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Chapter 12: Problem 83

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 (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): $$α = \frac{A}{I}$$
02

Calculate Reflectivity

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

Calculate Transmissivity

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

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. $$ε = α$$
05

Determine Values of Properties

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

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Most popular questions from this chapter

The reflectivity of aluminum coated with lead sulfate is \(0.35\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and \(0.95\) for radiation greater than \(3 \mu \mathrm{m}\). Determine the average reflectivity of this surface for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors?

Solar radiation is incident on a semi-transparent body at a rate of \(500 \mathrm{~W} / \mathrm{m}^{2}\). If \(150 \mathrm{~W} / \mathrm{m}^{2}\) of this incident radiation is reflected back and \(225 \mathrm{~W} / \mathrm{m}^{2}\) is transmitted across the body, the absorptivity of the body is (a) 0 (b) \(0.25\) (c) \(0.30\) (d) \(0.45\) (e) 1

Explain why surfaces usually have quite different absorptivities for solar radiation and for radiation originating from the surrounding bodies.

The human skin is "selective" when it comes to the absorption of the solar radiation that strikes it perpendicularly. The skin absorbs only 50 percent of the incident radiation with wavelengths between \(\lambda_{1}=0.517 \mu \mathrm{m}\) and \(\lambda_{2}=1.552 \mu \mathrm{m}\). The radiation with wavelengths shorter than \(\lambda_{1}\) and longer than \(\lambda_{2}\) is fully absorbed. The solar surface may be modeled as a blackbody with effective surface temperature of \(5800 \mathrm{~K}\). Calculate the fraction of the incident solar radiation that is absorbed by the human skin.

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.
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