Chapter 8

( Solidification and Grain Size Strengthening)

Solidification:

It requires two steps

1- Nucleation

2- Growth

Nucleation:

It occurs when a small piece (particle) of solid called (embryo) with a critical size forms from liquid.

Growth of the solid occurs when more atoms from liquid are attached to \

it until no more liquid remains.

Solidification takes place when the free energy of the solid is less than of the liquid at the freezing temperature.

- As the free energy difference becomes larger, the solid will be more stable. This energy difference is ΔFv (volume free energy)

In order for the solid to form, an interface must be created separating the solid from the liquid.

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A surface free energy σ is associated with this interface As the interface gets

larger, the greater the increase in σ.

The total change in free energy ΔG Produced when the embryo forms is:

Where

= volume of a spherical embryo of radius r.

= surface area of the spherical embryo.

σ = surface free energy

ΔFv = volume free energy (which is a negative change)

The total change in the free energy depends on the size of the embryo.

If the embryo is small, further growth will cause the free energy to increase, so it will remelt and causes the free energy to decrease.

No nucleation and no growth. So, the metal remains liquid

Since the liquid is present below the equilibrium freezing temperature, the liquid is undercooled.

If the embryo is large (greater the critical radius), the total energy

decreases when the size of the embryo increase, nucleation has occurred

and growth of the nucleus begins.

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Homogeneous Nucleation:

It occurs when the undercooling becomes large enough to cause formation of a stable nucleus. That’s when embryo to exceed the critical size..

Estimating the Size of the Stable Nucleus:

The size of the critical radius (r*) is given by:

ΔH f = the heat latent of fusion (it represents the heat that is given up during the liquid-solid transformation) of the metal

ΔT = Tm –T (is the undercooling when the liquid temperature is T)

Tm = the equilibrium freezing temperature in K

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Q) Calculate the size of the critical radius and the number of atoms in

the critical nucleus when solid copper forms by homogeneous nucleation.

Solution:

From the table:

The lattice parameter for FCC copper is:

a= 3.615Å = 3.615 x 10-8 cm

Vunit cell = (a)3 = (3.615 x 10-8 )3 = 47.24 x 10-24 cm3

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The number of unit cells in the critical nucleus is:

In FCC there are 4 atoms/unit cells.

So, No. of atoms in r* = (4 atoms/cell x 174 cells/nucleus) = 696 atoms

Heterogeneous Nucleation:

It occurs on the surface of impurities in the molten metal. Impurities provide a surface on which the solid can form.

Grain Size strengthening by Nucleation:

It occurs by adding impurities to the liquid metal.

It is called grain refinement or inoculation

Titanium or Boron added to Al liquid

- Very small particles of Al3Ti or TiB2 forms- These particles are sites for heterogeneous nucleation - Grain refining produces large number of grains.- So, more grain boundaries,- Grain size strengthening occurs.

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Glasses:

In certain cases of very rapid cooling, nucleation of the crystalline

solid never occurs. Instead an unstable amorphous (noncrystalline) solid

forms.

Growth:

After the formation of nuclei, growth occurs as atoms are attached

to the solid surface.

It depends on the removal of heat during solidification.

a) Specific heat of the liquid:

- It is the heat required to change the temperature of a unit weight of

the material by one degree.

- It must be removed either by radiation or conduction, until the

liquid cools to the freezing temperature.

b) Latent heat of fusion

It represents the energy that evolved as the disordered liquid

structure transforms to a more stable crystal structure.

Types of growth mechanisms:

The difference between the two types of growth mechanisms arises

because of the different sinks for the latent heat.

1) Planar Growth

It occurs by the movement of smooth solid-liquid interface into the

Liquid, where the container or mold must absorb the heat.

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2) Dendritic Growth:

When nucleation is poor, the liquid undercools to a temperature below

the freezing temperature before the solids forms. The undercooled

liquid absorbs the heat.

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Under this condition, a small solid protuberance (dendrite) forms at the interface.

Secondary and tertiary dendrite arms can also form on the primary

stalks to speed the evolution of the latent heat.

In pure metals, dendritic growth normally represents only a small

fraction of the total growth.

c=specific heat of the liquid, ΔT = heat that the undercooled liquid can absorb

ΔHf = latent heat represents the total heat that must be given up during

solidification.

As the undercooling ΔT increases, more dendritic growth occurs.

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Q76) Calculate the fraction of growth that occurs dendritically in copper which (a) nucleates homogeneously and (b) nucleates heterogeneously with 100C undercooling. Where latent heat of fusion for copper is 1628x10 6 J/m3, specific heat is 4.4x106 J/m3, undercooling for homogeneous nucleation is 2360C

Solution:

(a) For homogeneous nucleation

(b) For 100 C undercooling

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