Lecture 2 - Phase Diagrams - Part I [Compatibility Mode]

PROCESSING OF ENGINEERING ALLOYS

Lecture 2

Phase Diagrams – Part I

Our context

The understanding of phase diagrams for alloy systems is extremely important because:systems is extremely important because:

• There is a strong correlation between microstructure and mechanical propertiesp p

• The development of microstructure of an alloy is related to the characteristics of its phase diagram.– Even though most phase diagrams represent stable (or

equilibrium) states and structures

2

Learning Objectives

By the end of this lesson and associated studyyou will be able to:you will be able to:

• Schematically sketch and label simpleisomorphous and eutectic phase diagramsp p g

• Identify the composition of an alloy, its tempt etc.in a given a binary phase diagram.

• Assuming that the alloys are at equilibrium youwill be able to determine:– What phase(s) is (are) present– The composition of the phases– The mass fraction of the phases

3

•Alloy: Some definitions and basic concepts

A material that has metallic properties and is composed of two or more chemical elements of which at least one is a metal (i.e. steel is an alloy of carbon in iron; stainless steel is an alloy of carbon, chromium and sometimes nickel in iron). Components:•Components:

The elements or compounds which are present in the alloy(e.g., Cu and Zn in copper-zinc brass)

• Solute and solvent• Solute and solventSolvent represents the element or compound that is present in the greatest amount. Solute is used to denote an element or compound present in a minor concentration.co ce t at o

• Solid solutionA solid solution consists of atoms of at least two different types; the solute atoms occupy either substitutional or interstitial positions in the solvent lattice and the crystal lattice of the solvent is maintained.

• Systemi.System may refer to a specific body of material under consideration (e.g. a l dl f lt t l) O

4

ladle of molten steel). Or ii.a series of possible alloys consisting with the same components but without regard to alloy composition (e.g. iron-carbon system).

Phases• Phases:

A phase may be defined as a homogeneous portion of a system that has uniform physical and chemical characteristicsuniform physical and chemical characteristics.

Aluminum-Copper

β (lighter h )Copper

Alloy

α (darker

phase)

phase)Adapted from chapter-opening photograph, Chapter 9, Callister, Materials Science & Engineering: An

5

Engineering: An Introduction, 3e.

Solubility Limit• Solution – solid, liquid, or gas solutions, single phase• Mixture – more than one phase

• Solubility Limit:yMaximum concentration for which only a single phase solution exists.

Q tiQuestion: What is the solubility limit for sugar in water at 20ºC?

Sugar-water phase diagram Fig 9.1, Callister and Rethwisch, 8e 6

Effect of Temperature & Composition• Altering T can change # of phases: path A to B.• Altering C can change # of phases: path B to D.

DB

100

D (100ºC,C = 90)2 phases

B (100ºC,C = 70)1 phase

re (º

C)

L60

80 L (liquid)

+ water-

mpe

ratu

r L(liquid solution

i.e., syrup)40

60S

(solid sugar)

sugarsystem

70 80 1006040200

Tem 20

0Adapted from Fig. 9.1,

A (20ºC,C = 70)2 phases

70 80 1006040200C = Composition (wt% sugar)

7

p gCallister & Rethwisch 8e.

Criteria for Solid Solubility

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

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246Cu 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.

• Ni and Cu are totally soluble in one another for all proportions.

8

y p p

Phase equilibria

• Equilibrium – free energy is at a minimum undersome specified combination or temperature,some specified combination or temperature,pressure and composition

• A change in temperature, pressure and/org p , pcomposition for a system in equilibrium will resultin an increase in the free energy and in a possiblespontaneous change to another state wherebythe free energy is lowered.Ph ilib i f t ilib i it• Phase equilibrium - refers to equilibrium as itapplies to systems in which more than one phasemay existmay exist

9

Phase equilbiria in a liquid-solid system

Sugar-water syrup containedin a closed vessel and thesolution is in contact with solidsugar at 20°C. If the systemis at equilibrium theis at equilibrium thecomposition of the syrup is65wt % (C12 H22O11

_ H2O) and35 t% H O35wt% H2O

If the temperature of thet i dd l i dsystem is suddenly raised, say

to 100°C, this equilibrium istemporarily upset in that the

Sugar-water phase diagram Fig 9.1, Callister and Rethwisch, 8e

solubility limit is increased to80wt% sugar. 10

Phase equilbiria in solid-solid systems

• Free energy considerations - donot indicate the time periodpnecessary for the attainment of anew equilibrium state

• Often a state of equilibirum isnever completely achievedb th t f h tbecause the rate of approach toequilibrium is extremely slow;such a system is said to be in asuch a system is said to be in anonequilibrium or metastablestate.

11

One component (or unary) phase diagrams

Pressure-temperature diagram for H2OFig 9.2, Callister and Rethwisch, 8e

Binary Phase Diagrams

• Independent variables - Temperature andCompositionComposition

• Pressure held constant (1 atm almost alwaysused))

• If contain two components = Binary• Diagrams are “maps” that represent theDiagrams are maps that represent the

relationships between temperature and thecompositions and quantities of phases atequilibrium

Binary Isomorphous system

• Cu-Ni system• 2 phases: 1600

T(ºC)• 2 phases: L (liquid), α (FCC solid solution)

1500

1600

L (liquid)( )

• 3 different phase fields: 1300

1400

LL+ α

1100

1200 α(FCC solid

l ti )α

Note nomenclaturewt% Ni20 40 60 80 1000

1000solution)

Phase diagram for Cu-Ni• Note nomenclature gAdapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

Reading phase diagrams

If we know the composition andtemperature of our binary systemp y y(and at equilibrium), at least 3kinds of info are available,

1 the phases that are present1. the phases that are present,2. the composition of these

phases andp3. the fractions or percentages of

the phases

The copper-nickel phase diagram Fig. 9.3(a), Callister & Rethwisch 8e.

Phase Diagrams:Determination of phase compositions

Cu-Ni t

p p• Rule 2: If we know T and C0, then we can determine:

-- the composition of each phase.

1300

T(ºC)

L (liquid)

system• Examples:TA

A

At T = 1320ºC:Consider C0 = 35 wt% Ni tie line

1300 L (liquid)

α

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

At T = 1190ºC:

BTB

20

1200 (solid)

30 40 503532

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

DTD

43wt% Ni

35C0

32CLAt TB = 1250ºC:

Both α and L presentCL = C liquidus ( = 32 wt% Ni)

4Cα3

Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.),

S O ( )

16

qCα = C solidus ( = 43 wt% Ni)

ASM International, Materials Park, OH (1991).

Phase Diagrams:Determination of phase weight fractions

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

Determination of phase weight fractions

Cu-Ni • Examples:

At T : Only Liquid (L) present

T(ºC)

L (liquid)

system

TAA

tie lineConsider C0 = 35 wt% NiAt TA : Only Liquid (L) present

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

W 0 W 1 00

1300 L (liquid)

α

BTB R S

WL = 0, W α = 1.00

20

1200

α(solid)

30 40 503532

DTD

43

At TB : Both α and L present

3543 −Swt% Ni

20 30 40 5035C0

32CL

4Cα373.0

32433543

=−

=WL = SR + S

W RAdapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.),

17

= 0.27Wα = RR + S

ag a s o a y c e oys, as ( d ),ASM International, Materials Park, OH (1991).

The Lever Rule

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

What fraction of each phase?Think of the tie line as a lever

1300

T(ºC)

L (liquid)tie line

ML Mαα

( lid)

BTB

R S20

1200 (solid)

30 40 50C0CL Cα

SR

Mα x S = ML x R

LL CCRWCCSMW −−α 00

wt% Ni0 Cα

Adapted from Fig. 9.3(b), Callister & Rethwisch 8e.

18L

L

LL

LL CCSR

WCCSRMM

W−

=+

=−

=+

=+

=α

αα

α

α

00

Microstructur Ex: Cooling of a Cu-Ni Alloy

L (liquid)T(ºC) L: 35wt%Ni

Cu-Nit

• Phase diagram:Cu-Ni system.

130 0 Asystem• Consider

microstuctural changes that 4635

32Bα: 46 wt% Ni

L: 35 wt% Ni

C

120 0

changes that accompany the cooling of a

4332

α: 43 wt% Ni L: 32 wt% Ni

C

E

24 36D

120 0

α

C0 = 35 wt% Ni alloy EL: 24 wt% Ni

α: 36 wt% Ni

20 30 40 50110 0

(solid)

35

19wt% Ni

20 30 40 5035C0Adapted from Fig. 9.4,

Callister & Rethwisch 8e.

Cored vs Equilibrium Structures

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

L t t lidif h C 35 t% Ni

• Slow rate of cooling:Equilibrium structure

• Fast rate of cooling:Cored structure

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

q

First α to solidify:46 wt% Ni

Uniform Cα:35 wt% Ni

Last α to solidify:< 35 wt% Ni

20

Mechanical Properties: Cu-Ni System

• Effect of solid solution strengthening on:

-- Tensile strength (TS) -- Ductility (%EL)g ( ) y ( )

MP

a)

400 %E

L)

50

60

%EL for%EL for pure Cu

Stre

ngth

(

300

TS for pure Ni

gatio

n (%

40

50 %EL for pure Ni

Tens

ile S

Cu Ni0 20 40 60 80 100

200

TS for pure Cu

Elo

nCu Ni0 20 40 60 80 10020

30

Adapted from Fig. 9.6(a), Callister & Rethwisch 8e.

T

Composition, wt% Ni Composition, wt% NiCu Ni

Adapted from Fig. 9.6(b), Callister & Rethwisch 8e.

21

Binary-Eutectic Systems

2 componentshas a special compositionwith a min. melting T.

Cu-AgT(ºC)

• 3 single phase regions (L, α, β)

Ex.: Cu-Ag system system

L (liquid)1200T(ºC)

1000(L, α, β)• Limited solubility:

α: mostly Cu β: mostly Ag

α L + α L+β β

600

800TE 8.0 71.9 91.2779ºC

β: mostly Ag • TE : No liquid below TE

: Composition at t t T

• CE

α + β400

600

Adapted from Fig 9 7 Callister

temperature TE

C , wt% Ag20 40 60 80 1000

200CE• Eutectic reaction

L(C ) α(C ) + β(C )22

Adapted from Fig. 9.7, Callister & Rethwisch 8e.

Ag) wt%1.29( Ag) wt%.08( Ag) wt%9.71( β+αLcooling

heating

L(CE) α(CαE) + β(CβE)

EX 1: Pb-Sn Eutectic System

T(ºC)

• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine:-- the phases present Pb-Sn

systemAnswer: α + β T( C)

300 L (liquid)

systemAnswer: α + β-- the phase compositionsAnswer: Cα = 11 wt% Sn

C = 99 wt% SnL+ α

L+β20018.3

α 183ºC61.9 97.8

β-- the relative amountof each phase

150 SR

Cβ = 99 wt% Sn

Answer:

α + β100

150 SRW

α=

Cβ - C0Cβ - Cα

99 40 59

SR+S =

Answer:

C, wt% Sn20 60 80 1000 40

C011Cα

99Cβ

= 99 - 4099 - 11 = 59

88 = 0.67

Wβ =C0 - Cα

C C=RR+S Adapted from Fig 9 8 Callister

23

β Cβ - CαR+S

= 2988

= 0.33= 40 - 1199 - 11

Adapted from Fig. 9.8, Callister & Rethwisch 8e.

EX 2: Pb-Sn Eutectic System

T(ºC)

• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine:-- the phases present: Pb-Sn

systemAnswer: α + L

Answer: Cα = 17 wt% Sn-- the phase compositions

T( C)

300 L (liquid)L+α

systemAnswer: α + L

C = 46 wt% Sn

L+β200 α βL+α

183ºC

-- the relative amountof each phase

220SR

CL = 46 wt% Sn

A

α + β100

Wα =CL - C0

CL - Cα=

46 - 4046 - 17

Answer:

C, wt% Sn20 60 80 1000

L α

=629 = 0.21 40

C046CL

17Cα

Adapted from Fig 9 8 Callister

24WL =

C0 - Cα

CL - Cα=

2329 = 0.79

Adapted from Fig. 9.8, Callister & Rethwisch 8e.

Microstructural Developments in Eutectic Systems I

• For alloys for which

y

T(ºC) L: C0 wt% Sn For alloys for which C0 < 2 wt% Sn

• Result: at room temperaturepolycrystalline with grains of

( )

300 L

400

αL

0

-- polycrystalline with grains of α phase having composition C0

L+ α200

300 L

αT

(Pb-Snα: C0 wt% Sn

100 α + β

TE System)α: C0 wt% Sn

0C , wt% Sn

10

2

20C0

30Adapted from Fig. 9.11, C lli t & R th i h 8

25

2(room T solubility limit)

Callister & Rethwisch 8e.

Microstructural Developments in Eutectic Systems II

• For alloys for which

y

T(ºC) L: C0 wt% Sn For alloys for which 2 wt% Sn < C0 < 18.3 wt% Sn

• Result: t t t i + β

( )

300

L400

Lαat temperatures in α + β range

-- polycrystalline with α grainsand small β-phase particles

L + α

200

300

α

α

α: C0 wt% Sn

Pb Sn100

α + β

TEαβ

Adapted from Fig. 9.12, C lli t & R th i h 8

Pb-Snsystem

C wt% Sn10 200

C30

α β

2

26

Callister & Rethwisch 8e. C , wt% Sn18.3

C0

(sol. limit at TE)

2(sol. limit at T room )

Microstructural Developments in Eutectic Systems III

• For alloy of composition C0 = CE• Result: Eutectic microstructure (lamellar structure)

y

( )-- alternating layers (lamellae) of α and β phases.

Micrograph of Pb-Sn eutectic T(ºC)microstructure

Pb-Snsystem

300 LL+α

L: C0 wt% Sn

systemL + β200 α β183ºC

TE

Adapted from Fig. 9.14, Callister & Rethwisch 8e.

160µmα + β100

α: 18.3 wt%Snβ: 97.8 wt% Sn

27Adapted from Fig. 9.13, Callister & Rethwisch 8e.

C, wt% Sn20 60 80 1000 40

18.3 97.8CE61.9

Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister &

28

p g ,Rethwisch 8e.

Microstructural Developments in Eutectic Systems IV

• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% SnResult: h ti l d t ti i tit t

y

• Result: α phase particles and a eutectic microconstituent

Cα = 18.3 wt% Sn• Just above TE :T(ºC) L: C0 wt% Sn Lα

L

W (1 W ) 0 50

CL = 61.9 wt% SnS

R + SWα = = 0.50Pb-Snsystem

300 L

αL+ α

L

α

SR

SR

WL = (1- Wα) = 0.50

• Just below TE :C = 18 3 wt% Sn

L+β200

100

α β

β

TE

primary αeutectic α

eutectic β

Cα = 18.3 wt% SnCβ = 97.8 wt% Sn

SR + S

Wα = = 0.7320 60 80 1000

100

40

α + β

2918.3 61.9 97.8

R + SWβ = 0.27

Adapted from Fig. 9.16, Callister & Rethwisch 8e. C, wt% Sn

20 60 80 1000 40

Checking your learning

Question AnswerIn a solution the solvent is the element present in the smallest amount

True / Falsepresent in the smallest amountDoes equilibrium occurs when free energy is at a minimum or a maximum?Th 3 i d d t i blThere are 3 independent variables which may be used in a phase diagram. What are they?If we know the composition andIf we know the composition and temperature of a binary system, and can assume equilibrium, what are the 3 kinds of information available?Define a eutectic reactionA solidus line separates which phase fields?

30

Below is shown the lead-tin phase diagram

1. Using this diagram determine what are the h / hphases/phase

combinations for an alloy of composition 46 wt% Sn - 54 wt% Pb that is at equilibrium at 44°C?

2. For an alloy of composition 25 wt% Sn - 75 wt% Pb, calculate the phase(s) andcalculate the phase(s) and their composition(s) at 200°C

For two phase regions use the Lever rule1. Construct tie-line2. Note intersections of tie-

line and phase boundaries

3. Drop perpendiculars to get composition of the phases present (C andphases present (CL and Cα)

4 Use the lever rule to4. Use the lever rule to find the proportion of each phase CCS −each phase

LL CC

CCSR

SW−−

=+

=α

α 0

For point B