Phase Equilibrium

Phase Equilibrium

Engr 2110 – Chapter 9Dr. R. R. Lindeke

Topics for Study

Definitions of Terms in Phase studies Binary Systems

Complete solubility systems Multiphase systems

Eutectics Eutectoids and Peritectics Intermetallic Compounds The Fe-C System

Definitions

Component of a system: Pure metals and or compounds of which an alloy is composed, e.g. Cu and Ag. They are the solute(s) and solvent

System: A body of engineering material under investigation. e.g. Ag – Cu system

Solubility Limit: The maximum concentration of solute atoms that may dissolve in theSolvent to form a “solid solution” at some temperature.

Phases: A homogenous portion of a system that has uniform physical and chemical characteristics, e.g. pure material, solid solution, liquid solution, and gaseous solution, ice and water, syrup and sugar.

Microstructure: A system’s microstructure is characterized by the number of phases present, their proportions, and the manner in which they are distributed or arranged. Factors affecting microstructure are: alloying elements present, their concentrations, and the heat treatment of the alloy.

Single phase system = Homogeneous system Multi phase system = Heterogeneous system or mixtures

Definitions, cont

Phase Equilibrium: A stable configuration with lowest free-energy (internal energy of a system, and also randomness or disorder of the atoms or molecules (entropy).Any change in Temp., Comp., and Pressure = increase in free energy and away from Equilibrium. And move to another state

Equilibrium Phase Diagram: It is a diagram of the information about the control of microstructure or phase structure of a particular alloy system. The relationships between temperature and the compositions and the quantities of phases present at equilibrium are shown.

Definition that focus on “Binary Systems”

Binary Isomorphous Systems: An alloy system that contains two components that attain complete liquid and solid solubility of the components, e.g. Cu and Ni alloy. It is the simplest binary system to understand.

Binary Eutectic Systems: An alloy system that contains two components that has a special composition with a minimum melting temperature.

Definitions, cont

With these definitions in mind:ISSUES TO ADDRESS...

• When we combine two elements... what “equilibrium state” would we expect to get?• In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature (T ) and/or a Pressure (P)

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

Phase BPhase A

Nickel atomCopper atom

Lets consider a commonly observed System exhibiting: Solutions –liquid (solid) regions that contain a single phase Mixtures – liquid & solid regions contain more than one phase

• Solubility Limit: Max concentration for which only a single phase solution occurs (as we saw earlier).

Question: 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

Pure

Su

gar

Tem

pera

ture

(°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

Pure

W

ater

Adapted from Fig. 9.1, Callister 7e.

A 1-phase region

A 2-phase region

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


D (100°C,90)2 phases

B (100°C,70)1 phase

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

Gibbs Phase Rule: a tool to define the number of phases that can be found in a system at equilibrium

For the system under study the rule determines if the system is at equilibrium

For a given system, we can use it to predict how many phases can be expected

Using this rule, for a given phase field, we can predict how many independent parameters we can specify

Ex: Determine how many equilibrium phases that can exist in a ternary alloy? C = 3; N = 1 (temperature) P + F = 3+1 = 4 So, if F = 0 (all 3 elements compositions and the temperature is set) we could have (at most)

4 phases present

where:P is # phases at equil.F is # degrees of freedom of the system (independent parameters)C is # components (elements) in systemN is # "noncompostional" parameters in system (temp &/or Press

P F C N

ure)

Phase Diagrams: A chart based on a System’s Free Energy indicating “equilibuim” system structures

• Indicate ‘stable’ phases as function of T, Co, and P. • We will focus on: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used).

• PhaseDiagramfor Cu-Nisystem

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

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

• 3 phase fields are observed: LL +

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L + liquidus

solidusAn “Isomorphic” Phase System

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:

• Rule 1: If we know T and Co then we know the # and types of all phases present.

• Examples:A(1100°C, 60): 1 phase:

B(1250°C, 35): 2 phases: L +

Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991).

B (1

250°

C,3

5) A(1100°C,60)

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)L +

liquidus

solidus

30 40 50

L +

Cu-Ni system

Phase Diagrams:

• Rule 2: If we know T and Co 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

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM

International, Materials Park, OH, 1991.)

BTB

DTD

tie line

4C3

• 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

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of

Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)

Phase Diagrams:

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 +SNotice: as in a lever “the opposite leg” controls with a balance (fulcrum) at

the ‘base composition’ and R+S = tie line length = difference in composition limiting phase boundary, at the temp of interest

Tie line – a line connecting the phases in equilibrium with each other – at a fixed temperature (a so-called Isotherm)

The Lever Rule

How much of each phase?We can Think of it as a lever! So to balance:

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

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

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

• 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

Cored (Non-equilibrium) Cooling

Notice:The Solidus line is “tilted” in this non-equilibrium cooled environment

: Min. melting TE

2 componentshas a special compositionwith a min. melting Temp.


Binary-Eutectic (PolyMorphic) 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

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 (lever rule):

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


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 200°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

• 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 Temp. solubility limit)

TE

(Pb-SnSystem)

L

L: Co wt% Sn

: Co wt% Sn

• 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

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


Microstructures in Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.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

45.1% and 54.8%

Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister 7e.

• 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) = 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 = = 72.6 wt%

W = 27.4 wt%Adapted from Fig. 9.16, Callister 7e.

Pb-Snsystem

L+200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

L+

40

+

TE

L: Co wt% Sn LL

L+L+

+

200

Co, wt% Sn20 60 80 1000

300

100

L

TE

40

(Pb-Sn System)

Hypoeutectic & Hypereutectic Compositions

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 meutectic micro-constituent


hypereutectic: (illustration only)

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 m

hypoeutectic: Co = 50 wt% Sn


T(°C)

61.9eutectic

eutectic: Co = 61.9 wt% Sn

“Intermetallic” Compounds

Mg2Pb

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


An Intermetallic Compound is also an important part of the Fe-C system!

Eutectoid: solid phase in equilibrium with two solid phasesS2 S1+S3

+ Fe3C (727ºC)

intermetallic compound - cementite

coolheat

Peritectic: liquid + solid 1 in equilibrium with a single solid 2 (Fig 9.21)S1 + L S2

+ L (1493ºC)

cool

heat

Eutectic: a liquid in equilibrium with two solidsL +

Eutectoid & Peritectic – some definitions

coolheat

Eutectoid & Peritectic

Cu-Zn Phase diagram


Eutectoid transition +

Peritectic transition + L

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

-Eutectoid (B): +Fe3C

-Eutectic (A): L + Fe3C

Adapted from Fig. 9.24,Callister 7e.

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

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

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard) (ferrite-soft)

Max. C solubility in iron = 2.11 wt%

Hypoeutectoid Steel

Adapted from Figs. 9.24 and 9.29,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.)

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

Adapted from Fig. 9.30,Callister 7e.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)

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)

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 mHypereutectoid steel

pearlite

R S

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

wpearlite = wpearlite

sr

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

Fe3C

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 steelc) the amount of pearlite and proeutectoid ferrite ()

Solution:

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% C3

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

Solution, cont:c) 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

CCLooking at the Pearlite:11.1% Fe3C (.111*51.2 gm = 5.66 gm) & 88.9% (.889*51.2gm = 45.5 gm)

total = 45.5 + 48.8 = 94.3 gm

Alloying Steel with More Elements• Teutectoid changes: • Ceutectoid changes:

Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

Adapted from Fig. 9.35,Callister 7e. (Fig. 9.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

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

Looking at a Tertiary Diagram

When looking at a Tertiary (3 element) P. diagram, like this image, the Mat. Engineer represents equilibrium phases by taking “slices at different 3rd element content” from the 3 dimensional Fe-C-Cr phase equilibrium diagram

What this graph shows are the increasing temperature of the euctectoid and the decreasing carbon content, and (indirectly) the shrinking phase field

From: G. Krauss, Principles of Heat Treatment of Steels, ASM International, 1990

• 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 (cause) a range of microstructures.

Summary

Phase Equilibrium

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