Rotor EMF in squirrel cage induction motor

[cnh1995]

In a 3 phase induction motor,when stator rmf rotates, it cuts the rotor bars and induces an emf in them. Is this emf sinusoidal? I mean shouldn't it be a square wave since the rotor "bars" are being cut by the flux, like in case of concentrated stator windings of a synchronous generator? If the rotor emf is sinusoidal, how is the design of the bars?

 

[Babadag]

In my opinion, both magnetic fluxes: of stator and rotor rotate at the same speed-in a single wave.

This flux speed it is then stator flux speed n=f*60/p[rpm] where 2*p it is number of poles and f=supply frequency[Hz].

The EMF induced in a rotor bar-if we neglect harmonics-is of sinusoidal form following the magnetic flux form.

The number of phases of the rotor is then equal number of bars-and the phase to phase angle will

be 180/p/Zrotor where Z it is number of rotor bars.

The bar current is calculated starting from EMF and divided by rotor resistance -including skin effect

and proximity effect and the reactance-self reactance and mutual reactance between bar to bar.

Usually, each manufacturer has his own proceeding.

 

[Tero]

If I can add something to the discussion... perhaps correcting couble things. =)

If we are talking about steady state operation of traditiona induction machine (IM). First, the magnetic field produced by the stator is not sinusoidal unless you have toothless stator which are quite rare in "traditional" IM designs. The magnetic field is more like "stepped" form of sinusoidal (imagine sinusoidal wave and then make it blocky and stepped). This effect can be minimized a little bit with two level winding design, but still it is always full of slot harmonics but naturally there is fundamental wave which is sinusoidal.

In steady state when the stators rotating magnetic field creates a fundamental sinusoidalish current flow into the rotor bars. If we have i.e 50 Hz network voltage the current in stator is 50 Hz, but the rotor current is not same frequency as the network. This rotors own frequency is called slip frequency which is the result from the fact that IM rotor searches the steady state situation where flux lines passes the rotor bars in such maner that current produced in bars and in the end force produced by the whole squirrel cage matches the friction losses of the motor and possible load on the shaft. This slip depends on the design of the motor but for example 50Hz network two pole motors synchronous speed would be 3000rpm thus slip might be something like 50rpm, so motor would rotate at speed 2950 rpm. This 1,66% slip in frequency is little under 1 Hz and that is the frequency of the rotor. Funny fact is also that the magnetic field of the rotor rotates opposite direction vs the stator's field ie. if stator field rotates clockwise at speed of 50 Hz then in above example the rotor field would rotate counterclockwise at speed of little under 1 Hz. =)

Finally the phases in the rotor bars are not the amount of the bars as they don't exhibit phase angle between each bar. There is magnitude difference but still in traditional 3 phase motor the rotor has only 3 phases too. Naturally now we have not discussed about the effects of harmonics which again can create phase changes in the currents flowing in bars creating so called harmonic torques which can be observed as one of the many sources of vibration in motors operation. There are strict rule set how to select rotor bar count vs stator slot count. Rules can be found from general IM design books.

 

[cnh1995]

Funny fact is also that the magnetic field of the rotor rotates opposite direction vs the stator's field ie. if stator field rotates clockwise at speed of 50 Hz then in above example the rotor field would rotate counterclockwise at speed of little under 1 Hz. =)

I don't understand what you mean by this. As per my knowledge, rotor field and stator field both rotate in the same direction and rotor field is slower than the stator field (slip). If they were rotating in opposite directions, their speeds would add and we'd get rotor current of a frequency higher than the stator supply frequency. If the green part in your above post (colour changed by me) were true, the rotor current will have a frequency little less than 51 Hz.

 

[Salvador]

first of all the secondary rotor current is minus the losses also sinusoidal as its like the secondary winding of a transformer , the waveform follows the primary.

Secondly , @Tero why would the rotors magnetic field rotate counterclockwise? The whole point of an asynchronous aka induction motor atleast how I understand it is that the stator has a rotating magnetic field which induces current in the rotor windings , that current in turn creates a magnetic field which opposes the stator field , but since the stator field is moving the rotor currents are constantly changing and because they try to oppose those changes the rotor starts to turn as to follow the stator field so that the changes in it's current would be minimal.In other words the rotor's field tries to lock itself with the stator field as much as it can.When the load gets bigger the rotor slows and this creates larger currents in it's windings so it has more power to follow the stator field once again.

Nowhere have I heard that it's field runs counterclockwise to the stators' field.

A good quote from wikipedia about induction motors

The rotating magnetic flux induces currents in the windings of the rotor;[23] in a manner similar to currents induced in a transformer's secondary winding(s). The currents in the rotor windings in turn create magnetic fields in the rotor that react against the stator field. Due to Lenz's Law, the direction of the magnetic field created will be such as to oppose the change in current through the rotor windings. The cause of induced current in the rotor windings is the rotating stator magnetic field, so to oppose the change in rotor-winding currents the rotor will start to rotate in the direction of the rotating stator magnetic field. The rotor accelerates until the magnitude of induced rotor current and torque balances the applied load.

https://en.wikipedia.org/wiki/Induction_motor

 

[Babadag]

Let's say the rotating magnetic field is a ring of permanent magnet poles [N-S]. Let's say these poles do not rotate but only rotor is rotating with the difference of velocities. That means it rotates backward [counter clock wise] with the speed of v=slip*2*pi()*f/p*Rgap [Rgap=radius of the gap; p=number of pole pairs].

When a rotor bar of speed v passes a pole of B field an emf is produced E=B*leng of bar*v and since there are 2p poles [p pair of N-S] an emf wave will rotate with v velocity of the rotor.

A current will circulate through the bar but the current wave is delayed with the impedance angle and will produce a connected with rotor field wave.

This field wave rotates with the f*(1-s)*60/p rpm.

So what I said the both field waves rotate with the same velocity is wrong.

Since each rotor bar will pass the same pole at different time a different angle has to be applied for each current in each bar. You may consider then [if you want it] each bar as another phase [or not]. However, as you could see, no 3 phases are involved here but only number of poles. Of course, we give balanced 3 phase current (distributed in time) to balanced 3 phase winding (distributed in space) to produce the stator field.

 

[jim hardy]

In a 3 phase induction motor,when stator rmf rotates, it cuts the rotor bars and induces an emf in them. Is this emf sinusoidal? I mean shouldn't it be a square wave since the rotor "bars" are being cut by the flux, like in case of concentrated stator windings of a synchronous generator? If the rotor emf is sinusoidal, how is the design of the bars?

I owe you an apology, i never finished up your similar thread a couple months ago.
I was working up a step by step approach
Freeze frame your thinking for a moment.

Meantime to your question
imagine DC in the stator to make flux stand still
and rotor turning at slip speed
now it's just the familiar loop rotating in a stationary field. You already know it's a sinusoid.
As Babadaq said in post 6, how you make flux rotate is immaterial, it's relative motion between flux and rotor bars that makes torque.

At some instant we have flux stator flux B in blue, and rotor is rotating relative to it CCW .

Velocity of each bar relative to B changes from parallel to perpendicular back to parallel twice per revolution
is it clear that QVcrossB varies sinusoidally ?

Re your old thread....

I was starting to calculate current in each bar and resultant torque
assigned dimensions that included pi so as to make the expressions simple as possible and produce round numbers.
here's how far i got

Let's start here

Blue stator flux arrow rotates CCW at 3600 RPM, 60 hz

and for my thought experiment i'll give rotor these dimensions:

12.532 inch (1/pi meter) diameter by 19.685 inches (1/2 meter) long

Squirrel cage rotates CCW at 3540 RPM, 59 hz, slip = 60 RPM = 1.67%

i'll ride along with rotor and watch flux arrow pull ahead of me one rotation per second

motion of flux relative to squirrel cage is 1 revolution per second, counter clockwise

And i'll set flux density at 1 Tesla , which over cross sectional area of the rotor( L X W = 0.5 X 1/pi) is 1/2pi Webers.

Any pair of rotor bars forms a loop . The flux encircled by that loop varies siusoidally from -1/2pi to +1/2pi webers as B sweeps ahead of the rotor at 1 hz

so flux phi in red rotor loop = 1/2pi X sin(2pi t ), t = time in seconds

so flux = 1/2pi X sin(2pi t )

and voltage induced at any instant should be

2pi/2pi picos(2pit) = 1 volt peak at 1 hz

alternately, in any bar

e=blvsin(angle between velocity and B)

l was given as 1/2 meter,

radius = 1/2pi, so at 1 rps velocity = 2pir = 2pi/2pi = 1 meter /sec

each red bar has blv = 1T X 1/2 meter X 1m/sec = 1/2 volt, two bars make a loop agreeing with one volt by flux linkage method.

red bars e=blvsin90, =1 volt, top bar's current will flow out of page bottom one into page as shown by polarity sign

orange bars e=blvsin54 = 0.81

yellow bars e=blvsin18 = 0.31

green bars e=blvsin-18 = -0.31

brown bars e=blvsin-54 = -0.81

Next step is to estimate Land R of rotor turns and figure bar current, then torque on each set of bars
Maybe this thread will spur me to finish.
With much chagrin,

old jim

 

[cnh1995]

is it clear that QVcrossB varies sinusoidally ?

Meantime to your question
imagine DC in the stator to make flux stand still
and rotor turning at slip speed
now it's just the familiar loop rotating in a stationary field. You already know it's a sinusoid.
How you make flux rotate is immaterial, it's relative motion of flux and rotor bars that makes torque.

Thanks a lot! That is really helpful.

 

[Babadag]

I have to correct the "correction”. I was wrong in second place not in the first.

Explaining the induction phenomenon by locking the stator field is correct about the magnitude of the emf but it is not good explaining the rotation sense.

Let's say now the rotating magnetic field is a ring of permanent magnet poles [N-S].Let's say these poles rotate forward with the slip speed and the rotor is locked .That means the stator field rotates forward with the speed of v=slip*2*pi()*f/p*Rgap.

When a pole of B field and of speed v passes a rotor bar an emf is produced E=B*leng of bar*v.

and since there are 2p poles[p pair of N-S] an emf wave will rotate with v

velocity of the field B with respect the locked rotor.

A current-and a field generated by this- will circulate through the bar but the current wave is delayed with the impedance angle and will produce a field wave rotating with the same v with respect the rotor.

Now, since the rotor rotates with (vsynchron-v) the actual rotor field speed will be vsynchron-v+v.

That means in turn the rotor magnetic field and the stator magnetic field will rotate synchronously and the rotor field lag the stator field with a certain angle.

I have to apologize for my mistake!

 

[jim hardy]

That means in turn the rotor magnetic field and the stator magnetic field will rotate synchronously and the rotor field lag the stator field with a certain angle.

Thanks Mr Babadag
right there's the key to understanding the motor ...

think very simply for a moment
In my primitive picture up above
i aligned flux with red bars
were there any current in the red bars at that instant , red bar's flux would be perpendicular to blue flux and there'd be torque.
If red bars have zero resistance then current in them lags voltage in them by 90 degrees and is zero at instant shown, so there's no torque from red bars.
If red bars have some resistance then current lags by less than 90 degrees so there is torque from the red bars.

That's why adding resistance in the rotor gives more starting torque
not to mention it reduces starting current because the rotor at standstill looks less like a shorted secondary.
High starting current creates excess heat in the resistance of both stator and rotor windings .

Wound rotor induction motors bring the field outside via sliprings so the user can adjust rotor resistance. On something like a crane or elevator where you need high starting torque that's handy, you can start it as a high torque motor to get the load moving and after it's in motion decrease the resistance to make it low slip motor.
Not to mention the heat from rotor current is dissipated in the external field resistors not the rotor bars
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-13/wound-rotor-induction-motors/
https://www.copper.org/environment/...motor-rotor/pdf/Squirrel_Cage_Rotor_Slots.pdf

..................

sigh, sometimes i think I'm crazy.
Was so curious to tinker with a wound rotor motor i bid on one at government surplus sale.
To my astonishment i got it for $85.

My 600 lb gorilla

Oh well i can visit son in Jacksonville when i go pick it up.

 

[Babadag]

Thank you, very much Jim!

 

[jim hardy]

Thank you, very much Jim!

no, thanks go to you for that succinct statement about the fluxes.

I have to envision every term of an equation before i am comfortable with it.

This scary looking equation from http://www.electrical4u.com/torque-equation-of-three-phase-induction-motor/
(subscript 2 meaning in the rotor )

agrees with intuition
at any slip s
As rotor resistance approaches zero R₂ in numerator drives the fraction to zero , hence low starting torque of low slip motors with their low resistance rotors.
As rotor resistance approaches infinity R₂² in denominator takes over again driving the fraction to zero ,
and in between, s terms in both numerator and denominator shape the speed-torque curve.

That formula in a spreadsheet where you can vary rotor R and X and plot against s with just a keystroke would be fun.

...............

In my thought experiment described in post 7 it looks as if my torques(dipoles?) from brown, orange, yellow and green bars are cancelling one another by symmetry
but I've not yet ruled out algebra mistakes on my part
and Windows 10 hid my text file because it's in a pictures library
when i get all that straightened out i'll be back. (A friend from PF made a bona-fide Windows 7 disk appear )

but next several days are spoken for, got a concrete truck coming in the morning.

Thanks guys these exercises help me a lot.

Mr Babadag I've always admired your math skill. Thanks againold jim