4.11. ADVANCED RADIATION HEAT-TRANSFER PRINCIPLES

4.11A. Introduction and Radiation Spectrum

1. Introduction

This section will cover some basic principles together with some advanced topics on radiation that were not covered in Section 4.10. The exchange of radiation between two surfaces depends upon the size, shape, and relative orientation of these two surfaces and also upon their emissivities and absorptivities. In the cases to be considered the surfaces are separated by nonabsorbing media such as air. When gases such as CO2 and H2O vapor are present, some absorption by the gases occurs, which is not considered until later in this section.

2. Radiation spectrum and thermal radiation

Energy can be transported in the form of electromagnetic waves, which travel at the speed of light. Bodies may emit many forms of radiant energy, such as gamma rays, thermal energy, radio waves, and so on. In fact, there is a continuous spectrum of electromagnetic radiation. This electromagnetic spectrum is divided into a number of wavelength ranges, such as cosmic rays (λ < 10-13 m), gamma rays (λ = 10-13 to 10–10 m), thermal radiation (λ = 10-7 to 10-4 m), and so on. The electromagnetic radiation produced solely because of the temperature of the emitter is called thermal radiation and exists between the wavelengths of 10-7 and 10-4 m. This portion of the electromagnetic spectrum is of importance in radiant thermal heat transfer. Electromagnetic waves having wavelengths between 3.8 × 10-7 and 7.6 × 10-7 m, called visible radiation, can be detected by the human eye. This visible radiation lies within the thermal radiation range.

When different surfaces are heated to the same temperature, they do not all emit or absorb the same amount of thermal radiant energy. A body that absorbs and emits the maximum amount of energy at a given temperature is called a black body. A black body is a standard to which other bodies can be compared.

3. Planck's law and emissive power

When a black body is heated to a temperature T, photons are emitted from the surface which have a definite distribution of energy. Planck's equation relates the monochromatic emissive power E in W/m3 at a temperature T in K and a wavelength λ in m:

Equation 4.11-1


A plot of Eq. (4.11-1) is given in Fig. 4.11-1 and shows that the energy given off increases with T. Also, for a given T, the emissive power reaches a maximum value at a wavelength that decreases as the temperature T increases. At a given temperature the radiation emitted extends over a spectrum of wavelengths. The visible-light spectrum occurs in the low λ region. The sun has a temperature of about 5800 K and the solar spectrum straddles this visible range.

Figure 4.11-1. Spectral distribution of total energy emitted by a black body at various temperatures of the black body.


For a given temperature, the wavelength at which the black-body emissive power is a maximum can be determined by differentiating Eq. (4.11-1) with respect to λ at constant T and setting the result equal to zero. The result is as follows and is known as Wien's displacement law:

Equation 4.11-2


The locus of the maximum values is shown in Fig. 4.11-1.

4. Stefan-Boltzmann law

The total emissive power is the total amount of radiation energy per unit area leaving a surface with temperature T over all wavelengths. For a black body, the total emissive power is given by the integral of Eq. (4.11-1) at a given T over all wavelengths, or the area under the curve in Fig. 4.11-1:

Equation 4.11-3


This gives

Equation 4.11-4


The result is the Stefan-Boltzmann law with σ = 5.676 × 10-8 W/m2 · K4. The units of EB are W/m2.

5. Emissivity and Kirchhoff's law

An important property in radiation is the emissivity of a surface. The emissivity ε of a surface is defined as the total emitted energy of the surface divided by the total emitted energy of a black body at the same temperature:

Equation 4.11-5


Since a black body emits the maximum amount of radiation, ε is always <1.0.

We can derive a relationship between the absorptivity α1 and emissitivy ε1 of a material by placing this material in an isothermal enclosure and allowing the body and enclosure to reach the same temperature at thermal equilibrium. If G is the irradiation on the body, the energy absorbed must equal the energy emitted:

Equation 4.11-6


If this body is removed and replaced by a black body of equal size, then at equilibrium,

Equation 4.11-7


Dividing Eq. (4.11-6) by (4.11-7),

Equation 4.11-8


But α2 = 1.0 for a black body. Hence, since E1/EB = ε1,

Equation 4.11-9


This is Kirchhoff's law, which states that at thermal equilibrium α = ε of a body. When a body is not at equilibrium with its surroundings, the result is not valid.

6. Concept of gray body

A gray body is defined as a surface for which the monochromatic properties are constant over all wavelengths. For a gray surface,

Equation 4.11-10


Hence, the total absorptivity a and the monochromatic absorptivity αλ of a gray surface are equal, as are ε and ελ:

Equation 4.11-11


Applying Kirchhoff's law to a gray body, αλ = ελ and

Equation 4.11-12


As a result, the total absorptivity and emissivity are equal for a gray body even if the body is not in thermal equilibrium with its surroundings.

Gray bodies do not exist in practice; the concept of a gray body is an idealized one. The absorptivity of a surface actually varies with the wavelength of the incident radiation. Engineering calculations can often be based on the assumption of a gray body with reasonable accuracy. The a is assumed constant even with a variation in λ of the incident radiation. Also, in actual systems, various surfaces may be at different temperatures. In these cases, α for a surface is evaluated by determining the emissivity not at the actual surface temperature but at the temperature of the source of the other radiating surface or emitter, since this is the temperature the absorbing surface would reach if the absorber and emitter were at thermal equilibrium. The temperature of the absorber has only a slight effect on the absorptivity.

4.11B. Derivation of View Factors in Radiation for Various Geometries

1. Introduction

The concepts and definitions presented in Section 4.11A form a sufficient foundation so that the net radiant exchange between surfaces can be determined. If two surfaces are arranged so that radiant energy can be exchanged, a net flow of energy will occur from the hotter surface to the colder surface. The size, shape, and orientation of two radiating surfaces or a system of surfaces are factors in determining the net heat-flow rate between them. To simplify the discussion we assume that the surfaces are separated by a nonabsorbing medium such as air. This assumption is adequate for many engineering applications. However, in cases such as a furnace, the presence of CO2 and H2O vapor make such a simplification impossible because of their high absorptivities.

The simplest geometrical configuration will be considered first, that of radiation exchange between parallel, infinite planes. This assumption implies that there are no edge effects in the case of finite surfaces. First, the simplest case will be treated, in which the surfaces are black bodies, and then more complicated geometries and gray bodies will be treated.

2. View factor for infinite parallel black planes

If two parallel and infinite black planes at T1 and T2 are radiating toward each other, plane 1 emits radiation to plane 2, which is all absorbed. Also, plane 2 emits radiation to plane 1, which is all absorbed. Then for plane 1, the net radiation is from plane 1 to 2,

Equation 4.11-13


In this case all the radiation from 1 to 2 is intercepted by 2; that is, the fraction of radiation leaving 1 that is intercepted by 2 is F12, which is 1.0. The factor F12 is called the geometric view factor or simply view factor. Hence,

Equation 4.11-14


where F12 is fraction of radiation leaving surface 1 in all directions which is intercepted by surface 2. Also,

Equation 4.11-15


In the case of parallel plates, F12 = F21 = 1.0 and the geometric factor is simply omitted.

3. View factor for infinite parallel gray planes

If both of the parallel plates A1 and A2 are gray, with emissivities and absorptivities of ε1 = α1 and ε2 = α2, respectively, we can proceed as follows. Since each surface has an unobstructed view of the other, the view factor is 1.0. In unit time, surface A1 emits radiation to A2. Of this, the fraction ε2 (where α2 = ε2) is absorbed:

Equation 4.11-16


Also, the fraction (1 - ε2) or the amount (1 - ε2)() is reflected back to A1. Of this amount A1 reflects back to A2 a fraction (1 - ε1) or an amount (1 - ε1)(1 - ε2)(). The surface A2 absorbs the fraction ε2, or

Equation 4.11-17


The amount reflected back to A1 from A2 is (1 - ε2)(1 - ε1)(1 - ε2)(). Then A1 absorbs ε1 of this and reflects back to A2 an amount (1 - ε1)(1 - ε2)(1 - ε1)(1 - ε2) × (). The surface A2 then absorbs

Equation 4.11-18


This continues, and the total amount absorbed at A2 is the sum of Eqs. (4.11-16), (4.11-17), (4.11-18), and so on:

Equation 4.11-19


The result is a geometric series (M1):

Equation 4.11-20


Repeating the above for the amount absorbed at A1 which comes from A2,

Equation 4.11-21


The net radiation is the difference of Eqs. (4.11-20) and (4.11-21):

Equation 4.11-22


If ε1 = ε2 = 1.0 for black bodies, Eq. (4.11-22) becomes Eq. (4.11-13).

EXAMPLE 4.11-1. Radiation Between Parallel Planes

Two parallel gray planes which are very large have emissivities of ε1 = 0.8 and ε2 = 0.7; surface 1 is at 1100°F (866.5 K) and surface 2 at 600°F (588.8 K). Use English and SI units for the following:

  1. What is the net radiation from 1 to 2?

  2. If the surfaces are both black, what is the net radiation?

Solution: For part (a), using Eq. (4.11-22) and substituting the known values,


For black surfaces in part (b), using Eq. (4.11-13),


Note the large reduction in radiation when surfaces with emissivities less than 1.0 are used.


Example 4.11-1 shows the large influence that emissivities less than 1.0 have on radiation. This fact is used to reduce radiation loss or gain from a surface by using planes as a radiation shield. For example, for two parallel surfaces of emissivity ε at T1 and T2, the interchange is, by Eq. (4.11-22),

Equation 4.11-23


The subscript 0 indicates that there are no planes in between the two surfaces. Suppose that we now insert one or more radiation planes between the original surfaces. Then it can be shown that

Equation 4.11-24


where N is the number of radiation planes or shields between the original surfaces. Hence, a great reduction in radiation heat loss is obtained by using these shields.

4. Derivation of general equation for view factor between black bodies

Suppose that we consider radiation between two parallel black planes of finite size as shown in Fig. 4.11-2a. Since the planes are not infinite in size, some of the radiation from surface 1 does not strike surface 2, and vice versa. Hence, the net radiation interchange is less, since some is lost to the surroundings. The fraction of radiation leaving surface 1 in all directions which is intercepted by surface 2 is called F12 and must be determined for each geometry by taking differential surface elements and integrating over the entire surface.

Figure 4.11-2. Radiation between two black surfaces: (a) two planes alone, (b) two planes connected by refractory reradiating walls.


Before we can derive a general relationship for the view factor between two finite bodies, we must consider and discuss two quantities, a solid angle and the intensity of radiation. A solid angle ω is a dimensionless quantity which is a measure of an angle in solid geometry. In Fig. 4.11-3a the differential solid angle 1 is equal to the normal projection of dA2 divided by the square of the distance between the point P and area dA2:

Equation 4.11-25


Figure 4.11-3. Geometry for a solid angle and intensity of radiation: (a) solid-angle geometry, (b) intensity of radiation from emitting area dA.


The units of a solid angle are steradian or sr. For a hemisphere the number of sr subtended by this surface is 2π.

The intensity of radiation for a black body, IB, is the rate of radiation emitted per unit area projected in a direction normal to the surface and per unit solid angle in a specified direction as shown in Fig. 4.11-3b. The projection of dA on the line between centers is dA cos θ.

Equation 4.11-26


where q is in W and IB is in W/m2 · sr. We assume that the black body is a diffuse surface which emits with equal intensity in all directions, that is, I = constant. The emissive power EB which leaves a black-body plane surface is determined by integrating Eq. (4.11-26) over all solid angles subtended by a hemisphere covering the surface. The final result is as follows [see references (C3, H1, K1) for details]:

Equation 4.11-27


where EB is in W/m2.

In order to determine the radiation heat-transfer rates between two black surfaces, we must determine the general case for the fraction of the total radiant heat that leaves a surface and arrives on a second surface. Using only black surfaces, we consider the case shown in Fig. 4.11-4, in which radiant energy is exchanged between area elements dA1 and dA2. The line r is the distance between the areas, and the angles between this line and the normals to the two surfaces are θ1 and θ2. The rate of radiant energy that leaves dA1 in the direction given by the angle θ1 is IB1 dA cos θ1. The rate that leaves dA1 and arrives on dA2 is given by Eq. (4.11-28):

Equation 4.11-28


Figure 4.11-4. Area elements for radiation shape factor.


where 1 is the solid angle subtended by the area dA2 as seen from dA1. Combining Eqs. (4.11-25) and (4.11-28),

Equation 4.11-29


From Eq. (4.11-27), IB1 = EB1 /π. Substituting EB1 for IB1 into Eq. (4.11-29),

Equation 4.11-30


The energy leaving dA2 and arriving at dA1 is

Equation 4.11-31


Substituting for EB1 and for EB2 from Eq. (4.11-4) and taking the difference of Eqs. (4.11-30) and (4.11-31) for the net heat flow,

Equation 4.11-32


Performing the double integrations over surfaces A1 and A2 will yield the total net heat flow between the finite areas:

Equation 4.11-33


Equation (4.11-33) can also be written as

Equation 4.11-34


where F12 is a geometric shape factor or view factor and designates the fraction of the total radiation leaving A1 which strikes A2, and F21 represents the fraction leaving A2 which strikes A1. Also, the following relation exists:

Equation 4.11-35


which is valid for black surfaces and nonblack surfaces. The view factor F12 is then

Equation 4.11-36


Values of the view factor can be calculated for a number of geometrical arrangements.

5. View factors between black bodies for various geometries

A number of basic relationships between view factors are given below.

The reciprocity relationship given by Eq. (4.11-35) is

Equation 4.11-35


This relationship can be applied to any two surfaces i and j:

Equation 4.11-37


If surface A1 can only see surface A2, then F12 = 1.0.

If surface A1 sees a number of surfaces A2,A3, . . ., and all the surfaces form an enclosure, then the enclosure relationship is

Equation 4.11-38


If the surface A1 cannot see itself (surface is flat or convex), F11 = 0.

EXAMPLE 4.11-2. View Factor from a Plane to a Hemisphere

Determine the view factors between a plane A1 covered by a hemisphere A2 as shown in Fig. 4.11-5.

Figure 4.11-5. Radiant exchange between a flat surface and a hemisphere for Example 4.11-2.


Solution: Since surface A1 sees only A2, the view factor F12 = 1.0. Using Eq. (4.11-35),

Equation 4.11-35


The area A1 = πR2;A2 = 2πR2. Substituting into Eq. (4.11-35) and solving for F21,


Using Eq. (4.11-38) for surface A1, F11 = 1.0 - F12 = 1.0 - 1.0 = 0. Also, writing Eq. (4.11-38) for surface A2,

Equation 4.11-39


Solving for F22, F22 = 1.0 - F21 = 1.0 - = .


EXAMPLE 4.11-3. Radiation Between Parallel Disks

In Fig. 4.11-6 a small disk of area A1 is parallel to a large disk of area A2, with A1 centered directly below A2. The distance between the centers of the disks is R and the radius of A2 is a. Determine the view factor for radiant heat transfer from A1 to A2.

Figure 4.11-6. View factor for radiation from a small element to a parallel disk for Example 4.11-3.


Solution: The differential area for A2 is taken as the circular ring of radius x so that dA2 = 2πx dx. The angle θ1 = θ2. Using Eq. (4.11-36),


In this case the area A1 is very small compared to A2, so dA1 can be integrated to A1 and the other terms inside the integral can be assumed constant. From the geometry shown, r = (R2 + x2)1/2, cos θ1 = R/(R2 + x2)1/2. Making these substitutions into the equation for F12,


Integrating,


The integration of Eq. (4.11-36) has been performed for numerous geometrical configurations and values of F12 tabulated. Then,

Equation 4.11-34


where F12 is the fraction of the radiation leaving A1 which is intercepted by A2 and F21 the fraction reaching A1 from A2. Since the flux from 1 to 2 must equal that from 2 to 1, Eq. (4.11-34) becomes Eq. (4.11-35) as given previously:

Equation 4.11-35


Hence, one selects the surface whose view factor can be determined most easily. For example, the view factor F12 for a small surface A1 completely enclosed by a large surface A2 is 1.0, since all the radiation leaving A1 is intercepted by A2. In Fig. 4.11-7 the view factors F12 between parallel planes are given, and in Fig. 4.11-8 the view factors for adjacent perpendicular rectangles. View factors for other geometries are given elsewhere (H1, K1, P3, W1).

Figure 4.11-7. View factor between parallel planes directly opposed. (From W. H. McAdams, Heat Transmission, 3rd ed. New York McGraw-Hill Book Company, 1954. With permission.)


Figure 4.11-8. View factor for adjacent perpendicular rectangles. [From H. C. Hottel, Mech. Eng., 52, 699 (1930). With permission.]



4.11C. View Factors When Surfaces Are Connected by Reradiating Walls

If the two black-body surfaces A1 and A2 are connected by nonconducting (refractory) but reradiating walls as in Fig. 4.11-2b, a larger fraction of the radiation from surface 1 is intercepted by 2. This view factor is called . The case of two surfaces connected by the walls of an enclosure such as a furnace is a common example of this. The general equation for this case assuming a uniform refractory temperature has been derived (M1, C3) for two radiant sources A1 and A2 which are not concave, so they do not see themselves:

Equation 4.11-40


Also, as before,

Equation 4.11-41


Equation 4.11-42


The factor for parallel planes is given in Fig. 4.11-7 and for other geometries can be calculated from Eq. (4.11-36). For view factors F12 and for parallel tubes adjacent to a wall as in a furnace as well as for variation in refractory wall temperature, see elsewhere (M1, P3). If there are no reradiating walls,

Equation 4.11-43


4.11D. View Factors and Gray Bodies

A general and more practical case, which is the same as for Eq. (4.11-40) but with the surfaces A1 and A2 being gray with emissivities ε1 and ε2, will be considered. Nonconducting reradiating walls are present as before. Since the two surfaces are now gray, there will be some reflection of radiation, which will decrease the net radiant exchange between the surfaces below that for black surfaces. The final equations for this case are

Equation 4.11-44


Equation 4.11-45


where is the new view factor for two gray surfaces A1 and A2 which cannot see themselves and are connected by reradiating walls. If no refractory walls are present, F12 is used in place of in Eq. (4.11-41). Again,

Equation 4.11-46


EXAMPLE 4.11-4. Radiation Between Infinite Parallel Gray Planes

Derive Eq. (4.11-22) by starting with the general equation for radiation between two gray bodies A1 and A2 which are infinite parallel planes having emissivities ε1 and ε2, respectively.

Solution: Since there are no reradiating walls, by Eq. (4.11-43) becomes F12. Also, since all the radiation from surface 1 is intercepted by surface 2, F12 = 1.0. Substituting into Eq. (4.11-45), noting that A1/A2 = 1.0,


Then using Eq. (4.11-44),


This is identical to Eq. (4.11-22).


EXAMPLE 4.11-5. Complex View Factor for Perpendicular Rectangles

Find the view factor F12 for the configuration shown in Fig. 4.11-9 of the rectangle with area A2 displaced from the common edge of rectangle A1 and perpendicular to A1. The temperature of A1 is T1 and that of A2 and A3 is T2.

Figure 4.11-9. Configuration for Example 4.11-5.


Solution: The area A3 is a fictitious area between areas A2 and A1. Designate the area A2 plus A3 as A(23). The view factor F1(23) for areas A1 and A(23) can be obtained from Fig. 4.11-8 for adjacent perpendicular rectangles. Also, F13 can be obtained from Fig. 4.11-8. The radiation interchange between A1 and A(23) is equal to that intercepted by A2 and by A3:

Equation 4.11-47


Hence,

Equation 4.11-48


Solving for F12,

Equation 4.11-49



Methods similar to those used in this example can be employed to find the shape factors for a general orientation of two rectangles in perpendicular planes or parallel rectangles (C3, H1, K1).

EXAMPLE 4.11-6. Radiation to a Small Package

A small, cold package having an area A1 and emissivity ε1 is at temperature T1. It is placed in a warm room with the walls at T2 and an emissivity ε2. Using Eq. (4.11-45), derive the view factor for this and the equation for the radiation heat transfer.

Solution: For the small surface A1 completely enclosed by the enclosure A2, = F12 by Eq. (4.11-43), since there are no reradiating (refractory) walls. Also, F12 = 1.0, since all the radiation from A1 is intercepted by the enclosure A2 because A1 does not have any concave surfaces and cannot "see" itself. Since A2 is very large compared to A1, A1/A2 = 0. Substituting into Eq. (4.11-45),


Substituting into Eq. (4.11-44),


This is the same as Eq. (4.10-6) derived previously.


For solving complicated radiation problems involving more than four or five heat-transfer surfaces, matrix methods have been developed and are discussed in detail elsewhere (H1, K1).

4.11E. Radiation in Absorbing Gases

1. Introduction to absorbing gases in radiation

As discussed in this section, solids and liquids emit radiation over a continuous spectrum. However, most gases that are monatomic or diatomic, such as He, Ar, H2, O2, and N2, are virtually transparent to thermal radiation; that is, they emit practically no radiation and do not absorb radiation. Gases with a dipole moment and higher polyatomic gases emit significant amounts of radiation and also absorb radiant energy within the same bands in which they emit radiation. These gases include CO2, H2O, CO, SO2, NH3, and organic vapors.

For a particular gas, the width of the absorption or emission bands depends on the pressure and also the temperature. If an absorbing gas is heated, it radiates energy to the cooler surroundings. The net radiation heat-transfer rate between surfaces is decreased in these cases because the gas absorbs some of the radiant energy being transported between the surfaces.

2. Absorption of radiation by a gas

The absorption of radiation in a gas layer can be described analytically, since the absorption by a given gas depends on the number of molecules in the path of radiation. Increasing the partial pressure of the absorbing gas or the path length increases the amount of absorption. We define IλO as the intensity of radiation at a particular wavelength before it enters the gas and IλL as the intensity at the same wavelength after having traveled a distance of L in the gas. If the beam impinges on a gas layer of thickness dL, the decrease in intensity, dIλ, is proportional to Iλ and dL:

Equation 4.11-50


where Iλ is in W/m2. Integrating,

Equation 4.11-51


The constant αλ depends on the particular gas, its partial pressure, and the wavelength of radiation. This equation is called Beer's law. Gases frequently absorb only in narrow-wavelength bands.

3. Characteristic mean beam length of absorbing gas

The calculation methods for gas radiation are quite complicated. For the purpose of engineering calculations, Hottel (M1) has presented approximate methods for calculating radiation and absorption when gases such as CO2 and water vapor are present. Thick layers of a gas absorb more energy than do thin layers. Hence, in addition to specifying the pressure and temperature of a gas, we must specify a characteristic length (mean beam length) of a gas mass to determine the emissivity and absorptivity of a gas. The mean beam length L depends on the specific geometry.

For a black differential receiving surface area dA located in the center of the base of a hemisphere of radius L containing a radiating gas, the mean beam length is L. The mean beam length has been evaluated for various geometries, as presented in Table 4.11-1. For other shapes, L can be approximated by

Equation 4.11-52


Table 4.11-1. Mean Beam Length for Gas Radiation to Entire Enclosure Surface (M1, R2, P3)
Geometry of EnclosureMean Beam Length, L
Sphere, diameter D0.65D
Infinite cylinder, diameter D0.95D
Cylinder, length = diameter D0.60D
Infinite parallel plates, separation distance D1.8D
Hemisphere, radiation to element in base, radius RR
Cube, radiation to any face, side D0.60D
Volume surrounding bank of long tubes with centers on equilateral triangle, clearance = tube diameter D2.8D

where V is volume of the gas in m3, A is surface area of the enclosure in m2, and L is in m.

4. Emissivity, absorptivity, and radiation of a gas

Gas emissivities have been correlated and Fig. 4.11-10 gives the gas emissivity εG of CO2 at a total pressure of the system of 1.0 atm abs. The pG is the partial pressure of CO2 in atm and L is the mean beam length in m. The emissivity εG is defined as the ratio of the rate of energy transfer from the hemispherical body of gas to a surface element at the midpoint divided by the rate of energy transfer from a black hemisphere surface of radius L and temperature TG to the same element.

Figure 4.11-10. Total emissivity of the gas carbon dioxide at a total pressure of 1.0 atm. (From W. H. McAdams, Heat Transmission, 3rd ed. New York: McGraw-Hill Book Company, 1954. With permission.)


The rate of radiation emitted from the gas is in W/m2 of receiving surface element, where εG is evaluated at TG. If the surface element at the midpoint at T1 is radiating heat back to the gas, the absorption rate of the gas will be , where αG is the absorptivity of the gas for black-body radiation from the surface at T1. The αG of CO2 is determined from Fig. 4.11-10 at T1, but instead of the parameter pGL, the parameter pG L(T1/TG) is used. The resulting value from the chart is then multiplied by (TG/T1)0.65 to give αG. The net rate of radiant transfer between a gas at TG and a black surface of finite area A1 at T1 is then

Equation 4.11-53


When the total pressure is not 1.0 atm, a correction chart is available to correct the emissivity of CO2. Charts are also available for water vapor (H1, K1, M1, P3). When both CO2 and H2O are present the total radiation is reduced somewhat, since each gas is somewhat opaque to radiation from the other gas. Charts for these interactions are also available (H1, K1, M1, P3).

EXAMPLE 4.11-7. Gas Radiation to a Furnace Enclosure

A furnace is in the form of a cube 0.3 m on a side inside, and these interior walls can be approximated as black surfaces. The gas inside at 1.0 atm total pressure and 1100 K contains 10 mol % CO2 and the rest is O2 and N2. The small amount of water vapor present will be neglected. The walls of the furnace are maintained at 600 K by external cooling. Calculate the total heat transfer to the walls neglecting heat transfer by convection.

Solution: From Table 4.11-1, the mean beam length for radiation to a cube face is L = 0.60D = 0.60(0.30) = 0.180 m. The partial pressure of CO2 is pG = 0.10(100) = 0.10 atm. Then pGL = 0.10(0.180) = 0.0180 atm m. From Fig. 4.11-10, εG = 0.064 at TG = 1100 K.

To obtain αG, we evaluate αG at T1 = 600 K and pGL(T1/TG) = (0.0180) (600/1100) = 0.00982 atm · m. From Fig. 4.11-10, the uncorrected value of αG = 0.048. Multiplying this by the correction factor (TG/T1)0.65, the final correction value is


Substituting into Eq. (4.11-53),


For six sides, A = 6(0.3 × 0.3) = 0.540 m2. Then,



For the case where the walls of the enclosure are not black, some of the radiation striking the walls is reflected back to the other walls and into the gas. As an approximation when the emissivity of the walls is greater than 0.7, an effective emissivity ε' can be used:

Equation 4.11-54


where ε is the actual emissivity of the enclosure walls. Then Eq. (4.11-53) is modified to give the following (M1):

Equation 4.11-55


Other approximate methods are available for gases containing suspended luminous flames, clouds of nonblack particles, refractory walls and absorbing gases present, and so on (M1, P3).

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