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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoPlainText><span lang=EN-US>Dear Andrew<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>"</span> <span style='color:black'>As all the conduction through the TEG passes into the pot and the power output is 2W then how much of the 2kW or so passes through the TEG? <o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'>Either way the loss to the pot is only the 2W."<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>It is hard to know because we don't know the </span><span style='color:black'>Ä</span><span style='color:black'>T across the TEG. It might be a lot of that heat, but at a low temperature drop.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>The fact that adding cold water increases the power output noticeably inclines me to believe the drop is small and the flux high. Much safer in the long run to run the hot junctions cooler in case someone puts a cutting torch under the pot.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText><span style='color:black'>Regards<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'>Crispin<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p></div></body></html>