Eddy currents in case of synchronous and asynchronous motors

[PainterGuy]

Hi,

In an AC 3-phase induction motor (i.e. asynchronous motor) rotating magnetic field is produced by the stator and in this case rotor consists of simple cage having metallic bars or with coils wound around the rotor. The rotating magnetic field magnetizes the rotor and attracts it. As the rotor is attracted, it starts rotating physically. Now the rotor (or, rotor coils) is rotating and it produces eddy currents in the metallic bars (or, simple electric current in coils) which create their own magnetic field, and this magnetic field in turn induces voltage in stator coils. This induced voltage is opposite in polarity to that of AC supply and this is what controls the current flowing into stator coils.

When making a comparison of synchronous motor with asynchronous motor (let's assume unloaded one), it is said that in an induction motor, the mechanical speed of the rotor is less than the speed of the magnetic fields and the difference allows the stator to induce current in the rotor. My question is that why no eddy currents are generated in case of synchronous motor at all? I'm not able to visualize this. Could you please guide me? Thank you.

 

[cnh1995]

My question is that why no eddy currents are generated in case of synchronous motor at all?

Because stator magnetic field and rotor are rotating at the same (synchronous) speed (this is why it's called synchronous motor), unlike in an induction motor where the rotor speed is less than that of the stator magnetic field (slip).

 

[PainterGuy]

Thank you.

Okay. So, in case of synchronous motor there is no relative motion and that's why no eddy currents are produced. But even in case of asynchronous motor, there is no relative motion assuming unloaded motor. The rotor lags behind the stator field by a definite degree.

 

[cnh1995]

But even in case of asynchronous motor, there is no relative motion assuming unloaded motor.

Then there would be no eddy currents in the rotor. This is the ideal case. In practice however, rotor can't attain synchronous speed at no load. There is always some small slip present even at no load.

 

[PainterGuy]

Thank you.

In case of synchronous motor, both stator field and rotor field ideally get locked and there is no relative. In other words, ideally speaking, stator and rotor combines to make one rotating bar magnet.

But in case of asychronous motor, stator and rotor fields could not be combine to make a 'bar magnet' because they are not in straight line. The rotor lags behind stator field by a fixed amount, ideally. That lagging is slip and it's there even when the motor is unloaded. The slip increases with load.

Eddy currents require relative motion and I don't see any relative motion in both cases - synchronous and asynchronous motor. Where am I going wrong? Could you please guide me?

 

[cnh1995]

I don't see any relative motion in both cases - synchronous and asynchronous motor.

Eddy currents are induced in rotor body.
In a synchronous motor, the rotor body is moving at synchronous speed and hence, there is no relative motion between rotor body and rotating stator magnetic field.

In an asynchronous motor, rotor body is rotating slower than the stator magnetic field. Hence, there is a relative motion between the rotor body and stator magnetic flux. This is why emf is induced in the rotor bars (or windings) and eddy currents are geneated in the rotor core.

 

[gerbi]

@cnh1995
Very good explanation. That's the theory.

Hi,
[..] My question is that why no eddy currents are generated in case of synchronous motor at all?

Ideally there are no eddy currents in synchronous machine rotor body and no voltage induced in it's excitation winding.. However in practice there are almost always some amount of eddy currents being induced. This is due to non-ideal machine geometry, it's design as well as the voltage/current conditions it's experiencing. Those effects of unbalanced poly-phase systems are analyzed with application of symmetrical components. The one connected with the rotor eddy currents is the negative component. Rotating at double speed in the opposite direction to the positive one. For large synchronous generators there is a limit to both long and short term negative sequence current which is usually in 5-10% range (IEEE C50.13). The limit is due to thermal capabilities.

 

[anorlunda]

Several posters correctly said that an induction motor can never achieve synchronous speed.

But let's think what would happen if we supplied an external torque to speed it up slightly to achieve synchronous speed. (technically, we make is an induction generator with zero electric output). Then there would be zero currents induced in the rotor, and the stator field would appear to be DC as seen from the rotor.

In that state there is no difference between an induction machine and a synchronous machine with zero applied field voltage. .

 

[PainterGuy]

Thank you for your help, everyone.

Please have a look on the attachment. The metallic block is stationary and the magnet is moved toward left at the speed of 1m/sec. It will induce eddy currents in the block because it experiences a time-varying magnetic field. But if the block is also moved toward the left at the same speed as the magnet then no eddy currents would be produced because there is no relative motion between the block and magnet. Do you agree with me?

Slip is defined as the difference between synchronous speed and operating speed, at the same frequency, expressed in rpm, or in percentage or ratio of synchronous speed. Thus

where

is stator electrical speed,

is rotor mechanical speed. Slip, which varies from zero at synchronous speed and 1 when the rotor is at rest. [Wikipedia]

Let's say that if a point on stator field completes, let's say, 400 degrees in one second then a corresponding point on the rotor completes 360 degrees. Is the difference of 40 degrees between these two corresponding points constant? If it's constant, it'd mean no relative motion.

It looks like that the difference is not constant. In first second, stator point covers 400 degrees and in next second, it covers another 400 degrees. Meanwhile, the rotor point covers 360 degrees in first second and another 360 degrees in the next second. The difference between the two points is: (2x400)-(2x360)=80 degrees. It means that the relative motion exists between the corresponding points. Please let me know if I have it right. Thanks a lot.

 

[gerbi]

Let's say that if a point on stator field completes, let's say, 400 degrees in one second then a corresponding point on the rotor completes 360 degrees.

In this case there is a 40 deg/sec difference. Every second the relative position is changed by 40 deg. This means, that there is a relative motion.
So, again: at slip s≠0 there is a relative motion between stator electrical speed and rotor body. This relative motion between them is constant at constant slip.

 

[cnh1995]

Let's say that if a point on stator field completes, let's say, 400 degrees in one second then a corresponding point on the rotor completes 360 degrees. Is the difference of 40 degrees between these two corresponding points constant? If it's constant, it'd mean no relative motion.

This difference is 40 degrees per second.
Hence, relative angular velocity between stator and rotor is 40 deg/s and the slip is (400-360)×100/400=10%.

 

[gerbi]

Hence, relative angular velocity between stator and rotor is 40 deg/s and the slip is (400-360)×100/400=10%

The slip is in 0 - 1 range. So not a 10% but 0.1.

 

[PainterGuy]

Thank you.

The slip is in 0 - 1 range. So not a 10% but 0.1.

I might be wrong but I don't think that using 10% instead of 0.1 makes much of a difference. The Wikipedia article on induction motor does use a percentage value too. Thanks.

 

[gerbi]

Thank you.
I might be wrong but I don't think that using 10% instead of 0.1 makes much of a difference. The Wikipedia article on induction motor does use a percentage value too. Thanks.

Actually, this may be true. I see that some sources use %. Personally, I've never seen slip expressed in %.. so far. But those are old books, not in ENG. So - my bad.

 

[PainterGuy]

Hi,

I was reading further on eddy current and came across Wikipedia article on skin effect, https://en.wikipedia.org/wiki/Skin_effect. It says that the skin effect is due to opposing eddy currents induced by the changing magnetic field resulting from the alternating current. Its also says that at 60 Hz in copper, the skin depth is about 8.5 mm. The following table is an excerpt from the article.

Question 1:
Okay, at 60 Hz the skin depth in copper is about 8.5 mm but relevant to what? The skin depth is 8.5 mm if the copper conductor's radius is 20 mm, 100 mm or what? The article doesn't seem to mention this or I'm missing something obvious. Could you please guide me?

Question 2:
"Highly magnetic materials have a reduced skin depth owing to their large permeability μ_r as was pointed out above for the case of iron, despite its poorer conductivity. A practical consequence is seen by users of induction cookers, where some types of stainless steel cookware are unusable because they are not ferromagnetic." [Wikipedia]

I couldn't understand that how skin depth is related to the induction cooking. Then I reached the following section of Wikipedia article on Faraday cage: https://en.wikipedia.org/wiki/Faraday_cage#Faraday_cage

It says, "Faraday cages are Faraday shields which have holes in them and are therefore more complex to analyze. Whereas continuous shields essentially attenuate all wavelengths shorter than the skin depth, the holes in a cage may permit shorter wavelengths to pass through..."

This my understanding. In induction cooking, electromagnets generates EM waves which interact with a utensil. To full 'convert' those EM waves into eddy currents and hence heat, all the EM waves need to be captured by a utensil and none should pass through. If the utensil is made from a ferromagnetic material then it'll have shorter skin depth and therefore all EM waves are stopped and converted into heat. Do I make sense?

Thank you for your help.