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Dear Dr. Radiation,
I just went through your website descriptions of radiation physics. In general, it looks good, but a few things can use some clean up.
1. IR Equations: (I use Hudson's book Infrared System Engineering)
Planck's Law - OK (I use a simpler version, but this is OK (you might mention, this eq. generates the blackbody curves)
2. Stephan-Boltzman Law -
a. spelling errors on line 3
b. You should state that this gives the total energy emitted, but as the temperature rises, more of the energy is emitted at shorter wavelengths as given by Wein's Displacement Law and shown by the Planck blackbody curves. Thus, if you double the absolute temperature, you get 2 to the 4th or 16 times the TOTAL energy out, but more of this energy is output at shorter wavelengths and may not be felt as heat. (We normally consider the 8 to 12 micron region as thermal infrared).
3. Wein's Displacement Law - OK
Note: Our sun's energy at T=6000 K peaks at 0.5 microns (where our eye response also peaks). If our sun were cooler at 3000 K, its radiation would peak at 1.0 microns. (Our eyes, from evolution, would then peak at 1.0 microns, in the near infrared, but don't say this or you may catch Hell from some Born Again Creationist Christians who might be building Rumfords.)
4. Kirchhoff's Law : (Note spelling of Kirchhoff)
emissivity = absorbtance at any given wavelength and temperature
Thus "a good absorber is a good emitter" (which has to be true for the object to stay at the same temperature)
This can be expanded into the "common sense" equation I gave you that the total incident energy must equal the sum of the absorbed, reflected, and transmitted energy.
5. Absolute Zero in K=C+273.16
Absolute Zero in Rankin=F+459.69
(Note: There is a new energy source being researched that gets its energy from the non zero momentum of atoms even at absolute zero - See Aviation Week article recently)
6. Greenhouse
Good quality quartz glass can transmit out to about 2 microns. 0.5 microns is blue light. Objects at room temperature of about 80 deg. F (we call 300 K radiation) emit radiation which peaks at about 9.6 microns, which glass does not transmit. Thus the emitted energy from the inside objects in a greenhouse does not get out.
7. Glass doors.
Glass doors will not transmit the THERMAL infrared wavelengths from 8 to 12 microns, although they may transmit the short wavelengths around 1 micron, which we do not feel as heat. The glass will heat up, however, due to its absorptance and then will radiate out with an equal emissivity and at wavelengths determined by its temperature and Planck's Law.
Hope this is of some help.
If you want to chat about these items or some others, give me a call.
Regards to all,
Kimo
4/28/04 (Notes - not sent)
Kimo,
In trying to determine the radiant heat output of a masonry heater (masonry stove with flue channels) is it correct to assume that the total radiant energy of the source (in this case the wood fire) is represented by a Planck curve with a peak at 1000K or about one micron, and that the heat that gets through the masonry heater body is absorbed and re-radiated at about 500K so the radiant heat is no more than 1/16th of the original radiant energy of the fire? In other words an open fire produces far more radiant heat even though a fireplace may be less efficient due to the excess air loss and that a masonry heater radiates some but primarily heats air by conduction and convection.
Secondly, how do you deal with thermal storage? A masonry heater stores heat so that it continues to heat after the fire is out. Can that be described as a variation of Kirchhoff's law? Kirchhoff's law represents a stable situation but until equilibrium is reached more radiation as absorbed by the masonry than is emitted.
Finally, again in a masonry heater or masonry firebox, how would you calculate how much of the heat transferred to the masonry is due to radiation from the fire and how much from convection due to heated gasses impinging the masonry walls?
Jim
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