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Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

Short Answer

Expert verified
To show the given formula for the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace, we differentiate Planck's Spectral Distribution equation with respect to λ and T and then calculate the derivative. We get the ratio equation as: \[ \frac{d I_{\lambda}/I_{\lambda}}{d T/T} = \frac{C_2}{\lambda T}\frac{1}{1-\exp \left(-C_2 / \lambda T\right)} \] For given conditions \(0.65 \mu m\) and \(2000 K\), we can find the allowable variation in temperature, dT, by calculating the value of \(\frac{dT}{T}\) and then multiplying the result by the temperature T (2000 K). For \(10 \mu m \), the steps to calculate the allowable variation in temperature will be similar.

Step by step solution

01

Understand Planck's Spectral Distribution equation

Planck's Spectral Distribution equation is given by $$ I_{\lambda} = \frac{2 \pi h c^2 }{\lambda^5}\frac{1}{\exp \left(\frac{hc}{\lambda k_B T}\right) - 1} $$ Here, \(I_{\lambda}\) denotes the spectral intensity, and h, c, kB, λ, and T represent Planck's constant, the speed of light, Boltzmann constant, wavelength, and temperature, respectively.
02

Differentiate the equation with respect to λ and T

Differentiate Planck's Spectral Distribution equation with respect to λ and T. $$ \frac{d I_{\lambda}}{d T} =\frac{2 \pi h c^2 }{\lambda^5}\frac{d}{dT}\left(\frac{1}{\exp \left(\frac{hc}{\lambda k_B T}\right) - 1}\right) $$ Let \(C_2=\frac{hc}{k_B}\), then the above expression becomes $$ \frac{d I_{\lambda}}{d T} = \frac{2 \pi h c^2 }{\lambda^5}\frac{d}{dT}\left(\frac{1}{\exp \left(\frac{C_2}{\lambda T}\right) - 1}\right) $$
03

Calculate the derivative

Calculate the derivative of the above equation. $$ \frac{d I_{\lambda}}{d T} = \frac{2 \pi h c^2 }{\lambda^5}\frac{\frac{C_2}{\lambda T^2}}{\left(\exp \left(\frac{C_2}{\lambda T}\right) - 1\right)^2}\cdot\exp \left(\frac{-C_2}{\lambda T}\right) $$ Now, we can re-write the equation as: $$ \frac{d I_{\lambda}}{d T} = \frac{C_2}{\lambda T^2}\frac{I_{\lambda}^2}{1-\exp \left(-\frac{C_2}{\lambda T}\right)} $$
04

Formulate the given ratio

Formulate the ratio equation as mentioned in the problem. $$ \frac{d I_{\lambda}/I_{\lambda}}{d T/T} = \frac{C_2}{\lambda T}\frac{1}{1-\exp \left(-C_2 / \lambda T\right)} $$
05

Calculate the allowable variation in temperature for given conditions

For \(0.65 \mu m (6.5\times10^{-7}m)\) and \(2000 K\), we have: $$\frac{d I_{\lambda} / I_{\lambda}}{d T / T} = 0.005 \text{(for } 0.5 \% \text{ change)}$$ Plugging values into the ratio equation, $$0.005 =\frac{C_2}{(6.5\times10^{-7}) (2000)}\frac{1}{1-\exp \left(-\frac{C_2}{(6.5\times10^{-7})(2000)}\right)}$$ From this expression, we can find the value of \(\frac{dT}{T}\): $$\frac{dT}{T} = \left(\frac{d I_{\lambda} / I_{\lambda}}{0.005}\right)$$ Then, we can find the allowable variation in temperature, dT, by multiplying the result by T (2000 K). Now, for \(10 \mu m (10^{-5}m)\), the steps to calculate the allowable variation in temperature will be similar. Firstly, plug the given values into the ratio equation, and then compute the value of \(\frac{dT}{T}\). Finally, to find the allowable variation in temperature (dT), multiply the result by the temperature (2000 K).

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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 fundamental principle in the field of quantum mechanics and thermal radiation. It describes how electromagnetic radiation is emitted by a blackbody in thermal equilibrium. A blackbody is an idealized object that perfectly absorbs and emits all frequencies of radiation. Planck's Law provides the mathematical framework to understand this phenomenon.

At the heart of Planck's Law is the formula for the spectral intensity of radiation, which can be written as:
  • \[ I_{\lambda} = \frac{2 \pi h c^2 }{\lambda^5}\frac{1}{\exp \left(\frac{hc}{\lambda k_B T}\right) - 1} \]
Here, each variable represents a fundamental component:
  • \(I_{\lambda}\) is the spectral intensity at a wavelength \(\lambda\).
  • \(h\) is Planck's constant.
  • \(c\) is the speed of light.
  • \(k_B\) is the Boltzmann constant.
  • \(T\) is the absolute temperature of the blackbody.
This equation highlights how the intensity of radiation depends on the wavelength and the temperature of the blackbody. Understanding Planck's Law is crucial for dealing with topics in radiation heat transfer and calibrating radiometric devices.
Spectral Intensity
Spectral intensity is a measure of how much energy is emitted from a surface at a particular wavelength per unit time, area, and solid angle. It offers insights into how energy is distributed across different wavelengths.

In the context of blackbody radiation, spectral intensity is a key quantity described by Planck's Law. The function is complex but essential, allowing you to predict how a blackbody emits radiation depending on its temperature and the specific wavelength you are interested in.
  • Spectral intensity can increase dramatically with temperature, indicating more energy being emitted.
  • At longer wavelengths, intensity tends to decrease, resulting in fewer energy emissions.
  • Changes in intensity are influenced by temperature and wavelength, demonstrating their interrelatedness.
  • Engineers and scientists use spectral intensity to design systems like thermal cameras and radiometers, which require accurate thermal readings.
These properties underline the importance of mastering spectral intensity, as it impacts everything from environmental monitoring to the development of heating systems.
Temperature Variation
Temperature variation is a crucial concept in understanding and controlling blackbody radiation. It directly affects the spectral intensity of emitted radiation. When dealing with devices such as isothermal furnaces, precise temperature control is required to ensure accurate measurements and consistency.

To grasp the impact of temperature variation, consider how the spectral intensity changes when temperature shifts. Using part of the solution equation,
  • \[ \frac{d I_{\lambda}/I_{\lambda}}{d T/T} = \frac{C_{2}}{\lambda T}\frac{1}{1-\exp \left(-C_{2}/\lambda T\right)} \]
You can see the need to carefully calibrate temperature. This ratio tells you the sensitivity of intensity to small changes in temperature for a given wavelength. It is pivotal in ensuring that radiometric devices function accurately.
  • Small temperature variances can lead to significant spectral intensity changes.
  • Managing temperature variation helps in maintaining the integrity of measurements.
  • In calibration processes, keeping variations in check leads to reliable device outputs.
Understanding temperature variation is essential for applications needing high precision, such as optical measurements and scientific experiments.
Heat Flux Calibration
Heat flux calibration involves the accurate measurement and adjustment of heat transfer rates, often using blackbody sources. This process ensures that radiometric devices provide correct readings, critical in various industrial and scientific applications.

Calibration relies heavily on Planck's Law and the principles of spectral intensity and temperature control. Accurate heat flux measurements are essential when using instruments like heat flux gages and radiation thermometers.
Here are the steps and considerations in the calibration process:
  • The isothermal furnace's temperature must remain stable, as fluctuations can skew results.
  • Small apertures that mimic a blackbody are often used, as they provide a consistent radiation source.
  • Heat flux needs to be adjusted to match the desired levels, usually requiring iterative modifications.
  • Devices are calibrated over specific wavelengths, considering atmospheric conditions and material properties.
Such meticulous calibration practices ensure that the readings reflect true physical conditions, allowing for precision in subsequent analyses and applications.

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Most popular questions from this chapter

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 \(0.05 \leq \varepsilon_{x} \leq 0.35\).

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 \(P r=0.72\).

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 \(1 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\).

The 50 -mm peephole of a large furnace operating at \(450^{\circ} \mathrm{C}\) is covered with a material having \(\tau=0.8\) and \(\rho=0\) for irradiation originating from the furnace. The material has an emissivity of \(0.8\) and is opaque to irradiation from a source at room temperature. The outer surface of the cover is exposed to surroundings and ambient air at \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2}=\mathrm{K}\). Assuming that convection effects on the inner surface of the cover are negligible, calculate the heat loss by the furnace and the temperature of the cover.

Consider an opaque, gray surface whose directional absorptivity is \(0.8\) for \(0 \leq \theta \leq 60^{\circ}\) and \(0.1\) for \(\theta>60^{\circ}\). The surface is horizontal and exposed to solar irradiation comprised of direct and diffuse components. (a) What is the surface absorptivity to direct solar radiation that is incident at an angle of \(45^{\circ}\) from the normal? What is the absorptivity to diffuse irradiation? (b) Neglecting convection heat transfer between the surface and the surrounding air, what would be the equilibrium temperature of the surface if the direct and diffuse components of the irradiation were 600 and \(100 \mathrm{~W} / \mathrm{m}^{2}\), respectively? The back side of the surface is insulated.

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