1. Chapter 10: 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?
Phase B
Phase A
Nickel atom
Copper atom

2. Phase Equilibria: Solubility Limit
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase
• Solubility Limit:
Maximum concentration for
which only a single phase solution exists.
Sugar/Water Phase Diagram
Sugar
Temperature (°C) 20 40 60 80
100 C
= Composition (wt% sugar) L
(liquid solution
i.e., syrup)
Solubility
Limit
(liquid) + S
(solid
sugar) 4 6 8 10
Water
Adapted from Fig. 10.1,
Callister & Rethwisch 4e.
Question: What is the
solubility limit for sugar in
water at 20°C? 65
Answer: 65 wt% sugar.
At 20°C, if C < 65 wt% sugar: syrup
At 20°C, if C > 65 wt% sugar:
syrup + sugar

3. Components and Phases • Components: • Phases: b (lighter phase) a
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., a and b).
Aluminum-
Copper
Alloy b
(lighter
phase) a
(darker
phase)
Adapted from chapter-opening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e.

4. 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 (100°C,C = 70)
1 phase
D (100°C,C = 90)
2 phases 70 80
100 60 40 20
Temperature (°C) C
= Composition (wt% sugar) L (
liquid solution
i.e., syrup)
(liquid) + S
(solid
sugar)
water-
sugar
system
A (20°C,C = 70)
2 phases
Adapted from Fig. 10.1,
Callister & Rethwisch 4e.

5. Criteria for Solid Solubility
Simple system (e.g., Ni-Cu solution)
Crystal
Structure
electroneg
r (nm) Ni
FCC
1.9
0.1246 Cu
1.8
0.1278
Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility.
Ni and Cu are totally soluble in one another for all proportions.

6. Phase Diagrams • Indicate phases as a function of T, C, and P.
• For this course:
- binary systems: just 2 components.
- independent variables: T and C (P = 1 atm is almost always used).
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid) a
(FCC solid
solution) L +
liquidus
solidus
• 2 phases:
Phase
Diagram
for Cu-Ni
system L
(liquid) a
(FCC solid solution)
• 3 different phase fields: L
L + a a
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

7. Isomorphous Binary Phase Diagram
Cu-Ni system.
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid) a
(FCC solid
solution) L +
liquidus
solidus
• System is:
Cu-Ni
phase
diagram
-- binary
i.e., 2 components:
Cu and Ni.
-- isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

8. Phase Diagrams: Determination of phase(s) present
• Rule 1: If we know T and Co, then we know:
-- which phase(s) is (are) present.
wt% Ni 20 40 60 80
100
1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid) a
(FCC solid solution) L +
liquidus
solidus
Cu-Ni
phase
diagram
• Examples:
A(1100°C, 60 wt% Ni):
1 phase: a B
(1250°C,35) B
(1250°C, 35 wt% Ni):
2 phases: L + a
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
A(1100°C,60)

9. Phase Diagrams: Determination of phase compositions
• Rule 2: If we know T and C0, then we can determine:
-- the composition of each phase.
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 30 40 50
Cu-Ni system
• Examples: TA A
Consider C0 = 35 wt% Ni
tie line 35 C0
At TA
= 1320°C:
Only Liquid (L) present
CL = C0
( = 35 wt% Ni) B TB 32 CL 4 C 3
At TD
= 1190°C: D TD
Only Solid (a) present
C = C0
( = 35 wt% Ni)
At TB
= 1250°C:
Both 
and L present
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). CL
= C
liquidus
( = 32 wt% Ni) C
= C
solidus
( = 43 wt% Ni)

10. Phase Diagrams: Determination of phase weight fractions
• Rule 3: If we know T and C0, then can determine:
-- the weight fraction of each phase.
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5
Cu-Ni system TA A 35 C0 32 CL B TB D TD
tie line Ca R S
• Examples:
Consider C0 = 35 wt% Ni
At TA
: Only Liquid (L) present
WL = 1.00, Wa = 0
At TD :
Only Solid ( 
) present WL
= 0, W a
= 1.00
At TB :
Both 
and L present
= 0.27 WL = S R + Wa
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

11. The Lever Rule
Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm
wt% Ni 20
1200
1300
T(°C)
L (liquid) a
(solid) L +
liquidus
solidus 3 4 5 B T
tie line C0 CL C S R
What fraction of each phase? Think of the tie line as a lever
(teeter-totter) ML M R S
Adapted from Fig. 10.3(b), Callister & Rethwisch 4e.

12. Ex: Cooling of a Cu-Ni Alloy
• Phase diagram:
Cu-Ni system.
T(°C)
L (liquid)
L: 35wt%Ni
Cu-Ni
system
• Consider
microstuctural
changes that
accompany the
cooling of a
C0 = 35 wt% Ni alloy
a: 46 wt% Ni
L: 35 wt% Ni
130 a A + L B 46 35 C 43 32
a: 43 wt% Ni
L: 32 wt% Ni D 24 36
L: 24 wt% Ni
a: 36 wt% Ni a +
120 L E a
(solid)
110 20 3 35 4 5 C0
wt% Ni
Adapted from Fig. 10.4, Callister & Rethwisch 4e.

13. Cored vs Equilibrium Structures
• Ca changes as we solidify.
• Cu-Ni case:
First a to solidify has Ca = 46 wt% Ni.
Last a to solidify has Ca = 35 wt% Ni.
• Slow rate of cooling:
Equilibrium structure
• Fast rate of cooling:
Cored structure
Uniform Ca:
35 wt% Ni
First a to solidify:
46 wt% Ni
Last a to solidify:
< 35 wt% Ni

14. Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Tensile strength (TS)
-- Ductility (%EL)
Tensile Strength (MPa)
Composition, wt% Ni Cu Ni 20 40 60 80
100
200
300
400
TS for
pure Ni
TS for pure Cu
Elongation (%EL)
Composition, wt% Ni Cu Ni 20 40 60 80
100 30 50
%EL
for
pure Ni
for pure Cu
Adapted from Fig. 10.6(a), Callister & Rethwisch 4e.
Adapted from Fig. 10.6(b), Callister & Rethwisch 4e.

15. 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, a, b)
L (liquid)
1000
• Limited solubility:
a: mostly Cu
b: mostly Ag a L + a L + b
800
779°C b TE
8.0
71.9
91.2
600
• TE
: No liquid below TE a + b
: Composition at
temperature TE
• CE
400
200 20 40 60 CE 80
100
• Eutectic reaction
L(CE) (CE) + (CE) C ,
wt% Ag
Adapted from Fig. 10.7, Callister & Rethwisch 4e.
cooling
heating

16. EX 1: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine:
-- the phases present
Pb-Sn
system
Answer: a + b L + a b
200
T(°C)
18.3
C, wt% Sn 20 60 80
100
300
L (liquid)
183°C
61.9
97.8
-- the phase compositions
Answer: Ca = 11 wt% Sn
Cb = 99 wt% Sn
-- the relative amount of each phase
Answer:
150 R S 11 C 40 C0 99 C W  =
C - C0
C - C 59 88
= 0.67 S
R+S W  =
C0 - C
C - C R
R+S 29 88
= 0.33
Adapted from Fig. 10.8, Callister & Rethwisch 4e.

17. EX 2: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:
-- the phases present:
Pb-Sn
system
Answer: a + L L + b a
200
T(°C)
C, wt% Sn 20 60 80
100
300
L (liquid)
183°C
-- the phase compositions
Answer: Ca = 17 wt% Sn
CL = 46 wt% Sn
-- the relative amount of each phase
220 17 C R 40 C0 S 46 CL
Answer: W a =
CL - C0
CL - C 6 29
= 0.21
Adapted from Fig. 10.8, Callister & Rethwisch 4e. WL =
C0 - C
CL - C 23 29
= 0.79

18. Microstructural Developments in Eutectic Systems I
• For alloys for which
C0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of
a phase having
composition C0 L + a
200
T(°C) C ,
wt% Sn 10 2 20 C0
300
100 30 b
400
(room T solubility limit) TE
(Pb-Sn
System)
L: C0 wt% Sn a L
a: C0 wt% Sn
Adapted from Fig , Callister & Rethwisch 4e.

19. Microstructural Developments in Eutectic Systems II
L: C0 wt% Sn
• For alloys for which
2 wt% Sn < C0 < 18.3 wt% Sn
• Result:
at temperatures in a + b range
-- polycrystalline with a grains
and small b-phase particles
Pb-Sn
system L + a
200
T(°C) C ,
wt% Sn 10
18.3 20 C0
300
100 30 b
400
(sol. limit at TE) TE 2
(sol. limit at T
room ) L a
a: C0 wt% Sn a b
Adapted from Fig , Callister & Rethwisch 4e.

20. Microstructural Developments in Eutectic Systems III
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of a and b phases.
Adapted from Fig , Callister & Rethwisch 4e.
160 m
Micrograph of Pb-Sn
eutectic
microstructure
Pb-Sn
system
L
a
200
T(°C)
C, wt% Sn 20 60 80
100
300 L a b +
183°C 40 TE
L: C0 wt% Sn CE
61.9
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
Adapted from Fig , Callister & Rethwisch 4e.

21. Lamellar Eutectic Structure
Adapted from Figs & 10.15, Callister & Rethwisch 4e.

22. Microstructural Developments in Eutectic Systems IV
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result: a phase particles and a eutectic microconstituent WL
= (1- W a )
= 0.50 Ca
= 18.3 wt% Sn CL
= 61.9 wt% Sn S R + Wa =
• Just above TE :
Adapted from Fig , Callister & Rethwisch 4e.
Pb-Sn
system L + 
200
T(°C)
C, wt% Sn 20 60 80
100
300 a b  40 TE
L: C0
wt% Sn L  L 
18.3
61.9 S R
97.8 S R
primary a
eutectic b
• Just below TE : C 
= 18.3 wt% Sn 
= 97.8 wt% Sn S R + W =
= 0.73
= 0.27

23. Hypoeutectic & Hypereutectic
300 L
T(°C)
Adapted from Fig. 10.8,
Callister & Rethwisch 4e. (Fig adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) L + a a b
200 L + b
(Pb-Sn TE
System) a + b
100
(Figs and from Metals Handbook, 9th ed.,
Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
175 mm a
hypoeutectic: C0 = 50 wt% Sn
Adapted from
Fig , Callister & Rethwisch 4e.
hypereutectic: (illustration only) b
Adapted from Fig , Callister & Rethwisch 4e. (Illustration only)
C, wt% Sn 20 40 60 80
100
eutectic
61.9
eutectic: C0 = 61.9 wt% Sn
160 mm
eutectic micro-constituent
Adapted from Fig , Callister & Rethwisch 4e.

24. Intermetallic Compounds
Adapted from
Fig , Callister & Rethwisch 4e.
Mg2Pb
Note: intermetallic compound exists as a line on the diagram - not an area - because of stoichiometry (i.e. composition of a compound is a fixed value).

25. Eutectic, Eutectoid, & Peritectic
Eutectic - liquid transforms to two solid phases
L  +  (For Pb-Sn, 183C, 61.9 wt% Sn)
cool
heat
Eutectoid – one solid phase transforms to two other solid phases
S S1+S3
  + Fe3C (For Fe-C, 727C, 0.76 wt% C)
intermetallic compound - cementite
cool
heat
cool
heat
Peritectic - liquid and one solid phase transform to a second solid phase
S1 + L S2
 + L  (For Fe-C, 1493°C, 0.16 wt% C)

26. Eutectoid & Peritectic
Peritectic transformation  + L 
Cu-Zn Phase diagram
Eutectoid transformation   + 
Adapted from Fig , Callister & Rethwisch 4e.

27. Ceramic Phase Diagrams
MgO-Al2O3 diagram: 
Adapted from Fig , Callister & Rethwisch 4e.

28. Iron-Carbon (Fe-C) Phase Diagram
• 2 important
Adapted from Fig , Callister & Rethwisch 4e.
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a + d
(Fe)
C, wt% C
1148°C
T(°C)
727°C = T
eutectoid
points
- Eutectic (A): L Þ g +
Fe3C A
L+Fe3C
- Eutectoid (B): g Þ a +
Fe3C g
Result: Pearlite =
alternating layers of a
and Fe3C phases
120 mm
(Adapted from Fig , Callister & Rethwisch 4e.)
0.76 B
Fe3C (cementite-hard) a
(ferrite-soft)
4.30

29. Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite) a
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
(Fe)
C, wt% C
1148°C
T(°C)
727°C
(Fe-C
System) C0
0.76 g g
Adapted from Figs and 10.33,Callister & Rethwisch 4e.
(Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) g a a
pearlite
Adapted from Fig , Callister & Rethwisch 4e.
proeutectoid ferrite
pearlite
100 mm
Hypoeutectoid
steel

30. Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
(Fe)
C, wt% C
1148°C
T(°C)
727°C
(Fe-C
System) C0
0.76 g a
Adapted from Figs and 10.33,Callister & Rethwisch 4e.
(Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Wa = s/(r + s)
Wg =(1 - Wa) s r a
pearlite R S
Wpearlite = Wg
Wa’ = S/(R + S)
W =(1 – Wa’)
Fe3C
Adapted from Fig , Callister & Rethwisch 4e.
proeutectoid ferrite
pearlite
100 mm
Hypoeutectoid
steel

31. Hypereutectoid Steel T(°C) d L +L g g (austenite) Fe3C (cementite) g g
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a
L+Fe3C d
(Fe)
C, wt%C
1148°C
T(°C)
(Fe-C
System) g g
Adapted from Figs and 10.36,Callister & Rethwisch 4e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Fe3C g
pearlite C0
0.76
Adapted from Fig , Callister & Rethwisch 4e.
proeutectoid Fe3C
60 mm
Hypereutectoid
steel
pearlite

32. Hypereutectoid Steel Fe3C (cementite) L g (austenite) +L a d T(°C) g x
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+Fe3C a
L+Fe3C d
(Fe)
C, wt%C
1148°C
T(°C)
(Fe-C
System)
Fe3C g
Adapted from Figs and 10.36,Callister & Rethwisch 4e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
W =(1-Wg)
Wg =x/(v + x)
Fe3C x v
pearlite V X C0
0.76
Wpearlite = Wg
Wa = X/(V + X)
W =(1 - Wa)
Fe3C’
Adapted from Fig , Callister & Rethwisch 4e.
proeutectoid Fe3C
60 mm
Hypereutectoid
steel
pearlite

33. Example Problem
For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:
The compositions of Fe3C and ferrite ().
The amount of cementite (in grams) that forms in 100 g of steel.
The amounts of pearlite and proeutectoid ferrite () in the 100 g.

34. Solution to Example Problem
a) Using the RS tie line just below the eutectoid
Ca = wt% C CFe3C = 6.70 wt% C
Using the lever rule with the tie line shown
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
C , wt% C
1148°C
T(°C)
727°C C0 R S
CFe C 3 C
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g

35. Solution to Example Problem (cont.)
c) Using the VX tie line just above the eutectoid and realizing that
C0 = 0.40 wt% C Ca = wt% C Cpearlite = C = 0.76 wt% C
Fe3C (cementite)
1600
1400
1200
1000
800
600
400 1 2 3 4 5 6
6.7 L g
(austenite) +L
+ Fe3C a
L+Fe3C d
C, wt% C
1148°C
T(°C)
727°C C0 V X C C
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g

36. Alloying with Other Elements
• Teutectoid changes:
Adapted from Fig ,Callister & Rethwisch 4e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T
Eutectoid
(°C)
wt. % of alloying elements Ti Ni Mo Si W Cr Mn
• Ceutectoid changes:
Adapted from Fig ,Callister & Rethwisch 4e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
wt. % of alloying elements C
eutectoid
(wt% C) Ni Ti Cr Si Mn W Mo

37. 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.