EE 162: Electromechanical Energy Conversion and Transformers - KNUST PDF

KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY, KUMASI W SANE NO B A DEPARTMENT OF ELECTRICAL & ELECTRONIC ENGINEERING B.Sc. Electrical and Electronic Engineering, 1 EE 162: ELECTROMECHANICAL ENERGY CONVERSION AND TRANSFORMERS Credit: 3. P. Y. OKYERE SENIOR LECTURER KNUST SEPTEMBER 2012 (REVISED EDITION) Contact Address Dr. Philip Yaw OKYERE Department of Electrical & Electronic Engineering College of Engineering Kwame Nkrumah University of Science and Technology Kumasi, Ghana Phone: 0322063952 0208124340 E-mail: [email protected] [email protected] ii Course Author Dr. Philip Yaw Okyere graduated with a B.Sc. (First Class Honours) degree in electrical engineering in 1979. After one-year national service in the Department of Electrical & Electronic Engineering as Teaching Assistant, he proceeded to L’Institut National Polytechnique de Grenoble (INPG), France to pursue postgraduate studies in electrical engineering on a scholarship from the French Government. He graduated with DEA (“Mention: Bien”) and Docteur-Ingenieur degrees in 1982 and 1985 respectively. He returned to Ghana to join the Department of Electrical & Electronic Engineering of KNUST as a Lecturer in 1985 and was promoted to Senior Lecturer in 1995. Dr. Philip Yaw Okyere has taught courses in Electrical Machines, Electric Drives, Power Electronics and Power Systems and has rendered professional services to industry and the University. He has produced publications in diverse areas, including power system modelling, lightning protection, earth electrode resistance enhancement, application of artificial neural networks to power systems, and renewable energies. His current research interest is in integration of renewable energy sources into power systems. iii Course Outline EE 162 Electromechanical Energy Conversion and Transformers (3 0 3) Principles of Electromechanical Energy Conversion: Force and torque as rate of change of energy with position. Basic Transducers: Single Excitation, Alignment Force and Torque; Double Excitation, Alignment and Interaction Forces and torque Transformers: Construction; Basic theory; Phasor Diagram; Equivalent Circuits; No-load and Short-Circuit Tests; Voltage Regulation; Efficiency; Cooling methods; Polarity; Polyphase transformer Connections; Per-Unit Calculation; Parallel Operation of Transformers; Auto transformers; Tap-Changing transformers; Instrument Transformers. Recommended Textbooks 1. T. Wildi (1988): Electrical Power Technology, Prentice Hall Professional Technical Reference. 2. John Hindmarsh (1977):Electrical Machines and their Applications, Pergamon Press, Oxford 3. Robert Stein, William T. Hunt, Jr. (1979): Electric Power System Components, Transformers and rotating Machines, Van Nostrand Reinhold Company, New York 4. Richard A. Pearman (1994): Electrical Machinery & Transformers, Saunders College Publishing, Harcourt Brace College Publishers, New York. 5. B. A. Theraja, A. K. Theraja (2005): A Textbook of Electrical Technology, S. Chand & Company Ltd, New Delhi iv Contents Contact Address............................................................................................................................................ ii Course Author.............................................................................................................................................. iii Course Outline.............................................................................................................................................. iv Recommended Textbooks............................................................................................................................ iv CHAPTER ONE...............................................................................................................................................1 TRANSFORMERS............................................................................................................................................ 1 1 Introduction.......................................................................................................................................... 1 2 Principle of operation of a transformer................................................................................................ 1 3 Polarity and terminal markings of a transformer................................................................................. 2 4 Ideal transformer.................................................................................................................................. 2 4.1 Properties of an ideal transformer.............................................................................................. 2 4.1.1 Emf equation and voltage ratio (see Fig. 3).......................................................................... 2 4.1.2 Current ratio and power equation........................................................................................3 4.1.3 Phasor diagram of an ideal transformer (Fig. 4)................................................................... 4 4.1.4 Impedance ratio (see Fig. 5).................................................................................................. 5 4.1.5 Equivalent circuits of an ideal transformer........................................................................... 5 5 Practical single-phase transformer....................................................................................................... 7 5.1 Hysteresis loss.............................................................................................................................. 7 5.2 Eddy-current loss.......................................................................................................................... 7 5.3 Phasor diagram on no load.......................................................................................................... 9 5.4 Mutual and leakage fluxes in a transformer (see Fig. 9)........................................................... 10 5.5 Equivalent circuit of a practical transformer.............................................................................11 5.6 Equivalent circuit referred to the primary side (Fig. 12)........................................................... 12 5.7 Equivalent circuit referred to the secondary side (Fig. 13).......................................................13 5.8 Approximate equivalent circuits................................................................................................13 5.9 The complete phasor diagrams for loaded conditions.............................................................14 5.10 Rating of transformers............................................................................................................... 15 5.11 The turns ratio............................................................................................................................ 15 5.12 Definition of per-unit impedances.............................................................................................17 v 5.13 Voltage regulation......................................................................................................................17 5.14 Transformer output....................................................................................................................21 5.15 Efficiency..................................................................................................................................... 21 5.16 All-day efficiency........................................................................................................................23 5.17 Open-circuit and short-circuit tests on a transformer..............................................................24 5.17.1 Short-circuit test................................................................................................................. 24 5.17.2 Open-circuit test or no load test......................................................................................... 25 5.18 Construction of transformers.................................................................................................... 28 5.19 Polarity tests...............................................................................................................................28 5.19.1 Using a low-voltage ac source.............................................................................................29 5.19.2 Using dc source...................................................................................................................29 5.20 Parallel Operation of single-phase transformers...................................................................... 30 5.21 Load sharing of parallel-connected transformers..................................................................... 31 6 Three-phase transformers.................................................................................................................. 33 6.1 A three-phase transformer bank...............................................................................................33 6.2 A three-phase transformer unit.................................................................................................34 6.3 Winding arrangement................................................................................................................ 35 6.3.1 Star connection...................................................................................................................35 6.3.2 Delta connection................................................................................................................. 35 6.3.3 Zigzag (or interconnected star) connection........................................................................35 6.4 Vector groups............................................................................................................................. 35 6.5 Three-phase transformer connections......................................................................................37 6.5.1 Delta-Delta connection (Fig 29.a).......................................................................................37 6.5.2 Delta-Star connection (Fig. 29.b)........................................................................................37 6.5.3 Star-Delta connection.........................................................................................................38 6.5.4 Star-Star connection...........................................................................................................38 6.5.5 Star-Zigzag connection........................................................................................................38 6.6 Parallel operation of three-phase transformers (Fig. 30)......................................................... 42 7 Cooling methods.................................................................................................................................43 7.1 Air Cooling (Dry type transformers).......................................................................................... 44 7.1.1 AN........................................................................................................................................ 44 7.1.2 AF........................................................................................................................................ 44 vi 7.2 Oil-immersed, Oil cooling...........................................................................................................44 7.2.1 ONAN..................................................................................................................................44 7.2.2 ONAF...................................................................................................................................44 7.2.3 OFAF....................................................................................................................................44 7.3 Oil-immersed, Water cooling..................................................................................................... 45 7.3.1 ONWF..................................................................................................................................45 7.3.2 OFWF...................................................................................................................................45 8 Tap-changing Transformers................................................................................................................ 45 8.1 Changing the taps of transformers............................................................................................ 45 8.1.1 Off-load tap changing..........................................................................................................46 8.1.2 On-load tap changing..........................................................................................................46 9 Autotransformers................................................................................................................................48 9.1 Autotransformer equations....................................................................................................... 49 9.2 Advantages and disadvantages of autotransformer over two-winding transformer.............. 49 9.3 Two-winding transformer connected as an autotransformer..................................................50 9.4 Applications of autotransformers..............................................................................................52 10 Instrument transformers................................................................................................................ 52 10.1 The voltage or potential transformers (VTs or PTs)..................................................................53 10.2 Current transformers (CTs)........................................................................................................53 11 Further exercises.............................................................................................................................57 CHAPTER TWO............................................................................................................................................59 ELECTROMECHANICAL ENERGY CONVERSION........................................................................................... 59 1 Introduction........................................................................................................................................ 59 2 Forces and Torques Developed by Electromechanical Devices.......................................................... 59 3 Energy balance....................................................................................................................................59 4 Force in Singly Excited Electric Field System....................................................................................... 60 5 Force in Singly Excited Magnetic Systems.......................................................................................... 63 6 Useful relations for calculating forces and torques............................................................................64 7 Magnetic Field Energy and Coenergy.................................................................................................66 8 Torque in Singly-Excited Systems........................................................................................................70 9 Torques and Forces in Doubly Excited Magnetic Field Systems......................................................... 72 10 Alignment and Interaction Forces...................................................................................................74 vii 10.1 Alignment Forces........................................................................................................................74 10.2 Interaction forces.......................................................................................................................74 11 Motional emf..................................................................................................................................75 12 Examples of Electromechanical Devices......................................................................................... 76 12.1 Solenoid Relay (Fig. 10)..............................................................................................................76 12.2 Plunger relay (Fig. 11)................................................................................................................. 77 12.3 Horse shoe electromagnet.........................................................................................................78 12.4 Electrostatic voltmeter............................................................................................................... 79 12.5 Electrostatic loudspeaker...........................................................................................................81 13 Further exercises.............................................................................................................................83 viii CHAPTER ONE TRANSFORMERS 1 Introduction The transformer transfers electrical energy from one circuit to another via the medium of a pulsating magnetic field that links both circuits. The widespread development of ac power systems is principally due to the transformer. It enables us to produce and transmit power at economical voltages and to distribute it safely in factories and homes. In low-power low-current electronic and control circuits, it is used to provide impedance matching between a source and its load for maximum power transfer, to isolate one circuit from another, to isolate direct current while maintaining ac continuity between two circuits and to provide reduced ac voltages and currents for protection, metering, instrumentation and control. 2 Principle of operation of a transformer The transformer is a straight-forward application of Faraday’s Law of Electromagnetic Induction. Consider the general arrangement of a single-phase transformer shown in Fig. 1. An alternating voltage applied to coil 1, causes an alternating current to flow in the coil and this current produces an alternating flux in the iron core. A portion of the total flux links the second coil. The alternating flux induces a voltage in the second coil. If a load should be connected to the coil, this voltage would drive a current through it. Energy would then be transferred through the medium of magnetic field from coil 1 to coil 2. The combination of the two coils is called a transformer. The coil connected to the source is called the primary winding (or the primary) and the one connected to the load is called the secondary winding (or the secondary). Fig. 1 Elementary transformer 1 3 Polarity and terminal markings of a transformer Voltage E1 is induced in coil 1 and voltage E2 in coil 2. These voltages are in phase. Referring to Fig.2, suppose at any given instant when the primary terminal 1 is positive with respect to primary terminal 2, the secondary terminal 3 is also positive with respect to secondary terminal 4. Then terminals 1 and 3 are said to have the same polarity. To indicate that their polarities are the same, a dot is placed beside primary terminal 1 and secondary terminal 3. Alternatively, letters of the same suffix, A1 (for the high-voltage winding) and a1 (for the low-voltage winding) say can be used. We note that current I1 entering coil 1 through the dotted terminal 1 and current I2 entering coil 2 through the dotted terminal 3 create fluxes in the same direction. (a) Magnetic Circuit (b) Electric Circuit Fig. 2 Polarity and terminal markings 4 Ideal transformer An ideal transformer has no losses, no leakage flux and its core is infinitely permeable. An ideal transformer is shown in Fig. 3. The mutual flux Φm is confined to the iron. The primary current I1 is zero on no load. Fig. 3 Ideal transformer 4.1 Properties of an ideal transformer 4.1.1 Emf equation and voltage ratio (see Fig. 3) With the primary connected to an ac source V1, an alternating flux Φm is produced in the core. Let the flux be expressed as  m   max sin t webers (1) 2 The induced emf e1 as indicated in Fig. 1 is given by d d e1   N1 m   N1  max sin t   N1 max cos t dt dt e1  N 1 max sin t  90 (2) Hence N1 max 2fN1 E1    max 2 2 E1  4.44 fN 1 max (3) Similarly N 2  max 2fN 2  max E2    4.44 fN 2  max (4) 2 2 From equations (3) and (4), we obtain E1 N 1  (5) E2 N 2 The ratio a = N1/N2 is called the turns ratio. A step-up transformer has a < 1 and a step-down transformer has a > 1. In an ideal transformer, the applied voltage V1 and the induced voltage E1 must be identical. Hence we may write V1  E1 (6.a) and V1  4.44 fN 1 max (6.b) Equation (6.b) indicates that for a given frequency, number of turns and voltage, the peak flux  max must remain constant. 4.1.2 Current ratio and power equation On no load I1 = 0. Now if a load is connected across the secondary terminals (i.e. switch S is closed) current I2 flows through the load. This current produces mmf N2I2 which if it acted alone would by Lenz’s law, cause the mutual flux to reduce. Since when V1 is fixed the flux Φmax is also fixed, the primary develops mmf N1I1 which is such that N1 I1  N 2 I 2 (7.a) I1  I 2 a (7.b) In ideal transformer the secondary voltage V 2 E 2 (8) 3 The voltage V2 remains constant since E2 is fixed when the Φmax is fixed. It can be deduced from equations (5), (6.a), (7.a) and (8) that for an ideal transformer V1 I 1  V2 I 2 (9) That is there are no reactive and active losses in an ideal transformer. 4.1.3 Phasor diagram of an ideal transformer (Fig. 4) (a) No load (b) Load is resistive inductive Fig. 4 Phasor diagram of ideal transformer Example 1 An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200-V, 50-Hz source. The load across the secondary draws a current of 2 A at a power factor of 0.80 lagging. Calculate (a) the rms value of the primary current (b) the flux linked by the secondary winding (c) Draw the phasor diagram. Solution (a) N 1 I 1  N 2 I 2  90 I 1  2250  2  I 1  50 A V1 200 (b)  max    0.01 Wb 4.44 fN 1 4.44  50  90 N  (c) E 2   2  E1  5000 V  N1  The phase angle between V2 and I 2 is cos -1 (0.8)  36.9 4 Example 2 A 200-kVA, 6600-V / 400-V, 50-Hz 1-ph transformer has 80 turns on the secondary. Calculate (a) the approximate values of the primary and secondary full-load currents (b) the approximate number of primary turns and (c) the maximum value of the flux. Solution (a) Full - load primary current  (200  1000) 6600  30.3 A and full - load secondary current  (200  1000) 400  500 A (b) N 1  (80  6600) 400  1320 V2 400 (c)  max    0.0225 Wb 4.44 fN 2 4.44  50  80 4.1.4 Impedance ratio (see Fig. 5) The impedance seen by the source Z e  V1 I 1  E1 I 1  aE 2 I 2 a   a 2 E 2 I 2  a 2 Z Hence Ze  a2Z (10) Fig. 5 Impedance ratio 4.1.5 Equivalent circuits of an ideal transformer From equation (10), we can represent the transformer in Fig. 5 by equivalent circuit shown in Fig. 6.a. We may also write I 2 Z  E 2  E1 a  V1 a V1  I2Z (11) a and then represent the transformer by an equivalent circuit shown in Fig. 6.b 5 (a) Circuit referred to the primary side (b) Circuit referred to the secondary side Fig. 6 Equivalent circuits of ideal transformer Example 3 Calculate the voltage V and current I in the circuit of Fig. 7, knowing that the ideal transformer has a primary to secondary turns of 1:100 (i.e. a = 1/100). Fig. 7 See Example 3 Solution We shall shift all impedances to the primary side to obtain the equivalent circuit: V   aV Total impedance Z e  R 2   X L  X C   4 2  32  5  2 V1 10 Total current I    2A Ze 5 Voltage across the 4-ohm resistance V   IR  2  4  8 V. V The actual voltage V   100  8  800 V a 6 5 Practical single-phase transformer The windings of a practical transformer have both resistance and leakage inductance. The core is also imperfect: it has a core loss and finite permeability. The core loss has two components: hysteresis loss and eddy current loss. 5.1 Hysteresis loss When ferromagnetic material is subjected to alternating magnetization, the energy put into the magnetic field when the flux is increasing is not completely given back when the flux dies away but a certain portion is wasted. The area of the hysteresis loop gives the value of the energy loss per unit volume (WH) taking place for each complete cycle of magnetization. The loop area is x found experimentally to vary as Bmax up to moderate values of flux density (1.0 – 2.0 Wb/m2). The index x is named after Steinmetz and is about 1.6 though it may be higher. In practice the 2 hysteresis loss, for simplicity, is often taken as proportional to Bmax If f = frequency of magnetization, the power wasted in magnetic hysteresis Ph  fW H W/m 3 Ph  fBmax 1.6  fBmax 2 W/m 3 (12) The hysteresis constant η depends upon the magnetic material. 5.2 Eddy-current loss When a changing magnetic flux permeates any mass of metal, eddy currents are induced. The eddy currents cause the metal to heat. Eddy currents are also induced in a core revolving in a stationary or constant magnetic field. If iron or metal subjected to alternating magnetization is built up of laminations (or thin sheets) insulated from each other, the eddy current loss is reduced. The eddy current loss in a laminated core is given by Pe  (kf 2 t 2 Bmax 2 ) /  W/m 3 (13.a) where t = thickness of the laminations ρ = resistivity of the material k = a constant which depends on the waveform of the alternating flux At high frequencies where it is not practicable to make very thin laminations, core losses may be reduced by using ferrite cores or dust cores. Ferrite is a ceramic material having magnetic properties similar to silicon steel, and dust cores consist of fine particles of carbonyl iron or 7 permalloy (i.e. nickel and iron), each particle of which is insulated from its neighbour by a binding material. Such materials have a very high value of resistivity. For a given thickness and waveform, (13.a) reduces to Pe  k f 2 Bmax 2 W/m 3 (13.b) Example 4 The core of a transformer operating at 50 Hz has an eddy current loss of 100 W/m3 and the core laminations have a thickness of 0.50 mm. The core is redesigned so as to operate with the same eddy current loss but at a different voltage and at a frequency of 250 Hz. Assuming that at the new voltage the maximum flux density is one-third of its original value and the resistivity of the core remains unaltered, determine the necessary new thickness of the laminations. Solution From equation (13.a), we can write 2 2 2 Pe 2  f 2   t2   Bmax 2   1          Pe1  f1   t1   Bmax 1   2  B  1 Since eddy current loss and resistivity remain the same, and  max 2   , we have  Bmax 1  3 2 2 2  250   t 2   1  1        50   0.50   3   50   3  Therefore the new thickness t 2  0.50        0.30 mm  250   1  We note that if the new voltage had been given rather than Bmax2, then we would have used equation (6.b) to find Bmax2. From equation (6.b), we can write V1, 2  f 2  N 1, 2   max 2   f 2  N 1, 2  Bmax 2  A2             V1,1  f 1  N 1,1   max 1   f 1  N 1,1  Bmax 1  A1  If the number of turns and the cross section remain the same, then the equation reduces to V1, 2  f 2  Bmax 2      V1,1  f 1  Bmax 1  We can use the above equation to find the new voltage (V1,2) in example 4 if it is required. Example 5 In a transformer core of volume 0.16 m3 the total iron loss was found to be 2170 W at 50 Hz. The hysteresis loop of the core material, taken to the same maximum flux density, had an area of 9.0 cm2 when drawn to scales of 1 cm = 0.1 Wb/m2 and 1 cm = 250 AT/m. Calculate the total iron loss in the transformer core if it is energized to the same maximum flux density, but at a frequency of 60 Hz. 8 Solution Hysteresis loss WH = area x scale factors = 9 x 0.1 x 250 = 225 J/m3 At 50 Hz, hysteresis loss Ph  fWH  volume of core = 225 x 50 x 0.16 = 1800 W Therefore eddy-current loss = 2170 – 1800 = 370 W From equation (12), at 60 Hz, hysteresis loss = 1800 x (60/50) = 2160 W and from equation (13.b), eddy-current loss = 370 x (60/50)2 = 533 W Therefore total iron loss = 2160 + 533 = 2693 W Example 6 For the same maximum flux density, the total core loss in a core is 500 W at 25 Hz and 1400 W at 50 Hz. Find the hysteresis and eddy-current losses for both frequencies. Solution Since Bmax is constant, the losses can have the following forms: Ph  Af , Pe  Bf 2 and Pc  Ph  Pe  Af  Bf 2 For a frequency of 25 Hz, the core loss = 500 = A(25) + B(25)2 For a frequency of 50 Hz, the core loss = 1400 = A(50) + B(50)2 Solving the two equations, we obtain A = 12, B = 0.32 and the individual losses Ph = 300 W, Pe = 200 W at 25 Hz and Ph = 600 W, Pe = 800 W at 50 Hz 5.3 Phasor diagram on no load To furnish the power loss in the core, i.e. the core loss, a small current must be drawn from the source. This current Ip must be in phase with induced voltage E1. Also to create the mutual flux Φm, a magnetizing current Im in phase with Φm and lagging 90° behind E1 must be drawn to produce the required mmf. Fig. 8 Phasor diagram for practical transformer on no load We note that (i) the no-load current Io taken by the primary is the phasor sum of Ip and Im (ii) the difference between the value of the applied voltage V1 and that of the induced emf E1 is only about 0.05% when the transformer is on no load so the two can be considered to be equal (iii) Ip is very small compared with Im. Therefore the no-load power factor is very low. 9 Example 7 A 1-ph transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the iron core lc is 1.8 m and the joints are equivalent to an air gap lg of 0.1 mm. If the peak value of the flux density is to be 1.1 T when a voltage of 2200 V at 50 Hz is applied to the primary, find (a) the cross-sectional area of the core (b) the secondary voltage on no load (c) the primary current and power factor on no load. Assume the value of the magnetic field strength Hc for 1.1 T in iron to be 400 A/m, the corresponding iron loss to be 1.7 W/kg at 50 Hz and the density of the iron to be 7800 kg/m3. Solution The magnetic circuit of the transformer consists of iron core in series with the air gap. V1 2200 (a)  max    0.0206 Wb 4.44 fN 1 4.44  50  480  0.0206 Cross - sectional area of core Ac  max   0.0187m 2 Bmax 1.1 N 90 (b) Secondary voltage on no load V2  V1  2  2200   412.5 V N1 480 (c) Peak value of mmf required for the iron core Fg max  H c  l c  400  1.8  720 A Bg 1.1 Peak value of mmf for the air gap Fg max  H g l g  lg   0.0001  87.5 A o 4  10 7 Peak value of total mmf required to produce  max  Fc max  Fg max  720  87.5  807.5 A Peak value of total mmf 807.5 Peak value of magnetizing current    1.682 A N1 480 1.682 Its rms value assuming it to be sinusoidal I m   1.19 A 2 Volume of iron  l c  Ac  1.8  0.0187  0.0337 m 3 Mass of iron  volume  density  0.0337  7800  263 kg And iron loss  263  1.7  447 W I p  Iron loss V1  447 2200  0.203 A No load current  I o  1.19 2  0.203 2  1.21 A No load power factor  I p I o  0.203 1.21  0.168 lagging 5.4 Mutual and leakage fluxes in a transformer (see Fig. 9) The actual flux linking a coil can be considered to have two components: the mutual flux Φm linking both coils and the leakage flux Φl. The reluctance of the paths of the leakage flux is almost entirely due to the long air paths and is therefore practically constant. Consequently, the value of the leakage flux is proportional to the current in the coil. 10 Fig. 9 Transformer possessing two leakage fluxes and a mutual flux The secondary induced voltage Es is composed of two parts: (a) a voltage El2 induced by leakage flux Φl2 given by El2 = 4.44fN2Φl2,max (b) a voltage E2 induced by mutual flux Φm given by E2 = 4.44fN2Φmax Similarly, the primary induced voltage is composed of El1 =4.44fN1Φl1,max and El1 = 4.44fN1Φmax We can separate the four voltages E1, E2, El1 and El2 by rearranging the transformer circuit as shown in Fig. 10. The rearrangement of the transformer circuit makes it clear that El1 and El2 are voltage drops across reactances. These reactances called leakage reactances are given by X 1  El1 I 1 and X 2  El 2 I 2 Fig. 10 Separation of induced voltages 5.5 Equivalent circuit of a practical transformer The behaviour of a practical transformer may be conveniently considered by assuming it to be equivalent to an ideal transformer and then allowing for the imperfections of the actual transformer by means of additional circuits or impedances inserted between the supply and the primary winding and between the secondary winding and the load. The complete equivalent circuit of this transformer is shown in Fig. 11. R1 and R2 are resistances of the primary and secondary windings. X1 and X2 are the leakage reactances. The reactance Xm is such that it takes a reactive current Im (i.e. the magnetizing current) of the actual transformer. The core loss is accounted for by the resistance Rm which takes the component Ip of the primary current. 11 Fig. 11 Complete equivalent circuit of a practical transformer Ideal transformer equations still apply: I 2 N 1  N 2 I 2 and E1 E 2  N 1 N 2 5.6 Equivalent circuit referred to the primary side (Fig. 12) Fig. 12.Exact equivalent circuit referred to the primary side Let a  N 1 N 2 (14.a) Then X 2  ( N 1 N 2 ) 2 X 2  a 2 X 2 , R2  ( N 1 N 2 ) 2 R2  a 2 R2 , Z   ( N 1 N 2 ) 2 Z  a 2 Z (14.b)   V2  ( N 1 N 2 ) V2  aV2 , I 2  I 2 a (14.c)   It is worth noting that for a practical transformer, R2  R1 and X 2  X1 12 5.7 Equivalent circuit referred to the secondary side (Fig. 13) Fig.13 Exact equivalent circuit referred to the secondary side  R1  ( N 2 N 1 ) 2 R1  R1 a 2 , etc (15.a)   N 2 I p  N1 I p or I p  aI p etc (15.b) 5.8 Approximate equivalent circuits The exact equivalent circuit of the transformer is too exact for most practical engineering applications. Consequently, we can simplify it to make calculations easier. Approximate equivalent circuits commonly used for power transformer calculations are given in Fig. 13. (a) (b) 13 (c) Fig.14 Three approximate equivalent circuits The exciting or no load current Io causes voltage drop in the primary leakage impedance. In Fig. 14.a, this voltage drop is neglected. In Fig. 14.b, the exciting current is neglected entirely. Using this circuit may not cause serious errors in some applications such as voltage regulation calculations. In Fig. 14.c, Re is ignored. For transformers above 500 kVA where Xe is at least 5 times greater than Re, this circuit can be used to calculate say, voltage regulation. 5.9 The complete phasor diagrams for loaded conditions They are shown in Fig. 15 Fig. 15 Complete phasor diagrams for loaded conditions (drawn in two parts) 14 5.10 Rating of transformers To keep the transformer temperature at an acceptable level, limits are set to both the applied voltage (this determines the iron loss at a given frequency) and the current drawn by the load (this determines copper loss in the windings). The limits determine the rated voltage and rated current of transformers. The power rating of transformer Srated = rated voltage x rated current can be expressed in VA, kVA or MVA depending on the size of the transformer. The rated kVA (i.e. the rated power), frequency and voltage are always shown on the name plate. In large transformers, the corresponding currents are also shown. We note that Rated kVA = V1r x I1fl x10-3 = V2r x I2fl x 10-3 = V2n x I2fl x10-3 = E2n x I2fl x10-3 where V1r = rated primary voltage I1fl = rated primary current = primary full load current V2r = rated secondary voltage = V2n (no load secondary voltage corresponding to the rated primary voltage) = E2n (no load induced secondary voltage corresponding to primary rated voltage) I2fl = rated secondary current = secondary full load current 5.11 The turns ratio It is given by a  E1 E 2 E1  V1r on no load because I o Z1 is very small. Since the induced voltage in the secondary is equal to the secondary terminal voltage on no load a  V1r V 2 r (16.a) and I 1 fl  I 2 fl a (16.b) Example 8 A transformer is rated 10 kVA, 2400 / 240 V, 60 Hz. The parameters for the approximate equivalent circuit of Fig. 14.a are Rm = 80 kΩ, Xm = 35 kΩ, Re1 = 8.4 Ω and Xe1 = 13.7 Ω. Determine the voltage to be applied to the primary to obtain the rated current in the secondary when the secondary terminal voltage is 240 V. What is the input power factor? The load power factor is 0.8 lagging. 15 Solution  The secondary current when referred to the primary I 2  I 2 fl a  I 1 fl  10000 2400  4.17 A The load power factor angle   cos 1 0.8  36.9 If we choose the load current I2 as the reference phasor, then V 2  24036.9 The load voltage referred to the primary V   aV  240036.9  1920  j1440 volts 2 2 The voltage across the equivalent leakage impedance is Z I   8.4  j13.7 4.17  j 0  35  j 57 volts e1 2 The primary voltage required   V1  V 2  Z e1 I 2  1920  j1440  35  j 57   1955  j1497  246237.4 volts V1 1955  j1497 Ip    24.4375  j18.7125 mA Rm 80 k V 1955  j1497 Im  1   42.7714  j 55.8571 mA jX m j 35 k  I 1  I 2  I p  I m  4.17  j 0   0.0244  j 0.0187   0.0427  j 0.0558  4.2371  j 0.0371  4.237  0.50 A Phase angle between V1 and I 1  37.4   0.50  37.9 Input power factor  cos37.9  0.79 lagging Example 9 A 1-ph transformer operates from a 230-V supply. It has an equivalent resistance of 0.1 Ω and an equivalent leakage reactance of 0.5 Ω referred to the primary. The secondary is connected to a coil having a resistance of 200 Ω and a reactance of 100 Ω. Calculate the secondary terminal voltage. The secondary winding has four times as many turns as the primary. Solution Refer to approximate equivalent circuit of Fig. 14.b N 1 a  1  , Z  200  j100  , Z   a 2 200  j100  12.5  j 6.25  N2 4 Total impedance referred to the primary  Z e1  Z   0.1  j 0.5  12.5  j 6.25  12.6  j 6.75  2300  2300 I2  and V 2  12.5  j 6.25 12.6  j 6.75 12.6  j 6.75  230  12.5  j 6.25 230  13.9754 The magnitude of the voltage V 2    224.8719 volts 12.6  j 6.75 14.2941  V The actual secondary load voltage V2  2  4  224.8719  899 volts a 16 5.12 Definition of per-unit impedances The leakage impedances Z1 and Z2 on the primary and secondary side are expressed in per unit as follows: Z1 Zˆ1  (17.a) Z 1base Z2 Zˆ 2  (17.b) Z 2 base where V1r V2 V2 r V2 Z 1base   1r and Z 2 base   2r (17.c) I 1 fl S rated I 2 fl S rated The impedances are said to be expressed in per unit with reference to a base voltage and a base power which have been chosen in the case of transformers to be V1r, Srated for the primary circuit and V2r, Srated for the secondary circuit. The total impedance of the transformer in per unit  Zˆ1  Zˆ 2. Example 10 A single-phase transformer that is rated 3000 kVA, 69 kV / 4.16 kV, 60 Hz has an impedance of 8 per cent. Calculate the total impedance of the transformer referred to (a) the primary side (b) the secondary side Solution Vr2 69 2  10 6 (a ) Z1base    1587  S rated 3000  10 3 Z e1  Zˆ e1  Z1base  0.08 1587  127  V22r 4.16 2  10 6 (b) Z 2base    5.7685 S rated 3000  10 3 Z e 2  Zˆ e 2  Z 2 base  0.08  5.7685  0.46  2 2 N   4.16  Alternatively Z e 2   2  Z e1     127  0.46   N1   69  5.13 Voltage regulation With the primary voltage maintained constant, the secondary terminal voltage at no load differs from the secondary voltage under load. The voltage regulation or simply regulation of a transformer is the change in secondary voltage which occurs when the rated kVA output at a specified power factor is reduced to zero, with the primary voltage maintained constant. It is 17 usually expressed as a percentage (called percentage regulation), or a fraction of the rated or no- load terminal voltage (in per unit). The equivalent circuit given in Fig. 14.b is used to calculate voltage regulation. The circuit may be either referred to the primary or the secondary side. Therefore the circuit is represented in general form as shown in Fig. 16. If the circuit is referred to the primary side, then  E  V1 , V  V2 , Re  Re1 and X e  X e1 Re Xe E Load V Fig. 16 Circuit for computing voltage regulation  E  V V1  V2 Voltage regulation    pu (18.a) E V1 If the circuit is referred to the secondary side, then V E  V2 n  1 , V  V2 , Re  Re 2 and X e  X e 2 a V2 n  V 2 Voltage regulation   pu (18.b) V2 n The voltage regulation given by equations (18.a) and (18.b) are equal as shown below:   V  V2 V a  V2 a V2 n  V 2  1  1  (19) V1 V1 a V2 n When voltage regulation is known the terminal voltage can be calculated as follows: V2  1   V2 n (20) In general, let the load current be I lagging behind the load voltage V by . Then taking the load voltage as the reference phasor, we can write E  V 1  j 0  Re  jX e I cos   jI sin    V  IRe cos   IX e sin    j IX e cos   IRe sin   18 and hence E E  V  IRe cos   IX e sin  2  IX e cos   IRe sin  2 The second term under the root is usually negligible except at low leading power factors. Considering the first term only gives E  V  IRe cos   IX e sin  E  V  IRe cos   IX e sin  and (21.a) E  V IRe IX   cos   e sin  (21.b) E E E The angle  is negative when current is leading and positive when current is lagging. The voltage regulation is maximum for a given current or power when  =  where X R X tan   e or cos   e or sin   e (22.a) Re Ze Ze Proof: d X From E  V   IRe  sin    IX e cos   0 , we have tan   e  tan  d Re The maximum voltage regulation is given by R  X   R 2  X e2  E  V max  IRe cos   IX e sin   IRe  e   IX e  e   I  e    I Ze  (22.b)  Ze   Ze   Z e  where 2 2 Z e  Re  X e (22.c) From equation (21.b), if E is the rated voltage, then E  V  I fl  IR   I  IX    rated    cos    fl   sin  E rated I  E I  E   fl  rated   fl  rated  E rated  V  I  I fl R   I  I fl X      cos      sin  E rated  I  E I  E  fl  rated   fl  rated  E rated  V   I p.u. R p.u. cos   I p.u. X p.u. sin  (23) E rated where I p.u.  I I fl  current in per unit R p.u.  I fl R E rated  resistance in per unit , etc 19 We note the following: (i) Usually the quantities will be referred to the secondary side. (ii) Where E is the rated voltage and transformers parameters are given in per unit, equation (23) may be used. We should remember that with impedances in per unit, primary and secondary impedances can be added directly. (iii)The voltage regulation equations (21.a), (21.b), (22.b) and (23) are correct at any current or kVA. (iv)At rated current or kVA, Ip.u.=1. Example 11 A 100-kVA 1-ph transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.3 Ω and 0.01 Ω respectively, and the corresponding leakage reactances are 1.1 Ω and 0.035 Ω respectively. The supply voltage is 2200 V. Calculate (a) the equivalent impedance referred to the primary circuit and (b) the voltage regulation and the secondary terminal voltage for full load having a power factor of (i) 0.8 lagging and (ii) 0.8 leading (c) the maximum voltage regulation Solution (a) Re1  R1  ( N1 N 2 ) 2 R2  0.3  0.01  (400 80) 2  0.55  X e1  1.1  0.035  (400 80) 2  1.975   1 Z e1  0.552  1.9752 2  2.05  (b) (i ) cos   0.8  sin   0.6 I 1 fl  100  10 3 2200  45.45 A  R cos   X e1 sin   0.55  0.8  1.975  0.6    I 1 fl  e1   45.45   0.0336 pu  V1   2200  N  80 The secondary no load voltage V2 n  V1  2   2200   440 V  N1  400 The secondary terminal voltage V2  V2 n 1     4401  0.0336   425.2 V  R cos   X e1 sin   0.55  0.8  1.975  0.6  (ii )   I 1 fl  e1   45.45   0.0154 pu  V1   2200  The secondary terminal voltage V2  4401   0.0154  446.8 V I 1 fl Z e1 45.45  2.05 (c) Maximum voltage regulation  max    0.0424 pu. V1 2200 Example 12 The primary and secondary windings of a 30-kVA, 6000-V / 230-V transformer have resistances of 10 Ω and 0.016 Ω respectively. The total reactance of the transformer referred to the primary is 23 Ω. Calculate the percentage regulation of the transformer when supplying full load current at a power factor of 0.8 lagging. 20 Solution Re1  R1  ( N1 N 2 ) 2 R2  10  0.016  (6000 230) 2  20.89  X e1  23  I1 fl  30  10 3 6000  5 A  R cos   X e1 sin   20.89  0.8  23  0.6    I 1 fl  e1   5   0.0254 pu  V1   6000  Percent regulation  2.54% 5.14 Transformer output The transformer supply voltage and frequency are substantially constant. The heating therefore depends on the current taken by the load. Since the secondary voltage of the transformer is also substantially constant it means that the heating also depends on the load kVA. The transformer output is therefore usually quoted in kVA. The transformer load in kVA is given by S  V2 I 2  10 3 kVA (24) where V2 = actual load voltage and I2 = actual load current. When S is given, V2 is unknown and the load current is required, it can be estimated using the approximate equation S  V2 n I 2  10 3 kVA (25) 5.15 Efficiency The losses which occur in a transformer on load are composed of 2 2 (a) Copper losses in primary and secondary windings, namely I12 R1  I 22 R2  I 1 Re1  I 2 Re 2 (b) Iron losses in the core due to hysteresis and eddy currents. The iron losses depend on the peak value of the mutual flux Φm and frequency. It is therefore independent of load current if voltage and frequency are constant. Let Pi = the iron losses (fixed loss) in kW and Pc = the copper loss with full-load Srated kVA in kW Then the total loss at any load S = xSrated kVA at power factor cos  is Pi + x2 Pc and the efficiency is output xS rated cos  S rated cos     input xS rated cos   Pi  x Pc 2 P  S rated cos    i  xPc   x  21 For a given power factor, the efficiency is maximum when the expression in brackets is a minimum. Hence for a maximum efficiency, we have Pi x (26) Pc Proof: d  Pi  Pi   xPc   0 , we obtain  2  Pc  0 or Pi  x Pc 2 From dx  x  x According to equation (26), efficiency is maximum when the copper loss, x 2 Pc is equal to the fixed loss or iron losses Pc, or efficiency is maximum when the load is such that the resulting copper loss is equal to the fixed loss or iron losses. The efficiency of a transformer is calculated using this form of efficiency equation: losses Pi  x 2 Pc   1  1 (27) losses  output xS rated cos   Pi  x 2 Pc Example 13 The primary and secondary windings of a 500-kVA transformer have resistances of 0.42 Ω and 0.0011 Ω respectively. The primary and secondary voltages are 6600 V and 400 V respectively and the iron loss is 2.9 kW. Calculate the efficiency on full load at a power factor of 0.8. Solution 500  1000 Full - load secondary current I 2 fl   1250 A 400 1000 Full - load primary current I 1 fl  500   75.8 A 6600 Secondary copper loss on full load  1250 2  0.0011  1720 W Primary copper loss on full load  75.8 2  0.42  2415 W Total copper loss on full load , Pc  1720  2415  4135 W  4.135 kW Total loss on full load  Pc  Pi  4.135  2.9  7.035 kW Output power on full load at 0.8 pf  S rated cos   500  0.8  400 kW Pi  Pc 7.035 Efficiency on full load   1   1  0.9827  98.27 % S rated cos   Pi  Pc 400  7.035 Example 14 Find the output, at which the efficiency of the transformer of example 11 is maximum and calculate its value assuming the power factor of the load to be 0.8. 22 Solution Let the load S be a fraction x of the rated power, i.e. S = xSrated. Then Pi 2.9 x   0.837 Pc 4.135 Hence output at maximum efficiency  xS rated  0.837  500  418.5 kVA At maximum efficiency, total loss  Pi  x 2 Pc  Pi  Pi  2 Pi  2  2.9  5.8 kW and output power at 0.8 pf  xS rated cos   418.5  0.8  334.8 kW 5.8 Therefore, maximum efficiency  max  1   0.983  98.30 % 5.8  334.8 Example 15 A 400-kVA transformer has an iron loss of 2 kW and the maximum efficiency at 0.8 pf occurs when the load is 240 kW. Calculate (a) the maximum efficiency at unity power factor and (b) the efficiency on full load at 0.71 power factor Solution (a) Total loss at maximum efficiency  2 Pi  2  2  4 kW 240 Output kVA at maximum efficiency   300 kVA 0.8 Output power at maximum efficiency at unity pf  300  1  300 kW 4 Maximum efficiency at unity pf  max  1   0.9868  98.68 % 4  300 300 (b) The fraction x of full load kVA at which efficiency is maximum   0.75 400 P 2 Hence full load copper loss Pc  2i   3.56 kW x 0.75 2 Pi  Pc 2  3.56 Full load efficiency at 0.71 pf  1   1  98.08 % S rated cos   Pi  Pc 2  3.56  400  0.71 5.16 All-day efficiency Power transformers used in generating stations and transmission substations operate at substantially constant load. They are designed with minimal copper losses. Their maximum efficiency occurs at full load. Distribution transformers used in distribution networks on the other hand supplies load which varies widely over a 24-h period. They are always connected in circuit and operate less than full load for most of the time. Hence they are designed with minimal core losses. Their maximum efficiency occurs at 60 to 70 % of full load. The efficiency of these transformers is better 23 assessed on energy basis. The output and losses are calculated in kW hours over a 24-hour day. The all-day efficiency is defined as losses in kWh η allday  1  (28) losses in kWh  output in kWh Example 16 A 200-kVA 1-ph transformer has full load copper loss of 3.02 kW and iron loss of 1.6 kW. The transformer is in circuit continuously. For a total of 8 hours, it delivers a load of 160 kW at 0.8 pf. For a total of 6 hours, it delivers a load of 80 kW at unity power factor. For the remainder of the 24-h cycle, it is on no load. What is the all-day efficiency? Solution At 160 kW, 0.8 pf, kVA  200 kVA (full load), copper loss  3.02 kW iron loss  1.60 kW total loss  4.62 kW 2  80  At 80 kW, upf, kVA  80 kVA , copper loss     3.02  0.48 kW  200  iron loss  1.60 kW total loss  2.08 kW On no load (there is no copper loss on no load), Total loss  1.6 kW For 8 h output  160  8  1280 kWh loss  4.62  8  37 kWh For 6 h output  80  6  480 kWh loss  2.08  6  12.5 kWh For 10 h output  0  10  0 kWh loss  1.6  10  16 kWh In 24 h, total output  1760 kWh total loss  65.5 kWh 65.5 All day efficiency  1   0.9641  96.41 % 65.5  1760 5.17 Open-circuit and short-circuit tests on a transformer These two tests enable the efficiency and voltage regulation to be calculated without actually loading the transformer. 5.17.1 Short-circuit test This test is used to determine the leakage impedance. During this test, one winding is short- circuited and a reduced voltage Vsc applied to the other to cause rated current to flow. The test circuit and the equivalent circuit are shown in Fig. 18.a and Fig. 18.b respectively. The magnetizing branch is neglected because its current under this condition is less than 1 % of the total. 24 Voltage Vsc, current Isc and power Psc measured by the instruments are used to make the following calculations: V P (a ) Z e1  sc (b) Re1  sc2 (c) X e1  Z e21  Re21 (29) I sc I sc We note that the following: (i) Isc need not be the rated current since the equivalent circuit is linear. However, it is desirable that it should be near to the rated value so that stray losses (they are due to eddy currents set up in large section conductors, tank and metallic supports by leakage fluxes) are normal. (ii) The supply could be fed to either winding. It is often convenient on the high voltage transformers to supply the high-voltage winding, thus using a smaller current. Vsc which will be about 3-15 % of the rated value may also be more suitable for test facilities. (iii)In laboratory experiments using small transformers, the instruments positions shown minimize measurement errors. (iv)The circuit parameters obtained with equations (29.a) – (29.c) are referred to the side which the test voltage is applied. Fig. 18 Short-circuit test 5.17.2 Open-circuit test or no load test During this test, one winding is open-circuited and rated voltage at rated frequency is applied to the other. Quite often the low-voltage side is supplied to reduce the test voltage required for safety reasons. As with the short-circuit test, the equivalent circuit parameters will be referred to the side to which the test voltage is applied. The test circuit and equivalent circuit are shown in Fig. 19.a and b. The following calculations can be made: P 2 2 V V1 V1 (a ) I p  o ( b) I m  I o  I p (c ) R m  1 (d ) X m  ( e) a  (30) V1 Ip Im V2 Psc and Po represent the full load copper loss and the core loss (or iron losses) respectively. They can be used directly to calculate efficiency.. 25 Fig. 19 Open-circuit or no-load test Example 17 The circuit shown below was used in a test on a 3-kVA transformer. A variable voltage supply of fixed frequency was connected to terminals A and B and two tests were performed: (a) The voltage was raised to normal rated voltage and the meter reading were then 200 V, 24 W, 1.2 A (b) The terminals C and D were short-circuited and the voltage was raised until the transformer full-load current was flowing. The meter readings were then 6.4 V, 28 W, 15 A From the results of the tests, obtain: (i) the no-load current and its power factor (ii) the iron losses of the transformer at normal frequency and voltage (iii)the full-load copper loss (iv)the transformer resistances, Re1 and Rm and reactances Xe1 and Xm (v) the efficiency of the transformer at full load at a power factor of 0.8 (vi)the efficiency of the transformer at half full load and 0.8 power factor. C A A V B D Circuit diagram for Example 13 Solution (i) No - load current I o  1.2 A input power 24 No - load power factor    0.1 lag input apparent power 200  1.2 (ii) Iron losses Pi  Po  24 W 26 (iii) Full load copper loss Pc  Psc  28 W (iv) Open - circuit test calculations : P 24 2 2 Ip  o   0.12 A I m  I o  I p  1.2 2  0.12 2  1.19 A V1 200 V 200 V 200 Rm  1   1.67 kΩ Xm  1   168  I p 0.12 I m 1.19 Short - circuit test calculations : V 6.4 P 28 Z e1  sc   0.43  Re1  sc2  2  0.12  I sc 15 I sc 15 X e1  Z e21  Re21  0.432  0.12 2  0.41 Pi  x 2 Pc 24  28 (v)  fl  1   1  0.9788 pu  97.88 % xS rated cos   Pi  x Pc 2 (3000  0.8)  24  28 Pi  x 2 Pc 24  28( 12 ) 2 (vi)  1 fl  1   1   97.48 % 2 xS rated cos   Pi  x 2 Pc  12 (3000  0.8)  24  28( 12 ) 2 Example 18 A 10-kVA 1-ph transformer has a voltage ratio 1100 / 250 V. On no load and at normal voltage (1100 V) and frequency the input current is 0.75 A at a pf of 0.2 lagging. With the secondary short-circuited, full-load currents flow when the primary applied voltage is 77 V, the power input being 240 W. Calculate (a) the transformer equivalent resistance and reactance referred to the secondary side (b) the maximum value of the voltage regulation at full load and the load power factor at this regulation (c) the percentage of full-load current at which the transformer has maximum efficiency Solution (a) Referring the circuit to the secondary side, V 250 Vsc 2  sc1  77   17.5 V a 1100 S 10 000 I sc 2  I 2 fl  rated   40 A V2 n 250 V 17.5 P 240 Z e 2  sc 2   0.438  , Re 2  2sc  2  0.15  and X e 2  0.4382  0.152  0.41 I sc 2 40 I sc 2 40 (b) The maximum voltage drop at full load = I 2 fl Z e 2  17.5 V The maximum regulation occurs when    or when R 0.15 cos   cos   e 2   0.342 lag Z e 2 0.438 27 (c) Maximum efficiency occurs when x 2 Pc  Po. Therefore Po Vo I o cos  o 1100  0.75  0.2 x    0.829 or 82.9 % Pc Pc 240 5.18 Construction of transformers Power transformers are designed so that their characteristics approach those of an ideal transformer (a) To attain high permeability and consequently a small magnetizing current, the core is made of iron and the core forms a closed magnetic circuit (b) To keep the hysteresis loss down, high-grade grain-oriented steel having a narrow hysteresis loop is used (c) To keep the eddy current loss down, the core is laminated and 3 to 4% silicon is added to increase the resistivity of the steel. The laminations (about 0.4 mm thick) are insulated from each other by a thin layer of insulation, thus overall cross-sectional area is slightly greater than the actual cross-sectional area of the iron. (d) To keep the leakage fluxes low, the windings are arranged as shown in Fig. 20 in the case of 1-ph transformers. Two main forms of magnetic circuits are used: core- and shell-type arrangements. The core-type construction has the primary and secondary windings distributed over both core legs in order to reduce the amount of copper (i.e. length of turns is reduced). The transformer windings are carefully insulated from each other and from the core. Winding resistances R1 and R2 are also minimized to minimize copper losses. (a) Core - type constructi on with (b) Shell - type constructi on with concentric windings sandwiched windings Fig. 20 Common Single-phase transformer construction 5.19 Polarity tests The four terminals of a single-phase transformer may be mounted so that it has either additive or subtractive polarity as shown in Fig. 21. If it becomes necessary to determine whether a transformer has additive or subtractive polarity then either of the following polarity tests may be used: 28 (a) additive polarity (b) subtractive polarity Fig. 21 Two standard ways of mounting transformer terminals 5.19.1 Using a low-voltage ac source Referring to Fig. 22, (i) Connect the HV winding to ac source (ii) Connect a jumper J between any two adjacent HV and LV terminals (1, 3) (iii)Connect a voltmeter (V2) between the other two adjacent HV and LV terminals (2, 4) (iv)Connect a voltmeter (V1) across the HV winding. Reading of voltmeter V2 = E1 – E2 = V1 – E2 if 1 and 3 possess the same polarity = E1 + E2 = V1 + E2 if they do not Therefore if V2 is lower than V1 then the polarity is subtractive and if V2 is higher than V1 then the polarity is additive Fig. 22 Polarity test using ac source 5.19.2 Using dc source Referring to Fig. 23, (i) Connect a dc source to the LV winding (ii) Mark the terminal connected to the positive side of the source a2 (iii)Connect a dc voltmeter across the HV winding (iv)Close the switch and observe the movement of the pointer of the voltmeter (v) If the pointer moves upscale, the transformer terminal connected to the (+) terminal of the voltmeter is marked A2 and the other marked A1. Fig. 23 Polarity test using dc source 29 5.20 Parallel Operation of single-phase transformers Parallel connection of several transformers is widely used in electrical systems for the following reasons: (a) In many cases, the amount of power to be transformed is greater than that which can be built into one transformer (b) Frequently, the growth of load requires that the installed transformers supply an output greater than their total kVA capacity. Additional transformers are then installed to run in parallel with the existing transformer. (c) It is sometimes found desirable to supply a load through two or more units in order to reduce the cost of the spare unit required to ensure continuity of service in case of damage. The following conditions must be fulfilled when operating two or more single-phase transformers in parallel: (a) The polarity should be the same. The polarity can be either right or wrong. A wrong polarity results in a severe short circuit. Terminals of the same markings are connected together to ensure correct polarity. See Fig. 24. If the polarity markings are either incorrect or not present, the polarity of the incoming transformer can be checked by connecting a voltmeter across the paralleling switch (b) The voltage ratio should be the same. This is to avoid no-load circulating current and also over-loading on one transformer when the paralleled transformers are loaded. (c) The per-unit impedances should be equal in magnitude and have the same angle. When they are equal in magnitude, the transformers share kVA loads in proportion to their respective ratings. If both their magnitudes and angles are the same they will not only share kVA loads in proportion to their respective ratings but also the combined load kVA will be the algebraic sum of the kVA carried by each transformer. If the angles are different, the resultant kVA capacity of the paralleled group will be slightly smaller than the sum of their individual ratings if none should be overloaded. It is not very necessary that the angles should be the same Fig. 24 Connection ensuring correct polarity 30 5.21 Load sharing of parallel-connected transformers The equivalent circuit of two transformers in parallel feeding a common load ZL is shown in Fig. 25. The voltage ratios are supposed to be equal and the magnetizing branch is neglected. The circuit is referred to the secondary side but it may also be referred to the primary side. I1 Z1 I I2 Z2 V1 V ZL a Fig. 25 Equivalent circuit of two transformers in parallel Z1 is the impedance of transformer 1 referred to the secondary side and Z2 is the impedance of transformer 2 referred to the secondary side. Using the current divider rule, Z2 I1  I Z1  Z 2 Z1 I2  I Z1  Z 2 Multiplying by the terminal voltage V gives Z2 S1  S (31.a) Z1  Z 2 Z1 S2  S (31.b) Z1  Z 2 where S1  VI1  apparent power or load carried by transformer 1 (32.a) S 2  VI 2  apparent power or load carried by transformer 2 (32.b) S  VI  combined apparent power or load (32.c) The apparent powers S1, S2 and S are in complex notation. With the load voltage V as reference and the load current I lagging behind the voltage V by  , the combined load can be expressed as S  VI    VI  cos 1 (load pf ) and a transformer 1 load power factor say, obtained as cos 1 where 1 is the phase angle of S1. V  V  magnitude of the load voltage and I  I  magnitude of the load current. 31 Equations (31.a) and (31.b) hold for per-unit impedances provided that all are expressed with reference to a common base power. The following equation can be used to obtain a new per unit value with reference to a new base power. S base,new Ẑ new  Ẑ old  (33) S base,old Example 19 A 500-kVA transformer (Transformer 1) is connected in parallel with a 250-kVA transformer (Transformer 2). The secondary voltage of each is 400 V on no load. Find how they share a load of 750 kVA at power factor of 0.8 lagging if (a ) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and Z 2  0.015  j 0.04 pu  0.0427269.44 pu (b) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and Z 2  0.01  j 0.05 pu  0.0509978.69 pu (c) Z 1  0.01  j 0.05 pu  0.0509978.69 pu and Z 2  0.025  j 0.0444 pu  0.0509560.62 pu Solution If all the impedances are referred to a base power of 500 kVA, then only the impedances of transformer 2 will change. S base,new 500 Ẑ new  Ẑ old   Ẑ old   2 Ẑ old S base,old 250 Case (a) Z 1  0.01  j 0.05  0.05099 pu78.69 and Z 2  20.015  j 0.04  0.03  j 0.08  0.08544 pu69.44 Further Z 1  Z 2  0.01  j 0.05  0.03  j 0.08  0.04  j 0.13  0.1360 pu72.90 Total kVA load S  750  cos 1 0.8  750 kVA  36.9 Z2 0.0854469.44 S1  S   750  36.9   471 kVA  40.36 Z1  Z 2 0.13672.9  471 kVA at power factor of cos 40.36  0.762 lagging Similarly 0.0509978.69 S 2  750  36.9   281 kVA  31.11 0.13672.9  281 kVA at power factor of 0.856 lagging Remark: 32 Transformer 1 with larger per-unit impedance is under-loaded whereas transformer 2 with lower per-unit impedance is overloaded. Case (b) Z 2  20.01  j 0.05  0.02  j 0.1  0.10198 pu78.69 Z 1  Z 2  0.03  j 0.15  0.15297 pu78.69 0.1019878.69 S 1  750  36.9   500 kVA  36.9 0.1529778.69  500 kVA at power factor of 0.8 lagging 0.0509978.69 S 2  750  36.9   250 kVA  36.9 0.1529778.69  250 kVA at power factor of 0.8 laggging Remark: Load shared in proportion to transformer ratings. Arithmetic sum of loadings is equal to the combined load. A shorter approach can be used on recognizing that Z 1 pu  Z 2 pu Case (c) Z 2  20.025  j 0.0444  0.05  j 0.0888  0.1019 pu60.62 Z 1  Z 2  0.06  j 0.1388  0.1512 pu66.62 0.101960.62 S 1  750  36.9   505 kVA  42.9 0.151266.62  505 kVA at power factor of 0.732 lagging 0.0509978.69 S 2  750  36.9   253 kVA  24.82 0.151266.62  253 kVA at power factor of 0.91 lagging Remark: Transformers are slightly overloaded when the combined load is equal to the sum of individual kVAs 6 Three-phase transformers These are required to transform 3-phase power. The three-phase transformer may be either of the following: 6.1 A three-phase transformer bank This consists of three identical single-phase transformers having their windings externally connected for three-phase working. The single-phase transformers retain all their basic single- phase properties such as current ratio, voltage ratio and the flux in the core. The kVA capacity of the bank is the sum of their individual ratings. See Fig.26. 33 6.2 A three-phase transformer unit This is a single unit of special construction for three-phase working. Modern large transformers are usually of the three-phase three-legged core type shown in Fig.27. A leg carries the primary and secondary windings of a phase. The windings are internally connected. For a given total capacity, 3-phase units are much cheaper in capital cost, lighter, smaller and more efficient. Fig. 26 Delta-Star connection of single-phase transformers Fig. 27 Delta-Star connection of three-phase, three-legged core-type transformer 34 6.3 Winding arrangement The three windings, primary or secondary, can be connected in three different ways: 6.3.1 Star connection For this connection, phase voltage = line voltage/√3 which is about 58 % of the line voltage. This enables the insulation of the winding to be reduced to a minimum for a given supply voltage. For star connection line current = phase current. It is the most economical connection for a high- voltage winding. 6.3.2 Delta connection For this connection, phase voltage = line voltage. Therefore winding must be insulated for the full Line voltage. More turns are also required. With very high voltages a saving of 10 % may be achieved by using star-connection rather than delta connection on account of insulation. The saving is small, however, at voltages below 11 kV. For delta connection, phase current = line current/√3 so the winding cross-sectional area is 58 % of that required for the star connection. It it is the most economical for low-voltage winding. 6.3.3 Zigzag (or interconnected star) connection It is a modification of the star connection. Each phase winding is divided into two sections and placed on two different legs. The two sections are then connected in phase opposition. 15 % more turns are required for a given phase terminal voltage compared with a normal star. The three different winding arrangements give rise to several possible connection combinations: star-star, star-delta, star-zigzag, delta-star, delta-delta, etc 6.4 Vector groups The voltage induced in a primary phase winding is in phase with its corresponding secondary phase winding. However, the phase angles of the primary and secondary line voltages may differ depending on the type of winding connection. Three-phase transformers are given symbol or code which indicates the type of connection used for the high-voltage and the low-voltage winding and the phase displacement between the high-voltage side line to line voltage and its corresponding low-voltage side line to line voltage. The symbol is in the form Xxn where X (the capital letter) and x (the small letter) indicate the type of connection for the high-voltage and low-voltage windings respectively and n is a clock hour number which indicates the phase displacement. In this method of indicating the phase shift, the high-voltage side line to neutral voltage is represented by the minute hand of a clock always set at 12 and the corresponding low- voltage side line to neutral voltage represented by the hour hand. Thus when n = 11 it means the clock reads 11 O’clock and the low-voltage side line to line voltage or line to neutral voltage leads its corresponding voltage by 30o. In delta connection, the neutral point does not physically 35 Fig. 28 Vector symbols, diagrams and groups, winding connections, and phase displacement of 3-phase transformers 36 exist and it has to be found geometrically. Two-winding three-phase transformers are classified into four vector groups depending on the phase displacement as shown in the table below: Vector Group Phase Displacement Vector Symbol or Code 1 zero Yyo Ddo Dzo 2 180o Yy6 Dd6 Dz6 3 30o lag Dy1 Yd1 Yz1 4 30o lead Dy11 Yd11 Yz11 Fig. 28 above shows in addition vector diagrams and winding connections. Before connecting two or more transformers in parallel, their vector groups will have to be determined. 6.5 Three-phase transformer connections More common transformer connections are 6.5.1 Delta-Delta connection (Fig 29.a) This connection is economical for large low voltage transformer. Because there is no neutral point, a four-wire supply cannot be given. It is used in a 3-phase transformer bank but rarely in 3-phase transformer unit. It is possible to use this arrangement to provide 3-phase power with one transformer removed. This connection, known as open-delta or vee connection, can supply up to 57.7 % of the load capacity of the delta-delta connection. IL L1  L3 A1 I ph A2 a1 a2  L3 L1 C2 B1 c2 b1 VL V ph C1 B2 c1 b2  L2 L2 VL  V ph I L  3I ph Fig. 29.a Delta-delta connection 6.5.2 Delta-Star connection (Fig. 29.b) It is commonly used to step up alternator voltage to transmission line voltage. Another common application is in distribution service where as a step-down transformer, the windings are not the most economical. The secondary star point can be earthed and a four-wire supply given. 37 6.5.3 Star-Delta connection There is no secondary neutral and four-wire supplies cannot be given. The main use is as a step- down transformer at the load end of transmission line. Fig. 29.b Delta-star connection 6.5.4 Star-Star connection This is economical for high-voltage transformer. However, if the secondary load is unbalanced, the neutral point will be displaced and the line-to-neutral voltages will become unequal, a situation referred to as floating neutral. If the neutral of the primary and the neutral of the source are connected together usually by way of ground, the floating neutral problem can be eliminated. Again if there is no primary ground, the primary and secondary phase voltages will contain pronounced third harmonic voltages, which will cause the neutral point to oscillate above and below the earth at a voltage equal in magnitude to the third harmonic voltage. To eliminate the oscillation problem, a third delta-connected winding called tertiary winding is provided. Although it can be used to supply additional power, the tertiary winding generally has no external connection. 6.5.5 Star-Zigzag connection The star-zigzag arrangement reduces third harmonic voltages and its neutral point is not affected by unbalanced loading. The zigzag is restricted to comparatively low-voltage windings. Example 20 Three single-phase step-up transformers rated 40 MVA, 13.2 kV/80 kV are connected in delta- star on a 13.2 kV transmission line. If they feed a 90 MVA load, calculate the following: (a) the secondary line voltage (b) the currents in the transformer windings (c) the incoming and outgoing transmission line currents. Assume transformers are ideal. 38 Solution (a) Let the voltage applied to the primary of a transformer be Vph1 and the secondary terminal V ph 2 N 2 80 voltage be Vph2. Then considering the transformer to be ideal   V ph1 N 1 13.2 Since the primary connection is delta, the voltage applied to the primary of each single-phase transformer is equal to the transmission line voltage. Therefore N 80 V ph 2  2  V ph1   13.2  80 kV. N1 13.2 Since the secondary side is connected in star, the line voltage on the secondary side  80 3  138 kV V ph 2 V ph 1 13.2 kV I ph 1 I ph 2 90 (b) The load carried by each transformer   30 MVA 3 30 MVA From S  V ph1  I ph1 , current in the primary winding I ph1   2272 A 13.2 kV 30 MVA From S  V ph 2  I ph 2 , current in the secondary winding I ph2   375 A 80 kV (c) Current in incoming line  3I ph1  2272 3  3932 A Current in outgoing line  I ph 2  I L  375 A Alternatively, we can first calculate the line currents and use them to calculate the phase currents or currents in the windings. To calculate a line current we use the formula, S  3V L I L. Example 21 Three single-phase transformers have their primaries joined in delta to a 6600 V, three-phase, three-wire supply. Their secondaries are connected to give a three-phase, four-wire output at 415 V across lines. The total load on the transformers is a balanced load of 150 kW at 0.8 pf lag. If the voltage per turn on the primaries is 4, find (a) the number of turns on the primary winding and the secondary winding (b) the currents and voltages in all windings and lines, including the neutral wire on the secondary side 39 (c) kVA load on each transformer Assume transformers are ideal Solution V ph1 V ph 2 (a) Voltage per turn   4 N1 N2 For delta connection V ph1  V L1  6600 V V ph16600 Therefore N 1    1650 turns 4 4 V 415 For star connection V ph 2  L 2  3 3 V ph 2 415 Therefore N 2    60 turns 4 4 3 I L2 I L1 I ph1 415V 150 kW V ph 2 Load 0.8 pf lagging 6.6 kV V ph1 Power 150  10 3 (b) Secondary I L 2  I ph 2    261 A 3VL cos  3  415  0.8 IN  0 A because load is balanced  N 2 I ph 2  60  261 Primary phase current I ph1      9.5 A  N1  1650 Primary line current I L1  9.5 3  16.4 A. Example 22 A three-phase 415 V load takes a line current of 800 A from a 3300 / 415 V delta/star transformer. The 3300 V system is supplied from an 11000/3300 s

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