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</o:shapelayout></xml><![endif]--></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='color:#1F497D'>Dear Crispin,<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><snip><o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>></span><span lang=EN-CA style='color:black'>I was going to make a chart to post here showing the amount of extra energy involved but have been travelling.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Frank: ></span><span lang=EN-CA style='color:#1F497D'>>So how would you suggest doing this? <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Crispin: >></span><span lang=EN-CA style='color:#1F497D'>Just to calculate an excess of heat from the missing charcoal and it s heat value, minus the moisture evaporation needed to dry the fuel. As Dr Tom has said, it is nearly linear from 0-30% so that is easy. Just want to get a handle on the difference. Then look at why it is happening.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Frank: Think the combustion of the secondary just confuses and masks much of the info we want to know about what happens in the primary. Seems we just need the chemical components of the gases and particles reaching the secondary. So some questions:<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Measuring the chemicals, particles, volume of gas produced and moisture condensed in the collection bag:<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The chemicals should be CO2, CO, H and amount and ratio to determine the combustibility of the gases. <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The particles and their makeup determine the carbon and hydrogen that may not be combusted in the secondary.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The increased volume of gas produced from C, H and O from the wood turning into gas should indicate something and I think can be accurately measured.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>And the moisture in the collected gas that condensed upon cooling is water vapor and not the hydrogen that separated. This is the LHV calc. we talk about(?). <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>All this is for the TLUD stoves Paul and others use. Not for high temperature Rocket stoves. But what can be learned from the TLUD I would think can be used for other stoves. <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Is this how you look at it?<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Thanks<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Frank<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>Frank Shields<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>Control Laboratories, Inc.<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>42 Hangar Way<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>Watsonville, CA 95076<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>(831) 724-5422 tel<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>(831) 724-3188 fax<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><a href="http://www.biocharlab"><span style='color:blue'>www.biocharlab</span></a>.com<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'> <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'> <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>Not enough heat, hydrogen produced outside the Hot zone, some other chemical reaction sucking up the available heat (evaporating water), test equipment with a positive interference. <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>All possible. I think it may be that wood or coal (which is old biomass) has H2 driven from it when it is initially heated and this escapes from the fire zone into a chilled area inside the stove and gets to the chimney (sneakage as Tom Miles calls it). What a great word.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>Question that may seem unrelated: Does the ThermoWorks IR heat detector work through glass to measure the heat on the other side? <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>It will if the glass is not selective for IR and is transparent to IR. Maybe a quartz window would allow it. I did not check if a quartz window absorbs IR. Anyone know?<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>But you talk about ‘coal’ along with biomass. Wonder how all that differs. Can coal be used in a TLUD? <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The clean stoves being rolled out in Ulaan Baatar right now are TLUD’s. The fuel is raw lignite. They burn extremely cleanly right down to ash. There is a small amount of clinkering in the Silver stove because it tends to run with too much primary air and a very high temperature. The ash has a high melting temp so normally it is not a problem (Nalaikh coal). <o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>>…</span><span lang=EN-CA style='color:black'>The problem is at present the calculation methods do not consider that stove would every produce large amounts of charcoal so the calculated results are (erroneously) far from the real values.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>>So it seems the Stove needs be: ((WBT) + (biochar produced) = (wonderful rating))<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Well, the rating given at present is completely at odds with the actual fuel consumption and that is not tolerable. Simple as that. You can’t say a stove which saves no fuel saves 50% of the fuel because the calculation method has an error in it. Therefore, buy my stove! Crikey. You think the customer won’t notice?<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The issue it raises is that of the heat transfer efficiency and the efficiency as a system. IF the fuel is the consumption of raw fuel per burn cycle, then the system efficiency is that amount of fuel applied to the work done. If you want to chuck out char, that is your business, but that char is not going to be credited to the stove as a fuel saving. I just saw a kitchen with a separate charcoal stove used to burn the saved char from a wood fire. That is smart and is an efficiency, but it is not creditable to the stove that produces the char, only to the whole system of stoves in that kitchen. It is quite an interesting problem to define categorically.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>The two efficiencies (heat transfer and system) are not synonymous. In any case the way the WBT calculates the thermal efficiency is a proxy for the heat transfer efficiency, not the HTE so be careful how you state performance. The actual HTE is quite difficult to establish on a stove. Using an insulated pot gets a closer answer than most methods, but an insulated pot gives a misleading answer if you are trying to characterise the performance of the stove has a cooking device. People don’t cook with insulated pots. So one must decide what the desired metric is. Unfortunately at present the WBT’s calculate fuel consumption from the energy used in the heat transfer calculation instead of the other way round (which would be normal):<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Fuel consumed, work done => efficiency<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Not:<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Work done, efficiency => fuel consumed<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>because it allows unanticipated errors to be introduced.<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>We have much work to do. I wonder how efficiency we can be about it!<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'>Regards<br>Crispin<o:p></o:p></span></p><p class=MsoPlainText><span lang=EN-CA style='color:#1F497D'><o:p> </o:p></span></p></div></body></html>