LumaSense Technologies M67 User Manual
Page 8
M67, M67S Manual
7
Figure 2 - Transmission, Absorption, and Reflection of Infrared Energy.
Figure 2 shows the three modes by which the radiant energy striking an object may be dissipated:
a= absorption
t= transmission
r= reflection
The fractions of the total radiant energy, which are associated with each of the above modes of dissipation, are referred to as the
absorptivity (a) transmissivity (t) and the reflectivity (r) of the body. According to the theory of conservation of energy. The
extent to which materials reflect, absorb and transmit IR energy is know as the emissivity of the material.
Therefore,
1
t
a
r
The emissivity of a body is defined formally by the equation below, as the ratio of the radiant energy emitted by the body to the
radiation, which would be emitted by a blackbody at the same temperature. A blackbody is a theoretical surface, which absorbs
and re-radiates all the IR energy it receives. It does not reflect or transmit any IR energy. Perfect blackbody surfaces do not
exist in nature. The importance of emissivity in practical infrared thermometry is so important that it is discussed separately in
section 9, page 33 of this manual.
e
W
W
o
bb
Where,
W
o
= total radiant energy emitted by a body at a given temperature T.
W
bb
= total radiant energy emitted by a blackbody at the same temperature T.
If all energy falling on an object were absorbed (no transmission or reflection), the absorptivity would equal to 1. At a steady
temperature, all the energy absorbed could be re-radiated (emitted) so that the emissivity (e) of such a body would equal 1.
Thus,
a
e
(always): Kirchoff's Law
Objects whose emissivity equals 1 are called blackbodies.
The total radiant energy emitted by a body is proportional to the forth power of its absolute temperature:
W
e T
4
: Stefan-Boltzmann Law
where:
W = total radiant energy intensity emitted Watts/cm2.
T = absolute temperature of body, Kelvins
e = emissivity
= proportionality constant, called the Stefan-Boltzmann constant
(5.67 x 10
-12
Watts/cm
2
/K
4
)
Note that temperature in Kelvin = ºC + 273.15
= (ºF - 32)/1.8 + 273.15