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TEC/Peltier definitions, formulas, temp estimations

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lehpron
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2009/12/16 15:08:31 (permalink)
** This post contains a lot of math, prepare yourself **
 
TECs (ThermoElectric Chip), a.k.a. Peltiers, are heat pumps.  They are not coolers, they need to be cooled in order to work, otherwise they will burn out (or worse, burn your processor out).  Their ability to cool is dependant on the voltage they run at, the source load being pumped and the hotside's temperature.  If the voltage is too low, most of the specs are lost, we'll get to that in the formula section.  If either the source load or hotside temperature is too high, the unit will not cool, in fact it will begin to heat the source.
 
1.1 Definitions given by specs :
 
Vmax = max voltage the unit will run at
Imax = current in amps drawn at Vmax
DTmax = difference (Kelvin or Celcius, the same number) in temperature between both hot and cold sides when run at Vmax
Qmax = heat from the hotside when the cold side is insultated (high with load)
 
1.2 Definitions to be calculated from specs 
Preq = Imax × Vmax, power needed with running at maximum voltage
Qload = this is what the module will pump,  Preq - Qmax.
 
BTW, in metric, heat and power are both measured in watts, but they aren't the same things, the difference is efficiency, i.e. for TEC/Peltier, the efficency is Qload/Preq.
 
DT value varies per source load linearly, when source load = 0, DT=DTmax, when source load = Qload, DT=0.  When source exceeds Qload, DT is negative (in effect heating the source and not cooling it).  Going by this, TEC/Peltiers are best used if the source never exceeds Qload value.
 
1.3 Formulas: Let's say we run a TEC at a voltage less than Vmax, we'll call it V:
 
current draw, I, at voltage V = Imax × V ÷ Vmax
DT at voltage V = DTmax × (V÷Vmax)2
module heat, Q at voltage V = Qmax × (V÷Vmax)2
Power draw, P, at voltage V = I × V
pump limit, L, at voltage V =   P - Q  << this is what you factor in deciding which module to use in your experiments on CPU, GPU, chipsets or even RAM. 
 
As you can see, is the operating voltage is cut in half, the DT and heat/cooling terms cut into quarters.  This is why it is best to find a module where the Vmax isn't much higher than what you intend to run, so you get most of it back.  If you run at 12v, try to find a 14.4v module, as opposed to a 15.2 or 16.1.  You can also use multiple modules in an array, provided your cooling surface was large too.
 
For example, a 226W Qmax module has specs: Vmax = 15.2v, Imax= 24A, and DT=69K. Going by above formulas, the Qload is 134W if run at 15.2v dc.  So the source cannot be more than this to make use of the module.  But if run at 12v dc, the Qload drops to 86W -- most Intel 130W TDP CPUs will not qualify even at stock speeds.
 
FYI, TDP is a rating classification so that cooling companies target coolers to.  For Intel CPUs, it just means the CPU's stock heat value is between 130W and the next rating down at 95W.  It's a wide enough gap, but TDP isn't the actual value for each CPU. 
 
To estimate temperatures of a load, we require the Thermal Resistence (TR) of the coolers, ambient air in the case/room depending on where the cooling section is, and a source load.  Let's assume a 65W CPU stock, TR of 0.2 Celcius per Watt of heat for a high-end air cooler and 0.1 Celcius per Watt for midrange water LCS, 25C ambient air temp (that which goes into the cooler/rads).  For the preceeding examples, I'll use these TR's for air and water cooling.  Your specific coolers will have different values.
 
For a 226W Qmax module at 12v:
 
V = 12vdc
I = 24 × (12÷15.2) = 18.95A
P = 12 × 18.95 = 227.4W
Q = 226 × (12÷15.2)2 = 140.86W
DT = 69 × (12÷15.2)2 = 43K
L = 227.4 - 140.86 = 86.51W
 
Primary assumptions in following two examples:
  1. A cooler of specific TR is able to take any load.  They cannot in reality.  Air coolers are limited by number of heatpipes and fin surface area, water coolers are limited by their number of rads and which waterblocks or pumps are used.  Both cooling types are dependant on fan CFM ratings.
  2. Ambient air temperatures do not affect coldside going below ambient; they do.  A heatsink just transfers heat from a hotter region to a cooler region, i.e. CPU temps versus ambient.  But should this situation flip around, that ambient is warmer than the CPU, the heatsink's job will flip around too, to maintain a heat equilibrium.  However, if there was a proper seal, then heat isn't trying to get inside, neither is humidity and thus condensation.
The following example 1.4a, you can protect the cooler region with insulation, but example 1.4b you cannot as the nature of the cooling setup will literally keep temperatures near ambient.  Note, ambient isn't always room or case temperatures.  If you wanted, ambient could be exhaust of an AC blowing into a room, or an open winter's night window.
 
1.4a -- example by directly mounting module on the source load, with cooler on TEC/pelter
 
For a 95W load:
 
Qin = load into coldside = 95W
DTin (because DT varies with source load per load limit) = 43 - 43×(95÷86.51) = -4.22 C
Qout = hotside heat plus load = 140.86 + 95 = 235.86W
 
T hot (temperature of hotside with cooler on top) = T ambient + TR × Qout
= 25 + 0.2×235.86 = 72.17  (for air)
= 25 + 0.1×235.86 = 48.59  (for water)
 
T cold (temperature of coldside) = T hot - DTin
= 72.17 - (-4.22) = 76.39 (for air)
= 48.59 - (-4.22) = 52.8 (for water)
 
Note: these are not CPU or core temps, it's just the coldside of the module.  The copper base will be a bit warmer, as such the CPU and cores themselves, all by 1-2 deg.  Everything, including the TIM, absorbs heat along the way; can't remove it all.  Sure a better water cooling setup with a lower TR would bring down those high T hot temps, but it won't change the fact that 95W is too much for a 226W TEC/Peltier run at 12v.
If we used a 130W CPU instead, for our air and water examples, the coldside temps would rise to 100.8 for air, 74 for water.
 
1.4b -- example by placing module between two identical cooling sandwitches, then stacked onto source load
 
For the water LCS, this can be modeled as a "chiller", because you can't put two waterblocks on top of each other, essentially having two separate idential waterloops where they both share the TEC, placed upstream of the CPU block.
 
Again, using a 226W module and 95W load:
 
Qin = load into coldside, but in this case, source load × TR of first cooler
= 95W × 0.2 = 19W
= 95W × 0.1 = 9.5W
 
The reason this is because the first cooler takes some of the load away, the TEC just deals with the rest, that which is first cooler absorbed and raised it's own temps.  This impacts the Qout and DTin greatly.
 
DTin (because DT varies with coldside load per load limit)
= 43 - 43 × (19÷86.51) = 33.56 C (for air)
= 43 - 43 × (9.5÷86.51) = 38.28 C (for water)
 
Qout = hotside heat plus load
= 140.86 + 19 = 159.86W (for air)
= 140.86 + 9.5 = 150.36W (for water)
 
T hot (temperature of hotside with second cooler on top) = T ambient + TR × Qout
= 25 + 0.2 × 159.86 = 56.97 C (for air)
= 25 + 0.1 × 150.36 = 40.03 C (for water)
 
T cold (temperature of coldside) = T hot - DTin
= 56.97 - 33.56 = 23.41 (for air)
= 40.03 - 39.7 = 1.753 (for water)
 
Note: again, these aren't CPU or core temps, it's just the coldside of the module and thus IHS of the CPU (assuming perfect contact and heat transfer).  The copper base will be a bit warmer, as such the CPU and cores themselves, by 1-2 deg.  Everything, including the TIM, absorbs heat along the way. 
 
If we used a 130W CPU instead, for our air and water examples, the coldside temps would drop to 28.9 for air, 3.84 for water.
 
Also note the major differences in temperature using the TEC/Peltier differently.  With the water loop the coldside is below ambient for both loads, it is a chiller afterall.
 
Those are the basics, hope you enjoyed it, if you have questions, don't hesitate.  Not to throw a wrench into the wheel, but a real TEC will vary all it's specs by about 10-15% depending on the hotside's temperature (i.e. Qmax and Vmax will rise, but doesn't drop with lower temps than 25 C on the hotside), so it complicates the math a bit.  I didn't include it in this summary, I'm considering posting another write-up for it as well as a method for calibrating your cooler to figure it's TR value.
Editted for proper math symbols
post edited by lehpron - 2010/01/15 15:49:35

For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

Introduction to Thermoelectric Cooling
#1

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    mikotan
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    Re:TEC/Peltier definitions, formulas, temp estimations 2009/12/16 20:36:27 (permalink)
    re-sticky =P
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    overclocker333
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    Re:TEC/Peltier definitions, formulas, temp estimations 2009/12/16 21:29:03 (permalink)
    Ya got to love them... I have a 12 tec/5 waterblock tec chiller... I'm looking for some nice 40mm high Qmax tecs to stick in there as right now I have a mixture of 12706's and 12709's.... just make sure you have plenty of power to run these babies as you sill need a lot.   I have a PC Power & Cooling 1200 watt supplemented with a 650 watt Thermaltake graphics card psu and still wish I had more amperage!!!
     
    Thanks... and I vote Blue Ribbon anyway for the repost!

    5.2ghz highest recorded W3570 on chilled water in the world 
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    Another Cheap AC Chiller Project  
     Get It Colder II 
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    feathers632
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    Re:TEC/Peltier definitions, formulas, temp estimations 2009/12/31 10:43:40 (permalink)
    I used to do the peltier thing back in the P4 Northwood days.  I chilled my northwood below zero with peltier on cpu, cooled by water.

    I quit that after a few years because of the rising electricity bill.  A PC that consumes 600 watts+ is not a good idea for me.

    However, I do miss my peltier days and since I've recently upgraded to i7 860 on EVGA FTW 657, I'm wondering how it would be to use peltiers again.  The thing is, what I have this time is an extreme water system that chills the coolant via outside air.  Back in the old days my water rad used to dump waste heat back into the room and recycle it endlessly like most other noob water setups.  Having the water rad detached from the PC and cooled by outside air means my coolant can reach 8c or less.  So I'm thinking that would make a considerable difference to the hot-side of a peltier!  I'm thinking I could perhaps use a lower powered peltier (i.e. 75w) on CPU and cooled by my extreme water system?

    Then again I have tried building peltier water chillers in the past with varying degrees of success.  I know very well that the only really effective way to cool a peltier (especially above a certain power) is using water.

    So I have had this idea of a dual loop water system...

    The water chiller is made by sandwiching 6 or 8 peltiers between 2 hard disk waterblocks.  One waterblock is connected to the cold side of the peltiers and the other block cools the hot side.  The hot side hard disk waterblock will be cooled by my extreme water loop.  The cold side peltier hard disk block will feed peltier chilled coolant to the CPU waterblock.

    In other words, the hot side water loop is a full water loop whereas the cold side loop has no radiator, just the peltier chilling and a pump feeding cpu waterblock.  The peltiers will be  6 - 8 100w all fed with low voltage (perhaps 3 - 4 volts).

    ?????



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    #4
    lehpron
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    Re:TEC/Peltier definitions, formulas, temp estimations 2009/12/31 19:33:51 (permalink)
    feathers632
    I know very well that the only really effective way to cool a peltier (especially above a certain power) is using water.
    Not really, the key is getting a low thermal resistence on the hotside's cooling, this can be done easiest to impliment with water cooling, that's all.  With air cooling it can be done, but at the price of a loud fan pushing a large volume of air to compensate for heat removal.  It can be done with passive heatsink as well, given the TEC and source load properties, though the total surface area would have to be vast to compensate for lack of circulation (there is always ambient air flow).
    feathers632
    The water chiller is made by sandwiching 6 or 8 peltiers between 2 hard disk waterblocks.  One waterblock is connected to the cold side of the peltiers and the other block cools the hot side.  The hot side hard disk waterblock will be cooled by my extreme water loop.  The cold side peltier hard disk block will feed peltier chilled coolant to the CPU waterblock.   In other words, the hot side water loop is a full water loop whereas the cold side loop has no radiator, just the peltier chilling and a pump feeding cpu waterblock.  The peltiers will be  6 - 8 100w all fed with low voltage (perhaps 3 - 4 volts).  ?????
    To be honest, you would be blowing your money and effort into the wind.   TEC are useful only if the operating voltage is close enough to Vmax (per equations above) as it varies the DT, QLoad an Qmax exponentially. 
     
    For example, most high wattage modules have a Vmax=15v, so if you had a 100W unit run instead at 4v, you drop the Qmax to 100*(4/15)^2 = 7.1W -- this is the hotside waste heat from the module at 4v -- a 93% loss just for running it at a lesser voltage.  Qload -- being what you want taken from the CPU -- is about half this value, at 3W.  You would need a 5x5 array just to take care of your CPU at stock speeds, let alone overclocking.  It's better off you run your array at a higher voltage, with less modules, and stomach the powerbill.  At that point, I'm sure hardd rive blocks aren't meant or can efficiently remove more than 75W of heat, so just get a couple CPU blocks TEC sandwiches and arrainge them in a chainlink fashion.
     
    As I said earlier, watercooling definitely makes running a TEC easier considering the infrastructure and readily adaptable parts, but it isn't the only way.  You can use two aircoolers with a low wattage TEC sandwiched between, and it would have the same effect as a water chiller, just less expensive but with the cost of noise due to the needed airflow.  I've come up with a proposal a while back, but funding has been my Achilles' heel, so I can't show anyone yet.  The math is sound, I have no doubts of it working.  It shouldn't cost more than $200 (less if you have the fans) and can support loads of up to 220W at 25*C above ambient.  If you're (or anyone is) interested in the idea and a parts list, PM me.
    post edited by lehpron - 2009/12/31 19:37:54

    For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

    Introduction to Thermoelectric Cooling
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    feathers632
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/01/01 04:07:52 (permalink)
    Thanks for reply, Lephron.

    Fan noise is something I want to avoid which is why water cooling is the way to go for me. 

    Have you read this?

    http://watercooling.co.nz/node/57

    What is your opinion?




    i7 860 @ 4.3ghz
    FTW 657 A59 Bios.
    8gb GSKILL Ripjaw 1600 mhz 7 7 7 26 88 @ 1.65v.
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    CPU VTT 1.425
    DRAM 1.65
    Bclock 205
    QPI 4.2
    MCH Strap 1867 (this is critical at speeds past 4ghz must be no lower than 1867)
    CPU PWM Freq = 1019
    VTT PWM Freq = 634
    DDR PWM Freq =634
    CPU vcore 1.42
    Vdroop = Disabled.
    Manual memory mode (no XMP)
    PCIe 104mhz.
    Hyperthreading and all speedstep and turbo mode DISABLED!
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    lehpron
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/01/01 20:09:22 (permalink)
    That site pretty much sums up what I've talked about.  However, in place of exposing the actual math involved, there is a program that does it for you.  I'm more about teaching to make absolutely sure you know what is what, hence why I posted those forumulas; programs don't teach expect how to plug'n chug.  The old saying about "feeding a man fish feeds him for a day...", I just don't like giving out fish.
    post edited by lehpron - 2010/01/01 20:15:40

    For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

    Introduction to Thermoelectric Cooling
    #7
    nateman_doo
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/01/15 06:25:32 (permalink)
    VERY true saying.  I would rather a cheeseburger then fish though!  

    [sidetrack]
    Reminds me of a time I pulled over to help stranded motorists.  Two girls were pulled over at a rest stop on the Garden State Parkway, and i was just coming off duty.  I saw them both on their cell phones staring at a flat tire.  I decided the nice thing to do would be to help them.  Also I figured since I was in uniform I wouldn't get sprayed in the face with pepper spray.  Well of the 2 girls jibbering away on their cell phones the actual owner of the car would not get off the phone.  I was trying to show her HOW to change the tire in case she had it happen again, and she would be on her own-in an area with no reception.  Her friend put the phone down and watched the processess and I guided her as best I could, so all wasn't lost. 
    [/sidetrack]


    OK, back to post...

    Here is a suggestion: Special Alt Characters

    makes your formula's easier to read. 

    Makes this:

    For a 226W Qmax module at 12v:
     
    V = 12vdc
    I = 24 * (12/15.2) = 18.95A
    P = 12 * 18.95 = 227.4W
    Q = 226 * (12/15.2)^2 = 140.86W
    DT = 69 * (12/15.2)^2 = 43K
    L = 227.4-140.86 = 86.51W


    Look like this:

    For a 226W Qmax module at 12v:
     
    V =  12vdc                      
    I =   24 × (12÷15.2) = 18.95A
    P =  12 × 18.95 = 227.4W
    Q =  226 × (12÷15.2)² = 140.86W
    DT = 69 × (12÷15.2)² = 43K
    L =   227.4 – 140.86 = 86.51W

    Dummies like me fair much better with the math formula's looking like that     (← thats me with my dunce cap on)
    post edited by nateman_doo - 2010/01/15 06:38:14
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    lehpron
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/01/19 11:57:38 (permalink)
    Fixed for math characaters, and also recieved parts for a TEC-based cooling setup I'll share with everyone once I get it up and running.  Havn't decided if I want to post in here or elsewhere and just refer to this one for background info, latter might be better as the begining wall-of-text can be intimidating.

    For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

    Introduction to Thermoelectric Cooling
    #9
    nateman_doo
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/01/20 08:23:39 (permalink)
    lehpron

    ** This post contains a lot of math, prepare yourself **
     
    Editted for proper math symbols


    OMG!! HILARIOUS!!  a disclaimer!   for dummies like meeeee
    #10
    Halo_003
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    Re:TEC/Peltier definitions, formulas, temp estimations 2010/04/27 18:45:20 (permalink)
    Btw, my head almost exploded reading this thread.

    Great job!

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    humbleBe
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    Re:TEC/Peltier definitions, formulas, temp estimations 2012/04/17 00:26:25 (permalink)
    Hi
    My first post, I hope this quick question is in the right place.
    I have been reading up on peltiers and I have what might seem like a silly question or two just out of curiosity.
    1. What happens if you connect two peltiers directly together with both their cold sides facing each other?
    - Please assume that the current supplied and the heat dispersal is optimal and exactly the same for both and that the pelts are insulated against condensation.
    2. If you then put a copper water block between the two peltiers how thick would it need to be to achieve optimal cooling efficiency?
    i.e. Is the water block cooled more or do the pelts work against each other.
    Thanks
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    JaskarnSidhu
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    Re:TEC/Peltier definitions, formulas, temp estimations 2012/04/17 05:44:47 (permalink)
    Sandwiching a waterblock between the cold side of two TEC's will yield cooler/colder temps. You will be fine.
     
    As for thickness, i would say have the cold plates that touch each TEC at least 1/4" thick. So thats 1/2" in total for the cold plates, plus however thick you would have the water chambers/channels in your water block.


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    lehpron
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    Re:TEC/Peltier definitions, formulas, temp estimations 2012/04/17 07:43:49 (permalink)
    JaskarnSidhu has the right idea, I would only get a bit technical.
    humbleBe
    1. What happens if you connect two peltiers directly together with both their cold sides facing each other? - Please assume that the current supplied and the heat dispersal is optimal and exactly the same for both and that the pelts are insulated against condensation.
    If perfectly insulated, both coldsides would have the same cold temperature.  They are trying to pump heat out of each other and since there wouldn't be a separate coldplate, the ceramic plates connected would get cold.
     
    humbleBe
    2. If you then put a copper water block between the two peltiers how thick would it need to be to achieve optimal cooling efficiency?i.e. Is the water block cooled more or do the pelts work against each other.
    It isn't about thickness. per se, it is about how much surface area the circulating fluid has contact with; you can have more contact area (grooves, channels, fins) while keeping the water block low-profile thin. 
     
    Debuted back in fall of 2003 with the height of popularity in early 2004 and now discontinued, Swiftech made a block sandwiched between two 226W TEC's with waterblocks cooling on both sides called "MCW-CHILL 452"-- it was meant as an inline miniture chiller.  Swiftech used the baseplates of their CPU-waterblocks of the time to increase surface area.  Total of both bases would be approximately a 1/2" for reference.  At the time Swiftech charged $300 for what was basicaly an insulated stack of two TEC's and four CPU block bases.
     
    The same logic from heat removal for a waterblock also applies in the other direction for heat pumping, so it isn't just a matter of how thick it should be between TEC's.  I think it is simpler and easier to grab two modern CPU-waterblocks and place a TEC in between, and insulate the stack.  I plan to do that for another build in about 8-12 months. 

    For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

    Introduction to Thermoelectric Cooling
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    humbleBe
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    Re:TEC/Peltier definitions, formulas, temp estimations 2012/04/17 23:13:59 (permalink)
    Hi
    Thanks so much for the replies.
    I was curious if the two pelts were directly connected if one would pull the heat from the other ones hot side and overload or fail somehow.
    How far apart they should be - seems 1" is good.
    I found the link to the MCW quite interesting.
    One more general question:
    If you are feeding -10 celcius fluid to a modern intel CPU (eg: stock i7-2600k under load) at a typical 1 GPM what is the temperature of the water exiting the block?
     - is it still near ambient or is it much hotter requiring a radiator before going back to the pelt for cooling?
    Thanks
    post edited by humbleBe - 2012/04/18 18:17:14
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    lehpron
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    Re:TEC/Peltier definitions, formulas, temp estimations 2012/04/20 17:00:01 (permalink)
    humbleBe
    How far apart they should be - seems 1" is good.
    In a perfect world, all the heat transfers from one liquid to another as it passes through the heat pumps, meaning it isn't absorbing into anything in between, blocks included.  Alas, in real life heat is absorbed and not all transfers; that is why just about any heatsink from air to water maintains a temperature above ambient-- that is the equilibrium stored inthe heatsink itself.  For a TEC-based heat exchanger, your goal should be to have as little copper in between as possible.  Having one inch is bigger than the Swiftech MCW chiller.
     
    humbleBe
    One more general question:
    If you are feeding -10 celcius fluid to a modern intel CPU (eg: stock i7-2600k under load) at a typical 1 GPM what is the temperature of the water exiting the block? - is it still near ambient or is it much hotter requiring a radiator before going back to the pelt for cooling?
    We would need the specific heat of the coolant since all fluids retain some level of energy per temperature.  I would have used water since those tables are widely available online, but at -10 Celcius, it would have to be under pressure to keep it cold without solidifying. 
     
    For now, I'll use water anyway just so you see the effect CPU heat has and let's make the inlet temperature +10 Celcius.  The mathematical formula is know as the Heat Equation: 
     
    Qin = m-dot CP (T2-T1)

    where heat transfered (Qin) equals mass flow rate (m-dot) multiplied by the difference in temperatures (T2-T1) multipled by the specific heat at a constant pressure (CP).  For water CP = 4.184 kJ/kg K, for a coolant known as Ethylene Glycol commonly used in automobile anti-freeze, it is about half that of water.
     
    A stock 2600K uses approximately 93W of electricity, or about 80W TDP actual or 80 Joules per second of energy dissipated, this is the Qin.  A fluid flow of 1 gallon per minute converted to metric becomes about 0.06 kilograms per second.  Most waterblocks still absorb heat as mention above, the really nice ones absorb 5%, so 95% of the heat actually transfers into the fluid.  In this case, 76 joules per second at full load stock goes in the water.
     
    Doing the math and making sure the units are the same, you get a difference in temperature of 0.3 degrees celcius, twice that if using the Ethylene glycol coolant.  If you speed up the fluid flow, temperature difference cuts in half; if you double the load, you double the difference.
     
    Note: Liquid is just a medium and even radiators don't dump all the heat while they absorb some for themselves, your fluid temperature will slowly rise until it hits an equilibrium depending on the rad+fan+resevoir+insulation.  Not an easy calculation if you take all variables into account, but it gives you an idea of a simple case of fluid through a waterblock.
    post edited by lehpron - 2012/04/20 17:12:25

    For Intel processors, 0.122 x TDP = Continuous Amps at 12v [source].  

    Introduction to Thermoelectric Cooling
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