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Erschienen in:
24.07.2017
Distribution Ratios of Phosphorus Between CaO-FeO-SiO2-Al2O3/Na2O/TiO2 Slags and Carbon-Saturated Iron
Erschienen in: Metallurgical and Materials Transactions B | Ausgabe 5/2017
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Abstract
In order to effectively enhance the efficiency of dephosphorization, the distribution ratios of phosphorus between CaO-FeO-SiO2-Al2O3/Na2O/TiO2 slags and carbon-saturated iron (\( L_{\text{P}}^{\text{Fe-C}} \)) were examined through laboratory experiments in this study, along with the effects of different influencing factors such as the temperature and concentrations of the various slag components. Thermodynamic simulations showed that, with the addition of Na2O and Al2O3, the liquid areas of the CaO-FeO-SiO2 slag are enlarged significantly, with Al2O3 and Na2O acting as fluxes when added to the slag in the appropriate concentrations. The experimental data suggested that \( L_{\text{P}}^{\text{Fe-C}} \) increases with an increase in the binary basicity of the slag, with the basicity having a greater effect than the temperature and FeO content; \( L_{\text{P}}^{\text{Fe-C}} \) increases with an increase in the Na2O content and decrease in the Al2O3 content. In contrast to the case for the dephosphorization of molten steel, for the hot-metal dephosphorization process investigated in this study, the FeO content of the slag had a smaller effect on \( L_{\text{P}}^{\text{Fe-C}} \) than did the other factors such as the temperature and slag basicity. Based on the experimental data, by using regression analysis, \( \log L_{\text{P}}^{\text{Fe-C}} \) could be expressed as a function of the temperature and the slag component concentrations as follows:
$$ \begin{aligned} \log L_{\text{P}}^{\text{Fe-C}} & = 0.059({\text{pct}}\;{\text{CaO}}) + 1.583\log ({\text{TFe}}) - 0.052\left( {{\text{pct}}\;{\text{SiO}}_{2} } \right) - 0.014\left( {{\text{pct}}\;{\text{Al}}_{2} {\text{O}}_{3} } \right) \\ \, & \quad + 0.142\left( {{\text{pct}}\;{\text{Na}}_{2} {\text{O}}} \right) - 0.003\left( {{\text{pct}}\;{\text{TiO}}_{2} } \right) + 0.049\left( {{\text{pct}}\;{\text{P}}_{2} {\text{O}}_{5} } \right) + \frac{13{,}527}{T} - 9.87. \\ \end{aligned} $$