/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 129 Two plates, one with a black pai... [FREE SOLUTION] | 91Ó°ÊÓ

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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

A radiation thermometer is a device that responds to a radiant flux within a prescribed spectral interval and is calibrated to indicate the temperature of a blackbody that produces the same flux. (a) When viewing a surface at an elevated temperature \(T_{s}\) and emissivity less than unity, the thermometer will indicate an apparent temperature referred to as the brightness or spectral radiance temperature \(T_{\lambda}\). Will \(T_{\lambda}\) be greater than, less than, or equal to \(T_{s}\) ? (b) Write an expression for the spectral emissive power of the surface in terms of Wien's spectral distribution (see Problem 12.27) and the spectral emissivity of the surface. Write the equivalent expression using the spectral radiance temperature of the surface and show that $$ \frac{1}{T_{x}}=\frac{1}{T_{\lambda}}+\frac{\lambda}{C_{2}} \ln \varepsilon_{\lambda} $$ where \(\lambda\) represents the wavelength at which the thermometer operates. (c) Consider a radiation thermometer that responds to a spectral flux centered about the wavelength \(0.65 \mu \mathrm{m}\). What temperature will the thermometer indicate when viewing a surface with \(\varepsilon_{\lambda}(0.65 \mu \mathrm{m})=0.9\) and \(T_{x}=1000 \mathrm{~K}\) ? Verify that Wien's spectral distribution is a reasonable approximation to Planck's law for this situation.

Four diffuse surfaces having the spectral characteristics shown are at \(300 \mathrm{~K}\) and are exposed to solar radiation. Which of the surfaces may be approximated as being gray?

The energy flux associated with solar radiation incident on the outer surface of the earth's atmosphere has been accurately measured and is known to be \(1368 \mathrm{~W} / \mathrm{m}^{2}\). The diameters of the sun and earth are \(1.39 \times 10^{9}\) and \(1.27 \times 10^{7} \mathrm{~m}\), respectively, and the distance between the sun and the earth is \(1.5 \times 10^{11} \mathrm{~m}\). (a) What is the emissive power of the sun? (b) Approximating the sun's surface as black, what is its temperature? (c) At what wavelength is the spectral emissive power of the sun a maximum? (d) Assuming the earth's surface to be black and the sun to be the only source of energy for the earth, estimate the earth's surface temperature.

Solar radiation incident on the earth's surface may be divided into the direct and diffuse components described in Problem 12.9. Consider conditions for a day in which the intensity of the direct solar radiation is \(I_{\text {dir }}=210 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\) in the solid angle subtended by the sun with respect to the earth, \(\Delta \omega_{s}=6.74 \times 10^{-5} \mathrm{sr}\). The intensity of the diffuse radiation is \(I_{\text {dif }}=70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\). (a) What is the total solar irradiation at the earth's surface when the direct radiation is incident at \(\theta=30^{\circ}\) ? (b) Verify the prescribed value for \(\Delta \omega_{s}\), recognizing that the diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m}\) and the distance between the sun and the earth is \(1.496 \times 10^{11} \mathrm{~m}\) (1 astronomical unit).

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

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