Question

Skylights or "roof windows" are commonly used in homes and manufacturing facilities since they let natural light in during daytime 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 \(1-\mathrm{m}\)-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 \(0.5-\mathrm{cm}\)-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 $5.34 \mathrm{~m}^{2} . \mathrm{K} / \mathrm{W}\( ). Evaluate air properties at a film temperature of \)-7^{\circ} \mathrm{C}\( and \)1 \mathrm{~atm}$ pressure. Is this a good assumption?

Short answer

Answer: The primary advantage of using a motorized insulating cover for the skylight during cold weather and at night is to significantly reduce heat losses, as the heat loss through the skylight without a cover is much higher than the heat loss through the equivalent roof area. This helps in maintaining better insulation and energy efficiency.

Step-by-step solution

Calculate the Surface Area of the Skylight

To calculate the rate of heat loss, we need the surface area of the skylight. Given that the skylight is 1 m wide and 2.5 m long, the surface area can be calculated as: Area = width × length Area = 1 m × 2.5 m = 2.5 m²

Calculate the Temperature Difference

We have the interior temperature, exterior air temperature, and the effective sky temperature. We'll calculate the temperature difference between the inside and outside to use in the heat transfer calculations. ΔT = T_inside - T_outside = 20°C - (-10°C) = 30°C

Calculate the Heat Loss Due to Conduction

First, let's find the conductive heat transfer through the glass. Using the glass's thermal conductivity k and its thickness, we can calculate the conductive heat loss rate using Fourier's law: q_cond = (k * A * ΔT) / d q_cond = (0.78 W/m·K * 2.5 m² * 30°C) / 0.005 m = 11700 W

Calculate the Radiation Heat Loss

Next, we need to find the radiation heat loss. First, we need to calculate the emissive power of the glass using the Stefan-Boltzmann law: E = σ * T^4 E = 5.67e-8 W/m²·K^4 * (20 + 273.15)^4 = 448.59 W/m² The net radiation heat loss is calculated using the emissivity of the glass: q_rad = ε * A * (E1 - E2) Since E2 = σ * (-30 + 273.15)^4 = 195.42 W/m², q_rad = 0.9 * 2.5 m² * (448.59 - 195.42) W/m² = 631.72 W

Calculate the Total Heat Loss through the Skylight

Now we need to add the heat loss due to conduction and radiation: q_total = q_cond + q_rad = 11700 W + 631.72 W = 12331.72 W

Calculate the Heat Loss through the Equivalent Roof Area

Now we have to calculate the heat loss through the same surface area (2.5 m²) of the roof with an R-5.34 construction. Using the given R-value: q_roof = (A * ΔT) / R q_roof = (2.5 m² * 30°C) / 5.34 m²·K/W = 353.96 W

Compare the Results and Evaluate Assumption

Comparing the heat loss rates: Skylight heat loss: 12331.72 W Equivalent Roof heat loss: 353.96 W The heat loss through the skylight is significantly higher than the heat loss through the equivalent roof area. Therefore, it is clear that the skylight does not provide efficient insulation, and a motorized insulating cover would be necessary to reduce heat losses during cold weather and at night. The assumption of constant air properties is reasonable as temperature difference is relatively small and the air pressure is constant at 1 atm.