Binary Phase Diagrams
by
Azizan Aziz Ph.D
Azizan Aziz
School of Materials and Mineral Resources Engineering
Complete Solubility in the Solid state.
simplest binary (two component) system –
where the characteristics of the two elements are
so similar → form solid solutions at all composition
-the
Hume-Rothery Rule
1. Atomic radii differ < 15%
2. Electrochemical nature of elements must be similar
3. Other things being equal, a metal of lower valency is
more likely to dissolve a higher valency metal than
vice versa
4. The crystal structures are the same
Azizan Aziz
School of Materials and Mineral Resources Engineering
Complete Solid Solution
At high temp, A & B completely
soluble in each other in liq.
state
At low temp, there is a single,
solid-solution phase
Azizan Aziz
School of Materials and Mineral Resources Engineering
At given state pt (T,X) within the 2-phase
region an A-rich liquid exist in
equilibrium in B-rich solid solution
The composition of each phase is
determine by drawing line
connecting the liquidus and
solidus line- tie line
Composition of liq – intersection
with liquidus
Composition of solid –
intersection with solidus
Azizan Aziz
School of Materials and Mineral Resources Engineering
Application of Gibbs Phase Rule
Invariant point
Any chg in temp is
possible but composition
are not independent
Single-phase , F=2, T and X
can be change independently
Azizan Aziz
School of Materials and Mineral Resources Engineering
Fig. above summarizes the microstructures
characteristic of the various regions of this phase
diagram
Azizan Aziz
School of Materials and Mineral Resources Engineering
Cu – Ni system is a classic
example of binary diagram
with complete solid solution
A variety of commercial
copper-nickel alloys falls
within this system including a
super alloy called Monel
The Cu – Ni alloy is found in
numerous applications of
technology especially marine,
shipping , chemical plant etc
due to its corrosion resistance
Cu – Ni system is also identified as an isomorphous system
Azizan Aziz
School of Materials and Mineral Resources Engineering
Lever Rule
To determined the relative
amount of each phase in terms
of composition
α = mα/ mα+ mβ = xβ- x/ xβ- xα
β = mβ/ mα+ mβ = x- xα/ xβ- xα
Azizan Aziz
School of Materials and Mineral Resources Engineering
Determination of
Phase
composition
A tie line is drawn thru.
B at 1250oC.
Overall composition of Alloy = 35 wt%
Composition α = 43 wt% Ni
Composition L = 32 wt% Ni
Azizan Aziz
School of Materials and Mineral Resources Engineering
α = Co- CL/Cα – CL = 35 – 32 / 43 – 32 = 3/11 = 0.27
L = Co- Co/Cα – CL = 43 – 35 / 43 – 32 = 8/11 = 0. 73
α = 27% and L = 73%
Azizan Aziz
School of Materials and Mineral Resources Engineering
Microstructure development during solidification
of isomorphous alloys
Alloy composn:35 wt% Ni
At 1300oC compsn alloy:
35 wt% Ni – 65 wt% Cu
Point a : alloy completely
liquid
Pt b: the first α solid
begins to form
As cooling progresses (pt
c,d,e) the composition
and relative amount of
each phases will change
Azizan Aziz
School of Materials and Mineral Resources Engineering
Solidification of a Solid-Solution
Alloy
Azizan Aziz
School of Materials and Mineral Resources Engineering
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The change in
structure of a Cu40% Ni alloy
during
equilibrium
solidification. The
nickel and copper
atoms must
diffuse during
cooling in order to
satisfy the phase
program and
produce a uniform
equilibrium
structure.
Azizan Aziz
School of Materials and Mineral Resources Engineering
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The cooling curve for an isomorphous alloy during solidification. We
assume that cooling rates are small so as to allow thermal equilibrium
to take place. The changes in slope of the cooling curve indicate the
liquidus and solidus temperatures, in this case for a Cu-40% Ni alloy.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Nonequilibrium Solidification and
Segregation
 Coring - Chemical segregation in cast products, also
known as microsegregation or interdendritic
segregation.
 Homogenization heat treatment - The heat treatment
used to reduce the microsegregation caused during
nonequilibrium solidification.
 Macrosegregation - The presence of composition
differences in a material over large distances caused by
nonequilibrium solidification.
 Spray atomization - A process in which molten alloys or
metals are sprayed using a ceramic nozzle.
Azizan Aziz
School of Materials and Mineral Resources Engineering
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The change in
structure of a Cu40% Ni alloy
during
nonequilibrium
solidification.
Insufficient time
for diffusion in the
solid produces a
segregated
structure.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :Nonequilibrium Solidification of
Cu-Ni Alloys
Calculate the composition and amount of each phase in a
Cu-40% Ni alloy that is present under the nonequilibrium
conditions shown in Figure earlier at 1300oC, 1280oC,
1260oC, 1240oC, 1200oC, and 1150oC. Compare with the
equilibrium compositions and amounts of each phase.
Azizan Aziz
School of Materials and Mineral Resources Engineering
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The change in
structure of a Cu40% Ni alloy
during
nonequilibrium
solidification.
Insufficient time
for diffusion in the
solid produces a
segregated
structure.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :SOLUTION
We use the tie line upto the equilibrium solidus
temperature to calculate composition and percentages
of phases as per the lever rule. Similarly, the
nonequilibrium solidus temperature curve is used to
calculate percentages and concentrations of different
phases formed under nonequilibrium conditions.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :SOLUTION (Continued)
Azizan Aziz
School of Materials and Mineral Resources Engineering
(a) Microsegregation between
dendrites can be reduced by a
homogenization heat treatment.
Counterdiffusion of nickel and
copper atoms may eventually
eliminate the composition
gradients and produce a
homogeneous composition. (b)
Spray atomized powders of
superalloys. (c) Progression of
densification in low carbon
Astroalloy sample processed using
HIP. (Courtesy of J. Staite, Hann,
B. and Rizzo, F., Crucible
Compaction Metals.)
Azizan Aziz
School of Materials and Mineral Resources Engineering
NiO – MgO phase is a ceramic
exhibiting complete solid
solution
Notice that the composition
axis expressed in mole percent
rather wt percent.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Eutectic Diagrams With No Solid Solution
Features:
at relatively low temp, there is 2phase field pure solids A and B
solidus line is horizontal line that
corresponds to the eutectic
temperature
any material with eutectic
composition fully melted at
eutectic temp
Azizan Aziz
other than eutectic will not fully
melted at e.t – have to be heated
further thru a 2-phase region
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some representative microstructures for the binary eutectic
system are shown in Fig above
fine-grained eutectic microstructure in which there are alternating
layers of components, pure A and pure B
sharp solidification point of the eutectic composition generally leads
to the fine-grained nature of the eutectic microstructure.
Even slow cooling solidification has to be fast .Hence limited time
for diffusion – leading to fine-grained
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example is Al – Si eutectic
system
There is very small amt
of solid solubility
The aluminium-rich side
describe behaviour of
some important
aluminium alloys
While the silicon-rich
illustrates the limit of
aluminium doping in
producing p-type
semiconductor( Al has a
deficiency of valence
electron → electron hole
or positive charge
Azizan Aziz
carrier.
School of Materials and Mineral Resources Engineering
Eutectic Diagrams With limited Solid Solution
for many binary systems the two
components are partially soluble in each
other.
- the p.d intermediate between the two
cases mentioned earlier
solid solution region near each edge
the two SS phases α and β are
distinguishable
they frequently have different crystal
structures
crystal structure of α will be of A and β of B
α consist of B atoms In solid solution in the crystal lattice of A
Azizan Aziz
School of Materials and Mineral Resources Engineering
The Fig above represent the various microstructures
Azizan Aziz
School of Materials and Mineral Resources Engineering
The Eutectic Phase Diagram
 Solvus - A solubility curve that separates a single-solid
phase region from a two-solid phase region in the phase
diagram.
 Isopleth - A line on a phase diagram that shows constant
chemical composition.
 Hypoeutectic alloy - An alloy composition between that of
the left-hand-side end of the tie line defining the eutectic
reaction and the eutectic composition.
 Hypereutectic alloys - An alloy composition between that
of the right-hand-side end of the tie line defining the
eutectic reaction and the eutectic composition.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
The lead-tin equilibrium phase diagram.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Azizan Aziz
Solidification and
microstructure of a
Pb-2% Sn alloy.
The alloy is a
single-phase solid
solution.
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Figure 10.10 Solidification, precipitation, and microstructure
of a Pb-10% Sn alloy. Some dispersion strengthening occurs
as the β solid precipitates.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :Phases in the Lead–Tin (Pb-Sn)
Phase Diagram
Determine (a) the solubility of tin in solid lead at 100oC,
(b) the maximum solubility of lead in solid tin, (c) the
amount of β that forms if a Pb-10% Sn alloy is cooled to
0oC, (d) the masses of tin contained in the α and β
phases, and (e) mass of lead contained in the α and β
phases. Assume that the total mass of the Pb-10% Sn
alloy is 100 grams.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Solidification, precipitation, and microstructure of a Pb-10%
Sn alloy. Some dispersion strengthening occurs as the β solid
precipitates.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example : SOLUTION
(a) The 100oC temperature intersects the solvus curve
at 5% Sn. The solubility of tin (Sn) in lead (Pb) at 100oC
therefore is 5%.
(b) The maximum solubility of lead (Pb) in tin (Sn),
which is found from the tin-rich side of the phase
diagram, occurs at the eutectic temperature of 183oC
and is 97.5% Sn.
(c) At 0oC, the 10% Sn alloy is in a α + β region of the
phase diagram. By drawing a tie line at 0oC and
applying the lever rule, we find that:
%   10 - 2  100  8.2%
100 - 2
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example : SOLUTION (Continued)
(d) The mass of Sn in the α phase = 2% Sn  91.8 g
of α phase = 0.02  91.8 g = 1.836 g. Since tin (Sn)
appears in both the α and β phases, the mass of Sn
in the β phase will be = (10 – 1.836) g = 8.164 g.
(e) Mass of Pb in the α phase
= 98% Sn  91.8 g of α phase = 0.98  91.8 g =
89.964 g
Mass of Pb in the β phase
= 90 - 89.964 = 0.036 g.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Summary of calculations (for example).
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Solidification and microstructure of the eutectic alloy
Pb-61.9% Sn.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
The cooling curve for a eutectic alloy is a simple thermal
arrest, since eutectics freeze or melt at a single
temperature.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
(a) Atom redistribution during lamellar growth of a lead-tin
eutectic. Tin atoms from the liquid preferentially diffuse to
the β plates, and lead atoms diffuse to the α plates. (b)
Photomicrograph of the lead-tin eutectic microconstituent
(x400).
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :Amount of Phases in the
Eutectic Alloy
(a) Determine the amount and composition of each
phase in a lead-tin alloy of eutectic composition. (b)
Calculate the mass of phases present. (c) Calculate
the amount of lead and tin in each phase, assuming
you have 200 g of the alloy.
SOLUTION
(a) The eutectic alloy contains 61.9% Sn.
 : (Pb  19% Sn) %   97.5  61.9  100  45.35%
97.5  19.0
 : (Pb  97.5% Sn) %   61.9  19.0  100  54.65%
97.5  19.0
Azizan Aziz
School of Materials and Mineral Resources Engineering
SOLUTION (Continued)
(b) At a temperature just below the eutectic:
The mass of the α phase in 200 g of the alloy =
mass of the alloy  fraction of the a phase
= 200 g  0.4535 = 90.7 g
The amount of the β phase in 200 g of the alloy =
(mass of the alloy  mass of the a phase)
= 200.0 g  90.7 g = 109.3 g
Azizan Aziz
School of Materials and Mineral Resources Engineering
SOLUTION (Continued)
- Mass of Pb in the α phase = mass of the a phase in 200 g 
(concentration of Pb in α) = (90.7 g)  (1 – 0.190) = 73.467 g
- Mass of Sn in the α phase = mass of the a phase - mass of Pb
in the a phase = (90.7 – 73.467 g) = 17.233 g
- Mass of Pb in β phase = mass of the b phase in 200 g  (wt.
fraction Pb in β) = (109.3 g)  (1 – 0.975) = 2.73 g
- Mass of Sn in the β phase = total mass of Sn – mass of Sn in
the α phase = 123.8 g – 17.233 g = 106.57 g
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Summary of calculations (for Example).
Azizan Aziz
School of Materials and Mineral Resources Engineering
Hypoeutectic
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
The solidification and microstructure of a hypoeutectic
alloy (Pb-30% Sn).
Azizan Aziz
School of Materials and Mineral Resources Engineering
(a) A hypoeutectic lead-tin alloy. (b) A
hypereutectic lead-tin alloy. The dark
constituent is the lead-rich solid α, the light
constituent is the tin-rich solid β, and the fine
plate structure is the eutectic (x400).
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :Determination of Phases and
Amounts in a Pb-30% Sn Hypoeutectic
Alloy
For a Pb-30% Sn alloy, determine the phases
present, their amounts, and their compositions at
300oC, 200oC, 184oC, 182oC, and 0oC.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example : SOLUTION
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example :Microconstituent Amount and
Composition for a Hypoeutectic Alloy
Determine the amounts and compositions of each
microconstituent in a Pb-30% Sn alloy immediately after
the eutectic reaction has been completed.
Example 10.6 SOLUTION
At a temperature just above the eutectic—say, 184oC—the
amounts and compositions of the two phases are:
 :19% Sn %   61.9  30  100  74%  % Primary 
L : 61.9% Sn
Azizan Aziz
61.9  19
% L  30  19  100  26%  % eutectic at 182o C
61.9  19
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(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Figure 10.18 The cooling curve for a hypoeutectic Pb30% Sn alloy.
Azizan Aziz
School of Materials and Mineral Resources Engineering
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Nonequilibrium Freezing in the
Eutectic System
Figure 10.28
Nonequilibrium
solidification and
microstruture of a Pb15% Sn alloy. A
nonequilbrium
eutectic
microconstituent can
form if the
solidification is too
rapid.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Example of this type of p.d is Pb – Sn
- common solder alloy falls within this system
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their low melting ranges allow for joining of most metals
by convenient heating methods,with low risk of damage
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to heat-sensitive parts
solders with less than 5% tin are used for sealing
containers, coating and joining metals,and
applications with service temperatures exceeding
120oC
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School of Materials and Mineral Resources Engineering
solders between10% and 20% Sn -sealing cellular
automobile radiators and filling seams dents in
automobile bodies.
General purpose solder – 40 to 50% Sn –pastelike
during application associated with the two –phase liquid
plus solid region just above the eutectic temp.
Solders near eutectic composition –heat- sensitive
electronic components requiring minimum heat
application.
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School of Materials and Mineral Resources Engineering
Solvus line
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School of Materials and Mineral Resources Engineering
Eutectoid Diagram
the transformation of eutectic
liquid to relatively fine grained
microstructure of two solid
phases upon cooling – eutectic
reaction
L (eutectic) →α + β
some binary systems contain
a solid-state analog of the
eutectic reaction
eutectoid reaction is
γ ( eutectoid) → α + β
Azizan Aziz
School of Materials and Mineral Resources Engineering
Some representative microstructures are shown in Fig 9-18
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School of Materials and Mineral Resources Engineering
Azizan Aziz
School of Materials and Mineral Resources Engineering
The Fe-Fe3C system –most important commercial phase
diagram
Provides the major scientific basis for iron and steel
industry
The boundary bet. Iron & steel will be identified as a
carbon content of 2.0 wt%.
Roughly corresponds to the carbon solubility limit in the
austenite γ phase
The most important areas of interest on this diagram
around the eutectic and eutectoid rxns
Azizan Aziz
School of Materials and Mineral Resources Engineering
Fe – Fe3C phase diagram
Note that the composition axis
is given in wt% C eventhough
Fe3C and not carbon is a
component.
Eventhough graphite (C) is a
more stable precipitate than
Fe3C, the rate of graphite
precipitation is enormously
slower than that of Fe3C.
In common steels ( and
many cast iron) the Fe3C
phase is metastable
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School of Materials and Mineral Resources Engineering
Fe – C phase diagram
This diagram Identical to
Fe-Fe3C
However the
intermediate Fe3C does
not exist
Azizan Aziz
School of Materials and Mineral Resources Engineering
PERITECTIC DIAGRAM
On heating, solid transform to liquid and another solid
phase.
This is known as peritectic rxn
δ+L→ε
( on heating)
δ+L←ε
( on cooling)
The peritectic rxn is another invariant rxn involving 3
phases at equilibrium
Azizan Aziz
School of Materials and Mineral Resources Engineering
Referring to Cu-Zn system
A peritectic exist for this system (pt P) at 598oC and
78.6 wt%Zn – 21.4 wt%Cu.
There are altogether 5 peritectics in the Cu-Zn system
They involve intermediate solid solutions as low –temp
phases that transform upon heating.
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School of Materials and Mineral Resources Engineering
Point P
(peritectic)
Eutectoid pt
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This eg. A and B form stable
compound AB which does not
melt at a single temp
Compound AB undergo
incongruent melting i.e the liquid
formed has composition other
than AB
The peritectic reaction can be
written as
AB → L + B (on heating)
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School of Materials and Mineral Resources Engineering
This is representative microstructures for the peritectic
diagram
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School of Materials and Mineral Resources Engineering
Al2O3 – SiO2 system
Important binary
system to ceramics
industry (Fe-Fe3C in
steel ind)
Several important
ceramics fall within this
system.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Refractory silica bricks are nearly pure SiO2 having
0.1-0.6 mol% Al2O3.
For silica bricks operating at above 1600oC . Need to keep
Al2O3 content as low as possible to minimize the amount of
liq. phase
Common fireclay refractories having 16 – 32 mol% Al2O3
Their usefulness as structural elements in furnace
limited by the solidus(eutectic) temperatures at 1587
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School of Materials and Mineral Resources Engineering
Increase refractoriness occurs at composition of the
incongruently melting point of mullite ( 3Al2O3.2SiO2) 60 mol%
Care must be taken in producing mullite refractories –
overall composition > 72 wt% (60 mol%) Al2O3 to avoid twophase region.
This will ensure the
refractory remains
solid to the peritectic
temp. (1890oC)
High alumina
refractories 6090wt% Al2O3.(46-84
mol% )
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School of Materials and Mineral Resources Engineering
General Binary Diagrams
The formation of intermediate
compounds is a relatively
common occurrence in binary
system
Fig shows a case of intermediate
compound, AB which melts
congruently
This system is equivalent to two binary system
This is what we termed a general diagram - a composite of
two or more diagrams
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School of Materials and Mineral Resources Engineering
Analysing this complex diagrams is
quite simple
Simply deal with the smallest binary
system associated with the overall
composition and ignore all others.
Fig shown illustrate this.
It shows for an overall compsn bet. AB & B we can treat the
diagram as simple binary eutectic of AB and B.
For practical purposes A-AB does not exist
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School of Materials and Mineral Resources Engineering
temperature
Composition
The above Fig is a relatively complex general diagram with
4 intermediate compounds – A2B, AB, AB2 and AB4
And several eg. of individual binary diagrams.
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School of Materials and Mineral Resources Engineering
temperature
composition
But for the overall composition shown above, only AB2 – AB4
binary is relevant
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School of Materials and Mineral Resources Engineering
This MgO-Al2O3
system is simply a
composite two
binary systems but
limited solubility
This diagram include
important intermediate
compound spinel MgOAl2O3 (or MgAl2O4) –an
important family of
magnetic materials
Azizan Aziz
School of Materials and Mineral Resources Engineering
Eventhough spinel is distinct compound (50mol%Al2O350mol%MgO), it is presented as a single phase rather than
a single line.
Implies that spinel is stable over a range of composition. –
nonstoichiometric for other than normal composition
There is limited solubility of Al2O3 in MgO below 1400oC
(left-hand extremity)
Similarly the lack MgO solubility in Al2O3
Also two eutectics on either side and the stoichiometric
spinels melt congruently at about 2100oC
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School of Materials and Mineral Resources Engineering
Good eg. of general
diagram
This complex
diagram can be
analyzed as a
simple eutectic in
high – aluminium
region
Important age-hardenable are found near the κ phase
boundary
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School of Materials and Mineral Resources Engineering
Azizan Aziz
School of Materials and Mineral Resources Engineering
This is another eg.
of complex diagram
simple to analyzed.
For eg many
commercial brass
compositions lie in
the single-phase α
region
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School of Materials and Mineral Resources Engineering
Important binary ceramic
system ZrO2- CaO
There is one eutectic
eutectoid
Pure ZrO2 transform
monoclinic to tetragonal at
1000oC
Transformation involve
large change in volume
resulting in crack
Azizan Aziz
School of Materials and Mineral Resources Engineering
This problem is overcome by “stabilizing” the zirconia by
adding bet. 3-7wt% CaO (18 mol%)
As shown by the phase diagram, this addition of CaO
produces cubic and tetragonal crystal structure are retaine
from room to melting point 2500oC and crack formation
circumvented.
This partially stabilized zirconia (PSZ) is a very
refractory,structural material.
Higher calcia content cubic phase may be retained – fully
stabilized zirconia.
Azizan Aziz
School of Materials and Mineral Resources Engineering
An alloy in the A-B system described by the Fig above is formed by melting
equal parts of A and A2B. Qualitatively described the microstructural
development that will occur upon slow cooling
this melt.
Schoolof
of Materials
and Mineral Resources Engineering
Azizan Aziz
Microstructural development during
slow cooling
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Gradual solidification of the 50%A – 50%B composition
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The microstructural development binary eutectic is
straight forward and was illustrated earlier.
However the composition and relative amount of α and β
change slightly below eutectic temp . Microstructurally minor
change.
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Hypoeutectic Composition
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The gradual growth of β above eutectic temp is
comparable to complete solid solution diagram
With one difference- crystallite growth stops at the eutectic
temp with ~67% of the microstructure solidified
Remaining liq (w. eutectic compsn) transform suddenly to the
eutectic microstructure upon cooling thru theuetectic temp.
Cooling of hypereutectic composition -2 forms of β
phase
Large grains – (L + β) → proectectic β
Finer grain (in the lamellar eutectic) known as
eutectic β
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This Fig shows the development for hypoeutectic composition
Similar to hyper except
Large grain – proeutectic α
Finer grain lamellar eutectic α
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School of Materials and Mineral Resources Engineering
temperature
Two other types of
microstructural development
A. For overall composition 10% B
Development is quite similar to that for complete solution
binary system
i.e solidification to a single –phase solid solution that remains
stable upon cooling to low temp.
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School of Materials and Mineral Resources Engineering
temperature
Overall composition
20%B
Situation similar to 10%
except upon cooling, α
phase become
saturated with B
atoms
Further cooling – pptn of a small amt. of β phase
Pptn of second phase - occur along the grain boundaries (
some cases within the grain)
Such cases encounterd in Al-Cu system –preciptation
hardening
Azizan Aziz
School of Materials and Mineral Resources Engineering
This Fig. illustrate
the cooling path
of a white cast
iron (composition
3.0 wt % C
Azizan Aziz
School of Materials and Mineral Resources Engineering
The eutectoid rxn to produce pearlite. This composition (0.77
wt% C is close to that for 1080 plain carbon steel.
Azizan Aziz
School of Materials and Mineral Resources Engineering
Hypoeutectoid steel
(containing < 0.77 wt% C)
overall compsn 0.5 wt% C
Final microstructure
If austenite cools just
below 750oC, ferrite
nucleates and grow
usually at austenite
grain boundaries
Primary ferrite continues to grow until the temp falls to 727oC
Remaining austenite at that temp is now surrounded by ferrite
and has change composition 0.5 to 0.77 wt% C
Subsequent cooling below 727oC causes all remaining
austenite
School of rxn
Materials and Mineral Resources Engineering
Azizan Aziz to transform to pearlite by eutectoid
This Fig. shows the hyperectectoid steel
Similar to hypoectectoid except that the proectectoid phase is
cementite.
Azizan Aziz
School of Materials and Mineral Resources Engineering