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 for Metallic Systems Phase B Phase A Nickel atom Copper atom
Chapter 9: Phase Diagrams for Metallic Systems. 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... - PowerPoint PPT Presentation
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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 for Metallic Systems
Phase BPhase A
Nickel atomCopper atom
Phase Equilibria: Solubility LimitIntroduction
– Solutions – solid solutions, single phase– Mixtures – more than one phase
• Solubility Limit: Max concentration for which only a single phase solution occurs.
Question: What is the solubility limit at 20°C?
Answer: 65 wt% sugar.
If Co < 65 wt% sugar: syrup
If Co > 65 wt% sugar: syrup + sugar.
65
Sucrose/Water Phase Diagram
Pu
re
Su
gar
Tem
per
atu
re (
°C)
0 20 40 60 80 100Co =Composition (wt% sugar)
L (liquid solution
i.e., syrup)
Solubility Limit L
(liquid)
+ S
(solid sugar)20
40
60
80
100
Pu
re
Wat
er
• Components: The elements or compounds which are present in the mixture (e.g., Al and Cu)
• Phases: The physically and chemically distinct material regions that result (e.g., and ).
Aluminum-CopperAlloy
Components and Phases
(darker phase)
(lighter phase)
Effect of T & Composition (Co)• Changing T can change # of phases: path A to B.• Changing Co can change # of phases: path B to D.
D (100°C,90)2 phases
B (100°C,70)1 phase
A (20°C,70)2 phases
70 80 1006040200
Tem
pera
ture
(°C
)
Co =Composition (wt% sugar)
L (liquid solution
i.e., syrup)
20
100
40
60
80
0
L (liquid)
+ S
(solid sugar)
water-sugarsystem
Phase Equilibria
CrystalStructure
electro negativity
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.
• Ni and Cu are totally miscible in all proportions.
Simple solution system (e.g., Ni-Cu solution)
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
• 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 solid solution)
L + liquidus
solid
us
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(1100°C, 60): 1 phase:
B(1250°C, 35): 2 phases: L +
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
(FCC solid solution)
L +
liquidus
solid
us
Cu-Niphase
diagram
B (
1250
°C,3
5)
A(1100°C,60)
Phase Diagrams:composition of phases
• Rule 2: If we know T and Co, then we know: --the composition of each phase.
• Examples:
At TA = 1320°C:
Only Liquid (L) CL = Co ( = 35 wt% Ni)
At TB = 1250°C:
Both and L CL = C liquidus ( = 32 wt% Ni here)
C = C solidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( ) C = Co ( = 35 wt% Ni)
Co = 35 wt% Ni
wt% Ni
20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L +
Cu-Ni system
TAA
35Co
32CL
BTB
DTD
tie line
4C3
• 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
Phase Diagrams:weight fractions of phases
wt% Ni
20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L +
Cu-Ni system
TAA
35Co
32CL
BTB
DTD
tie line
4C3
R SAt TB: Both and L
% 733243
3543wt
= 27 wt%
WL S
R +S
W R
R +S
• Tie line – connects the phases in equilibrium with each other - essentially an isotherm
The Lever Rule
How much of each phase? Think of it as a lever
ML M
R S
RMSM L
L
L
LL
LL CC
CC
SR
RW
CC
CC
SR
S
MM
MW
00
wt% Ni
20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L + B
TB
tie line
CoCL C
SR
• 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; phase field extends from 0 to 100 wt% Ni.
• Consider Co = 35 wt%Ni.
Cooling in a Cu-Ni Binary
wt% Ni20
1200
1300
30 40 501100
L (liquid)
(solid)
L +
L +
T(°C)
A
35Co
L: 35wt%Ni
Cu-Nisystem
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
• 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
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
Te
nsile
Str
en
gth
(M
Pa
)
Composition, wt% NiCu Ni0 20 40 60 80 100
200
300
400
TS for pure Ni
TS for pure Cu
Elo
ng
atio
n (
%E
L)
Composition, wt% NiCu Ni0 20 40 60 80 10020
30
40
50
60
%EL for pure Ni
%EL for pure Cu
: Min. melting TE
Binary-Eutectic Systems
• Eutectic transition L(CE) (CE) + (CE)
• 3 single phase regions (L, ) • Limited solubility: : mostly Cu : mostly Ag • TE : No liquid below TE
• CE
composition
Ex.: Cu-Ag system
Co , wt% Ag
Cu-Agsystem
L (liquid)
L + L+
20 40 60 80 1000200
1200T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2779°C
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present:
EX: Pb-Sn Eutectic System (1)
+ --compositions of phases:
CO = 40 wt% Sn
--the relative amount of each phase:
L+L+
+
200
T(°C)
18.3
C, wt% Sn20 60 80 1000
300
100
L (liquid)
183°C 61.9 97.8
Pb-Snsystem
150
40Co
11C
99C
SR
C = 11 wt% SnC = 99 wt% Sn
W=C - CO
C - C
=99 - 4099 - 11
= 5988
= 67 wt%
SR+S
=
W =CO - C
C - C=R
R+S
=2988
= 33 wt%=40 - 1199 - 11
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find... --the phases present:
EX: Pb-Sn Eutectic System (2)
+ L--compositions of phases:
CO = 40 wt% Sn
--the relative amount of each phase:
W =CL - CO
CL - C=
46 - 40
46 - 17
= 6
29= 21 wt%
WL =CO - C
CL - C=
23
29= 79 wt%
L+
+
200
T(°C)
C, wt% Sn20 60 80 1000
300
100
L (liquid)
L+
183°C
Pb-Snsystem
40Co
46CL
17C
220SR
C = 17 wt% SnCL = 46 wt% Sn
• Co < 2 wt% Sn• Result: --at extreme ends --polycrystal of grains i.e., only one solid phase.
Microstructures in Eutectic Systems: I
0
L + 200
T(°C)
Co , wt% Sn10
2
20Co
300
100
L
30
+
400
(room T solubility limit)
TE
(Pb-SnSystem)
L
L: Co wt% Sn
: Co wt% Sn
• 2 wt% Sn < Co < 18.3 wt% Sn• Result:
Initially liquid + then alonefinally two phases
polycrystal fine -phase inclusions
Microstructures in Eutectic Systems: II
Pb-Snsystem
L +
200
T(°C)
Co , wt% Sn10
18.3
200Co
300
100
L
30
+
400
(sol. limit at TE)
TE
2(sol. limit at Troom)
L
L: Co wt% Sn
: Co wt% Sn
• Co = CE • Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of and crystals.
Microstructures in Eutectic Systems: III
160 m
Micrograph of Pb-Sn eutectic microstructurePb-Sn
systemL
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
L+ 183°C
40
TE
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
CE61.9
L: Co wt% Sn
Lamellar Eutectic Structure
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: crystals and a eutectic microstructure
Microstructures in Eutectic Systems: IV
WL = (1-W) = 50 wt%
C = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SW = = 50 wt%
• Just above TE :
• Just below TE :C = 18.3 wt% Sn
C = 97.8 wt% SnS
R + SW = = 73 wt%
W = 27 wt%18.3 61.9
SR
97.8
SR
primary eutectic
eutectic
Pb-Snsystem
L+200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
L+
40
+
TE
L: Co wt% Sn LL
L+L+
+
200
Co, wt% Sn20 60 80 1000
300
100
L
TE
40
(Pb-Sn System)
Hypoeutectic & Hypereutectic
hypereutectic: (illustration only)
175 m
hypoeutectic: Co = 50 wt% Sn
T(°C)
61.9eutectic
160 m
eutectic micro-constituent
eutectic: Co = 61.9 wt% Sn
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.
Eutectoid & Peritectic
• Eutectic - liquid in equilibrium with two solidsL + cool
heat
intermetallic compound - cementite
cool
heat
• Eutectoid - solid phase in equation with two solid phases
S2 S1+S3
+ Fe3C (727ºC)
cool
heat
• Peritectic - liquid + solid 1 solid 2
S1 + L S2
+ L (1493ºC)
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition +
Peritectic transition + L
Iron-Carbon (Fe-C) Phase Diagram
• 2 important points
-Eutectoid (B):
+Fe3C
-Eutectic (A): L + Fe3C
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
+Fe3C
+
L+Fe3C
(Fe) Co, wt% C
1148°C
T(°C)
727°C = Teutectoid
A
SR
4.30
Result: Pearlite = alternating layers of and Fe3C phases
120 m
R S
0.76
Ceu
tect
oid
B
Fe3C (cementite-hard) (ferrite-soft)
Chapter 9 - 27
Ferrite and Austenite
Ferrite Austenite
Hypoeutectoid Steel
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe3C
+ Fe3C
L+Fe3C
(Fe) Co , wt% C
1148°C
T(°C)
727°C
(Fe-C System)
C0
0.76
proeutectoid ferritepearlite
100 m Hypoeutectoidsteel
R S
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
r s
w =s/(r+s)w =(1- w)
Hypereutectoid Steel
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+Fe3C
+Fe3C
L+Fe3C
(Fe) Co , wt%C
1148°C
T(°C)
(Fe-C System)
0.7
6 Co
proeutectoid Fe3C
60 mHypereutectoid steel
pearlite
R S
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
sr
wFe3C =r/(r+s)w =(1-w Fe3C )
Fe3C
Example: Phase Equilibria
For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following
a) composition of Fe3C and ferrite ()
b) the amount of carbide (cementite) in grams that forms per 100 g of steel
c) the amount of pearlite and proeutectoid ferrite ()
Chapter 9 – Phase EquilibriaSolution:
g 3.94
g 5.7 CFe
g7.5100 022.07.6
022.04.0
100xCFe
CFe
3
CFe3
3
3
x
CC
CCo
b) the amount of carbide (cementite) in grams that forms per 100 g of steel
a) composition of Fe3C and ferrite ()
CO = 0.40 wt% C
C = 0.022 wt% C
CFe C = 6.70 wt% C3
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe3C
+ Fe3C
L+Fe3C
Co , wt% C
1148°C
T(°C)
727°C
CO
R S
CFe C3C
Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ()
note: amount of pearlite = amount of just above TE
Co = 0.40 wt% C
C = 0.022 wt% C
Cpearlite = C = 0.76 wt% C
Co CC C
x 100 51.2 g
pearlite = 51.2 gproeutectoid = 48.8 g
Fe 3
C (
cem
entit
e)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
(austenite)
+L
+ Fe3C
+ Fe3C
L+Fe3C
Co , wt% C
1148°C
T(°C)
727°C
CO
R S
CC
Alloying Steel with More Elements
• Teutectoid changes: • Ceutectoid changes:
TE
ute
cto
id (
°C)
wt. % of alloying elements
Ti
Ni
Mo SiW
Cr
Mn
wt. % of alloying elementsC
eut
ect
oid
(w
t%C
)
Ni
Ti
Cr
SiMn
WMo
• 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 for a range of microstructures.