# 2 weeks until the exam! Still feeling a little uncomfortable with Per Unit Analysis? Here is a detailed example broken down step by step



## Zach Stone P.E. (Oct 13, 2017)

I've been getting a lot of questions lately via email about solving questions using per unit analysis to solve for various system parameters so I put together an example with a detailed breakdown.

This example includes:


A reference voltage bus

Two transformers

One line impedance

One load impedance

A total of three voltage sections

All transformer ratios match

The reference bus voltage is equal to the nearest transformer's primary voltage ratio. 

Enjoy.

Per Unit Example for the Electrical PE Exam:

Using Per Unit Analysis and taking into account the transformer percent impedances, solve for the current in each part of the three-phase system shown below:




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:

______

I'm limited by the file size of images that I can upload in this post, so I've linked the article below that way you can still access the entire full-length solution that contains over 15 images of the circuit in various steps of the solution and the math worked out for every step.

Click here to see the full-length solution - Per Unit Example Tips, Tricks, and What to Watch Out for on the PE exam (Electrical PE Review)

With the PE exam right around the corner, I hope this helps!

Feel free to ask any questions on the article in this post, I would be happy to answer them and to help you work your way through the example if you get stuck or need a little extra help.


----------



## Omer (Oct 13, 2017)

Thanks for the question, it is good for practice.

However, for this specific problem we can directly calculate the current in the load side and then using the transformers ratio to obtain the current on the other zones, right?

Thanks again for the detailed explanation on the link.


----------



## Zach Stone P.E. (Oct 13, 2017)

Omer said:


> Thanks for the question, it is good for practice.
> 
> However, for this specific problem we can directly calculate the current in the load side and then using the transformers ratio to obtain the current on the other zones, right?
> 
> Thanks again for the detailed explanation on the link.




Yes, as long as you include all impedances.  This way you can validate your answers with the ones you get using the per unit method. 

The PE exam is likely to give you which base values to use such as "use the transformer ratings as base values" and then present the answers to you in per units.


----------



## rg1 (Oct 13, 2017)

Electrical PE Review said:


> Yes, as long as you include all impedances.  This way you can validate your answers with the ones you get using the per unit method.
> 
> The PE exam is likely to give you which base values to use such as "use the transformer ratings as base values" and then present the answers to you in per units.


Rarely we find a good explanation on use of pu system. The author has shown patience to explain it in nice steps, making sure that no gap is left while delivering the information/knowledge. Very excellent explanation of how to use pu system. My compliments to Zach and his team at @Electrical PE Review . Must be very helpful to many.Thanks a ton.


----------



## rg1 (Oct 13, 2017)

Omer said:


> Thanks for the question, it is good for practice.
> 
> However, for this specific problem we can directly calculate the current in the load side and then using the transformers ratio to obtain the current on the other zones, right?
> 
> Thanks again for the detailed explanation on the link.


While using actual impedances, make sure that they are transported at one end, either Gen end or Load end (at one Voltage to be more specific), through Transformers by that square formula else you get a wrong answer.


----------



## Omer (Oct 13, 2017)

Actually, transformers Impedances and TL impedance doesn't matter at all.

I just need to know the load resistance (which is provided 200 ohm) and the voltage at the load (provided 13.8 Kv)  to calculate the current

VLL/sqrt(3) / RL  =39.8 A (same answer)


----------



## Omer (Oct 13, 2017)

Of course, if I am not interested in the phase angle.


----------



## Zach Stone P.E. (Oct 13, 2017)

rg1 said:


> Rarely we find a good explanation on use of pu system. The author has shown patience to explain it in nice steps, making sure that no gap is left while delivering the information/knowledge. Very excellent explanation of how to use pu system. My compliments to Zach and his team at @Electrical PE Review . Must be very helpful to many.Thanks a ton.


Thanks rg1!


----------



## rg1 (Oct 13, 2017)

Omer said:


> Actually, transformers Impedances and TL impedance doesn't matter at all.
> 
> I just need to know the load resistance (which is provided 200 ohm) and the voltage at the load (provided 13.8 Kv)  to calculate the current
> 
> VLL/sqrt(3) / RL  =39.8 A (same answer)


You are right Omer. In this case it is true because it is a load current and the Thevenin Impedance is small compared to load Impedance and can be neglected. You will realise the difference when say, I give  you a load impedance of say 0.5 Ohms. Try and do it.


----------



## Zach Stone P.E. (Oct 13, 2017)

Omer said:


> Actually, transformers Impedances and TL impedance doesn't matter at all.
> 
> I just need to know the load resistance (which is provided 200 ohm) and the voltage at the load (provided 13.8 Kv)  to calculate the current
> 
> VLL/sqrt(3) / RL  =39.8 A (same answer)


Just want to say be careful here. 

The load is much larger than the line impedance and the reactance of both transformers so while you are getting the same rounded answer this approach could yield a wrong answer in the future.  For example, remove the load from the circuit and calculate the current due to just the line and transformer impedances.  

Think of a series circuit with several impedances.  The current drawn is due to the sum of impedances.


----------



## rg1 (Oct 13, 2017)

Electrical PE Review said:


> I've been getting a lot of questions lately via email about solving questions using per unit analysis to solve for various system parameters so I put together an example with a detailed breakdown.
> 
> This example includes:
> 
> ...


In this case the 13.8 KV left side bus supplying the power has no given Z, so it becomes infinite bus implying the bus with perfect constant Voltage or a source with Zero internal Impedance. Had the Z for bus been given, which sometimes come in form of Short Circuit capacity, that needs to be taken into account.


----------



## Omer (Oct 13, 2017)

I know what you are trying to say, and you are right.

Actually, what is going on is that, the secondary voltage of XFR T2 is no longer 13.8 Kv due to the voltage drop of the line and the transformers. and this becomes more evident as high currents flow in the system. The secondary voltage of T2 is even at zero voltage when a 3 phase short circuit occur at the secondary of T2. and the current will only be limited by the series impedances of the system.

however, if the voltage at the load was specified at 13.8 KV, I would use my previously mentioned approach.


----------

