CENG151 Introduction to Materials
Science and Selection
Tutorial 7
16th November, 2007
Reviewing Phase Diagram Terms

Phase - Any portion including the whole of a system, which is
physically homogeneous within it and bounded by a surface so
that it is mechanically separable from any other portions.

Component – The distinct chemical substance from which the
phase is formed.

Phase Diagram - A diagram (map) showing phases present
under equilibrium conditions and the phase compositions at
each combination of temperature and overall composition.

Gibbs Phase Rule - Used to determine the degrees of freedom
(F), the number of independent variables available to the
system:

F = C – P + 1 (for pressure at constant 1 atm.)
Solubility terms

Solubility - The amount of one material that will
completely dissolve in a second material without
creating a second phase.

Unlimited solubility - When the amount of one
material that will dissolve in a second material
without creating a second phase is unlimited.

Limited solubility - When only a maximum amount
of a solute material can be dissolved in a solvent
material.
In-Class Exercise: Revision
Apply the Gibbs
phase rule to a
sketch of the
MgO-Al2O3
phase diagram.
(Figure 9-26)
Find out for
each phase
compartment
the degrees of
freedom.
Solution 9.5
F=2-1+1=2
F=1-2+1=0
(F=2-2+1=1)
F=2-1+1=2
F=2-1+1=2
F=2-2+1=1
F=2-2+1=1
F=1-2+1=0
Phase diagram for unlimited solubility
The composition of the phases in a two-phase region
of the phase diagram are determined by a tie line
(horizontal line connecting the phase compositions)
at the system temperature.
Eutectic Diagram with No Solid Solution



Another binary system with
components that are so
dissimilar that their solubility
in each other in nearly
negligible.
There exists a 2-phase
region for pure solids (A+B).
Because A and B cannot
dissolve in each other!
The eutectic temperature
(eutektos greek for “easily
melted”) is the temperature
that eutectic composition is
fully melted.
Eutectic diagram with limited solid
solution



Many binary systems
are partially soluble 
intermediate phase
diagram of the
previous two!
α and  individual phases
are single solid solution
phases!
While α+ is a 2-phase
region for pure solids α
and .
Eutectic Diagram with Limited Solid Solution
Lead (Pb) – Tin (Sn) equilibrium phase diagram.
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Solidification and
microstructure
of a Pb-2% Sn
alloy. The alloy
is a single-phase
solid solution.
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Solidification, precipitation, and microstructure of a Pb-10%
Sn alloy. Some dispersion strengthening occurs as the β solid
precipitates.
The Lever Rule


Tells us the amount
of each phase there
are in the alloy 
wt%
C0 must be made up
of appropriate
amounts of α at
composition Cα and
of liquid at
composition CLiq.
The Lever Rule

Basically, the
proportion of the
phases present are
given by the relative
lengths of the tie line.
So, the proportions of
α and L present on
the diagram are:
x  x
m

m  m x  x
m
m  m

x  x
x  x
9.9
9.10
So, the left side of the tie line
gives the proportion of the
liquid phase, and the right side
of the tie line gives the
proportion of the alpha phase!
Example:
Application of Lever Rule

Calculate the amounts of
and L at 1250oC in the Cu-40% Ni
alloy shown in the figure below.
A tie line 1250°C
in the coppernickel system
that is used to
find the amount
of each phase.
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark
used herein under license.
Example: Solution
x = mass fraction of the alloy that is in
α phase.
Since we have only two phases, α and L. Thus, the
mass fraction of nickel in liquid will be 1 - x.
Total mass of nickel in 100 grams of the alloy = mass
of nickel in liquid + mass of nickel in α
So, 100  (% Ni in alloy = [(100)(1 – x)](% Ni in L) +
(100)[x](% Ni in α )
x = (40-32)/(45-32) = 8/13 = 0.62
If we convert from mass fraction to mass percent, the
alloy at 1250oC contains 62% α and 38% L.
Note that the concentration of Ni in α phase (at
1250oC) is 45% and concentration of nickel in liquid
phase (at 1250oC) is 32% (read from the diagram).
Example: Limited Solubility
Consider a 50:50 Pb –
Sn solder.
(a)For a temperature
of 200˚C,
determine (i) the
phases present, (ii)
their compositions,
and (iii) their
relative amounts
(expressed in
weight percent).
(b)Repeat part (a) for
100˚C.
Example: Solution
(a)Reading off the phase diagram at 50wt% and
200˚C,
(i) Phases present are  and liquid.
(ii)The composition of  is ~18wt% Sn and of L is
~ 54wt% Sn.
(iii)Using the Eqn 9.9 and 9.10, we have:
Lever Rule
x  x
m

m  m x  x
9.9
x  x

m  m x  x
9.10
m
xL  x
54  50
wt % 
 100% 
 100%
xL  x
54  18
 11.1%
wt % L 
x  x
50  18
 100% 
 100%
xL  x
54  18
 88.9%
Example: Solution (cont.)
(b) Similarly, at 100C, we obtain:
(i) Phases are  and .
(ii)  is ~5wt% Sn and  is ~ 99wt% Sn.
(iii)
x  x
99  50
wt % 
100% 
100%  52.1%
x  x
99  5
x  x
50  5
wt %  
100% 
100%  47.9%
x  x
99  5
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
The change in
structure of a
Cu-40% Ni
alloy during
nonequilibrium
solidification.
Insufficient
time for
diffusion in the
solid produces
a segregated
structure.
In-Class Exercise: Limited Solubility
Consider 1kg of an aluminum
casting alloy with 10wt%
silicon.
(a)Upon cooling, at what
temperature would the first
solid appear?
(b)What is the first solid phase
and what is its composition?
(c) At what temperature will the
alloy completely solidify?
(d)How much proeutectic phase
will be found in the
microstructure?
(e)How is the silicon distributed
in the microstructure at
576C?
In-Class Exercise: Solution
Follow microstructural development with the aid of
the phase diagram:
(a)For this composition, the liquidus is at ~595C.
(b) It is solid solution  with the composition of
~1wt% Si.
(c) At the eutectic temperature, 577C.
(d) Practically all of the proeutectic  will have
developed by 578C. Using the equation: (eqn9.9),
we obtain:
xL  x
12.6  10
1kg  
1kg 
m 
xL  x
12.6  1.6
 0.236kg  236 g
In-Class Exercise: Solution (cont.)
(e) At 576C, the overall microstructure is  + . The
amounts of each are:
x  x
1kg   100  10 1kg 
m 
x  x
100  1.6
 0.915kg  915 g
m 
x  x
1kg   10  1.6 1kg 
x  x
100  1.6
 0.085kg  85 g
But we found in (d) that 236g of the  is in the form of
relatively large grains of proeutectic phase, giving:
 eutectic   total   proeutectic
 915 g  236 g  679 g
In-Class Exercise: Solution (cont.)
The silicon distribution is then given by multiplying its
weight fraction in each microstructural region by the
amount of that region:
Si in proeutectic  = (0.016)(236g) = 3.8g
Si in eutectic  = (0.016)(679g) = 10.9g
Si in eutectic  = (1.000)(85g) = 85.0g
Finally, note that the total mass of silicon in the three
regions sums to 99.7g rather than 100.0g (=10wt% of
the total alloy) due to round-off errors.
Question 9.10
(c) (b)
(a)
Describe qualitatively the
microstructural
development during the
slow cooling of a melt
composed of the
following compositions.
(See Figure 9-16)
(a)10 wt% Pb – 90 wt% Sn,
(b)40 wt% Pb – 60 wt% Sn,
(c)50 wt% Pb – 50 wt% Sn.
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a
trademark used herein under license.
Solution 9.10
(a) The first solid to precipitate is solid solution, -Sn near
210˚C. At the eutectic temperature (183˚C), the
remaining liquid solidifies leaving a two phase
microstructure microstructure of solid solutions α-Pb and
-Sn.
(b) The first solid to precipitate is solid solution α-Pb near
188˚C. At the eutectic temperature (183˚C), the
remaining liquid solidifies leaving a two phase
microstructure of solid solution α-Pb and -Sn.
(c) The first solid to precipitate is solid solution α-Pb near
210˚C. At the eutectic temperature (183˚C), the
remaining liquid solidifies leaving a two-phase
microstructure of solid solution α-Pb and -Sn.
Question 9.13
Describe qualitatively
the microstructural
development that will
occur upon the slow
cooling of a melt
composed of the
following
compositions.
(See Figure 9-28)
(a) 20 wt% Mg – 80 wt% Al,
(b) 80 wt% Mg – 20 wt% Al.
(a)
(b)
Solution 9.13


(a) The first solid to precipitate is α. At
eutectic temperature (450˚C), the remaining
liquid solidifies leaving a two-phase
microstructure of solid solutions α and β.
(b) The first solid to precipitate is . At the
eutectic temperature (437˚C), the remaining
liquid solidifies leaving a two-phase
microstructure of solid solution  and .
Question 9.17
Calculate the amount
of each phase
present in 1kg of a 50
wt% Ni – 50 wt% Cu
alloy at the following
temperatures.
(See Figure 9-9)
(a)1400˚C
(b)1300˚C
(c)1200˚C
Solution 9.17
(a) In the single (L) region:
(c) In the single (α)
mL  1kg, m  0kg
(b) Two phases exist:
x  x
58  50
1kg  
1kg 
mL 
x  xL
58  46
 0.667 kg  667 g
x  xL
50  46
1kg  
1kg 
m 
x  xL
58  46
 0.333kg  333 g
phase region:
mL  0kg, m  1kg
Question 9.18
Calculate the amount of
each phase present in 1
kg of a 50 wt% Pb – 50
wt% Sn solder alloy at
the following
temperatures. (See Figure
9-16)
(a)300˚C
(b)200˚C
(c)100˚C
(d)0˚C
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a
trademark used herein under license.
Solution 9.18
(a) In the single (L) region:
mL  1kg, m  0kg
(b) Two phases exist (α-Pb + L):
x  x  Pb
50  18
1kg  
1kg 
mL 
xL  x  Pb
54  18
 0.889kg  889 g
m  Pb
xL  x
54  50
1kg  
1kg 

xL  x  Pb
54  18
 0.111kg  111g
Solution 9.18 (cont.)
(c) Two phases exist (α-Pb + -Sn):
m  Pb
x  Sn  x
99  50
1kg  
1kg 

x  Sn  x  Pb
99  5
 0.521kg  521g
x  x  Pb
50  5
1kg  
1kg 
m  Sn 
x  Sn  x  Pb
99  5
 0.479kg  479 g
m  Pb
(d) Two phases exist
(α-Pb + α-Sn):
x  Sn  x
100  50
1kg  
1kg 

x  Sn  x  Pb
100  1
 0.505kg  505 g
m  Sn
x  x  Pb
50  1
1kg  
1kg 

x  Sn  x  Pb
100  1
 0.495kg  495 g
End of Tutorial 7
Thank You!