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A small metal object, initially at \(T_{l}=1000 \mathrm{~K}\), is cooled by radiation in a low-temperature vacuum chamber. One of two thin coatings can be applied to the object so that spectral hemispherical emissivities vary with wavelength as shown. For which coating will the object most rapidly reach a temperature of \(T_{f}=500 \mathrm{~K} ?\)

Short Answer

Expert verified
To determine which coating will help the object most rapidly reach a temperature of 500 K, we should first calculate the effective emissivity for each coating by integrating the product of Planck's Law and the emissivity function for each coating over the entire wavelength range. Then, we can calculate the cooling rate for the object with each coating by applying the Stefan-Boltzmann equation. The coating that results in the higher cooling rate will help the object reach the target temperature faster. Since we don't have the detailed emissivity functions, we cannot determine the specific coating numerically, but the outlined solution steps can be applied with proper emissivity functions provided.

Step by step solution

01

Understand the Planck's Law and Stefan-Boltzmann equation

To begin, it's important to understand Planck's Law which describes the intensity of radiation emitted by a black body at different wavelengths. It can be expressed as: \(E_b(\lambda ,T)=\frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} -1}\) Where: - \(E_b(\lambda ,T)\) is the spectral radiance per unit wavelength - \(\lambda\) is the wavelength - \(T\) is the temperature - \(h\) is the Planck's constant - \(c\) is the speed of light - \(k_B\) is the Boltzmann constant Now, we can consider Stefan-Boltzmann equation for the total emissive power of a black body: \(E_b = \sigma T^4\) Where: - \(E_b\) is the total emissive power - \(\sigma\) is the Stefan-Boltzmann constant - \(T\) is the temperature
02

Relate Planck's Law and Stefan-Boltzmann equation to the problem

In our problem, the object is not a black body, so we need to consider its emissivity \((\varepsilon)\). The emissive power \(E\) can be related to \(E_b\) by: \(E=\varepsilon E_b\) Now, we can define the total emissivity for each coating as a function of wavelength and integrate this product over the entire range of wavelengths. This will give us the effective emissivity for each coating, which can be used to calculate the cooling rate.
03

Calculate the effective emissivity for each coating

We will integrate the product of Planck's Law and the emissivity for each coating over the entire wavelength range (0 to infinity) to find the effective emissivity for each coating. Denote emissivity function for coating 1 as \(\varepsilon_1(\lambda)\) and for coating 2 as \(\varepsilon_2(\lambda)\). Then the effective emissivity is: \(\varepsilon_{eff,1}=\int\limits_0^\infty \varepsilon_1(\lambda)E_b(\lambda ,T)d\lambda\) \(\varepsilon_{eff,2}=\int\limits_0^\infty \varepsilon_2(\lambda)E_b(\lambda ,T)d\lambda\)
04

Apply the effective emissivity to calculate cooling rate

Now, using the effective emissivities for each coating, we can calculate the cooling rate for the object with each coating by applying the Stefan-Boltzmann equation: \(E_{1}=\varepsilon_{eff,1}\sigma T^{4}\) \(E_{2}=\varepsilon_{eff,2}\sigma T^{4}\) The cooling rate is determined by the difference in emissive power. If \(E_1 > E_2\), then the object with coating 1 will cool faster. If \(E_2 > E_1\), then the object with coating 2 will cool faster. The coating with the higher cooling rate will help the object reach a temperature of 500 K most rapidly. Due to the missing detailed emissivity function, we cannot calculate the cooling rates numerically. However, the solution steps can be applied with proper emissivity functions given.

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Key Concepts

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

Planck's Law
Planck's Law is a cornerstone of understanding radiative cooling, as it describes the electromagnetic radiation emitted by a black body in thermal equilibrium. This law explains how the intensity of radiation varies with wavelength at a given temperature.
The core formula of Planck's Law is:
  • \(E_b(\lambda ,T)=\frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} -1}\)
Here, each variable plays a significant role:
  • \(E_b(\lambda ,T)\): Spectral radiance per unit wavelength is the amount of energy emitted at each wavelength.
  • \(\lambda\): The wavelength of emitted radiation. Shorter wavelengths generally involve higher energy emissions.
  • \(T\): The absolute temperature of the emitting body, in Kelvins. As temperature increases, the intensity and peak wavelength of radiation change.
  • \(h\): Planck's constant, a fundamental constant in quantum mechanics.
  • \(c\): The speed of light, crucial since radiative processes involve electromagnetic waves.
  • \(k_B\): Boltzmann constant, linking temperature to energy.
Planck's Law highlights that at higher temperatures, objects emit radiation more intensely and at shorter wavelengths. It is crucial for calculating how objects like the metal one in our exercise cool over time.
Stefan-Boltzmann Equation
The Stefan-Boltzmann Equation is integral to calculating the total amount of energy emitted by a black body as it cools. This provides a macroscopic view of radiative cooling by summing emissions over all wavelengths.
This equation is expressed as:
  • \(E_b = \sigma T^4\)
Where:
  • \(E_b\): Total emissive power or the power radiated per unit area by the object.
  • \(\sigma\): Stefan-Boltzmann constant, which ensures that units are consistent when relating temperature to energy.
  • \(T\): Temperature, raised to the fourth power, indicating how sensitive radiative power is to changes in temperature.
The equation shows that the energy emitted by a body increases with the fourth power of the temperature. So, small changes in temperature can significantly affect the energy output. In our problem, the Stefan-Boltzmann equation helps determine which coating will make the object cool faster, by comparing the effective emissive power with different coatings.
Emissivity
Emissivity is a crucial concept when dealing with real-world objects, like the metal in our exercise, since they aren't perfect black bodies. It represents the efficiency with which an object emits thermal radiation, compared to an ideal black body.
  • Definition: Emissivity \((\varepsilon)\) is a dimensionless value ranging from 0 to 1.
  • \(\varepsilon = 1\): The object behaves like a perfect black body, emitting maximum possible radiation at its temperature.
  • \(\varepsilon = 0\): The object is completely non-emissive, reflecting all radiation without absorbing any.
To understand cooling rates, we consider the effective emissivity, which is a product of spectral emissivity and Planck’s radiative distribution. This involves integrating emissivity across all wavelengths:
  • \(\varepsilon_{eff} = \int\limits_0^\infty \varepsilon(\lambda)E_b(\lambda ,T)d\lambda\)
Effective emissivity determines how much of the Planck's law predicted radiation is actually emitted by the coated metal object. The coating with higher effective emissivity will radiate energy more efficiently, thus cooling the object more rapidly. Understanding emissivity helps select the optimal coating for quicker temperature reduction, as seen in the objective of the exercise.

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Most popular questions from this chapter

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}

A diffuse, opaque surface at \(700 \mathrm{~K}\) has spectral emissivities of \(\varepsilon_{\lambda}=0\) for \(0 \leq \lambda \leq 3 \mu \mathrm{m}, \varepsilon_{\lambda}=0.5\) for \(3 \mu \mathrm{m}<\lambda \leq 10 \mu \mathrm{m}\), and \(\varepsilon_{\lambda}=0.9\) for \(10 \mu \mathrm{m}<\lambda<\infty\). A radiant flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), which is uniformly distributed between 1 and \(6 \mu \mathrm{m}\), is incident on the surface at an angle of \(30^{\circ}\) relative to the surface normal. Calculate the total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) area of the surface that reaches a radiation detector positioned along the normal to the area. The aperture of the detector is \(10^{-5} \mathrm{~m}^{2}\), and its distance from the surface is \(1 \mathrm{~m}\).

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

A cylinder of \(30-\mathrm{mm}\) diameter and \(150-\mathrm{mm}\) length is heated in a large furnace having walls at \(1000 \mathrm{~K}\), while air at \(400 \mathrm{~K}\) is circulating at \(3 \mathrm{~m} / \mathrm{s}\). Estimate the steady-state cylinder temperature under the following specified conditions. (a) The cylinder is in cross flow, and its surface is diffuse and gray with an emissivity of \(0.5\). (b) The cylinder is in cross flow, but its surface is spectrally selective with \(\alpha_{\lambda}=0.1\) for \(\lambda \leq 3 \mu \mathrm{m}\) and \(\alpha_{\lambda}=0.5\) for \(\lambda>3 \mu \mathrm{m}\). (c) The cylinder surface is positioned such that the airflow is longitudinal and its surface is diffuse and gray. (d) For the conditions of part (a), compute and plot the cylinder temperature as a function of the air velocity for \(1 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\).

One scheme for extending the operation of gas turbine blades to higher temperatures involves applying a ceramic coating to the surfaces of blades fabricated from a superalloy such as inconel. To assess the reliability of such coatings, an apparatus has been developed for testing samples under laboratory conditions. The sample is placed at the bottom of a large vacuum chamber whose walls are cryogenically cooled and which is equipped with a radiation detector at the top surface. The detector has a surface area of \(A_{d}=10^{-5} \mathrm{~m}^{2}\), is located at a distance of \(L_{\text {sl }}=1 \mathrm{~m}\) from the sample, and views radiation originating from a portion of the ceramic surface having an area of \(\Delta A_{c}=10^{-4} \mathrm{~m}^{2}\). An electric heater attached to the bottom of the sample dissipates a uniform heat flux, \(q_{b}^{\prime \prime}\), which is transferred upward through the sample. The bottom of the heater and sides of the sample are well insulated. Consider conditions for which a ceramic coating of thickness \(L_{c}=0.5 \mathrm{~mm}\) and thermal conductivity \(k_{c}=\) \(6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has been sprayed on a metal substrate of thickness \(L_{s}=8 \mathrm{~mm}\) and thermal conductivity \(k_{s}=\) \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The opaque surface of the ceramic may be approximated as diffuse and gray, with a total, hemispherical emissivity of \(\varepsilon_{c}=0.8\). (a) Consider steady-state conditions for which the bottom surface of the substrate is maintained at \(T_{1}=1500 \mathrm{~K}\), while the chamber walls (including the surface of the radiation detector) are maintained at \(T_{w}=90 \mathrm{~K}\). Assuming negligible thermal contact resistance at the ceramic- substrate interface, determine the ceramic top surface temperature \(T_{2}\) and the heat flux \(q_{b}^{\prime \prime}\). (b) For the prescribed conditions, what is the rate at which radiation emitted by the ceramic is intercepted by the detector?

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