<p>Accurately determining the thermodynamic data of the reactions between rare earth elements (REEs) and [O] and [S] in the liquid iron is an important foundation for the application of REEs in the steel industry. The difficulty in the flotation of rare earth inclusions and the side reactions between REEs and the crucible pose significant challenges in accurately measuring these thermodynamic data. This study used high-purity yttria crucibles as a smelting container, an induction furnace for smelting, and an increased scale of the experimental system to overcome the above problems. More reliable equilibrium constants and interaction coefficients for the Y-S and Y-O-S equilibrium under 1873&#xa0;K to 1973&#xa0;K were obtained. YS and Y<sub>2</sub>O<sub>2</sub>S were identified as stable equilibrium products corresponding to the Y-S and Y-O-S equilibria, respectively. The contents of [Y], [O], and [S] greatly influenced the reaction equilibrium in the system. As the [Y] content is less than 0.02 wt pct and the [S] content is higher than 0.0010 wt pct (S/O mass ratio &gt; 1), the equilibrium was dominated by the Y-S reaction, and its equilibrium constant and interaction coefficient can be expressed by <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11663_2025_3608_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="207" /> </InlineMediaObject> <EquationSource Format="TEX">\(\log K_{{{\text{YS}}}} = - 29437/T + 10.94\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>log</mo> <msub> <mi>K</mi> <mtext>YS</mtext> </msub> <mo>=</mo> <mo>-</mo> <mn>29437</mn> <mo stretchy="false">/</mo> <mi>T</mi> <mo>+</mo> <mn>10.94</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11663_2025_3608_Article_IEq2.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="177" /> </InlineMediaObject> <EquationSource Format="TEX">\(e_{{\text{S(YS)}}}^{{\text{Y}}} = - 8489/T + 3.14\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>e</mi> <mrow> <mtext>S(YS)</mtext> </mrow> <mtext>Y</mtext> </msubsup> <mo>=</mo> <mo>-</mo> <mn>8489</mn> <mo stretchy="false">/</mo> <mi>T</mi> <mo>+</mo> <mn>3.14</mn> </mrow> </math></EquationSource> </InlineEquation>, respectively. As the [Y] content ranges from 0.02 to 0.20 wt pct and the [S] content is less than 0.0010 wt pct (0.25 &lt; S/O mass ratio &lt; 1), the equilibrium was dominated by the Y-O-S reaction, and its equilibrium constant and interaction coefficient can be expressed by <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11663_2025_3608_Article_IEq3.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="219" /> </InlineMediaObject> <EquationSource Format="TEX">\(\log K_{{{\text{Y}}_{{2}} {\text{O}}_{{2}} {\text{S}}}} = - 43108/T + 9.59\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>log</mo> <msub> <mi>K</mi> <mrow> <msub> <mtext>Y</mtext> <mn>2</mn> </msub> <msub> <mtext>O</mtext> <mn>2</mn> </msub> <mtext>S</mtext> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mn>43108</mn> <mo stretchy="false">/</mo> <mi>T</mi> <mo>+</mo> <mn>9.59</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="11663_2025_3608_Article_IEq4.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="192" /> </InlineMediaObject> <EquationSource Format="TEX">\(e_{{{\text{S(Y}}_{{2}} {\text{O}}_{{2}} {\text{S)}}}}^{{\text{Y}}} = - 30681/T + 15\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msubsup> <mi>e</mi> <mrow> <mrow> <msub> <mtext>S(Y</mtext> <mn>2</mn> </msub> <msub> <mtext>O</mtext> <mn>2</mn> </msub> <mtext>S)</mtext> </mrow> </mrow> <mtext>Y</mtext> </msubsup> <mo>=</mo> <mo>-</mo> <mn>30681</mn> <mo stretchy="false">/</mo> <mi>T</mi> <mo>+</mo> <mn>15</mn> </mrow> </math></EquationSource> </InlineEquation>, respectively.</p>

错误:搜索内容不能为空,请输入英文关键词
错误:关键词超出字数限制,请精简
高级检索

Thermodynamic Assessment and Phase Equilibria of Yttrium and Sulfur in Liquid Iron

  • Jian Kang,
  • Hongpo Wang,
  • Zheyang Lin,
  • Juntong Shen,
  • Yu Wang,
  • Ke Chen

摘要

Accurately determining the thermodynamic data of the reactions between rare earth elements (REEs) and [O] and [S] in the liquid iron is an important foundation for the application of REEs in the steel industry. The difficulty in the flotation of rare earth inclusions and the side reactions between REEs and the crucible pose significant challenges in accurately measuring these thermodynamic data. This study used high-purity yttria crucibles as a smelting container, an induction furnace for smelting, and an increased scale of the experimental system to overcome the above problems. More reliable equilibrium constants and interaction coefficients for the Y-S and Y-O-S equilibrium under 1873 K to 1973 K were obtained. YS and Y2O2S were identified as stable equilibrium products corresponding to the Y-S and Y-O-S equilibria, respectively. The contents of [Y], [O], and [S] greatly influenced the reaction equilibrium in the system. As the [Y] content is less than 0.02 wt pct and the [S] content is higher than 0.0010 wt pct (S/O mass ratio > 1), the equilibrium was dominated by the Y-S reaction, and its equilibrium constant and interaction coefficient can be expressed by \(\log K_{{{\text{YS}}}} = - 29437/T + 10.94\) log K YS = - 29437 / T + 10.94 and \(e_{{\text{S(YS)}}}^{{\text{Y}}} = - 8489/T + 3.14\) e S(YS) Y = - 8489 / T + 3.14 , respectively. As the [Y] content ranges from 0.02 to 0.20 wt pct and the [S] content is less than 0.0010 wt pct (0.25 < S/O mass ratio < 1), the equilibrium was dominated by the Y-O-S reaction, and its equilibrium constant and interaction coefficient can be expressed by \(\log K_{{{\text{Y}}_{{2}} {\text{O}}_{{2}} {\text{S}}}} = - 43108/T + 9.59\) log K Y 2 O 2 S = - 43108 / T + 9.59 and \(e_{{{\text{S(Y}}_{{2}} {\text{O}}_{{2}} {\text{S)}}}}^{{\text{Y}}} = - 30681/T + 15\) e S(Y 2 O 2 S) Y = - 30681 / T + 15 , respectively.