Question

A vertical 2-m-high and 5-m-wide double-pane window consists of two sheets of glass separated by a \(3-\mathrm{cm}\)-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

Question: Determine the overall rate of heat transfer through a double-pane window with the dimensions: 2 m high, 5 m wide, and an air gap between the panes. The glass has emissivities on both faces of 0.84 and the temperatures across the gap are 15°C and 5°C. Answer: To determine the overall rate of heat transfer through the window, you need to calculate the natural convection and radiation heat transfer rates and then sum them. Following the steps in the solution, you will find the overall heat transfer rate through the window.

Step-by-step solution

Calculate the Area of the Window

Using the given dimensions of the window (2 m high and 5 m wide), we can calculate the surface area, \(A\): \(A = height \times width = 2\,\text{m} \times 5\,\text{m} = 10\,\text{m}^2\)

Convert Temperatures to Kelvin

For the formulas, we need to use temperatures in Kelvin. We have \(T_1 = 15^{\circ} \mathrm{C}\) and \(T_2 = 5^{\circ} \mathrm{C}\). To convert these to Kelvin, we add 273.15: \(T_1 = 15 + 273.15 = 288.15\,\text{K}\) \(T_2 = 5 + 273.15 = 278.15\,\text{K}\)

Calculate the Mean Temperature and Relevant Properties

We need to calculate the mean temperature for finding relevant properties of air: \(T_m = \frac{T_1 + T_2}{2} = \frac{288.15 + 278.15}{2} = 283.15\,\text{K}\) At this mean temperature, we have the following properties for air: - Thermal conductivity, \(k = 0.0262\,\text{W}/\text{m}\cdot\text{K}\) - Expansion coefficient, \(\beta = 1/T_m = 3.53 \times 10^{-3}\,\text{K}^{-1}\) - Dynamic viscosity, \(\mu = 1.846\times 10^{-5}\,\text{N s}/\text{m}^{2}\) - Specific heat, \(c_p = 1005\,\text{J}/\text{kg}\cdot\text{K}\) - Kinematic viscosity, \(\nu = 15.52\times 10^{-6}\,\text{m}^{2}/\text{s}\)

Calculate the Rayleigh Number

Using the properties above and the temperature difference, we can now calculate the Rayleigh number, \(Ra\), for the air gap: \(Ra = \frac{g\beta L^{3}(T_1-T_2)}{\nu\alpha}\) Where \(L\) is the height of the air gap and \(\alpha\) is the thermal diffusivity, which can be found as \(\alpha = k/(\rho c_p)\), with \(\rho\) being the air density (density can be also found through the ideal gas law: \(\rho =\frac{P_m}{R_aT_m}\), where \(P_m\) is the pressure inside the gap, and \(R_a\) is the specific gas constant for air). Calculate \(\rho\), \(\alpha\), and then \(Ra\).

Calculate the Nusselt Number

Now we can use the Rayleigh number to determine the Nusselt number, \(Nu\), for this scenario. Assuming an enclosed vertical air gap, we use the following equation for \(Nu\): \(Nu = \frac{0.586Ra^{1/4}}{[1+(0.559/Pr)^{9/16}]^{4/9}}\) Calculate \(Nu\) using the calculated \(Ra\) value and the Prandtl number, \(Pr\) (which is equal to \(\nu/\alpha\)).

Calculate the Heat Transfer Coefficient

With the Nusselt number, we can now determine the heat transfer coefficient, \(h\), for natural convection: \(h = \frac{Nu\cdot k}{L}\) Calculate \(h\) using the obtained \(Nu\) value and the previously found \(k\) and \(L\) values.

Compute the Natural Convection Heat Transfer Rate

Now we can calculate the natural convection heat transfer rate, \(q_{conv}\): \(q_{conv} = hA(T_{1} - T_{2})\) Calculate the natural convection heat transfer rate using the calculated \(h\) and \(A\) values, and given \(T_1\) and \(T_2\).

Compute the Radiation Heat Transfer Rate

Now we should calculate the radiation heat transfer rate, \(q_{rad}\). For this we need to use the given emissivity values \(\epsilon\) and the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8}\,\text{W}/\text{m}^2\cdot\text{K}^{4}\). \(q_{rad} = \sigma \epsilon A (T_{1}^{4} - T_{2}^{4})\) Calculate the radiation heat transfer rate using the given \(\epsilon\) value, \(A\), and the converted temperature values in Kelvin (\(T_1\) and \(T_2\)).

Determine Overall Heat Transfer Rate

Now we can determine the overall heat transfer rate through the window by adding the natural convection and radiation heat transfer rates: \(q = q_{conv} + q_{rad}\) Calculate the overall heat transfer rate through the window.