Chapter 9 - 1

ISSUES TO ADDRESS...• When we mix two elements... what equilibrium state do we get?

• In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature (T )

then... How many phases do we get? What is the composition of each phase? How much of each phase do we get (relative amount)?

Chapter 9: Phase Diagrams

Phase (A and B atoms)

Phase (A and B atoms)

Nickel atomCopper atom

StructurePropertiesProcessing

Chapter 9 - 2

Phase Equilibria: Review: Solubility Limit

Introduction

- Solutions – solid solutions, single phase (e.g. syrup)

- Mixtures – more than one phase (syrup and solid sugar)

Phase: Homogeneous portion of a system that has uniform physical and chemical properties

Solubility Limit:Max concentration for which only a single phase solution occurs.

Q: What is the solubility limit at 20°C?

Answer: 65 wt% sugar. If Co < 65 wt% sugar: syrup If Co > 65 wt% sugar: syrup + sugar.

65

Sucrose/Water Phase Diagram

Pu

re

Su

gar

Tem

per

atu

re (

°C)

0 20 40 60 80 100Co =Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid) + S

(solid sugar)

20

40

60

80

100

Pu

re

Wat

er

Chapter 9 - 3

• Components: The elements or compounds which are present in the mixture (e.g., Al and Cu)• Phases:The physically and chemically distinct material regions that result (e.g., and ).•EquilibriumNo change with time – fixed composition for each component

Al-Cu Alloy

Components and Phases

(darker phase)Includes Al and Cu

(lighter phase)Includes Al and Cu

Two phasesTwo Components

Chapter 9 - 4

Effect of T & Composition (Co)• Changing T can change # of phases:

D (100°C,90)2 phases

B (100°C,70)1 phase

path A to B.• Changing Co can change # of phases: path B to D.

A (20°C,70)2 phases

70 80 1006040200

Tem

pera

ture

(°C

)

Co =Composition (wt% sugar)

L (liquid solution

i.e., syrup)

20

100

40

60

80

0

L (liquid)

+ S

(solid sugar)

water-sugarsystem

Chapter 9 - 5

Phase Diagrams

Types of phase diagrams depend on Solubility • Isomorphous phase diagram (complete solubility) • Eutectic phase diagram (partial solubility) • Complex phase diagram (…….) How to obtain phase diagrams ?- From cooling curves for various starting solutions

(Thermal method)

• Thermodynamics: Phase Diagrams Indicate phases as function of T, Co, and P. • For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used).

Phase Diagram: A Plot of T versus C indicating various boundaries between phases

T

Time

? Phase Change

T

composition

Uses of Phase Diagrams: - No. of phases - Composition of each phase- Relative amount of each phase+ Predict the microstructure

Chapter 9 - 6

Phase Equilibria - Solutions

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

Both metals have •Nearly the same atomic radius•The same crystal structure (FCC) •similar electronegativities (W. Hume – Rothery rules) suggesting high mutual solubility.

Simple Solution System (e.g., Ni-Cu solution)Remember

Ni and Cu are totally miscible (soluble) in all proportions (compositions).

ONE PHASE ALWAYS – Isomorphous !

Chapter 9 - 7

Phase Diagrams- Metals

• 2 phases: L (liquid)

(FCC solid solution)

• 3 phase fields: LL +

Phase Diagram for Cu-Ni system

20 40 60 80 1000 wt% Ni1000

1100

1200

1300

1400

1500

1600

T(°C)

L (liquid)

(FCC solid Soln)

L + Liquidus line

Solidus lin

e

Isomorphous Phase Diagram Complete solubility - one phase field extends from 0 to 100 wt% Ni.

Chapter 9 - 8

Phase Change

• Phase boundary:Lines between different phases e.g. – Liquid – solid

• one phase • two phases …etc)

• Phase Change: – Solidification (or melting) – Phase Reaction: one phase gives DIRECTLY two phases e.g.

Liquid Solid phase 1 + Solid Phase 2 (both insoluble)• This occurs at

– Certain (fixed) temperature: TE

– Certain (fixed) composition: CE

• Eutectic Phase Reaction:L

• There are other types of phase reactions (Later)

Chapter 9 - 9

Cooling Curves

There are various types of cooling curves:

Depending on composition (characteristics)

1. Pure Element

2. Mixture of soluble 2 elements

3. Mixture of Partially Soluble 2 elements

4. Mixtures undergoing Phase Reactions

1. At the composition for phase reaction

2. Before the composition for phase reaction

Chapter 9 - 10

Types of Cooling Curves

2) Mixture of soluble 2 elements

Time

L

L

T

3) Phase Reaction at CE 3) Phase Reaction: Co > CE

Time

L

L

T

L

T

Time

L

L A or B)

1) Pure Element

Tmelting

Time

L

L

T

TE

Chapter 9 - 11

Constructing Phase Diagrams

Thermal Method• Start with several different solutions

• Each solution at certain co

• Prepare the solutions in liquid form (melting)• Cool each solution while recording T versus time• Plot the cooling curve for each solution • At each phase change: there are changes in the slope

– Latent Heat – Change in the properties (specific heat)– Phase reaction (later)

• Project cooling curves on T versus co plot.

• Connect the points from projection creating boundaries. • Label different obtained areas and boundaries.

Chapter 9 - 12

Constructing Isomorphous Phase Diagrams

Cooling Curves at Various Initial Co Phase Diagram T verus Co

T

CompositionTime

Chapter 9 - 13

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solid

us

Cu-Niphase

diagram

Using Phase Diagrams:# and types of phases

Rule 1: If we know T and Co, then we know: - Number and types of phases present.

• Examples:A(1100°C, 60): 1 phase:

B(1250°C, 35): 2 phases: L +

B (

1250

°C,3

5) A(1100°C,60)

Chapter 9 - 14

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

Using Phase Diagrams:composition of phases

Rule 2: If we know T and Co, then we know: - the composition of each phase.

• Examples:TA

A

35Co

32CL

At TA = 1320°C:

Only Liquid (L) CL = Co ( = 35 wt% Ni)

At TB = 1250°C:

Both and L CL = C liquidus ( = 32 wt% Ni here)

C = C solidus ( = 43 wt% Ni here)

At TD = 1190°C:

Only Solid ( ) C = Co ( = 35 wt% Ni)

Co = 35 wt% Ni

BTB

DTD

tie line

4C3

Chapter 9 - 15

Rule 3: If we know T and Co, then we know: --the relative amount of each phase (given in wt%).

• Examples:

At TA: Only Liquid (L)

W L = 1.0 , W = = 0At TD: Only Solid ( )

W L = 0, W = 1.0

Co = 35 wt% Ni

Using Phase Diagrams:weight fractions of phases

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

TAA

35Co

32CL

BTB

DTD

tie line

4C3

R S

At TB: Both and L

73.03243

3543

= 0.27

WL S

R +S

W R

R +S

Chapter 9 - 16

• Tie line – connects the phases in equilibrium with each other - essentially an isotherm

The Lever Rule

How much of each phase? ThinkThink of it as a lever (teeter-totter)

ML M

R S

RMSM L

L

L

LL

LL CC

CC

SR

RW

CC

CC

SR

S

MM

MW

00

wt% Ni

20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L + B

TB

tie line

CoCL C

SR

Chapter 9 - 17

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

35Co

L: 35wt%Ni

Cu-Nisystem

Observe the microstructure

while going down (cooling)

at the same composition.

• Consider Co = 35 wt%Ni.

Microstructure - Cooling in a Cu-Ni Binary

4635

4332

: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

: 36 wt% Ni

B: 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

Chapter 9 - 18

• C changes as we solidify.• Cu-Ni case:

• Fast rate of cooling: Cored structure

• Slow rate of cooling: Equilibrium structure

First to solidify has C = 46 wt% Ni.

Last to solidify has C = 35 wt% Ni.

Cored vs Equilibrium Phases

First to solidify: 46 wt% Ni

Uniform C:

35 wt% Ni

Last to solidify: < 35 wt% Ni

Chapter 9 - 19

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

--Tensile strength (TS)

- Peak as a function of Co

Te

nsile

Str

en

gth

(M

Pa

)

Composition, wt% NiCu Ni0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu

--Ductility (%EL,%AR)

- Min. as a function of Co

Elo

ng

atio

n (

%E

L)

Composition, wt% NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

Chapter 9 - 20

Eutectic Phase DiagramWith Eutectic Phase Reaction

L(CE) (CE) + (CE)The phase diagram is characterized with:

• 3 single phase regions (L, and )

• Partial Solubility – Solubility Limit of A in B forming B in A forming Occurs at

L (liquid)

L + L+

Wt% A: Co

1000

T

CE

TE

: Eutectic Composition • CE

• TE : Eutectic Temperature

Remember Phase Reaction at CE

Time

L

L

T

TE

Chapter 9 - 21

Eutectic Phase Diagram – Cooling Curves

L

L + L+

Wt% A: Co

1000

T(°C)

T

Time

L

L

1) Pure A or B

Tmelting

2) Giving or

Time

L

L

T

Time

L

L

T

L

Time

L

L

T

TE

What is the difference in this side

Chapter 9 - 22

Maximum solubility of A in B

L

L + L+

Wt% A: Co

1000

T(°C)

LiquidusSolidus

Solvus

Eutectic Reaction

Eutectic Phase Diagram

Maximum solubility of B in A

Boundaries - lines:•Liquidus•Solidus•Solvus

Now, apply previous rules ?On real systems

Chapter 9 - 23

Chapter 9 - 24

Chapter 9 - 25

Example-Eutectic System

: mostly Cu : mostly Ag

• TE : No liquid below TE

Ex.: Cu-Ag system

L (liquid)

L + L+

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE

8.0%

CE=71.9%

91.2%

TE=779°C

Given T and C Values are characteristics Of Cu-Ag system

Chapter 9 - 26

L+L+

+

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

183°C 61.9 97.8

• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (1)

+ --compositionscompositions of phases:

CO = 40 wt% Sn

--the relative amountrelative amount of each phase:

150

40Co

11C

99C

SR

C = 11 wt% SnC = 99 wt% Sn

W=C - CO

C - C

=99 - 4099 - 11

= 5988

= 0.67

SR+S

=

W =CO - C

C - C=R

R+S

=2988

= 0.33 (or 1 – 0.67)=40 - 1199 - 11

Chapter 9 - 27

L+

+

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

L+

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 200200°°CC, find... --the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (2)

+ L--compositionscompositions of phases:

CO = 40 wt% Sn

--the relative amountrelative amount of each phase:

W =CL - CO

CL - C=

46 - 40

46 - 17

= 6

29= 0.21

WL =CO - C

CL - C=

23

29= 0.79

40Co

46CL

17C

220SR

C = 17 wt% SnCL = 46 wt% Sn

Chapter 9 - 28

• Co < 2 wt% Sn• Result: --at extreme ends --polycrystal of grains i.e., only one solid phase.

Microstructures in Eutectic Systems: I

0

L+ 200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

30

+

400

(room T solubility limit)

TE

(Pb-SnSystem)

L

L: Co wt% Sn

: Co wt% Sn

Depends on the composition

Chapter 9 - 29

• 2 wt% Sn < Co < 18.3 wt% Sn• Result:

Initially liquid + then alonefinally two phases

polycrystal fine -phase inclusions

Microstructures in Eutectic Systems: II

Pb-Snsystem

L +

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

30

+

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

L

L: Co wt% Sn

: Co wt% Sn

Chapter 9 - 30

• Co = CE • Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of and crystals.

Microstructures in Eutectic Systems: III

160 m

Micrograph of Pb-Sn eutectic microstructure

Pb-Snsystem

L

200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L

L+ 183°C

40

TE

18.3

: 18.3 wt%Sn

97.8

: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

Chapter 9 - 31

Lamellar Eutectic Structure

Chapter 9 - 32

• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: crystals and a eutectic microstructure

Microstructures in Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary eutectic

eutectic

Pb-Snsystem

L+200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

L+

40

+

TE

L: Co wt% Sn LL

Chapter 9 - 33

• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: crystals and a eutectic microstructure

Microstructures in Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary eutectic

eutectic

WL = (1-W) = 0.50

C = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SW‘ = = 0.50

• Just above TE :

• Just below TE :

C = 18.3 wt% Sn

C = 97.8 wt% SnS

R + SW = = 0.73

W = 0.27

Pb-Snsystem

L+200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

L+

40

+

TE

L: Co wt% Sn LL

What is the fraction of Eutectic ?Pro-eutectic (primary)

Chapter 9 - 34

L+L+

+

200

Co, wt% Sn20 60 80 1000

300

100

L

TE

40

(Pb-Sn System)

Hypoeutectic & Hypereutectic

160 m

eutectic micro-constituent

hypereutectic: (illustration only)

175 m

hypoeutectic: Co = 50 wt% Sn

T(°C)

61.9eutectic

eutectic: Co = 61.9 wt% Sn

Chapter 9 - 35

Intermetallic Compounds

Note: intermetallic compound forms a line - not an area like ?

Mg2Pb

In Previous Phase diagramsα and β are terminal solid solutions

Can we obtain other intermediate phases ?

Now: Mg2Pb • is an intermetallic compound• with a ratio (2:1) (Mg:Pb)-Molar• melts at a fixed temperature M• Cooling curve ???

Because stoichiometry (i.e. composition) is exact: Calculate ?

Calculate Wt% Pb = 207 /(207+2(24.3)) X 100 = 81 %

Chapter 9 - 36

Same Story ?

• No. and type of each phase• Composition of each phase • Relative amount or fraction of each phase• Distinguish between

– primary phase– Eutectic phase

• Microstructure• Cooling Curves• Fill the regions with types of phases?

Chapter 9 - 37

Eutectoid & Peritectic Phase Reactions

• Eutectic - liquid in equilibrium with two solidsL + cool

heat

intermetallic compound - cementite

cool

heat

• Eutectoid - solid phase in equilibrium with two solid phases

S2 S1+S3

+ Fe3C (727ºC)

cool

heat

• Peritectic - liquid + solid 1 solid 2 (Fig 9.21)

S1 + L S2

+ L (1493ºC)

Chapter 9 - 38

Complex Phase Diagrams

α and η are terminal solid solutions - β, β’, γ, δ and ε are intermediate solid solutions.• new phenomena exist: ……… reaction ………. reaction

Chapter 9 - 39

Eutectoid & Peritectic

Cu-Zn Phase diagram

Eutectoid transition +

Peritectic transition + L

Chapter 9 - 40

Phase Diagram with Eutectoid Phase Reactions

The eutectoid (eutectic-like in Greek) reaction is similar to the eutectic

This phase diagram includes both:

Eutectic ReactionEutectoid Reaction

Our Interest In Fe-C phase diagram

Chapter 9 - 41

Iron-Carbon (Fe-C) Phase Diagram• 2 important points

-Eutectoid (B):

+Fe3C

-Eutectic (A): L + Fe3C

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austeniteaustenite)

+L

+Fe3C

+Fe3C

+

L+Fe3C

(Fe) Co, wt% C

1148°C

T(°C)

727°C = Teutectoid

A

SR

4.30

Result: Pearlite = alternating layers of and Fe3C phases

120 m

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hardhard) (ferriteferrite-softsoft)

Chapter 9 - 42

Development of Microstructure in Iron-Carbon alloys

PearlitePearlite Structure

By DiffusionDiffusion

Note: remember the names! Austenite, ferriteferrite, cemetitecemetite and pearlitepearlite

Chapter 9 - 43

Hypoeutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

+ Fe3C

L+Fe3C

(Fe) Co , wt% C

1148°C

T(°C)

727°C

(Fe-C System)

C0

0.76

proeutectoid ferritepearlite

100 m HypoeutectoidHypoeutectoidsteel

R S

w =S/(R+S)wFe3C =(1-w)

wpearlite = wpearlite

r s

w =s/(r+s)w =(1- w)

Chapter 9 - 44

Hypereutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

(Fe) Co , wt%C

1148°C

T(°C)

(Fe-C System)

0.7

6 Co

proeutectoid FeFe33CC

60 mHypereutectoid steel

pearlitepearlite

R S

w =S/(R+S)wFe3C =(1-w)

wpearlite = wpearlite

sr

wFe3C =r/(r+s)w =(1-w Fe3C )

Fe3C

Chapter 9 - 45

Phases in Fe–Fe3C Phase Diagram

α -ferrite - solid solution of C in …….. Fe• Stable form of iron at room temperature.• The maximum solubility of C is 0.022 wt% (interstitial solubility)• Soft and relatively easy to deform

+ Fe-C liquid solution

γ -austenite - solid solution of C in …….. FeThe maximum solubility of C is 2.14 wt % at 1147°C.Interstitial lattice positions are much larger than ferrite (higher C%)Is not stable below the eutectic temperature (727 °C) unless cooled rapidly (Chapter 10).

δ -ferrite solid solution of C in ……. FeThe same structure as α -ferriteStable only at high T, above 1394 °CAlso has low solubility for carbon (BCC)

Fe3C (iron carbide or cementite)This intermetallic compound is metastable, it remains as a compound indefinitely at room T, but decomposes (very slowly, within several years) into α -Fe and C (graphite) at 650 - 700 °C

Chapter 9 - 46

Example: Phase Equilibria

For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following

a) composition of Fe3C and ferrite ()

b) the amount of carbide (cementite) in grams that forms per 100 g of steel

c) the amount of pearlite and proeutectoid ferrite ()

Chapter 9 - 47

Example – Fe-C System

g 3.94

g 5.7 CFe

g7.5100 022.07.6

022.04.0

100xCFe

CFe

3

CFe3

3

3

x

CC

CCo

b) the amount of cementite in grams that forms per 100 g of steel

a) composition of Fe3C and ferrite ()

CO = 0.40 wt% C

C = 0.022 wt% C

CFe C = 6.70 wt% C3

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

+ Fe3C

L+Fe3C

Co , wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3C

Chapter 9 - 48

Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ()

note: amount of pearlitepearlite = amount of just above just above TE

Co = 0.40 wt% CC = 0.022 wt% CCpearlite = C = 0.76 wt% C

Co CC C

x 100 51.2 g

pearlite = 51.2 gproeutectoid = 48.8 g

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

+ Fe3C

L+Fe3C

Co , wt% C

1148°C

T(°C)

727°C

CO

R S

CC

Pearlite include ?

Eutectoid Fe3CFe3C

+Eutectoid ferrite ferrite

Chapter 9 - 49

Alloying Steel with More Elements

• Teutectoid changes: • Ceutectoid changes:

TE

ute

cto

id (

°C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

wt. % of alloying elements

Ce

ute

cto

id (

wt%

C)

Ni

Ti

Cr

SiMn

WMo

Chapter 9 - 50

• Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phase

for a given T and composition of the system.

• Alloying to produce a solid solution usually--increases the tensile strength (TS)--decreases the ductility.

• Binary eutectics and binary eutectoids allow for a range of microstructures.

Summary