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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US'>Dear Frank<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US'>Don’t let the other topics drop, OK?<o:p></o:p></span></p><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal><span style='color:#1F497D'>></span>A ) (Fuel Efficiency%) = ((task 1 + Task 2) / (dry wt biomass fuel) ) X 100<o:p></o:p></p></div><div><p class=MsoNormal>This is the dry fuel <u>taken from the forest</u> required to complete task 1 and task 2. Doesn’t matter the energy or time it takes to do it. As long as the stove(s) can complete the task(s) the stove(s) can be tested. There must be a very sharp endpoint telling when the task(s) are completed. Excess fuel left in the stove is wasted. <o:p></o:p></p></div><div><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>This takes a mass-based approach and it has little use for stove performance rating. The problem is there is more than one fuel and as they have different energy contents, comparing the mass consumed doesn’t tell us precisely the performance. It has to be done on an energy basis because the work done is based on energy (we have to use something). Similarly emissions per kg of fuel burned are not ideal for measuring stove performance. They are useful for air-shed modelling but are usually misconstrued because a fuel does not by itself, have ‘characteristic emissions’.<o:p></o:p></span></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal><span style='color:#1F497D'>></span>B ) (thermal efficiency%) = (((task 1 + task 2) / (total energy released)) X 100<o:p></o:p></p></div><div><p class=MsoNormal>This is the total energy in the dry fuel <u>added to the combustion chamber(s)</u> and used to complete task 1 and task 2. Doesn’t mater the energy going into the food or time it takes. Just that the two tasks can be completed. Anything left is wasted.<o:p></o:p></p></div><div><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>This is not the standard definition of ‘thermal efficiency’. Your (B) mixes two things. The thermal efficiency is normally the combustion efficiency multiplied by the heat transfer efficiency. This combined number is of interest to designers, as are both the contributing numbers. It is very important to understand that ‘thermal efficiency’ does not represent the fuel consumption for the reason that the fuel consumed and the energy in the fuel consumed are not part of the calculation of either the heat transfer efficiency or the combustion efficiency. Ergo, it can’t tell us the fuel consumption.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p></div><div><p class=MsoNormal><span style='color:#1F497D'>></span>C ) (energy efficiency%) = ((energy going into water for stove 1) + (energy going into water for stove 2) / (total energy from dry fuel released)) X 100<o:p></o:p></p></div><div><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>That is the classic Thermal Efficiency. The denominator is the heat that could theoretically be released from the fuel that is missing (turned into gases and particles). If you calculate the <i>actual</i> amount of heat released and put it in the denominator, you would get the heat transfer efficiency. By subtraction you can get the numbers needed to calculate the combustion efficiency.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p></div><div><p class=MsoNormal><span style='color:#1F497D'>></span>Are we on the same page?<o:p></o:p></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><div><blockquote style='margin-top:5.0pt;margin-bottom:5.0pt'><div><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>Well partly. You haven’t given a definition of the energy efficiency. The energy in the fuel consumed by the stove may or may not be released. Someone watching the fuel disappear from the forest can easily calculate how much energy is going into the stove per cooking session. A cook can easily determine how much cooking benefit they are getting. Obviously some of the energy is wasted in different forms: partially burned gases, unburned particulate matter, fuel dropped through the grate, carbon in the ash, char that is not useable in that stove. That is where the forest energy goes, and that is why I am using the term ‘energy efficiency’. It is best to discuss the performance in terms of the energy in the fuel taken from the source.<o:p></o:p></span></p></div></blockquote></div><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>I have another analogy: Plug in your electric hot plate. It has a long cord. The cord gets warm as the hot plate cooks. What is the electricity consumption of the hotplate? Is it all the energy that passes into the plug, flows along the wires to the hot plate, or is it only the energy that actually reaches the connections on the element? Should the heat lost from the long wire be included in the energy consumption or not? How does the electricity supply company view the matter?<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>If you want to know the thermal efficiency, you measure it at the element. If you want to know the energy efficiency of the whole system including the wire, you measure it at the plug. Heat lost from the long wire is wasted as far as cooking is concerned. It warms the room but not the food. <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>There is no point pretending that the heat lost from the wire is not involved in the electricity bill at the end of the month. In the same manner, there is no point pretending that unused char remaining after a cooking event does not come from the forest. A different stove with the same thermal efficiency producing less char remaining (waste) will draw less fuel from the forest. It is therefore more energy efficient.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>The problem stovers have created for donors is that on occasion the thermal efficiency has been reported as the fuel consumption, which clearly it is not if there is any unusable form of ‘leftover energy’. Ron and Paul are appealing for such leftovers to be counted as ‘fuel’ which leads one back to the thermal efficiency metric, and away from the energy efficiency metric. Stoves don’t produce fuel. Solid fuel burning stoves <i>process</i> it and inevitably waste some. <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>The UNFCCC, GEF, CDM and Gold Standard, World Bank, GIZ, and other donors galore asked for the relative fuel consumption of an improved stove measured against a quantified or default baseline (open fire). They never asked for the thermal efficiency. If they did, they did so by accident, by mistake, because thermal efficiency is not a measure of fuel consumption for a solid fuel stove, a fact recognised by, for example, India and China. <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>Stove projects that use thermal efficiency as a selection criterion are at huge risk of failure because these days, stoves might consume a heck of a lot more fuel than they completely burn. The enthusiastic promotion of TLUD pyrolysers and the thermal efficiency claims made for them is creating a crisis for foresters and funders as they gradually realise that the energy efficiency is only 2/3 of the thermal efficiency (or even less).<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>It’s not nice to fool Mother Nature.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>Regards<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>Crispin<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'><o:p> </o:p></span></p></div></div></body></html>