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Unformatted text preview: cen58933_ch11.qxd 9/9/2002 9:38 AM Page 561 CHAPTER F U N D A M E N TA L S O F T H E R M A L R A D I AT I O N o far, we have considered the conduction and convection modes of heat transfer, which are related to the nature of the materials involved and the presence of fluid motion, among other things. We now turn our attention to the third mechanism of heat transfer: radiation, which is characteristically different from the other two. We start this chapter with a discussion of electromagnetic waves and the electromagnetic spectrum, with particular emphasis on thermal radiation. Then we introduce the idealized blackbody, blackbody radiation, and blackbody radiation function, together with the Stefan–Boltzmann law, Planck’s law, and Wien’s displacement law. Radiation is emitted by every point on a plane surface in all directions into the hemisphere above the surface. The quantity that describes the magnitude of radiation emitted or incident in a specified direction in space is the radiation intensity. Various radiation fluxes such as emissive power, irradiation, and radiosity are expressed in terms of intensity. This is followed by a discussion of radiative properties of materials such as emissivity, absorptivity, reflectivity, and transmissivity and their dependence on wavelength, direction, and temperature. The greenhouse effect is presented as an example of the consequences of the wavelength dependence of radiation properties. The last section is devoted to the discussions of atmospheric and solar radiation because of their importance. S 11 CONTENTS 11-1 Introduction 562 11-2 Thermal Radiation 563 11-3 Blackbody Radiation 565 11-4 Radiation Intensity 571 11-5 Radiative Properties 577 11-6 Atmospheric and Solar Radiation 586 Topic of Special Interest: Solar Heat Gain Through Windows 590 561 cen58933_ch11.qxd 9/9/2002 9:38 AM Page 562 562 HEAT TRANSFER 11–1 Vacuum chamber Hot object Radiation FIGURE 11–1 A hot object in a vacuum chamber loses heat by radiation only. Person 30°C Fire 900°C Air 5°C Radiation FIGURE 11–2 Unlike conduction and convection, heat transfer by radiation can occur between two bodies, even when they are separated by a medium colder than both of them. I INTRODUCTION Consider a hot object that is suspended in an evacuated chamber whose walls are at room temperature (Fig. 11–1). The hot object will eventually cool down and reach thermal equilibrium with its surroundings. That is, it will lose heat until its temperature reaches the temperature of the walls of the chamber. Heat transfer between the object and the chamber could not have taken place by conduction or convection, because these two mechanisms cannot occur in a vacuum. Therefore, heat transfer must have occurred through another mechanism that involves the emission of the internal energy of the object. This mechanism is radiation. Radiation differs from the other two heat transfer mechanisms in that it does not require the presence of a material medium to take place. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. Also, radiation transfer occurs in solids as well as liquids and gases. In most practical applications, all three modes of heat transfer occur concurrently at varying degrees. But heat transfer through an evacuated space can occur only by radiation. For example, the energy of the sun reaches the earth by radiation. You will recall that heat transfer by conduction or convection takes place in the direction of decreasing temperature; that is, from a high-temperature medium to a lower-temperature one. It is interesting that radiation heat transfer can occur between two bodies separated by a medium colder than both bodies (Fig. 11–2). For example, solar radiation reaches the surface of the earth after passing through cold air layers at high altitudes. Also, the radiationabsorbing surfaces inside a greenhouse reach high temperatures even when its plastic or glass cover remains relatively cool. The theoretical foundation of radiation was established in 1864 by physicist James Clerk Maxwell, who postulated that accelerated charges or changing electric currents give rise to electric and magnetic fields. These rapidly moving fields are called electromagnetic waves or electromagnetic radiation, and they represent the energy emitted by matter as a result of the changes in the electronic configurations of the atoms or molecules. In 1887, Heinrich Hertz experimentally demonstrated the existence of such waves. Electromagnetic waves transport energy just like other waves, and all electromagnetic waves travel at the speed of light in a vacuum, which is C0 2.9979 108 m/s. Electromagnetic waves are characterized by their frequency or wavelength . These two properties in a medium are related by c (11-1) where c is the speed of propagation of a wave in that medium. The speed of propagation in a medium is related to the speed of light in a vacuum by c c0 /n, where n is the index of refraction of that medium. The refractive index is essentially unity for air and most gases, about 1.5 for glass, and about 1.33 for water. The commonly used unit of wavelength is the micrometer ( m) or micron, where 1 m 10 6 m. Unlike the wavelength and the speed of propagation, the frequency of an electromagnetic wave depends only on the source and is independent of the medium through which the wave travels. The frequency (the number of oscillations per second) of an electromagnetic wave can range cen58933_ch11.qxd 9/9/2002 9:38 AM Page 563 563 CHAPTER 11 from less than a million Hz to a septillion Hz or higher, depending on the source. Note from Eq. 11-1 that the wavelength and the frequency of electromagnetic radiation are inversely proportional. It has proven useful to view electromagnetic radiation as the propagation of a collection of discrete packets of energy called photons or quanta, as proposed by Max Planck in 1900 in conjunction with his quantum theory. In this view, each photon of frequency is considered to have an energy of e h hc λ (11-2) where h 6.6256 10 34 J · s is Planck’s constant. Note from the second part of Eq. 11-2 that the energy of a photon is inversely proportional to its wavelength. Therefore, shorter-wavelength radiation possesses larger photon energies. It is no wonder that we try to avoid very-short-wavelength radiation such as gamma rays and X-rays since they are highly destructive. λ , µm Electrical power waves 1010 109 108 107 Radio and TV waves 106 105 11–2 I 104 THERMAL RADIATION Although all electromagnetic waves have the same general features, waves of different wavelength differ significantly in their behavior. The electromagnetic radiation encountered in practice covers a wide range of wavelengths, varying from less than 10 10 m for cosmic rays to more than 1010 m for electrical power waves. The electromagnetic spectrum also includes gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, thermal radiation, microwaves, and radio waves, as shown in Figure 11–3. Different types of electromagnetic radiation are produced through various mechanisms. For example, gamma rays are produced by nuclear reactions, X-rays by the bombardment of metals with high-energy electrons, microwaves by special types of electron tubes such as klystrons and magnetrons, and radio waves by the excitation of some crystals or by the flow of alternating current through electric conductors. The short-wavelength gamma rays and X-rays are primarily of concern to nuclear engineers, while the long-wavelength microwaves and radio waves are of concern to electrical engineers. The type of electromagnetic radiation that is pertinent to heat transfer is the thermal radiation emitted as a result of energy transitions of molecules, atoms, and electrons of a substance. Temperature is a measure of the strength of these activities at the microscopic level, and the rate of thermal radiation emission increases with increasing temperature. Thermal radiation is continuously emitted by all matter whose temperature is above absolute zero. That is, everything around us such as walls, furniture, and our friends constantly emits (and absorbs) radiation (Fig. 11–4). Thermal radiation is also defined as the portion of the electromagnetic spectrum that extends from about 0.1 to 100 m, since the radiation emitted by bodies due to their temperature falls almost entirely into this wavelength range. Thus, thermal radiation includes the entire visible and infrared (IR) radiation as well as a portion of the ultraviolet (UV) radiation. What we call light is simply the visible portion of the electromagnetic spectrum that lies between 0.40 and 0.76 m. Light is characteristically no different than other electromagnetic radiation, except that it happens to trigger the 103 Microwaves 102 10 1 10–1 Thermal Infrared radiation Visible Ultraviolet 10–2 10–3 X-rays 10–4 10–5 10–6 γ-rays 10–7 10–8 10–9 Cosmic rays FIGURE 11–3 The electromagnetic wave spectrum. Plants Walls People Furniture FIGURE 11–4 Everything around us constantly emits thermal radiation. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 564 564 HEAT TRANSFER TABLE 11–1 The wavelength ranges of different colors Color Violet Blue Green Yellow Orange Red Wavelength band 0.40–0.44 0.44–0.49 0.49–0.54 0.54–0.60 0.60–0.67 0.63–0.76 FIGURE 11–5 Food is heated or cooked in a microwave oven by absorbing the electromagnetic radiation energy generated by the magnetron of the oven. m m m m m m sensation of seeing in the human eye. Light, or the visible spectrum, consists of narrow bands of color from violet (0.40–0.44 m) to red (0.63–0.76 m), as shown in Table 11–1. A body that emits some radiation in the visible range is called a light source. The sun is obviously our primary light source. The electromagnetic radiation emitted by the sun is known as solar radiation, and nearly all of it falls into the wavelength band 0.3–3 m. Almost half of solar radiation is light (i.e., it falls into the visible range), with the remaining being ultraviolet and infrared. The radiation emitted by bodies at room temperature falls into the infrared region of the spectrum, which extends from 0.76 to 100 m. Bodies start emitting noticeable visible radiation at temperatures above 800 K. The tungsten filament of a lightbulb must be heated to temperatures above 2000 K before it can emit any significant amount of radiation in the visible range. The ultraviolet radiation includes the low-wavelength end of the thermal radiation spectrum and lies between the wavelengths 0.01 and 0.40 m. Ultraviolet rays are to be avoided since they can kill microorganisms and cause serious damage to humans and other living organisms. About 12 percent of solar radiation is in the ultraviolet range, and it would be devastating if it were to reach the surface of the earth. Fortunately, the ozone (O3) layer in the atmosphere acts as a protective blanket and absorbs most of this ultraviolet radiation. The ultraviolet rays that remain in sunlight are still sufficient to cause serious sunburns to sun worshippers, and prolonged exposure to direct sunlight is the leading cause of skin cancer, which can be lethal. Recent discoveries of “holes” in the ozone layer have prompted the international community to ban the use of ozone-destroying chemicals such as the refrigerant Freon-12 in order to save the earth. Ultraviolet radiation is also produced artificially in fluorescent lamps for use in medicine as a bacteria killer and in tanning parlors as an artificial tanner. The connection between skin cancer and ultraviolet rays has caused dermatologists to issue strong warnings against its use for tanning. Microwave ovens utilize electromagnetic radiation in the microwave region of the spectrum generated by microwave tubes called magnetrons. Microwaves in the range of 102–105 m are very suitable for use in cooking since they are reflected by metals, transmitted by glass and plastics, and absorbed by food (especially water) molecules. Thus, the electric energy converted to radiation in a microwave oven eventually becomes part of the internal energy of the food. The fast and efficient cooking of microwave ovens has made them as one of the essential appliances in modern kitchens (Fig. 11–5). Radars and cordless telephones also use electromagnetic radiation in the microwave region. The wavelength of the electromagnetic waves used in radio and TV broadcasting usually ranges between 1 and 1000 m in the radio wave region of the spectrum. In heat transfer studies, we are interested in the energy emitted by bodies because of their temperature only. Therefore, we will limit our consideration to thermal radiation, which we will simply call radiation. The relations developed below are restricted to thermal radiation only and may not be applicable to other forms of electromagnetic radiation. The electrons, atoms, and molecules of all solids, liquids, and gases above absolute zero temperature are constantly in motion, and thus radiation is constantly emitted, as well as being absorbed or transmitted throughout the entire volume of matter. That is, radiation is a volumetric phenomenon. However, cen58933_ch11.qxd 9/9/2002 9:38 AM Page 565 565 CHAPTER 11 for opaque (nontransparent) solids such as metals, wood, and rocks, radiation is considered to be a surface phenomenon, since the radiation emitted by the interior regions can never reach the surface, and the radiation incident on such bodies is usually absorbed within a few microns from the surface (Fig. 11–6). Note that the radiation characteristics of surfaces can be changed completely by applying thin layers of coatings on them. Radiation emitted Gas or vacuum Solid 11–3 I BLACKBODY RADIATION A body at a temperature above absolute zero emits radiation in all directions over a wide range of wavelengths. The amount of radiation energy emitted from a surface at a given wavelength depends on the material of the body and the condition of its surface as well as the surface temperature. Therefore, different bodies may emit different amounts of radiation per unit surface area, even when they are at the same temperature. Thus, it is natural to be curious about the maximum amount of radiation that can be emitted by a surface at a given temperature. Satisfying this curiosity requires the definition of an idealized body, called a blackbody, to serve as a standard against which the radiative properties of real surfaces may be compared. A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation, regardless of wavelength and direction. Also, a blackbody emits radiation energy uniformly in all directions per unit area normal to direction of emission. (Fig. 11–7). That is, a blackbody is a diffuse emitter. The term diffuse means “independent of direction.” The radiation energy emitted by a blackbody per unit time and per unit surface area was determined experimentally by Joseph Stefan in 1879 and expressed as Eb(T ) T4 (W/m2) FIGURE 11–6 Radiation in opaque solids is considered a surface phenomenon since the radiation emitted only by the molecules at the surface can escape the solid. Uniform Nonuniform Blackbody Real body (11-3) where 5.67 10 8 W/m2 · K4 is the Stefan–Boltzmann constant and T is the absolute temperature of the surface in K. This relation was theoretically verified in 1884 by Ludwig Boltzmann. Equation 11-3 is known as the Stefan–Boltzmann law and Eb is called the blackbody emissive power. Note that the emission of thermal radiation is proportional to the fourth power of the absolute temperature. Although a blackbody would appear black to the eye, a distinction should be made between the idealized blackbody and an ordinary black surface. Any surface that absorbs light (the visible portion of radiation) would appear black to the eye, and a surface that reflects it completely would appear white. Considering that visible radiation occupies a very narrow band of the spectrum from 0.4 to 0.76 m, we cannot make any judgments about the blackness of a surface on the basis of visual observations. For example, snow and white paint reflect light and thus appear white. But they are essentially black for infrared radiation since they strongly absorb long-wavelength radiation. Surfaces coated with lampblack paint approach idealized blackbody behavior. Another type of body that closely resembles a blackbody is a large cavity with a small opening, as shown in Figure 11–8. Radiation coming in through the opening of area A will undergo multiple reflections, and thus it will have several chances to be absorbed by the interior surfaces of the cavity before any FIGURE 11–7 A blackbody is said to be a diffuse emitter since it emits radiation energy uniformly in all directions. Small opening of area A Large cavity T FIGURE 11–8 A large isothermal cavity at temperature T with a small opening of area A closely resembles a blackbody of surface area A at the same temperature. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 566 566 HEAT TRANSFER part of it can possibly escape. Also, if the surface of the cavity is isothermal at temperature T, the radiation emitted by the interior surfaces will stream through the opening after undergoing multiple reflections, and thus it will have a diffuse nature. Therefore, the cavity will act as a perfect absorber and perfect emitter, and the opening will resemble a blackbody of surface area A at temperature T, regardless of the actual radiative properties of the cavity. The Stefan–Boltzmann law in Eq. 11-3 gives the total blackbody emissive power Eb, which is the sum of the radiation emitted over all wavelengths. Sometimes we need to know the spectral blackbody emissive power, which is the amount of radiation energy emitted by a blackbody at an absolute temperature T per unit time, per unit surface area, and per unit wavelength about the wavelength . For example, we are more interested in the amount of radiation an incandescent light bulb emits in the visible wavelength spectrum than we are in the total amount emitted. The relation for the spectral blackbody emissive power Eb was developed by Max Planck in 1901 in conjunction with his famous quantum theory. This relation is known as Planck’s law and is expressed as Eb ( , T ) C1 [exp (C2 / T ) 5 1] (W/m2 · m) (11-4) where C1 C2 2 2 hc 0 3.742 108 W · m4/m2 hc0 /k 1.439 104 m · K Also, T is the absolute temperature of the surface, is the wavelength of the radiation emitted, and k 1.38065 10 23 J/K is Boltzmann’s constant. This relation is valid for a surface in a vacuum or a gas. For other mediums, it needs to be modified by replacing C1 by C1/n2, where n is the index of refraction of the medium. Note that the term spectral indicates dependence on wavelength. The variation of the spectral blackbody emissive power with wavelength is plotted in Figure 11–9 for selected temperatures. Several observations can be made from this figure: 1. The emitted radiation is a continuous function of wavelength. At any specified temperature, it increases with wavelength, reaches a peak, and then decreases with increasing wavelength. 2. At any wavelength, the amount of emitted radiation increases with increasing temperature. 3. As temperature increases, the curves shift to the left to the shorterwavelength region. Consequently, a larger fraction of the radiation is emitted at shorter wavelengths at higher temperatures. 4. The radiation emitted by the sun, which is considered to be a blackbody at 5780 K (or roughly at 5800 K), reaches its peak in the visible region of the spectrum. Therefore, the sun is in tune with our eyes. On the other hand, surfaces at T 800 K emit almost entirely in the infrared region and thus are not visible to the eye unless they reflect light coming from other sources. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 567 Violet Red 567 CHAPTER 11 1014 Visible light region 5800 K (Solar) 4000 K 1012 Locus of maximum power: λT = 2898 µm·K 2000 K 1010 Ebλ , W/ m2 ·µm 1000 K 500 K 108 300 K 106 100 K 104 102 1 0.01 0.1 1 10 100 1000 Wavelength λ, µm As the temperature increases, the peak of the curve in Figure 11–9 shifts toward shorter wavelengths. The wavelength at which the peak occurs for a specified temperature is given by Wien’s displacement law as ( T )max power 2897.8 m · K (11-5) This relation was originally developed by Willy Wien in 1894 using classical thermodynamics, but it can also be obtained by differentiating Eq. 11-4 with respect to while holding T constant and setting the result equal to zero. A plot of Wien’s displacement law, which is the locus of the peaks of the radiation emission curves, is also given in Figure 11–9. The peak of the solar radiation, for example, occurs at 2897.8/ 5780 0.50 m, which is near the middle of the visible range. The peak of the radiation emitted by a surface at room temperature (T 298 K) occurs at 9.72 m, which is well into the infrared region of the spectrum. An electrical resistance heater starts radiating heat soon after it is plugged in, and we can feel the emitted radiation energy by holding our hands facing the heater. But this radiation is entirely in the infrared region and thus cannot FIGURE 11–9 The variation of the blackbody emissive power with wavelength for several temperatures. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 568 568 HEAT TRANSFER Incident light Reflected Re d dw Re ello en Y re e G lu B w llo n Ye ree e G lu B Absorbed FIGURE 11–10 A surface that reflects red while absorbing the remaining parts of the incident light appears red to the eye. be sensed by our eyes. The heater would appear dull red when its temperature reaches about 1000 K, since it will start emitting a detectable amount (about 1 W/m2 · m) of visible red radiation at that temperature. As the temperature rises even more, the heater appears bright red and is said to be red hot. When the temperature reaches about 1500 K, the heater emits enough radiation in the entire visible range of the spectrum to appear almost white to the eye, and it is called white hot. Although it cannot be sensed directly by the human eye, infrared radiation can be detected by infrared cameras, which transmit the information to microprocessors to display visual images of objects at night. Rattlesnakes can sense the infrared radiation or the “body heat” coming off warm-blooded animals, and thus they can see at night without using any instruments. Similarly, honeybees are sensitive to ultraviolet radiation. A surface that reflects all of the light appears white, while a surface that absorbs all of the light incident on it appears black. (Then how do we see a black surface?) It should be clear from this discussion that the color of an object is not due to emission, which is primarily in the infrared region, unless the surface temperature of the object exceeds about 1000 K. Instead, the color of a surface depends on the absorption and reflection characteristics of the surface and is due to selective absorption and reflection of the incident visible radiation coming from a light source such as the sun or an incandescent lightbulb. A piece of clothing containing a pigment that reflects red while absorbing the remaining parts of the incident light appears “red” to the eye (Fig. 11–10). Leaves appear “green” because their cells contain the pigment chlorophyll, which strongly reflects green while absorbing other colors. It is left as an exercise to show that integration of the spectral blackbody emissive power Eb over the entire wavelength spectrum gives the total blackbody emissive power Eb: Eb(T) 0 Eb ( , T) d T4 (W/m2) (11-6) Thus, we obtained the Stefan–Boltzmann law (Eq. 11-3) by integrating Planck’s law (Eq. 11-4) over all wavelengths. Note that on an Eb – chart, Eb corresponds to any value on the curve, whereas Eb corresponds to the area under the entire curve for a specified temperature (Fig. 11–11). Also, the term total means “integrated over all wavelengths.” Ebλ Ebλ( λ, T ) Eb(T ) EXAMPLE 11–1 λ FIGURE 11–11 On an Eb – chart, the area under a curve for a given temperature represents the total radiation energy emitted by a blackbody at that temperature. Radiation Emission from a Black Ball Consider a 20-cm-diameter spherical ball at 800 K suspended in air as shown in Figure 11–12. Assuming the ball closely approximates a blackbody, determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 5 min, and (c) the spectral blackbody emissive power at a wavelength of 3 m. SOLUTION An isothermal sphere is suspended in air. The total blackbody emissive power, the total radiation emitted in 5 minutes, and the spectral blackbody emissive power at 3 mm are to be determined. Assumptions The ball behaves as a blackbody. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 569 569 CHAPTER 11 Analysis (a) The total blackbody emissive power is determined from the Stefan–Boltzmann law to be Eb T4 (5.67 10 8 W/m2 · K4)(800 K)4 23.2 103 W/m2 23.2 kW/m2 800 K That is, the ball emits 23.2 kJ of energy in the form of electromagnetic radiation per second per m2 of the surface area of the ball. 20 cm (b) The total amount of radiation energy emitted from the entire ball in 5 min is determined by multiplying the blackbody emissive power obtained above by the total surface area of the ball and the given time interval: D2 As (0.2 m)2 60 s (5 min) 1 min t EbAs t Qrad Ball 0.1257 m2 FIGURE 11–12 The spherical ball considered in Example 11–1. 300 s 1 kJ 1000 W · s (23.2 kW/m2)(0.1257 m2)(300 s) 876 kJ That is, the ball loses 876 kJ of its internal energy in the form of electromagnetic waves to the surroundings in 5 min, which is enough energy to raise the temperature of 1 kg of water by 50°C. Note that the surface temperature of the ball cannot remain constant at 800 K unless there is an equal amount of energy flow to the surface from the surroundings or from the interior regions of the ball through some mechanisms such as chemical or nuclear reactions. (c) The spectral blackbody emissive power at a wavelength of 3 m is determined from Planck’s distribution law to be 3.743 C1 Eb 5 exp C2 T 108 W · m4/m2 4 (3 m)5 exp 1 1.4387 10 m · K (3 m)(800 K) 1 3848 W/m2 · m The Stefan–Boltzmann law Eb(T ) T 4 gives the total radiation emitted by a blackbody at all wavelengths from 0 to . But we are often interested in the amount of radiation emitted over some wavelength band. For example, an incandescent lightbulb is judged on the basis of the radiation it emits in the visible range rather than the radiation it emits at all wavelengths. The radiation energy emitted by a blackbody per unit area over a wavelength band from 0 to is determined from (Fig. 11–13) Eb, 0– (T ) 0 Eb ( , T ) d (W/m2) Ebλ λ1 Eb, 0 – λ (T ) = Ebλ( λ, T ) d λ 1 0 Ebλ( λ, T ) (11-7) It looks like we can determine Eb, 0– by substituting the Eb relation from Eq. 11-4 and performing this integration. But it turns out that this integration does not have a simple closed-form solution, and performing a numerical integration each time we need a value of Eb, 0– is not practical. Therefore, we define a dimensionless quantity f called the blackbody radiation function as λ1 λ FIGURE 11–13 On an Eb – chart, the area under the curve to the left of the 1 line represents the radiation energy emitted by a blackbody in the wavelength range 0– 1 for the given temperature. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 570 570 HEAT TRANSFER TABLE 11–2 Blackbody radiation functions f T, m·K 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000 5200 5400 5600 5800 6000 T, m·K f 0.000000 0.000000 0.000000 0.000016 0.000321 0.002134 0.007790 0.019718 0.039341 0.066728 0.100888 0.140256 0.183120 0.227897 0.273232 0.318102 0.361735 0.403607 0.443382 0.480877 0.516014 0.548796 0.579280 0.607559 0.633747 0.658970 0.680360 0.701046 0.720158 0.737818 6200 6400 6600 6800 7000 7200 7400 7600 7800 8000 8500 9000 9500 10,000 10,500 11,000 11,500 12,000 13,000 14,000 15,000 16,000 18,000 20,000 25,000 30,000 40,000 50,000 75,000 100,000 f 0.754140 0.769234 0.783199 0.796129 0.808109 0.819217 0.829527 0.839102 0.848005 0.856288 0.874608 0.890029 0.903085 0.914199 0.923710 0.931890 0.939959 0.945098 0.955139 0.962898 0.969981 0.973814 0.980860 0.985602 0.992215 0.995340 0.997967 0.998953 0.999713 0.999905 Ebλ —— Eb fλ 1 – λ2 = f0 – λ – f0 – λ 2 1 Ebλ( λ, T ) ———–— Eb(T ) λ1 λ2 FIGURE 11–14 Graphical representation of the fraction of radiation emitted in the wavelength band from 1 to 2. λ 0 f (T) Eb ( , T ) d (11-8) T4 The function f represents the fraction of radiation emitted from a blackbody at temperature T in the wavelength band from 0 to . The values of f are listed in Table 11–2 as a function of T, where is in m and T is in K. The fraction of radiation energy emitted by a blackbody at temperature T over a finite wavelength band from 1 to 2 is determined from (Fig. 11–14) f 1– 2 (T ) f 2(T ) f 1(T ) (11-9) where f 1(T ) and f 2(T ) are blackbody radiation functions corresponding to 1T and 2T, respectively. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 571 571 CHAPTER 11 EXAMPLE 11–2 Emission of Radiation from a Lightbulb The temperature of the filament of an incandescent lightbulb is 2500 K. Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks. SOLUTION The temperature of the filament of an incandescent lightbulb is given. The fraction of visible radiation emitted by the filament and the wavelength at which the emission peaks are to be determined. Assumptions The filament behaves as a blackbody. Analysis The visible range of the electromagnetic spectrum extends from 0.4 m to 2 0.76 m. Noting that T 2500 K, the blackbody radia1 tion functions corresponding to 1T and 2T are determined from Table 11–2 to be 1T 2T (0.40 m)(2500 K) (0.76 m)(2500 K) 1000 m · K 1900 m · K → f → f 1 2 0.000321 0.053035 That is, 0.03 percent of the radiation is emitted at wavelengths less than 0.4 m and 5.3 percent at wavelengths less than 0.76 m. Then the fraction of radiation emitted between these two wavelengths is (Fig. 11–15) f 1– 2 f 2 f 1 0.053035 0.000321 Ebλ —— Eb f0.4 – 0.76 = f0 – 0.76 – f0 – 0.4 0.0527135 Therefore, only about 5 percent of the radiation emitted by the filament of the lightbulb falls in the visible range. The remaining 95 percent of the radiation appears in the infrared region in the form of radiant heat or “invisible light,” as it used to be called. This is certainly not a very efficient way of converting electrical energy to light and explains why fluorescent tubes are a wiser choice for lighting. The wavelength at which the emission of radiation from the filament peaks is easily determined from Wien’s displacement law to be ( T )max power 2897.8 m · K → max power 2897.8 m · K 2500 K 1.16 m Discussion Note that the radiation emitted from the filament peaks in the infrared region. 11–4 I RADIATION INTENSITY Radiation is emitted by all parts of a plane surface in all directions into the hemisphere above the surface, and the directional distribution of emitted (or incident) radiation is usually not uniform. Therefore, we need a quantity that describes the magnitude of radiation emitted (or incident) in a specified direction in space. This quantity is radiation intensity, denoted by I. Before we can describe a directional quantity, we need to specify direction in space. The direction of radiation passing through a point is best described in spherical coordinates in terms of the zenith angle and the azimuth angle , as shown in 0.4 0.76 1.16 λ, µ m FIGURE 11–15 Graphical representation of the fraction of radiation emitted in the visible range in Example 11–2. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 572 572 HEAT TRANSFER Figure 11–16. Radiation intensity is used to describe how the emitted radiation varies with the zenith and azimuth angles. If all surfaces emitted radiation uniformly in all directions, the emissive power would be sufficient to quantify radiation, and we would not need to deal with intensity. The radiation emitted by a blackbody per unit normal area is the same in all directions, and thus there is no directional dependence. But this is not the case for real surfaces. Before we define intensity, we need to quantify the size of an opening in space. z Emitted radiation I (θ , φ) θ dA y φ x FIGURE 11–16 Radiation intensity is used to describe the variation of radiation energy with direction. Plain angle, Arc length A slice of pizza of plain angle Solid Angle Let us try to quantify the size of a slice of pizza. One way of doing that is to specify the arc length of the outer edge of the slice, and to form the slice by connecting the endpoints of the arc to the center. A more general approach is to specify the angle of the slice at the center, as shown in Figure 11–17. An angle of 90˚ (or /2 radians), for example, always represents a quarter pizza, no matter what the radius is. For a circle of unit radius, the length of an arc is equivalent in magnitude to the plane angle it subtends (both are 2 for a complete circle of radius r 1). Now consider a watermelon, and let us attempt to quantify the size of a slice. Again we can do it by specifying the outer surface area of the slice (the green part), or by working with angles for generality. Connecting all points at the edges of the slice to the center in this case will form a three-dimensional body (like a cone whose tip is at the center), and thus the angle at the center in this case is properly called the solid angle. The solid angle is denoted by , and its unit is the steradian (sr). In analogy to plane angle, we can say that the area of a surface on a sphere of unit radius is equivalent in magnitude to the solid angle it subtends (both are 4 for a sphere of radius r 1). This can be shown easily by considering a differential surface area on a sphere dS r2 sin d d , as shown in Figure 11–18, and integrating it from 0 to , and from 0 to 2 . We get 2 S Solid angle, sphere Surface area, S A slice of watermelon of solid angle FIGURE 11–17 Describing the size of a slice of pizza by a plain angle, and the size of a watermelon slice by a solid angle. r 2 sin d dS 0 2 r2 0 sin d 4 r2 (11-10) 0 which is the formula for the area of a sphere. For r 1 it reduces to S 4 , and thus the solid angle associated with a sphere is 4 sr. For a hemisphere, which is more relevant to radiation emitted or received by a surface, it is 2 sr. The differential solid angle d subtended by a differential area dS on a sphere of radius r can be expressed as d dS r2 sin d d (11-11) Note that the area dS is normal to the direction of viewing since dS is viewed from the center of the sphere. In general, the differential solid angle d subtended by a differential surface area dA when viewed from a point at a distance r from dA is expressed as d dAn r2 d A cos r2 (11-12) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 573 573 CHAPTER 11 0 0 θ φ Radiation emitted into direction (θ ,φ) π/2 2π dS r 2 sin θ dθ dφ Ie (θ ,φ) r Solid angle: d dS/r 2 sin θ d θ dφ θ φ r dθ r sin θ dS dA dφ r sin θ dφ θ P dθ dφ Solid angle for a hemisphere: d Hemisphere π/2 2π sin θ d θ d φ r dφ 2π 0θ0φ where is the angle between the normal of the surface and the direction of viewing, and thus dAn dA cos is the normal (or projected) area to the direction of viewing. Small surfaces viewed from relatively large distances can approximately be treated as differential areas in solid angle calculations. For example, the solid angle subtended by a 5 cm2 plane surface when viewed from a point O at a distance of 80 cm along the normal of the surface is An r2 5 cm 2 (80 cm) 2 7.81 10 4 sr If the surface is tilted so that the normal of the surface makes an angle of 60˚ with the line connecting point O to the center of the surface, the pro(5 cm2)cos 60˚ 2.5 cm2, and the jected area would be dAn dA cos solid angle in this case would be half of the value just determined. Intensity of Emitted Radiation Consider the emission of radiation by a differential area element dA of a surface, as shown in Figure 11–18. Radiation is emitted in all directions into the hemispherical space, and the radiation streaming though the surface area dS is proportional to the solid angle d subtended by dS. It is also proportional to the radiating area dA as seen by an observer on dS, which varies from a maximum of dA when dS is at the top directly above dA ( 0˚) to a minimum of zero when dS is at the bottom ( 90˚). Therefore, the effective area of dA for emission in the direction of is the projection of dA on a plane normal to , which is dA cos . Radiation intensity in a given direction is based on a unit area normal to that direction to provide a common basis for the comparison of radiation emitted in different directions. The radiation intensity for emitted radiation Ie( , ) is defined as the rate · at which radiation energy dQ e is emitted in the ( , ) direction per unit area normal to this direction and per unit solid angle about this direction. That is, Ie( , ) ˙ d Qe d A cos d (r sin θ dφ )(r dθ ) ˙ dQe d A cos sin d d (W/m2 sr) (11-13) FIGURE 11–18 The emission of radiation from a differential surface element into the surrounding hemispherical space through a differential solid angle. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 574 574 HEAT TRANSFER The radiation flux for emitted radiation is the emissive power E (the rate at which radiation energy is emitted per unit area of the emitting surface), which can be expressed in differential form as ˙ dQ e dA dE Ie( , ) cos sin d d (11-14) Noting that the hemisphere above the surface will intercept all the radiation rays emitted by the surface, the emissive power from the surface into the hemisphere surrounding it can be determined by integration as 2 E /2 dE 0 hemisphere 0 Ie( , ) cos sin d d (W/m 2) (11-15) The intensity of radiation emitted by a surface, in general, varies with direction (especially with the zenith angle ). But many surfaces in practice can be approximated as being diffuse. For a diffusely emitting surface, the intensity of the emitted radiation is independent of direction and thus Ie constant. 2 /2 Noting that cos 0 sin d d , the emissive power relation in 0 Eq. 11-15 reduces in this case to A Projected area An A cos θ Diffusely emitting surface: n θ Solid angle, FIGURE 11–19 Radiation intensity is based on projected area, and thus the calculation of radiation emission from a surface involves the projection of the surface. (W/m2) Ie (11-16) Note that the factor in Eq. 11-16 is . You might have expected it to be 2 since intensity is radiation energy per unit solid angle, and the solid angle associated with a hemisphere is 2 . The reason for the factor being is that the emissive power is based on the actual surface area whereas the intensity is based on the projected area (and thus the factor cos that accompanies it), as shown in Figure 11–19. For a blackbody, which is a diffuse emitter, Eq. 11-16 can be expressed as Blackbody: Eb Ib (11-17) where Eb T4 is the blackbody emissive power. Therefore, the intensity of the radiation emitted by a blackbody at absolute temperature T is Blackbody: Incident radiation E Eb( T ) Ib(T ) T4 (W/m2 sr) (11-18) I i (θ , φ ) θ dA φ FIGURE 11–20 Radiation incident on a surface in the direction ( , ). Incident Radiation All surfaces emit radiation, but they also receive radiation emitted or reflected by other surfaces. The intensity of incident radiation Ii( , ) is defined as the rate at which radiation energy dG is incident from the ( , ) direction per unit area of the receiving surface normal to this direction and per unit solid angle about this direction (Fig. 11–20). Here is the angle between the direction of incident radiation and the normal of the surface. The radiation flux incident on a surface from all directions is called irradiation G, and is expressed as 2 G /2 dG hemisphere 0 0 Ii ( , ) cos sin d d (W/m2) (11-19) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 575 575 CHAPTER 11 Therefore irradiation represents the rate at which radiation energy is incident on a surface per unit area of the surface. When the incident radiation is diffuse and thus Ii constant, Eq. 11-19 reduces to Diffusely incident radiation: G (W/m2) Ii (11-20) Again note that irradiation is based on the actual surface area (and thus the factor cos ), whereas the intensity of incident radiation is based on the projected area. Radiosity Surfaces emit radiation as well as reflecting it, and thus the radiation leaving a surface consists of emitted and reflected components, as shown in Figure 11–21. The calculation of radiation heat transfer between surfaces involves the total radiation energy streaming away from a surface, with no regard for its origin. Thus, we need to define a quantity that represents the rate at which radiation energy leaves a unit area of a surface in all directions. This quantity is called the radiosity J, and is expressed as 2 /2 J 0 Ie r( , ) cos sin d d (W/m2) (11-21) 0 where Ie r is the sum of the emitted and reflected intensities. For a surface that is both a diffuse emitter and a diffuse reflector, Ie r constant, and the radiosity relation reduces to Diffuse emitter and reflector: J Ie r (W/m2) (11-22) For a blackbody, radiosity J is equivalent to the emissive power Eb since a blackbody absorbs the entire radiation incident on it and there is no reflected component in radiosity. Spectral Quantities So far we considered total radiation quantities (quantities integrated over all wavelengths), and made no reference to wavelength dependence. This lumped approach is adequate for many radiation problems encountered in practice. But sometimes it is necessary to consider the variation of radiation with wavelength as well as direction, and to express quantities at a certain wavelength or per unit wavelength interval about . Such quantities are referred to as spectral quantities to draw attention to wavelength dependence. The modifier “spectral” is used to indicate “at a given wavelength.” The spectral radiation intensity I ( , , ), for example, is simply the total radiation intensity I( , ) per unit wavelength interval about . The spectral intensity for emitted radiation I , e( , , ) can be defined as the rate at which · radiation energy dQ e is emitted at the wavelength in the ( , ) direction per unit area normal to this direction, per unit solid angle about this direction, and it can be expressed as I , e( , , ) ˙ dQe d A cos d dλ (W/m2 sr m) (11-23) Radiosity, J (Reflected irradiation) Irradiation, G Emissive power, E FIGURE 11–21 The three kinds of radiation flux (in W/m2): emissive power, irradiation, and radiosity. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 576 576 HEAT TRANSFER Then the spectral emissive power becomes 2 /2 E 0 Iλ, e Area 0 Iλ, e dλ (W/m2) Iλ, e ( , , )cos sin d d (11-24) Similar relations can be obtained for spectral irradiation G , and spectral radiosity J by replacing I , e in this equation by I , i and I , e r, respectively. When the variation of spectral radiation intensity I with wavelength is known, the total radiation intensity I for emitted, incident, and emitted reflected radiation can be determined by integration over the entire wavelength spectrum as (Fig. 11–22) Ie Iλ, e dλ 0 λ FIGURE 11–22 Integration of a “spectral” quantity for all wavelengths gives the “total” quantity. Ie 0 Iλ, e d λ , Ii 0 Iλ, i d λ , and Ie r 0 Iλ, e r dλ (11-25) These intensities can then be used in Eqs. 11-15, 11-19, and 11-21 to determine the emissive power E, irradiation G, and radiosity J, respectively. Similarly, when the variations of spectral radiation fluxes E , G , and J with wavelength are known, the total radiation fluxes can be determined by integration over the entire wavelength spectrum as E 0 Eλd λ, G 0 G λ d λ, and J 0 Jλ d λ (11-26) When the surfaces and the incident radiation are diffuse, the spectral radiation fluxes are related to spectral intensities as Eλ Iλ, e , Gλ Iλ, i , and Jλ Iλ, e r (11-27) Note that the relations for spectral and total radiation quantities are of the same form. The spectral intensity of radiation emitted by a blackbody at an absolute temperature T at a wavelength has been determined by Max Planck, and is expressed as Ib λ( λ , T ) 2 hc 2 0 λ5[exp(hc 0 / λ kT ) 1] (W/m2 sr m) (11-28) where h 6.6256 10 34 J s is the Planck constant, k 1.38065 10 23 J/K is the Boltzmann constant, and c0 2.9979 108 m/s is the speed of light in a vacuum. Then the spectral blackbody emissive power is, from Eq. 11-27, Eb ( , T ) A2 θ2 θ1 A1 T1 55° r 5 cm2 Ib ( , T ) (11-29) A simplified relation for Eb is given by Eq. 11-4. 40° 75 cm 3 cm2 600 K FIGURE 11–23 Schematic for Example 11–3. EXAMPLE 11–3 Radiation Incident on a Small Surface A small surface of area A1 3 cm2 emits radiation as a blackbody at T1 600 K. Part of the radiation emitted by A1 strikes another small surface of area A2 5 cm2 oriented as shown in Figure 11–23. Determine the solid angle subtended by A2 when viewed from A1, and the rate at which radiation emitted by A1 that strikes A2. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 577 577 CHAPTER 11 SOLUTION A surface is subjected to radiation emitted by another surface. The solid angle subtended and the rate at which emitted radiation is received are to be determined. Assumptions 1 Surface A1 emits diffusely as a blackbody. 2 Both A1 and A2 can be approximated as differential surfaces since both are very small compared to the square of the distance between them. Analysis Approximating both A1 and A2 as differential surfaces, the solid angle subtended by A2 when viewed from A1 can be determined from Eq. 11-12 to be An, 2 2–1 r 2 A2 cos r2 (5 cm2) cos 40 º (75 cm) 2 2 6.81 10 4 sr since the normal of A2 makes 40˚ with the direction of viewing. Note that solid angle subtended by A2 would be maximum if A2 were positioned normal to the direction of viewing. Also, the point of viewing on A1 is taken to be a point in the middle, but it can be any point since A1 is assumed to be very small. The radiation emitted by A1 that strikes A2 is equivalent to the radiation emitted by A1 through the solid angle 2–1. The intensity of the radiation emitted by A1 is I1 Eh(T1) T4 1 (5.67 10 8 W/m2 K4)(600 K)4 2339 W/m 2 sr This value of intensity is the same in all directions since a blackbody is a diffuse emitter. Intensity represents the rate of radiation emission per unit area normal to the direction of emission per unit solid angle. Therefore, the rate of radiation energy emitted by A1 in the direction of 1 through the solid angle 2–1 is determined by multiplying I1 by the area of A1 normal to 1 and the solid angle 2–1. That is, · Q 1–2 I1(A1 cos 1) 2–1 (2339 W/m2 sr)(3 2.74 10 4 W 10 4 cos 55˚ m2)(6.81 10 4 sr) Therefore, the radiation emitted from surface A1 will strike surface A2 at a rate of 2.74 10 4 W. · 4 Discussion The total rate of radiation emission from surface A1 is Q e A1 T1 2.204 W. Therefore, the fraction of emitted radiation that strikes A2 is 2.74 10 4/2.204 0.00012 (or 0.012 percent). Noting that the solid angle associated with a hemisphere is 2 , the fraction of the solid angle subtended by A2 is 6.81 10 4/(2 ) 0.000108 (or 0.0108 percent), which is 0.9 times the fraction of emitted radiation. Therefore, the fraction of the solid angle a surface occupies does not represent the fraction of radiation energy the surface will receive even when the intensity of emitted radiation is constant. This is because radiation energy emitted by a surface in a given direction is proportional to the projected area of the surface in that direction, and reduces from a maximum at 0˚ (the direction normal to surface) to zero at 90˚ (the direction parallel to surface). 11–5 I RADIATIVE PROPERTIES Most materials encountered in practice, such as metals, wood, and bricks, are opaque to thermal radiation, and radiation is considered to be a surface cen58933_ch11.qxd 9/9/2002 9:38 AM Page 578 578 HEAT TRANSFER phenomenon for such materials. That is, thermal radiation is emitted or absorbed within the first few microns of the surface, and thus we speak of radiative properties of surfaces for opaque materials. Some other materials, such as glass and water, allow visible radiation to penetrate to considerable depths before any significant absorption takes place. Radiation through such semitransparent materials obviously cannot be considered to be a surface phenomenon since the entire volume of the material interacts with radiation. On the other hand, both glass and water are practically opaque to infrared radiation. Therefore, materials can exhibit different behavior at different wavelengths, and the dependence on wavelength is an important consideration in the study of radiative properties such as emissivity, absorptivity, reflectivity, and transmissivity of materials. In the preceding section, we defined a blackbody as a perfect emitter and absorber of radiation and said that no body can emit more radiation than a blackbody at the same temperature. Therefore, a blackbody can serve as a convenient reference in describing the emission and absorption characteristics of real surfaces. Emissivity The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature. The emissivity of a surface is denoted by , and it varies between zero and one, 0 1. Emissivity is a measure of how closely a surface approximates a blackbody, for which 1. The emissivity of a real surface is not a constant. Rather, it varies with the temperature of the surface as well as the wavelength and the direction of the emitted radiation. Therefore, different emissivities can be defined for a surface, depending on the effects considered. The most elemental emissivity of a surface at a given temperature is the spectral directional emissivity, which is defined as the ratio of the intensity of radiation emitted by the surface at a specified wavelength in a specified direction to the intensity of radiation emitted by a blackbody at the same temperature at the same wavelength. That is, , ( , , , T) Iλ , e (λ, , , T ) Ibλ( λ , T ) (11-30) where the subscripts and are used to designate spectral and directional quantities, respectively. Note that blackbody radiation intensity is independent of direction, and thus it has no functional dependence on and . The total directional emissivity is defined in a like manner by using total intensities (intensities integrated over all wavelengths) as ( , , T) Ie( , , T ) Ib(T ) (11-31) In practice, it is usually more convenient to work with radiation properties averaged over all directions, called hemispherical properties. Noting that the integral of the rate of radiation energy emitted at a specified wavelength per unit surface area over the entire hemisphere is spectral emissive power, the spectral hemispherical emissivity can be expressed as cen58933_ch11.qxd 9/9/2002 9:38 AM Page 579 579 CHAPTER 11 Eλ( λ , T ) E b λ( λ , T ) ( , T) (11-32) Note that the emissivity of a surface at a given wavelength can be different at different temperatures since the spectral distribution of emitted radiation (and thus the amount of radiation emitted at a given wavelength) changes with temperature. Finally, the total hemispherical emissivity is defined in terms of the radiation energy emitted over all wavelengths in all directions as E(T ) Eb(T ) (T ) (11-33) Therefore, the total hemispherical emissivity (or simply the “average emissivity”) of a surface at a given temperature represents the ratio of the total radiation energy emitted by the surface to the radiation emitted by a blackbody of the same surface area at the same temperature. Ed Noting from Eqs. 11-26 and 11-32 that E and E ( , T ) 0 ( , T )Eb ( , T ), and the total hemispherical emissivity can also be expressed as E(T ) E b( T ) (T ) λ( λ , 0 T )Ebλ( λ , T )d λ (11-34) T4 since Eb(T ) T 4. To perform this integration, we need to know the variation of spectral emissivity with wavelength at the specified temperature. The integrand is usually a complicated function, and the integration has to be performed numerically. However, the integration can be performed quite easily by dividing the spectrum into a sufficient number of wavelength bands and assuming the emissivity to remain constant over each band; that is, by expressing the function ( , T) as a step function. This simplification offers great convenience for little sacrifice of accuracy, since it allows us to transform the integration into a summation in terms of blackbody emission functions. As an example, consider the emissivity function plotted in Figure 11–24. It seems like this function can be approximated reasonably well by a step function of the form constant, constant, constant, 1 • 0 ελ 1 (11-35) Then the average emissivity can be determined from Eq. 11-34 by breaking the integral into three parts and utilizing the definition of the blackbody radiation function as ε1 ε2 ε2 2 3 λ1 1 (T) 0 (T) 1 2 2 λ2 Eb λd λ 2 Eb 1 f0 1 2 λ1 Eb λd λ Eb f 1 2(T) 3 3 f 2 λ2 Eb λd λ Eb (T) ε1 ε3 ε3 λ1 (11-36) Radiation is a complex phenomenon as it is, and the consideration of the wavelength and direction dependence of properties, assuming sufficient data Actual variation λ2 FIGURE 11–24 Approximating the actual variation of emissivity with wavelength by a step function. λ cen58933_ch11.qxd 9/9/2002 9:38 AM Page 580 580 HEAT TRANSFER Real surface: εθ ≠ constant ελ ≠ constant Diffuse surface: εθ = constant Gray surface: ελ = constant Diffuse, gray surface: ε = ελ = εθ = constant FIGURE 11–25 The effect of diffuse and gray approximations on the emissivity of a surface. 1 Nonconductor εθ 0.5 Conductor 0 0° 15° 30° 45° θ 60° 75° 90° FIGURE 11–26 Typical variations of emissivity with direction for electrical conductors and nonconductors. exist, makes it even more complicated. Therefore, the gray and diffuse approximations are often utilized in radiation calculations. A surface is said to be diffuse if its properties are independent of direction, and gray if its properties are independent of wavelength. Therefore, the emissivity of a gray, diffuse surface is simply the total hemispherical emissivity of that surface because of independence of direction and wavelength (Fig. 11–25). A few comments about the validity of the diffuse approximation are in order. Although real surfaces do not emit radiation in a perfectly diffuse manner as a blackbody does, they often come close. The variation of emissivity with direction for both electrical conductors and nonconductors is given in Figure 11–26. Here is the angle measured from the normal of the surface, and thus 0 for remains radiation emitted in a direction normal to the surface. Note that nearly constant for about 40˚ for conductors such as metals and for 70˚ for nonconductors such as plastics. Therefore, the directional emissivity of a surface in the normal direction is representative of the hemispherical emissivity of the surface. In radiation analysis, it is common practice to assume the surfaces to be diffuse emitters with an emissivity equal to the value in the normal ( 0) direction. The effect of the gray approximation on emissivity and emissive power of a real surface is illustrated in Figure 11–27. Note that the radiation emission from a real surface, in general, differs from the Planck distribution, and the emission curve may have several peaks and valleys. A gray surface should emit as much radiation as the real surface it represents at the same temperature. Therefore, the areas under the emission curves of the real and gray surfaces must be equal. The emissivities of common materials are listed in Table A–18 in the appendix, and the variation of emissivity with wavelength and temperature is illustrated in Figure 11–28. Typical ranges of emissivity of various materials are given in Figure 11–29. Note that metals generally have low emissivities, as low as 0.02 for polished surfaces, and nonmetals such as ceramics and organic materials have high ones. The emissivity of metals increases with temperature. Also, oxidation causes significant increases in the emissivity of metals. Heavily oxidized metals can have emissivities comparable to those of nonmetals. ελ Eλ Blackbody, ε = 1 1 Blackbody, Ebλ Gray surface, ε = const. Gray surface, Eλ = ε Ebλ ε Real surface, ελ FIGURE 11–27 Comparison of the emissivity (a) and emissive power (b) of a real surface with those of a gray surface and a blackbody at the same temperature. T = const. 0 T = const. Real surface, Eλ = ελEbλ λ 0 (a) λ (b) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 581 581 CHAPTER 11 1.0 0.8 Silicon carbide, 1000 K 0.6 Tungsten 1600 K Aluminum oxide, 1400 K 0.4 Stainless steel, 1200 K heavily oxidized 0.2 Stainless steel, 800 K lightly oxidized 2800 K 0 0.1 0.2 0.4 0.6 1 2 46 Total normal emissivity, ε n Spectral, normal emissivity, ε λ, n 1.0 10 20 Heavily oxidized stainless steel 0.8 0.6 Aluminum oxide 0.4 Lightly oxidized stainless steel 0.2 Tungsten 0 40 60 100 0 500 0 1000 Wavelength, λ, µm 1500 2000 2500 3000 3500 Temperature, K (a) (b) FIGURE 11–28 The variation of normal emissivity with (a) wavelength and (b) temperature for various materials. Vegetation, water, skin Care should be exercised in the use and interpretation of radiation property data reported in the literature, since the properties strongly depend on the surface conditions such as oxidation, roughness, type of finish, and cleanliness. Consequently, there is considerable discrepancy and uncertainty in the reported values. This uncertainty is largely due to the difficulty in characterizing and describing the surface conditions precisely. EXAMPLE 11–4 Building materials, paints Rocks, soil Glasses, minerals Carbon Ceramics Oxidized metals Metals, unpolished Polished metals Emissivity of a Surface and Emissive Power The spectral emissivity function of an opaque surface at 800 K is approximated as (Fig. 11–30) 0.3, 0.8, 0.1, 1 2 3 0 3m 7m with wavelength is given. The average emissivity of the surface and its emissive power are to be determined. Analysis The variation of the emissivity of the surface with wavelength is given as a step function. Therefore, the average emissivity of the surface can be determined from Eq. 11-34 by breaking the integral into three parts, (T ) 0 2 Eb d 2 1 2 T4 1 f0– 1(T ) 2(f 2 2 f 1– 2 f 1) Eb d 3 1 T4 1f Eb d (T ) T4 3 3(1 0.4 0.6 0.8 1.0 ελ SOLUTION The variation of emissivity of a surface at a specified temperature 1 0.2 FIGURE 11–29 Typical ranges of emissivity for various materials. 3m 7m Determine the average emissivity of the surface and its emissive power. 1 0 f 2– (T ) f 2) 1.0 0.8 0.3 0.1 0 0 3 7 λ, µm FIGURE 11–30 The spectral emissivity of the surface considered in Example 11–4. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 582 582 HEAT TRANSFER where f 1 and f 2 are blackbody radiation functions corresponding to 2T. These functions are determined from Table 11–2 to be 1T 2T (3 m)(800 K) (7 m)(800 K) Note that f0– 1 f1 f0 since f 1. Substituting, 0.3 0.140256 0.521 2400 m · K 5600 m · K →f →f f 1, since f0 0, and f 0.8(0.701046 1 2 0.140256) and 0.140256 0.701046 f 2– 1T 0.1(1 f 2 1 f 2, 0.701046) That is, the surface will emit as much radiation energy at 800 K as a gray surface having a constant emissivity of 0.521. The emissive power of the surface is E T4 0.521(5.67 10 8 W/m2 · K4)(800 K)4 12,100 W/m2 Discussion Note that the surface emits 12.1 kJ of radiation energy per second per m2 area of the surface. Absorptivity, Reflectivity, and Transmissivity Incident radiation G, W/m 2 Reflected ρG Everything around us constantly emits radiation, and the emissivity represents the emission characteristics of those bodies. This means that every body, including our own, is constantly bombarded by radiation coming from all directions over a range of wavelengths. Recall that radiation flux incident on a surface is called irradiation and is denoted by G. When radiation strikes a surface, part of it is absorbed, part of it is reflected, and the remaining part, if any, is transmitted, as illustrated in Figure 11–31. The fraction of irradiation absorbed by the surface is called the absorptivity , the fraction reflected by the surface is called the reflectivity , and the fraction transmitted is called the transmissivity . That is, Absorptivity: Semitransparent material Absorbed αG Reflectivity: Transmissivity: Transmitted τG FIGURE 11–31 The absorption, reflection, and transmission of incident radiation by a semitransparent material. Gabs , G Gref , Reflected radiation G Incident radiation Transmitted radiation Gtr , G Incident radiation Absorbed radiation Incident radiation 0 1 (11-37) 0 1 (11-38) 0 1 (11-39) where G is the radiation energy incident on the surface, and Gabs, Gref, and Gtr are the absorbed, reflected, and transmitted portions of it, respectively. The first law of thermodynamics requires that the sum of the absorbed, reflected, and transmitted radiation energy be equal to the incident radiation. That is, Gabs Gref Gtr G (11-40) Dividing each term of this relation by G yields 1 For opaque surfaces, 0, and thus (11-41) cen58933_ch11.qxd 9/9/2002 9:38 AM Page 583 583 CHAPTER 11 1 (11-42) This is an important property relation since it enables us to determine both the absorptivity and reflectivity of an opaque surface by measuring either of these properties. These definitions are for total hemispherical properties, since G represents the radiation flux incident on the surface from all directions over the hemispherical space and over all wavelengths. Thus, , , and are the average properties of a medium for all directions and all wavelengths. However, like emissivity, these properties can also be defined for a specific wavelength and/or direction. For example, the spectral directional absorptivity and spectral directional reflectivity of a surface are defined, respectively, as the absorbed and reflected fractions of the intensity of radiation incident at a specified wavelength in a specified direction as λ, Iλ , abs ( λ , , ) (λ, , ) Iλ , i ( λ , , ) and λ, Iλ, ref (λ , , ) (λ, , ) Iλ, i ( λ , , ) (11-43) Likewise, the spectral hemispherical absorptivity and spectral hemispherical reflectivity of a surface are defined as and () G λ , ref ( λ) Gλ( λ) (11-44) where G is the spectral irradiation (in W/m2 m) incident on the surface, and G , abs and G , ref are the reflected and absorbed portions of it, respectively. Similar quantities can be defined for the transmissivity of semitransparent materials. For example, the spectral hemispherical transmissivity of a medium can be expressed as G λ , tr ( λ) The average absorptivity, reflectivity, and transmissivity of a surface can also be defined in terms of their spectral counterparts as Gd 0 Gd , 0 0 0 Incident ray Re fle cte dr (b) ay s (11-46) Gd 0 Normal Gd , Gd (a) (11-45) G λ( λ) Normal Reflected rays Gd 0 The reflectivity differs somewhat from the other properties in that it is bidirectional in nature. That is, the value of the reflectivity of a surface depends not only on the direction of the incident radiation but also the direction of reflection. Therefore, the reflected rays of a radiation beam incident on a real surface in a specified direction will form an irregular shape, as shown in Figure 11–32. Such detailed reflectivity data do not exist for most surfaces, and even if they did, they would be of little value in radiation calculations since this would usually add more complication to the analysis than it is worth. In practice, for simplicity, surfaces are assumed to reflect in a perfectly specular or diffuse manner. In specular (or mirrorlike) reflection, the angle of reflection equals the angle of incidence of the radiation beam. In diffuse reflection, radiation is reflected equally in all directions, as shown in Figure Incident ray (c) Normal θ ra y () Incident ray θ ted Gλ( λ) lec G λ , abs( λ) Re f () FIGURE 11–32 Different types of reflection from a surface: (a) actual or irregular, (b) diffuse, and (c) specular or mirrorlike. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 584 584 HEAT TRANSFER 1.0 7 9 Absorptivity, α 0.8 6 0.6 0.4 0.2 1. White fireclay 2. Asbestos 3. Cork 4. Wood 5. Porcelain 6. Concrete 7. Roof shingles 8. Aluminum 9. Graphite 5 2 3 4 8 1 0 300400 600 1000 2000 4000 6000 Source temperature, K FIGURE 11–33 Variation of absorptivity with the temperature of the source of irradiation for various common materials at room temperature. 11–32. Reflection from smooth and polished surfaces approximates specular reflection, whereas reflection from rough surfaces approximates diffuse reflection. In radiation analysis, smoothness is defined relative to wavelength. A surface is said to be smooth if the height of the surface roughness is much smaller than the wavelength of the incident radiation. Unlike emissivity, the absorptivity of a material is practically independent of surface temperature. However, the absorptivity depends strongly on the temperature of the source at which the incident radiation is originating. This is also evident from Figure 11–33, which shows the absorptivities of various materials at room temperature as functions of the temperature of the radiation source. For example, the absorptivity of the concrete roof of a house is about 0.6 for solar radiation (source temperature: 5780 K) and 0.9 for radiation originating from the surrounding trees and buildings (source temperature: 300 K), as illustrated in Figure 11–34. Notice that the absorptivity of aluminum increases with the source temperature, a characteristic for metals, and the absorptivity of electric nonconductors, in general, decreases with temperature. This decrease is most pronounced for surfaces that appear white to the eye. For example, the absorptivity of a white painted surface is low for solar radiation, but it is rather high for infrared radiation. Kirchhoff’s Law Sun α = 0.9 α = 0.6 Consider a small body of surface area As , emissivity , and absorptivity at temperature T contained in a large isothermal enclosure at the same temperature, as shown in Figure 11–35. Recall that a large isothermal enclosure forms a blackbody cavity regardless of the radiative properties of the enclosure surface, and the body in the enclosure is too small to interfere with the blackbody nature of the cavity. Therefore, the radiation incident on any part of the surface of the small body is equal to the radiation emitted by a blackbody at temT 4, and the radiation absorbed by the small perature T. That is, G Eb(T ) body per unit of its surface area is Gabs T4 G The radiation emitted by the small body is FIGURE 11–34 The absorptivity of a material may be quite different for radiation originating from sources at different temperatures. T4 Eemit Considering that the small body is in thermal equilibrium with the enclosure, the net rate of heat transfer to the body must be zero. Therefore, the radiation emitted by the body must be equal to the radiation absorbed by it: As T4 As T4 Thus, we conclude that (T ) (T ) (11-47) That is, the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature. This relation, which greatly simplifies the radiation analysis, was first developed by Gustav Kirchhoff in 1860 and is now called Kirchhoff’s law. Note that this relation is derived under the condition cen58933_ch11.qxd 9/9/2002 9:38 AM Page 585 585 CHAPTER 11 that the surface temperature is equal to the temperature of the source of irradiation, and the reader is cautioned against using it when considerable difference (more than a few hundred degrees) exists between the surface temperature and the temperature of the source of irradiation. The derivation above can also be repeated for radiation at a specified wavelength to obtain the spectral form of Kirchhoff ’s law: T T As , ε, α (T ) (T ) (11-48) This relation is valid when the irradiation or the emitted radiation is independent of direction. The form of Kirchhoff ’s law that involves no restrictions is the spectral directional form expressed as , (T ) , (T ). That is, the emissivity of a surface at a specified wavelength, direction, and temperature is always equal to its absorptivity at the same wavelength, direction, and temperature. It is very tempting to use Kirchhoff ’s law in radiation analysis since the relation together with 1 enables us to determine all three properties of an opaque surface from a knowledge of only one property. Although Eq. 11-47 gives acceptable results in most cases, in practice, care should be exercised when there is considerable difference between the surface temperature and the temperature of the source of incident radiation. G Eemit FIGURE 11–35 The small body contained in a large isothermal enclosure used in the development of Kirchhoff ’s law. Visible 1.0 0.8 The Greenhouse Effect You have probably noticed that when you leave your car under direct sunlight on a sunny day, the interior of the car gets much warmer than the air outside, and you may have wondered why the car acts like a heat trap. The answer lies in the spectral transmissivity curve of the glass, which resembles an inverted U, as shown in Figure 11–36. We observe from this figure that glass at thicknesses encountered in practice transmits over 90 percent of radiation in the visible range and is practically opaque (nontransparent) to radiation in the longer-wavelength infrared regions of the electromagnetic spectrum (roughly 3 m). Therefore, glass has a transparent window in the wavelength range 0.3 m 3 m in which over 90 percent of solar radiation is emitted. On the other hand, the entire radiation emitted by surfaces at room temperature falls in the infrared region. Consequently, glass allows the solar radiation to enter but does not allow the infrared radiation from the interior surfaces to escape. This causes a rise in the interior temperature as a result of the energy build-up in the car. This heating effect, which is due to the nongray characteristic of glass (or clear plastics), is known as the greenhouse effect, since it is utilized primarily in greenhouses (Fig. 11–37). The greenhouse effect is also experienced on a larger scale on earth. The surface of the earth, which warms up during the day as a result of the absorption of solar energy, cools down at night by radiating its energy into deep space as infrared radiation. The combustion gases such as CO2 and water vapor in the atmosphere transmit the bulk of the solar radiation but absorb the infrared radiation emitted by the surface of the earth. Thus, there is concern that the energy trapped on earth will eventually cause global warming and thus drastic changes in weather patterns. In humid places such as coastal areas, there is not a large change between the daytime and nighttime temperatures, because the humidity acts as a barrier τλ 0.6 Thickness 0.038 cm 0.318 cm 0.635 cm 0.4 0.2 0 0.25 0.4 0.6 1.5 3.1 4.7 6.3 7.9 0.7 Wavelength λ, µ m FIGURE 11–36 The spectral transmissivity of low-iron glass at room temperature for different thicknesses. Solar radiation Greenhouse Infrared radiation FIGURE 11–37 A greenhouse traps energy by allowing the solar radiation to come in but not allowing the infrared radiation to go out. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 586 586 HEAT TRANSFER on the path of the infrared radiation coming from the earth, and thus slows down the cooling process at night. In areas with clear skies such as deserts, there is a large swing between the daytime and nighttime temperatures because of the absence of such barriers for infrared radiation. 11–6 ^ n G0 = Gs cos θ Earth’s surface ys a ’s r Sun θ Gs , W/m2 ’s e rth er Eaosph atm FIGURE 11–38 Solar radiation reaching the earth’s atmosphere and the total solar irradiance. I ATMOSPHERIC AND SOLAR RADIATION The sun is our primary source of energy. The energy coming off the sun, called solar energy, reaches us in the form of electromagnetic waves after experiencing considerable interactions with the atmosphere. The radiation energy emitted or reflected by the constituents of the atmosphere form the atmospheric radiation. Below we give an overview of the solar and atmospheric radiation because of their importance and relevance to daily life. Also, our familiarity with solar energy makes it an effective tool in developing a better understanding for some of the new concepts introduced earlier. Detailed treatment of this exciting subject can be found in numerous books devoted to this topic. The sun is a nearly spherical body that has a diameter of D 1.39 109 m and a mass of m 2 1030 kg and is located at a mean distance of L 1.50 1011 m from the earth. It emits radiation energy continuously at a rate of Esun 3.8 1026 W. Less than a billionth of this energy (about 1.7 1017 W) strikes the earth, which is sufficient to keep the earth warm and to maintain life through the photosynthesis process. The energy of the sun is due to the continuous fusion reaction during which two hydrogen atoms fuse to form one atom of helium. Therefore, the sun is essentially a nuclear reactor, with temperatures as high as 40,000,000 K in its core region. The temperature drops to about 5800 K in the outer region of the sun, called the convective zone, as a result of the dissipation of this energy by radiation. The solar energy reaching the earth’s atmosphere is called the total solar irradiance Gs , whose value is Gs (4πL2) Gs (4πr 2)Eb Sun r 1m2 Eb = σT 4 sun Gs 1373 W/m2 (11-49) The total solar irradiance (also called the solar constant) represents the rate at which solar energy is incident on a surface normal to the sun’s rays at the outer edge of the atmosphere when the earth is at its mean distance from the sun (Fig. 11–38). The value of the total solar irradiance can be used to estimate the effective surface temperature of the sun from the requirement that (4 L2 )Gs 4 (4 r 2 ) Tsun (11-50) L Earth 1m2 FIGURE 11–39 The total solar energy passing through concentric spheres remains constant, but the energy falling per unit area decreases with increasing radius. where L is the mean distance between the sun’s center and the earth and r is the radius of the sun. The left-hand side of this equation represents the total solar energy passing through a spherical surface whose radius is the mean earth–sun distance, and the right-hand side represents the total energy that leaves the sun’s outer surface. The conservation of energy principle requires that these two quantities be equal to each other, since the solar energy experiences no attenuation (or enhancement) on its way through the vacuum (Fig. 11–39). The effective surface temperature of the sun is determined 5780 K. That is, the sun can be treated as a from Eq. 11-50 to be Tsun cen58933_ch11.qxd 9/9/2002 9:38 AM Page 587 587 CHAPTER 11 2500 5780 K blackbody Solar irradiation Spectral irradiation, W/ m2·µm blackbody at a temperature of 5780 K. This is also confirmed by the measurements of the spectral distribution of the solar radiation just outside the atmosphere plotted in Figure 11–40, which shows only small deviations from the idealized blackbody behavior. The spectral distribution of solar radiation on the ground plotted in Figure 11–40 shows that the solar radiation undergoes considerable attenuation as it passes through the atmosphere as a result of absorption and scattering. About 99 percent of the atmosphere is contained within a distance of 30 km from the earth’s surface. The several dips on the spectral distribution of radiation on the earth’s surface are due to absorption by the gases O2, O3 (ozone), H2O, and CO2. Absorption by oxygen occurs in a narrow band about 0.76 m. The ozone absorbs ultraviolet radiation at wavelengths below 0.3 m almost completely, and radiation in the range 0.3–0.4 m considerably. Thus, the ozone layer in the upper regions of the atmosphere protects biological systems on earth from harmful ultraviolet radiation. In turn, we must protect the ozone layer from the destructive chemicals commonly used as refrigerants, cleaning agents, and propellants in aerosol cans. The use of these chemicals is now banned in many countries. The ozone gas also absorbs some radiation in the visible range. Absorption in the infrared region is dominated by water vapor and carbon dioxide. The dust particles and other pollutants in the atmosphere also absorb radiation at various wavelengths. As a result of these absorptions, the solar energy reaching the earth’s surface is weakened considerably, to about 950 W/m2 on a clear day and much less on cloudy or smoggy days. Also, practically all of the solar radiation reaching the earth’s surface falls in the wavelength band from 0.3 to 2.5 m. Another mechanism that attenuates solar radiation as it passes through the atmosphere is scattering or reflection by air molecules and the many other kinds of particles such as dust, smog, and water droplets suspended in the atmosphere. Scattering is mainly governed by the size of the particle relative to the wavelength of radiation. The oxygen and nitrogen molecules primarily scatter radiation at very short wavelengths, comparable to the size of the molecules themselves. Therefore, radiation at wavelengths corresponding to violet and blue colors is scattered the most. This molecular scattering in all directions is what gives the sky its bluish color. The same phenomenon is responsible for red sunrises and sunsets. Early in the morning and late in the afternoon, the sun’s rays pass through a greater thickness of the atmosphere than they do at midday, when the sun is at the top. Therefore, the violet and blue colors of the light encounter a greater number of molecules by the time they reach the earth’s surface, and thus a greater fraction of them are scattered (Fig. 11–41). Consequently, the light that reaches the earth’s surface consists primarily of colors corresponding to longer wavelengths such as red, orange, and yellow. The clouds appear in reddish-orange color during sunrise and sunset because the light they reflect is reddish-orange at those times. For the same reason, a red traffic light is visible from a longer distance than is a green light under the same circumstances. The solar energy incident on a surface on earth is considered to consist of direct and diffuse parts. The part of solar radiation that reaches the earth’s surface without being scattered or absorbed by the atmosphere is called direct solar radiation GD. The scattered radiation is assumed to reach the earth’s surface uniformly from all directions and is called diffuse solar radiation Gd. 2000 Extraterrestrial 1500 O3 O2 1000 Earth’s surface H2O 500 O3 H2O H2O 0 0 0.5 H2O CO2 1.0 1.5 2.0 Wavelength, µ m 2.5 3.0 FIGURE 11–40 Spectral distribution of solar radiation just outside the atmosphere, at the surface of the earth on a typical day, and comparison with blackbody radiation at 5780 K. Mostly red White Sun Red Orange Yellow Blue Violet Air molecules Atmosphere Earth FIGURE 11–41 Air molecules scatter blue light much more than they do red light. At sunset, the light travels through a thicker layer of atmosphere, which removes much of the blue from the natural light, allowing the red to dominate. cen58933_ch11.qxd 9/9/2002 9:38 AM Page 588 588 HEAT TRANSFER Then the total solar energy incident on the unit area of a horizontal surface on the ground is (Fig. 11–42) Diffuse solar radiation D s ire ra olar ct di at io n ^ n θ Gd , W/m2 GD , W/m2 FIGURE 11–42 The direct and diffuse radiation incident on a horizontal surface at the earth’s surface. Gsolar GD cos Gd (W/m2) (11-51) where is the angle of incidence of direct solar radiation (the angle that the sun’s rays make with the normal of the surface). The diffuse radiation varies from about 10 percent of the total radiation on a clear day to nearly 100 percent on a totally cloudy day. The gas molecules and the suspended particles in the atmosphere emit radiation as well as absorbing it. The atmospheric emission is primarily due to the CO2 and H2O molecules and is concentrated in the regions from 5 to 8 m and above 13 m. Although this emission is far from resembling the distribution of radiation from a blackbody, it is found convenient in radiation calculations to treat the atmosphere as a blackbody at some lower fictitious temperature that emits an equivalent amount of radiation energy. This fictitious temperature is called the effective sky temperature Tsky. Then the radiation emission from the atmosphere to the earth’s surface is expressed as Gsky 4 Tsky (W/m2 ) (11-52) The value of Tsky depends on the atmospheric conditions. It ranges from about 230 K for cold, clear-sky conditions to about 285 K for warm, cloudy-sky conditions. Note that the effective sky temperature does not deviate much from the room temperature. Thus, in the light of Kirchhoff ’s law, we can take the absorptivity of a surface to be equal to its emissivity at room temperature, . Then the sky radiation absorbed by a surface can be expressed as Esky, absorbed Gsky 4 Tsky 4 Tsky (W/m2) (11-53) The net rate of radiation heat transfer to a surface exposed to solar and atmospheric radiation is determined from an energy balance (Fig. 11–43): Sun Atmosphere Gsolar Eabsorbed Esolar, absorbed s Gsolar s Gsolar Gsky Eemitted αs Gsolar · q net, rad ε Gsky Eabsorbed FIGURE 11–43 Radiation interactions of a surface exposed to solar and atmospheric radiation. Eemitted Esky, absorbed Eemitted 4 Tsky Ts4 4 (Tsky Ts4 ) (W/m2) (11-54) where Ts is the temperature of the surface in K and is its emissivity at room · temperature. A positive result for q net, rad indicates a radiation heat gain by the surface and a negative result indicates a heat loss. The absorption and emission of radiation by the elementary gases such as H2, O2, and N2 at moderate temperatures are negligible, and a medium filled with these gases can be treated as a vacuum in radiation analysis. The absorption and emission of gases with larger molecules such as H2O and CO2, however, can be significant and may need to be considered when considerable amounts of such gases are present in a medium. For example, a 1-m-thick layer of water vapor at 1 atm pressure and 100°C emits more than 50 percent of the energy that a blackbody would emit at the same temperature. In solar energy applications, the spectral distribution of incident solar radiation is very different than the spectral distribution of emitted radiation by cen58933_ch11.qxd 9/9/2002 9:39 AM Page 589 589 CHAPTER 11 the surfaces, since the former is concentrated in the short-wavelength region and the latter in the infrared region. Therefore, the radiation properties of surfaces will be quite different for the incident and emitted radiation, and the surfaces cannot be assumed to be gray. Instead, the surfaces are assumed to have two sets of properties: one for solar radiation and another for infrared radiation at room temperature. Table 11–3 lists the emissivity and the solar absorptivity s of the surfaces of some common materials. Surfaces that are intended to collect solar energy, such as the absorber surfaces of solar collectors, are desired to have high s but low values to maximize the absorption of solar radiation and to minimize the emission of radiation. Surfaces that are intended to remain cool under the sun, such as the outer surfaces of fuel tanks and refrigerator trucks, are desired to have just the opposite properties. Surfaces are often given the desired properties by coating them with thin layers of selective materials. A surface can be kept cool, for example, by simply painting it white. We close this section by pointing out that what we call renewable energy is usually nothing more than the manifestation of solar energy in different forms. Such energy sources include wind energy, hydroelectric power, ocean thermal energy, ocean wave energy, and wood. For example, no hydroelectric power plant can generate electricity year after year unless the water evaporates by absorbing solar energy and comes back as a rainfall to replenish the water source (Fig. 11–44). Although solar energy is sufficient to meet the entire energy needs of the world, currently it is not economical to do so because of the low concentration of solar energy on earth and the high capital cost of harnessing it. EXAMPLE 11–5 Selective Absorber and Reflective Surfaces TABLE 11–3 Comparison of the solar absorptivity s of some surfaces with their emissivity at room temperature Surface s Aluminum Polished Anodized Foil Copper Polished Tarnished Stainless steel Polished Dull Plated metals Black nickel oxide Black chrome Concrete White marble Red brick Asphalt Black paint White paint Snow Human skin (caucasian) Consider a surface exposed to solar radiation. At a given time, the direct and diffuse components of solar radiation are GD 400 and Gd 300 W/m2, and the direct radiation makes a 20° angle with the normal of the surface. The surface temperature is observed to be 320 K at that time. Assuming an effective sky temperature of 260 K, determine the net rate of radiation heat transfer for these cases (Fig. 11–45): (a) s 0.9 and 0.9 (gray absorber surface) (b) s 0.1 and 0.1 (gray reflector surface) (c) s 0.9 and 0.1 (selective absorber surface) (d) s 0.1 and 0.9 (selective reflector surface) 0.09 0.14 0.15 0.03 0.84 0.05 0.18 0.65 0.03 0.75 0.37 0.50 0.60 0.21 0.92 0.87 0.60 0.46 0.63 0.90 0.97 0.14 0.28 0.08 0.09 0.88 0.95 0.93 0.90 0.97 0.93 0.97 0.62 0.97 Winds Clouds Rain Reservoir Power lines Evaporation SOLUTION A surface is exposed to solar and sky radiation. The net rate of radiation heat transfer is to be determined for four different combinations of emissivities and solar absorptivities. Analysis The total solar energy incident on the surface is Gsolar GD cos Gd (400 W/m2) cos 20° 676 W/m2 (300 W/m2) Then the net rate of radiation heat transfer for each of the four cases is determined from: · q net, rad s Gsolar 4 (Tsky Ts4) HPP Solar energy FIGURE 11–44 The cycle that water undergoes in a hydroelectric power plant. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 590 590 HEAT TRANSFER ε (a) 0.9 s · q net, rad 0.9 and 0.9 (gray absorber surface): 0.9(676 W/m2) 0.9(5.67 10 8 W/m2 · K4)[(260 K)4 (320 K)4] 307 W/m2 λ (a) ε (b) s · q net, rad 0.1 and 0.1(676 W/m2) 34 W/m (c) s · q net, rad 0.1 0.1 (gray reflector surface): 0.1(5.67 10 8 W/m2 · K4)[(260 K)4 (320 K)4] 2 0.9 and 0.1 (selective absorber surface): 0.9(676 W/m2) 0.1(5.67 10 8 W/m2 · K4)[(260 K)4 (320 K)4] 575 W/m2 λ (b) ε 0.1 and 0.9 (selective reflector surface): 0.1(676 W/m2) 0.9(5.67 10 8 W/m2 · K4)[(260 K)4 (320 K)4] 234 W/m2 0.1 3 µm λ 3 µm Discussion Note that the surface of an ordinary gray material of high absorptivity gains heat at a rate of 307 W/m2. The amount of heat gain increases to 575 W/m2 when the surface is coated with a selective material that has the same absorptivity for solar radiation but a low emissivity for infrared radiation. Also note that the surface of an ordinary gray material of high reflectivity still gains heat at a rate of 34 W/m2. When the surface is coated with a selective material that has the same reflectivity for solar radiation but a high emissivity for infrared radiation, the surface loses 234 W/m2 instead. Therefore, the temperature of the surface will decrease when a selective reflector surface is used. λ 0.9 0.1 (d) s · q net, rad 0.9 (c) ε (d) FIGURE 11–45 Graphical representation of the spectral emissivities of the four surfaces considered in Example 11–5. TOPIC OF SPECIAL INTEREST* Solar Heat Gain Through Windows The sun is the primary heat source of the earth, and the solar irradiance on a surface normal to the sun’s rays beyond the earth’s atmosphere at the mean earth–sun distance of 149.5 million km is called the total solar irradiance or solar constant. The accepted value of the solar constant is 1373 W/m2 (435.4 Btu/h · ft2), but its value changes by 3.5 percent from a maximum of 1418 W/m2 on January 3 when the earth is closest to the sun, to a minimum of 1325 W/m2 on July 4 when the earth is farthest away from the sun. The spectral distribution of solar radiation beyond the earth’s atmosphere resembles the energy emitted by a blackbody at 5780°C, with about 9 percent of the energy contained in the ultraviolet region (at wavelengths between 0.29 to 0.4 m), 39 percent in the visible region (0.4 to 0.7 m), and the remaining 52 percent in the near-infrared region (0.7 to 3.5 m). The peak radiation occurs at a wavelength of about 0.48 m, which corresponds to the green color portion of the visible spectrum. Obviously a *This section can be skipped without a loss in continuity. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 591 glazing material that transmits the visible part of the spectrum while absorbing the infrared portion is ideally suited for an application that calls for maximum daylight and minimum solar heat gain. Surprisingly, the ordinary window glass approximates this behavior remarkably well (Fig. 11–46). Part of the solar radiation entering the earth’s atmosphere is scattered and absorbed by air and water vapor molecules, dust particles, and water droplets in the clouds, and thus the solar radiation incident on earth’s surface is less than the solar constant. The extent of the attenuation of solar radiation depends on the length of the path of the rays through the atmosphere as well as the composition of the atmosphere (the clouds, dust, humidity, and smog) along the path. Most ultraviolet radiation is absorbed by the ozone in the upper atmosphere, and the scattering of short wavelength radiation in the blue range by the air molecules is responsible for the blue color of the clear skies. At a solar altitude of 41.8°, the total energy of direct solar radiation incident at sea level on a clear day consists of about 3 percent ultraviolet, 38 percent visible, and 59 percent infrared radiation. The part of solar radiation that reaches the earth’s surface without being scattered or absorbed is called direct radiation. Solar radiation that is scattered or reemitted by the constituents of the atmosphere is called diffuse radiation. Direct radiation comes directly from the sun following a straight path, whereas diffuse radiation comes from all directions in the sky. The entire radiation reaching the ground on an overcast day is diffuse radiation. The radiation reaching a surface, in general, consists of three components: direct radiation, diffuse radiation, and radiation reflected onto the surface from surrounding surfaces (Fig. 11–47). Common surfaces such as grass, trees, rocks, and concrete reflect about 20 percent of the radiation while absorbing the rest. Snow-covered surfaces, however, reflect 70 percent of the incident radiation. Radiation incident on a surface that does not have a direct view of the sun consists of diffuse and reflected radiation. Therefore, at solar noon, solar radiations incident on the east, west, and north surfaces of a south-facing house are identical since they all consist of diffuse and reflected components. The difference between the radiations incident on the south and north walls in this case gives the magnitude of direct radiation incident on the south wall. When solar radiation strikes a glass surface, part of it (about 8 percent for uncoated clear glass) is reflected back to outdoors, part of it (5 to 50 percent, depending on composition and thickness) is absorbed within the glass, and the remainder is transmitted indoors, as shown in Figure 11–48. The conservation of energy principle requires that the sum of the transmitted, reflected, and absorbed solar radiations be equal to the incident solar radiation. That is, s s s 1 where s is the transmissivity, s is the reflectivity, and s is the absorptivity of the glass for solar energy, which are the fractions of incident solar radiation transmitted, reflected, and absorbed, respectively. The standard 3-mm- (1 -in.) thick single-pane double-strength clear window glass trans8 mits 86 percent, reflects 8 percent, and absorbs 6 percent of the solar energy incident on it. The radiation properties of materials are usually given for normal incidence, but can also be used for radiation incident at other Spectral transmittance 591 CHAPTER 11 1.00 1 0.80 2 0.60 3 0.40 0.20 0 0.2 2 0.4 0.6 1 Wave length, µm 3 45 1. 3 mm regular sheet 2. 6 mm gray heat-absorbing plate/float 3. 6 mm green heat-absorbing plate/float FIGURE 11–46 The variation of the transmittance of typical architectural glass with wavelength (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Fig. 11). Sun Direct radiation Window Diffuse radiation Reflected radiation FIGURE 11–47 Direct, diffuse, and reflected components of solar radiation incident on a window. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 592 592 HEAT TRANSFER Sun Incident solar radiation 100% 6-mm thick clear glass Transmitted 80% Reflected 8% Absorbed 12% Outward transfer of absorbed radiation 8% Inward transfer of absorbed radiation 4% FIGURE 11–48 Distribution of solar radiation incident on a clear glass. angles since the transmissivity, reflectivity, and absorptivity of the glazing materials remain essentially constant for incidence angles up to about 60° from the normal. The hourly variation of solar radiation incident on the walls and windows of a house is given in Table 11–4. Solar radiation that is transmitted indoors is partially absorbed and partially reflected each time it strikes a surface, but all of it is eventually absorbed as sensible heat by the furniture, walls, people, and so forth. Therefore, the solar energy transmitted inside a building represents a heat gain for the building. Also, the solar radiation absorbed by the glass is subsequently transferred to the indoors and outdoors by convection and radiation. The sum of the transmitted solar radiation and the portion of the absorbed radiation that flows indoors constitutes the solar heat gain of the building. The fraction of incident solar radiation that enters through the glazing is called the solar heat gain coefficient (SHGC) and is expressed as SHGC Solar heat gain through the window Solar radiation incident on the window · q solar, gain fi s s · q solar, incident (11–55) where s is the solar absorptivity of the glass and fi is the inward flowing fraction of the solar radiation absorbed by the glass. Therefore, the dimensionless quantity SHGC is the sum of the fractions of the directly transmitted ( s) and the absorbed and reemitted ( fi s) portions of solar radiation incident on the window. The value of SHGC ranges from 0 to 1, with 1 corresponding to an opening in the wall (or the ceiling) with no glazing. When the SHGC of a window is known, the total solar heat gain through that window is determined from · Q solar, gain SHGC Aglazing · q solar, incident (W) (11–56) · where Aglazing is the glazing area of the window and q solar, incident is the solar heat flux incident on the outer surface of the window, in W/m2. Another way of characterizing the solar transmission characteristics of different kinds of glazing and shading devices is to compare them to a wellknown glazing material that can serve as a base case. This is done by taking the standard 3-mm-(1 -in.)-thick double-strength clear window glass 8 sheet whose SHGC is 0.87 as the reference glazing and defining a shading coefficient SC as SC Solar heat gain of product Solar heat gain of reference glazing SHGC SHGC 1.15 SHGC SHGCref 0.87 (11–57) Therefore, the shading coefficient of a single-pane clear glass window is SC 1.0. The shading coefficients of other commonly used fenestration products are given in Table 11–5 for summer design conditions. The values for winter design conditions may be slightly lower because of the higher heat transfer coefficients on the outer surface due to high winds and thus cen58933_ch11.qxd 9/9/2002 9:39 AM Page 593 T ABLE 11–4 Hourly variation of solar radiation incident on various surfaces and the daily totals throughout the year at 40° latitude (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 15) Solar Radiation Incident on the Surface,* W/m2 Solar Time Direction of Surface 5 Jan. N NE E SE S SW W NW Horizontal Direct 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Apr. N NE E SE S SW W NW Horizontal Direct 0 0 0 0 0 0 0 0 0 0 July N NE E SE S SW W NW Horizontal Direct Oct. N NE E SE S SW W NW Horizontal Direct Date 8 9 10 11 12 noon 13 14 15 16 0 0 0 0 0 0 0 0 0 0 20 63 402 483 271 20 20 20 51 446 43 47 557 811 579 48 43 43 198 753 66 66 448 875 771 185 59 59 348 865 68 68 222 803 884 428 68 68 448 912 71 71 76 647 922 647 76 71 482 926 68 68 68 428 884 803 222 68 448 912 66 59 59 185 771 875 448 66 348 865 43 43 43 48 579 811 557 47 198 753 20 20 20 20 271 483 402 63 51 446 0 0 0 0 0 0 0 0 0 0 41 262 321 189 18 17 17 17 39 282 57 508 728 518 59 52 52 52 222 651 79 462 810 682 149 77 77 77 447 794 97 291 732 736 333 97 97 97 640 864 110 134 552 699 437 116 110 110 786 901 120 123 293 582 528 187 120 120 880 919 122 122 131 392 559 392 392 122 911 925 120 120 120 187 528 582 293 123 880 919 110 110 110 116 437 699 552 134 786 901 97 97 97 97 333 736 732 291 640 864 79 77 77 77 149 682 810 462 447 794 3 8 7 2 0 0 0 0 1 7 133 454 498 248 39 39 39 39 115 434 109 590 739 460 76 71 71 71 320 656 103 540 782 580 108 95 95 95 528 762 117 383 701 617 190 114 114 114 702 818 126 203 531 576 292 131 126 126 838 850 134 144 294 460 369 155 134 134 922 866 138 138 149 291 395 291 149 138 949 871 134 134 134 155 369 460 294 144 922 866 126 126 126 131 292 576 531 203 838 850 117 114 114 114 190 617 701 383 702 818 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 74 163 152 44 7 7 7 14 152 40 178 626 680 321 40 40 40 156 643 62 84 652 853 547 66 62 62 351 811 77 80 505 864 711 137 87 87 509 884 87 87 256 770 813 364 87 87 608 917 90 90 97 599 847 599 97 90 640 927 87 87 87 364 813 770 256 87 608 917 77 87 87 137 711 864 505 80 509 884 62 62 62 66 547 853 652 84 351 811 6 7 19 Daily Total 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 446 489 1863 4266 5897 4266 1863 489 2568 — 57 52 52 52 59 518 728 508 222 651 41 17 17 17 18 189 321 262 39 282 0 0 0 0 0 0 0 0 0 0 1117 2347 4006 4323 3536 4323 4006 2347 6938 — 103 95 95 95 108 580 782 540 528 762 109 71 71 71 76 460 739 590 320 656 133 39 39 39 39 248 498 454 115 434 3 0 0 0 0 2 7 8 1 7 1621 3068 4313 3849 2552 3849 4313 3068 3902 — 40 40 40 40 321 680 626 178 156 643 7 7 7 7 44 152 163 74 14 152 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 453 869 2578 4543 5731 4543 2578 869 3917 — 17 18 *Multiply by 0.3171 to convert to Btu/h · ft2. Values given are for the 21st of the month for average days with no clouds. The values can be up to 15 percent higher at high elevations under very clear skies and up to 30 percent lower at very humid locations with very dusty industrial atmospheres. Daily totals are obtained using Simpson’s rule for integration with 10-min time intervals. Solar reflectance of the ground is assumed to be 0.2, which is valid for old concrete, crushed rock, and bright green grass. For a specified location, use solar radiation data obtained for that location. The direction of a surface indicates the direction a vertical surface is facing. For example, W represents the solar radiation incident on a west-facing wall per unit area of the wall. Solar time may deviate from the local time. Solar noon at a location is the time when the sun is at the highest location (and thus when the shadows are shortest). Solar radiation data are symmetric about the solar noon: the value on a west wall before the solar noon is equal to the value on an east wall two hours after the solar noon. 593 cen58933_ch11.qxd 9/9/2002 9:39 AM Page 594 594 HEAT TRANSFER TABLE 11–5 Shading coefficient SC and solar transmissivity solar for some common glass types for summer design conditions (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 27, Table 11). Type of Glazing Nominal Thickness mm in. (a) Single Glazing Clear 3 6 10 13 Heat absorbing 3 6 10 13 (b) Double Glazing Clear in, 3a clear out 6 Clear in, heat absorbing outc 6 solar SC* 1 8 1 4 3 8 1 2 1 8 1 4 3 8 1 2 0.86 0.78 0.72 0.67 0.64 0.46 0.33 0.24 1.0 0.95 0.92 0.88 0.85 0.73 0.64 0.58 1 8 1 4 0.71b 0.88 0.61 0.82 1 4 0.36 0.58 *Multiply by 0.87 to obtain SHGC. a The thickness of each pane of glass. Combined transmittance for assembled unit. c Refers to gray-, bronze-, and green-tinted heat-absorbing float glass. b higher rate of outward flow of solar heat absorbed by the glazing, but the difference is small. Note that the larger the shading coefficient, the smaller the shading effect, and thus the larger the amount of solar heat gain. A glazing material with a large shading coefficient will allow a large fraction of solar radiation to come in. Shading devices are classified as internal shading and external shading, depending on whether the shading device is placed inside or outside. External shading devices are more effective in reducing the solar heat gain since they intercept the sun’s rays before they reach the glazing. The solar heat gain through a window can be reduced by as much as 80 percent by exterior shading. Roof overhangs have long been used for exterior shading of windows. The sun is high in the horizon in summer and low in winter. A properly sized roof overhang or a horizontal projection blocks off the sun’s rays completely in summer while letting in most of them in winter, as shown in Figure 11–49. Such shading structures can reduce the solar heat gain on the south, southeast, and southwest windows in the northern hemisphere considerably. A window can also be shaded from outside by vertical or horizontal architectural projections, insect or shading screens, and sun screens. To be effective, air must be able to move freely around the exterior device to carry away the heat absorbed by the shading and the glazing materials. Some type of internal shading is used in most windows to provide privacy and aesthetic effects as well as some control over solar heat gain. Internal shading devices reduce solar heat gain by reflecting transmitted solar radiation back through the glazing before it can be absorbed and converted into heat in the building. Draperies reduce the annual heating and cooling loads of a building by 5 to 20 percent, depending on the type and the user habits. In summer, they reduce heat gain primarily by reflecting back direct solar radiation (Fig. 11–50). The semiclosed air space formed by the draperies serves as an additional barrier against heat transfer, resulting in a lower U-factor for the window and thus a lower rate of heat transfer in summer and winter. The solar optical properties of draperies can be measured accurately, or they can be obtained directly from the manufacturers. The shading coefficient of draperies depends on the openness factor, which is the ratio of the open area between the fibers that permits the sun’s rays to pass freely, to the total area of the fabric. Tightly woven fabrics allow little direct radiation to pass through, and thus they have a small openness factor. The reflectance of the surface of the drapery facing the glazing has a major effect on the amount of solar heat gain. Light-colored draperies made of closed or tightly woven fabrics maximize the back reflection and thus minimize the solar gain. Dark-colored draperies made of open or semi-open woven fabrics, on the other hand, minimize the back reflection and thus maximize the solar gain. The shading coefficients of drapes also depend on the way they are hung. Usually, the width of drapery used is twice the width of the draped area to allow folding of the drapes and to give them their characteristic “full” or “wavy” appearance. A flat drape behaves like an ordinary window shade. A flat drape has a higher reflectance and thus a lower shading coefficient than a full drape. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 595 595 CHAPTER 11 External shading devices such as overhangs and tinted glazings do not require operation, and provide reliable service over a long time without significant degradation during their service life. Their operation does not depend on a person or an automated system, and these passive shading devices are considered fully effective when determining the peak cooling load and the annual energy use. The effectiveness of manually operated shading devices, on the other hand, varies greatly depending on the user habits, and this variation should be considered when evaluating performance. The primary function of an indoor shading device is to provide thermal comfort for the occupants. An unshaded window glass allows most of the incident solar radiation in, and also dissipates part of the solar energy it absorbs by emitting infrared radiation to the room. The emitted radiation and the transmitted direct sunlight may bother the occupants near the window. In winter, the temperature of the glass is lower than the room air temperature, causing excessive heat loss by radiation from the occupants. A shading device allows the control of direct solar and infrared radiation while providing various degrees of privacy and outward vision. The shading device is also at a higher temperature than the glass in winter, and thus reduces radiation loss from occupants. Glare from draperies can be minimized by using off-white colors. Indoor shading devices, especially draperies made of a closed-weave fabric, are effective in reducing sounds that originate in the room, but they are not as effective against the sounds coming from outside. The type of climate in an area usually dictates the type of windows to be used in buildings. In cold climates where the heating load is much larger than the cooling load, the windows should have the highest transmissivity for the entire solar spectrum, and a high reflectivity (or low emissivity) for the far infrared radiation emitted by the walls and furnishings of the room. Low-e windows are well suited for such heating-dominated buildings. Properly designed and operated windows allow more heat into the building over a heating season than it loses, making them energy contributors rather then energy losers. In warm climates where the cooling load is much larger than the heating load, the windows should allow the visible solar radiation (light) in, but should block off the infrared solar radiation. Such windows can reduce the solar heat gain by 60 percent with no appreciable loss in daylighting. This behavior is approximated by window glazings that are coated with a heat-absorbing film outside and a low-e film inside (Fig. 11–51). Properly selected windows can reduce the cooling load by 15 to 30 percent compared to windows with clear glass. Note that radiation heat transfer between a room and its windows is proportional to the emissivity of the glass surface facing the room, glass, and can be expressed as · Q rad, room-window glass 4 Aglass (Troom 4 Tglass) (11-58) Therefore, a low-e interior glass will reduce the heat loss by radiation in winter (Tglass Troom) and heat gain by radiation in summer (Tglass Troom). Tinted glass and glass coated with reflective films reduce solar heat gain in summer and heat loss in winter. The conductive heat gains or losses can be minimized by using multiple-pane windows. Double-pane windows are Summer Sun Winter Overhang Sun Window FIGURE 11–49 A properly sized overhang blocks off the sun’s rays completely in summer while letting them in in winter. Sun Drape Reflected by glass Reflected by drapes Window FIGURE 11–50 Draperies reduce heat gain in summer by reflecting back solar radiation, and reduce heat loss in winter by forming an air space before the window. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 596 596 HEAT TRANSFER Glass (colder than room) Sun · Qrad ~ ε No reflective film Low-e film (high infrared reflectivity) (a) Cold climates Glass (warmer than room) Sun · Qrad ~ ε Infrared Reflective film Visible Low-e film (b) Warm climates FIGURE 11–51 Radiation heat transfer between a room and its windows is proportional to the emissivity of the glass surface, and low-e coatings on the inner surface of the windows reduce heat loss in winter and heat gain in summer. usually called for in climates where the winter design temperature is less than 7°C (45°F). Double-pane windows with tinted or reflective films are commonly used in buildings with large window areas. Clear glass is preferred for showrooms since it affords maximum visibility from outside, but bronze-, gray-, and green-colored glass are preferred in office buildings since they provide considerable privacy while reducing glare. EXAMPLE 11–6 Installing Reflective Films on Windows A manufacturing facility located at 40° N latitude has a glazing area of 40 m2 that consists of double-pane windows made of clear glass (SHGC 0.766). To reduce the solar heat gain in summer, a reflective film that will reduce the SHGC to 0.261 is considered. The cooling season consists of June, July, August, and September, and the heating season October through April. The average daily solar heat fluxes incident on the west side at this latitude are 1.86, 2.66, 3.43, 4.00, 4.36, 5.13, 4.31, 3.93, 3.28, 2.80, 1.84, and 1.54 kWh/day · m2 for January through December, respectively. Also, the unit cost of the electricity and natural gas are $0.08/kWh and $0.50/therm, respectively. If the coefficient of performance of the cooling system is 2.5 and efficiency of the furnace is 0.8, determine the net annual cost savings due to installing reflective coating on the windows. Also, determine the simple payback period if the installation cost of reflective film is $20/m2 (Fig. 11–52). SOLUTION The net annual cost savings due to installing reflective film on the west windows of a building and the simple payback period are to be determined. Assumptions 1 The calculations given below are for an average year. 2 The unit costs of electricity and natural gas remain constant. Analysis Using the daily averages for each month and noting the number of days of each month, the total solar heat flux incident on the glazing during summer and winter months are determined to be Qsolar, summer Qsolar, winter 5.13 30 4.31 31 3.93 31 3.28 30 2.80 31 1.84 30 1.54 31 1.86 2.66 28 3.43 31 4.00 30 548 kWh/year 508 kWh/year 31 Then the decrease in the annual cooling load and the increase in the annual heating load due to the reflective film become Cooling load decrease Qsolar, summer Aglazing (SHGCwithout film SHGCwith film) (508 kWh/year)(40 m2)(0.766 0.261) 10,262 kWh/year Heating load increase Qsolar, winter Aglazing (SHGCwithout film SHGCwith film) (548 kWh/year)(40 m2)(0.766 0.261) 11,070 kWh/year 377.7 therms/year since 1 therm 29.31 kWh. The corresponding decrease in cooling costs and the increase in heating costs are cen58933_ch11.qxd 9/9/2002 9:39 AM Page 597 597 CHAPTER 11 Glass Decrease in cooling costs Increase in heating costs (Cooling load decrease)(Unit cost of electricity)/COP (10,262 kWh/year)($0.08/kWh)/2.5 $328/year Sun Air space (Heating load increase)(Unit cost of fuel)/Efficiency (377.7 therms/year)($0.50/therm)/0.80 $236/year Then the net annual cost savings due to the reflective film become Cost savings Decrease in cooling costs $328 $236 $92/year Increase in heating costs Reflected The implementation cost of installing films is Implementation cost Reflective film 2 2 ($20/m )(40 m ) $800 FIGURE 11–52 Schematic for Example 11–6. This gives a simple payback period of Simple payback period Transmitted Implementation cost Annual cost savings $800 $92/year 8.7 years Discussion The reflective film will pay for itself in this case in about nine years. This may be unacceptable to most manufacturers since they are not usually interested in any energy conservation measure that does not pay for itself within three years. But the enhancement in thermal comfort and thus the resulting increase in productivity often makes it worthwhile to install reflective film. SUMMARY Radiation propagates in the form of electromagnetic waves. The frequency and wavelength of electromagnetic waves in a medium are related by c/ , where c is the speed of propagation in that medium. All matter whose temperature is above absolute zero continuously emits thermal radiation as a result of vibrational and rotational motions of molecules, atoms, and electrons of a substance. Temperature is a measure of the strength of these activities at the microscopic level. A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation, regardless of wavelength and direction. The radiation energy emitted by a blackbody per unit time and per unit surface area is called the blackbody emissive power Eb and is expressed by the Stefan–Boltzmann law as Eb(T ) T4 where 5.670 10 8 W/m2 K4 is the Stefan–Boltzmann constant and T is the absolute temperature of the surface in K. At any specified temperature, the spectral blackbody emissive power Eb increases with wavelength, reaches a peak, and then decreases with increasing wavelength. The wavelength at which the peak occurs for a specified temperature is given by Wien’s displacement law as ( T)max power 2897.8 m K The blackbody radiation function f represents the fraction of radiation emitted by a blackbody at temperature T in the wavelength band from 0 to . The fraction of radiation energy emitted by a blackbody at temperature T over a finite wavelength band from 1 to 2 is determined from fλ 1 λ 2( T ) fλ 2( T ) f λ 1( T ) where fλ 1(T) and fλ 2(T) are the blackbody radiation functions corresponding to 1T and 2T, respectively. The magnitude of a viewing angle in space is described by solid angle expressed as d dAn/r2. The radiation intensity for emitted radiation Ie( , ) is defined as the rate at which radiation energy is emitted in the ( , ) direction per unit area normal to this direction and per unit solid angle about this direction. The radiation flux for emitted radiation is the emissive power E, and is expressed as 2 E /2 dE hemisphere 0 0 Ie( , ) cos sin d d For a diffusely emitting surface, intensity is independent of direction and thus cen58933_ch11.qxd 9/9/2002 9:39 AM Page 598 598 HEAT TRANSFER E Ie For a blackbody, we have Ib Eb and Ib(T) E b( T ) T4 The radiation flux incident on a surface from all directions is irradiation G, and for diffusely incident radiation of intensity Ii it is expressed as G Ii The rate at which radiation energy leaves a unit area of a surface in all directions is radiosity J, and for a surface that is both a diffuse emitter and a diffuse reflector it is expressed as J Ie The total hemispherical emissivity of a surface is the average emissivity over all directions and wavelengths. Radiation energy incident on a surface per unit surface area per unit time is called irradiation G. When radiation strikes a surface, part of it is absorbed, part of it is reflected, and the remaining part, if any, is transmitted. The fraction of incident radiation (intensity Ii or irradiation G) absorbed by the surface is called the absorptivity, the fraction reflected by the surface is called the reflectivity, and the fraction transmitted is called the transmissivity. Various absorptivities, reflectivities, and transmissivities for a medium are expressed as Iλ, abs (λ, , ) (, , ) , () r Iλ, i (λ, , ) Gλ, abs( λ) Gλ( λ) 0 Iλ, e d λ and E 0 Eλd λ The last relation reduces for a diffusely emitting surface and for a blackbody to I E ,e and Eb ( , T) Ib ( , T) The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature. Different emissivities are defined as Ie ( , , T ) Ib ( T ) Spectral hemispherical emissivity: E λ( λ, T ) ( , T) E bλ( λ, T ) Total hemispherical emissivity: (T) 0 T )E b λ( λ , T )d λ T4 Emissivity can also be expressed as a step function by dividing the spectrum into a sufficient number of wavelength bands of constant emissivity as (T ) 1 f0 λ1(T ) 2 fλ1 λ 2( T ) () Gλ, tr( λ) Gλ( λ) Gtr G The consideration of wavelength and direction dependence of properties makes radiation calculations very complicated. Therefore, the gray and diffuse approximations are commonly utilized in radiation calculations. A surface is said to be diffuse if its properties are independent of direction and gray if its properties are independent of wavelength. The sum of the absorbed, reflected, and transmitted fractions of radiation energy must be equal to unity, 1 For opaque surfaces, , λ( λ , Iλ, i ( λ, , ) 0, and thus Surfaces are usually assumed to reflect in a perfectly specular or diffuse manner for simplicity. In specular (or mirrorlike) reflection, the angle of reflection equals the angle of incidence of the radiation beam. In diffuse reflection, radiation is reflected equally in all directions. Reflection from smooth and polished surfaces approximates specular reflection, whereas reflection from rough surfaces approximates diffuse reflection. Kirchhoff ’s law of radiation is expressed as Total directional emissivity: E(T ) Eb( T ) Iλ, ref ( λ, , ) 1 Ib λ( λ, T ) ( , , T) , and Gref , and G Iλ, e ( λ, , , T ) ( , , , T) (, , ) Gλ( λ) Spectral directional emissivity: , , Gλ, ref ( λ) () Gabs , G where Ie r is the sum of the emitted and reflected intensities. The spectral emitted quantities are related to total quantities as Ie , and 3 fλ 2 (T ) (T) , (T), (T) (T), and (T) (T) That is, the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature. Gas molecules and the suspended particles in the atmosphere emit radiation as well as absorbing it. The atmosphere can be treated as a blackbody at some lower fictitious temperature, called the effective sky temperature Tsky that emits an equivalent amount of radiation energy. Then the radiation emitted by the atmosphere is expressed as cen58933_ch11.qxd 9/9/2002 9:39 AM Page 599 599 CHAPTER 11 Gsky q·net, rad 4 T sky The net rate of radiation heat transfer to a surface exposed to solar and atmospheric radiation is determined from an energy balance expressed as sGsolar 4 (T sky 4 Ts ) where Ts is the surface temperature in K, and emissivity at room temperature. is the surface REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and Air Conditioning Engineers, Handbook of Fundamentals, Atlanta, ASHRAE, 1993. 8. M. F. Modest, Radiative Heat Transfer. New York: McGraw-Hill, 1993. 9. M. Planck. The Theory of Heat Radiation. New York: Dover, 1959. 2. A. G. H. Dietz. “Diathermanous Materials and Properties of Surfaces.” In Space Heating with Solar Energy, ed. R. W. Hamilton. Cambridge, MA: MIT Press, 1954. 10. W. Sieber. Zeitschrift für Technische Physics 22 (1941), pp. 130–135. 3. J. A. Duffy and W. A. Beckman. Solar Energy Thermal Process. New York: John Wiley & Sons, 1974. 11. R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. 3rd ed. Washington, D.C.: Hemisphere, 1992. 4. J. P. Holman. Heat Transfer. 9th ed. New York: McGrawHill, 2002. 12. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul, MN: West, 1995. 5. H. C. Hottel. “Radiant Heat Transmission,” In Heat Transmission, 3rd ed., ed. W. H. McAdams. New York: McGraw-Hill, 1954. 13. Y. S. Touloukain and D. P. DeWitt. “Nonmetallic Solids.” In Thermal Radiative Properties. Vol. 8. New York: IFI/Plenum, 1970. 6. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. 14. Y. S. Touloukian and D. P. DeWitt. “Metallic Elements and Alloys.” In Thermal Radiative Properties, Vol. 7. New York: IFI/Plenum, 1970. 7. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. PROBLEMS* 11–1C What is an electromagnetic wave? How does it differ from a sound wave? 11–6C What is the cause of color? Why do some objects appear blue to the eye while others appear red? Is the color of a surface at room temperature related to the radiation it emits? 11–2C By what properties is an electromagnetic wave characterized? How are these properties related to each other? 11–7C Why is radiation usually treated as a surface phenomenon? 11–3C What is visible light? How does it differ from the other forms of electromagnetic radiation? 11–8C Why do skiers get sunburned so easily? Electromagnetic and Thermal Radiation 11–4C How do ultraviolet and infrared radiation differ? Do you think your body emits any radiation in the ultraviolet range? Explain. 11–5C What is thermal radiation? How does it differ from the other forms of electromagnetic radiation? *Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with an EES-CD icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 11–9C How does microwave cooking differ from conventional cooking? 11–10 Electricity is generated and transmitted in power lines at a frequency of 60 Hz (1 Hz 1 cycle per second). Determine the wavelength of the electromagnetic waves generated by the passage of electricity in power lines. 11–11 A microwave oven is designed to operate at a frequency of 2.8 109 Hz. Determine the wavelength of these microwaves and the energy of each microwave. 11–12 A radio station is broadcasting radio waves at a wavelength of 200 m. Determine the frequency of these waves. Answer: 1.5 106 Hz 11–13 A cordless telephone is designed to operate at a frequency of 8.5 108 Hz. Determine the wavelength of these telephone waves. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 600 600 HEAT TRANSFER Blackbody Radiation 11–14C exist? What is a blackbody? Does a blackbody actually 11–15C Define the total and spectral blackbody emissive powers. How are they related to each other? How do they differ? 11–16C Why did we define the blackbody radiation function? What does it represent? For what is it used? 11–17C Consider two identical bodies, one at 1000 K and the other at 1500 K. Which body emits more radiation in the shorter-wavelength region? Which body emits more radiation at a wavelength of 20 m? 11–18 Consider a 20-cm 20-cm 20-cm cubical body at 1000 K suspended in the air. Assuming the body closely approximates a blackbody, determine (a) the rate at which the cube emits radiation energy, in W, and (b) the spectral blackbody emissive power at a wavelength of 4 m. 11–19E The sun can be treated as a blackbody at an effective surface temperature of 10,400 R. Determine the rate at which infrared radiation energy ( 0.76–100 m) is emitted by the sun, in Btu/h · ft2. 11–20 The sun can be treated as a blackbody at 5780 K. Using EES (or other) software, calculate and plot the spectral blackbody emissive power Eb of the sun versus wavelength in the range of 0.01 m to 1000 m. Discuss the results. 11–21 The temperature of the filament of an incandescent lightbulb is 3200 K. Treating the filament as a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks. Reconsider Problem 11–21. Using EES (or other) software, investigate the effect of temperature on the fraction of radiation emitted in the visible range. Let the surface temperature vary from 1000 K to 4000 K, and plot fraction of radiation emitted in the visible range versus the surface temperature. of radiation transmitted through a 2-m 2-m glass window from blackbody sources at (a) 5800 K and (b) 1000 K. Answers: (a) 218,400 kW, (b) 55.8 kW Radiation Intensity 11–26C What does a solid angle represent, and how does it differ from a plane angle? What is the value of a solid angle associated with a sphere? 11–27C How is the intensity of emitted radiation defined? For a diffusely emitting surface, how is the emissive power related to the intensity of emitted radiation? 11–28C For a surface, how is irradiation defined? For diffusely incident radiation, how is irradiation on a surface related to the intensity of incident radiation? 11–29C For a surface, how is radiosity defined? For diffusely emitting and reflecting surfaces, how is radiosity related to the intensities of emitted and reflected radiation? 11–30C When the variation of spectral radiation quantity with wavelength is known, how is the corresponding total quantity determined? 11–31 A small surface of area A1 4 cm2 emits radiation as a blackbody at T1 800 K. Part of the radiation emitted by A1 strikes another small surface of area A2 4 cm2 oriented as shown in the figure. Determine the solid angle subtended by A2 when viewed from A1, and the rate at which radiation emitted by A1 that strikes A2 directly. What would your answer be if A2 were directly above A1 at a distance of 80 cm? A2 θ2 θ1 11–22 11–23 An incandescent lightbulb is desired to emit at least 15 percent of its energy at wavelengths shorter than 1 m. Determine the minimum temperature to which the filament of the lightbulb must be heated. 11–24 It is desired that the radiation energy emitted by a light source reach a maximum in the blue range ( 0.47 m). Determine the temperature of this light source and the fraction of radiation it emits in the visible range ( 0.40–0.76 m). 11–25 A 3-mm-thick glass window transmits 90 percent of the radiation between 0.3 and 3.0 m and is essentially opaque for radiation at other wavelengths. Determine the rate A1 T1 45° r 4 cm2 60° 80 cm 4 cm2 800 K FIGURE P11–31 11–32 A small circular surface of area A1 2 cm2 located at the center of a 2-m-diameter sphere emits radiation as a blackbody at T1 1000 K. Determine the rate at which radiation energy is streaming through a D2 1-cm-diameter hole located (a) on top of the sphere directly above A1 and (b) on the side of sphere such that the line that connects the centers of A1 and A2 makes 45˚ with surface A1. 11–33 Repeat Problem 11–32 for a 4-m-diameter sphere. 11–34 A small surface of area A 1 cm2 emits radiation as a blackbody at 1500 K. Determine the rate at which radiation energy is emitted through a band defined by 0 2 and 45 cen58933_ch11.qxd 9/9/2002 9:39 AM Page 601 601 CHAPTER 11 60˚ where is the angle a radiation beam makes with the normal of the surface and is the azimuth angle. determine the absorptivity and reflectivity of the filament at both temperatures. 11–35 A small surface of area A 1 cm2 is subjected to incident radiation of constant intensity Ii 2.2 104 W/m2 sr over the entire hemisphere. Determine the rate at which radiation energy is incident on the surface through (a) 0 45˚ and (b) 45 90˚, where is the angle a radiation beam makes with the normal of the surface. 11–45 The variations of the spectral emissivity of two surfaces are as given in Figure P11–45. Determine the average emissivity of each surface at T 3000 K. Also, determine the average absorptivity and reflectivity of each surface for radiation coming from a source at 3000 K. Which surface is more suitable to serve as a solar absorber? Radiation Properties ελ 11–36C Define the properties emissivity and absorptivity. When are these two properties equal to each other? 11–37C Define the properties reflectivity and transmissivity and discuss the different forms of reflection. 11–38C What is a graybody? How does it differ from a blackbody? What is a diffuse gray surface? 11–39C What is the greenhouse effect? Why is it a matter of great concern among atmospheric scientists? 11–40C We can see the inside of a microwave oven during operation through its glass door, which indicates that visible radiation is escaping the oven. Do you think that the harmful microwave radiation might also be escaping? 11–41 The spectral emissivity function of an opaque surface at 1000 K is approximated as 1 2 3 0.4, 0.7, 0.3, 0 2m 6m 2m 6m Determine the average emissivity of the surface and the rate of radiation emission from the surface, in W/m2. 1.0 0.9 0.8 (1) 0.5 0.2 0 0 0.1 (2) λ, µm 3 FIGURE P11–45 11–46 The emissivity of a surface coated with aluminum oxide can be approximated to be 0.2 for radiation at wavelengths less than 5 m and 0.9 for radiation at wavelengths greater than 5 m. Determine the average emissivity of this surface at (a) 5800 K and (b) 300 K. What can you say about the absorptivity of this surface for radiation coming from sources at 5800 K and 300 K? Answers: (a) 0.203, (b) 0.89 11–47 The variation of the spectral absorptivity of a surface is as given in Figure P11–47. Determine the average absorptivity and reflectivity of the surface for radiation that originates from a source at T 2500 K. Also, determine the average emissivity of this surface at 3000 K. Answers: 0.575, 32.6 kW/m2 11–42 The reflectivity of aluminum coated with lead sulfate is 0.35 for radiation at wavelengths less than 3 m and 0.95 for radiation greater than 3 m. Determine the average reflectivity of this surface for solar radiation (T 5800 K) and radiation coming from surfaces at room temperature (T 300 K). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors? 11–43 A furnace that has a 25-cm 25-cm glass window can be considered to be a blackbody at 1200 K. If the transmissivity of the glass is 0.7 for radiation at wavelengths less than 3 m and zero for radiation at wavelengths greater than 3 m, determine the fraction and the rate of radiation coming from the furnace and transmitted through the window. 11–44 The emissivity of a tungsten filament can be approximated to be 0.5 for radiation at wavelengths less than 1 m and 0.15 for radiation at greater than 1 m. Determine the average emissivity of the filament at (a) 2000 K and (b) 3000 K. Also, αλ 0.7 0.2 2 λ, µm FIGURE P11–47 11–48E A 5-in.-diameter spherical ball is known to emit radiation at a rate of 120 Btu/h when its surface temperature is 950 R. Determine the average emissivity of the ball at this temperature. 11–49 The variation of the spectral transmissivity of a 0.6-cm-thick glass window is as given in Figure P11–49. Determine the average transmissivity of this window for solar radiation (T 5800 K) and radiation coming from surfaces at room temperature (T 300 K). Also, determine the amount of cen58933_ch11.qxd 9/9/2002 9:39 AM Page 602 602 HEAT TRANSFER solar radiation transmitted through the window for incident solar radiation of 650 W/m2. Answers: 0.848, 0.00015, 551.1 W/m2 τλ 0 Answer: 413.3 R 11–60 The air temperature on a clear night is observed to remain at about 4°C. Yet water is reported to have frozen that night due to radiation effect. Taking the convection heat transfer coefficient to be 18 W/m2 · °C, determine the value of the maximum effective sky temperature that night. 0.92 0 0.3 into space at 0 R. If there is no net heat transfer into the spaceship, determine the equilibrium temperature of the surface. 3 λ, µm FIGURE P11–49 Atmospheric and Solar Radiation 11–50C What is the solar constant? How is it used to determine the effective surface temperature of the sun? How would the value of the solar constant change if the distance between the earth and the sun doubled? 11–61 The absorber surface of a solar collector is made of aluminum coated with black chrome ( s 0.87 and 0.09). Solar radiation is incident on the surface at a rate of 600 W/m2. The air and the effective sky temperatures are 25°C and 15°C, respectively, and the convection heat transfer coefficient is 10 W/m2 · °C. For an absorber surface temperature of 70°C, determine the net rate of solar energy delivered by the absorber plate to the water circulating behind it. 11–51C What changes would you notice if the sun emitted radiation at an effective temperature of 2000 K instead of 5762 K? Sun 11–52C Explain why the sky is blue and the sunset is yelloworange. Gsolar = 600 W/m2 11–53C When the earth is closest to the sun, we have winter in the northern hemisphere. Explain why. Also explain why we have summer in the northern hemisphere when the earth is farthest away from the sun. 11–54C What is the effective sky temperature? 11–55C You have probably noticed warning signs on the highways stating that bridges may be icy even when the roads are not. Explain how this can happen. T = 25°C Ts = 70°C 11–56C Unless you live in a warm southern state, you have probably had to scrape ice from the windshield and windows of your car many mornings. You may have noticed, with frustration, that the thickest layer of ice always forms on the windshield instead of the side windows. Explain why this is the case. 11–57C Explain why surfaces usually have quite different absorptivities for solar radiation and for radiation originating from the surrounding bodies. 11–58 A surface has an absorptivity of s 0.85 for solar radiation and an emissivity of 0.5 at room temperature. The surface temperature is observed to be 350 K when the direct and the diffuse components of solar radiation are GD 350 and Gd 400 W/m2, respectively, and the direct radiation makes a 30° angle with the normal of the surface. Taking the effective sky temperature to be 280 K, determine the net rate of radiation heat transfer to the surface at that time. 11–59E Solar radiation is incident on the outer surface of a spaceship at a rate of 400 Btu/h · ft2. The surface has an absorptivity of s 0.10 for solar radiation and an emissivity of 0.8 at room temperature. The outer surface radiates heat Tsky = 15°C Absorber plate Water tubes Insulation FIGURE P11–61 11–62 Reconsider Problem 11–61. Using EES (or other) software, plot the net rate of solar energy transferred to water as a function of the absorptivity of the absorber plate. Let the absorptivity vary from 0.5 to 1.0, and discuss the results. 11–63 Determine the equilibrium temperature of the absorber surface in Problem 11–61 if the back side of the absorber is insulated. Special Topic: Solar Heat Gain through Windows 11–64C What fraction of the solar energy is in the visible range (a) outside the earth’s atmosphere and (b) at sea level on earth? Answer the same question for infrared radiation. 11–65C Describe the solar radiation properties of a window that is ideally suited for minimizing the air-conditioning load. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 603 603 CHAPTER 11 11–66C Define the SHGC (solar heat gain coefficient), and explain how it differs from the SC (shading coefficient). What are the values of the SHGC and SC of a single-pane clear-glass window? 11–67C What does the SC (shading coefficient) of a device represent? How do the SCs of clear glass and heat-absorbing glass compare? Venetian blinds Double-pane window Light colored 11–68C What is a shading device? Is an internal or external shading device more effective in reducing the solar heat gain through a window? How does the color of the surface of a shading device facing outside affect the solar heat gain? 11–69C What is the effect of a low-e coating on the inner surface of a window glass on the (a) heat loss in winter and (b) heat gain in summer through the window? 11–70C What is the effect of a reflective coating on the outer surface of a window glass on the (a) heat loss in winter and (b) heat gain in summer through the window? 11–71 A manufacturing facility located at 32° N latitude has a glazing area of 60 m2 facing west that consists of doublepane windows made of clear glass (SHGC 0.766). To reduce the solar heat gain in summer, a reflective film that will reduce the SHGC to 0.35 is considered. The cooling season consists of June, July, August, and September, and the heating season, October through April. The average daily solar heat fluxes incident on the west side at this latitude are 2.35, 3.03, 3.62, 4.00, 4.20, 4.24, 4.16, 3.93, 3.48, 2.94, 2.33, and 2.07 kWh/day · m2 for January through December, respectively. Also, the unit costs of electricity and natural gas are $0.09/kWh and $0.45/therm, respectively. If the coefficient of performance of the cooling system is 3.2 and the efficiency of the furnace is 0.90, determine the net annual cost savings due to installing reflective coating on the windows. Also, determine the simple payback period if the installation cost of reAnswers: $53, 23 years flective film is $20/m2. 11–72 A house located in Boulder, Colorado (40° N latitude), has ordinary double-pane windows with 6-mm-thick glasses and the total window areas are 8, 6, 6, and 4 m2 on the south, west, east, and north walls, respectively. Determine the total solar heat gain of the house at 9:00, 12:00, and 15:00 solar time in July. Also, determine the total amount of solar heat gain per day for an average day in January. 11–73 Repeat Problem 11–72 for double-pane windows that are gray-tinted. 11–74 Consider a building in New York (40° N latitude) that has 200 m2 of window area on its south wall. The windows are double-pane heat-absorbing type, and are equipped with lightcolored venetian blinds with a shading coefficient of SC 0.30. Determine the total solar heat gain of the building through the south windows at solar noon in April. What would your answer be if there were no blinds at the windows? Heat-absorbing glass FIGURE P11–74 11–75 A typical winter day in Reno, Nevada (39° N latitude), is cold but sunny, and thus the solar heat gain through the windows can be more than the heat loss through them during daytime. Consider a house with double-door-type windows that are double paned with 3-mm-thick glasses and 6.4 mm of air space and have aluminum frames and spacers. The house is maintained at 22°C at all times. Determine if the house is losing more or less heat than it is gaining from the sun through an east window on a typical day in January for a 24-h period if the Answer: less average outdoor temperature is 10°C. Double-pane window Sun Solar heat gain 10°C 22°C Heat loss FIGURE P11–75 11–76 Repeat Problem 11–75 for a south window. 11–77E Determine the rate of net heat gain (or loss) through a 9-ft-high, 15-ft-wide, fixed 1 -in. single-glass window with 8 aluminum frames on the west wall at 3 PM solar time during a typical day in January at a location near 40° N latitude when the indoor and outdoor temperatures are 70°F and 45°F, Answer: 16,840 Btu/h gain respectively. 11–78 Consider a building located near 40° N latitude that has equal window areas on all four sides. The building owner is considering coating the south-facing windows with reflective film to reduce the solar heat gain and thus the cooling load. cen58933_ch11.qxd 9/9/2002 9:39 AM Page 604 604 HEAT TRANSFER But someone suggests that the owner will reduce the cooling load even more if she coats the west-facing windows instead. What do you think? 11–83 The surface in Problem 11–82 receives solar radiation at a rate of 820 W/m2. Determine the solar absorptivity of the surface and the rate of absorption of solar radiation. Review Problems 11–84 The spectral transmissivity of a glass cover used in a solar collector is given as 11–79 The spectral emissivity of an opaque surface at 1200 K is approximated as 1 2 3 0 for 0.85 for 2 0 for 2m 6m 6m Determine the total emissivity and the emissive flux of the surface. 11–80 The spectral transmissivity of a 3-mm-thick regular glass can be expressed as 0 0.85 0 1 2 3 for 0.35 m for 0.35 2.5 m for 2.5 m Determine the transmissivity of this glass for solar radiation. What is the transmissivity of this glass for light? 11–81 A 1-m-diameter spherical cavity is maintained at a uniform temperature of 600 K. Now a 5-mm-diameter hole is drilled. Determine the maximum rate of radiation energy streaming through the hole. What would your answer be if the diameter of the cavity were 3 m? 11–82 The spectral absorptivity of an opaque surface is as shown on the graph. Determine the absorptivity of the surface for radiation emitted by a source at (a) 1000 K and (b) 3000 K. αλ 0.8 0.1 0 0.3 FIGURE P11–82 1.2 λ, µm 1 2 3 0 0.9 0 for 0.3 m for 0.3 3m for 3m Solar radiation is incident at a rate of 950 W/m2, and the absorber plate, which can be considered to be black, is maintained at 340 K by the cooling water. Determine (a) the solar flux incident on the absorber plate, (b) the transmissivity of the glass cover for radiation emitted by the absorber plate, and (c) the rate of heat transfer to the cooling water if the glass cover temperature is also 340 K. 11–85 Consider a small black surface of area A 2 cm2 maintained at 600 K. Determine the rate at which radiation energy is emitted by the surface through a ring-shaped opening defined by 0 2 and 40 50˚ where is the azimuth angle and is the angle a radiation beam makes with the normal of the surface. Design and Essay Problems 11–86 Write an essay on the radiation properties of selective surfaces used on the absorber plates of solar collectors. Find out about the various kinds of such surfaces, and discuss the performance and cost of each type. Recommend a selective surface that optimizes cost and performance. 11–87 According to an Atomic Energy Commission report, a hydrogen bomb can be approximated as a large fireball at a temperature of 7200 K. You are to assess the impact of such a bomb exploded 5 km above a city. Assume the diameter of the fireball to be 1 km, and the blast to last 15 s. Investigate the level of radiation energy people, plants, and houses will be exposed to, and how adversely they will be affected by the blast. ...
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This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.

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