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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoPlainText><span style='color:#1F497D'>Dear Frank<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>There is no need to over-complicate the issues for HHV.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>There are three stages of calculation involved:<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>From liquid to boiling,<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>From boiling to evaporated water at boiling<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>From boiled vapour to hotter vapour<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>I realize for biomass the difference between J used in evaporation (2256<o:p></o:p></p><p class=MsoPlainText>j/g) and energy in water at 450c (2854 j/g) is well within the 'noise' of biomass fuel so doesn't matter which one we use.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:#1F497D'>The energy 'at 450' includes parts of all three segments of the calculation. The number 2854 is the sum of three things. I was showing how to calculate them separately.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>> And if we all are using the 2256 value that is what I will do. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:#1F497D'>2257 is probably the most accurate. Often people use 2260 for convenience being a rounder number.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>But it seems since the hydrogen in biomass never is in a liquid state or never even water vapor before going from solid state to water vapor at ~450 that we should be using 2854 ((2256<o:p></o:p></p><p class=MsoPlainText>+(1.72 X 350)). <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:#1F497D'>Hydrogen is in the solid state in the fuel (not the H2 in the moisture, the H2 in the biomass). It is liberated as a gas and most of it burns to water, but not all. There can be quite a lot of H2 in the exhaust. See this graph. The lower line is free H2(EF) which means it is the measured concentration multiplied by (Excess Air +100%). You can see there is nearly a fixed ratio between the CO and the H2. This is common though not universal. Even after 3 hours, there is still lots of free floating H2.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><img width=378 height=174 id="Picture_x0020_1" src="cid:image003.jpg@01CCCFBC.2264BB90"><o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>For the LHV calculation of methane; I see wiki says the HHV is product of water in liquid form <span style='color:#1F497D'>[and cooled to 0 degrees C] </span>and LLV is product of water in vapor form - same way you calculate biomass. <span style='color:#1F497D'>[It has nothing to do with the fuel type, it is how to calculate the heat content of combustion products] </span>But, of course water is never in the liquid form (until it completely condenses at below 100c.). Hydrogen is held by carbon or then water vapor at the combustion temp. so I am surprised they calculate (estimate) it the way they do. <o:p></o:p></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>In fact it is pretty reasonable. Lower heating value is what heating can be done by the fire. That is a pretty real world attitude. Yes, it has not been done exactly. But it is easy to calculate.<o:p></o:p></span></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>So I will not bother anyone with more questions, will calculate as everyone does (subtract energy from water liquid to water vapor) <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:#1F497D'>Yes. The HHV from a bomb calorimeter minus the latent heat of evaporation of the moisture from combustion of Hydrogen gives the useable dry fuel heat called LHV. So far no one has come up with a difference between the latent heat of evaporation and latent heat of condensation. If you can create one, you will have a perpetual motion heat engine!<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>Regards<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'>Crispin<o:p></o:p></span></p><p class=MsoPlainText><span style='color:#1F497D'><o:p> </o:p></span></p></div></body></html>