Question

Consider a \(1.2\)-m-high and 2-m-wide doublepane window consisting of two \(3-\mathrm{mm}\)-thick layers of glass $(k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( separated by a \)2.5$-cm-wide airspace. 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?

Short answer

Answer: The steady rate of heat transfer through the double-pane window is 444.4 W, and the temperature of its inner surface is 19.29°C. As the temperatures involved in this problem are relatively low, it is reasonable to assume that radiation has a minor effect compared to conduction and convection in this case.

Step-by-step solution

Calculate the resistances

First, we will compute the thermal resistance for each layer: the two layers of glass and the layer of air in between. For the glass layers, we have: $$R_{G} = \frac{\Delta x}{kA}$$ Where \(R_{G}\) is the thermal resistance of the glass, \(\Delta x = 3 \times 10^{-3}\,\text{m}\) is the thickness, \(k = 0.78 \frac{\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}}\) is the thermal conductivity, and A is the area. Considering the dimensions of the window, the area is: $$A = 1.2\,\text{m} \times 2\,\text{m} = 2.4\,\mathrm{m^2}$$ Computing the resistance for the glass layers, we have: $$R_{G} = \frac{3 \times 10^{-3}\,\text{m}}{0.78 \frac{\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}} \times 2.4\,\mathrm{m^2}} = 1.6026 \times 10^{-3}\,\mathrm{K/W}$$ Considering the air layer, the resistance is given by: $$R_{A} = \frac{1}{hA}$$ Where \(h = 10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the heat transfer coefficient and A is the area. Calculating the resistance, we have: $$R_{A} = \frac{1}{10 \frac{\mathrm{W}}{\mathrm{m}^{2}\cdot\mathrm{K}} \times 2.4\, \mathrm{m^2}} = 0.0417\,\mathrm{K/W}$$

Calculate the overall resistance and steady state heat transfer rate

Next, we will find the overall resistance by adding resistance of two glass layers and the air layer: $$R_{total} = 2R_{G} + R_{A}$$ $$R_{total} = 2(1.6026 \times 10^{-3}\,\mathrm{K/W}) + 0.0417 \,\mathrm{K/W} = 0.045\,\mathrm{K/W}$$ Now, we will compute the steady state heat transfer rate using the overall resistance and the given temperature difference: $$q = \frac{T_{1} - T_{2}}{R_{total}}$$ Where \(T_{1} = 20^{\circ} \mathrm{C}\) is the room temperature, \(T_{2} = 0^{\circ} \mathrm{C}\) is the outside temperature. Calculating the heat transfer rate, we have: $$q = \frac{20\,\mathrm{C} - 0\,\mathrm{C}}{0.045\, \mathrm{K/W}} = 444.4\,\mathrm{W}$$

Calculate the inner surface temperature

Finally, we will compute the temperature of the inner surface of the window. We will use the heat transfer rate and the resistance of the glass for this calculation: $$T_{surface} = T_{1} - q \times R_{G}$$ $$T_{surface} = 20^{\circ} \mathrm{C} - 444.4\,\mathrm{W} \times 1.6026 \times 10^{-3}\,\mathrm{K/W} = 19.29^{\circ} \mathrm{C}$$ Thus, the steady rate of heat transfer through the double-pane window is \(444.4\,\mathrm{W}\), and the temperature of its inner surface is \(19.29^{\circ} \mathrm{C}\). Regarding the assumption about disregarding radiation, since the temperatures involved in this problem are relatively low, it is reasonable to assume that radiation has a minor effect comparing to conduction and convection in this case.