Question 8 of 8: Carbide/Nitride Solubility in Austenite; AlN in Hot-Rolled Steel

Question 8: Carbide/Nitride Solubility in Austenite; AlN in Hot-Rolled Steel (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

8.1 — (a) Deriving Eq. (1) and its assumptions

For the reaction as printed, $[M]_\gamma + [X]_\gamma = [MX]$ (formation of the compound from dissolved solute), the standard free energy of reaction relates to the equilibrium constant by $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = -RT\ln K, \qquad K = \frac{a_{MX}}{a_M\,a_X}$$ where $a_M$, $a_X$ are the activities of dissolved M and X in austenite, $a_{MX}$ is the activity of the (solid) compound MX, and $\Delta H^{\circ}$, $\Delta S^{\circ}$ are for this formation reaction. Taking $\log_{10}$: $$\log_{10}\!\left(\frac{a_M\,a_X}{a_{MX}}\right) = \frac{\Delta H^{\circ}}{2.303\,R\,T} - \frac{\Delta S^{\circ}}{2.303\,R}$$ Matching this to the printed form $\log_{10}[\text{wt.\%M}][\text{wt.\%X}] = A - B/T$ requires three simplifying assumptions:

1. MX is a stoichiometric, essentially pure solid compound. Taking $a_{MX}=1$ removes it from the equilibrium expression, leaving the solubility product $a_M a_X = 1/K$ on the left-hand side — valid only if MX carries negligible solubility for other species and departs negligibly from ideal stoichiometry.
2. M and X obey Henry's law (dilute-solution behaviour) in austenite. At the low concentrations typical of microalloying/interstitial additions, the activities are taken proportional to concentration, $a_M=\gamma_M[\text{wt.\%M}]$ with $\gamma_M$ treated as a constant independent of composition; this lets the printed equation use raw weight-percent values in place of thermodynamic activities, with the (assumed T-independent, or only present through $A$) activity coefficients absorbed into the fitted constant $A$.
3. $\Delta H^{\circ}$ and $\Delta S^{\circ}$ of the reaction are independent of temperature over the fitted range (equivalent to neglecting any heat-capacity difference between products and reactants, $\Delta C_p\approx0$), which is what allows $A$ and $B$ to be reported as single constants rather than functions of $T$.

Under these three assumptions, $A=-\Delta S^{\circ}/2.303R$ and $B=-\Delta H^{\circ}/2.303R$ (both positive, because forming the nitride/carbide from solution is exothermic and lowers entropy; the Henry's-law activity coefficients are folded into $A$), i.e. Eq. (1) is exactly the van't-Hoff form of the solubility product, re-expressed in base-10 log and weight-percent units.

8.2 — (b) Maximum tolerable Al at 900 °C for 40 ppm N, without AlN precipitation

Given. $\log_{10}[\text{wt.\%Al}][\text{wt.\%N}] = A - B/T$ with $A=1.55$, $B=7060$; hot-rolling temperature $900\,{}^{\circ}\text{C} = 1173.15\ \text{K}$; nitrogen content $40\ \text{ppm} = 0.0040\ \text{wt.\%N}$.

Find. The maximum wt.%Al that keeps the steel just below the AlN solubility limit (i.e. the solubility PRODUCT, not exceeded) at 900 °C.

Approach. Evaluate the solubility product from Eq. (1) at $T=1173.15\ \text{K}$, then divide by the known wt.%N to isolate the maximum wt.%Al.

1. Evaluate the right-hand side of Eq. (1) at 900 °C. $$\log_{10}[\text{Al}][\text{N}] = 1.55 - \frac{7060}{1173.15} = 1.55 - 6.018 = -4.468$$
2. Convert to the solubility product. $$[\text{wt.\%Al}][\text{wt.\%N}] = 10^{-4.468} = 3.404\times10^{-5}$$
3. Solve for the maximum wt.%Al at the given nitrogen level. With $[\text{wt.\%N}]=0.0040$, $$[\text{wt.\%Al}]_{\max} = \frac{3.404\times10^{-5}}{0.0040} = 8.511\times10^{-3}\ \text{wt.\%}$$ $$\boxed{[\text{wt.\%Al}]_{\max} = 0.00851\ \text{wt.\%} \approx 85.1\ \text{ppm}}$$

Provided the steel's total aluminum content stays at or below about 85 ppm (0.0085 wt.%) at a 40 ppm nitrogen level, the product $[\text{Al}][\text{N}]$ remains below the Eq. (1) solubility limit at 900 °C and AlN will not precipitate during hot rolling; any aluminum content above this — common in fully aluminum-killed steels, which typically carry well over 200–300 ppm total Al — will drive AlN precipitation at this temperature, which is in fact the intended microalloying effect used for austenite grain-pinning in those grades, at the cost of the free (uncombined) nitrogen available for other reactions.