CHAPTER 3: DRIVING FORCES AND FLUXES FOR DIFFUSION 68 17. 18. 19. 20. 21. 22. 23. 24. 25. J. Hoekstra, A.P. Sutton, T.N. Todorov, and A.P. Horsfield. Electromigration of vacancies in copper. Phys. Rev. B, 62(13):8568-8571, 2000. P. Shewmon. Diffusion in Solids. The Minerals, Metals and Materials Society, War- rendale, PA, 1989. F.C. Larch6 and P.W. Voorhees. Diffusion and stresses, basic thermodynamics. Defect and Diffusion Forum, 129-130:31-36, 1996. J.P. Hirth and J. Lothe. Theory of Dislocations. John Wiley & Sons, New York, 2nd edition, 1982. F. Larch6 and J.W. Cahn. The effect of self-stress on diffusion in solids. Acta Metall., A.H. Cottrell. Dislocations and Plastic Flow. Oxford University Press, Oxford, 1953. L.S. Darken. Diffusion of carbon in austenite with a discontinuity in composition. Trans. AIME, 180:430-438, 1949. U. Mehmut, D.K. Rehbein, and O.N. Carlson. Thermotransport of carbon in two- phase V-C and Nb-C alloys. Metall. Trans., 17A(11):1955-1966, 1986. A.H. Cottrell and B.A. Bilby. Dislocation theory of yielding and strain ageing of iron. Proc. Phys. SOC. A, 49:49-62, 1949. 30 ( 10) : 1835-1845, 1982. EXERCISES 3.1 Component 1, which is unconstrained, is diffusing along a long bar while the temperature everywhere is maintained constant. Find an expression for the heat flow that would be expected to accompany this mass diffusion. What role does the heat of transport play in this phenomenon? Solution. The basic force-flux relations are - 1 J1 = -L11Vpi - Lig-VT T (3.85) TQ = -L~1Op1 - LQQ-VT 1 T Under isothermal conditions J; = -L11Vp1 TQ = -L~lVpi Therefore, using Eqs. 3.61 and 3.86, (3.86) (3.87) The heat flux consists of two parts. The first is the heat flux due to the flux of entropy, which is carried along by the mass flux in the form of the partial atomic entropy, S:. Beca_use 31 = +85'/8N1, a flux of atoms will transport a flux of heat given by JQ = TJs = TS1J1. The second part is a "cross effect" proportional to the flux of mass, with the proportionality factor being the heat of transport. 3.2 As shown in Section 3.1.4, the diffusion of small interstitial atoms (component 1) among the interstices between large host' atoms (component 2) produces a interdiffusivity, 5, for the interstitial atoms and host atoms in a V-frame D = c~OZD~ (3.88) given by Eq. 3.46, that is - EXERCISES 69 and therefore a flux of host atoms given by - dc:! dX Jv = -D- 2 (3.89) This result holds even though the intrinsic diffusivity of the host atoms is taken to be zero and the flux of host atoms across crystal planes in the local C-frame is therefore zero. Give a physical explanation of this behavior. Solution. When mobile interstitials diffuse across a plane in the V-frame, the material left behind shrinks, due to the loss of the dilational fields of the interstitials. This establishes a bulk flow in the diffusion zone toward the side losing interstitials and causes a compensating flow (influx) of the large host atoms toward that side even though they are not making any diffusional jumps in the crystal. The rate of loss of volume of the material (per unit area) on one side of a fixed plane in the V-frame due to a loss of interstitials is (3.90) In the V-frame this must be compensated for by a gain of volume due to a gain of host atoms so that -+-=o dV1 dV2 dt dt (3.91) where dVz/dt is the rate of volume gain due to the gain of host atoms corresponding to Substituting Eqs. 3.90 and 3.92 into Eq. 3.91 and using Eq. A.lO, (3.92) (3.93) 3.3 In a classic diffusion experiment, Darken welded an Fe-C alloy and an Fe- C-Si alloy together and annealed the resulting diffusion couple for 13 days at 1323 K, producing the concentration profile shown in Fig. 3.11 [23]. Initially, the C concentrations in the two alloys were uniform and essentially equal, whereas the Si concentration in the Fe-C-Si alloy was uniform at about 3.8%. After a diffusion anneal, the C had diffused “uphill” (in the direction of its concentration gradient) out of the Si-containing alloy. Si is a large substi- tutional atom, so the Fe and Si remained essentially immobile during the 6 0.6 e + 0.5 2 % 0.4 cu 0 C a, E 0-l 0.3 Ill1 -20 -10 0 10 20 Distance from weld (mm) Figure 3.11: Nonuniform concentration of C produced by diffusion from an initially uniform distribution. Carbon migrated from the Fe-Si-C (left) to the Fe-C alloy (right). From Darken [23]. 70 CHAPTER 3: DRIVING FORCES AND FLUXES FOR DIFFUSION diffusion, whereas the small interstitial C atoms were mobile. Si increases the activity of C in Fe. Explain these results in terms of the basic driving forces for diffusion. Solution. As the C interstitials are the only mobile species, Eq. 3.35 applies, and therefore J; = -L11Vp1 (3.94) (3.95) Using the standard expression for the chemical potential, p1 = py + kTlna1 where a1 = 71x1 is the activity of the interstitial C, (3.96) The coefficient L11 in Eq. 3.96 is positive and the equation therefore shows that the C flux will be in the direction of reduced C activity. Because the C activity is higher in the Si-containing alloy than in the non-Si-containing alloy at the same C concentration, the uphill diffusion into the non-Si-containing alloy occurs as observed. In essence, the C is pushed out of the ternary alloy by the presence of the essentially immobile Si. 3.4 Following Shewmon, consider the metallic couple specimen consisting of two different metals, A and B, shown in Fig. 3.12 [18]. The bonded end is at temperature TI and the open end is at T2. A mobile interstitial solute is kJ/mol in one leg and QFans = 0 in the other. Assuming that the interstitial concentration remains the same at the bonded interface at TI, derive the equation for the steady-state interstitial concentration difference between the two metal legs at Tz. Assume that TI > T2. present at the same concentration in both metals for which QYans = - 84 r 1 Figure 3.12: Metallic couple specimen made up of metals A and B. Solution. In the steady state, Eq. 3.60 yields CiQYans VT VCl = -~ kT2 Reducing to one dimension and integrating, Therefore, (3.97) (3.98) (3.99) EXERCISES 71 Therefore, for leg A, (3.100) while for leg B, cf(T2) = cf(T1). Finally, because cf(T1) = $(Ti) = cf(T2) ci(Ti), (3.101) 1 -l> -84000 (Ti - T2) Ac1 = cl(T1) exp { [ NokTiTz 3.5 Suppose that a two-phase system consists of a fine dispersion of a carbide phase in a matrix. The carbide particles are in equilibrium with C dissolved interstitially in the matrix phase, with the equilibrium solubility given by c1 = c,e o -AH/(kT) (3.102) If a bar-shaped specimen of this material is subjected to a steep thermal gradient along the bar, C atoms move against the thermal gradient (toward the cold end) and carbide particles shrink at the hot end and grow at the cold end, even though the heat of transport is negative! (For an example, see the paper by Mehmut et al. [24].) Explain how this can occur. 0 Assume that the concentration of C in the matrix is maintained in local equilibrium with the carbide particles, which act as good sources and sinks for the C atoms. Also, AH is positive and larger in magnitude than the heat of transport. Solution. Ea. 3.102. and therefore The local C concentration will be coupled to the local temperature by dcl - dci dT - AH dT I dx dT dx kT2 dx - - Substitution of Eq. 3.103 into Eq. 3.60 then yields Jl = D1cl (AH + Qtrans) dz dT kT2 (3.103) (3.104) Because (AH + Qtrans) is positive, the C atoms will be swept toward the cold end, as observed. 3.6 Show that the forces exerted on interstitial atoms by the stress field of an edge dislocation are tangent to the dashed circles in the directions of the arrows shown in Fig. 3.8. Solution. The hydrostatic stress on an interstitial in the stress field is given by Eq. 3.80 and the force is equal to = -0lVP. Therefore, (3.105) where A is a positive constant. Translating the origin of the (x’, y’) coord+inate system to a new position corresponding to (2’ = R,y’ = 0), the expression for Fl in the new (x, y) coordinate system is 72 CHAPTER 3: DRIVING FORCES AND FLUXES FOR DIFFUSION Converting to cylindrical coordinates, 1 r sin 8 $1 = -R1 AV [rZ+R2+2rRc~~B The gradient operator in cylindrical coordinates is d ld V =fir-+Ce dr T de (3.107) (3.108) Therefore, using Eq. 3.107 and Eq. 3.108 yields $1 = - 01 A {fir(Rz -r2)sin8+fie [(R2 +r2)cose+2Rr]} (3.109) The force on an interstitial lying on a cylinder of radius R centered on the origin where [RZ + r2 + 2Rr cos el2 r = R is then (3.110) The force anywhere on the cylinder therefore lies along -60, which is tangential to the cylinder in the direction of decreasing 0. 3.7 Consider the diffusional flux in the vicinity of an edge dislocation after it is suddenly inserted into a material that has an initially uniform concentration of interstitial solute atoms. (a) Calculate the initial rate at which the solute increases in a cylinder that has an axis coincident with the dislocation and a radius R. Assume that the solute forms a Henrian solution. (b) Find an expression for the concentration gradient at a long time when mass diffusion has ceased. Solution. (a) The diffusion flux is given by Eq. 3.83. Initially, the concentration gradient is zero and the flux is due entirely to the stress gradient. Therefore, hkT(1 - v) 1 r2 -' Now, integrate the flux entering the cylinder, noting that the B component con- tributes nothing: Rd9 = 0 2x Asin6 (3.112) where A = constant. Note that this result can be inferred immediately, due to the symmetry of the problem. (b) When mass flow has ceased, the flux in Eq. 3.83 is zero and therefore vc1= - 7:;;;; t yVlb [-1 ] (3.113) 3.8 The diffusion of interstitial atoms in the stress field of a dislocation was con- sidered in Section 3.5.2. Interstitials diffuse about and eventually form an sine cos0 ~ Cr + Tue r EXERCISES 73 equilibrium distribution around the dislocation (known as a Cottrell atmo- sphere), which is invariant with time. Assume that the system is very large and that the interstitial concentration is therefore maintained at a concentra- tion cy far from the dislocation. Use Eq. 3.83 to show that in this equilibrium atmosphere, the interstitial concentration on a site where the hydrostatic pressure, P, due to the dislocation is cyl = Cle 0 -nlp/(W (3.114) Solution. According to Eq. 3.83, (3.115) At equilibrium, = 0 and therefore lncyq + = a1 = constant (3.116) kT Because cyq = c? at large distances from the dislocation where P = 0, a1 = In&, Ceq 1- - C;e-%P/(kT) (3.117) 3.9 In the Encyclopedia of Twentieth Century Physics, R.W. Cahn describes A.H. Cottrell and B.A. Bilby's result that strain aging in an interstitial solid solu- tion increases with time as t213 as the coming of age of the science of quan- titative metallurgy [25]. Strain aging is a phenomenon that occurs when interstitial atoms diffuse to dislocations in a material and adhere to their cores and cause them to be immobilized. Especially remarkable is that the t213 relation was derived even before dislocations had been observed. Derive this result f0r an edge dislocation in an isotropic material. 0 Assume that the degree of the strain aging is proportional to the number of interstitials that reach the dislocation. 0 Assume that the interstitial species is initially uniformly distributed and that an edge dislocation is suddenly introduced into the crystal. 0 Assume that the force, -RlVP, is the dominant driving force for inter- stitial diffusion. Neglect contributions due to Vc. 0 Find the time dependence of the number of interstitials that reach the dislocation. Take into account the rate at which the interstitials travel along the circular paths in Fig. 3.8 and the number of these paths fun- neling interstitials into the dislocation core. Solution. The tangential velocity, u, of an interstitial tkaveling along a circular path of radius R in Fig. 3.8 will be proportional to the force F1 = -fIlVP exerted by the dislocation. In cylindrical coordinates, P is proportional to sinO/r, so (3.118) 74 CHAPTER 3: DRIVING FORCES AND FLUXES FOR DIFFUSION Therefore, v K F1 LX l/rz. As shown in Fig. 3.8, v at equivalent points on each circle will scale as l/r*, and because r at these points scales as R, 1 (3.119) The averagewelocity, (v), around each circular path will therefore scale as l/RZ. Since the distance around a path is 2nR, the time, tR, required to travel completely around (3.120) Therefore, at time, t, the circles with radii less than Rcrit K t1/3 (3.121) will be depleted of solute. During an increment of time dt, the average distance at which interstitials along the active flux circles approach the dislocation is equal (to a reasonable approximation) to ds = (v)dt. The total volume (per unit length of dislocation) supplying atoms during this period is then dV LX dt Jm (v) dR 0: Ldt Lit %it (3.122) where the integral is taken over only the active flux circles. Because the concentration was initially uniform, the number of interstitials reaching the dislocation in time t, des- ignated by N, is therefore proportional to the volume swept out. Therefore, substituting Eq. 3.121 in Eq. 3.122 and integrating, (3.123) More detailed treatments are given in the original paper by Cottrell and Bilby [25] and in the summary in Cottrell's text on dislocation theory [22]. 3.10 Derive the expression + DVCVPZ JA = kT for the electromigration of substitutional atoms in a pure metal, where Dv is the vacancy diffusivity and cv is the vacancy concentration. Assume that: There are two mobile components: atoms and vacancies. Diffusion occurs by the exchange of atoms and vacancies. There is a sufficient density of sources and sinks for vacancies so that the vacancies are maintained at their local equilibrium concentration everywhere. Solution. Vacancies are defects that scatter the conduction electrons and are therefore subject to a force which in turn induces a vacancy current. The vacancy current results in an equal and opposite atom current. The components are network constrained so that Eq. 2.21 for the vacancies, which are taken as the N,th component, is Because V~A = 0 (see Eq. 3.64) and pv = 0, EXERCISES 75 The vacancy current is therefore due solely to the_cross term arising from the current of conduction electrons (which is proportional to E). The coupling coefFicient for the vacancies is the off-diagonal coefficient Lvq which can be evaluated using the same procedure as that which led to Eq. 3.54 for the electromigration of interstitial atoms in a metal. Therefore, if (CV) is the average drift velocity of the vacancies induced by the current and Mv is the vacancy mobility, 3.11 (a) It is claimed in Section C.2.1 that the mean curvature, K, of a curved interface is the ratio of the increase in its area to the volume swept out when the interface is displaced toward its convex side. Demonstrate this by creating a small localized “bump” on the initially spherical interface illustrated in Fig. 3.13. I1 c L Figure 3.13: Circular cap (spherical zone) 011 a spherical interface. (b) Show that Eq. 3.124 also holds when the volume swept out is in the form of a thin layer of thickness dw, as illustrated in Fig. 3.14. Figure 3.14: with curvature K = (1/R1) + (1/&). Layer of thickness diu swept out by additioii of material at a11 interface 0 Construct the bump in the form of a small circular cap (spherical zone) by increasing h infinitesimally while holding r constant. Then show that dA dV /$=- (3.124) where dA and dV are, respectively, the increases in interfacial area and volume swept out due to the construction of the bump. 76 CHAPTER 3: DRIVING FORCES AND FLUXES FOR DIFFUSION Solution. (a) The area of the circular cap in Fig. 3.13 is A = 7r (T’ + h2) Here T and h are related to the radius of curvature of the spherical surface, R, by the relation R=?(l+$) 2h (3.125) The volume under the circular cap is given by 7r 7r V = -hr2 + -h3 2 6 If the bump is now created by forming a new cap of height h + dh while keeping T constant, dA = 27rhdh (3.126) (3.127) Therefore, using Eqs. 3.125, 3.126, and 3.127, and the fact that h2/r2 << 1, dA 2 dV-RZK - - (b) The increase in area is dA = (R1 + dw) dB1 (Rz + dw) dB2 - R1 dB1 R2 dBz = (RI + Rz) dw dB1 dB2 The volume swept out is dV = Ri dB1 Rz dBz dw Therefore, dA 1 1 _- - -+-=K dV Ri Rz CHAPTER 4 THE DIFFUSION EQUATION The diffusion equation is the partial-differential equation that governs the evolution of the concentration field produced by a given flux. With appropriate boundary and initial conditions, the solution to this equation gives the time- and spatial- dependence of the concentration. In this chapter we examine various forms assumed by the diffusion equation when Fick’s law is obeyed for the flux. Cases where the diffusivity is constant, a function of concentration, a function of time, or a function of direction are included. In Chapter 5 we discuss mathematical methods of obtaining solutions to the diffusion equation for various boundary-value problems. 4.1 FICK’S SECOND LAW If the diffusive flux in a system is f, Section 1.3.5 and Eq. 1.18 are used to write the diffusion equation in the general form dC + _- -n-V*J at where n is an added source or sink term corresponding to the rate per unit volume at which diffusing material is created locally, possibly by means of chemical reaction or fast-particle irradiation, and :is any flux referred to a V-frame. There frequently are no sources or sinks operating, and n = 0 in Eq. 4.1. When Fick’s law applies (see Section 3.1) and n = 0, Eq. 4.1 takes the general form Kinetics of Materials. By Robert W. Balluffi, Samuel M. Allen, and W. Craig Carter. 77 Copyright @ 2005 John Wiley & Sons, Inc. [...]... coefficients must be rank-two tensors Therefore, (4.77) (4.78) or in a matrix representation involving rank-one tensors (vectors) and rank-two tensors, (4.79) (a) When written in full, Eq 4.79 becomes - J? ff11 J? - 59 3 ff 13 p11 pl2 p3 1 ff22 ff 23 p2l p22 p 23 a31 a32 ff 33 p31 p32 p 33 71 1 5" 1 J,4 ff12 ff2l JZQ 712 7 13 711 v12 7 13 721 722 7 23 721 731 732 733 731 722 v32 v 33 77 23 (b) Onsager's relation... thermal conductivity is K1 = 35 5 J m-' s-' K-* , nearly that 1 of carbon-diamond Between the graphite layers, where the bonding is very weak, the conductivity is much lower, K l = 89 .3 J m - l s - l K-l Figure 4.6 is therefore representative of single-crystal graphite, where the basal plane is parallel to the layers shown In general, the properties of crystals and other types of materials, such as composites,... corresponding flux vectors will then be, respectively, 23 -g[ -9 0 13 [ -9 0 23 [ I,I] (4.84) Because of the cubic crystal symmetry, the fluxes parallel t o the directions o f the gradients in the three cases (i.e., - g D 1 1 , - g D 2 2 , and - 9 0 3 3 ) must be equal Therefore, 0 1 1 (4.85) = D22 = 0 3 3 Preservation o f fourfold symmetry along the 2 1 , 2 2 , and 2 3 axes requires that any flux components in... D ) Let elements of such a transformed system be identified by a “hat.” Then B,, 0 0 (4.60) =[ i3] ; The diagonal elements of b are the eigenvalues of D, and the coordinate system of b defines the principal axes 21,22, 23 (the eigensystem) In the principal axes coordinate system, the diffusion equation then has the relatively simple form det Dll D12 Dl3 Dl2 022 0 23 X Dl3 0 23 033 - =0 (4.62) 90 CHAPTER... interdifFusion profile for this system 30 0 -0 .6 -0 .4 -0 .2 0 0.2 Distance (cm) Figure 4.8: Silva and Mehl [8] Concentration profile observed in Cu -3 0 at % Zn diffusion couple From Da 4 3 The Kirkendall effect can be studied by embedding an inert marker in the original step-function interface (x = 0) of the diffusion couple illustrated in Fig 3. 4 Show that this marker will move in the V-frame or, equivalently, with... diffusion is therefore isotropic Forms of the diffusivity tensor D for other crystal systems are tabulated in Nye's text [7] Bibliography 1 I.M Gelfand and S.V Fomin Calculus of Variations Prentice-Hall, Englewood Cliffs, NJ 19 63 2 W.C Carter, J.E Taylor, and J.W Cahn Variational methods for microstructural evolution J O M , 49(12) :3 0 -3 6 1997 3 P.M Morse and H Feshbach Methods of Theoretical Physics, Vols 1... McGrawHill, New York, 19 53 4 J Crank The Mathematics of Diffusion Oxford University Press, Oxford, 2nd edition 1975 5 C hlatano On relation between diffusion coefficients and concentrations of solid metals (the nickel-copper system) Jpn J Phys., 8 (3) :10 9-1 13, 1 933 6 P Shewmon Diffusion in Solids The Minerals, Metals and Materials Society, Warrendale, PA, 1989 7 J F Nye Physical Properties of Crystals Oxford... charge in a planar half-space where the plane bounding the half-space is held at zero potential The negative image charge outside the half-space allows superposition and satisfies the boiindary condition at the plane bounding the half-space [3] 85 4 3 DIFFUSIVITY AS A FUNCTION OF CONCENTRATION Each step function evolves according to an error-function solution of the type given by Eq 4 .31 and their superposition... the nondiffused ends of the specimen, according to x , where cy = at1I2 is a constant Solution According t o Eq 3. 23, the instantaneous velocity of any marker is given by (4.68) 93 EXERCISES Now, in general, c1 = C I ( ~ ) ,where q = x/& Because ( D l - 0 2 ) is a function of 7, we can represent it as [Dl - 0 2 1 = h(q) Similarly, we can write d c l / d x = (dcl/dq)t-’/’ = f ( 7 ) t-’l2 Putting these... profile generally contains information about the concentration dependence of the diffusivity For constant D , stepfunction initial conditions have the error function (Eq 4 .31 ) as a solution to dc/dt = Dd2c/dx2.When the diffusivity is a function of concentration, d C d2c dD(c) - = D(c )- + -( 8c)2 dt dx2 (4. 43) dc For identical initial conditions, the difference between a measured profile and the error-function . {fir(Rz -r2 )sin8+fie [ (R2 +r2 )cose+2Rr]} (3. 109) The force on an interstitial lying on a cylinder of radius R centered on the origin where [RZ + r2 + 2Rr cos el2 r = R is then. 3. 5 Suppose that a two-phase system consists of a fine dispersion of a carbide phase in a matrix. The carbide particles are in equilibrium with C dissolved interstitially in the matrix. dislocation are tangent to the dashed circles in the directions of the arrows shown in Fig. 3. 8. Solution. The hydrostatic stress on an interstitial in the stress field is given by Eq. 3. 80