CENG151 Introduction to Materials Science and Selection

Phase Diagrams Homework Solutions

2007

Homework Problem: 9.3

(a) (b) (c) Calculate the degrees of freedom for a 50:50 copper nickel alloy at: 1400 ˚C where it exists as a single, liquid phase.

1300 ˚C where it exists as a two-phase mixture of liquid and solid solution.

1200 ˚C where it exists as a single, solid-solution phase.

Assume a constant pressure of 1 atm above the alloy in each case.

Solution

9.3

Using Gibbs Phase Rule:  F = C – P + 1 = 2 – 1 + 1 (b) (c) = 2 F = 2 – 2 + 1 = 2 F = 2 – 1 + 1 = 2 b c a

Homework Problem:

9.5

Apply the Gibbs phase rule to a sketch of the MgO-Al 2 O 3 phase diagram.

(Figure 9-26) Find out for each phase compartment the degrees of freedom.

Solution 9.5

F=1-2+1=0 F=2-1+1=2 (F=2-2+1=1) F=2-1+1=2 F=2-2+1=1 F=2-1+1=2 F=2-2+1=1 F=1-2+1=0

Homework Problem 9.8

Describe qualitatively the microstructural development that will occur upon slow cooling of a melt composed of 50wt% Al and 50wt% Si (see Figure 9-13)

Solution 9.8

 The first solid to precipitate is a solid solution,  , near 1045 ˚C. At the eutectic temperature (577˚C), the remaining liquid solidifies leaving a two-phase microstructure of solid solutions α and  .

Phase change Phase change

Homework Problem 9.12

Describe qualitatively the microstructural development that will occur upon slow cooling of an alloy with equal parts (by weight) of aluminum and  phase (Al 2 Cu) (see Figure 9-27)

Solution 9.12

  Equal parts of Al and  is about 26.25wt% Cu.

The first solid to precipitate is  near 570 ˚C. At the eutectic temperature (548.2˚C), the remaining liquid solidifies leaving a two-phase microstructure of solid solutions  and  .

  Phase change Phase change 

Homework Problem 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:

m L

 1

kg

,

m

  0

kg

(b) Two phases exist ( α-Pb + L) :

m L

  0

x x L

 

x



x

  

Pb Pb

.

889

kg

 889

g



m

 

Pb

 

x L x L

 0 .

111

kg



x x

 

Pb

 111

g

50  18 54  18  54  50 54  18

Solution 9.18 (cont.)

(c) Two phases exist ( α-Pb +  -Sn) :

m

 

Pb x



x

 

Sn Sn

 

x



x Pb

   99 99   50 5    0 .

521

kg

 521

g m

 

Sn

 

x



x

 

Sn x

  

Pb x

 

Pb

0 .

479

kg

 479

g

 50  5 (d) Two phases exist ( α-Pb + α-Sn) : 99  5

m

 

Pb

 0 .



x



x

 

Sn



Sn

 505

kg

 

x x

 

Pb

505

g

 100 100  50  1

m

 

Sn

 

x



x

 

Sn

0 .

495

kg x

  

x



Pb



Pb

 495

g

 50  1 100  1

Homework Problem 9.23

Calculate the amount of proeutectoid α present at the grain boundaries in 1 kg of a common 1020 structural steel (0.20wt% C). (See Figure 9-19) 728 ˚C

Solution 9.23

In effect, we need to calculate the equilibrium amount of ferrite at 728˚C.

m

  0 .

77  0 .

20 0 .

77  0 .

02  0 .

76

kg

=760g

End of Homework Solutions

Thank You!