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Unformatted text preview: FIND: When a phase transfonnation 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 to solid, there
are two steps involved in the process. They are: 1. nucleation of the new phase, and
2. growth of the phase. Nucleation relates to the formation of the new phase and the development of the
interface separating the two phases. Nucleation can either occur randomly throughout the structure  ter'med homogeneous automation or at speciﬁc sites such as interfaces  termed
heterogeneous nucIearion. Growth  Once the phases has nucleated, it begins to grow. The growth process is
controlled by diffusion and undercooh'ng. As in the nucleation step, there my be
competing processes that lead to a maximum growth rate at an intermediate temperature. FIND: We presented a derivation in Section 8.2.3 showing that the barrier for
nucleation, AG“, decreases with increasing undercooiing following the proportionality 1
AG* “ D By starting with an expression for the free energy of a distribution of spherical particles
of radius r, derive equation 8.29a. Explain each step in the derivation. Explain any
assumptions that are made. SOLUTION: To determine the barrier to the nucleation process, AG‘ we begin by
noting that the free energy as a function of particle size for homogeneous nucleation has
two terms, one that increases with r“, the interfaeiai energy per unit volume term, and one
that decreases with r’. A maximum occurs in Athr) at some r‘. These graphical
relationships are sketched below. .36 all?
lulcﬂ‘ueiul energy 0: r: fin The dependence of the various free energy terms associated with nucleation as a function
of temperature: (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). as?
The change in free energy can be written as: AG(r) = (4mg) yﬂ+[%rtr3] :16“ In this equation we assume that the nuclei can he considered as a random distribution of
spheres. To locate the maximmn of a ﬁanction we equate the ﬁrst derivative of the function with respect to the parameter which is the variable to zem. Here we assume that
r is the only variable. The is 15L is indep'eudent'of size and orentation. (JACK?) _ __ 2 1*:
d, If u. .. D  Surva+41tr AG”
0?
r1 : inn. AGll'
:1 Using equation 8.24 for 11:35“ we have: r" = 927‘“) TEKAHf' on Inwriting
AGf' = AHHIT E)ﬁi" we 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 undereoolings, i.e.
small ﬁT‘s, at the large undercoolings that are typical for homogeneous nucleatiou that approximation may not be 1«alid and a more complex tenn is required. But for a ﬁrst
order approximation this assumption is reasonable.) In order to detennine the value of AG{1‘) at 1", we introduce the expression
r * = (2gSL) TE!(AH:"M) into
_ 2 4 3 :—
AG(r)  4m 15L+3—1tr AG, "
mm H. = 4nlc—2yﬂwgme'Mfrﬂ
4 _
+[§)n[(—ZYE)TEI(AH: “DP'10,
3
§4u(4yﬂT;IAH:AT2 r ﬂ+_1:(— ByﬂTngHEATﬁAq
48 T1 2 32 3 3 3
“1:533me AT)——j—1tyﬂEVTIAH or AG
43 a 2 2_ 1 32 3 3. . AH .
371:1“? “AH“ AT ——3—1t[1&TEI(AH‘, AT!” [ T; or
43 32
=[_3._ 3)::1'31 T21(AHfAT2) If all the terms in parentheses are constant then, 1 AG " m ____t.
at" FIND: Explain the simultaneous inﬂuence that undercooling has on the barrier to
nucleation and the atomic rearrangements necessary to initiate the transfonnation. Show how these competing effects lead to classical Ccurve behavior in the nucleation of
diﬂilsional transformations. SOLUTION: With larger undercoolings, both r" and AG“ decrease, suggesting 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 arrangements 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. The
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 Ccarve. Tﬂnsfmmaticn temperlure Net Mlcleuinn me I npmduct {If two
heavy duller: lineal Tcnmerellrre Start of nucleation Nucleumnmelsll in: (a) (b) 4. The inﬂuence oftemperature on the mobility term and the nucleation barrier term. The opposing processes result in a maximum in the nucleation rate at an intermediate
temperature. Since the time for nucleation is inversely related to the nucleation rate, the time curve exhibits a minimmn at an intermediate temperature. Because of its shape, this curve is
often referred to as a Ceurw. FIND: Explain how the value 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 ﬁmction of 1 when a
liquid transforms to a solid, for example. In the development of that equation it was assumed that the transforming phase was spherical and the interfacial energy, 'st was
isotropic. That equation consists of two terms on the right hand side, he AG(1') = (41:14) Ya. + 43 m3 (AG...) Since YsL > O and AG“ < D for AT > 0, the ﬁrst term increases with radius, and the second
decreases. Figure 8.23 illustrates that there is a maximum that occurs at some 1' we
designated as r“ and a corresponding AG, we designated as AG“ where I" = (“ZYSLyAGv 2
151:2"Jr 3 1 so =  chew
* may): 13" (M)2 Both the critical radius, r‘, and the barrier to the nucleation process contain 'm in the
numerator. ”thinking in terms of the barrier to nucleation, :16", there is a cubic
dependence on interfacial energy. The larger the SL 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 irrterfacial energies to stimulate
nucleation. FIND: Compare homogeneous 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 at speciﬁc .. sites. In the case of solidiﬁcation they can be at mold walls, unintentional additions such 7” as ceramic inclusions from crucibles or it may occur at nucleating agents which are
intentionally added to control the solidiﬁcation microstructure. ...
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 Summer '08
 Tannebaum

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