物理化學より見たる平爐操業諸過程の關連性について VIII

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Under acid slag-SiO_2 FeO, MnO system-the fundamental equilibrium reactions are thought as follows, that is : - In slag In steel bath equations equilibrium constants equations equilibrium constants (FeO)=(Fe)+(O) K_1(=(Fe)(O)/(FeO)) [FeO]=[Fe]+[O] k_1 (MnO)=(Mn)+(O) K_2 [MnO]=[Mn]+[O] k_2 (SiO_2)=(Si)+2(O) K_3 [SiO_2]=[Si]+2[O] k_3 (CO)=(C)+(O) K_4 [CO]=[C]+[O] k_4 Furthermore distribution constants distribution constants [FeO]e/(FeO)e=L_1 [Fe]e/(Fe)e=l_1 [MnO]e/(MnO)e=L_2 [Mn]e/(Mn)e=l_2 [SiO_2]e/(SiO_2)e=L_3 [Si]e/(Si)e=l_3 [CO]e/(CO)e=L_4 [O]e/(C)e=l_4 [O]e/(O)e=l_5 As an example, Mn reaction is thought. From K_1 and K_2 equations above (FeO) [Mn]/(MnO)=KMn・(Mn)e/(Mn)・(Fe)/(Fe)e is derived. When slag is saturated with SiO_2 and (Mn)e/(Mn)・(Fe)/(Fe)e=1 (ΣFeO) [Mn]/(ΣMnO)=K^S_<Mn> is derived. It is interesting that this equations derived with only three conditions as above is quite same form with the equation of complete equilibrium which is derived with seven conditions (K_1, K_2, k_1, k_2 any three of L_1, L_2, l_1 l_2 l_5) When k_1 and k_2 reactions are in equilibrium (ΣFeO)e[Mn]/(ΣMnO)=K^S_<Mn>・[MnO]/[MnO]e is derived, and when distribution equilibrium of MnO is assumed, [FeO][Mn]/(ΣMnO)=K^S_<Mn> L^S_<FeO> is obtained. That is, Mn reaction should be discussed also from MnO distribution between slag and bath about Si reaction it is quite the same.
社団法人日本鉄鋼協会の論文
1949-10-25