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I just did some calculations, and to me it seems that the induction motor only works when the rotors coils have finite resistance:

I will show that for a single conducting loop, whose current is driven by change of flux due to the exterior magnetic field, the torque on the loop will be zero when the current is zero, and when the voltage is zero as well.

enter image description here

I assume that the stators coils produce a rotating magnetic field. For the sake of the question, I I'm switchting to a rotating frame of reference, that rotates with the same frequency, by which also the rotor rotates. In this system, the only thing that is moving is the rotor (unless it's rotating with the same frequency. I assume that there is a "slip", so in the reference frame of the outer magnetic field, the rotor rotates to the right). I'm aware that a rotor doesn't consist of only one loop, but of many, and that they are arranged in different angles. I still think that for the sake of the question, it's enough to draw one loop.

In the 2nd picture, the voltage induced in the loop BY THE OUTER ROTATING MAGNETIC FIELD ALONE is at max, because the change in magnetic flux reaches an extreme point when the magnetic field lines are perpendicular to plane of the loop. The voltage drives the current in the loop. Because the loop is an inductor itself (it creates a magnetic field as well), the current is behind the voltage by 90 degrees. That means the current is at max in reverse direction in the first picture. However, In the situation of the first picture, there isn't any torque acting on the rotor, because the torque generated by Lorentz forces acting on the charges sums up to zero. In the 2nd picture, current would generate the maximum possible torque. But here (since current is behind voltage by 90°) the current is zero.

Question: Is it wrong to just look at the 2 cases that I described here? Or do we in general need to introduce resistors into the rotors circuit, to change the phase-shift between voltage and current?

Quantumwhisp
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  • Look at ... for the sake of argument, the 30 degree case, where the angle of the coil is 30 degrees behind the LH image (i.e. pointing to 10 o'clock) I think you will find torque. –  Aug 01 '20 at 14:39
  • *All* motor windings have finite resistance, unless you cool them with liquid helium to make the copper superconductive. Is that what you intend to do? – Mike Waters Aug 01 '20 at 15:24
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    @MikeWaters God I hope so. – DKNguyen Aug 01 '20 at 15:50
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    I don't understand the downvote. @MikeWaters ideal inductors are usually modelled without resistance. For somebody who doesn't have mutch clue, my first guess was that coppers resistance of the copper isn't big compared to inductive reactance of the moter. Besides that, the question here deals with the question wether an induction motor with lossless windings can work in principle. – Quantumwhisp Aug 01 '20 at 16:07
  • @BrianDrummond Indeed, there I have torque. But 30 degrees ahead of the LH image (pointing at 2 o'clock). I have torque in the oposite direction, of the same amount (because the current's direction shifts in the moment of the LH image). – Quantumwhisp Aug 01 '20 at 16:45
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    @MikeWaters one of the degrees of freedom in designing induction machines is the rotor resistance. Machines can have rotor resistance introduced deliberately, because it gives higher starting torque, at the cost of more slip and reduced efficiency. I think it's valid to ask what happens if you should go all the way the other way -- even if it's just in theory. – TimWescott Aug 01 '20 at 20:08
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    _"I assume that there is a "slip""_ - why? – Bruce Abbott Aug 01 '20 at 20:30
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    @BruceAbbott because it's the usual mode of operation for an induction motor. The other steady-state possible is that there is no slip at all, but this not an interesting steady state, because nothing happens. – Quantumwhisp Aug 01 '20 at 22:18
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    Slip is proportional to rotor resistance. What do you think happens when the resistance is zero? – Bruce Abbott Aug 01 '20 at 23:29
  • @BruceAbbott that's what I'm trying to find out at the moment. Can you tell me why slip should be solely proportional to resistance? How mutch slip there is is determined by the torques the motor and the load acts on the rotor (and at least the load can vary extremely, independent of the rotor resistance). – Quantumwhisp Aug 02 '20 at 06:45
  • It seems to me that the slip *frequency* would be 0, but the slip *phase* would be whatever is required to produce torque (e.g. to the 10 o'clock position). The 2 o'clock position, as expected, produces torque in the opposite direction because you are using your induction machine as a generator. –  Aug 03 '20 at 17:21

2 Answers2

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Superconductors are impervious to magnetic fields, so if you have a superconducting rotor no fields lines will enter it. (or pass through the apertures of the squirrel cage)

With no field lines in the rotor there will be no torque.

you could possibly get it running synchrounosly as a variable reluctance motor. but you'd need a way to start it, and there wouldn't be much torque available to keep it running.

Jasen Слава Україні
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Using the usual equivalent circuit, the electrical power converted to mechanical power is calculated as the power dissipated in R2x(1-s)/s where R2 is the rotor resistance. That would seem to require a finite rotor resistance. The rotor resistance is quite small in most induction motors, but the other impedances in the equivalent circuit are also quite small, so the rotor resistance is not small in comparison.

If you look at the torque vs speed equation Torque-Speed Equation for Induction Motor, it looks to me that it does not fail if R2 = 0.

You may also want to look at When load increases in rotor of induction motor how does stator draws more current? for more detail about the equivalent circuit.

As commented by @MikeWaters, there is not much reason to pursue this unless you are interested in a superconductor rotor.

Re Comments

Increasing the rotor resistance results in higher slip at a given torque as shown below. Resistance can be increased and decreased by changing the rotor bar size and material and be connecting external resistors through slip rings. The highest slip without external resistance that is commonly offered is about 13% at rated torque. Slip in the area of 1.5 to 2 percent seems to be available with aluminum rotor bars. Copper rotor bars must be able to provide slip a bitbelow 1.5%.

The equivalent circuit is used in motor design, but the design process is a lot more than that. With most motors, more than one equivalent circuit or more branches in the rotor circuit are needed to model changes in the rotor resistance when the slip is high. The simple circuit works only for rotor bars that do nor extend very far into the rotor from the surface. In order to provide higher torque at high slip, the rotor bars are extended more deeply or split into a surface part and a deep part.

If the motor is intended only for use with a VFD, the motor could be designed with big copper rotor bars close to the surface. That seems to be the strategy used for the early model Tesla cars.

An induction motor is not supposed to be able to develop torque without slip, but I don't know what happens as the torque vs. speed curve becomes vertical. That seems to be what will happen as R2 approaches zero.

Calculations

The following shows the results of progressively reducing R2. T1 represents a curve for the equivalent circuit that I believe represents a motor that was actually constructed. I don't remember where I got the data. T2 shows the "normal" effect of reducing rotor resistance, the peak torque remains the same but occurs at a lower slip. T3 shows a slight rise in peak torque but a more drastic reduction in torque on the high-slip side of the peak. T4 and T5 show the peak torque falling with reduced R2. Per unit values of R2 are 1, .7, .23, .14 and.02 for T1, T2, T3, T4 and T5 respectively. The equivalent circuit used omitted the magnetization branch.

enter image description here

  • Your 2nd paragraph is misleading to me: Did you mean to write " "it does not fail if R2 = 0"? In general your answer seems to be that for R2 = 0, there isn't any torque. Do I understand you right at this point? – Quantumwhisp Aug 01 '20 at 17:04
  • Yes, I corrected that. Thank you. –  Aug 01 '20 at 19:14
  • What do you mean by "doesn't fail". The equation still makes sense, but the outcome would be that T= 0. I'm nitpicking on this, because that's still a part of your question which otherwise would be in dissagreement with the previous paragraph. If this is understood, then my question is answered. – Quantumwhisp Aug 01 '20 at 19:20
  • I believe that you get something other than T = 0. I am not sure if the value is reasonable, but I think it is. –  Aug 01 '20 at 19:34
  • The Enumerator will be zero if R2 = 0, that's why I thought that T = 0 as well. This value would line up with what you wrote previously, and with my personal findings. – Quantumwhisp Aug 01 '20 at 19:57
  • Yes. I see that. I was looking at a simplified version in which I may have made an error. I believe that it holds together as R2 approaches zero, but the power factor will get lower. I think that an induction motor with a super conducting rotor may work, but the current may need to be regulated by a variable current source. I am not inclined to spend very much time on this. –  Aug 01 '20 at 20:14
  • Maybe the motor becomes synchronous as R2 goes to zero. S also goes to zero, so the term R2/S becomes 1. I can't believe it would just stop working. Makes no sense. – user57037 Aug 01 '20 at 20:30
  • I don't think that equation is intended to be used in motor design. It is an equation that describes torque for a given motor. It does not consider that the shape of the torque vs slip curve will change as R2 changes. But as you move from a NEMA type D rotor (stall torque is peak torque) to a Type A or Type B (peak torque happens closer to sychronous speed) you get a sense of how the curve changes. – user57037 Aug 01 '20 at 20:33
  • https://electrical-engineering-portal.com/wp-content/uploads/typical-motor-speed-torque-curves.gif – user57037 Aug 01 '20 at 20:36
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    @mkeith you're right, it won't stop working, instead times Tc become infinite. Superconductor coils are perfect EM shields in that their enclosed net flux cannot change. A superconductor ring should behave like a PM magnet: synchronous motor with torque and lag-angle, and zero slip. But also I suspect that the starting torque will be zero, and the zero-ohm rotor will never start up (it might just vibrate back and forth, a net zero torque over time.) – wbeaty Aug 01 '20 at 21:31
  • @wbeaty might need a starter resistor with a centrifugal switch. LOL. – user57037 Aug 01 '20 at 23:57