# How do I calculate the Cr concentration in an Fe-C-Cr alloy?

Say, I have a 0.1 % C and 18 % Cr in a Fe-alloy. At a particular temperature, we have precipitate formation (of Cr23C6), which requires many Cr atoms. At a given temperature , how do I evaluate the remaining Cr atoms in the alloy ?

Any pointers for the analytical calculation would be greatly appreciated!

• What is the chemical reaction "eating" ? Not one I came across in school or university. Oct 4, 2021 at 5:09
• to my understanding ( a metallurgist can comment on this) the remaining Cr atoms (and the phases of the steel also depend on the rate of cooling.
– NMech
Oct 4, 2021 at 5:25
• This is a topic for a graduate seminar, not so much stack exchange. Estimate the relative proportions of each of several carbides that are present; and then estimate the chemical composition ( relative to Cr, Fe) for each one, I have seen similar work on Cr;Mo pressure vessel steels ( 3 Cr ;1 Mo) and it gets to complicated for me, Oct 4, 2021 at 21:09

Assume your initial concentrations are in kg/100 kg (wt%).

Find the initial amounts. Take a basis of 100 kg of total material. This gives you the initial masses $$m_j$$ of C, Cr, and Fe in your sample.

Prepare to do the chemical reaction. Convert the initial masses $$m_j$$ to initial moles $$n_j$$ using the molar mass $$M_j$$ with $$n_j = m_j/M_j$$.

Determine which component is limiting the final product. The reaction consumes 23 moles of Cr per 6 moles of C. Assume that the reaction goes to completion. Take the ratio of $$n_{Cr}/n_C$$ from the above. If it is greater than $$23/6$$, C is limiting. Otherwise, Cr is limiting.

Do the chemical reaction. Assume that the reaction has a yield of $$y$$ as a relative amount. A value $$y = 0$$ is no reaction and $$y = 1$$ is complete reaction. Use the moles of the limiting reactant (either C or Cr). The moles of the precipitate that form $$n_{Cr23C6}$$ is found by the expression $$n_{Cr23C6} = y\ n_C/6$$ if C is limiting or $$n_{Cr23C6} = y\ n_{Cr}/23$$ if Cr is limiting. The moles of C left is $$n_{C,f} = n_C - 6\ n_{Cr23C6}$$. The moles of Cr left is $$n_{Cr,f} = n_{Cr} - 23\ n_{Cr23C6}$$.

You now have the final composition in moles for C, Cr, Cr$$_{23}$$C$$_6$$, and Fe (unchanged). Determine the relative molar composition or convert the final moles back to masses and determine the relative mass composition.

You are asking the wrong question. You want to know if there is Cr depletion ( reduced concentration) near the Cr carbides which have effectively removed Cr from solution. There are corrosion tests to evaluate Cr depletion, known as sensitization. It can be remedied with thermal homogenization; stabilization. Much more a concern with austenitics than your 430SS

Either by an electron microprobe or by small-area xray diffraction.

• Sorry I f my question was misleading. I meant to calculate the Cr concentration analytically, not measurements. Oct 4, 2021 at 4:21