Chap 9 Solns
CHAPTER 10
PHASE DIAGRAMS
PROBLEM SOLUTIONS
Development of Microstructure in Eutectic Alloys 10.16 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. Solution Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.
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10.17 Is it possible to have a magnesium–lead alloy in which the mass fractions of primary α and total α are 0.60 and 0.85, respectively, at 460°C (860°F)? Why or why not? Solution In order to make this determination we need to set up the appropriate lever rule expression for each of these quantities. From Figure 10.20 and at 460C, C = 41 wt% Pb, CMg Pb = 81 wt% Pb, and Ceutectic = 67 wt% Pb. 2
For primary
W' =
Ceutectic C0 Ceutectic C
=
67 C0 = 0.60 67 41
Solving for C0 gives C0 = 51.4 wt% Pb. Now the analogous expression for total
W =
CMg P b C0 2
CMg P b C 2
=
81 C0
81 41
= 0.85
which yields a value of 47 wt% Pb for C0. Therefore, since these two C0 values are different, this alloy is not possible.
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10.18 For a lead–tin alloy of composition 80 wt% Sn–20 wt% Pb and at 180°C (355°F) do the following: (a) Determine the mass fractions of α and β phases. (b) Determine the mass fractions of primary β and eutectic microconstituents. (c) Determine the mass fraction of eutectic β. Solution (a) This portion of the problem asks that we determine the mass fractions of and phases for an 80 wt% Sn-20 wt% Pb alloy (at 180C). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the + phase field. From Figure 10.8 and at 180C, C = 18.3 wt% Sn, C = 97.8 wt% Sn, and Ceutectic = 61.9 wt% Sn. Therefore, the two lever-rule expressions are as follows:
W =
W =
C C0 97.8 80 = = 0.224 C C 97.8 18.3
C0 C C C
80 18.3 = 0.776 97.8 18.3
=
(b) Now it is necessary to determine the mass fractions of primary and eutectic microconstituents for this same alloy. This requires that we utilize the lever rule and a tie line that extends from the maximum solubility of Pb in the phase at 180C (i.e., 97.8 wt% Sn) to the eutectic composition (61.9 wt% Sn). Thus
W' =
We =
C0 Ceutectic C Ceutectic C C0 C Ceutectic
=
=
80.0 61.9 = 0.504 97.8 61.9
97.8 80.0 = 0.496 97.8 61.9
(c) And, finally, we are asked to compute the mass fraction of eutectic , We. This quantity is simply the difference between the mass fractions of total and primary as We = W – W' = 0.776 – 0.504 = 0.272
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10.21 For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 950°C (1740°F), 860°C (1580°F), 800°C (1470°F), and 600°C (1100°F). Label all phases and indicate their approximate compositions. Solution The illustration below is the Cu-Zn phase diagram (Figure 10.19). A vertical line at a composition of 52 wt% Zn-48 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement (i.e., 950C, 860C, 800C, and 600C).
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:
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Congruent Phase Transformations Binary Eutectic Systems Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions 10.24 Figure 10.40 is the tin–gold 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 tin-gold phase diagram (Figure 10.40).
There are two eutectics on this phase diagram. One exists at 10 wt% Au-90 wt% Sn and 217C. The reaction upon cooling is
L +
The other eutectic exists at 80 wt% Au-20 wt% Sn and 280C. This reaction upon cooling is
L +
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There are three peritectics. One exists at 30 wt% Au-70 wt% Sn and 252C. Its reaction upon cooling is as follows:
L +
The second peritectic exists at 45 wt% Au-55 wt% Sn and 309C. This reaction upon cooling is
L +
The third peritectic exists at 92 wt% Au-8 wt% Sn and 490C. This reaction upon cooling is
L +
There is one congruent melting point at 62.5 wt% Au-37.5 wt% Sn and 418C. Its reaction upon cooling is
L
No eutectoids are present.
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Ceramic Phase Diagrams 10.26 From Figure 10.24, the phase diagram for the MgO–Al2O3 system, it may be noted that the spinel solid solution exists over a range of compositions, which means that it is nonstoichiometric at compositions other than 50 mol% MgO–50 mol% Al2O3. (a) The maximum nonstoichiometry on the Al2O3-rich side of the spinel phase field exists at about 2000°C (3630°F) corresponding to approximately 82 mol% (92 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. (b) The maximum nonstoichiometry on the MgO-rich side of the spinel phase field exists at about 2000°C (3630°F) corresponding to approximately 39 mol% (62 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. Solution (a) For this portion of the problem we are to determine the type of vacancy defect that is produced on the Al2O3-rich side of the spinel phase field (Figure 10.24) and the percentage of these vacancies at the maximum nonstoichiometry (82 mol% Al2O3). On the alumina-rich side of this phase field, there is an excess of Al3+ ions, which means that some of the Al3+ ions substitute for Mg2+ ions. In order to maintain charge neutrality, Mg2+ 3+
vacancies are formed, and for every Mg2+ vacancy formed, two Al ions substitute for three Mg2+ ions. Now, we will calculate the percentage of Mg2+ vacancies that exist at 82 mol% Al2O3. Let us arbitrarily choose as our basis 50 MgO-Al2O3 units of the stoichiometric material, which consists of 50 Mg2+ ions and 100 Al3+ ions. Furthermore, let us designate the number of Mg2+ vacancies as x, which means that 2x Al3+ ions have been added and 3x Mg2+ ions have been removed (two of which are filled with Al3+ ions). Using our 50 MgOAl2O3 unit basis, the number of moles of Al2O3 in the nonstoichiometric material is (100 + 2x)/2; similarly the number of moles of MgO is (50 – 3x). Thus, the expression for the mol% of Al2O3 is just
100 2x 2 mol% Al2O3 = 100 100 2x (50 3x) 2 If we solve for x when the mol% of Al2O3 = 82, then x = 12.1. Thus, adding 2x or (2)(12.1) = 24.2 Al3+ ions to the original material consisting of 100 Al3+ and 50 Mg2+ ions will produce 12.1 Mg2+ vacancies. Therefore, the percentage of vacancies is just
% vacancies =
12.1 100 = 8.1% 100 50
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(b) Now, we are asked to make the same determinations for the MgO-rich side of the spinel phase field, for 39 mol% Al2O3. In this case, Mg2+ ions are substituting for Al3+ ions. Since the Mg2+ ion has a lower charge than the Al3+ ion, in order to maintain charge neutrality, negative charges must be eliminated, which may be accomplished by introducing O2- vacancies. For every 2 Mg2+ ions that substitute for 2 Al3+ ions, one O2- vacancy is formed. Now, we will calculate the percentage of O2- vacancies that exist at 39 mol% Al2O3. Let us arbitrarily choose as our basis 50 MgO-Al2O3 units of the stoichiometric material which consists of 50 Mg2+ ions 100 Al3+ ions. Furthermore, let us designate the number of O2- vacancies as y, which means that 2y Mg2+ ions have been added and 2y Al3+ ions have been removed. Using our 50 MgO-Al2O3 unit basis, the number of moles of Al2O3 in the nonstoichiometric material is (100 – 2y)/2; similarly the number of moles of MgO is (50 + 2y). Thus, the expression for the mol% of Al2O3 is just
100 2 y 2 mol% Al2O3 = 100 100 2 y (50 2 y) 2 If we solve for y when the mol% of Al2O3 = 39, then y = 7.91. Thus, 7.91 O2- vacancies are produced in the original material that had 200 O2- ions. Therefore, the percentage of vacancies is just
% vacancies =
7.91 100 = 3.96% 200
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10.29 Consider 3.5 kg of austenite containing 0.95 wt% C, cooled to below 727°C (1341°F). (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 0.95 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 =
CFe C C0 3
CFe C C 3
=
6.70 0.95 = 0.86 6.70 0.022
which, when multiplied by the total mass of the alloy, gives (0.86)(3.5 kg) = 3.01 kg of total ferrite. Similarly, for total cementite,
C0 C 0.95 0.022 WFe C = = = 0.14 3 CFe C C 6.70 0.022 3
And the mass of total cementite that forms is (0.14)(3.5 kg) = 0.49 kg. (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. Applying Equation 10.22, in which C1' = 0.95 wt% C
Wp =
6.70 C 1' 6.70 0.95 = = 0.97 6.70 0.76 6.70 0.76
which corresponds to a mass of (0.97)(3.5 kg) = 3.4 kg. Likewise, from Equation 10.23
WFe C' = 3
C1' 0.76 5.94
=
0.95 0.76 = 0.03 5.94
which is equivalent to (0.03)(3.5 kg) = 0.11 kg of the total 3.5 kg mass. (d) Schematically, the microstructure would appear as:
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10.30 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy containing 0.35 wt% C. Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0.35 wt% C iron-carbon alloy are considered in this problem. From Equation 10.20
Wp =
C0' 0.022 0.35 0.022 = = 0.44 0.74 0.74
And, from Equation 10.21 (for proeutectoid ferrite)
W' =
0.76 C0' 0.76 0.35 = = 0.56 0.74 0.74
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PHASE DIAGRAMS
PROBLEM SOLUTIONS
Development of Microstructure in Eutectic Alloys 10.16 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. Solution Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.
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10.17 Is it possible to have a magnesium–lead alloy in which the mass fractions of primary α and total α are 0.60 and 0.85, respectively, at 460°C (860°F)? Why or why not? Solution In order to make this determination we need to set up the appropriate lever rule expression for each of these quantities. From Figure 10.20 and at 460C, C = 41 wt% Pb, CMg Pb = 81 wt% Pb, and Ceutectic = 67 wt% Pb. 2
For primary
W' =
Ceutectic C0 Ceutectic C
=
67 C0 = 0.60 67 41
Solving for C0 gives C0 = 51.4 wt% Pb. Now the analogous expression for total
W =
CMg P b C0 2
CMg P b C 2
=
81 C0
81 41
= 0.85
which yields a value of 47 wt% Pb for C0. Therefore, since these two C0 values are different, this alloy is not possible.
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10.18 For a lead–tin alloy of composition 80 wt% Sn–20 wt% Pb and at 180°C (355°F) do the following: (a) Determine the mass fractions of α and β phases. (b) Determine the mass fractions of primary β and eutectic microconstituents. (c) Determine the mass fraction of eutectic β. Solution (a) This portion of the problem asks that we determine the mass fractions of and phases for an 80 wt% Sn-20 wt% Pb alloy (at 180C). In order to do this it is necessary to employ the lever rule using a tie line that extends entirely across the + phase field. From Figure 10.8 and at 180C, C = 18.3 wt% Sn, C = 97.8 wt% Sn, and Ceutectic = 61.9 wt% Sn. Therefore, the two lever-rule expressions are as follows:
W =
W =
C C0 97.8 80 = = 0.224 C C 97.8 18.3
C0 C C C
80 18.3 = 0.776 97.8 18.3
=
(b) Now it is necessary to determine the mass fractions of primary and eutectic microconstituents for this same alloy. This requires that we utilize the lever rule and a tie line that extends from the maximum solubility of Pb in the phase at 180C (i.e., 97.8 wt% Sn) to the eutectic composition (61.9 wt% Sn). Thus
W' =
We =
C0 Ceutectic C Ceutectic C C0 C Ceutectic
=
=
80.0 61.9 = 0.504 97.8 61.9
97.8 80.0 = 0.496 97.8 61.9
(c) And, finally, we are asked to compute the mass fraction of eutectic , We. This quantity is simply the difference between the mass fractions of total and primary as We = W – W' = 0.776 – 0.504 = 0.272
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10.21 For a 52 wt% Zn–48 wt% Cu alloy, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 950°C (1740°F), 860°C (1580°F), 800°C (1470°F), and 600°C (1100°F). Label all phases and indicate their approximate compositions. Solution The illustration below is the Cu-Zn phase diagram (Figure 10.19). A vertical line at a composition of 52 wt% Zn-48 wt% Cu has been drawn, and, in addition, horizontal arrows at the four temperatures called for in the problem statement (i.e., 950C, 860C, 800C, and 600C).
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:
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Congruent Phase Transformations Binary Eutectic Systems Equilibrium Diagrams Having Intermediate Phases or Compounds Eutectoid and Peritectic Reactions 10.24 Figure 10.40 is the tin–gold 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 tin-gold phase diagram (Figure 10.40).
There are two eutectics on this phase diagram. One exists at 10 wt% Au-90 wt% Sn and 217C. The reaction upon cooling is
L +
The other eutectic exists at 80 wt% Au-20 wt% Sn and 280C. This reaction upon cooling is
L +
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There are three peritectics. One exists at 30 wt% Au-70 wt% Sn and 252C. Its reaction upon cooling is as follows:
L +
The second peritectic exists at 45 wt% Au-55 wt% Sn and 309C. This reaction upon cooling is
L +
The third peritectic exists at 92 wt% Au-8 wt% Sn and 490C. This reaction upon cooling is
L +
There is one congruent melting point at 62.5 wt% Au-37.5 wt% Sn and 418C. Its reaction upon cooling is
L
No eutectoids are present.
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Ceramic Phase Diagrams 10.26 From Figure 10.24, the phase diagram for the MgO–Al2O3 system, it may be noted that the spinel solid solution exists over a range of compositions, which means that it is nonstoichiometric at compositions other than 50 mol% MgO–50 mol% Al2O3. (a) The maximum nonstoichiometry on the Al2O3-rich side of the spinel phase field exists at about 2000°C (3630°F) corresponding to approximately 82 mol% (92 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. (b) The maximum nonstoichiometry on the MgO-rich side of the spinel phase field exists at about 2000°C (3630°F) corresponding to approximately 39 mol% (62 wt%) Al 2O3. Determine the type of vacancy defect that is produced and the percentage of vacancies that exist at this composition. Solution (a) For this portion of the problem we are to determine the type of vacancy defect that is produced on the Al2O3-rich side of the spinel phase field (Figure 10.24) and the percentage of these vacancies at the maximum nonstoichiometry (82 mol% Al2O3). On the alumina-rich side of this phase field, there is an excess of Al3+ ions, which means that some of the Al3+ ions substitute for Mg2+ ions. In order to maintain charge neutrality, Mg2+ 3+
vacancies are formed, and for every Mg2+ vacancy formed, two Al ions substitute for three Mg2+ ions. Now, we will calculate the percentage of Mg2+ vacancies that exist at 82 mol% Al2O3. Let us arbitrarily choose as our basis 50 MgO-Al2O3 units of the stoichiometric material, which consists of 50 Mg2+ ions and 100 Al3+ ions. Furthermore, let us designate the number of Mg2+ vacancies as x, which means that 2x Al3+ ions have been added and 3x Mg2+ ions have been removed (two of which are filled with Al3+ ions). Using our 50 MgOAl2O3 unit basis, the number of moles of Al2O3 in the nonstoichiometric material is (100 + 2x)/2; similarly the number of moles of MgO is (50 – 3x). Thus, the expression for the mol% of Al2O3 is just
100 2x 2 mol% Al2O3 = 100 100 2x (50 3x) 2 If we solve for x when the mol% of Al2O3 = 82, then x = 12.1. Thus, adding 2x or (2)(12.1) = 24.2 Al3+ ions to the original material consisting of 100 Al3+ and 50 Mg2+ ions will produce 12.1 Mg2+ vacancies. Therefore, the percentage of vacancies is just
% vacancies =
12.1 100 = 8.1% 100 50
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(b) Now, we are asked to make the same determinations for the MgO-rich side of the spinel phase field, for 39 mol% Al2O3. In this case, Mg2+ ions are substituting for Al3+ ions. Since the Mg2+ ion has a lower charge than the Al3+ ion, in order to maintain charge neutrality, negative charges must be eliminated, which may be accomplished by introducing O2- vacancies. For every 2 Mg2+ ions that substitute for 2 Al3+ ions, one O2- vacancy is formed. Now, we will calculate the percentage of O2- vacancies that exist at 39 mol% Al2O3. Let us arbitrarily choose as our basis 50 MgO-Al2O3 units of the stoichiometric material which consists of 50 Mg2+ ions 100 Al3+ ions. Furthermore, let us designate the number of O2- vacancies as y, which means that 2y Mg2+ ions have been added and 2y Al3+ ions have been removed. Using our 50 MgO-Al2O3 unit basis, the number of moles of Al2O3 in the nonstoichiometric material is (100 – 2y)/2; similarly the number of moles of MgO is (50 + 2y). Thus, the expression for the mol% of Al2O3 is just
100 2 y 2 mol% Al2O3 = 100 100 2 y (50 2 y) 2 If we solve for y when the mol% of Al2O3 = 39, then y = 7.91. Thus, 7.91 O2- vacancies are produced in the original material that had 200 O2- ions. Therefore, the percentage of vacancies is just
% vacancies =
7.91 100 = 3.96% 200
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10.29 Consider 3.5 kg of austenite containing 0.95 wt% C, cooled to below 727°C (1341°F). (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 0.95 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 =
CFe C C0 3
CFe C C 3
=
6.70 0.95 = 0.86 6.70 0.022
which, when multiplied by the total mass of the alloy, gives (0.86)(3.5 kg) = 3.01 kg of total ferrite. Similarly, for total cementite,
C0 C 0.95 0.022 WFe C = = = 0.14 3 CFe C C 6.70 0.022 3
And the mass of total cementite that forms is (0.14)(3.5 kg) = 0.49 kg. (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. Applying Equation 10.22, in which C1' = 0.95 wt% C
Wp =
6.70 C 1' 6.70 0.95 = = 0.97 6.70 0.76 6.70 0.76
which corresponds to a mass of (0.97)(3.5 kg) = 3.4 kg. Likewise, from Equation 10.23
WFe C' = 3
C1' 0.76 5.94
=
0.95 0.76 = 0.03 5.94
which is equivalent to (0.03)(3.5 kg) = 0.11 kg of the total 3.5 kg mass. (d) Schematically, the microstructure would appear as:
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10.30 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy containing 0.35 wt% C. Solution The mass fractions of proeutectoid ferrite and pearlite that form in a 0.35 wt% C iron-carbon alloy are considered in this problem. From Equation 10.20
Wp =
C0' 0.022 0.35 0.022 = = 0.44 0.74 0.74
And, from Equation 10.21 (for proeutectoid ferrite)
W' =
0.76 C0' 0.76 0.35 = = 0.56 0.74 0.74
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