Transcript INDUSTRIAL MATERIALS - B Tech Mechanical Engineering

INDUSTRIAL MATERIALS
Instructed by:
Dr. Sajid Zaidi
PhD in Advanced Mechanics, UTC, France
MS in Advanced Mechanics, UTC, France
B.Sc. in Mechanical Engineering, UET, Lahore
B.TECH Mechanical Technology
IQRA COLLEGE OF TECHNOLOGY (ICT)
INTERNATIONAL ISLAMIC UNIVERSITY, ISLAMABAD
Formation of Alloys
For most manufacturing applications, metals are not
used in their pure form.
 Engineering materials tend to be ALLOYS, materials
composed of two or more different elements, and they
tend to exhibit their own characteristic properties.
 An ‘‘alloy’’ can also be defined as a material that
exhibits properties of a metallic material and is made
from multiple elements.
 There are different ways in which a metal might respond
to the addition of another element.
 The most popular way is when the two elements exhibit
some degree of solubility in the solid state. The two
materials form a SOLID SOLUTION, where the ALLOY
ELEMENT dissolves in the BASE METAL.
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Formation of Alloys
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The solid solution can be
◦ Substitutional or Interstitial
In the substitutional solutions, some atoms of the alloy
elements occupy lattice sites normally filled by atoms of
the base metal. The replacement is totally random in
nature, with the alloy atoms being distributed throughout
the base lattice.
In the interstitial solutions, the alloy element atoms
squeeze into the empty spaces between the atoms of the
base metal lattice.
Substitutional
Interstitial
Phases
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Formation of Alloys
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Alloys can be classified as:
◦ Single – Phase Alloys
◦ Multiple – Phase Alloys
A phase can be defined as a homogeneous portion of a
system that has uniform physical and chemical
characteristics.
A phase has the following characteristics:
◦ same structure or atomic arrangement throughout
◦ roughly the same composition and properties
throughout
◦ definite interface between the phase and any
surrounding or adjoining phases.
Phases
Water has three phases:
liquid water, solid ice, and steam.
 Phases do not always have
to be solid, liquid, and gaseous forms
of a material.
 An element, such as iron (Fe), can exist in FCC and
BCC crystal structures. These two solid forms of iron
are two different phases of iron that will be stable at
different temperatures and pressure conditions.
 Similarly, ice, itself, can exist in several crystal
structures.
 Carbon can exist in many forms (e.g., graphite or
diamond). These are only two of the many possible
phases of carbon.
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Formation of Alloys
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Phase Rule
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Formation of Alloys
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Josiah Willard Gibbs developed the phase rule in 1875–
1876. It describes the relationship between the number
of components and the number of phases for a given
system and the conditions that may be allowed to change
(e.g., temperature, pressure, etc.)
The general form of Phase Rule is:
2+C=F+P
C = number of chemically independent components
F = number of degrees of freedom
P = number of phases present
2 = both temperature and pressure are allowed to
change
Phase Rule
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Formation of Alloys
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Unary phase diagram
Melting point
Boiling point
Phase Rule
◦ Point A
◦ Point B
◦ Point X
Triple Point (point X)
 Condition of sublime
 P-T diagram
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Schematic unary phase diagram
for magnesium
Solubility
INDUSTRIAL MATERIALS
Formation of Alloys
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How much of each material or component we can
combine without producing an additional phase i.e., the
solubility of one material into another (e.g., sugar in
water, copper in nickel, phosphorus in silicon, etc.).
Unlimited Solubility
Limited Solubility
No Solubility
Unlimited Solubility
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Formation of Alloys
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Water and Alcohol have unlimited solubility and when
combined only one phase is produced.
Similarly, if we were to mix any amounts of liquid copper
and liquid nickel, only one liquid phase would be
produced. This liquid alloy has the same composition and
properties everywhere, because nickel and copper have
unlimited liquid solubility.
Unlimited Solubility
INDUSTRIAL MATERIALS
Formation of Alloys
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If the liquid copper-nickel alloy solidifies and cools to
room temperature while maintaining thermal equilibrium,
only one solid phase is produced. After solidification, the
copper and nickel atoms do not separate but, instead, are
randomly located within the FCC crystal structure. Within
the solid phase, the structure, properties, and composition
are uniform and no interface exists between the copper
and nickel atoms. Therefore, copper and nickel also have
unlimited solid solubility. The solid phase is a solid
solution of copper and nickel.
Unlimited Solubility
A solid solution is not a mixture. A mixture contains more
than one type of phase whose characteristics are retained
when the mixture is formed. In contrast to this, the
components of a solid solution completely dissolve in one
another and do not retain their individual characteristics.
Conditions of Unlimited Solubility
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Formation of Alloys
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In order for an alloy system to have unlimited solid solubility,
certain conditions must be satisfied. These conditions, the
Hume-Rothery rules, are as follows:
 Size factor: The atoms or ions must be of similar size, with
no more than a 15% difference in atomic radius, in order to
minimize the lattice strain
Unlimited Solubility
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Formation of Alloys
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Crystal structure: The materials must have the same crystal
structure; otherwise, there is some point at which a transition
occurs from one phase to a second phase with a different
structure.
Valence: The ions must have the same valence; otherwise,
the valence electron difference encourages the formation of
compounds rather than solutions.
Electronegativity: The atoms must have approximately the
same electronegativity. Electronegativity is the affinity for
electrons. If the electronegativities differ significantly,
compounds form as when sodium and chlorine atoms
combine to form sodium chloride.
Limited Solubility
Salt or sugar have a limited solubility in water.
 If we add a small amount of liquid zinc to liquid copper, a
single liquid solution is produced. When that copper-zinc
solution cools and solidifies, a single solid solution having
an FCC structure results, with copper and zinc atoms
randomly located at the normal lattice points.
 However, during the solidification of the liquid solution
containing more than about 30% Zn, some of the excess
zinc atoms combine with some of the copper atoms to
form a CuZn compound.
 Two solid phases now coexist: a solid solution of
 copper saturated with about 30% Zn plus a CuZn
compound.
 The solubility of zinc in copper is limited.
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Formation of Alloys
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Formation of Alloys
Limited Solubility
Solid – Solution Strengthening
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Formation of Alloys
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In metallic materials, one of the important effects of solidsolution formation is the resultant solid-solution
strengthening.
This strengthening is caused by increased resistance to
dislocation motion.
This is one of the important reasons why brass (Cu-Zn
alloy) is stronger than pure copper.
Similarly small levels of carbon strengthen iron.
Jewelry could be made out of pure gold or silver.
However, pure gold and pure silver are extremely soft and
malleable and the jewelry pieces made will not retain their
shape. That’s why jewelers add copper to gold or silver.
INDUSTRIAL MATERIALS
Formation of Alloys
Solid – Solution Strengthening
The effects of several
alloying elements on the
yield strength of copper.
Nickel and zinc atoms are
about the same size as
copper atoms, but beryllium
and tin atoms are much
different from copper
atoms. Increasing both
atomic size difference and
amount of alloying element
increases solid-solution
Strengthening.
Solid – Solution Strengthening
Degree of Solid-Solution Strengthening
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The degree of solid-solution strengthening depends on
two factors.
First, a large difference in atomic size between the
original (host or solvent) atom and the added (guest or
solute) atom increases the strengthening effect.
A larger size difference produces a greater disruption of
the initial crystal structure, making slip more difficult.
Second, the greater the amount of alloying element added,
the greater the strengthening effect. A Cu-20% Ni alloy is
stronger than a Cu-10% Ni alloy.
Solid – Solution Strengthening
Effect of Solid-Solution Strengthening on Properties
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Formation of Alloys
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The yield strength, tensile strength, and hardness of alloys
are greater than those of the pure metals. For example,
small concentrations of Mg are added to aluminum to
provide higher strength to the aluminum alloys used in
making aluminum beverage cans.
Almost always, the ductility of the alloy is less than that
of the pure metal. Only rarely, as in copper-zinc alloys,
does solid-solution strengthening increase both strength
and ductility.
Solid – Solution Strengthening
Effect of Solid-Solution Strengthening on Properties
Electrical conductivity of the alloy is much lower than that of
the pure metal. This is because electrons get more scattered off
the atoms of the alloying elements. Solid-solution
strengthening of copper or aluminum wires used for
transmission of electrical power is not recommended because
of this pronounced effect. Electrical conductivity of many
alloys, although lower than that of pure metals, is often more
stable as a function of temperature.
 The resistance to creep, or loss of strength at elevated
temperatures, is improved by solid-solution strengthening.
High temperatures do not cause a catastrophic change in the
properties of solid-solution-strengthened alloys. Many hightemperature alloys, such as those used for jet engines, rely
partly on extensive solid-solution strengthening.
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Formation of Alloys
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Formation of Alloys
Solid – Solution Strengthening
The effect of additions of zinc
to copper on the properties of
the solid-solution strengthened
Alloy.
Phase Diagram
A phase diagram shows the phases and their compositions
at any combination of temperature and alloy composition.
 When only two elements or two compounds are present in
a material, a binary phase diagram can be constructed.
 Isomorphous binary phase diagrams are found in a
number of metallic and ceramic systems. In the
isomorphous systems, which include the copper-nickel
systems, only one solid phase forms; the two components
in the system display complete solid solubility.
 In phase diagram, x-axis represents mole%, weight%,
atomic% or mole fraction of one of the components. The
y-axis represents the temperature.
 A ternary phase diagram is a phase diagram for systems
consisting of three components.
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Formation of Alloys
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Formation of Alloys
Phase Diagram
Liquidus
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Transition in structure appear as characteristic points in
the temperature – time plot of the cooling history, known
as cooling curve. It is obtained when a fixed composition
material is heated and subsequently cooled at a uniformly
slow rate
Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
Cooling curves for Cu-Ni alloy form different Ni
composition
Phase Diagram
Liquidus temperature is the
temperature above which a material
is completely liquid. The upper
curve represents the liquidus
temperatures for copper-nickel
alloys for different compositions.
 We must heat an alloy above
the liquidus temperature to produce
a completely liquid alloy that can
then be cast into a useful shape. The
liquid alloy begins to solidify when the temperature
decreases to the liquidus temperature. For the Cu-40%Ni
alloy in figure the liquidus temperature is 1280ºC.
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Formation of Alloys
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Phase Diagram
Solidus temperature is the
temperature below which the alloy
is 100% solid. The lower curve in
figure represents the solidus
temperatures for Cu-Ni alloys for
different compositions.
 An alloy is not completely solid
until the material cools below the
solidus temperature.
 If we use a copper-nickel alloy at high temperatures, we
must be sure that the service temperature is below the
solidus so that no melting occurs. For the Cu-40% Ni
alloy in figure, the solidus temperature is 1240ºC.
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Formation of Alloys
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Phase Diagram
Alloys melt and freeze over a
range of temperatures between the
liquidus and the solidus. The
temperature difference between the
liquidus and the solidus is the
freezing range of the alloy.
 Within the freezing range, two
phases coexist: a liquid and a solid.
For Cu-Ni, the solid is a solution of
copper and nickel atoms and is designated as the α phase.
 For the Cu-40% Ni alloy in figure, the freezing range is
1280 – 1240 = 40ºC.
 Note that pure metals solidify at a fixed temperature i.e.,
the freezing range is zero degrees.
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Formation of Alloys
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Phase Diagram
Often we are interested to know the phases present in an
alloy at a particular temperature.
 If we plan to make a casting, we must be sure that the
metal is initially all liquid.
 If we plan to heat treat an alloy component, we must be
sure that no liquid forms during the process.
 Different solid phases have different properties. For
example, BCC Fe (indicated as α phase on the iron carbon
phase diagram) is magnetic. However, FCC iron
(indicated as γ phase on the Fe-C diagram) is not.
 The phase diagram can be treated as a road map; if we
know the temperature and alloy composition we can
determine the phases present.
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Formation of Alloys
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Phase Diagram
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Formation of Alloys
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From the phase diagram for the NiO-MgO binary system
[figure below], describe a composition that can melt at
2600ºC but will not melt when placed into service at
2300ºC. Often we are interested to know the phases
present in an alloy at a particular temperature.
Phase Diagram
One method to improve the
fracture toughness of a ceramic
material is to reinforce the
ceramic matrix with ceramic
fibers. A materials designer has
suggested that Al2O3 could be
reinforced with 25% Cr2O3 fibers, which would interfere
with the propagation of any cracks in the alumina. The
resulting composite is expected to operate under load at
2000ºC for several months. Criticize the appropriateness of
this design.
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Formation of Alloys
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Phase Diagram
Gibbs Rule for Isomorphous Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
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We keep the pressure fixed (e.g., one atmosphere), which
is normal for binary phase diagrams.
The phase rule can be rewritten as:
1 + C = F + P (for constant pressure)
Determine the degrees of freedom in a Cu-40% Ni alloy at
(a) 1300ºC, (b) 1250ºC, and (c) 1200ºC.
◦ At 1300ºC, P = 1, C = 2,
1+C=F+P
→
◦ At 1250ºC, P = 2, C = 2,
1+C=F+P
→
◦ At 1200ºC, P = 1, C = 2,
1+C=F+P
→
F=2
F=1
F=2
Phase Diagram
Gibbs Rule for Isomorphous Phase Diagram
As there is only one degree of freedom in a two-phase region of a
binary phase diagram, the compositions of the two phases are always
fixed when we specify the temperature. Therefore, we can use a tie
line to determine the composition of the two phases.
 A tie line is a horizontal line
within a two-phase region drawn at
the temperature of interest (see
figure). In an isomorphous system,
the tie line connects the liquidus and
solidus points at the specified
temperature. The ends of the tie line
represent the compositions of the
two phases in equilibrium.
 Tie lines are not used in
single-phase regions because we do not have two phases to ‘‘tie’’ in.
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Formation of Alloys
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Phase Diagram
Use of Tie Line
Determine the
composition of each
phase in a Cu-40% Ni
alloy at 1300ºC,
1270ºC, 1250ºC, and
1200ºC (see fig.).
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Formation of Alloys
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Phase Diagram
Lever Rule (amount of each phase)
INDUSTRIAL MATERIALS
Formation of Alloys
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To calculate the amounts of liquid and solid, we construct
a lever on our tie line, with the fulcrum of our lever being
the original composition of the alloy. The leg of the lever
opposite to the composition of the phase, whose amount
we are calculating, is divided by the total length of the
lever to give the amount of that phase.
The lever rule in general can be written as:
Phase percent = opposite arm of lever x 100
total length of tie line
Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
Lever Rule (amount of each phase)
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Fraction of the solid phase = (Wo – Wl)/(Ws-Wl)
Phase Diagram
Lever Rule (amount of each phase)
INDUSTRIAL MATERIALS
Formation of Alloys
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Determine the amount of each phase in the Cu-40% Ni
alloy at 1300ºC, 1270ºC, 1250ºC, and 1200ºC.
at
1300ºC: There is only one phase, so 100% L.
at
1270ºC: %L = 50 – 40 x 100 = 77%
50 – 37
%α = 40 – 37 x 100 = 23%
50 – 37
Eutectic
A eutectic or eutectic mixture is a mixture of two or more
phases at a composition that has the lowest melting point.
 It is where the phases simultaneously crystallize from
molten solution.
 The proper ratios of phases to obtain a eutectic is
identified by the eutectic point on a binary phase diagram.
 The eutectic point is the point where the liquid phase
borders directly on the solid (α + β) phase.
INDUSTRIAL MATERIALS
Formation of Alloys
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Eutectic point
Eutectic
The eutectic point represents the minimum melting
temperature of any possible A-B alloy.
 The temperature that corresponds to this point is known as
the eutectic temperature.
 Not all binary system alloys have a eutectic point: those
that form a solid solution at all concentrations, such as the
Cu-Ni system, have no eutectic.
 An alloy system that has a eutectic is often referred to as a
eutectic system, or eutectic alloy.
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Formation of Alloys
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Cu – Ag Ectectic Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
Binary – Eutectic Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
Binary – Eutectic Phase Diagram
Cu –Ag Eutectic Phase Diagram
 Three single-phase regions can be found on the diagram:
α, β, and liquid.
 The α phase is a solid solution rich in copper; it has silver
as the solute component and an FCC crystal structure.
 The β phase solid solution also has an FCC structure, but
copper is the solute.
Cooling curve of
a eutectic alloy
 No liquid below TE
 CE is the composition at TE
 The intersection of CE and TE is called the
eutectic point. If the alloy has the eutectic
composition, it will simultaneously crystallize from molten
solution i.e., without having any freezing range.
INDUSTRIAL MATERIALS
Formation of Alloys
Binary – Eutectic Phase Diagram
Cu –Ag Eutectic Phase Diagram
 The solubility in each of the solid phases is limited.
 The solubility limit for the α phase corresponds to the
boundary line, CBA, between the α/(α+β) and α/(α+L)
phase regions.
 It increases with temperature to a maximum [8.0 wt% Ag
at 779ºC] at point B, and decreases back to zero at the
melting temperature of pure copper, point A [1085ºC].
 At temperatures below 779ºC, the solid solubility limit
line separating the α and α+β phase regions is termed as
solvus line (CB).
 The boundary AB between the α and α+L regions is the
solidus line.
INDUSTRIAL MATERIALS
Formation of Alloys
Binary – Eutectic Phase Diagram
Cu –Ag Eutectic Phase Diagram
 For the β phase, both solvus and solidus lines also exist,
HG and GF, respectively.
 There are also three two-phase regions found for the
copper–silver system: α+L, β+L, and α+β.
 The α and β phase solid solutions coexist for all
compositions and temperatures within the α+β phase field.
 The α+liquid and β+liquid phases also coexist in their
respective phase regions.
 An important reaction occurs for an alloy of composition
CE as it changes temperature in passing through TE; this
reaction may be written as follows:
INDUSTRIAL MATERIALS
Formation of Alloys
Binary – Eutectic Phase Diagram
Cu –Ag Eutectic Phase Diagram
 Upon cooling, a liquid phase is transformed into the two
solid α and β phases at the temperature TE; the opposite
reaction occurs upon heating. This is called a eutectic
reaction; CαE and CβE are the respective compositions of
the α and β phases at TE. Thus, for the copper–silver
system, the eutectic reaction may be written as follows:
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The eutectic reaction, upon cooling, is similar to
solidification for pure components, however, the solid
product of eutectic solidification is always two solid
phases, whereas for a pure component only a single phase
forms.
Binary – Eutectic Phase Diagram
INDUSTRIAL MATERIALS
Formation of Alloys
Pb –Sn Eutectic Phase Diagram
(Ex) Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, determine:
INDUSTRIAL MATERIALS
Formation of Alloys
-- the phases present
T(°C)
Answer: a + b
-- the phase compositions
300
Answer: Ca = 11 wt% Sn
Cb = 99 wt% Sn
-- the relative amount
of each phase
Answer:
200
150
C - C0
W = b
a
Cb - Ca
=
=
Pb-Sn
system
a
L+a
18.3
L +b b
183°C
61.9
100
99 - 40
59
=
= 0.67
99 - 11
88
Wb = C0 - Ca
Cb - Ca
40 - 11
29
=
= 0.33
99 - 11
88
L (liquid)
97.8
a + b
0 11 20
Ca
40
C0
60
80
C, wt% Sn
99100
Cb
(Ex contd. )
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:
INDUSTRIAL MATERIALS
Formation of Alloys
-- the phases present:
Answer: a + L
-- the phase compositions
T(°C)
Answer: Ca = 17 wt% Sn
CL = 46 wt% Sn
-- the relative amount
of each phase
Answer:
300
L+a
220
200
CL - C0
46 - 40
=
Wa =
CL - Ca
46 - 17
6
=
= 0.21
29
C0 - Ca
23
=
= 0.79
WL =
CL - Ca
29
L (liquid)
a
L +b b
183°C
100
a + b
0
17 20
Ca
40 46 60
80
C0 CL C, wt% Sn
100
(Ex contd.)
INDUSTRIAL MATERIALS
Formation of Alloys
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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.
(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.
Hypoeutectic and Hypereutectic
Alloys
INDUSTRIAL MATERIALS
Formation of Alloys
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A hypoeutectic alloy is an alloy whose composition will
be between that of the left-hand-side end of the tie line
defining the eutectic reaction and the eutectic
composition. As a hypoeutectic alloy cools, the liquid
begins to solidify at the liquidus temperature, producing
solid α. However, solidification is completed only after
going through the eutectic reaction.
An alloy composition between that of the right-hand-side
end of the tie line defining the eutectic reaction and the
eutectic composition is known as a hypereutectic alloy.
INDUSTRIAL MATERIALS
Formation of Alloys
Hypoeutectic and Hypereutectic
Alloys
The cooling curve for a hypoeutectic Pb-30% Sn alloy