7/24/2019 230 F14 HW9 SOLS.pdf

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MSE 230 Solutions for Assignment 9 Fall 2014

1. 185C 181Ci) 25 wt.% Sn

Phases: !+L !+

Phase Composition C!= 18.3 wt.% Sn C!= 18.3 wt.% SnCL= 61.9 wt.% Sn C= 97.8 wt.% Sn

Phase Amounts: W! =

61.9 " 25

61.9 " 18.3= 0.85 W

! =

97.8 " 25

97.8 "18.3= 0.917

WL =

25!18.3

61.9!18.3= 0.15 W" =

25#18.3

97.8#18.3= 0.084

ii) 60 wt.% Sn

Phases: 185C 181C

!+L !+

Phase Composition: C!= 18.3 wt.% Sn C!= 18.3 wt.% Sn

CL= 61.9 wt.% Sn C= 97.8 wt.% Sn

Phase Amounts: W! =

61.9 " 60

61.9 " 18.3= 0.04 W" =

97.8# 60

97.8#18.3= 0.475

WL =

60 !18.3

61.9!18.3= 0.96 W =

60"18.3

97.8"18.3= 0.525

iii) 90 wt.% Sn

185C 181C

Phases: +L !+

Phase Composition: C= 97.8 wt.% Sn C!= 18.3 wt.% Sn

CL= 61.9 wt.% Sn C= 97.8 wt.% Sn

Phase Amounts: W =

90!

61.9

97.8 ! 61.9= 0.78 W

! =

97.8 " 90

97.8"18.3= 0.10

WL =

97.8! 90

97.8! 61.9= 0.22 W

=

90 !18.3

97.8 ! 18.3= 0.90

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Things to Notice:

1) For the 25 wt.% Sn alloy primary !forms before the eutectic reaction while for the 90 wt.% Sn

alloy primary forms before the eutectic reaction.

2) As we get closer to the eutectic alloy composition less primary

!or primary forms before theeutectic reaction occurs.

3) Notice that as the alloy composition changes the phase compositions stay the same, but the

amount of each phase changes. For example, at 181C much more of the !phase is observed forthe 20 wt.% Sn alloy than for the 90 wt.% Sn alloy.

b) With part a) in mind, let's sketch the microstructures we would expect to observe at 181C for

each alloy composition.

25 wt.% Sn: The microstructure should have mostly primary !(W!= 0.85 at 185C) with the

remainder being eutectic !and .

60 wt.% Sn: The microstructure should have mostly eutectic !and because there is very little

primary !(W!= 0.04 at 185C).

90 wt.% Sn: The microstructure should have mostly primary (W= 0.78 at 185C) with the

remainder being eutectic !and .

! "

Primary!

! " ! "

Primary

"

Pb/Sn Microstructures at 181C

Co=25 wt.% Sn Co=90 wt.% SnCo=60 wt.% Sn

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Problem 2

a) 90 wt. % Pb and 10 wt. % Sn. Upon cooling from 350C changes in slope occur at the following

transitions: L to !+L at 300C, !+L to !at 280C and !to !+"at 150C. Note that unlike parts b)

and c) there is no flat region in the cooling curve. Since this composition does not cross the eutectic line

during cooling there are no junctions where there are zero degrees of freedom.

b) 60 wt. % Pb and 40 wt. % Sn. Upon cooling from 350C changes in slope occur at the

following transitions: L to !+L at 230C, !+L to !+"at 183C. The latter transition is the

eutectic reaction involving three phases: L, !and", so the slope becomes flat because there are

zero degrees freedom (see part d). Until all of the liquid has become !and", the temperature

stays fixed at the eutectic temperature of 183C.

!"#$"%&'(%"*+,-

.#"

/00

100

200

120+,

/32+,

!"#$"%&'(%"*+,-

.#"

/00

100

200

130+,

/40+,

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Problem 3

a) The following table has the relevant data starting with r, the atomic radius, as listed in the inside frontcover of Callister. The rest of the data is calculated in the same way as other XRD problems from

previous assignments so I will leave the details to you. Please recall that a is the lattice parameter and

d is the interplanar spacing. Copper and silver both have a FCC crystal structure.

r (nm) a(nm) d(111) d(200) 2#(111) 2#(200)

Cu 0.128 0.362 0.2090 0.181 43.2 50.4

Ag 0.144 0.407 0.235 0.207 38.3 44.4

The important issue to note is the mixture of silver and copper powders at room temperature will resultin a two-phase mixture so the XRD pattern will have two sets of peaks, one for the pure silver powder,

the other for the pure copper powder.

b) Please refer to the Ag-Cu phase diagram in Callister. An alloy of 50 wt.% Cu and 50 wt.% silver

heated to 1000C will be in the liquid single phase region. When cooled to and held at 800C, the alloy

resides in the !+L two-phase region. The !phase is predominantly copper, but also contains

approximately 10 wt. % silver in solid solution with the copper. There is no "phase (which ispredominantly silver) present at 800C for this composition, so the silver peaks will no longer be present

in the XRD pattern. The Cu peaks from part a) will be shifted to slightly lower values of 2#to

accommodate the larger sized silver atoms that are now in solid solution. The liquid phase is not

crystalline and will not generate peaks on the XRD pattern. The lever rule shows that you expect to havemuch more of the liquid phase than the !phase, so you might expect a decrease in the peak intensities

as well, but we are not concerned about that for this problem.

To summarize, the new XRD pattern will display one set of peaks corresponding to the copper-rich !

phase. Therefore, the copper peaks remain, but shift to slightly lower values of 2#due to some silver in

solid solution. The peaks associated with pure silver are no longer observed due to the liquid phase.

!

#!$% &' &% %'$'

() *+++,

-. *+++, ()*#'',-.*#'',

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Problem 4

Hypoeutectoid steel of composition Fe with 0.45 wt.% C.

a) W!(total)=

CFe 3C

" Co

CFe3C "C!=

6.7 " 0.45

6.7 " 0.022

=0.936

WFe 3C

= 1!W" = 0.064

b) W!(proeutectoid) $the !phase formed just above 727C in the !+%two phase region

W!(proeutectoid) =0.77! 0.45

0.77 ! 0.022= 0.42

W!(eutectoid) = W!(total) - W!(proeutectoid)= 0.936 - 0.42

W!(eutectoid) = 0.516

c)

! Fe3C

! "

Above 727 C Below 727C

!

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Problem 5*

Eutectoid Fe3C = Total Fe3C - Proeutectoid Fe3C

0.1 =C0" 0.022

6.7" 0.022"

C0" 0.76

6.7" 0.76

! ! total Fe3C Proeutectoid Fe3C

0.1 =C0! 0.022

6.68!

C0! 0.76

5.94

(6.68) (5.94) (0.1) = 5.94 (C0- 0.022) - 6.68 (C0- 0.76)

Solving for Cogives Co= 1.34 wt.% C

Checking your answer:1.34! 0.022

6.7! 0.022!

1.34! 0.76

6.7! 0.76= 0.1

!

!+""

!+Fe3C

"+Fe3C

Fe3C

10 wt. %

eutectoid Fe3C

C0=?

0.022 0.77 6.7

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Problem 6*

a)

b) The formation of the #precipitates requires solid state diffusion. If we cooled to a low

temperature very quickly (a process called "quenching") we could prevent precipitate formation fromoccurring because diffusion would be too slow. This would trap the copper in solid solution eventhough not all of it wants to be there. In other words, there is a large driving force for atomic

rearrangement to form the #precipitates, but the diffusivity is too low for the atoms to respond to the

driving force.

!

"

" Particles Dispersed in an !Matrix