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A spherical aluminum shell of inside diameter \(D=2 \mathrm{~m}\) is evacuated and is used as a radiation test chamber. If the inner surface is coated with carbon black and maintained at \(600 \mathrm{~K}\), what is the irradiation on a small test surface placed in the chamber? If the inner surface were not coated and maintained at \(600 \mathrm{~K}\), what would the irradiation be?

Short Answer

Expert verified
In conclusion, the irradiation on a small test surface placed in the spherical aluminum shell when the inner surface is coated with carbon black and maintained at 600 K is approximately \(7.35 * 10^3 Wm^{-2}\). If the inner surface were not coated and maintained at 600 K, the irradiation would be approximately 735 Wm^{-2}.

Step by step solution

01

Using the Stefan-Boltzmann Law formula for irradiation, \[E = K * T^4\] We have, \(E = (5.67 * 10^{-8} Wm^{-2}K^{-4})*(600 K)^4\) #Step 2: Solve for E when the surface is coated with Carbon Black#

Simply plug the values into the formula and solve for E: \[ E = (5.67 * 10^{-8} Wm^{-2}K^{-4})*(600 K)^4\] \[ E = (5.67 * 10^{-8} Wm^{-2}K^{-4})* 129600000000 K^4\] \[ E = 7.35 * 10^3 Wm^{-2}\] The irradiation on the small test surface is approximately \(7.35 * 10^3 Wm^{-2}\) when the surface is coated with carbon black and maintained at 600 K. Now we need to find the irradiation when the surface is not coated. #Step 3: Calculate Emissivity for Uncoated Aluminum at 600K# When the surface is not coated we need to consider the emissivity of the aluminum. The emissivity of a material is a measure of how effectively it radiates energy, with 0 being no radiation and 1 being the ideal blackbody. The emissivity of aluminum at 600 K is approximately 0.1. We will use this value to find the irradiation on the test surface when the inner surface is not coated. #Step 4: Calculate the Irradiation for Non-Coated Surface with 600 K and Emissivity 0.1# We will modify the Stefan-Boltzmann Law formula to include the emissivity value: \[E = \varepsilon * K * T^4\] Here, \(\varepsilon\) represents the emissivity and the other variables remain the same.
02

Now, we need to plug the values into the formula and solve for E: \[E = (0.1) * (5.67 * 10^{-8} Wm^{-2}K^{-4}) * (600 K)^4\] #Step 5: Solve for E when the inner surface is not coated#

Plugging the values into the formula and solving for E: \[ E = (0.1)*(5.67 * 10^{-8} Wm^{-2}K^{-4})*(600 K)^4\] \[ E = (0.1)*(5.67 * 10^{-8} Wm^{-2}K^{-4})*129600000000K^4\] \[ E = 735 Wm^{-2}\] The irradiation on the small test surface is approximately \(735 Wm^{-2}\) when the inner surface is not coated and maintained at 600 K. In conclusion, the irradiation on a small test surface placed in the chamber when the inner surface is coated with carbon black and maintained at 600 K is \(7.35 * 10^3 Wm^{-2}\). If the inner surface were not coated and maintained at 600 K, the irradiation would be 735 Wm^{-2}.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in the realm of thermodynamics that relates the energy radiated by a black body to the fourth power of its temperature. This law is mathematically represented as \( E = \text{σ} \times T^4 \), where \( E \) is the radiant energy per unit area per unit time (also known as radiative flux or irradiation), \( \text{σ} \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} Wm^{-2}K^{-4} \)), and \( T \) is the absolute temperature in kelvins (K).

In a practical situation, such as a radiation test chamber, the Stefan-Boltzmann Law enables the calculation of the energy radiated by the chamber's walls, assuming they behave as a black body. A black body is an idealized physical body that absorbs all incident electromagnetic radiation and reflects none. The thermal radiation emitted by a black body is defined solely by its temperature, making the Stefan-Boltzmann Law a crucial component in calculating emitted thermal radiation.

Understanding this law is essential when dealing with systems that involve heat transfer by radiation, as it gives a quantitative measure of the energy radiated and allows for a comparison between different temperatures and materials.
Emissivity
Emissivity is a complex but crucial concept when analyzing thermal radiation. It is a measure of how efficiently a surface emits thermal radiation relative to a perfect black body. Emissivity values range between 0 and 1, where a perfect black body, which is an ideal emitter, has an emissivity of 1, and a perfect reflector, which does not emit radiation at all, has an emissivity of 0.

Most real materials fall somewhere in between these two extremes, and the emissivity will affect the actual radiation output according to the modified Stefan-Boltzmann Law: \( E = \text{ε} \times \text{σ} \times T^4 \), where \( \text{ε} \) represents the emissivity of the material. It's important to note that emissivity is not only dependent on the material but also on the surface texture and the temperature. A rough or oxidized surface typically has a higher emissivity than a shiny, polished surface.

For example, in the case of an aluminum shell used in a radiation test chamber, if the shell's inner surface is coated with carbon black, the emissivity is close to 1, vastly increasing the thermal radiation compared to the native aluminum surface, which has a much lower emissivity. The difference in emissivity dramatically changes the thermal radiation environment within the chamber, influencing any experiments or tests conducted.
Radiation Test Chamber
A radiation test chamber is a specialized environment used to study the effects of radiation on various materials or to simulate the conditions that an object might experience in outer space or other extreme environments. The key to replicating these conditions is precise control over the radiation levels within the chamber. By using emission properties of materials and understanding how they interact with Stefan-Boltzmann Law and emissivity, engineers and scientists can tailor the radiation environment inside the chamber.

A radiation test chamber is often designed to mimic a vacuum, reducing heat loss through conduction or convection. This focus on radiative heat transfer allows for accurate simulations and experiments. When the inner surface of such a chamber is coated with carbon black, it simulates a near-ideal black body, maximizing the irradiation. Such conditions are essential for calibrating instruments, testing thermal protection materials, or studying the behavior of systems intended for use in space.

A thorough understanding of the fundamental concepts like Stefan-Boltzmann Law and emissivity is hence essential for designing and interpreting the results from radiation test chambers.

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Most popular questions from this chapter

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.

Consider Problem \(4.51 .\) (a) The students are each given a flat, fot-surface silver mirror with which they collectively irradiate the wooden ship at location B. The reflection from the mirror is specular, and the silver's reflectivity is \(0.98\). The solar irradiation of each mirror, perpendicular to the direction of the sun's rays, is \(G_{S}=1000 \mathrm{~W} / \mathrm{m}^{2}\). How many students are needed to conduct the experiment if the solar absorptivity of the wood is \(\alpha_{w}=0.80\) and the mirror is oriented at an angle of \(45^{\circ}\) from the direction of \(G_{S}\) ? (b) If the students are given second-surface mirrors that consist of a sheet of plain glass that has polished silver on its back side, how many students are needed to conduct the experiment? Hint: See Problem \(12.62 .\)

A small disk \(5 \mathrm{~mm}\) in diameter is positioned at the center of an isothermal, hemispherical enclosure. The disk is diffuse and gray with an emissivity of \(0.7\) and is maintained at \(900 \mathrm{~K}\). The hemispherical enclosure, maintained at \(300 \mathrm{~K}\), has a radius of \(100 \mathrm{~mm}\) and an emissivity of \(0.85\). Calculate the radiant power leaving an aperture of diameter \(2 \mathrm{~mm}\) located on the enclosure as shown.

Solar irradiation of \(1100 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a large, flat, horizontal metal roof on a day when the wind blowing over the roof causes a convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outside air temperature is \(27^{\circ} \mathrm{C}\), the metal surface absorptivity for incident solar radiation is \(0.60\), the metal surface emissivity is \(0.20\), and the roof is well insulated from below. (a) Estimate the roof temperature under steady-state conditions. (b) Explore the effect of changes in the absorptivity, emissivity, and convection coefficient on the steady-state temperature. 12.108 Neglecting the effects of radiation absorption, emission, and scattering within their atmospheres, calculate the average temperature of Earth, Venus, and Mars assuming diffuse, gray behavior. The average distance from the sun of each of the three planets, \(L_{s p}\), along with their measured average temperatures, \(\bar{T}_{p}\), are shown in the table below. Based upon a comparison of the calculated and measured average temperatures, which planet is most affected by radiation transfer in its atmosphere? \begin{tabular}{lcc} \hline Planet & \(L_{x-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\ \hline Venus & \(1.08 \times 10^{11}\) & 735 \\ Earth & \(1.50 \times 10^{11}\) & 287 \\ Mars & \(2.30 \times 10^{11}\) & 227 \\ \hline \end{tabular}

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 \(\lambda \geq 3 \mu \mathrm{m}\).

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