Question

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and \(3200 \mathrm{~W} / \mathrm{m}^{2}\), respectively. If the temperature of the side surfaces is \(485 \mathrm{~K}\), the emissivity of the side surfaces is (a) \(0.37\) (b) \(0.55\) (c) \(0.63\) (d) \(0.80\) (e) \(0.89\)

Short answer

Answer: The emissivity of the side surfaces is approximately 0.63.

Step-by-step solution

Understand and write down the Stefan-Boltzmann law

The Stefan-Boltzmann law is given by the formula: \(R = \epsilon \sigma T^4 + (1 - \epsilon)R_{ref}\), where \(R\) is the radiosity, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}\)), \(T\) is the temperature, and \(R_{ref}\) is the radiosity of the reflected radiation.

Identify the unknowns and the knowns

We are given: - Temperature of the side surfaces: \(T_{side} = 485 \mathrm{~K}\) - Radiosity of the top surfaces: \(R_{top} = 7500 \mathrm{~W} / \mathrm{m}^{2}\) - Radiosity of the side surfaces: \(R_{side} = 3200 \mathrm{~W} / \mathrm{m}^{2}\) We need to find the emissivity of the side surfaces, \(\epsilon_{side}\).

Calculate the radiosity of the reflected radiation

The base surface (black) has a temperature of \(400 \mathrm{~K}\), so its radiosity is given by \(R_{base} = \sigma (400)^4 = 5.67 \times 10^{-8} \times (400)^4 \approx 57913.6 \mathrm{~W} / \mathrm{m}^{2}\). Assuming that all the radiation emitted by the base surface is reflected by the side surfaces, we have \(R_{ref} = R_{base}\).

Use the Stefan-Boltzmann law to find the emissivity of the side surfaces

Plug the knowns into the equation for the side surfaces: \(3200 = \epsilon_{side} \times (5.67 \times 10^{-8}) \times (485)^4 + (1 - \epsilon_{side}) \times 57913.6\). Now, solve for \(\epsilon_{side}\): \(\epsilon_{side} \approx 0.63\) The correct answer is (c) \(0.63\).

Key concepts

Emissivity Calculation

Understanding how to calculate emissivity is key to solving thermal radiation problems. Emissivity (\(\epsilon\)) is a measure of a surface's ability to emit thermal radiation compared to a perfect black body. It ranges from 0 to 1, with 1 indicating 100% efficiency in emitting radiation. To find emissivity, we use the Stefan-Boltzmann law equation, which factors in both emitted and reflected radiation.For example, with the side surfaces in our exercise:- Radiosity (\(R_{side} = 3200\; \mathrm{W/m}^2\))- Temperature (\(T_{side} = 485\; \mathrm{K}\))- Known Stefan-Boltzmann constant (\(\sigma = 5.67 \times 10^{-8} \mathrm{W}/\mathrm{m}^2\mathrm{K}^4\))The calculation involves finding the portion of radiosity attributed to thermal emission versus reflection, giving us the emissivity value. By setting up the Stefan-Boltzmann equation stepwise and replacing the variables with known values, we isolate and solve for \(\epsilon_{side}\).This method clearly showcases how the interactions of emission and reflection determine a surface's emissive properties.

Radiosity

Radiosity is an essential concept in understanding thermal radiation, especially for determining how surfaces interact in a thermal environment. Radiosity (\(R\)) refers to the total radiation leaving a surface, which includes both emitted and reflected radiation. To calculate radiosity, you need to consider both these factors. In the exercise:- The base surface emits its radiation entirely, as it is black, and contributes to the radiosity of the side surfaces.The side surfaces, however, combine this incoming reflected radiation with their thermal emission, leading to a lower radiosity of \(3200\; \mathrm{W/m}^2\) compared to the base. Understanding how emitted and reflected energies contribute to the total radiosity helps us in calculating other factors like emissivity and understanding energy balance in thermal systems.

Thermal Radiation

Thermal radiation is energy emitted by matter as electromagnetic waves due to its temperature. It is a fundamental method of heat transfer, alongside conduction and convection. Every object with a temperature above absolute zero emits thermal radiation, the intensity and wavelength of which depend on the object's temperature.Key points include:- **Emission**: Higher temperature equals more intense radiation, as seen with the hotter side surfaces in the furnace.- **Reflection**: Not all incident radiation is absorbed; some is reflected, influencing the radiosity calculations.- **Surface Interaction**: Different surfaces exchange thermal radiation based on their temperatures and emissivities. In the given exercise, understanding thermal radiation helps us grasp why the side and top surfaces have distinct radiosity values due to differences in temperature and \(\epsilon\), illustrating energy transfer within the furnace environment.