Concept of slip speed in Self Excited Induction Generator (SEIG)

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Discussion Overview

The discussion revolves around the concept of slip speed in Self Excited Induction Generators (SEIGs), particularly focusing on how these generators develop output power when the rotor speed increases and the relationship between rotor speed and stator frequency. Participants explore the implications of negative slip and the role of external capacitors in maintaining excitation flux.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Shahvir questions how a SEIG can develop output power when the rotor speed increases to the point where the slip between the rotor and air-gap flux becomes zero, suggesting that this would imply zero generated output power.
  • Uart argues that the excitation air-gap flux in a SEIG rotates at a frequency determined by the rotor current and the rotor speed, indicating that the slip must be negative for the generator to produce net electrical power.
  • Shahvir seeks clarification on the physics of negative slip in SEIGs, requesting a detailed explanation of the stages a SEIG goes through from zero speed to higher speeds, and how to compute the output frequency of a SEIG.
  • Shahvir expresses confusion about the relationship between rotor speed, stator frequency, and the role of capacitors in sourcing excitation amps, noting that this makes the concept of slip less clear.
  • Another participant provides a reference to a paper that may help clarify the topic.

Areas of Agreement / Disagreement

Participants exhibit differing views on the relationship between rotor speed and air-gap flux in SEIGs, with some asserting that negative slip is necessary for power generation while others question the implications of zero slip. The discussion remains unresolved with multiple competing views present.

Contextual Notes

Participants highlight the dependence of the excitation process on rotor speed and the role of capacitors, which introduces complexity in understanding slip in SEIGs. There are unresolved questions regarding the computation of output frequency and the stages of operation for SEIGs.

b.shahvir
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Hi Guys, :smile:

In case of an Induction Generator (IG) connected to grid, the rotor runs at super synchronous (above synchronous) speed exhibiting -ve slip. This is possible since the grid frequency is fixed at say, 50Hz. However, in case of a Self Excited Induction Generator (SEIG), it starts off as a synchronous generator at low speeds due to residual magnetism. As the rotor speed eventually increases, the external capacitor injects sufficient reactive power into the stator to maintain air-gap excitation flux.

But the stator current frequency of SEIG is determined by rotor speed and hence the excitation air-gap flux would rotate at the same speed as that of the spinning rotor (if a 3 phase SEIG is considered). In this case, how will the SEIG develop output power if slip between rotor and air-gap flux is zero?.. as at every step the increment in rotor speed will cause a proportional increment in stator frequency and at each step, the air-gap excitation flux will be in synchronism with the rotor speed resultiing in zero slip. Zero slip would mean zero generated output power! :confused:

Can someone pls. clear the above concept for me? Also can this concept be applied to a single phase SEIG? Pls. help :frown:

Kind regards,
Shahvir
 
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b.shahvir said:
But the stator current frequency of SEIG is determined by rotor speed and hence the excitation air-gap flux would rotate at the same speed as that of the spinning rotor (if a 3 phase SEIG is considered).

I don't think that's correct shahvir. That would apply if the rotor currects were DC but of course they're not. With 3-phase AC rotor current the excitation air-gap flux will rotate at the frequency of the rotor current (divide pole pairs) plus that of the spinning rotor.

Since the slip must be negative in order to produce net electrical power then the rotor currents will actually have 3-pase currents which produce a rotating field (relative to the rotor) in the opposite direction to that of the rotors physical motion. So the air-gap flux will rotate at a frequency which is less than that of the rotors physical motion.

Hope that helps.
 
uart said:
Since the slip must be negative in order to produce net electrical power then the rotor currents will actually have 3-pase currents which produce a rotating field (relative to the rotor) in the opposite direction to that of the rotors physical motion. So the air-gap flux will rotate at a frequency which is less than that of the rotors physical motion.

Hope that helps.

Thanks Uart, although I've somewhat got the idea of what you are trying to say, but the above part is a bit unclear to me...especially the the rotor field in opposition to stator field. So, will it be possible for you to explain the physics of -ve slip in a SEIG in a bit more detail? The explanation might include the stages which a SEIG goes thru. from zero speed to, say, any desired speed. Also, since SEIG starts off as a synchronous generator due to residual magnetism, I consider rotor to be 2 pole.

How does one compute output frequency of a SEIG?

Thanks very much & Kind regards,
Shahvir
 
Last edited:
Hey b.shahvir,

This "www.naun.org/journals/circuitssystemssignal/cssp-62.pdf"[/URL] paper might be of some help.

Regards
 
Last edited by a moderator:
My interest in Induction Generators started off out of curiosity of Seimens make electric traction system used in my city which employ inverter fed 3 ph induction motors also used for regenerative braking. This curiosity eventually led me to an interest in SEIGs as they are independent source of power supply.

However, slip speed is something easily exhibited by grid connected IGs (as grid frequency is fixed and hence speed of rotating stator magnetic field remains constant), but SEIGs use capacitors which themselves depend on rotor speed for sourcing excitation amps. Hence the frequency at which the capacitors operate itself depends on rotor speed so the concept of slip between rotor and stator rotating magnetic field becomes a bit sketchy to understand as stated in my earlier post. I wish some clarity could be provided in this regard. :frown:

Thanks & Kind regards,
Shahvir
 

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