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Wear and Dissolution of MgO–C Refractory Lining in Belly and Top Cone Regions of BOF Vessel

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Abstract

The MgO–carbon bricks of varying carbon concentration are used as refractory lining material in BOF. Laser profiling of the lining is done at regular intervals to keep track of refractory wear in different parts of vessels. The main factors which affect the kinetics of dissolution of graphite flakes lying between the MgO grains in the belly region are attack by CO2 in gas and FeO in slag, and temperature. FeO can easily penetrate the MgO grains along grain boundaries and reach those places where graphite flakes are present. Kinetic models of refractory wear are analyzed on the basis of data obtained from actual laser profile measurements. The bricks salvaged from the top cone region at the end of the campaign have been subjected to scanning electron microscope (SEM) and electron probe micro analyzer (EPMA) investigations. The possible cause of wear in top cone region is also the oxidation of carbon in the brick by CO2 gas and direct attack by FeO thrown from the jet impact region.

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Correspondence to Brahma Deo.

Appendix

Appendix

1.1 Derivation of Rate Equation and Comparison of Actual and Estimated Wear Rates

From Eq. (18) we get

$$\frac{{\partial n_{c} }}{\partial t}\frac{{(1 + K_{e} a_{\text{FeO}} )}}{{k_{s} Aa_{\text{FeO}} }} = - P_{\text{CO}}^{b} \left( {K_{e} a_{FeO} - \frac{{P_{{{\text{CO}}_{2} }}^{b} }}{{P_{\text{CO}}^{b} }}} \right)$$
(21)

Since \(P_{\text{CO}}^{b} = \, 1 - P_{{{\text{CO}}_{2} }}^{b}\) (atm)

$$\frac{{\partial n_{c} }}{\partial t}\frac{{(1 + K_{e} a_{\text{FeO}} )}}{{k_{s} Aa_{\text{FeO}} }} = - (1 - P_{{{\text{CO}}_{2} }}^{b} )\left[ {K_{e} a_{\text{FeO}} - \frac{{P_{{{\text{CO}}_{2} }}^{b} }}{{(1 - P_{{{\text{CO}}_{2} }}^{b} )}}} \right]$$
(22)

Or,

$$\frac{{\partial n_{c} }}{\partial t}\frac{{(1 + K_{e} a_{\text{FeO}} )}}{{k_{s} Aa_{\text{FeO}} }} = - \left[ {K_{e} a_{\text{FeO}} - P_{{{\text{CO}}_{2} }}^{b} (1 + K_{e} \, a_{\text{FeO}} )} \right]$$
(23)

Or, calculating \(P_{{{\text{CO}}_{2} }}^{b}\)

$$P_{{{\text{CO}}_{2} }}^{b} = \frac{{\partial n_{c} }}{\partial t}\frac{1}{{k_{s} Aa_{\text{FeO}} }} + \frac{{K_{e} a_{\text{FeO}} }}{{ ( 1+ K_{e} a_{\text{FeO}} )}}\;\;({\text{atm}})$$
(24)

From Eq. (7)

$$P_{{{\text{CO}}_{2} }}^{i} = - \frac{{\partial n_{c} }}{\partial t}\frac{1}{{Ak_{c} }}\;\; \left( {\text{atm}} \right)$$
(25)

From Eqs. (14), (24) and (25)

$$\frac{{\partial n_{c} }}{\partial t} = - \frac{{Ak_{g} 1.01 \times 10^{5} }}{{RT_{f} }}\left( {\left\{ {\frac{{\partial n_{c} }}{\partial t}\frac{1}{{k_{s} Aa_{\text{Feo}} }} + \frac{{K_{e} a_{\text{FeO}} }}{{(1 + K_{e} a_{\text{FeO}} )}}} \right\} + \frac{{\partial n_{c} }}{\partial t}\frac{1}{{Ak_{c} }}} \right)$$
(26)
$$\frac{{dn_{c} }}{A\partial t} = \frac{ - 1}{{\left( {\frac{1}{{K_{e} a_{\text{FeO}} }} + 1} \right)\left[ {\left( {\frac{{RT_{f} }}{{1.01 \times 10^{5} k_{g} }}} \right) + \left( {\frac{1}{{k_{s} a_{\text{FeO}} }}} \right) + \left( {\frac{1}{{k_{c} }}} \right)} \right]}}$$
(27)

Constants:

  • Total pressure P = 1 atm

  • ρ = 2250 kg/m3

  • ks = 2.27 mol m−2 s−1 atm−1

  • Ke = 0.17453

  • kg = 0.2154 m s−1

  • kc = 30.6 mol m−2 s−1 atm−1

  • aFeo = 0.4

  • R = 8.314 Pa m3 mol−1 K−1

  • T = 1900 K

1.2 Sample calculations

From Table 1, if thickness eroded is 19 mm in 237 heats, then the average wear rate in each heat is

$$X = \frac{\Delta }{\Delta }\frac{x}{heat}\;{\text{m}}/{\text{heat}}$$
$${\text{X}} = \frac{19}{237 \times 1000} = 8.01 \times 10^{ - 5} \;{\text{m}}/{\text{heat}}$$

Wear rate (k mol s−1), assuming that brick contains 0.1 mass fraction of graphite flakes, total reaction time in one heat is 10 min (=60 * 10 s), reaction area of brick is 1 m2, and density of MgO-10% carbon brick is 2250 kg m−3

$$= \frac{X\rho }{12\Delta t} \times 0.1$$
$$= \frac{{8.01 \times 10^{ - 5} \times 2250 \, }}{12 \times 60 \times 10} \times 0.1$$
$$= 2.50527 \times 10^{ - 6} \;{\text{kg}}\;{\text{mol}}\;{\text{s}}^{ - 1}$$
(28)

Now for reaction area of 1 m2, when both r1 and r2 are rate controlling,

$$\frac{{\partial n_{c} }}{\partial t} = \frac{ - 1}{{\left( {\frac{1}{{K_{e} a_{\text{FeO}} }} + 1} \right)\left[ {\frac{{RT_{f} }}{{1.01 \times 10^{5} k_{g} }} + \frac{1}{{k_{s} a_{\text{FeO}} }}} \right] \times 1000}}$$
$$\frac{{\partial n_{c} }}{\partial t} = \frac{ - 1}{{\left( {\frac{1}{0.17453 \times 0.4} + 1} \right)\left[ {\frac{8.314 \times 1900}{{1.01 \times 10^{5} \times .2154}} + \frac{1}{2.27 \times 0.4}} \right] \times 1000}}$$
$$\frac{{\partial n_{c} }}{\partial t} = 3.57 \times 10^{ - 5} \;{\text{kg}}\;{\text{mol}}\;{\text{s}}^{ - 1}$$

The graphite is present in the MgO bricks in the form of flakes. The flakes are oriented in all possible directions. It is difficult to estimate accurately the fractional surface area of graphite available for reaction. Further, the refractory surface is not a planar surface, and at some places the refractory surface may be covered by solid/mushy slag. If the area of reaction with carbon, only for the sake of simplicity, is assumed to be 10% of the nominal surface of MgO–C refractory (i.e. directly proportional to mass percent of carbon in the brick), then the predicted wear rate will be

$$3.57 \times 10^{ - 5} \times 0.1 = \, 3.57 \times 10^{ - 6} \;{\text{kg}}\;{\text{mol}}\;{\text{s}}^{ - 1}$$
(29)

From Eqs. (28) and (29), the ratio of actual to predicted wear rate is therefore = 2.50/3.57 = 0.70.

It shows that the predicted wear rate (theoretical value from model) is slightly higher than the actual wear rate.

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Deo, B. Wear and Dissolution of MgO–C Refractory Lining in Belly and Top Cone Regions of BOF Vessel. Trans Indian Inst Met 70, 1965–1971 (2017). https://doi.org/10.1007/s12666-016-1018-1

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