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Change of Base Values in Per-Unit Systems, Part 3a.
In the previous topics, we introduced per-unit systems and discussed their advantages in detail. In this Topic, we will be discussing a crucial aspect of per-unit systems that is the changes in base values. In this video Part 3a, we will derive a formula that will come handy in complex analysis of power systems.
Until now, we have looked into examples and problems that involve a single base value for the entire power system. But what if the different regions in a power system have varying base values? For example, the generating side would be working on 200MVA base power, the transmission line has base power of 50MVA and the base power for load side is 100MVA. Such types of power systems require the selection of one base value for the whole power system and also require conversion of different bases to the selected one. And same goes for the other quantities involved.
We know that when we convert a power system into its per-unit equivalent, we find out per-unit impedances according to the selected base value. Now, we will derive the formula for calculating a new per-unit impedance if the selected base value is different from the given one.
The general formula that is used to calculate per-unit impedance is:
We know that the actual impedance remains the same no matter what the base is. So, by rearranging the above equation we get:
Zactual = Zp.u × Zbase
For a given base value we can write this equation as:
Zactual = Zp.u,given × Zbase,given
If we want to calculate the per-unit impedance according to a new base value, we can then write the formula as:
From our previous videos, we know that:
Putting these equations to find out Z_(p.u,new):
Rearranging the above equation, we get:Z_(p.u,new)= Z_(p.u,given)×〖V_(base,given)〗^2/〖V_(base,new)〗^2 ×S_(base,new)/S_(base,given)
So, finally we have derived a complete formula that will aid us in tackling varying bases in a power system. We can use this formula to calculate new per-unit impedances and hence make a per-unit equivalent model of a power system.
In the next video we will solve an example from the book that incorporates the above formula so that we can have a clear concept in our minds.
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