Question

A vertical double-pane window consists of two sheets of glass separated by a \(1.5-\mathrm{cm}\) air gap at atmospheric pressure. The glass surface temperatures across the air gap are measured to be \(278 \mathrm{~K}\) and \(288 \mathrm{~K}\). If it is estimated that the heat transfer by convection through the enclosure is \(1.5\) times that by pure conduction and that the rate of heat transfer by radiation through the enclosure is about the same magnitude as the convection, the effective emissivity of the two glass surfaces is (a) \(0.47\) (b) \(0.53\) (c) \(0.61\) (d) \(0.65\) (e) \(0.72\)

Short answer

Answer: The effective emissivity of the two glass surfaces is approximately 0.53.

Step-by-step solution

Write the heat transfer equations.

We write the equations for conduction, convection, and radiation heat transfer. Conduction: \(q_{cond} = k\frac{T_1-T_2}{L}\), where: - \(k\) is the thermal conductivity of air - \(T_1\) is the temperature of the first glass surface - \(T_2\) is the temperature of the second glass surface - \(L\) is the air gap thickness Convection: \(q_{conv} = h_{conv}A(T_1-T_2)\), where: - \(h_{conv}\) is the convective heat transfer coefficient - \(A\) is the surface area Radiation: \(q_{rad} = \epsilon\sigma A(T_1^4 - T_2^4)\), where: - \(\epsilon\) is the effective emissivity of the glass surfaces - \(\sigma\) is the Stefan-Boltzmann constant

Write the relationship between heat transfer rates.

We are given that: - \(q_{conv} = 1.5q_{cond}\) - \(q_{rad} \approx q_{conv}\) From these relationships, we can write: \(q_{conv} \approx q_{rad} = 1.5q_{cond}\)

Substitute the heat transfer equations into the relationship.

We substitute the heat transfer equations into the relationship to find the effective emissivity. \(h_{conv}A(T_1-T_2) \approx \epsilon\sigma A(T_1^4 - T_2^4) = 1.5k\frac{T_1-T_2}{L}\)

Solve for the effective emissivity epsilon.

Rearrange the equation to solve for \(\epsilon\). \(\epsilon = \frac{1.5k(T_1-T_2)}{\sigma(T_1^4 - T_2^4)L}\) Given values: - \(T_1 = 278\) K - \(T_2 = 288\) K - \(L = 0.015\) m (1.5 cm in meters) - \(\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}\) (Stefan-Boltzmann constant) - \(k = 0.026 W m^{-1} K^{-1}\) (thermal conductivity of air at normal room temperature and pressure) Now plug in the given values into the expression for \(\epsilon\). \(\epsilon = \frac{1.5 \times 0.026(288 - 278)}{(5.67 \times 10^{-8})(288^4 - 278^4)(0.015)}\) Calculate the value of \(\epsilon\): \(\epsilon \approx 0.53\) The correct option is (b) [0.53].

Key concepts

Thermal Conductivity

Thermal conductivity is a property of materials that indicates their ability to conduct heat. It is often denoted by the symbol \( k \). In the context of our double-pane window problem, the thermal conductivity refers to the measure of how well the air trapped between the two panes of glass can transport heat from the hot side to the cold side. The greater the thermal conductivity, the more efficient the material is at transferring heat.

Mathematically, the rate of heat transfer due to conduction through a material can be expressed by Fourier’s law: \( q_{cond} = k\frac{T_1-T_2}{L} \), where \( T_1 \) and \( T_2 \) are the temperatures on either side of the material and \( L \) is the thickness of the material separating them. In our scenario, air's thermal conductivity plays a vital role in determining the overall heat transfer by conduction through the air gap.

Convective Heat Transfer

Convective heat transfer is the mode of heat transfer that occurs when a fluid (liquid or gas) moves past a solid boundary. This movement can either be natural, driven by the buoyancy effects due to temperature differences, or forced, due to mechanical means like a pump or fan. Convective heat transfer is crucial for understanding how heat is transferred within the air gap of the double-pane window.

The rate of heat transfer by convection can be calculated using the formula: \( q_{conv} = h_{conv}A(T_1-T_2) \), where \( h_{conv} \) represents the convective heat transfer coefficient, \( A \) is the surface area in contact with the fluid, and \( T_1-T_2 \) symbolizes the temperature differential driving the heat transfer. In cases where the convection is enhanced by some intrinsic movement within the fluid, heat is transferred more efficiently, thus the convective heat transfer coefficient is higher.

Radiation Heat Transfer

Radiation heat transfer is the energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules. This type of heat transfer does not require a medium, which means it can occur in a vacuum. For our window, thermal radiation plays a role in transferring heat across the air gap without directly heating the air itself.

The heat transfer due to radiation is described by the Stefan-Boltzmann law, given by the equation: \( q_{rad} = \epsilon\sigma A(T_1^4 - T_2^4) \), where \( \epsilon \) is the emissivity of the material surfaces, \( \sigma \) is the Stefan-Boltzmann constant, and \( T_1^4 - T_2^4 \) is the difference in temperatures raised to the fourth power. Emissivity is a measure of how effectively a surface emits thermal radiation, and it is dimensionless, with a value between 0 and 1. A perfect black body, which is an idealized physical body that absorbs all incident electromagnetic radiation, has an emissivity of 1.

Effective Emissivity

Effective emissivity is an essential concept when multiple surfaces are involved in the radiation heat transfer process, as is the case in the double-pane window problem. It accounts for the fact that the emissivity of a system involving several surfaces can differ from that of the individual surfaces due to reflections and interactions between the surfaces.

The effective emissivity is a combined measure that simplifies the overall emissive properties of the two glass surfaces. It can be lower than the individual emissivities of the panes due to the reflections of radiative heat between the panes, which can lead to some of the heat being absorbed or re-emitted.

In solving our exercise, to find the effective emissivity, we compare the rate of heat transfer by convection to that by radiation, considering that they are approximately equal and both contribute to the overall heat transfer. This comparison allows us to solve for the effective emissivity, which ultimately helps us to understand how efficient the glass panes are in their role as barriers to not just conductive or convective heat transfer, but radiative as well.