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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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07-09-2004, 07:57 AM | #1 |
Cooling Neophyte
Join Date: Apr 2004
Location: Warwickshire, england
Posts: 10
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Can someone explain this to me?
I heard from many forums that 2 X single heatercore in parallel > heatercore dual
However, I never got a detailed explanation about this. They all say it works better but I've got many different explainations. Some said it's because the water is slower, it stay in the radiator longer and therefore cools better. Some say it gets a better flow rate? I'm really confused, someone help me out please. |
07-09-2004, 01:23 PM | #2 |
Cooling Neophyte
Join Date: Jun 2003
Location: denmark
Posts: 73
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better flowrate
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07-09-2004, 01:43 PM | #3 | |
CoolingWorks Tech Guy Formerly "Unregistered"
Join Date: Dec 2000
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Quote:
NOT predictable if the rads really have low flow resistance (fitting loss can be greater) ah yes, more testing but when the changed flow rate is factored into the C/W the effect either way is very small, 0.004°C/W gain using our MCR80s in parallel the biggest challenge with rads is finding the space to mount them, don't sweat their connection |
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07-09-2004, 07:14 PM | #4 |
Cooling Neophyte
Join Date: Apr 2004
Location: Warwickshire, england
Posts: 10
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What do you think?
That's how I'm doing it in my radbox. |
07-09-2004, 07:17 PM | #5 |
CoolingWorks Tech Guy Formerly "Unregistered"
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looks ok by me
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07-09-2004, 07:24 PM | #6 |
Cooling Savant
Join Date: Jan 2004
Location: Alberta
Posts: 631
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I think 2X singles would be better than 1X dual since the area is much larger... A double is what, 10*6 when a single is 6*6?
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07-09-2004, 07:34 PM | #7 |
Cooling Savant
Join Date: Jun 2004
Location: Florida
Posts: 414
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um
I hope that second rad isn't pulling air from the first |
07-09-2004, 07:38 PM | #8 | |
Cooling Neophyte
Join Date: Apr 2004
Location: Warwickshire, england
Posts: 10
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Quote:
Unfortunately, that's something that I can't avoid happening. I do have an opened side window on the radbox though so it shouldn't be too bad. Any ideas though? |
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07-09-2004, 07:47 PM | #9 |
Cooling Savant
Join Date: Jan 2004
Location: Alberta
Posts: 631
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Why? What's the big issue with that? The only explanation I've seen is that the air is warmer, but with 2240LPM of air, the increase would be marginal if measurable...
It also seems to me that that would increase the airflow through both radiators by a certain amount as there is twice the pressure being offered, although there is twice the back pressure... Nevermind that wouldn't make a difference... |
07-09-2004, 09:01 PM | #10 | |
Cooling Savant
Join Date: Jun 2004
Location: Florida
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are you joking look into delta t, and how important it is to cooling if rad inlet air is 3C warmer, CPU die WILL be 3C warmer how exactly did you calculate 2240LPM? what value(s) did you use as the static backpressure for the radiator? please note a fan's airflow is rated under no restriction conditions, at original poster make a duct to side window. |
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07-09-2004, 09:58 PM | #11 |
CoolingWorks Tech Guy Formerly "Unregistered"
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need to be in series if plumbed that way, warm air one first
should be ok I never looked at the fans, bleh |
07-09-2004, 10:26 PM | #12 |
Cooling Savant
Join Date: Jan 2004
Location: orlando FL
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Turn the one on the left around so it blows out of the case, cut a hole in the center so they both pull air from the side or top of the case. That way they both have fresh air.
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07-09-2004, 11:03 PM | #13 |
Cooling Savant
Join Date: Jan 2004
Location: Alberta
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I assumed 80CFM but that's horribly wrong. It'll be around half that.
Now, 40CFM is 1.13267386 m^3/minute or 1132.7lpm, and the heat of a system is around 12000J/minute The heat capacity of air is 1.005kJ/kgC and the density is 1.29kg/1000L so the mass of the momentary flow is 1.46kg. 1.46*1.005 = 1.47kJ for each degree of temperature rise. 12000/1470 = 8.2C Umm... That's a useless number. Ok, let's try it again. The volume of a heater core is 1.18L and the air is recycled 960 times per minute. 8.2C/960 times = a temperature rise of 0.00854C across the radiator. The second radiator is 0.0085C warmer. Did I do that wrong or does anyone have any theory or testing showing otherwise? I think series is fine if that's correct. |
07-10-2004, 05:27 AM | #14 |
Cooling Neophyte
Join Date: Apr 2004
Location: Warwickshire, england
Posts: 10
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So, should I change the rads to run in series?
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07-10-2004, 02:32 PM | #15 |
Cooling Savant
Join Date: Nov 2003
Location: Suffolk, UK
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Like Bill said at the start of the thread. It makes very little differance.
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07-11-2004, 02:58 PM | #16 | |
Cooling Neophyte
Join Date: Apr 2004
Location: Warwickshire, england
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From ocuk forums:
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Should be better now. |
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07-11-2004, 08:32 PM | #17 |
Thermophile
Join Date: Sep 2002
Location: Melbourne, Australia
Posts: 2,538
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Air-flow series hypothetical scenario:
Using BillA's ThermoChill radiator review here, and Panaflo fan specs here. Using AA's figures above: The heat capacity of air is 1005 J/kgC and the density is 1.29 kg/m³. So 1m³/min (~35CFM) of air-flow has a thermal resistance of 21.6W/C (or 0.0463C/W) Let's use the ThermoChill 120.1 performance to crunch some math, and let's use a 5C water-air inlet differential into the first radiator. Just rough eye-balling of data curves. Let's consider the Panaflo L1A 12cm fan (~69CFM rated, 30dBA, 38mm thick fan) For a single 120.1, the L1A would push around an estimated 0.75m³/min through the core. The radiator would have an estimated C/W of around 0.07. If we now stuck two of the rads in air-flow series with the same fan, we'd see around about 0.5m³/min of air-flow through the rads. At a 5C water-air differential, the first radiator would have a C/W of around 0.10. Now to work out the efficiency of the second radiator, we'll assume that our system is in equilibrium with a 5C air-water delta into the first rad. This means that the first radiator is dissipating 50W of heat into the air flowing through it. At 0.5m³/min, the air is therefore warming up by 50 / (21.6 * 0.5) =~ 4.6C as it flows through the first radiator. We'll ignore that the coolant will have cooled off somewhat after this as it flows into the second radiator. So the second radiator now has a water-air differential of just 0.4C, or at a C/W of 0.10 with a 0.5m³/min air-flow rate, is providing just 4W of additional cooling effect after the air got warmed by the first radiator. So overall with a 5C initial differential, the system is providing ~54W of cooling potential, or a total system C/W of around 0.09, which is definitely worse than just having a single radiator by itself. Alternately, if we had the radiators in air-flow parallel, the same fan could push around 1.25m³/min through the pair of radiators, and offer a C/W of around 0.05 or so. So with the L1A: Single 12cm radiator: C/W =~ 0.07 Dual 12cm radiators in air-flow series: C/W =~ 0.09 Dual 12cm radiators in air-flow parallel: C/W =~ 0.05 The issue here being that at very low air-flow rates, these radiators are above 90% efficient in terms of water-air thermal transfer, and so the addition of a second radiator just chokes the air-flow, while being fed with air that has closely approached the coolant temperature. If you can push the air-flow rates up high enough where the radiator's thermal transfer efficiency drops away, then the second radiator starts to become more effective, but by that stage you do need fairly powerful and high pressure fans to push enough air-flow through the radiators for that to happen. |
07-11-2004, 09:57 PM | #18 | |
Cooling Savant
Join Date: Jan 2004
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07-11-2004, 10:21 PM | #19 | |
Thermophile
Join Date: Sep 2002
Location: Melbourne, Australia
Posts: 2,538
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So given two fans, the example would then be saying that if you had a fan mounted on each side of a single radiator, and then compared that to two fans pulling though two radiators in air-flow series, then the two fans on the single radiator would still offer better performance than the two fans on the two radiators in air-flow series. Munching through the math, you need some very strong fannage in air-flow series to get the efficiency of the first radiator (in comparison to the thermal capacity of the air) low enough such that the second radiator starts to do more than next to nothing. Coupled with the drop in air-flow rates due to the in-series operation, for most radiator setups, we're talking ear-deafening powerful fans to get to the level where two radiators in air-flow series are working better than a single radiator. The greater the radiator thickness that the air has to get through, the stronger the air-pressure needs to be to make it worth-while, and air-pressure is not what axial fans are good at. After spending some time looking at heater-cores and radiators, I started to understand exactly why most heater-cores are two-row only and only as thick as they are. It truly is about the best trade-off in terms of pressure-drop vs performance against the sorts of air-pressures that the fans used on them operate at, and that even included heater-core blowers. If he put the radiators in air-flow parallel, with one fan sucking on each radiator, in isolation to the other, that'd be the "ideal" way that he could set that system up. Last edited by Cathar; 07-11-2004 at 10:29 PM. |
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07-11-2004, 11:22 PM | #20 |
Cooling Neophyte
Join Date: May 2004
Location: Little Rock, Arkansas, USA
Posts: 27
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Dammit.
I was working on a front and top mounted radiator system. This prooves that three months of research may not even be enough. Thank you Cathar.... I guess... for prooving with math how my setup will suck with two radiators. Now I am left with two choices: 1) scrap the front radiator or 2) make a shroud where the air from the front radiator is exhausted out somewhere and turn the back fans into intakes instead of exhaust. . . . . . . I'll tell you why watercooling as an industry is being hurt--and it's not because of cheap low quality kits--it's because 90% of what you read about it is total BS. The low quality kits are secondary to this IMHO. See, right now, I want to pull my f'n hair out. My air trap solution turned out to not be so hot, my radiator selection and placement turned out to look nice, but not much else. My choice to add a second radiator to compensate turned out to be kinda dumb given that they are in air flow series instead of air flow parallel. Had I simply had access to a list of facts (FACTS! with math to back them up!) three months ago, then I might have done a better job of things. As it is now, I feel disheartened and pissed off. It's not because I didn't research things, it's because it's taken me over three months to sift out the horseshit and get to the truth of things. THAT's why WCing can't be mainstream (that and the maintenance aspects). At least not now. And I still have a thousand questions, for instance--Cathar preferably: My top mounted radiator is going to be exhausting air that is already in the case. Air that is going to be heated by a processor that likely puts out around 89 watts max as well as two 160gb samsung spinpoint hard drives. The fan is a Panaflow 12cm--the one you were talking about earlier. The radiator is a crappy single BIX. Given this, is there a way you could just guesstimate how much I am going to be hurt by not mounting the crappy BIX where outside air flows through it instead? I don't know, maybe this should be a new thread--although the "Can you please explain to me" title is kind of a catch all. These are my parts: 1/2" ID, MCP600, TDX (not lucite) or Cathar's new one, Danger Den GPU WB maybe, Single Bix Top mounted, Tygon. What should I change? You can get more details in this thread. . . . dammit. Bruce |
07-12-2004, 12:25 AM | #21 | |
Cooling Neophyte
Join Date: Nov 2003
Location: Hong Kong
Posts: 81
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07-12-2004, 12:55 AM | #22 | |
Thermophile
Join Date: Sep 2002
Location: Melbourne, Australia
Posts: 2,538
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To me, the moment some statement is made without some mathematics to provably back it up, then it is really just an opinion, not a given fact. Problem is, even if someone did go to all the trouble to lay it all out and "give the facts", people will still make the wrong decision for a few reasons: 1) They don't know who to trust. What with many sites just regurgitating garbage that originated in deep dark history from some guy who contemplated his navel instead of mathematically figuring something out, and so gave birth to the internet equivalent of a "wive's tale", most people just go with the majority voice, rather than the more difficult to understand hard-line factual/experimentally derived analysis. Most people's eyes do just glaze over the moment a ² symbol appears in some text, and for some people it doesn't even take that much. You can only dumb down and generalise some concepts so much before they start to lose meaning and become just as bad as the more general "untruths" out there, and in some ways worse, because people see some simplified maths, forgot to read the caveats that were applied to get it that simple, and all of a sudden some specific scenario's proof becomes "gospel" for a far wider range of scenarios that the original statement was ever intended to cover. 2) No one is right all the time. Especially not me. I'm so embarrassed about various statements I've made in the past. I'd hate to be origin of some of what I talk about above, because I, too, am learning. It's also why I'm very hesitant to disclose any mathematical theory I have about block design. I know that it'd most likely get shot down by the true theorists (something which I am not), but more dangerously may mislead people. The more I learn, the more that I feel that I don't know anything. I'll answer you other questions in a bit - forgive my soap-box speech - I know I do that a bit too. |
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07-12-2004, 01:02 AM | #23 | ||
Thermophile
Join Date: Sep 2002
Location: Melbourne, Australia
Posts: 2,538
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1m³/min of air-flow is therefore 1.29kg of air Thermal capacity of air is 1005 J/kgC Therefore, thermal capacity of 1m³ of air is 1.29 x 1005 =~ 1296 J/C 1 Watt = 1 Joule per Second (Units: J/s) Need to convert flow rate into "per second" units => 1 m³ per minute is 0.01667 m³ per second Therefore, thermal capacity of 1m³/min of air = 0.01667 x 1296 = 21.6 J/(C*s) or 21.6 W/C |
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07-12-2004, 01:27 AM | #24 | ||
Thermophile
Join Date: Sep 2002
Location: Melbourne, Australia
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Last edited by Cathar; 07-12-2004 at 01:44 AM. |
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07-12-2004, 01:35 AM | #25 | |
Cooling Savant
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