Does compressor cooling consume a lot of electricity?
I'm thinking of trying it but does it consume a lot of electricity? Its easier for me if I wanted because I have a friend who deals with air-conditioners. How much would an old compressor cost? It would'nt be nice to ask for something very expensive from him.
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less then pelts if you compare it by cooling power.
[This message has been edited by Freakyfrank (edited 03-10-2001).] |
Well yes and no. Pelt cooling is less efficient then compressor cooling, but compressor cooling can eat more power then a pelt any day.
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I'm starting to question that. . .vapochills unit only consumes 50watts. . .that's a far higher cooling to power consumption ratio than any pelt I've seen.
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isnt that 50Watt at 120V?
------------------ C-ya Joe - Owner/Editor www.ProCooling.com Where the Completely Addicted Come to Cool Off Somebody set up us the bomb. |
nope, that's 50watts at 12volts. . .it plugs into any standard computer PSU. you hook it up to the same PSU that the rest of your computer components plug into. it draws only 50watts of power from the PSU.
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Hmm wow, well it does make sense. The Freon Vapor phase process is vastly more efficient then Pelt cooling.
------------------ C-ya Joe - Owner/Editor www.ProCooling.com Where the Completely Addicted Come to Cool Off Somebody set up us the bomb. |
Ok I just asked my friend about it and he says it will waste a lot of electricity. Thus I don't think he'll be happy to pass me one.
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Well just cuz it eats a lot of electricity doesn't mean that it is expensive... or does it?
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but power is a work measurement, in electricity its P=IxE, if you know the power and the voltage, then its I = P/E. so the variable is the I or the Amps, 50 watts is 4.166 amps at 12 volts or .4166 amps at 120 volts. but either way, 50 watts is still 50 watts. as you increase the voltage the current will drop. This is the primary reason the power companies run such high voltages on there transmission lines.
Ron |
Not exactly... They use such high voltages to reduce energy losses. It's formula to calculate lost energy is W=R*I^2 so you see you want a very high voltage and a very low I.
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