HMWK 5 Solutions

HMWK 5 Solutions - HW 5 Due on Dec. 2 5.1 Briefly explain...

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HW 5 Due on Dec. 2 5.1 Briefly explain the difference between self-diffusion and interdiffusion 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. 5.6 The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m 2 at 500 ° C. Assume a diffusion coefficient of 1.0 × 10 -8 m 2 /s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained. 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 × 10 -8 m 2 /s )( 0.20 m 2 ) (3600 s/h) 0.6 - 2.4 kg/m 3 5 × 10 - 3 m = 2.6 × 10 -3 kg/h 5.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200 C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 10 -11 m 2 /s, and the diffusion flux is found to be 1.2 10 -7 kg/m 2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m 3 . How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m 3 ? Assume a linear concentration profile. 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
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Assume x A is zero at the surface, in which case x B = 0 + ( 6 × 10 -11 m 2 /s ) 4 kg/m 3 - 2 kg/m 3 1.2 × 10 - 7 kg/m 2 -s = 1 × 10 -3 m = 1 mm 5.8 A sheet of BCC iron 1 mm thick was exposed to a carburizing gas atmosphere on one
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HMWK 5 Solutions - HW 5 Due on Dec. 2 5.1 Briefly explain...

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