CAP5 callister

5-8

5.8 This problem calls for computation of the diffusion coefficient for a steady-state diffusion situation.
Let us first convert the carbon concentrations from weight percent to kilograms carbon per meter cubed using
Equation 4.9a. For 0.015 wt% C
CC

CC" =

CC ρC +

C Fe

× 10 3

ρ Fe

0.015

=

0.015
2.25 g /cm3

+

99.985

× 10 3

7.87 g /cm3

1.18 kg C/m3

Similarly, for 0.0068 wt% C
CC" =

0.0068
0.0068
2.25 g /cm3

+

99.9932

× 10 3

7.87 g /cm 3

= 0.535 kg C/m3

Now, using a rearranged form of Equation 5.3
⎡x − x ⎤
B⎥
D = − J⎢ A
⎢⎣ CA − C B ⎥⎦
⎡
⎤
− 2 × 10−3 m
⎥
= − (7.36 × 10-9 kg/m2 - s)⎢
⎢⎣ 1.18 kg / m3 − 0.535 kg / m3 ⎥⎦

= 2.3 x 10-11 m2/s

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5-14

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 C0 = 0.55 wt% C. From Equation 5.5
C x − C0
⎛ x ⎞
0.25 − 0.55
=
= 0.5455 = 1 − erf ⎜
⎟
⎝ 2 Dt ⎠
0 − 0.55
Cs − C0

Thus,
⎛ x ⎞ erf ⎜
⎟ = 0.4545
⎝ 2 Dt ⎠

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
0.4545 − 0.4284
=
0.45 − 0.40 0.4755 − 0.4284

And, z = 0.4277

Which means that x = 0.4277
2 Dt

And, finally

x = 2(0.4277) Dt = (0.8554)

(4.3 × 10−11 m2 /s)( 3.6 × 10 4 s)

= 1.06 x 10-3 m = 1.06 mm
Note: this problem may also be solved using the “Diffusion” module in the VMSE software. Open the “Diffusion” module, click on the “Diffusion Design” submodule, and then do the following:

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