Equilibria in Materials

009 fall Advanced Physical Metallurgy Phase Equilibria in Materials 10. 13. 009 Eun Soo Park Office: 33-316 Telephone: 880-71 Email: espark@snu.ac.kr Office hours: by an appointment 1

Contents for previous class - Binary Phase Diagrams: Reactions in the Solid State * Eutectoid reaction: α β + γ * Monotectoid reaction: α1 β+ α * Peritectoid reaction: α + β γ - Allotropy of the Components * SYSTEMS IN WHICH ONE PHASE IS IN EQUILIBRIUM WITH THE LIQUID PHASE * SYSTEMS IN WHICH TWO PHASES ARE IN EQUILIBRIUM WITH THE LIQUID PHASE * Metatectic reaction: β L + α Ex. Co-Os, Co-Re and Co-Ru

Contents for today s class Solidification: Liquid Solid < Nucleation > & Nucleation in Pure Metals Homogeneous Nucleation Heterogeneous Nucleation Nucleation of melting < Growth > Equilibrium Shape and Interface Structure on an Atomic Scale Growth of a pure solid 1) Continuous growth : Atomically rough or diffuse interface ) Lateral growth : Atomically flat of sharply defined interface Heat Flow and Interface Stability 3

Solidification: Liquid Solid 4 Fold Symmetric Dendrite Array 4

Solidification: Liquid Solid 4 Fold Anisotropic Surface Energy/ Fold Kinetics, Many Seeds 5

Solidification: Liquid - casting & welding - single crystal growth - directional solidification - rapid solidification 4.1. Nucleation in Pure Metals T m : G L = G S - Undercooling (supercooling) for nucleation: 50 K ~ 1 K <Types of nucleation> Solid - Homogeneous nucleation - Heterogeneous nucleation 6

4.1.1. Homogeneous Nucleation Δ H -T ΔS ΔG =0= Δ H-T m ΔS Driving force for solidification G L = H L TS L ΔG = G S = H S TS S L : ΔH = H L H S ΔS=Δ H/T m =L/T m (Latent heat) T = T m - ΔT ΔG =L-T(L/T m ) (LΔT)/T m ΔG = LΔT T m, 7

8 4.1.1. Homogeneous Nucleation L L S G G ) ( 1 + = SL SL L L S S A G G G γ + + = L S G G, SL SL S L S A G G G G G γ + = = Δ ) ( 1 SL r r G r G γ π π 3 4 3 4 + Δ = Δ : free energies per unit volume for spherical nuclei (isotropic) of radius : r

4.1.1. Homogeneous Nucleation Why r * is not defined by ΔG r = 0? r < r* : unstable (lower free E by reduce size) r > r* : stable (lower free E by increase size) r* : critical nucleus size r* dg=0 Unstable equilibrium Fig. The free energy change associated with homogeneous nucleation of a sphere of radius r. 9

Gibbs-Thompson Equation Δ G = ΔG of a spherical particle of radius, r ΔG of a supersaturated solute in liquid in equilibrium with a particle of radius, r r( s) 4 π r γ 3 4π r Δ Gr() l = ΔG 3 Equil. condition for open system Δμ should be the same. Δ μ = π γ = π Δ 8 r 4 r G r γ m r /mole or γ / per unit volume r ΔG = γ / r SL * r*: in (unstable) equilibrium with surrounding liquid r = γ SL ΔG 10

The creation of a critical nucleus ~ thermally activated process ΔG = LΔT T m r γ γ T 1 = = ΔG L ΔT * SL SL m of atomic cluster Fig. 4.5 The variation of r* and r max with undercooling ΔT The number of clusters with r * at < ΔT N is negligible. 11

Formation of Atomic Cluster At the T m, the liquid phase has a volume -4% greater than the solid. Fig. A two-dimensional representation of an instantaneous picture of the liquid structure. Many close-packed crystal-like clusters (shaded) are instantaneously formed. 1

Formation of Atomic Cluster When the free energy of the atomic cluster with radius r is by 4 3 Δ Gr = π r Δ G + 4 πr γsl, 3 how many atomic clusters of radius r would exist in the presence of the total number of atoms, n 0? A A A A L A A 1 3 4 m 1 exp ΔG n = n1 kt exp ΔG n3 = n kt exp ΔG n4 = n3 kt 1 3 3 4 M m 1 m exp ΔG m = nm 1 n n m kt Δ G +Δ G + L+ΔG = n kt 1 3 m 1 m exp 1 m n n m r = n ΔG kt 1 m exp 1 exp ΔGr = n0 kt 13

Formation of Atomic Cluster Compare the nucleation curves between small and large driving forces. Gibbs Free Energy large driving force small driving force r γ γ T 1 = = ΔG L ΔT * SL SL m r* r* radius 14

Formation of Atomic Cluster n o : total # of atoms. ΔG r : excess free energy associated with the cluster # of cluster of radius r n r = n 0 exp ΔG RT r - holds for T > T m or T < T m and r r* -n r exponentially decreases with ΔG r Ex. 1 mm 3 of copper at its melting point (no: 10 0 atoms) ~10 14 clusters of 0.3 nm radius (i.e. ~ 10 atoms) ~10 clusters of 0.6 nm radius (i.e. ~ 60 atoms) effectively a maximum cluster size, ~ 100 atoms ~ 10-8 clusters mm -3 or 1 cluster in ~ 10 7 mm 3 15

The creation of a critical nucleus ~ thermally activated process r γ γ T 1 = = ΔG L ΔT * SL SL m Fig. 4.5 The variation of r* and r max with undercooling ΔT The number of clusters with r * at < ΔT N is negligible. 16

4.1.. The homogeneous nucleation rate - kinetics How fast solid nuclei will appear in the liquid at a given undercooling? C 0 : atoms/unit volume C * : # of clusters with size of C* ( critical size ) C = C 0 exp( ΔG kt * hom ) clusters / m 3 The addition of one more atom to each of these clusters will convert them into stable nuclei. N hom = f 0 C o exp( ΔG kt * hom ) nuclei / m 3 s 3 16πγ T f o ~ 10 11 s -1 : frequency vibration frequency energy SL m ΔG * = of diffusion in liquid surface area 3 L ( T ) C o ~ 10 9 atoms/m 3 Δ N hom -3 1 * 1 cm s when ΔG ~ 78 kt 1 17

4.1.. The homogeneous nucleation rate - kinetics N hom f 0 C o exp{ A } ( ΔT ) where 3 16πγ SLT A = 3 L kt : insensitive to Temp m How do we define ΔT N? N hom ~ 1 Δ T critical value for detectable nucleation - critical supersaturation ratio - critical driving force - critical supercooling for most metals, ΔT N ~0. T m (i.e. ~00K) Fig. 4.6 The homogeneous nucleation rate as a function of undercooling T. T N is the critical undercooling for homogeneous nucleation. 18

Real behavior of nucleation Under suitable conditions, liquid nickel can be undercooled (or supercooled) to 50 K below T m (1453 o C) and held there indefinitely without any transformation occurring. In the refrigerator, however, water freezes even ~ 1 K below zero. In winter, we observe that water freezes ~ a few degrees below zero. Why this happens? What is the underlying physics? Which equation should we examine? 3 3 16πγ SL 16πγSLT m 1 Δ G* = = 3 ( Δ G ) 3 L ( Δ T ) N hom = f 0 C o exp( ΔG kt * hom ) Normally undercooling as large as 50 K are not observed. The nucleation of solid at undercooling of only ~ 1 K is common. 19

4.1.3. Heterogeneous nucleation 3 From 16πγ T SL m 1 ΔG * = 3L ( ΔT ) Nucleation becomes easy if by forming nucleus from mould wall. Fig. 4.7 Heterogeneous nucleation of spherical cap on a flat mould wall. γ SL γ = γ cosθ + γ ML SL ML SM SM cosθ = ( γ γ ) / γ SL Δ G = Δ G + A γ + A γ A γ het S v SL SL SM SM SM ML In terms of the wetting angle (θ ) and the cap radius (r) (Exercies 4.6) 4 3 Δ Ghet = π r Δ G + 4 πr γsl S( θ) 3 where S θ θ θ ( ) = ( + cos )(1 cos ) / 4 0

S(θ) has a numerical value 1 dependent only on θ (the shape of the nucleus) Δ G = S() θ ΔG * * het hom 3 γ SL 16πγ r * = and ΔG * = ΔG 3ΔG SL S ( θ ) S( θ ) θ = 10 S(θ) ~ 10-4 θ = 30 S(θ) ~ 0.0 θ = 90 S(θ) ~ 0.5 1

S(θ) has a numerical value 1 dependent only on θ (the shape of the nucleus) Δ G = S() θ ΔG * * het hom 3 γ SL 16πγ r * = and ΔG * = ΔG 3ΔG SL S ( θ ) Fig. 4.8 The excess free energy of solid clusters for homogeneous and heterogeneous nucleation. Note r* is independent of the nucleation site.

The Effect of ΔT on G* het & G* hom? Plot G* het & G* hom vs ΔT and N vs ΔT. N hom 1 cm -3 1 s Fig. 4.9 (a) ariation of G* with undercooling ( T) for homogeneous and heterogeneous nucleation. (b) The corresponding nucleation rates assuming the same critical value of G* 3

Barrier of Heterogeneous Nucleation ΔG * = 16 πγ 3 3 16πγ ( 3cosθ ( θ ) + SL SL S = 3ΔG 3ΔG 4 cos 3 θ ) θ Δ G = S( θ ) ΔG * * het hom A 3 * * 3cosθ + cos θ Δ Gsub = ΔGhomo 4 B A + A B 3 3cosθ + cos θ = = 4 S( θ ) How about the nucleation at the crevice or at the edge? 4

Nucleation Barrier at the crevice What would be the shape of nucleus and the nucleation barrier for the following conditions? contact angle = 90 groove angle = 60 1 6 ΔG * homo 3 * 4 ΔG * homo homo 1 ΔG 1 * 4 ΔG homo 5

How do we treat the non-spherical shape? A Substrate B Good Wetting A Substrate Bad Wetting B * * A Δ Gsub = ΔGhomo A + B Effect of good and bad wetting on substrate 6

Nucleation inside the crevice In both of the nucleation types considered so far it can be shown that 1 Δ G* = * Δ G * : volume of the critical nucleus (cap or sphere) Nucleation from cracks or crevices should be able to occur at very small undercoolings even when the wetting angle θ is relatively large. However, that for the crack to be effective the crack opening must be large enough to allow the solid to grow out without the radius of the solid/liquid inteface decreasing below r*. Inoculants ~ low values of θ low energy interface, fine grain size 7

4.1.4. Nucleation of melting Although nucleation during solidification usually requires some undercooling, melting invariably occurs at the equilibrium melting temperature even at relatively high rates of heating. Why? γ + γ < γ SL L S (commonly) In general, wetting angle = 0 No superheating required! 8

Growth We have learned nucleation. Once nucleated, the nucleus will grow. How does it grow? In order to see the details of growth, we need to know the structure of the surface on an atomic scale. How does it look like? Do liquid and solid nuclei differ in growth mechanism? 9

Water Drops Natural Minerals Topaz ( 황옥 ) Stibnite ( 휘안광 ) 30

Equilibrium Shape and Interface Structure on an Atomic Scale How do you like to call them? rough interface singular (smooth) interface What about the dependence of surface energy on crystal directions? isotropic γ anisotropic γ Compare the kinetic barrier for atomic attachment. Which has a low growth barrier? 31

Equilibrium Shape and Interface Structure on an Atomic Scale atomically-disordered atomically-flat Apply thermodynamics to this fact and derive more information. Entropy-dominant weak bonding energy stable at high T Enthalpy-dominant strong bonding energy stable at low T 3

Thermal Roughening singular (smooth) interface rough interface Heating up to the roughening transition. 33

Equilibrium shape of NaCl crystal Thermal Roughening 60 710 J.C. Heyraud, J.J. Metois, J. Crystal Growth, 84, 503 (1987) 34

Atomic iew Ideal Surfaces singular vicinal More realistic surface Realistic surfaces of crystals typically look like this at low temperature At sufficiently high temperature, the structure becomes atomically rough 35

4.. Growth of a pure solid Two types of solid-liquid interface 1. Continuous growth : Atomically rough or diffuse interface. Lateral growth : Atomically flat of sharply defined interface 36

4..1 Continuous growth The migration of a rough solid/liquid interface can be treated in a similar way to the migration of a random high angle grain boundary. - Driving force for solidification ΔG = L T m ΔT i L: latent heat of melting Δ T i : undercooling of the interface Liquid Solid - Net rate of solidification ν = k 1 ΔT i k 1 : properties of boundary mobility Reference (eq. 3.1) ν = M ΔG / m The solidification of metals is usually a diffusion controlled process. - Pure metal grow at a rate controlled by heat conduction. - Alloy grow at a rate controlled by solute diffusion. 37

4.. Lateral growth Materials with a high entropy of melting prefer to form atomically smooth, closed-packed interfaces. For this type of interface the minimum free energy also corresponds to the minimum internal energy, i.e. a minimum number of broken solid bonds. Two ways in which ledges and jogs (kinks) can be provided. 1) Surface (-D) nucleation ) Spiral growth Condition for Atomic Attachment Suppose the building unit (atom) has 6 bonds to be saturated 5 4 site ΔE / atom 3 1 1 3 4 5 4φ φ 0φ +φ +4φ stable stable stable : unstable unstable 38 kink

How many unsaturated bonds are there if they are epitaxial to the underneath atomic layer? L + 4 φ /atom + 4 φ /atom + 8 φ /4 atoms + φ /atom + 1 φ / 9 atoms + 16 φ /16 atoms Δf = -ktln(p/p e ) Δf 9Δf 16Δf Δf Δf 4 + φ / atom + 1 φ /atom 3 Draw the plot showing how the free energy varies with the number of atoms in the presence of supersaturation (driving force) for growth. Δf Δf Gibbs Free Energy no. of atoms - Dimensional Nucleation - if large # of atoms form a disc-shaped layer, - self-stabilized and continue to grow. - ΔT becomes large, r*. - v exp (-k /ΔT i ) 39

) Spiral growth: Growth by Screw Dislocation Crystals grown with a low supersaturation were always found to have a growth spirals on the growing surfaces. - addition of atoms to the ledge cause it to rotate around the axis of screw dislocation - If atoms add at an equal rate to all points along the step, the angular velocity of the step will be initially greatest nearest to the dislocation core. - the spiral tightens until it reaches a min radius of r* - v (growth rate) = k 3 (ΔT i ) 40

Growth by Screw Dislocation Burton, Cabrera and Frank (BCF, 1948) elaborated the spiral growth mechanism, assuming steps are atomically disordered... Their interpretation successfully explained the growth velocity of crystals as long as the assumption is valid 3) Growth from twin boundary - another permanent source of steps like spiral growth not monoatomic height ledge but macro ledge 41

Kinetic Roughening Rough interface - Ideal Growth diffusion-controlled dendritic growth Singular interface - Growth by Screw Dislocation Growth by -D Nucleation The growth rate of the singular interface cannot be higher than ideal growth rate. When the growth rate of the singular Interface is high enough, it follows the ideal growth rate like a rough interface. kinetic roughening 4