Liquid steel contains initially 0.05 mass% P and this has to be reduced to 0.01 mass% using a basic slag. The equilibrium distribution ratio of P between slag and metal is Lp = ( % P )slag( % P )metal, = 80; Assuming that in itially the slag does not contain any phosphorous (P), then the minimum weight of slag (ton) required per ton of steel is

This question belongs to Metallurgical Engineering Metallurgical Thermodynamics And Kinetics

Metallurgical Thermodynamics and Kinetics

Liquid steel contains initially 0.05 mass% P and this has to be reduced to 0.01 mass% using a basic slag. The equilibrium distribution ratio of P between slag and metal is $$Lp = \frac{{\left( {\% P} \right){\text{slag}}}}{{\left( {\% P} \right){\text{metal}}}}, = 80;$$ Assuming that in itially the slag does not contain any phosphorous (P), then the minimum weight of slag (ton) required per ton of steel is

A) 0.025 B) 0.05 C) 0.075 D) 0.10

Answer & Solution

Correct Answer: Option B

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