Chapter 9: Phase Diagrams

ISSUES TO ADDRESS...• When we combine two elements... what is the resulting equilibrium state?

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

then... How many phases form? What is the composition of each phase? What is the amount of each phase? What is the structure of each phase

Chapter 9: Phase Diagrams

Phase BPhase A

Nickel atomCopper atom

Phase BehaviorLaughlin p. 67

• Components: The elements or compounds which are present in the alloy (e.g., Al and Cu)

• Phases: The physically and chemically distinct material regions that form (e.g., α and β).

Aluminum-Copper Alloy

Components and Phases

β

(darker phase)

α (lighter phase)

1µm

Optical Mettalography

Solubility Limit

Question: What is the solubility limit for sugar in water at 20ºC?

65

• Solubility Limit: Maximum concentration for which only a single phase solution exists.

Sugar/Water Phase Diagram

Tem

per

atu

re (

ºC)

0 20 40 60 80 100

Composition (wt% sugar)

L

(syrup)

Solubility Limit L

(liquid)

+ S

(solid sugar)20

40

60

80

100

• Solution – solid, liquid, or gas solutions, single phase• Mixture – more than one phase

Problem 9.2: At 170°C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in Pb?The lead-tin phase diagram is shown in the Animated Figure 9.8.

Isomorphous Binary Phase Diagram

• System is:-- binary 2 components: Cu and Ni.

-- isomorphous i.e., complete solubility of one component in another

Cu-Ni phase diagram

wt% Ni20 40 60 80 1000

1000

1100

1200

1300

1400

1500

1600

T(º

C)

L (liquid)

α

solid solution

L + αliquidus

solid

us

Criteria for Solid Solubility: Hume–Rothery rules

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

• Same crystal structure• Similar electronegativities • Similar atomic radii

• Ni and Cu are totally soluble in one another for all proportions.

Structure

BCC

FCC

Periodic Table

Cu-Ni phase diagram

wt% Ni20 40 60 80 1000

1000

1100

1200

1300

1400

1500

1600

T(º

C)

L

α L + α

Determination of phases present

• If we know T and Co, then we know which phases are present.

• Examples:A(1100ºC, 60 wt% Ni): 1 phase: α

B(1250ºC, 35 wt% Ni): 2 phases: L + α B

(12

50ºC

,35)

A(1100ºC,60)

Determination of phase compositions: Lever Rule

What fraction of each phase? Think of the tie line as a lever

(teeter-totter)

ML Mα

R S

M x S ML x R

L

L

LL

LL CC

CC

SR

RW

CC

CC

SR

S

MM

MW

00

wt% Ni

20

1200

1300

T(ºC)

L

α

30 40 50

L + α B

TB

tie line

C0CL Cα

SR

Tie line –– also sometimes called an isotherm

wt% Ni20

1200

1300

30 40 50110 0

L

α

T(ºC)

35C0

L: 35wt%Ni

C0 = 35 wt% Ni alloy

Slow Cooling of a Cu-Ni Alloy

4635

4332

a: 43 wt% Ni

L: 32 wt% Ni

α: 46 wt% NiL: 35 wt% Ni

L: 24 wt% Ni

a: 36 wt% Ni

24 36

Last a to solidify:< 35 wt% Ni

Cored vs Equilibrium Structures

Uniform Ca: 35 wt% Ni

First α to solidify:46 wt% Ni

wt% Ni20

1200

1300

30 40 501100

L

α

T(ºC)

35C0

4635

4332

24 36

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

-- Tensile strength (TS) -- Ductility (%EL)

Ten

sile

Str

eng

th (

MP

a)

Composition, wt% NiCu Ni0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu

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

2 components=low melting

Binary-Eutectic Systems

• 3 single phase regions (L, α, β)

Ex.: Cu-Ag system

L (liquid)

α L + α L+β β

α + β

wt% Ag20 40 60 80 1000

200

1200

400

600

800

1000

CE

TE 8.0 71.9 91.2779ºC

Ag) wt%1.29( Ag) wt%.08( Ag) wt%9.71( Lcooling

heating

Eutectic Phase Reaction:

Cu-Ag system

• For alloys for which C0 < 2 wt% Sn• Result: at room temperature -- polycrystalline with grains of a phase having

composition C0

Microstructural Development in Eutectic Systems I

0

L+ α200

T(ºC)

wt% Sn10

2

20C0

300

100

L

α

30

α +β

400

(room T solubility limit)

TE

Pb-Sn

αL

L: C0 wt% Sn

α: C0 wt% Sn

• For alloys for which 2 wt% Sn < C0 < 18.3 wt% Sn• Result: at temperatures in α + β range -- polycrystalline with a grains and small β-phase particles

Microstructural Development in Eutectic Systems II

L + α

200

T(ºC)

C, wt% Sn10

18.3

200C0

300

100

L

α

30

α + β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

L: C0 wt% Sn

αβ

a: C0 wt% Sn

• For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of α and β phases.

Microstructural Development in Eutectic Systems III

160 μm

Micrograph of Pb-Sn eutectic microstructure

200

T(ºC)

wt% Sn

20 60 80 1000

300

100

L

α β

L+ α

183ºC

40

TE

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE61.9

L: C0 wt% Sn

Lamellar Eutectic Structure

• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn• Result: α phase particles and a eutectic microconstituent

Microstructural Development in Eutectic Systems IV

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1- Wα) = 0.50

Cα = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SWα = = 0.50

• Just above TE :

• Just below TE :C

α = 18.3 wt% Sn

Cβ = 97.8 wt% Sn

SR + S

Wα = = 0.73

Wβ = 0.27

200

T(ºC)

wt% Sn

20 60 80 1000

300

100

L

αL+

α

40

TE

L: C0 wt% Sn

Lα

L+αL+β

α + β

200

C, wt% Sn20 60 80 1000

300

100

L

α β

TE

40

System)

Hypoeutectic & Hypereutectic

160 mm

eutectic micro-constituent

hypereutectic: (illustration only)

β

ββ

ββ

β

175 mm

α

α

α

αα

α

hypoeutectic: C0 = 50 wt% Sn

T(ºC)

61.9eutectic

eutectic: C0 = 61.9 wt% Sn

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound exists as a line - not an area – because stoichiometry (i.e. composition of a compound) is fixed.

• Eutectoid –all solid phases

S2 S1+S3

ϒ α+ Fe3C (For Fe-C, 727ºC, 0.76 wt% C)

intermetallic compound - cementite

cool

heat

Eutectic, Eutectoid, & Peritectic

• Eutectic - liquid transforms to two solid phases

L S1+S3 (For Pb-Sn, 183ºC, 61.9 wt% Sn) cool

heat

cool

heat

• Peritectic - liquid and one solid phase transform to a second solid phase S1 + L S2

δ + L ϒ (For Fe-C, 1493ºC, 0.16 wt% C) • Peritectoid – all solid phases

S1 + S2 S3

Eutectoid & Peritectic

Eutectoid transformation δ γ + ε

Peritectic transformation γ + L δ

Cu-Zn Phase diagram

Iron-Carbon (Fe-C) Phase Diagram

- Eutectoid (B):

γ → α +Fe3C

- Eutectic (A): L → γ + Fe3C

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ+Fe3C

α+Fe3C

α+γ

δ

(Fe) wt% C

1148ºC

T(ºC)

α 727ºC = Teutectoid

4.30

Result: Pearlite = alternating layers of α and Fe3C phases

120 mm0.76

B

γ γγγ

A L+Fe3C

Fe3C (cementite-hard)α (ferrite-soft)

Hypereutectoid Steel1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ +Fe3C

α +Fe3C

L+Fe3C

δ

(Fe) C, wt%C

1148ºC

T(ºC)

a

0.76

C0

Fe3C

γ

γγ

γγγ γ

proeutectoid Fe3C

60 μm

pearlite

pearlite

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

(Fe) C, wt% C

1148ºC

T(ºC)

α727ºC

C0

0.76

Hypoeutectoid Steel

proeutectoid ferritepearlite

100 μmγ γγγ

α

pearlite

γγ γ

γα

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

a) The compositions of Fe3C and ferrite (α).

b) The amount of cementite (in grams) that forms in 100 g of steel.

c) The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.

Solution to Example Problem

WFe3C R

R S

C0 C

CFe3C C

0.40 0.0226.70 0.022

0.057

b) Wight Fraction of cementite

a) The compositions of Fe3C and ferrite (α).RS tie line just below the eutectoidCα = 0.022 wt% C

CFe3C = 6.70 wt% C

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

C , wt% C

1148ºC

T(ºC)

727ºC

C0

R S

CFe C3Cα

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

pearlite

The amounts of pearlite in the 100 g.

c) Using the VX tie line just above the eutectoid and realizing thatC0 = 0.40 wt% C

Cα = 0.022 wt% C

Cpearlite = Cγ = 0.76 wt% C

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ(austenite)

γ+L

γ + Fe3C

+ Fe3C

L+Fe3C

δ

C, wt% C

1148ºC

T(ºC)

727ºC

C0

V X

CγCα

Wpearlite V

V X

C0 C

C C

0.40 0.0220.76 0.022

0.512

Amount of pearlite in 100 g

= (100 g)Wpearlite

= (100 g)(0.512) = 51.2 g

α

pearlite

Alloying with Other Elements

• Teutectoid changes:

TE

ute

cto

id (

ºC)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

• Ceutectoid changes:

wt. % of alloying elements

Ce

ute

cto

id (

wt%

C)

Ni

Ti

Cr

SiMn

WMo

VMSE: Interactive Phase Diagrams

Change alloy composition

Microstructure, phase compositions, and phase fractions respond interactively

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

given the temperature and composition of the system.

• The microstructure of an alloy depends on -- its composition, and -- whether or not cooling rate allows for maintenance of equilibrium.

• Important phase diagram phase transformations include eutectic, eutectoid, and peritectic.

Summary

70 80 1006040200

Tem

pera

ture

(ºC

)

C = Composition (wt% sugar)

L (liquid solution

i.e., syrup)

20

100

40

60

80

0

L (liquid)

+ S

(solid sugar)

Effect of Temperature & Composition

• Altering T can change # of phases: path A to B.• Altering C can change # of phases: path B to D.

D (100ºC,C = 90)

2 phases B (100ºC,C = 70)

1 phase

A (20ºC,C = 70)

2 phases

L+αL+β

α + β

200

T(ºC)

18.3

wt% Sn20 60 80 1000

300

100

L (liquid)

α 183ºC 61.9 97.8

β

• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC

Pb-Snsystem

EX 1: Pb-Sn Eutectic System

-- the relative amount of each phase

150

40C0

11Cα

99Cβ

SR

Cα = 11 wt% Sn

Cβ = 99 wt% Sn

Wα

=Cβ - C0

Cβ - Cα

=99 - 4099 - 11

= 5988

= 0.67

SR+S

=

Wβ

=C0 - Cα

Cβ - Cα

=RR+S

=2988

= 0.33=40 - 1199 - 11

Ca = 17 wt% Sn

L+β

a + b

200

T(ºC)

wt% Sn20 60 80 1000

300

100

L

α β

L+ α

183ºC

• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC

EX 2: Pb-Sn Eutectic System

the relative amount of each phase

Wa =CL - C0

CL - Cα

=46 - 40

46 - 17

= 6

29= 0.21

WL =C0 - Cα

CL - Cα

=23

29= 0.79

40C0

46CL

17Cα

220SR

CL = 46 wt% Sn