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Chapter 9 - 11
ISSUES TOADDRESS... When we combine two elements...
what equilibrium state do weget? In particular, if we specify...
--a composition (e.g., wt% Cu - wt% Ni),and
--a temperature (T)then...How many phases do we get?What is the composition of each
phase?How much of each phase do we get?
Chapter 9: Phase Diagrams
Phase BPhase A
Nickel
atomCopperatom
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Chapter 9 - 22
Phase Equilibria: Solubility LimitIntroduction
Solutions solid solutions, single phase Mixtures more than one phase
Solubility Limit:
Max concentration forwhich only a single phase
solution occurs.
Question: What is the
solubility limit at 20C?
Answer: 65 wt% sugar.
If Co < 65 wt% sugar: syrupIf Co > 65 wt% sugar: syrup + sugar.
65
Sucrose/Water PhaseDiagram
P
u
r
e
S
u
g
ar
Temperature
(C
)
0 20
40
60
80
10
0
C
o
=Composition (wt%
sugar)
L(liquid
solutioni.e.,syrup)
SolubilityLimit
L(liqu
id)+S(sol
idsugar)
20
40
60
80
10 0
P
u
r
e
Wa
te
r
Adapted from Fig. 9.1,Callister 7e.
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Chapter 9 - 33
Components:
The elements or compounds which are present in the mixture (e.g., Al and Cu) Phases:
The physically and chemically distinct material regionsthat result (e.g., and ).
Aluminum-Copper
Alloy
Components and Phases
(darker
phase)
(lighter
phase)
Adapted fromchapter-openingphotograph,Chapter 9,Callister 3e.
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Chapter 9 - 44
Effect ofT& Composition (Co) Changing T can change # of phases:
Adapted from Fig.9.1,Callister 7e.
D(100C,90)2 phases
B(100C,70)
1 phase
pathA to B.
Changing Co can change # of phases: path B to D.
A(20C,70)2 phases
70
80
100
60
40
20
0
Temperature
(C
)
Co
=Composition (wt%sugar)
L( liquid
solutioni.e.,
syrup)20
10
0
4
0
60
80
0
L(liquid)+
S(sol
idsugar)
water-sugarsystem
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Chapter 9 - 55
Phase Equilibria
CrystalStructure
electroneg r (nm)
Ni FCC 1.9 0.1246Cu FCC 1.8 0.1278
Both have the same crystal structure (FCC) and havesimilar electronegativities and atomic radii (W. Hume Rothery rules) suggesting high mutual solubility.
Simple solution system (e.g., Ni-Cu solution)
Ni and Cu are totally miscible in all proportions.
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Chapter 9 - 66
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) (FCC solidsolution) 3 phase fields:
L
L+
wt%20
40
60
80
10
01000
1100
1200
1300
1400
1500
16
00
T(
C)L(liquid)
(FCC solid
solution
)
L+
liq
uid
ussolid
us
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Chapter 9 - 77
wt%
Ni
20
40
60
80
100
01000
1100
1200
1300
1400
1500
1600
T(C)
L
(liquid)
(FCC
solidsolution)
L+
liquid
ussolid
us
Cu-Niphasediagram
Phase Diagrams:# and types of phases
Rule 1: If we know T and Co, then we know:--the # and types of phases present.
Examples:
A(1100C,
60):1 phase: B(1250C,
35):2 phases: L +
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 2 5 0 C , 3 5 )
A(1100C,60)
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Chapter 9 - 88
wt%Ni
20
1200
13
00
T(C)
L
(liquid)
(solid)
L +
liquid
us
solid
us
30
40
50
L+
Cu-Nisystem
Phase Diagrams:composition of phases
Rule 2: If we know T and Co
, then we know:--the composition of each phase.
Examples:
TAA
35Co
32CL
At
T
A =
1320C:Only Liquid(L)
CL =
Co ( = 35 wt%
Ni)
AtT
B = 1250C:
Both and
LCL =C
liquidus
( = 32 wt% Ni here)
C =
C
solidus
( = 43 wt% Ni
here)
AtT
D =1190C:Only Solid
()
C =
Co ( = 35 wt%
Ni
)
Co = 35 wt%Ni
Adapted from Fig. 9.3(b), Callister 7e.(Fig. 9.3(b) is adapted from Phase Diagramsof Binary Nickel Alloys, P. Nash (Ed.), ASMInternational, Materials Park, OH, 1991.)
BTB
DTD
tieline
4C3
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Chapter 9 - 99
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
=0At
TD: Only Solid
()
W
L= 0,W
= 100 wt%
Co = 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.)
Phase Diagrams:weight fractions of phases
wt%Ni
20
1200
13
00
T(C)
L
(liquid)
(solid)
L +
liquid
us
solid
us
30 40 50
L +
Cu-Nisystem
TAA
35Co
32CL
BTB
DTD
tieline
4C3
R S
AtT
B:Both
andL
%773377
3377 wt=
=
= 27 wt%
WL
= SR+ S
W
=R
R+ S
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Chapter 9 - 1010
Tie line connects the phases in equilibrium with eachother - essentially an isotherm
The Lever Rule
How much of each phase?Think of it as a lever (teeter-totter)
ML
M
R S
RMSM L =
L
L
LL
L
LCC
CC
SR
RW
CC
CC
SR
S
MM
MW
=
+
=
=
+
=
+
=
77
wt%Ni
20
1200
1300
T(C)
L
(liquid)
(solid)
L +
liquid
us
solid
us
30 40 50
L
+
BTB
tieline
CoCL C
SR
Adapted from Fig. 9.3(b),Callister 7e.
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Chapter 9 - 1111
wt%
Ni
20
120
0
130
0
30 40 50110
0
L(liquid)
(solid)
L+
L +
T(
C)
A
35Co
L: 35wt
%Ni Cu-Nisystem
Phase diagram:
Cu-Ni system.
System is:
--binaryi.e., 2 components:Cu and Ni.
--isomorphousi.e., completesolubility of onecomponent in
another; phase
field extends from0 to 100 wt% Ni.
Adapted from Fig. 9.4,
Callister 7e.
Consider
Co = 35 wt%Ni.
Ex: Cooling in a Cu-Ni Binary
46
35 4
3
3
2
: 43 wt%Ni
L: 32 wt%Ni
L: 24 wt%Ni: 36 wt%
Ni
B: 46 %
L: 35 wt%Ni
C
D
E
24
36
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Chapter 9 - 1212
Nonequilibrium Cooling
In application not sufficienttime for diffusionResult is cored structureSolidus curve shifts to higherNi contents
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Chapter 9 - 1313
C changes as we solidify. Cu-Ni case:
Fast rate of cooling:
Cored structure Slow rate of cooling:
Equilibrium structure
First to solidify has C = 46 wt%Ni.
Last to solidify has C = 35 wt%Ni.
Cored vs Equilibrium Phases
First
tosolidify:46 wt%
Ni
UniformC
:
35 wt%Ni
Last
tosolidify:< 35 wt%
Ni
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Chapter 9 - 1414
Mechanical Properties:Cu-Ni System
Effect of solid solution strengthening on:
--Tensile strength (TS) --Ductility (%EL,%AR)
--Peak as a function of Co --Min. as a function of Co
Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.
TensileStren
gth
(MPa)
Composition,wt% Ni
Cu
Ni
0 2
0
4
0
6
0
8
0
1
00
200
300
400 TS
forpure Ni
TS forpure Cu
Elongatio
n
(%EL)
Composition, wt% Ni
Cu
Ni
0 20
40
60
80
100
2
0
30
40
5
0
60
%
EL
for
pureNi
%EL
for pureCu
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Chapter 9 - 1515
: Min. meltingTE
2components
has a special composition
with a min. melting T.
Adapted from Fig. 9.7,Callister 7e.
Binary-Eutectic Systems
Eutectic transition
L(CE) (C E) + (C E)
3 single phaseregions
(L,
,
)
Limitedsolubility:: mostly
Cu : mostlyAg
TE: No liquid belowTE
CEcomposition
Ex.: Cu-Ag system
Cu-Agsystem
L (liquid)
L + L+
+
Co
,
wt% Ag20
40
60
80
100
0200
1200
T(C)
4
00
600
800
10
00
CE
TE
8.0
71.9
91.2
779C
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Chapter 9 - 1616
L+ L+
+
200
T(C)
18.3
C, wt%Sn
20
60
80
100
0
300
10
0
L
(liquid) 183C
61.9
97.8
For a 40 wt% Sn-60 wt% Pb alloy at 150C, find...
--the phases present: Pb-Snsystem
EX: Pb-Sn Eutectic System (1)
+ --compositions of phases:
CO = 40 wt% Sn
--the relative amountof each phase:
150
40Co
11C
99C
SR
C = 11 wt%SnC = 99 wt%
Sn
W=C -
COC -7
C
=99 -4099 -11
=5988
= 67 wt%
SR+S
=
W =CO -
CC
=R
R+S
=2
988
= 33 wt
%
=40 -
1199 -11
Adapted from Fig. 9.8,Callister 7e.
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Chapter 9 - 1717
L+
+
200
T(C)
C, wt%Sn
20
60
80
100
0
300
10
0
L
(liquid)
L+
183C
For a 40 wt% Sn-60 wt% Pb alloy at 220C, find...
--the phases present: Pb-Snsystem
Adapted from Fig. 9.8,Callister 7e.
EX: Pb-Sn Eutectic System (2)
+ L--compositions of phases:
CO = 40 wt% Sn
--the relative amountof each phase:
W=CL -COCL -
C7
=46 -4046 -17
= 629
= 21 wt%
WL =CO -
C7CL -C
=2329
= 79 wt%
40Co
46CL
17C
7
220
SR
C = 17 wt%SnCL = 46 wt%
Sn
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Chapter 9 - 1818
Co < 2 wt% Sn Result:
--at extreme ends--polycrystal of grains
i.e., only one solid phase.
Adapted from Fig. 9.11,Callister 7e.
Microstructuresin Eutectic Systems: I
0
L+ 200
T(C)
Co
,
wt%Sn
10
2
20C
o
300
1
00
L
30
+
400
(room T solubilitylimit)
TE
(Pb-SnSystem)
L
L: Co wt%Sn
: %
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Chapter 9 - 1919
2 wt% Sn < Co < 18.3 wt% Sn Result:
Initially liquid + then alone7
finally two phases polycrystal fine -phase
inclusions
Adapted from Fig. 9.12,Callister 7e.
Microstructuresin Eutectic Systems: II
Pb-Snsystem
L +
200
T(C)
Co
,
wt%Sn
10
18
.3
20
0Co
3
00
100
L
30
+
400
(sol. limit atTE)
TE
2(sol. limit atT
roo
m
)
L
L: Co wt%Sn
: %
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Chapter 9 - 2020
Co = CE Result: Eutectic microstructure (lamellar structure)--alternating layers (lamellae) of and crystals.
Adapted from Fig. 9.13,Callister 7e.
Microstructuresin Eutectic Systems: III
Adapted from Fig. 9.14, Callister 7e.
160
Micrograph of Pb-Sn
eutecticmicrostructure
Pb-Snsystem
L
+
+
200
T(C)
C, wt%Sn
20
60
80
10
0
0
300
100
L
L+
183C
40
TE
18
.3
: 18.3
%
97
.8
: 97.8 %
C
E61.9
L: Co wt%
Sn
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Chapter 9 - 2121
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister7e.
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Chapter 9 - 2222
18.3 wt% Sn < Co < 61.9 wt% Sn Result: crystals and a eutectic microstructure
Microstructuresin Eutectic Systems: IV
18
.3
61
.9
SR
97
.8
SR
primary
eutectic
eutectic
WL
=(1-
W) = 50 wt%
C = 18.3 wt%Sn
CL
= 61.9 wt%Sn S
R+
SW
== 50 wt%
Just aboveTE :
Just below
TE :C = 18.3 wt%SnC = 97.8 wt%Sn S
R+
SW
=
= 73 wt%
W = 27 wt
%Adapted from Fig. 9.16,
Callister 7e.
Pb-Snsystem
L+200
T(C)
Co, wt%Sn
20
60
80
10
0
0
300
100
L
L+
40
+
TE
L:Co
wt%Sn
L
L
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Chapter 9 - 2323
L+ L+
+
200
Co, wt%Sn
20
60
80
100
0
3
00
100
L
TE
40
(Pb-SnSystem)
Hypoeutectic & Hypereutectic
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
meutectic micro-constituentAdapted from Fig. 9.14,
Callister 7e.
hypereutectic: (illustrationonly)
Adapted from Fig. 9.17,Callister 7e. (Illustration only)
(Figs. 9.14 and 9.17from MetalsHandbook, 9th ed.,Vol. 9, Metallographyand Microstructures,
American Society forMetals, Materials Park,OH, 1985.)
175
hypoeutectic: Co = 50 wt%Sn
Adapted fromFig. 9.17, Callister 7e.
T(C)
61.9
eutectic
eutectic: Co=61.9wt%Sn
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Chapter 9 - 2424
An Example of Complex Binary Alloy Systems
Liquid
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Chapter 9 - 2525
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a line - not an area - becausestoichiometry (i.e. composition) is exact.
Adapted fromFig. 9.20, Callister 7e.
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Chapter 9 -
2626
Eutectoid & Peritectic
Eutectic - liquid in equilibrium with two solidsL +cool
heat
intermetallic compound -cementitecool
heat
Eutectoid - solid phase in equation with two solid phases
S2 S1+S3
+ Fe3C (727C)
cool
heat
Peritectic - liquid + solid 1 solid 2 (Fig 9.21)S1 + L S2
+ L (1493C)
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Chapter 9 -
2727
Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted fromFig. 9.21, Callister 7e.
Eutectoid transition +
Peritectic transition + L7 7
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Chapter 9 -
2828
Iron-Carbon (Fe-C) Phase Diagram
2 important
points
-Eutectoid
(B): + Fe3C
-Eutectic(A):L + Fe3
C
Adapted from Fig. 9.24,Callister 7e.
Fe3C
(cementite)
160014001200100080060
0400
0 1 2 3 4 5 6 6.7
L
(austeni
te)
+L
+Fe3C
+Fe
3C
+
L+Fe3C
(Fe)
Co, wt% C
1148C
T(
C)
727C= T
eutectoid
A
SR
4.30
Result: Pearlite =
alternating layers of
and Fe3C phases
120
(Adapted from Fig. 9.27, Callister 7e.)
R S
0.76C
eut
ect
oid
B
Fe3C (cementite-hard)
(ferrite-soft)
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Chapter 9 -
2929
Hypoeutectoid Steel
Adapted from Figs. 9.24
and 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
(cementite)
1600
1400
1200
10
008006004
00
0 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
+Fe3C
L+Fe3C
(Fe)
Co, wt%C
1148C
T(C)
727C
(Fe-CSystem)
C0 0 . 7 6
Adapted from Fig. 9.30,Callister 7e.
proeutectoid ferritepearlite
100
Hypoeutectoid ste
el
RS
w = S/(
R+ S)
wFe3C
= (1-
w)
w pearlite
=
wpearlite
r s
w =s /(r +s)
w = (1-
w)
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Chapter 9 -
3030
Hypereutectoid Steel
Fe3C
(cementi
te)
1600140012001
0008006004
00
0 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
+Fe
3C
L+Fe3C
(Fe)
Co, wt%C
1148C
T(C)
Adapted from Figs. 9.24
and 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-CSystem)
0.
7 6
Co
Adapted from Fig. 9.33,Callister 7e.
proeutectoid Fe3C
60
Hypereutectoid ste
elpearlite
R S
w =S/(
R+S)
wFe3C
= (1-
w)
wpearlite
=
wpearlite
sr
wFe3C
=r /(r+s)
w =(1-
w Fe3
C
)
Fe3C
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Chapter 9 -
3131
Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperaturejust below the eutectoid, determine thefollowing
a) composition of Fe3C and ferrite ( )
b) the amount of carbide (cementite) in gramsthat forms per 100 g of steel
c) the amount of pearlite and proeutectoidferrite ( )7
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Chapter 9 -
3232
Chapter 9 Phase EquilibriaSolution:
g7.77
g.77CFe
g7.7777333.77.7
777.77.7
777xCFeCFe
7
CFe7
7
7
=
=
=
=
=
+
x
CCCCo
b) the amount of carbide (cementite) ingrams that forms per 100 g of steel
a) composition of Fe3C and ferrite ( )
CO = 0.40 wt% C
C = 0.022 %
= 6.70 %
3
Fe3C(cem
entite)
16
0014001200100
0800600400
0 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3
C
+Fe3C
L+Fe3C
Co, wt% C
1148C
T(C)
727C
CO
R S
CFeC
3C
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Chapter 9 -
3333
Chapter 9 Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ( )
note: amount of pearlite = amount ofjust above TECo = 0.40 wt% C
C = 0.022 wt% CCpearlite = C = 0.76 wt% C
+=CoC
CC
x777=77.7g
pearlite = 51.2 g proeutectoid = 48.8 g
Fe3C(cem
entite)
16
0014001200100
0800600400
0 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3
C
+Fe3C
L+Fe3C
Co, wt% C
1148C
T(C)
727C
CO
RS
CC
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Chapter 9 -
3434
Alloying Steel with More Elements
Teutectoidchanges:
Ceutectoidchanges:
Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34 fromEdgar C. Bain, Functions of the Alloying Elements inSteel, American Society for Metals, 1939, p. 127.)
Adapted from Fig. 9.35,Callister 7e. (Fig. 9.35 fromEdgar C. Bain, Functions of the Alloying Elements inSteel, American Society for Metals, 1939, p. 127.)
TEute
cto
id
(C)
wt. % of alloying
elements
Ti
Ni
Mo
Si W
CrMn
wt. % of alloying
elements
Ceute
ctoi
d
(wt
%C)
Ni
Ti
Cr
Si M
nW
Mo
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Chapter 9 -
3535
Phase diagrams are useful tools to determine:
--the number and types of phases,--the wt% of each phase,--and the composition of eachphasefor a given T and composition of the system.
Alloying to produce a solid solution usually
--increases the tensile strength (TS)--decreases the ductility.
Binary eutectics and binary eutectoids allow fora range of microstructures.
Summary
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Chapter 9
3636
CoreProblems:
Self-helpProblems:
ANNOUNCEMENTS
Reading: