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.

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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.

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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 -4p/3 r3 ?GV 4pr2 g.
• Differentiate to find the stationary point (at
  which the rate of change of free energy turns
  negative).

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Critical radius, free energy
Not at ?Gr0!!!

• d(?Gr) 0 -4p/ r2 ?G 8prg.
• From this we find the critical radius and
  critical free energy. r 2g/?GV ?G
  16pg3/3?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 -4p/3 r3 (?GV - ?GS) 4pr2
  g. d(?Gr) 0 -4p r2 ?G 8prg. r 2g /
  (?GV - ?GS) ?G 16pg3 / 3(?GV - ?GS)2.

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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!

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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.

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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.

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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, 110bcc in ferrite and
  111fcc in austenite will favor plate-like
  growth of ferrite. For hcp Ti precipitating in
  bcc Ti, a good atomic match is found between the
  0001hcp (low temperature allotrope) and the
  110bcc (high temperature allotrope).
• Note coherent vs. incoherent interfaces are to
  be discussed in more detail later.

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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 PE), and w is of the
  same order as the atomic jump frequency f w
  exp -(?Gbulk/kT).

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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.

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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.

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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.

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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.

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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.

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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.

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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
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Griffith Eq., contd.

• In order to compute the energy balance as the
  crack lengthens we need, (a) the elastic
  energy released ps2c2t/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 -ps2c2t/E 4ctg

low stress
Courtney
high stress
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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.

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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.

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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.

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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, phrg, exceeds
  this maximum pressure, the film will burst and
  the water will flow out of the hole.

h
d
p
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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.

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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
  factor modifies the critical free energy ?G
  16pgab3/3(?GV-?GS)2 S(q) 16pgab3/3(?GV-?GS)2
  0.5(2 cosq)(1 - cosq)2

gab
a
Grainboundary in alpha
q
gaa
b
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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.

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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 PE.

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Vacancies

• The role of vacancies is discussed in PE 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!).

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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.

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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 0C, 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 -40C and saturated air, or
  relative humidity of several hundred percent.
• http//cimss.ssec.wisc.edu/wxwise/class/dewfog.htm
  l

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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.