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How much heat comes from an electric motor 2

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marko333

Mechanical
Aug 6, 2003
21
Hello.

If I have a 45kW 3-phase electric motor, can I estimate how much heat is generated from it?

For example, if I know that the power factor is 0.78, is it reasonable to assume all the inefficiencies eventually translate to heat? So that I can say about 10kW of heat is generated?

Thankyou.

Marko.
 
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Suggestion: The motor manufacturer tech support may provide the motor heat dissipation value (in watts).
The estimate of the motor heat dissipation from the motor efficiency can be derived too; however, the whole loss calculated over the efficiency is not RI^2, watts. Some of losses causing the efficiency to be lower than 1 are covering the windage, and rotor dynamics.
E.g., if the motor is 95% efficient, 45kW is used as a motor output, then 45kW/0.95 - 45kW=2.368kW are losses. If windage and dynamics consumes 2/3 than 1/3 is RI^2 heat dissipation, i.e. about .79kW.
 
marko333: You do realize that 45kW will be dumped in the "area" of the motor. If the motor is running a leather stacker then all of the motor's consumed energy will be dumped in the immediate area. This is often messed up when people are doing air conditioning calculations.

 
For a 45 kW motor, the output is 45 kW.If the efficiency is app. 82%, the input is 50 kW. 5 kW is the loss.

For identifying the actual value of the following losses , testing and calculation is requred.
Windage & friction losses
No load magnetic & copper loss
Laod copper loss (stator & rotor)
 
60hzhmmmm, i do not fully understand what you mean. could you please explain a little more? Are you saying that in the area of the motor, there will be 45kW of heat? I am trying to estimate a heat load for a HVAC calc

I agree that a 45 kW motor will turn that 45kW of energy into some mechanical form.

For a 80% efficient motor, it will draw a total 55kW of energy.

So, that 10kW of energy left over after 45kW are converted to mechanical energy is the result of friction, resistance, slip etc. So generally, does most of this energy loss degrade to heat/noise? Or is only about 1/3 of it converted to heat? Is there a general 'rule of thumb' to estimate this heat loss?

Thankyou
 
Basically, all losses end up as heat, including friction, windage, core and winding. So just use the efficiency of the motor. A 100 hp motor that has an efficiency of 90% requires about 111 hp of input energy (at full load). It will deliver 100 hp to the shaft and dissipate about 11 hp (8.2kW)as heat. It gets a little more complicated at partial load - you would need a curve of efficiency versus load to be highly accurate.

The power factor doesn't play a role in heat loss determination because it doesn't directly relate to efficiency. In fact, extremely high efficiency motors have poor power factors.
 
All of the used electrical energy is converted into heat:
The loss in the motor itself, the useful part of the energy
gets converted into mechanical energy first and finally --
in the external mechanism -- into heat.


<nbucska@pcperipherals.com>
 
Suggestion to the previous posting: Normally, RI**2 in watts is considered as a heat source. Although, there is some heat developed in bearings due to bearing friction. Watts that consumed by motor windage losses and motor rotational dynamics do not significantly contribute heat to the room environment.
 
Suggestion to the previous posting: All losses end up as heat. Some losses are greater than others, but since the original question related to the heat rejected by the motor, the breakdown of losses is immaterial.
 
All the input to the motor is finally converted into heat.if the driven equipment is doing the work within the room, for eg a grinding wheel,all the energy is converted into heat with in the room.If the driven equipment is a pump,the mechanical input to pump is converted into pressure energy of the pumped fluid.If the fluid leaves the room,it gets converted into heat at some other place along the pipeline.In this case the heat dissipation in the room is due to inefficiency of the motor only.

Since the most stable form of energy is heat, all energy eventually gets converted into heat energy.

So froma HVAC load estimation point of view,the heat dissipation into the room is very much dependent on the system configuaration as explained in above example
 
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