chapter 9: phase diagrams
DESCRIPTION
Chapter 9: Phase Diagrams. Phase b ( A and B atoms). Phase a ( A and B atoms). Nickel atom. Copper atom. ISSUES TO ADDRESS. • When we mix two elements ... what equilibrium state do we get?. Structure Properties Processing. • In particular, if we specify... - PowerPoint PPT PresentationTRANSCRIPT
Chapter 9 - 1
ISSUES TO ADDRESS...• When we mix 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 (relative amount)?
Chapter 9: Phase Diagrams
Phase (A and B atoms)
Phase (A and B atoms)
Nickel atomCopper atom
StructurePropertiesProcessing
Chapter 9 - 2
Phase Equilibria: Review: Solubility Limit
Introduction
- Solutions – solid solutions, single phase (e.g. syrup)
- Mixtures – more than one phase (syrup and solid sugar)
Phase: Homogeneous portion of a system that has uniform physical and chemical properties
Solubility Limit:Max concentration for which only a single phase solution occurs.
Q: 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
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 regions that result (e.g., and ).•EquilibriumNo change with time – fixed composition for each component
Al-Cu Alloy
Components and Phases
(darker phase)Includes Al and Cu
(lighter phase)Includes Al and Cu
Two phasesTwo Components
Chapter 9 - 4
Effect of T & Composition (Co)• Changing T can change # of phases:
D (100°C,90)2 phases
B (100°C,70)1 phase
path A to B.• Changing Co can change # of phases: path B to D.
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
Chapter 9 - 5
Phase Diagrams
Types of phase diagrams depend on Solubility • Isomorphous phase diagram (complete solubility) • Eutectic phase diagram (partial solubility) • Complex phase diagram (…….) How to obtain phase diagrams ?- From cooling curves for various starting solutions
(Thermal method)
• Thermodynamics: 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).
Phase Diagram: A Plot of T versus C indicating various boundaries between phases
T
Time
? Phase Change
T
composition
Uses of Phase Diagrams: - No. of phases - Composition of each phase- Relative amount of each phase+ Predict the microstructure
Chapter 9 - 6
Phase Equilibria - Solutions
CrystalStructure
electroneg r (nm)
Ni FCC 1.9 0.1246
Cu FCC 1.8 0.1278
Both metals have •Nearly the same atomic radius•The same crystal structure (FCC) •similar electronegativities (W. Hume – Rothery rules) suggesting high mutual solubility.
Simple Solution System (e.g., Ni-Cu solution)Remember
Ni and Cu are totally miscible (soluble) in all proportions (compositions).
ONE PHASE ALWAYS – Isomorphous !
Chapter 9 - 7
Phase Diagrams- Metals
• 2 phases: L (liquid)
(FCC solid solution)
• 3 phase fields: LL +
Phase Diagram for Cu-Ni system
20 40 60 80 1000 wt% Ni1000
1100
1200
1300
1400
1500
1600
T(°C)
L (liquid)
(FCC solid Soln)
L + Liquidus line
Solidus lin
e
Isomorphous Phase Diagram Complete solubility - one phase field extends from 0 to 100 wt% Ni.
Chapter 9 - 8
Phase Change
• Phase boundary:Lines between different phases e.g. – Liquid – solid
• one phase • two phases …etc)
• Phase Change: – Solidification (or melting) – Phase Reaction: one phase gives DIRECTLY two phases e.g.
Liquid Solid phase 1 + Solid Phase 2 (both insoluble)• This occurs at
– Certain (fixed) temperature: TE
– Certain (fixed) composition: CE
• Eutectic Phase Reaction:L
• There are other types of phase reactions (Later)
Chapter 9 - 9
Cooling Curves
There are various types of cooling curves:
Depending on composition (characteristics)
1. Pure Element
2. Mixture of soluble 2 elements
3. Mixture of Partially Soluble 2 elements
4. Mixtures undergoing Phase Reactions
1. At the composition for phase reaction
2. Before the composition for phase reaction
Chapter 9 - 10
Types of Cooling Curves
2) Mixture of soluble 2 elements
Time
L
L
T
3) Phase Reaction at CE 3) Phase Reaction: Co > CE
Time
L
L
T
L
T
Time
L
L A or B)
1) Pure Element
Tmelting
Time
L
L
T
TE
Chapter 9 - 11
Constructing Phase Diagrams
Thermal Method• Start with several different solutions
• Each solution at certain co
• Prepare the solutions in liquid form (melting)• Cool each solution while recording T versus time• Plot the cooling curve for each solution • At each phase change: there are changes in the slope
– Latent Heat – Change in the properties (specific heat)– Phase reaction (later)
• Project cooling curves on T versus co plot.
• Connect the points from projection creating boundaries. • Label different obtained areas and boundaries.
Chapter 9 - 12
Constructing Isomorphous Phase Diagrams
Cooling Curves at Various Initial Co Phase Diagram T verus Co
T
CompositionTime
Chapter 9 - 13
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
Using Phase Diagrams:# and types of phases
Rule 1: If we know T and Co, then we know: - Number and types of phases present.
• Examples:A(1100°C, 60): 1 phase:
B(1250°C, 35): 2 phases: L +
B (
1250
°C,3
5) A(1100°C,60)
Chapter 9 - 14
wt% Ni20
1200
1300
T(°C)
L (liquid)
(solid)L +
liquidus
solidus
30 40 50
L +
Cu-Ni system
Using Phase Diagrams:composition of phases
Rule 2: If we know T and Co, then we know: - the composition of each phase.
• Examples:TA
A
35Co
32CL
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
BTB
DTD
tie line
4C3
Chapter 9 - 15
Rule 3: If we know T and Co, then we know: --the relative amount of each phase (given in wt%).
• Examples:
At TA: Only Liquid (L)
W L = 1.0 , W = = 0At TD: Only Solid ( )
W L = 0, W = 1.0
Co = 35 wt% Ni
Using Phase Diagrams:weight fractions of phases
wt% Ni20
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 S
At TB: Both and L
73.03243
3543
= 0.27
WL S
R +S
W R
R +S
Chapter 9 - 16
• Tie line – connects the phases in equilibrium with each other - essentially an isotherm
The Lever Rule
How much of each phase? ThinkThink of it as a lever (teeter-totter)
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
Chapter 9 - 17
wt% Ni20
1200
1300
30 40 501100
L (liquid)
(solid)
L +
L +
T(°C)
A
35Co
L: 35wt%Ni
Cu-Nisystem
Observe the microstructure
while going down (cooling)
at the same composition.
• Consider Co = 35 wt%Ni.
Microstructure - 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
Chapter 9 - 18
• 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 - 19
Mechanical Properties: Cu-Ni System
• Effect of solid solution strengthening on:
--Tensile strength (TS)
- Peak 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
--Ductility (%EL,%AR)
- Min. as a function of Co
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
Chapter 9 - 20
Eutectic Phase DiagramWith Eutectic Phase Reaction
L(CE) (CE) + (CE)The phase diagram is characterized with:
• 3 single phase regions (L, and )
• Partial Solubility – Solubility Limit of A in B forming B in A forming Occurs at
L (liquid)
L + L+
Wt% A: Co
1000
T
CE
TE
: Eutectic Composition • CE
• TE : Eutectic Temperature
Remember Phase Reaction at CE
Time
L
L
T
TE
Chapter 9 - 21
Eutectic Phase Diagram – Cooling Curves
L
L + L+
Wt% A: Co
1000
T(°C)
T
Time
L
L
1) Pure A or B
Tmelting
2) Giving or
Time
L
L
T
Time
L
L
T
L
Time
L
L
T
TE
What is the difference in this side
Chapter 9 - 22
Maximum solubility of A in B
L
L + L+
Wt% A: Co
1000
T(°C)
LiquidusSolidus
Solvus
Eutectic Reaction
Eutectic Phase Diagram
Maximum solubility of B in A
Boundaries - lines:•Liquidus•Solidus•Solvus
Now, apply previous rules ?On real systems
Chapter 9 - 23
Chapter 9 - 24
Chapter 9 - 25
Example-Eutectic System
: mostly Cu : mostly Ag
• TE : No liquid below TE
Ex.: Cu-Ag system
L (liquid)
L + L+
Co , wt% Ag20 40 60 80 1000
200
1200T(°C)
400
600
800
1000
CE
TE
8.0%
CE=71.9%
91.2%
TE=779°C
Given T and C Values are characteristics Of Cu-Ag system
Chapter 9 - 26
L+L+
+
200
T(°C)
18.3
C, wt% Sn20 60 80 1000
300
100
L (liquid)
183°C 61.9 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)
+ --compositionscompositions of phases:
CO = 40 wt% Sn
--the relative amountrelative amount of each phase:
150
40Co
11C
99C
SR
C = 11 wt% SnC = 99 wt% Sn
W=C - CO
C - C
=99 - 4099 - 11
= 5988
= 0.67
SR+S
=
W =CO - C
C - C=R
R+S
=2988
= 0.33 (or 1 – 0.67)=40 - 1199 - 11
Chapter 9 - 27
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 200200°°CC, find... --the phases present: Pb-Sn
system
EX: Pb-Sn Eutectic System (2)
+ L--compositionscompositions of phases:
CO = 40 wt% Sn
--the relative amountrelative amount of each phase:
W =CL - CO
CL - C=
46 - 40
46 - 17
= 6
29= 0.21
WL =CO - C
CL - C=
23
29= 0.79
40Co
46CL
17C
220SR
C = 17 wt% SnCL = 46 wt% Sn
Chapter 9 - 28
• 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
Depends on the composition
Chapter 9 - 29
• 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
Chapter 9 - 30
• 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 microstructure
Pb-Snsystem
L
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
Chapter 9 - 31
Lamellar Eutectic Structure
Chapter 9 - 32
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: crystals and a eutectic microstructure
Microstructures in Eutectic Systems: IV
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
Chapter 9 - 33
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: crystals and a eutectic microstructure
Microstructures in Eutectic Systems: IV
18.3 61.9
SR
97.8
SR
primary eutectic
eutectic
WL = (1-W) = 0.50
C = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SW‘ = = 0.50
• Just above TE :
• Just below TE :
C = 18.3 wt% Sn
C = 97.8 wt% SnS
R + SW = = 0.73
W = 0.27
Pb-Snsystem
L+200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
L+
40
+
TE
L: Co wt% Sn LL
What is the fraction of Eutectic ?Pro-eutectic (primary)
Chapter 9 - 34
L+L+
+
200
Co, wt% Sn20 60 80 1000
300
100
L
TE
40
(Pb-Sn System)
Hypoeutectic & Hypereutectic
160 m
eutectic micro-constituent
hypereutectic: (illustration only)
175 m
hypoeutectic: Co = 50 wt% Sn
T(°C)
61.9eutectic
eutectic: Co = 61.9 wt% Sn
Chapter 9 - 35
Intermetallic Compounds
Note: intermetallic compound forms a line - not an area like ?
Mg2Pb
In Previous Phase diagramsα and β are terminal solid solutions
Can we obtain other intermediate phases ?
Now: Mg2Pb • is an intermetallic compound• with a ratio (2:1) (Mg:Pb)-Molar• melts at a fixed temperature M• Cooling curve ???
Because stoichiometry (i.e. composition) is exact: Calculate ?
Calculate Wt% Pb = 207 /(207+2(24.3)) X 100 = 81 %
Chapter 9 - 36
Same Story ?
• No. and type of each phase• Composition of each phase • Relative amount or fraction of each phase• Distinguish between
– primary phase– Eutectic phase
• Microstructure• Cooling Curves• Fill the regions with types of phases?
Chapter 9 - 37
Eutectoid & Peritectic Phase Reactions
• Eutectic - liquid in equilibrium with two solidsL + cool
heat
intermetallic compound - cementite
cool
heat
• Eutectoid - solid phase in equilibrium with two solid phases
S2 S1+S3
+ Fe3C (727ºC)
cool
heat
• Peritectic - liquid + solid 1 solid 2 (Fig 9.21)
S1 + L S2
+ L (1493ºC)
Chapter 9 - 38
Complex Phase Diagrams
α and η are terminal solid solutions - β, β’, γ, δ and ε are intermediate solid solutions.• new phenomena exist: ……… reaction ………. reaction
Chapter 9 - 39
Eutectoid & Peritectic
Cu-Zn Phase diagram
Eutectoid transition +
Peritectic transition + L
Chapter 9 - 40
Phase Diagram with Eutectoid Phase Reactions
The eutectoid (eutectic-like in Greek) reaction is similar to the eutectic
This phase diagram includes both:
Eutectic ReactionEutectoid Reaction
Our Interest In Fe-C phase diagram
Chapter 9 - 41
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
(austeniteaustenite)
+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-hardhard) (ferriteferrite-softsoft)
Chapter 9 - 42
Development of Microstructure in Iron-Carbon alloys
PearlitePearlite Structure
By DiffusionDiffusion
Note: remember the names! Austenite, ferriteferrite, cemetitecemetite and pearlitepearlite
Chapter 9 - 43
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 HypoeutectoidHypoeutectoidsteel
R S
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
r s
w =s/(r+s)w =(1- w)
Chapter 9 - 44
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 FeFe33CC
60 mHypereutectoid steel
pearlitepearlite
R S
w =S/(R+S)wFe3C =(1-w)
wpearlite = wpearlite
sr
wFe3C =r/(r+s)w =(1-w Fe3C )
Fe3C
Chapter 9 - 45
Phases in Fe–Fe3C Phase Diagram
α -ferrite - solid solution of C in …….. Fe• Stable form of iron at room temperature.• The maximum solubility of C is 0.022 wt% (interstitial solubility)• Soft and relatively easy to deform
+ Fe-C liquid solution
γ -austenite - solid solution of C in …….. FeThe maximum solubility of C is 2.14 wt % at 1147°C.Interstitial lattice positions are much larger than ferrite (higher C%)Is not stable below the eutectic temperature (727 °C) unless cooled rapidly (Chapter 10).
δ -ferrite solid solution of C in ……. FeThe same structure as α -ferriteStable only at high T, above 1394 °CAlso has low solubility for carbon (BCC)
Fe3C (iron carbide or cementite)This intermetallic compound is metastable, it remains as a compound indefinitely at room T, but decomposes (very slowly, within several years) into α -Fe and C (graphite) at 650 - 700 °C
Chapter 9 - 46
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 - 47
Example – Fe-C System
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 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 - 48
Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite ()
note: amount of pearlitepearlite = amount of just above just above TE
Co = 0.40 wt% CC = 0.022 wt% CCpearlite = 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
Pearlite include ?
Eutectoid Fe3CFe3C
+Eutectoid ferrite ferrite
Chapter 9 - 49
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 elements
Ce
ute
cto
id (
wt%
C)
Ni
Ti
Cr
SiMn
WMo
Chapter 9 - 50
• 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.
Summary