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02-05-2004, 05:22 AM | #26 |
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[quote=lolito_fr]
1)I was also thinking of heat produced in the block due to friction, and conversion of potential energy into heat. For the Cascade, 2.5m @ 7.5lpm turns out to be roughly 3W unless I'm missing anything? Not entirely negligeable, but not even close to 25W. Plus this would be a quadratic variation, rather than the quasi-linear one that we're observing. 2) I find it hard to believe that secondary losses could amount to 25% variation in total power dissipation, considering that the sink temperature is apparently not varying by more than 0.5°C, and even the core is only changing by 2°C. (Cascade data) Nothing can be that closely coupled to the waterblock for such a small change in temperature to produce such a large increase in power? 3) Non uniform mixing of flow, or sensor temps influenced by pipe surface temps seems like a good possibility. (QUOTE] 1) Your sums are correct ,3 watt is about right: Using Watts ~ m(H2O)*LPM/6 2) Just dunno. Simple modelling is nigh impossible with the complexity of the thermal enviroment.May be possible to split losses/gains into the component which enters/leaves the system through mb/die path and that through Non-die/wb path. The former being possibly only "C/W" dependant and the latter being "Flow-rate and wb" design dependant.Just possibly,this is evident in the "watt v C/W" plot.Some attempt to split these is in Incoherent's work. However consider it to be extreme speculation and not worth pursuing until thoughts are more crystallised 3) Again dunno,but would have thought Bill(in his "'Heat Source Efficacy' measurements) would have encountered this problem if it exists. |
02-05-2004, 05:26 AM | #27 |
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A little physics.
Pressure is just potential energy. So to pump 7.5litres of water up to 2.5m, you need m*g*h Joules of energy. m=mass in kg (or volume in litres, as 1litre of water=1kg) g=9.8m/s^2 h=height in meters 7.5*9.8*2.5=184J If you pump 7.5 litres in one minute, that’s 184J/minute. Divide by 60 to get J/s: 184/60=3.07J/s=3.07W Derived formula: P=0.16Q.H P in Watts Q in lpm H in m |
02-05-2004, 05:32 AM | #28 |
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sorry! Les beat me to it.
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02-05-2004, 05:57 AM | #29 | |
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Quote:
I only gave the result, not the derivation. It is the derivation which is important. |
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02-05-2004, 06:59 AM | #30 |
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I know it's probably me who doesn't understand the goal of this exercice, but wouldn't calculation cpu wattage be easier and more reliable using a formula based on Vcore and cpu specs ?
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02-05-2004, 09:26 AM | #31 |
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This whole thing about being able to calculate CPU wattage using flow rate and delta T sounds fishy to me. All you are measuring is how much power (thermal) is being transfered to the water. Not the power output of the CPU. What you are seeing must be secondary heat loss.
There could also be problems with the equation. Last edited by murray13; 02-05-2004 at 09:34 AM. |
02-05-2004, 09:29 AM | #32 |
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pHaestus, a very small offset in one of your water temp probes could give this result (the strange increase in power as flowrate increases).
I am talking about an offset of perhaps less than 0.05 degrees. A simple way to test this would be to reverse the flow though the waterblock you have in your setup now, so that inlet T becomes outlet T and vice versa. The block performance would of course suffer but, if there is in fact an offset, we could work out the exact amount. If you can find the time to take many measurements at a few flowrates I think that this would be rather valuable. I am not questioning your setup (your data shows that it is actually stunningly accurate, the noise is unbelievably low) but an offset between the thermistors might be very small and difficult to detect. If you have been interchanging inlet T and outlet T then this is of course irrelevant. It proves that the setup is excruciatingly accurate. Cheers Incoherent |
02-05-2004, 09:47 AM | #33 |
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I'm suspecting 'h' needs to be addressed
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02-05-2004, 10:02 AM | #34 | |
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Quote:
I am guessing "head loss"? or "heat capacity"? Sorry Bill, you are losing me. (not too difficult ) Cheers Incoherent |
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02-05-2004, 10:13 AM | #35 |
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'h', as in convection coefficient
in pHaestus' graph #2 the 'apparent' wattage goes from low to high along with the flow rate - one would observe that the value of 'h' is also going from low to high (relatively speaking) I'm guessing the equation has been simplified a bit much no answers here, just a suspicion |
02-05-2004, 10:31 AM | #36 |
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I'll see about just swapping the thermistors tonight incoherent
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02-05-2004, 11:05 AM | #37 |
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yea, sorry, washers not needed
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02-05-2004, 04:58 PM | #38 |
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I Still don't get what it is the result is supposed to be, C/W of the system is easy to calculate using other much easier to get values. This program can help do i very easy, I know the page is in Danish, but just press Download and the program is in english.
http://www.overclocking.dk/download....de=show&id=324 |
02-05-2004, 06:42 PM | #39 |
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GIGO, Myth
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02-05-2004, 06:58 PM | #40 |
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Damn hopefully beeing Danish can excuse my poor english.. GIGO? don't understand.. ( I know i prob. misunderstood the topic of this discussion.. but i thought he wanted to calculate the C/W of his system..)
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02-05-2004, 07:10 PM | #41 |
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The more I think about it, the more I'm convinced that temp gradient in the hose is responsible for the large difference in wattage between the cascade and maze4 at the same cpu temp. The flow through the outer 2 channels on each side of the maze are problably picking up little or no heat while the center channel may be picking up 50% or more. The cascade with all the flow starting at the core combined with the high turbulence should definitely see a better mixing than the maze.
If the above is true then changing the depth of the sensor ( if possible) would yield a large change in the maze wattage (50 watts perhaps), while yielding a small change in the cascade (0 to 10watts) If this also proves true, then blocking off the flow to the maze's outer channels (filling in with silicone for instance) should give a nice performance boost by increasing the velocity of the inner channels for the same flow rate. I'm guessing as high as 1.5C at 1 gpm. Edit: spelling Last edited by freeloadingbum; 02-05-2004 at 08:24 PM. |
02-05-2004, 07:31 PM | #42 |
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GIGO is an old programmer acronym. Short for Garbage In, Garbage Out. Those CPU Watt calculators aren't grounded in reality. Sure I can calculate a C/W, but it'll be crap.
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02-05-2004, 08:37 PM | #43 |
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pHaestus
Were the tests performed with the TBredB 1700+ at 2200MHz and 1.85V ? Don't think MHz and V were specified in article. Info would be useful for cross-ref to Incoherent's thread. BTW Benchtest com Calculator indicates 113Watts (100% usage) for a TBred 1700+ at 2200MHz and 1.85V. Last edited by Les; 02-05-2004 at 08:46 PM. |
02-05-2004, 09:34 PM | #44 |
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Beginning to wonder if splitting atoms wouldnt be easier :shrug:
A great thread just wish I understood what all them pretty lines ment. |
02-06-2004, 05:00 AM | #45 |
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#42 Ahh!
I know the eventhough we have spent a lot of time geting the calculations i the program as correct as can be.. it will always be more of a theretical result than a real life one, and it we do get similar results compared to the BTW benchtest calculator. So we prob. use the same formulas. The question realy is how far the result calculated this way is from the real life result which he is trying to calculate using the formula.. But i get your point, and think i will wait and see if we get some workable formulaes, and perhaps they could be added to the program.. |
02-06-2004, 09:43 AM | #46 |
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An observation pHaestus
If you are using anti-freeze in your system, the value for heat capacity should be less than 4186J/kg*°C. Unfortunately I don't know how much but I would guess that it would be in the region of 10% less with the 1:4 mix you are using. Not particularly significant when comparing waterblocks but something to remember if we want to compare data from different power measurement techniques. It does bring your W results down to a perhaps more realistic level but doesn't explain the slope. Again, I suspect an offset. A bit of qualification for this statement in these two charts. This a bit colourful maybe but it is showing the calculated delta T curves vs flowrate for various power levels. Also shown are the Ts from your data for all the blocks. As you can see the delta T's adjusted by an offset of -0.07 deg fit very closely to a 70W heat source. The second chart shows the new power calculation vs flow rate. This is a bit ideal but might point in the direction of a (small) probe error. Another thing, I think that pressure loss needs to be measured for these blocks at different flowrates. I think that presented as it is the Cascade seems to be massively ahead of the competion (a conclusion I believe is not incorrect) but for a given pump/system the flow will be less through the Cascade, and hence the performance delta will not be as much as you'd conclude from simply looking at flow rate. I think this has been discussed elsewhere. Not to take ANYTHING away from your review/testing methods mind you, or from the Cascade. Both are fantastic. Cheers Incoherent Last edited by Incoherent; 02-06-2004 at 09:52 AM. |
02-06-2004, 10:17 AM | #47 |
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Incoherent
You have been busy. Good thinking. Never thought of introducing a known error. I use Scott Gamble's data for anti-freeze correction. Yep 10% seems about right. Graphs look eminently sensible. Will check your Excel manipulations for stupidity - will only post if find. |
02-06-2004, 10:28 AM | #49 |
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Using pH's but will cross-check.
Thanks. |
02-06-2004, 11:17 AM | #50 |
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Incoherent you are 100% right about needing pressure drop vs. flow rate data. My digital pressure gauge was acting strangely when I was testing that MCWChill; I can see if maybe it was the low temperatures that were messing it up and if so generate. Otherwise I bid on ebay periodically but haven't yet gotten lucky.
I took the day off today and just went downstairs to look at the test system. The pump is running but the PC isn't on. There is a deltaT of 0.06C across the Maze4 wb at 1.5GPM; raising the flow rate to ~3gpm raises the delta T to 0.07C. Guess I'll swap the inlet and outlet and see if that offset persists (in the opposite direction); if so I'll just go back and adjust the outlet temps by the appropriate amounts. Winder where that came from; I specifically remember putting both these probes into water and they read exactly the same as water temp changed...
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