Chapter 9: Phase Diagrams

Chapter 10: Phase Diagrams
Show figures 10-1 and 10-3, and discuss the difference between a component and a phase. A
component is a distinct chemical entity, such as Cu, Ni, NiO or MgO. A phase is a chemically
and structurally homogeneous portion of the microstructure. A phase often contains more
than one component. Figure 10-1 (below) illustrates a two-component (sugar, water), two-phase
(sugar solution, solid sugar) system. Figures 10-3 illustrates a system with two-components (Cu,
Ni) and two phases (liquid, solid solution). In some cases, the two different phases could have the
same crystal structure, but the two materials are immiscible because of differing atomic radii,
electronegativity, valence, etc. Clearly, solid, liquid and gas also represent different phases.
Phase diagrams:
Phase diagrams map the number and types of phases of phases that are present, the composition of
each phase, and the microstructures that exist. For practical reasons, this course will only cover
binary (two-component) phase diagrams, although of course many materials of interest contain
more than two components. Type 316 stainless steel has the following composition (weight %): C
(0.10), Mo (2.5), Ni (11), Cr (18), Fe (68.4). The phase diagram of this material is 5-dimensional,
and beyond the scope of this course.
The Cu-Ni and binary phase diagram (Figure 10.3) is the simplest type of binary phase diagrams for
two metals. Figure 10.3 shows that Cu and Ni are mutually soluble at room temperature throughout
the entire range of compositions. Remember from Chapter 5 that we discussed the Hume-Rothery
rules for deciding when two metals will be soluble, and we concluded that Cu and Ni form a perfect
solid solution. The detailed comparison was:
RCu = 0.128 nm, RNi = 0.125 nm,
2.4% difference.
Electronegativities of Cu and Ni are
1.9 and 1.8, respectively.
Both have a FCC crystal structure.
Valences are +1, +2.
Our conclusion (Chapter 5) that Cu
and Ni form a complete solid
solution is represented graphically
by Figure 10.3 (at the right). When
two metals do not form a complete
solid solution, the phase diagram
will become more complex.
The Al2O3- Cr2O3 phase diagram
(Figure 10.23) is the simplest type of
binary phase diagram for two
ceramic materials. Figure 10.23 is
completely analogous to Figure
10.3. While we did not explicitly
discuss Hume-Rothery-type rules for
predicting ceramic solubility, similar
guidelines do exist. We expect a
new
criterion,
the
same
stoichiometric ratio of anions to
cations, to replace the criterion of
having similar valence. From the
inside cover of the text, the ionic
radii of Al and Cr 0.053 and 0.063
nm.
The borderline between the liquid
phase and two-phase (liquid + solid)
region is termed the liquidus line,
while the borderline between the solid phase and two-phase (liquid + solid) region is termed the
solidus line. In the two-phase region, given the temperature, we can determine the composition of
each of the phases by drawing a tie line, as illustrated in figure 10.3b.
Example Problem:
At point B (35 wt% Ni, 1250ºC) on figure 10.3b, what are the mass fraction liquid and the mass
fraction solid?
This must be solved using the lever rule, or the inverse lever rule, I have heard both terminologies
used. As described in the text:
WL =
C − C0
S
= α
R + S Cα − C L
WL =
42.5 − 35
= 0.68
42.5 − 31.5
The reason that some people call this the inverse lever rule is the line segment (S) chosen to
calculate the mass fraction liquid is on the solid side of the phase diagram, not the liquid side.
Similarly:
Wα =
C − CL
R
= 0
R+ S
Cα − C L
Wα =
35 − 31.5
= 0.32
42.5 − 31.5
Obviously, these must add up to 1.0.
Example Problem:
Calculate the amount of each phase present in 1 kg of a 50 wt.% Ni- 50 wt.% Cu alloy at a) 1400°C,
b) 1300°C and c) 1200°C.
a)
For a 50 wt.% Ni- 50 wt.% Cu alloy at 1400°C, we are in the liquid (L) region of the phase
diagram. Therefore, we have 1 kg of liquid (L).
b)
For a 50 wt.% Ni- 50 wt.% Cu alloy at 1300°C, we are in the solid + liquid (α+L) region of
the phase diagram. Here we must use the lever rule to calculate the mass fraction of each
phase. Unfortunately, figure 10.2 does not contain an insert enlarging the α+L region of the
phase diagram, so accurate reading of this figure is difficult. However, if we draw a tie line
across the α+L region at 1300°C, the endpoints are at about 45 wt.% Ni and 60 wt.% Ni.
Therefore, the mass fractions are:
WL =
Cα − C 0
(1 kg ) = 60 − 50 (1 kg ) = 0.67 kg
60 − 45
Cα − C L
Wα =
C0 − C L
(1 kg ) = 50 − 45 (1 kg ) = 0.33 kg
60 − 45
Cα − C L
Note that when determining the mass fraction of the liquid (solid) phase, the numerator
contains the composition of the solid (liquid) phase. This seems backwards to some people,
and they refer to the lever rule as the inverse lever rule to remind themselves of this. To
ensure that you get the correct answer, you should always look at the problem visually and
determine which mass fraction is larger. Then you can check the answer. In addition, the
two mass fractions (0.67 and 0.33) must add to 1.
c)
For a 50 wt.% Ni- 50 wt.% Cu alloy at 1200°C, we are in the solid (α) region of the phase
diagram. Therefore, we have 1 kg of solid (α).
Show figures 10.4 and 10.5. What happens during cooling from high temperature (liquid) to room
temperature (solid)? As soon as the liquidus line is crossed, the solid solution phase begins to
precipitate from the liquid. What are the composition of the precipitating solid and the remaining
liquid? The composition of the solid and liquid at every step along the way can be determined by
drawing a tie line at that temperature. Eventually, the temperature drops below the solidus line, and
precipitation of the solid is complete. No further significant change occurs in the microstructure
upon cooling to room temperature.
Show Figure 10.6, which illustrates strain hardening for both pure Cu and pure Ni, when small
amounts of the other are added.
Eutectic phase diagrams:
What happens when two materials are not completely soluble, the more typical situation? In the
simplest case, a binary eutectic phase diagram is observed. Show figures 10.7 and 10.8.
For Ag and Cu, we can apply the Hume-Rothery “rules” as follows:
RCu = 0.128 nm, RAg = 0.144 nm, 12.5% difference.
Electronegativities of Cu and Ag are both 1.9.
Both have a FCC crystal structure.
Valences are +1, +2.
Although Cu and Ag seem to follow the Hume-Rothery rules, Figure 10.7 shows that they do not
form a complete solid solution. The critical difference is that their atomic radii are not that similar.
This shows that the Hume-Rothery rules are really guidelines, not rules.
We can also apply the Hume-Rothery “rules” to Sn and Pb as follows:
RSn = 0.151 nm, RPb = 0.175 nm, 15.9% difference.
Electronegativities of Sn and Pb are both 1.8.
Sn has a BCT crystal structure, while Pb has a FCC crystal structure
Valences of both are +2, +4.
Since they have different crystal structures, the relative insolubility of Sn and Pb is expected, and is
shown graphically in Figure 10.8. The eutectic reaction is described by L → α + β, occurring at the
eutectic composition. Any phase transformation during which a liquid transforms simultaneously to
two solid phases is termed a eutectic reaction. The binary eutectic phase diagram has two liquidus
lines separating L from (L + α) and (L + β) and several solidus lines separating α from (α + L), β
from (β + L), (α + β) from (α + L), and (α + β) from (β + L). Note that the borderline between the
α phase and the (α + β) and (α + L) regions defines the T-dependent solubility limit of Sn in Pb.
Show figure 10.8, the Pb-Sn phase diagram. Does the eutectic reaction only occur when you
happen to be cooling an alloy of exactly the eutectic composition (61.9 wt.% Sn)? No. When the
Sn composition is between 19 and 97.5 wt.% Sn, liquid of the eutectic composition exists when the
alloy crosses the horizontal line during cooling from high temperature (liquid) to room temperature.
Explain this in detail with two examples. Show figures 10.9 through 10.11, which illustrate the
microstructures that exist at various compositions and temperatures. When a liquid of eutectic
composition solidifies, a eutectic microstructure is formed with alternating layers of α and β phases.
These thin layers form because the solidification reactions occurs all at once, rather than gradually
as in the case of complete solid solubility (Cu-Ni for example). Diffusion is not rapid enough to
completely separate the α and β phases that form.
Note that Pb-Sn materials were widely used as solders, since the melting temperature (183°C) of the
eutectic is considerably below that of pure Sn (232.0°C) and pure Pb (327.5°C). Due to
environmental concerns about Pb, they are being replaced by other solder materials, such as Ag-Sn.
This is an extremely good example of using phase diagrams to engineer the properties of
materials!! All soldier materials have phase diagrams that look like Figure 10.8.
Show Figures 10.11 through 10.17 and explain the development of the microstructure of Sn-Pb
alloys upon cooling.
The Figure on the left side illustrates phase changes for a Sn composition below the solubility
limit in Pb, at all temperatures. The Figure on the right side illustrates the phase changes for
a Sn composition below the solubility limit in the α phase at elevated temperature, but above
that solubility limit at room temperature.
Gradual cooling at the eutectic composition (Figure above) results in simultaneous conversion
from 100% liquid to 100% solid. Since phase separation is instantaneous, but solid diffusion
is slow, the striped eutectic microstructure in the Figure below is formed.
In the Figure above, the Sn composition is 40 wt%, higher than the solubility limit of Sn in the
α phase, but less than the eutectic composition. The final microstructure is shown in the
figure below, which illustrated the primary α surrounded by stripes of the eutectic
microstructure.
Example Problem:
Calculate the amount of each phase present in 1 kg of a 50 wt.% Pb- 50 wt.% Sn alloy at a) 300°C,
b) 200°C and c) 100°C. Also calculate the amount of each microstructure.
a)
For a 50 wt.% Pb- 50 wt.% Sn alloy at 300°C, we are in the liquid (L) region of the phase
diagram. Therefore, we have 1 kg of liquid (L). The liquid is also the microstructure.
b)
For a 50 wt.% Pb- 50 wt.% Sn alloy at 200°C, we are in the solid + liquid (α+L) region of
the phase diagram. Here we must use the lever rule to calculate the mass fraction of each
phase. If we draw a tie line across the α+L region at 200°C, the endpoints are at about 17
wt.% Sn and 54 wt.% Sn. Therefore, the mass fractions are:
WL =
C 0 − Cα
(1 kg ) = 50 − 17 (1 kg ) = 0.89 kg
54 − 17
C L − Cα
Wα =
C L − C0
(1 kg ) = 54 − 50 (1 kg ) = 0.11 kg
C L − Cα
54 − 17
Again, note that the numerator of each quantity contains the composition of the opposite
phase, but this must be true in order to obtain mostly liquid (L) phase. Looking at the tie
line, the composition of interest (50 wt.% Sn) is much closer to the liquid (L) than to the
solid (α) phase, so mL >> mα. In addition, the two mass fractions (0.89 and 0.11) sum to 1.
Compare the equations given above for mL and mα to those employed in the previous
problem for mL and mα. They are not the same!! In the current problem, the liquid is
on the right-hand side of the α phase, but in the previous problem, the liquid is on the
left-hand side of the α phase. If you memorize formulas instead of understanding the
use of the lever rule, you will not be able to do this type of problem correctly!!
Note that the two microstructures that exist here are the same as the two phases, liquid and α
solid solution.
c)
For a 50 wt.% Pb- 50 wt.% Sn alloy at 100°C, we are in the α + β region of the phase
diagram. Drawing a tie line across this region of the phase diagram, the endpoints are at
about 98 and 5 wt.% Sn. Therefore, the mass fractions are:
Wα =
Wβ =
Cβ − C0
C β − Cα
(1 kg ) = 98 − 50 (1 kg ) = 0.52 kg
98 − 5
C 0 − Cα
(1 kg ) = 50 − 5 (1 kg ) = 0.48 kg
98 − 5
C β − Cα
Here the microstructure differs from the phase, and this is determined by what happens just
above the eutectic temperature, 183ºC. First, note that we on the left side of the eutectic
composition, 61.9 wt% Sn, so the primary phase is α. Remember, the primary phase is the
one that precipitates out prior to the eutectic reaction. The remaining liquid is transformed
into the eutectic microstructure. Therefore:
W primary α (100°C ) = Wα (183°C + ) =
W eutectic (100°C ) = W L (183°C + ) =
C L − C0
(1 kg ) = 61.9 − 50 (1 kg ) = 0.27 kg
C L − Cα
61.9 − 18.3
C 0 − Cα
(1 kg ) = 50 − 18.3 (1 kg ) = 0.73 kg
C L − Cα
61.9 − 18.3
The distinction between the mass (mass fraction) of each phase and the mass (mass fraction)
of each microstructure is an extremely important one. About 1/3 of the class will get this
wrong on the next exam.
Example Problem
The microstructure of a Cu-Ag alloy at 775ºC consists of primary α and eutectic microstructures. If
the mass fractions of these two micro-constituents are 0.73 and 0.27, respectively, determine the
alloy composition.
First, you need to realize that primary α refers to the α phase that precipitates out prior to the
eutectic reaction. Thus, the weight fraction primary α is equal to the weight fraction α just above
the eutectic temperature, 779ºC. Similarly, the weight fraction eutectic microstructure is equal to
the weight fraction L just above this temperature.
W primary α (775°C ) = Wα (779°C + ) =
W primary α (775°C ) =
0.73 =
C eut − C 0
C eut − C α
71.9 − C 0
71.9 − 8.0
71.9 − C 0
71.9 − 8.0
C 0 = 25.3 wt %
Show figures 10.19, 20.20 and 10.22 (Cu-Zn, Mg-Pb and Ni-Ti). These are much messier phase
diagrams, but every region in each phase diagram contains either one phase or two phases. In the
two-phase regions, the mass fraction of each phase can be obtained from the (inverse) lever rule,
just as we have done above for more complex phase diagrams.
Some of these phase diagrams contain intermetallic compounds, which behave as though they are a
chemically distinct component. These show up as something like a vertical line in a phase diagram.
Examples include Mg2Pb in figure 10.20 and the vertical line at 44.9 wt% Ti in figure 10.22.
An additional reaction is also introduced, the peritectic reaction δ + L → γ ε, where a solid and a
liquid phase react to form a different solid phase.
Example Problem:
Consider 1 kg of brass with a composition of 35 wt.% Zn- 65 wt.% Cu. This phase diagram is
shown in figure 10.21.
a)
Upon cooling, at which temperature does the first solid appear?
Drawing a vertical line at 35 wt.% Zn, 930°C.
b)
What is the first solid phase to appear, and what is its composition?
On the same vertical line, the sample crosses from the L phase to the two-phase α + L
region. Thus, the first solid that forms is the Cu-rich α solid solution, which probably has a
FCC structure. Drawing a tie line at 930°C, the composition of the first α to precipitate is
about 25 wt.% Zn.
c)
At which temperature will the alloy completely solidify?
From the same vertical line, complete solidification occurs at about 900°C.
d)
Over which temperature range will the microstructure be completely in the α phase?
The alloy is 100% α phase from 670°C down to about 200°C, where this alloy enters the α
+ β two-phase region.
Show a series of binary ceramic phase diagrams(Al2O3- Cr2O3, MgO- Al2O3, ZrO2-CaO), and
discuss the analogies to binary metal phase diagrams:
Eutectoid phase diagrams:
Show figure 10.28. Obviously, the Fe-Fe3C phase diagram is of enormous commercial importance,
since the primary ingredients of most steels are Fe and C. The addition of small amounts of C (0.21.0 wt.%) increases the ductility of Fe tremendously and prevents brittle failure. Figure 10.28 is not
a true phase diagram, since Fe3C is metastable, gradually degrading to form Fe and C (graphite).
However, this reaction is nearly infinitely slow at room temperature, so common practice is to treat
Fe3C as though it were a stable compound. Note that pure Fe has a room-temperature BCC α phase
(ferrite) and a higher temperature FCC γ phase (austenite). This is not uncommon, particularly
among ceramic materials.
The Fe-Fe3C phase diagram shows a eutectic reaction, but it also shows a eutectoid reaction, during
which γ → α + Fe3C. Any phase transformation during which a solid phase transforms
simultaneously to two other solid phases is referred to as a eutectoid reaction. Figures 10.30
through 10.37 illustrate the microstructures that are present at different compositions and
temperatures. The transformations associated with the eutectoid reaction are similar to those
associated with the eutectic reaction. The main difference is that for the eutectoid reaction, the
initial reactant is a solid, whereas for the eutectic reaction, the initial reactant is a liquid. This
explains the difference between Figure 10.33 (Fe3C precipitates along the grain boundaries of γ) and
Figure 10.12 (α precipitates as solid chunks out of liquid L).
Example Problem
Consider 2.5 kg of austenite containing 0.65 wt% C, cooled to below 727ºC.
a)
What is the pro-eutectoid phase?
The pro-eutectoid phase forms prior to the eutectoid reaction, so that is the α (ferrite) phase.
b)
How many kg each form of ferrite and cementite?
This involves using the inverse lever rule:
Wα =
C Fe C − C 0
3
C Fe C − Cα
(2.5 kg )) =
3
W Fe C =
3
C 0 − C alpha
C Fe C − Cα
3
c)
6.70 − 0.65
(2.5 kg ) = 2.3 kg
6.70 − 0.022
(2.5 kg )) = 0.65 − 0.022 (2.5 kg ) = 0.2 kg
6.70 − 0.022
How many kg each form of pearlite and the pro-eutectoid phase?
Remember, the pro-eutectoid α precipitates above 727ºC, while the remaining γ is
transformed at that temperature into pearlite. Therefore:
W pro−α (727°C − ) = Wα (727°C + ) =
C eutectoid − C 0
(2.5 kg )) = 0.76 − 0.65 (2.5 kg ) = 0.4 kg
0.76 − 0.022
C eutectoid − Cα
W pearlite (727°C − ) = Wγ (727°C + ) =
C 0 − Cα
(2.5 kg )) = 0.65 − 0.022 (2.5 kg ) = 2.1 kg
C eutectoid − C α
0.76 − 0.022
Once again, a large percentage of the class will confuse the answers to parts b and c.
Example Problem:
Calculate the amount of the proeutectoid phase present in 1 kg of steel containing 1.20 wt.% C.
The proeutectoid phase refers to the phase that precipitates out of solid solution prior to the
eutectoid reaction. When the C content is less than 0.77 wt.% C, the proeutectoid phase is α ferrite.
When the C content is greater than 0.77 wt.% C, the proeutectoid phase is Fe3C.
From section 10.19 and figure 10.26, this problem is really asking what is the amount of Fe3C that
exists just above the eutectoid temperature (727°C+). This can be determined using the lever rule:
W Fe C =
3
C γ − C eut
C Fe C − C eut
3
(1 kg ) =
1.20 − 0.76
(1 kg ) = 0.07 kg
6.70 − 0.76
You should realize that other alloying elements beyond C are commonly used in many types of
steel, especially stainless steels. Show figures 10.38 and 10.39, which illustrate the effect of some
alloying elements on the eutectoid temperature and eutectoid composition of steel.