1、材料科学基础课后习题答案7 SOLUTION FOR CHAPTER 71. FIND: Label the phase fields in Figure HP7-1.SOLUTION: There are two regions of single phase equilibrium separated by a region of two-phase equilibrium. The line separating the single phase liquid from the two phase liquid and solid region is called the liquidu
2、s, and the line separating the two phase liquid and solid from the single phase solid is called the solidus.SKETCH:2. FIND: The name given to the type of equilibrium diagram shown in Fig. HP7-1.SOLUTION: The diagram is termed isomorphous. Above the liquidus there is only one phase, namely the liquid
3、, and it has the same structure regardless of composition. Below the solidus is the solid phase, and as well it has the same structure regardless of composition.3. FIND: What is the crystal structure of component B, why?SOLUTION: There are no phase boundaries below the solidus consequently the phase
4、 and therefore the structure must be the same. If A is FCC, then B and all compositions between A and B must be the same phase and hence same structure. If A is FCC, then B must also be FCC.4. FIND: Sketch equilibrium cooling curves for alloy Xo and pure component B. Explain why they have different
5、shapes.SOLUTION: The slope of the temperature versus time behavior for the alloy in the single phase region is controlled by the cooling rate and the heat transferred from the liquid. At the liquidus temperature a small amount of solid is formed, releasing an amount of latent heat of fusion related
6、to the volume of solid transformed. Hence, at the liquidus temperature there will be a change in slope. Since heat is being generated the slope will be less than that above the liquidus. As the temperature is reduced through the two-phase region a small amount of solid is formed and a corresponding
7、heat released; once the solidus temperature is reached no additional transformation takes place and there is again a change in slope. The slope is greater as the solid cools than as the liquid and solid cooled. For the pure component, solid and liquid are in equilibrium at TB, and hence, a horizonta
8、l line when liquid transforms to solid at TB.SKETCH:5. FIND: The liquidus temperature and the solidus temperature of alloy XoSOLUTION: From figure HP7-1 the liquidus temperature is approximately 1110oC and the solidus temperature is approximately 1070oC.6. FIND: Determine compositions and phase frac
9、tions of each phase in equilibrium at 1100oC for alloy Xo.SOLUTION: At equilibrium, the temperature of two phases must be the same, and the composition of the solid is found where the tie line intersects the solidus and the liquid is at the intersection of the tie line with the liquidus. From the ph
10、ase diagram the composition of solid is 0.35B and that of the liquid is 0.55B. Using the lever rule the fraction of liquid, fL, and fraction of solid, fs is determined.7. FIND: Sketch fL and fS for alloy Xo as the alloy is cooled under equilibrium conditions from 1200oC to room temperature.SOLUTION:
11、 At the liquidus temperature the amount of liquid is almost 100% with only a very small amount of solid. Conversely at the solidus temperature, the amount of solid is almost 100% with only a very small amount of liquid. In a two-phase field: fL + fs = 1.8. FIND: Changes in the compositions of the li
12、quid and solid phases during quilibrium cooling of alloy Xo through the two-phase field.SOLUTION: At the liquidus temperature, the composition of the solid is approximately 0.30B. As equilibrium solidification progresses, the composition of the solid increases to 0.5 B, the maximum value it can reac
13、h. The liquid on the other hand is initially the composition of the alloy Xo, and as equilibrium solidification progresses the composition increases in B to the maximum composition in the liquid of approximately 0.7 B.9. FIND:From the following data construct a plausible equilibrium phase diagram. C
14、omponent A melts at 800C and B melts at 1000C; A and B are completely soluble in one another at room temperature; and if solid containing 0.3 B is heated under equilibrium conditions, the solid transforms to liquid having the same composition at 500C.GIVEN: Melting temperatures for the two component
15、s, congruent melting temperature and composition.SOLUTION: The sketch shown below satisfies all of the requirements stated in the problem. The melting temperature of pure A is 800C and that for pure B is 1000C. The congruent melting temperature is at 500C at a composition containing 0.3 B. Below the
16、 congruent melting temperature there is a continuous solid solution. 10. FIND: For alloys containing 10, 22, 25, 27, and 40 wt% Ni, determine the number of phases present and the composition of the phases in equilibrium at 1200oC.SOLUTION:Alloy composition in % NiNumber of phases in equilibriumCompo
17、sition ofPhases (% Ni)101 (Liquid)10%221 (Liquid)22%252 (Liquid + Solid)liquid:22%, solid: 32%272 (Liquid + Solid)liquid: 22%, solid: 32%401 (Solid)40%11. FIND: Beginning with a statement of mass balance, derive the lever rule in a two-phase system.SOLUTION: In a two phase field for an alloy of some
18、 over all composition Xo, the solute is distributed in the two phases: xo = f x + f x where the compositions are expressed in terms of one of the components, f + f = 1. Then, f = 1 - f. Substituting: xo = (1 - f) x + f x xo = x - x f + xf xo - x = (x - x) f12. FIND: Discuss each of the factors that
19、permit the Cu-Ni system to be isomorphous over the temperature range 350-1000oC.SOLUTION: The empirical rules of Hume-Rothery identify the characteristics that two elements must have in common for extensive solubility. This should require thata. the two components must have the same crystal structur
20、eb. the atomic radii of the two atoms must be similar c. the two components have the comparable electro negativities, andd. the two components have the similar valence.13. FIND: What does the temperature 322OC represent in Figure HP7-2?SOLUTION: 322oC is called the critical temperature. At the criti
21、cal temperature there is a corresponding critical composition. For an alloy of this composition, cooling under equilibrium conditions from above the critical temperature to below this temperature results in the formation of two phases from one phase of the critical composition. As cooling continues
22、the two phases that form have different compositions.14. FIND: In Figure HP7-2, are 1 and 2 different crystal structures?SOLUTION: 1 and 2 are two phases having the same structure but different compositions. The composition of the two phases are determined as in any two phase system by using the tie
23、-line. Where the tie-line at the equilibrium temperature intersects the phase boundaries determines the composition of the two phase in equilibrium.15. FIND: Location of the equilibrium phase boundary at temperature T.GIVEN: Alloy 1 containing 30%B and alloy 2 containing 50% B, when equilibrated at
24、temperature T are in the same two-phase (L + S) region. The fraction of liquid in alloy 1 is 0.8 and the fraction of liquid in alloy 2 is 0.4. SKETCH:SOLUTION: Since the fraction of liquid in alloy 1 is greater than that of solid, the liquidus is to the left of 0.3, and since the fraction of solid i
25、n alloy 2 is greater than that of liquid, the solidus must lie to the right of alloy 2. Using the lever rule:for alloy 1for alloy 2 0.8XL - 0.8Xs = Xs - 0.3 0.4XL - 0.4Xs = Xs - 0.5Solve for Xs and XL, 0.8Xs - 0.8XL = Xs - 0.3 0.8Xs - 0.8XL = 2Xs - 1.0 _ 0 = -Xs + 0.7, or Xs = 0.7BTo find XL, substi
26、tute Xs into one of the equations: 0.4 - 0.56 = -0.8XL -0.16 = -0.8XL, or XL = 0.2BAs a check, use the other equation to calculate the fraction of liquid from the compositions.16. FIND: Label all regions of the phase diagram and the boundaries of monovariant equilibrium for the diagram shown in Figu
27、re HP7-3.SKETCH/SOLUTION:17. FIND: Sketch an equilibrium cooling curve from above the eutectic to room temperature for an alloy of eutectic composition.SKETCH/SOLUTION:18. FIND: Explain why the equilibrium cooling curves for alloys on either side of the eutectic composition will be different than th
28、e equilibrium cooling curve for a eutectic alloy.SOLUTION: At the eutectic temperature, liquid of composition XE is in equilibrium with two solids, X and X . For a hypoeutectic alloy (composition to the left of the eutectic), the first phase to form at the liquidus temperature of the alloy is . When
29、 the eutectic temperature is reached the liquid of the eutectic composition is in equilibrium with the two solids, one of composition X and the other X. Consequently, depending upon composition, the closer the overall composition of the alloy is to XE, the less proeutectic and the more eutectic. Sim
30、ilarly, for alloys to the right, proeutectic will form. In terms of the phase rule at constant pressure, for the eutectic reaction F = C - P + 1 = 2 - 3 + 1 = 0. The eutectic is invariant, and solidifies under equilibrium conditions at one temperature, TE. For alloys on either side of the eutectic c
31、omposition but between X and X, the proeutectic phase forms and cooling occurs as it does in any two-phase s-l region. Once the eutectic isotherm is reached, the remaining liquid of eutectic composition solidifies isothermally.19. FIND: The maximum solid solubility of B in A and of A in B in Figure
32、HP7-3.SOLUTION: The maximum solubility of B in A occurs at 0.15 B and the maximum solubility of A in B is 0.1 A.20. FIND: For an alloy of eutectic composition in Figure HP7-3, determine the composition of the solid phases in equilibrium with the liquid.SOLUTION: The composition of the two solid phase and in equilibrium with liquid at T_ occurs at 0.15 B and 0.9 B.21. FIND:
