Induction motor flux variations

[gilver]

Dear shahvir

You're wright when one moves a shorted coil in a magnetic field (residual in this case) so that it cuts the magnetic field an EMF is generated and a Induction current will flow. Because of this current a counter force (torque) will be produced (braking action as you mentioned). But finally it is the prime mover who has to compensate the counter torque due to induction current or later the load that will be connected. But he has to compensate all the losses in the machine as well. So if the load (current) is to big for the prime mover his rotational speed will decrease and at a certain point the reactive power build up between the machine coils and capacitors will be to small and the emf will drop. (The voltage versus Ireactive current looks like a U/Iload curve of a dc serie generator) because selfexciting of this induction generator depend on the capacitor as well as on the rotational speed. so if one or both becomes to small the induction motor can't build up his reactive power. and the emf will be minimum

best regards
gilver

 

[b.shahvir]

Although the reply is technically correct, to visualize self-excited induction generator generating power just by a capacitor connected in parallel is difficult. I tried finding info on the net but it is presented in such a way as to just accept it blindly!

My query arose because the induction motor basically behaves as a generalized transformer and as such to imagine a self-excited transformer generating it's own power (just by connecting a capacitor in parallel) seems a bit unpalatable, although technically correct.
Also, we must remember that in an induction motor there is no flux cutting (dynamically induced EMF) as discussed earlier. Hence, to imagine a generator which suplies power to a load without flux cutting seems odd.

Thanx &
Shahvir

 

[gilver]

When an induction motor is been driven by a prime mover and nothing connected to the coils. You'll measuring a small voltage on the coils (about 3 v ac). At that instance the machine is working as a synchronous generator with the rotor as rotating field. The rotor can be seen as a permanent magnet due to its residual flux. When one connect capacitors on the coils, the coils work here as as power source and charge the capacitors (reactive power is being exchanged). next the capacitors will discharge and reactive power is delivered to the stator coils. The armature reaction of the stator coil will increase the total flux so the emf increases and the capacitors will be charged more than before etc.. more flux more emf more charge ... This kind of snowball effect is limited due to magnetisation curve and capacitor curve see attachement

The emf is delivering a reactive current to the capacitors and next he is drawing capacitive current from the capacitors to build and maintain his magnetic field

But if the Capacitor or rotational speed are to small you can't start this system

When you start such a system you will see the progressive rising of the alternator voltage

 

[b.shahvir]

Thanx very much

I'll do my own little analysis and get back if doubts persist.


Shahvir

 

[b.shahvir]

Ok, so now I’ve come to understand that a self-excited induction generator initially acts as an AC synchronous generator as the rotor and stator cores are permanently magnetized with N and S poles. The rotating permanent magnet N and S rotor poles now cut the stator conductors (dynamically induced EMF) inducing AC voltage in them, which in turn charges the capacitor and the cycle continues with the capacitor providing reactive current to the stator to sustain magnetic flux in the machine. Plz correct me if I’m wrong.

Now, let us concentrate on the part wherein the stator of an Induction Generator has already been excited by reactive amp-turns. This reactive amp-turns can be supplied by the local grid or parallel capacitor.

I want to understand the following pertaining the same;

1) In case of a self-excited Induction Generator, on what factors does the generator frequency and speed of synchronously revolving air-gap flux depend on? Does it depend on the speed of rotor or the LC time constant of the oscillatory circuit formed by the external capacitor and the stator winding inductance?

2) Assume an Induction Motor is connected to a finite grid and running above synchronous speed. The motor now acts as an Induction Generator. In this case, is the generated voltage or power reflected back to the finite grid by the induction generator due to ‘transformer action’ (statically induced EMF) or due to ‘generator action’ (flux cutting, dynamically induced EMF)?
This question is analogous to my earlier query on induced EMF in stator winding…. only thing now I’ve considered an Induction Motor acting as an Induction Generator connected to a 'finite grid system'!

3) Since I’ve have assumed a ‘finite grid system’, will the Induction Generator increase grid frequency as the rotor is spun ‘above’ synchronous speed?

I did be very grateful for a suitable reply to my questions.

Thanks &
Shahvir

 

[b.shahvir]

Ok, so now I’ve come to understand that a self-excited induction generator initially acts as an AC synchronous generator as the rotor and stator cores are permanently magnetized with N and S poles. The rotating permanent magnet N and S rotor poles now cut the stator conductors (dynamically induced EMF) inducing AC voltage in them, which in turn charges the capacitor and the cycle continues with the capacitor providing reactive current to the stator to sustain magnetic flux in the machine. Plz correct me if I’m wrong.

Now, let us concentrate on the part wherein the stator of an Induction Generator has already been excited by reactive amp-turns. This reactive amp-turns can be supplied by the local grid or parallel capacitor.

I want to understand the following pertaining the same;

1) In case of a self-excited Induction Generator, on what factors does the generator frequency and speed of synchronously revolving air-gap flux depend on? Does it depend on the speed of rotor or the LC time constant of the oscillatory circuit formed by the external capacitor and the stator winding inductance?

2) Assume an Induction Motor is connected to a finite grid and running above synchronous speed. The motor now acts as an Induction Generator. In this case, is the generated voltage or power reflected back to the finite grid by the induction generator due to ‘transformer action’ (statically induced EMF) or due to ‘generator action’ (flux cutting, dynamically induced EMF)?
This question is analogous to my earlier query on induced EMF in stator winding…. only thing now I’ve considered an Induction Motor acting as an Induction Generator connected to a 'finite grid system'!

3) Since I’ve have assumed a ‘finite grid system’, will the Induction Generator increase grid frequency as the rotor is spun ‘above’ synchronous speed?

I did be very grateful for a suitable reply to my questions.

Thanks &
Shahvir

Can someone please guide me in this regard? I did be very much grateful.

Thanks & regards,
Shahvir

 

[gilver]

Daer shahvir

when an induction generator is spun above his synchronous frequency, the IG will draw reactive power from the grid , just as it will do as motor but now it'll be capacitive, and the IG will injected active power in the grid the amount is depending on machine constants and slip. the frequency of this currents will be grid frequency. Because the IG has to build up his field, he needs to draw magnetising current from the grid (grid frequency). So can an IG change grid frequency, not because he generates a current/voltage with higher frequency, because it doesn't. Maybe because the Ig ask for to much reactive power, but if the grid can't deliver it the IG will fall flat. So no matter what the magnetude of the negative slip may be, the primary current will have the frequency of that correspions to the speed of the rotating magnetic field (stator) and this is determined by the frequency of the grid. Remenber an IG can only deliver active power and will only draw reactive power. The grid has to deliver all the reactive power to all his loads including the reactive power to the IG

in case of the self excited IG once started The L C combination detemine the frequency, and if the speed is increased the frequency remaine quasi constant. So slip will occur between rotor and stator rotating field, rotor current will flow and the residual flux will increase and so on (see motor action)



best regards

 

[b.shahvir]

Dear Gilver,

Thanx for reply, very much grateful. The 1st part is understood by me, in that, as per your explanation although the rotor spins above synchronous speed, the grid frequency will remain constant. This is because the voltage induced in stator of IG depends on slip of IG rotor and revolving mahnetic field. Plz correct me if I'm wrong!

in case of the self excited IG once started The L C combination detemine the frequency, and if the speed is increased the frequency remaine quasi constant. So slip will occur between rotor and stator rotating field, rotor current will flow and the residual flux will increase and so on (see motor action)
best regards

If the frequency of IG depends on LC combination due to external capacitor and IG stator inductance, then how can it be tuned at 50 or 60HZ? I had come across an article in which it was mentioned that frequency of self excited IG reaches as high as 183HZ!

Then how can frequency of IG be adjusted to give output frequency of 50HZ? Also, once external capacitor starts developing reactive current and maintains a magnetic field, what is relation beetween speed of revolving stator magnetic field as compared to rotor speed of IG? as now, the effect of residual magnetic field becomes ineffective. Finally, is there a definite formula for calculating frequency of self excited IG? I hope you understand what I'm trying to convey!

Thanx very much &
Shahvir

 

[b.shahvir]

I did be grateful if someone can guide me!

Kind regards,
Shahvir