Emissivity and Absorption
Suppose two objects are in a hollow (Fig. 4.30): object A, which is black, and object B, which is gray (a body that does not absorb all the incoming radiation). The energy and mass are in balance when the temperatures of A, B, and C are equal. In the balanced state the radiation emitted by the bodies is equal to the radiation received.
The radiation density in C is in constant balance at all points and in all directions (for a given frequency). The radiation density is
FIGURE 4.29 Radiation intensity of a blackbody as a function of wavelength (temperature parameter).
When the radiant energy d<t> passes through a surface element dA in the direction of its normal vector, in a space angle element dio.
For a black object,
Lv Lmv dv=>Lv = Lmv. (4.206)
A gray object absorbs only part of the incoming radiation: incoming = aLv dv= outgoing = eLmv dv.
From Kirchhoff’s law,
A( v, T, ft) = e( v, T, ft) (4.207)
Absorption ratio = emissivity ft = direction angle to the surface normal
The radiation intensity from a surface to a semispace is temperature dependent on
«Pemis = *Mm = je(v)Mmv dv (4.208)
Mm = vT4 -8 W
(T = 5.67032 x 10 —j—4 j the Stefan-Boltzmann constant, m K
The radiation intensity absorbed by the surface is
<Pabs = aE = |a( v)E„ dv (4.209)
E = radiation intensity = incoming 4>M.
Emissivity is strongly dependent on the surface quality. The emissivity of a rough surface is greater than that of a smooth surface, increasing the rate of absorption. Emissivity values are found in textbooks. Care must be taken when using these values, as they usually denote total emissivities. The emissivity is considered constant in the spectrum, and this may be a poor approximation.
For example, the emissivity of white paper is high ( = 0.93 ) at room temperature and, according to Kirchhoff’s law, the absorption factor at the same temperature is also high. At relatively low temperature, as in this case, the radiation is concentrated in the long wavelengths, according to Wien’s law. However, when the same paper receives radiation from the sun, at a radiation temperature of 6000 K, when the absorption factor is small, this radiation has a short wavelength. A white object is a good reflector; this is why white clothing is used in the tropics.
When a low-temperature heating radiator is painted, the color is selected according to heat radiation. At this relatively low temperature, the radiation lies almost outside the visible region. The color may be deceptive. Snow is a good reflector of visible radiation; still, its total emissivity and therefore its absorption in normal conditions is as high as 0.98. Visible radiation passes through glass; its emissivity at the temperature 20 °C is 0.98. Glass radiation lies within the 300-2800 nm range. An important feature of glass is that it is opaque to longwave radiation, which is produced by low-temperature emitters. It is this phenomenon that is termed the greenhouse effect.
The radiation emitted by the sun, due to its high temperature, has a short wavelength. Glass is transparent at this wavelength, allowing the radiation to pass through into the interior of the building. This energy is absorbed by the room surfaces, causing them to rise in temperature and to become low — temperature emitters. The radiation from these low-temperature surfaces is longwave, to which glass is opaque, and thus the radiation cannot escape through the glass to the outdoors, resulting in a rise in the space temperature. The transmission of radiation through the glass depends on the spectral characteristics of the nature of the glass.
Silica glass transmits 92% of the radiation in the wavelength region of 0.3-2.7 jim, and it is impervious to other radiation. Determine the wavelength of the sun’s radiation that the glass transmits when the sun is treated as a blackbody, T = 5600 K. What happens for a blackbody at a temperature of 295 K?
According to Eq. (4.204), at the median point of the spectral energy, As0, T = 4107 |j, m K, and therefore the median point of the spectral energy of the sun is
A50 = 4107 jim K/5600 K = 0.733
Calculations using Planck’s radiation law show w’hich part of the radiation energy remains in the wavelength range:
A50/A, = 0.733/0.35 = 2.095 —» ^ [ ^MАdA = 0.06
A50/A2 = 0.733/2.7 = 0.2716 —> ^(*MAdA = 0.97
Thus 97% — 6 % = 91 % of the radiation energy lies in the range, and the silica glass transmits 91 % • 92% = 83.7% of the radiation energy.
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