Discussion in 'General Electronics Chat' started by jinksung, Jul 22, 2012.

1. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
I need help from you, electrical guyes.

I am a materials guy and I am working on inductor materials, soft magnetic materials.

I tired to understand how induction motor works. And I found lots of difficulties.

My questions are;
1. Is induction motors a constant power machine or not?
Since the resistance of stator winding is fixed and the applied voltage is constant to the stator, so it looks to me that current flow is fixed regardless of loading. But I found that this is not the case. Why is that?

If the rotor draws current by magnetization, still the resistance of the stator is same. Could you please explain it not by an electrical circuit but by reasoning? Say
- current in stator winding induce magnetic field,
- stator tooth are magnetized by the magnetic field,
- magnetic field in the stator magnetizes the rotor,
- EMF is induced in a rotor bar,
- magnetic field of opposite direction forms inthe rotor bar,
- rotation of rotor
- current in stator winding induce magnetic field,
- stator tooth are magnetized by the magnetic field,
- magnetic field in the stator magnetizes the rotor,
- EMF is induced in a rotor bar,
- magnetic field of opposite direction forms inthe rotor bar,
- slip in the rotor due to load
- rotation of rotor

The only difference is slip.
How does slip influence more current at stator winding?

What is the meaning of efficiency of induction motors?
People mention that the efficiency of induction motors is 75-90%.
And I found that the magnetizing current for inductor is about 20~60% (average 33% of FLC). Magnetizing current means current for no load case. (No mechanical work at all.)
If the magnetizing current is 33%, does this mean that sum of loading current and loss current is 67%? If this is the case, the efficiency of , say, 90% is actually 67%x0.9?

According to www.lmphotonics.com/energy.htm

The current flowing into an induction motor comprises three major components, magnetizing current, loss current and load current. The magnetizing current is essentially constant, being dependent only on the applied voltage....For a large motor, the magnetizing current can be as low as 20% of the rated full load current of the motor.

3. If high permeability materials were developed, how will it affect the efficiency of motor?
For example, to magnetze the stator core to 1.5T, an existing material needs 1000A/m. And the new material needs 500A/m.
Does the above case mean that with the new material, magnetizing current is reduce by half? So we can save lots of energy?
Is there any effect on load current?

2. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
There are many different types of AC motors.

For instance synchronuous, and asynchronous.

The different current is explained very simple.

If the motor must produce more torque, then this will result in higher back EMF.

The magnetic force to overcome it must be higher. And so the current!

AC electrical circuits, and DC, are not the same. If you do not fully understand the fundamental difference then I suggest to buy a book about this topic.

Motors are most likely very effective, not 100%, but often 70 to 95%.

For instance a hand drill having nominal rating of 700W will heat up after a while, but no way near to 700W. If it is heavily loaded, then it will heat up more.

3. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Thanks takao21203. Let's say a squirrel-cage induction motor.
And also assume using 1.5T magnetic field for stator.
Then, what will happen?

4. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Also, if we use the new material with higher permeability in a DC motor, then what will happen? How much energy will be saved?

5. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
I don't know motors enough to know what will happen (or not) at 1.5T, eventually using new high-perm materials.

I have read some related books, but this is more than 10 years ago.

Currently I only have one book here about motors, enough to understand different types of motors.

Small DC motors often use magnets for instance inside hard drives.

It is an example where a motor is used to rotate an object, but otherwise it is not loaded. But it is not exactly a classical DC motor. Brushless commutation...

The biggest loss will be caused by mechanical friction.

On the linked website there is a lot of explanation however no actual motors, or energy saving devices can be seen.

6. ### crutschow Expert

Mar 14, 2008
20,471
5,797
1. An induction motor power input is proportional to the mechanical power output (neglecting magnetizing current and other losses).

The rotor and stator act as a transformer. The frequency of the induced current in the rotor is determined by the slip frequency between the rotating field and the rotor speed. As the mechanical load increases, the rotor slows down, increasing the slip frequency and thus the amount of current induced in the rotor. By transformer action the stator current is also increased. Thus the stator current is proportional to the mechanical load (and amount of slip).

2. Magnetizing current is reactive current and requires no power (other than that dissipated by the current through the winding resistance). Since the magnetizing current is 90 degrees out of phase with the voltage, you add that to the load current using vectors (phasors). It does not add linearly.

3. A higher permeability material should lower the magnetizing current but that is only a small factor in the motor efficiency at full load. Most of the wasted power comes from load current through the winding resistance in the stator and rotor. A motor with superconducting windings, for example, would have very high efficiency.

Gex91 and strantor like this.
7. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Thanks crutschow, but still I have a question on magnetizing current.
When people mention about magnetizing current, they are talking about no load current (almost no slip condition) and it is about average 33% of full load current. So, if a high permeability material is used(say 2000 vs 1000), can I save 50% of the magnetizing current to get the same magnetic flux density? In other words, is no load current reduced down to 16.5%?
And some people say that magnetizing current is almost constant during operation of the motors with any load. If this is the case, can we save 16.5% of electric power to get the same mechanical work?

8. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Let's change my question as follows;
Generally materials scientists consider only loss component.
Iron loss, copper loss, windage loss, stray current loss.
And we believe the rest of energy is for mechannical work.
So we are trying to reduce Iron loss without sacrificing magnetic induction.

Let's assume we have materials with same level of iron loss.
However, its magnetization behavior is different.
For example, permeability of 1000 and 2000 for the same 1.5T magnetization.

Iron loss and windage loss appears to be same.
The major difference lies in copper loss since magnetizing current is different.

At this point, what I have noticed is that the reduction in copper loss is due to reduction in exciting power.

Now, how will the reduction in exciting power affect the electric power consumption? or efficiency?

If we only consider loss components, except for the copper loss everythig is same.
In full load condition, how much portion can be attributed to the magnetizing current?

It's really puzzling!

9. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
As crutschow said, if magnetizing current is reactive current and requires no power, then maybe what appears to be important is only loss components. Is it really so?

10. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
Small motors with high torque are built using high-permability materials, if I remember right.

A transformer is somehow analogy to motor even if not a very good one. Transformer does not have moving parts.

And single toroid inductor is not much different from a Transformer, indeed in principle, a full transformer can be built on a toroid. The physics for only a coil are the same.

Indeed permability, and magnetization losses are important properties for toroid cores. There are many different materials, even if appearance is similar. Granularity is different I think.

11. ### crutschow Expert

Mar 14, 2008
20,471
5,797
Magnetizing current is due to the inductance of the motor and is thus inductive (reactive) current which only causes a power loss from that current going through the winding resistance. Thus for a 16.5% reduction in magnetizing current you would get a 16.5% reduction in the magnetizing losses due to this resistance, not a 16.5% reduction in full load motor power.

12. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
If this is the case, copper loss by magnetising current will be reduced.
So, 33%(magnetizing current)x(3/4) of copper loss will be reduced.
(I^2R Loss, since current becomes 1/2, its copper loss is 1/4)
Load current is also transferred through magntic circuit.
If there is more current in stator windings, then it will contribute to the copper loss.
So, when rotor bar draws more current by loading, there will be more current in stator windings. Because of this, more copper loss in the stator windings during loading.

Load current cannot be separated form magnetizing process, What I mean is any type of energy transffered to the rotor should use magnetic circuit.
My concept is not clear now. On this point, I will post it later.

13. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
If a motor has 95% efficiency, and copper loss are 2%, then 16% reduction of copper loss are:

2% * 16%

which roughly correlates to 0.02 * 0.16

After changing units to percents again, you get 0.32%

So it would change to 1.68%

Neglible!

14. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Thanks takao21203
If copper loss is 2% and magnetizing current is 33%. then what I can save is
2%x0.33x3/4 = 0.495%

But what I am thinking is, as I posted above, load current.
Load current cannot be transferred sepatately.
What I mean is; it's not something like
'magnetization is one thing and load carrying is another.'
The same process, the magnetization, should be involved in energy transfer, even for the load current.

Originally, what I have thought was
2% copper loss x (3/4) = 1.5% improvement in motor efficiency,
(in case if we are using the same magnetization value, say 1.5T.)

And Actually, in an induction motor, motor efficiency is about 75~90%.
Therefore, loss components cover about 10~25%.
And copper loss at stator is about 30~40% of the whole loss.

So, we can save a lot of energy using high permeability materials.

But still question remains. How much energy?
In an induction motor, is it
3/4 of whole copper loss induced by total current or
3/4 of copper loss by so called magnetizing current?

15. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
If you for instance consider an electric train, big losses are caused by braking.

And mechanical friction between wheels/iron tracks.

Old Tram cars actually used Rheostat systems built on the car roof.

Modern trains (locomotive etc.) use motor/generator combination, or electronic converters.

Using expensive high-permability materials would increase system cost considerably.

For the Euro Tunnel (for instance), very powerful locomotives are used, rated at 7MW.

So if you save just one % from 7MW, that's 70 kW already.

The system is nuclear powered, so it won't really make a difference to use 70 kW more or less, even if it might seem a lot of electric power.

16. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Let's think differently.
In a transformer,
if there is no load in a secondary side,
then there is only magnetizing current in the primary.
However, if there is a load,
then more current will flow through the primary windings.

My question is how does more current in the primary winding affect magnetization of the stator core.

Since there is a more current, more magnetization?

But this might not be the case???
Because many tranformer guyes treat iron loss (loss induced by magnetization) as a constant regardless of loading.
They treat transformers as a constant magnetic flux density machine.

17. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
Thanks takao21203
But as a materials engineers, I need to stick to the efficiency of materials, regardless of how small it is. That is a goal of magnetic materials engineers.

And what I want to know is
how can we achieve that goal?

But I found that, as you see in our discussion, it is not clear at all.

I know low core loss is important, I mean everybody knows about it.
The meaning of core loss is straight forward.

However, meaning of permeability(magnetization) is much more complex.

And I would like to clearfy it.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Simply assuming high permeability materials as the key issue is an oversimplification. one needs to keep in mind the important issue of the air gap reluctance.

Induction machine manufacturers also use magnetic wedges in the stator winding slots to reduce magnetizing current.

It is common practice in machine analysis to distinguish load & magnetizing currents.

19. ### jinksung Thread Starter New Member

Jul 20, 2012
22
0
If we assume the same geometry, is the air gap a constant rather than a variable?
- Same magnetic field in stator with different current flows.
- Same air gap.
- Same toque
So, with different current flows at the stator windings, we may have same torque.
Am I right? I mean no load case.

Last edited: Jul 23, 2012
20. ### takao21203 AAC Fanatic!

Apr 28, 2012
3,785
502
I have uploaded above mentioned document to my web domain (since it is 11MB too large for attachment).

Starting at page 102, there are explanations, formulae and definitions covering the topic. They cover magnetic cores however the topology is very similar to electric motors.

http://pic.hitechworld.org/data/siemens_matsushita_coils.pdf

I hope it can be of use to you.