Induction motor contribution to a non bolted fault. 1

[waross]

For discussion:
A motor acts as an induction generator when the grid frequency is lower than the motor speed converted to frequency.
With a bolted fault, a motor that normally turns at 1750 RPM with a slip speed of 50 RPM will be seeing a slip speed of about 1750 RPM.
Is it valid to use the starting current at 50 RPM for the contribution to a fault?

For a non bolted fault; Does the motor acting as an induction generator feed any reactive current into the fault.
This may be a lower frequency such as 55 Hz back fed into a 60 Hz fault.
The question is not the actual frequency but whether and lower frequency current is back fed into a fault.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[cranky108]

I sort of have the same question, but about an induction generator, with a connected capacitor bank, for var support.

I suspect it can contribute the normal output, depending on the local voltage seen at the generator, sort of like a limited output generator.
Like a faster transition to steady-state than a sync-generator.

But I think it has mostly to do with the terminal voltage

 

[dpc]

Bill,

I've never thought about it in those terms. When a fault occurs, the motor excitation is lost and the motor speed starts to decrease a bit. The contribution from the motor is a function of the motor's subtransient reactance, just as for a generator. We generally assume something in the 16% to 25% range for the X". But current decays rapidly due to lack of excitation - generally in a few cycles. The fault impedance will impact the phase angle of the current provided by the generator, just as any other impedance in the circuit would impact it.

When motor contributions for faults are calculated, we almost always ignore the frequency decay of the motor contribution, because the overall impact is small and a more precise calculation would be a serious PITA.

Motor contribution is very real, but it's also very brief.

Cheers,

Dave

 

[krisys]

In my own view, it is logical to assume that the short circuit contribution should be based on the induction motor characteristics, mainly the induction motor short circuit contribution is a function of motor locked rotor current. Same should hold good when it operates as induction generator as well.

 

[waross]

Any comments, information or speculation on the frequency of the fault current generated by the induction motor?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[aussiejohn2]

Hi Bill,
Interesting question. If we consider a very low resistance fault (not quite bolted) then I think it is clear that there will be two different frequency contributions. At the other extreme with a very high resistance, the motor will remain synchronised. There must be a transition at some intermediate resistance? I have never noticed this behaviour in simulations, but haven't particularly looked and think it must be there.
I don't have a good transient package installed at the moment but am tempted to dig out the license and reinstall one to confirm. I have a spice dq motor model - I think that may actually suffice.
Regarding starting current at 50 rpm, I normally just use the starting current (at 0 rpm), i.e. x" as others say above. This is endorsed by standards, and if you consider the shape of the current-speed curve it makes negligible difference.
John.

 

[waross]

Another issue concerning the motor starting current curve and the contribution to fault currents:
A widely accepted value for locked rotor current is 6 times full load current.
Depending on the point-on-wave instant of energization, There may be a first quarter cycle transient that is considerably higher than the accepted starting current typically shown on starting current curves.
In the event of a sudden bolted fault, is it possible to have a similar current transient?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[aussiejohn2]

Hi Bill,
I think of that brief high current as magnetising inrush, just like for a transformer. I know that back-to-back transformers can "outrush", and it seems reasonable to think your faulted motor can too. For the motor case, though, the extra energy/fault current is not much (Breakers and busbars should already be rated for the peak current of a reasonable x:r).
JD.

 

[jraef]

As to spinning reserve contribution to a fault, a prof I once had used the acronym "BBS" on the chalkboard* without explaining it, so that someone in the room would ask. Turned out he meant "Brief But Spectacular". 2-3 cycles, but because that is shorter than the shortest clearing time of any sort of protective device, it must be factored in.[pre][/pre]

*Yes, I'm that old


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[dpc]

Bill,

The asymmetrical inrush current on motor starting depends on motor X/R ratio and the voltage phase angle when energized (so it's different in each phase). When considering fault contribution from a running motor, there can be similar dc offset current, but as a function of the X/R of the equivalent impedance of the entire system. Where multiple sources exist, each parallel path may have a different X/R ratio so the amount of asymmetry can vary from each source. So for any short circuit, there can be dc offset current of some degree depending on X/R ratio and voltage phase angle. Again it will be different in each phase (for three-phase fault).

Cheers,

Dave