Heat load calculation?

Dieses Thema im Forum "Overclocking & Cooling" wurde erstellt von Arctucas, 28. Mai 2009.

  1. cdawall

    cdawall Guest

    no the fan would dissipate 5w of heat still it would just transfer some of the heat to the air.
  2. zAAm

    zAAm Guest

    Once again, if you know the power the component draws you can calculate the amount of BTU's it will dissipate. Just multiply the power in watts with 3.412.

    Uhm, no. Does the heat then make the air move?? *Wtf :wtf:
  3. yogurt_21

    yogurt_21 Guest

    an aquarium chiller seems like overkill being that they're rated for much more volume. personally I took a vaopchill classic and attached it to a water block on my loop. it comes after the radiator in the loop. temps went from 30c idle on the cpu with a straight water loop to 5c idle.

    granting that is just the cpu and I didn't test it for long. (had to dismantle it when mwe moved and havent put it back together)

    but being that you're looking for more extreme cooling an aquarium unit may be just what the doctor ordered. lol

    edit: and theres not way you could even get close to 6000btu heat load form your components. cpu at a high average of 150 = 511btu + (80w) gpu's 273 btu each +chipset (30w) 102 btu + whatever esle you want to cool hard drive (20w) 68 btu. so say a high estimate of 4 gpu's in quad sli or quadfire plus an oced core i7 (150w) plus the chipset north and south and 8 hard disks.

    that monster rig would come to 2331 btu. 1/3 of what that chiller can handle.
  4. cdawall

    cdawall Guest

    actually heat can make air move but thats against the point. the motor producing the mechanical spining still puts out still uses 5w still produces friction etc.
  5. zAAm

    zAAm Guest

    Let's go back to basics then. Conservation of power. If a closed system draws 5W of power, it must dissipate all of it so none is destroyed. Now, a closed system is something that does not have any influence on outside objects and is not influenced by any outside objects. Clearly a fan isn't a closed system since it interacts with the air OUTSIDE. If a fan was a closed system it would blow the heck out of the air inside it but would not disturb the air around it.

    Now, since power is given to the air in terms of speed and momentum, power is subtracted from the system. Once again the total power dissipated must equal the power drawn. Since 5W is drawn, 5W should be dissipated. But since some power is transferred to the air, the power converted to heat MUST be less than 5W. Input (5W) = Heat(?W) + Power given to Air(?W). Since the power given to the air is not zero, the power converted to heat cannot be 5W.

    Essentially you can see it as a system that draws 5W, converts maybe 2W of that energy into heat and the remaining 3W of energy (still in electrical form) is converted to mechanical energy and transferred to the air to make it move.

    Hope this settles it. *Stick Out Tongue Heat load calculation? :p
  6. Bo$$

    Bo$$ Guest

    W1zzard speaks the truth, some of you guys need to read up on some basic physics *Smile Heat load calculation? :) *Toast :toast:

    couldnt have said it better
  7. There is no constant BTU.Not all components even behave the same.. But each component will have a TDP (Thermal design power) that gives you a threshold for stock thermal output. You probably wont find anything for chipset or GPU online, but CPU has it. The rest is probably guesswork.