Transformer Circulation Currents 3

[eglaude]

I have a power system infrastructure that effectively puts two 3000kVA, 13kV/2.4kV, Z1=5.55% and Z2=6.33%, in parallel via basically a main-tie-main configuration. It's not as "clean" an arrangement, due to several feeder breakers that allow the tie between the two transformers. Old infrastructure flexibility.

Breaker coordination and selectivity is in general satisfactory, and typically the parallelling of these transformers is not done. We tie the two togehter when either of the transformers needs to be taken out for service.

We tripped one of the smaller feeder breakers when the busses were teffectively ied togehter even though the total currrent on both transformers were within the trip settings of the breaker. Typically our procedure calls to look for a split in current draw between the two transformers when a tie is made. Our systems are typically double ended and matched.

The mismatched transformer impedance is what I believe is causing the circulation current and resulting in a breaker trip of one of the in-line breakers that makes up the bus tie arrangement.

Can anyone give me some information on calculating circulation currents in mismatched transformers?

Thanks.

 

[electricpete]

Mismatch of transformer impedance affects load sharing.

Mismatch of voltage ratio creates circulating current. Check to see if transformers are both on the same tap. If tap were 1% different, than circulating current would be
1% / (5.5% + 6.5%) = 1 / (12) ~ 8% of base.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[stevenal]

Good info at the Beckwith site.

http://www.beckwithelectric.com/

 

[mpparent]

I hope you are braced for that type of switching procedure.

Mike

 

[eglaude]

mpparent,

When you say braced, you are refering to the BIL ratings of the Tx, and the MVA rating of the switchgear. Yes, the gear can handle it.

Good info electricpete and stevenal, thanks guys.

 

[RalphChristie]

Like stevenal said, good info at Beckwith.

Just a few comments:

On what did the feeder tripped? If you temporary parallel these two transformers I can't see why would a feeder trip due to a circulating current, except if (like electricpete noted) the transformers where on totally different taps. When you parallel them, each transformer will draw plus/minus half of the load (one a little more than the other due to different impedances) + the circulating current. The circulating current can just be that high if the transformers where on totally different taps.

mpparent makes a very important note:
It is not just your MVA-rating of your switchgear that must be able to handle the current during paralleling. (Note: fault-current) Everything downstream, cables, outgoing feeders etc. will see double the normal fault-current if there occur such a fault during paralleling. But, if you do it temporarily, the chances of such a fault is not high - however, be aware of it.

Another thing, makes sure that both LV and HV breakers of each transformer is tripped during a fault condition inside the transformer. That is to ensure there is no back-feed from one transformer into the other transformer (during paralleling) if there occurs a fault. (Diff operation, REF operation, Buchholtz operation, etc)

Regards
Ralph

 

[eglaude]

Thanks Ralph,

Good info.You are correct, upon further investigation, the taps on the transformers were changed to compensate for an undervoltage occurance at a large refrigeration unit. The two transformers had approximately 200 V difference, i.e. T1=2500V, T2=2300V, or 9% difference. Using pete's calcs, that equates to 75% of the base current or 540A in circulation current. A breaker from a switchgear line-up, typically used to feed a small line-up can be used to parallel the Tx's through a disconnect used as a tie.This breaker is set to a low trip setting, 300A, which tripped out after 5 seconds, consistant with the associated TCC.

Thanks
Eric

 

[RalphChristie]

Eric

I am glad you've found the problem. One of our procedures before we put some of our transformers in parallel is to ensure both transformers is on the same tap. I do not know how your tapchangers look like, but we have OLTCs, that makes tapping easy. If you have off-circuit tapchangers, please ensure the transformer is NOT energized before you change taps.

See also the reply of ScottyUK in thread238-121895.



You can also determine the circulating current by:

(Va-Vb)/(Za-Zb)

where:
Va - secondary line terminal voltage of trsf A having the lower ratio, i.e. the higher secondary voltage
Vb - secondary line terminal voltage of trsf B having the higher ratio, i.e. the lower secondary voltage
Za, Zb - Ohmic impedances of trsfs A and B respectively, and are obtained from

Za = (Vza x Va)/(100 x Ia)
Zb = (Vzb x Vb)/(100 x Ib)

Where:
Vza, Vzb - percentage impedance voltage drop at full-load ratings of trsfs A and B respectively
Ia, Ib - Full-load line currents in amperes of trsfs A and B respectively

Must admit, Pete's equation looks much easier

Regards
Ralph

 

[eglaude]

Thanks Ralph, however there is a big difference in Pete's equation and the equation that you indicate. From our incident, we have data that gets us close to what Pete's equation generates.

Can you clarify Vza, and Vzb for me, as I believe I'm not applying the impedances correctly.

Thanks

Thanks.

 

[mpparent]

Ralph,

Thanks for beating me to the punch!

Mike

 

[electricpete]

I'm not sure I understand Ralph's equation either.

"Vza, Vzb - percentage impedance voltage drop at full-load ratings of trsfs A and B respectively"

The p.u. voltage drop at full load is Za and Zb. It looks like maybe you are trying to use those other quantities to convert to a non-p.u. quantity? In that case why are you using an actual voltage rather than rated. Not sure I follow.

To my way of thinking there is a voltage in the circulating loop circuit equal to Va-Vb (secondary side voltages). The impedance limiting the circulating current is Za+Zb. The current is voltage over impedance (Va-Vb)/(Za+Zb)... how did you end up with Za-Zb in the denominator.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[RalphChristie]

These equations come from the J&P Transformer Book. (12th edition, Martin Heathcote) After doing the calculations myself, I noted the difference. However, if you change the sign in (Za-Zb) to (Za+Zb) the answer seems more reasonable.

From Pete's equation:
200 of 2400 = 8.33%
8,33/12 = 69.44%
Circulating current = 501A

From J&P Book's equation (with sign changed):
Za = (Vza x Va)/(100 x Ia)
= (5.55 x 2400)/(100 x 722)
= 0.1845
Zb = (Vzb x Vb)/(100 x Ib)
= (6.33 x 2400)/(100 x 722)
= 0.2104
Circulating current = (Va-Vb)/(Za[red]+[/red]Zb)
= (2500 - 2300)/(0.1845 + 0.2104)
= 506A

But, I am not sure if this is correct, let me first try to confirm it before you apply it.
Sorry for any inconvenience.

Regards
Ralph