Radiant Emittance
163
Example 10.5. Find the average radiant emittance of the earth and the
sun.
Solution. The earth approximates a blackbody radiator emitting at
288 K. The average emittance of the earth is therefore (5.67 x
W m-2 K - ~ x (288 K ) ~ = 390 w/m2. The sun emittance is approximately that of a blackbody at 6000 K. Using Eq. (10.7) again, the
energy emitted is therefore 73 M W / ~ ' at the surface of the sun.
The energy emitted by nonblackbodies is given by:
@ = / &(A)Eb(h, t) dh
(10.8)
where &(A) is the spectral distribution of emissivity and Eb(h, T) is from
Eq. (10.5). A gray body is one which has no wavelength dependence of
the emissivity so the integration produces:
Natural surfaces are not perfect gray bodies, and the result of applying
Eq. (10.8), is a power of T that is not exactly 4 in the Stefan-Boltzmann
equation. In practice, for the range of normal terrestrial temperatures, all
bodies can be treated as gray bodies and Eqs. (10.7) and (10.9) can be
used with an appropriate average emissivity. This approach also works
for computing atmospheric emittance, even though the atmosphere is far
from a gray body. In the next chapter we show that the emissivity of most
natural surfaces is between 0.95 and 1 .O. For most of our calculations we
assume a value of 0.97. The emissivity of a clear atmosphere, however,
is much lower, as can be seen in Fig. 10.6. Clouds increase the emissivity
of the atmosphere and the emissivity of a completely overcast sky with a
low cloud base is near unity.
Several empirical formulae are available for computing estimates
of clear sky emissivity. One with reasonable theoretical justification is
(Brutsaert, 1984):
where e, is the vapor pressure (Ha) measured at height of one to two
meters and T, is the air temperature (kelvins). The reasoning behind this
formula is that atmospheric thermal radiation is primarily a function of
the water vapor concentration in the first few kilometers of the atmosphere
and is most strongly dependent on the vapor concentration in the first few
hundred meters. Thus a measurement of vapor concentration at a height
of one to two meters, combined with estimates of vapor and temperature
profiles to 5 km, can be used to estimate emissivity.
Since vapor pressure and minimum temperature are strongly correlated, correlations have also been made between temperature at a height
of one to two meters and clear sky emissivity. Swinbank (1963) suggests
163
Example 10.5. Find the average radiant emittance of the earth and the
sun.
Solution. The earth approximates a blackbody radiator emitting at
288 K. The average emittance of the earth is therefore (5.67 x
W m-2 K - ~ x (288 K ) ~ = 390 w/m2. The sun emittance is approximately that of a blackbody at 6000 K. Using Eq. (10.7) again, the
energy emitted is therefore 73 M W / ~ ' at the surface of the sun.
The energy emitted by nonblackbodies is given by:
@ = / &(A)Eb(h, t) dh
(10.8)
where &(A) is the spectral distribution of emissivity and Eb(h, T) is from
Eq. (10.5). A gray body is one which has no wavelength dependence of
the emissivity so the integration produces:
Natural surfaces are not perfect gray bodies, and the result of applying
Eq. (10.8), is a power of T that is not exactly 4 in the Stefan-Boltzmann
equation. In practice, for the range of normal terrestrial temperatures, all
bodies can be treated as gray bodies and Eqs. (10.7) and (10.9) can be
used with an appropriate average emissivity. This approach also works
for computing atmospheric emittance, even though the atmosphere is far
from a gray body. In the next chapter we show that the emissivity of most
natural surfaces is between 0.95 and 1 .O. For most of our calculations we
assume a value of 0.97. The emissivity of a clear atmosphere, however,
is much lower, as can be seen in Fig. 10.6. Clouds increase the emissivity
of the atmosphere and the emissivity of a completely overcast sky with a
low cloud base is near unity.
Several empirical formulae are available for computing estimates
of clear sky emissivity. One with reasonable theoretical justification is
(Brutsaert, 1984):
where e, is the vapor pressure (Ha) measured at height of one to two
meters and T, is the air temperature (kelvins). The reasoning behind this
formula is that atmospheric thermal radiation is primarily a function of
the water vapor concentration in the first few kilometers of the atmosphere
and is most strongly dependent on the vapor concentration in the first few
hundred meters. Thus a measurement of vapor concentration at a height
of one to two meters, combined with estimates of vapor and temperature
profiles to 5 km, can be used to estimate emissivity.
Since vapor pressure and minimum temperature are strongly correlated, correlations have also been made between temperature at a height
of one to two meters and clear sky emissivity. Swinbank (1963) suggests