PowerPoint 프레젠테이션 - SNU OPEN COURSEWARE

재 료 상 변 태

Phase Transformation of Materials

2008. 11. 06.

박 은 수

서울대학교 재료공학부

Contents for previous class

• Coherency Loss

• Type of Interfaces

• Solid/Liquid Interfaces

• Interface Migration

Faceted interface > L_f/T_m≈ 4R > Diffuse interface^{: 대부분의 금속} ~ R

At interface,

Contents for today’s class

• Nucleation in Pure Metals

• Homogeneous Nucleation

- formation of atomic cluster - homogeneous nucleation rate

• Heterogeneous Nucleation

• Nucleation of melting

4. Solidification

: Liquid Solid

- casting

- single crystal growth - directional solidification - rapid solidification

4.1. Nucleation in Pure Metals T_m : G_L = G_S

- Undercool (supercool) for nucleation: 250 K ~ 1 K

<Types of nucleation>

- Homogeneous nucleation - Heterogeneous nucleation

Driving force for solidification (ch.1.2.3)

4.1.1. Homogeneous Nucleation

T

m

T G L Δ

=

Δ

^{L : ΔH = H}

L – H^S (Latent heat) T = T_m - ΔT

,

G^L = H^L – TS^L G^S = H^S – TS^S

4.1.1. Homogeneous Nucleation

L V L

S V G

V

G₁ = ( + ) G₂ = V_SG_V^S +V_LG_V^L + A_SLγ_SL

L V S

V G

G ,

SL SL

S V L

V

S

G G A

V G

G

G = − = − − + γ

Δ

_{2 1}

( )

SL V

r

r G r

G π

³

4 π

²

γ

3

4 Δ +

−

= Δ

: free energies per unit volume

for spherical nuclei (isotropic) of radius : r

Gibbs-Thompson Equation

3 ( )

4

r l 3 V

G

π

r G

Δ = × Δ

8 r 4 r 2 G_V

μ π γ π

Δ = = Δ

ΔG of a spherical particle of radius, r ²

( ) 4

Gr s

π γ

r

Δ =

ΔG of a supersaturated solute in liquid in equilibrium with a particle of radius, r

Equil. condition for open system Δμ should be the same.

r r

V_m γ 2

/mole or

r γ

2 / per unit volume

* /

2 r

G

_V

= γ

_SL

Δ

V SL

r G

= Δ

∗ 2γ

r*: in (unstable) equilibrium with surrounding liquid

4.1.1. Homogeneous Nucleation

* SL SL m

V V

2 2 T 1

r G L T

γ ^⎛ γ ^⎞

= Δ = ⎜⎝ ⎟⎠ Δ

Fig. 4.2 The free energy change associated with homogeneous nucleation of a sphere of radius r.

2 2

2 3 2

3

) (

1 3

16 )

( 3

* 16

T L

T G G

V m SL V

SL

⎟⎟ Δ

⎠

⎜⎜ ⎞

⎝

= ⎛

= Δ

Δ π γ πγ

r < r* : unstable (lower free E by reduce size) r > r* : stable (lower free E by increase size) r* : critical nucleus size

Why r^* is not defined by ΔG_r= 0?

Δ ≅ Δ

m

G L T T

r* & ΔG ↓ as ΔT ↑

Formation of Atomic Cluster

At the T_m, the liquid phase has a volume 2-4% greater than the solid.

Fig. 4.4

A two-dimensional representation of an

instantaneous picture of the liquid structure.

Many close-packed crystal-like clusters (shaded) are instantaneously formed.

Formation of Atomic Cluster

3 2

4 4 ,

r 3 V SL

G πr G π γr

Δ = − Δ +

When the free energy of the atomic cluster with radius r is by

how many atomic clusters of radius r would exist in the presence of the total number of atoms, n₀?

1 2 3 4 m 1 m

A →A →A →A →L→A ₋ → A

1 2

2 1exp G

n n

kT

⎛ Δ → ⎞

= ⎜− ⎟

⎝ ⎠

2 3

3 2 exp G

n n

kT

⎛ Δ → ⎞

= ⎜− ⎟

⎝ ⎠

3 4

4 3 exp G

n n

kT

⎛ Δ → ⎞

= ⎜− ⎟

⎝ ⎠

M ¹

1 exp

m m

m m

n n G

kT

− →

−

⎛ Δ ⎞

= ⎜− ⎟

⎝ ⎠

1 2 2 3 1

1exp

m m

m

G G G

n n

kT

→ → − →

⎛ Δ + Δ + + Δ ⎞

= ⎜− ⎟

⎝ ⎠

L

1 1 exp

m m

n n G

kT

⎛ Δ → ⎞

= ⎜− ⎟

⎝ ⎠

0 exp ^r

r

n n G

kT

⎛ Δ ⎞

= ⎜⎝− ⎟⎠

radius

small driving force

large driving force

Gibbs Free Energy

Compare the nucleation curves

between small and large driving forces.

Formation of Atomic Cluster

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Δ−

= RT

n G

n_{r 0} exp ^r n_o : total # of atoms.

ΔG_r : excess free energy associated with the cluster

# of cluster of radius r

- holds for T > T_m or T < T_m and r ≤ r*

- n_r exponentially decreases with ΔG_r

Ex. 1 mm³ of copper at its melting point (no: 10²⁰ atoms)

→ ~10¹⁴ clusters of 0.3 nm radius (i.e. ~ 10 atoms)

→ ~10 clusters of 0.6 nm radius (i.e. ~ 60 atoms)

→ effectively a maximum cluster size, ~ 100 atoms

~ 10^-8 clusters mm^-3 or 1 cluster in ~ 10⁷ mm³

Formation of Atomic Cluster

Fig. 4.5 The variation of r* and r_max with undercooling ΔT The number of clusters with r^* at < ΔT_N is negligible.

4.1.2. The homogeneous nucleation rate - kinetics

How fast solid nuclei will appear in the liquid at a given undercooling?

C₀ : atoms/unit volume

C^* : # of clusters with size of C* ( critical size )

) exp(

* hom

0 kT

C G

C^∗ = − Δ

clusters / m³

The addition of one more atom to each of these clusters will convert them into stable nuclei.

) exp(

* hom 0

hom kT

C G f

N = _o − Δ

nuclei / m³∙s

f_o ~ 10¹¹ : frequency ∝ vibration frequency energy of diffusion in liquid surface area

C_o ~ 10²⁹ atoms/m³

2 2

2 3

) (

1 3

* 16

T L

G T

V m SL

⎟⎟ Δ

⎠

⎜⎜ ⎞

⎝

= ⎛

Δ πγ

) } exp{ (

₂

0

hom

T

C A f

N

_o

− Δ

≈

^A _{L kT}^T

V

m SL 2

2 3

3 16πγ

=

: insensitive to Temp

where

4.1.2. The homogeneous nucleation rate - kinetics

Fig. 4.6 The homogeneous nucleation rate as a function of undercooling ∆T. ∆T_N is the critical undercooling for homogeneous nucleation.

→ critical value for detectable nucleation - critical supersaturation ratio

- critical driving force - critical supercooling

- hom

* 3 1

1 cm s when G ~ 78 k

N ≈ ⁻ Δ T

How do we define ΔT_N?

Under suitable conditions, liquid nickel can be undercooled (or supercooled) to 250 K below T_m (1453^oC) and held there indefinitely without any transformation occurring.

Normally undercooling as large as 250 K are not observed.

The nucleation of solid at undercooling of only ~ 1 K is common.

In the refrigerator, however, water freezes even ~ 1 K below zero.

In winter, we observe that water freezes ~ a few degrees below zero.

Which equation should we examine?

) exp(

* hom 0

hom kT

C G f

N _o Δ

−

=

* ( ) ( )

3 3 2

SL SL m

2 2 2

V V

16 16 T 1

G 3 G 3 L T

πγ ^⎛ πγ ^⎞

Δ = Δ =⎜⎝ ⎟⎠ Δ

Why this happens? What is the underlying physics?

Real behavior of nucleation

Nucleation becomes easy if γ _SL ↓ by forming nucleus from mould wall.

From

Fig. 4.7 Heterogeneous nucleation of spherical cap on a flat mould wall.

4.1.3. Heterogeneous nucleation

2 2

2 3

) (

1 3

* 16

T L

G T

V m SL

⎟⎟ Δ

⎠

⎜⎜ ⎞

⎝

= ⎛

Δ πγ

SM SL

ML γ θ γ

γ = cos +

SL SM

ML γ γ

γ

θ ( )/

cos = −

het S v SL SL SM SM SM ML

G V G A γ A γ A γ

Δ = − Δ + + −

3 2

4 4 ( )

het 3 V SL

G ^⎧ πr G π γr ^⎫S θ

Δ = −⎨ Δ + ⎬

⎩ ⎭

where S( )θ = (2 cos )(1 cos ) / 4+ θ − θ ²

In terms of the wetting angle (θ ) and the cap radius (r)

S(θ) has a numerical value ≤ 1 dependent only on θ (the shape of the nucleus)

* *

( )

hom

G

het

S θ G

Δ = Δ

Fig. 4.8 The excess free energy of solid clusters for homogeneous and heterogeneous nucleation. Note r* is independent of the nucleation site.

) 3 (

* 16

* 2 ₂

3 θ

πγ

γ S

G G G and

r

V SL V

SL ⋅

= Δ Δ Δ

=

( ) S θ

* *

( )

hom

G

het

S θ G

Δ = Δ

^{θ = 10 → S(θ) ~ 10-4}

θ = 30 → S(θ) ~ 0.02 θ = 90 → S(θ) ~ 0.5

4.1.3. Heterogeneous nucleation

The Effect of ΔT on ∆G*_het & ∆G*_hom?

Fig. 4.9 (a) Variation of ∆G* with undercooling (∆T) for homogeneous and heterogeneous nucleation.

(b) The corresponding nucleation rates assuming the same critical value of ∆G*

-3 1

hom 1 cm

N ≈ s⁻

Plot ∆G*_het & ∆G* _hom vs ΔT and N vs ΔT.

θ

VA

VB

Barrier of Heterogeneous Nucleation

4

) cos

cos 3 2 ( 3

) 16 3 (

* 16

3 2

3 2

3 θ πγ θ θ

πγ ⋅ − +

= Δ Δ ⋅

= Δ

V SL V

SL

S G G G

θ θ

⎛ − + ⎞

Δ = Δ ⎜ ⎟

⎝ ⎠

3

* 2 3 cos cos

4

*

sub homo

G G

2 3 cos cos3

4 ( )

A

A B

V S

V V

θ θ θ

− +

= =

+

How about the nucleation at the crevice or at the edge?

* *

( )

hom

G

het

S θ G

Δ = Δ

* homo

3

4 ΔG

homo^*

1

2 ΔG 1

_homo^*

4 ΔG

Nucleation Barrier at the crevice

contact angle = 90 groove angle = 60

What would be the shape of nucleus and the nucleation barrier for the following conditions?

* homo

1

6 ΔG

How do we treat the non-spherical shape?

⎛ ⎞

Δ = Δ ⎜ ⎝ + ⎟ ⎠

* * A

sub homo

A B

G G V

V V

Substrate

Good Wetting

VA

VB Substrate

Bad Wetting

VA

VB

Effect of good and bad wetting on substrate

Nucleation inside the crevice

GV

V

G = Δ

Δ *

2

* 1

V* : volume of the critical nucleus(cap or sphere) Wetting angle이 크더라도nucleation이 가능 할 수 있다.

(그러나 crack의 입구는 critical radius 보다 커야 함.)

Although nucleation during solidification usually requires some undercooling, melting invariably occurs at the equilibrium melting temperature even at relatively high rates of heating.

Why?

SV LV

SL

γ γ

γ + <

In general, wetting angle = 0 No superheating required!

4.1.4. Nucleation of melting

(commonly)