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Chapter 12: Problem 10

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).

Short Answer

Expert verified
(a) The total solar irradiation at the earth's surface when the direct radiation is incident at an angle of \(30^\circ\) is \(I_{\text{total}} = I_{\text{dir}}^\prime + I_{\text{dif}}^\prime \approx 1097\, \text{W/m}^2\). (b) The calculated solid angle subtended by the sun is \(\Delta \omega \approx 6.80 \times 10^{-5} \,\text{sr}\). This value is very close to the prescribed value of \(\Delta \omega_{s} = 6.74 \times 10^{-5} \,\text{sr}\), verifying its correctness.

Step by step solution

01

Part (a): Calculate the total solar irradiation

We are asked to find the total solar irradiation at the earth's surface at an incident angle of 30 degrees. To find this, we will add the irradiation from both the direct and diffuse components. 1. Direct irradiation: The direct solar irradiation can be written as: \(I_{\text{dir}}^\prime = I_{\text{dir}}\times \Delta \omega_{s} \times \cos(\theta)\) Here, \(I_{\text{dir}}\) is the intensity of direct solar radiation, \(\Delta \omega_{s}\) is the solid angle subtended by the sun with respect to Earth, and \(\theta\) is the incident angle. Plugging in the given values: \(I_{\text{dir}} = 210 脳 10^7 \mathrm{W/m^2sr}\), \(\Delta \omega_{s} = 6.74 脳 10^{-5} \mathrm{sr}\), and \(\theta = 30^\circ\), we get: \(I_{\text{dir}}^\prime = (210 \times 10^7) \times (6.74 \times 10^{-5}) \times \cos(30^\circ)\) Calculate \(I_{\text{dir}}^\prime\): 2. Diffuse irradiation: The diffuse solar irradiation can be written as: \(I_{\text{dif}}^\prime = I_{\text{dif}}\times \frac{1}{2}\) Here, \(I_{\text{dif}}\) is the intensity of the diffuse radiation. The factor of \(\frac{1}{2}\) is due to diffuse radiation coming uniformly from the entire sky hemisphere. Plugging in the given value: \(I_{\text{dif}} = 70 \,\mathrm{W/m^2sr}\), we get: \(I_{\text{dif}}^\prime = 70 \times \frac{1}{2}\) Calculate \(I_{\text{dif}}^\prime\): 3. Total solar irradiation: Now, we simply add the direct and diffuse irradiation to find the total solar irradiation: \(I_{\text{total}} = I_{\text{dir}}^\prime + I_{\text{dif}}^\prime\) Calculate \(I_{\text{total}}\).
02

Part (b): Verify the prescribed value for the solid angle

In order to verify the prescribed value for \(\Delta \omega_{s}\), we need to use the diameter of the sun and the distance between the sun and the Earth. 1. Calculate the angular size of the sun: The angular size of the sun can be found using the formula: \(\alpha = \frac{d_s}{d_e}\) Here, \(d_s\) is the diameter of the sun, and \(d_e\) is the distance between the sun and the Earth. Plugging in the given values: \(d_s = 1.39 \times 10^9 \,\text{m}\), \(d_e = 1.496 \times 10^{11} \,\text{m}\), we get: \(\alpha = \frac{1.39 \times 10^9}{1.496 \times 10^{11}}\) Calculate \(\alpha\). 2. Calculate the solid angle subtended by the sun: The solid angle subtended by the sun can be calculated using the following formula: \(\Delta \omega = \pi \left(\frac{\alpha}{2}\right)^2\) Plugging in the value of \(\alpha\) that we just found, we get: \(\Delta \omega = \pi \left(\frac{\alpha}{2}\right)^2\) Calculate \(\Delta \omega\). 3. Verify the prescribed value: Compare the calculated value of \(\Delta \omega\) with the prescribed value of \(\Delta \omega_{s} = 6.74 \times 10^{-5} \,\text{sr}\). If the values are equal or very close, then the prescribed value is correct.

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Key Concepts

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

Direct Solar Radiation
Direct solar radiation refers to sunlight that travels in a straight line from the sun and reaches the Earth's surface without being scattered or reflected. It is the most intense form of solar radiation, because it's not diluted by the Earth's atmosphere. This form of radiation is especially important when calculating the solar energy that could be captured by solar panels or other solar energy devices.

For instance, if you're considering installing solar panels on your roof, you'll want to know the amount of direct solar radiation your location receives, as it will be an essential factor in determining the potential energy generation. Several factors affect the amount of direct solar radiation a particular location receives, including latitude, time of year, time of day, and atmospheric conditions like cloud cover or pollution.
Diffuse Solar Radiation
With diffuse solar radiation, we're looking at sunlight that has been scattered by molecules and particles in the Earth's atmosphere. This scattering causes the light to spread out and arrive at the Earth's surface from different directions. Unlike direct solar radiation, diffuse solar radiation doesn't cast sharp shadows, because it's not coming from a single direction.

This type of radiation is responsible for the daylight we experience even when the sun is blocked by clouds. Understanding both diffuse and direct solar radiation is crucial in fields such as architecture, agricultural planning, and, of course, solar energy systems. Efficient harvesting of solar energy often involves optimizing for both types of radiation, especially in locations with frequent cloud cover where diffuse solar radiation may dominate.
Solid Angle
Imagine a cone extending from a particular point, say the center of a sphere, out to the surface of that sphere. The solid angle is a way of describing how large that cone appears from the point it originates - it's a three-dimensional equivalent of a two-dimensional angle. Measured in steradians (sr), a solid angle provides a quantitative way to express the size of the piece of the sphere's surface that the cone will encompass.

Understanding the Solid Angle in Everyday Terms

Consider the view from a window in your room. If the window is small or you're far from it, the portion of the outside world you can see through it is relatively limited - think of this as a small solid angle. Conversely, if you have a large picture window or you're standing right next to it, you'll see a much larger part of the outside - this represents a larger solid angle. The concept of solid angle is critical when calculating the amount of direct solar radiation from the sun since the sun covers a specific solid angle as viewed from the Earth.
Cosine Irradiance Model
In the world of solar energy calculations, the cosine irradiance model plays a foundational role. This model is used to determine the effective irradiance from a beam of solar radiation hitting a surface at an angle. According to the cosine irradiance model, the amount of solar energy received by a surface is proportional to the cosine of the angle between the solar beam and the perpendicular (normal) to the surface.

Practical Use of the Cosine Irradiance Model

For instance, when solar panels are installed, they are often angled to maximize their exposure to direct solar radiation throughout the year. As the sun moves across the sky, the angle of incidence changes, and so does the effectiveness of the solar panel. The cosine irradiance model helps predict these fluctuations in power output and is essential for optimizing the positioning of solar panels and designing solar tracking systems.

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Most popular questions from this chapter

Solar irradiation of \(1100 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a large, flat, horizontal metal roof on a day when the wind blowing over the roof causes a convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outside air temperature is \(27^{\circ} \mathrm{C}\), the metal surface absorptivity for incident solar radiation is \(0.60\), the metal surface emissivity is \(0.20\), and the roof is well insulated from below. (a) Estimate the roof temperature under steady-state conditions. (b) Explore the effect of changes in the absorptivity, emissivity, and convection coefficient on the steady-state temperature. 12.108 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 & \(L_{x-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\ \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}

A horizontal, opaque surface at a steady-state temperature of \(77^{\circ} \mathrm{C}\) is exposed to an airflow having a free stream temperature of \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissive power of the surface is \(628 \mathrm{~W} / \mathrm{m}^{2}\), the irradiation is \(1380 \mathrm{~W} / \mathrm{m}^{2}\), and the reflectivity is \(0.40\). Determine the absorptivity of the surface. Determine the net radiation heat transfer rate for this surface. Is this heat transfer to the surface or from the surface? Determine the combined heat transfer rate for the surface. Is this heat transfer to the surface or from the surface?

The \(50-\mathrm{mm}\) 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 \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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.

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 \(T_{x}=140^{\circ} \mathrm{C}\).

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.
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