Chapter 9: Phase Diagrams
ISSUES TO ADDRESS...
• When we combine two elements...
what is the resulting equilibrium state?
• In particular, if we specify...
-- the composition (e.g., wt% Cu - wt% Ni), and
-- the temperature (T)
then...
How many phases form?
What is the composition of each phase?
What is the amount of each phase?
What is the structure of each phase
Phase A
Phase Behavior
Laughlin p. 67
Phase B
Nickel atom
Copper atom
Components and Phases
• Components:
The elements or compounds which are present in the alloy
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that form (e.g., α and β).
Optical Mettalography
α (lighter
phase)
β (darker
phase)
1µm
Aluminum-Copper Alloy
Solubility Limit
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase
Sugar/Water Phase Diagram
• Solubility Limit:
Question: What is the
solubility limit for sugar in
water at 20ºC?
Temperature (ºC)
Maximum concentration for
which only a single phase
solution exists.
100
Solubility
Limit
80
L
(liquid)
60
40
L
(syrup)
20
0
20
+
S
(solid
sugar)
40
6065 80
100
Composition (wt% sugar)
Problem 9.2: At 170°C, what is the maximum solubility (a) of Pb in Sn? (b) of Sn in
Pb? The lead-tin phase diagram is shown in the Animated Figure 9.8.
Isomorphous Binary Phase Diagram
Cu-Ni phase diagram
1600
• System is:
1500
-- binary
-- isomorphous
i.e., complete solubility of
one component in another
1400
T(ºC)
2 components: Cu and Ni.
L (liquid)
1300
α
1200
1100
1000
solid solution
0
20
40
60
wt% Ni
80
100
Criteria for Solid Solubility: Hume–Rothery rules
•Same crystal structure
•Similar electronegativities
•Similar atomic radii
Crystal
Structure
electroneg
r (nm)
Ni
FCC
1.9
0.1246
Cu
FCC
1.8
0.1278
• Ni and Cu are totally soluble in one another for all proportions.
Structure
BCC
FCC
Periodic Table
Determination of phases present
• If we know T and Co, then we know which phases are present.
Cu-Ni phase diagram
• Examples:
B(1250ºC, 35 wt% Ni):
2 phases: L + α
L
1400
T(ºC)
A(1100ºC, 60 wt% Ni):
1 phase: α
1500
1300
B (1250ºC,35)
1600
α
1200
1100
1000
A(1100ºC,60)
0
20
40
60
wt% Ni
80
100
Determination of phase compositions: Lever Rule
Tie line –– also sometimes called an isotherm
T(ºC)
What fraction of each phase?
Think of the tie line as a lever
(teeter-totter)
tie line
1300
L
B
TB
Mα
ML
α
1200
R
20
30CL
S
C0 40 Cα
50
wt% Ni
C  C0
ML
S
WL 


ML  M R  S C  CL

R
S
M x S  ML x R
C0  CL
R
W 

R  S C  CL
Slow Cooling of a Cu-Ni Alloy
C0 = 35 wt% Ni alloy
T(ºC) L
L: 35wt%Ni
130 0
L: 35 wt% Ni
α: 46 wt% Ni
35
46
32
24
43
L: 32 wt% Ni
36
: 43 wt% Ni
120 0
L: 24 wt% Ni
: 36 wt% Ni
α
110 0
20
30
35
C0
40
50
wt% Ni
Cored vs Equilibrium Structures
Uniform C:
35 wt% Ni
T(ºC)
L
130 0
35
46
32
24
43
36
120 0
First α to solidify:
46 wt% Ni
α
Last  to solidify:
< 35 wt% Ni
110 0
20
30
35
C0
40
50
wt% Ni
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Ductility (%EL)
400
TS for
pure Ni
300
TS for pure Cu
200
0 20 40
Cu
60 80 100
Ni
Composition, wt% Ni
Elongation (%EL)
Tensile Strength (MPa)
-- Tensile strength (TS)
60
%EL for pure Cu
%EL for
pure Ni
50
40
30
20
0 20
Cu
40
60
80 100
Ni
Composition, wt% Ni
Binary-Eutectic Systems
=low melting
2 components
Cu-Ag system
Ex.: Cu-Ag system
• 3 single phase regions
(L, α, β)
1200
L (liquid)
1000
TE
800
α
L+α
L+β β
779ºC
8.0
71.9 91.2
600
α +β
400
200
0
20
40
60 CE 80
wt% Ag
Eutectic Phase Reaction:
L(71.9 wt% Ag)
cooling
heating
(8.0 wt% Ag)  (91.2 wt% Ag)
100
Microstructural Development in Eutectic Systems I
Pb-Sn
• For alloys for which
C0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of
a phase having
composition C0
T(ºC)
L: C0 wt% Sn
400
L
α
L
300
200
TE
100
α
L+ α
α: C0 wt% Sn
α+β
0
10
20
30
C0
wt% Sn
2
(room T solubility limit)
Microstructural Development in Eutectic Systems II
• For alloys for which
400
2 wt% Sn < C0 < 18.3 wt% Sn
• Result:
at temperatures in α + β range 300
-- polycrystalline with a grains
and small β-phase particles 200
L: C0 wt% Sn
T(ºC)
L
L
α
L +α
a: C0 wt% Sn
α
TE
α
β
100
α+ β
0
10
20
30
C0
C, wt%
2
(sol. limit at T room )
18.3
(sol. limit at TE)
Sn
Microstructural Development in Eutectic Systems III
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of α and β phases.
T(ºC)
Micrograph of Pb-Sn
eutectic microstructure
L: C0 wt% Sn
300
α
200
L+ α
L
β
183ºC
TE
100
β: 97.8 wt% Sn
α: 18.3 wt%Sn
0
20
18.3
40
60
CE
61.9
80
100
97.8
wt% Sn
160 μm
Lamellar Eutectic Structure
Microstructural Development in Eutectic Systems IV
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result: α phase particles and a eutectic microconstituent
• Just above TE :
L: C0 wt% Sn
L
T(ºC)
300
L
Cα = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
= 0.50
=
Wα
R+S
WL = (1- Wα ) = 0.50
α
L+ α
α
200
R
TE
S
S
R
• Just below TE :
100
primary α
eutectic α
eutectic β
0
20
18.3
40
60
61.9
80
wt% Sn
100
97.8
Cα = 18.3 wt% Sn
Cβ = 97.8 wt% Sn
Wα = S = 0.73
R+S
Wβ = 0.27
Hypoeutectic & Hypereutectic
300
L
T(ºC)
α
200
L+ α
L+β β
TE
System)
α+β
100
0
20
40
hypoeutectic: C0 = 50 wt% Sn
α
α
α
60
80
eutectic
61.9
C, wt% Sn
hypereutectic: (illustration only)
eutectic: C0 = 61.9 wt% Sn
β
β
α α
β
α
175 mm
100
160 mm
eutectic micro-constituent
β β
β
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound exists as a line - not an area – because
stoichiometry (i.e. composition of a compound) is fixed.
Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L cool S1+S3
(For Pb-Sn, 183ºC, 61.9 wt% Sn)
heat
• Eutectoid –all solid phases
intermetallic compound - cementite
S2
S1+S3
cool
ϒ heat α+ Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
• Peritectic - liquid and one solid phase transform to a
second solid phase
S1 + L
S2
δ +L
cool
heat
ϒ
(For Fe-C, 1493ºC, 0.16 wt% C)
• Peritectoid – all solid phases
S1 + S2
S3
Eutectoid & Peritectic
Cu-Zn Phase diagram
Peritectic transformation γ + L
Eutectoid transformation δ
γ+ε
δ
Iron-Carbon (Fe-C) Phase Diagram
T(ºC)
1600
δ
L → γ + Fe3C
- Eutectoid (B):
γ → α + Fe3C
L
1400
1200
γ +L
γ
(austenite)
γ γ
γ γ
1000
800
α
120 mm
Result: Pearlite =
alternating layers of
α and Fe3C phases
L+Fe3C
γ +Fe3C
727ºC = T eutectoid
B
600
400
0
(Fe)
A
1148ºC
α +Fe3C
1
0.76
2
3
4
4.30
5
6
wt% C
Fe3C (cementite-hard)
α (ferrite-soft)
Fe3C (cementite)
- Eutectic (A):
6.7
Hypereutectoid Steel
T(ºC)
1600
δ
γγ
γ γ
γ γ
Fe3C
1200
proeutectoid Fe3C
L
1400
γ +L
γ
(austenite)
L+Fe3C
1148ºC
1000
γ +Fe3C
800
60 μm
a
600
400
0
(Fe)
pearlite
pearlite
α +Fe3C
0.76
γ
1 C0
2
3
4
5
6
6.7
C, wt%C
T(ºC)
1600
δ
L
1400
γ γ
γ γ
1200
γ +L
γ
(austenite)
γ γ
γ γ
α
γ + Fe3C
800
pearlite
proeutectoid ferrite
727ºC
α
600
400
0
(Fe) C0
pearlite
100 μm
L+Fe3C
1148ºC
1000
α + Fe3C
1
0.76
α
Hypoeutectoid Steel
2
3
4
5
6
6.7
C, wt% C
For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the
following:
a)
The compositions of Fe3C and ferrite (α).
b)
The amount of cementite (in grams) that forms in 100 g of steel.
c)
The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.
Solution to Example Problem
a) The compositions of Fe3C and ferrite (α).
RS tie line just below the eutectoid
Cα = 0.022 wt% C
CFe3C = 6.70 wt% C
pearlite
b) Wight Fraction of cementite
1600
δ

R
C  C
 0
R  S CFe3C  C
0.40  0.022
 0.057
6.70  0.022
T(ºC)
1200
γ
γ +L
1000
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g
L+Fe3C
1148ºC
(austenite)
Fe C (cementite)
WFe3C 
L
1400
γ + Fe3C
800
727ºC
R
S
α + Fe3C
600
400
0
Cα C0
1
2
3
4
C, wt% C
5
6
6.7
CFe
3C
The amounts of pearlite in the 100 g.
c) Using the VX tie line just above the eutectoid and
realizing that
C0 = 0.40 wt% C
Cα = 0.022 wt% C
Cpearlite = Cγ = 0.76 wt% C
α
pearlite
1600


V
C  C
 0
V  X C  C
0.40  0.022
 0.512
0.76  0.022
Amount of pearlite in 100 g
= (100 g)Wpearlite
T(ºC)
1200
γ
γ +L
L+Fe3C
1148ºC
(austenite)
1000
γ + Fe3C
800
727ºC
VX
 + Fe3C
600
400
0
1
Cα C0 Cγ
= (100 g)(0.512) = 51.2 g
2
3
4
C, wt% C
5
6
Fe C (cementite)
Wpearlite 
L
1400
6.7
Alloying with Other Elements
Ti
Mo
Si
W
Cr
Mn
Ni
wt. % of alloying elements
• Ceutectoid changes:
Ceutectoid (wt% C)
T Eutectoid (ºC)
• Teutectoid changes:
Ni
Cr
Si
Ti Mo
W
Mn
wt. % of alloying elements
VMSE: Interactive Phase Diagrams
Microstructure, phase compositions, and phase fractions respond interactively
Change alloy composition
Summary
• Phase diagrams are useful tools to determine:
-- the number and types of phases present,
-- the composition of each phase,
-- and the weight fraction of each phase
given the temperature and composition of the system.
• The microstructure of an alloy depends on
-- its composition, and
-- whether or not cooling rate allows for maintenance of
equilibrium.
• Important phase diagram phase transformations include
eutectic, eutectoid, and peritectic.
Effect of Temperature & Composition
• Altering T can change # of phases: path A to B.
• Altering C can change # of phases: path B to D.
B (100ºC,C = 70) D (100ºC,C = 90)
1 phase
Temperature (ºC)
100
2 phases
L
80
(liquid)
60
L
(liquid solution
40
i.e., syrup)
+
S
(solid
sugar)
A (20ºC,C = 70)
20
2 phases
0
0
20
40
60 70 80
100
C = Composition (wt% sugar)
EX 1: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC
Cα = 11 wt% Sn
Pb-Snsystem
T(ºC)
Cβ = 99 wt% Sn
300
-- the relative amount
of each phase
200
=
40 - 11
29
=
= 0.33
99 - 11
88
α
L+ α
100
L+β β
183ºC
18.3
150
Cβ - C0
S
=
W =
R+S
Cβ - Cα
α
99 - 40
59
=
=
= 0.67
99 - 11
88
C0 - Cα
R
Wβ =
=
Cβ - Cα
R+S
L (liquid)
61.9
R
97.8
S
α+β
0 11 20
Cα
40
C0
60
80
wt% Sn
99100
Cβ
EX 2: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC
Ca = 17 wt% Sn
CL = 46 wt% Sn
the relative amount
of each phase
CL - C0
46 - 40
=
Wa =
CL - Cα
46 - 17
6
=
= 0.21
29
C0 - Cα
23
=
= 0.79
WL =
CL - Cα
29
T(ºC)
300
α
220
200
L+ α
R
L
L+β β
S
183ºC
100
a +b
0
17 20
Cα
40 46 60
C0 CL
80
wt% Sn
100