John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 82
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The gas consists of a small fraction of an absorbingspecies (say CO2 ) mixed into a nonabsorbing species (say N2 ). If the absorbing gas has a partial pressure pa and the mixture has a total pressurep, the correlation takes this form:εg = fn pa L, p, Tg(10.51)The parameter pa L is a measure of the layer’s optical depth; p and Tgaccount for changes in the absorption bands with pressure and temperature.Radiative heat transfer570§10.5Hottel and Sarofim [10.12] provide such correlations for CO2 and H2 O,built from research by Hottel and others before 1960. The correlationstake the form (10.52)εg pa L, p, Tg = f1 pa L, Tg × f2 p, pa , pa Lwhere the experimental functions f1 and f2 are plotted in Figs. 10.22 and10.23 for CO2 and H2 O, respectively.
The first function, f1 , is a correlation for a total pressure of p = 1 atm with a very small partial pressureof the absorbing species. The second function, f2 , is a correction factorto account for other values of pa or p. Additional corrections must beapplied if both CO2 and H2 O are present in the same mixture.To find the net heat transfer between the gas and the walls, we mustalso find the total absorptance, αg , of the gas.
Despite the equality of themonochromatic emittance and absorptance, ελ and αλ , the total values,εg and αg , will not generally be equal. This is because the absorbedradiation may come from, say, a wall having a much different temperaturethan the gas with a correspondingly different wavelength distribution.Hottel and Sarofim show that αg may be estimated from the correlationfor εg as follows:6Tg 1/2Twαg =· ε pa L, p, Tw(10.53)TwTgFinally, we need to determine an appropriate value of L for a givenenclosure. The correlations just given for εg and αg assume L to bea one-dimensional path through the gas.
Even for a pair of flat platesa distance L apart, this won’t be appropriate since radiation can travelmuch farther if it follows a path that is not perpendicular to the plates.For enclosures that have black walls at a uniform temperature, we canuse an effective path length, L0 , called the geometrical mean beam length,to represent both the size and the configuration of a gaseous region. Thegeometrical mean beam length is defined asL0 ≡4 (volume of gas)boundary area that is irradiated(10.54)Thus, for two infinite parallel plates a distance apart, L0 = 4A/2A =2. Some other values of L0 for gas volumes exchanging heat with allpoints on their boundaries are as follow:6Hottel originally recommended replacing the exponent 1/2 by 0.65 for CO2 and0.45 for H2 O.
Theory, and more recent work, both suggest using the value 1/2 [10.13].Figure 10.22 Functions used to predict εg = f1 f2 for watervapor in air.571Figure 10.23 Functions used to predict εg = f1 f2 for CO2 inair. All pressures in atmospheres.572Gaseous radiation§10.5• For a sphere of diameter D, L0 = 2D/3• For an infinite cylinder of diameter D, L0 = D• For a cube of side L, L0 = 2L/3• For a cylinder with height = D, L0 = 2D/3For cases where the gas is strongly absorbing, better accuracy can beobtained by replacing the constant 4 in eqn. (10.54) by 3.5, lowering themean beam length about 12%.We are now in position to treat a problem in which hot gases (saythe products of combustion) radiate to a black container. Consider anexample:Example 10.11A long cylindrical combustor 40 cm in diameter contains a gas at1200◦ C consisting of 0.8 atm N2 and 0.2 atm CO2 .
What is the netheat radiated to the walls if they are at 300◦ C?Solution. Let us first obtain εg . We have L0 = D = 0.40 m, a totalpressure of 1.0 atm, pCO2 = 0.2 atm, and T = 1200◦ C = 2651◦ R.Then Fig. 10.23a gives f1 as 0.098 and Fig. 10.23b gives f2 1, soεg = 0.098. Next, we use eqn. (10.53) to obtain αg , with Tw = 1031◦ R,pH2 O LTw /Tg = 0.031:αg =1200 + 273300 + 2730.5(0.074) = 0.12Now we can calculate Qnetg-w . For these problems with one wallsurrounding one gas, the use of the mean beam length in findingεg and αg accounts for all geometrical effects, and no view factor isrequired. The net heat transfer is calculated using the surface area ofthe wall:4Qnetg-w = Aw εg σ Tg4 − αg σ Tw= π (0.4)(5.67 × 10−8 ) (0.098)(1473)4 − (0.12)(573)4= 32 kW/mTotal emissivity charts and the mean beam length provide a simple,but crude, tool for dealing with gas radiation.
Since the introduction573Radiative heat transfer574§10.6of these ideas in the mid-twentieth century, major advances have beenmade in our knowledge of the radiative properties of gases and in thetools available for solving gas radiation problems. In particular, bandmodels of gas radiation, and better measurements, have led to betterprocedures for dealing with the total radiative properties of gases (see,in particular, References [10.11] and [10.13]).
Tools for dealing with radiation in complex enclosures have also improved. The most versitileof these is the previously-mentioned Monte Carlo method [10.4, 10.7],which can deal with nongray, nondiffuse, and nonisothermal walls withnongray, scattering, and nonisothermal gases. An extensive literaturealso deals with approximate analytical techniques, many of which arebased on the idea of a “gray gas” — one for which ελ and αλ are independent of wavelength. However, as we have pointed out, the gray gasmodel is not even a qualitative approximation to the properties of realgases.7Finally, it is worth noting that gaseous radiation is frequently lessimportant than one might imagine.
Consider, for example, two flames: abright orange candle flame and a “cold-blue” hydrogen flame. Both havea great deal of water vapor in them, as a result of oxidizing H2 . But thecandle will warm your hands if you place them near it and the hydrogenflame will not. Yet the temperature in the hydrogen flame is higher. Itturns out that what is radiating both heat and light from the candle is soot— small solid particles of almost thermally black carbon.
The CO2 andH2 O in the candle flame actually contribute relatively little to radiation.10.6Solar energyThe sunThe sun continually irradiates the earth at a rate of about 1.74×1014 kW.If we imagine this energy to be distributed over a circular disk with theearth’s diameter, the solar irradiation is about 1367 W/m2 , as measuredby satellites above the atmosphere. Much of this energy reaches theground, where it sustains the processes of life.7Edwards [10.11] describes the gray gas as a “myth.” He notes, however, that spectralvariations may be overlooked for a gas containing spray droplets or particles [in arange of sizes] or for some gases that have wide, weak absorption bands within thespectral range of interest [10.3].
Some accommodation of molecular properties can beachieved using the weighted sum of gray gases concept [10.12], which treats a real gasas superposition of gray gases having different properties.§10.6Solar energyThe temperature of the sun varies from tens of millions of kelvin in itscore to between 4000 and 6000 K at its surface, where most of the sun’sthermal radiation originates. The wavelength distribution of the sun’senergy is not quite that of a black body, but it may be approximated assuch. A straightforward calculation (see Problem 10.49) shows that ablack body of the sun’s size and distance from the earth would producethe same irradiation as the sun if its temperature were 5777 K.The solar radiation reaching the earth’s surface is always less thanthat above the atmosphere owing to atmospheric absorption and theearth’s curvature and rotation.
Solar radiation usually arrives at an angleof less than 90◦ to the surface because the sun 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 varieswith latitude, time of day, and day of year. Trigonometry and data forthe 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 of atmospheric conditions.
In fact, whenthe sun passes through the atmosphere at a low angle (near the horizon), the path of radiation through the atmosphere is longer, providingrelatively more opportunity for atmospheric absorption and scattering.Additional moisture in the air can increase the absorption by H2 O, and,of course, clouds can dramatically reduce the solar radiation reachingthe ground. The consequence of these various effects is that the solarradiation received on the ground is almost never more than 1200 W/m2and is often only a few hundred W/m2 . Extensive data are available forestimating the ground level solar irradiation at a given location, time, anddate [10.14, 10.15].The distribution of the Sun’s energy and atmosphericirradiationFigure 10.24 shows what becomes of the solar energy that impinges onthe earth if we average it over the year and the globe, taking account ofall kinds of weather.
Only 45% of the sun’s energy actually reaches theearth’s surface. The mean energy received is about 235 W/m2 if averagedover the surface and the year. The lower left-hand portion of the figureshows how this energy is, in turn, all returned to the atmosphere and tospace.The solar radiation reaching the earth’s surface includes direct radiation that has passed through the atmosphere and diffuse radiation thathas been scattered, but not absorbed, by the atmosphere.
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