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A radiation detector has an aperture of area \(A_{d}=10^{-6}\) \(\mathrm{m}^{2}\) and is positioned at a distance of \(r=1 \mathrm{~m}\) from a surface of area \(A_{3}=10^{-4} \mathrm{~m}^{7}\). The angle formed by the normal to the detector and the surface normal is \(\theta=30^{\circ}\). The surface is at \(500 \mathrm{~K}\) and is opaquc, diffuse, and gray with an emissivity of \(0.7\). If the surface irtadiation is \(1500 \mathrm{~W} / \mathrm{m}^{2}\), what is the rate at which the detector intercepts radiation from the surface?

Short Answer

Expert verified
The detector intercepts radiation at a rate of approximately \(1.33 \times 10^{-8} \mathrm{~W}\).

Step by step solution

01

Understand the Given Parameters

The exercise provides several key parameters: the area of the detector aperture, \(A_d = 10^{-6} \) \( \mathrm{m}^2\), the distance from the surface, \(r = 1 \mathrm{~m}\), the area of the surface, \(A_3 = 10^{-4} \mathrm{~m}^2\), the angle \( \theta = 30^\circ \), surface temperature \(500 \mathrm{~K}\), emissivity \(\varepsilon = 0.7\), and the surface irradiation \(1500 \mathrm{~W} / \mathrm{m}^2\).
02

Calculate the Angle Factor \( F_{d3} \)

First, calculate the angle factor. Since the detector and the surface form an angle \( \theta \), we use the relation: \[ F_{d3} = \frac{\cos\theta}{\pi r^2} A_d \] Substitute \( \theta = 30^\circ \), \( A_d = 10^{-6} \mathrm{~m}^2\), and \( r = 1 \mathrm{~m} \): \[ F_{d3} = \frac{\cos 30^\circ}{\pi (1)^2} \times 10^{-6} \approx \frac{0.866}{3.14159} \times 10^{-6} \approx 2.755 \times 10^{-7} \].
03

Calculate the Blackbody Emission from the Surface

Calculate the blackbody emissive power, \( E_b \), using Stefan-Boltzmann's law:\[ E_b = \sigma T^4 \] where \( \sigma = 5.67 \times 10^{-8} \mathrm{~W/m}^2\cdot \mathrm{K}^4 \) and \( T = 500 \mathrm{~K} \):\[ E_b = 5.67 \times 10^{-8} \times (500)^4 \approx 2830 \mathrm{~W/m}^2 \].
04

Calculate the Emitted Power from Surface

The actual emitted power per unit area from the surface, \( E \), considering emissivity \( \varepsilon \):\[ E = \varepsilon E_b = 0.7 \times 2830 \approx 1981 \mathrm{~W/m}^2 \].
05

Determine Net Radiation Flux from Surface

The net radiation flux \( q_{rad} \) received by the detector from the surface is the difference between emitted power \( E \) and the given irradiation of the surface. Calculate it as:\[ q_{rad} = E - G_3 = 1981 - 1500 \approx 481 \mathrm{~W/m}^2 \].
06

Calculate Radiation Intercepted by the Detector

Finally, the rate at which the radiation detector intercepts radiation from the surface is:\[ Q_{d} = F_{d3} A_3 \times q_{rad} \]Substitute \( F_{d3} \approx 2.755 \times 10^{-7} \), \( A_3 = 10^{-4} \mathrm{~m}^2 \), and \( q_{rad} \approx 481 \mathrm{~W/m}^2 \):\[ Q_{d} = 2.755 \times 10^{-7} \times 10^{-4} \times 481 \approx 1.33 \times 10^{-8} \mathrm{~W} \].

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Key Concepts

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

Emissivity
Emissivity is a measure of an object's ability to emit thermal radiation compared to a perfect blackbody. It is a dimensionless quantity ranging from 0 to 1. A perfect blackbody has an emissivity of 1, meaning it emits the maximum possible radiation at a given temperature. In the context of the problem, the emissivity is crucial because it directly influences how much radiation the surface actually emits.
A real surface will always have an emissivity less than 1, which means it does not emit as much thermal radiation as a perfect blackbody. The emissivity affects the calculation of emitted power using the formula: \[ E = ext{emissivity} imes E_b \] where \(E_b\) is the blackbody emissive power, calculated using the Stefan-Boltzmann law. Understanding the role of emissivity helps in accurately calculating the thermal radiation leaving the surface.
Stefan-Boltzmann Law
The Stefan-Boltzmann law states that the power emitted by a blackbody per unit area is proportional to the fourth power of its absolute temperature. This law is represented by the equation: \[ E_b = ext{Stefan-Boltzmann constant} imes T^4 \] where the Stefan-Boltzmann constant \( \sigma \) is approximately \( 5.67 \times 10^{-8} \mathrm{~W/m}^2\cdot \mathrm{K}^4 \), and \( T \) is the temperature in Kelvin.
This law is fundamental for determining the total power radiated by an object due to its temperature. In applied scenarios, like the given exercise, we often consider emissivity to adjust for real materials that are not perfect blackbodies. Therefore, multiplying \( E_b \) with emissivity gives us a more accurate measure of the actual emitted radiation.
Understanding this law is vital for solving heat transfer problems involving radiation.
Radiation Detector
A radiation detector is a device that captures and measures radiation energy emanating from a source. In the exercise, the detector is characterized by its aperture area, which determines how much radiation can potentially be collected. The detector's ability to measure radiation accurately depends on several factors such as distance from the source and the angle at which it is positioned.
The detector's position relative to the source surface, described by the distance and angle, directly impacts the calculation of the rate of intercepted radiation. Only the portion of radiation that passes through the detector's aperture and falls within its field of view can be measured.
Thus, understanding how detectors work helps in setting them correctly in practical applications where detecting radiation accurately and efficiently is needed.
Angle Factor
The angle factor, also known as the view factor or configuration factor, represents how much of the radiation emitted by one surface reaches another surface. It is affected by the geometrical arrangement and orientation between both surfaces. In the equation used in the step-by-step solution: \[ F_{d3} = \frac{\cos\theta}{\pi r^2} A_d \] Here, \( \theta \) is the angle between the normal of the surface emitting radiation and the detector, \( r \) is the distance between the two surfaces, and \( A_d \) is the detector's aperture area.
The cosine of the angle (cosine law) adjusts the view factor to account for the actual effective area of the surface as seen from the detector.
Understanding angle factors is crucial for determining the net radiation exchange between surfaces since they encompass the geometrical dependency of radiation heat transfer.

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Most popular questions from this chapter

A diffuse, opoque surface at \(700 \mathrm{~K}\) has spectral emissivities of \(\varepsilon_{\mathrm{i}}=0\) for \(0 \leq \lambda \leq 3 \mu \mathrm{m}\), \(\varepsilon_{\mathrm{a}}=0.5\) for \(3 \mu \mathrm{m}<\lambda \leq 10 \mu \mathrm{m}\), and \(\varepsilon_{\mathrm{x}}=0.9\) for \(10 \mu \mathrm{m}<\lambda<\infty\). A radiant flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), which is uniformly distributed between 1 and \(6 \mu \mathrm{m}\), is incident on the surface at an angle of \(30^{\circ}\) relative to the surface normal. Calculate the total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) arca of the surface that reaches a radiation detector pusitioned along the normal to the area. The aperture of the detector is \(10^{-5} \mathrm{~m}^{2}\), and its distance from the surface is \(1 \mathrm{~m}\).

A roof-cooling system, which operates by maintaining a thin film of water on the roof surface, may be used to reduce air-conditioning costs or to maintain a cooler environment in nonconditioned buildings. To determine the effectiveness of such a system, consider a sheet metal roof for which the solar absorptivity \(\alpha_{s}\) is \(0.50\) and the hemispherical emissivity \(\varepsilon\) is \(0.3\). Representative conditions correspond to a surface convection coefficient \(h\) of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a solar irradiation \(G_{S}\) of \(700 \mathrm{~W} / \mathrm{m}^{2}\), a sky temperature of \(-10^{\circ} \mathrm{C}\), an utmospheric temperature of \(30^{\circ} \mathrm{C}\), and a relutive humidity of \(65 \%\). The roof may be assumed to be well insulated from below. Determine the roof surface temperature without the water film. Assuming the film and roof surface temperatures to be cqual, determine the surface temperature with the film. The solar absorptivity and the hemispherical emissivity of the film-surface combination are \(\alpha_{s}=0.8\) and \(\varepsilon=0.9\), respectively.

An opaque surface, \(2 \mathrm{~m}\) by \(2 \mathrm{~m}\), is maintained at \(400 \mathrm{~K}\) and is simultaneously exponed to solar irradiation with \(G=1200 \mathrm{~W} / \mathrm{m}^{2}\). The surface is diffuse and its spectral absorptivity is \(\alpha_{n}=0,0.8,0\), and \(0.9\) for \(0 \leq \lambda \leq 0.5 \mu m, 0.5 \mu m<\lambda \leq 1 \mu m .1 \mu m<\lambda \leq\) \(2 \mu \mathrm{m}\), and \(\lambda>2 \mu \mathrm{m}\), respectively. Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface.

A radiator on a proposed satellite solar power station must dissipate beat being generated within the satellite by madiating it into space. The radiator surface has a solar absorptivity of \(0.5\) and an emissivity of \(0.95\). What is the equilibrium surface temperature when the solar irradiation is \(1000 \mathrm{~W} / \mathrm{m}^{2}\) and the required heat dissipation is 1500 W/m \({ }^{2}\) ?

A horizontal, opaque surface at a steady-sate temperature of \(77^{\circ} \mathrm{C}\) is exposed in an airflow having a free stream temperature of \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissive power of the surface is \(628 \mathrm{~W} / \mathrm{m}^{2}\), the irradiation is \(1380 \mathrm{~W} / \mathrm{m}^{2}\), and the reflectivity is \(0.40\). Determine the absorptivity of the surface. Determine the act ractiation heat transfer rate for this surface. Is this heat transfer to the wurface or from the surface? Determine the combined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

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