L-28(TB)(ET) ((EE)NPTEL) - Module 7 Transformer Version 2...

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Module 7 Transformer Version 2 EE IIT, Kharagpur
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Lesson 28 Problem solving on Transformers Version 2 EE IIT, Kharagpur
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Contents 28 Problem solving on Transformer (Lesson-28) 4 28.1 Introduction ……………………………………………………………………. 4 28.2 Problems on 2 winding single phase transformers ……………………………. 4 28.3 Problems on 3-phase ideal transformer ………………………………………. .. 10 28.4 Problems on ideal auto transformers …………………………………………. .. 14 Version 2 EE IIT, Kharagpur
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28.1 Introduction In this lesson some typical problems on transformer are solved with emphasis on logical steps involved. For a practical two winding transformer, the knowledge of approximate equivalent circuit is of utmost importance in order to predict its performance. Equivalent circuit parameters are either supplied directly or indirectly in terms of O.C and S.C test data. The first problem enumerates in detail how to get the equivalent circuit parameters from test data. The importance of the side (LV or HV) in which calculations are carried out is highlighted. The second problem, in fact, is an extension of the first problem. Calculation of regulation, efficiency and maximum efficiency are dealt with in these problems. Next few problems highlight the basic calculation steps involved in ideal 3-phase transformer and ideal auto transformer since the equivalent circuit of these transformers are outside the scope of first year electrical technology course. 28.2 Problems on 2 winding single phase transformers 1. The O.C and S.C test data are given below for a single phase, 5 kVA, 200V/400V, 50Hz transformer. O.C test from LV side : 200V 1.25A 150W S.C test from HV side: 20VV 12.5A 175W Draw the equivalent circuit of the transformer (i) referred to LV side and (ii) referred to HV side inserting all the parameter values. Solution Let us represent LV side parameters with suffix 1 and HV side parameters with suffix 2. Computation with O.C test data Let us show the test data in the approximate equivalent circuit (Figure 28.1) of the transformer as given below. Due to the fact that the HV side is open circuited, there will be no current in the branch e1 e1 + rj x . So entire power of 150W is practically dissipated in R cl 1 . The no load current I 01 = 1.25 A is divided into: magnetizing component I m 1 and core loss component I cl 1 as depicted in the phasor diagram figure 28.1. 1.25 A 200 V Figure 28.1: O.C equivalent circuit and phasor diagram. r e1 x e1 I m1 X m1 I c1 R c11 Open circuit I c11 I m1 I o1 θ o θ o V 1 Version 2 EE IIT, Kharagpur
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No load (or O.C) power factor cos θ o = 150 200 1.25 × = 0.6 θ o = cos -1 0.6 = 53.13º Hence, sin θ o = 0.8 After knowing the value of cos θ o and sin θ o and referring to the no load phasor diagram, I m 1 and I cl 1 can be easily calculated as follows.
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L-28(TB)(ET) ((EE)NPTEL) - Module 7 Transformer Version 2...

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