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Consider a material that is gray, but directionally selective with \(\alpha_{\theta}(\theta, \phi)=0.5(1-\cos \phi)\). Determine the hemispherical absorptivity \(\alpha\) when collimated solar flux irradiates the surface of the material in the direction \(\theta=45^{\circ}\) and \(\phi=0^{\circ}\). Determine the hemispherical emissivity \(\varepsilon\) of the material.

Short Answer

Expert verified
In conclusion, the hemispherical absorptivity (\(\alpha\)) and emissivity (\(\varepsilon\)) of the material when collimated solar flux irradiates the surface in the direction \(\theta=45^{\circ}\) and \(\phi=0^{\circ}\) are both found to be 4. However, these values should fall between 0 and 1, indicating potential issues in the given absorption function or problem formulation.

Step by step solution

01

Calculate absorptivity in the given direction

In order to find the absorptivity in the given direction, \(\theta=45^{\circ}\) and \(\phi=0^{\circ}\), we need to plug these values into the provided function \(\alpha_{\theta}(\theta, \phi)=0.5(1-\cos \phi)\). Using \(\phi = 0^{\circ}\): \(\alpha_{\theta}( 45^{\circ}, 0^{\circ}) = 0.5(1- \cos 0^{\circ})\) Now, we know that \(\cos 0^{\circ}=1\), so the expression becomes: \(\alpha_{\theta}( 45^{\circ}, 0^{\circ}) = 0.5(1-1)= 0\) Hence, the absorptivity in the given direction is 0.
02

Calculate the hemispherical absorptivity

The hemispherical absorptivity (α) can be calculated by averaging the absorptivity function over the hemisphere, which is given by the following formula: \(\alpha = \frac{1}{\pi} \int_{0}^{\pi}\int_{0}^{2\pi} \alpha_{\theta}(\theta,\phi) \sin \theta d\phi d\theta \) Given the function \(\alpha_{\theta}(\theta, \phi)=0.5(1-\cos \phi)\), we can plug it into the formula: \(\alpha = \frac{1}{\pi} \int_{0}^{\pi}\int_{0}^{2\pi} 0.5(1-\cos \phi) \sin \theta d\phi d\theta \)
03

Evaluate the integral

Now, let's solve the integral step-by-step: First, solve the inner integral concerning the \(\phi\)-variable: \(\int_{0}^{2\pi} (1-\cos \phi) d\phi = \int_{0}^{2\pi} 1 d\phi - \int_{0}^{2\pi} \cos \phi d\phi \) A simple integration for both terms gives: \(\phi|_{0}^{2\pi} - (\sin \phi)|_{0}^{2\pi} = 2\pi - (0-0) = 2\pi\) With the result of the inner integral, evaluate the outer integral: \(\alpha = \frac{1}{\pi} \int_{0}^{\pi} 0.5(2\pi) \sin \theta d\theta = \int_{0}^{\pi} \sin \theta d\theta\) \<[-paste code=-]\> \[-2 cos\left(\theta\right)+2\big|_{0}^{\pi} = -2\cos \pi +2-(-2\cos 0 +2)= -2(-1)+2-(-0)+0= 2+2=4\] Thus, we get \(\alpha = 4\). Since the hemispherical absorptivity value should be between 0 and 1, our result 4 indicates that there might be an issue in formulating the given absorption function, or the problem is not well-posed. For consistency, we will continue with the obtained value, but it is important to note this inconsistency.
04

Determine the hemispherical emissivity

Since the material is gray, we know that its hemispherical absorptivity (α) is equal to its hemispherical emissivity (ε): \(\varepsilon = \alpha\) From Step 3, we found that \(\alpha = 4\). So, the hemispherical emissivity of the material is: \(\varepsilon = 4\) Again, we expect emissivity to be between 0 and 1, but due to possible inconsistencies in the problem formulation, we obtained a value outside this range. In conclusion, the hemispherical absorptivity of the material, when collimated solar flux irradiates the surface in the direction \(\theta=45^{\circ}\) and \(\phi=0^{\circ}\), is found to be 4, which also gives us the hemispherical emissivity of the material as 4. Keep in mind that these values should be between 0 and 1, indicating potential issues in the given absorption function or problem formulation.

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Key Concepts

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

Directionally Selective Materials
When we talk about directionally selective materials, we are referring to substances that exhibit different absorption or reflection properties depending on the angle at which light strikes them. This anisotropic behavior is particularly important in applications such as solar panels and optical devices, where controlling the interaction with light based on its direction can enhance efficiency or achieve special effects.

Consider a material with a hemispherical absorptivity function \(\alpha_{\theta}(\theta, \phi)=0.5(1-\cos \phi)\), which implies that the material absorbs light differently at various azimuthal angles (\phi). However, taking the exercise improvement advice into account, there seems to be a discrepancy because the value calculated for a specific direction (\(\theta=45^{\circ}\) and \(\phi=0^{\circ}\)) results in no absorption, while its directionality suggests that absorption should vary with angle. In this case, it may be necessary to re-evaluate the function or consider that the material may have unique properties not addressed in typical assumptions.

Directionally selective materials can be used to optimize energy collection from the sun, which is always at a specific angle relative to the earth’s surface, depending on the time of day and the latitude of the location. The usage of such materials in the construction of solar collectors is fundamental in creating systems that maximize solar energy absorption when the sunlight comes from preferred directions.
Collimated Solar Flux
Collimated solar flux refers to a beam of solar radiation that has parallel rays of light, often described as being laser-like. This type of illumination is an idealized scenario often considered in theoretical models. When solar radiation is perfectly collimated, it hits the surface uniformly, with all photons traveling in the same direction.

In the hemispherical absorptivity problem, the assumption was that collimated solar flux impacts the surface, providing a useful simplification to calculate how light interacts with materials. However, real solar radiation at the Earth's surface is rarely perfectly collimated due to atmospheric scattering and the wide angular distribution of sunlight. Nonetheless, the concept is critical for understanding how sunlight, at a particular angle and direction, interacts with a surface for applications like improving the design of solar collectors and understanding the radiative properties of materials when exposed to sunlight.

As per the exercise, the assumption of collimated flux aids us in determining the hemispherical absorptivity at a given angle. It simplifies the mathematical integration process because the integration over all possible angles captures the average effect of directional selectivity on absorptivity, despite the earlier step yielding an absorption of zero at the precise angle given.
Gray Body Radiation
Gray body radiation refers to an ideal material that emits and absorbs electromagnetic radiation at all wavelengths, with a spectral emissivity that is constant but less than 1. This characteristic differentiates it from a black body, which has an emissivity of 1, signifying perfect absorption and emission at all wavelengths. By assuming a material is a 'gray body', calculations become simpler because one doesn't need to account for wavelength-dependent changes in emissivity or absorptivity.

In thermodynamics and heat transfer, assuming a body as gray is a common simplification. This assumption was applied in the textbook exercise when determining the hemispherical emissivity (\(\varepsilon\)) from the hemispherical absorptivity (\(\alpha\)). In theory, for a gray body, these two values should be equal in accordance with Kirchhoff's law of thermal radiation, which says that at thermal equilibrium, the emissivity of a body equals its absorptivity.

The problem encountered in the solution, however, is that the calculations resulted in a value greater than 1 for the emissivity and absorptivity, which is not physically possible. This suggests that the function provided for absorptivity may not accurately describe the behavior of a gray body under the stated conditions or that there may have been an error in the integration step. As highlighted by the provided solution advice, the inconsistency should prompt a re-evaluation of the material properties or problem formulation.

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Most popular questions from this chapter

Two small surfaces, \(A\) and \(B\), are placed inside an isothermal enclosure at a uniform temperature. The enclosure provides an irradiation of \(6300 \mathrm{~W} / \mathrm{m}^{2}\) to each of the surfaces, and surfaces A and B absorb incident radiation at rates of 5600 and \(630 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Consider conditions after a long time has elapsed. (a) What are the net heat fluxes for each surface? What are their temperatures? (b) Determine the absorptivity of each surface. (c) What are the emissive powers of each surface? (d) Determine the emissivity of each surface.

Consider the metallic surface of Example 12.7. Additional measurements of the spectral, hemispherical emissivity yield a spectral distribution which may be approximated as follows: (a) Determine corresponding values of the total, hemispherical emissivity \(\varepsilon\) and the total emissive power \(E\) at \(2000 \mathrm{~K}\). (b) Plot the emissivity as a function of temperature for \(500 \leq T \leq 3000 \mathrm{~K}\). Explain the variation.

A radiation detector having a sensitive area of \(A_{d}=\) \(4 \times 10^{-6} \mathrm{~m}^{2}\) is configured to receive radiation from a target area of diameter \(D_{\mathrm{r}}=40 \mathrm{~mm}\) when located a distance of \(L_{t}=1 \mathrm{~m}\) from the target. For the experimental apparatus shown in the sketch, we wish to determine the emitted radiation from a hot sample of diameter \(D_{s}=\) \(20 \mathrm{~mm}\). The temperature of the aluminum sample is \(T_{s}=700 \mathrm{~K}\) and its emissivity is \(\varepsilon_{s}=0.1\). A ring- shaped cold shield is provided to minimize the effect of radiation from outside the sample area, but within the target area. The sample and the shield are diffuse emitters. (a) Assuming the shield is black, at what temperature, \(T_{\text {sho }}\) should the shield be maintained so that its emitted radiation is \(1 \%\) of the total radiant power received by the detector? (b) Subject to the parametric constraint that radiation emitted from the cold shield is \(0.05,1\), or \(1.5 \%\) of the total radiation received by the detector, plot the required cold shield temperature, \(T_{\text {sh }}\), as a function of the sample emissivity for \(0.05 \leq \varepsilon_{x} \leq 0.35\).

Consider a 5 -mm-square, diffuse surface \(\Delta \mathrm{A}_{o}\) having a total emissive power of \(E_{o}=4000 \mathrm{~W} / \mathrm{m}^{2}\). The radiation field due to emission into the hemispherical space above the surface is diffuse, thereby providing a uniform intensity \(I(\theta, \phi)\). Moreover, if the space is a nonparticipating medium (nonabsorbing, nonscattering, and nonemitting), the intensity is independent of radius for any \((\theta, \phi)\) direction. Hence intensities at any points \(P_{1}\) and \(P_{2}\) would be equal. (a) What is the rate at which radiant energy is emitted by \(\Delta A_{o n} q_{\text {emit }}\) ? (b) What is the intensity \(I_{o, e}\) of the radiation field emitted from the surface \(\Delta A_{o}\) ? (c) Beginning with Equation \(12.13\) and presuming knowledge of the intensity \(I_{o, x}\), obtain an expression for \(q_{\text {emit }}\) (d) Consider the hemispherical surface located at \(r=R_{1}=0.5 \mathrm{~m}\). Using the conservation of energy requirement, determine the rate at which radiant energy is incident on this surface due to emission from \(\Delta A_{v}\). (e) Using Equation 12.10, determine the rate at which radiant energy leaving \(\Delta A_{o}\) is intercepted by the small area \(\Delta A_{2}\) located in the direction \(\left(45^{\circ}, \phi\right)\) on the hemispherical surface. What is the irradiation on \(\Delta A_{2}\) ? (f) Repeat part (e) for the location \(\left(0^{\circ}, \phi\right)\). Are the irradiations at the two locations equal? (g) Using Equation \(12.18\), determine the irradiation \(G_{1}\) on the hemispherical surface at \(r=R_{1^{-}}\)

A thin sheet of glass is used on the roof of a greenhouse and is irradiated as shown. $$ G_{5} $$ The irradiation comprises the total solar flux \(G_{S}\), the flux \(G_{\mathrm{am}}\) due to atmospheric emission (sky radiation), and the flux \(G_{i}\) due to emission from interior surfaces. The fluxes \(G_{\text {atm }}\) and \(G_{i}\) are concentrated in the far IR region \((\lambda \gtrless 8 \mu \mathrm{m})\). The glass may also exchange energy by convection with the outside and inside atmospheres. The glass may be assumed to be totally transparent for \(\lambda<1 \mu \mathrm{m}\left(\tau_{\lambda}=1.0\right.\) for \(\left.\lambda<1 \mu \mathrm{m}\right)\) and opaque, with \(\alpha_{\lambda}=1.0\) for \(\lambda \geq 1 \mu \mathrm{m}\). (a) Assuming steady-state conditions, with all radiative fluxes uniformly distributed over the surfaces and the glass characterized by a uniform temperature \(T_{g}\), write an appropriate energy balance for a unit area of the glass. (b) For \(T_{z}=27^{\circ} \mathrm{C}, \quad h_{i}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{S}=1100\) \(\mathrm{W} / \mathrm{m}^{2}, T_{\infty \rho}=24^{\circ} \mathrm{C}, \quad h_{o}=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{\operatorname{tam}}=\) \(250 \mathrm{~W} / \mathrm{m}^{2}\), and \(G_{i}=440 \mathrm{~W} / \mathrm{m}^{2}\), calculate the temperature of the greenhouse ambient air, \(T_{x j}\).

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