Solving the Concentration Gradient: A Calculation Challenge

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SUMMARY

The forum discussion centers on calculating the nitrogen flux through a 1-mm sheet of FCC iron in a heat exchanger at 1200℃. The concentrations of nitrogen at the two surfaces are 0.04 atomic percent and 0.005 atomic percent. The correct calculation for the concentration gradient involves using the atomic volume of the FCC structure, leading to a flux of 7.57 x 1012 N atoms/cm2-s. The user identified an error in the solution manual regarding the volume of the cell in the denominator of the calculation.

PREREQUISITES
  • Understanding of concentration gradient calculations
  • Familiarity with FCC (Face-Centered Cubic) crystal structures
  • Knowledge of diffusion equations in materials science
  • Basic thermodynamics, particularly the Arrhenius equation
NEXT STEPS
  • Study the derivation of the diffusion equation in materials science
  • Learn about the properties of FCC iron and its atomic structure
  • Explore the application of the Arrhenius equation in diffusion processes
  • Investigate methods for calculating concentration gradients in solid materials
USEFUL FOR

This discussion is beneficial for materials scientists, chemical engineers, and students studying diffusion processes in solid-state physics or materials science.

tkuehl
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I have a problem with calculating the concentration gradiant. Here is the question and the solution from the solution manual and the numbers don't add up.

A 1-mm sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200℃. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. Determine the flux of nitrogen through the foil in atoms/cm2−s.

solution
(a) Ac/Ax = [.00005 - .0004]*4 atoms/cell(3.589E-8)^3
--------------------------------------------
.1cm
=-3.03 x 1020 N atoms/cm3-cm

(b) J=-D(Ac/Ax) = -0.0034 exp[-34,600/(1.987)(1473)](-3.03 x 1020)
= 7.57 x 1012 N atoms/cm2-s

I would like to know how they get the answer in part a.
 
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I figured it out. the volume of the cell needs to be in the denominator. Our professor takes everything out of the solutions manual and never checks it.