No Load Speed of Induction Motor

Wikipedia can explain it to anyone!

thx!!!!^^

Is that your interpretation or the one from Wiki?

Spare me the insults.

I would spare you some humor if it does not brush off!

Our friend here is up to his ears in homework!

You got that right! What surprises me is that the answers are so intuitive.

It's intuitive only if you understand that an induction motor uses two magnets to produce torque. You have to know how a magnet works and the difference between force and torque. The tricky part that does take a near epiphany revelation is how the rotor magnet is formed and modulated. This introduces the cryptic term of slip.

That should be enough bread crumbs to complete the assignment or for Hansel to find the oven in the gingerbread house.

It's the rotatey-most of the carrier throom all loony-loony and the fallolloper all slippy-sliders and deep joy in the turn-it.

The movement (rotation) of an induction motor rotor is created by the rotating rotor magnetic field lagging the rotating stator (winding coils) magnetic field. - Slip is the speed difference between the two rotating fields usually expressed in percent.

As load is applied to the rotor the rotor speed decreases creating a higher difference in speed of rotation between the stator field and the rotor field. (Higher slip factor.)

If the motor is running with no load attached to the rotor slip will be the minimum value allowed by motor design efficiency.

It is one of those things that is easy once you "get it".

Imagine two stationary electromagnets with AC powered windings such that their opposite pole ends are pointed toward each other - and that a single/multi coil of wire (a rotor) is placed on an axle between these so that the sides were close to one of the magnetic poles and the whole assembly can spin as in an electric motor.

Over time the poles would be S-N and then N-S with the direction of the AC current and the magnetic flux between these would cut back and forth across the coils of wire and induce a current in the coil - first one way then the other. The coil will sit there trying to move first one way and then the other as the current induced in the coil by the AC magnetic field reacts with the magnetic field that induced it - this is left/right hand rule stuff and in maths is a vector cross product. Physically, in an induction motor, this is the high-current high-slip stall or starting condition.

Mathematically, a pulsing field situation like the above is the same as two magnetic fields rotating in opposite directions - that is why there is no sustained rotation.

Now add another two poles at 90 degrees to the first - and the result is nothing much different from before and still with no rotation. There will be a resultant single pulsing field from the two contributing pole pairs, with a centre line in space between the two sets of poles - and still no sustained movement, as before

NOW THE IMPORTANT BIT. However, add a little extra like including an inductor in line with the field winding for one of the magnetic pairs, or a capacitor across the windings for one of the magnetic pairs, or have more turns in the winding on one of the magnet pairs (which makes it more inductive and is the same as adding an inductor), and this causes the current and hence the magnetic field in of the magnetic pairs to lead or lag the other.

In mathematical and in physical terms this situation is the same as having the average of the two fields stay at a constant strength and rotate (from the leading field pair toward the lagging field pair) - and the result of this is that there is a net dragging motion on the coils in the direction of the apparently rotating field - SO the coil moves continually in one direction instead of pulsing back and forth.

And in relation to the original question - the higher the torque stopping the rotor from turning, the greater the drag needed to keep it turning, and so the greater the current required to get the torque, and the greater the slip to generate the current - no load is only enough slip to cover losses.

The dragging effect relationship is instantaneously like in in a DC generator where motion of the rotor in a magnetic field induces a back emf that causes current and a resultant torque or in a DC motor where current in a rotor in a magnetic field causes torque and the resulting motion induces a back emf.

Did you also write the explanation for the Turbo Encabulator?

Turbo Encabulator - YouTube

Because if you didn't, you could have.

I still don't "get it'.

I beg your pardon if my demonstration will boring you.

The power transferred from stator to rotor [gap power] -see Steinmetz model of an induction motor- it is:

Pg=3*Rrot*Irot^2/slip [for three-phase induction motor].

since Xrot=Xrot0*slip[negligible now] Xrot0=rotor reactance at start.

Irot=Vrot0*slip/Rrot -the voltage induced in rotor is proportional with rotor current frequency that means with the slip. Then:

Pg=3*Vrot0^2/Rrot*slip or:

slip=Pg/3/Vrot0^2*Rrot

Pg=Pmec-rotorLosses No-load=Pmec=0

rotorLosses=copperLosses+mechanicalLosses

If mechanicalLosses=[approx.]constant[ventilation+friction] then Pg>0 and then a small current in rotor has to be and this will provoke a slight copperLosses too.

Briefly Pg is small [but not nil] then

slip[small]=Pg[small]*Rrot/3/Vrot0^2

Rrot=[approx.]constant[slightly depends on slip-due to skin effect- and on temperature-of course]

Because at no load, there is no additional burden for the the motor to drive or carry that will cause it slow down and slip / lag behind!

Sorry vsar slip=(synchronous velocity-rotor velocity)/ synchronous velocity[s=(nsynch-nrot)/nsynch nrot=nsynch*(1-s)].That means small slip= elevated rotor velocity, big slip[as at start s=1, nrot=0] slow rotor.Pmech=0 Pg small, slip small, elevated rotor speed.

7anoter4 -

There is no argument nor disagreement on what you stated! That's the theoretical description or definition of a motor slip, which is also theoretically inverse to it's synchronous speed!

My response to the OP was based on how his question was constructed or posted... it is based more on the natural, physical being, its own mechanical mass that will always be present and can never be overcome by an induction motor..

I'm with you on this, vsar. The question the OP is asking is not for an equation that correlates with the observation. The question is why is the equation true. IMHO TrevorM started down that path but lost his point along the way. A good rewrite of that short essay might help but without knowing the schooling level of the OP it will likely either not enlighten or drone on after making the point.

The one pivotal insight that nobody has yet brought up here is that the rotor is a spinning transformer. The frequency of the signal that gets coupled through this transformer is correlated with the slip. If an outside torque spins the rotor precisely at synchronous speed (the outside force is needed to overcome mechanical and other losses) then the frequency through this transformer is zero. As anyone working with AC electricity should know a transformer will not couple power with a zero frequency AC signal.

i run faster when i am carrying nothing. With a full backpack, i run slower, if at all

I should agree with all opinion: big load small speed for squirrel cage induction motor and for series excitation D.C. motor. But shunt and separate excitation D.C. motors are not sensible to load [in admissible limits].The same a synchronous a.c. motor.

Also the wound rotor induction motor could keep a relative constant velocity by changing the resistance rotor connected with.