Percentage Impedance vs Percentage Voltage Regulation

[TAFEDout]

With regard to calculating load sharing when paralleling transformers, and calculating short-circuit fault currents, how similar is Percentage Voltage Regulation (given with respect to full-load voltage) to Percentage Impedance?

I know how both quantities can be found by experiment/calculation; however, it would seem that for some (if not all) practical purposes the two quantity values might be interchangeable.

I assume that due to the much lower voltage during the short-circuit test to find/confirm the %Z, the Fe losses are negligible and therefore are not included when calculating %Z. Are there any other differences between Percentage Voltage Regulation and Percentage Impedance?

(Hopefully, I haven’t lost the plot - as all my reference texts apparently keep the two quantities very separate)

 

[Marmite]

The regulation that occurs at the secondary terminals of a transformer when
a load is supplied consists of voltage drops due to the resistance of the windings and voltage drops due to the leakage reactance
between the windings. These two voltage drops are in quadrature with one another, the resistance drop being in phase with the load current. The value is always positive and indicates a voltage drop with load.
The voltage regulation is significantly influenced by the power factor of the load, in particular for transformers with high impedance.
Regards
Marmite

 

[TAFEDout]

True – the percentage regulation is heavily dependent on the load power factor. Whereas the load for the short-circuit test for percentage impedance is unity. So the %Reg and %Z would be different if the load was not unity.

During the short-circuit test for finding the percentage impedance, rated current is flowing, so the effect of winding resistance and leakage reactance would be accounted for (?)

So assuming we are talking about a percentage regulation value for a load at unity pf, how similar is this value to the percentage impedance?

 

[waross]

The transformer has a combination of resistance and reactance. There will be a voltage drop across this combination whenever current flows in the transformer. The phase angle and power factor of this current depends on the X:R ratio and the power factor is very low.
Any current through the transformer will cause a voltage drop which may be calculated by Current times Impedance. However reactance predominates and the power factor of this voltage drop will be quite low.
To determine regulation a vector representation may be subtracted from the open circuit voltage vector. Because of the phase angle of this drop in relation to the phase angle of the load voltage of a unity PF load, the resultant load voltage drop is less than the simple arithmetic difference.

Look at it another way:
Draw a vector to represent the voltage of a unity PF load.
Draw a vector at 5% of the load voltage and at 80 degrees lagging. (representing 5% impedance voltage)
Add these vectors. The resultant is the open circuit voltage of the transformer.
The regulation is the percentage difference between the open circuit voltage and the voltage at the load. Because the impedance voltage drop is mostly quadrature, it has a small effect on the overall voltage drop of the load voltage.
Now repeat this exercise with load vectors at other than unity power factor to see how the load power factor affects the regulation.
There may be a few errors here but the purpose is to illustrate the relationship between regulation and impedance, and how this relation is affected by the load power factor.

Bill
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"Why not the best?"
Jimmy Carter

 

[TAFEDout]

Thanks Bill and Marmite, it is starting to ‘gel’! I’m back in my office on Friday, where I’ll get into the phasor diagrams! Cheers