Loss due to Circulating Current in Parallel Transformers 1

[CKent]

I remember participating in a discussion sometime in Aug 2003 regarding paralleling transformer (see thread238-68183)...I'm doing a study now on technical loss...I have come to the issue of circulating current in parallel XFs...circulatin current (Ic) arises due to difference in the turns ration of the XFs in parallel...Ic adds up to the current in the XF with lower turns ration while it subtracts to the current in the XF with higher turns ration...this effectively changes the capacity of the XFs in parallel...however, in general I'm being confused on how this affects the losses of the combined parallel combination...of course the one with the lower turns ration will have a resulting additional loss due to this circulating current...however, isn't it that the other XF with higher turns ration effectively will have reduce loss because the current it its wdg will be lesser than the actual load...so assuming that the 2 XFs have equal resistance, isn't it that the effective loss of the parallel combination is actually equal as in the condition wherein there is no circulating current (in cases with equal turns ratio)...

In the thread I mentioned above, one of the respondents has this to say in paralleling XFs...
"But an advantage is the reduced losses, especially near full load of one transformer. This applies equally to I^2X and I^2R loss - on a 25MVA transformer, times about one hundred substations, it adds up. Halving the current quarters the loss. But the primary advantage is supply availability - it was this that was the primary driver for us doing it."

How could this be?

 

[itsmoked]

CKent; You would do much better to read this entire thread carefully as waross wrote a "manifesto" on the subject.


thread238-143892

 

[jghrist]

If you work out an example, you would see that the sum of the losses of two identical parallel transformers increases as the split of the current varies from 50-50.

 

[waross]

Hi jghrist
I think that the lossses will depend on the resistive component of the transformer impedance. This can be different in transformers of equal ratings. Even though they may be sharing the load 50-50.
Apart from that, I agree with you completely. It's the I² part of I²R that does it.

 

[stevenal]

It's an oversimplification to say the circulating current adds to one and subtracts from the other. In phase and out of phase components must be treated separately, and the circulating current is primarily inductive. With no load or a load with unity pf, the resultant current on both transformers will increase as one goes lagging while the other goes leading. Bypassing the load, this current does no useful work at all, it simply heats the transformer windings and the conductors and connections in the circulating current path. It's all loss. With a lagging load, a little circulating current may offset the vars supplied by one transformer resulting in less loss in that unit. Overall, though, losses have increased because of the presence of circulating current.

 

[waross]

Hi stevenal
I agree with you if the circulating currents are caused by harmonics, or distorted wave forms. With different wave forms the instanteneous voltage can be higher for one transformer at one point in the cycle and lower than the other at another point in the cycle. As the voltage relationship is changing several times a cycle, the resultant current will be a higher frequency. The current is actually going from "A" to "B", and then from "B" to "A" several times a cycle. This is one cause of harmonics.
I suspect that you are more familiar with harmonics than I am, and I respect your posts in regards to harmonic currents.
My submissions are not to do with harmonics but with mismatched voltages and/or turns ratios. After you consider the effects of the mismatched characteristics, you can add the effects of the harmonics.
yours
In the instance where the transformer characteristics and wave forms are similar, but the turns ratios or the percent impedances are different, the resulting current is actually trying to transfer energy from one system to another. Failing to do that the resulting voltage and the resulting division of load will be dependant on the impedance of the transformers and their turns ratios.
Calculating the actual current in a transformer may require considering the power factor of the various loads, but once the current is determined, at what -ever power factor or phase angle, the losses are I²R, and there is no place in the this formula for phase angle.

 

[stevenal]

Waross,

The circulating currents I'm speaking of are fundamental. There is nothing in the circulating current circuit that is non-linear to cause harmonics. It is conceivable that a non-linear load might move the source voltage enough to cause harmonic circulating current, but I am not considering that circumstance. I stand by my earlier posting. Superposition is the theory that allows the circulating component of current to be calculated for a no-load condition and then applied to the loaded condition. Great references at Beckwithelectric.com.

 

[dpc]

Stevenal is right - the circulating currents (fundamental)caused by a turns ratio mismatch will increase the overall system losses. And they must be added as phasors, taking the phase angles into account.

 

[veritas]

Gentlemen

I think it is more correct to say that circulating currents are due to the difference in open circuit secondary voltages of trfrs in parallel (this is the bottomline for circulating current - not merely just due to different turns ratios). The latter is most commonly the cause but we had a case where two trfrs were paralelled with the HV bus being about 5km upstream from the trfrs in question. The HV cables feeding the trfrs had different impedances resulting in the HV terminal voltages being different. With the trfrs having the same turns ratio the open circuit secondary voltages were different resulting in circulating current.

Also, all else being equal, impedance mismatching does not result in circulating current but in disproportionate sharing of the load current.

Finally, another interesting phenomenon with circulating current is that with different turns ratios, the LV circulating current magnitude is the same for both trfrs but not the HV. This results in an imbalance, the difference being supplied by the source. The source current is primarily reactive. This fact is used by some utilities to sink excess vars on the system during low load conditions.

Hope this helps.

Regards.

 

[waross]

I agree with you veritas. You just said what I was trying to say.
Given the condition you described, as you start to load the transformer pair, one transformer will initially take all the load plus the circulating current. The second transformer will carry only the circulating current.
As the load increases, the circulating current will drop on both transformers. As the load is increased further the circulating current will eventually drop to zero. At that point, 100% of the load will still be carried by one transformer and the other will have virtually zero secondary current.
Further increases in the load will be carried by both transformers.
Parallel transformers of different percent impedance ratings will not share the load in the proper proportion. In an emergency you can get a better sharing of current between transformers of different percent impedences by simulating a higher voltage on the transformer with the higher percent impedance by setting it's tap down. (If the tap is set low, then the actual voltage is more than the transformer is expecting.)
The capacity of a transformer bank may have to be de-rated drastically in the event of poor load sharing.
You can dramatically improve the proportional load sharing at heavy loads on mismatched transformers by changing tap settings.
The down side is the circulating currents at light load conditions.
A caveat. Transformers with equal percent impedances will share the load in proportion to their KVA ratings.
Transformers with UNEQUAL percent impedances may be adjusted to share the load proportionately at one load level only. The load sharing will be disproportionate at higher or lower loads.
A rule of thumb is that the effective percent impedance at full load will be changed by the percentage of voltage change. Setting a tap 2.5% low is equivalent to numerically reducing its percent impedance 2.5%. (7% impedance becomes 4.5% impedance.
This is a combination of "It was a dark and stormy night!" and "Whatever gets you through the night!"
A note to you gentlemen with the high priced modelling software. Can you set up a model of two parallel transformers with different pecent impedances and see what the software says about load sharing? Then try changing the applied voltage a percent or so and see what the loading curves look like.
yours

 

[jghrist]

stevenal's point is that the circulating current is out of phase with the load current, so you have to add and subtract vectorially. Circulating current is mostly reactive. If the load is resistive, there is no amount of current 90° lagging current that you can subtract to make the vectorial sum zero.

 

[waross]

Hi jghrist
I will accept that statement. However loading reduces the voltage difference that causes the circulating currents and this is what reduces them to zero, not the load current.
If you change my wording to "the load current adds vectorilly to the circulating current" then I believe my description is correct.
Please work a couple of examples.
The circulating currents we are discussing (I think) are 60 Hz., not harmonics. They are caused by unequal voltages. Loading will change the secondary voltages and so change the magnitude of the circulating currents.
yours

 

[stevenal]

I also agree with what veritas (good handle) said. He did not, however, claim that downstream loading affected the circulating current component. The only way load will reduce the circulating current is if the load causes one of the transformers to go off line, or the source impedance ahead of the transformers is significant enough that load moves the primary voltage. If one assumes an infinite bus at the primary, open circuit voltage difference (ideal source voltage behind the impedance) drives the circulating current though the constant impedance of the transformers and bus work that make up the circulating current path.

This is what the math says. Look at the Beckwith Electric Application notes on the subject. But my experience is not just in working the math. I operate actual transformers in parallel, and have observed the affects of loading and out of step LTC positions on circulating current as well as on the individual transformer loading. The infinite bus primary, constant impedance, variable turns ratio model works very well since LTC position has a negligible affect on impedance.

 

[dpc]

waross,

OK - I modeled two transformers in parallel in EasyPower. When impedance are unequal, the load is shared per the impedance ratio, with the lower impedance xfmr taking more of the load. So if one transformer has 6% Z and the other has 3%, 2/3 of the load is supplied by the 3%Z unit and 1/3 by the 6%Z unit. Changing the primary voltage has no effect on the load sharing split.

Was that the question you were asking??

 

[waross]

dpc
Thanks.
Now if it's possible, change the applied primary voltage on one transformer.
Put the primary voltage on the transformer with 6%Z up one percent. You should see a circulating current at no load, and better sharing of the load at higher load levels. The circulating current should also be gone at the high load level.
If you put the primary voltage up 3 % you should see close to equal load sharing at full load and close to zero circulating current at full load.
This is assuming an infinite supply.
Thanks for the help.
yours

 

[dpc]

waross

Are you talking about changing the actual voltage at the transformer primary or changing the turns ratio of the transformer. If the transformers are truly in parallel, the primary voltage is the same.

 

[waross]

dpc
Good point. Change a tap setting please.
In the example you have chosen, 3%Z and 6%Z we should get interesting results if you set the tap on the 6%Z transformer down 2.5%
Thanks
yours

 

[stevenal]

dpc
I suspect if Easypower is like the software I use, it won't give you circulating current directly. Begin with no load to see circulating current directly. Then use phasor addition/ subtraction to see what part of the transformer current goes to load and what part goes to circulating current. To make it easy, use a purely resistive load and purely reactive transformer impedances. Prediction: load current will remain proportional to the inverse of the transformer impedance while circulating current will be constant.

 

[dpc]

If primary tap on the 3%Z transformer is lowered to .975 of nominal, its share of the load (load is purely resistive) increases to about 41% of the total, with the 6%Z transformer providing 59%. But there is a substantial amount of circulating current that is purely reactive. kVAR flow is about 42% of the real power delivered to the load.

Total losses are higher when taps are not matched, due to the circulating current and the unavoidable losses it creates.

The circulating current is a result of the turns ratio difference between the two transformers. When the two transformers have identical turns ratios, there is no circulating current.

 

[waross]

Fellas, are you telling me that two partly loaded transformers with resistive loads and identical terminal voltages may not always be paralleled without causing circulating currents?
I understood that the phase angle of the secondary current was determined by the load. What would introduce a phase shift in one transformer that is different than the other?
The load sharing in dpc's model is about what I expected it to be, but I'm having a problem with the circulating current.
respectfully