Per Unit System – Practice Problem Solved For Easy Understanding

Let’s understand the concept of per unit system by solving an example. In the one-line diagram below, the impedance of various components in a power system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base.

Power System Oneline - PEguru.com - Power Systems Engineering
Figure 1: Oneline Diagram Of A Power System

Now that you have carefully examined the system and its parameters, the equivalent impedance diagram for the above system would look something like the following.

 

Impedance Diagram - PEguru.com - Power Systems Engineering
Figure 2: Impedance Diagram Of A Power System

Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance. Capacitive effects between lines and to ground are ignored as well.

To obtain the new normalized per unit impedances, first we need to figure out the base values (Sbase, Vbase, Zbase) in the power system. Following steps will lead you through the process.

Step 1: Assume a system base

Assume a system wide S_{base} of 100MVA. This is a random assumption and chosen to make calculations easy when calculating the per unit impedances.

So, S_{base} = 100MVA

Step 2: Identify the voltage base

Voltage base in the system is determined by the transformer. For example, with a 22/220kV voltage rating of T1 transformer, the V_{base} on the primary side of T1 is 22kV while the secondary side is 220kV. It does not matter what the voltage rating of the other components are that are encompassed by the V_{base} zone.

See figure below for the voltage bases in the system.

Voltage Base - PEguru.com - Power Systems Engineering
Figure 3: Voltage Base In The Power System

Step 3: Calculate the base impedance

The base impedance is calculated using the following formula:

Z_{base}=\frac{{kV_{base}}^2}{S_{base MVA}} Ohms……………………………………………………(1)

For T-Line 1: Z_{base}=\frac{(220)^2}{100}= 484 Ohms

For T-Line 2: Z_{base}=\frac{(110)^2}{100}= 121 Ohms

For 3-phase load: Z_{base}=\frac{(11)^2}{100}= 1.21 Ohms

Step 4: Calculate the per unit impedance

The per unit impedance is calculated using the following formulas:

Z_{p.u.}=\frac{Z_{actual}}{Z_{base}} ……………………………………………………………………………..(2)

Z_{p.u._{new}}=Z_{p.u._{old}}(\frac{S_{base_{new}}}{S_{base_{old}}})(\frac{V_{base_{old}}}{V_{base_{new}}})^2 ……………………………….(3)

The voltage ratio in equation (3) is not equivalent to transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.

For T-line 1 using equation (2): X_{l1_{p.u.}}=\frac{48.4}{484}= 0.1 pu

For T-line 2 using equation (2): X_{l2_{p.u.}}=\frac{65.43}{121}= 0.5 pu

For 3-Phase load:

Power Factor: \cos^{-1}(0.6)=\angle{53.13}

Thus, S_{3\phi}(load)=57\angle{53.13}

Z_{act}=\frac{(V_{rated})^2}{\overline{S}^*}= \frac{10.45^2}{57\angle{-53.13}}= 1.1495+j1.53267 Ohms

Per unit impedance of 3-phase load using equation (2)= \frac{1.1495+j1.5326}{1.21} = 0.95+j1.2667 pu

For generator, the new per unit reactance using equation (3)

X_{sg}= 0.18(\frac{100}{90})(\frac{22}{22})^2 = 0.2 pu

For transformer T1: X_{t1}= 0.1(\frac{100}{50})(\frac{22}{22})^2 = 0.2 pu

For transformer T2: X_{t2}= 0.06(\frac{100}{40})(\frac{220}{220})^2 = 0.15 pu

For transformer T3: X_{t3}= 0.064(\frac{100}{40})(\frac{22}{22})^20.16 pu

For transformer T4: X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^20.2 pu

For Motor, X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^20.25 pu

The equivalent impedance network with all the impedances normalized to a common system base and the appropriate voltage base is provided below.

Impedance Diagram - PEguru.com - Power Systems Engineering
Per Unit Impedance Diagram

 

 

If you think you learnt something today then you will love the eBook I prepared for you. It has 10 additional and unique per unit problems. Go ahead, preview it. Get the complete version for only $1.99. Thanks for supporting this blog.

Per Unit Systems: Ten Total Problems Solved With Detailed Explanation

  • Per unit systems is an important concept in power system analysis. So much so, it shows up in all exams pertaining to power. Besides being taught in higher education, it is quite commonly used by engineers in the power industry. Equipment impedances are furnished in per unit. Source impedance for fault analysis is furnished in per unit. In this book, you will find enough problems solved for you to get the hang of this subject.

To view full load amps due to motor load and inductive load at Bus 2, see this post.

Summary:

  1. Assume a Sbase for the entire system.
  2. The Vbase is defined by the transformer and any off-nominal tap setting it may have.
  3. Zbase is derived from the Sbase and Vbase.
  4. The new per unit impedance is obtained by converting the old per unit impedance on old base values to new ones. See equations (2) and (3).

* * * * *

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kaushik vastarpara

its really bcoz by reading this my confusion abut selection of base nd other is very clear…sommust read it frend /…thank u

Pavan

This is really helpful. I didn’t really got it when reading through this, but when I saw the below comment by Mike, it seems like a question worth answering. However the content is really clear and understandable. Keep up the good work!
Thanks,

mike

I don’t understand one part: When calculating Xtl2 you are using (22/22) which is reflected from where? Vbase in T2 is 220 primary and 11 secondary, so where does 22 come from?

The same for Xtl4.

mark

will the impedance or p.u impedance in each line will be like in series? will the current for the PRIMARY AND SECONDARY of the transformer now be equal??? how will i find the actual line current for each line and for the whole system…

chris

A single phase ,350 kva, 1380v generator has an internal impedance Zg of j6 ohms. The generator is used to supply a load of 250kva/440v at power factor 0.78 lagging. determine: the turns ratio of the transfomer, the impedance per km if the line between the generator and the transformer is 5km, the voltage regulation of the system.
Using the ratings of the generator as base values determine the generated per unit voltage that is required to produce a full load current under short circuit condition.
CAN SOMEONE HELP ME WIT THESE CALCULATION PLZ!!!

kam

sorry but i didnt get my reply yet, so could you pls help me out???????? thanks a lot

karthik

very well explained but could you pls show me how to calculate voltage and current in both lines, will be very greatful , thanks a lot……….

kam

It is really well explained but could you pls show me how to calculate voltage at but 3 and current in both lines, will be very greatful , thanks a lot

manish

what if transformers are connected in star and delta connection?

Anayat

i am very new to Power side , so i really dont know abt all these concepts , what we only have T1 and T2 , and all the rating given are three phase line to line ? how we ll solve it then?

very nicely explained….to the point and complete..thanks a lot 🙂

Sanket

VERY NICELY EXPLAINED THANK-YOU ………
I WILL VISIT WEBSITE AGAIN FOR FURTHER REFERENCES.

BABULS RAJ

Thank u so much…..after searching for a proper explanation for the same in so many sites, i got it finally from your site. Clear explanation with proper diagrams with multi colour…….very nice …..

Renjith M

Commendable work. But there is a small error. The per-unit system is the ratio of two quantities of the same units. Therefore it is unitless. Well that is what I know. So accordingly we specify the per-unit quatities as just ‘P.U’. So you need to remove the ‘Ohms’ from the text and insert ‘p.u’

Alfredo

It was very useful, but it is short because is necessary to get the complete solution, any way I liked.

Abdul Rauff

Very Good Info About PSA.Thanks Alot

Hilary

Protection engieering, i have been give the reactance as Xd’ to calculate faults on a system do i convert to Xd” how do i do this

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