chapter 9: phase diagrams · chapter 9 - 11 0 20 40 60 80 100 wt% ni 1000 1100 1200 1300 1400 1500...
TRANSCRIPT
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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
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Chapter 9 - 2
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
ga
r
Tem
pera
ture
(°C
)
0 20 40 60 80 100C
o=Composition (wt% sugar)
L(liquid solution
i.e., syrup)
Solubility Limit L
(liquid)
+ S
(solid sugar)20
40
60
80
100
Pu
re
Wa
ter
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 regions
that result (e.g., α and β).
Aluminum-CopperAlloy
Components and Phases
α (darker
phase)
β (lighter
phase)
Adapted from chapter-opening photograph, Chapter 9, Callister 3e.
A phase maybe defined as a homogeneous portion of a system that has uniform physical and chemical characteristics.
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Chapter 9 - 4
Effect of T & Composition (Co)• Changing T can change # of phases:
Adapted from Fig. 9.1, Callister 7e.
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
Te
mp
era
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
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Chapter 9 - 5
Phase Equilibrium
Equilibrium: minimum energy state for a given T, P, and composition
(i.e. equilibrium state will persist indefinitely for a fixed T, P and
composition).
Phase Equilibrium: If there is more than 1 phase present, phase
characteristics will stay constant over time.
Phase diagrams tell us about equilibrium phases as a function of T,
P and composition (here, we’ll always keep P constant for simplicity).
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Chapter 9 - 6
Unary Systems
Triple point
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Chapter 9 - 7
Phase Equilibria
0.12781.8FCCCu
0.12461.9FCCNi
r (nm)electronegCrystalStructure
• Both have the same crystal structure (FCC) and have similar 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 - 8
Unary Systems
Single component system
Consider 2 metals:
Cu has melting T = 1085oCNi has melting T = 1453oC (at standard P = 1 atm)
1085oC
Cu
solid
liquid
T
1453oC
Ni
solid
liquid
T
What happens when Cu and Ni are mixed?
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Chapter 9 - 9
Binary Isomorphous Systems
2 componentsComplete liquid and solid solubility
1085oC
Cu
solid
liquid
T
1453oC
Ni
solid
liquid
T
wt% Ni0 100
Expect Tm of solution to lie in between Tm of two pure components
For a pure
component, complete melting
occurs before T
increases (sharp phase transition).
But for
multicomponent
systems, there is usually a
coexistence of L and S.
L
S
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Chapter 9 - 10
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
Diagramfor 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)
α (FCC solid solution)
• 3 phase fields: LL + α
α
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
α
(FCC solid solution)
L+ αliq
uidus
solid
us
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Chapter 9 - 11
wt% Ni20 40 60 80 10001000
1100
1200
1300
1400
1500
1600T(°C)
L (liquid)
α(FCC solid solution)
L+ α
liquidus
solid
us
Cu-Niphase
diagram
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: α
B(1250°C, 35): 2 phases: L + α
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
250°C
,35)
A(1100°C,60)
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Chapter 9 - 12
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L + α
liquidus
solidus
30 40 50
L + α
Cu-Ni system
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 α and L
CL = C liquidus ( = 32 wt% Ni here)
Cα = Csolidus ( = 43 wt% Ni here)
At TD = 1190°C:
Only Solid ( α)
Cα = Co ( = 35 wt% Ni)
Co = 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.)
BTB
DTD
tie line
4Cα3
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Chapter 9 - 13
Determining phase composition in 2-phase region:
1. Draw the tie line.
2. Note where the tie line intersects the liquidus and solidus lines (i.e.
where the tie line crosses the
phase boundaries).
3. Read off the composition at the boundaries:
Liquid is composed of CL amount
of Ni (31.5 wt% Ni).
Solid is composed of Cαααα amount of
Ni (42.5 wt% Ni).
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Chapter 9 - 14
• 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)
WL = 100 wt%, Wα = 0
At TD: Only Solid ( α)
WL = 0, Wα = 100 wt%
Co = 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.)
Phase Diagrams:weight fractions of phases
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L + α
liquidus
solidus
30 40 50
L + α
Cu-Ni system
TAA
35Co
32CL
BTB
DTD
tie line
4Cα3
R S
At TB: Both α and L
% 733243
3543wt=
−
−=
= 27 wt%
WL= S
R +S
Wα= R
R +S
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Chapter 9 - 15
Lever Rule: Derivation
Since we have only 2 phases:
1=+ αWWL
Conservation of mass requires that:
Amount of Ni in α-phase + amount of Ni in liquid phase = total amount of Nior
oLLCCWCW =+αα
(1)
(2)
From 1st condition, we have: L
WW −= 1α
Sub-in to (2):oLLL
CCWCW =+− α)1(
Solving for WL
and Wα gives :
L
o
L
CC
CCW
−
−=
α
α
L
Lo
CC
CCW
−
−=
α
α
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Chapter 9 - 16
• 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 (teeter-totter)
ML
Mα
R S
RMSM L ⋅=⋅α
L
L
LL
LL
CC
CC
SR
RW
CC
CC
SR
S
MM
MW
−
−=
+=
−
−=
+=
+=
α
α
α
α
α
00
wt% Ni
20
1200
1300
T(°C)
L (liquid)
α
(solid)L + α
liquidus
solidus
30 40 50
L + αB
TB
tie line
Co
CL Cαααα
SR
Adapted from Fig. 9.3(b), Callister 7e.
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Chapter 9 - 17
wt% Ni20
1200
1300
30 40 501100
L (liquid)
α
(solid)
L+ α
L+ α
T(°C)
A
35Co
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 in
another; α phasefield extends from0 to 100 wt% Ni.
Adapted from Fig. 9.4,
Callister 7e.
• ConsiderCo = 35 wt%Ni.
Ex: Cooling in a Cu-Ni Binary
4635
4332
α: 43 wt% Ni
L: 32 wt% Ni
L: 24 wt% Ni
α: 36 wt% Ni
Bα: 46 wt% NiL: 35 wt% Ni
C
D
E
24 36
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Chapter 9 - 18
Microstructures in Isomorphous Alloys
Microstructures will vary on the cooling rate (i.e. processing conditions)
1. Equilibrium Cooling: Very slow cooling to allow phase equilibrium to
be maintained during the cooling process.
a (T>1260oC): start as
homogeneous liquid solution.
c (T= 1250oC): calculate composition
and mass fraction of each phase.
Cα = 43 wt% Ni; CL = 32 wt% Ni
d (T~ 1220oC): solidus line reached.
Nearly complete solidification.
Cα = 35 wt% Ni; CL = 24 wt% Ni
e (T<1220oC): homogeneous solid
solution with 35 wt% Ni.
b (T ~ 1260oC): liquidus line
reached. α phase begins tonucleate. Cα = 46 wt% Ni; CL = 35 wt% Ni
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Chapter 9 - 19
Non-equilibrium cooling
Since diffusion rate is especially low in solids, consider case where:
Cooling rate >> diffusion rate in solidCooling rate << diffusion rate in liquid
(equilibrium maintained in liquids phase)
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Chapter 9 - 20
Non-equilibrium coolinga’ (T>1260oC): start as
homogeneous liquid solution.
c’ (T= 1250oC): solids that formed at pt b’
remain with same composition (46wt%) and
new solids with 42 wt% Ni form around the
existing solids (Why around them?).
d’ (T~ 1220oC): solidus line reached. Nearly
complete solidification.
•Previously solidified regions maintain original
composition and further solidification occurs at 35
wt% Ni.
e (T<1220oC): Non-equilibrium solidification
complete (with phase segregation).
b’ (T ~ 1260oC): liquidus line
reached. α phase begins to nucleate. Cα = 46 wt% Ni; CL = 35 wt% Ni
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Chapter 9 - 21
• Cα changes as we solidify.• Cu-Ni case:
• Fast rate of cooling:Cored structure
• Slow rate of cooling:Equilibrium structure
First α to solidify has Cα = 46 wt% Ni.Last α to solidify has Cα = 35 wt% Ni.
Cored vs Equilibrium Phases
First α to solidify:
46 wt% Ni
Uniform Cα:
35 wt% Ni
Last α to solidify:
< 35 wt% Ni
Cored structure can be eliminated by a homogenization heat treatment at a temperature below the solidus point
for the particular alloy composition.
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Chapter 9 - 22
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.
Te
nsile
Str
eng
th (
MP
a)
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 80 100
20
30
40
50
60
%EL for pure Ni
%EL for pure CuAt RT
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Chapter 9 - 23
: Min. melting TE
2 componentshas a special compositionwith a min. melting T.
Adapted from Fig. 9.7, Callister 7e.
Binary-Eutectic Systems
• Eutectic transition
L(CE) α(CαE) + β(CβE)
• 3 single phase regions (L, α, β)
• Limited solubility: α: mostly Cu
β: mostly Ag
• TE : No liquid below TE
• CE
composition
Ex.: Cu-Ag system
Cu-Agsystem
L (liquid)
α L + αL+β β
α + β
Co , wt% Ag20 40 60 80 1000
200
1200T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2779°C
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Chapter 9 - 24
Eutectic Point
Eutectic point: Where 2 liquidus lines
meet (pt. E).
Sometimes also referred to as invariant
point.
Eutectic Reaction:
cool
L(CE) α(CαE) + β(CβE)
similar to one component (pure) system
except 2 solid phases.
Eutectic Isotherm: Horizontal solidus
line at T = TE.
heat
Cu-Ag phase diagram
Single phase regions: α, β, L-phase
2-Phase coexistence regions: α+β, α+L and β+L
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Chapter 9 - 25
L+αL+β
α + β
200
T(°C)
18.3
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α183°C
61.9 97.8β
• 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)
α + β--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase:150
40Co
11Cα
99Cβ
SR
Cα = 11 wt% Sn
Cβ = 99 wt% Sn
Wα=Cβ - CO
Cβ - Cα
=99 - 4099 - 11
=5988
= 67 wt%
SR+S
=
Wβ =CO - Cα
Cβ - Cα
=R
R+S
=29
88= 33 wt%=
40 - 11
99 - 11
Adapted from Fig. 9.8, Callister 7e.
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Chapter 9 - 26
L+β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L (liquid)
α β
L+α
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)
α + L--compositions of phases:
CO = 40 wt% Sn
--the relative amount
of each phase:
Wα =CL - CO
CL - Cα
=46 - 40
46 - 17
=6
29= 21 wt%
WL =CO - Cα
CL - Cα
=23
29= 79 wt%
40Co
46CL
17Cα
220SR
Cα = 17 wt% SnCL = 46 wt% Sn
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Chapter 9 - 27
• Co < 2 wt% SnAdapted from Fig. 9.11,
Callister 7e.
Microstructures in Eutectic Systems: I
0
L+ α200
T(°C)
Co, wt% Sn10
2
20Co
300
100
L
α
30
α+β
400
(room T solubility limit)
TE
(Pb-SnSystem)
αL
L: Co wt% Sn
α: Co wt% Sn
1. One component rich
composition.a: start with homogeneous liquid.
b: α-phase solids with liquid.Compositions and mass fractions
can be found via tie lines and lever
rule.
c: α-phase solid solution only.
Net result: polycrystalline α solid.
Cooling at this composition is similar to binary isomorphous systems.
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Chapter 9 - 28
• 2 wt% Sn < Co < 18.3 wt% Sn
Adapted from Fig. 9.12, Callister 7e.
Microstructures in Eutectic Systems: II
Pb-Snsystem
L + α
200
T(°C)
Co , wt% Sn10
18.3
200Co
300
100
L
α
30
α+ β
400
(sol. limit at TE)
TE
2(sol. limit at Troom)
L
α
L: Co wt% Sn
αβ
α: Co wt% Sn
2. One-component rich but cooling to α α α α + ββββ coexistence.d: homogeneous liquid.
e: α + L phase (same as previous but at different compositions and mass fractions).
f: all α-phase solid solution.
g: α + β phase (passing through solvus line leads to exceeding solubility limit and βphase precipitates out).
Net result: polycrystalline α-solid with fine
β crystals.
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Chapter 9 - 29
• Co = CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of α and β crystals.
Adapted from Fig. 9.13, Callister 7e.
Microstructures in Eutectic Systems: III
Adapted from Fig. 9.14, Callister 7e.
160µm
Micrograph of Pb-Sneutectic microstructure
Pb-Snsystem
L + β
α + β
200
T(°C)
C, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
183°C
40
TE
18.3
α: 18.3 wt%Sn
97.8
β: 97.8 wt% Sn
CE61.9
L: Co wt% Sn
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Chapter 9 - 30
Lamellar Eutectic Structure
Adapted from Figs. 9.14 & 9.15, Callister
7e.
L
Sn
Pb
β
α Pb rich
Sn rich
In order to achieve large
homogeneous regions, long diffusion lengths are required.
Lamellar structure forms because
relatively short diffusion lengthsare required.
α
β
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Chapter 9 - 31
• 18.3 wt% Sn < Co < 61.9 wt% Sn• Result: α crystals and a eutectic microstructure
Microstructures in Eutectic Systems: IV
18.3 61.9
SR
97.8
SR
primary α
eutectic αeutectic β
WL = (1-Wα) = 50 wt%
Cα = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SWα= = 50 wt%
• Just above TE :
• Just below TE :
Cα = 18.3 wt% Sn
Cβ = 97.8 wt% SnS
R + SWα= = 73 wt%
Wβ = 27 wt%Adapted from Fig. 9.16,
Callister 7e.
Pb-Snsystem
L+β200
T(°C)
Co, wt% Sn
20 60 80 1000
300
100
L
α β
L+α
40
α+β
TE
L: Co wt% Sn LαLα
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Chapter 9 - 32
L+αL+β
α + β
200
Co, wt% Sn20 60 80 1000
300
100
L
α βTE
40
(Pb-SnSystem)
Hypoeutectic & Hypereutectic
Adapted from Fig. 9.8, Callister 7e. (Fig. 9.8 adapted from Binary Phase
Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)
160 µm
eutectic micro-constituentAdapted from Fig. 9.14,
Callister 7e.
hypereutectic: (illustration only)
β
ββ
ββ
β
Adapted from Fig. 9.17,
Callister 7e. (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 µm
α
α
α
αα
α
hypoeutectic: Co = 50 wt% Sn
Adapted from
Fig. 9.17, Callister 7e.
T(°C)
61.9
eutectic
eutectic: Co =61.9wt% Sn
Microconstituent: an
element of a microstructure with
identifiable and characteristic structure (at
pt. m there are 2 microconstituents: primary
α and eutectic structures)
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Chapter 9 - 33
Intermediate phases
Intermediate solid solutions (intermediate phases): Solid solutions that do
not extend to pure components in the phase diagram.
Cu-Zn
Terminal solid
solutions: α and η.
Intermediate solid
solutions: β, γ, δ and ε.
Tie lines and lever rule can
be used to determine compositions and wt% of
phases.
e.g. at 800oC with 70 wt%
Zn
CL = 78 wt% ZnCγ = 67 wt% Zn
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Chapter 9 - 34
Intermetallic Compounds
Mg2Pb
Note: intermetallic compound forms a discrete line - not an area - because stoichiometry (i.e. composition) is exact.
Adapted from
Fig. 9.20, Callister 7e.
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Chapter 9 - 35
Eutectoid & Peritectic
• Eutectic - liquid in equilibrium with two solids
L α + βcool
heat
intermetallic compound - cementite
cool
heat
• Eutectoid - solid phase in equation with two solid phases
S2 S1+S3
γ α + Fe3C (727ºC)
cool
heat
• Peritectic - liquid + solid 1 � solid 2 (Fig 9.21)
S1 + L S2
δ + L γ (1493ºC)
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Chapter 9 - 36
Eutectoid & Peritectic
Cu-Zn Phase diagram
Adapted from Fig. 9.21, Callister 7e.
Eutectoid transition δ γ + ε
Peritectic transition γ + L δ
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Chapter 9 - 37
Congruent phase transformation
Congruent transformation: no change in composition upon phase
transformation. (for example: allotropic transformation)
Incongruent transformation: phase transformation where at least one of the
phases go through composition change. (for example: isomorphous, eutectic and
eutectoid system)
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Chapter 9 - 38
Ceramic phase diagrams
Al2O3-Cr2O3 MgO-Al2O3
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Chapter 9 - 39
Gibbs Phase Rule
P + F = C + NNumber of
phases present
Degree of freedom (externally controllable
parameters: i.e. T, P, and C)
Number of components
Number of non-
compositional variables (Temperature & Pressure)
e.g. Cu-Ag phase diagram
Cu and Ag are the only components-> C = 2
Temperature is the only non-compositional variable here (i.e. fixed pressure).-> N = 1 (but in general N = 2)
When 2 phases are present-> P = 2 which leads to F = C+N-P = 2+1-2 = 1
When only 1 phase is present.-> P = 1 which leads to F = 2
What does this mean? Why should you care?
A criterion for the number of phases that will coexist within a system at equilibrium
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Chapter 9 - 40
Gibbs Phase Rule
In the previous example of Cu-Ag phase diagram, when F = 1, only one
parameter (T or C) needs to be specified to completely define the system.
e.g. (for α+L region)If T is specified to be 1000oC, compositions are already determined
(Cα and CL).
Or
If composition of the a phase is
specified to be Cα then both T and CL
are already determined.
Cα CL
The nature of the phases is important,
not the relative phase amounts.
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Chapter 9 - 41
Gibbs Phase Rule
When F = 2, both T and C have to be specified to completely define the
state of the system.
e.g.(for α region)If T is specified to be 800oC, Cα can be any where between 0 to ~8 wt% Ag)
Or
If composition of the a phase is
specified to be Cα = 3 wt%, then T and
can be any where between ~600 to 1100oC.
Cα
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Chapter 9 - 42
Gibbs Phase Rule
Where in the Cu—Ag diagram, is there a 0 degree of freedom?
(i.e. T, P, and C are all fixed)---eutectic isotherm
F=3-P=0
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Chapter 9 - 43
Concept Check
• Question: For a ternary system, three
components are present; temperature is also
a variable. What is the maximum number of
phases that may be present for a ternary
system, assuming that pressure is held
constant?
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Chapter 9 - 44
TVR Tuscan Speed 6, high-performance sports
car with an austempered ductile iron crankshaft.
The world's first bridge made of iron in 1779. The entire structure is made of cast iron. (near Broseley,
UK)
Iron-Carbon Systemstructural material
Ferrite Magnets
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Chapter 9 - 45
Iron-Carbon System
The Akashi Kaikyo bridge, a 3-span 2-
hinged truss-stiffened suspension bridge.
completed in 1998. It connects Kobe with
Awaji Island. It is the world's longest
suspension bridge, with a span between the towers of 1.9 km.
Millau Viaduct in France, the
highest bridge in the world.
Golden Gate Bridge
Steel bridges
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Chapter 9 - 46
• Iron-carbon (Fe-C) system is one of the most important binary systems due to the versatile uses of the iron-based structural alloys.
• This phase diagram is so important in understanding the equilibrium structure, and in the design of heat treatment process of iron alloys.
• The most important part of the phase diagram is the region below6.7 w% carbon. All practical iron-carbon alloys contain C below 6.7 w%. This part is of the phase diagram is thus the most analyzed part of the iron carbon phase diagram.
Iron-Carbon System
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Chapter 9 - 47
Iron-Carbon System
IronTypical metal (e.g. Cu)
Solid
Liquid
Tm
TT(oC)
912
1394
1538
Ferrite (BCC)
Austenite (FCC)
Ferrite (BCC)
Liquid
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Chapter 9 - 48
Iron-Carbon SystemIron-Iron Carbide Diagram
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Chapter 9 - 49
Remarks on Fe-C system
• C is an interstitial element in Fe matrix.
• C has limited solid solubility in the alpha BCC phase (narrow region close to pure iron). Max solid solubility of C in alpha iron is 0.022 w% at 727 °C.
• Alpha iron an be made magnetic below 768°C.
• Austenite phase is not stable below at 727 °C. Max solid solubility of C into austenite is 2.14 w% at 1147°C, much larger that that in alpha phase.
• Austenite is a non-magnetic phase, and heat treatment of Fe-C alloys involving austenite is so important.
• Cementite phase (intermetallic, Fe3C) forms over a large region of the Fe-C phase diagram, but it is a metastable phase (heating above 650 °C for years decomposes this phase into alpha iron and graphite). Cementite is very hard and brittle.
• δ ferrite is stable only at relatively high T, it is of no technological importance and is not discussed further.
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Chapter 9 - 50
Ferrite (90x) Austenite (x325)Relatively softDensity: 7.88g/cm3
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Chapter 9 - 51
Classification Scheme of Ferrous Alloys
• IronPure ion contains less than 0.008wt% C
• Steel
0.008-2.14 wt% C, in practice,<1.0 wt% C
• Cast Iron
2.14-6.70 wt% C, in practice, <4.5 wt% C
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Chapter 9 - 52
Iron-Carbon (Fe-C) Phase Diagram
• 2 important points
-Eutectoid (B):γ ⇒ α +Fe3C
-Eutectic (A):L ⇒ γ +Fe3C
Adapted from Fig. 9.24,Callister 7e.
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ
(austenite)
γ+L
γ+Fe3C
α+Fe3C
α+γ
L+Fe3C
δ
(Fe) Co, wt% C
1148°C
T(°C)
α 727°C = Teutectoid
A
SR
4.30
Result: Pearlite = alternating layers of α and Fe3C phases
120 µm
(Adapted from Fig. 9.27, Callister 7e.)
γ γγγ
R S
0.76
Ce
ute
cto
idB
Fe3C (cementite-hard)
α (ferrite-soft)
γ (0.76 wt% C)
cool
heat
α (0.022 wt% C) + Fe3C (6.7 wt% C)
Eutectoid cooling:
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Chapter 9 - 53
Pearlite
Pearl-microscope picture
colonies
(X500)
Mechanically, pearlite has properties intermediate between the soft, ductile ferrite and the hard, brittle cementite.
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Chapter 9 - 54
Hypoeutectoid Steel
Adapted from Figs. 9.24 and 9.29,Callister 7e. (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
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ + Fe3C
α+ Fe3C
L+Fe3C
δ
(Fe) Co, wt% C
1148°C
T(°C)
α727°C
(Fe-C System)
C0
0.7
6
Adapted from Fig. 9.30,Callister 7e.
proeutectoid ferritepearlite
100 µmHypoeutectoid
steel
R S
α
wα =S/(R+S)
wFe3C=(1-wα)
wpearlite = wγpearlite
r s
wα =s/(r+s)wγ =(1- wα)
γ
γ γ
γα
αα
γγγ γ
γ γγγ
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Chapter 9 - 55
Hypereutectoid Steel
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ +Fe3C
α +Fe3C
L+Fe3C
δ
(Fe) Co, wt%C
1148°C
T(°C)
α
Adapted from Figs. 9.24 and 9.32,Callister 7e. (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)
0.7
6 Co
Adapted from Fig. 9.33,Callister 7e.
proeutectoid Fe3C
60 µmHypereutectoid steel
pearlite
R S
wα =S/(R+S)
wFe3C=(1-w α)
wpearlite = wγpearlite
sr
wFe3C=r/(r+s)
wγ =(1-w Fe3C)
Fe3C
γγ
γ γ
γγγ γ
γγγ γ
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Chapter 9 - 56
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 (α)
b) the amount of carbide (cementite) in grams
that forms per 100 g of steel
c) the amount of pearlite and proeutectoid
ferrite (α)
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Chapter 9 - 57
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
=α
=
=−
−=
−
−=
α+ α
α
x
CC
CCo
b) the amount of carbide (cementite) in grams that forms per 100 g of steel
a) composition of Fe3C and ferrite (α)
CO = 0.40 wt% CCα = 0.022 wt% CCFe C = 6.70 wt% C
3
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
Co , wt% C
1148°C
T(°C)
727°C
CO
R S
CFe C3Cαααα
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Chapter 9 - 58
Chapter 9 – Phase Equilibriac. the amount of pearlite and proeutectoid ferrite (α)
note: amount of pearlite = amount of γ just above TE
Co = 0.40 wt% CCα = 0.022 wt% CCpearlite = Cγ = 0.76 wt% C
γ
γ + α=
Co −Cα
Cγ −Cα
x 100 = 51.2 g
pearlite = 51.2 g
proeutectoid α = 48.8 g
Fe
3C
(ce
me
ntite
)
1600
1400
1200
1000
800
600
4000 1 2 3 4 5 6 6.7
L
γ (austenite)
γ+L
γ + Fe3C
α + Fe3C
L+Fe3C
δ
Co , wt% C
1148°C
T(°C)
727°C
CO
R S
CγγγγCαααα
![Page 59: Chapter 9: Phase Diagrams · Chapter 9 - 11 0 20 40 60 80 100 wt% Ni 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid) α (FCC solid solution) L + α l i q u i d u s s o l i d](https://reader033.vdocuments.us/reader033/viewer/2022042018/5e767469e2972b63f8301f92/html5/thumbnails/59.jpg)
Chapter 9 - 59
Effect of alloying elements on Fe-C phase diagram
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Chapter 9 - 60
Alloying Steel with More Elements
• Teutectoid changes: • Ceutectoid changes:
Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34 from 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. 9.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
TE
ute
cto
id(°
C)
wt. % of alloying elements
Ti
Ni
MoSi
W
Cr
Mn
wt. % of alloying elements
Ce
ute
cto
id(w
t%C
)
Ni
Ti
Cr
SiMn
WMo
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Chapter 9 - 61
• 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