Chapter 6
Solidification
Almost all processing of technologically important metals includes a step
where liquid material is cooled to form a solid. It is very common, for example, for the components of metallurgical alloys to be mixed in the liquid state
so that, upon solidification, an alloy with the desired composition is formed.
In order to form a solid from an undercooled melt we need formation of crystalline nuclei and growth of these nuclei to form a solid. Transformations
which occur by these two processes, nucleation and growth, are examples of
first order phase transitions. Before beginning our discussion of nucleation
and growth of pure materials, we make some general comments about phase
transitions.
The condition of thermodynamic equilibrium implies minimization of
some free energy. For processes at constant pressure P , we minimize the
Gibbs free energy, G = H − T S, while for processes at constant volume V
we minimize the Helmholtz free energy F = U − T S, where H and U are the
enthalpy and internal energy respectively and are related by H = U + P V ,
and S is the entropy and T the temperature. For most solid-liquid or solidsolid transitions it is usually more practical to work at constant pressure,
so G is the appropriate thermodynamic potential. However, at atmospheric
pressure, the changes in the P V term for condensed phase reactions are usually small compared to changes in the other terms, so either F or G can be
used.
A first order phase transition involves two distinct phases which are physically separated by a sharp1 interface, across which there is usually a sym1
It is often impossible to distinguish the phase of the atoms right at the interface, so
157
158
CHAPTER 6. SOLIDIFICATION
metry change. Each phase is represented by a free energy which is a function
of temperature, pressure (or volume), and composition. In equilibrium, the
system will configure itself in the phase which has the lower free energy.
Points where the free energies of the two phases cross are transition points.
For example, consider the free energy as a function of temperature for solid
and liquid phases shown in Fig. 6.1. The free energies of the liquid and solid
cross at the melting temperature Tm . In equilibrium, the free energy of the
system is a continuous curve following the free energy of the liquid above Tm
and that of the solid below Tm . There is a slope change in the free energy at
the Tm , so entropy given by:
∂F ∂G S=−
=−
∂T P
∂T V
will have a discontinuity at the transition point:
∆SF = S
liquid
−S
solid
T =Tm
=
∂Gsolid ∂Gliquid −
∂T P
∂T P
T =Tm
A liquid which is cooled infinitely slowly will transform to the solid phase
at the melting temperature. However real transitions occur at finite rates,
and thus involve excursions from equilibrium which provide the driving force
for the transition. A liquid cooled below Tm will have a free energy higher
than that of the solid by an amount:
∆G = Gliquid − Gsolid
Since solidification will decrease the free energy of the system, ∆G is said to
be the driving force for the transformation. This is exactly analogous to the
driving force for diffusion being provided by gradients in external potentials
or gradients in the non-ideal part of the chemical potential (Sec. 4.3).
We can estimate this driving force with a very simple expression if we are
willing to assume that there is no difference in the heat capacities of the two
phases. Since G = H − T S we find for the temperature derivatives:
∂G ∂H ∂S = −S =
−T
−S
∂T P
∂T P
∂T P
even sharp interfaces may be spread out over a few atomic distances.
159
G
∆G
solid
liquid
TM
T
liquid
S
∆S F
solid
TM
T
Figure 6.1: Schematic of free energy versus temperature for solid and liquid
phases.
160
So that:
CHAPTER 6. SOLIDIFICATION
∂H ∂S ∂ 2 G =T
= −T
CP ≡
∂T P
∂T P
∂T 2 P
where CP is the heat capacity at constant pressure, which is the change
in enthalpy per change in temperature. We can then find the difference in
enthalpy between the liquid and solid states by integrating from the transition
temperature:
∆H = H
liquid
−H
solid
T
= ∆HF +
Tm
∆CP dT
where ∆HF is the heat of fusion which is the heat released by solidification
given by:
∆HF = H liquid − H solid T =Tm
and ∆CP = CPliquid − CPsolid . Similarly we can find the entropy difference:
∆S = S
liquid
−S
solid
T
= ∆SF +
Tm
∆CP
dT
T
Since ∆G = ∆H − T ∆S, and at T = Tm ∆G = 0 we see that:
∆HF = Tm ∆SF
(6.1)
If we now assume that ∆CP = 0, that is the curvature of the free energies of
the two phases are zero or equal, then we see that:
∆G = ∆HF − T ∆SF
Inserting Eqn. 6.1 we find:
∆HF
Tm
Tm − T
= ∆HF
Tm
∆G = ∆HF − T
(6.2)
Equation 6.2 is known as the Turnbull extrapolation. It is valid for most
practical solidification situations, except in cases of extremely rapid cooling
where large undercoolings can be present.
In contrast to the behavior of a first order phase transition, a second
order phase transition has both a continuous free energy and a continuous
161
entropy (free energy slope). The transition usually occurs over a temperature
range of several degrees rather than at one point. The free energy is usually
represented as a single continuous curve, while the entropy has separate
curves for the two phases. There is a discontinuity in slope of S, which
results in a discontinuity in heat capacity, as shown in Fig. 6.2.
Some structural transitions are second order phase transitions, for example the FCC to FCT transition in BaTi03 . Order-disorder transitions,
for example the β to β transition in CuZn, can also be second order phase
transitions.
G
solid
liquid
TM
T
liquid
S
solid
TM
T
liquid
cP
solid
TM
T
Figure 6.2: Schematic of free energy, entropy and heat capacity as a function
of temperature for a second order phase transition.
162
6.1
CHAPTER 6. SOLIDIFICATION
Nucleation
We now return to solidification of a pure undercooled melt, which, of course,
is an example of a first order phase transition. The thermodynamic driving force provided by the undercooling, must accomplish two processes; nucleation and growth. In considering nucleation, we first establish that the
surface or interface energy will result in an energy barrier which destabilizes
nuclei under a certain critical size. We then examine the population of nuclei
and the rate at which critical nuclei form to find the nucleation rate.
6.1.1
Nucleation Barrier
We consider the free energy change associated with formation of a solid
particle from an undercooled liquid, as shown schematically in Fig. 6.3. There
are two terms, the first is the negative, volume term associated with the bulk
condensation energy, and the second is the positive, surface term associated
with formation of the new interfacial area between the solid and the liquid.
Thus, if ∆GT is the total free energy change associated with formation of a
solid particle, we can write:
∆GT = −
V
∆gA + AγSL
VA
where V is the volume of the solid particle, A is the solid particle surface area,
VA is the atomic volume, ∆gA is the free energy change per atom associated
with the liquid-solid transformation, and γSL is the surface energy. The free
energy change per atom ∆gA is related to the free energy change per volume
∆GV through:
∆gA = ∆GV VA
If we take N to be the number of atoms in the solid particle, we can write:
∆GT = −N ∆gA + ηN 2/3 γSL
(6.3)
where η is a shape factor defined by:
η=
A
N 2/3
Since the negative, volume term of Eqn. 6.3 is linear in n while the positive, surface term varies as N 2/3 , the sum of the two terms will exhibit a
6.1. NUCLEATION
163
Liquid
Liquid
Solid
Figure 6.3: Schematic of undercooled liquid and an undercooled liquid with
a solid particle.
peak, as shown in Fig. 6.4. The critical number of atoms, N ∗ , corresponding
to this can be found by setting:
∂∆GT =0
∂N N =N ∗
and solving for N ∗ , we find:
∗
N =
2ηγSL
3∆gA
3
(6.4)
This expression for the critical number of atoms in a solid particle is valid
for arbitrarily-shaped particles.
The free energy, ∆G∗ , of a critical size solid particle can be found by
inserting N ∗ from Eqn. 6.4 into the expression for the free energy change as
a function of number of atoms Eqn. 6.3. We find:
∆G∗ = −N ∗ ∆gA + ηγSL (N ∗ )2/3
3
4 η 3 γSL
=
27 ∆gA2
γSL η(N ∗ )2/3
=
3
(6.5)
If our solid particles are spherical in shape, the number of atoms in a
particle with radius r is:
4
N=
πr3
3VA
and the shape factor becomes:
η=
A
4πr2
2/3
=
= (36π)1/3 VA
2/3
3
N 2/3
[4πr /(3VA )]
CHAPTER 6. SOLIDIFICATION
Free Energy, ∆GT
164
N*
surface
volume
total
N
Figure 6.4: ∆GT versus N showing volume and surface contributions resulting in a peak at N = N ∗ .
so that the free energy change associated with forming a particle of radius r
is:
∆gA
4
4
+ 4πr2 γSL = − πr3 ∆GV + 4πr2 γSL
∆GT = − πr3
3
VA
3
∗
We can find the critical radius r by differentiating this directly or by substituting the above expressions for η and n into Eqn. 6.4:
r∗ =
2γSL VA
2γSL
=
∆gA
∆GV
The energy change associated with formation of a particle of this radius is:
∆G∗ =
3
3
16πVA2 γSL
16πγSL
=
3∆gA2
3∆G2V
(6.6)
If we consider the process of adding an one additional atom to a solid
particle of a given size, we see that for solid particles with N < N ∗ the addition of one more atom will increase the free energy, while for solid particles
with N > N ∗ , addition of an atom will decrease the free energy. Hence, solid
particles with N < N ∗ tend to shrink and disappear. These solid particles
are known as embryos. Solid particles with N > N ∗ will grow, since this
6.1. NUCLEATION
165
decreases the free energy. The process of adding one additional atom to a
solid particle with N = N ∗ is the nucleation process, since this results in a
nuclei, the growth of which will decrease the free energy.
6.1.2
Distribution of Solid Particle Sizes
The equilibrium distribution of solid particles is described by xN which is
the number fraction of solid particles with a given size. The concentration of
a particle of a given size can be found in a manner exactly analogous to that
used in finding the equilibrium concentration of vacancies. The formation of
clusters in the collection of atoms increases the number of ways the system
can arrange itself, and hence increases the entropy. Thus, even though the
enthalpy will increase if a sub-critical cluster is formed, there will be some
concentration of clusters of a given size which will minimize the free energy.
Thus xN is given by:
∆GT
xN = exp −
kB T
where ∆GT is given by Eqn. 6.3. The actual number fraction of solid particles
of a given size will be different from the equilibrium distribution since solid
particles with N > N ∗ will grow and will be removed from the distribution.
The Volmer-Webber theory of nucleation assumes that the actual number
fraction of solid particles with N atoms, ZN , is given by:
ZN =
xN if N < N ∗
0
if N > N ∗
(6.7)
In this theory, the number of critically sized nuclei is given by:
Z ∗ = exp −
∆G∗
kB T
= xN ∗
with ∆G∗ given by Eqn. 6.5 or 6.6.
The Becker-Döring theory of nucleation goes beyond the unrealistic assumptions of Eqn. 6.7 and considers the effect of the nucleation process on
the population of particles. By considering the evolution of the population
distribution via the reactions:
N + 1 ⇐⇒
N+1 − 1 ⇐⇒
N+1
N
166
CHAPTER 6. SOLIDIFICATION
they find (after some mathematical shenanigans which are beyond the scope
of this course) the number of critical sized nuclei:
1
∆G∗
∆G∗ 1/2
exp
−
N ∗ 3πkB T
kB T
∗ 1/2
∆G
1
=
xN ∗
N ∗ 3πkB T
Z∗ =
The factor:
1
N∗
∆G∗
3πkB T
1/2
≡ Γz
is known as the Zelldovich factor and was first derived by Farkas. A comparison of the population of particles predicted by the Volmer-Webber and
Becker-Döring theories is shown in Fig. 6.5.
ZN
Volmer-Webber
xN
Becker-Doring
0
1
N/N*
Figure 6.5: Schematic of number of nuclei as a function of number of atoms
for Volmer-Webber and Becker-Döring nucleation theories. Dotted line is the
equilibrium number assuming no depletion due to growth.
6.1. NUCLEATION
6.1.3
167
Nucleation Rate
A nuclei will be formed each time an atom sticks to a critical sized particle.
Hence the nucleation rate, I, per volume is given by:
I = RσA∗
Z∗
VA
where R is the sticking rate per site, σ is the sites per area on the critical
nucleus, and A∗ is the area of a critical nucleus, and VA is the atomic volume.
In the case of a liquid-solid transformation, the attachment rate is exactly
analogous to the exchange frequency in the diffusion process, and can be
expressed as:
∆GM
R = ν0 exp −
kB T
where ν0 is some vibration frequency, and ∆GM is the activation energy for
the jump process. Thus we have for the nucleation rate:
σA∗ ν0
I=
VA N ∗
∆G∗
3πkB T
1/2
∆G∗ + ∆GM
exp −
kB T
The temperature dependence is dominated by ∆G∗ which is given by Eqn. 6.5
or for a sphere, Eqn. 6.6. We can use the Turnbull extrapolation to find ∆gA
for the liquid-solid case:
∆T
∆gA = ∆hF
Tm
where ∆hF = hL − hS is the difference in enthalpy per atom between the
liquid and solid states. For spherical nuclei we find:
3
16πγSL
3∆G2V
3
Tm2
16πVA2 γSL
=
3∆h2F (∆T )2
∆G∗ =
Relative to this 1/(∆T )2 dependence in the exponent the other terms are
nearly constant, and Turnbull has suggested that homogeneous nucleation of
the solid from an undercooled liquid can be approximated by:
∆G∗
1
39
I
≈ 10 exp −
m3 s
kB T
(6.8)
168
6.1.4
CHAPTER 6. SOLIDIFICATION
Heterogeneous Nucleation
In practice, nucleation of the solid phase rarely takes place homogeneously
in the bulk of the liquid. In most situations, nucleation initiates on the
walls of the containment vessel or on particles of high melting point oxides
inadvertently dispersed in the melt. These inhomogeneous nucleation sites
reduce the nucleation barrier by lowering the surface energy cost of forming
a nucleus. In fact, only by dispersing the liquid into droplets small enough
so that there is an appreciable chance of not having any nucleation sites in
a droplet, can homogeneous nucleation be achieved.
To examine how heterogeneities effect the energy of forming a solid phase
particle in an undercooled liquid, we consider formation of a spherical solid
cap of radius r and contact angle θ on a mold wall, as shown in Fig. 6.6. Here
we have three different interfacial energies to consider: γSL is the surface
energy between the solid and liquid, γSM is that between the solid and the
mold, and γLM is that between the liquid and the mold. The contact angle
θ is determined by the surface energy balance:
γLM = γSM + γSL cos θ
This expression, derived in Appendix C can be viewed as a force balance
considering each of the surface energies as a surface stress, although strictly
speaking γ is the surface energy and not the surface stress. It is straightforward geometry to show that the volume VS of the spherical cap is:
VS =
πr3 2 − 3 cos θ + cos3 θ
3
and the area ASL of the solid liquid contact is:
ASL = 2πr2 (1 − cos θ)
and the area ASM of solid mold contact is:
ASM = π(r sin θ)2
The formation of this spherical cap involves: changing the liquid to the
solid, creating new interface area ASL and changing the interface energy from
γLM to γSM in the contact area ASM . Hence the free energy of formation is:
∆GT = −VS ∆GV + ASL γSL + ASM (γSM − γLM )
6.1. NUCLEATION
169
γSL
liquid
θ
γLM
solid
γSM
r
mold
Figure 6.6: Schematic of a solid spherical cap with radius r and contact angle
θ forming on a mold wall.
which can be shown to be given by:
!
4
∆GT = − πr3 ∆GV + 4πr2 γSL S(θ)
3
where S(θ) is a contact angle dependent function:
S(θ) = (2 + cos θ)(1 − cos θ)2 /4
which is shown plotted in Fig. 6.7.
Apart from the factor of S(θ), this is the same expression we obtained for
homogeneous nucleation. The energy of formation will again have a peak at
a the critical radius r∗ given by:
r∗ =
2γSL
∆GV
but the energy barrier will be reduced by a factor of S(θ):
∆G∗ =
3
16πγSL
S(θ)
3∆G2V
This is shown schematically in Fig. 6.8.
Since ∆G∗het is much less than ∆G∗homo , nucleation can occur at much
smaller undercoolings. This is a result of the increased ratio of volume to
surface energy for the spherical cap compared to a sphere.
170
CHAPTER 6. SOLIDIFICATION
S(θ)
1.0
0.5
0.0
0.0
0.5
1.0
θ/π
Figure 6.7: Plot of the reduction factor S(θ) for nucleation barrier for heterogeneous nucleation.
∆GT
Homogeneous
Heterogeneous
0
1
2
r/r*
Figure 6.8: Formation free energy as a function of particle radius both homogeneous and heterogeneous nucleation of a solid particle in an undercooled
liquid.
6.1. NUCLEATION
171
Recall that the probability of a critical nucleus occurring on a site is:
∆G∗
kT
so that even though the number of possible sites for nucleation is much
smaller for heterogeneous nucleation, the probability of it happening on a
given site is so much higher that heterogeneous nucleation usually dominates.
exp −
6.1.5
Numerical Estimation of Nucleation Rates
So far we have discussed nucleation with only passing reference to any real
numbers. We can make some good headway toward estimating nucleation
rates by using a couple of simple assumptions which are in the spirit of
those often made this course. The result of this simple analysis is a good
starting point for estimations of nucleation rates and helps to solidify our
understanding of the underlying materials science.
We first consider the critical radius r∗ , given by
r∗ =
2γ
∆GV
where γ is the interface energy, in this case between liquid and solid, and
∆GV is the volume driving force for the transition, given by
∆GV ≈ ∆HF
∆hF ∆T
∆T
=
Tm
VA Tm
where VA is the atomic volume, ∆HF and ∆hF are the enthalpy of fusion per
volume and per atom respectively, and ∆T and Tm are the undercooling and
melting temperature respectively. The interface energy γ can be estimated
as
γ ≈ FBB ∆hF σ
where FBB is the fraction of broken bonds at the interface and σ is the atom
surface density. Using ∆hF = ∆HF VA and
σ ≈ (1/VA )2/3
and taking FBB ≈ 1/2 as found for liquid/solid interfaces in our chapter on
surfaces, we can find
1/3 Tm
r∗ ≈ VA
∆T
172
CHAPTER 6. SOLIDIFICATION
for a fcc with lattice parameter a this becomes
r∗ ≈
a Tm
Tm
= 0.63a
1/3
(4) ∆T
∆T
For Cu with a lattice parameter of a ≈ 3.6 Å and a melting temperature of
Tm ≈ 1358 K, we find a critical radius of about 23 Å for an undercooling of
10% which is about 136 K.
Now let’s turn to the nucleation rate. Starting with the nucleation event
being defined as the process of adding an atom to a critically sized nucleus
we found:
Z∗
I = RA∗ σ
VA
∗
where A is the area of a critically sized nucleus, R is the attachment rate,
σ is the number of atomic sites per area, Z ∗ is the number fraction of critically sized clusters and VA is the atomic volume. Using The Becker-Döring
expression for Z ∗ we find
I=
A∗ ν0
5/3
VA N ∗
∆G∗
3πkB T
1/2
∆G∗ + ∆GM
exp −
kB T
where N ∗ is the number of atoms in a critically-sized nucleus and we have
2/3
again used σ ≈ 1/VA . ∆G∗ is the nucleation barrier given (for spherical
nuclei) by
3
∆hF
16πγ 3
16πFBB
∆G∗ =
=
2
3∆GV
3(∆T /Tm )2
Taking
4
and
N∗ =
π(r∗ )3
A∗ = 4π(r∗ )2
3VA
and again using the above expression for the interface energy γ we find
2ν0
I=
VA
FBB ∆hF
∆G∗ + ∆GM
exp −
kB T
kB T
Further, we recognize that nucleation takes place close to the melting point
so ∆hF /(kB T ) ≈ ∆hF /(kB Tm ). Earlier in the course we noted that the
enthalpy of fusion scaled with the melting point, and that for a large number
of materials
∆hF
≈ 1.1
kB Tm
6.1. NUCLEATION
173
Using again FBB ≈ 1/2 we find
1.5 ν0
∆G∗ + ∆GM
I≈
exp −
VA
kB T
To make further progress we must estimate the migration energy. This is
different from the activation barrier for diffusion in that it need not contain
the cost of forming a defect. Hence our earlier scaling relations won’t be
much help. We instead just assume that the energy cost for migration from
liquid to solid of the order of the difference in bond strength between solid
and liquid. Hence we find
2
∆GM ≈ ≈ ∆hF
z
Hence
∆GM
exp −
kB T
2Tm ∆hF
≈ exp −
zT kB Tm
≈ 0.8
where we have assumed we are near T = Tm . Hence it appears that this
migration energy cost does not play a big role in nucleation at small undercoolings2 .
Lastly we must make some estimate of ν0 . This is likely close to but less
than a typical atomic vibration, and in absence of other information we take
ν0 ≈ 1012 /s
Putting this all together we find
1.2 × 1012
∆G∗
I ≈
exp −
VA
kB T
12
1.2 × 10
Tm 2 Tm
≈
exp −2.3
VA
∆T
T
Using the volume per atom of Cu we find a prefactor of 1 × 1041 /m3 s, which
is a factor of 100 times that obtained by Turnbull, who assumed a a factor
of 10 higher frequency, larger atomic volume and a larger migration energy
cost (taken to be that of viscous flow so exp(−∆GM /kB T ) ≈ 10−2 ). This
uncertainty in prefactor is endemic to nucleation theory and and variations
2
As the temperature decreases, nucleation will be curtailed by this migration energy
cost, and with rapid cooling, can sometimes be suppressed completely, resulting in an
amorphous glass.
174
CHAPTER 6. SOLIDIFICATION
by a factor ±2 in the exponent are common. As discussed below, this is not
as bad as it seems as the undercooling where copious nucleation is observed
is relatively insensitive to the prefactor. In fact to a good approximation the
nucleation rate is just given by
I ≈ const.
ν0
∆G∗
exp −
VA
kB T
where the constant is in the range of 10−2 − 1.
We can plot the nucleation rate per atom site as a function of undercooling as shown in Figure 6.9. The sudden onset of copious nucleation occurs
somewhere between 20% and 25% undercooling. For example, a nucleation
rate of one per cubic micron per second corresponds to IVA ≈ 1.2 × 10−11 /s,
which occurs at an undercooling of about 23%. This figure also shows that
the undercooling at which copious nucleation occurs is relatively insensitive
to the prefactor, since prefactors which differ by a factor of 100 only change
the onset of copious nucleation by a fraction of a percent.
The onset of copious nucleation at 23% is at a rather higher undercooling
than is usually achieved in practice3 . In most practical situations nucleation
occurs at undercoolings below 10%. This is due to the dominance of heterogeneous nucleation on the mold wall or on refractory particles in the melt.
Heterogeneous nucleation takes place on the interface between the melt
and the solid surfaces in which the melt is in contact (the mold walls or
inclusions in the melt). We can apply the same analysis to find the nucleation
rate per area for this case with only a slight modification of our starting
expression
I het = RσA∗ σZ ∗het
where the only difference is that now Z ∗het is the number fraction of critically
sized particles, so σZ ∗het is the number of critically sized particles per unit
interface area. We modify the Becker-Döring expression for the homogeneous
case to find:
1/2
1 ∆G∗het
∆G∗het
∗het
Z
= ∗
exp −
N
3πkB T
kB T
where ∆G∗het is the heterogeneous nucleation barrier given by
∆G∗het = ∆G∗ S(θ)
3
in fact one could even question the validity of using the Turnbull free energy extrapolation down this far
6.1. NUCLEATION
175
-12
100x10
5000
80
4000
60
3000
0.5
0.45
0.40
0.35
40
N*
I VA (1/s)
FBB
2000
20
0
1000
0
5
10
15
20
25
0
∆T/Tm (%)
Figure 6.9: Nucleation rate per atom volume as a function of fractional undercooling calculated using the assumptions detailed above. The dashed line
corresponds to the nucleation rate of 1 per second per cubic micron. The
dotted line is the same expression with just a factor of 100 smaller prefactor.
Also shown is the number of atoms in the critical nucleus. Numbers correspond to values of FBB . We see that a change of 10% in the estimation of
the interface energy has a change in predicted nucleation rate comparable to
a change of a factor of 100 in the prefactor.
176
CHAPTER 6. SOLIDIFICATION
Hence the heterogeneous nucleation rate is given by:
I
het
=
A∗ ν0
4/3
VA N ∗
∆G∗het
3πkB T
1/2
∆G∗het + ∆GM
exp −
kB T
2/3
where we have again made the substitution σ = 1/VA . Using
∗
∗ 2
a = 2π(r ) (1 − cos θ)
we find
I het =
2ν0
2/3
VA
4π(r∗ )3
N =
S(θ)
3VA
∗
and
FBB ∆hF
∆G∗het + ∆GM
exp −
kB T (2 + cos θ)
kB T
It is interesting to compare our estimations for homogeneous and heterogeneous nucleation rates. To do so we consider a collection of N atoms
with volume V = N VA . Homogeneous nucleation takes place throughout the
sample volume so the total number of homogeneous nucleation events per
unit time is
Ṅ hom = V I hom
However heterogeneous nucleation takes place only on the area of the sample
in contact with the mold. Hence the total number of heterogeneous nucleation events per unit time is
Ṅ het = AI het
A is the area of the mold wall where heterogeneous nucleation will occur. If,
for simplicity, we assume that nucleation takes place at the container walls,
we can say A = V 2/3 . To compare these we can multiply each by VA /V to
put it on a per atomic site basis, so the quantities we compare are
Ṅ hom VA
= I hom VA
V
FBB ∆hF
∆G∗ + ∆GM
= 2ν0
exp −
kB T
kB T
and
Ṅ het VA
VA
= I het 1/3
V
V
VA 1/3
FBB ∆hF
∆G∗het + ∆GM
= 2ν0
exp −
V
kB T (2 + cos θ)
kB T
6.2. GROWTH
177
We can see right away that the prefactor for the heterogeneous case is smaller
by the factor4
VA 1/3
V
Taking a typical volume of 1 liter = 1×10−3 m3 , and using the atomic volume
of Cu, we find
VA 1/3
= 2.3 × 10−9
V
Hence the prefactor for the heterogeneous nucleation rate is much smaller.
However, the lowering of the nucleation barrier by a factor of S(θ) more
than compensates. We can estimate the undercooling range over which heterogeneous nucleation dominates. Setting Ṅ het = Ṅ hom and solving for the
undercooling gives:
∆T ≈
Tm c
3
16πFBB
[1 − S(θ)]
ln(V /VA )
∆hF
kB Tm
Tm
T
1/2
Using numbers as above, assuming a volume of 1 liter, and taken θ = 60◦
corresponding to the case where γSL ≈ γLM ≈ 2γSM , so that S(θ) ≈ 0.16, we
can find the crossover undercooling or about 40%. Hence, we predict that
heterogeneous nucleation dominates for the most undercoolings of practical
interest. This is shown in Figure 6.10, where we show both the homogeneous
and heterogeneous nucleation rate for a volume of 1 liter.
6.2
6.2.1
Growth
Atom Attachment Rate
Once a nucleus has formed, the transformation proceeds with the process of
growth, where atoms are transported across the interface from one phase to
the other. The undercooling which has paid the thermodynamic price for
the formation of the interface surrounding the nucleus, provides the thermodynamic driving force for this process as well. The process of an atom
jumping from the liquid to the solid is a diffusive jump with a driving force,
just as we saw in diffusion with a chemical concentration gradient. (Sec. 4.3)
The energy as a function of position during this jump is shown in Fig. 6.11
4
√
Ignoring the factor 1/ 2 + cos θ.
178
CHAPTER 6. SOLIDIFICATION
-12
100x10
I VA (1/s)
80
60
heterogeneous
homogeneous
40
20
0
0
5
10
15
20
25
∆T/Tm (%)
Figure 6.10: Homogeneous and heterogeneous nucleation rate as a function
of fractional undercooling for a volume of 1 liter.
which is similar to Fig. 4.3. The activation energy ∆GM is the formation
free energy associated with the activated complex where the atom is halfway
between the liquid and solid. The driving force ∆gN is the the net lowering
in free energy per atom in going from liquid to solid, and is the fraction of
the total driving force of the transformation ∆gT remaining after the energy
costs of interface formation are paid.
We can calculate the rate at which an atom will jump from liquid to a
given site on the solid:
RLS
−∆GM
∆gN
= nL aL νL exp
+
kB T
2kB T
where nL is the number of liquid atoms available per site on the solid, aL is
the fraction of them jumping in the correct direction, and νL is the vibration
frequency in the liquid. We can write a similar expression for the rate at
which atoms jump from a given site in the solid to the liquid:
RSL
−∆GM
∆gN
= aS νS exp
−
kB T
2kB T
6.2. GROWTH
179
Energy
Activated complex
Liquid
∆G’M
Solid
∆gN
Reaction Coordinate
Figure 6.11: Schematic of the energy of an atom during a jump from an
undercooled liquid to a solid.
where as and νS are the corresponding factors in the solid. We know that at
no driving force, RSL = RLS so that:
aS νS = nL aL νL ≡ ν0
The net rate RN which atoms leave the liquid to attach at a given site on
the solid is:
RN = RLS − RSL
−∆GM
= ν0 exp
kB T
∆gN
exp
2kB T
−∆gN
− exp
2kB T
!
For small driving forces e±∆gN /2kB T ≈ 1 ± ∆gN /2kB T and we find:
−∆GM
ν0 ∆gN
exp
RN =
kB T
kB T
so that the rate of atom attachment to the solid is proportional to the net
driving force.
180
6.2.2
CHAPTER 6. SOLIDIFICATION
Uniform Growth
The fraction of the driving force available to encourage atom attachment is
a function of the site to which the atom is attaching. For example, an atom
attaching to a smooth surface creates an adatom site (Fig. 5.7) which costs
energy ∆g+1 to create more surface area, or equivalently, more broken bonds.
(Recall that by just considering the nearest neighbor bonds in a simple cubic
material, and ignoring the formation entropy, we have previously estimated
that ∆g+1 ≈ ∆2hT /3, where ∆hT is the enthalpy per atom associated with
the transformation.) (Table 5.2, Section 5.2). Hence the net driving force is:
∆gN = ∆gT − ∆g+1
The growth will only occur if the driving force is sufficient to create the new
surface defect and still have drive the atom attachment.
In contrast, an atom attaching at a kink creates no new surface energy
or broken bonds. Hence for kink attachment:
∆gN = ∆gT
the whole driving force is available for atomic attachment kinetics, and attachment can occur at arbitrarily small undercoolings. Therefore, a surface
which has a large concentration of kinks (a rough surface), can grow more
easily than a smooth surface.
In the case of a very rough surface or for large undercoolings, growth can
occur by attachment of atoms at random sites. This is known as uniform
growth since the growth occurs at a uniform rate everywhere on the surface.
In this case, the velocity of growth will be given by:
aν0 ∆gN
∆GM
vU =
exp −
kB T
kB T
where a is the distance gained per layer of atoms, ν0 is an effective attachment attempt frequency, ∆gN is the net thermodynamic driving force for
the transition, and ∆GM is the activation energy for the jump process. This
uniform growth rate is proportional to the undercooling and is constant in
time for a fixed undercooling.
6.2.3
Nonuniform Growth Modes
In the case where the surface is smooth and the undercooling is insufficient to
allow attachment at random sites, growth can occur by attachment of atoms
6.2. GROWTH
181
to kink sites. This attachment results in motion of the kink along the face of
the step. This motion in turn results in motion of the step along the surface,
and this step motion results in addition of a layer of atoms, which gives a
growth velocity. This is shown schematically in Fig. 6.12.
v
vk
vs
Figure 6.12: Schematic showing the relationship between kink, step, and
surface velocity.
We have seen that the rate of attachment of atoms at a kink site is given
by:
ν0 ∆gT
−∆GM
Rk =
exp
kB T
kB T
where ∆gT is the total thermodynamic driving force for the transition, which
is used here since, in the case of kink attachment, there is no formation of new
surfaces so the net driving force is the total thermodynamic driving force.
This will result in a kink velocity, vk , given by:
vk = aRk
The kinks will be spaced by
yk = a/x0k
where a is the size of an atom and x0k is the equilibrium fraction of step sites
which have kinks on them. The time τk required to advance the step one
atom is given by:
yk
a
τk =
= 0
vk
xk vk
Hence the step velocity will be given by the distance moved divided by this
time, or:
a
= x0k vk
vs =
τk
182
CHAPTER 6. SOLIDIFICATION
The time τs required to advance the surface one atom is given by the step
spacing ys divided by the step velocity vs :
τs =
ys
vs
and the surface growth velocity, v, will then be given by the distance advanced
divided by this time:
a
avs
ax0 vk
v=
=
= k
τs
ys
ys
Inserting the above expression for vk we find:
v=
a2 x0k ν0 ∆gT −∆G /kB T
M
e
y s kB T
(6.9)
There are several limiting cases for the growth behavior as governed by
Eqn. 6.9. Each has characteristic behavior as a function of undercooling.
Step Motion Limited
Growth Velocity, v
If there are many steps, the growth will just be limited by the motion of
steps. At small undercoolings, the growth rate will be less than that for
uniform growth, since ax0k /ys < 1. As the undercooling increases, uniform
growth will begin to occur, so that the growth rate will approach that for
uniform growth. This behavior is shown schematically in Fig. 6.13.
Uniform Growth
Step Motion Limited
Non-Uniform Growth
Undercooling, ∆T
Figure 6.13: Schematic of growth rate versus undercooling for the case of
nonuniform growth limited by the step velocity.
6.2. GROWTH
183
Step Nucleation Limited
If there are few steps on the surface, the growth will be limited by the supply
of steps rather than their motion. One source of steps is the two dimensional
nucleation of islands on the surface. This is just the two dimensional analogue
of the nucleation we have previously covered. The islands are assumed to be
cylinders one atom high on the surface, as shown schematically in Fig. 6.14.
The free energy change associated with formation of an island will have two
counteracting components. The first is the volume term, which is just the
bulk transformation energy times the volume of the island. The second is
the free energy associated with the increase of surface area. We can write:
2
∆G2D
T = −πr a
∆gT
+ 2πraγSL
VA
where r is the island radius, VA is the atomic volume, and γSL is the surface
energy. The first term is negative and the second term is positive, since
γSL > 0. The surface term will dominate at small r, while the volume term
will dominate at large r. Hence there will be a maximum in ∆GT at some
r = r∗ . This maximum will act as an effective energy barrier to formation of
islands. We find the maximum by setting:
∂∆G2D
T =0
∂r r=r∗2D
r
a
Figure 6.14: Schematic of island nucleating on the surface of a solid.
We find:
r∗2D =
γSL VA
∆gT
184
CHAPTER 6. SOLIDIFICATION
The energy associated with this maximum, ∆G∗2D is found to be:
2
VA
πaγSL
∆gT
∆G∗2D =
1/3
2
πVA γSL
≈
∆GV
2
πFBB
∆hf
≈
(∆T /Tm )
1/3
where we have taken a ≈ VA , and in the last step made approximations
similar to those made in section 6.1.5.
For the two-dimensional nucleation of steps on a crystal surface the nucleation rate is given by
I 2D = RσA∗2D σZ ∗2D
where A∗2D is the perimeter area of the island. For this two dimensional case
Becker and Döring found the number density of critically sized nuclei to be
Z ∗2D =
∆G∗2D
∗2D
e−∆G /kB T
∗2D
N kB T
For the two dimensional pill box nuclei we find
∗2D
A
= 2πr
∗
1/3
VA
and
N
∗2D
π(r∗ )2 VA
=
VA
1/3
1/3
where we have taken the height of the island to be VA . Inserting these into
the expression for nucleation rate per unit surface area we find
I
2D
=
≈
≈
2πFBB ν0
2/3
VA
πν0
2/3
VA
πν0
2/3
VA
∆hF
kB T
∆G∗2D + ∆GM
exp −
kB T
∆G∗2D + ∆GM
exp −
kB T
2
VA
πaγSL
exp −
∆gT kB T
(6.10)
where we have taken FBB = 1/2. Since ∆gT in the denominator of Eqn. 6.10
is proportional to ∆T , the nucleation rate will climb rapidly as ∆T increases,
as shown in Fig. 6.15.
185
Nucleation Rate, I
6.2. GROWTH
Undercooling, ∆T
Figure 6.15: Schematic of island nucleation rate as a function of undercooling.
Growth Velocity, v
As the nucleation increases, the growth rate will also rise, since the nucleation of islands will provide steps on which to grow. As ∆T increases, the
growth rate will approach that of uniform growth as the nucleation of islands
provides sufficient steps. Hence the growth rate as a function of undercooling
will behave as shown in Fig. 6.16.
Uniform Growth
Island Nucleation Limited
Non-Uniform Growth
Undercooling, ∆T
Figure 6.16: Schematic of growth rate as a function of undercooling for the
situation where the growth is limited by the supply of steps, and the steps
are a result of island nucleation.
Growth Limited by Supply of Screw Dislocation Steps
A screw dislocation intersecting the surface will leave a ledge ending at the
position of the dislocation. This will provide a location for growth at kink
186
CHAPTER 6. SOLIDIFICATION
sites. Since one end of this ledge is pinned at the position of the dislocation,
growth along this ledge will result in a spiral configuration. This situation is
shown schematically in Fig. 6.17. The center of this spiral configuration will
be unstable if it is smaller than the critical radius, r∗ found in the previous
section. This limits the radial spacing of the spiral steps to be ys = 2πr∗ .
Inserting this into Eqn. 6.9 we find:
a2 x0k
ν0
−∆GM
v=
(∆gT )2
exp
2πγSL VA
kB T
kB T
Thus the growth velocity is quadratic in ∆T , and as ∆T increases, the
growth will again become uniform, as shown in Fig. 6.18.
a)
b)
c)
Figure 6.17: Schematic of a) screw dislocation ledge at surface and b) development of spiral surface step by addition of atoms, and c) the spiral feature
which results from growth along this ledge.
6.2.4
Interface Stability and Dendrite Formation
Solid phase growth takes place by the motion of the solid-liquid interface. The
interface velocity increases monotonically with undercooling, and is dependent on the structure, crystallographic orientation, and defect concentration
Growth Velocity, v
6.2. GROWTH
187
Uniform Growth
Screw Dislocation Step Limited
Non-Uniform Growth
Undercooling, ∆T
Figure 6.18: Schematic of growth velocity as a function of undercooling for
the case where growth is limited by screw dislocation spiral steps.
of the interface. The motion of the interface results in release of the enthalpy
of melting ∆HF sometimes called the latent heat of fusion. To achieve steady
state growth with velocity v, this heat must be removed from the interface by
heat flow into the liquid or solid. We can express this as a heat conservation
equation:
n̂·v∆HF = KS n̂·∇T |S − KL n̂·∇T |L
(6.11)
where KS and KL are the thermal conductivities of the solid and liquid state
respectively, and n̂ is the surface normal pointing from the solid to liquid.
The first term on the right is the heat flow away from the interface into the
solid, while the second term is the heat flow away from the interface into the
liquid. (Note that a negative gradient ∇T |L in the liquid will give a positive
heat flow out from the interface to the liquid.) Basically Eqn. 6.11 just states
that the faster the latent heat is removed from the interface, the faster the
surface will grow.
In order to examine growth morphology, we consider the stability of a
small deviation from a planar interface. In general interface shape stability
is a complicated problem requiring the use of perturbation methods. Local
thermal fluctuations and impurities will always be present and result in small
local fluctuations in growth velocity. The question is, will these fluctuations
grow or shrink? Here we use simple intuitive arguments to determine under
what conditions a planar interface will be stable.
We consider two possibilities. In the first case, shown schematically in
Fig. 6.19, the liquid is superheated relative to the melting point. This is a
188
CHAPTER 6. SOLIDIFICATION
situation which might exist in a melt which nucleated on a cool mold wall, or
a melt which has been heated by rejection of latent heat. There is a positive
gradient into the liquid so heat flows from the liquid into the interface. A
protuberance will increase the positive temperature gradient into the liquid,
and increase the surface area for heat flow. Hence the heat flow into the
interface will increase and the growth will slow down. The protuberance will
then shrink as the surrounding interface regions advances at a faster rate.
In contrast, consider growth of a interface into an undercooled liquid
(Fig. 6.20). This is a situation which might exist if nucleation occurred in
the central region of an undercooled melt. Now there is a negative temperature gradient from the interface (which is near the melting temperature)
into the liquid, so heat flows from the interface into the liquid. This negative
gradient is larger in front of a protuberance so the heat flow away from the
interface is larger here and the protuberance will grow faster than the surrounding interface. The protuberance will grow and eventually branch again
resulting in dendritic or tree-like branches as shown in Fig. 6.21. Dendrites
forming during solidification of a pure melt are known as thermal dendrites
to distinguish them from those resulting from solidification of an alloy.
Dendrites tend to grow in definite crystallographic directions, corresponding to the fastest growing surfaces, Typically these are the (100) surface in
FCC metals and (11̄00) in HCP metals. Individual dendrite arms can have
a misorientation of up to several degrees, resulting in a solid with strong
texture but many small angle grain boundaries.
6.2.5
Gibbs-Thompson Effect
One aspect of growth of both small particles and dendrite arms is the effect
of the curved interface. We see that the free energy inside a curved interface
is higher than that in an infinite solid. This is known as the Gibbs-Thompson
effect. Consider a solid sphere of radius r in an undercooled liquid. The free
energy associated with forming this sphere is
4
∆GT = −∆GV πr3 + γ4πr2
3
Adding one atom will increase the volume by VA the atomic volume. The
resulting increase in radius can be found through
VA = δV =
dV
δr = 4πr2 δr
dr
6.2. GROWTH
a)
189
Solid
b)
Liquid
Solid
Liquid
T
Heat
Flow
v
Heat
Flow
z
c)
Solid
Liquid
d)
Solid
Liquid
T
Heat
Flow
v
Heat
Flow
z
Figure 6.19: Schematic of stable liquid-solid interface growing with velocity
v into a superheated liquid. a) Planar interface (heavy line) and isotherms
(fine lines) in solid and liquid phases. Heat flow is represented by double
arrows. b) Temperature as a function of position in liquid and solid phases.
c) Interface with protuberance. d) Temperature as a function of position
in front of and behind the protuberance. The isotherms in the solid are
more widely spaced behind the protuberance leading to a decreased gradient
and less heat flow away from the interface into the solid. The isotherms
in the liquid are closer together in front of the protuberance leading to a
larger positive gradient and increased heat flow into the interface from the
liquid. Hence the protuberance grows more slowly than the surrounding
planar interface, and the planar interface is stable.
190
CHAPTER 6. SOLIDIFICATION
a)
Solid
b)
Liquid
Solid
Liquid
T
Heat
Flow
Heat
Flow
v
z
c)
Solid
Liquid
d)
Solid
Liquid
T
Heat
Flow
Heat
Flow
v
z
Figure 6.20: Schematic of an unstable liquid-solid interfacegrowing with velocity v into an undercooled liquid. a) Planar interface (heavy line) and
isotherms (fine lines) in solid and liquid phases. Heat flow is represented
by double arrows. b) Temperature as a function of position in liquid and
solid phases. c) Interface with protuberance. d) Temperature as a function
of position in front of and behind the protuberance. Again, the isotherms
in the solid are more widely spaced behind the protuberance leading to a
decreased gradient and less heat flow away from the interface into the solid.
And again, the isotherms in the liquid are closer together in front of the
protuberance. However, in this case, this leads to a larger negataive gradient
away from the interface and increased heat flow away from the interface into
the liquid. Hence the protuberance grows more quickly than the surrounding
planar interface, and the planar interface is unstable.
6.2. GROWTH
191
Figure 6.21: Schematic of dendrite growth sequence.
The corresponding change in energy is
δGT = −∆GV 4πr2 δr + γ8πr δr
2γ
δV
= −∆GV +
r
We see an effective increase in the free energy per volume of:
∆GrV =
2γ
r
Hence the free energy inside any curved interface will be higher than that of
an infinite solid. We can use the Turnbul extrapolation:
∆GV = ∆HF
∆T
Tm
to relate this free energy difference to an undercooling:
∆T r =
2γ Tm
r ∆HF
The undercooling must be greater than ∆T r to drive the growth of a solid
bounded by a curved interface.
6.2.6
Dendrite Growth Rate
Lets consider the growth rate of a dendrite tip into an undercooled liquid
which is at T∞ far from the tip. We assume that the growth rate is limited by
192
CHAPTER 6. SOLIDIFICATION
the removal of latent heat, so the tip temperature is held at T r = Tm − ∆T r .
The thermal profile away from the tip looks roughly as in Fig. 6.22 where
the temperature difference ∆Tc between the tip and the liquid far from the
tip is:
∆Tc = ∆T0 − ∆T r
2γ Tm 1
= ∆T0 1 −
r ∆HF ∆T0
r∗
= ∆T0 1 −
r
where ∆T0 is the undercooling relative to the melting point of the solid, and
r∗ is the same critical radius we found for homogeneous nucleation. The
thermal gradient in the liquid near the tip is approximately:
∆Tc
r
and for simplicity we assume that all conduction of heat is in the liquid
(∇T |S = 0). Equation 6.11 then becomes:
∇T |L ∼ −
KL ∆Tc
r∗
KL ∆T0
KL
∇T |L ≈
1−
=
v=−
∆HF
∆HF r
∆HF r
r
(6.12)
The velocity falls off at large r due to the decrease in the thermal gradient
to remove the latent heat, and it goes to zero at r = r∗ since all the driving
force (undercooling) is used to push atoms across a curved interface into a
region of higher free energy (chemical potential).
We can find the radius of the fastest growing dendrite by differentiating
Eqn. 6.12:
dv = 0 → rmax = 2r∗
dr r=rmax
The maximum velocity is then:
vmax
6.3
KL ∆T0
r∗
=
1−
∆HF rmax
rmax
=
KL ∆T0
KL
=
(∆T0 )2
∗
4∆HF r
8γTm
Alloy Solidification
So far we have considered the solidification of a single component liquid. The
addition of a second component introduces a new level of complexity to the
problem. We now deal with some of these complications.
6.3. ALLOY SOLIDIFICATION
193
Tm
∆T
r
Tr
∆T0
∆Tc
T∞
z
Figure 6.22: Schematic of temperature profile near a growing dendrite tip.
We consider solidification of a alloy in a region of the phase diagram as
shown in Fig. 6.23, which shows the temperature boundaries as a function of
solute mole fraction x. We define a partition coefficient k to be the ratio of
equilibrium mole fractions in the liquid and solid, that is:
k=
xS
xL
where xL and xS are the mole fraction of solute in a liquid and solid in equilibrium at a given temperature (for example T2 in Fig. 6.23). For simplicity,
we assume that the partition coefficient is independent of temperature, which
means that the liquidus and solidus will be straight lines.
For illustration purposes, we ignore the complications of geometry and
consider one dimensional solidification of a bar as shown in Fig. 6.24, where
heat is extracted out one end and the solidification front moves down the
bar. We consider three limiting cases.
• Infinitely slow equilibrium solidification
• No diffusion in the solid but perfect mixing in the liquid.
• No diffusion in the solid and diffusional mixing in the liquid.
194
CHAPTER 6. SOLIDIFICATION
T
Solidus
T1
T2
T3
xL
Liquidus
xS
kx0
x0
k
x0 xmax
xE
x
Figure 6.23: Simplified region of a phase diagram.
Solid
Liquid
Heat
Figure 6.24: Schematic of a bar for one-dimensional solidification
6.3. ALLOY SOLIDIFICATION
195
Equilibrium Solidification
In this case, as we cool the melt of initial mole fraction x0 (Fig. 6.23), the
solidification begins at temperature T1 with formation of a solid of mole
fraction kx0 . As the cooling continues, the mole fraction of the solid remains
uniform across the solid portion of the bar, and the mole fraction follows
the solidus. The liquid mole fraction also remains uniform and follows the
liquidus. For example, at temperature T2 , the bar is partially solidified with
a solid of mole fraction xS and the remaining liquid has mole fraction xL as
shown in Figs. 6.23 and 6.25. The position of the interface, that is the fraction
of solid in the bar, is determined by the lever rule. At the temperature T3
the last liquid solidifies with mole fraction x0 /k.
xL
x0
xS
Figure 6.25: Partially solidified bar at temperature T2 under the conditions
of equilibrium solidification. The bottom axis is position. The cross hatched
regions are equal in area since the solute rejected in the solid is added to
the liquid. The mole fractions xS and xL move, following the solidus and
liquidus as the temperature drops and solidification continues.
No Diffusion in Solid - Perfect mixing in Liquid
Frequently there will be insufficient time for solute redistribution in the solid,
but there will be very good mixing in the liquid, either by intentional mixing
or by convective flows. In this case, the first solid again forms at temperature
T1 with mole fraction kx0 , and again the liquid will be enriched in solute since
k < 1. The next solid will be slightly richer in solute, since local equilibrium
is maintained at the interface (xS /xL = k). However, now the solute is not
redistributed in the solid, so the solid concentration will vary with position
along the bar. We can define an average mole fraction in the solid xS ,
which will vary with temperature as shown in Fig. 6.26. The position of the
interface at a given temperature will be determined by the lever rule between
196
CHAPTER 6. SOLIDIFICATION
xS and xL . To find xS as a function of position along the bar, we equate
the solute rejected into liquid with increase in liquid solute concentration:
(xL − xS ) dfS = (1 − fS ) dxL
where fS is the fraction of the bar which is solid, and (1 − fS ) is the fraction
which is liquid. In a homework problem, you show that:
xS = kx0 (1 − fS )k−1 and xL = x0 fLk−1
You also show that the last solidification will take place at the eutectic temperature TE at the eutectic mole fraction xE , and one can also show that the
average concentration is given by:
xS =
+
x0 *
1 − (1 − fs )k
fS
T
T1
xL
T2
xS
kx0
x0 xmax
x0
k
xE
x
Figure 6.26: Phase diagram region illustrating solidification in the case of
no diffusion in the solid and perfect mixing in the liquid. The average mole
fraction xS in the solid is shown as a line.
No Diffusion in Solid - Diffusional Mixing in Liquid
In this case, we have no stirring or convection in the liquid, and the solute
rejected is transported away into the liquid by diffusion. As the liquid is
6.3. ALLOY SOLIDIFICATION
a)
197
xL
x0
kx0
b)
xL
x0
kx0
c)
xE
x0
xmax
kx0
Figure 6.27: Schematic illustrating steps in solidification for the case of no
diffusion in the solid and perfect mixing in the liquid. a) Initial solidification
at mole fraction kx0 . b) At an intermediate temperature. c) Final concentration profile, showing variation of mole fraction along the length of the bar
and region of eutectic solid.
198
CHAPTER 6. SOLIDIFICATION
cooled, solid first forms with mole fraction kx0 at temperature T1 . This
results in a rapid buildup of solute in the liquid near the interface, so that
there is a rapid increase in the mole fraction of the solid which is subsequently
formed. If a steady-state is imposed, with a constant interface velocity v, the
interface temperature becomes T3 , the mole fraction in the liquid climbs to
x0 /k, and the solid formed has mole fraction x0 . There is a bow wave of
rejected solute in the liquid ahead of the solidification front. As the end of
the bar is reached, this solute in this bow wave is compressed into a small
volume resulting in a large concentration, and the final solid to form is of the
eutectic composition as shown in Fig. 6.28.
a)
x0
kx0
b)
x0
k
x0
kx0
c)
xE
x0
xmax
kx0
Figure 6.28: Schematic illustrating steps in solidification for the case of no
diffusion in the solid and diffusional mixing in the liquid. a) Rapid buildup
of solute in the liquid and increase in concentration of the solid. b) The
solidification at an intermediate temperature showing steady-state region,
bow-wave in liquid and initial transient in the solid. c) Final solute distribution along bar showing initial and final transients.
We can find a relationship between the flux and interface velocity by
6.4. CONSTITUTIONAL UNDERCOOLING
199
equating the flux excluded by the moving interface to the flux down the
concentration gradient in the liquid.
dcL [cL (zS ) − cS (zS )]v = −D
dz z=zS
(6.13)
where cL (zS ) and cS (zS ) are the concentrations in the solid and liquid respectively at the solid-liquid interface (z = zS ), and D is the diffusivity in
the liquid. Equation 6.13 represents a boundary condition on the diffusion
equation in the liquid. This is exactly analogous to our treatment of the
moving interface problem, but with no diffusion on one side (solid) of the
interface. This is just Eqn. 1.13 from the moving interface problem in section 1.6. It is also analogous to solidification of a pure material, as in section
6.2.4 Eqn. 6.11, where the interface motion is limited by the removal of the
latent heat. Here the interface motion is limited by the rate of removal of
the excess solute.
In a homework problem we see that the concentration in the liquid is
given by:
1−k
zv
xL = x0 1 +
exp −
k
D
where z = z − zS , so that the bow wave has a characteristic width of ∼ D/v.
6.4
Constitutional Undercooling
In absence of liquid phase mixing, the rate of growth during alloy solidification is determined by the rate of solute diffusion into the liquid. This
is analogous to growth of a pure metal into an undercooled melt, where
the growth rate is determined by the rate of heat transport into the liquid.
Hence, dendrite formation would seem likely.
Consider steady-state solidification of a planar interface. The variation
of solute concentration ahead of interface (Fig. 6.28) results in a variation
of local liquidus temperature TL (z). We define −m to be the slope of the
liquidus line, given by:
m=
T1 − T3
x0
1
k
−1
=
k
1−k
T1 − T3
x0
200
CHAPTER 6. SOLIDIFICATION
where k is the partition coefficient, and T1 and T3 are the temperatures where
equilibrium solidification begins and ends for a liquid of mole fraction solute
x0 (Fig. 6.23). We then find for the local liquidus temperature:
zv
k
+ exp −
TL (z) = Tm − mxL (z) = Tm − (T1 − T3 )
1−k
D
(6.14)
which has the limits:
TL (0) = Tm − (T1 − T3 )
1
x0
= Tm − m = T3
1−k
k
and:
TL (∞) = Tm − (T1 − T3 )
k
= Tm − mx0 = T1
1−k
This is shown plotted in Fig. 6.29
T(z)
T
T
TL(z)
T
z
Figure 6.29: Local liquidus temperature for steady-state solidification with
a planar interface.
We assume that the actual temperature near the front of the interface
increases linearly with distance:
dT T (z) = T3 + z
dz zS
6.5. EUTECTIC SOLIDIFICATION
201
is the temperature gradient in the liquid near the interface. This
where dT
dz zS
actual temperature is also shown in Fig. 6.29 for the case where there is a
region where:
T (z) < TL (z),
(6.15)
that is, the local temperature is lower than the liquidus temperature. This
condition is known as constitutional undercooling. In this case, a protrusion
which sticks out in front of the interface will have a lower temperature than
the local liquidus, and so will grow. Hence if the region in front of the
interface satisfies condition 6.15, dendrite formation will occur.
If the temperature gradient into the liquid is large enough, then condition 6.15 will never be met, and a planar interface will be stable. Mathematically, this stability condition is:
dT dTL >
dz zS
dz zS
Differentiating Eqn. 6.14 we find:
dT v
>
(T1 − T3 )
dz zS
D
The temperature range T1 − T3 is known as the equilibrium freezing range
of the alloy. We see that systems with a large equilibrium freezing range
will have a greater tendency for constitutional undercooling and formation
of dendritic growth. The development of dendrites is shown in Fig. 6.30.
6.5
Eutectic Solidification
We next consider eutectic solidification, represented by:
L→α+β
There is a diverse medley of possible two-phased microstructures which can
be produced by eutectic solidification, even in the relatively simple case of solidification along one direction at a constant velocity. These include lamella
or plate structures, rods, discontinuous structures, and irregular distribution
of phases. The behavior is determined by a variety of factors including the interface energies, diffusion rate in the melt, volume fraction of the two phases,
preferred orientations for growth, and relative growth rates of the phases.
202
CHAPTER 6. SOLIDIFICATION
a)
b)
c)
d)
e)
Figure 6.30: Schematic showing development of dendrites in alloy solidification. a) Planar interface. b) Protrusion develops, grows rejecting more
solute into the surrounding region. This slows growth near protuberance. c)
Dendrite grows more surrounding regions slow. d) Dendrites grow. e) Final
cellular microstructure.
For the purposes of illustration, we consider the unidirectional solidification at velocity v of a two-component eutectic liquid, resulting in a lamellar
structure comprised of alternating plates of A-rich α-phase and B-rich βphase. This process is illustrated in Fig. 6.31. Solidification of α causes
buildup of B in the liquid in front of the growing interface. This excess of
B must diffuse laterally in the liquid to the β-phase solidification front. A
similar process occurs for the excess A in front of the β-phase regions. The
resulting atomic fluxes are shown schematically in Fig. 6.31.
One important aspect of two-phase solidification is the presence of interfaces between the phases. The energy contributions of these interfaces affect
growth morphology and rate. For example, consider the interfacial tension
equilibrium at the growing solid-liquid interface, as shown schematically in
Fig. 6.32. If we assume that the surface tension with the liquid is the same
for both α and β-phases, and we assume that the interfaces are in force
equilibrium, we find:
γαβ = 2γL cos θ
For many cases, the solid-liquid interface energy is much higher than that
between the two phases, that is γL γαβ , so that θ ∼ π/2. If however,
the α-β interface energy is larger than twice the solid-liquid interface en-
6.5. EUTECTIC SOLIDIFICATION
203
ergy (γαβ > 2γL ), then interfacial tension equilibrium is not possible, and a
discontinuous microstructure will form.
v
α
β
JA
JB
α
λ
β
α
l
d β
Figure 6.31: Schematic of solidification of eutectic liquid into lamellar microstructure with lamella spacing λ, showing the flux of A atoms into the
growing α-phase region and the flux of B atoms into the growing β-phase
region.
Even when the interface force equilibrium allows formation of lamellar
structures, the growth rate and lamellar spacing reflect the energy cost of
the α-β interface. The shaded volume element in Fig. 6.31, which has length
l, depth d and height λ, has interface energy 2ldγαβ , so the energy per volume
is:
2ldγαβ
2γαβ
=
ldλ
λ
Hence the formation of this eutectic lamella from the melt results in a change
in free energy per volume of:
∆G(λ) = Gsolid − Gliquid = ∆G(∞) +
2γαβ
λ
204
CHAPTER 6. SOLIDIFICATION
γ αβ
γL
α
β
Figure 6.32: Schematic of interface
tension balance in eutectic growth.
θ
L
γL
Here ∆G(∞) is the bulk free energy change associated with the solidification
of the liquid to the eutectic solid, which can be approximated by the Turnbull
extrapolation:
∆HF ∆T
∆G(∞) = −
TE
where ∆HF is the (positive) heat of fusion for melting of the eutectic, TE is
the equilibrium eutectic temperature, and ∆T is the undercooling given by
∆T = TE − T . Note that at temperatures below TE , ∆G(∞) is negative,
but for small undercoolings ∆G(λ) may not be, since the energy cost of the
interfaces is a positive quantity. For a given lamellar spacing λ, the minimum
undercooling ∆T1 required for eutectic solidification is found by setting
∆G(λ)|∆T =∆T1 = 0
Solving for ∆T1 we find:
2γαβ TE
∆HF λ
At undercoolings greater than ∆T1 , where the temperature is below TE −∆T1 ,
the free energy change ∆G(λ) is negative and eutectic solidification with a
lamellar spacing λ can occur.
Conversely, by similar arguments, we can find the minimum spacing λmin
which is stable for a given level of undercooling. This is:
∆T1 =
λmin =
2γαβ TE
∆HF ∆T
(6.16)
This can be represented schematically in a couple of ways. If we plot the free
energy as a function of composition at a given temperature, we see that the
free energy of the solid phases is a function of the lamellar spacing. This is
shown in Fig. 6.33, where the two extremes are shown. For infinite lamella
spacing, the free energy curves of the solid phases have a tangent line lying
6.5. EUTECTIC SOLIDIFICATION
205
below the liquid free energy curve by an amount ∆G(∞). For the case where
λ = λmin the solid free energy curves share a tangent with the liquid free
energy curve. This raising of the solid phase free energy curves has the
effect of lowering the liquidus lines on the equilibrium phase diagram, since
solidification with a finite lamellar spacing will only occur at temperatures
below TE − ∆T1 . This is shown schematically in Fig. 6.34.
Gα(λ min)
Gα(∞)
GL
Gβ(λ min)
∆G(∞)
Gβ(∞)
x
Figure 6.33: Schematic of free energy representation for eutectic solid. Shown
are the free energy for the liquid and bulk solid phases as a function of
composition. For a lamellar spacing of λmin , the free energy cost of the
interface formation raises the free energy curves of the solid phases to have
a common tangent with the liquid free energy curve.
During actual growth the undercooling ∆T will be greater than ∆T1
which is the minimum required for interface formation. If we write:
∆T = ∆T1 + ∆T2
then the excess undercooling ∆T2 is that which is available to drive the other
growth processes, such as the atomic fluxes mentioned earlier and shown
schematically in Fig. 6.31. For growth at lamella spacing greater than the
minimum allowed λmin , the free energy curves for the solid phases will be
raised only part of the way to the point of tangency with the liquid free
energy curve, as shown in Fig. 6.35a. The free energy driving force |∆G(∞)|
provides an amount 2γαβ /λ for interface formation and an amount:
2γαβ
2γαβ 2γαβ
1
1
|∆G(∞)| −
=
= 2γαβ
−
−
λ
λmin
λ
λmin λ
206
CHAPTER 6. SOLIDIFICATION
λ=∞
λ<∞
T
∆T
1
x
Figure 6.34: Schematic phase diagram for a eutectic system showing the
lowering of the liquidus lines for finite lamellar spacing.
for other growth processes, including driving the diffusion flux. By using the
Turnbull extrapolation we can associate this driving force to the undercooling
∆T2 (= ∆T − ∆T1 ):
∆HF ∆T2
1
1
= 2γαβ
−
TE
λmin λ
Solving for ∆T2 and inserting λmin from Eqn. 6.16 we find:
∆T2
2γαβ TE
=
∆HF
1
1
−
λ
λmin
λmin
= ∆T 1 −
λ
(6.17)
To deal with the complicated problem of atomic flux in the concentration
profiles in front of the growing eutectic interface we make some simplifying
assumptions. We first assume that the liquid directly in front of the interface
is in equilibrium with the growing phase, that is, the atomic fraction of B
atoms in front of the β-phase is xLβ and that in front of the α-phase is xLα ,
as shown in Fig. 6.35a. These have corresponding atomic concentrations
cLβ and cLα , which are related to the atomic fraction by a factor of the
atomic volume. Hence there is a composition change along the interface of
∆c = cLα − cLβ . Further, we assume that the liquidus lines near the eutectic
are linear with slopes which are equal in magnitude but opposite in sign. This
allows us to deduce a simple relationship between the undercooling ∆T2 and
the composition difference. If we take m to be the magnitude of the slope of
6.5. EUTECTIC SOLIDIFICATION
207
GL
Gβ (λ min )
Gβ (λ)
2γ αβ
λ min
∆G(∞)
2γαβ
λ
xαL
xLβ
x
Gβ(∞)
xLα
xβL
λ=
∞
λ<∞
T
∆T
∆T1
∆T2
xLβ
x
xLα
Figure 6.35: Schematic of the a) free energy and b) phase diagram for conditions of eutectic growth .
208
CHAPTER 6. SOLIDIFICATION
the liquidus line, and VA to be the (constant) atomic volume, then we have
(referring to Fig. 6.35b):
∆T2 =
m Lα
m
x − xLβ = VA ∆c
2
2
Solving for ∆c we find:
2∆T2
(6.18)
mVA
Hence there exists a concentration gradient along the interface which is
approximately the concentration difference ∆c divided by the distance over
which it occurs. That is:
dc
∆c
≈
dy
λ/2
where y is measured along the interface. There will be a flux of atoms along
the interface in response to this gradient with magnitude given by:
∆c =
Jy =
dc −DL
dy =
2DL ∆c
λ
(6.19)
where DL is the diffusivity in the liquid in front of the growing interface. For
a given undercooling, the growth rate of the eutectic phase will be determined
by this flux. During a time increment dt, the flux will carry 2Jy dt B atoms
per area away from the front of the growing α-phase region. During this
time, the interface will advance a distance dz which will expel (cE − cαL )
from the growing region, where cαL is the concentration corresponding to
xαL in Fig. 6.35a and cE is the eutectic concentration. Equating these two
B atom amounts we find:
cE − cαL dz = 2Jy dt
where the factor of 2 accounts for the flux to adjacent slabs of β-phase on
either side of the growing α-phase. Solving for the growth velocity and using
Eqn. 6.19 for Jy , Eqn. 6.18 for ∆c, and Eqn. 6.17 for ∆T2 we find:
2Jy
− cαL
4DL ∆c
=
λ (cE − cαL )
8DL
λmin
∆T 1
=
1−
(cE − cαL ) mVA λ
λ
v =
cE
(6.20)
6.5. EUTECTIC SOLIDIFICATION
209
We can see that, for a given undercooling, there will be a peak in growth
velocity as a function of lamellar spacing. For large λ the diffusion is slow
since the composition difference is spread out over a large distance so the
composition gradient is small. Conversely, as λ approaches λmin the driving
force is entirely consumed by interface formation.
By differentiating Eqn. 6.20 we can find the lamellar spacing which corresponds to the fastest growth, which will be the most likely lamellar spacing
to occur. We find:
dv =0
dλ λ=λ∗
→
λ∗ = 2λmin
Inserting this into Eqn. 6.20 above we find maximum growth velocity:
2DL ∆T 1
− cαL mVA λmin
DL ∆HF (∆T )2
=
γαβ mVA TE (cE − cαL )
2
1
16DL γαβ TE
=
E
αL
∆HF mVA (c − c ) λ∗
vmax =
cE
so we see that the maximum velocity is proportional to the square of the
undercooling or to the square of the reciprocal of the lamellar spacing.