Phase Diagrams [9]

1>

Brazil’s map

Cu-Ag phase diagram

Some important definitions…

2>

• Component - chemically recognizable species (Fe and C in carbon steel, H2O and NaCl in salted water). A binary alloy contains two components, a ternary alloy – three, etc.

• Phase – a portion of a system that has uniform physical and chemical characteristics. Two distinct phases in a system have distinct physical or chemical characteristics(e.g. water and ice) and are separated from each other by definite phase boundaries. A phase may contain one or more components.

• A single-phase system is called homogeneous, systems with two or more phases are mixtures or heterogeneous systems.

• Solvent - host or major component in solution.• Solute - minor component

Solubility Limit of a component in a phase is the maximum amount of the componentthat can be dissolved in it (e.g. alcohol has unlimited solubility in water, sugarhas a limited solubility, oil is insoluble). The same concepts apply to solid phases: Cu and Ni are mutually soluble in any amount (unlimited solid solubility), while C has a limited solubility in Fe.

Some important definitions…

3>

Solubility Limit:

after Callister & Rethwisch (2014)

Some important definitions…

4>

• Microstructure:The properties of an alloy depend not only on proportions of the phases but also on how they are arranged structurally at the microscopic level. Thus, the microstructure is specified by the number of phases, their proportions, and their arrangement in space.

This is an alloy of Fe with 4 wt.% C. There are several phases. The long gray regions are flakes of graphite. The matrix is a fine mixture of BCC Fe and Fe3C compound.

graphite

matrix

Some important definitions…

5>

• Equilibrium – thermodynamic state where a system is stable at constant temperature, pressure and composition, not changing with time.

Gibb’s Phase Rule

6>

A tool to define the number of phases and/or degrees of phase changes that can be found in a system at equilibrium

P + F = C + NP: # of phases presentF: degrees of freedom (temp., pressure, composition)C: components or compoundsN: noncompositional variables (usually 1)

For the Cu-Ag system @ 1 atm for a single phase P:N=1 (temperature), C = 2 (Cu-Ag), P= 1 (a, b, L)F = 2 + 1 – 1= 2This means that to characterize the alloy within a single phase field, 2 parameters must be given: temperature and composition.

If 2 phases coexist, for example, α+L , β+L, α+β, then according to GPR, we have 1 degree of freedom: F = 2 + 1 – 2= 1. So, if we have temperature or composition, then we can completely define the system.

If 3 phases exist (for a binary system), there are 0 degrees of freedom. This means the composition and temperature are fixed. This condition is met for a eutectic system by the eutectic isotherm.

Unary phase diagram

7>

water

pure iron

Binary Systems

8>

CrystalStructure

Electro-negativity

r (nm)

Ni FCC 1.9 0.1246Cu FCC 1.8 0.1278

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

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

Cu-Ni Isomorphous phase diagram

9>

Information:

Phases,

Compositions,

Weight fractions.

Lever Rule

Determination of phase(s) present

10>

Rule 1: If we know T and Co, then we know:- which phase(s) is (are) present.

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(ºC)

L (liquid)

α(FCC solidsolution)

L + α

liquidu

s

solidu

sCu-Niphase

diagram

• Examples:A (1100ºC, 60 wt% Ni):

1 phase: αB (1250ºC, 35 wt% Ni):

2 phases: L + α

B(1

250º

C,3

5)A(1100ºC,60)

Determination of phase compositions

11>

Rule 2: If we know T and C0, then we can determine:- the composition of each phase.

wt% Ni20

1200

1300

T(ºC)

L (liquid)

α(solid)L +

α

liquidus

solidus

30 40 50

L + α

Cu-Ni system• Examples:

TAA

35C0

32CL

At TA = 1320ºC: Only Liquid (L) present CL = C0 ( = 35 wt% Ni)

At TB = 1250ºC: Both α and L presentCL = C liquidus ( = 32 wt% Ni) Cα = Csolidus ( = 43 wt% Ni)

At TD = 1190ºC: Only Solid (α) presentCα = C0 ( = 35 wt% Ni)

Consider C0 = 35 wt% Ni

DTD

tie line

4Cα3

BTB

Determination of phase weight fractions

12>

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, Wα = 0

At TD : Only Solid ( α) present WL = 0, Wα = 1.00

wt% Ni20

1200

1300

T(ºC)

L (liquid)

α(solid)L +

α

liquidus

solidus

30 40 50

L + α

Cu-Ni system

TA A

35C0

32CL

BTB

DTD

tie line

4Cα3

R S

At TB : Both α and L present

73.032433543

=−−

=

= 0.27

WL =S

R +S

Wα =R

R +S

Consider C0 = 35 wt% Ni

The Lever Rule: A Proof

13>

Sum of weight fractions:

Conservation of mass:Combine above equations:

WL + Wα = 1

Co = WLCL + WαCα

= RR + S

Wα =Co − CLCα − CL

= SR + S

WL =Cα − CoCα − CL

• A geometric interpretation:Co

R S

WαWL

CL Cαmoment equilibrium:

1− Wαsolving gives Lever Rule

WLR = WαS

Development of the microstructure

14>

Solidification in the solid + liquid phase occurs gradually upon cooling from the liquidusline.

The equilibrium compositionof the solid and the liquid change gradually during cooling (as can be determined by the tie-line method.)

Nuclei of the solid phase form and they grow to consume all the liquid at the solidus line.

Development of the microstructureWhen cooling is too fast for atoms to diffuse and produce equilibrium conditions, nonequilibrium concentrationsare produced. The nonuniformcomposition produced by nonequilibrium solidification is known as segregation.Microsegregation can cause hot shortness which is the melting of the material below the melting point of the equilibrium solidus.Homogenization, which involves heating the material just below the non-equilibrium solidus and holding it there for a few hours, reduces the microsegregation by enabling diffusion to bring the composition back to equilibrium. 15>

Cored vs Equilibrium Structures

16>

• Slow rate of cooling:Equilibrium structure

• Fast rate of cooling:Cored structure

First α to solidify:46 wt% NiLast α to solidify:< 35 wt% Ni

• Cα changes as we solidify.• Cu-Ni case: First α to solidify has Cα = 46 wt% Ni.

Last α to solidify has Cα = 35 wt% Ni.

Uniform Cα:35 wt% Ni

Mechanical Properties of the Cu-Ni System

17>

Effect of solid solution strengthening on:

Three-phase Reactions in Phase Diagrams

18>

Eutectoid & Peritectic

19>

Cu-Zn Phase diagram:

Eutectoid transition δ γ + ε

Peritectic transition γ + L δ

Intermetallic Compounds

20>

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

Mg2Pb

Microstructures in Eutectic Systems

21>

Pb-Snsystem

Sn) wt%7.89( Sn) wt%.38(1 Sn) wt%9.61( βα +Lcooling

heating

Lamellar Eutectic Structure

22>

A 2-phase microstructure resulting from the solidification of a liquid having the eutectic composition where the phases exist as a lamellae that alternate with one another.Formation of eutectic layered microstructure in the Pb-Sn system during solidification at the eutectic composition. Compositions of α and β phases are very different. Solidification involves redistribution of Pb and Sn atoms by atomic diffusion.

Pb‐rich

Sn‐rich

Hypoeutectic & Hypereutectic Compositions

23>

L+αL+β

α + β

200

Co, wt% Sn20 60 80 1000

300

100

L

α βTE

40

(Pb-SnSystem)

160 μmeutectic micro-constituent

hypereutectic: (illustration only)

β

ββ

ββ

β

175 μm

α

α

α

αα

α

hypoeutectic: Co = 50 wt% Sn

T(°C)

61.9eutectic

eutectic: Co =61.9wt% Sn

Iron-Carbon (Fe-C) Phase Diagram

24>

• 2 important points

-Eutectoid (B):γ ⇒ α +Fe3C

-Eutectic (A):L ⇒ γ +Fe3C

Fe3C

(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

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α 727°C = Teutectoid

ASR

4.30Result: Pearlite = alternating layers of α and Fe3C phases

120 μm

γ γγγ

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard)α (ferrite-soft)

Max. C solubility in γ iron = 2.11 wt%

Iron-Carbon (Fe-C) Phase Diagram

25>

Fe3C

(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

(Fe) Co, wt% C

1148°C

T(°C)

α727°C

(Fe-C System)

C0

0.76

proeutectoid ferritepearlite

100 μm Hypoeutectoidsteel

R S

α

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

r s

wα =s/(r+s)wγ =(1- wα)

γγ γ

γαα

α

γγγ γ

γ γγγ

Hypoeutectoid Steel

Iron-Carbon (Fe-C) Phase Diagram

26>

Hypereutectoid Steel

Fe3C

(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

(Fe) Co, wt%C

1148°C

T(°C)

α

(Fe-C System)

0.76 Co

proeutectoid Fe3C

60 μmHypereutectoid steel

pearlite

R S

wα =S/(R+S)wFe3C =(1-w α)

wpearlite = wγpearlite

sr

wFe3C =r/(r+s)wγ =(1-w Fe3C)

Fe3C

γγγ γ

γγγ γ

γγγ γ

Application of the Lever Rule (I)

27>

Example: For a AISI 1040 carbon steel (0.40 wt% C) at a temperature just below the eutectoid, determine the following:

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

forms per 100 g of steelc) the amount of pearlite and proeutectoid ferrite (α)

Application of the Lever Rule (I)

28>

g 3.94g 5.7 CFe

g7.5100 022.07.6022.04.0

100xCFe

CFe

3

CFe3

3

3

=α

=

=−−

=

−−

=α+ α

α

x

CCCCo

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

a) composition of Fe3C and ferrite (α)CO = 0.40 wt% CCα = 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

γ (austenite)

γ+L

γ + Fe3C

α + Fe3C

L+Fe3C

δ

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3Cα

Solution:

Application of the Lever Rule (I)

Solution (cont.):

29>

c) the amount of pearlite and proeutectoid ferrite (α) note: amount of pearlite = amount of γ just above TE

Co = 0.40 wt% CCα = 0.022 wt% CCpearlite = Cγ = 0.76 wt% C

γγ + α

=Co −CαCγ −Cα

x 100 = 51.2 g

pearlite = 51.2 gproeutectoid α = 48.8 g

Fe3C

(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

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CγCα

Application of the Lever Rule (II)

30>

For Pb-Sn system at 150°C, calculate relative amounts of each phase by (a) Mass fraction (b) Volume fraction

Pb-Sn system

Application of the Lever Rule (II)

31>

0.6711994099

CC

CCW

gm/cm7.3:ρgm/cm11.2:ρ

:Given

αβ

1βα

3β

3α

=−−=

−

−=

αβ

α1β 0.331199

1140CCCC

W =−−=

−−

= 1-0.67=0.33

Solution (a):

Application of the Lever Rule (II)

32>

33

33

cm4.52gm/cm7.3

gm33)(v

cm5.98gm/cm11.2

gm67)(v

FractionVolume(b) Solution

==

==

β

α

0.574.525.98

5.98vv

vVβα

αα =+

=+

=

Vβ = 1 − 0.57 = 0.43

Ternary Phase Diagrams

33>A

Hypothetical ternary phase diagrams where binary phase diagrams are present at the three faces.

Many alloy systems are based on three or more elements.To describe the changes in structure with temperature, a three-dimensional phase diagram is required, which is somewhat complicated.In the plot below the shaded area is the temperature at which freezing begins.

Ternary Phase Diagrams

34>

The isothermal plot of a ternary phase diagram is usually constructed using an equilateral triangle with the composition of each pure element (or compound) at each corner of the triangle to predict the phases at a particular temperature.

An isothermal plot at room temperature for the hypothetical ternary phase diagram.

20A-20B-60C

CALLISTER JR, W. D. AND RETHWISCH, D. G. Materials Science and Engineering: An Introduction, 9th edition.

John Wiley & Sons, Inc. 2014, 988p. ISBN: 978-1-118-32457-8.

ASHBY, M. and JONES, D. R. H. Engineering Materials 1: An Introduction to Properties, Applications and Design.

4th Edition. Elsevier Ltd. 2012, 472p. ISBN 978-0-08-096665-6.

ASKELAND, D. AND FULAY, P. Essentials of Materials Science & Engineering, 2nd Edition.

Cengage Learning. 2009, 604p. ISBN 978-0-495-24446-2

SMALLMAN, R. E. and NGAN, A.H.W. Physical Metallurgy and Advanced Materials, 7th Edition.

Elsevier Ltd. 2007, 650p. ISBN 978-0-7506-6906-1.

References

35Nota de aula preparada pelo Prof. Juno Gallego para a disciplina Ciência dos Materiais de Engenharia.® 2016. Permitida a impressão e divulgação. http://www.feis.unesp.br/#!/departamentos/engenharia-mecanica/grupos/maprotec/educacional/