Upload
rose-hines
View
218
Download
0
Embed Size (px)
Citation preview
8/10/2019 MSE 2001
1/40
1
Chapter 7 Phase Equilibria and
Phase Diagrams
Introduction
The one-component phase diagram
Phase equilibria in a two-component system The eutectic phase diagram
The peritectic phase diagram
The monotectic phase diagram
Complex diagrams
8/10/2019 MSE 2001
2/40
2
Phase
A chemically and structurally homogeneous region of a material
A part of a system, physically distinct, macroscopically homogeneous,and of fixed or variable composition.
It is mechanically separable from the rest of the system.
A phase is a region within which all the intensive variables vary
continuously, whereas at least some of them have discontinuities at the
borders between phases.
ice water
ice + water2 phases = solid phase + liquid phase
I want to drink 2-phase water consisting
of solid phase and liquid phase.
8/10/2019 MSE 2001
3/40
3
Phase diagramGraphical representation of the combination of temperature, pressure,
composition, or other variables for which specific phases exist at equilibrium.
Phase diagram of Water (H2O)Phase diagram of Carbon dioxide (CO2)
State point: a position on the phase diagram
0.47 g/cm3
@304.25 K & 72.9 atm
8/10/2019 MSE 2001
4/40
4
One-Component Phase Diagrams
Gibbs Phase Rule for systems in equilibrium
Component: a chemical species whose concentration in a phase can be varied
independently of the other species concentration
Number of degrees of freedom in equilibrium is the number of variables (p, T, or
composition) that can be independently adjusted without disturbing equilibrium.
water
F = C
P + 2C- Components
P- Number of phases
F- Degrees of freedomF = 2
F = 1F = 1
F = 0F = 0
F = 2F = 2
8/10/2019 MSE 2001
5/40
5
Example of One-Component Phase Diagrams
iron SiO2
8/10/2019 MSE 2001
6/40
6
Two-Component Phase Diagrams
Temperature
Pressure
Composition for materials A and B
(Composition of one-component
system?)
For one state point in closed system we
need three variables (p, T, & X).
If we fix the pressure,
(T, X(composition))2-D space
(p, T, X(composition))3-D space
F = CP + 1
F = 2
F = 1T
Xs Xl
XA+XB=Xtotal
F = C P + 2
C- Components
P- Number of phases
F- Degrees of freedomF = 2
F = 1F = 1
F = 0F = 0
F = 2
F = 2
One-component
system
8/10/2019 MSE 2001
7/40
7
Two-Component Phase Diagrams
Specification of composition
Atomic percentage (=atomic fraction=atomic number fraction=mole fraction)
Weight percentage (=weight fraction)
100
/%/%
/%%%
BofwtatomicBofwtAofwtatomicAofwt
AofwtatomicAofwtAofatomicAofmole
100
%%%% BofwtatomicBofatomicAofwtatomicAofatomic
AofwtatomicAofatomicAofwt
Question1) Calculate the atomic fraction of copper in aluminum for a two-
component alloy containing 5 wt % copper. Atomic mass is 63.55 for Cu and 26.98
for Al.
8/10/2019 MSE 2001
8/40
9
Two-Component Phase Diagrams
isomorphous system
has complete solubilityof one component in another.
Non-isomorphous system?
Cu-Ni phase diagram
Both Cu and Ni have the
same crystal structure, FCC,
similar radii,electronegativity and
valence.
F = CP + 1
8/10/2019 MSE 2001
9/40
10
Two-Component Phase Diagrams
In a single-phase field, the composition
of the phase is the composition of the
alloy
In a two-phase field, the amount of
each phase and the composition of
each phase can be determined using
a tie line and the lever rule
In a single-phase field, the composition
of the phase is the composition of the
alloy
Phase boundaries
liquidus boundary and solidus boundary
Xo
8/10/2019 MSE 2001
10/40
8/10/2019 MSE 2001
11/40
12
If we know T and Co(initial composition),then we know thecomposition of each phase.
Examples:
Cu-Ni
system
C0= 35 wt% Ni
At 1300 C:
Only liquid (L)
CL= C0(= 35 wt% Ni)
At 1150 C:
Only solid (a)
Ca= C0(= 35 wt% Ni)
At TB:
Both aand L
CL= Cliquidus(= 32 wt% Ni)
Ca= Csolidus(=43 wt% Ni)
Two-Component Phase Diagrams
8/10/2019 MSE 2001
12/40
13
If we know T and Co, then we know the amount of each phase
(given in wt%).
C0= 35 wt% Ni
At 1300 C:
Only liquid (L)
WL
= 100 wt%, Wa
= 0 wt%
At 1150 C:
Only solid (a)
WL= 0 wt%, Wa= 100 wt%
At TB:
Both aand L WL= S/(R+S) =
(43-35)/(43-32) = 73 wt%
Wa= R/(R+S) =
(35-32)/(43-32) = 27 wt%
The lever rule
Two-Component Phase Diagrams
Cu-Ni
system
8/10/2019 MSE 2001
13/40
14
The Lever Rule in a Two-Component System
LLS
S
total
L fXX
XX
M
M 0
SLSL
total
S fXX
XX
M
M 0
8/10/2019 MSE 2001
14/40
15
Hamsters ?
Type A Type B
Assumption: Type B hibernates earlier than Type A.
Type A starts hibernation in colder day
than Type B does.
B is solidified at higher T than A.
Type A Type B
They hibernate.
8/10/2019 MSE 2001
15/40
16
When Types A and B got together, they had a party.
In summer, they mix and
happily play together.
Closed system Closed system
As the weather gets colder and colder.
They start hibernating
One phase Two phases
Z Z Z Z..
No More Active (solid) phase.
Composition of type B:
XB=80 %
Still Active (liquid) phase.
Composition of type B:
XB=28.6 %
Total composition is not changed (XB, total=50 %).
Total composition: XB, total=50 %
WL=58.3 % WS=41.7 %
8/10/2019 MSE 2001
16/40
18
Two-Component Phase Diagrams
1
1
1
l s
o l l s s
l s
o l s s s
o l l s s s
o l s s l
o l
s
s l
f fX X f X f
f f
X X f X f
X X X f X fX X f X X
X Xf
X X
s o
l
s l
X XfX X
Composition, XB
Temp
erature
8/10/2019 MSE 2001
17/40
19
The Lever Rule in a Two-Component System
LLS
S
total
L fXX
XX
M
M 0
SLSL
total
S fXX
XX
M
M 0
8/10/2019 MSE 2001
18/40
20
Effect of Cooling Rate
Fast cooling rate
More local heterogeneity
Slow cooling rate
More homogeneous structure
Just slow Very slow
8/10/2019 MSE 2001
19/40
21
Two-Component Phase Diagrams
Composition of the
liquid for each alloy
Composition of the
solid for each alloy
Alloy 1: 0.2
Alloy 2: 0.3Alloy 3: 0.5
Alloy 4: 0.6
Alloy 5: 0.8
Alloy 1: 0.8
Alloy 2: 0.8
Alloy 3: 0.8
Alloy 4: 0.8
Alloy 5: 0.8
Composition X0for each alloy
Alloy 1: 0.2
Alloy 2: 0.2
Alloy 3: 0.2
Alloy 4: 0.2
Alloy 5: 0.2
For Alloy 2,
83.02.08.0
3.08.0
Lf 17.02.08.0
2.03.0
Sf0.3Composition, XB
Temperature
8/10/2019 MSE 2001
20/40
22
At temperature T2
0.77 0.60
0.77 0.50
0.63
l
l
f
f
0.60 0.50
0.77 0.50
0.37
s
s
f
f
At temperature T3
0.72 0.60
0.72 0.45
0.44
l
l
f
f
0.60 0.45
0.72 0.45
0.56
s
s
f
f
Analysis of an Isomorphous Phase Diagram
X0=0.60
Temperature f l xl fs Xs
T1 1 0.6 0 0.85
T2 0.63 0.5 0.37 0.77
T3 0.44 0.45 0.56 0.72
T4 0.29 0.4 0.71 0.68
T5 0 0.3 1 0.6
8/10/2019 MSE 2001
21/40
23
Two-Component Phase Diagrams
isomorphous systemhas complete solubilityof one component in
another.
Cu-Ni phase diagramAn isomorphous system is only possible
for substitutional solid solution.
The substitution occurs randomlyon their
FCC lattice sites because the Cu and Ni
atoms are so similar.
Cu-Ni alloy
Excellent corrosion resistance
Used for water-cooled heat exchangers
The size difference between the solute and solvent must be no greater than ~15%.
The electronegativities of the two atomic species must be comparable.
The valence of the two species must be similar.
The crystal structures of the two species must be the same.
Hume-Rothery Rules
24
8/10/2019 MSE 2001
22/40
24
Two-Component Phase Diagrams
Four isomorphous systems
Formation of substitutional solid solution
All the Hume-Rothery rules are satisfied.
Cu-Ni Ge-Si
Ag-Au NiO-MgO
25
8/10/2019 MSE 2001
23/40
25
Free Energy and Phase Diagram
At High T At Low TAt intermediate T
Xo
A
B
C
A B C
26
8/10/2019 MSE 2001
24/40
26
When Types A and B got together, they had a party.
The free energy determines the composition!
The composition is adjusted to minimize the free energy.
In summer, they mix and
happily play together.
Closed system Closed system
As the weather gets colder and colder.
They start hibernating
One phase Two phases
Z Z Z Z..
No More Active (solid) phase.
Composition of type B:
XB=80 %
Still Active (liquid) phase.
Composition of type B:
XB=28.6 %
Total composition is not changed (XB, total=50 %).
Total composition: XB, total=50 %
WL=58.3 % WS=41.7 %
27
8/10/2019 MSE 2001
25/40
27
Free Energy and Phase Diagram
How to use the chemical potential?Liquidus and solidus line is determined at the points
where the chemical potentials of the phases are the
same.
How to minimize the Gibbs free energy?
Constraint 1:
Constraint 2:
Constraint 3: solidsolidliquidliquidtotal fGfGG
solidBsolidliquidBliquidB fXfXX ,,,0 1 solidliquid ff
Chemical potential, (definition):
TpiX
G
,
We use chemical potential to describe the phenomenon.
G: free energy
Xi: composition (number of molecule or atom of material i)p: pressure
T: temperature
28
8/10/2019 MSE 2001
26/40
28
1. A:B=50:50
X0,B=0.50
A A
A A
A
A
AAA A
B
B
B
BB B
BB
B B
BB
AA
A A
A A
A
A
AAA A
B
B
B
BB B
BB
B B
BB
AA
2. In the 2-phase region, the system will have
solid phase immediately in the liquid phase.
But they want to minimize the free energy by changing the composition (XB).
Liquid phase wants to increase the XB.
Solid phase wants to decrease the XB.
Solid phase
Liquid phase
Liquid phase
29
8/10/2019 MSE 2001
27/40
29
Gmin,S
Gmin,L
A A
A A
A
A
AAA A
B
BB
BB B
BB
B B
BB
AA
3. This could be done by moving B from solid
phase to liquid phase.
B
Liquid phase increases the XB.Solid phase decreases the XB.
0
,
Tpi
liquid
liquidX
G
0
,
Tpi
solidsolid
XG
As long as the free energy can be reduced, the solid phase will keep giving B to the liquid
phase.
For a while, XB,Solidand XB,Liquid will approach the minimum points, Gmin,Sand Gmin,L, respectively.
(highly) Negative slope
(slightly) Positive slope
Solid phase
Liquid phase
The liquid phase want to get more B.
The solid phase want to give out more B.
30
8/10/2019 MSE 2001
28/40
30
Gmin,S
Gmin,L
A A
A A
A
A
AAA A
B
BBB B
BB
B B
BB
AA
4. Although XB,Solid arrives at Gmin,S, XB,Liquid is still
far from Gmin,L.
B
Furthermore, the magnitude of chemical potentialliquid, is still large in comparison to solid(=0).
0
,
Tpi
liquid
liquidX
G
0
,
Tpi
solidsolid
X
G
Therefore, the liquid phase still get B from the solid phase.
Due to this, XB,Solidwill pass through Gmin,Sand go up.
For a while, XB,Liquid will keep going down towards Gmin,L.
(Still) negative slope
zero slope
B
B
0
,
Tpi
liquid
liquidX
G
0
,
Tpi
solidsolid
X
G
Please note that the chemical potentials are all negative If the solid composition passesthrough the Gmin,S.Both phases want to get more B.
Solid phase
Liquid phase
The situation is like Tug of War.
31
8/10/2019 MSE 2001
29/40
31
Gmin,S
Gmin,L
A A
A A
A
A
AAA A
B
BB
BB
B B
BB
AA
5. Since the solid phase has lost B too much,
Gsolidbecomes larger (which means that the solid
phase becomes unstabler).
B
Gsolidand Gliquidarrive at the points (4) and (5),respectively.
Now, the solid phase and the liquid phase have the same chemical potential, which means
that they need B component to the same extent.
The B component cannot move any more between those phases.
Therefore, XB,Solid and XB,Liquidare determined at the points (4) and (5).
If we do this for other X0,Bfor various temperature, we will get various pairs of (4) and (5).
a collection of (4): solidus line
a collection of (5): liquidus line
B
B
0
,
Tpi
liquid
liquidX
G
0
,
Tpi
sol idsolid
X
G
solidliquid
Solid phase
Liquid phase
8/10/2019 MSE 2001
30/40
33) h 0 ( ) h h h h h h
8/10/2019 MSE 2001
31/40
33
A
B
Q) We have a composition XB0 (at T=T0) which is in the two-phase region as shown in the
following phase diagram and the corresponding free energy plot. The solid phase and the
liquid phase have different compositions: XB
Sfor the solid phase and XB
Lfor the liquid phase.
(3) At the points A and B, the chemical potentials have (positive, negative) value.
What is the meaning of this?
Both phases want to get more B.
(4) At T=T0, the tie line meets points A and B with the solidus line and the liquidus
line, respectively, which means that the composition of the solid phase and liquid
phase are XB
S(point 4) and XB
L(point 5). Please explain how to use the chemical
potential to determine the points 4 and 5.
To obtain phase equilibrium, each phase needs to have the same chemical potential
for material. The chemical potential is the slope of the free energy plot, so that thefree energy curves for solid phase and liquid phase have the common tangent line
that has the same slope.
34
8/10/2019 MSE 2001
32/40
34
Eutect ic Phase Diagrams
Composition, XB
Temperature
TA
TB
Composition, XB
Temperature
TA
TB
Feeling?
eutektos, meaning 'easily melted.'
8/10/2019 MSE 2001
33/40
Eutectic Phase Diagrams
Composition, XB
TA
TB
X1 XE X2A B
Temperature
Xa
Xb
T
Xs
Xl
TXs
Xl
TXa
Xb
One-phase liquid
Two-phase solid
One-
phase
solid
One-phase
solid
two-
phase two-
phase
8/10/2019 MSE 2001
34/40
Eutectic Phase Diagrams
F = 2, specify
temperature
and composition
F = 1, specify
temperature or thecomposition of one
of the phases
F = 0, temperatureand compositions
of the phase are
fixed.
Composition, XB
TA TB
X1 XE X2A B
Temperature
F = 2
F = 2
F = 2
Eutectic point
F = 1
F = 1
F = 1
F = 0
38
E t ti Ph Di
8/10/2019 MSE 2001
35/40
Eutectic Phase Diagrams
Symbolic expression: Liquid Solid 1 + Solid 2
L a+ b
Eutectic Reaction
Through eutectic point in eutectic
isotherm, one-phase Liquid state
becomes two-phase Solid state.
39
8/10/2019 MSE 2001
36/40
Eutectic Phase Diagrams
Question) from the following eutectic phase diagram
(1) Your system is located in two- phase
region (L+ ) with X0=0.27. If you cool itdown slowly (equilibrium cooling), what
would be the fraction of primary solid ()
at the eutectic temperature?
588.02037
2737
27.0
0
0
a
aXX
XXf
X
L
LP
(2) From the same system, what would be the fraction of liquid at the eutectic
temperature?
412.02037
20270
a
a
XX
XXf
L
eut
L
40
8/10/2019 MSE 2001
37/40
Eutectic Phase Diagrams
Question) from the following eutectic phase diagram
(3) From the liquid (eutectic composition) in
question (2), what amount of and bwill
be formed just below the eutectic isotherm?
679.02073
3773
ab
b
aXX
XXf
eut
L
321.02073
2037
ab
ab
XX
XXf
eut
L
Therefore, the fraction of alloy composed of the eutectic andbobtained from
the eutectic liquidin question (2)is
280.0679.0412.0 aa fff eut
L
eut
132.0679.01412.0 bb
fff eut
L
eut
412.02037
20270
a
a
XX
XXf
L
eut
L
8/10/2019 MSE 2001
38/40
42
8/10/2019 MSE 2001
39/40
Peritectic Phase Diagrams
Symbolic expression: Liquid + Solid 1 Solid 2
L + a b
1. Peritectic Point and Peritectic
isotherm!
2. Through peritectic point, Liquid+solid becomes solid.
43
8/10/2019 MSE 2001
40/40
Peritectic Phase Diagram
X1=0.125 wt %
X2=0.170 wt %
X3=0.350 wt %
Question) Determine the composition and relative amounts of each phase
present just above and below the peritectic isotherm for each of the three
alloy compositions indicated.
fL
f
f
fL
f
f
fL
f
f
X1=0.125
X2=0.170
X3=0.350
just above just below
X1 X2 X3
0.09 0.17 0.53
+ L+
L+
L