It looks like I have a few new ideas for chilling. - Page 2

Pump->Cold Side TEC->CPU->120MM Rad->HOT Side TEC->480MM Rad->Res->Pump

This does not make sense.

Why not place the TEC cold side directly onto the CPU, then have a waterblock on the warm side to remove the heat to the radiator which will reject it to the air?

Wiki also states these TEC coolers are only like 5-10% efficient, so they will produce a lot more heat then what they remove, just like any other refrigeration system.
Edited by ginger_nuts - 10/9/12 at 4:26am

Who are you going to believe ? People that actually use Tec`s or a supposed knowledgable site. Using a TEC at 40-60 % of it`s Vmax will provide the best eff. it will typically be 70 - 90% eff. around there not 5 - 10. Mounting the TEC on the cpu is possible but the IHS on the die is not totally flat and you will run into frost and condensation problems. Dual sided chiller unit is best solution, please take a good look at chiller designs and TEC`s in general so that you can get a better grasp of how they work.

The second law of thermodynamics specifies that it is statistical impossible creating a machine that works the way you want it to work:

A TEC is a heat-pump. Transfers heat from one of its sides to the other at the expense of electricity that gets converted to...more heat!
That means that if the "cold side" can effectively "pump-away" 100W of heat, dissipated off the hot side will be those 100W of "work" the TEC produces, plus the extra heat generated doing so, which will sum up to much more than 100W total.

Adding both the cold and the hot side of a TEC in the same loop, will end up you heating up the water with the TEC's own heat produced every second ontop of the heat the elements being "cooled" would produce. You don't just cancel out the hot and cold effects, you seriously penalize the performance in comparison to having no TEC to begin with.

The TEC - if used at all - should be either interfered between the CPU and the CPU block, dumping the hot side's heat into the cooling loop, or indirectly cooling your primary loop's water, with a secondary loop or huge air heatsink dissipating the heat out of it.

why don't you try it and see?

it does not matter where you put your rad, tec, gpu, cpu, pump, etc. it does not matter if you put cpu first then the then the pump second, etc, etc, it's a closed loop, there is no first or last, everything you dump in that loop will get MIXED/EQUALIZED, because the pump circulates its contents.

so to calculate you temps, you add all your heat output and subtract it to the cold output, so it doesn't really makes sense to mix the cold and hot output of the TEC, you might as well not use it.

Bit of a pointless comment, most pumps will take 50-60 degrees Celsius fluid.

The fluid would not reach that unless the rads are seriously undersized.

yes, you are right, that theory has been tried and tested by many people, ask the experts and they will agree with that theory.

as i said, it does not matter how you place your loop, you can place you pump before or after the cpu, it does not matter, the temperature across your loop is all the same, maybe 1c or 2c difference but that's about it.
Edited by ghostrider85 - 10/9/12 at 11:50pm

In Theory I think you could do this but it would be very impractical.

If I got my numbers right, it takes 4186 joules to raise or lower 1 liter of water 1c.

Watts = joules/sec so if you wanted to cool 1 liter of water in one second it would take 4186 watts to do it. Thats just 1c.

if we are trying to flash the water in a chiller block I'd say we are looking closer to .10 liters or ~410 watts to do 1c.

So if we wanted to flash our water 10c then it would take 4100 watts to do so.

You would need at least 10 400 watt tecs to flash the water, then run that to the cpu then take that water and cool the tec.

This would work because we are flash removing heat in the loop to the cpu and then putting in right back into the loop as it exits the cpu.

Now you would have to have a hell of a radiator to cool the 4000 watts, plus the waste heat of another 6000 watts + the cpu of 250 watts.

your radiator would need to be able to dissipate 10,000 watts.

I'd love to see this in action.. I think the radiator would take up half your room though.


Now the reason we don't normally do it this way and use 2 loops is because we can use multiple turns of the water to lower the temp as it runs through the chiller as long as we don't add more heat back into this loop.

Example we can chill one liter of water 10c in 100 seconds (4186 joules/liter * 10degrees / 400 watt tec) using a single 400 watt tec.

We use the second loop to dissipate the heat from the first loop as well as the waste heat of the tec.
Edited by Mindchi|l - 10/10/12 at 10:40am

My summer rog box can dissipate about 2000w of heat @ 10c delta t.

This is what is looks like
Warning: Spoiler! (Click to show)
That's an 18x12 inch heat exchanger with two 10" fans pushing a ton of air (600cfm/ea.).

So you would need five of those to flash cool... That's a lot of radiators!

I wanna see this happen!