Estimate transformer loss from nameplate impedance?

[SacramentoFred]

Is it possible for me to make a rough estimate of the loss that a three-phase 360 kVA power transformer would have just from the percent impedance listed on the nameplate? Any insight appreciated! Thanks, Fred

 

[dpc]

No, not really. The transformer impedance is just the leakage reactance.

Transformers are very efficient. If you assumed 2% loss at full load, that would probably be conservative, unless there is something special about this transformer.

 

[jghrist]

IEEE Std 141 Red Book has a graph of typical X/R ratios. This would be a rough estimate only because transformer losses can vary a lot; if losses were evaluated at purchase, expect them to be lower than typical.

R% = Z%·cos(atan(X/R))
FL loss = rating·R%

 

[prc]

Designers can only make a good guess from impedance value. Higher the impedance from normal value, more copper loss and less iron loss.If less impedance,higher iron loss and less copper loss.If it is a unit of recent times, iron loss may be 0.2 % and copper loss about 1.2 %

 

[waross]

You would get a closer figure under load using the regulation but still not completely accurate. Impedance propedominates under short circuit conditions. Regulation is more important under normal loading.

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

 

[jghrist]

You would get a closer figure under load using the regulation but still not completely accurate. Impedance propedominates under short circuit conditions. Regulation is more important under normal loading.

How would you get losses from the regulation. Regulation is a calculated value and is not shown on the nameplate anyway. It is generally shown on the transformer test report, but if you had that, you would have the losses also.

You need the load loss and the impedance to calculate regulation - See IEEE C57.12.90 Section 14, Calculated Data.

 

[SacramentoFred]

Wow, thanks, all, for the insight. I am a bit confused because I found after making my post that the Blume text on transformers states that Percentage Impedance is approximately equal to 100x the ratio of KVA output vs KVA input. Maybe that's how it was in days of yore, and the newer NEMA TP1 high efficiency models change that rule of thumb?

It sounds that from the consensus, although this transformer has a Percentage Impedance of almost 7%, I can expect an efficiency of better than 96%.

I really appreciate the feedback. Fred

 

[dpc]

There is not a direct relationship between impedance and efficiency.

I would expect the efficiency to be higher than 96%.

Also, the efficiency varies with the load on the transformer.

 

[waross]

Percentage Impedance is approximately equal to 100x the ratio of KVA output vs KVA input.

I think that you have misread or misremembered the original statement.
Replace the word impedance with the word efficiency.
The regulation or PU voltage drop (or rise, to carry the accuracy to the 4th or 5th significant figure) will approximate the load losses.
The PU current deficiency at no load will approximate the no load losses.
A catalog or brochure may list the regulation and/or the efficiency. The test report confirms that the transformer meets the published figures.

Is it possible for me to make a rough estimate of the loss that a three-phase 360 kVA power transformer would have just from the percent impedance

No. The regulation figure will give a much closer rough estimate than the impedance, but still ignores lo load losses.

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