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White Paper WP009001EN Effective January 2015

Abstract

Transformers used in commercial and industrialapplications typically show impedances marked inpercentages.

While the technical definition of this percentageor per unit impedance is that this value is apercentage of the base impedance of the trans-former, there is a more descriptive way of thinkingof it.

The per-unit impedance describes that percentageof the rated voltage required to produce full load

current while the transformer output is shorted.The lower the impedance, the lower the voltagerequired to produce full load current. Lower imped-ance transformers allow higher fault currents toflow at a particular voltage.

Measuring impedance in units of percentagegreatly simplifies the calculation of currents andvoltages in a power system. While absoluteimpedance (measured in ohms) could certainlyalso be used, doing so greatly complicates thecalculations.

Per unit impedances

Consider a transformer with a 5% impedance.A voltage is applied to the primary with the

secondary winding shorted (faulted).

At 5% input voltage 100% FLA producedHow much current flows at higher voltages?Ignoring non-linear effects such as core saturation,if 5% voltage produces 100% FLA, then:

At 10% input voltage 200% FLA isproduced

At 50% input voltage 1000% FLA isproduced

At 100% input voltage 2000% FLA isproduced

Note that at nominal full voltage on the primarywinding, the fault currents, both primary andsecondary, increase to 20 times FLA.

Knowing this, one advantage of publishing per uimpedances becomes evident. Virtually by inspetion, a first approximation of this transformersavailable fault current is to multiply the FLA x1/%Z.

ISC

= IFL

. 1%Z

Equation 1:

Absolute impedances

While that seems like a useful rule of thumb,wouldnt the solution be just as easy if the manfacturers published absolute ohms? While labelpercent impedance is independent of the trans-former voltage (or tap), absolute impedance is n

In other words, separate absolute impedanceswould need to be calculated not only for everyvoltage winding, but also for every tap settingof every winding! Having to deal with only onepercent value simplifies things.

Using Transformer Impedance

Consider an ideal transformer (lossless, zeroimpedance) connected to a source V1 and proviing current to a load Z

L.

V1

I1

V1

V2

ZL=

1

4160 : 480

V2

I2

Transformer nameplate impedance:

Per unit or absolute?David G. Loucks

Eaton

Figure 1

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Eaton

1000 Eaton BoulevardCleveland, OH 44122United StatesEaton.com

2015 EatonAll Rights ReservedPrinted in USAPublication No.: WP009001EN / CSSC-1412-809February 2015

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All other trademarks are property

of their respective owners.

Transformer Nameplate ImpedanceWhite Paper WP009001EN

Effective January 2015

Figure 1 shows input and output voltages along with the imped-ance of the load. This provides enough information to calculate thesecondary current.

Equation 2: I2= =

V2

ZL

480V1

= 480 amps

Looking first at our lossless (to simplify the math) transformer,consider that the product of primary voltage and current equals the

product of the secondary voltage and the secondary current:

Equation 3: V1I

1= V

2I

2

For the example values given in Figure 1:

Equation 4: 4160I1= 480480

or rewriting to solve for I1:

Equation 5: I1=

480.4804160

= 55.4 amps

Since impedance is defined as the ratio of voltage divided bycurrent, using the known voltage and the derived current, theimpedance looking through the transformer to the load can befound.

V1

4160V

55.4 A

I1

Zeq

V2

ZL=

1V

2

I2

Figure 2

Performing the math to solve for this effective impedance Zeq

:

= = 75.1 V

1 416055.4

Zeq

=I1

Equation 6:

Notice that this lossless transformer was assumed to have noimpedance. The impedance looking through the transformer actu-ally is the impedance of the 1 ohm secondary, but somehow modi-fied by the transformer to appear as 75 ohms.

When performing a power systems analysis 75 ohms is the actualcircuit impedance that limits the 4160V to only 55.4 amperes.However, It is clear that this value could not be published on thetransformer since its value is dependent on the load and one otherfactor.

That other factor is turns ratio.

Note that this multiplier, 75.1 is equal to the turns ratio squared:

1=75.14160480(

(2Equation 7:

It is evident, then, that the secondary load impedance observedat the primary terminals of the transformer changes with the ratioof the primary to secondary voltage. That means, changing tapssettings will change the apparent impedance looking through thetransformer!

Consider how this complexity multiplies when looking at a systemwith transformers feeding transformers. With absolute valuesof impedance, it would be necessary to perform this turns ratio

squared calculation on every transformer and for every tap setting.Since upstream settings would affect downstream settings, theproblem scales geometrically. Every one of those voltage ratioswould need to be calculated before determining fault currents!

Because of this complexity, it is clear why power systems engi-neers began to favor per unit calculations instead of using absolutemeasurements.

Example calculations: Absolute vs per unit

Consider a 1000 kVA, 5.75%Z, 12.47 kV to 480Y/277 transformer.

The constant of proportionality used to convert per unit impedanceto absolute impedance is the base impedance.

Equation 8: Zohms

= Zpu

*Zbase

Beginning with the 480V secondary, the base impedance is calcu-lated as:

Equation 9: Zbase

=V

baseLL2

Sbase_3

= 4802

1000000= 0.23

With base impedance calculated, solve for actual impedance.

Equation 10: Zohms

= Zpu

*Zbase

= 0.0575*0.23

= 0.0132

Looking into the 480V transformer from the secondary side, thattransformer would appear to place 0.0132 impedance in serieswith the source.

To show how this value is not const ant, we recall the statementthat absolute impedance changes with voltage. Solving Eq 9 againbut replacing 480V with 12.47 kV:

Equation 11: Zbase

=V

baseLL2

Sbase_3

= 124702

1000000= 156

With the new primary base impedance, the absolute impedance iscalculated:

Equation 12: Zohms

= Zpu

*Zbase

= 0.0575*156 = 8.9

This means that measured on the 12.47 kV circuit, the transformerappears to place 8.9 impedance in series with the load, not 0.0132.

Which is correct?

They both are. The difference just depends on which voltagereference you are calculating the impedance.

Conclusion

While transformers could be marked with either per unit or absoluteimpedances, the per unit calculations are preferred by powersystems engineers.

Per unit values do not change with voltage. This simplifiescalculations.