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CHAPTER 14
14.1 The gap between two parallel black plates at T1and T2, respectively, is filled with a particle-laden gas.
Radiative equilibrium prevails, the particle loading is a fixed volume fraction, with particles manufactured
from two different materials (one a specular reflector, the other a diffuse reflector, both having the same ǫ).
Sketch the nondimensional heat flux Ψ = q/σ(T4
1−T4
2) vs. particle size (but keeping volume-fraction constant).
Solution
If we assume the particles to be large, we have from equations (12.29) and (12.66)
4a,and τL=βL=2fvL
4a∝1
a.
For small particles Rayleigh scattering applies, and from equations (12.29) and (12.46), or from equation (12.53),
very small particles. Since specu-
lar spheres have a forward scattering
peak (see Fig. 12-13) more energy is
transmitted from plate to plate than for
diffuse spheres, which have a strong
backward scattering peak. Since acan-
not increase beyond a single large par-
ticle, the heat flux cannot increase be-
yond a certain point.
0
q/
σ
(T
1
4
−
T
4
)
2
amax
diffuse
specular
a
360
CHAPTER 14 361
14.2 Consider radiative equilibrium in a one-dimensional, gray, nonscattering, plane-parallel medium bounded
by isothermal black plates at temperatures T1and T2. To make a simple closed-form solution possible to
determine the heat flux between the plates, it has been proposed to replace the radiative-equilibrium slab by a
constant temperature slab, with its temperature evaluated at T4
av =1
2(T4
1+T4
2).Under what optical conditions
is this a good idea, if ever?
0
or, following Example 14.1:
0 1 2 3 4 5
τL
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Ψ
Exact radiative equilibrium
Constant temperature slab at Tav
362 RADIATIVE HEAT TRANSFER
14.3 Consider a space enclosed by infinite, diffuse-gray parallel plates filled with a gray nonscattering medium.
The surfaces are isothermal (both at Tw), and there is uniform and constant heat generation within the
medium per unit volume, ˙
Q′′′. Conduction and convection are negligible so that ∇·q=˙
Q′′′. Set up the
integral equations describing temperature and heat flux distribution in the enclosure, i.e., show that
Φ(τ)=σT4−Jw
˙
Q′′′/4κ=1+1
2ZτL
0
Φ(τ′)E1(|τ−τ′|)dτ′,
Ψ(τ)=q
˙
Q′′′/κ =τ−τL
2.
2.
Once Φhas been determined the radiosity is related to wall temperature through equation (14.48), or
Q′′′
τL
364 RADIATIVE HEAT TRANSFER
14.4 A semi-infinite, gray, isotropically scattering medium, originally at a temperature of 0 K, is subjected to
collimated irradiation with a constant heat flux q0normal to its nonreflecting surface. Set up the integral
relationships governing steady-state temperature and radiative heat flux within the medium, assuming
radiative equilibrium. Hint: Collimated irradiation with heat flux q0has the radiative intensity
I0(θ, ψ)=
q0
2πsin θ δθ,0≤θ < δθ, 0≤ψ≤2π,
0,elsewhere.
0
366 RADIATIVE HEAT TRANSFER
14.5 Two large, isothermal gray plates at T1=2000 K, T2=1000 K with ǫ=ǫ1=ǫ2=0.5 are separated by a gap
of width L=1 m filled with purely (isotropically) scattering particles. If the heat flux between the plates has
been measured as 223.3 kW/m2, what is the medium’s scattering coefficient?
CHAPTER 14 367
14.6 Consider two parallel, black, isothermal plates spaced 1 m apart with T1=2000 K and T2=1000 K. The
medium between the plates is gray and at radiative equilibrium with a nonconstant absorption coefficient of
κ=κ0+κ′
1z;κ0=10−2cm−1, κ′
1=2×10−4cm−2.
The medium does not scatter.
(a) What is the heat flux between the plates?
(b) What is the temperature at the medium’s center (z=1
2L)?
368 RADIATIVE HEAT TRANSFER
14.7 An infinite, black, isothermal plate bounds a semi-infinite space filled with black spheres. At any given
distance, z, away from the plate the particle number density is identical, namely NT=6.3662 ×108m−3.
However, the radius of the suspended spheres diminishes monotonically away from the surface as
a=a0e−z/L;a0=10−4m,L=1 m.
(a) Determine the absorption coefficient as a function of z(you may make the large-particle assumption).
(b) Determine the optical coordinate as a function of z. What is the total optical thickness of the semi-infinite
space?
(c) Assuming that radiative equilibrium prevails, determine the heat loss from the plate.
CHAPTER 14 369
14.8 The radiative heat transfer between two isothermal, black plates at temperatures T1and T2and separated
by a nonparticipating gas is to be minimized. Enough of a black material is available to place a 1 mm thick
radiation shield between the plates. Alternatively, the same amount of material could be used in the form of
small spheres of 0.1 mm radius to be suspended between the plates. Which possibility results in lower heat
flux, assuming conduction and convection to be negligible?
370 RADIATIVE HEAT TRANSFER
14.9 Two infinite, isothermal plates at temperatures T1and T2are separated by a cold, gray medium of optical
thickness τL=κL(no scattering).
(a) Calculate the radiative heat flux at the bottom plate, the top plate, and the net radiative energy going
into the gray medium, assuming that both plates are black.
(b) Repeat (a), but assume that both plates have the same temperature T, and that both plates are gray with
equal emittance ǫ(diffuse emission and reflection).
or
q(0) =σT4
1−2σT4
2E3(τL),
q(L)=−hσT4
2−2σT4
1E3(τL)i.
q(0) =−q(L)=Jw[1−2E3(τL)].
The radiosity is determined from equation (14.48) as
q(0) =−q(L)=1
Q′′ =[1−2E3(τL)]σT4
w
CHAPTER 14 371
14.10 A semi-infinite, absorbing-emitting, nonscattering medium at uniform temperature is in contact with a gray-
diffuse wall at Twand with emittance ǫw.
(a) The medium is gray, and has a constant absorption coefficient. Determine the net radiative heat flux at
the wall.
(b) Let the medium be nongray with nonconstant absorption coefficient κλ, and the wall be nongray and
nondiffuse with spectral, directional emittance ǫ′
λ. How would this affect the wall heat flux?
372 RADIATIVE HEAT TRANSFER
14.11 A 1 m thick, isothermal slab bounded by two cold black plates has a temperature of 3000 K, and a nongray
absorption coefficient that can be approximated by
κλ=(0, λ < 2µm,
0.20 cm−1, λ > 2µm.
Calculate the total heat loss by radiation from the slab (in W/cm2).
=0.20×100
The total heat loss from top and bottom is
2|q(0)|=2×5.670 ×10−8×30004W
m2[1 −0.73778][1 −2×0] =2.41 ×106W/m2
CHAPTER 14 373
14.12 Consider (a) two parallel plates, (b) two concentric spheres, (c) two concentric cylinders. The bottom/inner
surface needs to dissipate a heat flux of 30 W/cm2and has a gray-diffuse emittance ǫ1=0.5. The top/outer
surface is at T2=1000 K with ǫ2=0.8. The medium in between the surfaces is gray and nonscattering
(κ=0.1 cm−1), has a thickness of L=5 cm and is at radiative equilibrium. Determine the temperature at the
bottom/inner surface necessary to dissipate the supplied heat for the three different cases (the radii of the
inner cylinder and sphere are R1=5 cm). Discuss the results.
T1=1972 K.
(b) Concentric Spheres: In the absence of heat generation we have, from equation (14.75) and Table 14.2 for
τL=2(τ2−τ1)=1.0 and R1/R2=1
5.670 ×10−12 ×0.4594 =1.252 ×1013 K4
T1=1881 K.
5.670 ×10−12 ×0.4339 =1.319 ×1013 K4,
T1=1906 K.
374 RADIATIVE HEAT TRANSFER
14.13 Consider a very hot sphere of a nongray gas of radius R=1 m in 0 K surroundings that have been evacuated.
The gas has a single absorption-emission band in the infrared, with an absorption coefficient
κη=(0, η < 3000 cm−1=η0,
κ0e−(η−η0)/ω, η > 3000 cm−1,
where κ0=1 cm−1, ω =200 cm−1. During cool-down the sphere is always isothermal, and remains of constant
size (i.e., constant density ρ=1000 g/m3). The heat capacity of the gas is cp=1 kJ/kg K. Determine the time
required to cool the gas from Ti=6000 K to Te=1000 K. Sketch qualitatively the behavior of Ψ = q/σT4vs. T.
Hint: To make an analytical solution possible, you may make the following assumptions:
(a) Ein(x)=R1
0(1−e−xξ)dξ/ξ =E1(x)+ln x+γE≃ln x+γE(for sufficiently large x; see also Appendix E).
(b) Wien’s distribution may be used.
Solution
and τ0=κ0R=1 cm−1×1 m =100. Integrating equation (14.13-A) over the spectrum (assuming the band to
be narrow enough to extract Ibηfrom the integral) leads to
I(θ)=Z∞
0
Iηdη=Ibη0Z∞
η01−e−2τηcos θdη.
Changing the integration variable to x=τη/τ0we obtain
dx =−1
q=2πZπ/2
0
I(θ) cos θsin θdθ=2πZ1
0
ωIbη0Ein(2τ0µ)µdµ.
Since τ0µ≫1 wherever µis appreciable, we may replace this by
0
=π×200 cm−1[ln 200 +0.57721 −0.5]Ibη0=3378 cm−1Ibη0.
From a heat balance, assuming radiation to be the sole mechanism for the sphere to lose heat:
−ρcpVdT
dt =Aq
CHAPTER 14 375
or
Ebη0=Ebλ0/η2
0=C1η3
0e−C2η0/T,
so that
×e−1.4388 cmK×3000 cm−1/T=−3.258 ×103K
4.316 =(1.4135 −1) e4.316/4.316 =7.175,
0.7193 =1.1203 −e0.7193
0.7193 =−1.734,
and
t=7.175 +1.734
0.7549 =11.8 s.
376 RADIATIVE HEAT TRANSFER
14.14 It is proposed to construct a high-temperature heating element by guiding hot combustion gases through a
silicon carbide tube. The outside of the SiC tube then radiates heat toward the load. Such devices are known
as “radiant tubes.” For the design of such a radiant tube you may make the following assumptions:
(i) the combustion gas inside the radiant tube is essentially gray and isothermal with κ=0.2 cm−1and
Tgas =2000 K;
(ii) the silicon carbide tube wall is essentially isothermal and of negligible thickness, with a gray-diffuse
emittance of 0.8 on both sides,
(iii) the long tube is contained in a large furnace with a background temperature of 1000 K.
Determine the necessary tube diameter to achieve a radiant heating rate of 100 kW per m length of tube.
