I can't tell if you guys are making fun of me or putting yourself down, just ask questions, don't hesitate, anything. But you have to do it, I can't initiate as I can't read minds.
__NAZ, when you get back in here I'd like to know your stock Vcore, I'm still have a hard time with how you figured your 860 to be dissipating 270 Watts, seems almost double what it could be.
Going by the formula:
(OC TDP÷stock TDP) = (19÷21) x (200÷133) x (1.3÷stock Vcore)
2 if assuming 95W TDP as your stock (which it isn't btw), and 270 for the OC TDP, then your stock Vcore was way down at 0.89v...
This was done by taking 270÷95 = 2.84, then divide the multi ratio and divide the bclk ratio, then take the square root. Lastly, divide 1.3 from this number and you get the value for stock Vcore. When you do the TDP equation, do the division first, then raise the voltage exponent, them multiply everything together.
Fact is, the typical i7 CPU will have a stock value between 1.1 and 1.3, OC TDP will not be higher than 160W at your settings for 3.8GHz -- you could use an H50 with two high-flow fans and call it a day.
The impact of hyperthreading according to many reviews have shown that enabling it increases temperature by 5%; all things equal, this accounts for a 5% increase in heat wattage.
As for estimating 860's TDP, I set the fastest of a series/type as the hottest part and scale down by clockspeed (or process tech or cache). There is an i7 880 rumored to appear, so I'll tentatively assume 880 is 95W TDP to qualify for the rating. 860 has the 21x stock multi, 880 is supposed to get the 23x stock multi, so 860 is 9% slower. Assuming the same, or very similar stock voltage, then that's 9% less heat, or about 86W -- this is closer to true i7 860 stock heat load. Redoing the OC TDP equation and using
this review's reference stock Vcore (since I don't know yours):
(19÷21) x (200÷133) x (1.3÷ 1.247)
2 = 1.478
86W x 1.478 = 127W --- this is closer to your heat load, not 270.
Above I stated best to have a gap between what you need and what you have, don't get minimum. So multiplied by 2 or 3 and we get a radiator set from 260 - 390W, any dual 120 radiator will suffice, more than that will bring temperatures lower but with diminishing returns.
If you pruchased those dual rads from Swiftech, MCR220, that's a thermal resistence factor of 0.033
oC/W supporting 325W at 2gal/min and a pair of 80CFM fans. A 2gal/min pump is a their 350/355 pump and if you choose
Swiftech's Apogee GT CPU block which is rated for about 0.05
oC/W at 2 gal/min. Total thermal resistence factor is the sum, 0.05+0.033 =
0.083oC/W.
What does this mean and why should you care? At a load of 127W, a resistence of 0.083
oC/W means the cooler as a whole will raise temperatures by 127W x 0.083
oC/W = 10.5
oC above ambient. Which ambient, air? Not necessarily, the liquid in the loop doesn't act as just a medium, not all the heat is transfered from the block to the radiator, its the whole reason the liquid heats up. For Swiftech's radiator tests, they maintained a 10
oC difference between air temeprature and the liquid temperature throughout. So the 10.5
oC actually adds to the liquid temperature.
Your theoretical
minimum full load temperature with dual rads would then be 10 + 10.5 = 20.5
oC above the reference air temperature, which is the air entering the radiator (not air measured anywhere inside or outside your case or general room temperature). If the air temperature entering your dual 120's is 35
oC, then the CPU IHS full load temperatures could be 55.5
oC. If you went for Swiftech's triple 120 rad instead your full load temps dip into the upper 40's.
Here's another example:
When K|ngp|n ran his QX6700 to a WR 4.9GHz back in 2006, I memorized his numbers: 15 multi from stock 10, 327Mhz FSB from stock 266, and Vcore of 1.792 from stock 1.25. His full load temps were -191
oC centigrade, used LN
2. His TDP ratio was = (15÷10) x (327÷266) x (1.792÷1.25)
2 = 3.79 times higher than stock, or almost 500W of heat from a 65nm quad! LN
2 boils at -196
oC, so the thermal resistence of his copper pot was the change in temperature divide by heat wattage or (-196 - (-191))÷ ~500 = 0.01
oC/W.
post edited by lehpron - 2010/06/27 15:57:02