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Download Week 1 - Transformer
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ELECTRICAL MACHINE DET 204/3 JIMIRAFIZI BIN JAMIL CHAPTER 1 Transformer Introduction 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. The most important transformers are: tasks performed by • 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). 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. Ideal Transformer An ideal transformer is a lossless device with an input winding and output winding. v p (t ) v s (t ) Np Ns a a = turns ratio of the transformer a = N2 = V2 N1 V1 Sp SS 1 lossless N p i p ( t ) N s is ( t ) i p (t ) i s (t ) 1 a Power in ideal transformer Output Power Pout Pin V p I p cos Reactive Power Qout Qin V p I p sin Apparent Power S out S in V p I p sin 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: ZL VL IL ZL' a2ZL The equivalent circuit of a transformer The major items to be considered in the construction of such a model are: • Copper (I2R) 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 ΦLP and ΦLS 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. Nonideal or actual transformer Mutual flux Nonideal or actual transformer Ep = primary induced voltage Es = secondary induced voltage Vp = primary terminal voltage Vs = secondary terminal voltage Ip = primary current Is = secondary current Ie = excitation current IM = magnetizing current XM = magnetizing reactance IC = core current RC = core resistance Rp = resistance of primary winding Rs = resistance of the secondary winding Xp = primary leakage reactance Xs = secondary leakage reactance Exact equivalent circuit of the actual transformer a) The transformer model referred to primary side b) The transformer model referred to secondary side Approximate equivalent circuit of the actual transformer a) The transformer model referred to primary side b) The transformer model referred to secondary side Per unit System The per unit value of any quantity is defined as Actual Quantity Per Unit, pu Base value of quantity Quantity – may be power, voltage, current or impedance Two major advantages in using a per unit system: 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. Per Unit (pu) in Single Phase System Pbase , Sbase ,Qbase VbaseI base Vbase Z base I base Ybase I base Vbase ( Vbase )2 Z base Sbase 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 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 = 9,600 V1 = 9,600 2,400 = 4A I2 = 9,600 V2 = 9,600 120 @ = 80 A I2 = I1 a 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 236.9 o . 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 !