MSE 230 Assignment 7 Solutions Fall 2013 Problems with an by them will not be graded but solutions will be provided 1 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 0 6 2 4 kg m3 1 0 x 10 m s 0 2 m 3600 s h 5 x 10 3 m 8 2 2 2 6 x 10 2 3 kg h Carbon should diffuse through iron more rapidly than chromium and chromium should have a higher activation energy for diffusion Carbon is much smaller than iron and may diffuse via interstitial sites Chromium is close in size to iron and diffuses by a vacancy diffusion mechanism Therefore for chromium to diffuse it must exchange lattice positions with a vacant lattice site The extra requirement for vacancy formation results in a higher activation energy for chromium 3 On the same graph sketch qualitatively correct plots of ln D vs 1 T for i carbon diffusing through BCC iron and ii radioactive iron diffusing through BCC iron Please provide reasons for any differences between the two plots The plot reflects the fact that carbon should diffuse more rapidly through iron than iron diffuses through itself This is because carbon diffuses through iron via interstitial sites while iron needs vacancies to diffuse The higher absolute value of the carbon diffusivity represents the more rapid diffusion rate while the lower slope for carbon diffusivity represents the lower activation energy required for interstitial diffusion relative to vacancy diffusion 4 In order to solve this problem we must first compute the value of D from the data given at 1000 K o this requires the combining of both Equations 5 3 and 5 8 Solving for D from these expressions o gives J D exp o C x 5 4 x 10 10 kg m2 s 125000 J mol exp 8 31 J mol K 1000 K 4 350 kg m 5 26 x 10 6 2 m s The value of the diffusion flux at 1300 K may be computed using these same two equations as follows J D 5 26 x 10 6 Q C exp d o x RT 125000 J mol 8 31 J mol K 1300 K 2 4 m s 350 kg m exp 1 74 x 10 8 2 kg m s 5 5 15 This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0 45 wt at a point 5 mm from the surface From Equation 5 6b 2 x constant Dt But since the temperature is constant so also is D constant and 2 x constant t or x t 2 1 x 1 t 2 2 2 Thus 2 5 mm 10 h 2 5 0 mm t 2 2 from which t 40 h 2 6 This is a nonsteady state diffusion situation thus it is necessary to employ Equation 6 5 utilizing the following values for the concentration parameters C 0 0025 wt N o C 0 45 wt N s C 0 12 wt N x Therefore C C x o C C s o 0 12 0 0025 0 45 0 0025 x 0 2626 1 erf 2 Dt And thus x 0 7374 erf 2 Dt Using linear interpolation and the data presented in Table 6 1 z erf z 0 75 0 7112 y 0 7374 0 80 0 7421 0 7374 0 7112 y 0 75 0 7421 0 7112 0 80 0 75 From which y x 2 Dt 0 7924 The problem stipulates that x 0 45 mm 4 5 x 10 4 m Therefore 4 5x10 4 m 0 7924 2 Dt Which leads to Dt 8 063 x 10 8 m 2 Furthermore the diffusion coefficient depends on temperature according to Equation 6 8 and as 7 2 stipulated in the problem D 3 x 10 m s and Q 76 150 J mol Hence o d Qd 8 2 Dt D exp o RT t 8 063 x 10 m 3 0 x 10 7 2 m s exp 76150 t 8 063 x 10 8 m2 8 31 J mol K T And solving for the time t t in s 0 269 9163 7 exp T Thus the required diffusion time may be computed for some specified temperature in K Below are temperatures that lie within the range stipulated in the problem tabulated t values for three different The higher the temperature the higher the diffusivity and the less time taken to reach a concentration of 0 12 wt nitrogen at a depth of 0 45 mm below the surface Temperature C K Time s h 500 773 37 861 10 52 550 823 18 425 5 12 600 873 9 738 2 70 7 Recrystallization Time Recrystallization Temperature The time is takes for recrystallization to occur decreases with increasing temperature This is because at higher temperatures the rate of diffusion is higher Since atoms move around by self diffusion during recrystallization the higher the temperature as long as it stays below the melting temperature the shorter the recrystallization time
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