114 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION An estimate of the penetration distance for the error-function solution (Eq. 5.23) is the distance where c(x, t) = ~0/8, or equivalently, erf[x/(2m)] = -3/4: which corresponds to x RZ 1.6fi (5.71) A reasonable estimate for the penetration depth is therefore again 2m. To estimate the time at which steady-state conditions are expected, the required penetration distance is set equal to the largest characteristic length over which diffusion can take place in the system. If L is the characteristic linear dimension of a body, steady state may be expected to apply at times r >> L2/Dmin, where Dmin is the smallest value of the diffusivity in the body. Of course, there are many physical situations where steady-state conditions will never arise, such as when the boundary conditions are time dependent or the system is infinite or semi-infinite. Bibliography 1. P.M. Morse and H. Feshbach. Methods of Theoretical Physics, Vols. 1 and 2. McGraw- 2. J. Crank. The Mathematics of Diffusion. Oxford University Press, Oxford, 2nd edition, 3. H.S. Carslaw and J.C. Jaeger. Conduction of Heat in Solids. Oxford University Press, Hill, New York, 1953. 1975. Oxford, 2nd edition, 1959. EXERCISES 5.1 A flat bilayer slab is composed of layers of material A and B, each of thickness L. A component is diffusing through the bilayer in the steady state under conditions where its concentration is maintained at c = co = constant at one surface and at c = 0 at the other. Its diffusivity is equal to the constants DA and DB in the two layers, respectively. No other components in the system diffuse significantly. Does the flux through the bilayer depend on whether the concentration is maintained at c = co at the surface of the A layer or the surface of the B layer? Assume that the concentration of the diffusing component is continuous at the A/B interface. Solution. Solve for the difFusion in each layer and match the solutions across the A/B interface. Assume that c = co at the surface of the A layer and let c = c~/~ be the concentration at the A/B interface. Using Eq. 5.5, the concentration in the A layer in the interval 0 < x < L is (5.72) (5.73) For the B slab in the interval L 5 x 5 2L, (5.75) EXERCISES 115 Setting JA = JB and solving for cAIB, DA DA + DB co CAIB = The steady-state flux through the bilayer is then c0 D~D~ J=- L DA+DB (5.76) (5.77) J is invariant with respect to switching the materials in the two slabs, and therefore it does not matter on which surface c = CO. 5.2 Find an expression for the steady-state concentration profile during the radial diffusion of a diffusant through a cylindrical shell of thickness, AR, and inner radius, R'", in which the diffusivity is a function of radius D(r). The boundary conditions are C(T = R'") = c'" and c(r = R'" + AR) = coUt. Solution. The gradient operator in cylindrical coordinates is d 16 d dr rd0 dz V = -Cr + go + -Cz The divergence of a flux J'in cylindrical coordinates is - 1 d(rJr) 1 dJ0 dJ, V. J= + +- r dr r d0 dz Therefore, the steady-state radially-symmetric difFusion equation becomes which can be integrated twice to give (5.78) (5.79) (5.80) (5.81) The integration constant a1 is determined by the boundary condition at R'" + AR: (5.82) 5.3 Find the steady-state concentration profile during the radial diffusion of a diffusant through a bilayer cylindrical shell of inner radius, R'", where each layer has thickness AR/2 and the constant diffusivities in the inner and outer layers are Din and DOut. The boundary conditions are c(r = R'") = c'" and C(T = R'" + AR) = coUt. Will the total diffusion current through the cylinder be the same if the materials that make up the inner and outer shells are exchanged? Assume that the concentration of the diffusant is the same in the inner and outer layers at the bilayer interface. Solution. The concentration profile at the bilayer interface will not have continuous derivatives. Break the problem into separate difFusion problems in each layer and then impose the continuity of flux at the interface. Let the concentration at the bilayer interface be Inner region: R'" 5 r 5 Ri" + 116 CHAPTER 5. SOLUTIONS TO THE DIFFUSION EQUATION Using Eq. 5.82, The flux at the bilayer interface is Outer region: R'" + 5 r 5 Rin + AR r + $0 C~ut - ci/o R'"+AR In ( R1"+AR/2) In ( Rin + AR/2 COUt(T) = The flux at the bilayer interface is Setting the fluxes at the interfaces equal and solving for ci/O yields Cy~ut out + ainCin aout + Cyin ci/o = c where R'" + A R/ 2 (5.83) (5.84) (5.85) (5.86) (5.87) (5.88) Putting Eq. 5.87 into Eqs. 5.83 and 5.85 yields the concentration profile of the entire cylinder . The total current diffusing through the cylinder (per unit length) is Using Eq. 5.87, aout (Gout - Gin) ci/o - - - @out + (5.89) (5.90) If everything is kept constant except Din and DOut, use of Eq. 5.90 in Eq. 5.89 shows that nin nout uu IK CY~ DOut + CY~ Din (5.91) where a1 and a2 are constants. Clearly, I will be different if the materials making up the inner and outer shells are exchanged and the values of DOut and Din are therefore ex- changed. This contrasts with the result for the two adjoining flat slabs in Exercise 5.1. 5.4 Suppose that a very thin planar layer of radioactive Au tracer atoms is placed between two bars of Au to produce a thin source of diffusant as illustrated in Fig. 5.8. A diffusion anneal will cause the tracer atoms to spread by self-diffusion as illustrated in Fig. 5.3. (A mathematical treatment of this spreading out is presented in Section 4.2.3.) Suppose that the diffusion ex- EXERCISES 117 Thin source Figure 5.8: Thin planar tracer-atom source between two long bars. periment is now carried out with a constant electric current passing through the bars along x. (a) Using the statement of Exercise 3.10, describe the difference between the way in which the tracer atoms spread out when the current is present and when it is absent. (b) Assuming that DVCV is known, how could you use this experiment to determine the electromigration parameter p for Au? Solution. (a) The electric current produces a flux of vacancies in one direction and an equal flux of atoms in the reverse direction, so that 4 + JA = -Jv (5.92) Using the statement of Exercise 3.10, this will result in an average drift velocity for each atom, given by (5.93) The tracer atoms will spread out as they would in the absence of current: however, they will also be translated bodily by the distance Ax = (VA)t relative to an embedded inert marker as illustrated in Fig. 5.9. Inert em bedded marker (a) t=O 1 1 4 I I Figure 5.9: (a) The initially thin distribution of tracer atoms that, subsequently, will spread due to diffusion and drift due to electromigration. (b) The electromigration has caused the distribution to spread out and to be translated bodily by Ax = (VA)t relative to the fixed marker. 118 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION This may be shown by choosing an origin at the initial position of the source in a coordinate system fixed with respect to the marker. The diffusion equation is then (5.94) where (VA)C is the flux due to the drift. Defining a moving (primed) coordinate system with its origin at x = (‘UA)t, 2’ = X - (WA)t (5.95) Using [a( )/ax], = [a( )/ad]t, the drift velocity does not appear in the resulting diffusion equation in the primed coordinate system, which is d *c * a2 *c at ax‘2 -= D- The solution in this coordinate system can be obtained from Table 5.1; nd e-e’2/(4*Dt) *c(x/, t) = - dm (5.96) (5.97) The distribution therefore spreads independently of (wA), but is translated with velocity (VA) with respect to the marker. (b) The velocity (V)A can be measured experimentally and then p can be obtained through use of Eq. 5.93 if DVCV is known. It will be seen in Chapter 8 that DVCV can be determined by use of Eq. 8.17 if *D is known. *D can be determined from the measured distribution illustrated in Fig. 5.96 using Eq. 5.97 and the method outlined in Section 5.2.1. 5.5 Obtain the instantaneous plane-source solution in Table 5.1 by representing the plane source as an array of instantaneous point sources in a plane and integrating the contributions of all the point sources. Solution. Assume an infinite plane containing m point sources per unit area each of strength nd. The plane is located in the (y~) plane at x = 0. All the point sources in the plane lying within a thin annular ring of radius r and thickness dr centered on the z-axis will contribute a concentration at the point P located along the x-axis at a distance, x, given by nd e-(z2+r2)/(4Dt) (47rDt)3I2 dc = m27rr dr (5.98) where the point-source solution in Table 5.1 has been used. The total concentration is then obtained by integrating over all the point sources in the plane, so that where M = mnd is the total strength of the planar source per unit area. 5.6 Consider an infinite bar extending from -cc to +cc along x. Starting at t = 0, heat is generated at a constant rate in the x = 0 plane. Show that the temperature distribution along the bar is (5.100) EXERCISES 119 where P = power input at x = 0 (per unit area) and cp = specific heat per unit volume. Next, show that Finally, verify that this solution satisfies the conservation condition x T(z, t) cp dx = Pt Solution. The amount of heat added (per unit area) at z = 0 in time dt is Pdt. Using the analogy between problems of mass diffusion and heat flow (Section 4.1), each added amount of heat, Pdt, spreads according to the one-dimensional solution for mass diffusion from a planar source in Table 5.1: dT=-[ 1 Pdt ] e-z2/(4nt) CP 2(7rKt)1/2 (5.102) Because the term in brackets represents an incremental energy input per unit volume, the factor (cp)-’ must be included to obtain an expression for the corresponding incremental temperature rise, dT. Let Then 2 X2 a2=lG t=y- P a1 = - 2CP J J:/” exp (-my2) T(x,y) = -2a1 dY Y2 (5.103) (5.104) Integrating by parts and converting back to the variables (z, t) yields ~(z, t) = 2a1 e-azltd2 + 4a16 eCCZ d< (5.105) Substituting for a1 and a2, we finally obtain (5.106) Note that the solution given by Eq. 5.106 holds for z 2 0 because the positive root of & was used. The symmetric solution for z 5 0 is easily obtained by changing the sign of z. All the heat stored in the specimen at the time t is represented by the integral Q = 2 [w T(a, t) cp dx The first bracketed term in Eq. 5.107 has the value 2Pt. The second term can be integrated by parts and has the value Pt. Therefore, Q = 2Pt - Pt = Pt and the stored heat is equal to the heat generated during the time t, given by Pt. 120 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION 5.7 Consider the following boundary-value problem: dC -(z = 4m, t) = 0 ax 0 Use the superposition method to find the time-dependent solution. Show that when 26 >> a, the solution in (a) reduces to a standard in- stantaneous planar-source solution in which the initial distribution given by Eq. 5.108 serves as the source. 0 Use the following expansions for small E: (5.109) 2E 2 erf(z + E) = erf(z) + -e-' + * *. e' = 1 +E+ J;; Solution. (a) The concentration of diffusant located between 6 and 5 + d< in the initial dis- tribution acts as a planar source of thickness, d<, and produces a concentration increment at a distance, 2, given by (5.110) The total concentration produced at x is then obtained by integrating over the distribution. Therefore, Using the relations oe-u2 du = - J;; [erf(P) - erf(cu)] (5.112) 2 The solution is (5.113) x+a x-a c(2, t) =% 2a { erf ( -) x - erf ( T) EXERCISES 121 (b) Expanding Eq. 5.114 for small values of a/A = a/m produces the result c(x,t) = - (5.115) This is just the solution for a planar source of strength nd corresponding to the content per unit area of the original distribution given by Eq. 5.108. 5.8 (a) Find the solution c(z, y, z, t) of the constant-D diffusion problem where the initial concentration is uniform at CO, inside a cube of volume u3 centered at the origin. The concentration is initially zero outside the cube. Therefore, if 1x1 5 4 and Iy1 5 $ and 121 5 4 otherwise c(z, y, z, t = 0) = and c(z = fm, y = fml z = Am, t) = 0 (b) Show that when 2m >> a, the solution reduces to a standard instan- taneous point-source solution in which the contents of the cube serve as the point source. Use the erf(z + E) expansion in Eq. 5.109. Solution. (a) The method of superposition of point-source solutions can be applied to this problem. Taking the number of particles in a volume dV = dXdqdC equal to dN = co dXdqdC as a point source and integrating over all point sources in the cube using the point-source solution in Table 5.1, the concentration at x, y, z is c(x, Yl 2, t) co dX dqdC e-[(~-~)2+(y-q)2+(z-C)z]/(4Dt) (5.116) (4~Dt)~l~ The integral can be factored J-a/2 ' J-a/z The integrals all have similar forms. Consider the first one. Let u = (x-x)/a; then 122 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION Therefore, the solution can be written x - a/2 (b) Expansion of Eq. 5.119 using Eq. 5.109 produces the result cga3 -r2/(4Dt) (47rDt)3/2 C= (5.119) which is just the solution for a point source containing the contents of the cube corresponding to cga3 particles. 5.9 Determine the temperature distribution T = T(z, y, 2, t) produced by an ini- tial point source of heat in an infinite graphite crystal. Plot isothermal curves for a fixed temperature as a function of time in: (a) The basal plane containing the point source (b) A plane containing the point source with a normal that makes a 60" (c) A plane containing the c-axis and the point source angle with the c-axis The thermal diffusivity in the basal plane is isotropic and the diffusivity along the c-axis is smaller than in the basal plane by a factor of 4. Solution. Using Eq. 4.61 and the analogy between mass diffusion and thermal diffusion, the basic differential equation for the temperature distribution in graphite can be written (5.120) where 21 and 22 are the two principal coordinate axes in the basal plane and 23 is the principal coordinate along the c-axis. In order to make use of the point-source solution for an isotropic medium as in Sec- tion 4.5, rescale the axes Then Eq. 5.120 becomes dT at - = (RiRL) The solution of Eq. 5.122 for the point source in (Table 5.1) 53 = 1/6 6 (5.121) (qh) (5.122) these coordinates then has the form T - e-(€:+Eg+E~)/[4(kini)1/3tl (5.123) t3/2 where cy is a constant. Converting back to the principal axis coordinates yields (5.124) EXERCISES 123 (a) Isotherms in the basal plane: In the basal plane passing through the origin, 3% = 0 (5.125) Q and T 22, %3 = 0, t) = - ,-(*?+*P)/(4kllt) t3/2 Isotherms for a fixed temperature at increasing times are shown in Fig. 5.10. They are circles, as expected, because the thermal conductivity is isotropic in the basal plane. Initially, the isotherms spread out and expand because of the heat conduction but they will eventually reverse themselves and contract toward the origin, due to the finite nature of the initial point source of heat. h i Figure 5.10: passes through the origin. Isotherms for a fixed temperature at increasing times in a basal plane that (b) Isotherms in a 60" inclined plane: The isotherms on a plane with a normal in- clined 60" with respect to the c-axis can be determined by expressing the solution (Eq. 5.124) in a new coordinate system rotated 60" about the 21 axis. The new (primed) coordinates are 0 ( ) = ( ;os60" sin60' ) ( ii ) 0 -sin60" cos60" In the new coordinates, with zb = 0, the temperature profile in the inclined plane passing through the origin is Figure 5.11 shows the isotherms as a function of time. Again the curves expand and contract with increasing time. However, the isotherms are elliptical because the thermal conductivity coefFicient is different along the c-axis and in the basal plane. [...]... c1 c1(-Co,t) = C1(-Z,O) = cf C2(Co,t) c2(-Co,t) R = C 2 ( Z , O ) = c2 L = c2(-x,O) = c2 (6.57) and therefore R c2 L c2 = a1 + + + a4 =a1 - a2 a2 a3 + a3 - a4 Solving the four equations above for the four coefficients, (6.59) a4 = (022 - Dl1 + A)(c,R - c;) + 2D21(~ ?- c?) 4A Substituting Eq 6.59 into Eq 6.56, the final solution for c2 cz" c2 is + ck = _ 2 + + (011 (D22 -0 22 - Dli + A)(c,R - c;) -2 021(~p... ;I (6.32) and c, and cp are the two concentrations given by [;] ; ;] ; =A-l[ (6.33) These quantities therefore have the forms - D22 + A J , = -J i + A 2A D2 1 D l l - D22 - A Jp = -J1 A 2A -0 21 Dii J2 (6. 34) 52 and -0 21 c, = - A cp D2 1 A =-c1 c1 + - Dll- D22 +A c2 2A Dll - D22 - A 2A (6.35) c2 The flux problem can now be easily solved in the diagonalized system using Eq 6.31; the solution can then be... a1 = 0: co - + u2 e P (5. 145 ) - m m transform the boundary condition given by Eq 5. 141 t o obtain -D (g) = x=o -a E(0,p) (5. 146 ) Therefore, putting Eq 5. 145 into Eq 5. 146 yields CY Cn (5. 147 ) and (5. 148 ) where h = a / D Find the desired solution by taking the inverse transform using a table o f transforms t o obtain c(x,t ) = co [erf (m)+ X eh=+h2gt erfc (& +h m ) ] (5. 149 ) (b) From Eq 5. 149 the concentration... following series expansions of erf(z): For small z , (5. 142 ) and for large z , 1 1 1 x 3 - - - + 1 X 3 X 5 erfc(z) = 1- erf(z) = 2 3 9 + ) (5. 143 ) (c) Give a physical interpretation of the results in (b) Solution (a) Use the Laplace transform method Transforming the diffusion equation along with the initial condition given by Eq 5. 141 yields the same result as Eq 5. 64: (5. 144 ) The solution is therefore... are step-function concentration profiles and hence error-function profiles having the form (6. 54) are tried as solutions t o the uncoupled diffusion equations given by Eq 6 .44 Using Eqs 6.26 and 6.37, the relationships between the concentrations c1 and c2 and the concentrations ca and cp in the diagonalized system are DII - 0 2 2 - A ‘Or 2021 cz = co cg + c1 = + + DII - DZZ A cp 2021 (6.55) 142 CHAPTER... State Institute of Metals, London, 1987 2 A Vignes and J.P Sabatier Ternary diffusion in Fe-Co-Ni 245 :179 5-1 802, 1969 alloys Trans TMS-AIME, 3 C Cserhati, U Ugaste, M.J.H van Dal, N.J.H.G.M Lousberg, A.A Kodentsov, and F.J.J van Loo On the relationship between interdiffusion and tracer diffusion coefficients in ternary solid solutions Defect and Diffusion Forum, 19 4- 1 99:18 9-1 94, 2001 4 M.E Glicksman... independent of concentration EXERCISES 143 Solution Assume solutions in the diagonalized system having the standard point- source form = C, (-w) r2 cp = (4. iixa+Rt)3/2 exp (6.61) Using Eqs 6.26 and 6.37, c1 = Dii - 0 2 2 - A 2021 c2 = c, cg + I -t Dll - D~~+ A CP 2021 (6.62) Conservation o f components 1 and 2 requires that W 47 ~ 4n r c l r 2 d r =N1 = constant (6.63) W c2 7-2 dr = N ~ constant = Jo By substituting... and requiring that Eq 6.63 be satisfied, Dl1 - 0 2 2 - A 2021 + + Dl1 - 0 2 2 A a g = N1 2021 a, a p = N2 + (6. 64) By solving Eq 6. 64 for a, and a g and putting the results into Eq 6.62 with the help of Eq 6.61, the final solution for c2 is + + exp (1 + (Dl1 - + A)N2 - 2021N c2 = 2D21N1 ( A 0 2 2 - D11)Nz 16A(7 ~ X + t ) 3 / 2 022 16A(rX- t)3/2 exp &) (-A) (6.65) A similar expression may be found for... c;) -2 021(~p ~ - + A)($ + 2 0 2 1 ( ~ ?- c?) erf 4A - ck) 4A f ) A similar expression holds for c1 6.2 Point sources of components 1 and 2, containing N1 and Nz atoms, respec- tively, are located at the origin r = 0 in a large piece of pure component 3 Solve for the resulting three-dimensional diffusion field, assuming that the diffusivities D l l , 0 2 1 , and 0 2 2 , are independent of concentration... ternary case, (6 .43 ) Time-Dependent Solutions In the time-dependent case, the diffusion equations given by Eq 6.23 are coupled However, they can be uncoupled by again using the diagonalizat ion met hod Using the transformation matrix A on Eq 6.23 gives for the ternary case, a -[ at and for the general case, I.' = [ ; :+Iv2[ - cg A1 a 0 - at : 0 0.'' x 0 2 0 :] ; (6 .44 ) 0 0 * 0 : (6 .45 ) 0 AN-1 where the . gradients are present and all other driving forces-such as thermal gradients or electric fields- 'Such defects, if present, will be assumed to be in local thermal equilibrium at very. Thin source Figure 5.8: Thin planar tracer-atom source between two long bars. periment is now carried out with a constant electric current passing through the bars along x. (a) Using. (c) We can rewrite hm as am. Therefore, as cy becomes small, or at short times t, or as D increases, c(0, t) approaches CO. For small a, surface desorption is compensated