Dephosphorization Reaction - Steel Making - Lecture Notes, Study notes of Metallurgy

Download Dephosphorization Reaction - Steel Making - Lecture Notes and more Study notes Metallurgy in PDF only on Docsity! Lecture 8: Dephosphorization Reaction Contents Preamble Equilibrium considerations How low γP2O5 should be? Effect of FeO and CaO on dephosphorization Illustration Conditions for dephosphorization Conditions for simultaneous removal of C and P Key words: Dephosphorization, steelmaking reactions, BOF steelmaking Preamble Phosphorus removal from hot metal is the most important refining reaction. Phosphorus has atomic number 15 and it can give up all 5 electrons from its outermost shell to become P5+ or accept 3 electrons to become P3− to attain stable configuration. This means that phosphorus can be removed both under oxidizing as well as reducing conditions. But removal of phosphorus under reducing conditions is not practical since its removal is highly hazardous. Thus P removal is practised mostly under oxidizing conditions. Equilibrium Considerations: Phosphorus removal reaction 2 [P] + 5 [O] = (P2O5) (1) ∆ G° = −740375 + 535.365T J/mol At T > 1382K, ∆ G° becomes positive which results in decomposition of P2O5 to P and O. Thus removal of phosphrous requires that aP2O5 must be reduced. KP = aP 2O 5 [wt % P]2[wt % O]5 (2) Now [wt% O] = aFeO [wt% O]sat . and (3) Docsity.com log[wt% O]sat . = − 6320 T + 2.734 (4) By equation 2 and 3 and replacing aP2O5by using Raoult’s Law and after rearrangement N(P 2O 5) [wt % P]2 = KP (aFeO ) 5[wt % O]sat . γP 2O 5 (5) γP2O5 is activity coefficient of P2O5 in slag. The LHS of equation 5 is index of dephosphorization and denotes distribution of phosphorus between slag and metal. Higher value of LHS demands low γP2O5 in a slag of a given composition. How low γP2O5should be? Consider dephosphorization in a slag of aFeO = 0.31 at 1773K. Initial %P in metal is 0.1 and mole fraction of NP2O5 in slag = 0.01. Let us calculate γP2O5which will allow dephosphorization. log KP = 38668 T − 27.96 (6) At 1773 K, KP = 7.06 × 10−7 [wt% O] can be determined by equation 3 and 4. We substitute the values in equation 5. We get γP2O5 = 4.16 × 10 −16 Now the question before us: how to attain such a low value of γP2O5 in a slag of given composition? Such a low value of γP2O5can be attained when we use basic oxides which have a very strong tendency to form a stable chemical compound. The different basic oxides have different ability to lower γP2O5 . The following expression describes the relative effects of basic oxides on γP2O5 . log γP2O5 = −24.64 �NCaO + 0.682NMgO + 0.591NMnO + 0.545NFeO − 0.091NSi O2� − 42000 T + + 23.58 (7) Alkaline oxides Na2O and BaO are stronger than CaO but they are corrosive to the refractory lining and hence not used. Consider a slag NCaO = 0.56, NMgO = 0.12, NMnO = 0.06, NFeO = 0.1 andNSi O2 = 0.6 We calculate γP2O5at different temperatures T (K) γP2O5 1773 1.74 × 10−18 1823 0.778 × 10−18 1873 0.324 × 10−18 Decrease in temperature increases γP2O5which favours dephosphorization reaction. Docsity.com