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Chapter 12: Problem 89

A two-color pyrometer is a device that is used to measure the temperature of a diffuse surface, \(T_{s}\). The device measures the spectral, directional intensity emitted by the surface at two distinct wavelengths separated by \(\Delta \lambda\). Calculate and plot the ratio of the intensities \(I_{\lambda+\Delta, e}\left(\lambda+\Delta \lambda, \theta, \phi, T_{s}\right)\) and \(I_{\lambda, e}\left(\lambda, \theta, \phi, T_{s}\right)\) as a function of the surface temperature over the range \(500 \mathrm{~K} \leq T_{s} \leq 1000 \mathrm{~K}\) for \(\lambda=5 \mu \mathrm{m}\) and \(\Delta \lambda=0.1,0.5\), and \(1 \mu \mathrm{m}\). Comment on the sensitivity to temperature and on whether the ratio depends on the emissivity of the surface. Discuss the tradeoffs associated with specification of the various values of \(\Delta \lambda\). Hint: The change in the emissivity over small wavelength intervals is modest for most solids, as evident in Figure 12.17.

Short Answer

Expert verified
The intensity ratio of a two-color pyrometer is given by \(\dfrac{I_{\lambda+\Delta, e}}{I_{\lambda, e}} = \dfrac{\dfrac{1}{e^{\frac{hc}{(\lambda+\Delta\lambda) kT}}-1}}{\dfrac{1}{e^{\frac{hc}{\lambda kT}}-1}}\), where we have assumed the change in emissivity over small wavelength intervals is modest. Calculate the intensity ratio as a function of surface temperature over the range 500K to 1000K for 位 = 5 渭m and 螖位 = 0.1, 0.5, and 1 渭m. Plot the intensity ratio against the surface temperature and analyze the sensitivity to temperature and the role of emissivity. Note that tradeoffs exist for smaller and larger 螖位 values in terms of sensitivity to temperature change, device usability, and assumptions regarding emissivity.

Step by step solution

01

Setup the Intensity Ratio Expression

For the given problem, we are given the spectral intensity at the two wavelengths: \(I_{\lambda, e}\left(\lambda, \theta, \phi, T_{s}\right)\) and \(I_{\lambda+\Delta, e}\left(\lambda+\Delta \lambda, \theta, \phi, T_{s}\right)\). Using the Plank's law equation for both intensities, we get the expression for the intensity ratio as: \(\dfrac{I_{\lambda+\Delta, e}}{I_{\lambda, e}} = \dfrac{\epsilon(\lambda+\Delta\lambda) \dfrac{1}{e^{\frac{hc}{(\lambda+\Delta\lambda) kT}}-1}}{\epsilon(\lambda) \dfrac{1}{e^{\frac{hc}{\lambda kT}}-1}}\)
02

Apply the Hint

The hint suggests that the change in emissivity over small wavelength intervals is modest for most solids. Therefore, for our calculations, we can assume that the emissivities are approximately equal at the two wavelengths (\((蔚(\lambda+\Delta\lambda) \approx 蔚(\lambda)\))): \(\dfrac{I_{\lambda+\Delta, e}}{I_{\lambda, e}} = \dfrac{\dfrac{1}{e^{\frac{hc}{(\lambda+\Delta\lambda) kT}}-1}}{\dfrac{1}{e^{\frac{hc}{\lambda kT}}-1}}\)
03

Calculate the Ratio for Each 螖位

Now that we have the intensity ratio expression, plug in the values of 位 and 螖位 to find the intensity ratio. Calculate the intensity ratio as a function of the surface temperature over the range 500K to 1000K for 位 = 5 渭m and 螖位 = 0.1, 0.5, and 1 渭m.
04

Plot the Intensity Ratio

Plot the intensity ratio against the surface temperature for each value of 螖位. This will help much in the discussion we will do in step 5.
05

Discuss the Results

Based on the plots created in Step 4, we can discuss the sensitivity of the intensity ratio to temperature and whether the ratio depends on the emissivity of the surface. 1. Sensitivity to temperature: Compare the slopes of the intensity ratio as a function of temperature for the three values of 螖位. A steeper slope indicates higher sensitivity to temperature change, whereas a flatter slope indicates lower sensitivity. 2. Dependence on emissivity: As the change in emissivity is assumed to be modest, the ratio should not significantly depend on the emissivity of the surface. 3. Tradeoffs associated with specification of the various values of 螖位: Discuss the influence of having smaller or larger values for 螖位. A smaller 螖位 value might lead to a lower sensitivity to temperature change, while a larger 螖位 value might cause a higher sensitivity to temperature change. However, a larger 螖位 might also have implications on the device's usability and accuracy, as well as the assumption made about the emissivity. It is important to balance sensitivity to temperature change with practically choosing a 螖位 value. By completing these steps and discussing the results, we can better understand the relationships between the intensity ratio, temperature, and emissivity, as well as the trade-offs associated with choosing different 螖位 values.

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Key Concepts

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

Spectral Intensity Ratio
In two-color pyrometry, the spectral intensity ratio is a critical element used to determine the surface temperature. This ratio involves measurements of the emitted light at two different wavelengths close to each other. By using Planck's Law, we find the expression for the intensity ratio:
  • \[\dfrac{I_{\lambda+\Delta, e}}{I_{\lambda, e}} = \dfrac{\epsilon(\lambda+\Delta\lambda) \dfrac{1}{e^{\frac{hc}{(\lambda+\Delta\lambda) kT}}-1}}{\epsilon(\lambda) \dfrac{1}{e^{\frac{hc}{\lambda kT}}-1}}\]
In practical terms, this formula helps us know the intensity ratio as a function of surface temperature, using wavelengths
given as \(\lambda+\Delta\lambda\) and \(\lambda\). By assuming identical emissivity at both wavelengths due to small ranges,
this ratio simplifies our calculations, reducing errors potentially caused by changes in emissivity.
Temperature Sensitivity
Temperature sensitivity is crucial when assessing the effectiveness of a two-color pyrometer. It describes how responsive the spectral intensity ratio is to changes in temperature. When plotting the intensity ratio over a temperature range, the slope of the plot reveals sensitivity:
  • A steeper slope indicates high sensitivity.
  • A flatter slope suggests low sensitivity.
This sensitivity varies with the selection of \(\Delta\lambda\):
  • Smaller \(\Delta\lambda\) values may yield reduced temperature sensitivity.
  • Larger \(\Delta\lambda\) values enhance sensitivity.
However, practical constraints need consideration; a larger \(\Delta\lambda\) might not always be usable due to assumed emissivity consistency, impacting overall accuracy. Balancing these factors is key to maximizing effectiveness.
Emissivity Dependence
In the realm of spectral intensity measurement, emissivity is a measure of a material's ability to emit energy compared to a perfect black body. Emissivity can vary with wavelength, but within small intervals, this variation is often negligible. Leveraging this characteristic assists in assuming that emissivity
at \(\lambda\) and \(\lambda+\Delta\lambda\) is approximately equal, simplifying calculations:
  • The simplified intensity ratio becomes:\[\dfrac{I_{\lambda+\Delta, e}}{I_{\lambda, e}} = \dfrac{\dfrac{1}{e^{\frac{hc}{(\lambda+\Delta\lambda) kT}}-1}}{\dfrac{1}{e^{\frac{hc}{\lambda kT}}-1}}\]
This approach minimizes the influence of emissivity on the ratio, allowing the two-color pyrometer to focus more on temperature variations rather than dealing with differences in emissivity. Thus, assuming the change in emissivity is modest aids in achieving a more reliable temperature measurement.

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Most popular questions from this chapter

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 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.

An opaque, horizontal plate has a thickness of \(L=21 \mathrm{~mm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Water flows adjacent to the bottom of the plate and is at a temperature of \(T_{x, w}=25^{\circ} \mathrm{C}\). Air flows above the plate at \(T_{x, a}=260^{\circ} \mathrm{C}\) with \(h_{a}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The top of the plate is diffuse and is irradiated with \(G=1450 \mathrm{~W} / \mathrm{m}^{2}\), of which \(435 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The steady-state top and bottom plate temperatures are \(T_{t}=43^{\circ} \mathrm{C}\) and \(T_{b}=35^{\circ} \mathrm{C}\), respectively. Determine the transmissivity, reflectivity, absorptivity, and emissivity of the plate. Is the plate gray? What is the radiosity associated with the top of the plate? What is the convection heat transfer coefficient associated with the water flow?

Solar radiation incident on the earth's surface may be divided into the direct and diffuse components described in Problem 12.9. Consider conditions for a day in which the intensity of the direct solar radiation is \(I_{\text {dir }}=210 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\) in the solid angle subtended by the sun with respect to the earth, \(\Delta \omega_{s}=6.74 \times 10^{-5} \mathrm{sr}\). The intensity of the diffuse radiation is \(I_{\text {dif }}=70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\). (a) What is the total solar irradiation at the earth's surface when the direct radiation is incident at \(\theta=30^{\circ}\) ? (b) Verify the prescribed value for \(\Delta \omega_{s}\), recognizing that the diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m}\) and the distance between the sun and the earth is \(1.496 \times 10^{11} \mathrm{~m}\) (1 astronomical unit).

Solar flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\) is incident on the top side of a plate whose surface has a solar absorptivity of \(0.9\) and an emissivity of \(0.1\). The air and surroundings are at \(17^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the plate and air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the bottom side of the plate is insulated, determine the steady-state temperature of the plate.
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