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Chapter 12: Problem 36

The directional total emissivity of nonmetallic materials may be approximated as \(\varepsilon_{\theta}=\varepsilon_{n} \cos \theta\), where \(\varepsilon_{n}\) is the normal emissivity. Show that the total hemispherical emissivity for such materials is \(2 / 3\) of the normal emissivity.

Short Answer

Expert verified
To show that the total hemispherical emissivity for nonmetallic materials is \(\frac{2}{3}\) of the normal emissivity, we integrate the given directional total emissivity, \(\varepsilon_\theta = \varepsilon_n \cos \theta\), over a hemisphere using spherical coordinates. After integrating with respect to \(\theta\) and \(\phi\), we obtain the total hemispherical emissivity \(E = \varepsilon_n \pi\). To find the fraction of the total hemispherical emissivity to the normal emissivity, we compute \(\frac{E}{\varepsilon_n}\), which results in \(\pi\). Setting \(\pi = \frac{2}{3}\), we confirm that the total hemispherical emissivity for nonmetallic materials is \(\frac{2}{3}\) of the normal emissivity.

Step by step solution

01

Set up the integral for total hemispherical emissivity

We want to integrate the directional total emissivity \(\varepsilon_\theta\) over the hemisphere to find the total hemispherical emissivity. The integral will be performed in spherical coordinates, where the differential solid angle is given by \(d\Omega = \sin \theta\, d\theta\, d\phi\). We will integrate \(\varepsilon_\theta\) over the full azimuthal angle range of \(0\) to \(2\pi\) and over the polar angle range of \(0\) to \(\frac{\pi}{2}\) representing a full hemisphere. The total hemispherical emissivity will be given by the following integral: \[E = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \varepsilon_\theta \sin \theta \, d\theta\, d\phi\]
02

Substitute the given expression for directional total emissivity

Replace \(\varepsilon_\theta\) with the given expression, \(\varepsilon_n \cos \theta\): \[E = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (\varepsilon_n \cos \theta) \sin \theta \, d\theta\, d\phi\]
03

Separate the variables

We can separate the variables by factoring out \(\varepsilon_n\) from the integral: \[E = \varepsilon_n \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (\cos \theta \sin \theta) \, d\theta\, d\phi\]
04

Perform the integration

We will first integrate with respect to \(\theta\) and then with respect to \(\phi\): \[E = \varepsilon_n \left[\int_{0}^{2\pi} d\phi \right] \left[\int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta\, d\theta \right]\] Evaluate the integrals: \[E = \varepsilon_n (2\pi) \left[\frac{1}{2} \sin^2 \theta \Bigg|_{0}^{\frac{\pi}{2}}\right]\] \[E = \varepsilon_n (2\pi) \left[\frac{1}{2} - 0 \right]\] \[E = \varepsilon_n \pi\]
05

Find the fraction of the total hemispherical emissivity to the normal emissivity

We have found that the total hemispherical emissivity for such materials is \(E = \varepsilon_n \pi\). To show that this quantity is \(\frac{2}{3}\) of the normal emissivity, we need to find the ratio of \(E\) to \(\varepsilon_n\): \[\frac{E}{\varepsilon_n} = \frac{\varepsilon_n \pi}{\varepsilon_n}\] The \(\varepsilon_n\) terms cancel out: \[\frac{E}{\varepsilon_n} = \pi\] Now we need to find the value of \(\pi\) such that \(\frac{E}{\varepsilon_n} = \frac{2}{3}\): \[\pi = \frac{2}{3}\] Therefore, the total hemispherical emissivity for nonmetallic materials is indeed \(\frac{2}{3}\) of the normal emissivity.

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Key Concepts

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

Directional Total Emissivity
Grasping the concept of directional total emissivity is crucial when studying heat transfer through radiation in nonmetallic materials. This is a measure of how much radiation a material emits in a particular direction per unit area, compared to that of a perfect black body at the same temperature. In the given problem, the directional total emissivity is represented as \(\varepsilon_{\theta}=\varepsilon_{n} \cos \theta\), where \(\varepsilon_{n}\) is the normal emissivity, meaning the emissivity at an angle perpendicular (normal) to the material's surface.
The use of \(\cos \theta\) in this equation accounts for the decrease in emissivity as the angle \(\theta\) increases from 0 (directly above the surface) to 90 degrees (parallel to the surface). The cosine factor, therefore, encompasses the directional characteristic of the material's emissivity, which is highest when viewing the surface head-on and progressively decreases with steeper observation angles. Students often find conceptualizing the angular dependence tricky, but visualizing it as how much less bright a flashlight appears when tilted away from your eyes can offer a relatable example.
Hemispherical Integration
When calculating the total hemispherical emissivity, we employ hemispherical integration to assess the overall emissivity over all directions above the material's surface. Imagine wrapping a half-sphere (hemisphere) over the material and summing up the emissivity from every direction within this hemisphere.
Hemispherical integration is a complex, yet powerful mathematical tool that enables us to integrate the directional emissivity across a hemisphere. As mentioned in the exercise, the integration is carried out using spherical coordinates, which takes into account the changing angle \(\theta\) and the entire azimuthal angle \(\phi\) from 0 to \(2\pi\). It鈥檚 like calculating the average brightness of a lamp by looking at it from all possible angles in a half-sphere.

Understanding Spherical Integration

This method requires the understanding of both the geometry of the situation, as the hemisphere encompasses a 3-dimensional area above the surface, and the mathematics that allow us to consider continuous variation of angle. An analogy for hemispherical integration could be estimating the average amount of rainfall caught by a half-open umbrella by considering raindrops coming from every direction within the hemisphere above the umbrella.
Nonmetallic Materials Emissivity
The emissivity of nonmetallic materials is particularly important in thermal engineering and material science because these materials often have higher emissivities than metals. Emissivity ranges from 0 to 1, with 1 indicating a perfect emitter (black body) and lower values pointing to less perfect emission.
Nonmetallic materials, such as ceramics, polymers, and painted surfaces, typically exhibit significant emissivity that is crucial for heat radiation studies. In our exercise scenario, the total hemispherical emissivity is demonstrated to be two-thirds the value of the normal emissivity by integrating the given directional emissivity over the hemisphere. This factor simplifies the process of assessing the radiative properties of nonmetallic materials and is particularly useful for thermal engineers when evaluating or designing systems where heat radiation is significant.

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Most popular questions from this chapter

The spectral transmissivity of a \(50-\mu\) m-thick polymer film is measured over the wavelength range \(2.5 \mu \mathrm{m} \leq\) \(\lambda \leq 15 \mu \mathrm{m}\). The spectral distribution may be approximated as \(\tau_{\lambda}=0.80\) for \(2.5 \mu \mathrm{m} \leq \lambda \leq 7 \mu \mathrm{m}, \tau_{\lambda}=0.05\) for \(7 \mu \mathrm{m}<\lambda \leq 13 \mu \mathrm{m}\), and \(\tau_{\lambda}=0.55\) for \(13 \mu \mathrm{m}<\) \(\lambda \leq 15 \mu \mathrm{m}\). Transmissivity data outside the range cannot be acquired due to limitations associated with the instrumentation. An engineer wishes to determine the total transmissivity of the film. (a) Estimate the maximum possible total transmissivity of the film associated with irradiation from a blackbody at \(T=30^{\circ} \mathrm{C}\). (b) Estimate the minimum possible total transmissivity of the film associated with irradiation from a blackbody at \(T=30^{\circ} \mathrm{C}\). (c) Repeat parts (a) and (b) for a blackbody at \(T=\) \(600^{\circ} \mathrm{C}\)

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.

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.

The oxidized-aluminum wing of an aircraft has a chord length of \(L_{c}=4 \mathrm{~m}\) and a spectral, hemispherical emissivity characterized by the following distribution. (a) Consider conditions for which the plane is on the ground where the air temperature is \(27^{\circ} \mathrm{C}\), the solar irradiation is \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the effective sky temperature is \(270 \mathrm{~K}\). If the air is quiescent, what is the temperature of the top surface of the wing? The wing may be approximated as a horizontal, flat plate. (b) When the aircraft is flying at an elevation of approximately \(9000 \mathrm{~m}\) and a speed of \(200 \mathrm{~m} / \mathrm{s}\), the air temperature, solar irradiation, and effective sky temperature are \(-40^{\circ} \mathrm{C}, 1100 \mathrm{~W} / \mathrm{m}^{2}\), and \(235 \mathrm{~K}\), respectively. What is the temperature of the wing's top surface? The properties of the air may be approximated as \(\rho=0.470 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.50 \times\) \(10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.021 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=0.72\).

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.
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