Consider a \(1.2\)-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass \((k=\) \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separated by a \(3-\mathrm{cm}\)-wide air space. Determine the steady rate of heat transfer through this window and the temperature of its inner surface for a day during which the room is maintained at \(20^{\circ} \mathrm{C}\) while the temperature of the outdoors is \(0^{\circ} \mathrm{C}\). Take the heat transfer coefficients on the inner and outer surfaces of the window to be \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and disregard any heat transfer by radiation. Evaluate air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

To find the total thermal resistance, we will consider all the thermal resistances in series: 1. Convection resistance from the room to the inner glass surface 2. Conduction resistance through the first glass layer 3. Convective resistance from the first air layer to the second glass layer 4. Conduction resistance through the second glass layer 5. Convective resistance from the second glass layer to the outside The convection resistance is given by: $$R_{conv} = \frac{1}{hA}$$ The conduction resistance is given by: $$R_{cond} = \frac{L}{kA}$$ where: \(h\) = heat transfer coefficient \(A\) = surface area \(L\) = thickness of the conduction material \(k\) = thermal conductivity Calculating the given values, we have: Area of the window, \(A = 1.2 * 2 = 2.4 \ \mathrm{m^2}\) Thickness of glass, \(L_{glass} = 3 * 10^{-3} \ \mathrm{m}\) Width of air space, \(L_{air} = 3*10^{-2} \ \mathrm{m}\) Now let's calculate individual thermal resistances: 1. \(R_{conv1} = \frac{1}{(10)(2.4)} = 0.04167 \ \mathrm{m^2 K/W}\) 2. \(R_{cond1} = \frac{3*10^{-3}}{(0.78)(2.4)} = 0.001282 \ \mathrm{m^2 K/W}\) 3. \(R_{conv2}\) of air layer is missing. We will address this in the assumption evaluation part. 4. \(R_{cond2} = \frac{3*10^{-3}}{(0.78)(2.4)} = 0.001282 \ \mathrm{m^2 K/W}\) 5. \(R_{conv3} = \frac{1}{(25)(2.4)} = 0.01667 \ \mathrm{m^2 K/W}\) The total thermal resistance (excluding the air layer) is the sum of all individual resistances: $$R_{total} = R_{conv1} + R_{cond1} + R_{cond2} + R_{conv3} = 0.04167 + 0.001282 + 0.001282 + 0.01667 = 0.0609 \ \mathrm{m^2 K/W}$$ Now let's evaluate the assumption part and address the missing air layer.

The problem asks if it's a good idea to disregard the heat transfer by radiation and evaluate air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. The heat transfer through the air layer is mainly by conduction, which is insignificant compared to conduction through the glass and convection, and therefore this assumption is reasonable. The air properties can also be evaluated at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Since radiation heat transfer is neglected, we only need to consider the convective heat transfer coefficient inside the air gap. However, we don't have this value in the given information. Hence, this step remains incomplete.

We know the total thermal resistance calculated earlier (excluding the air layer) and the temperature difference between the room and outside. Now we can find the steady heat transfer rate using the formula: $$Q = \frac{T_{room} - T_{outside}}{R_{total}}$$ Substitute the given values: $$Q = \frac{20 - 0}{0.0609} = 327.9 \ \mathrm{W}$$

Using the convective heat transfer equation: $$Q = h_{1}A(T_{room} - T_{surface1})$$ Rearranging the equation and solving for \(T_{surface1}\): $$T_{surface1} = T_{room} - \frac{Q}{h_{1}A} = 20 - \frac{327.9}{(10)(2.4)} = 18.64^{\circ} \mathrm{C}$$ The steady heat transfer through the double-pane window is \(327.9 \ \mathrm{W}\), and the temperature of the inner surface is \(18.64^{\circ} \mathrm{C}\). However, this solution doesn't consider the air layer convective resistance, which would possibly alter the results slightly.