CENG151 Introduction to Materials Science and Selection

Phase DiagramsHomework Solutions

2007

Homework Problem: 9.3

Calculate the degrees of freedom for a 50:50 copper-nickel alloy at:

(a) 1400˚C where it exists as a single, liquid phase.

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

(c) 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

= 2

(b) F = 2 – 2 + 1 = 2

(c) F = 2 – 1 + 1 = 2

a

b

c

Homework Problem: 9.5

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

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 (Al2Cu) (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:

(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

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.

=760g

kgkgm 76.0102.077.0

20.077.0

End of Homework Solutions

Thank You!