9.5
Cite the phases that are present and the
phase compositions for the following alloys:
(a) 90 wt% Zn–10 wt% Cu at 400_C(750_F).
(b) 75 wt%Sn–25wt%Pb at 175_C (345_F).
(c) 55 wt% Ag–45 wt% Cu at 900_C(1650_F).
(d) 30 wt% Pb–70 wt% Mg at 425_C(795_F).
(e) 2.12 kg Zn and 1.88 kg Cu at 500_C(930_F).
(f ) 37 lbm Pb and 6.5 lbm Mg at 400_C(750_F).
(g) 8.2 mol Ni and 4.3 mol Cu at 1250_C (2280_F).
(h) 4.5 mol Sn and 0.45 mol Pb at 200_C(390_F).
9.5 This problem asks that we cite the phase or phases present for several
alloys at specified temperatures.
(a) For an alloy composed of 15 wt% Sn-85 wt% Pb and at 100C, from
Figure 9.7,  and  phases are present, and
C = 5 wt% Sn-95 wt% Pb
C = 98 wt% Sn-2 wt% Pb
(b) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, from
Figure 9.18, only the  phase is present; its composition is 25 wt% Pb-75
wt% Mg.
(c) For an alloy composed of 85 wt% Ag-15 wt% Cu and at 800C, from
Figure 9.6,  and liquid phases are present, and
C = 92 wt% Ag-8 wt% Cu
CL = 77 wt% Ag-23 wt% Cu
(d) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, from
Figure 9.17,  and  phases are present, and
C = 51 wt% Zn-49 wt% Cu
C = 58 wt% Zn-42 wt% Cu
(e) For an alloy composed of 1.25 kg Sn and 14 kg Pb and at 200C, we
must first determine the Sn and Pb concentrations, as
CSn =
1.25 kg
 100 = 8.2 wt%
1.25 kg  14 kg
CPb =
14 kg
 100 = 91.8 wt%
1.25 kg  14 kg
From Figure 9.7, only the  phase is present; its composition is 8.2 wt%
Sn-91.8 wt% Pb.
(f) For an alloy composed of 7.6 lbm Cu and 114.4 lbm Zn and at 600C,
we must first determine the Cu and Zn concentrations, as
CCu =
C Zn =
7.6 lb m
 100 = 5.0 wt%
7.6 lb m  144.4 lbm
144.4 lb m
 100 = 95.0 wt%
7.6 lb m  144.4 lb m
From Figure 9.17, only the L phase is present; its composition is 95.0
wt% Zn-5.0 wt% Cu
(g) For an alloy composed of 21.7 mol Mg and 35.4 mol Pb and at 350C,
it is first necessary to determine the Mg and Pb concentrations, which we
will do in weight percent as follows:
'
mPb
= nm APb = (35.4 mol)(207.2 g/mol) = 7335 g
Pb
'
mMg
= nm AMg = (21.7 mol)(24.3 g/mol) = 527 g
Mg
CPb =
7335 g
 100 = 93 wt%
7335 g  527 g
CMg = 100 wt%  93 wt% = 7 wt%
From Figure 9.18, L and Mg2Pb phases are present, and
CL  94 wt % Pb  6 wt% Mg
CMg Pb  81 wt% Pb  19 wt% Mg
2
(h) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900C, it
is first necessary to determine the Cu and Ag concentrations, which we
will do in weight percent as follows:
'
mCu
= nm
Cu
ACu = (4.2 mol)(63.55 g/mol) = 266.9 g
'
mAg
= nm AAg = (1.1 mol)(107.87 g/mol) = 118.7 g
Ag
CCu =
266.9 g
 100 = 69.2 wt%
266.9 g  118.7 g
C Ag =
118.7 g
 100 = 30.8 wt%
266.9 g  118.7 g
From Figure 9.6,  and liquid phases are present; and
C = 8 wt% Ag-92 w% Cu
CL = 45 wt% Ag-55 wt% Cu
9.10 Below is a portion of the H2O–NaCl phase diagram:
(a) Using this diagram, briefly explain how spreading salt on ice that is at a
temperature below 0oC (32oF) can cause the ice to melt.
(b) What concentration of salt is necessary to have a 50% ice–50% liquid brine at 10oC (14oF)?
(a)
Spreading salt on ice will lower the melting temperature, since the
liquidus line decreases from 0C to the eutectic temperature at about 21C. Thus, ice at a temperature below 0C (and above -21C) can be
made to form a liquid phase by the addition of salt.
(b) We are asked to compute the concentration of salt necessary to have
a 50% ice-50% brine solution at -10C (14F). At -10C,
Cice = 0 wt% NaCl-100 wt% H2O
C
= 13 wt% NaCl-87 wt% H O
brine
2
Thus,
Wice = 0.5 =
Cbrine  Co
Cbrine  Cice
=
13  Co
13  0
Solving for Co (the concentration of salt) yields a value of 6.5 wt% NaCl93.5 wt% H2O.
9.13
Consider a specimen of ice I which is at -10oC and 1 atm pressure. Using Figure 9.33,
the pressure–temperature phase diagram for H2O, determine the pressure to which the
specimen must be raised or lowered to cause it (a) to melt, and (b) to sublime.
9.13 This problem asks us to consider a specimen of ice I which is at -10C
and 1 atm pressure.
(a) In order to determine the pressure at which melting occurs at this
temperature, we move vertically at this temperature until we cross the Ice
I-Liquid phase boundary of Figure 9.33. This occurs at approximately 570
atm; thus the pressure of the specimen must be raised from 1 to 570
atm.
(b) In order to determine the pressure at which sublimation occurs at this
temperature, we move vertically downward from 1 atm until we cross the
Ice I-Vapor phase boundary of Figure 9.33. This intersection occurs at
approximately 0.0023 atm.
9.30
A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature from an
elevated temperature in such a way that the high-temperature microstructure is
preserved. This microstructure is found to consist of the  phase and Mg2Pb, having
respective mass fractions of 0.65 and 0.35.
Determine the approximate temperature from which the alloy was quenched.
9.30 We are asked to determine the approximate temperature from which a
Pb-Mg alloy was quenched, given the mass fractions of  and Mg2Pb
phases. We can write a lever-rule expression for the mass fraction of the
 phase as
W = 0.65 =
CMg Pb  Co
2
CMg Pb  C 
2
The value of Co is stated as 45 wt% Pb-55 wt% Mg, and CMg Pb is 81
2
wt% Pb-19 wt% Mg, which is independent of temperature (Figure 9.18);
thus,
0.65 =
81  45
81  C

which yields
C = 25.6 wt% Pb
The temperature at which the -( + Mg2Pb) phase boundary (Figure
9.18) has a value of 25.6 wt% Pb is about 360C (680F).
9.62
Is it possible to have an iron–carbon alloy for which the mass fractions of total ferrite
and pearlite are 0.860 and 0.969, respectively? Why or why not?
9.62 This problem asks if it is possible to have an iron-carbon alloy for which
W = 0.860 and Wp = 0.969. In order to make this determination, it is
necessary to set up lever rule expressions for these two mass fractions in
terms of the alloy composition, then to solve for the alloy composition of
each; if both alloy composition values are equal, then such an alloy is
possible. The expression for the mass fraction of total ferrite is
W =
CFe C  C o
3
CFe C  C
3
=
6.70  Co
6.70  0.022
= 0.860
Solving for this Co yields Co = 0.95 wt% C.
Equation (9.20) as
Now for Wp we utilize
Wp =
6.70  C1'
5.94
= 0.969
This expression leads to C1' = 0.95 wt% C. Since Co = C1' , this alloy is
possible.
9.63
Compute the maximum mass fraction of eutectoid cementite in an iron–carbon alloy
that contain 1.00 wt% C.
9.63 This problem asks that we compute the mass fraction of eutectoid
cementite in an iron-carbon alloy that contains 1.00 wt% C. In order to
solve this problem it is necessary to compute mass fractions of total and
proeutectoid cementites, and then to subtract the latter from the former.
To calculate the mass fraction of total cementite, it is necessary to use the
lever rule and a tie line that extends across the entire  + Fe3C phase
field as
Co  C
1.00  0.022
WFe C =
=
= 0.146
CFe C  C
6.70  0.022
3
3
Now, for the mass fraction of proeutectoid cementite we use Equation
(9.21)
WFe C' =
3
C1'  0.76
5.94
=
1.00  0.76
= 0.040
5.94
And, finally, the mass fraction of eutectoid cementite WFe C'' is just
3
WFe C  WFe C  WFe C   0.146  0.040  0.106
3
3
3
9.65
The mass fraction of eutectoid ferrite in an iron–carbon alloy is 0.71. On the basis of
this information, is it possible to determine the composition of the alloy? If so, what is
its composition? If this is not possible, explain why.
9.65 This problem asks whether or not it is possible to determine the
composition of an iron-carbon alloy for which the mass fraction of
eutectoid ferrite is 0.71; and if so, to calculate the composition. Yes, it is
possible to determine the alloy composition; and, in fact, there are two
possible answers. For the first, the eutectoid ferrite exists in addition to
proeutectoid ferrite. For this case the mass fraction of eutectoid ferrite
(W'') is just the difference between total ferrite and proeutectoid ferrite
mass fractions; that is
W '' = W  - W '
Now, it is possible to write expressions for W and W' in terms of Co,
the alloy composition. Thus,
W'' =
=
CFe C  Co
3
CFe C  C 
3

0.76  C o
0.74
6.70  C o
0.76  C o

= 0.71
6.70  0.022
0.74
And, solving for Co yields Co = 0.61 wt% C.
For the second possibility, we have a hypereutectoid alloy wherein
all of the ferrite is eutectoid ferrite. Thus, it is necessary to set up a lever
rule expression wherein the mass fraction of total ferrite is 0.71.
Therefore,
W =
CFe C  C o
3
CFe C  C
3
=
6.70  Co
6.70  0.022
And, solving for Co yields Co = 1.96 wt% C.
= 0.71