Question

Problem 5.10

A sheet of BCC iron 4.9-mm thick was exposed to a carburizing atmosphere on one side and a decarburizing atmosphere on the other side at 725°C. After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces were determined to be 0.011 and 0.0074 wt%. Calculate the diffusion coefficient if the diffusion flux is 2.6 × 10^-8 kg/m²-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm³, respectively.

Hint: Use Equation 4.9a—that is, for this problem

to convert concentrations from weight percent carbon to kilograms of carbon per cubic meter or iron.

        

m2/s

Answer 1

Solution Let us first convert the carbon concentrations from weight percent to kilograms Carbon pere Meter cubed. For 0.011 Wtc. to kilograms Canne Carbon concent co" = x103 Cc ce + Cfe Sc ffe 0. ON to3 0.011 + 99.989 2.25 9/m3 . 7.87 %/cm3 = 0.865 kg/m3 For 0.0074 wt%C: 0.0074 ce" = 0.0074 99.9926 + 1 2.25 g/m3 . 7-879/43 = 0.5822 kgspot. Now using a rearanged form of equations 5.2, D = -J 0:-Jan en Where J= diffusion Flux 2-1 [ and incorporating values of plate thickness and diffusion FLUX provided in the problem 14ete ment, as well as the two concentrations of carbon determined above, the diffusion coefficient is computed as follows:

D = -(2.6*10-8 c o r 4.9 10-3 m. /m?s ) [ 0.5822 kg - 0.865 kg] 0.5822 0-865 D = 450-495 * 10-12 mºs or, D= . 4.50*10-10 m/ Hope This Helps. Thumbs Up please. Thank You.