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**To**:**<nfc at actwin_com>****Subject**:**Re: NFC: Peltier Junction Cooling****From**:**"Andrew S Dalton" <ASDALTON at prodigy_net>**- Date: Fri, 8 Jan 1999 19:05:24 -0500

-----Original Message----- From: Mark Binkley <mbinkley at earthling_net> To: nfc at actwin_com <nfc at actwin_com> Date: Friday, January 08, 1999 1:07 PM Subject: 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. If there's no heat leakage (yeah, right), then any amount of cooling, no matter how small, could reach the temperature that you want. The difference is how long it would take. Let's say that the Peltier junction has a heat transfer capacity of only 50 watts (171 Btu/hr). A Btu is defined as the amount of heat needed to raise 1 pound of water by one degree Fahrenheit, and there are about 8.5 lbs of water in a gallon. If you have a 20 gallon tank at 80 deg F and want to lower it to 60 deg F, the amount of heat you need to remove is the following: (20 gal H2O)(8.5 lb H2O/gal)(1 Btu/lb H2O/deg F)(80 deg F - 60 deg F) = 3400 Btu. The amount of time needed to go from 60 to 80 deg F would be: 3400 Btu/(171 Btu/hr) = 19.9 hr. Ouch! And that's assuming that no heat leaks into the aquarium, which we know isn't the case. This doesn't tell us how much power we would need to _keep_ the tank at the lower temperature, but it gives a good indication that 50 watts of cooling won't be enough. If you want to know how much power is needed to maintain a low temperature, it is necessary to take into account the heat leakage into the tank from the outside air. To do this, we can take advantage of the fact that the rate of heat transfer is directly proportional to the surface area, and also directly proportional to the difference between the temperatures of the aquarium and its surroundings. Mathematically: Q = U*A*(To - Ti). Q is the rate of heat transfer, A is the surface area of the tank, To is the temperature of the outside air, and Ti is the temperature inside the aquarium. U is an experimentally determined number called the overall heat transfer coefficient, and it is usually a constant. (If I remember correctly, the R-value of home insulation is the reciprocal of this number, or 1/U.) For our purposes, it is not necessary to separate the specific heat transfer and area terms, so we will combine them into a single parameter UA, which has units of watts/deg C or Btu/hr/deg F. If you know UA for your aquarium, then the rate of heat transfer required is easy to determine. I have to work out a few equations first, but I'll post another message explaining how you can do a simple experiment to determine the UA of your aquarium. ------------------------------------------------------------ [\ Andrew Dalton [\[\ [\ [\ [\[\[\[\ [\ [\ [\[\ [\[\ [\ [\ [\ [\ [\[\[\[\[\[\[\[\ [\ [\ [\[\ [\[\ [\ [\ [\ [\ [\[\[\[\ [\[\[\[\ [\ [\ [\ [\ [\[\ [\[\ [\[\ [\[\ [\ [\ [\ [\ [\ [\ [\ [\ [\[\[\[\[\[\[\[\[\[\[\[\[\[\[\[\ ------------------------------------------------------------

**Re: NFC: Peltier Junction Cooling***From*: Sajjad Lateef <sajjad at uic_edu>

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