phasediagram

PHASE DIAGRAMS

Prof. H. K. Khaira

Professor

Deptt. of MSME

M.A.N.I.T., Bhopal

Contents

• 1. Phase diagram• 2. Phase Rule• 3. Lever Rule• Microstructure Development

Phase Diagrams

Phase

• Phase is a homogenous, physically distinct and mechanically separable portion of the material with a given chemical composition and structure.

Phase diagrams

Properties of a materials depend on 1. Number of phases present2. Type of phases present3. Amount of phases present and 4. Form of the phases present

The Properties can be changed by altering these quantities. In order to make these changes, it is essential to know the conditions under which these quantities exist and the conditions under which a change in phase will occur.

Phase diagrams

• The best method to record the data related to phase changes in many alloy systems is in the form of phase diagrams, also known as equilibrium diagrams or constitutional diagrams.

Phase diagrams

• In order to specify completely the state of a system in equilibrium, it is necessary to specify three independent variables.

• These variables, which are externally controllable, are temperature, pressure and composition.

• Phase diagram is the graphical presentation of the phases present in a system under different conditions of pressure, temperature and composition.

Phase diagrams

• In metallurgical systems, the pressure is usually taken as atmospheric pressure. Thus the phase diagram shows the phases present at different compositions and temperatures.

Phase diagrams

• With pressure assumed to be constant at atmospheric pressure, the equilibrium diagram indicates the structural changes due to variation of temperature and composition.

• Phase diagrams show the phases present under equilibrium conditions, that is, under conditions in which there will be no change with time.

• Equilibrium conditions may be approached by extremely slow heating and cooling, so that if a phase change is to occur, sufficient time is allowed.

Phase diagrams

The phase diagram in Figure displays an alloy of two metals which forms a solid solution at all relative concentrations of the two species

Phase diagrams

• Phase diagrams are usually plotted with composition on X axis and the temperature on Y axis,.

Types of Equilibrium Diagrams

• 1. Complete solid solution type• 2. Eutectic type• 3. Peritectic type

Complete solid solution type

Complete solid solution type

• The phase diagram in which both the constituents are soluble in each other in solid state at all compositions, is known as complete solid solution type phase diagram.

Complete solid solution Diagram

The phase diagram in Figure displays an alloy of two metals which forms a solid solution at all relative concentrations of the two species

Nickel-Copper Phase Diagram

Germanium-Silicon PhaseDiagram

Eutectic Type

Eutectic Type

• It is a phase diagram containing eutectic reaction.

• Eutectic reaction is : L = α (s) + β (s)

In this reaction, a liquid phase (L) decomposes into a mixture of two solid phases (α and β) on cooling.

Eutectic Diagram

Tin-Lead Phase Diagram

Gold-Germanium Phase Diagram

Copper-Silver Phase Diagram

Peritectic Diagram

• It is a phase diagram containing peritectic reaction.

• peritectic reaction : L + α (s) = β (s)

• In peritectic reaction, a liquid (L) and a solid (α) transform in to another solid (β) on cooling.

Phase diagram for Fe–C system (dotted lines represent iron-graphite equilibrium).

Application of Phase Diagrams

• Phase diagram gives us – Overall Composition– Solidus– Liquidus– Limits of Solid Solubility– Chemical Composition of Phases at any temperature– Amount of Phases at any temperature– Invariant Reactions– Development of Microstructure

Overall Composition

• Concentration: Relative amounts of each constituent

• It is the horizontal axis in all binary phase diagrams

• The scale can be in weight %, atomic % or mole %

Solidus and Solidus

• Solidus– Temperature up to which alloy is completely solid– Temperature at which liquefaction begins

• Liquidus– Temperature up to which alloy is completely liquid– Temperature at which solidification begins

Chemical Composition of Phases

• It is the chemical composition of each phase in the system

• In a system having more than one phase, each phase will have a unique chemical composition which will be different from each other, and will also be different from the overall composition

• Not to be confused with overall composition

Invariant Reactions

• Eutectic: L = α (s) + β (s); e.g., Pb-Sn• Peritectic: α (s) + L = β (s); e.g., Pb-In• Monotectic: L1 = α (s) + L2; e.g., Cu-Pb• Syntectic: L1 + L2 = α (s); e.g., Na-Zn• Metatectic: β (s) + α (s) = L1 e.g., U-Mn

Cooling Curve

• A cooling curve is a graphical plot of the changes in temperature with time for a material over the entire temperature range through which it cools.

Cooling Curve for Pure Metals

Cooling Curve

• This is by far the most widely used experimental method.

• It relies on the information obtained from the cooling diagrams.

• In this method, alloys with different compositions are melted and then the temperature of the mixture is measured at a certain time interval while cooling back to room temperature.

Cooling Curve

• A cooling curve for each mixture is constructed and the initial and final phase change temperatures are determined. Then these temperatures are used for the construction of the phase diagrams

Cooling Curve

• Under equilibrium conditions, all metals exhibit a definite melting or freezing point. If a cooling curve is plotted for a pure metal. It will show a horizontal line at the melting or freezing temperature.

Cooling curve for the solidification of a pure metal.

Cooling curve for a pure metal showing possible

undercooling.

Cooling curve for a solid solution.

Phase Diagram of alloy A+B

Series of cooling curves for different alloys in a completely soluble system. The dotted lines indicate the form of the phase

diagram

Cooling Curves

The Phase Rule

43

The Phase RuleDegree of freedom or Variance (f): the number of intensive variables that can be changed independently without disturbing the number of phases in equilibrium.

The phase rule: a general relation among the variance f, the number of components c and the number of phases p at equilibrium for a system of any composition.

f = c – p + 2

45

The Phase Rule

(a) (b)

The difference between (a) a single-phase solution, in which the composition is uniform on a microscopic scale, and (b) a dispersion, in which regions of one component are embedded in a matrix of a second component.

Physical Chemistry

Chapter 7

46

The Phase Rule

Phase: a state of matter that is uniform throughout in chemical composition and physical state. (Gibbs)

Number of phase (p): Gas or gaseous mixture – single phaseLiquid – one, two and three phases

two totally miscible liquids – single phasea slurry of ice and water – two phases

Solid – a crystal is a single phasean alloy of two metals – two phases (immiscible)- one phase (miscible)

Physical Chemistry

Chapter 7

Simple Example

51

H2O phase diagram: P — TRegion (s, l, g):

f=2, one phase

t/℃

A

D C

0.00611

0.01

solid

gas

liquid

O

P /

10 5

Pa

374.2

218 atm

99.974

1 atm

0.0024

I

R

S

Y

Tf TbT3

Line (OA, AD, AC):

f=1, two phases in equilibrium

Point (A):

f=0, three phases in equilibrium

Tc

Physical Chemistry

Chapter 7

52

f＝ 1－ p＋ 2＝ 3－ p， (C＝ 1）

f ≥0, p ≥1, 3≥p≥1

p＝ 1， f＝ 2

p＝ 2， f＝ 1

p＝ 3， f＝ 0

One-component phase equilibrium

For a one-component system (pure water)

Physical Chemistry

Chapter 7

Lever Rule

Lever Rule

• The compositions of the two coexisting phases at a point in a two phase region is given by the points of intersection of the tie line from that point with the boundaries of the respective phases.

• The relative amounts of two coexisting phases at a point are INVERSELY proportional to the distances of the point from intersection points of the tie line from the point with the phase boundaries.”

54

A + B

Translating This Statement

• “two coexisting phases”Means you are in a 2 phase region.

A B

A + LB + L

L

C

• Pick an arbitrary point C.• There are two phases at point C. • These phases are A an B• Hence Phases A and B will be in equilibrium at point C

P Q

Tie Lines

The line from A to B through C is a tie line.

A tie line is,• An isothermal line• That connects two equilibrium phases

A + B

Translating This Statement

The tie line from point C intersect boundaries of phase A and B at 0% A and 0% B respectively.

Hence the composition of phase A will be 0% B or pure A.

Similarly, the composition of B will be pure B or 0% A.

A B

A + LB + L

L

C

The distances are AC and AB

A + B

Translating This Statement

“The intercepts of the tie line with the phase boundaries A and B are PA and CQ respectively.

A B

A + LB + L

L

CP Q

A + B

Translating This Statement

The intercepts of the tie line with the phase boundaries A and B are CA and CB respectively.

A B

A + LB + L

L

C

A + B

Translating This Statement

• “The amounts . . . Are inversely proportional”Means AC / AB is the fraction of B

A B

A + LB + L

L

C

And, CB / AB is the fraction of A

B =

A =

A + B

A Sample Calculation

A B

A + LB + L

L

C

• Draw a horizontal tie line through the point.

D

• Measure its length

• Measure the length of each side

AD = 1 cm

AC = .75 cm CD = .25 cm

• Calculate the amounts A & L

%25cm1

cm25.A %75

1

75.L

• Identify the phases.

A (solid)

L (liquid)

A more complex system . . .

• Identify the phases.

• Measure the line lengths.

• Draw a tie line.

• Calculate the amounts of each phase present.

C

A B

A = Li2O-B2O3

B = Li2O-2B2O3

AB = .75 cm

AC = .188 cm

CB = .562 cm

%7575.

562.A %25

75.

188.B

One Last Note.

If a point is in a single phase region (including a solid solution), NO tie line is used. There is 100% of that phase.

C

Summary

• The lever rule is used to calculate the relative percents of each phase when 2 or more phases are present.

• The first step in lever rule calculation is to draw a tie line through the composition.

• Next one measures the lengths of the tie line, and the distance from the composition to each phase.

• The relative concentration of a phase is proportional to the distance from the other phase to the composition, divided by the length of the tie line. (Opposite length / total length)

65

66

Cu-Ni Phase Diagram: T vs. c (wt% or at%)

• Tell us about phases as function of T, Co, P. • For this course:

--binary systems: just 2 components.

--independent variables: T and Co (P = 1atm is always used).

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

a

(FCC solid

solution)

L + aliquidus

solid

us

Contains 2 Phases1. L2. a

There are 3 phase fields1. L2. L + a3. a

-- isomorphousi.e., complete

solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.

67

Phase Diagrams: # and types of phases

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

• Examples:

A(1100°C, 60 wt% Ni): 1 phase: a

B(1250°C, 35 wt% Ni): 2 phases: L + a

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

a

(FCC solidsolution)

L + aliquidus

solid

us

B (

1250

°C,3

5) A(1100°C,60)

Cu-Niphase

diagram

68

Phase Diagrams: composition of phases

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

wt% Ni20

1200

1300

T(°C)

L (liquid)

a(solid)L + a

liquidus

solidus

30 40 50

L + a

Cu-Ni system

TAA

35C0

32CL

DTD

tie line

4C

3

BTB

• Examples:

At TA = 1320°C:

Only Liquid (L) present CL = C0

( = 35 wt% Ni)

At TB = 1250°C:

Both and L present CL

= C liquidus ( = 32 wt% Ni)

C = C solidus ( = 43 wt% Ni)

At TD = 1190°C:

Only Solid (a) present C = C0 ( = 35 wt% Ni)

Consider C0 = 35 wt% Ni

69

Phase Diagrams: wt. fraction of phases

• Rule 3: If we know T and C0, then can determine: -- the weight fraction of each phase.

• Examples:

At TA : Only Liquid (L) present

WL = 1.00, Wa = 0

At TD : Only Solid ( ) present

WL = 0, Wa

= 1.00

At TB : Both and L present

73.03243

3543

= 0.27

WL= S

R + S

Wa= R

R + S

Consider C0 = 35 wt% Ni

wt% Ni20

1200

1300

T(°C)

L (liquid)

a(solid)L + a

liquidus

solidus

30 40 50

L + a

Cu-Ni system

TAA

35C0

32CL

BTB

DTD

tie line

4Ca

3

R S

W = wt. fraction of phase out of whole.

70

The Lever Rule• Sum of weight fractions:

• Conservation of mass (Ni):

• Combine above equations:

W

LW

1

C

oW

LC

LW

C

RR S

W Co CLC CL

SR S

WLC Co

C CL

moment equilibrium:

1 Wsolving gives Lever Rule

WLR WS

wt% Ni

20

1200

1300

T(°C)

L (liquid)

a(solid)L + a

liquidus

solidus

30 40 50

L + aB

TB

tie line

C0CLC

SR

71

The Lever Rule: an interpretation

ML M

R S

M x S ML x R

• Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm

wt% Ni

20

1200

1300

T(°C)

L (liquid)

a(solid)L + a

liquidus

solidus

30 40 50

L + aB

TB

tie line

C0CLC

SR

What fraction of each phase?

Think of tie line as a lever

L

L

LL

LL CC

CC

SR

RW

CC

CC

SR

S

MM

MW

00

72

Phase and Microstructure (equilibrium)Example: Cooling in Cu-Ni Binary

wt% Ni20

1200

1300

30 40 50110 0

L (liquid)

a (solid)

L + a

L + a

T(°C)

A

35C0

L: 35wt%Ni

4635

4332

a: 43 wt% Ni

L: 32 wt% Ni

Ba: 46 wt% NiL: 35 wt% Ni

C

EL: 24 wt% Ni

a: 36 wt% Ni

24 36D

• Consider microstuctural changes that accompany the cooling of a C0 = 35 wt% Ni alloy

• From liquid, solid phase nucleates.

• From solid, other phases can nucleate.

• Like ice, many grains of solid form.

• wt% of SOLUTE given by line dropped from boundaries

• Fraction of PHASES present given by the “lever rule”.

• Microstructure different depending on cool slowly or quench.

73

Binary-Eutectic Systemshas a special compositionwith a minimum melting T.

• Ex: Cu-Ag 3 single-phase regions (L, , )• Limited solubility : mostly Cu

: mostly Ag• TE: no liquid below TE.• cE: composition for min.

melting T (Eutectic).

Eutectic: direct from liquid to 2-phase solid upon cooling: L +

Cu-Agsystem

L (liquid)

a L + aL+b b

a + b

C , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

Ag) wt%1.29( Ag) wt%.08( Ag) wt%9.71( bL cooling

heating

74

Example 2: Pb-Sn Eutectic System

Ca = 17 wt% Sn-- phase compositions

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:

-- phases present:

-- relative amt of phases

Wa =CL - C0

CL - C=

46 - 40

46 - 17= 0.21

WL =C0 - C

CL - C=

23

29= 0.79

+ L

CL = 46 wt% Sn

40C0

46CL

17C

L+b

a + b

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a b

L + a

183°C

220SR

75

Example 2: Pb-Sn Eutectic System

Ca = 17 wt% Sn-- phase compositions

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:

-- phases present:

-- relative amt of phases

Wa =CL - C0

CL - C=

46 - 40

46 - 17= 0.21

WL =C0 - C

CL - C=

23

29= 0.79

+ L

CL = 46 wt% Sn

40C0

46CL

17C

L+b

a + b

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a b

L + a

183°C

220SR

76

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

--phases present: + b --compositions of phases:

Example 1: Lead-Tin (Pb-Sn) Eutectic DiagramSolder for electronics

Use the “Lever Rule”

Ca = 11 wt% Sn

Cb = 99 wt% SnL + a

L+b

a + b

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a183°C

61.9 97.8

b

150

40C0

11C

99C

SR

Adapted from Fig. 10.8, Callister & Rethwisch 3e.

-- relative amount of each phase:

W

=C - C0

C - C

=99 - 4099 - 11

= 5988

= 0.67

SR+S

=

W

=C0 - C

C - C=R

R+S

=2988

= 0.33=40 - 1199 - 11

77

Example 2: Pb-Sn Eutectic System

Ca = 17 wt% Sn-- phase compositions

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:

-- phases present:

-- relative amt of phases

Wa =CL - C0

CL - C=

46 - 40

46 - 17= 0.21

WL =C0 - C

CL - C=

23

29= 0.79

+ L

CL = 46 wt% Sn

40C0

46CL

17C

L+b

a + b

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

a b

L + a

183°C

220SR

78

• For alloys for 18.3wt%Sn < Co < 61.9wt%Sn

• Result: a crystals and a eutectic microstructure

Microstructure “below” Eutectic (hypoeutectic)

18.3 61.9

SR

97.8

SR

primary aeutectic a

eutectic b

Adapted from Fig. 10.16, Callister & Rethwisch 3e.

Pb-Snsystem

L+ 200

T(°C)

C, wt% Sn

20 60 80 1000

300

100

L

a b

L+

40

+

TE

L: C0 wt% Sn LL

WL = (1-Wa) = 0.50

Ca = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SWa = = 0.50

• Just above TE :

• Just below TE :C

= 18.3 wt% Sn

C = 97.8 wt% Sn

SR + S

W = = 0.73

W = 0.27

79

Solder: Lead-Tin (Pb-Sn) microstructure

For 50 wt% Pb alloy:• Lead-rich phase (dark)• Lamellar eutectic structure of Sn-rich phase (light).

* Why is Liquid-phase ~62.9wt%Sn and -phase ~16.3wt%Sn at 180 C?* For fraction of total phase (both eutectic and primary), use the Lever Rule.

+

L +

L

80

T(°C)

(Pb-Sn System)

L + 200

Co, wt% Sn20 400

300

100

L

60

+ b

TE b

080 100

L + b

18.361.9

97.8

Cohypoeutectic

Cohypereutectic

eutectic

hypereutectic: (illustration only)

160m

eutectic: Co=61.9wt%Sn

175m

b

bb

bb

b

hypoeutectic: Co=50wt%Sn

eutectic micro-constituent

Hypoeutectic & Hypereutectic

Adapted from Fig. 10.8, Callister & Rethwisch 3e.

(Figs. 10.14 and 10.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

81

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 ().

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

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

Example Problem Steel

82

Solution to Problem

WFe3C R

R S

C0 C

CFe3C C

0.40 0.0226.70 0.022

0.057

b) Use lever rule with the tie line shown

a) Use RS tie line just below Eutectoid

Ca = 0.022 wt% C

CFe3C = 6.70 wt% C

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 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

C , wt% C

1148°C

T(°C)

727°C

C0

R S

CFe C3C

83

Solution to Problem

c) Using the VX tie line just above the eutectoid and realizing that

C0 = 0.40 wt% C

Ca = 0.022 wt% C

Cpearlite = C = 0.76 wt% C

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

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

C, wt% C

1148°C

T(°C)

727°C

C0

V X

CC

Example Problem

• One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following:

• a) Weight % of liquid and α at 250ºC• b) Chemical composition of the liquid and α at 250ºC• c) Weight % of the liquid and α just above the eutectic

temperature• d) Chemical composition of the liquid and α at just above the

eutectic temperature

90

92

• Lever Rule is useful to determine:

- the composition of each phase,- and the wt% of each phase

•

SummaryLever Rule

Microstructure Development

• The microstructure developed depends on the overall composition and the cooling rate

Composition dependence ofmicrostructure

Composition dependence ofmicrostructure

Composition dependence ofmicrostructure

Composition dependence ofmicrostructure

Composition dependence ofmicrostructure

Phase diagram for Pb–Sn system. Alloy 1: 63Sn–37Pb, Alloy 2: 70Pb–30Sn, Alloy 3: 70Sn–30Pb.