Question

A horizontal opaque flat plate is well insulated on the edges and the lower surface. The top surface has an area of \(5 \mathrm{~m}^{2}\), and it experiences uniform irradiation at a rate of \(5000 \mathrm{~W}\). The plate absorbs \(4000 \mathrm{~W}\) of the irradiation, and the surface is losing heat at a rate of \(500 \mathrm{~W}\) by convection. If the plate maintains a uniform temperature of \(350 \mathrm{~K}\), determine the absorptivity, reflectivity, and emissivity of the plate.

Short answer

Question: Determine the plate's absorptivity, reflectivity, and emissivity. Answer: The absorptivity of the plate is 0.8, its reflectivity is 0.2, and its emissivity is approximately 0.89.

Step-by-step solution

Calculate the total energy balance of the plate

First, let's calculate the total energy balance of the plate. The plate receives irradiation, absorbs some of that energy, and loses energy by convection. The remaining energy should be equal to the energy emitted by the plate as radiation. Given that the plate absorbs \(4000 \mathrm{~W}\) of the irradiation and loses \(500 \mathrm{~W}\) by convection, we can calculate the energy emitted by the plate as: Energy emitted = \(4000 \mathrm{~W}\) (absorbed) - \(500 \mathrm{~W}\) (lost by convection) = \(3500 \mathrm{~W}\)

Determine the plate's absorptivity

Absorptivity is defined as the fraction of incident radiation that is absorbed by a surface. We know that the plate absorbs \(4000 \mathrm{~W}\) of the irradiation and the total irradiation is \(5000 \mathrm{~W}\). Therefore, the absorptivity \((\alpha)\) can be calculated as: Absorptivity \((\alpha)\) = \(\frac{\text{absorbed energy}}{\text{incident energy}}\) \(\alpha\) = \(\frac{4000 \mathrm{~W}}{5000 \mathrm{~W}}\) = \(0.8\)

Determine the plate's reflectivity

Reflectivity \((\rho)\) is defined as the fraction of incident radiation that is reflected by a surface. Since an opaque surface either absorbs or reflects all the incident radiation, and we have already calculated absorptivity, we can simply subtract absorptivity from 1 to find the reflectivity: Reflectivity \((\rho)\) = \(1 - \alpha\) \(\rho\) = \(1 - 0.8\) = \(0.2\)

Determine the plate's emissivity

Emissivity \((\varepsilon)\) is the effectiveness of a surface in emitting energy as thermal radiation and can be found using the Stefan-Boltzmann law: Emitted energy = \(\varepsilon \sigma A T^4\) Where \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature in Kelvin. We know the emitted energy is \(3500 \mathrm{~W}\), the surface area is \(5 \mathrm{~m}^{2}\), and the temperature is \(350 \mathrm{~K}\). The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}\). Rearrange the Stefan-Boltzmann law to solve for emissivity: \(\varepsilon\) = \(\frac{\text{emitted energy}}{\sigma A T^4}\) \(\varepsilon\) = \(\frac{3500 \mathrm{~W}}{(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}) (5 \mathrm{~m}^2) (350 \mathrm{~K})^4}\) \(\varepsilon \approx 0.89\) Thus, the absorptivity of the plate is \(0.8\), its reflectivity is \(0.2\), and its emissivity is approximately \(0.89\).

Key concepts

Absorptivity

Absorptivity is a measure of how much of the incident radiation is absorbed by a surface. It is an important concept in thermal radiation, as it determines how much of the incoming energy is retained by the object. Absorptivity is denoted by the symbol \(\alpha\) and is expressed as a fraction or a percentage.

• An absorptivity of \(0\) means that the surface absorbs none of the incoming radiation.
• An absorptivity of \(1\) means that the surface absorbs all the incoming radiation completely, acting as a perfect absorber.

In the context of the given problem, the absorptivity is calculated by taking the ratio of the absorbed energy to the incident energy. Given that the plate absorbs \(4000 \mathrm{~W}\) out of the total \(5000 \mathrm{~W}\) of irradiation, its absorptivity \((\alpha)\) is given by:\[\alpha = \frac{4000 \mathrm{~W}}{5000 \mathrm{~W}} = 0.8\]This means the plate absorbs 80% of the radiation it receives.

Reflectivity

Reflectivity is the fraction of incident radiation that is reflected by a surface and is denoted by the symbol \(\rho\). It deals with the portion of radiation that does not get absorbed but instead bounces off the surface. Reflectivity is crucial in applications where minimizing heat gain is desirable, as in reflective coatings and thermal insulation.

• A reflectivity of \(0\) implies that all radiation is absorbed with none being reflected.
• A reflectivity of \(1\) signifies that all radiation is reflected, equivalent to a perfect mirror.

For an opaque object, the sum of absorptivity and reflectivity should equal 1, as the object neither transmits energy through it nor allows it to pass. In the exercise, having calculated an absorptivity \((\alpha)\) of \(0.8\), the reflectivity \((\rho)\) can be found simply as:\[\rho = 1 - \alpha = 1 - 0.8 = 0.2\]This indicates that the plate reflects 20% of the incident radiation.

Emissivity

Emissivity, symbolized as \(\varepsilon\), indicates the efficiency with which a surface emits thermal radiation. It is a critical factor in thermal analysis, particularly for materials that release heat via radiation. Emissivity is especially relevant in cooling technologies and the thermal design of spacecraft.

• An emissivity of \(0\) means that the surface emits no thermal radiation.
• An emissivity of \(1\) indicates a perfect emitter, often referred to as a "blackbody."

To evaluate emissivity, the Stefan-Boltzmann law is used, which provides the relationship between emitted energy, area, temperature, and emissivity. In this scenario, the emitted energy is \(3500 \mathrm{~W}\), and we know the area \(A\), temperature \(T\), and Stefan-Boltzmann constant \(\sigma\). Plugging in values:\[\varepsilon = \frac{3500 \mathrm{~W}}{(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}) \times (5 \mathrm{~m}^2) \times (350 \mathrm{~K})^4}\]Solving this gives an emissivity \(\varepsilon \approx 0.89\), demonstrating the plate's efficiency in emitting energy as thermal radiation. This means the plate emits about 89% of the energy it is capable of emitting as if it were a perfect blackbody at the same temperature.