How electric motors consume energy?

[Charles123]

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

 

[russ_watters]

In an ideal motor, power out is EQUAL to power in.

 

[truesearch]

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.

 

[Charles123]

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.

 

[truesearch]

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.)

 

[russ_watters]

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.

 

[truesearch]

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

 

[Charles123]

Sure, but what causes the voltage drop? (in an ideal motor)

 

[truesearch]

Sorry ! don't understand what you mean !

 

[russ_watters]

Sure, but what causes the voltage drop? (in an ideal motor)

What about the explanations you already got did you not understand?

 

[CWatters]

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.

 

[Charles123]

"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

 

[russ_watters]

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?

 

[Charles123]

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?

 

[russ_watters]

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.

 

[Charles123]

So the voltage drop would be solely due to resistance in the wires of the motor?

 

[nikhil khatri]

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 .

 

[Lsos]

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.

 

[CWatters]

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.

 

[sophiecentaur]

"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.

 

[Dadface]

(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)

 

[Charles123]

Dadface, thank you for your exposition.
Thank you all for your inputs/answers.
Regards

 

[cavinmozarmi]

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