# Re: NFC: Peltier Junction Cooling

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OK, 1 Joule=0.24 cal.  Can you get Joules from all this mess?
(Can't get to my physics book right now).

> >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
>
>
>

Prost,

Martin

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