Chap 05 Solns-6E - CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS...

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CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS 5.1 Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal. 5.2 Self-diffusion 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 Steady-state 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 species--i.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 steady-state 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 = - ∆Ατ ∆Χ ∆ξ = - 1.0ξ10 -8 μ 2 ( 29 0.2μ 2 ( 29 (3600σ/η29 0.6 - 2.4 κγ /μ 3 5 ξ 10 -3 μ 85
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= 2.6 x 10 -3 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 = - ∆ Χ Α - Χ Β ξ Α - ξ Β 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 - Χ Β ϑ Assume x A is zero at the surface, in which case x B = 0 + 6 x 10 -11 m 2 /s ( 29 4 κγ /μ 3 - 2 κγ /μ 3 ( 29 1.2 ξ 10 -7 κγ /μ 2 = 1 x 10 -3 m = 1 mm 5.8 This problem calls for computation of the diffusion coefficient for a steady-state 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 ρ Χ + Χ Φε ρ Φε ξ10 3 = 0.012 0.012 2.25 g/ cm 3 + 99.988 7.87 γ / χμ 3 ξ10 3 0.944 kg C/m 3 86
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Similarly, for 0.0075 wt% C C C '' = 0.0075 0.0075 2.25 g/ cm 3 + 99.9925 7.87 γ / χμ 3 ξ10 3 = 0.590 kg C/m 3 Now, using a form of Equation (5.3) D = - ϑ ξ Α - ξ Β Χ Α - Χ Β = - 1.40ξ10 -8 κγ/μ 2 ( 29 - 10 -3 μ 0.944 κγ /μ 3 - 0.590 κγ / μ 3 = 3.95 x 10 -11 m 2 /s 5.9
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This homework help was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona- Tucson.

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Chap 05 Solns-6E - CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS...

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