Ch09 Modified

8/2/2019 Ch09 Modified

1/36

Chapter 9 - 11

ISSUES TOADDRESS... When we combine two elements...

what equilibrium state do weget? In particular, if we specify...

--a composition (e.g., wt% Cu - wt% Ni),and

--a temperature (T)then...How many phases do we get?What is the composition of each

phase?How much of each phase do we get?

Chapter 9: Phase Diagrams

Phase BPhase A

Nickel

atomCopperatom


2/36

Chapter 9 - 22

Phase Equilibria: Solubility LimitIntroduction

Solutions solid solutions, single phase Mixtures more than one phase

Solubility Limit:

Max concentration forwhich only a single phase

solution occurs.

Question: What is the

solubility limit at 20C?

Answer: 65 wt% sugar.

If Co < 65 wt% sugar: syrupIf Co > 65 wt% sugar: syrup + sugar.

65

Sucrose/Water PhaseDiagram

P

u

r

e

S

u

g

ar

Temperature

(C

)

0 20

40

60

80

10

0

C

o

=Composition (wt%

sugar)

L(liquid

solutioni.e.,syrup)

SolubilityLimit

L(liqu

id)+S(sol

idsugar)

20

40

60

80

10 0

P

u

r

e

Wa

te

r

Adapted from Fig. 9.1,Callister 7e.


3/36

Chapter 9 - 33

Components:

The elements or compounds which are present in the mixture (e.g., Al and Cu) Phases:

The physically and chemically distinct material regionsthat result (e.g., and ).

Aluminum-Copper

Alloy

Components and Phases

(darker

phase)

(lighter

phase)

Adapted fromchapter-openingphotograph,Chapter 9,Callister 3e.


4/36

Chapter 9 - 44

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

Adapted from Fig.9.1,Callister 7e.

D(100C,90)2 phases

B(100C,70)

1 phase

pathA to B.

Changing Co can change # of phases: path B to D.

A(20C,70)2 phases

70

80

100

60

40

20

0

Temperature

(C

)

Co

=Composition (wt%sugar)

L( liquid

solutioni.e.,

syrup)20

10

0

4

0

60

80

0

L(liquid)+

S(sol

idsugar)

water-sugarsystem


5/36

Chapter 9 - 55

Phase Equilibria

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246Cu FCC 1.8 0.1278

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

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

Ni and Cu are totally miscible in all proportions.


6/36

Chapter 9 - 66

Phase Diagrams

Indicate phases as function of T, Co, and P. For this course:

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

2 phases:

L

(liquid) (FCC solidsolution) 3 phase fields:

L

L+

wt%20

40

60

80

10

01000

1100

1200

1300

1400

1500

16

00

T(

C)L(liquid)

(FCC solid

solution

)

L+

liq

uid

ussolid

us


7/36

Chapter 9 - 77

wt%

Ni

20

40

60

80

100

01000

1100

1200

1300

1400

1500

1600

T(C)

L

(liquid)

(FCC

solidsolution)

L+

liquid

ussolid

us

Cu-Niphasediagram

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(1100C,

60):1 phase: B(1250C,

35):2 phases: L +

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

B

( 1 2 5 0 C , 3 5 )

A(1100C,60)


8/36

Chapter 9 - 88

wt%Ni

20

1200

13

00

T(C)

L

(liquid)

(solid)

L +

liquid

us

solid

us

30

40

50

L+

Cu-Nisystem

Phase Diagrams:composition of phases

Rule 2: If we know T and Co

, then we know:--the composition of each phase.

Examples:

TAA

35Co

32CL

At

T

A =

1320C:Only Liquid(L)

CL =

Co ( = 35 wt%

Ni)

AtT

B = 1250C:

Both and

LCL =C

liquidus

( = 32 wt% Ni here)

C =

C

solidus

( = 43 wt% Ni

here)

AtT

D =1190C: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 Diagramsof Binary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH, 1991.)

BTB

DTD

tieline

4C3


9/36

Chapter 9 - 99

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

=0At

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 ofBinary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH, 1991.)

Phase Diagrams:weight fractions of phases

wt%Ni

20

1200

13

00

T(C)

L

(liquid)

(solid)

L +

liquid

us

solid

us

30 40 50

L +

Cu-Nisystem

TAA

35Co

32CL

BTB

DTD

tieline

4C3

R S

AtT

B:Both

andL

%773377

3377 wt=

=

= 27 wt%

WL

= SR+ S

W

=R

R+ S


10/36

Chapter 9 - 1010

Tie line connects the phases in equilibrium with eachother - essentially an isotherm

The Lever Rule

How much of each phase?Think of it as a lever (teeter-totter)

ML

M

R S

RMSM L =

L

L

LL

L

LCC

CC

SR

RW

CC

CC

SR

S

MM

MW

=

+

=

=

+

=

+

=

77

wt%Ni

20

1200

1300

T(C)

L

(liquid)

(solid)

L +

liquid

us

solid

us

30 40 50

L

+

BTB

tieline

CoCL C

SR

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


11/36

Chapter 9 - 1111

wt%

Ni

20

120

0

130

0

30 40 50110

0

L(liquid)

(solid)

L+

L +

T(

C)

A

35Co

L: 35wt

%Ni Cu-Nisystem

Phase diagram:

Cu-Ni system.

System is:

--binaryi.e., 2 components:Cu and Ni.

--isomorphousi.e., completesolubility of onecomponent in

another; phase

field extends from0 to 100 wt% Ni.

Adapted from Fig. 9.4,

Callister 7e.

Consider

Co = 35 wt%Ni.

Ex: Cooling in a Cu-Ni Binary

46

35 4

3

3

2

: 43 wt%Ni

L: 32 wt%Ni

L: 24 wt%Ni: 36 wt%

Ni

B: 46 %

L: 35 wt%Ni

C

D

E

24

36


12/36

Chapter 9 - 1212

Nonequilibrium Cooling

In application not sufficienttime for diffusionResult is cored structureSolidus curve shifts to higherNi contents


13/36

Chapter 9 - 1313

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

tosolidify:46 wt%

Ni

UniformC

:

35 wt%Ni

Last

tosolidify:< 35 wt%

Ni


14/36

Chapter 9 - 1414

Mechanical Properties:Cu-Ni System

Effect of solid solution strengthening on:

--Tensile strength (TS) --Ductility (%EL,%AR)

--Peak as a function of Co --Min. as a function of Co

Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.

TensileStren

gth

(MPa)

Composition,wt% Ni

Cu

Ni

0 2

0

4

0

6

0

8

0

1

00

200

300

400 TS

forpure Ni

TS forpure Cu

Elongatio

n

(%EL)

Composition, wt% Ni

Cu

Ni

0 20

40

60

80

100

2

0

30

40

5

0

60

%

EL

for

pureNi

%EL

for pureCu


15/36

Chapter 9 - 1515

: Min. meltingTE

2components

has a special composition

with a min. melting T.


Binary-Eutectic Systems

Eutectic transition

L(CE) (C E) + (C E)

3 single phaseregions

(L,

,

)

Limitedsolubility:: mostly

Cu : mostlyAg

TE: No liquid belowTE

CEcomposition

Ex.: Cu-Ag system

Cu-Agsystem

L (liquid)

L + L+

+

Co

,

wt% Ag20

40

60

80

100

0200

1200

T(C)

4

00

600

800

10

00

CE

TE

8.0

71.9

91.2

779C


16/36

Chapter 9 - 1616

L+ L+

+

200

T(C)

18.3

C, wt%Sn

20

60

80

100

0

300

10

0

L

(liquid) 183C

61.9

97.8

For a 40 wt% Sn-60 wt% Pb alloy at 150C, find...

--the phases present: Pb-Snsystem

EX: Pb-Sn Eutectic System (1)

+ --compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

150

40Co

11C

99C

SR

C = 11 wt%SnC = 99 wt%

Sn

W=C -

COC -7

C

=99 -4099 -11

=5988

= 67 wt%

SR+S

=

W =CO -

CC

=R

R+S

=2

988

= 33 wt

%

=40 -

1199 -11



17/36

Chapter 9 - 1717

L+

+

200

T(C)

C, wt%Sn

20

60

80

100

0

300

10

0

L

(liquid)

L+

183C

For a 40 wt% Sn-60 wt% Pb alloy at 220C, find...

--the phases present: Pb-Snsystem


EX: Pb-Sn Eutectic System (2)

+ L--compositions of phases:

CO = 40 wt% Sn

--the relative amountof each phase:

W=CL -COCL -

C7

=46 -4046 -17

= 629

= 21 wt%

WL =CO -

C7CL -C

=2329

= 79 wt%

40Co

46CL

17C

7

220

SR

C = 17 wt%SnCL = 46 wt%

Sn


18/36

Chapter 9 - 1818

Co < 2 wt% Sn Result:

--at extreme ends--polycrystal of grains

i.e., only one solid phase.


Microstructuresin Eutectic Systems: I

0

L+ 200

T(C)

Co

,

wt%Sn

10

2

20C

o

300

1

00

L

30

+

400

(room T solubilitylimit)

TE

(Pb-SnSystem)

L

L: Co wt%Sn

: %


19/36

Chapter 9 - 1919

2 wt% Sn < Co < 18.3 wt% Sn Result:

Initially liquid + then alone7

finally two phases polycrystal fine -phase

inclusions


Microstructuresin Eutectic Systems: II

Pb-Snsystem

L +

200

T(C)

Co

,

wt%Sn

10

18

.3

20

0Co

3

00

100

L

30

+

400

(sol. limit atTE)

TE

2(sol. limit atT

roo

m

)

L

L: Co wt%Sn

: %


20/36

Chapter 9 - 2020

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


Microstructuresin Eutectic Systems: III

Adapted from Fig. 9.14, Callister 7e.

160

Micrograph of Pb-Sn

eutecticmicrostructure

Pb-Snsystem

L

+

+

200

T(C)

C, wt%Sn

20

60

80

10

0

0

300

100

L

L+

183C

40

TE

18

.3

: 18.3

%

97

.8

: 97.8 %

C

E61.9

L: Co wt%

Sn


21/36

Chapter 9 - 2121

Lamellar Eutectic Structure

Adapted from Figs. 9.14 & 9.15, Callister7e.


22/36

Chapter 9 - 2222

18.3 wt% Sn < Co < 61.9 wt% Sn Result: crystals and a eutectic microstructure

Microstructuresin 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 S

R+

SW

== 50 wt%

Just aboveTE :

Just below

TE :C = 18.3 wt%SnC = 97.8 wt%Sn S

R+

SW

=

= 73 wt%

W = 27 wt

%Adapted from Fig. 9.16,

Callister 7e.

Pb-Snsystem

L+200

T(C)

Co, wt%Sn

20

60

80

10

0

0

300

100

L

L+

40

+

TE

L:Co

wt%Sn

L

L


23/36

Chapter 9 - 2323

L+ L+

+

200

Co, wt%Sn

20

60

80

100

0

3

00

100

L

TE

40

(Pb-SnSystem)

Hypoeutectic & Hypereutectic

Adapted from Fig. 9.8,Callister 7e. (Fig. 9.8adapted from Binary PhaseDiagrams, 2nd ed., Vol. 3,T.B. Massalski (Editor-in-Chief), ASM International,Materials Park, OH, 1990.)

160

meutectic micro-constituentAdapted from Fig. 9.14,

Callister 7e.

hypereutectic: (illustrationonly)

Adapted from Fig. 9.17,Callister 7e. (Illustration only)

(Figs. 9.14 and 9.17from MetalsHandbook, 9th ed.,Vol. 9, Metallographyand Microstructures,

American Society forMetals, Materials Park,OH, 1985.)

175

hypoeutectic: Co = 50 wt%Sn

Adapted fromFig. 9.17, Callister 7e.

T(C)

61.9

eutectic

eutectic: Co=61.9wt%Sn


24/36

Chapter 9 - 2424

An Example of Complex Binary Alloy Systems

Liquid


25/36

Chapter 9 - 2525

Intermetallic Compounds

Mg2Pb

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



26/36

Chapter 9 -

2626

Eutectoid & Peritectic

Eutectic - liquid in equilibrium with two solidsL +cool

heat

intermetallic compound -cementitecool

heat

Eutectoid - solid phase in equation with two solid phases

S2 S1+S3

+ Fe3C (727C)

cool

heat

Peritectic - liquid + solid 1 solid 2 (Fig 9.21)S1 + L S2

+ L (1493C)


27/36

Chapter 9 -

2727

Eutectoid & Peritectic

Cu-Zn Phase diagram


Eutectoid transition +

Peritectic transition + L7 7


28/36

Chapter 9 -

2828

Iron-Carbon (Fe-C) Phase Diagram

2 important

points

-Eutectoid

(B): + Fe3C

-Eutectic(A):L + Fe3

C


Fe3C

(cementite)

160014001200100080060

0400

0 1 2 3 4 5 6 6.7

L

(austeni

te)

+L

+Fe3C

+Fe

3C

+

L+Fe3C

(Fe)

Co, wt% C

1148C

T(

C)

727C= T

eutectoid

A

SR

4.30

Result: Pearlite =

alternating layers of

and Fe3C phases

120

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

R S

0.76C

eut

ect

oid

B

Fe3C (cementite-hard)

(ferrite-soft)


29/36

Chapter 9 -

2929

Hypoeutectoid Steel

Adapted from Figs. 9.24

and 9.29,Callister 7e.(Fig. 9.24 adapted fromBinary Alloy PhaseDiagrams, 2nd ed., Vol. 1,T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)

Fe3C

(cementite)

1600

1400

1200

10

008006004

00

0 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

(Fe)

Co, wt%C

1148C

T(C)

727C

(Fe-CSystem)

C0 0 . 7 6


proeutectoid ferritepearlite

100

Hypoeutectoid ste

el

RS

w = S/(

R+ S)

wFe3C

= (1-

w)

w pearlite

=

wpearlite

r s

w =s /(r +s)

w = (1-

w)


30/36

Chapter 9 -

3030

Hypereutectoid Steel

Fe3C

(cementi

te)

1600140012001

0008006004

00

0 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3C

+Fe

3C

L+Fe3C

(Fe)

Co, wt%C

1148C

T(C)

Adapted from Figs. 9.24

and 9.32,Callister 7e.(Fig. 9.24 adapted fromBinary Alloy PhaseDiagrams, 2nd ed., Vol. 1,T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)

(Fe-CSystem)

0.

7 6

Co


proeutectoid Fe3C

60

Hypereutectoid ste

elpearlite

R S

w =S/(

R+S)

wFe3C

= (1-

w)

wpearlite

=

wpearlite

sr

wFe3C

=r /(r+s)

w =(1-

w Fe3

C

)

Fe3C


31/36

Chapter 9 -

3131

Example: Phase Equilibria

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

a) composition of Fe3C and ferrite ( )

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

c) the amount of pearlite and proeutectoidferrite ( )7


32/36

Chapter 9 -

3232

Chapter 9 Phase EquilibriaSolution:

g7.77

g.77CFe

g7.7777333.77.7

777.77.7

777xCFeCFe

7

CFe7

7

7

=

=

=

=

=

+

x

CCCCo

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

a) composition of Fe3C and ferrite ( )

CO = 0.40 wt% C

C = 0.022 %

= 6.70 %

3

Fe3C(cem

entite)

16

0014001200100

0800600400

0 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3

C

+Fe3C

L+Fe3C

Co, wt% C

1148C

T(C)

727C

CO

R S

CFeC

3C


33/36

Chapter 9 -

3333

Chapter 9 Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ( )

note: amount of pearlite = amount ofjust above TECo = 0.40 wt% C

C = 0.022 wt% CCpearlite = C = 0.76 wt% C

+=CoC

CC

x777=77.7g

pearlite = 51.2 g proeutectoid = 48.8 g

Fe3C(cem

entite)

16

0014001200100

0800600400

0 1 2 3 4 5 6 6.7

L

(austenite)

+L

+Fe3

C

+Fe3C

L+Fe3C

Co, wt% C

1148C

T(C)

727C

CO

RS

CC


34/36

Chapter 9 -

3434

Alloying Steel with More Elements

Teutectoidchanges:

Ceutectoidchanges:

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

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

TEute

cto

id

(C)

wt. % of alloying

elements

Ti

Ni

Mo

Si W

CrMn

wt. % of alloying

elements

Ceute

ctoi

d

(wt

%C)

Ni

Ti

Cr

Si M

nW

Mo


35/36

Chapter 9 -

3535

Phase diagrams are useful tools to determine:

--the number and types of phases,--the wt% of each phase,--and the composition of eachphasefor 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 fora range of microstructures.

Summary


36/36

Chapter 9

3636

CoreProblems:

Self-helpProblems:

ANNOUNCEMENTS

Reading:

Ch09 Modified

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