Question

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.

Short answer

Answer: The total rate of heat transfer through the window by both natural convection and radiation is approximately 144.48 W.

Step-by-step solution

Calculate the temperature difference between the glass surfaces

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.

Calculate the heat transfer by radiation

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: \(q_r = (5.67 × 10^{-8}) \times 10 \times (288.15^4 - 278.15^4)\) \(q_r = 51.38 \mathrm{ \space W}\)

Calculate the Grashof and Prandtl numbers

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) \(Pr = \frac{1029 \times 1.406 \times 10^{-5}}{0.0247}\) \(Pr = 0.7131\)

Calculate the Nusselt number

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. \(Nu = \frac{1}{2} (0.890) (1.2306 \times 10^{9})^{1/4} (0.7131)^{1/3}\) \(Nu \approx 75.45\)

Calculate the heat transfer by natural convection

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: \(q_c = 0.931 \times 10 \times (15 - 5)\) \(q_c \approx 93.1 \mathrm{ \space W}\)

Determine the total heat transfer

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 \(144.48 \mathrm{\space W}\).

Key concepts

Natural Convection

Natural convection is a mode of heat transfer that occurs in fluids (such as air or water) due to the presence of a temperature gradient. When a fluid adjacent to a hot surface gets heated, it becomes less dense and rises, while the cooler, denser fluid sinks. This creates a circulatory motion known as convection currents. In the case of the double-pane window in the exercise, the air gap between the glass panes exhibits natural convection due to the temperature difference between the two surfaces.

In calculating natural convection heat transfer in the step by step solution, understanding the Grashof number, which represents the ratio of the buoyant to viscous force in a fluid, is crucial. It is an indication of the strength of the natural convection. Together with the Prandtl number, which reflects the fluid’s thermal diffusivity, these dimensionless numbers are essential for estimating the convective heat transfer coefficient using empirical correlations such as the Nusselt number.

Radiative Heat Transfer

Radiative heat transfer is another method by which heat can be transferred through the window. It involves the emission of infrared radiation, a form of energy emitted by all objects based on their temperature. The amount of heat transferred by radiation between the two glass panes of the window depends on their temperatures and emissivities.

The Stefan-Boltzmann law, as employed in the exercise solution, quantifies the amount of heat transferred per unit area and is crucial for calculating radiative heat transfer. It requires the conversion of temperatures to an absolute scale (Kelvin), and it considers the fourth power of the absolute temperatures, indicating that radiative heat transfer is significantly sensitive to temperature changes.

Thermal Conductivity

Thermal conductivity is the property of a material to conduct heat. It is largely dependent on the material's composition and structure and measures the amount of heat, in watts, that can move through a square meter of the material when there is a temperature gradient (measured in Kelvin) across the length of the material (measured in meters). In the exercise, the thermal conductivity of air plays a pivotal role in calculating the convective heat transfer coefficient, which is then used to find the amount of heat transferred by natural convection.

Nusselt Number

The Nusselt number, a dimensionless number used in the exercise solution, represents the enhancement of heat transfer through a fluid layer as a consequence of convection relative to conduction. A higher Nusselt number indicates a more efficient convection process. To determine the Nusselt number for natural convection between vertical plates, as in our window scenario, specific empirical correlations are used. These correlations involve the other dimensionless numbers such as the Grashof and Prandtl numbers, incorporating them into an equation to yield the Nusselt number.

Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the power radiated per unit area of the surface of a black body is directly proportional to the fourth power of the black body's thermodynamic temperature. In the context of the exercise, this law allows us to calculate the rate of heat transfer by radiation through the window. The law involves a constant, the Stefan-Boltzmann constant \( \sigma \), and requires that temperatures are expressed in Kelvin to accurately represent the thermal radiation emitted from the glass surfaces of the window, adjusted for their non-black body characteristics through the emissivity value.