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Irradiation on a semi-transparent medium is at a rate of \(520 \mathrm{~W} / \mathrm{m}^{2}\). If \(160 \mathrm{~W} / \mathrm{m}^{2}\) of the irradiation is reflected from the medium and \(130 \mathrm{~W} / \mathrm{m}^{2}\) is transmitted through the medium, determine the medium's absorptivity, reflectivity, transmissivity, and emissivity.

Short Answer

Expert verified
Based on the given information, calculate the absorptivity, reflectivity, transmissivity, and emissivity of the semi-transparent medium.

Step by step solution

01

Calculate Absorptivity

To find the medium's absorptivity, we need to calculate the difference between incident irradiation which is absorbed by the medium and the irradiation that is reflected and transmitted. Absorptivity (伪) could be determined by dividing the absorbed irradiation by the incident irradiation. Incident irradiation: \(I = 520 \mathrm{W/m^2}\) Reflected irradiation: \(R = 160 \mathrm{W/m^2}\) Transmitted irradiation: \(T = 130 \mathrm{W/m^2}\) First, find the absorbed irradiation: $$A = I - (R+T)$$ Now, divide the absorbed irradiation by the incident irradiation to find the absorptivity (伪): $$伪 = \frac{A}{I}$$
02

Calculate Reflectivity

Reflectivity (蟻) is simply the ratio of the reflected irradiation to the incident irradiation, which is: $$蟻 = \frac{R}{I}$$
03

Calculate Transmissivity

Transmissivity (蟿) is similarly the ratio of the transmitted irradiation to the incident irradiation: $$蟿 = \frac{T}{I}$$
04

Calculate Emissivity

Based on Kirchhoff's law of thermal radiation, for an opaque medium, absorptivity (伪) equals emissivity (蔚). However, here we have a semi-transparent medium. Since the problem doesn't provide any additional information for the emissivity calculation, let's assume it as an opaque medium for the sake of the exercise and use Kirchhoff's law to find emissivity. $$蔚 = 伪$$
05

Determine Values of Properties

Now we have all equations needed to find the values of the 4 properties. Calculate them as follows: $$A = I - (R+T) = 520 - (160 + 130) = 230 \mathrm{W/m^2}$$ $$伪 = \frac{A}{I} = \frac{230}{520} = 0.442$$ $$蟻 = \frac{R}{I} = \frac{160}{520} = 0.308$$ $$蟿 = \frac{T}{I} = \frac{130}{520} = 0.25$$ $$蔚 = 伪 = 0.442$$ So, the semi-transparent medium has an absorptivity of 0.442, reflectivity of 0.308, transmissivity of 0.25, and an emissivity of 0.442 (assuming an opaque medium).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Irradiation
Irradiation refers to the amount of thermal energy that hits a surface per unit area. It's often measured in watts per square meter (W/m虏). For example, in the exercise, the surface of the medium is being impacted by irradiation at a rate of 520 W/m虏. Understanding irradiation is crucial because it forms the basis on which we measure how much energy a medium receives from external sources. In many scientific and engineering contexts, irradiation values help determine the potential energy available for absorption, reflection, or transmission. It's a key component in calculating thermal energy flow and plays a significant role in assessing energy efficiency in various materials. These insights are critical when analyzing the radiative properties of materials, especially in fields such as thermodynamics and thermal engineering.
Absorptivity
Absorptivity, denoted by the Greek letter alpha (伪), is a property that measures the fraction of incident irradiation absorbed by a material. The formula to calculate absorptivity is:\[伪 = \frac{A}{I}\]where \(A\) represents the absorbed irradiation and \(I\) is the total incident irradiation. In our exercise, we had to first determine the absorbed irradiation, which was calculated to be 230 W/m虏. Thus, by substituting into the formula, the absorptivity was found to be 0.442. Absorptivity varies depending on the material's nature and surface characteristics. Understanding absorptivity helps in determining how much of the thermal energy is stored within a medium, an important factor in designing systems for thermal insulation and solar energy harnessing.
Reflectivity
Reflectivity, symbolized by rho \((蟻)\), is a property that quantifies how much incident irradiation is reflected off a surface. Reflectivity is expressed as a ratio of the reflected power to the incident power. The formula is:\[蟻 = \frac{R}{I}\]where \(R\) is the reflected irradiation and \(I\) is the total incident irradiation. In the given exercise, we calculated reflectivity as 0.308, meaning 30.8% of the incident irradiation was reflected. Reflectivity depends on factors such as the surface texture, color, and angle of incident irradiation. It is a crucial parameter for products and structures which are exposed to solar radiation, as it affects their temperature regulation. Low reflectivity indicates that more energy is either absorbed or transmitted, which can lead to increased heat retention in the material.
Kirchhoff's Law
Kirchhoff's Law of thermal radiation establishes a relationship between absorptivity and emissivity. According to the law, for a body in thermal equilibrium, its emissivity \((蔚)\) is equal to its absorptivity \((伪)\). This principle assumes that the material is opaque, meaning no light is transmitted through. In the case of semi-transparent materials, the direct application can be complicated as they transmit part of the incident irradiation. In the exercise, it was assumed for simplicity that the medium behaves like an opaque body, thus treating emissivity \((蔚)\) as equivalent to absorptivity \((伪)\). For engineering applications, Kirchhoff's Law provides an essential tool for predicting the thermal behavior of materials. Ensuring materials have optimal absorptivity and emissivity characteristics help in creating efficient thermal systems in industries ranging from manufacturing to building design.

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Most popular questions from this chapter

Solar radiation is incident on the front surface of a thin plate with direct and diffuse components of 300 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. The direct radiation makes a \(30^{\circ}\) angle with the normal of the surface. The plate surfaces have a solar absorptivity of \(0.63\) and an emissivity of \(0.93\). The air temperature is \(5^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The effective sky temperature for the front surface is \(-33^{\circ} \mathrm{C}\) while the surrounding surfaces are at \(5^{\circ} \mathrm{C}\) for the back surface. Determine the equilibrium temperature of the plate.

A semi-transparent plate \(\left(A_{1}=2 \mathrm{~cm}^{2}\right)\) has an irradiation of \(500 \mathrm{~W} / \mathrm{m}^{2}\), where \(30 \%\) of the irradiation is reflected away from the plate and \(50 \%\) of the irradiation is transmitted through the plate. A radiometer is placed \(0.5 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. If the temperature of the plate is uniform at \(350 \mathrm{~K}\), determine the irradiation that the radiometer would detect.

What changes would you notice if the sun emitted radiation at an effective temperature of \(2000 \mathrm{~K}\) instead of \(5762 \mathrm{~K}\) ?

What is the greenhouse effect? Why is it a matter of great concern among atmospheric scientists?

Describe the solar radiation properties of a window that is ideally suited for minimizing the air-conditioning load.

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