### There is a lot of confusion surrounding the per unit system, and rightfully so. If you’ve never used it before, it is easy to become confused with all of the individual pieces that make the system work.

### To help clear things up, here is an in-depth break down on an example question that is similar to what you can expect to see on the NCEES Electrical Power PE Exam.

### Per Unit System Example for the Electrical PE Exam

Using the Per Unit system and taking into account the transformer percent impedances, solve for the current in each part of the three-phase system shown below. Assume both transformers are either delta delta or wye wye connected and that there is no phase shift between primary and secondary current and voltage.

### Solution:

The usefulness of the per unit system is in converting all system impedances to per unit impedances and re-drawing the circuit without having to worry about the different voltage levels from each transformer.

In this example, there are two transformers that divide the system into three sections of different voltages. The first step is to divide the system into different sections based on voltage levels. I do this by typically drawing a straight line below each transformer:

The first step is to divide the system into different sections based on voltage levels.

I typically do this by drawing a straight line directly below each transformer:

The next step is to choose the base values for power and voltage.

The base power will be the same in for each section, but each section will have a different base voltage.

The problem might tell you to use specific values for base power and voltage. If it does, use them accordingly as the answer choices will most likely still be in per units and using a different base will change the resulting per unit system values.

If the problem does not assign the base values for you, then it is advantageous to pick one of the existing MVA values in the system such as the apparent power rating of one of the machines.

Similar for voltage, it is advantageous to pick one transformer and use either the primary or secondary voltage as the base voltage in a particular section, and then use the remaining transformer ratios to step up or down the voltage base for each neighboring section accordingly. Why this is advantageous will become clear when we run the math.

If all transformer ratio’s match, then the secondary voltages of all upstream transformers are equal to the primary voltage all downstream transformers and vice versa, then your voltage base in each section will be equal to the primary and secondary voltages of each transformer. That is the case for this example.

For base power, I’ve arbitrary picked T1’s MVA rating for the system and the voltage ratios of the transformers for the base voltage in each section:

Now that we have the assigned base values for power and voltage, we’ll need to calculate the base impedance for each section so that we can use it to calculate the per unit impedances later on.

The formula for base impedance is:

We can solve for impedance using voltage and power, and if we use the base voltage and base power in each section, then the resulting impedance will be the base impedance for each section as shown below:

Now that we have calculated the base impedance for each section, we can start calculating the per unit impedances of each system element using the following formula:

We will start by calculating the per unit line impedance and per unit load impedance first before we tackle the transformers.

The line per unit impedance and load per unit impedance are calculated as shown below:

Now that we know the per unit line and load impedances, let’s drawn them on the per unit circuit diagram:

In order to complete the above diagram, we’ll need to calculate the per unit impedances of both transformers T1 and T2.

The catch here is that we don’t know either of the transformer impedances in ohms, we are instead given their percent impedances.

When you are given the percent impedance of a machine such as a transformer, generator, or motor, the percent impedance is given in the base ratings of the machine itself.

This means that in order to calculate the percent impedances of the transformers, we actually need to perform a per unit base change using the following formula:

In this case, the old base will be the ratings of each transformer, and the new base will be the new bases we have chosen.

Since we are working with impedances, we will need to plug in the impedance formula that uses the square of the voltage divided by power:

In the above formula, SB and VB will be the chosen bases of our system and section, and S and V will be the transformer voltage and power rating.

Transformers have one power rating, so choosing that is easy. However, they have two voltages, a primary and secondary.

If we use the transformer’s primary voltage for V we will need to use the base voltage on the primary side of the transformer for VB in the formula above.

If we use the transformer’s secondary voltage for V then we will need to use the base voltage on the secondary side of the transformer for VB instead.

This will be more clear when run the math below.

Let’s base change each transformer’s percent impedance one at a time starting with transformer T1 using the primary voltage rating as V and the voltage base on the primary side of the transformer for VB:

Notice that since we originally chose 45MVA as the base power for the system, both power terms are equal and cancel. This will not be true for T2.

Also, notice, that both voltage values are also equal and cancel since the base voltage in each section is equal to each transformer’s voltage ratio.

It is important to be aware that the voltage values in the above formula **will not cancel** when the actual voltage in a section is not equal to the transformer’s voltage ratios such as when a generator is connected that supplies a voltage slightly higher or slightly lower than the primary voltage ratio of the transformer it is connected to, or, when all transformer ratios do not match as mentioned in the first several paragraphs.

Notice that since our base power and base voltage for T1 equal the base values we were changing to, there is no actual change in the percent impedance. We started with 5% and ended with 5%.

Last bit of importance is to notice the imaginary “j” term placed in front of the final per unit impedance value for T1.

Transformers are made up of large inductive coils and their impedance is mostly reactive.

**The PE exam may not remind you of this and it will be up to you to not only recognize that this is actually a reactive impedance but to also add the imaginary j term to it.**

Let’s move on and calculate the per unit impedance for transformer T2:

Now that we have both per unit impedances let’s fill in the remaining values in our per unit impedance diagram. again, notice both

Again, notice that both transformer per unit impedances are reactive terms and we had to add the “j” multiplier:

Now that we have all impedances converted to per unit values of the same base, it’s time to calculate the per unit current using Ohm’s law:

However, in order to calculate the per unit impedance, we will need to calculate the per unit voltage.

Luckily for us, the voltage on our input bus is equal to the base voltage selected in that zone which simplifies the math since both terms cancel:

That’s right, the per unit voltage is equal to 1.

This greatly simplifies solving for the per unit current which will now equal the inverse of the total series per unit impedance of the system:

The per unit current will be the same value for each voltage section.

Now that we know the per unit current, we can finally calculate the actual current in each voltage section by multiplying by the per unit current by each base current:

However, this requires us to know what the base current in each voltage section is equal to.

We can calculate it using the chosen base power and the chosen base voltage in each section as shown below.

Notice the √3 multiplier since it is a three phase system:

Now that we have each of the base current values we can complete the problem and solve for the actual current in each section in amps:

We can now fill in the current in amps for each section to the original system diagram that we started with:

For a quick check of our work, we can use the transformer voltage ratios to verify our currents are equal in each zone depending on voltage level.

If we did our work correctly we should get almost the exact same answer for each current by starting with one and using the transformer ratios to solve for the rest

Let’s start with current I3 and work our way back to I2 and I1:

Looks good!

Here is our final answer solving for the individual currents in amps of the original system using the per unit method:

Ria says

Hi Zach,

I got familiar with your website and the course you provide on YouTube just two weeks ago. I absolutely love listening to your lectures and solving the the quiz problems (the free courses). I wish I knew about your website at least 2 months ago! Never the less, I highly recommend people to go over your course, as you go over concepts and basics and not just to pass the exam which I appreciate a lot. I am a first time taken and have benefited from the free videos you have posted.

The PU problem was very well and neatly explained with all steps. I do not have a problem with PU, but I was still glad you posted something. Keep doing the good work! You are a great instructor. Thanks for your help. Fingers crossed I pass.

Zach Stone, P.E. says

Thanks for the complements Ria! Good luck on the October exam. If for some reason you don’t pass this time around and need some extra help, you know right where to find me.

Maysam Radvar says

Hi Zach,

Thank you so much for the thorough explanations on this topic. PU calculations are one of the most important and fundamental concepts in Power System modeling and Protection Systems. We were running Transient Stability Analysis utilizing PSS/E the other day at our office and we were given the machine’s data in machine’s base. However, the line data provided from transmission planner was given in system’s base. It was easy to enter the values in program and move on with the simulation but we verified the simulation results with the technique you explained above and the outcome brought smile to our face! Thanks for your help and keep up with awesome work!

Zach Stone, P.E. says

That’s a neat story Maysam thanks for sharing, I’m glad the information in the article helped you and your colleagues!

Patrick says

Hi Zach,

Great article on the PU method! The way you broke down the process is both simple and effective. Following you methods has helped mitigate any small errors that could be difference between getting the question right versus getting the question wrong!

Zach Stone, P.E. says

Thanks Patrick, I’m glad it helped!

Seth Newbold says

Zach,

Thanks for posting this. Per Unit analysis has definitely been one of my most challenging concepts to consistently perform correctly. I’m definitely going to use this during my last couple weeks of preparation before the test to hopefully solidify my understanding of per unit problems.

Zach Stone, P.E. says

Hi Seth,

I’m glad the article was helpful!

Anmar Hiassat says

This is wonderful. It was the easiest way to understand one of the most complected topics for me.

Zach Stone, P.E. says

Glad it was helpful Anmar!

Steven says

Thanks for posting Zach! The way you break down the different steps based on the different sections of the system is very helpful! It simplifies per unit analysis tremendously.

Egis says

Thanks for posting. Since PU method is not my favorite, but with this example makes easier to understand and follow the steps to solution. what a great example and hopefully able to solve PU problems during exam.

Melanie Parker says

What a great article! Per unit can be very intimidating, but I’m feeling confident after these examples. Thanks!

Rania Genidy says

Thank you for posting this Zach, you simplified the PU and going step by step with all the explanation made it very easy. I feel very confident that any question in Per unit will be a piece of cake. it could have not been described better. thanks again

Sam says

Very thoughtful and precise! Helped me to divulge my understanding of PU calculation, very detailed explanation so that I can easily follow.

Orlando says

This is just the beginning of what else ElectricalPEReview.com offers. Everything has been very helpful and help me understand PU analysis and much more. Thanks Zach!

Zach Stone, P.E. says

Thanks Orlando!

Jason says

Excellent article on per unit analysis. Between this article and your live webinar I feel much more confident solving per unit analysis problems.

Jeff Beer says

Hi Zach.

Once again, my course work in the Electrical PE Review has helped to clear up my confusion in problem solving. This time regarding the PU System performed on a complex transmission line consisting of two series transformers, step up and step down configurations. The solution for this problem was presented step by step in a smooth and coherent fashion. Specifically, the technique shown of drawing a vertical line below each transformer, separating out the base voltages and power, allowed me to see more clearly what information the problem had presented me and what information was left for me to solve. I will use the technique going forward both in the PE Exam and my own future circuit analysis. Thank you again!

Zach Stone, P.E. says

You are welcome, thank you Jeff!

Eklas Hossain says

Very helpful and complete discussion. I was always confused about the different bases, and struggled to pick one; but now it has become a whole lot clear. I specially like the discussion on the MVA method. Thank you for doing such a great job!

Will says

I really like how you’ve laid out this problem; though I can’t imagine a problem this long on the actual PE test. I’m following well up to the point where you simplify Ipu. That step could use some explanation. It might be good to add a step where you multiply by the complex conjugate and convert to polar notation to get the simplified answer for Ipu.

Thank You!

Zach Stone, P.E. says

Hi Will, thanks for the comment.

I agree that this is not a 6-minute problem that you’d likely expect on the actual PE exam.

However, you could definitely expect to see any number of quicker to solve but still challenging per unit problems that are designed to be explained and demonstrated by this single, more in-depth, example. It was a way for me to compact a series of helpful lessons for per unit analysis contained within a single article.

Do you mind pointing out which exact step you are referring to with simplifying Ipu? If you do, I’d be happy to expand on it in both the article and here in the comment section as an update.

Thanks!

Justin says

Toward the end of the article, you indicate your currents are off due to rounding when checking your currents in each section using the transformer voltage ratios. The error is actually due to using the incorrect current 2 base value in your current 2 calculation (you used 367.5A instead of 376.5A). This leads you to calculate I2 as 7.8A as opposed to 7.96A.

Very helpful, concise, easy to understand article on how per unit analysis works; greatly helped me. Thank you!

Zach Stone, P.E. says

Hi Justin,

you are absolutely correct! Great catch, that was bothering me for a long time when I made the article because it should have been identical. Thanks for taking the time to point this out, I went and corrected the values. I’m glad you enjoyed the article on per unit analysis.