Question for Cathar

[jaydee]

Originally posted by Cathar


My educated guess, remembering that GPU's and NB's are much lower heat emitting devices than CPU's, is that you'd see less than a 0.2C difference on a GPU block, and less than a 0.1C difference on a NB block as a result of the slightly lower flow rates.

Is this still true? Seems to me the new Radeons 9700 and the newer GForces are putting out 50C-70C loaded from what they been saying. Thats pretty toasty. Seems to me about like adding a second CPU. :shrug:

[SysCrusher]

Cathar is correct in what he states. You have to think about it on a molecular level. The water temp will rise maybe .2C. Which means it allows more water molecules(alot considering 6lpm) to come in contact with the heated surface. This is why turbulance and velocity are just important as flow rate. Narrowing the slot even more will be a benefit to cooling performance but depends on how well the pump can push it. A stronger pump can use a even narrower slot. Depending on what pump head, flow, pressure is how narrow the slot "can" be up until the slot is so narrow it decreases the flow and effects the block itself and other blocks in the loop. Then you have to think of added heat from the pump from it working harder to push that water through a narrower slot.

You only can make it so small till you see deminishing returns. That's true with making the slot bigger too. It's a very fine balancing act and Cathar put alot of thought and work into it. LOL I'v tried it doing it the home brewed way and it's not easy.

[Gulp35]

I'm confused

<sarcastic>
All this time I was thinking that when a waterblock gives better temperatures It's because more heat is going into the water.
</sarcastic>


take for example that if a computer creates 100Joules /sec. (100W) in the perfect waterblock all 100W would go into the cooling meduim and make the CPU be at ammient temps. So as waterblock efficientcy increases the water would get warmer.

So in real life say your CPU is 5C above amient, I am guessing that the die of silicon is about 10g (I have no idea about this i am guessing completely) that would mean that it would take 35.5J to raise the temp that much which leaves 64.4 J to go into the water. If the temp was 5C higher then only 29J would go into the water. I hope you get my drift.

But cathar is right that the amount of temp rise is minimal because for every liter of water in the system it takes 4184 Joules to warm it 1C. this would leave a whole lot of time for the water temp to rise and for the rad to remove temps.

I don't mean to sound critical, but I am trying to clear some things up that I feel haven't been.

[Cathar]

Originally posted by Gulp35
take for example that if a computer creates 100Joules /sec. (100W) in the perfect waterblock all 100W would go into the cooling meduim and make the CPU be at ammient temps. So as waterblock efficientcy increases the water would get warmer.

So in real life say your CPU is 5C above amient, I am guessing that the die of silicon is about 10g (I have no idea about this i am guessing completely) that would mean that it would take 35.5J to raise the temp that much which leaves 64.4 J to go into the water. If the temp was 5C higher then only 29J would go into the water. I hope you get my drift.

If only 29J/s (W) is going into the water, then where's the other 71W going?

[SysCrusher]

Your on the right track Gulp. I think what your tried to explain in the first part of your post was depending upon if the water is standing still. In this case the water is constantly being cooled. More heat is going into the water but is also being cooled just as fast by the rad so the block receives an endless supply of cool water. Plus the water is moving fast enough through the block that it doesn't heat up enough than if the water was moving though the block at a slower rate (we're talking awfully slow). Turbulance and velocity just allow more water molecules to absorb more heat. With a 70W heat load and a 6lpm flow rate the water absorbs the same amount of heat in the same given amount of time regardless of how good the block is. The White Water just minipulates the water to it's advantage better than other blocks do. This is just my laymans terms of explaining since Im no PHD in fluid dynamics.

I can't even detect a temperature difference between my inlet and outlet water of my block. If one has a temp difference of 1.5-2C between their water inlet and outlet, they should be nearly running at ambient temps. lol

[bigben2k]

Originally posted by theetruscan
Under what controlled conditions were such tests taken? This is total trash.

I'm sorry to be so blunt, but such statements really are total crap.

The water won't be warmer. Why? Because the CPU is pumping the same amount of heat into the water. There is no "extra heat". If the CPU has clocked at a certain speed, and runs at a certain voltage, running a certain program, it emits the same amount of heat energy.
__________________________________________

Are you sure? I mean obviously the CPU releases the same amount of heat, so some of the reasoning may be faulty, but it seems reasonable that the reduction in flow rate from the additional pressure drop could result in something on the order of 1C temperature increase. I could be completely off base here, but it was my understang that lower flow rates result in higher water temps.

The jet inpingement throws out (to some extent) the lower-flow-is-worse theorem, because of the (non-recoverable) pressure drop being created by the jet. It's still valid, but not by the same equation.

The temperature gradient is what makes all blocks different. If you think about it, given the same flow rate, different blocks will perform differently, because they all allow a different amount of energy to remain within the baseplate, or the die contact area.

Cathar succesfully balanced a very thin baseplate, an optimal fin configuration, and an excellent flow configuration. If you can see that, then you know why WW works so well.

If you're "almost there", I can tell you that simply making the baseplate thinner doesn't always work, and that it can make the performance worse, if it's not balanced with fins and flow. In other words, if you're going to expose the water to a higher temp (same wattage, but higher temp), then you better make sure that the water can take that heat away.

[Cathar]

The technical answer is that a one waterblock may offer a lower thermal resistance than another, through whatever means available (turbulence, jet impingement, fine-channels, whatever). The whole point being that the water is manipulated in such a way that for the time that the water is in contact with the hot sections of the block (which has been heated up by the CPU) that the water achieves a higher coefficient of convection, also referred to as h. The units of h are Watts per (Area * dT), where dT is the temperature difference between the surface (of the waterblock) and the liquid flowing over that surface.

Now each water-block offers a fixes amount of area for cooling, so if we multiple h by the area of the waterblock, we are left with units of W/dT, or how many watts of heat energy will move into the water based on the temperature difference between the waterblock surface temperature and the water flowing over that surface. T is often measured in Celcius, and so the term is commonly summarised at W/C, or even some-times inverted to C/W.

Now some waterblocks will have a higher W/C value than others. Higher is better. For example, one block might have a W/C value of 20, while another might have just 10.

For the one with a W/C value of 20, and a heat source of 100W, the surface of the block coming into contact with the water must be 100/20=5C warmer than the water before 100W of heat energy is transferred into the water.

For the block with a W/C value of 10 and a heat source of 100W, the surface of the block coming into contact with the water must be 100/10=10C warmer than the water before 100W of heat energy is transferred into the water.

Now since the waterblocks are attached to the CPU (and ignoring the thermal paste junction for now) this means that the CPU will be 5C cooler for the block with the W/C of 20, because it is more efficient at transferring the heat into the water. The block with a W/C of 20 is more efficient because it doesn't need to get as hot before it's able to transfer 100W of heat into the water, unlike the block with the W/C of 10, which has to get much hotter. Since the CPU is attached to the blocks and is the source of heat, the CPU is correspondingly hotter as well for the second block.

This is what's referred to as thermal resistance. The watts (heat energy) transferred into the water is the same, it's just a question of how hot does the block need to get before it transfers the heat energy that the CPU is emitting.

Please also note that the above is am extremely simplistic explanation and does not attempt to take into account concepts like varying thermal gradients through the metal itself.

[pippin88]

Originally posted by Cathar
The technical answer is that a one waterblock may offer a lower thermal resistance than another, through whatever means available (turbulence, jet impingement, fine-channels, whatever). The whole point being that the water is manipulated in such a way that for the time that the water is in contact with the hot sections of the block (which has been heated up by the CPU) that the water achieves a higher coefficient of convection, also referred to as h. The units of h are Watts per (Area * dT), where dT is the temperature difference between the surface (of the waterblock) and the liquid flowing over that surface.

Now each water-block offers a fixes amount of area for cooling, so if we multiple h by the area of the waterblock, we are left with units of W/dT, or how many watts of heat energy will move into the water based on the temperature difference between the waterblock surface temperature and the water flowing over that surface. T is often measured in Celcius, and so the term is commonly summarised at W/C, or even some-times inverted to C/W.

Now some waterblocks will have a higher W/C value than others. Higher is better. For example, one block might have a W/C value of 20, while another might have just 10.

For the one with a W/C value of 20, and a heat source of 100W, the surface of the block coming into contact with the water must be 100/20=5C warmer than the water before 100W of heat energy is transferred into the water.

For the block with a W/C value of 10 and a heat source of 100W, the surface of the block coming into contact with the water must be 100/10=10C warmer than the water before 100W of heat energy is transferred into the water.

Now since the waterblocks are attached to the CPU (and ignoring the thermal paste junction for now) this means that the CPU will be 5C cooler for the block with the W/C of 20, because it is more efficient at transferring the heat into the water. The block with a W/C of 20 is more efficient because it doesn't need to get as hot before it's able to transfer 100W of heat into the water, unlike the block with the W/C of 10, which has to get much hotter. Since the CPU is attached to the blocks and is the source of heat, the CPU is correspondingly hotter as well for the second block.

This is what's referred to as thermal resistance. The watts (heat energy) transferred into the water is the same, it's just a question of how hot does the block need to get before it transfers the heat energy that the CPU is emitting.

Please also note that the above is am extremely simplistic explanation and does not attempt to take into account concepts like varying thermal gradients through the metal itself.

Excellent explanation.

I knew that the water wouldn't get hotter on a different block, but had not come to grips with the reasons for this.

This put it all pretty clear.

[8-Ball]

They way I consider the efficiency of waterblocks/loops and so on is as follows.

As Cathar has stated, the CPU is giving out a heat load, lets say 90W. For a given flow rate, the rad will be able to dissipate 90W for a particular temperature differential between water and air. The water should stabilise at this temperature.

Now we can assume (roughly) that we have water at this temperature flowing through the block. The block has a C/W ratio. Lets say 0.2 for a good block. This implies that for every 0.2 of a degree difference in temperature, 1 watt of heat will be transferred to the water. So for 90 Watts, you would need a temp differential of 18 degrees between the water and the block.

A more efficient block would be able to transfer 90W from the die with a lower temp differential, lets say 0.17 C/W ratio.

There would now need to be 15.3 degrees difference between the water and the die.

However, you are still only getting 90W of heat transferred into the water, which if the flow rate is the same as before, will raise the temp of the water by the same amount, and for the same flow rate, the rad should still be performing the same.

I'm sure it then gets complicated because the cpu will most probably clock higher, introducing more heat into the water raising it's temperature. But you would have changed the conditions for comparison.

Now consider raising the flow rate. The rad will perform slightly more efficiently, (lower C/W ratio between water and air for rad), thus for the same heatload, the water temp should be slightly lower, but only very slightly, as the C/W ratio will not change much. The block should now also perform with a lower C/W ratio, thus decreasing the temp difference between the water and the block required to remove 90W.

It is my understanding that people often think of this the wrong way round.

If you want to transfer 90W from the cpu to the air (via water cooling), you need to consider the following stages, in this order!

(Consider a constant flow rate - constant C/W ratios in block and rad)

1. How much warmer wil the water have to be than the air blown through the rad, for 90W to be transferred.

90 x C/W for rad = dTaw

(dTaw is temp difference between air and water).

2. For the same flow rate and given C/W ratio for block, how much higher will the temp of the cpu have to be than the water in the block, for 90W of heat to be transferred to the water.

90 x C/W for block = dTwc

(dTwc is temp difference between water and cpu).

So if the ambient temperature is A, then for a given flow rate, the temp of the cpu should be A + dTaw + dTwc.

This is very approximate as it assumes no other sources of heat in the loop, and assumes that the temp of the water doesn't vary throughout the loop, which for a high flow system is not far off.

However, the general principle still stands, If you are working from a given ambient air temp with a given heatload, you need to work back fro the air to consider how much hotter each stage will have to be to allow the transfer of 90W, until you reach the cpu. Add these up and you can get a rough estimate of the expected CPU temp.

The main point of this however, aws to explain why a more efficient waterblock, does NOT raise the temps of the water, merely the temp difference between the water and the cpu. IE, lowers the CPU temp.

8-ball

[8-Ball]

Oops.

Sorry for the long post.

8-ball

[Gulp35]

the 71J go to warming the silicon die 10C over ambient.
Specific Heat (amount of energy required to raise 1g of substance 1C or K) of silicon is .71J/g*C so for a 10g piece of silicon to be rasied 10C ((.71J/g*c)*10g*10C)=71J.
there is my 71J

I was not describing the water temp in the enitre system but the temp of the water as it is exiting the block (i.e. warmed water that has not passed through the radiator to be cooled).

Yes the water temperature goes up if we are measuring the exit temperature
but No the water inlet temp will not change dramaticly if a large enough radiator is given (Heatecores are designed to handle a lot more than 100W of heat , I think i heard that 1 horsepower is equal to ~700W).

[Cathar]

Originally posted by Gulp35
the 71J go to warming the silicon die 10C over ambient.
Specific Heat (amount of energy required to raise 1g of substance 1C or K) of silicon is .71J/g*C so for a 10g piece of silicon to be rasied 10C ((.71J/g*c)*10g*10C)=71J.
there is my 71J

I was not describing the water temp in the enitre system but the temp of the water as it is exiting the block (i.e. warmed water that has not passed through the radiator to be cooled).

It's 71J per second, not just 71J.

If 71J/s were heating up just the silicon die, it would reach 120C in less than 10 seconds and fry shortly thereafter.

Suggest that you read some of the posts above. The water is not heated up more anywhere. The difference is that some blocks needs to get hotter (and hence allow the CPU to get hotter) before they transfer all of the CPU's heat into the water. The amount of heat transferred does not change depending on the block.

[Since87]

Originally posted by Cathar
The amount of heat transferred does not change depending on the block.

[NITPICKY]

Actually with a better block, a smaller fraction of the CPU's heat output is going to conduct down the CPU pins. Therefore the amount of heat conducted through the block will be higher.

[/NITPICKY]

Under no circumstances, is this a DISADVANTAGE of a high performance block though.

[SysCrusher]

Originally posted by Cathar
The amount of heat transferred does not change depending on the block.

This is where some are getting it confused and that other website introduced the myth. I lack the proper education in this area to explain it correctly though. It's a matter of lowering the thermal resistance and using turbulance,velocity plus flow to take advantage of the water as much as possible as it passes through the block - same thing with a rad. Water only heats up so much in a given amount of time. Some water blocks have to heat up more before it can transfer the same amount of heat to the water as a block that doesn't have to heat up as much. This is why copper is far superior to aluminum but they both transfer the same amount of heat to the water. The heat just travels through copper and into the water faster than aluminum. This is one reason why copper is used more often in electrical applications. Electrical resistance has pecular effect - Heat. Copper can deal with this heat better allowing more electrical current to pass through it faster with less resitance.

[Since87]

Originally posted by SysCrusher
The heat just travels through copper and into the water faster than aluminum. This is one reason why copper is used more often in electrical applications. Electrical resistance has pecular effect - Heat. Copper can deal with this heat better allowing more electrical current to pass through it faster with less resitance.

No. The primary reason copper is used in electrical applications is that it has a low electrical resistance. Because of the low electrical resistance, for a given current, wire length, and wire gauge, less power will be dissipated in a copper wire than in an aluminum wire. The heat spreading ability of copper is a tertiary concern.

[Cathar]

Originally posted by Since87
[NITPICKY]

Actually with a better block, a smaller fraction of the CPU's heat output is going to conduct down the CPU pins. Therefore the amount of heat conducted through the block will be higher.

[/NITPICKY]

Under no circumstances, is this a DISADVANTAGE of a high performance block though.

Actually I didn't want to confuse the issue by saying that as I was trying to keep it all relatively straight-forward.

If you're interested in a discussion on this, I started exactly such a thread on this topic about 2 months ago over here:

http://forum.oc-forums.com/vb/showth...hreadid=148310

[SysCrusher]

Originally posted by Since87
No. The primary reason copper is used in electrical applications is that it has a low electrical resistance. Because of the low electrical resistance, for a given current, wire length, and wire gauge, less power will be dissipated in a copper wire than in an aluminum wire. The heat spreading ability of copper is a tertiary concern.

I figured I get something wrong. Does heat effect electrical conductivity? Know that I got it way off topic. I stand corrected though and I didn't look at that way.

[Cathar]

Originally posted by SysCrusher
I figured I get something wrong. Does heat effect electrical conductivity? Know that I got it way off topic. I stand corrected though and I didn't look at that way.

Now this is something that has often played across my mind.

Electrical resistance rises as a conductor/semi-conductor gets warmer. Electrical resistance is the main source of heat from CPU's.

If a more efficient water-block (or heatsink, or whatever) keeps a CPU cooler, then will that CPU dissipate less heat due to the lower electrical resistances due to it being cooler?

[8-Ball]

This is something I too have considered. Not exactly as discussed but more with testing rigs such as that used by BillA.

The heating elements work by joule heating which is dependent on the resistance. Again would a better block reduce the resistance thus reducing the power output.

8-ball

[8-Ball]

Also, I know it was a long post, but did what I said above make sense, or have I missed the mark?

8-ball

[Cathar]

Originally posted by 8-Ball
Also, I know it was a long post, but did what I said above make sense, or have I missed the mark?

8-ball

Seemed fine to me.

[Rayman2k2]

cathar, do you have an email addy cause i gots a few questions.

[Since87]

Originally posted by SysCrusher
I figured I get something wrong. Does heat effect electrical conductivity? Know that I got it way off topic. I stand corrected though and I didn't look at that way.

Yes. Copper has a TCR (Temperature Coefficient of Resistance) of about 0.4%/C.

So if you raise the temperature of a copper wire by 10C the resistance of the wire will increase 4%.

Originally posted by Cathar
Now this is something that has often played across my mind.

Electrical resistance rises as a conductor/semi-conductor gets warmer. Electrical resistance is the main source of heat from CPU's.

If a more efficient water-block (or heatsink, or whatever) keeps a CPU cooler, then will that CPU dissipate less heat due to the lower electrical resistances due to it being cooler?

I don't know the TCR of CMOS transistors and in any case I suspect it varies greatly with the doping of the particular IC. (A quick google didn't help.) I did find a datasheet for an N-channel MOSFET that indicated the TCR for the transistor was 1%/C but I have no idea how representative that is.

The TCR of copper will get you a 2% lower resistance in the copper interconnects if block 'W' gives you a 5C lower temp than block 'X'. Not a big difference. Also the copper resistances will be small compared to the semiconductor resistances, so I think the copper resistance can be ruled out as a significant factor. The semiconductor resistance may well be much more significant.

Another area where better cooling may have an impact is transistor leakage. As the transistors used in CPU's get smaller and smaller, they tend to leak more current even when they are shut off. Leakage in MOSFET's generally increases with temperature, so a cooler processor is probably consuming less power due to lower leakage current. Leakage tends to be very nonlinear with temperature though, so I don't have any prediction of how significant this may be.

It would be easy enough to test this. Leakage current is idle current. Just connect the proper supply across the VCC and GND pins of a sacrificial processor, ground the clock input, and measure the current draw as you vary the processor temperature. (If one had a CPU socket off a MOBO, it wouldn't even be necessary to sacrifice a processor.) This may be listed in the manufacturer's data. I've never looked.

[Since87]

Originally posted by 8-Ball
This is something I too have considered. Not exactly as discussed but more with testing rigs such as that used by BillA.

The heating elements work by joule heating which is dependent on the resistance. Again would a better block reduce the resistance thus reducing the power output.

8-ball

Yes, a better block would have an effect on the heat emitted by a die simulator IF you did not vary the supply to the heating element in order to maintain a constant power dissipation.

To the best of my knowledge, Bill varies the input voltage to his heating elements to keep the power at the right level.

[8-Ball]

I kind of guessed he would've already thought of this.

8-ball