Question Heat Transfer 8 of 12 = * Your answer is incorrect. 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 Tih -Surface coating Copper rod, * = 400 Wim-K, a=10m?! -Super insulator If the rod is initially at 25°C and gas flow for which h = 300 W/m2.K and To =300°C is initiated, what is the melting point temperature, in °C. of the coating if melting is observed to occur att =400 s? TS = i 293.36 °C

SolutionGiver dota:\begin{array}{l}T_{i}=25^{\circ} \mathrm{C} \\T_{\infty}=300^{\circ} \mathrm{C} \\h=300 \mathrm{w} / \mathrm{m}^{2} \cdot \mathrm{k} \\t_{m}=400 \mathrm{sec} . \\\alpha=10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{k}\end{array}Assumptions:-The rod is one dimentional conduction solid. (losses ane negligible to insulation).Negligible termal resistance of coatingThe properties of rod are constant.The rod approximated as semi-infinite medium.Thermal response corresponding to transient Conduction in a semi-infinate solid is given by\begin{array}{l}\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}\right]\right] \\*\left[\operatorname{erfc}\left[\frac{x}{2(\alpha t)^{1 / 2}}+\frac{n(\alpha t)^{1 / 2}}{k}\right]\right] \\\end{array}for x=0 \quad T(x, t)=T(0, t)=T_{s} and \operatorname{ertc}(0)=1\begin{array}{l} \therefore \frac{T_{S}-T_{i}}{T_{\infty}-T_{i}}=1-\exp \left[\frac{h^{2} \alpha t_{m}}{k^{2}}\right] \operatorname{ertc}\left[\frac{h\left(\alpha t_{n}\right)^{1 / 2}}{k}\right] \\\frac{T_{S}-T_{i}}{T_{\infty}-T_{i}}=1-\exp \left[\frac{(300)^{2}\left(10^{-4}\right)(400)}{(400)^{2}}\right] \operatorname{ertc}\left[\frac{(300)\left(10^{-4} \times 400\right)^{1 / 2}}{400}\right] \\\frac{T_{S}-T_{i}}{T_{\infty}-T_{i}}=1-\exp [0.0225] \operatorname{erfc}[0.15] \\\Rightarrow T_{S}-T_{i}=\left(T_{\infty}-T_{i}\right)[1-\exp [0.0225] \operatorname{erfc}[0.15]] \\\Rightarrow T_{S}=T_{i}+\left[T_{\infty}-T_{i}\right][1-\exp [0.025] \operatorname{ertc}[0.15]] \\T_{S}=25+[300-25][1-(1.0253)(0.832)] \\T_{S}=25+40.411 \\T_{S}=65.411^{0} \mathrm{C}\end{array}The melting point of coating is T_{S}=65.411^{\circ} \mathrm{C} ...