Phase Transformations Workshop David Laughlin Materials Science and Engineering Carnegie Mellon University Pittsburgh, PA 15213
Outline What is a Phase? Heterogeneous / Homogeneous Transformations Nucleation Thermodynamics Free Energy Phase Diagram Phase Rule Lever Rule Order parameter
Kinetics Arrhenius s law Diffusion TTT curves Nucleation Rate Johnson / Mehl / Avrami Driving Force Al Cu System Coarsening Spinodal Decomposition Disorder / Order Combined SD and Order The CuNiSn System
Definition A phase is a physically distinct region of a system which has attained thermodynamic equilibrium, and which has a specific set of parameters (h 1, h 2,..) which specify its physical properties. Such parameters are called order parameters and include composition, structure, atomic order, magnetic order, etc. A change of phase occurs when one or more of its order parameters changes discontinuously.
Types of Solid State Phase Transformations Reconstructive (grain boundary reaction) Replacive (atomic ordering) Displacive (martensite)
There are Two Basic Kinds of Transformations Homogeneous and Heterogeneous Homogeneous Transformations are those which occur at every part of the system and at the same time. Usually the process occurs in degrees Heterogeneous Transformations occur at specific sites in the system leaving the rest of the system untransformed.
Precipitation is a Heterogeneous Transformation α α + β The b phase forms at some places and not others
Heterogeneous Transformations Homogeneously Nucleated or Heterogeneously Nucleated In homogeneous nucleation, all sites of the system have the same probability of transforming Not so in heterogenous nucleation: some sits are favored
Heterogeneous Nucleation is more frequently encountered Dislocations Special Sites include: Grain boundaries Surfaces Structural defects like stacking faults
Tools used to understand Phase Transformations include: Thermodynamics Kinetics Crystallography Microstructure Properties electrical resistivity
Thermodynamics Why is a certain phase stable? Lowest Energy? So if atoms in a gas are brought together to form a condensed phase, their energy is lowered. What about the other direction? Condensed phases do become gases. Is the energy lowered in this case?
No! Why do gases form form condensed phases? ENTROPY! That famous word which cause undergraduate students (and others as well) to quake! The ENTROPY of the Universe is always increasing! (But not everywhere all the time!)
A Phase Change occurs as a compromise between the ENERGY and the ENTROPY of the system. At low Temperatures: ENERGY At High Temperatures: ENTROPY The trasnsformation Temperature is where they are equally important>
Gibbs pointed this out and defined a function he called Free Energy A U - TS G H - TS A is called the Helmholtz Free Energy G is called the Gibbs Free Energy
Precipitation from solid solution can be understood from this point of view. At high temperature entropy dominates, and equilibrium is single phase with all solute atoms dispersed throughout the matrix. At low temperatures energy dominates, so atoms come out of solution and form phases with lower configurational entropy.
This fact helps to explain allotropic transformations as well. Usually the high temperature phase is the one with higher entropy, which often correlates with the crystal structure Ti HCP and BCC Co is FCC at high temperature (high symmetry) and HCP at low T Fe BCC FCC BCC here the low T BCC is ferromagnetic!
Gibbs Free Energy G G = G (T, P, composition) G Liquid Solid Tm T G T = -S 0 The phase with the greatest slope has the highest entropy and is therefore the HIGH TEMPERATURE Phase
Free energy versus Temperature Curves for solid, liquid and gas phases of a material at one composition. Note that the envelope of the lowest free energy determines the equilibrium phase. S to L at T m L to G at T v
First note the curvature of the Free Energy curves Positive implies stability 2 G 2 c 0
if negative, the phase would decompose!
Now look at two or more phases BASIC PRINCIPLE: Equilibrium is obtained by the lowest free energy may be one phase or it may be two phases.
Gibbs Equilibrium Phase Rule P + F = C + 2 where P is the number of phases C is the number of components F is the number of degrees of freedom this is derived by equating the chemical potentials of all the phases in equilibrium
Application of the Phase Rule 1+F=2+1; F=2 X and T T L A α α + β β A % B B T E 2 + F = 2+1 F = 1 either x or T 3+F = 2 + 1 F = 0
The Lever Rule T A L Alloy of Co at T 1 How much β? α α + β β T B T E C Co C β α A % B B w/o β = 100 (C ο -C α )/(C β -C α )
A series of free energy curves showing how a phase diagram is developed from free energy curves at different temperatures
Phase Diagram with an Intermediate Phase Note the T 0 curves
Order Parameters Another thermodynamic parameter (besides composition) exists in many alloys, which measures the degree of order within a phase. The order may be atomic ordering or it may be magnetic ordering etc. For perfect order, all A atoms at 000 and all B atoms at the center of cell
But this is not always the case. Sometimes only 72% of the A atoms may be at 000 and 28% of them in the center of the cell. The crystal is still ordered, but not completely so η = r F A 1 F A A
Two kinds of Order- Disorder Transitions First and Higher Order Sometimes called discontinuous and Continuous Transitions
Free Energy vs. Order Parameter for a First Order transition Note that away from equilibrium the transition can become continuous!
Free Energy vs. order parameter and composition. Note that at some compositions the reaction is continuous and others it is not.
Precipitation of an ordered phase showing regions of instability. On cooling, a disordered phase becomes unstable with respect to ordering below T i
Now let us look at the Free Energy of the b phase Up to now we assumed the phase was in equilibrium with the a phase. This means that we assumed the effects of surface energy and strain energy were neglected! G b = g b V If we include the surface energy: G b = g b V + s A where g b free energy unit vol. V volume A surface area s surface energy per unit area
For a spherical particle V = 4/3 p r 3 A = 4 p r 2 Thus, since G b = g b V + s A G b / V = g b + s 3/r For large r same as before For small r, the free energy is larger
Let us look at how this affects phase equilibria β coh G α W β inc A B C inc C coh
Thus smaller particles have a higher solubility The phase diagram can look like:
Now the same kind of thing happens for the addition of the strain energy term G b = g b V + s A + W V where W is the strain energy per unit volume This also raises the free energy curve of the precipitate phase A strained phase has greater solubility in the matrix than an unstrained one.
Kinetics How fast does a phase form This is often more important than what phase is the equilibrium one! I = K exp( -Dg * /kt) I is the rate of nucleation Dg * is barrier to nucleation (all precipitation reactions have a barrier to their initiation)
Let us look at the form of this equation rate = K exp( -Q/kT) as T increases, the rate increases or as Q decreases, the rate increases Q is called activation energy The equation is Arrhenius law
Typical plots are as shown below The slope is -Q/R Rate 1/T
Another important equation that has this form is the one for the temperature dependence of the diffusion coefficient D = D O exp( - Q RT D ) Here, Q D is the activation energy for diffusion which in substitutional solid solutions is usually the sum of the activation energies of the formation of vacancies and the motion of vacancies
Time-Temperature-Transformation Time T No transformation Transformation nearly complete The lower region follows Arrhenius law. Why not the upper?
Look at the nucleation rate equation I = K exp( -Dg * /kt) As the temperature approaches the transition temperature, Dg * gets larger and larger because it is equal to Dg * = 16 p s 3 / 3 Dg v 2 and Dg v goes to zero at the transition temperature
Time-Temperature-Transformation Time T No transformation Transformation nearly complete Importance of quench rate Knee of the curve, etc
X = 1 - exp( -kt n ) This equation is sometimes called the Johnson/Mehl/ Avrami equation
X = 1 - exp( -kt n ) Thus dx dt = nkt n-1 (1 - X) Also d 2 dt X 2 = (n - 1)nkt n-1 (1 - X) + knt n-1 (- dx ) dt Set equal to zero to find the inflection point t n = n - 1 nk or X = 1 - exp( - n - 1 ) n
Back to the Nucleation rate equation Dg * = 16 p s 3 / 3 Dg v 2 Note the importance of the surface energy term, s and the driving force term, Dg v Let us look at Dg v How do we obtain this value? From the Free Energy Curves!
Note that the value of Dg v is largest for the more stable phase. At first sight it looks like this means that the barrier to nucleation is smallest for the stable phase. BUT we must look at the surface energy term! This term comes in as a cubic. This is the secret to why less stable phases form faster than stable ones! It is almost always because the surface energy term of the less stable is smaller than that of the stable phase. Hence the value of the barrier to nucleation, Dg * is smaller!
The Al-Cu system exemplifies this very well
Different crystal structures of metastable phases in Al-Cu alloys. Note the similar lattice parameters of the metastable phases to the Al matrix. This gives small surface energy.
Note how the least stable phase (GP) is he one which forms first. This is because it is most like the Al matrix.
How do we get smaller values of the surface energy? Coherent interfaces usually have smaller values of s Ordered phases bases on the matrix phase can be coherent and therefore have smaller surface energies. In CuNiSn the phases L1 2 and DO 22 are crystallographic derivatives of the FCC solid solution. Thus they have small surface energies and can precipitate faster,
We have assumed up to now homogeneous nucleation. What happens if nucleation occurs on a special site? There is a decrease in the effective s term which increase the kinetics! In general the equation: G b = g b V + s A + W V shows that we can lower the barrier by either nucleating on a special site and thereby lower the surface energy or by nucleating on a site with strain energy and lower the overall strain energy of the system. Hence dislocations and grain boundaries act as special sites!
Such nucleation is called heterogeneous nucleation. Grain Boundary reactions are heterogeneous Spinodal Decomposition on the other hand is homogeneous, occurring within the grains at all sites.
Crystallography The crystallography of a phase greatly affects its thermodynamics and the kinetics of formation Example: d in Al-Li alloys d has the L1 2 structure. Since it is a crystallographic derivative of FCC, it can form coherently on all faces with the Al matrix. Thus its surface energy is low and hence it will nucleate very fast! The same is true for (CuNi) 3 Sn L1 2 phase
Coarsening Early Theories Deal With the Kinetics (r) 3 -(r o ) 3 = kt k D g X e
Note the proper way to plot this! 3 3 r - r o Time NOT r vs ln(t), because the term r o may be important
This equation was derived under severe assumptions: zero volume fraction of second phase ideal solution thermodynamics spherical precipitates no strain effects random distribution BUT it seems to hold well for many cases.
Free Energy Construction for the Driving Force for Coarsening
Note that the gradient that exists which gives rise to the diffusion of solute comes from the differences between the composition of phases due to their size: For two particles of size r 1 < r 2 we have X X 2 X1 X 2gVm = 1+ RTr 2gVm 1 = 1+ ( - RT r 1 1 1 r 2 ) Thus for r 1 < r 2 X 1 > X 2 so solute will flow from the smaller particle to the larger one!
Three parameters that control rate of coarsening D, the diffusion constant of solute in the matrix g the surface energy of the precipitates X e the solubility of the solute in the matrix
Homogenization During solidification the solute is not able to maintain equilibrium with respect to the matrix. This is due to the small values of diffusion in the solid state relative to diffusion in the liquid. There are two kinds of segregation: positive and negative. These are defined below
For this case k = X S /X L < 1.
If the solidus and solvus have the opposite slope, k > 1. T k =X S /X L X L X S
Homogenize Within One Hour? X Dt X X = (10 6mm -14 m s 2 3600s) Now if D is changed to 10 X = 60mm -12 To make D larger we go to higher T
1 D D T T if 1 D D T T if )) T 1 T 1 ( R Q exp( D D ) RT Q exp( D D ) RT Q exp( D D 2 1 2 1 2 1 2 1 T T 1 2 T T 1 2 2 1 T T 2 o T 1 o T > < < > - - = - = - =
Spinodal Decomposition Early History Gibbs On the Equilibrium of Heterogeneous Substances (1874) CuNiFe, Daniel & Lipson, 1943,4 Hillert, MIT thesis (1956) Cahn and Hilliard (1958)
The negative curvature signals instability with respect to decomposition!
Cahn s Linear Theory Predicts Microstructure 1. Periodicity 2. Alignment in Elastically anisotropic materials Anisotropy 1 Y = (c 2 where 11 + c 12 )B B = Ø Œ3 - º c 11 + 2( 2c 44 - c 11 c11 + 2c12 2 + c )(l m 12 2 + 2 m n 2 + l 2 n 2 ø ) œ ß For most cubic materials, <100> is minimum
Isotropic Spinodal Decomposition, FeCrCo alloy and simulation
Cu-Ni-Fe with 25 nm modulations
Grain Boundary, note that SD occurs right up to the boundary
Fe-Be Alloy having undergone SD
Disorder / Order Transformations
The DO 3 structure has a sites Sn and the other three sites random dispersion of Ni and Cu
Ground State (T=0) Diagrams for BCC Derivative Structures
Ground State (T=0) Diagrams for FCC Derivative Structures
Can Spinodal Decomposition and Ordering Occur in the Same Alloy? It would seem not. Regular solution thermodynamics allows the interaction constant to be plus or minus. When plus, clustering When minus, atomic ordering
But this implies only near neighbors are important. There are some ordered phases which have the same number of opposite first neighbors as the disordered parent phase. (e..g. CuPt) Thus we should include second neighbors as well Look at CsCl 8 opposite near neighbors 6 same next near neighbors
Most nn are opposite but most nnn are not the same. To get all nn opposite and all nnn the same we put all the dark atoms in one section of the system (cluster) and then order!
The Cu Ni Sn System Cellular Decomposition Grain boundary precipitation Spinodal Decomposition Atomic Ordering
Ternary Phase Diagrams Location of the alloy A 0.6 B 0.3 C 0.1 on a ternary phase diagram
The atomic percentages, r i are obtained from weight percentages w i by the relation: r 1 w1 100 = Ł A1 ł n wi 1 Ł Ai ł For Cu 15 w/o Ni and 8 w/o Sn we get Cu 16.7 a/o Ni and 4.4 a/o Sn For Cu 7.5w/o Ni and 5w/o Sn we get Cu 8.3 a/o Ni and 2.7a/o Sn atomic mass Cu 63.546 (A i ) Ni 58.7 Sn 118.71
FCC L1 2 Cu 3 Au L1 0 CuAu II
DO 22 DO 3
Isopleth for 15 w/o Ni CuNiSn Diagram (from Zhao et al.)
Look at the Discontinuous Transformation at 799 K (526 C) it takes 100 s to start at 680 K (407 C) it takes 1000 s to start 1 Q = K exp[ - 100 2 1 Q = K exp[ - 10000 2 Solving, we get Q = 42,050 cal What T for 1000 s? 1 1000 1 10000 1 799 1 ] 680 42050 = exp[ - ][ 2 ] 1 T - 1 ] 680 T = 735K or 462 C The Q is close to activation energy for grain boundary diffusion
Copper 7.5 w/o Ni and 5 w/o Sn, after Zhao et al.
Isopleth for 7.5 w/o Ni CuNiSn Diagram (from Zhao et al.)
L1 2 and DO 22 have similar Energies
The mean electron concentration, N e (electrons per atom) is an accurate indicator of which of the two phases will be stable. If Ne > 8.5 DO 22 if Ne < 8.5 L1 2 Co 3 Ti: (3x9+4)/4 < 8 L1 2 Ni 3 V: (3x10+5)/4 =8.75 DO 22 (see previous slide)
Look at (CuNi) 3 Sn Cu has 11 electrons Ni has 10 electrons Sn has 2 electrons For (Cu 1.5 Ni 1.5 )Sn (1.5 X (11 + 10) +2) / 4 = 33.5/4 = 8.375 Close to 8.5! Favors L1 2
To stabilize the L1 2 phase we want to decrease the number of electrons in the phase. More Ni than there presently or Add Co in place of Cu or Ni (9 e s) Add a Group IIIA in place of Sn
Recent Results from CMU on CuNiSn Alloys will be presented DSC Scans TEM Bright Field Dark Field Diffraction Also Results from BrushWellman will be discussed