480GTX Heat Load Test - Page 5 - Overclock.net

**ira-k** · 09-18-07

Quote:

Originally Posted by Thundergod989

i'll let you know how it stands up to 2 8800ultra's and 1 q6600 -- all overclocked.

It should handle that with-out breaking a sweat, I had 487W on it and my water out to room ambient was only a 4.06C delta...So just going by that this should easily handle 1000W I would think...We just really want to know, I asked HW Labs and they told me 2200W but wouldn't give up the room to W/O delta so were doing this to see what it really is....I'm going to guess its really 1100W with a 10C delta, with 4-100+ cfm fans,why they wouldn't want to say that I have no idea because just that is a killer rad...And like Martin said it gives us something to do...

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Zeraphil · 09-19-07

Did you get to test it, Ira? I was going to suggest a rather simple alternative to the whole "water boiler and stuff" thing, but if you went ahead and got it then I guess it wouldn't be of too much help.

Basically consisted of boiling the water in a kettle to a specific temperature denominated as "tf". Stick a thermometer in it and let the temperature stabilize, then pour it in your test loop. Yeah, basically what you have to control is the volume of water you pour to fill the loop, the intial temperature, and the final temperature. Just measure the time it takes for the water to drop to ambient assisted by the radiator. You can calculate the initial energy (j0) of the water and the final energy(jf). The difference jf-j0, and the time it took to achieve jf, minus the heat input by the pump, should be enough to calculate how much heat dissipation the rad can handle....assuming that the loss of heat by passive diffusion in th eloop is minimal. I'm sure I might be missing something, so correct me if I'm wrong, and if I'm making too many assumptions, then pay this no heed >.<

**ira-k** · 09-19-07

Quote:

Originally Posted by Zeraphil

Did you get to test it, Ira? I was going to suggest a rather simple alternative to the whole "water boiler and stuff" thing, but if you went ahead and got it then I guess it wouldn't be of too much help.

Basically consisted of boiling the water in a kettle to a specific temperature denominated as "tf". Stick a thermometer in it and let the temperature stabilize, then pour it in your test loop. Yeah, basically what you have to control is the volume of water you pour to fill the loop, the intial temperature, and the final temperature. Just measure the time it takes for the water to drop to ambient assisted by the radiator. You can calculate the initial energy (j0) of the water and the final energy(jf). The difference jf-j0, and the time it took to achieve jf, minus the heat input by the pump, should be enough to calculate how much heat dissipation the rad can handle....assuming that the loss of heat by passive diffusion in th eloop is minimal. I'm sure I might be missing something, so correct me if I'm wrong, and if I'm making too many assumptions, then pay this no heed >.<

I did on the first page with 487W of heat dump...Another one of my pumps died and I ordered another one, I'll have to make a res for this new pump its pretty big, then I'll just Pelt up my 320W TEC on my back-up comp...That way I'll have pretty well know values, it will put me up around 487W + 320W or so....If it stays under a 10C delta I would guess its a 1K rad for sure...Even looking at the 4.06C Delta I had on it with a 487W load I'm sure it is but I just want to play with it some more...

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Zeraphil · 09-19-07

Quote:

Originally Posted by ira-k

I did on the first page with 487W of heat dump...Another one of my pumps died and I ordered another one, I'll have to make a res for this new pump its pretty big, then I'll just Pelt up my 320W TEC on my back-up comp...That way I'll have pretty well know values, it will put me up around 487W + 320W or so....If it stays under a 10C delta I would guess its a 1K rad for sure...Even looking at the 4.06C Delta I had on it with a 487W load I'm sure it is but I just want to play with it some more...

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YEah, though I think you could test it for its 2k capacity with what I told ya.
Like, for example, if you calculate that your loop holds about 2 L worth of water, that would be 2kg (assuming density at 100 C is 1, it's really .95 but for simplicities sake), that means that if you -just- bring water to a boil, your thermometer reads 100 C, and say ambient temperature was 23 C where you're at, then you calculate how much joules you got in that water, defined by Q = specific heat x mass x delta T; holy shoot that means you have 4.186 joule/g*C * 2000 grams * 77. That means you have 644644 Joules in that steam pot! So, if you turn on the loop right then and there, and obviate passive convection through the tubes... IF and only IF the rad has a 2k Watt heat dispersion, it would have to disperse 644644 Joules / 2000 Joules per s, or achieve ambient water temperature (23 C in this case) in 322.22 seconds, or 5.37 minutes, obviating, again, passive convection. How does that sound for an afternoon, eh? Even I like the idea.

EDIT: Again, I might be obviating some important detail, so don't kill me, I just like fluid thermodynamics, but then again, I'm just a biologist >.<

**ira-k** · 09-20-07

Quote:

Originally Posted by Zeraphil

YEah, though I think you could test it for its 2k capacity with what I told ya.
Like, for example, if you calculate that your loop holds about 2 L worth of water, that would be 2kg (assuming density at 100 C is 1, it's really .95 but for simplicities sake), that means that if you -just- bring water to a boil, your thermometer reads 100 C, and say ambient temperature was 23 C where you're at, then you calculate how much joules you got in that water, defined by Q = specific heat x mass x delta T; holy **** that means you have 4.186 joule/g*C * 2000 grams * 77. That means you have 644644 Joules in that steam pot! So, if you turn on the loop right then and there, and obviate passive convection through the tubes... IF and only IF the rad has a 2k Watt heat dispersion, it would have to disperse 644644 Joules / 2000 Joules per s, or achieve ambient water temperature (23 C in this case) in 322.22 seconds, or 5.37 minutes, obviating, again, passive convection. How does that sound for an afternoon, eh? Even I like the idea.

That just gave me an instant head-ache...

...I'll try it out or adding a 320W TEC to it when I get my new pump in...I think its at best a 1k or 1.1k rad so I will start from there and work up...Thanks......

EDIT: I don't care if you are a biologist it still gave me a head-ache...

Zeraphil · 09-20-07

Quote:

Originally Posted by ira-k

That just gave me an instant head-ache...

...I'll try it out or adding a 320W TEC to it when I get my new pump in...I think its at best a 1k or 1.1k rad so I will start from there and work up...Thanks......

lol >.< My bad Ira