91Ó°ÊÓ

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 \(0.7 \leq \lambda_{c} \leq 5 \mu \mathrm{m}\).

Step by step solution

01

Determine Hemispherical Absorptivity for Both Wavelengths

Given, the spectral, directional absorptivity for \(\lambda < \lambda_c\) is: \[\alpha_{\lambda, \theta}(\lambda, \theta) = \alpha_1 \cos\theta\] And for \(\lambda > \lambda_c\): \[\alpha_{\lambda, \theta}(\lambda, \theta) = \alpha_2\] Now, we need to find the total, hemispherical absorptivity \(\alpha_S\), which is the integration of the spectral, directional absorptivity over all wavelengths and solid angles.

02

Integrate Absorptivity over Solid Angle and Wavelength

To find \(\alpha_S\), we integrate \(\alpha_{\lambda, \theta}(\lambda, \theta)\) over solid angle and wavelength. Using the data given and considering \(\theta=45^{\circ}\): \[\alpha_S = \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\lambda_c} [\alpha_1 \cos\theta] \sin\theta \, d\lambda \, d\theta \, d\phi + \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{\lambda_c}^{\infty} [\alpha_2] \sin\theta\, d\lambda \, d\theta \, d\phi\] Plug in the given values \(\alpha_1 = 0.93\), \(\alpha_2 = 0.25\), and \(\lambda_c = 2\mu m\) and solve for \(\alpha_S\). #b) Calculate total, hemispherical emissivity - Expression and Evaluation.#

03

Determine Stefan-Boltzmann Law for Emissivity

We will use the Stefan-Boltzmann Law for emissivity to find an expression for the total, hemispherical emissivity \(\varepsilon\): \[\varepsilon \sigma T_p^4 = \int_{0}^{\infty} \varepsilon_{\lambda}(\lambda,T_p) E_b(\lambda, T_p) \, d\lambda\] Where \(\sigma\) is the Stefan-Boltzmann constant, \(E_b(\lambda, T_p)\) is the blackbody radiation curve, and \(\varepsilon_{\lambda}(\lambda, T_p)\) is the spectral hemispherical emissivity.

04

Kirchhoff's Law of Radiation

Using Kirchhoff's Law of Radiation, which states that absorptivity equals emissivity for opaque materials: \[\varepsilon_{\lambda}(\lambda, T_p) = \alpha_{\lambda, \theta}(\lambda, \theta)\] Substitute this into the Stefan-Boltzmann Law for emissivity and obtain an expression for the total, hemispherical emissivity \(\varepsilon\). Then evaluate it for \(T_p = 60^{\circ} C = 333.15 K\) and the given values of \(\alpha_1\), \(\alpha_2\), and \(\lambda_c\). #c) Calculate net radiant heat flux to the plate.#

05

Determine the Absorbed Radiation Flux

Using the given solar flux of \(q_s^{\prime\prime} = 1000 W/m^2\), find \(q_{abs}^{\prime\prime} = \alpha_S q_s^{\prime\prime}\).

06

Determine the Emitted Radiation Flux

Using the emissivity found in part (b), determine the emitted radiation flux as \(q_{em}^{\prime\prime} = \varepsilon \sigma T_p^4\).

07

Calculate Net Radiant Heat Flux

Subtract the emitted radiation flux from the absorbed radiation flux to find the net radiant heat flux: \(q_{\text {net }}^{\prime \prime} = q_{abs}^{\prime\prime} - q_{em}^{\prime\prime}\). #d) Explore the effect of cutoff wavelength.#

08

Use Radiation/Band Emission Factor Option

Using the prescribed conditions and the Radiation/ Band Emission Factor option in the Tools section of IHT, evaluate the effect of \(F_{(0\rightarrow \lambda_j)}\) on \(\alpha_S\), \(\varepsilon\), and \(q_{\text {net }}^{\prime \prime}\) for the wavelength range \(0.7\leq \lambda_c \leq 5\mu m\).

09

Analyze and Plot the Results

Plot the values of \(\alpha_S\), \(\varepsilon\), and \(q_{\text {net }}^{\prime \prime}\) against \(\lambda_c\) to visualize the effect of varying the cutoff wavelength in the given range.

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Key Concepts

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

Spectral Absorptivity

Spectral absorptivity refers to how effectively a material absorbs radiation at different wavelengths. It's a critical property for the efficiency of solar collectors, as they rely on absorbing solar energy effectively.

For the solar collector absorber plate, we have formulas that define its absorptivity based on wavelengths. For

• Wavelengths less than a critical value \(\lambda_c\): \[ \alpha_{\lambda} = \alpha_1 \cos \theta \] where \(\theta\) is the angle formed by the sun's rays, and \(\alpha_1\) is a constant.
• For wavelengths greater than \(\lambda_c\): \[ \alpha_{\lambda} = \alpha_2 \] where \(\alpha_2\) is another constant.

Understanding how these values change with wavelength helps in designing and optimizing the solar collector's performance.

By integrating these spectral absorptivity functions over all relevant angles and wavelengths, we obtain the overall, or total, absorptivity of the collector. This comprehensive measure determines how well the solar collector can harness energy from the sun.

Hemispherical Emissivity

Hemispherical emissivity is a measure of how effectively a material emits energy as thermal radiation across all wavelengths and directions. It’s crucial for energy balance in solar collectors.

We can calculate hemispherical emissivity using the Stefan-Boltzmann Law:\[\varepsilon \sigma T_p^4 = \int_{0}^{\infty} \varepsilon_{\lambda}(\lambda,T_p) E_b(\lambda, T_p) \, d\lambda\]Here, the constant \(\sigma\) is the Stefan-Boltzmann constant, \(E_b(\lambda, T_p)\) shows the blackbody radiation curve, and \(\varepsilon_{\lambda}\) is the spectral emissivity. Hemispherical emissivity tells us how much of the absorbed energy is emitted back.

When evaluating for a specific temperature, such as \(T_p = 60^{\circ} C\), Kirchhoff's Law (further explained below) helps us link absorbed and emitted energy, ensuring that the absorber plate emits energy according to its absorptive properties.

Kirchhoff's Law of Radiation

Kirchhoff's Law of Radiation is a foundational principle that connects absorptive and emissive properties of materials. It states that for opaque materials, spectral absorptivity equals spectral emissivity at thermal equilibrium. This means \(\alpha_{\lambda} = \varepsilon_{\lambda}\).

This law is particularly useful as it simplifies calculations. It allows direct use of absorptive data to determine emissivity, reducing complexities in thermal radiation analysis.

The implications for solar collectors are significant. By ensuring that the materials are finely tuned to this balance, a solar collector can efficiently absorb solar energy while minimizing energy loss by emission.

In practical applications, understanding Kirchhoff's Law helps in optimizing aspects of solar collector design, such as coating and materials, to maximize energy retention and efficiency.