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Chapter 12: Problem 6

According to its directicnal distribution, solar radiation incident on the earth's surface may be divided into two components. The direct component consists of parallel rays incident at a fixed renith angle \(\theta\), while the diffuse component consists of radiation that may be approximated as being diffusely distributed with \(\theta\). Consider clear sky conditions for which the direct radiation is incident at \(\theta=30^{\circ}\), with a total flux (based on an area that is normal to the rays) of \(q_{\text {dre }}^{*}=\) \(1000 \mathrm{~W} / \mathrm{m}^{2}\), and the total intensity of the diffuse radiation is \(l_{\text {ar }}=70 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{sr}\). What is the total solar irradiation at the earth's surface?

Short Answer

Expert verified
The total solar irradiation is 936 W/m虏.

Step by step solution

01

Understand the Given Information

We have:- Direct radiation with a zenith angle \( \theta = 30^{\circ} \) and a flux of \( q_{\text{dre}}^{*} = 1000 \, \text{W/m}^2 \), based on an area normal to the rays.- Diffuse radiation with an intensity of \( l_{\text{ar}} = 70 \, \text{W/m}^2\cdot\text{sr} \).
02

Calculate the Direct Radiation Flux on the Surface

Since the direct radiation is incident at an angle, we must consider the angle of incidence when calculating the flux on the horizontal surface. The actual direct radiation on the surface is given by:\[q_{\text{direct}} = q_{\text{dre}}^{*} \cdot \cos(\theta)\]Substituting \( \theta = 30^{\circ} \):\[q_{\text{direct}} = 1000 \, \text{W/m}^2 \cdot \cos(30^{\circ}) \approx 1000 \, \text{W/m}^2 \cdot 0.866 \approx 866 \, \text{W/m}^2\]
03

Consider the Diffuse Radiation Contribution

The diffuse radiation is assumed to be distributed uniformly across the hemisphere, so its contribution to the irradiation on the horizontal surface is the given intensity: \( l_{\text{ar}} = 70 \, \text{W/m}^2 \). Since this is diffuse, you do not need to adjust for angle.
04

Calculate Total Solar Irradiation

To get the total solar irradiation on the surface, sum the direct and diffuse components.\[q_{\text{total}} = q_{\text{direct}} + l_{\text{ar}}\]Substitute the calculated values:\[q_{\text{total}} = 866 \, \text{W/m}^2 + 70 \, \text{W/m}^2 = 936 \, \text{W/m}^2\]
05

Final Answer

The total solar irradiation at the earth's surface under the given conditions is \( 936 \, \text{W/m}^2 \).

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Key Concepts

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

Direct Radiation
Direct radiation is one of the two major components of solar radiation reaching the Earth's surface. It refers to sunlight that travels in a straight line from the sun and is neither scattered nor absorbed by the atmosphere. This type of radiation is highly directional, meaning that it impacts surfaces at specific angles depending on the position of the sun. Direct radiation's effectiveness or intensity on a horizontal surface can be affected by the angle of incidence. For instance, when the sun is directly overhead, the radiation strikes the surface perpendicularly, maximizing its intensity. However, as the sun's position changes throughout the day or with seasons, the angle of incidence varies, consequently altering the intensity of direct radiation received.To find the component of direct radiation that impacts a horizontal surface, you need to adjust for this angle, typically calculated using the formula:- \[ q_{\text{direct}} = q_{\text{dre}}^{*} \cdot \cos(\theta) \] Where:
  • \(q_{\text{dre}}^{*}\) is the flux of direct radiation at a 90-degree angle.
  • \(\theta\) is the zenith angle or the angle of incidence.
Diffuse Radiation
Diffuse radiation is another important component of solar radiation. Unlike direct radiation, it does not have a clear path from the sun to the surface. Instead, it is sunlight that has been scattered by molecules and particles in the earth's atmosphere. This scattering causes the light to reach the surface from all directions, rather than a single, direct path. Because diffuse radiation is scattered across the sky, it can illuminate shaded areas where direct sunlight doesn't easily reach. This contribution is critical for solar energy systems, especially during cloudy days when direct sunlight is obstructed. The intensity of diffuse radiation is typically considered uniform across a horizontal surface, meaning no adjustment for angle is necessary. It is given in terms of watts per square meter. For calculations involving total solar radiation, diffuse radiation is simply added to the calculated value of direct radiation.
Solar Irradiation
Solar irradiation is the measure of solar power received per unit area at a surface. It combines both direct and diffuse radiation contributions to determine the total solar energy available at the Earth's surface.Solar irradiation is a crucial concept for applications that depend on solar energy, such as photovoltaic systems. Knowing the total irradiation helps estimate the potential energy yield of solar panels, informing system design and energy predictions.In clear sky conditions, solar irradiation can be computed by summing the contributions from direct and diffuse radiation:- \[ q_{\text{total}} = q_{\text{direct}} + l_{\text{ar}} \] Where:
  • \(q_{\text{direct}}\) is the adjusted direct radiation considering the angle of incidence.
  • \(l_{\text{ar}}\) is the diffuse radiation intensity.
This sum provides a comprehensive view of total solar energy available, essential for optimizing solar energy collection.

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Most popular questions from this chapter

A radiation thermometer is a madiometer calibrated to indicate the temperature of a blackbody. A steel billet having a diffuse, gray surface of emissivity \(0.8\) is heated in a furnace whose walls are at \(1500 \mathrm{~K}\). Estimate the temperature of the billet when the radiation thermometer viewing the billet through a small hole in the furnace indicates \(1160 \mathrm{~K}\).

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\right)\) of 30 -mm diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A temperature sensor imbedded in the tip of a small tube having a diffuse, gray surface with an emissivity of \(0.8\) is centrally positioned within a large air-conditioned toom whose walls and air temperature are 30 and \(20^{\circ} \mathrm{C}\), respectively. (a) What temperature will the sensor indicate if the convection coefficient between the sensot tube and the air is \(5 \mathrm{~W} / \mathrm{m}^{2}\) - K? (b) What would be the effect of using a fan to induce airflow over the tube? Plot the sensor temperature as a function of the convection cocfficient for \(2 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and values of \(\mathrm{e}=0.2,0.5\). and \(0.8\).

A roof-cooling system, which operates by maintaining a thin film of water on the roof surface, may be used to reduce air-conditioning costs or to maintain a cooler environment in nonconditioned buildings. To determine the effectiveness of such a system, consider a sheet metal roof for which the solar absorptivity \(\alpha_{s}\) is \(0.50\) and the hemispherical emissivity \(\varepsilon\) is \(0.3\). Representative conditions correspond to a surface convection coefficient \(h\) of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a solar irradiation \(G_{S}\) of \(700 \mathrm{~W} / \mathrm{m}^{2}\), a sky temperature of \(-10^{\circ} \mathrm{C}\), an utmospheric temperature of \(30^{\circ} \mathrm{C}\), and a relutive humidity of \(65 \%\). The roof may be assumed to be well insulated from below. Determine the roof surface temperature without the water film. Assuming the film and roof surface temperatures to be cqual, determine the surface temperature with the film. The solar absorptivity and the hemispherical emissivity of the film-surface combination are \(\alpha_{s}=0.8\) and \(\varepsilon=0.9\), respectively.

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat ftux gages, radiation thermometens, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation 12.24, show that the ratio of the fractional change in the spectral intensity to the fractional change in the termperature of the furnace has the form $$ \frac{d I_{2} I_{2}}{d T I T}=\frac{C_{2}}{\lambda T} \frac{1}{\operatorname{cxp}\left(-C_{2} \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furmace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation al \(10 \mu \mathrm{m}\) ?
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