# Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

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solutions MANUAL FOR
Thermal
Radiation Heat
Transfer
by
John R. Howell
Robert Siegel
M. Pinar Menguc
solutionS MANUAL FOR
Thermal Radiation
Heat Transfer
by
John R. Howell
Robert Siegel
M. Pinar Menguc
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
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Version Date: 20150326
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Solutions to Homework
Problems
Table of Contents
Chapter 1
Introduction to Radiative Transferโฆโฆโฆโฆ..โฆโฆโฆโฆโฆ.โฆ1.1
Chapter 2
Radiative Properties at Interfacesโฆโฆโฆโฆ.โฆโฆ….โฆโฆ.โฆ.2.1
Chapter 3
Radiative Properties of Opaque Materialsโฆโฆ..โฆ..โฆโฆโฆ.3.1
Chapter 4
Configuration Factors for Diffuse Surfaces with
Uniform Radiosityโฆ.โฆ…โฆโฆโฆ.โฆโฆ..โฆโฆโฆโฆโฆโฆโฆโฆ…4.1
Chapter 5
Radiation Exchange in Enclosures Composed of Black
and/or Diffuse-Gray Surfacesโฆโฆโฆโฆโฆโฆโฆ.โฆโฆโฆโฆโฆ.5.1
Chapter 6
Exchange of Thermal Radiation among Nondiffuse
Nongray Surfacesโฆ..โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ….โฆโฆโฆโฆ6.1
Chapter 7
Radiation Combined with Conduction and Convection
at Boundariesโฆ…..โฆโฆโฆโฆโฆโฆโฆ…โฆ..โฆโฆโฆโฆ..โฆโฆโฆ.7.1
Chapter 8
Inverse Problems in Radiative Heat Transferโฆโฆ…….โฆ.โฆ8.1
Chapter 9
Properties of Absorbing and Emitting Mediaโฆโฆโฆโฆโฆ.โฆ.9.1
Chapter 10 Fundamental Radiative Transfer Relationsโฆ..โฆโฆ……โฆ.10.1
Chapter 11 Radiative Transfer in Plane Layers and
Multidimensional Geometriesโฆโฆโฆโฆโฆโฆ…โฆโฆ..โฆโฆ…11.1
Chapter 12 Solution Methods for Radiative Transfer in
Participating Mediaโฆ.โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..โฆโฆ. .12.1
Chapter 13 Conjugate Heat Transfer in Participating Mediaโฆโฆโฆ.โฆ.13.1
Chapter 14 Electromagnetic Wave Theoryโฆโฆโฆโฆโฆ…โฆ.โฆโฆโฆ..โฆ.14.1
Chapter 15 Absorption and Scattering by Particles
and Agglomeratesโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ…โฆ.โฆโฆโฆ…โฆ.15.1
Chapter 16 Near-Field Thermal Radiationโฆโฆโฆโฆโฆโฆโฆโฆ..โฆ..โฆโฆ16.1
Chapter 17 Radiative Effects in Translucent Solids,
Windows, and Coatingsโฆโฆโฆโฆโฆโฆ..โฆโฆโฆ.โฆโฆโฆโฆ…17.1
1.Introduction to Radiative Transfer
SOLUTIONS- CHAPTER 1
1.1
What are the wave number range in vacuum and the frequency range for the
visible spectrum (0.4 to 0.7 ๏ญm)? What are the wave number and frequency values at
the spectral boundaries between the near and the far infrared regions?
SOLUTION: c0 = 2.99792458 x 108 m/s; ๏ฌ1 = 0.4 x 10-6 m, ๏ฌ2 = 0.7 x 10-6 m,
๏จ1 = 1 / ๏ฌ1 = 2.5 x 106 m-1, ๏จ2 = 1 / ๏ฌ2 = 1.428571 x 106 m-1,
๏ฎ1 = c0 / ๏ฌ1 = 2.99792458 x 108 / 0.4 x 10-6 = 7.494811 x 1014 s-1,
๏ฎ2 = c0 / ๏ฌ2 = 2.99792458 x 108 / 0.7 x 10-6 = 4.282749 x 1014 s-1,
๏ฌboundary = 25 x 10-6 m, ๏จboundary = 1 / ๏ฌboundary = 4 x 104 m-1,
๏ฎboundary = c0/๏ฌboundary = 2.99792458 x 108 / 25 x 10-6=1.19917 x 1013 s-1
Answer: 2.5 x 106 to 1.4286 x 106 m-1; 4.2827 x 1014 to
7.4948 x 1014 s-1; 4 x 104 m-1; 1.1992 x 1013 s-1.
1.2
Radiant energy at a wavelength of 2.0 ๏ญm is traveling through a vacuum. It then
enters a medium with a refractive index of 1.24.
(a) Find the following quantities for the radiation in the vacuum: speed,
frequency, wave number.
(b) Find the following quantities for the radiation in the medium: speed,
frequency, wave number, and wavelength.
SOLUTION:
(a) c0 = 2.99792458 x 108 m/s; ๏ฌo = 2.0 x 10-6 m, n = 1.24,
so speed in the vacuum = 2.99792458 x 108 m/s;
๏ฎ0 = c0 / ๏ฌ0 = 2.99792458 x 108 / 2.0 x 10-6 = 1.49896229 x 1014 s-1.
๏จ0 = 1 / ๏ฌ0 = 5 x 105 m-1;
(b) Speed in the medium c m = c0 /n = 2.99792458 x 108 /1.24
= 2.417681113 x 108 m/s;
๏ฎm = ๏ฎ0 = 1.49896229 x 1014 s-1;
๏ฌm = cm / ๏ฎm = 2.417681113 x 108 / 1.49896229 x 1014 = 1.6129032 x 10-6 m;
๏จm = 1 / ๏ฌm = 6.20 x 105 m-1.
Answer: (a) 2.9979 x108 m/s; 1.4990 x1014s-1; 5×105 m-1
(b) 2.4176 x 108 m/s; 1.4990 x 1014 s-1; 6.20 x 105 m-1; 1.6129 x 10-6 m
1.1
1.Introduction to Radiative Transfer
1.3
Radiation propagating within a medium is found to have a wavelength within the
medium of 1.570 ๏ญm and a speed of 2.500×108 m/s.
(a) What is the refractive index of the medium?
(b) What is the wavelength of this radiation if it propagates into a vacuum?
SOLUTION:
C0 = 2.99792458×108 m/s; ๏ฌm = 1.570×10-6 m; cm = 2.500×108 m/s.
(a) nm = c0/cm = 1.199
(b) ๏ฌ0= nm๏ฌm = 1.882×10-6 m = 1.882 ๏ญm.
Answer: (a) 1.199; (b) 1.882 ๏ญm.
1.4
What range of radiation wavelengths are present within a glass sheet that has a
wavelength-independent refractive index of 1.33 when the sheet is exposed in vacuum
to incident radiation in the visible range ๏ฌ0 =0.4 to 0.7 ๏ญm?
SOLUTION: nm = 1.33; ๏ฌm1 = 0.4/nm = 0.301 ๏ญm;
๏ฌm2 = 0.7/nm = 0.526 ๏ญm.
Answer: 0.301 to 0.526 ๏ญm.
1.5
A material has an index of refraction n(x) that varies with position x within its
thickness. Obtain an expression in terms of co and n(x) for the transit time for radiation
to pass through a thickness L. If n(x) = ni(1 + kx), where ni and k are constants, what is
the relation for transit time? How does wave number (relative to that in a vacuum) vary
with position within the medium?
SOLUTION: n(x) = co / c(x), so c(x) = co / n(x) = dx/d๏ด. Then
t๏ฝ
ni L
ni ๏ฆ
1 L
kL2 ๏ถ
n
(
x
)
dx
๏ฝ
(
1
๏ซ
kx
)
dx
๏ฝ
L
๏ซ
๏ง
๏ท
c0 ๏ฒx ๏ฝ0
c0 ๏ฒx ๏ฝ0
c0 ๏จ
2 ๏ธ
c = ๏ฌm๏ฎ ; ๏ฎ = c / ๏ฌm = co / ๏ฌo
co / c = n = ๏ฌo / ๏ฌm; ๏ฌm / ๏ฌo = 1/n so
๏จm / ๏จo = n = ni(1 + kx).
Answer :
2
ni
kL
L+
co
2 ; ni(1 + kx).
1.2
1.Introduction to Radiative Transfer
Derive Equation 1.26 by analytically finding the maximum of the E๏ฌb/T5 vs. ๏ฌT
1.6
relation ( Equation 1.20).
SOLUTION: Take the derivative of E๏ฌb/T5 with respect to ๏ฌT. To simplify notation, let
๏ญ ๏ฝ ๏ฌT .
๏น
d ( E๏ฌ b / T 5 )
2๏ฐ C1
d ๏ฉ
d
๏ญ5
๏ฝ
๏ช
๏บ ๏ฝ 2๏ฐ C1
๏จ ๏ญ ๏ฉ / ๏exp(C2 / ๏ญ ) ๏ญ 1๏
5
d(๏ญ )
d ( ๏ญ ) ๏ช ๏จ ๏ญ ๏ฉ ๏exp(C2 / ๏ญ ) ๏ญ 1๏ ๏บ
d(๏ญ)
๏ซ
๏ป
2
๏ฌ๏ฏ
exp(C2 / ๏ญ ) ๏ผ๏ฏ
๏ญ6
๏ญ5 ๏ญC2 / ( ๏ญ )
๏ฝ 2๏ฐ C1 ๏ญ[ ๏ญ5 ๏จ ๏ญ ๏ฉ ] / ๏exp(C2 / ๏ญ ) ๏ญ 1๏ ๏ญ ๏จ ๏ญ ๏ฉ
๏ฝ
2
๏exp(C2 / ๏ญ ) ๏ญ 1๏ ๏ฏ๏พ
๏ฏ๏ฎ
๏ป
๏จ
๏ฝ
๏ฝ
๏ฉ
๏ฌ๏ฏ
2๏ฐ C1
C2 / ๏ญ
๏ฏ๏ผ
๏ญ[ ๏ญ5 ๏ซ
๏ฝ
๏ญ ๏exp(C2 / ๏ญ ) ๏ญ 1๏ ๏ฏ๏ฎ
๏1 ๏ญ exp(๏ญC2 / ๏ญ )๏ ๏ฏ๏พ
6
Setting the result = 0 to find the maximum,
C2 / ๏จ ๏ฌT ๏ฉmax
C2 / ๏ญmax
๏ฝ
๏ฝ5
1 ๏ญ exp(๏ญC2 / ๏ญmax ) 1 ๏ญ exp ๏ฉ ๏ญC2 / ๏จ ๏ฌT ๏ฉmax ๏น
๏ซ
๏ป
Clearly, the product (๏ฌT)max at the maximum of the Planck curve is equal to a constant,
which must be found by iteration. Using iteration or a root-finding program gives
๏จ ๏ฌT ๏ฉmax ๏ฝ C3 ๏ฝ 2897.8๏ญm ๏ K .
1.7
A blackbody is at a temperature of 1250 K (1250 K), and is in air.
(a) What is the spectral intensity emitted in a direction normal to the black
surface at ๏ฌ๏ = 3.75 ๏ญm?
(b) What is the spectral intensity emitted at ๏ฑ๏ = 45ยฐ with respect to the normal of
the black surface at ๏ฌ๏ = 3.75 ๏ญm?
(c) What is the directional spectral emissive power from the black surface at ๏ฑ =
45ยฐ and ๏ฌ๏ ๏ = 3.75 ๏ญm?
(d) At what ๏ฌ๏ is the maximum spectral intensity emitted from this blackbody, and
what is the value of this intensity?
(e) What is the hemispherical total emissive power of the blackbody?
SOLUTION:
(a) At ๏ฌT = 4687.5 ๏ญm K, I๏ฌ b,n ๏ฝ
2C1
๏ฌ ๏จe
5
C2 / ๏ฌT
๏ฉ
๏ญ1
= 7823 W/m2โข๏ญmโขsr.
(b) Because the intensity from a blackbody is independent of angle of emission,
the result is the same as part (a).
(c) E๏ฌb(3.75 ๏ญm, 45ยฐ) = ๏ I๏ฌb cos 45ยฐ = 7823 x 0.7071 = 5532 W/m2โข๏ญmโขsr
(d) From Equation 1.26, ๏ฌmaxT = C3, and from Table A.4,
C3 = 2897.8 ๏ญm K. Thus, ๏ฌmax = 2318.2 / 1250 = 2.3182 ๏ญm. At ๏ฌmax,
1.3
1.Introduction to Radiative Transfer
I๏ฌ b,max ๏ฝ
2C
1
C2 / ๏ฌmaxT
5
๏ฌmax
๏จe
๏ฉ
๏ญ1
= 12499.08 W/m2โข๏ญmโขsr
(e) From Equation 1.32 and Table A-4, Eb = ๏ณT4
= 5.67040×10-8 W/m2โขK4 x (1250)4 K4 = 138,437.5 W/m2
Answers: (a) 7,823W/(m2ยท๏ญmยทsr); (b) 7,823W/(m2ยท๏ญmยทsr);
(c) 5532 W/(m2ยท๏ญmยทsr); (d) 2.3182 ๏ญm, 12499 W/(m2ยท๏ญmยทsr); (e) 138,437 W/m2 .
1.8
Plot the hemispherical spectral emissive power E๏ฌb for a blackbody in air
[W/(m2ยท๏ญm)] as a function of wavelength (๏ญm) for surface temperatures of 2000 and
6250K.
SOLUTION:
E๏ฌb = 2๏ฐC1/ { ๏ฌ5[e(C2/๏ฌT) – 1]} For a wavelength range of 0.01 to 5 ๏ญm,
following figures are obtained for different surface temperatures .
1.4
1.Introduction to Radiative Transfer
1.9
For a blackbody at 2250 K that is in air, find :
(a) the maximum emitted spectral intensity (kW /m2โข๏ญmโขsr ).
(b) the hemispherical total emissive power (kW /m2 ).
(c) the emissive power in the spectral range between ๏ฌo = 2 and 8๏ญm.
(d) the ratio of spectral intensity at ๏ฌo = 2 ๏ญm to that at ๏ฌo = 8 ๏ญm.
SOLUTION:
(a) I๏ฌb,max = C4 T5 = 4.09570×10-12 x 22505 = 236.17 kW / ๏ญmโขm2โขsr
(b) Eb = ๏ณT4 = 5.67040×10-8(W/m2โขK4) x (2250)4 (K4) = 1453.26 kW/m2
(c) Using Equation 1.37, for ๏ฌ2T = 18,000 ๏ญmโขK, F0๏ฎ ๏ฌ2T = 0.97766 and for
๏ฌ1T = 4500 ๏ญmโขK, F0๏ฎ ๏ฌ1T = 0.56429
F๏ฌ2T ๏ฎ ๏ฌ1T = 0.41336
Eb(๏ฌ1๏ฎ๏ฌ2) = 0.3482 ๏ณT4 = 600.730 kW/m2
2C
(d) Using I ๏ฌb,n ๏ฝ 5 C / ๏ฌ1T
,
๏ฌ ๏จ e 2 ๏ญ 1๏ฉ
[ I๏ฌb(๏ฌ=2) ] / [I๏ฌb(๏ฌ=8) ] = 1.5861×105/ 2.9695×103 = 53.4139
Answers: (a) 236.17 kW/m2โข๏ญmโขsr; (b) 1453.26 kW/m2;
(c) 600.730 kW/m2; (d) 53.4139
1.5
1.Introduction to Radiative Transfer
1.10 Determine the fractions of blackbody energy that lie below and above the peak of
the blackbody curve.
SOLUTION: At the peak, ๏ฌmaxT = 2893 ๏ญmK. Using Equation 1.37 with
C
14388( ๏ญ m ๏ K )
๏ธ๏บ 2 ๏ฝ
๏ฝ 4.965 gives F0๏ญ2897.8 ๏ฝ 0.25005 . Therefore about 25 percent of
๏ฌT 2897.8( ๏ญ m ๏ K )
the blackbody energy is at wavelengths below the peak, and 75 percent is at longer
wavelengths.
1.11
A blackbody at 1250K is radiating in the vacuum of outer space.
(a) What is the ratio of the spectral intensity of the blackbody at
๏ฌ๏ = 2.0๏ญm to the spectral intensity at ๏ฌ๏ = 5๏ญm?
(b) What fraction of the blackbody emissive power lies between the wavelengths
of ๏ฌ๏ = 2.0๏ญm and ๏ฌ๏ = 5๏ญm?
(c) At what wavelength does the peak energy in the radiated spectrum occur for
this blackbody?
(d) How much energy is emitted by the blackbody in the range 2.0 ๏ฃ ๏ฌ๏ ๏ฃ 5 ๏ญm?
SOLUTION:
(a) ๏ฌ1T = 2.0 x 1250 = 2500 ๏ญm K; ๏ฌ2T = 5 x 1250 = 6250 ๏ญm K
(Note that these values bracket the peak in the blackbody curve.)
2C
Using I๏ฌb ๏ฝ 5 C / ๏ฌ1T
,
๏ฌ ๏จ e 2 ๏ญ 1๏ฉ
I๏ฌ๏ฑb /I๏ฌ๏ฒb = 11823.5 / 4237.4 = 2.7903 (b) From Equation 1.37, F0๏ฎ๏ฌ1T = F0๏ฎ2500
= 0.16136,
F0๏ฎ๏ฌ2T = F0๏ฎ7250 = 0.75792. The fraction between ๏ฌ1T and ๏ฌ2T is then=
((0.75792- 0.16136) = 0.59656.
(c) From Table A.4, C3 = 2897.77 ๏ ๏ญm K, so ๏ฌmax = 2897.77 / 1250 =
2.31824๏ ๏ญm.
(d) From Table 1.2, the band emission is
( F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T ) ๏ณT4 = 0.59656 x 5.67040×10-8(W/m2โขK4) x (1250)4 (K4)
= 82587.2 W/m2
Answer: (a) 2.7903; (b) 0.59656.; (c) 2.31824๏ญm; (d) 82587.2 W/m2.
1.12 Solar radiation is emitted by a fairly thin layer of hot plasma near the sun’s surface.
This layer is cool compared with the interior of the sun, where nuclear reactions are
occurring. Various methods can be used to estimate the resulting effective radiating
temperature of the sun, such as determining the best fit of a blackbody spectrum to the
observed solar spectrum. Use two other methods (below), and compare the results to
the oft-quoted value of Tsolar = 5780 K.
a) Using Wien’s Law and taking the peak of the solar spectrum as 0.50 ๏ญm,
estimate the solar radiating temperature.
1.6
1.Introduction to Radiative Transfer
b) Given the measured solar constant in Earth orbit of 1368 W/m2, and using the
“inverse square law” for the drop in heat flux with distance, estimate the solar
temperature. The mean radius of the Earth’s orbit around the sun as 149×106 km and
the diameter of the sun as 1.392×106 km.
SOLUTION:
a)
๏จ ๏ฌT ๏ฉmax ๏ฝ C3 ๏ฝ 2898๏ญm ๏ K; Tsolar ๏ฝ
2898๏ญm ๏ K
๏ฝ 5796K
0.50๏ญm
1/ 4
๏ฆ
๏ถ
๏ง
๏ท
2
2
RSun
RSun
qsolar
2
4
๏ง
๏ท
qsolar ๏ฝ 1353(W / m ) ๏ฝ qSun 2
๏ฝ ๏ณTSun 2
: TSun ๏ฝ
2
๏ง
๏ท
REarth orbit
REarth orbit
R
๏ง๏ง ๏ณ 2 Sun ๏ท๏ท
๏จ REarth orbit ๏ธ
b)
๏ฆ
๏ง
๏ง
๏ง
1368(W / m2 )
๏ฝ๏ง
2
1.392 ๏ด 105 / 2 km2
๏ง
๏ญ8
2 4
๏ง 5.6704 ๏ด 10 (W / m K )
2
๏ง
149 ๏ด 106 km2
๏จ
๏จ
๏ฉ๏จ ๏ฉ
๏ฉ๏จ ๏ฉ
๏จ
1/ 4
๏ถ
๏ท
๏ท
๏ท
๏ท
๏ท
๏ท
๏ท
๏ธ
๏ฝ 5766K
Answers: a) 5796K; b) 5766 K.
1.13
K.
The surface of the sun has an effective blackbody radiating temperature of 5780
(a) What percentage of the solar radiant emission lies in the visible range ๏ฌ = 0.4
to 0.7 ๏ญm?
(b) What percentage is in the ultraviolet?
(c) At what wavelength and frequency is the maximum energy per unit
wavelength emitted?
(d) What is the maximum value of the solar hemispherical spectral emissive
power?
SOLUTION:
(a) ๏ฌ1T = 5780 K x 0.4๏ ๏ญm = 2312 ๏ญm K ;
๏ฌ2T = 5780 K x 0.7๏ ๏ญm = 4046 ๏ญm K
๏F = F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T = 0.48916 – 0.12240 = 0.36676, or 36.7 %
(b) From Fig. 1.2, the UV range is taken as 0.01 to 0.4 ๏ญm.
๏ฌ1T = 5780 K x 0.01๏ ๏ญm = 57.8 ๏ญm K; ๏ฌ2T = 5780 K x 0.4๏ ๏ญm = 2312 ๏ญm K.
๏F = F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T = 0.12240 – 0 = 0.12240, or 12.2 %
1.7
1.Introduction to Radiative Transfer
(c) The maximum energy is at ๏ฌ๏ max, where, from Table A.4, C3 = 2897.8 ๏ ๏ญm
K, so ๏ฌmax = 2897.8 / 5780 = 0.50134 ๏ ๏ญm.
The corresponding frequency is
๏ฎmax = co / ๏ฌ๏ = 2.9979×108 (m/s) / 0.50134 x 10-6 (m) = 5.9798×1014 Hz
(d) At ๏ฌmax T = C3 = 2897.8 ๏ ๏ญm K, using
E๏ฌ max b ๏ฝ
2๏ฐC1
5
๏ฌmax
๏จe
C2 / ๏ฌmax T
๏ฉ
๏ญ1
gives E๏ฌ๏ฌmax,b) =8.301×107 W/(m2ยท๏ญm).
Answer: (a) 36.7%; (b) 12.2%; (c) 0.5013 ๏ญm, 5.98×1014 Hz;
(d) 8.301×107 W/(m2ยท๏ญm).
1.14 A blackbody radiates such that the wavelength at its maximum emissive power
is 2.00 ๏ญm. What fraction of the total emissive power from this blackbody is in the range
๏ฌ๏ = 0.7 to ๏ฌ๏ = 5 ๏ญm?
SOLUTION: Wien’s law gives ๏ฌmax T = C3 = 2897.8 ๏ ๏ญm K, so
T = C3/๏ฌmax= 2897.8 / 2= 1448.9 K. Thus, ๏ฌ1T = 1448.9 K x 0.7๏ ๏ญm = 1014.2 ๏ญm K;
๏ฌ2T = 1448.9 K x 5๏ ๏ญm = 7244.5 ๏ญm K. Using Equation 1.37, ๏F = F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T
=
0.82137 – 0.00038= 0.82099, or 82.09 %
Answer: 0.821.
1.15 A blackbody has a hemispherical spectral emissive power of 0.03500W/(m2ยท๏ญm)
[0.0400W/(m2ยท๏ญm)] at a wavelength of 90 ๏ญm. What is the wavelength for the maximum
emissive power of this blackbody?
SOLUTION: With E๏ฌb and ๏ฌ given, solve for T from Planck’s equation,
E๏ฌb = 2๏ฐC1/ { ๏ฌ5[e(C2/๏ฌT) – 1]} which becomes T = C2/{๏ฌ๏ ln [1+(2๏ฐC1/๏ฌ5 E๏ฌb)]}
Substituting numerical values gives T = 154.7 K, and Wien’s displacement law, ๏ฌmaxT
= C3, is used to find ๏ฌmax = 18.73 ๏ญm.
Answer: 18.73 ๏ญm.
1.16 A radiometer is sensitive to radiation only in the interval 3.6 ๏ฃ ๏ฌ๏ ๏ ๏ฃ 8.5๏ญm. The
radiometer is used to calibrate a blackbody source at 1000K. The radiometer records
that the emitted energy is 4600 W/m2. What percentage of the blackbody radiated
energy in the prescribed wavelength range is the source actually emitting?
1.8
1.Introduction to Radiative Transfer
SOLUTION:
๏ ๏ ๏ฌ๏ฌ ๏ญm
3.6
8.5
๏ ๏ ๏ฌT, ๏ญm K
3600
8500
F 0-๏ฌT
0.4036
0.87417
F๏ฌ1T๏ฎ๏ฌ2T = 0.87417 – 0.4036 = 0.47058
Eb,๏๏ฌ = F๏ฌ1T๏ฎ๏ฌ2T ๏ณT4 = 0.47058 x 5.67040×10-8 x 10004 = 26,684 W/m2
Percentage sensed = (4600/26684) x 100 = 17.239%
Answer: 17.24 %.
1.17
What temperature must a blackbody have for 25 percent of its emitted energy to
be in the visible wavelength region?
SOLUTION: There are two solutions to this problem- one for the temperature at which
the bulk of the energy is in the IR, and one for a higher temperature where the major
portion is in the UV. We can assume a temperature and, by trial and error or by using a
root solver, find the fraction in the visible (0.4 ๏ฃ ๏ฌ๏ ๏ฃ 0.7), and continue until the fraction is
0.25.
For example, for the lower temperature, try T = 4600 K. Then ๏ฌ1T = 4600 K x
0.4๏ ๏ญm = 1840 ๏ญm K; ๏ฌ2T = 4600 K x 0.7๏ ๏ญm = 3220 ๏ญm K. Using Equation 1.37, ๏F =
F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T = 0.32253 – 0.04422 = 0.27831, or 27.8 %. Try a lower
temperature and proceed to convergence.
Using a root solver in a computational software package, find T such that
0.7
๏ฒ๏ฌ E๏ฌ (T )d ๏ฌ ๏ญ 0.25 ๏ฝ 0
๏ฝ0
b
๏ณT 4
1.9
1.Introduction to Radiative Transfer
This gives the two results T = 4,343 and 12,460 K.
Answer: 4,343 K, 12,460 K. (Note, two solutions are possible!)
1.18
Show that the blackbody spectral intensity I๏ฌb increases with T at any fixed
value of ๏ฌ๏ ๏ฎ๏
SOLUTION: From Equation 1.15, I๏ฌb = 2C1/{๏ฌ5 [ e C2/๏ฌT – 1 ] }. The change in I๏ฌb
with T for a fixed ๏ฌ is:
C
2
๏ฆ C ๏ถ
C2
๏ญ2C1e ๏ฌT ๏ง ๏ญ 22 ๏ท
๏ฌT
๏ถ
I
2
C
C
e
๏ฆ ๏ฌb ๏ถ
๏จ ๏ฌT ๏ธ ๏ฝ 1 2
๏ง
๏ท ๏ฝ
2
2
6
C2
C2
๏ฌ
๏จ ๏ถT ๏ธ๏ฌ
๏น
๏น
5 ๏ฉ ๏ฌT
2 ๏ฉ ๏ฌT
๏ฌ ๏ชe ๏ญ 1๏บ
T ๏ชe ๏ญ 1๏บ
๏ซ
๏ป
๏ซ
๏ป
The values of C1, C2, and ๏ฌ are all positive, so that (๏ถI๏ฌb / ๏ถT)๏ฌ๏ is always positive.
Hence, I๏ฌb must increase with T for every ๏ฌ๏ฎ
There are other ways to carry out this proof, such as plotting I๏ฌb or showing that
I๏ฌb itself increases as T increases by noting that the denominator of Planck’s spectral
distribution of intensity decreases with increasing T at any fixed wavelength.
1.10
1.Introduction to Radiative Transfer
1.19
Blackbody radiation is leaving a small hole in a furnace at 1200 K (see the
figure.) What fraction of the radiation is intercepted by the annular disk? What fraction
passes through the hole in the disk?
1.5 cm
1 cm
3 cm
Hole in
furnace wall
SOLUTION: From Table 1.2, the fraction of radiation to an annular disk is:
1.5
2๏ฐ sin2 ๏ฑ 2 ๏ญ sin2 ๏ฑ1
๏ฝ sin2 ๏ฑ 2 ๏ญ sin2 ๏ฑ1
๏ฐ
2
1
.
๏ฑ
52
๏ฑ1
3
where sin ๏ฑ1 = 1/(32 + 12)1/2 ; sin ๏ฑ2 = 1.5/(32 + 1.52)1/2.
Then Fannular disk = 1.52/(32 + 1.52) – 1/(32 + 12) = 0.2000 – 0.1000 = 0.1000.
F hole = sin2๏ฑ1 = 1/(32 + 12) = 0.1000
Answer: 0.1000; 0.1000.
1.20
A sheet of silica glass transmits 85% of the radiation that is incident in the
wavelength range between 0.38 and 2.7๏ญm, and is essentially opaque to radiation
having longer and shorter wavelengths. Estimate the percent of solar radiation that the
glass will transmit. (Consider the sun as a blackbody at 5780 K.)
If the garden in a greenhouse radiates as a black surface and is at 40 ยฐC, what
percent of this radiation will be transmitted through the glass?
1.11
1.Introduction to Radiative Transfer
Solar
radiation
Glass
Garden at
40๏ฐ C
SOLUTION: Part 1; ๏ฌ1T = 0.38๏ ๏ญm x 5780 K = 2196.4 ๏ญm K; ๏ฌ2T = 2.7๏ ๏ญm x 5780 K=
15606๏ญm K. Using Equation 1.37, ๏F = F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T = 0.96951 – 0.10022 =
0.86929. The percent transmitted is then 0.86929 x 0.85 x 100 = 73.9 %.
Part 2: ๏ฌ1T = 0.38๏ ๏ญm x (40 + 273) K = 118.94 ๏ญm K; ๏ฌ2T = 2.7๏ ๏ญm x 313 K=
845.1๏ญm K. Using Equation 1.37, ๏F = F0๏ฎ๏ฌ2T – F0๏ฎ๏ฌ1T = 0.000037 – 0.00000 =
0.000037. The percent transmitted is then 0.000037 x 0.88 x 100 = 0.003 %.
Answer: 73.9%; 0.003%
1.21 Derive Wien’s displacement law in terms of wave number by differentiation of
Planck’s spectral distribution in terms of wave number, and show that T/๏จ๏ max = 5099.4
๏ญm K.
SOLUTION: From Equation 1.17, E๏จb = 2๏ฐC1๏จ3/[eC2๏จ/T – 1 ]. To obtain the maximum,
differentiate with respect to ๏จ,
C2๏จ
๏ฆ CT2๏จ
๏ถ
3
6๏ฐ C1๏จ ๏ง e ๏ญ 1๏ท ๏ญ 2๏ฐ C1๏จ ๏จC2 / T ๏ฉ e T
๏ถ
E
๏ฆ ๏จb ๏ถ
๏จ
๏ธ
๏ง
๏ท ๏ฝ
2
๏ฆ CT2๏จ
๏ถ
๏จ ๏ถ๏จ ๏ธT
๏ง e ๏ญ 1๏ท
๏จ
๏ธ
2
C2๏จmax
๏ฆ C2๏จmax
๏ถ C ๏จ
๏ง
๏ท
2
max
๏ญ 1๏ท ๏ฝ
e T
Setting this equal to zero gives 3๏ง e T
T
๏จ
๏ธ
for which the solution is (T/๏จmax ) = constant. Substitution of
(T/๏จmax) = 5099.4 ๏ญm K shows this solution to be valid.
1.12
1.Introduction to Radiative Transfer
1.22 A student notes that the peak emission of the sun according to Wien’s
displacement law is at a wavelength of about ๏ฌmax = C3/5780 K
= 2897.8/ 5780 = 0.501 ๏ญm. Using ๏จmax = 1/0.501 ๏ญm, the student solves again for the
solar temperature using the result derived in Homework Problem 1.21. Does this
computed temperature agree with the solar temperature? Why? (This is not trivial– put
some thought into why.)
SOLUTION: Solving for T using ๏จmax = 1/0.501 ๏ญm in (T / ๏จmax) = 5099.4 ๏ญm K gives
T = 5099.4 x 0.501 = 2554.8 K, which is much too small for the solar radiating
temperature. The reason is that E๏ฌ๏ is the energy per unit ๏ฌ๏ interval, while E๏จ is the
energy per unit ๏จ interval. The peaks in the curves of E๏ฌ vs ๏ฌ and E๏จ vs ๏จ do not occur
at corresponding values of ๏จ and ๏ฌ as related by ๏ฌ๏ ๏ฝ๏ 1/๏จ. (See Homework Problem
1.24 for the correct relation.)
1.23 Derive the relation between the wave number and the wavelength at the peak of
the blackbody emission spectrum. (You may use the result of Homework Problem 1.22.)
SOLUTION: At a given temperature, Wien’s law in terms of wavelength (in SI units) is T
= 2897.8 / ๏ฌ๏ max and, from the result of Homework Problem 1.18, in terms of wave
number, T = 5099.4 ๏จmax. At a given T, these may be equated to give ๏จ max = 2897.8
/ ( 5099.4 ๏ฌ๏ max ) = 0.56826 / ๏ฌmax where ๏จmax will be in ๏ญm-1. For wave number in
cm-1 and wavelength in ๏ญm, the relation is
๏จmax (cm-1) = 5682.6 / ๏ฌmax (๏ญm)
Answer: ๏จmax (cm-1) = 5682.6 / ๏ฌmax (๏ญm).
1.24 A solid copper sphere 3 cm in diameter has on it a thin black coating. Initially,
the sphere is at 850 K, and it is then placed in a vacuum with very cold surroundings.
How long will it take for the sphere to cool to 200K? Because of the high thermal
conductivity of copper, it is assumed that the temperature within the sphere is uniform at
any instant during the cooling process. (Properties of copper: density, ๏ฒ = 8950 kg/m3;
specific heat, c = 383 J/kgโขK.)
dT
๏ฝ ๏ณT 4 A s , or
SOLUTION: An energy balance on the sphere gives ๏ญ ๏ฒcV
d๏ด
๏ฒcV dT
d๏ด ๏ฝ ๏ญ
. Integrating and using V=๏ฐD3/6 and As = ๏ฐD2,
4
๏ณA s T
๏ด๏ฝ
๏ฒ cD ๏ฆ 1
1๏ถ
๏ง 3 ๏ญ 3 ๏ท ๏ฝ 12430 s ๏ฝ 3.453 hr .
18๏ณ ๏จ Tf Ti ๏ธ
Answer: 3.453 hr.
1.13
1.Introduction to Radiative Transfer
1.25 A black rectangular sheet of metal, 6 cm by 10 cm in size, is heated uniformly
with 2350 Watts by passing an electric current through it. One face of the rectangle is
well insulated. The other face is exposed to vacuum and very cold surroundings. At
thermal equilibrium, what fraction of the emitted energy is in the wave number range
from 0.33 to 2 ๏ญm-1?
SOLUTION: The energy emitted must be equal to the electrical energy supplied:
Qelec = ๏ณTA4A, or TA = (Qelec/A๏ณ)1/4 = [2350/(6x10x10-4×5.6704×10-8)]1/4 =
1621.2 K. The wavelengths corresponding to the wave numbers are ๏ฌ1 = 1/2 = 0.5 ๏ญm;
๏ฌ2 = 1/0.33 = 3.03 ๏ญm. Then ๏ฌ1T = 0.5×1621.2 = 810.58 ๏ญmโขK and ๏ฌ2T = 2×1621.2 =
4912.6๏ญmโขK. Using the series expression Equation 1.37 for F0๏ฎ๏ฌT gives the fraction of
emitted energy as 0.6225.
Answer: 0.6225.
1.26 Spectral radiation at ๏ฌ๏ = 2.445 ๏ญm and with intensity 7.35 kW/(m2โข๏ญmโขsr) enters
a gas and travels through the gas along a path length of 21.5 cm. The gas is at uniform
temperature 1000 K and has an absorption coefficient ๏ซ2.445 ๏ญm = 0.557 m-1. What is
the intensity of the radiation at the end of the path? Neglect scattering, but include
emission by the gas.
SOLUTION: Taking into account both absorption and emission gives
I๏ฌ (S ) ๏ฝ I๏ฌ (0)e๏ญ๏ซ๏ฌ S ๏ซ I๏ฌb 1 ๏ญ e ๏ญ๏ซ๏ฌ S and e๏ญ๏ซ๏ฌS ๏ฝ e๏ญ0.557๏ด0.215 ๏ฝ 0.8871
๏จ
๏ฉ
For I๏ฌb at 2.445 ๏ญm and 1000 K, Equation 1.15 gives I๏ฌb=3.803 kW/(m2 ๏ญm sr)
Substituting into the original equation gives
I๏ฌ ๏ฝ 7.35 ๏ด 0.8871 ๏ซ 3.803(1 ๏ญ 0.8871) ๏ฝ 6.950kW / (m2 ๏ ๏ญm ๏ sr ) .
Answer: 6.950 kW/( m2โข๏ญmโขsr ).
1.27 Radiation from a blackbody source at 2000 K is passing through a layer of air at
12,000 K and 1 atm. Considering only the transmitted radiation (that is, not accounting
for emission by the air), what path length is required to attenuate by 35% the energy at
the wavelength corresponding to the maximum emission by the blackbody source? (At
this ๏ฌ, take ๏ซ๏ฌ๏ = 1.200 x 10-1 cm-1 for air at 12,000 K and 1 atm.
SOLUTION: From Wien’s displacement law,
๏ฌmax ๏ฝ
C3 2897๏ญm ๏ K
๏ฝ
๏ฝ 1.4489๏ญm
T
2000K
I๏ฌ ๏จS ๏ฉ
๏ฝ 0.65 ๏ฝ e ๏ญ๏ซ๏ฌS ๏ฝ e ๏ญ0.12S
I๏ฌ ๏จ 0 ๏ฉ
Solving for S yields S = 3.589 cm.
Answer: 3.59 cm.
Then
1.14
1.Introduction to Radiative Transfer
1.28 A gas layer at constant pressure P has a linearly decreasing temperature across
the layer, and a constant mass absorption coefficient ๏ซm (no scattering). For radiation
passing in a normal direction through the layer, what is the ratio I2/I1 as a function of
T1, T2, and L? The temperature range T2 to T1 is low enough that emission from the
gas can be neglected. The gas constant is R.
SOLUTION: For a linear temperature
distribution,
T1
T ๏ญT
dT ๏ฝ ๏ญ 1 2 dx , and for
L
I1
T(x)
I2
x
T2
L
absorption, dI ๏ฝ ๏ญ๏ซ ( x )I ( x )dx ๏ฝ ๏ญ๏ซ m ๏ฒ ( x )I ( x )dx ๏ฝ ๏ญ๏ซ m
P
I ( x )dx Integrating from x = 0 to
RT ( x )
T2 dT
๏ซ P L dx
๏ซ mPL
dI
๏ฝ๏ญ m ๏ฒ
๏ฝ
๏ฒ
I1 I ( x )
R 0 T ( x ) R ๏จT1 ๏ญ T2 ๏ฉ T1 T ( x )
x = L, ๏ฒ
I2
๏ซ PL
๏ฆT ๏ถ
๏ญ m
ln๏ง 1 ๏ท
๏ฆI ๏ถ
๏ฆT ๏ถ
๏ซ mPL
I
R T ๏ญT
T
ln ๏ง 2 ๏ท or 2 ๏ฝ e ๏จ 1 2 ๏ฉ ๏จ 2 ๏ธ
Then, ln ๏ง 2 ๏ท ๏ฝ
I1
๏จ I1 ๏ธ R ๏จT1 ๏ญ T2 ๏ฉ ๏จ T1 ๏ธ
๏ซ m PL
๏ฆ T1 ๏ถ
๏ญ
ln๏ง ๏ท
I
R T ๏ญT
T
Answer: 2 ๏ฝ e ๏จ 1 2 ๏ฉ ๏จ 2 ๏ธ
I1
.
1.30 Hawking (1974) predicts that black holes should emit a perfect blackbody
radiation spectrum in proportion to the surface gravity of the black hole. Further
predictions are that the surface gravity of a black hole is inversely proportional to its
mass. Thus, very small black holes should emit very large blackbody radiation. If a
black hole with 1 Earth mass has a predicted blackbody temperature of 10-7 K, what
fraction of an Earth mass black hole will emit at T = 2.7K (equal to the background
temperature of space) and thus possibly be detectable?
SOLUTION: The relation between black hole mass and its emitted blackbody
temperature is, from the problem statement,
๏ฆ 10๏ญ7 ๏ถ
T1 m2
๏ฝ 3.7 ๏ด 10๏ญ8 mEarth
๏ฝ
, so m2 ๏ฝ mEarth ๏ด ๏ง
๏ท
T2 m1
๏จ 2.7 ๏ธ
(The Earth mass is about 5.97 ๏ด 1024 kg, so the smaller black hole will have an effective
mass of 2.2 ๏ด 1017 kg.)
1.15
1.Introduction to Radiative Transfer
1.30 Observers at NASA noticed an unexpected small deceleration of the early deep
space probes Pioneer 10 and 11 as they moved away from Earth. This was known as
โthe Pioneer Anomaly.โ After many speculations and false starts, researchers
determined that thermal radiation pressure exerted by radiation from thermoelectric
generators heated by small nuclear power sources plus radiated waste heat from the
electronics was the culprit. Assuming a projected radiating area of 0.2 m2 facing along
the spacecraft trajectory, a spacecraft mass of 258 kg, and a mean radiating
temperature on a the outward facing surface (effective emissivity = 0.3) of 375 K, what
deceleration was exerted on the spacecraft? [Much more detailed analysis can be found
in Turyshev et al. (2012). The detailed thermal model predicts a deceleration of 7 to 10
x10-10 m/s2.]
SOLUTION: Taking a = F/m =(PA)/m, the radiation pressure due to emission from the
surface is needed. The radiation pressure by emission is (Equation 1.65): P =
2ฮต๏ณT4/3c=[3×0.3×5.67×10-8 x 3754(W/m2)/3×2.998×108 (m/s)]
=1.12×10-6 Pa.
Thus, the deceleration is
a = [1.12×10-6 (Pa) x 0.2 (m2)/ 258 (kg)] = 8.68 x 10-10 m/s.
1.16
2. Radiative Properties at Interfaces
SOLUTIONS-CHAPTER 2
2.1
A material has a hemispherical spectral emissivity that varies considerably with
wavelength but is fairly independent of surface temperature (see, for example, the
behavior of tungsten in Figures 3.31 and 3.32. Radiation from a gray source at Ti is
incident on the surface uniformly from all directions. Show that the total absorptivity for
the incident radiation is equal to the total emissivity of the material evaluated at the
source temperatureTi .
SOLUTION: For a gray source, the incident radiation is proportional to blackbody
radiation at the source temperature; that is, CE๏ฌb(Ti) d๏ฌ in the spectral range d๏ฌ.
From Equation 2.31 the total hemispherical absorptivity is
๏ฅ
๏ฒ ๏ก๏ฌ (TA )dQ๏ฌ,id๏ฌ
๏ก ๏ฝ ๏ฌ๏ฝ0 ๏ฅ
๏ฒ๏ฌ๏ฝ0 dQ๏ฌ,id๏ฌ
Substituting for the incident energy, ๏ก ๏ฝ
๏ฒ
๏ฅ
๏ฌ๏ฝ0
๏ก ๏ฌ (TA )CE๏ฌb ๏จ Ti ๏ฉ d๏ฌ
๏ฒ
๏ฅ
๏ฌ๏ฝ0
CE๏ฌb ๏จ Ti ๏ฉ d๏ฌ
๏ฅ
๏ฒ ฮต E ๏จ T ๏ฉ d๏ฌ
From Table 2.2, ๏ก ๏ (T ) = ฮต ๏ (T ) = ฮต๏ฌ, so that ๏ก ๏ฝ
๏ฌ
A
๏ฌ
๏ฌ๏ฝ0
A
๏ฌ
๏ฌb
๏ณTi
i
4
๏ฅ
๏ฒ ฮต E ๏จ T ๏ฉ d๏ฌ ๏ฝ ๏ก ๏
For properties independent of surface temperature,๏ ฮต ๏ฝ
๏ฌ๏ฝ0
๏ฌ
๏ฌb
๏ณTi
2.2
K.
i
4
Using Figure 3.31, estimate the hemispherical total emissivity of tungsten at 2600
๏ฅ
SOLUTION: Use a numerical or graphical integration to find ฮต ๏ฝ
๏ฒ๏ฌ ฮต ๏ฌ E๏ฌ d ๏ฌ
๏ฝ0
b
๏ณT
4
A careful numerical integration with ฮต๏ฌ (๏ฌ๏ >2.65 ๏ญm ) = 0.1 gives ฮต = 0.284.
Answer: 0.284.
2.3
Suppose that ฮต๏ฌ๏ is independent of ๏ฌ๏ (gray-body radiation). Show that F0๏ฎ๏ฌT
represents the fraction of the total radiant emission of the gray body in the range from 0
to ๏ฌT.
2.1
2. Radiative Properties at Interfaces
SOLUTION: The emission in a wavelength interval from ๏ฌ = 0 to ๏ฌ๏ is
๏ฌ
๏ฒ๏ฌ
* ๏ฝ0
ฮต ๏ฌ E๏ฌb d ๏ฌ * and, for all wavelengths the emission is ๏ฒ
๏ฅ
๏ฌ ๏ฝ0
ฮต ๏ฌ E๏ฌb d ๏ฌ .
The fraction of energy emitted for the range 0 ๏ฎ ๏ฌ๏ is, if ฮต ๏ฅ๏ฌ๏ is independent of ๏ฌ,
Fraction (0 ๏ฎ ๏ฌ ) ๏ฝ
ฮต๏ฌ ๏ฒ
๏ฌ
๏ฌ * ๏ฝ0
ฮต๏ฌ ๏ฒ
E๏ฌb d ๏ฌ *
๏ฅ
๏ฌ ๏ฝ0
E๏ฌb d ๏ฌ
๏ฌ
๏ฝ
๏ฒ๏ฌ E๏ฌ d ๏ฌ *
๏ฒ๏ฌ E๏ฌ d ๏ฌ
* ๏ฝ0
๏ฅ
b
๏ฝ0
b
From Equation 1.33, this is F0๏ฎ๏ฌT.
2.4
For a surface with hemispherical spectral emissivity ฮต ๏ฌ , does the maximum of
the E๏ฌ distribution occur at the same ๏ฌ๏ as the maximum of the E๏ฌb distribution at the
same temperature? (Hint: examine the behavior of dE๏ฌ๏ /d๏ฌ.) Plot the distributions of
E๏ฌ๏ as a function of ๏ฌ for the data of Figure 2.9 at 600 K and for the property data at 700
K. At what ๏ฌ๏ is the maximum of E๏ฌ? How does this compare with the maximum of E๏ฌb?
SOLUTION: E๏ฌ ๏ฝ ฮต ๏ฌ E๏ฌb
๏ฌ๏ฏ
๏ผ๏ฏ
2๏ฐ C1ฮต ๏ฌ
d๏ญ 5
๏ฝ
๏ผ๏ฏ
dE๏ฌ
d ๏ฌ๏ฏ
๏ฌ ๏ญ5
๏ฏ ๏ฌ ๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป ๏พ๏ฏ
๏ฎ
๏ฝ
๏ฝ 2๏ฐ C1ฮต t
๏ญ
๏ฝ
d๏ฌ
d๏ฌ
d ๏ฌ ๏ฎ๏ฏ ๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป ๏พ๏ฏ
๏ฌ
๏ญC2 / ๏ฌ 2 ๏ฉ exp ๏จC2 / ๏ฌ ๏ฉ ๏ผ๏ฏ
๏ญ5๏ฌ ๏ญ6
๏ฏ
๏ญ5 ๏จ
=2๏ฐ C1ฮต t ๏ญ
๏ญ๏ฌ
2 ๏ฝ
๏ฉ
๏น
exp
C
/
๏ฌ
T
๏ญ
1
๏จ
๏ฉ
๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป ๏ฏ๏พ
2
๏ฏ๏ฎ ๏ซ
๏ป
=
๏ฌ๏ฏ
๏ผ๏ฏ
๏จC2 / ๏ฌ ๏ฉ
2๏ฐ C1ฮต t
๏ญ
5
๏ซ
๏ญ
๏ฝ
๏ฉ๏ซ1 ๏ญ exp ๏จC2 / ๏ฌT ๏ฉ๏น๏ป ๏ฏ๏พ
๏ฌ 6 ๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป ๏ฏ๏ฎ
where ฮต t is the total hemispherical emittance.
(C2 / ๏ฌ )
๏ฝ5
1 ๏ญ exp(๏ญC2 / ๏ฌT )
(ฮปT)max=2897.8 ๏ญm.K. So for the same temperature, the maximum occurs at the same
point as Eฮปb. Only the intensity of this power is reduced by the numerical value of
emittance. For data in Figure 2.9, the following figure is obtained
Setting the result = 0 to find the maximum;
2.2
2. Radiative Properties at Interfaces
2.3
2. Radiative Properties at Interfaces
Eฮป is maximum at 2897.8/600=4.83 ๏ญm which overlaps with the plot.
Eฮป is maximum at 2897.8/700=4.14 ๏ญm which overlaps with the plot.
2-5
Find the emissivity at 400 K and the solar absorptivity of the diffuse material with
the measured spectral emissivity shown in the figure.
SOLUTION:
The emissivity using Equation 2.10 is
๏ฅ
๏ฒ ฮต ๏ฌ ๏จT ๏ฉ E๏ฌ ๏จT ๏ฉ d ๏ฌ
ฮต (T ) ๏ฝ ๏ฌ
b
๏ฝ0
๏ฝ
0.83 ๏ฒ
1.90
๏ฌ ๏ฝ0
๏ณT
4
E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏ณT
4
๏ซ
0.50 ๏ฒ
2.80
๏ฌ ๏ฝ1.93
E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏ณT
4
๏ซ
0.17 ๏ฒ
๏ฅ
๏ฌ ๏ฝ 2.80
E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏ณT 4
๏ฝ 0.83F0๏ญ1.90T ๏ซ 0.50F1.90T ๏ญ2.80T ๏ซ 0.17F2.80T ๏ญ๏ฅ
๏ฝ 0.83F0๏ญ1.90T ๏ซ 0.50 ๏จ F0๏ญ2.80T ๏ญ F0๏ญ1.90T ๏ฉ ๏ซ 0.17 ๏จ1 ๏ญ F0๏ญ2.80T ๏ฉ
so only two blackbody fractions need be found. Using Equation 1.37 for the F values,
ฮต (T ๏ฝ 5780K ) ๏ฝ 0.83F0๏ญ1.90๏ด5780 ๏ซ 0.50 ๏จ F0๏ญ2.80๏ด5780 ๏ญ F0๏ญ1.90๏ด5780 ๏ฉ ๏ซ 0.17 ๏จ1 ๏ญ F0 ๏ญ2.80๏ด5780 ๏ฉ
๏ฝ 0.83 ๏ด 0.9316 ๏ซ 0.50 ๏ด ๏จ 0.9745 ๏ญ 0.9316 ๏ฉ ๏ซ 0.17 ๏ด (1๏ญ 0.9745) ๏ฝ 0.7990.
Answer: ฮต (T=400 K) = 0.17; ๏กs(T=5780 K) = 0.7990
2.4
2. Radiative Properties at Interfaces
2.6
The surface temperature-independent hemispherical spectral absorptivity of a
surface is measured when it is exposed to isotropic incident spectral intensity, and the
results are approximated as shown below. What is the total hemispherical emissivity of
this surface when it is at a temperature of 1000 K?
1.0
0.75
๏ก๏ฌ
0.50
0.25
0
1.2
2.0
(2.8)
๏ฅ
๏ก ๏ฌE๏ฌb ๏จ TA ๏ฉ d๏ฌ
๏ฒ
SOLUTION: ฮต ๏ฌ = ๏ก๏ฌ from Table 2.2 gives ฮต ๏ฝ 0 ๏ฅ
so that
E
T
d
๏ฌ
๏ฒ ๏ฌb ๏จ A ๏ฉ
๏ฌ๏ฌ๏ ๏ญm
0
ฮต = 0.75 F0๏ฎ1200 + 0.5 F 1200๏ฎ2000
= 0.75 x 0.002134 + 0.5 x (0.06673 – 0.002134)
= 0.03390.
Answer: 0.0339.
2.7 (a) Obtain the total absorptivity of a diffuse surface with properties given in the
figure for incident radiation from a blackbody with a temperature of 6200 K.
(b) What is the total emissivity of the diffuse surface with properties given in the figure if
the surface temperature is 500 K?
1.0
0.95
๏ก๏ฌ=๏ ฮต๏ฌ
0.15
0
1.5
๏ฌ๏ฌ๏ ๏ญm
2.5
2. Radiative Properties at Interfaces
SOLUTION:
๏ฅ
(a)
Using ๏ก ๏ฝ
๏ฒ ๏ก ๏ฌ q๏ฌ d ๏ฌ with q๏ฌ,i = E๏ฌb(6200K) and Equation 1.37 for the blackbody
๏ฒ q๏ฌ d ๏ฌ
0
,i
๏ฅ
0
,i
fractions gives ๏ก=0.95 F0๏ฎ1.5×6200+0.15 (1-F0๏ฎ1.5×6200)
= 0.95 x 0.8975+0.15(1-0. 0.8975)= 0.8680
(b) Similarly, for emission, ฮต=0.95 F0๏ฎ1.5×500+0.15 (1-F0๏ฎ1.5×500)
= 0.95 x (๏ป0)+0.15[1-(๏ป0)] = 0.15.
Answer: (a) 0.8680; (b) 0.15.
2.8 For the spectral properties given in the figure for a diffuse surface:
(a) what is the solar absorptivity of the surface (assume the solar temperature is 5800
K)?
(b) what is the total hemispherical emissivity of the surface if the surface temperature is
700 K?
๏ฅ
1.0
0.9
๏ก๏ฌ = ฮต๏ฌ
0.1
0
1.2
3
Wavelength, ๏ฌ๏ฌ๏ ๏ญm
๏ฅ
SOLUTION: The definitions of ฮต and ๏ก in terms of ฮต ๏ฌ and ๏ก๏ฌ are used as in Problems
2.5 and 2.6. Using Equation1.37,
(a) ๏กฮป = 0.9F0-1.2×5800 + 0.1(F0-3×5800 โ F0-1.2×5800) + 1(1 โ F0-3×5800 )
= 0.72505 + 0.01704 + 0.02399 = 0.76608
For a diffuse surface, Table 2.2 gives ฮต ๏ฌ = ๏ก๏ฌ๏ฎ๏ Then
(b) ฮต = 0.9F0-1.2×700 + 0.1(F0-3×700 โ F0-1.2×700) + 1(1-F0-3×700)
= 0 + 0.00830 + 0.91695 = 0.92528.
Answer: (a) 0.766; (b) 0.925.
2.6
2. Radiative Properties at Interfaces
2.9 A white ceramic surface has a hemispherical spectral emissivity distribution at 1600
K as shown. What is the hemispherical total emissivity of the surface at this surface
temperature?
1.0
0.8
0.6
ฮต ๏ฌ๏
0.4
0.2
0
0
2
4
6
8
10
12
๏ฌ, ๏ญm
SOLUTION: Numerical or graphical integration is necessary, as no analytical
integration appears possible even for this simple variation in spectral emissivity. From
Equation 2,10 with Eฮปb=๏ฐIฮปb
๏ฅ
๏ฒ ฮต ๏ฌ ๏จ1600K ๏ฉ E๏ฌ ๏จ1600K ๏ฉ d ๏ฌ
ฮต ๏จ1600K ๏ฉ ๏ฝ ๏ฌ
๏ฝ0
b
๏ณ ๏จ1600 ๏ฉ
4
Now,
ฮต ๏จ1600K ๏ฉ ๏ฝ
๏ซ
๏ฆ ๏ฌ ๏ญ ๏ฌ1 ๏ถ ๏น
1 ๏ฌ1
1 ๏ฌ2 ๏ฉ
ฮต E d๏ฌ ๏ซ
ฮต ๏ชฮต 1 ๏ซ ๏จฮต 2 ๏ญ ฮต 1 ๏ฉ ๏ง
๏ท ๏บE๏ฌb d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ0 1 ๏ฌ b
4 ๏ฒ๏ฌ ๏ฝ๏ฌ
๏ณTA
๏ณTA
๏จ ๏ฌ2 ๏ญ ๏ฌ1 ๏ธ ๏ป
๏ซ
1 ๏ฌ3
ฮต E d๏ฌ
๏ณTA4 ๏ฒ๏ฌ ๏ฝ๏ฌ2 2 ๏ฌb
where ฮต 1 =0.2, ฮต 2 =0.8, ฮป1=3 ๏ญm, ฮป2=7 ๏ญm, ฮป3= 10 ๏ญm
Numerical Romberg integration of the ฮต (1600 K) equation gives 0.28128.
Answer: 0.281
2.7
2. Radiative Properties at Interfaces
2.10 A surface has the following values of hemispherical spectral emissivity at a
temperature of 800K.
ฮต ๏ฌ(800 K)
๏ฌ๏ฌ ๏ญm
8
0
0
0.2
0.4
0.6
0.8
0.8
0.8
0.7
0.6
0.4
0.2
0
0
(a) What is the hemispherical total emissivity of the surface at 800 K?
(b) What is the hemispherical total absorptivity of the surface at 800 K if the
incident radiation is from a gray source at 1800 K that has an emissivity of 0.815? The
incident radiation is uniform over all incident angles.
SOLUTION:
(a)
0.8
0.6
ฮต ๏ฌ(750K)
0.4
0.2
0
0
1
2
3
4
๏ฌ๏ฌ๏ ๏ญm
From Equation 2.10,
2.8
5
6
7
8
2. Radiative Properties at Interfaces
๏ฅ
๏ฒ๏ฌ ฮต ๏ฌ ๏จ800K ๏ฉ E๏ฌ ๏จ800K ๏ฉ d ๏ฌ
ฮต (800K ) ๏ฝ
b
๏ฝ0
๏ณT
4
A
๏ฝ
๏ฌ ๏ญ ๏ฌ1 ๏น
1 ๏ฌ2 ๏ฉ
ฮต ๏ซ ๏จฮต 2 ๏ญ ฮต 1 ๏ฉ
E ๏จ 800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 1
1
๏ฌ2 ๏ญ ๏ฌ1 ๏บ๏ป ๏ฌb
๏ณ TA
๏ซ
๏ฝ
๏ฌ ๏ญ ๏ฌ2 ๏น
1 ๏ฌ3 ๏ฉ
ฮต ๏ซ ๏จฮต 3 ๏ญ ฮต 2 ๏ฉ
E ๏จ 800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 2
2
๏ฌ3 ๏ญ ๏ฌ2 ๏บ๏ป ๏ฌb
๏ณ TA
๏ซ
๏ฝ
๏ฌ ๏ญ ๏ฌ3 ๏น
1 ๏ฌ4 ๏ฉ
ฮต ๏ซ ๏จฮต 4 ๏ญ ฮต 3 ๏ฉ
E ๏จ 800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 3
3
๏ฌ4 ๏ญ ๏ฌ3 ๏บ๏ป ๏ฌb
๏ณ TA
๏ซ
where TA =
K, ฮต 1=0, ฮต 2=0.8, ฮต 3=0.8, ฮต 4=0, ๏ฌ1=1 ๏ญm, ๏ฌ2 = 3 ๏ญm, ๏ฌ3 = 4 ๏ญm, ๏ฌ4 = 8
๏ญm. Accurate numerical integration of the ฮต (
K) equation gives 0.43826.
(b) From Equation 2.25
๏ฅ
๏ฅ
๏ฒ ๏ก๏ฌ ๏จ800K ๏ฉ 0.815E๏ฌ ๏จ1800K ๏ฉ d ๏ฌ ๏ฝ ๏ฒ๏ฌ ฮต ๏ฌ ๏จ800K ๏ฉ E๏ฌ ๏จ1800K ๏ฉ d ๏ฌ
๏ก (800K ) ๏ฝ ๏ฌ
๏ณ 1800
๏ฒ 0.815E๏ฌ ๏จ1800K ๏ฉ d ๏ฌ
b
๏ฝ0
๏ฝ0
๏ฅ
b
4
b
0
๏ฝ
๏ฌ2 ๏ฉ
๏ฌ ๏ญ ๏ฌ1 ๏น
1
ฮต ๏ซ ๏จฮต 2 ๏ญ ฮต 1 ๏ฉ
E ๏จ1800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 1
1
๏ฌ2 ๏ญ ๏ฌ1 ๏บ๏ป ๏ฌb
๏ณ 1800
๏ซ
๏ฝ
๏ฌ3 ๏ฉ
๏ฌ ๏ญ ๏ฌ2 ๏น
1
ฮต ๏ซ ๏จฮต 3 ๏ญ ฮต 2 ๏ฉ
E ๏จ1800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 2
2
๏ฌ3 ๏ญ ๏ฌ2 ๏บ๏ป ๏ฌb
๏ณ 1800
๏ซ
๏ฝ
๏ฌ4 ๏ฉ
๏ฌ ๏ญ ๏ฌ3 ๏น
1
ฮต ๏ซ ๏จฮต 4 ๏ญ ฮต 3 ๏ฉ
E ๏จ1800K ๏ฉ d ๏ฌ
4 ๏ฒ๏ฌ ๏ฝ๏ฌ ๏ช 3
3
๏ฌ4 ๏ญ ๏ฌ3 ๏บ๏ป ๏ฌb
๏ณ 1800
๏ซ
Numerical integration of the ๏ฒ
๏ฅ
๏ฌ ๏ฝ0
ฮต ๏ฌ E๏ฌbd ๏ฌ gives 0.42709.
Answer: (a) 0.438; (b) 0.427.
2.11 Find the emissivity at 950 K and the solar absorptivity of the diffuse material with
the measured spectral emissivity shown in the figure. This will require numerical
integration.
2.9
2. Radiative Properties at Interfaces
SOLUTION:
The emissivity in the various ranges can be expressed as
0 ๏ฃ ๏ฌ ๏ผ 0.70 : ฮต ๏ฌ ๏ฝ 0.83
๏ฆ ๏ฌ ๏ญ 0.70 ๏ถ
0.70 ๏ฃ ๏ฌ ๏ผ 1.90 : ฮต ๏ฌ ๏ฝ 0.83 ๏ญ 0.33 ๏ง
๏ท
๏จ 1.20 ๏ธ
1.90 ๏ฃ ๏ฌ ๏ผ 2.80 : ฮต ๏ฌ ๏ฝ 0.50
๏ฆ ๏ฌ ๏ญ 2.80 ๏ถ
2.80 ๏ฃ ๏ฌ ๏ผ 3.50 : ฮต ๏ฌ ๏ฝ 0.50 ๏ญ 0.33 ๏ง
๏ท
๏จ 0.70 ๏ธ
๏ฌ ๏ณ 3.50;ฮต ๏ฌ ๏ฝ 0.17
The total emissivity is then
๏ฅ
๏ฒ ฮต ๏ฌ ๏จT ๏ฉ E๏ฌ ๏จT ๏ฉ d ๏ฌ
ฮต (T ) ๏ฝ ๏ฌ
b
๏ฝ0
๏ณT
4
๏ฉ
๏ฆ ๏ฌ ๏ญ 0.70 ๏ถ ๏น
๏ช0.83 ๏ญ 0.33 ๏ง 1.20 ๏ท ๏บ E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏จ
๏ธ๏ป
๏ซ
๏ฌ ๏ฝ0
๏ฝ
๏ซ
4
4
๏ณT
๏ณT
3.50 ๏ฉ
๏ฆ ๏ฌ ๏ญ 2.80 ๏ถ ๏น
2.80
0.50 ๏ญ 0.33 ๏ง
๏ท ๏บ E๏ฌb ๏จT ๏ฉ d ๏ฌ
0.50 ๏ฒ
E๏ฌb ๏จT ๏ฉ d ๏ฌ ๏ฒ๏ฌ ๏ฝ2.80 ๏ช๏ซ
0.70
๏จ
๏ธ๏ป
๏ฌ ๏ฝ1.90
๏ซ
๏ซ
4
4
๏ณT
๏ณT
๏ซ
0.83 ๏ฒ
0.70
0.17 ๏ฒ
๏ฅ
E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏ฌ ๏ฝ3.50
1.90
๏ฒ๏ฌ
๏ฝ0.70
E๏ฌb ๏จT ๏ฉ d ๏ฌ
๏ณT 4
The blackbody fractions can be used to evaluate the first, third and fifth integrals, but
the second and third probably require numerical integration. The results using numerical
integration are ฮต (950 K) = 0.0000+0.0684+0.1185+0.0556+0.0823 = 0.3249; ฮต (5780
K) = 0.4060+0.3225+0.0215+0.0041+0.0024 = 0.7564.
Answer: ฮต (950 K) = 0.3248; ฮต (5780 K) =๏ ๏ ๏กs(5780 K) = 0.7564.
2.10
2. Radiative Properties at Interfaces
2.12 A diffuse surface at 1000 K has a hemispherical spectral emissivity that can be
approximated by the solid line shown.
(a) What is the hemispherical-total emissive power of the surface? What is the
total intensity emitted in a direction 60ยฐ from the normal to the surface?
(b) What percentage of the total emitted energy is in the wavelength range 5 <๏ ๏ฌ
< 10 ๏ญm? How does this compare with the percentage emitted in this wavelength range
by a gray body at 1000 K with an emissivity ฮต = 0.611?
1.0
ฮต ๏ฌ๏
0.8
0.7
0.55
0.45
๏ฅ
0.2
0
0 1
4
5
7
๏ฅ
10
๏ฌ, ๏ญm
SOLUTION:
๏ฅ
๏ฒ ฮต ๏ฌ ๏จT ๏ฉ E๏ฌ ๏จT ๏ฉ d ๏ฌ ; the blackbody fractions for each spectral
(a): ฮต ๏จT ๏ฉ ๏ฝ ๏ฌ
๏ฝ0
A
b
A
๏ณT
band are obtained from Equation 1.37 as given in the table below:
A
๏ฌ๏ ๏ range
ฮต๏ฌ
4
A
๏ฌT range
F0๏ฎ๏ฌT – F0๏ฎ๏ฌT
๏F0๏ฎ๏ฌT
ฮต ๏ฌ ๏F0๏ญ๏ฌT ๏
0.000032
0.00032
๏ป0
0.48086-0.00032
0.48054
0.21624
0.63371-0.48086
0.15285
0.10699
0.80793-0.63371
0.17422
0.13937
0.9134-0.80793
0.10547
0.05801
1 – 0.9134
0.08660
0.017321
ฮต ๏ ๏ฝ๏ ๏๏ ฮต ๏ฌ๏ ๏F0๏ฎ๏ฌT = 0.53800
= ฮต ๏ณT4 = 0.53800x 5.6704×10-8 x 9004 = 30506.9 W/m2
0-1
1-4
4-5
5-7
7 – 10
10 – ๏ฅ
0.2
0.45
0.7
0.8
0.55
0.2
0 – 1000
1000-4000
4000-5000
5000-7000
7000-10000
10000 – ๏ฅ
E๏ฌb
I๏ฌb =E๏ฌb/๏ฐ = 9710.7 W/m2โขsr
(b) For the range 5<๏ฌ<10, the percentage is
[(0.13937+0.05801)/ 0.53800]x100=36.69%. For a gray surface, the percentage is the
same as for a black surface, or
100 (F0๏ฎ9000 – F0๏ฎ4500) = 100 (0.91339 – 0.63371) = 27.97 %.
Answer: (a) 30507 W/m2; 9711 W/m2โขsr (b) 36.69 %; 27.97 %
2.11
2. Radiative Properties at Interfaces
2.13 The ฮต ๏ฌ for a metal at 1000 K is approximated as shown, and it does not vary
significantly with the metal temperature. The surface is diffuse.
1.0
ฮต ๏ฌ๏
0.6
0.35
๏ฅ
0.15
0
0
2
๏ฌ, ๏ญm
4
6
๏ฅ
(a) What is ๏ก for incident radiation from a gray source at 1200 K with ฮตsource =
0.822?
(b) What is ๏ก๏ for incident radiation from a source at 1200 K made from the same
metal as the receiving plate?
SOLUTION:
๏ฅ
๏ฒ ๏ก๏ฌ ๏จ ๏ฝ ฮต ๏ฌ ๏ฉ 0.822E๏ฌ ๏จ1200K ๏ฉ d ๏ฌ ๏ฝ
(a) ๏ก ๏ฝ ๏ฌ
๏
๏ฌ๏ ๏ range
ฮต๏ฌ
0-2
2-4
4-๏ฅ
๏
0.6
0.35
0.15
๏
0.822๏ณ 12004
๏ฅฮต ๏ฌ ๏F
๏ฌTi range
F0๏ฎ๏ฌTi – F0๏ฎ๏ฌTi
b
๏ฝ0
๏F
ฮต ๏ฌ๏ ๏F
0-2400
0.14026- 0
0.14026
0.08415
2400-4800
0.60753-0.14026
0.46727
0.16355
1 – 0.60753
0.39247
0.05887
4800-๏ฅ
๏
๏
๏
๏
๏ ๏ ๏ ๏ ๏ ๏ ๏ ๏ ๏ก = ๏ ฮต ๏ฌ๏ ๏F = 0.30657
(b) ๏ก = ๏๏ ๏ก๏ฌ (๏ก๏ฌ๏F) / (๏๏ ๏ก๏ฌ๏ ๏F)
= [ 0.62 x 0.14026 + 0.352 x 0.46727 + 0.152 x 0.39247 ] / 0.30657 = 0.38022
Answer: (a) 0.30657; (b) 0.38022.
2.14 The directional total absorptivity of a gray surface is given by the expression ๏ก(๏ฑ)
= 0.450cos2๏ฑ๏ ๏ where ๏ฑ is the angle away from the normal to the surface.
(a) What is the hemispherical total emissivity of the surface?
(b) What is the hemispherical- hemispherical total reflectivity of this surface for
diffuse incident radiation (uniform incident intensity)?
2.12
2. Radiative Properties at Interfaces
(c) What is the hemispherical-directional total reflectivity for diffuse incident
radiation reflected into a direction 75ยฐ from the normal?
SOLUTION:
ฮต ๏ฝ๏ก ๏ฝ
(a)
1
4๏ฐ
๏ก ๏จ๏ฑ ๏ฉ cos ๏ฑ d๏ท
๏ฐ ๏ฒ๏ท
2๏ฐ
๏ฝ0
๏ฐ /2
๏ญ0.9 cos4 ๏ฑ
๏ฝ
0.450
cos
๏ฑ
d
๏ท
๏ฝ
๏ฝ 0.2250
๏ฐ ๏ฒ๏ท ๏ฝ0
4
0
4๏ฐ
3
(b)
๏ฒ๏ ๏ฝ๏ ๏ฑ๏ ๏ญ๏ ๏ก๏ ๏ฝ๏ ๏ฑ๏ ๏ญ๏ ฮต ๏ ๏ฝ๏ 1 – 0.2250= 0.7750.
(c) Using Equation 2.43, ๏ฒ(๏ฑr,๏ฆr) =๏ฒ(๏ฑ,๏ฆ) = 1 – ๏ก(๏ฑ) = 1 – 0.450 cos2 (75ยฐ) = 0.9699.
Answer: (a) 0.225; (b) 0.775; (c) 0.97.
2.15 Using Figure 3.24, estimate the total absorptivity of typewriter paper for normally
incident radiation from a blackbody source at 1200 K.
SOLUTION: Use Equations 2.81 and 2.39 for a gray surface and assume that there is
no dependence on circumferential angle ๏ฆ. From the plot we have symmetry for the
values of
.
๏ก ๏จ๏ฑ ๏ฝ 0,TA ๏ฉ ๏ฝ 1 ๏ญ ๏ฒ ๏จ๏ฑ ๏ฝ 0,TA ๏ฉ ๏ฝ 1 ๏ญ 2๏ฐ ๏ฒ
๏ฐ /2
๏ฑ r ๏ฝ0
๏ฒ ๏จ๏ฑ ๏ฝ 0,๏ฑr ,TA ๏ฉ cos ๏ฑr sin๏ฑr d๏ฑr ๏ฝ 1 ๏ญ 0.1079 ๏ฝ
๏ฝ 0.8921
Evaluation requires numerical integration.
Answer: 0.8921.
2.16 A gray surface has a directional emissivity as shown in the figure. The
properties are isotropic with respect to circumferential angle ๏ฆ.
(a) What is the hemispherical emissivity of this surface?
(b) If the energy from a blackbody source at 650 K is incident uniformly from all
directions, what fraction of the incident energy is absorbed by this surface?
(c) If the surface is placed in a very cold environment, at what rate must energy
be added per unit area to maintain the surface temperature at 1000 K?
2.13
2. Radiative Properties at Interfaces
SOLUTION:
(a) From Equation 2.9,
ฮต ๏ฝ 2๏ฒ ฮต ๏จ ๏ฑ ๏ฉ cos ๏ฑ sin ๏ฑ d๏ฑ ๏ฝ 2 ๏ฉ ๏ฒ
๏ฐ /2
1/2
๏ฑ๏ฝ0
๏ซ๏ช sin ๏ฑ๏ฝ0
0.9 sin ๏ฑ d ๏จ sin ๏ฑ ๏ฉ ๏ซ ๏ฒ
1
sin ๏ฑ๏ฝ1/2
0.5 sin ๏ฑ d ๏จ sin ๏ฑ ๏ฉ ๏น
๏ป๏บ
๏ฉ
๏ฆ1
๏ถ
1 ๏ถ๏น
๏ฆ
1/2
1
๏ญ 0๏ท
1๏ญ ๏ท๏บ
๏ง
2
2
๏ช
๏ฉ
๏น
๏ง
sin ๏ฑ
sin ๏ฑ
4
๏ธ ๏ซ 0.5
4 ๏บ ๏ฝ 0.600
๏บ ๏ฝ 2 ๏ช0.9 ๏จ
๏ฝ 2 ๏ช0.9
๏ซ 0.5
๏ง
๏ท
2
2
2
2
๏ช
๏ช๏ซ
๏ง
๏ท๏บ
sin ๏ฑ๏ฝ0
sin ๏ฑ๏ฝ1/2 ๏บ
๏ป
๏ช๏ซ
๏จ
๏ธ ๏บ๏ป
(b) Because the surface is gray, from Table 2.2, ๏ก๏ ๏ฝ๏ ฮต ๏ฎ Hence, ๏ก๏ = 0.600 = fraction
absorbed.
(c) Emissive power = ฮต ๏ณT4 = 0.600 x 5.67040×10-8 x 10004 = 34,022W/m2.
Answer: (a) 0.600; (b) 0.600; (c) 34,022 W/m2 .
2.17 Using Figure 3.44, estimate the ratio of normal total solar absorptivity to
hemispherical total emissivity for aluminum at a surface temperature of 650 K with a
coating of 0.1-๏ญm dendritic lead sulfide crystals. Assume the surface is diffuse. (The
solar temperature can be taken as 5780 K.)
SOLUTION: From the figure, approximate the normal hemispherical spectral reflectivity
as shown below. Using ๏ก๏ฌ๏ฌn = 1 – ๏ฒ๏ฌ๏ฌn, the spectral normal absorptivity is approximated
as shown below:
2.14
2. Radiative Properties at Interfaces
0.9
0.8
๏ฅ
๏ก๏ฌ,n
๏ฒ๏ฌ,n
๏ฅ
0.2
0.1
0
0
3
3
๏ฌ๏ฌ๏ ๏ญm
๏ฌ๏ฌ๏ ๏ญm
For ๏กn,solar, use Tsolar = 5780 K: then (๏ฌT)cutoff = 3×5780
= 17340 ๏ญmโขK, and F0๏ฎ(๏ฌT)cutoff = 0.97880 [Equation 1.37].
Then ๏กn,solar = 0.9 F0๏ฎ (๏ฌT)cutoff + 0.2 (1 – F0๏ฎ (๏ฌT)cutoff)
= 0.9 x 0.97880 + 0.2 ( 1 – 0.97880 ) = 0.8852
ฮต (T = 650 K) ๏ป ฮต n(T = 650 K) = 0.9 F0๏ฎ1950 + 0.2 (1 – F0๏ฎ1950)
= 0.9 x 0.05920 + 0.2 (1 – 0.05920 )= 0.2414 (NOTE: This assumes that the
hemispherical ฮต ๏ป ฮต n.) Thus, ๏กn,solar /๏ฅ(T = 650 K) = 0.8852/0.2414 = 3.6662.
Answer: 3.666.
2.18 A gray surface has a directional total emissivity that depends on angle of incidence
as ฮต ๏จ๏ฑ๏ฉ = 0.788 cos ๏ฑ. Uniform radiant energy from a single direction normal to the
cylinder axis is incident on a long cylinder of radius R. What fraction of energy striking
the cylinder is reflected? What is the result if the body is a sphere rather than a
cylinder?
Qincident
R
SOLUTION: ๏ก(๏ฑ) = ฮต (๏ฑ) = 0.788 cos ๏ฑ. ๏ฒ(๏ฑ) = 1 – ๏ก(๏ฑ) = 1 – 0.788 cos ๏ฑ.
For the cylinder, the intercepted energy on dA per unit length of cylinder is
q dA cos ๏ฑ = q (Rd๏ฑ๏ฉ cos ๏ฑ, and the reflected energy per unit length is
q (Rd๏ฑ๏ฉ cos ๏ฑ ( 1 – 0.788 cos ๏ฑ ). The ratio is
๏ฐ /2
reflected q ๏ฒ๏ฑ ๏ฝ0 R cos ๏ฑ ๏จ1 ๏ญ 0.788 cos ๏ฑ ๏ฉ d๏ฑ
0.788๏ฐ
๏ฝ
๏ฝ 1๏ญ
๏ฝ 0.3811
๏ฐ /2
incident
4
q
R cos ๏ฑ d๏ฑ
๏ฒ๏ฑ
๏ฝ0
2.15
2. Radiative Properties at Interfaces
For the sphere, the intercepted energy by a ring element is
R sin๏ฑ
dA = R d๏ฑ
๏ฑ๏
R
q 2๏ฐR sin๏ฑ Rd๏ฑ cos๏ฑ, and the ratio of reflected to incident energy is
๏ฐ /2
reflected q ๏ฒ๏ฑ ๏ฝ0 2๏ฐ R sin ๏ฑ cos ๏ฑ ๏จ1 ๏ญ 0.788 cos ๏ฑ ๏ฉ d๏ฑ
๏ฝ
๏ฐ /2
incident
q
2๏ฐ R 2 sin ๏ฑ cos ๏ฑ d๏ฑ
2
๏ฒ๏ฑ
๏ฝ0
๏ฐ /2
๏ฝ
๏ฆ sin2 ๏ฑ 0.788 cos3 ๏ฑ ๏ถ
๏ซ
๏ง
๏ท
3
๏จ 2
๏ธ0
๏ฐ /2
sin2 ๏ฑ
2 0
Answer:
;
1 0.788๏ฐ
๏ญ
2
3
๏ฝ
๏ฝ 0.4747
1
2
.
2.19 A flat metal plate 0.1 m wide by 1.0 m long has a temperature that varies only
along the long direction. The temperature is 900 K at one end, and decreases linearly
over the one meter length to 350 K. The hemispherical spectral emissivity of the plate
does not change significantly with temperature but is a function of wavelength. The
wavelength dependence is approximated by a linear function decreasing from ฮต ๏ฌ =
0.85 at ๏ฌ = 0 to ฮต ๏ฌ = 0.02 at ๏ฌ = 10 ๏ญm. What is the rate of radiative energy loss from
one side of the plate? The surroundings are at a very low temperature.
SOLUTION: The rate of energy loss is given by the total emissive power integrated
over the plate length, or
Q ๏ฝ 0.1(m2 )๏ฒ
๏ฝ 0.1๏ฒ
1
1
๏ฒ
10
x ๏ฝ0 ๏ฌ ๏ฝ0
10
๏ฆ
ฮต ๏ฌ E๏ฌb ๏จT ๏ฉ d ๏ฌdx
2๏ฐ C1
0.83๏ฌ ๏ถ
๏ฒ๏ฌ ๏ง๏จ 0.85 ๏ญ 10 ๏ท๏ธ
d ๏ฌ dx ๏ฝ 416.6W .
๏ฌ
๏ผ
๏ฉ
๏น
C
๏ฏ
๏ฏ
2
๏ฌ 3 ๏ญexp ๏ช
๏บ ๏ญ 1๏ฝ
๏ฏ๏ฎ
๏ซ ๏ฌ ๏จ 350 ๏ซ 550 x ๏ฉ ๏ป ๏ฏ๏พ
where the final result is obtained by numerical integration of the double integral.
Answer: 416.6 W.
x ๏ฝ0
๏ฝ0
2.16
2. Radiative Properties at Interfaces
2.20 A thin ceramic plate, insulated on one side, is radiating energy from its exposed
side into a vacuum at very low temperature. The plate is initially at 1200 K, and is to
cool to 300 K. At any instant, the plate is assumed to be at uniform temperature across
its thickness and over its exposed area. The plate is 0.25 cm thick, and the surface
hemispherical-spectral emissivity is shown in the figure and is independent of
temperature. What is the cooling time? The density of the ceramic is 3200 kg/m3, and
its specific heat is 710 J/(kgโขK).
๏ฅ
1.0
0.85
ฮต ๏ฌ๏
๏ฅ
0
1.0
10
๏ฌ๏ฌ๏ ๏ญm
SOLUTION: For an area A, the energy equation is
๏ฅ
2๏ฐC1
dT
๏ญ V๏ฒc
๏ฝ qA ๏ฝ A ๏ฒ ฮต ๏ฌ
d๏ฌ
๏ฌ๏ฝ 0
dt
๏ฆ C2 ๏ถ ๏น
5๏ฉ
๏ฌ ๏ชexp ๏ง
๏ท ๏ญ 1๏บ
๏จ ๏ฌT ๏ธ ๏ป
๏ซ
0.85 ๏จ ๏ฌ ๏ญ 1๏ฉ
๏ฌ๏ฝ 0
9
๏ฝ A๏ฒ
10
๏ฅ
2๏ฐC1
2๏ฐC1
d๏ฌ ๏ซ A ๏ฒ
0.85
d๏ฌ
๏ฌ๏ฝ10
๏ฆ C2 ๏ถ ๏น
๏ฆ C2 ๏ถ ๏น
5๏ฉ
5๏ฉ
๏ฌ ๏ชexp ๏ง
๏ฌ ๏ชexp ๏ง
๏ท ๏ญ 1๏บ
๏ท ๏ญ 1๏บ
๏จ ๏ฌT ๏ธ ๏ป
๏จ ๏ฌT ๏ธ ๏ป
๏ซ
๏ซ
Integrating gives
๏ฒ cV 1200
t๏ฝ
A ๏ฒ
T ๏ฝ300
10
0.85 ๏จ ๏ฌ ๏ญ 1๏ฉ
๏ฝ1
9
๏ฒ๏ฌ
dT
๏ฅ
2๏ฐ C1
2๏ฐ C1
d
๏ฌ
๏ซ
0.85 5
d๏ฌ
๏ฒ
5
๏ฌ
๏ฝ
10
๏ฌ ๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป
๏ฌ ๏ฉ๏ซexp ๏จC2 / ๏ฌT ๏ฉ ๏ญ 1๏น๏ป
Numerical integration is required, and results in๏ t = 1796 s = 29.9 min. (Note: V/A =
0.25×10-2 m.)
Answer: ๏ t๏ = 29.9 min.
2.17
3. Radiative Properties of Opaque Materials
SOLUTIONS-CHAPTER 3
3.1
An electrical insulator has a refractive index of n = 1.332 and has a smooth
surface radiating into air. What is the directional emissivity for the direction normal to the
surface? What is it for the direction 60o away from the normal?
SOLUTION: From Equation 3.12 with n1 = 1, n2 = 1.332,
ฮตn = 1 – [(n2 – 1)/(n2+1)]2 = 1 – [(1.332- 1)/( 1.332+ 1)] 2 = 0.9797.
Find ๏ฃ from Equation 3.3 using n1 = 1, n2 = 1.332, ๏ฑ = 60ยฐ.
sin ๏ฃ = (n1/n2 ) sin ๏ฑ = (1/1.332) sin 60ยฐ = 0.6502
Then ๏ฃ๏ = sin -1(0.6502) = 40.55ยฐ.
ฮต(60ยฐ) = 1 – ๏ฒ(60ยฐ); using Eq. 3.7a
ฮต(70ยฐ) = 1 – (1/2)[sin2(๏ฑ๏ ๏ญ๏ ๏ฃ)/sin2(๏ฑ๏ ๏ซ๏ ๏ฃ)] {1 + [cos2(๏ฑ๏ ๏ซ๏ ๏ฃ)/cos2(๏ฑ๏ ๏ญ๏ ๏ฃ)]}
= 1 – (1/2)[sin2(19.45ยฐ)/ sin2(100.55ยฐ)] {1+[cos2(100.55ยฐ)/cos2(19.45ยฐ)]}
= 1 – (1/2)(0.1108/0.9665)2 [1 + (-0.1831/0.8891)2] = 0.9405.
Answer: 0.9797; 0.9405.
3.2
A smooth hot ceramic dielectric sphere with an index of refraction n = 1.43 is
photographed with an infrared camera. Calculate how bright the image is at locations B
and C relative to that at A. (Camera is distant from sphere.)
Camera
A
1.5 cm
1.9 cm
B
C
2 cm
3.1
3. Radiative Properties of Opaque Materials
SOLUTION: A blackbody has the same intensity for all directions. Hence, to compare
intensities (brightness), compare the directional emissivities.
๏ฑ๏
1.5
๏ฑ๏
B
2
๏
๏ฑB = sin-1(1.5/2) = 48.590ยฐ; ๏ฃB = sin-1(1.5/2n) = 31.633ยฐ
๏ฑC = sin-1(1.9/2) = 71.805ยฐ; ๏ฃC = sin-1(1.9/2n) = 41.631ยฐ
ฮต(๏ฑ=0) = ๏ฅA = 1 – [(n-1) / (n+1)]2 = 1 – (0.43/2.43)2 = 0.968.
ฮต ๏ฝ 1๏ญ
2
2
1 sin ๏จ๏ฑ ๏ญ ๏ฃ ๏ฉ ๏ฉ cos ๏จ๏ฑ ๏ซ ๏ฃ ๏ฉ ๏น
1
๏ซ
๏ช
๏บ
2 sin2 ๏จ๏ฑ ๏ซ ๏ฃ ๏ฉ ๏ซ cos2 ๏จ๏ฑ ๏ญ ๏ฃ ๏ฉ ๏ป
ฮตB ๏ฝ 1๏ญ
2
2
1 sin ๏จ16.957 ๏ฉ ๏ฉ cos ๏จ 80.223 ๏ฉ ๏น
1
๏ซ
๏ช
๏บ ๏ฝ 0.9548
2 sin2 ๏จ 80.223 ๏ฉ ๏ซ cos2 ๏จ16.957 ๏ฉ ๏ป
2
2
1 sin ๏จ 30.174 ๏ฉ ๏ฉ cos ๏จ113.436 ๏ฉ ๏น
ฮตC ๏ฝ 1 ๏ญ
๏ช1 ๏ซ
๏บ ๏ฝ 0.8182
2 sin2 ๏จ113.436 ๏ฉ ๏ซ
cos2 ๏จ 30.174 ๏ฉ ๏ป
Thus, ฮตB/ฮตA = 0.9548/0.968 = 0.986; ฮตC/ฮตA = 0.8182/0.968 = 0.845.
Answer: 0.986; 0.845.
3.3
A particular dielectric material has a refractive index of n = 1.350. For a smooth
radiating surface, estimate:
(a) the hemispherical emissivity of the material for emission into air.
(b) the directional emissivity at ๏ฑ = 60ยฐ into air.
(c) the directional hemispherical reflectivity in air for both components of
polarized reflectivity. Plot both components for n=1.350 on a graph similar to Figure 3.5.
Let ๏ฑ be the angle of incidence.
SOLUTION:
(a) From Equation 3.11,
ฮต๏ ๏ = (1/2) – [ (3 x 1.350 + 1) x (1.350 -1)]/[6 x (1.350 +1 )2] – {1.3502x(1.350 2 1)2/(1.350 2 + 1)3} ln[(1.350 – 1)/( 1.350 + 1)]
+ 2 x 1.350 x (1.350 2 + 2×1.350 -1)/[ (1.350 2 + 1) x (1.350 4 – 1)]
3.2
3. Radiative Properties of Opaque Materials
– {8 x 1.350 4x (1.350 4 + 1)/[( 1.350 2 + 1) x (1.350 4 – 1)2]} ln (1.350)
= 0.9309
The result may also be found from Figure 3.3.
(b) From Figure 3.2, ฮต(๏ฑ๏ = 60ยฐ) = 0.85. Could also use Equation 3.6 to obtain
๏ฃ๏ฝ39.9ยฐ, substitute ๏ฃ into Equation 3.9 for๏ ๏ฒ( ๏ฑ )=0.0629, and use ฮต = 1 – ๏ฒ. This gives
ฮต(๏ฑ๏ = 60ยฐ) = 0.9371.
(For this particular set of parameters, the hemispherical emissivity and the directional
emissivity at ๏ฑ = 60ยฐ are nearly the same.)
Answer: (a) 0.9309, (b) 0.9371.
3.4
A smooth dielectric material has a normal spectral emissivity of ฮต๏ฌ,n = 0.725 at a
wavelength in air of 6 ๏ญm. Find or estimate values for:
(a) the hemispherical spectral emissivity ฮต๏ฌ๏ at the same wavelength.
(b) the perpendicular component of the directional hemispherical spectral
reflectivity ๏ฒ ๏ฌ ,๏ (๏ฑ) at the same wavelength and for incidence at ๏ฑ = 40ยฐ.
SOLUTION: From Equation 3.12
ฮตn = 4n/(n+1)2 ; n2 + 2n + 1 = 4n/0.725 ; so n = 3.2053 (or 0.3120, but n cannot
be < 1).
a) From Figure 3.3: ฮต/ฮตn = 0.965; so
๏ฅ = 0.965×0.725 = 0.70. Equation 3.8 gives 0.7034
b) From Equation 3.8, ๏ฒ ๏ฌ ,๏ (๏ฑ)
= {[(3.2053 2-0.4132)1/2 – 0.7660]/[( 3.2053 2-0.4132]1/2 + 0.7660]}2 = 0.3694.
Answer: (a) 0.703; (b) 0.369.
3.5
An inventor wants to use a light source and some Polaroid glasses to determine
when the wax finish is worn from her favorite bowling alley. She reasons that the wax
will reflect as a dielectric with n = 1.40, and that the parallel component of light from the
source will be preferentially absorbed and the perpendicular component strongly
reflected by the wax. When the wax is worn away, the wood will reflect diffusely. At
what height should the light source be placed to maximize the ratio of perpendicular to
parallel polarization from the wax as seen by the viewer?
h=?
2m
20 m
3.3
3. Radiative Properties of Opaque Materials
SOLUTION: To make ๏ฒ ๏ / ๏ฒII a maximum, use Brewster's angle (Equation 3.7 et seq.):
๏ฑ๏ = tan-1(n2/n1) = tan-1(1.40)
๏ฑr
2m
h=?
๏ฑ๏
20 – x
x
20 m
Since reflection is specular, ๏ฑr = ๏ฑ๏ฎ๏ Thus, x/2 = tan ๏ฑ๏ = 1.40 = (20-x)/h,
so x = 2.80 m, h = 12.286 m (a high ceiling!!)
Answer: 12.3 m.
3.6
A smooth ceramic dielectric has an index of refraction n = 1.58, that is
independent of wavelength. If a flat ceramic disk is at 1100 K, how much emitted energy
per unit time is received by the detector when it is placed at ๏ฑ = 0ยฐ or at ๏ฑ = 60ยฐ? Use
relations from electromagnetic theory.
0.6 cm
Detector
0.6 cm
0.3 m
n
Detector
๏ฑ๏
0.3 m
Disk
0.5 cm
SOLUTION: Neglect reflected energy from the surroundings. Then the energy reaching
the detector is I cos ๏ฑ d๏ dA, where dA = (๏ฐ x 0.0052/4); d๏ = (๏ฐ x 0.0062/4) (1/0.32).
In the normal direction, ฮตn = 4n/(n+1)2 = 4×1.58/(2.58)2 = 0.9495.
Then the energy for (๏ฑ = 0)
= [๏ณ 11004/๏ฐ] (๏ฐ x 0.0062/4) (1/0.32) (๏ฐ x 0.0052/4) x 0.9495
= 1.5477×10-4 W.
At ๏ฑ = 60ยฐ, ๏ฃ = sin-1 (sin 60ยฐ/1.58) = 33.24ยฐ, so
3.4
3. Radiative Properties of Opaque Materials
๏ฒ ๏จ๏ฑ ๏ฝ 75๏ฐ ๏ฉ ๏ฝ
2
2
1 sin ๏จ๏ฑ ๏ญ ๏ฃ ๏ฉ ๏ฉ cos ๏จ๏ฑ ๏ซ ๏ฃ ๏ฉ ๏น
1
๏ซ
๏ช
๏บ ๏ฝ 0.1021
2 sin2 ๏จ๏ฑ ๏ซ ๏ฃ ๏ฉ ๏ซ cos2 ๏จ๏ฑ ๏ญ ๏ฃ ๏ฉ ๏ป
ฮต(๏ฑ) = 1 – ๏ฒ(๏ฑ) = 0.8979. The energy rate at ๏ฑ =60ยฐ is then
=[๏ณ11004/๏ฐ] cos(60ยฐ) (๏ฐ๏ x 0.0062/4) (1/0.32) (๏ฐ๏ x 0.0052/4) x 0.8979= 7.3183×10-5 W.
Answer: 15.48×10-5 W at ๏ฑ๏ = 0ยฐ; 7.318×10-5 W at ๏ฑ = 60ยฐ.
3.7
At a temperature of 300 K these metals have the resistivities (Table 3.2):
Copper
Gold
Aluminum
1.72 x 10-6 Ohm-cm
2.27 x 10-6 Ohm -cm
2.73 x 10-6 Ohm -cm
What are the theoretical normal total emissivities and hemispherical total emissivities of
these metals, and how do they compare with tabulated values for clean, unoxidized,
polished surfaces?
SOLUTION: Using Equation 3.40 for the normal total emissivity :
ฮตn๏ = 0.578 ( reT)1/2 – 0.178 reT + 0.0584 (reT)3/2
and Eq. 3.45 for the total hemispherical emissivity:
ฮต๏ = 0.776 (reT)1/2 – [0.309 – 0.0889 ln(reT)]reT – 0.0175 (reT)3/2
(reT)1/2
Material
reT
Copper
516×10-6 22.7×10-3
681×10-6 26.1×10-3
819×10-6 28.6×10-3
Gold
Aluminum
ฮตn
๏ ๏ ๏ ฮต๏
11.72×10-6
17.7×10-6
0.0130
0.0171 0.02
0.0150
0.0196 0.018
23.4 x10-6
0.0164
0.0214 0.04
(reT)
3/2
๏ ๏ ๏ ๏ ๏ ๏ ฮตn๏
(Table B.1)
Alternatively, using Equation 3.39 gives for the normal value ฮตcopper = 0.0131;
ฮตgold = 0.0150; ฮตaluminum = 0.0164.
Answer: ฮตn: 0.0131; 0.0157; 0.0168; ฮต: 0.0170. 0.0202, 0.0217.
3.8
A highly-polished metal disk is found to have a measured normal spectral
emissivity of 0.095 at a wavelength of 12 ๏ญm. What is:
(a) the electrical resistivity of the metal (Ohm -cm) ?
(b) the normal spectral emissivity of the metal at ๏ฌ = 10 ๏ญm?
(c) the refractive index n of the metal at ๏ฌ = 10 ๏ญm?
(Note any assumptions that you make in obtaining your answers.)
SOLUTION:
(a) From Equation 3.35 (assumes n = ๏ซ for large wavelength.),
ฮต๏ฌn = 0.095 = 36.5 (re/๏ฌo)1/2 – 464 (re/๏ฌ0); ๏ฌ0 = 12 ๏ญm
Solve for re = 8.7161 x 10-5 Ohm๏ญcm.
3.5
3. Radiative Properties of Opaque Materials
(b) Substitute the re from part (a) into Equation 3.35 with ๏ฌ0 = 10 ๏ญm to obtain
ฮต๏ฌn ( ๏ฌ๏ = 10 ๏ญm) = 0.1037.
(c) Can solve using Equation 3.33 or Figure 3.6:
Using the figure with n = ๏ซ, n = about 18.2. Using Equation 3.33 with
ฮต๏ฌn(๏ฌ๏ = 10 ๏ญm) = 0.1037 and solving for n gives n = 18.26.
Answer: (a) 8.7161 x 10-5 Ohm -cm; (b) 0.104; (c) 18.3.
3.9 Show using Equation 3.7 that the parallel component of reflectivity becomes zero
๏ฆn ๏ถ
when ๏ฑ ๏ฝ tan ๏ญ1 ๏ง 2 ๏ท .
๏จ n1 ๏ธ
SOLUTION:
2
1/2
2
๏ฌ
๏ฉ๏ฆ n ๏ถ 2
๏น ๏ผ
๏ฆ
๏ถ
n
๏ฏ 2 cos ๏จ๏ฑ ๏ฉ ๏ญ ๏ช 2 ๏ญ sin2 ๏ฑ ๏บ ๏ฏ
๏ง ๏ท
i
i
๏ฏ ๏ง๏จ n1 ๏ท๏ธ
๏ช๏ซ๏จ n1 ๏ธ
๏บ๏ป ๏ฏ๏ฏ
๏ฏ
๏ฒ ๏จ๏ฑ i ๏ฉ ๏ฝ ๏ญ
1/2 ๏ฝ
2
๏ฉ๏ฆ n ๏ถ 2
๏น ๏ฏ
๏ฏ ๏ฆ n2 ๏ถ
2
๏ฏ ๏ง ๏ท cos ๏จ๏ฑ i ๏ฉ ๏ซ ๏ช๏ง 2 ๏ท ๏ญ sin ๏ฑ i ๏บ ๏ฏ
๏ช๏ซ๏จ n1 ๏ธ
๏บ๏ป ๏ฏ๏พ
๏ฏ๏ฎ ๏จ n1 ๏ธ
To make ๏ฒ ๏จ๏ฑ ๏ฉ ๏ฝ 0, the numerator must be = 0, so must prove that
1/2
2
๏ฉ๏ฆ n ๏ถ 2
๏น
๏ฆ n2 ๏ถ
2
2
๏ช
๏บ
cos
๏ฑ
๏ญ
๏ญ
sin
๏ฑ
๏จ ๏ฉ ๏ง ๏ท
๏ง ๏ท
n
n
๏ช
๏บ๏ป
๏จ 1๏ธ
๏ซ๏จ 1 ๏ธ
๏ฆn ๏ถ
๏ฝ 0 . Substituting ๏ง 2 ๏ท ๏ฝ tan ๏ฑ gives
๏จ n1 ๏ธ
1/2
2
๏ฉ๏ฆ n ๏ถ 2
๏น
๏ฆ n2 ๏ถ
2
2
๏ช
๏บ
cos
๏ฑ
๏ญ
๏ญ
sin
๏ฑ
๏จ
๏ฉ
๏ง ๏ท
๏ง ๏ท
๏ช๏ซ๏จ n1 ๏ธ
๏บ๏ป
๏จ n1 ๏ธ
1/2
๏ฉ sin2 ๏ฑ
๏น
sin2 ๏ฑ
2
๏ฝ
๏ญ๏ช
๏ญ
sin
๏ฑ
๏บ
cos ๏จ๏ฑ ๏ฉ ๏ซ cos2 ๏จ๏ฑ ๏ฉ
๏ป
๏ฝ tan2 ๏ฑ cos ๏จ๏ฑ ๏ฉ ๏ญ ๏ฉ๏ซ tan2 ๏ฑ ๏ญ sin2 ๏ฑ ๏น๏ป
1/2
1/2
๏ฉ
๏น
sin2 ๏ฑ
1
๏ฝ
๏ญ sin ๏ฑ ๏ช
๏ญ
1
๏บ
2
cos ๏จ๏ฑ ๏ฉ
๏ซ cos ๏จ๏ฑ ๏ฉ ๏ป
1/2
1/2
๏ฌ
๏ฌ
๏ฉ
๏น ๏ผ๏ฏ
๏ฉ 1 ๏ญ cos2 ๏จ๏ฑ ๏ฉ ๏น ๏ผ๏ฏ
1
๏ฏ
๏ฏ
๏ฝ sin ๏ฑ ๏ญ tan ๏ฑ ๏ญ ๏ช
๏ญ 1๏บ ๏ฝ ๏ฝ sin ๏ฑ ๏ญ tan ๏ฑ ๏ญ ๏ช
๏บ ๏ฝ
2
2
๏ซ cos ๏จ๏ฑ ๏ฉ ๏ป ๏ฏ๏พ
๏ซ cos ๏จ๏ฑ ๏ฉ ๏ป ๏ฏ๏พ
๏ฏ๏ฎ
๏ฏ๏ฎ
1/2
๏ฌ
๏ฉ sin2 ๏จ๏ฑ ๏ฉ ๏น ๏ผ๏ฏ
๏ฏ
๏ฝ sin ๏ฑ ๏ญ tan ๏ฑ ๏ญ ๏ช
๏บ ๏ฝ ๏ฝ sin ๏ฑ ๏จ tan ๏ฑ ๏ฑ tan ๏ฑ ๏ฉ ๏ฝ 0
2
๏ซ cos ๏จ๏ฑ ๏ฉ ๏ป ๏ฏ๏พ
๏ฏ๏ฎ
(Must choose negative sign, or reflectivity can be < 0.)
3.10 A clean metal surface has a normal spectral emissivity of ฮต๏ฌ,n = 0.06 at a
wavelength of 12 ๏ญm. Find the value of the electrical resistivity of the metal.
SOLUTION: From Equation 3.23b
3.6

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