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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link="#0563C1" vlink="#954F72"><div class=WordSection1><p class=MsoPlainText>Dear Andrew<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Most interesting range of topics - I hope not to expand it.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Here we go:<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>It would bond most easily with H2 to make water vapour and this is of course detectable using an NDIR cell (or other).<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>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.<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>I think that can't be correct.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Suppose: 50% C and 5% Hydrogen (by mass)<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Carbon at 32.8 and Hydrogen at 120 MJ/kg<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'>Carbon 16.4 MJ Hydrogen 6.0 MJ = 22.4<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Something like that. A few losses here and there, the Hydrogen matter.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>>If there is no carbon burning a great deal of the combustion can take place with no outside air at all. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>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?<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>>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%. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>Again this ties in with the equivalence ratio and Lambda being the same, so lambda=er=stoichiometric air plus excess air.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That is how to think about it.<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>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 <o:p></o:p></p><p class=MsoPlainText>the primary reaction.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>However, Honda <a href="http://www.hondata.com/techlambda.html">claims</a> 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”<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>And for that things work well. Taking that equipment and trying to analyse pyrolysers is very misleading because it will just say O<sub>2</sub> = 0.0.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>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.<o:p></o:p></p><p class=MsoPlainText> <o:p></o:p></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>...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.<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>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...<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Didn't we discuss the conditions necessary to create town gas from burning coal with water sprayed onto the fire?<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>...either way it is highly endothermic and I doubt a high moisture content fuel can either reach the required temperature of supply the energy.<o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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. <o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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%?<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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).<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Very, very interesting.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>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.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Regards<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'>Crispin<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p></div></body></html>