Phase DiagramsBinary Eutectoid Systems

Iron-Iron-Carbide Phase DiagramSteels and Cast Iron

Weeks 7 - 8

1

What is Phase?• The term ‘phase’ refers to a separate and identifiable

state of matter in which a given substance may exist.• Applicable to both crystalline and non-crystalline

materials• An important refractory oxide silica is able to exist as

three crystalline phases, quartz, tridymite and cristobalite, as well as a non-crystalline phase, silica glass, and as molten silica

• Every pure material is considered to be a phase, so also is every solid, liquid, and gaseous solution

• For example, the sugar–water syrup solution is one phase, and solid sugar is another

2

Introduction to Phase Diagram• There is a strong correlation between

microstructure and mechanical properties, and the development of microstructure of an alloy is related to the characteristics of its phase diagram

• It is a type of chart used to show conditions at which thermodynamically distinct phases can occur at equilibrium

• Provides valuable information about melting, casting, crystallization, and other phenomena

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4

• 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?

ISSUES TO ADDRESS...

Phase BPhase A

Nickel atomCopper atom

Solubility Limit• At some specific temperature, there is a

maximum concentration of solute atoms that may dissolve in the solvent to form a solid solution, which is called as Solubility Limit

• The addition of solute in excess of this solubility limit results in the formation of another compound that has a distinctly different composition

• This solubility limit depends on the temperature

5

Solubility Limit Sugar-Water

6

Microstructure• the structure of a prepared surface of material as

revealed by a microscope above 25× magnification• The microstructure of a material can strongly

influence properties such as strength, toughness, ductility, hardness, corrosion resistance, high/low temperature behavior, wear resistance, etc

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8

• 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 ).

Aluminum-CopperAlloy

Components and Phases

(darker phase)

(lighter phase)

9

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

PHASE EQUILIBRIA• Free Energy -> a function of the internal energy of a

system, and also the disorder of the atoms or molecules (or entropy)

• A system is at equilibrium if its free energy is at a minimum under some specified combination of temperature, pressure, and composition

• A change in temperature, pressure, and/or composition for a system in equilibrium will result in an increase in the free energy

• And in a possible spontaneous change to another state whereby the free energy is lowered

10

Unary Phase Diagram• Three externally controllable parameters that

will affect phase structure: temperature, pressure, and composition

• The simplest type of phase diagram to understand is that for a one-component system, in which composition is held constant

• Pure water exists in three phases: solid, liquid and vapor

11

Pressure-Temperature Diagram (Water)• Each of the phases will exist under equilibrium conditions

over the temperature–pressure ranges of its corresponding area

• The three curves (aO, bO, and cO) are phase boundaries; at any point on one of these curves, the two phases on either side of the curve are in equilibrium with one another

• Point on a P–T phase diagram where three phases are in equilibrium, is called a triple point

12

Binary Phase Diagrams• A phase diagram in which temperature and

composition are variable parameters, and pressure is held constant—normally 1atm

• Binary phase diagrams are maps that represent the relationships between temperature and the compositions and quantities of phases at equilibrium, which influence the microstructure of an alloy.

• Many microstructures develop from phase transformations, the changes that occur when the temperature is altered

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Phase Equilibria

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246

Cu FCC 1.8 0.1278

• 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.

15

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

• PhaseDiagramfor Cu-Nisystem

• 2 phases: L (liquid) (FCC solid solution)

• 3 phase fields: LL +

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L + liquidus

solidus

16

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solidusCu-Niphase

diagram

Phase Diagrams:# and types of phases

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

• Examples:A(1100°C, 60): 1 phase:

B(1250°C, 35): 2 phases: L +

B (1

250°

C,3

5) A(1100°C,60)

17

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

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

18

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

At TA: Only Liquid (L) W L = 100 wt%, W = 0

At TD: Only Solid ( ) W L = 0, W = 100 wt%

Co = 35 wt% Ni

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

TA A

35Co

32CL

BTB

DTD

tie line

4C3

R S

At TB: Both and L

% 7332433543 wt

= 27 wt%

WL SR +S

W R

R +S

19

• 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 M

R S

RMSM L

L

L

LL

LL CC

CCSR

RWCCCC

SRS

MMMW

00

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L + B

TB

tie line

CoCL C

SR

20

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

35Co

L: 35wt%Ni

Cu-Nisystem

• Phase diagram: Cu-Ni system.• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100 wt% Ni.• Consider Co = 35 wt%Ni.

Ex: Cooling in a Cu-Ni Binary

46354332

: 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

21

• 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

22

Mechanical Properties: Cu-Ni System• Effect of solid solution strengthening on:

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

--Peak as a function of Co --Min. as a function of Co

Tens

ile S

treng

th (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% Ni

Cu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

Eutectic SystemA eutectic system is a mixture of chemical compounds or elements that has a single chemical composition that solidifies at a lower temperature than any other composition

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: Min. melting TE

2 componentshas a special compositionwith a min. melting T.

Binary-Eutectic Systems

• Eutectic transitionL(CE) (CE) + (CE)

• 3 single phase regions (L, ) • Limited solubility: : mostly Cu : mostly Ag • TE : No liquid below TE

• CE

composition

Ex.: Cu-Ag system Cu-Agsystem

L (liquid)

L + L+

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

25

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)

+ --compositions of phases:

CO = 40 wt% Sn

--the relative amount of each phase:

150

40Co

11C

99C

SR

C = 11 wt% SnC = 99 wt% Sn

W=C - CO

C - C

= 99 - 4099 - 11 = 59

88 = 67 wt%

SR+S =

W =CO - C

C - C=R

R+S

= 2988

= 33 wt%= 40 - 1199 - 11

26

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 220°C, find... --the phases present: Pb-Sn

system

EX: Pb-Sn Eutectic System (2)

+ L--compositions of phases:

CO = 40 wt% Sn

--the relative amount of each phase:

W =CL - CO

CL - C=

46 - 4046 - 17

= 629 = 21 wt%

WL =CO - C

CL - C=

2329 = 79 wt%

40Co

46CL

17C

220SR

C = 17 wt% SnCL = 46 wt% Sn

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• 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

28

• 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

29

• 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

30

Lamellar Eutectic Structure

31

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

Microstructures in Eutectic Systems (Pb-Sn): IV

18.3 61.9

SR

97.8

SR

primary eutectic

eutectic

WL = (1-W) = 50 wt%

C = 18.3 wt% Sn CL = 61.9 wt% Sn

SR + S

W = = 50 wt%

• Just above TE :

• Just below TE :C = 18.3 wt% SnC = 97.8 wt% Sn

SR + S

W = = 73 wt%

W = 27 wt%

Pb-Snsystem

L+200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

L+

40

+

TE

L: Co wt% Sn LL

32

L+L+

+

200

Co, wt% Sn20 60 80 1000

300

100

L

TE

40

(Pb-Sn System)

Hypoeutectic & Hypereutectic

160 meutectic micro-constituent

hypereutectic: (illustration only)

175 m

hypoeutectic: Co = 50 wt% Sn

T(°C)

61.9eutectic

eutectic: Co = 61.9 wt% Sn

33

Eutectoid & Peritectic• Eutectic - liquid in equilibrium with two solids

L + coolheat

intermetallic compound - cementite

coolheat

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

+ Fe3C (727ºC)

coolheat

• Peritectic - liquid + solid 1 solid 2 • S1 + L S2

+ L (1493ºC)

34

Eutectoid & PeritecticCu-Zn Phase diagram

Eutectoid transition +

Peritectic transition + L

35

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

-Eutectoid (B): +Fe3C

-Eutectic (A): L + Fe3C

Fe3C

(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 = Teutectoid

ASR

4.30Result: Pearlite = alternating layers of and Fe3C phases

120 m

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard) (ferrite-soft)

36

Hypoeutectoid Steel

Fe3C

(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 Hypoeutectoidsteel

R S

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

wpearlite = wpearlite

r s

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

(Ferrite) + Fe3C (cementite)

+

37

Hypereutectoid Steel

Fe3C

(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)

0.76 Co

proeutectoid Fe3C

60 mHypereutectoid steel

pearlite

R S

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

wpearlite = wpearlite

sr

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

Fe3C

+

38

Case StudyFor 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 ()

39

Case StudySolution:

g 3.94g 5.7 CFe

g7.5100 022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

x

CCCCo

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

a) composition of Fe3C and ferrite ()

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

3

Fe3C

(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

40

Case Studyc. the amount of pearlite and proeutectoid ferrite ()

note: amount of pearlite = amount of 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

Fe3C

(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

41

Alloying Steel with More Elements

• Teutectoid changes: • Ceutectoid changes:

T Eut

ecto

id (°

C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

wt. % of alloying elementsC

eute

ctoi

d (w

t%C

)

Ni

Ti

Cr

SiMnWMo

Self-Study Task• Applications of Phase Diagrams of Alloys

in Mechanical and Manufacturing Engineering

42

43

• 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