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Unread 02-15-2003, 09:55 PM   #16
Since87
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Join Date: Sep 2002
Location: Indiana
Posts: 834
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Quote:
Originally posted by unregistered

Blackeagle, you are exactly the audience I like to 'hit'
re the heat load:
I test at 70W ACTUAL, this is very much equal to 100W per Radiate or one of those other smoke-and-mirrors 'calculators'
- believe me, you will not be able to apply much more heat than I am doing
AND the "C/W" does NOT change according to the applied heat
(read some of the articles on the site in my sig)
I think Blackeagle was saying that he doesn't understand the relevance of "C/W" numbers. (It's kind of implicit that "C/W" doesn't change with applied heat.)

Blackeagle

To expand on what Bill was saying:

It doesn't matter whether Bill tests at 1 Watt or 1000 Watts of applied heat. (Actually, for practicality reasons it matters greatly, but that's another issue.) The "C/W" values he determines in testing is a constant value over the entire range of heat loads we could possibly care about.

The "C/W" of a waterblock (Rt [for thermal resistance]) indicates that for any applied heat (H) the temperature difference (dT) between the inlet water and the die will be:

dT = H * Rt

So if the power dissipated in the die is 125 Watts and the "C/W" of the waterblock is 0.2 at the flowrate of the system, then:

dT = 125 * 0.2 = 25C

It's that simple.

If you assume that the temperature of the water is the same everywhere in the loop... (Not true, but the temperature differences are small enough to leave to the masochistic among us.)

Anyway, if you assume the water is at a constant temperature, you can just add the "C/W" of the wateblock and the "C/W" of the radiator and use the same equation to calculate the difference in temperature between radiator intake air and die.
Using a fairly reasonable "C/W" value for a radiator of 0.05:

dT = 125 * (0.2 + 0.05) = 31.25C

So if the ambient air flowing into the radiator is 20C, the die will be at 51.25C.

Now there's a horde of issues I'm not bringing up here, because I wanted to focus on the relevance of "C/W". Most of the math involved in designing a good system is not much more complicated than that though.

Hopefully it's obvious now, that it doesn't much matter what heatload Bill tests at, because the "C/W" numbers he measures are just as useful at 125 Watts as they are at 70 Watts.
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