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Chapter 7 Phase Equilibria and

Phase Diagrams

Introduction

The one-component phase diagram

Phase equilibria in a two-component system The eutectic phase diagram

The peritectic phase diagram

The monotectic phase diagram

Complex diagrams

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Phase

A chemically and structurally homogeneous region of a material

A part of a system, physically distinct, macroscopically homogeneous,and of fixed or variable composition.

It is mechanically separable from the rest of the system.

A phase is a region within which all the intensive variables vary

continuously, whereas at least some of them have discontinuities at the

borders between phases.

ice water

ice + water2 phases = solid phase + liquid phase

I want to drink 2-phase water consisting

of solid phase and liquid phase.

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Phase diagramGraphical representation of the combination of temperature, pressure,

composition, or other variables for which specific phases exist at equilibrium.

Phase diagram of Water (H2O)Phase diagram of Carbon dioxide (CO2)

State point: a position on the phase diagram

0.47 g/cm3

@304.25 K & 72.9 atm

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One-Component Phase Diagrams

Gibbs Phase Rule for systems in equilibrium

Component: a chemical species whose concentration in a phase can be varied

independently of the other species concentration

Number of degrees of freedom in equilibrium is the number of variables (p, T, or

composition) that can be independently adjusted without disturbing equilibrium.

water

F = C

P + 2C- Components

P- Number of phases

F- Degrees of freedomF = 2

F = 1F = 1

F = 0F = 0

F = 2F = 2

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Example of One-Component Phase Diagrams

iron SiO2

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Two-Component Phase Diagrams

Temperature

Pressure

Composition for materials A and B

(Composition of one-component

system?)

For one state point in closed system we

need three variables (p, T, & X).

If we fix the pressure,

(T, X(composition))2-D space

(p, T, X(composition))3-D space

F = CP + 1

F = 2

F = 1T

Xs Xl

XA+XB=Xtotal

F = C P + 2

C- Components

P- Number of phases

F- Degrees of freedomF = 2

F = 1F = 1

F = 0F = 0

F = 2

F = 2

One-component

system

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Specification of composition

Atomic percentage (=atomic fraction=atomic number fraction=mole fraction)

Weight percentage (=weight fraction)

100

/%/%

/%%%

BofwtatomicBofwtAofwtatomicAofwt

AofwtatomicAofwtAofatomicAofmole

100

%%%% BofwtatomicBofatomicAofwtatomicAofatomic

AofwtatomicAofatomicAofwt

Question1) Calculate the atomic fraction of copper in aluminum for a two-

component alloy containing 5 wt % copper. Atomic mass is 63.55 for Cu and 26.98

for Al.

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isomorphous system

has complete solubilityof one component in another.

Non-isomorphous system?

Cu-Ni phase diagram

Both Cu and Ni have the

same crystal structure, FCC,

similar radii,electronegativity and

valence.

F = CP + 1

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In a single-phase field, the composition

of the phase is the composition of the

alloy

In a two-phase field, the amount of

each phase and the composition of

each phase can be determined using

a tie line and the lever rule

In a single-phase field, the composition

of the phase is the composition of the

alloy

Phase boundaries

liquidus boundary and solidus boundary

Xo

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If we know T and Co(initial composition),then we know thecomposition of each phase.

Examples:

Cu-Ni

system

C0= 35 wt% Ni

At 1300 C:

Only liquid (L)

CL= C0(= 35 wt% Ni)

At 1150 C:

Only solid (a)

Ca= C0(= 35 wt% Ni)

At TB:

Both aand L

CL= Cliquidus(= 32 wt% Ni)

Ca= Csolidus(=43 wt% Ni)


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If we know T and Co, then we know the amount of each phase

(given in wt%).

C0= 35 wt% Ni

At 1300 C:

Only liquid (L)

WL

= 100 wt%, Wa

= 0 wt%

At 1150 C:

Only solid (a)

WL= 0 wt%, Wa= 100 wt%

At TB:

Both aand L WL= S/(R+S) =

(43-35)/(43-32) = 73 wt%

Wa= R/(R+S) =

(35-32)/(43-32) = 27 wt%

The lever rule


Cu-Ni

system

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The Lever Rule in a Two-Component System

LLS

S

total

L fXX

XX

M

M 0

SLSL

total

S fXX

XX

M

M 0

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Hamsters ?

Type A Type B

Assumption: Type B hibernates earlier than Type A.

Type A starts hibernation in colder day

than Type B does.

B is solidified at higher T than A.

Type A Type B

They hibernate.

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When Types A and B got together, they had a party.

In summer, they mix and

happily play together.

Closed system Closed system

As the weather gets colder and colder.

They start hibernating

One phase Two phases

Z Z Z Z..

No More Active (solid) phase.

Composition of type B:

XB=80 %

Still Active (liquid) phase.


XB=28.6 %

Total composition is not changed (XB, total=50 %).

Total composition: XB, total=50 %

WL=58.3 % WS=41.7 %

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1

1

1

l s

o l l s s

l s

o l s s s

o l l s s s

o l s s l

o l

s

s l

f fX X f X f

f f

X X f X f

X X X f X fX X f X X

X Xf

X X

s o

l

s l

X XfX X

Composition, XB

Temp

erature

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The Lever Rule in a Two-Component System

LLS

S

total

L fXX

XX

M

M 0

SLSL

total

S fXX

XX

M

M 0

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Effect of Cooling Rate

Fast cooling rate

More local heterogeneity

Slow cooling rate

More homogeneous structure

Just slow Very slow

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Composition of the

liquid for each alloy

Composition of the

solid for each alloy

Alloy 1: 0.2

Alloy 2: 0.3Alloy 3: 0.5

Alloy 4: 0.6

Alloy 5: 0.8

Alloy 1: 0.8

Alloy 2: 0.8

Alloy 3: 0.8

Alloy 4: 0.8

Alloy 5: 0.8

Composition X0for each alloy

Alloy 1: 0.2

Alloy 2: 0.2

Alloy 3: 0.2

Alloy 4: 0.2

Alloy 5: 0.2

For Alloy 2,

83.02.08.0

3.08.0

Lf 17.02.08.0

2.03.0

Sf0.3Composition, XB

Temperature

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At temperature T2

0.77 0.60

0.77 0.50

0.63

l

l

f

f

0.60 0.50

0.77 0.50

0.37

s

s

f

f

At temperature T3

0.72 0.60

0.72 0.45

0.44

l

l

f

f

0.60 0.45

0.72 0.45

0.56

s

s

f

f

Analysis of an Isomorphous Phase Diagram

X0=0.60

Temperature f l xl fs Xs

T1 1 0.6 0 0.85

T2 0.63 0.5 0.37 0.77

T3 0.44 0.45 0.56 0.72

T4 0.29 0.4 0.71 0.68

T5 0 0.3 1 0.6

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isomorphous systemhas complete solubilityof one component in

another.

Cu-Ni phase diagramAn isomorphous system is only possible

for substitutional solid solution.

The substitution occurs randomlyon their

FCC lattice sites because the Cu and Ni

atoms are so similar.

Cu-Ni alloy

Excellent corrosion resistance

Used for water-cooled heat exchangers

The size difference between the solute and solvent must be no greater than ~15%.

The electronegativities of the two atomic species must be comparable.

The valence of the two species must be similar.

The crystal structures of the two species must be the same.

Hume-Rothery Rules

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Four isomorphous systems

Formation of substitutional solid solution

All the Hume-Rothery rules are satisfied.

Cu-Ni Ge-Si

Ag-Au NiO-MgO

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Free Energy and Phase Diagram

At High T At Low TAt intermediate T

Xo

A

B

C

A B C

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When Types A and B got together, they had a party.

The free energy determines the composition!

The composition is adjusted to minimize the free energy.

In summer, they mix and

happily play together.

Closed system Closed system

As the weather gets colder and colder.

They start hibernating

One phase Two phases

Z Z Z Z..

No More Active (solid) phase.


XB=80 %

Still Active (liquid) phase.


XB=28.6 %

Total composition is not changed (XB, total=50 %).

Total composition: XB, total=50 %

WL=58.3 % WS=41.7 %

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Free Energy and Phase Diagram

How to use the chemical potential?Liquidus and solidus line is determined at the points

where the chemical potentials of the phases are the

same.

How to minimize the Gibbs free energy?

Constraint 1:

Constraint 2:

Constraint 3: solidsolidliquidliquidtotal fGfGG

solidBsolidliquidBliquidB fXfXX ,,,0 1 solidliquid ff

Chemical potential, (definition):

TpiX

G

,

We use chemical potential to describe the phenomenon.

G: free energy

Xi: composition (number of molecule or atom of material i)p: pressure

T: temperature

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1. A:B=50:50

X0,B=0.50

A A

A A

A

A

AAA A

B

B

B

BB B

BB

B B

BB

AA

A A

A A

A

A

AAA A

B

B

B

BB B

BB

B B

BB

AA

2. In the 2-phase region, the system will have

solid phase immediately in the liquid phase.

But they want to minimize the free energy by changing the composition (XB).

Liquid phase wants to increase the XB.

Solid phase wants to decrease the XB.

Solid phase

Liquid phase

Liquid phase

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Gmin,S

Gmin,L

A A

A A

A

A

AAA A

B

BB

BB B

BB

B B

BB

AA

3. This could be done by moving B from solid

phase to liquid phase.

B

Liquid phase increases the XB.Solid phase decreases the XB.

0

,

Tpi

liquid

liquidX

G

0

,

Tpi

solidsolid

XG

As long as the free energy can be reduced, the solid phase will keep giving B to the liquid

phase.

For a while, XB,Solidand XB,Liquid will approach the minimum points, Gmin,Sand Gmin,L, respectively.

(highly) Negative slope

(slightly) Positive slope

Solid phase

Liquid phase

The liquid phase want to get more B.

The solid phase want to give out more B.

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Gmin,S

Gmin,L

A A

A A

A

A

AAA A

B

BBB B

BB

B B

BB

AA

4. Although XB,Solid arrives at Gmin,S, XB,Liquid is still

far from Gmin,L.

B

Furthermore, the magnitude of chemical potentialliquid, is still large in comparison to solid(=0).

0

,

Tpi

liquid

liquidX

G

0

,

Tpi

solidsolid

X

G

Therefore, the liquid phase still get B from the solid phase.

Due to this, XB,Solidwill pass through Gmin,Sand go up.

For a while, XB,Liquid will keep going down towards Gmin,L.

(Still) negative slope

zero slope

B

B

0

,

Tpi

liquid

liquidX

G

0

,

Tpi

solidsolid

X

G

Please note that the chemical potentials are all negative If the solid composition passesthrough the Gmin,S.Both phases want to get more B.

Solid phase

Liquid phase

The situation is like Tug of War.

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Gmin,S

Gmin,L

A A

A A

A

A

AAA A

B

BB

BB

B B

BB

AA

5. Since the solid phase has lost B too much,

Gsolidbecomes larger (which means that the solid

phase becomes unstabler).

B

Gsolidand Gliquidarrive at the points (4) and (5),respectively.

Now, the solid phase and the liquid phase have the same chemical potential, which means

that they need B component to the same extent.

The B component cannot move any more between those phases.

Therefore, XB,Solid and XB,Liquidare determined at the points (4) and (5).

If we do this for other X0,Bfor various temperature, we will get various pairs of (4) and (5).

a collection of (4): solidus line

a collection of (5): liquidus line

B

B

0

,

Tpi

liquid

liquidX

G

0

,

Tpi

sol idsolid

X

G

solidliquid

Solid phase

Liquid phase

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A

B

Q) We have a composition XB0 (at T=T0) which is in the two-phase region as shown in the

following phase diagram and the corresponding free energy plot. The solid phase and the

liquid phase have different compositions: XB

Sfor the solid phase and XB

Lfor the liquid phase.

(3) At the points A and B, the chemical potentials have (positive, negative) value.

What is the meaning of this?

Both phases want to get more B.

(4) At T=T0, the tie line meets points A and B with the solidus line and the liquidus

line, respectively, which means that the composition of the solid phase and liquid

phase are XB

S(point 4) and XB

L(point 5). Please explain how to use the chemical

potential to determine the points 4 and 5.

To obtain phase equilibrium, each phase needs to have the same chemical potential

for material. The chemical potential is the slope of the free energy plot, so that thefree energy curves for solid phase and liquid phase have the common tangent line

that has the same slope.

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Eutect ic Phase Diagrams

Composition, XB

Temperature

TA

TB

Composition, XB

Temperature

TA

TB

Feeling?

eutektos, meaning 'easily melted.'

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Eutectic Phase Diagrams

Composition, XB

TA

TB

X1 XE X2A B

Temperature

Xa

Xb

T

Xs

Xl

TXs

Xl

TXa

Xb

One-phase liquid

Two-phase solid

One-

phase

solid

One-phase

solid

two-

phase two-

phase

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F = 2, specify

temperature

and composition

F = 1, specify

temperature or thecomposition of one

of the phases

F = 0, temperatureand compositions

of the phase are

fixed.

Composition, XB

TA TB

X1 XE X2A B

Temperature

F = 2

F = 2

F = 2

Eutectic point

F = 1

F = 1

F = 1

F = 0

38

E t ti Ph Di

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Symbolic expression: Liquid Solid 1 + Solid 2

L a+ b

Eutectic Reaction

Through eutectic point in eutectic

isotherm, one-phase Liquid state

becomes two-phase Solid state.

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Question) from the following eutectic phase diagram

(1) Your system is located in two- phase

region (L+ ) with X0=0.27. If you cool itdown slowly (equilibrium cooling), what

would be the fraction of primary solid ()

at the eutectic temperature?

588.02037

2737

27.0

0

0

a

aXX

XXf

X

L

LP

(2) From the same system, what would be the fraction of liquid at the eutectic

temperature?

412.02037

20270

a

a

XX

XXf

L

eut

L

40

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Question) from the following eutectic phase diagram

(3) From the liquid (eutectic composition) in

question (2), what amount of and bwill

be formed just below the eutectic isotherm?

679.02073

3773

ab

b

aXX

XXf

eut

L

321.02073

2037

ab

ab

XX

XXf

eut

L

Therefore, the fraction of alloy composed of the eutectic andbobtained from

the eutectic liquidin question (2)is

280.0679.0412.0 aa fff eut

L

eut

132.0679.01412.0 bb

fff eut

L

eut

412.02037

20270

a

a

XX

XXf

L

eut

L

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Peritectic Phase Diagrams

Symbolic expression: Liquid + Solid 1 Solid 2

L + a b

1. Peritectic Point and Peritectic

isotherm!

2. Through peritectic point, Liquid+solid becomes solid.

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Peritectic Phase Diagram

X1=0.125 wt %

X2=0.170 wt %

X3=0.350 wt %

Question) Determine the composition and relative amounts of each phase

present just above and below the peritectic isotherm for each of the three

alloy compositions indicated.

fL

f

f

fL

f

f

fL

f

f

X1=0.125

X2=0.170

X3=0.350

just above just below

X1 X2 X3

0.09 0.17 0.53

+ L+

L+

L

mse 2001

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