MSE 104: Phase Metallurgy: David Dye (2010-11)

Phase Diagrams Question Sheet 2

Comments and corrections to david.dye@imperial.ac.uk

Problems and Answers

Presented below is the aluminium-rich end of the Al-Cu binary phase diagram, which is the basis ofmany high-strength aluminium alloys. You may assume that all the phase boundaries may be describedby straight lines (not necessarily the case in reality).

0

100

200

300

400

500

600

700

0 10 20 30 40 50

L + α

α + θ

Composition (wt.% Cu)

Tem

pe

ratu

re (

oC

)

5.5 wt.% Cu 33 wt.% Cu

θ - Al2Cu

L + θL

α

660 oC

Al

548 oC

1 wt.% Cu

1. For an alloy containing 4wt.% Cu,

(a) Calculate, assuming slow cooling, the volume fraction and composition of the solid at 600◦C.

(b) Repeat the calculation for rapid cooling conditions (Scheil).

(c) For slow cooling, at what temperature does solidification complete?

(d) For fast cooling, what is the fraction eutectic?

(e) Sketch the solidified microstructure you would expect to observe for each assumption.

Answers:(a) First, we need to do some small calculations. The solidus line drops 112◦C in 5.5 wt.% Cu,or 20.36◦Cwt.%−1. So at 600◦C, or 60◦C below the freezing point of pure Al, the compositionof the solid will be 60/20.36 = 2.94 wt.%. The volume fraction will be given by the lever rule,which requires us to know the composition of the liquid at 600◦C. By a similar argument, wefind that its composition will be CL = 60/3.39 = 17.7 wt.%. So using the lever rule we find thatVα = (17.7− 4)/(17.7− 2.9) = 13.7/14.8 = 92.5%.(b) The partition coefficient k = Cα/CL = 5.5/33 = 1/6. The solid composition will vary from thatfound at the start of solidification = kC = 4/6 = 0.67 wt.% to the composition at the solidifyinginterface at 600◦C, which we have already found, 2.94 wt%. We then use the Scheil equation tofind Vs (Eq 84 in the notes). Rearranging, we find

Vs = 1− explnCs/kC

k − 1= 1− exp

{

−6

5ln

2.94

4/6

}

= 83.1%

so at any given temperature, more liquid is present for Scheil solidification than for equilibriumcooling. This makes sense, firstly because there is less Cu in the solid, so there must necessarily bemore liquid and secondly because for fast cooling we don’t have time for solidification to occur sothere will be a supersaturation of liquid waiting to solidify.

2 MSE 104: Phase Metallurgy: David Dye (2010-11)

(c) For equilibrium cooling, solidification completes when the alloy composition intersects the solidusline. This is at 660− 20.36× 4 = 578.6◦C.(d) For Scheil cooling, this is a repeat of (b) but where the liquid composition is the eutecticcomposition, or equivalently where the solid composition at the interface is 5.5 wt.%. So again

Vs = 1− explnCs/kC

k − 1= 1− exp

{

−1.2 ln5.5

4/6

}

= 92.1%

and so the fraction eutectic is 7.9%.(e) For slow cooling, we will just have grain of solid alpha phase. For Scheil cooling we will havegrain of cored alpha surrounded by a eutectic mixture of α and θ.

2. The SnO2–TiO2 system shows complete solid solution at high temperatures and a solvus at lowertemperatures. Thermodynamic data indicate that the value ofWH is 28.3 kJmol−1 in the expressionfor the enthalpy of mixing ∆Hmix = xSxTWH , where xS is the mole fraction SnO2 and xT is themole fraction TiO2.

(a) Calculate ∆Hmix at xS = 0, 0.1, 0.3, 0.5, 0.7, 0.9 and 1 and plot your values on a graph of∆Hmix against composition.

(b) Calculate the entropy of mixing at the same compositions and add a curve for −∆Smix to yourgraph (∆Smix = −R(xS lnxS + xT lnxT ) for random mixing of Sn and Ti on their sites).

(c) Calculate the Gibbs energy of mixing for these compositions at temperatures T of 1550◦C,1300◦C and 1000◦C. Make another graph of these values.

(d) From this graph, estimate the compositions of the co-existing phases in the two-phase regionand hence plot the solvus on a binary phase diagram.

Answer:We can make the following table:xS 0 0.1 0.3 0.5 0.7 0.9 1xT = 1− xS 1 0.9 0.7 0.5 0.3 0.1 0

(a) ∆Hmix kJmol−1 0 2.54 5.94 7.08 5.94 2.54 0

(b) −∆Smix Jmol−1 0 -2.70 -5.08 -5.76 -5.08 -2.70 0

(c) ∆Gmix(1823K) Jmol−1 0 -2380 -3320 -3430 -3320 -2380 0

(d) ∆Gmix(1573K) Jmol−1 0 -1700 -2050 -1990 -2050 -1700 0

(e) ∆Gmix(1273K) Jmol−1 0 -156 864 1312 864 -156 0

Note that we need to remember to do our calculations in K and mole fraction.

The graphs look like this;

-8

-6

-4

-2

0

2

4

6

8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

H

-S

-4000

-3500

-3000

-2500

-2000

-1500

-1000

-500

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1823K

1573K

1273K xs

H (kJ mol-1)

-S (J mol-1)

G (J mol-1)

xs

So at high temperatures, we notice that there is only on minimum and therefore there is no wayto draw a common tangent between a low xS and a high xT variant of the phase to lower theoverall energy of an alloy. However, at low temperatures there are two minima with a maxima inbetween. Therefore an alloy at (say) xS = 0.3 can lower its overall energy by decomposing into twovariants of the phase. At 1000◦C, these minima are at about xS = 0.1 and 0.9. So we can draw aphase boundary so that in between 0.1 and 0.9, we have a two phase region. If we do this at everytemperature, we generate the phase diagram in Figure 49.

So the point of this question is that is shows us how to calculate a phase diagram from expressionsfor the Gibbs energy / energies of the phase or phases.

MSE 104: Phase Metallurgy: David Dye (2010-11) 3

3. The MgO–FeO system shows complete solubility in both the solid and liquid phases. Values for thed420 interplanar spacing of the solid phase obtained by X-ray diffraction are given below.

(a) MgO and FeO are cubic with a = 4.213 A and 4.307 A, respectively. Using dhkl = a/√h2 + k2 + l2,

calculate d420 for the end-member (0 and 100) compositions.

(b) Vegard postulated that the lattice parameter (unit cell size) of a material varies linearly withcomposition. Thereby determine the unknown compositions of the solid phase in the Table(ignore the possible effect of thermal expansion).

(c) Hence plot the equilibrium phase diagram for the system.

(d) The Figure below shows the Gibbs energy curves of the solid and liquid phases for the sys-tem. By comparison with your equilibrium diagram, determine the temperature to which itcorresponds.

Liquid Composition Temp. Solid Solutionmol% MgO ◦C d420 A mol% MgO

0 1350 020 1900 0.95140 2200 0.94760 2500 0.94480 2700 0.943100 2800 100

G

FeO MgOmol % MgO

Solid

Liquid

65% 92%

Answer:(a) Plugging the numbers into the equation, d420(MgO)= 0.942 A and d420(FeO)= 0.963 A.(b) We can see that for mol%MgO=0, d420 = 0.963 in the solid, and for mol%MgO=1, d420 = 0.942,from (a). So d420 decreases by 0.021 A as the amount of MgO in solution varies from 0 to 1. Whenwe plot the graph, we can then just ask what mol% would give us the d-spacing found in the table,because those d-spacings must lie on that straight line.This gives us the graph below, at left.(c) We have therefore found the compositions of the solid corresponding to the given liquid compo-sitions and can plot the phase diagram;

Liquid Comp Temp. Solid Compositionmol% MgO ◦C d420 A mol% MgO

0 1350 0.963 020 1900 0.951 5740 2200 0.947 7660 2500 0.944 9080 2700 0.943 95100 2800 0.942 100

0

10

20

30

40

50

60

70

80

90

100

0.94 0.945 0.95 0.955 0.96 0.965

mo

l% M

gO

d_420 (A)

0

500

1000

1500

2000

2500

3000

0 20 40 60 80 100

Te

mp

era

ture

(o

C)

mol% MgO

L

L+S

S

(d) If we plot the common tangent on the diagram given, we find the liquid and solid compositionsare 65% and 90%, which corresponds on the phase diagram to a temperature of around 2600 ◦C.

4 MSE 104: Phase Metallurgy: David Dye (2010-11)

4. You are provided below with a partial phase diagram for the TiO2–Nb2O5 system.

(a) Designate each of the single phase fields (some have very limited solid solubility) and determinethe nominal compositions of the phases.

(b) Label the two phase fields and note the compositions and temperatures of the eutectic points.

(c) Draw a schematic diagram of the variation of Gibbs Energy against Composition for all thephases at a temperature of 1466◦C, that is, between the lower two eutectic temperatures.Label the known compositions of the phases and draw common tangents where appropriate[hint: draw the liquid first].

Answer:

(a) There are five phases; the liquid, α-Nb2O5, two intermediate compounds which we can denote βand γ, and finally δ-TiO2. β has around 25% TiO2, so the Nb2O5 and TiO2 are in the ratio 3:1. Sothis compound has a content of Nb6TiO17. Similarly, γ is at 50:50 composition, so its compositionis Nb2TiO7.

(b) There are eutectic poinst at about 1464◦C and 20%, 1468 ◦C and 37%, and 1476◦C and 58%.

(c) So the real trick here is to understand what the shapes of the Gibbs energy curves for thephases must look like. The liquid has a huge solubility range, so its Gibbs energy must vary quitegradually. β and γ, in contrast, only exist at very tightly defined compositions, so hypothetical β orγ at compositions away from that actually observed must be very unfavourable and therefore havevery high energy. The α and δ solid solutions are the intermediate case: they exist over a range ofcompositions but not all compositions; so their Gibbs Energy curves will vary fairly sharply.

The second thing we know from the phase diagram are the ranges over which the phases coexist.So at this temperature there are four two phase regions: α+L, β+L, β + γ and γ + δ. And thereare three single phase regions α, L and δ.

So then we draw the Gibbs energy curves for the five phases such that they form common tangentscorresponding to the two phase regions, and are unconditionally the lowest energy for a single phasefor the single phase solution regions. There are several ways to do this! One is given overleaf.

So in this question, we have gone through the reverse process - we have related the phase diagramto the form of the Gibbs energy curves. Finally, note that the common tangent goes through pointsof the same gradient on the Gibbs energy curves of different phases, NOT the minima.

MSE 104: Phase Metallurgy: David Dye (2010-11) 5

Tem

pe

ratu

re (

oC

)

Nb2O5 TiO2mol% TiO2

0 10080604020

1460

1480

1500

1520

αss

δssβ

γ

α+ββ+γ γ+δ

γ+L

α+Lβ+L

β+L γ+L

δ+L

L

Nb2O5 TiO2mol% TiO2

0 80604020

G (

kJ

mo

l-1)

αss δssβ+γ γ+δα+L β+LL

αss

δssβ

γ

L

6 MSE 104: Phase Metallurgy: David Dye (2010-11)

5. CMSX-4 is a nickel superalloy with composition Ni-6.4Cr-9.7Co-0.6Mo-1.0Ti-5.6Al-6.5Ta-6.4W-0.1Hf-3.0Re (wt.%). It is composed of two phases, an fcc γ-Ni phase and a primitive cubic γ′ phaseof nominal composition Ni3Al, where the Al atoms are placed on the cube face centres and the Niatoms are placed on the cube corners. You may assume that the solubility of Al, Ta and Ti in theγ is nil and that these atoms are only found on the Al site in the γ′ phase. Further, assume thatRe has no solubility in the γ′ and that all the remaining atomic species have complete solubilityand partition evenly in the γ phase and on the Ni site in the γ′. Therefore, calculate the volumefraction of γ′ and the compositions of each phase in at.%.

To do this problem, we need to figure out and then go through a series of logical steps.(a) First, we convert to atomic % to find out the fractions of atoms in the system, following Equation 11in the notes.(b) Then, we sum the atomic fractions of the Al, Ta and Ti. The γ′ phase has an composition X3Y, andthe question tells us that Y = Al, Ta or Ti, whilst X = Ni, Cr, Co, Mo, W, or Hf. Also, the γ phasedoesn’t contain any Al, Ta or Ti. So, the content of these atoms sets the amount of γ′ that must bepresent. The total of them is 16.08 at.%, so the γ′ fraction is 4× that, or 64.1%.(c) For the γ′ Al site, we can normalise these values to obtain their ratio’s on that site.(d) For all the species that occur on the γ′ Ni site, we now copy their compositions into a row; the sumof their contents is 82.9.(e) We then normalise each of these to get their ratios on the Ni site by dividing by 82.9%.So now we have the γ′ composition - it is (Ni75.8Cr9.0Co12.1Mo0.5W2.6Hf0.003)3(Al78.5Ta13.6Ti7.9).(f) Alternatively, the overall γ′ composition is given in (f) - multiply the Ni-site components by 0.75 andthe Al-site component by 0.25.(g) Now for the γ-“Ni” phase, we need to find out what we have left. For each species, we have theamount we started off with (62.91% for Ni) less the amount that is in the γ′ (for Ni, 56.9% × 64.3%).This leaves us with 26.3 at% for Ni. The sum of these fractions is the fraction of γ phase, or 35.7%.(h) Finally, we find the ratios of those by dividing by 35.7% to find the γ composition.

Ni Cr Co Mo Ti Al Ta W Hf Re sumat. mass 58.69 52.00 58.93 95.94 47.87 26.98 180.95 183.84 178.49 186.21wt% 60.7 6.4 9.7 0.6 1 5.6 6.5 6.4 0.1 3 100cx/mx 1.03 0.12 0.16 0.01 0.02 0.21 0.036 0.035 0.00056 0.016 1.6

(a) at.% 62.91 7.49 10.01 0.38 1.27 12.62 2.19 2.12 0.034 0.98 100(b) γ′ Al site 1.27 12.62 2.19 16.08(c) γ′ Al % 7.9 78.5 13.6 100(d)γ′ Ni site 62.9 7.5 10.0 0.4 2.1 0.034 82.9(e) γ′ Ni % 75.8 9.0 12.1 0.5 2.6 0.041 100(f) γ′ at% 56.9 6.8 9.1 0.3 2.0 19.6 3.4 1.9 0.031 100

(g) γ Ni site 26.3 3.1 4.2 0.2 0.0 0.0 0.0 0.9 0.014 1.0 35.7(h) γ at% 73.8 8.8 11.7 0.4 2.5 0.040 2.7 100

So, this is actually rather a tricky question. We can make some observations;(i) it often helps to make a spreadsheet to do the arithmetic. (ii) If we make some assumptions, we canmake an educated guess at the phase compositions of even quite complicated alloys.This alloy is used very heavily by Rolls-Royce to make single crystal turbine blades in jet engines - it isone of the best high temperature materials known, with oxidation and creep resistance, toughness andstrength. The γ′ phase is a very stable, very strong intermetallic that forms a fine dispersion of 10–200 nmprecipitates in the alloy, imparting almost all of the high temperature strength found in the alloy. Thematrix is just there to give the material some ductility, because intermetallics on their own tend to bebrittle. When designing these sorts of alloys it helps to have an idea of which atoms go where. In practice,the Al, Ti and Ta elements that promote the γ′ don’t partition this strongly, nor do the γ-formers likeMo partition so evenly. In fact, the alloy has a γ′ fraction of about 69%; but our estimate here isn’t toobad!Another thing to notice is the dramatic effect of atomic mass. Despite there only being 5.6 wt.% Al inthe alloy, it account for more than 1/8th of the atoms. Conversely, only 1 in 50 of the atoms are Ta.But, because Ta goes to the γ′ Al site, it still accounts for about 1/8th of the atoms on that site, so itseffect is quite dramatic.One final thing to notice is that although the Ni content looks quite low at around 60 wt.%, on its atomicsite it is actually quite high, at around 3/4.