to enhance the knowledge about phase transformation, and clear the basics of concept.

1. Phase Transformation - Basic Concepts

2. Figure 9-2 An interface is created when a solid
forms from the liquid.

3. Figure 9-3 The total free energy of the solid-liquid system
changes with the size of the solid.
SL

4. Phase Transformation - Basic Concepts
• Homogeneous nucleation
– a sphere of radius r of a solid forms in a
liquid
GL: free energy per volume of liquid
GS: free energy per volume of solid
γ: solid-liquid Interface free energy per area of the A=4r2
V= 4/3 r3

5. Phase Transformation - Basic Concepts

6. Phase Transformation - Basic Concepts
In addition, the phase transformation process often
involves the diffusion of atoms which has its own
activation energy, Qd.

7. -The total nucleation rate (nuclei per volume per second)
Phase Transformation - Basic Concepts
N

8. Phase Transformation - Basic Concepts

9. Driving Forces - Examples
Driving Force Equation Approx. Value Force
(MPa)
Stored energy of
deformation
1/2rmb2 r~1015 10
Grain boundaries 2sb/R sb~0.5 J/m2 10-2
Surface energy 2Dss/d D~10-3m 10-4
Chemical driving
force
R(T1-
T0)c0lnc0
5%Ag in Cu @
300°C
10+2
Magnetic field m0H2Dc/2 H~107A/m 10-4
Elastic energy t2(1/E1-
1/E2)/2
t~10 MPa 10-4
Temperature gradient DS
2lgradT/Wa
l~5x10-10 10-5

10. Quantitative Approach
• we can quantify the driving force for solidification (and allotropic
phase change).
• Assume that the free energies of the solid and liquid are
approximately linear with temperature.
Temperature
Molar
Free
Energy
GS
GL
TMelt
∆G
∆T

11. Driving force: solidification
• Writing the free energies of the solid and liquid as:
GS = HS - TSS
GL = HL - TSL
 ∆G = ∆H - T∆S
• At equilibrium, i.e. Tmelt, then the ∆G=0, so we can
estimate the melting entropy as:
∆S = ∆H/Tmelt = L/Tmelt
where L is the latent heat (enthalpy) of melting.
• Ignore the difference in specific heat between solid and
liquid, and we estimate the free energy difference as:
∆G  L-T(L/Tmelt) = L∆T/Tmelt

12. Driving force: Precipitation
• The other case of interest is
precipitation.
• This is more complicated because
we must consider the
thermodynamics of solids with
variable composition.
• Consider the chemical potential of
component B in phase alpha
compared to B in beta. This
difference, labeled as ∆Gn on the
right of the lower diagram is the
driving force (expressed as energy
per mole, in this case).
• To convert to energy/volume, divide
by the molar volume for beta: ∆GV =
∆Gn/Vm.
R
S
T
U
N
P
Q
Te

13. Driving force for nucleation
• It is important to realize the difference between the
driving force for the reaction as a whole, which is given
by the change in free energy between the
supersaturated solid solution and the two-phase mixture,
shown as ∆G0 in 5.3b (the distance N-S).
• Why a different free energy for nucleation? Because the
first nuclei of beta to appear do not significantly change
the composition of the parent material. Thus the free
energy change for nucleation is the rate of change of
free energy for the new, product phase (beta).

14. Driving force, contd.
• Based on the expression for the chemical potential of a
component (P&E 1.68):
µB = GB + RT lnBXB
• We can identify the chemical potential for B in the
supersaturated phase as µB
0, and the chemical potential
for B in precipitated beta in equilibrium with the alpha
phase as µB
e. Then the molar free energy difference,
∆Gn, (distance P-Q) is given by
∆Gn = µB
0 - µB
e = RT ln(B
0XB
0/B
eXB
e).
• We can simplify this for ideal solutions (=1) or for dilute
solutions (=constant) to:
∆Gn = RT ln(X0/Xe).

15. Details
• A more detailed approach considers the ratios of the
compositions of the phases in addition to the differences in
chemical potential [Martin et al., p10-11].
• Consider fig. 5.3b again. The triangle R-P-Q is similar
(mathematically) to the triangle R-T-U.
• The decrease in free energy per mole of precipitate is the
distance P-Q, which is smaller than the difference in
chemical potential for B in the supersaturated material at X0
and in the beta precipitate at Xbeta.
∆Gn = (µB
0 - µB
e){Xbeta-Xe}/{1-Xe}
= RT ln(B
0XB
0/B
eXB
e) {Xbeta-Xe}/{1-Xe} .
∆Gn = RT ln(X0/Xe) {Xbeta-Xe}/{1-Xe}.
• However, for precipitate that is pure B (or close to that) the
fraction in curly brackets, {Xbeta-Xe}/{1-Xe}, is close to one,
hence the result given on the previous slide.

16. Examples
• The driving force is very sensitive to the mole fractions of
the precipitate, Xbeta,, supersaturated matrix, X0, and
equilibrium matrix, Xe.
• For a ratio Xe/X0 = 10-3, Xbeta=1, and a temperature of
300K (i.e. very low solubility at a low enough
temperature), then the driving force is estimated to be -
17,000 J/mole. For the precipitation of pure Cu from Al,
for example, this is a feasible estimate.
• For many precipitation reactions, the volume fraction of
the precipitate, Vbeta= {X0 -Xe}/ {Xbeta-Xe}, is of order 0.1,
which means that the driving force is much less, ~ -1,700
J/mole. If the precipitate is not pure B, but incorporates
a significant amount of A, as for example in Al6Mn, then
there is a further reduction in the driving force.

17. Solid Solubility
• Before we can proceed, however, we need to look at the
variation in solubility with temperature. This variation can be
quite complex, so we will simplify the situation (considerably)
by considering a regular solid solution. See section 1.5.7 in
P&E. In short, the solubility can often be approximated by an
Arrhenius expression, where Q is the heat absorbed
(enthalpy) when I mole of beta dissolves in A as a dilute
solution.
mB

 GB

W 1 XB
 
2
 RT ln XB
From fig.1.36b,
GB

 mB

 DGB
XB  XB
e
DGB  RT ln XB
e
W 1 XB
e
 
2
XB
e
1
XB
e
 exp
DGB W
RT






 exp
DSB
R
exp
DHB W
RT






 A exp
Q
RT







18. Driving force as a function of temperature
• Despite the complications of the previous analysis, the
net result is startlingly simple.
• The driving force is, to first order, proportional to the
undercooling!
DGn  RTln
X0
Xe
 RT ln
Aexp
DHB
RT
e










Aexp
DHB
RT






 RTln exp
DHB
RTe

















  RT ln exp
DHB
RT












 RT
DHB
RTe

DHB
RT







  DHBT
T  Te
TTe








DGn  
DHB
Te
DT

19. Nucleation Rate
• Armed with this information
about driving force (note that
the proportionality to
undercooling is the same as
for solidification), we can
now look at how the
nucleation rate varies with
temperature.
• The energy barrier for
nucleation varies as the
square of the driving force:
the nucleation rate is
exponentially dependent on
the barrier height.

20. Elastic energy effect
• Elastic effects have
the effect of
offsetting the
undercooling by
∆GS.
• Diffusion limits the
rate at which atoms
can form a new
phase and hence
limits the nucleation
rate at low
temperatures.

21. Allotropic Phase Transformations
• These are, for example, eutectoid transformations (or
peritectoid).
• The calculation of driving force is similar:
– Notation (e.g. Fe-C system):
X0:= alloy composition (e.g. C in Fe)
X:= equilibrium austenite composition
X:= equilibrium ferrite composition
– Driving force:
∆Gn = {X - X}/X RT ln{(1-X/(1-X0)}.
• For a typical case of T=1000K, X = 0.04, X = 0.001,
X0:= 0.02 (equivalent to 0.4 weight % C), ∆Gn = -170
J/mole. This is much smaller than the previous
estimates for precipitation. Homogeneous nucleation is
essentially impossible!

22. Summary
• For solidification, the driving force is
proportional to the undercooling.
• For the case of precipitation where the
solid solubility is described by an ideal or
regular solution, the driving force is also
proportional to the undercooling.
• The driving force is the information that we
need in order to predict the nucleation
rate.

23. Thanks!
Any questions?
You can find me at
● @hamzaahmed
● hamzaahmed0696@mail.me