þ ¼ C0 þ ðCs  C0 Þ 1  erf pffiffiffiffiffiffi ; 2 Dt As it must, this is the same expression as that obtained in question 1.3. 3. Application to cementation 3.1. The latter equation can be re-expressed as C ðu Þ ¼ Cs þ ðC0  Cs Þerf ðu Þ. C Thus erf ðu Þ ¼ CCssC . We know that Cs = 1% and C0 = 0.1%. We can then 0 deduce that erf ðu Þ ’ 0.61 to get the expected value C = 0.45%. As erf ðu Þ  u for small values of u (see table) thus the approximate solution is u ’ 0.61. We can now derive the time needed to reach 0.45% of carbon at the desired depth (0.05 cm), we find t = 1.2 × 104 s that gives approximately t = 3.33 h.

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3.2. Because t  x 2 , the time required to get twice the previous depth is four times larger so that t  13.3 h. 02

x . If 3.3. It is easy to see that the time required to get a depth x′ is t ¼ D0 ð20;61 Þ2

we want to get x′ twice that previously reached (x) for the same time t, Q

then we need D0 ¼ 4D. As D0 ¼ D0 eRT 0 and D ¼ 0; 27e17400=T cm2s−1, we can deduce that T 0 ¼ Q=R  1330 K. As the initial temperature is D lnð4D0 Þ 930 °C (1203 K), the temperature increase is 127 °C.

2.2.5

Diffusion in Thin Sandwich Films

We consider the diffusion of a very thin sandwich layer (thickness e) composed of B atoms in a pure A sample. Let the x axis be perpendicular to the film, placed in x ¼ 0. We will study the diffusion of B atoms along this direction x. 1. Represent this sandwich layer (ABA) and the concentration profile of B atoms C ðx; 0Þ at t ¼ 0. The latter is modelled as a rectangle function of width e centred in x = 0. Show schematically, without any calculation, how the concentration profile C ðx; t Þ will vary for three successive times t1 , t2 , t3 . pffiffiffiffiffiffiffiffiffi 2. Remind the expression of the diffusion distance (x ¼ hx 2 i) as a function of D, the diffusion coefficient of B atoms (in A), and time t for one-dimensional diffusion. What is the shape of C ðx; t Þ at a given time t, and what statistical distribution would you use to model this profile? How does the width of the profile L vary with time? 3. In order to solve the second Fick equation, let us write C ðx; t Þ ¼ KP ðx; t Þ with C the concentration of B atoms per unit of volume. K is the number of atoms deposited in the layer per unit area. What are the units of the statistical distribution P ðx; t Þ? Is it a probability? What is the value of the integral function of P ðx; t Þ over all the space (x). Deduce that of C ðx; t Þ. 4. P ðx; t Þ is modelled by a Gaussian distribution G ðx; rÞ, whose standard deviation is r: x 2 1 G ðx; rÞ ¼ pffiffiffiffiffiffiffiffiffiffi e 2r2 2pr2

R þ1 This distribution is normalized so that: 1 G ðx Þdx ¼ 1. Examine the evolution of G ðx; rÞ for decreasing value of σ. How does σ vary with time t. Show that G ðx; rÞ tends towards the Dirac distribution when t tends towards zero? What are the conditions on the film thickness (ε) for this distribution to be realistic? It is reminded that the Dirac distribution dðx Þ ¼ 0 for any x 6¼ 0 (δ is the so-called Kronecker symbol). dðx Þ can be seen as a very high peak of vanishing thickness centred in x = 0.

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5. Show that C ðx; t Þ ¼ KG ðx; rÞ is a solution of the second Fick equation. Deduce the expression of the standard deviation r as a function of D, and time t. Compare r to the previous expression of the diffusion distance x. What is your conclusion? 6. Find, in another way, the solution of the second Fick equation using Fourier transforms (FT). It is reminded that:  2  @ C FT ¼ 4p2 k 2 FT ðC Þ @x 2   @C @FT ðC Þ FT ¼ @t @t  px 2  2 2 FT e a2 ¼ aepk a

7. 8. 9.

10. 11.

Give the expression of the parameter a as a function of the standard deviation of the Gauss distribution (r) and show that the solution is the same as the one previously established. Represent qualitatively the concentration profile C ðx; t Þ for successive times (t0 very small, t1 , t2 …). Show the evolution of both r(t) and C ð0; t Þ with time. What is the property of the product rC ð0; t Þ? What is your interpretation? We shall now consider a thin sandwich film of boron (B atoms) in nickel (A atoms). The activation energy of the diffusion of boron in nickel is Q = 1.15 eV and the pre-exponential factor is D0 = 2 10–7 m2/s. Calculate the time that is required for the film to diffuse over a width L = 10 lm at a temperature T = 800 °C. We will take here L ¼ 4r. What is your interpretation of L? What is the proportion of B atoms in this region of width L? There is now a stress gradient in the film that is perpendicular to the boron-rich film. Boron atoms are in interstitial position in the FCC structure of nickel. As B atoms are bigger than the interstice size, B atoms lead to the dilatation of the lattice (positive size effect). The applied stress leads to the migration of boron atoms from the most compressed areas (x\0) to less compressed regions (x [ 0). This leads to an additional transport term (C~ v) in the first Fick equation governing the boron concentration C . It is assumed that the rate of migration ~ v due to stress remains constant whatever x. Give the expression of the flux density of boron atoms J. Deduce that of @C @t . ðxx0 Þ2

K 12. Show that C 0 ðx; t Þ ¼ pffiffiffiffiffiffiffi e 2r2 is the solution of this new equation. Give the 2pr2 expression for x0 ðt Þ. What is your interpretation of this term? Represent qualitatively the shape of the concentration profile C ðx; t Þ for increasing times (t ¼ 0, t1 , t2 …).

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Solution to the problem 1. We can schematically represent the sandwich film and the temporal evolution of the concentration profile of B atoms as follows:

2. The distance travelled by B atoms in one dimension for a diffusion time t is pffiffiffiffiffiffiffiffi x ¼ 2Dt . The concentration profile has the shape of a Gauss distribution that spreads out in proportion as B atoms diffuse. The width of the profile Lðt Þ thus pffiffiffiffiffiffiffiffi evolves as x and is proportional to 2Dt .

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3. As C ðx; t Þ ¼ KP ðx; t Þ, the concentration of B atoms being expressed in atoms per unit volume and K being the number of B atoms per unit area, P ðx; t Þ is therefore a probability density in m−1 (in one dimension). By definition: Z þ1 P ðx; t Þdx ¼ 1 1

and therefore,

R þ1 1

C ðx; t Þdx ¼ K. x2

1 4. We consider here that P ðx; t Þ ¼ G ðx; rÞ ¼ pffiffiffiffiffiffiffi e2r2 . 2pr2 The maximum of G ðx; rÞ in x = 0 increases when decreasing σ. The width of this distribution, proportional to its standard deviation σ, increases with time t whereas the distribution height decreases. It is clear that G ðx; rÞ gets narrower as σ decreases, concomitantly the maximum of the distribution increases. Giving the definition of δ(x), G ðx; rÞ tends towards the Dirac distribution δ(x) when t tends toward zero. G ðx; rÞ ! dðx Þ when t ! 0. Hence, according to this Gaussian model, P ðx; t ¼ 0Þ ¼ dðxÞ for t = 0. The real distribution at t = 0 gets close to the Dirac distribution for very thin layers. This is a realistic approximation when the thickness of the film is close to that of a single atomic layer. 5. When the diffusion coefficient is independent of concentration, the second Fick equation is given by:

@C ðx; t Þ @ 2 C ðx; t Þ ¼D @t @x 2 We shall check that C ðx; t Þ ¼ KG ðx; rÞ is solution of this equation. Replacing this expression into the second Fick equation leads to:    2  dr x 2 1 x 1 C ðx; t Þ  ¼ DC ðx; t Þ 4  2 dt r3 r r r D As a result: dr dt ¼ r . pffiffiffiffiffiffiffiffi The standard deviation is therefore: r ¼ 2Dt . As a result, r ¼ x. 6. C ðx; t Þ can be found using the Fourier transform of the second Fick equation which is written:

@FT ðC Þ ¼ 4p2 k 2 D FT ðC Þ @t It is easy to derive the solution of this differential equation in the Fourier space: 2 2 FT ðC Þ ¼ Ae4p k Dt . A is an integration constant. Let us take the inverse Fourier  px 2  2 2 transform, since FT e a2 ¼ aepk a , then

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Diffusion, Segregation and Solid-State Phase Transformations

C ðx; t Þ ¼

A px22 e a a

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffiffiffi with a ¼ 4pDt , that can be re-written as a ¼ 2pr with r ¼ 2Dt . Thus, C ðx; t Þ is proportional to the Gauss distribution: C ðx; t Þ ¼ AG ðx; rÞ. The constant A can be derived from limit conditions. Hence, when t ! 0 we have shown that the concentration profile tends toward the Dirac distribution, then: Adðx Þ ¼ Kdðx Þ. We can thus deduce that A ¼ K and we get the expected solution C ðx; t Þ ¼ KG ðx; rÞ as obtained previously: x2 K C ðx; t Þ ¼ pffiffiffiffiffiffiffiffiffiffi e2r2 2 2pr 7. We can now represent the concentration profile C ðx; t Þ for different times:

K ffi The temporal evolution of C ð0; t Þ ¼ pffiffiffiffiffiffiffi and r ¼ 4pDt

pffiffiffiffiffiffiffiffi 2Dt is shown below:

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8. We notice that the product rC ð0; t Þ ¼ pKffiffiffiffi remains constant. This is simply the 2p result of the matter conservation. Consequently, the height of the concentration peak in x = 0 decreases in inverse proportion as the profile width. pffiffiffiffiffiffiffiffi L2 9. We find that D = 8 × 10–13 m2/s. As L ¼ 4r ¼ 4 2Dt then t ¼ 32D = 3.9 s. 10. The concentration profile is a Gaussian distribution. One well-known property of G(x) is that the probability that x remains in the [−2σ, +2σ] interval is 95% (error function as given in the table provided in problem 2.2.4). When defining the distribution width as L ¼ 4r we have only 5% of B atoms outside this width. 11. The flux density of boron atoms is now written: ! ~ J ¼ Dr C þ c~ v As

@C @t

¼ div~ J , we have: @C @2C @C ¼D 2 v : @t @x @x

K 12. The partial derivative of C 0 ðx; t Þ ¼ pffiffiffiffiffiffiffi e 2pr2

ðxx0 Þ2 2r2

leads to:

! 2 @C 0 dr 1 ð x  x Þ dx0 ðx  x0 Þ 0  þ ¼ C0 þ C0 dt r r2 @t r3 dt The latter term is equal to the sum of both the Fick and transport term: ! @2C 0 @C 0 1 ðx  x 0 Þ2 v ðx  x 0 Þ 0 ¼C D  2 þ D v þ C0 r r2 @x 2 @x r4 The term-by-term identification of the last two equations implies that C 0 is 0 solution if and only if: v ¼ dx it is necessary dt . As v ¼ cste then x0 ¼ vt. Moreover, pffiffiffiffiffiffiffiffi D ¼ , that leads to the expected expression r ¼ x ¼ 2Dt . The maximum that dr dt r of the concentration profile in x ¼ x0 moves towards positive distance x with a 0 velocity v ¼ dx dt . At the same time, the peak amplitude decreases, and its width increases.

100

2.2.6

Diffusion, Segregation and Solid-State Phase Transformations

Kirkendall Effect and Diffusion in CuZn

It was in brass (CuZn), in 1947, that Ernest Kirkendall discovered the effect bearing his name. The melting point of copper is 1083 °C while that of zinc is close to 420 °C. Zinc has a high solubility in copper (FCC) exceeding 30% over a wide range of temperature so that it is easy to prepare a single-phase CuZn solid solution. A diffusion couple is formed by depositing copper on the surface of a CuZn brass sample whose Zn content is below the solubility limit (FCC solid solution of Zn in copper). The thickness of the diffusion couple is of the order of a few cm. The diffusion problem is treated in one dimension. 1. Sketch the brass sample, on which pure copper is deposited. The coordinate axis (x) is taken as perpendicular to the interface. The latter will be taken as the origin of the reference frame (x ¼ 0, we take x positive in CuZn). 2. In your opinion, which element diffuses the fastest and why? 3. If voids were to form, where would they be observed relative to the Cu/CuZn interface and why? 4. Inert molybdenum wires are placed as markers to follow the movement of the crystal lattice. These inert wires follow the atomic planes without reacting with the alloy. The Kirkendall effect is then observed. Indicate without calculation in which direction the planes and the molybdenum markers migrate. 5. Consider a sample of brass with 25% at. Zn treated at a temperature T = 800 °C for a given time t. Represent qualitatively, one below the other, the concentration profiles Ci ðx Þ of copper and zinc (i = Cu, Zn), the corresponding fluxes of atoms Ji ðx Þ and the vacancy flux JV ðx Þ, as well as the rate of vacancy creation/annihilation sðx Þ ¼ @CV =@t. Comment on these curves. Define the various regions of sðx Þ and indicate both the direction of the lattice movement in the diagram and the wind of vacancies. In which position (x) is the vacancy velocity v maximum?

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6. After 30 days at 800 °C, microscopy images reveal that distance (d) travelled by the molybdenum wires is 52 μm. Calculate the lattice velocity v. 7. The Zn concentration profile obtained using microanalysis techniques allows us to estimate the partial diffusion coefficients of Cu and Zn at 800 °C, which we assume to be independent of the local composition. The atomic fraction of Zn measured at a given distance x0 is XZn = 20% and the gradient of Zn atomic fraction is estimated to be @XZn =@x = 0.17/mm. The interdiffusion coefficient deduced from the observed concentration profiles is estimated to be D = 1.13 10–13 m2/s. Using the expressions for the interdiffusion coefficient D and the lattice displacement ! velocity v previously estimated, express DZn and DCu as a function of D,v, r XZn and XZn . Calculate the values of the diffusion coefficients DZn and DCu . 8. How does the melting temperature of the CuZn alloy vary with the Zn content? What is the consequence on the partial diffusion coefficients DCu and DZn ? Sketch, on the same graph, how DCu , DZn and the interdiffusion coefficient D evolve with XZn . The Zn concentration increases with x in this experiment. Solution to the problem 1. We consider a Cu/CuZn diffusion couple, which can be represented schematically as follows:

2. The melting point of copper is 1083 °C, that of zinc is close to 420 °C, cohesion of Zn is therefore weaker. It is easier to “break” the Zn–Zn bonds, than those linking

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Diffusion, Segregation and Solid-State Phase Transformations

copper atoms. The enthalpies of migration and formation of vacancies are therefore lower. Zn is therefore the fastest diffuser because of its lower melting temperature (“hot” diffuser). 3. The higher mobility of Zn makes the flux of Zn diffusing to Cu greater than the reverse flux of Cu. It appears therefore a flux of vacancies from Cu to brass (see previous scheme). The vacancies will then agglomerate and lead to the formation of voids in the Zn-rich region, under the Cu/CuZn interface in the region x [ 0. 4. The vacancy flux towards x [ 0 results in the displacement of atomic planes in the same direction. Therefore, the Mo wires migrate from copper towards the brass substrate. 5. The situation is as follows:

As Ji ðx Þ ¼ Di @C@xi ðx Þ, JCu is positive whereas JZn is negative. The vacancy flux is Zn given by JV ¼ ðDCu  DZn Þ @C @x . As ðDCu  DZn Þ is negative, the wind of

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vacancies is heading towards increasing values of x, thus, towards the brass substrate. The vacancy creation/annihilation rate is given by: @ CZn v sðx Þ ¼ @Cv =@t ¼  @J @x ¼ ðDCu  DZn Þ @x 2 . 2

As ðDCu  DZn Þ is negative, this rate is of opposite sign as compared with the 2 curvature of CZn ðx Þ and it vanishes at the inflection point (@@xC2Zn ¼ 0), as shown in the previous figure. v with q the atomic density (i.e. Finally, the vacancy flux is written with ~ J v ¼ q~ the inverse of the atomic volume) and ~ v the vacancy velocity. We thus have ! ~ v ¼ ðDZn  DCu Þ rqCZn . Once again, we can see that the vacancy wind is directed towards the brass region (x > 0). This velocity is maximum when the concentration gradient of Zn is maximum, at the inflection point. 6. The Mo wires have travelled over a distance d of 52 lm after t = 30 days at 800 °C. In a steady state, the lattice displacement speed v is constant. Writing d ¼ vt leads to v = 0.02 nm/s. 7. We know that XZn = 0.2 and that @XZn =@x = 0.17/mm in x0 and D = 1.13 10–13 m2/s. The interdiffusion coefficient is given by D ¼ DZn XCu þ DCu XZn ! and ~ v ¼ ðDZn  DCu Þr XZn . Combining these two expressions gives: DZn ¼ D þ XZn rXv Zn  1.36.10–13 m2/s and DCu ¼ DZn  rXv Zn  0.18.10–13 m2/s. 8. Since the melting temperature of Zn is lower than that of copper, the melting temperature Tm of CuZn alloy decreases when increasing the Zn content. In addition, atomic mobility is known to increase when T/Tm increases, i.e. when Tm decreases (“hotter atoms” diffuse more rapidly). This implies that both DZn and DCu increase with the Zn content as shown below (i.e. for increasing distances from the initial interface). Moreover, DCu \D\DZn . The diffusion coefficients therefore vary as follows:

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2.2.7

Short-Circuits of Diffusion: Grain Boundaries and Dislocations

1. Diffusion along grain boundaries 1.1. The polycrystalline material is modelled by a periodic network of cubic grains of side d that are separated by grain boundaries (GBs) of width e. Cubes are parallel to the three directions of an orthogonal frame (xyz). Only diffusion along the grain boundaries that are parallel to the diffusion flow (z) are considered. Let ~ J V and ~ J GB be the diffusion fluxes in volume and along GBs in the z direction and DV , DGB the corresponding bulk and GB diffusion coefficients. Sketch the structure in the plane xOy that is perpendicular to the diffusion flow (z). Show that the total flux ~ J along z is J V whose respective weights are expressed as a weighted sum of ~ J GB and ~ expressed as a function of e and d. Give a simplified expression for e  d. This approximation is generally observed as e  0.5 nm and d > 1 μm. 1.2. It is assumed that there is a concentration gradient along z. Show that one can express the diffusion fluxes as a function of an apparent diffusion   coefficient D ¼ DV þ DGB . Express DV and DGB as a function of the volume and GB diffusivities DV , DGB and of the ratio e=d. It is assumed that e\\d. 1.3. Let us assume that the pre-exponential factor D0 is common to both volume and GB diffusion. Sketch the Arrhenius diagram showing the  evolution of DV and DGB with the inverse of temperature. The activation energies for bulk and GB diffusion are, respectively, QV and QGB ¼ 0:5QV . Indicate on this diagram the temperature T  for which both bulk and GB contributions to diffusion are equal. What are the related diffusion modes when T \T  and T [ T  respectively? 1.4. Calculate the expression of T  as a function of e, d, and QV . What are the diffusion modes for a material containing very fine grains (nanomaterials) or on the contrary for a material with very large grains? 1.5. Copper diffusion in BCC iron is considered (a ferrite), D0 = 0.57 cm2/s, and QV = 238 kJ/mol. The ferrite–austenite transition temperature is Tt = 910 °C. The grain size is d = 100 μm and e  0.5 nm. Calculate T  and compare it to Tt . What are your conclusions?

2. Diffusion along dislocations 2.1. We will perform a similar calculation for dislocations that we model by a periodic lattice of tubes parallel to the diffusion flux (“pipe diffusion”) and separated by a distance L. We shall assume that dislocations behave as high diffusivity pipes of radius r = 1 nm. Sketch the microstructure. The pre-exponential factor of the bulk diffusion coefficient D0 is equal to that related to diffusion along dislocations. The related diffusivities are DV for

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bulk diffusion and Dd for diffusion along the dislocations. Let us assume that activation energies are still given by QV and Qd ¼ QV =2. Calculate the expression of the apparent diffusion coefficient (D) as a function of the dislocation density per unit area, which can be written d ¼ 1=L2 . Express D in the following way: D ¼ DV þ Dd . In an annealed undeformed iron, the average distance L between dislocations is close to 10 μm. Show that under these conditions, DV  DV . 2.2. Calculate the expression of temperature T  for which both bulk and dislocation contributions to diffusion are equal. Calculate T  for the diffusion of copper in ferrite. 2.3. Compare this temperature to that calculated for grain boundaries. For what dislocation density would high diffusivity pathways along dislocations have an influence comparable (in terms of T*) to that along grain boundaries? Compare the related distance between dislocations to the previous distance L = 10 μm. Is this a highly deformed (strain-hardened) material? Solution to the problem 1. Diffusion along grain boundaries 1.1. The structure can be represented as follows:

The diffusion of atoms has 2 contributions ~ J V and ~ J GB . The diffusion flux being given by the number of atoms migrating per unit area, we can express total flux ~ J as the sum of both fluxes, weighted by the respective surface fractions, i.e. d 2 =ðd þ eÞ2 for bulk diffusion and ð2de þ e2 Þ=ðd þ eÞ2 for diffusion along GBs. Thus,

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Diffusion, Segregation and Solid-State Phase Transformations

~ J ¼

2de þ e2 ~ ~ þ J J GB : V ðd þ e Þ2 ðd þ e Þ2 d2

When e\\d, we have: ~ ~ J ¼~ J V þ 2e d J GB .

! 1.2. When a concentration gradient r C occurs along z, there is a diffusion flux ! that can be written ~ J ¼ Dr C with D ¼ D  þ D . It is easy to show V

GB

  ¼ 2e that approximate expressions are DGB d DGB and DV ¼ DV if e\\d. QV

QV =2

RT . The logarithm of each 1.3. We can therefore write D ¼ D0 e RT þ 2e d D0 e contribution (in volume and along GBs) can then be represented in an Arrhenius diagram as follows:

When T [ T  , the bulk diffusion dominates whereas when T \T  (low temperatures), diffusion occurs mainly along grain boundaries. 1.4. Temperature T  is such that DV ¼ 2e d DGB so that: QV

D0 e RT  ¼

QV =2 2e D0 e RT  d

This leads to T  ¼ 2R QlnV d . For large grains (d ! þ 1), this temperature ð2eÞ tends to 0, so the bulk diffusion is always dominant. However, for a nanomaterial (but still in the limit e\\d), T  gets very large and diffusion is dominated by the high diffusivity pathways along GBs. 1.5. Considering the diffusion of copper in a iron and grains of 100 µm in size, one gets T   970 °C that is higher than Tt above which ferrite transforms into austenite. Bulk diffusion is never predominant in ferrite whatever the temperature. The diffusion in ferrite along grain boundaries is therefore always preponderant whatever the temperature.

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2. Diffusion along dislocations 2.1. The modelled microstructure can be represented as follows:

As before, the diffusion coefficient can be expressed as the sum of the bulk and dislocation diffusion coefficients DV and Dd weighted by the respective areas: D¼

L2  pr 2 pr 2 D þ Dd V L2 L2

The latter equation can be re-expressed as D ¼ DV þ Dd . When the density of dislocations (d) is relatively low, such that pr 2 \\L2 , the approximated weighted diffusivities reads DV ¼ DV and Dd ¼ pr 2 dDd . 2.2. By definition T  is such that DV ¼ Dd . As a result, T  ¼ QV L2  556 °C. 2R ln

pr 2

The diffusion along dislocations is preponderant at low temperatures when T \T  .   2.3. We have found previously that Tdislocation \TJG . These two temperatures qffiffiffiffi 2 L d dp would be equal if pr 2 ¼ 2e. We then get L ¼ r 2e  560 nm. Hardening in materials results in an increase in the dislocation density. Since the dislocation density is quite low, the material can be considered as not very strain hardened.

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2.2.8

Diffusion, Segregation and Solid-State Phase Transformations

Driven Diffusion in a Force Field and Spinodal Decomposition

Let us consider a binary alloy AB and the diffusion of B atoms under the action of a chemical potential gradient (lB ). Using a simple atomistic model in one dimension, we will demonstrate that the velocity of B atoms vB0 is proportional to the chemical ! potential gradient: ~ v 0B ¼ MB r lB where MB is the mobility of B atoms whose expression will be given as a function of the diffusion coefficient DB . 1. The concentration of B atoms is CB and XB represents the atomic fraction of B atoms. Let us consider a one-dimensional lattice of periodicity a along x (x ¼ na with integer n). Let us approximate the potential configuration energy of B atoms WB ðx Þ as a sine like function of the position x. Represent schematically WB ðx Þ: (i) in the absence of any chemical potential gradient and indicate the height of the diffusion barrier DHD and (ii) in the presence of a ! constant negative chemical potential gradient r lB . 2. Let us consider an ideal solution (no chemical interactions between A and B atoms) and the presence of a concentration gradient. Calculate the norm of the gradient rlB as a function of the concentration gradient. DHD

3. Let us first consider the situation where rlB = 0. Let C0 ¼ t0 e kT be the jump frequency of B atoms towards one of their first two neighbours. t0 is a constant proportional to the vibration frequency of atoms. Let us consider for instance the atom in x = 0. The effective jump frequency C of this atom is the difference between that towards positive coordinates C þ (x [ 0) and that in the other direction C (x\0). What is the effective jump frequency C and the related diffusion flux under these conditions? We shall now examine the diffusion in the presence of a chemical potential gradient. Let lB ðx Þ be the chemical potential of a B atom at position x. In the presence of a non-zero and negative gradient, the diffusion barrier DH is decreased by a value equal to dlB (0) in the opposite direction. Give the expressions of C þ and C as function of C0 , DHD and dlB . Show that the net jump frequency C ¼ C þ  C can be expressed as C ¼ K C0 . We assume that dlB \\kT . Give an approximate expression for K as a function of rlB and a. 4. Give the expression for the migration velocity of B atoms vB0 as a function of C and ! a. Demonstrate that ~ v 0B ¼ MB r lB . This expression, demonstrated in 1D, remains valid in 3D. Give the expression of MB . Knowing that DB ¼ C0 a 2 in cubic crystals, deduce the well-known Darken equation giving MB as a function of DB . 5. This migration of B atoms under the action of the chemical potential gradient, leads to a flux of B atoms (JB ) that can be easily written as a function of vB0 and CB . Show that for an ideal solution, the expression of JB is identical to that of the first Fick law. Again, this demonstration remains valid in 3D. 6. Write the chemical potential lB as the sum of an ideal part liB and an additional term uB ðx Þ accounting for the non-ideality of the alloy. Demonstrate that JB is

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7.

8.

9.

10.

11.

109

the sum of a Fickian term and a transport term being proportional to a force ! ~B ðx Þ ¼ r uB ðx Þ. Find the Nernst–Einstein equation giving the transport F velocity vB (different from vB0 ) as a function of FB ðx Þ. The AB alloy presents a miscibility gap leading to the precipitation of a coherent phase b that has the same BCC structure than that of the parent phase a (e.g. FeCr alloys). Thus, for high supersaturation in B atoms, a spinodal decomposition regime can be observed. Using the previous expressions, show that in the presence of a chemical force, the flux density JB can be put in the form of Fick’s first law with an “apparent” diffusion coefficient DB that accounts for the thermodynamics of the system. Give the expression of DB ðXB Þ as a function of DB , XB and the second derivative of free enthalpy G ðXB Þ of the alloy. For that purpose, use the Gibbs–Duhem relation. We remind that G ¼ XA lA þ XB lB in a binary alloy. Show that one finds the classical Fick law when the solid solution is ideal, with DB = DB . Give the shape of free enthalpy G ðXB Þ when the alloy decomposes into two phases a and b of same structure. Give the sign of the diffusion coefficient DB ðXB Þ between the inflection points of G ðXB Þ and outside. Compare the related decomposition regimes according to the curvature of G ðXB Þ using Fick’s laws. Thermal fluctuations of concentrations occur at t = 0. These are modelled by a sinusoidal function of amplitude dCB0 and periodicity k. Describe the temporal evolution of these concentration fluctuations dCB ðt Þ. Assuming that DB varies little with CB and that it can thus be considered as constant, find the expression of dCB ðt Þ that is solution of the second Fick equation. Represent dCB ðt Þ for DB < 0 and compare to the homogenization problem when DB [ 0. The alloy AB is modelled by a regular solution. Calculate the expression of the second derivative of free enthalpy G ðXB Þ and of the spinodal line TS ðXB Þ corresponding to the inflection points of G ðXB Þ. Show that TS tends towards the critical temperature TC of the system when XB ¼ 1=2. TC is the temperature above which A and B atoms are miscible in all proportions. Deduce the expression of DB ðXB Þ as a function of the ratio TS =T . What is the sign of DB when T [ TS and T \TS , respectively? When the alloy is concentrated with significantly different mobilities Mi (i = A, B), the Kirkendall effect must be considered. As a consequence, a transport term Ci~ v taking into account the lattice velocity ~ v is to be added to the previous equations for the fluxes Ji (i = A, B). By writing that the sum of fluxes of A and B atoms is zero in the fixed frame of reference related to the sample, determine the expression of ~ v as a function of chemical potential gradients of A and B species. Substituting ~ v by its expression show that we find the classical form for the flux of B atoms in a chemical potential gradient lB : ! ~ J B ¼ MCB r lB Give the expression of the inter-mobility M as a function of MA and MB . Using the Gibbs–Duhem relation show that the latter equation can be re-written in the form of the following generalized flux equation:

110

Diffusion, Segregation and Solid-State Phase Transformations ! ~ J B ¼ M  C0 r ðlB  lA Þ

with C0 the concentration of atoms per unit volume. Give the expression of M  as a function of M , XA and XB . Show that M can simply be expressed as a function of the interdiffusion coefficient D. Demonstrate that the generalized equation for the flux can be put in the form of a Fick equation involving an apparent diffusion coefficient DB . Express the latter as a function of D, XB and of the second derivative of G ðXB Þ. Solution to the problem 1. Below are represented the configuration potential energies of B atoms WB(x) in a 1D crystal of periodicity a. WB(x) is shown in the absence of any chemical potential gradient (top diagram) and in the presence of a constant negative gradient (lower diagram, δ μB < 0).

2. For an ideal solution, we have lB ¼ GB þ RT ln XB , thus

dlB dx

B ¼ RT X1B dX dx

dCB ¼ RT CB dx . The chemical potential gradient is proportional to the concentration gradient. ! 3. When r lB ¼ 0, the height of the diffusion barrier is identical whatever the jump directions (x [ 0 and x\0). The latter is DHD (previous figure). The atoms DHD

migrate in both directions with the same frequency C0 ¼ t0 e kT . Thus, the net jump frequency is given by C ¼ C þ  C ¼ 0. The diffusion flux is also zero in the absence of chemical gradient.

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Let us consider the atom located at x = 0. In the presence of a negative gradient (dlB \0), the diffusion barrier is lower for increasing x (x [ 0): DH ¼ DHD þ dlB with δ μB < 0. Conversely, in the opposite direction (x\0), the barrier is higher: DH ¼ DHD  dlB . Related jump frequencies read: C þ ¼ t0 e C  ¼ t0 e

ðDHD þ dlB Þ kT

ðDHD dlB Þ kT

It is clear that C þ [ C0 and C \C0 when dlB \0. This leads to a net jump frequency C ¼ C þ  C that is written:    dlB  þ dlB dlB kT kT e C ¼ C0 e ¼ 2C0 sinh kT   B This equation can be expressed in the form C ¼ K C0 with K ¼ 2 sinh dl kT .   dlB dlB a B For dlB \\kT we have sinh dl kT  kT . Moreover dlB ¼ dx 2 because the difB a fusion barriers are centred in  a2. We can then conclude that C ¼ C0 dl dx kT and

a dlB K ¼  kT dx (>0). 4. It is natural to write for the migration velocity vB0 ¼ Ca. We deduce that:

vB0 ¼ C0

dlB a 2 C0 a 2 ¼ rlB dx kT kT

The writing of the latter equation in 3D leads to the well-known relationship: ! ~ v 0B ¼ MB r lB 2 B with MB ¼ D kT (the Darken equation) and DB ¼ C0 a for cubic lattices. 5. The migration of B atoms under the action of the chemical potential gradient leads to a flux JB ¼ CB vB0 . Considering an ideal solution, we can write lB ¼ GB þ kT ln XB (J/atom) and thus:

kT ! ! r CB : r lB ¼ CB We finally find the classical Fick law: kT ! ! ! ~ r CB ¼ MB kT r CB ¼ DB r CB ; J B ¼ CB MB CB with again DB ¼ MB kT .

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Diffusion, Segregation and Solid-State Phase Transformations

6. Let us write lB ¼ liB þ uB . We have: ! ! ! ! ~ J B ¼ MB CB r lB ¼ DB r CB  MB CB r uB ¼ DB r CB þ CB~ vB with: duB DB ¼ FB ðx Þ dx kT ! B ~ In 1D, FB ðx Þ ¼  du dx . Generalization in 3D gives: F B ¼ r uB . We then end up ~B . with the so-called Nernst–Einstein equation: ~ v B ¼ DkTB F ! dlB dCB B 7. Let us write ~ J B ¼ MB CB r lB . We can write dl dx ¼ dCB dx for the gradient component along x. Expressing the gradient components along y and z in similar ! way leads to ~ J B ¼ D  r CB with an apparent diffusivity that reads: vB ¼ MB ruB ¼ MB

B

DB ¼ MB CB

dlB dl ¼ M B XB B : dCB dXB

Let us express the chemical potential as a function of the free enthalpy. As dG G ðXB Þ ¼ lB XB þ lA XA then dX ¼ lB  lA . Using the Gibbs–Duhem equation, B XA dlA þ XB dlB ¼ 0, we obtain:   d 2 G dlB dlA dlB XB 1 dlB ¼  ¼ 1 þ : ¼ 1  XB dXB 1  XB dXB2 dXB dXB dXB Substituting this expression of the first derivative of lB into that of the apparent diffusivity leads to: d2G DB ¼ MB XB ð1  XB Þ dX 2. B

The apparent diffusivity can therefore be negative if the second derivative of G is d2G 1  negative. For an ideal solution, dX 2 ¼ kT X ð1X Þ and we recover DB ¼ MB kT ¼ B B B

DB that is nothing but the so-called Darken equation. 8. We consider phase separation into two phases of same structure a and b. Consequently G ðXB Þ can be represented as follows:

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2

d G We can see in this figure that dX 2 \0 in the concave region between both inflexion B

points. In this domain we have therefore DB \0. According to the first Fick ! equation, ~ J B ¼ DB r CB , the flux of B atoms goes up the concentration gradient. This so-called up-hill diffusion of B atoms leads to the amplification of concentration fluctuations. The system in unstable. This is the regime of spinodal d2G decomposition. On the other hand, outside the instability zone dX 2 [ 0 and B

DB [ 0. Concentration fluctuations are damped, if supersaturated the system is in a metastable state. This corresponds to the nucleation and growth regime. 9. We have shown before that dCB ðt Þ increases with time when DB \0. Considert ing13 that DB ¼ cste, it is immediate to demonstrate that dCB ðt Þ ¼ dCB0 ðx Þes is 2 2  d CB k B solution of the second Fick equation dC dt ¼ DB dx 2 with s ¼ 4p2 jD  j the relaxation B time. The amplitude increases exponentially with time. In the homogenization t problem for which DB [ 0, we get dCB ðt Þ ¼ dCB0 ðx Þe s . The amplitude decreases exponentially with time. We can represent dCB ðt Þ in the following way:

2

d G 10. The spinodal line TS ðXB Þ is such that dX 2 ¼ 0. Here, the free enthalpy (in J/mol) B

is written: G ¼ ð1  XB ÞGA þ XB GB þ NZ eXB ð1  XB Þ þ RT ½ð1  XB Þ lnð1  XB Þ þ XB ln XB

where N is the Avogadro constant, Z is the coordination number and e the order energy (chapter 1). Thus,

d2G 1 1 ¼ RT þ  2NZ e 1  XB XB dXB2

In reality the calculation of the linear theory of Cahn is more complex because DB ¼ f ðCB Þ, as discussed in problem 3.2.10. 13

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114

By definition

d2G dXB2

¼ 0 at T ¼ TS so that:

1 1 RTS þ ¼ 2NZ e; 1  XB XB

Then, TS ¼ 2Zk e XB ð1  XB Þ. For XB ¼ 1=2, TS ¼ Z2ke ¼ TC . TC is the critical temperature above which A and B are miscible in all proportions (chapter 1). Above this temperature, there is no more inflection points, and no more possible phase separation. Combing the previous expression of TS with that of the apparent diffusivity DB ¼ 2

d G MB XB ð1  XB Þ dX 2 results in the following equation: B     TS TS DB ¼ MB RT 1  ¼ DB 1  T T

As expected, the apparent diffusion coefficient is positive when T [ TS and negative for T \TS (spinodal regime). In the above equation for DB , units of MB (m2s−1J−1 mol) are different from those of question 7 (m2s−1J−1), as we have written G in J/mol here and not per atom. 11. In the presence of the Kirkendall effect, the diffusion fluxes with respect to the sample reference frame are written: ! ~ J B ¼ MB CB r lB þ CB~ v ! ~ v J A ¼ MA CA r lA þ CA~ ~ v is the displacement velocity of the lattice with respect to the immobile reference of the solid. In the immobile reference frame of the sample, we have: ~ JA þ~ J B ¼ 0: ! ! We then get ~ v ¼ MA XA r lA þ MB XB r lB . Injecting this expression of the velocity into the expression of ~ J B and using the Gibbs–Duhem relation, XA dlA þ XB dlB ¼ 0, gives: ! ~ J B ¼ MCB r lB with M ¼ XA MB þ XB MA , that is the inter-mobility. ! ! We again use Gibbs–Duhem to show that r lB ¼ XA r ðlB  lA Þ so that: ! ~ J B ¼ M  C0 r ðlB  lA Þ with M  ¼ MXA XB . C0 is the atomic density, i.e. the inverse of the atomic volume.

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Di As Mi ¼ kT (i = A,B), the previous equation for M can be rewritten in the form: D M ¼ kT with D ¼ XA DB þ XB DA the interdiffusion coefficient. The expression for M is the generalized Darken equation. We can now express the diffusion flux as a function of the second derivative of G. dG As dX ¼ lB  lA , it is immediate to write: B

d G! ~ J B ¼ M  r CB : dXB2 2

The latter equation can be re-expressed as: ! ~ J B ¼ DB r CB D d G We finally get DB ¼ kT ð1  XB ÞXB dX 2 . The apparent diffusivity again has the 2

B

sign of the second derivative of free enthalpy. DB is negative in the spinodal domain and positive outside.

2.2.9

A New Fick Law According to Howe

Andrew Howe (Sheffield, UK) published in 2002, in Scripta Materialia, an article [22] that proposed a new law which, he thinks, is more general than Fick’s law. In this problem, we propose to examine his demonstration. The modification introduced with respect to the classical Fick diffusion is that the jump frequency depends on the local concentration in the considered interstitial plane (C0 ðC Þ, see figure). We consider the interstitial diffusion of B atoms in a solid A. 1. Preamble: Darken’s equation (1953) 1.1. Show that, in the presence of a chemical potential gradient, the diffusion flux can be written in the form of Fick’s equation provided that the diffusion coefficient has the following expression: ~ ¼ D ð1 þ K Þ D

ð2:1Þ

with D the diffusion coefficient of B in A. Give the expression of the constant K as a function of the activity coefficient c and of the atomic fraction X in B atoms and that of D as a function of the mobility M (Darken’s equation). We remind that chemical potential is given by l ¼ l0 þ RT lnðcX Þ. 1.2. Under which condition can the second law be written in the following form? @C ~ ¼ DDC @t

ð2:2Þ

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Diffusion, Segregation and Solid-State Phase Transformations

2. The new diffusion law (Howe, 2002) 2.1. We treat the problem of the diffusion of B atoms in an AB alloy in one dimension along the direction Ox (figure below). The distance between planes is b. All concentrations will hereafter refer to B atoms. We consider two successive planes (planes 1 and 2) of concentrations C and C þ dC , respectively. The jump frequencies depend on x and C and are respectively C1 ¼ C0 and C2 ¼ C0 þ dC0 . By expressing the gradients (@=@x) of both C and C0 , demonstrate the Howe equation: J ¼

@ ðDC Þ @x

ð2:3Þ

In the calculation of J , we will neglect the second order term dC dC0 . Give the expression of D as a function of b, C0 and Z  defined as the number of atomic jumps in a given direction (i.e. x [ 0) that have a non-zero projected distance on the Ox axis. Compare with the classical expression.

2.2. Show that we find the first classical Fick equation provided that C0 does not depend on the local concentration C ðx Þ. Give the expression of D as a function of C0 , b and Z  . Show that the diffusion coefficient related to the diffusion of B atoms in simple cubic lattices writes D ¼ C0 a 2 and remains valid for BCC and FCC structures (a is the lattice parameter). 3. Diffusion in a chemical force field 3.1. The diffusion is no longer Brownian. There is a driving force “pushing” B atoms to migrate towards low chemical activity regions (i.e. low chemical

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potential). We now assume that D is constant (C0 is not a function of C and thus D does not depend on x). We shall now show that the new equation of Howe14 (question 2.1) is valid provided that D ¼ cD with c the activity coefficient. D is the diffusion coefficient for an ideal solution (c ¼ 1). Show that we find a similar expression as Fick’s equation with an apparent diffusion coefficient whose expression is similar to that given in (2.1): ~ ¼ D  ð1 þ K Þ D

ð2:4Þ

! Thus, one can write the flux as follows (2.3): J ¼ r ðD C Þ. Let DGD be the height of the migration barrier (free enthalpy) when there is no chemical force (ideal solution). 3.2. Give the classical expression of D as a function of temperature T , ν and Z*. Now consider that D ¼ cD and show that this introduces a diffusion barrier depending on the local chemical potential l (l1 and l2 in planes 1 and 2). Represent schematically the diffusion barrier between two spatial configurations where the potentials on the left and on the right of the barrier are respectively l1 and l2 . Show that the net jump frequency depends on the difference of chemical potentials between planes 1 and 2 ! and that the diffusion flux J ¼ r ðD C Þ is proportional to the chemical potential gradient. 3.3. Let us demonstrate that we can write D ¼ cD in the general case. Re-express the classical expression of the diffusion flux (C0 6¼ C0 ðC Þ) in a chemical force field in the following form: J ¼ D

@C  f ðD; cÞ @x

ð2:5Þ

We consider that this equation remains valid if we replace D by D ¼ cD. Compare term by term this equation (2.5) with that given by Howe (2.3). Deduce that D is necessarily proportional to the activity coefficient c. Solution to the problem 1. Preamble: Darken’s equation (1953) 1.1. The diffusion flux of B atoms can be written as: @l J ¼ MC @x C is the concentration of B atoms given by C ¼ X=Vat, with Vat the atomic volume. M is the mobility of B atoms. Let us use the general expression of the chemical potential l ¼ l0 þ RT lnðcX Þ. c is the activity coefficient. Injecting this expression into the equation of the diffusion flux leads to:

This problem relies on the postulate made by A. Howe, i.e. D ¼ cD, whose validity could be uncertain. 14

118

Diffusion, Segregation and Solid-State Phase Transformations   d ln c @C ~ @C ¼ D J ¼ MRT 1 þ d ln X @x @x ~ ¼ D ð1 þ K Þ (1) Writing the diffusivity in the form: D d ln c leads to: K ¼ d ln X . The diffusion coefficient of B atoms in A is given by the well-known Darken equation: D ¼ MRT . 1.2. The second Fick equation reads:   @C @ ~ @ ¼ D C @t @x @x ~ depends on the activity that is generally a function of x and cannot thus D be “taken out” of the divergence term. Moreover, D can also be a function of x. Consequently, we can write the usual Fick law that follows only if c and D are both constant: @C ~ ¼ DDC @t

2. Establishment of the new diffusion law (Howe, 2002) 2.1. Let us sketch the diffusion process as follows:

ð2Þ

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Writing the difference of fluxes from interstitial plane 1 and plane 2 (J1!2 ) and inversely from 2 to 1 (J2!1 ) gives: J ¼ J1!2  J2!1 ¼ Z  bðC1 C1  C2 C2 Þ with Z  the number of jumps along Ox that have a non-zero projection and b the jump length. We have C1 ¼ C and C2 ¼ C þ dC . Similarly, C1 ¼ C0 and C2 ¼ C0 þ dC0 . Neglecting the second order term dC0 dC leads to: J ¼ Z  bðC0 dC þ C dC0 Þ: Since the distance between planes is very small, we can write dC ¼ bð@C =@xÞ. This can be seen as the first order Taylor expansion of 0 concentration in x þ b. Similarly, dC0 ¼ b @C @x . Thus,   @C @C0 þC J ¼ Z  b2 C0 : @x @x Writing D ¼ Z  b2 C0 leads to the expected Howe equation: @ ðDC Þ @x

J ¼

ð3Þ

! This equation can be generalised in 3D: ~ J ¼ r ðDC Þ. It is worth mentioning that the latter is also the equation of Christian (1965). If D is constant, we find again the classical first Fick equation.  @C0 0 2.2. Writing J ¼ Z  b2 C0 @C as demonstrated before, @C @x þ C @x @x ¼ 0 gives the expected Fick law: J ¼ Z  b2 C0

@C @C ¼ D : @x @x

In a simple cubic, we have a single jump (Z* = 1) in the direction x [ 0 (or x\0) and the projected jump distance on Ox is equal to the lattice parameter: b = a. As a result, D ¼ C0 a 2 . Let us calculate D for both BCC and FCC structures. The number of jumps of non-zero projection toward x > 0 is Z  ¼ 4 in both BCC and FCC structures and b ¼ a=2. As a result, the previous equation D ¼ C0 a 2 remains valid for all the cubic structures. 3. Diffusion in a chemical force field 3.1. Let us write Howe’s equation in 1D: J ¼

@ ðD  C Þ ; @x

120

Diffusion, Segregation and Solid-State Phase Transformations Substituting D ¼ cD in the latter equation and keeping in mind that d ln X ¼ d ln C gives:   d ln c @C ~ @C ¼ D J ¼ cD 1 þ d ln X @x @x  ~ ¼ D 1 þ with D

d ln c d ln X



that leads to the expected form given in (1): ~ ¼ D  ð1 þ K Þ D

ð4Þ

with K ¼ ddlnlnXc . When a chemical driving force occurs, then D is to be replaced by D ¼ cD in Howe’s equation. Γ and consequently D  depend on concentration and on position x. DGD

3.2. We know that C0 ¼ te RT when no chemical driving force occurs. DGD is the diffusion barrier and t a constant depending on the vibration frequency of atoms. Previous question enables to write that: D ¼ tZ  b2 e

DGD RT

:

Let us calculate γ first as a function of chemical potential of an ideal solution (c ¼ 1) so as to calculate D ¼ cD. As l ¼ l0 þ RT lnðcX Þ one can write: c¼e

llid RT

with lid ¼ l0 þ RT lnðX Þ. Thus, one gets: lid

D  ¼ tZ  b2 e RT e

ðDGD lÞ RT

:

When c ¼ 1, l ¼ lid and one recovers D  ¼ D. This latter equation shows that the migration barrier for a given species is lowered by its chemical potential l and the net jump frequency Γ from site 1 to site 2 now depends on the difference between related chemical potentials in these sites 1 and 2, as shown in the figure below. Jump frequencies C1 and C2 in sites 1 and 2 are related to the respective diffusivities D1 ¼ Z  b2 C1 and D2 ¼ Z  b2 C2 . As l1 \l2 in the figure below, the diffusion barrier in sites 1 is larger than in sites 2. Consequently C2 [ C1, there is a net flow of atoms from sites 2 toward sites 1, where the chemical potential is smaller. In addition, calculating J ¼ @ ðD C Þ=@x using the latter expression of D* shows that J is

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proportional to chemical gradient @l=@x as expected. It can be shown   that J ¼  DRTC @l @x with D ¼ cD. By identification we get C1 ¼ c1 C0 with c1 the activity in site 1 and C2 ¼ c2 C0 with c2 the activity in site 2.

3.3. Let us write: J ¼ D with v ¼ DRTF and F ¼  @/ @x .

@C þ Cv @x

ð5Þ



d ln c  d ln c As @/ @x ¼ RT dx , we have v ¼ D dx and we recover the following expression for the flux:   d ln c @C ; J ¼ D 1 þ d ln X @x

122

Diffusion, Segregation and Solid-State Phase Transformations that we have derived in question 3.1 from the Howe equation when imposing D ¼ cD. This necessary condition can also be evidenced considering Howe’s equation in the following form: J ¼

@ ðD  C Þ @C @D  ¼ D C : @x @x @x 

 d ln c Identification of the two latter equations gives: @D @x ¼ D dx . Integrating the latter equation gives: D ¼ cK 0 with K 0 a constant to be determined. For γ = 1 (no chemical force), D ðc ¼ 1Þ ¼ K 0 and D ¼ D, therefore K 0 ¼ D and consequently D ¼ cD. Thus, Howe’s equation is equivalent to the classical equation of the flux in a force field (as given in the course) if we replace D by D ¼ cD.

2.2.10

Driven Diffusion in a Force Field: Application to Nabarro-Herring Creep

When a material is subjected to a tensile stress r at a temperature for which diffusion is important (T [ 0:3Tf in metals, with Tf the melting temperature), we observe an elongation of the sample in the direction of stress r that increases with time. This is called creep. Quickly, after a first stage called primary creep where dislocations multiply, a stationary regime is reached during which the density of dislocations reaches a plateau and remains approximately constant. This secondary stage of creep results in an elongation dl (l being the dimension of the specimen) increasing linearly with time (secondary creep). For low stresses, creep is essentially dl driven by diffusion and not by dislocations and the strain rate e0 ¼ 1l dt is proportional to the stress r. We speak of Nabarro-Herring creep or Newtonian creep. The viscosity is defined as the ratio g ¼ r=e0 (expressed in poise). We will limit ourselves to diffusion creep in this problem. We will thus establish the expression of viscosity g. It is a problem of vacancy diffusion in a force field that can be solved using the Nernst–Einstein equation. The force to be considered here is the tensile stress r applied to the specimen surfaces. 1. We consider a cylindrical specimen of length l subjected to a tensile stress r along an axis (z) that is parallel to its length. The volume of the specimen is V ¼ lS with S its top surface area. The action of the stress results in a flux of vacancies from the ends of the specimen (from the top and bottom surfaces) to its lateral edges. This leads to an elongation of the specimen dl and a decrease in its diameter that will be neglected in the following calculations. Make a diagram showing the flux of vacancies from the extremities to the edges of the specimen. Since surfaces are perfect vacancy sinks and sources, we shall assume that the gradient of vacancy concentration near surfaces is zero.

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We can therefore assume that, in the steady state, the diffusive term is negligible compared to the transport term. dl We will calculate the expression of the flow velocity e0 ¼ 1l dt as a function of the atomic volume X and the vacancy flux JV . For that purpose, calculate the variation of volume dV at the ends of the specimen as a function of the elongation dl and S. Derive the expression of e0 . Then express dV as a function of the number of vacancies dn moving from one of both ends to the specimen sides as a function of the atomic volume X. Deduce the expression of e0 as a function of n 0 ¼ dn=dt, S, l and X. Then express n 0 as a function of the vacancy flux (JV ). From the Nernst–Einstein equation linking the velocity v and the force F (expressed by atom), give JV as a function of r. Using the expression for the vacancy diffusion coefficient DV as a function of the self-diffusion coefficient D, give the final expression of e0 as a function of D, X l, kT and r. Deduce the expression of viscosity g. How does it vary with temperature T ? We speak of Newtonian creep when viscosity g is constant (stationary creep). Is it the case here? 2. Nickel-based superalloys are key materials in aeronautics for aircraft turbojet engines. Turbine blades, in particular, are subject to considerable stresses (centrifugal forces due to blade rotation) and to temperatures that can approach 1000 °C. In some aircraft such as the Rafale (Dassault), the blades are moreover monocrystalline. The absence of grain boundaries, which are weak points under stress, allows for better performance than polycrystalline blades. We will apply the previous calculations to these materials. We shall consider pure nickel for simplicity. The crystal parameter of nickel (FCC) is a = 0.36 nm. The data for the self-diffusion of nickel are D0 = 1.9 cm2/s and Q = 68 kcal/mol (1 cal = 4.18 J). Calculate the viscosity g at 1000 °C for the blade length l = 10 cm. Deduce the flow rate e0 (per hour) under a stress of 100 MPa. Deduce the blade elongation dl after 20 years of service for aircraft flying 10 h/week in average. What do you conclude? 3. We will now apply these calculations to polycrystalline nickel (civil jet engines) whose grain boundaries constitute defect sinks or sources just like surfaces. It is considered that the previous calculations remain valid for a polycrystalline material. The length to be considered in calculations is now the grain size l 0 , and we take l 0 = 100 µm in average. Calculate the new viscosity. All things being equal, compare to the previous value. What is the ratio of viscosities between monocrystalline and polycrystalline alloys? What do you deduce from this? Calculate the flow rate (per hour) for r = 100 MPa. The grains of a material often have a dispersion in size. What will be the “response” of grains to creep according to their size?

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Solution to the problem 1. The volume change at each specimen end is given by dV ¼ S dl2 (figure below).

dl The flow velocity e0 ¼ 1l dt is then written:

e0 ¼

2 dV : Sl dt

If dn is the related number of vacancies migrating from surface area S of the specimen, then we can write that dV ¼ Xdn with X the atomic volume. As a result, we get: e0 ¼

2X 0 n Sl

with n 0 ¼ dn=dt. The number of vacancies migrating to one end is written: n 0 ¼ JV S The vacancy flux JV can be expressed using the Nernst–Einstein equation. Neglecting the diffusive term leads to: JV ¼ CV v

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with CV the vacancy concentration. The vacancy velocity v reads: v¼

DV F kT

F is the force per atom that is exerted: F ¼ rS na and na the number of atoms in the specimen: na ¼ Sl=X. DV is the diffusion coefficient of the vacancies: DV ¼ D=XV with Xv ¼ Cv X and D the self-diffusion coefficient. Substituting these expressions in those of the flux leads to: JV ¼

D r kT l

D r Substituting the latter equation in that of n 0 gives n 0 ¼ kT l S and finally we get 0 for e :

e0 ¼

2XD r kT l 2

that can be expressed as a function of the viscosity: r e0 ¼ g 2

Q

kT with g ¼ l2XD the viscosity. The latter varies with temperature as T =D  T =e RT . At a given temperature, the viscosity is constant. It is the Newtonian creep. The viscosity decreases with temperature because the exponential term prevails over the linear term. Q 2. At 1000 °C, we have D  D0 e RT = 4.07 10–16 m2/s. Nickel is FCC and the elementary cell (volume = a 3 ) contains 4 atoms, as a result X ¼ a 3 =4 = 1.16 10–29 m3. For l = 10 cm, we find g = 1.85 1022 poises. For a stress of 100 MPa, we get e0 = 1.95 10–11 h−1. Thus, after 20 years of service (10 h per week) the elongation is about dl  20 nm. This value is very small and has no significant influence. 3. Let g0 be the viscosity of the polycrystalline material. We have g0 =g ¼ ðl 0 =l Þ2 = 10–6. So e0 is 106 times greater in polycrystalline alloys. The elongation of grains is increased in the same proportion. It is thus clear that small grains and consequently material with nanograins will be more affected. Creep will result in profound changes of the microstructure in terms of grain size distribution and grain shape.

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2.2.11

Diffusion, Segregation and Solid-State Phase Transformations

Driven Diffusion of Impurities in a Force Field, the Formation of Cottrell Atmospheres

The diffusion and segregation of impurities to dislocation lines in alloys lead to the formation of Cottrell-Bilby clouds. This segregation phenomenon, predicted theoretically by Cottrell and Bilby [23], is encountered in many materials, whether metallic (carbon in steels, boron in FeAl alloys, see figure below), ionic crystals (silver in NaCl) or even semiconductors (dopants B, As or P in silicon). These Cottrell atmospheres play an important role in plasticity in metals by hindering the dislocation displacement or even by pinning them during deformation (structural hardening effect).

Three-dimensional atomic tomography image showing a boron-enriched Cottrell atmosphere (large black dots) in an ordered alloy (B2) FeAl [24]. Only Al-rich atomic planes (001) are shown here (small black dots, iron is not represented). The Burgers circuit shows the presence of an edge dislocation along which boron segregates (image E. Cadel, GPM). The dislocation line is perpendicular to the figure.

In order to calculate the spatial distribution of impurities around an edge dislocation line, we shall study their diffusion in the force field due to their elastic interaction with an edge dislocation. Let us consider a dilute solid solution AB where B is an impurity interacting with the stress field of the line defect. The dislocation line is set along axis Oz that is perpendicular to the figure. Its Burgers vector is parallel to x and its norm is equal to b. The elastic interaction energy (/) of a B atom with the dislocation at a given distance r is expressed in cylindrical coordinates (r; h) as follows:

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/ðr; hÞ ¼ K

sin h r

lb where K ffi 3p dV . μ is the shear modulus and dV the change in volume introduced by the impurity in solid solution.

1. A dislocation behaves like a mechanical +/− dipole. The region noted + is in compression above the line and that noted − is in expansion below. The impurity B will therefore interact differently according to the size effect that it produces in the host lattice of the solid solution. According to the sign of the volume variation (that is called size effect), B atoms will segregate either above the dislocation (dV \0) or below (dV [ 0). What is the sign of / in both cases? Interstitial atoms dilate the lattice when their size is larger than that of the interstice, so that dV [ 0. This is the most frequent case. dV \0 otherwise. If B is in substitutional position in the lattice, the volume difference between B and A atoms is to be considered, dV ¼ VB  VA . Show on a diagram (x, y) the regions corresponding to / ¼ þ /0 [ 0 and / ¼ /0 for interstitial impurities (for which dV [ 0). 2. Show that both equipotential lines (/ ¼ /0 ) are circles in the figure plane (r; h). Give the coordinates of the circle centre and its radius as a function of /0 and K. Represent these equipotentials in the plane (r; h) for dV [ 0. What is the shape of the interaction iso-surfaces in 3D? How do these equipotential change for dV \ 0? 3. Fickian diffusion term is not considered. Express the velocity vector ~ v (transport term) as well as the only related transport flux ~ J B as a function of the potential energy of interaction of B with the dislocation /, of DB (diffusion coefficient of B in A) and C the concentration in B atoms. ! ~ ¼ r By analogy with electric field lines (E V ), and without calculation, deduce from the shape of the equipotentials the shape of the transport flux lines ~ J B . For this purpose, draw qualitatively various equipotentials for increasing values of /0 , in particular for a very small and a very large value. 4. We will now calculate the shape of the transport flux lines. Let M be the point representing the position of the considered B atom. Write that the displacement ! vector d ðOM Þ is colinear to the gradient of / using vector product d ðOM Þ ^ r /. Deduce the shape of flux lines. Let us remind that the components of the gradient vector in cylindrical coordinates (r; h) are given by: rr ¼

@ 1 @ and rh ¼ : @r r @h

5. We now consider the Fick term so as to study the equilibrium state. Write the flux of B atoms ~ J B considering both the diffusive and the transport terms. Deduce its components Jr and Jh in cylindrical coordinates. Write thermodynamics equilibrium. Deduce the equation of iso-concentration curves C ðr; hÞ

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related to B atoms. For this purpose, make the change in variable u ¼ sinr h. Let C0 be the concentration of B far from the dislocation. /ðr;hÞ

Show that C ðr; hÞ can be written as: C ðr; hÞ ¼ C0 e kT . What is the shape of the iso-concentration lines? Let us consider that B atoms lead to a positive size effect (dV [ 0). This is the case of carbon in interstitial position in steels. From the expression of C ðr; hÞ, show in which region (i.e. domain of h) there is enrichment. 6. Let us estimate the “size” of this cloud by taking it equal to the distance r0 for which concentration C is significantly larger than that far from the line defect (C0 ). Let us take C ¼ 10C0 . Calculate r0 in nm for carbon in iron with l = 200 GPa, b = 0.2 nm. We will take T = 1000 K, h ¼ p=2 and dV ¼ 0; 9b3 . 7. The expression found for C ðr; hÞ diverges when K sin h\0 and r ! 0. This does not make physical sense. Imagine how the actual interaction energy / should vary in proportion as the B impurities migrate and segregate along the dislocation line D. Explain why the concentration in this region is greatly overestimated when using the previous classical expression for /ðr; hÞ.

Solution to the problem 1. We consider the following situation:

lb The elastic interaction energy is written / ¼ K sinr h with K ffi 3p dV . The sign of / is that of dV sin h. For B atoms to segregate along the dislocation line, / must be negative. /\0 for dV \0 when sin h [ 0 (h 2 ½0; p , compression region) and /\0 for dV [ 0 when sin h\0 (h 2 ½0; p , expansion region). Substitutional B atoms will segregate in the compression region (+ region) when dV \0 (small substitutional B atoms as compared to solvent A atoms) and in the expansion region (region), when dV [ 0 (big atoms). For interstitial atoms most often dV [ 0 (e.g. carbon in steels) so that segregation takes place in the expansion region (/0 \ 0, diagram below).

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2. We have /0 ¼ K sinr h with /0 [ 0. Equipotential lines are such that /0 ¼ cste. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Let us write r ¼ x 2 þ y 2 and y ¼ r sin h. Substituting these expressions in that of /0 leads  to the equation below that shows that þ /0 ðr; hÞ are circles whose K centre is in x ¼ 0; y ¼ 2/

0

jK j : and whose radius is r ¼ 2/ 0

    K 2 K 2 x2 þ y  ¼ 2/0 2/0

  K In the same way, equipotential lines /0 ðr; hÞ are circles centred in 0;  2/ 0

jK j with the same radius r ¼ 2/ (see diagram below for dV [ 0 and thus 0 jK j ¼ þ K ). As a result, the larger /0 , the smaller the circle radius. All these properties remain valid whatever Z . The equipotentials in 3 dimensions are therefore cylinders that are all tangent to the dislocation line. When dV \0, K \0 so that /\0 above the dislocation line (y [ 0) and / [ 0 below. This is the reverse situation compared to the diagram below.

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Diffusion, Segregation and Solid-State Phase Transformations

! B ~ 3. We know that ~ v ¼ MB r U with MB ¼ D kT (Darken’s equation). Writing J B ¼ ! BC r /. The flux lines are therefore the same as those of C~ v leads to ~ J B ¼  DkT force that are perpendicular to the equipotentials (analogy with the electric field lines that are perpendicular to voltage equipotentials). We can anticipate that flux lines are circles.

4. We shall calculate the equation of flux lines as defined by the gradient of the interaction energy written in cylindrical coordinates:  K  !  r 2 sin h : r/ ¼ K r 2 cos h r;h ! ! As the flux  lines  are perpendicular to r /, thus d ðOM Þ ^ r / ¼ 0 with dr . This leads to: d ðOM Þ ¼ rdh r;h rdh sin h þ dr cos h ¼ 0 It is straightforward to see that r ¼ A cos h is solution with A a positive constant. Since r 2 ¼ x 2 þ y 2 and r ¼ x= cos h, we have r 2 ¼ Ax and we end up with x 2 þ y 2 ¼ Ax. This last equation is that of a circle of radius A=2 and centre ðA=2; 0Þ:    2 A 2 A x þ y2 ¼ 2 2

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131

The transport flux lines are therefore circles of radius A=2 whose centre is in x ¼ A=2 (see previous diagram). 5. The sum of both the diffusion and transport fluxes ~ J B is written: DB C ! ! ~ r/ J B ¼ DB r C  kT The polar components of this flux are therefore given by: Jr ¼ DB

@C DB C K þ sin h @r kT r 2

and Jh ¼ DB

1 @C DB C K  cos h r @h kT r 2

At equilibrium, the flux is zero, as a result Jh and Jr are both zero. Let us write u ¼ sin h=r. The two latter equations lead to the single following equation: dC CK þ ¼0 du kT Ku

Integrating the latter equation gives: C ¼ Be kT with B a constant. When r ! 1, u ! 0 and the concentration tends toward C ¼ C0 , then B ¼ C0 . Moreover, it is easy to see from the definition of the elastic energy that Ku ¼ /. Consequently /

C ¼ C0 e kT and we recognize the Boltzman distribution, as expected. We see that / ¼ cst implies C ¼ cst. The iso-concentrations are therefore of the same form as the equipotentials. The dislocation is enriched if C [ C0 and thus if /\0. B atoms that lead to a positive size effect (dV [ 0) will segregate in regions where u ¼ sin h=r\0, that is below the dislocation line. This is the case of most of interstitial elements (B,C,N). 6. For C ¼ 10C0 we find r0  1 nm. 7. In proportion as the dislocation line enriches in B atoms with time, the increase in B concentration entails a partial relaxation of the deformation field, which in turn will modify the interaction energy /. In contrast to the simple equation of /0 given previously, the real energy gain associated with the segregation of one B atom should be lower when the dislocation line is already enriched. Thus, in order not to overestimate the concentration of impurities in Cottrell atmospheres, the expression of / should be modified so that it decreases with the concentration of B atoms along the line defect.

Diffusion, Segregation and Solid-State Phase Transformations

132

2.2.12

Diffusion in Semiconductors: The Generation of an Internal Nernst Field

1. Doping of silicon by diffusion 1.1. Silicon (valence 4) is doped with arsenic (valence 5) in order to obtain a N-type doping. Arsenic is an electron donor. At the operating temperature of microelectronic devices (300 K) dopants are ionized. Each arsenic atom releases an electron: As → As+ + e−. In this experiment, the concentration (C ) of dopants is uniform along x so that the concentration of electrons per unit volume (n) is also uniform. Under the action of an external ~ these electrons acquire an average velocity that is expressed electric field E ~ with le the electric mobility of electrons in the as follows: ~ v ¼ le E semiconductor. Find the expression of the diffusivity of electrons De as a function of le . The measured resistivity of a silicon rod is q = 0.005 Ωm. Using the expression for the charge flux Je , calculate the concentration of electrons (n) given that le = 0.13 m2 V−1 s−1. All dopants are assumed to be ionized and therefore activated, i.e. they generate conduction electrons. Calculate the atomic fraction of N-dopants (As). Silicon has a diamond structure with 8 atoms per unit cell. The crystal parameter is a = 0.543 nm. 1.2. We now wish to make a diode that is nothing else than a PN junction (see diagram below). Boron, of valence 3, is an electron acceptor that will be used to get a P-doped region through the generation of positive charged holes. Once ionized, each boron atom gives a positively charged hole h+: B → B− + h+. All boron atoms will be assumed to be ionised in the following. The concentration of holes h+ is then equal to that of boron atoms. In order to get a P doped region where holes are the majority charge carriers in the previous N doped sample, the sample surface is exposed to a boron-enriched atmosphere at a given temperature T. Boron atoms thus diffuse from the surface to the bulk. The continuous supply of boron-enriched gas, at constant partial pressure, maintains constant the surface concentration of boron (C ð0; t Þ ¼ Cs ,8t). The partial pressure is such that Cs is higher than the arsenic concentration of the already N-doped sample so that holes are majority carriers (compared to electrons) near the surface (P zone). The heat treatment is stopped before equilibrium is reached so that the boron concentration decreases with x. We assume that the diffusion coefficient D of boron in silicon does not depend on x. By analogy with cementation (carbon diffusion from the surface, problem 2.2.4), give without any calculation the expression of  pffiffiffiffiffiffi C ðx; t Þ as a function of the “error” function erf ðu Þ with u ¼ x= 2 Dt . x!1

We will take C ðx; 0Þ ¼ 0,8x [ 0 and C ðx; t Þ ! 0. Represent schematically C ðx; t Þ, for increasing times t [ 0.

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1.3. Diffusion of boron is performed at 1150 °C. Calculate the diffusion coefficient of boron (D0 = 2.64 cm2/s and Q = 3.6 eV/atom). 1.4. It is reminded that the hole concentration (p) is equal to the boron concentration C because dopants are all ionised. We want that at a given depth x0 , p becomes equal to the concentration of electrons (n). We thus realize a PN junction at this depth. For x\x0 , this is a P-region where holes are majority charge carriers. Conversely, for x [ x0 , electrons are majority carriers, and we get a N region. Make a diagram representing the concentration profiles of boron C ðx Þ ¼ pðx Þ for t [ 0 (p is the hole concentration) and that of arsenic C ¼ n. The surface concentration of boron being Cs = 5.1024 atoms/m3, calculate the time required (in hours) for the PN junction to be localized at depth x0 = 2.7 µm. We give erf ð2:2Þ  0:998. 2. Implantation doping and Nernst field in silicon In this problem, independent of the previous one, the N doping of a silicon sample is performed by ion implantation of arsenic.15 This results in a dopant concentration profile C ðx Þ showing a maximum localised at a distance x ¼ d from the surface. The coordinate axis Ox is perpendicular to the surface. C ðx Þ has the 15

The implantation (atomic jet of the dopants towards the surface of the substrate) gives rise to the creation of “irradiation” defects in the crystal (Si self-interstitials, vacancies), and leads to a bell-shaped profile of the dopants, which are for a large part in interstitial position in the lattice after implantation. A thermal activation treatment of the dopants at a sufficient temperature (e.g. “flash” treatment at 1000 °C) is then necessary after implantation in order to replace the dopants in substitutional position on the silicon sites. At the operating temperature of semiconductors (300 K), we consider that all dopants are ionized. The dopants are electron donors if they are N dopants, like arsenic of valence 5, or electron acceptors (creation of holes h+) if they are P dopants like boron of valence 3 (silicon is of valence 4).

134

Diffusion, Segregation and Solid-State Phase Transformations

shape of a Gaussian that is truncated on the left side of the maximum. Let D be the diffusion coefficient of arsenic and De that of electrons. 2.1. The concentration gradients of dopants when they are all ionized give rise to concentration gradients of electrons @n=@x along x. These space charges will give rise to an electric field E ðx Þ called the Nernst field that also depends on x. The ionized dopants (As+) will be subjected to this field E. Give the expression of the flux of dopants J along x as a function of D, C ðx Þ, T and E. 2.2. Let us now calculate the expression of E ðx Þ. In the same way as before, write the electron flux (Je ) as a function of the electron concentration n. Writing that, in the absence of an external electric field, Je is necessarily zero, deduce the expression of the internal field E ðx Þ as a function of @n=@x. 2.3. It is assumed that all dopants are ionized. Replacing E ðx Þ by its expression in the dopant flux equation J , show that we find a Fick equation with an effective diffusion coefficient D ¼ hD with D the diffusion coefficient of dopants. Give the value of h. What is your conclusion on the influence of the internal Nernst field E on diffusion? 2.4. The arsenic concentration profile (shown by atom probe tomography) reveals, as expected, a dome shape after implantation (figure below). The arsenic concentration is 1019 at/cm3 near the surface (Cs ) and that of the maximum (CM ) is 1021 at/cm3. The latter peak is observed at a depth xM ¼ 10 nm. Calculate the atomic fraction at the surface in ppm and that in xM in %. How does E ðx Þ vary with x? Sketch E ðx Þ and C ðx Þ below. Can the Nernst field become zero? What is its influence on the diffusion of arsenic ions As+?

Atom probe tomography image and arsenic concentration profile after implantation (S. Duguay, GPM).

2.5. Calculate an order of magnitude of the average Nernst field E (in V/μm) in the region comprised between x ¼ 0 and xM , and at the service

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135

temperature of the device (300 K). C ðx Þ will be assumed to vary linearly with x in the considered region. Do you think that the field E is important? Deduce the corresponding potential difference dV . 2.6. A large part of implanted dopants is not activated after implantation because they are not in the substitutional lattice sites. In other words, they do not provide electrons because they do not link with neighbour silicon atoms via the expected covalent bounds. An activation treatment of dopants at 1000 K is therefore required for these dopants to be replaced in lattice sites. This treatment also decreases the concentration of point defects induced by implantation. Explain why the diffusion is faster than expected during the early stages of heat treatment. Why is it interesting to use very rapid laser-assisted flash treatments? Solution to the problem 1. Doping of silicon by diffusion 1.1. The average velocity of electrons under the action of an external electric ~ reads ~ ~ with le the electric mobility of electrons in the field E v ¼ le E ~ with F ~ ¼ eE. ~ Thus semiconductor. By definition, we have ~ v ¼ Me F le ¼ eMe . e is the absolute value of the charge of the electron ( [ 0) and De (Darken’s equation). The diffusion coefficient of the electrons is Me ¼ kT thus given by: l kT De ¼ e : e The electric charge flux Je is written: ~ ~ v ¼ nele E: J e ¼ ne~ The diffusive term proportional to @n=@x is zero because the electron ~ leads to the expression of concentration n is constant. Writing ~ J e ¼ rE electrical conductivity r ¼ nele. As the resistivity is given by q ¼ 1=r, we get: 1 n¼ ¼ 961  1021 m3 ; qele with ρ = 0.005 Ωm and le = 0.13 m2 V−1s−1. All dopants, whose concentration is C , are ionized and thus activated, therefore n ¼ C . Because silicon has 8 atoms per cell, the atomic volume is given by Vat ¼ a 3 =8  20 angstroms3 with a = 0.543 nm (lattice parameter). The concentration per unit of volume writes C ¼ VXat . As dopants are all ionised, thus X ¼ nVat  0.19 ppm (1.9 × 10–7). 1.2. Let us use the solution of the second Fick law for the diffusion of impurities from the sample surface (course reminder and problem 2.2.4). We then have: C ðx; t Þ ¼ Cs  ðCs  C0 Þerf ðu Þ  pffiffiffiffiffiffi with u ¼ x= 2 Dt , t [ 0.

136

Diffusion, Segregation and Solid-State Phase Transformations For x ¼ 0 (u ¼ 0), erf ðu Þ ¼ 0 and as expected we get C ð0; t Þ ¼ Cs . The bulk concentration far from the surface remains equal to the initial conx!1 centration C0 that is zero here: C ðx; t Þ ! C0 ¼ 0. The depth profile thus obeys the following equation (see figure below): C ¼ Cs ð1  erf ðu ÞÞ

1.3. D0 = 2.64 cm2/s and Q = 3.6 eV/atom. The diffusivity at 1150 °C is thus: Q

D ¼ D0 e kT  4:82  1017 m2 =s 1.4. For x\x0 , the zone is P-doped and holes h+ are majority charge carriers. Conversely, for x [ x0 , electrons are majority carriers and the region is N-doped. The concentration profiles related to the PN junction can be represented schematically as follows:

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137

The PN junction is located at x ¼ x0 where the hole concentration (i.e. the boron concentration C ) is equal to the initial electron concentration (n): C ¼ n ¼ Cs ð1  erf ðu ÞÞ We can deduce that erf ðu Þ ¼ 1  n=Cs . For Cs = 5.1024 atoms/m3, this  pffiffiffiffiffiffi gives u = 2.2 knowing that erf ð2:2Þ  0:998. By definition, u ¼ x= 2 Dt . We can then calculate the time required for the PN junction to be localized at the depth x0 = 2.7 µm and we find t = 2,17 h. 2. Implantation doping and Nernst field in silicon 2.1. The flux of ionised dopants (As+) J is written: J ¼ D

@C DC þ eE @x kT

2.2. On the other hand, in the absence of an external electric field, the flux of electrons (Je ) is zero, so that: Je ¼ De

@n De  neE ¼ 0 @x kT

We can therefore derive the expression of the Nernst electric field: E¼

kT @n : ne @x

2.3. All dopants are ionized, thus C ¼ n. Injecting the previous expression of E into the flux equation related to As+ leads to: J ¼ 2D

@C @x

Writing that D  ¼ hD leads to h ¼ 2. The internal Nernst electric field doubles the diffusion coefficient. 2.4. As Cs = 1019 at/cm3 and CM = 1021 at/cm3 in x ¼ xM = 10 nm, writing that atomic fractions are Xs ¼ CVat (Vat = 20 angstroms3) leads to Xs = 200 ppm and XM = 2.104 ppm or 2 at.%. @C ~ Because C ¼ n, then E ¼  kT ne @x . This shows that electric field E opposes to concentration gradient. As a consequence, the force exerted on As+ ions ~ ¼ þ eE ~ also opposes to concentration gradient in the same way as F Fick’s term. The electric field thus reinforces diffusion as shown previously ~ is negative for x\xM , zero in ðD ¼ 2DÞ. As shown in the figure below, E

Diffusion, Segregation and Solid-State Phase Transformations

138

x ¼ xM and positive in x [ xM . The internal field enhances the diffusion of arsenic and the spreading of the concentration profile. The influence of E on the concentration profile and the electric field profile E(x) can be sketched as follows:

@C 2.5. Let us calculate the electric field E ¼  kT Ce @x . The linear approximation of CM Cs the concentration profile in the first region (x\xM ) gives @C @x ¼ xM



CM xM   C2M . 6

1035 m−4. The average concentration in this region is close to

C We thus obtain a mean electric field equal to |E|  5 × 10 V/m = 5 V/µm and then dV ¼ E=xM  0.05 V for xM = 10 nm. The Nernst field is not negligible. 2.6. Diffusion is enhanced during the early stages of heat treatment at 1000 K because of the high amount of point defects (vacancies, self-interstitials)

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139

induced by implantation. These defects are known to enhance diffusion of dopants. The activation treatment at 1000 K will decrease the concentration of these point defects. The diffusion coefficient of the dopants will therefore decrease with time down to its expected value. A flash treatment, which by definition takes place over a short period, will limit transient enhanced diffusion.

2.2.13

Driven Diffusion in a Force Field: Oxidation of Nickel

The oxidation of nickel leads to the formation of an oxide Ni1-δO that grows at the sample surface. It is non-stoichiometric and is described as a crystal composed of two embedded simple cubic sublattices related to nickel and oxygen, respectively. A fraction d of the nickel sites are occupied by structural vacancies that accommodate the deviation from ideal stoichiometry (atomic fraction of nickel lower than 50at.%). The thermal vacancies are neglected. Oxidation is assumed to be controlled by the self-diffusion of nickel on its sublattice in the form of Ni2+ cations that migrate from the internal Ni/NiO interface to the surface (figure below). Nickel diffuses via a vacancy mechanism leading to the diffusion of vacancies v 2 in the opposite direction. Oxidation of nickel at the surface requires electrons to move towards the surface. We have thus the diffusion of charged holes h þ from the surface towards the Ni/NiO interface in the same direction as v 2 (diagram below). In the present problem, oxidation will be studied by considering the diffusion of vacancies v 2 and holes h+. This is equivalent to considering the diffusion of nickel and electrons.

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Let z be the distance from the Ni/NiO interface that we take as origin (z ¼ 0). The high mobility of holes (or electrons) compared to that of the vacancies creates a ~ that affects the space charge. This results in an internal Nernst electric field E þ 2 migration of holes h and that of doubly charged vacancies v . In the first part of the problem, we consider oxidation in a general way. A metal M giving rise to the formation of an oxide (MO) is considered. The valence a of metal M will not be given explicitly in order to preserve the generality of the established expressions. In the second part, theory will be applied to the oxidation of nickel. 1. Modelling diffusion and the Nernst field 1.1. Let a be the valence of the metal (M) and e the absolute value of the electron charge. Give the expression of the charge of vacancies qv and holes qh as a function of e and a. Give the expression of the concentration of holes Ch as a function of the concentration of vacancies Cv and a. Give the expression of the atomic fraction of vacancies Xv in the oxide as a function of d. Note that the vacancy concentration at the MO/M interface, close to the equilibrium fraction in the metal, is lower than that at the surface. The concentration of vacancies therefore increases with z, and leads to their Fickian diffusion, from the surface towards the MO/M interface (previous figure) 1.2. Write the 1D expressions of fluxes ~ J v and ~ J h in the presence of a Nernst ~ Let Dv and Dh be the diffusion coefficients of vacancies and electric field E. holes, respectively. Considering metal valence a, write the required relaJ h so as to ensure that there is no net electric tionship between ~ J v and ~ ~ Show that it is proportional to current. Deduce the expression for E. ! Dh  Dv and to the concentration gradient of vacancies r Cv . 1.3. Holes are much more mobile than vacancies. Give the approximate ~ is proportional to 1=Cv . ~ when Dh [[ Dv . Show that E expression of E 1.4. Give the sign of E ðz Þ. What is its influence on the charged species h þ and v a ? Represent a diagram showing the fluxes of vacancies and holes in the oxide, and below this diagram, in correspondence, sketch E ðz Þ and Cv ðz Þ. 1.5. Deduce the expression of Jv as a function of a, Dv and rCv when Dh [[ Dv . Show that the flux has the form of the first Fick equation with an apparent diffusion coefficient Dv . Give the expression of Dv as a function of Dv . 1.6. Let us solve the second Fick equation in steady state conditions. Show that the vacancy concentration Cv ðz Þ grows linearly with distance z. Give the integration constants as a function of the vacancy concentrations at the surface (Cvs ) and at the M/MO interface (Cvi ). How does E evolve with Cv and with z? Give the expression of E as a function of the atomic fractions of vacancies Xvs and Xvi at the surface and at the metal-oxide interface respectively.

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2. Oxidation of nickel 2.1. Let us now apply the previous model to the oxidation of nickel at 1000 °C in steady state condition. The NiO oxide has a cubic structure (a = 0.418 nm). The lattice contains 8 atoms (4Ni, 4O). The concentration of vacancies at the oxide surface in equilibrium with the imposed partial pressure of oxygen (PO2 ) and that at the NiO/Ni interface have been measured. The atomic fraction of vacancies in the nickel sub-lattice measured near the surface is ds = 10–4. That at the Ni/NiO interface is di = 4.10–6, a value that is close to twice the equilibrium atomic fraction of vacancies in nickel. We will take a = 2 (Ni2+). Give the approximate expression of E ðz Þ when ds [[ di . Calculate the Nernst field (E in V/μm) in steady state in z ¼ 0 and in z0 = 20 µm. 2.2. Since electric field E is not constant, space charges develop. The related ~ ¼ q=e with e ¼ e0 er charge density q is given by the Poisson equation: divE the electric permittivity in the oxide. The relative permittivity of NiO is er = 9 whereas the permittivity in vacuum is e0 = 1/(36π109). Demonstrate that, in steady state condition, q is proportional to 1=Xv2 . What is its sign? 2.3. We define the concentration of space charges per unit volume Cq by the expression jqj ¼ Cq e. Calculate its expression as a function of d. Compare Cq at the surface and at the Ni/NiO interface. Calculate both related space charge fractions Xq (per atomic site) and compare them to the vacancy fractions Xv at the surface and at the Ni/NiO interface. Is the neutrality respected? 2.4. We are now interested in the kinetics of nickel oxidation. It is assumed that the Ni/NiO interface reaction rate is very fast compared to bulk diffusion of nickel. Diffusion can be shown to control the growth kinetics except in the early stages where the oxide is very thin (problem 3.2.6). Show that the oxidized thickness z0 increases with time according to the following law: z02 ¼ ct. Give the expression of c as a function of Dv and Xvs when ds [[ di . Show that c is proportional to the diffusion coefficient DNi of Ni in NiO. 2.5. Calculate the thickness of the oxide (in microns) after 8 h given that DNi = 2.8 10–14 cm2/s at 1000 °C.

Solution to the problem 1. Modelling of diffusion and of the Nernst field 1.1. It is immediate to see that qv ¼ ae with a the valence of the metal and e the absolute value of the electron charge. The charge of holes is qh ¼ þe. The ionisation process of vacancies obeys the following surface reaction: v ! v a þ ah þ . As a result, the concentration of holes is Ch ¼ aCv . We will assume that this equility is true whatever z.

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The Ni1-δO oxide is composed of two sublattices. Vacancies are located only in the metal sublattice. Their atomic fraction reduced to this sublattice is d. Because it is assumed that there are no vacancies in the other oxygen sublattice (thermal vacancies are neglected), the overall vacancy fraction as calculated over both sublattices is half of d: Xv ¼ d=2. J h in the direction perpendicular to the oxide surface gives: 1.2. Writing ~ J v and ~ Jv ¼ Dv

@Cv Dv þ Cv qv E @z kT

Jh ¼ Dh

@Ch Dh þ Ch qh E @z kT

and

As there is no electric current, the sum of the fluxes of charge is zero and necessarily Jh ¼ aJv. On the other hand, qh ¼ þe and qv ¼ ae. Using the relation Ch ¼ aCv in both flux expressions, gives the Nernst electric field ~ E: ~ ¼ Dh  Dv kT ! r Cv : E Dh þ aDv eCv As expected, the electric field is zero if the mobility of the holes is equal to that of the vacancies. 1.3. For Dh [[ Dv , we have the approximate expression of the field: ~ ’ kT ! r Cv : E eCv The norm of the internal electric field decreases as the concentration of vacancies increases (when the vacancy concentration gradient varies little). ! 1.4. The sign of E ðz Þ is therefore that of r Cv , the latter being positive because the concentration of vacancies increases with distance z from the M/MO interface to the surface. As E ðz Þ [ 0 in the considered region, the field accelerates the diffusion of vacancies from the surface to the MO/M ~ opposes to the movement interface (F ¼ qv E\0 because qv ¼ ae) but E of holes (F ¼ þeE [ 0, see diagram below). Electric field E decreases with z because the concentration of vacancies increases with distance z. The concentration gradient of vacancies varies little with z and is even constant in the steady state (Jv ¼ cste). It is therefore the term 1=Cv that controls the variation of the field E.

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kT @Cv 1.5. Replacing the expression of the field E ’ eC into the equation related v @z to the vacancy flux gives:

Jv ¼ Dv

@Cv @z

with Dv ¼ Dv ð1 þ aÞ. The Nernst electric field therefore increases the diffusion rate of vacancies. The increase is proportional to the metal valence. When the valence is α = 1, we have a factor 2 whereas we get a factor 3 for a ¼ 2 (as for nickel in the following).

144

Diffusion, Segregation and Solid-State Phase Transformations @Jv v ~ 1.6. In the steady state @C @t ¼ divJ v ¼ 0. Thus, @z ¼ 0 and consequently v Dv @C @z ¼ cst. The vacancy concentration Cv ðz Þ is thus:

Cv ðz Þ ¼ kz þ k0 : We shall determine both λ and k0 integration constants. As the vacancy concentration in z ¼ 0 is equal to that at the M/MO interface Cvi , thus k0 ¼ Cvi . On the other hand, as the surface concentration writes Cv ðz0 Þ ¼ Cvs ¼ Cvi þ kz0 , therefore k ¼ Cv ðz Þ ¼ Cvi þ

Cvs Cvi z0 .

This leads to:

Cvs  Cvi z z0 C s C i

The concentration gradient is therefore rCv ¼ v z0 v . Replacing this equation in the approximate expression of the field leads to: E’

kT Cvs  Cvi 1 e Cv z0

This latter equation indicates that in steady state conditions, the field is  1 proportional to 1=Cv ¼ Cvi þ kz . E ðz Þ has a hyperbolic dependence on z (as in previous figure). As Cv ¼ Xv =Vat with Vat the atomic volume, E ðz Þ is written: E’

kT Xvs  Xvi  e Xvi z0 þ Xvs  Xvi z

2. Oxidation of nickel 2.1. Let us express the latter equation as a function of δ: E’

kT ds  di e di z0 þ ðds  di Þz

As ds = 10–4 at the surface and di = 4.10–6 at the Ni/NiO interface thus ds [[ di . We then obtain the approximated expression: E’

kT ds : e di z0 þ ds z

kT ds In z ¼ 0 we have E ’ ez . The field for z0 = 20 µm and T = 1273 K is 0 di

kT close to 0.137 V/µm. In z ¼ z0 we find E ’ ez = 5.5.10–3 V/µm. 0

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~ ¼ q=e in 1D gives q ¼ e dE . The 2.2. Writing the Poisson equation divE dz expression of E is: E’

kT dXv : eXv dz

In steady state conditions, we get:   kT 1 dXv 2 q ¼ e e Xv2 dz ds di v Thus, we have q\0. Note that dX dz ¼ k ¼ 2z0 and therefore q ¼ 0 only if ds ¼ di . 2.3. We have q ¼ Cq e. Thus, the charge concentration is written:   kT 1 dXv 2 Cq ¼ e 2 2 e Xv dz ds di v Let us remind that XV ¼ d=2 and dX dz ¼ 2z0 . At the interface (z ¼ 0), 19 3 d ¼ di , therefore Cq  8.5.10 charges/m whereas in z ¼ z0 , d ¼ ds , then the average surface concentration is Cq  3.4.1018 charges/m3. The concentration of charges at the surface is low compared to that at the Ni/NiO interface. The electrons in excess at the Ni/NiO interface will migrate towards the surface and “ionize” the vacancies. The holes, conversely, migrate from the surface to the Ni/NiO interface. The atomic fraction of electric charges is given by Xq ¼ Cq Vat with Vat ¼ a 3 =8. Thus, Xq  7.8.10–10 in z ¼ 0 (Ni/NiO interface). This is very low compared to the vacancy atomic fraction Xv = 2.10–6. The neutrality is almost respected. We have Xq  3.1.10–11 in z ¼ z0 whereas the vacancy fraction at the interface is 0.5.10–4. This guarantees a much better neutrality. Note that Cq tends toward zero when dXv =dz ! 0, that is for large oxyde thickness. 2.4. The increase in volume of the material consecutive to oxidation can be easily expressed as a function of the flux of vacancies:

dV ¼ jJv jSdt2Vat : S is the surface area of the sample. Note that each vacancy gives rise to the diffusion of one Ni atom of atomic volume Vat . As each Ni atom reacts with one oxygen atom (Ni þ O ! NiO), the volume increases by 2Vat. Writing Cs that dV ¼ Sdz0 and Jv ¼ Dv z0v gives: z0 dz0 ¼ 2Dv Xvs dt

146

Diffusion, Segregation and Solid-State Phase Transformations Because z0 ðt ¼ 0Þ ¼ 0, integrating the latter equation leads to: z02 ¼ ct with c ¼ 4Dv Xvs ð1 þ aÞ. As Ni only diffuses in the Ni sublattice of the oxide, in which the atomic fraction is d ¼ 2Xv , thus, DNi ¼ 2Dv Xv , and pffiffiffiffiffiffiffiffiffiffi therefore c ¼ 2DNi ð1 þ aÞ ¼ 6DNi (a ¼ 2). The thickness z0 grows as DNi t (parabolic growth), a value that is proportional to the mean square diffusion distance, as expected for bulk diffusion. The growth rate dz0 =dt pffiffiffiffiffiffiffiffiffiffiffiffiffiffi increases with the valence α as ð1 þ aÞ. 2.5. We find z0 = 0.7 µm.

Chapter 3 Kinetics of Formation of a New Phase 3.1 3.1.1

Course Reminder Continuous and Discontinuous Precipitation

Let us consider a binary alloy AB supersaturated in B atoms, subjected to an isothermal treatment at a temperature T0 leading to the precipitation of a new phase b within the parent phase a (phase diagram in figure 3.1) [25–27]. The precipitation of a coherent b phase (i.e. with the same structure as the parent phase α) can be continuous or discontinuous depending on the difference δa between the lattice parameters (a) of both a and b phases. The continuous precipitation process is the most frequent. It is in particular observed when β has a parameter close to α (small relative misfit δa/a). The supersaturation in B atoms in matrix α decreases continuously with time in proportion as β precipitation develops. This is what is observed in CuCo, FeCu, NiAl alloys and in many other systems, in particular light aluminium alloys for aeronautics (figures 3.2 and 3.3). When the misfit between lattice parameters of both coherent phases is large, the new b phase may prefer to nucleate to grain boundaries (GBs) because this minimises the overall interface energy. The growth of the new β phase then proceeds via a discontinuous precipitation mechanism. This is also called cellular precipitation because of the observed cellular microstructure. In the course of kinetics, the growth of b induces the migration of GBs (figure 3.4). There is a discontinuity of compositions at the mobile a=a0 interface. a0 designates the supersaturated parent phase in front of GB whose composition is that of the alloy (X0 ). a is the transformed parent phase, with composition Xa , in equilibrium with b precipitates (Xb ). This phase grows behind the transformation front at the expense of a0 whose composition is not yet affected and remains that of the alloy (X0 ). The concentration profile is therefore discontinuous across the moving a=a0 . GBs sweep the supersaturated a0 phase and collect the B atoms required for the growth of the b phase behind the front. Solute atoms then diffuse along moving GB towards

DOI: 10.1051/978-2-7598-2743-5.c003 © Science Press, EDP Sciences, 2022

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FIG. 3.1 – Schematic equilibrium phase diagram showing the presence of a two-phase domain in which one phase b is in equilibrium with the parent solid solution a. Xa and Xb are the equilibrium atomic fractions of B atoms in a phases and b phases at T ¼ T0 , respectively, and X0 is the nominal fraction of B atoms.

FIG. 3.2 – Atom probe tomography analysis (28 × 28 × 90 nm3) showing the precipitation of small MgZn2 precipitates (grey envelopes, 5–25 nm in size) in an AlZnMgCu alloy subjected to two successive ageing treatments at 120 °C and 150 °C [28]. Dots represent solute atoms (Zn, Mg, Cu). The image shows the iso-concentration surfaces at 10% zinc (W. Lefebvre, GPM).

growing precipitate lamellae. This mode of transformation is more often observed at low temperatures for which diffusion along GBs prevails over that in bulk. This discontinuous precipitation mechanism is observed, in particular, in CuAg alloys for which the misfit da=a between the almost pure copper parent phase and the silver precipitates reaches 12% (figure 3.5).

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149

FIG. 3.3 – High resolution transmission electron microscopy image (ARM 200F operated at 200 kV in STEM-HAADF mode along direction) showing the structure of a MgZn2 precipitate in a 7000 series aluminium-based alloy (W. Lefebvre, GPM).

FIG. 3.4 – Discontinuous transformation process. (a) b precipitates nucleate at the GB

(t ¼ t1 ). (b) The subsequent growth of β lamellae at t2 [ t1 gives rise to migration of the GB. The equilibrium α phase develops at the expense of the supersaturated a0 parent phase (X0 > Xα). B atoms in excess in a0 migrate towards β along GBs and lead to the growth of β.

Lamellae of b phase observed during discontinuous precipitation are spaced quasi-periodically along GBs with a minimum period km driven by thermodynamics and that depends on temperature T . For k ¼ km , the gain of volume free enthalpy associated with β precipitation is just compensated by the interface energy α/β, the energy balance is then zero. For low values of dT , a simple calculation leads to: km ¼

K dT

with K ¼ 2rVDHm TS . TS is the solvus temperature of the a phase above which the alloy is single-phase, dT ¼ T  TS is sometimes called the supercooling (by abuse of language), r is the interfacial energy, Vm the molar volume, and DH ([ 0) the enthalpy gain due to the transformation. The microstructure is refined by lowering the temperature (km decreases when the supercooling dT increases).

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Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.5 – Electron microscopy image showing discontinuous precipitation of silver lamellae (in light contrast) in a Cu5at.%. Ag [29] alloy – in light the Ag precipitates growing on one side of the GB (X. Sauvage, M. Bonvalet GPM). Similar equations are obtained in eutectoid reactions (problem 3.2.1). However, the mechanism of eutectoid transformations (one solid c gives two solids a and b) is different from that of discontinuous transformation. First, because it involves three different phases with a singularity in the phase diagram, the eutectoid point located at TE (eutectoid plateau). Moreover, in the eutectoid transformation, the lamellae of the new phase b nucleate at grain boundaries (GB) as in the discontinuous transformation, but GBs do not migrate. This is the well-known example of the pearlitic transformation in steels where the FCC c phase (austenite) is transformed into ferrite a (BCC) that is depleted in carbon. The latter, by rejecting carbon, causes the precipitation of cementite Fe3C that is a phase rich in carbon. The expression of the minimum spacing km between cementite lamellae is the same as before (see problem 3.2.1) by exchanging TS for TE . The microstructure is finer and the transformation more rapid as the supercooling “dT ” increases.

3.1.2

Instability and Metastability

Consider a supersaturated binary alloy and the coherent precipitation of a b phase with the same structure as the parent phase a. Compositional fluctuations of thermal origin always exist at non-zero temperature in materials. Let us examine how these fluctuations evolve with time. Considering an atomic fraction of B atoms (X) close to the nominal content of solutes B (X0 ), one can make a Taylor expansion of free enthalpy G ðX Þ:

2
dG

1 2d G
þ  G ðX Þ ¼ G ðX0 Þ þ DX
þ ðDX Þ dX X0 2 dX 2
X0

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151

Derivatives are taken in X0 and DX ¼ X  X0 . By averaging G(X) in space ðhG ðX ÞiÞ, the odd terms cancel out because there are “as many” positive and negative DX fluctuations of same amplitudes. Limiting the Taylor expansion to second order leads to: E 2

1D 2 d G
ðDX Þ hG ðX Þ  G ðX0 Þi ¼ 2 dX 2
X0

The average gain of free enthalpy hG ðX Þ  G ðX0 Þi should be negative for the composition fluctuations to develop. Writing that hG ðX Þ  G ðX0 Þi is close to kT , we end up with [30]: D E 2kT ðDX Þ2 ¼ d 2 G
:
2 dX

X0

FIG. 3.6 – Free enthalpy diagram G ðX Þ of a regular solid solution giving rise to phase separation. Both α and β phases have the same structure and their equilibrium compositions are Xa and Xb as given by the common tangent. The free enthalpy gain is negative (dG\0) between both inflexion points. This is favourable to the amplification of small fluctuations of concentrations. This is the so-called spinodal regime. A zoom of a part of G ðX Þ close to the solubility limit Xa is shown (low supersaturation). On both sides of inflexion points, the curvature of G(X) is positive and the system is outside the spinodal domain. The positive energy balance (dG [ 0) is unfavourable to the development of composition fluctuations. The free enthalpy of the system can however decrease by nucleation of the b phase. In classical nucleation theory, nuclei are assumed to have the equilibrium composition of the β phase Xb , in the early stages of precipitation (figure 3.9).

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Diffusion, Segregation and Solid-State Phase Transformations

D E The mean square amplitude of fluctuations ðDX Þ2 is necessarily positive. This is only achieved if d 2 G=dX 2 \0. The system is then in an unstable state (chapter 1). This is the spinodal regime. The free enthalpy G ðX Þ shows a negative curvature between both inflection points of G ðX Þ (figure 3.6). The spinodal domain is located inside the miscibility gap for which the supersaturation is high. The apparent diffusion coefficient, proportional to the second derivative of G ðX Þ gets negative (see chapter 2 on diffusion). The flux of solutes goes up the concentration gradient, this is the so-called up-hill diffusion. Small thermal fluctuations grow spontaneously, with no barrier to cross (figure 3.6). The a=b interfaces are diffuse and the concentration of solutes in the b phase increases gradually with time until reaching its equilibrium value. Various studies carried out by atom probe nanoanalysis have revealed the existence of the spinodal regime in many systems from these two criteria. The spinodal decomposition of highly supersaturated FeCr alloys is a typical example (figure 3.7).

FIG. 3.7 – 3D image (10*10*30 nm3) obtained by atom probe tomography showing the spinodal decomposition of a FeCr ferritic steel (BCC structure). The iso-concentration surfaces at 20at.% of chromium reveal the interfaces between both a (Fe-rich) and a0 (Cr-rich) interconnected phases [31] (courtesy F. Danoix, GPM).

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153

This instability regime only makes sense for iso-structural transformations for which only the chemical composition distinguishes the new phase from the parent phase and for which we have a unique G ðX Þ function. In some systems, such as FeCr, one observes a complex interconnected microstructure in which the b phase surrounds a but a also surrounds b (figure 3.7). Outside the spinodal domain, the concavity of G ðX Þ is positive (figure 3.6). The thermal fluctuations are then necessarily damped for two reasons: (i) the diffusion coefficient is positive (chapter 2), the solutes go down the concentration gradients according to the classical Fick law; (ii) the fluctuations result in a free enthalpy higher than that of the non-decomposed alloy (figure 3.6). In both regions located outside the spinodal lines of the phase diagram (i.e. outside the inflexion points), the enthalpy of the system can only decrease through nucleation of the b phase. For small supersaturations, the system is then in a metastable state and must

FIG. 3.8 – (a) Nucleation and growth of the b phase in a supersaturated phase (a). The atomic fractions of B atoms (X) at α/β interfaces are the equilibrium fractions (Xa and Xb ) of both a and b phases as given by the phase diagram. A local equilibrium occurs at the interface. (b) Spinodal decomposition of a highly supersaturated solid solution. The amplitude of the thermal concentration fluctuations initially present in the solid solution increases with time until equilibrium compositions (t ¼ t3 ) are reached.

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cross a barrier for the nucleation of the new phase b. There is then an incubation period before nuclei form. Nuclei are assumed to have the equilibrium composition of the b phase (Xb ) as soon as β nucleates (classical nucleation theory). One then speaks of the nucleation and growth regime (figure 3.8). The β phase first nucleates and then grows in proportion as the supersaturation in B atoms (X0  Xa ) decreases. In this precipitation regime, a=b interfaces are abrupt contrary to the spinodal decomposition regime (figure 3.8), for which interfaces are diffuse.

3.1.3

Homogeneous Nucleation

Various authors have contributed to the early developments of the so-called classical theory of nucleation. Let us mention the pioneers, Volmer [32] and Weber (1926) who studied the nucleation of water drops in supersaturated steam but also Becker and Döring (1935) [33], Turnbull (1948) [34] and Zeldovich [35]. John W. Cahn (1961) [36] eventually proposed a theory of nucleation called the non-classical nucleation theory in which α/β interfaces are diffuse and where nuclei may not have the equilibrium composition of the β phase. We shall now present the fundamental principles of the classical nucleation and growth theory. Let us consider a diluted but supersaturated binary alloy AB in which nucleate spherical precipitates of a new phase (b), whose structure is a priori different from that of the parent phase (a). The corresponding G ðX Þ diagram is given in figure 3.9. We will assume that the density of nuclei per unit volume is low so that we can (i) ignore the interactions between them, (ii) consider that the quantity of solutes B in nuclei is so low that the composition of the parent phase a remains unchanged and equal to the initial composition X0 . We can simply express the free enthalpy variation DG resulting from the formation of a spherical embryo of radius R as follows: 4 DG ¼  pR3 ðDgn  Dge Þ þ 4pR2 r 3 DG should be negative for the nucleation of the β phase to be energetically favourable. Dgn ð[ 0Þ is the driving force of nucleation (figure 3.9), to which the elastic energy of coherence Dge opposes. The latter is generally smaller (otherwise nucleation cannot occur). The elastic term Dge finds its origin in the elastic deformation of the lattice caused by the misfit (δa/a) that may occur between the lattice parameters of the a and b phases. This is notably the case for the coherent precipitation where a and b phases have the same structure. The second term, proportional to R2 , is related to interface energy and is positive. It then opposes to nucleation (as Dge ). r is the interfacial energy per unit area. This physical quantity is also often called interfacial tension (J/m2 are equivalent to N/m). The equilibrium concentration of embryos of size R at a temperature T is given by Boltzmann’s law: Cn ðRÞ ¼ Cn0 e

DG ðRÞ kT

:

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155

FIG. 3.9 – Schematic representation of G ðX Þ related to two phases a and b that have different structures. X0 is the nominal composition of the alloy, Xα and Xβ are the equilibrium atomic   fractions of B atoms in both phases (as given by the common tangent). GP  GQ =Vb ¼ Dgn is the driving force for nucleation of the β phase. Vb is the molar volume of b phase.

Cn0 is the concentration of nucleation sites. We will take it equal to the concentration of B atoms (X0 =vat with vat the atomic volume). This term is sometimes taken equal to the concentration of lattice sites (1=vat ) which is much larger. What is really important in the latter equation is however the exponential term. The previous equation of DGðRÞ expresses the competition occurring between the driving force term that is negative (in general Dgn [ Dge ) and proportional to the volume of nuclei (  R3 ) and the positive interface term proportional to the surface of the nuclei (  R2 ). This competition leads to a maximum of DGðRÞ. As shown in figure 3.10, the maximum of DG ðRÞ, denoted as DG  , occurs when the nuclei radius reaches a critical value (R ¼ R ). R* is the so-called critical radius of nucleation (figure 3.10). Nuclei of the new β phase have to cross the nucleation barrier DG  to enter into the growth regime. When R > R*, the free enthalpy decreases which means that growth of the embryo is energetically favourable. R* and the corresponding maximum DG  of DGðRÞ can be determined by writing that dG=dR ¼ 0 in R ¼ R , this gives: R ¼

2r Dgn  Dge

156

Diffusion, Segregation and Solid-State Phase Transformations 4 DG  ¼ prðR Þ2 3

The height of the nucleation barrier (DG  ) therefore decreases when the driving force Dgn increases (i.e. when the supersaturation increases). In the classical theory of nucleation, the composition of nuclei is assumed to be that of the equilibrium phase (Xb ) as given by the common tangent of G(X). As already discussed, there is a local equilibrium at the α/β interface, but this equilibrium is that of the phase diagram and does not account for capillarity. It is worth mentioning that another criterion, based on the maximization of the driving force, could either be used to determine the composition of nuclei. This is the rule of the parallel tangents to Gα(X) and Gβ(X) instead of the common tangent. When the curvature of Gb ðX Þ is important (figure 3.9), both common and parallel tangents are very close to each other. The composition of nuclei is then close to the equilibrium composition of the β phase. Otherwise, there is a small difference. The so-called Gibbs–Thomson effect (due to interface energy) modifies the local equilibrium at the interface and leads to a shift in phase compositions. This is examined later in the section dealing with coarsening. However, when the curvatures of Gb ðX Þ and Ga ðX Þ are close to each other, it can be shown that the influence of Gibbs–Thomson effect is of the order of the supersaturation and therefore remains small. Consequently, deviation from the common tangent will not be considered hereafter.

FIG. 3.10 – Free enthalpy ΔG(R) due to the formation of a spherical nuclei as function of its radius R. Embryos that have a radius greater than R* can grow. The balance is however favourable (ΔG(R) < 0) to the development of the β phase only when R [ 3R =2. The Zeldovich zone of width d is the region where DG ðRÞ is close to the top of the barrier DG  within kT .

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157

As shown in figure 3.10, the energy balance is really favourable to the development of β phase only if DG\0. This is realized when R [ 3R =2. However, thermal fluctuations allow subcritical embryos (R\R ) to cross the barrier DG  . In classical nucleation theory, it is assumed that nuclei can grow if R [ R . In this case, DG ðRÞ decreases each time an additional B atom is added to the nucleus. The stationary nucleation rate is expressed as the number of supercritical embryos per unit of time and volume and is given by the Boltzman distribution: In ¼ I0 e

QD kT

e

DG  kT

I0 ¼ AtCn0 with t the vibration frequency of atoms and Cn0 the concentration of nucleation sites. QD is the activation energy related to the diffusion of B atoms towards the nuclei. A is a constant proportional to the surface of the critical nuclei. The expression of this quantity is not really crucial and will not be explicited. The main factor that really drives the nucleation rate is the exponential term that depends on both the mobility of B atoms and the nucleation barrier DG  . In is stationary (constant during the nucleation process) if DG  remains constant. The previous equation of DG  shows that the driving force Dgn should then be constant. This condition is only fulfilled in the early stages of nucleation during which the supersaturation remains nearly constant (low volume fraction of precipitates). The expression of the nucleation rate reflects the competition between the nucleation barrier (DG  ðT Þ) and the mobility of B atoms (QD =kT ) which varies differently with temperature. At high temperatures, the low supersaturation in B atoms produces a lower density of precipitates (low nucleation rate). On the other hand, they grow quickly because the atomic mobility is high (figure 3.1). At low temperatures it is the opposite. The precipitates are numerous because the supersaturation is high and they are generally smaller. The equilibrium volume fraction of precipitates as given by the lever rule (chapter 1) is more important at low temperatures (lower solubility of B atoms in the parent α phase). However, diffusion is slower at low temperatures. The calculation of the nucleation rate In and of the critical radius R requires the determination of thethree terms  Dgn , Dge and r. Let us start with the driving force of nucleation Dgn ¼ GP  GQ =Vb . The latter is the gain of free enthalpy by unit of volume (Vb is the molar volume of phase b) consecutive to the formation of a nuclei that have the equilibrium composition of phase β. For a dilute regular solution and for nuclei that are almost pure in B atoms (Xb  1), we obtain the following simplified expression (see problem 3.1.3):   RT X0 Dgn ¼ ln Vb Xa Atomic fractions (X) refer to solutes B. X0 is the nominal fraction of solutes in the alloy and Xa that in the parent phase (α) at equilibrium. Dgn increases when X0 is increased, that is when the supersaturation ðX0  Xa ðT ÞÞ is larger. The latter decreases for increasing T (higher solubility Xa ðT Þ, figure 3.1). The nucleation rate and the density of nuclei therefore decreases at the highest temperatures, when

158

Diffusion, Segregation and Solid-State Phase Transformations

approaching the solvus temperature. R increases in proportion as we get closer to the solvus line in phase diagram (solubility limit) where the supersaturation becomes zero. The second term Dge is the coherency elastic energy (per unit of volume). As mentioned earlier, for coherent nuclei, this term results from the difference between the lattice parameters of the nucleus and of the surrounding α phase (aβ and aα, respectively). The lattice misfit leads to an elastic strain and consequently to a stress. According to Eshelby’s [37] theory, the related elastic energy density Dge is proportional to the square of the relative lattice misfit (δa/a) between nuclei and the surrounding a phase: Dge ¼ E

d2 1v

E is the Young modulus of rigidity and v the poisson coefficient (  0.3). According to Vegard’s law, the crystalline parameter is proportional to the atomic fraction, so we can write for coherent phases:  ab  aa  d¼2 Xb  Xa ab þ aa For a coherent nuclei, the interfacial tension r originates from the existence of numerous unfavourable AB bounds at the a=b interface, which give rise to an excess energy proportional to the order energy e. In this simplified approach, it can be shown that r is proportional to e and is written: ! 2 r ¼ e rX The interfacial energy generally varies from 10 to 200 mJ/m2 for coherent a=b interfaces for which there is continuity of the lattice across the interface (figure 3.11). r can reach 500 mJ/m2 for semi-coherent interfaces (figure 3.12). For these interfaces, the coherence stress is relaxed by dislocations (figure 3.13) spaced by a distance d ¼ b=d. b is the norm of the Burgers vector of the dislocations. b ¼ a=2 for (001) edge dislocations in BCC (a is the lattice parameter). Interfacial tension r can reach 1 J/m2 for incoherent interfaces (figure 3.13), a value that is close to that of general grain boundaries (r  0.78 J/m2 in iron). This makes the homogeneous nucleation of incoherent phases very difficult. They often nucleate heterogeneously on extended defects (grain boundaries, dislocations). Coherent precipitates have a low interfacial energy σ, but the misfit leads to an elastic energy density Dge rather important. Figure 3.11 shows schematically the situation for a β phase that has a larger lattice parameter than parent phase a (B atoms are bigger than A atoms). This elastic contribution decreases drastically when precipitates are semi-coherent (figure 3.12) or incoherent (figure 3.13). However, at the same time, this has a detrimental effect on interfacial energy σ that may drastically increase. Consequently, homogeneous nucleation of such phases is often more difficult. In reality, nuclei may shrink when their radius is too close to the critical radius. They may become subcritical because of thermal fluctuations (kT ). A more refined

Kinetics of Formation of a New Phase

159

FIG. 3.11 – Coherent a=b interface. β phase is rich in B atoms that are bigger than A atoms that are mainly located in the α phase. The lattice parameter of b phase is then larger than that of parent phase a. This introduces a distortion of the lattice close to the interface that results in a coherency elastic energy that is proportional to the volume of the precipitate.

FIG. 3.12 – α/β semi-coherent interface. Periodic dislocations accommodate the misfit (da=a) between α and β phases. Distance (d) between dislocations is such that da=a ¼ b=d. b is the modulus of the Burgers vector of dislocations and d their spacing. The elastic energy is partially relaxed by these dislocations. theory considering these regression effects has been developed by Turnbull and Zeldovich. When the energy gain is close to the top of the nucleation barrier (ΔG*), i.e. DG ðRÞ [ DG   kT (figure 3.10), the nucleus may indeed shrink and become subcritical. We thus have to consider the heterophase fluctuations around R . R may then decrease below R . According to this theory, nuclei can only grow irreversibly when they leave the so-called Zeldovich zone of width d such as DG   DG [ kT (figure 3.10).

Diffusion, Segregation and Solid-State Phase Transformations

160

FIG. 3.13 – α/β incoherent interface. The lattices of both α and β phases are decoupled with almost no deformation at the interface. The elastic energy is almost zero, but the interfacial energy is rather large. In this Zeldovich zone around R , for which DG   DG\kT , embryos have a size that “oscillates” in a stochastic way around R* during some time (s). In contrast to the classical nucleation theory, the Zeldovich model introduces the notion of incubation time s. This is the time that is needed for embryos to leave the Zeldovich zone of width d (such as R [ R þ d =2). Growth of nuclei is then observed for t > τ. The Zeldovich width can easily be calculated using a Taylor expansion of DG ðRÞ to second order near the critical size (see problem 3.2.11): 2 d ¼ pffiffiffi : Z p Z is the Zeldovich factor:

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 d 2 DG

Z¼ 2pkT dn 2
n¼n

n is the number of atoms in a nucleus of radius R: n ¼ 3v4at pR3 . n  is therefore the number of atoms in a critical nucleus of size R*. The latter expression of Z shows that the Zeldovich
factor is larger when the absolute value of the curvature of ΔG is 2
larger. As ddnDG \0, we speak in term of absolute curvature hereafter. As 2
 n¼n

expected (figure 3.10), the Zeldovich width increases when its curvature decreases. Hence, the nucleation rate can be written as a function of time: I ðt Þ ¼ Z b Cn0 e

DG  kT

s

et:

b is the frequency at which B atoms aggregate to nuclei. The latter can be written (problem 3.2.12):

Kinetics of Formation of a New Phase

b ¼

161

4pðR Þ2 DC0 : a4

C0 is the nominal concentration of B atoms and a is the interatomic distance. Note that other expressions of b are proposed in the literature. Previous equation of I(t) shows that the nucleation rate I(t) is proportional to b . The expression of I(t) shows that there is again a competition between the mobility   D  and the driving force for nucleation. of B atoms D ¼ D0 exp Q kT Incubation time s is obtained by expressing b as a diffusion coefficient in the framework of Einstein’s diffusion theory (random walk of nuclei around R* in the “phase space” of their size R): b ¼

ð d Þ 2 2s

This leads to: s ¼ pZ22 b . As shown previously, the Zeldovich factor increases with the second derivative of ΔG(R) (in absolute value). Consequently, incubation time s decreases when the curvature of ΔG increases. A simple calculation shows that its curvature close to the critical size increases with the driving force. Hence, as expected, the incubation time decreases when the driving force increases, that is when the supersaturation increases (larger nominal concentration X0).

3.1.4

Heterogeneous Nucleation

When the b phase does not have the same structure as the parent phase a (incoherent phases), homogeneous nucleation in the bulk of grains is very difficult or even impossible. One then observes the preferential nucleation of the b phase to extended defects, namely grain boundaries (GBs), stacking faults or dislocations (figures 3.14 and 3.15). This mode of precipitation is predominant at high temperatures because the supersaturation and the driving force of nucleation decrease. This may lead to a zone around GBs that is depleted in both solutes and vacancies (GBs are point defect sinks) giving rise to a deserted zone where there is no precipitation (PFZ, precipitation free zone). We then observe the decoration of extended defects by precipitates (b) whose growth is also favoured as GBs and dislocation are both short circuits of diffusion. When both homogeneous and heterogeneous nucleation cohabit, the precipitates that nucleate on GBs or dislocations are bigger than those formed in the grain. Let us consider GBs. The latter generate an interfacial energy raa that can be relaxed partly by the heterogeneous nucleation of the b phase that develops on each side of the GB (figure 3.16). The balance of interfacial tensions at the triple point of GBs (figure below) leads to the equality: raa ¼ 2rab cos h

162

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.14 – Electron microscopy image (STEM HAADF) showing in dark contrast the heterogeneous precipitation of incoherent precipitates at GBs (Mg-rich b phase) in AlMgZr alloys containing 5at.%. Mg and a low Zr content. In bright contrast, some precipitates of “dispersoids” rich in Zr, that is a low-grade element, also appear (X. Sauvage, GPM).

h (  p=2) is the wetting angle and rab the interfacial energy between a and b  phases. A simple calculation (problem 3.2.4) shows that the new barrier DGGB is lowered by a factor S(θ) that depends on the wetting angle θ (figure 3.17):  DGGB ¼ DG  S ðhÞ

with S ðhÞ ¼ 12 ð2 þ cos hÞð1  cos hÞ2 . When the interfacial energy of GBs (raa ) is very low compared to rab , cos h ¼ raa =2rab ! 0. Wetting angle h then tends towards p=2 and the nucleus becomes spherical. As a result, S(θ) tends towards 1 (figure 3.17). Hence, when the  interfacial energy of GBs is very low, DGGB ! DG  and the heterogeneous nucleation of the β phase on GBs is no more favourable compared to homogeneous nucleation. Increasing raa leads to a decrease in S ðhÞ and subsequently to a smaller  nucleation barrier DGGB . The latter tends towards zero if raa ¼ 2rab . Then, h ! 0 and we have a perfect wetting of GBs by the b phase. This looks similar to segregation phenomenon (chapter 1). However, this is a distinct phenomenon. In contrast to precipitation, segregation does not require the alloy to be supersaturated. It is worth mentioning that when h ! 0, β then becomes two-dimensional. Thus, previous calculations do not hold any more. The limit case h ! 0 has therefore no real physical meaning.

Kinetics of Formation of a New Phase

163

FIG. 3.15 – Electron microscopy image showing the homogeneous and heterogeneous precipitation of Ag particles (in dark contrast) in Cu5at.%. Ag alloy having undergone a precipitation treatment at 440 °C for 30 min. Cu and Ag have the same structure (FCC) but with a strong parametric deviation (12%) giving rise to both homogeneous precipitation and heterogeneous nucleation on dislocations (dark contrast lines) of silver precipitates (X. Sauvage and M. Bonvalet, GPM).

FIG. 3.16 – Heterogeneous nucleation of a b precipitate on the grain boundary between grains of the parent phase a. Balance of interfacial tensions at the two triple points sets the wetting angle h.

164

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.17 – Representation of the function S ðhÞ to which the heterogeneous nucleation barrier  DGGB is proportional to. Nucleation of the b phase is easier for small wetting angles h.

3.1.5

Nucleation of Metastable Phases

When the precipitated phase (β) has a different structure from that of the parent phase, its nucleation gets difficult. The incubation time becomes very long. One can then observe the nucleation of a coherent metastable phase (b0 ) that has a different structure than that of the equilibrium phase (β). Such a metastable phase nucleates easier because of its lower interfacial energy. This is very often observed in aluminium-based alloys. Consider AlCu alloys for which the equilibrium phase h is Al2Cu (incoherent tetragonal ordered phase). There is then the nucleation of pre-precipitation zones in the form of copper platelets of thickness equal to an atomic plane (001), called Guinier–Preston (GP) zones (figures 3.18, 3.19 and 3.20), named after the researchers who first identified them by X-ray diffraction. This metastable phase has of course a higher free energy than the equilibrium phase (figure 3.21), which leads to a higher metastable solubility limit Xa=GP (figures 3.21 and 3.22). Thus, these GP zones are only observed if the copper content is sufficient and higher than this limit Xa=GP . In AlCu alloys, other intermediate metastable phases called h00 (coherent Al2Cu phase, figure 3.21), and h0 (tetragonal, partially coherent) may precipitate. They have lower free enthalpies than the GP zones but larger than the equilibrium h phase. The metastable solubility limits related to these metastable phases decrease in proportion as phases are more stable (GP ! h00 ! h0 ) by approaching the solubility limit Xa=# (figure 3.21) of the stable phase. For large supersaturation, one then has a cascade of metastabilities leading to the successive nucleation of GP zones, then of h00 , h0 and h phases during the precipitation kinetics. The kinetic paths of such successive transformations are complex. GP zones can grow to finally form the h00 phase. A part of GPs is then transformed into h00 while the other GP zones dissolve. The h0 phase nucleates to dislocations while

Kinetics of Formation of a New Phase

165

FIG. 3.18 – Atom probe tomography image (16 × 16 × 50 nm3) revealing the presence of copper-rich GP zones, parallel to the (001) planes in Al4at.%.Cu alloys (A. Bigot, GPM) [38].

FIG. 3.19 – Guinier–Preston (GP) zones appear as pure copper monolayer platelets in AlCu alloys. Copper atoms are smaller and lead to a distortion of the crystal lattice.

the heterogeneous nucleation of h is observed at the a=h0 interfaces. These metastability cascades are encountered in many other alloys (AlZnMg, AlMgSi, niobium steels [40]). In aluminium-based alloys, numerous vacancies are trapped during quenching from the annealing temperature. This supersaturation of vacancies accelerates the diffusion of solute atoms and leads to nucleation and growth of precipitates even at room temperature. Hence, GP zones form and develop in AlCu alloys aged at room temperature, which leads to a structural hardening in the early stages of precipitation (historical experiments of Alfred Wilm in Düren in 1906, giving rise to the first duralumin alloys whose names come from the town Düren).

Diffusion, Segregation and Solid-State Phase Transformations

166

FIG. 3.20 – Field ion microscopy [39] image (50 × 50 nm) showing Cu-rich h00 precipitates in an Al alloy containing 1.7at.%. of Cu and aged at 150 °C for 24 h. The precipitates appear as platelets a few nm thick and are oriented along the 3 equivalent directions of the aluminium FCC lattice (W. Lefebvre, GPM).

3.1.6

Growth

Once precipitates reach their critical size, the nuclei of b phase can grow. The growth rate depends on the crystallographic nature of a=b interfaces. Incoherent interfaces are generally more mobile. Semi-coherent interfaces, composed of periodically spaced dislocations accommodating the lattice misfit, migrate by a slower wedge growth mechanism (terrace edges on the interface a=b). Growth leads to precipitates in form of platelets whose large semi-coherent faces migrate slowly and whose lateral incoherent faces are very mobile (figure 3.23). The anisotropic shape is here driven by the migration kinetics of interfaces. The thickness to width ratio of the platelets tends to decrease during their growth. This effect is at the origin of the so-called Widmanstätten morphology. Anisotropy of the surface energy may also lead to platelets or rods, and more generally to non-spherical shapes. In the growth stage of precipitation, bulk diffusion of solutes in the supersaturated parent phase (a) towards a=b interfaces lead to the migration of interfaces. Let us examine the migration of planar α/β interfaces for which the rearrangement of the interface and the related reaction are rapid (e.g. incoherent interfaces) compared to the bulk diffusion of solute atoms. It is then the diffusion process that controls the growth kinetics because it is the slowest process. It can then be shown that the

Kinetics of Formation of a New Phase

167

FIG. 3.21 – Free enthalpy G ðX Þ of (i) the parent solid solution (a), (ii) the metastable phases (GP, h00 , h0 ) and (iii) the stable h phase. The solubility limits related to metastable phases are larger than the equilibrium solubility related to the equilibrium phase θ. The lower free enthalpy of phases, the more stable, and the smaller the solubility limit. The compositions of phases h00 and h0 in equilibrium with a are here very close to that of the stable h phase (Xb ).

migration of planar interfaces follows a parabolic law that is characteristic of the bulk diffusion of solutes (Zener model, problem 3.2.9): pffiffi x ¼ fm D t : This equation shows that there is a competition between the equilibrium mole fraction of b phase (fm ) and the diffusion coefficient (D) when the temperature increases. fm decreases when increasing temperature T whereas diffusion coefficient D increases “exponentially”. Hence, it is easy to understand that the growth rate v ¼ dx=dt increases with T up to a maximum before decreasing up to solvus temperature TS where the growth rate cancels (the solubility limit related to the given nominal composition alloy is reached at T ¼ TS ). For spherical precipitates (radius R) whose growth rate is controlled by diffusion, Fick’s laws solved in spherical coordinates lead to a slightly different parabolic law (problems 3.2.5 and 3.2.7):

168

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.22 – Metastable phase diagram showing the metastable solubility limits of solutes in the α parent phase (dotted line) relative to metastable phases (GP, h00 , h0 ). The latter are larger than the equilibrium solubility limit Xa=# related to the equilibrium h phase.

FIG. 3.23 – Temporal evolution of the morphology of a precipitate with highly mobile incoherent lateral interfaces and semi-coherent planar interfaces.

R¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi 2fm Dt :

This law is valid only for low densities of precipitates (low precipitated mole fraction fm ) without interactions, and growing thus independently in a solid solution of quasi-constant composition (figure 3.8). It is worth mentioning that, in reality, the solute concentration of the parent phase α decreases with time because of the growth of precipitates and the subsequent increase of fm with time.

3.1.7

The Spinodal Decomposition

Spinodal decomposition is observed in many systems (figures 3.24 and 3.25). Let us mention the concentrated alloys FeCr, AlNiCo, and CuNiFe, but also polymers (polystyrene-polyvinyl methylene). Spinodal decomposition is of great importance, especially for stainless steels used in the primary cooling pipes of nuclear power plants, as it can lead to a decrease in resilience. It is an important problem of safety of power plants not only in France but also in Europe and elsewhere. The theory of spinodal decomposition with be detailed in problem 3.2.10.

Kinetics of Formation of a New Phase

169

FIG. 3.24 – Spinodal decomposition of Fe20at.%.Cr alloys aged at 450 °C. The material decomposes into two coherent BCC phases that are the chromium-depleted phase (α) and the chromium-enriched phase (a0 ). This atomic tomography image (30 × 30 nm2 section) reveals the interconnected and percolated a=a0 structure. The 30at.%.Cr concentration isosurfaces (bright contrast) and the 31at.%.Cr isosurfaces (darker contrast) exhibit the exterior and the interior of the Cr-rich a0 domains, respectively (F. Danoix, GPM). As for the nucleation regime where the size of nuclei has to exceed a critical size to grow, in spinodal decomposition, and for the same reasons, the concentration fluctuations can only grow if the driving force of decomposition outweighs the opposing terms (interfacial and elastic energies). This introduces a critical wavelength kc above which concentration fluctuations can develop. kc can be calculated by writing the energy balance. The free enthalpy associated with the development of compositional fluctuations dX ¼ X  X0 around the nominal atomic fraction X0 of the alloy is written:  Z 
!
2

ðg ðX Þ  g ðX0 ÞÞ þ Ki
r X
þ Ke ðdX Þ2 dV : DG ðX Þ ¼ G ðX Þ  G ðX0 Þ ¼ V

All terms in the integral, taken in the entire volume, are energy densities per unit volume. The first term g ðX Þ  g ðX0 Þ is the chemical energy gain. It is negative and
!
2

opposes the interfacial (Ki
r X
) and elastic (Ke ðdX Þ2 ) terms. The latter terms are both positive and oppose the decomposition. Let us consider

sinusoidal fluctuations
!
whose wavelength is k. The concentration gradient
r X
is maximum in x ¼ 0 (modulo λ) and is written 2pdX=k. A Taylor expansion to second order gives for the driving force term:

170

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.25 – Time evolution of Cr concentration profiles in ferrite (α) of a nuclear power plant austeno-ferritic (α + γ) duplex steel aged at 350 °C (up to 30 000 h). The Cr profiles show that the amplitude of concentration fluctuations increases with ageing time. The concentration amplitude distributions (on the left) deduced from concentration profiles reflect this increase of the amplitude of fluctuations with time (F. Danoix, GPM) [41].


2
1 2d g
g ðX Þ  g ðX0 Þ ¼ ðdX Þ : 2 dX 2
X¼X0 The first order term of the Taylor expansion is not considered in this expression because statistically it cancels in the integral as there are as many fluctuations with positive and negative amplitudes (dX). Fluctuations can develop if DG ðX Þ\0, this is fulfilled if the interior of the integral is negative:
 2 1 d 2 g

2p þ Ki þ Ke \0
2 2 dX X¼X0 k

Kinetics of Formation of a New Phase

171

and therefore necessarily for k [ kc with: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 2Ki
: kc ¼ 2pu t d2g
2Ke þ dX 2
X¼X0

In a mean field approach, with interactions limited to first neighbours, the interface term can be written as Ki ¼ Ae=vat with e the order energy and A a constant proportional to the square of the interaction distances. Note that Ki is an energy per unit length (vat is the atomic volume).   E da 2 Let us remind (section on nucleation) that Dge ¼ 1v a . This equation can be   2 E d ln a 2 rewritten in the following form: Dge ¼ Ke ðDX Þ with Ke ¼ 1v . dX The expression of kC has a physical meaning only if the interior of the square root is positive. This imposes that:
d 2 g

[ 2Ke :  dX 2
X¼X0 When the coherence stress is zero (Ke ¼ 0) or negligible, we find the necessary condition for spinodal decomposition to occur: d 2 g=dX 2 \0, as given in chapter 1. This defines the so-called incoherent spinodal. This is an abuse of language because in reality the two phases remain coherent in spinodal decomposition. Let us remind that indeed the concept of spinodal decomposition only makes sense when both co-existing phases have the same structure and can be described by a single free enthalpy curve G(X). The term “incoherent spinodal” simply expresses that the elastic energy of coherence is not accounted for (negligible misfit between phases), as for incoherent interfaces for which this contribution is zero. In the more general case, in the presence of an elastic term due to the coherence misfit, the driving force d 2 g=dX 2 must be even more negative. This requires the system to be even more in the core of the miscibility gap (i.e. larger supersaturations). The above condition then defines the coherent spinodal domain. The latter is inside the coherent miscibility gap that itself is inside the incoherent gap (figure 3.26). The temporal evolution of the concentration C ðr; t Þ is controlled by the divergence of the diffusion flux (second Fick’s law). The latter is proportional to the chemical potential gradient and thus to the second derivative of g ðX Þ whose expression is given in chapter 2 devoted to diffusion. In the early stages of spinodal decomposition, the amplitude of fluctuations is small. Cahn [42] and Hilliard have thus proposed a theory that holds in the framework of this hypothesis. This theory, called linear, predicts that the amplitude of concentration fluctuations Aðk Þ increases exponentially with time provided that λ > λC or k\kc with k ¼ 2p=k the wave number and kc ¼ 2p=kc : 

 2

Aðk Þ ¼ A0 eRðk Þt

with Rðk Þ ¼ 2M0 Vm Ki kc2  k k 2 .

172

Diffusion, Segregation and Solid-State Phase Transformations

FIG. 3.26 – Phase diagram T ðX Þ showing (i) the coherent miscibility gap that is narrower than the incoherent gap because of the additional elastic term due to lattice misfit, (ii) the related coherent and incoherent spinodal lines.

M0 is the inter-mobility (chapter 2) that is proportional to the interdiffusion coefficient D of solutes. M0 also depends on the nominal atomic fraction of solutes X0 (problem 3.2.10) that writes: M0 ¼ DX0 ð1  X0 Þ=RT . The dependence of M0 with X0 is sometimes neglected in literature. During the kinetics of spinodal decomposition, there is a gradual damping of concentration fluctuations of small wavelengths k\kc and amplification of large wavelength Fourier components of the concentrapffiffiffi tion signal (k [ kc ). It can be shown that Rðk Þ is maximum when kM ¼ kc = 2. This is shown for instance using small angle neutron or X-ray diffusion. This theory is limited to the early stages of decomposition for which the amplitude of the fluctuations is small and the assumptions reasonable. Another theory (LBM) developed by Langer, Bar-On, Miller [43], allows to predict the evolution of the wavelengths of the decomposed structure. This so-called LBM theory predicts the following temporal law evolution k ¼ k0 t n with n  0.2. Experimental exponents often differ from this theoretical value.

3.1.8

Coarsening

Let us now consider the advanced stages of growth of precipitates. The supersaturation in the parent phase has now decreased by a large amount. The driving force is thus not any more the supersaturation but the reduction of the overall interface energy. Smaller precipitates will dissolve to the profit of larger ones for which the surface to volume ratio is smaller. During the precipitation kinetics, the particles do not all have the same history from their nucleation stage up to their growth stage. Some particles of the new phase are growing while others have just been formed (i.e. nucleated). The precipitates formed at the beginning of the heat treatment are at a given time t, larger than those that appear later. During the growth stage, supersaturation decreases

Kinetics of Formation of a New Phase

173

and the molar fraction of precipitates tends towards its equilibrium value, as given by the lever rule. The driving force for growth, driven by supersaturation, gradually disappears. However, a=b interfaces lead to an excess of free energy that favours large particles whose ratio between the driving force term (in R3 for spherical precipitates) and the surface energy (in R2 ) is more favourable. We will then observe a competitive growth, leading to the dissolution of the smallest particles, that are expensive in interfacial energy, to the benefit of large ones (figures 3.27 and 3.28). The density of precipitates will then decrease while the average size of precipitates will continue to increase. This regime of competitive growth is called coarsening or Oswald ripening.

FIG. 3.27 – Scanning electron microscopy micrograph (backscattered electrons) illustrating the highly ordered microstructure observed in an advanced stage of coarsening in a model cobalt-based superalloy. The alloy (Co9at.%.Al7at.%.W) was aged at 900 °C for 200 h. The micrograph reveals a very large volume fraction of cuboidal Al and W enriched c0 precipitates oriented along directions. The average particle size is 160 nm (A. Azzam and X. Sauvage, GPM) [44].

The excess free enthalpy due to capillary effects is given by: DG ¼ Vm Dp Vm is the molar volume. Dp ¼ pb  pa is the pressure difference between phases at the a=b interface. The latter is given by the Laplace–Young equation Dp ¼ 2r=R

174

Diffusion, Segregation and Solid-State Phase Transformations

with r the interfacial energy per unit area and R the radius of curvature of the α/β interface16. Replacing this equation in that of DG leads to: DG ¼

2rVm : R

This excess of free enthalpy decreases with R and vanishes for planar interfaces. It gives rise to an increase in the solubility limit of B atoms at the curved interface. Under certain approximations, one can demonstrate that (problem 3.2.7): XaR ¼ Xa1 e

2rVm 1 R Rg T

:

XaR is the atomic fraction of B atoms in the parent phase α close to the curved α/β interface. Rg is the gas constant. Xa1 is the equilibrium solubility limit for R ! 1 (planar interfaces) as given by the phase diagram. The exponent is generally small compared to one, so that we can linearize the previous equation (first order Taylor expansion):  q XaR ¼ Xa1 1 þ R m with q ¼ 2rV Rg T , the capillary length. This equation is known as the Gibbs–Thomson equation. The new growth law taking into account this Gibbs–Thomson effect predicts a temporal evolution of the mean radius hRi of the particles in t 1=3 . Coarsening is thus slower than pure growth that is in t 1=2 . hRi is given by (problem 3.2.7):

hRi3 ¼ hR0 i3 þ KR ðt  t0 Þ: t0 denotes the coarsening onset time and R0 the initial radius at t = t0. The classical LSW theory of coarsening gives for pure precipitates in B: KR ¼

8 rVm2 DCa1 9 NRg T

with N the Avogadro number. The concentration is expressed in at/m3 in this equation During the coarsening regime, precipitates smaller than a certain critical size Rc dissolve in favour of larger ones. Rc has the same physical origin (interfacial energy) as that of the critical radius in nucleation theory and its expression is identical (problem 3.2.7). We can also show that Rc is equal to the average size hRi of the population of precipitates (figure 3.28). Both Rc and hRi increase with time. This competitive growth results in a decrease in the precipitate density NV (number of particles/m3). The temporal evolution of NV ðt Þ can be deduced from 16

The Laplace-Young equation can be demonstrated as follows. The increase in free energy dF related to the increase in the volume of a precipitate dVb is dF ¼ pa dVa  pb dVb þ rdAb with dAb the related increase in the precipitate area. Moreover, dVa ¼ dVb . At equilibrium dF ¼ 0. dA Consequently, pb  pa ¼ dVbb . Thus, Dp ¼ 2r R for spherical precipitates of radius R.

Kinetics of Formation of a New Phase

175

FIG. 3.28 – Coarsening of precipitates. There is a competitive growth between particles: the precipitates smaller than the average size hRi dissolve while the larger ones grow. There is a concomitant inversion of the concentration gradients at a=b interfaces. that of the average size of precipitates. As a first approximation, we can write the volume fraction as: 4 fv ¼ phRi3 NV : 3 The volume fraction (fv ) remaining approximately constant during coarsening and as hRi3 is proportional to t, the density of precipitates decreases in t 1 . The calculation (problem 3.2.7) shows however that there remains a residual supersaturation in the parent phase that decreases in t 1=3 . The volume fraction, very close to the equilibrium one, continues to grow very slightly during the coarsening stage.

3.1.9

Overall Kinetics

The temporal evolution of the nucleation rate I ðt Þ, the matrix composition Ca ðt Þ, the number density NV ðt Þ and the average radius of precipitates hRðt Þi is schematically represented in figure 3.29. They summarize the kinetic evolution of a supersaturated alloy from the early stages of nucleation up to the coarsening stage. The three stages namely, nucleation, growth and coarsening can be identified. After incubation time s (Zeldovich–Turnbull theory), nucleation rate I ðt Þ increases rapidly (nucleation stage), then remains constant (stationary nucleation rate) and eventually decreases at the beginning of the growth stage. Number density NV first increases during nucleation, then grows linearly with time when I ðt Þ get constant. NV reaches a maximum at the beginning of the growth regime. During the third stage of coarsening, NV eventually decreases as t 1 . In the very first stages of precipitation, the concentration of B atoms in the parent phase Ca ðt Þ remains approximately constant, i.e. during stationary nucleation provided that the number density of nuclei remains small. Then supersaturation decreases quickly during the growth regime and the concentration in the parent phase will finally tend asymptotically towards the equilibrium solubility, as mentioned earlier, Ca ðt Þ decreases in t 1=3 during the coarsening stage.

176

Diffusion, Segregation and Solid-State Phase Transformations

In the early stages of precipitation (nucleation), the average size of precipitates remains approximately constant and close to the critical radius. hRi then increases in t 1=2 (growth) and finally in t 1=3 in the coarsening regime. This theoretical scenario is however schematic. In reality, there is often an overlap of the nucleation and growth stages. New nuclei continue to be formed while the precipitates that nucleated earlier have already entered the growth regime. Similarly, there is quite often an overlap between the growth and coarsening stages and the pure growth regime is almost never observed. One then observes a decrease in the density of precipitates before growth ends, as coarsening operates, while the content of B solutes in the parent phase is still significantly above the solubility limit.

FIG. 3.29 – Overall kinetics of precipitation with the three consecutive nucleation, growth

and coarsening stages, showing the temporal evolution of (a) the nucleation rate I ðt Þ (b) the concentration of B atoms in the parent solid solution Ca ðt Þ, (c) the number density of precipitates per unit of volume NV ðt Þ and (d) the mean radius of precipitates hRðt Þi.

Kinetics of Formation of a New Phase

3.2 3.2.1

177

Problems Eutectoid Transformations

The eutectoid transformations are of great importance in steels (basically FeC) in which austenite c (FCC phase) transforms into pearlite which is a mixture of the α ferrite (BCC FeC solid solution containing a small amount of carbon) and cementite Fe3C. The cementite, that is a carbide, is hereafter called the b phase. In the early stages of transformation, colonies of b carbides nucleate at grain boundaries (GBs). Carbide lamellae alternate with ferritic phase α and form a quasi-periodic microstructure. These a=b colonies grow and eventually invade the whole grain, making the parent phase c disappear. We will model this phase transformation by considering the eutectoid transformation c ! a þ b in a binary alloy AB (A = Fe and B = C in steels). Lamellae of the b phase are assumed to all have the same width, which remains constant during the eutectoid reaction. Lamellae nucleate at GBs and then grow with periodicity λ. The three phases are assumed to have the same molar volume Vm . The formation of the β phase creates α/β interfaces that cost energy and opposes to the phase transformation. The contribution of β/γ and α/γ interfaces will not be considered hereafter. 1. Let us consider an austenitic steel that has the eutectoid composition. Let XE be the atomic fraction of solute B in parent phase c corresponding to the eutectoid point. DG (>0) is the molar free enthalpy gain related to the eutectoid transformation c ! a þ b. Let TE be the eutectoid temperature. Sketch G ðX Þ for the three phases α, β, γ at T ¼ TE and T \TE . Represent DG in this scheme and the equilibrium atomic fraction of carbon in both α and β phases (Xa , Xb ). Represent schematically the phase diagram. 2. The distance between b lamellae (i.e. the periodicity k of the microstructure) depends on the thermodynamics of the system, in particular on the energy per unit area of a=b interfaces (σ). Only α/β interfaces are considered in the following. We will study this problem as a function of the “supercooling” dT ¼ TE  T (abuse of language referring to the problem of crystal nucleation in liquid) for T \TE . Let us define Dg as the effective driving force of transformation (J/m3) decreased by the interface term (σ) that opposes nucleation. Express Dg as a function of both DG (J/mole) and the surface-to-volume ratio A=V of the a þ b microstructure (A is the area of a=b interfaces). Express A=V as a function of k. Deduce the expression of Dg as a function of k, Vm and DG. Sketch the microstructure for T \TE . 3. What is the condition on the sign of Dg for the eutectoid transformation to be possible? Deduce the minimum spacing between lamellae km that is required. 4. We will calculate DG ðT Þ. What is the value of DG at TE ? Deduce the expression of DS as a function of the related gain of enthalpy DH. It is assumed that DS and DH are constants for small supercooling dT . Deduce the approximate expression of DG ðT Þ and then that of km as a function of the enthalpy gain per unit volume

178

5.

6.

7.

8.

Diffusion, Segregation and Solid-State Phase Transformations

DHV , dT , σ and TE . Represent km as a function of dT . How does the microstructure change with the supercooling dT ? The growth of both α and β phases is driven by the bulk diffusion of solutes (carbon in steels) in the c phase from the supersaturated parent phase α to the new phase b. It is assumed that the migration of solutes takes place in the c phase close to a=c and b=c interfaces. There is no important diffusion along interfaces. Make a schematic drawing showing the diffusion fluxes. Let D be the diffusion coefficient of solutes. Write that the solute flux J as given by the Fick equation is equal to Cc v with v the migration velocity of the transformation front and Cc the nominal concentration of solute atoms in the parent phase c. Justify this result. Let dC ¼ Ca=c  Cb=c be the concentration difference in the c phase between that near the a=c interface and that near the b=c interface for T ¼ TE  dT . The concentration gradient is then approximately equal to dC =k. Deduce the expression of v as a function of D, dC and k. We will derive an approximate expression of dC from the phase diagram. Let us assume that there is a local equilibrium near a=c and b=c interfaces. Xb=c and Xa=c are the atomic fractions of solutes near b=c and a=c interfaces, respectively. Represent these quantities in the schematic representation of G ðX Þ for the three phases α, β, γ when T ¼ TE  dT . Show on a generic phase diagram T ðX Þ that Xa=c and Xb=c can be obtained by extrapolation in the a þ b domain of both c/c þ a and c/c þ b equilibrium lines. Under which conditions (on dT and on the interface geometry) can we write that dX1 ¼ Xa=c  Xb=c ¼ kX dT with kX a constant that can be deduced from the phase diagram? Microstructural observations show that both a=c and b=c interfaces are curved. This leads to a Gibbs–Thomson effect that modifies dX, which now varies as 1=R (see section on coarsening). For the sake of simplicity, a unique mean radius of curvature R is considered, with no distinction between the a=c and b=c interfaces. Besides, it is reasonable to set R as proportional to the distance (k) between lamellae of b. dX is therefore proportional to 1=k. The driving force Dg vanishes for k ¼ km (minimum critical distance). This implies that dX ¼ 0 when k ¼ km . It is then natural to write: dX ¼ dX1 ð1  km =kÞ that cancels when k ¼ km . Give the expression of the transformation velocity v in substituting dX by its expression. Deduce the value kM of k for which v is maximum (vM ). Represent v as a function of k. Show that the maximum velocity vM is proportional to ðdT Þ2 . We shall now apply the previous model to the pearlitic transformation of a eutectoid steel (0.77 wt.% carbon, TE = 727 °C) subjected to an isothermal treatment at T = 707 °C. The interfacial energy of α/β interfaces is σ = 700 mJ/m2 and the transformation enthalpy per unit volume is DHV = 8.108 J/m3. Calculate the critical value km . Electron microscopy observations show a fairly wide dispersion of the distances between cementite lamellae with an average value near 0.18 μm. Compare this value to the distance kM for which the migration rate is maximum. Conclude.

Kinetics of Formation of a New Phase

179

9. Examination of the FeC phase diagram indicates that dX1 = 1at.% at 707 °C and the eutectoid composition (Xc ) is 3.6at.% (0.77 wt.%). The diffusion data for carbon in austenite are D0 = 0.15 × 10−4 m2/s and Q = 143 kJ/mol. Calculate the maximum velocity vM . Compare this result to the measured velocity that is close to 1 μm/s. Diffusion along a=c and a=b interfaces was not considered in the previous calculation. What would be the influence on vM ? 10. The material is now treated at 687 °C. Calculate the new value of dT and compare to the previous value. Calculate again km , kM and vM . Compare to previous values. What is the role of the temperature? The average grain size is d = 100 μm. What is the order of magnitude of the time required for the austenitic grains (γ phase) to be entirely transformed into pearlite?

Solution to the problem 1. The three phases α, β, γ coexist at the eutectoid point (XE , TE). The related phase diagram T(X) can be sketched as follows:

At the eutectoid point (XE ) at T ¼ TE , the three α, β, γ phases are at thermodynamic equilibrium, the tangent of the three G(X) curves is thus common.

180

Diffusion, Segregation and Solid-State Phase Transformations

When T \TE the parent phase c transforms into a þ b. The related gain of free enthalpy DG [ 0 is shown in the following diagram.

Xa and Xb are the atomic fractions of B atoms when α and β phases are at equilibrium at the considered temperature T \TE . 2. The development of the eutectoid transformation creates new α/β interfaces. The latter leads to an interfacial energy σ that opposes to the formation of the a þ b lamellar structure. Thus, the effective driving force of transformation is given by: Dg ¼

DG 2A r Vm V

Factor 2 in this equation is due to the presence of two a=b interfaces per period k. 2r It is easy to see that approximately VA ¼ 1k. Therefore, Dg ¼ DG Vm  k .

Kinetics of Formation of a New Phase

181

3. In order for the transformation to take place, it is necessary that Dg [ 0. This requires the periodicity λ to be larger than a minimum value: λ > λm with m km ¼ 2rV DG . 4. At the eutectoid point, the energy gain associated with the transformation is necessarily zero, so that DG ðT ¼ TE Þ ¼ 0. Now, we know that DG ðT Þ ¼ DH  T DS. Thus, DS ¼ DH TE . For T close to TE , DH and DS vary little and are assumed to be constant for small undercooling. When dT ¼ TE  T is small, we can thus write: DG ðT Þ ¼ DH

dT TE

We deduce that: km ¼

2rTE DHV dT

with DHV ¼ DH =Vm . The latter equation for λm shows that the microstructure becomes finer (smaller k) when the supercooling dT increases (lower temperature). The dependence of λm with δT is hyperbolic as represented below.

182

Diffusion, Segregation and Solid-State Phase Transformations

5. Only bulk diffusion is considered, Fick’s diffusion flux (in absolute value) writes DrC ¼ Cc v. It is assumed that rC ¼ dC =k, thus the velocity reads: v ¼ Dk dC Cc .

6. Both a=c and b=c interfaces are in local equilibrium during the phase transformation. Consequently, the atomic fractions Xb=c and Xa=c at these interfaces are given by the common tangents between G(X) curves related to b and c phases and α and c phases, respectively, as shown in the following diagram. Note that dX1 ¼ Xa=c  Xb=c [ 0.

Let us consider the phase diagram represented below. The equilibrium lines separating the c and a þ c regions on one hand, and the c and b þ c regions on other hand, give both Xa=c and Xb=c values, respectively, by linear extrapolation at T < TE .

Kinetics of Formation of a New Phase

183

We can write dX1 ¼ Xa=c  Xb=c as being proportional to the supercooling provided that dT is small. From the phase diagram, we have dX1 ¼ kX dT with kX a constant that depends on the slope of equilibrium lines. The phase diagram gives the equilibrium concentrations for planar interfaces. Therefore, the previous expression of dX1 is only valid when the radius of curvature of the lamellae is infinite. A correction is thus required when interfaces are curved. 7. We start from the equation dX ¼ dX1 ð1  km =kÞ that considers the Gibbs– Thomson effect (curvature). As v ¼ Dk dC Cc and C = X/Vat with Vat the atomic volume, consequently:

v¼

  D dX D dX1 km ¼ 1 : k Xc k Xc k

The extrapolation of equilibrium lines of the phase diagram gives dX1 ¼ kX dT . We then end up with: v¼

DkX k  km dT : Xc k2

184

Diffusion, Segregation and Solid-State Phase Transformations

The velocity is maximum when have:

@v @k

¼ 0 and consequently for kM ¼ 2km . Thus, we

vM ¼

DkX dT 4Xc km

2 2rTE X DHV As km ¼ DH , thus, vM ¼ Dk 8Xc rTE ðdT Þ . V dT 8. We find km = 0.0875 μm and kM = 0.175 µm, that is very close to the average value observed separating the lamellae (0.18 μm). Lamellae separated by kM , whose growth fronts are the fastest, prevail over the others. 9. The steel being annealed at 707 °C, then dT = 20 K. We find D = 3.5 × 10−13 m2/s. The weight fraction of carbon in the austenite is 0.77 wt.% (eutectoid point). As the atomic mass of iron (  56 amu) and carbon (  12 amu), the atomic fraction of carbon in the austenite is therefore approximately 56/12 times larger, that gives a value close to Xγ = 3.6at.%. Using the previous expression, we find vM = 0.28 μm/s. This value is of the same order of magnitude as the measured velocity (1 μm/s). Nevertheless, the calculated value is lower. This discrepancy can be due to both γ/α and γ/β interfaces that are high diffusivity pathways. The moving interfaces can drag carbon atoms that will rapidly diffuse along interfaces towards the cementite lamellae. This will accelerate the migration of the transformation front and thus increase vM . 10. When the steel is annealed at 687 °C, dT (40 K) is twice the previous value (20 K). As a result, kM and km are twice as small leading to a finer scale microstructure. The maximum velocity that increases in ðdT Þ2 contributes to an increase in vM by factor four. However, the diffusion coefficient decreases with decreasing temperature (D = 2.5 × 10−13 m2/s at 687 °C compared to D = 3.5 × 10−13 m2/s at 707 °C). We then find vM  0.8 μm/s that is a little bit smaller than four times the previously calculated value (0.28 μm/s at 707 °C). A rough estimate of the time required for the austenitic grains to entirely transform into pearlite is given by: t = d/2 vM . For d = 100 μm, t is close to 1 min.

Kinetics of Formation of a New Phase

3.2.2

185

Nucleation of Precipitates in Ni-Al Alloys

Ni-Al binary alloys are a model system for nickel based superalloys used in the aerospace industry (jet engines). Nickel-based superalloys present excellent mechanical properties at high temperatures (up to 1200 °C), thanks to the presence of a high-volume fraction of ordered Ni3Al c0 precipitates (L12 structure) forming after heat treatment in the c FCC parent phase. We will study the nucleation of c0 Ni3Al precipitates in a supersaturated Ni-Al alloy whose atomic fraction in Al is X0 = 13at.%. For the sake of simplicity, we will not consider the chemical order in the c0 phase. We shall assume that the c Al-depleted parent phase can be treated as a regular solution. The crystalline parameter of nickel (FCC) is a = 0.354 nm. The molar volumes of both c and c0 are very close and will be considered hereafter as equal and noted Vm . We will establish the expression of the nucleation driving force Dgn (J/m3) and deduce the related critical radius. 1. Give the expression for the chemical potentials of Ni and Al in the c phase as a function of the activities ai (i = Ni, Al) of Ni and Al. Express these same chemical potentials for a regular solution as a function of X ¼ NZ e. N is the Avogadro number, Z the coordination number, and ε the order energy. Deduce the expression of the ratio ai =Xi (activity coefficient) as a function of both X and atomic fractions Xi (i = Ni, Al). 2. Represent schematically the free enthalpy diagrams G ðX Þ of the c solid solution of Al in Ni and that of the Ni3Al phase. To simplify the writing, we note X the atomic fraction of aluminium, the nominal atomic fraction X0 and Xc that in the c phase at equilibrium (Xc0 is the Al equilibrium atomic fraction in c0 ). Plot on this diagram the gain in free enthalpy of nucleation (in J/mol), also known as the driving force, DGn ¼ Vm Dgn for X ¼ X0 . Calculate Dgn as a function of chemical potentials l0i in X ¼ X0 , lci in X ¼ Xc , and of Xc0 . Deduce the expression of Dgn as a function of the related activities ai0 in X ¼ X0 and aic in X ¼ Xc . 3. The solubility limit of Al in c at T = 625 °C is Xc = 11at.% and X0 = 13at.%. Show that since X0