Normally, to get the total radiative heat flux (qr) in W/m2 from the spectral radiative intensity (Ilamda) in W/(m2*um*str), Ilamda is integrated twice--with respect to solid angles and wavelengths.
In case Ilamda represents only solar irradiance (e.g., from the AM1.5G spectrum) incident on a surface or volume, how is the integral with respect to solid angles going to be carried (if any)?
For example, will integration with respect to solid angle (if any) in this case result in a factor of 4*pi (or 2*pi or pi)?
In other words, is there a need to integrate with respect to solid angle when we want to get the total radiative heat flux (in W/m2) from the AM1.5G spectrum?
The sun is an isotropic radiator for all practical purposes. Its irradiance is typically given in (W/m^2/nm) since the distance sun-earth is reasonably constant. One obtains the total flux received by a surface through integration over the wavelength range of interest and simply multiplying by the projected area of the surface normal to the direction of the incoming irradiance.
We can consider the Sun as a Lambertian source of light, with a spectral radiance L(lambda) expressed in W/(m2*um*sr). Pay attention: here, m2 refers to the Sun surface, not the Earth surface. Moreover, by definition of Lambertian surface, the spectral radiance L(lambda) of the Sun is independent on the direction of emission. To obtain the total flux emitted by the Sun, you have to integrate the spectral radiance L(lambda) over the spectrum, obtaining the radiance L in W/(m2*sr), after over the solid angle, obtaining the emittance M in W/(m2), and finally over the surface, obtaining the total flux Ftot in W: Ftot(W)=S*M=S*pi*L= 3.84*1026W. The radiance L can be derived as:
L=INT[dlambda*L(lambda)]=Ftot(W)/(pi*S)= Ftot(W)/[pi*( pi*D2)]= 3.84*1026/[pi2*(1.392*109)2]=2*107W/(m2*sr),
where D is the Sun diameter. The irradiance on the Earth, outside the atmosphere, will be:
Eext= Ftot(W)/(4pi*d2)= 3.84*1026/[4pi*(1.496*1011)2]=1367W/m2, where d is the Sun-Earth distance.
The irradiance on the Earth ground is attenuated by the atmosphere and becomes (at AM1.5): E=1000W/m2=INT[dlambda*E(lambda)], where E(lambda) is the spectral irradiance in W/(m2*um), that is the AM1.5G spectrum.
When you ask: “is there a need to integrate with respect to solid angle when we want to get the total radiative heat flux (in W/m2) from the AM1.5G spectrum?”, I answer: “No, because the AM1.5G spectrum does not contain the information on the solid angle”. The integration on the solid angle is needed, and in fact it was already made, but on the spectral radiance L(lambda) of the Sun, not starting from the quantity that you call Ilamda.
Thank you, Dr. Parretta, for the beautifully put explanation.
My problem was assuming that AM1.5 is a measure of spectral radiative intensity (W/(m2*str*um)) while it is actually a measure of spectral radiative heat flux (W/(m2*um)) since it is readily integrated over the sun's surface area.
I was probably confusing AM1.5 with the case of considering the sun as a blackbody emitter where the blackbody spectral radiative intensity (W/(m2*str*um)) is multiplied by the solid angle of the sun as seen from earth to obtain the extraterrestrial spectral radiative heat flux (W/(m2*um)) on earth.
For more information you can refer Principles of Solar Engineering,
by D. Yogi Goswami , Frank Kreith , Jan F. Kreider
- Badges
- Science topic
- Similar topics
- Engineering
- Thermal Engineering
- Thermofluid
- Heat Transfer