Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
[tex]\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big][/tex]
Using [tex]D_o[/tex] and [tex]E_a[/tex] values obtained from the graph:
Thus;
[tex]\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr[/tex]
[tex]B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr[/tex]
So, the initial time required to grow oxidation is expressed as:
[tex]t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)[/tex]
[tex]where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719[/tex]
∴
[tex]2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)[/tex]
[tex]2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr[/tex]
NOW;
[tex]1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453[/tex]
[tex]d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}[/tex]
[tex]d_o =0.02609 \ OR \ -0.0939[/tex]
Thus; since we will consider the positive sign, the initial thickness [tex]d_o[/tex] is ;
≅ 0.261 μm
