Sorry if this has all been covered before - I searched the forums, and didn't find a similar post.
My impression is that every wood stove I've ever used emits most of its heat under most conditions by convection - like the so-called "radiators" in homes and cars. Even with the big window on my current insert, it's only under exceptional conditions that radiation equals or exceeds blower-air heat. This has significant consequences: for example, there is no way that this unit will, as the dealer had said, keep my pipes from freezing in the event of a power outage - much less keep the house comfortably warm.
The only time I get much radiation is with an extremely hot fire, which I've been unable to maintain for more than an hour or two. After that, the box fills up with glowing coals, which don't deliver much heat. I got to wondering about that.
As some people here know, energy emitted from a so-called "black body" (a scientific ideal fairly well approximated by coals or the specks of glowing soot in a flame) goes like the fourth power of absolute temperature. For example, a fire at 1500F (ca. 1100K) would emit about twice as much total radiant energy as one at 1200F (ca. 900K).
That's an impressive change, but my sense is that the increased radiation output during those peak bursts is even more dramatic. Also, I noticed that a 6" bed of coals sends very little heat out the blower, and not much out the front door either. If I open the door half-way, the heat outside the glass is still moderate - but just inside, I can't even leave my hand for a few seconds.
So - I think that the glass is blocking much of the radiation. To dig deeper, we need to consider the wavelength spectrum of emitted heat.
A useful number is Wien's displacement constant, ca. 2898 kelvin-microns. Simply put, the emission peak of a black body will be at a wavelength equal to the that constant divided by the absolute temperature. For the sun, the emission peak is at about 5760 degrees. Divided into 2898, that gives .503 microns = 503 nanometers, conveniently near the middle of the human eye's sensitivity. For wood fires, emission peaks in the mid-infrared: about 2.7 microns at 1500F, 3.1 microns at 1200F.
What about the window? Most matter, even "clear" stuff, absorbs pretty strongly at wavelengths longer than about 3 microns. Here's a relevant spectral transmission plot:
www.sinclairmfg.com/datasheets/borosilicatecurve.htm
(I'm assuming that the faceplate is a Pyrex-class glass.) It looks like at least half of the radiation coming from inside the stove will be absorbed by the window. At higher temperatures, not a little but a LOT more will get out past that absorption edge. It's kind of like turning up the dimmer on an incandescent bulb: even at low settings, there's plenty of output. We just can't see it, because it's almost all in the infrared. The filament has to get very hot before the short-wavelength "tail" of the radiation curve gets bright to our eyes, and the shift in peak wavelength is more important than than the increase on overall power.
What happens to the absorbed radiation? It's not lost; just heats up the window. Some will be re-radiated back into the stove, and some radiated outward. The problem is that it's being re-radiated at a much lower temperature. (Note that the window itself never glows.) And per the fourth-power law, there will be very little radiation at that lower temperature. So some heat will still get out, but mainly by convection off the faceplate.
If this reasoning is correct, very hot fires send more radiation from the window partly by greatly enhanced short-wavelength radiation passing through the window, and partly by more-efficient re-radiation of heat absorbed by the window.
Corrections and refinements welcome.
-Theo
My impression is that every wood stove I've ever used emits most of its heat under most conditions by convection - like the so-called "radiators" in homes and cars. Even with the big window on my current insert, it's only under exceptional conditions that radiation equals or exceeds blower-air heat. This has significant consequences: for example, there is no way that this unit will, as the dealer had said, keep my pipes from freezing in the event of a power outage - much less keep the house comfortably warm.
The only time I get much radiation is with an extremely hot fire, which I've been unable to maintain for more than an hour or two. After that, the box fills up with glowing coals, which don't deliver much heat. I got to wondering about that.
As some people here know, energy emitted from a so-called "black body" (a scientific ideal fairly well approximated by coals or the specks of glowing soot in a flame) goes like the fourth power of absolute temperature. For example, a fire at 1500F (ca. 1100K) would emit about twice as much total radiant energy as one at 1200F (ca. 900K).
That's an impressive change, but my sense is that the increased radiation output during those peak bursts is even more dramatic. Also, I noticed that a 6" bed of coals sends very little heat out the blower, and not much out the front door either. If I open the door half-way, the heat outside the glass is still moderate - but just inside, I can't even leave my hand for a few seconds.
So - I think that the glass is blocking much of the radiation. To dig deeper, we need to consider the wavelength spectrum of emitted heat.
A useful number is Wien's displacement constant, ca. 2898 kelvin-microns. Simply put, the emission peak of a black body will be at a wavelength equal to the that constant divided by the absolute temperature. For the sun, the emission peak is at about 5760 degrees. Divided into 2898, that gives .503 microns = 503 nanometers, conveniently near the middle of the human eye's sensitivity. For wood fires, emission peaks in the mid-infrared: about 2.7 microns at 1500F, 3.1 microns at 1200F.
What about the window? Most matter, even "clear" stuff, absorbs pretty strongly at wavelengths longer than about 3 microns. Here's a relevant spectral transmission plot:
www.sinclairmfg.com/datasheets/borosilicatecurve.htm
(I'm assuming that the faceplate is a Pyrex-class glass.) It looks like at least half of the radiation coming from inside the stove will be absorbed by the window. At higher temperatures, not a little but a LOT more will get out past that absorption edge. It's kind of like turning up the dimmer on an incandescent bulb: even at low settings, there's plenty of output. We just can't see it, because it's almost all in the infrared. The filament has to get very hot before the short-wavelength "tail" of the radiation curve gets bright to our eyes, and the shift in peak wavelength is more important than than the increase on overall power.
What happens to the absorbed radiation? It's not lost; just heats up the window. Some will be re-radiated back into the stove, and some radiated outward. The problem is that it's being re-radiated at a much lower temperature. (Note that the window itself never glows.) And per the fourth-power law, there will be very little radiation at that lower temperature. So some heat will still get out, but mainly by convection off the faceplate.
If this reasoning is correct, very hot fires send more radiation from the window partly by greatly enhanced short-wavelength radiation passing through the window, and partly by more-efficient re-radiation of heat absorbed by the window.
Corrections and refinements welcome.
-Theo
