Transformers in parallel in abnormal situation, no LTC 1

[Sideswiper]

I'm a newbie protection engineer and I have a mentor who will look over this, so don't panic over how ignorant I am. I'm getting a whole week of protection engineer training next week.
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We have a substation with two transformers that are usually run independantly. There is a tie breaker on the low side that is normally open.

The high side voltage is 115kV and the low side voltage is 13.09kV. Right now the high side of both transformers is tapped at 110kV.

If the tie breaker closed, would there be circulating current?

If we change the high side tap on one transformer to 130kV, and the tie breaker closed, what would the circulating current be? Since it is a temporary condition to have the tie breaker closed, would it be a big deal?

Transformer A
Impedance: 10.43% @ 18 MVA, angle 88.4 degrees

Transformer B
Impedance: 9.11% @ 20 MVA, angle 87.34 degrees

So if I try to use this formula,
i(circulating) = Vdiff/(Z1+Z2)

Vdiff is the calculated secondary voltages minus transformer loss before the tie breaker closes? Or is it just the difference in the high side tapping?

I guess I calculate the low side voltage for both transformers before the tie breaker closes (high side voltage divided by high side tap times low side voltage), then subtract transformer loss (current from peak MVA load divided by voltage with load angle time the transformer impedance). The difference in the two is my Vdiff and then I can get i(circulating).

 

[7anoter4]

Hi Sideswiper
The attached pdf. may be useful.

 

[burnt2x]

Which one will you change-tap, A or B?

 

[Sideswiper]

It addresses the low side taps being different, but I have different high side taps. Also the impedance is a whole percentage point different between the two transformers.

Maybe I just divide 20kV (the difference in the high side taps) by the sum of the transformer impedances? Seems like I have to translate to the secondary voltages like I mentioned in the original post to get the Vdiff I need, but I can't justify why at the moment.

Maybe the answers are staring me in the face and I just can't see it yet.

 

[Sideswiper]

We'd be increasing the tap on transformer A. It's loaded twice as heavily as transformer B and that is causing problems somehow (I'll get clarification on that when I can, it always helps to understand the problem...). The distribution side doesn't seem to want to simply balance the feeder load which would be the best solution... then if we had to close the tie breaker we wouldn't have to worry so much about circulating current.

 

[stevenal]

Look at application notes 11 and 13 at:

http://www.beckwithelectric.com/infoctr/index.htm

 

[burnt2x]

IMHO,I can't understand the rationale by tap-changing TR "A" to 130kV! Raising the tap of TR "A" with Z=10.43% will make the situation worse! More winding, more Z! By just looking at the numbers, you should lower the tap on "A".
Try using a loadflow software and simulate. The best setup I had in mind is to lower "A" tap to 113kV (if possible) and you get a very good sharing of load between the two trafos! Max. total load of 35MVA can be achieved by this setup.
Another thing is that you should make sure you will be above 95% of the secondary voltage to avoid violations.
Please check because I just did a quickie on this problem.

My $.02

 

[waross]

You can't or shouldn't numerically add impedances. impedance is the vector sum of reactance and resistance. Use your respective phase angles or the X/R ratio to determine the respective R and X. Add the Rs, add the Xs. Calculate the new impedance.
As I understand circulating currents they are independent of load currents, but will add to the load current of the transformer with the higher open circuit voltage and subtract from the transformer with the lower open circuit voltage.
Remember that normally the load impedance is much higher than the transformer impedance. For a rigorous solution you must determine the effective resistance and reactance of the load and combine with the transformer R and X.

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

 

[Sideswiper]

burnt: The transformers are normally run as independant and not in parallel. Closing the tie breaker on the low end would not be done often but I'd like to know the circulating current if it is. Whatever is driving the decision to increase the tap is not how they would operate in parallel but rather how it is working now. Or are you saying it is strange to raise the tap on a heavily loaded transformer? Maybe it is, I don't know.

waross: Yes I tend to just enter the Z impedance vectors in my calculator and add them, but I'm familiar with using the Rs and Xs. Same thing. I have the peak load resistances and reactances and calculated the low side voltages during those peaks (which I clumsily tried to describe in the original post). Thanks for the clear answer, I think I'm on the right track.

 

[slavag]

Hi.
As I remember, max permissible difference between impedance
+/- 10%.
In this case is more then +/- 10%. I'm not sure, but I think right solution is additional reactor to transformer with lower impedance. But in this case haven't any circulating current.
Check load sharing for full load.
resultant impedance will:
uk= ( 18+20)/ ((18/10.43)+( 20/9.11))=9.69%
S1= 18^10.43/9.69=19.2MVA ( overload)
S2= 20^9.11/9.69=18.7MVA (underload)
S= 19.2+18.7= 37.9MVA I think is not "bad"
Voltage difference isn't recommended. Check possible circ. current.
Ir1= 18MVA/1.73^13.09kV= 795A
Ir2= 20MVA/1.73^13/09kV= 883A
dU=3%
Icir= 3/((10.43/795)+(9.11/883))= 130A
Regards.
Slava

 

[cranky108]

Your question on changing the high side tap, well it's the same as changing the low side tap. In both cases you are changing the transformer winding ratio.
It's the ratio that is important, along with the impedance. Off nominal transformers are not uncommon, but to look at the impedances correctly you need to have them on the same base. I personally like 100 Mva for everything.
If you have load-flow software, use it to complete the calculations.

 

[jghrist]

Consider the following:

    +-----------+----> IL
    |           |
    Z1 ^        Z2 ^
    |  | I1     |  | I2
    |  |        |  |
    |           |
  + |         + |
    V1          V2
  - |         - |
    +-----------+

V1 is the secondary no-load voltage of transformer 1. V2 is the secondary no-load voltage of transformer 2. IL is the load current. Z1 is the impedance,in ohms referenced to the low side, of transformer 1. Z2 is the impedance,in ohms referenced to the low side, of transformer 2.

V1 - I1·Z1 = V2 - I2·Z2
I1 + I2 = IL

You have two equations with two unknowns, I1 and I2. All quantities are complex.

If IL = 0, this reduces to the calculation of circulating current in Section 3.2 of 7anoter4's Beckwith attachment.

You can see that increasing the primary tap of T1 (decreasing the secondary voltage), all other things being equal, will decrease the current through T1. As Burnt2x notes, increasing the tap will increase Z1 also, which will tend to decrease I1. The increase in Z1 with higher taps is not easily determined except by test, but it won't be enough to offset the higher secondary voltage.

 

[waross]

I apologize. I may have misspoken. The load current can be expected to be at a reasonably good power factor. The circulating current will be determined by the X and R of the transformers and will be mostly reactive. It will add and subtract to the load currents a little but not much.

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

 

[burnt2x]

Okay, you might need visuals to understand your case very well!
The open circuit voltages of your TR A sec = 11.58kV and TR B sec = 13.09 kV (see attached graphics.

Q: How could you close the tie breaker with this condition?

 

[burnt2x]

Ooops!

 

[7anoter4]

Hi
As usual I am late.
I think the way Slava showed is correct if :
S1= 18/10.43*9.69=16.7 MVA (underload)
S2= 20/9.11/9.69=21.3 MVA ( overload)
If the supply [HV] is different [from tap changer] the way indicated by Jghrist seems to be more appropriate.
Also I think if both trfx are in the same tap changer position the circulated current should be the minimum.
See attached xls. If it is still interesting ,of course.
Best regards,

 

[slavag]

7another4, you are never late.
Thanks for your great work.
Best Regards.
Slava

 

[7anoter4]

Thank you Slava!