1

Phase equilibria: solubility limit

Introduction– Solutions – solid solutions, single phase– Mixtures – more than one phase

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

Question: What is the solubility limit at 20°C?

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

Sucrose/Water Phase Diagram

Pure

Su

gar

Tem

pera

ture

(°C

)

0 20 40 60 80 100Co =Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid) +

S

(solid sugar)20

40

60

80

10 0

Pure

W

ater

Adapted from Fig. 9.1,Callister 7e.

2

Components and phases

• Components: The elements or compounds which are present in the mixture (e.g., Al and Cu)• Phases: The physically and chemically distinct material regions that result (e.g., α and β).

Aluminum-CopperAlloy

α (darker

phase)

β (lighter

phase)

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

3

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)α (FCC solid solution)

• 3 phase fields: LL + αα

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α

(FCC solid

solution)

L + αliqu

idus

solidu

s

4

Phase diagrams

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

α

(FCC solid solution)

L + α

liquidu

s

solidu

s

Cu-Niphase

diagram

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

• Examples:A(1100°C, 60): 1 phase: α

B (1250°C, 35): 2 phases: L + α

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

250°

C,3

5) A(1100°C,60)

5

Phase diagrams

wt% Ni20

1200

1300

T(°C)

L (liquid)

α

(solid)L + α

liquidus

solidus

30 40 50

L + α

Cu-Nisystem

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

• Examples:T A

A

35C o

32C L

At TA = 1320°C:

Only Liquid (L) C L = C o ( = 35 wt% Ni)

At TB = 1250°C:

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

C α = C solidus ( = 43 wt% Ni here)

At TD = 1190°C:

Only Solid ( α ) C α = C o ( = 35 wt% Ni)

C o = 35 wt% Ni

Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH, 1991.)

BT B

DT D

tie line

4C α

3

6

Phase diagrams

• 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 α = 0At TD : Only Solid (α )

W L = 0, Wα = 100 wt%

C o = 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.)

wt% Ni20

1200

1300

T(°C)

L (liquid)

α

(solid)L + α

liquidus

solidus

30 40 50

L + α

Cu-Nisystem

T AA

35C o

32C L

BT B

DT D

tie line

4C α

3

R S

At TB : Both α and L

% 733243

3543wt=

!

!=

= 27 wt%

WL =S

R + S

Wα =R

R + S

7

The lever rule

Tie line – connects the phases in equilibrium with each other - essentially anisotherm

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

ML Mα

R S

RMSML!=!"

L

L

LL

L

LCC

CC

SR

RW

CC

CC

SR

S

MM

MW

!

!=

+=

!

!=

+=

+=

"

"

"

"

"

00

wt% Ni

20

1200

1300

T(°C)

L (liquid)

α

(solid)L + α

liquidus

solidus

30 40 50

L + α

BT B

tie line

CoCL Cα

SR

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

8

Cooling

wt% Ni20

1200

1300

30 40 501100

L (liquid)

α (solid)

L + α

L + α

T(°C)

A

35Co

L: 35wt%NiCu-Ni

system

• Phase diagram: Cu-Ni system.

• System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; α phase field extends from 0 to 100 wt% Ni.

Adapted from Fig. 9.4,Callister 7e.

• Consider Co = 35 wt%Ni.

46354332

α: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Niα: 36 wt% Ni

Bα: 46 wt% NiL: 35 wt% Ni

C

D

E

24 36

9

Binary eutectic systems

: Min. melting TE

2 componentshas a special compositionwith a min. melting T.

Adapted from Fig. 9.7,Callister 7e.

• Eutectic transitionL(CE) α(CαE) + β(CβE)

• 3 single phase regions (L, α, β )

• Limited solubility: α : mostly Cu

β : mostly Ag

• TE : No liquid below TE

• CE

composition

Ex.: Cu-Ag system

Cu-Agsystem

L (liquid)

α L + α L+β β

α + β

Co , wt% Ag20 40 60 80 1000

200

1200T(°C)

400

600

800

1000

CE

TE 8.0 71.9 91.2779°C

10

Binary eutectic systems

L+αL+β

α+ β

200

T(°C)

18.3

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α183°C

61.9 97.8β

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

Pb-Snsystem

α + β

--compositions of phases:

CO = 40 wt% Sn

--the relative amount of each phase:

150

40Co

11Cα

99Cβ

SR

Cα = 11 wt% Sn

Cβ = 99 wt% Sn

W α =Cβ - COCβ - Cα

=99 - 4099 - 11

=5988

= 67 wt%

SR+S

=

W β =CO - CαCβ - Cα

=RR+S

=29

88= 33 wt%=

40 - 1199 - 11

Adapted from Fig. 9.8,Callister 7e.

11

Binary eutectic systems

L+β

α+ β

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L (liquid)

α β

L+ α

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present:

Pb-Snsystem

Adapted from Fig. 9.8,Callister 7e.

α + L

--compositions of phases:

CO = 40 wt% Sn

--the relative amount of each phase:

W α =CL - COCL - Cα

=46 - 40

46 - 17

= 6

29= 21 wt%

W L =CO - CαCL - Cα

=23

29= 79 wt%

40Co

46CL

17Cα

220SR

Cα = 17 wt% SnCL = 46 wt% Sn

12

Microstructures in eutectic systems

• Co < 2 wt% Sn• Result: --at extreme ends --polycrystal of α grains i.e., only one solid phase.

Adapted from Fig. 9.11,Callister 7e.

0

L+ α200

T(°C)

Co, wt% Sn10

2

20Co

300

100

L

α

30

α+β

400

(room T solubility limit)

TE(Pb-SnSystem)

αL

L: Co wt% Sn

α: Co wt% Sn

13

Microstructures in eutectic systems

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

Initially liquid + α then α alone finally two phases

α polycrystal fine β-phase inclusions

Adapted from Fig. 9.12,Callister 7e.

Pb-Snsystem

L + α

200

T(°C)

Co , wt% Sn10

18.3

200Co

300

100

L

α

30

α+ β

400

(sol. limit at TE)

TE

2(sol. limit at Troom)

Lα

L: Co wt% Sn

αβ

α: Co wt% Sn

14

Microstructures in eutectic systems

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

Adapted from Fig. 9.13,Callister 7e.

Adapted from Fig. 9.14, Callister 7e.160 µm

Micrograph of Pb-Sn eutectic microstructure

Pb-Snsystem

L + β

α + β

200

T(°C)

C, wt% Sn20 60 80 1000

300

100

L

α β

L+ α183°C

40

TE

18.3

α: 18.3 wt%Sn

97.8

β: 97.8 wt% Sn

CE61.9

L: Co wt% Sn

15

Microstructures in eutectic systems

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

18.3 61.9

SR

97.8

SR

primary αeutectic α

eutectic β

WL = (1- W α ) = 50 wt%

C α = 18.3 wt% Sn

CL = 61.9 wt% SnS

R + SW α = = 50 wt%

• Just above TE :

• Just below TE :

C α = 18.3 wt% Sn

C β = 97.8 wt% SnS

R + SW α = = 73 wt%

W β = 27 wt%Adapted from Fig. 9.16,Callister 7e.

Pb-Snsystem

L+β200

T(°C)

Co, wt% Sn

20 60 80 1000

300

100

L

α β

L+α

40

α+β

TE

L: Co wt% Sn LαLα

16

Hypoeutectic & hypereutectic

L+αL+β

α+β

200

Co, wt% Sn20 60 80 1000

300

100

L

α β

TE

40

(Pb-Sn System)

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 µmeutectic micro-constituent

Adapted from Fig. 9.14,Callister 7e.

hypereutectic: (illustration only)

β

ββ

ββ

β

Adapted from Fig. 9.17,Callister 7e. (Illustrationonly)

(Figs. 9.14 and 9.17from MetalsHandbook, 9th ed.,Vol. 9,Metallography andMicrostructures,American Society forMetals, MaterialsPark, OH, 1985.)

175 µm

α

α

α

αα

α

hypoeutectic: Co = 50 wt% Sn

Adapted fromFig. 9.17, Callister 7e.

T(°C)

61.9eutectic

eutectic: Co = 61.9 wt% Sn

17

Intermetallic compounds

Mg2Pb

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

Adapted fromFig. 9.20, Callister 7e.

18

Peritectic & eutectoid

• Cu-Zn Phase diagram

Adapted fromFig. 9.21, Callister 7e.

Eutectoid transition δ γ + ε

Peritectic transition γ + L δ

19

Fe-C phase diagram

• 2 important

points

-Eutectoid (B):

γ ⇒ α + Fe3C

-Eutectic (A):

L ⇒ γ + Fe3C

Adapted from Fig. 9.24,Callister 7e.

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

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

γ γγγ

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard)α (ferrite-soft)

20

Hypoeutectoid steel

Adapted from Figs. 9.24and 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

(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

Adapted from Fig. 9.30,Callister 7e.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

21

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

Adapted from Figs. 9.24and 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-C System)

0.76

Co

Adapted from Fig. 9.33,Callister 7e.

proeutectoid Fe3C

60 µmHypereutectoid steel

pearlite

R S

wα =S/(R+S)wFe3C =(1-wα)

wpearlite = wγpearlite

sr

wFe3C =r/(r+s)wγ =(1-w Fe3C)

Fe3C

γγγ γ

γγγ γ

γγγ γ