Per Unit System - Practice Problem Solved for Easy Understanding _ Power Systems Engineering

9/13/2017

Per Unit System - Practice Problem Solved For Easy Understanding | Power Systems Engineering

Power Pow er System Sys temss Engi E nginee neerri ng

Per Unit Un it System Sys tem – Prac P racti tice ce Probl Pro blem em Solve So lved d For Fo r Easy E asy Under Un dersta stand nding ing

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Let ’s understand the concept of per unit s yst em by solvi so lvi ng an examp ex amp le. In the one -line diagram below, the impedance of o f various components in a power

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sys tem, typically derived from their nameplates, name plates, are presented. The task now is to n ormalize these values using a common commo n base.

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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

Per Unit System -

the following.

Practice Problem Solved For Easy Understanding Power Circuit Breaker - Operation and Control Scheme Wha t D o Symm S ymm etr etrica ica l, Asy mme tri cal , Momentary, Interrupting, Close & Latch Ratings Mean? Direct Transfer Trip Scheme

http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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Figure 2: Impedance Diagram Of A Power System

Guide

Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance.

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Capacitive eects between lines and to ground are ignored as well.

Substation Sub-

To obtain the new normalized per unit impedances, rst we need to gure out

Surface Engineering:

the base values (Sbase, Vbase, Zbase) in the power system. Following steps will

An Overvi ew of Wha t a

lead you through the process.

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Step 1: Assume a system base Assume a s ystem wid e

of 100MVA. Thi s i s a r andom ass umptio n a nd chose n

to make calculations easy when calculating the per unit impedances. So,

= 100MVA

Step 2: Identify the voltage base Vol tage base in the sys tem is determined by the tra nsf ormer. For exa mpl e, wit h a 22/220kV voltage rating of T1 transformer, the

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

zone.

See gure below for the voltage bases in the system.


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Figure 3: Voltage Base In The Power System

Ste p 3: C alc ula te the bas e i mpedan ce

The base impedance is calculated using the following formula: Ohms…………………………………………………………………..(1)

For T-Line 1:

= 484 Ohms

For T-Line 2:

= 121 Ohms

For 3-phase load:

= 1.21 Ohms

Step 4: Calculate the per unit impedance The per unit impedance is calculated using the following formulas: ……………………………………………………………………………..(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): For T-line 2 using equation (2):

= 0.1 pu = 0.5 pu

For 3-Phase load: Power Factor: http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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Thus, = 1.1495+j1.53267 Ohms Per unit impedance of 3-phase load using equation (2)=

=

0.95+j1.2667 pu For generator, the new per unit reactance using equation (3) = 0.2 pu For transformer T1:

= 0.2 pu

For transformer T2:

= 0.15 pu

For transformer T3:

= 0.16 pu

For transformer T4:

= 0.2 pu

For Motor,

= 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 dened by the transformer and any o-nominal tap setting it may have. 3. Zbase is derived from the Sbase and Vbase. http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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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). ***** bas e v alu es

per unit

per unit impedances

per unit system

per unit value

Power

transformers

Per Unit Systems (41%, 71 Votes) Power System Protection (39%, 68 Votes) Power System Equipment Like Breakers, Transformers Etc (18%, 32 Votes) Other (2%, 3 Votes)

Total Voters:

53 Responses to Per Unit System – Practice Problem Solved For Easy Understanding says: Can someone help please how can i calculate R and X from p.u. to base uni ts?

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?, Yconnected resistors. Choose a base of 75 MVA, 66 kV for the high vol tage side of the tra nsforme r and specify the base for the low vol tage side. Deter mine the per-u nit resistanc e of the load on the base for the low-vol tag e side. T hen, det ermine the load


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resistance R_L i n ohms referred to the high-voltage side and the per-unit value of this resistance on the chosen base.

says: please add more

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

says: Love it.. A job well done.

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 equival ent impedan ce when they are conne cted in three phase at 30 kva.

says: @sayonsom What we dev eloped in this exercise is a positive sequenc e network. A transformer’s winding conguration does not aect 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 aected by transformer winding conguration. Delta and ungrounded Y have a big impact in the design of zero sequence impedance network. You will need a zero seque nce netwo rk for unsym metrical fault current analysis like L-G or L-L-G etc. I am getting into symmetrical components with this discussion.

says: http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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good thing, to see a dedicated power system website.

says: How does the calculations change if the transformers are Y-Y or Y-Delta connected in dierent parts of the net work?

says: Awesome !!! Best explanation ever.. thank u

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

says: Thanks a lot.. Nice elaboration..

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 dierent values for Vba se4.

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

says:


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

says: Hi, thanks for the nice sharing, I wana know how to model delta-grounded y transformer for load ow calculation.thanks

says: nice job dude, highly comprehensible and presise, thumbs up

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

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

says: is that posible to calculate each parameter without giving a base vol tage?

says: good job

says: what do you mean by… .. s*-??


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says: Can we nd the short circuit current at each end?

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 vol tage of the synch ronous machine? (Please help me solve thi s question) thanks

says: ver y useful thank s

says: How can we determine the voltage on the bus 1

says: Do you know how to nd the voltage at the bus? Thanks

says: Hi how we can nd the voltage in bus1 in PU and in volte

says: tanks alot save me alot of stress

says: . Obtain the per unit impedance(reactance) diagram of the power system shown in the g G1 : 30MVA , 10.5KV, X?=1.6 ? G2 : 15MVA , 6.6KV, X?=1.2 ? http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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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

says: how to convert ohms value to per unit value

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 Vra ted = 4.1 6kV , S = 2MV A <-3 6.87. Can you help?!

says: awesome

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

says: @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.

says:


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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,

says: I don’t understand one part: When calculating X tl2 you are using (22/22) which is reected from where? Vbase in T2 is 220 primary and 11 secondary, so where does 22 come from? The same for Xtl4.

says: will the impedan ce or p.u impedan ce 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 nd the actual line current for each line and for the whole system…

says: A single phase ,350 kva , 1380v generator has an interna l impedance Zg of j6 ohms. The generator is used to supply a load of 250kva/440v at power factor 0.78 l agging. 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!!!

says: Kam, Once you have the impedance network, use the current division rule to determine the current owing 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. http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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I will solve one for the currents in the future but for now, you will have to learn how to reduc e a circuit (us ing KVL and KCL ) to determine the currents.

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

says: ver y well explained but could y ou pls show me how to calculate vol tage and c urrent in both lines, will be ver y greatful , thank s a lot……….

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

says: what if trans formers are conne cted in star and delta connection?

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?

says: ver y nicely explained…. to the point and complete. .th anks a lot

says: VER Y NICELY EXPLAINED THANK -YOU ……… I WILL VISIT WEBSITE AGAIN FOR FURTHER REFERENCES. http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

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

says: Nice catch. Fixed it. Thanks.

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’

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

says: Ver y Good Info Abo ut PSA.Thank s Alot

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

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Per Unit System - Practice Problem Solved for Easy Understanding _ Power Systems Engineering

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