Heat Transfer

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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} ...