Chap 05 Solns-6E

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

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CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS 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 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 85

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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 10 -3 m = 1.06 mm 5.13 This problem asks us to compute the nitrogen concentration ( C x ) at the 1 mm position after a 10 h diffusion time, when diffusion is nonsteady- state. From Equation (5.5) C x - C o C s - C o = C x - 0 0.1 - 0 = 1 - erf x 2 Dt = 1 - erf 10 - 3 m (2) 2.5 x 10 - 11 m 2 /s ( 29 (10 h)(3600s /h) = 1 - erf (0.527)
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## This note was uploaded on 08/30/2011 for the course EGN 3365 taught by Professor Staff during the Spring '11 term at University of Florida.

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

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