[Stoves] Excess air

[ajheggie at gmail.com]

[Default] On Mon, 26 Aug 2013 11:04:31 -0400,"Crispin
Pemberton-Pigott" <crispinpigott at gmail.com> wrote:

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

Good, having now read some of the earlier posts on the subject I see
some has been discussed recently

>>>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.

Actually I don't think O2 becomes available, some O radicals may exist
for a very short period.
>
> 
>
>>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).

Yes but I don't know what an NDIR cell is nor how it could be
distinguished from moisture in the wood.
>
> 
>
>>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.

You removed the calculation and it was just a first approximation to
show that the energy from wood was nearly the same as the energy from
burning the carbon content in that wood. In fact your calculation
below shows it to be a better case as you have upped the energy
available to the carbon and the energy from the 1 spare hydrogen to be
even nearer the wood cv, remember for my approximation I assume the
H7O3 represents 1 spare hydrogen and 3H2O (already completely oxidised
but still losing the enthalpy of evaporation in the reaction).
> 
>
>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

From the above I say 2nd approximation

 0.52*32.8+0.009*120=18.1MJ
>
> 
>
>Something like that. A few losses here and there, the Hydrogen matter.

Yes but only one of them makes a net contribution.
>
> 
>
>>>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 proof to me is that even if one uses bone dry wood and heats it up
to 330C not a lot of heat is given up as the pyrolysis takes place and
as far as I know no noticeable oxygen is present as flaming the offgas
is initially difficult although later there is a CO diffuse flame.

> 
>
>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.

Yes that's a way of looking on it.


> 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.

Probably and because it's a cell the output is not linear with
changing O2 content.
>
> 
>
>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"

Actually I don't read it that their meter does read in the range they
say, it may well be that they are measuring something like HC and CO
and using a look up table to deduce conditions outside a normal
zirconium cell can. As you say they may have a different type of
sensor.

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

Not quite so easy, what happened was the coke was blasted by air as an
updraught gasifier, this raised the coke temperature to 1100C and a
mixture of CO and predominantly nitrogen having a free ride, as Philip
Lloyd says, (and we have often discussed this nitrogen problem in the
past). The offgas was then used to raise steam and superheat it, this
very hot steam was then used instead of air to blast the coke and
produce the water shift reaction and H2 and CO was produced, if clean
enough this is a true syngas. As this process was highly endothermic
the coke gradually cooled down even with the superheated steam so the
process was repeated with an air blast.

>
>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.

Yes I find I can burn woodchip with 66% moisture but its only
smouldering (i.e nascent char burning) and the offgas is too diluted
by excess air and steam to support a flame, flue temperatures hover
below 100C and hardly any heat gets into the water for circulation to
the building.
>
> 
>
>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.


Yes Oxygen as well as all the other elements in the input will be
conserved.

I really cannot comment on the instruments as I have seldom had access
to them and even then it was simple testo type devices and as I said
the cells on this  use derivatives to deduce gas content from, if used
in a gas boiler one has to tell them whether it is natural gas or lpg
because the lookup table is different. Wood is a complex mixture so I
expect it is easy to confuse them.

AJH (possibly offline now till next Saturday though I will read the
first few lines of messages)