Home-work #9_solutions
9.2 At 500C (930F), what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?
Solution
(a) From Figure 9.7, the maximum solubility of Cu in Ag at 500C corresponds to the position of
the –( + ) phase boundary at this temperature, or to about 2 wt% Cu.
(b) From this same figure, the maximum solubility of Ag in Cu corresponds to the position of the
–( + ) phase boundary at this temperature, or about 1.5 wt% Ag.
9.5 Consider a specimen of ice that is at 210C and 1 atm pressure. Using Figure 9.2, 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.
Solution
The figure below shows the pressure-temperature phase diagram for H2O, Figure 10.2; a vertical
line has been constructed at -10C, and the location on this line at 1 atm pressure (point B) is also noted.
(a) Melting occurs, (by changing pressure) as, moving vertically (upward) at this temperature, we
cross the Ice-Liquid phase boundary. 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-Vapor phase boundary. This intersection occurs at
approximately 0.0023 atm.
9.8 Cite the phases that are present and the phase compositions for the following alloys:
(a) 90 wt% Zn-10 wt% Cu at 400C (750F)
(b) 75 wt% Sn-25 wt% Pb at 175C (345F)
(c) 55 wt% Ag-45 wt% Cu at 900C (1650F)
(d) 30 wt% Pb-70 wt% Mg at 425C (795F)
(e) 2.12 kg Zn and 1.88 kg Cu at 500C (930F)
(f) 37 lbm Pb and 6.5 lbm Mg at 400C (750F)
(g) 8.2 mol Ni and 4.3 mol Cu at 1250C (2280F)
(h) 4.5 mol Sn and 0.45 mol Pb at 200C (390F)
Solution
This problem asks that we cite the phase or phases present for several alloys at specified
temperatures.
(a) That portion of the Cu-Zn phase diagram (Figure 9.19) that pertains to this problem is shown
below; the point labeled ―A‖ represents the 90 wt% Zn-10 wt% Cu composition at 400C.
As may be noted, point A lies within the and phase field. A tie line has been constructed at 400C; its
intersection with the + phase boundary is at 87 wt% Zn, which corresponds to the composition of the
phase. Similarly, the tie-line intersection with the + phase boundary occurs at 97 wt% Zn, which is
the composition of the phase. Thus, the phase compositions are as follows:
C = 87 wt% Zn-13 wt% Cu
C = 97 wt% Zn-3 wt% Cu
(b) That portion of the Pb-Sn phase diagram (Figure 9.8) that pertains to this problem is shown
below; the point labeled ―B‖ represents the 75 wt% Sn-25 wt% Pb composition at 175C.
As may be noted, point B lies within the + phase field. A tie line has been constructed at 175C; its
intersection with the + phase boundary is at 16 wt% Sn, which corresponds to the composition of the
phase. Similarly, the tie-line intersection with the + phase boundary occurs at 97 wt% Sn, which
is the composition of the phase. Thus, the phase compositions are as follows:
C = 16 wt% Sn-84 wt% Pb
C = 97 wt% Sn-3 wt% Pb
(c) The Ag-Cu phase diagram (Figure 9.7) is shown below; the point labeled ―C‖ represents the
55 wt% Ag-45 wt% Cu composition at 900C.
As may be noted, point C lies within the Liquid phase field. Therefore, only the liquid phase is present; its
composition is 55 wt% Ag-45 wt% Cu.
(d) The Mg-Pb phase diagram (Figure 9.20) is shown below; the point labeled ―D‖ represents the
30 wt% Pb-70 wt% Mg composition at 425C.
As may be noted, point D lies within the phase field. Therefore, only the phase is present; its
composition is 30 wt% Pb-70 wt% Mg.
(e) For an alloy composed of 2.12 kg Zn and 1.88 kg Cu and at 500C, we must first determine
the Zn and Cu concentrations, as
CZn 2.12 kg
2.12 kg 1.88 kg 100 53 wt%
CCu 1.88 kg
2.12 kg 1.88 kg 100 47 wt%
That portion of the Cu-Zn phase diagram (Figure 9.19) that pertains to this problem is shown below; the
point labeled ―E‖ represents the 53 wt% Zn-47 wt% Cu composition at 500C.
As may be noted, point E lies within the + phase field. A tie line has been constructed at 500C; its
intersection with the + phase boundary is at 49 wt% Zn, which corresponds to the composition of the
phase. Similarly, the tie-line intersection with the + phase boundary occurs at 58 wt% Zn, which is
the composition of the phase. Thus, the phase compositions are as follows:
C = 49 wt% Zn-51 wt% Cu
C = 58 wt% Zn-42 wt% Cu
(f) For an alloy composed of 37 lbm Pb and 6.5 lbm Mg and at 400C, we must first determine the
Pb and Mg concentrations, as
CPb 37 lbm
37 lbm 6.5 lbm
100 85 wt%
CMg 6.5 lbm
37 lbm 6.5 lbm
100 15 wt%
That portion of the Mg-Pb phase diagram (Figure 9.20) that pertains to this problem is shown below; the
point labeled ―F‖ represents the 85 wt% Pb-15 wt% Mg composition at 400C.
As may be noted, point F lies within the L + Mg2Pb phase field. A tie line has been constructed at 400C;
it intersects the vertical line at 81 wt% Pb, which corresponds to the composition of Mg2Pb. Furthermore,
the tie line intersection with the L + Mg2Pb-L phase boundary is at 93 wt% Pb, which is the composition of
the liquid phase. Thus, the phase compositions are as follows:
CMg2Pb
= 81 wt% Pb-19 wt% Mg
CL = 93 wt% Pb-7 wt% Mg
(g) For an alloy composed of 8.2 mol Ni and 4.3 mol Cu and at 1250C, it is first necessary to
determine the Ni and Cu concentrations, which we will do in wt% as follows:
nNi' nmNi
ANi (8.2 mol)(58.69 g/mol)= 481.3 g
nCu' nmCu
ACu (4.3 mol)(63.55 g/mol)= 273.3 g
CNi 481.3 g
481.3 g + 273.3 g 100 63.8 wt%
CCu 273.3 g
481.3 g + 273.3 g 100 36.2 wt%
The Cu-Ni phase diagram (Figure 9.3a) is shown below; the point labeled ―G‖ represents the 63.8 wt% Ni-
36.2 wt% Cu composition at 1250C.
As may be noted, point G lies within the phase field. Therefore, only the phase is present; its
composition is 63.8 wt% Ni-36.2 wt% Cu.
(h) For an alloy composed of 4.5 mol Sn and 0.45 mol Pb and at 200C, it is first necessary to
determine the Sn and Pb concentrations, which we will do in weight percent as follows:
nSn’ nmSn
ASn (4.5 mol)(118.71 g/mol)= 534.2 g
nPb' nmPb
APb (0.45 mol)(207.2 g/mol)= 93.2 g
CSn 534.2 g
534.2 g + 93.2 g 100 85.1 wt%
CPb 93.2 g
534.2 g + 93.2 g 10014.9 wt%
That portion of the Pb-Sn phase diagram (Figure 9.8) that pertains to this problem is shown below; the
point labeled ―H‖ represents the 85.1 wt% Sn-14.9 wt% Pb composition at 200C.
As may be noted, point H lies within the + L phase field. A tie line has been constructed at 200C; its
intersection with the L + Lphase boundary is at 74 wt% Sn, which corresponds to the composition of
the L phase. Similarly, the tie-line intersection with the + Lphase boundary occurs at 97.5 wt% Sn,
which is the composition of the phase. Thus, the phase compositions are as follows:
C = 97.5 wt% Sn-2.5 wt% Pb
CL = 74 wt% Sn-26 wt% Pb
9.13 For an alloy of composition 74 wt% Zn-26 wt% Cu, cite the phases present and their compositions at
the following temperatures: 850C, 750C, 680C, 600C, and 500C.
Solution
This problem asks us to determine the phases present and their concentrations at several
temperatures, for an alloy of composition 74 wt% Zn-26 wt% Cu. From Figure 9.19 (the Cu-Zn phase
diagram), which is shown below with a vertical line constructed at the specified composition:
At 850C, a liquid phase is present; CL = 74 wt% Zn-26 wt% Cu
At 750C, and liquid phases are present; C = 67 wt% Zn-33 wt% Cu; CL = 77 wt% Zn-23 wt%
Cu
At 680C, and liquid phases are present; C = 73 wt% Zn-27 wt% Cu; CL = 82 wt% Zn-18 wt%
Cu
At 600C, the phase is present; C = 74 wt% Zn-26 wt% Cu
At 500C, and phases are present; C = 69 wt% Zn-31 wt% Cu; C = 78 wt% Zn-22 wt% Cu
9.17 A 90 wt% Ag-10 wt% Cu alloy is heated to a temperature within the + liquid phase region. If the
composition of the liquid phase is 85 wt% Ag, determine:
(a) The temperature of the alloy
(b) The composition of the phase
(c) The mass fractions of both phases
Solution
(a) In order to determine the temperature of a 90 wt% Ag-10 wt% Cu alloy for which and liquid
phases are present with the liquid phase of composition 85 wt% Ag, we need to construct a tie line across
the + L phase region of Figure 9.7 that intersects the liquidus line at 85 wt% Ag; this is possible at about
850C.
(b) The composition of the phase at this temperature is determined from the intersection of this
same tie line with solidus line, which corresponds to about 95 wt% Ag.
(c) The mass fractions of the two phases are determined using the lever rule, Equations 9.1 and
9.2 with C0 = 90 wt% Ag, CL = 85 wt% Ag, and C = 95 wt% Ag, as
W =C0 CL
C CL
=90 85
95 85= 0.50
WL =C C0
C CL
=95 90
95 85= 0.50
9.27 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 α phase and Mg2Pb, having respective mass fractions of 0.65 and 0.35. Determine
the approximate temperature from which the alloy was quenched.
Solution
We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% 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 =CMg2Pb C0
CMg2Pb C
The value of C0 is stated as 45 wt% Pb-55 wt% Mg, and CMg2Pb is 81 wt% Pb-19 wt% Mg, which is
independent of temperature (Figure 9.20); thus,
0.65 =81 45
81 C
which yields
C = 25.6 wt% Pb
The temperature at which the –( + Mg2Pb) phase boundary (Figure 9.20) has a value of 25.6 wt% Pb is
about 360C (680F).
9.33 The microstructure of a lead-tin alloy at 180C (355F) consists of primary β and eutectic structures.
If the mass fractions of these two microconstituents are 0.57 and 0.43, respectively, determine the
composition of the alloy.
Solution
Since there is a primary microconstituent present, then we know that the alloy composition, C0
is between 61.9 and 97.8 wt% Sn (Figure 9.8). Furthermore, this figure also indicates that C = 97.8 wt%
Sn and Ceutectic = 61.9 wt% Sn. Applying the appropriate lever rule expression for W'
W' =C0 Ceutectic
CÊ Ceutectic
=C0 61.9
97.8 61.9= 0.57
and solving for C0 yields C0 = 82.4 wt% Sn.
9.37 For a 30 wt% Zn-70 wt% Cu alloy, make schematic sketches of the microstructure that would be
observed for conditions of very slow cooling at the following temperatures: 1100C (2010F), 950C
(1740F), 900C (1650F), and 700C (1290F). Label all phases and indicate their approximate
compositions.
Solution
The illustration below is the Cu-Zn phase diagram (Figure 9.19). A vertical line at a composition
of 30 wt% Zn-70 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures
called for in the problem statement (i.e., 1100C, 950C, 900C, and 700C).
On the basis of the locations of the four temperature-composition points, schematic sketches of the four
respective microstructures along with phase compositions are represented as follows:
9.39 The room-temperature tensile strengths of pure lead and pure tin are 16.8 MPa and 14.5 MPa,
respectively.
(a) Make a schematic graph of the room-temperature tensile strength versus composition for all
compositions between pure lead and pure tin. (Hint: you may want to consult Sections 9.10 and 9.11, as
well as Equation 9.24 in Problem 9.64.)
(b) On this same graph schematically plot tensile strength versus composition at 150°C.
(c) Explain the shapes of these two curves, as well as any differences between them.
Solution
The (a) and (b) portions of the problem ask that we make schematic plots on the same graph for
the tensile strength versus composition for lead-tin alloys at both room temperature and 150C; such a
graph is shown below.
(c) Upon consultation of the Pb-Sn phase diagram (Figure 9.8) we note that, at room temperature
(20C), about 1.5 wt% of Sn is soluble in Pb (within the -phase region at the left extremity of the phase
diagram). Similarly, only about 1 wt% of Pb is soluble in Sn (within the -phase region at the left
extremity). Thus, there will a solid-solution strengthening effect on both ends of the phase diagram—
strength increases slightly with additions of Sn to Pb [in the phase region (left-hand side)] and with
additions of Pb to Sn [in the phase region (right-hand side)]; these effects are noted in the above figure.
This figure also shows that the tensile strength of pure lead is greater than pure tin, which is in agreement
with tensile strength values provided in the problem statement.
In addition, at room temperature, for compositions between about 1.5 wt% Sn and 99 wt% Sn,
both and phase will coexist, (Figure 9.8), Furthermore, for compositions within this range, tensile
strength will depend (approximately) on the tensile strengths of each of the and phases as well as their
phase fractions in a manner described by Equation 9.24 for the elastic modulus (Problem 9.64). That is, for
this problem
(TS)alloy (TS)V + (TS)V
in which TS and V denote tensile strength and volume fraction, respectively, and the subscripts represent
the alloy/phases. Also, mass fractions of the and phases change linearly with changing composition
(according to the lever rule). Furthermore, although there is some disparity between the densities of Pb and
Sn (11.35 versus 7.27 g/cm3), weight and volume fractions of the and phases will also be similar (see
Equation 9.6).
At 150C, the curve will be shifted to significantly lower tensile strengths inasmuch as tensile
strength diminishes with increasing temperature (Section 6.6, Figure 6.14). In addition, according to Figure
9.8, solubility limits for both and phases increase—for the phase from 1.5 to 10 wt% Sn, and for the
phase from 1 to about 2 wt% Pb. Thus, the compositional ranges over which solid-solution strengthening
occurs increase somewhat from the room-temperature ranges; these effects are also noted on the 150C
curve above. Furthermore, at 150C, it would be expected that the tensile strength of lead will be greater
than that of tin; and for compositions over which both and phases coexist, strength will decrease
approximately linearly with increasing Sn content.
9.42 Figure 9.36 is the aluminum-neodymium phase diagram, for which only single-phase regions are
labeled. Specify temperature-composition points at which all eutectics, eutectoids, peritectics, and
congruent phase transformations occur. Also, for each, write the reaction upon cooling.
Solution
Below is shown the aluminum-neodymium phase diagram (Figure 9.36).
There are two eutectics on this phase diagram. One exists at 12 wt% Nd-88 wt% Al and 632C.
The reaction upon cooling is
LAl Al11Nd3
The other eutectic exists at about 97 wt% Nd-3 wt% Al and 635C. This reaction upon cooling is
L AlNd3 + Nd
There are four peritectics. One exists at 59 wt% Nd-41 wt% Al and 1235C. Its reaction upon
cooling is as follows:
L + Al2Nd Al11Nd3
The second peritectic exists at 84 wt% Nd-16 wt% Al and 940C. This reaction upon cooling is
L + Al2Nd AlNd
The third peritectic exists at 91 wt% Nd-9 wt% Al and 795C. This reaction upon cooling is
L + AlNd AlNd2
The fourth peritectic exists at 94 wt% Nd-6 wt% Al and 675C. This reaction upon cooling is
L + AlNd2 AlNd3
There is one congruent melting point at about 73 wt% Nd-27 wt% Al and 1460C. Its reaction
upon cooling is
L Al2Nd
No eutectoids are present.
9.43 Figure 9.37 is a portion of the titanium-copper phase diagram for which only single-phase
regions are labeled. Specify all temperature-composition points at which eutectics, eutectoids, peritectics,
and congruent phase transformations occur. Also, for each, write the reaction upon cooling.
Solution
Below is shown the titanium-copper phase diagram (Figure 9.37).
There is one eutectic on this phase diagram, which exists at about 51 wt% Cu-49 wt% Ti and
960C. Its reaction upon cooling is
L Ti2Cu + TiCu
There is one eutectoid for this system. It exists at about 7.5 wt% Cu-92.5 wt% Ti and 790C.
This reaction upon cooling is
+ Ti2Cu
There is one peritectic on this phase diagram. It exists at about 40 wt% Cu-60 wt% Ti and
1005C. The reaction upon cooling is
L Ti2Cu
There is a single congruent melting point that exists at about 57.5 wt% Cu-42.5 wt% Ti and
982C. The reaction upon cooling is
L TiCu
.50 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F).
(a) What is the proeutectoid phase?
(b) How many kilograms each of total ferrite and cementite form?
(c) How many kilograms each of pearlite and the proeutectoid phase form?
(d) Schematically sketch and label the resulting microstructure.
Solution
(a) The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid
composition (0.76 wt% C).
(b) For this portion of the problem, we are asked to determine how much total ferrite and
cementite form. Application of the appropriate lever rule expression yields
W =CFe3C C0
CFe3C C
=6.70 1.15
6.70 0.022= 0.83
which, when multiplied by the total mass of the alloy (1.0 kg), gives 0.83 kg of total ferrite.
Similarly, for total cementite,
WFe3C =C0 C
CFe3C C
=1.15 0.022
6.70 0.022= 0.17
And the mass of total cementite that forms is (0.17)(1.0 kg) = 0.17 kg.
(c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite)
form. Applying Equation 9.22, in which
C1' = 1.15 wt% C
Wp =6.70 C 1
'
6.70 0.76=
6.70 1.15
6.70 0.76= 0.93
which corresponds to a mass of 0.93 kg. Likewise, from Equation 9.23
WFe3C' =C1
' 0.76
5.94=
1.15 0.76
5.94= 0.07
which is equivalent to 0.07 kg of the total 1.0 kg mass.
(d) Schematically, the microstructure would appear as: