Transition from Internal to External Oxidation of Mn Steel Alloys

Abstract

The transition from internal to external oxidation of Mn-steels depends on temperature, oxygen partial pressure or dew point and alloy composition. A description for this transition includes, besides the diffusion coefficients of Mn and oxygen, also on the critical volume fraction oxide precipitates. This volume fraction for Mn-steel alloys has been determined experimentally and equals 0.2, whereas usually a value of 0.3 is adopted. A low critical volume fraction oxide implies that the alloy is more prone to external oxidation. Oxidation experiments with 1.7, 3.5 and 7.0 wt% Mn-steel alloys at 950 °C, i.e. in the austenitic region, at different oxygen partial pressure or dew points in the range of −45 to +10 °C showed excellent agreement with the oxidation mode predicted using the data obtained for the diffusivities and critical volume fraction oxide.

Appendices

Appendix 1

The relation between dew point (DP) and partial pressure of water vapor follows from [28]:

$$ \log p_{{H_{2} O}} = \frac{9.8DP}{273.8 + DP} - 2.22\quad \quad DP \le 0^\circ {\text{C}} $$

(1-1)

$$ \log p_{{H_{2} O}} = \frac{7.58DP}{240 + DP} - 2.22\quad \quad DP > 0^\circ {\text{C,}} $$

(1-2)

where the units of partial pressure of water vapor ($p_{{H_{2} O}}$) and dew point (DP) are in atmosphere and degree Celsius, respectively. Next, the corresponding partial pressure of oxygen can be calculated with the Gibbs free energy of formation ($\Updelta G_{{H_{2} O}}^{0}$) of water vapor [29]:

$$ H_{2 \, (g)} + \frac{1}{2}O_{2 \, (g)} = H_{2} O_{ \, (g)} ,\;{\text{with}}:\;\Updelta G_{{H_{2} O}}^{0} = - 245519 + 53.35T\;\left( {{\text{J}}/{\text{mol}}} \right), $$

(1-3)

in which T is the absolute temperature in Kelvin.

Appendix 2

For the oxygen solubility in Fe it holds that [25]:

$$ \log \left( {N_{O}^{(s)} } \right) = \log \left( {\frac{{p_{{H_{2} O}} }}{{p_{{H_{2} }} }}} \right) - \frac{5000}{T} - 0.67\quad \quad {\text{bcc}} - {\text{Fe}} $$

(2-1)

$$ \log \left( {N_{O}^{(s)} } \right) = \log \left( {\frac{{p_{{H_{2} O}} }}{{p_{{H_{2} }} }}} \right) - \frac{4050}{T} - 1.52\quad \quad {\text{fcc}} - {\text{Fe}} $$

(2-2)

Here $N_{O}^{(s)}$denotes the surface concentration of oxygen in mole fraction. $p_{{H_{2} O}}$ and $p_{{H_{2} }}$ are the partial pressure of water vapor and hydrogen in atm., respectively, and T is the absolute temperature in Kelvin.

Appendix 3

In order to obtain the spatial size distribution of internal oxide precipitates from the area observed in the cross-sections, the Saltykov area method [13] was adopted. This method is based on the relative section areas of the precipitates, i.e. A/A _max , where A _max is the area of the largest precipitate. When assuming that the precipitates are spherical, then the area A can be related to its diameter D. Next, a series of size groups j is obtained by taking the logarithm of the ratio pertaining to the upper boundary of the subsequent size groups equal to 10⁻¹, i.e. $\log \left( {\frac{{D_{j + 1} }}{{D_{j} }}} \right) = 10^{ - 1}$

Here, 12 size groups were considered. Next, the number of internal oxide particles divided by the total test area (IOZ) for all of the size groups $(N_{A} )_{i}$ was counted (using ImageJ software [12]). These values are substituted in the so-called working equation, which reads [13]:

$$ \left( {N_{v} } \right)_{j} = \frac{1}{{D_{j} }}\left[ \begin{gathered} 1.6461\left( {N_{A} } \right)_{i} - 0.4561\left( {N_{A} } \right)_{i - 1} - 0.1162\left( {N_{A} } \right)_{i - 2} - 0.0415\left( {N_{A} } \right)_{i - 3} \hfill \\ - 0.0173\left( {N_{A} } \right)_{i - 4} - 0.0079\left( {N_{A} } \right)_{i - 5} - 0.0038\left( {N_{A} } \right)_{i - 6} - 0.0018\left( {N_{A} } \right)_{i - 7} \hfill \\ - 0.0010\left( {N_{A} } \right)_{i - 8} - 0.0003\left( {N_{A} } \right)_{i - 9} - 0.0002\left( {N_{A} } \right)_{i - 10} - 0.0002\left( {N_{A} } \right)_{i - 11} \hfill \\ \end{gathered} \right], $$

(3-1)

where $(N_{v} )_{j}$ represents the number of particles per unit volume in the j ^th class interval, where $j \in \left\{ {1,12} \right\}$. The largest particle size corresponds to j = 1. For each selected value of j, i is set equal to j, which determines the number of terms used inside the brackets. $D_{j}$ is the maximum diameter of each size group concerned. The total number of particles per unit volume, $N_{V}$, equal the sum of the $(N_{v} )_{j}$ values, hence:

$$ N_{V} = \sum\limits_{i}^{j} {\left( {N_{v} } \right)_{j} } $$

(3-2)

The mean diameter of the precipitates is obtained from [13], i.e.:

$$ \overline{D} = \frac{1}{{N_{V} }}\sum\limits_{i}^{j} {\left( {N_{v} } \right)_{j} D_{j} } $$

(3-3)