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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.
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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.
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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
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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)
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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
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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
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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.
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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.
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18
Peritectic & eutectoid
• Cu-Zn Phase diagram
Adapted fromFig. 9.21, Callister 7e.
Eutectoid transition δ γ + ε
Peritectic transition γ + L δ
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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)
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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
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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