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The way to avoid fans and radiators in your loop - Page 5

post #41 of 57
Quote:
Originally Posted by httuner View Post
what about the algae buildup that'll happen?
http://www.koolance.com/water-coolin...product_id=944

As I mentioned. Run a closed loop for your pc, run whatever non-corrosive fluid you want for the heat disposal medium, (oil, water, Sex Panther by Odeon, etc).
 
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post #42 of 57
Quote:
Originally Posted by nowwater View Post
Please! For the sake of my research! Please, tell me how did you get the "2.47 watt/hours per gallon per degree"!
Look up specific heat capacity of water and change units.

4.184 joules/gC, ie 4.184 joules of heat to raise 1 gram of water by 1 degree C.

1 joule = 1 watt applied for 1 second
3600 seconds in 1 hour
1000 grams h20 = 1 liter h20 = .264 gallons


I get 4,184/3600= 1.16 watt applied x 1 hour to heat 1 liter h20 by 1C or

1.16 * 3.78 = 4.39 watts applied x 1 hour to heat 1 gallon h20 by 1c

For a 20 gallon aquarium, 439 watts applied for 10 hour to heat 20 gallons of water by 50C
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post #43 of 57
I was going to say it is 527 watts to raise 20 galls of water by 50 deg in 10 hours assuming no heat losses then I realised that your US gallons are different to my imperial ones

From this though anyone can see that for some this may indeed be a solution of sorts.

If you can genuinely say that you never exceed a given number of hours using your pc
If you can genuinely say that your processor runs cool enough to accept a rise in temps of "x" degrees
Then you could calculate the size of tank/mass of water you would need to run silent and putting a rad with fans in the loop would enable a fast cool down if the situation arose.

For 24/7 use this is a fail unless the tank has enough surface area to lose heat to the surroundings faster that you are adding it above a certain temp differential. I believe this would obviate the use of a fish tank
post #44 of 57
Quote:
Originally Posted by u3b3rg33k View Post
http://www.koolance.com/water-coolin...product_id=944

As I mentioned. Run a closed loop for your pc, run whatever non-corrosive fluid you want for the heat disposal medium, (oil, water, Sex Panther by Odeon, etc).
I was thinking in terms of having fish/plants in the aquarium, how fishes poop and algae can buildup quite fast due to living things in the aquarium thus creating all sorts of mess in the aquarium that'll eventually contaminate the water =] Then this water will be pump throughout the water cooling system causing ill-effects.

lol, I mean whats the point of having an aquarium if you're not gonna put fish in it =] But that heat-exchanging is really neat
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post #45 of 57
Quote:
Originally Posted by nowwater View Post
Why does the water has to be evaporated?
This is exactly like the water in the reservoir. The reservoir is sealed up.
No evaporation can happen.
1. Water evaporates from reservoirs...give it time

2. The aquarium idea will be fine for 20-30 minutes. It just delays the inevitability of water heating up. Think of it as boiling water. A small pot of water boils quickly...a large pot takes longer.
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post #46 of 57
Quote:
Originally Posted by nowwater View Post
Please! For the sake of my research! Please, tell me how did you get the "2.47 watt/hours per gallon per degree"!
I just did a quick search and found that here:

http://wiki.answers.com/Q/How_many_w...ter_one_degree

Didn't check using specific heat formulas or anything. Thermo was one of my weak subjects in college, and that was about 15 years ago now..
    
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post #47 of 57
One fairly easy to understand way is calculate in calories first, convert to joules, then divide by time in seconds

This is for imperial gallons
If 1 calorie raises 1 cc water 1 degree C... multiply by number of cc in 20 galls
90800 calories raises 90800 cc water 1 deg C Then multiply for target temp
4540000 calories raises 90800 ccwater 50 deg C then multiply by 4.184 for joules
18977200 joulesraises 90800 cc water 50 deg C then divide by target time
527.14 watts raises 90800 cc water 50 deg C in 36000 seconds
post #48 of 57
how much would the water cool off if circulated by a normal fishtank filter???

just seems like no one is putting that factor in.
post #49 of 57
Quote:
Originally Posted by AlaskaFox View Post
how much would the water cool off if circulated by a normal fishtank filter???

just seems like no one is putting that factor in.
lolwhut?

fishtank filters don't cool anything off, if anything they're known to add heat to the water because they've got a few watt pump in them.... If you mean evaporation... it depends on the surface area... I had a 4'x2' surface area with constant disruption but no fans over the top and it evaporated a gallon a day with the hood closed. I can imagine it evaporating 5-8 gallons a day if you ran a fan over it, and that certainly would reduce temps... But much more effective ways of utilizing this principal are with bong coolers which are surprisingly well documented in this forum, just do a search.
post #50 of 57
Stirring via circulating through filter isnt going to get rid of heat.

80+% heat loss of an open tank is through evaporation. OP was talking about a sealed tank, which will minimize evaporation, hence minimize heat loss. Radiation and transmission through walls, the other 2 ways of heat loss of tank, are too small heat loss to waste time approximating heat loss, which is why no one bothered to approximate heat loss.
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