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Question

The rear window of an automobile is defogged by attaching a thin, transparent, film-type heating element to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. (a) For 4-mm-thick window glass, determine the electrical power required per unit window area to maintain an inner surface temperature of 15C15^{\circ} \mathrm{C} when the interior air temperature and convection coefficient are T,i=25CT_{\infty, i}=25^{\circ} \mathrm{C} and hi=10W/m2Kh_{i}=10 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, while the exterior (ambient) air temperature and convection coefficient are T,o=10CT_{\infty, o}=-10^{\circ} \mathrm{C} and ho=65W/m2Kh_{o}=65 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}.

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Known:\textbf{Known:}

  • window glass thickness, L=4L=4mm=0.004= 0.004m
  • inner surface of the window temperature, Ts,i=15T_{s,i}=15^\circC
  • interior air temperature and convection factor, T,i=25T_{\infty,i}=25^\circC, hi=10Wm2Kh_i=10\frac{\text{W}}{\text{m}^2\text{K}}
  • exterior ambient air temperature and convection factor, T,o=10T_{\infty,o}=-10^\circC, ho=65Wm2Kh_o=65\frac{\text{W}}{\text{m}^2\text{K}}
  • b)\textbf{b)} convection factors for plotting the graphs, h0=2,20,65,100Wm2Kh_0=2, 20, 65, 100 \frac{\text{W}}{\text{m}^2\text{K}}

Since there are no other parameters involved, electrical power can be found as the heat flux on the plane wall. Heat flux expression can be set as energy balance on the inner surface of the window:

q=T,iTs,i1h+Ts,iT,o1ho+Lkq=2515110+15+10165+0.0041.4q=1270.48Wm2\begin{align*} &q''=-\frac{T_{\infty,i}-T_{s,i}}{\frac{1}{h}}+\frac{T_{s,i}-T_{\infty,o}}{\frac{1}{h_o}+\frac{L}{k}}\\ &q''=-\frac{25-15}{\frac{1}{10}}+\frac{15+10}{\frac{1}{65}+\frac{0.004}{1.4}}\\ &\boxed{q''=1270.48 \frac{\text{W}}{\text{m}^2}}\\ \end{align*}

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