Huge Radiators: Roundup

[kinghong1970]

The RX480 I got is literally flawless. The fins are perfectly spaced, the finish is smooth and even, all the threads are in the right place etc.
If my RX480 represents all the current RX rads then there is really nothing to worry about. But of course I can't really know since I only have that one radiator here.

i was playing around with my RX480 and i realized that my has one flaw... the 4 corners of the radiator does not lay flat against a flat surface.
it's a bit "twisted" to give appx 1/8" wobble at 2 opposing corners.

meh... so long as it hold water and cools it...

and now... for this part...

A radiators performance shows in it's temperature difference "Water - Air". i.e. how much hotter does the water get than the ambient temperature? The smaller this difference, the better the rad performs.

This difference is dependant on the radiators efficiency at transferring heat (in short: it's performance) and the amount of heat it has to deal with. The more heat is being dumped, the higher the temperature difference is going to be, obviously.

Now, C/W is exactly that measurement - temperature difference Water - Air - but adjusted to how much heat was dumped into the loop during testing. The heaters don't deliver a completely constant heat load, so just comparing the absolute temp differences isn't quite fair.

So if your data shows that radiator X produced a temp difference Water - Air of, say, 7.5° at a given fan speed, that's one part of the equation. Let's say that the average heatload during the test was 290W.

This results in a C/W of 7.5/290 = 0.025862

ok, so according to your chart, you have temps for air in and air out as well as temps for water in and water out...

which figures do you use for the temp difference of "Water -Air" ?

To translate C/W values back to tangible water temperatures, just multiply them with the estimated heatload in your LC loop and add your room temperature to that.

as for estimation of your heatload, where and how?

With the above example, I would say my system dumps maybe 180W of heat into the loop so it's 0.025862*180 = 4.66° plus my room temperature 24° = 28.66° water temperature.
So I could roughly estimate that I'd get a water temp of about 29° with a radiator that has a C/W of 0.025862

So much for my explanation. I hope it helps (and I also hope I didn't make any mistakes...).

aaah, i get this part... you're basically taking the CW value and plugging it backwards to get appx water temps in your loop when exiting the radiator...

i think my brain cell just split!!!
cell count hittin double digits now!!!

watch out you 1337 testers!!! evolution is in action!