Introduction to Engineering Materials ENGR2000 Chapter 9…engineering.armstrong.edu/cameron/ENGR2000_phasedi… · · 2012-10-26Introduction to Engineering Materials ENGR2000 Chapter

Introduction to Engineering MaterialsENGR2000

Chapter 9: Phase Diagrams I

Dr. Coates

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Definitions and Basic Concepts• A component is a chemically distinct and

essentially indivisible substance.– Fe, Si, C are single component systems– H2O is a single component system– window glass (SiO2, Na2O & CaO) is a three-

component system




9.3 Phases

• A phase is a homogeneous portion of a systemthat has uniform physical and/or chemicalcharacteristics.– Liquid– Solid– Gaseous– Crystal structure– Pure metal




Phase and Component

• Pure FCC Al is a– single component, single phase system

• A mixture of pure ice and pure water is– a single component, two phase system

• A mixture of BCC Fe and FCC Fe is a– single component, two phase system.

• A solid solution of Cu and Ni is a– two component, single phase system

• A solid solution of NiO and MgO is a– two component, single phase system




9.2 Solubility Limit

• Maximum concentration of solute atoms that maydissolve in the solvent to form a solid solution– At a specific temperature– For a specific alloy system

• How would temp. affect solubility limit?• Why?

– Atoms in solvent are able to vibrate with greateramplitude and frequency, allowing more spaces per unittime for solute atoms to occupy




Sugar-water solution + solid sugar at 20 C in equilibrium




9.4 Microstructure (Review 4.10 polishing and etching)

• Microstructure– subject to direct observation using optical or electron

microscopes• Metal alloys characterized by

– # of phases present– Proportions of phases– Distribution or arrangement of phases

• Microstructure depends on– Alloying elements– Concentrations– Heat treatment




9.5 Phase Equilibrium

• Free energy = function (internal energy,randomness or disorder of the atoms)

• A system is at equilibrium if its free energy is at aminimum under some specified combination oftemperature, pressure and composition.

• Phase equilibrium is reflected by a constancy withtime in the phase characteristics of a system.(applies where more than one phase exists in thesystem)




Sugar-water solution + solid sugar at 20 C in equilibrium

If suddenly increase temp from 200 C to 800 C, what happens?




9.5 Phase Equilibria

• Free energy considerations provide informationabout the equilibrium characteristics of aparticular system– No indication of time periods needed for attainment of new

equilibrium state

• Often equilibrium never achieved because rate ofapproach to equilibrium is extremely slow-Systemstate is said to be in non-equilibrium or metastable




9.8 Binary Isomorphous Systems

• Binary– 2 components

• Isomorphous– Complete liquid and solid solubility

Phase diagrams are maps that present the relationship between temperature and composition and the quantities of phases at equilibrium

•presented here are for a constant pressure of 1 atm




Substitutional solid solutions

• Cu & Ni solution - Isomorphous– Atomic radii for Cu & Ni are 0.128 nm & 0.125 nm.– Both have FCC crystal structure– EN values are 1.9 & 1.8– Valences are +1 & +2.




• The liquid L is a homogenous liquid solutioncomposed of both copper and nickel

• The a phase is a substitutional solid solutionconsisting of both Cu and Ni atoms, having anFCC crsytal structure.

• At temp. below 10800 C, Cu and Ni are mutuallysoluble in each other in the solid state for allcompositions. Why? (See section 4.3)




pure

com

pone

ntC

u

pure

com

pone

ntN

i

MeltingTemperature of pure Cu




9.8 Interpretation of Phase DiagramsPhases Present

60 wt % Ni – 40 wt % Cu alloyat 1100 C





Determination of Phase Compositions


NiwtC .%60a

The composition at (11000 C) is given by

SINGLE PHASE!




• Within a two phase system, how is thecomposition of each phase calculated?

• Ex. Suppose we wanted to know what % nickelwas present in the a portion within the a +L phase

NiwtCNiwtC

L .%5.31.%5.42

a

Note-these are the equilibrium concentrations of the two phases




Determination of Phase Amounts


%100aW

What (mass fraction) percentage of the a phase is present

1. in the a region2. in the L region3. in the a+L region?

%0aW




Determination of Phase Amounts- Lever rule


1 :Note

68.0

32.0

.%35.%5.31.%5.42

0

0

0

L

LL

L

L

L

WWCCCCW

CCCCW

NiwtCNiwtCNiwtC

a

a

a

aa

a




Reading Assignments

• Section 9.8– Tie line– Lever rule– Example 9.1– Volume fractions & mass fractions




9.9 Development of Microstructure in Isomorphous Alloys

• Equilibrium cooling– Cooling occurs very slowly– Phase equilibrium is continuously maintained







9.8 Development of Microstructure in Isomorphous Alloys

• Non-equilibrium cooling– Rapid cooling rates (in most practical situations)– Sufficient time must be allowed at each temperature for

compositional readjustment, otherwise– Equilibrium is not always maintained




Because diffusion in the solid a phase is relatively slow!




At a give temp in the two phase region, will more or less liquid be present for nonequilibrium cooling vs equilibrium cooling?

How does Ni content compare along solidus line for nonequilibrium vs equilibrium cooling?

How much solidus deviation would you expect with increasing cooling rate?




• Distribution of grains will be non-uniform(segregation)

• Center of each grain is rich in high meltingelement/concentration, the concentration of lowmelting element increases from this regiontowards grain boundary-coring

• CORING will affect temperature response/suddenloss of mechanical integrity as structure isreheated




9.10 Mechanical Properties of Isomorphous Alloys

• Each component experiences solid-solution strengthening– At temp below lowest-melting component




•Three phases, a, b, liquid

•a -Solid solution rich in Cu, silver as solute FCC crystal structure

•b-Solid solution, FCC, Cu is solute

•L-liquid•Three two phase regions

•a+b•a+L•a+b

•Tie lines and lever rule still apply

9.11 Binary Eutectic Systems




9.11 Binary Eutectic Systems

Invariant point

Melting temperaturePure copper

Max solubility Ag ina phase

Max solubility Cu inb phase




• Which line represents lowest temperature at whicha liquid phase may exist for any Cu-Ag alloy atequilibrium?

• What happens to temp at which alloys becometotally liquid (melting temp)as Ag is added to Cu?Vice versa?

• What is special about point E?

BEG

AE EF

Eutectic Reaction-upon cooling, liquid is transformed to two solid aand b phases




Eutectic (easily melted) point

re temperatueutectic - ncompositio eutectic -

:n compositio ofalloy an for reaction Eutectic

E

E

EEE

E

TC

CCCLC

ba ba

L(71.9 wt% Ag)

a(8.0wt% Ag)+b(91.2 wt% Ag)




60-40 Solder (60 wt. % Sn – 40 wt. % Pb)

• Alloy iscompletelymolten at~185 C

• Easily melted• Low-temperature

solder




Example 9.2

• For a 40 wt % Sn – 60 wt % Pb alloy at 150 C– What phases are present?– What are the compositions of the phases?




SnwtCSnwtC

%98%11

b

a

ba :nscompositio Phase

& :Bpoint at Phases




Example 9.3

• For a 40 wt % Sn – 60 wt % Pb alloy at 1500 C,calculate the relative amount of each phase presentin terms of– mass fraction and– volume fraction.At 1500 C take the densities of Pb and Sn to be

11.23g/cm3 and 7.24 g/cm3 respectively.




33.0

67.0

.%40%98%11

1

1

1

ab

ab

ab

ba

b

a

CCCCW

CCCC

W

SnwtCSnwtCSnwtC

:fractions) (mass amounts Phase

:nscompositio Phase




3

3

3

3

/29.7

23.112

24.798

100100

/59.10

23.1189

24.711

100100

/24.7/23.11

cmg

PbC

SnC

cmg

PbC

SnC

cmgSncmgPb

PbSn

PbSn

bbb

aaa

:phase each of Density

:densities Given




43.0

58.0

b

b

a

a

b

b

b

b

b

a

a

a

a

a

WW

W

V

WW

W

V

:fractions) (volume amounts Phase




9.11 Development of Microstructure in Eutectic Alloys

• Eutectic alloys– Eutectic composition

• Hypoeutectic alloys– Alloy composition < eutectic composition

• Hypereutectic alloys– Eutectic composition < alloy composition

Several different types of microstructures are possible for the slow cooling of eutectic alloys!




SnwtCSnwt

C

%2%0 1

1

:system Pb-Sn theFor

C) 20(~ etemperatur roomat component of solubility solidmax

component pure

20




SnwtCSnwt

C

%3.18%2:system Pb-Sn For the

re temperatueutecticat component of solubilitymax

re temperaturoomat component of solubilitymax

2

2




SnwtSnwtSnwtL

SnwtC

C

.%8.97.%3.18.%9.61:reaction Eutectic

%9.61:system Pb-Sn For the

ncompositio eutectic

3

3

ba




Eutectic structure

• Microstructure of solid that results from theeutectic reaction

• Alternating layers (lamellae) of a and b phasesthat form simultaneously







SnwtCSnwt

C

%9.61%3.18:system Pb-Sn For the

isotherm. eutectic thecross cooled, when that,eutectic the

other than nscompositio all

4

4







1

3.189.619.61

3.189.613.18

'4

'4

'

PeutecticL

eutecticLe

WW

:Check

CQP

QW

:primary the of fraction) (mass amount Phase

CQP

PWW

:ituentmicroconst eutectic the of fraction) (mass amount Phase

a

a

a

The eutectic microconstituent always forms from the liquid having the eutectic composition

We, Wa’, are mass fractions of eutectic

microconstituent and primary a




1

3.188.973.18

3.188.978.97

'4

'4

totaltotal

total

total

WW :Note

CRPQ

PW

CRQP

RQW

: and total the of fraction) (mass amounts Phase

ba

b

a

ba




Intermediate Phases or Compounds

• Terminal Solid Solution– The two solid phases exist over composition ranges

near the concentration extremes• Intermediate Solid Solution

– Intermediate phases found at other than the twocomposition extremes

• Intermetallic Compound– Discrete intermediate compounds exist (rather than

solid solutions) in metal-metal solutions




9.12 Equilibrium Diagrams having Intermediate Phases or Compounds

Intermediate solid solutions

Terminal solid solution(commercial brass)




51

Intermetallic Compounds

Mg2Pb

Note: intermetallic compound exists as a line on the diagram - not an area -because of stoichiometry (i.e. composition of a compound is a fixed value).

Adapted from Fig. 9.20, Callister & Rethwisch 8e.

Mg-Mg2Pb eutectic system

Mg2Pb-Pb eutectic system




• Mg2Pb (magnesium plumbide) exists by itselfonly at the precise composition (81 wt% Pb)

• What is the melting temperature of Mg2Pb?

• How does the solubility of magnesium in leadcompare to the solubility of lead in magnesium?

• Note that the diagram could be separated into twoeutectic diagrams at Mg2Pb line




53

• Eutectoid – one solid phase transforms to two other solid phasesS2 S1+S3

a + Fe3C (For Fe-C, 727ºC, 0.76 wt% C)

intermetallic compound -cementite

coolheat

Eutectic, Eutectoid, & Peritectic• Eutectic - liquid transforms to two solid phases

L a + b (For Pb-Sn, 183ºC, 61.9 wt% Sn)coolheat

coolheat

• Peritectic - liquid and one solid phase transform to a second solid phaseS1 + L S2

+ L (For Fe-C, 1493ºC, 0.16 wt% C)




54

Eutectoid & Peritectic

Cu-Zn Phase diagram


Eutectoid transformation +

Peritectic transformation + L

A eutectoid reaction is found in the iron-carbon system, very important in heat treating of steels!




9.15 Congruent Phase Transformations

• Transformations with no change in composition• Melting of pure materials• Allotropic transformations

• Incongruent– At least one of the phases will experience a change in

composition• Eutectic, eutectoid• Melting of an alloy that belongs to an isomorphous system




Congruent melting for solid solution of 44.9 wt.% Ti

Where is peritectic reaction?

Congruent/incongruent?

Portion of Ni-Ti Phase Diagram




57

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

points

- Eutectoid (B): a +Fe3C

- Eutectic (A):L + Fe3C

Adapted from Fig. 9.24,Callister & Rethwisch 8e.

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a+Fe3C

(Fe) C, wt% C

1148ºC

T(ºC)

a 727ºC = T eutectoid

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

120 mm

(Adapted from Fig. 9.27, Callister & Rethwisch 8e.)

0.76

B

A L+Fe3C

Fe3C (cementite-hard)a (ferrite-soft)




58Fe

3C (c

emen

tite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

a+ Fe3C

L+Fe3C

(Fe) C, wt% C

1148ºC

T(ºC)

a727ºC

(Fe-C System)

C0

0.76

Hypoeutectoid Steel

Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (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.)

Adapted from Fig. 9.30, Callister & Rethwisch 8e. proeutectoid ferritepearlite

100 mm Hypoeutectoidsteel

a

pearlite

a

aa




59Fe

3C (c

emen

tite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

a+ Fe3C

L+Fe3C

(Fe) C, wt% C

1148ºC

T(ºC)

a727ºC

(Fe-C System)

C0

0.76

Hypoeutectoid Steel

a

aa

srWa’ = s/(r + s)W =(1 – Wa’)

R Ta

pearlite

Wpearlite = W

Watotal = T/(R + T)W =(1 – Watotal)Fe3C


Adapted from Fig. 9.30, Callister & Rethwisch 8e. proeutectoid ferritepearlite

100 mm Hypoeutectoidsteel

The eutectoid microconstituent pearlite will form from the solid having the eutectic composition




60

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

a +Fe3C

L+Fe3C

(Fe) C, wt%C

1148ºC

T(ºC)

a


(Fe-C System)

0.76

C0

Fe3C


proeutectoid Fe3C

60 mm Hypereutectoid steel

pearlite

pearlite




61Fe

3C (c

emen

tite)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

a +Fe3C

L+Fe3C

(Fe) C, wt%C

1148ºC

T(ºC)

a

Hypereutectoid Steel

(Fe-C System)

0.76

C0

pearlite

Fe3C

xv

V X

Wpearlite = W

Watotal = X/(V + X)

W =(1 - Watotal)Fe3C’

W =(1-W)W =x/(v + x)

Fe3C


proeutectoid Fe3C

60 mm Hypereutectoid steel

pearlite





62

Example Problem

For a 99.6 wt% Fe-0.40 wt% C steel at a temperaturejust below the eutectoid, determine the following:

a) The compositions of Fe3C and ferrite (a).b) The amount of cementite (in grams) that forms in

100 g of steel.c) The amounts of pearlite and proeutectoid ferrite

(a) in the 100 g.




63

Solution to Example Problem

WFe3C R

R S

C0 Ca

CFe3C Ca

0.40 0.0226.70 0.022

0.057

b) Using the lever rule with the tie line shown

a) Using the RS tie line just below the eutectoid

Ca = 0.022 wt% CCFe3C = 6.70 wt% C

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

a+ Fe3C

L+Fe3C

C, wt% C

1148ºC

T(ºC)

727ºC

C0

R S

CFe C3C

Amount of Fe3C in 100 g = (100 g)WFe3C

= (100 g)(0.057) = 5.7 g




64

Solution to Example Problem (cont.)c) Using the VX tie line just above the eutectoid and realizing

thatC0 = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

(austenite)

+L

+ Fe3C

a+ Fe3C

L+Fe3C

C, wt% C

1148ºC

T(ºC)

727ºC

C0

V X

CC

Wpearlite V

V X

C0 Ca

C Ca

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




65

Alloying with Other Elements• Teutectoid changes:

Adapted from Fig. 9.34,Callister & Rethwisch 8e. (Fig. 9.34 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

T Eut

ecto

id(º

C)

wt. % of alloying elements

Ti

Ni

Mo SiW

Cr

Mn

• Ceutectoid changes:

Adapted from Fig. 9.35,Callister & Rethwisch 8e. (Fig. 9.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

wt. % of alloying elementsC

eute

ctoi

d(w

t% C

)

Ni

Ti

Cr

SiMnWMo




66

• Phase diagrams are useful tools to determine:-- the number and types of phases present,-- the composition of each phase,-- and the weight fraction of each phase given the temperature and composition of the system.

• The microstructure of an alloy depends on -- its composition, and -- whether or not cooling rate allows for maintenance of

equilibrium.• Important phase diagram phase transformations include

eutectic, eutectoid, and peritectic.

Summary




67

9.11, 9.13, 9.17, 9.20, 9.32

Core Problems:

Self-help Problems:

ANNOUNCEMENTSReading:

9.1 thru 9.20, will not be tested on 9.16, 9.17




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Introduction to Engineering Materials ENGR2000 Chapter 9…engineering.armstrong.edu/cameron/ENGR2000_phasedi… · · 2012-10-26Introduction to Engineering Materials ENGR2000 Chapter