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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.

Short Answer

Expert verified
The surface absorbs 3840 W, emits 5848 W, has a radiosity of 6328 W, and loses 2488 W by net radiation.

Step by step solution

01

Calculate the Absorbed Irradiation

The absorbed irradiation can be calculated using the relation \( q_{abs} = G \cdot A \cdot \bar{\alpha} \), where \( \bar{\alpha} \) is the average absorptivity over the spectrum of solar irradiation. To find \( \bar{\alpha} \), find the solar spectrum distribution which predominantly lies between 0.5 to 1.0 \( \mu m \) making \( \alpha = 0.8 \). Then,\[ q_{abs} = 1200 \mathrm{~W/m^2} \times 4 \mathrm{~m^2} \times 0.8 = 3840 \mathrm{~W} \]
02

Calculate the Emissive Power

The emissive power is given by the Stefan-Boltzmann law, \( E = \epsilon \sigma T^4 \), where \( \epsilon \) is the total hemispherical emissivity. Assuming \( \epsilon = \alpha(\lambda > 2 \mu m) = 0.9 \) for thermal radiation from the surface at 400K. \[ E = 0.9 \times 5.67 \times 10^{-8} \mathrm{~W/m^2\cdot K^4} \times (400 \mathrm{~K})^4 = 1462 \mathrm{~W/m^2} \] For the total area, the emissive power is:\[ E_{total} = 1462 \mathrm{~W/m^2} \times 4 \mathrm{~m^2} = 5848 \mathrm{~W} \]
03

Determine Radiosity

Radiosity \( J \) is the sum of emitted and reflected radiation, \( J = E + \rho G \). Here \( \rho = 1 - \epsilon = 0.1 \). So,\[ J = 1462 \mathrm{~W/m^2} + 0.1 \times 1200 \mathrm{~W/m^2} = 1582 \mathrm{~W/m^2} \] Over the total area, radiosity is:\[ J_{total} = 1582 \mathrm{~W/m^2} \times 4 \mathrm{~m^2} = 6328 \mathrm{~W} \]
04

Calculate Net Radiation Heat Transfer

The net radiation heat transfer \( q_{net} \) is the difference between absorbed irradiation and radiosity.\[ q_{net} = q_{abs} - J_{total} = 3840 \mathrm{~W} - 6328 \mathrm{~W} = -2488 \mathrm{~W} \]A negative value indicates that the surface loses heat through net radiation.

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Key Concepts

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

Absorbed Irradiation
Absorbed irradiation refers to the portion of incoming radiant energy that an object takes in, rather than reflecting or transmitting. In the context of surfaces exposed to solar energy, it's essential to know the amount of radiation absorbed for energy calculations. For a given surface, it can be calculated by the formula: \( q_{abs} = G \times A \times \bar{\alpha} \), where
  • \( G \) is the incoming solar irradiation in W/m²,
  • \( A \) is the area of the surface in m², and
  • \( \bar{\alpha} \) is the average absorptivity of the surface.
Absorptivity (\( \alpha \)) is a measure of how effective a surface is at absorbing radiation, varying with the wavelength of the incoming radiation. In the example provided, due to the spectral characteristics of the irradiation and the surface, the average absorptivity (\( \bar{\alpha} \)) is determined based on the spectral distribution of solar energy. It primarily lies between 0.5 to 1.0 \( \mu m \), leading to an absorptivity value of 0.8, capturing much of the expected irradiation effectively.
Emissive Power
Emissive power (\( E \)) represents the radiation emitted by a surface due to its thermal energy. According to the Stefan-Boltzmann law, the emissive power of a surface at a given temperature is calculated with the formula: \( E = \epsilon \sigma T^4 \). Here,
  • \( \epsilon \) is the emissivity of the surface,
  • \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \text{ W/m}^2\cdot \text{K}^4) \), and
  • \( T \) is the absolute temperature in Kelvin.
Emissivity is the efficiency at which a surface emits thermal radiation compared to that of an ideal black body, which has an emissivity of 1. In typical calculations, selectively absorbed and emitted wavelengths are considered. For instance, in this case, the emissive power assumes an emissivity of 0.9, indicative of the surface's ability to emit thermal radiation in the spectrum beyond 2 \( \mu m \). This lets us calculate the energy emitted per unit area, and subsequently, for the entire surface area, which is often necessary for comprehensive thermal management analyses.
Radiosity
Radiosity involves the total radiant energy leaving a surface, which includes both emitted and reflected radiation. It's a comprehensive energy balance on a surface and can be represented by the formula: \( J = E + \rho G \), where
  • \( E \) is the emissive power of the surface,
  • \( \rho \) is the reflectivity (\( \rho = 1 - \epsilon \), since \( \epsilon + \rho = 1 \) for opaque surfaces), and
  • \( G \) is the incoming irradiation.
Radiosity is a crucial metric when conducting thermal radiation studies of surfaces in environments exposed to external sources of heat, such as the Sun. This value helps in understanding how much total energy flows away from a surface. It includes both what the surface itself emits and what it reflects from external sources, offering insights into various practical applications like thermal insulation efficacy and energy flow in systems.
Net Radiation Heat Transfer
Net radiation heat transfer reflects the balance between absorbed radiation and the radiosity of a surface. It is characterized by the formula: \( q_{net} = q_{abs} - J_{total} \). If \( q_{net} \) is positive, the surface absorbs more energy than it emits and reflects, consequently heating up. If it’s negative, as seen in the example, the surface radiates or reflects more energy than it absorbs, meaning the surface is losing heat. Understanding net radiation heat transfer is vital in practical scenarios like heating earth structures, energy surplus or deficit in spacecraft systems, and evaluating the thermal stress in architectural components due to radiative energy transfers.

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Most popular questions from this chapter

The neighborhood cat likes to sicep on the roof of our shed in the backyard. The roofing surface is weathered galvanized sheet metal \(\left(\varepsilon=0.65, \alpha_{S}=0.8\right)\). Consider a cool spring day when the ambient air temperature is \(10^{\circ} \mathrm{C}\) and the convection coefficient can be estimated from an empirical correlation of the form \(\bar{h}_{1}=1.0 \Delta T^{4 / 3}\), where \(\mathrm{T} T\) is the difference between the surface and ambient temperatures. Assume the sky temperature is \(-40^{\circ} \mathrm{C}\). (a) Assuming the backside of the roof is well insulated, calculate the roof temperarure when the solar irradiation is \(600 \mathrm{~W} / \mathrm{m}^{2}\). Will the cat enjoy sleeping under these conditions? (b) Consider the case when the hackside of the roof is not insulated, but is exposed to ambient air with the same convection cocfficient relation and experiences radiation exchange with the ground, also at the ambient air temperarure. Calculate the roof tempernture and comment on whether the roof will be a comfortable place for the cat to snooze.

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.

According to its directicnal distribution, solar radiation incident on the earth's surface may be divided into two components. The direct component consists of parallel rays incident at a fixed renith angle \(\theta\), while the diffuse component consists of radiation that may be approximated as being diffusely distributed with \(\theta\). Consider clear sky conditions for which the direct radiation is incident at \(\theta=30^{\circ}\), with a total flux (based on an area that is normal to the rays) of \(q_{\text {dre }}^{*}=\) \(1000 \mathrm{~W} / \mathrm{m}^{2}\), and the total intensity of the diffuse radiation is \(l_{\text {ar }}=70 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{sr}\). What is the total solar irradiation at the earth's surface?

Sheet steel emerging from the hot roll section of a steel mill has a temperature of \(1200 \mathrm{~K}\), a thickness of \(\delta=3 \mathrm{~mm}\), and the following distribution for the spectral, hemispherical emissivity. The density and specific heat of the steel are 7900 \(\mathrm{kg} / \mathrm{m}^{3}\) and \(640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. What is the total, hemispherical emissivity? Accounting for emission from beth sides of the sheet and neglecting conduction, convection, and radiation from the surroundings. determine the initial time rate of change of the sheet temperature \((d T / d t)\). As the steel cools. it oxidizes and its total, hemispherical emissivity increases. If this increase maty be correlated by an expression of the form \(\varepsilon=\varepsilon_{120}[1200 \mathrm{~K} T(\mathrm{~K})]\). how long will it take for the steel to cool from 1200 to \(600 \mathrm{~K}\) ?

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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