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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>You raise an important objection and it is not, in my experience, correct. It is only correct when 'things go normally' such as in a constant state furnace burning a low oxygen content fuel such as kerosene or coal.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>" I disagree. Stoichiometric air is the amount of air needed to combine with the fuel if combustion goes to completion and there is no fuel or oxygen left over."<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Actually is it not as simple as that for a fuel with any O2 in it. How do you know if the O2 came from within the fuel or from air supplied to the fuel? Thus the constant burning condition is different from the evolving fuel composition condition - the latter being exactly what we have in a pyrolyser.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>"Excess air is that air over and above the stoichiometric air that is supplied in order for all the fuel to be able to "meet" and combine with oxygen from the air. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Only by definition, not in practise. In practise conditions were being regularly encountered when the calculated excess air ratio was -120,000% or -185,000% (a negative value) when shortly before it had been at 2000%, a positive value. The problem was traced to the formula itself - it does not treat the combustion in a chemically balanced manner - that is, it makes assumptions about the availability of O<sub>2</sub> as if there was none in the fuel. It is an incorrect assumption though it holds for gasoline and therefore most vehicles.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That is why I developed a new way of calculating the 'effective EA' meaning what the EA would have to be to get the conditions we are measuring, as if there was no O<sub>2</sub> in the fuel.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>"It thus follows that there will be oxygen in the exhaust that is equal to the excess air supplied."<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Well that is the root of the problem. It is only true under certain sets of conditions. What happens when the reading for O<sub>2</sub> plus the reading for CO/2 (which is to turn the O in the CO into O<sub>2</sub> that supposedly came from the air supplied) sums to a number larger than 21%? The denominator in the formula turns negative.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That was the source of the impossibly high negative values for EA in a stove that had a fire burning in it and obviously high excess air.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The 'correct' values for the place held by the number 21 in the standard formula should actually be the total O<sub>2</sub> in the emitted species, including combustion moisture which id at present ignored completely.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The formula (<i>HTP</i> version) then replaces the CO/2 with a detected species that still have oxidation potential so as to represent that potential correctly. It is subtracted from the numerator's O<sub>2</sub> measured value on top and on the bottom.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><u> [(O2-(CO/2)]<o:p></o:p></u></p><p class=MsoPlainText>[21-(O2-(CO/2)] x 100 = EA% is the standard formula, in case anyone forgot<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Here is the alternative<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><u> [(O<sub>2</sub>-(CO/2 plus all other oxidising potential)]<o:p></o:p></u></p><p class=MsoPlainText>[All detected O-containing gases factored to O<sub>2</sub>)- [(O<sub>2</sub>-(CO/2 plus all other oxidising potential)] <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>With all values in %.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The result is the chemically balanced correct EA value (still pretending O<sub>2</sub> is actually from real air) no matter what the conditions in the fire are and no matter where the O<sub>2</sub> came from. It works for rocket engines with excess air (but consuming no air at all) and for stoves in which O<sub>2</sub> is being manufactured from the available water in the fuel. Because we know the O<sub>2</sub> in the fuel and the O<sub>2</sub> in the air does not mean we know the O<sub>2</sub> that will be manufactured in the fire.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I measured stoves with a burning wood+coal fire that had O<sub>2</sub> stack measurements above 21%, even as high as 24%, <i>while the fire was burning</i>. That O<sub>2</sub> above the ambient level was coming from the fuel and the water in the fire. So what then is the EA under such conditions?<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>"The fact that the fuel is already partially oxidised makes no difference to the excess air value."<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Actually it makes a measurable difference.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>"It does make a difference to the total oxygen in the exhaust constituents but that is an entirely different matter."<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>It is part of the same matter. EA is an expression of the availability of O<sub>2</sub>, expressed in a way that assumes all O<sub>2</sub> comes from the air. The 'standard' formula only works in certain cases. We have been using the alternative in the <i>HTP</i> for at least 5 years for that reason. For those cases for which the old version was intended, they give the same answer. Otherwise they do not.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I know most people are not interested in this level of chemical detail which is why Profs Lloyd and Annegarn and I talk about it at conferences and dinner, not here. But the error is real and the effect on reported values is real and there is a solution.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I believe that with the right equipment we can tell when the water evaporates from the fuel which would lead us to the point where we can tell in real time what the fuel composition is as it burns. That will give us much more accurate firepower values in real time.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>This is stuff fascinating.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Regards<o:p></o:p></p><p class=MsoPlainText>Crispin<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p></div></body></html>