91Ó°ÊÓ

To find the Grashof number, we need to know the properties of the air at the average temperature between the inner and outer glass layers. We can find this average temperature as: $$T_{avg} = \frac{T_{room} + T_{inner}}{2}$$ Plugging in the given values, we find the average temperature: $$T_{avg} = \frac{26^{\circ} \mathrm{C} + 18^{\circ} \mathrm{C}}{2} = 22^{\circ} \mathrm{C}$$ At this temperature, we can look up the air properties of kinematic viscosity (\(\nu\)) and thermal expansion coefficient (\(\beta\)) from standard air tables. For this case, we can use: $$\beta = 3.39 \times 10^{-3} \,\mathrm{K}^{-1}$$ $$\nu = 15.89 \times 10^{-6} \,\mathrm{m}^{2}/\mathrm{s}$$ Now we can find the Grashof number using the formula: $$Gr = \frac{g \beta (T_{room} - T_{inner}) L^3}{\nu^2}$$ Plugging in the given values and the values we found for air properties: $$Gr = \frac{9.81 \, \mathrm{m/s^2} \cdot 3.39 \times 10^{-3} \,\mathrm{K}^{-1} \cdot (26 - 18) \,\mathrm{K} \cdot (0.02 \, \mathrm{m})^3}{(15.89 \times 10^{-6} \,\mathrm{m}^{2}/\mathrm{s})^2}$$ Calculating the Grashof number, we get: $$Gr = 192,165$$

Now we need to find the Nusselt number, which can be correlated as a function of the Grashof number and Prandtl number for natural convection. For a vertical air gap, we can use the following correlation: $$Nu = 0.138 \cdot (Gr \cdot Pr)^{1/3}$$ We need to find the Prandtl number, which is defined as the ratio of momentum diffusivity to thermal diffusivity: $$Pr = \frac{\nu}{\alpha}$$ For air at \(22^{\circ} \mathrm{C}\), we can find the thermal diffusivity (\(\alpha\)) from standard air tables. For this case, we can use: $$\alpha = 22.49\times 10^{-6} \,\mathrm{m}^{2}/\mathrm{s}$$ Plugging in the value for kinematic viscosity (\(\nu\)) found earlier, we can calculate the Prandtl number: $$Pr = \frac{15.89 \times 10^{-6} \, \mathrm{m}^{2}/\mathrm{s}}{22.49 \times 10^{-6} \, \mathrm{m}^{2}/\mathrm{s}}$$ $$Pr = 0.707$$ Now substituting Gr and Pr in the Nusselt number correlation: $$Nu = 0.138 \cdot (192,165 \cdot 0.707)^{1/3}$$ Calculating the Nusselt number, we get: $$Nu = 14.312$$

Now that we have the Nusselt number, we can find the heat transfer coefficient (\(h\)) using the following formula: $$h = \frac{Nu \cdot k}{L}$$ At \(22^{\circ} \mathrm{C}\), we can find the thermal conductivity (\(k\)) of air from standard air tables: $$k = 0.026 \, \mathrm{W/m} \cdot \mathrm{K}$$ Plugging in the given air gap thickness (\(L\)) and the values for \(Nusselt \, number\) and \(thermal \, conductivity\), we can calculate the heat transfer coefficient: $$h = \frac{14.312 \cdot 0.026 \, \mathrm{W/m} \cdot \mathrm{K}}{0.02 \, \mathrm{m}}$$ $$h = 18.808 \, \mathrm{W/m}^2 \cdot \mathrm{K}$$

Finally, we can find the temperature of the outer glass layer (\(T_{outer}\)) by rearranging the formula for heat transfer and solving for \(T_{outer}\): $$q = hA(T_{room} - T_{outer})$$ $$T_{outer} = T_{room} - \frac{q}{h \cdot A}$$ We can find the heat loss per unit area (\(q'\)) using the formula: $$q' = h(T_{inner} - T_{outer})$$ $$q = q' \cdot A$$ $$T_{outer} = T_{room} - \frac{h(T_{inner} - T_{outer}) \cdot A}{h \cdot A}$$ $$T_{outer} = T_{room} - h(T_{inner} - T_{outer})$$ Plugging in the given values and the values we found for \(h\) and \(T_{room}\), we can find the temperature of the outer glass layer: $$18^{\circ} \mathrm{C} = 26^{\circ} \mathrm{C} - 18.808 \, \mathrm{W/m}^2 \cdot \mathrm{K}(T_{outer} - 18^{\circ} \mathrm{C})$$ $$T_{outer} = 16.42^{\circ} \mathrm{C}$$ Next, we can find the heat loss per unit area (\(q'\)) from the heat transfer formula: $$q' = 18.808 \, \mathrm{W/m}^2 \cdot \mathrm{K}(18^{\circ} \mathrm{C} - 16.42^{\circ} \mathrm{C})$$ $$q' = 29.72 \, \mathrm{W/m}^2$$ Now, we can find the total heat loss (\(q\)) by multiplying the heat loss per unit area by the total area of the window: $$q = 29.72 \, \mathrm{W/m}^2 \times (1.5\,\mathrm{m} \times 2.8 \,\mathrm{m})$$ $$q = 124.53 \, \mathrm{W}$$ So, the temperature of the outer glass layer is \(16.42^{\circ} \mathrm{C}\) and the rate of heat loss through the window by natural convection is \(124.53 \mathrm{W}\).