A simple procedure for measuring surface convection heat transfer coefficients involves coating the surface with a thin layer of material having a precise melting point temperature. The surface is then heated and, by determining the time required for melting to occur, the convection coefficient is determined. The following experimental arrangement uses the procedure to determine the convection coefficient for gas flow normal to a surface. Specifically, a long copper rod is encased in a super insulator of very low thermal conductivity, and a very thin coating is applied to its exposed surface. If the rod is initially at 25°C and gas flow for which h = 200 W/m2 ∙ K and T∞ = 300°C is initiated, what is the melting point temperature of the coating if melting is observed to occur at t = 400 s?

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Calculate the melting point temperature of the surfaced coating at a distance x=0 if the melting observed to occur at time 400 \mathrm{~s}.\begin{array}{l}\frac{T(x, t)-T_{i}}{T_{\infty}-T_{i}}=\operatorname{erf}\left(\frac{x}{2 \sqrt{\alpha t}}\right)-\left[\exp \left(\frac{h x}{k}+\frac{h^{2} \alpha t}{k^{2}}\right)\right]\left[\operatorname{erfc}\left(\frac{x}{2 \sqrt{\alpha t}}+\frac{h \sqrt{\alpha t}}{k}\right)\right] \\\frac{T(0, t)-25}{300-25}=\left\{\begin{aligned}\operatorname{erfc}\left(\frac{0}{2 \sqrt{10^{-4} \times 400}}\right) \\-\left\{\begin{array}{l}{\left[\exp \left(\frac{(200)(0)}{400}+\frac{(200)^{2}\left(10^{-4}\right)(400)}{(400)^{2}}\right)\right]} \\\times\left[\operatorname{erfc}\left(\frac{0}{2 \sqrt{10^{-4} \times 400}}+\frac{\left.200 \times \sqrt{10^{-4} \times 400}\right)}{400}\right)\right]\end{array}\right\}\end{aligned}\right\} \\\frac{T_{s}-25}{300-25}=\left\{\operatorname{erfc}(0)-\left\{\left[\exp \left(0+\frac{(4)(400)}{160000}\right)\right] \times\left[\operatorname{erfc}\left(0+\frac{0.2}{2}\right)\right]\right\}\right\} \\T_{s}-25=275 \times\{\operatorname{erfc}(0)-\{[\exp (0.01)] \times[\operatorname{erfc}(0.1)]\}\} \\T_{s}=25+275 \times\{\operatorname{erfc}(0)-\{[\exp (0.01)][\operatorname{erfc}(0.1)]\}\} \\\end{array}Obtain the Gaussian error function value at \operatorname{erf}(0) from the Table Gaussian Error Function at the value w=0. \operatorname{erf}(0)=0Calculate the complementary error function value at w=0. \operatorname{erfc} w=1-\operatorname{erf} w \operatorname{erfc}(0)=1-\operatorname{erf}(0.00)=1-0=1Obtain the Gaussian error function (\operatorname{erf} w) value from the Table Gaussian Error Function at the value w=0.10 : \operatorname{erf}(0.1)=0.11246Calculate the complementary error function value at w=0.10. \operatorname{erfc} w=1-\operatorname{erf} w\operatorname{erfc}(0)=1-\operatorname{erf}(0.10)=1-0.11246=0.88754Plug the values in equation (1).\begin{aligned}T_{s} & =25+275 \times\{\operatorname{erfc}(0)-\{[\exp (0.01)][\operatorname{erfc}(0.1)]\}\} \\& =25+275 \times\{1-1.01 \times 0.88754\} \\& =25+275 \times 0.10354 \\& =53.5^{\circ} \mathrm{C}\end{aligned} ...