Question

A sheet of Fe 1.0 mm thick is exposed to an oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m^2/sec) if the diffusion flux is 1.4×10^-8kg/m^2-sec.

(a) 9.87*10^-12

(b) 9.87*10^-13

(c) 9.87*10^-11

(d) 9.87*10^-10

The question was posed to me during an interview for a job.

I would like to ask this question from Fick’s Law-2 topic in section Modes and Diffusion of Mass Transfer

Answer 1

Correct answer is (a) 9.87*10^-12

For explanation: Convert from wt% to kg/m^3 .

                          Concentration in kg/m^3=[C1/(C1/ρ1+C2/ρ2)]*1000

                          0.012% and 0.075% in kg/m^3 are 0.270 and 1.688 respectively.

                          Putting all the values in the equation JA=(-)DABdC/dx.