# Heat load calculation?



## Arctucas (May 28, 2009)

How would I determine the heat load (in BTU) of various PC components; e.g. CPU, GPU, etc.?


----------



## zAAm (May 28, 2009)

I'm guessing you need this to determine the size of an aircon?

Firstly, find out how much power each component draws. Then if this component does not convert that energy into something like mechanical energy (i.e. hard drives and fans), then the heat energy is the power draw. So if you have a CPU with a 100W power draw it will output 100W of heat (give or take). The same can't be said for hard drives and fans since most of their energy is converted into mechanical energy. Thus a harddrive with a 20W power draw can dissipate maybe 10W of heat and a 5W fan can dissipate maybe 2W or so.

Then you just sum all the power values (I think you can use 50% of hard drive and fan ratings for heat) to get a total power draw.
When you have this value (in watts), you can convert it to BTU's by the following calculation:

*1 Watt = 3.412 BTU's/hour.*

So if you keep a system with a 300W power draw (would probably be about 280W of heat or so) online for an hour it would generate 280 x 3.412 = 955 BTU/hour worth of heat.

BTW, it is weird to convert from a unit of power (Watt) to a unit of energy (BTU)...


----------



## Arctucas (May 28, 2009)

zAAm said:


> I'm guessing you need this to determine the size of an aircon?
> 
> Firstly, find out how much power each component draws. Then if this component does not convert that energy into something like mechanical energy (i.e. hard drives and fans), then the heat energy is the power draw. So if you have a CPU with a 100W power draw it will output 100W of heat (give or take). The same can't be said for hard drives and fans since most of their energy is converted into mechanical energy. Thus a harddrive with a 20W power draw can dissipate maybe 10W of heat and a 5W fan can dissipate maybe 2W or so.
> 
> ...



Actually, I was considering a chilled water cooling system, and wanted to know how to size it considering the various components of my system.


----------



## Dippyskoodlez (May 28, 2009)

Arctucas said:


> Actually, I was considering a chilled water cooling system, and wanted to know how to size it considering the various components of my system.



The efficiency of the A/C/Compressor would change I imagine(The cooling capacity, anyways), if you used a custom condensor/evap, and used it to chill a reservoir or such.

A normal A/C compressor should be enough to chill an average system. (I think 2500/5000BTU?)

You could simply submerge the condenser in the resevoir, or if you're able to re-gas, and braze copper, you can create an evap. custom designed for your res.

Keep in mind a compressor will run 24/7(It has a constant heat load), whereas in a normal air conditioner, it switches off and on, so overall power consumption goes much much higher.

Regardless, chilled, a system should have enough power to bring components to negative temperatures, or else its simply not worth the hassle (Electricity bill/Temperatures for components..) I don't think you can "design a system for a heatload unless you're talking a multiphase cascade for Large heatloads (stage cooling).

A single compressor will depend the most on your cooling method- Direct Die, or Chilled water (How well designed your water setup is, and how you handle the sub zero temperatures will change everything, down to the pumps and waterblocks being used.) 

Its a whole new realm of cooling. And mighty cool


----------



## Arctucas (May 28, 2009)

Dippyskoodlez said:


> The efficiency of the A/C/Compressor would change I imagine(The cooling capacity, anyways), if you used a custom condensor/evap, and used it to chill a reservoir or such.
> 
> A normal A/C compressor should be enough to chill an average system. (I think 2500/5000BTU?)
> 
> ...



Thanks, I am evaluating which route to take; buy a pre-built aquarium chiller, or build one using an air conditioner compressor and plate heat exchanger.

I am not wanting sub-zero temperatures, just to maintain 20°-30° below ambient.

My question is how to size the chiller; basically how to determine the heat load that will be dumped into the water loop, so I am able to correctly size the chiller in BTUs.


----------



## Dippyskoodlez (May 28, 2009)

Arctucas said:


> Thanks, I am evaluating which route to take; buy a pre-built aquarium chiller, or build one using an air conditioner compressor and plate heat exchanger.
> 
> I am not wanting sub-zero temperatures, just to maintain 20°-30° below ambient.



I'm not sure how heavy duty aquarium chillers are, a common route is to just use an air conditioner... they're usually pretty cheap (I picked one up at wally-world for like $54).

Building an apparatus to dip the cold end of the A/C while still pushing the heat outside is a painful task sometimes, but will probably give you some extremely nice temperatures.

Taking the thermostat off, gives some really fun frost on it 

Just make sure you have a drill, dremel, some wood, and a suitable container for a resevoir, and a way to seal it all off.

After building the resevoir, you probably want to let it run for a bit, come up with how you're going to handle the thermostat (dipping it in might work...), but you will probably still need to insulate the tubing, because lower than ambient temperatures will still probably condense water/ice on it.

Just get creative!

Xtremesystems.org forums might be a slightly better resource for technical questions regarding putting a condenser submerged in a resevoir, and proper coolant mixtures to prevent a frozen loop!



> My question is how to size the chiller; basically how to determine the heat load that will be dumped into the water loop, so I am able to correctly size the chiller in BTUs.



If you want just below ambient temperatures, it's not really worth the hassle TBH. Use a Peltier. 

Using a "Chiller" you dont really need to calculate the load, as long as you're chilling the water... controlling how much/how fast the resevoir of coolant is chilled would simply be determined via thermostat... Most people would probably just disable it, because people generally want it as cool as possible.

And since you should be insulating it anyways, why not go "REALLY COLD"? 





For the technical "Heat load", you could use the TDP of the specific devices you are cooling from tech docs- Most CPU's "Stock" will give you about 75-110w of heat. If you're cooling the NB, and GPU, that will change things.

Overclocked, different voltages and different CPU's will behave differently, and there isn't really a good way to give thermal output numbers (Trial/error perhaps?).. It's also not consistent. My a64 3700+ is a Toledo, but 90% of all other 3700+'s are San Diego CPU's. Mine runs exceptionally cool, because it has 2 CPU's on the package.

Also, load will effect output- Idle a CPU being clocked down is like 10w of power, even less of heat. Under full load it can sky rocket, and most computers aren't under 100% load 24/7/365.


----------



## Arctucas (May 28, 2009)

Dippyskoodlez said:


> I'm not sure how heavy duty aquarium chillers are, a common route is to just use an air conditioner... they're usually pretty cheap (I picked one up at wally-world for like $54).
> 
> Building an apparatus to dip the cold end of the A/C while still pushing the heat outside is a painful task sometimes, but will probably give you some extremely nice temperatures.
> 
> ...



I see several that are rated for 6000+ BTU, and will cool a 200 gallon aquarium up to 30°F below ambient. If I am using a reasonable size loop (a couple of gallons or so), I would believe that 30°C drop might be possible.

On the other hand, building a chiller would not only be cheaper, but I would have the satisfaction of doing it myself. That is the way I would prefer to go, but again, I would like to size the chiller properly.

I will check out XS.


----------



## Dippyskoodlez (May 28, 2009)

Arctucas said:


> I see several that are rated for 6000+ BTU, and will cool a 200 gallon aquarium up to 30°F below ambient. If I am using a reasonable size loop (a couple of gallons or so), I would believe that 30°C drop might be possible.
> 
> On the other hand, building a chiller would not only be cheaper, but I would have the satisfaction of doing it myself. That is the way I would prefer to go, but again, I would like to size the chiller properly.
> 
> I will check out XS.


aquarium chillers would probably do the trick if that's what you're looking at. (Not a fish guy, sorry  )

But yeah, Thermal output of a computer is a rollercoaster, and you can't really hit a "Target temperature".

In most scenarios you don't want a "target temperature" like you need in an aquarium either, colder is better.

and 6000BtU is most DEFINITELY enough to cool a computer 

A CPU is at most a 2-3cmx2-3cm piece of silicon pumping out heat, whereas a 200 gallon aquarium will have fish inside, the raw surface area and heat transfer going in/out are radically different.


----------



## W1zzard (May 28, 2009)

zAAm said:


> So if you have a CPU with a 100W power draw it will output 100W of heat (give or take). The same can't be said for hard drives and fans since most of their energy is converted into mechanical energy.



i disagree. a hdd that draws 20W will dissipate 20W. if that weren't the case the HDD would store energy ---time--> lots of energy, infinite amounts, *poof* black hole. which isnt the case


----------



## zAAm (May 28, 2009)

W1zzard said:


> i disagree. a hdd that draws 20W will dissipate 20W. if that weren't the case the HDD would store energy ---time--> lots of energy, infinite amounts, *poof* black hole. which isnt the case



Remember I'm talking about HEAT ENERGY and not total energy. I don't believe all the power would be dissipated into heat. Some of that energy goes toward overcoming the rotational friction of the bearings to keep the drive spinning at a constant speed. Since the platters aren't mounted on frictionless bearings in a vacuum they can't be sustained with no energy. Also, because you have a disk spinning you have potential energy being stored in the drive as well (not in the sense of a black hole though ) ... The head is moved by an electromagnet that converts some of its power into heat and also because of a change in flux in an inductor converts electric energy into magnetic energy which translates into a moving head. Thus all the energy can NOT be converted into heat or else nothing would move in the drive


----------



## W1zzard (May 28, 2009)

friction = heat. explain where the energy goes. you can not destroy energy (and i doubt you are creating mass)

you put energy into the platter which you get back in form of heat when the platter stops. you need to add more energy to the platter to keep it running to overcome friction on the spindle which is converted into heat energy


----------



## zAAm (May 29, 2009)

Ok, granted. The hard drive at full speed would almost be a closed system in this regard (By taking ALL the components into account at once). Essentially it would convert electrical energy into just enough mechanical energy to overcome friction which in turn generates heat energy. During spin-up however you are not counteracting friction but rather moving mass so then most of the mechanical energy is converted to potential energy and not heat. At spin-down you're once again converting the stored potential energy into electrical and heat energy. I think we had two opposite ends of the same straw here...

A fan is another story because it's not a closed system. (In essence the buck does not stop with it ) The fan converts electrical energy into mechanical energy which pushes against air and does work doing so. So energy would be transferred into the air so to speak. So then a 5W fan would not dissipate 5W of heat or else it would not be able to transfer any energy into moving the air.


----------



## Arctucas (May 29, 2009)

All well and good, but I still need to know how to determine the total amount of heat (in BTU) of my components?


----------



## cdawall (May 29, 2009)

zAAm said:


> Ok, granted. The hard drive at full speed would almost be a closed system in this regard (By taking ALL the components into account at once). Essentially it would convert electrical energy into just enough mechanical energy to overcome friction which in turn generates heat energy. During spin-up however you are not counteracting friction but rather moving mass so then most of the mechanical energy is converted to potential energy and not heat. At spin-down you're once again converting the stored potential energy into electrical and heat energy. I think we had two opposite ends of the same straw here...
> 
> A fan is another story because it's not a closed system. (In essence the buck does not stop with it ) The fan converts electrical energy into mechanical energy which pushes against air and does work doing so. So energy would be transferred into the air so to speak. So then a 5W fan would not dissipate 5W of heat or else it would not be able to transfer any energy into moving the air.



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


----------



## zAAm (May 29, 2009)

Arctucas said:


> All well and good, but I still need to know how to determine the total amount of heat (in BTU) of my components?



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.



cdawall said:


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



Uhm, no. Does the heat then make the air move??


----------



## yogurt_21 (May 29, 2009)

Arctucas said:


> I see several that are rated for 6000+ BTU, and will cool a 200 gallon aquarium up to 30°F below ambient. If I am using a reasonable size loop (a couple of gallons or so), I would believe that 30°C drop might be possible.
> 
> On the other hand, building a chiller would not only be cheaper, but I would have the satisfaction of doing it myself. That is the way I would prefer to go, but again, I would like to size the chiller properly.
> 
> I will check out XS.



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.


----------



## cdawall (May 29, 2009)

zAAm said:


> 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??



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.


----------



## zAAm (May 29, 2009)

cdawall said:


> 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.



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.


----------



## Bo$$ (May 29, 2009)

W1zzard said:


> i disagree. a hdd that draws 20W will dissipate 20W. if that weren't the case the HDD would store energy ---time--> lots of energy, infinite amounts, *poof* black hole. which isnt the case




W1zzard speaks the truth, some of you guys need to read up on some basic physics  




> 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.




couldnt have said it better


----------



## Dippyskoodlez (May 29, 2009)

Arctucas said:


> All well and good, but I still need to know how to determine the total amount of heat (in BTU) of my components?



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.


----------

