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SolidificationMost metals are melted and then cast into semifinished or finished
shape.
Solidification of a metal can be divided into the following steps:
•Formation of a stable nucleus
•Growth of a stable nucleus.
Formation of stable
nuclei
Growth of crystals Grain structure
Polycrystalline Metals
In most cases, solidification begins from multiple sites, each
of which can produce a different orientation
The result is a “polycrystalline” material consisting of many
small crystals of “grains”
Each grain has the same crystal lattice, but the lattices are
misoriented from grain to grain
Driving force: solidification
• Writing the free energies of the solid and liquid as:
GVS = HS - TSS
GVL = HL - TSL
∴∴∴∴ ∆GV = ∆H - T∆S
• At equilibrium, i.e. Tmelt, then the ∆GV = 0, so we can estimate
the melting entropy as:
∆S = ∆H/Tmelt
where -∆H 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:
SL AA ⇒ For the reaction to proceed to the right ∆GV
must be negative.
MeltMelt
VT
THH
T
THG
∆×∆=∆−∆≅∆
NUCLEATION
The two main mechanisms by which nucleation of a solid particles in
liquid metal occurs are homogeneous and heterogeneous nucleation.
Homogeneous Nucleation
Homogeneous nucleation occurs when there are no special objects
inside a phase which can cause nucleation. For instance when a pure
liquid metal is slowly cooled below its equilibrium freezing
temperature to a sufficient degree numerous homogeneous nuclei are
created by slow-moving atoms bonding together in a crystalline
form.
Consider the free energy changes when some atoms in the liquid
collapse and agglomerate to form a solid of radius r.
The energy changes involve two terms:
(a)The chemical free energy change associated with the transfer of
atoms from liquid to solid state (∆∆∆∆Gv);
(b)the interfacial energy (γγγγ)due to the creation of new interface
(liquid-solid interface)!
Assume that ∆∆∆∆Gv is the change in free energy per unit volume and
∆∆∆∆GT is the total Free energy change, r is the radius of the nucleus
γππ 2
3
43
4rG
rG VT +∆=∆⇒
084 0
2
0
0
=+∆=∆
=
γππ rGrdr
GdV
rr
T
VGr
∆−=⇒
γ20
2
32
00
3
16
3
4)(
V
TG
rrG
∆==∆⇒
πγπγ
r0 : critical radius;
* for r < r0 : the growth of the droplet ⇒ ∆∆∆∆GT ↑↑↑↑ ⇒ the embryos
should shrink and disappear!
* for r > r0 : the growth of the droplet ⇒ ∆∆∆∆GT ↓↓↓↓ ⇒ the nuclei
could steadily grow!
γ
TH
Tr
f
m
∆∆=⇒
γ20
m
f
VT
THG
∆∆=∆
Where:
-∆∆∆∆Hf = Latent Heat of fusion
∆∆∆∆T = amount of undercooling at which
the nucleus form
Heterogeneous NucleationHeterogeneous nucleation is the nucleation that occurs in a liquid
on the surfaces of its container, insoluble impurities or other
structural material (catalyst) which lower the critical free energy
required to form a stable nucleus.
ro
γγγγL-C
γγγγL-S
γγγγS-C
θθθθ
Catalyst
Solid
Liquid γγγγL-C = γγγγS-C + γγγγL-S Cosθθθθ
Supersaturated Solutions
• If the liquid is just at the freezing point, only a few molecules stick, because they have comparatively high energy
• As the liquid is cooled, more molecules can form into nuclei.• When the nucleus is big enough (because of undercooling) the
supercooled liquid suddenly changes to a solid.
• Metals often experience undercooling of 50 to 500 oC
• Homogeneous nucleation usually only occurs in the lab.
• Impurities provide a “seed” for nucleation
• Solidification can start on a wall.
• It’s like cloud seeding, or water condensing on the side of a glass.
• Adding impurities on purpose is called inoculation
• Nucleation begins
• Chill Zone
• Columnar Zone (a) There may be dendrites in the columnar zone (b)
Grains grow in preferred directions
• Equiaxed Zone
Imperfections in Solids
Materials are often stronger when they have defects. The mechanical
and electrical properties of a material are affected by the presence of
defects. The study of defects is divided according to their dimension:
Defects in Solids
Gemstones – Hope Diamond blue color due to boron impurities (ppm)
• Metals - ductility, stiffness, brittleness, etc. drastically affected
Examples of the large impact of defects:
Bonding
+
Structure
+
Defects
Properties
The defects have a profound
effect on the macroscopic
properties of materials
The processing determines the defects
Chemical
Composition
Type of Bonding
Crystal Structure
Thermomechanical
Processing
Micro
structure
0D – point defects: vacancies and interstitials impurities.
1D – linear defects: dislocations (edge, screw, mixed)
2D – planar defects: grain boundaries, surfaces.
3D – extended defects: pores, cracks.
Point Defects (0D)
Vacancies and Self-Interstitials
A vacancy is a lattice position that is vacant because the atom is
missing. It is created when the solid is formed.
They occur naturally as a result of thermal vibrations.
An interstitial is an atom that occupies a place outside the normal
lattice position.
It may be the same type of atom as the others (self interstitial) or an
impurity atom.
In the case of vacancies and interstitials, there is a change in the
coordination of atoms around the defect. This means that the forces
are not balanced in the same way as for other atoms in the solid
(lattice distortion).
Vacancy - a lattice position that is vacant because the atom is missing.
Interstitial - an atom that occupies a place outside the normal lattice
position. It may be the same type of atom as the others (self
interstitial) or an impurity interstitial atom.
The number of vacancies formed by thermal agitation follows an
Arrhenius type of equation:
where NA is the total number of atoms in the solid, QV is the energy
required to form a vacancy (per atom or per mole), kB is Boltzmann
constant, R is the gas constant and T the temperature in Kelvin.
Note that kT(300 K) = 0.025 eV (room temperature) is much smaller
than typical vacancy formation energies.
For instance, QV(Cu) = 0.9 eV/atom.
This means that NV/NA at room temperature is exp(-36) = 2.3 × 10-16,
an insignificant number.
Thus, a high temperature is needed to have a high thermal
concentration of vacancies. Even so, NV/NA is typically only about
0.0001 at the melting point.
−=
Tk
QNN
B
VAV exp
−=RT
QNN V
AV exp
Calculate equilibrium number of vacancies per cubic meter for
copper at 1000oC
Given:
Activation Energy per vacancy = 0.9 eV/atom
atomic weight of copper = 63.5 g/mol
and density at 1000oC = 8.40 g/cm3
Boltzmann’s constant kB=1.38 × 10-23 J/atom-K=8.62 × 10-5 eV/atom-K
325
5
328
0
328
336323
1022
127310628
901008
12731000
1008
563
10408100236
mvacanciesXN
KKeVX
eVmatomsXN
kT
QNN
toequalisKCatvacanciesofnumbertheThus
matomsXN
molg
mcmcmgmolatomsX
A
NN
V
V
vv
Cu
a
/.
))(/.(
).(exp)/.(
exp
)(,
/.
/.
)/)(/.)(/.(
=
=
−=
=
==
−
ρ
Solution:
Determine N, number of atomic sites per cubic meter for Cu
-
+
Point defects in ionic crystals are charged. Coulombic forces are
large and any charge imbalance wants to be balanced. Charge
neutrality --> several point defects created:
Frenkel defect: a cation vacancy and a cation interstitial or an
anion (negative ion) vacancy and anion interstitial. (Anions are
larger so it is not easy for an anion interstitial to form).
Schottky defect: pair of anion and cation vacancies
Imperfections in Ceramics
Frenkel defectSchottky defect
• Frenkel or Schottky defects: no change in cation to anion ratio →→→→compound is stoichiometric
• Non-stoichiometry (composition deviates from the one predicted
by chemical formula) may occur when one ion type can exist in
two valence states, (e.g. Fe2+, Fe3+). In FeO, usual Fe valence state
is 2+. If two Fe ions are in 3+ state, then a Fe vacancy is
required to maintain charge neutrality →→→→ fewer Fe ions →→→→ non-
stoichiometry
Imperfections in Ceramics
� Impurity atoms can exist as either substitutional or interstitial
solid solutions
�Substitutional ions substitute for ions of like type
� Interstitial ions are small compared to host structure (anion
interstitials are unlikely).
Impurities in Ceramics
Interstitial impurity atom
Substitutional impurity ions
�Solubility high if
ion radii and charges
match
�Incorporation of
ion with different
charge state requires
compensation by
point defects.
Impurities in Solids
All real solids are impure. A very high purity material, say
99.9999% pure (called 6N – six nines) contains ~ 6 × 1016
impurities per cm3.
Impurities are often added to materials to improve the properties.
For instance, carbon added in small amounts to iron makes steel,
which is stronger than iron. Boron impurities added to silicon
drastically change its electrical properties.
Solid solutions are made of a host, the solvent or matrix) which
dissolves the solute (minor component). The ability to dissolve is
called solubility. Solid solutions are:
•homogeneous
•maintain crystal structure
•contain randomly dispersed impurities (substitutional or
interstitial)
Solid SolutionSolid Solution Solids dissolve other solids
Solid solutions are made of a host (the solvent or matrix) which
dissolves the minor component (solute). The ability to dissolve is
called solubility.
So, most of engineering materials are solid solutions, i.e., alloys:
solvent and solutes
Two ways: depending on the size and host structure
Substitutional
Ni/Cu
- solvent: usually the element
present in greatest amount
(sometimes referred to as “host
atoms”)
- solute: usually the element
present in minor concentration.
Interstitial
C/Fe
For fcc, bcc, hcp structures the voids (or interstices) between the
host atoms are relatively small ⇒ atomic radius of solute should be
significantly less than solvent.
Normally, max. solute concentration ≤ 10%, (2% for C-Fe)
Second Phase: as solute atoms are added, new compounds/structures
are formed, or solute forms local precipitates. Whether the addition of
impurities results in formation of solid solution or second phase
depends the nature of the impurities, their concentration and
temperature, pressure…
•Similar atomic size (to within 15%)
•Similar crystal structure
•Similar electronegativity (otherwise a compound is formed)
•Similar valence
Composition can be expressed in weight percent, useful when
making the solution, and in atomic percent, useful when trying to
understand the material at the atomic level.
Example
Ni is completely miscible in Cu (all rules apply)
Zn is partially miscible in Cu (different valence, different crystal
structure)
Factors for high solubility in Substitutional alloys
(Hume-Rothery Solubility Rules)
Ni - Cu binary isomorphous
alloyNi Cu
crystal structure FCC FCC
atomic radius 0.125 0.128
2.4%
electronegativities 1.8 1.8
valence 2 + 2 +
Solubility Cu in Ni 100%
Pb Cu
FCC FCC
0.175 0.128
36.7%
1.6 1.8
2+, 4+ 2 +
Limited solubility
(eutectic) alloys
Solubility Cu in Pb 0.1%
Solid Solution:
homogeneous
maintain crystal structure
contain randomly dispersed impurities
(substitutional or interstitial)
Second Phase: as solute atoms are added, new compounds /
structures are formed, or solute forms local precipitates
Examples
(A)Calculate the critical radius (in nanometers) of a
homogeneous nucleus that forms when pure liquid copper
solidifies. Assume ∆∆∆∆T(undercooling) = 0.2Tmelt.
For Cu Tm=1083oC; Heat of fusion (∆∆∆∆Hf) = 1826J/cm
3; Surface
Energy (γγγγ) = 177x10-7 J/cm2; Lattice parameter of FCC copper
= a=0.361nm
Calculate the number of atoms in the critical-sized nucleus at this
undercooling.
(A)
We make use of the equation for a spherical nucleus to calculate
the size of the critical nucleus
( )( )( )
nmcmxTcmJ
TcmJx
TH
Tr
m
m
f
m 969.01069.92.0.1826
.1017722 8
3
27
0 ===∆∆
= −−
−−γ
Then, the volume of the critical nucleus is
The volume of an FCC cell is
The number of cells in the critical nucleus is
As the number of atoms in an FCC cell is 4 then the total number of
atoms is
( ) 333 82.397.03
4
3
4nmrnucleuscritVol ===−− ππ
( ) 333 047.0361.0 nmnmacellVol ===−
cellsnm
nm
cellVol
nucleusVol34.81
047.0
82.33
3
==−
−
atomsxrofAtomsTotalNumbe 32534.814 ==
(B) Calculate (a) the equilibrium number of vacancies per cubic
meter in pure magnesium at 450oC. Assume that the energy of
formation of a vacancy in pure magnesium is 0.89eV. (b) What is
the vacancy fraction at 600oC?
(Boltzmann’s constant kB=8.62x10-5eV/K)
Atomic weight of Mg = 24.31g.mol-1
Density of Magnesium = 1.74g.cm-3
(B)
First, we calculate the number of magnesium atoms in a cubic
meter
322
15
328
0
328
336323
10712
72310628
89010314
723450
10314
3124
10741100236
−
−−−
−
=
−=
−=
=
==
mvacanciesxN
KKeVx
eVmatomsxN
Tk
QNN
toequalisKCatvacanciesofnumbertheThus
matomsxN
molg
mcmcmgmolatomsX
A
NN
V
V
B
vv
Mg
a
..
))(..(
).(exp)..(
exp
)(,
/.
/.
).)(/.)(/.(ρ
Vacancy fraction at 600oC
6
151036.7
)873)(.1062.8(
)89.0(exp −
−−=
−= x
KKeVx
eV
N
NV