Influence of Process Parameters on Countercurrent Reactor Reduction of Oxidized Mill Scale Waste and Its Co-relationship with Mathematical Model

Abstract

Mill scale is treated as a waste product by the steel industry, so it is essential to recycle and reuse this waste for recovery of metallic iron and its single oxide derivative. Most of the study in the field of mill scale reduction focuses on the static mill scale reduction. Optimizing the process in dynamic conditions becomes more difficult because of the complexity involved inside the process. In the present method, mill scale having particle size ranges from 106 to 53 µm was converted to single-phase oxide Fe₂O₃ by blowing pure oxygen at 1100 °C and then it was subjected to a reducing atmosphere of 50% H₂ and 50% N₂ (1:1) at 875 ± 5 °C at a varying reactor angle from 2° to 5° and different mass flow rates varying from 0.037 to 0.148 kg/min. A generalized experimental model (GEM) has been formulated for the degree of reduction of oxide, which encompasses twelve parameters consisting of reactor design and process parameters. A Buckingham theorem was used to obtain dimensionless parameter for the degree of reduction, and a logical correlation using the proposed model has been established.

Graphical Abstract

Appendix 1

Derivation of Prediction of Dimensionless Degree of Reduction Model

$$ \begin{gathered} {\text{R}} = {\text{f}}({\text{D}},{\text{ L}},\beta ,{\text{ g}},\omega_{{\text{r}}} {\text{,T}}_{{\text{r}}} {\text{,Q}}_{{\text{vg,}}} ,\nu ,{\text{d}}_{{\text{P}}} ,\rho_{{\text{g}}} ,\rho_{{\text{s}}} ) \, = \, 0 \hfill \\ {\text{f}}({\text{D}},{\text{ L}},\beta ,{\text{ g}},\omega_{{\text{r}}} {\text{,T}}_{\rm r} {\text{,Q}}_{{{\text{vg}}}} ,\nu ,{\text{d}}_{{\text{P}}} {\text{,R}}, \, \rho_{{\text{g}}} ,\rho_{{\text{s}}} ) = \, 0 \hfill \\ \end{gathered} $$

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Total no. variables (n) = 12.

Total no. fundamental dimensions (m) = 4.

No. of dimensionless (π) term = n − m = 12 − 4 = 8.

Repeating variables: (1) Characteristic length (L), (2) flow rate of gas (Q_vg), (3) temperature of reduction (T_r), (4) density of H₂ gas (ρ_g).

Finding out the expression for degree of reduction in terms of the above-mentioned parameters

$$ {\text{f }}\left( {\pi_{{1}} , \, \pi_{{2}} ,\pi_{{3}} ,\pi_{{4}} ,\pi_{{5}} ,\pi_{{6}} ,\pi_{{7}} ,\pi_{{8}} } \right) = 0 $$

Term can be calculated as follows:

$$ (1)\quad \phi_{1}\;\text{term\; calculations:} \;(\text{Non-repeating\; variables}: \text{Diameter\; of\; CCR\; of\; D})$$

$$ \begin{gathered} \pi_{{1}} = {{L}}^{{{a}}} {{Q}}_{{{\text{vg}}}}^{{{b}}} {{T}}_{{\text{r}}}^{{{c}}} \rho_{{\text{g}}}^{{{d}}} {{D}} \hfill \\ \left[ {{{M}}^{0} {{L}}^{0} {{T}}^{0} \theta^{0} } \right] = \, \left[ {{{M}}^{0} {{L}}^{{1}} {{T}}^{0} \theta^{0} } \right]^{{{a}}} \left[ {{{M}}^{0} {{L}}^{{3}} {{T}}^{{ - {1}}} \theta^{0} } \right]^{{{b}}} \left[ {{{M}}^{0} {{L}}^{0} {{T}}^{0} \theta^{{1}} } \right]^{{{c}}} \left[ {{{M}}^{{1}} {{L}}^{{ - {3}}} {{T}}^{0} \theta^{0} } \right]^{{{d}}} \left[ {{{M}}^{0} {{L}}^{{1}} {{T}}^{0} \theta^{0} } \right] \hfill \\ \end{gathered} $$

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For M, d = 0.

For L, a + 3b − 3d + 1 = 0, a =  − 1.

For T, b = 0,

For θ, c = 0

$$ \begin{aligned} \pi_{{1}} & = {{L}}^{{ - {1}}} {{Q}}_{{{\text{vg}}}}^{0} {{T}}_{{\text{r}}}^{0} \rho_{{\text{g}}}^{0} {{D}} \hfill \\ {\pi}_{1} & = \frac{D}{L}\end{aligned} $$

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Similarly,

$$ {\pi }_{2} = \frac{{\beta }}{{{{T}}_{{\text{r}}} }},\;{\pi }_{3} = \frac{{{{L}}^{3} {\omega }_{\rm r} }}{{{{Q}}_{{{\text{vg}}}} T_{\rm r} }},{\pi }_{4} = \frac{{L^{5} {{g}}}}{{{{Q}}_{\rm vg}^{2} }},{\pi }_{5} = \frac{\nu }{{L^{5} {\rho }_{\rm g} }},{\pi }_{6} = \frac{{{{ d}}_{\rm p} }}{{{L}}},{\pi }_{7} = {{R}}, \;{\pi}_{8} = \frac{{{\rho }_{\rm s} }}{{{\rho}_{\rm g} }}. $$

$$ f\left( {{\pi }_{1} ,{\pi }_{2} ,{\pi }_{3} ,{\pi }_{4} ,{\pi }_{5} ,{\pi }_{6} ,{\pi }_{7} ,{\pi }_{8} } \right) = 0 $$

$$ f\left( {\frac{{{D}}}{{{L}}}\;\frac{{\beta }}{{T_{\rm r} }}\;\frac{{{{L}}^{3} {\omega }_{\rm r} }}{{{{Q}}_{\rm vg} T_{\rm r}}},\;\frac{{L^{5} {{g}}}}{{{{Q}}_{\rm vg}^{2} }}\;\frac{\nu }{{L^{5} {\rho }_{\rm g} }}\;\frac{{{{D}}_{\rm p} }}{{{L}}}\;{{R}},\;\frac{{{\rho}_{\rm s}}}{{{\rho}_{\rm g} }}}\right) = 0 $$

$$ f\left( {{{R}},\;\frac{{{D}}}{{{L}}}\;\frac{{\beta }}{{T_{\rm r} }}\;\frac{{{{L}}^{3} {\omega }_{\rm r} }}{{{{Q}}_{\rm vg} T_{\rm r} }},\frac{{L^{5} {{ g}}}}{{{{ Q}}_{\rm vg}^{2} }}\;\frac{\nu}{{L^{5} {\rho }_{\rm g} }}\;\frac{{{{D}}_{\rm p} }}{{{L }}}\;\frac{{{\rho }_{\rm s} }}{{{\rho }_{\rm g} }}} \right) = 0. $$

$$ R =\Upphi \left( {\frac{{{D}}}{{{L}}} \times { }\frac{{\beta }}{{T_{\rm r} }} \times \frac{{{{L}}^{3} {\omega }_{\rm r} }}{{{{Q_{\rm vg}}}T_{\rm r} }} \times \frac{{{{Q}}_{\rm vg} }}{{{{L}}^{3} }} \times \frac{{L^{5} {{g}}}}{{{{Q}}_{\rm vg}^{3} }} \times \frac{\nu }{{L^{5} {\rho }_{\rm g} }} \times \frac{{{{D}}_{\rm p} }}{{{L}}} \times \frac{{{\rho }_{\rm s} }}{{{\rho }_{\rm g} }}} \right)^{A} $$

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