Correct waterblock sequence

A

aphexddb

Weaksauce
Joined
Jun 10, 2004
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I'm a complete watercooling n00b, just about to order my first kit form Danger Den. I will be cooling the cpu, gpu, and mobo. When i get everything whats the best order to link the water blocks together from the cold water input? I'm guessing first the CPU, then GPU, then the mobo. That seems to make sense, but do GPU's get really hot, like hotter than the cpu? btw The cpu is a Athlon FX-53 on the Abit AV8 if that matters..

And yes I will be overclocking :D
 
i'd guess pump -> cpu -> gpu -> chipset -> rad.

i might put chipset before gpu if it helped clean up the lines.
 
aphexddb said:
I'm a complete watercooling n00b, just about to order my first kit form Danger Den. I will be cooling the cpu, gpu, and mobo. When i get everything whats the best order to link the water blocks together from the cold water input? I'm guessing first the CPU, then GPU, then the mobo. That seems to make sense, but do GPU's get really hot, like hotter than the cpu? btw The cpu is a Athlon FX-53 on the Abit AV8 if that matters..

And yes I will be overclocking :D
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No need to worry about order at all. Just keep which ever is convenient in order :) (now if you got res... it goes right before pump usually for the priming purposes... other than that... nope)
 
flak said:
i'd guess pump -> cpu -> gpu -> chipset -> rad.
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That's probably the most commonly used, if you add a reservoir just through that in at the end right after the rad.

Some people prefer putting the rad directly after the pump to remove the heat put in the circuit by the pump, but it has it's advantages and disadvantages. Pick something that works the best for you and your case.
 
stumpy said:
That's probably the most commonly used, if you add a reservoir just through that in at the end right after the rad.

Some people prefer putting the rad directly after the pump to remove the heat put in the circuit by the pump, but it has it's advantages and disadvantages. Pick something that works the best for you and your case.
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It's a common misconception that putting the radiator after the pump cools water to go to components after it. The laws of physics and thermodynamics, however, say that this is a load of steaming BS. The water when flowing in a closed loop reaches an equlibrium temperature where the water is the EXACT same temperature EVERYWHERE in the loop, and doesn't have "cool" paths after the rad and "hot" paths before.

As long as you go directly from your reservoir (or T line or whatever you use for filling/bleeding) into your pump inlet, your flowrates will be optimal and your temps good.

-OMP
 
Reservoir into pump inlet = optimum flow, gotcha.

Follow up question, since I'm gonna use a reservoir does the type of it matter much or is it completely asthetic? I heard somewhere that drive bay reservoirs are not great, is this because of proximity to hot drives? Or should this this be a new thread ;)
 
Technically you don't need a reservoir or a Tline or anything at all to hold water in a closed loop. Reservoirs are both asthetic and useful for some systems, in that they can show off the watercooling goodness, and they allow for easier filling and bleeding. They have absolutly no impact on cooling/heating unless you actually place something hot against the side and ensure good thermal contact (don't do that :p ).

Drive bays get a bad rap because idiots tend to install them incorrectly and they leak. You are ordering a bayres from Dangerden correct? Make sure there is a good ammount of teflon tape on the barbs and caps when you screw them in, and do a complete leak test before installing the 2600$ worth of FX53 holyness.

-OMP
 
It's a common misconception that putting the radiator after the pump cools water to go to components after it. The laws of physics and thermodynamics, however, say that this is a load of steaming BS. The water when flowing in a closed loop reaches an equlibrium temperature where the water is the EXACT same temperature EVERYWHERE in the loop, and doesn't have "cool" paths after the rad and "hot" paths before.
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Could you send me a link that states this or explains this? I know physics can be confusing at times, but this seems just wrong from what I would normally think.

For example when you boil a pot of water the water boils from the hottest part, the bottom and rise's to the top. From what your saying the water should turn to vapor at once. (which would be quite an amazing site)

Not to mention the whole deal with water being the least dense at 4degree's celcius.

If you ask me pump->rad->cpu->gpu->res->repeat, makes most sense to me.
 
After a little reading this is my understanding of a water cooled setup based on the laws of thermodynamics. Some physics info here

First law of thermodynamics
The change in internal energy of a system is the sum of the heat transferred to or from the system and the work done on or by the system.

Second law of thermodynamics
The entropy -- a measure of the unavailability of a system's energy to do useful work -- of a closed system tends to increase with time.

Third law of thermodynamics
For changes involving only perfect crystalline solids at absolute zero, the change of the total entropy is zero.

Zeroth law of thermodynamics
If two bodies are each in thermal equilibrium with a third body, then all three bodies are in thermal equilibrium with each other.

So it seems to me that:
First law - basically the heat in a closed loop radiator comes from the water blocks.
Second law - the entropy (heat) increases with time.
Zeroth law - eventually all the parts in the water loop reach the same temp.

This makes sense to me, if any physics majors out there wanna whip out some corrections it would be appreciated
 
Your wrong on your intrepretation of the zeroth law, it means if you have object A in equilibrium with object C, and object B is in equilibrium with object C, then object A is in eqilibrium with object B.

Which is usefull if object A and B cant be directly compared.

But it says nothing of a closed loop being in equilibrium everywhere in the loop.

I know that water will transfer heat quickly, so my general thought is that the water will get warmer the closer to the hotspot, and cooler the closer to the radiator. And would never reach an equilibrium.
 
Lord of Shadows said:
Could you send me a link that states this or explains this? I know physics can be confusing at times, but this seems just wrong from what I would normally think.

For example when you boil a pot of water the water boils from the hottest part, the bottom and rise's to the top. From what your saying the water should turn to vapor at once. (which would be quite an amazing site)

Not to mention the whole deal with water being the least dense at 4degree's celcius.

If you ask me pump->rad->cpu->gpu->res->repeat, makes most sense to me.
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http://scienceworld.wolfram.com/physics/ThermalEquilibrium.html

This is a pretty basic definition of thermal equilibrium from wolfram, but it is pretty good for what I am trying to show.

From your boiling water example: Should the water evaporate instantly? No. When the water encounters a heat source (thermal energy from the burner), according to the 4 laws (including the aptly named Zeroth law) of thermodynamics, heat energy will be deposited into the water until the exchange rate of energy is zero. When this happens, the system is said to be in equilibrium, so the water can not be or get any hotter. This is verified by the fact that water boils at 212degrees F (100 C) and while it boils it is no hotter and no colder. Steam and other such things occur when excess energy being input by the burner can not "over heat" the boiling water, so energy is released and steam is formed, thus maintaining the equilibrium state.

Now we can apply this to a watercooling loop:

As the water encounters the heat source (waterblocks for the CPU, GPU, NB, and a little bit from the pump in this case) the water absorbs heat energy. This energy is transfered to the radiator where fins and waterchanels dissapate the heat energy. Is the water entering the radiator hotter? The answer is of course yes, but only until the system reaches equilibrium. According to what has been stated above, when a hotter area and a cooler area come in contact (I.E. before and after blocks and radiator), the system will exchange energy until at thermal equilibrium, making the water the same temperature. I know its hard to understand at first...Thermo isn't really intuitive, but the laws of physics don't lie, every part of the closed loop is the same temperature once the system has been flowing enough for the input of heat into the water can be balanced by the radiator.

Enough about thermo now...I spent enough time before I got my degree (B.S. in Physics and Astronomy) studying this sorta crap :rolleyes:

Here are some things that can help or hurt your temps depending on which route you go:

Tubing length.: Keeping the tubing as short as possible will help you a great deal. The more water you push through a system the fastest the better. This gives the radiator to dissapate more heat that the water absorbs from heat sources. If you have a decision to go pump -> rad -> blocks or pump -> blocks -> rad, and one option is a foot of tubing shorter, you want to go with the shorter option. Don't go crazy measuring and being anal retentive about tubing lengths, but it does make a difference.

Airflow: Make sure you get the radiator the coldest and freshest air possible. According to thermo, colder air = lower equlibrium temps, so you do yourself no good by using air from inside your case (which still has lots of components producing heat even though you watercool) and passing it over the radiator. Cooler outside air is the way to go. Some people like to put fans over their pump to limit the heat they put into the water. If you can do this without using the air to/from the radiator, great. If not, then whatever energy dissapated by the airflow over the pump will go back into cooling the radiator and not help.

Water additives: Some water additives lower surface tension and raise thermal capactity in water which will allow you to dissapate more heat through the radiator. I'm not an expert in this area, so do some research. Water wetter and Zerex are popular choices.

I'm tired of typing so thats all I'm going to say now. Maybe more tomorrow unless someone else has some questions.

-OMP
 
Lord of Shadows said:
Your wrong on your intrepretation of the zeroth law, it means if you have object A in equilibrium with object C, and object B is in equilibrium with object C, then object A is in eqilibrium with object B.

Which is usefull if object A and B cant be directly compared.

But it says nothing of a closed loop being in equilibrium everywhere in the loop.

I know that water will transfer heat quickly, so my general thought is that the water will get warmer the closer to the hotspot, and cooler the closer to the radiator. And would never reach an equilibrium.
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Good catch. The Zeroth law does not say anything about equilibrium in closed systems, but the law of thermal equilibrium is subject to the zeroth law. It's basically saying that upon reaching an equilibrium point, all parts in contact with the loop are in equilibrium.
 
Alright, so the rate heat enters the system soon matches the rate heat exits the system when in a steady state.

I just dont understand how you can add heat in one place and remove it from another some distance away and still have a equal temp in both areas. Wouldnt there be a small temp difference that is proportional to the distance of the two sources; due to heat needing to be transfered within the water?
 
Lord of Shadows said:
Alright, so the rate heat enters the system soon matches the rate heat exits the system when in a steady state.

I just dont understand how you can add heat in one place and remove it from another some distance away and still have a equal temp in both areas. Wouldnt there be a small temp difference that is proportional to the distance of the two sources; due to heat needing to be transfered within the water?
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Ok...Just think of it this way. 1cm^3 of block of water will pass through the water block in what? less than 1/100th of second? How much can 100watt heater heat water by in 1/100th of second? Not even noticeble amount. (just look at wattage of hair drier, etc) You are talking about such a minor difference which is immesurable by switching the order. Only reason I said reservoir first or t-line first before pump is to make sure it doesn't break by having not primed... nothing else.
 
Lord of Shadows said:
Alright, so the rate heat enters the system soon matches the rate heat exits the system when in a steady state.

I just dont understand how you can add heat in one place and remove it from another some distance away and still have a equal temp in both areas. Wouldnt there be a small temp difference that is proportional to the distance of the two sources; due to heat needing to be transfered within the water?
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Kinda, but this is getting into some advanced thermodynamics that really isn't interesting or applicable for this discussion. The long answer is yes with a long explination about entropy and differentials, and the short answer is no, simply because the distances between components and the rate of the water flowing in the system is so great. Even if there is a difference between the blocks and radiator in temperature, it will be too small to measure correctly without a large percent error.

Jinu117 hints at the physics behind the problem, and the principle is the same, the water simply doesn't absorb heat like you think.
 
I think it is hilarious that the dude asks for the best waterblock sequence and the thread turns into a Physics lecture! :p
 
I think it is hilarious that the dude asks for the best waterblock sequence and the thread turns into a Physics lecture!
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Especially in one of utter nonsense. The whole trick with the laws of thermodynamics is that they refer to closed systems. A cooling loop is not a closed system, since you're permanently adding and removing heat. And the whole way the systems works is by having temperature gradients, aka cold water goes into block, warm water comes out, warm water goes into radiator, cold water comes out. And how boiling points should come into play in this connection is also rather unclear, unless you're refering to the state of the system just before the loud "Oh S..T".
Oh, and if I would skip the chipset block at first and measure the water temperature after the GPU. Chances are the water is warmer then the chipset is anyway.
 
In summary, there is a tiny, possibly insignificant differenec in water temp anywhere in your loop. Arrange the parts so that there is the least amount of tubing being used and so that bends flow smoothly. This will have more of an impact on temps than block position.
 
whitewale said:
Especially in one of utter nonsense. The whole trick with the laws of thermodynamics is that they refer to closed systems. A cooling loop is not a closed system, since you're permanently adding and removing heat. And the whole way the systems works is by having temperature gradients, aka cold water goes into block, warm water comes out, warm water goes into radiator, cold water comes out. And how boiling points should come into play in this connection is also rather unclear, unless you're refering to the state of the system just before the loud "Oh S..T".
Oh, and if I would skip the chipset block at first and measure the water temperature after the GPU. Chances are the water is warmer then the chipset is anyway.
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The NASA Technical Dictionary defines a closed system as:
"In thermodynamics, a system so chosen that no transfer of mass takes place across its boundaries"


So if a watercooling loop isn't a closed system, then why don't you have to refill the loop every time water passes through the radiator? The exchange of heat energy somehow goes somewhere else besides the water or the waterblock?

Think about it.

The debate over open vs. closed for a watercooling loop is long and complicated and the overall answer is one such that a loop retains a closed-state while all else remains constant, and a temporary open-state during what you call a gradient change.

Example:

Take a watercooling loop operating while the computer is off. The only heat added is by the pump, which is constant, so no gradient and no "warm spots" before or after the pump/radiator. Then you turn the computer on. Suddenly there are more heat sources from waterblocks. The loop becomes "open" in that there are new heat sources that the radiator has to deal with. When the computer boots and is at idle in windows or whatever, the heat from the processor becomes constant, along with the NB and GPU, so the system reaches a "closed" equilibrium.

Now, open a program or a game, or Prime, or whatever that taxes the processor, causing to heat up more from its idle state. The system loop "opens" again until it can reach a stable equilibrium with the added heat and the radiator where it again becomes "closed".

Now, stop playing the game or whatever and bring the processor/GPU back to idle and reverse the above process. This is just what happens when you use your computer. The loop in its rest state is closed and in equilibrium, so when you add a heat gradient (like playing a game or using the CPU) the system adapts in order to reach an equilibrium state.

These gradients you speak of do exist, but not to the extent you think they do. Considering the volume of water and changes in thermal energy added to water between an idle processor and a load one (not much, certainly less that a few watts), the ultimate result is a tiny tiny differential in heat between heat sources. Enough theory, tests have been done over at procooling.com to measure these differentials between radiators and blocks, and the big answer was that there was at most a half of a degree in water temperature. A half degree in water temperature make literally no difference in overall thermal capacity and dissapation rates in a pseudo-closed loop.

Questions? Bueller? Bueller?

-OMP
 
OneMadPoptart said:
It's a common misconception that putting the radiator after the pump cools water to go to components after it. The laws of physics and thermodynamics, however, say that this is a load of steaming BS. The water when flowing in a closed loop reaches an equlibrium temperature where the water is the EXACT same temperature EVERYWHERE in the loop, and doesn't have "cool" paths after the rad and "hot" paths before.

-OMP
Click to expand...


Yeah, I should've realized that before, I AM a physics student after all....oh well. But yes that is absolutely true, The only time that it would matter is if you had mixed temperatures in the water right before you turned it on , and then it would be slightly different until the water reached equilibrium.
 
Good argument OneMadPoptart,

However... :)

Whilst functionally you may be very close to correct in saying that the thermal load on the water travelling through the cpu block is only 1/2 c or so (no idea, not interested whether it's 1/2 c, 5c or 1 million c - actually the last would be kind of cool), that is a very different thing from thermal equilibrium. Whilst NASA may define a thermally closed system as one where mass isn't transferred, (haven't checked) you have to remember that mass is energy, so when they say mass, they will include such delights as photons, and other energy carrying vectors, including thermal radiatiom. To 'close' the system under discussion over a given period, the 'closed' system would have to include all the power the system would convert to heat, all the radiated heat released by the system into the environment, and everything in between. Certainly over any time period entropy will increase, as more of the electricity is converted to heat energy in the environment, but equally certainly the system is not in equilibrium, clearly there are non reversible changes over time. The CPU/GPU will do most of the work, converting electricity into heat, with the watercooling system simply transferring the heat waste from the comparatively low entropy concentration on the chip(s) to higher entropy distributed heat in the wider environment via the radiator.

In summary: The water cooling system cannot be treated as a closed system, as energy is being transferred both into and out of the system, which is after all how it works, and as previously stated energy is mass, hence not closed according to your NASA definition.

As I said at the start, for all practical purposes you're probably right about the thermal gradient, but not for the reasons you initially stated, and you seemed to regard the science as pretty important, so thought I'd try and help out. Any similarity between different parts of the system is a function of the imprecision of your measuring tools, not because the different areas are in equilibrium.

Oh and regarding the initial question; given that there's only a 1/2c or so taken on at the CPU, with I would guess similar at the GPU (maybe slightly more for a high end card), and quite a lot less at the chipset, then the maximum impact of chip order should be no more than 1c or so. I would suggest that it is far more important to ensure the clearest flow possible, as reduced flow should have a far greater impact. That is with the caveat of the 1/2c figure being correct, as I said, no idea, not really bothered.

Cheers
 
john_with_a_J said:
Good argument OneMadPoptart,

Whilst NASA may define a thermally closed system as one where mass isn't transferred, (haven't checked) you have to remember that mass is energy, so when they say mass, they will include such delights as photons, and other energy carrying vectors, including thermal radiatiom. To 'close' the system under discussion over a given period, the 'closed' system would have to include all the power the system would convert to heat, all the radiated heat released by the system into the environment, and everything in between. Certainly over any time period entropy will increase, as more of the electricity is converted to heat energy in the environment, but equally certainly the system is not in equilibrium, clearly there are non reversible changes over time. The CPU/GPU will do most of the work, converting electricity into heat, with the watercooling system simply transferring the heat waste from the comparatively low entropy concentration on the chip(s) to higher entropy distributed heat in the wider environment via the radiator.

In summary: The water cooling system cannot be treated as a closed system, as energy is being transferred both into and out of the system, which is after all how it works, and as previously stated energy is mass, hence not closed according to your NASA definition.

As I said at the start, for all practical purposes you're probably right about the thermal gradient, but not for the reasons you initially stated, and you seemed to regard the science as pretty important, so thought I'd try and help out. Any similarity between different parts of the system is a function of the imprecision of your measuring tools, not because the different areas are in equilibrium.

Oh and regarding the initial question; given that there's only a 1/2c or so taken on at the CPU, with I would guess similar at the GPU (maybe slightly more for a high end card), and quite a lot less at the chipset, then the maximum impact of chip order should be no more than 1c or so. I would suggest that it is far more important to ensure the clearest flow possible, as reduced flow should have a far greater impact. That is with the caveat of the 1/2c figure being correct, as I said, no idea, not really bothered.

Cheers
Click to expand...

Yes, I realize all of this, but for watercooling loops it really makes no difference. Let me see if I can tell you all why:

You say that mass is energy. This is true, and everyone (almost) knows the equation E = MC^2, saying that energy in a system is equal to the product of the mass of the system times the speed of light squared.

Why doesn't this matter for us? Lets find out how much mass is really being transfered:
Solving for M, we divide both sides by the speed of light squared so we get the equation M = E / C^2. Lets plug in a value of around 100 Watts for heat from a processor (this is possibly a little high, and I dont remember the values for intel and amd standards). A Watt is a Joule (energy) per second so a value of 100 (J/s) / 300,000,000 (m/s) (speed of light in meters per second) we get a grand total of 0.000000333 kilograms per second radiated by the cooling loop. Do you know how much mass that is? NOTHING. Maybe if you leave your watercooling loop up for a million years you will get a noticable mass exchange. At this level, the mass transfer is amazingly insignificant so we can conclude that no mass transfer is occuring and the system is basically closed in this respect. In relation to the above argument, entropy differentials and deltas really mean nothing when dealing in the macroscopic world.

The simple fact remains however that energy is being dumped and released into and from the system. Think of a watercooling loop without a radiator. What happens? The water gets hotter and hotter and the processor does as well because of the continuios dump of energy into the system...duh... now add the raditor. What happens? The radiator dissapates heat so the water does not get any hotter with the added thermal energy, so it basically sets an upper limit on the ammount of energy that can be added into the water. So we've now got a system that both adds and dissapates heat, so what? They balance each other out so you don't get any net difference in input vs. output of thermal energy, which is defined as a thermally and basically closed system.

I never said that watercooling loops are 100% closed, but the fact ramains that we treat them as such because we can ignore some of the finer points of quantum mechanics because the added effect to the system is negligable. My own instincts tell me that the system isnt closed classically because the definitions are so exacting to what a closed system actually is, but we still treat a loop as such because even though the system is "open" to whatever small degree that it is, the differentials are so small that it makes life a hell of a lot easier to call it a closed system.

Phew... I like this argument. This is fun :) Of course don't take my words as absolute truth here, not until I can dig my old thermo books out of storage... :rolleyes: Anyone else want to add something to this discussion?

-OMP
 
Wow, I never thought a post on [H] would reference (and use) the famous E= M*C^2 equation. I guess you never know what to expect. Keep at it , I'm enjoying reading this stuff.
 
Why not just calculate how much the water warms/cools?

Take 1 gallon/min (a typical flow rate), convert it and multiply by water's density (1 gram/cm^3) to get 63.1 grams/second. Next, multiply by the specific heat of water (4.184 Joules/gram/°C) to get 264 W/°C (one Watt equals one Joule/second). So, your 100 watt CPU will result in the water leaving the block 0.38 °C hotter (100/264).

Flow order doesn't much matter, unless you have loads of heat to deal with (peltiers) or really low flow.
 
HeThatKnows said:
Why not just calculate how much the water warms/cools?

Take 1 gallon/min (a typical flow rate), convert it and multiply by water's density (1 gram/cm^3) to get 63.1 grams/second. Next, multiply by the specific heat of water (4.184 Joules/gram/°C) to get 264 W/°C (one Watt equals one Joule/second). So, your 100 watt CPU will result in the water leaving the block 0.38 °C hotter (100/264).

Flow order doesn't much matter, unless you have loads of heat to deal with (peltiers) or really low flow.
Click to expand...

This is assuming 100% efficiency heat/energy transfer which really doesn't happen between the copper in waterblocks and the processor...Still it does put an upper limit on the temperature rise after a waterblock, which confirms what has been said above.

Thanks for the quick calculation.

-OMP
 
HeThatKnows said:
Why not just calculate how much the water warms/cools?

Take 1 gallon/min (a typical flow rate), convert it and multiply by water's density (1 gram/cm^3) to get 63.1 grams/second. Next, multiply by the specific heat of water (4.184 Joules/gram/°C) to get 264 W/°C (one Watt equals one Joule/second). So, your 100 watt CPU will result in the water leaving the block 0.38 °C hotter (100/264).

Flow order doesn't much matter, unless you have loads of heat to deal with (peltiers) or really low flow.
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Even dual peltier pumping out approximately 400W plus didn't show any temp diff when I tried while back :p
 
jinu117 said:
Even dual peltier pumping out approximately 400W plus didn't show any temp diff when I tried while back :p
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Exactly. I think it's safe to say that there isn't any noticable temp differences before or after the radiator and waterblocks.
 
Yes this is kind of fun. I think we've possibly drifted off the original topic a bit with the science, but the quality of discussion is probably high enough, and the people involved nice enough that we'll get away with it :)

I think the discussion is probably ending up with us all rather agreeing, and can I just say - OneMadPopTart - your later posts are excellent examples of clear thinking, backed up with solid knowledge and understanding, and very well stated. I realise that probably sounds rather patronising, but it's intended to be a genuine compliment, something I don't think gets done enough on these sorts of forums.

Rather my post was tackling your earlier posts, specifically your first post where you claimed that the temperatures were exactly identical throughout the system, indeed you capitalised exact and everywhere to stress the point. Although by the time I posted you had already begun to drift away from this position I couldn't let it go unchallenged could I?

Now does anyone fancy a game of 'The Earth is flat, and was made in 4004 BC'?

Cheers
 
:D
john_with_a_J said:
Yes this is kind of fun. I think we've possibly drifted off the original topic a bit with the science, but the quality of discussion is probably high enough, and the people involved nice enough that we'll get away with it :)

I think the discussion is probably ending up with us all rather agreeing, and can I just say - OneMadPopTart - your later posts are excellent examples of clear thinking, backed up with solid knowledge and understanding, and very well stated. I realise that probably sounds rather patronising, but it's intended to be a genuine compliment, something I don't think gets done enough on these sorts of forums.

Rather my post was tackling your earlier posts, specifically your first post where you claimed that the temperatures were exactly identical throughout the system, indeed you capitalised exact and everywhere to stress the point. Although by the time I posted you had already begun to drift away from this position I couldn't let it go unchallenged could I?

Now does anyone fancy a game of 'The Earth is flat, and was made in 4004 BC'?

Cheers
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Thank you for the nice words. I try to help people as much as possible when I post (that is, the nice and somewhat open minded ones). I'm happy an agreement was reached eventually too, I think my fingers were getting tired of typing that much. I'm glad you noticed my deviation from my original posts...I did make the first couple intending to make a point, and then we started this whole thing so a little more science and tiny corrections were necessary. ...Good times...

-OMP
 
aphexddb Entropy is not 'Heat', it is the state of disorder, I believe you may have meant 'Enthalpy'.

OneMadPoptart is entirely correct in every one of his posts. john_with_a_j, though it may be said that 'matter is energy' the mass actually being transfered is inversly proportional to c^2 (aka its incredibly small, so small in fact that the mass of the water transfering this mass far far far exceeds the mass being transfered itself. As for types of systems, there are 3 types,

http://www.engineersedge.com/thermodynamics/types_systems.htm

A watercooling loop is infact a closed system, the quantum properties of mass/energy relationships don't really apply here.

OMP, I commend you for trying to explain a thermodynamic concept that is really quite difficult for the laymen to grasp, I think its worth noting though that equilibrium is based on a differential taken over an INFANTESIMAL amount of time (dt) so when you look at the situation from this perspective, though at one infantesimal time you may see a temperature of x degrees in the loop, and at another infantesimal time you would see temp y, when integrated over all time you can see that the system is in fact in a state of equilibrium, a stranger and more difficult point of science to address, but a correct one none the less :)

HiTech

PS: Physics SUCKS, Chemistry RULES (specifically Physical Chemistry, aka thermodynamical, kinetical, and quantum chemistry )!!! :)
 
So back to the question...

i think pump-res-waterblocks-radiator makes the most sense b/c theres no chance of bubbles getting to the waterblocks, thus best cooling.
 
Sobek said:
I think it is hilarious that the dude asks for the best waterblock sequence and the thread turns into a Physics lecture! :p
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2 days later and we're still at it. :p It's an interesting discussion and I'm learning new things so I don't mind.
 
ATH25 said:
So back to the question...

i think pump-res-waterblocks-radiator makes the most sense b/c theres no chance of bubbles getting to the waterblocks, thus best cooling.
Click to expand...

This would quite possibly be the WORST cooling loop for watercooling systems.

Going pump directly into the res will leave you with completly horrible water pressure (thus less force and flow) and pretty much be a waste of a pump. Go pump -> blocks -> rad -> res -> pump or pump -> rad -> blocks -> res -> pump depending on which way allows for the shortest tubing.
 
HiTech-Hate said:
A watercooling loop is infact a closed system, the quantum properties of mass/energy relationships don't really apply here.

...I think its worth noting though that equilibrium is based on a differential taken over an INFANTESIMAL amount of time (dt) so when you look at the situation from this perspective, though at one infantesimal time you may see a temperature of x degrees in the loop, and at another infantesimal time you would see temp y, when integrated over all time you can see that the system is in fact in a state of equilibrium, a stranger and more difficult point of science to address, but a correct one none the less :)
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I couldn't have said it any better. Thanks for the great explanation.

-OMP

P.S. Chemistry SUCKS. Quantum mechanics and high-energy astrophysics RULE.
 
I just put my first wc setup in my comp last night, I went Pump -> Rad -> Block -> Pump,
i'll let you know if it works well or not in this setup :). As OMP said though, shortest tubing is best, not only for water pressure and cooling capacity, but also for aesthetics (with all this tubing in my case, I at least want it to look half decent). And to all you wcers out there that are ecing in a Lian Li PC65 I tip my hat to you, it was a BITCH to get all this stuff crammed in there (especially when your drive cage is full of 4hds, and you cant mount your heatercore on the bottom).
 
Here's a question to bring back the original spirit of the post . . .

I've been wondering if placing the GPU block directly after the CPU block is the best idea. Running hot water over your GPU seems to defeat the purpose of water-cooling it in the first place. I guess the ultimate question is this: will the post-CPU water be hotter than the temperature of the GPU, or cooler? Also, if it's cooler, by how much? If it's cooler, there is still room for the GPU's heat to be absorbed. But will the amount temperature difference allow for more heat absorption than simple air-cooling?

I propose using a Y to send the cooled water from the radiator directly to both the GPU and CPU simultaneously. This will obviously decrease the amount of water that gets to each block by 1/2, but how will this affect the overall cooling of each?

Has anybody experimented with this? If not, I'll probably end up doing so.
 
Vellicator said:
Here's a question to bring back the original spirit of the post . . .

I've been wondering if placing the GPU block directly after the CPU block is the best idea. Running hot water over your GPU seems to defeat the purpose of water-cooling it in the first place. I guess the ultimate question is this: will the post-CPU water be hotter than the temperature of the GPU, or cooler? Also, if it's cooler, by how much? If it's cooler, there is still room for the GPU's heat to be absorbed. But will the amount temperature difference allow for more heat absorption than simple air-cooling?

I propose using a Y to send the cooled water from the radiator directly to both the GPU and CPU simultaneously. This will obviously decrease the amount of water that gets to each block by 1/2, but how will this affect the overall cooling of each?

Has anybody experimented with this? If not, I'll probably end up doing so.
Click to expand...

I'm thinking of doing that in my setup but haven't decided yet.
 
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