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ISSUES TO ADDRESS...• When we combine two elements...

• what equilibrium state do we get?• 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 muchof each phase do we get?

CHAPTER 9: PHASE DIAGRAMS

Chapter 9- 1

• Solubility Limit:Max concentration for which only a solution occurs.

• Ex: Phase Diagram: Water-Sugar System

Question: What is thesolubility limit at 20C?

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

• Solubility limit increases with T:e.g., if T = 100C, solubility limit = 80wt% sugar.

Pu

re

Su

ga

r

Tem

pe

ratu

re(°

C)

0Co=Composition (wt% sugar)

20 40 6065 80 100

20

40

60

80

100

Pu

re

Wa

ter

THE SOLUBILITY LIMITحد حالليت

SolubilityLimit L

(liquid)

L +(liquid solution S

i.e., syrup) (solid

sugar)

Chapter 9- 2

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• Components:The elements or compounds which are mixed initially

(e.g., Al and Cu)• Phases:

The physically and chemically distinct material regions that is produced in the composition (e.g., and ).

Aluminum-CopperAlloy

(darker phase)

(lighter phase)

COMPONENTS AND PHASES

Chapter 9- 3

• Changing T can change number of phases: path A to B.• Changing Co can change number of phases: path B to D.

• water-sugarsystem

20 40 60 70 80 100Co=Composition (wt% sugar)

00

Tem

pe

ratu

re(°

C)

B(100,70)1 phase

40

20

100

D(100,90)2 phases

60

80

EFFECT OF TEMPERATURE & COMPOSITION

L(liquid)

+L S

(liquid solution (solid i.e., syrup) sugar)

A(70,20)2 phases

Chapter 9- 4

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• Phase diagram for one-component system (H2O)

EFFECT OF PRESSURE & TEMPERATURE

Chapter 9- 5

• For binary systems: just 2 components.--independent variables: T and C0 (P = 1atm is always used).

•PhaseDiagramfor Cu-Nisystem

• 2 phases:L (liquid)

(FCC solid solution)

• 3 phase fields: LL +

wt% Ni

(FCC solid solution)

0 20 40 60 80 1001000

1100

1200

1300

1400

1500

1600

L (liquid)

PHASE DIAGRAMS

T(°C)

Chapter 9- 6

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• Rule 1: If we know T and C0, then we know:--the number and types of phases present.

wt% Ni0 20 40 60 80 1001000

1100

1200

1300(FCC solidsolution)

A(1100,60)B

(12

50,3

5)

• Examples:A(1100, 60):

1 phase:

B(1250, 35):2 phases: L +

1400

1500

1600T(°C)

L (liquid)

Cu-Ni phase

diagram

NUMBER AND TYPES OF PHASES

Chapter 9- 7

C wt% Ni20

1200TD

L (liquid)

(solid)

303235CLC0

4043 50

D

TBB

• Examples: At TA:

C0 = 35wt%NiCL = C0 ( = 35wt% Ni)

Only Liquid (L)

At TD:C = C0 ( = 35wt% Ni)

Only Solid ()

At TB:CL = Cliquidus ( = 32wt% Ni here)C = Csolidus ( = 43wt% Ni here)Both and L

TA1300

T(°C)A

tie line

PHASE DIAGRAMS: composition of phases

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

Chapter 9- 8

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Lever Ruleقانون اهرم

L

L

LL

CC

CC

SR

RW

CC

CC

SR

SW

0

0

PHASE DIAGRAMS: weight fractions of phases

• Rule 3: If we know T and Co, then we know:--the amount of each phase (given in wt%).

Chapter 9- 9

Cu-Nisystem

• Examples:C0 = 35wt%Ni

At TA: Only Liquid (L)

WL = 100wt%, W = 0At TD: Only Solid ()

WL = 0, W = 100wt%At TB: Both and L

W R

R S

WL R S

43 32

73wt %S 43 35

= 27wt%

wt% Ni20

1200TD

TA1300 L (liquid)

T(°C)

(solid)

303235CLC0 C

4043 50

A

D

TBB

tie line

R S

PHASE DIAGRAMS: weight fractions of phases

Chapter 9- 10

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•Sum of weight fractions: WL W 1

• Conservation of mass (Ni): C0 WLCL WC• Combine above equations:

C0 CL RW

R SC CL S

RSWL C C0

C CL

• A geometric interpretation:C0CL C moment equilibrium:

1W

solving gives Lever Rule

WLR WS

THE LEVER RULE: A PROOF

R S

WL W

Chapter 9- 11

--isomorphousi.e., completesolubility of one component inanother; phasefield extends from0 to 100wt% Ni.

• ConsiderC0 = 35wt%Ni.

50

wt% Ni

110020 30

1200

1300

40

T(°C) L (liquid)

(solid)

A

D

B C

35C0

L: 35wt%Ni

E

L: 35wt%Ni

464332

24

35

36 L: 32wt%Ni

: 43wt%Ni

L: 24wt%Ni

: 36wt%Ni

Cu-Nisystem

EX: COOLING IN A Cu-Ni SYSTEM

Chapter 9- 12

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Fast coolingSlow coolingChapter 9- 13

• C changes as we solidify.• Cu-Ni case:

• Fast rate of cooling:Cored structure

• Slow rate ofcooling:

Equilibrium structure

Uniform C:

35wt%Ni

First to solidify has C = 46wt%Ni.Last to solidify has C = 36wt%Ni.

First to solidfy: 46wt%Ni

Last to solidfy:< 35wt%Ni

CORED VS EQUILIBRIUM PHASES

Chapter 9- 14

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• Effect of solid solution strengthening on:

--Ultimate Tensile strength (UTS) --Ductility (%EL,%AR)

--Peak as a function of C0 --Min. as a function of C0

MECHANICAL PROPERTIES: Cu-Ni System

Elo

ng

ati

on

(%E

L)

Composition, wt%NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

Ten

sile

Str

en

gth

(MP

a)

Composition, wt%NiCu Ni

TS for pure Cu

0 20 40 60 80 100200

300

400

TS for pure Ni

Chapter 9- 15

2 componentshas a special compositionwith a min. melting T.

• TE:No liquid below TE• CE:Min. melting

temperature composition

T(°C)Ex.: Cu-Ag system• 3 single phase regions

(L, )• Limited solubility:

: mostly Cu: mostly Ni

Composition, wt% Ag

L (liquid)

L + 779°C

800 L+

2000 20 40 60 CE80 100

1200

400

600

1000

TE 8.0 71.9 91.2

Cu-Agsystem

BINARY-EUTECTIC SYSTEMS

21 PhaseSolidPhaseSolidLiquid

L

Chapter 9- 16

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• For a 40wt%Sn-60wt%Pb alloy at 150C, find...--the phases present:

+ --the compositions

of the phases:

Pb-Snsystem

EX: Pb-Sn EUTECTIC SYSTEM (1)

L + L+

200

150

100

T(°C)

18.3

60 80 100

Comp., wt% Sn0 20 40

C0

300L (liquid)

183°C61.9 97.8

Chapter 9- 17

• For a 40wt%Sn-60wt%Pb alloy at 150C, find...--the phases present: + --the compositions

of the phases:C = 11wt%SnC = 99wt%Sn

--the relative amounts of each phase:

W 59 67wt%8829

W 88

33wt%

Pb-Snsystem

EX: Pb-Sn EUTECTIC SYSTEM (2)

L + L+

200

T(°C)

18.3

Comp., wt% Sn0 11 20 40 60 80 99100

C0

300

150

100

L (liquid)

183°C61.9 97.8

R S

Chapter 9- 18

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L + 200

T(°C)

20 30Comp., wt% Sn

10

2

0C0

300

100

L

: C0wt%Sn

L: C0wt%Sn

L

+

400

(room T solubility limit)

TE Pb-SnSystem

• C0 < 2wt%Sn• Result:

--polycrystal of grains.

MICROSTRUCTURES IN EUTECTIC SYSTEMS-I

Chapter 9- 19

• 2wt%Sn < C0 < 18.3wt%Sn• Result:

-- polycrystal with fine crystals.

: C0wt%SnL +

200

T(°C)

Comp., wt% Sn0 10

18.3

20 30C0

300

100

L

L: C0wt%Sn

+

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

L

Pb-Snsystem

MICROSTRUCTURES IN EUTECTIC SYSTEMS-II

TE Eutectic Temperature

Chapter 9- 20

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MICROSTRUCTURE 2wt%Sn < C0 < 18.3wt%Sn

فاز زمينه فاز •پخش ھمراه با رسوبات ذرات فاز •

شده در زمينه

Chapter 9- 21

L + 200

T(°C)

Comp., wt% Sn

0 20 40

300

100

L

60

L: Cowt%Sn

+

TE

: 18.3wt%Sn

L +

080 100

CE 61.9

18.3 97.8

183°C

: 97.8wt%Sn160m

Micrograph of Pb-Sn eutectic microstructure

• C0 = CE• Result: Eutectic microstructure

--alternating layers of and crystals.

Pb-Snsystem

MICROSTRUCTURES IN EUTECTIC SYSTEMS-III

Chapter 9- 22

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MICROSTRUCTURES IN EUTECTIC SYSTEMS-III

Directions of diffusion of tin and leadatoms are indicated by blue and redarrows, respectively.

• Just below TE: C = 18.3wt%SnC = 97.8wt%Sn

• C0= 61.9wt%Sn

= SR + S

W =45.15wt%

W = 54.85wt%

Chapter 9- 23

L + 200

T(°C)

Comp., wt% Sn

0 20 40

300

100

L

60

L: C0wt%Sn

+

TE

080 100

L +

C018.3 61.9 97.8

L

L

primary

S

S

RR

eutectic

Pb-Snsystem

• 18.3wt%Sn < C0 < 61.9wt%Sn• Result: crystals and a eutectic microstructure

WL = (1-W) =50wt%

• Just above TE:C = 18.3wt%Sn

CL = 61.9wt%Sn

W =R + S =50wt%S

• Just below TE: C = 18.3wt%SnC = 97.8wt%Sn

= SR + S

W =72.7wt%

W = 27.3wt%

MICROSTRUCTURES IN EUTECTIC SYSTEMS-IV

eutectic

Chapter 9- 24

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T(°C)

(Pb-Sn System)

L + 200

tic100 Co, wt% Sn0 20 40 60

300

100

L

+

TE

080

L +

18.3

hypoeutectic: Co=50wt%Sn 61.997.8

hypereutectic: (illustration only)

Co ectic hypereutechypoeutCo

eutectic

160m

eutectic micro-constituent

eutectic: Co=61.9wt%Sn

175m

HYPOEUTECTIC & HYPEREUTECTIC

Chapter 9- 25

Phase Transformations استحاله فازي

21 PhaseSolidPhaseSolidLiquid

L

EUTECTIC

321 PhaseSolidPhaseSolidPhaseSolid

EUTECTUID

21 PhaseSolidPhaseSolidLiquid

L

PERITECTIC

PERITECTUID321 PhaseSolidPhaseSolidPhaseSolid

Chapter 9- 26

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Cu-Zn PHASE DIAGRAM

Chapter 9- 27

IRON-CARBON (Fe-C) PHASE DIAGRAM

5/12/2017

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120mResult: Pearlite =alternating layers of and Fe3C phases.

• 2 important points-Eutectic (A):

L Fe3C-Eutectoid (B):

Fe3C

Fe3

C(c

em

en

tite

)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

(Fe) 0.77 Comp., wt% C4.30

727°C = Teutectoid

1148°C

T(°C)

A

B

SR

R S

Fe3C (cementite-hard) (ferrite-soft)

Ce

ute

cto

id

IRON-CARBON (Fe-C) PHASE DIAGRAM

Chapter 9- 29

Microstructure of a eutectoidsteel (pearlite microstructure)consisting of alternating layersof -ferrite (the light phase)and Fe3C (thin layers most ofwhich appear dark).

Chapter 9- 30

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PEARLITE PHASE

Chapter 9- 31

w =S/(R+S)wFe3C =(1-w)

HYPOEUTECTOID STEEL

(Fe-CSystem)

0 Co

Fe3C

(ce

me

nti

te)

800

600

4001 2 3 4 5

1600

1400

1200

1000

6 6.7Co, wt% C

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

0.7

7

727°C

1148°C

T(°C)

R S

w =s/(r+s)w =(1-w)

r s

pearlitewpearlite = w

100m Hypoeutectoid steel

Chapter 9- 32

5/12/2017

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Chapter 9- 33

HYPOEUTECTOID STEEL

0.38 wt% C steel Chapter 9- 34

5/12/2017

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(Fe-CSystem)

1 Co 2

Fe3

C(c

em

en

tite

)

1600

1400

1200

1000

800

wFe3C =r/(r+s)600

4000 3 4 5 6 6.7

Co, wt% C

L

austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

0.7

7

1148°C

T(°C)

R S

pearlitewpearlite = w

w =S/(R+S)wFe3C =(1-w)

60m Hypereutectoidsteel

s

w =(1-wFe3C)

r

Fe3C

HYPEREUTECTOID STEEL

Chapter 9- 35

HYPEREUTECTOIDSTEEL

Chapter 9- 36

5/12/2017

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HYPEREUTECTOIDSTEEL

Chapter 9- 37

TE

ute

cto

id(°

C)

0 4 8 12wt. % of alloying elements

Ti

Ni600

800

1000

1200 Mo SiW

Cr

Mn

wt. % of alloying elements

Ce

ute

cto

id(w

t%C

)

0 4 8 120

NiCr

Ti0.2

0.4

0.6

0.8

SiMnW

Mo

• Teutectoid changes: • Ceutectoid changes:

ALLOYING STEEL WITH MORE ELEMENTS

Chapter 9- 38

5/12/2017

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• Need a material to use in high temperature furnaces.• Consider Silica (SiO2) - Alumina (Al2O3) system.• Phase diagram shows:

mullite, alumina, and crystobalite (made up of SiO2)tetrahedra as candidate refractories.

25Composition (wt% alumina)

T(°C)

1400

1600

1800

2000

2200

20 40 60 80 1000

alumina +

mullite

mullite + L

mulliteLiquid

(L)

mullite + crystobalite

crystobalite + L

alumina + L

3Al2O3-2SiO2

APPLICATION:  REFRACTORIES

Chapter 9- 39

• Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phasefor 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

Chapter 9- 40