Per Unit System - Practice Problem Solved For Easy Understanding | Power Systems Eng...Page 1 of 11 Power Systems Engineering Per Unit System – Practice Problem Solved For Easy Understanding Enter your email address to subscribe to this blog and receive notifications of new 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 posts by email. Join 429 other subscribers system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base. Email Address Subscribe Try Google AdWords™ google.com/AdWords Get $75 Worth of Advertising When You Spend $25 With AdWords 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. Recent Posts Tags Choosing Between Resistor and Reactor for Neutral Ground Impedance Power Cable Neutral and System Grounding Substation SCADA System - Design Guide Cable and Conduit Installation in Substations - Best Practices Substation SubSurface Engineering: http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi... 10/10/2014 Power Cable Neutral To obtain the new normalized per unit impedances. Momentary.Per Unit System .Practice Problem Solved For Easy Understanding | Power Systems Eng. Following steps will lead you through the process. Page 2 of 11 An Overview of What a Geotech Survey Reveals Per Unit System Practice Problem Solved For Easy Understanding Power Circuit Breaker . So. first we need to figure out and System Grounding the base values (Sbase. It does not matter what the voltage rating of the other components are that are encompassed by the zone.Operation and Control Scheme What Do Symmetrical. Zbase) in the power system. See figure below for the voltage bases in the system.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi. For example. Asymmetrical... Vbase. How to Pass the Power System PE exam Capacitive effects between lines and to ground are ignored as well. Step 1: Assume a system base Assume a system wide of 100MVA. Interrupting. 10/10/2014 . Close & Figure 2: Impedance Diagram Of A Power System Latch Ratings Mean? Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance... This is a random assumption and chosen to make calculations easy when calculating the per unit impedances. with a 22/220kV voltage rating of T1 transformer. http://peguru. = 100MVA Step 2: Identify the voltage base Voltage base in the system is determined by the transformer. the on the primary side of T1 is 22kV while the secondary side is 220kV. .5 pu For 3-Phase load: Power Factor: Thus.1 pu = 0.(3) The voltage ratio in equation (3) is not equivalent to transformers voltage ratio. Page 3 of 11 Figure 3: Voltage Base In The Power System Step 3: Calculate the base impedance 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. 10/10/2014 .21 Ohms Step 4: Calculate the per unit impedance The per unit impedance is calculated using the following formulas: …………………………………………………………………………….. For T-line 1 using equation (2): For T-line 2 using equation (2): = 0..Per Unit System ..Practice Problem Solved For Easy Understanding | Power Systems Eng.(2) ……………………………….com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi. 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. http://peguru... .2 pu For transformer T2: = 0.. Zbase is derived from the Sbase and Vbase.Practice Problem Solved For Easy Understanding | Power Systems Eng. = 0. the new per unit reactance using equation (3) = 0.2 pu For transformer T1: = 0. 2.. 3.Per Unit System ..15 pu For transformer T3: = 0.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi.25 pu The equivalent impedance network with all the impedances normalized to a common system base and the appropriate voltage base is provided below. 4.2667 pu For generator. 10/10/2014 .16 pu For transformer T4: = 0. The Vbase is defined by the transformer and any off-nominal tap setting it may have. Per Unit Impedance Diagram Summary: 1.95+j1. 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). Assume a Sbase for the entire system.53267 Ohms Per unit impedance of 3-phase load using equation (2)= = 0. Page 4 of 11 = 1.1495+j1.2 pu For Motor. ***** http://peguru. . Choosing Between Resistor and Reactor for Neutral Ground Impedance Faults involving ground produce high fault current magnitude especially when the transformer (s) neutral is solidly grounded....0%Transformer No. You will find it on the .8 power factor lagging is taken from the 33kv. s*-?? an says: Can we find the short circuit current at each end? samuel says: A load of 50mw at 0. Determine How To Load Two Paralleled Transformers Two three phase transformers with ratings shown below are paralleled on their secondary side.. 10/10/2014 . calculate the terminal voltage of the synchronous machine? (Please help me solve this question) thanks Nikhil says: very useful thanks alshaia says: http://peguru. You may also like: Book Review: Protective Relaying . ( taking a base MVA of 100mva). 1 ... Z = 5. Mounting Provisions for Pole Mounted Transformers Pole mounted transformers for station service in substations or to feed residential or commercial buildings are fitted with mounting lugs that are sized per . It is hard to miss this book if you have been in the power industry for a while.Transformer No.Per Unit System .. It is time to step back and understand this concept from ......com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi. Symmetrical Components Demystified Learning symmetrical components can be overwhelming if you consider its mathematical aspect only. 32 Responses to Per Unit System – Practice Problem Solved For Easy Understanding IJAJ says: what do you mean by…..Principles and Application by Lewis Blackburn and Thomas . Page 5 of 11 base values per unit per unit impedances per unit system per unit value Power transformers Please subscribe to this blog to receive informative articles. 2 ..Practice Problem Solved For Easy Understanding | Power Systems Eng... The neutral ground circuit in the transformer provides . 4160V/480V.1000kVA. 2 ? per phase on the high tension side T2 (3 phase): 15MVA . 33/6. X= 16 ? per phase on the high tension side Transmission line : 20.. 6.6KV. 0. 6..5KV...Practice Problem Solved For Easy Understanding | Power Systems Eng. X= 15. Obtain the per unit impedance(reactance) diagram of the power system shown in the fig G1 : 30MVA .f lagging Load B : 40MW .com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi.Per Unit System .5 ohm per phase Load A : 15MW . 33/11KV .6KV. great article – thanks very much! I have a similar problem to http://peguru.2KV . 10. 6. 10/10/2014 .6 ? G2 : 15MVA .85 p. X?=1.9 p. 0.2 ? G3 : 25MVA . 11KV . X?=0.6KV .f lagging 5 abi says: how to convert ohms value to per unit value Lee Taylor says: Hi. X?=1.56 ? T1 (3 phase): 15MVA . Page 6 of 11 How can we determine the voltage on the bus 1 BRian says: Do you know how to find the voltage at the bus? Thanks tahseen says: Hi how we can find the voltage in bus1 in PU and in volte noa says: tanks alot save me alot of stress abi says: . Page 7 of 11 solve but I am struggling with the Zact calculation.. Correct values are now shown in the calculations.16kV. My inputs are Vrated = 4.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… http://peguru. Appreciate the feedback. Since the ratio of Vbase_old/Vbase_new is the same.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi. Pavan says: This is really helpful.Per Unit System . does not change. Keep up the good work! Thanks. it seems like a question worth answering. therefore.87... I didn’t really got it when reading through this.Practice Problem Solved For Easy Understanding | Power Systems Eng. S = 2MVA <-36. Can you help?! bhanu says: awesome 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 Admin says: @Pavan @Mike: That’s a typo.. 10/10/2014 . mark says: will the impedance or p. However the content is really clear and understandable. the end result. so where does 22 come from? The same for Xtl4. but when I saw the below comment by Mike. 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. determine: the turns ratio of the transfomer. the voltage regulation of the system.. so could you pls help me out???????? thanks a lot karthik says: very well explained but could you pls show me how to calculate voltage and current in both lines.Practice Problem Solved For Easy Understanding | Power Systems Eng. Page 8 of 11 chris says: A single phase . the impedance per km if the line between the generator and the transformer is 5km. thanks a lot………. 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.78 lagging. 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 . kam says: sorry but i didnt get my reply yet. you will have to learn how to reduce a circuit (using KVL and KCL) to determine the currents. 10/10/2014 .. CAN SOMEONE HELP ME WIT THESE CALCULATION PLZ!!! Admin says: Kam. use the current division rule to determine the current flowing each line. thanks a lot http://peguru.350 kva. will be very greatful . Once you have the impedance network. The generator is used to supply a load of 250kva/440v at power factor 0. I will solve one for the currents in the future but for now.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi.. I am not sure I understand “voltage at 3″.Per Unit 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. 1380v generator has an internal impedance Zg of j6 ohms. .. so i really dont know abt all these concepts . The per-unit system is the ratio of two quantities of the same units.Per Unit System . Clear explanation with proper diagrams with multi colour…….com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi..thanks a lot Sanket says: VERY NICELY EXPLAINED THANK-YOU ……… I WILL VISIT WEBSITE AGAIN FOR FURTHER REFERENCES...to the point and complete. Thanks. Well that is what I know.. Therefore it is unitless.U’. and all the rating given are three phase line to line ? how we ll solve it then? richa says: very nicely explained…. i got it finally from your site.u’ http://peguru. But there is a small error. So you need to remove the ‘Ohms’ from the text and insert ‘p.Practice Problem Solved For Easy Understanding | Power Systems Eng..after searching for a proper explanation for the same in so many sites. BABULS RAJ says: Thank u so much…. what we only have T1 and T2 . Renjith M says: Commendable work. So accordingly we specify the per-unit quatities as just ‘P. Page 9 of 11 manish says: what if transformers are connected in star and delta connection? Anayat says: i am very new to Power side . 10/10/2014 .very nice …. Fixed it. Admin says: Nice catch. Abdul Rauff says: Very Good Info About PSA. Page 10 of 11 Alfredo says: It was very useful.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi.. any way I liked.. Yamayee AB O UT CO N T ACT TE R MS & C O ND IT I O NS PR IV ACY PO L IC Y SI T E M A P http://peguru. Bala Lewis Blackburn Protective Relaying AC system W ELCO M E T O P E G U RU flash test asymmetrical current lockout relays modern power system momentary rating NCEES Professional NEC 2011 Handbook Power System Analysis and Design neutral grounding NFPA 70E OSHA 1910 Pass PE Exam permissive over reaching transfer trip per unit Engineering system PE Sample Problems pilot relaying power blackout power flow power system Power transformers PPE protective relaying short circuit rating steady state reactive power support stability substation conduit details symmetrical components symmetrical current system frequency thermal limitation transformers voltage drop limitation Zia A..Practice Problem Solved For Easy Understanding | Power Systems E.Thanks Alot Hilary says: Protection engieering. but it is short because is necessary to get the complete solution.. i have been give the reactance as Xd’ to calculate faults on a system do i convert to Xd” how do i do this Leave a Reply A R CH I V ES January 2013 December 2012 September 2012 June 2012 May 2012 March 2012 January 2012 November 2011 August 2011 July 2011 June 2011 May 2011 April 2011 March 2011 September 2010 July 2010 June 2010 TA G C LO U D B O O K S O N MY SH ELF arc flash study arc base values cable shield grounding direct buried conduits direct transfer trip FACT generator swing Ground grid HVDC IEEE 1584 John Camara Juan L.Per Unit System . 10/10/2014 . com/2011/06/per-unit-system-practice-problem-solved-for-easy-understandi. 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