re: Input electrical power = fan heat

[TurbineBlade]

Hi,

I read about that when a fan + motor runs inside an air stream, the fan heat into the conditional space is equal to the electrical power input. I am trying to understand this concept. If the electrical power input is 10W, then

say, 1 W goes to heat as motor loss, and
9 W goes to produce the flow at certain velocity. However, `some` energy turns into heat due to the friction from the fan blade.

As the air stream is still moving, why all the energy become heat input into the space.

thanks

Turbineblade

 

[IRstuff]

Where else would it go? It's simple conservation of energy. If the fan consumes 10W, part of it goes into the work of moving the air, part goes into heating the air (but, that minuscule unless the blades are really cranking), part of it goes into electrical and frictional losses in the fan. The former two already go into the air stream. The latter is expressed as heat, which is usually cooled by convection, but not by the air stream. When the entire fan is in the air stream, then the heat is removed by the same air stream, so all of the 10 W goes into the air stream.

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[TurbineBlade]

Hi,

Thanks. When I tried to break it down mathematically, I run into the problem that

Pelect = Qloss (motor+fan) + (diff. pressure)*flow

The (diff. pressure)*flow is the work done by the fan to move the air, say 9W. However, we are also counting this same amount of energy (9W) as heat input into the air stream? Is the explanation that K.E transfers into friction due to the air resistance, which become heat into the space eventually?

 

[Drazen]

That is say "dynamic" energy of air that also have to be relieved in cooling application, that's why it is counted.

 

[EnergyProfessional]

ultimately everything becomes low grade heat. The universe started with a lot of exery. In the end everything will be entropy.

but in the AHU a fraction of electricity will become pressure. That pressure will be reduced o heat in the ductwork and heats up air in the system. It depends on what you need that information for. Rule of thumb is fan adds 2°F in AHU.

 

[TurbineBlade]

Hi,

Use the example above, the 9W that produces the pressure rise at a given flow is to overcome the friction in the ductwork. The friction, which is an irreversible pressure loss through the duct, eventually results to heat and transfer to the air stream and the space. If I have an ACU with ducting that recirculate air inside a space. The heat generated from the fan = the electrical power to the fan. Is this correct concept?

 

[Compositepro]

Correct. The law of conservation of energy: Energy cannot be created or destroyed. It can only be converted from one form to another. If you want to get into nuclear physics it can also be converted int mass, E=MC^2.

 

[Drazen]

it is a little more complex.

if you want to get it completely, it is best to use analogy with refrigeration cycle - compression makes adiabatic change of temperature. the similar is with fan, though air density change is not considered in low pressure ventilation equations.

that is, mechanical energy input raises air temperature. air does not receive any additional heat through this, but your air loses cooling potential by this temperature rise, so you need to apply additional cooling (equal to mechanical energy input balance) to maintain cooling capacity of your supply air.

this is where complexity of thermodynamic principles becomes visible, as temperature level is not directly bound with energy level - which is also the case with many other thermodynamic processes.

for practical purposes you can take formulas as they are or invest time to grasp all principles, but the most important thing is to not try to simplify it by force.

I don't know what kind of background your nuclear specialty provides you with, but for understanding of adiabatic or isothermic processes you really need some study of thermodynamics.

 

[Compositepro]

Drazen, you completely lost me on that one. Why is it any more complicated than simple conservation of energy?

The trick to solving most problems is to not make them more complicated than they are.

 

[EnergyProfessional]

TurbineBlade: maybe tell us what exactly you want to accomplish or find out? Ultimately everything becomes heat. teh question is at what location. Mot inefficiency and fan efficiency will heat up the air in the fan plenum. So X5 of electricity become heat right there.
the remainder of electricity became potential energy and kinetic energy (static and velocity pressure). Part of that get's converted to heat in the ductwork. (energy doesn't get lost, it just get transformed). After leaving the diffuser you have some velocity pressure left... ultimately this becomes heat within the room.

Is any of that noticeable? Only the part in the fan since this may be 50% or more of that energy in a small space. the rest is spread over such a large area, that you don't notice an increase in temp.

 

[Drazen]

compositepro, you are right that good knowledge most frequently leads to simplest solutions. however, to reach good knowledge it is not good to try to simplify things by force only because they are felt as complex.

thermodynamics is really not simple, but there are many engineering shortcuts that work if you apply them punctually, taking lot of care about limitations.

if you want good understanding, things are different though. i am not better than others, though i am practicing thermodynamics for more than 20 years, there are still things which i don't fully understand. when i reach those zones, I rely on proven advice, without trying to reach my simplified explanation. (sometimes i try, but when noticing holes, i give up and leave it to better times when someone would fund my deeper study .)

here, i tried to direct op in right direction. he is likely experienced engineer and it wouldn't be good that he explains to himself adiabatic compression by figuring out friction.

 

[willard3]

Consider basic laws of physics, ie, conservation of energy and the second law of thermo.

 

[TurbineBlade]

Hi,

Thank you for the responses.

Energyprofessional,

I have a fan that takes electrical input of about 50 kW. The fan, fan motor and the duct work are all inside a conditioned space. It is a huge fan and blow air through a cooling coil downstream. I suggest not to run the fan when the cooling supply water is out of service as it will heat up the space (500m3) enclosed by concrete walls. The question comes up why an electrical input of 50kW to the fan will indeed become heat when the work done by the fan is to move the air, which should be the kinetic energy. By now, I can say that the kinetic energy used to move the air is small and the kinetic energy will also transfer into heat since the air discharge out from the grille will eventually stop. Most of the energy from the fan is used to overcome the friction in the duct.

 

[LittleInch]

do you have your units right? 500m3 = a room 13m square by 3 m high and you have a 50kW fan motor!

That will certainly heat up your room pretty fast if you've got no cooling

"Direct" heat from the motor everyone can see and feel the motor getting hot, but it's the spread out low temperature rise of the air due to friction that is difficult to comprehend for many people, but just think of energy and its inability to be destroyed.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[dbill74]

TurbineBlade,
I think you're over thinking this. With the fan in ductwork/cabinet in a room, the fan motor is going to heat the air in the duct before the room itself, because the air is moving it will effectively move the heat from the fan out of the room thereby minimizing the amount of heat entering the room due to the fan motor to be negligible. This is true if the cooling coil is in operation or not.

To determine the amount of heat entering the airstream from the fan motor, I suggest downloading Loren Cook's Engineering Cookbook from

http://www.lorencook.com/downloads.asp.

Page 44 lists approximate amount of heat gained by an airstream based on motor size (Hp).

It is perfectly acceptable to run the fan to provide ventilation air when there is no need for cooling. You need to review the system's current sequence of operation and verify that the fan running with the cooling coil off is the way it is intended to operate. Changing the sequence of operation as you suggest can have unintended consequences besides a hot mechanical room.

Bill

 

[TurbineBlade]

Hi,
Correction. The room is about 1500m3.

If I were to size tempoeary acu, I would need to include the 50kw as cooling load if the fan were to turn on.

Bill,

The fan with ductwork recirculates air inside the room.

 

[EnergyProfessional]

so just circulating in that room? Are you cooling something else in that room or what is that fan for? It sounds like you need to size cooling for he actual motor power (not sure if 50kW is the rated or actual power) + whatever else you need to cool.

If you use Trace or other load software the fan heat would be added as well.

 

[dbill74]

What is the point of the fan then? Where is the cooling coil? How are you getting heat out of the space?
Regardless, the info in the book I referenced is still applicable. Not all the electrical input power is going to be converted to heat energy, some of it is converted to work energy to rotate the fan and move the air.

 

[Compositepro]

Eventually (within seconds), all of that energy is converted to heat. This is one of the most basic and important concepts in any field of science or engineering.

 

[EnergyProfessional]

Exergy (availability) will be converted to entropy irreversibly. This one rule will literally explain all HVAC questions.