CHAPTER 10
10.1 A semi-infinite medium 0 ≤z<∞consists of a gray, absorbing-emitting gas that does not scatter, bounded
by vacuum at the interface z=0. The gas is isothermal at 1000 K, and the absorption coefficient is κ=1 m−1.
The interface is nonreflecting; conduction and convection may be neglected.
(a) What is the local heat generation that is necessary to keep the gas at 1000 K?
(b) What is the intensity distribution at the interface, that is, I(z=0, θ, ψ), for all θand ψ?
(c) What is the total heat flux leaving the semi-infinite medium?
2.
Since the medium is gray, this expression is valid on a spectral and on a total basis. For θ > π/2 the radiation
is coming from a semi-infinite space. For this case it is more convenient to integrate away from the point of
interest (along τ′′
sin the sketch) rather than toward it: Changing the integration variable in equation (10.29)
τs→∞ Zτs
0
0
2< θ ≤π.
Thus, a large isothermal body radiates like a blackbody at that temperature. Using azimuthal symmetry we
268 RADIATIVE HEAT TRANSFER
where the last integral cannot be solved explicitly, but has been tabulated extensively under the name
“exponential integral E2(x)” (cf. Appendix E). Similarly,
q=2πZπ
0
I(τ, θ) cos θsin θdθ
=2πIb“Zπ/2
cos θsin θdθ#
˙
Q′′′ =2πIbκE2(τ)=2σT4κE2(τ)
=2×5.670 ×10−8×10004W/m2×1 m−1E2(κz),
˙
Q′′′ =1.1340 ×105E2(κz) W/m3.
(b) It is clear from the above discussion, or by setting z=0 in the intensity relation:
0,0≤θ < π
CHAPTER 10 269
10.2 Reconsider the semi-infinite medium of Problem 10.1 for a temperature distribution of T=T0e−z/L,T0=
1000 K,L=1 m. What are exiting intensity and heat flux for this case? Discuss how the answer would change
if κwould vary between 0 and ∞.
270 RADIATIVE HEAT TRANSFER
10.3 Repeat Problem 10.1 for a medium of thickness L=1 m. Discuss how the answer would change if κwould
vary between 0 and ∞.
2.
Since the medium is gray, this expression is valid on a spectral and a total basis. For θ > π/2 the intensity is
readily found with τs=−(τL−τ)/cos θ(since cos θ < 0) as
I(τ, θ)=Ib1−e+(τL
−τ)/cos θ,π
0
0
=2πIb[2−E2(τ)−E2(τL−τ)],
where the two integrals cannot be solved explicitly, but have been tabulated extensively under the name
“exponential integral E2(x)” (cf. Appendix E). Similarly,
q=2πZπ
0
I(τ, θ) cos θsin θdθ
˙
Q′′′ =dq
dz =κ(4πIb−G)=κ4πIb1−1+1
2E2(τ)+1
2E2(τL−τ)
=κ2πIb[E2(τ)+E2(τL−τ)]=2κσT4[E2(τ)+E2(τL−τ)]
=2×1 m−1×5.670×10−8×10004W
m2[E2(τ)+E2(τL−τ)],
q(0) =−4.426 ×104W
m2.
272 RADIATIVE HEAT TRANSFER
10.4 A semi-infinite, gray, nonscattering medium (n=2, κ =1 m−1) is irradiated by the sun normal to its surface
at a rate of qsun =1000 W/m2. Neglecting emission from the relatively cold medium, determine the local heat
generation rate due to absorption of solar energy.
Hint: the solar radiation may be thought of as being due to a radiative intensity which has a large value Io
over a very small cone of solid angles δΩ, and is zero elsewhere, i.e.,
I(ˆs)=Ioover δΩalong ˆn,
0 elsewhere,
and
qsun =Z4π
I(ˆs)ˆn ·ˆs dΩ = IoδΩ.
CHAPTER 10 273
10.5 A 1 m thick slab of an absorbing/emitting gas has an approximately linear
temperature distribution as shown in the sketch. On both sides the medium is
bounded by vacuum with nonreflecting boundaries.
(a) If the medium has a constant and gray absorption coefficient of κ=1 m−1,
what is the intensity (as function of direction) leaving the hot side of the
slab?
(b) Give an expression for the radiative heat flux leaving the hot side.
T1 = 1000 K
1 m
T2 = 2000 K
1=6.633 ×5.670 ×10−8×10004=3.76 ×105W/m2.
CHAPTER 10 275
10.6 A semitransparent sphere of radius R=10 cm has a parabolic temperature profile T=Tc(1 −r2/R2), Tc=
2000 K. The sphere is gray with κ=0.1 cm−1,n=1.0, does not scatter, and has nonreflective boundaries.
Outline how to calculate the total heat loss from the sphere (i.e., there is no need actually to carry out
cumbersome integrations).
276 RADIATIVE HEAT TRANSFER
10.7 Repeat Problem 10.6, but assume that the temperature is uniform at 2000 K. What must be the local production
of heat if the sphere is to remain at 2000 K everywhere? Note: The answer may be left in integral form (which
must be solved numerically). Carry out the integration for r=0 and r=R.
=3.629 ×108e−1W/m3=1.335 ×107W/m3
τr=τR:
˙
Q′′′(R)=2κσT4Z+1
−1
e−τRµ−τR|µ|dµ=2κσT4“Z0
−1
dµ+Z1
0
e−2τRµdµ#
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