1、材料科学基础课后习题答案8 SOLUTIONS FOR CHAPTER 81. FIND: When a phase transformation occurs such as a liquid phase transforming to a solid below its melting temperature, what are the two steps involved in the process? Briefly describe each.SOLUTION: During a phase transformation such as a liquid transforming t
2、o solid, there are two steps involved in the process. They are:1. nucleation of the new phase, and2. growth of the phase.Nucleation relates to the formation of the new phase and the development of the interface seperating the two phases. Nucleation can either occur randomly throughout the structure
3、- termed homogeneous nucleation or at specific sites such as interfaces - termed heterogeneous nucleation.Growth - Once the phases has nucleated, it begins to grow. The growth process is controlled by diffusion and undercooling. As in the nucleation step, there may be competing processes that lead t
4、o a maximum growth rate at an intermediate temperature.2. FIND: We presented a derivation in Section 8.2.3 showing that the barrier for nucleation, G*, decreases with increasing undercooling following the proportionalityBy starting with an expression for the free energy of a distribution of spherica
5、l particles of radius r, derive equation 8.2-9a. Explain each step in the derivation. Explain any assumptions that are made.SOLUTION: To determine the barrier to the nucleation process, G* we begin by noting that the free energy as a function of particle size for homogeneous nucleation has two terms
6、, one that increases with r2, the interfacial energy per unit volume term, and one that decreases with r3. A maximum occurs in G(r) at some r*. These graphical relationships are sketched below. The dependence of the various free energy terms associated with nucleation as a function of temperature: (
7、a) the relationship between cluster radius and surface energy of a growing spherical solid phase in liquid, (b) the relationship between the cluster radius and (c) the sum of (a) and (b). The change in free energy can be written as:In this equation we assume that the nuclei can be considered as a ra
8、ndom distribution of spheres. To locate the maximum of a function we equate the first derivative of the function with respect to the parameter which is the variable to zero. Here we assume that r is the only variable. The is SL is independent of size and orentation.Using equation 8.2-4 for we have:I
9、n writingwe have assumed that the heat capacity difference between the liquid and solid phases is zero. (Note: Although this may be a reasonable assumption at small undercoolings, i.e. small Ts, at the large undercoolings that are typical for homogeneous nucleation that approximation may not be vali
10、d and a more complex term is required. But for a first order approximation this assumption is reasonable.)In order to determine the value of G(r) at r*, we introduce the expressionintoRearranging,If all the terms in parentheses are constant then,3. FIND: Explain the simultaneous influence that under
11、cooling has on the barrier to nucleation and the atomic rearrangements necessary to initiate the transformation. Show how these competing effects lead to classical C-curve behavior in the nucleation of diffusional transformations.SOLUTION: With larger undercoolings, both r* and G* decrease, suggesti
12、ng that simply lowering the temperature of the system allows nucleation to occur ever more readily. However, there are practical kinetic limits to this effect. For example, with decreased temperature there is a corresponding reduction in atomic mobility. The random fluctuations in the local arrangem
13、ents of atoms is the process that provides the clusters. Since the formation of the clusters depends on atomic mobility, a reduction in the temperature reduces the rate of clustering. Thus, as shown in Figure (a) below, the overall nucleation rate exhibits a maximum at an intermediate temperature. T
14、he maximum in the nucleation rate leads to a minimum in the time required to nucleate a phase, as shown in Figure b. Because of its shape, this curve is known as a C-curve.(a) The influence of temperature on the mobility term and the nucleation barrier term. The opposing processes result in a maximu
15、m in the nucleation rate at an intermediate temperature. (b) Since the time for nucleation is inversely related to the nucleation rate, the time curve exhibits a minimum at an intermediate temperature. Because of its shape, this curve is often referred to as a C-curve.4. FIND: Explain how the value
16、of interfacial energy between the parent phase and the transforming phase affects the critical radius and the barrier to nucleation.SOLUTION: Equation 8.25 gives the change in free energy as a function of r when a liquid transforms to a solid, for example. In the development of that equation it was
17、assumed that the transforming phase was spherical and the interfacial energy, SL, was isotropic. That equation consists of two terms on the right hand side, i.e. G(r) = (4r2) SL + 4/3 r3 (Gv)Since SL 0 and GV 0, the first term increases with radius, and the second decreases. Figure 8.2-3 illustrates
18、 that there is a maximum that occurs at some r we designated as r* and a corresponding G, we designated as G* where r* = (-2SL)/Gvand Both the critical radius, r*, and the barrier to the nucleation process contain SL in the numerator. Thinking in terms of the barrier to nucleation, G*, there is a cu
19、bic dependence on interfacial energy. The larger the S-L interfacial energy, the larger the barrier to nucleation and hence the more difficult. The use of nucleating agents is based on the principle of introducing particles with lower interfacial energies to stimulate nucleation.5. FIND: Compare hom
20、ogeneous nucleation and heterogeneous nucleation.SOLUTION: The process of homogeneous nucleation occurs at random locations in the parent phase. The distribution of the transforming phase occurs without regard to specific sites, such as mold walls in solidification. Heterogeneous nucleation occurs a
21、t specific sites. In the case of solidification they can be at mold walls, unintentional additions such as ceramic inclusions from crucibles or it may occur at nucleating agents which are intentionally added to control the solidification microstructure.6. FIND: What is the difference between the fol
22、lowing interfaces?a. coherentb. partially coherent, andc. incoherentSOLUTION: A coherent interface is one in which there is a one-to-one correspondence of atomic planes across the interface. This type of interface occurs when the lattice parameters in the two phases are the same or very close. When
23、the lattice parameters are different in the two phases, the increase in strain energy that will occur as the particle grows necessitates the periodic insertion of dislocations at the interface to accommodate the misfit. This type of interface is referred to as partially coherent. An incoherent inter
24、face occurs between two phases of different crystal structures and sufficiently different atomic spacings that can not be accommodated by dislocations. 7. FIND: How does interfacial energy vary with coherency?SOLUTION: Interfacial energy is sensitive to the nature of the interface separating the two
25、 phases. The interfacial energy increases going from coherent to partially coherent to incoherent, i.e. incoh. part.coh. . coh.8. FIND: Based upon your answer to question 7, explain how the probability for heterogeneous nucleation changes as the type of interface changes from coherent to partially c
26、oherent to incoherent.SOLUTION: Since the barrier to nucleation, G*, is related to / in the following way: G* /3increasing / would increase the barrier to homogeneous nucleation. Consequently, the probability for heterogeneous nucleation would increase as interfacial energy increases.9. FIND: Figure
27、 5.3-5 contains a schematic illustration of a twin boundary in a crystal. From the point of view of coherency, what is the nature of the type of twin boundary illustrated in the figure? Comment on the relative energy of a twin boundary compared with a random grain boundary in a polycrystalline mater
28、ial.GIVEN: Figure 5.3-5 illustrates a schematic of a twin in a matrix showing the two twin boundaries separating the matrix from the twin and Figure 8.2-10 illustrates an incoherent interface separating the matrix from a precipitate. Schematic of a twinSchematic of an incoherent boundarySOLUTION: At
29、oms that are on the twin plane are part of the stacking sequence in the matrix above the twin plane as well as the stacking of atoms below the twin plane. Since a coherent interface is an interface that occurs when there is a one-to-one correspondence across the interface, then the twin illustrated
30、in this figure would be classified as a coherent twin boundary. The incoherent boundary illustrated above occurs in a system when there is not a match across the boundary separating two phases. Since a general grain boundary represents a situation where the orientation of two grains across a boundar
31、y are not the same, we would therefore expect that there would not be a match of atoms across the boundary. Thus, a coherent twin plane would have lower interfacial energy than the interfacial energy associated with grain boundary separating two randomly oriented grains.10. FIND: In certain nickel-b
32、ase superalloys, a second phase can precipitate coherently from the matrix during aging because the lattice parameters of the two phases are very close and both phases are cubic. For a coherent precipitate in this system, what is the most likely relationship between the crystallographic axes in the matrix phase and that of the precipitate? Explain, using sketches.GIVEN: The matrix and precipitate are both cubic with similar lattice parameters.SOLUTION: The best match
