[Stoves] Excess air

Mon Aug 26 10:04:31 CDT 2013

Dear Andrew

Most interesting range of topics - I hope not to expand it.

Here we go:

>>There is almost enough O2 in biomass to burn all the Hydrogen but not
quite. If there is no carbon burning a great deal of the combustion can take
place with no outside air at all. 

>I think you are looking at this the wrong way round, the wood (which Tom
Reed suggested a surrogate formula for of CH1.4O0.6) is a mixture of
cellulose, lignin and hemicellulose, these are derived from the simple
sugars that are made during photosynthesis, the oxygen is always bonded to
hydrogen or carbon atoms , i.e the fuel is already partially oxidised. 

I agree with the proxy formula. Saying it is partially oxidised, yes well...
to burn it has to come apart (and does) making O2 available as you explain.

>Working out the bond energies involved needs a younger more agile brain
than mine but I somehow doubt that any of the oxygen originally bonded to
the wood survives as free oxygen in the flue gas.

It would bond most easily with H2 to make water vapour and this is of course
detectable using an NDIR cell (or other).

>What we can easily show is that the energy available in wood is almost
exactly the same as the energy available in burning the same mass of carbon
as is in the wood, which again suggests that it is carbon oxidation that is
providing the energy for the combustion.

I think that can't be correct.

Suppose: 50% C and 5% Hydrogen (by mass)

Carbon at 32.8 and Hydrogen at 120 MJ/kg

Carbon 16.4 MJ Hydrogen 6.0 MJ = 22.4

Something like that. A few losses here and there, the Hydrogen matter.

>>If there is no carbon burning a great deal of the combustion can take
place with no outside air at all. 

>Well the case in a retort is that no carbon burns, can free oxygen other
than that which occupies interstitial spaces in the wood be detected in the
offgas of a retort?

Under certain conditions I think It would, but how to prove it? One proof is
to do it dry and wet to see what the result is. Steam is really reactive,
meaning chemically active and could be involved.

>>The term Lambda is used loosely. Testo uses it incorrectly in their
computer outputs to describe Excess Air. Lambda is strictly the total air
demand, not the excess air only. Suppose the EA is 250%. Then Lambda is
350%. 

>Again this ties in with the equivalence ratio and Lambda being the same, so
lambda=er=stoichiometric air plus excess air.

That is how to think about it.

>The point I was trying to make is that a lambda sensor as used in stoves
and boilers never detects lambda of less than 1 because there is no oxygen
to exchange electros with in the detector, all the oxygen has been consumed
in 

the primary reaction.

So what is really another case of misnaming is that it is not really a
'Lambda detector' it is an excess air detector with 1 added to the dial. If
it can't detect the fact that the combustion was only getting 30% of the
stoichiometric air it needed, then it is not a lambda detector. It just
detected excess air, and probably at low values it is not even doing a good
job of that because it is using a simplified version of the formula that
only really works well at higher values. We get used to the numbers and we
use them, but the thing is a bit of a mess viewed as a chemist would.

However, Honda claims <http://www.hondata.com/techlambda.html>  their
"Lambda detectors" measure values less than 1. At full power they want the
Lambda value to be 0.88-0.92. They claim it is an O2 meter. So if EA is 0%,
what are they detecting?? See down the page at "Tuning mixture"

>Most of the big boilers I dealt with burning woodchip with a moisture
content of more than 30% seemed to still showing 5-10% oxygen in the flue
gas, so an excess air of 50-100% and lambda of 1.5-2.

And for that things work well. Taking that equipment and trying to analyse
pyrolysers is very misleading because it will just say O2 = 0.0.

>S Varunkumar in his thesis discussed here recently further breaks down this
small amount of primary air and its behaviour into separate modes of
pyrolysis and offgas evolution depending on its superficial velocity.

The superficial velocity has other value as Dr Tom pointed out for years and
I have been keeping that in mind when looking at these devices. Because
quantifying the input air volume is so difficult (because it is such a small
amount) working out what is going on in the fire or pyrolysis zone depends
on a lot of guesswork and misunderestimation.

>>The manufacture of O2 in the fire is the Water Gas Shift Reaction. It
would happen using combustion moisture I suppose but I have not seen that
definitively.    I have definitive measurements though from wet fuel fires. 

>...the trouble with wet fuels is that they need so much primary air just to
burn them that lots of this subsequently travels through the bed and if it
doesn't get consumed in a secondary flame will appear in the flue gases.

Ah, but this is not how to create the conditions. First, light a wood fire
with perhaps 2 kg of wood and a little coal  - maybe 1 kg or less on the
sides to get those nuggets hot. The stove is heated and the chimney get
working.  When the wood is reaching the char-mostly stage, put on wet
lignite with 50% volatiles. This smothers the fire placing wet (25%
moisture) coal on a very hot wood fire that is already burning. The bottom
will immediately dry producing a lot of steam and immediately thereafter,
volatiles that are not burned because there is too much cooling by the steam
immediately above the hot charcoal and wood.

This creates a huge quantity of condensed volatiles which are the major air
pollution problem in Ulaanbaatar. Between that point and the time when
flames emerge through the top of the coal pile (at which point the PM and CO
immediately almost disappear) there are conditions suitable for cracking
water - either combustion moisture or fuel moisture.

>I thought the water shift reaction was an equilibrium reaction and needed
high temperature, I think water vapour only dissociates above 1500C in the
absence of a catalyst...

Well this is where theory gets in the way of reality. Yes it takes place in
isolation at 1500 but town gas producers did not operate at 1500 C, correct?
The hot carbon is there and there is heat from burning some of the H2. 

Didn't we discuss the conditions necessary to create town gas from burning
coal with water sprayed onto the fire?

>...either way it is highly endothermic and I doubt a high moisture content
fuel can either reach the required temperature of supply the energy.

So it is endothermic and it absorbs a lot of heat which is why the early
fire, with high O2 in it, is not effective at raising the temperature of the
stove. As a heater, it sucks, literally.

So this is really quite easy to demonstrate: all you have to do is add up
the Oxygen found in all the species that were detected in the stack.

Free O2 + the O2 in CO2 plus the CO divided by 2 to make O2, most people
have that much. Then the H2O from combustion of the hydrogen (which you can
get by measuring free H2 and working out how much was available) and SO2 (in
the case of coal) and NO/2 and NO2 if there is any, and N2O/2. 

Just add them up for each measurement interval. If the entire spectrum of
gases were measured, then all the O2 should be accounted for, right? There
is the O2 in the fuel to begin with which is probably burned to water, and
there is O2 from the air. What is the explanation if the total is above 22%
of the emitted parts per million? Or 24%?

The first thing to suspect is the detectors, right? So check them and do it
again. Same result. It is not for long, it is not in all stoves, but it is a
reproducible effect in certain cases of having hot, wet lignite sitting on
top of a hot wood fire that is being smothered, in a high excess air
condition, in a stove with a lot of space above the fuel pile (ensuring
rapid chilling).

Very, very interesting.

The upside of the story (even if it is all wrong) is that the chemically
balanced approach to the calculation automatically corrects for instrument
drift in that the resulting answer, which is not a direct measure, remember,
is closer to the truth than the effect of the same instrument drift using
the simplified formula that is applied in combustion analysers. For that
reason we never use any calculated result from a measuring device, only the
raw data. There are too many shortcuts applied.

Regards

Crispin

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