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Does high flow make radiator placement less important?

post #1 of 12
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Scenario: 2xQuad Radiator -> CPU -> GPU. Assume the radiators have more than enough capacity to remove all the heat from the loop on the return trip and bring the coolant back to room temp.

Now we take into account the pump. If it were moving water really slow, the liquid would pick up a lot of heat from the CPU before going into the GPU. This would seem to indicate that you would have to have a radiator between the CPU and GPU just to get the GPU's cold water. However if your water was moving really fast, all parts of loop would always be ~ room temperature?

In a nutshell I'm wondering if I can throw a lot of rad (2x RX480) and a lot of pump (2 x MCP655) at my loop, and then not have to worry about single vs dual loops or using Y splitters to feed the CPU and GPU in parallel from a single loop. I am however a bit worried that 2 of those pumps on 5 will blow my res up.
post #2 of 12
Water works like air, so more flow means more heat transferred. Someone did some fanless radiator testing (and I did as well) and the results are not good. Even with a dual 140mm and a dual 120mm, the water just got hotter and hotter.
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post #3 of 12
it's been said over and over and over again.
Contrary to nit picky people out there, the order of your loop doesn't matter past the standard reservoir before/above pump part.
even in lower flow situations, the water coming out of your cpu block MIGHT be 0.5c warmer than it went in usually not that much though...
the water in your loop flows through your loop so many times before actually getting a chance to warm up that it's also gone through the radiators enough times to nullify it.

Either your entire loop warms up over time, or it doesn't. If it does, get more fans and/or radiator.
post #4 of 12
You just need one rx480 rad. One mcp655 will handle it just fine.
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post #5 of 12
Stock phrase time!

It takes ~260W to heat water at 1GPM by 1°C

Ahhh, much better.


It is a linear relationship, so if you half the flow rate you double the temperature rise, half the power half the temperature rise and so on.

Even at 0.5GPM you would be looking at something in the region of ~2°C water temperature rise through your CPU block, the GPU isn't going to care much about that.

You are right that as the flow rate increases the temperature rise through the components decreases, and the temperature difference between the "hot" and "cold" parts of your loop decreases. The opposite is true that as the flow rate drops the temperature difference increases.

However for the flow rate to be low enough for the temperature rise to be important you would have other problems. As the flow rate drops below ~0.7GPM the heat transfer in the CPU block starts to drop off rapidly. This is due to increased lamina flow in the CPU block which inhibits heat transfer to some extent.


Tl:Dr - Keep the flow rate at about 1GPM and you will have no problems with loop order.
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post #6 of 12
Thread Starter 
Awesome guys. Thanks for the help. I will stick with my quad rad and single pump and see how everything goes. Part of my concern is from people on water still getting 60C or more CPU temps. I presume this is because there is a "lag" in heat transfer between the IHS and waterblock, so even if you had a million pumps the CPU temperature would not go down. Not unless you increase the delta T by using chilled water...
post #7 of 12
Quote:
Originally Posted by dr/owned View Post

Awesome guys. Thanks for the help. I will stick with my quad rad and single pump and see how everything goes. Part of my concern is from people on water still getting 60C or more CPU temps. I presume this is because there is a "lag" in heat transfer between the IHS and waterblock, so even if you had a million pumps the CPU temperature would not go down. Not unless you increase the delta T by using chilled water...

Pretty much.

The other thing is that a lot of people get water cooling and then crank their v-core and clock speeds.

Then they come back and say "hey, why is my CPU so hot?"
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post #8 of 12
Quote:
Originally Posted by GingerJohn View Post

Stock phrase time!

It takes ~260W to heat water at 1GPM by 1°C

Ahhh, much better.


It is a linear relationship, so if you half the flow rate you double the temperature rise, half the power half the temperature rise and so on.

Even at 0.5GPM you would be looking at something in the region of ~2°C water temperature rise through your CPU block, the GPU isn't going to care much about that.

You are right that as the flow rate increases the temperature rise through the components decreases, and the temperature difference between the "hot" and "cold" parts of your loop decreases. The opposite is true that as the flow rate drops the temperature difference increases.

However for the flow rate to be low enough for the temperature rise to be important you would have other problems. As the flow rate drops below ~0.7GPM the heat transfer in the CPU block starts to drop off rapidly. This is due to increased lamina flow in the CPU block which inhibits heat transfer to some extent.


Tl:Dr - Keep the flow rate at about 1GPM and you will have no problems with loop order.

this is confusing..
GPM=gallon per minute? do you mean galon?
because
specific heat capacity of water= 4.185j/g
1j=1w
4.185*1000*3.78541=15,843.833555j/galon= 15,843.833555 watts/galon

DeltaT*amount of water(in galons)*specific heat capacity= energy required
since both of those numbers are 1, the wattage required would be 15,844 watts wouldnt it?
 
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post #9 of 12
Quote:
Originally Posted by AznDud333 View Post

this is confusing..
GPM=gallon per minute? do you mean galon?
because
specific heat capacity of water= 4.185j/g
1j=1w
4.185*1000*3.78541=15,843.833555j/galon= 15,843.833555 watts/galon

DeltaT*amount of water(in galons)*specific heat capacity= energy required
since both of those numbers are 1, the wattage required would be 15,844 watts wouldnt it?

Check my sig, or just follow this link

Edit: Yes, gallon. I know, mixing units, bad me!

4.185J/gK

1W = 1J/s

1GPM = 3.785 L/m = 3785g/m = 63.08g/s

dT = Energy / (SHC * mass)

To do it on a time basis
dT = Power / (SHC*mass flow rate)

Power = dT * SHC * mass flow rate

for dT of 1°C
Power = 1* 4.185 * 63.08 = 264W
Edited by GingerJohn - 3/4/13 at 4:16pm
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post #10 of 12
Quote:
Originally Posted by GingerJohn View Post

Check my sig, or just follow this link

hmm quite advanced, when i first saw your post i was curious what you were saying exactly, i guess i misinterpreted the 260w for 1c of water at 1gpm sentence, sounded a lot like it takes 260watts to heat 1 gallon of water to a degree higher
 
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