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ELECTRICAL MACHINE DET 204/3 JIMIRAFIZI BIN JAMIL Transformer CHAPTER 1.

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Presentation on theme: "ELECTRICAL MACHINE DET 204/3 JIMIRAFIZI BIN JAMIL Transformer CHAPTER 1."— Presentation transcript:

1 ELECTRICAL MACHINE DET 204/3 JIMIRAFIZI BIN JAMIL Transformer CHAPTER 1

2 A transformer is a device that changes ac electric energy at one voltage level to ac electric energy at another voltage level through the action of a magnetic field. Introduction

3 The most important tasks performed by transformers are: changing voltage and current levels in electric power systems. matching source and load impedances for maximum power transfer in electronic and control circuitry. electrical isolation (isolating one circuit from another or isolating dc while maintaining ac continuity between two circuits).

4 It consists of two or more coils of wire wrapped around a common ferromagnetic core. One of the transformer windings is connected to a source of ac electric power – is called primary winding and the second transformer winding supplies electric power to loads – is called secondary winding.

5 An ideal transformer is a lossless device with an input winding and output winding. Ideal Transformer lossless a = turns ratio of the transformer a = N2 = V2 N1 V1

6 Power in ideal transformer Output Power Reactive Power Apparent Power

7 Impedance transformation through the transformer The impedance of a device – the ratio of the phasor voltage across it in the phasor current flowing through it:

8 The equivalent circuit of a transformer Copper (I 2 R) losses: Copper losses are the resistive heating in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings. Eddy current losses: Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer. Hysteresis losses: Hysteresis losses are associated with the arrangement of the magnetic domain in the core during each half cycle. They are complex, nonlinear function of the voltage applied to the transformer. Leakage flux: The fluxes Φ LP and Φ LS which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self inductance in the primary and secondary coils, and the effects of this inductance must be accounted for. The major items to be considered in the construction of such a model are:

9 Nonideal or actual transformer Mutual flux

10 Nonideal or actual transformer E p = primary induced voltageE s = secondary induced voltage V p = primary terminal voltageV s = secondary terminal voltage I p = primary currentI s = secondary current I e = excitation currentI M = magnetizing current X M = magnetizing reactance I C = core current R C = core resistanceR p = resistance of primary winding R s = resistance of the secondary windingX p = primary leakage reactance X s = secondary leakage reactance

11 Exact equivalent circuit of the actual transformer a)The transformer model referred to primary side b)The transformer model referred to secondary side

12 Approximate equivalent circuit of the actual transformer a)The transformer model referred to primary side b)The transformer model referred to secondary side

13 Per unit System The per unit value of any quantity is defined as Quantity – may be power, voltage, current or impedance

14 1.It eliminates the need for conversion of the voltages, currents, and impedances across every transformer in the circuit; thus, there is less chance of computational errors. 2.The need to transform from three phase to single phase equivalents circuits, and vise versa, is avoided with the per unit quantities; hence, there is less confusion in handling and manipulating the various parameters in three phase system. Two major advantages in using a per unit system:

15 Per Unit (pu) in Single Phase System

16 Example An ideal transformer is rated at 2400/120 V, 9.6 kVA and has 50 turns on the secondary side. Calculate; a)Turns ratio b)The number of turns on the primary side c)The current ratings for the primary and secondary windings

17 Solution a)This is step-down transformer, since V1 = 2,400 V > V2 = 120 V a = V2 = 2,400 = 20 V1 120 b) N1 = ? a = N2 N1 20 = 50 N1 N1 = 50 20 = 2.5 turns c) S = V1 I1 = V2 I2 = 9.6kVA. I1 = 9,600I2 = 9,600@I2 = I1 V1 V2 a = 9,600 = 4A = 9,600 = 80 A 2,400 120

18 Assignment 2 Consider an ideal, single phase 2400/240 V transformer. The primary is connected to a 2200 V source and the secondary is connected to an impedance of. a) Find the secondary output current and voltage. b) Find the primary input current. c) Find the load impedance as seen from the primary side. d) Find the input and output apparent powers. e) Find the output power factor. Try It !


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