Solutions HW Assignments - Ch 5-7

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

This preview shows pages 1–5. Sign up to view the full content.

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

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

View Full Document
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
= 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

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

View Full Document
5.31
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online