Motor winding connection error 1

[MakingAComeback]

Folks,

We started for the first time recently an 18000 hp induction motor: 13.8kV, 900 rpm. The Multilin 469 relay took it out immediately on short circuit. In round numbers phase C current was about 8000 amps, while phases A and B currents were the same at 4500 amps or so. Normal starting current is less than 3000 amps. We spent a lot of time checking the current transformer circuity and concluded that the problem wasn't there. After much head scratching and a surge test of the stator we concluded that the T3 and T6 leads had been marked wrong at the factory. Reversed them and all was well on the next start.

Conceptually I can grasp what caused the symptoms we observed, but it would be educational to see a mathematical expression that represents our situation. Can you help?

 

[waross]

First read thread237-179569

Your voltages and windings may be represented as equal magnitude vectors originating at a point.
There will be a vector at zero degrees, one at 120 degrees and one at 240 degrees in both cases. If you reverse the connections of the winding represented by the 240 degree vector, you will now have the windings represented by vectors at zero degrees, 60 degrees (The reversed 240 degree winding) and 120 degrees. The vector relationship of the applied voltage remains 0, 120, and 240 degrees. Not a good match.
respectfully

 

[electricpete]

Interesting question.

If you looked at the phases as independent impedances in each of the three legs, it wouldn't matter if you reversed one pair of leads.

So it would seem that it must depends on mutual flux

The normal flux pattern would be:
A B' C A' B C' A B' C A' B C' etc (60 degrees apart)


If you reverse B phase, you have:

A B C A' B' C' A B C A' B' C' (120 degrees apart)

You went from 60 degrees apart from adjacent coil to 120 degrees apart from adjacent groups.

The motor must build up enough flux within a group to counter the applied voltage. It's easier to do that when the groups are 60 degrees apart than 120 degrees apart.

Look at group AA'. It is overlapped by the next group over approx 2/3 of it's span.

If the next group is B'B (normal configuration), then that group is 60 electrical degrees away and adds to the AA' flux.

If the next group is BB' (miswired), then that next group is 120 electrical degrees away and subtracts from the AA' flux.

In the miswired configuration, phase AA' developes less flux (for a given current), and must draw more current in order to create enough flux to balance the applied voltage.

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[electricpete]

I changed the scenario above to involve a miswire of phase A.

http://home.houston.rr.com/electricpete/eng-tips/Miswire2.pdf

Hopefully the approach is clear if you read the comments, but please ask if it's not clear.

In the normal configuration, the contribution from the overlapping groups doubles the flux linking group AA'. (The overlapping groups contribute flux equal to the orignal AA'). This is shown in attached vector calcuation.

It stands to reason that if you reverse AA' and wire it as A'A, the contribution of the overlapping groups is now equal/opposite and the flux is now 0 (if the system remained balanced). This is also shown in attached vector calculation.

Since phase A cannot generate any flux to cancel the applied voltage, it's current will go through the roof.


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[electricpete]

Actually, if we assumed and unbalance such that phases B and C continued to contribute their normal LR current and phase A increased to develop the same flux it would during a normal start, then phase A should triple.

Normally we get 1 from A plus 1 from overlapping groups to get 2.

For the miswire, we get 1 from the overlapping groups, A, is adding in the negative direction, we need to add -3 from A in order to reach the same magnitude of flux as the normal case -3+1=-2... magnitude = 2.

There is of course a current limiting effect from your supply, and obviously a lot of assumptions built into the above discussion.

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[electricpete]

In my vector analaysis I was also making the assumption that the neutral voltage of the machine wouldn't change... another not-so-good assumption... so the results are a very approximate answer which should be viewed qualitatively.

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[MakingAComeback]

Dear Mr. Pete,

Thanks for giving my question so much thought. I can follow your reasoning fairly well, and feel that yours is a valid approach. Very enlightening.

A clarification please . . . the (evalf) item in your pdf explanation . . . what is that? A software result or something similar?

Thinking further about the situation and employing some of your thoughts made me think that the winding reversal in our motor led to something akin to an open delta situation in a transformer. With little or no back emf on the C winding (your A) we would have had full line to line voltage on the A and B windings thus giving us phase currents sqrt.(3) higher than normal starting current. That would have resulted in phase C current being nearly 3 times greater than normal starting current. In round numbers we had such a relationship among the currents.

Another term might apply also. We could think of the reversed leads' effectively producing a "virtual short circuit" on the T3-T6 winding.

Thanks again for sharing your insight, knowledge, and experience.