Azizan Aziz School of Materials and Mineral Resources Engineering Binary Phase Diagrams Azizan Aziz Ph.D by

Azizan Aziz School of Materials and Mineral Resources Engineering

Binary Phase Diagrams

Azizan Aziz Ph.D

by


Complete Solubility in the Solid state.

-the simplest binary (two component) system –where the characteristics of the two elements are so similar → form solid solutions at all compositionHume-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


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


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


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


Fig. above summarizes the microstructures characteristic of the various regions of this phase diagram


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


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


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


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%


Microstructure development during solidification of isomorphous alloys

At 1300oC compsn alloy:

35 wt% Ni – 65 wt% Cu

Alloy composn:35 wt% Ni

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


Solidification of a Solid-Solution Alloy


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The change in structure of a Cu-40% 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.


©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.


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.


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The change in structure of a Cu-40% Ni alloy during nonequilibrium solidification. Insufficient time for diffusion in the solid produces a segregated structure.


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.

Example :Nonequilibrium Solidification of Cu-Ni Alloys


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The change in structure of a Cu-40% Ni alloy during nonequilibrium solidification. Insufficient time for diffusion in the solid produces a segregated structure.


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.


Example :SOLUTION (Continued)


(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.)


NiO – MgO phase is a ceramic exhibiting complete solid solution

Notice that the composition axis expressed in mole percent rather wt percent.


Eutectic Diagrams With No Solid Solution

at relatively low temp, there is 2-phase 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

other than eutectic will not fully melted at e.t – have to be heated further thru a 2-phase region

Features:


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


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 carrier.


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


The Fig above represent the various microstructures


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.

The Eutectic Phase Diagram


(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.


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Solidification and microstructure of a Pb-2% Sn alloy. The alloy is a single-phase solid solution.



Figure 10.10 Solidification, precipitation, and microstructure of a Pb-10% Sn alloy. Some dispersion strengthening occurs as the β solid precipitates.


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.

Example :Phases in the Lead–Tin (Pb-Sn) Phase Diagram



Solidification, precipitation, and microstructure of a Pb-10% Sn alloy. Some dispersion strengthening occurs as the β solid precipitates.


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:

%2.81002 - 1002 - 10 %


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.



Summary of calculations (for example).



Solidification and microstructure of the eutectic alloy Pb-61.9% Sn.



The cooling curve for a eutectic alloy is a simple thermal arrest, since eutectics freeze or melt at a single temperature.



(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).


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.

%65.541000.195.970.199.61 % ) %5.97(:

%35.451000.195.979.615.97 % ) %19(:

SnPb

SnPb


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


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



Summary of calculations (for Example).



The solidification and microstructure of a hypoeutectic alloy (Pb-30% Sn).

Hypoeutectic


(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).


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.


Example : SOLUTION


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:

Co182at eutectic %%26100199.61

1930 %Sn 61.9% :

Primary %%74100199.61309.61 %Sn 19% :

LL



Figure 10.18 The cooling curve for a hypoeutectic Pb-30% Sn alloy.


Nonequilibrium Freezing in the Eutectic System

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Figure 10.28 Nonequilibrium solidification and microstruture of a Pb-15% Sn alloy. A nonequilbrium eutectic microconstituent can form if the solidification is too rapid.


Example of this type of p.d is Pb – Sn

- common solder alloy falls within this system their low melting ranges allow for joining of most metals by convenient heating methods,with low risk of damage 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


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.


Solvus line


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


Some representative microstructures are shown in Fig 9-18



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


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


Fe – C phase diagram

This diagram Identical to Fe-Fe3C

However the intermediate Fe3C does not exist


PERITECTIC DIAGRAMOn 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


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.


Point P (peritectic)

Eutectoid pt


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)


This is representative microstructures for the peritectic diagram


Al2O3 – SiO2 system

Important binary system to ceramics industry (Fe-Fe3C in steel ind)

Several important ceramics fall within this system.


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


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 two-phase region.

This will ensure the refractory remains solid to the peritectic temp. (1890oC)

High alumina refractories 60-90wt% Al2O3.(46-84 mol% )


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


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


tem

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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.


tem

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composition

But for the overall composition shown above, only AB2 – AB4

binary is relevant


This MgO-Al2O3 system is simply a composite two binary systems but limited solubility

This diagram include important intermediate compound spinel MgO-Al2O3 (or MgAl2O4) –an important family of magnetic materials


Eventhough spinel is distinct compound (50mol%Al2O3-50mol%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


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



This is another eg. of complex diagram simple to analyzed.

For eg many commercial brass compositions lie in the single-phase α region


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


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.


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 of this melt.


Microstructural development during slow cooling


Gradual solidification of the 50%A – 50%B composition


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.


Hypoeutectic Composition


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


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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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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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


This Fig. illustrate the cooling path of a white cast iron (composition 3.0 wt % C


The eutectoid rxn to produce pearlite. This composition (0.77 wt% C is close to that for 1080 plain carbon steel.


Hypoeutectoid steel (containing < 0.77 wt% C)

overall compsn 0.5 wt% C

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 to transform to pearlite by eutectoid rxn

Final microstructure


This Fig. shows the hyperectectoid steel

Similar to hypoectectoid except that the proectectoid phase is cementite.

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Azizan Aziz School of Materials and Mineral Resources Engineering Binary Phase Diagrams Azizan Aziz Ph.D by