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.

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.

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.

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

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})^2$0.16 pu

For transformer T4: $X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^2$0.2 pu

For Motor, $X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^2$0.25 pu

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

Per Unit Impedance Diagram

### 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).

* * * * *

### 53 Responses to Per Unit System – Practice Problem Solved For Easy Understanding

1. Tom says:

Can someone help please how can i calculate R and X from p.u. to base units?

2. samantha says:

please help: Three transformers each rated 25 MVA, 38. 1 /3.81 kV are connected star-delta with a balanced load of three 0.6?, Y-connected resistors. Choose a base of 75 MVA, 66 kV for the high-voltage side of the transformer and specify the base for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then, determine the load resistance R_L i n ohms referred to the high-voltage side and the per-unit value of this resistance on the chosen base.

3. daran says:

4. Rusty says:

Shouldn’t the 3-ph load be multiplied by (100/57)?

5. Mian Faizan says:

Love it.. A job well done. 🙂

6. PAUL PEREZ says:

HELP! How do determine the equivalent three phase impedance of three single phase transformer? Example. Three each of 10 kva transformer, 7.52kv-240V, wiith an individual impedance of 4%. What is the equivalent impedance when they are connected in three phase at 30 kva.

@sayonsom

What we developed in this exercise is a positive sequence network. A transformer’s winding configuration does not affect it. The positive sequence network is good for calculating balanced load current and 3 phase faults (not involving ground).

The zero sequence network however does get affected by transformer winding configuration. Delta and ungrounded Y have a big impact in the design of zero sequence impedance network.

You will need a zero sequence network for unsymmetrical fault current analysis like L-G or L-L-G etc.

I am getting into symmetrical components with this discussion.

8. Vanjinathan P says:

good thing, to see a dedicated power system website.

9. sayonsom says:

How does the calculations change if the transformers are Y-Y or Y-Delta connected in different parts of the network?

10. Shalini says:

Awesome !!! Best explanation ever.. thank u

11. Naz says:

good job. its a comprehensive article. Appreciate your hard work

12. Jathin says:

Thanks a lot.. Nice elaboration..

13. Ashwin says:

Can someone tell me what would happen if for T2 and T4, the Primary and Secondary were swapped i.e. T2 would be 11/220 and T4 would be11/110?

I have a simialr problem and I am getting two different values for Vbase4.

14. narayan says:

IJAJ :what do you mean by….. s*-??
ans: it mean conjugate of S i.e changing sign of angle only
mike: When calculating Xtl2 using (22/22) . Vbase in T2 is 220 primary and 11 secondary, 220 come from?
ans: in all transformer we are allowed only take primary or secondary as reference. here primary is taken

15. Harsh says:

sir how to choose base kv value transformer? some time you take (22/22) nd sometime (10.45/11)

16. Aslam says:

Hi, thanks for the nice sharing,

I wana know how to model delta-grounded y transformer for load flow calculation.thanks

17. tuafi says:

nice job dude, highly comprehensible and presise, thumbs up

18. CJ says:

Good example of the method but pay attention the generator can supply only 90MVA the loads absorb 57 + 66.5 =123.5 MVA the network will in reality overload or under power the loads

19. Amey says:

if the transformer’s secondary is grounded by a neutral impedance then how to proceed with the calculations please suggest with an example

20. demis tesfaw says:

is that posible to calculate each parameter without giving a base voltage?

good job

22. IJAJ says:

what do you mean by….. s*-??

23. an says:

Can we find the short circuit current at each end?

24. samuel says:

A load of 50mw at 0.8 power factor lagging is taken from the 33kv.( taking a base MVA of 100mva), calculate the terminal voltage of the synchronous machine? (Please help me solve this question) thanks

25. Nikhil says:

very useful thanks 🙂

26. alshaia says:

How can we determine the voltage on the bus 1

27. BRian says:

Do you know how to find the voltage at the bus?
Thanks

28. tahseen says:

Hi
how we can find the voltage in bus1 in PU and in volte

29. noa says:

tanks alot save me alot of stress

30. abi says:

. Obtain the per unit impedance(reactance) diagram of the power system shown in the fig
G1 : 30MVA , 10.5KV, X?=1.6 ?
G2 : 15MVA , 6.6KV, X?=1.2 ?
G3 : 25MVA , 6.6KV, X?=0.56 ?
T1 (3 phase): 15MVA , 33/11KV , X= 15.2 ? per phase on the high tension side
T2 (3 phase): 15MVA , 33/6.2KV , X= 16 ? per phase on the high tension side
Transmission line : 20.5 ohm per phase
Load A : 15MW , 11KV , 0.9 p.f lagging
Load B : 40MW , 6.6KV , 0.85 p.f lagging
5

31. abi says:

how to convert ohms value to per unit value

32. Lee Taylor says:

Hi, great article – thanks very much! I have a similar problem to solve but I am struggling with the Zact calculation. My inputs are Vrated = 4.16kV, S = 2MVA <-36.87. Can you help?!

33. bhanu says:

awesome

34. kaushik vastarpara says:

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

@Pavan @Mike: That’s a typo. Correct values are now shown in the calculations. Since the ratio of Vbase_old/Vbase_new is the same, the end result, therefore, does not change. Appreciate the feedback.

36. Pavan says:

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,

37. mike says:

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.

38. mark says:

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…

39. chris says:

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,
Once you have the impedance network, use the current division rule to determine the current flowing each line. I am not sure I understand “voltage at 3”, if bus 3 is faulted (3ph) then it is zero otherwise it should be the same as nominal voltage as seen on the secondary side of the transformer.
I will solve one for the currents in the future but for now, you will have to learn how to reduce a circuit (using KVL and KCL) to determine the currents.

41. kam says:

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

42. karthik says:

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……….

43. kam says:

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

44. manish says:

what if transformers are connected in star and delta connection?

45. Anayat says:

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?

46. richa says:

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

47. Sanket says:

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

48. BABULS RAJ says:

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 …..

Nice catch. Fixed it. Thanks.

50. Renjith M says:

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’

51. Alfredo says:

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

52. Abdul Rauff says:

Very Good Info About PSA.Thanks Alot

53. Hilary says:

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