
What is Blackbody Radiation and how is it calculated?
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Post Author
espadmin
1. What is a black body?
A [GLOSS]blackbody[/GLOSS] is a perfect absorber of incident radiation. Whatever the wavelength of incident radiation, no thermal radiation is reflected from the surface of a black body. A blackbody also emits the maximum energy possible for any given temperature.
Only a few substances such as carbon black approach the properties of a blackbody. Most surfaces, even many that appear black, are not perfect absorbers or emitters of thermal radiation.
2. How much radiation is emitted from a black body?
The energy emitted from a blackbody surface varies with wavelength. As the body gets hotter, more energy is emitted at shorter wavelengths. Above about
600°C, there is sufficient energy emission within the visible waveband of the spectrum (0.380.76
mm) for the surface to appear red. Planck first derived the relationship between energy emission and wavelength in 1900 from quantum atomic theory. The monochromatic [GLOSS]emissive power[/GLOSS] of a black surface (E_{b},_{l})
at temperature
T(°Kelvin) is given by:
where
c_{1} = 3.74×10^{8} Wmm^{4}m^{2}
c_{2} = 1.4388×10^{4}
mm.K
and l is in micron
(mm) and E_{b},_{l} is in
Wm^{2}mm^{1}
The [GLOSS]spectral emission[/GLOSS] from a black body is plotted in Figure 1 for two temperatures. For a body between 500 and
1500°C more than 98% of the radiant energy falls in the industrially useful near infrared waveband
(0.7625mm). The wavelength corresponding to the peak radiant flux is inversely proportional to temperature T(K) and is given by Wien’s displacement law:
Figure 1. Radiant flux from a black body as a function of wavelength.
3. The StefanBoltzmann equation
Integration of the Planck equation over all wavelengths leads to the well known Stefan Boltzmann equation for [GLOSS]total radiation[/GLOSS] exchange from a black body surface:
where,
s
= the Stefan Boltzmann constant = 5.6687×10^{8} Wm^{2}K^{4}
E_{b} is the total energy emission per unit area over all wavelengths from a point on a black surface to all directions in the hemisphere above it (i.e. into
2p[GLOSS]steradians[/GLOSS] of solid angle)
4. An example
The outside surface of a furnace wall is at 120°C. What is the rate of heat loss from the surface assuming it approximates to a black body? Be careful first to convert temperature to degrees Kelvin ( T(K) =
T(°C)+273.0)
E_{b} = 5.6687×10^{8} x (120+273)^{4} = 1352W/m^{2}
Sources
Thermal Radiation Heat Transfer, Robert Siegel & John R.Howell, 3rd Edition,
1992, Hemisphere, ISBN 0891162712