A vertical 2-m-high and 5-m-wide double-pane window consists of two sheets of glass separated by a 3 -cm-thick air gap. In order to reduce heat transfer through the window, the air space between the two glasses is partially evacuated to \(0.3 \mathrm{~atm}\) pressure. The emissivities of the glass surfaces are 0.9. Taking the glass surface temperatures across the air gap to be \(15^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C}\), determine the rate of heat transfer through the window by natural convection and radiation.

First, we need to find the temperature difference between the glass surfaces: ΔT = T1 – T2 = 15°C – 5°C = 10°C Here, T1 = 15°C and T2 = 5°C.

To calculate the heat transfer by radiation, we can use the formula: \(q_r = \sigma \space A \space (T_1^4 - T_2^4)\) where: - \(q_r\) is the heat transfer by radiation (W) - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 × 10^{-8} \space \mathrm{W/ m^2 \cdot K^4}\)) - \(A\) is the window area (m²) - \(T_1\) and \(T_2\) are the temperatures of the glass surfaces in Kelvin First, convert the temperatures to Kelvin: \(T_1\) = 15°C + 273.15 = 288.15 K \(T_2\) = 5°C + 273.15 = 278.15 K Calculate the area of the window: A = height × width = 2m × 5m = 10 m² Now, we can calculate the heat transfer by radiation: \(q_r = (5.67 × 10^{-8}) \times 10 \times (288.15^4 - 278.15^4)\) \(q_r = 51.38 \mathrm{ \space W}\)

To calculate the Grashof and Prandtl numbers, we need the properties of air, such as thermal conductivity, kinematic viscosity, thermal expansion coefficient, and specific heat at constant pressure, at the average temperature between the glass surfaces, which is \((T_1 + T_2)/2\) = (15 + 5) / 2 = 10°C. Properties of air at 10°C can be found from standard tables or online sources: - Thermal conductivity, \(k = 0.0247 \mathrm{\space W/ m \cdot K}\) - Kinematic viscosity, \(\nu = 1.406 \mathrm{ \times} 10^{-5} \mathrm{\space m^2/s}\) - Thermal expansion coefficient, \(\beta = 0.00328 \mathrm{\space K^{-1}}\) - Specific heat at constant pressure, \(c_p = 1029 \mathrm{\space J/ kg \cdot K}\) Next, we need to calculate the Grashof number, \(Gr\), which is given by: \(Gr = \frac{g \beta ΔT L^3}{\nu^2}\) where: - \(g\) is the acceleration due to gravity (\(9.81 \mathrm{ \space m/s^2}\)) - \(\beta\) is the thermal expansion coefficient (K¯¹) - \(ΔT\) is the temperature difference between the glass surfaces (K) - \(L\) is the height of the window (m) - \(\nu\) is the kinematic viscosity (m²/s) \(Gr = \frac{9.81 \times 0.00328 \times 10 \times 2^3}{(1.406 \times 10^{-5})^2}\) \(Gr = 1.2306 \times 10^{9}\) Now, we need to calculate the Prandtl number, \(Pr\), which is given by: \(Pr = \frac{c_p \nu}{k}\) where: - \(c_p\) is the specific heat at constant pressure (J/kg·K) - \(\nu\) is the kinematic viscosity (m²/s) - \(k\) is the thermal conductivity (W/m·K) \(Pr = \frac{1029 \times 1.406 \times 10^{-5}}{0.0247}\) \(Pr = 0.7131\)

We can use the appropriate correlation for natural convection between vertical parallel plates to calculate the Nusselt number, \(Nu\). A common correlation is: \(Nu = \frac{1}{2} C_1 Gr^{1/4} Pr^{1/3}\) where \(C_1 = 0.890\) is a constant. \(Nu = \frac{1}{2} (0.890) (1.2306 \times 10^{9})^{1/4} (0.7131)^{1/3}\) \(Nu \approx 75.45\)

Next, we need to calculate the heat transfer by natural convection, which can be calculated using the formula: \(q_c = h A (T_1 - T_2)\) where: - \(h\) is the convective heat transfer coefficient (W/m²·K) - \(A\) is the window area (m²) - \(T_1\) and \(T_2\) are the temperatures of the glass surfaces in °C First, we need to calculate the convective heat transfer coefficient, \(h\): \(h = \frac{k Nu}{L}\) where: - \(k\) is the thermal conductivity (W/m·K) - \(Nu\) is the Nusselt number - \(L\) is the height of the window (m) \(h = \frac{0.0247 \times 75.45}{2}\) \(h = 0.931 \mathrm{ \space W/ m^2 \cdot K}\) Now, we can calculate the heat transfer by natural convection: \(q_c = 0.931 \times 10 \times (15 - 5)\) \(q_c \approx 93.1 \mathrm{ \space W}\)

Finally, we can determine the total heat transfer by adding the heat transfer by radiation and natural convection: Total heat transfer = \(q_r\) + \(q_c\) = 51.38 W + 93.1 W = 144.48 W Therefore, the total rate of heat transfer through the window by both natural convection and radiation is approximately \(144.48 \mathrm{\space W}\).