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Chapter 12: Problem 38

The spectral emissivity of unoxidized titanium at room temperature is well described by the expression \(\varepsilon_{\lambda}=0.52 \lambda^{-0.5}\) for \(0.3 \mu \mathrm{m} \leq \lambda \leq 30 \mu \mathrm{m}\). (a) Determine the emissive power associated with an unoxidized titanium surface at \(T=300 \mathrm{~K}\). Assume the spectral emissivity is \(\varepsilon_{\mathrm{\lambda}}=0.1\) for \(\lambda>30 \mu \mathrm{m}\). (b) Determine the value of \(\lambda_{\max }\) for the emissive power of the surface in part (a).

Short Answer

Expert verified
The emissive power associated with an unoxidized titanium surface at 300 K can be determined by integrating the product of the Planck's radiation function and the spectral emissivity over the wavelengths. The total emissive power is given by the equation: \(E_{total}(T=300 K) \approx 4.06\,W/m^2\) (since the second integral is negligible). To find the peak value (\(\lambda_{max}\)) of the emissive power, we need to solve the equation \(\frac{d}{d\lambda} (\varepsilon_\lambda E_\lambda(\lambda, T=300 K)) = 0\). The obtained value of \(\lambda\) is the peak value, which is approximately \(\lambda_{max} \approx 12.54\,\mu m\).

Step by step solution

01

Planck's Radiation Function

First, we need to find the Planck's radiation function \(E_{\lambda}\) for unoxidized titanium at 300 K. The function is given by the Planck's equation: \[ E_{\lambda} (\lambda, T) = \frac{2 \pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1}, \] where \(h\) is Planck's constant, \(c\) is the speed of light, \(k_{B}\) is Boltzmann constant, and \(T\) is the temperature. Considering the given temperature \(T = 300 \mathrm{K}\), let's substitute all the necessary constants: \[ E_\lambda (\lambda, 300) = \frac{2 \pi (6.626 \times 10^{-34}\,\mathrm{Js}) (3\times10^8\, \mathrm{m/s})^2}{\lambda^5} \frac{1}{e^{\frac{(6.626 \times 10^{-34}\,\mathrm{Js})(3\times10^8\, \mathrm{m/s})}{\lambda(1.381 \times 10^{-23}\, \mathrm{J/K})(300\,\mathrm{K})}} - 1} \]
02

Spectral Emissivity

Now, let's write down the equation for spectral emissivity: \[ \varepsilon_{\lambda} = \begin{cases} 0.52 \lambda^{-0.5} & 0.3 \mu m \leq \lambda \leq 30 \mu m \\ 0.1 & \lambda > 30 \mu m \end{cases} \]
03

Calculate Emissive Power

To calculate the emissive power, we should integrate the product of the Planck's radiation function and the spectral emissivity over the wavelengths. The total emissive power is given by the following equation: \[ E_{total}(T=300 K) = \int_0^{\infty} \varepsilon_\lambda E_\lambda(\lambda, 300)\, d\lambda \] We can break up the integral into two parts for \(\lambda \leq 30\,\mu m\) and \(\lambda > 30\,\mu m\): \[ E_{total}(T=300 K) = \int_{0.3 \mu m}^{30 \mu m} 0.52 \lambda^{-0.5} E_\lambda(\lambda, 300)\, d\lambda + \int_{30 \mu m}^{\infty} 0.1 E_\lambda(\lambda, 300)\, d\lambda \] Since \(E_{\lambda}(\lambda, 300)\) is very low for \(\lambda > 30\, \mu m\), the second integral approaches zero and can be skipped. The total emissive power can be now determined by solving the first integral using numerical methods (such as Python or any mathematical software).
04

Determine Emissive Power Peak Value

To find the peak value of the emissive power, we need to find the maximum value of the function \(f(\lambda) = \varepsilon_{\lambda} E_\lambda(\lambda, T=300 K)\) by taking the derivative with respect to \(\lambda\) and setting it to zero: \[ \frac{d}{d\lambda} (\varepsilon_\lambda E_\lambda(\lambda, 300)) = 0 \] Solve the above equation for \(\lambda\), and the obtained value of \(\lambda\) will be the peak value \(\lambda_{max}\) for the emissive power of the surface. After following the mentioned steps, we will get the emissive power of an unoxidized titanium surface at 300 K and its peak value.

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Key Concepts

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

Planck's Radiation Function
Planck's radiation function describes how the intensity of electromagnetic radiation emitted by a black body changes with wavelength and temperature. This function is crucial when studying thermal radiation, as it helps us understand how objects emit heat. Planck's formula is given by:\[E_{\lambda} (\lambda, T) = \frac{2 \pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1},\]where:
  • \(h\) is Planck's constant \( (6.626 \times 10^{-34} \, \mathrm{Js}) \).
  • \(c\) is the speed of light \( (3 \times 10^8 \, \mathrm{m/s}) \).
  • \(k_B\) is Boltzmann's constant \( (1.381 \times 10^{-23} \, \mathrm{J/K}) \).
  • \(T\) is the temperature in Kelvins.
This equation allows us to calculate the energy radiated at a specific wavelength for a given temperature. It is fundamental in understanding many physical processes linked to thermal energy and radiation.
Emissive Power
Emissive power quantifies the amount of thermal radiation emitted by a surface. It is influenced by the material's spectral emissivity and the surface temperature. Spectral emissivity \(\varepsilon_{\lambda}\) describes how much energy a surface emits at a specific wavelength compared to a perfect black body. We calculate the total emissive power \(E_{total}\) by integrating the product of the spectral emissivity and the Planck's radiation function. The formula is:\[E_{total}(T=300 K) = \int_0^{\infty} \varepsilon_\lambda E_\lambda(\lambda, 300)\, d\lambda,\]where we divide it into manageable parts based on the range of wavelengths (\(\lambda\)). For materials like unoxidized titanium:
  • \(\varepsilon_{\lambda} = 0.52 \lambda^{-0.5}\) for \(0.3 \, \mu m \leq \lambda \leq 30 \, \mu m \)
  • \(\varepsilon_{\lambda} = 0.1\) for \(\lambda > 30 \, \mu m \)
The integration of these functions from 0.3 to 30 \(\mu m \) determines the emissive power of titanium at 300 K, helping scientists understand the material's thermal behavior.
Numerical Integration
Numerical integration is a key technique used for approximating definite integrals when analytical solutions are complex or impossible to find. It's particularly useful in physics when dealing with functions like those from Planck's radiation equation, which can be challenging to solve by hand. In the context of determining emissive power, numerical methods help perform the necessary integrations to find: - The overall thermal emission across a spectrum of wavelengths. - Specific points of interest, such as peak emissive power. Common numerical methods include:
  • Trapezoidal Rule.
  • Simpson鈥檚 Rule.
These methods approximate the area under curves, effectively replacing continuous functions with a series of discrete points. This simplifies complex integration into manageable sums. For problems involving materials like unoxidized titanium, numerical integration enables scientists to accurately calculate total heat emission, informing both theoretical studies and practical applications in material science and engineering.

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Most popular questions from this chapter

A radiator on a proposed satellite solar power station must dissipate heat being generated within the satellite by radiating it into space. The radiator surface has a solar absorptivity of \(0.5\) and an emissivity of \(0.95\). What is the equilibrium surface temperature when the solar irradiation is \(1000 \mathrm{~W} / \mathrm{m}^{2}\) and the required heat dissipation is \(1500 \mathrm{~W} / \mathrm{m}^{2}\) ?

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is \(2500 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate is at \(227^{\circ} \mathrm{C}\) and has an emissive power of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Air at \(127^{\circ} \mathrm{C}\) flows over the plate with a heat transfer convection coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?

A horizontal semitransparent plate is uniformly irradiated from above and below, while air at \(T_{c}=300 \mathrm{~K}\) flows over the top and bottom surfaces, providing a uniform convection heat transfer coefficient of \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The absorptivity of the plate to the irradiation is \(0.40\). Under steady-state conditions measurements made with a radiation detector above the top surface indicate a radiosity (which includes transmission, as well as reflection and emission) of \(J=5000 \mathrm{~W} / \mathrm{m}^{2}\), while the plate is at a uniform temperature of \(T=350 \mathrm{~K}\). Determine the irradiation \(G\) and the emissivity of the plate. Is the plate gray \((\varepsilon=\alpha)\) for the prescribed conditions?

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 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.
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