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**To**:**nfc at actwin_com****Subject**:**Re: NFC: Peltier Junction Cooling****From**:**mbinkley at earthling_net (Mark Binkley)**- Date: Fri, 8 Jan 1999 13:59:36 -0400

>Electrical work: W = VA = 12*8.5 = 102 watts. >W/Q >= 1 - Tc/Th, so Q <= W/(1 - Tc/Th). >Tc = (0 + 273) K = 273 K. Th = (50 + 273) K = 323 K. >Therefore, Q <= 102 watts/(1 - 273/323) = 659 watts. > >So this particular device could transfer _at most_ 659 watts (2249 Btu/hr). >The actual amount is probably a lot less. > >If we knew the actual heat transfer rate, we could compare the efficiency of >the Peltier junction with that of an ordinary refrigerator or air >conditioner. (Just convert the heat transfer from Btu/hr to watts.) OK, now calculate that for a gallon of water (assuming no heat is lost through glass or elswhere). How many degrees temp can a gallon of water be lowered with that much power? I have no clue how to calculate this. Mark Binkley Columbus Ohio USA <))>< mbinkley at earthling_net

**Re: NFC: Peltier Junction Cooling***From*: "D. Martin Moore" <archimedes at master_localink4.com>

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