Chapter 9 Radiation Heat Transfer

Forsberg Heat Transfer
Chapter 9
Radiation Heat Transfer




Electromagnetic Radiation
Electromagnetic radiation wavelengths:
Thermal radiation 0.1 to 100m (microns)
Visible light 0.4 to 0.8m
X-rays 10 −11 to 2 x 10 −8 m
Microwaves 1 mm to 10 m
Radio waves 10 m to 30 km
co =  
co = speed of light in vacuum = 2.9979 x 10 8 m / s
 = wavelength of radiation, m
 = frequency of radiation, / s




1

, Blackbody Emission
A "blackbody" emits radiation at the maximum
possible rate. The emission is given by Planck's Law:
2 hco2  −5
Eb ( ,T ) =
 hc 
exp  o  − 1
  kT 
Eb = spectral emissive power of a blackbody, W / (m2 m)
h = Planck's constant k = Boltzmann's constant
C1  −5
Eb ( ,T ) =
exp ( C2 / T ) − 1
C1 = 3.7417 x 108 (W/m2 )  (m)4
C2 = 1.4388 x 10 4 m  K




Blackbody Spectral Emissive Power




9
Spectral E
8
5800
10
7
6
10
534
1000
10
2 K K
1−10
10300
0−12
−3
0.1




K 1 10 100




10
10Wavele 2

, Stefan-Boltzmann Law
Integrating the Spectral Emissive Power over all
wavelengths, we get the Stefan-Boltzmann Law:

Eb (T ) = 0 E b ( ,T ) d  =  T 4
 = Stefan-Boltzmann constant = 5.670 x 10 −8 W / m2 K4

Terminology:
"Spectral" = parameter depends on wavelength
"Total" = parameter is independent of wavelength




Wien’s Displacement Law

Looking at the spectral-emissive-power figure,
it is seen that the curves have maximums. As the
temperature increases, the maximums move to
a higher wavelength. This is Wien's Displacement
Law: maxT = 2898 m  K
max = wavelength of maximum emission
T = absolute temperature of blackbody, K




3

, For a 5800 K blackbody, max = = 0.500 m
For a 1000 K blackbody, max = = 2.90 m
For a 300 K blackbody, max = = 9.66 m


The sun can be approximated as a blackbody at 5800 K.
The space inside a car or in a room is about 300 K.
Glass has a high transmission at low wavelengths,
but a low transmission at higher wavelengths.
Solar radiation enters the space, but radiation in the
space has difficulty leaving. The greenhouse effect.




Blackbody Radiation Function
Let's say we want the total emissive power for a
blackbody at temperature T for a wavelength
range of 1 to 2 . We want

Eb (1 → 2 ,T ) = 12 E b ( ,T ) d 
The fraction of radiation emitted by a blackbody
in wavelength range 1 to 2 is
 
 Eb ( ,T ) d  1 E b ( ,T ) d 
2 2

F1 →2 = 1 =
0 Eb ( ,T ) d  T 4

 2 C1  −5
 d

1 exp ( C2 / T ) − 1
F1 →2 =
T 4




4