College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Exercise Eleven – Fundamentals of Radiation
1. The variation of spectral
transmissivity of a 0.6-cm thick
glass window is shown in the figure
at the right. Determine the average
transmissivity for this window for
solar radiation (T = 5,800 K) and
radiation coming from surfaces in
the room (t = 300 K). Also,
determine the amount of solar
radiation transmitted through the
window for incident solar radiation
of 650 W/m2. (Problem and figure
P12.51 from Çengel, Heat and Mass
Transfer.)
Here we have to evaluate the average transmissivity, , which is given by the general equation
(which is the first integral on the right) that is subsequently applied to the profile in this problem.


1
1
 E d  4
4   b ,
T 0
T
0.3 m
 0Eb, d 
0
1
T 4
3 m
 0.92Eb, d 
0.3 m
1
T 4

 0E
b ,
d
3 m
The non-zero part of this equation is simply 0.92 times the fraction of total radiation between the
wavelengths of 1 = 0.3 m and 2 = 3 m. We can find this from the Table 12-2 on page 672 of
the text which gives the fraction f as a function of the product T.

3 m
0.92
Eb, d  0.92 f  2T   f  1T 
T 4 0.3m
For T = 5800 K, 1T = (0.3 m)(5800 K) = 1740mK and 2T = (3 m)(5800 K) = 17400mK.
Interpolation in Table 12-2 gives
f  1T   f  1740 m  K   0.019718 
f  2T   f  17400 m  K   0.019718 
.039341  .019718
1740  1600  0.033454
1800  1600
.980860  .973814
17400  160  0  0.978746
18000  16000
Thus the total transmissivity of the surface for radiation with a source temperature of 5800 K is
0.978746 – 0.033454 = 0.870 .
If the temperature is 300 K, 1T = (0.3 m)(300 K) = 90mK and 2T = (3 m)(300 K) =
900mK. Interpolation in Table 12-2 gives
f  1T   f  90 m  K   0.0 
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
.0.0  0.0
90  0  0.0
200  0
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise eleven solutions
ME 375, L. S. Caretto, Spring 2007
f  1T   f  900 m  K   0.000016 
Page 2
0.000321  0.000016
900  800  0.00016
1000  800
Thus the total transmissivity of the surface for radiation with a source temperature of 5800 K is
0.00016 – 0.000000 = 0.00016 .
The amount of incoming solar radiation transmitted through the glass uses the transmissivity for a
radiation source at 5,800 K. So, if the incident radiation is 650 W/m 2, then the transmitted
radiation is 0.870 times this or 566 W/m2 .
2. The variations of spectral emissivity of
two surfaces are shown in the figure at
the right. Determine the average
emissivity of each surface at T = 3000 K.
Also determine the average absorptivity
and reflectivity of each surface for
radiation coming from a source at 3000
K. Which surface is more suitable to
serve as a solar absorber? (Problem and
figure P12.47 from Çengel, Heat and
Mass Transfer.)
For both surfaces 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.

3 m

1
1
1
  4    Eb, d  4   1 Eb, d  4   2 Eb, d 
T 0
T 0
T 3 m
 1 f  3 m T    2 1  f  3 m T    2   1   2  f  3 m T 
For T = 3000 K, (3 m)T = (3 m)(3000 K) = 9000 mK; at this value of T, f = 0.890029 in
Table 12-2 on page 672 of the text. Thus we have the following computational equation.
   2   1   2  f  3 mT    2  0.890029 1   2 
For the surface, with 1 = 0.2 and 2 = 0.9
   2  0.8900291   2   0.9  0.8900290.2  0.9 = 0.28
By Kirchoff’s law  =  so  = 0.28 for this surface.
To determine the reflectivity for this surface we use the relationship that  +  +  = 1. We do not
have any data on the transmissivity so we cannot really apply this equation. We can get an
estimate of the maximum reflectivity by assuming that the surface is opaque; that is  = 0. In this
case  +  = 1, so  = 1 –  = 1 – 0.28 or  = 0.72 .
For the surface, with 1 = 0.8 and 2 = 0.1
   2  0.8900291   2   0.1  0.8900290.8  0.1 = 0.72
By Kirchoff’s law  =  so  = 0.72 for this surface.
Exercise eleven solutions
ME 375, L. S. Caretto, Spring 2007
Page 3
To determine the reflectivity for this surface we again assume that  = 0.to get a maximum value
for . For this surface,  = 1 –  = 1 – 0.72 or  = 0.28 .
The surface with 1 = 0.8 and 2 = 0.1 is a better surface for a solar collector. It has a high
absorptivity in the low wavelength region where solar energy is concentrated and a low emissivity
at longer wavelengths where most of the radiation emitted by the solar collector would occur.
3. A solar collector has a surface whose solar absorptivity is 0.87 and whose emissivity at
terrestrial temperatures is 0.09. At a time when the air temperature is 25oC and the
effective sky temperature is 15oC, the collector is shut off so there is no heat transfer out
of the back of the collector. Determine the equilibrium temperature of the collector
surface if the incident solar radiation is 600 W/m2 and the convective heat transfer
coefficient is 10 W/m2oC.
If there is no heat transfer out of the back of the collector the net heat transfer into the front of the
collector must be zero. This consists of the solar irradiation absorbed by the collector minus the
heat lost by convection and radiation to the environment. Setting this heat balance gives the
following relationship among heat fluxes entering the collector.
4
 s Gsolar  hT  Ts    Tsky
 Ts4   0
Substituting the given problem data into this equation, using absolute temperatures for the
radiation calculations, gives
4
 s Gsolar  hT  Ts    Tsky
 Ts4   0
0.87 6002W 
m


10 W
5.670 x10 8 W




288.15 K 4  Ts4  0
298
.
15
K

T

0
.
09
s
2
2
4
m K
m K
This equation for Ts cannot be solved explicitly. Using numerical software such as the goal seek
method of Excel gives Ts = 346 K .