8.11 Equation of transfer or radiative transfer equation (RTE) 283
Converting I to q with R replacing x, we have
σ TW4 |
e |
(−κ R) |
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σ Tg4 |
(1− e |
(−κ R) |
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π |
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π |
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q(R) = σ Tg4 (1− e(−κ R) ) |
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For optically thin gases, κ R 1 |
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q(R) = σTg4κ R |
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q(R) can be written as εgσ Tg4 where εg can be defined as the gas emissivity. |
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From the above definition, it is clear that εg = κ R.
So, a gas emissivity can be defined as the product of two quantities. The first one k is the thermal part, and the second one R is the geometric part. R is, in fact, the mean beam length Lm for the gas. The key achievement thus far has been our ability to separate out the geometric part from the thermal part. That in gas radiation, the gas emissivity εg is a function of the geometry is unsettling for us. However, the thin gas approximation and the accompanying simplifications lets us introduce emissivity εg and proceed with an engineering treatment of gas radiation which is often adequate if our problem of interest is a rocket nozzle or a combustion chamber.
For a few commonly encountered geometries, the mean beam lengths are given in Table 8.5. The product of k and Lm will then give εg, which gives us a handle in solving the problem of gas radiation.
The foregoing framework presented basic ideas in gas radiation with a “clever” strategy to obtain gas emissivity. Please note that in this development, both εg and αg
Table 8.5 Mean beam lengths Lm for different gas geometries.
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Sl. No. |
Geometry |
length |
Le |
1. |
Hemisphere radiating to an element at the |
Radius R |
R |
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center of the base |
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2. |
Sphere (radiation to surface) |
Diameter D |
0.65D |
3. |
Infinite circular cylinder (radiation to the |
Diameter D |
0.95D |
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curved surface) |
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Circular cylinder of equal height and diam- |
Diameter D |
0.60D |
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eter (radiation to the entire surface) |
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5. |
Infinite parallel planes (radiation to planes) |
Spacing between |
1.80L |
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planes L |
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6. |
Cube (radiation to any surface) |
Side L |
0.66L |
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Arbitrary shape of volume V (radiation to |
Volume to area ratio |
3.6V/A |
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the surface of area A) |
V/A |
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Adapted from Howell et al. (2011).
284 CHAPTER 8 Thermal radiation
are constructs we introduced to make the problem more tractable. The basic properties in gas radiation are the monochromatic emission coefficient, ελ, and monochromatic absorption coefficients αλ, both of which are not dimensionless and have units of m−1.
8.11.2 Enclosure analysis in the presence of an absorbing or emitting gas
We now extend the theory of evacuated enclosures to problems involving radiation in optically thin gases. Consider two walls of an enclosure with areas A1 and A2 (see Fig. 8.31). Consider two elemental areas in them. Let the distance between them be R. Let the unit vectors be n1 and n2, and the angles subtended by them be θ1 and θ2, respectively.
Let all the surfaces in the enclosure be gray and diffuse. Now let us consider an optically thin gas filling the enclosure.
The radiation leaving dA1 that falls on dA2 is given by
J1dA1dA2 cos(θ1 )cos(θ2 ) |
e(−κ R) |
(8.176) |
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π R2 |
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e(−κ R) ≈ 1−κ R |
(8.177) |
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using Eq. (8.177), Eq. (8.176) can be expressed as
J1dA1dA2 cos(θ1 )cos(θ2 ) |
−κ |
J1dA1dA2 cos(θ1 )cos(θ2 ) |
(8.178) |
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π R2 |
π R |
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FIGURE 8.31
Enclosure analysis in the presence of an absorbing and emitting gas.
8.11 Equation of transfer or radiative transfer equation (RTE) 285
Assuming a uniform radiosity of J1, the irradiation of A2 due to radiation from A1 is
J1κ ∫A |
∫A |
dA1dA2 cos(θ1 )cos(θ2 ) |
− J1κ ∫A |
∫A |
dA1dA2 cos(θ1 )cos(θ2 ) |
(8.179) |
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π R |
2 |
π R |
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The first term within the integral is A1F12, which is also A2F21 (Please recall the definition of view factor Fij). In the second term the integrand should also be something similar, and let us say it can be represented as A1L12 = A2 L21.
Let us say it is A1L12, where L12 is the mean beam length.
Therefore, the irradiation on A2 due to radiation coming from A1 is given by
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J A F −κ J |
A L = A G+ |
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12 |
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Using the reciprocal rule the above equation can be written |
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A2G2+ |
= J1 A2 F21 −κ J1 A2 L21 |
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(8.181) |
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G+ |
= J F |
1− κ L21 |
= |
J F |
1− κ L12 |
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By induction, |
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2 12 |
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F21 |
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Substituting for |
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of the gas, we have |
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G1+ |
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(8.184) |
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L12 and L21 represent the mean beam length
Lij also follows the reciprocal rule, as in Eq. (8.185)
τ12 |
= 1 − |
κ L21 |
= 1 − |
κ L12 |
(8.185) |
F |
F |
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21 |
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The irradiation falling on the ith surface in the enclosure can then be written as
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N |
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Gi = εgσTg4 + ∑Fij J jτij |
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(8.186) |
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j=1 |
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κ Lij |
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where τ ij = 1 |
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Fij |
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Fji |
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Please note that the irradiation Gi on the ith surface of the enclosure includes the gas radiation (first term) and the radiation falling from the other surfaces that is attenuated by the gas as seen by the presence of the term τij. The radiosity of the ith surface is given by
Ji = εiσTi4 + (1 − εi )Gi |
(8.188) |
286CHAPTER 8 Thermal radiation
Finally, the net radiation from the ith surface is given by
qi = Ji − Gi |
(8.189) |
The simplification afforded by the above development is too profound to be missed. The above development clearly shows that with a workaround, the original theory of radiosity for evacuated enclosures can be applied for an optically thin, gray
case. One can easily see that for k = 0, that is, the gas is transparent εg = 0 and τij=1 and |
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N |
Eq. (8.190) reduces to Gi = ∑ Fij J j , which is nothing but Eq. (8.120) that gave an |
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J = 1 |
expression for irradiation, G |
i |
on the ith surface in a N surface evacuated enclosure. |
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8.11.3 Calculation of emissivities and absorptivities for a mixture of gases
Mixtures of gases are encountered in furnaces and in internal combustion chambers. Burning hydrocarbons results in water vapor and carbon dioxide. Both are radiatively participative and they interact such that they have overlapping spectral bands.
A detailed method of solving would involve solving the RTE at millions of lines in the EM spectrum. This is very cumbersome.
For a quick engineering solution, we need a simplified approach. The gas emissivity εg is a function of the following five quantities. Lm = mean beam length
Pg = partial pressure of the gas P = total pressure of the gas
C = concentration of other gases Tg = temperature of the gas
This can be written as εg = f(Lm , Pg , P, C, Tg ). It is intuitive to see that emissivity is a function of temperature. The concentration of other gases needs to be included as the gas emissivity can depend on the interaction of gases and overlap of absorption bands. Hottel prepared charts for a total mixture pressure of 1 atmosphere, and a correction factor is to be applied for other pressures (Hottel and Hoyt, 1954).
The total emissivity of CO2, εc, is a function of the product of the partial pressure of CO2 and mean beam length and its gas temperature.
εC = f1 (PC Lm ,Tg )
The total emissivity of water vapor, water vapor, mean beam length, and gas
εw, is a function of the partial pressure of temperature.
εw = f2 (Pw Lm ,Tg )
A mixture of carbon dioxide and water vapor is a nongray gas. In view of this, εg =/ αg.
Leckner correlations:
Leckner gave correlations for calculation of the total emissivity and absorptivity of a mixture of CO2 and H2O (Leckner, 1972).
8.11 Equation of transfer or radiative transfer equation (RTE) 287
The emissivity for zero partial pressure condition is given by
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M N |
T |
j |
p L |
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ε0 |
( pa L, p = 1bar,Tg ) = exp ∑∑cji |
g |
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log |
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i=0 j=0 |
0 |
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a |
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where T0 = 1000 K and (paL)0 = 1 bar cm.
The emissivity for different pressure conditions is given by
ε ( pa L, p,Tg ) |
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(a − 1)(1 |
− P ) |
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( p |
L) |
m |
2 |
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= 1 |
− |
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E |
exp |
−c log10 |
a |
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ε0 ( pa L,1bar,Tg ) |
a + b − 1 + PE |
pa L |
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(8.190)
(8.191)
In Eq. (8.191), PE is an effective pressure and a, b, c, and (paL)m are correlation parameters.
As aforementioned, a correction factor must be added to take care of the overlap of both carbon dioxide and water vapor. The constants in the above equation are given in Tables 8.6 and 8.7. The correction factor for emissivity is given by
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ζ |
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1 |
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( pH2O + pCO2 ) |
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2.76 |
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ε = |
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− 0.0089ζ |
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log10 |
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Finally, to calculate the emissivity and absorptivity of the gas mixture containing H2O and CO2, the following equations can be used.
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εi ( pi L, p,Tg ) = ε0,i ( pi L,1bar,Tg ) |
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( pi L, p,Tg ), i = CO2 |
or H2O |
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Table 8.6 Correlation constants for the determination of the total emissivity for water vapor.
M,N |
2,2 |
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C00 . . . CN0 |
−2.2118 |
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−1.1987 |
0.035596 |
. . . |
0.85667 |
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0.93048 |
−0.4391 |
C0M. . . CNM |
−0.10838 |
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t )/p |
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0.045915 |
P |
(p + 2.56p |
/ |
o |
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a |
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(paL)m/(paL)0 |
13.2 t2 |
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A |
2.144 |
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1.888–2.053log10t, |
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1.1/t1.4 |
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0.5 |
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T0 = 1000K, p0 = 1 bar, t = T/T0, (paL)0 = 1 bar cm
288 CHAPTER 8 Thermal radiation
Table 8.7 Correlation constants for the determination of the total emissivity for carbon dioxide.
M,N |
2,3 |
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C00. . .CN0 |
−3.9893 |
2.7669 |
−2.1081 |
0.39163 |
. . . |
1.271 |
−1.1090 |
1.0195 |
−0.21897 |
C0M. . . CNM |
−0.23678 |
−0.19731 |
−0.19544 |
0.044644 |
PE |
(p + 0.28pa)/po |
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paL)0 |
0.054/t2, |
t < 0.75 |
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0.225 t2, |
t > 0.75 |
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A1 + 0.1 / t1.45
B0.23
C1.47
T0 = 1000 K, p0 = 1 bar, t = T/T0, (paL)0 = 1 bar cm
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αi ( pi L, p,Tg ,Ts ) = |
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Ts |
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(8.197) |
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Example 8.14: A furnace is in the shape of a sphere and is 0.6 m in diameter. It is filled with a gas mixture at a temperature of 1500 K and a pressure of 1.8 atmosphere. The gas mixture consists of H2O at a partial pressure of 0.8 atm and CO2 with a partial pressure of 0.4 atm. The cavity wall is black and is maintained at 600 K. Determine the heat transfer to the wall.
Solution:
Given that:
Diameter of the spherical cavity = 0.6 m Pressure of the gas mixture = 1.8 atm Temperature of the gas mixture = 1500 K Partial pressure of carbon dioxide = 0.4 atm Partial pressure of water vapour = 0.8 atm Wall temperature of the cavity = 600 K
Lm = Mean beam length = 0.65D = 0.39 m |
(8.198) |
Emissivity calculation of CO2
pa L = 0.4 × 0.39 = 0.156 atm m |
(8.199) |
8.11 Equation of transfer or radiative transfer equation (RTE) 289
From Table 8.7,
PE = 1.912, t = 1.5,a = 1.056 b = 0.23,c = 1.47 and ( pa L)m = 0.506 bar cm
Substituting the above values in Eq. (8.190), we get ε0 = 0.11. From Eq. (8.191), we get εCO2 = 0.111 .
Emissivity calculation of H2O:
pa L = 0.8 × 0.39 = 0.312 atm m
From Table 8.6,
PE = 3.47, t = 1.5,a = 1.53b = 0.62,c = 0.5 and ( pa L)m = 29.7 bar cm
Substituting the above values in Eq, 8.190, we get ε0 = 0.195. From Eq. (8.191), we get εH2O
ζ = |
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Total emissivity of CO2 and H20 = 0.326
In a similar fashion, absorptivity can be calculated using Eq. (8.196).
Absorptivity of CO2 = 0.131
Absorptivity of H2O = 0.297
From Eq. (8.197),
Total absorptivity of CO2 and H20 = 0.131 + 0.297 − 0.035 = 0.393 Heat transfer to the wall having a temperature of 600 K,
q1 = J1 − G1
Q1 = (J1 − G1 )4π R2
ε1 = 1
J1 = ε1σ T14 + (1 − ε1 )G1 = σ T14
G1 = εgσ Tg4 + τ g (σ T14 )
(8.200)
(8.201)
(8.202)
(8.203)
(8.204)
290 CHAPTER 8 Thermal radiation
αg +τ g + ρg = 1; ρg = 0 αg +τ g = 1; τ g = 1−αg
G1 = εgσTg4 + (1 − αg )(σT14 )
q1 = J1 − G1 = σT14 − εgσTg4 − σT14 + αgσT14
q1 = αgσT14 − εgσTg4
For the problem under consideration,
q1 = 0.393× 5.67 ×10−8 × 6004 −0.326 × 5.67 ×10−8 ×15004
q1 = −90.69 kWm2
Q1 = q1 4π R2 = −102.57 kW
Q1 = −102.57 kW
(8.205)
(8.206)
(8.207)
(8.208)
(8.209)
(8.210)
(8.211)
(8.212)
Problems
8.1Determine the wavelength that corresponds to the maximum emission from the following surfaces.
a.Tungsten filament at 2900 K.
b.Heated metal at 1100 K.
c.Human skin at 306 K.
d.A metal surface cooled cryogenically and maintained at 90 K.
8.2The spectral, hemispherical emissivity of tungsten may be approximated by the distribution shown in Fig. 8.32. Consider a cylindrical tungsten filament that is of diameter D = 0.8 mm and length L = 20 mm. The filament is enclosed in an evacuated bulb and is heated by an electrical current to a steady-state temperature of 2900 K. What is the total, hemispherical emissivity when the filament temperature is 2900 K?
8.3An opaque surface has a hemispherical spectral reflectivity, as shown in Fig. 8.33A. It is subjected to spectral irradiation, as shown in Fig. 8.33B.
a.Sketch the spectral hemispherical absorptivity distribution.
b.Determine the total irradiation on the surface.
c.Determine the radiant flux that is absorbed by the surface.
d.Determine the total hemispherical absorptivity of the surface.
8.4An opaque surface 2 m × 2 m is maintained at 400 K and is simul-
taneously exposed |
to solar irradiation with G = |
1200 W/m2. The sur- |
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face |
is diffuse and |
its spectral absorptivity is αλ |
= 0, 0.8, 0, and 0.9 |
for |
0 m ≤ λ ≤ 0.5 m,0.5 m ≤ λ ≤ 1 m,0 m ≤ λ ≤ 0.5 m,λ ≥ 2 m, respectively. |
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8.11 Equation of transfer or radiative transfer equation (RTE) 291
FIGURE 8.32
Spectral emissivity.
FIGURE 8.33
Variation of (A) hemispherical spectral reflectivity and (B) spectral irradiance with wavelength.
Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface.
8.5Consider a two-dimensional evacuated rectangular enclosure shown in Fig. 8.34. Determine all the view factors.
8.6Consider a two-dimensional evacuated regular hexagonal duct of side a, as shown in Fig. 8.35. Determine all the view factors.
8.7Consider two very large parallel plates with diffuse gray surfaces as shown in Fig. 8.36. Details are given in the figure. Determine the net radiation heat transfer between the surfaces.
8.8An oven is in the shape of a triangle, which is infinitely deep in the direction perpendicular to the paper. The heated surface is maintained at 1200 K, and
292 CHAPTER 8 Thermal radiation
FIGURE 8.34
Rectangular enclosure.
FIGURE 8.35
Two-dimensional regular hexagonal duct of side a.
FIGURE 8.36
Infinite parallel plates considered in exercise problem 8.35.
one surface is insulated. The third surface is at 490 K. The enclosure is evacuated and is 1 m wide on all the three sides. The emissivity of two sides of the enclosure is 0.8 (surface at 1200 K and the insulated one), while that of the third surface is 0.4. Determine the rate at which heat energy must be supplied to the heated side per unit length at 1200 K. What is the temperature of the insulated surface?
8.9The space between two very long concentric tubes with diameter 25.4 mm and 50.8 mm is evacuated. The inner surface of the outer tube and the outer surface of the inner tube are diffuse and gray and have emissivities of 0.03 and 0.07,