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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='color:#1F497D'>Dear Ron<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'>I am answering your post first because of its currency but I am not forgetting that I need to respond to Paal and Paul from the weekend and Vetle’s last very information-filled message. I have been extremely busy over the past week preparing for the COP 17 demo and beating Old Man Winter with some brickwork and also today’s demonstration of three burning modes and fuels for the visiting delegation from China (which went well).<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><div><div><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>I like the computations you have provided, but think there are additional efficiencies needing computation. Yours below gives zero benefit to the (very sizeable) charcoal production.- which obviously will entail some howls from those (like me) interested in Biochar.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I went through the spreadsheet first and it seems you have captured the numbers well. The column D numbers are only very slightly different from the ones I used and they are because I had to round slightly to give the mass data. <o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I do immediately have a question as to whether the ‘efficiency of char production’ has anything to do with ‘thermal efficiency. Yes they can bot he viewed as efficiencies but they are not related, of course. <o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:black'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>First, I think this sort of efficiency measurement should be done without a pot lid. </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Why?<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>The 22% number which is now measuring something about the tightness of the lid (and not related to the stove) should jump to above 25%, I guess.<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Of this I am absolutely sure: tightness of a lid fit has nothing to do with heating efficiency of boiling temperature or anything other than limiting heat loss. The ‘tight lid’ theory is an old saw that does not bear repeating. In fact I recall a discussion on this list in which I showed that putting a very heavy 1 kg perfectly fitting gas tight lid on a pot compared with a loose lumpy lid was the equivalent of moving the stove 5 floors up a building. It is irrelevant, literally. The pot was a good quality stainless steel one with a good lid fit (loos but sealing well).<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>I will send an Excel spread sheet to Erin and Tom on this – so others can play with it. </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I received it, thanks.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>That SS shows that about 50% of the initial energy remains in the produced char</span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I am not immediately clear what SS means. I agree that the remaining char contains roughly 50% of the initial energy available in the fuel. I also immediately caution everyone to realise and remember and repeat that this is a highly conditioned statement. It is only true because of the moisture content of that particular fuel with that particular analysis. I have to revise the calculation because the ask content is apparently 3% or so (confirmed by Meredith from REAP Canada this morning. We were on a tour of switchgrass production and a pelleting plant of vast capacity.)<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>>…</span><span style='font-family:"Arial","sans-serif";color:black'>I say that makes the overall stove efficiency, making no other changes in Crispin's work, about 75% (or 25% "loss" – not 75% loss). </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Well we have to be careful what is claimed here. The boiling efficiency of the pot from beginning to end was indeed 22% of the available heat. It was not 22% of the heat offered to the pot. That is for sure. It is also not correct to make the calculation of the char-subtracted value without pointing out the implications of a slightly different moisture content because in these cases where the char mass is high, a change in moisture gives a disproportionately large increase in the apparent efficiency. In fact, I tossed in the 22% because I did not have a high heat efficiency which I would much have preferred. I will try to get one in the next few days if anyone thinks it is important. It is certainly much higher than 22%.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>That is, I claim that just adding the boiling and charcoal-making efficiencies together is valid. The missing 25% is energy that was not captured usefully.<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The char-making efficiency is related to the theoretical maximum char possible and has nothing to do with heat transfer efficiency. I am surprise you would want to add them together.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The engineering approach would be to take the heat offered to the pot and evaluate how much of it was retained within the pot in the form of hotter water or evaporated water. That is not the heat transfer efficiency, it is the system efficiency. The system efficiency heat plus the heat passing into the pot and lost through radiation, conduction and convection added together would constitute the total heat transferred. No one on this list that I have ever read (which is not the whole history) has offered a heat transfer efficiency for consideration. Maybe we should give some thought to that.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>But, even without adding the charcoal – making efficiency, one can approximately double (as opposed to the above-stated tripling) Crispin's calculated 22% “boil-away” efficiencies by dividing the useful output by the net energy consumed in the boiling task, not by the total amount available . </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Correct.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Heat retained by the pot (as described above as system efficiency heat) <i>H</i><sub>1</sub> divided by the Heat potential, minus heat value of the char remaining <i>H</i><sub>2</sub> = system efficiency.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><i><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>H</span></i><sub><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>1</span></sub><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> / <i>H</i><sub>2</sub> x 100 = Eff %<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:black'>To repeat - I calculate that about half the initial available switchgrass energy is still available in the produced char, and that this MUST be accounted for when comparing a pyrolysis stove with a combusting stove. </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I do not see that it gets special treatment. The stove type is not affecting ordinary engineering calculations. If a stove make 5 g of 500 g of char, determining the system efficiency requires that the heat value of the remaining char be deducted.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>When it comes to evaluation the performance of a stove in the field, in other words a practical evaluation of fuel consumption, one has to consider whether or not the char can be ‘used’. It might be for fuel or agriculture. If it cannot be re-used in the same stove, it is lost to the stove and fuel and is a loss. I have no problem with this being reflected in a note about the performance. But I think it is a stretch to say that from and energy use point of view, a stove that produces char is ‘twice as efficient because it does not burn the whol fuel’ which is what I understand from your calculation where you add the % char to the system efficiency.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>This comes closer to the efficiency on the water boiling side, but still does not fully capture the value of the char. </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The system efficiency is not intended to capture the value of the char because it has nothing to do with the combustion or the heat transfer. It is a completely separate subject with its own rules and understandings. It would by quite legitimate to point out how much of the carbon in the original fuel was in the char, for example, because someone might want to optimise the retention of carbon. <o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Would you agree that it is reasonable to report the amount of heat that was useful to the pot as a % of the potential of the fuel, as well as the % of the heat offered to the pot? In fairness, the combustion efficiency of the fire should be taken into account when making this calculation of system efficiency, and should be demanded if calculating the heat transfer efficiency. The heat transfer efficiency is a very narrow examination of something taking place within a chain of events – one link of it, so to speak.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>The lost energy is about half of the non-char energy (not 3/4 of the input energy - as implied with a 22 or 25% efficiency computation).<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I did not state any losses or describe them. You have inferred that I said the char as a loss. It is left over and I said so. It might be fuel, it might be fertiliser. In the light of the comments I presented about the thermal efficiency having to overcome the carbon subtracted from the fuel-stove energy chain, I think it is relevant to present all the information so the user can interpret the test in a way that is applicable to their intended purpose.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>To restate, with simple numbers that might result with a re-test without a pot lid, that hypothetically finds that 25% of the energy was productively used in boiling the water and also found that half the energy remained in the char:<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I do not see how taking off the lid would do anything other than increase the radiation losses (which are not counted in a standard test) and increase the evaporation (which is). The result of taking off a pot ALWAYS results in a lower system efficiency number. The reason is that water has a emissivity coefficient of about 0.99, similar to black oil. If you take off the pot, a lot of heat is lost by radiation to the sky/ceiling. This loss is not countable by present methods and appears to be an inefficiency, that is it lowers the system efficiency number in all cases. There is no case where you can increase the uncounted loss, and increase the system efficiency number.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>>…</span><span style='font-family:"Arial","sans-serif";color:black'>My preference: eff3 = eff1 +eff2 = eff1+ 2*eff1 = 3* eff1 = ¾ </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>You cannot add efficiencies that if they are producing the same thing like heated water, but not if they do not have the same units. Steam and char are not the same units.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>(I believe that having the same denominator when calculating efficiencies is essential. </span><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Exactly. The denominator of one is MJ and the denominator of the other is g.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>></span><span style='font-family:"Arial","sans-serif";color:black'>Of the ¾ useful output 2/3 is in the char and 1/3 is in the boiled water; one quarter of the input energy was not captured.)<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>It is important to note that char is not a ‘useful output’ from heat generation point of view. It is unburned carbon. It went in unburned and came out unburned. That is not an efficiency.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:black'>Combined (EPA-like) eff4 = eff1/(1-eff2) = eff1/(1-1/2) = 2 eff1 = ½ (I think this undervalues the char – and has no theoretical basis when computing efficiencies. It does a better job of bringing in the char, but not the correct calculation. This says that about half of the available energy passing the pot is captured. It does not give enough credit to the char.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>But about half the energy offered to the pot DID enter the pot. How well it made char is irrelevant to the system efficiency. I am tired of repeating that so I will end now.<o:p></o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-family:"Arial","sans-serif";color:#1F497D'>The spreadsheet:<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The char content was expressed (as normal) as a % of the dry fuel mass, not the wet fuel mass because if expressed as a % of the wet fuel mass, one cannot make correct comparisons between different stoves and fuels. Yes, I know people usually report the wet fuel number but it is not very helpful.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I have corrected all the numbers relating to ash content and derived therefrom.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>It was 97 degree rise – sorry if there was some mistake. It was 3-100 heating.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I have added the correct heat value for dry switchgrass and calculated the heat available.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The line about pot lid off has been removed. [It occurs to me that an understanding that the lid coming off would increase the thermal efficiency may be rooted in the error in the WBT3 that calculates an ‘efficiency’ for simmering which is itself a conceptual error. Increasing the evaporation during a simmering test appears to increase the system efficiency which is impossible. It is an artefact of the math error. Any WBT3 efficiency number is incorrect for this reason. Only the high power heating efficiency is valid.]<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>As the char heat content is included ‘above’ in the sheet I have removed all the lower section with estimates and the combining of thermal efficiency and char making realised.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The Average system efficiency is 56.66%. It also produced char which could theoretically have been 100% of the carbon and 0.176 as much ash and volatiles (which is 15% of the total mass). That char could have been 617.6 which I calculated as 1050g x 50% Carbon / 0.85 = 617.6 g.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The char realised was 365g so it is 59% of the char that might have been produced.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Some may jump in here with a ‘fixed carbon’ angle. Well, there is carbon in the volatile materials and it is no possible to know exactly how much because there is no exact value for what is ‘fixed’ save a different number for each and every temperature. The best we can say scientifically is that is it 59% of what might have been.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Now, it is clear that adding 56.66% thermal efficiency to 59% char production realised does not make sense. It is not a stove with ‘115.8% efficiency’. I hope that is obvious to everyone. It is an efficient stove that makes char. If that is what you want, maybe use it. As mentioned in previous posts, is it constructed based on the formulas used to make a Vesto. The main difference between this and other TLUDs like the Anderson gasifier and Peko Pe is there is a lot more flame space which provides more draft to the secondary air which provides better mixing and turbulence. It does not have a pot skirt or even a proper top deck the size of the pot, both of which might increase the efficiency.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>As I said, this is a pretty efficient little stove! It cost less than $1 to make and cooks for about 2 hours per kg. I will try to get a better number for different stages of the burn. The power drops slowly during the burn. There was no un-pyrolysed fuel at the end.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>The edited spreadsheet is attached.<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Regards<o:p></o:p></span></p><p><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Crispin<o:p></o:p></span></p></div></div></div></body></html>