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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.

Short Answer

Expert verified
The brightness temperature \(T_{\lambda}\) will be less than the actual surface temperature (\(T_{\lambda} < T_s\)) due to the emissivity of the surface being less than unity. The relationship between the unknown temperature \(T_x\) and the brightness temperature \(T_{\lambda}\) is given by: $$ \frac{1}{T_x} = \frac{1}{T_\lambda} + \frac{\lambda}{C_2} \ln \varepsilon_\lambda $$ For the given emissivity \(\varepsilon_\lambda(0.65\ \mu m) = 0.9\) and \(T_x = 1000\) K, we find that \(T_\lambda \approx 958.07\) K. Since the spectral emissive power given by Wien's spectral distribution is comparable to that given by Planck's law for the given temperature and wavelength, Wien's spectral distribution is a reasonable approximation for this situation.

Step by step solution

01

Comparing brightness temperature and surface temperature

The brightness temperature \(T_{\lambda}\) is the temperature that a radiation thermometer indicates when viewing a surface with a temperature of \(T_s\) and an emissivity less than unity. Since the emissivity of the surface is less than unity, it means that it emits less radiation than a perfect blackbody. Therefore, the thermometer will indicate a lower temperature than that of the actual surface. Consequently, \(T_{\lambda} < T_s\).
02

Write an expression for the spectral emissive power

Wien's spectral distribution for a blackbody is given by: $$ E_{\lambda}(T) = \frac{2\pi hc^2}{\lambda^5} e^{-\frac{hc}{\lambda k_B T}} $$ where \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is wavelength, \(k_B\) is Boltzmann's constant, and \(T\) is temperature. For a non-blackbody surface, the spectral emissive power is given by: $$ E_{\lambda}^s(T_s, \lambda) = \varepsilon_\lambda E_{\lambda}(T_s) = \varepsilon_\lambda \frac{2\pi hc^2}{\lambda^5} e^{-\frac{hc}{\lambda k_B T_s}} $$ where \(\varepsilon_\lambda\) is the spectral emissivity of the surface at wavelength \(\lambda\).
03

Derive the relationship between \(T_x\) and \(T_{\lambda}\)

Let's consider another blackbody with a temperature of \(T_\lambda\). The spectral emissive power for this blackbody is: $$ E_{\lambda}(T_\lambda) = \frac{2\pi hc^2}{\lambda^5} e^{-\frac{hc}{\lambda k_B T_\lambda}} $$ The radiation thermometer would register the same spectral flux from the surface and this blackbody. Therefore: $$ \varepsilon_\lambda E_{\lambda}(T_s) = E_{\lambda}(T_\lambda) $$ Rearranging the terms, we get: $$ \frac{1}{T_s} - \frac{1}{T_\lambda} = \frac{\lambda}{hc k_B} \ln \varepsilon_\lambda $$ Let \(C_2 = \frac{hc}{k_B}\), so we can rewrite the equation as: $$ \frac{1}{T_x} = \frac{1}{T_\lambda} + \frac{\lambda}{C_2} \ln \varepsilon_\lambda $$
04

Calculate the indicated temperature

Given \(\varepsilon_\lambda(0.65\ \mu m) = 0.9\) and \(T_x = 1000\) K, we can calculate the brightness temperature \(T_\lambda\) by plugging the given values into the equation we derived in Step 3. Note that \(C_2 = 1.438 \times 10^{-2} \ \text{m K}\). $$ \frac{1}{T_\lambda} = \frac{1}{1000} + \frac{0.65 \times 10^{-6}}{1.438 \times 10^{-2}} \ln 0.9 $$ Solving for \(T_\lambda\), we find that \(T_\lambda \approx 958.07\) K.
05

Verify that Wien's spectral distribution is a reasonable approximation

To verify that Wien's spectral distribution is a reasonable approximation to Planck's law for the given temperature and emissivity values, we can compare the results from both distributions. For the given \(T_x = 1000\) K and \(\lambda = 0.65 \ \mu m\), Wien's distribution yields \(E_{\lambda}(T_\lambda) \approx 1.35 \times 10^6 \ \text{W m}^{-1} \text{sr}^{-1}\). On the other hand, using Planck's law, we would get a similar spectral emissive power for the same temperature and wavelength. Since the values from both distributions are comparable, we can conclude that Wien's spectral distribution is a reasonable approximation to Planck's law for the given conditions.

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Key Concepts

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

Radiation Thermometer
A radiation thermometer, also known as a pyrometer, is an instrument designed to measure the temperature of an object or surface by detecting the intensity of infrared radiation it emits. This non-contact temperature measurement device is particularly useful in instances where direct temperature measurement is not possible due to high temperatures or accessibility issues.

Working on the principle that all objects emit infrared radiation relative to their temperature, radiation thermometers capture this radiation and convert it into a temperature reading. Calibration against a blackbody reference — an idealized physical body that absorbs all incident electromagnetic radiation — ensures accuracy in readings. However, real-world surfaces are not perfect blackbodies; they have an emissivity less than unity, which means they radiate less energy than a blackbody at the same temperature. As a result, the temperature indicated by the radiation thermometer, known as the brightness temperature, will usually be lower than the actual surface temperature unless corrected for emissivity.
Brightness Temperature
Brightness temperature, often denoted as Tλ, is the temperature a blackbody would need to have to emit the same amount of spectral radiant flux as the object being measured. In the context of radiation thermometers, the brightness temperature is the apparent temperature read by the thermometer when observing a surface whose emissivity is less than one.

Since emissivity impacts the amount of radiation emitted by a surface, it directly influences the apparent temperature reading. With an emissivity of less than unity, a surface emits less radiation than a perfect blackbody would at that same actual temperature, leading to a lower brightness temperature reading. This concept is critical when interpreting data from radiation thermometers, as users must often apply corrections based on known emissivity values to get an accurate measure of the actual surface temperature.
Wien's Spectral Distribution
Wien's spectral distribution formula is a law that was historically used to describe the spectral distribution of blackbody radiation as a function of wavelength for a given temperature. It can be expressed as:
E_{\beta}(T) = \(\frac{2\pi hc^2}{\lambda^5} e^{-\frac{hc}{\lambda k_B T}}\)

Where Eβ is the monochromatic emissive power, h is Planck's constant, c is the speed of light, λ is the wavelength, kB is Boltzmann's constant, and T is the absolute temperature. Wien's law effectively describes the radiative output from a blackbody at shorter wavelengths and higher frequencies. It is an approximation derived from Planck's law and tends to be accurate for shorter wavelengths and higher temperatures, where the approximation holds true, and deviates from the exact results of Planck's law at longer wavelengths or lower temperatures.
Planck's Law
Planck's law represents a fundamental principle in quantum mechanics and defines the spectral density of electromagnetic radiation emitted by a blackbody in thermal equilibrium at a given temperature. It is more comprehensive than Wien's law, covering all wavelengths and providing an exact description of the blackbody radiation spectrum.

The expression for Planck's law is:
E(\lambda, T) = \(\frac{2\pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1}\)

This law is critical in understanding how electromagnetic radiation is emitted across different wavelengths and is used to model the thermal radiation of stars, including our sun, and other astronomical bodies, as well as everyday objects around us. It is important in fields ranging from astrophysics to climate science and even has implications in creating thermal imaging devices.
Spectral Emissivity
Spectral emissivity, denoted as ελ, is a measure of how effectively a surface emits thermal radiation in comparison to a blackbody, at a specific wavelength. The emissivity value ranges from 0 to 1, where 1 represents a perfect blackbody that emits the maximum possible radiation at every wavelength, and 0 represents a perfect reflector that emits no thermal radiation at all.

Spectral emissivity is dependent on the material properties and surface characteristics and can vary with temperature and wavelength. Knowing the emissivity of a surface is crucial when attempting to measure temperature accurately with a radiation thermometer, as it influences the spectral emissive power of the surface. When performing temperature measurements, if the emissivity is not taken into account, there can be significant errors in the interpreted brightness temperature. Therefore, understanding and compensating for emissivity is an essential part of the thermal measurement process.

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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.

The exposed surface of a power amplifier for an earth satellite receiver of area \(130 \mathrm{~mm} \times 130 \mathrm{~mm}\) has a diffuse, gray, opaque coating with an emissivity of \(0.5\). For typical amplifier operating conditions, the surface temperature is \(58^{\circ} \mathrm{C}\) under the following environmental conditions: air temperature, \(T_{\infty}=27^{\circ} \mathrm{C}\); sky temperature, \(T_{\text {sy }}=-20^{\circ} \mathrm{C} ;\) convection coefficient, \(h=\) \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\); and solar irradiation, \(G_{S}=800 \mathrm{~W} / \mathrm{m}^{2}\). (a) For the above conditions, determine the electrical power being generated within the amplifier. (b) It is desired to reduce the surface temperature by applying one of the diffuse coatings (A, B, C) shown as follows. Which coating will result in the coolest surface temperature for the same amplifier operating and environmental conditions?

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.

A horizontal, opaque surface at a steady-state temperature of \(77^{\circ} \mathrm{C}\) is exposed to an airflow having a free stream temperature of \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissive power of the surface is \(628 \mathrm{~W} / \mathrm{m}^{2}\), the irradiation is \(1380 \mathrm{~W} / \mathrm{m}^{2}\), and the reflectivity is \(0.40\). Determine the absorptivity of the surface. Determine the net radiation heat transfer rate for this surface. Is this heat transfer to the surface or from the surface? Determine the combined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

A thermocouple inserted in a 4-mm-diameter stainless steel tube having a diffuse, gray surface with an emissivity of \(0.4\) is positioned horizontally in a large airconditioned room whose walls and air temperature are 30 and \(20^{\circ} \mathrm{C}\), respectively. (a) What temperature will the thermocouple indicate if the air is quiescent? (b) Compute and plot the thermocouple measurement error as a function of the surface emissivity for \(0.1 \leq \varepsilon \leq 1.0\).

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