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Chapter 12: Problem 54

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 \(1 \mathrm{~m}\).

Short Answer

Expert verified
The total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) area of the surface that reaches the radiation detector positioned along the normal to the area can be calculated using the following formula: \(P_{\mathrm{detected}} =\frac{ P_{\mathrm{emitted}} \times P_{\mathrm{incident}} \times 10^{-5} \mathrm{~m}^{2}}{1^{2} \mathrm{~m}^{2} }\) where \(P_{\mathrm{emitted}}\) is the power emitted by the surface, and \(P_{\mathrm{incident}}\) is the power incident on the surface. After calculating these values using the given information and plugging them into the formula, you can solve for the total radiant power, \(P_{\mathrm{detected}}\).

Step by step solution

01

Calculate the blackbody spectral radiance emitted by the surface at different wavelengths

For an opaque surface with temperature T = 700 K and given spectral emissivities, we can calculate the blackbody spectral radiance, B位, for different wavelengths using Planck's law: \(B_{\lambda}(\lambda, T) = \frac{2 \pi h c^2 }{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1}\) where, h = 6.626 x 10^{-34} J路s (Planck's constant) c = 3 x 10^8 m/s (speed of light) kB = 1.381 x 10^{-23} J/K (Boltzmann's constant) 位 = wavelength T = 700K
02

Calculate the radiative energy emitted by the surface at different wavelengths

We know the blackbody spectral radiance emitted at different wavelengths. Now, we multiply it with the given spectral emissivities 蔚(位) to calculate the radiative energy emitted. Emissivities: \(0\leq\lambda\leq3\mu m\) We have 蔚(位) = 0 \(3\mu m<\lambda\leq10\mu m\) We have 蔚(位) = 0.5 \(10\mu m<\lambda<\infty\) We have 蔚(位) = 0.9 We also know the area of the surface A = 10^{-4} m^2. So, the total emitted power P_emitted at different wavelengths will be: \(P_{\mathrm{emitted}}(\lambda) = \varepsilon_{\lambda}B_{\lambda}A\)
03

Calculate the power incident on the surface

We are given that a radiant flux of 1000 W/m虏 is uniformly distributed between 1 and 6 渭m and incident on the surface at an angle 30掳 relative to the surface normal. So, the power incident on the surface, P_incident, can be calculated as: \(P_{\mathrm{incident}} = F_{\mathrm{incident}} \times A \times \cos(\theta) = 1000 \mathrm{~W} / \mathrm{m}^{2} \times 10^{-4} \mathrm{~m}^{2} \times \cos(30^{\circ})\)
04

Calculate the total emitted power reaching the detector

The power detected by the detector can be calculated by taking the ratio of the product of emitted and incident power to the power emitted by the surface. Now we multiply by the aperture area of the detector, 10^{-5} m虏, and divide it by the square of the distance between the surface and the detector, 1虏 m虏. \(P_{\mathrm{detected}} =\frac{ P_{\mathrm{emitted}} \times P_{\mathrm{incident}} \times 10^{-5} \mathrm{~m}^{2}}{1^{2} \mathrm{~m}^{2} }\) Now, we just need to plug all the calculated values for P_emitted and P_incident from Steps 2 and 3 into the above formula to find the total emitted power that reaches the detector and solve for P_detected.

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Key Concepts

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

Emissivity
Emissivity is a measure of how effectively a surface emits thermal radiation. It is symbolized by \( \varepsilon \) and ranges from 0 to 1. A value of 1 indicates a perfect emitter, known as a blackbody, while 0 represents a non-emissive surface. In practice, most surfaces have emissivity values between these extremes.

For the given surface in the exercise, the spectral emissivity is defined across different wavelength intervals. For example:
  • \(0 \leq \lambda \leq 3 \mu m\): \(\varepsilon_{\lambda} = 0\) - The surface does not emit radiation in this range.
  • \(3 \mu m < \lambda \leq 10 \mu m\): \(\varepsilon_{\lambda} = 0.5\) - The surface emits at half the blackbody rate in this range.
  • \(10 \mu m < \lambda < \infty\): \(\varepsilon_{\lambda} = 0.9\) - The surface emits at 90% of the blackbody rate.
These values are crucial for determining how much heat the surface emits, impacting calculations of radiant heat transfer.
Planck's Law
Planck's Law is fundamental in understanding blackbody radiation. It describes the distribution of electromagnetic radiation emitted by a blackbody in thermal equilibrium at a definite temperature. The formula is given by:\[B_{\lambda}(\lambda, T) = \frac{2 \pi h c^2 }{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_BT}} - 1}\]Here,
  • \(h\) is Planck鈥檚 constant (\(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\)).
  • \(c\) is the speed of light (\(3 \times 10^8 \text{ m/s}\)).
  • \(\lambda\) is the wavelength.
  • \(T\) is the temperature in Kelvin.
  • \(k_B\) is Boltzmann's constant (\(1.381 \times 10^{-23} \text{ J/K}\)).
This law helps to compute the spectral radiance \(B_\lambda\), which is the power emitted per unit area, per unit wavelength, and per unit solid angle. For surfaces that aren't perfect blackbodies, emissivity is applied to adjust the radiance according to the material鈥檚 properties.
Spectral Radiance
Spectral radiance, denoted \(B_\lambda\), measures the power emitted by a surface per unit area per unit wavelength per unit solid angle. It provides insight into how much energy is radiated at particular wavelengths and is crucial for understanding radiant heat transfer.

By using Planck's Law combined with emissivity, the calculation of spectral radiance helps predict how real materials emit radiation compared to ideal blackbodies. For surfaces with known emissivities, the emitted power at various wavelengths can be found by multiplying Planck's equation's output with the respective emissivity. This gives a complete picture of the energy distribution across different wavelengths, which is integral in fields like thermal imaging and energy efficiency.
Blackbody Radiation
Blackbody radiation is the theoretical emission of electromagnetic radiation by an object that is a perfect emitter and absorber, known as a blackbody. Blackbodies are idealized objects, where the radiation emitted solely depends on the object鈥檚 temperature, not its material or surface properties.

The significance of blackbody radiation lies in its ability to establish baseline emission levels, against which real surfaces鈥攖hose with emissivity less than one鈥攃an be compared. By understanding blackbody radiation, one can apply modifications using emissivity to determine the actual emitted power from real surfaces.

In practice, no objects are true blackbodies. However, blackbody models can approximate the behavior of stars, heated metals, or any other opaque and non-reflective object, simplifying complex calculations in thermal and optical physics.

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Most popular questions from this chapter

A furnace with a long, isothermal, graphite tube of diameter \(D=12.5 \mathrm{~mm}\) is maintained at \(T_{f}=2000 \mathrm{~K}\) and is used as a blackbody source to calibrate heat flux gages. Traditional heat flux gages are constructed as blackened thin films with thermopiles to indicate the temperature change caused by absorption of the incident radiant power over the entire spectrum. The traditional gage of interest has a sensitive area of \(5 \mathrm{~mm}^{2}\) and is mounted coaxial with the furnace centerline, but positioned at a distance of \(L=60 \mathrm{~mm}\) from the beginning of the heated section. The cool extension tube serves to shield the gage from extraneous radiation sources and to contain the inert gas required to prevent rapid oxidation of the graphite tube. (a) Calculate the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) on the traditional gage for this condition, assuming that the extension tube is cold relative to the furnace. (b) The traditional gage is replaced by a solid-state (photoconductive) heat flux gage of the same area, but sensitive only to the spectral region between \(0.4\) and \(2.5 \mu \mathrm{m}\). Calculate the radiant heat flux incident on the solid-state gage within the prescribed spectral region. (c) Calculate and plot the total heat flux and the heat flux in the prescribed spectral region for the solidstate gage as a function of furnace temperature for the range \(2000 \leq T_{f} \leq 3000 \mathrm{~K}\). Which gage will have an output signal that is more sensitive to changes in the furnace temperature?

Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front side is exposed to ambient air that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), solar irradiation of \(600 \mathrm{~W} / \mathrm{m}^{2}\), and an effective sky temperature of \(-40^{\circ} \mathrm{C}\). What is the electrical power \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) required to maintain the plate surface temperature at \(T_{s}=60^{\circ} \mathrm{C}\) if the plate is diffuse and has the designated spectral, hemispherical reflectivity?

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

The absorber plate of a solar collector may be coated with an opaque material for which the spectral, directional absorptivity is characterized by relations of the form $$ \begin{array}{ll} \alpha_{\lambda, \dot{\theta}}(\lambda, \theta)=\alpha_{1} \cos \theta & \lambda<\lambda_{c} \\ \alpha_{\lambda, \theta}(\lambda, \theta)=\alpha_{2} & \lambda>\lambda_{c} \end{array} $$ The zenith angle \(\theta\) is formed by the sun's rays and the plate normal, and \(\alpha_{1}\) and \(\alpha_{2}\) are constants. (a) Obtain an expression for the total, hemispherical absorptivity, \(\alpha_{S}\), of the plate to solar radiation incident at \(\theta=45^{\circ}\). Evaluate \(\alpha_{5}\) for \(\alpha_{1}=0.93, \alpha_{2}=\) \(0.25\), and a cut-off wavelength of \(\lambda_{c}=2 \mu \mathrm{m}\). (b) Obtain an expression for the total, hemispherical emissivity \(\varepsilon\) of the plate. Evaluate \(\varepsilon\) for a plate temperature of \(T_{p}=60^{\circ} \mathrm{C}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}\), and \(\lambda_{c}\). (c) For a solar flux of \(q_{s}^{\prime \prime}=1000 \mathrm{~W} / \mathrm{m}^{2}\) incident at \(\theta=45^{\circ}\) and the prescribed values of \(\alpha_{1}, \alpha_{2}, \lambda_{c}\), and \(T_{p}\), what is the net radiant heat flux, \(q_{\text {net }}^{\prime \prime}\), to the plate? (d) Using the prescribed conditions and the Radiation/ Band Emission Factor option in the Tools section of \(I H T\) to evaluate \(F_{\left(0 \rightarrow \lambda_{j}\right)}\), explore the effect of \(\lambda_{c}\) on \(\alpha_{S}, \varepsilon\), and \(q_{\text {net }}^{N}\) for the wavelength range \(0.7 \leq \lambda_{c} \leq 5 \mu \mathrm{m}\).

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 & \(\boldsymbol{L}_{s-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\\ \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}
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