3-Winding Transformer Impedance Calcualtion

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Andrew445

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Hello!

I am attempting to model 3-winding step-up transformers in SKM PowerTools. This transformer is 1260/1386kVA, 34500V/380V/380V with delta on the high side and ungrounded-wye on the secondary/tertiary. The percent impedances are given as H-X=3.82, H-Y=3.72, X-Y=8.86.

The problem is that SKM needs all this information entered as %RPositive, %Xpositive, %RNegative, %XNegative. How do I calculate the required inputs from the nameplate data?
 
Note carefully that although the NEC recognizes ungrounded delta, there are strict requirements for grounding the neutral for almost all wye configurations.
 
If you don't have the transfomer test data, you will have to make some assumptions.

The data you have is the %Z for H-X, H-Y & X-Y. Verify that the values are all on the same base, usually the H winding KVA (1260 kVA).

%Z = %R + j %X, vectorially. Or: %Z = sqrt( %R^2 + %X^2).

If you have the X/R ratio for the transformer you can calculate R & X using algebra.

If you have the transformer test results and have the kw load loss value at the rated load you can calculate %R= 100% x KW loss/KVA rating. (Do not include the no load loss in the kw loss number).

%X = sqrt( %Z^2 - %R^2).

If you don't have the loss data or the X/R ratio, take a guess at it. 20-30 might be in the right range. You will notice it doesn't make much of a difference in the symmetrical fault current calculations. It might affect the asymmetrical values.

Assume negative sequence %Z is equal to the positive sequence values.
 
Thanks for the help!!

Unfortunately I could not readily get the X/R ratio so I just had to make some assumptions that the %Z is largely reactive. Your calculations help me understand this better.

This is for photovoltaic GSU transformers with special grounding requirements, so indeed it is ungrounded-wye.
 
In IEEE 141, Figure 4A-1, there is a chart that provides typical X/R ratios for transformers. The information is based on IEEE C37.010.

For your transformer, assuming each winding is 1/2 the total KVA, the base for H-X is 630KVA, and the same for H-Y; the typical X/R would then be is around 5.

The OP did not state the base for the impedance, however for three winding transformers it is typical that the H-X impedance, and H-Y is on the low LV winding base, so in this case 630KVA. It is given this way because it is not uncommon to have two different ratings for the windings.

Knowing the Z and X/R, the R and X components can then be calculated.
 
change base ratings on 3W

change base ratings on 3W

Quick follow up... to change the base rating of the H-X and H-Y windings (X-Y also), do we simply multiply by (kVAbase new/ kVA base old)?

specifically here,

(Z% @ H-X base 1260) = (1260/630) * (Z% @ H-X base 630)
 
Phil Corso said:
Andrew445,

Hopefully you are dealing with an hypothetical excercise. The arrangement under consideration will result in serious problems!

Regards, Phil Corso
Click to expand...

Now if that isn't a leading response. OK, I'll be the the sacrificial lamb, why?

Wait for it.........................................
 
James5280 said:
Quick follow up... to change the base rating of the H-X and H-Y windings (X-Y also), do we simply multiply by (kVAbase new/ kVA base old)?

specifically here,

(Z% @ H-X base 1260) = (1260/630) * (Z% @ H-X base 630)
Click to expand...

You can have three-wdg xfmrs that are rated the full rating on both secondary wdgs. This prevents restriction of using the full rating of the xfmr if one of the wdgs is not used.

I would rather see a screenshot or picture of the nameplate. Got one?
 
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