CHAPTER 7
7.1
φ
qsun
d
d
A1
A2
Two identical circular disks of diameter d=1 m are connected at one
point of their periphery by a hinge. The configuration is then opened
by an angle φ. Surface 1 is a diffuse reffector, but emits and absorbs
according to
ǫ′
λ=0.95 cos θ, λ ≤3µm,
0.5, λ > 3µm.
Disk 2 is black. Both disks are insulated. Assuming the opening angle
to be φ=60◦, calculate the average equilibrium temperature for each of
the two disks, with solar radiation entering the configuration parallel to
Disk 2 with a strength of qsun =1000 W/m2.
1=0.5 and ǫ(1)
1as
ǫ(1)
(0.95 cos θ) sin θcos θdθ=20.95 −1
π/2
=0.633.
ǫ(1)
1
ǫ(1)
1
q(1)
1=−αo1Ho1=−αo1qsun cos θsun
=−0.8231000 cos 30◦=−712.5 W/m2.
1
ǫ(1)
1−1
q(1)
2=
ǫ(1)
1−1
F2−1q(1)
1.
A1
0.633 −11
3(−712.5) =−137.7 W/m2.
Spectral range 2 follows from equation (7.5b) as
i=1 : q(2)
1
ǫ(2)
=Eb1−F1−2Eb2,
1
=1565.7 W/m2
Eb2=
1
F2−1−
1
216 RADIATIVE HEAT TRANSFER
7.2 Reconsider 7.1 for the case that surfaces A1and A2are long, rectangular plates.
Solution
Using the semigray approximation we employ equation (7.5), with ǫ(2)
1=0.5 and ǫ(1)
1as
ǫ(1)
(0.95 cos θ) sin θcos θdθ=20.95 −1
π/2
=0.633.
ǫ(1)
1
ǫ(1)
1
q(1)
1=−αo1Ho1=−αo1qsun cos θsun
=−0.8231000 cos 30◦=−712.5 W/m2.
1
ǫ(1)
1−1
q(1)
2=
ǫ(1)
1−1
F2−1q(1)
1.
q(1)
2=1
0.633 −11
2(−712.5) =−206.5 W/m2.
Spectral range 2 follows from equation (7.5b) as
i=1 : q(2)
1
ǫ(2)
=Eb1−F1−2Eb2,
1
=1800.2 W/m2
Eb2=
1
F2−1−
1
218 RADIATIVE HEAT TRANSFER
7.5 Repeat Problem 5.36 for the case that the top of the copper shield is coated with white epoxy paint (Fig. 3-33).
q(2)
1=ǫ(2)
1Eb1,
i.e., the net heat ffux going into the shield from the top is
q1=q(1)
(ρcp)cδdTc
dt =−q1−q2
=α1qsun cos ϕ+FǫσT4
w−(ǫ1+Fǫ)σT4
c
or
ZTc
(ρcp)cδdTc
CHAPTER 7 219
7.6 A cubical enclosure has five of its surfaces maintained at 300 K, while the sixth is isothermal at 1200 K. The
entire enclosure is coated with a material that emits and reffects diffusely with
ǫλ=0.2,0≤λ < 4µm,
0.8,4µm<λ<∞.
Determine the net radiative heat ffuxes on the surfaces.
m=2,i=1 : −
6
X
Fi−jEbc =−Ebc =q(2)
h
c
6
X
Fi−j
qh=50.4355
6−0.43555.67010−812004−50.7987
6−0.79875.67010−83004
=46,359 W/m2
qc=−1
5qh=−9272 W/m2
220 RADIATIVE HEAT TRANSFER
2
ǫ(m)−1
ǫ(m)−1q(m)
2
E(m)
bh −E(m)
bc =q(m)
h
ǫ(m)−1
ǫ(m)−1q(m)
c.
In general, conservation of energy can only be applied to the entire spectrum, not just a band. However, if
ǫλ=0 outside the single band presently considered, then all heat ffuxes over the remainder of the spectrum
must be zero. Therefore, energy conservation may be applied to each individual band, and q(m)
h+5q(m)
c=0.
This is readily verified by applying the governing equation to a second surface, say i−2. Thus,
E(m)
qh=50.2
12004f(4800)
=0.60753
−3004f(1200)
=0.00213
+50.8
6−0.85.67010−8h12004(1−0.60753) −3004(1−0.00213)i=47,458 W/m2
7.8 Repeat Problem 6.26 for the case that the corner is coated with a diffusely emitting, specularly reffecting layer
whose spectral behavior may be approximated by
ǫλ=0.8,0≤λ < 3µm,
0.2,3µm<λ<∞.
The line source consists of a long filament at 2500 K inside a quartz tube, i.e., the source behaves like a gray
body for λ < 2.5µm but has no emission beyond 2.5µm.
alizing that an observer situated
S´
S´
(2)
A2
W=1.
A(1)=W ! per unit depth
id(1)=1 ! T specified
A(2)=W ! per unit depth
id(2)=1 ! T specified
T(2)=500. ! T in K
! View Factors; since configuration is open (iclsd=2), diagonal terms are also needed
iclsd=2
! Range 1&2 are the same in this problem (no specular reflected energy hits another surface)
Fs(1,1,1)=0.; Fs(1,1,2)=1.-.5*sqrt(2.); Fs(1,2,2)=0.;
W=1.
A(1)=W ! per unit depth
T(1)=500. ! T in K
! Surface 2 (left)
A(2)=W ! per unit depth
T(2)=500. ! T in K
! View Factors; since configuration is open (iclsd=2), diagonal terms are also needed
iclsd=2
! Range 1&2 are the same in this problem (no specular reflected energy hits another surface)
224 RADIATIVE HEAT TRANSFER
7.10 Repeat Problem 5.26, for the case that A1is coated with a material that has a spectral, directional emittance of
ǫ′
λ=(0.9 cos θ,λ < 4µm,
0.3, λ > 4µm. 0≤θ < π
2.
CHAPTER 7 225
7.11
Consider the configuration shown, consisting of a conical cavity A1and an opposing
circular disk with a hole at the center, as shown (d=1 cm). Defocussed laser
radiation at 10.6 µm enters the configuration through the hole in the disk as shown,
the beam having a strength of qL=103W/cm2. The disk A2is a gray, diffuse
material with ǫ2=0.1, and is perfectly insulated. Surface A1is kept at a constant
temperature of 500 K. No other external surfaces or sources affect the heat transfer.
(a) Assuming surface A1to be gray and diffuse with ǫ1=0.3 determine the
amount of heat that needs to be removed from A1(Q1).
(b) If A1were coated with the material of Problem 3.12, how would you determine
Q1? Set up any necessary equations and indicate how you would solve them
(no actual solution necessary). Would you expect Q1to increase/decrease/stay
the same (and why)?
(c) What other simple measures can you suggest to improve the accuracy of the
solution (to either (a) or (b))?
1=0.16 (=αfrom Problem 3.12). Then
i=1 : 0 =
1
1
1=0.32 (ǫfrom Problem 3.12) :
i=1 : (1 −F1−1)Eb1−F1−2Eb2=
1
1
1−1
2,(7.11-C)
1
2
1=0.32 > ǫ1grey. Both effects will make Q1smaller (in absolute value).
Using semigrayxchdf for this problem returns
Q1=−237W, T2=1509 K
1=0.1, ǫ(2)
1=0.4, ǫ(3)
1=0.16 yields spectrally “exact” results
except of diffusely reffected and reabsorbed laser irradiation. Along those lines, the 1
CHAPTER 7 227
7.13
R
4R
6R
R
concentrator
oil
tube
qsun=103W/m2
In the solar energy laboratory at UC Merced parabolic concentrators are
employed to enhance the absorption of tubular solar collectors as shown
in the sketch. Assume that solar energy enters the cavity normal to the
opening, with a strength of qsun =1000 W/m2(per unit area normal
to the rays). The parabolic receiver is coated with a highly reffective
gray, diffuse material with ǫ1=0.05, and is kept cold by convection (i.e.,
emission from it is negligible). Calculate the collected solar energy as a
function of tube outer temperature (say, for 300 K, 400 K, 500 K),
(a) assuming the tube to be gray with emittance ǫ2=0.90,
(b) assuming the tube to be covered with black nickel, using the 2-band
approach;
It is sufficient to treat tube and concentrator each as single zones.
Solution