When other gases are involved, especially at hightemperatures, as in furnaces, or when long distances are involved, as inthe atmosphere, gas radiation can become an important part of the
Trang 1564 Radiative heat transfer §10.5
even emit additional photons The result is that fluids can play a role inthe 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 hightemperatures, as in furnaces, or when long distances are involved, as inthe atmosphere, gas radiation can become an important part of the heatexchange process
How gases interact with photons
The photons of radiant energy passing through a gaseous region can beimpeded in two ways Some can be “scattered,” or deflected, in variousdirections, and some can be absorbed into the molecules Scattering is
a fairly minor influence in most gases unless they contain foreign ticles, such as dust or fog In cloudless air, for example, we are aware
par-of the scattering par-of sunlight only when it passes through many miles par-ofthe 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 anglefor hundreds of miles Radiation in the blue wavelengths has all beenscattered out before it can be seen Thus, we see only the unscatteredred 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 tain soot or in pulverized coal combustion Mie scattering has a strongangular variation that changes with wavelength and particle size [10.8].The absorption or emission of radiation by molecules, rather thanparticles, will be our principal focus The interaction of molecules withradiation — photons, that is — is governed by quantum mechanics It’shelpful at this point to recall a few facts from molecular physics Each
con-photon has an energy hc o /λ, where h is Planck’s constant, c ois 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 thanvisible photons, which are in turn more energetic than infrared photons
It is not surprising that hotter objects emit more visible photons
Trang 2§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 availallow-able 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
Trang 3566 Radiative heat transfer §10.5
molecular structure A knowledge of the energy levels in a molecule, inconjunction with quantum theory, allows specific atoms and bonds to be
identified This is called spectroscopy (see [10.9, Chpt 18 & 19] for anintroduction; see [10.10] to go overboard)
At the wavelengths that correspond to thermal radiation at typicaltemperatures, it happens that transitions in the vibrational and rotationmodes of molecules have the greatest influence on radiative absorptance.Such transitions can be driven by photons only when the molecule hassome asymmetry.4 Thus, for all practical purposes, monatomic and sym-metrical diatomic molecules are transparent to thermal radiation Themajor components of air—N2 and O2—are therefore nonabsorbing; so,too, are H2and such monatomic gases as argon
Asymmetrical molecules like CO2, H2O, CH4, O3, NH3, N2O, and SO2,
on the other hand, each absorb thermal radiation of certain wavelengths.The first two of these, CO2and H2O, are always present in air To under-stand how the interaction works, consider the possible vibrations of CO2and H2O shown in Fig 10.19 For CO2, the topmost mode of vibration
is symmetrical and has no interaction with thermal radiation at normalpressures The other three modes produce asymmetries in the moleculewhen they occur; each is important to thermal radiation
The primary absorption wavelength for the two middle modes of CO2
is 15 µm, which lies in the thermal infrared The wavelength for the tommost mode is 4.3 µm For H2O, middle mode of vibration interactsstrongly with thermal radiation at 6.3 µm The other two both affect2.7 µm radiation, although the bottom one does so more strongly In ad-dition, H2O has a rotational mode that absorbs thermal radiation havingwavelengths of 14 µm or more Both of these molecules show additionalabsorption 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
bot-CO2 and at 1.9 and 1.4 µm for H2O) Additional absorption bands canappear at high temperature or high pressure
Absorptance, transmittance, and emittance
Figure10.20shows 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 ment A vibration of the molecule must create a fluctuating dipole moment in order
mo-to interact with phomo-tons A rotation interacts with phomo-tons only if the molecule has a permanent dipole moment.
Trang 4§10.5 Gaseous radiation 567
Figure 10.20 The attenuation of
radiation through an absorbing (and/orscattering) gas
ness dx, the intensity will be reduced by an amount di λ:
where ρ is the gas density and κ λ is called the monochromatic
absorp-tion coefficient If the gas scatters radiaabsorp-tion, 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 λare all m2/kg.
If ρκ λis constant through the gas, eqn (10.43) can be integrated from
an initial intensity i λ0 at x = 0 to obtain
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:
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.
5All 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.
Trang 5568 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, whileallowing radiation with somewhat higher or lower wavelengths to passalmost unhindered Figure10.21shows the absorptance of water vapor
as a function of wavelength for a fixed depth We can see the absorptionbands at wavelengths of 6.3, 2.7, 1.9, and 1.4 µm that were mentionedbefore
A comparison of Fig 10.21 with Fig 10.2 readily shows why tion from the sun, as viewed from the earth’s surface, shows a number
radia-of spikey indentations at certain wavelengths Several radia-of those tions occur in bands where atmospheric water vapor absorbs incomingsolar radiation, in accordance with Fig.10.21 The other indentations inFig 10.2 occur where ozone and CO2 absorb radiation The sun itselfdoes not have these regions of low emittance; it is just that much of theradiation in these bands is absorbed by gases in the atmosphere before
indenta-it can reach the ground
Just as α λ and ε λare equal to one another for a diffuse solid surface,they are equal for a gas We may demonstrate this by considering anisothermal gas that is in thermal equilibrium with a black enclosure thatcontains 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:
Now, if radiation from some other source is transmitted through
a nonscattering isothermal gas, we can combine the absorption from
Trang 6§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 λ0 e −ρκ λ L
=τ λ +i λ b
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 ghave 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 CO2) mixed into a nonabsorbing species (say N2) If the
ab-sorbing gas has a partial pressure p aand 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
Trang 7570 Radiative heat transfer §10.5
Hottel and Sarofim [10.12] provide such correlations for CO2and H2O,built from research by Hottel and others before 1960 The correlationstake the form
tion for a total pressure of p = 1 atm with a very small partial pressure
of the absorbing species The second function, f2, is a correction factor
to account for other values of p a or p Additional corrections must be
applied if both CO2 and H2O 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 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 α g may be estimated from the correlation
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, 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 as
L0≡ 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:
6 Hottel originally recommended replacing the exponent 1/2 by 0.65 for CO 2 and 0.45 for H O Theory, and more recent work, both suggest using the value 1/2 [ 10.13 ].
Trang 8Figure 10.22 Functions used to predict ε g = f1f2 for water
vapor in air
571
Trang 9Figure 10.23 Functions used to predict ε g = f1f2 for CO2 inair All pressures in atmospheres.
572
Trang 10§10.5 Gaseous radiation 573
• 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/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 N2 and 0.2 atm CO2 What is the net
heat radiated to the walls if they are at 300◦C?
Solution. Let us first obtain ε g We have L0 = D = 0.40 m, a total
pressure 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, so
ε g = 0.098 Next, we use eqn (10.53) to obtain α g , with T w = 1031 ◦R,
Now we can calculate Qnetg-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
Total emissivity charts and the mean beam length provide a simple,
but crude, tool for dealing with gas radiation Since the introduction
Trang 11574 Radiative heat transfer §10.6
of 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 diation in complex enclosures have also improved The most versitile
ra-of 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 are
based on the idea of a “gray gas” — one for which ε λ and α λ are pendent of wavelength However, as we have pointed out, the gray gas
inde-model is not even a qualitative approximation to the properties of real
gases.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 have
a 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 hydrogen
flame will not Yet the temperature in the hydrogen flame is higher It
turns out that what is radiating both heat and light from the candle is soot
— small solid particles of almost thermally black carbon The CO2 and
H2O in the candle flame actually contribute relatively little to radiation
The sun
The sun continually irradiates the earth at a rate of about 1.74×1014kW
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 measured
by satellites above the atmosphere Much of this energy reaches theground, where it sustains the processes of life
7 Edwards [ 10.11 ] describes 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 of gray gases having different properties.
Trang 12§10.6 Solar energy 575
The temperature of the sun varies from tens of millions of kelvin in its
core to between 4000 and 6000 K at its surface, where most of the sun’s
thermal radiation originates The wavelength distribution of the sun’s
energy is not quite that of a black body, but it may be approximated as
such A straightforward calculation (see Problem 10.49) 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 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 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
Figure10.2shows the reduction of solar radiation by atmospheric
ab-sorption for one particular set of atmospheric conditions In fact, when
the sun passes through the atmosphere at a low angle (near the
hori-zon), the path of radiation through the atmosphere is longer, providing
relatively more opportunity for atmospheric absorption and scattering
Additional moisture in the air can increase the absorption by H2O, and,
of course, clouds can dramatically reduce the solar radiation reaching
the ground The consequence of these various effects is that the solar
radiation received on the ground is almost never more than 1200 W/m2
and is often only a few hundred W/m2 Extensive data are available for
estimating the ground level solar irradiation at a given location, time, and
date [10.14,10.15]
The distribution of the Sun’s energy and atmospheric
irradiation
Figure10.24 shows what becomes of the solar energy that impinges on
the earth if we average it over the year and the globe, taking account of
all kinds of weather Only 45% of the sun’s energy actually reaches the
earth’s surface The mean energy received is about 235 W/m2if averaged
over the surface and the year The lower left-hand portion of the figure
shows how this energy is, in turn, all returned to the atmosphere and to
space
The solar radiation reaching the earth’s surface includes direct
radi-ation that has passed through the atmosphere and diffuse radiradi-ation that
... class="text_page_counter">Trang 115 74< /b> Radiative heat transfer< /i> §10.6
of these ideas in the mid-twentieth century, major advances... 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... 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,