Does Water Cooling your PC Also Cool Down Your Room? - The Workshop

[SaperPL]

2 hours ago, TheWoollyMammoth said:

You didn't factor in the higher power consumption of the water cooler,  which generates more heat than saves at the heat source. 

Quick search on the internet says a pump draws 12-18W so it's still less than the difference on the charts I linked and not all of that energy is dissipated as heat since some of it is used to move the fluids around.

2 hours ago, LeapFrogMasterRace said:

This not a controlled experiment and that is well within the margin of error and those cpus aren't even clocked the same in the graph they are using more watt energy because of that P=IV.

Those are charts from test on the same cpu with two different locked clocks and I'm not comparing the two charts together... Try to understand what are we talking about and please read the axis descriptions before responding

48 minutes ago, LeapFrogMasterRace said:

At least I know the difference between a forum post and an article .

I know you are new but perhaps you should look at our community standards, on this forum we try to be nice to each other. 

I posted this forum thread since it was the closest use case I could find to the current processors. I had all this stuff over my university courses and If I were to show you example on some random non-PC chip you wouldn't treat that seriously, I think.

 

Thanks for short and precise explanation Martin86.

 

I love Linus's understanding of the market and manufacturing costs, that he shares with everyone, but whenever he tries to do science, it's all over the place

[vanished]

I'm going to ignore the whole "of course it doesn't help" topic since that's obvious and has been beaten to death already; I've got something different I'd like to see!

 

I wonder about if you placed the PC on the ground, vs placing it on a desk roughly half the height of the room.  Heat rises, and if there was very little mixing in the room, in theory, the computer on the desk would have less air to "play with" since it would heat up the top half of the room without heating the bottom half much, while the computer on the floor could heat the entire room.  So, what I'd like to see tested is if a computer on a desk will heat up the "room*" faster than a computer on the floor?

 

* I don't mean the whole room, I mean if you placed the temperature probe like 5 feet of the ground or something, would it go up faster with the PC on the desk than if it was on the floor?

[officialanal]

[andrew_nyr]

22 hours ago, LinusTech said:

Amazon: http://geni.us/BvX
NCIX: http://bit.ly/1RNNgXu

 

We water cool the CPU and video card of a computer to determine if that'll make your gaming environment more comfortable.. FOR SCIENCE!

 

 

 

Not unless you do whole room water cooling with the radiator on the outside https://www.youtube.com/watch?v=b8bLtg9J1Oc

[andrew_nyr]

hi linus

 

[PepeSilvia]

It's a shame you didn't do the half hour intervals for the second test. I think the most interesting thing that the video could have shown is whether water cooling, by dissipating heat more effectively, heats up the room faster. I'm assuming that it wouldn't be a huge difference, but it would be interesting to see how it played out.

[vanished]

11 hours ago, PepeSilvia said:

It's a shame you didn't do the half hour intervals for the second test. I think the most interesting thing that the video could have shown is whether water cooling, by dissipating heat more effectively, heats up the room faster. I'm assuming that it wouldn't be a huge difference, but it would be interesting to see how it played out.

I dont see how it would be any different.  You have the same energy/heat output, which is why it was at the same final equilibrium temperature, so the heating rate would be identical.

 

Edit: I just remembered that they did not reach equilibrium in the test due to time constraints but my point stands: in both cases the same temperature increase happens in the same duration, thus in both cases heat output in watts was the same (as expected) and thus intermediate readings would also all be the same. In fact, since they did not reach equilibrium, you can consider their "final" reading as an intermediate reading itself.

Edited March 7, 2016 by Ryan_Vickers

[SaperPL]

15 hours ago, Ryan_Vickers said:

I dont see how it would be any different.  You have the same energy/heat output, which is why it was at the same final equilibrium temperature, so the heating rate would be identical.

 

Edit: I just remembered that they did not reach equilibrium in the test due to time constraints but my point stands: in both cases the same temperature increase happens in the same duration, thus in both cases heat output in watts was the same (as expected) and thus intermediate readings would also all be the same. In fact, since they did not reach equilibrium, you can consider their "final" reading as an intermediate reading itself.

If linus did the interval tests while running the first scenario and didn't do that in the second one this means he could've let out bits of hot air just by opening the door on those intervals which could make a slight difference of few degrees.

[PepeSilvia]

15 hours ago, Ryan_Vickers said:

I dont see how it would be any different.  You have the same energy/heat output, which is why it was at the same final equilibrium temperature, so the heating rate would be identical.

 

Edit: I just remembered that they did not reach equilibrium in the test due to time constraints but my point stands: in both cases the same temperature increase happens in the same duration, thus in both cases heat output in watts was the same (as expected) and thus intermediate readings would also all be the same. In fact, since they did not reach equilibrium, you can consider their "final" reading as an intermediate reading itself.

I agree with your second point more or less but the rate that the heat was exhausted into the room was the variable i'm interested in, not the wattage/heat from the components (which was the same). The components using the same amount of power but being significantly different temperatures suggests that the rate of transfer of heat produced is greater with a water cooled system (obvs). The component temps recorded stabilize at a lower temperature, presumably in about the same time-span using the same amount of power on water. So are they generating less heat output or is more heat being transferred through the cooling loop? Does the rad expel the same amount as the heatsink? Is the difference held within the loop itself? If the rad is expelling more heat than the heatsink then there's a fair chance the room temp will increase more quickly to begin with using a water loop. Even if both begin to plateau at about the same room temp the time it takes to hit the plateau could be different. Eventually the heat escaping the room will either stabilize the room temperature or present a similar curve in both scenarios.

[vanished]

5 hours ago, PepeSilvia said:

I agree with your second point more or less but the rate that the heat was exhausted into the room was the variable i'm interested in, not the wattage/heat from the components (which was the same). The components using the same amount of power but being significantly different temperatures suggests that the rate of transfer of heat produced is greater with a water cooled system (obvs). The component temps recorded stabilize at a lower temperature, presumably in about the same time-span using the same amount of power on water. So are they generating less heat output or is more heat being transferred through the cooling loop? Does the rad expel the same amount as the heatsink? Is the difference held within the loop itself? If the rad is expelling more heat than the heatsink then there's a fair chance the room temp will increase more quickly to begin with using a water loop. Even if both begin to plateau at about the same room temp the time it takes to hit the plateau could be different. Eventually the heat escaping the room will either stabilize the room temperature or present a similar curve in both scenarios.

The heat generated by the part must match the heat released by the cooler attached or the part will continuously heat up or cool down until they do match. The chip can sit at a lower temperature on liquid cooling because the resistance to heat transfer posed by the liquid cooler is less than that of the air cooler - the liquid cooler can transfer heat to the room at X watts when at 50 deg while the smaller air cooler has to be at 80 deg to also transfer at X watts ( just example numbers)

[Martin86]

I wanted to add a few points to my previous post in this thread.

 

1. When a chip is operating at a cooler temperature the core voltage can be lowered to achieve the same core frequency as a higher temp/voltage would require. Because the resistance in the copper decrease and less voltage is requied to push current through it. Thus it would produce less heat then on a higher operating temperature. Modern CPU's does Auto-regulate voltage, but I am not sure if it does so linearly with frequency or temperature or some other mechanism. It can be the case that the extra water pump would require more watts then what is saved on the CPU.

 

2. For a system on a lower temperature, it gives the CPU room to increase its frequency. The higher the frequency the faster a task gets completed, and for a few nanoseconds the core can idle at a lower voltage, which produce less heat. This is how mobile devices are developed, get the task done as fast as possible and idle to save power. It is better to be at 95W for 10% of a millisecond, then at 45W for 90% of a millisecond, it works kinda like a duty-cycle.

 

3. The CPU can move a Thread around to different physical cores (scheduling) to prevent one core to heat up more then the others. This is to "spread out" the heat generation on a larger silicon surface. The higher the CPU temperature, the more the scheduler moves the thread around, resulting in less performance. L1 Caches and core contexts has to be copied around on the chip which takes time (more duty-cycle), more watts and results in less performance.

 

4. Simply by testing the system (outside in a room with plenty of ventilation) with both Air and Water cooling and reading the wattage off a WattMeter would give all the answers for this experiment. If the wattage consumed is greater with water cooling then air; it would output more heat because wattage = heat generated. It would be interesting to test water cooling with the pump connected to a different power supply to see the drop in watts needed due to lower temps! The room only becomes interesting as the ambient-temperature-rise would set new conditions for what the core temperatures would stabilize at. But remember to compare apples to apples (voltages, frequencies, and bechmark results).

 

5. A system operating on a high temperature would also emmit more Infrared Radiation from the case itself; because the case has a higher temperature. Remember experienced heat is not only from convective heat transfers, but also conductive (touching) and radiating (photons). Sitting next to a huge case with a temperature of 40+ celsius would expose you to more IR photons, and it will heat up more the surface the case is sitting on (rubber feet vs metal). The case would also irradiate the walls, ceiling and floor which would also heat up the room which in turn would make those surfaces emmit more radiation towards the user. A water cooled system would put more of the heat into the air which a good ventilation system can toss out pretty quickly, but the radiated and conducted heat would "stick around" more.

 

6. Benchmark stats would also need to be recorded. What is the point of proving for a given situation that Air and Water cooling gives the same room temperature if the benchmark stats for one is lower then the other? Again apples to apples.