How to base change percent impedance values (and why the formula looks backward compared to the base change formula) - Power Exam Sub Forum - Engineer Boards
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# How to base change percent impedance values (and why the formula looks backward compared to the base change formula)

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Hi Everyone, I just published two new videos that I'd like to share to help address one of the questions that I get asked the most lately about base changing the percent impedance values of machines (transformers, generators, motors, etc).

Have you ever noticed that the percent impedance base change formula below:

seems to have the old and new power bases on the wrong side of the fraction compared to the standard per unit base change formula:

Or, have you ever seen this version of the per unit impedance base change formula and were not sure where it came from:

Click play to see why all three of formulas are actually the same as long as a very specific condition is met that we will discuss in the video:

How about a worked out practice problem to try this out and apply the formulas from the video above?

Let's base change the 5.9% impedance of the generator shown in the one line diagram below using transformer T2's ratings as the new base values of the system:

First, we'll start by dividing the one line diagram into three separate voltage zones created by each transformer and determine the new power base and new voltage base in each of the three voltage zones.

Next, we'll use the base changing formula with the old per unit impedance of the generator (Zpu old) using the generator's ratings for the old base values (Vb old, and Sb Old), and the new base values located in zone 1 at the generator (Vb new, and Sb new).

Last, we will cover how to tell with the old voltage base is equal to the new voltage base (Vb Old = Vb new) by looking at the voltage ratings of each device in the online line diagram and matching all primary and secondary voltages so that we know when we can use the shortened version of the per unit impedance base change formula.

You can work out this practice problem along with me by clicking play below:

I hope you enjoyed the new videos and learned something new that you can take with you to the up coming October 2019 electrical PE exam.

Edited by Zach Stone, P.E.

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Zach,

Thank you for posting this video. I was able to follow and understand it.  I tried turning this into a fault current analysis problem to see if I could solve it. What if there was a 3-phase fault on the right side of the bus? If I use the MVA method, I get 18404 amps.  If I use the Per Unit method, I get 18265 amps.  Is that a big enough difference to be concerned?

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14 hours ago, MEtoEE said:

Zach,

Thank you for posting this video. I was able to follow and understand it.  I tried turning this into a fault current analysis problem to see if I could solve it. What if there was a 3-phase fault on the right side of the bus? If I use the MVA method, I get 18404 amps.  If I use the Per Unit method, I get 18265 amps.  Is that a big enough difference to be concerned?

I would check your rounding. Both methods should yield the exact same number. You can try storing values in your calculator as variables to get a more precise answer.

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