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Watercooling using only a water reservoir for cooling water?

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zist

New Member
Joined
Aug 11, 2003
I seem to remember someone once showing me a pic of a water setup using only blocks/pump and a large water reservoir.

I know you can use a big passively cooled radiator, but this requires a very powerful pump(or so ive been told) and you need to buy the radiator. I was thinking that if you just used a (very) large (closed) plastic tank of water that you might still achieve some decent cooling without stressing the pump(the tank could also be pretty flat so it wouldnt have to pump the water to large heights).

Anyone have an idea about how much water this requires, to manage 2-3 waterblocks(CPU(AMD)+GPU and possibly NB)?

Would a Hydor Pro(L20) pump be sufficient for this(they cost about 1/3 of the eheims 1046/1048 here).

My primary concern here is keeping everything silent, not OC'ing(though i would like to run my Barton 2500+@ 11x200Mhz, which worked fine even with my old budget water cooling kit( which was sold as a closed circuit(ie I didnt put it together myself) and leaked! )).

Sorry if these are stupid questions, but im a bit of newbie to watercooling.
 
You may be able to use a fully passive resevoir system, but it would still work with evaporation, meaning that you would have to have the res be open air, and it would require frequent refilling. Here is an instance of this at work. But it will only be effective for a few hours of usage, and I highly doubt that it would be able to handle all those waterblocks. I would strongly suggest a heatercore, coupled with a fan like this undervolted to 12V for absolute silence. In the long run, it will be a cheaper solution, and overall far superior.
 
Ooops, this post was supposed to be in Watercooling....

Anyway, the system will need to be closed, i dont want to have to refill water frequently. But water in a tank looses heat to its surrondings due to exchange with the tank itself( i guess even thin plastic will limit this alot ) but also through radiating heat, so with a large enough tank it definetly would work(assuming that the surrondings can absorb the heat without raising the ambient temp significantly). The question is whether or not the required tank would be too large to be feasible.

I can easily have a tank with ~50L of water behind my computer(possibly even 100L, considering that 50cmx50cmx50cm is 125L).
 
First, Welcome to the forums!

Now... I don't think you could have a tank large enough to sufficiently shed off enough heat to overclock your CPU for extended periods. Your water would just constatnly get hotter and hotter until it becomes useless as a cooling mechanism. You have to have some way of shedding the heat, simply increasing the volume of the water doesn't help much. It will lengthen the time it takes for the body of water to heat up, but it *will* eventually heat up. The heat source is still there and it's going to require the same level of cooling as it did before you started this cooling plan.
 
It will work, but you will need to refill your water every few hours. If this is feasable for you, it should work fine. You may want to look into constructing an evaporative cooler(bong). This won't need refilling for far longer periods of time, give you silence, and actually superior performance to a radiator. This may give you some ideas.
 
It won't work in a closed system, as the others hinted at too. There is a very good reason for the design of heatercore/radiators, being densely finned and preferrably made out of copper- that reason is heat transfer. There is no way a closed bucket, even with say an aluminium wall would be able to achieve even a sizeable portion of the radiator's heat exchange rate.
You may realize how problematic that is if you think about how even a good radiator needs dedicated airflow for proper operation... your bucket would be without airflow too.
 
I fully understand your scepsis(sp?), and i do know that it will be hugely inefficent(poor surface area for one as pointed out, and no airflow), but even so i think you underestimate the heat-loss that will take place.
Try taking a (closed) bottle of water from the fridge, and place it in your living room, it doesnt take long for the water to get from cold to lukewarm pretty quickly (or more similar a hot cup of coffee/tea).
But 50L might not be enough, if we heat the water by 300W(as an upper bound), it takes 7*50 = 350 seconds ~ 6 minutes for the 50L to increase its temp by 1C. The water will loose heat to the surrondings in proportion to the temp difference between the water and the surronding air. So when the water is 1C higher than its surrondings the exchange will be minimal, but when the difference is 10C it will be alot more. This and alot of missing variables makes its very hard to calculate how much heat 50L of water can handle, and at which temperature it will stabilize.

Im considering putting this to the test by simply boiling 1.5L of water and then putting it in a plastic bottle and see how fast the temp decreases. That way i can calculate backwards how many watts it losses.
 
Right... if you took a SEALED plastic jug of hot water into a room at 70°, it wouldn't take long for that jug to cool down to room temperature. However, you aren't talking about a stationary source of heat, you are talking about continually ADDING to that heat. Even if that were the only factor here, it still wouldn't be possible. The main thing however is the fact that you are trying to COOL with this water. Every hour that passes, you'll see the water temp get hotter and your CPU cooling will suffer as a result. The only way to shed off the effects of that 'constant' heat source is to either periodically shut it off, ergo defeating the purpose of a 'real' cooling system, or implement some sort of active cooling, ie a radiator loop.

A solid body of sealed liquid will not shed it's heat by itself. It *will* get hotter eventually, unless *some* other method of heat dissipation is introduced. Evaporation... radiation... chilling... Otherwise, your only 'cooling' feature is the temperature of your surroundings, which without lots of airflow and/or surface area, mean little to a contained body of water.

sorry, I just don't think this is possible.
 
Some physics guy once told me that a cup of coffee looses most of its heat due to heat radiation, which shouldnt be too limited in this case.(i would have assumed that it lost the most to evaporation though)

And yes i know that i will be adding to that heat, but lets say that at 35C the water looses 300W by itself(probably unlikely without huge amounts of water, but thats what im trying to find out), then the temperature will *stay* at 35C when heated by 300W.

I've seen a watercooling shop sell a thin 60cmx60cm "mini" radiator(of the standard type you would use to heat an office etc), and claim that it can dissipate 550W of heat. Thats of course with a slightly better surface area, and better material(iron or copper), but it does give an idea of how much heat its possible to dissipate. Of course the water might need to become pretty hot compared to the surrondings to achieve this dissipation but i dont mind a full load temp of 55-60C on a hot summer day.

I will test it with 1.5L of water and see how much heat it dissipates at different temperatures. You guys are probably right, but i think its worth it to be sure :)
 
A jar of coffee may cool quickly with no airflow, but now put that coffee in a radiator and blow a little air through...

I think the tank would have to be sufficiently massive - my guess would be about the size of an ATX case.
 
My guess would also be something like a standard ATX case, but i might actually be able/willing to have so large a tank.
 
if would be fine if your res was a fish tank =d the huge glass panels and the massive amount of water should be enough to radiate the heat and keep it more or less at room temp..

you can keep the pump in the tank too =d
 
A 1 liter sealed plastic container will not work, period.

However, the concept is not one that will never work under any circumstances- if only you take it far enough.

Now, there are a couple of ways to go about this; a smart way and a way that is less so. I'd like to call the smart way the one where you increase the surface area and use superior materials (Cu, Alu) for it. This is a concept vividly embodied in a radiator.
Alternatively, you could add a huge buffer and increase the surface area only a bit, using an insulating material... in short, the sealed bucket theory. :)

PS The coffee and fridge water stories have some unnoticed aspects to them; the steam coming off the coffee = evaporation, which adds a lot to the liquid's temperature loss. As for the cold water; yes it becomes warmer (though it can take long for a bottle that remains sealed)- but the temperature delta between ambient and 3C fridge temp is rather big too.
 
Of course 1L(or 1.5L) of water wont work, i just dont have anything larger handy that i can easily test with. I just want to get an idea of how much heat 1.5L can dissipate(and yes i know i cant just multiply it up to say 50L, because the surface area ratio will change).

I *would* prefer using a normal radiator(not normal like in watercooling but normal like in heating houses :), but i suspect it requires a *very* powerful pump, and such a pump is likely to be noisy(not to mention expensive). Still such a radiator might be just as easy to find as a large water container, perhaps even more so.

I think ill start looking out for people that are throwing away big water containers/radiators :) Ill get back to you on my sealed bottle experiment ;)

cherryp00t:
If i had a fish-tank, trust me, gold fishes would already be swimming past my CPU :)
 
How about using a really large radiator from a car, preferably with a V8. Inlets are usually on the top with outlets on the bottom so Gravity is assisting in making the water flow through the radiator.
 
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