Solutions HW Assignments - Ch 5-7

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 10-11 m 2 /s ( 29 4 kg / m 3- 2 kg / m 3 ( 29 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- 0.55 = 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 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 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 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 ( 29 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 ( 29 exp- 80,000 J/ mol (8.31 J/ mol - K)(900 + 273 K) = 1.69 x 10= 1....
View Full Document

Page1 / 13

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online