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Two plates, one with a black painted surface and the other with a special coating (chemically oxidized copper) are in earth orbit and are exposed to solar radiation. The solar rays make an angle of \(30^{\circ}\) with the normal to the plate. Estimate the equilibrium temperature of each plate assuming they are diffuse and that the solar flux is \(1368 \mathrm{~W} / \mathrm{m}^{2}\). The spectral absorptivity of the black painted surface can be approximated by \(\alpha_{\lambda}=0.95\) for \(0 \leq \lambda \leq \infty\) and that of the special coating by \(\alpha_{\lambda}=0.95\) for \(0 \leq \lambda<3 \mu \mathrm{m}\) and \(\alpha_{\lambda}=0.05\) for \(\lambda \geq 3 \mu \mathrm{m}\).

Short Answer

Expert verified
The estimated equilibrium temperatures for the two plates are approximately \(364.5\mathrm{K}\) for the black painted surface and \(336.7\mathrm{K}\) for the special coated surface.

Step by step solution

01

Calculate the absorbed solar radiation for each plate.

Both plates are exposed to the same solar flux. However, due to the angle of incidence, the absorbed solar radiation differs. We will use the product of the incident angle and the solar flux to calculate the absorbed solar radiation. For this, we can use the equation: \[A = F_\text{solar} \cdot \cos(\theta)\] Where: \(A\) = absorbed solar radiation (W/m²), \(F_\text{solar}\) = solar flux given by \(1368 \mathrm{~W} / \mathrm{m}^2 \), \(\theta\) = angle between the solar rays and the normal to the plate, given by \(30°\).
02

Calculate the absorptivity-integrated power for each plate.

For each plate, we first need to calculate the absorptivity-integrated power using the given spectral absorptivities (\(\alpha_\lambda\)): Black painted surface: Since the absorptivity is constant and given by \(\alpha_{\lambda} = 0.95 \) for all wavelengths, the absorptivity-integrated power for this plate is simply 0.95. Special coating: Here, the spectral absorptivity is different for different wavelength ranges. We need to first find the fraction of the total power of solar radiation for each of these wavelength ranges, and then use the given spectral absorptivities accordingly. From the solar spectrum and the Planck's law, about 51% of the total solar radiation is below \(3\mathrm{~\mu m}\) and the remaining 49% is above \(3\mathrm{~\mu m}\). So, absorptivity-integrated power for the special coated plate is: \(\alpha_\text{integrated} = 0.95 \times 0.51 + 0.05 \times 0.49 \approx 0.51\)
03

Calculate total absorbed power for each plate.

Next, we calculate the total absorbed power for each plate using the absorbed solar radiation and absorptivity-integrated power: Black painted surface: \[P_\text{absorbed} = A \times \alpha_\text{integrated} = F_\text{solar} \cdot \cos(\theta) \times 0.95\] Special coating: \[P_\text{absorbed} = A \times \alpha_\text{integrated} = F_\text{solar} \cdot \cos(\theta) \times 0.51\]
04

Calculate the emitted heat for each plate using Stefan-Boltzmann law.

The emitted heat can be calculated using Stefan-Boltzmann law, which is given by: \[P_\text{emitted} = \sigma T^4\] Where: \(P_\text{emitted}\) = power emitted per unit area, \(\sigma\) = Stefan-Boltzmann constant, given by \(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\), \(T\) = temperature of the surface (in Kelvin).
05

Set the absorbed power equal to the emitted power to find equilibrium temperature.

At equilibrium, the absorbed power equals the emitted power. Therefore, Black painted surface: \[P_\text{absorbed} = P_\text{emitted} \Rightarrow F_\text{solar} \cdot \cos(\theta) \times 0.95 = \sigma T^4\] Special coating: \[P_\text{absorbed} = P_\text{emitted} \Rightarrow F_\text{solar} \cdot \cos(\theta) \times 0.51 = \sigma T^4\]
06

Solve for the equilibrium temperatures for both plates.

Finally, we can solve the equations for both plates to find their respective equilibrium temperatures: Black painted surface: \[ T_\text{black} = \left(\frac{F_\text{solar} \cdot \cos(\theta) \times 0.95}{\sigma}\right)^{1/4} \approx 364.5\mathrm{K} \] Special coating: \[ T_\text{special} = \left(\frac{F_\text{solar} \cdot \cos(\theta) \times 0.51}{\sigma}\right)^{1/4} \approx 336.7 \mathrm{K} \]

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Key Concepts

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

Stefan-Boltzmann Law
Understanding the Stefan-Boltzmann law is essential in the study of thermal radiation. This principle is rooted in the field of thermodynamics and quantifies the power radiated from a black body in terms of its temperature. Specifically, the law states that the total energy radiated per unit surface area of a black body across all wavelengths is directly proportional to the fourth power of the black body's temperature.

The mathematical expression for the Stefan-Boltzmann law can be written as:
\[ P_{\text{emitted}} = \sigma T^4 \]
where:\( P_{\text{emitted}} \) is the energy emitted per unit area,\( \sigma \) represents the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)), and \( T \) is the absolute temperature of the black body in Kelvin. This law provides a foundational understanding for how objects at different temperatures emit energy. It is crucial when solving problems that involve the equilibrium temperature between absorbed and emitted thermal radiation.

When applying this law, it's important to note that only perfect black bodies strictly follow the relation. However, for practical purposes, it is often applied to real objects by considering the emissivity factor, which accounts for how closely an object approximates a black body in terms of radiation.
Solar Radiation Absorption
The solar radiation absorption by any surface is a critical consideration in calculating equilibrium temperatures, as demonstrated in the exercise involving plates in orbit. Absorbed solar radiation depends on factors like the angle of incidence, surface properties, and the intensity of incoming solar radiation.

The solar flux, \( F_{\text{solar}} \), is the constant flow of energy from the sun per unit area, given for Earth's vicinity as approximately \(1368 \mathrm{W/m^2}\). However, not all this flux is absorbed—some is reflected or passes through. The actual amount of absorbed solar energy (\( A \)) can be calculated using the formula:
\[ A = F_{\text{solar}} \cdot \cos(\theta) \]
where \( \theta \) is the angle of incidence measured from the normal to the surface. The cosine factor reduces the effective area of the plate exposed to radiation, a concept known as 'projection effect'. In our exercise, the angle is \(30^\circ\), leading to a cosine factor of \(\cos(30^\circ)\), which results in less energy absorption as compared to if the sunlight were hitting the plate directly at normal incidence (\(0^\circ\)).

Solar radiation absorption directly influences the thermal energy balance of the surface, impacting the equilibrium temperature calculation. This concept is essential in various fields, such as solar panel design, climate modeling, and space engineering.
Spectral Absorptivity
Spectral absorptivity, \( \alpha_{\lambda} \), is an integral concept regarding surfaces absorbing electromagnetic radiation at different wavelengths. This property varies across materials and influences how much radiation is absorbed at specific wavelengths, crucial for calculating an object's temperature in space.

In the textbook exercise, spectral absorptivity signifies how the black painted surface and special coating respond to solar radiation. A black painted surface is described with a constant spectral absorptivity of \( \alpha_{\lambda} = 0.95 \) across all wavelengths, indicating nearly complete absorption and minimal reflection. This quality makes black surfaces adept at harvesting solar energy.

The special coating, however, has a variable spectral absorptivity: \( \alpha_{\lambda} = 0.95 \) for wavelengths less than \(3 \mu m\) and \( \alpha_{\lambda} = 0.05 \) beyond that. This differentiation helps the material respond differently to various parts of the solar spectrum.

By understanding the spectral absorptivity of materials, engineers can design surfaces to achieve a desired thermal behavior, such as maximizing absorption in solar panels or reducing heat gains in thermal control systems for satellites. It's a critical factor in controlling the temperature and energy efficiency of systems exposed to radiation.

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Most popular questions from this chapter

Photovoltaic materials convert sunlight directly to electric power. Some of the photons that are incident upon the material displace electrons that are in turn collected to create an electric current. The overall efficiency of a photovoltaic panel, \(\eta\), is the ratio of electrical energy produced to the energy content of the incident radiation. The efficiency depends primarily on two properties of the photovoltaic material, (i) the band gap, which identifies the energy states of photons having the potential to be converted to electric current, and (ii) the interband gap conversion efficiency, \(\eta_{\mathrm{bg}}\), which is the fraction of the total energy of photons within the band gap that is converted to electricity. Therefore, \(\eta=\eta_{\mathrm{bg}} F_{\mathrm{bg}}\) where \(F_{\mathrm{bg}}\) is the fraction of the photon energy incident on the surface within the band gap. Photons that are either outside the material's band gap or within the band gap but not converted to electrical energy are either reflected from the panel or absorbed and converted to thermal energy. Consider a photovoltaic material with a band gap of \(1.1 \leq B \leq 1.8 \mathrm{eV}\), where \(B\) is the energy state of a photon. The wavelength is related to the energy state of a photon by the relationship \(\lambda=1240 \mathrm{eV} \cdot \mathrm{nm} / B\). The incident solar irradiation approximates that of a blackbody at \(5800 \mathrm{~K}\) and \(G_{S}=1000 \mathrm{~W} / \mathrm{m}^{2}\). (a) Determine the wavelength range of solar irradiation corresponding to the band gap. (b) Determine the overall efficiency of the photovoltaic material if the interband gap efficiency is \(\eta_{\text {bog }}=0.50\) (c) If half of the incident photons that are not converted to electricity are absorbed and converted to thermal energy, determine the heat absorption per unit surface area of the panel.

The exposed surface of a power amplifier for an earth satellite receiver of area \(130 \mathrm{~mm} \times 130 \mathrm{~mm}\) has a diffuse, gray, opaque coating with an emissivity of \(0.5\). For typical amplifier operating conditions, the surface temperature is \(58^{\circ} \mathrm{C}\) under the following environmental conditions: air temperature, \(T_{\infty}=27^{\circ} \mathrm{C}\); sky temperature, \(T_{\text {sy }}=-20^{\circ} \mathrm{C} ;\) convection coefficient, \(h=\) \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); and solar irradiation, \(G_{S}=800 \mathrm{~W} / \mathrm{m}^{2}\). (a) For the above conditions, determine the electrical power being generated within the amplifier. (b) It is desired to reduce the surface temperature by applying one of the diffuse coatings (A, B, C) shown as follows. Which coating will result in the coolest surface temperature for the same amplifier operating and environmental conditions?

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is \(2500 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate is at \(227^{\circ} \mathrm{C}\) and has an emissive power of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Air at \(127^{\circ} \mathrm{C}\) flows over the plate with a heat transfer convection coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

An enclosure has an inside area of \(100 \mathrm{~m}^{2}\), and its inside surface is black and is maintained at a constant temperature. A small opening in the enclosure has an area of \(0.02 \mathrm{~m}^{2}\). The radiant power emitted from this opening is \(70 \mathrm{~W}\). What is the temperature of the interior enclosure wall? If the interior surface is maintained at this temperature, but is now polished, what will be the value of the radiant power emitted from the opening?

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