A Technical Question

when i first got interested in water cooling, i had the idea of bringing my ambient room temp down. i know that sounds odd. but my comp was heating up the room like a space heater. (tis a very small room heh). anyway, after thinking about it, water cooling shouldnt affect the ambient room temp at all should it. i mean the cpu is still dissipating the same amount of energy into the same amount of air. actual cpu temp shouldnt matter, unless im thinking incorrectly. well lower ambient temps or not, water cooling has certainly provided me with many hours of enjoyment, besides the fact that i have yet to put a complete setup together after like 2.5 years. :-\ all my projects r too extreme and just fizzle out.

Unless you water cooling is loosing mass p), it shouldnt contribute anymore heat to the system than you previous air setup (Not accounting for pump heat, extra fans, and a higher overclock). It just does it more efficiently.

Hm, I would have thought that with the pump generating heat as well, it will output slightly more heat than with a conventional air setup, however the difference is minute and would be unnoticable.

The pump does add a small amount of heat to the "system" of the room.

Couple of thoughts before I make my point though:

If your cpu temp is lower with water than it was with air... where do you think that heat went?

Water cooling is more efficient, and thus allows you to move more heat during a given period of time. Where are we moving that heat?

When talking about heat between two different "zones" or "areas" or what ever suits your fancy; heat, much like electric charges, will (all things being equal) tend to equalize.

Since theres still the same amount of heat, and your "disipating" it more effectively (quickly), the "system" of the room will equalize temps much more rapidly. The AC isn't able to keep up quite as well, and on top of that, I'd bet your dumping more air out of your case with your new setup. (either that or your case temp is getting ungodly high). All of this is assuming you have let it run long enough for the loop temp to equalize.

so... long story even longer... yes water cooling can raise your ambient.

.....unless your heatercore is in a different room.

Something that watercooling makes easier, is the transport of said heat to another location.
My heatercore is in the basement, for example. It does little to change ambient temps in the basement, due to the fact that it's a stone cellar that equalizes quickly with the moist ground on the otherside of the stone walls.
With it also goes my pump heat, and the NB heat too, for they are both watercooled.

It hardly makes lanning easy though...

well actually if you get lower temps with water than you will be adding more heat to your room, more heat removed from cpu = more heat into room
+ the heat from the pump
but this should be such a small amount that i doubt you would ever notice it

Diggrr said:

.....unless your heatercore is in a different room.

Something that watercooling makes easier, is the transport of said heat to another location.
My heatercore is in the basement, for example. It does little to change ambient temps in the basement, due to the fact that it's a stone cellar that equalizes quickly with the moist ground on the otherside of the stone walls.
With it also goes my pump heat, and the NB heat too, for they are both watercooled.

It hardly makes lanning easy though...

Click to expand...

you can make box cooler for lan partys with quick disconects to make it easer

Guderian said:

The pump does add a small amount of heat to the "system" of the room.

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Yes.

Guderian said:

If your cpu temp is lower with water than it was with air... where do you think that heat went?

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Into air, eventually.

Guderian said:

Water cooling is more efficient, and thus allows you to move more heat during a given period of time. Where are we moving that heat?

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Same air you'd move it with a heatsink

Guderian said:

so... long story even longer... yes water cooling can raise your ambient.

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Only by the heat of the pump.

heat transfer efficency increases as delta T increases.

energy cannot be created or destroyed. The CPU die temp may be cooler, but the same watts of energy are being disipated per minute as a heatsink

If you don't believe me, I'll back up with simple physics and formulas...

energy cannot be created or destroyed.

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Agreed.

The CPU die temp may be cooler, but the same watts of energy are being disipated per minute as a heatsink.

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Incorrect. If the die is cooler.... where did that heat/energy go? We both agree it didn't disappear, and it doesn't just hang out in your loop. Its in the room, or the case.

Hopefully I'm misunderstanding you, because I can't believe you would say a heatsink disipates the same amount of energy as a decent WC Rig.

Guderian said:

Hopefully I'm misunderstanding you, because I can't believe you would say a heatsink disipates the same amount of energy as a decent WC Rig.

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you're only misunderstanding the facts

same energy in, same energy out.

V=I*R
I=V/R

watts=V*I

watts=V^2/R

Vcore and resistance are constant (at a given CPU load)

so, watts is constant

so, heat is constant

so, heat dissipated is constant

it's just that a watercooling rig doesn't need as high a delta T to reach equlibrium

could call in some people from procooling, or use more physics if needed

Guderian said:

Agreed.



Incorrect. If the die is cooler.... where did that heat/energy go? We both agree it didn't disappear, and it doesn't just hang out in your loop. Its in the room, or the case.

Hopefully I'm misunderstanding you, because I can't believe you would say a heatsink disipates the same amount of energy as a decent WC Rig.

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WHEN AT STEADY STATE both dissipate the same amount of heat.
Q=AU(Tcpu-Tfluid) ----> Equation that governs heat transfer
A= area available to heat trasner
U=heat trasnfer coefficient (function of fluid physical properties and geometry)
Tcpu=die temperature
Tfluid= cooling fluid temperature, be it air or water

A thermal equilibrium will be established for both air coolin and watercooling. Air-cooling has more heat transfer resistances because air is being used as a cooling fluid. The die temperature will continue to rise until the temperature difference between the die and the surrounding air can dissipate the heat output of the processor.

Because water cooling is more efficient heat transfer ( The heat transfer coefficient in the above equation is higher) a lower temperature difference is required between the die temperature and the water temperature to transfer the same amount of heat.

EDIT: The at steady state part is very important. When the temperatures are transient (heatsink temperature is increasing or water temperature is increasing, the above equation gets much more complicated). Let me know if you're interested - I can reason out the situation, but its got a lot more differential equations.

heat transfer efficency increases as delta T increases

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The bigger the temp difference, the faster the heat will move.

A thermal equilibrium will be established for both air coolin and watercooling. Air-cooling has more heat transfer resistances because air is being used as a cooling fluid. The die temperature will continue to rise until the temperature difference between the die and the surrounding air can dissipate the heat output of the processor.

Because water cooling is more efficient heat transfer ( The heat transfer coefficient in the above equation is higher) a lower temperature difference is required between the die temperature and the water temperature to transfer the same amount of heat.

Click to expand...

So basically, since the loop equalizes faster, and is disipating the same amount of heat, the die temp doesn't have time to rise to levels you get with a standard HSF.

That does make sense.

EDIT: Why didn't you tell me I was misspelling Dissipate.

rudnik68 said:

Air-cooling has more heat transfer resistances because air is being used as a cooling fluid.

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aircooling usually doesn't have as much surface area, too, but yes

Guderian: glad you now understand, now go stop some misinformaiton out there

greenman100 said:

you're only misunderstanding the facts

same energy in, same energy out.

V=I*R
I=V/R

watts=V*I

watts=V^2/R

Vcore and resistance are constant (at a given CPU load)

so, watts is constant

so, heat is constant

so, heat dissipated is constant

it's just that a watercooling rig doesn't need as high a delta T to reach equlibrium

could call in some people from procooling, or use more physics if needed

Click to expand...

i think your slightly rong becuse youre cooler with water cooling the semiconducters are cooler moving more watts a cold wire will move more elect than a hot one this internal restance.
cooler more flo

Guderian said:

... So basically, since the loop equalizes faster, and is disipating the same amount of heat, the die temp doesn't have time to rise to levels you get with a standard HSF. ...

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Not quite... the die temperatures will be lower for the cooling solution that provides the least amount of thermal resistance. This allows the same amount of heat to be transfered (remember that we're talking about an equilibrium case here) with a smaller temperature gradient (i.e. smaller DT between the CPU die and the water).

Also note that in both systems all of the heat (ignoring secondary losses at the waterblock, etc) is eventually dispersed into air... the difference is that the watercooling setup can generally have a larger surface area than a heat sink can.

lclark2074 said:

i think your slightly rong becuse youre cooler with water cooling the semiconducters are cooler moving more watts a cold wire will move more elect than a hot one this internal restance.
cooler more flo

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you're looking at a very minute current difference, if at all.

But in the overall scheme of the thread... you are not going to raise your ambient more with water, than with air (pump heat not withstanding).

I was originally thinking lower temps on die = more heat removed = more heat moved into the ambient ( a very simple and logical chain of thought, with absolutely no basis in science). when in fact the lower temp is due to the "heat transfer efficency increase as delta T increased".

the difference is that the watercooling setup can generally have a larger surface area than a heat sink can.

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Isn't this the very reason that WC is more effecient, and thus able to move X amount of heat with lower Temp Delta?

Guderian said:

... Isn't this the very reason that WC is more effecient, and thus able to move X amount of heat with lower Temp Delta?

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I'm not so sure that's the only factor. Keep in mind that there are two temperature gradients to be concerned with in water cooling: both CPU - water and the water - air. I think that the real bonus to water cooling is the efficiency of the first step... though the large heat exchanger is definitely one of the reasons water cooling works as a system.

I'm gonna jump in with some physics here as well

E=mCdT

E is energy (J). m is mass (kg). C is a constant for the material concerned (J per K per kg). dT is the change in temperature (K).

Let us say a certain CPU produces 50 Watts of heat (using easy numbers here).

Thus in one second 50 Joules of energy in the form of heat, as per the relationship Power = Energy/Time is released from the CPU. To make life easier I am going to assume that the heatsink absorbs all this heat and thus our heatsink has a temperature rise, dependant on E, m and C.

Now, 1 mole of air is 29grams (I think). 22.4litres of air is 29 grams at STP. 1 mole of water is 18 grams. Thus 0.018 litres of water is 18 grams.

1 litre of air is 0.00129 kgrams, Specfic Heat capacity (C) is 1 J/kg/K
1 litre of water is 1 kgram, Specific Heat capacity (C) is 4.184 J/kg/K

Thus when removing 50J of energy:

E = m.C.dT => dT = E/m.C

For Water dTw = 50/1x4.184 = 11.95K
For Air dTa = 50/0.00129x1 = 38760K

Hence 1 litre of water will heat up a LOT less than 1 litre of air when given the same amount of heat energy.

Don't know if this is relevant just felt like revising some physics in the hope that it is of use to someone.