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Unformatted text preview: CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS 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 5.12 This problem asks that we determine the position at which the carbon concentration is 0.25 wt% after a 10h heat treatment at 1325 K when C o = 0.55 wt% C. From Equation (5.5) C x C o C s C o = 0.25 0.55 0.55 = 0.5455 = 1  erf x 2 Dt Thus, erf x 2 Dt = 0.4545 Using data in Table 5.1 and linear interpolation z erf (z ) 0.40 0.4284 z 0.4545 0.45 0.4755 z 0.40 0.45 0.40 = 0.4545 0.4284 0.4755 0.4284 And, z = 0.4277 Which means that x 2 Dt = 0.4277 And, finally x = 2(0.4277) Dt = (0.8554) 4.3 x 10 11 m 2 / s ( 29 3.6 x 10 4 s ( 29 = 1.06 x 103 m = 1.06 mm 5.24 This problem asks that we compute the temperature at which the diffusion flux is 6.3 x 10 10 kg/m 2s. Combining Equations (5.3) and (5.8) yields J = D o ∆ C ∆ x exp Q d RT Solving for T from this expression leads to T = Q d R 1 ln D o ∆ C J ∆ x = 80, 000 J / mol 8.31 J / mol K 1 ln 6.2 x 10 7 m 2 / s ( 29 0.45 kg / m 3 ( 29 6.3 x 10 10 kg / m 2 s ( 29 10 2 m ( 29 = 900 K = 627 ° C 5.30 In order to compute the diffusion time at 900 ° C to produce a carbon concentration of 0.75 wt % at a position 0.5 mm below the surface we must employ Equation (5.6b) with position constant; that is Dt = constant Or D 600 t 600 = D 900 t 900 In addition, it is necessary to compute both D 600 and D 900 using Equation (5.8). From Table 5.2, for the diffusion of C in α Fe, Q d = 80,000 J/mol and D o = 6.2 x 107 m 2 /s. Therefore, D 600 = 6.2 x 107 m 2 /s ( 29 exp 80,000 J/ mol (8.31 J/ mol  K)(600 + 273 K) = 1.01 x 1011 m 2 /s D 900 = 6.2 x 107 m 2 /s ( 29 exp 80,000 J/ mol (8.31 J/ mol  K)(900 + 273 K) = 1.69 x 10= 1....
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This note was uploaded on 05/03/2010 for the course ME 250750 taught by Professor Signer during the Summer '10 term at Wichita State.
 Summer '10
 Signer

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