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Solutions for Thermal Radiation Heat Transfer, 7th Edition Howell (All Chapters included)

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Complete Solutions Manual for Thermal Radiation Heat Transfer, 7th Edition by John R. Howell, M. Pinar Mengüc, Kyle Daun, Robert Siegel ; ISBN13: 9780367347079. (Full Chapters included Chapter 1 to 19).... 1. Introduction to Radiative Transfer. 2. Radiative Properties at Interfaces. 3. Radiative ...

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Thermal Radiation Heat Transfer
7th Edition by John R. Howell



Complete Chapter Solutions Manual
are included (Ch 1 to 19)




** Immediate Download
** Swift Response
** All Chapters included

, 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–
0.7 μm)? What are the wave number and frequency values at the spectral boundary (25 m)
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 .4 x 10-6 = 7.494811 x 1014 s-1,
2 = c0  2 = 2.99792458 x .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 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 Radiation propagating within a medium is found to have a wavelength within the medium of 1.633
μm and a speed of 2.600 × 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 = 2.99792458x108 m/s; m = 1.633x10-6 m; cm = 2.600x108 m/s.
(a) nm = c0/cm = 1.153
(b) 0= nmm = 1.883x10-6 m = 1.883 m.
Answer: (a) 1.153; (b) 1.883 m.

1.3 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
ni  kL2 
 
L L
1 ni
t= n( x )dx = (1 + kx )dx =
L + 
c0 x =0 c0
x =0 c0  2 
c = m; = c / m = co / o; co / c = n = o / m; m / o = 1/n so m / o = n = ni(1 + kx).
2
ni kL
Answer : L+ ; ni(1 + kx).
co 2

1.4 Derive Equation 1.30 by analytically finding the maximum of the Eλb/T 5 versus λT relation (Equation
1.24).




1.1

, 1.Introduction to Radiative Transfer


SOLUTION: Take the derivative of Eb/T5 with respect to T. To simplify notation, let  = T .

d  
d (E  b / T 5 )
d( )
= 
2 C1
 = 2 C1
d (  )  (  )  exp(C2 /  ) − 1 
5
d
d( )
 
(  ) / exp(C2 /  ) − 1
−5


 

 (
= 2 C1 [ −5 (  ) ] /  exp(C2 /  ) − 1 − (  )
−6 −5 −C2 / (  )
2
)
exp(C2 /  ) 

exp(C2 /  ) − 1 
2

2 C1  C2 /  
= 6 [−5 + 
  exp(C2 /  ) − 1  1 − exp( −C2 /  ) 
C2 / max C2 / ( T )max
Setting the result = 0 to find the maximum, = =5
1 − exp( −C2 / max ) 1 − exp −C2 / ( T ) 
 max 

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 ( T )max = C3 = 2897.8  m  K
.

1.5 A blackbody is at a temperature of 1350 K and is in air.
(a) What is the spectral intensity emitted in a direction normal to the black surface at λ = 4.00 μm?
(b) What is the spectral intensity emitted at θ = 50o with respect to the normal of the black surface at
λ = 4.00μm?
(c) What is the directional spectral emissive power from the black surface at θ = 50o and λ = 4.00
μm?
(d) At what λ 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:
2C1
(a) At T = 5400 m K, Ib,n = = 8706 W/m2•m•sr.
 e
5
( C2 / T
−1 )
(b) Because the intensity from a blackbody is independent of angle of emission, the result is the
same as part (a).
(c) Eb(4 m, 50°) = Ib cos 50° = 8706 x 0.6428 = 5596 W/m2•m•sr
(d) From Equation 1.30, maxT = C3 = 2897.8 m K. Thus, max = 2897. = 2.1465 m.
2C
At max, Ib,max = 1
= 18,365 W/m2•m•sr
5
max (e
C2 / maxT
)
−1

(e) From Equation 1.35 and Table A-4, Eb = T4 = 5.67040x10-8 W/m2•K4 x (1350)4 K4
= 1.883x105 W/m2
Answers: (a) 8706 W/(m2·m·sr); (b) 8706 W/(m2·m·sr); (c) 5596 W/(m2·m·sr);
(d) 2.1465 m, 18365 W/(m2·m·sr); (e) 188.34 kW/m2.

1.6 For a blackbody at 2450 K that is in air, find
(a) The maximum emitted spectral intensity (kW/m 2·μm·sr)
(b) The hemispherical total emissive power (kW/m 2)
(c) The emissive power in the spectral range between λo = 1.5 and 7 μm
(d) The ratio of spectral intensity at λo = 1.5 μm to that at λo = 7 μm




1.2

, 1.Introduction to Radiative Transfer

SOLUTION:
(a) Ib,max = C4 T5 = 4.09570x10-12 x 24505 = 361.54 kW / m•m2•sr
(b) Eb = T4 = 5.67040x10-8(W/m2•K4) x (2450)4 (K4) = 2043.1 kW/m2
(c) Using Equation 1.41, for 2T = 17,150 m•K, F0- 2T = 0.97815 and for 1T = 3675 m•K,
F0- 1T = 0.41877; F2T - 1T = 0.55938; Eb(1-2) = 0.55938 T4 = 1142.8 kW/m2
2C1
(d) Using Ib,n = , Ib(=1.5)] / [Ib(=7)] = 3.1911x105/ 5.3935x103 = 59.165
 e
5 C2 / T
(−1 )
Answers: (a) 361.54 kW/m2•m•sr; (b) 2043.1 kW/m2; (c) 1142.8 kW/m2; (d) 59.17

1.7 Determine the fractions of blackbody energy that lie below and above the peak of the blackbody
curve.
C 14388(  m  K )
SOLUTION: At the peak, maxT = 2893 mK. Using Equation 1.41 with   2 = = 4.965
T 2897.8(  m  K )
gives F0−2897.8 = 0.25005 . Therefore, about 25 percent of the blackbody energy is at wavelengths below
the peak, and 75 percent is at longer wavelengths.

1.8 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 λ = 0.4–0.7 μm?
(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. 1.4, 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 %
(c) The maximum energy is at  max, where, from Table A.4, C3 = 2897.8 m K, so
max = 2897. = 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 C1
E max b = gives Emax,b =8.301x107 W/(m2·m).

5
max (e
C2 / maxT
)
−1

Answer: (a) 36.7%; (b) 12.2%; (c) 0.5013 m, 5.98x1014 Hz;
(d) 8.301x107 W/(m2·m).

1.9 A blackbody has a hemispherical spectral emissive power of 0.0390 W/(m2·μm) at a wavelength of
85 μm. 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,




1.3

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