phase equilibrium
DESCRIPTION
phase equilibrium xTRANSCRIPT
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Phase Equilibria
Reading: West 11-12
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PHASE DIAGRAMS• Also called equilibrium or constitutional diagramsAlso called equilibrium or constitutional diagrams• Plots of temperature vs. pressure, or T or P vs.
composition, showing relative quantities of phases at equilibriumequilibrium
• Pressure influences phase structure– Remains virtually constant in most applicationsy pp– Most solid-state phase diagrams at 1 atm
• Note: metastable phases pdo not appear on equilibrium phase diagrams
Fe–Fe3C phase diagramdiagrams g
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PHASESA phase is a homogeneous portion of a system with uniform physical and A phase is a homogeneous portion of a system with uniform physical and
chemical characteristics, in principle separable from the rest of the system
A difference in either physical or chemical properties constitutes a phase
gaseous state
• seemingly only one phase occurs (gases always mix)
lt t lt N O SiO
liquid state• often only one phase occurs (homogeneous solutions)
e.g., salt water, molten Na2O-SiO2
• two immiscible liquids (or liquid mixtures) count as two phases
solid state• crystalline phases: e.g., ZnO and SiO2 = two phases
• polymorphs: e g wurtzite and sphalerite ZnS are different phases• polymorphs: e.g., wurtzite and sphalerite ZnS are different phases
• solid solutions = one phase (e.g., Al2O3-Cr2O3 mixtures)25
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PHASE EQUILIBRIA
The equilibrium phase is always the one with the lowest free energy
G = H – TS
The driving force for a phase change is the minimization of free
Equilibrium → state with minimum free energy under some g m m z f f
energym m m f gy mspecified combination of temperature, pressure, and composition e.g., meltingg g
unstable
metastable
G
metastable
equilibrium state 26
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GIBBS PHASE RULE
P + F = C + 21870s
F C P + 2GibbsP: number of phases present at equilibrium
C: number of components needed to describe the system
F = C – P + 2
C: number of components needed to describe the systemF: number of degrees of freedom, e.g. T, P, composition
The number of components (C) is the minimum number of chemically independent constituents needed to describe the composition of the phases present in the system.
e.g., • salt water. C = 2 (NaCl and water)g , ( )• solid magnesium silicates. C = 2 (MgO and SiO2)• solid MgAl silicates. C = 3. (MgO, Al2O3, SiO2)
The degrees of freedom (F) is the number of independent variables that must be specified to define the system completely.
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ONE COMPONENT PHASE DIAGRAMS
P + F = C + 2with C = 1
F=2with C = 1
P + F = 3F 0 F=1
Composition is fixed,Only T and P can vary
F=0 F=1
y y
Three possibilities:• P = 1 … F = 2 (bivariant phase field)
P 2 F 1 ( i i h )• P = 2 … F = 1 (univariant phase curve)• P = 3 … F = 0 (invariant phase point)
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EXAMPLE: BOILING WATER
P + F = C + 2
C = 1 (water)C = 1 (water)P = 2 (vapor + liquid)
F 1 ( i h T PF = 1 (either T or P,but not both)
→ coexistence curve
F=1
*once we specify either T or P of our boiling water, the other g ,variable is defined as well
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CLAUSIUS-CLAPEYRON EQUATIONExpresses the pressure dependence of phase transitions
From a to b, starting from Gibbs-Duhem equation:
Expresses the pressure dependence of phase transitions as a function of temperature (gives slopes of coexistence curves)
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dPdTddPvdTsdg
From a to b, starting from Gibbs Duhem equation
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222
dgdgdPvdTsdg
12
12
vvss
dTdP
12
THS
HdP T VTdT
derived ~1834
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Slope of the coexistence curves:
HdP
VTdT
H positive along arrows(melt, sublime, vaporize)
V negative only for melting
*Ice less dense than water31
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ONE COMPONENT PHASE DIAGRAMS
Carbon
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More than 100 tons of synthetic diamonds are yproduced annually worldwide by firms like Diamond Innovations (previously part of General Electric), Sumitomo Electric, and De Beers.
Gemesis, GE, Sumitomo Electric, and De Beers 33
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ONE COMPONENT PHASE DIAGRAMS
SiO2
4.287 g/cm3C2/c
2.93 g/cm3
2.65 g/cm3
P42/mnmP31212.65 g/cm
there are also many metastable phases (not shown) 34
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OTHER EXAMPLES
Ice
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OTHER EXAMPLES
CO2
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OTHER EXAMPLES
Sulfur
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TWO COMPONENT (BINARY) DIAGRAMS
P + F C + 2with C = 2
Composition is now variable: T, P, andcomposition can vary
P + F = C + 2
When vapor pressures are negligible:
P + F = C + 1Condensed phase rule
Pressure is no longer a variable: only T and composition matter
P + F = 3Three possibilities (as before):
• P = 1 … F = 2 (bivariant phase field)( p )• P = 2 … F = 1 (univariant phase curve)• P = 3 … F = 0 (invariant phase point) 38
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SIMPLE EUTECTIC SYSTEMSsimplest binary diagram for solids P + F = 3p y g
(liquid)P=1F=2
P=2P=2
P 2
P 2F=1
P 2F=1
P=2F=1
(solid)
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• no compounds/solid solutions in solid state• only single phase liquid at high temperatures• partial melting at intermediate temperatures• partial melting at intermediate temperatures
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First, specify overall composition of the systemSecond, pick a temperature.
→ The compositions of the phases (1 or 2) are then fixedhe compos t ons of the phases ( or ) are then f xed
isotherme.g., start with liquid
Liquid1490°C
qat a certain composition
70% A30% B
isoplethisopleth
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now, slowly cool liquidcool liquid
crystals of Abegin to form
Liquidus curve: specifies the maximum temperature at which crystals can co-exist with the melt in thermodynamic equilibrium 42
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Maximum T at which crystals can exist
LIQUIDUS CURVEMaximum T at which crystals can exist.aka Saturation Solubility Curve
Melting point depression: the effect of a soluble impurity f p yon the melting point of pure compounds.
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For example, consider salty ice:
add salt, freezing point falls… 44
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keep cooling
more crystals of A form, depleting melt of Aof A
P=2F=1
the system is now a mixture of crystals of pure A and the melt45
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YX f YX f
Al h i h XfY h l i f h h A d l Along the isotherm XfY, the relative amounts of the phases A and melt vary but the composition of the individual phases does not vary. Along XfY, the two phases are pure A and melt of ~60% A & 40% B.
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tie line
YX f
tie line
YX f
Th l i f h h (“ h i i ”) b
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The relative amounts of the two phases (“phase composition”) can be determined by using the level rule along the tie line XfY: Amount of A = fY/XY and amount of liquid = Xf/XY.
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PHASE COMPOSITION AND LEVER RULEL l th f ti l t f t h i l Lever rule: the fractional amounts of two phases are inversely proportional to their distances along the tie line (isotherm) from the bulk composition axis
“balance the teeter-totter”
1 1 2 2f L f L1 Phase 2
X Yf
Phas
e 1
L1 = Xf L2 = fY
1 1 2f f L fYL1 1 2
2 1 11f f Lf f L
2
11 2
fYXY
LfL L
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Ph 2
X Yf
hase
1
Phase 2
Ph f‘’f‘
Overall Composition Fraction of liquid (phase 2)
f‘ff’’
Xf’ /XY = 10%Xf/XY = 60%Xf’’ /XY = 85%f Xf /XY = 85%
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cool more
note that liquid becomes richer in Bbecomes richer in B
We can use the phase diagram to determine the phase composition, the relative amounts of A and melt at a certain T and bulk comp.
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cool some more
last of the liquid solidifies as crystals just above solidus:
of A and Bfraction A: fe/Xemelt fraction: Xf/Xe
X fpassing through the passing through the solidus is called the eutectic reaction:
Liq e + A A + BLiq. e + A A + B
Solidus curve: gives the lowest temperature at which liquids can exist in equilibrium over a given compositional range 51
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cool some more
sample is now mixture of A and B below solidus:
crystalsfraction A constantat fY/XY
f ti B t tfraction B constantat Xf/XY
ffX Y
P=2F=1F=1
not much interesting happens below the solidus line. The solid just further cools with no change in composition. 52
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crystals of A precipitate first
crystals of B
precipitate first (A primary)
0% Bprecipitate first (B primary)
~0% B~100% liq.
~0% A0% A~100% liq.
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slow heating is the reverse of slow cooling
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EUTECTIC COMPOSITION (“easily melted”)system with the eutectic composition will melt/freeze congruentlysystem with the eutectic composition will melt/freeze congruently(i.e., completely, with no change in composition) and at the lowest T
composition with the
eutectic point is invariant
composition with the lowest melting point
eutectic point is invariant
P + F = 3
- 3 phases present
F = 0
eutectic point
p p(A, B, liquid e)
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VaporVapor--LiquidLiquid--Solid Nanowire GrowthSolid Nanowire Growth
metal catal sts
alloy
vapornanowire
800 deg. In-situ TEM
catalysts liquid
I II III
Alloying
NucleationI
GrowthIIIII
Unidirectional growth is the consequence of an anisotropy in solid liquid interfacial energyUnidirectional growth is the consequence of an anisotropy in solid-liquid interfacial energy.Y. Wu et al. J. Am. Chem. Soc. 2001, 123, 3165
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BINARY SYSTEMS WITH COMPOUNDS
T 1 i l d AB th t lt tl Type 1: single compound AB that melts congruently
• AB melts congruently• AB melts congruently(no change in composition)
t i l t ti t• two simple eutectic systemsstitched together
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BINARY SYSTEMS WITH COMPOUNDS
T 2 i l d AB th t lt i tl
• AB melts incongruently at T2
Type 2: single compound AB that melts incongruently
g y 2(liquid composition is x,liquid fraction is zy/zx)
• x is a peritectic point(3 phases: A, AB, liquid x)
• AB melts completely at T3
li id i i b• liquid compositions between x and m precipitate AB first
Peritectic point: an invariant point (liquid & two solids) in which the composition of the liquid phase is not between that of the two solids (not a minimum in the intersection between two liquidus lines) 59
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let’s cool liquid of composition n:
1) crystal of A form at T3
2) A f til li id 2) more A forms until liquid reaches composition x
3) At T peritectic reaction 3) At T2, peritectic reaction occurs:Liq. x + A Liq. x + AB
(crystalline phase changes from A to AB, often very slowly)
4) Cooling to T1, more AB forms, liquid composition moves to m eutectic reaction occurs
5) below T1, we have AB and B 60
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BINARY SYSTEMS WITH COMPOUNDS
T 3 i l d AB ith li it f t bilitType 3: single compound AB with upper limit of stability
• AB decomposes before melting (it disproportionates to A and B)
• AB has upper limit of stability
• simple eutectic system above T• simple eutectic system above T1
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THE CaO – SiO2 BINARY DIAGRAM 11 compounds!
• Ca2SiO4 and CaSiO3 meltcongruently
11 compounds!
immiscibleliquids y
• Ca3SiO5 and Ca3Si2O7melt incongruently
• Ca3SiO5 has a lower limit of stability(decomposes to CaO and (decomposes to CaO and Ca2SiO4 below 1250°C)
• liquid immiscibility domeliquid immiscibility domeat the silica-rich end ofthe diagram
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BINARY SYSTEMS WITH IMMISCIBLE LIQUIDS
pp upper consolutetemperature
immiscibility domeBelow T3, the liquid phase separates into two liquids of different comp
cool this comp.
LA+L 2 Liq. invariant monotectic points
T3into two liquids of different comp.
A +L B +L
A+La
c invariant monotectic pointsat points a and c
P + F = 3T2
A +B- 3 phases present
(liquid a, liquid c, A)
A B
Liq a + Liq c Liq c + AAt T2, monotectic reaction occurs:
Liq. a Liq. c Liq. c A
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THERMODYNAMICS OF LIQUID IMMISCIBILITYEnthalpy of mixing determines whether two liquids A and B mixEntha py of m ng t rm n s wh th r two qu s an m
Gmix = H – TS • S > 0 for mixing• if H < 0 (stronger A‐B attraction), the( g ),liquids are miscible at all temps
• if H > 0 (stronger A‐A, B‐B attraction):ibl i ibl if H < TS (hi h T)• possibly miscible if H < TS (high T)
• immiscible when H > TS (low T)
when a liquid enters the immiscibility dome the unmixing process when a liquid enters the immiscibility dome, the unmixing process (phase separation) occurs by one of two major mechanisms:
1) nucleation and growth 1) nucleation and growth (large compositional fluctuations required for ΔG<0)
2) i d l d i i2) spinodal decomposition(even infinitesimal compositional fluctuations give ΔG<0) 64
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SOLID SOLUTIONSa solid-state solution; a crystalline material with variable compositiony p
Two types:b tit ti l ( l t t /i l l t t /i )• substitutional (solute atom/ion replaces solvent atom/ion)
• interstitial (solute atom/ion occupies interstitial site)
Formation of solid solutions favored by:Formation of solid solutions favored by:• similar atomic/ionic radii (15%)• same crystal structure
i il l t ti iti
Ex m l s:
• similar electronegativities• same valency
Examples:• Al2O3 – Cr2O3 (substitutional, complete) • Si-Ge (substitutional, complete)
F C (i i i l i l)• Fe-C (interstitial, partial)
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BINARY SYSTEMS WITH SOLID SOLUTIONSsimplest solid solution system has complete miscibility in both solid and liquid states
P=1F=2
• copper melts at 1085 C
• nickel melts at 1455 C
• above liquidus: single phase liquid
P=1F=2
• between liq. and sol.: two phases
• below solidus: single phase solid α-phase is a substitutional solidconsisting of both Cu and Ni and on an FCC lattice
g p
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the composition of the solid solution has to change continuously CRYSTALLIZATION
• at T3, s.s. of composition s1 forms
during crystallization in order to maintain equilibrium
cool this comp.
liquidusL
3, p 1(lever rule shows ~100% liquid
of composition x1)
T2
T3• at T2, s.s. is composition s2, richer
in A, and liquid is composition x2, depleted of B(lever rule shows 33% s s and
tie line
solidus
(lever rule shows 33% s.s. and 67% liquid) T1
A Bx
s.s.
ssx
• at T1, s.s. is composition x1 and the final liquid is composition x3(lever rule shows ~100% s.s.)
xA Bx1 s1s2x2x3
The amount of s.s. and it’s A content continuously increases during cooling.
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Equilibrium solidification requires very slow coolingmi t t d i li
• No microstructural changes until theliquidus line is reached at b
microstructure during cooling
liquidus line is reached at b
• From b to d: -phase increases asdictated by lever rule and compositionchanges as dictated by tie lines andtheir intersection with solidus/liquidus
maintaining equilibrium during cooling results in a homogeneouscooling results in a homogeneous
alpha phase
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Somewhat faster cooling results in fractional crystallizationmi t t d i limicrostructure during cooling
• diffusion in solids is very slow, requires time• non-equilibrium (fast) cooling causes:q g
1) cored structures (inhomogeneous s.s.)2) metastable phases3) undercooled liquid
Consequences of non-equilibrium solidification: segregation, cored structure.upon reheating: grain boundaries will
3) undercooled liquid
upon reheating: grain boundaries willmelt first causing loss of structural integrity
coring…“onions”
coring in metals can be eliminated by heating just below the solidus for sufficient time 69
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CRYSTALLIZATION
Very slow cooling rate: crystallization to the same composition
Slightly faster: fractional crystallizationg y y
Very fast (quenched): no time for crystallization, glass forms
A glass is a non-crystalline (amorphous) solid, such as glassy silica glassy silica
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METALLIC GLASSESh t lli ll
small samples often produced by rapid cooling (millions of degrees a second) of
amorphous metallic alloys
cooling (millions of degrees a second) of molten metal, e.g., by melt spinning
bulk metallic glasses (BMG), 1990s, often f d f ll ith h l iti lformed from alloys with much lower critical cooling rates. These alloys contain many elements (Fe, Ti, Cu, Mg, Pd, etc.), thus “frustrating” the elements in their ability to form an ordered lattice.
properties: • better wear/corrosion resistance (no grain boundaries)l h l d (l h f h)• lower thermal conductivity (low phonon mean free path)
• higher strength (no line defects, no grain boundaries)• easy processing (they soften and flow with heating)
Ti40Cu36Pd14Zr10, believed to be noncarcinogenic, is about 3 times stronger than titanium potential artificial bone
e.g.,
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BINARY SOLID SOLUTIONS WITH THERMAL MAX/MIN
Indifferent point (P = 2, F = 1), NOT invariant point
There are NEVER three phasesThere are NEVER three phases.
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BINARY EUTECTIC WITH PARTIAL SOLID SOLUTIONvery commonly crystalline phases have only partial solubility in very commonly crystalline phases have only partial solubility in each other (immiscible solid solutions)
• 3 two phase regions• one invariant point
iti l f • compositional range of the two s.s. depends on temperature
ll th m xim m
e
• usually, the maximumsolubility of B in A (and vice versa) occurs at solidus tempat solidus temp
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EXAMPLE OF PARTIAL SOLID SOLUTION EUTECTIC
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FCC i l i h i A
Melting points of pure metals
FCC, copper is solute, rich in Agmax. 8.8% Cu
Ag in Cu8.8% Cu –maximumAg in Cu
solubility8%
maximum
Solid solution rich in copperppsilver as solute (FCC) – max of8% Ag at 779C
Solvus line – line between single phase solid and multi-phase solid
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Eutectic isothermEutectic isotherm
E t ti i thEutectic isotherm
coolingLiq E (71 9% Ag) α (8 0% Ag) + β (91 2% Ag)
Eutectic reaction:
heatingLiq. E (71.9% Ag) α (8.0% Ag) + β (91.2% Ag)
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Pb-Sn (solder):
63/37 solder – eutectic comp.
150C isotherm (40% Sn – 60% Pb)
Given a 40% Sn – 60% Pb alloy at 150C:(a) What phase(s) are present?(b) What is (are) the composition(s) of the phase(s)?( ) ( ) p ( ) p ( )(c) What is (are) the relative amount(s) of the phase(s)?
use tie line and Lever Rule!77
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composition of the two phases: use tie line
• 40% Sn – 60% Pb alloy at 1500CPh t• Phases present• Temperature – composition
point in + region ( and ill i t)will coexist)
• C (10% Sn – 90% Pb)• C (98% Sn – 2% Pb)
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phase composition: use Lever Rule
Relative amounts of each phase (weight)
4098CC66.0
109840981
CC
CCW
34.0109810401
CC
CCW
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MICROSTRUCTURAL CHANGES IN BINARY EUTECTICSPure element to maximum solubility at
Cooling composition C1:
Pure element to maximum solubility atroom temperature
Alloy is liquid until it passes through liquidus at b: -phase begins to form
More is formed while passing through +L region – compositional differences dictated by tie lines and boundaries
All liquid crystallizes to -phase with C1
Results in polycrystalline solid with uniform p y y fcomposition
Note crystal maintains -structure all the way to room temperatures (at equilibrium)way to room temperatures (at equilibrium).
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MICROSTRUCTURAL CHANGES IN BINARY EUTECTICSComposition range between room
C li iti C
Composition range between roomtemperature solubility and max. solidsolubility at eutectic temperature
Cooling composition C2:
Changes are similar to previous case as we move to solvus line
Just above solvus line, at point f, microstructure consists of grains with composition Ccomposition C2
Upon crossing the solvus line, the -solidsolubility is exceeded formation of y-phase particles
With continued cooling, particles willcontinue to grow in size because the continue to grow in size because the mass fraction of increases slightly with decreasing temperature
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EUTECTIC MICROSTRUCTURES
Solidification of eutectic composition:
L (61 9% Sn) (18 3% Sn) + (97 8% Sn) redistribution of tin andL (61.9% Sn) (18.3% Sn) + (97.8% Sn) – redistribution of tin and lead at eutectic temperature
Redistribution is accomplished by diffusion – microstructure is lamellae orRedistribution is accomplished by diffusion microstructure is lamellae or columnar structure. This lamellae/columnar configuration: atomic diffusion of lead and tin need only occur over relatively short distances 82
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During the eutectic transformation from liquid to solid ( + ), a redistribution of and tin is necessary because and and have different
i i i h f hi h i hcompositions, neither of which is the same as the liquid. Redistribution is through diffusion.
Liquid
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lamellar rodlike
globular acicular
Each eutectic alloy has its own characteristic microstructure
spheroidal, nodular, or globular; acicular (needles) orspheroidal, nodular, or globular; acicular (needles) or rod; and lamellar (platelets, Chinese script or dendritic, or filigreed) 84
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Compositions other than eutectic that, when cooled, cross the eutectic point
As eutectic temperature line is crossed, liquid phase, which is of the eutectic p , q p ,composition, will transform to eutectic structure (often lamellae)
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Anteater inclusionu
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cool this comp.
e
30
Relative amounts of microconstituents just below solidus:Primary microconstitutent, f’ = Q/(p + Q) = (61.9 – 30)/(61.9 – 18.3) = 73% is primary alpha
Eutectic microconstituent, fe = same as liquid phase from which it was transformed, fefe = p/(p + Q) = (30 – 18.3)/(61.9 – 18.3) = 27% is liquid
The eutectic liquid fractionates into alpha and beta solid solutions:The eutectic liquid fractionates into alpha and beta solid solutions:f = R/(p + Q + R) = (97.8 – 61.9)/(97.8 – 18.3) = 45% alpha
Total α = 73% + 27% × 0.45 = 85% ; 73% is primary α, 12% is eutectic α, 15% is β
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longer time longer time available to phase separate yields larger yields larger phase domains
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BINARIES WITH SOLID‐SOLID PHASE TRANSITIONSin systems without solid solutions solid-solid phase transitions are in systems without solid solutions, solid-solid phase transitions are indicated by isotherms
T1: αB βB
T2: αA βA2 β
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BINARIES WITH SOLID‐SOLID PHASE TRANSITIONSin systems with complete solid solubility the solid-solid phase in systems with complete solid solubility, the solid-solid phase transition temperature usually varies with composition
west 11.18two phase region containsone solid and one liquid phase
two phase region containstwo solid phases
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ALL SOLID INVARIANT POINTS
Liq. E βA + βBEutectic point E:
Eutectoid point B: βB βA + αB disproportionationreactions
Peritectoid point Y: βA + αB αAreactions
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BINARIES: THE BIG PICTURE
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Note: “oid” analogues of each these reactions may exist
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THREE‐COMPONENT CONDENSED SYSTEMS
with C = 3P + F = C + 1
P + F = 4
If i t t th i d d t i blIf pressure is constant, three independent variables: T & the concentrations of two components
The presence of four phases: F = 0, invariant point(usually three crystalline phases and liquid)
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EUTECTIC SYSTEM WITHOUT COMPOUNDS
Take 3 simple binary eutectic systems …
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… combine to form a triangular prism phase space
TemperatureTemperature
Composition projection onto b f t i l
p(in plane) base of triangle
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FINDING COMPOSITION ON A TERNARY DIAGRAM
Two Line MethodDraw two lines through Composition 1parallel to any two sides of the triangle –here AB and AC The intersection of
Two Line Method
here AB and AC. The intersection of these two lines with the third side (BC) divides this side into three line segments. The lengths of the individual line segments are proportional to the relative amounts of the three components A, B, and C.
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subliquidus equilibriasubliquidus equilibria
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CRYSTALLIZATION PATHWAYFor a general composition:
1) start at primary phase field2) proceed along univariant boundary curve3) finish t ut ti p int
For a general composition:
3) finish at eutectic point
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Anorthite
CRYSTALLIZATION PATHWAY ‐ EXAMPLE
Anorthite –Wollastonite –Sphene System
Anorthite – CaAl2Si2O8Wollastonite – CaSiO3Sphene – CaSiTiOSphene – CaSiTiO5
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EUTECTIC SYSTEM WITH ONE COMPOUNDsimplest case is one binary compound on a binary joinsimplest case is one binary compound on a binary join
• binary join A-BC(only A and BC along this line)
• two eutectic trianglesg(A-B-BC and A-BC-C)
ternary diagrams with multiple compounds solid solutions etcternary diagrams with multiple compounds, solid solutions, etc.become very complicated very quickly
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