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'[EE]:BTU's of a system'
2002\04\11@125824 by

changing topics from my current ICD problems...

If you have a 300W power supply, and *assume* (however bad it is) that it
will be drawing a total of 300W (lets pretend its a perfect world and no
power factor correction, or the supply is PF corrected)then if you want to
calculate the BTU's to determine the cooling requirements, do you simply use
the 300W figure?

What I am getting at can you safely approximate the fact that the entire
300W will at some point be dissipated as heat?  Or only a certain
percentage?

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Usually the 300W rating is what the load will dissipate.  If the supply
is 75% efficient, the load gets 300W and the supply has to dissipate
100W.

If the load is in the same box as the supply, such as in a PC, and the
supply fan is doing all the cooling, then the fan in the above case
would have to cool the 100W of the supply plus the 300W of the load.

Sherpa Doug

> {Original Message removed}
Well, the 300W power supply, in an ideal world, will radiate NO heat and
have no cooling requirements.

The components connected to it, however, typicially radiate radiation.
This is in the form of heat, light, EM, RF, and other fields.

-Adam

Micro Eng wrote:

{Quote hidden}

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surely only a percentage. Some energy must be going into whatever you
are powering with the power supply. I don't know your situation. But
if it was driving a motor. The electric power you put in, minus the
mechanical power you get out, is the power that must be dissipated
else where mainly heat!

Ed

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>What I am getting at can you safely approximate the fact that the entire
>300W will at some point be dissipated as heat?  Or only a certain
>percentage?

The 300W is what it can deliver as output.  You will draw, on the
input, somewhat more.  The difference is the heat that you will
get.  If it were 100% efficient, it could deliver the power and
there would be no heat.   In the power supply, that is.   Your
300W will show up in the load, in the thing you connected to
the power supply, and if it's in the same box, then you'll get
to dissipate that anyway.

Power factor effects are reactive, and don't become heat.  That's
why they're called "imaginary".

Barry

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> If you have a 300W power supply, and *assume* (however bad it is) that it
> will be drawing a total of 300W (lets pretend its a perfect world and no
> power factor correction, or the supply is PF corrected)then if you want to
> calculate the BTU's to determine the cooling requirements, do you simply
use
> the 300W figure?

It depends on what is connected to the supply.  300W is the maximum the
supply is guaranteed to provide over its operating range.  Properly designed
systems make sure that the worst case load never exceeds that.  Most systems
would run below that during normal operation.

Another issue to consider is that the supply is not 100% efficient.  80% is
probably safe for a normal switching power supply.  So if the system is
drawing the full 300W from the supply, figure the supply is drawing 340W
from the power line.

> What I am getting at can you safely approximate the fact that the entire
> 300W will at some point be dissipated as heat?  Or only a certain
> percentage?

If it is a computer system, then it all eventually ends up as heat.  If the
power is used to run a mechanical shaft, for example, then it is transferred
somewhere else.

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2002\04\12@031900 by
>If you have a 300W power supply, and *assume* (however bad it is) that
>it will be drawing a total of 300W (lets pretend its a perfect world and
>no power factor correction, or the supply is PF corrected)then if you
>want to calculate the BTU's to determine the cooling requirements, do
>you simply use the 300W figure?

No, you take into account the efficiency. Pin - Ploss = Pout. Ploss
becommes heat for sure. A modern SMPSU will be at least 70% efficient, so
you are left with at most 100W of heat from the supply. However, if the
device powered is also considered, then you need to remove most of the Pin
as heat because Pout turns into heat while running circuits. Exceptions to
this are electromechanical machines and electrochemical syestems and RF
transmitters where a significant part of Pout does not become heat.

>What I am getting at can you safely approximate the fact that the entire
>300W will at some point be dissipated as heat?  Or only a certain
>percentage?

If you know nothing about the system and only have the PSU (f.ex. if you
are a cabinet maker and make cabinets with built-in PSUs - to be filled
with systems by your prospective clients), then you do need to account for
all the input power, as heat, in your cabinet and provide suitable
ventilation or forced cooling arrangements.

If you do know everything about your project (say a circuit that consumes
8W and is mounted in a plastic handheld box) then you also need to make
sure that the box does not become too hot to handle (literally).
Temperature over 45C anywhere outside a box is not good. A super-miniature
circuit with umpteen watts will require kitchen mitts to operate and a
non-flammable support unless you take care of this from start.

Peter

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1 Watt = 3.41 BTUH. Only a percentage of the power will be dissipated as
heat energy.
This is what the term "efficiency" is all about. You should be able to
determine the amount of energy lost as heat from specifications. If not, you
can always back into it by measuring heat sink temperature if you know the
number of square inches of surface area.

Chris
{Quote hidden}

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> Only a percentage of the power will be dissipated as
> heat energy.
> This is what the term "efficiency" is all about.

Oh, and where does the rest of it go?  In a computer system, effectively all
of the input power goes to heat.  Unless you are deliberately transmitting
power to somewhere else, like driving a shaft or emitting electromagnetic
waves, virtually everything ends up as heat.

Of course it does depend on what the "enclosure" is you are considering.  If
you are only care about the power supply itself, then figure about 20% to
25% of the input power heats the supply.  If you are looking at the whole
computer system, then the remaining 75 to 80% ends up as heat too.

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It's all a matter of perspective. I work on systems with multi-kW heater
loads, and typically am called on to supply values for make up air for plant
cooling systems. The amount of heat dissipated into the local environ from
the electronic controls is miniscule compared to the overall, and is
certainly covered in my safety margin, or "fudge factor". I do, however
consider 100% of the control power as heat input.

{Quote hidden}

So then, electrically speaking, computing systems are really pretty
inefficient then.....

> Of course it does depend on what the "enclosure" is you are
> considering.  If
> you are only care about the power supply itself, then figure
> about 20% to
> 25% of the input power heats the supply.  If you are looking
> at the whole
> computer system, then the remaining 75 to 80% ends up as heat too.

As regards the smaller heat contribution from the supply (which if I'm not
mistaken, is the area in which this thread started...) wouldn't the "drop
out" of the regulation actually allow you to determine supply waste heat?

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> As regards the smaller heat contribution from the supply (which if I'm not
> mistaken, is the area in which this thread started...) wouldn't the "drop
> out" of the regulation actually allow you to determine supply waste heat?

In a linear supply, the wasted power is proportional to the voltage being
dropped by the supply because the input and output currents are the same.
This is not true of switching supplies.

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(978) 742-9014, http://www.embedinc.com

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