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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoNormal>Dear Testers<o:p></o:p></p><p class=MsoNormal>The presentation on accessing the Gold Standard carbon financing mechanism given at ETHOS on the weekend mentions that credit can be given for relative performance on the basis of thermal efficiency.<o:p></o:p></p><p class=MsoNormal>There are quite a number of people testing these days and it would be helpful to have a discussion to bring forth issues to do with the calculation of thermal efficiency, what it means and how to measure it.<o:p></o:p></p><p class=MsoNormal>This need is separate from the UN Foundation’s Global Alliance which is a US-based group managed by Timothy Wirth <a href="http://www.unfoundation.org/about-unf/our-leadership/timothy-e-wirth.html">http://www.unfoundation.org/about-unf/our-leadership/timothy-e-wirth.html</a> <o:p></o:p></p><p class=MsoNormal>It seems the PCIA will merge into the GA’s stove programme (can someone confirm this impression?) with resulting pressures to find some common ground for performance testing. <o:p></o:p></p><p class=MsoNormal>Reviewing tests recently performed by David Kamemia (SeTAR Centre) using the WBT 3.01 I was pleasantly surprised to find a number of corrections in the mathematics including a separate reporting of the thermal efficiency for the boiling and simmering sections. I am not sure how far the corrections have gone but it is definitely better than the v3.0 that has been used until recently by many testers.<o:p></o:p></p><p class=MsoNormal>With that frequency of use in mind, I am proposing that a thermal efficiency formula error in the simmering portion of the test be recognized and possibly corrected.<o:p></o:p></p><p class=MsoNormal>The simmering efficiency in WBT3.01 is given by adding the change in heat content in the simmering water (usually it is a negative number because the temperature drops) to the energy needed to boil off the missing water. Issues that can be discussed separately are the pointlessness of making this calculation in the first place, the use of lids significantly changing the apparent thermal efficiency and the reporting of the boiling power divided by the simmering power as a ‘turn down ratio’ when clearly it is not the turn down ratio of a stove. These issues are ignored bellow.<o:p></o:p></p><p class=MsoNormal>The formula calculating the thermal efficiency uses the average mass of water in the pot during the simmering test. I am pretty sure I agreed with this formula at some time in the past to state the ‘mass of water simmered’. It is indeed the average mass of water simmered, but it is not the right mass to use for calculating things related to heat. Here is the average mass formula:<o:p></o:p></p><p class=MsoNormal>[1] (M1+M2)/2 = average of the initial and final mass when M1 is the initial mass and M2 is the final mass.<o:p></o:p></p><p class=MsoNormal>Tf = final temperature and Ti = the initial temperature. The average mass is next multiplied by (4.186 * (Tf-Ti)) yielding the average mass of water simmered times heat capacity times Final temp minus Initial temp. The answer is in Joules.<o:p></o:p></p><p class=MsoNormal>The mass of water evaporated during simmering, M3, is multiplied by 2260 to give the quantity of heat needed to boil it away.<o:p></o:p></p><p class=MsoNormal>It can all be written as:<o:p></o:p></p><p class=MsoNormal>[2] (M1+M2)/2 * 4.186 * (Tf-Ti)° + M3 * 2260 = Joules net. <o:p></o:p></p><p class=MsoNormal>It is assumed by WBT3.x to mean the number of Joules of absorbed during simmering which obviously it is not but that is another conversation. My point is that the formula does not give the right answer anyway.<o:p></o:p></p><p class=MsoNormal>The thermal efficiency is taken to be the number Joules as calculated above divided by the heat applied to the pot (mass of fuel consumed * LHV): Joules absorbed/Joules offered = efficiency.<o:p></o:p></p><p class=MsoNormal>So, what to do?<o:p></o:p></p><p class=MsoNormal>Assume we are at sea level and water boils at 100 C. Assume the simmering temperature is 95 C (final). Assume the initial mass of water is 1000 g and the final mass is 900 g.<o:p></o:p></p><p class=MsoNormal>The WBT3.01 formula, as is, gives <o:p></o:p></p><p class=MsoNormal>[3] (1000+900)/2 * 4.186 * (95-100)° + 100 * 2260 = 206,117 Joules<o:p></o:p></p><p class=MsoNormal>Initially the mass is 1000 g. It is 100 C to start. 100 g will be boiled. It took 100 * 2260 = 226,000 Joules to boil off the 100 g. All that heat came from pot, whether it was stored in the water or received it from the fire.<o:p></o:p></p><p class=MsoNormal>The heat remaining in the water not evaporated (the “enthalpy” of the water) dropped. But the water that was boiled off did not participate in this dropping, only the 900 g of water that was left behind. So the enthalpy of the water in the pot dropped by: 900 g remaining * 4.186 * (Tf-Ti which is -5 degrees):<o:p></o:p></p><p class=MsoNormal>[4] 900 * 4.186 * -5° = -18,837 Joules (Enthalpy change is negative because the temperature dropped).<o:p></o:p></p><p class=MsoNormal>If you observed that some of the water evaporated when the temperature is below 100, it makes no difference as it takes more heat to evaporate water below the boiling point. The increase is nearly linear going from 2257 at 100 C to 2501 at 0 C.<o:p></o:p></p><p class=MsoNormal>So the final answer is -18,837 + 226,000 = 207,163 Joules which is 0.5% higher. Not much, right? Right. But it is right.<o:p></o:p></p><p class=MsoNormal>Here is the formula:<o:p></o:p></p><p class=MsoNormal>[5] M2 * 4.186 * (Tf-Ti) + M3 * 2260 = Joules<o:p></o:p></p><p class=MsoNormal>Let’s make some tomato sauce for tortillas in an open pot:<o:p></o:p></p><p class=MsoNormal>Assume we are at sea level and water boils at 100 C. Assume the simmering temperature is 95 C (final). Assume the initial mass of water is 1000 g and the final mass is 400 g.<o:p></o:p></p><p class=MsoNormal>[6] (1000+400)/2 * 4.186 * -5° + 600 * 2260 = 1,341,349<o:p></o:p></p><p class=MsoNormal>[7] 400 * 4.186 * -5° + 600 * 2260 = 206,117 = 1,347,628<o:p></o:p></p><p class=MsoNormal>The difference is 6,279 Joules but only 0.47%<o:p></o:p></p><p class=MsoNormal>Lastly, use a pressure cooker to boil some tough beef:<o:p></o:p></p><p class=MsoNormal>Assume we are at sea level and water boils at 110 C. Assume the simmering temperature is 105 C (final). Assume the initial mass of water is 1000 g and the final mass is 995 g.<o:p></o:p></p><p class=MsoNormal>[8] (1000+995)/2 * 4.186 * -5° + 5 * 2260 = -9,578 J<o:p></o:p></p><p class=MsoNormal>[9] 400 * 4.186 * -5° + 600 * 2260 = 206,117 = -9,525 J<o:p></o:p></p><p class=MsoNormal>The difference is 52 Joules but still only 0.55% - not much, but consistently different.<o:p></o:p></p><p class=MsoNormal>If you apply the last result to an efficiency calculation, dividing it by the Joules of heat released by the fire, the answer will (normally) be a small negative number. I leave you to think about the implications of <i>that</i> when reporting that the simmering performance of the stove is negative.<o:p></o:p></p><p class=MsoNormal>Regards<o:p></o:p></p><p class=MsoNormal>Crispin<o:p></o:p></p><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><o:p> </o:p></p></div></body></html>