Question 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. Gas flow T., -Surface coating Copper rod, k = 400 W/m.k, a = 10 m/s -Super insulator If the rod is initially at 25°C and gas flow for which h = 150 W/m2.K and T =300°C is initiated, what is the melting point temperature, in °C. of the coating if melting is observed to occur at t =400 s? T = i °C

SolutionGiven data\begin{aligned}k & =400 \mathrm{w} / \mathrm{m} \cdot \mathrm{k} ; \alpha=10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\h & =150 \mathrm{w} / \mathrm{m}^{2} \cdot \mathrm{k} ; \quad T_{\infty}=300^{\circ} \mathrm{C} ; T_{i}=25^{\circ} \mathrm{C} \\t=t_{m} & =400 \mathrm{sec} .\end{aligned}The termal response of transient conduction in a semiinfinite solid is\begin{aligned}\frac{T_{(x, t)}-T_{i}}{T_{\infty}-T_{i}}=\operatorname{erfc}\left[\frac{x}{2(\alpha t)^{1 / 2}}\right]- & {\left[\exp \left[\frac{h x}{k}+\frac{h^{2} \alpha t}{k^{2}}\right]\right] } \\& *\left[\operatorname{erfc}\left[\frac{x}{2(\alpha t)^{1 / 2}}+\frac{h(\alpha t)^{1 / 2}}{k}\right]\right]\end{aligned}when \left.x=0 ; T^{T}(x, t)=T_{(0, t)}\right)=T_{s} and \operatorname{erfc}(0)=1.\begin{array}{l}\frac{T_{s}-T_{i}}{T_{\infty}-T_{i}}=1-\exp \left[\frac{h^{2} \alpha t}{k^{2}}\right] \operatorname{ertc}\left[\frac{h(\alpha t)^{1 / 2}}{k}\right] \\\frac{T_{s}-T_{i}}{T_{\infty}-T_{j}}=1-\exp \left[\frac{(150)^{2}\left(10^{-4}\right)(400)}{(400)^{2}}\right] \operatorname{erfc}\left[\frac{150\left(10^{-4} \times 400\right)^{1 / 2}}{400}\right] \\\frac{T_{s}-25}{300-25}=1-[\exp (0.005625) * \operatorname{erfc}(0.075)] \\T_{s}-25=275[1-(1.00564)(0.9155)]\end{array}\begin{aligned}& { }^{s}-25=21.817 \\\Rightarrow & T_{s}=21.817+25 \\& T_{s}=46.817^{\circ} \mathrm{C}\end{aligned}\therefore The melting point of Coating is T_{S}=46.87^{\circ} \mathrm{C} ...