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Jacaranda (Engineering) 3333 Mail Code Phone: 818.677. E-mail: lcaretto@csun.edu 8348 Fax: 818.677.
College of Engineering and Computer Science Mechanical Engineering Department
Mechanical Engineering 37 5 Heat Transfer Spring 2007 Number 17629 Instructor: Larry Caretto
May 2 Homework Solutions
12 - 14 A cordless telephone is designed to operate at a frequency of 8.5x10^8 Hz. Determine the wavelength of these telephone waves.
We can use the relationship between wavelength, frequency, and the speed of light, c = 2.998x10^8 m/s.
Hz s
x Hz
s
x m c 1
- 5 10
8
8
0.353 m
12 - 20 Consider a cube, 20 cm on a side, that is suspended in air and emitting radiation, that closely approximates a black body, at T = 750 K. Find (a) the rate at which the cube emits energy in W, (b) the spectral black-body emissive power at a wavelength of 4 m.
We can use the Stefan-Boltzmann formula for the emissive power of a black-body: E = AT^4 , where the area of the cube is six times the area of each side = 6(0.2 m)^2 = 0.24 m^2. The radiation then is
2 4 2 4
8 4
- 24 750
m K m K
x W
E AT 4306 W
The spectral radiation is found from the following formula
(^51). 43878 4 750
2
4
5
1 b , C 2 T mK m K m e
m
W m
e
C
E
3045 W/m^2 m
May 2 homework solutions ME 3 75 , L. S. Caretto, Spring 200 7 Page 2
12 - 48 The emissivity of a surface coated with aluminum oxide can be approximated to be 0.
for radiation at wavelengths less than 5 m and 0.9 for radiation greater than 5 m. Determine the average emissivity of this surface at (a) 5800 K and (b) 300 K. What can you say about the absorptivity of this surface for radiation coming from sources at 5800 K and 300 K?
For both temperatures the average emissivity, , is given by the general equation (which is the first integral on the right) that is subsequently applied to the profile in this problem. In this equation 1 = 0.15 and 2 = 0.9.
f ^ m T ^ ^ f ^ m T ^ ^ ^ f ^ m T
E d T
E d T
E d T (^) m
b
m
b b
1 2 2 1 2
5
4 2 ,
5
0
4 1 , 0
4 ,
(^)
For T = 5800 K, ( 5 m)T = (5 m)( 58 00 K) = 29,000 mK; at this value of T, f = 0.994715 from interpolation in Table 12-2 on page 672 of the text. Thus we have the following result for T = 5800 K.
5800 K 2 0. 994715 1 2 0. 9 0. 994715 0. 15 0. 9 = 0. 15
For T = 300 K, (5 m)T = (5 m)(300 K) = 1500 mK; at this value of T, f = 0.013754 from interpolation in Table 12-2. Thus we have the following result for T = 300 K.
(^300) K 2 0. 013574 1 2 0. 9 0. 013574 0. 15 0. 9 = 0.
From Kirchoff’s law we know that the spectral absorptivity equals the spectral emissivity (assuming a diffuse surface): = l. Since we have integrated over the entire black-body radiation spectrum here we know that the total absorptivity will equal the total emissivity. Thus
= 0. 15 for radiation at 5800 K and = 0.89 for radiation at 300 K.
12 - 50 E A 5 - in diameter spherical ball is known to emit radiation at a rate of 550 But/h when its surface temperature is 950 R. Determine the average emissivity of the ball at this temperature.
We can use the formula for the emissive power of a body with a given emissivity : E = AT^4 , where the surface area of the ball equals D^2. Solving for the emissivity and substituting values gives the result as follows.
4
2
2 4
(^4248)
950 12
R
in
ft in h ft R
x Btu
h
Btu
DT
E
AT
E
