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Re: NFC: Peltier Junction Cooling

>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