Consider a 1.2-m-high and 2-m-wide glass window with a thickness of \(6 \mathrm{~mm}\), thermal conductivity \(k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.9\). The room and the walls that face the window are maintained at \(25^{\circ} \mathrm{C}\), and the average temperature of the inner surface of the window is measured to be \(5^{\circ} \mathrm{C}\). If the temperature of the outdoors is \(-5^{\circ} \mathrm{C}\), determine \((a)\) the convection heat transfer coefficient on the inner surface of the window, \((b)\) the rate of total heat transfer through the window, and \((c)\) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case?

Using the Fourier's law of heat conduction, we can find the conduction heat transfer through the glass window. The formula for heat conduction is: \(Q_{cond} = k \cdot A \cdot \frac{ΔT}{d}\) Where: - k is the thermal conductivity of the glass (\(0.78 \,\mathrm{W/m \cdot K}\)) - A is the area of the window (\(1.2\,\mathrm{m} \times 2\,\mathrm{m} = 2.4\, \mathrm{m^2}\)) - ΔT is the temperature difference between the room and the outer surface of the window (\(25^\circ \mathrm{C} - (-5^\circ \mathrm{C}) = 30\,\mathrm{K}\)) - d is the thickness of the glass window (\(6\,\mathrm{mm} = 0.006\,\mathrm{m}\)) Now, calculate the conduction heat transfer: \(Q_{cond} = 0.78 \times 2.4 \times \frac{30}{0.006} = 936\,\mathrm{W}\)

The formula for convection heat transfer is: \(Q_{conv} = h_{in} \cdot A \cdot (T_{room} - T_{avg_{in}})\) Since we have already calculated the conduction heat transfer (Q_cond), we can assume that Q_cond ≈ Q_conv. Therefore, we can solve for the convection heat transfer coefficient (h_in): \(h_{in} = \frac{Q_{conv}}{A \cdot (T_{room} - T_{avg_{in}})} = \frac{936\,\mathrm{W}}{2.4\,\mathrm{m^2} \times (25^\circ \mathrm{C} - 5^\circ \mathrm{C})} = 13\,\mathrm{W/m^2 \cdot K}\)

As we mentioned before, in this case, we can assume that Q_cond ≈ Q_conv. Therefore, the total heat transfer (Q_total) through the window is: \(Q_{total} = Q_{conv} = 936\,\mathrm{W}\)

The formula for the combined natural convection and radiation heat transfer is: \(Q_{conv+rad} = h_{out} \cdot A \cdot (T_{avg_{in}} - T_{out})\) Again, we can assume that Q_conv+rad ≈ Q_total. Therefore, we can solve for the combined natural convection and radiation heat transfer coefficient (h_out): \(h_{out} = \frac{Q_{total}}{A \cdot (T_{avg_{in}} - T_{out})} = \frac{936\,\mathrm{W}}{2.4\,\mathrm{m^2} \times (5^\circ \mathrm{C} - (-5^\circ \mathrm{C})} = 26\,\mathrm{W/m^2 \cdot K}\) In conclusion: (a) The convection heat transfer coefficient on the inner surface of the window (h_in) is \(13\,\mathrm{W/m^2 \cdot K}\). (b) The rate of total heat transfer through the window (Q_total) is \(936\,\mathrm{W}\). (c) The combined natural convection and radiation heat transfer coefficient on the outer surface of the window (h_out) is \(26\,\mathrm{W/m^2 \cdot K}\). Considering that the glass thickness is only \(6\mathrm{~mm}\) and the heat transfer coefficients (h_in and h_out) have much larger values, it can be reasonable to neglect the thermal resistance of the glass in this case. This is because the relatively thin glass and high heat transfer coefficients make the contribution of the glass to the overall heat transfer negligible.