278 RADIATIVE HEAT TRANSFER
10.8 Repeat Problem 10.6, but assume that the temperature is uniform at 2000 K. Also, there is no heat production,
meaning that the sphere cools down. How long will it take for the sphere to cool down to 500 K (the heat
capacity of the medium is ρc=1000 kJ/m3K and the conductivity is very large, i.e., the sphere is isothermal
at all times).
CHAPTER 10 279
10.9 A relatively cold sphere with a radius of Ro=1 m consists of a nonscattering gray medium that absorbs with
an absorption coefficient of κ=0.1 cm−1and has a refractive index n=2. At the center of the sphere is a
small black sphere with radius Ri=1 cm at a temperature of 1000 K. On the outside, the sphere is bounded
by vacuum. What is the total heat flux leaving the sphere? Explain what happens as κis increased from zero
to a large value.
is seen is so small and cos θ≃1, we can write
q(Ro)≃I(Ro, θ =0) ×1×δΩ≃Ib(Ri)e−κRo×πR2
i
=0.8889 ×4π10−4m2×22×5.670 ×10−8W
m2K41012K4e−0.1 cm−1100cm,
280 RADIATIVE HEAT TRANSFER
10.10 A laser beam is directed onto the atmosphere of a (hypothetical) planet. The planet’s atmosphere contains
0.01% by volume of an absorbing gas. The absorbing gas has a molecular weight of 20 and, at the laser
wavelength, an absorption coefficient κη=10−4cm−1/(g/m3). It is known that pressure and temperature
distribution of the atmosphere can be approximated by p=p0e−2z/Land T=T0e−z/L, where p0=0.75 atm,
T0=400 K are values at the planet surface z=0, and L=2 km is a characteristic length. What fraction of the
laser energy arrives at the planet’s surface?
=0.04570 e−z/Lg/m3.
0
=4.570 ×10−6cm−1−Le−z/L
∞
=0.9140.
Fraction on planet′s surface =e−τη=40%.
CHAPTER 10 281
282 RADIATIVE HEAT TRANSFER
L=15.076/0.15 cm−1=100 cm
CHAPTER 10 283
10.12 Repeat Problem 10.11 for a medium with refractive index n=2, bounded by vacuum (i.e., a slab with reflecting
surfaces).
Hint: (1) Part of the laser beam will be reflected when first hitting the slab, part will penetrate into the slab.
Part of this energy will be absorbed by the layer, part will hit the rear face, where again a fraction will be
reflected back into the slab, and the rest will emerge from the slab, etc. Similar multiple internal reflections
will take place with the emitted energy before emerging from the slab. (2) To calculate the slab–surroundings
reflectance, show that the value of the absorptive index is negligible.
I(0) =Q
e−(r/R)2
=6.3662 ×1010e−(r/R)2W
m2sr0.1764 =1.1230 ×1010e−(r/R)W/m2sr.
284 RADIATIVE HEAT TRANSFER
0
=Ib(1 −t)=n2σT4
π(1 −t),
where the subscripts ηwere omitted since the medium is gray, and the integration is easily carried out since
Ib=const. Of this, the fraction
Ib(1−t)(1+ρt)1−ρ
1−ρ2t2=IbAslab =Ib
1−(1.1111 ×0.2231)2
=0.708122×5.670×10−8×10004
W
6.3662 ×1010Tslab =22×5.670 ×104
πAslab
6.3662 ×1010 8
t=1.2757 ×10−6(1−0.8889t−0.1111t2)=1.2757 ×10−6
κη
CHAPTER 10 285
10.13 A thin column of gas of cross section δAand length Lcontains a uniform suspension of small particles that
absorb and scatter radiation. The scattering is according to the phase function (a)Φ = 1 (isotropic scattering),
(b)Φ = 1+A1cos Θ(linear anisotropic scattering, A1is a constant), and (c)Φ = 3
4(1 +cos2Θ) (Rayleigh
scattering), where Θis the angle between incoming and scattered directions. A laser beam hits the column
normal to δA. What is the transmitted fraction of the laser power? What fraction of the laser flux goes through
an infinite plane at Lnormal to the gas column? What fraction goes back through a plane at 0? What happens
to the rest?
288 RADIATIVE HEAT TRANSFER
10.15 Show that, by setting βη=0 and Iw=J/π, the radiosity integral equation (5.25) can be recovered from
equation (10.69) for a nonparticipating medium surrounded by diffusely reflecting walls.
Hint: break up the heat flux in equation (10.69) into two parts, incoming radiation Hand exiting radiation
J. For the latter assume rto be an infinitesimal distance above the surface and evaluate the integral in
equation (10.69).