Microstructure-Properties: II Nucleation Rates

1. Microstructure-Properties: IINucleation Rates 27-302 Lecture 2 22 October Fall, 2002 Prof. A. D. Rollett

2. Materials Tetrahedron Processing Performance Properties Microstructure

3. Objective • The objective of this lecture is to describe the nucleation rate for diffusive phase transformations in terms of the driving force, the interfacial energy and the elastic properties of the phases.

4. References • Phase transformations in metals and alloys, D.A. Porter, & K.E. Easterling, Chapman & Hall. • Materials Principles & Practice, Butterworth Heinemann, Edited by C. Newey & G. Weaver.

5. Nucleation Rate • The rate at which nucleation of a new phase occurs is critical to the prediction of phase transformation behavior. • We will contrast homogeneous nucleation (extremely rare!) with heterogeneous nucleation (typical) rates. • Why study homogeneous nucleation? Useful foundation and simplest to understand. • Bottom line: the quickest transformation wins!

6. Thermodynamics of nucleation • How should we understand nucleation? • The crucial point is to understand it as a balance between the free energy available from the driving force, and the energy consumed in creating new interface (between parent and product phases). Once the rate of change of free energy becomes negative, then an embryo can grow. • Parallel to the Griffith analysis: once the rate of (free) energy change becomes negative with crack length increase, then the crack can grow without limit.

7. Nucleation paths • It is important to remember that the actual outcome is always the process that leads most rapidly to the change for which a (thermodynamic) driving force exists. • Different transformations are observed for the same material because different types of transformation occur most rapidly for different undercoolings. In carbon steels, for example, austenite decomposes to pearlite most rapidly for small undercoolings but to martensite for large undercoolings. • Anisotropy in the interfacial energy forces growing grains to adopt anisotropic shapes in order to minimize high energy orientations of the interface. • Anisotropy in growth rates has a similar effect. • Heterogeneous nucleation on surfaces, pre-existing interfaces (grain boundaries), dislocations etc. is very important. • Elastic energy plays a major role in constraining nucleation.

8. Homogeneous Nucleation • Assume that the new, product phase appears as spherical particles. • Free energy released by transformation is proportional to the volume. • Free energy consumed by creation of interface is proportional to the surface area of particle and the interfacial energy, g. • Net change in free energy per particle, ∆Gr:∆Gr = -4π/3 r3 ∆GV + 4πr2g. • Differentiate to find the stationary point (at which the rate of change of free energy turns negative).

9. Critical radius, free energy Not at ∆Gr=0!!! • d(∆Gr) = 0 = -4π/ r*2 ∆G* + 8πr*g. • From this we find the critical radius and critical free energy.r* = 2g/∆GV∆G* = 16πg3/3∆GV2 • Crucial difference from solidification: the role of elastic energy!

10. Elastic energy • Why does elastic energy play such an important role in solid state phase transformations? • Volume changes on transformation of order a few % are typical. Elastic energy is symmetric: net (hydrostatic) tension or compression leads to an increase in elastic energy. This elastic energy cost for creation of a new phase, ∆GS (= Ee2/2), must be subtracted in proportion to the volume of new phase.∆Gr = -4π/3 r3 (∆GV - ∆GS) + 4πr2g.d(∆Gr) = 0 = -4π r2 ∆G* + 8πr*g.r* = 2g / (∆GV - ∆GS);∆G* = 16πg3 / 3(∆GV - ∆GS)2.

11. Homogeneous Nucleation: examples • Two examples of homogeneous nucleation in the solid state are known.1) Cu with 1-3% Co can be heat treated to precipitate Co homogeneously. We will examine this case in a homework aimed at predicting TTT diagrams.2) Ni superalloys will precipitate Ni3Al homogeneously at small undercoolings because of the small lattice misfit and small interfacial energy. • Why only these cases? Small interfacial energy, and small elastic energy difference. • Everything else: heterogeneous!

12. Elastic Anisotropy • Remember that most crystalline solids are elastically anisotropic: this means that the shape of a new phase is likely to be anisotropic. • If either the parent or the product phase is more compliant in a particular direction, larger dimensions parallel to this direction will be favored over stiffer directions. This is offset by the interfacial energy term which must increase as the surface-to-volume ratio increases. • Example: Guinier-Preston zones in the Al-Cu system, which are platelets on {100}.

13. Interfacial Energy • Even in solidification, the anisotropy of the interfacial energy matters. The energy of the solid-liquid interface varies depending on which crystallographic surface is involved. {111} surfaces tend to have the lowest energy in fcc metals. • In solid state precipitation, the anisotropy of the interface matters even more! This is because there are two crystalline surfaces involved in the interface. If the crystal structures are different (often the case) then low energy interfaces require good atomic matching between the two planes. Sometimes this results from combining close-packed interfaces.

14. Interfacial Anisotropy • If the interfacial energy varies with direction, i.e. is anisotropic, then the shape of the nucleus is also likely to be anisotropic. Low energy facets will be favored over high energy ones. • Example: Widmanstätten precipitation in steels, titanium alloys. Good atomic matching (coherent interface) between, say, {110}bcc in ferrite and {111}fcc in austenite will favor plate-like growth of ferrite. For hcp Ti precipitating in bcc Ti, a good atomic match is found between the {0001}hcp (low temperature allotrope) and the {110}bcc (high temperature allotrope). • Note: coherent vs. incoherent interfaces are to be discussed in more detail later.

15. Nucleation rate • 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. • Population (concentration) of critical embryos, C*, is given by a Boltzmann factor, where C0 is the number of atoms per unit volume:C* = C0 exp -(∆G*/kT) • The rate at which a critical embryo is formed, f, depends on the migration of atoms, i.e. diffusion, which is again given by a Boltzmann factor, where ∆Gbulk is the activation energy for (bulk) diffusion (∆Gm in P&E), and w is of the same order as the atomic jump frequency:f = w exp -(∆Gbulk/kT).

16. Nucleation rates, contd. • Based on this approach, we can now understand the extremely strong dependence of nucleation rate on undercooling. • Note that the net effect of elastic energy is to offset (decrease) the equilibrium transformation temperature.

17. Effect of undercooling • The effect of undercooling on the nucleation rate is drastic, because of the non-linear relation between the two quantities. • By incorporating the previous expression into the nucleation rate we obtain the following:C* = C0 exp -(∆G*/kT) = • Finally the nucleationrate is the productof C* and f: N = f C* • Note that N has 2 exponential terms in it.

18. Nucleation Rate • The combined equations are as follows. • The nucleation rate is the product of C* and f. Note that the product of w and C0 is a large number because w is of the order of the atomic vibration frequency, and C0 is the number of atoms per unit volume.

19. 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.

20. Analogies: climbing over a mountain pass • The simplest analogy for nucleation is that of climbing over a mountain pass on a bicycle. As long as you are climbing up the hill, you have to pedal and expend energy. Once you get to the top, you can then coast down the other side. It does not matter how high the pass is, you always get to coast once you reach the top.

21. Analogies: Griffith Eq. • The second analogy is with the Griffith equation. • The Griffith equation is the result of examining the rate at which energy is added to and subtracted from a system. Energy is added to the system (energy cost) in order to create new crack surface area. Energy is removed from the system (energy released) as the elastic strain energy is decreased. • The crack will propagate catastrophically when the rate of energy input becomes negative, i.e. as soon as energy starts to be released. • The tensile stress on the body plays the role of the driving force for crack advance.

22. Griffith Eq., contd. • The Griffith equation is an energy-based criterion for (catastrophic) crack propagation. • As a crack advances, so a region of radius ~ crack length unloads. The elastic energy of this region is now available to do work, i.e. create new surface area. • The elastic energy of the body goes down as the square of the crack length whereas the energy consumed in creating a longer crack is proportional (linear) to the crack length. Note that the relevant length, c, is the half-length of en elliptical crack. Therefore the derivation is for a crack that is fully contained within a material (even though many practical examples involve a crack that starts at a free surface). c s c dc

23. Griffith Eq., contd. • In order to compute the energy balance as the crack lengthens we need, (a) the elastic energy released = πs2c2t/Eand (b) the energy consumed = 4ctg. • To understand the balance point, consider the energy of the system, UTOT, as a function of crack length (similar to the nucleation criterion): UTOT = -πs2c2t/E + 4ctg low stress [Courtney] high stress

24. Griffith Eq., contd. • To find the “peak of the curve”, apply standard calculus and find the value of the crack length for which the derivative is zero. • Notice that the “break-even point” for getting more elastic energy back than you have to put into the crack surface is stress dependent: higher stress, shorter crack required.

25. Analogies, continued: a dam burst • A good analogy for driving pressure is water pressure. The higher a water column, the more pressure there is at the bottom to push water out (or drive it through pipes). • An analogy for nucleation is a dam failure. Consider a reservoir that is very full (as in the Johnstown flood, May 31st, 1889). The weight of water is equivalent to a (huge!) driving pressure for release of the water. If the dam has any weak points where water can seep through, the seepage itself tends to weaken the dam. The failure of a dam is, in some sense, auto-catalytic. Once seepage starts, it weakens the seepage sites; the flow rate increases, thereby accelerating the damage. Eventually, the damage propagates and the dam fails catastrophically.

26. A more precise analogy? • Consider the following situation. A box with perforations in a line on one side is filled with water. Obviously, the water flows out through the perforations. If, however, a thin plastic sheet (cling-film, for example) is used to line the perforated side of the box. As the box is filled up with water, the pressure on each perforation will increase. If the plastic film is flexible (think of a balloon), it will bulge out. If it bulges too far, however, it will burst (again, like a balloon). Once this happens, the water will run out and the pressure is released. The height of the water column that is required to burst the plastic is equivalent to the driving pressure required for nucleation to occur.

27. Water Column with Hole • The pressure of the water in the column is resisted by the surface tension, s, of the plastic film lining the column. The maximum pressure that the film can withstand is 2s/d. If the hydrostatic head pressure, p=hrg, exceeds this maximum pressure, the film will burst and the water will flow out of the hole. h d p

28. Heterogeneous Nucleation • Heterogeneous nucleation must occur on some substrate: grain boundaries triple junctions dislocations (existing) second phase particles • Consider a grain boundary: why is it effective? Answer: by forming on a grain boundary, an embryo can offset its “cost” in interfacial energy by eliminating some grain boundary area.

29. Grain boundary nucleation • The semi-angle, q = cos-1gaa/2gab • As for solidification, the radius of the spherical caps depends only on the interfacial energy, so:r* = 2gab/(∆GV-∆GS)but a shape factormodifies the critical free energy:∆G* = 16πgab3/3(∆GV-∆GS)2S(q) = 16πgab3/3(∆GV-∆GS)20.5(2 + cosq)(1 - cosq)2 gab a Grainboundary in alpha q gaa b

30. Other heterogeneous sites • Other sites for heterogeneous nucleation have been listed. • For the same contact angle, grain corners (quadruple points) are more effective than grain edges (triple lines), which are more effective than grain boundaries.

31. Heterogeneous nucleation rate • The rate of heterogeneous nucleation, Nhet, is described by a very similar equation as previously described for homogeneous nucleation, Nhomo. The critical difference is in the critical free energy, ∆G*, and the density of sites, C1. • Homogeneous:Nhomo = w exp -(∆Gbulk/kT)C0 exp -(∆Ghomo*/kT) • Heterogeneous:Nhet = w exp -(∆Gbulk/kT)C1 exp -(∆Ghet*/kT) • For grain boundary nucleation, for example, the ratio of site densities, C1/C0 = d/D, where D is the grain size, and d is the boundary thickness: see Table 5.1 in P&E.

32. Vacancies • The role of vacancies is discussed in P&E and, essentially, dismissed in terms of heterogeneous nucleation sites. • There is, however, an interesting and practical aspect of vacancies in nucleation that relates to diffusion. • Normally when you cool a material, the vacancy concentration changes only rather slowly from the high temperature value: quenching essentially preserves it. This quenched-in vacancy concentration can allow nucleation to occur at lower-than-expected temperatures. • Up-quenching can eliminate the excess vacancy concentration, and block nucleation (practiced in Al alloys!).

33. 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.

34. Another view of nucleation: clouds! • CONDENSATION: DEW, FOG AND CLOUDS • At night, the surface and the air near the surface cools radiatively. Air that comes in contact with the cold surface cools by conduction. If the air and the surface cools to the dew point, the temperature at which saturation occurs, dew forms. If the air temperature drops below freezing, the dew will freeze becoming frozen dew. If the dew point is less that 0°C, then we refer to it as the frost point, and frost forms by deposition. Dew, frozen dew and frost form in shallow layers near the surface. They are a form of precipitation, a moisture sink of the atmosphere. • Condensing water to form a cloud is not quite simple. • HOMOGENEOUS NUCLEATION • Homogeneous nucleation occurs when the water vapor molecules condense and form a cloud droplet. To do this requires an environmental temperature of -40°C and saturated air, or relative humidity of several hundred percent. • http://cimss.ssec.wisc.edu/wxwise/class/dewfog.html

35. Clouds, contd. • HETEROGENEOUS NUCLEATION • It turns out that saturating the air is not always enough to form a cloud. The water vapor molecules need a site to condense on. This site is called a Condensation Nuclei and the process referred to as heterogeneous nucleation. Cloud condensation nuclei (CCN) are about 1 micron in size. There are two types of condensation nuclei: • HYGROSCOPIC - water-seeking • HYDROPHOBIC - water-repelling • Most particles are released into the atmosphere near the ground, this is where largest concentrations of CCN are. There are generally more CCN over land than over oceans. CCNs are one type of aerosol in the atmosphere. • HAZE: haze contains a large concentration of nuclei - dust or salt particles. • Dry haze • Wet haze - condensation can occur at relative humidity of 80%. • As the relative humidity increases and approaches 100% the haze particles grow larger and condensation beings on the less-active nuclei. When the visibility lowers to less than 1 km (.62 mi.) and the air contains water droplets we have a FOG.