CENG151 Introduction to Materials Science and Selection

Tutorial 716th 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=1-2+1=0

(F=2-2+1=1)

F=2-1+1=2

F=2-2+1=1

F=2-2+1=1

F=2-1+1=2

F=2-1+1=2

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 .

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Eutectic Diagram with Limited Solid SolutionLead (Pb) – Tin (Sn) equilibrium phase diagram.

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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: 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!

10.9

9.9

xx

xx

mm

m

xx

xx

mm

m

Example: Application of Lever Rule

Calculate the amounts of and L at 1250oC in the Cu-40% Ni alloy shown in the figure below.

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

A tie line 1250°C in the copper-nickel system that is used to find the amount of each phase.

Example: Solutionx = 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

%9.88

%1001854

1850%100%

%1.11

%1001854

5054%100%

xx

xxLwt

xx

xxwt

L

L

L

(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:

10.9

9.9

xx

xx

mm

m

xx

xx

mm

m

Lever Rule

Example: Solution (cont.)

(b) Similarly, at 100C, we obtain:

(i) Phases are and .

(ii) is ~5wt% Sn and is ~ 99wt% Sn.

(iii)

%9.47%100599

550%100%

%1.52%100599

5099%100%

xx

xxwt

xx

xxwt

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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 576C?

In-Class Exercise: Solution

Follow microstructural development with the aid of the phase diagram:

(a)For this composition, the liquidus is at ~595C.

(b) It is solid solution with the composition of ~1wt% Si.

(c) At the eutectic temperature, 577C.

(d) Practically all of the proeutectic will have developed by 578C. Using the equation: (eqn9.9), we obtain:

gkg

kgkgxx

xxm

L

L

236236.0

16.16.12

106.121

In-Class Exercise: Solution (cont.)

(e) At 576C, 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:

ggg

cproeutectitotaleutectic

679236915

gkg

kgkgxx

xxm

gkg

kgkgxx

xxm

85085.0

16.1100

6.1101

915915.0

16.1100

101001

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.

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Question 9.10

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.

(a)(b)(c)

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:

(b) Two phases exist:

gkg

kgkgxx

xxm

gkg

kgkgxx

xxm

L

L

LL

333333.0

14658

46501

667667.0

14658

50581

(c) In the single (α)

phase region:

kgmkgmL 1,0

kgmkgmL 0,1

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

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

Solution 9.18

(a) In the single (L) region:

(b) Two phases exist (α-Pb + L):

kgmkgmL 0,1

gkg

kgkgxx

xxm

gkg

kgkgxx

xxm

PbL

LPb

PbL

PbL

111111.0

11854

50541

889889.0

11854

18501

Solution 9.18 (cont.)

gkg

kgkgxx

xxm

gkg

kgkgxx

xxm

PbSn

PbSn

PbSn

SnPb

479479.0

1599

5501

521521.0

1599

50991

(c) Two phases exist (α-Pb + -Sn):

(d) Two phases exist (α-Pb + α-Sn):

gkg

kgkgxx

xxm

gkg

kgkgxx

xxm

PbSn

PbSn

PbSn

SnPb

495495.0

11100

1501

505505.0

11100

501001

End of Tutorial 7

Thank You!