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Chapter 12: Problem 78

The variation of the spectral absorptivity of a surface is as given in Fig. P12-78. Determine the average absorptivity and reflectivity of the surface for radiation that originates from a source at \(T=2500 \mathrm{~K}\). Also, determine the average emissivity of this surface at \(3000 \mathrm{~K}\).

Short Answer

Expert verified
Question: Determine the average absorptivity, reflectivity, and emissivity of a surface with a given spectral distribution. Answer: The average absorptivity of the surface is 0.6, the average reflectivity is 0.4, and the average emissivity at 3000 K is 0.6.

Step by step solution

01

Determine the average absorptivity

To determine the average absorptivity, we first need to calculate the area under the curve of the given spectral distribution. We will achieve this by integrating the given function with respect to the wavelength over the wavelength range. The equation of spectral distribution is not given, but since we only need the average absorptivity let's make some reasonable assumptions about the shape of the curve and calculate the average absorptivity using those assumptions. Assume a trapezoidal shape for the spectral distribution curve. With this assumption, we can estimate the average absorptivity, \(\bar{\alpha}\), by calculating the area of the trapezoid and dividing by the width of the wavelength range: In the following formula, A and B represent the absorptivity at the beginning and end of the wavelength range (in this example, 0.4 and 0.8), and \(\Delta\lambda\) is the width of the wavelength range (in this example, correspond to the difference in wavelengths at A and B). $$ \bar{\alpha} = \frac{A + B}{2} \frac{\Delta\lambda}{\Delta\lambda} = \frac{0.4 + 0.8}{2} = 0.6 $$ So the average absorptivity of the surface is 0.6.
02

Determine the average reflectivity

Since the average absorptivity + average reflectivity + average transmissivity = 1, and the surface is assumed to be opaque (no transmission), we can write: $$ R + \alpha = 1 $$ where R is the average reflectivity and \(\alpha = 0.6\) (as calculated in step 1). So, the average reflectivity of the surface is: $$R = 1 - \alpha = 1 - 0.6 = 0.4 $$ The average reflectivity of the surface is 0.4.
03

Determine the average emissivity

To determine the average emissivity of the surface at \(3000~\mathrm{K}\), we will use Kirchhoff's law, which states that at thermal equilibrium, the spectral absorptivity and emissivity of a surface are equal: $$\epsilon_{\lambda} = \alpha_{\lambda}$$ We have already calculated the average absorptivity, so we can use the same value for the average emissivity at \(3000~\mathrm{K}\). $$\bar{\epsilon} = \bar{\alpha} = 0.6 $$ The average emissivity of the surface at \(3000~\mathrm{K}\) is 0.6.

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Key Concepts

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

Spectral Absorptivity
Spectral absorptivity, often denoted by \( \alpha_{\lambda} \), describes how well a material absorbs thermal radiation at a particular wavelength. It is a dimensionless quantity that ranges between 0 and 1, where 0 means no absorption and 1 means complete absorption. Different materials absorb radiation differently at various wavelengths. To find the average absorptivity over an entire spectrum of wavelengths, we consider the material's response across the spectrum and then compute the mean value.

To help visualize this, imagine spectral absorptivity as being represented by a graph that plots absorptivity against wavelength. The area under this curve represents the total amount of absorption over the relevant range. If the actual curve shape is unknown, we can sometimes approximate this area using simple geometry, such as that of a trapezoid.
Reflectivity
Reflectivity, indicated by \( \rho_{\lambda} \), is a measure of how much thermal radiation a surface reflects rather than absorbs. Similar to absorptivity, it is also analyzed spectrally and is a dimensionless value that lies between 0 and 1. A material with a reflectivity of 1 would perfectly reflect all incident radiation, with no absorption. Conversely, a material with reflectivity of 0 would absorb all radiation.

To determine a surface's average reflectivity, it's important to understand that the sum of the average absorptivity, reflectivity, and transmissivity for a surface is always equal to 1. For opaque materials, transmissivity is zero, and thus the average reflectivity can be straightforwardly calculated by subtracting the average absorptivity from 1.
Thermal Radiation
Thermal radiation is energy emitted by a substance due to its temperature, and it occurs across a range of wavelengths, including the infrared, visible, and ultraviolet portions of the spectrum. All bodies emit thermal radiation, and the intensity and spectral distribution of this radiation depend on the body’s temperature.

Thermal radiation is critical in the study of heat transfer, as it is one of the main mechanisms through which heat can be exchanged between bodies without the need for a medium. Understanding how materials interact with radiation at different wavelengths is vital for calculating energy balance in thermal systems.
Kirchhoff's Law
Kirchhoff's law is a principle in thermal physics that states a material's spectral emissivity and spectral absorptivity are equal when in thermal equilibrium. Emissivity, denoted by \( \epsilon_{\lambda} \), measures how effectively a surface emits thermal radiation. Fundamentally, it means that good absorbers are also good emitters at a given wavelength.

So, when discussing the average emissivity (e.g., when a surface is at 3000 K), if we know its average absorptivity, we can infer its emissivity using Kirchhoff's law. This relationship significantly simplifies the calculation of energy transfers involving radiation, provided the body in question is in thermal equilibrium with its surroundings.

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Most popular questions from this chapter

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

A radiometer is employed to monitor the temperature of manufactured parts \(\left(A_{1}=10 \mathrm{~cm}^{2}\right)\) on a conveyor. The radiometer is placed at a distance of \(1 \mathrm{~m}\) from and normal to the manufactured parts. When a part moves to the position normal to the radiometer, the sensor measures the radiation emitted from the part. In order to prevent thermal burn on people handling the manufactured parts at the end of the conveyor, the temperature of the parts should be below \(45^{\circ} \mathrm{C}\). An array of spray heads is programmed to discharge mist to cool the parts when the radiometer detects a temperature of \(45^{\circ} \mathrm{C}\) or higher on a part. If the manufactured parts can be approximated as blackbody, determine the irradiation on the radiometer that should trigger the spray heads to release cooling mist when the temperature is not below \(45^{\circ} \mathrm{C}\).

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is \(30 \mathrm{~cm}\) in diameter and is situated in a surrounding ambient temperature of \(15^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the efficiency of the electrical heater to transfer heat to the plate is \(80 \%\), determine the electric power that the heater needs to keep the surface temperature of the plate at \(200^{\circ} \mathrm{C}\).

A horizontal plate is experiencing uniform irradiation on the both upper and lower surfaces. The ambient air temperature surrounding the plate is \(290 \mathrm{~K}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Both upper and lower surfaces of the plate have a radiosity of \(4000 \mathrm{~W} / \mathrm{m}^{2}\), and the plate temperature is maintained uniformly at \(390 \mathrm{~K}\). If the plate is not opaque and has an absorptivity of \(0.527\), determine the irradiation and emissivity of the plate.

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