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8/6/2019 Ch09 Lecture Note
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Chapter 9 - 1
ISSUES TO ADDRESS... When we combine two elements...
what equilibrium state do we get? 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 atomCopper atom
Chapter 9 - 2
Phase Equilibria: Solubility LimitIntroduction
Solutions solid solutions, single phase Mixtures more than one phase
Solubility Limit :Max concentration forwhich only a single phasesolution occurs.
Question: What is thesolubility limit at 20 C?
Answer: 65 wt% sugar .If C o < 65 wt% sugar: syrupIf C o > 65 wt% sugar: syrup + sugar.
65
Sucrose/Water Phase Diagram
P u r e
S u g a r
T e m p e r a t u r e
( C )
0 20 40 60 80 100C o =Composition (wt% sugar)
L(liquid solution
i.e., syrup)
SolubilityLimit L
(liquid)+S
(solidsugar)20
40
60
80100
P u r e
W a t e r
Adapted from Fig. 9.1,Callister 7e.
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Chapter 9 - 3
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-CopperAlloy
Components and Phases
(darkerphase)
(lighterphase)
Adapted fromchapter-openingphotograph,Chapter 9,Callister 3e.
Chapter 9 - 4
Effect of T & Composition ( C o ) Changing T can change # of phases:
Adapted fromFig. 9.1,Callister 7e.
D (100 C,90)2 phases
B (100 C,70)1 phase
path A to B . Changing C o can change # of phases: path B to D .
A (20 C,70)2 phases
70 80 1006040200
T e m p e r a
t u r e
( C )
C o =Composition (wt% sugar)
L(liquid solution
i.e., syrup)
20
100
40
60
80
0
L(liquid)+S
(solidsugar)
water-sugarsystem
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Chapter 9 - 5
Phase Equilibria
0.12781.8FCCCu
0.12461.9FCCNi
r (nm)electronegCrystalStructure
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.
Chapter 9 - 6
Phase Diagrams Indicate phases as function of T , C o , and P . For this course:
-binary systems: just 2 components.-independent variables: T and C o (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 Phase Diagrams of Binary Nickel Alloys , P. Nash(Ed.), ASM International, Materials Park,OH (1991).
2 phases:L (liquid) (FCC solid solution)
3 phase fields:LL +
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T (C)
L (liquid)
(FCC solidsolution)
L + l i q u
i d u s
s o l i d u
s
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Chapter 9 - 7
Note:The Ni-Cu alloy system shown in the previous slide is a binary isomorphous system;
Note: Isomorphous : Having the same structure. In the phase diagram sense,isomorphicity means having the same crystal structure or complete solid solubility for all compositions.
Chapter 9 - 8
wt% Ni20 40 60 80 10001000
1100
1200
1300
14001500
1600T (C)
L (liquid)
(FCC solidsolution)
L +
l i q u i d u s
s o l i d u
sCu-Niphase
diagram
Phase Diagrams :# and types of phases
Rule 1: If we know T and C o , then we know:--the # and types of phases present.
Examples:A(1100 C, 60):
1 phase: B (1250 C, 35):
2 phases: L +
Adapted from Fig. 9.3(a), Callister 7e.(Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys , P. Nash(Ed.), ASM International, Materials Park,OH, 1991).
B ( 1 2 5 0 C
, 3 5 )
A (1100
C,60)
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Chapter 9 - 9
wt% Ni20
1200
1300
T (C)
L (liquid)
(solid) L
+
l i q u i d u s
s o l i d u s
30 40 50
L +
Cu-Nisystem
Phase Diagrams :composition of phases
Rule 2: If we know T and C o , then we know:--the composition of each phase.
Examples:T A A
35C o32C L
At T A = 1320 C:Only Liquid ( L)C L = Co ( = 35 wt% Ni)
At T B = 1250 C:Both and LC L = C liquidus ( = 32 wt% Ni here)C = C solidus ( = 43 wt% Ni here)
At T D = 1190 C:Only Solid ( )C = C o ( = 35 wt% Ni )
C o = 35 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.)
B T B
D T D
tie line
4C 3
Chapter 9 - 10
Rule 3: If we know T and C o , then we know:--the amount of each phase (given in wt%).
Examples:
At T A: Only Liquid (L)W L = 100 wt%, W = 0
At T D : Only Solid ( )W L = 0, W = 100 wt%
C o = 35 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.), ASMInternational, Materials Park, OH, 1991.)
Phase Diagrams :weight fractions of phases
wt% Ni20
1200
1300
T (C)
L (liquid)
(solid) L
+
l i q u i d u s
s o l i d u s
30 40 50
L +
Cu-Nisystem
T A A
35C o
32C L
B T B
D T D
tie line
4C 3
R S
At T B : Both and L
%7332433543 wt =
=
= 27 wt%
W L =S
R + S
W =R
R + S
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Chapter 9 - 11
Tie line connects the phases in equilibrium witheach other - essentially an isotherm
The Lever Rule
How much of each phase?Think of it as a lever (teeter-totter)
M L M
R S
R M S M L=
L
L
LL
LL C C
C C S R
R W
C C C C
S R S
M M M
W
=+
=
=+
=+
=
00
wt% Ni
20
1200
1300
T (C)
L (liquid)
(solid) L
+
l i q u i d u s
s o l i d u s
30 40 50
L +
B T B
tie line
C o C L C
S R
Adapted from Fig. 9.3(b),Callister 7e.
Chapter 9 - 12
wt% Ni20
1200
1300
30 40 501100
L (liquid)
(solid)
L +
L +
T (C)
A
35C o
L: 35wt%Ni
Cu-Nisystem
Phase diagram:Cu-Ni system.
System is:--binary
i.e. , 2 components:
Cu and Ni.--isomorphousi.e., completesolubility of onecomponent inanother; phasefield extends from0 to 100 wt% Ni.
Adapted from Fig. 9.4,Callister 7e.
ConsiderC o = 35 wt%Ni .
Ex: Cooling in a Cu-Ni Binary
4635
4332
: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
: 36 wt% Ni
B : 46 wt% NiL: 35 wt% Ni
C
D
E
24 36
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Chapter 9 - 13
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 to solidify:46 wt% Ni
Uniform C :35 wt% Ni
Last to solidify:< 35 wt% Ni
Chapter 9 - 14
Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on:
--Tensile strength ( TS ) --Ductility (% EL,% AR )
--Peak as a function of C o --Min. as a function of C o
Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e.
T e n s
i l e S t r e n g t h
( M P a )
Composition, wt% NiCu Ni0 20 40 60 80 100
200
300
400
TS forpure Ni
TS for pure Cu E l o n g a
t i o n
( % E L )
Composition, wt% NiCu Ni0 20 40 60 80 10020
30
40
50
60
%EL for
pure Ni
%EL for pure Cu
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Chapter 9 - 15
: Min. melting T E
2 componentshas a special compositionwith a min. melting T.
Adapted from Fig. 9.7,Callister 7e.
Binary-Eutectic Systems
Eutectic transitionL(C E ) (C E ) + (C E )
3 single phase regions(L, , )
Limited solubility: : mostly Cu: mostly Ag
T E : No liquid below T E C E
composition
Ex.: Cu-Ag systemCu-Agsystem
L (liquid)
L + L+
+
C o , wt% Ag20 40 60 80 1000
200
1200T (C)
400
600
800
1000
C E
T E 8.0 71.9 91.2779 C
Chapter 9 - 16
L+ L+
+
200
T (C)
18.3
C , wt% Sn20 60 80 1000
300
100
L (liquid)
183 C61.9 97.8
For a 40 wt% Sn-60 wt% Pb alloy at 150 C, find...--the phases present: Pb-Sn
system
EX: Pb-Sn Eutectic System (1)
+ --compositions of phases:
C O = 40 wt% Sn
--the relative amountof each phase:
150
40C o
11C
99C
S R
C = 11 wt% SnC = 99 wt% Sn
W =C - C O C - C
= 99 - 4099 - 11 =5988 = 67 wt%
S R +S =
W =C O - C C - C
=R R +S
= 2988
= 33 wt%= 40 - 1199 - 11
Adapted from Fig. 9.8,Callister 7e.
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Chapter 9 - 17
L+
+
200
T (C)
C , wt% Sn20 60 80 1000
300
100
L (liquid)
L+
183 C
For a 40 wt% Sn-60 wt% Pb alloy at 200 C, find...--the phases present: Pb-Sn
system
Adapted from Fig. 9.8,Callister 7e.
EX: Pb-Sn Eutectic System (2)
+ L--compositions of phases:
C O = 40 wt% Sn
--the relative amountof each phase:
W =C L - C O C L - C
=46 - 4046 - 17
=6
29 = 21 wt%
W L =C O - C C L - C
=2329 = 79 wt%
40 C o
46C L
17C
220S R
C = 17 wt% SnC L = 46 wt% Sn
Chapter 9 - 18
C o < 2 wt% Sn Result:
--at extreme ends
--polycrystal of grainsi.e., only one solid phase.
Adapted from Fig. 9.11,Callister 7e.
Microstructuresin Eutectic Systems: I
0
L+ 200
T (C)
C o , wt% Sn10
2
20C o
300
100
L
30
+
400
(room T solubility limit)
T E (Pb-SnSystem)
L
L: C o wt% Sn
: C o wt% Sn
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Chapter 9 - 19
2 wt% Sn < C o < 18.3 wt% Sn Result:
Initially liquid + then alonefinally two phases
polycrystalfine -phase inclusions
Adapted from Fig. 9.12,Callister 7e.
Microstructures
in Eutectic Systems: II
Pb-Snsystem
L +
200
T (C)
C o , wt% Sn10
18.3
200C o
300
100
L
30
+
400
(sol. limit at T E )
T E
2(sol. limit at T room )
L
L: C o wt% Sn
: C o wt% Sn
Chapter 9 - 20
C o = C E 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 m
Micrograph of Pb-Sneutectic
microstructurePb-Snsystem
L +
+
200
T (C)
C , wt% Sn
20 60 80 1000
300
100
L
L+183 C
40
T E
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
C E 61.9
L: C o wt% Sn
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Chapter 9 - 21
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister 7e.
Chapter 9 - 22
18.3 wt% Sn < C o < 61.9 wt% Sn Result: crystals and a eutectic microstructure
Microstructuresin Eutectic Systems: IV
18.3 61.9
S R
97.8
S R
primary eutectic
eutectic
W L = (1- W ) = 50 wt%
C = 18.3 wt% Sn
C L = 61.9 wt% SnS
R + S W = = 50 wt%
Just above T E :
Just below T E :C = 18.3 wt% SnC = 97.8 wt% Sn
S R + S
W = = 73 wt%
W = 27 wt%Adapted from Fig. 9.16,Callister 7e.
Pb-Snsystem
L+200
T (C)
C o , wt% Sn
20 60 80 1000
300
100
L
L+
40
+
T E
L: C o wt% Sn LL
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Chapter 9 - 23
L+ L+
+
200
C o , wt% Sn20 60 80 1000
300
100
L
T E
40
(Pb-SnSystem)
Hypo eutectic & Hyper eutectic
Adapted from Fig. 9.8,Callister 7e. (Fig. 9.8adapted from Binary Phase Diagrams , 2nd ed., Vol. 3,T.B. Massalski (Editor-in-Chief), ASM International,Materials Park, OH, 1990.)
160 meutectic micro-constituent
Adapted from Fig. 9.14,Callister 7e.
hypereutectic: (illustration only)
Adapted from Fig. 9.17,Callister 7e. (Illustrationonly)
(Figs. 9.14 and 9.17from Metals Handbook , 9th ed.,Vol. 9,Metallography and
Microstructures ,American Society forMetals, MaterialsPark, OH, 1985.)
175 m
hypoeutectic: C o = 50 wt% Sn
Adapted fromFig. 9.17, Callister 7e.
T (C)
61.9eutectic
eutectic: C o =61.9wt% Sn
Chapter 9 - 24
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a line - not an area -because stoichiometry (i.e. composition) is exact.
Adapted fromFig. 9.20, Callister 7e.
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Chapter 9 - 25
Eutectoid & Peritectic Eutectic( ) - liquid in equilibrium with two solids
L + coolheat
intermetallic compound- cementite
coolheat
Eutectoid( ) - solid phase in equation with twosolid phasesS 2 S 1+S 3
+ Fe 3C (727C)
coolheat
Peritectic( ) - liquid + solid 1 solid 2 (Fig 9.21)
S 1 + L S 2 + L (1493C)
Chapter 9 - 26
Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted fromFig. 9.21, Callister 7e.
Eutectoid transition +
Peritectic transition + L
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Chapter 9 - 27
Iron-Carbon (Fe-C) Phase Diagram 2 important
points
-Eutectoid ( B ): + Fe 3C
-Eutectic ( A):L + Fe 3C
Adapted from Fig. 9.24, Callister 7e .
F e 3
C ( c e m e n
t i t e )
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe 3C
+Fe 3C
+
L+Fe 3C
(Fe) C o , wt% C
1148 C
T (C)
727 C = T eutectoid
AS R
4.30Result: Pearlite =alternating layers of and Fe 3C phases
120 m
(Adapted from Fig. 9.27, Callister 7e .)
R S
0.76
C e u
t e c t o i
d
B
Fe 3C (cementite-hard) (ferrite-soft)
Chapter 9 - 28
Hypo eutectoid Steel
Adapted from Figs. 9.24and 9.29, Callister 7e .(Fig. 9.24 adapted fromBinary Alloy Phase Diagrams , 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)
F e 3
C ( c e m e n t
i t e )
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe 3C
+ Fe 3C
L+Fe 3C
(Fe) C o , wt% C
1148 C
T (C)
727 C
(Fe-CSystem)
C 0 0
. 7 6
Adapted from Fig. 9.30, Callister 7e .proeutectoid ferritepearlite
100 m Hypoeutectoidsteel
R S
w =S /(R +S )w Fe 3C =(1- w )
w pearlite = w pearlite
r s
w =s /(r +s )w =(1- w )
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Chapter 9 - 29
Hyper eutectoid Steel
F e 3
C ( c e m e n t
i t e )
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe 3C
+Fe 3C
L+Fe 3C
(Fe) C o , wt%C
1148 C
T (C)
Adapted from Figs. 9.24and 9.32, Callister 7e .(Fig. 9.24 adapted fromBinary Alloy Phase Diagrams , 2nd ed., Vol.1, T.B. Massalski (Ed.-in-Chief), ASM International,Materials Park, OH,1990.)
(Fe-CSystem)
0 . 7 6 C o
Adapted from Fig. 9.33, Callister 7e .
proeutectoid Fe 3C
60 mHypereutectoidsteel
pearlite
R S
w =S /(R +S )w Fe 3C =(1- w )
w pearlite = w pearlite
s r
w Fe 3C =r /( r +s )w =(1- w Fe 3C )
Fe 3C
Chapter 9 - 30
Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine thefollowing
a) composition of Fe 3C and ferrite ( )b) the amount of carbide (cementite) in grams
that forms per 100 g of steelc) the amount of pearlite and proeutectoid
ferrite ( )
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Chapter 9 - 31
Chapter 9 Phase EquilibriaSolution:
g3.94g5.7CFe
g7.5100022.07.6022.04.0
100xCFe
CFe
3
CFe3
3
3
=
=
=
=
=+
x
C C C C o
b) the amount of carbide(cementite) in grams thatforms per 100 g of steel
a) composition of Fe 3C and ferrite ( )C O = 0.40 wt% CC = 0.022 wt% CC Fe C = 6.70 wt% C3
F e 3
C ( c e m e n
t i t e )
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe 3C
+ Fe 3C
L+Fe 3C
C o , wt% C
1148 CT (C)
727C
C O
R S
C Fe C3C
Chapter 9 - 32
Chapter 9 Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ( )
note: amount of pearlite = amount of just above T E
C o = 0.40 wt% CC = 0.022 wt% CC
pearlite= C
= 0.76 wt% C
+
=C o C C C
x 100 = 51.2 g
pearlite = 51.2 gproeutectoid = 48.8 g
F e 3
C ( c e m e n
t i t e )
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe 3C
+ Fe 3C
L+Fe 3C
C o , wt% C
1148 CT (C)
727 C
C O
R S
C
C
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Chapter 9 - 33
Alloying Steel with More Elements
T eutectoid changes: C eutectoid changes:
Adapted from Fig. 9.34, Callister 7e . (Fig. 9.34from Edgar C. Bain, Functions of the Alloying Elements in Steel , American Society for Metals,1939, p. 127.)
Adapted from Fig. 9.35, Callister 7e . (Fig. 9.35from Edgar C. Bain, Functions of the Alloying Elements in Steel , American Society for Metals,1939, p. 127.)
T E u t e c
t o i d ( C )
wt. % of alloying elements
Ti
Ni
Mo Si W
Cr
Mn
wt. % of alloying elements
C e u
t e c t o i
d ( w t % C )
Ni
Ti
Cr
SiMn
WMo
Chapter 9 - 34
Phase diagrams are useful tools to determine:--the number and types of phases,--the wt% of each phase,--and the composition of each phase
for 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 - 35
Core Problems:
Self-help Problems:
ANNOUNCEMENTSReading: