Re: Brown's gas??
- From: Tim Wescott <tim@xxxxxxxxxxxxxxxx>
- Date: Thu, 08 Dec 2005 17:39:58 -0800
Andy Dingley wrote:
Did I say "competent"? Allow me to appeal to your scrabble bag analogy and remind you that I said "partially". If you were interested in the radiant heat from the flame then it probably _does_ radiate the same amount as a similar-sized black thing at that temperature -- at least in a room-temperature environment.On Thu, 08 Dec 2005 14:52:46 -0800, Tim Wescott <tim@xxxxxxxxxxxxxxxx> wrote:
- There's some babble about a "cool flame". This is simply ridiculous. The idea that it "radiates at 129°C" is particularly silly. It's a hydrogen-oxygen flame - they're hot, that's how they work. It's not magic though, nor are normal rules of physics or chemistry suspended for this magic gimmick.
That may be the one (partially) true claim, actually.
None of their claim, as I paraphrased it, is at all true. I don't just shake words out of a Scrabble bag, I use them very carefully.
Hydrogen remains largely transparent as it burns, so it emits very little light.
Indeed. But although it radiates _little_ light, the spectrum of that which it does radiate follows the same Planck distribution as for glowing soot.
There is no competent way in which you can equate the dim light of a hydrogen flame with "radiation at 129°C"
Of course it's a meaningless distinction, because you're going to want to point the flame at something and as soon as you do that thing is going to heat up conductively. Since it's most likely a good emitter then there's going to be heat radiated all over the place.
Yes, but I bet they used the el-cheapo one from Harbor Freight.
So if you looked at the flame with an infrared pyrometer it would show up as not much warmer than it's surroundings.
That's a fault with infrared bolometers. Better modern pyrometers (such as those used for measuring gas temperatures) don't measure total fluxes, they measure the ratios between fluxes at a number of different wavelengths. Older ones (disappearing wire etc.) simply used a target of a known heated material (usually a lump of firebrick) and looked for the peak emission wavelength / colour.
Red ocher paint is largely gray; there are very few things that have a really pure color -- and those things emit at the wavelengths that they absorb when they're cool. In fact, researchers (I think in Japan) have designed filaments for light bulbs that are only emissive in the optical band -- they're "white" at UV and IR. The bulbs are significantly more efficient, because they don't radiate useless wavelengths.
bright yellow flames get that way because soot is black,
"Black body" radiation has _nothing_ to do with the fact that cold soot looks black to human vision. Red ochre paint is a good simulation of a black body radiator at gas flame temperatures and it will return to looking just the same red colour when it's cool.
Were soot clear (or white) at temperature it would not glow so strongly. And it's "emissive" (black) at visible wavelengths at ambient, as well.
Look at 'clear' glass out of a furnace for a counterexample -- it's glowing, but not nearly as strongly as it would if it were really emissive at visible wavelengths. Take the same size chunk of iron at the same temperature and you'll light up the room.
--
Tim Wescott Wescott Design Services http://www.wescottdesign.com .
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