CHAPTER 3 47
3.20 Determine the total, normal emittance of copper, silver and gold for a temperature of 1500 K. Check your
results by comparing with Fig. 3-8.
Gold 4.10 ×1050.820 ×1051.8293 ×10−20.1353 0.075
These ǫncoincide with the dashed line of Fig. 3-8. Experimental data show considerable scatter for such
48 RADIATIVE HEAT TRANSFER
3.21 Determine the total, hemispherical emittance of copper, silver and gold for a temperature of 1500 K. Check
your results by comparing with Fig. 3-11.
Gold 0.1353 0.6687 0.091
Again, these ǫcoincide with the dashed line of Fig. 3-11.
CHAPTER 3 49
3.22 A polished platinum sphere is heated until it is glowing red. An observer is stationed a distance away, from
where the sphere appears as a red disk. Using the various aspects of electromagnetic wave theory and/or
Fig. 3-9 and Table 3.3, explain how the brightness of emitted radiation would vary across the disk, if observed
with (a) the human eye, (b) an infrared camera.
0.2 µm [this may also be shown by evaluating ε′′/ε′from equation (3.64), substituting into equation (3.62)].
Using the Drude theory for the visible, however, results in still appreciable values for ε′′ as well as k, and Pt
should appear to the human eye more or less like a metal [see (b) below]. In reality, Pt starts deviating from
50 RADIATIVE HEAT TRANSFER
3.23 Two aluminum plates, one covered with a layer of white enamel paint, the other polished, are directly facing
the sun, which is irradiating the plates with 1000 W/m2. Assuming that convection/conduction losses of the
plates to the environment at 300 K can be calculated by using a heat transfer coefficient of 10 W/m2K, and
that the back sides of the plates are insulated, estimate the equilibrium temperature of each plate.
0.08, λ > 2µm,
keeping in mind that 90% of the sun’s energy is below 1.5µm, while virtually all emission from a surface with
Tc<500 K is beyond 4 µm (i.e., the emittance over the range 1.5µm<λ<4µm is unimportant). Then
=0.9−0.6×0.93962 =0.336
ǫn=0.9−0.6f(2 ×Tc)=0.9 for Tc<500 K
ǫ≃0.85 from Fig. 3-19.
aluminum:
αn=0.08 +(0.25 −0.08) ×0.93962 =0.240
σT4
∞
=1000 W/m2K
5.670×10−8×3004W/m2K=2.177
Tc
T∞4
+6.532
ǫTc
T∞=αn
ǫ2.177 +1+6.532
ǫ
52 RADIATIVE HEAT TRANSFER
3.24 Consider a metallic surface coated with a dielectric layer.
(a) Show that the fraction of energy reflected at the vacuum-dielectric interface is negligible (n1=1.2; k1=0).
(b) Develop an expression for the normal, spectral emittance for the metal substrate, similar to the Hagen-
Rubens relationship.
(c) Develop an approximate relation for the directional, spectral emittance of the metal substrate for large
wavelengths and moderate incidence angles, say θ < 75◦.
CHAPTER 3 53
3.25
vacuum, n0 = 1
dielectric, n1, k1
metal, n2, k2
A plate of metal with n2=k2=100 is covered with a dielectric as shown.
The dielectric has an absorption band such that n1=2, and k1=1 for
0.2µm< λ < 2µm and k1=0 elsewhere. The dielectric is thick enough,
such that any light travelling through it of wavelengths 0.2µm< λ < 2µm
is entirely absorbed before it reaches the metal.
(a) What is the total, normal emittance of the composite if its temperature is 400 K?
(b) What is the total, normal absorptance if sun shines perpendicularly onto the composite?
For 0.2µm< λ < 2µm the composite is opaque, and
ρnλ=ρ01 =(n1−1)2+k2
1
(n1+1)2+k2
1
=12+12
32+12=0.2; 0.2µm<λ<2µm.
0.0390,elsewhere.
(a)
ǫn=1
σT4Z∞
0
ǫnλEbλ(400 K) dλ=0.0390 ×f(0.2T)+1−f(2T)
54 RADIATIVE HEAT TRANSFER
3.26 Estimate the total, normal emittance of α-SiC for a temperature of (i) 300 K, (ii) 1000 K. You may assume the
spectral, normal emittance to be independent of temperature.
2.2µm< λ < 9.3µm (for 1000 K) with maximum Ebλ(i.e., weight factor for emittance evaluation) near 9.5µm
ǫnλ=
0.9, λ < 10 µm,
0,10 µm<λ<13 µm,
0.65, λ > 13 µm,
one obtains
0.85, λ > 6µm,
=0.85 −0.05 f(6000) =0.85 −0.05 ×0.73778 =0.813
CHAPTER 3 55
3.27 Estimate the total, hemispherical emittance of a thick slab of pure silicon at room temperature.
Solution
For room temperature (≃300 K) the important part of the spectrum for the evaluation of the emittance is
0.1<f(λT)<0.9, or 7.3µm<λ<31 µm (from Appendix C). It is difficult, for this range, to make an accurate
0.20, λ > 8µm,
one obtains
ǫn=0.75 f(8×300 µmK) +0.20 1−f(2400)
56 RADIATIVE HEAT TRANSFER
3.28 Estimate and compare the total, normal emittance of room temperature aluminum for the surface finishes
given in Fig. 3-25.
0.1<f(λT)<0.9, or 7.3µm< λ < 31 µm (from Appendix C), with particular emphasis around λ≃9.5µm
0.1, λ > 7.5µm,
polished : ǫnλ,pl =0.06, λ < 7.5µm,
0.04, λ > 7.5µm,
evaporated : ǫnλ,ev =0.015,all λ.
Thus
3.29 A satellite orbiting earth has part of its (flat) surface coated with spectrally selective “black nickel,” which is
a diffuse emitter and whose spectral emittance may be approximated by
ǫλ=0.9, λ < 2µm,
0.25, λ > 2µm.
Assuming the back of the surface to be insulated, and the front exposed to solar irradiation of qsol =1367 W/m2
(normal to the surface), determine the relevant α/ǫ-ratio for the surface. What is its equilibrium temperature?
What would be its equilibrium temperature if the surface is turned away from the sun, such that the sun’s
rays strike it at a polar angle of θ=60◦?
σT4Z∞
0
=0.250 for moderate T,say T<700 K.
Thus αn
58 RADIATIVE HEAT TRANSFER
3.30 Repeat Problem 3.29 for white paint on aluminum, whose diffuse emittance may be approximated by
ǫλ=0.1, λ < 2µm,
0.9, λ > 2µm.
=0.9−0.8×0.93962 =0.149.
The hemispherical emittance is
σT4Z∞
0
=0.900 for moderate T,say T<700 K.
Thus αn
CHAPTER 3 59
3.31 Estimate the spectral, hemispherical emittance of the grooved materials shown in Fig. 3-36. Repeat Problem
3.29 for these materials, assuming them to be gray.
0
0=0.372
αnλ=ǫ′
λ(θ=0) =0.9
60 RADIATIVE HEAT TRANSFER
3.32 Repeat Problem 1.7 for a sphere covered with the grooved material of Fig. 3-36, whose directional, spectral
0.0,40◦< θ < 90◦.
Assume the material to be gray.
0
0=0.9 sin240◦
Therefore,
CHAPTER 3 61
3.33 A solar collector consists of a metal plate coated with “black nickel.” The collector is irradiated by the sun
with a strength of qsol =1000 W/m2from a direction which is θ=30◦from the surface normal. On its top the
surface loses heat by radiation and by free convection (heat transfer coefficient h1=10 W/m2K), both to an
atmosphere at Tamb =20◦C. The bottom surface delivers heat to the collector fluid (h2=50 W/m2K), which
flows past the surface at Tfluid =20◦C. What is the equilibrium temperature of the collector plate? How much
energy (per unit area) is collected (i.e., carried away by the fluid)? Discuss the performance of this collector.
Assume black nickel to be a diffuse emitter.
0.25, λ > 2µm,
it follows that
ǫ=1
ǫλEbλdλ=0.9f(2 µmT)+0.25 1−f(2 µmT)
(h1+h2)Ta
=0.860×1000×cos 30◦W/m2
(10 +50) ×293 W/m2=4.2365 ×10−2
62 RADIATIVE HEAT TRANSFER
3.34 Make a qualitative plot of temperature vs. the total hemispherical emittance of
(a) a 3 mm thick sheet of window glass,
(b) polished aluminum, and
(c) an ideal metal, which obeys the Hagen-Rubens relation.
0.3µm< λ < 2.5µm.
(b) From Fig. 3-25 ǫλincreases with decreasing λ, i.e., the λ-range important for increasing T. Therefore, ǫ
increases with T.
CHAPTER 3 63
3.35 A horizontal sheet of 5 mm thick glass is covered with a 2 mm thick layer of water. If solar radiation is incident
normal to the sheet, what are the transmissivity and reflectivity of the water/glass layer at λ1=0.6µm and
λ2=2µm? For water mH2O(0.6µm) =1.332 −1.09 ×10−8i,mH2O(2 µm) =1.306 −1.1×10−3i; for glass
mglass(0.6µm) =1.517 −6.04 ×10−7i,mglass(2 µm) =1.497 −5.89 ×10−5i.
mw=1.332−1.09×10−8i,m1=1.517−6.04×10−7i
ρ1=1.332−1
1.332+12
=0.02027, ρ2=1.517−1.332
1.517+1.332 2
=0.00422
ρ3=1.517−1
=0.04219 since all k2≪1.
τ1=e−4π×6.04×10−7
64 RADIATIVE HEAT TRANSFER
Thus, for scheme (i)
R−
w=0.02027 +0.00422 ×(1−0.02027)2
1−0.02027 ×0.00422 =0.02432,
T=0.97568 ×(1−0.04219) ×0.93871
1−0.04219 ×0.02432 ×0.938712=0.87803 ≃88%.
Equivalently, from scheme (ii)
mw=1.306−1.1×10−3i,m1=1.497−5.89×10−5i
τw=e−4π×1.1×10−3
×2 mm/2µm=10−6≃0
CHAPTER 3 65
3.36
Glass
Glass
Collector
T
T
TR
2
T
R
2
T
A solar collector plate of spectral absorptance αcoll =0.90 is fitted with
two sheets of 5 mm thick glass as shown in the adjacent sketch. What
fraction of normally incident solar radiation is absorbed by the collector
plate at a wavelength of 0.6µm? At 0.6µmmglass =1.517−6.04×10−7i.
and (3.101), with
ρ1=1.517 −1
1.517 +12
=0.04219, τ1=e−4π×6.04×10−7
×5 mm/0.6µm=0.93871,
1
Ac=0.90 ×0.74831
1−(1−0.90) ×0.13348 =0.68259 ≃68%.