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