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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link="#0563C1" vlink="#954F72"><div class=WordSection1><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>Dear Saloop<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'>I have look again at the formulas used for calculating the efficiency and you should be careful in drawing conclusions about the stove and its various forms until you have corrected the calculations.<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'>This formula (copied below) does <b>not</b> give the correct efficiency value. Because you are interpreting the calculated results to make conclusions about the design, it is really important that you use a slightly modified test method and correct the formula.<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><img width=378 height=71 id="Picture_x0020_1" src="cid:image002.jpg@01D02520.89168500"><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'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>There are two problems with this formula. <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>The first is that you have counted the heat gained by the water before it started boiling. No problem. Then you evaporated water, no problem. Later the water cooled and some of it evaporated. That is a problem. The heat driving that evaporation did not come from the fire (which is out) but from the cooling of the water. Energy was transformed from hot water into hot steam at a conversion rate of at least 540 kCal/kg (more, if the temperature at the time was below the boiling point).<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'>I mentioned this problem before. <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'>Here is your formula broken down:<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'>{2 (99.8-31.5) [I left out the x1]<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'>That is: water mass times Delta T. The result is the enthalpy change in the water.<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'>+(0.58 x 540}<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'>That is: the enthalpy change from water to steam for 0.58 kg of water.<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'>However you note that the test was finished and the water mass missing was measured at the end of a cooling period. Is that correct? Your description does not state explicitly that the water mass was determined at the end of the gasifier experiment, though it does say so for the Chula experiment. If so there is a problem. <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'>You have not calculated the change in enthalpy for the pot cooling between flame out and the time when you determined the mass of water evaporated. This change in enthalpy is negative, not positive and is missing from the formula.<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 is the same as the first portion but with a negative number:<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'>{? (36-99.8) [in the case of Test 1]<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'>{? (36-99.8) What number should be in place of the ‘?’. It should be the final mass.<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'>1.134 (36-99.8) <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>= 1.134 x -63.8 <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>= -72.34 Kcal heat lost from the hot water and turned into evaporation and the heating of the local environment directly from the pot surface (which is an unknown value).<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'>Thus the reasonably correct total heat gained number is:<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'>{2 (99.8-31.5) - 1.134 (36-99.8) + (0.58 x 540}<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'>Water heating – water cooling + total water evaporated.<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'>Second problem with the calculation:<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>You have allowed the pot to cool long after the fire has been removed from under it. The pot loses heat into the air, not only from evaporation. You do not know the amount of heat gained by the pot and water that was subsequently lost to the surrounding air. You do not know how much heat was lost from the pot when the water temperature was rising and or when it was boiling. Letting the pot cool for so long has allowed a large amount of heat to be lost from the pot ‘uncounted’ meaning, counting the lost heat that was not turned into evaporation. The longer the wait (without fire) the greater the unknown and the greater the error and the less confidence we can have in the final number.<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'>To correct this you need to know the mass of the water at the moment when the gasification stops and stop the experiment at that time. You have recorded the time, and the temperature, but not the water mass remaining. If you know the mass of water remaining when the gasification is completed, you can get a pretty good figure for the cold start system 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><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 are two problems with the test procedure.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>The first is that it is not clear what ‘efficiency’ you are trying to determine. An efficiency is a ratio and there are many ratios that can be reported. As I read the description, it seems to me you want to know the system efficiency, that is, the % of useful heat available in the fuel that is detected in the pot.<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 would be a very unusual cook who was ‘cooking’ while the pot reduced in temperature to 36 C. I think this is a reasonable assumption. So if you could report the mass of water either when the flame goes out or after it cools to 95 C (whether the flame is out or not – otherwise known as the Chinese Method) you can use the extended formula above that accounts for the enthalpy of cooling.<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'>To get the best answer, record the mass of water when the gasification is finished, or when water drops to 95 C (you decide and report which you have done).<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 second problem with the procedure is that the pot was allowed to cool while sitting on the stove which allows some of the heat retained by the stove body to be transferred to the pot. This can be quite misleading when trying to determine whether or not the change in architecture is affecting the performance. The cooling energy calculation has such large unknowns in it that decisions about results may be misguided because of something taking place long after the real cooking has stopped. That should be avoided.<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'>Regarding the secondary air inlets, my previous comments still apply. I won’t repeat them here. The number of holes it too large and the diameter too small to impel the air into the centre of the gas chamber. <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'>I agree with all Julien’s comments. I believe a better seal between the riser and the body would prevent hot air which is now rising past the outside of the riser and into the flaming area at the top edge of the riser. That air is surely cooling the flame and pot surface and interfering with the transfer of heat to the pot.<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><p>Respected sir,<br> Please find the resized documents attached with this mail. It has been sized down to 800 KB. Hope it is possible to post it.<o:p></o:p></p><p>Thanking you,<o:p></o:p></p></div></body></html>