Skylights or "roof windows" are commonly used in homes and manufacturing facilities since they let natural light in during day time and thus reduce the lighting costs. However, they offer little resistance to heat transfer, and large amounts of energy are lost through them in winter unless they are equipped with a motorized insulating cover that can be used in cold weather and at nights to reduce heat losses. Consider a 1 -m-wide and \(2.5\)-m-long horizontal skylight on the roof of a house that is kept at \(20^{\circ} \mathrm{C}\). The glazing of the skylight is made of a single layer of \(0.5\)-cm-thick glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\varepsilon=0.9)\). Determine the rate of heat loss through the skylight when the air temperature outside is \(-10^{\circ} \mathrm{C}\) and the effective sky temperature is \(-30^{\circ} \mathrm{C}\). Compare your result with the rate of heat loss through an equivalent surface area of the roof that has a common \(R-5.34\) construction in SI units (i.e., a thickness-to-effective-thermal- conductivity ratio of \(\left.5.34 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\right)\). Evaluate air properties at a film temperature of \(-7^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

To find the heat loss through the skylight caused by conduction and radiation, we apply the following equations: - For conduction: \(\dot{Q}_{cond} = \frac{kA \Delta T}{d}\), where \(k\) is the thermal conductivity, \(A\) is the area of the skylight, \(\Delta T\) is the temperature difference, and \(d\) is the thickness of the glazing. - For radiation: \(\dot{Q}_{rad} = \varepsilon A \sigma (T_{sky}^4 - T_{glass}^4)\), where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_{sky}\) and \(T_{glass}\) are the sky and glass temperatures in Kelvin. First, convert the temperatures into Kelvin: \(T_{sky} = -30^{\circ}C + 273.15 = 243.15\,K\), \(T_{glass} = 20^{\circ}C + 273.15 = 293.15\,K\). Next, calculate the area of the skylight: \(A = 1\,m \times 2.5\,m = 2.5\,m^2\). Now, calculate the heat loss through conduction and radiation: \(\dot{Q}_{cond} = \frac{0.78\frac{W}{m \cdot K} \cdot 2.5\,m^2 \cdot (20^{\circ}C + 10^{\circ}C)}{0.5\,cm} = \frac{0.78\frac{W}{m \cdot K} \cdot 2.5\,m^2 \cdot 30\,K}{5 \times 10^{-3} m} = 702\,W\) \(\dot{Q}_{rad} = 0.9 \cdot 2.5\,m^2 \cdot 5.67 \times 10^{-8}\frac{W}{m^2 \cdot K^4} (243.15^{\,4} - 293.15^{\,4}) = -401\,W\) As the radiation heat loss is negative, we only consider the conduction heat loss, which is the dominating heat transfer mechanism. So, \(\dot{Q}_{skylight} = 702\,W\).

We first find the overall heat transfer coefficient (U-value) for the R-5.34 roof: \(U = \frac{1}{R}\), where \(R\) is the thermal resistance. \(U = \frac{1}{5.34\,\frac{m^2 \cdot K}{W}} = 0.187\,\frac{W}{m^2 \cdot K}\) Now, we can calculate the heat loss through an equivalent surface area using the equation: \(\dot{Q}_{roof} = UA\Delta T\): \(\dot{Q}_{roof} = 0.187\,\frac{W}{m^2 \cdot K} \cdot 2.5\,m^2 \cdot 30\,K = 140.6\,W\)

Now we have the heat losses through the skylight and the equivalent roof surface, so we can compare them: \(\dot{Q}_{skylight} = 702\,W\) \(\dot{Q}_{roof} = 140.6\,W\) The heat loss through the skylight is significantly higher than that of the equivalent roof surface. This illustrates the importance of proper insulation for skylights to reduce energy consumption. The given air properties - film temperature of -7°C and 1 atm pressure - represent a reasonable assumption as it stands between the outside and inside temperatures and is under normal atmospheric pressure conditions.