19 2 HETEROGENEOUS NUCLEATION 481 Homogeneous 0.3 3i 0 0.5 1 1.5“ 2 yaa/yap = 2 cos l/l Figure 19.14: Re imes in which grain corner, edge. boundary, and homogeneous nucleation are predictecf to be dominant. From Cahn [18]. 19.2.2 Nucleation on Dislocations Dislocations in crystals have an excess line energy per unit length that is associated with the elastic strain field of the dislocation and the bad material in its core. In many cases, the formation of a particle of the new phase at the dislocation can reduce this energy, enabling it to act as a favorable site for heterogeneous nucleation. The original treatment of heterogeneous incoherent nucleation on dislocations was by Cahn [19]. The general topic, including coherent nucleation on dislocations, has been reviewed by Larch6 [20]. Incoherent Nucleation. Consider first incoherent nucleation on dislocations [ 191. For linearly elastic isotropic materials, the energy per unit length El inside a cylin- der of radius T having a dislocation at its center is given by and EL = -I.(&) Pb2 (screw dislocation) 4n Pb2 (edge dislocation) 47~ (1 - v) In (k) El = (19.50) (19.51) where b is the Burgers vector and R, is the usual effective core radius. Poisson’s ratio v is approximately 0.3 for many solids, so to a fair approximation, the energy difference between edge and screw dislocations can be ignored. Following Cahn. El=-ln(&) Bb 2 (19.52) where B x pbl(2n). Allowing the entire region inside a radius T to transform to incoherent @ will allow essentially all of the dislocation energy originally inside the transformed region to be “released.” Thus, the dislocation catalyzes incoherent nucleation by eliminating some of the dislocation’s total energy. It is important to note that the dislocation will still effectively exist in the material along with its strain energy outside the transformed region, even though the incoherent @ has replaced the core region. For example, a Burgers circuit around the dislocation in the matrix material surround- ing the incoherent @-phase cylinder will still have a closure failure equal to b. On 482 CHAPTER 19: NUCLEATION forming the incoherent cylinder of radius r, the total free energy change per unit length is (terms independent of r) (19.53) Bb 2 AG’(r) =m2Ag~+2.1rry- -lnr+ Extreme values of AG’(r) are given by the condition Bb br 2r = 2./r(rAg~ + 7) - - = 0 8 AG‘ (r ) (19.54) Plotting AG’(r) vs. r in Fig. 19.15, two types of behavior are evident, depending on the value of the parameter, a, where (19.55) For a > 1, nucleation is barrierless-i.e., the transformation is controlled solely by growth kinetics. However, for a c 1, a barrier exists. The local minimum of AG’(r) at point A in the plot corresponds to a metastable cylinder of p of radius ro forming along the dislocation line. (In a sense, this is analogous to the Cottrell atmosphere described in Section 3.5.2.) In Eq. 19.54, the metastable cylinder’s radius is (19.56) The nucleation barrier for a c 1 is then related to the difference in AG’(r) between the states A and B in Fig. 19.15, where the radius rc corresponding to the unstable state at B is given from Eq. 19.54 as (19.57) However, the dislocation is practically infinitely long compared to the size of any realistic critical nucleus. If the nucleus were of uniform radius along a long length of the dislocation, AGc would be very large. A critical nucleus will form from a local fluctuation in the form of a “bulge” of the cylinder associated with the metastable state A, as illustrated in Fig. 19.16. The problem is thus to find the particular bulged-out shape that corresponds to a minimum activation barrier for nucleation. tl B r0 rc r Fi ure 19.16: cyfndrical precipitate along the core of a dislocation. From Cahn [lQ]. Possible free energy vs. size behavior for the formation of an incoherent 19 2. HETEROGENEOUS NUCLEATION 483 Figure 19.16: dislocation. Possible shape for incoherent critical nucleus forming along the core of a Let the function r(t) specify the shape of the nucleus. The energy to go from the metastable state A to the unstable state B (see Fig. 19.15) can be expressed AG = [AG' (r) - AG' (TO)] dt (19.58) J From earlier equations, 7rAgB(r2-Tz) -"I.(:) 2 +2ry [r/q-rO]} dt (19.59) The unknown shape r( t) is determined by minimizing AG using variational calcu- lus techniques. The solution to the Euler equation for this problem is somewhat complicated, requiring some substitutions and lengthy algebra [19]. From the re- sulting equations, one can plot the ratio of the activation barrier for nucleation on dislocations AGp to that for homogeneous nucleation As: vs. a, in a manner analogous to the plot given in Fig. 19.13, which compared nucleation on various sites in polycrystals. The resulting plot in Fig. 19.17 shows a dramatic decrease in the relative value of AGf as cy -, 1. Cahn also considered briefly the nucleation kinetics and showed that for reason- able values of the parameters in the theory, nucleation on dislocations in solids can be copious [19]. Typically, this occurs when a is in the range 0.4-0.7. " 0 0.2 0.4 0.6 0.8 1.0 a Figure 19.17: at dislocations with increasing values of the parameter a (see Eq. 19.55). From Cahn [19]. Lowering of the activation barrier for heterogeneous incoherent nucleation 484 CHAPTER 19. NUCLEATION Coherent Nucleation. The elastic interaction between the strain field of the nucleus and the stress field in the matrix due to the dislocation provides the main catalyzing force for heterogeneous nucleation of coherent precipitates on dislocations. This elastic interaction is absent for incoherent precipitates. For coherent particles with dilational strains, there is a strong interaction with the elastic stress field of edge dislocations [20]. If a particle has a positive dilational transformation strain (& + E&, + &FZ > 0), it can relieve some of the dislocation’s strain energy by forming in the region near the core that is under tensile strain. Conversely, when this strain is negative, the particle will form on the compressive side. Interactions with screw dislocations are generally considerably weaker, but can be important for transformation strains with a large shear component. Deter- minations of the various strain energies use Eshelby’s method of calculating these quantities [20]. Bibliography 1. F.K. LeGoues, H.I. Aaronson, Y.W. Lee, and G.J. Fix. Influence of crystallography upon critical nucleus shapes and kinetics of homogeneous f.c.c f.c.c. nucleation. I. The classical theory regime. In International Conference on Solid-Solid Phase Transfor- mations, pages 427-431, Warrendale, PA, 1982. The Minerals, Metals and Materials Society. 2. D.T. Wu. Nucleation theory. Solid State Phys., 50:37-187, 1997. 3. J.W. Christian. The Theory of Transformations in Metals and Alloys. Pergamon 4. K.C. Russell. Linked flux analysis of nucleation in condensed phases. Acta Metall., 5. K.C. Russell. Grain boundary nucleation kinetics. Acta Metall., 17(8):1123-1131, 6. K.C. Russell. Nucleation in solids: The induction and steady-state effects. Adv. 7. J.D. Eshelby. On the determination of the elastic field of an ellipsoidal inclusion, and 8. F.R.N. Nabarro. The influence of elastic strain on the shape of particles segregating 9. F. Bitter. On impurities in metals. Phys. Rev., 37(11):1527-1547, 1931. 10. D.M. Barnett, J.K. Lee, H.I. Aaronson, and K.C. Russell. The strain energy of coherent ellipsoidal precipitates. Scnpta Metall., 8(12):1447-1450, 1974. 11. S.M. Allen and J.C. Chang. Elastic energy changes accompanying the gamma-prime rafting in nickel-base superalloys. J. Mater. Res., 6(9):1843-1855, 1991. 12. J.D. Eshelby. Elastic inclusions and inhomogeneities. In I.N. Sneddon and R. Hill, editors, Progress in Solid Mechanics, volume 2, pages 89-140, Amsterdam, 1961. Nort h-Holland. 13. A.J. Ardell and R.B. Nicholson. On the modulated structure of aged Ni-A1. Acta Metall., 14(10):1295-1310, 1966. 14. J.K. Lee, D.M. Barnett, and H.I. Aaronson. The elastic strain energy of coherent ellipsoidal precipitates in anisotropic crystalline solids. Metall. Trans. A, 8(6):963- 970, 1977. 15. H.I. Aaronson and F.K. LeGoues. An assessment of studies on homogeneous diffusional nucleation kinetics in binary metallic alloys. Metall. Tk-ans. A, 23(7):1915-1945, 1992. Press, Oxford, 1975. 16( 5):761-769, 1968. 1969. Colloid Interface Sci., 13(3-4):205-318, 1980. related problems. Proc. Roy. SOC. A, 241(1226):376-396, 1957. in an alloy. Proc. Phys. SOC., 52(1):90-104, 1940. EXERCISES 485 16. 17. 18. 19. 20. J.W. Cahn and J.E. Hilliard. F'ree energy of a non-uniform system-111. Nucleation in a two-component incompressible fluid. J. Chem. Phys., 31(3):688-699, 1959. H.I. Aaronson and J.K. Lee. The Kinetic Equations of Solid-rSolid Nucleation The- ory and Comparisons with Experimental Observations, pages 165-229. The Minerals, Metals and Materials Society, Warrendale, PA, 2nd edition, 1999. J.W. Cahn. Acta Metall., 4(5):449459, 1956. J.W. Cahn. Nucleation on dislocations. Acta Metall., 5(3):16+172, 1957. F.C. Larch& Nucleation and precipitation on dislocations. In F.R.N. Nabarro, editor, Dislocations in Solids, volume 4, pages 137-152, Amsterdam, 1979. North-Holland. The kinetics of grain boundary nucleated reactions. EXE RClS ES 19.1 An equilibrium temperaturecomposition diagram for an A-B alloy is shown in Fig. 19.18a. A nucleation study is carried out at 800 K using an alloy of 30 at. % B. The alloy is initially homogenized at 1200 K, then quenched to 800 K where the steady-state homogeneous nucleation rate is determined to be 10' m-3 s-'. Since this rate is so small as to be barely detectable, it is desired to change the alloy composition (i-e., increase the supersaturation) so that with the same heat treatment the nucleation rate is increased to 1021 m-3 s-l. Estimate the new alloy composition required to achieve this at 800 K. Use the free energy vs. composition curves in Fig. 19.18b, and assume that the interphase boundary energy per unit area, 7, is 75 mJ m-2. List important assumptions in your analysis. - 3 1200 - f ElOOO- c" 800- v - B I I1 I I1 I Ill 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Atomic fraction B (4 Atomic fraction B (b) Figure 19.18: AgB, vs. atomic fraction of component B at T = 800 K. (a) Equilibrium diagram for A-B alloy. (b) Plot of free-energy density, Solution. Important assumptions include that the interfacial free energy is isotropic, that elastic strain energy is unimportant, and that the nucleation rates mentioned are for steady-state nucleation. The critical barrier to nucleation, Ap,, can be calculated for the 0.3 atomic fraction B alloy using the tangent-to-curve construction on the curves in Fig. 19.18b to provide the value AgB = -9 x 10' Jm-3 for the chemical driving force for this supersaturation at 800 K. AgC is given for a spherical critical nucleus by 486 CHAPTER 19: NUCLEATION Note that at this temperature, kT = 1.38 x x 800 = 1.10 x lo-", 50 that at 800 K and XB = 0.3, AG, % 79kT. Based on the criterion that for significant nucleation AG, 5 76kT (Section 19.1.7), it is reasonable that the nucleation rate is "barely detectable" in the alloy with XB = 0.3. The steady-state nucleation rate will be proportional to exp[-AG,/(kT)] 50 we know that at 800 K and XB = 0.3, lo6 = ~'exp(-79) (19.61) where the constant C' is equal to NP.2 in the classical theory for steady-state nucleation. We need to find the critical nucleation barrier necessary to achieve the nucleation rate of 10". and this will be or 1 o6 exp( -79) loz1 exp[ -AG,/( kT)] -= or -34.54+79= - AGc kT In 10-l~ = -79 + - kT (19.62) (19.63) and thus for the higher nucleation rate we must have AG, *: 44.5kT = 4.91 x lO-"J. Next, solve for the chemical driving force required to get AG, down to this value, as follows: Finally, use the free-energy density vs. composition curves and work the tangent-to- curve construction in reverse. Using the result that AgB = -12 x 107Jm-3, the corresponding tangent to the a-phase curve will be at about 33 at. % B. This calculation serves as a good example of the high sensitivity of nucleation rate to the degree of supersaturation. 19.2 The data below are typical for a metal solid solution that can precipitate a phase 0 from a matrix phase a. Assume that the structures of both phases are such that 0 could form by coherent homogeneous nucleation or, alternatively, by incoherent homogeneous nucleation. Also, assume that strain energy can be neglected during incoherent nucleation but must be taken into account during coherent nucleation. Using the data below, answer the following: (a) Below what temperature does incoherent nucleation become thenody- (b) Below what temperature does coherent nucleation become thenody- (c) Which type of nucleation, coherent or incoherent, do you expect to occur Data namically possible? namically possible? at 510 K? Justify your answer. -yc = 160 mJ m-2 7' = 800 mJ m-2 AgE = 2.6 x lo9 J m-3 AgB = 8 x lo6 (T - 900K) J m -3 K-' (coherent interface) (incoherent interface) (coherent particle) (driving force for precipitation) Solution. (a) Nucleation becomes thermodynamically possible if the thermodynamic driving For sufficiently large volumes nuclt force for the transformation is negative. EXERCISES 487 ated incoherently in the absence of strain energy, and where the interfacial energy has become unimportant, the total energy change will be negative if AgB < 0. Therefore, we need AgB = 8 x lo6 (T - 900 K) J m-3 K-’ < 0, or T < 900 K. (b) For coherent nucleation to be thermodynamically possible, AgB + AgE < 0. Therefore, we need 8 x lo6 (T - 900 K) + 2.6 x lo9 < 0, or T < 575 K. (c) Assuming that the number of available sites for nucleation is the same for both coherent and incoherent mechanisms, the nucleation mechanism one expects to observe will be determined by the critical free-energy barrier, AG,. Because the nucleation rates are proportional to exp[-AG,/(kT)], the mechanism with the lowest value of AG, will dominate and be observable if AG, 5 76kT, approxi- mately. Assuming spherical nuclei, A& = 167ry3/[3(Ags + Ag,)’], where (19.65) y = yi, AgE = 0 y = yc, AgE # 0 (incoherent nucleation) (coherent nucleation) Using the given data, at T = 510K, AgB = -3.12 x lo9 Jm-3. With this, AGc = 8.81 x lO-”J (incoherent nucleation) (19.66) ( AG, = 2.54 x lO-”J (coherent nucleation) and, similarly, AGC/(76kT) = 1.65 > 1 AGC/(76kT) = 0.475 < 1 (incoherent nucleation) (coherent nucleation) (19.67) Consequently, coherent nucleation is expected. 19.3 Martensitic transformations involve a shape deformation that is an invariant- plane strain (simple shear plus a strain normal to the plane of shear). The elastic coherency-strain energy associated with the shape change is often min- imized if the martensite forms as thin plates lying in the plane of shear. Such a morphology can be approximated by an oblate spheroid with semiaxes (r, r, c), with T >> c. The volume V and surface area S for an oblate spheroid are given by the relations (19.68) V = -r c and S = 2rr2 47r 2 3 The coherency strain energy per unit volume transformed is Ac ASE = 1- (19.69) (a) Find expressions for the size and shape parameters for a coherent critical nucleus of martensite. Use the data below to calculate values for these parameters. (b) Find the expression for the activation barrier for the formation of a coherent critical nucleus of martensite. Use the data below to calculate the value of this quantity. (c) Comment on the likelihood of coherent nucleation of martensite under these conditions. 488 CHAPTER 19: NUCLEATION (d) Make a sketch of the free-energy surface AG(r, c) and indicate the loca- tion of the critical nucleus configuration (T~, c,) on the surface. Data AgB = -170 MJ m-3 y = 150 mJ m-' A = 2.4 x lo3 MJ m-3 (chemical driving force at observed transformation temperature) (interphase boundary energy per unit area) (strain energy proportionality factor) Solution. (a) Write the free energy to form a nucleus in the usual way as the sum of a bulk free-energy term, a strain-energy term, and an interfacial-energy term so that ( 19.70) Now AG = AG(c, T) and the critical values of c and T are then found by applying the simultaneous conditions 4 4 3 3 AG = -TT'CA~B + -TTC~A + 2~r'r Substituting Eq. 19.70 into Eqs. 19.71 and solving for rC and cc yields (19.71) (19.72) Using the data provided, these quantities evaluate to (19.73) C T rC = 50nm cc = 1.5nm - = 0.035 (b) Substituting Eqs. 19.72 into Eq. 19.70 then yields 32~ A2y3 AgC = -~ (AgB)4 Using the data provided, this quantity is equal to AG, = 7.8 x J (19.74) (19.75) (c) Nucleation would proceed at observable rates if A& 5 76kT. Assuming a nucle- ation temperature of 350 K, = 1.6 x 105 (19.76) 7.8 x J - A Gc kT 1.38 x J K-l x 350K which is huge compared to 76! So homogeneous nucleation would be very unlikely. Note that the size parameter rC is particularly large and thus the critical nucleus volume is large, consistent with the large value of A&. (d) The saddle point on the free-energy surface, (rC,cc), is indicated in Fig. 19.19. EXERCISES 489 Figure 19.19: Saddle point on free-energy surface. 19.4 Derive Eq. 19.22; i.e., 1 1 a2A(7~ -= (w) 62 8kT N=Nc Solution. Approximate the curve of AGN vs. n/ in Fig. 19.6 by a circle of radius R. Then kT = R - (19.77) Expanding Eq. 19.77 and neglecting the higher-order terms, 1 11 P-mz The standard expression for the curvature, 1/R, is (19.78) (19.79) Combining Eqs. 19.78 and 19.79, the desired result is then obtained. 19.5 Derive Eq. 19.36 for the free-energy change due to the annihilation of excess vacancies at nucleating incoherent clusters during precipitation. Hint: The chemical potential of excess vacancies is given by Eq. 3.66. Solution. First, calculate the energy change contributed by the excess vacancies which are eliminated to relieve the strain due to the dilatation $1. If V is the cluster volume, AV = 3€TIV. The number of vacancies required is then N = 3ETlV/n and the free-energy change due to the removal of these vacancies is therefore (19.80) Next, calculate the free-energy change due to the destruction of the additional vacancies which are removed to the point where the rate of buildup of elastic strain due to their annihilation is just equal to the rate at which energy is given up by the vacancy annihilation. If N vacancies are destroyed in this fashion, the volume of matrix removed 490 CHAPTER 19: NUCLEATION is Nn and the dilational strain that is induced is then Nn/3V. Using Eq. 19.25, the strain energy that is created is (19.81) The energy released by the annihilated vacancies is As; = NkTln(Xv/XGq), and the total energy change is then (19.82) AG" is minimized when aAG"/aN = 0, and carrying out this operation, the minimum value is (19.83) Adding Eqs. 19.80 and 19.83, the total energy change (per unit cluster volume) is then finally 2 (19.84) 3zn 9(1 - Y) Ag = kTln n 19.6 Figure 19.20 shows a cross section through the center of a critical nucleus that has cylindrical symmetry around the vertical axis EF. AB and CD are the traces of flat facets that possess the interfacial energy (per unit area) yf, and AC and BD are the traces of the spherical portion of the interface that possesses the corresponding energy y. E / F Figure 19.20: Critical nucleus shape. (a) Construct a Wulff plot that is consistent with the critical nucleus shape (b) Show that the free energy to form this critical nucleus can be written (Wulff shape) in Fig. 19.20. 1 AQ,(sphere) 2 A& = - (3 cos a - coS3 a) (19.85) where Ag,(sphere) is the free energy to form a critical nucleus for the same transformation but which is spherical and possesses the interfacial energy y. Assume the classical model for the nucleus. [...]... CHAPTER 20: GROWTH OF PHASES IN CONCENTRATION AND THERMAL FIELDS where the relation v,": = [,lye]" / P - v:yP has been used and where = flux of i into the a phase at the interface measured in the a-phase V-frame [ J y ' ] " l P = flux of i into the @ phase at the interface measured in the @-phase V-frame t~a";c,= velocity of the interface measured in the a-phase V-frame vfll"p = velocity of the interface... is invoked The solutions are of the error-function type and are given by cg - cgw CEO - cagoo - 1 - erf ( x/diZ) 1 - erf ( A / G ) (20.18) (20.19) with x ( t )= A& (20.20) where A = constant A is now found by substituting Eqs 20.18, 20.19, and 20.20 into Eq 20 .12 for the velocity of the interface (in the VP-frame) with the result +Q P fi G (cpB” - c p ) Cp-CgP ,-A2/(46@) l+erf(A/G) (20.21) Therefore,... nucleus is Arect = bc c c = y l O / v + ylO/sub - yv/sub 2y10/" A94 &a (19.109) T h e two areas are equal when (yll/sub-yv/sub)2 = 2y10/v [ylO/wb -& - p u b + (Jz- l)yV/SUb] (19.110) 19.11 We wish t o prove by means of the Wulff construction (Section C.3.1) that the equilibrium shape of the grain boundary nucleus in Fig 19 .12 is indeed composed of two spherical-cap-shaped interfaces The nucleus has cylindrical... yields dc where a1 "-a1 - -e v2 - 2 (20.41) dq = constant Integrating again, (20.42) Carrying out the integration and determining a1 by applying the boundary conditions, cam - cff (1/7) e-q2 - fierfc(7) (20.43) cam - Cap ( I / ~ R )e-7; - J?Ferfc(vR) Next, V R is determined in the usual way by invoking the Stefan condition at the interface, which has the form dt (20.44) r=R Use of Eq 20.43 in Eq 20.44... surface of a casting of Cu-Ni-Mn alloy made visible when solidification shrinkage causes liquid to retreat into the casting interior (a) Low magnification (b) Higher-magnification image showing side-branch evolution near dendrite tips Micrographs courtesy of J Feuchtwanger 20.3.1 Stability of Liquid/Solid Interface during Solidification of a Unary System Consider first the solidification of a body of superheated... Use of Eq 20.43 in Eq 20.44 then produces the desired relation for - CffP 3 - e-& 7 R - 2 (cP" - cap 1 (l/vR) e-qg - J;rerfc(rlR) VR: (20.45) Of special interest is the current of B atoms into the particle, which is given by I = 47rR25j" Using Eq 20.43 in Eq 20.46, (g) r=R (20.46) The last approximation is obtained by an expansion of the denominator to first order and is usually valid since T]R is... interfacial equilibrium The volume of the nucleus is then the volume of the cone of height d plus the volume of the spherical cap of radius R and is given by The area of the cap, A , is given by 2rRh, where h is its height Therefore, A = 2rRh = 2rR2[ 1- COS(CY/~)] (19.90) The free energy t o form a nucleus as in Fig 19.22 is then + 2r~3 AG = -[ 1 - C O S ( C Y / ~ ) ]2rR2[ 1- C O S ( C Y / ~ ) ] ~ ~ ~ (19.91)... interface measured in the a-phase V-frame vfll"p = velocity of the interface measured in the @-phase V-frame v" ,: = velocity of the a-phase V-frame measured in the P-phase V-frame (" concentration of i at the a / @interface on the @ side ( p therefore = precedes a in the superscript), crP = concentration of i at the a / @interface on the precedes 3 in the superscript) ! , CY side ( a therefore b The... and simple linear profile, as shown, and that the atomic volume of each component is the same in each phase so that no changes in overall volume occur Employing the same interfacial-reaction rate-constant model used to write Eq 13.24, the following five equations will apply under general conditions: J p = K1c( :: - c” ’) (20.25) J p = - ( C p a - c$) dX1 dt (20.26) J p = K2 (c” )::c - (20.27) (20.28)... critical parameter determining the mode of growth The magnitude of & , in turn, depends directly upon the magnitude of the rate constant Ki Determining the magnitude of Ki requires the construction of a detailed model for the source action of the interface based on one (or more) of the mechanisms described in Section 13.4 Experimental results for the growth of layers has been reviewed [6, 71 It is . mirror symmetry across the grain-boundary plane. Figure 19.27 shows a cross section of the nucleus centered in a patch of boundary of constant circular area, A,. The area of the nucleus projected. Conversely, when this strain is negative, the particle will form on the compressive side. Interactions with screw dislocations are generally considerably weaker, but can be important for transformation. distinguished from purelv conservative growth that occurs when no such long-range transport is required. -1 In the treatment of this growth it is necessary to determine the concentration