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EET 103 Chapter 5 (Lecture 1) Single Phase Transformer 1 Introduction to Transformer 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. 2 Introduction to Transformer • Transformers are constructed of two coils or more placed around the common freeomagnetic core so that the charging flux developed by one will link to the other. • The coil to which the source is applied is called the primary coil. • The coil to which the load is applied is called the secondary coil. 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 5 Mutual Inductance • Mutual inductance exits between coils of the same or different dimensions. • Mutual inductance is a phenomenon basic to the operation of the transformer. 6 Mutual Inductance • A transformer is constructed of 2 coils placed so that the changing flux developed by one will link the other. • The coil to which the source is applied is called primary • The coil which the load is applied is called secondary. 7 Mutual Inductance (Cont…) 8 9 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 i p (t ) 1 i s (t ) a N pi p ( t ) N sis ( t ) a = turns ratio of the transformer 10 Power in ideal transformer Pout Pin V p I p cos Qout Qin V p I p sin S out Sin Vs I s V p I P Where is the angle between voltage and current 11 Impedance transformation transformer through the 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 12 Non-ideal or actual transformer Mutual flux 13 Losses in the transformer • 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. 14 Losses in 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. 15 Transformer equivalent circuit Ie Ic Im Ep = primary induced voltage Vp = primary terminal voltage Ip = primary current Ie = excitation current XM = magnetizing reactance RC = core resistance Rs = resistance of the secondary winding Xs = secondary leakage reactance Es = secondary induced voltage Vs = secondary terminal voltage Is = secondary current IM = magnetizing current IC = core current Rp = resistance of primary winding Xp = primary leakage reactance 16 Dot convention 1. If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core. 2. If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding. 17 Exact equivalent circuit the actual transformer a. The transformer model referred to primary side b. The transformer model referred to secondary side 18 Approximate equivalent circuit the actual transformer a. The transformer model referred to primary side b. The transformer model referred to secondary side 19 Exact equivalent circuit of a transformer refer to primary side Ep = primary induced voltage Vp = primary terminal voltage Ip = primary current Ie = excitation current XM = magnetizing reactance RC = core resistance winding Rs = resistance of the secondary winding Xs = secondary leakage reactance Es = secondary induced voltage Vs = secondary terminal voltage Is = secondary current IM = magnetizing current IC = core current Rp = resistance of primary Xp = primary leakage reactance 20 The Equation: Primary side Secondary side I p Ie I s / a ES I s ( Rs jX s ) Vs Ie IC I M Vs I s Z L V p I p ( R p jX p ) E p E p I C RC Vp Ep Is N p a Vs Es I p N s E p I M ( jX M ) E p I e ( RC // jX M ) 21 Exact equivalent circuit of a transformer referred to primary side Rp Ip a2Xs Xp Is/a a2Rs Ie Vp Ep aVs Exact equivalent circuit of a transformer referred to secondary side Rp/a2 aIp Xp/a2 Rs Is Xs aIe Vp/a aIc Rc/a2 aIm Ep/a = Es Vs XM/a2 22 Approximate equivalent circuit of a transformer referred to primary side Ip Reqp jXeqp Is/a + + Reqp=Rp+a2Rs Vp Rc aVs jXM - Xeqp=Xp+a2Xs - Approximate equivalent circuit of a transformer referred to secondary side aIp Reqs + jXeqs Is + Reqs=Rp / a2+Rs Vp/a Rc/a2 - jXM/a2 Vs Xeqs=Xp / a2+Xs 23 Example 1 A single phase power system consists of a 480V 60Hz generator supplying a load Zload=4+j3W through a transmission line ZLine=0.18+j0.24W. Answer the following question about the system. a) If the power system is exactly as described below (figure 1(a)), what will be the voltage at the load be? What will the transmission line losses be? b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line (figure 1(b)). What will the load voltage be now? What will the transmission line losses be now? 24 ILine IG ZLoad=0.18+j0.24W + ILoad VLoad V=48000V ZLoad=4+j3W - Figure 1 (a) T1 IG 1:10 ILine T2 ILoad ZLine=0.18+j0.24W + 10:1 ZLoad=4+j3W V=48000V VLoad Figure 1 (b) - 25 Solution 5.1 (a) From figure 1 (a) shows the power system without transformers. Hence IG = ILINE = ILoad. The line current in this system is given by I line V Z line Z load 4800V (0.18W j 0.24W) ( 4W j 3W) 4800 4800 4.18 j 3.24W 5.2937.8 90.8 37.8 26 Therefore the load voltage is Vload I lineZ load (90.8 37.8 A)( 4W j 3W) (90.8 37.8 A)(536.9W) 454 0.9 and the line losses are Ploss I line Rline 2 (90.8 A) (0.18W) 2 1484W 27 (b) From figure 1 (b) shows the power system with the transformers. To analyze the system, it is necessary to convert it to a common voltage level. This is done in two steps i) Eliminate transformer T2 by referring the load over to the transmission’s line voltage level. ii) Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the transmission line’s voltage over to the source side. The value of the load’s impedance when reflected to the transmission system’s voltage is Z 'load a 2 Z load 10 2 ( ) ( 4W j 3W) 1 400W j 300W 28 The total impedance at the transmission line level is now Z eq Z line Z 'load 400.18W j 300.24W 500.336.88W The total impedance at the transmission line level (Zline+Z’load) is now reflected across T1 to the source’s voltage level Z 'eq a 2 Z eq a 2 ( Z line Z 'load ) 1 2 ( ) (0.18W j 0.24W 400W j 300W) 10 (0.0018W j 0.0024W 4W j 3W) 5.00336.88W 29 Notice that Z’’load = 4+j3 W and Z’line =0.0018+j0.0024 W. The resulting equivalent circuit is shown below. The generator’s current is 4800V IG 95.94 36.88 A 5.00336.88W a) System with the load referred to the transmission system voltage level b) System with the load and transmission line referred to the generator’s voltage level 30 Knowing the current IG, we can now work back and find Iline and ILoad. Working back through T1, we get N p1 I G N s1 I line I line N p1 N S1 IG 1 (95.94 36.88 A) 10 9.594 36.88 A 31 Working back through T2 gives N p 2 I line N s 2 I load I load N p2 Ns2 I line 10 ( )(9.594 36.88 A) 1 95.94 36.88 A It is now possible to answer the questions. The load voltage is given by Vload I load Z load (9.594 36.88 A)(536.87W) 479.7 0.01V 32 the line losses are given by Ploss ( I line ) 2 Rline (9.594 A) (0.18W) 2 16.7W Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers compared to the system without transformers. 33 Parameter determination of the transformer Open circuit test Provides magnetizing reactance and core loss resistance Obtain components are connected in parallel 34 Experiment Setup In the open circuit test, transformer rated voltage is applied to the primary voltage side of the transformer with the secondary side left open. Measurements of power, current, and voltage are made on the primary side. Since the secondary side is open, the input current IOC is equal to the excitation current through the shunt excitation branch. Because this current is very small, about 5% of rated value, the voltage drop across the secondary winding and the winding copper losses are neglected. 35 Admittance Yoc I oc Voc Open circuit Power Factor Poc PF cos Voc I oc Open circuit Power Factor Angle cos 1 Poc Voc I oc Angle of current always lags angle of voltage by I oc 1 1 Yoc GC jBM j Voc RC XM 36 Short circuit test – Provides combined leakage reactance and winding resistance – Obtain components are connected in series 37 Experiment Setup In the short circuit test, the secondary side is short circuited and the primary side is connected to a variable, low voltage source. Measurements of power, current, and voltage are made on the primary side. The applied voltage is adjusted until rated short circuit currents flows in the windings. This voltage is generally much smaller than the rated voltage. 38 Impedances referred to the primary side Z sc Vsc I sc Power Factor of the current Psc PF cos Vsc I sc Angle Power Factor 1 Psc cos Vsc I sc Therefore Z sc Vsc0 0 Vsc 0 0 I sc I sc Z sc Req jX eq R p a Rs j X p a X s 2 2 39 The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60-Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the primary side of the transformer and the following data were taken: Open- circuit test (on Short- circuit test (on primary) primary) Voc = 8000 V Vsc = 489 V Ioc = 0.214 A Isc = 2.5 A Poc = 400 W Psc = 240 W Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit 40 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 41 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. 42 Per Unit (pu) in Single Phase System Pbase ,Qbase , Sbase VbaseI base Vbase Z base I base I base Ybase Vbase 2 ( Vbase ) Z base Sbase 43 Voltage Regulation (VR) The voltage regulation of a transformer is defined as the change in the magnitude of the secondary voltage as the current changes from full load to no load with the primary held fixed. VR At no load, VS VP Vp VR a VS , fl Req a VS , fl VS , fl VS ,nl VS , fl X 100% Is X 100% Xeq + + Vp/a Vs - - 44 Phasor Diagram Vp/a jIsXeq Vs IsReq Is Vp/a Lagging power factor jIsXeq Is Vs IsReq Unity power factor 45 Vp/a jIsXeq Is IsReq Vs Leading power factor 46 Efficiency The efficiency of a transformer is defined as the ratio of the power output (Pout) to the power input (Pin). Pout X 100% Pin Pout Pout P losses X 100% Vs I s cos X 100% Vs I s cos Pcu Pcore Pcore = Peddy current + Physteresis And Pcu= Pcopper losses 47 Pcu = Copper losses are resistive losses in the primary and secondary winding of the transformer core. They are modeled by placing a resistor Rp in the primary circuit of the transformer and resistor Rs in the secondary circuit. PCORE = Core loss is resistive loss in the primary winding of the transformer core. It can be modeled by placing a resistor Rc in the primary circuit of the transformer. 48 Example A 15 kVA, 2400/240-V transformer is to be tested to determine its excitation branch components, its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer Open- circuit test Short- circuit test Voc = 2400 V Vsc = 48 V Ioc = 0.25 A Isc = 6.0 A Poc = 50 W Psc = 200 W 49 The data have been taken by using the connections of open circuit test and short circuit test a. Find the equivalent circuit of this transformer referred to the high voltage side. b. Find the equivalent circuit of this transformer referred to the low voltage side. c. Calculate the full load voltage regulation at 0.8 lagging power factor and 0.8 leading power factor. d. What is the efficiency of the transformer at full load with a power factor of 0.8 lagging? 50