Determine the \(U\)-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions. The glazings are made of clear glass having an emissivity of \(0.84\). Take the average air space temperature at design conditions to be \(10^{\circ} \mathrm{C}\) and the temperature difference across the air space to be \(15^{\circ} \mathrm{C}\).

We will first calculate the resistance for conduction through the glass, which can be determined by the equation: \(R_\text{cond}=\frac{d}{kA}\), where \(d\) is the thickness of the glass, \(k\) is the thermal conductivity of the glass, and \(A\) is the area of the glass. We know the thermal conductivity of glass to be \(k=1 \,\text{W}\,(\text{m} \cdot \text{K})^{-1}\). Assuming the glass thickness to be \(4 \,\text{mm}\) and considering unit area \(A=1 \,\text{m}^{2}\), we have: \(R_\text{cond}=\frac{0.004\,\text{m}}{1\,\text{W}\,(\text{m} \cdot \text{K})^{-1} \cdot 1\,\text{m}^{2}}=0.004\,\text{m}^{2}\,(\text{K}\,\text{W})^{-1}\). The heat transfer coefficient for conduction through glass can be calculated as: \(U_\text{cond}=\frac{1}{R_\text{cond}}=\frac{1}{0.004\,\text{m}^{2}\,(\text{K}\,\text{W})^{-1}}=250\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).

We will now calculate the radiation heat transfer coefficient (\(U_\text{rad}\)) within the air space using the following equation: \(U_\text{rad}=\frac{\sigma\epsilon(Th^{4}-Tc^{4})}{Th-Tc}\), where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8}\,\text{W}\,(\text{m}^{2}\cdot\text{K}^{4})^{-1}\)), \(\epsilon\) is the emissivity of the glass, \(Th\) is the hot surface temperature and \(Tc\) is the cold surface temperature. We are given the emissivity of the glass as \(0.84\), the average air space temperature as \(10^{\circ}\mathrm{C}\), and the temperature difference across the air space as \(15^{\circ}\mathrm{C}\). First, determine the hot and cold surface temperatures: \(Th=10+7.5=17.5^{\circ}\mathrm{C}=290.65\,\text{K}\), \(Tc=10-7.5=2.5^{\circ}\mathrm{C}=275.65\,\text{K}\). Now, calculate the radiation heat transfer coefficient: \(U_\text{rad}=\frac{5.67\times10^{-8}\,\text{W}\,(\text{m}^{2}\cdot\text{K}^{4})^{-1} \cdot 0.84 \cdot (290.65^{4}-275.65^{4})}{290.65-275.65} \approx 8.12\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).

We will now calculate the convection heat transfer coefficient (\(U_\text{conv}\)) within the air space using the following equation: \(U_\text{conv}=1.32\Delta T^{1/3}\), where \(\Delta T\) is the temperature difference across the air space. Given the temperature difference across the air space as \(15^{\circ}\mathrm{C}\), we have: \(U_\text{conv}=1.32\cdot15^{1/3}\approx 4.36\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).

Finally, we will calculate the overall U-factor by summing up the heat transfer coefficients from the above steps. Since the window has two glass panes, we need to consider the conduction through both panes. Therefore, \(U_\text{total}=U_\text{cond}+U_\text{rad}+U_\text{conv}+U_\text{cond} = 250 + 8.12 + 4.36 + 250 \approx 512.48\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\). The U-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions is approximately \(512.48\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).