STUDY OF IDEAL TRANSFORMER AND PRACTICAL TRANSFORMER


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1 STUDY OF IDEAL TRANSFORMER AND PRACTICAL TRANSFORMER RUBY DHANKAR, SAPNA KAMRA,VISHAL JANGRA Abstract This paper proposes the study of real and ideal transformer. It also explains load, noload conditions and phasor diagrams of ideal and practical transformer. For a better understanding and an easier explanation of a practical transformer, certain idealizing assumptions are made which are close approximations for a practical transformer. Keywords transformer, flux, reactance, magnetizing current, primary and secondary windings. I. INTRODUCTION Michel faraday introduced the principle of electro magnetic induction in 1831.it estates that a voltage appear across the terminals of an electric coil when the flux linked with the coil changes. The magnitude of the induced voltage is corresponding to the rate of change of flux linkages.this finding forms the basic for many magneto electric devices.the primitive use of this phenomenon was in the development of induction coils. These coils were used to develope high voltage pulses to ignite the explosive charges in the mines. As the d.c. power arrangement was in use at that time, very little of transformer principle was made use of. In the d.c. supply system the generating station and the load centre have to be necessarily close to each other due to the demand of economic transmission of power. Also the d.c. generators cannot be scaled up due to the drawback of the commutator. 2. Discussion As discussed earlier the transformer is a static device working on the principle of faraday s law of induction. It estates that a voltage appears across the terminals of an electric coil when the flux correlated with the same changes.this emf is proportional to the rate of variation of flux linkages. Establishing mathematically. Here the common constuctional aspects alone are explained. 1. Ideal transformer 2. practical transformer 2.1. ideal transformer Earlier it is seen that a voltage is induced in a coil when the flux correlated with the same changes. It posses certain essential features of a real transformer but some details of minor significance are ignored which will be introduced stepbystep while analyzing a transformer. The idealizing assumptions made are as follows 1) No winding resistance 2) No magnetic leakage 3) No iron loss and zero magnetizing current Ideal transformer equations: using faraday s law of induction.. (1) IJIRT INTERNATONAL JOURNAL OF INNOVATIVE RESEARCH IN TECHNOLOGY 1258
2 ... (2) Combining ratio of (1) & ( 2) turns ratio will be equal to... (3) Where for stepdown transformers, a > 1 for stepup transformers, a < 1 By law of Conservation of Energy, apparent, real and reactive power are each conserved in the input and output Winding resistances in transformer Primary and secondary windings made up of copper wire. Every conductor has its own resistance. So primary and secondary side both have resistance. Primary resistance R 1 and secondary resistance R 2 are in series with the respect to the windings. Figure 1 shows that the resistance of the windings on both sides. Due to winding resistance when current passes through the windings there will voltage drop IR and creates power loss. thus the E 1 < V 1 and V 2 < E (4) Combining (3) & (4) with this endnote ideal transformer identity yields the... (5) By Ohm's Law and ideal transformer identity... (6) Apparent load impedance Z' L (Z L referred to the primary) 2.2. practical transformer... (7) Leakage reactance in transformer Mutual Flux Φ Links With Both The Primary And Secondary Side. Primary Current I 1 Creates Individual Flux Φ 1 In Primary Side And Secondary Current I 2 Produces Flux Φ 2 In Secondary Side, Those Two Fluxes Is Not Common In Both Sides. Thus The Flux Φ 1 And Flux Φ 2 Are Known As Leakage Flux in Transformer. Figure 2 Shows Leakage Fluxes Ideal transformer has no losses although practical transformer have Iron loss Magnetic leakage Winding resistances Iron losses in transformer Alternating flux Φ passes through iron kernel. It produces eddy current and hysteresis loss in it. Two of these losses called either iron loss or core loss. Iron loss rely upon the core volume, supply frequency, maximum flux density etc. Magnitude of iron loss is small in practical transformer. The way of leakage flux is through the air mainly. The effect of primary leakage flux Φ 1 generates an inductive reactance X 1 series in primary winding and secondary leakage flux Φ 2 introduces an IJIRT INTERNATONAL JOURNAL OF INNOVATIVE RESEARCH IN TECHNOLOGY 1259
3 inductive reactance X 2 in series with the secondary winding shown in figure 1. Primary leakage inductance, L 1 = primary leakage flux linkages / primary current = N 1 Φ 1 / I 1 Primary leakage reactance, is X 1 = 2πf L 1 In the same manner secondary leakage inductance, L 2 = N 2 Φ 2 / I 2 Secondary leakage reactance, is X 2 = 2πf L 2.There is No power loss occurs due to leakage reactance. But it changes the power factor as well there is voltage loss due to IX drop. Flux leakage is absolutely small about 5% of mutual flux Φ in a transformer. But it cannot be avoided or ignored No load practical transformer The component I w is admitted as iron loss or active or working component. It is in phase with the enforced voltage V 1. It provides a very small primary copper loss and iron loss. I w = I 0 cos Φ 0 Here It is clear that I 0 is the phasor sum of I m and I w. I 0 = (I m 2 + I w 2 ) No load power factor, cos Φ 0 = I w / I 0 At no load practical transformer primary copper loss I 0 2 R is quite small and this loss may be ignored. Hence, primary no load input power of practical transformer is equal to the iron loss in transformer. No load input power is, W 0 = Iron loss As primary loss in practical transformer is quite small so it may be written at no load, V 1 = E 1. Here is no load in secondary so E 2 = V 2. Practical transformer on no load phasor diagram Figure 4 is the phasor of practical transformer on no load condition. Primary small current I 0 is phasor sum of I m and I w. A practical transformer diagram is shown in figure 3, there is no load in secondary terminal it is open circuited. When ac source is coupled in primary a small current I 0 flows through the primary. It occurs a very small extent of copper loss and iron loss in the primary. In order that the primary no load current I 0 is not 90 behind the applied voltage V 1 but lags it by angle Φ 0 < 90. Primary no load current I 0 lags by V 1 voltage by an angle Φ 0 < 90. We can deduce the magnetizing and iron loss current using above equations now we will solve a math. Here No load input power, W 0 = V 1 I 0 cos Φ 0 In primary side for I 0 we get two components I w and I m. The component I m is admitted as magnetizing component. This component creates mutual flux Φ in the core. I m lagging behind V 1 by 90. I m = I 0 sin Φ 0 IJIRT INTERNATONAL JOURNAL OF INNOVATIVE RESEARCH IN TECHNOLOGY 1260
4 For practical transformer on load we will take two cases Case 1: when the transformer has no winding resistance and leakage flux is also zero. Case 2: in this when the transformer has winding resistance and leakage flux Case 1: No winding resistance and no leakage flux Phasor diagram of practical transformer on load Figure 1 :represent a practical transformer on load. We have to suppose it has no winding resistance and no leakage flux. Another word winding resistance and leakage flux is neglected. For this supposition V 1 = E 1 and V 2 = E 2. Take the load at secondary is inductive load which causes the secondary current I 2 to lag the secondary voltage V 2 by Φ 2. Primary current I 1 must meet two conditions i) Primary current must supply the no load current I 0 to accommodate the iron losses in the transformer and to provide flux in the core. ii) Primary current must supply a current I 2 to counteract the demagnetizing influence of secondary current I 2. The magnitude of I 2 will be Figure 2 : represent the phasor diagram of practical transformer on load for inductive load. Here E 1 and E 2 are lagging behind by common flux Φ by 90. Phasor sum of I 0 and I 2 is the primary current I 1. I 2 is anti phase with I 2. The value of K is supposed to be unity so primary phasor is equal to secondary phasor. Primary power factor = cos Φ 1 Secondary power factor = cos Φ 2 Primary input power = V 1 I 1 cos Φ 1 Secondary input power = V 2 I 2 cos Φ 2 Case 2: Transformer with resistance and leakage reactance N 1 I 2 = N 2 I 2 Or I 2 = I 2 N 2 /N 1 = K I 2 The phasor sum of I 2 and I 0 is the total primary current I 1. I 1 = I 2 + I 0 Whereas I 2 = K I 2 I 2 is 180 out of phase with I 2 current. IJIRT INTERNATONAL JOURNAL OF INNOVATIVE RESEARCH IN TECHNOLOGY 1261
5 Figure 3 :shows a practical transformer with winding resistance and leakage resistance. This is real condition which exists in a practical transformer. Voltage drop R 1 and X 1 occurs in primary side so V 1 > E 1 and voltage drop R 2 and X 2 occurs in secondary side so V 2 < E 2. Assume an inductive load which causes the secondary current I 2 to lag behind the secondary voltage V 2 by Φ 2. Primary current I 1 must follow the two conditions i) Primary current must supply no load current I 0 to meet the iron losses in the transformer and to provide flux in the core. ii) It must supply a current I 2 to counteract the demagnetizing influence of secondary current I 2. The magnitude of I 2 will be N 1 I 2 = N 2 I 2 Primary input power = V 1 I 1 cos Φ 1 Secondary input power = V 2 I 2 cos Φ 2 I 1 = I 2 + I 0 Where I 2 = K I 2 3. Conclusion This paper concludes that An ideal transformer is the one whose windings do not have any ohmic resistance and whose core does not have any leakage flux and eddy current losses. A real transformer is the one whose windings do have some amount of ohmic resistance and the core also have some leakage flux and eddy current losses. References 1. J.B. Gupta Electromechanical energy conversion, S.K. kataria & Sons. 2. ece.colorado.edu/bart/book Or I 2 = I 2 N 2 /N 1 = K I 2 The phasor sum of I 2 and I 0 is the total primary current I 1. Phasor diagram of practical transformer resistance and reactance: with Figure 4: represents the phasor diagram of a practical transformer for the usual case of inductive load. Here E 1 and E 2 are lagging behind by mutual flux Φ by 90. Primary power factor = cos Φ 1 Load power factor = cos Φ 2 IJIRT INTERNATONAL JOURNAL OF INNOVATIVE RESEARCH IN TECHNOLOGY 1262
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