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Report CopyRight/DMCA Form For : Notes For An Introductory Course On Electrical Machines

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Contents, Preface ix,1 Three Phase Circuits and Power 1. 1 1 Electric Power with steady state sinusoidal quantities 1. 1 2 Solving 1 phase problems 5, 1 3 Three phase Balanced Systems 6. 1 4 Calculations in three phase systems 9,2 Magnetics 15. 2 1 Introduction 15, 2 2 The Governing Equations 15. 2 3 Saturation and Hysteresis 19, 2 4 Permanent Magnets 21.

2 5 Faraday s Law 22, 2 6 Eddy Currents and Eddy Current Losses 25. 2 7 Torque and Force 27,3 Transformers 29, 3 1 Description 29. 3 2 The Ideal Transformer 30, 3 3 Equivalent Circuit 32. 3 4 Losses and Ratings 36, 3 5 Per unit System 37. v, vi CONTENTS, 3 6 Transformer tests 40, 3 6 1 Open Circuit Test 41.

3 6 2 Short Circuit Test 41, 3 7 Three phase Transformers 43. 3 8 Autotransformers 44,4 Concepts of Electrical Machines DC motors 47. 4 1 Geometry Fields Voltages and Currents 47,5 Three phase Windings 53. 5 1 Current Space Vectors 53, 5 2 Stator Windings and Resulting Flux Density 55. 5 2 1 Balanced Symmetric Three phase Currents 58, 5 3 Phasors and space vectors 58.

5 4 Magnetizing current Flux and Voltage 60,6 Induction Machines 63. 6 1 Description 63, 6 2 Concept of Operation 64, 6 3 Torque Development 66. 6 4 Operation of the Induction Machine near Synchronous Speed 67. 6 5 Leakage Inductances and their Effects 71, 6 6 Operating characteristics 72. 6 7 Starting of Induction Motors 75, 6 8 Multiple pole pairs 76. 7 Synchronous Machines and Drives 81, 7 1 Design and Principle of Operation 81.

7 1 1 Wound Rotor Carrying DC 81, 7 1 2 Permanent Magnet Rotor 82. 7 2 Equivalent Circuit 82, 7 3 Operation of the Machine Connected to a Bus of Constant Voltage. and Frequency 84, 7 4 Operation from a Source of Variable Frequency and Voltage 88. 7 5 Controllers for PMAC Machines 94, 7 6 Brushless DC Machines 95. 8 Line Controlled Rectifiers 99, 8 1 1 and 3 Phase circuits with diodes 99.

8 2 One Phase Full Wave Rectifier 100, 8 3 Three phase Diode Rectifiers 102. 8 4 Controlled rectifiers with Thyristors 103, CONTENTS vii. 8 5 One phase Controlled Rectifiers 104, 8 5 1 Inverter Mode 104. 8 6 Three Phase Controlled Converters 106, 8 7 Notes 107. 9 Inverters 109, 9 1 1 phase Inverter 109, 9 2 Three phase Inverters 111.

10 DC DC Conversion 117, 10 1 Step Down or Buck Converters 117. 10 2 Step up or Boost Converter 119, 10 3 Buck boost Converter 122. Preface, The purpose of these notes is be used to introduce Electrical Engineering students to Electrical. Machines Power Electronics and Electrical Drives They are primarily to serve our students at. MSU they come to the course on Energy Conversion and Power Electronics with a solid background. in Electric Circuits and Electromagnetics and many want to acquire a basic working knowledge. of the material but plan a career in a different area venturing as far as computer or mechanical. engineering Other students are interested in continuing in the study of electrical machines and. drives power electronics or power systems and plan to take further courses in the field . Starting from basic concepts the student is led to understand how force torque induced voltages. and currents are developed in an electrical machine Then models of the machines are developed in. terms of both simplified equations and of equivalent circuits leading to the basic understanding of. modern machines and drives Power electronics are introduced at the device and systems level and. electrical drives are discussed , Equations are kept to a minimum and in the examples only the basic equations are used to solve. simple problems , These notes do not aim to cover completely the subjects of Energy Conversion and Power.

Electronics nor to be used as a reference not even to be useful for an advanced course They are. meant only to be an aid for the instructor who is working with intelligent and interested students . who are taking their first and perhaps their last course on the subject How successful this endeavor. has been will be tested in the class and in practice . In the present form this text is to be used solely for the purposes of teaching the introductory. course on Energy Conversion and Power Electronics at MSU . E G STRANGAS,E Lansing Michigan and Pyrgos Tinos, ix. A Note on Symbols, Throughout this text an attempt has been made to use symbols in a consistent way Hence a script. letter say v denotes a scalar time varying quantity in this case a voltage Hence one can see. v 5 sin t or v v sin t, The same letter but capitalized denotes the rms value of the variable assuming it is periodic . Hence , v 2V sin t,The capital letter but now bold denotes a phasor . V V ej , Finally the script letter bold denotes a space vector i e a time dependent vector resulting from.

three time dependent scalars , v v1 v2 ej v3 ej2 , In addition to voltages currents and other obvious symbols we have . B Magnetic flux Density T , H Magnetic filed intensity A m . Flux Wb with the problem that a capital letter is used to show a time. dependent scalar , flux linkages of a coil rms space vector . s synchronous speed in electrical degrees for machines with more than. two poles , o rotor speed in electrical degrees for machines with more than two poles . m rotor speed mechanical speed no matter how many poles . r angular frequency of the rotor currents and voltages in electrical de . grees , T Torque Nm , Real and Imaginary part of .

x, 1, Three Phase Circuits and Power,Chapter Objectives. In this chapter you will learn the following , The concepts of power real reactive and apparent and power factor. The operation of three phase systems and the characteristics of balanced loads in Y and in . How to solve problems for three phase systems, 1 1 ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES. We start from the basic equation for the instantaneous electric power supplied to a load as shown in. figure 1 1, i t , , , v t , , Fig 1 1 A simple load. p t i t v t 1 1 , 1, 2 THREE PHASE CIRCUITS AND POWER.

where i t is the instantaneous value of current through the load and v t is the instantaneous value. of the voltage across it , In quasi steady state conditions the current and voltage are both sinusoidal with corresponding. amplitudes i and v and initial phases i and v and the same frequency 2 T 2 f . v t v sin t v 1 2 , i t i sin t i 1 3 , In this case the rms values of the voltage and current are . s, Z, 1 T 2 v , V v sin t v dt 1 4 , T 0 2, s, Z, 1 T 2 i . I i sin t i dt 1 5 , T 0 2, 6 v 6 i, and these two quantities can be described by phasors V V and I I . Instantaneous power becomes in this case , p t 2V I sin t v sin t i .

1, 2V I cos v i cos 2 t v i 1 6 , 2, The first part in the right hand side of equation 1 6 is independent of time while the second part. varies sinusoidally with twice the power frequency The average power supplied to the load over. an integer time of periods is the first part since the second one averages to zero We define as real. power the first part , P V I cos v i 1 7 , If we spend a moment looking at this we see that this power is not only proportional to the rms. voltage and current but also to cos v i The cosine of this angle we define as displacement. factor DF At the same time and in general terms i e for periodic but not necessarily sinusoidal. currents we define as power factor the ratio , P, pf 1 8 . VI, and that becomes in our case i e sinusoidal current and voltage . pf cos v i 1 9 , Note that this is not generally the case for non sinusoidal quantities Figures 1 2 1 5 show the cases.

of power at different angles between voltage and current . We call the power factor leading or lagging depending on whether the current of the load leads. or lags the voltage across it It is clear then that for an inductive resistive load the power factor is. lagging while for a capacitive resistive load the power factor is leading Also for a purely inductive. or capacitive load the power factor is 0 while for a resistive load it is 1 . We define the product of the rms values of voltage and current at a load as apparent power S . S VI 1 10 , ELECTRIC POWER WITH STEADY STATE SINUSOIDAL QUANTITIES 3. 1, 0 5, i t , 0, 0 5, 1, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 2, 1, u t , 0, 1, 2, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 1 5, 1, p t , 0 5, 0, 0 5, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. Fig 1 2 Power at pf angle of 0o The dashed line shows average power in this case maximum. 1, 0 5, i t , 0, 0 5, 1, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1.

2, 1, u t , 0, 1, 2, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 1 5, 1, p t , 0 5, 0, 0 5, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. Fig 1 3 Power at pf angle of 30o The dashed line shows average power. and as reactive power Q, Q V I sin v i 1 11 , Reactive power carries more significance than just a mathematical expression It represents the. energy oscillating in and out of an inductor or a capacitor and a source for this energy must exist . Since the energy oscillation in an inductor is 1800 out of phase of the energy oscillating in a capacitor . 4 THREE PHASE CIRCUITS AND POWER, 1, 0 5, i t , 0. 0 5, 1, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1, 2.

1, u t , 0, 1, 2, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 1, 0 5, p t , 0, 0 5, 1, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. Fig 1 4 Power at pf angle of 90o The dashed line shows average power in this case zero. 1, 0 5, i t , 0, 0 5, 1, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 2, 1, u t , 0, 1, 2, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1. 0 5, 0, p t , 0 5, 1, 1 5, 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1.

Fig 1 5 Power at pf angle of 180o The dashed line shows average power in this case negative the opposite. of that in figure 1 2, the reactive power of the two have opposite signs by convention positive for an inductor negative for. a capacitor , The units for real power are of course W for the apparent power V A and for the reactive power. V Ar , SOLVING 1 PHASE PROBLEMS 5, Using phasors for the current and voltage allows us to define complex power S as . S VI 1 12 , 6 v 6 i, V I 1 13 ,and finally, S P jQ 1 14 . For example when, p , v t 2 120 sin 377t V 1 15 , 6.

p , i t 2 5 sin 377t A 1 16 , 4, then S V I 120 5 600W while pf cos 6 4 0 966 leading Also . 6 6 6 4, S VI 120 5 579 6W j155 3V Ar 1 17 , Figure 1 6 shows the phasors for lagging and leading power factors and the corresponding complex. power S , P, S jQ, jQ, S, P I, V, V, I, Fig 1 6 a lagging and b leading power factor. 1 2 SOLVING 1 PHASE PROBLEMS, Based on the discussion earlier we can construct the table below . Type of load Reactive power Power factor, Reactive Q 0 lagging.

Capacitive Q 0 leading, Resistive Q 0 1, 6 THREE PHASE CIRCUITS AND POWER. We also notice that if for a load we know any two of the four quantities S P Q pf we can. calculate the other two e g if S 100kV A pf 0 8 leading then . P S pf 80kW, q, Q S 1 pf 2 60kV Ar or, sin v i sin arccos 0 8 . Q S sin v i , Notice that here Q 0 since the pf is leading i e the load is capacitive . Generally in a system with morePthan one loads P or sources real and P. reactive power balance but, not apparent power i e Ptotal i Pi Qtotal i Qi but Stotal 6 i Si . In the same case if the load voltage were VL 2000V the load current would be IL S V. 100 103 2 103 50A If we use this voltage as reference then . 6, V 2000 0 V, 6 6 o, I 50 i 50 36 9 A, 6 6 36 9o.

S V I 2000 0 50 P jQ 80 103 W j60 103 V Ar,1 3 THREE PHASE BALANCED SYSTEMS. Compared to single phase systems three phase systems offer definite advantages for the same power. and voltage there is less copper in the windings and the total power absorbed remains constant rather. than oscillate around its average value , Let us take now three sinusoidal current sources that have the same amplitude and frequency but. their phase angles differ by 1200 They are , , i1 t 2I sin t . 2 , i2 t 2I sin t 1 18 , 3, 2 , i3 t 2I sin t , 3. If these three current sources are connected as shown in figure 1 7 the current returning though node. n is zero since , 2 2 , sin t sin t sin t 0 1 19 , 3 3.

Let us also take three voltage sources , , va t 2V sin t . 2 , vb t 2V sin t 1 20 , 3, 2 , vc t 2V sin t , 3. connected as shown in figure 1 8 If the three impedances at the load are equal then it is easy. to prove that the current in the branch n n0 is zero as well Here we have a first reason why. THREE PHASE BALANCED SYSTEMS 7, i1, i2 n, i3, Fig 1 7 Zero neutral current in a Y connected balanced system. , v1, , n , n, v2 v3 , Fig 1 8 Zero neutral current in a voltage fed Y connected balanced system . The purpose of these notes is be used to introduce Electrical Engineering students to Electrical Machines Power Electronics and Electrical Drives They are primarily to serve our students at MSU they come to the course on Energy Conversion and Power Electronics with a solid background

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