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#128817 01/26/04 03:22 PM
Joined: Oct 2003
Posts: 10
The per unit impedance listed on a transformer, as one such N/P read says,
"the per unit resistance 1 % the reactance per unit is 5 %."

Is this referred to the primary or the secondary or the whole transformer ?
Second if an Open circuit test is performed on the LV side is the per unit ratings on the N/P can they be used directly on the Low voltage side for example if as above it says resistance per unit 1 % does this mean the equivalent circuits resistance is 0.01 ohms as seen by the secondary.

Any help would be appreciated

#128818 01/28/04 06:06 PM
Joined: Dec 2002
Posts: 110
If I understand your question you want to relate the % per unit impedance to the actual impedance.
First, this is a way to express the Series impedance of the T/F as a per cent.
Second, To convert the % per unit impedance to its actual value it is first necessary to divide the 1 % per unit and the 5 % per unit by 100, this would give .01 + j.05.
the nex thing to do is to find the Z-base of the primary side That is system used for T/Fs where the standard SI units are not used but are referred to their bases.
That is Per Unit value = Actual value/base value.
For example, single phase T/F 50KVA 8000v/240 has a % per unit Name plate of say 1.2 and 7 for resistance and reactance respectively.
1. convert the per cent to decimal, or (.013 + j.070 then it is necessary to find the Z-base.
That is the total impedance base for the transformer referred to the primary.
That is this the base voltage squared/base KVA or 8000(Squared)/50KVA and this is 1280 ohms.
This Z-base in now multiplied by the converted % per units.
Or (.012 + j.070) * 1280 = (15.36 +j89.6)
the Series Impedance referred to the primary is Req = 15.36 ohms and Xeq = 89.6 ohms.
Now in order to find the series impedance referred to the secondary all that is needed is to divided the two Primaries values by the turns ration squared.

Or from the secondary find the Z-base as referred to the secondary.
Since Pin = Pout the S-base is still = 50KVA and the V-base is 240 volts and the result will be related to the turns ration squared.

Hope this helps.


#128819 01/30/04 04:10 AM
Joined: Oct 2000
Posts: 2,722
Broom Pusher and
Per Impedance Data, Resistances and Reactances of Transformers tend to follow normal patterns according to the ratings.

The Graphs I have seen for %R and %X, are determined by measurement of load loss on Z test.
This would relate to the entire Transformer - including the Core, not just a single winding.
(I mention the Core for its losses and etc.).

Quantities expressed in Per-Unit or % are the same regardless of whether they are are referred to the Primary side or the Secondary side.

Per the second question, Per-Unit quantities are one thing - as listed above, but the Transformer's Regulation will result in situations where if the load current and Secondary voltage are at rated values, the Primary voltage must exceed rated value.
This is found by using this Equation:

Ep = Es+Is(Rs+jX)+IpRp

* Ep = RMS volts at primary terminal (phasor);
* Es = RMS volts at secondary terminal (phasor);
* Ip = RMS amperes in primary (phasor);
* Is = RMS amperes in secondary (phasor);
* Rp = Ohms resistance of primary winding;
* Rs = Ohms resistance of secondary winding;
* X = Ohms reactance of Transformer.

The excess is the Regulation.

Regulation for Lagging Power Factor is:

Gr = [(Rr+Pr)²+(Xr+Qr)²]¹/² -1


Go = 100[PrRr+QrXr + (( (PrXr-QrRr)²÷2 ))]

* Go = regulation, in percent;
* Gr = regulation, in per-unit;
* Pr = Load Power Factor in per-unit;
* Qr = (1-Pr²)¹/²;
* Rr = resistance of transformer in per-unit;
* Xr = reactance of transformer in per-unit.

Then there's Efficiency ratings!

Hope the information from Wocolt and myself are sufficient.
If you need additional data, feel free to post away!


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128820 01/30/04 03:16 PM
Joined: Apr 2002
Posts: 2,527
Nota Bene: Who says Nobody ever really uses Ohm’s Law and Why should I have to learn it?

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