## PER UNIT CONSTANTS IN POWER SYSTEM CALCULATIONS BASICS AND TUTORIALS

Per unit quantities greatly simplify comparisons between items of power apparatus and aid in solving fault calculations. Per unit is a method of normalizing the characteristics of elements in a power electronics system so they can be represented independent of the particular voltage at that point in the system. Their characteristics are translated relative to a common base so that extended calculations can be made easily.

In its simplest form, a per unit quantity is merely the percent quantity divided by 100. It spares one the nonsense of 50% voltage times 50% current equals 2500% power. In per unit notation, 0.5 pu voltage times 0.5 pu current equals 0.25 pu power as it should be. A transformer with 6% impedance would have a per unit impedance of 0.06 pu.

Although not described as such, this impedance is based on the rated voltage and current of the transformer. It accommodates the differences in primary and secondary voltages by describing the percent rated voltage in either winding required to produce rated current in that winding with the other winding shorted.

When other elements are added to a system, however, there will be a whole set of different ratings of the various elements. A 500-kVA transformer at 4160 V with 6% reactance may serve a 50-kVA transformer at 480 V with 4% reactance that, in turn, serves a 5-kVA lighting transformer at 120 V with 3% reactance.

It is a real nuisance to chase the various voltages and currents back through the system to find, for example, the short circuit current at the final transformer. Per unit quantities make it easy.

First, one must choose a particular power level as a base quantity. The selection is completely arbitrary but is usually related to the rating of one of the component items. In this case, the 50-kVA transformer will be used as the base, and its leakage impedance will be 4% on that base, 0.04 pu.

To relate the 5-kVA lighting transformer to this quantity, one simply multiplies the 5-kVA impedance of 0.03 pu on its own base by the power ratio of 50 kVA/5 kVA = 10. With the two in current, 28.9 pu on the 5-kVA base. At the 50-kVA transformer, this fault will result in 1/0.346 pu = 2.89 pu current on a 50 kVA base, and at the 500-kVA transformer the fault is 1/3.46 = 0.289 pu on a 500-kVA base.

At any point in the system, one can define a base impedance as
Zbase = VLL^2/VA
or
Zbase =VLL^2/(1000×kVA) where VA or kVA is a three-phase rating.

Then, in ohms, Zohms = Zbase Zpuat that base. The base impedance is the impedance that, when connected to each line of a three-phase system at rated voltage, will draw rated load current and develop rated voltamperes.

It is worth the effort to develop a familiarity with the per unit system, because it greatly eases conversations with utility engineers, motor designers, transformer designers, and others associated with power electronics. It is universally used. cascade, the total impedance is now 0.04 + 0.03 × 10 = 0.34 pu.

The 500-kVA transformer by the same procedure becomes 0.06 50/500 = 0.006 pu on the 50-kVA base. The series string impedance is then 0.006 + 0.04 + 0.30 = 0.346 pu on the 50 kVA base. This total series impedance is 0.0346 pu on a 5-kVA base, and a fault on the secondary of the 5-kVA transformer will result in 1/0.0346 = 28.9 times rated.