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Topic 3: Change of Base Values in Per-Unit Systems, Part 3b.
In the previous part, Part 3a, we derived a formula that was used in the cases where the base values are changed.
In this video Part 3b, we will explain how the formula is employed using a simple example.
Now this is example 1.5 from page 30 of Power System Analysis by John Grainger (3rd Edition).
In this example the reactance of the generator is given as: X” is equal to 0.25 pu.
And the rating of the generator is specified as: V equals 18kV and S is equal to 500MVA.
The problem asks us to find the reactance of the generator in per-units according to the new base values given.
And the new base values are this: V base new is equal to 20kV and S base new is equal to 100MVA.
Now we can see that the reactance X’’ is specified in per-units in this question.
We can understand that this value which is the per unit value can be written as this:
X” per unit given is equal to 0.25 pu.
And we have to calculate the new value for this reactance.
We also know that the rating of the generator given
are taken from the base values as we learned in earlier topics of this series.
So we understand that the given X’’ value is calculated
using the given base values so we can write them as this:
V base given is equal to 18kV and S base given is equal to 500MVA.
Now for calculating the new value of X’’ in per-units
we use the formula that we derived in the previous Part 3a which is:
Z per unit new is equal to Z per unit given times V base given squared divided by V base new squared times S base new over S base given.
Or another way of writing this
is X” per unit new is equal to X” per unit given times V base given squared over V base new squared times S base new over S base given.
Now entering in the values from the problem statement:
X” per unit new is equal to 0.25 times 18kV squared over 20kV squared times 100MVA over 500MVA.
And solving this equation what we get is very very simple.
X” per unit new is equal to 0.25 times 0.81 times 0.2 or it is equal to 0.0405pu.
So that is the value of new per unit impedance.
So if we compare the new impedance value, we can clearly see that it’s quite smaller than the previous one.
We are comparing 0.0405 per unit value with our original 0.25 per unit.
This makes total sense because previously, the 0.25 per unit impedance was calculated with a considerably larger MVA rating (500 MVA)
so when we converted the impedance to a smaller MVA rating (100 MVA) the effective impedance value decreased considerably.
And that makes total sense.
Hence we have the answer to the simple example we solved.
The method is employed everywhere where there is a need of calculating new per-unit values
if the base values have been changed; whether it is simple power system analysis or a complex one.
This video marks the end of the series of “Per-Unit Systems in Power Systems”.
Let’s summarize what we have learned in the 3 topics of this series:
1. Basics of per-unit systems and their usage.
2. Formulas for converting systems into per-unit systems.
3. Advantages of using per-unit systems in various power system applications.
4. Per-unit systems for varying base values.
We hope you find this content useful and enlightening.
We also hope that you have enjoyed watching this series and understanding the concepts and topics as a student or professional.
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