Quantitative Approach for Determining the Critical Volume Fraction for the Transition from Internal to External Oxidation

Abstract

A theoretical approach is presented to calculate the critical volume fraction f ^* _v for the transition from internal to external oxidation. The focus was on the evolution of internal-oxide volume fraction during the initial stage of oxidation. An equation to calculate f ^* _v was developed by considering an effective diffusion coefficient of oxygen in the alloy (D ^eff _O ) which takes into account the evolving blocking effect by forming internal oxide precipitates. Based on this physical process, three types of oxidation behavior can result, depending on the properties of the alloy and exposure atmosphere. Further discussions on the remaining discrepancy between predicted and experimentally-determined f ^* _v values are provided for refinement on the theoretical prediction of this parameter.

Appendices

Appendix 1

Determining volume fraction (f _v ′) from mole fraction (N _BOv ′)

$$\begin{gathered} f_{v}^{'} = \frac{{V_{{BO_{\upsilon } }} }}{{V_{{BO_{\upsilon } }} + V_{matrix} }}\;where\;V_{i} \;is\;volume\;of\;i \hfill \\ = \frac{{n_{{BO_{\upsilon } }} V_{m}^{oxide} }}{{n_{{BO_{\upsilon } }} V_{m}^{oxide} + (n_{total} - n_{{BO_{\upsilon } }} )V_{m}^{alloy} }}\;where\;n_{i} \;is\;mole\;of\;i \hfill \\ = \frac{1}{{1 + \left( {\frac{{n_{total} }}{{n_{{BO_{\upsilon } }} }} - 1} \right)\frac{{V_{m}^{alloy} }}{{V_{m}^{{BO_{\upsilon } }} }}}}\;note\;that\;\frac{{n_{{BO_{v} }} }}{{n_{total} }} = N_{{BO_{v} '}} \hfill \\ = \frac{{V_{m}^{{BO_{\upsilon } }} /V_{m}^{alloy} }}{{V_{m}^{{BO_{\upsilon } }} /V_{m}^{alloy} + \left[ {\frac{{\pi N_{O}^{S} D_{O} }}{{\nu D_{B} (N_{B}^{0} )^{2} }}\frac{{1 - f_{v} }}{{2 + f_{v} }} - 1} \right]}} \hfill \\ = \frac{{V_{m}^{{BO_{\upsilon } }} /V_{m}^{alloy} }}{{V_{m}^{{BO_{\upsilon } }} /V_{m}^{alloy} + \left[ {Z\frac{{1 - f_{v} }}{{2 + f_{v} }} - 1} \right]}} \hfill \\ \end{gathered}$$

where V ^oxide _m , V ^alloy _m are molar volumes of BO_ν and alloy and $Z \equiv \frac{{\pi N_{O}^{S} D_{O} }}{{vD_{B} \left( {N_{B}^{0} } \right)^{2} }}$

Appendix 2

Mathematical derivation to determine f ^* _v (after Leblond [6]).

Letting f ^′_v  = f _v, the Eq. (12) becomes:

$$f_{v} = \frac{{V_{m}^{oxide} /V_{m}^{alloy} }}{{V_{m}^{oxide} /V_{m}^{alloy} + \left[ {Z\frac{{1 - f_{v} }}{{2{ + }f_{v} }} - 1} \right]}}$$

This can be converted to a quadratic equation:

$$\left( {R - Z - 1} \right) \cdot f_{v}^{2} + \left( {R + Z - 2} \right)f_{v} - 2R = 0$$

(15)

where R = V ^oxide _m /V ^alloy _m .

For the critical transition point, the quadratic equation has a unique solution, which means the discriminant is zero:

$$\varDelta = \left( {R + Z - 2} \right)^{2} + 8R\left( {R - Z - 1} \right) = 0$$

From that, we can determine Z as a function of R:

$$Z_{1} = 3R + 2 + 2\sqrt {6R}\; {\text{or}}\; Z_{2} = 3R + 2 - 2\sqrt {6R}$$

When the quadratic Eq. (15) has a double solution, this solution is

$$f_{v} = - \frac{R + Z - 2}{{2\left( {R - Z - 1} \right)}}$$

(16)

Substituting Z ₁ into the Eq. (16), it is found that f _v is negative, which is not realistic for a volume fraction.

Therefore, the only solute is obtained by substituting Z ₂ into Eq. (16), which is:

$$\begin{gathered} f_{v}^{*} = \frac{{2\sqrt {V_{m}^{oxide} /V_{m}^{alloy} } }}{{\sqrt 6 + 2\sqrt {V_{m}^{oxide} /V_{m}^{alloy} } }} \hfill \\ \hfill \\ \end{gathered}$$