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Eutectic Diagram with No Solid Solution ... The eutectic temperature (eutektos greek for 'easily melted') is the temperature ... Eutectic diagram with limited ... – PowerPoint PPT presentation

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Title: CENG151 Introduction to Materials Science and Selection


1
CENG151 Introduction to Materials
Science and Selection
  • Tutorial 7
  • 16th November, 2007

2
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.)

3
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.

4
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.
5
Solution 9.5
F1-210
6
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.
7
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
    (AB). 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.

8
Eutectic diagram with limited solid solution
  • Many binary systems are partially soluble ?
    intermediate phase diagram of the previous two!
  • a and ? individual phases are single solid
    solution phases!
  • While a? is a 2-phase region for pure solids a
    and ?.

9
Eutectic Diagram with Limited Solid SolutionLead
(Pb) Tin (Sn) equilibrium phase diagram.
10
Solidification and microstructure of a Pb-2 Sn
alloy. The alloy is a single-phase solid
solution.
11
Solidification, precipitation, and microstructure
of a Pb-10 Sn alloy. Some dispersion
strengthening occurs as the ß solid precipitates.
12
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 a at
    composition Ca and of liquid at composition CLiq.

13
The Lever Rule
  • Basically, the proportion of the phases present
    are given by the relative lengths of the tie
    line. So, the proportions of a and L present on
    the diagram are

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!
14
Example
Application of Lever Rule
A tie line 1250C in the copper-nickel system
that is used to find the amount of each phase.
15
Example Solution
x mass fraction of the alloy that is in a
phase. Since we have only two phases, a 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 a So, 100 ? ( Ni in alloy (100)(1
x)( Ni in L)
(100)x( Ni in a ) x (40-32)/(45-32) 8/13
0.62 If we convert from mass fraction to mass
percent, the alloy at 1250oC contains 62 a and
38 L. Note that the concentration of Ni in a
phase (at 1250oC) is 45 and concentration of
nickel in liquid phase (at 1250oC) is 32 (read
from the diagram).
16
Example Limited Solubility
  • Consider a 5050 Pb Sn solder.
  • For a temperature of 200C, determine (i) the
    phases present, (ii) their compositions, and
    (iii) their relative amounts (expressed in weight
    percent).
  • Repeat part (a) for 100C.

17
Example Solution
  • Reading off the phase diagram at 50wt and 200C,
  • Phases present are ? and liquid.
  • The composition of ? is 18wt Sn and of L is
    54wt Sn.
  • Using the Eqn 9.9 and 9.10, we have

Lever Rule
18
Example Solution (cont.)
  • (b) Similarly, at 100?C, we obtain
  • (i) Phases are ? and ?.
  • (ii) ? is 5wt Sn and ? is 99wt Sn.
  • (iii)

19
  • The change in structure of a Cu-40 Ni alloy
    during nonequilibrium solidification.
    Insufficient time for diffusion in the solid
    produces a segregated structure.

20
In-Class Exercise Limited Solubility
  • Consider 1kg of an aluminum casting alloy with
    10wt silicon.
  • Upon cooling, at what temperature would the first
    solid appear?
  • What is the first solid phase and what is its
    composition?
  • At what temperature will the alloy completely
    solidify?
  • How much proeutectic phase will be found in the
    microstructure?
  • How is the silicon distributed in the
    microstructure at 576?C?

21
In-Class Exercise Solution
  • Follow microstructural development with the
    aid of the phase diagram
  • For this composition, the liquidus is at 595?C.
  • It is solid solution ? with the composition of
    1wt Si.
  • At the eutectic temperature, 577?C.
  • Practically all of the proeutectic ? will have
    developed by 578?C. Using the equation (eqn9.9),
    we obtain

22
In-Class Exercise Solution (cont.)
(e) At 576?C, the overall microstructure is ?
?. The amounts of each are But we found in
(d) that 236g of the ? is in the form of
relatively large grains of proeutectic phase,
giving
23
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.
24
Question 9.10
(a)
(b)
(c)
  • Describe qualitatively the microstructural
    development during the slow cooling of a melt
    composed of the following compositions. (See
    Figure 9-16)
  • 10 wt Pb 90 wt Sn,
  • 40 wt Pb 60 wt Sn,
  • 50 wt Pb 50 wt Sn.

25
Solution 9.10
  • (a) The first solid to precipitate is solid
    solution, ?-Sn near 210C. At the eutectic
    temperature (183C), the remaining liquid
    solidifies leaving a two phase microstructure
    microstructure of solid solutions a-Pb and ?-Sn.
  • (b) The first solid to precipitate is solid
    solution a-Pb near 188C. At the eutectic
    temperature (183C), the remaining liquid
    solidifies leaving a two phase microstructure of
    solid solution a-Pb and ?-Sn.
  • (c) The first solid to precipitate is solid
    solution a-Pb near 210C. At the eutectic
    temperature (183C), the remaining liquid
    solidifies leaving a two-phase microstructure of
    solid solution a-Pb and ?-Sn.

26
Question 9.13
(a) 20 wt Mg 80 wt Al, (b) 80 wt Mg 20
wt Al.
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)
(b)
27
Solution 9.13
  • (a) The first solid to precipitate is a. At
    eutectic temperature (450C), the remaining
    liquid solidifies leaving a two-phase
    microstructure of solid solutions a and ß.
  • (b) The first solid to precipitate is ?. At the
    eutectic temperature (437C), the remaining
    liquid solidifies leaving a two-phase
    microstructure of solid solution ? and ?.

28
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)
  • 1400C
  • 1300C
  • 1200C

29
Solution 9.17
(c) In the single (a) phase region
  • (a) In the single (L) region
  • (b) Two phases exist

30
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)
  • 300C
  • 200C
  • 100C
  • 0C

31
Solution 9.18
  • (a) In the single (L) region
  • (b) Two phases exist (a-Pb L)

32
Solution 9.18 (cont.)
(c) Two phases exist (a-Pb ?-Sn)
(d) Two phases exist (a-Pb a-Sn)
33
End of Tutorial 7
  • Thank You!
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