Chap 9 Solns
CHAPTER 10
PHASE DIAGRAMS
PROBLEM SOLUTIONS
Solubility Limit 10.1 Consider the sugar–water phase diagram of Figure 10.1. (a) How much sugar will dissolve in 1000 g of water at 80°C (176°F)? (b) If the saturated liquid solution in part (a) is cooled to 20°C (68°F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20°C? (c) How much of the solid sugar will come out of solution upon cooling to 20°C? Solution (a) We are asked to determine how much sugar will dissolve in 1000 g of water at 80C. From the solubility limit curve in Figure 10.1, at 80C the maximum concentration of sugar in the syrup is about 74 wt%. It is now possible to calculate the mass of sugar using Equation 5.6 as
Csugar(wt%) =
74 wt% =
msugar msugar mwater msugar
msugar 1000 g
100
100
Solving for msugar yields msugar = 2846 g (b) Again using this same plot, at 20C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. (c) The mass of sugar in this saturated solution at 20C (m' sugar ) may also be calculated using Equation 5.6 as follows:
64 wt% =
mÕsugar
mÕsugar 1000 g
100
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which yields a value for m' sugar of 1778 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that precipitated out of the solution upon cooling m"sugar; that is
m"sugar = msugar m' sugar = 2846 g 1778 g = 1068 g
One-Component (or Unary) Phase Diagrams
10.3 Consider a specimen of ice that is at –15°C and 10 atm pressure. Using Figure 10.2, 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 -15C, and the location on this line at 10 atm pressure (point B) is also noted.
(a) Melting occurs, (by changing pressure) as, moving vertically (upward) along this line, we cross the Solid-Liquid phase boundary. This occurs at approximately 1,000 atm; thus, the pressure of the specimen must be raised from 10 to 1,000 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, we move vertically downward along this line from 10 atm until we cross the Solid-Vapor phase boundary. This intersection occurs at approximately 0.003 atm.
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Interpretation of Phase Diagrams (Binary Isomorphous Systems) (Binary Eutectic Systems) (Equilibrium Diagrams Having Intermediate Phases or Compounds) 10.5 Cite the phases that are present and the phase compositions for the following alloys: (a) 25 wt% Pb–75 wt% Mg at 425°C (800°F) (b) 55 wt% Zn–45 wt% Cu at 600°C (1110°F) (c) 7.6 lbm Cu and 144.4 lbm Zn at 600°C (1110°F) (d) 4.2 mol Cu and 1.1 mol Ag at 900°C (1650°F) Solution (a) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, from Figure 10.20, only the phase is present; its composition is 25 wt% Pb-75 wt% Mg. (b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, from Figure 10.19, and phases are present, and C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu (c) For an alloy composed of 7.6 lb m Cu and 144.4 lbm Zn and at 600C, we must first determine the Cu and Zn concentrations (using Equation 5.6), as
CCu =
7.6 lbm 100 = 5.0 wt% 7.6 lbm 144.4 lbm
CZn =
144.4 lbm 100 = 95.0 wt% 7.6 lbm 144.4 lbm
From Figure 10.19, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu (d) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900C, it is necessary to determine the Cu and Ag concentrations in weight percent. However, we must first compute the masses of Cu and Ag (in grams) using a rearranged form of Equation 5.7:
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' = n mCu mCu 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
Now, using Equation 5.6, concentrations of Cu and Ag are determined as follows:
CCu =
266.9 g 100 = 69.2 wt% 266.9 g 118.7 g
CAg =
118.7 g 100 = 30.8 wt% 266.9 g 118.7 g
From Figure 10.7, and liquid phases are present; and C = 8 wt% Ag-92 w% Cu CL = 45 wt% Ag-55 wt% Cu
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10.7 A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400°C (2550°F) to 1200°C (2190°F). (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? (c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? Solution Shown below is the Cu-Ni phase diagram (Figure 10.3a) and a vertical line constructed at a composition of 50 wt% Ni-50 wt% Cu.
(a) Upon cooling form 1400C, the first solid phase forms at the temperature at which this vertical line intersects the L–( + L) phase boundary--i.e., at about 1320C. (b) The composition of this solid phase corresponds to the intersection with the L–( + L) phase boundary, of a tie line constructed across the + L phase region at 1320C--i.e., C = 62 wt% Ni-38 wt% Cu. (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Ni with the (+ L)– phase boundary--i.e., at about 1270C.
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(d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the intersection with the L–(+ L) boundary, of the tie line constructed across the + L phase region at 1270C--i.e., CL is about 37 wt% Ni-63 wt% Cu.
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10.8
Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and
temperatures given in Problem 10.5. Solution (a) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, only the phase is present; therefore W = 1.0. (b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, compositions of the and phases are C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu And, since the composition of the alloy, C0 = 55 wt% Zn-45 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent zinc
W =
W =
C C0 58 55 = = 0.43 C C 58 51 C0 C C C
=
55 51 = 0.57 58 51
(c) For an alloy composed of 7.6 lbm Cu and 144.4 lbm Zn (95.0 wt% Zn-5.0 wt% Cu) and at 600C, only the liquid phase is present; therefore, WL = 1.0
(d) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag (30.8 wt% Ag-69.2 wt% Cu) and at 900C, compositions of the and liquid phases are C = 8 wt% Ag-92 w% Cu CL = 45 wt% Ag-55 wt% Cu And, since the composition of the alloy, C0 = 30.8 wt% Ag-69.2 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent silver
W =
CL C0 45 30.8 = = 0.38 CL C 45 8
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WL =
C0 C CL C
=
30.8 8 = 0.62 45 8
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10.10 A 40 wt% Pb–60 wt% Mg alloy is heated to a temperature within the α + liquid-phase region. If the mass fraction of each phase is 0.5, then estimate: (a) The temperature of the alloy (b) The compositions of the two phases Solution (a) We are given that the mass fractions of and liquid phases are both 0.5 for a 40 wt% Pb-60 wt% Mg alloy and are asked to estimate the temperature of the alloy. Using the appropriate phase diagram, Figure 10.20, by trial and error with a ruler, a tie line within the + L phase region that is divided in half for an alloy of this composition exists at about 540C. (b) We are now asked to determine the compositions of the two phases. This is accomplished by noting the intersections of this tie line with both the solidus and liquidus lines. From these intersections, C = 26 wt% Pb, and CL = 54 wt% Pb.
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10.12 Is it possible to have a copper–silver alloy of composition 20 wt% Ag–80 wt% Cu that, at equilibrium, consists of α and liquid phases having mass fractions Wα = 0.80 and WL = 0.20? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why. Solution Yes, it is possible to have a Cu-Ag alloy of composition 20 wt% Ag-80 wt% Cu which consists of mass fractions W = 0.80 and WL = 0.20. Using the appropriate phase diagram, Figure 10.7, by trial and error with a ruler, the tie-line segments within the + L phase region are proportioned such that
W = 0.8
CL C 0 CL C
for C0 = 20 wt% Ag. This occurs at about 800C.
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10.14 Determine the relative amounts (in terms of volume fractions) of the phases for the alloys and temperatures given in Problems 10.5a and b. Given here are the approximate densities of the various metals at the alloy temperatures:
Metal
Temperature (°C)
Density (g/cm3)
Cu
600
8.68
Mg
425
1.68
Pb
425
10.96
Zn
600
6.67
Solution This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problems 10.5a and b. This is accomplished by using the technique illustrated in Example Problem 10.3, and also the results of Problems 10.5 and 10.8. (a) This is a Pb-Mg alloy at 425C, wherein only the phase is present. Therefore, V = 1.0.
(b) This is a Zn-Cu alloy at 600C, wherein C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu W = 0.43 W = 0.57 3
Zn = 6.67 g/cm
3
Cu = 8.68 g/cm
Using these data it is first necessary to compute the densities of the and phases using Equation 5.13a. Thus
=
100 CZn () Zn
CCu() Cu
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100
=
51
6.67 g /cm 3
=
= 7.52 g/cm3
49 8.68 g /cm 3
100 CZn () Zn
CCu() Cu
100
=
58 6.67 g /cm 3
= 7.39 g/cm3
42 8.68 g /cm 3
Now we may determine the V and V values using Equation 10.6. Thus,
W V =
W
W
0.43
=
7.52 g /cm 3 = 0.43 0.43 0.57 7.52 g /cm 3 7.39 g /cm 3
W V =
W
W
0.57 =
7.39 g /cm 3 = 0.57 0.43 0.57 7.52 g /cm 3 7.39 g /cm 3
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PHASE DIAGRAMS
PROBLEM SOLUTIONS
Solubility Limit 10.1 Consider the sugar–water phase diagram of Figure 10.1. (a) How much sugar will dissolve in 1000 g of water at 80°C (176°F)? (b) If the saturated liquid solution in part (a) is cooled to 20°C (68°F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20°C? (c) How much of the solid sugar will come out of solution upon cooling to 20°C? Solution (a) We are asked to determine how much sugar will dissolve in 1000 g of water at 80C. From the solubility limit curve in Figure 10.1, at 80C the maximum concentration of sugar in the syrup is about 74 wt%. It is now possible to calculate the mass of sugar using Equation 5.6 as
Csugar(wt%) =
74 wt% =
msugar msugar mwater msugar
msugar 1000 g
100
100
Solving for msugar yields msugar = 2846 g (b) Again using this same plot, at 20C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. (c) The mass of sugar in this saturated solution at 20C (m' sugar ) may also be calculated using Equation 5.6 as follows:
64 wt% =
mÕsugar
mÕsugar 1000 g
100
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which yields a value for m' sugar of 1778 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that precipitated out of the solution upon cooling m"sugar; that is
m"sugar = msugar m' sugar = 2846 g 1778 g = 1068 g
One-Component (or Unary) Phase Diagrams
10.3 Consider a specimen of ice that is at –15°C and 10 atm pressure. Using Figure 10.2, 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 -15C, and the location on this line at 10 atm pressure (point B) is also noted.
(a) Melting occurs, (by changing pressure) as, moving vertically (upward) along this line, we cross the Solid-Liquid phase boundary. This occurs at approximately 1,000 atm; thus, the pressure of the specimen must be raised from 10 to 1,000 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, we move vertically downward along this line from 10 atm until we cross the Solid-Vapor phase boundary. This intersection occurs at approximately 0.003 atm.
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Interpretation of Phase Diagrams (Binary Isomorphous Systems) (Binary Eutectic Systems) (Equilibrium Diagrams Having Intermediate Phases or Compounds) 10.5 Cite the phases that are present and the phase compositions for the following alloys: (a) 25 wt% Pb–75 wt% Mg at 425°C (800°F) (b) 55 wt% Zn–45 wt% Cu at 600°C (1110°F) (c) 7.6 lbm Cu and 144.4 lbm Zn at 600°C (1110°F) (d) 4.2 mol Cu and 1.1 mol Ag at 900°C (1650°F) Solution (a) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, from Figure 10.20, only the phase is present; its composition is 25 wt% Pb-75 wt% Mg. (b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, from Figure 10.19, and phases are present, and C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu (c) For an alloy composed of 7.6 lb m Cu and 144.4 lbm Zn and at 600C, we must first determine the Cu and Zn concentrations (using Equation 5.6), as
CCu =
7.6 lbm 100 = 5.0 wt% 7.6 lbm 144.4 lbm
CZn =
144.4 lbm 100 = 95.0 wt% 7.6 lbm 144.4 lbm
From Figure 10.19, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu (d) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900C, it is necessary to determine the Cu and Ag concentrations in weight percent. However, we must first compute the masses of Cu and Ag (in grams) using a rearranged form of Equation 5.7:
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' = n mCu mCu 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
Now, using Equation 5.6, concentrations of Cu and Ag are determined as follows:
CCu =
266.9 g 100 = 69.2 wt% 266.9 g 118.7 g
CAg =
118.7 g 100 = 30.8 wt% 266.9 g 118.7 g
From Figure 10.7, and liquid phases are present; and C = 8 wt% Ag-92 w% Cu CL = 45 wt% Ag-55 wt% Cu
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10.7 A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400°C (2550°F) to 1200°C (2190°F). (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? (c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? Solution Shown below is the Cu-Ni phase diagram (Figure 10.3a) and a vertical line constructed at a composition of 50 wt% Ni-50 wt% Cu.
(a) Upon cooling form 1400C, the first solid phase forms at the temperature at which this vertical line intersects the L–( + L) phase boundary--i.e., at about 1320C. (b) The composition of this solid phase corresponds to the intersection with the L–( + L) phase boundary, of a tie line constructed across the + L phase region at 1320C--i.e., C = 62 wt% Ni-38 wt% Cu. (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Ni with the (+ L)– phase boundary--i.e., at about 1270C.
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(d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the intersection with the L–(+ L) boundary, of the tie line constructed across the + L phase region at 1270C--i.e., CL is about 37 wt% Ni-63 wt% Cu.
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10.8
Determine the relative amounts (in terms of mass fractions) of the phases for the alloys and
temperatures given in Problem 10.5. Solution (a) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, only the phase is present; therefore W = 1.0. (b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, compositions of the and phases are C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu And, since the composition of the alloy, C0 = 55 wt% Zn-45 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent zinc
W =
W =
C C0 58 55 = = 0.43 C C 58 51 C0 C C C
=
55 51 = 0.57 58 51
(c) For an alloy composed of 7.6 lbm Cu and 144.4 lbm Zn (95.0 wt% Zn-5.0 wt% Cu) and at 600C, only the liquid phase is present; therefore, WL = 1.0
(d) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag (30.8 wt% Ag-69.2 wt% Cu) and at 900C, compositions of the and liquid phases are C = 8 wt% Ag-92 w% Cu CL = 45 wt% Ag-55 wt% Cu And, since the composition of the alloy, C0 = 30.8 wt% Ag-69.2 wt% Cu, then, using the appropriate lever rule expressions and taking compositions in weight percent silver
W =
CL C0 45 30.8 = = 0.38 CL C 45 8
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WL =
C0 C CL C
=
30.8 8 = 0.62 45 8
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10.10 A 40 wt% Pb–60 wt% Mg alloy is heated to a temperature within the α + liquid-phase region. If the mass fraction of each phase is 0.5, then estimate: (a) The temperature of the alloy (b) The compositions of the two phases Solution (a) We are given that the mass fractions of and liquid phases are both 0.5 for a 40 wt% Pb-60 wt% Mg alloy and are asked to estimate the temperature of the alloy. Using the appropriate phase diagram, Figure 10.20, by trial and error with a ruler, a tie line within the + L phase region that is divided in half for an alloy of this composition exists at about 540C. (b) We are now asked to determine the compositions of the two phases. This is accomplished by noting the intersections of this tie line with both the solidus and liquidus lines. From these intersections, C = 26 wt% Pb, and CL = 54 wt% Pb.
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10.12 Is it possible to have a copper–silver alloy of composition 20 wt% Ag–80 wt% Cu that, at equilibrium, consists of α and liquid phases having mass fractions Wα = 0.80 and WL = 0.20? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why. Solution Yes, it is possible to have a Cu-Ag alloy of composition 20 wt% Ag-80 wt% Cu which consists of mass fractions W = 0.80 and WL = 0.20. Using the appropriate phase diagram, Figure 10.7, by trial and error with a ruler, the tie-line segments within the + L phase region are proportioned such that
W = 0.8
CL C 0 CL C
for C0 = 20 wt% Ag. This occurs at about 800C.
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10.14 Determine the relative amounts (in terms of volume fractions) of the phases for the alloys and temperatures given in Problems 10.5a and b. Given here are the approximate densities of the various metals at the alloy temperatures:
Metal
Temperature (°C)
Density (g/cm3)
Cu
600
8.68
Mg
425
1.68
Pb
425
10.96
Zn
600
6.67
Solution This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problems 10.5a and b. This is accomplished by using the technique illustrated in Example Problem 10.3, and also the results of Problems 10.5 and 10.8. (a) This is a Pb-Mg alloy at 425C, wherein only the phase is present. Therefore, V = 1.0.
(b) This is a Zn-Cu alloy at 600C, wherein C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu W = 0.43 W = 0.57 3
Zn = 6.67 g/cm
3
Cu = 8.68 g/cm
Using these data it is first necessary to compute the densities of the and phases using Equation 5.13a. Thus
=
100 CZn () Zn
CCu() Cu
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100
=
51
6.67 g /cm 3
=
= 7.52 g/cm3
49 8.68 g /cm 3
100 CZn () Zn
CCu() Cu
100
=
58 6.67 g /cm 3
= 7.39 g/cm3
42 8.68 g /cm 3
Now we may determine the V and V values using Equation 10.6. Thus,
W V =
W
W
0.43
=
7.52 g /cm 3 = 0.43 0.43 0.57 7.52 g /cm 3 7.39 g /cm 3
W V =
W
W
0.57 =
7.39 g /cm 3 = 0.57 0.43 0.57 7.52 g /cm 3 7.39 g /cm 3
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