A Gibbs Energy Minimization Approach for Modeling of Chemical Reactions in a Basic Oxygen Furnace

Abstract

In modern steelmaking, the decarburization of hot metal is converted into steel primarily in converter processes, such as the basic oxygen furnace. The objective of this work was to develop a new mathematical model for top blown steel converter, which accounts for the complex reaction equilibria in the impact zone, also known as the hot spot, as well as the associated mass and heat transport. An in-house computer code of the model has been developed in Matlab. The main assumption of the model is that all reactions take place in a specified reaction zone. The mass transfer between the reaction volume, bulk slag, and metal determine the reaction rates for the species. The thermodynamic equilibrium is calculated using the partitioning of Gibbs energy (PGE) method. The activity model for the liquid metal is the unified interaction parameter model and for the liquid slag the modified quasichemical model (MQM). The MQM was validated by calculating iso-activity lines for the liquid slag components. The PGE method together with the MQM was validated by calculating liquidus lines for solid components. The results were compared with measurements from literature. The full chemical reaction model was validated by comparing the metal and slag compositions to measurements from industrial scale converter. The predictions were found to be in good agreement with the measured values. Furthermore, the accuracy of the model was found to compare favorably with the models proposed in the literature. The real-time capability of the proposed model was confirmed in test calculations.

Appendix A: Initial Values for Pair Fractions

Before starting the algorithm for the solution of the pair fractions, a reasonably good initial guess should be generated. This can be done by inserting Eq. [37] into Eq. [36]. After reorganizing, this produces three equations

$$ \begin{aligned}& X_{AB}^{2} \left( {1 - K_{AB} } \right) + {\text{X}}_{AB} \left( {{\text{X}}_{AC} + {\text{X}}_{BC} - 2{\text{Y}}_{A} - 2{\text{Y}}_{B} } \right)\\ &\quad + 4{\text{Y}}_{A} {\text{Y}}_{B} - 2\left( {{\text{X}}_{BC} {\text{Y}}_{A} + {\text{X}}_{AC} {\text{Y}}_{B} } \right) + {\text{X}}_{AC} {\text{X}}_{BC} = 0, \\ & X_{AC}^{2} \left( {1 - K_{AC} } \right) + {\text{X}}_{AC} \left( {{\text{X}}_{AB} + {\text{X}}_{BC} - 2{\text{Y}}_{A} - 2{\text{Y}}_{C} } \right) \\ &\quad + 4{\text{Y}}_{A} {\text{Y}}_{C} - 2\left( {{\text{X}}_{BC} {\text{Y}}_{A} + {\text{X}}_{AB} {\text{Y}}_{C} } \right) + {\text{X}}_{AB} {\text{X}}_{BC} = 0, \\ & X_{BC}^{2} \left( {1 - K_{BC} } \right) + {\text{X}}_{BC} \left( {{\text{X}}_{AB} + {\text{X}}_{AC} - 2{\text{Y}}_{B} - 2{\text{Y}}_{C} } \right)\\ &\quad + 4{\text{Y}}_{B} {\text{Y}}_{C} - 2\left( {{\text{X}}_{AC} {\text{Y}}_{B} + {\text{X}}_{AB} {\text{Y}}_{C} } \right) + {\text{X}}_{AB} {\text{X}}_{AC} = 0, \\ \end{aligned} $$

(A1)

where $K_{AB} = { \exp }\left( {\frac{{{{\Delta }}g_{AB} }}{RT}} \right)$, etc. These equations are only functions of (X _AB , X _AC , X _BC ). By subtracting the second row equation from the first row equation, subtracting the third row equation from the first row equation and subtracting the second row equation from the third row equation again three equations are obtained:

$$ \begin{aligned} & X_{AB}^{2} \left( {1 - K_{AB} } \right) + 2{\text{X}}_{AB} \left( {2{\text{Y}}_{C} - 1} \right) + 4{\text{Y}}_{A} {\text{Y}}_{B} \\ &\quad - X_{AC}^{2} \left( {1 - K_{AC} } \right) - 2{\text{X}}_{AC} \left( {2{\text{Y}}_{B} - 1} \right) - 4{\text{Y}}_{A} {\text{Y}}_{C} = 0, \\ & X_{AB}^{2} \left( {1 - K_{AB} } \right) + 2{\text{X}}_{AB} \left( {2{\text{Y}}_{C} - 1} \right) + 4{\text{Y}}_{A} {\text{Y}}_{B} \\ &\quad - X_{BC}^{2} \left( {1 - K_{BC} } \right) - 2{\text{X}}_{BC} \left( {2{\text{Y}}_{A} - 1} \right) - 4{\text{Y}}_{B} {\text{Y}}_{C} = 0 \\ & X_{BC}^{2} \left( {1 - K_{BC} } \right) + 2{\text{X}}_{BC} \left( {2{\text{Y}}_{A} - 1} \right) + 4{\text{Y}}_{B} {\text{Y}}_{C} \\ &\quad - X_{AC}^{2} \left( {1 - K_{AC} } \right) - 2{\text{X}}_{AC} \left( {2{\text{Y}}_{B} - 1} \right) - 4{\text{Y}}_{A} {\text{Y}}_{C} = 0. \end{aligned} , $$

(A2)

The advantage of this manipulation is that now each equation is only dependent on two unknown pair fractions. Also in Eq. [A2] there are no terms where two pair fractions are multiplying each other like in Eq. [A1]. By applying the solution of second degree polynomial equation to the first row equation, we get

$$ \begin{aligned} {\text{X}}_{AB} & = \frac{{ - B_{AB} \pm \sqrt {B_{AB}^{2} - 4A_{AB} \left( {C_{AB} - A_{AC} X_{AC}^{2} - B_{AC} {\text{X}}_{AC} - C_{AC} } \right)} }}{{2A_{AB} }} \\ A_{AB} & = \left( {1 - K_{AB} } \right), \quad A_{AC} = \left( {1 - K_{AC} } \right) \\ B_{AB} & = 2\left({2{\text{Y}}_{C} - 1}\right), \quad B_{AC} = 2\left( {2{\text{Y}}_{B} - 1} \right) \\ C_{AB} & = 4 {\text{Y}}_{A} {\text{Y}}_{B} , \quad C_{AC} = 4 {\text{Y}}_{A} {\text{Y}}_{C} \end{aligned} $$

(A3)

where the coefficients are the terms multiplying the pair fractions in Eq. [A2]. So that X _AB could have any real solutions the determinant should be either zero or positive:

$$ 4A_{AB} \left( {A_{AC} X_{AC}^{2} + B_{AC} {\text{X}}_{AC} + C_{AC} - C_{AB} } \right) + B_{AB}^{2} \ge 0. $$

(A4)

Using Eq. [A4] limits can be acquired for pair fraction X _AC . In Figure 10, all possible parabolas are presented, which give limiting values for the pair fraction and have positive values for Eq. [A4] between X = [0, 1]. If product A _AB A _AC is negative, the parabola is opening down (Figure 10(a) through (c)). If the product is positive, the parabola is opening up (Figure 10(d) through (f)). For clarification, there are more possibilities for parabolas which give positive values when X = [0, 1], but these will not give new limits that are between [0, 1]. It is now assumed that Eq. [A4] has two solutions X ₁ and X ₂ as well as X ₁ < X ₂. If the parabola is opening down, the limiting values for X _AC must be between the two solutions.

Fig. 10

All possible parabolas from (a) to (f) for Eq. [A4] that can give limiting values for pair fraction

Since fractions can have values only between zero and one, the new limiting values for X _AC can be calculated as follows:

$$ \begin{array}{*{20}c} {L_{AC} = { \hbox{max} }\left( {X_{1} ,0} \right)} & {U_{AC} = { \hbox{min} }\left( {X_{2} ,1} \right)} \\ \end{array} , $$

(A5)

where L _AC is the lower limit and U _AC is the upper limit. The upward opening parabola can give two ranges for limits that are between [0, 1]. This happens when 0 ≤ X ₁, X ₂ ≤ 1. The corresponding situation is in Figure 10(d). Then the first limits are

$$ \begin{array}{*{20}c} {L_{AC}^{1} = 0} & {U_{AC}^{1} = X_{1} } \\ \end{array} , $$

(A6)

and the second limits are

$$ \begin{array}{*{20}c} {L_{AC}^{2} = X_{2} } & {U_{AC}^{2} = 1.} \\ \end{array} $$

(A7)

The correct solution has to be between the first lower and upper limits or the second lower and upper limits. Using Eq. [A2] similarly as above limits can be acquired for the three pair fractions (X _AB , X _AC , X _BC ). Additional improvements in the initial guesses can be achieved by inserting the limiting values for X _AC in Eq. [A3] and studying whether the solutions for X _AB are even more limiting than the previous limits. This second step was noticed to usually decrease the limits from previous values. Finally, the initial guesses are calculated as averages of the lower and upper limits:

$$ \begin{array}{*{20}c} {X_{AC} = 0.5\left( {L_{AC} + U_{AC} } \right),} & {X_{AB} = 0.5\left( {L_{AB} + U_{AB} } \right),} & {X_{BC} = 0.5\left( {L_{BC} + U_{BC} } \right).} \\ \end{array} $$

(A8)

Initial values for the three other pair fractions can now be calculated from Eq. [37].