SAMPLE EXAM PROBLEM PROTECTION (6 OF 80 PROBLEMS)

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Chad Wood
5 years ago
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1 SAMPLE EXAM PROBLEM PROTECTION (6 OF 80 PROBLEMS) SLIDE In this video, we will cover a sample exam problem for the Power PE Exam. This exam problem falls under the topic of Protection, which accounts for 6 of 80 problems on the PE exam. The question reads, Two generators are connected in parallel and each serves their own transformer. The outputs of these transformers are connected to a common bus and then a common transmission line. If a three phase fault occurs on the transmission line, then what will be the fault current? The ratings of the equipment are shown on the next slide. You can assume the circuit is three phase and the neutrals are solidly grounded. SLIDE 2 The properties of the generators and transformers are shown on this slide. The voltages, MVAs and sub-transient reactances are provided for the generator. The voltages, MVAs and percent impdeances are provided for the transformers. Finally the rated voltage and impedance per phase of the transmission line is provided. This problem only provides the necessary information to complete the problem, but the PE exam may also include extra information like transformer configurations, transient reactances as opposed to sub-transient reactances and negative or zero components. Where as in this problem only positive components are shown. Generator : 500 MVA, 3.8 kv, X = 0.5 Transformer : 500 MVA, 3.8/230 kv, X = 0.3 Generator 2:,200 MVA, 3.8 kv, X = 0.08 Transformer 2:,200 MVA, 3.8/230 kv, X = 0.08 Transmission Line: 230 kv, X = 0 Ω per phase On the exam, you should be able to construct a simple one line diagram to show which components are in parallel and which components are in series, if the PE exam does not provide a diagram for you. Protection-

2 SLIDE 3 There are multiple methods to complete this problem, like the MVA method and the Per Unit Method. However, this video focuses on the MVA method and if you would like more information on the Per Unit method, please see the website in the description link. SLIDE 4 The first step in the MVA method is to convert all components properties to equivalent short circuit MVA. This is the maximum amount of power that can flow through the component during a short circuit. Each component has its own way of finding its equivalent short circuit MVA which will be covered in the next slide. SLIDE 5 For the generator the short circuit MVA value is found by dividing the full load MVA of the generator by the generator s sub transient reactance. Do not use the transient reactance or steady state reactance, since the transient and steady state reactance values do not give you the instantaneous short circuit current. SLIDE 6 ; Next, the transformer short circuit MVA value describes the total amount of apparent power through a transformer during a 3-phase fault. This value is found by dividing the full load MVA of the transformer by the percent impedance of the transformer. 00% % ; Protection-2

3 SLIDE 7 The short circuit current in a transmission line is simply a function of the voltage in the line and the impedance in the line. The short circuit MVA value is found by taking the rated voltage in kilo-volts squared and dividing the value by the per phase impedance. The units will result in MVA for the short circuit MVA of the transmission line. SLIDE 8 ; Ω The previous calculations will result in the following short circuit MVA values as shown in this figure. The next step is to combine each of these MVA values with the equations for combining components in parallel and series on the next slide. Protection-3

4 SLIDE 9 The following equations can be used to combine each component. You can notice from these equations that when parallel components are used, then the MVA is increased and when series components are used the resulting total MVA is reduced. SLIDE 0 First you want to work from left to right to simplify the circuit. First, combine the generator and transformer in series for both sets. Notice, that the series components will increase impedance to short circuit current which decreases the MVA short circuit value., ,500 Protection-4

5 SLIDE Then combine these two circuits in parallel. Notice that parallel lines provide additional paths for short circuit current to flow, which increases the MVA short circuit value. & 9,286 SLIDE 2 Finally combine the previous value in series with the transmission line, which results in the value as shown. 3,370 & SLIDE 3 Protection-5

6 Finally the last step is to use the resulting MVA combined value for the entire circuit from the power source to the fault location and the apparent power equation to find the short circuit current or fault current for a three phase fault. The equation is simply P = I*V*Root 3. Make sure you use the voltage at the fault location. The resulting answer is (C) 8,640 A. Thank you! 3, , 8, Protection-6

BE Semester- VI (Electrical Engineering) Question Bank (E 605 ELECTRICAL POWER SYSTEM - II) Y - Y transformer : 300 MVA, 33Y / 220Y kv, X = 15 %

BE Semester- VI (Electrical Engineering) Question Bank (E 605 ELECTRICAL POWER SYSTEM - II) Y - Y transformer : 300 MVA, 33Y / 220Y kv, X = 15 % BE Semester- V (Electrical Engineering) Question Bank (E 605 ELECTRCAL POWER SYSTEM - ) All questions carry equal marks (10 marks) Q.1 Explain per unit system in context with three-phase power system and