The Muppet’s Guide to:
The Structure and Dynamics of Solids
Phase Diagrams
Phase Diagrams
• A phase diagram is a graphical
representation of the different phases
present in a material.
• Commonly presented as a function of
composition and temperature or pressure
and temperature
Applies to elements, molecules etc. and can also be used
to show magnetic, and ferroelectric behaviour (field vs.
temperature) as well as structural information.
Components and Phases
• Components:
The elements or compounds which are present in the mixture
(e.g., Al and Cu)
• Phases:
The physically and chemically distinct material regions
that result (e.g., a and b).
AluminumCopper
Alloy
b (lighter
phase)
a (darker
phase)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Unary Phase Diagrams
A pressure-temperature plot showing the different phases present in
H2O.
Phase
Boundaries
Crossing
any line
results in a
structural
phase
transition
Upon crossing one of these boundaries the phase abruptly
changes from one state to another. Latent heat not shown
Reading Unary Phase Diagrams
Melting Point (solid → liquid)
Triple Point
(solid + liquid + gas)
Boiling Point
(liquid→ gas)
Sublimation (solid → gas)
As the pressure falls, the boiling point reduces, but the melting/freezing
point remains reasonably constant.
Reading Unary Phase Diagrams
P=1atm
Melting Point: 0°C
Boiling Point: 100°C
P=0.1atm
Melting Point: 2°C
Boiling Point: 68°C
Water Ice
http://images.jupiterimages.com/common/detail/13/41/23044113.jpg,
http://www.homepages.ucl.ac.uk/~ucfbanf/ice_phase_diagram.jpg
Binary Phase Diagrams
Phase B
Phase A
Nickel atom
Copper atom
• 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?
How much of each phase do we get?
Phase Equilibria: Solubility Limit
– Solutions – solid solutions, single phase
– Mixtures – more than one phase
• Solubility Limit:
Answer: 65 wt% sugar.
If Co < 65 wt% sugar: syrup
If Co > 65 wt% sugar: syrup + sugar.
L
(liquid)
60
L
40
(liquid solution
i.e., syrup)
20
0
+
S
(solid
sugar)
20
40
6065 80
100
Co =Composition (wt% sugar)
Pure
Sugar
solubility limit at 20°C?
Solubility
Limit
80
Pure
Water
Question: What is the
100
Temperature (°C)
Max concentration for
which only a single phase
solution occurs.
Sucrose/Water Phase Diagram
Salt-Water(ice)
http://webserver.dmt.upm.es/~isidoro/bk3/c07sol/Solution%20properties_archivos/image001.gif
Phase Diagrams
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
T(°C)
• 2 phases:
1600
• Phase
Diagram
for Cu-Ni
at P=1 atm.
L (liquid)
a (FCC solid solution)
L (liquid)
1500
1400
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
Figure adapted from Callister, Materials science and engineering, 7th Ed.
80
• 3 phase fields:
L
L+a
a
100
wt% Ni
Phase Diagrams
• Indicate phases as function of T, Co, and P.
• For this course:
-binary systems: just 2 components.
-independent variables: T and Co (P = 1 atm is almost always used).
T(°C)
1600
• Phase
Diagram
for Cu-Ni
at P=1 atm.
Liquidus:
Separates the liquid
from the mixed L+a
phase
L (liquid)
1500
1400
Solidus:
Separates the
mixed L+a phase
from the solid
solution
1300
a
(FCC solid
solution)
1200
1100
1000
0
20
40
60
Figure adapted from Callister, Materials science and engineering, 7th Ed.
80
100
wt% Ni
Number and types of phases
• Rule 1: If we know T and Co, then we know:
- the number and types of phases present.
• Examples:
T(°C)
1600
A(1100°C, 60):
1 phase: a
B (1250°C,35)
L (liquid)
1500
1400
B(1250°C, 35):
2 phases: L + a
1300
a
(FCC solid
solution)
1200
A(1100°C,60)
1100
1000
Cu-Ni
phase
diagram
0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
40
60
80
100
wt% Ni
Composition of phases
• Rule 2: If we know T and Co, then we know:
--the composition of each phase.
• Examples:
T(°C)
Cu-Ni
system
A
TA
Co = 35 wt% Ni
1300 L (liquid)
At T A = 1320°C:
Only Liquid (L)
B
TB
CL = Co ( = 35 wt% Ni)
At T D = 1190°C:
1200
D
Only Solid ( a)
TD
Ca = Co ( = 35 wt% Ni)
20
3032 35
At T B = 1250°C:
CLCo
Both a and L
CL = C liquidus ( = 32 wt% Ni here)
Ca = C solidus ( = 43 wt% Ni here)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
tie line
a
(solid)
4043
50
Ca wt% Ni
Cooling a Cu-Ni Binary - Composition
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(°C) L (liquid)
130 0
L: 35 wt% Ni
a: 46 wt% Ni
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
35
32
--isomorphous
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
L: 35wt%Ni
B
C
46
43
D
24
L: 32 wt% Ni
36
120 0
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
USE LEVER RULE
The Lever Rule – Weight %
• Tie line – connects the phases in equilibrium
with each other - essentially an isotherm
T(°C)
How much of each phase?
Think of it as a lever
tie line
1300
L (liquid)
B
TB
a
(solid)
1200
R
20
30C
S
40 C
a
L Co
R
50
wt% Ni
WL 
Ma
ML
C  C0
ML
S

 a
ML  M a R  S Ca  CL
Figure adapted from Callister, Materials science and engineering, 7th Ed.
S
M a S  M L R
Wa 
C  CL
R
 0
R  S Ca  CL
Weight fractions of phases – ‘lever rule’
• Rule 3: If we know T and Co, then we know:
--the amount of each phase (given in wt%).
• Examples:
Cu-Ni
system
T(°C)
C o = 35 wt% Ni
At T A : Only Liquid (L)
W L = 100 wt%, W a = 0
At T D : Only Solid ( a)
W L = 0, W a = 100 wt%
At T B : Both a and L
43  35
 73 wt %
43  32
WL 
S
R +S

Wa 
R
R +S
= 27 wt%
Figure adapted from Callister, Materials science and engineering, 7th Ed.
A
TA
1300
TB
1200
TD
20
tie line
L (liquid)
B
S
R
D
3032 35
C LC o
a
(solid)
40 4 3
50
C a wt% Ni
Cooling a Cu-Ni Binary – wt. %
• Phase diagram:
Cu-Ni system.
• System is:
--binary
i.e., 2 components:
Cu and Ni.
T(°C) L (liquid)
130 0
L: 92 wt%
a: 8 wt%
i.e., complete
solubility of one
component in
another; a phase
field extends from
0 to 100 wt% Ni.
• Consider
Co = 35 wt%Ni.
Cu-Ni
system
A
32
--isomorphous
L: 35wt%Ni
34
B
C
46
43
D
24
L: 73 wt%
36
120 0
a: 27 wt%
E
L: 8 wt%
a: 92 wt%
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
Equilibrium cooling
• Multiple freezing sites
– Polycrystalline materials
– Not the same as a single crystal
• The compositions that freeze are a
function of the temperature
• At equilibrium, the ‘first to freeze’
composition must adjust on further cooling
by solid state diffusion
Diffusion is not a flow
Concept behind mean free path in
scattering phenomena conductivity
Our models of diffusion are based on a random walk approach and
not a net flow
http://mathworld.wolfram.com/images/eps-gif/RandomWalk2D_1200.gif
Diffusion in 1 Dimension
• Fick’s First Law
J = flux – amount of material per unit area per unit time
C = concentration
dC
J   D T 
dx
Diffusion coefficient which we expect
is a function of the temperature, T
Diffusion cont….
• Requires the solution of the continuity equation:
The change in concentration as a function of time in a
volume is balanced by how much material flows in per time
unit minus how much flows out – the change in flux, J:
C
J

0
t
x
•
BUT
dC
J   D T 
dx
giving Fick’s second law (with D being constant):
C
  C 
2C

D
  D T  2
t
x  x 
x
Solution of Ficks’ Laws
t=0
Co
For a semi-infinite sample
the solution to Ficks’ Law
gives an error function
distribution whose width
increases with time
C
C
t=t
x
Consider slabs of Cu and Ni.
Interface region will be a
mixed alloy (solid solution)
Interface region will grow as a
function of time
Slow Cooling in a Cu-Ni Binary
T(°C) L (liquid)
L: 35wt%Ni
Co = 35 wt%Ni.
Enough time is
allowed at each
temperature
change for atomic
diffusion to occur.
–
Thermodynamic
ground state
Each phase is
homogeneous
130 0
L: 35 wt% Ni
a: 46 wt% Ni
Cu-Ni
system
A
B
C
D
120 0
L: 32 wt% Ni
a: 43 wt% Ni
E
L: 24 wt% Ni
a: 36 wt% Ni
a
(solid)
110 0
20
Figure adapted from Callister, Materials science and engineering, 7th Ed.
30
35
Co
40
50
wt% Ni
L
Non –
equilibrium
cooling
No-longer in the
thermodynamic
ground state
Reduces the
melting
temperature
Figure adapted from Callister, Materials science and engineering, 7th Ed.
α+L
α
Cored vs Equilibrium Phases
• Ca changes as we solidify.
• Cu-Ni case:
First a to solidify has Ca = 46 wt% Ni.
Last a to solidify has Ca = 35 wt% Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure
First a to solidify:
46 wt% Ni
Last a to solidify:
< 35 wt% Ni
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Uniform C a:
35 wt% Ni
Binary-Eutectic Systems – Cu/Ag
2 components
has a special composition
with a min. melting temperature
a
phase:
b
phase:
Mostly
copper
Solvus line
– the
solubility
limit
Mostly
Silver
• Limited solubility – mixed phases
• 3 phases regions, L, a and b and 6 phase fields - L, a and b, L+a, L+b, ab
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Binary-Eutectic Systems
Cu-Ag system
T(°C)
The Eutectic point
1200
L (liquid)
TE, Eutectic temperature, 779°C
1000
CE, eutectic composition, 71.9wt.%
a
TE
• TE
: No liquid below TE
Min. melting TE
800
E
779°C
CaE=8.0
600
CE=71.9
L +b b
CbE=91.2
ab
400
200
• Eutectic transition
L(CE)
L+a
0
a(CaE) + b(CbE)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Co
20
40
60 CE 80
wt% Ag in Cu/Ag alloy
100
Any other composition, Liquid
transforms to a mixed L+solid phase
Pb-Sn (Solder) Eutectic System (1)
• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...
--the phases present: a + b
--compositions of phases:
CO = 40 wt% Sn
Ca = 11 wt% Sn
Cb = 99 wt% Sn
--the relative amount
of each phase:
Wa =
S
R+S
C -C
= b O
Cb - Ca
99 - 40
59
=
= 67 wt%
99 - 11
88
CO - Ca
Wb = R
=
Cb - Ca
R+S
40 - 11
29
=
= 33 wt%
=
99 - 11
88
Pb-Sn
system
T(°C)
300
200
L (liquid)
a
L+a
18.3
150
L +b b
183°C
61.9
R
97.8
S
100
a+ b
=
0 11 20
Ca
40
Co
60
80
C, wt% Sn
99100
Cb
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Microstructures in Eutectic
Systems: II
L: Co wt% Sn
T(°C)
• 2 wt% Sn < Co < 18.3 wt% Sn
• Result:
400
 Initially liquid → liquid + a
 then a alone
 finally two phases
 a poly-crystal
 fine b-phase inclusions
L
300
L +a
a
200
TE
a: Co wt% Sn
a
b
100
a+ b
0
10
20
Co
Co ,
2
(sol. limit at T room )
18.3
(sol. limit at TE)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
L
a
Pb-Sn
system
30
wt% Sn
Microstructures
in Eutectic Systems: Co=CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
T(°C)
Micrograph of Pb-Sn
eutectic
microstructure
L: Co wt% Sn
300
Pb-Sn
system
a
200
L+a
L
TE
100
ab
0
Lb b
183°C
20
18.3
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
100
97.8
C, wt% Sn
Figures adapted from Callister, Materials science and engineering, 7th Ed.
160m
Microstructures
in Eutectic Systems: Co=CE
• Result: Eutectic microstructure (lamellar structure)
--alternating layers (lamellae) of a and b crystals.
T(°C)
L: Co wt% Sn
300
Pb-Sn
system
a
200
L+a
Wa 
Lb b
183°C
TE
Pb
rich
L
Wb 
100
ab
0
20
18.3
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
100
97.8
C, wt% Sn
Figures adapted from Callister, Materials science and engineering, 7th Ed.
97.8  61.9
 45.2%
97.8  18.3
61.9  18.3
 54.8%
97.8  18.3
Sn Rich
Lamellar Eutectic Structure
At interface, Pb moves to a-phase
and Sn migrates to b- phase
Lamellar form to minimise diffusion
distance – expect spatial extent to
depend on D and cooling rates.
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Sn
Pb
Microstructures IV
• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
L: Co wt% Sn
T(°C)
300
Pb-Sn
system
a
200
L+a
L
a
L
R
a L
L+b b
S
TE
a+ b
100
0
20
18.3
40
60
61.9
80
Co, wt% Sn
Figure adapted from Callister, Materials science and engineering, 7th Ed.
100
• Just above TE :
C a = 18.3 wt% Sn
CL = 61.9 wt% Sn
Wa= S
= 50 wt%
R +S
WL = (1- W a) = 50 wt%
Microstructures IV
• 18.3 wt% Sn < Co < 61.9 wt% Sn
• Result: a crystals and a eutectic microstructure
L: Co wt% Sn
T(°C)
300
Pb-Sn
system
a
200
L+a
TE
R
100
a+ b
L
a
L
R
a L
L+b b
S
• Just below TE :
C a = 18.3 wt% Sn
C b = 97.8 wt% Sn
Wa = S
= 73 wt%
R +S
W b = 27 wt%
S
Primary, a
0
20
18.3
40
60
61.9
80
Co, wt% Sn
Figure adapted from Callister, Materials science and engineering, 7th Ed.
100
97.8
Eutectic, a
Eutectic, b
Intermetallic Compounds
a
phase:
Mostly
Mg
Mg2Pb
b
phase:
Mostly
Lead
Note: intermetallic compound forms a line - not an area because stoichiometry (i.e. composition) is exact.
Figure adapted from Callister, Materials science and engineering, 7th Ed.
Eutectoid & Peritectic
Peritectic transition  + L
Cu-Zn Phase diagram
Eutectoid transition 
+
Figure adapted from Callister, Materials science and engineering, 7th Ed.

mixed liquid and solid to
single solid transition
Solid to solid ‘eutectic’ type
transition
Iron-Carbon (Fe-C) Phase Diagram
L
 + Fe3C
-Eutectoid (B):

a + Fe3C
T(°C)
1600

L
1400
1200
 +L

(austenite)
 
 
1000
800
a
600
Result: Pearlite =
alternating layers of
a and Fe3C phases
S
 +Fe3C
727°C = T eutectoid
R
S
1
0.76
C eutectoid
120 m
0
(Fe)
Figure adapted from Callister, Materials science and engineering, 7th Ed.
L+Fe3C
R
B
400
A
1148°C
Fe3C (cementite)
• 2 important
points
-Eutectic (A):
2
3
a+Fe3C
4
5
6
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
a (ferrite-soft)
Iron-Carbon
http://www.azom.com/work/pAkmxBcSVBfns037Q0LN_files/image003.gif