Solutions HW Assignments - Ch 5-7

Solutions HW Assignments - Ch 5-7 - CHAPTER 5 DIFFUSION...

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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 10 -11 m 2 /s () 4 kg/m 3 2 kg/m 3 ( ) 1.2 x 10 7 kg/m 2 -s = 1 x 10 -3 m = 1 mm 5.12 This problem asks that we determine the position at which the carbon concentration is 0.25 wt% after a 10-h 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 0.55 = 0.5455 = 1 erf x 2D t Thus, erf x t = 0.4545
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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 = 0.4545 0.4284 0.4755 0.4284 And, z = 0.4277 Which means that x 2D t = 0.4277 And, finally x = 2(0.4277) Dt = (0.8554) 4.3 x 10 11 m 2 /s ( ) 3.6 x 10 4 s ( ) = 1.06 x 10 -3 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 2 -s. Combining Equations (5.3) and (5.8) yields J = D o Δ C Δ x exp Q d RT Solving for T from this expression leads to T = Q d R 1 ln D o Δ C J Δ x
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= 80,000 J /mol 8.31 J /mol- K 1 ln 6.2 x 10 7 m 2 /s () 0.45 kg/m 3 6.3 x 10 10 kg/m 2 -s 10 2 m = 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 10 -7 m 2 /s. Therefore, D 600 = 6.2 x 10 -7 m 2 /s exp 80,000 J/mol (8.31 J/mol - K)(600 + 273 K) = 1.01 x 10 -11 m 2 /s D 900 = 6.2 x 10 -7 m 2 /s exp (8.31 J/mol - K)(900 + = 1.69 x 10 -10 m 2 /s Now, solving for t 900 gives t 900 = D 600 t 600 D 900 1.01 x 10 11 m 2 (100 min) 1.69 x 10 10 m 2 = 5.98 min
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5.31
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Solutions HW Assignments - Ch 5-7 - CHAPTER 5 DIFFUSION...

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