Phase Diagrams:
composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
T(°C)
Cu-Ni
system
A
TA
Co = 35 wt% Ni
tie line
1300 L (liquid)
At T A = 1320°C:
Only Liquid (L)
B
TB
CL = Co ( = 35 wt% Ni)
α
At T D = 1190°C:
(solid)
1200
D
Only Solid ( α)
TD
Cα = Co ( = 35 wt% Ni)
20
3032 35 4043
50
At T B = 1250°C:
CLCo
Cα wt% Ni
Both α and L
Adapted from Fig. 9.3(b), Callister 7e.
9.3(b) is adapted from Phase Diagrams
CL = C liquidus ( = 32 wt% Ni here) (Fig.
of Binary Nickel Alloys, P. Nash (Ed.), ASM
Cα = C solidus ( = 43 wt% Ni here) International, Materials Park, OH, 1991.)
2
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%).
• Examples:
Co = 35 wt% Ni
At T A : Only Liquid (L)
W L = 100 wt%, W α = 0
At T D: Only Solid ( α)
W L = 0, Wα = 100 wt%
At T B : Both α and L
WL =
Wα =
43 − 35
S
=
= 73 wt %
R + S 43 − 32
R
= 27 wt%
R +S
Cu-Ni
system
T(°C)
A
TA
tie line
L (liquid)
1300
B
R S
TB
1200
D
TD
20
3 032 35
CLCo
α
(solid)
4 0 43
50
Cα 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.), ASM
International, Materials Park, OH, 1991.)
3
The Lever Rule
• Tie line – connects the phases in equilibrium with each
other - essentially an isotherm
How much of each phase?
Think of it as a lever (teeter-totter)
T(°C)
tie line
1300
L (liquid)
B
TB
α
(solid)
1200
R
20
S
R
3 0C C
40 C
α
L o
wt% Ni
WL =
Mα
ML
50
S
M α ⋅S = M L ⋅R
Adapted from Fig. 9.3(b),
Callister 7e.
C − C0
ML
S
=
= α
M L + M α R + S C α − CL
Wα =
C − CL
R
= 0
R + S C α − CL
4
Ex: Cooling in a Cu-Ni Binary
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(°C) L (liquid)
130 0
L: 35 wt% Ni
α: 46 wt% Ni
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
35
32
--isomorphous
i.e., complete
solubility of one
component in
another; α phase
field extends from
0 to 100 wt% Ni.
L: 35wt%Ni
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
α: 43 wt% Ni
E
L: 24 wt% Ni
α: 36 wt% Ni
α
(solid)
110 0
20
30
Adapted from Fig. 9.4,
Callister 7e.
35
Co
40
50
wt% Ni
5
Cored vs Equilibrium Phases
• Cα changes as we solidify.
• Cu-Ni case: First α to solidify has Cα = 46 wt% Ni.
Last α to solidify has Cα = 35 wt% Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First α to solidify:
46 wt% Ni
Last α to solidify:
< 35 wt% Ni
Uniform C α:
35 wt% Ni
6
Binary-Eutectic Systems
has a special composition
with a min. melting T.
2 components
Cu-Ag
system
T(°C)
Ex.: Cu-Ag system
1200
• 3 single phase regions
L (liquid)
1000
(L, α, β)
α L + α 779°C
• Limited solubility:
L +β β
800
T
α: mostly Cu
8.0
71.9 91.2
E
β: mostly Ag
600
• TE : No liquid below TE
α+β
400
• CE : Min. melting TE
composition
200
• Eutectic transition
L(CE)
0
α(CαE) + β(CβE)
20
40
60 CE 80
100
Co , wt% Ag
Adapted from Fig. 9.7,
Callister 7e.
7
EX: Pb-Sn Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: α + β
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
--the relative amount
of each phase:
Wα =
C - CO
S
= β
R+S
Cβ - Cα
Pb-Sn
system
300
200
L (liquid)
α
L+ α
18.3
150
100
99 - 40
59
=
= 67 wt%
99 - 11
88
C - Cα
Wβ = R = O
Cβ - C α
R+S
L +β β
183°C
61.9
R
97.8
S
α+β
=
=
40 - 11
29
=
= 33 wt%
99 - 11
88
0 11 20
Cα
40
Co
60
80
C, wt% Sn
99100
Cβ
Adapted from Fig. 9.8,
Callister 7e.
8
EX: Pb-Sn Eutectic System (2)
• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...
--the phases present: α + L
T(°C)
--compositions of phases:
CO = 40 wt% Sn
Cα = 17 wt% Sn
CL = 46 wt% Sn
--the relative amount
of each phase:
CL - C O
46 - 40
=
Wα =
CL - Cα
46 - 17
6
=
= 21 wt%
29
Pb-Sn
system
300
α
220
200
L (liquid)
L+ α
R
L +β β
S
183°C
100
CO - C α
23
=
WL =
= 79 wt%
29
CL - C α
α+β
0
17 20
Cα
40 46 60
Co CL
80
100
C, wt% Sn
Adapted from Fig. 9.8,
Callister 7e.
9
Microstructures
in Eutectic Systems: I
• Co < 2 wt% Sn
• Result:
--at extreme ends
--polycrystal of α grains
i.e., only one solid phase.
T(°C)
L: Co wt% Sn
400
L
α
L
300
α
200
(Pb-Sn
System)
α: Co wt% Sn
TE
α+ β
100
Adapted from Fig. 9.11,
Callister 7e.
L+ α
0
Co
10
20
30
Co, wt% Sn
2
(room T solubility limit)
10
Microstructures
in Eutectic Systems: II
L: Co wt% Sn
• 2 wt% Sn < Co < 18.3 wt% Sn 400T(°C)
• Result:
 Initially liquid + α
 then α alone
 finally two phases
 α polycrystal
 fine β-phase inclusions
L
300
L+α
α
200
TE
α: Co wt% Sn
α
β
100
α+ β
0
Adapted from Fig. 9.12,
Callister 7e.
(sol.
L
α
10
20
Pb-Sn
system
30
Co
Co , wt%
2
limit at T room )
18.3
(sol. limit at TE)
Sn
11
Microstructures
in Eutectic Systems: III
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of α and β crystals.
T(°C)
L: Co wt% Sn
300
Pb-Sn
system
α
200
L+ α
L
L+β β
183°C
TE
100
α+β
0
20
18.3
Adapted from Fig. 9.13,
Callister 7e.
Micrograph of Pb-Sn
eutectic
microstructure
40
β: 97.8 wt% Sn
α: 18.3 wt%Sn
60
CE
61.9
80
160 µm
Adapted from Fig. 9.14, Callister 7e.
100
97.8
C, wt% Sn
12
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
7e.
13
Microstructures
in Eutectic Systems: IV
• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: α crystals and a eutectic microstructure
L: Co wt% Sn
T(°C)
300
Pb-Sn
system
α
200
L+ α
TE
L
α
L
R
α L
L+β β
S
S
R
primary α
eutectic α
eutectic β
0
20
18.3
Adapted from Fig. 9.16,
Callister 7e.
40
60
61.9
Cα = 18.3 wt% Sn
CL = 61.9 wt% Sn
Wα = S = 50 wt%
R+S
WL = (1- Wα) = 50 wt%
• Just below TE :
α+β
100
• Just above TE :
80
Co, wt% Sn
100
97.8
Cα = 18.3 wt% Sn
Cβ = 97.8 wt% Sn
Wα = S = 73 wt%
R+S
Wβ = 27 wt%
14
Hypoeutectic & Hypereutectic
300
L
T(°C)
Adapted from Fig. 9.8,
Callister 7e. (Fig. 9.8
adapted from Binary Phase
Diagrams, 2nd ed., Vol. 3,
T.B. Massalski (Editor-inChief), ASM International,
Materials Park, OH, 1990.)
α
200
L+ α
α+β
20
40
hypoeutectic: Co = 50 wt% Sn
α
α
(Pb-Sn
System)
100
0
(Figs. 9.14 and 9.17
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
L+β β
TE
α
60
80
eutectic
61.9
hypereutectic: (illustration only)
β
β
α
Adapted from
Fig. 9.17, Callister 7e.
Co, wt% Sn
eutectic: Co = 61.9 wt% Sn
α α
175 µm
100
β
β β
β
160 µm
eutectic micro-constituent
Adapted from Fig. 9.14,
Callister 7e.
Adapted from Fig. 9.17,
Callister 7e. (Illustration
only)
15
Intermetallic Compounds
Adapted from
Fig. 9.20, Callister 7e.
Mg2Pb
Note: intermetallic compound forms a line - not an area because stoichiometry (i.e. composition) is exact.
16
•
Eutectoid
&
Peritectic
Eutectic - liquid in equilibrium with two solids
L
cool
α+β
heat
• Eutectoid - solid phase in equation with
two solid phases
intermetallic compound
S2
- cementite
S1+S3
γ
cool
heat
α + Fe3C
(727ºC)
• Peritectic - liquid + solid 1  solid 2 (Fig
9.21)
S1 + L
S2
δ+L
cool
γ
(1493ºC)
heat
17
Eutectoid & Peritectic
Peritectic transition γ + L
δ
Cu-Zn
Phase
diagram
Eutectoid transition δ
γ+ε
Adapted from
Fig. 9.21, Callister 7e.
18
Iron-Carbon (Fe-C) Phase Diagram
T(°C)
1600
δ
1200
γ ⇒ α + Fe3C
γ +L
γ
(austenite)
γ γ
γ γ
1000
α
800
600
120 µm
Result: Pearlite =
alternating layers of
α and Fe3C phases
(Adapted from Fig. 9.27, Callister 7e.)
S
γ +Fe3C
727°C = Teutectoid
R
S
1
0.76
L+Fe3C
R
B
400
0
(Fe)
A
1148°C
2
3
α+Fe3C
4
5
6
Fe3C (cementite)
L ⇒ γ + Fe3C
-Eutectoid (B):
L
1400
C eutectoid
• 2 important
points
-Eutectic (A):
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
α (ferrite-soft)
Adapted from Fig. 9.24,Callister 7e.
19
Hypoeutectoid Steel
T(°C)
1600
δ
L
γ γ
γ γ
α
γ
γ
α
γ αγ
γ +L
γ
1200
(austenite)
1000
800
γ + Fe3C
r s
727°C
αRS
w α =s /(r +s) 600
w γ = (1- wα )
400
0
α
(Fe)
pearlite
L+Fe3C
1148°C
α + Fe3C
1
C0
w pearlite = w γ
2
3
4
5
Adapted from Figs. 9.24
and 9.29,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
6.7
100 µm
w α = S/(R + S)
w Fe3 =(1- w α )
C
6
(Fe-C
System)
Co , wt% C
0.76
γ γ
γ γ
Fe3C (cementite)
1400
pearlite
Hypoeutectoid
steel
proeutectoid ferrite
Adapted from Fig. 9.30,Callister 7e.
20
Hypereutectoid Steel
T(°C)
1600
δ
L
Fe3C
γ
γ
γ +L
γ
1200
(austenite)
γ
γ
1000
γ γ
γ γ
r
800
w Fe3C =r /( r +s)
w γ =(1- w Fe3C )
α R
600
400
0
(Fe)
pearlite
L+Fe3C
1148°C
γ +Fe3C
0.76
γ
γ
γ
γ
(Fe-C
System)
s
S
1 Co
w pearlite = w γ
w α =S/( R +S)
w Fe3C =(1- w α )
α +Fe3C
2
3
4
5
6
Fe3C (cementite)
1400
Adapted from Figs. 9.24
and 9.32,Callister 7e.
(Fig. 9.24 adapted from
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-inChief), ASM International,
Materials Park, OH,
1990.)
6.7
Co , wt%C
60 µmHypereutectoid
steel
pearlite
proeutectoid Fe3C
Adapted from Fig. 9.33,Callister 7e.
21
Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature
just below the eutectoid, determine the
following
a) composition of Fe3C and ferrite (α)
b) the amount of carbide (cementite) in grams
that forms per 100 g of steel
c) the amount of pearlite and proeutectoid
ferrite (α)
22
Chapter
9
–
Phase
Equilibria
a) composition of Fe C and ferrite (α)
Solution:
3
b) the amount of carbide
(cementite) in grams that
forms per 100 g of steel
CO = 0.40 wt% C
Cα = 0.022 wt% C
CFe C = 6.70 wt% C
3
1600
δ
1200
0.4 − 0.022
x 100 = 5.7g
=
6.7 − 0.022
L
γ
α = 94.3 g
L+Fe3C
1148°C
(austenite)
1000
γ + Fe3C
800
Fe3C = 5.7 g
γ +L
Fe C (cementite)
Co − Cα
Fe3C
1400
=
x100 T(°C)
Fe3C + α CFe3C − Cα
727°C
R
S
α + Fe3C
600
400
0
Cα CO
1
2
3
4
Co , wt% C
5
6
6.7
CFe
3C
23
c.
Chapter
9
–
Phase
Equilibria
the amount of pearlite and proeutectoid ferrite (α)
note: amount of pearlite = amount of γ just above TE
1600
δ
L
1400
T(°C)
γ +L
Co − Cα
γ
γ
x 100 = 51.2 g 1200
=
(austenite)
γ + α Cγ − Cα
L+Fe3C
1148°C
1000
γ + Fe3C
800
727°C
RS
pearlite = 51.2 g
proeutectoid α = 48.8 g
α + Fe3C
600
400
0
1
Cα CO Cγ
2
3
4
5
6
Co , wt% C
24
Fe C (cementite)
Co = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 0.76 wt% C
6.7