“Phase Transformation in Materials”

“Phase Transformation in Materials”

Eun Soo Park

Office: 33-313

Telephone: 880-7221 Email: espark@snu.ac.kr

Office hours: by an appointment

2018 Fall

09.06.2018

1

2

Microstructure-Properties Relationships

Microstructure Properties

Alloy design &

Processing Performance

“ Phase Transformation ”

“Tailor-made Materials Design”

down to atomic scale Contents for previous class

(Ch1) Thermodynamics and Phase Diagrams (Ch2) Diffusion: Kinetics

(Ch3) Crystal Interface and Microstructure

(Ch4) Solidification: Liquid  Solid

(Ch5) Diffusional Transformations in Solid: Solid  Solid (Ch6) Diffusionless Transformations: Solid  Solid

Background to understand

phase

transformation

Representative Phase

transformation

Contents of this course_Phase transformation

Contents for previous class

4

Chapter 1

Thermondynamics and Phase Diagrams

Contents for today’s class I

- Equilibrium

- Single component system

Gibbs Free Energy

as a Function of Temp. and Pressure

- Driving force for solidification

- Classification of phase transition

5

Q1: “thermodynamic equilibrium”?

Lowest possible value of Gibb’s Free Energy

→ The main use of thermodynamics in metallurgy is to allow

the prediction of whether an alloy is in equilibrium.

Equilibrium

Mechanical equilibrium

:

Thermal equilibrium

: absence of temperature gradients in the system

Chemical equilibrium

: no further reaction occurs between the reacting substances i.e. the forward and reverse rates of reaction are equal.

Thermodynamic equilibrium

: the system is under mechanical, thermal and chemical equilibrium

The properties of the system-P, T, V, concentrations-do not change with time.

Chapter 1

6 matchbox

7

PV E

H  

System Condensed

E

H 

Relative Stability of a System Gibbs Free Energy

H : Enthalpy ; Measure of the heat content of the system

E : Internal Energy, Kinetic + Potential Energy of a atom within the system Kinetic Energy :

Atomic Vibration (Solid, Liquid)

Translational and Rotational Energy in liquid and gas.

Potential Energy : Interactions or Bonds between the atoms within the system

T : The Absolute Temperature

S : Entropy, The Randomness of the System Chapter 1.1

Gibbs free energy : G=E +PV-TS=H-TS

Useful when P is constrained during thermodynamic process.

Mixture of one or more phases

8

 0 dG

2 1 0

G G G

   

Equilibrium

Phase Transformation Chapter 1.1

metastable

unstable

Activation E

stable

9

Q2: What is single component system?

Different elements or chemical compounds

10

1.2 Single component system

One element (Al, Fe) One type of molecule (H

O)

Fe C

1.2.1 Gibbs Free Energy as a Function of Temp.

TS H

G  

‐ Allotropic forms?

‐ How is phase stability measured?

11

Q3: C _v vs. C _p ?

Specific heat

(the quantity of heat (in joules) required to raise the temperature of substance by 1K)

at constant volume VS. at constant pressure

* What is the role of temperature on equilibrium?

12

When V is constant,

V V

V T

E T

C ( Q) ( )



 



  E C dT

T C E

V V

V ^{ (}_^{ ) or }



Experimentally, it is easy to keep constant P (Cp) than constant V because it is difficult to make constant V (Cv)  pressure ex) 1 atm,

When pressure is constant,

dV P

Q

dE    

PV E

H  

dT P dV dT

dE dT

Q  



0

VdP Q

VdP PdV

PdV Q

VdP PdV

w Q

VdP PdV

dE dH



































dT V dP dT

Q dT

dH   

constant is

when

0 P

dTdP 

P P

P C

dT Q dT

dH )  ( ) 

(  ^{H }



PV E

H  

dV P Q dE  

In alloy work one usually consider systems at constant pressure so that enthalpy changes are more important then energy changes.

13

C_P ; tempeature-dependent function

2





 a bT CT C

(empirical formula above room temp)

Molecules have internal structure because they are composed of atoms that have different ways of moving within molecules. Kinetic energy stored in these internal degrees of freedom contributes to a substance’s

specific heat capacity and not to its temperature.

14

Q4: How is C _p related with H and S?

* What is the role of temperature on equilibrium?

15

Draw the plots of (a) C

vs. T, (b) H vs. T and (c) S vs. T.

 

   

P

P

C H

T

 

T

H C dT

 

T

C

S dT

T

How is C_p related with H and S?

H = 0 at 298K for a pure element in its most stable state.

S = ?

H = ?

  

     

P

P

C S

T T

Enropy : q S  T

: When considering a phase transformation or a chemical reaction, the important thing is not the value of the thermodynamic function, but the amount of change.

16

Q5: How to draw the plots of H vs.T and G vs. T in single component system?

* What is the role of temperature on equilibrium?

 ^ 









1 0

1 0

) , ( )

, ( )

, ( )

,

(

P

T

T

S P T dT dP

P T V T

P G T

P G

SdT VdP

dG

P dP dT G

T dG G

P T G G

T P



 







 



 







 

 ( , )

SdT VdP

SdT TdS

VdP PdV

PdV TdS

dG

TS d PV

d dE TS

d dH dG

TS H

G































) ( )

( )

(

P V S G

T G

T P

 



 







 

 



 







 ,

Compare the plots of H vs.T and G vs. T.

18

Q6: G ^S vs G ^L as a function of temperature?

* What is the role of temperature on equilibrium?

19

• Which is larger, H

or H

?

• H

> H

at all temp.

• Which is larger, S

or S

?

• S

> S

at all temp.

→

decreases more rapidly with increasing temperature than that of the solid.

• Which is larger, G

or G

at low T?

• G

> G

G

> G

1.2.1 Gibbs Free Energy as a Function of Temp.

Fig. 1.4 Variation of enthalpy (H) and free energy (G) with temperature for the Solid and liquid phases of a pure metal. L is the latent heat of melting, Tm the

Equilibrium melting temperature.

20

Considering P, T

 ^ 











1 0

1 0 0

0

0

, ) ( , ) ( , )

( )

, (

)

, (

P P

T

T

S P T dT dP

P T V T

P G T

P G

SdT VdP

dG

P T G G

G

T (K)

800 1300

2.9 kJ

300

P = 1 bar

S(graphite) = 5.74 J/K, S(diamond) = 2.38 J/K,

G

T(K)

0

phase

transformation

T S G

P



 



 









S(water) = 70 J/K S(vapor) = 189 J/K

21

V T

H dT

dP

T S H

V S V

V S S dT

dP

eq eq

eq eq



 



 





 





 



 



 





이므로 여기서









(applies to all coexistence curves)

Q7: What is the role of pressure on equilibrium?

* Clausius‐Clapeyron Relation :

22

When the two phases with different molar volumes are in equilibrium, if the pressure changes, the equilibrium temperature T should also change with pressure.

If α & β phase are equilibrium,

dT S dP V dG

dT S dP V dG





















At equilibrium,



 dG

dG 

V T

H dT

dP

T S H

V S V

V S S dT

dP

eq eq

eq eq



 



 





 





 



 



 





이므로 여기서









For, α→γ ; ∆V (-), ∆H(+) 0



 



 





V T

H dT

dP

eq

For, γ →liquid; ∆V (+), ∆H(+)

0



 



 





V T

H dT

dP

eq

* Clausius‐Clapeyron Relation :

(applies to all coexistence curves)

Fcc (0.74)

Bcc (0.68)

1.2.2 Pressure Effects

hcp (0.74)

Fig. 1.5 Effect of pressure on the equilibrium phase diagram for pure iron

P V G

T

 



 









Here,

23

Q8: How to classify phase transition?

“First order transition” vs “Second order transition”

24

The First-Order Transition

G

T S

T

S = L/T

T C_P

N P

P T

T S C

,



 







  Latent heat

Energy barrier

Discontinuous entropy, heat capacity

The Second Order Transition

G

T S

T

S=0 Second-order transition

T C_P

No Latent heat Continuous entropy



 



 







 

N P

P T

T S C

,

26

Q9: What is the driving force for

“Solidi ication: Liquid → Solid”?

27

T

T G L 





L : ΔH = H^L – H^S (Latent heat) T = T_m - ΔT G^L = H^L – TS^L

G^S = H^S – TS^S

ΔG = Δ H -T ΔS ΔG =0= Δ H-T_mΔS ΔS=Δ H/T_m=L/T_m

ΔG =L-T(L/T_m)≈(LΔT)/T_m

(eq. 1.17)

1.2.3 Driving force for solidification

4 Fold Anisotropic Surface Energy/2 Fold Kinetics, Many Seeds 28

4. Solidification : Liquid Solid

29

4.1.1. Homogeneous Nucleation

L V L

S

V G

V

G

 (  ) G

 V

G

 V

G

 A



L V S

V G

G ,

SL SL

S V L

V

S

G G A

V G

G

G       



( )

SL V

r

r G r

G 

4 



3

4  







: free energies per unit volume

for spherical nuclei (isotropic) of radius : r

30

4.1.1. Homogeneous Nucleation

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

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 G_r = 0?

r* dG=0

Unstable equilibrium

*

* 2

G _ r 



SL V

r r G r

G  ³ 4 ² 

3

4  







31

1.2.3 Driving force for solidification

*

2 r T

T

G L ^SL

m

 

 



Liquid

Solid

SL V

r r G r

G  ³ 4 ² 

3

4  







Electrostatic levitation in KRISS

+

PSD (x) PSD (y)

HV (z-axis)

HV (x-axis)

HV (y-axis)

He-Ne laser He-Ne laser

Heating laser

Feedback

Feedback

T : ~3000 ^oC P : ~ 10^-7Torr

KRISS material : Dr. G.W.Lee

T

34

400 600 800 1000

Recalescence Recalescence

ΔT = 90℃ ΔT = 84℃

Temperature (℃)

Time (sec)

T_m = 666℃

Cyclic cooling curves of Zr

Ti

Cu

Ni

Be

35

: insensitive to Temp.

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 How do we define nucleation temperature, T_N ?

* The homogeneous nucleation rate - kinetics

hom 2

~ 1 N T



→ for most metals, ΔT_N~0.2 T_m (i.e. ~200K)

36

* Relationship between Maximum Supercoolings and T

- casting & welding - single crystal growth - directional solidification

- rapid solidification Nucleation in Pure Metals

T_m : G_L = G_S

- Undercooling (supercooling) for nucleation: 250 K ~ 1 K

<Types of nucleation>

- Homogeneous nucleation - Heterogeneous nucleation

Solidification: ^{Liquid Solid}

38

Q10: What is the driving force for

“Melting: Solid → Liquid”?

39

T

T G L 





L : ΔH = H^L – H^S (Latent heat) T = T_m - ΔT G^L = H^L – TS^L

G^S = H^S – TS^S

ΔG = Δ H -T ΔS ΔG =0= Δ H-T_mΔS ΔS=Δ H/T_m=L/T_m

ΔG =L-T(L/T_m)≈(LΔT)/T_m

(eq. 1.17)

* Driving force for melting

ΔG

T ΔT

40

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

 

  

In general, wetting angle = 0 No superheating required!

Nucleation of melting

(commonly)

Liquid Undercooled Liquid Solid

<Thermodynamic>

Solidification:

• Interfacial energy ΔT_N

Melting:

• Interfacial energy

SV LV

SL

 

  

No superheating required!

No ΔT_N T_m

vapor

Melting and Crystallization are Thermodynamic Transitions

42

Chapter 1

 0 dG

Contents for today’s class

- Equilibrium

- Driving force for solidification

- Classification of phase transition

2 1 0

G G G

   

Phase Transformation

Tm

T G  L



First order transition: CDD/Second order transition: CCD

Thermondynamics and Phase Diagrams

Gibbs Free Energy as a Function of Temp. and Pressure

- Single component system

P V S G

T G

T P

 



 







 

 



 







 ,

V T

H dT

dP

eq

eq 

  ) (

Clausius-Clapeyron Relation

Lowest possible value of G No desire to change ad infinitum