Questions regarding transformer's percent impedance in other than nominal voltages

[Reinserkmo]

Hello,

I'd like some help to consider the calculation of a current in a transformer's SC test. Suppose i have this theoretical transformer :

Vpri = 230 kV
Srated = 100 MVA
Z% = 5 %

For testing purposes, i'll use a 440 V generator to energize the transformer's primary side with its secondary side shorted. In order to calculate the current which will circulate in the primary side during the test, should i perform a p.u base change to 440V ? how would you guys calculate the current in A using p.u to solve the problem ?

Thanks in advance!

 

[waross]

Percent impedance voltage;
The supply voltage needed to drive rated current through a shorted secondary.

Vpri = 230 kV

5% of 230 kV = 11.5 kV
0.44 kV is only about 0.04% of 11.5 kV.
I would not depend on a test at such a low percentage of standard conditions.

 

[Reinserkmo]

Percent impedance voltage;
The supply voltage needed to drive rated current through a shorted secondary.

5% of 230 kV = 11.5 kV
0.44 kV is only about 0.04% of 11.5 kV.
I would not depend on a test at such a low percentage of standard conditions.

Thanks for the reply,

Let me give more context, the test's objective is to measure the differential relay's current in order to confirm correct CT's polarity and collect oscilograms during comissioning. I need to rate the diesel generator's correctly for it to be able to whitstand the test current without tripping its breaker.

Knowing the transformer's percent impedance, i should be able to calculate the current that will circulate during the test and then calculated the rater power of the generator to be used in the test.

 

[System Protection]

You should be able to convert from PU to actual impedance (ohms) and calculate from there. It's been a long time since I've tested a transformer, but I'd often have to do that in reverse. Start with an actual impedance and convert to PU to check against the nameplate.

 

[Reinserkmo]

Greetings,

If i convert the z% from p.u to ohms, considering the transformers nominal voltage, wouldnt the impedance be off for my test? as with the generators voltage being significantlly different than rated voltage, the leakage reactance would be way off than that calculated under nominal settings ?

 

[bacon4life]

Is that a common field test for transformer commissioning?

 

[Reinserkmo]

Is that a common field test for transformer commissioning?

When its possible to do this test, its great because we can confirm that the CTs wiring is correctly executed and we can measure the differential current before the energization date, allowing us to prevent unwanted tripping by the 87 and REF protections during the assets energization.

 

[waross]

I still think that your test voltage is much too low.
Again:

5% of 230 kV = 11.5 kV
0.44 kV is only about 0.04% of 11.5 kV.
I would not depend on a test at such a low percentage of standard conditions.

At such a low percentage of rated voltage, I suspect that hysteresis effects will be significant and you will be in a non linear range of excitation current

 

[System Protection]

I should clarify my comments as the transformers I used to test were autos.

I could get a decently accurate reading on a short circuit impedance test with very little voltage. Just using 120VAC and a variac.

The excitation requirements are less than a regular transformer though.

There was obviously never enough current flowing to test the CTs.

 

[Reinserkmo]

I should clarify my comments as the transformers I used to test were autos.

I could get a decently accurate reading on a short circuit impedance test with very little voltage. Just using 120VAC and a variac.

The excitation requirements are less than a regular transformer though.

There was obviously never enough current flowing to test the CTs.

Yea, i've done this test 2 times before. First time it didn't work because the test current was too great for the generator to handle, then it would trip its breaker. Second time, it worked fine because i had a large generator and it was able to withstand the test current.

Sadly, i don't have a variac avaible at the moment, so i will have to do the test using a generator but i need to size it correctly in order to be able to do the test at the first place or it will be just like the first time.

I've got hold on a sheet that calculates said current, i've performed some tests with it comparing the results with an oscillogram that i have from a real test and the results were equal. Apparently, the sheet just calculates the p.u voltage using the generator's voltage divided by the transformer's rated voltage and then calculates the current dividing that per unit voltage by transformer's rated z%.

What i don't understand is why the sheet's calculations uses the rated z% for a condition other than transformer's nominal values.

 

[Lucas_Sterquino]

Greetings,

If i convert the z% from p.u to ohms, considering the transformers nominal voltage, wouldnt the impedance be off for my test? as with the generators voltage being significantlly different than rated voltage, the leakage reactance would be way off than that calculated under nominal settings ?

Hey mate, two things to consider:
1) I think there are CT testers in the market that you could use to confirm polarity and CT ratings. Generally, the differential protection could be triggered and tested through secondary injection or use a proper tester and running the cables through the CTs, simulating faults accordingly. Obviously, I have no idea of what type of CTs and the voltage level you are working at, so take this comment as a heads up.

2) Regarding the circuit calculation - assuming the transformer % impedance is your only load:
2.1 - First you define your base. In this case, take the transformer rated values as your base: Vb = 230kV, Sb = 100MVA
2.2 - Calculate your Generator voltage in the per-unit: Vg = 440/230k = 0.00191 pu
2.3 - Calculate the current in per-unit: Ig = Vg/Z% = 0.00191/0.05 = 0.03826 pu
2.4 - Calculate the current in amps: Ig' = Ig x (Sb/(sqrt(3)*Vb)) = 0.03826 x 100M/(sqrt(3)*230k) = 9.6A

You can calculate the other way around:
- From the base values, obtain the transformer ohms: Zt = Z% x Vb^2/Sb = 0.05 x 230k^2/100M = 26.45 ohms
- Calculate the generator current: Ig = Vg/(sqrt(3)*Zt) = 440/(sqrt(3)*26.45) = 9.6A

Now, be aware of the transformer inrush current and ensure your gen will stand with it. And the above calcs are reference only to show the per-unit method.

Thanks,

Lucas.