564 Radiative heat transfer §10.5 even emit additional photons. The result is that fluids can play a role in the thermal radiation to the the surfaces that surround them. We have ignored this effect so far because it is generally very small, especially in air and if the distance between the surfaces is on the or- der of meters or less. When other gases are involved, especially at high temperatures, as in furnaces, or when long distances are involved, as in the atmosphere, gas radiation can become an important part of the heat exchange process. How gases interact with photons The photons of radiant energy passing through a gaseous region can be impeded in two ways. Some can be “scattered,” or deflected, in various directions, and some can be absorbed into the molecules. Scattering is a fairly minor influence in most gases unless they contain foreign par- ticles, such as dust or fog. In cloudless air, for example, we are aware of the scattering of sunlight only when it passes through many miles of the atmosphere. Then the shorter wavelengths of sunlight are scattered (short wavelengths, as it happens, are far more susceptible to scattering by gas molecules than longer wavelengths, through a process known as Rayleigh scattering). That scattered light gives the sky its blue hues. At sunset, sunlight passes through the atmosphere at a shallow angle for hundreds of miles. Radiation in the blue wavelengths has all been scattered out before it can be seen. Thus, we see only the unscattered red hues, just before dark. When particles suspended in a gas have diameters near the wave- length of light, a more complex type of scattering can occur, known as Mie scattering. Such scattering occurs from the water droplets in clouds (of- ten making them a brilliant white color). It also occurs in gases that con- tain soot or in pulverized coal combustion. Mie scattering has a strong angular variation that changes with wavelength and particle size [10.8]. The absorption or emission of radiation by molecules, rather than particles, will be our principal focus. The interaction of molecules with radiation — photons, that is — is governed by quantum mechanics. It’s helpful at this point to recall a few facts from molecular physics. Each photon has an energy hc o /λ, where h is Planck’s constant, c o is the speed of light, and λ is the wavelength of light. Thus, photons of shorter wave- lengths have higher energies: ultraviolet photons are more energetic than visible photons, which are in turn more energetic than infrared photons. It is not surprising that hotter objects emit more visible photons. §10.5 Gaseous radiation 565 Figure 10.19 Vibrational modes of carbon dioxide and water. Molecules can store energy by rotation, by vibration (Fig. 10.19), or in their electrons. Whereas the possible energy of a photon varies smoothly with wavelength, the energies of molecules are constrained by quantum mechanics to change only in discrete steps between the molecule’s allow- able “energy levels.” The available energy levels depend on the molecule’s chemical structure. When a molecule emits a photon, its energy drops in a discrete step from a higher energy level to a lower one. The energy given up is car- ried away by the photon. As a result, the wavelength of that photon is determined by the specific change in molecular energy level that caused it to be emitted. Just the opposite happens when a photon is absorbed: the photon’s wavelength must match a specific energy level change avail- able to that particular molecule. As a result, each molecular species can absorb only photons at, or very close to, particular wavelengths! Often, these wavelengths are tightly grouped into so-called absorption bands, outside of which the gas is essentially transparent to photons. The fact that a molecule’s structure determines how it absorbs and emits light has been used extensively by chemists as a tool for deducing 566 Radiative heat transfer §10.5 molecular structure. A knowledge of the energy levels in a molecule, in conjunction with quantum theory, allows specific atoms and bonds to be identified. This is called spectroscopy (see [10.9, Chpt. 18 & 19] for an introduction; see [10.10] to go overboard). At the wavelengths that correspond to thermal radiation at typical temperatures, it happens that transitions in the vibrational and rotation modes of molecules have the greatest influence on radiative absorptance. Such transitions can be driven by photons only when the molecule has some asymmetry. 4 Thus, for all practical purposes, monatomic and sym- metrical diatomic molecules are transparent to thermal radiation. The major components of air—N 2 and O 2 —are therefore nonabsorbing; so, too, are H 2 and such monatomic gases as argon. Asymmetrical molecules like CO 2 ,H 2 O, CH 4 ,O 3 ,NH 3 ,N 2 O, and SO 2 , on the other hand, each absorb thermal radiation of certain wavelengths. The first two of these, CO 2 and H 2 O, are always present in air. To under- stand how the interaction works, consider the possible vibrations of CO 2 and H 2 O shown in Fig. 10.19. For CO 2 , the topmost mode of vibration is symmetrical and has no interaction with thermal radiation at normal pressures. The other three modes produce asymmetries in the molecule when they occur; each is important to thermal radiation. The primary absorption wavelength for the two middle modes of CO 2 is 15 µm, which lies in the thermal infrared. The wavelength for the bot- tommost mode is 4.3 µm. For H 2 O, middle mode of vibration interacts strongly with thermal radiation at 6.3 µm. The other two both affect 2.7 µm radiation, although the bottom one does so more strongly. In ad- dition, H 2 O has a rotational mode that absorbs thermal radiation having wavelengths of 14 µm or more. Both of these molecules show additional absorption lines at shorter wavelengths, which result from the superpo- sition of two or more vibrations and their harmonics (e.g., at 2.7 µm for CO 2 and at 1.9 and 1.4 µm for H 2 O). Additional absorption bands can appear at high temperature or high pressure. Absorptance, transmittance, and emittance Figure 10.20 shows radiant energy passing through an absorbing gas with a monochromatic intensity i λ . As it passes through an element of thick- 4 The asymmetry required is in the distribution of electric charge — the dipole mo- ment. A vibration of the molecule must create a fluctuating dipole moment in order to interact with photons. A rotation interacts with photons only if the molecule has a permanent dipole moment. §10.5 Gaseous radiation 567 Figure 10.20 The attenuation of radiation through an absorbing (and/or scattering) gas. ness dx, the intensity will be reduced by an amount di λ : di λ =−ρκ λ i λ dx (10.43) where ρ is the gas density and κ λ is called the monochromatic absorp- tion coefficient. If the gas scatters radiation, we replace κ λ with γ λ , the monochromatic scattering coefficient. If it both absorbs and scatters ra- diation, we replace κ λ with β λ ≡ κ λ + γ λ , the monochromatic extinction coefficient. 5 The dimensions of κ λ , β λ , and γ λ are all m 2 /kg. If ρκ λ is constant through the gas, eqn. (10.43) can be integrated from an initial intensity i λ 0 at x = 0 to obtain i λ (x) = i λ 0 e −ρκ λ x (10.44) This result is called Beer’s law (pronounced “Bayr’s” law). For a gas layer of a given depth x = L, the ratio of final to initial intensity defines that layer’s monochromatic transmittance, τ λ : τ λ ≡ i λ (L) i λ 0 = e −ρκ λ L (10.45) Further, since gases do not reflect radiant energy, τ λ +α λ = 1. Thus, the monochromatic absorptance, α λ ,is α λ = 1 − e −ρκ λ L (10.46) Both τ λ and α λ depend on the density and thickness of the gas layer. The product ρκ λ L is sometimes called the optical depth of the gas. For very small values of ρκ λ L, the gas is transparent to the wavelength λ. 5 All three coefficients, κ λ , γ λ , and β λ , are expressed on a mass basis. They could, alternatively, have been expressed on a volumetric basis. 568 Radiative heat transfer §10.5 Figure 10.21 The monochromatic absorptance of a 1.09 m thick layer of steam at 127 ◦ C. The dependence of α λ on λ is normally very strong. As we have seen, a given molecule will absorb radiation in certain wavelength bands, while allowing radiation with somewhat higher or lower wavelengths to pass almost unhindered. Figure 10.21 shows the absorptance of water vapor as a function of wavelength for a fixed depth. We can see the absorption bands at wavelengths of 6.3, 2.7, 1.9, and 1.4 µm that were mentioned before. A comparison of Fig. 10.21 with Fig. 10.2 readily shows why radia- tion from the sun, as viewed from the earth’s surface, shows a number of spikey indentations at certain wavelengths. Several of those indenta- tions occur in bands where atmospheric water vapor absorbs incoming solar radiation, in accordance with Fig. 10.21. The other indentations in Fig. 10.2 occur where ozone and CO 2 absorb radiation. The sun itself does not have these regions of low emittance; it is just that much of the radiation in these bands is absorbed by gases in the atmosphere before it can reach the ground. Just as α λ and ε λ are equal to one another for a diffuse solid surface, they are equal for a gas. We may demonstrate this by considering an isothermal gas that is in thermal equilibrium with a black enclosure that contains it. The radiant intensity within the enclosure is that of a black body, i λ b , at the temperature of the gas and enclosure. Equation (10.43) shows that a small section of gas absorbs radiation, reducing the inten- sity by an amount ρκ λ i λ b dx. To maintain equilibrium, the gas must therefore emit an equal amount of radiation: di λ = ρκ λ i λ b dx (10.47) Now, if radiation from some other source is transmitted through a nonscattering isothermal gas, we can combine the absorption from §10.5 Gaseous radiation 569 eqn. (10.43) with the emission from eqn. (10.47) to form an energy bal- ance called the equation of transfer di λ dx =−ρκ λ i λ +ρκ λ i λ b (10.48) Integration of this equation yields a result similar to eqn. (10.44): i λ (L) = i λ 0 e −ρκ λ L    =τ λ +i λ b  1 − e −ρκ λ L     ≡ε λ (10.49) The first righthand term represents the transmission of the incoming intensity, as in eqn. (10.44), and the second is the radiation emitted by the gas itself. The coefficient of the second righthand term defines the monochromatic emittance, ε λ , of the gas layer. Finally, comparison to eqn. (10.46) shows that ε λ = α λ = 1 − e −ρκ λ L (10.50) Again, we see that for very small ρκ λ L the gas will neither absorb nor emit radiation of wavelength λ. Heat transfer from gases to walls We now see that predicting the total emissivity, ε g , of a gas layer will be complex. We have to take account of the gases’ absorption bands as well as the layer’s thickness and density. Such predictions can be done [10.11], but they are laborious. For making simpler (but less accurate) estimates, correlations of ε g have been developed. Such correlations are based on the following model: An isothermal gas of temperature T g and thickness L, is bounded by walls at the single temperature T w . The gas consists of a small fraction of an absorbing species (say CO 2 ) mixed into a nonabsorbing species (say N 2 ). If the ab- sorbing gas has a partial pressure p a and the mixture has a total pressure p, the correlation takes this form: ε g = fn  p a L, p, T g  (10.51) The parameter p a L is a measure of the layer’s optical depth; p and T g account for changes in the absorption bands with pressure and temper- ature. 570 Radiative heat transfer §10.5 Hottel and Sarofim [10.12] provide such correlations for CO 2 and H 2 O, built from research by Hottel and others before 1960. The correlations take the form ε g  p a L, p, T g  = f 1  p a L, T g  ×f 2  p,p a ,p a L  (10.52) where the experimental functions f 1 and f 2 are plotted in Figs. 10.22 and 10.23 for CO 2 and H 2 O, respectively. The first function, f 1 , is a correla- tion for a total pressure of p = 1 atm with a very small partial pressure of the absorbing species. The second function, f 2 , is a correction factor to account for other values of p a or p. Additional corrections must be applied if both CO 2 and H 2 O are present in the same mixture. To find the net heat transfer between the gas and the walls, we must also find the total absorptance, α g , of the gas. Despite the equality of the monochromatic emittance and absorptance, ε λ and α λ , the total values, ε g and α g , will not generally be equal. This is because the absorbed radiation may come from, say, a wall having a much different temperature than the gas with a correspondingly different wavelength distribution. Hottel and Sarofim show that α g may be estimated from the correlation for ε g as follows: 6 α g =  T g T w  1/2 ·ε  p a L T w T g ,p,T w  (10.53) Finally, we need to determine an appropriate value of L for a given enclosure. The correlations just given for ε g and α g assume L to be a one-dimensional path through the gas. Even for a pair of flat plates a distance L apart, this won’t be appropriate since radiation can travel much farther if it follows a path that is not perpendicular to the plates. For enclosures that have black walls at a uniform temperature, we can use an effective path length, L 0 , called the geometrical mean beam length, to represent both the size and the configuration of a gaseous region. The geometrical mean beam length is defined as L 0 ≡ 4 (volume of gas) boundary area that is irradiated (10.54) Thus, for two infinite parallel plates a distance  apart, L 0 = 4A/2A = 2. Some other values of L 0 for gas volumes exchanging heat with all points on their boundaries are as follow: 6 Hottel originally recommended replacing the exponent 1/2 by 0.65 for CO 2 and 0.45 for H 2 O. Theory, and more recent work, both suggest using the value 1/2 [10.13]. Figure 10.22 Functions used to predict ε g = f 1 f 2 for water vapor in air. 571 Figure 10.23 Functions used to predict ε g = f 1 f 2 for CO 2 in air. All pressures in atmospheres. 572 §10.5 Gaseous radiation 573 • For a sphere of diameter D, L 0 = 2D/3 • For an infinite cylinder of diameter D, L 0 = D • For a cube of side L, L 0 = 2L/3 • For a cylinder with height = D, L 0 = 2D/3 For cases where the gas is strongly absorbing, better accuracy can be obtained by replacing the constant 4 in eqn. (10.54) by 3.5, lowering the mean beam length about 12%. We are now in position to treat a problem in which hot gases (say the products of combustion) radiate to a black container. Consider an example: Example 10.11 A long cylindrical combustor 40 cm in diameter contains a gas at 1200 ◦ C consisting of 0.8 atm N 2 and 0.2 atm CO 2 . What is the net heat radiated to the walls if they are at 300 ◦ C? Solution. Let us first obtain ε g . We have L 0 = D = 0.40 m, a total pressure of 1.0 atm, p CO 2 = 0.2 atm, and T = 1200 ◦ C = 2651 ◦ R. Then Fig. 10.23a gives f 1 as 0.098 and Fig. 10.23b gives f 2  1, so ε g = 0.098. Next, we use eqn. (10.53) to obtain α g , with T w = 1031 ◦ R, p H 2 O LT w /T g = 0.031: α g =  1200 + 273 300 + 273  0.5 (0.074) = 0.12 Now we can calculate Q net g-w . For these problems with one wall surrounding one gas, the use of the mean beam length in finding ε g and α g accounts for all geometrical effects, and no view factor is required. The net heat transfer is calculated using the surface area of the wall: Q net g-w = A w  ε g σT 4 g −α g σT 4 w  = π(0.4)(5.67 × 10 −8 )  (0.098)(1473) 4 −(0.12)(573) 4  = 32 kW/m Total emissivity charts and the mean beam length provide a simple, but crude, tool for dealing with gas radiation. Since the introduction [...]... surface includes direct radiation that has passed through the atmosphere and diffuse radiation that 575 Radiative heat transfer 576 45 % reaches the earth’s surface §10.6 45 % is transmitted to the earth directly and by diffuse radiation 33 % is reflected back to space 22% is absorbed in the atmosphere Sensible heat transfer to atmosphere Radiation that reaches the outer atmosphere from the sun Net radiation...Radiative heat transfer 5 74 §10.6 of these ideas in the mid-twentieth century, major advances have been made in our knowledge of the radiative properties of gases and in the tools available for solving gas radiation problems In particular, band models of gas radiation, and better measurements, have led to better procedures for dealing with the total radiative properties of gases (see, in particular,... is rarely directly overhead We have seen that a radiant heat flux arriving at an angle less than 90◦ is reduced by the cosine of that angle (Fig 10 .4) The sun’s angle varies with latitude, time of day, and day of year Trigonometry and data for the earth’s rotation can be used to find the appropriate angle Figure 10.2 shows the reduction of solar radiation by atmospheric absorption for one particular set... shows that a black body of the sun’s size and distance from the earth would produce the same irradiation as the sun if its temperature were 5777 K The solar radiation reaching the earth’s surface is always less than that above the atmosphere owing to atmospheric absorption and the earth’s curvature and rotation Solar radiation usually arrives at an angle of less than 90◦ to the surface because the... several materials Among these, we identify several particularly selective solar absorbers and solar reflectors The selective absorbers have a high absorptance for solar radiation and a low emittance for infrared radiation Consequently, they do not strongly reradiate the solar energy that they absorb The selective solar reflectors, on the other hand, reflect solar energy strongly and also radiate heat efficiently... solar energy in a lower temperature space The atmospheric greenhouse effect and global warming The atmosphere creates a greenhouse effect on the earth’s surface that is very similar to that caused by a pane of glass Solar energy passes through the atmosphere, arriving mainly on wavelengths between about 0 .3 and 3 µm The earth’s surface, having a mean temperature of 15 ◦ C or so, radiates mainly on infrared... evaporation, CO2 by animal respiration, CH4 through plant decay and digestion by livestock, and so on Human activities, however, have significantly increased the concentrations of all of the gases Fossil fuel combustion increased the CO2 579 Radiative heat transfer 580 §10.6 0.8 Temperature Anomaly, ˚C 0.6 0 .4 0.2 0.0 -0 .2 Annual mean 5-year mean -0 .4 -0 .6 1880 1900 1920 1 940 1960 1980 2000 Year Figure... What is F1–2 for the squares shown in Fig 10.29? Configuration for Chapter 10: Radiative heat transfer 586 Figure 10.29 Prob 10.12 Configuration for 10. 13 10. 14 Figure 10 .30 Prob 10. 14 A particular internal combustion engine has an exhaust manifold at 600◦ C running parallel to a water cooling line at 20◦ C If both the manifold and the cooling line are 4 cm in diameter, their centers are 7 cm apart, and... infrared wavelengths longer than 5 µm Certain atmospheric gases have strong absorption bands at these longer wavelengths Those gases absorb energy radiated from the surface, and then reemit it toward both the surface and outer space The result is that the surface remains some 30 K warmer than the atmosphere In effect, the atmosphere functions as a radiation shield against infrared heat loss to space The gases... the gray gas as a “myth.” He notes, however, that spectral variations may be overlooked for a gas containing spray droplets or particles [in a range of sizes] or for some gases that have wide, weak absorption bands within the spectral range of interest [10 .3] Some accommodation of molecular properties can be achieved using the weighted sum of gray gases concept [10.12], which treats a real gas as superposition . The geometrical mean beam length is defined as L 0 ≡ 4 (volume of gas) boundary area that is irradiated (10. 54) Thus, for two infinite parallel plates a distance  apart, L 0 = 4A / 2A = 2. Some other values. solar radiation reaching the earth’s surface is always less than that above the atmosphere owing to atmospheric absorption and the earth’s curvature and rotation. Solar radiation usually arrives. through the gas. Even for a pair of flat plates a distance L apart, this won’t be appropriate since radiation can travel much farther if it follows a path that is not perpendicular to the plates. For