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General Liquid/Water Cooling Discussion For discussion about Full Cooling System kits, or general cooling topics. Keep specific cooling items like pumps, radiators, etc... in their specific forums. |
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11-17-2004, 07:12 PM | #176 | |
Cooling Neophyte
Join Date: Oct 2004
Location: France
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Quote:
We need same graph but with a constant power as a CPU will put the same power in water in a high or a flow rate. So the coolant/air differential will increase when flow decrease... And the EHEIM1048 and 1046 are used for low flow since they are quiet and resistant, but their PQ curve aren't really good for high pressure. Try the same test with the Laing DDC and D4 (maybe results will be the same, but I dont think EHEIM are representative of good pumps for high drop pressure) EDIT: sorry if this point was already discussed, I realize I havent read half of the topic yet ! EDIT2 : IMO, most users mix "low flow" with "µchannel". In commercial blocks which use "µchannel", the pressure drop is very high so the "low flow" is a direct effect. But it's not the "low flow" which is efficient but the "µchannel". As you all know, same block will always be more efficient with higher flow (without dealing with the pump and his power). If you look at the tests of roscal's blocks (named "proto") you can see an example of a real "µchannel" (0.3 mm) whith a PQ curve very different of all other µchannel blocks ( http://www.cooling-masters.com/image...pdc_blocs1.png ) To my mind, the good way is in this direction : µchannel to increase the exchange surface and decrease the layer (in a 0.3mm channel, the layer can't be more than 0.15 mm ) but without adding restriction like the nexxos or 1a blocks with "stupid" way for water. Last edited by Uncle`BuZZ; 11-17-2004 at 07:44 PM. |
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11-17-2004, 11:17 PM | #177 | ||
Thermophile
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Location: Melbourne, Australia
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Quote:
Radiators do improve in efficiency with increasing flow rates. This fact invalidates the following section of your response. Quote:
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11-18-2004, 03:20 AM | #178 |
Cooling Savant
Join Date: Dec 2003
Location: France
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The pink line on BillA's graph gives you this, if you keep power constant:
btw, would really like to see pHaestus test the 0.3mm proto |
11-18-2004, 04:56 AM | #179 | |
Cooling Savant
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Quote:
Undoubtedly... Why? Muse... ...ignore me. |
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11-18-2004, 05:42 AM | #180 | |
Cooling Neophyte
Join Date: Oct 2004
Location: France
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Quote:
All graphs on the HE 120 series test page are based on a 5°C coolant/air differential. As input is equal to coolant/air differential * specific heat capacity of coolant * mass flow rate when you modify the flow with a constant coolant/air differential you modify the input, so with more flow you introduce more heat in the rad, so the results is evident : rad dissipated more heat, but it doesnt mean than a rad with the same input but with more flow will disspated more heat (this appears to be right as shown by the graph dt vs flow at 71 W) 1 Watt = 1 J/s cp = 4185 J/kg/°C (depends of temperature, but very few changes) lpm/60 gives liters per second 1 liter of water is about 1 kg so lpm/60 * cp * delta T gives kg/s * J/kg/°C * °C = J/s = W On the graph you posted : at 2 lpm, you have 2/60 * 4185 * 5 = 697.5 W in input at 10 lpm, you have 10/60 * 4185 * 5 = 3487.5 W in input (hum, it's amazing, maybe i missed something because they are great values, am I wrong ?) Ok, I got it , I keep my wrong reasoning to show you how I am stupid and to let you understand what I meant. After reading of the article, the delta T isn't between inlet air and inlet coolant but between inlet air and the average temp of coolant resulting of dissipation, so it doesn't mean the input is directly proportional to flow as I believed. But I dont understand your quote : my sentence "We need same graph but with a constant power as a CPU will put the same power in water in a high or a flow rate. So the coolant/air differential will increase when flow decrease..." is true to my mind, and it's exactly the graph posted above. The answer is evident on this graph, not on your where we don't know nothing about input. |
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11-18-2004, 05:49 AM | #181 | |
Cooling Neophyte
Join Date: Oct 2004
Location: France
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Quote:
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11-18-2004, 06:12 AM | #182 | |
Thermophile
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Location: Melbourne, Australia
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Quote:
The graph shows exactly what the input is to sustain the 5C air-water temperature delta at equilibrium. All the information that is on the graph that I presented from BillA's excellent article is enough to formulate a C/W graph for the radiator's performance with respect to flow rate - simply divide 5C by the input power (watts) shown to arrive at the C/W for that flow rate. Multiple the resulting C/W axis by the CPU watts (71W) and you would arrive at the exact same graph as lolito_fr presented above. I don't believe you are stupid. I am glad that you are thinking about it deep enough to get yourself perhaps a little confused, but that's a good thing. A least you're thinking! |
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11-18-2004, 06:31 AM | #183 |
Cooling Savant
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Buzz, not sure who is more confused after reading your post- you or me?
You cannot say that Bills' (not Cathars) graph is wrong and mine is correct - as my graph is just a different interpretation of this one: http://thermal-management-testing.com/Thermo12.gif Saying that BillA or Cathar is incorrect is heresy - you had better have some concrete evidence if you were to make such a claim, lol. edit: too slow. Cathar, thx for explaining the math for me |
11-18-2004, 07:02 AM | #184 | |
Cooling Neophyte
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Quote:
What I don't see is how the average temp of coolant is representative of the input... I don't see the input in the graph. It's the average term who confuses me... I dont see what it represents... the temp on outlet coolant is talking to me, as the delta T between inlet and outlet, but not the average... because you can have the same average with 10 and 4 °C and 8 and 6°C inlet coolant temp and flow give the input, delta T between inlet and oultlet with flow give the power dissipated, but I dont see what represents the average temp... |
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11-18-2004, 10:34 AM | #185 |
Cooling Savant
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Hmmmm. I think I see now...
I think you were saying that it looked as though the increased power at higher flow rates was because the liquid was not cooling down as much, so the average fluid temp was higher? So the increased rad efficiency as flow increases would be clearer if the difference between the inlet and outlet fluid temp was very small (ie negligeable)? Edit: Oops. The fact that the average fluid temp is higher for increased flow rates is one reason why the rad is more efficient. Bleh. So my question now: how does Bill regulate the average fluid temp in the rad? Last edited by lolito_fr; 11-18-2004 at 11:20 AM. |
11-18-2004, 12:26 PM | #186 |
CoolingWorks Tech Guy Formerly "Unregistered"
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avg temp NOT regulated
the 5°C deltaT is air inlet -> coolant inlet the LMTD (Tm) is calculated per the std eq |
11-18-2004, 01:09 PM | #187 |
Cooling Savant
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ok, that makes sense.
It does now make me wonder if we're using your graphs correctly to predict overall system temps, seeing as the liquid side dT across the rad will be adding to the rad dT ?? maybe I'm just looking too close... |
11-18-2004, 02:38 PM | #188 |
CoolingWorks Tech Guy Formerly "Unregistered"
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all my testing prior to Swiftech was the bench testing of components, and my focus was on identifying a procedure that would distinguish the capability differences
the data is adaptable to system's characterization, but may need some adjustment I must admit to not spending any time on systems performance prediction, I just test the components together to arrive at a system C/W N.B. systems C/Ws must be measured at the max heat load when you fellows think that you have a viable prediction scheme -> work out what the number might be for a Swiftech kit; then you can compare to the measured value on the site this should be fun BTW, I have values for both the MCW650 and the other unannounced pump so this pump heat business can be objectively assessed - and I'm not adverse to helping Bruce, I could test the MCP600 if needed |
11-18-2004, 02:44 PM | #189 | |
Cooling Savant
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Quote:
dT(cpu-air) = dT(cpu-water) + W(wb+pump)x"C/W"(rad) - W(wb+pump)/(69.5466x flow(lpm)) Bloody hard work using the beermat method. Edit: Changed + for - sign to account for direction of heat flow. Last edited by Les; 11-18-2004 at 03:46 PM. |
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11-18-2004, 02:56 PM | #190 |
Cooling Savant
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Bill, that sounds like fun indeed. Nice little homework project
Les, thats not what I wanted to hear (the W(wb+pump)/(69.5466x flow(lpm)) bit) I was thinking that the 2 LMTD's (rad & waterblock) "cancelled" each other out. (lets just forget pump heat here for a minute!) I will draw a diagram I must be missing something!! edit: just using simplified LMTDs for the example edit2: replaced "LMTD" by "average". suspect this has something to do with it(?) Last edited by lolito_fr; 11-18-2004 at 03:21 PM. |
11-18-2004, 03:37 PM | #191 |
Cooling Savant
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Thanks for diagram; have changed a "+" for a "- " sign
Last edited by Les; 11-18-2004 at 03:54 PM. |
11-18-2004, 03:47 PM | #192 |
Cooling Savant
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Is W*wb/(69.5466x flow(lpm)) not already accounted for in the dT(cpu-water) ?
Edit: No, because pH gives the dT between CPU and water inlet Damn. so my graphs are wrong. Last edited by lolito_fr; 11-18-2004 at 03:58 PM. |
11-18-2004, 03:59 PM | #193 |
Cooling Savant
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Do not think so.
All Bill's "C/W"s are with respect to inlet temperature. pHaestus's dTs are with respect to inlet temperature. Will have a rethink(easy to get confused), but think I am right Edit: Yes; I think your edit is correct Last edited by Les; 11-18-2004 at 04:05 PM. |
11-18-2004, 04:07 PM | #194 |
Big PlayerMaking Big Money
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I can provide water avgs (or LMTD) if needed.
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11-18-2004, 04:10 PM | #195 |
Cooling Savant
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All Bill's "C/W"s are with respect to inlet temperature.
Now I am confused. "the LMTD (Tm) is calculated per the std eq" Does this not mean the graph is corrected for 5°C LMTD? pH, Thx but I think we can manage as long as we know exactly what is measured/reported |
11-18-2004, 04:17 PM | #196 | |
Cooling Savant
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Quote:
dT(cpu-air) = dT(cpu-water) + W(wb+pump)x"C/W"(rad) - W(wb+pump)/(69.5466x flow(lpm)) gives system dT from the existing data. However. with my continual insistence of complicating, I am not happy with using a constant "wb extracted" heat or pump heat. Still believe "wb extracted heat" changes with "C/W" - something like one of these curves image Last edited by Les; 11-18-2004 at 04:23 PM. |
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11-18-2004, 04:30 PM | #197 |
Cooling Savant
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ugh. I think the pump heat thing we can just about cope with (input power vs flow?) but the cpu secondary losses...
edit: the cpu is not part of the cooling system - the system C/W is independent. Last edited by lolito_fr; 11-19-2004 at 03:45 AM. Reason: meant cpu not wb secondary losses |
11-18-2004, 04:34 PM | #198 |
Cooling Savant
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I am just being awkward
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11-18-2004, 04:39 PM | #199 |
Cooling Neophyte
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the zern PQ+ could be a good target for a futur test to see a µchannel block without useless drop pressure.
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11-19-2004, 06:47 AM | #200 | |
Cooling Savant
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yep. pity that Rosco and pH only have 2 hands
Quote:
the 676 looks like a stretched Miracle Midget Wonder (on this page) the MCR120 looks like a BIX, but thinner? |
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