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 of
many high-strength aluminium alloys. You may assume that all the phase boundaries may be described
by straight lines (not necessarily the case in reality).
700
660 oC
L
Temperature (oC)
600
L+θ
L+α
548 oC
α
500
33 wt.% Cu
5.5 wt.% Cu
400
α+θ
300
200
100
1 wt.% Cu
0
0
10
Al
20
30
Composition (wt.% Cu)
40
50
θ - Al2Cu
1. For an alloy containing 4wt.% Cu,
(a)
(b)
(c)
(d)
(e)
Calculate, assuming slow cooling, the volume fraction and composition of the solid at 600◦ C.
Repeat the calculation for rapid cooling conditions (Scheil).
For slow cooling, at what temperature does solidification complete?
For fast cooling, what is the fraction eutectic?
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◦ C wt.%−1 . So at 600◦ C, or 60◦ C below the freezing point of pure Al, the composition
of 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, we
find that its composition will be CL = 60/3.39 = 17.7 wt.%. So using the lever rule we find that
Vα = (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 that
found at the start of solidification = kC = 4/6 = 0.67 wt.% to the composition at the solidifying
interface at 600◦C, which we have already found, 2.94 wt%. We then use the Scheil equation to
find Vs (Eq 84 in the notes). Rearranging, we find
6 2.94
ln Cs /kC
= 83.1%
= 1 − exp − ln
Vs = 1 − exp
k−1
5
4/6
so at any given temperature, more liquid is present for Scheil solidification than for equilibrium
cooling. This makes sense, firstly because there is less Cu in the solid, so there must necessarily be
more liquid and secondly because for fast cooling we don’t have time for solidification to occur so
there 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 solidus
line. 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 eutectic
composition, or equivalently where the solid composition at the interface is 5.5 wt.%. So again
5.5
ln Cs /kC
= 92.1%
= 1 − exp −1.2 ln
Vs = 1 − exp
k−1
4/6
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 have
grain 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 lower
temperatures. Thermodynamic data indicate that the value of WH is 28.3 kJmol−1 in the expression
for the enthalpy of mixing ∆Hmix = xS xT WH , where xS is the mole fraction SnO2 and xT is the
mole 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 your
graph (∆Smix = −R(xS ln xS + xT ln xT ) 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 region
and hence plot the solvus on a binary phase diagram.
Answer:
We can make the following table:
xS
0
0.1
xT = 1 − xS
1
0.9
0
2.54
(a) ∆Hmix kJ mol−1
0 -2.70
(b) −∆Smix J mol−1
(c) ∆Gmix (1823 K) J mol−1 0 -2380
(d) ∆Gmix (1573 K) J mol−1 0 -1700
-156
(e) ∆Gmix (1273 K) J mol−1 0
0.3
0.7
5.94
-5.08
-3320
-2050
864
0.5
0.5
7.08
-5.76
-3430
-1990
1312
0.7
0.3
5.94
-5.08
-3320
-2050
864
0.9
0.1
2.54
-2.70
-2380
-1700
-156
1
0
0
0
0
0
0
Note that we need to remember to do our calculations in K and mole fraction.
The graphs look like this;
H (kJ mol-1) 8
-S (J mol-1)
xs
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
6
H
-500
-S
4
-1000
2
-1500
0
1823K
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-2000
xs
-2500
-4
-3000
-6
-3500
-8
1573K
1273K
-2
G (J mol-1)
-4000
So at high temperatures, we notice that there is only on minimum and therefore there is no way
to draw a common tangent between a low xS and a high xT variant of the phase to lower the
overall energy of an alloy. However, at low temperatures there are two minima with a maxima in
between. Therefore an alloy at (say) xS = 0.3 can lower its overall energy by decomposing into two
variants of the phase. At 1000◦C, these minima are at about xS = 0.1 and 0.9. So we can draw a
phase boundary so that in between 0.1 and 0.9, we have a two phase region. If we do this at every
temperature, 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 expressions
for 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 the
d420 interplanar spacing of the solid phase obtained by X-ray diffraction are given below.
√
(a) MgO and FeO are cubic with a = 4.213 Å and 4.307 Å, respectively. Using dhkl = a/ h2 + k 2 + 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 with
composition. 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 system. By comparison with your equilibrium diagram, determine the temperature to which it
corresponds.
Liquid Composition
mol% MgO
0
20
40
60
80
100
Temp.
◦
C
1350
1900
2200
2500
2700
2800
Solid Solution
d420 Å mol% MgO
0
0.951
0.947
0.944
0.943
100
G
Solid
Liquid
65%
FeO
mol % MgO
92%
MgO
Answer:
(a) Plugging the numbers into the equation, d420 (MgO)= 0.942 Å and d420 (FeO)= 0.963 Å.
(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 Å as the amount of MgO in solution varies from 0 to 1. When
we 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 compositions and can plot the phase diagram;
Liquid Comp
mol% MgO
0
20
40
60
80
100
Temp.
◦
C
1350
1900
2200
2500
2700
2800
Solid Composition
d420 Å mol% MgO
0.963
0
0.951
57
0.947
76
0.944
90
0.943
95
0.942
100
3000
100
2500
L
Temperature (oC)
90
80
mol% MgO
70
60
50
40
30
L+S
2000
1500
S
1000
500
20
10
0
0
0.94
0.945
0.95
0.955
0.96
0.965
0
20
40
60
mol% MgO
80
100
d_420 (A)
(d) If we plot the common tangent on the diagram given, we find the liquid and solid compositions
are 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 –Nb2 O5 system.
(a) Designate each of the single phase fields (some have very limited solid solubility) and determine
the 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 the
phases 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-Nb2 O5 , two intermediate compounds which we can denote β
and γ, and finally δ-TiO2 . β has around 25% TiO2 , so the Nb2 O5 and TiO2 are in the ratio 3:1. So
this compound has a content of Nb6 TiO17 . Similarly, γ is at 50:50 composition, so its composition
is Nb2 TiO7 .
(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 the
phases must look like. The liquid has a huge solubility range, so its Gibbs energy must vary quite
gradually. β and γ, in contrast, only exist at very tightly defined compositions, so hypothetical β or
γ at compositions away from that actually observed must be very unfavourable and therefore have
very high energy. The α and δ solid solutions are the intermediate case: they exist over a range of
compositions 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 there
are 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 tangents
corresponding to the two phase regions, and are unconditionally the lowest energy for a single phase
for 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 diagram
to the form of the Gibbs energy curves. Finally, note that the common tangent goes through points
of the same gradient on the Gibbs energy curves of different phases, NOT the minima.
MSE 104: Phase Metallurgy: David Dye (2010-11)
5
L
1500
γ
δ+L
β
1480
αss
1460
α+L
γ+L
β+L
β+L
γ+δ
β+γ
α+β
0
Nb2O5
δss
γ+L
20
40
mol% TiO2
G (kJ mol-1)
Temperature (oC)
1520
60
80
100
TiO2
γ
αss
δss
β
L
αss
0
Nb2O5
α+L L β+L
20
γ+δ
β+γ
40
mol% TiO2
60
δss
80
TiO2
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-6.4W0.1Hf-3.0Re (wt.%). It is composed of two phases, an fcc γ-Ni phase and a primitive cubic γ ′ phase
of nominal composition Ni3 Al, where the Al atoms are placed on the cube face centres and the Ni
atoms are placed on the cube corners. You may assume that the solubility of Al, Ta and Ti in the
γ is nil and that these atoms are only found on the Al site in the γ ′ phase. Further, assume that
Re has no solubility in the γ ′ and that all the remaining atomic species have complete solubility
and partition evenly in the γ phase and on the Ni site in the γ ′ . Therefore, calculate the volume
fraction 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 11
in the notes.
(b) Then, we sum the atomic fractions of the Al, Ta and Ti. The γ ′ phase has an composition X3 Y, and
the question tells us that Y = Al, Ta or Ti, whilst X = Ni, Cr, Co, Mo, W, or Hf. Also, the γ phase
doesn’t contain any Al, Ta or Ti. So, the content of these atoms sets the amount of γ ′ that must be
present. 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 sum
of 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 (Ni75.8 Cr9.0 Co12.1 Mo0.5 W2.6 Hf0.003 )3 (Al78.5 Ta13.6 Ti7.9 ).
(f) Alternatively, the overall γ ′ composition is given in (f) - multiply the Ni-site components by 0.75 and
the 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 the
amount 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.
at. mass
wt%
cx/mx
(a) at.%
(b) γ ′ Al site
(c) γ ′ Al %
(d)γ ′ Ni site
(e) γ ′ Ni %
(f) γ ′ at%
(g) γ Ni site
(h) γ at%
Ni
58.69
60.7
1.03
62.91
Cr
52.00
6.4
0.12
7.49
Co
58.93
9.7
0.16
10.01
Mo
95.94
0.6
0.01
0.38
62.9
75.8
56.9
26.3
73.8
7.5
9.0
6.8
3.1
8.8
10.0
12.1
9.1
4.2
11.7
0.4
0.5
0.3
0.2
0.4
Ti
47.87
1
0.02
1.27
1.27
7.9
2.0
0.0
Al
26.98
5.6
0.21
12.62
12.62
78.5
19.6
0.0
Ta
180.95
6.5
0.036
2.19
2.19
13.6
3.4
0.0
W
183.84
6.4
0.035
2.12
Hf
178.49
0.1
0.00056
0.034
Re
186.21
3
0.016
0.98
2.1
2.6
1.9
0.9
2.5
0.034
0.041
0.031
0.014
0.040
1.0
2.7
sum
100
1.6
100
16.08
100
82.9
100
100
35.7
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 can
make 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 is
one of the best high temperature materials known, with oxidation and creep resistance, toughness and
strength. The γ ′ phase is a very stable, very strong intermetallic that forms a fine dispersion of 10–200 nm
precipitates in the alloy, imparting almost all of the high temperature strength found in the alloy. The
matrix is just there to give the material some ductility, because intermetallics on their own tend to be
brittle. 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 like
Mo partition so evenly. In fact, the alloy has a γ ′ fraction of about 69%; but our estimate here isn’t too
bad!
Another thing to notice is the dramatic effect of atomic mass. Despite there only being 5.6 wt.% Al in
the 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 its
effect is quite dramatic.
One final thing to notice is that although the Ni content looks quite low at around 60 wt.%, on its atomic
site it is actually quite high, at around 3/4.