CHAPTER 9.1 - PHASE DIAGRAMS FOR METALLIC SYSTEMS.pdf

Chapter 9 -
Chapter 9: Phase Diagrams
1

Chapter 9 - 2
Schematically sketch simple isomorphous and eutectic phase diagram and label the
various phase regions, label the liquidus, solidus and solvus line.
Given a binary phase diagram (isomorphous and eutectic systems), the composition
of an alloy, its temperature, and assuming that the alloy is at equilibrium, determine:
What phase(s) is (are) present,
The composition(s) of the phase(s), and
The mass fraction(s) of the phase(s)
For some given binary phase diagram, do the following:
Locate the temperatures and compositions of all eutectic, eutectoid, and
peritectic, phase transformations,
Write reactions for all of these transformations for either heating or cooling.
Given the composition of an iron-carbon alloy containing between 0.022 wt% and
2.14 wt% C, be able to:
Specify whether the alloy is hypoeutectoid or hypereutectoid,
Name the proeutectoid phase,
Compute the mass fraction of proeutectoid phase and pearlite, and
Make a schematic diagram of the microstructure at a temperature just
below the eutectoid.
Learning Outcomes

Chapter 9 - 3
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

Chapter 9 - 4
Phase Equilibria: Solubility Limit
Question: What is the
solubility limit for sugar in
water at 20ºC?
Answer: 65 wt% sugar.
At 20ºC, if C < 65 wt% sugar: syrup
At 20ºC, if C > 65 wt% sugar:
syrup + sugar
65
• Solubility Limit:
Maximum concentration for
which only a single phase
solution exists.
Sugar/Water Phase Diagram
Sugar
Temperature
(ºC)
0 20 40 60 80 100
C = Composition (wt% sugar)
L
(liquid solution
i.e., syrup)
Solubility
Limit L
(liquid)
+
S
(solid
sugar)
20
40
60
80
100
Water
Adapted from Fig. 9.1,
Callister & Rethwisch 8e.
• Solution – solid, liquid, or gas solutions, single phase
• Mixture – more than one phase

Chapter 9 - 5
• 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., a and b).
Aluminum-
Copper
Alloy
Components and Phases
a (darker
phase)
b (lighter
phase)
Adapted from chapter-
opening photograph,
Chapter 9, Callister,
Materials Science &
Engineering: An
Introduction, 3e.

Chapter 9 - 6
70 80 100
60
40
20
0
Temperature
(ºC)
C = Composition (wt% sugar)
L
(liquid solution
i.e., syrup)
20
100
40
60
80
0
L
(liquid)
+
S
(solid
sugar)
Effect of Temperature & Composition
• Altering T can change # of phases: path A to B.
• Altering C can change # of phases: path B to D.
water-
sugar
system
D (100ºC,C = 90)
2 phases
B (100ºC,C = 70)
1 phase
A (20ºC,C = 70)
2 phases

Chapter 9 - 7
Criteria for Solid Solubility
Crystal
Structure
electroneg r (nm)
Ni FCC 1.9 0.1246
Cu FCC 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.
Simple system (e.g., Ni-Cu solution)
• Ni and Cu are totally soluble in one another for all proportions.

Chapter 9 - 8
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).
Phase
Diagram
for Cu-Ni
system
Adapted from Fig. 9.3(a), Callister &
Rethwisch 8e. (Fig. 9.3(a) is adapted from
Phase Diagrams of Binary Nickel Alloys,
P. Nash (Ed.), ASM International,
Materials Park, OH (1991).
• 2 phases:
L (liquid)
a (FCC solid solution)
• 3 different phase fields:
L
L + a
a
wt% Ni
20 40 60 80 100
0
1000
1100
1200
1300
1400
1500
1600
T(ºC)
L (liquid)
a
(FCC solid
solution)

Chapter 9 - 9
Cu-Ni
phase
diagram
Isomorphous Binary Phase Diagram
• 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; a phase
field extends from
0 to 100 wt% Ni.
wt% Ni
20 40 60 80 100
0
1000
1100
1200
1300
1400
1500
1600
T(ºC)
L (liquid)
a
(FCC solid
solution)

Chapter 9 -
wt% Ni
20 40 60 80 100
0
1000
1100
1200
1300
1400
1500
1600
T(ºC)
L (liquid)
a
(FCC solid
solution)
Cu-Ni
phase
diagram
10
Phase Diagrams:
Determination of phase(s) present
• Rule 1: If we know T and Co, then we know:
-- which phase(s) is (are) present.
• Examples:
A(1100ºC, 60 wt% Ni):
1 phase: a
B(1250ºC, 35 wt% Ni):
2 phases: L + a
B
(1250ºC,35) A(1100ºC,60)

Chapter 9 - 11
wt% Ni
20
1200
1300
T(ºC)
L (liquid)
a
(solid)
30 40 50
Cu-Ni
system
Phase Diagrams:
Determination of phase compositions
• Rule 2: If we know T and C0, then we can determine:
-- the composition of each phase.
• Examples:
TA
A
35
C0
32
CL
At TA = 1320ºC:
Only Liquid (L) present
CL = C0 ( = 35 wt% Ni)
At TB = 1250ºC:
Both a and L present
CL = Cliquidus ( = 32 wt% Ni)
Ca = Csolidus ( = 43 wt% Ni)
At TD = 1190ºC:
Only Solid (a) present
Ca = C0 ( = 35 wt% Ni)
Consider C0 = 35 wt% Ni
D
TD
tie line
4
Ca
3
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).
B
TB

Chapter 9 - 12
• Rule 3: If we know T and C0, then can determine:
-- the weight fraction of each phase.
• Examples:
At TA : Only Liquid (L) present
WL = 1.00, Wa = 0
At TD : Only Solid ( a) present
WL = 0, Wa = 1.00
Phase Diagrams:
Determination of phase weight fractions
wt% Ni
20
1200
1300
T(ºC)
L (liquid)
a
(solid)
30 40 50
Cu-Ni
system
TA
A
35
C0
32
CL
B
TB
D
TD
tie line
4
Ca
3
R S
At TB : Both a and L present
73
.
0
32
43
35
43




= 0.27
WL
 S
R +S
Wa
 R
R +S
Consider C0 = 35 wt% Ni
Phase Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International, Materials
Park, OH (1991).

Chapter 9 - 13
• Tie line – connects the phases in equilibrium with
each other – also sometimes called an isotherm
The Lever Rule
What fraction of each phase?
Think of the tie line as a lever
(teeter-totter)
ML Ma
R S

Ma x S  ML x R
L
L
L
L
L
L
C
C
C
C
S
R
R
W
C
C
C
C
S
R
S
M
M
M
W












a
a
a
a
a
0
0
wt% Ni
20
1200
1300
T(ºC)
L (liquid)
a
(solid)
30 40 50
B
TB
tie line
C0
CL Ca
S
R
Adapted from Fig. 9.3(b),

Chapter 9 - 14
wt% Ni
20
1200
1300
30 40 50
110 0
L (liquid)
a
(solid)
T(ºC)
A
35
C0
L: 35wt%Ni
Cu-Ni
system
• Phase diagram:
Cu-Ni system.
• Consider
microstuctural
changes that
accompany the
cooling of a
C0 = 35 wt% Ni alloy
Ex: Cooling of a Cu-Ni Alloy
46
35
43
32
a: 43 wt% Ni
L: 32 wt% Ni
B
a: 46 wt% Ni
L: 35 wt% Ni
C
E
L: 24 wt% Ni
a: 36 wt% Ni
24 36
D

Chapter 9 -
• Slow rate of cooling:
Equilibrium structure
• Fast rate of cooling:
Cored structure
First a to solidify:
46 wt% Ni
Last a to solidify:
< 35 wt% Ni
15
• 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.
Cored vs Equilibrium Structures
Uniform Ca:
35 wt% Ni

Chapter 9 - 16
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
-- Tensile strength (TS) -- Ductility (%EL)
Adapted from Fig. 9.6(a),
Tensile
Strength
(MPa)
Composition, wt% Ni
Cu Ni
0 20 40 60 80 100
200
300
400
TS for
pure Ni
TS for pure Cu
Elongation
(%EL) Composition, wt% Ni
Cu Ni
0 20 40 60 80 100
20
30
40
50
60
%EL for
pure Ni
%EL for pure Cu
Adapted from Fig. 9.6(b),

Chapter 9 - 17
2 components
has a special composition
with a min. melting T.
Binary-Eutectic Systems
• 3 single phase regions
(L, a, b)
• Limited solubility:
a: mostly Cu
b: mostly Ag
• TE : No liquid below TE
: Composition at
temperature TE
• CE
Ex.: Cu-Ag system
Cu-Ag
system
L (liquid)
a L + a
L+bb
a  b
C, wt% Ag
20 40 60 80 100
0
200
1200
T(ºC)
400
600
800
1000
CE
TE 8.0 71.9 91.2
779ºC
Ag)
wt%
1.2
9
(
Ag)
wt%
.0
8
(
Ag)
wt%
9
.
71
( b

a
L
cooling
heating
• Eutectic reaction
L(CE) a(CaE) + b(CbE)

Chapter 9 - 18
L+a
L+b
a + b
200
T(ºC)
18.3
C, wt% Sn
20 60 80 100
0
300
100
L (liquid)
a 183ºC
61.9 97.8
b
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine:
-- the phases present Pb-Sn
system
EX 1: Pb-Sn Eutectic System
Answer: a + b
-- the phase compositions
-- the relative amount
of each phase
150
40
C0
11
Ca
99
Cb
S
R
Answer: Ca = 11 wt% Sn
Cb = 99 wt% Sn
W
a
=
Cb - C0
Cb - Ca
=
99 - 40
99 - 11
=
59
88
= 0.67
S
R+S
=
Wb =
C0 - Ca
Cb - Ca
=
R
R+S
=
29
88
= 0.33
=
40 - 11
99 - 11
Answer:

Chapter 9 - 19
Answer: Ca = 17 wt% Sn
-- the phase compositions
L+b
a + b
200
T(ºC)
C, wt% Sn
20 60 80 100
0
300
100
L (liquid)
a b
L+a
183ºC
• For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine:
-- the phases present: Pb-Sn
system
EX 2: Pb-Sn Eutectic System
-- the relative amount
of each phase
Wa =
CL - C0
CL - Ca
=
46 - 40
46 - 17
=
6
29
= 0.21
WL =
C0 - Ca
CL - Ca
=
23
29
= 0.79
40
C0
46
CL
17
Ca
220
S
R
Answer: a + L
CL = 46 wt% Sn
Answer:

Chapter 9 - 20
• For alloys for which
C0 < 2 wt% Sn
• Result: at room temperature
-- polycrystalline with grains of
a phase having
composition C0
Microstructural Developments
in Eutectic Systems I
0
L+ a
200
T(ºC)
C, wt% Sn
10
2
20
C0
300
100
L
a
30
a+b
400
(room T solubility limit)
TE
(Pb-Sn
System)
a
L
L: C0 wt% Sn
a: C0 wt% Sn

Chapter 9 - 21
• 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
in Eutectic Systems II
Pb-Sn
system
L + a
200
T(ºC)
C, wt% Sn
10
18.3
20
0
C0
300
100
L
a
30
a+ b
400
(sol. limit at TE)
TE
2
(sol. limit at Troom)
L
a
L: C0 wt% Sn
a
b
a: C0 wt% Sn

Chapter 9 - 22
• For alloy of composition C0 = CE
• Result: Eutectic microstructure (lamellar structure)
-- alternating layers (lamellae) of a and b phases.
in Eutectic Systems III
160m
Micrograph of Pb-Sn
eutectic
microstructure
Pb-Sn
system
Lb
a  b
200
T(ºC)
C, wt% Sn
20 60 80 100
0
300
100
L
a b
L+a
183ºC
40
TE
18.3
a: 18.3 wt%Sn
97.8
b: 97.8 wt% Sn
CE
61.9
L: C0 wt% Sn

Chapter 9 - 23
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
& Rethwisch 8e.

Chapter 9 - 24
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn
• Result: a phase particles and a eutectic microconstituent
in Eutectic Systems IV
18.3 61.9
S
R
97.8
S
R
primary a
eutectic a
eutectic b
WL = (1-Wa) = 0.50
Ca = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
R + S
Wa = = 0.50
• Just above TE :
• Just below TE :
Ca = 18.3 wt% Sn
Cb = 97.8 wt% Sn
S
R + S
Wa= = 0.73
Wb = 0.27
Pb-Sn
system
L+
b
200
T(ºC)
C, wt% Sn
20 60 80 100
0
300
100
L
a b
L+a
40
a+b
TE
L: C0 wt% Sn L
a
L
a

Chapter 9 - 25
L+a
L+b
a + b
200
C, wt% Sn
20 60 80 100
0
300
100
L
a b
TE
40
(Pb-Sn
System)
Hypoeutectic & Hypereutectic
(Fig. 10.8 adapted from
Binary Phase Diagrams,
2nd ed., Vol. 3, T.B.
Massalski (Editor-in-Chief),
ASM International,
Materials Park, OH, 1990.)
160 m
eutectic micro-constituent
hypereutectic: (illustration only)
b
b
b
b
b
b
(Illustration only)
(Figs. 9.14 and 9.17
from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
175 m
a
a
a
a
a
a
hypoeutectic: C0 = 50 wt% Sn
Adapted from
Fig. 9.17, Callister &
Rethwisch 8e.
T(ºC)
61.9
eutectic
eutectic: C0 =61.9wt% Sn

Chapter 9 - 26
Intermetallic Compounds
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).
Adapted from
Fig. 9.20, Callister &
Rethwisch 8e.

Chapter 9 - 27
• Eutectoid – one solid phase transforms to two other
solid phases
S2 S1+S3
 a + Fe3C (For Fe-C, 727ºC, 0.76 wt% C)
intermetallic compound
- cementite
cool
heat
Eutectic, Eutectoid, & Peritectic
• Eutectic - liquid transforms to two solid phases
L a + b (For Pb-Sn, 183ºC, 61.9 wt% Sn)
cool
heat
cool
heat
• Peritectic - liquid and one solid phase transform to a
second solid phase
S1 + L S2
 + L  (For Fe-C, 1493ºC, 0.16 wt% C)

Chapter 9 - 28
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transformation   + 
Peritectic transformation  + L 

Chapter 9 - 29
Iron-Carbon (Fe-C) Phase Diagram
• 2 important
points
- Eutectoid (B):
  a +Fe3C
- Eutectic (A):
L   +Fe3C
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
+Fe3C
a+Fe3C

(Fe) C, wt% C
1148ºC
T(ºC)
a 727ºC = Teutectoid
4.30
Result: Pearlite =
alternating layers of
a and Fe3C phases
120 m
(Adapted from Fig. 9.27,
Callister & Rethwisch 8e.)
0.76
B
 


A L+Fe3C
Fe3C (cementite-hard)
a (ferrite-soft)

Chapter 9 - 30
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 +Fe3C
a+Fe3C
L+Fe3C

(Fe) C, wt% C
1148ºC
T(ºC)
a
727ºC
(Fe-C
System)
C0
0.76
Hypoeutectoid Steel
Adapted from Figs. 9.24
and 9.29,Callister &
Rethwisch 8e.
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
Adapted from Fig. 9.30, Callister & Rethwisch 8e.
proeutectoid ferrite
pearlite
100 m
Hypoeutectoid
steel
a
pearlite

 

a
a
a


 
 



Chapter 9 - 31
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 +Fe3C
a+Fe3C
L+Fe3C

(Fe) C, wt% C
1148ºC
T(ºC)
a
727ºC
(Fe-C
System)
C0
0.76
Hypoeutectoid Steel

 

a
a
a
s
r
Wa = s/(r +s)
W =(1 - Wa)
R S
a
pearlite
Wpearlite = W
Wa’ = S/(R +S)
W =(1 – Wa’)
Fe3C
Rethwisch 8e.
Binary Alloy Phase
Diagrams, 2nd ed., Vol.
1, T.B. Massalski (Ed.-in-
Chief), ASM International,
Materials Park, OH,
1990.)
proeutectoid ferrite
pearlite
100 m
Hypoeutectoid
steel

Chapter 9 - 32
Hypereutectoid Steel
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 +Fe3C
a +Fe3C
L+Fe3C

(Fe) C, wt%C
1148ºC
T(ºC)
a
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)
(Fe-C
System)
0.76
C0
Fe3C


 


 


 
proeutectoid Fe3C
60 mHypereutectoid
steel
pearlite
pearlite

Chapter 9 - 33
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 +Fe3C
a +Fe3C
L+Fe3C

(Fe) C, wt%C
1148ºC
T(ºC)
a
Hypereutectoid Steel
(Fe-C
System)
0.76
C0
pearlite
Fe3C


 
x
v
V X
Wpearlite = W
Wa = X/(V +X)
W =(1 - Wa)
Fe3C’
W =(1-W)
W =x/(v + x)
Fe3C
proeutectoid Fe3C
60 mHypereutectoid
steel
pearlite
Rethwisch 8e. (Fig. 9.24
adapted from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B. Massalski
(Ed.-in-Chief), ASM
International, Materials
Park, OH, 1990.)

Chapter 9 - 34
Example Problem
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 (a).
b) The amount of cementite (in grams) that
forms in 100 g of steel.
c) The amounts of pearlite and proeutectoid
ferrite (a) in the 100 g.

Chapter 9 - 35
Solution to Example Problem
WFe3C 
R
R  S

C0 Ca
CFe3C Ca

0.40 0.022
6.70 0.022
 0.057
b) Using the lever rule with
the tie line shown
a) Using the RS tie line just below the eutectoid
Ca = 0.022 wt% C
CFe3C = 6.70 wt% C
Fe
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 + Fe3C
a+ Fe3C
L+Fe3C

C, wt% C
1148ºC
T(ºC)
727ºC
C0
R S
CFe C
3
Ca
Amount of Fe3C in 100 g
= (100 g)WFe3C
= (100 g)(0.057) = 5.7 g

Chapter 9 - 36
Solution to Example Problem (cont.)
c) Using the VX tie line just above the eutectoid and
realizing that
C0 = 0.40 wt% C
Ca = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
Fe
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L

(austenite)
+L
 + Fe3C
a+ Fe3C
L+Fe3C

C, wt% C
1148ºC
T(ºC)
727ºC
C0
V X
C
Ca
Wpearlite 
V
V  X

C0 Ca
C Ca

0.40 0.022
0.76 0.022
 0.512
Amount of pearlite in 100 g
= (100 g)Wpearlite
= (100 g)(0.512) = 51.2 g

Chapter 9 -
VMSE: Interactive Phase Diagrams
37
Change alloy composition
Microstructure, phase compositions, and phase fractions respond interactively

Chapter 9 - 38
Alloying with Other Elements
• Teutectoid changes:
Adapted from Fig. 9.34,Callister & Rethwisch 8e.
(Fig. 9.34 from Edgar C. Bain, Functions of the
Alloying Elements in Steel, American Society for
Metals, 1939, p. 127.)
T
Eutectoid
(ºC)
wt. % of alloying elements
Ti
Ni
Mo
Si
W
Cr
Mn
• Ceutectoid changes:
Adapted from Fig. 9.35,Callister & Rethwisch 8e.
(Fig. 9.35 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

Chapter 9 - 39
• 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.
Summary

Chapter 9 - 40
Core Problems:
Self-help Problems:
ANNOUNCEMENTS
Reading:

CHAPTER 9.1 - PHASE DIAGRAMS FOR METALLIC SYSTEMS.pdf

More Related Content

What's hot (20)

Similar to CHAPTER 9.1 - PHASE DIAGRAMS FOR METALLIC SYSTEMS.pdf (20)

Recently uploaded (20)

CHAPTER 9.1 - PHASE DIAGRAMS FOR METALLIC SYSTEMS.pdf