Dear Tom R
So when we provide very hot secondary air into a gas stream at about 700 C it
is (mostly) burning C, not CO? Maybe?
Thanks
Crispin
Dear Richard and All combustors, gasifiers and Pyrolysers:
Good question, "Why is the match flame blue near the base, but turns yellow in
the outer region?"
<><><>
As the wood in the match heats, the cellulose in the wood breaks down to form
CO and H2. (The lignin in the wood turns to charcoal.)
I could write a long equation using cellulose (C6H10O5)n, but the chemistry is
easier if I approximate it for Carbo-Hydrates, "CH2O", an approximate formula
for all sugars, starch and cellulose.
CH2O ===> CO + H2
Solids emit black body radiation mostly proportional to their surface
temperature. But burning gases emit arbitrary colors due to the flame
reactions. So the flame close to the match shows these blue flame colors.
<><><>
So why does the flame give off black body radiation at a greater distance from
the burning match?
Carbon monoxide is a queer duck. It is THE MOST STABLE molecule at high
temperatures. However, at low temperatures (below ~700C as the flame gas
cools) it disproportionates to carbon and CO2.
2CO ===> C(soot) + CO2
This is known as the Reverse Boudouard reaction. However, it takes a while for
the soot to form.
The same thing is true if you look very closely at a candle flame.
<><><>
I started to answer your question this morning, but did not have a good answer.
When I returned to the question in the afternoon my subconscious must have
accessed "Reverse Boudouard". Funny how the mind works.
Your flame chemist,
Tom Reed
Thomas B Reed
280 Hardwick Rd
Barre, MA 01005
508 353 7841
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