Chapter P. Phase diagrams.
After studying this chapter, you will be able to
P1.
P2.
P3.
P4.
Sketch the phase diagram of a solid solution and analyze this diagram: determine
the composition and amount of each phase for a given overall composition and
temperature.
Sketch the phase diagram of a simple eutectic alloy and define the eutectic point,
liquidus, solidus and solvus lines. Analyze this diagram in terms of number,
composition and amount of the phases.
Sketch a simple phase diagram with an intermediate compound and analyze the
diagram in terms of number, composition and amount of phases.
Analyze the phase diagram of the iron-carbon system that underlies steel.
In Chapter 5, we have seen that metals are strengthened by introducing foreign atoms,
either randomly dispersed in the host metal, or in the form of precipitates. In Chapter 4,
we have found that a high electric resistance is obtained by the introduction of
“impurities”. The deliberate introduction of foreign atoms is called alloying.
In order to process the alloys or utilize alloys effectively, it is important to know what
one obtains when one melts two metals together and lets them solidify to form an alloy.
This is the topic of this chapter.
For simplicity, we will restrict ourselves to binary alloys, which are composed of
two metals. When two metals are molten together and solidified, the alloy can take three
principal forms: a solid solution, a eutectic or a compound. This is illustrated in Figure
P.1.
In a solid solution, the atoms of both elements A and B occupy random positions
in the crystal. The classic example is the copper – nickel alloy. The solid solution is
obtained because the strength of the Cu – Ni bond is intermediate between those of the
Cu – Cu and the Ni – Ni bonds.
A eutectic is formed when the A – B bond is weaker than the A – A and the B –
B bond and each component has a low solubility in the other. (It is the solid equivalent of
a water-oil mixture). In this case, the liquid solution solidifies by separating into two
distinct phases, one rich in element A and one rich in element B. This will be illustrated
by the lead – tin or the bismuth – tin alloys, which are the materials used as solder.
Materials that form eutectics are commonly used for precipitation strengthening.
A compound is obtained when the A – B bond is stronger than the A - A and the
B – B bonds. This will be illustrated by the semiconductor Gallium Arsenide.
1
Figure P.1. Schematic illustration of differing configurations of A and B atoms in binary
alloys. (C ). Solid solution, (D) Segregation in a eutectic. (E) Compound formation.
The behavior of such alloys is codified in the phase diagrams. A phase diagram
is a drawing that shows what phases an alloy will form at all temperatures and
compositions in thermodynamic equilibrium, which is the state that nature aspires to
and is obtained if sufficient time is allowed. When the alloy is cooled rapidly, the
equilibrium may not be obtained and another form of alloy will be formed. Such rapid
cooling is an important tool of the engineer, as we shall see in chapter K. In this section,
we concentrate on thermodynamic equilibrium.
P.1.
Phases, Components and Phase Diagrams.
Let us take a glass of water and pour some sugar into it and add ice cubes. We
now have two components: water and sugar. (Ice and water are the same component).
We also have three phases: water, ice and solid sugar. Now let us stir until the sugar is
dissolved in the water. We now have tow components: water and sugar and two phases:
the water-sugar solution and the ice cubes. If we wait until all the ice melts, we obtain
one phase and two components.
A phase is a homogeneous, physically distinct, portion of matter that is present in
a non homogeneous system. It may be a single component or a mixture.
A component is an ingredient of a chemical system the concentration of which in
the different phases is capable of independent variation. (Webster’s International
Dictionary). Notice that we have used water and sugar as the components. Water is H2O and sugar is a
complex hydrocarbon molecule. Why did we not use hydrogen, oxygen and carbon as components?
2
Because that is not necessary: neither the water nor the sugar decomposes in this system and the
concentration of both is capable of independent variation in our glass).
The composition of a phase is the relative amount of the components it contains.
(The composition of the sweet water is the amount of sugar in it).
We will also be concerned about the amount of a phase, which is the fraction of
the mixture that is in the particular phase. (How many percent ice and how many percent
liquid?).
In a binary phase diagram (illustrated in figures P.2, P3, and P.8 for example) we
plot the compositions on the horizontal scale and the temperature on the vertical scale.
The overall composition represents the relative amounts of the components that are
present in the alloy. (The amounts we melted together).
P.2.
Solid solutions.
Copper – nickel alloy forms solid solutions. No matter how much nickel we add
to the copper, we obtain a solid in which the copper and nickel atoms are dispersed at
random. The phase diagram of copper – nickel alloys is shown in figure 6.2. As in all
phase diagrams, the x – scale represents the composition c of the alloy. In the present
example, it shows the percentage of nickel contained in the alloy. We have a choice of
representing the composition in atomic percent or weight percent. The weight fractions
Figure 6.2. The phase diagram of copper – Nickel alloys. This is figure 5-12 of Ohring.
are more convenient for engineering purposes; these are usually represented in the bottom
scale. Atomic percentages, which are of scientific interest, are commonly shown in the
top scale. These scales are, of course, different because of the different atomic weights of
the elements.
3
The first thing we note is that pure nickel has a well-defined melting point of
1455 C; it is liquid above this temperature and solid below. The same is true for pure
copper, at 1083 oC. This is not the case for the alloys. Take the alloy containing 35 %
nickel shown as the solid vertical line (a – b) in the figure. When we mix 35 pounds of
nickel and 65 pounds of copper, we form the overall composition C0 (=35 %). When
this alloy is cooled from point (a), namely 1400 oC, it starts solidifying at 1360 degrees
and becomes completely solid at 1310 degrees only. Between these two temperatures,
we obtain two phases, one liquid and one solid. This phenomenon is described in the
diagrams by the liquidus line above which the alloy is completely liquid and the solidus
line below which it is completely solid. The region between these two lines, (shaded in
the figure) is a two-phase region: it contains solid and liquid.
o
P.3.
Analysis of binary phase diagrams
a. Single Phase In a single phase field (in this or any other phase diagram) the chemical
and physical analysis of what is present in thermodynamic equilibrium is very simple.
The percent by weight or by atoms of each constituent component is equal to the overall
composition. Let us consider a 35wt% Cu and 65wt. %Ni alloy (close in composition to
Monel metal which is used in petroleum refineries) and heat it to 1400C (state a, Fig.P.2).
The alloy would melt and if we could make a chemical analysis at this temperature it
would contain 65wt. %Ni and 35wt.% Cu. Physically, we know that 100% of what is
present is liquid. Similarly, if the original alloy were cooled to1100C (state b) we know
the chemical analysis is identical (i.e.,65wt. %Ni and 35wt.% Cu), and that physically,
100% of what is present is solid.
In any single phase field of any binary equilibrium phase diagram the rules are:
1. The chemical composition is the same as that of the original alloy.
2. Physically, the alloy is either 100% homogeneous liquid or 100% homogeneous
solid.
3. Single phase alloy liquids and solids are generally stable over some range of
composition and temperature.
b. Two Phase Mixture, Tie Line Construction and Chemical Composition
If the system state point falls in a two phase (L + S) field a two phase physical
mixture exists. The chemical and physical analyses follow different rules from those of
the single phase field. Both analyses center on the construction of a tie line. This is a
horizontal line (constant temperature) drawn through the state point (c) so that it extends
to the left and right until it ends at the single phase boundaries on either side. Such a tie
line is shown at 1340C for the initial 35wt% Cu - 65wt. %Ni alloy.
Within the L+S field, the composition of the solid is given by the intersection
point of the tie line with the solidus curve (read off on the horizontal axis). Similarly, the
corresponding composition of the liquid is given by the intersection of the tie line with
the liquidus curve. For example, in the 65wt. %Ni - 35wt% Cu alloy at 1340C the solid
has a composition of 27 wt. % Cu-73 wt% Ni, while the liquid has a composition of 40
4
wt. %Cu-60 wt% Ni.
c. Two Phase Mixture, Lever Rule and Relative Phase Amounts
We are only half way through our analysis of the two phase mixture. Now that we
know the chemical composition of each phase let us determine physical composition,
namely the relative proportions or weights of each phase present. Consider a portion of
the phase diagram containing an alloy of A and B atoms with an initial composition Co
expressed in terms of wt.% B. Assume that two phases α and β coexist in equilibrium
and the chemical composition of phase α is Cα and the composition of phase β is Cβ. (It
is common to identify solid phases by Greek letters.) Thus, phase α contains Cα wt.% of
component B and [100 - Cα] wt. % of component A. The problem is to determine the
fraction fα of the mixture by weight that is phase α and the fraction fβ  = (1 - fα) that is
phase β. If the total alloy weighs W grams, conservation of the mass of B atoms requires
W Co = WCαfα + W Cβ (1 - f α).
(P.1)
(i.e. the amount of copper in the alloy is the amount in the solid + the amount in the
liquid). The term on the left represents the total weight of B atoms in the alloy, and the
terms on the right the weight of B atoms partitioned, respectively to phases α and β.
Solving for fα and fβ yields the Lever Rule, which is expressed by
fα= (Cβ - Co) / (Cβ- Cα) and fβ = (Co - Cα) / (Cβ - Cα).
(P.2)
Note that, in order to obtain fα, one measures the distance from the overall composition
C0 to the composition of the β phase and divides the result by the length of the tie line.
The Lever Rule is so named because the fraction of a phase (fα) multiplied by its “lever
arm” (Co - Cα) is equal to the product of fβ and lever arm (Cβ - Co). Thus, whichever
phase ( boundary) the initial composition is closest to, is present in the greatest amount.
When applied to our 65wt. %Ni - 35wt.% Cu alloy at 1340C, the fraction of solid is (4035)/(40-27) = 0.385 and that of liquid is (35-27)/(40-27) = 0.615.
In summary, for any two - phase field of any binary equilibrium phase diagram
the rules are:
1. Draw a horizontal tie line corresponding to the chosen temperature and extending
to the left and right phase limits from the overall composition C0. (Tie lines are only
drawn in two-phase fields. They make no sense in a one phase field).
2. The chemical composition of the two phases is given by the ends of the tie line,
5
extended vertically down to, and read off the horizontal axis.
3. The weight fraction of each phase within the two phase mixture is given by the
Lever Rule (Eq. P.2).
4. Quantitative phase analyses can not be made if the system state lies exactly on a
line or boundary that separates phases. One must either be slightly above or slightly
below the line in order to analyze the system by the above rules.
Figure P.3. The germanium-silicon phase diagram.
P.3.1. Analysis of an equilibrium – cooled alloy.
We now have all the tools required to analyze a binary alloy system. Rather than
the Cu-Ni system which has a restricted L + S field (narrow solidification range) making
tie line constructions difficult to read, let us consider the Ge-Si system of Fig. P.3. which
has a similar but wider two - phase field.
EX.5- 3
An alloy containing 6X1022 atoms of Ge and 4X1022 atoms of Si is
heated to 1300C and cooled to 950C.
a. What is the alloy composition in atomic and weight percent?
b. At temperatures of 1300C, 1230C, 1200C, 1150, 1100C, 1070C and 950C: what
phases are present? what are their chemical compositions?; how much of each phase
is physically present? Assume equilibrium is attained at each temperature.
ANS. a. The atomic percent of Ge is 6X1022 / (6X1022 + 4X1022)X100 = 60 at.% .
Therefore, 40 at.% Si is present.
The weight of the Ge atoms is
6X1022 MGe / NA = 6X1022X72.6 / 6.02X1023 = 7.24 g.
6
The weight of the Si atoms is
4X1022 MSi / NA = 4X10 22 X28.1/ 6.02X1023 = 1.87 g.
The wt% Ge = 7.24 / (7.24 + 1.87)X100 =79.5. Therefore there is
20.5 wt% Si as noted on the upper weight percent silicon scale. Both weight and
atomic percent notations are widely used as measures of composition.
b. The included table is self explanatory and has the required answers.
Temp C
Phases
Chemical
Composition
(at%Si)
Phase amounts
( atomic fraction)
Comment
a. 1300
L
40
1.0
all liquid
b. 1230
L
S
~40
72
~1.0
~0
solidification
beginning
c. 1200
L
S
34
67
(67-40)/ (67-34) = 0.82
18.2
d. 1150
L
S
25
56
(56-40)/(56-25) = 0.52
0.48
e. 1100
L
S
47.5 (47.5-40)/(47.5-16) =0.24
16
0.76
f. 1070
L
S
12
~40
~0
~1
solidification
almost complete
g. 950
S
40
1.0
all solid
from Eq.P.3
“
“
“
“
“
An idealized picture of the solidification events during slow cooling of this alloy
would first start with an amorphous melt. At 1230C tiny Si - rich particles would begin to
form at the mold wall and perhaps in the bulk of the melt as well. As the temperature
drops, the particles continue to grow by thickening and lengthening. Surrounding the
particles is the remaining Ge - rich liquid which continually shrinks in amount until it
effectively disappears at 1070C, leaving grain boundaries behind.
P.3.2. Coring.
It is important to know that the composition of the solid particles changes, during
cooling, from 72 to 40 % Si. This occurs by diffusion of the Si atoms in the solid, which
takes time. The equilibrium values of the above example are reached only if cooling is
slow enough to permit the diffusion to proceed to completion.
If cooling is too rapid, the center of the growing solid particles is too rich in Si;
7
this silicon is missing in the liquid which becomes poorer in Si than indicated in figure
P.3. The final result is a cored structure in which the centers of the grains are too rich
in Si and the outside of the grains is too poor in Si. Such non-equilibrium structures are
undesirable and are corrected by subsequent thermal treatment, such as a lengthy heating
in the solid state.
P. 4 The binary eutectic phase diagram.
Many pairs of elements (e.g., Bi-Cd, Sn-Zn, Ag-Cu, Al-Si etc) do not mix. The
bond between the two elements A-B is weaker than the A-A and the B-B bond and each
of them can dissolve a limited amount of the other. These pairs solidify in a manner
similar to Pb-Sn whose eutectic phase diagram is shown in Figure P.4. The eutectic is so
named for the critical eutectic point corresponding to the eutectic temperature (183oC
in Sn-Pb) and eutectic composition (61.9% Sn in Sn-Pb). Above the eutectic
temperature, it is as though two independent solid solution systems exist - one on the Pb
side yielding Pb rich solid solutions (the α phase), the other on the Sn side giving rise to
Sn rich solid solutions (the β phase). Their liquidus lines converge at the eutectic point
(61.9 wt. %Sn and 183C). This means that the liquid phase of every alloy containing
between 19.2 to 97.5 wt% Sn, will eventually reach the eutectic composition (61.9 wt.%
Sn) upon cooling from elevated temperature to slightly above 183oC.
solidus
liquidus
Eutectic
point
Solvus line
Figure B.4. Pb-Sn phase diagram showing eutectic reaction. 2-phase regions are colored.
Liquid at this eutectic composition decomposes at 183C into a mixture of two solid
phases: α, a Pb rich solid solution, and β, a Sn rich solid solution. The reaction is
invariant; both the temperature and phase compositions remain constant during the
8
transformation of eutectic liquid to α + β. Consider now the cooling behavior of an alloy
of exactly eutectic composition (61.9 wt.% Sn-30.1 wt.% Pb). Until the melt reaches
183C there is the usual temperature drop. But then the temperature remains at the eutectic
temperature until the eutectic reaction is complete. This is followed by cooling to room
temperature of the two - phase solid. The two - phase eutectic microstructure that
develops is generally characteristic of the alloy system. It often appears as a lamellar
array of alternating plates of the two phases, but sometimes as a dispersion of one phase
in a matrix of the other. This is shown in figure P.5.A.
A
B
Figure P.5. (A) Microstructure of a Pb-Sn eutectic with 61.9 % Sn. (B)
Microstructure of a Pb-Sn alloy of composition 50 wt% Sn; with proeutectic α
phase and eutectic. The dark regions are Pb- rich α phase; the bright areas are Sn
– rich β phase. (From Callister).
The rules of phase analysis in single- and two - phase fields given previously will
be applied once again in the example below. First, however, a few additional details of
eutectic solidification should be noted. The solid α phase (in equilibrium with L) that
forms above the eutectic temperature is known as the proeutectic or primary α phase.
Similarly, solid proeutectic β phase separates from the melt in Sn - rich alloys at T >
183C. As the eutectic reaction proceeds, the eutectic mixture of α + β adds its presence
to the already solidified larger crystals of proeutectic α (or β). See figure P.5.B.
The solvus curve in Figure P.4. on the Pb – rich side shows that the solubility of
Sn in Pb decreases as the temperature is lowered. When the alloy cools below the
eutectic temperature, solute is rejected from proeutectic phases because of the lower
solubility.
EX.P-2.
a. For the alloys, 5wt.% Sn - 95 wt.% Pb, 40 wt% Sn - 60 wt.% Pb,
and 70 wt% Sn - 30 wt% Pb, identify the phases present and determine their
compositions and relative amounts at 330C, 184C and at 25C.
b. For a 1 kg specimen of the 70wt% Sn-30 wt% Pb alloy, what weight
of Pb is contained within the eutectic microstructure at 25C?
9
ANS. The answers are tabulated below.
Temp C
Phases
a. 1.
330
184
25
L
α
α
β
Chemical
Phase amounts
Composition
(wt. fraction)
(wt% Sn)
5wt.%Sn - 95wt.%Pb
5
5
~2
 100
~
1.0
1.0
(100-5)/(100-2) = 0.97
(5-2)/(100-2) = 0.03
Comment
all liquid
all solid
no eutectic
microstructure
present
_______________________________________________________
2.
40 wt.%Sn-60 wt.% Pb
330
184
L
40
1.0
all liquid
α
19.2
(61.9-40) / (61.9-19.2) = 0.51
proeutectic α
L
61.9
(40 - 19.2)/ (61.9-19.2) =0.49
eutectic liquid
25
α
~2
(100-40)/(100-2) = 0.61
β
~100
(40-2)/(100-2) = 0.39
or alternatively,
182
α
~2
(61.9-40)/(61.9-2) = 0.37
α+β
61.9
(40-2)/(61.9- 2) = 0.63
eutectic mixture
_________________________________________________________
3.
70wt% Sn-30 wt% Pb
330
184
L
70
1.0
all liquid
β
97.5 (70-61.9)/(97.5-61.9) = 0.23
proeutectic β
L
61.9 (97.5-70)/ (97.5-61.9) = 0.77
eutectic liquid
25
α
~2
(100-70)/(100-2) = 0.31
β
~100 (70-2)/ (100-2) = 0.69
or alternatively,
182
β
~100
(70-61.9)/(100-61.9) = 0.21
α+β
61.9
(100-70)/(100-61.9) = 0.79 eutectic mixture
_________________________________________________________
b.
In 1 kg of the 70wt.%Sn-30 wt% Pb alloy the eutectic mixture
weighs 0.79 kg. Of this, 61.9% is Sn and 38.1% is Pb. Therefore, the amount of Pb
= (0.79)(.381) = 0.30 kg.
Schematic microstructures of these alloys are sketched in Fig.P.6 for the indicated
alloys. They should be compared with actual room temperature microstructures of several
10
alloys in Pb-Sn system (Fig P.7). A noteworthy feature of eutectic systems is the low
melting temperature at the eutectic point, which is lower than the melting points of either
component. For this reason, the alloy containing 63 wt.%Sn-37wt.%Pb is common
electrical solder, while the 50 wt.%Sn-50 wt% Pb alloy was used in plumbing. (Recently,
the use of Pb in plumbing has been outlawed).
Figure P.6. Schematic microstructures of three alloys in the Pb-Sn system at temperatures
of 330, 184 and 25oC, The eutectic microstructure is composed of alternate α and β
layers.
P.4.1. The formation of precipitates.
The analysis of the diagram for 5% Sn in examples P.2. and illustrated in figure
P.6. shows the use of the limited solubility in eutectics for precipitation strengthening of
alloys. For any alloy containing less than 19.2 wt% Sn, the material is a solid solution (α
phase). As the temperature is lowered, the composition Co reaches the solubility limit (it
crosses the solvus line) and β phase is precipitated and presents an obstacle to dislocation
motion. We shall visit this concept again in chapter K where we will examine the use of
cooling rates to control the size of the precipitates.
P.5.
Intermediate Compounds and Intermediate Phases.
Consider the Ga-As phase diagram depicted in Fig. P.8. A dominant feature in
the phase diagram is the central compound GaAs, containing equiatomic amounts of Ga
and As. Compounds often have well defined stoichiometries, in this case 50 at.% Ga- 50
at.% As, and are therefore denoted by vertical lines in phase diagrams. The strong bond
11
Figure P. 7 Microstructures of Pb-Sn alloys at room temperature. (A) 30 wt% Sn – 70
wt% Pb. Dark proeutectic α - phase surrounded by eutectic. (B) 50 wt% Sn- 50 wt% Pb.
Dark proeutectic α phase within the light eutectic (C) 63 wt% Sn – 37 wt% Pb. Twophase eutectic microstructure. (D) 70 wt% Sn – 30 wt% Pb. Light proeutectic β phase
surrounded by eutectic.
12
between gallium and arsenic atoms in GaAs is reflected in the melting temperature of the
compound, which is higher than that of either component. The interpretation of this
phase diagram follows the rules we have examined before. Let us assume that we have a
melt that is slightly rich in Ga, say 40 atom % As and 60 % Ga. Solidification will start
around 1230 oC. At 1200 oC, the system contains solid GaAs and a liquid phase
containing 32 % Ga. There will be (40 – 32)/(50 – 32) = 44.5% GaAs and 55 % liquid.
At 30 oC we have (50 – 40)/(50 – 0) = 20 % almost pure Ga and 80 % GaAs. Below
29.5 oC, which is the melting point of gallium, we find 20% solid gallium and 80 %
GaAs. On the As –rich side of the diagram, the analysis is similar, the arsenic solidifies
at 810 oC.
Figure P.8.
GaAs.
The phase diagram of the Ga-As system with formation of the compound
13
Figure P. 9. The SiO2 – Al2O3 phase diagram.
The SiO2 – Al2O3 system of figure P.9. is instructive. The components are not
elements, but the compounds SiO2 (silica) and Al2O3 (alumina). These are chosen
because neither substance decomposes at the temperatures shown. The system forms the
intermediate compound Mullite with nominal composition 3Al2O3.2SiO2.
This
compound, marked as γ phase, exists over a broader stoichiometric range than does
GaAs, namely from 71 to 73 wt% alumina. Alloy mixtures of these oxides are used to
make refractory bricks that line high temperature industrial furnaces.
Mullite can be considered as the end compound of the SiO2 - Mullite eutectic
system to its left and of the Mullite - alumina eutectic system to its right. The analysis of
the phase diagram on either side of mullite is identical to that of a eutectic system.
An alloy system can also contain intermediate phases. These are shown in the CuZn diagram figure P.10. The α,β,ψ, ε and η phases are actually solid solutions with
different crystal structures. The system includes several commercially important coppers
and brasses. For example α brass, or cartridge brass contains 30% zinc, Muntz metal or
β brass contains 40% zinc. The transition from liquid to solid takes place by peritectic
reactions which we discuss next.
14
Figure P.10. Cu-Zn phase diagram. α brass contains 30% Zn, β brass contains 40% Zn.
P.6. Peritectic Solidification.
Not uncommon in the catalog of solidification sequences is the invariant
peritectic reaction which can be written in the generic form
L + β → α.
(P.3)
Peritectic reactions often occur when there is a large difference in melting point between
the two components. The liquid reacts with the higher-temperature solid to form a new
solid phase. On the phase diagram the peritectic reaction is characterized by the
prominent isotherm, and appears as an upside down eutectic. There are few examples of
alloy systems that exhibit only peritectic behavior over the complete temperature range.
One is the platinum-rhenium system, whose phase diagram is shown in Fig. P.11. Above
the peritectic temperature of 2450C, solid solution solidification (β) prevails in Re - rich
alloys. Below this temperature down to the melting point of Pt, liquid exists with Pt - rich
compositions which also solidify as solid solutions α.
15
Figure P.11. Peritectic phase diagram of Pt-Re.
Only alloys with compositions between ~43%wt.% Re to ~54 wt.% Re participate
in the peritectic reaction. For any alloy in this range a liquid containing ~43%wt.% Re
and a solid β containing ~54 wt.% Re coexist just above 2450C. The proportions depend
on the initial composition and can be calculated from the Lever Rule. Given sufficient
time the β phase will react with the liquid and produce the α phase as Equation P.3
suggests. But, as the temperature drops, will be rejected from α and both phases will
persist down to lower temperatures. Unlike eutectics, peritectics have no particularly
distinguishing microstructural features.
The issue of equilibrium is an important one in peritectic solidification. Since
peritectic reactions are sluggish they don’t readily proceed to completion unless
maintained at high temperature for long times. During practical cooling rates the
peritectic reaction may be bypassed yielding nonequilibrium phases that may not appear
on the phase diagram.
16
P.7.
The Iron – Carbon system.
The best known and arguably most important phase diagram in metallurgy is the
iron – carbon diagram shown in Figure P.12. Note that the diagram does not extend to
pure carbon because all reactions of interest occur at concentrations below that of the
compound Fe3C that contains 6.67 wt. % carbon. From this diagram, much can be
learned about plain carbon steels and cast irons, two of the most widely utilized classes of
structural metals. Figure P.12 is not a true equilibrium diagram, however, because Fe3C
can decompose into graphite (C) and Fe at elevated temperatures, but it describes what is
happening in the processing of steel.
At the extreme left of the diagram, pure iron is polymorph; it has the BCC crystal
structure in the lower temperature range, then it transforms to the FCC structure at 914 oC
and back to the BCC structure at 1394oC. We will not be interested in the latter
transformation because it is of not practical use. This polymorphism is an extraordinary
present nature has offered us: it is responsible for the wide range of properties of steel,
from very ductile steels used in buildings and bridges for their safety to the hard tool
steels used in manufacturing. FCC iron (the γ phase, also called Austenite) can dissolve
Pearlite = α + Fe3C
Figure P.12. The phase diagram of the iron – carbon system. The diagram is shown only
up to 6.67% carbon, which corresponds to the compound Fe3C.
appreciable amounts of carbon (up to 2.14 wt% = 8 at% at 1147oC) but the BCC structure
(the α phase called Ferrite) can not dissolve more than 0.022 wt% (= 0.08 at%) carbon
17
and practically none at room temperature. Therefore the Austenite solid solution
containing 0.76 wt% carbon undergoes, when cooled below 727oC, a eutectoid reaction
into almost pure α iron (ferrite) and Fe3C. The compound Fe3C is known as Cementite.
The eutectoid that is formed is called Pearlite; it consists of alternating layers of ferrite
and cementite and is shown in figure P. 13. Note that ferrite, being pure α iron, is very
soft and that cementite is a hard ceramic. This combination provides steel with strength
and ductility.
(We also note in the phase diagram a eutectic point at 4.3 wt% carbon and
1147 C, but this is of minor technical importance).
o
Figure P.13. The microstructure of pearlite. The dark lines are cementite layers, the
bright areas represent ferrite.
The Fe- Fe3C phase diagram exhibits all of the solidification and transformation
behaviors already discussed in addition to the eutectoid transformation that we have just
examined.
Let us analyze the phases produced from melts containing three different amounts
of carbon, namely 0.4 0.76 and 1.2 wt% carbon. A look at figure P.12 reveals that, at
900oC, say, all three alloys are Austenite solid solutions.
We now examine what happens when these alloys are cooled to room
temperature. Let us start with the steel that contains 0.76 wt% C; this is a eutectoid steel.
Cooling from the melt, it will form an Austenitic solid solution all the way down to
727oC. At this temperature, it undergoes the eutectoid reaction and forms pearlite (Figure
P.13), as we have already seen.
Now let us examine a hypoeutectoid steel, namely one that contains less than
0.76 wt% C. For this purpose, we reproduce in figure P.14 the iron-carbon phase
diagram only in the vicinity of the eutectoid point (0.76wt%C and 727oC).
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Figure P.14. Equilibrium phase transformations of hypoeutectic and hypereutectic steels
together with schematic microstructures.
At 900oC , the alloy containing 0.4 wt% carbon is Austenite. The microstructure,
sketched in a circle, consists of γ - phase grains. As the temperature passes the solvus
line at 800 oC, the γ phase is enriched in carbon and rejects pure α iron. This segregation
usually occurs at grain boundaries. As we cool the alloy to just above the eutectoid
temperature, more α iron is rejected and the carbon concentration in the γ phase
approaches 0.76 wt%. The α iron formed above the eutectoid point is called
proeutectoid α. We can estimate the amount of proeutectoid α by means of the lever
rule: fα = (0.4 – 0.022)/(0.76 – 0.022) = 51%. Now we cool to just below the eutectoid
temperature, the proeutectoid α is not affected, but the remaining austenite (γ) undergoes
the eutectoid transformation into pearlite (which is α + Fe3C). How much proeutectoid
α? 51%, and 49% pearlite. The microstructure of the steel is shown in Figure P. 15.
Since the α iron is soft, hypeutectoid steels are softer and more ductile than eutectoid
steels.
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Figure P.15. Microstructure of a hypoeutectoid carbon
steel. The bright areas are ferrite. (From Callister).
The hypereutectoid steel at 1.2 % is analyzed in the
same fashion. At 900oC, it is austenite (single phase
solid solution). As we cool to 800oC, the solubility
limit is exceeded, and cementite is precipitated to the
grain boundaries, the amount of the latter increases as
we cool to, say, 728 degrees. The lever rule can again
compute the amount of proeutectoid Fe3C (7.5%).
When we cool through the eutectoid point, the
remaining austenite (containing 0.76wt% C)
undergoes the eutectoid transformation and becomes
pearlite. The microstructure is sketched in figure P.14
and shown in figure P.16.
Figure P.16. Hypereutectoid steel.
Cementite is bright.
With all three materials, cooling down to room temperature does not change the
microstructure but the ferrite rejects the little carbon it contained and becomes pure BCC
iron. Since cementite is a hard material, hypereutectoid steels are harder and less ductile
than pearlite. We see that, in steels, the higher the carbon concentration, the harder the
material.
This, however, is not the whole story. So far, we have examined equilibrium
phase diagrams, namely materials that are cooled very slowly. All three steels we have
examined can be made much harder by rapid cooling. These will be examined in the nest
chapter where we will study the kinetics of phase transformations.
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Glossary
A phase is a homogeneous, physically distinct, portion of matter that is present in a non
homogeneous system. It may be a single component of a mixture.
A component is an ingredient of a chemical system.
The composition of a phase (C) is the relative amount of the components it contains.
The amount of a phase (f) , is the fraction of the mixture that is in the particular phase.
In a solid solution, the components occupy random positions in the crystal.
The liquidus line represents the lowest temperatures of a liquid as a function of
composition. It forms the boundary between the liquid phase and the two-phase system.
The solidus line represents the highest temperature of solids as a function of
composition. It is the border between the solid phase and the two-phase region.
The solvus line represents the limit of solubility on one component in the other. It is
generally the limit between a solid solution phase and a two-phase eutectic region.
A eutectic is an alloy of immiscible components. At the eutectic point the liquid
transforms into the eutectic, which consists of two distinct phases arranged in alternating
layers.
A proeutectic phase is formed above the eutectic temperature.
A eutectoid reactions is similar to a eutectic reaction; the difference is that a solid phase
(instead of a liquid) reacts to produce two phases. The result is called a eutectoid.
A eutectoid steel contains the eutectoid concentration (0.76 wt%) of carbon.
A hypoeutectoid steel contains less than 0.76 wt% carbon.
A hypereutectoid steel contains more than 0.76 $ carbon.
Austenite is the FCC solid solution in steel, its other name is γ phase.
Ferrite is the BCC form of iron in steel. It contains very little carbon. It is the α phase.
Cementite is the compound Fe3C containing 6.67 wt% carbon in steel.
Pearlite is the eutectoid phase composed of ferrite and cementite in steel.
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Recapitulation.
1). Phase diagrams show the phases one obtains in an alloy system for every
composition and temperature. They represent the thermodynamic equilibrium situation
which may requires long times to be reached, especially at the lower temperatures.
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
Binary alloys can form solid solutions, eutectics, or compounds, depending on
the relative strengths of the bonds between the atoms.
In alloys, a two-phase region separates the liquid phase from the solid in the
diagram. In this region, the liquid and the solid have different compositions.
The composition of the liquid is given by the intersection of the liquidus curve
with a tie line representing the temperature of interest; the composition of the
solid is the intersection of the tie line with the solidus curve.
The amounts of phases can be computed by the lever rule. The amount of
solid is the ratio of the length of tie line between the overall composition and
the liquidus curve to the length of the tie line between the liquidus and
solidus. The amount of liquid phase is the ratio of the line between the solidus
and the overall composition divided by the length between liquidus and
solidus. (Note: for the liquid phase, go to the solidus line and vice versa!).
Solid solutions are one phase systems. They occur over the whole
composition range for solid solution solids; they occur at both ends of the
diagram for eutectics.
Eutectics form when the two components are poorly soluble in each other. At
the eutectic point (a specific composition and temperature), the liquid
separates into two solid phases which form the eutectic). Solids formed above
the eutectic point are called proeutectic.
Compounds form when A-B bond is stronger than the A-A or B-B bond. They
can be considered as the end phases of eutectic regions.
Steel is an alloy of iron with 0.2 to 1.2 wt% carbon.
The phase diagram of steel has a eutectoid point at which the hightemperature FCC solid solution (Austenite) decomposes into a eutectoid
(pearlite) formed of BCC iron(ferrite) and the compound Fe3C (cementite).
At hypoeutectoid compositions, a proeutectoid ferrite forms in addition to the
pearlite. Such steels are softer than eutectoid steels.
At hypereutectoid compositions, a proeutectoid cementite forms in addition to
the pearlite, resulting in a harder steel than pearlite.
Problems.
P.1 A solder manufacturer wishes to make a batch of solder having the eutectic composition. On
hand is a supply of 250 kg of electrical solder scrap containing 60 Wt % Sn-40 Wt% Pb, and
1250 kg of plumbing solder scrap containing 60 wt% Pb-40 wt% Sn. If all of the scrap is to be
melted how much of which metal must be added to achieve the desired 61.9 Wt% Sn
composition?
P.2 Gallium arsenide crystals are grown from Ga rich melts. Can you mention one reason why
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As rich melts are not used for this purpose?
P.3 A 90 Wt% Sn-10 Wt% Pb alloy is cooled from 300C to 0C.
a. What is the composition of the alloy in terms of atomic or molar percent?
b. Draw a cooling curve for this alloy.
d. In what ways does the equilibrium microstructure of this alloy differ from that of the 90
Wt%Pb-10 Wt% Sn alloy.
P.4 The Al2O3 - Cr2O3 phase diagram resembles that of Cu-Ni. Al2O3 melts at 2040C and
Cr2O3 melts at 2275C.
a. Sketch the phase diagram.
b. How many components are there in this system?
d. It is desired to grow a ruby single crystal with the composition 22 wt% Cr2O3. On your phase
diagram indicate the melt composition and temperature for growth.
P.5 Suppose you have a large quantity of silver amalgam (Ag-Hg alloy) scrap. The alloy is
liquid at low temperatures and has much less value than the isolated pure metals. Design a
physical process to separate these metals.
P.6a. In the Ga-As binary system perform total chemical composition and phase amount analyses
for an 80 At% Ga-20 At% As alloy at 1200C, 1000C, 200C and 29C.
b. What is different about the two regions that are labeled Liquid + GaAs?
P.7 It is desired to pull an alloy single crystal of composition 78 At%Ge-22 at% Si from a binary
Ge-Si melt.
a. At what temperature should the crystal be pulled?
b. What melt composition would you recommend?
c. As the crystal is pulled what must be done to ensure that its stoichiometry is maintained
constant?
P.8 Consider the 50 wt% Pt - 50 wt% Re alloy.
a. This alloy is heated to 2800C and cooled to 800C. Sketch an equilibrium cooling curve for this
alloy.
b. Perform complete equilibrium chemical and physical phase analyses at 2800C, 2452C, 2448C,
and 1000C.
P.9. Make enlarged sketches of the phase regions that surround the two highest isotherms in the
Cu-Zn phase diagram (Fig.5-27).
a. What is the name of the solidification behavior displayed in both cases.
b. Which line(s) on this phase diagram represent the solubility of Zn in the phase?
c. Which line(s) on this phase diagram represent the solubility of Zn in the phase?
d. Which line(s) on this phase diagram represent the solubility of Cu in the phase?
e. Is it possible for the solubility of a component to drop as the temperature is raised? If possible,
give an example in this phase diagram.
P.10 Perform quantitative phase analyses at temperatures (C and f) corresponding to the four
states indicated for the 0.4%C and 1.2 %C steels in the Fe-Fe3C system of Fig.P.14. Do your
analyses correspond to the sketched microstructures?
5-23 Up until the middle ages iron was made by reducing iron ore (e.g., Fe2O3) in a bed of
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charcoal (carbon). Steel objects were shaped by hammering the resulting solid sponge iron that
was reheated to elevated temperatures using the same fuel. As the demand for iron and steel
increased, furnaces were made taller and larger. They contained more iron ore and charcoal that
were heated for longer times. An interesting and extremely important thing occurred then! Liquid
iron was produced. Why?
(Molten iron issuing from the bottom of the furnace could be readily cast, a fact that helped to
usher in the Machine Age).
P.11 Give a good physical reason for the following statements:
a. Elements that are soluble in all proportions in the solid state have the same crystal structure.
b. Unlike substitutional solid solutions, interstitial solid solutions do not range over all
compositions.
P.12 Distinguish between intermediate phases and compounds.
Contrast metals and nonmetallic inorganic materials with respect to their tendency to form
intermediate phases and compounds in binary phase diagrams.
P.13a. A 40 wt % Al2O3 - 60 wt % SiO2 melt is slowly cooled to 1700C. What is the proeutectic
phase and how much of it is there?
b. Furnace bricks are made from this composition. Is the melting point of these bricks higher or
lower than pure silica bricks?
c. What is the highest melting temperature attainable for bricks containing both SiO2 and
Al2O3?
P.14 A 10 gram ball bearing sphere is made of a steel containing 1.1Wt %C. Suppose the ball is
austenized and slowly cooled to room temperature.
a. What is the weight of pearlite present?
b. What is the total weight of ferrite present?
c. What is the total weight of cementite present?
d. What is the weight of cementite present in the pearlite?
e. What is the weight of cementite present as the proeutectoid phase?
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