How electric motors consume energy?

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SUMMARY

The discussion focuses on the energy consumption of electric motors, specifically addressing how electrical energy is converted into mechanical energy and heat. Key factors include heat dissipation due to resistance in the motor windings, counter-electromotive force (back EMF), and losses from friction in bearings and brushes. Real-world motors experience energy losses, with some brushless DC permanent magnet motors achieving efficiencies over 95%. Understanding these mechanisms is crucial for optimizing motor performance and efficiency.

PREREQUISITES
  • Basic understanding of electric motor operation
  • Familiarity with concepts of voltage, current, and resistance
  • Knowledge of electromagnetic principles, including back EMF
  • Understanding of energy conversion processes in electrical systems
NEXT STEPS
  • Research the principles of back EMF in electric motors
  • Study the impact of resistance in motor windings on efficiency
  • Explore the design and efficiency of brushless DC permanent magnet motors
  • Investigate methods to reduce friction losses in motor components
USEFUL FOR

Electrical engineers, motor designers, and anyone involved in optimizing electric motor performance and efficiency will benefit from this discussion.

  • #31
pgardn:
"I don't understand which fields are doing work and at what points in time in the motor/battery setup. And that is the essence of understanding the energy situations in your very first post that I did not realize, for me. Work. I personally get myself to understand most of this stuff with fields and a reference object and what is doing work on that object through a series of events. I cannot manage that with this problem. Maybe you understand what fields are doing work on electrons, or the electrons within the lattice, during the different phases of a motor turning?" - I am sorry, but I don’t follow.
"As for case I. I was already assuming the wires had the same resistance. Therefore there are two diff ranges of energy allowed for the electrons. 1. in the battery 2. In the wire. No motor necessary." - I don`t understand either. The motor has to have some effect, doesn't it? If we are holding the motor still, we are applying a force equal to the one applied in the rotor trough electromagnetism. Therefore mechanical work is not being done, so all energy has to be dissipated as heat. Am I wrong?
Lsos:
You just gave your input, an important one to the discussion, thank you for that!
In regards to your second post, see my second comment above…
CWatters:
"In an ideal motor with zero losses there is not voltage drop and it allways runs flat out regardless of the load." - If this was accurate then power going in the motor would equal power going out, so you would not be transforming electrical energy in mechanical energy, but creating mechanical energy maintaining all the electrical one. You could have as much motors in series as you wanted…
"Back to the real world motor...Consider what happens when you stall the motor. The only thing limiting the current is the coil and brush resistance. All the voltage drop appears across these." - again, If we are holding the motor still, we are applying a force equal to the one applied in the rotor trough electromagnetism. Therefore mechanical work is not being done, so all energy has to be dissipated as heat. There has to be an extra effect besides the normal resistance of the wires.
sophiecentaur:
" There is no essential difference between the electrical energy being transferred to thermal in the resistance and into 'bulk' kinetic as the motor moves. The thermal energy that turns up in the hot resistor is only another manifestation of kinetic energy - it's just randomised." -Agree.
Can you give some input to how we "loose" the electrical energy in the motor. Power in must be higher that power going out the motor, regardless of the fact that we let the rotor turn or not. By what mechanisn(s)?
Thank you all for your inputs!
 
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  • #32
Power in needs to be greater than power out because there are energy losses.
The useful power out from an electric motor is mechanical kinetic energy. The armature wires get hot because of their resistance and hi is a form of kinetic energy but it would be misleading to include this in the total KE output .
There are several ways by which energy is wasted in an electric motor
1) heat produced by current flowing through the resistance of the armature
2) eddy current in the iron core
3) hysteris losses in the iron
4) resistance at the commutator contacts
5) windage and bearing friction...larger motors have a cooling fan
6) distortion of the applied magnetic field due to the armature field
All of these need to be taken into account to understand what happens to energy in electric motors.
 
  • #33
truesearch, you shot down my theory that back emf is how the kinetic energy output of an electric motor manifests itself as a load on the generator/ battery. I guess that since that time I (and I imagine other readers) was hoping that you would offer a better explanation. In the above, you explain how energy is wasted in an electric motor. But how is it NOT wasted?

Let me reiterate my understanding of what's happening:
Electrons flow through the wire, creating a magnetic field which turns the motor. If there is a load on the motor, then that load will push back on the motor, which will push back on the magnetic field, which will push back on the electrons flowing through the wire...and you lose electrical energy.

Is there much more to it than this? Is what I'm describing above not "back emf" but something more...perhaps Lenz's Law?
 
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  • #34
I'm confused here because it looks to me like the question has been thoroughly answered from multiple directions. But for clarity, Charles, there are two categories of energy dissipation in a motor:
1. Conversion to mechanical.
2. Losses.

Your question is about #1, right?

Then through Lenz's law and back EMF, do you get that the motion of the electrons creates an EMF that literally pushes back against the motion of the electrons? That's voltage (drop). So what else is there that you don't get?
 
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  • #35
Charles123 said:
pgardn:
"I don't understand which fields are doing work and at what points in time in the motor/battery setup. And that is the essence of understanding the energy situations in your very first post that I did not realize, for me. Work. I personally get myself to understand most of this stuff with fields and a reference object and what is doing work on that object through a series of events. I cannot manage that with this problem. Maybe you understand what fields are doing work on electrons, or the electrons within the lattice, during the different phases of a motor turning?" - I am sorry, but I don’t follow.
"As for case I. I was already assuming the wires had the same resistance. Therefore there are two diff ranges of energy allowed for the electrons. 1. in the battery 2. In the wire. No motor necessary." - I don`t understand either. The motor has to have some effect, doesn't it? If we are holding the motor still, we are applying a force equal to the one applied in the rotor trough electromagnetism. Therefore mechanical work is not being done, so all energy has to be dissipated as heat. Am I wrong?
Lsos:
You just gave your input, an important one to the discussion, thank you for that!
In regards to your second post, see my second comment above…
CWatters:
"In an ideal motor with zero losses there is not voltage drop and it allways runs flat out regardless of the load." - If this was accurate then power going in the motor would equal power going out, so you would not be transforming electrical energy in mechanical energy, but creating mechanical energy maintaining all the electrical one. You could have as much motors in series as you wanted…
"Back to the real world motor...Consider what happens when you stall the motor. The only thing limiting the current is the coil and brush resistance. All the voltage drop appears across these." - again, If we are holding the motor still, we are applying a force equal to the one applied in the rotor trough electromagnetism. Therefore mechanical work is not being done, so all energy has to be dissipated as heat. There has to be an extra effect besides the normal resistance of the wires.
sophiecentaur:
" There is no essential difference between the electrical energy being transferred to thermal in the resistance and into 'bulk' kinetic as the motor moves. The thermal energy that turns up in the hot resistor is only another manifestation of kinetic energy - it's just randomised." -Agree.
Can you give some input to how we "loose" the electrical energy in the motor. Power in must be higher that power going out the motor, regardless of the fact that we let the rotor turn or not. By what mechanisn(s)?
Thank you all for your inputs!

charles at this point I don't know what you are looking for. Truesearch went through a multitude of engineering points about what happens at every single step if you DONT consider a the held motor a single wire. That would require going back to the original applet and looking at all the details that really exist. Of course in case I when you start to loop the wire, the brushes, etc... this is detailed engineering. What one would do to make sure a motor worked as efficiently as possible... This is patently not what I was trying to accomplish. And not what I thought you were after.

To get to the meat of the matter which is what I think you are after... for me, you have gotten to the point of asking the equivalent of: by what mechanism does gravity work? I can't answer by what mechanism masses attract each other. The world about us has told us that they do. You can model fields, you can model distorted space, at some point you reach a set of observations and math that are just not mechanistic anymore and we are left with math that describes the phenomena accurately but the mechanistic modeling only goes so far, they are flawed. So force carrier particles is a better model to explain why masses attract each other. The particles, we will call them gravitons... I am cooked by this time...
 
  • #36
truesearch, thank you for this answer.
I agree with all that you said, but all those things are due to a real motor not being an ideal one. But in an ideal one Power out still has to be lower that power in. Why?
regards
 
  • #37
In an ideal motor, power out is EQUAL to power in.
 
  • #38
If you include ALL forms of energy then power out always equals power in.
Usually we are interested in USEFUL power out, the rest is called 'losses'
Electric motors can be very efficient, less than 10% of the energy supplied by the battery is wasted.
 
  • #39
I am sorry I put the question the wrong way. Of course in an ideal motor power out equals power in, but power in is electric power and power out is mechanical power. So electrical energy is fully transformed to mechanical work, rotor rotation. So electrical power out, in an ideal motor is zero. Why? That was what I meant before.
 
  • #40
Electric motors are machines that convert electrical energy to mechanical energy, they do not produce electrical energy.
The only other thing that I can think of that may shed some light for you is a battery connected to a bulb... all of the electrical energy from the battery is converted to heat and light by the bulb, the bulb does not produce any electrical energy !
Hope this is some help.
(I feel that it is back emf that is causing you problems... but I am not certain.)
 
  • #41
Charles123 said:
So electrical power out, in an ideal motor is zero. Why? That was what I meant before.
This represents a common but basic misunderstanding of electricity. You appear to believe that power flows through wires regardless of load, thus any device that doesn't consume all of that power let's some pass through. That is not correct. An electrical energy consuming device has no electrical energy output, only input.
 
  • #42
All machines are devices that convert an input of energy into an output of energy.
Transformers convert electrical energy input to a different electrical energy output.
Levers convert mechanical energy into mechanical energy.
motors do not produce electrical energy as an output.
etc
 
  • #43
Sure, but what causes the voltage drop? (in an ideal motor)
 
  • #44
Sorry ! don't understand what you mean !
 
  • #45
Charles123 said:
Sure, but what causes the voltage drop? (in an ideal motor)
What about the explanations you already got did you not understand?
 
  • #46
Ok try this...

If you put a permanant magnet close to an electro magnet that's off nothing happens. When you turn the electromagnet on a repelling or attracting force is created between them. This force causes the permanant magnet to move.

Work = force x distance

The permanant magnet does negative work on the winding...or the winding does work on the permanant magnet which ever way you prefer to look at it.

When th permanant magnet has moved away from the first winding the commutator switches off the first winding and applies power to another winding just as as the permanant magnet arrives in the right position. So the permanant magnet does negative work on each winding in turn until it gets back to the first one.

Energy is therefore convertde from electrical to mechanical form.
 
  • #47
"What about the explanations you already got did you not understand?" - The only explanation given that would happen in an ideal motor is back emf, that if you read the first posts was actually suggested by me, while asking if it was it, all the others are loses related to the functioning of a real motor. What I mean is that it has to be a voltage drop in the motor, or as I said before you could have all the motors you wanted in series, and they would all perform the same way.
CWatters, I understand how electrical energy, using electromagnetism produces mechanical work. That is not my question, I you read the first posts you will see that I am not questioning how an electrical motor works. I am asking how is that the electrical energy going in the motor, and use to produce mechanical work (and heat in a real motor), is no longer available in the wires going out of the motor.
Thank you all
Regards
 
  • #48
Then back emf really is the answer you are looking for. The wording of your post implies you don't like that answer though. Is there something about back emf you don't like/get?
 
  • #49
I have no problem with the answer, actually it was my initial intuitive answer, but I wanted to be sure it was the only factor (in an ideal motor). But in the example given of one holding the motor so the rotor doesn`t turn, what happens there? There is no back emf, but there has to be a voltage drop, how does it happen?
 
  • #50
Charles123 said:
But in the example given of one holding the motor so the rotor doesn`t turn, what happens there? There is no back emf, but there has to be a voltage drop, how does it happen?
That's potentially dangerous. The lack of back emf means much higher current until the source voltage is dragged down or other losses increase to dissipate the energy as heat...potentially burning out the motor.
 
  • #51
So the voltage drop would be solely due to resistance in the wires of the motor?
 
  • #52
we say motor consumes power or energy in contrast with transformer,that transformer does not consume power it utilisese it. While motor which works on the same principle of transformer consumes power.This is because in motor work is to be done to rotate the rotor so that much amount of energy gets consume in motor and hence the rating of motor is in kilowatt while rating of transformer is in KVA.
hope it may help you .
 
  • #53
Charles123 said:
So the voltage drop would be solely due to resistance in the wires of the motor?

In a stalled motor, I would say yes.

If you broke apart the motor and threw away the casing/ magnets/ armature/ everything except the wire...the effect on the voltage drop would probably not be much different.
 
  • #54
Charles123 said:
So the voltage drop would be solely due to resistance in the wires of the motor?

In a stalled motor yes. In the wire and the brushes if it's a brushed motor.

In a running motor there are other sources of loss. See page one of this thread.
 
  • #55
Charles123 said:
"What about the explanations you already got did you not understand?" - The only explanation given that would happen in an ideal motor is back emf, that if you read the first posts was actually suggested by me, while asking if it was it, all the others are loses related to the functioning of a real motor. What I mean is that it has to be a voltage drop in the motor, or as I said before you could have all the motors you wanted in series, and they would all perform the same way.
CWatters, I understand how electrical energy, using electromagnetism produces mechanical work. That is not my question, I you read the first posts you will see that I am not questioning how an electrical motor works. I am asking how is that the electrical energy going in the motor, and use to produce mechanical work (and heat in a real motor), is no longer available in the wires going out of the motor.
Thank you all
Regards

This shows that you still haven't got this right. Which wire is the one "coming out of the motor"? 'Positive and negative' are only arbitrary names, given to them. The energy is used in the motor (the potential drop across the terminals). No energy is needed for the current to pass through the wires, either the one from the positive or the negative terminal of the battery.
If one or both of the wires were resistive, then energy would be transferred in them and there would be less potential available for the motor.
It's the same situation when you connect three resistors in series, according to their resistance, different proportions of the supply volts are dropped across each one. This sharing out is established in the first nanosecond or so after the supply is switched on.
 
  • #56
(Let the supply voltage=V and the motor resistance=R)

When the motor is turned on it starts to turn and because the coil is rotating in a magnetic field it acts like a generator and a back emf (E) builds up.As it picks up speed E increases and the current (I) decreases.E reaches a maximum value which depends on the speed reached which ,in turn, depends on how heavily the motor is loaded.We can write:

V-E=IR

Multiplying by I we can write:

VI=EI+I^2R

VI=input power,EI=mechanical output power(work done per second by the motor)and
I^2R=power losses due to electrical heating.

(This is a simplified treatment and ignores other losses)
 
  • #57
Dadface, thank you for your exposition.
Thank you all for your inputs/answers.
Regards
 
  • #58
1) heat produced by current flowing through the resistance of the armature
2) eddy current in the iron core
3) hysteris losses in the iron
4) resistance at the commutator contacts
5) windage and bearing friction...larger motors have a cooling fan
6) distortion of the applied magnetic field due to the armature field
 

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