Question

Consider a gray and opaque surface at \(0^{\circ} \mathrm{C}\) in an environment at \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(0.8\). If the radiation incident on the surface is \(240 \mathrm{~W} / \mathrm{m}^{2}\), the radiosity of the surface is (a) \(38 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(132 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(240 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(300 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(315 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

Answer: The radiosity of the surface is approximately 362.724 W/m².

Step-by-step solution

Find the emitted energy by the surface considering its temperature and emissivity

To find the emitted energy, we'll use the Stefan-Boltzmann Law. The emitted energy by the surface can be given as \(E_{emit} = εσT_{surface}^{4}\), where ε is the emissivity, σ is the Stefan-Boltzmann constant (approximately \(5.67 x 10^-8 W/m^2K^4\)), and \(T_{surface}\) is the surface temperature in Kelvin. The surface temperature is given as \(0° \mathrm{C}\), which is equivalent to \(273 \mathrm{K}\) in Kelvin. Now, calculate the emitted energy: \(E_{emit} = (0.8)(5.67 x 10^{-8} \mathrm{W/m^2K^4})(273 \mathrm{K})^4 = 314.724 \mathrm{~W} / \mathrm{m}^{2}\)

Calculate the reflected energy from the incident radiation

To find the reflected energy, we need to consider the surface's reflectivity. Reflectivity is equal to (1- ε), where ε is the emissivity of the surface. Reflectivity = 1- ε = 1 - 0.8 = 0.2 Now, multiply the reflectivity by the incident radiation to find the reflected energy: \(E_{reflected} = (0.2)(240 \mathrm{~W} / \mathrm{m}^{2}) = 48 \mathrm{~W} / \mathrm{m}^{2}\)

Calculate the radiosity of the surface by adding the emitted and reflected energy

To find the radiosity, simply add the emitted energy and the reflected energy that we calculated in steps 1 and 2, respectively. Radiosity = \(E_{emit} + E_{reflected} = 314.724 \mathrm{~W} / \mathrm{m}^{2} + 48 \mathrm{~W} / \mathrm{m}^{2} = 362.724 \mathrm{~W} / \mathrm{m}^{2}\) However, none of the given options matches the calculated value, so it is necessary to verify the data provided in the exercise or possible errors during the resolution process.

Key concepts

Understanding the Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a cornerstone in the study of thermal radiation, which is the emission of energy from the surface of an object due to its temperature. This law states that the energy emitted by a blackbody per unit area is directly proportional to the fourth power of the body's absolute temperature. Mathematically, it is expressed as \
\( E_{emit} = \$\sigma T^{4} \$\), where \( E_{emit} \) is the emitted energy per unit area, \( \sigma \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} \mathrm{W/m^2K^4} \)), and \( T \) is the absolute temperature in Kelvin.

In practical scenarios, surfaces are often not ideal blackbodies, which is where emissivity, denoted by \( \epsilon \), becomes crucial. For non-ideal surfaces (or real-world objects), the emitted energy is scaled by their emissivity, as \( E_{emit} = \epsilon\sigma T^{4} \). Emissivity values range from 0 to 1, with 1 representing a perfect blackbody.

• To apply this law, you must know the temperature and emissivity of the surface.
• Ensure all temperatures are converted to Kelvin.
• Use precise values for emissivity and the Stefan-Boltzmann constant.

Emissivity and Its Role in Thermal Radiation

Emissivity, represented by the Greek letter \( \epsilon \), is a measure of a material's ability to emit thermal radiation compared to that of a perfect blackbody. A blackbody, which is an idealized physical body, absorbs all incident electromagnetic radiation and re-emits energy with a characteristic spectrum dependent on temperature. Real materials emit less radiation than a blackbody and hence have emissivities less than 1.

The emissivity of a surface affects how it exchanges thermal radiation with its environment, including radiation that is absorbed, reflected, and emitted. When calculations are made for heat transfer or radiation based on the Stefan-Boltzmann Law, the emissivity value must be factored in to predict accurate results.

• The concept of emissivity is central when relating a real object's radiation to the idealized Stefan-Boltzmann Law.
• When emissivity is accounted for, the formula becomes \( E_{emit} = \epsilon\sigma T^{4} \), making it applicable to real-world materials.
• Materials with high emissivity are effective radiators, while materials with low emissivity reflect more than they emit.

Thermal Radiation and How It Applies to Surfaces

Thermal radiation is the process by which energy is released by the movements of charged particles within atoms in the form of electromagnetic waves. All bodies at a temperature above absolute zero emit thermal radiation. The Stefan-Boltzmann Law quantifies the emitted thermal radiation for ideal blackbodies, and emissivity adjusts this for real materials.

However, the thermal radiation that a surface exchanges with its surroundings also consists of reflected and absorbed radiation. The term 'radiosity' describes the total amount of radiation leaving a surface, which includes both the emitted and reflected portions. To calculate it effectively, one must combine the effects of emission and reflection according to the material's properties: \( Radiosity = E_{emit} + E_{reflected} \).

• To understand heat transfers and thermal effects, it's vital to address all forms of radiative exchange.
• The sum of emitted and reflected radiation provides us with a comprehensive understanding of a surface's thermal interaction with its environment.
• An accurate assessment of radiosity is necessary for thermal management in a wide range of applications, from building insulation to spacecraft design.