Question

Consider a flat ceiling that is built around $38-\mathrm{mm} \times 90-\mathrm{mm}\( wood studs with a center-to-center distance of \)400 \mathrm{~mm}\(. The lower part of the ceiling is finished with \)13-\mathrm{mm}$ gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2},{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a 13 -mm plywood layer, a layer of felt $\left(R=0.011 \mathrm{~m}^{2},{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\(, and linoleum \)\left(R=0.009 \mathrm{~m}^{2},{ }^{\circ} \mathrm{C} / \mathrm{W}\right)$. Both sides of the ceiling are exposed to still air. The airspace constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide airspace between the studs \((a)\) does not have any reflective surface, \((b)\) has a reflective surface with \(\varepsilon=0.05\) on one side, and \((c)\) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the airspace.

Short answer

Answer: The R-value for the ceiling with reflective surfaces with an emissivity of 0.05 on both sides is 7.210 m²K/W, and the U-factor is 0.139 W/m²K.

Step-by-step solution

1. Calculate the R-value of individual layers

To find the R-value of the ceiling, we first need to find the R-value for individual layers: - Wood subfloor: \(R=0.166\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) - Gypsum wallboard: Thickness is 13 mm, we will need the thermal conductivity (\(k\)) of the wallboard to calculate the R-value. - Plywood: Thickness is 13 mm, we will need the thermal conductivity (\(k\)) of plywood to calculate the R-value. - Felt: \(R=0.011\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) - Linoleum: \(R=0.009\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) For the calculation, we assume the following thermal conductivity values: - Gypsum wallboard: \(k_{gw} = 0.17\,\mathrm{W}/\mathrm{m}\mathrm{}^{\circ}\mathrm{C}\) - Plywood: \(k_{p} = 0.13\,\mathrm{W}/\mathrm{m}\mathrm{}^{\circ}\mathrm{C}\) Using the formula for R-value, \(R=\frac{thickness}{k}\) - Gypsum wallboard: \(R_{gw}=\frac{0.013\,\mathrm{m}}{0.17\,\mathrm{W}/\mathrm{m}\mathrm{}^{\circ}\mathrm{C}} = 0.076\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) - Plywood: \(R_{p}=\frac{0.013\,\mathrm{}}{0.13\,\mathrm{W}/\mathrm{m}\mathrm{}^{\circ}\mathrm{C}} = 0.100\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) Now, we have the R-value for every layer.

2. Calculate the R-value of the airspace with different scenarios

To find the R-value of the airspace, we will analyze three different scenarios \((a)\), \((b)\), \((c)\) with different reflective surfaces: (a) No reflective surface: Without a reflective surface, the R-value of the airspace can be found using the formula: \(R_{air}=\frac{1}{k_{air}*velocity}\), where \(k_{air}\) is the thermal conductivity of air and velocity is the velocity of air. However, we need more information about the velocity of air in this scenario. (b) Reflective surface with \(\varepsilon=0.05\) on one side: Let's assume that the required emissivity is given by \(\varepsilon=0.05\), and the temperature difference across airspace is \(\Delta T=5.6\,\mathrm{}^{\circ}\mathrm{C}\). The formula for calculating the R-value of airspace with reflective surface on one side is given by: \(R_{air\,b}=\frac{1}{hr}=\frac{1}{\frac{\varepsilon\sigma(T_{mean}^{2}+T_{inside}^{2})(T_{mean}+T_{inside})}}\) For this formula, we need: - \(T_{mean}=10^{\circ} \mathrm{C}\) - \(\sigma = 5.67\cdot10^{-8}\, \mathrm{W}/\mathrm{m}^2\mathrm{K}^4\) Calculation: \(R_{air\,b} =\frac{1}{0.05\cdot5.67\cdot10^{-8}\cdot((10+273)^2+(10+5.6+273)^2)((10+273)+(10+5.6+273))} = 4.357\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) (c) Reflective surfaces with \(\varepsilon=0.05\) on both sides: For this scenario, we use this formula with two reflective surfaces: \(R_{air\,c}=\frac{1}{2hr}=\frac{1}{2\frac{\varepsilon\sigma(T_{mean}^{2}+T_{inside}^{2})(T_{mean}+T_{inside})}}\) Calculation: \(R_{air\,c}=\frac{1}{2\times0.05\times5.67\times10^{-8}\times((10+273)^2+(10+5.6+273)^2)((10+273)+(10+5.6+273))} = 8.713\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\)

3. Calculate the R-value and U-factor of the ceiling for each scenario

Finally, we will calculate the R-value and U-factor of the ceiling by considering the airspace constitutes 82% of the heat transmission area and the studs and headers constitute 18%. We will add the R-value of the layers to the R-value of the airspace obtained in each scenario and obtain the U-factor with the formula \(U = \frac{1}{R}\). (a) No reflective surface: Incomplete information. (b) Reflective surface with \(\varepsilon=0.05\) on one side: \(R_{total\,b}=0.82\cdot R_{air\,b}+ 0.18\cdot(R_{gw}+R_{wood\,sub}+R_{p}+R_{felt}+R_{lin})\) \(R_{total\,b}=0.82\cdot 4.357 + 0.18\cdot(0.076+0.166+0.100+0.011+0.009)= 3.573+0.065=3.638\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) Now, the U-factor: \(U_{total\,b}=\frac{1}{R_{total\,b}}= \frac{1}{3.638} = 0.275\,\mathrm{W}/\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}\) (c) Reflective surfaces with \(\varepsilon=0.05\) on both sides: \(R_{total\,c}=0.82\cdot R_{air\,c}+ 0.18\cdot(R_{gw}+R_{wood\,sub}+R_{p}+R_{felt}+R_{lin})\) \(R_{total\,c}=0.82\cdot 8.713 + 0.18\cdot(0.076+0.166+0.100+0.011+0.009) = 7.145+0.065 = 7.210\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) Now, the U-factor: \(U_{total\,c}=\frac{1}{R_{total\,c}}= \frac{1}{7.210} = 0.139\,\mathrm{W}/\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}\) The R-value and U-factor for each scenario are: - (b) Reflective surface with \(\varepsilon=0.05\) on one side: \(R_{total\,b} = 3.638\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) and \(U_{total\,b} = 0.275\,\mathrm{W}/\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}\). - (c) Reflective surfaces with \(\varepsilon=0.05\) on both sides: \(R_{total\,c} = 7.210\,\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}/\mathrm{W}\) and \(U_{total\,c} = 0.139\,\mathrm{W}/\mathrm{m}^{2}\mathrm{}^{\circ}\mathrm{C}\).