Short URL of this page: https://gpac.link/2T1tzkA

This video was brought to you by GeneralPAC.com, making power systems Intuitive, Open and Free for Everyone, Everywhere. Consider subscribing and supporting through patreon.com/GeneralPAC. This is a mechanism for you to support us financially so we can continue making high quality power system video tutorials. Our corporate sponsor for this topic is AllumiaX.com from Seattle, Washington. Contact them for industrial and commercial power system studies.

Why we use Per-Unit in Power Systems, Part 2c.

In this video, we will demonstrate another advantage of using Per-unit in Power systems.

Let’s start with the basics. In order to perform power system analysis of a system having one or more than one Transformers, which usually is the case, we have to convert a transformer into its equivalent circuit referred to either the primary side or the secondary side. There are formulas to achieve this, such as turns ratio formula, voltage formula, impedance formula etc. Impedance on primary side should be multiplied by the square of turns ratio to give the secondary side impedance. We have to find the leakage reactance and magnetizing currents as well.

However, when a transformer is transformed into its per unit equivalent, the need of tedious calculations performed to refer transformers to either the HV side or the LV side is eliminated. The magnetizing currents are neglected and even there is no requirement to calculate impedances on different sides individually. Let’s see how it is made easy. Consider a transformer:

(Draw a transformer like shown)

We know that to find Z2 referred to primary side, the formula is:

Simplifying:

**Z
I
**

_{2,p.u}= z_{1}x_{1,base}V

_{1,base}

Hence, we prove that the per unit impedance of primary and secondary sides are equal in magnitude. This means that calculating per unit impedance of either side is enough to analyze a transformer.

Let’s take an example from the book Power System Analysis by John J. Grainger (International Edition 1994)

Example 2.5, pg. 56, Power System Analysis by John J. Grainger (International Edition 1994) Solution:

On the low-voltage side:

**Z
( 110 V )
= 4.84 Ω
**

_{base}=^{2}2 x 10

^{3}

**Z
0.06
4.84
= 0.0124 p.u
**

_{p.u}=By doing this two-step calculation we know that Zpu on both sides i.e. the low voltage side and high voltage side, is equal to 0.0124 p.u. And because of our derivation earlier in this video, we know that if we calculate it by using the impedance referred from the high-voltage side, the calculation would be lengthy but result is exactly the same.

This concludes that if we want to analyze a transformer, we could simply calculate the per unit impedance on the either side of the transformer. In the next video, we will see the case of multiple transformers in power systems.

We hope, you have a continued interest in this topic and series as a student or professional. We also hope you find this content useful and enlightening. Please consider subscribing to GeneralPac.com and becoming a patron on patreon.com/generalpac.

Thankyou.