Thread: Brown's gas??
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Tim Wescott
 
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Default Brown's gas??

Andy Dingley wrote:
On Thu, 08 Dec 2005 14:52:46 -0800, Tim Wescott
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"

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.

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.

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.

Yes, but I bet they used the el-cheapo one from Harbor Freight.

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.

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.

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