Induction Generator at no load

[b.shahvir]

Hi all,

Assume a standalone synchronous generator feeding power to a single induction machine which is made to run as induction generator at super-synchronous speed without any external electrical load connected to the machine busbars.

What will be the behavior of the induction machine under these conditions at super synchronous speeds? If zero active power flows into the busbars (no load), will the induction generator rotor current also be zero? (transformer action). But how is this possible, especially since the rotor conductors are being continuously cut by the stator magnetic field above synchronous speed?! please explain

Thanks,
Shahvir

 

[anorlunda]

Your question is self contradictory. You say that the syncronous generator is "feeding power" but later you say no load and zero power. If there is no friction and no load, then why would the induction machine run at supersynchronous speed? Please specify your question better.

 

[b.shahvir]

Initially I have considered the induction m/c to be running as induction motor. Then slowly the shaft speed of induction motor is increased to super-synchronous speed. The induction motor now acts as a generator but the only problem is there is no electrical load connected to the finite busbars for the induction generator to supply active power to. If zero active power flows into the busbars, then logically the induction generator rotor current must also be zero due to transformer theory. But this cannot be possible, especially since the rotor conductors are being continuously cut by the stator magnetic field above synchronous speed. Hence rotor current has to be circulating in the rotor conductors even at no load, herein lies the contradiction & my doubt.

 

[anorlunda]

If zero active power flows into the busbars, then logically the induction generator rotor current must also be zero

Do you mean zero instantaneously, or zero averaged over a whole AC cycle?

You could also consider an induction machine running with exactly zero slip. Because of piecewise linearity, the parameters at +epsilon slip and -epsilon slip should be arbitrarily close to those at zero slip. Motor/generator, at zero slip they are the same thing.

Why do you pose the question with a synchronous generator connected. Why not an ideal AC voltage source?

 

[b.shahvir]

An ideal AC voltage source is infinite grid. On infinite grid there is always a load for the induction generator active power to sink into. But in the above case of standalone sync generator on finite busbars, there is no electrical load connected. In such a condition how will the ind generator behave at super-synchronous speeds. zero active power over a whole cycle (anyways it doesn't matter as there is no electrical load on the busbars)

 

[anorlunda]

You are still self contradictory. Why would the induction generator run supersynchrounously if there is no power? If it does run supersynchronously, why won't power flow between the IM and the other source or other generator?

You do realize that no load in the middle is needed. Power can flow from one machine to another in either direction. One acts as a generator and the other as a motor. The motor one is the load.

 

[b.shahvir]

You do realize that no load in the middle is needed. Power can flow from one machine to another in either direction. One acts as a generator and the other as a motor. The motor one is the load.

This would have been the case if both machines were sync gens as they are not dependent on each other to create magnetizing current, but not so in case of ind gen. the stator of ind gen (slave m/c) depends on sync gen (boss m/c) to magnetize it's stator coils. The ind gen is always slave to the sync gen & cannot drive the sync gen into motoring mode

 

[anorlunda]

The ind gen is always slave to the sync gen & cannot drive the sync gen into motoring mode

That's tricky. It also ignores reactive power. Suppose I put mechanical energy into an IG and excitation into a SG and connect the two, what would you expect to happen? The energy has to go somewhere. Either they run as generator-motor or the speeds run away until something breaks.

You didn't answer my earlier question about zero current. Instantaneously zero or average over an AC cycle zero?

 

[b.shahvir]

That's tricky. It also ignores reactive power. Suppose I put mechanical energy into an IG and excitation into a SG and connect the two, what would you expect to happen? The energy has to go somewhere. Either they run as generator-motor or the speeds run away until something breaks.

That is precisely my doubt, but the above is true only in case of 2 SG in parallel. The IG is a slave m/c as it depends on SG to magnetize itself. So it cannot generate it's own reactive power to magnetize the SG & push it into motoring mode

You didn't answer my earlier question about zero current. Instantaneously zero or average over an AC cycle zero?

both

 

[anorlunda]

If current is zero instantaneously, then it is the same as an open circuit; meaning no connection at all. Your question is self-contradictory.

 

[b.shahvir]

If current is zero instantaneously, then it is the same as an open circuit; meaning no connection at all. Your question is self-contradictory.

It will act like an open circuit as there is no external electrical load connected to the finite bus & so I come back to my initial question

 

[anorlunda]

Assume a standalone synchronous generator feeding power to a single induction machine

It is not feeding power. It is not even connected. You have an isolated IM spinning open circuit.

 

[b.shahvir]

It is not feeding power. It is not even connected. You have an isolated IM spinning open circuit.

From the standpoint of SG yes! But i think you have not yet understood the concept. An IG cannot feed power to another SG as it is dependent on the SG for it's magnetic excitation

 

[anorlunda]

But i think you have not yet understood the concept.

I certainly don't understand what you want. I'm out of this thread.

 

[b.shahvir]

I certainly don't understand what you want. I'm out of this thread.

My question is quite direct, you can re-read it if you have not understood