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I have a closed area and I need to calculate the heat load for HVAC purposes.

In this system I have lots of electrical motors that move a mass back and forth on linear rails.

It's my understanding via the law of conservation of energy, that ALL energy put into this area, will turn into heat (via losses or via kinetic energy), unless it's somehow stored as potential energy inside the area, for use later on (when it still will be turned into thermal energy when converted back to kinetic).

The question is if I should assume that all energy I put into the motors measured in Watt, should be used for calculating the need for air cooling / HVAC in this space. Or do I use the efficiency of the motor output for this calculation?

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So long as your entire apparatus is in the room, then yes, ultimately all of the energy ends up as heat. You could make the case that if, e.g., you were crashing things together and thus bending or reshaping stuff, that some small percentage went into that reforming of molecular bonds, but from an HVAC standpoint that's in the noise. Same would go for, say machines generating sound, some of which is re-emitted outside the room.

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  • $\begingroup$ The answer I was coming here to write. Also negligible is electromagnetic noise from the motors (unless the system is intentionally a radio antenna). $\endgroup$ – rob May 12 at 13:40
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Your considerations are pretty much right.

If you don't carry out the energy in some other form (chemical, loaded springs, shafts, belts, compressed air going to the outside, etc...) the whole power input ends up as heat.

On the other hands, be aware that motors not always work at their rated power and the heat output will probably average over hours (or more) because of the heat capacity of every material involved.

If you have the system in hand, a watt-meter at the power socket will do the trick. That's how we estimate the HVAC for servers.

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Except for energy which is buried and never retrieved, all energy on earth ends up being radiated into space (at a rate which is pretty much equal to the rate that we receive it from the sun).

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