Solidification Most 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
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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