1473 K で CaO-MgO-CaCl₂ フラックスと平衡する Ti-Cu 系合金中の酸素とカルシウムの溶解度

抄録

  Solubility of Ca and O in molten Ti-Cu alloys equilibrated with a CaO-MgO-CaCl₂ ternary flux in CaO or MgO crucibles at 1473 K was analytically investigated. The Equilibrium constant of deoxidization reaction by dissolved Ca can be determined as follows:
  CaO_{(s)in CaO crucible}=Ca_{in Metal}+O_{in Metal}, −3.25=logK, K=(a_Ca•a_O)/(a_CaO)
Moreover, from the investigated partial pressure of O around the reaction cell, an equilibrium constant of dissoluble reaction of O in the molten alloys can be determined as follows:
  1/2O₂=O_{in Metal}, 16.08=logK, K=[a_{O in Metal}]/(P_O2^1/2)
If it is possible to use an equilibrium constant of calcia formation published as a thermodynamic data book, an equilibrium constant of dissolution reaction of calcium into molten alloys can be determined as follows,
  Ca_{in Flux}=Ca_{in Metal}, −2.29=logK, K=[a_{Ca in Metal}]/(a_{Ca in Flux})
The activity of CaO in the CaO-MgO-CaCl₂ ternary flux equilibrated with a MgO crucible is determined as 0.76 compared to the normal state (a_CaO=1) using a CaO crucible. This is obtained from the difference of the equilibrium constants of CaO formation using CaO and MgO crucibles, as expressed by the relation: logK=logK′+loga_CaO.