Phase Diagrams
ISSUES TO ADDRESS...
• When we combine two elements...
what equilibrium state do we get?
• In particular, if we specify...
-- a composition (e.g., wt% Cu - wt% Ni), and
-- a temperature (T )
• Then...
How many phases do we get?
What is the composition of each phase?
What is the amount of each phase?
Phase B
Phase A
Nickel atom
Copper atom
2
Phase Equilibria: Solubility Limit
– Solutions – solid solutions, single phase
– Mixtures – more than one phase
Phase: A homogeneous portion of a system that have
uniform physical and chemical characteristics.
Question: What is the
solubility limit at 20°C?
Answer: 65 wt% sugar.
100
If Co < 65 wt% sugar: syrup
If Co > 65 wt% sugar: syrup + sugar.
Solubility
Limit
80
60
L
40
(liquid solution
i.e., syrup)
20
0
Pure
Water
Max concentration for
which only a single phase
solution occurs.
L
(liquid)
+
S
(solid
sugar)
20
40
6065 80
100
Co =Composition (wt% sugar)
Pure
Sugar
Solubility Limit:
Temperature (°C)
•
Sucrose/Water Phase Diagram
3
Effect of T & Composition (Co)
• Changing T can change # of phases:
path A to B.
• Changing Co can change # of phases:
path B to D.
B (100°C,70)
1 phase
watersugar
system
Adapted from
Fig. 9.1,
Callister 7e.
Temperature (°C)
100
D (100°C,90)
2 phases
80
L
60
(liquid)
+
L
(liquid solution
40
i.e., syrup)
20
0
0
S
(solid
sugar)
A (20°C,70)
2 phases
20
40
60 70 80
100
Co =Composition (wt% sugar)
4
Components and Phases
• Components:
The elements or compounds which are mixed initially
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that result (e.g., a and b).
AluminumCopper
Alloy
Adapted from
Fig. 9.0,
Callister 3e.
Phase Equilibria
Simple solution system (e.g., Ni-Cu solution)
Crystal
Structure
Electro neg
r (nm)
Ni
FCC
1.9
0.1246
Cu
FCC
1.8
0.1278
• Both have the same crystal structure (FCC) and have
similar electronegativities and atomic radii (W. HumeRothery rules) suggesting high mutual solubility.
• Ni and Cu are totally miscible in all proportions (complete
solid solubility).
6
• It can be considered as a map that
determine the phases depending on
composition and temperature.
Temperature
• Equilibrium phase diagrams represent
the relationship between temperature
and the compositions and the quantities
of phases present at equilibrium.
100% A
Map of
phases
present
Wt% of B
•
Solid solution: It exists when different atoms take part in
building a crystal lattice (single phase).
•
Type of solid solutions:
•
•
Substitutional solid solution
•
Interstitial solid solution
In both cases solute atoms completely dissolve in the host
metal forming a single phase.
B 100%
• Pure metals
solidifies at a
constant
temperature
which is known
as the melting
temperature.
• Binary alloys
solidifies over a
range of
temperatures.
Temperature
Cooling curves:
Time
Time
Solid solution phase diagram:
• How to construct a
phase diagram
from cooling
curves?
• Draw the cooling
curves of different
composition
alloys, the upper
inflection point
form a loci for the
liquidus line, the
lower inflection
Melting
point form a loci
temp. of
for the solidus
Cu
line.
Melting
temp. of
Ni
Some notes about the phase diagram:
• The liquid L is a homogeneous liquid solution of both copper
and nickel.
• The 𝛼 phase is a substitutional solid solution consisting of
both copper and nickel atoms, and having FCC crystal
structure.
• Below 1085°C there is complete solubility between copper
and nickel (same crystal structure FCC, nearly equal atomic
radii, and electronegativity and similar valency).
• Melting temperatures for pure nickel is 1085°C and for pure
nickel is 1453°C.
• Copper-nickel system is termed isomorphous because of this
complete liquid and solid solubility.
• For each particular composition and temperature the number
and type of phases can be pointed out.
Phase diagrams: number and types of phases
• Rule 1: If we know T and Co, then we know the number and
types of phases present.
• Examples:
Cu-Ni
phase
diagram
Adapted from Fig. 9.2(a), Callister 6e.
(Fig. 9.2(a) is adapted from Phase
Diagrams of Binary Nickel Alloys, P.
Nash (Ed.), ASM International,
Materials Park, OH, 1991).
Phase diagrams: composition of phases
• Construct a tie line across the two phase region
• Intersection of the tie line and the phase boundaries on either
sides are noted.
• Perpendiculars are dropped from these intersection to the
horizontal composition axis
For an alloy of
composition 35%Ni,
65%Cu, determine the
phase composition at
1250°C
If only one phase is present, for example a composition of 60%Ni40%Cu at 1100°C (point A), only the 𝛼 phase is present. The
composition of this phase is the same as the composition of the
alloy, which mean that the composition of a is 60% Ni-40 %Cu .
Phase diagrams: composition of phases
Rule 2: If we know T and Co, then we know the composition of
each phase.
Cu-Ni system
• Examples:
Phase diagrams: weight fractions of phases
Rule 3: If we know T and Co, then we know the amount of each
phase.
Cu-Ni
system
• Examples:
=
43- 35
= 73wt%
43- 32
= 27wt%
𝑚!
𝑊! =
𝑚! + 𝑚"
𝑚"
𝑊" =
𝑚! + 𝑚"
The lever rule: Proof
• Sum of weight fractions: 𝑊! + 𝑊" = 1
• Conservation of mass (Ni): 𝐶# = 𝑊! 𝐶! + 𝑊" 𝐶"
• Combine above equations:
• A geometric interpretation: moment equilibrium:
WLR = WaS
1- Wa
solving gives Lever Rule
1 − 𝑊! 𝑅 = 𝑊! 𝑆
𝑅 = 𝑊! 𝑆 + 𝑊! 𝑅 = 𝑊! 𝑅 + 𝑆
Development of microstructure in isomorphous
(solid solution) alloys:
•Solidification start at point b
at 1260°C.
•Solidification is complete at
point d, 1220°C.
•In between the alloy is
composed of two phases, 𝛼
and L.
•The number written beside
each phase is the chemical
composition of that phase.
Mechanical properties: Cu-Ni System
Effect of solid solution strengthening on:
--Tensile strength (TS)
Adapted from Fig. 9.5(a), Callister 6e.
--Ductility (%EL,%AR)
Adapted from Fig. 9.5(b), Callister 6e.
Example:
Example:
A 65 wt% Ni-35 wt% Cu alloy
is heated to a temperature
within the (𝛼 + 𝐿) phase
region. If the composition of
the 𝛼 phase is 70 wt% Ni,
determine:
a) The temperature of the
alloy.
b) The composition of the
liquid phase.
c) The mass fractions of
both phases.
1340
𝐶# 𝐶"𝐶!
Binary-Eutectic Systems
has a special composition with a
minimum melting temperature.
2 components
Cu-Ag system
Cooling
curve
Ag
a is a solid solution phase of
Ag dissolved in Cu.
b is a solid solution phase of
Cu dissolved in Ag.
Cu
Wt% Ag
TE = 779 oC
Ag
Eutectic reaction
Liquidus line
Solidus line
An important reaction occurs
for an alloy of composition CE
, 71.9 wt%Ag (eutectic point)
as the alloy passes through
the eutectic temperature TE.
Upon cooling the liquid is
transformed to two solid
phases a and b
Solvus line
Cu
Eutectic reaction: L àS1+S2
For copper silver system
Wt% Ag
Ag
Eutectic means easily melted, it corresponds to the lowest melting
temp. composition, and its cooling curve is the same as of pure
metals where solidification happens at a constant temperature.
General notes:
• A general rule is that single-phase regions are always
separated from each other by a two-phase region that consist of
the two single phases that it separates.
• The solvus line is analogous to the solubility limit line in the
sugar-water system. The solubility of Ag in Cu is limited at room
temperature and increases with increasing temp.
• These diagrams can also be termed as partial solid solubility
phase diagram.
• This phase diagram can also be constructed using cooling
curves for different composition alloys.
Example: Pb-Sn eutectic system
For a 40wt%Sn-60wt%Pb alloy at 150C, find:
a) the phases present.
b) the compositions of the phases.
c) the relative amounts of each phase.
Development of microstructure in binary eutectic
system
Co < 2wt%Sn
Result:
• polycrystal of a grains.
• The minimum solubility
limit is not exceeded.
That’s why we end with a
single solid solution phase
2wt%Sn < Co <18.3wt%Sn
Result:
• a polycrystal with fine b
crystals.
• b phase is considered
to be precipitates.
Co = CE =61.9 wt% Sn
Result: Eutectic microstructure in the form of alternating layers of
a and b crystals (lamellar structure)
18.3wt%Sn < Co < 61.9wt%Sn
Result: primary a crystals and an eutectic microstructure ae
and be
Upon cooling and at the liquidus line primary a starts to appear and
grows with cooling until the eutectic temperature (TE). Upon
reaching TE the primary a is completely formed and the remaining
liquid transforms into the eutectic microconstituent (ae and be)
according to the eutectic reaction.
TE+DT
TE-DT
Pb-Sn system
𝑊"! = 𝑊"$ + 𝑊""
Weight fraction of eutectic microconstituent (𝑊!! +𝑊#! ) is the
same as the fraction of liquid from which it transforms just
above TE (TE +DT) :
𝐿 → 𝑊!! + 𝑊#!
thus 𝑊$ = 𝑊!! + 𝑊#!
Above TE (TE +DT) the phases are a and L so the tie line is drawn
between the solidus (for 100% a) and liquidus (for 100% L) line at
TE (between 18.3% and 61.9%)
Weight fraction of primary a (𝑊!% ) is the fraction of a phase
just before the eutectic transformation (TE+DT)
Weight fraction of total a (primary
and eutectic) and total b can be
found at (TE-DT), where the only
phases present is a and b, by
drawing a tie line that extends
from the left solvus line (for a) and
the right solvus line (for b) at TE.
Weight fraction of eutectic a:
𝑊"" = 𝑊"! − 𝑊"$
Example:
For a copper–silver alloy of composition 25 wt%Ag–75wt%Cu and at 775°C:
(a) Determine the mass fractions of 𝛼 and phases 𝛽.
(b) Determine the mass fractions of primary 𝛼 and eutectic microconstituents.
(c) Determine the mass fraction of eutectic 𝛼.
Hypoeutectic & Hypereutectic
Variation of mechanical properties with alloying
Maximum strength is achieved at the eutectic composition
due to the large number of boundary layers in the lamellar
eutectic microstructure.
Eutectoid and Peritectic reactions
Eutectoid
reaction: a solid
phase 𝛿
transforms into
two other solid
phases 𝛾 + 𝜖.
Peritectic
reaction: one solid
phase transforms
into a liquid phase
and another solid
phase
Allotropy: The ability of the material to exist in more than
one crystal structure depending on the temperature. Iron has
three allotropic form.
Cooling curve
of pure iron
Curies Temp.
liquid
1540
d - iron
1394 BCC
g-iron
FCC
912
Non-magnetic a-iron
BCC
768
Magnetic
time
Iron – Carbon system
The phase diagram will be studied up to 6.7 wt% C, where
an intermediate compound iron carbide (Fe3C) is formed,
because in practice all steels and cast iron have carbon
contents less than 6.7 wt% C.
Steel
Cast iron
(Fe3C)
Phases present:
• Austenite (g) : solid solution of carbon in FCC g iron.
Maximum solubility of carbon in austenite is 2.14wt%,
occurs at 1147oC. Austenite is very soft and malleable.
Steel can be at one time100% austenite and so it can be
easily forged.
• Ferrite (a): solid solution of carbon in BCC a iron.
Maximum solubility of carbon is 0.022 wt%, occur at
727oC. Ferrite is soft and ductile. (B.H.N. : 70)
• Ferrite (d) : Similar to a ferrite, except for the range of
temperature over which it exist.
• Cementite (Fe3C) : Chemical compound of iron carbide.
Extremely hard and brittle (BHN : 800)
Important reactions:
Eutectoid reaction: at 0.76wt% C and temp.: 727°C
Pearlite
Pearlite: It is the eutectoid structure à mixture of two
phases (a + Fe3C). Pearlite is strong and tough (BHN: 160).
Microstructure is lamellar (alternating layers of a and Fe3C).
Eutectic reaction: at 4.3 wt%C and temp.:1147°C
Iron-Carbon (Fe-C) Phase Diagram
Hypoeutectoid Steel (0.022wt%C < %C < 0.76wt%C)
• The proeutectoid phase
(the phase that appears
beside g above the
eutectoid temp. 727°C)
is a ferrite and it forms
along the g boundaries.
• When austenite g
reaches the eutectoid
temp. 727°C transforms
completely to pearlite.
Hypereutectoid Steel (0.76wt%C < %C < 2.14wt%C)
• The proeutectoid phase
(the phase that appears
beside g above the
eutectoid temp. 727°C) is
Fe3C cementite and it
forms along the g
boundaries.
• When austenite g
reaches the eutectoid
temp. 727°C transforms
completely to pearlite.
Weight fractions steel phases (hypoeutectoid)
Weight fraction of
pearlite is the same as
the weight fraction of
austenite from which it
is formed (just above
the eutectoid line)
Weight fraction of
proeutectoid a
Weight fractions steel phases (hypereutectoid)
Weight fraction of
proeutectoid Fe3C
Consider 1.0 kg of austenite containing 1.15 wt%C, cooled to below
727°C.
a) What is the proeutectoid phase?
b) How many kilograms each of total ferrite and cementite form?
c) How many kilograms each of pearlite and the proeutectoid
phase form?
d) Schematically sketch and label the resulting microstructure.
Effect of alloying elements
• Teutectoid changes:
• Ceutectoid changes: