Evolution of Chemical Composition and Modeling of Growth Nonmetallic Inclusions in Steel Containing Yttrium

Abstract

The programs WYK_Stal and Bi-Growth, developed at AGH-UST, Kraków, Poland, were used for simulating the refining process, the formation of non-metallic inclusions, and their growth. The Fe-Y-Al-O-S-Ca system in pre-oxidized steel was analyzed, where yttrium formed precipitates from both O and S. When first Al and second Y were added to steel, the proportion of Al2O3 inclusions remained constant. This resulted in higher yttrium losses for oxide formation, whereas the sulfur content promoted sulfide phase formation. The introduction of yttrium at the end of refining contributed to reducing the consumption of this element in the non-metallic phase formation. The addition of aluminum and then calcium were sufficient to achieve a high degree of deoxidation and desulfurization. Calculations performed with WYK_Stal for both (a) and (c) versions of the model showed that the sulfide phase was constituted by CaS and FeS (model c) and CaS (model (a)). The participation of the calcium sulfide phase turned out to be dominant in the inclusions. Their presence was also identified in the slag phase. Simulations of the growth of complex oxide and oxo-sulfide inclusions using the Bi_Growth program showed that the yttrium content of the steel has a decisive role in the formation of complex oxide inclusions and the final oxygen content of the steel. In contrast, for the growth of oxide-sulfide inclusions, the character of growth is determined by the sulfur content of steel.

Rem-O-S Compound inclusions growth model

“…Calculations were performed, and the following data were used for the calculations: number of nuclei in 1 cm 3 of steel equals to 10 6 , concentration: S = 0.008%, Y = 0.09%, O = 0.01%, equilibrium concentration for Y = 0.005%, molar volume v1 of sulfide 20 (cm 3 •mol −1 ), molar volume v2 of oxide 20 (cm 3 •mol −1 ) [1]. and 5 show the growth of a bi-component nucleus.…”

“…Non-metallic inclusions in steel affect the microstructure and properties of steel [1]. They are usually considered a detrimental part of the refining process.…”

Research of Injection Methods for Y2o3 Nanoparticles Into Nickel- Free Stainless Steel During Induction Vacuum Remelting

“…Another important condition for the success of ceramic particles injection is a good wetting of their surface with the iron melt, otherwise we should expect that after the carrier dissolution yttrium oxide will form clusters or conglomerates, which will not allow it to spread evenly in the melt volume, • Blowing of Y 2 O 3 powders into melt volume using inert gas carrier will be effective only if wetting of Y 2 O 3 particles with melt on the border of their contact will be intensive, otherwise gas bubbles will rise to the surface and will carry away the entire volume of powder that did not pass into melt. • The possibility of formation of Y 2 O 3 particles in the volume of Y-doped Fe-13Cr melt is justified by the high activity of yttrium toward oxygen [22,23] and the solubility of oxygen in Fe-Cr melt [24]. The first series of experiments was to analyze the Fig.…”

Effect of Rare Earth Yttrium on Inclusion Characteristics of Grain-Oriented Silicon Steel

“…The thermal expansion coefficient of rare earth inclusions are similar to that of steel substrates, avoiding the generation of cracks at the interface between rare earth inclusions and steel substrates during steel processing, thereby improving the impact resistance of the steel. With the addition of rare earth yttrium, since the affinity of Y to O and S is stronger than that of Al, rare earth yttrium has a metamorphic effect on Al 2 O 3 inclusions [37]. The formation reaction formula and standard Gibbs free energy of the formation of Al 2 O 3 are as follows: From the calculation results, it is known that the Gibbs free energy of formation of the above formula is −292,574.63 J/mol, and the reaction is going forward, that is, with the addition of rare earth Y in molten steel, Y can transform Al 2 O 3 into Y 2 O 2 S. However, it is still limited by the reaction equilibrium, and a small amount of Al 2 O 3 remains in the steel.…”