Motor Winding with Superconductor

Motors consist of winding on a stucture that is free to rotate within a magnetic field,

There are several sources of power loss firstly in generating the magnetic field, if super conductors are used for this apart from refrigeration power the losses fall to zero.

Then we come to the rotating part if super conductors are to be used here there are two problems those of refrigeration and the poor performance of super conductors in varying magnetic field also there are the losses involved in feeding the current into the moving conductors.

Some super conducting motors have been built but I don't know if any are a commercial sucess

Shyprum,

How refrigeration does comes in picture. Poster is just talking about the superconductor not of how it is made.

Cos θ= resistance/Impedance,

And if Resistance is Zero, Cos θ is Zero.

And input power is always zero if Cos θ is Zero.

May be his assessment is fair, still I am not sure.

I am puzzled by all this talk of power factors if the motor is spinning quite freely with no mechanical load the phase angle is about 90° and the power factor zero but as soon as a mechanical load is put on the shaft the phase angle moves towards zero and the power factor towards 1 .

Refrigeration is relevent to supercondutors!

I think the poster is confusing I^2R losses with actual power generated...

Poster, the actual power generated does not depend on the amount of losses in the generator

I think the poster is also confusing a load's power factor with a generator's output...

John.

For me nothing confusing, but I don't know answer.

When poster is talking about motor, how does generator comes in light.

Who is confused?

Mark,

motor / generator = same thing, in this analysis...

Fine lets accepted,

But I don't see anything confusing. I heard that motors are made with super conductors,

But there must be some answer to the poster's question. Just think of how one can calculate the power factor for a super motor. Every thing is ideal.

You are right as you load motor, power factor run toward unity, and this is because of denominator (impedance) decreases.

What you said is based on the common sense. But I am thinking of mathematical answer.

Since numerator is zero, how power factor can approaches to unity. Even in case of zero value for denominator (impedance) power factor can be 0/0, = in-determinant,

Motors may be made with supper conductor. But question is "what is replacing zero value of numerator"

Free running motor will run at a power factor of 0, and that's fine, But what about loaded motor? Either zero or in-determinant,

when the power factor is zero it means that no power is adsorbed from the power supply, now when some load is put on the motor shaft power is adsorbed, you dont need resistence in the windings it is only a nuisense.

ya you are correct resistance means not only mention the conductors ohmic losses, its also include load and miss allignment of shafts....no one can make a moter with 100% perfect design... so what there is some miss alignment always present in the moor it will produce resistance and give power factor...

So you mean something like misalignment etc is essential to get the power factor.

Ok, fine there will be certain losses and that why you will get power factor.

Now just consider a super bearings supper magnetic materials and everything is super.

So can we get some power factor now? If yes then how? I guess this for what poster is looking for.

The power factor isn't relevant here I think...

What is relevant is that with an ideal motor, i.e. no losses, friction etc... the rotor will follow the rotating magnetic field perfectly and in synch with the rotating field, so no power is consumed as there are no losses.

As soon as a load is put on the motor the rotor will lag the rotating magnetic field to use the energy stored in the field to turn the output shaft... This means the rotor is lagging the rotating field by a bit more... Increase the load and the rotor will lag some more... Causing the apparent power factor to change...

Eventually the load on the shaft will increase to the point where the rotor can't keep up with the rotating field and will stall, it will over load the motor...

Whether the motor has additional losses is similar to having a load on the output shaft... Whether the motor is superconducting or whatever makes no difference... Superconducting coils will only reduce the losses a bit, there will still be friction and core losses to overcome...

That's my take of the question... John.

R-S,

Probably your greatest concern when using SC windings might be very high inrush currents that could likely quench your SCM's SC properties. As far as the mains are concerned, your motor's SC windings represent a perfect short-circuit. Your only salvation will be the winding inductance. Apart from that there is very little to differentiate an AC motor built this way than one build with ordinary copper windings. Now why you'd actually want to build an AC SCM motor is a matter you may wish to discuss with a physician.

-e

Everybody commented in his way and thanks for that, but I request to all please see the following,

My only question is "what is replacing resistance in the equation, which makes a SC motor to perform with output.

It's very simple,

Cos θ= base / hypotenuse

In case of a (resistive +inductive) circuit

Base is =Resistance,

Hypotenuse = (2*3.14*frequency*inductance + resistance)^(1/2)

So now we can say Cos θ= Resistance/(2*3.14*frequency*inductance + resistance)^(1/2)

Now think of Ideal motor, (when no resistive loss is )

Cos θ= 0/ (2*3.14*frequency*inductance+0)^(1/2) = 0

So input power = V*I*Cos θ= V*I*0=0

And if input is Zero Output can't be more then zero.

I am sure there must be some answer for it, but I am not getting it so far.

Rakesh, all the mathematics in the world are not going to help you if you don't understand what the mathematics mean. Your slavish obsession with the resistance term in these equations is evidence enough for that. It's one thing to follow a recipe; but quite another to bake the cake. What you're doing here is equivalent to re-arranging the deck chairs on the Titanic.

Sir, Lets say if you are a motor designer and get an opportunity to design, can you deny mathematics?

I hope no,

BTW I got the answer and I shell share it later with all participated.

Not denying mathematics at all! The point is this: the "what" of the mathematics is of little use if one does not first understand the "why" of it - and the "when" (as in knowing when it applies, and when it does not). If the principles are well understood, the mathematics naturally follow. If the principles are not well understood, no amount of mathematics will help you. If it's the forest you wish to understand, does it help to fixate on one tree? Folks here have been trying to show you the forest, but you've missed their point every time and persist in banging your head against the same tree. If you're going to learn how to bake, Rakesh, you've got to get your nose out of that recipe book.

GUD 2 see such a amazing thinking....

there r some more pbms, would arise further as i read abt that matter............