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(1)

“Phase Equilibria in Materials”

Eun Soo Park

Office: 33-313

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

Office hours: by an appointment

2017 Fall

09.11.2017

1

(2)

Microstructure-Properties Relationships

Microstructure Properties

Alloy design &

Processing Performance

“ Phase Transformation ”

“Tailor-made Materials Design”

down to atomic scale

2

Contents for previous class

(3)

3

Important!!!

Understanding and Controlling

Phase Transformation of Materials

Contents for previous class

(4)

Contents in Phase Transformation

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

(Ch3) Crystal Interface and Microstructure

(Ch4) Solidification: Liquid  Solid

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

Basic concept

for understanding Phase Transformation

Representative

Phase Transformation

D.A. Porter and K.E. Eastering, "Phase Transformations in Metals and Alloys"

4

Contents for previous class

(5)

Phase Equilibria in Materials Thermodynamics

Phase diagrams

Binary, Ternary , Quarternary phase diagram

5

Contents for previous class

(6)

Gibbs Free Energy as a Function of Temp. and Pressure

- Single component system - Equilibrium

Contents for today’s class

- Classification of phase transition

First order transition/Second order transition Driving force for solidification

- Solidification : Liquid → Solid - Phase Transformation

CHAPTER 1 Introductory Thermondynamics

6

(7)

7

Q1: “thermodynamic equilibrium”?

Lowest possible value of Gibb’s Free Energy

(8)

Equilibrium

Mechanical equilibrium

: total potential energy of the system is a minimum.

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

8 matchbox

(9)

' ' A const const

TS E

X     

Thermodynamic free energy , the energy in a physical system that can be converted to do work,

When no useful work is being extracted from the sub-system, The energy X reaching a minimum at equilibrium, when dX=0.

If no chemical species can enter or leave the sub-system, then ∑ μiR Nican be ignored.

If furthermore the temperature of the sub-system is such that T is always equal to TR, then

in particular:

Helmholtz free energy, the energy that can be converted into work at a constant temperature and volume

 0 dX

If the volume V is constrained to be constant, then

where F is the thermodynamic potential called Helmholtz free energy, F=E−TS.

Under constant volume conditions therefore, dF < 0 if a process is to go forward;

and dF=0 is the condition for equilibrium.

XETSconst '  Aconst ' Helmholtz free energy : F=E−TS.

Useful when V is constrained during thermodynamic process.

9

(10)

If the sub-system pressure P is constrained to be equal to the external reservoir pressure PR,

,where G is the Gibbs free energy, G=E−TS+PV. Therefore under constant pressure conditions, if dG ≤ 0, then the process can occur spontaneously, because the change in system energy exceeds the energy lost to entropy. dG=0 is the condition for equilibrium.

This is also commonly written in terms of enthalpy, where H=E+PV. G=H-TS

const G

const PV

TS E

X      

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

Useful when P is constrained during thermodynamic process.

Gibbs free energy, the energy that can be converted into work at a uniform temperature and pressure throughout a system

10

(11)

11

TS H

G  

PV E

H  

System Condensed

E

for

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

(12)

12

 0 dG

2 1

0

G G G

   

Equilibrium Lowest possible value of Gibb’s Free Energy No desire to change ad infinitum

Phase Transformation Chapter 1.1

metastable

unstable

Activation E

stable

(13)

13

Q2: What is single component system?

(14)

14

1.2 Single component system

One element (Al, Fe)

One type of molecule (H

2

O)

Fe C

1.2.1 Gibbs Free Energy as a Function of Temp.

TS H

G  

Allotropic form?

(15)

15

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?

(16)

16

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 C

p

dT

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.

(17)

17

CP

;

tempeature-dependent function

2

a bT CT C

P

(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.

(18)

18

Table of specific heat capacities

Substance Phase Cp

J / g·K

Cp

J / mol·K

Cv

J / mol·K

Volumetric heat capacity

J / cm3·K

Aluminium Solid 0.897 24.2 2.422

Copper solid 0.385 24.47 3.45

Diamond solid 0.5091 6.115 1.782

Gold solid 0.1291 25.42 2.492

Graphite solid 0.710 8.53 1.534

Iron solid 0.450 25.1 3.537

Lithium solid 3.58 24.8 1.912

Magnesium solid 1.02 24.9 1.773

Silver solid 0.233 24.9

Water liquid (25 °C) 4.1813 75.327 74.53 4.184

Zinc solid 0.387 25.2

All measurements are at 25 °C unless otherwise noted.

(19)

19

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

* What is the role of temperature on equilibrium?

(20)

20

Draw the plots of (a) C

p

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

 

   

P

P

C H

T

 

298

T

H C dT

P

 

0

T

C

P

S dT

T

How is Cp 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.

(21)

21

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?

(22)

22

1 0

1 0

) , ( )

, ( )

, ( )

,

(

0 0 P 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.

(23)

23

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

* What is the role of temperature on equilibrium?

(24)

24

• Which is larger, H

L

or H

S

?

• H

L

> H

S

at all temp.

• Which is larger, S

L

or S

S

?

• S

L

> S

S

at all temp.

Gibbs free energy of the liquid

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

• Which is larger, G

L

or G

S

at low T?

• G

L

> G

S

(at low Temp) and G

S

> G

L

(at high Temp)

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.

(25)

25

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

(26)

26

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 :

(27)

27

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,

(28)

28

Q8: How to classify phase transition?

“First order transition” vs “Second order transition”

(29)

29

The First-Order Transition

G

T S

T

S = L/T

T CP

N P

P T

T S C

,

Latent heat

Energy barrier

Discontinuous entropy, heat capacity

(30)

The Second Order Transition

G

T S

T

S=0 Second-order transition

T CP

No Latent heat Continuous entropy

N P

P T

T S C

,

(31)

31

Q9: What is the driving force for

“Solidi ication: Liquid → Solid”?

(32)

Phase Transformation

G

Metastable

Unstable

Stable barrier Metastable

Change to another phase metastable state structure or composition or order

2 1

0

G G G

   

32

(33)

33

T

m

T G L

L : ΔH = HL – HS (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

m

ΔS 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

Liquid →Solid transformation?

(34)

Turnbull’s Insight for Supercooling

How does the liquid metals resist on the crystallization?

What is the origin of high nucleation barrier against the crystallization?

What is the structure of interface between liquids and crystals?

What is the local order of liquid metals?

34

Liquid Crystal

Easy nucleation

Microcrystalline structure for liquid metals –Same short range order with crystals!

2 ,

*

3 g

= 16 W

s l

3

T T

= H

gl s f

(35)

Turnbull’s solid-liquid interface

Solid state physics v47, 1 (1994)

Supercooling

Turnbull’s insight : Liquid orders substantially near a crystal surface

due to entropy decreasing caused by the crystal-melt interfacial tension

35

(36)

4 Fold Anisotropic Surface Energy/2 Fold Kinetics, Many Seeds 36

4. Solidification : Liquid Solid

(37)

37

4.1.1. Homogeneous Nucleation

L V L

S

V G

V

G

1

 (  ) G

2

V

S

G

VS

V

L

G

VL

A

SL

SL

L V S

V G

G ,

SL SL

S V L

V

S

G G A

V G

G

G       

2 1

( )

SL V

r

r G r

G

3

4 

2

3

4  

: free energies per unit volume

for spherical nuclei (isotropic) of radius : r

(38)

38

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 Gr = 0?

r* dG=0

Unstable equilibrium

*

* 2

G r

SL

SL V

r r G r

G3 42

3

4

(39)

39

1.2.3 Driving force for solidification

*

2 r T

T

G L

SL

m

 

 

Liquid

solidification

Solid

SL V

r r G r

G3 42

3

4

(40)

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

(41)

41 NANO STRUCTURED MATERIALS LAB.

(42)

42

400 600 800 1000

Recalescence Recalescence

ΔT = 90℃ ΔT = 84℃

Temperature (℃)

Time (sec)

Tm = 666℃

Cyclic cooling curves of Zr

41.2

Ti

13.8

Cu

12.5

Ni

10

Be

22.5

(43)

43

: insensitive to Temp.

The homogeneous nucleation rate as a function of undercooling ∆T.

∆TN 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, TN ?

* The homogeneous nucleation rate - kinetics

hom 2

~ 1 N T

→ for most metals, ΔTN~0.2 Tm (i.e. ~200K)

(44)

44

* Relationship between Maximum Supercoolings and T

m

(45)

- casting & welding - single crystal growth - directional solidification - rapid solidification

Nucleation in Pure Metals

Tm : GL = GS

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

<Types of nucleation>

- Homogeneous nucleation - Heterogeneous nucleation

Solidification: Liquid Solid

45

(46)

46

Q10: What is the driving force for

“Melting: Solid → Liquid”?

(47)

47

T

m

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

m

ΔS 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 Solid Liquid transformation?

(48)

48

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)

(49)

Liquid Undercooled Liquid Solid

<Thermodynamic>

Solidification:

Liquid Solid

• Interfacial energy ΔTN

Melting:

Liquid Solid

• Interfacial energy

SV LV

SL

 

  

No superheating required!

No ΔTN Tm

vapor

Melting and Crystallization are Thermodynamic Transitions

(50)

Gibbs Free Energy as a Function of Temp. and Pressure

- Single component system

 0 dG

P V S G

T G

T P

 

 

 

 

 

 ,

- Equilibrium

Contents for today’s class

Lowest possible value of G No desire to change ad infinitum

V T

H dT

dP

eq

eq

  ) (

Clausius-Clapeyron Relation

- Classification of phase transition

First order transition: CDD/Second order transition: CCD

Driving force for solidification

Tm

T GL

- Solidification : Liquid → Solid

Themodynamic transition

2 1

0

G G G

   

- Phase Transformation

CHAPTER 1 Introductory Thermondynamics

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