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Microstructure of a lead-tin alloy
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Pratt and Whitney F119 Engine
Why study Phase Diagrams?
• Provide valuable information about: – Melting
– Casting
– Microstructure
– Crystallization
– Temperature
– Pressure
– Composition
– Phase transformations
Phase Diagrams
• Component: a chemical constituent of an alloy, which may be used to specify its composition – Example: In a copper-zinc brass, the
components are Cu and Zn
• Solid solution (Chapter 4): – Consists of atoms of at least two different types
– Solute atoms occupy either substitutional or interstitial positions
– Crystal structure of solvent is maintained
Solubility Limit
• Maximum concentration of solute atoms that may dissolve in the solvent, at a specific temperature, to form a solid solution
• Adding solute past the solubility limit yields another solid solution (different crystal structure) or compound that has a different composition
Phases • Phase: a homogeneous portion of a system that
has uniform physical and chemical characteristics • The following are considered phases:
– Solid solution – Liquid solution – Gaseous solution
• If more than one phase is present in a system, each will have its own distinct properties
• Phases exist over a range of compositions, temperatures and pressures
• Example: Water and Ice two phases (physically dissimilar, but chemically the same)
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Stability of Phase
• Phase stability depends on temperature, pressure and composition,
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Phases
For each of the following situations how many phases are present?
1. Ice cubes floating in a glass of water.
2. Water at room temperature containing 20 wt% sugar.
3. A bottle of 7-Up.
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Components vs. Phases • Components:
– The elements (or compounds) that are mixed to make the alloy
• Al and Cu
• MgO and Al2O3
• styrene and butadiene
– The composition is the concentration of each component present in an alloy
• Phase: – Physically and/or chemically
distinct regions in the material
Phase β Phase α
Aluminum atom
Copper atom
Al-Cu alloy
Sugar- Water Demo
• Initially, sugar is added to water and there is a sugar- water solution or syrup
• Solution becomes more concentrated with sugar until the solubility limit is reached (saturated)
• System cannot dissolve any more sugar (at the specified temperature) and further additions settle to the bottom of the container
• System now contains two phases; syrup and solid (undissolved) sugar crystals
Solid Solubility
• We define solubility as the maximum equilibrium concentration of a solute in a host material.
Answer: 65 wt% sugar
ALS: What is the solubility of sugar in water at 20oC?
ALS: If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?
a) 650 g of sugar
b) 2000 g of sugar
c) 1350 g of sugar
Solid Solubility
• We define solubility as the maximum equilibrium concentration of a solute in a host material Previous Answer: 65 wt% sugar
ALS: If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?
Solid Solubility
• We define solubility as the maximum equilibrium concentration of a solute in a host material Previous Answer: 65 wt% sugar
Answer: The solution is 65% (about 2/3) sugar and 1/3 water. The weight of the sugar is twice that of the water, so 2000 g is the answer.
If you leave a saturated water/sugar solution open to the air and the water starts to evaporate, what will happen?
Solubility Limits
Answer: At some point there won’t be enough water to hold all of the sugar. The solubility for sugar will be exceeded, and some of it will come out of the solution in crystal form (i.e. precipitate) to compensate.
Phase Diagrams
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One-Component (Or Unary) Phase Diagrams
• 3 externally controllable parameters that affect phase structure:
– Temperature
– Pressure (usually scaled logarithmically)
– Composition
• Unary phase diagrams only deal with pressure-temperature graphs composition remains constant
• Along the phase boundaries, the phases on either side are in equilibrium
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(H20) Each of the phases will exist in equilibrium
conditions over the corresponding temperature
• Triple point: all three states are in equilibrium (at 273.16 K at a pressure of 6.04x10-3 atm in the example)
Binary Phase Diagrams
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Binary Phase Diagrams
• Temperature and composition are variable parameters
• Binary alloys – contain two components
• Pressure held constant normally ≈ 1 atm
• Maps that represent the relationship between temperatures, compositions and quantities of phases at equilibrium
• Accurately predict phase transformations and resulting microstructures
Binary Isomorphous Systems
• Easiest binary phase diagram to understand – Example: copper-nickel system
• Temperature is plotted along the vertical axis
• Composition of the alloy is plotted along the horizontal axis – Bottom- weight %
– Top – atom %
– Composition ranges from 0 wt% Ni (100 wt% Cu) on the left of the horizontal axis to 100 wt% Ni (0 wt% Cu) on the right of the horizontal axis
Binary Isomorphous Systems
• Three phase regions or fields: – Alpha(α)
• Substitutional solid solution consisting of Cu and Ni and both have an FCC crystal structure
– Liquid (L) • Homogeneous liquid solution
composed of both copper and nickel
– Two phase α + L
• The copper-nickel system is termed isomorphous due to the complete liquid and solid solubility of the two components
Phase Boundary
Interpretation of Phase Diagrams
• For a binary system of known composition and temperature that is at equilibrium, you can identify:
1. Phases present 2. Compositions of phases 3. Percentages or fractions of phases
• Phases Present:
– Locate the temperature-composition point on the diagram and one can identify the phase(s) within the phase field that are/ is present
Interpretation of Phase Diagrams
• Phase compositions: – One phase present:
• composition of phase is the same as the overall composition of the alloy
– Two phases present: • A Tie line is constructed across two-phase region at
a specified temperature of the alloy
• Intersections of tie lines and phase boundaries on both sides are noted
• The compositions corresponding to these points are read from the x-axis.
Interpretation of Phase Diagrams
• Determining Phase Amounts: – One phase present:
• Alloy is composed entirely of that phase; phase fraction is 1.0 or percentage is 100%
– Two phases present: • Use the lever rule; tie line is constructed across the two-phase
region at a specified temperature of the alloy • Overall alloy composition is located on the tie line • Fraction of one phase is determined by taking the length of the tie
line from the overall alloy composition to the phase boundary for the other phase, and dividing by the total lie line length
• The fraction of the other phase is determined in the same manner • Segment lengths can be determined either by direct measurement
from the phase diagram using ruler or by subtracting compositions from the composition axis
• Liquid mass fraction:
• α-Phase mass fraction:
Alternate Approach - Phase diagrams
L
LCC
CC
SR
SW
0
L
L
CC
CC
SR
RW
0
Developing Microstructures – Isomorphous Alloys
• Equilibrium cooling:
– Cooling occurs very slowly
Mechanical Properties of Isomorphous Alloys
• Solute Strengthening
TS for pure Cu
TS for pure Ni
%EL for pure Cu
%EL for pure Ni
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Binary Eutectic Systems
One common type of alloy system is called the eutectic system. Binary eutectic systems occur when 2 elements completely dissolve into each other in the liquid state, but have only limited solubility in the solid state.
More than one solid phase exists in these systems. There are multiphase regions between single-phase regions. Which phases are present depends, as before, on the temperature and the composition of the material.
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Binary Eutectic Systems
The copper-silver alloy system is binary eutectic.
• Consider Cu-Ag alloys:
− How many phases are there in the system?
− What are the two solid phases?
− Where is the eutectic point?
3 (L, , b)
: mostly Cu
b: mostly Ag
eutectic temperature: TE eutectic composition: CE
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ALS: Pb-Sn Eutectic System
• Consider an alloy which contains 40 wt% Sn at 150oC.
• What phases are present? What are their compositions?
• Answer:
There are 2 phases: a and b.
Ca = 11 wt% Sn
Cb = 99 wt% Sn
The lead-tin system is also binary eutectic
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ALS: Pb-Sn Eutectic System
• How much of each phase is present?
%6788
59
W
1 33%W Wb
• Answer: From the lever rule:
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Adapted from Fig. 9.9,
Callister 6e.
Microstructures in Eutectic Systems
• Consider a eutectic system (e.g. Pb-Sn) with an initial concentration (Co) less than 6 wt% Sn.
• The result is... – … a single phase
polycrystal with grains.
10% 30% 20% 0%
100
300
400
200
%Sn
Te
mp
era
ture
oC
Co
L
L
b
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Pb-Sn
system
Adapted from Fig. 9.10,
Callister 6e.
Microstructures in Eutectic Systems
• Consider a eutectic system with an initial concentration (Co) between 2 and 18.3 wt% Sn.
• The result is... –… a 2-phase
polycrystal with... • … and β grains. • … a fine grained
structure.
: C o wt%Sn L +
2 0 0
T(°C)
C o , wt% Sn 1 0
18.3
2 0 0 C o
3 00
1 00
L
3 0
L: C o wt%Sn
+ b
4 00
(sol. limit at TE)
T E
2 (sol. limit at Troom)
L
b
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Pb-Sn
system
Adapted from Fig. 9.11, Callister 6e.
Adapted from Fig. 9.12, Callister 6e. (Fig. 9.12 from Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
Microstructures in Eutectic Systems • When the initial concentration (Co) equals the eutectic
composition (CE) the result is...
– … alternating layers of and β crystals
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L +
2 0 0
T(°C)
wt% Sn
2 0 4 0 0
3 00
1 00
L
6 0
L: C o wt%Sn
+ b
T E b
0 8 0 100
L + b
C o 18.3 61.9
L
L
primary
97.8
S
S
R
R
eutectic eutectic b
Pb-Sn
system
Adapted from Fig. 9.14,
Callister 6e.
Microstructures in Eutectic Systems
• Co between 18.3 and 61.9 wt% Sn results in... – … a combination of crystals and eutectic regions
Just above TE:
W L = (1- W ) = 50 wt%
C = 18.3 wt% Sn
C L = 61.9 wt% Sn
S
R + S W = = 50 wt%
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Pb-Sn
system
Adapted from Fig. 9.14,
Callister 6e.
Microstructures in Eutectic Systems
• Co between 18.3 and 61.9 wt% Sn results in... – … a combination of crystals and eutectic regions
Just below TE:
C = 18.3 wt%Sn
C b = 97.8 wt%Sn
S
R + S W = = 73 wt%
W b = 27 wt%
Eutectoid and Peritectic Reactions
• Eutectoid reaction: a reaction wherein upon cooling, one solid phase transforms isothermally and reversibly into two new solid phases that are intimately mixed
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cooling
heating
cooling
heating
L
• Peritectic reaction: a reaction wherein upon cooling, a solid and a liquid phases transform isothermally and reversibly to a solid phase having a different composition
Copper-zinc phase diagram
The Iron-Carbon System
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The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram
• When heated, pure iron experiences two changes in its crystal structure before it reaches its melting temperature:
– At room temperature, exists as ferrite (α iron) and has a BCC structure
– Ferrite experiences a polymorphic transformation to FCC austenite (γ iron) at 912°C
– Experiences a second transformation to BCC phase known as δ ferrite at 1394°C, which finally melts at 1538°C
The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram
• Cementite, (Iron Carbide - Fe3C). It is 25at% C or 6.7 wt% C.
The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram
Euctectic Point
Eutectoid Point
Eutectic reaction for the iron-iron carbide
system:
Eutectoid reaction for the iron-iron carbide system:
)%7.6()%022.0()%76.0( 3 CwtCFeCwtCwtcooling
heating
CFeLcooling
heating3
Development of Microstructure in Iron-Carbon Alloys
• The microstructure depends on: – Concentration of carbon – Heat treatment
• Pearlite: a two-phase microstucture results from the transformation of austenite and consists of alternating layers ( or lamellae) of -ferrite and cementite
43 Pearlite
Hypoeutectoid Alloys
• Hypoeutectoid alloy : (less than eutectoid) between 0.022 and 0.76wt% C
• Proeutectoid ferrite: that formed above the eutectoid temperature
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Hypereutectoid Alloys • Hypereutectoid alloys:
containing between 0.76 and 2.14 wt% C
• Proeutectoid cementite: that which forms before the eutectoid reaction cementite composition remains constant as temperature changes
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Non-Equilibrium Cooling
• In most situations, the cooling rates are impractically slow and unnecessary
• Nonequilibrium effects of practical importance: – Phase changes or transformations at temperatures other than those
predicted in phase diagrams – The existence of non-equilibrium phases that do not appear on the phase
diagram
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Chapter 9 Review
What did you learn?
•
•
•
•
•
•
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Chapter 9
Practice Problems
Practice Problems
1. Salt is often spread on roads during the winter in order to depress the freezing point of water. A binary phase diagram for water and salt (NaCl) is shown below.
Liquid (Brine)
Field “Z”
Practice Problems
a) What are the two phases that exist in equilibrium in the field “Z” region.
b) At -10°C determine the overall composition for which
equal amounts of these phases exist in equilibrium. c) What are the compositions of each phase under these
conditions? d) What is the lowest temperature at which adding salt
can prevent water from freezing completely?
Practice Problems
Answer:
a)Liquid (Brine) + Ice
b)8 wt% NaCl
c)Ice – 0.5 wt% NaCl
Salt + Brine – 16 wt% NaCl
d) -21°C
Practice Problems
2. Using the Fe-Fe3C phase diagram, compute the weight fractions of proeutectoid ferrite and pearlite, respectively, that form in an iron-carbon alloy containing 0.25 wt% C if the alloy is slowly cooled to a temperature just below the eutectoid temperature.
Practice Problems
a) Wα = 0.31, Wp = 0.69 b) Wα = 0.69, Wp = 0.31 c) Wα = 0.56, Wp = 0.44 d) Wα = 0.44, Wp = 0.56
Practice Problems
Answer:
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b) isanswer theTherefore,
31.0022.076.0
022.025.0
69.0022.076.0
25.076.0
pW
W
Practice Problems
c)
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% 12.11or 1211.07317.08528.0
:reaction eutectic during formed
8528.03.188.97
308.97
:reaction eutecticAfter
7317.03.189.61
309.61
:reaction eutectic Before
W
W
W