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</o:shapelayout></xml><![endif]--></head><body bgcolor=white 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 Paul<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'>Thanks for that, and I will add a correction to the part you highlight:<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><div><p class=MsoNormal><b><span lang=EN-US style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>></span></b>So the difference in the TEMPERATURE OF THE WATER is of little consequence, being about 3 deg C. <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'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D'>The energy required to evaporate water at a temperature other than 100 degrees (at sea level pressure) is higher. The difference is 2.44 Joules per degree under 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'>Thus the idea that it takes a different amount of energy to lose water at a different temperature is not correct.<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 calculate the energy gained by a pot that is evaporating water (and ignore the losses from the pot itself, which are fixed) the calculation is not evaporated mass x 2257 Joules/gram.<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 boiling temperature minus temperature at which the water evaporated x 2.44) + mass evaporated x 2257 = Joules.<o:p></o:p></span></p></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'>If you don’t know the temperature at which the water went missing, and you only have the final temperature (which is the case in the WBT) the calculation 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'>Missing water mass x 2257 + missing water mass x (boiling temperature minus final temperature) x 2.44. That is the heat that was gained by the pot plus the heat that was lost be a reduction of temperature (and that actually heat turns into evaporated 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'>The correct accounting for this energy is straight forward and not novel. The WBT formula for the heat gained by the pot is almost correct. The HTP does it correctly. Other methods do it incorrectly (Indian and Chinese methods). All these sorts of things are picked up during external review. If the methods were reviewed by experts, these mistakes would have been eliminated long ago (we can assume).<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 a more fundamental problem with this ‘heat gained’ problem however. Knowing how much heat (net) is gained by the pot does not mean we can calculate ‘the efficiency of simmering’. There is no ‘efficiency of simmering’ that can be expressed on a thermal basis. This fundamental fact means the metric is wrong to start with. As a measurement, it is not in the IWA because by then the EPA and others had accepted the fact it was not useful. Continuing to talk about the thermal efficiency of simmering only indicates that one does not understand what it means. Simmering is an arbitrary task that does not have a thermal efficiency component because it does not require ‘work’ to be performed, and ‘work’ has a definition.<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 hold a bucket of water at a constant distance above the floor for 30 minutes, you have not accomplished any work, but you will have expended a lot of energy to hold it there. You can measure that energy used, not you cannot express that energy in terms of having accomplished the task of ‘doing nothing’ that qualifies as ‘work’. You could just as well hang it on a nail. The nail isn’t doing and ‘work’ either and the ‘energy efficiency’ of hanging a bucket on a nail is mathematical nonsense.<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'>Hanging a bucket of hot water on the nail of 97 degrees is not ‘work’. Therefore there is no ‘thermal efficiency of simmering’. It is an oxymoron. The metric is invalid. Changing the calculation method doesn’t make it valid. It fails at the conceptual stage.<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'>Overcoming losses of heat from the pot by convection and radiation is ‘work’, but no one is measuring that and you can’t report the efficiency of doing something you have not measured.<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 am glad we are getting onto the same page, which is normal science and engineering.<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></div></body></html>