Transformers in parallel in abnormal situation, no LTC
Transformers in parallel in abnormal situation, no LTC
(OP)
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).
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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).






RE: Transformers in parallel in abnormal situation, no LTC
The attached pdf. may be useful.
RE: Transformers in parallel in abnormal situation, no LTC
RE: Transformers in parallel in abnormal situation, no LTC
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.
RE: Transformers in parallel in abnormal situation, no LTC
RE: Transformers in parallel in abnormal situation, no LTC
http://www.beckwithelectric.com/infoctr/index.htm
RE: Transformers in parallel in abnormal situation, no LTC
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
RE: Transformers in parallel in abnormal situation, no LTC
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
RE: Transformers in parallel in abnormal situation, no LTC
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.
RE: Transformers in parallel in abnormal situation, no LTC
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
RE: Transformers in parallel in abnormal situation, no LTC
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.
RE: Transformers in parallel in abnormal situation, no LTC
CODE
| |
Z1 ^ Z2 ^
| | I1 | | I2
| | | |
| |
+ | + |
V1 V2
- | - |
+-----------+
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.
RE: Transformers in parallel in abnormal situation, no LTC
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Transformers in parallel in abnormal situation, no LTC
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?
RE: Transformers in parallel in abnormal situation, no LTC
RE: Transformers in parallel in abnormal situation, no LTC
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,
RE: Transformers in parallel in abnormal situation, no LTC
Thanks for your great work.
Best Regards.
Slava
RE: Transformers in parallel in abnormal situation, no LTC