what does this phrase mean "a transformer with 5% impedance'

[andrewward]

Hi all,

what does this phrase mean "a transformer with 5% impedance'? Is it just the p.u. impedance? Also, what are the formulae for calculating that percentage impedance?

Thanks,
Andrew

 

[alehman]

Z % = Z p.u. x 100
Zbase = kV^2 / MVA
Z p.u. = Z in ohms / Zbase

Please see any power engineering text for more detail on per-unit method.

 

[4654]

for a simple understanding, this is percentage of voltage. The rated current will be obtained when the this much % voltage applies on the primary. Basically for the fault current calculations, this will be useful.

 

[dotcomm]

Just to expand explanation by 4654 -

It is the percentage voltage applied on the primary to produce a full rated load current in the 'short circuited secondary'.

 

[WvH]

Another angle: It is the voltage loss in %pu, on the trf base, across the trf when rated current flows through the trf.

 

[jbartos]

Suggestion: Additionally, the transformer impedance is a complex number:
Zpu=Rpu+jXpu
pu=per unit
It may also be presented as impedance in series with a voltage source.

 

[jbartos]

Addition to alehman (Electrical) Oct 21, 2003 marked ///\\Z % = Z p.u. x 100
Zbase = kV^2 / MVA
Z p.u. = Z in ohms / Zbase
///kV is the secondary rated voltage since the primary voltage must be higher for the transformer secondary to deliver the rated kVA, Voltage and Current.\\

 

[boomsystems]

Summary is. It is the % of terminal voltage required to be applied on one side ( Primary) of the winding in order to circulate full load current on other ( Secondary) when other side is solidly short circuited.

 

[rbulsara]

Let me jump in as well.

%Z is not calculated, but rather measured by test, as indicated above, by shorting the seconday terminals and applying just enough voltage that will create the rated full load current in the secondary. The ratio of this voltage to per rated primary voltage times 100= %Z.

The meaning or signinficance of the Z% impedance is that it helps determine the maximum short circuit current the transformer can deliver. Assuming that the primary is connected to a infinite source, the maximum Isc= (rated full load amps*100/ percent impedance)

or Isc = FLA/Z in per unit , 5%=0.05 per unit.

For example, a 500kVA,480V secondary tranformer (600A FLA)with 5% Z will can deliver maximum of 600/.05 =12000A of SCC.

Hope this helps.

 

[andrewward]

Thanks everyone for the help. Most appreciated. I now realise that I was after the definition that the Z% is the % terminal voltage required to obtain the rated current for a short-circuit secondary. We have been working on short-circuit currents.

 

[alehman]

Another wording of wvh's comment - it is roughly the % voltage change at the load terminals going from no load to full load, assuming a zero impedance source at the primary. This is called voltage regulation.

Clarification to jbartos's comment - per-unit or % impedance is the same from either the primary or secondary. Calculate Zbase for the voltage rating of the winding being considered. The measured impedance in ohms divided by Zbase for the side you are measuring will give you the same impedance regardless of which winding you measure from.

 

[jbartos]

Suggestion: The transformer ratings are on the transformer output, e.g. MVA, kV, kW, etc. It means that the transformer input will have to have higher kV than the rated kV primary obtained by the transformer turns ratio. There is always the transformer impedance that is necessary to be taken into account for obtaining correct results and the correct transformer parameters. However, by measurements there has to be one transformer per unit impedance only since the transformer electrical equivalent circuit impedance can be located on the primary side or on the secondary side (the parallel magnetizing branch in the transformer electrical equivalent circuit is assumed negligible).