Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

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solutions MAnuAl FoR
byThermal
radiaTion heaT
Transfer
John r. howell
robert siegel
m. Pinar menguc

Solutions to Homework
Problems
Table of Contents
Chapter 1 Introduction to Radiative Transfer…………..…………….…1.1
Chapter 2 Radiative Properties at Interfaces………….……....…….….2.1
Chapter 3 Radiative Properties of Opaque Materials……..…..……….3.1
Chapter 4 Configuration Factors for Diffuse Surfaces with
Uniform Radiosity….…...……….……..……………………...4.1
Chapter 5 Radiation Exchange in Enclosures Composed of Black
and/or Diffuse-Gray Surfaces………………….…………….5.1
Chapter 6 Exchange of Thermal Radiation among Nondiffuse
Nongray Surfaces…..……………………………....…………6.1
Chapter 7 Radiation Combined with Conduction and Convection
at Boundaries….....…………………...…..…………..……….7.1
Chapter 8 Inverse Problems in Radiative Heat Transfer…….......….…8.1
Chapter 9 Properties of Absorbing and Emitting Media…………….….9.1
Chapter 10 Fundamental Radiative Transfer Relations…..……......….10.1
Chapter 11 Radiative Transfer in Plane Layers and
Multidimensional Geometries………………...……..……...11.1
Chapter 12 Solution Methods for Radiative Transfer in
Participating Media….………………………………..……. .12.1

Chapter 13 Conjugate Heat Transfer in Participating Media……….….13.1
Chapter 14 Electromagnetic Wave Theory……………...….………..….14.1
Chapter 15 Absorption and Scattering by Particles
and Agglomerates…………………………...….………...….15.1
Chapter 16 Near-Field Thermal Radiation……………………..…..……16.1
Chapter 17 Radiative Effects in Translucent Solids,
Windows, and Coatings………………..……….…………...17.1

1.Introduction to Radiative Transfer
1.1
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 m)? 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; 1 = 0.4 x 10-6 m, 2 = 0.7 x 10-6 m,
1 = 1 / 1 = 2.5 x 106 m-1, 2 = 1 / 2 = 1.428571 x 106 m-1,
1 = c0 / 1 = 2.99792458 x 108 / 0.4 x 10-6 = 7.494811 x 1014 s-1,
2 = c0 / 2 = 2.99792458 x 108 / 0.7 x 10-6 = 4.282749 x 1014 s-1,
boundary = 25 x 10-6 m, boundary = 1 / boundary = 4 x 104 m-1,
boundary = c0/boundary = 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 m 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; o = 2.0 x 10-6 m, n = 1.24,
so speed in the vacuum = 2.99792458 x 108 m/s;
0 = c0 / 0 = 2.99792458 x 108 / 2.0 x 10-6 = 1.49896229 x 1014 s-1.
0 = 1 / 0 = 5 x 105 m-1;
(b) Speed in the medium cm = c0 /n = 2.99792458 x 108 /1.24
= 2.417681113 x 108 m/s;
m = 0 = 1.49896229 x 1014 s-1;
m = cm / m = 2.417681113 x 108 / 1.49896229 x 1014 = 1.6129032 x 10-6 m;
m = 1 / m = 6.20 x 105 m-1.
Answer: (a) 2.9979 x108 m/s; 1.4990 x1014s-1; 5x105 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.Introduction to Radiative Transfer
1.2
1.3 Radiation propagating within a medium is found to have a wavelength within the
medium of 1.570 m and a speed of 2.500x108 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 = 2.99792458x108 m/s; m = 1.570x10-6 m; cm = 2.500x108 m/s.
(a) nm = c0/cm = 1.199
(b) 0= nmm = 1.882x10-6 m = 1.882 m.
Answer: (a) 1.199; (b) 1.882 m.
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 0 =0.4 to 0.7 m?
SOLUTION: nm = 1.33; m1 = 0.4/nm = 0.301 m;
m2 = 0.7/nm = 0.526 m.
Answer: 0.301 to 0.526 m.
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. Then 
 
     
 
  2
0 0
0 0 0
1 1 2
( ) ( )
L L
i i
x x
n n kL
t n x dx kx dx L
c c c
c = m ;  = c / m = co / o
co / c = n = o / m; m / o = 1/n so
m / o = n = ni(1 + kx).
Answer :ni
co
L + kL2
2 ; ni(1 + kx).

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1.Introduction to Radiative Transfer
1.3
1.6 Derive Equation 1.26 by analytically finding the maximum of the E
b/T5 vs.
T
relation ( Equation 1.20).
SOLUTION: Take the derivative of E
b/T5 with respect to
T. To simplify notation, let.T
         
       
 
   
5 51
1 25
2
2
6 5 2 2
1 2 2
2
1 2
6
2 2
( / ) 2 2 / exp( / ) 1
( ) ( ) ( )exp( / ) 1
/ ( ) exp( / )
2 [ 5 ] / exp( / ) 1 exp( / ) 1
2 /
[ 5
exp( / ) 1 1 exp( / )
bd E T Cd d
C C
d d dC
C C
C C C
C C
C C

 
 
   
  
  
    
  
  
   
    









 











Setting the result = 0 to find the maximum, 
 
22 max max
2 max 2 max
// 5
1 exp( / ) 1 exp /
C TC
C C T

 

 
     
Clearly, the product (
T)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  3max 2897.8T C m 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 
= 3.75 m?
(b) What is the spectral intensity emitted at 
= 45° with respect to the normal of
the black surface at = 3.75 m?
(c) What is the directional spectral emissive power from the black surface at  =
45° and 
= 3.75 m?
(d) At what 
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 T = 4687.5 m K, 2
1
, /5
2
1
b n C T
C
I e
 


 = 7823 W/m2•m•sr.
(b) Because the intensity from a blackbody is independent of angle of emission,
the result is the same as part (a).
(c) Eb(3.75 m, 45°) = Ib cos 45° = 7823 x 0.7071 = 5532 W/m2•m•sr
(d) From Equation 1.26, maxT = C3, and from Table A.4,
C3 = 2897.8 m K. Thus, max = 2318.2 / 1250 = 2.3182 m. At max,

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1.Introduction to Radiative Transfer
1.4 2 max
1
,max /5
max
2
1
b C T
C
I e



 
= 12499.08 W/m2•m•sr
(e) From Equation 1.32 and Table A-4, Eb = T4
= 5.67040x10-8 W/m2•K4 x (1250)4 K4 = 138,437.5 W/m2
Answers: (a) 7,823W/(m2·m·sr); (b) 7,823W/(m2·m·sr);
(c) 5532 W/(m2·m·sr); (d) 2.3182 m, 12499 W/(m2·m·sr); (e) 138,437 W/m2 .
1.8 Plot the hemispherical spectral emissive power Eb for a blackbody in air
[W/(m2·m)] as a function of wavelength (m) for surface temperatures of 2000 and
6250K.
SOLUTION: Eb = 2C1/ { 5[e(C2/T) - 1]} For a wavelength range of 0.01 to 5 m,
following figures are obtained for different surface temperatures .

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1.Introduction to Radiative Transfer
1.5
1.9 For a blackbody at 2250 K that is in air, find :
(a) the maximum emitted spectral intensity (kW /m2•m•sr ).
(b) the hemispherical total emissive power (kW /m2 ).
(c) the emissive power in the spectral range between o = 2 and 8m.
(d) the ratio of spectral intensity at o = 2 m to that at o = 8 m.
SOLUTION:
(a) Ib,max = C4 T5 = 4.09570x10-12 x 22505 = 236.17 kW / m•m2•sr
(b) Eb = T4 = 5.67040x10-8(W/m2•K4) x (2250)4 (K4) = 1453.26 kW/m2
(c) Using Equation 1.37, for 2T = 18,000 m•K, F0 2T = 0.97766 and for
1T = 4500 m•K, F0 1T = 0.56429
F2T  1T = 0.41336
Eb(12) = 0.3482 T4 = 600.730 kW/m2
(d) Using 2
1
, /5
2
1



 
b n C T
C
I e ,
[ Ib(=2) ] / [Ib(=8) ] = 1.5861x105/ 2.9695x103 = 53.4139
Answers: (a) 236.17 kW/m2•m•sr; (b) 1453.26 kW/m2;
(c) 600.730 kW/m2; (d) 53.4139

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1.Introduction to Radiative Transfer
1.6
1.10 Determine the fractions of blackbody energy that lie below and above the peak of
the blackbody curve.
SOLUTION: At the peak, maxT = 2893 mK. Using Equation 1.37 with2 14388( ) 4.965
2897.8( )
C m K
T m K

 


  

gives0 2897.8 0.25005F   . Therefore about 25 percent of
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
= 2.0m to the spectral intensity at 
= 5m?
(b) What fraction of the blackbody emissive power lies between the wavelengths
of 
= 2.0m and 
= 5m?
(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  
 5 m?
SOLUTION:
(a) 1T = 2.0 x 1250 = 2500 m K; 2T = 5 x 1250 = 6250 m K
(Note that these values bracket the peak in the blackbody curve.)
Using 2
1
5
2
1/b C T
C
I e



  ,
Ib /Ib = 11823.5 / 4237.4 = 2.7903 (b) From Equation 1.37, F01T = F02500
= 0.16136,
F02T = F07250 = 0.75792. The fraction between 1T and 2T is then=
((0.75792- 0.16136) = 0.59656.
(c) From Table A.4, C3 = 2897.77 m K, so max = 2897.77 / 1250 =
2.31824m.
(d) From Table 1.2, the band emission is
( F02T - F01T ) T4 = 0.59656 x 5.67040x10-8(W/m2•K4) x (1250)4 (K4)
= 82587.2 W/m2
Answer: (a) 2.7903; (b) 0.59656.; (c) 2.31824m; (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 m,
estimate the solar radiating temperature.

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1.Introduction to Radiative Transfer
1.7
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 149x106 km and
the diameter of the sun as 1.392x106 km.
SOLUTION:
a)   
      

3 solarmax
2898 m K
T C 2898 m K; T 5796K
0.50 m
b)   
   
1/ 4
2 2
2 4Sun Sun solar
solar Sun Sun Sun2 2 2
Earth orbit Earth orbit Sun
2
Earth orbit
1/ 4
2
25 2
8 2 4
26 2
R R q
q 1353(W / m ) q T : T
R R R
R
1368(W / m ) 5766K
1.392 10 / 2 km
5.6704 10 (W / m K )
149 10 km

 
 
       
  
 
 
 
 
 
  
 
 
 
 
Answers: a) 5796K; b) 5766 K.
1.13 The surface of the sun has an effective blackbody radiating temperature of 5780
K.
(a) What percentage of the solar radiant emission lies in the visible range  = 0.4
to 0.7 m?
(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) 1T = 5780 K x 0.4m = 2312 m K ;
2T = 5780 K x 0.7m = 4046 m K
F = F02T - F01T = 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 m.
1T = 5780 K x 0.01m = 57.8 m K; 2T = 5780 K x 0.4m = 2312 m K.
F = F02T - F01T = 0.12240 - 0 = 0.12240, or 12.2 %

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1.Introduction to Radiative Transfer
1.8
(c) The maximum energy is at max, where, from Table A.4, C3 = 2897.8 m
K, so max = 2897.8 / 5780 = 0.50134 m.
The corresponding frequency is
max = co / = 2.9979x108 (m/s) / 0.50134 x 10-6 (m) = 5.9798x1014 Hz
(d) At max T = C3 = 2897.8 m K, using  

  2 max
1
max b C / T5
max
2 C
E e 1
gives Emax,b) =8.301x107 W/(m2·m).
Answer: (a) 36.7%; (b) 12.2%; (c) 0.5013 m, 5.98x1014 Hz;
(d) 8.301x107 W/(m2·m).
1.14 A blackbody radiates such that the wavelength at its maximum emissive power
is 2.00 m. What fraction of the total emissive power from this blackbody is in the range
= 0.7 to 
= 5 m?
SOLUTION: Wien's law gives max T = C3 = 2897.8 m K, so
T = C3/max= 2897.8 / 2= 1448.9 K. Thus, 1T = 1448.9 K x 0.7m = 1014.2 m K;
2T = 1448.9 K x 5m = 7244.5 m K. Using Equation 1.37, F = F02T - F01T
=
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/(m2·m)
[0.0400W/(m2·m)] at a wavelength of 90 m. What is the wavelength for the maximum
emissive power of this blackbody?
SOLUTION: With Eb and  given, solve for T from Planck's equation,
Eb = 2C1/ { 5[e(C2/T) - 1]} which becomes T = C2/{ln [1+(2C1/5 Eb)]}
Substituting numerical values gives T = 154.7 K, and Wien's displacement law, maxT
= C3, is used to find max = 18.73 m.
Answer: 18.73 m.
1.16 A radiometer is sensitive to radiation only in the interval 3.6 
 8.5m. 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?

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1.Introduction to Radiative Transfer
1.9
SOLUTION:
 m T, m K F 0-T
3.6 3600 0.4036
8.5 8500 0.87417
F1T2T = 0.87417 - 0.4036 = 0.47058
Eb, = F1T2T T4 = 0.47058 x 5.67040x10-8 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   0.7), and continue until the fraction is
0.25.
For example, for the lower temperature, try T = 4600 K. Then 1T = 4600 K x
0.4m = 1840 m K; 2T = 4600 K x 0.7m = 3220 m K. Using Equation 1.37, F =
F02T - F01T = 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 that0.7
0
4
( ) 0.25 0
bE T d
T
  






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1.Introduction to Radiative Transfer
1.10
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b increases with T at any fixed
value of 

SOLUTION: From Equation 1.15, Ib = 2C1/{5 [ e C2/T - 1 ] }. The change in Ib
with T for a fixed  is:2
2
2 2
2
1 2
1 2
2 26
5 2
2 2
1 1
C
CT
T
b
C C
T T
C
C e
I C C eT
T e T e






 

 
       
     
    
   
The values of C1, C2, and  are all positive, so that (Ib / T)is always positive.
Hence, Ib must increase with T for every 
There are other ways to carry out this proof, such as plotting Ib or showing that
Ib 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.

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1.Introduction to Radiative Transfer
1.11
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?
SOLUTION: From Table 1.2, the fraction of radiation to an annular disk is:1
2
2
21
2
2
2
sinsin
2
sinsin2 



where sin 1 = 1/(32 + 12)1/2 ; sin 2 = 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1 = 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m, 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 °C, what
percent of this radiation will be transmitted through the glass?
1.5 cm
Hole in
furnace wall
1 cm
3 cm
2
1
1.5
1
.
5
3

Loading page 15...

1.Introduction to Radiative Transfer
1.12
SOLUTION: Part 1; 1T = 0.38m x 5780 K = 2196.4 m K; 2T = 2.7m x 5780 K=
15606m K. Using Equation 1.37, F = F02T - F01T = 0.96951 - 0.10022 =
0.86929. The percent transmitted is then 0.86929 x 0.85 x 100 = 73.9 %.
Part 2: 1T = 0.38m x (40 + 273) K = 118.94 m K; 2T = 2.7m x 313 K=
845.1m K. Using Equation 1.37, F = F02T - F01T = 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/
max = 5099.4
m K.
SOLUTION: From Equation 1.17, Eb = 2C13/[eC2/T - 1 ]. To obtain the maximum,
differentiate with respect to , 
2 2
2
2 3
1 1 2
2
6 1 2 /
1
C C
T T
b
C
T T
C e C C T e
E
e
 
  
    
     
 




 
 

Setting this equal to zero givesT
C
max2T
C max2max2
e
T
C
1e3
 









for which the solution is (T/max ) = constant. Substitution of
(T/max) = 5099.4 m K shows this solution to be valid.
Garden at
40 C
Solar
radiation
Glass

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Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

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