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Unformatted text preview: CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS 5.1 Selfdiffusion is atomic migration in pure metalsi.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal. 5.2 Selfdiffusion may be monitored by using radioactive isotopes of the metal being studied. The motion of these isotopic atoms may be monitored by measurement of radioactivity level. 5.3 (a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism. (b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom. 5.4 Steadystate diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing speciesi.e., the diffusion flux is independent of time. 5.5 (a) The driving force is that which compels a reaction to occur. (b) The driving force for steadystate diffusion is the concentration gradient. 5.6 This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to employ both Equations (5.1a) and (5.3). Combining these expressions and solving for the mass yields M = JAt =  DAt C x =  1.0 x 108 m 2 /s ( 29 0.2 m 2 ( 29 (3600 s/h) 0.6  2.4 kg / m 3 5 x 10 3 m 85 = 2.6 x 103 kg/h 5.7 We are asked to determine the position at which the nitrogen concentration is 2 kg/m 3 . This problem is solved by using Equation (5.3) in the form J =  D C A C B x A x B If we take C A to be the point at which the concentration of nitrogen is 4 kg/m 3 , then it becomes necessary to solve for x B , as x B = x A + D C A C B J Assume x A is zero at the surface, in which case x B = 0 + 6 x 1011 m 2 /s ( 29 4 kg / m 3 2 kg / m 3 ( 29 1.2 x 10 7 kg / m 2 s = 1 x 103 m = 1 mm 5.8 This problem calls for computation of the diffusion coefficient for a steadystate diffusion situation. Let us first convert the carbon concentrations from wt% to kg C/m 3 using Equation (4.9a). For 0.012 wt% C C C '' = C C C C C + C Fe Fe x 10 3 = 0.012 0.012 2.25 g / cm 3 + 99.988 7.87 g / cm 3 x 10 3 0.944 kg C/m 3 86 Similarly, for 0.0075 wt% C C C '' = 0.0075 0.0075 2.25 g / cm 3 + 99.9925 7.87 g/ cm 3 x 10 3 = 0.590 kg C/m 3 Now, using a form of Equation (5.3) D =  J x A x B C A C B =  1.40 x 108 kg/m 2 s ( 29 10 3 m 0.944 kg / m 3 0.590 kg / m 3...
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This homework help was uploaded on 04/17/2008 for the course MATL 2100 taught by Professor Kim during the Spring '08 term at Auburn University.
 Spring '08
 KIM

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