Please check my math

[crownonfire]

The 980x has a TDP of 130w
Anandtech's 980x setup at 4.13Ghz had a power consumption of 205w

http://www.anandtech.com/show/2961/i...z-with-6-cores

So lets say the heat load is 205 watts.


Lets say we use the 19911-5M31-28CZ
peltier from
http://www.customthermoelectric.com/tecs_Qcmax.html

Specs

Qmax 400watts
dTmax 62 C
Vmax 24.8vv
Imax 28 amps

Lets say we run the peltier at 75% power, 300 watts.

The difference between the hotside and coldside of the peltier will be

Delta T = (1 - (Heat Load/Max Cooling Power)) * Max Temp Difference

Delta T = (1 - (205/300))*62 = 19.6 deg C

The peltier hotside will be

cpu watts+ tec watts = 205+300 = 505 watts

---------------------------------
Lets assume a perfect world for simplicity

Ambient would be 22C
CPU surface would be 22C-19.6 = 0.4C
Then a rad setup would need to handle 505watts


First, are these calcs correct?
Second, I know I said a perfect world, but how far is this math from reality? Assuming I use a cpu peltier waterblock.

[Ultrasonic2]

your maths is wrong.
your making a number of assumptions that are incorrect

[=[PULSAR]=]

the dTmax is 62C that would be pretty hard to keep 500W cooled even with watercooling.

[Serpentarius]

here's a TEC calculator ..... from Ultrasonic2

http://www.overclock.net/peltiers-te...alculator.html

[Ultrasonic2]

The 980x has a TDP of 130w
Anandtech's 980x setup at 4.13Ghz had a power consumption of 205w

http://www.anandtech.com/show/2961/i...z-with-6-cores

So lets say the heat load is 205 watts.


Lets say we use the 19911-5M31-28CZ
peltier from
http://www.customthermoelectric.com/tecs_Qcmax.html

Specs

Qmax 400watts
dTmax 62 C
Vmax 24.8vv
Imax 28 amps

Lets say we run the peltier at 75% power, 300 watts.
( with 75% of vmax applied does not = 75% of Qmax or dTmax)

The difference between the hotside and coldside of the peltier will be

Delta T = (1 - (Heat Load/Max Cooling Power)) * Max Temp Difference

Delta T = (1 - (205/300))*62 = 19.6 deg C

The peltier hotside will be

cpu watts+ tec watts = 205+300 = 505 watts
(nope the 300 you are referring to is Qc not the amount of electricity used to move the 300 watts of heat)

---------------------------------
Lets assume a perfect world for simplicity

Ambient would be 22C
CPU surface would be 22C-19.6 = 0.4C
Then a rad setup would need to handle 505watts
( If the ambient is 22 then water temp might be 25 and the hot side temp of the tec being 30. then the cold side of the TEC would be 30 -19.6 = 10.4 however there is 2 layers of TIM and the heat speared in the way so when a load is applied to it it's thermal resistance needs to be taken into account ( im just making up numbers))

First, are these calcs correct?
Second, I know I said a perfect world, but how far is this math from reality? Assuming I use a cpu peltier waterblock

[zipdogso]

Delta T = (1 - (Heat Load/Max Cooling Power)) * Max Temp Difference

In addition to what has been said before you cannot say 75% power = 75% Qmax (it doesn't.) and then use the max temp difference of the TEC.
The max difference of the TEC is only at max power with zero heat transfer inefficienceies....in the real world...unobtenarium. Judging from the charts for this TEC at 75% power the max Dt is more like 55º but there again it is in ideal conditions not real world ones. (Your thermal transfer inefficiencies are your TIM, Hot/Cold plates, the pressure you apply to the TEC...)
If it worked this formula would only work at full power which nobody does anyway and it is simply a crappy formula, used badly, by people who don't know what they are talking about.
I personally wish I could press a button and instantly destroy all versions of this formula on the net

Second, I know I said a perfect world, but how far is this math from reality? Assuming I use a cpu peltier waterblock.

...it is absolute *%$£^*&"!.

If anyone needs anymore proof - of all the formula touted around by TEC manufacturers you will never find this one it was circulated on the net by the "Old School" TEC users and ballooned from there. Indeed I even looked at it myself when I started looking at TECs 2 years ago but once you understand how TECs work it is obvious that it is a very bad formula especially when used by people who don't understand it.
There are too many variables in a running TEC to simplify it in this manner but to top it all the very core thinking of it is incorrect because the Dt does not vary in a linear fashion.

Just for the record:- looking at the charts for the TEC you linked.... if you powered it at 75% power your Q would be somewhere around 125-175w (Incidently that's only at a hotside temp of 27ºC.) rather short of your 300w estimation It does seem a bit low...but I double checked and was being fair with the Dt. It is unlikely your Dt will be less than 30º at 75% power.
So if the heatload of 205w is the max for standard usage this TEC will probably be Ok for the most part, will probably struggle when the loads go up BUT using it would defy the simple logic of using a TEC with greater power than the max you might encounter.

As Ultrasonic2 said :- You should also note that correct usage of Dt (to find the temp of the coldside of the TEC.) is subtract it from the WORKING temp of the hotside. This can easily be 10-15ºC or more higher than CASE ambient which of course depending on your fan arrangement will be higher than your room ambient. With a theoretical Dt in the range of 30º or more you should go below ambient (seems to be all everyone wants these days.) but you would not have what I would call good cooling - you might get the coldside to 15º and your CPU a few degrees more than that but only with good hotside cooling and A1 case cooling. When your CPU idles and the load on the coldside reduces you will gain Dt in compensation and your coldside temp will fall towards zeroº. But what's the point of idle temps ??? I can't see why people like quoting them - It is bit like saying your car is efficient when parked in your yard with the engine running.

[crownonfire]

Thank you to everyone for the replies!

I have taken time to read thoroughly and understand the mistakes I made. Sorry for the short reply, I currently have a case of dysentery.