Heterogeneous nucleation ! undercooling of a few K sufficient lt surface Nheterogeneous > Nhomogeneous due to reduced nucleus/me
Solid Liquid
Fig. 5-4. Sketch of homogeneous and
heterogeneous Tm nucleation. The latter occurs at existing surfaces.
heterogeneous homogeneous (5.8) "Gc = f"Gc f = f (cos#) cos#: wetting angle Nucleation and Growth Topic 4 M.S Darwish MECH 636: Solidification Modelling
ned in eq. (5.8) as function of Fig. 5-5. (Left) Nucleation at a wall and wetting angle #. (Right) f defi elt wets the solid substrate. cos # for a flat substrate. As expected nucleation is facilitated if the m Heterogeneous nucleation facilitated by: - similar crystal structure (low misfitstrain) - chemical affinity - rough surface (reduced melt/nucleus surface) Grafting of melt with small particles ! fine grains. 5-4 Objectives
By the end of this lecture you should be able to: Explain the term homogeneous as applied to nucleation events Understand the concept of critical size and critical free energy Differentiate between unstable cluster (embryos) and stable nuclei
Derive expressions for (r*,N, ...) in terms of ∆Gv & ∆T. List typical heterogeneous nucleation sites for solidification Understand the term wetting or contact angle, θ Explain why the wetting angle is a measure of the efficiency of a particular nucleation site Write an expression relating critical volumes of heterogeneous and homogeneous nuclei. Introduction
During Solidification the atomic arrangement changes from a random or short-range order to a long range order or crystal structure.
Nucleation occurs when a small nucleus begins to form in the liquid, the nuclei then grows as atoms from the liquid are attached to it.
The crucial point is to understand it as a balance between the free energy available from the driving force, and the energy consumed in forming new interface. Once the rate of change of free energy becomes negative, then an embryo can grow. Energy Of Fusion
ΔGV = GL − GS = ΔHV − TΔS G Stable Stable solid liquid ⎛ V ⎞ liquid ΔHV = LV = ⎜ ⎟ hm !G ⎝ ρs ⎠ solid H h V h V GS Δ V m S m Tm = = Δ = ρ T !T G Δ S ρsΔS s m L T Tm V h V Temperature ΔG = h − T m V m ρ ρ T s s m ⎛ ⎞ ⎛ ⎞ hmV T V ΔT = ⎜1 − ⎟ = ⎜ ⎟ hm ρs ⎝ Tm ⎠ ⎝ ρs ⎠ Tm L ΔT = V Tm
Homogeneous Nucleation
2 Liquid Liquid ASL = 4πr γSL
4 3 VS = πr Solid 3
G G = G + !G 1 2 1 L S L G1 = (VS + VL )GV G2 = VSGV + VLGV + ASLγ SL
L ΔT ΔG = V V T m ΔG = G − G 2 1 -ve +ve 4 S L 3 2 = V G − G + A γ ΔG = − πr ΔGV + 4πr γ SL S ( V V ) SL SL 3 1. When r is smaller than some r* an increase = −VSΔGV + ASLγ SL in r leads to an increase of ∆G -> unstable 2. When r is larger than some r* an increase in r leads to a decrease of ∆G -> stable Critical radius
Not at ∆G=0!!!
Differentiate to find the stationary point (at which interfacial energy ! r2 the rate of change of free energy turns negative). !G
d(ΔG) = 0 dr
2 !G* ∗ ∗ 0 −4π r ΔGV + 8πr γ = 0 r ( ) * r !G From this we find the critical radius and critical r
free energy. Volume free-energy !r3"T ∗ 2γ ⎛ 2γ T ⎞ 1 r = SL = ⎜ SL m ⎟ ΔGV ⎝ LV ⎠ ΔT 4 3 2 ΔG = − πr ΔGV + 4πr γ SL 3 ⎛ 3 2 ⎞ 3 ∗ 16πγ SL 16πγ SLTm 1 ΔG = 2 = ⎜ 2 ⎟ 2 3ΔG ⎝ 3LV ⎠ ΔT V ( )
Cluster and Nuclei
interfacial energy ! r2 !G if r Unstable solid particles with r * Volume free-energy Stable solid particles with r>r are !r3"T referred to as nuclei Since ∆G = 0 when r = r* the critical nuclei is effectively in (unstable) equilibrium with the surrounding liquid Effect of Undercooling interfacial energy ! r2 At r* the solid sphere is at equilibrium with its !G G Stable Stable surrounding thus the solid sphere and the liquid solid liquid have the same free energy liquid " " r2 > r1 2γ !G SL 2" ΔG = SL V ∗ !G* # solid r1 r 0 2" SL r * # r! r2 GS How r* and ∆G* decrease with undercooling ∆T !Gr! !T GL ! Volume free-energy !r3"T 500 T Tm Temperature Nuclei are stable , ˚C 300 T in this region ! 2 ∗ ∗ −4π(r ) ΔGV + 8πr γ = 0 Embryos form in this 100 region and may redissolve 5x10-7 10-6 1.5x10-6 Critical radius of particle, r* (cm) Variation of r* and rmax with ∆T • Although we now know the critical values for an embryo to become a nucleus, we do not know the rate at which nuclei will appear in a real system. • To estimate the nucleation rate we need to know the population density of embryos of the critical size and the rate at which such embryos are formed. • The population (concentration) of critical embryos is r r* given by ΔG − r kT rmax nr = noe 0 !TN !T k is the Boltzmann factor, no is the total number of atoms in the system ∆Gr is the excess of free energy associated with the cluster Homogeneous Nucleation Rate Nhom taking a ∆G equal to ∆G*, then the concentration of clusters to reach the critical size can be written as: ΔG∗ − hom ∗ kT 3 C = Coe clusters/m !TN !T The addition of one more atom to each of these clusters would convert them into stable nuclei r r* If this happens with a frequency fo, ∗ ΔGhom − r kT 3 max Nhom = foCoe nuclei/m A − 3 2 T 2 (Δ ) 3 16πγ SLTm N f C e nuclei/m 0 !T !T hom = o o A = 2 N 3LV kT The effect of undercooling on the nucleation rate is drastic, because of the non-linear relation between the two quantities as is shown in the plot Heterogeneous Nucleation 3 ⎛ 3 2 ⎞ ∗ 16πγ SL 16πγ SLTm 1 ΔG = 2 = ⎜ 2 ⎟ 2 3ΔGV ⎝ 3LV ⎠ (ΔT) it is clear that for nucleation to be facilitated the interfacial energy term should be reduced Liquid " SL Solid " " ML Nucleating agent ! " SM ! ! ! γ ML = γ SM + γ SL cosθ (γ ML − γ SM ) cosθ = γ SL Heterogeneous Nucleation L S L G1 = (VS + VL )GV + (AML ′ + AML )γ ML G2 = VSGV + VLGV + AML ′ γ ML + ASLγ SL + ASM γ SM Liquid Liquid " SL Solid " ML " " ML Nucleating agent " SM Nucleating agent ! " SM ! ! ! ! ! ΔG = G2 − G1 = −VSΔGV + ASLγ SL + ASM γ SM − AMLγ ML ⎧ 4 3 ⎫ ΔGhet = ⎨− πr ΔGv + 4πγ SL ⎬ S(θ) ⎩ 3 ⎭ 2 ∆G (2 + cosθ)(1− cosθ) hom S(θ) = <1 4 Critical r and ∆G 3 ∗ 2γ SL 16πγ r ∗ SL !G = ΔG = 2 S(θ) ΔGV 3ΔGV !G* # # "Ghet "Ghom # "Ghom "G# 0 het r r* ! ! Critical value !G! ! r for nucleation !Gr !T N Nhet Nhom θ =10˚→ S(θ) =10−4 θ = 30˚→ S(θ) = 0.02 !T model does not work for θ = 0˚ Heterogeneous Nucleation Rate ΔG∗ Mould walls not flat − het ∗ kT n = n1e number of atoms in contact with nucleating agent surface Critical radius for solid ΔG∗ − het kT 3 Nhet = f1C1e nuclei /m number of atoms in contact with nucleating agent surface per unit volume Nucleation in cracks occur with very little undercooling for cracks to be effective the crack Exercise show that opening should be large enough to allow 1 the solid to grow out without the radius ΔG∗ = V ∗ΔG 2 v of the solid/liquid interface decreasing below r* Nucleation of Melting While nucleation during solidification requires some undercooling, melting invariably occurs at the equilibrium temperature even at relatively high rates of heating. this is due to the relative free energies of the solid/vapour, solid/liquid and liquid/vapour interfaces. It is always found that γ SL + γ LV < γ SV Therefore the wetting angle θ = 0 and no superheating is required for nucleation of the liquid Growth of a Pure Solid Solid Liquid Solid Liquid T T T m x m In a pure metal solidification is v v controlled by the rate at which the latent heat of solidification Solid Liquid Solid Liquid can be conducted away from the solid/liquid interface. Heat Heat Solid Liquid Solid Liquid Tm Tm dTS dTL k = k + vL Solid Liquid Solid Liquid S dx L dx V Development of Thermal Dendrites dT dT k S = k L + vL S dx L dx V dT dTL ΔTc S ≈ 0 ≈ dx dx r Tm !Tr !To TS dTL 1 kL ΔTc v ≈ −kL ≈ − !Tc dx L L r V V TL,far 2γTm ∗ 2γT Solid Liquid m r ΔTr = ⇒ r = LV r LV ΔTr x k 1⎛ r∗ ⎞ v ≈ L ⎜1 − ⎟ LV r ⎝ r ⎠ r = 2r∗ Alloy Solidification Limited Diffusion in Solid and Liquid T k=cS/cL Solid Liquid cL(x) cS c cL 1 T1 T2 T3 co critical DL/v gradient T c TL(x) T (c ) A kco co cmax co/k ceut B 1 o TL (cL) T3(c1) constitutional undercooling x Solid Liquid xL 0 Constitutional undercooling and solidification morphology Constitutional undercooling and solidification morphology Fig. 5-9. How con- stitutional under- cooling affects solidification mor- phology. Crucial parameters: • Local solidification rate: if low, solute has time to diffusFei g. 5-9. How con- away from interface into bulk lsitqituuitdio nal under- ! planar growth cooling affects • grad T: > critical value ! no constitutional undercoolinsgolidification mor- (cf. Fig. 5-9). phology. Steep grad T + low SL interface velocity ! planar growth (e.g. growth of Si single crystals) Planar growth " cell growth " dendrite growth Crucial pa#rameters: # # • Locanol usndoelrciodoilifnigcationm ordaetraete: unidfe rlcooowlin,g solutest rhonag sun tdiemrcoeol intgo diffuse aw5-7ay from interface into bulk liquid ! planar growth • grad T: > critical value ! no constitutional undercooling (cf. Fig. 5-9). Steep grad T + low SL interface velocity ! planar growth (e.g. growth of Si single crystals) Planar growth " cell growth " dendrite growth # # # no undercooling moderate undercooling strong undercooling 5-7 Summary • By considering the balance between the release of free energy by transformation and the cost of creating new interface, the critical free energy for nucleation and the critical size of the nucleus can be derived. • The exponential dependence of nucleation rate on undercooling means that, in effect, no nucleation will be observed until a minimum undercooling is achieved. • The undercooling required for nucleation is increased by volume changes on transformation, but decreased by the availability of heterogeneous nucleation sites. Units • Consider the units of the various quantities that we have examined. • For driving force, the units are either Joules/mole (∆Gm) or Joules/m3 (∆Gv); dimensions = energy/mole, energy/volume. • For interfacial energy, the units are Joules/m2; dimensions = energy/area. • For critical radius, the units are m (or nm, to choose a more practical unit); dimensions = length. • For nucleation rate, the units are number/m3/s; dimensions are number/volume/ time. • For critical free energy, the units are Joules; dimensions are energy. What is less obvious is how to scale the energy against thermal energy. When one calculates a value for ∆G*, the values turn out to be of the order of 10-19J, or 1eV. This is reasonable because we are calculating the energy associated with an individual cluster or embryonic nucleus, I.e. energies at the scale of atoms. Therefore the appropriate thermal energy is kT (not RT). • For the activation energy (enthalpy) of diffusion, in the equation for nucleation rate, the units depend on the source of the information. If the activation energy for diffusion is specified in Joules/mole, then the appropriate thermal energy is RT, for example. Latent Heat of fusion (J/ Typical undercooling for Metal Freezing Temp. (°C) Surface enrgy (J/cm2) cm3) Hv Ga 30 488 56 10-7 76 Bi 271 543 54 10-7 90 Pb 327 237 33 10-7 80 Ag 962 965 126 10-7 250 Cu 1085 1628 177 10-7 236 Ni 1453 2756 255 10-7 480 Fe 1538 1737 204 10-7 420 N2O 0 40