# Some limiting solutions for radiative transfer

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## The Diffusion Solution

Figure 1: Geometry for diffusion solution

If the participating medium is optically thick, then the solution of the RTE reduces considerably in form. Optically thick means that the mean free path for radiation, 1/ ( κλ + σs) = 1/βλ, is much smaller than the dimensions of the system being studied. In that case, radiative transfer becomes a diffusion process. This is analogous to the reduction of a molecular energy transport analysis to a conduction (diffusion) analysis at high molecular densities.

Consider a 1-dimensional system of infinite parallel plates containing a participating, isotropically scattering medium with wavelength dependent attenuation coefficient βλ(Fig. 1). If the plates are separated by a distance H, the RTE, eq. (68), reduces in this one-dimensional case to

$\frac{{\cos \theta }}{{{\beta _\lambda }}}\frac{{d{I_\lambda }(x,\theta )}}{{dx}} = \frac{{\cos \theta }}{{H{\beta _\lambda }}}\frac{{d{I_\lambda }(x,\theta )}}{{d(x/H)}} = {i_\lambda }(x/H) - {I_\lambda }(x/H,\theta )\qquad \qquad(1)$

If 1 / H βλ << 1 (optically thick), then the intensity can be expanded in a series,

${I_\lambda }\left( {x,\theta } \right) = {I_\lambda }^{\left( 0 \right)} + \frac{1}{{{\beta _\lambda }H}}{I_\lambda }^{\left( 1 \right)} + {\left( {\frac{1}{{{\beta _\lambda }H}}} \right)^2}{I_\lambda }^{\left( 2 \right)} + ......\qquad \qquad(2)$

Substituting the expansion into eq. (1) and using

${i_\lambda }\left( {{\tau _\lambda }} \right) \equiv (1 - {\omega _\lambda }){I_{\lambda b}}({\tau _\lambda }) + \frac{{{\omega _\lambda }}}{{4\pi }}\int_{\theta = 0}^\pi {\int_{\phi = 0}^{2\pi } {{I_\lambda }\left( {\theta ,\phi } \right)} } \sin \theta d\theta d\phi$

from The Radiative Transfer Equation for the source function iλ(x / H),

$\begin{array}{l} \frac{{\cos \theta }}{{H{\beta _\lambda }}}\left[ {\frac{{\partial I_\lambda ^{\left( 0 \right)}(x,\theta )}}{{\partial (x/H)}} + \frac{1}{{H{\beta _\lambda }}}\frac{{\partial I_\lambda ^{\left( 1 \right)}(x,\theta )}}{{\partial (x/H)}} + ...} \right] \\ = (1 - {\omega _\lambda }){I_{\lambda b}}({\tau _\lambda }) + \frac{{{\omega _\lambda }}}{{4\pi }}\int_{\theta = 0}^\pi {\int_{\phi = 0}^{2\pi } {\left[ {{I_\lambda }^{\left( 0 \right)} + \frac{1}{{{\beta _\lambda }H}}{I_\lambda }^{\left( 1 \right)} + {{\left( {\frac{1}{{{\beta _\lambda }H}}} \right)}^2}{I_\lambda }^{\left( 2 \right)} + ......} \right]} } d{\Omega _i} \\ - \left[ {{I_\lambda }^{\left( 0 \right)} + \frac{1}{{{\beta _\lambda }H}}{I_\lambda }^{\left( 1 \right)} + {{\left( {\frac{1}{{{\beta _\lambda }H}}} \right)}^2}{I_\lambda }^{\left( 2 \right)} + ......} \right] \\ \end{array}\qquad \qquad(3)$

Collecting all terms of zero order in 1 / Hβλ gives

$I_\lambda ^{\left( 0 \right)} = \left( {1 - {\omega _\lambda }} \right){I_{\lambda b}}\left( {{\tau _\lambda }} \right) + \frac{{{\omega _\lambda }}}{{4\pi }}\int_{{\Omega _1} = o}^{4\pi } I _\lambda ^{(0)}d{\Omega _i}\qquad \qquad(4)$

Because neither term on the right-hand side depends on Ωi, it follows that $I_\lambda ^{\left( 0 \right)}$ also cannot depend on Ωi. In that case, the integral in eq. (4) can be evaluated, and the result is

$I_\lambda ^{\left( 0 \right)} = {I_{\lambda b}}\left( {{\tau _\lambda }} \right)\qquad \qquad(5)$

This identifies the first term in the series in eq. (2). Now the first order terms are collected in eqs. (3), giving

$\cos \theta \left[ {\frac{{\partial I_{\lambda b}^{}(x)}}{{\partial (x/H)}}} \right] = \frac{{{\omega _\lambda }}}{{4\pi }}\int_{d{\Omega _i} = 0}^{4\pi } {{I_\lambda }^{\left( 1 \right)}} d{\Omega _i} - {I_\lambda }^{\left( 1 \right)}\qquad \qquad(6)$

Multiply through by dΩi = 2πsinθdθ and integrate over dΩi. The result is

$0 = \int_{{\Omega _i} = 0}^{4\pi } {I_\lambda ^{\left( 1 \right)}d{\Omega _i} - {\omega _\lambda }\int_{{\Omega _i} = 0}^{4\pi } {I_\lambda ^{\left( 1 \right)}d{\Omega _i}} } \qquad \qquad(7)$

Because ωλ ≠0, the integral in eq. (7) must be equal to zero. In that case, eq. (6) reduces to

${I_\lambda }^{\left( 1 \right)} = - \cos \theta \left[ {\frac{{\partial I_{\lambda b}^{}(x)}}{{\partial (x/H)}}} \right]\qquad \qquad(8)$

Now, the first two terms in the series for intensity [eq. (2)] are known. If the series is truncated after these two terms, the result is

${I_\lambda }\left( {x,\theta } \right) = {I_{\lambda b}}\left( x \right) - \frac{1}{{{\beta _\lambda }}}\cos \theta \frac{{\partial I_{\lambda b}^{}(x)}}{{\partial x}}\qquad \qquad(9)$

For the diffusion case, then, the local intensity depends only on the local blackbody intensity and its gradient. Now the local spectral radiative flux is found by substituting eq. (9) into

${{q''}_{rad, \lambda}} = \int_{\Omega = 0}^4 \pi {I_{\lambda}} s X n d \Omega = \int_{\Omega = 0}^4 \pi {I_{\lambda}}(\theta)cos \theta d \Omega$
$\begin{array}{l} {{q''}_{rad,\lambda }}\left( x \right) = \int_{{\Omega _i} = 0}^{4\pi } {{I_\lambda }} \left( \theta \right)\cos \theta d\Omega = \int_{{\Omega _i} = 0}^{4\pi } {\left[ {{I_{\lambda b}} - \frac{1}{{{\beta _\lambda }}}\cos \theta \frac{{\partial I_{\lambda b}^{}(x)}}{{\partial x}}} \right]} \cos \theta d{\Omega _i} \\ {\rm{ }} = \int_{\theta = - \pi /2}^{\pi /2} {\int_{\varphi = 0}^{2\pi } {{I_{\lambda b}}\cos \theta \sin \theta d\theta d\phi } } - \int_{\theta = - \pi /2}^{\pi /2} {\int_{\phi = 0}^{2\pi } {\frac{1}{{{\beta _\lambda }}}\frac{{\partial I_{\lambda b}^{}(x)}}{{\partial x}}} } {\cos ^2}\theta \sin \theta d\theta d\phi \\ {\rm{ }} = 0 - \frac{{4\pi }}{{3{\beta _\lambda }}}\frac{{\partial I_{\lambda b}^{}(x)}}{{\partial x}} = - \frac{4}{{3{\beta _\lambda }}}\frac{{\partial E_{\lambda b}^{}(x)}}{{\partial x}}{\rm{ }}\left( {{\rm{1}}0.{\rm{146}}} \right) \\ \end{array}$

For the diffusion solution, the local spectral radiant heat flux is thus dependent on only the local gradient in the spectral emissive power. This is known as the Rosseland diffusion equation, and was first derived from studies of radiative transfer through the Sun's upper atmosphere. If the properties of the participating medium are gray, then integrating over all wavelengths gives

${q''_{rad}} = - \frac{4}{{3\beta }}\frac{{dE_b^{}(x)}}{{dx}} = - \frac{{4\sigma }}{{3\beta }}\frac{{d{T^4}(x)}}{{dx}} = - \frac{{16\sigma {T^3}}}{{3\beta }}\frac{{dT(x)}}{{dx}}\qquad \qquad(11)$

This is the same form as the Fourier equation for heat conduction, and the radiative conductivity can be defined as

${k_{rad}} = \frac{{16\sigma {T^3}}}{{3\beta }}\qquad \qquad(12)$

If the medium is non-gray, then eq. (10) can be integrated over wavelength to find the total heat flux:

${q''_{rad}} = - \frac{4}{3}\int_{\lambda = 0}^\infty {\frac{1}{{{\beta _\lambda }}}\frac{{\partial E_{\lambda b}^{}(x)}}{{\partial x}}d\lambda \equiv - \frac{4}{{3{\beta _R}}}\frac{{d{E_b}}}{{dx}}} \qquad \qquad(13)$

Here, βR is an averaged attenuation coefficient known as the Rosseland mean attenuation coefficient. If attenuation is due only to absorption (nonscattering medium), the value of κR is found from

$\begin{array}{l} \frac{1}{{{\kappa _R}}} = \frac{{\int_{\lambda = 0}^\infty {\frac{1}{{{\kappa _\lambda }}}\frac{{\partial E_{\lambda b}^{}(x)}}{{\partial x}}d\lambda } }}{{\frac{{d{E_b}}}{{dx}}}} \\ = \int_{\lambda = 0}^\infty {\frac{1}{{{\kappa _\lambda }}}\frac{{dE_{\lambda b}^{}}}{{d{E_b}}}d\lambda = } \frac{1}{\sigma }\int_{\lambda = 0}^\infty {\frac{1}{{{\kappa _\lambda }}}\frac{{dE_{\lambda b}^{}}}{{d{T^4}}}d\lambda } \\ \end{array}\qquad \qquad( 14)$

The derivative $dE_{\lambda b}^{}/d{T^4}$ is found by taking the required derivative of the Planck blackbody relation with respect to T4. The use of the Rosseland mean absorption coefficient has some obvious problems; if there are regions of the spectrum that are transparent, then eq. (14) will become indeterminate. This is overcome in practice by finding the Rosseland mean for those parts of the spectrum that meet the optically thick criterion, and applying the diffusion solution in those regions. Where the medium is transparent, the wavelength dependent solutions for no participating medium are applied.

## The Slip Boundary Conditions

In the absence of conduction or convection, radiation problems have unfamiliar behavior near boundaries. The medium temperature does not approach the boundary temperature as a solid boundary with fixed temperature is approached. This can be seen for the case of a gray medium near a gray boundary (Fig. 2). An energy equation can be written on a plane very near the boundary.

Figure 2: Derivation of slip boundary condition

If the boundary is gray-diffuse and at temperature Tw, then the radiative flux (radiosity) leaving the boundary in the +x direction is ${q''_{ + x}} = {\rm{ }}\varepsilon \sigma T_w^4 + (1 - \varepsilon ){q''_{ - x}}$. The flux in the negative x direction comes from the optically thick medium and can be found by integrating eq. (9) over the hemisphere of directions with intensity having negative x components. This gives

$\begin{array}{l} {{q''}_{ - x}} = \int_{\phi = 0}^{2\pi } {\int_{\theta = \pi }^{ - \pi /2} {I(x,\theta )\cos \theta \sin \theta d\theta d\phi } } \\ = \int_{\phi = 0}^{2\pi } {\int_{\theta = \pi }^{ - \pi /2} {\left[ {{I_b}\left( {x = 0} \right) - \frac{1}{\beta }\cos \theta \frac{{\partial I_b^{}(x = 0)}}{{\partial x}}} \right]\cos \theta \sin \theta d\theta d\phi } } \\ = \pi {I_b}\left( {x = 0} \right) - \frac{{2\pi }}{\beta }\frac{{\partial I_b^{}(x = 0)}}{{\partial x}}\left. {\left[ {\frac{{{{\cos }^3}\theta }}{3}} \right]} \right|_{\theta = 0}^{ - \pi /2} = {E_b}\left( {x = 0} \right) + \frac{{2\pi }}{{3\beta }}\frac{{\partial I_b^{}(x = 0)}}{{\partial x}} \\ = {E_b}\left( {x = 0} \right) + \frac{2}{{3\beta }}\frac{{\partial E_b^{}(x = 0)}}{{\partial x}} = {E_b}\left( {x = 0} \right) - \frac{{q''(x = 0)}}{2} \\ \end{array}\qquad \qquad(15)$

Substituting into an energy balance across the area dA adjacent to the boundary,

$\begin{array}{l} q''\left( {x = 0} \right) = {{q''}_{ + x}} - {{q''}_{ - x}} = \left[ {\varepsilon \sigma T_w^4 + \left( {1 - \varepsilon } \right){{q''}_{ - x}}} \right] - {{q''}_{ - x}} = \varepsilon \left( {\sigma T_w^4 - {{q''}_{ - x}}} \right) \\ {\rm{ }} = \varepsilon \left[ {\sigma T_w^4 - \left( {\sigma {T^4}(x = 0) - \frac{{q''\left( {x = 0} \right)}}{2}} \right)} \right] \\ \end{array}\qquad \qquad(16)$

Substituting eq. (15),

${T^4}(x = 0) = T_w^4 - \frac{{q''\left( {x = 0} \right)}}{\sigma }\left( {\frac{1}{\varepsilon } - \frac{1}{2}} \right)\qquad \qquad(17)$

Equation (17) shows that the temperature of the medium at the boundary is less than the boundary temperature when the heat transfer is positive in the +x direction.

Now we can find an expression for the radiative heat flux between two infinite parallel plates at T1, ε1 and T22 separated by a gray medium of optical thickness τ using the diffusion approximation.

The diffusion solution, eq. (11) is

${q''_{rad}} = - \frac{{4\sigma }}{{3\beta }}\frac{{d{T^4}(x)}}{{dx}} = - \frac{{4\sigma }}{3}\frac{{d{T^4}(x)}}{{d\tau }}$

For the one-dimensional case, the heat flux is constant, so the equation can be integrated between the medium boundaries to give

${q''_{rad}}\int_{\tau = 0}^\tau {d{\tau ^*}} = - \frac{{4\sigma }}{3}\int_{T = T(\tau = 0)}^{T\left( \tau \right)} {d{T^4} = - } \frac{{4\sigma }}{3}\left[ {{T^4}\left( \tau \right) - {T^4}\left( 0 \right)} \right] = {q''_{rad}}\tau$

Using the slip boundary condition eq. (17) to eliminate the medium temperatures at the boundaries, T4(0) and T4(τ) in terms of the boundary temperatures,

$\begin{array}{l} {{q''}_{rad}} = \frac{{4\sigma }}{{3\tau }}\left[ {{T^4}\left( 0 \right) - {T^4}\left( \tau \right)} \right] \\ {\rm{ }} = \frac{{4\sigma }}{{3\tau }}\left\{ {\left[ {T_1^4 - \frac{{{{q''}_{rad}}}}{\sigma }\left( {\frac{1}{{{\varepsilon _1}}} - \frac{1}{2}} \right)} \right] - \left[ {T_2^4 + \frac{{{{q''}_{rad}}}}{\sigma }\left( {\frac{1}{{{\varepsilon _2}}} - \frac{1}{2}} \right)} \right]} \right\} \\ \end{array}$

or, finally,

${q''_{rad}} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{{3\tau }}{4} + \frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}$

Comparing with the result of the results with out participating media, ${q''_{rad}} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}$, it is seen that for the radiation diffusion case, the presence of the participating medium adds a radiative resistance of 3τ / 4.

## The Nearly Transparent Gas

If the medium has a small (but non-zero) absorption coefficient, then it will still emit from a volume element in accordance with

de = 4κPσT4dV

from Properties of Participating Media. However, minimal attenuation due to absorption will occur as the emitted intensity travels along a path, and absorption by and within the volume element can be neglected. This observation is often used in analyzing radiation loss from nonscattering small flames such as for laboratory burners, since the radiation energy loss from the element becomes

${q_{rad}} = 4{\kappa _P}\sigma T_{dV}^4dV\qquad \qquad(18)$

and this is easily included in an energy balance that includes the combustion energy release for determining flame temperature, and often serves as a first-order radiative loss correction to the adiabatic flame temperature.

## Isothermal Media

Figure 3: Black cold hemisphere enclosing a participating isothermal nonscattering medium with uniform absorption coefficient a.

Consider a high-temperature volume of participating medium contained in an enclosure with cold black walls. If the medium is well-mixed, then its temperature and properties are uniform within the enclosure. Given the temperature and composition of the medium, the Planck mean absorption coefficient can be computed from

$de = 4dV\int_{\lambda = 0}^\infty {{\kappa _\lambda }{E_{\lambda b}}} d\lambda$

from Properties of Participating Media using the spectral absorption coefficient.

Consider a hemisphere containing a nonscattering uniform isothermal medium at temperature T (Fig. 3). At the center of the base of the hemisphere is an element dA. The intensity striking dA from the direction of the ring element on the surface of the hemisphere is given by

${I_\lambda }({\tau _\lambda }) = {I_\lambda }(0)\exp \left( { - {\tau _\lambda }} \right) + \int_{\tau {*_\lambda } = 0}^{{\tau _\lambda }} {{i_\lambda }} \left( {\tau {*_\lambda }} \right)\exp \left[ { - \left( {{\tau _\lambda } - \tau {*_\lambda }} \right)} \right]d\tau {*_\lambda }$
${I_\lambda }({\kappa _\lambda }R) = {I_\lambda }(0)\exp \left( { - {\kappa _\lambda }R} \right) + \int_{\kappa {*_\lambda } = 0}^{{\kappa _\lambda }} {{\kappa _\lambda }{i_\lambda }} \left( {{\kappa _\lambda }r} \right)\exp \left[ { - {\kappa _\lambda }\left( {R - r} \right)} \right]dr\qquad \qquad(19)$

Here, r is the distance along R from the hemisphere surface, and for this geometry and a nonscattering medium, τλ has been replaced with κλR and τλ * with κλr. Since the hemisphere boundary is cold and black, Iλ(0) = 0. For the nonscattering isothermal medium, the source function iλ(κλr) reduces to:

${i_\lambda }\left( {{\kappa _\lambda }r} \right) = {I_{\lambda b}}({\kappa _\lambda }r) = {I_{\lambda b}}(T)\qquad \qquad(20)$

and eq. (19) becomes

$\begin{array}{l} {I_\lambda }({\kappa _\lambda }R) = {I_{\lambda b}}\left( T \right)\int_{{a_\lambda }r = 0}^{{a_\lambda }R} {{\kappa _\lambda }\exp \left[ { - {\kappa _\lambda }\left( {R - r} \right)} \right]} dr \\ {\rm{ = }}{I_{\lambda b}}\left( T \right)\exp ( - {\kappa _\lambda }R)\left[ {\exp ({\kappa _\lambda }R) - 1} \right] = {I_{\lambda b}}\left( T \right)\left[ {1 - \exp ( - {\kappa _\lambda }R)} \right] \\ \end{array}\qquad \qquad(21)$
Table 1 Mean Beam Lengths for Common Geometries
 Geometry Characterizing Dimension Mean Beam Length Le for Finite Optical Thickness Sphere radiating to surface Diameter D 0.65D Circular cylinder of infinite lengthradiating to interior surface DiameterD 0.95D Circular cylinder of height equal to twodiameters radiating to:Plane endConcave surfaceEntire surface Diameter D 0.60D0.76D0.73D Infinite slab of medium radiating to:Element on one faceBoth bounding planes Slab thickness H 1.8H1.8H Rectangular parallelepiped:1×1×4 volume radiating to:1×4 face1×1 faceAll faces Shortest side X 0.82X0.71X0.81X

Note that the term containing the exponential is the spectral emittance ελ(R) of the medium is now found from:

${{q''}_{rad,\lambda }} = \int_{\omega = 0}^{4\pi } {{I_\lambda }} \left( \theta \right)\cos \theta d\Omega = \int_{\omega = 0}^{4\pi } {{\varepsilon _\lambda }(R){I_{\lambda b}}\left( T \right)} \cos \theta d\Omega$
= ${\varepsilon _\lambda }(R){I_{\lambda b}}\left( T \right)\int_{\omega = 0}^{4\pi } {\cos \theta d\Omega }$
= ${\varepsilon _\lambda }(R){I_{\lambda b}}\left( T \right)\int_{\phi = 0}^{2\pi } {\int_{\theta = 0}^{\pi /2} {\cos \theta } \sin \theta d\theta d\phi } = {\varepsilon _\lambda }(R)\pi {I_{\lambda b}}\left( T \right)$
= ${\varepsilon _\lambda }(R){E_{\lambda b}}\qquad \qquad(22)$

Equation (22) is clearly a very simple equation for determining the heat flux for this particular geometry. For more complex geometries, the quantity Iλ(θ) inside the integral will vary with Ω, and the integration to find the radiative flux at a location on the boundary will become quite complex. For any given geometry, however, there is some value of a fictitious length, called the mean beam length, Le, that can be substituted for R in the hemisphere relation, eq. (22), and will give the correct value for heat flux that is found from the detailed integration. The spectral radiative heat flux is then found from

${q''_{rad,\lambda }} = {\varepsilon _\lambda }({L_e}){E_{\lambda b}}\qquad \qquad(23)$

Integrating over all wavelengths and substituting the Planck mean emittance

$\varepsilon \left( {{L_e},T} \right) = \frac{{\int_{\lambda = 0}^\infty {{\varepsilon _\lambda }({L_e}){E_{\lambda b}}\left( {T,\lambda } \right)d\lambda } }}{{\sigma {T^4}}}$

gives the total heat flux as

${q''_{rad}} = \varepsilon ({L_e})\sigma {T^4}\qquad \qquad(24)$

Values for the mean beam length have been computed for many common geometries. Some of these are given in Table 1.

## General geometries

The following equation from Properties of Participating Media

de = 4κPσT4dV

indicates that the total energy emitted by an isothermal volume element is

$de = 4{a_P}\sigma {T^4}dV\qquad \qquad(25)$

If the medium is optically thin, then there is no attenuation of this energy before striking the cold black boundary, so the average radiative flux over the entire boundary is

${q''_{rad}} = \frac{e}{A} = \frac{{4{\kappa _P}\sigma {T^4}V}}{A}\qquad \qquad(26)$

Also, for the optically thin case, the emittance of the gas can be rewritten using a series expansion of the exponential term to give

$\varepsilon \left( {{L_e}} \right) = 1 - \exp ( - \kappa {L_e}) = 1 - \left[ {1 - \kappa {L_e} + \frac{{{{\left( {\kappa {L_e}} \right)}^2}}}{{2!}} - ...} \right] \approx \kappa {L_{e,0}}\qquad \qquad(27)$

Substituting this result into eq. (24) gives

${q''_{rad}} = \kappa {L_{e,0}}\sigma {T^4}\qquad \qquad(28)$

Comparing eqs. (28) and (26) shows that for an optically thin medium radiating to a cold black boundary, the mean beam length is

${L_{e,0}} = \frac{{4V}}{A}\qquad \qquad(29)$

By comparison of the result of eq. (29) with values of mean beam length computed using complete integration over the volume of a gas that is not optically thin, it is found that to correct the optically thin approximation for absorbing gases, reasonable values for the mean beam length for most geometries are

${L_e} = 0.9{L_{e,0}} = \frac{{3.6V}}{A}\qquad \qquad(30)$

## Gas Emittance Values

Participating media have extremely varying wavelength dependence, and the assumption of a gray gas is much less applicable than the assumption of gray surfaces.

The wavelength dependence occurs because gases absorb radiation through converting radiative energy into internal energy. For gases the conversion processes are by absorption of photons that causes energy transitions between quantized energy states. At lower gas temperatures, these transitions occur between vibrational-rotational states, and at higher temperatures electronic transitions and transitions from so-called bound electronic states to free electron states through dissociation and ionization can occur. Monatomic gases (argon, krypton, neon, etc.) and most homopolar diatomic molecules (N2,O2, etc) do not have transitions that are activated by radiation in the wavelength regions of engineering interest except at very high temperatures such as in spacecraft re-entry. Thus, the atmospheric gases are essentially transparent and do not participate in radiative transfer. Polyatomic molecules such as H2O,CO2, and hydrocarbons have transitions that can be activated (that is, absorption can occur) in the infrared. These gases are of particular engineering interest in the design of utility and chemical process furnaces, and in the study of global warming.

The absorption of radiation results in a transition between a pair of quantized energy states Ei and Ej, and this appears as an absorption line in the absorption spectrum. The wavelength of absorption that causes this transition is given by

${E_j} - {E_i} = h{\nu _{i - j}} = \frac{{hc}}{{{\lambda _{i - j}}}}\qquad \qquad(31)$

where vij is the frequency of the absorbed photon and λij is its wavelength. The line is very narrow; in fact it would have zero spectral width if the lower and upper quantized energy states were separated by an exact energy interval. Absorption of a photon then would occur only for photons that have exactly the frequency or wavelength given by eq. (31). However, the energy difference between the upper and lower energy states is not exact due to several effects that perturb the values of the energy states in eq. (31) and cause the lines observed for absorption by a group of molecules to have finite spectral width. These include the uncertainty principal, the velocity distribution of the gas molecules relative to the speed of light (Doppler broadening), the interactions with nearby molecules that distort the energy states (collision broadening), the effects of electric fields from free electrons or related effects (Stark broadening) and others.

Because there are very many vibrational-rotational states, there are many lines in the absorption spectrum. For CO2 and H2O, the lines are closely spaced in certain regions of the IR spectrum, and indeed the line widths overlap to the extent that in low resolution spectra, the lines coalesce into absorption bands.

Given the computational difficulty of integrating the spectrum over thousands of individual lines within the bands to find total heat transfer (called a line-by-line calculation), much effort has been expended to develop higher-level engineering approximations to the spectral properties that can be used in practical calculations. Some approaches are based on finding averaged properties over narrow spectral bands based on either experimental measurement or line-by-line calculation (narrow band correlations); others are based on correlation of the characteristics of the wide absorption bands (wide-band approximations), and some correlate the properties over the entire spectrum to provide the total gas emittance. Finding useful approaches for correlating the absorbing properties of gases for use in engineering calculations is an active research area (Modest and Zhang, 2002; Denison and Webb, 1995).

Because the emittance of a participating gas mixture (say CO2 and H2O as present in combustion products) depends on gas temperature, total pressure, and concentration, correlation is not simple. An additional complication for mixtures of participating gases is that the absorption lines and bands of the individual gases may overlap, and this must be accounted for in addition to predicting the properties of the individual gases.

Hottel (1954) presented curves of the emittance εLe for CO2 and H20 which have been widely used in the mean beam length approach. Hottel's curves were based on experimental data with extrapolations to some temperature and Le-partial pressure regions based on theory. These curves have been updated based on more recent data, and the approach remains very useful for the isothermal gas mean beam length approximation used in this section. Rather than using Hottel's original curves, it is useful to have analytical expressions for the emittance. For water vapor in air, Cess and Lian (1976) give the correlation

Table 2: Constants for use in eq. (32)
 T(K) a0 a1(m-1/2atm-1) 300 0.683 1.17 600 0.674 1.32 900 0.700 1.27 1200 0.673 1.21 1500 0.624 1.15
${\varepsilon _{{H_2}O}}({L_e},T) = {a_0}\left[ {1 - \exp ( - {a_1}\sqrt X } \right]\qquad \qquad(32)$

where $X = {p_{{H_2}O}}{L_e}\left( {{p_{air}} + b{p_{{H_2}O}}} \right)\left( {300/T} \right)$ and $b = 5.0{\left( {300/T} \right)^{1/2}} + 0.5$. In these relations, T is in K, p in atm, and Le in m. The constants in eq. (32) are in Table 2.

Leckner (1972) gives empirical expressions for the total emittance derived from expressions for narrow band behavior summed over the spectrum for both water vapor and CO2. In these correlations and equations, p is in bar, and Le is in cm. The most accurate expressions from Leckner agree within five percent to experimental data for T>400K. The correlation equation is

$\varepsilon = \exp \left\{ {{a_0} + \sum\limits_{i = 1}^M {{a_i}} {{\left[ {\log (p{L_e})} \right]}^i}} \right\}\qquad \qquad(33)$
Table 3: Coefficients cji for Water Vapor and CO2 Emittance in eq. (33)
 i c0i cli c2i c3i c4i Water Vapor, T > 400 K, M=2, N=2 0 -2.2118 -1.1987 0.035596 1 0.85667 0.93048 -0.14391 2 -0.10838 -0.17156 0.045915 Carbon Dioxide, T > 400K, M=3, N=4 0 -3.9781 2.7353 -1.9822 0.31054 0.015719 1 1.9326 -3.5932 3.7247 -1.4535 0.20132 2 -0.35366 0.61766 -0.84207 0.39859 -0.063356 3 -0.080181 0.31466 -0.19973 0.046532 -0.0033086

where ${a_i} = {c_{0i}} + \sum\limits_{j = 1}^N {{c_{ji}}} {\left( {T/1000} \right)^j}$ and the values of cji are in Table 3 for water vapor and CO2. A plot of the emittance predicted by eq. (33) is in Fig. 4. Observe that the emittance increases with the pressure-path length product as expected. The trend with temperature is that emittance generally decreases with increasing temperature for water vapor. In contrast, CO2 tends to go through a peak in emittance at about 1200 K.

The individual emittances for H2O and CO2 in air must be modified when both gases are present in a mixture, which is commonly the case. This is because the individual spectral lines and absorption bands for the two gases overlap in some spectral regions, and simple addition of the emittance will overpredict the effect of the mixture. In some cases, a simple addition can predict a gas absorptance and emittance that is greater than unity at certain wavelengths. The overlap correction equation is

Figure 4: Computed emittance of water vapor from eq. (33) using the coefficients of Table 10.3. Units of pL are bar-cm.
$\varepsilon (p{L_e}) = {\varepsilon _{{H_2}O}}({p_{{H_2}O}}{L_e}) + {\varepsilon _{C{O_2}}}({p_{C{O_2}}}{L_e}) - \Delta \varepsilon \qquad \qquad(34)$

Hottel (1954) presents a graph of the approximate band overlap correction Δε. An empirical expression for the band overlap correction that is in good agreement with the Hottel chart (Leckner, 1972) valid for 1000 < T < 2200 K and all pressures is

$\Delta \varepsilon = \left( {\frac{\zeta }{{10.7 + 101\zeta }} - 0.0089{\zeta ^{10.4}}} \right){\left[ {{{\log }_{10}}(p{L_e})} \right]^{2.76}}\qquad \qquad(35)$

where $\zeta = {p_{{H_2}O}}/\left( {{p_{{H_2}O}} + {p_{C{O_2}}}} \right)$, p is in bars, and Le in cm.

The correlations and overlap corrections are for properties of the absorbing gases mixed with air at a total pressure of one atmosphere. If the total pressure differs considerably from one atm, then a pressure correction must be applied to the predicted one atm emittance because of increased pressure broadening of the individual lines that make up the bands that are summed to obtain the total emittance. Hottel (1954) has also presented graphs for this correction, and Leckner (1972) has provided an algebraic expression. Because most engineering equipment where radiation is important operates at near to one atm, the pressure correction is not given here.

The mean beam length approximation for radiation in an enclosure of isothermal gases provides a straightforward tool for radiative transfer estimates. The restrictions to isothermal gases with cold black boundaries can be relaxed; however, these approximations are often met quite well in utility boilers and furnaces where combustion gases are highly turbulent and well mixed, giving near isothermal gas mixtures; the surfaces are covered with ash and soot, making them near-black; and the boundaries are water-walls at relatively low temperatures compared with the combustion gases.

## References

Cess, R.D. and Lian, M.S., 1976, “A Simple Parameterization for the Water Vapor Emissivity,” J. Heat Transfer, Vol. 98, pp. 676-678.

Denison, M.K. and Webb, B.W., 1995, “The Spectral-Line Weighted-Sum-of-Gray-Gases Model for H2O/CO2 Mixtures,” J. Heat Transfer, Vol. 117, pp. 788-798.

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.

Hottel, H.C, 1954, “Radiant Heat Transmission,” Chap. 4, in W.H. McAdams (Ed.), Heat Transmission, 3rd ed., McGraw-Hill, New York, NY.

Leckner, B., 1972, “Spectral and Total Emissivity of Water Vapor and Carbon Dioxide,” Combustion and Flame, Vol. 19, pp. 33-48.

Modest, M.M. and Zhang, H., 2002, “The Full-Spectrum Correlated-k Distribution for Thermal Radiation From Molecular Gas-Particulate Mixtures,” J. Heat Transfer, Vol. 124, pp. 30-38.

Siegel, R. and Howell, J.R., 2002, Thermal Radiation Heat Transfer, 4th ed., Taylor and Francis, New York, NY.