Question

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

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

Step-by-step solution

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 (α) 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 (α): $$α = \frac{A}{I}$$

Calculate Reflectivity

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

Calculate Transmissivity

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

Calculate Emissivity

Based on Kirchhoff's law of thermal radiation, for an opaque medium, absorptivity (α) equals emissivity (ε). 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. $$ε = α$$

Determine Values of Properties

Now we have all equations needed to find the values of the 4 properties. Calculate them as follows: $$A = I - (R+T) = 520 - (160 + 130) = 230 \mathrm{W/m^2}$$ $$α = \frac{A}{I} = \frac{230}{520} = 0.442$$ $$ρ = \frac{R}{I} = \frac{160}{520} = 0.308$$ $$τ = \frac{T}{I} = \frac{130}{520} = 0.25$$ $$ε = α = 0.442$$ 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).

Key concepts

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/m²). For example, in the exercise, the surface of the medium is being impacted by irradiation at a rate of 520 W/m². 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 (α), is a property that measures the fraction of incident irradiation absorbed by a material. The formula to calculate absorptivity is:\[α = \frac{A}{I}\]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/m². 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 \((ρ)\), 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:\[ρ = \frac{R}{I}\]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 \((ε)\) is equal to its absorptivity \((α)\). 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 \((ε)\) as equivalent to absorptivity \((α)\). 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.