1

Chapter 9 - 1

ISSUES TO ADDRESS...

• When we combine 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?

Phase Diagrams

Phase BPhase A

Nickel atomCopper atom

Chapter 9 - 2

Phase Equilibria: Solubility Limit

• Introduction– Solutions – solid solutions, solute atoms dissolve in the

solvent to form a solid solution – a single phase– Mixtures – more than one phase

• Solubility Limit:Max concentration forwhich only a single phase solution occurs.

Question: What is thesolubility limit at 20°C?

Answer: 65 wt% sugar.If Co < 65 wt% sugar at 20°C : syrupIf Co > 65 wt% sugar at 20°C : syrup + sugar.

The solubility of sugar in a sugar-water syrup

2

Chapter 9 - 3

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

• Phases:A homogeneous portion of a system that has uniform physical

and chemical characteristics (e.g., a and b).

Aluminum-Copper Alloy

Components and Phases

a (darker phase)

b (lighter phase)

Chapter 9 - 4

Effect of T & Composition (Co)

• Changing T can change number of phases: path A to B.

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

D (100°C,90)2 phases

B (100°C,70)1 phase

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

3

Chapter 9 - 5

Phase Equilibria

0.12781.8FCCCu

0.12461.9FCCNi

r (nm)electronegCrystalStructure

• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume –Rothery rules) suggesting high mutual solubility.

Simple solution system (e.g., Ni-Cu solution)

• Ni and Cu are totally miscible in all proportions.

Chapter 9 - 6

Binary Phase Diagrams

• Indicate phases as function of T, Co, and P (pressure). • For this course:

- binary systems: just 2 components.- independent variables: T and Co (P = 1 atmosphere is almost

always used).

• Phase Diagram for Cu-Ni system

• 2 phases:L (liquid)

a (FCC solid solution)

• 3 phase fields: LL + aa

(wt% Ni)20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

a

(FCC solid solution)

L + aliquidus

solidus

4

Chapter 9 - 7

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

a(FCC solid solution)

L + a

liquidus

solidus

Cu-Niphase

diagram

Phase Diagrams: numbers and types of phases

• Rule 1: If we know T and Co, then we know:-- the numbers and type of phases present.

• Examples:A (1100°C, 60):

1 phase: a

B(1250°C, 35):2 phases: L + a

B(1

250°

C,3

5)A(1100°C,60)

Chapter 9 - 8

Phase Diagrams: composition of phases

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

• Examples:

At TA = 1320°C: Only Liquid (L) CL = Co ( = 35 wt% Ni)

At TB = 1250°C:

Both a and LCL = C liquidus ( = 32 wt% Ni here)

Ca = Csolidus ( = 43 wt% Ni here)

At TD = 1160°C:

Only Solid ( a) Ca = Co ( = 35 wt% Ni)

C0 = 35 wt% Ni

A

D

5

Chapter 9 - 9

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

• Examples (9.1 p263):

wt% Ni

20

1200

1300

T(°C)

L (liquid)

a

(solid)L + a

liquidus

solidus

3 0 4 0 5 0

L + a

T AA

35C o

32C L

BT B

DT D

tie line

4C a3

R S

At TA: Only Liquid (L)

WL = 100 wt%, Wa = 0At TD: Only Solid ( a)

WL = 0, Wa = 100 wt%

Co = 35 wt% Ni

At TB: Both a and L

% 7332433543

wt=-

-=

= 27 wt%

WL =S

R +S

Wa= R

R +S

Phase Diagrams: weight fractions of phases

Cu-Ni system

Chapter 9 - 10

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

The Lever Rule

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

ML Ma

R S

RMSM L=a

L

L

LL

LL CC

CCSR

RW

CCCC

SRS

MMM

W-

-=

+=

-

-=

+=

+=

a

a

a

a

a

00

wt% Ni

20

1200

1300

T(°C)

L (liquid)

a(solid)L + a

liquidus

solidus

30 40 50

L + aB

TB

tie line

CoCL Ca

SR

Adapted from Fig. 9.3(b), Callister 7e.

6

Chapter 9 - 11

• Phase diagram:Cu-Ni system.

• System is:--binary

i.e., 2 components:Cu and Ni.

--isomorphousi.e., completesolubility of onecomponent inanother; a phasefield extends from0 to 100 wt% Ni.

• ConsiderCo = 35 wt%Ni.

Equilibrium Cooling in a Cu-Ni Binary System

Chapter 9 - 12

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

• Fast rate of cooling:Cored structure

• Slow rate of cooling:Equilibrium structure

First a to solidify has Ca = 46 wt% Ni.Last a to solidify has Ca = 35 wt% Ni.

Cored vs Equilibrium Phases

First a to solidify: 46 wt% Ni

Uniform Ca:

35 wt% Ni

Last a to solidify: < 35 wt% Ni

7

Chapter 9 - 13

nonequilibrium Cooling in a Cu-Ni Binary System

Schematic representation of the development of microstructure during the nonequilibrium solidification of a 35wt%Ni-65wt%Cu alloy

Chapter 9 - 14

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

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

Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.

Ten

sile

Str

engt

h (M

Pa)

Composition, wt% NiCu Ni0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu

Elo

ngat

ion

(%E

L)

Composition, wt% NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

8

Chapter 9 - 15

: Min. melting TE

2 componentshas a special compositionwith a minimum melting T.

Binary-Eutectic Systems

• Eutectic transitionL(CE) a(CaE) + b(CbE)

• 3 single phase regions (L, a, b)

• Limited solubility: a: mostly Cu b: mostly Ag

• TE : No liquid below TE

• CE

composition

Ex.: Cu-Ag systemCu-Ag system

L (liquid)

a L + a L+bb

a + b

Cowt% Ag

20 40 60 80 1000200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

Chapter 9 - 16

9

Chapter 9 - 17

The tin-bismuth phase diagram

Composition of the Bi-Sn soldier: 57wt% Bi

Questions:

1. Give the names for the single phases in the Sn-Bi system

2. To fíll the name for each phase region

3. The eutectic temperature and the eutectic composition

4. Tm for pure Sn and Bi

Chapter 9 - 18

For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, (a) What phases are present, (b) What are the composition of the phases; (c) Calculate the relative amount of each phase present; (d) The same questions as in (a-c) if the alloy is at 220ºC

10

Chapter 9 - 19

L+aL+b

a +b

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a 183°C61.9 97.8

b

• 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)

a + b--compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

150

40Co

11Ca

99Cb

SR

Ca = 11 wt% SnCb = 99 wt% Sn

Wa=Cb - CO

Cb - Ca

=99 - 4099 - 11

= 5988

= 67 wt%

SR+S

=

Wb=CO - Ca

Cb - Ca=R

R+S

=2988

= 33 wt%=40 - 1199 - 11

Chapter 9 - 20

L+b

a +b

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a b

L+a

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...--the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (2)

a + L--compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

Wa =CL - CO

CL - Ca

=46 - 40

46 - 17

=6

29= 21 wt%

WL =CO - Ca

CL - Ca=

23

29 = 79 wt%

40Co

46CL

17Ca

220SR

Ca = 17 wt% SnCL = 46 wt% Sn

11

Chapter 9 - 21

• Co < 2 wt% Sn• Result:

--at extreme ends--polycrystal of a grains

i.e., only one solid phase.

Microstructures in Eutectic Systems: I

0

L+ a200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

a

30

a+b

400

(room T solubility limit)

TE

(Pb-SnSystem)

aL

L: Co wt% Sn

a: Co wt% Sn

Chapter 9 - 22

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

Initially liquid + athen a alonefinally two phases

a polycrystalfine b-phase inclusions

Microstructures in Eutectic Systems: II

Pb-Snsystem

L + a

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

a

30

a+ b

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

La

L: Co wt% Sn

ab

a: Co wt% Sn

12

Chapter 9 - 23

• Co = CE• Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of a and b crystals.

Microstructures in Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.

160mm

Micrograph of Pb-Sneutectic microstructure

Pb-Snsystem

L + b

a + b

200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L

a b

L+a

183°C

40

TE

18.3

a: 18.3 wt%Sn

97.8

b: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

Chapter 9 - 24

Lamellar Eutectic Structure

The eutectic structure forms in the alternating layers:

fi atomic diffusiom of lead and tin need only occur over relatively short distances

13

Chapter 9 - 25

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

Microstructures in Eutectic Systems: IV

18.3 61.9

SR

97.8

SR

primary aeutectic a

eutectic b

WL = (1- Wa) = 50 wt%

Ca = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SWa= = 50 wt%

• Just above TE :

• Just below TE :Ca = 18.3 wt% Sn

Cb = 97.8 wt% SnS

R + SWa= = 73 wt%

Wb = 27 wt%Adapted from Fig. 9.16, Callister 7e.

Pb-Snsystem

L+b200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

a b

L+a

40

a+b

TE

L: Co wt% Sn LaLa

Chapter 9 - 26

L+aL+b

a +b

200

Co, wt% Sn20 60 80 1000

300

100

L

a bTE

40

(Pb-SnSystem)

Hypoeutectic & Hypereutectic

Adapted from Fig. 9.8, Callister 7e. (Fig. 9.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)

160 mm

eutectic micro-constituentAdapted from Fig. 9.14, Callister 7e.

hypereutectic: (illustration only)

b

bb

bb

b

Adapted from Fig. 9.17, Callister 7e. (Illustration only)

(Figs. 9.14 and 9.17 from Metals Handbook, 9th ed.,Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

175 mm

a

a

a

aa

a

hypoeutectic: Co = 50 wt% Sn

Adapted from Fig. 9.17, Callister 7e.

T(°C)

61.9eutectic

eutectic: Co =61.9wt% Sn

14

Chapter 9 - 27

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.

1. Give the name for each phase field

2. To point out the eutectic reactions in the Mg-Pb system, give the temperature TEand the composion CE for each reaction.

Chapter 9 - 28

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.

15

Chapter 9 - 29

Eutectoid & Peritectic

• Eutectic - liquid in equilibrium with two solidsL a + bcool

heat

intermetallic compound - cementite

cool

heat

• Eutectoid - solid phase in equation with two solid phasesS2 S1+S3

g a + Fe3C (727ºC)

cool

heat

• Peritectic - liquid + solid 1 solid 2 (Fig 9.21)S1 + L S2

d+ L g (1493ºC)

L fi S1 + S2

Chapter 9 - 30

Eutectoid & Peritectic

Cu-Zn Phase diagram

Adapted from Fig. 9.21, Callister 7e.

Eutectoid transition d g+ e

Peritectic transition g+ L d

16

Chapter 9 - 31

The copper-zinc phase diagram

Question: Try to find the eutectoid and the peritectic reactions in the Cu-Zn system

Chapter 9 - 32

Microstructural and property changes in iron-carbon alloys

1. The Iron-iron carbide (Fe-Fe3C) Phase diagram

2. Development of microstructure in Fe-C alloys

3. Phase transformation and TTT diagram

17

Chapter 9 - 33

d-phaseAusteniteFerriteCementite

PealiteBainiteMartensite

Fe-Fe3C phase diagramEutectic reactionEutectoid reaction

Maximum solubility

Chapter 9 - 34

The Iron-ironcarbide (Fe-Fe3C) Phase diagram

18

Chapter 9 - 35

Iron-Carbon (Fe-C) Phase Diagram

• 2 important points

-Eutectoid (B):g⇒ a +Fe3C

-Eutectic (A):L ⇒ g+Fe3C

Adapted from Fig. 9.24,Callister 7e.

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g (austenite)

g+L

g+Fe3C

a+Fe3C

a+g

L+Fe3C

d

(Fe) Co, wt% C

1148°C

T(°C)

a 727°C = Teutectoid

ASR

4.30

Result: Pearlite = alternating layers of a and Fe3C phases

120 mm

(Adapted from Fig. 9.27, Callister 7e.)

g ggg

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard)a (ferrite-soft)

Chapter 9 - 36

Hypoeutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g (austenite)

g+L

g+ Fe3C

a+ Fe3C

L+Fe3C

d

(Fe) Co, wt% C

1148°C

T(°C)

a727°C

(Fe-C System)

C0

0.76

Adapted from Fig. 9.30,Callister 7e.proeutectoid ferritepearlite

100 mm Hypoeutectoidsteel

R S

a

wa =S/(R+S)wFe3C

=(1-wa)

wpearlite = wgpearlite

r s

wa =s/(r+s)wg=(1- wa)

g

g g

ga

aa

ggg g

g ggg

19

Chapter 9 - 37

Hypereutectoid Steel

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g (austenite)

g+L

g+Fe3C

a +Fe3C

L+Fe3C

d

(Fe) Co, wt%C

1148°C

T(°C)

a

Adapted from Figs. 9.24 and 9.32,Callister 7e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

(Fe-C System)

0.76

Co

Adapted from Fig. 9.33,Callister 7e.

proeutectoid Fe3C

60 mmHypereutectoid steel

pearlite

R S

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

wpearlite = wgpearlite

sr

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

Fe3C

ggg g

ggg g

ggg g

Chapter 9 - 38

Example: Phase Equilibria

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

a) composition of Fe3C and ferrite (a)b) the amount of carbide (cementite) in grams that forms per

100 g of steelc) the amount of pearlite and proeutectoid ferrite (a)

20

Chapter 9 - 39

Chapter 9 – Phase Equilibria

Solution:

g 3.94

g 5.7 CFe

g7.5100 022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

=a

=

=-

-=

-

-=

a+ a

a

x

CCCCo

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

a) composition of Fe3C and ferrite (a)

CO = 0.40 wt% CCa = 0.022 wt% CCFe C = 6.70 wt% C

3

Fe 3

C (

cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

g (austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3Ca

Chapter 9 - 40

Chapter 9 – Phase Equilibria

c. the amount of pearlite and proeutectoid ferrite (a) note: amount of pearlite = amount of g just above TE

Co = 0.40 wt% CCa = 0.022 wt% CCpearlite = Cg= 0.76 wt% C

g

g+a=

Co -CaCg-Ca

x 100 = 51.2 g

pearlite = 51.2 gproeutectoid a = 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

g (austenite)

g+L

g + Fe3C

a + Fe3C

L+Fe3C

d

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CgCa

21

Chapter 9 - 41

The Gibbs phase rule:

P + F = C + N

P: the number of phases present

F: the number of degrees of freedom or the number of externally controlled variables (T, P and conposition etc.)

C: the number of components in the system

N: noncompositional variables (T and P)

Cu-Ag systemIn the a+L phase region: N=1, C=2, P=2, fi F=1In the a phase region: N=1, C=2, P=1, fi F=2

Chapter 9 - 42

• 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 fora range of microstructures.

Summary

Homework: 9.8, 9.14, 9.24, 9.33, 9.37, 9.38, 9.42, 9.43, p304-307