Power System Dynamics with Computer-Based Modeling and Analysis 9781119487463, 9781119487449, 9781119487456, 1119487463

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Power System Dynamics with Computer-Based Modeling and Analysis

Power System Dynamics with Computer-Based Modeling and Analysis Yoshihide Hase Power Systems Engineering Consultant Japan

Tanuj Khandelwal ETAP – Operation Technology, Inc. USA

Kazuyuki Kameda Eltechs Engineering & Consulting Co., Ltd Japan

This edition first published 2020 © 2020 John Wiley & Sons Ltd All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by law. Advice on how to obtain permission to reuse material from this title is available at http://www.wiley.com/go/permissions. The right of Yoshihide Hase, Tanuj Khandelwal, and Kazuyuki Kameda to be identified as the authors of this work has been asserted in accordance with law. Registered Offices John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, USA John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, UK Editorial Office The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, UK For details of our global editorial offices, customer services, and more information about Wiley products visit us at www.wiley.com. Wiley also publishes its books in a variety of electronic formats and by print-on-demand. Some content that appears in standard print versions of this book may not be available in other formats. Limit of Liability/Disclaimer of Warranty While the publisher and authors have used their best efforts in preparing this work, they make no representations or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties, including without limitation any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives, written sales materials or promotional statements for this work. The fact that an organization, website, or product is referred to in this work as a citation and/or potential source of further information does not mean that the publisher and authors endorse the information or services the organization, website, or product may provide or recommendations it may make. This work is sold with the understanding that the publisher is not engaged in rendering professional services. The advice and strategies contained herein may not be suitable for your situation. You should consult with a specialist where appropriate. Further, readers should be aware that websites listed in this work may have changed or disappeared between when this work was written and when it is read. Neither the publisher nor authors shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Library of Congress Cataloging-in-Publication Data Names: Hase, Yoshihide, 1937– author. | Khandelwal, Tanuj, 1977– author. | Kameda, Kazuyuki, 1947– author. Title: Power System Dynamics with Computer-Based Modeling and Analysis / Yoshihide Hase (power systems engineering consultant, Japan), Tanuj Khandelwal (ETAP–Operation Technology, Inc., USA), Kazuyuki Kameda (Eltechs Engineering & Consulting Co., Ltd, Japan). Description: Hoboken, NJ : Wiley, 2020. | Includes bibliographical references and index. | Identifiers: LCCN 2019010507 (print) | LCCN 2019012442 (ebook) | ISBN 9781119487463 (Adobe PDF) | ISBN 9781119487449 (ePub) | ISBN 9781119487456 (hardcover) Subjects: LCSH: Electric power systems. | Electric power systems–Reliability. Classification: LCC TK3001 (ebook) | LCC TK3001 .H635 2020 (print) | DDC 621.319–dc23 LC record available at https://lccn.loc.gov/2019010507 Cover Design: Wiley Cover Images: Courtesy of Chubu Electric Power Company Inc., Courtesy of ETAP – Operaton Technology, Inc., Courtesy of TEPCO, © Martin Capek/Shutterstock Set in 10/12pt Warnock by SPi Global, Pondicherry, India

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v

Contents About the Authors xxix Preface xxxi Acknowledgments xxxiii

Part A 1

1.1 1.2 1.3 1.4 1.5 2

2.1 2.2 2.2.1 2.2.2 2.2.3 2.3 2.4 2.4.1 2.4.2 2.4.2.1 2.4.2.2 2.5 3

3.1 3.1.1 3.1.1.1 3.1.1.2 3.1.1.3 3.1.2 3.1.3 3.2 3.2.1

Power Systems Theories and Practices

3 Overview 3 Voltage, Current, Electric Power, and Resistance Electromagnetic Induction (Faraday’s Law) 4 Self Inductance and Mutual Inductance 6 Mutual Capacitance 7

1

Essentials of Electromagnetism

3

11 Euler’s Formula 11 Complex Number Notation of Electricity Based on Euler’s Formula 12 Case 1 13 Case 2 13 Case 3 13 LR Circuit Transient Calculation Using Complex Number Notation and the Laplace Transform LCR Circuit Transient Calculation 16 Case 1 17 Case 2 18 Calculation of the Steady-State Current Component 18 Calculation of the Transient Current Component 20 Resistive, Inductive, and Capacitive Load, and Phasor Expressions 21 Complex Number Notation (Symbolic Method) and the Laplace Transform

25 Overhead Transmission Lines with Inductive LR Constants 25 Three-Phase Single-Circuit Line Without an Overhead Grounding Wire (OGW) Voltage and Current Equations, and Equivalent Circuits 25 Measurement of Line Impedances Zaa, Zab, Zac 27 Working Inductance (Laa–Lab) 27 Three-Phase Single Circuit Line with OGW, OPGW 28 Three-Phase Double-Circuit Line with LR Constants 29 Overhead Transmission Lines with Capacitive C Constants 30 Stray Capacitance of Three-Phase Single-Circuit Line 30 Transmission Line Matrices and Symmetrical Components

25

14

vi

Contents

3.2.2 3.2.3 3.3 3.3.1 3.3.2 3.3.3 3.3.3.1 3.3.3.2 3.3.3.3 3.4 3.5 3.5.1 3.5.2 3.5.3 3.5.4 3.6 3.6.1 3.6.2 3.7

Three-Phase Single-Circuit Line with OGW 31 Three-Phase Double-Circuit Line 31 Symmetrical Coordinate Method (Symmetrical Components) 32 Fundamental Concepts of Symmetrical Components 32 Definition of Symmetrical Components 34 Implications of Symmetrical Components 36 Transformation from abc Quantities to 012 Quantities 36 Inverse Transformation from 012 Quantities to abc Quantities 38 Balanced Three-Phase Condition 38 Conversion of a Three-Phase Circuit into a Symmetrical Coordinated Circuit 39 Transmission Lines by Symmetrical Components 39 Single-Circuit Line with LR Constants 39 Double-Circuit Line with LR Constants 41 Single-Circuit Line with Stray Capacitance C 43 Double-Circuit Line with C Constants 45 Generator by Symmetrical Components (Simplified Description) 47 Simplified Symmetrical Equations 47 Reactance of a Generator 48 Description of a Three-Phase Load Circuit by Symmetrical Components 49

4

Physics of Transmission Lines and Line Constants

4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.1.4.1 4.1.4.2 4.1.4.3 4.1.5 4.2 4.2.1 4.2.2 4.2.3 4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.4 4.5 4.5.1 4.5.2

51 Inductance 51 Self-Inductance L11 of a Straight Conductor 51 Two Parallel Conductors in Space, and the Working Inductance L11 −L12 53 Inductance of n-Parallel Conductors in Space 54 Inductance of an Overhead Single-Phase Line Conductor 56 ψ 11 Caused by Conductor 1 57 ψ 12 Caused by Conductor 2 (Imaginary Conductor, Radius H) 57 ψ 11 − ψ 12 Total Flux Linkage of Conductor 1 57 Inductances of Three-Phase Line Conductors 57 Capacitance and Leakage Current 59 Potential Voltages and Capacitance of a Single Overhead Conductor 59 Potential Voltages and Capacitance of a Three-Phase Overhead Line 61 Stray Capacitance and Leakage Current of Phase-Balanced Lines 64 Actual Configuration of Overhead Transmission Lines 66 Structure 66 Equivalent Radius req of Multi-Bundled Conductors 66 Line Resistance 67 Typical Transmission Line Constants and Summary of Impedance/Capacitance Matrices Special Properties of Working Inductance and Working Capacitance 68 MKS Rational Unit System 71 Fundamental Concepts 71 Practical MKS Units for Electrical Engineering Physics 72

5

The Per-Unit Method

5.1 5.2 5.3 5.3.1 5.3.2 5.4 5.5

77 Fundamental Concepts of the PU Method 77 PU Method for a Single-Phase Circuit 77 PU Method for Three-Phase Circuits 79 Base Quantities with the PU Method for Three-Phase Circuits 79 Unitization of Three-Phase Circuit Equations 80 Base Quantity Modification of Unitized Impedance 80 Unitized Symmetrical Circuit: Numerical Example 81

67

Contents

6.1 6.1.1 6.1.2 6.1.3 6.1.4 6.2 6.2.1 6.2.2 6.2.3 6.2.4 6.3 6.3.1 6.3.2 6.3.3 6.3.3.1 6.3.3.2 6.3.3.3 6.3.4 6.4 6.5 6.6 6.6.1 6.6.2 6.7 6.7.1 6.7.2 6.8 6.9 6.9.1 6.9.2 6.9.3 6.9.4 6.9.5

91 Single-Phase Three-Winding Transformer 91 Fundamental Equations Before Unitization 91 Determining Base Quantities for Unitization 91 Unitizing the Original Equation 92 Introducing the Unitized Equivalent Circuit 93 − − Δ-Connected Three-Phase, Three-Winding Transformer 95 Fundamental Equations Before Unitization 95 Determining Base Quantities for Unitization 97 Using the Original Equation 98 Symmetrical Equations and the Equivalent Circuit 99 Three-Phase Transformers with Various Winding Connections 101 Core Structure and Zero-Sequence Excitation Impedance 101 Various Winding Models 101 Delta Windings and Special Properties 102 Case 1: Balanced Three-Phase nth Harmonic Currents 103 Case 2: nth Harmonic Current Flow in Phase-a 104 Case 3 : nth Harmonic Current of Synchronized Delay with Power Frequency 105 % IZ of a Three-Winding Transformer 105 Autotransformers 105 On-Load Tap-Changing Transformer (LTC Transformer) 107 Phase-Shifting Transformer 109 Fundamental Equations 110 Application for Loop-Circuit Lines 112 Woodbridge Transformers and Scott Transformers 113 Woodbridge Transformers 113 Scott Transformers 115 Neutral Grounding Transformer 116 Transformer Magnetic Characteristics and Inrush Current Phenomena 118 Hysteresis Characteristics 118 Flux DC Offset Magnetization 121 Residual Flux 121 Transformer Inrush Current 122 Transformer Inrush Current-Limiting Switching Control 123

7

Fault Analysis Based on Symmetrical Components

6

7.1 7.2 7.2.1 7.2.2 7.2.3 7.2.4 7.2.5 7.3 7.4 7.4.1 7.4.2 7.5 7.5.1 7.5.2

Transformer Modeling

127 Fundamental Concepts of Fault Analysis Based on the Symmetrical Coordinate Method 127 Line-to-Ground Fault (Phase-a to Ground Fault: 1ϕG) 127 Condition Before the Fault 129 Phase-a to Ground Fault 129 Voltages and Currents at Virtual Terminal Point f in the 012-domain 130 Voltages and Currents at an Arbitrary Point under Fault Conditions 131 Fault Under No-Load Conditions 132 Fault Analysis at Various Fault Modes 132 Conductor Opening 137 Single-Phase (Phase a) Conductor Opening 137 Two-Phase (Phase-b, -c) Conductor Opening 139 Visual Vector Diagrams of Voltages and Currents under Fault Conditions 139 Three-Phase Fault: 3ϕS, 3ϕG (Solidly Neutral Grounding System, High-Resistance Neutral Grounding System) 139 Phase b to c Fault: 2ϕS (for Solidly Neutral Grounding System, High-Resistance Neutral Grounding System) 141

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Contents

7.5.3 7.5.4 7.5.5 7.5.6 7.6 7.6.1 7.6.2

Phase-a to Ground Fault: 1ϕG (Solidly Neutral Grounding System) 143 Double Line-to-Ground (Phases b and c) Fault: 2ϕG (Solidly Neutral Grounding System) 146 Phase-a Line-to-Ground Fault: 1ϕG (High-Resistance Neutral Grounding System) 148 Double Line-to-Ground (Phases b and c) Fault: 2ϕG (High-Resistance Neutral Grounding System) Three-Phase-Order Misconnections 151 Phase-abc to acb Misconnection 152 Phase-abc to bca Misconnection 153

8

Fault Analysis with the αβ0-Method 155 αβ0-Method (Clarke-Components) 155 Definition of the αβ0-Coordinate Method (αβ0-Components) 155 The Transformation of Arbitrary Waveform Quantities 156 Interrelation Between αβ0-Components and Symmetrical Components 159 Circuit Equation and Impedance with the αβ0-Coordinate Method 160 Single-Circuit Transmission Lines 162 Double-Circuit Transmission Lines 163 Generators 164 Transformer Impedances and Load Impedances in the αβ0-Domain 166 Fault Analysis with αβ0-Components 166 Line-to-Ground Fault (Phase a to Ground Fault: 1G) 166 The Phase-bc Line to Ground Fault 168 Other Mode Short-Circuit Faults 170 Open-Conductor Mode Faults 170 Advantages of the αβ0-Method 171 Fault-Transient Analysis with Symmetrical Components and the αβ0-Method 171 Transmission-Line Equations for Transient Analysis 171 Comparison of Transient Analysis with Symmetrical Components and the αβ0-Method

8.1 8.1.1 8.1.2 8.1.3 8.1.4 8.1.5 8.1.6 8.1.7 8.1.8 8.2 8.2.1 8.2.2 8.2.3 8.2.4 8.3 8.4 8.4.1 8.4.2 9

9.1 9.1.1 9.1.2 9.1.2.1 9.1.2.2 9.1.2.3 9.1.2.4 9.1.2.5 9.2 9.2.1 9.2.2 9.3 9.3.1 9.3.2 9.3.3 9.3.4 9.3.5 10

10.1 10.1.1 10.1.2 10.1.3 10.1.4 10.1.4.1

Power Cables 175 Structural Features of Power Cables 175 Structures of CV(XLPE) Cable and OF Cable 175 Features of Power Cables 179 Insulation 179 Production Processes 179 Environmental Layout Conditions and Stresses 180 Metallic Sheath Circuits and Outer-Covering Insulation 181 Electrical Specifications and Factory Testing Levels 181 Circuit Constants of Power Cables 183 Cable Inductance 183 Cable Capacitance and Surge Impedance 187 Metallic Sheaths and Outer Coverings 190 Role of Metallic Sheaths and Outer Coverings 190 Double Sheath-End Terminal Grounding Method (Solid-Sheath-Bonding Method) 190 Single Sheath-End Terminal Grounding Method (Single-Sheath-Bonding Method) 191 Cross-Bonding Metallic-Shielding Method 191 Surge Protection at Jointing Boxes 193 Synchronous Generators, Part 1: Circuit Theory 195 Generator Model in a Phase abc-Domain 195 Stator and Rotor Windings Structure 195 Relative Angular Position Between Rotor and Stator 197 Three-Phase abc-Domain Circuit Equations and Inductance Matrices 197 Introduction of Stator Inductance Matrix: labc(t) 200 Introduction of the Stator and Rotor Mutual Inductance Matrix labc–F(t), lF–abc(t)

202

172

150

Contents

10.2 10.2.1 10.2.2 10.2.3 10.3 10.3.1 10.3.2 10.3.3 10.3.4 10.4 10.4.1 10.4.2 10.5 10.5.1 10.5.1.1 10.5.1.2 10.5.1.3 10.5.2 10.5.2.1 10.5.2.2 10.5.2.3 10.5.2.4 10.6 10.7 10.8 10.8.1 10.8.2 10.8.2.1 10.8.2.2 10.8.2.3 10.9 10.9.1 10.9.1.1 10.9.1.2 10.9.1.3 10.9.1.4 10.9.2 10.9.3 10.9.4 10.9.5 10.10 10.10.1 10.10.2

dq0 Method (dq0 Components) 203 Definition of the dq0 Method and Its Physical Meaning 203 Mutual Relation of the dq0-, abc-, and 012-Domains 204 Characteristics of dq0-Domain Quantities 205 Transformation of Generator Equations from the abc-Domain to the dq0-Domain 206 Transformation of Stator Voltage Equations to the dq0-Domain 206 Transformation of the Rotor Voltage Equation 208 Transformation of the Stator Flux Linkage Equation 208 Transformation of the Rotor Flux Linkage Equation 209 Physical Meanings of Generator Equations in the dq0-Domain 210 Main Fluxes and Leakage Fluxes 210 Self-Inductance Ld, Lq 211 Generator dq0-Domain Equations 213 Setting Base Quantities for s-Coils, f-Coils, and k-Coils 213 Capacity Base (VA or MVA Base) 214 Voltage Base and Current Base of the Stator d-, q-, 0-Coils 214 Unitization of Time t to Radians 214 Unitization of Generator dq0-Domain Equations 214 Unitization of the Stator Voltage edq0 Equation 214 Unitization of the Rotor Voltage Equation 215 Unitization of the Stator Flux Linkage ψ dq0 Equation 215 Unitization of the Rotor Flux Linkage Equation 217 Generator dq0-Domain Equivalent Circuit 218 Generator Operating Characteristics and Vector Diagram on the d- and q-Axes Plane 220 Generator Transient Reactance 223 Initial Condition Just Before Sudden Disturbance 223 Assorted d-axis and q-axis Reactance for Transient Phenomena 224 Time Interval t = 0–3 Cycles (50 ms) 224 Time Interval t = 3 to Approximately 60 Cycles (50 ms to 1 sec) 224 Time Interval t = 1 sec to Steady-State Condition 225 Symmetrical Equivalent Circuits of Generators 225 Positive-Sequence Circuit 225 Subtransient Period: t = 0–3 cycles (0–50 ms) 226 Transient Period: t = 3–60 cycles (50 ms to 1 sec) 227 Steady-State Period: After 1 sec 228 Evaluation of the Positive-Sequence Equivalent Circuit 228 Negative-Sequence Circuit 229 Positive-Sequence Current Behavior in the d-axis Circuit 229 Negative-Sequence Current Behavior on the d-axis Circuit 229 Zero-Sequence Circuit 231 Laplace-Transformed Generator Equations and Time Constants 231 Laplace-Transformed Equations 231 Open-Circuit Transient Time Constants T d0 , T d0 , T q0 , T q0 232

10.10.3 10.10.4 10.11 10.11.1 10.11.2 10.11.3 10.12 10.13 10.13.1 10.13.2

Short-Circuit Transient Time Constants T d ,T d ,T q , T q 233 Short-Circuit Time Constant of the Stator (Armature) Winding T a 234 Measuring Generator Reactance 235 Measuring d-axis Reactance xd and Short-Circuit Ratio SCR 235 Measuring Negative-Sequence Reactance X 2 237 Measuring Zero-sequence Reactance X 0 238 Relations Between the dq0-Domain and αβ0-Domain 239 Calculating Generator Short-Circuit Transient Current Under Load 239 Calculating Transient Current with Laplace Transform 239 Transient Short-Circuit Current Calculation for a No-Load Generator 243

ix

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Contents

Synchronous Generators, Part 2: Characteristics of Machinery 251 Apparent Power P + jQ in the abc-, 012-, dq0-Domains 251 Definition of Apparent Power 251 Expanded Apparent Power for Arbitrary Waveform Voltages and Currents 252 Apparent Power of a Three-Phase Circuit in the 012-Domain 253 Apparent Power in the dq0-Domain 255 Mechanical (Kinetic) Power and Generating (Electrical) Power 257 Mechanical Input Power and Electrical Output Power 257 The First Term for Steady-State Power 258 The Second Term for Transient Power 258 The Third Term for Joule Loss Power 258 Steady-State Condition 259 Transient Condition Due to Sudden Disturbance 259 Kinetic Equation for Generators 259 Dynamic Characteristics of Generator (Kinetic Motion Equation) 259 Dynamic Equation for Generators as an Electrical Expression 261 Power Conversion between Rotor Mechanical Power and Stator Electrical Power 261 Step 1: Power Transmission from Prime Mover to Rotor Shaft 262 Step 2: Power Transmission from Generator Rotor to Stator Coil 262 Step 3: Power Transmission from the Stator Coil to the Impedance Load 262 Speed Governors 267 Generator Excitation System 267 Generator Operating Characteristics with P-Q (or p-q) Coordinates 269 Generator Ratings and Capability Curves 271 Upper-Limit Curve of Apparent Power P + jQ or p + jq (Curve ②−③) 272 Upper-Limit Curve of Excitation Voltage Efd (Equivalent to if) (Curve ①–②) 272 Stability-Limit Curve (Curve ⑤−⑦) 274 Limit Curve against Extraordinary Local Heating on the Stator Coil End (Curve ③−④) 274 Generator’s Locus in the pq-Coordinate Plane under Various Operating Conditions 275 The Locus under Fixed Excitation Efd 275 Locus under Fixed Terminal Voltage eG 275 Locus under Fixed Effective Power P 276 Locus under Fixed Terminal Current iG 276 Leading Power-Factor (Under-Excitation Domain) Operation, and UEL Function by AVR 277 Generator as a Reactive Power Generator 277 Under-Excitation (Leading Power-Factor Operation) and the Problem of Overheating at the Stator Core End 277 11.7.3 Under-Excitation Limit Protection by AVR 282 11.8 Operation at Over-Excitation (Lagging Power-Factor Operation) 282 11.9 Thermal Generators’ Weak Points (Negative-Sequence Current, Higher Harmonic Current, Shaft-Torsional Distortion) 282 11.9.1 Generator Volume Size and Unit Capacity 282 11.9.2 Critical I2-Withstanding Capability 283 11.9.3 Rotor Overheating Caused by DC and Higher Harmonic Currents 285 11.9.3.1 nth-Order Harmonic Current Flowing into a Phase-a Coil of a Generator 285 11.9.3.2 DC Current Flow 285 11.9.3.3 Three-Phase nth-Order Current Flow 286 11.10 Transient Torsional Twisting Torque of a TG Coupled Shaft 287 11.10.1 Transient Torsional Torque Caused by a Sudden Network Disturbance 287 11.10.2 Amplification of Torsional Torque 288 11.10.3 Subsynchronous Resonance 290 11.11 General Description of Modern Thermal/Nuclear TG Units 290 11.11.1 ST Unit for Thermal Generation 290

11

11.1 11.1.1 11.1.2 11.1.3 11.1.4 11.2 11.2.1 11.2.1.1 11.2.1.2 11.2.1.3 11.2.2 11.2.3 11.3 11.3.1 11.3.2 11.3.3 11.3.3.1 11.3.3.2 11.3.3.3 11.3.4 11.3.5 11.4 11.5 11.5.1 11.5.2 11.5.3 11.5.4 11.6 11.6.1 11.6.2 11.6.3 11.6.4 11.7 11.7.1 11.7.2

Contents

11.11.2 11.11.3

Combined Cycle System with Gas and Steam Turbines ST Unit for Nuclear Generation 295

12

297 P-δ Curves and Q-δ Curves 297 Power Transfer Limits of Grid-Connected Generators (Steady-State Stability) 299 Apparent Power of Generators 299 Power-Transfer Limits of Generators (Steady-state Stability) 301 Power Cycle Diagram 302 Mechanical Analogy to Steady-State Stability 304 Transient Stability 306 Definitions of Steady-State Stability, Transient Stability, and Dynamic Stability 306 Steady-State Stability 306 Transient Stability 306 Dynamic Stability 306 Mechanical Acceleration Equation for a Two-Generator System 306 Transient Stability under Fault Condition (Equal-Quadrant Method) 308 Case Study 1: Transient Stability Is Successfully Maintained 308 Case Study 2: The System Condition Exceeds the Transient Stability Limit 309 Case Study 3: High-Speed Reclosing Is Conducted 309 Dynamic Stability 309 Quick Excitation Control with an AVR 309 Quick Driving-Power Adjustment with a Speed-Governor Control 310 Four-Terminal Circuit and the P − δ Curve under Fault Conditions 310 Four-Terminal Circuits 310 Four-Terminal Circuit in a Transmission Line Before Fault 311 Four-Terminal Circuit in a Transmission Line under Fault 311 P-δ Curve under Various Fault-Mode Conditions 312 Three-Phase Fault Mode (3ϕS) 312 Double-Phase Fault Mode (2ϕS) 312 Single-Phase Fault Mode (1ϕG) 313 Double-Phase Opening Mode (2ϕOp) 313 Single Phase Opening Mode(1ϕOp) 313 PQV Characteristics and Voltage Instability (Voltage Avalanche) 313 Apparent Power at the Sending Terminal and Receiving Terminal 313 Voltage Sensitivity Characteristics with a Small Disturbance ΔP, ΔQ 314 Circle Diagram for Apparent Power 315 PQV Characteristics, and PV and QV Curves 316 PQV Characteristics of Load 316 PV Mode Voltage Collapse (PV Avalanche) 318 QV Mode Voltage Collapse (QV Avalanche) 318 PQV Steady-State Stability 319 Generator Characteristics with an AVR 319 VQ Control (Voltage and Reactive Power Control) of Power Systems 319 Generator Transfer Function 320 Transfer Function of a Generator Plus Load 322 Transfer Function of a Generator under Special Load Conditions 323 Duties of an AVR 324 Transfer Function of a Generator Plus an AVR 325 Transfer Function of the Total Power System Including an AVR and Load 327 Generator Operation Limit With and Without an AVR in PQ Coordinates 330 Generator Operation without an AVR 330 Generator Operation with an AVR 330

12.1 12.2 12.2.1 12.2.2 12.2.3 12.2.4 12.3 12.3.1 12.3.1.1 12.3.1.2 12.3.1.3 12.3.2 12.3.3 12.3.3.1 12.3.3.2 12.3.3.3 12.4 12.4.1 12.4.2 12.5 12.5.1 12.5.2 12.5.3 12.6 12.6.1 12.6.2 12.6.3 12.6.4 12.6.5 12.7 12.7.1 12.7.2 12.7.3 12.7.4 12.7.5 12.7.6 12.7.7 12.7.8 12.8 12.8.1 12.8.2 12.8.3 12.8.4 12.8.5 12.8.6 12.8.7 12.9 12.9.1 12.9.2

Steady-State, Transient, and Dynamic Stability

292

xi

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Contents

12.9.3 12.9.3.1 12.9.3.2 12.10 12.10.1 12.10.2

Charging Transmission Lines with a Generator with and without an AVR 331 Charging Lines with a Generator without an AVR 331 Charging Lines with a Generator with an AVR 331 VQ (Voltage and Reactive Power) Control with an AVR 332 Reactive Power Distribution for Multiple Generators and Cross-Current Control Pf Control and VQ Control 334

13

Induction Generators and Motors (Induction Machines)

332

13.1 13.2 13.2.1 13.2.2 13.2.3 13.2.4 13.2.5 13.3 13.3.1 13.3.2 13.3.3 13.3.4 13.3.5 13.3.6 13.3.7 13.4

337 Introduction to Induction Motors and Generators 337 Doubly Fed Induction Generators and Motors 337 abc-Domain Voltages and Currents Equations 337 dq0-Domain Transformed Equations 342 Phasor Expressions for dq0-domain Transformed Equations 349 IM Driving Power and Torque 351 Steady-State Operation 354 Squirrel-Cage Induction Motors 355 Circuit Equations 355 Torque-Speed Characteristics Equation for a Squirrel-Cage Induction Machine 358 Reverse, Startup, and Ordinary Running and Overspeed Operating Ranges 359 Torque, Air-Gap Flux, Speed, and Power as the Basis of Power-Electronic Control 360 Startup Operation 365 Rated-Speed Operation 365 Overspeed Operation and Braking Operation 365 Proportional Relations of Mechanical Quantities and Electrical Quantities as a Basis of Power-Electronic Control 367

14

Directional Distance Relays and R–X Diagrams

14.1 14.1.1 14.1.2 14.2 14.2.1 14.2.2 14.2.3 14.3 14.3.1 14.3.2 14.3.3 14.3.4 14.4 14.4.1 14.4.2 14.4.2.1 14.4.2.2 14.4.3 14.4.4 14.5 14.6

371 Overview of Protective Relays 371 The Missions and Duties of Protective Relays 371 Classification of Relays 372 Directional Distance Relays (DZ-Ry) and R–X Coordinate Plane 372 Fundamental Algorithms of Directional Distance Relays 373 R–X Coordinates (R–X Diagram) and P–Q Coordinates 374 R–X Characteristics and Algorithms of Distance Relays 374 R–X Diagram Locus under Fault Conditions 375 Directional Distance Relay(44S-1, 2, 3 Relays) for Line-to-Line Fault Detection 375 Directional-Distance Relay(44G-Relays) for Line-to-Ground Fault Detection 378 Behavior of 44G-Relays against Line-to-Line Short-Circuit Faults 381 Directional-Grounding Relay (67G-Relays) for a High-Impedance Neutral Grounded System Impedance Locus under Ordinary Load Conditions and Step-Out Conditions 381 Impedance Locus under Ordinary Load Conditions 381 Impedance Locus under Transient Conditions 383 The Circle Locus When k = Er/Es Is Changed Under Fixed δ (the k-circles) 384 The Circle Locus When δ Is Changed from 0 to 360 under Fixed k (the δ-circles) 384 The Impedance Locus under Step-Out Conditions 384 Step-Out Detection with Directional Distance Relays 385 Impedance Locus Under Faults with Load-Flow Conditions 385 Loss of Excitation Detection by Distance Relays (40-Relay) 386

15

Lightning and Switching Surge Phenomena and Breaker Switching

15.1 15.1.1 15.1.2 15.1.3

Traveling Wave on a Transmission Line, and Equations 391 Traveling-Wave Equations 391 The Ideal (No-Loss) Line 393 The Distortion-Less Line 394

391

381

Contents

15.1.4 15.2 15.3 15.3.1 15.3.2 15.3.3 15.4 15.4.1 15.4.2 15.5 15.6 15.7 15.7.1 15.7.2 15.7.3 15.8 15.8.1 15.8.2 15.9 15.9.1 15.9.2 15.9.3 15.10 15.11 15.11.1 15.11.2 15.12 15.12.1 15.12.2 15.12.3 15.12.4 15.12.5 15.12.6 15.12.7 15.12.8 15.12.9 15.12.10 15.12.11 15.12.12 15.12.13 15.13 15.14 15.15 15.16 15.17

Laplace Transformed Solution of Voltage and Current 396 Four-Terminal Network Equations between Two Arbitrary Points 398 Examination of Line Constants 399 Overhead Transmission-Line Constants 399 Power-Cable Constants 400 Approximation of Distributed-Constants Circuits and Accuracy of Concentrated-Constants Circuits 401 Behavior of Traveling Waves at Transition Points 401 Incident Wave, Transmitted Wave, and Reflected Wave at a Transition Point 401 Voltage and Current Traveling Waves at Typical Transition Points 403 Surge Overvoltages and Their Three Different, Confusing Notations 404 Behavior of Traveling Waves at a Lightning-Strike Point 406 Traveling Wave Phenomena of Three-Phase Transmission Lines 408 Surge Impedance of Three-Phase Lines 408 Symmetrical Coordinate Analysis of Lightning Strikes 409 Line-to-Ground and Line-to-Line Traveling Waves 410 Reflection Lattices and Transient Behavior Modes 413 Reflection Lattices 413 Oscillatory and Non-oscillatory Convergence 414 Switching Surge Phenomena Caused by Breakers Tripping 415 Calculating Fault-Current Tripping (Single-Phase, Single-Source Circuit) 416 Calculating Fault-Current Tripping (Double Power Source Circuit) 419 Breaker TRV and RRRV 423 Breaker Phase Voltages and Recovery Voltages after Fault Tripping 424 Three-Phase Breaker TRVs across Independent Poles 426 First Pole Tripping 426 Second- and Third-Pole Tripping 429 Circuit Breakers and Switching Practices 432 Fundamentals of Breakers 432 Terminology of Switching Phenomena 433 Short-Circuit Current (Lagging Power-Factor Current) Tripping 435 Leading Power-Factor Small-Current Tripping 436 Short-Distance Line Fault Tripping (SLF) 441 Current-Chopping Phenomena with Small Current Tripping and Lagging Power Factor Step-Out Tripping 444 Current-Zero Missing 444 Overvoltages Caused by Breaker Closing (Close-Switching Surge) 445 Resistive Tripping and Resistive Closing by Circuit Breakers 447 Standardized Switching Surge Level Requested by EHV/UHV Breakers 447 Overvoltage Reduction with Resistive Tripping 448 Overvoltage Reduction with Resistive Closing 450 Switching Surge Caused by Line Switches (Disconnecting Switches) 452 Surge Phenomena Caused on Power Cable Systems 454 Lightning Surge Caused on Cable Lines 456 Switching Surge Caused on Cable Lines 458 Surge Voltages Caused on Cables and GIS Jointed Points 459

16

Overvoltage Phenomena 463

16.1 16.2 16.3 16.4 16.4.1 16.4.2

Neutral-Grounding Methods 463 Arc-Suppression Coil (Petersen Coil) Neutral-Grounded Method 467 Overvoltages Caused by a Line-to-Ground Fault 467 Other Low-Frequency Overvoltage Phenomena (Non-resonant Phenomena) Ferranti Effect 469 Overvoltage Due to Transmission Line Charging 469

469

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16.4.3 16.4.4 16.5 16.5.1 16.5.2 16.5.3 16.5.4 16.6 16.7 16.7.1 16.7.2 16.7.3 16.8 16.8.1 16.8.2 16.8.3 16.8.4 16.8.5

Self-Excitation of Generators 471 Sudden Load Tripping or Load Failure 471 Lower-Frequency Resonant Overvoltages 472 Positive-Sequence Series Resonance 472 Series Resonance under Temporary Conditions (Faults, Phase Opening, Reclosing Time, etc.) Transformer Winding Resonant Oscillation Triggered by Switching Surges 474 Ferroresonance Caused by Core Saturation 474 Interrupted Ground Fault of a Cable Line in a Neutral-Ungrounded System 475 Switching Surge Overvoltages 475 Overvoltages Caused by Breakers Closing (Breaker-Closing Surge) 476 Overvoltages Caused by Breakers Tripping (Breaker-Tripping Surge) 476 Switching Surges Caused by Line Switches 477 Overvoltage Phenomena Caused by Lightning Strikes 477 Direct Strike on Phase Conductors (Direct Flashover) 477 Direct Strike on an OGW or Tower Structure (Inverse Flashover) 478 Induced Strokes (Electrostatic-Induced Strokes and Electromagnetic-Induced Strokes) 479 Capacitive Induced Lightning Surges 479 Inductive Induced Lightning Surges 480

17

Insulation Coordination 481 Overvoltages as Insulation Stresses 481 Classification of Overvoltages 483 Maximum Continuous (Power-Frequency) Overvoltages: Us 483 TOVs and Representative TOVs: Urp 484 Single Line-to-Ground Faults 484 Load Rejection 484 Loss of Neutral Grounding 484 Slow-Front Overvoltages 485 Fast-Front Overvoltages 485 Very Fast-Front Overvoltages 485 Fundamental Process of Insulation Coordination 486 Insulation Coordination 486 Specific Principles of Insulation Strength and Breakdown 486 Insulation Design Criteria for Overhead Transmission Lines 486 Insulation Design Criteria for Substation and Substation Apparatus 487 Insulation Design Criteria for Power Cable Line 487 Countermeasures on Transmission Lines to Reduce Overvoltages and Flashover 487 Using Plural OGWs and OPGWs 487 Reasonable Allocation and Air Clearances for Conductors and Grounding Wires 488 Reduction of Tower Surge Impedance 488 Using Arcing Horns (Arcing Rings) 488 Tower-Mounted Arrester Devices 489 Using Unequal Circuit Insulation (Double-Circuit Lines) 490 Using High-Speed Reclosing 490 Overvoltage Protection with Arresters at Substations 491 Surge Protection Using Metal–Oxide Surge Arresters 491 Metal–Oxide Arresters 493 Arrester Ratings, Classification, and Selection 496 Separation Effects of Station Arresters 497 Station Protection Using OGWs and Reduced Grounding Resistance 499 Direct Lightning Strikes on Substations 499 OGWs in the Station Area 499 Reduction of Station Grounding Resistance and Surge Impedance 499

17.1 17.2 17.2.1 17.2.2 17.2.2.1 17.2.2.2 17.2.2.3 17.2.3 17.2.4 17.2.5 17.3 17.3.1 17.3.2 17.3.2.1 17.3.2.2 17.3.2.3 17.4 17.4.1 17.4.2 17.4.3 17.4.4 17.5 17.6 17.7 17.8 17.8.1 17.8.2 17.8.3 17.8.4 17.9 17.9.1 17.9.2 17.9.3

473

Contents

17.10 17.10.1 17.10.2 17.10.3 17.10.4 17.10.5 17.10.6 17.10.7 17.10.8 17.11 17.11.1 17.11.2 17.11.3 17.11.4 17.11.5 17.12 17.12.1 17.12.2 17.12.3 17.13

Insulation Coordination Details 501 Definitions and Principal Topics in the Standards 501 Insulation Configuration 503 Insulation Withstanding Level and BIL, BSL 503 Standard Insulation Levels and Principles 504 Insulation Levels for Power Systems under 245 kV 504 Insulation Levels for Power Systems over 245 kV 508 Evaluating the Degree of Insulation Coordination 509 Insulation of Power Cables 509 Transfer Surge Voltages through Transformers, and Generator Protection 509 Electrostatic Transfer Surge Voltage (Single-Phase Transformers) 510 Electrostatic Transfer Surge Voltage (Three-Phase Transformers) 512 Transfer Voltage Caused at the Generator Terminal Side 516 Transfer-Surge Protection 517 Electromagnetic Transfer Voltage 518 Transformer Internal High-Frequency Voltage Oscillation Phenomena 518 Equivalent Circuit of a Transformer in the HF Domain 518 Transient Oscillatory Voltages Caused by Incident Surges 518 Reducing Internal Oscillatory Voltages (Non-oscillatory Windings) 523 Oil-Filled Transformers Versus Gas-Filled Transformers 524

18

Harmonics and Waveform Distortion Phenomena 527 Classification of Harmonics and Waveform Distortion 527 Impact of Harmonics 527 Generators and Industrial Motors 527 Nonlinear Loads 529 Harmonic Phenomena Caused by Power Cable Line Faults 529 Transient Current Equation 529 Step 1 529 Step 2 530 Transient Fault Current 532 Waveform Distortion and the Impact on Protective Relays 534

18.1 18.2 18.2.1 18.2.2 18.3 18.3.1 18.3.1.1 18.3.1.2 18.3.2 18.3.3 19

19.1 19.2 19.3 19.4 19.5 19.5.1 19.5.2 19.5.3 19.5.4 19.5.5 19.5.6 19.5.7 19.5.8 19.6 19.6.1 19.6.2 19.6.3 19.6.4 19.6.5

Power Electronic Applications, Part 1: Devices 535 Fundamental Concepts of Power Electronics 535 Power Switching with Power Devices 535 Snubber Circuit 539 Voltage Conversion with Switching 540 Power Electronics Devices 542 Classification and Features of Power Semiconductors 542 Diodes 542 Thyristors 542 Gate Turn-Off (GTO) Thyristor 544 Bipolar Junction Transistor (BJT) or Power Transistor 544 Power Metal Oxide Semiconductor Field Effect Transistors (MOSFET) 546 Insulated Gate Bipolar Transistors (IGBT) 546 Intelligent Power Modules (IPM) 547 Mathematical Background for Analyzing Power Electronics Applications 547 Fourier Series Expansion 547 Averaged Value and Effective Value of Arbitrary Waveform Quantities 548 Power, Power Factor, and Distortion Factor of Arbitrary Waveforms 548 Repetitive On-Off Switching of DC-Quantities 549 Alternate Rectangular Waveform 549

xv

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19.6.6 Alternate Rectangular Waveform with Switch-On at α and Switch-Off at β 19.6.7 Power of Waveform Distorted Voltage and Current 552

20.1 20.1.1 20.1.2 20.1.3 20.1.4 20.1.5 20.1.6 20.1.7 20.1.8 20.2 20.2.1 20.2.2 20.2.3 20.2.4 20.2.5 20.3 20.3.1 20.3.2 20.3.3 20.3.4 20.3.5 20.3.6 20.4 20.4.1 20.4.2 20.4.3 20.5 20.5.1 20.5.2 20.6

553 AC-to-DC Conversion: A Rectifier with a Diode 553 Single-Phase Rectifier with Pure Resistive Load R 553 Inductive Load and the Role of Series-Connected Inductance L 554 Roles of Freewheeling Diodes and Current-Smoothing Reactors 556 Single-Phase Diode Bridge Full-Wave Rectifier 557 Roles of Voltage-Smoothing Capacitors 558 Three-Phase Half-Bridge Rectifiers 559 Current Overlapping 560 Three-Phase Full-Bridge Rectifiers 561 AC-to-DC Controlled Conversion: Rectifier with a Thyristor 562 Single-Phase Half-Bridge Rectifier with a Thyristor 562 Single-Phase Full-Bridge Rectifier with a Thyristor 565 Three-Phase Full-Bridge Rectifier with Thyristors 567 Higher Harmonics and the Ripple Ratio 568 Commutating Reactance: Effects of Source-Side Reactance 570 DC-to-DC Converters (DC-to-DC Choppers) 571 Voltage Step-Down Converters (Buck Choppers) 571 Step-Up (Boost) Converters (Boost Choppers) 573 Buck-Boost Converters (Step-Down/Step-Up Converters) 575 Two- and Four-Quadrant Converters (Composite Choppers) 576 Pulse-Width Modulation Control (PWM) of a DC-DC Converter 577 Multiphase Converters 578 DC-to-AC Inverters 579 Overview of Inverters 579 Single-Phase Inverters 579 Three-Phase Inverters 582 PWM Control of Inverters 583 Principles of PWM Control (Triangle Modulation) 584 Another PWM Control Scheme (Tolerance Band Control) 586 AC-to-AC Converters (Cycloconverters) 587

21

Power Electronics Applications, Part 3: Control Theory

20

21.1 21.2 21.2.1 21.2.2 21.2.3 21.2.4 21.2.5 21.2.6 21.3 21.3.1 21.3.2 21.3.3 21.3.4 21.4 21.4.1 21.4.2 21.4.3

Power Electronics Applications, Part 2: Circuit Theory

589 Introduction 589 Driving Motors 589 Induction Motor (IM) Driving Control 589 Volts per Hertz (V/f) Control (or AVAF Inverter Control) 591 Constant Torque and Constant Speed Control 593 Space Vector PWM Control of IMs (Sinusoidal Control Method) 593 Space Vector PWM Control (Rotor-Flux Oriented Control) 596 dq-Sequence Currents PWM Control (Sinusoidal Control) 597 Static Var Compensators (SVC: A Thyristor-Based Approach) 597 SVCs 597 TCR and TCC 598 Asymmetrical Control Method with PWM Control for SVC 600 Statcom or SVG 600 Active Filters 603 Basic Concepts of Active Filters 603 Active Filtering with the dq Method 605 Vector PWM Control Based on the dq Method 607

551

Contents

21.4.4 21.4.5 21.5 21.6 21.7 21.7.1 21.7.2 21.8 21.9 21.10 21.11 21.11.1 21.11.2 21.12 21.12.1 21.12.2 21.13

Converter Modeling as a dq Coordinates Laplace Transfer Function Active Filter Using the PQ Method or αβ Method 609 Generator Excitation Systems 609 Adjustable-Speed Pumped-Storage Generator-Motor Units 610 Wind Generation 615 Wind Generators 618 Gearless Wind Turbine with Double-Fed IGs 618 Small Hydro Generation 618 Solar Generation (Photovoltaic Generation) 619 High-Voltage DC Transmission (HVDC Transmission) 621 FACTS Technology 625 Overview of FACTS 625 Thyristor Controlled or Protected Series Capacitors (SCs) 625 Railway Applications 627 Railway Substation System 627 Electric Train Engine Motor-Driving System 628 Uninterruptible Power Supplies 628

Appendix A Mathematical Formulae 631 Appendix B Matrix Equation Formulae 635

Part B 22

22.1 22.2 22.2.1 22.2.1.1 22.2.1.2 22.2.1.3 22.2.2 22.2.3 22.3 22.3.1 22.3.1.1 22.3.1.2 22.3.1.3 22.3.2 22.3.3 22.3.4 22.3.4.1 22.3.4.2 22.3.5 22.4 22.5 22.5.1 22.5.2 22.5.2.1 22.6 22.6.1 22.6.2 22.7 22.7.1

Digital Computation Theories

639

641 Introduction 641 Network Types 642 Active Elements and Passive Elements 642 Independent Sources 642 Dependent Sources 643 Source Transformation 644 Linear Elements and Nonlinear Elements 644 Bilateral Elements and Unilateral Elements 645 Circuit Elements 645 Resistors 645 Current-Divider Rule 646 Voltage-Divider Rule 646 Resistivity 647 Inductors 648 Capacitors 649 R–L–C Networks 650 Parallel Connection of a Resistor and Inductor (RL Network) 651 Parallel Connection of a Resistor and Capacitor (RC Network) 651 Circuit with Lumped Elements 653 Ohm’s Law 653 Kirchhoff’s Circuit Laws 655 Kirchhoff’s Current Law 655 Kirchhoff’s Voltage Law 655 Limitations 656 Electrical Division Principle 656 Current Division 656 Voltage Division 657 Instantaneous, Average, and RMS Values 657 Root Mean Square 658 Digital Computation Basics

607

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22.8 22.8.1 22.8.2 22.8.3 22.9 22.9.1 22.9.1.1 22.9.1.2 22.10 22.10.1 22.10.2 22.11 22.11.1 22.11.1.1 22.11.1.2 22.11.1.3 22.12 22.12.1 22.12.2 22.12.3 22.12.4 22.12.5 22.12.6 22.12.7 22.12.8 22.13 22.13.1 22.13.1.1 22.13.2 22.13.2.1 22.13.2.2 22.13.2.3 22.13.2.4 22.13.2.5 22.13.3 22.13.4 22.13.5 22.13.5.1 22.13.5.2 22.13.6 22.13.7 22.13.8 22.13.9 22.14 22.14.1 22.14.1.1 22.14.2 22.14.2.1 22.14.3 22.14.3.1 22.14.4 22.14.5 22.14.6 22.14.6.1

Nodal Formulation 658 Superposition Theorem 658 Nodal Analysis 660 Mesh Analysis 661 Procedure for Mesh Analysis 662 Equivalent Circuits 663 Series Equivalent Circuits 663 Parallel Equivalent Circuits 664 Norton’s and Thévenin’s Equivalents 664 Thévenin’s Theorem 664 Norton’s Theorem 666 Maximum Power Transfer Theorem 668 Proof of the Maximum Power Transfer Theorem 668 Condition for Maximum Power Transfer 668 Maximum Power Transfer Value 668 Efficiency of Maximum Power Transfer 669 Linear System Mathematics 670 Matrix Algebra 670 Matrix Types 670 Matrix Addition and Subtraction 671 Matrix Multiplication 671 Matrix Determinant 672 Triangle Matrix – Lower (L) and Upper (U) 672 Matrix Inversion 673 Adjugate and Cofactor of a Matrix 674 Network Topology 675 Basic Terminology of Network Topology 675 Graphs 675 Types of Graphs 676 Connected Graphs 676 Unconnected Graphs 676 Directed Graph 677 Undirected Graphs 677 Subgraphs and Their Types 677 Tree 677 Co-Trees 677 Matrix Operations 678 Cramer’s Rule 678 Gaussian Elimination 679 Gauss–Jordan Elimination 680 LU Factorization 680 LU Factorization with Partial Inversion or Pivoting 681 LU Factorization with Complete Inversion 681 Power System Matrices 681 Incidence Matrices 682 Procedure to Find the Incidence Matrix 682 Fundamental Loop Matrices 682 Procedure to Find the Fundamental Loop Matrix 683 Fundamental Cut-Set Matrices 683 Procedure to Find the Fundamental Cut-Set Matrix 684 Algorithms for Formation of Network Matrices 684 Singular Transformation 686 Bus Admittance Matrix Using Nodal Analysis 688 Building the YBUS Matrix 690

Contents

22.14.7 22.15 22.15.1 22.15.2 22.15.3 22.15.4 22.16 22.16.1 22.16.2 22.16.3 22.16.4 22.16.5

Sparse Matrix Techniques 691 Transformer Modeling 692 An Ideal Transformer 692 Three-Phase Transformer Model 693 Scott-T Connected Transformer 694 Le-Blanc Connected Transformer 695 Transmission Line Modeling 696 Modeling Long Transmission Lines 697 Lumped Parametric Π Equivalent Circuits of Transmission Lines Short-Length Line 698 Medium-Length Line 699 Load Modeling 699

23

23.1 23.2 23.3 23.4

Power-Flow Methods 701 Newton–Raphson Method 701 Gauss–Seidel Method 702 Adaptive Newton–Raphson Method Fast-Decoupled Method 703

24

Short-Circuit Methods 705

24.1 24.1.1 24.1.2 24.1.3 24.1.4 24.1.4.1 24.1.4.2 24.1.5

698

703

ANSI/IEEE Calculation Methods 705 ½ Cycle Network 705 1.5–4 Cycle Network 705 30 Cycle Network 706 ANSI Multiplication Factor 707 Local and Remote Contributions 707 No AC Decay Ratio 707 Momentary (1/2 Cycle) Short-Circuit Current Calculation (Buses and High-Voltage Circuit Breakers (HVCB)) 707 24.1.6 High-Voltage Circuit Breaker Interrupting Duty Calculation 708 24.1.6.1 Contact Parting Time 708 24.1.6.2 S Factor 708 24.1.7 Low-voltage Circuit Breaker Interrupting Duty Calculation 710 24.1.8 Fuse Interrupting Short-circuit Current Calculation 711 24.1.9 Center-Tap Transformer Impedance Model for a One-Phase Short Circuit 711 24.1.10 Short-Circuit Calculations Using Constant-Current Sources 712 24.1.10.1 Current-Limiting Functionality 712 24.1.10.2 FRT Model Page – Reactive Current Control 717 24.1.11 SC Model for Constant-Current Sources During Unbalanced Faults 718 24.2 IEC Calculation Methods 719 24.2.1 General Description of Calculation Methodology 719 24.2.2 Definition of Terms 720 24.2.2.1 Initial Symmetrical Short-Circuit Current (I k) 720 24.2.2.2 Peak Short-Circuit Current (Ip) 720 24.2.2.3 Symmetrical Short-Circuit Breaking Current (Ib) 720 24.2.2.4 Steady-State Short-Circuit Current (Ik) 720 24.2.2.5 Subtransient Voltage (E ) of a Synchronous Machine 720 24.2.2.6 Far-from-Generator Short-Circuit 720 24.2.2.7 Near-to-Generator Short-Circuit 720 24.2.2.8 Subtransient Reactance (Xd ) of a Synchronous Machine 720 24.2.2.9 Minimum Time Delay (Tmin) of a Circuit Breaker 720 24.2.2.10 Voltage Factor c 721 24.2.3 Initial Symmetrical Short-Circuit Current Calculation 721

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24.2.4 24.2.5 24.2.6 24.2.7 24.2.8 24.2.9 24.2.10 24.2.11 24.2.12 24.2.13 24.2.14 24.2.15 24.2.15.1 24.2.16 24.2.17

Peak Short-Circuit Current Calculation 721 Symmetrical Short-Circuit Breaking Current Calculation 722 DC Component of Short-Circuit Current Calculation 722 Asymmetrical Short-Circuit Breaking Current Calculation 722 Steady-State Short-Circuit Current Calculation 722 Meshed and Non-Meshed Networks 723 Adjustment of Ib 723 Modeling Power-Station Units 723 Network Bus, Connecting Bus, and Auxiliary System Bus for a Power-Station Unit 723 Wind Power-Station Units 723 Power-Station Units with Full-Size Converters 724 IEC Short-Circuit Mesh Determination Method 725 Meshed/Non-Meshed Systems 725 Comparison of Device Rating and Short-Circuit Duty 726 Calculating IEC Device Capability 726

25

729 Problem Formulation 729 Characteristic Harmonic Currents 731 Interharmonics 732 Subharmonics 732 Methodology and Standards 733 Ideal Current Source 734 Thévenin/Norton Equivalent Sources 734 Harmonic Indices 735 Harmonic Factor 735 Individual Harmonic Distortion (IHD) 736 Arithmetic Summation (ASUM) 736 Telephone Influence Factor 736 I∗T Product 737 kVT Product 737 C Message 737 Telephone Form Factor (TFF) 737 Distortion Index (DIN) 738 Total Interharmonic Distortion (TIHD) 738 Total Subharmonic Distortion (TSHD) 738 Group Total Harmonic Distortion (THDG) 738 Subgroup Total Harmonic Distortion (THDS) 739 Harmonic Power Factor 739 Harmonic Component Modeling 740 Harmonic Current Sources 740 Harmonic Voltage Sources 741 Power System Components 741 Long Transmission Lines and Cables 741 Short-Line Model 742 Synchronous Generators 742 Induction Machines 742 Conventional Loads 743 System Resonance 743 Harmonic Mitigation 744 Passive Filters 745 Zigzag Grounding Transformers 746 Multiple Rectifier Bridges and Transformer Phase Shifting

25.1 25.1.1 25.1.2 25.1.3 25.2 25.2.1 25.2.2 25.3 25.3.1 25.3.2 25.3.3 25.3.4 25.3.5 25.3.6 25.3.7 25.3.8 25.3.9 25.3.10 25.3.11 25.3.12 25.3.13 25.3.14 25.4 25.4.1 25.4.1.1 25.5 25.5.1 25.5.2 25.5.3 25.5.4 25.5.5 25.6 25.7 25.7.1 25.7.2 25.7.3

Harmonics

747

Contents

25.7.4 Three-Rectifier Arrangement 748 25.7.4.1 Four-Rectifier Arrangement 748

26.1 26.2

749 Methodology and Standards 749 Performance Indices 752

27

Numerical Integration Methods 755

27.1 27.2 27.3 27.4 27.5

Accuracy 755 Stability 755 Stiffness 757 Predictor–Corrector Runge–Kutta 758

28

761 Power-Flow Injections 761 Voltage Magnitude Constraints 762 Line-Flow Thermal Constraints 762 Line-Flow Constraints as Current Limitations 763 Line-Flow Constraints as Voltage Angle Constraints

26

28.1 28.2 28.3 28.4 28.5

Reliability

757

Optimization

Part C

Analytical Practices and Examples using ETAP

763

765

29

Introduction to Power System Analysis 767

29.1 29.2 29.3 29.4 29.5 29.6 29.7 29.7.1 29.7.2 29.7.3 29.7.4 29.7.5 29.7.6

Planning Studies 767 Need for Power-System Analysis 768 Computers in Power Engineering 768 Study Approach 768 Operator Training 772 System Reliability and Maintenance 772 Electrical Transient Analyzer Program (ETAP) 772 Virtual Reality Operation 772 Total Integration of Data 772 One-Line Diagrams 773 Simplicity in Data Entry 773 Multidimensional Database 773 Other ETAP Analysis Modules 775

30

One-Line Diagrams

30.1 30.2 30.3 30.4 30.4.1 30.4.2 30.4.3 30.5 30.6

Introduction 777 Engineering Parameters 777 One-Line Diagram Symbols 778 Power-System Configurations 780 Transmission and Distribution Substation Configurations 783 Primary Distribution Configurations 784 Secondary Distribution Configurations 787 Network Topology Processing 787 Illustrative Example – Per-Unit and Single-Line Diagram 790

31

Load Flow 791 Introduction 791 Study Objectives 791

31.1 31.2

777

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31.3 31.3.1 31.3.2 31.3.2.1 31.3.2.2 31.3.2.3 31.4 31.4.1 31.4.1.1 31.4.1.2 31.4.1.3 31.4.1.4 31.4.1.5 31.4.1.6 31.5 31.6 31.7 31.8 31.9 31.9.1 31.9.2 31.9.3 31.9.4 31.10 31.11 31.11.1 31.11.1.1 31.11.1.2 31.11.2 31.11.2.1 31.11.2.2 31.11.3 31.11.3.1 31.11.3.2 31.11.4 31.11.4.1 31.11.4.2 31.11.4.3 31.11.4.4 31.11.5 31.11.5.1 31.11.5.2 31.11.5.3 31.11.5.4 31.11.5.5 31.11.6 31.11.6.1 31.11.6.2 31.11.6.3 31.11.6.4 31.11.7 31.11.7.1 31.11.7.2

Problem Formulation 792 Generation and Load Bus Modeling 792 Modeling of Loads (ZIP) 793 Constant Power (P) Load 793 Constant Impedance (Z) Load 793 Constant Current (I) Loads 793 Calculation Methodology 794 Load-Flow Convergence 795 Negative Impedance 796 Negative Reactance 796 Zero or Very Small Impedance 796 Widely Different Branch Impedance Values 796 Long Radial System Configurations 796 Bad Bus Voltage Initial Values 796 Required Data for ETAP 796 Data Collection and Preparation 797 Model Validation 797 Study Scenarios 799 Contingency Analysis 800 Bus Voltage Security Index (PIv/vsp) 800 Real Power Flow Change Index (PIΔP) 800 Reactive Power Flow Change Index (PIΔQ) 800 Branch Overloading Security Index (PIS/Ssp) 801 Optimal or Optimum Power Flow 801 Illustrative Examples 803 Example – Load-Flow Study for a Transmission System 803 System Modeling 803 Load-Flow Study 805 Example – Load-Flow Study for an Industrial System 808 System Modeling 808 Load-Flow Study 812 Example – Load-Flow Study for a Renewable Energy System 817 System Modeling 817 Load-Flow Study 821 Example – Impedance and Ampacity of Transmission Lines 824 Impedance of Transmission Lines 824 Results of Impedance Calculation 824 Ampacity of Transmission Lines 824 Results of Ampacity Study 830 Example – Cable Ampacity and Cable Sizing for an Industrial System 830 Standards 830 Ampacity of Feeder Cables 831 Results of the Ampacity Study 831 Cable-Sizing Study 831 Results of Cable Sizing Study 832 Example – Underground Cables for an Industrial System 834 Steady-State Temperature Calculation 834 Configuration of the Raceway (Duct Bank) 834 Required Data 835 Results of the Steady-State Temperature Study 835 Example – OPF for a Transmission System 835 Objective Conditions and Constraints for Optimization 837 Results of the OPF Study 838

Contents

32

32.1 32.2 32.2.1 32.3 32.3.1 32.3.2 32.3.2.1 32.3.2.2 32.3.2.3 32.3.2.4 32.3.3 32.3.3.1 32.3.3.2 32.3.3.3 32.3.4 32.3.4.1 32.3.4.2 32.3.4.3 32.3.4.4 32.3.4.5 32.3.4.6 32.3.4.7 32.3.4.8 32.3.4.9 32.3.4.10 32.3.4.11 32.4 32.4.1 32.4.2 32.5 32.5.1 32.5.2 32.5.3 32.5.4 32.5.5 32.6 32.6.1 32.6.2 32.6.3 32.6.3.1 32.6.3.2 32.6.3.3 32.6.3.4 32.6.3.5 32.6.3.6 32.6.4 32.6.4.1 32.6.4.2 32.6.4.3 32.6.4.4 32.6.4.5 32.6.4.6

Short-Circuit/Fault Analysis 841 Introduction 841 Analysis Objectives 841 Problem Formulation 842 Methodology and Standards 846 Methods 846 Standards 846 ANSI/IEEE Standard Methodology 846 IEC 60909 Standard Methodology 847 IEC 61363 Transient Short-Circuit Calculation 848 Open-Phase Fault or Series Fault 849 Calculation Procedure 850 Balanced-System Fault 850 Comparison of Device Rating and Short-Circuit Duty 852 Unbalanced System Fault 852 Required Data for Short-Circuit Analysis 852 Bus Data 853 Branch Data 853 Power Grid Data 853 Synchronous Machine Data 853 Induction Motor Data 853 Lumped Load Data 853 Inverter/UPS/VFD Data 854 High-Voltage Circuit-Breaker Data 854 Low-Voltage Circuit-Breaker Data 854 Fuse Data 854 Other Data 855 Study Scenarios 855 SC Contributions 855 Switch/Protective Device Topology 855 Results and Reports 856 ANSI Standard 856 IEC Standard 856 Bus Alert 856 Protective Device Alert 857 Short Circuit Result Analyzer 858 Illustrative Examples 858 Sizing Buses and Breakers per ANSI Standard 858 Example – IEC Short Circuit 865 Example – Short Circuit/Fault Analysis for a Transmission System 865 Analysis Objectives 865 Standards 865 Required Data 867 Study Scenarios 867 Study Conditions 867 Results of SC/Fault Analysis 867 Example – Short Circuit/Fault Analysis for an Industrial System 868 Analysis Objectives 868 Standards 868 Required Data 868 Study Scenarios 868 Study Conditions 868 Results of SC/Fault Analysis 869

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Contents

32.6.5 32.6.5.1 32.6.5.2 32.6.5.3 32.6.5.4 32.6.5.5 32.6.5.6 32.6.6 32.6.6.1 32.6.6.2 32.6.6.3 32.6.6.4

Example – Short Circuit/Fault Analysis for a Renewable Energy System Analysis Objectives 870 Standards 870 Required Data 870 Study Scenarios 871 Study Conditions 871 Results of Short Circuit/Fault Analysis 872 Example – Grounding Grid System for an Industrial System 873 Analysis Objective 873 Methodology and Standards 873 Required Data 876 Results of Grounding Grid Study 877

33

Motor Starting 881 Methods 881 Motor Fundamentals 881 Induction Motors 881 Synchronous Motors 882 Difference Between Synchronous and Induction Motors 883 Motor Rated Power 883 Torque – Speed 883 Motor-Connected Load Types 884 Motor Inrush 886 Motor-Starting Methods 886 Direct Across-the-Line Start or Direct Online Starting (DOL) 887 Star-Delta Starting 887 Rotor Resistance Starting 887 Reactor Starting 888 Auto-Transformer Starting 888 Comparison Table 889 Motor Inertia 889 Motor Nameplate 890 Electric Motor Standard Comparison 891 Motor and System Frequency Impact 892 Analysis Objectives 893 Voltage Drop 893 Methodology and Standards 894 Static Motor Starting or Nondynamic Motor Model 894 Illustrative Example – Motor Starting 895 Illustrative Example – Motor Sizing 896 Illustrative Example – Static Starting 896 Illustrative Example – Static Starting with an Autotransformer 897 Dynamic Motor Starting 898 Dynamic Motor Acceleration with Motor Circuit Models 899 Torque Slip Characteristic Curve Model 901 Control of Motor Characteristic Curves 901 Transient Stability or Full Network Dynamics 901 Required Data 902 Additional Data for Starting Motors 903 Illustrative Examples 903 Static Motor Starting with Load Change 903 Dynamic Motor Starting 904 Motor Starting with VFD 906 Synchronous Motor Starting 908

33.1 33.1.1 33.1.1.1 33.1.1.2 33.1.1.3 33.1.2 33.1.3 33.1.4 33.1.5 33.1.6 33.1.6.1 33.1.6.2 33.1.6.3 33.1.6.4 33.1.6.5 33.1.6.6 33.1.7 33.1.8 33.1.9 33.1.10 33.2 33.2.1 33.3 33.3.1 33.3.1.1 33.3.1.2 33.3.1.3 33.3.1.4 33.3.2 33.3.2.1 33.3.2.2 33.3.3 33.3.4 33.4 33.4.1 33.5 33.5.1 33.5.2 33.5.3 33.5.4

870

Contents

33.5.5 33.5.5.1 33.5.5.2 33.5.5.3 33.5.5.4 33.5.5.5 33.6 33.7

Example – Motor-Starting Study for an Industrial System 911 Analysis Objectives 911 Methodology and Standards 911 Required Data 912 Dynamic Modeling of Motor Starting (IM-3, 5.45 MW, 4-Pole) 913 Results of the Motor-Starting Study 913 Motor-Starting Plots and Results 913 Motor-Starting Alerts 916

34

Harmonics

34.1 34.2 34.3 34.3.1 34.3.2 34.4 34.5 34.5.1 34.5.1.1 34.5.1.2 34.5.1.3 34.5.2 34.5.3 34.5.4

917 Introduction 917 Analysis Objectives 919 Required Data 921 Harmonic Spectrum Data 922 Effect of kVA and Source Impedance 922 Harmonic Load Flow and Frequency Scan 923 Illustrative Examples 924 Example – Harmonic Study for an Industrial System 924 Analysis Objectives 924 Required Data 924 Results of the Harmonic Study 924 Effect of Capacitors on System Resonance and Distortion 927 Harmonic Mitigation – Passive Filters 931 Harmonic Cancelation Using Transformer Phase Shift 933

35

Transient Stability

35.1 35.2 35.3 35.3.1 35.4 35.4.1 35.4.2 35.4.2.1 35.4.2.2 35.4.2.3 35.4.2.4 35.4.2.5 35.4.2.6 35.4.2.7 35.4.2.8 35.4.2.9 35.4.2.10 35.4.2.11 35.4.3 35.4.3.1 35.4.3.2 35.4.3.3 35.4.3.4 35.4.3.5 35.4.3.6 35.4.3.7 35.4.4 35.4.4.1

939 Introduction 939 Analysis Objectives 940 Basic Concepts of Transient Stability 942 Stability Limits 943 Dynamic Models 944 Power Grid 944 Synchronous Machines 944 Notations and Symbols 944 General Concept of Modeling Synchronous Machines Transient Model for Round Rotors 945 Subtransient Model for Round Rotors 945 Subtransient Model for Salient-pole (IEEE 2.1) 945 Subtransient Model for Salient-pole (IEEE 2.2) 946 Frequency-Dependent Model 947 Excitation System 947 Governor-Turbines 948 Power System Stabilizers 949 Shaft Torsion 949 Induction Machines 949 Single1 Model 949 Single2 Model 950 Double-Cage Model (DBL1) 951 Double-Cage Model (DBL2) 953 Frequency-Dependent Model 953 Shaft Torsion Model 954 Load Model 955 Wind Turbine Generators 955 Type 1 – Induction Generator 955

945

xxv

xxvi

Contents

35.4.4.2 35.4.4.3 35.4.4.4 35.4.5 35.4.6 35.4.7 35.4.8 35.4.9 35.5 35.6 35.6.1 35.6.2 35.6.3 35.7 35.8 35.9 35.9.1 35.9.2 35.9.3 35.10 35.11 35.11.1 35.11.2 35.12 35.12.1 35.12.1.1 35.12.1.2 35.12.1.3 35.12.1.4 35.12.1.5 35.12.1.6 35.12.2 35.12.2.1 35.12.2.2 35.12.2.3 35.12.2.4 35.12.2.5 35.12.2.6 35.12.2.7 35.12.3 35.12.3.1 35.12.3.2 35.12.3.3 35.12.4 35.12.4.1 35.12.4.2 35.12.4.3

Type 2 – Induction Generator with Variable Slip 956 Type 3 – Doubly Fed Induction Generator 957 Type 4 – Full-Scale Power Converter 961 Inverter (PV Array Inverter) 966 Uninterruptible Power Supplies 966 Variable Frequency Drives 966 Dynamic Lumped Motor Load Model 966 Protection Relays 966 User-Defined Models 967 Parameter Tuning 967 Inertial Components 969 Integral Components 970 Inertial-Differential Components 970 Single-Generator Power System Model 971 Data Collection and Preparation 973 Study Scenarios 974 Islanding Study 974 Bus Transfers 975 Load Shedding 975 Stability Improvement 977 System Simulation 977 Events 977 Results, Reports, and Plots 979 Illustrative Examples 979 Determine Critical Fault-Clearing Time (IEEE 9-Bus System) 979 Second Run 980 Third Run 980 Fourth Run 980 Fifth Run 980 Conclusions 984 Final Run 984 Stability Analysis of an Industrial Facility 984 Steps 984 Run the TS Study (First Time) 985 Rerun the TS Study (Second Run) 985 Modify the Study Case to Simulate a Bus Transfer 985 Rerun the TS Study (Third Run) 985 Use Frequency Relay to Perform Load Shedding 987 Rerun the TS Study (Fourth Run) 988 Example – Transient-Stability Analysis for a Transmission System 988 Analysis Objectives 989 Required Data 989 Results of Transient-Stability Analysis 990 Example – Transient-Stability Analysis for an Industrial System 995 Analysis Objectives 995 Required Data 996 Results of Transient-Stability Analysis 997

36

Reliability Assessment

36.1 36.2 36.3 36.3.1 36.3.2

1003 Introduction 1003 Analysis Objectives 1003 Problem Formulation 1004 Generation Reliability Assessment 1004 Transmission Reliability Assessment 1004

Contents

36.3.3 36.4 36.5 36.5.1 36.5.2 36.5.3 36.5.4 36.5.4.1 36.5.4.2 36.5.4.3 36.5.4.4

Distribution Reliability Assessment 1005 Required Data 1005 Illustrative Examples 1005 Simple Radial System 1005 Single and Double Contingency 1008 Reliability Index Calculation 1011 Example – Reliability Assessment for a Transmission System Analysis Objectives 1013 Required Data 1014 Study Scenarios 1014 Results of Reliability Assessment 1016

37

Protective Device Coordination 1019 Introduction 1019 Study Criteria 1020 Study Objectives 1020 Low-Voltage Circuit Breakers 1020 Relays 1022 Electromechanical Relays 1022 Static Relays 1025 Digital Relays 1025 Numerical Relays 1025 Relay Types/Functions 1027 Phase Fault Relays 1027 Instantaneous Overcurrent Protection (50) 1027 Inverse-Time Relays (51) 1027 Ground Fault Relays 1027 Differential Relays (87) 1028 Undervoltage and Overvoltage Relays 1028 Directional Relays 1028 Methodology 1028 Discrimination by Time 1029 Discrimination by Current 1030 Discrimination by Current and Time 1030 Scale Selection 1031 Log–Log Plot 1034 Required Data 1035 Principle of Protection 1036 Principle of Selectivity/Coordination 1037 Selectivity Time Margins 1039 Art of Protection and Coordination >600 V 1040 Bus Relays 1041 Feeder Relays 1041 Induction Motors 1041 Motor Protection 1042 Cables 1042 Capacitors 1042 Power Transformers 1044 Transformer Primary-Phase Fault 1045 Transformer Ground Fault 1045 Transformer Primary Ground 1046 Transformer Primary Fuse Phase 1046 Transformer Secondary, Low-Resistance Grounded Transformer Secondary, Solid Grounded 1047

37.1 37.1.1 37.1.2 37.1.3 37.2 37.2.1 37.2.2 37.2.3 37.2.4 37.2.5 37.2.5.1 37.2.5.2 37.2.5.3 37.2.5.4 37.2.5.5 37.2.5.6 37.2.5.7 37.3 37.3.1 37.3.2 37.3.3 37.3.4 37.3.5 37.4 37.5 37.6 37.6.1 37.7 37.7.1 37.7.2 37.7.3 37.7.4 37.7.5 37.7.6 37.7.7 37.7.7.1 37.7.7.2 37.7.7.3 37.7.7.4 37.7.7.5 37.7.7.6

1047

1013

xxvii

xxviii

Contents

37.7.7.7 37.7.7.8 37.8 37.8.1 37.8.2 37.8.3 37.8.4

Primary Neutral (YD) 1048 Directional Overcurrent 1048 Illustrative Examples 1048 Basic Operation and Phase Protection 1048 Illustrative Example – Ground Fault Protection 1063 Evaluating Phase and Ground Settings Using a Sequence of Operation Illustrative Example – Star Auto-Evaluation 1068

Appendix C

Standards, Regulations, and Best Practice

Further Reading Index 1085

1083

1071

1064

xxix

About the Authors Yoshihide Hase was born in Gifu Prefecture, Japan, in 1937. After graduating in electrical engineering from Kyoto University, he joined the Toshiba Corporation in 1960 and took charge of various power system projects, both at home and abroad, including the engineering of generating station equipment, substation equipment, as well as power system control and protection, until 1996. During that time, he held the positions of general manager, senior executive of technology for the energy systems sector, and chief fellow. In 1996, he joined Showa Electric Wire & Cable Company as the senior managing director and representative director and served on the board for eight years. He was a lecturer at Kokushikan University for five years from 2004. He was the vice president of the IEEJ (1995–1996) and has been bestowed as an honorary member. He was also the representative officer of the Japanese National Committee of CIGRE (1987–1996) and has been bestowed as a distinguished member of CIGRE. He is also a member of the Institute of Electrical and Electronics Engineers (IEEE). He is the author of two books from Wiley: Handbook of Power Systems Engineering (first ed. 2007, second ed. 2013), and of three Japanese books from Maruzen. —Yoshihide Hase: [email protected] Tanuj Khandelwal was born in Mumbai, India, in 1977. After graduating with his bachelor of engineering from the University of Mumbai in 1995 with an emphasis on electronics and telecommunications, he completed his master of science in electrical engineering in 2001 from California State University, Long Beach, USA. He joined ETAP thereafter, and is responsible for software research and development and engineering consulting projects, and he has conducted numerous power system analyses for industrial, renewable, commercial, and utility power systems. He is currently chief technology officer (CTO) at ETAP and a senior member of the Institute of Electrical and Electronics Engineers (IEEE). —Tanuj Khandelwal: [email protected]

Kazuyuki Kameda was born in Tochigi Prefecture, Japan, in 1947. After graduating in electrical engineering from Gunma Technical College, he joined the Toyo Engineering Corporation in 1968 and took charge of engineering and construction for various chemical plants and power system projects worldwide. In 1989, he established Eltechs Engineering and Consulting Co., Ltd., which executed power system engineering and power system analyses. He currently serves in sales and marketing at ETAP as a representative of OTI (Operation Technology, Inc.) in Japan. —Kazuyuki Kameda: [email protected]

xxxi

Preface In the last three decades of the twentieth century, most analytical engineering calculations were specialized tasks and were usually conducted by full-time power system experts who were familiar with mainframe and general-purpose computers such as IBM PCs. Circumstances have changed rapidly in the past two decades of the twenty-first century. We are now living in an era when anyone can learn and conduct engineering simulations at their desk using powerful computers and sophisticated engineering software platforms. The rapid changes in the electrical power engineering discipline are no exception. Many engineers who are conducting simulations have unrelated expertise and spend the majority of their time working on different engineering activities. Their background may be mechanical engineering, computers, civil engineering, physics, or mathematics instead of electrical engineering. However, they are able to conduct analytical tasks by themselves at their desks as needed, because easy-to-use analytical platforms are available as desktop applications. In addition, the capabilities of electrical application platforms have expanded far beyond just performing calculations: now they can perform a variety of engineering work. ETAP, as a typical example of the prevailing worldwide power system analytical engineering platforms, has been utilized by electrical power engineers with various expertise as a robust, multipurpose, essential engineering support tool and even as a powerful communication tool in the fields of commercial businesses and engineering societies. However, a powerful engineering tool is not a magic box. Essential factors required for a good power engineer are good background knowledge with regard to power system behavior and engineering practices, as well as a high level of skill to utilize engineering tools to maximum effect. In its 37 chapters, divided into three parts, this book brings together critical contributions from international engineers and researchers to provide an outstanding balance of power system engineering theory, simulation, and practical applications that can motivate the reader to simulate electrical systems rather than just to perform analysis. Part A of the book covers power systems background theories and practices, Part B discusses power circuit computation theories for power systems, and Part C demonstrates how to utilize computation platforms in order to perform effective simulation and analysis. This book is intended for students in electrical and other engineering disciplines. In addition, it was designed to be useful as a reference and self-study guide for the professional dealing with electrical energy systems. The book’s broad coverage together with the practical use of computer-based power system analysis makes it ideal for use in a number of ways, including service courses taught to non-electrical engineering majors. The organization and details of the material in this book should enable instructors to select topics to include in courses within the modern engineering curriculum with sufficient flexibility. Part A, “Power Systems Theories and Practices,” is a revision of Handbook of Power Systems Engineering with Power Electronics Applications, second ed., by Yoshihide Hase, published in 2013 by Wiley. The original book’s 28 chapters have been totally revised and recast into new 21 chapters, including additional new content. The following is from the preface for the original book: This book deals with the art and science of power systems engineering for those engineers who work in electricityrelated industries such as power utilities, manufacturing enterprises, engineering companies, or for students of electrical engineering in universities and colleges. Each engineer’s relationship with power system engineering is extremely varied, depending on the types of companies they work for and their positions. We expect readers to study the characteristics of power systems theoretically as a multi-dimensional concept by means of this book, regardless of readers’ business roles or specialties.

xxxii

Preface

We have endeavored to deal with the following three points as major features of the book: First, the book covers the theories of several subsystems, such as generating plants, transmission lines and substations, total network control, equipment-based local control, protection, and so on, as well as phenomena ranging from power (fundamental) frequency to lightning and switching surges, as the integrally unified art and science of power systems. Any equipment in a power system network plays its role by closely linking with all other equipment, and any theory, technology or phenomenon of one power system is only a viewpoint of the profound dynamic behavior of the system. This is the reason why we have covered different categories of theories combined in a single hierarchy in this book. Secondly, readers can learn about the essential dynamics of power systems mostly through mathematical approaches. We explain our approach by starting from physically understandable equations and then move on to the final solutions that illustrate actual phenomena, and never skip explanations or adopt half-measures in the derivations. Thirdly, the book deals with scientific theories of power system networks that will essentially never change. We intentionally excluded descriptions of advanced technologies, expecting such technologies to continue to advance year by year. In this book, most of the theoretical explanation is based on typical simple circuits with one or two generators and one or two transmission lines. Precise understanding of the phenomena in such simple systems must always be the basis of understanding actual large systems and the incidents that may occur on them. Once the reader has a firm grasp of theoretical power system concepts and a good understanding of electrical energy components, Part B of this book (Chapters 22–28) should be referenced to learn about the use of component models and efficient computational techniques in the development of computer programs representing the steady and dynamic states of electrical power systems. Note that power systems are a very diverse field, and there are multiple books that delve into the complex math associated with numerical methods. The authors have tried to present adequate numerical computation methods and their overall usage in order to prepare readers to arm themselves with sufficient understanding before tackling power systems analysis using computer-based software. Some basic knowledge of power system theory, matrix analysis, and electrical engineering laws is presented. After the introduction of various numerical methods, specific method applications are discussed for power flow, short-circuit, harmonics, and reliability analysis. Part C (Chapters 29–37) combines knowledge from Part A and Part B to demonstrate in a practical manner the use of a sophisticated and user-friendly power system analysis software platform: Electrical Transient Analyzer Program (ETAP), designed and developed by Operation Technology, Inc., Irvine, CA, USA. ETAP is a handy tool for solving numerical problems with a large number of buses from commercial power systems all the way to utility electrical networks. Part C includes new benchmark examples from industrial, renewable, and utility power systems to illustrate best practice for engineers to conduct power system analysis. Keeping in mind current developments and future requirements, the book has a planned preference toward computer simulation of power systems and the application of ETAP for analyses and solutions of complex problems. The information in Part C integrates, through examples, several hand calculations that are solved later by ETAP to illustrate the differences between estimated/shorthand calculations as well as solutions to practical problems and mitigation techniques. This will not only help readers to comprehend the ETAP solutions to practical problems but will also encourage readers to modify and even develop their own simulation results. The reader is strongly encouraged to read the ETAP user guide for additional information on program options that are not replicated in this book. While the explanations of Part C and Part B are based on ETAP, they are also good references for readers who utilize other, similar platforms. Part A was drafted by Yoshihide Hase, Part B and Part C were drafted by Tanuj Khandelwal, and the Part C industry benchmark examples were drafted and solved by Kazuyuki Kameda. The authors strongly hope that young engineers in the field study this book and use it to contribute to society’s future. The book contains a large volume of pages, but it is just the tip of the iceberg; a considerable amount of material was left on the proverbial cutting-room floor. In conclusion, the objective in writing this text and the accompanying ETAP package was to provide a student-oriented, comprehensive, and up-to-date reference book, with consistent notation and necessary detailed explanation at the level for which it is intended. Should a second edition be required, the authors would be grateful to receive your comments and any suggestions for improvements on any aspect of this book. Yoshihide Hase Tanuj Khandelwal Kazuyuki Kameda September, 2019

xxxiii

Acknowledgments The authors express deep thanks to Toshiba, TEPCO, CEPCO and Showa Wire & Cable Co., who granted some courtesy figures and photo materials for the Part A chapters. Further for Part B and part C, words may not be enough to express our gratitude to the management at ETAP who permitted the use of the ETAP software to bring life to our theoretical examples. In particular, we would like to acknowledge our gratitude to our publisher, our reviewers, and the staff who provided invaluable constructive criticism. Last but not the least, we are indebted to the countless faculty members and young readers for their eager acceptance of our work.

1

Part A Power Systems Theories and Practices

3

1 Essentials of Electromagnetism 1.1

Overview

Electricity and magnetism were considered two separate forces until James Clerk Maxwell’s revolutionary work in 1873, A Treatise on Electricity and Magnetism. Maxwell showed that the interactions of positive and negative charges were mediated by one force. Four main effects resulting from the interaction between electricity and magnetism may be experimentally demonstrated and observed in our world today: i) Electric charges attract or repel each another with a force inversely proportional to the square of the distance between them: i.e. dissimilar charges attract, and similar ones repel. ii) Magnetic poles attract or repel one another in a manner similar to positive and negative charges and always exist as pairs. iii) Electric current inside a wire creates a corresponding magnetic field outside the wire. Its direction (clockwise or counterclockwise) depends on the direction of the current in the wire. iv) Current is induced in a loop of wire when it is moved toward or away from a magnetic field, or a magnet is moved toward or away from it; the direction of current depends on that of the movement. Electromagnetic force is one of the four known fundamental forces including weak nuclear, strong nuclear, and gravitational force. Electromagnetic force is responsible for practically all phenomena one encounters in daily life above the atomic scale, with the exception of gravity. For the most part, all forces involved in interactions between atoms can be explained by the electromagnetic force acting between the electrically charged atomic nuclei and electrons of the atoms. Electromagnetic forces also explain how these particles carry momentum by their movement. In addition, electromagnetic force is involved in all forms of chemical phenomena. Without these profound, significant theoretical and practical discoveries, power generation, transmission, and utilization would not exist as we know them today.

1.2

Voltage, Current, Electric Power, and Resistance

If a conductor loop circuit exists and an electrical source (or electromotive force EMF) is series-connected to the loop circuit, free electrons in the conductor begin to travel. Such a phenomenon is usually explained as follows: “Electric charge q is moved by electromotive force v.” The current is defined as moving velocity (dq/dt) of the electric charge q. Following is the equation definition of current: it =

d q t dt t

q t =

i t dt

11

0

Next, Ohm’s law states that “If a voltage v is given to an object element, current i will flow. The voltage v and the current i are proportional to each other.” The voltage and current are time-changing variables v(t) and i(t) with the parameter t.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

4

1 Essentials of Electromagnetism

Switch

Source voltage e(t)

L i S

i(t)

R

VR(t)

V (a)

(b)

Figure 1.1 Power and resistance.

Following is the equation that defines electrical resistance: v t =R i t R=

vt it

12

Electrical resistance R by our definition means the proportional constants of the voltage v(t) and the current i(t). As shown in Figure 1.1a, electrical resistance R is a characteristic of an individual object material. There are conductive materials and nonconductive (resistive) materials, all of which have different electrical resistance values. Figure 1.1b illustrates the power consumption of the load, where terminals a and b of the load material are connected to a battery. If the voltage difference v across the two terminals is a constant value, the current i is also a constant value. The total electric charge dq move from terminal a to terminal b during a small period of time dt, and the charge that moves dq must be equal to i dt. Further, the original electrical potential energy value U decreases by a small amount of energy dU. Decreased energy: d q t 13 dt This means the electrical mode potential energy dU is converted into another type of energy: thermal energy. dU is called electric energy loss or Joule loss. The conversion speed dU(t)/dt of the electrical energy dU to thermal energy is the definition of electric power P(t): dU = dq v t = i t dt v t

d U t =i t v t dt

it =

14

P t

From Eqs. (1.3) and (1.4), we obtain the definition of a watt P t =i t

2

R

P t =

vt 2 R

15

and the definition of a watt hour: ∞

U t =

P t dt

16

0

1.3

Electromagnetic Induction (Faraday’s Law)

Figure 1.2a illustrates that no current will be generated in the loop circuit if the magnet is at a standstill, whereas current will be generated if the magnet is moved. Further, the current’s magnitude is proportional to the moving velocity of the magnet. This is the Faraday’s law of electromagnetism. Faraday’s law can be interpreted as follows from the viewpoint of physics. The magnetic flux lines emitting from the N pole of the magnet gradually spread. Therefore, if the magnet is brought closer to the loop circuit, the number of magnetic lines that pass through the loop increases, and magnetic energy is supplied to the loop conductor, so that free electrons are forced to move; in other words, current is caused by the

1.3 Electromagnetic Induction (Faraday’s Law)

S N

S

(a)

(b)

Figure 1.2 Magnetic induction.

induction of EMF. If the magnet is at a standstill, the current and EMF disappear, because the number of magnetic lines does not change. As shown in Figure 1.2b, if the switch is not closed, current cannot flow through the loop circuit, and magnetic flux is not generated: no magnetic field is created. Magnetic flux is produced by loop current that begins to flow when the switch closes. If the current increases, the magnetic flux also increases proportionally, so that some of the induced flux passes through the adjacent second loop conductor in the figure, and current and EMF are induced in the second loop conductor. The second loop’s current and EMF are also proportional to the moving velocity of the magnet. If the first loop’s current is a constant magnitude, the magnetic flux that passes through the second loop will also become a constant magnitude, and the second loop’s current and EMF will disappear. This is the principle of mutual electromagnetic induction that was discovered by Michael Faraday and independently by Joseph Henry. It was experimentally proven that electricity and magnetism always affect each other and thus should be combined as a single concept: electromagnetism. This discovery suggested an important fact: electricity can be produced in a stable manner if a magnet is moved in a stable way by mechanical driving power. Faraday’s law is as follows, using a coil that has N turns: d Φt dt d ε t = −N Φ t dt ε t =−

17 18

where ε(t): the induced electromotive force (EMF) Φ(t): the linking flux. In the case of an N-turn solenoid coil with no flux leaks, EMF obviously is proportional to the number of turns N. Just after Faraday’s discovery, Heinrich Lenz formulated Lenz’s law in 1833. It states that an induced electric current flows in a direction such that the current opposes the change that induced it. This was later explained as a special case of the law of conservation of energy. Figure 1.3 illustrates Lenz’s law. Whenever the N pole of the magnet is moved close to the loop conductor, current is induced such that the original magnetic field is canceled. A new magnetic field is produced whose NS pole’s direction is inverse to the original magnetic field.

S

μ→ z

Figure 1.3 Lenz’s law.

N

N

S

5

6

1 Essentials of Electromagnetism

1.4

Self Inductance and Mutual Inductance

Suppose we have an N-turn solenoid coil that is rigidly wound with no flux leaks. If a power source is connected to the coil, current i(t) will flow and flux Φ(t) will be induced. In this case, all of the flux Φ(t) is linking flux (leakage flux is zero) that links with the current N times. The induced linking flux (or flux linkage) Φ(t) is proportional to current i(t) and the number of turns. The definition of self-inductance is L i t =N Φ t ∴ L=

N Φt it

19

The total magnetic flux linkage is N Φt The current i(t) and magnetic flux linkage N Φ(t) are proportional, and the proportional coefficient N Φ(t)/i(t) is the definition of inductance L. The physical unit of L is the henry H = (T m2/A). The inductance L is the magnetic flux linkage per 1 amp. If current flows to any coil, voltage v(t) (or EMF) is induced that is proportional to the changing speed di(t)/dt of the current i(t) (Faraday’s law). Also, from Eqs. (1.8) and (1.9), di t Φt = −N dt dt where N Φ t = L i t , based on the definition of inductance L ε t = −L

1 10a 1 10b

Following is a popular equation for the voltage between the coil terminals where EMF ε(t) is called by voltage v(t) and the minus sign is removed. v t =L

di t Φt =N dt dt

1 10c

In Figure 1.2b, coil 2 (N2 turns) is magnetically coupled with coil 1 (N1 turns). If current i1(t) flows into coil 1, flux Φ1(t) is induced; this can be divided into the linking flux ΦM(t), which links with coil 2 as well as coil 1, and the leakage flux Φleak1(t), which links only with coil 1 and does not link with coil 2. The flux ΦM(t) produces i2(t) in coil 2. The flux ΦM(t) that couples coil 1 and coil 2 is called mutual linking flux or mutual flux linkage, and may also be called main flux as a technical engineering term. The equations are as follows: For coil 1: ε1 t = N1

ΦM t + Φleak1 t ΦM t Φleak1 t = N1 + N1 dt dt dt

ε1 t = − L 1

di1 t dΦ1 t = − N1 dt dt

Φ1 t = ΦM t + Φleak1 t

1 11a 1 11b

For coil 2, the definition of mutual inductance M is M i2 t = N2 ΦM t

1 11c

If the current i1(t) changes, ΦM(t) also changes proportionally, and the EMF ε2(t) is induced in coil 2 (secondary voltage). By the derivative, ε2 t = − M

d dΦM t i2 t = − N2 dt dt

1 11d

Eq. (1.11d) is the derivative form of Eq. (1.11c). The minus sign indicates that ε2(t) is induced in the opposed direction of dΦM(t)/dt according to Lenz’s law.

1.5 Mutual Capacitance

Coil 1 and Coil 2 should be reciprocal theoretically and in the form of equations, where their leakage fluxes are taken into account. The source voltage is ε1 t = N 1

ΦM t + Φleak1 t ΦM t Φleak1 t = N1 + N1 = v1 t + vleak1 t dt dt dt

ε2 t = N2

ΦM t + Φleak2 t ΦM t Φleak2 t = N2 + N2 = v2 t + vleak2 t dt dt dt

1 12a

where Terminal voltages

v1 t = N1

ΦM t dt

v2 t = N2

ΦM t dt

Leakage voltages

1 12b

vleak1 t = N1

Φleak1 t dt

vleak2 t = N2

Φleak2 t dt

1 12c

and from Eq. (1.12b), v1 t ΦM t v2 t = = N1 dt N2

1 12d

Further, the power transferred between the two coils must be preserved: v 1 t i 1 t = v2 t i 2 t

1 12e

From Eqs. (1.12d) and (1.12e), v1 t N1 i2 t = = N2 i1 t v2 t and i1 t N1 = i2 t N2 equal ampere − turn

1.5

1 12f

Mutual Capacitance

In Figure 1.4a, two electrodes a and b are facing each other and are connected to a battery. The potential difference or voltage vab(t) between a and b is zero in the initial condition with switch S in an open position. If switch S is closed, vab(t) immediately increases from zero until it is the same voltage value as the battery. This phenomenon can be physically interpreted that if switch S is closed, electrons at terminal a are forced to move to terminal b by a chemical process. As a result, the number of electrons at terminal a decreases (positive electric charge +q is stored), and the number of electrons at terminal b increases (electric charge −q is stored). Now, look at the parallel facing plain plate electrodes in Figure 1.4a. If voltage is charged to the electrodes, a uniform electric field E is caused, and electric charges +q and −q at both plates will increase until the voltage V is the same value as that of the battery. The value of q is obviously proportional to the intensity of the electric field E and the effective area A of the plates (A is called the Gaussian surface), so that the following equation is satisfied: q = ε0 A E

1 13

The flux ΦE of the electric field is proportional to A and E and has the following equation: ΦE = A E

1 14

7

8

1 Essentials of Electromagnetism

l h

Switch C + Battery

–

+

V

–

B

Current I

Capacitor C

S (a)

(b)

+ + + + + + + + + + +q

Gaussian surface

d A –q – – – – – – – – – –

A

+q B

A

–q

Integration directional pass (c)

(d)

(e)

Figure 1.4 Displacement current according to Maxwell and capacitance.

The electric field E is uniform from distance 0 to the plate-to-plate distance d so that the potential difference (voltage) V across the electrodes is +

V=

−

d

Eds =

ds = E d

1 15

0

From Eqs. (1.13) and (1.15), by eliminating E, ε0 A V d q=C V q C= V ε0 A farad F = coulomb volt where C d q=

1 16a 1 16b 1 16c 1 16d

Capacitance (C) is physically the proportional coefficient of q and V and is defined by Eq. (1.16c). The capacitance C is proportional to the area A and is in inverse proportion to the gap distance d as written in Eq. (1.16d). The voltage V and other quantities need not be constant values; they are written as time-dependent variables: t

q t = it =

−∞

i t dt = C v t

vt =

d d q t =C v t dt dt

1 C

t −∞

i t dt

1 17a 1 17b

Next, the loop circuit in Figure 1.4b includes a capacitor with facing disc plates and a time-changing current i(t) (typically, AC current or transient current) is flowing through the circuit. All of Eqs. (1.13)–(1.17) are satisfied at any arbitrary time t, so the proportional relationships of all the variable quantities are preserved. Then the equations are given as follows, where ε0 is called permittivity. q t = ε0 A E t d d i t = q t = ε0 A E t dt dt +

vt =

−

1 18a 1 18b

d

E t ds =

E t ds = E t d 0

1 18c

1.5 Mutual Capacitance

The voltage v(t) is given by integrating the electric field E(t) from distance 0 to d. Now, let’s further examine the capacitor gap. James Clerk Maxwell focused on the total flux of the electric field ΦE(t). He assumed, as shown in Figure 1.4c, that the current id(t) flows through the gap across distance d and that id(t) is proportional to the changing speed of the flux of the electric field ΦE(t)/dt and to the permittivity constant of the gap ε0. He named the current id(t) the displacement current. The total flux of the electrical field is ΦE t = A E t

1 19a

and the current is id t = ε0

d ΦE t dt

1 19b

Or, from Eqs. (1.19a) and (1.19b), id t = ε0

d d ΦE t = ε0 A E t dt dt

1 19c

Equation (1.19c) of the displacement current id(t) that by Maxwell’s definition flows across the gap and Eq. (1.18b) of the conductor current i(t) are of the same form. Therefore, id(t) and i(t) can be handled as the same current: the same current flows through the capacitor gap: id t = i t

1 20

Figure 1.4d shows the shows the stream passes of the magnetic fields in the inner part of the disc and the outer edges. Figure 1.4e is a special case with two pointed electrodes. By the way, Maxwell’s hypothesis of displacement current was a great step in modern scientific history, because the theory indicated that electricity or electrical power (and probably light and light power) can be transferred across space with arbitrary permittivity (including a vacuum) without the “existence of ether.” The concept of capacitance can be summarized as follows. Assuming an arbitrary point m on a surface of a material and another point n on a surface of the same material or a different material (regardless of whether the materials are conductors or insulators), a small point-to-point capacitance ΔC leakage exists between points m and n. Also, capacitance C exists between the two surfaces, because it is an integrated part of the point-to-point capacitance. If an electrical potential difference between the surfaces (voltage across the surfaces) v(t) exists, capacitive leakage current i(t) flows through the capacitance, where i(t) can be calculated using Eq. (1.17b). Capacitance between two surfaces and across a gap is usually called leakage capacitance, by which they can be easily distinguished with solid-state capacitors. The leakage capacitance C is given by Eq. (1.16d). It is proportional to the surface area A and to the specific permittivity ε0 of the gap, and it is in inverse proportion to the distance d across the two surfaces.

9

11

2 Complex Number Notation (Symbolic Method) and the Laplace Transform There are some essential mathematical tools that are indispensable for electrical power engineering in order to handle three-phase AC electrical quantities qualitatively as well as quantitatively. These are typically (i) complex numbers notation (originally referred to as the symbolic method), (ii) symmetrical components (012-transform) analysis, (iii) dq0-transform analysis, (iv) the per-unit method, (v) Laplace transform analysis, (vi) matrices equational calculations, and (vii) transfer-functional analysis. In addition, (viii) αβ0-transform analysis (Clarke components) may sometimes be used. This chapter briefly introduces essential aspects of these tools, including mathematical definitions and logical fundamentals. So, individual explanations may be looked up here as in a dictionary, and the methods to use the tools are covered in succeeding chapters.

2.1

Euler’s Formula

Leonhard Euler (1707–1783), who was born in Switzerland, was the greatest mathematician in modern mathematics history. Many different formulas bear Euler’s name, but the most famous and important formula in the field of engineering is the following: e jx = cos x + jsin x e

± jx

2 1a

= cosx ± jsin x

2 1b

And the special case of x = π/2 is π

e j2 = e j90 = j

2 1c

Euler’s formula can be illustrated in the form of a complex plane as shown in Figure 2.1. Electrical engineers have to handle various electrical quantities using complex number notations, and Eq. (2.1) gives the essential basis of such notations. This section explains the basic mathematical background of this formula. In the modern algebra, the arbitrary function f(x) of variable x can be expanded as the infinitive series expanded form, that is known as Taylor Series Expansion: f x =f a +

f

1

1

a

x −a +

f

2

2

a

x− a 2 +

f

3

a

x −a

3

f

3

n

n

In the equation, f (k)(a) means that f (x) is differentiated k times and the x of Or as of Maclaurin’s Series Expansion theorem by putting a 1

2

3

a

x−a

n

dn f x is put into a, namely x dx n

2 2a a.

0.

n

0 f 0 2 f 0 3 f 0 n x+ x + x x 2 2b 1 2 3 n Therefore, the functions ex, ejx, sin x, and cos x can be easily expanded as follows using the Maclaurin’s theorem: f x =f 0 +

f

∞

1 n 1 1 1 1 x = 1 + x + x2 + x3 + x4 + x5 + n 2 3 4 5 n=0 1 1 1 1 jx 2 + jx 3 + jx 4 + jx 5 + e jx = 1 + jx + 2 3 4 5

ex =

2 3a 2 3b

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

12

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

Im i

eiφ = cos φ + i sin φ

sin φ φ 0 cos φ

1 Re

Figure 2.1 Complex number coordinates.

1 2 1 4 1 6 1 8 x + x − x + x 2 4 6 8 1 3 1 5 1 7 1 9 sin x = x − x + x − x + x 3 5 7 9 cos x = 1 −

2 4a 2 4b

From Eqs. (2.4a) and (2.4b), cos x + j sin x = 1 −

1 2 1 4 1 6 1 8 x + x − x + x 2 4 6 8

+ j x−

1 3 1 5 1 7 1 9 x + x − x + x 3 5 7 9

25

1 1 1 1 jx 2 + jx 3 + jx 4 + jx 5 + = 1 + jx + 2 3 4 5 The right side of Eq. (2.5) is equal to the right side of Eq. (2.3b), deriving Eq. (2.1a). Further, putting x ± x yields Eq. (2.1b). In addition, putting x π in the equation yields Eq. (2.6). This formula is famous because the unique, important mathematical values π, e, and j are all included in this one simple equation: e jπ = − 1

26

By the way, complex number notation based on the Euler’s Formula is not the sole possession of electrical engineers. For example, it is also the most practical tool for working with theories of modern physics such as the theory of elemental particles and electromagnetic theory.

2.2

Complex Number Notation of Electricity Based on Euler’s Formula

In Figure 2.2, the resistance R and inductance L are connected in series via a switch S. Let’s examine the transient phenomenon of the circuit at time t ≥ 0 just after the switch is closed at time t = 0. The following functional relations of Eq. (2.7a) are maintained throughout the duration of t ≥ 0: vR t = R i t d vL t = L i t dt e t = vR t + v L t ∴ e t =R i t +L

2 7a

d it dt

2 7b

2.2 Complex Number Notation of Electricity Based on Euler’s Formula

So, the current i (t) can be found as the solution of the differential Eq. (2.7b). 2.2.1

Case 1

In the case where the AC source voltage e(t) = E cos(ωt + θ) with frequency f = ω/2π is given, d it 2 8a dt And the solution i(t) is given by the following equation (the solution process will be explained later):

Switch

Source voltage e (t)

i (t)

R

VR (t)

L

VL (t)

e t = Ecos ωt + θ = R i t + L

it =

Figure 2.2 LR circuit.

R E cos ωt + θ − φ − e − L t cos θ − φ Z

2 8b

where R2 + ωL

Circuit impedance: Z = cosφ =

R Z

cos φ =

ωL Z

φ = tan − i

L Time constant: T = Z = R 2.2.2

2

ωL (angle of power factor) R

R2 + ωL

ωL

2

R

Case 2

In the case where the AC source voltage e(t) = E sin(ωt + θ) with frequency f = ω/2π is given, the differential equation and the solution are e t = Esin ωt + θ = R i t + L i t =

d i t dt

2 9a

R E sin ωt + θ −φ − e − L t sin θ −φ Z

2 9b

Note that Case 2 is obtained by replacing θ from Case 1 with θ − 90 , because we know cos ωt + θ − 90 = sin ωt + θ

2.2.3

Case 3

Now, imagine a complex number source voltage in the form of composite form of Eq. (2.8a) + j Eq. (2.9a), where the current solution must be in the form of Eq. (2.8b) + j Eq. (2.9b): e t + je t = Ecos ωt + θ + jEsin ωt + θ = R i t + ji t

+L

d i t + ji t dt

2 10a

i t = i t + ji t =

R E − t cos ωt + θ − φ − e L cos θ − φ Z

+j

R E − t sin ωt + θ −φ −e L sin θ − φ Z

2 10b

The Eq. (2.10a) and Eq. (2.10b) can be simply reformed as follows by referring Euler’s formula. e t = Ee j ωt + θ = R i t + L it =

d it dt

E j ωt + θ −φ E j ωt + θ − φ E − Rt j θ − φ R − e − Lt e j θ −φ = − e e e L e Z Z Z

e t and i t with the upper dot indicate the complex number voltage and current.

2 10c 2 10d

13

14

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

Obviously, if the voltage source is given by complex number form e t = Ee j ωt + θ , then the circuit equation can be given by Eq. (2.10c) with the complex number voltage and current variables. Further the current solution can be given by Eq. (2.10d) also as the complex number current i t , where the first term and the second term of the right side are the steady state term and the transient term respectively. Of course the real part of Eq. (2.10c)(2.10d) correspond with Eq. (2.8a)(2.8b), and the imaginary part of that correspond with Eq. (2.9a)(2.9b). In other words, the source voltage can be given by complex number Eej(ωt + θ) instead by real number E cos(ωt + θ) or E sin(ωt + θ), then the complex number solution is derived, and finally the real part or imaginary part of the solution gives the actual engineering solution. In conclusion, circuit transient calculation as well as steady state calculation can be conducted either by real number calculation or by complex number calculation. However complex number calculation is always far easier than real number calculation. This is because complex number calculation can release lengthy real number addition and multiplication calculation of trigonometric functions. This is the base concept of complex number notation. By the way, if we need only a steady-state current solution, then omitting the transient term of Eq. (2.10d), ist t =

E j ωt + θ − φ e Z

2 11a

Then from Eqs. (2.10c) and (2.11a), e t = E e j ωt + θ = R i t + L

d E j ωt + θ −φ E E j ωt + θ − φ i t =R e +L jωe j ωt + θ − φ = R + jωL e dt Z Z Z

2 11b

∴ e t = E e j ωt + θ = R + jωL ist t or e t = E e j ωt + θ = Z ist t where Z = R + jωL

2 11c

If we give the source voltage using complex number Eq. (2.11a), then the steady-state current calculation can be obtained as the solution of complex number algebraic Eq. (2.11b), instead of solving the original differential Eq. (2.8a) or (2.9a). Eq. (2.11b) can be written by replacing d/dt from Eq. (2.10b) with jω as a formality. Now the concept of circuit impedance Z = R + jωL is introduced, and the steady-state calculation can be done quite easily with a complex number algebraic calculation.

2.3 LR Circuit Transient Calculation Using Complex Number Notation and the Laplace Transform Transient phenomena cannot be calculated by utilizing jωL or 1/jωC reactive impedance; we have to solve the differential equations for individual circuits. Solving differential equations is generally quite difficult, and applying the Laplace transform in combination with complex number notation is usually a great help. This is due to special properties of the exponential constant e, as are shown in Eq. (2.12): e ± jx = cos x ± sin x = 1 d jx d j ωt + θ e = j e jx , e = jω e j ωt + θ dx dx

2 12a 2 12b

Let’s solve the differential Eq. (2.10b) by using the Laplace transform. The differential equation is again d 2 13a E e j ωt + θ = R i t + L i t dt The Laplace transform method replaces d/dt with s and replaces ejθ with 1/(s − jω) when switching is caused at t = 0 (refer to Appendix A): d 1 1 , e j ωt + θ = e iθ e jωt s, e iωt 2 13b e iθ dt s − jω s− jω So, Eq. (2.13a) in the t-domain can be transformed into the Laplace s-domain as follows: 1 = R i s + L si s 2 13c Ee jθ s −jω

2.3 LR Circuit Transient Calculation Using Complex Number Notation and the Laplace Transform

Then, the current solution for the s-domain is ∴ is =

Ee jθ L

1 s − jω

s+

2 13d

R L

Now we need to conduct the Laplace inverse transform from s-domain to t-domain. Referring to the Laplace transform (Appendix A), 1 s+α s+β Then, with α

it =

=

1 e −αt −e − βt β−α

− jω and β

2 14a

R/L,

R R Ee jθ 1 Ee jθ Ee jθ R −jωL − t − t e jωt −e L = e jωt −e L = 2 L R R + jωL Z + jω L

Ee jθ cosϕ − jsin ϕ Z

∴ it =

E Z

e jωt − e R

R − t L

e j ωt + θ −ϕ − e − L t e j θ −ϕ

= =

Ee j θ − ϕ Z

e jωt −e

R − t L

=

E Z

e jωt − e

R − t L

e j ωt + θ −ϕ −e

R − t j θ −ϕ L e

E j ωt + θ −ϕ E − R t j θ −ϕ e − e Le Z Z Steady-state term Transient term

where Z = R + jωL Z =

T=

R2 + ωL tan φ =

ωL R cos φ = R Z

sin φ =

ωL Z

L attenuation time constant R

2 14b

or ∴it =

E Z

R

cos ωt + θ − φ − e − L t cos θ − φ

+j

Solution in case of source voltage Ecos ωt + θ

E Z

R

sin ωt + θ −φ −e − L t sin θ − φ

Solution in case of source voltage Esin ωt + θ

2 14c

Equations. (2.14b) and (2.14c) are the same as Eqs. (2.10d) and (2.10b), respectively; and the real part and imaginary parts of Eq. (2.14c) are the same as Eqs. (2.8b) and (2.9b) respectively. Figure 2.3 shows a typical transient current waveform that was calculated in Excel:

1.5 1 0.5 0 –0.5 –1 –1.5 –2 Figure 2.3 Transient current of an LR circuit.

Steady-state term Transient term Transient current

ω = 50Hz R = 10Ω L = 2H 2π ωL = 89 φ = tan − 1 θ = 75 R f=

T=

L = 0 2 sec R

15

16

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

2.4

LCR Circuit Transient Calculation

Next, let’s examine an LCR series-connected circuit, as shown in Figure 2.4. The circuit can be written with the following equations: e t = vR t + vL t + vC t Switch

i (t)

or

R

Source voltage e (t)

e t =R i t +L

VR (t)

L

VL (t)

C

VC (t)

d 1 it + dt C

t −∞

i t dt

where

vR t = R i t , v L t = L

Capacitor C

d 1 i t , vC t = dt C

t −∞

i t dt

2 15

If the source voltage is given by the complex number form e t = Ee j ωt + θ

Figure 2.4 LCR circuit.

e t = Ee j ωt + θ = R i t + L

d 1 it + dt C

t −∞

2 16a

i t dt

then the Laplace-transform equation is Ee jθ

1 1 1 1 = R i s + L si s + i s = R + Ls + s −jω C s Cs

2 16b

is

Then

is =

Ee jθ L

s R 1 s2 + s + L LC

s − jω

=

Ee jθ L

s s −jω

R s+ 2L

2

1 R − + LC 2L

2

2 17

So our last task is to find the Laplace inverse-transformed current i t of the previous equation. The following Laplace formula can be used: jωe jωt

s s − jω Or, replacing α

s − jω

2

s −α − β

2

2

α −jω − β

2

+ e αt

α2 − β2 −jωα sinhβt − jωβcosh βt β α− jω 2 −β2

2 18a

− α, s s + α 2 −β2

jωe jωt + e −αt α + jω 2 − β2 Steady-state term

α2 − β2 + jωα sinhβt − jωβcosh βt β α + jω 2 −β2 Transient term

2 18b

2.4 LCR Circuit Transient Calculation

2.4.1

Case 1

1 R 2 ≥ . LC 2L Refer to Eqs. (2.17) and (2.18b), and define ωdαβ as follows:

Let’s consider the case where

1 R − LC 2L

2

ωd

R R , α2 = 2L 2L

2

α=

, β = j ωd

1 R − LC 2L

j

, −β2 = ω2d =

1 R − LC 2L

2

2

> 0, α2 − β2 = α2 + ω2d =

1 LC

2 19a

The first thing to do is to calculate the steady-state current component ist t . The term {(α + jω)2 − β2} included in the denominators of the steady-state component and transient component is α + jω 2 −β2 = − ω2 + α2 −β2 + j2ωα = − ω2 + =

where φ = tan

jω 1 R + j ωL − L ωC ωL −

−1

R

1 ωC

,

Z =

=

1 R + j2ω LC 2L

jω jφ e Z L

R2 + ωL −

1 ωC

2

2 19b

So, the steady-state component of ist t is jωe jωt Ee jθ jωe jωt E j ωt + θ −φ = = e 2 2 Z L jω jφ α + jω − β e Z L Then the current solution in complex number form is E j ωt + θ − φ i t st = e Z or E E cos ωt + θ −φ + j sin ωt + θ − φ i t st = Z Z it

st

=

Ee jθ L

2 19c

2 19d

The real part and imaginary parts of Eq. (2.19d) are the steady-state current solutions for the source voltage E cos(ωt + θ) or E sin(ωt + θ), respectively. Next, we can calculate the transient current component itr t . Refer to Euler’s formula in Appendix A. cosh βt =

e jωd t + e − jωd t e βt −e − βt = cos ωd t, sinh βt = = jsin ωd t 2 2

Then it

tr =

α2 − β2 + jωα sinh βt −jωβcosh βt

Ee jθ − αt e L

= e − αt

β α + jω 2 −β2 1 R + jω jsin ωd t −jω jωd cos ωd t LC 2L jω jφ e Z jωd L

jθ

Ee L

jθ

R 1 sin ωd t + j sin ωd t 2L ωLC ω jφ e Z − ωd L R 1 sin ωd t−cos ωd t −j sin ωd t 2ωd L ωd LωC

ω ωd cosωd t −

= e − αt

Ee L

= e − αt

Ee j θ −φ Z

2 19e

17

18

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

f=

2

R = 10Ω L = 2H C = 0 00001F

Steady-state term

1 0

Transient term

–1

Transient current

ω = 50Hz 2π 1 R − LC 2L

–2

2

= 5 × 104 > 0

ωd =

1 R − LC 2L

Z =

R2 + ωL−

2

= 224 ≥ 0 1 ωC

2

= 312Ω

Figure 2.5 Transient current of an LCR circuit: Case 1.

2L = 0 1 sec φ = tan − 1 T= R

∴ it

R

− 2Lt tr = e

Ee j θ − φ Z

ωL− R

1 ωC

= 82 6

R 1 sin ωd t− cosωd t − j sin ωd t 2ωd L ωd LωC

θ = 75

2 19g

2L (attenuation time constant) R This is the complex number current solution. Here is the a + jb form:

where T =

∴ it

tr

=e

−

R t 2L

+ je

−

E Z

R t 2L

cos θ − φ E Z

R 1 sin ωd t− cos ωd t + sin θ −φ sin ωd t 2ωd L ωd LωC

2 19h

R 1 sin ωd t− cos ωd t −cos θ − φ sin ωd t 2ωd L ωd LωC

sin θ − φ

The real part and imaginary parts are the current solutions in the case where the source voltage is given by E cos(ωt + θ) or E sin(ωt + θ), respectively. Figure 2.5 shows a typical transient current waveform calculated in Excel: Case 1 is summarized in Table 2.1.

2.4.2

Case 2

Next, let’s look at the case

α=

R R , α2 = 2L 2L

1 R ≤ LC 2L

2

, − β2 = ω2d =

2

1 R − LC 2L

2

> 0, β = ωd =

R 2L

2

−

1 1 , α2 − β2 = α2 + ω2d = LC LC

2 20a

2.4.2.1 Calculation of the Steady-State Current Component

The term in nent is

α + jω 2 −β2 included in the denominators of the steady-state component and the transient compo-

Table 2.1 Switching transient current of an LCR series circuit.

t=0

To solve the transient current equation where R2 < 4 L/C with initial conditions i = 0, vc = 0 at t = 0, L

i(t)

L

di 1 + Ri + idt = 0 dt c

The solution for the transient current:

e(t)

#1

C

R

vc(t)

Steady-state term of the current

Source voltage by cosine function e(t) = E cos(ωt + θ)

#2

Source voltage by sine function e(t) = E sin(ωt + θ)

#3

Source voltage by complex number function (symbolic e t = e t + je t method) = E e j ωt + θ

i t s= it

s

=

E cos ωt + θ −φ Z E sin ωt + θ −φ Z

i t s = i t s + ji t s

Transient term of the current

it

T

R , ωd = 2L

Z =

ω20 − α2 =

R2 + ωL−

1 ωC

1 R − LC 2L 2

2

, ω0 =

, φ = tan − 1

E Z E Z

it

T

=

it

T

=i t

E j ωt + θ − φ e = Z

ω = 2π f: angular velocity of power frequency f

α=

=

1 LC

1 ωL− ωC R

Notes: Euler’s formula: e±jφ = cos φ ± j sin φ. The symbolic relation (#1) + j(#2) = (#3) is preserved for all quantities of the circuit. See Section 2.4 for the calculation method using Laplace transforms.

E = Z

T

Circuit current

e − αt cos θ − φ

α α2 + ω2d sinωd t−cos ωd t − cos θ − φ + 90 sin ωd t ωd ωωd

e − αt cos θ − φ

α α + sinωd t−cos ωd t − sin θ −φ + 90 sinωd t ωd ωωd

+ ji t e

− αt

2

ω2d

e

S

+i t

T

=i t

S

+i t

T

it

i t =i t

T j θ−φ

i t =i t

α α + sin ωd t−cos ωd t − e ωd ωωd 2

ω2d j θ − φ + 90

sin ωd t

S

+i t

= i t + ji t

T

20

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

α + jω 2 −β2 = − ω2 + α2 −β2 + j2ωα = − ω2 + =

jω 1 R + j ωL − L ωC

1 R + j2ω LC 2L

jω jδ e Z L

=

Then the steady-state current component is it

st

=

Ee jθ L

jωe jωt α + jω 2 − β2

=

Ee jθ jωe jωt E j ωt + θ −φ = e Z L jω jφ e Z L

2 20b

2.4.2.2 Calculation of the Transient Current Component

Recall the following hyperbolic equations: e ωd t + e − ωd t e ωd t − e − ωd t sin βt = sin ωd t = cosh βt = cosh ωd t = 2 2 Ee jθ −αt e i t tr = L

−

β α + jω 2 − β2

Ee j θ −φ Z

= e −αt =e

α2 −β2 + jωα sinh βt − jωβcosh βt

1 + jωα sinh ωd t − jωωd cosh ωd t Ee jθ −αt LC = e jω jφ L e Z ωd L

α 1 sinh ωd t sinhωd t−coshωd t − j ωd ωd ωLC

Ee j θ − φ Z

R t 2L

2 20c

2 20d

R 1 sinh ωd t−cosh ωd t − j sinhωd t ωd 2L ωd ωLC

Then, in the form of a complex number, it

tr

=e

−

R t 2L

+ je where T =

−

E Z R t 2L

cos θ − φ E Z

R 1 sinh ωd t− cosh ωd t + sin θ − φ sin ωd t ωd 2L ωd ωLC

sin θ − φ

R 1 sinh ωd t− cosh ωd t −cos θ −φ sin ωd t ωd 2L ωd ωLC

2L time constant R

2 20e

The real part and imaginary parts are the current solutions in the case where the source voltage is given by E cos(ωt + θ) or E sin(ωt + θ), respectively. R By the way, the attenuating factor of the Eq. (2.20d) for an LCR circuit is unexpectedly e − 2L , whereas that of Eq. (2.14b) R for an LR circuit is e − L . This is true because Eq. (2.20d) with the condition C + ∞ (1/ωC is negligibly small) coincides with Eq. (2.14b). Putting C ∞ in Eq. (2.20d), then α=

R 2L

β = ωd ,

α2 −β2 =

1 = 0, LC

R =1 ωd 2L

2 21a

Recall sinhωd t−cosh ωd t = − e ωd t , so that Eq. (2.20) is simplified as follows: it

R

tr

= e − 2Lt

∴ it

R R Ee j θ − φ Ee j θ −φ Ee j θ −φ sinh ωd t− cosh ωd t = e − 2Lt −e ωd t = e − 2Lt Z Z Z

R

− Lt tr = − e

Ee j θ −φ Z

R

R

− e − 2Lt = − e − Lt

Ee j θ −φ Z

2 21b

2.5 Resistive, Inductive, and Capacitive Load, and Phasor Expressions

f=

ω = 50Hz 2π

1

R = 40Ω, L = 2H, C = 0 01F, 2

1 R − LC 2L R 2L

ωd =

Steady-state term

–1

Transient term Transient current

= − 50 < 0, 2

1 − = 7 07, LC

R2 + ωL −

Z =

0

1 ωC

–2

2

Figure 2.6 Transient current of an LCR circuit: Case 2.

= 629Ω,

2L = 0 1 sec, φ = tan − 1 T= R

ωL − R

1 ωC

= 86 3 , θ = 75

Eq. (2.21b) is the same as Eq. (2.14b) for the LR circuit. Figure 2.6 shows a typical transient current waveform calculated in Excel. As a final conclusion, all the equations demonstrated in Sections 2.3 and 2.4 are true for linear calculation of any voltage and current values v t ,i t , any impedance Z , or any frequency f = ω/2π. This means complex number notation as well as the Laplace transform can be adopted for any linear calculation of transient phenomena and surge phenomena. The transient current calculation of the LCR circuit (Case 2) is summarized in Table 2.1; the source voltage is expressed in three different forms:

2.5

Resistive, Inductive, and Capacitive Load, and Phasor Expressions

Let’s again examine the steady-state phenomenon of an LCR series load as shown in Figure 2.4. Referring to Eq. (2.19d) and setting θ 0 in the equation, the steady-state source voltage e(t) and current are as follows: e t = Ee jωt it

st

=

E j ωt −φ e complex number solution Z

2 22a

or e t = E cosωt it

st

=

E cos ωt −φ real number expression Z

where φ = tan− 1

1 ωL − ωC

R

Z =

R2 + ωL −

1 ωC

2 22b

2

2 22c

21

22

2 Complex Number Notation (Symbolic Method) and the Laplace Transform

Also, the terminal voltages of LCR elements are: vR t = R i t = R vL t = L vC t =

E j ωt −φ e Z

di t d E j ωt −φ E j ωt −φ =L e = jωL e = jωL i t dt Z Z dt

1 C

t −∞

i t dt =

1 C

t −∞

E j ωt −φ 1 E j ωt −φ 1 e dt = −j e = −j it Z ωC Z ωC

Namely vR t = R i t vL t = jωL i t 1 it vC t = −j ωC And e t = Ee jωt = vR t + vL t + vC t

2 23

π

where e j2 = e j90 = j Eq. (2.23) shows that the voltage vR t of the resistor terminal is in phase with the current i t and the voltage vL t of the inductor is leaded by 90 from i t , whereas vC t of the capacitor terminal is lagged by 90 from i t . The amplitude of the source voltage across all three components in a series LCR circuit is made up of the three individual component voltages vL t , vC t , and vR t , with the current common to all three components. The vector diagrams therefore have the current vector as their reference, with the three voltage vectors plotted with respect to this reference. Figures 2.7a and b show such situations.

(b) VR t

VL t (a) VC

VL VC

t

E = VR + VL + VC

E = VR + VL + VC VL + VC VR

t IR

IR VC Figure 2.7 LCR series-connected circuit.

t

2.5 Resistive, Inductive, and Capacitive Load, and Phasor Expressions

We cannot simply add vL t , vC t , and vR t to find the supply voltage e(t) across all three components, because all three voltage vectors point in different directions with regard to the current vector. Therefore, we have to find the supply voltage e(t) as the phasor sum of the three voltages. The phasor diagram for a series LCR circuit is produced by combining the three individual phasors and adding the vector voltages. Since the current flowing through the circuit is common to all the circuit elements, we can use this as the reference vector, with the three voltage vectors drawn relative to it at their corresponding angles.

23

25

3 Transmission Line Matrices and Symmetrical Components

In order to understand fully the nature of power systems, we need to study the nature of transmission lines as the first step. In this chapter, we examine the characteristics and basic equations of three-phase overhead transmission lines. However, the actual quantities of the constants are described in Chapter 4.

3.1

Overhead Transmission Lines with Inductive LR Constants

3.1.1

Three-Phase Single-Circuit Line Without an Overhead Grounding Wire (OGW)

3.1.1.1

Voltage and Current Equations, and Equivalent Circuits

A three-phase single-circuit line between a point m and a point n with only L and R and without an OGW can be written as shown in Figure 3.1a. In the figure, rg and Lg are the equivalent resistance and inductance of the earth, respectively. The outer circuits I and II connected at points m and n can theoretically be three-phase circuits of any kind. All the voltages Va, Vb, Vc and currents Ia, Ib, Ic are vector quantities; and the symbolic arrows show the measuring directions of the three-phase voltages and currents, which have to be written in the same direction for the three phases as a basic rule to describe the electrical quantities of three-phase circuits. In Figure 3.1, the currents Ia, Ib, Ic in each phase conductor flow from left to right (from point m to point n). Accordingly, the composite current Ia + Ib + Ic has to return from right to left (from point n to m) through the earth–ground pass. In other words, the three-phase circuit has to be treated as the set of “three-phase conductors + one earth circuit” pass. In Figure 3.1a, the equations of the transmission line between m and n can be easily described as follows. Here, voltages V and currents I are complex-number vector values: m Va − n Va

= ra + jωLaag Ia + jωLabg Ib + jωLacg Ic − mn Vg

①

m Vb − n Vb

= jωLbag Ia + rb + jωLbbg Ib + jωLbcg Ic − mn Vg

②

m Vc − n Vc

= jωLcag Ia + jωLcbg Ib + rc + jωLccg Ic − mn Vg

③

where mn Vg = rg + jωLg Ig = − rg + jωLg Ia + Ib + Ic Substituting ④ into ①, and then eliminating m Va − n Va

31

④

mnVg, Ig,

= ra + rg + jωLaag + Lg Ia + rg + jωLabg + Lg Ib + rg + jwLacg + Lg Ic

⑤

Substituting ④ into ② and ③ in the same way, m Vb − n Vb

= rg + jωLbag + Lg Ia + rb + rg + jωLbbg + Lg Ib + rg + jωLbcg + Lg Ic

⑥

m Vc − n Vc

= rg + jωLcag + Lg Ia + rg + jωLcbg + Lg Ib + rc + rg + jωLccg + Lg Ic

⑦

32

Now, the original Eq. (3.1) and the derived Eq. (3.2) are the equivalent of each other, so Figure 3.1b, showing Eq. (3.2), is also the equivalent of Figure 3.1a.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

26

3 Transmission Line Matrices and Symmetrical Components

Point m

Point n

nVa

I c rb mVb rc

mVc

Earth surface

nVb

Lbcg

Ig = – (Ia + Ib + Ic)

raa Laa Zab

mVa

Zac

Ib Zaa

mVb Ic Zbb m Vc

nVc

rg

Ia

Point n

nVa

nVb Zbc n Vc

Zcc

Outer-circuit II

Ib ra Laag

Outer-circuit I

mVa

Labg Lacg

Outer-circuit II

Outer-circuit I

Ia

Point m

Ideal ground (equal potential) zero impedance

Lg

Ig = – (Ia + Ib + Ic)

mnVg

(a)

(b)

Figure 3.1 Single-circuit line with LR constants.

Eq. (3.2) can be expressed in the form of a matrix equation, and the following equations are derived accordingly (refer to Appendix B for the matrix equations): m Va

n Va

m Vb − m Vc

n Vb n Vc

ra + rg + jωLaag + Lg rg + jωLbag + Lg = rg + jωLcag + Lg

rg + jωLabg + Lg rb + rg + jωLbbg + Lg rg + jωLcbg + Lg

rg + jωLacg + Lg rg + jωLbcg + Lg rc + rg + jωLccg + Lg

Ia Ib Ic 33

raa + jωLaa rba + jωLba rca + jωLca

rab + jωLab rbb + jωLbb rcb + jωLcb

Zaa Zba Zca

Zac Zbc Zcc

Zab Zbb Zcb

rac + jωLac rbc + jωLbc rcc + jωLcc

Ia Ib Ic

Ia Ib Ic

where Zaa = raa + jωLaa = ra + rg + jω Laag + Lg 34

Zbb , Zcc are written in similar equation forms and Zac , Zbc are also written in similar forms Now, we can apply symbolic expressions for this matrix equation as follows: m V abc − n V abc

= Z abc I abc

35

where m Va m V abc

=

m Vb m Vc

n Va

,

n V abc

=

n Vb n Vc

,

Zaa Z abc = Zba Zca

Zab Zbb Zcb

Zac Zbc , Zcc

Ia I abc = Ib Ic

36

Summarizing the previous equations, Figure 3.1a can be described as Eqs. (3.3) and (3.6) or Eqs. (3.5) and (3.6), in which the resistance rg and inductance Lg of the earth return pass are already reflected in all four equations, although

3.1 Overhead Transmission Lines with Inductive LR Constants

Ig and mnVg are eliminated in Eqs. (3.5) and (3.6). We can consider Figure 3.1b as the equivalent circuit of Eqs. (3.3) and (3.4) or Eqs. (3.5) and (3.6). In Figure 3.1b, earth resistance rg and earth inductance Lg are already included in the line constants Zaa, Zab, etc. so the earth in the equivalent circuit of Figure 3.1b is “the ideal earth” with zero impedance. Therefore the earth can be expressed in the figure as the equal-potential (zero-potential) earth plane at any point. It is clear that the mutual relation between the constants of Figure 3.1a and Figure 3.1b is defined by Eq. (3.4). It should be noted that the self-impedance Zaa and mutual impedance Zab of phase a, for example, involve the earth resistance rg and earth inductance Lg. Generally, in actual engineering tasks, Figure 3.1b and Eqs. (3.3) and (3.4) or Eqs. (3.5) and (3.6) are applied instead of Figure 3.1a and Eqs. (3.1) and (3.2); in other words, the line impedances are given as Zaa, Zab, etc. instead of Zaag, Zabg. The line impedances Zaa, Zbb, Zcc are called the self-impedances of the line including the earth–ground effect, and Zab, Zac, Zbc, etc. are called the mutual impedances of the line including the earth–ground effect. Measurement of Line Impedances Zaa, Zab, Zac

3.1.1.2

Let’s consider how to measure the line impedances, taking the earth effect into account. As we know from Figure 3.1b and Eqs. (3.3) and (3.4), the impedances Zaa, Zab, Zac, etc. can be measured by the circuit connection shown in Figure 3.2a. The conductors of the three phases are grounded to earth at point n, and the phase b and c conductors are opened at point m. Accordingly, the boundary conditions nVa = nVb = nVc = 0, Ib = Ic = 0 can be adopted for Eq. (3.3): m Va

−

m Vb m Vc

∴

m Va

0 0 0

=

Ia = Zaa ,

Zaa Zba Zca

Zab Zbb Zcb

m Vb

Zac Zbc Zcc

Ia = Zba ,

m Vc

Ia 0 0

①

Ia = Zca

②

37

Therefore the impedances Zaa, Zab, Zac can be calculated from the measurement results of mVa, mVb, mVc, and Ia. All the impedance elements in the impedance matrix Zabc of Eq. (3.7) can be measured in the same way. Working Inductance (Laa–Lab)

3.1.1.3

Figure 3.2b shows the case where current I flows along the phase a conductor from point m to n and comes back from n to m only through the phase b conductor on the return pass. The equation is with boundary conditions Ia = –Ib = I, Ic = 0, nVa = nVb: m Va

n Va

−

m Vb m Vc

n Va n Vc

=

Zaa Zba Zca

Zab Zbb Zcb

Zac Zbc Zcc

I −I 0

3 8a

Ia

Ia Ib = 0 Ic = 0 A

mVa

mVb

mVc

Ic = 0 A Ig = 0

Ig = –Ia (a) Figure 3.2 Measuring circuit of line impedance.

Ib = –Ia

V

(b)

27

28

3 Transmission Line Matrices and Symmetrical Components

Therefore m Va − n Va

= Zaa − Zab I

m Vb − n Vb

= − Zbb − Zba I

V = m Va − m Vb = V I=

voltage drop of the phase a conductor between points m and n voltage drop of the phase b conductor between points m and n

Zaa −Zab + Zbb − Zba I

m Va − m Vb

I = Zaa −Zab + Zbb − Zba

=

twice values of working impedance

① 3 8b ②

Equation (3.8b)① indicates the voltage drop of the parallel circuit wires a, b under the condition of the go-and-returncurrent connection. The current I flows out at point m on the phase a conductor and returns to m only through the phase b conductor, so any other current flowing does not exist on the phase c conductor or earth–ground pass. In other words, Eq. (3.8b)① is satisfied regardless of the existence of the third wire or earth–ground pass. Therefore, the impedance (Zaa–Zab) as well as (Zbb–Zba) should be specific values that are determined only by the relative condition of the phase a and b conductors, and they are not affected by the existence or absence of the third wire or earth–ground pass. (Zaa–Zab) is called the working impedance, and the corresponding (Laa–Lab) is called the working inductance of the phase a conductor with the phase b conductor. Furthermore, as the conductors a and b are generally of the same specification (the same dimension, same resistivity, etc.), the impedance drop between m and n of the phase a and b conductors should be the same. Accordingly, the working inductances of both conductors are clearly the same: (Laa–Lab) = (Lbb–Lba). The value of the working inductance can be calculated from the following well-known equation, which is derived by an electromagnetic analytical approach as a function only of the conductor radius r and the parallel distance sab between the two conductors: sab Laa − Lab = Lbb − Lba = 0 4605 log10 39 + 0 05 mH km r This is the equation for the working inductance of the parallel conductors a and b, whose derivation is shown in Chapter 4 as the theory of electromagnetism. The equation shows that the working inductance Laa–Lab for the two parallel conductors is determined only by the relative distance between the two conductors sab and the radius r, so it is not affected by any other conditions such as other conductors or the distance from the earth surface. The working inductance can also be measured as the value (1/2) V/I by using Eq. (3.8b)②. 3.1.2

Three-Phase Single Circuit Line with OGW, OPGW

Most high-voltage transmission lines are equipped with overhead grounding wires (OGWs) and/or OGWs with optical fibers (OPGW, for communication use). See Figure 4.8 for more detail. In the case of a single-circuit line with single OGW, the circuit includes four conductors; the fourth conductor (x in Figure 3.3) is earth-grounded at all the transmission towers. Therefore, using the figure for the circuit, Eq. (3.3) has to be replaced by the following equation: m Va

n Va

m Vb − m Vc m Vx = 0

Point m

m Va m Vb mVc

Zaa V Zba n b = Zca n Vc Zxa n Vx = 0 Ia Ib Ic Ix

a b c

m Vx

– (Ia + Ib + Ic + Ix) Overhead grounding wire

Zab Zbb Zcb Zxb

Zac Zbc Zcc Zxc

Zax Zbx Zcx Zxx

Ia Ib Ic Ix

Point n

3 10a

x a nVa

nVx

nVb nVc

Earth grounded at every tower

Figure 3.3 Single-circuit line with Overhead Grounding Wire (OGW).

c b

3.1 Overhead Transmission Lines with Inductive LR Constants

Extracting the fourth row, 1 Zxa Ia + Zxb Ib + Zxc Ic Zxx

Ix = −

3 10b

Substituting Ix into the first, second, and third rows of Eq. (3.10a), m Va m Vb

n Va

−

m Vc

n Vb n Vc

Zaa = Zba Zca

Zab Zbb Zcb

Zac Zbc Zcc

Ia Zax Ix Ib + Zbx Ix Ic Zcx Ix

Zaa −

Zax Zxa Zxx

Zab −

Zax Zxb Zxx

Zac −

Zax Zxc Zxx

= Zba −

Zbx Zxa Zxx

Zbb −

Zbx Zxb Zxx

Zbc −

Zbx Zxc Zxx

Zca −

Zcx Zxa Zxx

Zcb −

Zcx Zxb Zxx

Zcc −

Zcx Zxc Zxx

Zaa Zba Zca

Zab Zbb Zcb

Zac Zbc Zcc

where Zax = Zxa ,

Ia Ib Ic 3 11

Ia Ib Ic Zbx = Zxb ,

Zcx = Zxc

Zaa = Zaa −δaa , Zab = Zab − δab Zax Zxa Zax Zxb δaa = , δab = Zxx Zxx This is the fundamental equation of the three-phase single-circuit line with OGW in which Ix has already been eliminated and the impedance elements of the grounding wire are slotted into the three-phase impedance matrix. Equation (3.11) is obviously of the same form as Eq. (3.3), while all the elements of the rows and columns in the impedance matrix have been revised to smaller values with corrective terms δax = ZaxZxa/Zxx, etc. These equations indicate that the three-phase single-circuit line with OGW can be expressed as a 3 × 3 impedance matrix equation in the form of Eq. (3.11) regardless of the existence of OGW, as was the case with Eq. (3.3). Also, we can see that OGW plays a role not only to shield lines against lightning but also to reduce the self- and mutual reactance of transmission lines.

3.1.3

Three-Phase Double-Circuit Line with LR Constants

The three-phase double-circuit line can be written as in Figure 3.4 and Eq. (3.12) regardless of the existence or absence of OGW: m Va

n Va

m Vb

n Vb

m Vc m VA m VB m VC

−

n Vc n VA n VB n VC

Zaa Zba Zca = ZAa ZBa ZCa

Zab Zbb Zcb ZAb ZBb ZCb

Zac Zbc Zcc ZAc ZBc ZCc

ZaA ZbA ZcA ZAA ZBA ZCA

ZaB ZbB ZcB ZAB ZBB ZCB

ZaC ZbC ZcC ZAC ZBC ZCC

Ia Ib Ic IA IB IC

3 12

In addition, if the line is appropriately phase balanced, the equation can be expressed by Eq. (3.36a), as seen later in Section 3.5.2.

29

30

3 Transmission Line Matrices and Symmetrical Components

m

mVA

mVB

n

IA IB IC Ia Ib Ic

mVC

nVC

m Va V m b mVc

n Vc

nVb

nVB nVA

nVa

Figure 3.4 Three-phase double-circuit line with LR constants.

3.2

Overhead Transmission Lines with Capacitive C Constants

3.2.1

Stray Capacitance of Three-Phase Single-Circuit Line

Referring to Figure 3.5a, leakage currents Ia, Ib, Ic of a three-phase transmission line are given by Eq. (3.13a) or (3.13b): Ia = jωCaa Va + jωCab Va − Vb + jωCac Va − Vc Ib = jωCbb Vb + jωCbc Vb − Vc + jωCba Vb − Va

3 13a

Ic = jωCcc Vc + jωCca Vc − Va + jωCcb Vc −Vb Or, with a small modification, Ia Caa + Cab + Cac Ib = jω − Cba Ic − Cca

− Cab Cba + Cbb + Cbc − Ccb

− Cac −Cbc Cca + Ccb + Ccc

Va Vb Vc

3 13b

This is the fundamental equation for stray capacitances of a three-phase single-circuit transmission line whose conductor layout is shown in Figure 3.5a. A well-balanced transmission line capacitance can be expressed by Eq. (3.14), and the circuit can be expressed as shown in Figure 3.5b or 3.5c. Cs + 2Cm −Cm = jω −Cm

Ia Ib Ic

−Cm Cs + 2Cm −Cm

I abc

−Cm −Cm Cs + 2Cm

Va Vb Vc

3 14

V abc

C abc

∴ I abc = jωC abc V abc where Cs ≡ Caa

C ab jω a jω C

(V a

Cbb ≡ Ccc Cm ≡ Cab

) b – Vb

Cab

Cbc Cca

(V – a V c) Caa Cbb

c

Vb

ac

Va

Cbc ≡ Cca

Ccc

Vc

a b c

a

Cm Cm

Cm

Cs Cs Cs

b c Cs Cs Cs 3Cm 3Cm 3Cm

jω Caa Va (a)

(b) Single-circuit line

Figure 3.5 Stray capacitance of a three-phase single-circuit line.

(c) Single-circuit line

3.2 Overhead Transmission Lines with Capacitive C Constants

Incidentally, Figure 3.5b is equivalent to Figure 3.5a, and it can be reconfigured to Figure 3.5c where delta-connected Cm capacitors are transformed into star-connected 3Cm branches. Further, the total capacitance of one phase C ≡ Cs + 3Cm in Figure 3.4c is called the working capacitance of a pair of two conductors. The detailed theory of these leakage currents, the capacitance matrix, and calculating the equations of these capacitive elements will be discussed in Chapter 4, Section 4.2.

3.2.2

Three-Phase Single-Circuit Line with OGW

Four conductors with phase names a, b, c, x exist in this case, so the following equation can be derived as an extended form of Eq. (3.13b): Ia = jωCaa Va + jωCab Va − Vb + jωCac Va −Vc + jωCax Va −Vx

3 15a

where Vx = 0, because OGW is earth-grounded at every tower. Accordingly, Ia Caa + Cab + Cac + Cax Ib = jω − Cba Ic −Cca

− Cab Cba + Cbb + Cbc + Cbx −Ccb

− Cac − Cbc Cca + Ccb + Ccc + Ccx

Va Vb Vc

3 15b

This matrix equation is again in the same form as Eq. (3.13b). However, the phase-to-ground capacitance values (diagonal elements of the matrix C) are increased (the value of Cax is increased for the phase a conductor, from Caa + Cab + Cac to Caa + Cab + Cac + Cax).

3.2.3

Three-Phase Double-Circuit Line

Six conductors with phase names a, b, c, A, B, C exist in this case, as shown in Figure 3.6, so the following equation can be derived as an extended form of Eqs. (3.15a) and (3.15b): Ia = jω Caa Va + Cab Va − Vb + Cac Va − Vc + CaA Va −VA + CaB Va −VB + CaC Va −VC

3 16a

Then

Ia Ib Ic = jω IA IB IC

Caa + Cab + Cac + CaA + CaB + CaC

− Cab

−Cac

−CaA

− CaB

− CaC

− Cba

Cba + Cbb + Cbc + CbA + CbB + CbC

− Cbc

− CbA

− CbB

−CbC

− Cca

−Ccb

Cca + Ccb + Ccc + CcA + CcB + CcC

− CcA

−CcB

− CcC

− CAa

−CAb

− CAc

CAA + CAB + CAC + CAa + CAb + CAc

−CAB

− CAC

− CBa

− CBb

− CBc

− CBA

CBA + CBB + CBC + CBa + CBb + CBc

− CBC

− CCa

−CCb

− CCc

− CCA

−CCB

CCA + CCB + CCC + CCa + CCb + CCc

Va Vb Vc VA VB VC

3 16b

31

32

3 Transmission Line Matrices and Symmetrical Components

C´m a

A Cm

Cm B b C´m c

C C´m

Cs

Cs

Cs

Cs Cs

Cs

Figure 3.6 Stray capacitance of a double-circuit line (well balanced).

It is obvious that the double-circuit line with OGW can be expressed in the same form. The case of a well-transposed double-circuit line is shown by Eq. (3.16b) and Figure 3.6:

Ia Ib Ic = jω IA IB IC

Cs

Cs + 2Cm + 3Cm

− Cm

− Cm

−Cm

− Cm

− Cm

− Cm

Cs + 2Cm + 3Cm

− Cm

−Cm

− Cm

− Cm

− Cm

− Cm

Cs + 2Cm + 3Cm

−Cm

− Cm

− Cm

− Cm

− Cm

− Cm

Cs + 2Cm + 3Cm

− Cm

− Cm

− Cm

− Cm

− Cm

−Cm

Cs + 2Cm + 3Cm

− Cm

− Cm

− Cm

− Cm

−Cm

− Cm

Cs + 2Cm + 3Cm

Caa ≒ Cbb ≒ Ccc ≒ CAA ≒ CBB ≒ CCC

Va Vb Vc VA VB VC

3 17

one phase-to-ground capacitance

Cm

Cab ≒ Cbc ≒

≒ CAB ≒ CBC ≒

capacitance between two conductors of the same circuit

Cm

CaA ≒ CbC ≒

≒ CAa ≒ CBb ≒

capacitance between two conductors of a different circuit

Here, we have studied the fundamental equations and circuit models of transmission lines and the actual calculation method for the L, C, R constants. Concrete values of the constants are investigated in Chapter 4.

3.3

Symmetrical Coordinate Method (Symmetrical Components)

3.3.1

Fundamental Concepts of Symmetrical Components

Now we will examine three-phase circuit quantities. Electrical quantities in one phase cannot be treated independently of another phase, because not only self-inductance Laa, Lbb, Lcc or self-capacitance Caa, Cbb, Ccc, but also mutual inductance Lab, Lbc, Lca or mutual capacitance Cab, Cbc, Cca exists between one phase and another. Therefore, three phase voltages va(t), vb(t), vc(t), three phase currents ia(t), ib(t), ic(t), and three phase fluxes φa(t), φb(t), φc(t) at an arbitrary one point in a three-phase circuit must always be treated as the abc-phase set values. Furthermore, in case of equipment with winding coils, mutual inductance and mutual capacitance exist not only between different phases at one circuit point but also across the longitudinal multiple arbitrary points in the same phase

3.3 Symmetrical Coordinate Method (Symmetrical Components)

or in different phases. Rotating machines as well as transformers have a complicated circuit structure with complex distributed inductance and capacitance. For this reason, generators, motors, and transformers cannot be well specified or described in the form of abc-phase domain circuit characteristic values or in the form of circuit constants. In other words, we do not have a straightforward way to specify the circuit characteristics of three-phase equipment or to calculate three-phase circuit steady-state or transient phenomena. The symmetrical coordinate method (symmetrical components) provides a definite solution with which this critical difficulty can be solved. A three-phase power system cannot be described in the form of an abc-domain circuit for most analytical purposes, whereas it can be described by adopting the symmetrical coordinate method in the form of a 012-domain circuit. The symmetrical coordinate method is the ultimate basis for a three-phase circuit description that includes specifications for various three-phase equipment as well as transmission lines, the combined circuit configuration, and all kinds of calculations for circuits including steady-state, transient, and surge phenomena analysis. The symmetrical coordinate method is often referred to as the analytical tool for traditional fault analysis, probably for historical reasons; however, this method is a very powerful mathematical tool for three-phase circuit theory and calculations. This section first examines the essential concepts of the symmetrical coordinate method, followed by a circuit description of three-phase transmission lines and other equipment by symmetrical components. The three-phase circuit generally has four electric conducting passes (phase a, b, c passes and an earth pass), and these four electric passes are closely coupled by mutual inductances L as well as by mutual capacitances C. Therefore, phenomena on any pass of a three-phase circuit cannot be independent of phenomena on the other passes. The symmetrical coordinate method (symmetrical components) is a kind of variable-transformation technique from a mathematical viewpoint. That is, three electrical quantities on abc phases are always handled as one set of variables in the abc domain, and these three variables are then transformed into another set of three variables named positive (1), negative (2), and zero (0) sequence quantities in the newly defined 012 domain. An arbitrary set of three variables in the abc domain and the transformed set of three variables in the 012 domain are mathematically in one-to-one correspondence with each other. Therefore, the phenomena of abc-phase quantities in any frequency zone can be transformed into the 012 domain and can be observed, examined, and solved from the standpoint of the defined 012 domain. Then the behavior or solution in the 012 domain can be retransformed into the original abc domain. It can be safely said that the symmetrical coordinate method is an essential analytical tool for any kind of three-phase circuit phenomenon, and it is inevitably utilized in every kind of engineering work with power systems. Only symmetrical components provide a way to obtain the large, precise analytical circuits for integrated power systems including generators, transmission lines, station equipment, and loads. Figure 3.7 shows such a transformation between the two domains in one-to-one correspondence. One set of abcphase currents Ia, Ib, Ic (or phase voltages Va, Vb, Vc) at an arbitrary point in the three-phase network based on the

Analytical subject Ia Ib Ic 0-1-2 method Circuit equation f(I0 I1 I2)

rm]

fo ans

[Tr

Circuit equation f(Ia Ib Ic)

[Transform] α-β-0 method

rm]

fo ans

[Tr

Circuit equation f (Iα Iβ I0)

d-q-0 method Circuit equation f(id iq i0)

Calculation

Calculation

Calculation

Calculation

Solution

Solution Ia Ib Ic

Solution

Solution id iq i0

I0 I1 I2 [Inverse-transform]

Figure 3.7 Concept of variable transformation.

Iα Iβ I0 [Inverse-transform] [Inverse-transform]

33

34

3 Transmission Line Matrices and Symmetrical Components

abc domain is transformed to another set of three variables named I0, I1, I2 (or V0, V1, V2) in the 012 domain, using the defined transformation rule. The equations of the original abc domain will be changed into new equations in the 012 domain, with which three-phase power systems can be described as precise, simple circuits. Therefore, complex subjects in the abc domain can be treated and resolved easily in the 012 domain, and the solution in this domain is easily inversetransformed into the correct solution in the original abc domain. There are two other important transformation methods: a) α β 0 transformation method, (Ia, Ib, Ic) (Iα, Iβ, I0): This is also useful as a complementary analytical tool of symmetrical components. In some special circuits, α β 0 components provide easier solutions for problems for which symmetrical components may not give good solutions. (Id, Iq, I0): This is a very powerful transformation specialized for the treatb) dq0 transformation method, (Ia, Ib, Ic) ment of generators and other rotating machinery. Rotating machines can be described as precise, simple circuits only with the dq0 method. Due to the precise description of generator characteristics with the dq0 method, dynamic system behavior can be analyzed. We will learn more about these methods in later chapters. 3.3.2

Definition of Symmetrical Components

Let’s imagine AC voltages and currents at an arbitrary point of a three-phase circuit and name these quantities with time-dependent complex-number variables Va(t), Vb(t), Vc(t) and Ia(t), Ib(t), Ic(t). In association with this set of voltages and currents, we introduce a new set of complex-number voltages and currents V0(t), V1(t), V2(t) and I0(t), I1(t), I2(t) and define them in the following equations, where the notation (t) that indicates the time-dependency of the variables is omitted for simplicity. Note that all the V, I, φ variables and impedance Z values are complex number values in this section, because the symmetrical coordinate method uses complex number notation for the application of both steady-state phenomena and transient phenomena. 1 Va + Vb + Vc 3 1 V1 = Va + aV b + a2 Vc 3 1 V2 = Va + a2 Vb + aV c 3 V0 =

or

V0 V1 V2

1 1 1 = 3 1

3 18

1 a a2 a

V 012 =

1 a2 a

Va Vb Vc

∴ V 012 = a V abc

V abc

1 I a + Ib + Ic 3 1 I1 = Ia + aI b + a2 Ic 3 1 I2 = Ia + a2 Ib + aI c 3 I0 =

or

I0 I1 I2 I 012 =

1 1 1 = 3 1

1 a a2 a

1 a2 a

3 19 Ia Ib Ic I abc

∴ I 012 = a I abc

3.3 Symmetrical Coordinate Method (Symmetrical Components)

a and a2 are called vector operators and are defined as follows: 1 a= − +j 2 1 a2 = − − j 2

3 = e j120 = 120 = cos120 + jsin 120 2 3 = e −j120 = − 120 = cos120 − jsin 120 2

3 20a

2

1 3 = 120 = − + j 2 2

where 120 = 2π/3 [rad]. The vector operators a, a2 can be modified into the following equations and written as vectors, as shown in Figure 3.8: a2 + a + 1 = 0

a2 = e − j120 = −120 = + 120 a3 −1 = a −1 a2 + a + 1 = 0

a3 = 1

a2 + a = − 1

a2 + 1 = − a

a + 1 = − a2

a4 = a3 a = a

a5 = a3 a2 = a2

a = e j120 = 120

a

−1

=a

−1

3

a =a

2

a = a

2

a

1 −a = a 1 − a = a

=a

−2

3 20b

3

a =a

a−a = j 3 2

=1

3

−2

a −a2 = a2 j 3

2

a2 −1 = a + 1 a −1 = − a2 a −1 = − a2 −a2 j 3 = a j 3 where j = e j90 = 90 , −j = e −j90 = 90 The defined set of voltages V0, V1, V2 are named zero (0), positive (1), and negative (2) sequence voltages, respectively; and the set of currents I0, I1, I2 are also named zero (0), positive (1), and negative (2) sequence currents in the newly defined 012 domain. As Va, Vb, Vc, Ia, Ib, Ic are expressed as complex-number quantities (effective valued or peak valued) in the a–b–c domain, then V0, V1, V2, I0, I1, I2 are consequently complex-number quantities (effective valued or peak valued) in the 012 domain. The inverse matrix equations of Eqs. (3.18) and (3.19) can be easily introduced as follows: Va = V0 + V1 + V2 Vb = V0 + a2 V1 + aV 2

or

Vc = V0 + aV 1 + a2 V2

Va Vb Vc

=

V abc =

1 1 1

1 a2 a

1 a a2

V0 V1 V2

3 21

V 012

a −1

a – a2 = j 3 a=–

1 +j 3 2 2

aV

120° 120°

1·V = V

1

a –1 = – a2j 3

120°

a2 = –

–1

1 –j 3 2 2

a 2V

(a)

(b)

a2 – 1 = aj 3

1

–a

a2

a2 – a = – j 3 (c)

2

Figure 3.8 Vector operators a, a .

1 – a2 = –aj 3

– a2

a

1– a = a2j 3

35

36

3 Transmission Line Matrices and Symmetrical Components

Ia = I0 + I1 + I2 2

Ib = I0 + a I1 + aI 2

Ia Ib Ic

or

2

Ic = I0 + aI 1 + a I2

=

1

1

1

1

a

2

a

1

a

a2

I abc =

a −1

I0 I1 I2

3 22

I 012

Equations (3.18) and (3.19), for transformation to the 012 domain, and Eqs. (3.21) and (3.22) for inverse transformation to the abc domain, are the basic definitions of the symmetrical components transformation. Incidentally, the vector operator matrices a and a−1 are inverse matrices of each other: 1 1 1 a a−1 = 3 1

1

1

1

1

1

a

2

a

1

a

2

a

2

a

1

a

a2

a

a

=

a −1

1+1+1

1 + a2 + a

1 + a + a2

1 1 + a + a2 3 1 + a2 + a

1 + a3 + a3

1 + a2 + a4

1 + a4 + a2

1 + a3 + a3

3 23 1 = 0 0

0 1 0

0 0 =1 1

a−1 × a = a a−1 = 1 All the quantities in these equations are assigned as complex-number quantities; however, any assignment does not exist in the definition with regard to frequency or waveforms. In other words, the quantities may contain DC and/or higher harmonics. It should be noted that the symmetrical coordinate transformation can be applied not only for power frequency steady-state phenomena but also for transient phenomena of any kind or even for lightning, switching, and surge phenomena. The voltage and current quantities are assigned as complex numbers in these definitions, so the corresponding realnumber equations (or imaginary-number equations) can be extracted from them, which indicates the real behavior of the actual voltage and current quantities in the abc domain as well as in the 012 domain. Finally, needless to say, all the electrical quantities in the abc domain, such as electric charge q, electric lines E, flux φ, etc., can be transformed into the 012 domain using the same definitions with the previous vector operators.

3.3.3

Implications of Symmetrical Components

We need to examine additional aspects of the symmetrical components defined by previous above equations. The explanation in this section focuses on the current I, and obviously the same analogy can be applied to all other quantities. 3.3.3.1 Transformation from abc Quantities to 012 Quantities

Equation (3.25) can be transformed into Eq. (3.24) by multiplying both sides of the equations by 3: 3I0 = 3I1 = 3I2 =

•• •

Ia Ia Ia

+ + +

Ib aI b

+

Ic

+

2

+

a 2 Ic aI c

a Ib

3 24

For the first term: the same current components For the second term: counterclockwise-balanced current components For the third term: clockwise-balanced current components.

Whenever the current quantities are composed only of power-frequency components (sinusoidal waveforms), they can be visualized by drawing them as vectors in complex-number domain coordinates. Figure 3.9a shows the composition process of I0, I1, I2 from Ia, Ib, Ic.

3.3 Symmetrical Coordinate Method (Symmetrical Components)

3I0 = Ia + Ib + Ic

Zero-sequence

2

3I1 = Ia + aIb + a Ic

Positive-sequence

2

3I2 = Ia + a Ib + aIc

Negative-sequence

=

Ia

Ib

Ia

aIb

+

Ia

3 (I0 + I1 + I2)

Ic

+

2

a Ib

3Ia

0

a2Ic aIc 0 Ic

Ib

Ic aIc

2

I0

a Ib I2

Ia I1

2

a Ib

Ia

+ Ib 3I 0 = I a Ia aIb 1

2

3I2 = Ia + a Ib + aIc

Ia

a Ic

=I

a

+a Ib

Symmetrical sequence currents

aIc

Ib

Ia 2

3I

aIb

+ Ic

Ia

2

a Ic

Phase-b current

+a

2

Phase-c current

Phase-a current

Ic (a)

a phase

Ia = I0 + I1 + I2 a2I

b phase

Ib = I0 +

c phase

Ic = I0 + aI1 + a2I2

1

+ aI2

=

I0

a2I1

+

I2

+

aI1 0

3I0

aI2 a2I2 0

I1 + a 2I 2

I0 aI1

Ic = I0 + a

aI1

I1

I0

Ia + Ib + Ic

a2I2

I0

a2I2

Ia = I0 + I

a2I1

1

aI2

I0

Phase-a, -b, -c currents

+I

I2

2

I1

Ib = I0 + a2I1 + aI2

I0 I0 I0

aI2

a2I1

I2

I1

I0

Zero-sequence current

Positive-sequence current

Negative-sequence current

(b) Figure 3.9 abc-domain vectors and 012-domain vectors. (a) Composition of I0, I1, I2 from Ia, Ib, Ic. (b) Composition of Ia, Ib, Ic from I0, I1, I2.

37

38

3 Transmission Line Matrices and Symmetrical Components

3.3.3.2 Inverse Transformation from 012 Quantities to abc Quantities

Equation (3.22) is modified as follows: Ia = Ib = Ic =

• • •

I0 I0 I0

+ + +

I1 a2 I1 aI 1

+ + +

I2 aI 2 a2 I2

3 25

Clockwise-balanced complex-number currents I1, a2I1, aI1 are the components of the phase a, phase b, phase c currents, respectively (positive-sequence components). Counterclockwise-balanced complex-number currents I2, aI2, a2I2 are the components of the phase a, phase b, phase c currents, respectively (negative-sequence components). The three same-value quantities I0, I0, I0 are the components of the phase a, phase b, phase c currents, respectively (zero-sequence components).

Figure 3.9b shows the composition process of Ia, Ib, Ic from I0, I1, I2. (Figure 3.9a and b are drawn as a mutually paired case; however, the vectors in Figure 3.13 are drawn in half-dimensional length.) Again, the quantities of the abc and 012 domains are bilaterally transformable using the earlier definition. 3.3.3.3 Balanced Three-Phase Condition

Figure 3.10 shows the special case where three-phase currents are balanced with a sinusoidal waveform. As Ia, Ib, Ic are clockwise phase balanced, then Ib = a 2 Ia ,

Ia = Ia ,

Ic = aI a

3 26a

and 1 1 Ia + Ib + Ic = Ia 1 + a 2 + a = 0 3 3 1 1 I1 = Ia + aI b + a2 Ic = Ia 1 + a a2 + a2 a = Ia 3 3 3 26b 1 1 I2 = Ia + a2 Ib + aI c = Ia 1 + a2 a2 + a a 3 3 1 = Ia 1 + a + a 2 = 0 3 Under the balanced three-phase condition, the zero-sequence current I0 and negative-sequence current I2 are zero (or do not exist), and only positive-sequence current I1 exists with the same vector value as Ia. The three-phase quantities Ia, Ib, Ic or the transformed I0, I1, I2 under steady-state conditions (i.e. including only power frequency terms) can be visualized as vectors in Gauss coordinates, whether balanced or unbalanced. Although transient quantities or multi-frequency quantities may not be visualized simply, the equational relations between the abc and 012 domains are always justified. I0 =

Ic = aIa 120° 240°

Ia

I1 = Ia I0 = I2 = 0

2 I b = a Ia

(a) Three-phase balanced current

(b) Current with symmetrical components

Figure 3.10 Symmetrical components of balanced three-phase currents.

3.5 Transmission Lines by Symmetrical Components

Note that the currents Ia, Ib, Ic at an arbitrary point in a three-phase circuit and the corresponding currents I1, I2, I0 in positive, negative, and zero-sequence circuits should be marked by arrows ( ) in the same direction as the symbolic rule. The arrows of voltage polarities have to be selected analogously.

3.4

Conversion of a Three-Phase Circuit into a Symmetrical Coordinated Circuit

As the first step in studying the symmetrical components transformation, we need to examine how the equations and the related drawn circuits in the abc domain are transformed into those in the symmetrical 012 domain. Let’s try to transform Eqs. (3.3), (3.5), and (3.6) of the transmission line in Figure 3.1b into symmetrical components. The equations for the transmission line (between points m and n), Eqs. (3.5) and (3.6), are written again here: m V abc − n V abc

= Z abc I abc

3 27

We also have transformation equations with regard to voltages and currents at points m and n. For point m: m V 012

=a

m V abc

m V abc

I 012 = a I abc

= a−1

I abc = a

−1

m V 012

3 28

I 012

And for point n: n V 012

= a n V abc

= a − 1 n V 012 I abc = a −1 I 012 n V abc

I 012 = a I abc

3 29

Because the currents at points m and n are assumed to be equal (because leakage current through the stray capacitance of the line is ignored), the suffix m or n is omitted from the symbol Iabc. Multiplying by a at the top (i.e. left-multiplying) of both sides in Eq. (3.33), it can easily be changed into a symmetric equation: a

m V abc

−a n V abc = aZ abc I abc

m V 012n

∴ ie

a −1 I 012

V 012

= a Z abc a −1 I 012 m V 012 − n V 012 = Z 012 I 012 Z 012 = a Z abc a −1 m V 012 − n V 012

Z 012 I 012

3 30

Equation (3.27) was transformed into Eq. (3.30) by symmetrical coordinates. The abc impedance matrix Zabc was transformed into the 012 sequence impedance matrix Z 012, which is defined by Eq. (3.30).

3.5

Transmission Lines by Symmetrical Components

3.5.1

Single-Circuit Line with LR Constants

Assuming that the transmission line in Figure 3.1b is well phase transposed, Zaa ≒ Zbb ≒ Zcc

Zs

Zab ≒ Zba ≒ Zbc ≒ Zcb ≒ Zca ≒ Zac

3 31

Zm

Then Eqs. (3.5) and (3.6) can be simplified as follows: m Va m Vb m Vc m V abc

n Va

−

n Vb n Vc

− n V abc

Zs = Zm Zm

Zm Zs Zm

=

Z abc

Zm Zm Zs

Ia Ib Ic I abc

3 32

39

40

3 Transmission Line Matrices and Symmetrical Components

Accordingly, Z012 of Eq. (3.30) is a−1

Z abc

Z 012 = a Z abc a

−1

Zs = a Zm Zm

Zm Zs Zm

Zm Zm Zs

1

1

1

1

2

a

a

1

a

a2

Zs + 2Zm

Zs + a2 + a Zm

Zs + a + a2 Zm

= a Zs + 2Zm

a2 Zs + 1 + a Zm

aZ s + 1 + a2 Zm

Zs + 2Zm

aZ s + 1 + a2 Zm

a2 Zs + 1 + a Zm 3 33

1 1 1 = 3 1

1

1

Zs + 2Zm

Zs − Zm

Zs −Zm

a

a

2

Zs + 2Zm

a Zs − Zm

a Zs −Zm

2

a

Zs + 2Zm

a Zs − Zm

a2 Zs − Zm

a

a

2

Z abc a −1

Zs + 2Zm 0 = 0

0 Zs − Zm 0

0 0 Zs −Zm

Zs + 2Zm 0 0

0 Zs −Zm 0

0 0 Zs − Zm

Namely, m V0

n V0

m V1 − m V2

n V1 n V2

mV 012

nV 012

=

Z 012

or

I0 I1 I2 I 012

Z0 0 0

0 Z1 0 Z 012

0 0 Z1

I0 I1 I2 I 012

3 34

m V0 − n V0

= Zs + 2Zm I0 = Z0 I0

m V1 − n V1

= Zs − Zm I1 = Z1 I1

m V2 − n V2 = Zs − Zm I2 = Z1 I2 where Z0 = Zs + 2Zm , Z1 = Zs − Zm

This is the equation of a single-circuit transmission line in the symmetrical components domain. Z012 is a simple diagonal matrix in which all the off-diagonal elements vanish (become zero). This means the positive-, negativeand zero-sequence equations are mutually independent of each other because mutual impedances do not exist among them. Now we can conclude that, if the original three-phase circuit is phase balanced (this assumption is acceptable for most cases, with only small errors), positive (1) sequence, negative (2) sequence, and zero (0) sequence circuits can be independently handled. Figure 3.11 shows the equivalent circuit of a three-phase (single-circuit) transmission line by symmetrical components, which is drawn from Eq. (3.34). Symmetrical impedances Z0, Z1, Z2 are defined by Eq. (3.34) in relation to the original impedances Zs, Zm, while from the relation Z0 > Z1 = Z2

3 35

That is, for the transmission line, the positive-sequence impedance Z1 and the negative-sequence impedance Z2 are equal and smaller than the zero-sequence impedance Z0. Note that, as the transmission line is not perfectly phase balanced, very small off-diagonal elements may exist in the impedance matrix Z012, so that positive-, negative- and zero-sequence circuits are mutually linked by small mutual inductances. If necessary, we can examine the strict impedance matrix by calculating equation Z012 = a Zabc a−1 without any assumption such as Eq. (3.31).

3.5 Transmission Lines by Symmetrical Components

Point m

Point n

Outer circuit

Outer circuit

Z1 = Zs – Zm Positive-sequence circuit

I1

mV1

nV1

Z2 = Z1 = Zs – Zm Negative-sequence circuit

I2 nV2

m V2

Z0 = Zs + 2Zm

n Zero-sequence circuit

I0

m V0

nV0

Figure 3.11 The equivalent circuit of a three-phase single-circuit transmission line (impedances).

3.5.2

Double-Circuit Line with LR Constants

Let’s examine a double-circuit transmission line, as shown in Figure 3.4, assuming that the first and second circuits are well phase transposed. The symbols 1V, 1I and 2V, 2I shown here refer to the voltages and currents of the first and second circuits: 1 m Va 1 m Vb 1 m Vc 2 m Va 2 m Vb 2 m Vc

−

1 m V abc 2 m V abc

−

1 n Va 1 n Vb 1 n Vc 2 n Va 2 n Vb 2 n Vc

Zs Zm Zm = Zm Zm Zm

Zm Zs Zm Zm Zm Zm

Zm Zm Zs Zm Zm Zm

Zm Zm Zm Zs Zm Zm

Zm Zm Zm Zm Zs Zm

1

Zm Zm Zm Zm Zm Zs

Ia Ib 1 Ic 2 Ia 2 Ib 2 Ic 1

3 36a

or 1 n V abc 2 n V abc

=

Z sm Zm

Zm Z sm

1 2

I abc Z sm 1 I abc + Z m = I abc Z m 1 I abc + Z sm

2 2

I abc I abc

where we assume Zs

Zaa ≒ Zbb ≒ Zcc ≒ ZAA ≒

self-impedance

Zm

Zab ≒ Zbc ≒ Zca ≒ ZAB ≒ ZBC ≒

mutual impedance between the conductors of the same circuit

Zm

ZaA ≒ ZaB ≒ ZaC ≒ ZbA ≒

mutual impedance between the conductors of another circuit

3 36b

Symmetrical quantities of double-circuit line 1m V 012 , 1n V 012 , 2m V 012 , 2n V 012 , 1 I 012 , 2 I 012 are introduced in conjunction with abc domain quantities 1m V abc , 1n V abc , 2m V abc , 2n V abc , 1 I abc , 2 I abc : 1 m V 012

=a

1 m V abc ,

1 n V 012

=a

1 n V abc ,

1

I 012 = a 1 I abc

2 m V 012

=a

2 m V abc ,

2 n V 012

=a

2 n V abc ,

2

I 012 = a 2 I abc

1 m V abc

= a −1

1 1 m V 012 , n V abc

= a−1

1 n V 012 ,

1

I abc = a − 1

1

I 012

2 m V abc

= a −1

2 2 m V 012 , n V abc

= a−1

2 n V 012 ,

2

I abc = a − 1

2

I 012

① 3 37 ②

41

42

3 Transmission Line Matrices and Symmetrical Components

The equation of circuit 1 in Eq. (3.36a) can be transformed to the 012 domain: 1 1 1 2 m V abc − n V abc = Z sm I abc + Z m I abc ∴ a −1 1m V 012 − a −1 1n V 012 = Zsm a −1 1 I 012

+ Z m a −1

2

I 012

Left-multiplying by a and recalling that a a−1 = 1, Z sm a −1

for circuit1 1m V 012 − 1n V 012 = a

1

I 012 + a Z m a −1

2

I 012 3 38

and for circuit 2 analogously 2 1 m V 012 − n V 012

= a Zm a

−1

1

I 012 + a Z sm a

−1

2

I 012

a Zsm a−1 in this equation is equal to Z012 of the single-circuit line in Eq. (3.33), so a Z sm a

Zs + 2Zm 0 = 0

−1

0 Zs − Zm 0

0 0 Zs −Zm

and a Zm a 1 1 1 = 3 1

−1

=a

1 a a2

Zm Zm Zm 1 a2 a

a

Zm Zm Zm

Zm Zm Zm

3Zm 3Zm 3Zm

0 0 0

1 1 1 0 0 0

1 a2 a

1 a a2

3Zm 0 = 0

0 0 0

0 0 0

Zm a − 1

Accordingly, 1 m V 012 2 m V 012

or

1 mV 0 1 mV 1 1 mV 2 2 mV 0 2 mV 1 2 mV 2

1 n V 012 2 n V 012

−

−

1 nV 0 1 nV 1 1 nV 2 2 nV 0 2 nV 1 2 nV 2

a Z sm a − 1 = a Z sm a − 1 Zs + 2Zm 0 0 = 3Zm 0 0 Z0 0 0 = Z0M 0 0

0 Z1 0 0 0 0

a Z sm a − 1 a Z sm a − 1 0 Zs − Zm 0 0 0 0 0 0 Z1 0 0 0

where Z1 = Zs − Zm ,

Z0 = Zs + 2Zm ,

Z0M = 3Zm

I 012 Z 012 = 2 I 012 Z 0M 1

0 0 Zs −Zm 0 0 0

Z0M 0 0 Z0 0 0

0 0 0 0 Z1 0

3Zm 0 0 Zs + 2Zm 0 0 0 0 0 0 0 Z1

1

I0 I1 1 I2 2 I0 2 I1 2 I2 1

Z 0M Z 012

0 0 0 0 Zs − Zm 0

1 2

I 012 I 012

0 0 0 0 0 Zs −Zm

1

I0 I1 1 I2 2 I0 2 I1 2 I2 1

3 39a

3.5 Transmission Lines by Symmetrical Components 1I

1

2I

1

Z1=Zs – Zm #1

Z1=Zs – Zm 1 n V1

1 mV1 2 mV1

Positive-sequence circuit

#2

2 n V1 1I

Z2=Z1=Zs – Zm

2I

Z2=Z1=Zs – Zm

2 2

#1 1 n V2

1 mV2 2 mV2

Negative-sequence circuit

#2

2 n V2

Z0M = 3Z´ m Point m

Point n

1I

0

2I

0

Z0=Zs + 2Zm

Circuit #1

Z0=Zs + 2Zm

Zero-sequence circuit

1 mV0

1 n V0

#2

2 n V0

2 mV0

Figure 3.12 The equivalent circuit of a three-phase double-circuit transmission line (impedance).

Equation (3.39a) can be recasted into the following equation: 1 m V0 2 m V0

−

1 n V0 2 n V0

=

Z0 Z0M

1 m V1 2 m V1

1 n V1 2 n V1

=

Z1 0

0 Z1

1

−

Z1 = 0

0 Z1

1

−

1 n V2 2 n V2

1 m V2 2 m V2

where Z0 = Zs + 2Zm ,

1

Z0M Z0

2

2

2

I0 I0

I1 I1

3 39b

I2 I2

Z0M = 3Zm ,

Z1 = Zs − Zm

Figure 3.12 shows the equivalent circuit of the three-phase double-circuit transmission line by symmetrical components, which is drawn from Eq. (3.39a) or (3.39b). The positive-, negative-, and zero-sequence circuits are independent of each other (mutual inductances do not exist). In the positive- and negative-sequence circuits, the mutual inductances do not exist between lines 1 and 2. However, in the zero-sequence circuit, lines 1 and 2 are mutually coupled together by Z0M = 3Zm . Zs and Zm are actually the averaged values of Zaa, Zbb, Zcc and Zab, Zbc, Zca, respectively, so that Zs and Zm can be calculated by using Eq. (4.17) from Chapter 4. The positive-sequence impedance Z1 = Zs–Zm is derived from the working inductance Laa–Lab given by Eq. (3.9), which we will discuss later in detail. 3.5.3

Single-Circuit Line with Stray Capacitance C

The stray capacitances of a well-phase-transposed single-circuit line are shown by Figure 3.6a and b and Eq. (3.14). Eq. (3.14) is repeated here: Ia Ib = jω Ic I abc = jω ×

Cs + 2Cm −Cm −Cm

−Cm Cs + 2Cm −Cm C abc

−Cm −Cm Cs + 2Cm

Va Vb Vc × V abc

3 40

43

44

3 Transmission Line Matrices and Symmetrical Components

The transformation of this equation into symmetrical components is as follows: I012 = a I abc = a jω Cabc Vabc = jω a where C 012 = a C abc a

Cabc a −1 V012

Cs = 0 0

1 a a2

a −1

C abc

a

1 1 1 C012 = 3 1

jωC012 V012

−1

1 a2 a

Cs + 2Cm −Cm −Cm

0 Cs + 3Cm 0

−Cm Cs + 2Cm −Cm

−Cm −Cm Cs + 2Cm

1 1 1

1 a2 a

1 a a2

3 41a

0 0 Cs + 3Cm

Then I0 Cs I1 = jω 0 I2 0

0 Cs + 3Cm 0

0 0 Cs + 3Cm

C 012

I 012 = jω × where C0 = Cs ,

V 012

×

C1 = C2 = Cs + 3Cm

0 C1 0

V0 C0 V1 = jω 0 V2 0

0 0 C1

C 012

V0 V1 V2

3 41b

V 012

working capacitance

C1

This is the equation for stray capacitances of a single-circuit transmission line in the symmetrical components domain. C012 is a simple diagonal matrix in which all the off-diagonal elements vanish (become zero). Figure 3.13a shows the equivalent circuit of a three-phase (single-circuit) transmission line by symmetrical components, which is drawn from Eq. (3.41b). The positive (1), negative (2), and zero (0) sequence circuits of the phasebalanced three-phase transmission line are obviously independent of each other.

I1 Positive-sequence circuit

C1 = Cs + 3Cm C1: working capacitance

V1

I1

a2I1

Cs

Cs

V1 2 a V1 aV1 Cs

aI1

3Cm

I2 Negative-sequence circuit

V2

I2

C2 = C1 = Cs + 3Cm

V2

aI2

n

a2I2

aV2 a2V2

Cs

Cs

Cs

3Cm

3Cm

3Cm n

0

I0 Zero-sequence circuit

I0

C0 = Cs

V0

V0

V0 V0 Cs

0 I0 Cs

0

I0 Cs

3Cm

3Cm

3Cm n

3I0

(a)

3Cm

3Cm

0

(b)

Figure 3.13 The equivalent circuit of a three-phase single-circuit transmission line (capacitance).

0

3.5 Transmission Lines by Symmetrical Components

Symmetrical capacitances C0, C1, C2 are defined by Eq. (3.41b) as the related values Cs, Cm that will be discussed in Chapter 4. The physical meaning of the relations C0 = Cs, C1 = Cs + 3Cm can be understood from Figure 3.13b, where zero-sequence current cannot flow in the circuit branch of 3Cm because point n is not earth grounded.

3.5.4

Double-Circuit Line with C Constants

The stray capacitances of a well-phase-transposed double-circuit line are shown in Figure 3.6. The symbols 1V, 1I and 2 V, 2I are adopted as quantities of circuits 1 and 2, respectively, here. Concerning the phase a current of circuit 1, 1

Ia = jωCs 1 Va + jωCm 1 Va − 1 Vb + jωCm 1 Va − 1 Vc + jωCm 1 Va − 2 Va + jωCm 1 Va − 2 Vb + jωCm 1 Va − 2 Vc

①

∴ 1

3 42a 1

Ia = jω Cs + 3Cm + 3Cm Va −jωCm 1 Va + 1 Vb + 1 Vc − jωCm 2 Va + 2 Vb + 2 Vc 2

1

= jω Cs + 3Cm + 3Cm Va −jω3Cm 1 V0 −jω3Cm V0

②

Similar equations are derived for the phase b and c currents. Accordingly, 1

1

1

2

1

1

1

2

Ia Ib = jω Cs + 3Cm + 3Cm 1 Ic

Va Vb − jω3Cm 1 Vc

V0 V0 − jω3Cm 1 V0

V0 V0 2 V0

3 42b

This equation is easily transformed into symmetrical components: 1

2

1

I0 = jω Cs + 3Cm + 3Cm V0 −jω3Cm 1 V0 −jω3Cm V0 2

1

= jω Cs + 3Cm V0 −jω3Cm V0 = jωCs 1 V0 + jω3Cm

1

V0 − 2 V0 3 43a

1

1

I1 = jω Cs + 3Cm + 3Cm V1

1

I2 = jω Cs + 3Cm + 3Cm V2

1

Accordingly, 1

I0 Cs + 3Cm 1 I1 0 1 I2 0 = jω 2 I0 − 3Cm 2 I1 0 2 I2 0

0 Cs + 3Cm + 3Cm 0 0 0 0

0 0 Cs + 3Cm + 3Cm 0 0 0

− 3Cm 0 0 Cs + 3Cm 0 0

0 0 0 0 Cs + 3Cm + 3Cm 0

0 0 0 0 0 Cs + 3Cm + 3Cm

1

V0 V1 1 V2 2 V0 2 V1 2 V2 1

3 43b

45

46

3 Transmission Line Matrices and Symmetrical Components

Namely, 1 2

1 2

I0 Cs + 3Cm = jω I0 − 3Cm = jω

Cs Cs

jω

C0 C0

2

2

C1 0

1 2

V0 + C0 1 V0 − 2 V0 V0 + C0 2 V0 − 1 V0

C1 0

2

V0 V0

0 C1

1 2

0 C1

1 2

= jω

− C0 C0 + C0

C0 + C0 − C0

0 Cs + 3Cm + 3Cm

1 2

1 2

V0 V0

①

V1 V1

V1 V1

②

I2 Cs + 3Cm + 3Cm = jω I2 0 jω

1

V0 + 3Cm 1 V0 − 2 V0 V0 + 3Cm 2 V0 − 1 V0

I1 Cs + 3Cm + 3Cm = jω I1 0 jω

1

1

− 3Cm Cs + 3Cm

0 Cs + 3Cm + 3Cm

1 2

3 43c

V2 V2

V2 V2

③

where positive-sequence capacitance

C1 = C2 = Cs + 3Cm + 3Cm

zero-sequence capacitance

C0 = Cs , C0 = 3Cm

④

This is the equation of the stray capacitances of double-circuit lines, from which the equivalent circuit in Figure 3.14 is derived.

#2

Positive-sequence

2V

1

C1

1V

2V

2

1

1V 2

C1

C1

#2 C′0 Zero-sequence

2V

0

C1 = Cs + 3Cm + 3C ′m (C1: working capacitance)

C1

#2

Negative-sequence

#1

1V 0

C0

C0

#1 C1 = C2 = Cs + 3Cm + 3C′m

Circuit #1

C0 = Cs C′0 = 3C′m

Figure 3.14 The equivalent circuit of a three-phase double-circuit transmission line (capacitance).

3.6 Generator by Symmetrical Components (Simplified Description)

3.6

Generator by Symmetrical Components (Simplified Description)

3.6.1

Simplified Symmetrical Equations

A synchronous generator (or synchronous motor) may be considered a machine containing balanced three-phase ideal power sources and balanced three-phase leakage impedances, so that the generator is simply expressed as an approximate circuit as shown in Figure 3.14. (Detailed approaches are discussed in Chapter 10.) Now, we have the following from Figure 3.15: Ea Eb = a2 Ea Ec = aE a

Va Vb Vc

−

=

− V abc

E abc

Zs Zm Zm

Zm Zs Zm

Zm Zm Zs

Ia Vn Ib − V n Ic Vn I abc

Z abc

=

Vn = − Zn Ia + Ib + Ic = −Zn 3I0 = −3Zn I0

① 3 44

−V n

②

Ea , Eb ,Ec the generated source voltages of three-phase-balanced design Equation (3.44) can be transformed into symmetrical components by left-hand multiplication of the symmetric operator a: a E abc −a V abc = aZ abc I abc − a V n

3 45a

∴ E 012 − V 012 = aZ abc a −1 I 012 −a V n where E0 1 1 1 E 012 = E1 = a E abc = 3 E2 1

1 a a2

1 a2 a

Ea 0 2 a Ea = Ea aE a 0

The first term on the right (a Zabc a−1) is the same form as in Eq. (3.33). The second term on the right (a Vn) is 1 1 1 a Vn = 3 1

1 a a2

1 a2 a

Virtual generating source terminal

Vn Vn Vn = 0 = Vn 0

Generator terminal Ia

Zs Ec=aEa

Ea Zs Eb=a2Ea

Vn

Zs

− 3Zn I0 0 0

Zm Zm

Ib Zm

Ic

Va Vb

Zn

Ia + Ib + Ic = 3I0

Vc

Figure 3.15 Generator (easy concept with the abc-phase domain).

47

48

3 Transmission Line Matrices and Symmetrical Components

Accordingly, 0 V0 Z0 Ea − V1 = 0 0 V2 0

0 Z1 0

0 0 Z2

I0 3Zn I0 I1 + 0 I2 0

or

3 45b −V0 = Z0 + 3Zn I0 Ea −V1 = Z1 I1 −V2 = Z2 I2

This is the transformed symmetrical equation of the generator; Figure 3.16 shows the symmetrical equivalent circuits of Eq. (3.45b). The figure shows that a power source exists only in the positive-sequence circuit, and the negative- and zero-sequence circuits consist only of passive impedances. A generator may be theoretically referred to as a “positivesequence power generator.” 3.6.2

Reactance of a Generator

Equation (3.45b), derived from Figure 3.15 or Eq. (3.44), shows that the generator has time-independent constant symmetrical reactance, and the positive- and negative-sequence reactance are the same quantities. However, this is not correct. The generator reactance will change from time to time under transient conditions, and, moreover, the positive- and negative-sequence reactance as well as the zero-sequence reactance are different. The generator can strictly be treated only by the d–q–0 transformation method in which the new concept of direct-axis reactance (xd , xd , xd) and quadratureaxis reactance (xq , xq , xq) are introduced. Now, by applying the reactance (xd , xd , xd) as positive-sequence reactance, Figure 3.16 can be treated as the mostly correct equivalent circuit for the generator, while the positive-sequence reactance will change from time to time as shown in Chapter 10, Table 10.1, under transient conditions. For most analyses of primarily power frequency phenomena (fault analysis, for example), Eq. (3.45b) and Figure 3.16 can be applied as the satisfactory equivalent circuit of the synchronous generator, while the following reactance is used in equivalent circuits (details are examined in Chapter 10). For the positive-sequence reactance, x1 =

xd direct-axis subtransient reactance 0 −3 cycle time, 0 45 or 60 ms xd direct-axis transient reactance 3 − 50 or 60 cycle time , 1 sec xd direct-axis steady-state reactance 1 sec I1 Z1 = jx1

Positive-sequence circuit

V1

x1 =

Ea I2

Negative-sequence circuit

V2

Z2 = jx2

I0 Zero-sequence circuit

V0

Z0 = jx0 3Zn

Figure 3.16 Generator (easy concept with the 012-phase domain).

{

xd′′ (0 ~3 cycles time) xd′ (3 ~60 cycles time) xd (1 sec ~)

3.7 Description of a Three-Phase Load Circuit by Symmetrical Components

The time in parentheses means duration just after a sudden change of circuit condition. For the negative-sequence reactance, x2 can be treated as constant for most cases, although it may change slightly just after a sudden change of circuit condition. And for the zero-sequence reactance, x0 can always be treated as constant. The values of xd , xd , xd, x2, x0 for a synchronous generator are given on the name-plate in terms of ratings. Table 10.1 in Chapter 10 shows typical values of generator reactance.

3.7

Description of a Three-Phase Load Circuit by Symmetrical Components

In power-receiving substations, feeder lines are connected to the HV, MV, or LV bus. Some of them are connected to other generating stations and substations through the lines, and others to load stations. The equation for the total load is approximately written as follows: Va Vb Vc

=

Zs Zm Zm

V abc =

Zm Zs Zm

Zm Zm Zs

Ia Ib Ic

Z abc

3 46

I abc

or by symmetrical components V0 Z0 V1 = 0 V2 0

0 Z1 0

0 0 Z2

I0 I1 I2

3 47

where Z1 = Z2 = Zs – Zm, Z0 = Zs + 2Zm > Z1 = Z2. It is obvious that the 1-, 2-, 0-sequence networks are mutually independent, and the load can be approximately expressed simply by Z1, Z2, Z0, respectively.

49

51

4 Physics of Transmission Lines and Line Constants We studied circuit equations of transmission lines in the previous chapter, including self-inductance Ls, mutual inductance Lm, self-capacitance Cs, and mutual capacitance Cm, as well as working inductance Ls − Lm, working capacitanceC ≡ Cs + 3Cm, and the symmetrically transformed L0 L1 L2 and C0 C1 C2. Now we will examine what transmission line inductance and capacitance mean in terms of electromagnetism. Further, we will introduce equations and the typical values of various LC-line constants with regard to the physics of electromagnetism.

4.1

Inductance

Equation (3.9) for working inductance Laa − Lab was briefly shown in the previous chapter. In this section, we examine what transmission line inductance means in terms of electromagnetism. Self-Inductance L11 of a Straight Conductor

4.1.1

As is shown in Figure 4.1a, conductor 1 (radius r) is laid out straight in an area of permeability μ = μs μ0 (μ0 is permeability in vacuum space, μs is relative permeability, and μs = 1.0 in vacuum space). If current i flows through conductor 1, concentric circular magnetic paths are composed in a conductor section as well as in the space outside of the conductor surface, and the central point O of conductor 1 is also the central point of induced concentric magnetic paths. The concentric magnetic paths in the outer space of conductor 1 are examined first. A thin concentric magnetic ring path at point x from O with length 2πx and width dx can be imaged. The magnetic resistance R of the ring path is proportional to the length of the ring path 2πx[m], and is inversely proportional to the sectional area 1 dx [m2]. Namely, R=

2πx A turn Wb μdx

where x ≥ r

4 1a

where: μ = μs μ0 is the permeability of the ring path. μ0: permeability in vacuum space (μ0 = 4π × 10−7 in the MKS rational unit system) μs: relative permeability (μs = 1.0 in vacuum space and almost 1.0 in air space) The reason μ0 is 4π × 10−7 in the MKS rational unit system is discussed later, in Section 4.5. If current i[A] flows through the conductor, or if electromotive force i[A turn] is charged in the conductor (where the transmission line has one turn), flux dφ is produced through the ring path with sectional depth dx, and dφ per unit length is proportional to i and inverse proportional to R: dφ =

i μ i = dx R 2πx

Wb m

4 1b

The transmission line is essentially a one-turn conductor, and the linking flux value per unit length dψ is dψ = 1 dφ =

μ i dx 2πx

4 1c

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

52

4 Physics of Transmission Lines and Line Constants

y y S1 Conductor1 Radius r

dx dx r

D = 2h

g

x o

S1

S2 +q

S2

OV

S3 Sn

i1

S –q Conductor 2 (a) Conductor 1

i2

i3 in

(b) Conductor 1 and 2

radius r

(c) Multi-conductors

Figure 4.1 Current and flux linkage of parallel conductors.

Therefore the total linking flux ψ out of the space from the conductor surface (radius r[m]) to point S is S

ψ out =

r

S

dψ out =

1 dφ =

r

S r

μ i μ i dx = loge x 2πx 2π

μ i S μ μ S loge = s 0 loge = 2π r r 2π

s r

42

i

Next, we examine the linking flux ψ in in the conductor section. If current i[A] flows through the conductor with uniform density, the current within a space of diameter x[m] is ix = i

2πx2 x2 =i 2 2 2πr r

where r ≥ x ≥ 0

A

4 3a

The intensity of magnetic field H at the ring path with length 2πx and width dx, which is x distant from point O in the radial direction, is H=

ix A turn m 2πx

4 3b

The flux density B is μcond μ0 ix μcond μ0 i x = 2πr 2 2πx

B = μcond μ0 H =

Wb m2

4 3c

where μcond is the relative permeability of the conductor. The flux at the x distant ring path with width dx and ring area 1 × dx is dφ = B 1 × dx = Bdx =

μcond μ0 i x dx Wb m 2πr 2

4 3d

The turn number of the conductor within a circle of radius x is x2/r2, assuming that current i is uniformly distributed in the conductor section. Then the linking flux value is ψ in =

r 0

r

dψ in =

μ μ = cond 4 0 2πr

μcond μ0 i x x2 dx = 2πr 2 r2 0 r

ix dφ = 0

1 4 x 4

r 0

μ μ i = cond 4 0 2πr

1 4 r 4

μcond μ0 i x3 dx 4 0 2πr r

μ μ i = cond 0 i 8π

44

4.1 Inductance

As the result of Eqs. (4.2) and (4.4), the total linking flux produced by current i of conductor 1 and interlink with current i within the circular sectional area with radius r from the conductor center point O to the outer point S is ψ total = ψ out + ψ in =

μs μ0 S μ μ loge + cond 0 r 2π 8π

45

i

The definition of inductance is L = ψ/i and dψ/dt = L di/dt. This means L is the proportional constants of flux ψ and current i (or flux ψ per i = 1A). Then L11 =

ψ total μs μ0 S μ μ = loge + cond 0 r i 2π 8π

4 6a

S S or L11 = 0 4605μs log10 + 0 05μcond × 10 − 6 H m = 0 4605μs log10 + 0 05μcond mH km r r where μ0 permeability of vacuum space ,and μ0 = 4π × 10

−7

4 6b

by MKS rational unit system

This is the self-inductance of conductor 1, and the equation corresponds with Eq. (3.9).

4.1.2

Two Parallel Conductors in Space, and the Working Inductance L11 −L12

In Figure 4.1b, two conductors 1 and 2 (radius r) are laid out in parallel at distance D. Current i[A] goes out on conductor 1 and comes back from 2, or current i[A] flows in conductor 1 and current –i [A] flows in conductor 2 in the same direction. Now we imagine an arbitrary point y(S1, S2), which is S1 distant from 1 and S2 distant from 2; point y is far distant from both conductors 1 and 2, such that S1 S2 D. Current i[A] of conductor 1 produces concentric flux in conductor 1, and this flux interlinks with current i of conductor 1, so that the linking flux value is given by Eq. (4.5). That is, ψ 11 =

μs μ0 S1 μ μ loge + cond 0 2π r 8π

i where μ = μcond μ0

4 7a

Next, current –i[A] of conductor 2 produces concentric flux in conductor 2. The linking flux to which current i of conductor 1 links is ψ 12, which can be calculated by accumulating dψ 12 = (−i)dφ from D to S2: − ψ 12 =

S2

S2

dψ 12 =

D

− i dφ =

D

μ μ S2 = S 0 loge 2π D

S2 D

μS μ0 − i μ μ −i loge x dx = S 0 2πx 2π

S2 D

−i

The total linking flux of current i of conductor 1 is the sum of ψ 11 and −ψ 12, recalling that S ≡ S1 ψ 11 −ψ 12 =

4 7b

μ S μ μ loge + cond 0 2π r 8π

μ D μ μ log + cond 0 = 2π e r 8π

i+

μS μ0 S loge D 2π

S2

D

r:

−i 4 7c

i

Then, the general equations of L11, L12 and the working inductance L11 − L12 are L11 =

ψ 11 μ S μ μ S μ μ = log + cond 0 = loge + 2π r 8π 2π e r i 8π

4 8a

L12 =

ψ 12 μ S μ S = loge = loge 2π D 2π D i

4 8b

L11 −L12 =

ψ 11 − ψ 12 μs μ0 D μ μ = loge + cond 0 r i 2π 8π

4 8c

53

54

4 Physics of Transmission Lines and Line Constants

where μ = μs μ0 S ≡ S1

S2

D

r

μ0: permeability of vacuum space, and μ0 = 4π × 10−7 in the MKS rational unit system μs: relative permeability of air, and μs = 1.0000004 1.0 μcond: relative permeability of copper, and μcond = 0.999994 1.0 Equations (4.8a), (4.8b), and (4.8c) can be modified as follows in the form of a common logarithm and with μ0 = 4π × 10−7 in the MKS rational unit system: L11 − L12 = 0 4605μs log10

D + 0 05μcond mH km working inductance r

S1 + 0 05μcond mH km self − inductance r S2 mH km mutual inductance L12 = 0 4605μs log10 D

L11 = 0 4605μs log10

4 9a 4 9b 4 9c

where μs μcond 1.0 S ≡ S1 S2 D r Obviously, the working inductance L11 − L12 is derived only as a function of the distance D between the two parallel conductors and the radius r, so that the actual value is dependent only on the allocation of the two conductors, whereas L11 , L12 are the uncertain values affected by parameters S1, S2, respectively. Note that the working inductance L11− L12 of parallel conductors 1 and 2 is not affected by the existence or nonexistence of a third or other conductors. L11− L12 is based on the calculation where conductors 1 and 2 are loop connected, and all of conductor 1’s current i1 = i comes back as conductor 2’s current i2 =− i simultaneously, so that it is not affected by the currents and fluxes of other conductors. Also note that the term μcond μ0/8π or 0.05μcond is quite small and can be ignored for most actual engineering purposes. 4.1.3

Inductance of n-Parallel Conductors in Space

Next, Figure 4.1c shows a space where parallel conductors 1, 2, 3 n (conductor radius r) lie in a vacuum (or uniform) space, and currents i1, i2, i3, in are flowing in the same direction through these conductors, respectively. We will look at the total flux linkages given an arbitrary point p whose distances from the conductors are S1, S2, S3, respectively. The total linking flux to which current i1 of conductor 1 links is ψ 1 = loge

Sj S1 μcond μ0 S2 + loge + i2 + loge ij + r 8π D12 D1j

i1 + i2 + i3 +

+ loge

Sn in D1n

in = 0, Djk is the distance between conductors j and k

ψ1

L11 L12 L13

L1n

i1

ψ2

L21 L22 L23

L2n

i2

ψ 3 = L31 L32 L33

L3n

i3

ψn

Lnn

in

Ln1 Ln2 Ln3

4 10a 4 10b

4 10c

Sj μs μ0 S1 μ μ μ μ loge + cond 0 , Ljj = s 0 loge 2π r 8π 2π r Sj μs μ0 S2 μs μ0 L12 = loge loge , L1j = 2π D12 2π D1j L11 =

4 10d

and L11 − L12 =

μs μ0 D12 loge , 2π r

L11 − L1j =

D1j μs μ0 loge 2π r

4 10e

The inductance L11 − L12 given by Eq. (4.10e) is called the working inductance between the conductors 1 and 2. It is interesting that working inductance L11 − L12 is given as the fixed value whose parameters are the conductor radius r and conductor-to-conductor distance Djk, whereas the self and mutual inductance elements given by Eq. (4.10d) are mathematically nondeterminant values because the unknown distance S is included.

4.1 Inductance

By the way, we are interested in the potential voltages v1, v2, v3 of these conductors. The potential voltage v1 of conductor 1 is defined as the voltage difference from zero volts at the neutral voltage point, which is far from the conductors. So, we can satisfy the following equation S

S1

S2

Sn

4 11

Dj k

Then Eq. (4.10a) can be modified as follows: ψ1 =

μs μ0 S μ μ loge + cond 0 r 2π 8π

i1 +

μs μ0 S S loge i2 + loge i3 + D12 D13 2π

=

μs μ0 1 μ μ μ μ 1 1 loge + cond 0 i1 + s 0 loge i2 + loge i3 + r D12 D13 2π 8π 2π

∴ ψ1 =

μs μ0 1 μ μ μ μ 1 1 loge + cond 0 i1 + s 0 loge i2 + loge i3 + r D12 D13 2π 8π 2π

+

μs μ0 loge S i1 + i2 + i3 + 2π

4 12 Therefore the matrix equation of the flux linkage in Eqs. (4.10c) and (4.10d) can be replaced by the following equation: ψ1

L11 L12 L13

L1n

i1

ψ2

L21 L22 L23

L2n

i2

ψ 3 = L31 L32 L33

L3n

i3

ψn

Lnn

in

Ln1 Ln2 Ln3

i1 + i2 + i3 +

4 13a

in = 0

where μs μ0 1 μ μ μ μ 1 loge + cond 0 , Ljj = s 0 loge r r 2π 8π 2π μ μ 1 μ μ 1 L12 = s 0 loge , L1j = s 0 loge D12 D1j 2π 2π

L11 =

4 13b

Or, using the MKS rational unit system, 1 L11 = 0 4605 log10 + 0 05, r L12 = 0 4605 log10

1 , D12

Ljj = 0 4605 log10 L1j = 0 4605 log10

1 r

1 D1j

4 13c

As mentioned earlier, the self and mutual inductance elements Ljj, Ljk in the inductance matrix in Eqs. (4.10c, d) are mathematically nondeterminant values because the unknown distance S is included. However, those equations can be replaced by Eqs. (4.13a, b), where the self and mutual inductance elements Ljj, Ljk are written as determinant values without S. Now we can use Eqs. (4.13a, b, c) from an engineering viewpoint, where the equational relation between ψ and i is preserved. Further, the self and mutual inductance elements can be calculated using Eq. (4.13b) or (4.13c) as fixed values whose parameters are conductor radius r and conductor-to-conductor distance Djk. The working inductance between conductor 1 and j is given by the following equation, where the small term μcond μ0/ 8π or 0.05 is omitted: L11 −L1j =

D1j D1j μs μ0 loge = 0 4605 log10 working inductance 2π r r

4 13d

Next, we will examine the potential voltages v1, v2, v3 of these conductors. Faraday’s law shows that the changing speed of linking flux dψ/dt and the caused voltage v are proportional to each other; in other words, we know v = dψ/dt, using the MKS unit system.

55

56

4 Physics of Transmission Lines and Line Constants

Then the following v − i matrix equation is derived: v1 v2 v3

d = dt

vn

ψ1

L11 L12 L13

L1n

i1

ψ2

L21 L22 L23

L2n

ψ 3 = L31 L32 L33

L3n

i2 d i3 dt

ψn

Lnn

Ln1 Ln2 Ln3

4 14

in

Obviously, if the diameter size 2r and the physical allocation Djk of the conductors are fixed, the inductance-matrix is also fixed. Then the potential voltages v1, v2, vn can be calculated if we know i1, i2, in. Next, we will examine the voltage-sag Δvj = vj − vj that is caused between the two points on the conductors whose longitudinal distance is Dlength. The equation is

4.1.4

Δv1

v1

v1

L11 L12 L13

L1n

i1

Δv2

v2

v2

L21 L22 L23

L2n

Δv3 = v3 − v3 = Dlength

L31 L32 L33

L3n

i2 d i3 dt

Δvn

Ln1 Ln2 Ln3

Lnn

vn

vn

4 15

in

Inductance of an Overhead Single-Phase Line Conductor

Now we will examine an overhead single-phase line whose conductor is laid at height h m from the ground surface, as shown in Figure 4.2; current i(t) goes through the line and come back through the earth return pass. A hypothesis may be set up that is analogous to Figure 4.1(b) and Eqs. (4.9a, b, and c), that if conductor 2 is replaced by a single imaginary earth conductor (radius r) laid at a depth of h m, then distance D is replaced by 2h. With the hypothesis, L11 − L12, L11, L12 can be calculated using Eqs. (4.9a, b, and c), respectively, where D is replaced by 2h. However, it has been experimentally observed that the calculated values are always a little smaller in comparison with the field measured values, and, furthermore, that the calculated values and the measured values are almost the same if we assume that the imaginary earth conductor is laid at depth H0 (=600~1000 m) instead the depth h, as shown in Figure 4.2a.

+q

+q h

h Ground surface H0 ≒ 600~900 m Imaginary datum plane

H0

Imaginary earth conductor H ≒ 2H0 + h

–q

–q

(a) Figure 4.2 Equivalent earth conductor.

(b)

4.1 Inductance

Next, let’s examine the interpretation of these facts in Figure 4.2b. We assume that the returned earth current −i(t) is uniformly flowing through the big circular area of radius H = 2H0 + h. This is the special case from Figure 4.1b in that conductor 1 (radius r) and conductor 2 (radius H = 2H0 + h) are laid in parallel through uniform air space, and distance D is replaced by H + h ( 2H0). The flux linkages and inductance of overhead conductor 1 in this case are calculated next. 4.1.4.1 ψ 11 Caused by Conductor 1

The flux linkage ψ 11 caused by current +i of conductor 1 is the same as Eq. (4.7a): ψ 11 = ψ out + ψ in =

μs μ0 S1 μ μ loge + cond 0 2π r 8π

4 16a

i

where μ = μs μ0 (permeability of air space). 4.1.4.2 ψ 12 Caused by Conductor 2 (Imaginary Conductor, Radius H)

The flux linkage ψ 12 of conductor 2 caused by the earth current −i is given by Eq. (4.7b), where the earth conductor radius is H instead of r, so D is replaced by H + h ( 2H0) : μs μ0 S2 loge 2π H +h

−ψ 12 = ψ out + ψ in =

−i

4 16b

4.1.4.3 ψ 11 − ψ 12 Total Flux Linkage of Conductor 1

Then with the condition S ψ 11 −ψ 12 =

S1, S2

r, the total flux linage ψ 11 − ψ 12 of conductor 1 is

H

μs μ0 S μ μ loge + cond 0 r 2π 8π

μs μ0 S loge H +h 2π

i+

μ μ H + h μcond μ0 + = s 0 loge r 2π 8π

i

−i

μs μ0 H +h loge r 2π

4 16c i

The second term is of course negligible: L11 −L12 =

ψ 11 − ψ 12 i

μ H +h H +h log = 0 4605log10 2π e r r

4 16d

where H 2H0 + h and H0 is 600~900 m μ = μs μ0 (permeability of air space) In conclusion, the calculated value of ψ 11 − ψ 12 in that the earth conductor of radius H 2H0 + h is assumed and that of the measured values become almost coincident each other. Therefore, the assumption that the ground current flows through the circle with radius H ( 2H0) is reasonable. H0( H/2) is called the equivalent depth of the earth conductor. 4.1.5

Inductances of Three-Phase Line Conductors

Now we will investigate an actual three-phase transmission line, which is installed along the earth ground surface. The average heights of the phase-a,b,c conductors are ha, hb, hc [m], respectively. Consider phase-a,b,c conductors and one virtual ground conductor g (radius rg) that is buried in the depth 2H0. We assume 2H0 Ha + ha Hb + hb Hc + hc (see Figure 4.3). The equation of potential voltage is va vb vc vg

=

Laa Lab Lac Lg

ia

Lba Lbc Lbc Lg

d ib dt ic

Lca Lcb Lcc Lg Lg

Lg

ia + ib + ic + ig = 0

Lg Lgg

ig

4 17

57

58

4 Physics of Transmission Lines and Line Constants

b

Dab a

c

+ib

+ic

hb

hc ≒ 6~80 m

+ia ha ha = H0

Ground surface H0 = 600–900m Imaginary ground plane

ha = H0

hb+H0

Daβ

hc+H0 ≒ H0

Dbα

α

–ia

–ib

–ic γ

β Figure 4.3 Three-phase line-inductance model.

The equations can be modified as follows by eliminating ig = − (ia + ib + ic): va

Laa − Lag Lab − Lag Lac − Lag

vb

= Lba − Lbg Lbb − Lbg Lbc − Lbg

vc

Lca − Lcg Lcb − Lcg Lcc − Lcg

vg = Lgg − Lg

ia d ib dt ic

Laa Lab Lac Lba Lbb Lbc Lca Lcb Lcc

d ia + ib + ic dt

ia d ib dt ic 4 18a

where Laa Lab

μs μ0 1 1 loge = 0 4605 log10 r r 2π μs μ0 1 1 loge Lab −Lag = = 0 4605 log10 Dab Dab 2π Laa −Lag =

μs μ0 Dab Dab loge = 0 4605 log10 2π r r μ μ h + 2H0 Lgg − Lg = s 0 loge 2π H Laa − Lab =

4 18b

Next, the equation for the voltage sag caused on the conductors with length Dlength is our main interest. It is as follows, referring to Eq. (4.15): Δva

va

va

Laa Lab Lac

Δvb = vb − vb = Dlength

Lba Lbb Lbc

Δvc

Lca Lcb Lcc

vc

vc

ia d ib dt ic

4 19a

4.2 Capacitance and Leakage Current

and 0 4605 log10

ha + 2H0 ha + 2H0 ha + 2H0 0 4605 log10 0 4605 log10 r Dab Dac

Lba Lbb Lbc = 0 4605 log10

ha + 2H0 hb + 2H0 ha + 2H0 0 4605 log10 0 4605 log10 Dba r Dbc

Laa Lab Lac

Lca Lcb Lcc 0 4605 log10

4 19b

ha + 2H0 ha + 2H0 hc + 2H0 0 4605 log10 0 4605 log10 Dca Dcb r

We know that H 2H0 (the radius of the imaginary earth conductor) is less than a few hundred meters from a practical physics viewpoint. In the equation, the inductance and resistance of earth ground are included in the 3 × 3 inductance matrix. The working inductance between conductor a and b is Laa − Lab =

μs μ0 Dab Dab loge = 0 4605log10 2π r r

4 19c

By the way, the equivalent depth of earth conductor H0 depends on the geological strata. It is typically rather shallow (say, 600–800 m) in regions of younger strata from the Quaternary period or later, but it is generally deep (800–900 m) in regions of older strata from the Tertiary period or earlier. Finally, all the equations in this section are in the form of instantaneous quantities, so they can be used for transient analysis. However, in the case of steady-state calculations, d/dt in the equations can be replaced by jω, and the derived equations correspond to Eqs. (3.3) and (3.4), respectively.

4.2

Capacitance and Leakage Current

4.2.1

Potential Voltages and Capacitance of a Single Overhead Conductor

Figure 4.4a shows one long conductor 1 (radius r) laid in a uniform space of permittivity ε, and electric charge with the value +q per unit length is applied to the conductor. An electrical line of force proportional to +q is generated and emitted uniformly in all radial directions on the orthogonal plane. Now imagine a concentric circle with radius x. All line of forces emitted by +q pass through the circle (this is called a Gaussian plane). The intensity of the electric field (or electric line of force) E(x) at a point on the circle is proportional to +q and inversely proportional to 2πx. Then E x =

+q 2πεx

4 20

Let’s suppose that +q [C/m](coulomb per length) is to be moved at point x in the field of force E(x), work to oppose E(x) (in physics terminology) is required, and the accumulated value of the “work” is the potential energy of charge +q required by the movement work. Next, in Figure 4.4b, conductor 1 (radius r) and conductor 2 (radius r) are laid in parallel, where the distance between the two conductors is D12, and the electric charge per unit length is +q and −q, respectively. Imagine an arbitrary point y(S1, S2) whose distance from conductor 1 and 2 is S1 and S2, respectively. The intensity of the electric fields E1(S1) and potential voltage v+q produced by +q are E1 S =

+q 2πεS1

r1

v+q =

E1 S dS

4 21a

S1

Also, E2(S2) and v−q produced by −q are E2 S =

−q 2πεS2

r

v−q =

E2 S dS S2

4 21b

59

60

4 Physics of Transmission Lines and Line Constants

Gaussian plane Flux linkage line vy y

Electric line of force 2πx r

+q

Equipotential line

S1

+q v1

x

S2

h

D = 2h

Electric line of force

g OV Ground surface

h

Radius r –q (b) Two conductors laid in parallel

(a) One conductor laid in uniform space Figure 4.4 Potential voltage in space with one overhead conductor.

Then the potential voltage at point y is, by the theorem of superposition, r

vy = v + q + v −q =

r

E1 S dS +

E2 S dS =

S1

S2

r

1 dS − S S1

q 2πε

r

1 dS S S2

4 22a

The contents of the brackets {} can be modified as follows: r

1 dS + S1 S

−1 dS = S2 S r

r D

1 dS + 2S

D 2 S1

1 dS − S

r D

1 dS + 2S

D 2 S2

1 dS = S

D 2 S1

1 dS − S

D 2 S2

1 S2 dS = loge S S1

Then ∴ vy S1 , S2 =

q 2πε

D 2 S1

1 dS − S

D 2 S2

1 dS S

=

q S2 loge 2πε S1

The potential voltage v1g of conductor 1 to the ground voltage 0 V is given as the special case of S1 D12 = 2h, v1g =

q D12 q 2h log log = −v2g = 2πε e r 2πε e r

And v2g is given by putting S1 v2g = − v1g = −

D12, S2

q D12 q 2h loge loge = 2πε 2πε r r

4 22b r and S2

4 23a r, +q

− q: 4 23b

The voltage difference between conductor 1 and the virtual conductor 2 v12 when +q and −q are given to them respectively is v12 = v1g − v2g = 2v1g =

q D12 log πε e r

4 23c

Then the per-unit length capacitance of the overhead conductor 1, the capacitance between conductor 1 and the ground, is from Eq. (4.23a)

4.2 Capacitance and Leakage Current

q 1 1 2πε 0 02413εs = μF km = μF km 4 24 = = 1 D12 1 2h 2h 2h v1g loge loge loge log10 2πε r 2πε r r r 1 = 9 × 109 where ε = εs ε0, 4πε0 Now, we have derived the capacitance of a single conductor to the earth ground C1g. Equation (4.22b) also shows that the planes y(S1, S2), given fixed S2/S1 values, give equipotential voltage planes of the electric line of force. The voltage of the entire ground surface, including the midpoint of the straight line O1O2 (S1 = S2 =D/2), become 0 V because S2/S1 = 1. Conductors 1 and 2 are the conformal transposition of the ground surface. C1g

4.2.2

Potential Voltages and Capacitance of a Three-Phase Overhead Line

Now, consider the three-phase line in Figure 4.5. Our purpose is to find the following equations with regard to the figure: paa pab pac

qa

vb = pba pbb pbc

qb

vc

pca pcb pcc

qc

qa

kaa kab kac

va

qb = kba kbb kbc

vb

qc

vc

va

kca kcb kcc

vabc = pabc qbc

4 25a

qabc = kabc vbc

4 25b

Where pabc qabc = 1

4 25c

p [m/F] are the coefficients of the potential, and k [F/m] are the electrostatic coefficients of static capacity. Equation (4.25a) and the matrix pabc can be easily found using the analogy of the explanation for conductors 1 and 2 in the previous section: paa , pbb ,pcc self coefficients of the potential Putting qb = qc = 0 in Eq. (4.25a), we have va = paa qa va is calculated assuming ±qa are given to conductors a and α, respectively. This is the same case as Figure 4.4, but the names of conductors 1 and 2 are replaced by a and α. Then Eq. (4.23a) is cited: paa

q 1 1 0 02413εs = F m = μF km = 1 D 1 2h 2ha vaa aα a loge loge log10 2πε r 2πε r r 4 26a

pbb and pcc are of the same form by analogy:

b

+qb

Dab

a +qa

Dbc

ha

Dca

Radius r c +qc

hb

hc

pab,pbc,pca (mutual coefficients of the potential)

Ground surface

Putting qa = qc = 0 in Eq. (4.25a), we have va = pab qb va is calculated assuming ±qb are given to conductors b and β respectively. This is the same case as Figure 4.4b and Eq. (4.22b), where the point y(S1, S2) is replaced by y(Dab, Daβ). Then, q 1 0 02413εs pab = F m = μF km 1 Daβ 2haβ vab loge log10 2πε Dab Dab 4 26b

hb

ha D bα

hc

Daβ –qa

γ

–qc

α β

–qb

Figure 4.5 Three-phase overhead line.

61

62

4 Physics of Transmission Lines and Line Constants

pbc and pca are of the same form by analogy. Now we have found the following pabc matrix equation: loge

paa pab pac pba pbb pbc =

Daα Daβ Daγ loge loge r Dab Dac

1 Dbα Dbβ Dbγ loge loge loge 2πε Dba r Dbc

pca pcb pcc loge

m F

Dcα Dcβ Dcγ loge loge Dca Dcb r

4 26c

2ha Daβ Daγ loge loge loge r Dab Dac = 2 × 9 × 109 loge

Dbα 2hb Dbγ loge loge Dba r Dbc

loge

Dcα Dcβ 2hc loge loge Dca Dcb r

Dab 2 − ha −hb

where Daβ = Dbα =

2

+ ha + hb

2

=

km μF

Dab 2 + 4ha hb

4 26d

This is the pabc matrix of Eq. (4.25a). Our next goal is to find Eq. (4.25b) and the kabc matrix, which is obviously the inverse matrix of the pabc matrix. Then kaa = pbb pcc − p2bc Δ kbb = kcc =

pcc paa − p2ca paa pbb − p2ab

F m

Δ

F m

Δ

F m

kab = kba = − pab pcc − pac pbc Δ

F m

kbc = kac = − pbc paa −pba pca Δ

F m

kca = kac = − pca pbb −pcb pab Δ

F m

Δ = paa pbb pcc + 2pab pbc pac − paa p2bc + pbb p2ca + pcc p2ab

m3 F 3

4 27

The kabc matrix has been mathematically found as the parameters of the pabc matrix elements, which are quite lengthy. However, kabc can be modified to a simple form as a matrix with parameters of Cabc matrix elements. Modifying Eq. (4.25b), qa = kaa va + kab vb + kac vc = kaa + kab + kac va + − kab va −vb + −kac va − vc

C m

qb = kba + kbb + kbc vb + − kbc vb − vc + −kba vb − va

C m

qc = kca + kcb + kcc vc + −kca vc −va + − kca + vc − vb

C m

4 28a

Then qa = Caa va + Cab va − vb + Cac va − vc

C m

qb = Cbb vb + Cbc vb − vc + Cca vb −va

C m

qc = Ccc vc + Cca vc − va + Ccb vc −vb

C m

4 28b

4.2 Capacitance and Leakage Current

And in the form of derivatives, dqa d d d va − vb + Cac va − v c = Caa va + Cab dt dt dt dt dq d d d vb − vc + Cca vb − v a ib = b = Cbb vb + Cbc dt dt dt dt dq d d d vc −va + Ccb vc − v b ic = c = Ccc vc + Cca dt dt dt dt ia =

A m 4 28c

where Caa = kaa + kab + kac

F m

Cab = −kab

F m

Cbb = kba + kbb + kbc

F m

Cbc = − kbc

F m

Ccc = kca + kcb + kcc

F m

Cca = − kca

F m

Cac = − kac

F m

Cba = − kba

F m

Ccb = −kcb

F m,

4 28d

Further, Eq. (4.28b) can be modified as follows: Caa + Cab + Cac

− Cab

−Cac

va

qb =

−Cba

Cba + Cbb + Cbc

− Cbc

vb

qc

− Cca

− Ccb

Cca + Ccb + Ccc

vc

qa

4 28e

And in the form of derivatives, the leakage current equation is as follows: qa Caa + Cab + Cac − Cab − Cac d = qb = − Cba Cba + Cbb + Cbc −Cbc dt qc −Cca − Ccb Cca + Ccb + Ccc

ia ib ic

va d vb dt vc

4 28f

Now we have found kabc matrix of Eq. (4.25b). In conclusion, if we know the radius and allocation of all the line conductors, all the constants can be calculated in the kabc matrix (Eq. (4.27)) Cabc (Eq. (4.28d)). order pabc matrix (Eq. 4.26c) Note that Figures 4.6a and b as well as Figure 4.5 are based on the variables q and v. In contrast, Figure 4.6c is based on Eq. (4.28c), which may be more familiar as the v and i circuital expression. The current i flowing through the capacitance elements is called the leakage current.

(a)

a

(b)

b

Radius r

Cab

qb

vab

c qc

vca va

vb

qa va

vc

Figure 4.6 Stray capacitance of a single-circuit line.

Caa

(c)

b

(V a b

Cbc c

a

vbc

Charge qa

qb

Cca Cbb

vb

qc Ccc

Ca jω a jω C

Cab

Cbc Cca

c

(V – a V c) Caa Cbb

Ccc

Vb

ac

Va vc

) b – Vb

jω Caa Va

Vc

63

64

4 Physics of Transmission Lines and Line Constants

4.2.3

Stray Capacitance and Leakage Current of Phase-Balanced Lines

As illustrated in Figure 4.5, if a three-phase transmission line is well-phase-balanced, probably by transposition, the matrix equations of pabc, kabc, Cabc can be simplified: ha ≒ hb ≒ hc ,

h

Dll = Dab = Dba ≒ Dbc = Dcb ≒ Dca = Dac

Daβ ≒ Dbα ≒ Daγ ≒ Dcα ≒ Dbγ ≒ Dcβ paa ≒ pbb ≒ pcc

ps

4 29b

pab = pba ≒ pac = pca ≒ pbc = pcb

pm

4 29a

Accordingly, the p and k coefficients of Eqs. (4.26a, b) and (4.27) are simplified as follows: Δ = p3s + 2p3m − 3ps p2m = ps − pm

2

ps + 2pm

kaa ≒ kbb ≒ kcc ≒ p2s − p2m

ks

Δ=

ps + pm ps − pm ps + 2pm

kab = kba ≒ kac = kca ≒ kbc = kcb = − pm ps −p2m

km

Δ

−pm ps − pm ps + 2pm

=

ks + 2km =

1 ps + 2pm

Caa ≒ Cbb ≒ Ccc = ks + 2km =

Cs

1 ps + 2pm

Cab = Cba ≒ Cac = Cca ≒ Cbc = Ccb = − km

Cm =

2h r sbα 9 pab ≒ pbc ≒ pca = 2 × 9 × 10 loge Dll

pm

4 29d

pm pm = Cs ps −pm ps + 2pm ps −pm

paa ≒ pbb ≒ pcc = 2 × 9 × 109 loge

ps

4 29c

D2ll + 2h

≒ 2 × 9 × 109 loge 9

= 2 × 9 × 10 loge

m F

①

m F

2

Dl l 2h 1+ Dl l

2

1 2

m F

②

where generally h > sl l ,

2h Dl l

2

1

and ∴ pm ≒ 2 × 9 × 109 loge

2h Dl l

m F

②

4 29e

4.2 Capacitance and Leakage Current

Cs =

1 ≒ ps + 2pm

1

1 = 2h 2h 8h3 2 × 9 × 109 loge + 2loge 2 × 9 × 109 loge 2 r sl l rDl l 0 02413 0 02413 = × 10 − 9 F m = μF km ① 8h3 8h3 log10 2 log10 2 rDl l rDl l zero-sequence capacitance

while 2h 2h loge log10 pm Dll Dll ≒ = Dl l 2h 2h ps − pm log10 loge − loge r r Dl l 2h 2h log10 log10 pm 0 02413 Dl l Dl l ∴ Cm = Cs ≒ Cs = μF km ② Dl l Dl l ps −pm 8h3 log10 log log10 2 10 r r rDl l

4 29f

Then the stray capacitance and the leakage current equations of a well-balanced three-phase line can be written as follows in Eqs. (4.30a, b) and is shown in Figure 4.7: −Cm Cs + 2Cm −Cm

Cs + 2Cm − Cm = jω − Cm

Ia Ib Ic I abc

−Cm −Cm Cs + 2Cm

Va Vb Vc

4 30a

V abc

C abc

∴ I abc = jωC abc V abc Incidentally, Figure 4.7a can be modified to Figure 4.7b, where the total capacitance of one phase C ≡ Cs + 3Cm is called the working capacitance of single-circuit transmission lines, and can be calculated by the following equation: Cs + 3Cm = ks + 2km + 3 − km = ks −km =

C

1

=

2 × 9 × 109 loge

2h 2h −loge r Dll

1 ps − pm

1

=

2 × 9 × 109 loge

Dl l r

F m

0 02413 working capacitance ① μF km Dl l log10 r In the case of multi-bundled n conductor lines , the radius r is replaced by the equivalent radius reff , m ② req = r 1 n × w n −1 n where w is the geometrical averaged distance between bundled conductors =

4 30b

Note that Eq. (4.30a) is the same as Eq. (3.40) in Chapter 3. The equivalent radius req of ② is discussed in the next section. a b c

a

Cm Cm

Cm

Cs Cs Cs

(a) Single-circuit line

b c

Cs Cs Cs 3Cm 3Cm 3Cm

(b) Single-circuit line

Figure 4.7 Stray capacitances of a single-circuit overhead line (well-balanced).

65

66

4 Physics of Transmission Lines and Line Constants

4.3

Actual Configuration of Overhead Transmission Lines

4.3.1

Structure

Figure 4.8 shows the typical structure of an extra high voltage (EHV) double-circuit transmission line composed of multiple bundled conductors. Let’s look back at the inductive and capacitive circuit equations of a transmission line. Table 4.2 and Figure 4.9 show a typical advanced aluminum conductor steel reinforced (ACSR) conductor for overhead transmission lines. Due to recent advanced metal–alloy production and wire-drawing technology, lightweight, largecurrent-capacity conductors with high-temperature-withstanding characteristics even at 230 C have been realized, as shown in the table. Furthermore, aluminum conductor carbon fiber reinforced (ACFR) conductors (where the tension member in steel twisted wires is replaced by carbon fiber twisted wires) have been experimentally adopted in order to realize lightweight conductors. 4.3.2

Equivalent Radius req of Multi-Bundled Conductors

For most recent large-capacity transmission lines, multi-bundled conductor lines (n = 2–8 per phase) are being utilized, as shown in Figure 4.8. In the case of n-bundled conductors (the radius of each conductor is r), the L and C constants can be calculated by replacing the radius r of the transmission lines with the equivalent radius req = r1/n × w(n − 1)/n. OGW or OPGW Formed aluminum-clad steel wire Aluminum pipe Optical fiber cable R

S

T

T

Suspension insulator (pin insulators)

S

Fin (low audible noise, low wind noise type)

Power conductor TACSR Multi-bundled conductor

R

Zinc-coated steel core Press-formed thermalwithstandable Aluminum alloy Spacer

Figure 4.8 Overhead double-circuit transmission line.

e

or

c el

te

ds

y allo

m

te oa

nd

ta iths

c-c

l-w ma

n

Zi

m inu

alu

Courtesy of Exsym corporation

r

ed

the

m for

ss-

Pre

Figure 4.9 High-temperature-withstanding aluminum-clad steel wire (TACSR).

4.3 Actual Configuration of Overhead Transmission Lines

Laa − Lag from Eq. (4.19) and C ≡ Cs + 3Cm from Eq. (4.30b), for example, can be modified as follows, by adopting the equivalent radius req instead of the individual conductor radius r: Laa C

ha + 2H0 + 0 1 mH km req 1 0 02413 F m = Cs + 3Cm = D Dll ll 2 × 9 × 109 loge log10 req req Laa − Lag = 0 4605log10

req = r 1 n × w n −1

4 31a μF km

n

4 31b 4 31c

w[m] is the geometrical averaged distance of bundled conductors. The mutual inductance Lab and mutual capacitance Cab are not affected by the adoption of multi-bundled conductor lines because the radius r is not included in the equations. The theoretical proof of req = r1/n × w(n − 1)/n is shown in Supplement 1 at the end of this chapter. Numerical check: Using TACSR = 810 mm2 (see Chapter 2), 2r = 40 mm and four bundled conductors (n = 4), with the square allocation w = 50 cm averaged distance: w = w12 w13 w14 w23 w24 w34 = 50 50 2 50 50 50 2 50 req = r 1

n

w n− 1

n

= 2 01

4

1 6 1 6

= 57 24cm

4 32

57 253 4 = 24 7 cm

The equivalent radius req = 24.7 cm is 12.4 times r = 2.0 cm, so the line self-inductance Laa can also be reduced by the application of bundled conductors. 4.3.3

Line Resistance

Earth resistance rg in Figure 3.1a is negligibly small. Accordingly, the mutual resistances rab, rbc, rca in Eq. (3.4) become zero. Therefore, the specific resistances of conductors ra, rb, rc are actually equal to the resistances raa, rbb, rcc in the impedance matrix of Eq. (3.11). In addition to the power loss caused by the linear resistance of conductors, nonlinear losses called the skin-effect loss and corona loss occur on the conductors. These losses become progressively larger in higher-frequency zones, so they are influential factors for the attenuation of traveling waves in surge phenomena. However, they can usually be ignored for power frequency phenomena, because they are smaller than the linear resistive loss and, further, very much smaller than the reactance value of the line, at least for power frequency. With regard to bundled conductors, due to the result of the enlarged equivalent radius req, the dielectric strength around the bundled conductors is somewhat relaxed, so corona losses can also be reduced. Skin-effect losses of bundled conductors are obviously far smaller than for a single conductor whose aluminum cross-section is the same as the total sections of the bundled conductors. Further, in addition to the R, L, C constants, the leakage resistance G is the fourth line constant. This is typically the creepage resistance of insulators of transmission lines or station equipment that is a parallel resistance with stray admittance jωC, and usually has an extraordinarily large ohmic value. G is an important constant that is significantly affected by the insulation characteristics of individual high-voltage insulators, attenuation ratio of surge phenomena, and so on. However, G can be ignored for most ordinary circuit analysis (except for surge analysis), because it has a large resistance on the order of, say, megaohms 4.3.4

Typical Transmission Line Constants and Summary of Impedance/Capacitance Matrices

The circuit equations with LR constants in the form of a-b-c-domains are given by Eqs. (3.11) and (3.12), and the impedance matrix and elements can be calculated with Eq. 4.19b. Further, the working inductance is calculated with Eq. (4.19). The circuit equations and impedance matrix by 012-transformed (symmetrical) domain are given in Eq. (3.34) and Figure 3.11. The circuit matrix equations with capacitive C constants in the form of a-b-c-domains are given by Eq. (4.30a). Further, the capacitive elements Cs and Cm can be calculated using Eq. (4.29f).

67

68

4 Physics of Transmission Lines and Line Constants

By the way, the circuit impedance equations and the impedance matrix with a 012-transformed (symmetrical) domain are given by Eqs. (3.40) and (3.41) and Figure 3.13. L, C constant values of individual overhead transmission lines are different because the allocations of conductors (in other words, the physical length r, h, Sll, etc., of the tower design) and the geological characteristics of the earth ground vary. However, the line constants are not much different for lines of similar voltage classes, because the physical dimensions of the conductors are not very different. In addition, the constants are not much different even for lines of different voltages, because the physical lengths of the variables r, h, Sll, etc. based on the conductor allocation of the tower design are included in the logarithmic terms of the equations for L, C. Table 4.1 shows typical L, C values of single-circuit lines and double-circuit lines. Numerical check of Laa and Lab : from Eq. 4.18b, Ls = 0 4605 log10

ha + 2H0 req

ha + 2H0 Dab Dl l Ls −Lm = 0 4605 log10 req Lm = 0 4605 log10

••

Assuming H0

800m and ha + 2H0

4 33 1600m, then

Ls= 2.39, 2.26, 1,94, 1,47mH/km for req = 0.01m, 0.02m, 0.1m, 1.0m, respectively. Lm= 1.66,1.52,1.20, 1.05 mH/km for Dl l = 0.5m, 1m, 5m, 10m, respectively. Numerical check of Cs and 3Cm: from Eq. (4.29e), Cs =

0 02413 μF km 8h3 log10 req D2l l

① 4 34

2h log10 0 02413 Dl l Cm = μF km ② Dl l 8h3 log10 log10 r req D2l l Assuming req = 0.02m and Dl l = 1m: For h = 10m

For h = 25m

For h = 50m

Cs = 0.0043μF/km

Cs = 0.0034μF/km

Cs = 0.0031μF/km

3Cm = 0.0099μF/km

3Cm = 0.0102μF/km

3Cm = 0.0111μF/km

All of these calculated values are very close to the typical values of the individual constants given in Table 4.1. Table 4.2 shows ACSR conductors for overhead transmission lines.

4.4

Special Properties of Working Inductance and Working Capacitance

The equations for working inductance and working capacitance were introduced in the previous section. These are shown again from Eqs. (4.8c) and (4.30b), in vacuum space: ψ 11 −ψ 12 μ0 D μ μ = loge + cond 0 working inductance r i 2π 8π q 1 working capacitance C1 = = 1 2h v1 loge 2πεs ε0 r L11 − L12 =

4 35a 4 35b

Also permeability and permittivity were explained through the deriving process. These are, again, in the MKS rational unit system,

Table 4.1 Typical LRC constants of overhead transmission lines. Single-circuit line

Circuit equation

Positive-sequence

Zero-sequence

Positive-sequence

Zero-sequence

Z1 = Zs − Zm

Z0 = Zs + 2Zm

Z1 = Zs − Zm

Z0 = Zs + 2Zm Z0M = 3Z’m

Z1

Z0

Eq. (2.15)

Eq. (2.15)

jY1 = jωC1

jY0 = jωC0

C 1= C s + 3Cm

C0 = Cs Eq. (2.22)

Eq. (2.22) Typical value

Double-circuit line

Z1

Z0

Z0M

Eq. (2.20)

Eq. (2.20)

C1 = Cs + 3Cm + 3C’m jY1 = jωC1

jY0 = jωC0 = jωCs jY 0 = jωC 0 = jω3C’m

C1

C1

C0

Zero-sequence (First lane)

Zero-sequence (Second lane)

Z00 = Z0 + Z0M = Zs + 2Zm + 3Z’m

Z01 = Z0 − Z0M = Z0 + 2Zm − 3Z’m

Z00

jY00 = jωC0 = jωCs

C′0 = 3C′m C 0 = Cs

C0= Cs

Eq. (2.24)

Surge impedance L C

Z01

jY01 = jω(C0 + 2C 0) = jω(Cs + 6C’m)

C0 + 2C´0 =Cs + 6C´m

Eq. (2.24)

Inductance [mH/km] Ls = 2 mH/km Lm = 1 mH/km 3 L’m = 2.5 mH/km

L1 = Ls − Lm L1 = 2–1 = 1 mH/km (= j0.314 Ω/km)

L0 = Ls + 2Lm L0 = 2 + 2 × 1 = 4 mH/km (= j1.26 Ω/km)

L 1 = Ls − Lm L1 = 2–1 = 1 mH/km (= j0.314 Ω/km)

L0 = Ls + 2Lm L0 = 2 + 2 × 1 = 4 mH/km (= j1.26 Ω/km)

L0M = 3 L’m L0 = 2.5 mH/km (= j0.78 Ω/km)

L00 = Ls + 2Lm + 3 L’m L00 = 2 + 2 + 2.5 = 6.5 mH/km (= j2.04 Ω/km)

Z01 = Ls + 2Lm − 3 L’m L01 = 2 + 2–2.5 = 1.5 mH/km (= j0.47 Ω/km)

Single line Ω

Capacitance [μF/km] C0 = 0.005 μF/km 3Cm = 0.01 μF/km 3C’m = 0.0025 μF/km

C1 = Cs + 3Cm C1 = 0.005 + 0.010 = 0.015 μF/km

C0 = Cs C0 = 0.005 μF/km

C1 = Cs + 3Cm + 3C’m C1 = 0.005 + 0.010 +0.0025 = 0.0175 μF/km

C0 = Cs C0 = 0.005 μF/km

C 0 = 3C’m C 0 = 3C’m = 0.0025 μF/km

C0 = Cs C0 = 0.005 μF/km

C0 + 2Cm = Cs + 6C m = 0.005 + 2 × 0.0025 = 0.010 μF/km

Double line 239 Ω

(Continued)

Table 4.1 (Continued) Single-circuit line

Example 1

Example 2

Example 3

Example 4

500 kV. double circuit line STACSR (810mm2) × 4

275 kV. double circuit line STACSR (810 mm2) × 2

154 kV, double circuit line ACSR (610 mm2) × l

66 kV. double circuit line ACSR (330 mm2) × 1

Double-circuit line

Positive-sequence

Zero-sequence

Positive-sequence

Zero-sequence

(0.86 mH/km) 0.019 + j0.27 Ω/km

(3.34 mH/km) 0.30 + jl.05 Ω/km

(0.86 mH/km) 0.019 + j0.27 Ω/km

(3.34 mH/km) 0.30 + j1.05 Ω/km

0.013 μF/km 1.18A/km

0.0085 μF/km

0.016 μF/km 1.41 A/km

(0.73 mH/km) 0.0085+ j0.23 Ω/km

0.26+ j1.10 Ω/km

(0.73 mH/km) 0,0085 + j0.23 Ω/km

0.0156 μF/km 0.778 A/km

0.0054 μF/km

0.0 l85 μF/km 0.92 A/km

(1.21 mH/km) 0.005 + j0.38 Ω/km

0.30+ jl.43 Ω/km

(1.21 mH/km) 0.005+ j0.38 Ω/km

0.0094 μF/km 0.262 A/km

0.0055 μF/km

0.0109 μF/km 0.148 A/km

(1.15 mH/km) 0.05 + j0.36 Ω/km

0.31 + j1.47 Ω/km

(1.15 mH/km) 0.05 + j0.36 Ω/km

0.0101 μF/km 0.120 A/km

0.0055 μF/km

0.0122 μF/km 0.065 A/km

0.26 + j1.10 Ω/km

0.30 + j1.43 Ω/km

0.31 + j1.47 Ω/km

Zero-sequence (First lane)

Zero-sequence (Second lane)

Surge impedance

(2.20 mH/km) 0.28+ j0.69 Ω/km

(5.54 mH/km) 0.58 + j1.74 Ω/km

(1.15 mH/km) 0.02+ j0.36 Ω /km

S 257 Ω

0.0029 μF/km

0.0062 μF/km

0.012 μF/km

D 232 Ω

(2.42 mH/km) 0.25+ j0.76 Ω/km

(5.92 mH/km) 0.51 + j1.86 Ω/km

(0.18 mH/km) 0.10+ j0.34 Ω /km

S 216 Ω

0.0029 μ/km

0.0058 μF/km

0.0116 μF/km

D 199 Ω

(2.90 mH/km) 0.25 + j0.91 Ω/km

(7.45 mH/km) 0.55 + j2.34 Ω/km

(1.66 mH/km) 0.05 + j0.52 Ω/km

S 358 Ω

0.0015 μF/km

0.0044 μF/km

0.0074 μF/km

D 333 Ω

(3.22 mH/km) 0.25+ j1.01 Ω/km

(7.90 mH/km) 0.56+ j0.48 Ω/km

(1.46 mH/km) 0.06+ j0.46 Ω /km

S 337 Ω

0.0021 μF/km

0.0042 μF/km

0.0082 μF/km

D 307 Ω

Note 1: The reactances are the derived values from j2πfL = 2π 0 50 L on a 5 0 Hz base, so that 1 mH/km corresponds to jX = 2π (50) × 10−3 = 0314 Ω/km. Therefore jX for 60 Hz systems should he modified to 1.2 times larger values. Note 2: The charging current is derived from I = j2π fC (1 / 3) V based on 50 Hz as RMS values. Accordingly, the charging current I for Example 1 is calculated from I = 2π 50 (0.015 10−6)(l/ 3) 500 103 = 1.36 A/km per phase. The charging current for 60 Hz systems should be modified to 1.2 times larger values. Note 3: The theoretical details of zero-sequence first-/second-lane impedance of double circuit line can be seen in chapter 4 of “Power Systems Engineering: second ed. by Y. Hase (Wiley 2013).

4.5 MKS Rational Unit System

Table 4.2 ACSR conductors for overhead transmission lines (typical examples). Continuous

Temporary

Maximum temperature [ C]

Maximum current [A]

Maximum temperature [ C]

Maximum current [A]

ACSRa

90

829

120

1125

TACSR

150

1323

180

150S

ZTACSR

210

1675

240

1831

XTACIR

230

1715

290

2004

μ0 ε0

permeability of vacuum space ,and μ0 = 4π × 10 −7 1 permittivity of vacuum space ,and = 9 × 109 4πε0

4 36

Now, let’s further examine these equations. The right-side second term μcond μ0/8π of Eq. (1.44a) is the linking flux in a narrow conductor section, so it can be ignored when phenomena in wide spaces are examined. The working inductance L11 − L12 and working capacitance C1 relate to each other as follows: 1 Laa − Lab

Ca

=

1 μ0 ε0

c0

the constant value

4 37a

In the MKS unit system c0

1 = μ0 ε0

1 4π × 10 −7

1 4π × 9 × 109

= 3 × 108 m sec 4 37b

= 300, 000 km sec It was found that 1 Laa − Lab Ca always becomes 1 μ0 ε0 , which takes the constant value c0 unconditionally. From the viewpoint of physics, if current flows through a straight conductor laid out in three-dimensional vacuum space, it is accompanied by a magnetic line with permeability μ0 and an electric line of force with permittivity ε0. Furthermore, 1 μ0 ε0 takes the constant value c0 unconditionally. Equations (4.35a), (4.35b), (4.36), (4.37a), and (4.37b) are the climax of the conclusion presented by James C. Maxwell in 1873 in his famous paper. The constant c0 has a value with a dimension of distance/time or velocity. With these conclusive equations, Maxwell presumed as follows: i) An electromagnetic wave exists, and it can propagate through vacuum space without “ether.” Energy can also be propagated through vacuum space. ii) The propagating velocity of the wave is always the constant value 1 μ0 ε0 = c0 , and it is 300 000 km/sec if measured by the MKS rational unit system. This was Maxwell’s theoretical discovery of the electromagnetic wave. He also proved by analogy that light from the sun must also be a kind of wave, with the same velocity of 300 000 km/sec. It is interesting that the working inductance and working capacitance given by Eqs. (4.10e) and (4.30b) are coincident with the positive-sequence inductance L1 = Ls − Lm (Eq. (3.34) for the form of impedance Z1 = Zs − Zm in Chapter 3) and the positive-sequence capacitance C1 = Cs + 3Cm (Eq. (3.41b)), respectively.

4.5

MKS Rational Unit System

4.5.1

Fundamental Concepts

We will discuss the fundamentals of the MKS rational unit system as the last subject of this chapter. The/velocity of an electromagnetic wave c0 is a universal, unchanging constant, and its value is c0 ≡ 1 μ0 ε0 = 3 × 108 [m/sec] if measured by the MKS unit system. c0 is the unchanging value 300 000 km/sec, so μ0 ε0 is also an

71

72

4 Physics of Transmission Lines and Line Constants

unchanging value. In other words, we can create an arbitrary unit system in which either μ0 or ε0 can be freely determined, whereas the other is dependently defined, because c0 and μ0 ε0 are unchangeable fixed values. If either μ0 or ε0 is defined with a value, the other should be defined dependently to satisfy the equation. Therefore, μ0 and ε0 are defined as follows by the MKS rational unit system: c0

1 = 3 × 108 m sec μ0 ε0

4 38a

μ0 = 4π × 10 − 7 H m 1 1 = ε0 = −7 2 4π × 10 c0 4π × 9 × 109

4 38b 4 38c

Now, let’s go back to the historical story of the MKS rational unit system. Coulomb’s laws for electric force and magnetic force can be described by the Gaussian unit system and by the MKS rational unit system as follows: Coulomb’s law by electric charge q1 ,q2 Coulomb’s law by magnetic pole m1 , m2 q1 q2 m1 m2 Gaussian unit system F= 2 F= r r2 1 q1 q2 1 m1 m2 MKS rational unit system F = F= V m2 4πε0 r 2 4πμ0 r2 4 39 where ε0 and μ0 are defined by Eqs. (4.48b and c) with the MKS rational unit system. In order to compare both unit systems, imagine a hollow sphere as shown in Figure 4.10. If the radius is r, the surface area is 4πr2 regardless of the unit system. If electric charge +q = 1.0 is placed at the center point of the sphere, the electric line of force will radiate uniformly toward the sphere surface. Now, we are free to count the total number of the radiated lines of force. The number is counted as 1 (one) by the Gaussian unit system, as 4π by the CGS (cm, gr, sec) rational unit system, and as 4π × 10−7 by the MKS (m, kg, sec) rational unit system. The MKS rational unit system is based on the same concept as the CGS rational unit system, except that it uses m instead of cm and kg instead of g. Using the Gaussian unit system, the expression of Coulomb’s law is simple; however, the number of lines of force per unit area becomes 1/4πr2. With the CGS unit system, the total number is 4π, and the number per unit area at the surface is 1/r2, which means we can count the lines of force per unit area using the equation without 4π. Generally, with the CGS rational unit system, we can escape the inconvenience of 4π or 2 π by removing them in the related equations for counting various physical quantities; on the other hand, 4π or 2 π is always included in equations based on the Gaussian unit system. In conclusion, light speed c0 is a fixed value of 3 × 108 m/sec or300 m/μsec in the MKS rational unit system. Then, subordinately, μ0 and ε0 are defined by Eq. (4.48b and c) in the MKS rational unit system. Radius r Following is a comparison of the MKS and CGS rational unit systems with Sphere Surface 4πr2 regard to force F and energy: ΔS

1 newton = 105 dyne energy = force

distance =

kg m sec2 newton

m =

g cm sec2 dyne

cm 107

4 40 5

+q

7

The number differ by 10 for force and by 10 for energy.

4.5.2 Practical MKS Units for Electrical Engineering Physics

Figure 4.10 Hollow sphere.

A summary of various electrical practical units is explained in brief as the last part of this chapter. The meter unit system was established in 1875, and the unit system based on the three fundamental units meter, kilogram, and second became popular

4.5 MKS Rational Unit System

Table 4.3 Fundamental units of the International Unit System (SI). Quantity

Name

Symbol

Distance

Meter

m

Weight

Kilogram

kg

Time

Second

sec

Current

Ampere

A

Thermodynamic temperature

Kelvin

K

Molecule volume

Mol

mol

Light intensity

Candela

cd

Table 4.4 Definition of various derived units in electrical physics. newton = m kg/sec2 pascal = newton/m2 joule = newton m watt = joule/sec = newton m/sec volt = watt/ampere ohm = volt/ampere weber = volt/sec tesla = weber/m2 henry = weber/ampere coulomb = ampere sec farad = coulomb/volt

all over the world. In 1951, the ampere was added as the fourth fundamental unit, and the expanded MKSA unit system was authorized, which meant various units for electrical physics were officially combined with various units for Newtonian physics. After that, the kelvin (K) for temperature and candela (cd) for light intensity were added. Then, in 1960, the International Unit System (SI: International System of Units) was established; it includes the seven fundamental units shown in Table 4.3. This is today’s expanded MKS unit system. All other units are defined dependently as derived units from these seven fundamental units. In addition, useful derived units are defined with proper unit names. For example, the unit for electric charge ±q is the time-integration of ampere having a unit value of Ampere sec. The new unit name coulomb was defined for the derived unit ampere sec. In other words, C = A sec is a derived unit defined with a proper unit name. Table 4.4 shows various derived units with proper defined unit names in electrical physics.

Supplement 1 Proof of Equivalent Radius req = r1/n wn–1/n for a Multi-Bundled Conductor The equivalent radius req = r1/n wn–1/n of a multi-bundled conductor in Eqs. (4.30b) and (4.31c) can be proved as follows. Calculation of Equivalent Radius for Inductance Assume that we have a one-phase n-bundled conductor where n: number of conductors, r: radius of each conductor, w: averaged distance between two conductors, and h: height above ground level. If all the elemental conductors are well balanced, the following equation is derived analogously to Eq. (4.30a):

73

74

4 Physics of Transmission Lines and Line Constants

r v1

s v1

Ls Lm

Lm

i1

r v2

s v2

Lm Ls

Lm

i2

−

= jω

r vn

1

s vn

Lm Lm

Ls

in

If the voltage and current of the bundled conductor are v and i, the voltage and current of each elemental conductor is v and i/n: rv rv

sv sv

−

rv

= jω

sv

Ls Lm

Lm

i n

Lm Ls

Lm

i n

Lm Lm

Ls

i n

2

Then we have r v − s v = jω

1 n

Ls + n −1 Lm

where Ls = 0 4605 log10 Lm = 0 4605 log10

3

i

h+H + 0 05 r

4a

h+H + 0 05 w

4b

If the bundled conductor is equivalent to a single conductor with radius req, is arranged at the same height h, and is charged with the same v and i, r v − s v = jωLeq

5

i

where Leq = 0 4605 log10

h+H + 0 05 req

6

Because Eqs. (3) and (5) should be equal, then Leq = Ls + n− 1 Lm

1 n

7

and therefore 0 4605 log10

h+H + 0 05 = req

0 4605 log10

h+H h+H + 0 05 = 0 4605 log10 1 n n− 1 req w r

0 4605 log10

h+H + 0 05 + n −1 r

0 4605 log10

h+H + 0 05 w

1 n

Then, n

+ 0 05

8

and therefore req = r 1

n

w n −1

n

Now, Eq. (4.30b②) has been proven using Eqs. (8) and (9).

9

4.5 MKS Rational Unit System

Calculation of Equivalent Radius of Capacitance If the voltage and charge of an n-bundled conductor are v and + q, the charge of each elemental conductor is +q/n. Then the following equation is derived analogously to Eqs. (4.25a) and (4.26a, b, c): v = 2 q n loge

= 2q loge = 2q loge

r1

n−1 2h 2h 2 q n loge × 9 × 109 × 9 × 109 + r w 1 1 n

2h r n

+ loge 2h w n−1

n

2h w

n−1 n

× 9 × 109

10

× 9 × 109

If the n-bundled conductor is equivalent to a single conductor with radius req and the same height h, and it is charged with the same v and + q, v = 2qloge

2h × 9 × 109 req

11

Comparing both equations, the following equation is derived: req = r 1

n

w n−1

n

Now Eqs. (4.31b, c) have been proven by Eqs. (10), (11), and (12).

12

75

77

5 The Per-Unit Method 5.1

Fundamental Concepts of the PU Method

The per-unit (PU) method (or % method) is a technique for handling any kind of quantity with its particular dimensions as quantities of a dimensionless ratio value based on 1.0 pu or 100%. This practice is a very useful approach and is applied widely in many engineering fields, eliminating the troublesome handling of several different kinds of quantities. However, in power system engineering, the PU method has various meanings, such as a technique for describing electrical circuits, far exceeding the simple meaning of the only convenient method to remove troublesome dimensions. Many individual structures that form power systems can be combined as one circuit (instead of a connection diagram) only by using the PU method. Furthermore, transformers can be handled by PU expressions as equipment in which Kirchhoff’s law is applied. The value of the PU method for electricians is summarized as follows: a) Kirchhoff’s law is satisfied among the currents of a transformer’s primary, secondary, and tertiary windings, so that transformers can be described as very simple circuits. b) Generators can also be described as accurate, simple circuits (see Chapter 10). c) Transmission lines, generators, transformers, loads, and other equipment of different types and ratings can all be combined together as one circuit. This can only be done by applying the PU method. d) The PU method provides relief from troublesome handling of practical dimensions (V, A, MVA, Ω, Wb, etc.). Items a), b), and c) are very important, and item d) is supplementary for electrical engineers.

5.2

PU Method for a Single-Phase Circuit

Let’s first examine the PU method for a single-phase circuit. The basic equations of voltage, current, and apparent power are V volt = Z ohm I ampere VA volt ampere = P + jQ volt ampere = V volt I ∗ ampere where V , I, Z,VA or S

51

complex-number quantities; I ∗ the conjugate of I

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

78

5 The Per-Unit Method

Now, in order to unitize the V, Z, I, VA quantities, the base quantities with signs Vbase, Ibase, Zbase, VAbase are introduced. All the base quantities are scalars (real numbers, or a vector of ∠0 ) and have to satisfy the following equations: Vbase volt = Zbase ohm Ibase ampere 5 2a VAbase volt ampere = Vbase volt Ibase ampere or Ibase ampere =

VAbase volt ampere Vbase volt 5 2b

Zbase

V2 Vbase volt = base ohm ohm = Ibase ampere VAbase

We can select any arbitrary value for base voltage Vbase and base capacity VAbase, but the base current Ibase and base impedance Zbase must be determined, depending on Vbase and VAbase, to satisfy Eqs. (5.2a, b). Equation (5.1) can be unitized with the base quantities of Eq. (5.2a) as follows: V Z I = Vbase Zbase Ibase VA P + jQ P Q V I∗ = = +j = VAbase VAbase VAbase VAbase Vbase Ibase

53

Using an overbar as the symbol for unitized quantities, V =Z I VA = P + jQ = V I where V = VA =

∗

V Z I∗ V ∗ , Z= , I = , V= Vbase Zbase Vbase Ibase

54

VA P Q , P= , Q= VAbase VAbase VAbase ∗

The unitized quantities V , Z, I ,VA, P + jQ are nondimensional complex numbers. Equation (5.4) is the same as the original Eq. (5.1), and the vector phase relations in Eq. (5.1) are preserved in Eq. (5.4) because all the base quantities are selected as scalars (a vector of ∠0 ). Unitized quantities can obviously be changed into actual values with individual dimensions using the following equations: V volt = V Vbase , Z = Z Zbase , I = I Ibase VA = VA VAbase , P = P VAbase , Q = Q VAbase Figure 5.1 summarizes the PU method.

55

5.3 PU Method for Three-Phase Circuits

System condition with practical unit values V, I, Z, VA

Per-unitization V=

V Vbase

Analysis

Conversion to practical unit values V = V · Vbase

Base quantities VAbase, Vbase Ibase, Zbase Figure 5.1 The concept of the per-unit (PU) method.

5.3

PU Method for Three-Phase Circuits

Now we will introduce the PU method for three-phase circuits. 5.3.1

Base Quantities with the PU Method for Three-Phase Circuits

When using the PU method for three-phase circuits, the line-to-line (l–l) base quantities and line-to-ground (l–g) base quantities are defined. Both must be strictly distinguished as the premise of three-phase circuit analysis for any investigation purpose. These base quantities are defined as follows: VA3ϕbase = 3 VA1ϕbase = 3 Vl −gbase Il − gbase = 3 Vl −lbase Il −lbase = 3 Vl −lbase Il −gbase ① Vl − lbase = 3 Vl −gbase 3 Il − lbase = Il −gbase

②

56

③

The base capacities (VA or MVA) and voltage (V or kV) are defined first, and then the base current [A], impedance Z [ohm], admittance [mho], etc. are dependently defined as follows: VA1ϕbase VA3ϕbase = Vl −gbase 3 Vl −lbase kVA3ϕbase MVA3ϕbase = = × 103 3 KV l − lbase 3 KV l −lbase

Il − gbase =

Zl − gbase =

④

Vl −gbase Vl − lbase 2 kV l −lbase 2 = = × 103 Il −gbase VA3ϕbase kVA3ϕbase

kV l −lbase 2 MVA3ϕbase 1 MVA3ϕbase = Yl −gbase = Zl − gbase kV l − lbase 2 =

Vl −lbase Il − lbase 1 Il −lbase = Yl − lbase = Zl −lbase Vl − lbase Zl −lbase =

⑤

57

⑥ ⑦ ⑧

The values of all the unitized quantities based on the l–l bases are written as variables with the suffix l – l, and those based on the l–g bases are written with the suffix symbol l – g according to the description rule.

79

80

5 The Per-Unit Method

5.3.2

Unitization of Three-Phase Circuit Equations

We know Eqs. (3.44) and (3.45b) for a simple generator model. These equations can be per-unitized as follows: Unitized generator equation Ea

Zs

Zm

Zm

Ia

E b = a2 E a − V b = Z m

Zs

Zm

Ib − V n

Ec = a Ea

Zm

Zs

Ic

Va

Zm

Vc

Vn Vn

V n = Z n I a + I b + I c = −Z n 3I 0 = −3Z n I 0 Ea = Zs = Ia =

Ea Vl −gbase

, Eb =

Eb Vl −gbase

,E c =

Ec Vl −gbase

,V n =

Vn

58

Vl − gbase

Zs Zm Zn ,Z m = ,Z n = Vl − gbase Zl − gbase Zl − gbase Ia Il − gbase

0

,I b =

V0

Ea − V 1 = 0 V2

Ib Il −gbase

, Ic =

Ic Il − gbase

Z0

0

0

I0

0

Z1

0

I1 +

0

0

Z2

I2

3Z n I 0 0 0

− V 0 = Z 0 I 0 + 3Z n I 0

or

59

E a −V 1 = Z 1 I 1 − V 2 = Z2 I 2 Obviously, the forms of the equations are preserved with unitization. A remarkable effect of the PU method can be seen in Chapter 6, when we explain the transformer circuit expression.

5.4

Base Quantity Modification of Unitized Impedance

In practical engineering, the quantities (MVA capacity, voltage, current, impedance, etc.) of individual members of a power system network (generators, transformers, transmission lines, etc.) are generally dictated by ohmic values or by PU values with different individual PU bases. On the other hand, in order to obtain a total combined system circuit for these members, MVAbases must be unified for the entire system first; furthermore, base voltages for each section must be selected to satisfy the turn ratio of the transformers (or, typically, to adopt rated voltages for each section). Accordingly, base impedances of individual equipment often must be changed to another base value. Therefore, we need to examine how to change the base quantities of impedances. There is an impedance element Z [Ω], which can be written as unitized impedance with two different base quantities: Z Ω = Z old Zold base = Z new Z new base where Zold base Ω =

Voldl −lbase 2 kV oldl −lbase 2 = VAold3ϕbase MVAold3ϕbase

Znew base Ω =

Vnewl −lbase 2 kV newl −lbase 2 = VAnew3ϕbase MVAnew3ϕbase

5 10

5.5 Unitized Symmetrical Circuit: Numerical Example

Accordingly, the formula to change both the base capacity and base voltage is Z new = Z old

Zold base = Z old Znew base

VAnew3ϕbase VAold3ϕbase

= Z old

MVAnew3ϕbase MVAold3ϕbase

kV oldl − lbase kV newl − lbase

Voldl −lbase Vnewl −lbase

2

2

5 11

The formula to change only the base capacity is Z new = Z old

VAnew3ϕbase = Z old VAold3ϕbase

MVAnew3ϕbase MVAold3ϕbase

5 12

And the formula to change only the base voltage is Z new = Z old

Voldl −lbase Vnewl −lbase

2

= Z old

kV oldl −lbase kV newl −lbase

2

5 13

As a general practice, one unified value of MVAbase must be selected, and then kVbase values of individual sections across each transformer must be determined. Through these processes, Zbase as well as Ibase are dependently determined for each section. Then, the derived base impedance is adopted as Znew base to obtain the unified circuit.

5.5

Unitized Symmetrical Circuit: Numerical Example

The basic theory of power system circuit analysis was been examined in Chapters 3 and 4. Now, we will demonstrate a numerical calculation in which a symmetrical equivalent circuit of a model system will be derived using the previously studied theories. Table 5.1 contains the diagram of a model system including overhead lines, cable lines, generating station, substations, and some loads. The subject of the exercise is to derive the symmetrical equivalent circuit of the model system shown in Table 5.1, given 1000 MVA is assigned as the base MVA quantity and the rated voltages of each section are to be selected as the base voltages of each section. The result of this exercise is shown in Table 5.2, and the calculation process is described next. 1) Determining PU base quantities MVAbase = 1000 MVA is given, and the rated voltages 22, 500, 66, 154, and 66 kV of Sections A, B, C, D, and E, respectively, are selected as the Vbase for each section. Then, the base quantities of each section of the model system are calculated by applying Eqs. (5.13a, b), and the result is summarized in Table 5.2. 2) Generator G1 The given reactance values are the percentage reactance based on the 625 MVA rated capacity of the generator. Accordingly, they should be modified into PU values (i.e. [pu]) based on 1000 MVA:

jx1 =

jxd = j0 25 ×

1000 = j0 400 pu 625

jxd = j0 29 ×

1000 = j0 464 pu 625

jxd = j1 56 ×

1000 = j2 495 pu 625

1000 = j0 368 pu 625 1000 = j0 192 pu x0 = j0 12 × 625 jx2 = j0 23 ×

81

5 The Per-Unit Method

Table 5.1 The model power system. Goal: To obtain a symmetrical equivalent circuit with the assigned base quantities of 1000 MVA and rated base voltages of each section.

tio LT

Section A G1

ine le l m) b a 5k :c L3 4kV 1 Open (15

nC

Sec

Section B L1 500 kV 90 km

Tr1

Tertiary 66 kV

Tr2

Power frequency: 50Hz

L2: overhead line (154kV 30 km) Tr3 66 kV

LD1 Load

RG1

Secondary NL1 Section D

Primary NGR1

L1 overhead line (90 km double circuits), nominal square 810 mm2 (diameter 38.4 mm) × 4 bundled conductors (averaged distance 20 cm), grounding wire OGW 410 mm2 (diameter 28.6 mm)

54

23.

57

A

.45

30

13.5

3

23.

18

TR3; transformer 250 MVA 154/66 kV %XP-s = 12% %X0 = 10%

NGR1; grounding resistance rating 200A

L3; cable line 15 km, leakage current 210A per phase, single-core cable (inductance ignored)

NL1; neutral compensating reactor rating 15 000 kVA

LT; 66 kV voltage regulator (reactor), 80 MVA (neutral ungrounded)

B

43.5

54

a

25.58

8

1

LD1; 66 kV load 200 MW pf = 0.8 where Z1 ≒ Z2

NGR2; grounding resistance rating 100A

3 38.0

13.

Tr1; transformer 1000 MVA 22 kV/500 kV, % impedance 13%

13.5

18

20

Tr2; transformer 750 MVA 500/154/66 k v % impedance P-to-S; 23% (750 base MVA) P-to-T; 18% (250 base MVA) S-to-T; 9% (250 base MVA)

C

c

b

NGR2 Section E

9 13 .45 10

x

RG1; neutral-ground resistance 100A

32

G1; generator rating 625MVA, 22 kV xd = 156% x’d = 29% x d = 25% x2 = 23% x0 = 12%

13.

82

Assuming He = 300 m Stray capacitances are to be taken into account.

L2; 154 kV 30 km Self-inductance Ls = 2.4 [mH/km] Mutual-inductance Lm = 1.1 [mH/km] Self-capacitance Cs = 0.0052 [μF/km] Mutual-capacitance Cm = 0.003 [μF/km]

3) Neutral resistance of the generator RG1 The neutral earthing with 100 A resistance means the resistance value for which a current of 100 A would flow when the phase voltage 22 3kV is charged to the generator’s neutral point: 3 kV × 103 = R × 100 127 pu = 262 pu 3R = 786 pu ∴ R = 127 Ω = 0 484

22

4) Transformer Tr1 See Chapter 6, Sections 6.2 and 6.3, for the calculation processes. The impedance of 13% is based on the rated 1000 MVA, 22/500 kV base, so the base quantities need not be changed. The transformer is solidly grounded at the high-tension neutral point. Accordingly, the neutral resistance in Table 6.1, Figure d, is zero. Then jx1 = jx2 = j0 13 pu jx0 = j0 13 pu P Zn

= 0 Z ex0 = ∞

Table 5.2 Equivalent circuit in symmetrical components. Base quantities for each section Section A

Section B

Section C

Section D

Section E

Base capacity

MVA3ϕbase

Base voltage

kVl-lbase

22 kV

500 kV

66 kV

154 kV

66 kV

kVl-gbase

22

500

66

154

66

Base current

Il − gbase =

Base impedance

MVA3ϕbase = 1000 MVA, MVA1ϕbase = 333.3 MVA

MVA3ϕbase × 103 3kV l −lbase

kV l − lbase 2 Zl − gbase = MVA3ϕbase

Section B Section A

Positive-seq.

E1

j0.13

jx′′d=j0.4 (jx′d=j0.464)

– j9.2

j0.368

– j9.2

j0.13 j0.192 786

Zex0 ⇋∞ – j25.0

3kV = 38 kV

3kV = 89 kV

3kV = 38 kV

1000 × 103 = 1155 A 3 × 500

1000 × 103 = 8748 A 3 × 66

1000 × 103 = 3749 A 3 × 154

1000 × 103 = 8748 A 3 × 66

222 = 0 484 Ω 1000

5002 = 250 Ω 1000

662 = 4 36 Ω 1000

1542 = 23 7 Ω 1000

662 = 4 36 Ω 1000

j0.115

j0.334

C Section j0.52

– j0.027

j0.387

j0.115 – j9.2

j0.334

j0.48

j12.49

– j17.9

– j0.027 j0.52

– j492 j0.48

j0.387

j0.115 – j9.2

j12.49 –j17.9

j0.303

– j0.027 j66.6 j0.334

j0.303 j0.387 – j31.6 Zex0 – j25.0 ⇋∞

Section E 3.20 + j2.40

Section D

Reactor

j0.162 Zero-seq.

3kV = 289 kV

1000 × 103 = 26240 A 3 × 22

j0.115

j0.13 Negative-seq.

3kV = 12 7kV

j1.83

–j321

j0.40 112.5

56.4 j∞

– j17.9

– j861

3.20 + j2.40

84

5 The Per-Unit Method

The high-tension side of the zero-sequence circuit is earth-grounded through jx0, while the low-tension side is open. 5) 500 kV double-circuit transmission line L1 Each phase of this line consists of four bundled conductors. Because all the conductor sizes and allocations are given, the inductance L and capacitance C can be calculated by applying the equations from Chapter 4. The equivalent radius of the four bundled conductors is as follows (see Eq. (4.31c)) r = 0 0192 m, 1 req = r n

n −1 r n =0

w=0 2 m 1

3

01924 × 0 204 = 0 1113 m

where w is the averaged distance of the multiple-bundled conductors. The radius of the overhead grounding wire is rx = 0 0143 m The height of the conductors from the imaginary datum plane (see Figure 4.3) is H0 = 300 m ha + 2H0 ≒ hb + 2H0 ≒

≒ hA + 2H0 ≒

≒ hx + 2H0

2H0 = 600 m

The averaged phase-to-phase distance within the same circuit is Dll = Dab Dbc Dca

1 3=

1

13 54 × 13 54 × 27 3 = 17 04 m

The averaged distance between one phase of circuit 1 and one phase of circuit 2, DlL, is DlL = =

1 DaA DaB DaC 3

1 DbA DbB DbC 3

18 × 23 3 × 32 45

1 3

= 23 87 × 22 14 × 23 87

1 DcA DcB DcC 3

23 3 × 20 × 23 3 1 3 = 23

1 3

1 3

18 × 23 3 × 32 45

1 3

1 3

28 m

The averaged distance between the phase conductor and the overhead grounding wire (OGW) is Dlx = Dax Dbx Dcx

1 3=

1

13 54 × 25 53 × 38 08 3 = 23 55 m

Finally, the height of the OGW is hx = 67 m a) Calculating inductance and impedance i) Self-inductance and impedance, including the earth grounding effect (but before modification with the OGW effect) Referring to Eqs. (4.6b) and (4.31a), and with req = 0.1113 and ha + 2H0 600 m, Ls = 0 4605log10

ha + 2H0 1 600 1 + 0 05 1 + + 0 05 1 + = 0 4605 log10 n 0 1113 4 req

= 1 781 mH km ∴ Zs ≒ Zaa ≒ Zbb ≒ ZAA ≒ ZBB ≒ = jX S = j2π 50 1 781 × 10 −3 = j0 559 Ω km

where f = 50 Hz

ii) Mutual inductance and impedance between phases of the same circuit, including the earth grounding effect (but before modification with the OGW effect) Referring to Eqs. (4.6b) and (4.18), and with Dll = 17.04 [m],

5.5 Unitized Symmetrical Circuit: Numerical Example

Lm = 0 4605log10

ha + 2H0 600 + 0 05 = 0 762 mH km + 0 05 = 0 4605log10 17 04 Dll

∴ Zm ≒ Zab ≒ Zbc ≒ ZAB ≒

= jX m = j2π 50 × 0 762 × 10 − 3 = j0 239 Ω km

iii) Mutual inductance and impedance between one phase of circuit 1 and one phase of circuit 2, including the earth grounding effect (but before modification with the OGW effect) With DlL = 23.28, Lm = 0 4605log10

ha + 2H0 600 + 0 05 = 0 700 mH km + 0 05 = 0 4605log10 23 28 DlL

∴ Zm ≒ ZaA ≒ ZaB ≒ ZaC ≒ ZbA ≒

= jX m = j2π 50 × 0 700 × 10 −3 = j0 220 Ω km

iv) Corrected impedance ZS, Zm, Z m with the OGW effect Referring to Eqs. (3.10a, b) and (3.11), the correction factor δax with the OGW effect must be subtracted from each impedance matrix element in the case of a single-circuit line. But for double circuits, the equation before the correction with the OGW effect is given by Eq. (1.19). Furthermore, the following equation is obtained as the OGW effect: Ix = −

1 Zxa Ia + Zxb Ib + Zxc Ic + ZxA IA + ZxB IB + ZxC IC Zxx

Thus, the correction factor δax is subtracted from all the elements of the 6 × 6 impedance matrix, in the same way as for the single-circuit line. Referring to Eq. (4.13c), and using rx = 0.0143 m and hx + 2H0 600 m, Lxx = 0 4605log10

hx + 2H0 600 + 0 1 = 2 22 mH km + 0 1 ≒ 0 4605log10 0 0143 rx

∴ Zxx = j2π 50 × 2 18 × 10 −3 = j0 697 Ω km Referring to Eq. (4.13c) and using hx + 2H0

600 m and DlL = 23.55 m,

Llx = Lax ≒ Lbx ≒ Lcx ≒ LAx ≒ ≒ Lxa ≒ hx + 2H0 600 + 0 05 = 0 698 mH km + 0 05 = 0 4605log10 = 0 4605log10 23 55 Dlx ∴ Zlx = j2π 50 × 0 698 × 10 −3 = j0 219 Ω km The correction factor with OGW is δ=

Zlx Zlx j0 219 2 = j0 069 Ω km = Zxx j0 697

Accordingly, all the self- and mutual-impedance elements must have the same correction factor subtracted, δ = j 0.069 Ω/km: Zs , Zm , Zm after correction by OGW effect Zs = j0 559− j0 069 = j0 490 Ω km Zm = j0 239− j0 069 = j0 170 Ω km Zm = j0 220− j0 069 = j0 151 Ω km This is the final calculated result for the transmission line self−/mutual-impedance matrix. v) Symmetrical impedance: Z1, Z2, Z0, Z0M Referring to Eq. (3.39a), Z1 = Z2 = ZS −Zm = j0 489− j0 169 = j0 320 Ω km Z0 = ZS + 2Zm = j0 492 + 2 × j0 169 = j0 830 Ω km Z0M = 3Zm = 3 × j0 151 = j0 450 Ω km

85

86

5 The Per-Unit Method

vi) Unitizing symmetrical impedance Zl–gbase = 250 Ω, line length l = 90 km: j0 320 × 90 = j0 115 pu 250 j0 830 × 90 = j0 303 pu Z0 = 250 j0 453 × 90 = j0 162 pu Z 0M = 250 Z1 = Z2 =

b) Calculating stray capacitance i) Capacitances Cs, Cm, C m before correction with the OGW effect The averaged height of the phase conductors is 1

1

h = ha hb hc 3 = 30 × 43 5 × 57 3 = 42 05 m The averaged distance between phases of the same circuit is Dll = 17 04 m The averaged distance between one phase of circuit 1 and one phase of circuit 2 is DlL = 23 28 m The equivalent conductor radius is req = 0 1113 m Referring to Eq. (4.29f), 0 02413 0 02413 = 0 00566 μF km 8h3 8 × 42 053 log10 log10 req D2ll 0 1113 × 17 042 2h 2 × 42 05 log10 log10 Dll 17 04 0 00180 μF km = 0 00566 × Cm = Cs 17 04 Dll log10 log10 0 1113 req Cs =

2h log10 0 02413 0 02413 SlL Cm = = DlL 8h3 8 × 42 053 log log10 log 10 10 req req D2lL 0 1113 × 23 282

2 × 42 05 23 28 0 00145 μF km 23 28 log10 0 1113

log10

Then, the capacitance matrix elements of Eq. (3.17) are Cs + 2Cm + 3Cm = 0 0129 μF km − Cm = − 0 00180 μF km − Cm = − 0 00145 μF km ii) Effect of the OGW on the capacitances Referring to Figure 3.5, one OGW (symbol x in the following equation) is to be added to the scenario in the figure. Given this condition, Eq. (3.13a) for Ia is to be modified with the addition of the term Cax (Va – Vx), where Vx = 0. The modified equation is Ia = jω Caa Va + Cab Va −Vb + Cac Va −Vc + CaA Va − VA + CaB Va − VB + CaC Va −Vc + Cax Va − Vx = jω Caa + Cab + Cac + CaA + CaB + CaC + Cax Va − Cab Vb − Cac Vc −CaA VA − CaB VB −CaC Vc ≒ jω

Cs + Cax + 2Cm + 3Cm Va −Cm Vb − Cm Vc − Cm VA − Cm VB −Cm VC

5.5 Unitized Symmetrical Circuit: Numerical Example

Comparing both equations, the resulting modification is Caa Caa + Cax. That is, it can be said that the selfcapacitance Caa becomes a little larger with the addition of OGW. In other words, referring to Eqs. (3.16b) and (3.17) and Eqs. (3.43b, c), the value of Cs would probably become a few percent larger with the modification of Cs Cs + Clx in comparison to the line with the same conductor allocation and without the OGW. (Details of this calculation are omitted in this book.) iii) Symmetrical capacitance The modification of the value of Cs with the addition of the OGW is ignored here (Clx ≒ 0) because it is only a few percent. Referring to Eq. (3.43c) and Figure 3.14, C1 = C2 = CS + 3Cm + 3Cm = 0 00566 + 3 × 0 00180 + 3 × 0 00145 = 0 0154 μF km C0 = CS = 0 00566 μF km C0 = 3Cm = 3 × 0 00145 = 0 00435 μF km ∴

− jX c1 =

1 −j = = − j206 × 103 Ω km jY 1 2π 50 × 0 0154 × 10 −6

= jX c2 =

1 jY 2

− jX c0 =

1 −j = = − j562 × 103 Ω km jY 0 2π 50 × 0 00566 × 10 −6

−jXc0 =

1 −j = = −j712 × 103 Ω km j Y0 2π 50 × 0 00435 × 10 −6

Incidentally, assuming a modification effect of 3% of Cs with the OGW in the previous calculation, C0 = Cs is also modified by 1.05 times, whereas C1 = C2 = CS + Clx + 3Cm + 3Cm is modified by 1.01 times. In other words, the positive- and negative-sequence capacitances C1, C2 are almost not affected by the existence of the OGW, whereas the zero-sequence capacitance C0 has a slightly larger value. iv) Unitizing symmetrical capacitance Unitizing the ohmic values with the 250 Ω base, − jX c1 = −j206 × 103 250 = −j824 pu km − jX c0 = −j562 × 103 250 = −j2248 pu km − jX c0 = −j712 × 103 250 = −j2848 pu km From the concentrated constants of the 90 km length, − jX c1 = −j864 × 90 = − j9 2 pu − jX c0 = −j2248 90 = −j25 0 pu − jX

c0

= − j2848 90 = −j31 6 pu km

Accordingly, the charging current per circuit per phase under normal conditions is I = 1 9 2 = 0 109 pu = 0 109 × 1155 A 90 km = 126 A 90 km 6) The transformer Tr2r (see Sections 6.2 and 6.3 for the calculation process) P − S Z = j0

23 pu 750 MVA base

P − T Z = j0

18 pu 250 MVA base

S − T Z = j0

09 pu 250 MVA base

These %IZ values are written in two different bases, where 750 MVA is the rated capacity of the primary and secondary windings, and 250 MVA is the rated capacity of the tertiary winding. The PU values must be modified into those for the 1000 base MVA:

87

88

5 The Per-Unit Method

1000 = j0 307 pu 1000 MVA base 750 1000 = j0 72 pu 1000 MVA base P −Δ Z = j0 18 × 250 1000 = j0 36 pu 1000 MVA base S −Δ Z = j0 09 × 250 P −S Z = j0

23 ×

Then j 0 307 + 0 72 − 0 36 = j0 334 pu 1000 MVA base 2 2 j 0 307 + 0 36 − 0 72 P −S Z + S − Δ Z − P − Δ Z = = − j0 027 pu 1000 MVA base SZ = 2 2 j 0 72 + 0 36 − 0 307 P − Δ Z + S − Δ Z − P −S Z = = j0 387 pu 1000 MVA base ΔZ = 2 2 PZ =

P −S Z + P −Δ Z − S − Δ Z

=

The equivalent circuit from Table 5.1 has been obtained. The derived impedance element SZ has a minus sign, so the element in the equivalent circuit is a series capacitive unit (condenser). 7) Neutral resistance NGR1 and neutral reactor NL1 For NGR1, the resistance value for 154/ 3 kV and 200 A is r0 =

154 3 × 103 = 445 Ω 200

For NL1, the reactance value for 154/ 3 kV and 15 000 kVA is jx0 = j

154

2

3 × 103 = j527 Ω 15000

Unitizing with the base impedance 23.7 Ω, 445 = 18 8 pu , 3r0 = 56 4 pu 1000 MVA base 23 7 j527 = j22 2 pu , j3x0 = j66 6 pu 1000 MVA base jx0 = 23 7

NGR1 r0 = NL1

8) 66 kV, 80 MVA reactor bank for voltage regulation LT The single-phase capacity of the reactor is 80/3 MVA. Then

jx1 = jx2 = j

Vl −g VA1ϕ

2

66 × 103 3 =j 80 × 106 3

2

= j54 45 Ω

Unitizing with the base impedance 4.36 Ω, jx1 = jx2 =

j54 45 = j12 49 pu 1000 MVA base 4 36

jx0 = ∞ The zero-sequence impedance is infinitely large because the neutral reactor is open. 9) 154 kV single-circuit transmission line L2 The constants Ls, Lm, Cs, Cm are given. The impedance for a 30 km length is jx1 = jx2 = j2πf Ls −Lm = j2π × 50 2 4 − 1 1 × 10 −3 × 30 12 3 = j0 52 pu = j12 3 Ω = j 23 7

5.5 Unitized Symmetrical Circuit: Numerical Example

jx0 = j2πf Ls + 2Lm = j2π × 50 2 4 + 2 × 1 1 × 10 − 3 × 30 43 3 = j43 3 Ω = j = j0 1 83 pu 23 7 and the capacitance for 30 km is jyc1 = jyc2 = j2πf Cs + 3Cm = j2π × 50 0 0052 + 3 × 0 003 × 10 −6 × 30 = j133 8 × 10 − 6 Ω −1 jyc0 = j2πfC s = j2π × 50 0 0052 × 10 −6 × 30 = j49 0 × 10 −6 Ω − 1 1 − j7474 = − j321 pu = −j7474 Ω = 23 7 j133 8 × 10 −6 1 − j20 408 = − j861 pu = − j20 408 Ω = − jxc0 = 23 7 j49 0 × 10 −6

∴ − jxc1 = − jxc2 =

The leakage current per phase for 30 km is 1 = 0 003 pu = 0 003 × 3749 A ≒ 11 2 A 321 10) 154 kV, 15 km power cable line (three single-core cables) L3 The line consists of three single-core cables, so that C0 = C1 = C2. The leakage current for 15 km is 210 A per phase. Then 154 × 103 3 − jxc1 = −jxc2 = − jxc0 = − j423 Ω 210 423 ∴ −jxc1 = − jxc2 = − jxc0 = −j = −j17 9 pu 23 7 Inductance is ignored. 11) The transformer Tr3 and NGR2 Modifying the base capacity from 250 to 1000 MVA, 1000 = j0 48 pu 1000 MVA base 250 1000 = j0 4 pu 1000 MVA base jx0 = j0 10 × 250

jP − S x = j0 12 ×

There is also the relation P − Sx > x0, because jx0 includes the parallel effect of Z ex 0 in the zero-sequence equivalent circuit of Case d in Table 6.1. For NGR2 at 100 A, 154 3 × 103 = 889 Ω 100 889 = 37 5 pu ∴ 3r0 = 112 5 pu r0 = 23 7 r0 =

12) 66 kV, 200 MW load At 200 MW with the power factor cos φ = 0.8,

= 4 0∠37 pu 1000 MVA base

250

0

0

φ 20

15

1 02 200 − j150 pu = P − jQ = 200 −j150 MVA = 1000 Z1 = 0 20 − j0 15 pu 1000 MVA base 1 = 3 20 + j2 40 pu ∴ Z1 = 0 20 − j0 15

89

90

5 The Per-Unit Method

Z1 ≒ Z2 is assumed for the load, although the impedance of a rotating load would generally be Z1 Z2. Table 5.2 shows the symmetrical equivalent circuit of the power system from Table 5.1, obtained by combining the results from steps 1–12. It is a circuit, instead of a connection diagram, for the given power system, which includes all the effective LRC constant elements. This is of course the first step of any system analysis.

91

6 Transformer Modeling 6.1

Single-Phase Three-Winding Transformer

Our purpose is to find the per-unit (PU) equivalent circuit of this transformer. The process may be divided into four steps. We will carry them out, step by step.

6.1.1

Fundamental Equations Before Unitization

A single-phase three-winding transformer can be written using the circuit shown in Figure 6.1a for power-frequency phenomena, where PN SN T N are the numbers of turns of the primary (P), secondary (S), and tertiary (T) windings, respectively. The transformer excitation current under a no-load condition can usually be ignored (the excitation impedance is large enough), except when core saturation is caused by abnormally high charging voltages. Therefore, the relation of voltages and currents in this transformer may be described by the following equation, in which the leakage impedances of three windings are taken into consideration: PV SV TV

ZPP = ZSP ZTP

ZPS ZSS ZTS

ZPT ZST ZTT

PI SI

①

TI

61 PI PN

+ SI SN + T I

TN

=0

ZPS = ZSP , ZPT = ZTP , ZST = ZTS

② ③

ZPP, ZSS, ZTT are the self-impedances of the primary (P), secondary (S), and tertiary (T) windings, and ZPS, ZPT, ZST are the mutual impedances between the three windings.

6.1.2

Determining Base Quantities for Unitization

Now we need to per-unitize the equation, sticking to the rule of per-unitization described in Chapter 5. Each base quantity for unitization of this transformer must be determined so that the following equations are satisfied: VAbase = P Vbase P Ibase = S Vbase S Ibase = T Vbase P Vbase S Vbase T Vbase = = PN SN TN P Ibase P N = S Ibase S N = T Ibase T N

T Ibase

① ②

62

③

In other words: ∗

1 Base capacities (VAbase) of the primary (P), secondary (S), and tertiary (T) windings are selected to have equal values (Equation ①).

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

92

6 Transformer Modeling PI

SI SN

PI

SV

Secondary PV

TI

TN

SI

SZ

TI

TZ

PN

Primary PV

PZ

NV

T V Tertiary

(a)

SV

TV

(b) Equivalent circuit, PU basis PZ

= ( P–S Z + P–TZ – S–TZ )/2 Z S = (P–SZ + S–TZ – P–TZ )/2 TZ = (P–TZ + S–TZ – P–SZ)/2 P–SZ, P–TZ, S–TZ, are given on the name-plate

Figure 6.1 Single-phase, three-winding transformer. ∗

2 Base voltages of the primary (P), secondary (S), and tertiary (T) windings are proportional to the turns ratio (transformation ratio) of the three windings (Equation ②). ∗ 3 Base amperes of the primary (P), secondary (S), and tertiary (T) windings are dependently determined as values of base capacity (∗1) divided by each base voltage (∗2). That is, the base ampere-turns of the primary (P), secondary (S), and tertiary (T) windings have the same value (Equation ③).

6.1.3

Unitizing the Original Equation

Let’s unitize the original Eq. (6.1) using the base quantities from Eq. (6.2) (recall that unitized quantities are indicated by an overbar, e.g. P V P V ): PV

= PV

P Vbase , S V

= SV

S Vbase , T V

= TV

T Vbase

VAbase VAbase , S I = S I S Ibase = S I P Vbase S Vbase

PI

= PI

P Ibase

= PI

TI

= TI

T Ibase

= TI

63

VAbase T Vbase

The equation for PV from Eq. (6.1) ① can be unitized as shown here: + ZPS S I S Ibase + ZPT T I T Ibase VAbase VAbase VAbase = ZPP P I + ZPS S I + ZPT TI V V P base S base T Vbase VAbase VAbase VAbase ∴ P V = ZPP P I + ZPS S I + ZPT 2 V V V V P base P base s base P base T Vbase PV

= PV

Z PP

P Vbase

PI

= ZPP

P I P Ibase

+ Z PS S I + Z PT

6 4a TI

TI

SV , T V

can be unitized analogously. Next, Eq. (6.1) ② can be unitized by Eq. (6.2) ③, as follows: PI PN P Ibase P N

+

SI SN S Ibase S N

∴ PI + SI + T I = 0

+

TI TN T Ibase T N

=0

6 4b

6.1 Single-Phase Three-Winding Transformer

Accordingly, VAbase 2 P Vbase

ZPP PV SV TV

=

ZPS

ZSP

VAbase P Vbase SVbase

ZTP

VAbase Z P base T Zbase

ZPP = ZSP ZTP

ZPS ZSS ZTS

ZPT ZST ZTT

VAbase V P base S Vbase

ZPT

VAbase V P base T Vbase

VAbase 2 S Vbase

ZST

VAbase S Vbase T Vbase

ZSS

VAbase V S base T Vbase

ZTS

ZTT

TI

65 ①

SI TI

②

+ SI + T I = 0 VAbase where Z PP = 2 P Vbase ZPS

SI

PI

PI

Z PS

VAbase 2 T Vbase

PI

VAbase = ZSP P Vbase S Vbase

VAbase P Vbase S Vbase

Z SP etc

In conclusion, Eq. (6.5) is the unitized equation of Eq. (6.1) with the base quantities of Eq. (6.3). In Eq. (6.5), the summation of the unitized vector currents of the primary (P), secondary (S), and tertiary (T) windings is zero. In other words, the unitized transformer circuit equations are able to satisfy Kirchhoff’s law. 6.1.4

Introducing the Unitized Equivalent Circuit

We have introduced Eq. (6.5) as the unitized fundamental equations of a transformer in which the vector sum of the currents is zero. Therefore, it would be useful if the equation could be written as the one-to-one corresponding equivalent circuit shown in Figure 6.1b. We can do that. It is clear that Figure 6.1b satisfies Eq. (6.5) ②. Then, if we define the impedances P Z, S Z, T Z in the figure so that the circuit satisfies Eq. (6.5) ①, the figure is the perfect equivalent circuit of the transformer that satisfies Eq. (6.5). Now let’s find such a condition. i) Given the condition TĪ = 0 (with the tertiary terminal opened), Figure 6.1b and Eq. (6.5) must coincide (with the tertiary terminal opened). Using TĪ = 0 in Eq. (6.5), we have PV

− S V = Z PP

PI

+ Z PS S I − Z SP

= Z PP + Z SS − 2Z PS

PI

+ Z SS S I

PI

On the other hand, using TĪ = 0 in Figure 6.1b, we have PV

− SV =

PI

PZ + SZ

PI

= P − S Z P I,

PZ + SZ = P−SZ

+ SI = 0

The following equation must be satisfied in order for the two equations to coincide with each other under the tertiaryterminal-open condition: P −S Z = P Z + S Z = Z PP

+ Z SS − 2Z PS

In the same way, the following conditions must be satisfied. ii) The required condition that Figure 6.1b and Eq. (6.5) coincide under the secondary-terminal-open condition (SĪ = 0) is P −T Z = P Z + T Z = Z PP

+ Z TT − 2ZPT

93

94

6 Transformer Modeling

iii) The required condition that Figure 6.1b and Eq. (6.5) coincide with each other under the primary-terminal-open condition (PI = 0) is S −T Z = S Z + T Z = Z SS

+ ZTT − 2Z ST

Summarizing, Figure 6.1b can be the precise equivalent circuit of the transformer by satisfying the previous three equations for the impedances. Accordingly, for the transformer equations, Tertiary terminal open P V − S V =

PZ + SZ

Secondary terminal open P V − T V =

PZ + T Z

Primary terminal open S V − T V = S Z + T Z

= P − S Z P I,

PI

PI SI

= P −T Z P I,

= S − T Z S I,

PI

+ SI = 0

PI

+ TI = 0

SI

+ TI = 0

66

where the definitions of impedances are Leakage impedance between P and S under the condition T I = 0 P −S Z = P Z + S Z = Z PP

+ Z SS − 2Z PS

Leakage impedance between P and T under the condition S I = 0 P −T Z = P Z + T Z = Z PP

+ Z TT − 2Z PT

6 7a

Leakage impedance between S and T under the condition P I = 0 S −T Z = S Z + T Z = Z SS

+ Z TT − 2Z ST

or, using the definition of P Z, S Z, T Z in the equivalent circuit in Figure 6.1b, PZ = SZ = TZ =

P −S Z + P −T Z − S −T Z

2 P −S Z + S − T Z − P −T Z

2 P −T Z + S −T Z − P −S Z

2

= Z PP + Z ST − Z PS −Z PT = Z SS + Z PT − Z PS −Z ST

6 7b

= Z TT + Z PS − Z PT − Z ST

Figure 6.1b with the impedances P Z, S Z, T Z becomes the unitized equivalent circuit of the transformer by defining the impedances as in Eq. (6.7b). The equivalent circuit of course satisfies Kirchhoff’s law with unitization. The impedances Z PP, Z PS, etc. are the self- and mutual-impedance (actually reactance), so the physical concept can be visualized from the winding structures, and the values can be estimated by engineers in transformer designs. The impedance P–SZ can be measured as the leakage reactance between the primary and secondary terminals under the tertiary-winding-open condition, and P–TZ, S–TZ can be measured similarly. On the other hand, PZ, SZ, TZ are the impedances defined by Eqs. (6.7a, b) to obtain the equivalent circuit in Figure 6.1b, and we cannot explain that they are the impedances of this or that part of the transformer from the viewpoint of transformer structure. However, transformers can be treated as a kind of black box by utilizing the previously defined equivalent circuits for power-frequency phenomena in power system networks. Incidentally, the resistances of the transformer windings are negligibly small, so Z can be replaced by j X or j ω L. ZPP = jXPP, ZPS = jXPS, as well as jP–SX, jP–TX, jS–TX, etc. and have positive values (reactance). However, one of jPX, jSX, jTX could have a negative value, just like a series capacitive element in the equivalent circuit. With regard to practical engineering, the percentage impedance drop voltages (%IZ) of individual transformers are indicated on their nameplates and are the percentage expression of leakage reactance P − S X, P − T X, S − T X Accordingly, utilizing these values, P Z, S Z, T Z can be derived from Eq. (6.7b). In practical engineering, the percentage value P–SX is usually given by the base MVA of the primary winding side, while P–SX, S–TX may be given by the base MVA of the tertiary winding side on a nameplate, so that the base value conversion is required to derive the equivalent circuit. This will be discussed in Sections 6.2.2 and 6.3.4. A transformer without tertiary winding can be treated by omitting T Z in the equivalent circuit of the three-winding transformer.

6.2

6.2 6.2.1

− − Δ-Connected Three-Phase, Three-Winding Transformer

− − Δ-Connected Three-Phase, Three-Winding Transformer Fundamental Equations Before Unitization

Figure 6.2b shows a typical three-phase three-winding transformer with − − Δ-connected windings, whose connection diagram with the terminal code names is printed on the nameplate as shown in Figure 6.2a. This connection is called the tertiary 30 lagging connection, because the phase angle of the low-tension bushing terminal a is 30 lagging in comparison with the bushing U and u terminals. The code names of all the bushing terminals have been changed in Figure 6.2b because the special names of the terminals are used only for analytical purposes, as shown here: primary (U, V, W

R, S, T), secondary (u, v, w

r, s, t), tertiary (a, b, c

b, c, a)

The tertiary terminal names a, b, c are intentionally changed by a 120 rotation, so the vector directions of newly named a, b, c terminals (the original c, a, b terminals, respectively) are rectangular to the phases R, S, T and r, s, t, respectively. Moreover, the quantities inside each tertiary winding (with suffix Δ) and the quantities outside each tertiary bushing (with suffix T) must be strictly distinguished from each other. There are three single-phase, three-winding transformers with the same ratings, whose winding connection is shown in Figure 6.1. These three single-phase transformers can be composed as one bank of three-phase transformer, as shown in Figure 6.2b, by simply connecting the bushing terminals. Accordingly, Eq. (6.8) is introduced as the fundamental equation of the three-phase transformer in Figure 6.2b: P Va

P Vn

S Va

S Vn

Δ Va P Vb S Vb − Δ Vb P Vc S Vc Δ Vc

0 P Vn S Vn 0 P Vn S Vn 0

ZPP ZSP ZΔP 0 = 0 0 0 0 0

ZPS ZSS ZΔS 0 0 0 0 0 0

ZPΔ ZSΔ ZΔΔ 0 0 0 0 0 0

0 0 0 ZPP ZSP ZΔP 0 0 0

0 0 0 ZPS ZSS ZΔS 0 0 0

0 0 0 ZPΔ ZSΔ ZΔΔ 0 0 0

0 0 0 0 0 0 ZPP ZSP ZΔP

0 0 0 0 0 0 ZPS ZSS ZΔS

0 0 0 0 0 0 ZPΔ ZSΔ ZΔΔ

P Ia S Ia Δ Ia P Ib S Ib

①

Δ Ib P Ic S Ic Δ Ic

where ZPS = VSP ,ZPΔ = ZΔP P Vn

= P Zn P In = P Zn P Ia + P Ib + P Ic = P Zn 3P I0

②

S Vn

= S Zn S In = S Zn S Ia + S Ib + S Ic = S Zn 3S I0

③

T Ia T Ib

Δ Ic

=

T Ic

Δ Vc

Δ Ib

−

Δ Ia

T Vb

=

T Vc T Va

④

Δ Ic

Δ Ib

Δ Va Δ Vb

Δ Ia

68

T Vc

−

⑤

T Va T Vb

P Ia P N

+ S Ia S N + Δ Ia

ΔN

=0

P Ib P N

+ S Ib S N + Δ Ib

ΔN

=0

P Ic P N

+ S Ic S N + Δ Ic

ΔN

⑥

=0

The submatrix of Equation ① is equal to Eq. (6.1). Equations ② and ③ correspond to the neutral connection of the primary and secondary windings; equations ④ and ⑤ correspond to the delta connection of the tertiary windings. Equation ⑥ is a base physical concept for any transformer.

95

96

6 Transformer Modeling

H-tension

M-tension

U

u

O

W

V

a c

o

w

L-tension

v b

(a) Phase name on name-plate (Phase-a of the tertiary winding is 30° lagging) Primary

Secondary S Ic

P Ia

S Ia

Tertiary TIa

S Ib

b SN turn

PN turn P Va

Vc

SVc

PVc

PVb

S Zn

TIc

ΔN turn

TVa

TVc TVb

(The tertiary Phase-a is 90° leading towards the primary phase-a)

= 3SI0

= 3PI0

ΔI b

c

SVn

SIn = SIa + SIb + SIc

PIn = PIa + PIb + PIc

ΔV b

SVb

SVa

PVn

PZn

ΔVa

ΔI a

a

PIb

I Δ c

Δ

P Ic

TIb

(b) The winding connection with the phase name for analytical purposes

PI1

PZ

SI1

SZ ΔZ

Positive- V P 1 seq.

SV1

× (–j) ΔI1=

– jTI1

ΔV1=

– jTV1

TI1 PZ = (P– SZ

–j

ΔZ = (P– ΔZ

PI2

Negativeseq.

SI2

SZ

ΔI2 = jTI2

ΔZ

PV2

PI 0

Zeroseq.

PZ

SV2

3PZn

PZ

PVn

PV0

PV0 – PVn

SZ

3SZn

ΔZ

SVn

SV0 – SVn ΔI0

ΔV0 =

ΔV2 = jTV2

×j TI 2 TV2

j

SI0 TI0 =

0

SV0

0

T I0

TV0

Zex0 ; Zero-sequence excitation impedance (c) The equivalent circuit in the symmetrical domain Figure 6.2

+ P– ΔZ –

S– ΔZ) /2

TV1 SZ = (P– SZ + S– ΔZ – P– ΔZ) /2

− − Δ transformer (low-tension winding 30 lagging connection).

TV0

+ S– ΔZ – P– SZ) /2

6.2

6.2.2

− − Δ-Connected Three-Phase, Three-Winding Transformer

Determining Base Quantities for Unitization

The base quantities for the unitization of this transformer are determined so that the following equations are satisfied: 1 VA3ϕbase = VA1ϕbase = P Vl − gbase P Il −gbase = S Vl −gbase S Il −gbase 3 = Δ Vl −lbase

Δ Il − lbase

= T Vl −gbase

T Il − gbase

=

3

1 3

T Vl − gbase

T Il −gbase

①

k1 k2

ΔVl −lbase = 3

T Vl −gbase

②

1 3

T Il − gbase

③

ΔVl −lbase = P Vl −gbase

=

PN

P Il − gbase P N

=

1 3

P Zl − gbase

=3

SN

=

Δ Vl − lbase ΔN

=

=

=

= S Il −gbase S N = Δ Il −lbase

T Il − gbase Δ N

S Zl −gbase =

Δ Zl − lbase

S Vl − gbase

P Vl −gbase

3

T Vl − gbase ΔN

④

k1

69

ΔN

⑤

k2

2

⑥

VA1ϕbase S Vl −gbase

2

⑦

VA1ϕbase Δ Vl −gbase

VA1ϕbase

T Zl −gbase

2

=

3 T Vl −gbase VA1ϕbase

2

=3

T Vl − gbase

2

VA1ϕbase ⑧

In other words: ∗

1 Base capacities (VAbase) of the primary (P), secondary (S), and tertiary (T) windings have the same value (Equation ①). ∗ 2 Base voltages of the primary (P), secondary (S), and tertiary (T) windings are proportional to the turn ratio (transformation ratio) of H(high)−/M(medium)−/L(low)-tension windings (Equation ④). This condition is satisfied simply by applying the rated voltages of each winding as base voltages. ∗ 3 Base amperes of the primary (P), secondary (S), and tertiary (T) windings are dependently determined as values of the base capacity (∗1) divided by each base voltage (∗2). In other words, the ampere-turn bases of the primary (P), secondary (S), and tertiary (T) windings are to have equal values (Equations ②③④⑤). ∗ 4 Base impedances are dependently determined (Equations ⑥⑦⑧).

97

98

6 Transformer Modeling

6.2.3

Using the Original Equation

Equation (6.8) can be divided by the appropriate base quantities, which are defined by Eq. (6.9), so that the unitized equations are derived as follows: PV a

Z PP Z SP V S n Z ΔP 0 V P n SV n = 0 V P n SV n 0

Z PS Z SS Z ΔS

PV n

SV a ΔV a PV b SV b − ΔV b PV c SV c ΔV c

Z PΔ Z SΔ Z ΔΔ

PIa SIa ΔI a

Z PP Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PIb

①

SIb ΔI b

Z PP Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PIc SIc ΔI c

where VA1ϕbase

ZPP

P Vl − g base

Z PP Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ = Z ΔΔ

ZPS

2

VA1ϕbase V P l − g base S Vl − g base

ZSP

VA1ϕbase P Vl − g base Δ Vl − g base

ZΔP

VA1ϕbase P Vl − g base S Vl − g base

ZSS

VA1ϕbase S Vl − g base

ZΔS

2

VA1ϕbase S Vl − g base Δ Vl − g base

and

ZPΔ

VA1ϕbase P Vl − g base Δ Vl − g base

ZSΔ

VA1ϕbase V S l − g base Δ Vl − g base

ZΔΔ

VA1ϕbase Δ Vl − g base

2

②

Z PS = Z SP ,Z PΔ = Z ΔP ,etc , + P I b + P I c = P Z n 3P I 0

③

+ S I b + S I c = S Z n 3S I 0

④

PV n

= P Zn P I n = P Zn

PIa

SV n

= S Zn S I n = S Zn

SIa

3 T Ia 3 T Ib = 3 T Ic 3 ΔV a 3 ΔV b = 3 ΔV c

ΔI c ΔI a

ΔI b

−

ΔI c

ΔI b TVb TVc TVa

PIa

+ S I a + ΔI a = 0

PIb

+ S I b + ΔI b = 0

PIc

+ S I c + ΔI c = 0

=

ΔI a TVc

−

TVa TVb

−1 0 1

0 1 −1

=

0 −1 1

1 −1 0 1 0 −1

ΔI a

⑤

ΔI b ΔI c

−1 1 0

TvV a TVb

⑥

TVc

⑦

6 10

− − Δ-Connected Three-Phase, Three-Winding Transformer

6.2

6.2.4

Symmetrical Equations and the Equivalent Circuit

The fundamental equations from Eq. (6.10) for this transformer can be transformed into the symmetrical domain to derive Eq. (6.11). The process of transformation is shown in the Supplement at the end of this chapter. PV 0

PV n

SV 0

SV n

ΔV 0 PV 1 SV 1 − ΔV 1 PV 2 SV 2 ΔV 2

0 0 0 0 0 0 0

Z PP Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PI0 SI0 ΔI 0

Z PP Z SP Z ΔP

=

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PI1 SI1

①

ΔI 1

Z PP Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PI2 SI2 ΔI 2

PV n

= P Z n P I n = 3P Z n P I 0

②

SV n

= S Z n S I n = 3S Z n S I 0

③

T I0 T I1

=

T I2

0 TV1 = TV2

0 jΔ I 1 − jΔ I 2

T I0

or

ΔV 0

jΔ V 1 − jΔ V 2

PI0

+ S I 0 + ΔI 0 = 0

PI1

+ S I 1 + ΔI 1 = 0

PI2

+ S I 2 + ΔI 2 = 0

or

− jT I 1 jT I 2

0 = ΔI 1 ΔI 2

0 − jT V 1 = jT V 2

6 11

④

ΔV 0

⑤

ΔV 1 ΔV 2

⑥

It is clear from Eq. (6.11) that mutual inductances do not exist between positive-, negative-, and zero-sequence quantities. These equations can be recast as follows, and the positive-, negative-, and zero-sequence quantities can be treated independently. For the positive sequence: PV 1

Z PP

Z PS

Z PΔ

PI1

SV 1

= Z SP

Z SS

Z SΔ

SI1

Z ΔΔ

ΔI 1

Z PΔ Z SΔ Z ΔΔ

PI2

ΔV 1

Z ΔP

Z ΔS

PI1

①

+ S I 1 + ΔI 1 = 0 ②

ΔI 1 = −jT I 1 ΔV 1 = − jT V 1

6 12a ③

For the negative sequence: PV 2 SV 2 ΔV 2

Z PP = Z SP Z ΔP

Z PS Z SS Z ΔS

SI2 ΔI 2

PI2

①

+ S I 2 + ΔI 2 = 0 ②

ΔI 2 = jT I 2 ΔV 2 = jT V 2

③

6 12b

99

100

6 Transformer Modeling

And for the zero sequence: PI0 PV 0 SV 0 ΔV 0

PV n

−

SV n ΔV n

Z PP = Z SP Z ΔP

Z PS Z SS Z ΔS

Z PΔ Z SΔ Z ΔΔ

PI0 SI0

①

T I0

ΔI 0

+ S I 0 + ΔI 0 = 0

②

= 0, Δ V 0 = 0

6 12c

P V n = 3P Z n P I 0

③ SV n

= 3S Z n S I 0

These are the unitized equations in the symmetrical coordinate domain. Positive-sequence Eq. (6.12a) is the same as Eq. (6.5) for a single-phase transformer, so the equivalent circuit must be the same as in Figure 6.1b by the same analogy described in Section 6.1. The negative- and zero-sequence quantities can be treated the same way. The symmetrical equivalent circuits corresponding to Eqs. (6.12a–c) can be written as shown in Figure 6.2c, where impedances P Z, S Z, Δ Z are defined by the equation PZ =

SZ =

ΔZ =

P −S Z + P −Δ Z − S − Δ Z

2 P −S Z + S − Δ Z − P − Δ Z

2 P − Δ Z + S − Δ Z − P −S Z

6 13

2

P −S Z = P Z + S Z = Z PP

+ Z SS − 2Z PS

P −Δ Z = P Z + Δ Z = Z PP

+ Z ΔΔ −2Z PΔ

S −Δ Z = S Z + Δ Z = Z SS

+ Z ΔΔ −2Z SΔ

The expression on the right side and the neutral grounding terminal in Figure 6.2c are strictly in a one-to-one correspondence to Equations ②③ of Eqs. (6.12a–c). Numerical check: we will use as a typical example a 1000 MVA, 500 kV transformer for substation use, with

••

Rated capacity H : 1000 MVA, M : 1000 MVA, L : 300 MVA Rated voltage 500 kV/275 kV/63 kV

With the percentage impedances P–SX = 14% (1000 MVAbase), P–ΔX = 44% (1000 MVAbase), and S–ΔX = 26% (1000 MVAbase), the equivalent circuit reactance of the transformer can be calculated as follows using Eq. (6.13), where SX takes capacitive values: PZ

= jP X = 16% = j0 16pu, S Z = jS X = 2% = −j0 02 pu , Δ Z = jΔ X = 28% = j0 28pu

Now let’s consider Eq. (6.12c) and the corresponding zero-sequence equivalent circuit. Because we have the equations = 0 and TĪ0 = 0, the Δ terminal is earth grounded, and the tertiary (T) terminal is open. This means the zerosequence current from the tertiary (T) outside circuit cannot flow into the (delta windings of the) transformer, although the zero-sequence current from the primary (P) or secondary (S) outside circuit can flow into the (delta windings of the) transformer. On the other hand, the equations for PV n and SV n in Eq. (6.12c) require us to insert 3PZ n and 3SZ n into the primary and secondary branches, respectively. Therefore, if the primary and secondary neutral terminals are solidly earth grounded, the zero-sequence current inflow from outside to the primary terminal flows partly into the delta winding (as the circulating current) and partly ΔV 0

6.3 Three-Phase Transformers with Various Winding Connections

out through the secondary terminal. If the neutral terminal on the secondary side is opened or highly resistive grounded (SZ n = ∞), all the zero-sequence inflow current from the primary side circulates through the delta windings. Equations (6.12a–c) and the equivalent circuit in Figure 6.2 as the expression for the three-phase, three-winding transformer in the symmetrical sequence domain are important because: 1 The 1–2–0 sequence circuits are mutually independent. 2 The unitized simple circuits allow the use of Kirchhoff’s law. 3 They contain common reactance for 1–2–0 sequence circuits. We know that an actual large power system can be expressed as a precise single large circuit, combining both lines and equipment with various rated capacities and voltages. It is satisfying to think that the most significant key factor of such a technique owes much to Eqs. (6.12a–c) and the equivalent circuit in Figure 6.2 for three-phase transformers, realized by the symmetrical coordinate transformation and appropriate unitization. The zero-sequence excitation impedance Z ex0 in Figure 6.2 will be discussed later.

6.3

Three-Phase Transformers with Various Winding Connections

Three-phase transformers with various winding connections and their unitized equations and equivalent circuits are shown in Table 6.1. Figure a in the table is the case of Figure 6.2. The equations and the equivalent circuits for transformers of other winding connections can be described in the same way as in Figure a. Autotransformers can be expressed by the same equivalent circuits.

6.3.1

Core Structure and Zero-Sequence Excitation Impedance

Table 6.2 shows a typical core structure of a transformer bank. Say we want to impose balanced three-phase voltages (i.e. positive- or negative-sequence voltages) from primary terminals. The induced fluxes from the balanced voltage charging are also balanced three-phase, so that any flux pass on the laminate steel core will not be saturated under normal voltage operation. This is why the excitation impedance Z ex can be ignored as very large impedance values under the condition of balanced three-phase voltages and currents. Next, let’s impose zero-sequence voltages from the primary terminals shown in the figures in Table 6.2. In the case of the transformers in Figures b and c, the created zero-sequence flux ϕ0 may be saturated because the return pass of ϕ0 is almost absent or of high magnetic reluctance (ϕ0 may be partly forced to flow into the air gap and tank), so flux saturation would be caused, and an abnormal temperature increase on the saturated flux pass would occur, if saturation by ϕ0 were to continue for a long time. Of course, the excitation current increases under saturation phenomena, which means the excitation impedance for zero-sequence voltage Z ex0 as a part of the equivalent circuit would have a smaller value. The transformers in Figure D have the auxiliary fourth magnetic pass of the laminated steel core by which zerosequence flux saturation can be prevented. The transformers in Figure a (one bank of three single-phase transformers) do not have such a limitation. Although the excitation impedance Zex can be ignored in most cases, Zex0 may occasionally need to be taken into account in the zero-sequence circuit as shown in Figure 6.2c. Figure 6.3 shows a typical example of no-load excitation current of a large-capacity transformer (1000 MVA, 500kv/ 275kv/63kv). Each phase current on the tertiary side is only a few amperes, which is 0.1% or less of the rated current and is negligibly small.

6.3.2

Various Winding Models

Figures (a)–(e) in Table 6.1 show typical transformer winding constructions and the symmetrical equivalent circuits. Due to the existence of the delta windings, the zero-sequence circuit of the delta-winding side is isolated from the Y winding-side circuits. Moreover, ΔZ is earth grounded in the zero-sequence equivalent circuit. These are the two important reasons to explain the role of the delta windings.

101

102

6 Transformer Modeling

Table 6.1 The equivalent circuit of various transformer windings.

Three-phase three-winding transformer Figure-a

PZn

_

positive-seq.

PZ

_

PZ

negative-seq.

zero-seq.

_ 3PZ n _ Z _P Z ex0

Figure-b

_ ΔZ

_

_

SZ

_ ΔZ

_

SZ

PZ

_ 3 SZ n

_ ΔZ

PZ

_ 3PZ n _ Z _P Z ex0

_

SZ

_ ΔZ

_

SZ

_ ΔZ

_

SZ

_ ΔZ

_

PZ

_

PZ

_ 3PZ n _ Z _P Z ex0

Figure-d

_

PZ n

_

_

Figure-c

_

SZn

SZ

Three-phase two-winding transformer

SZ n

_

SZ

_ ΔZ

_

SZ

_ ΔZ _ 3SZ n

_

SZ

_ ΔZ

Figure-e

_

PZ n

_

_

PZ

_ P–SZ

_

PZ

SZ

_

SZ

_ P–SZ

_ 3PZ n _ PZ _ Z ex0

_

SZ

_

PZ

_

SZ

_ P–SZ

_

PZ

_

SZ

_ P–SZ

_

PZ

_ Z ex0

_

SZ

Note: The reactances PX, SX, ΔX are calculated by Eq. (5.21), in which P-SX, P-ΔX, S-ΔX are given on the nameplate of the transformer as % IZ. Base MVA of P-SX, P-ΔX, S-ΔX should be unified before calculating with the equation.

Main (step-up) transformers for large power stations contain the windings in Figure b or d, and generators as well as local power station circuits are connected to the delta windings. High-voltage substation transformers have the windings shown in Figure a, b, c, or d. The transformer in Figure c has − connected windings and the third winding without an output terminal, although these windings are not fed out by the bushing terminals. 6.3.3

Delta Windings and Special Properties

Most three-phase transformers today have set-in delta windings. The advantages gained by adopting delta windings as general practice are summarized as follows; all of them can be explained by previously mentioned unique characteristics of the zero-sequence domain: a) To isolate the zero-sequence circuit of the delta-winding side from that of the -winding side. In other words, to intercept zero-sequence current I0 flowing across the different winding-side circuits. b) To reduce the largely zero-sequence reactance of the transformer, thus reducing and stabilizing neutral point voltages, and then reducing phase-over voltages under normal conditions or temporary overvoltage (TOV) during faults or any other power system disturbances. c) To reduce zero-sequence impedance and to stabilize the neutral (zero-sequence) voltages under normal conditions or during system faults. d) To intercept the through pass of DC or harmonic currents from the high-tension (HT) to medium-tension (MT) side equipment, in particular to protect generators or motors against abnormal operation or damage. e) To protect the transformer from damage that may be caused by zero-sequence in-flow current or by DC or 3nth (3, 6, 9,…) harmonic currents (overheating, vibration, over voltages, waveform distortion, etc.) in the cores, insulated windings, yokes, clamps, or any other structural part. Transformer delta-winding connections can intercept the through pass of DC or 3nth (3, 6, 9,…) harmonic currents. Let’s examine the behavior of harmonic currents flowing through delta windings.

6.3 Three-Phase Transformers with Various Winding Connections

Table 6.2 Typical structures of three-phase transformers.

3 × Single-phase transformers Figure-A I0 I0

ϕ

Shell type Figure-B I0

I0

ϕ

I0

Core type(three-poles)

I0

Figure-C I0

I0

I0

Core type(four poles) Figure-D I0

I0

I0

ϕ

ϕ

ϕ

ϕ

ϕ

*1

*2

*3

ϕ

ϕ

*1

Note: *1 The zero-sequence flux ϕ0 induced by the zero-sequence current I0 passes only through the core. Accordingly, the zero-sequence excitation impedance Z ex0 is usually ignored as quite a large impedance value. *2 The hatched part is apt to be saturated by zero-sequence current I0. Accordingly, the zero-sequence excitation impedance Z ex0 is relatively small, say Z ex0 ≒ 1.0–5.0 per unit. *3 When the zero-sequence current I0 is forced to flow into the transformer windings, the induced flux return pass could be through clamps, the inner wall of the tank, air gaps, r rather than the laminated core. The counter-electromotive force and the flux ϕ0 induced by the zero-sequence current I0 are small (because the magnetic reluctance of the return pass in air gaps is large). Accordingly, Z ex0 in the zero-sequence equivalent circuit is small, say Z ex0 ≒ 0.3–1.0 per unit.

Voltage phase a-b

Current phase-a

Current phase-b Current phase-c

Excitation current at tertiary bushing is a few amperes (average value) When a transformer is energized from one side where another side is in an open condition, power with the same amount of closed area in the hysteresis characteristic curve is required in order to establish flux through the core, which is lost as thermal consumption loss of the core. Figure 6.3 Excitation current of a three-phase transformer. Autotransformer 1000 MVA, 500 kv/275 kv/63 kv (core-type five core legs; measured value with no-load condition).

6.3.3.1

Case 1: Balanced Three-Phase nth Harmonic Currents

Balanced three-phase nth harmonic currents are Ia = Iejnωt n

Ib = Ie jn ωt − 120 = e −j120 Ic = Ie

jn ωt + 120

= e

j120

n

Ia = a2n Ia n

Ia = a Ia

6 14

103

104

6 Transformer Modeling

Stationary harmonics caused by solid-state power conditioners (inverters, converters, rectifiers, etc.) or very small waveform distortion from generators may be classified in this category. Transforming Eq. (6.14) into the 1–2–0 domain, I0 1 1 I1 = 1 3 I2 1

1 a a2

1 a2 a

Ia 2n

a Ia a n Ia

1 + a2n + a n 1 1 + a2n + 1 + a n + 2 = 3 1 + a2n + 2 + a n + 1

Ia

6 15a

Then, for the case of n = 3 m (0, 3, 6, 9, …) I0 Ia I1 = 0 I2 0

6 15b

and of n = 3 m + 1 (1, 4, 7, …) I0 0 I1 = Ia I2 0

6 15c

and of n = 3 m + 2 (2, 5, 8,…) I0 0 I1 = 0 I2 Ia

6 15d

That is, the behavior of balanced three-phase nth harmonic currents has the following characteristics: n = 1, 4, 7, …: behaves as positive-sequence currents n = 2, 5, 8, …: behaves as negative-sequence currents n = 0, 3, 6, 9, …: behaves as zero-sequence currents This is why transformers with delta windings can intercept 0 (DC), third, sixth, … harmonic currents. 6.3.3.2 Case 2: nth Harmonic Current Flow in Phase-a

Ia = Iejnωt

6 16

Ib = Ic = 0 A waveform-distorted single-phase load may cause harmonic currents of this type: 1 I0 = I1 = I2 = Iejnωt 3 In this case, negative- and zero-sequence currents of the nth order flow through the circuit.

6 17

6.4 Autotransformers

Case 3 : nth Harmonic Current of Synchronized Delay with Power Frequency

6.3.3.3

Here, Ia = Iejnωt Ib = Ie j nωt − 120 = a2 Ia Ic = Ie

j nωt + 120

6 18

= a Ia

Then I0 1 1 I1 = 1 3 I2 1

1 a a2

1 a2 a

Ia 0 a 2 Ia = Ia a Ia 0

6 19

The harmonic current in Eq. (6.18) consists of only positive-sequence components. In other words, the current including harmonics in Eq. (6.18) is the positive-sequence current (120 phase-balanced current) with harmonic distortion. 6.3.4

% IZ of a Three-Winding Transformer

Primary and secondary MVA ratings of typical substations using three-winding transformers are usually the same, while tertiary MVA ratings may be smaller (say, 30% or 35%). The %IZ described on the nameplate is usually given by the base MVA, as follows:

•• •

Between the primary and secondary P–SZ: primary and secondary base MVA Between the primary and tertiary P–TZ: tertiary base MVA Between the secondary and tertiary S–TZ: tertiary base MVA

In order to find the equivalent circuit of the transformer, the earlier PU impedance values must be modified into new PU values based on a single, common base MVA. In other words, P–TZ, S–TZ are modified to new PU values based on primary and secondary MVA capacity, and then the new P–TZ, S–TZ, as well as P–SZ, are put into Eq. (6.13) to find PZ, SZ, T Z. (The calculation is demonstrated in the next section.)

6.4

Autotransformers

Figure 6.4a shows a single-phase three-winding transformer, in which one terminal of the primary and one terminal of the secondary winding are connected. A transformer with this type of connection is called an autotransformer. The related equations of this transformer connection are written as follows: Pv

ZPP S v = ZSP ZTP Tv

ZPS ZSS ZTS

where Pv = PV − SV Sv = SV Tv = TV

ZPT ZST ZTT

Pi Si

①

Ti

6 20

Pi = PI

②

Si = PI

+ SI

③

Ti = TI

Equation (6.20) ① corresponds to Eq. (5.6), except that the symbols of variables V, I are replaced by v, i.

105

106

6 Transformer Modeling

P

I P

Primary

Pv

P

i

T

L

PV

T

i

T

L

S

i

L S

I

P

Tertiary TV

Tv

Pv

S

Sv

I

Secondary S

P

i

P

L

Series coil

I S

PV

Shunt coil

V

(a)

S

S

i

L

Sv

S

I

V

(b)

Figure 6.4 Autotransformer.

By substituting Eq. (6.20) ② ③ into ① and modifying it, the following equation is derived: PV SV TV

ZPP + ZSS + ZPS + ZSP ZSP + ZSS = ZTP + ZTS

ZPS + ZSS ZSS ZTS

ZPT + ZST ZST ZTT

PI SI

6 21

TI

Equation (6.21) is of the same form as Eq. (6.1), which means a transformer with rated values Pv, Sv, Tv, and Pi, Si, Ti can be applied as a transformer with new rated values PV, SV, TV and PI, SI, TI under the condition that the voltage insulation of the primary winding can withstand the voltage of the primary side network PV = Pv + Sv, and the current capacity of the secondary winding is for the secondary side SI = Pi + Si. In other words, Eq. (6.21) and Eq. (6.1) are equivalent only when the impedance matrix of Eq. (6.1) is replaced by that of Eq. (6.21). Accordingly, all the theory already described in this chapter for ordinal transformers can be applied to autotransformers. Further, the following equations with regard to MVA capacity are derived for an autotransformer whose MVA capacities on the primary and secondary sides are the same: Self-winding transformer capability MVAself = P v P i = S v S i where Pv Sv Tv = = =k N N P S TN

① 6 22

Auto-transformer capability MVAauto = P V

PI

=

Pv + Sv

Si = Sv

Pi + Si

= SV

where co-ratio α

PV

− SV

PV

SI

② =

Pv Pv + Sv

Comparing the ratio of the MVA capacities, MVAauto P V P I = = MVAself Pv Pi

Pv + Sv Pv Pi

Pi

=

Pv + Sv Pv

=

1 α

β

6 23

The MVA capacity with an autotransformer connection can be enlarged β times, but of course with appropriate design of the insulation and current capacity of the windings. The primary winding for Pv, Pi is called the series coil because the current from the primary side flows directly to the secondary side through this coil. The series winding

6.5 On-Load Tap-Changing Transformer (LTC Transformer)

coil is not earth grounded and is required to have an insulation level for the rated value of PV. The secondary winding (shunt coil) is required to have a current capacity of SI = Pi + Si. As a numerical check, Autotransformer 500 kV 275 kV 66 kV 500− 275 = 0 45 α= 500 β = 1 0 45 = 2 2 The capability of an autotransformer is equal to that of an ordinary transformer. Autotransformers are usually adopted for EHV or UHV substation uses mainly due to the economic benefit. The weight of an autotransformer can generally be reduced, compared to an ordinal transformer of the same MVA capacity. However, it must be noted that the percentage impedance PZ, SZ, TZ will become quite small without appropriate countermeasures during design. It can be determined by recalculating the equivalent impedance with Eq. (6.7a) based on the new impedance matrix from Eq. (6.21) instead of Eq. (6.20), although a description is omitted here. The tertiary winding may be omitted depending on the specification, as shown by Figure 6.4b. The equation for the single-phase autotransformer is the same as Eq. (6.1) for the ordinal transformer, and the situation is the same for a three-phase transformer. Therefore, the same symmetrical equivalent circuit can be adopted for a network calculation.

6.5

On-Load Tap-Changing Transformer (LTC Transformer)

The LTC transformer has an on-load tap-changer that is connected in series to the neutral-side terminal of primary (HT) windings, as shown in Figure 6.5a. The number of turns of the primary windings (tap ratio) can be changed (typically 0 to ±10%) under on-load operation. LTC transformers are typically installed at larger power-receiving substations in large cities or industrial areas with higher load densities. They are operated as key transformers to regulate voltages as well as reactive power around the station area, which are generally automatically controlled by AVR or AQR equipment in combination with shunt reactor banks and/or capacitor banks. As the turn ratio of the transformer is changed under tap-changing operation, the PU method explained in Section 6.2 should be modified. We will study in this section how we can treat the network, including the LTC transformer, using the PU method. The three-winding transformer with turn numbers pN:sN:tN shown in Figure 6.1a is described by Eq. (6.7a) using the base quantities in Eq. (6.2). Now, imagine a new condition where the primary turn is modified as pN pN . This condition can be expressed by the following equation: For turn numbersp N p Vbase pN

=

s Vbase sN

p Ibase p N pN

k

pN

sN

=

tN

t Vbase tN

and base quantities =

p

Vbase pN

= s Ibase s N = t Ibase t N = p Ibase

=

pN

①

Vbase p Ibase = = 0 9 to 1 1 p Vbase p Ibase

p

the primary voltage and current p V, p I are, for the PU value by the base quantities p Vbase , p Ibase

pN

turns base ,

pV

=

pV pI I = V I p basep p base

②

and for the PU value by the base quantities p Vbase , p Ibase p N turns base pV

=

pV p Vbase

pI

=

pI I p base

③

6 24

107

6 Transformer Modeling

HV MV LV

Core

108

On-load tap changer LTC Tap selector Pole selector

(a) on-load tap changer (b) Equivalent circuit

pI

Ideal-Tr

pIʹ

sI sZʹ

pZʹ

tZʹ

pVʹ

pV

pN

Turn numbers

pNʹ

Base quantities

pV

= k · pVʹ 1 · pI = pIʹ k

tI sV

pNʹ: sN

ʹ pVbase

tV

: tN

/ pN ʹ= sVbase /sN = tVbase /t N = t Ibase·t N

ʹ ·pNʹ= sIbase.sN pIbase

Figure 6.5 On-load tap-changing transformer.

The equivalent circuit for this condition is written on the right side of Figure 6.5b. The equivalent impedance pZ , sZ , may be slightly changed from the original pZ, sZ, tZ. The relation between both base quantities is

tZ

p Vbase pN

=

p

Vbase pN

,

p Ibase p N

= p Ibase

pN

6 25

Then pN p Vbase pI = k= = = 0 9−1 1 p Ibase pN p Vbase

6.6 Phase-Shifting Transformer

275 : 66kV

pZ

Tr1

sZ tZ

pI

pZʹ

pI

Tr2

sZʹ

pV pV

tZʹ

275kkV : 66kV k = 0.9–1.1

1:k

Ideal Tr

1 · pV k

k · pI

(b) Equivalent circuit

(a) Loop circuit Figure 6.6 Loop circuit.

Accordingly, pV

=

pI =

pV p Vbase pI

p Ibase

=

=

Vbase p Vbase

pV

p

p

Vbase

p Ibase

pI

p Ibase

p Ibase

=

1 k

=k

pV

6 26 pI

Equation (6.26) is the equation for the ideal transformer with turn numbers pN:pN , which can also be written as on the left side of Figure 6.5b. In other words, this figure is the equivalent circuit of the LTC transformer under the condition of the primary turn pN , because the figure satisfies all the equations in Eqs. (6.24)–(6.26). The tap ratio is generally specified to within 10%, so k = 0:9–1.1. The reactance p–sX, p–tX, s–tX for each tap number is generally indicated on the nameplate of the individual transformer. In conclusion, Figure 6.5 or Eqs. (6.24)–(6.26) with the ideal transformer must be adopted for circuit analysis of power-frequency phenomena such as power-flow analysis or P–Q–V stability analysis. Figure 6.6a is a typical example of a system including a loop circuit, in which transformer Tr2 with the tap changer may be operated with a different winding ratio than Tr1. The power-frequency phenomena of this system can be calculated using the circuit in Figure 6.6b. All the necessary equations can be derived from this equivalent circuit.

6.6

Phase-Shifting Transformer

In networks with loop circuits, the current flow for each line is distributed proportionally to the inverse ratio of the line impedances, which may not necessarily be an appropriate distribution in regard to the rated capacity limit of each line, P–Q–V regulation/stability, load flow economy, and so on. The phase-shifting transformer (or simply phase-shifter) can change the current flow distribution actively for this loop system. Needless to say, the phase-shifter can control the power flow of the meshed circuit network system as well as of the parallel loop circuit system if the installed location and the MVA bank capacity are planned appropriately. Figure 6.7a shows a typical connection diagram for a phase-shifter installed at a location where the primary- and secondary-side rated voltages are the same. As shown in the figure, the phase-shifting transformer bank consists of one three-phase regulating transformer and one three-phase series-connected transformer, whose windings are connected in series directly through the outer bushings or oil ducts.

109

110

6 Transformer Modeling

1Va 1 Vb 1V c

1Ia

2Ia

Nʹ

1Ib

1Ia - 2Ia

N

Vʹc

1Ic

vb

vc

vʹa iʹa

ia

va

iʹc vʹc

ib

ng

2Ic

Iʹa

Vʹb

n

2Ib

Vʹa

1Va - Vaʹ

= 2Va

1Vb - Vbʹ

= 2Vb

1Vc - Vcʹ

= 2Vc

nʹ iʹb vʹb

ic ig

Regulating Tr

Series Tr

(a)

1 Vc

1I 0

2Vc

2 Va

k (1Vb - 1Vc)

2I0

ig

1V0

1Va

j 3 k •1V1

1I1 1Vb

2Vb

Zerosequence circuit

2V0

1V1

1I1

2I1

Positivesequence circuit

2V1

j 3 k 2I1 •

1I c

2I c

j 3 k •1V2

1I2 1V2

2I2

2V2

2I a

1I b

k (2Vb - 1Vc)

1I a

2I b

Negativesequence circuit

j 3 k •2I2

(b)

(c) Equivalent circuit

Figure 6.7 Phase-shifting transformer (for a solidly grounded system).

6.6.1

Fundamental Equations

The following equations can be derived from Figure 6.7a: 1 V abc

N

=

vabc n

N 1 I abc − 2 I abc + niabc + ng ig = 0

V abc vabc = ' N' n

ig where ig = ig ig

V abc = 1 Vabc − 2 V abc

6 27 N

1 −1 1

−1 ϕ

−1 1

va vb = vc vabc

va vb vc

ia ib ic

vabc

iabc

2 I abc + n iabc = 0

−1

ia ib ic

ϕ

iabc

1 =

−1 1

−1 1

6.6 Phase-Shifting Transformer

Accordingly, 1 −kϕ

1 Vabc

= 2 V abc ,

k=

n N 1, N n

I abc + kg ig = 1 + kϕ

2 I abc ,

kg =

ng N

6 28

or 2 Va

= 1 Va −k 1 Vb − 1 Vc

1 Ia

+ kg i g = 2 Ia + k 2 Ib − 2 Ic

2 Vb

= 1 Vb −k 1 Vc − 1 Va

1 Ib

+ kg ig = 2 Ib + k 2 Ic − 2 Ia

2 Vc

= 1 Vc −k 1 Va − 1 Vb

1 Ic

+ kg i g = 2 Ic + k 2 Ia − 2 Ib

6 29

The regulating transformer generally has taps and polarity switches (typically of the on-load-changing mechanism) so that k can be specified as a plus or minus value; typically. k = +0.1 to 0 to −0.1. The vector diagrams for the phase-shifter under the balanced three-phase condition are shown in Figure 6.7b. Now we transform Eq. (6.28) into symmetrical components: 1 − k a × ϕ × a −1 1 I 012

+ kg

1 V 012

= 2 V 0012

ig 0 = 1 + k a × ϕ × a −1 0

a2 − a

a × ϕ × a −1 =

2 I 012

6 30 −j 3

= a −a

2

j 3

Then 1 V0

1 I0

= 2 V0

1 + j 3k

1 V1

= 2 V1

1 −j 3k

1 V2

= 2 V2

①

+ K g i g = 2 I0

1 I1

= 1 −j 3k

2 I1

1 I2

= 1 + j 3k

2 I2

②

6 31

From this equation, the symmetrical equivalent circuit in Figure 6.7c is derived, which can be used to analyze powerfrequency phenomena. Figure 6.8 is another example of a phase-shifting transformer, installed at a location where the HT windings are solidly grounded and the LT windings are resistive grounded. The transformer can control power flow as well as change the voltages.

a

a

b

b

c

c

Figure 6.8 Phase-shifting transformer (for impedance neutral grounded system).

111

112

6 Transformer Modeling

6.6.2

Application for Loop-Circuit Lines

Figure 6.9 shows a positive-sequence equivalent circuit of a loop network in which the previously described phaseshifter is applied at the receiving terminal r. The positive-sequence equations are Route 2

Route 1

V s1 − V rr1 = I 1 Z 1 ②

V s1 −V rr1 = I 1 Z 1 V rr1 − V r1 = − j 3k V rr1 1 V r1 = I 1 Z 1 ∴ V s1 − 1 + j 3k

Z 1 ≒ jx1

①

6 32

③

Z 1 = jx1

For the case where a phase-shifter exists in route 1, I1 = S s1 = Ps1 + jQs1 = V s1 I1∗

1 = ∗ Z1

1 1 V r1 V s1 − Z1 1 + j 3k V 2s1 −

6 33

1 V s1 V ∗r1 1 −j 3k

For the case where a phase-shifter does not exist in route 1 (equivalent to the case of k = 0), S s1, k = 0 = Ps1 + jQs1 =

1 V 2s1 −V s1 V ∗r1 Z ∗1

6 34

The difference in the apparent power in these two cases is ΔS s1 = S s1 − Ss1, k = 0 = ≒ 3k 2 − j 3k

∗ − j 3k V s1 V ∗r1 3k 2 −j 3k V s1 Vr1 = 2 +1 Z1∗ 3 k↖ Z ∗1 1 −j 3k

V s1 V ∗r1 − jx1

3k + j3k 2 3k V s1 V ∗r1 ≒ V s1 V ∗r1 x1 x1

=

①

6 35

With V s = Vs ∠δ ,V r = Vr ∠0 then ΔPs1 + ΔjQs1 ≒

3k V s1 V ∗r1 x1

cosδ + j sin δ

②

The approximation in the process is based on k = +0.1 to −0.1. The power flow of Eq. (6.35) on route 1 can be controlled by the existence of the phase-shifter. In the network, it can be assumed that line 1 is longer than line 2, so the reactance of line 1 is larger than that of line 2 (jx1 > jx2), whereas the line capacity of line 1 is far larger than that of line 2. The power flow of this circuit may not be appropriate, because the power-flow distribution of line 2 increases despite the far smaller current capacity.

s

route-2

r

Zʹ1 = jxʹ1 ·

route-1 ·

Vs1

j 3k · Vrr1

Z1 = jx1 Vrr1

Vr1

load

Figure 6.9 Application for a loop circuit (positive-sequence circuit).

6.7 Woodbridge Transformers and Scott Transformers

The phase-shifter can solve the inadequate current distribution by changing the current-flow distribution with time. Furthermore, adequate current-distribution control can reduce the angular difference across points s and r so that the phase-shifter can improve system stability to some extent.

6.7

Woodbridge Transformers and Scott Transformers

Electric train loads are typically fed by a single phase circuit, so they are fatally unbalanced loads from the viewpoint of a three-phase network. Woodbridge-type and Scott-type transformers have special windings in order to mitigate the load imbalance.

6.7.1

Woodbridge Transformers

The connection diagram of a Woodbridge transformer is shown in Figure 6.10(a). This is a three-phase, three-winding transformer whose S (secondary) and T (tertiary) windings are delta connections and have the same design, except that their winding terminal connection is specially arranged. The voltage-transformed power is fed out from the A terminal as well as from the B terminal. The voltage boost-up transformer Tr2 (turn ratio 1: 3) is also connected to the IB pIa

3 IB HT

pVa

pVc

s Ic pIc

p Vn

sVc sVb

sVa

tVa

s Ib pVb

s Ia

pIb

IA

LT-B

t Ib tVb

VB (60 kV)

60 (kV) 3

tVc

B-terminal load (single phase)

t Ic

1: 3

t Ia

VA (60 kV)

(a) Three-phase connection

LT-A A-terminal load (single phase) –IB

p Ic

=

– 3 IA

1 · (– 3 I – I ) A B 3k

IB

jIA

(b) Vector diagram p Ib =

1 · ( 3 I –I ) A B 3k

IA

–IB Figure 6.10 Woodbridge transformer.

p Ia =

3 IA

1 3k

· (2 IB)

113

114

6 Transformer Modeling

B terminal. The A and B terminals would typically be fed out to feeding sections 1 and 2 (the up- and down-train lines, for example), both of which have a similar load in terms of capacity as well as load pattern over time. The equations are pn

p Ia

+ s n s Ia + t Ia = 0

∴k

pn

p Ib

+ s n s Ib + t Ib = 0

k

p Ib

+ s Ib + t Ib = 0

p Ic

+ s n s Ic + t Ic = 0

k

p Ic

+ s Ic + t Ic = 0

pn

p Ia

+ s Ia + t Ia = 0

where pn

turns of the primary winding

sn

turns of the secondary and tertiary windings , k =

6 36

pn sn

A-terminal current IA = s Ic − s Ib = t Ic − t Ib

①

6 37

B-terminal current 3IB = s Ic − s Ia + t Ib − t Ia

②

Eliminating tIb, tIc from Eqs. (6.36) and (6.37), s Ib − s Ic

=

−k 2

p Ib − p Ic

6 38

Using these equations, the current IA, IB can be written as a function of the primary current as follows: IA =

k 2

p Ib − p Ic

3IB = s Ic − s Ia + t Ib − t Ia =

k 2p Ia − p Ib − p Ic 2

namely IA k 0 = 3IB 2 2

IA k 0 = 3IB 2 2

1 −1

1 −1

−1 −1 −1 −1

p Ia p Ib p Ic

1 1 1

1 a2 a

1 a a2

6 39a p I0 p I1 p I2

=

k 0 2 0

−j 3 3

−j 3 3

p I0 p I1 p I2

Accordingly, IA 3 0 k = IB 0 2

−j 1

j 1

p I0 p I1 p I2

6 39b

6.7 Woodbridge Transformers and Scott Transformers

Then IA = − j

3 k 2

p I1 − p I2

①

3 k p I1 + p I2 IB = 2 Inverse equations 1 IB + jI A p I1 = 3k

••

1 IB − jI A 3k

②

p I2

=

I p 0

=0

p Ia

=

1 2IB 3k

p Ib

=

1 3k

I p c

=

1 − 3IA −IB 3k

3IA − IB

6 39c

③

The equations indicate the following: = 0: The zero-sequence current is always zero regardless of the load conditions. = 0: This is with the condition jIA = IB. That is, the negative-sequence currents can be reduced to zero if the magnitudes of the A-terminal load and B-terminal load are equal, or it can be reduced if both load values are similar over time, whereas pI2 would become a large value of pI1 = pI2 if IA = 0 or IB = 0. pI0 pI2

6.7.2

Scott Transformers

The connection diagram for a Scott transformer is shown in Figure 6.11, where the turn ratios of the primary windings are n1:n2:n3 = 1:2: 3 and of the secondary windings are nT:nM = 1:1. This transformer is equivalent to a Woodbridgetype transformer from an application’s viewpoint. Referring to Figure 6.11, the following equation is justified.

p Ia pVa

HT

IB

p Ib

LT-B

pVb

VB

n2

p Ic

n1

n3

n3

pVc

VA LT-A

A-terminal

Figure 6.11 Scott transformer.

IA

B-terminal

115

116

6 Transformer Modeling

nM IM = n3 Ib – n3 Ic nT IT = n2 Ia – n1 Ib + Ic where n1 n2 n3 = 1 2

6 40a

3 and nT = nM

Then nM IM = Ib − Ic 3n1 nT IT = 2Ia − Ib + Ic n1

6 40b

Therefore k 0 = 2 2

1 −1

−1 −1

IM 3 0 k = IT 0 2

−j 1

j 1

IM 3IT

where

Ia k 0 Ib = 2 0 Ic

−j 3 3

−j 3 3

p I0 p I1

6 41a

p I2

p I0 p I1

6 41b

p I2

3 n1 k = ,nT = nM nM 2

Equations (6.41a, b) are entirely equal to Eqs. (6.39a, b). So, the functional characteristics of the Scott transformer are equivalent to those of the Woodbridge transformer, while the coil structure is asymmetrical and quite different. The neutral point of the Scott transformer may be solid- or resistive-grounded. The Scott three-phase transformer was invented by a Westinghouse engineer, C. F. Scott, in the late 1890s, and was pervasive in industrial applications of single-phase loads – in particular, in rail load applications. The Woodbridge transformer became an alternative in the 1970s. The typical applications of these transformers are railroad substations, where single-phase loads are usually divided into two groups of similar load capacity and load patterns (typically, to and from lines) in order to minimize the negative-sequence current component.

6.8

Neutral Grounding Transformer

Larger industrial factories in fields such as chemicals, metal furnaces, steel, electrical appliances, etc. often adopt powerreceiving transformers with delta windings for lower-voltage sides, in order to isolate the factories’ zero-sequence circuits from that of the power system network. However, the factories’ in-house networks must be neutral grounded to obtain in-house insulation coordination. Neutral grounding transformers are adopted in such cases as special transformers to earth–ground the zero-sequence circuit of the in-house networks. Figure 6.12a shows the winding connection diagram. Figure 6.12c is the equivalent circuit marking the first core pole, where all equations are unitized. The related equations are va − va = Z S I A vb −vb = ZS IB

va = Zex IA − IC ①

vc −vc = ZS IC

vc = Zex IC − IB

V A = va − vb VB = vb −vc VC = vc −va

vb = Zex IB − IA

③

② 6 42

6.8 Neutral Grounding Transformer

vaʹ IA

–vbʺ

–vcʺ

IB

VA VB VC

IC

vcʹ vʹa

vʹb

vaʺ

vʹc

vbʺ

vbʹ VA

–vbʺ vʹa

vcʺ

VB

IC

vʹc g

–vaʺ

vʹb

–vcʺ

–vaʺ

R

VC (a)

IA

vaʹ

Zs ⇋ jxs IA – IC

(b)

IC

Zex ⇋ ∞

vaʺ

Zs : short-circuit impedance Zex : excitation impedance

(c) Figure 6.12 Neutral grounding transformer.

VA = ZS IA + Zex 2IA − IB −IC = ZS IA + 3Zex IA − I0

6 43a

= ZS + 3Zex IA − 3Zex I0 ∴ VA = ZS + 3Zex IA − 3Zex I0 VB = ZS + 3Zex IB −3Zex I0

6 43b

VC = ZS + 3Zex IC − 3Zex I0 Transforming this into symmetrical components, V0 = ZS I0 V1 = ZS + 3Zex I1

6 44

V2 = ZS + 3Zex I2 and V0 = ZS = jxS I0 V1 Z1 = Z2 = = ZS + 3Zex ≒ ∞ I1 Zex ≒ ∞ Z0 =

6 45

117

118

6 Transformer Modeling

This equation shows that jx0 = jxs; but on the other hand, jx1 = jx2 ≑ ∞ because the excitation reactance values are quite large. This is the principle of the neutral grounding transformer. Figure 6.12b is the vector diagram explaining the function of the neutral grounding transformer. If such a transformer is installed and is earth grounded through resistance R, the zero-sequence impedance becomes Z0 = 3R + jXs. Recall that third-harmonic quantities behave like zero-sequence quantities, so this transformer can also bypass the third-harmonic current to earth. Incidentally, if a generator exists in the in-house network, the generator is generally neutral earth grounded through resistance (the resistive value of 3 R = 50–200 Ω is typically selected so that the phase-to-grounding fault current becomes 100–200 A). Accordingly, a neutral grounding transformer may be omitted.

6.9

Transformer Magnetic Characteristics and Inrush Current Phenomena

6.9.1

Hysteresis Characteristics

Now we will discuss a transformer as a machine based on electromagnetism with the variable quantities v(t), i(t), φ(t). The flux loop pass structure of a transformer or any rotating machine is composed of laminated steel plate cores (see Figure 6.13). Nothing is as precious as the core material, because of its remarkable characteristics: ① it has high flux density with a magnetic azimuth for the direction of production; ② it experiences small steel core loss (eddy current loss); ③ it has wide, stable temperature characteristics; ④ it is easy to process; and ⑤ it is relatively cheap. In Figure 6.14a, if the power source is connected at the primary coil such that excitation current iex(t) flows and v1(t) is established, flux φ(t) will be induced in the core, and the values of v1(t), iex(t), φ(t) will not be affected by the presence or non-presence of the secondary-side disconnected winding. In the core, an eddy current is caused by flux φ(t), and power is consumed: this is called core loss or eddy current loss Pex = i2ex Rex . The current iex(t) is called the excitation current of the core. Core loss caused by the flux is the nature of the core: in other words, excitation current iex(t) is flowed in order that the voltage v1(t) is established by connecting the power source. The core-loss power is typically 1% or less of the rated kVA capacity. However, if the power source is too weak to supply the core-loss, the primary winding voltage cannot be built up to 1.0 pu. The following equation is for a two-winding transformer based on Faraday’s law, which says that voltage is proportional to the changing speed of the flux: v1 t dφ t v2 t = = n1 dt n2

6 46a

We can assume that the primary source-side voltage v1(t) is sinusoidal, so v1 t = V1 cos ωt φt =

t −∞

v1 t dt = Φsin ωt + Φk

6 46b

The equations explain that if v1(t) is sinusoidal, then φ(t)/dt and v2(t) are sinusoidal; φ(t) is also sinusoidal, but the angular time is 90 delayed to v1(t), and DC offset Φk may be included. However, the excitation current iex(t) cannot be sinusoidal, as shown in Figure 6.14b–6.14c. Under the normal operation of a transformer, the primary current i1(t) consists of primary load current i1lord(t) plus excitation current iex(t), or i1(t) = i1lord(t) + iex(t). However, the excitation current iex(t) corresponding to the core-loss is quite small, so it will be negligible for most of the analysis. Next, we will study conditions where the secondary winding is disconnected or absent, as in the case of a reactor. If an alternate power source of an arbitrary angular speed ω = 2πf is connected to the primary winding, φ(t)−i(t) characteristics as shown in Figure 6.15a can be measured. This is called the hysteresis curve or i- φ saturation curve of the core. The instantaneous operating point (φ(t), i(t)) moves one round per cycle of the alternate current. The flux density and the total flux values are saturated to a specific value. Further, if the source voltage frequency increases, the looped curve becomes larger. One-cycle power loss caused in the core is calculated as the looped area size of the hysteresis curve, as explained by Eq. (6.47). So, if the source frequency increases, the core-loss also increases.

6.9 Transformer Magnetic Characteristics and Inrush Current Phenomena

Center of core

Center of core

LT-coil Barrier (paper press-board) HT-coil

Barrier (paper press-board)

LT-coil

HT-coil

(a) Multilayer cylindrical winding method

(b) Disc-winding method

Bus duct (LT-side) Steel-core clamp plate Upper yoke Conservator Neutral bushing Transformer tank (inside wall magnetically shielded)

Cable head (HT-side)

Cooler units

Binding tape HT-coil LT-coil Side core Main core (laminated silicon steel plate) Core support plate (c) Structure Figure 6.13 Three-phase transformer.

Let’s calculate the thermal loss energy consumed in the core per cycle: 2π

Ploss θ dθ =

0

∴

2π

2π

v1 t 0

2π

Ploss θ dθ =

0

iex t dt =

n1 0

2π

dφ t dt

2π

iex t dt =

n1 iex φ dφ

0

n1 iex φ dφ looped area size of hysteresis curve

6 47

0 2π

The left side 0 n1 iex φ dφ is the looped area size of the hysteresis curve, because it is the one-cycle integration of n1 iex(φ). Then, the equation shows the reason that core loss is given by the area of the hysteresis loop. As shown in Figure 6.165b, during the time when v1(t) and iex(t) have the same polarity, supplied power from the source is consumed

119

φ(t)

v(t) iex(t) (a) Core and coil structure φ(t) v(t)

Charging voltage v(t)

iex(t)

iex(t) φm

residual flux Φresidual

t

(b) Waveform of v(t), φ(t), iex(t)

(b) Residual flux

Figure 6.14 Wave forms of source voltage v1(t), core flux φ(t), and excitation current iex(t).

φ

vi + iex

Wb/m3

d

d e

e

1.0 v

25 Hz 50 Hz 150

100

50

50

100

dφ c

150 AT/m

1.0 a

(a) Hysteresis curve

b

φ

iex

Flux Linkage

Power Transformer Inrush Current Magnetizing Current

(d) Typical waveform of inrush current Figure 6.15 Hysteresis curve (φ − i curve) and inrush current.

g

c

f

b a g

(b) Voltage, flux, and excitation current versus iron loss

Core Characteristics

(c) Hysteresis v(t)-i(t) characteristics

t

f

6.9 Transformer Magnetic Characteristics and Inrush Current Phenomena

0 π/2 π

0 π/2 π

in the core; and on the other hand, during the time when v1(t)and ϕ iex(t) have a different polarity, energy is supplied from the core to K1 the source. ϕC C The excitation current iex(t) is typically within 1% of the rated current whenever a transformer is operated under ordinal vol- ν, ϕ ν ϕB B ϕA iC tages1.0 ± 0.1. It may be a few amperes even for large transformers such as 1000 MVA, 500 kV class. Such small excitation 3π/2 loss is due to a significantly smaller hysteresis loss, or, in other ϕA A i O 2π ωt 0 π /2 π words, to the thin hysteresis curve characteristics of the core. Then, the excitation impedance Zex vrms/iex under nominal volϕdc iA tages is quite large, so it can be ignored for most commercial frequency circuit analysis, including transient calculations. This is why Zex is ignored in the positive/negative-sequence equivalent circuit given in Table 6.1. Figure 6.3 shows typical waveforms K2 for excitation current, including fifth, seventh, and eleventh harmonic current components. The third, sixth, and ninth harmonics are not included for a transformer with a delta-winding connection. Figure 6.16 DC bias flux of a transformer with DC current Hysteresis characteristics may be explained as a v(t)-i(t) satura- component. tion curve instead of a φ(t)-i(t) curve. The reason is as follows. If the source voltage at the primary winding v(t) is sinusoidal, then using Eqs. (6.46a, b), φ(t)/dt is also sinusoidal; and φ(t) is sinusoidal while the phase angle of φ(t) is 90 degrees ahead of v(t). In other words, we have the equations φ t = 2Φrms sin ωt and v t = 2Vrms sin ωt − 90 . Φrms and Vrms have a one-to-one correspondence, so the hysteresis curve can be written as a φ-i curve or v- i curve. The hysteresis v-i curve may be assumed to be a composite of three straight lines, as shown in Figure 6.16, because the core-loss is reasonably small so the width is quite thin. The knee-point voltage of a transformer is typically around 1.2 PU based on the rated voltage. If the source voltage exceeds the knee point, the excitation current iex becomes quite large and has a sharp-pointed waveform. Consequently, the core temperature will rise quickly due to increased core-loss v(t) i(t)dt. Such continuous overvoltage operation should be avoided. 3π/2 2π

3π/2 2π

ωt

ωt

iC

iA

6.9.2

Flux DC Offset Magnetization

Figure 6.16 shows simplified core-saturation characteristics of a transformer core flux φ versus current i. The knee points K1, K2 of a transformer core hysteresis curve are generally designed at 115–120% of the rated flux values as part of the design philosophy. In the figure, the core flux range φA and the current waveform iA show normal operation without a DC current component idc, where the flux locus is symmetrical at the central point A. If the DC current component idc appears and is superposed, the center point may shift from point A to B. Further, if idc increases or if the flux AC component is increased, the operating zone will exceed the saturation knee point K1 or K2 due to the large excitation current and cause a steep, distorted waveform. φC and iC in the figure show such a condition. The DC bias flux is typically caused by the following: i) ii) iii) iv)

Transformer inrush current Transformer ferroresonance Generator self-excitation Phenomena caused in the thyristor (single-phase half-bridge rectifier)

6.9.3

Residual Flux

If a pure iron bar is magnetized by wound coil with an AC or DC current source, it becomes magnetized iron, and the magnetized flux will not disappear even if the current source is removed. Such flux is called residual flux. For the same reason, whenever a transformer is charged, residual fluxes with DC constant values Φra, Φrb, Φrc remains after the transformer is discharged. Now we will examine three-phase residual flux in detail.

121

122

6 Transformer Modeling

Various loads in the adjacent wide local power system Line impedance

Transformer Br1

Surge absorber

Br2 Load Stray capacitance

Power source VT1

VT2

Figure 6.17 Typical single-line diagram of a substation.

In Figure 6.17, whenever a charged transformer with a load condition is disconnected at time top0 due to the associated HV-side breaker tripping, transient phenomena of secondary-winding (LV) side currents i2a(t) , i2b(t) , i2c(t) , voltages va(t), vb(t), vc(t), and core fluxes φa(t) , φb(t), φc(t) are caused in the existing loop circuit of the LV-side even though the primary-winding (HV) side currents i1a(t), i1b(t), i1c(t) became zero at top0 when the breaker tripped. The phenomena continue for time duration Δt until top1 (Δt = top1 − top0) when the transient voltages and currents disappear and the fluxes become constant values Φra, Φrb, Φrc (∵the flux is given by the voltage integration, which becomes a constant value when the voltage becomes zero). Such free-attenuating transient phenomena are also caused when a transformer with a no-load condition (the load side breaker Br2 is kept open and the conductor is in dead-end condition in Figure 6.17) is disconnected due to the associated load-side breaker Br1 tripping. There is secondary-side stray capacitance of the load-side dead-end three-phase conductors (and capacitance of the surge absorber or power cable), so the loop circuit with outer-side total capacitance C and the leakage inductance of the transformer LV-windings still exists after the breaker trips. Therefore, the residual fluxes are the standstill fluxes that remain at the end of the transient phenomenon. 6.9.4

Transformer Inrush Current

When a transformer with its LV-side (secondary-side) circuit open is switched in from the HV-side (primary source side), a transient current 3–10 times larger than the rated transformer current is caused for several cycles or for up to several seconds. This is called transformer inrush current (see Figure 6.15d). Inrush current is caused when instantaneous fluxes exceed the knee point of the core hysteresis curve for some length of time just after initial charging, as shown by Figure 6.15c or Figure 6.16. The currents can be quite large and unbalanced three-phase with sharp-pointed (harmonic-rich) waveforms. Because saturation occurs only every half cycle, each phase’s current increases during one polarity. Once the core saturates, however, the winding inductance (the definition of φ/i) is greatly reduced, and only the resistance of the primary-side windings and the impedance of the power line limit the current. Inrush currents generated in a transformer HV-winding are widely distributed into all the HV-side neighborhood circuit branches due to the law of circuit inverse-ratio impedance, and voltage sag of probably 10% to 25% will also be caused. Consequently, the inrush phenomena can cause damage or problems for primary-side substation equipment, HV-side connected loads, and the transformer. Typical damages or problems include the following:

•• ••

The transformer: Mechanical buckling or mechanical looseness of the winding structures. Motors and generators: Torsional torque, twisting, and eccentricity/rotor and stator insulation damages caused by thermal heating. Load breakers and fuses: Overcurrent tripping and/or under-voltage tripping. Fuse cutoff breaking. Protective relays and control systems: Malfunction and/or abnormal operation caused by overcurrent or undervoltage.

Now we will study the case where a transformer with residual fluxes Φra, Φrb, Φrc is charged at angular time θcl by the associated breaker’s three-phase simultaneous closing operation. At time ωt = θcl+, the source voltages are

6.9 Transformer Magnetic Characteristics and Inrush Current Phenomena

va t = V cos ωt vb t = V cos ωt − 2π 3

6 48

vc t = V cos ωt + 2π 3 And the core fluxes are φa t =

t θcl va

t dt = Φ0 sin ωt−sin θcl + Φra

φb t =

t θcl vb

t dt = Φ0 sin ωt − 2π 3 − sin θcl − 2π 3

+ Φrb

φc t =

t θcl vc

t dt = Φ0 sin ωt + 2π 3 − sin θcl + 2π 3

+ Φrc

6 49a

Or, by modification, φa t =

t θcl va

t dt = Φ0 sin ωt + Φra − Φ0 sin θcl

φb t =

t θcl vb

t dt = Φ0 sin ωt −2π 3 + Φrb −Φ0 sin θcl − 2π 3

φc t =

t θcl vc

t dt = Φ0 sin ωt + 2π 3 + Φrc −Φ0 sin θcl + 2π 3

6 49b

As shown by Eq. (6.49a), φa(t), φb(t), φc(t) are the standstill values Φra, Φrb, Φrc before the breaker closing time θcl, but so-called prospective fluxes created by the source voltages are added at t = θcl+. Then, using the modified Eq. (6.49b), DC-bias components appear as the initial values of the inrush transient term. As a result, fluxes of one or two phases become quite large and exceed the knee point of the hysteresis curve and cause severe inrush currents. The transient flux has a theoretical maximum of 2Φ0 + Φr. This is the reason for transformer inrush current phenomena. Obviously, the residual fluxes Φra, Φrb, Φrc and the associated angular time θcl of the breaker closing are the two primary parameters that affect the transient phenomena, and the inrush current values as well as the waveforms can differ due to chance.

6.9.5

Transformer Inrush Current-Limiting Switching Control

Transformer inrush current-limiting technologies based on the breaker three-pole simultaneous switch control (TPSC) method have been considered an unsolved issue for a many years. The author established a new TPSC technology that can significantly limit transformer inrush current, and transformer inrush current-limiting devices (TPSCD) based on the new technology have been successfully adopted in over 100 commercial transformer banks in Japan since 2013. The principle of the new technology is introduced here. We will first examine a current breaker. Three-phase load currents or fault currents are tripped by a breaker due to natural current-zero tripping. The electrical current-zero times of currents ia, ib, ic are different from each other, so the electrical trip times of the breaker’s three poles are also different from each other, whereas the mechanical disconnection time of each pole is the same. But when a no-load transformer is tripped, very small excitation currents (a few amperes or less) are flowing before the trip. Therefore, the breaker can cut off the small three-phase excitation currents iaex, ibex, icex simultaneously just after the Ltotal breaker’s mechanical breaking time top0. Such tripping is Tr Br2 Br1 called chop-mode current breaking. So, we can conclude that no-load transformer breaking is actually three-pole simultaCtotal R neous tripping, instead of cascade tripping. Now the transformer and the connected power grid are balanced three-phase, and the three-pole current break by V V the associated breaker happens at a single time top0, so the transient behavior caused after the breaker trips must be balanced three-phase. Thus the transient voltages va(t), vb(t), vc(t), currents ia(t), ib(t), ic(t), and fluxes φ(t), φb(t), φc(t) are always balanced three-phase. Then, the transient phenomenon can be interpreted as a Single attenuation mode Oscillatory attenuation mode positive-sequence loop circuit break, as shown in Figure 6.18. The waveforms of the fluxes may be in free Figure 6.18 A transformer secondary-winding side-loop oscillatory attenuating mode or simple attenuating mode, circuit model.

123

124

6 Transformer Modeling

depending on the L, C, R constants of the loop circuit. If the stray capacitance Ctotal becomes relatively larger, they are apt to become oscillatory mode waveforms (see Figure 6.18). Figure 6.19 shows the transient waveforms of voltages and fluxes measured at a 220(Y)/110(Δ)/11 kV(Δ) transformer of a hydropower station. The voltages are the measured waveforms of the station three-phase v-trs, and the fluxes are the calculated waveforms with the built-in digital calculation processing function of the TPSCD, based on the mathematical continuous integration of the three-phase voltages. The instantaneous quantities va(t), vb(t), vc(t) and φa(t), φb(t), φc(t) are also shown with three-phase phasor notation, and they become rotating equilateral triangle phasors over transient time top0 to top1. The voltage equilateral phasors attenuate while rotating at the free angular velocity and finally disappear at top1, or at θop1 in angular time. The flux equilateral phasors also attenuate while rotating; the rotating speed gradually slows down and finally stops at top1 or θop1 when the voltages disappear, because the fluxes are always synchronized with the integration of the voltages. In this case, the equilateral flux phasors turned twice and the polarity changed twice while maintaining the balanced three-phase conditions, and the residual flux phasors shrank to values of 30%. The transient time interval Δt = top1 − top0 was approximately 30 ms. The author conducted similar field tests for more than 100 commercial banks of 6–275 kV transformers. All of the recorded data showed that the transient time duration Δt = top1 − top0 is distributed to the value within 3–100 ms, and the phasor Φr of the residual fluxes is distributed to the value within 0.25–0.8 pu. These figures for the distributed bands of Δt = top1 − top0 and Φr may be concluded to be common values, as the result of many recorded commercial field experiences. In conclusion, transformer residual fluxes are always balanced three-phase if the source-side grid is balanced threephase. This conclusion has been statistically Table 2.2 proven based on many commercial trials as well as theoretical experiments by the author and his colleague.

va

φc

vb

vc

φa

φb

Voltage (Measured wave form)

1pu

Flux (Calculated wave form)

1pu

top0

va

top1

vb

Voltage phasor

1pu

Flux phasor

1pu

vc

φc

φa φb

0

10

20

30

40

50

Time [ms] Figure 6.19 Transient phenomenon caused just after an unloaded transformer tripped (recorded voltages and flux waveforms at a 60 MVA, 220(Y)/110(Δ)/11 kV(Δ) transformer of a hydropower station).

6.9 Transformer Magnetic Characteristics and Inrush Current Phenomena

Now, the three-phase residual fluxes are written by the below equilateral triangle phasors. Φra = Φr ∠θop1 Φrb = Φr ∠ θop1 − 2π 3

①

Φr =

Φrc = Φr ∠ θop1 + 2π 3

2 Φra 2 + Φ2rb + Φ2rc ② 3

6 50

Figure 6.20 illustrates this TPSC technology. The method can be referred to as synchronization of three-phase residual flux-phasors to three-phase source flux-phasors. During the many commercial field trials conducted by the author and his colleague, inrush currents disappeared at all the transformer banks where the TPSC technology was adopted. Figure 6.21 shows one of commercial trials of a TPSC device installed in a transformer (24,7 kVA,11 kV/543 V) at a chemical company. When the control was switched out of service, a maximum of 9 pu of inrush current and 23%u of voltage sag were recorded. When the switch control was in service, the currents and voltage sag were successfully limited within 2 pu and 1%, respectively.

The flux phasors at the time of breaker arc extinction

The residual flux phasors (standstill)

The prospective (initial charging) flux phasors

10

30 Inrush current Voltage sag

8

24 18

6 Limiter in service 4

12

2

6

0

0

30

60

90

120

150

Voltage sag [%]

Inrush current [pu] (1pr-1510 Ams)

Figure 6.20 Three-pole simultaneous switch control based on three-phase phasors.

0 180

Phase angular difference of the residual flux and the initial energizing fluxes [degrees]

Δθ = θopi – θclose Figure 6.21 Transformer inrush phenomena and a switch control: measured data (transformer 24,7 kVA, 11 kV/543 V for a chemical factory).

125

126

6 Transformer Modeling

Supplement Transformation from Equation 6.10 to Equation 6.11 a) Transformation from Equation 6.10 ① to Equation 6.11 ① PV a PV b

PV n

−

PV c

or

PV n

PIa PIb

= Z PP

PV n

P V abc − P V n

SIa

+ Z PS

PIc

= Z PP

P I abc

ΔI a

SIb

+ Z PΔ

SIc

ΔI b

1

ΔI c

+ Z PS S I abc + Z PΔ

Δ I abc

Multiplying the left-hand side by a, a P V abc − a P V n = Z PP a P I abc + Z PS a S I abc + Z PΔ a

Δ I abc

PV n

0 0

P V 012

PV 0

∴

PV 1

P I 012

PI0

PV n

−

0 0

PV 2

S I 012

= Z PP

PI1

SI0

+ Z PS

PI2

SI1

2

Δ I 012

ΔI 0

+ Z PΔ

SI2

ΔI 1 ΔI 2

This is the symmetrical equation in regard to primary voltages. The secondary and tertiary side equations are derived analogously. b) Transformation from Equation 6.10 ③④ to Equation 6.11 ②③. This is self-explanatory. c) Transformation from Equation 6.10 ⑤ to Equation 6.11 ④ Equation 6.10 ⑤ is

3T I abc =

0 1 −1

−1 0 1

1 −1 0

3

Δ I abc

Then

1 a T I 012 = a T I abc = 3

0 1 −1

−1 0 1

1 −1 0

a

−1

Δ I 012

1 = 3

0 ΔI0 a − a2

ΔI1 =

a2 − a Δ I 2

0 ΔI 0 j ΔI 0 − j ΔI 0

This is Equation 6.11 ④. Equation 6.11 ⑤ can be derived from Equation 6.10 ⑥ analogously.

4

127

7 Fault Analysis Based on Symmetrical Components 7.1 Fundamental Concepts of Fault Analysis Based on the Symmetrical Coordinate Method Three-phase circuit analysis including load-flow analysis and short-circuit analysis is conducted using the symmetrical coordinates method (symmetrical components), both to calculate small circuits or computational simulations of large-scale power systems, or for steady-state analysis or transient analysis. This is because the circuital constants (impedance, capacitance, etc.) of generators, transformers, load circuits, overhead transmission lines, power cables, and buses – in other words, the impedance and capacitance of all components of large and small power systems – can be specified and given as symmetrical (012-domain) constants, or as dq0-domain constants. The circuital characteristics of all individual lines and equipment can be specified with symmetrical 012-constants or with dq0-constants, which may be a kind of derivatives of symmetrical 012-constants. Therefore, three-phase circuits that become the bases of circuit analysis can be written using the symmetrical components. The αβ0 (Clarke components) method may be partly utilized for very special circuit analysis (typically, overvoltage calculations across a breaker second pole tripping), but the circuital bases of such special analysis are symmetrical constant circuits. The symmetrical components method is essential not only as a circuital calculation tool, but also as a tool to specify electrical characteristics of lines and equipment. Figure 7.1 shows the process flow of fault analysis using symmetrical components. The first step is to transform the power system connection and fault condition into the 012-domain circuit. The second step is to find the circuit solution in the 012-domain. The last step is to inverse-transform the solution into the abc-domain.

7.2

Line-to-Ground Fault (Phase-a to Ground Fault: 1ϕG)

Recall that a three-phase power system can be drawn as a connection diagram in the abc-domain and not as a circuit in the abc-domain. Let’s examine phase-a to ground short-circuit fault (say, phase-a 1ϕG) at an arbitrary point f on a transmission line. Figure 7.2a shows the partial connection diagram of the power system including point f, where virtual abc-terminals branch out at fault point f. The power system before the fault at point f can be drawn as a symmetrical circuit in the 012-domain, as shown within the dashed lines in Figure 7.2b, where the corresponding virtual terminals branch out at point f.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

128

7 Fault Analysis Based on Symmetrical Components

a1) System equation Sabc (Vabc, Iabc, Zabc)

b1) Fault condition at point f f fabc ( f Vabc , f Iabc)

transform ; × a

transform ; × a b2) Fault condition f f012 ( f V012 , f I012)

a2) System equation S012 (V012, I012, Z012)

c)

a-b-c domain

0-1-2 domain

Solution on 0-1-2 domain Inverse-transform ; × a –1

d)

Solution on a-b-c domain

a-b-c domain

Figure 7.1 Procedure for fault analysis.

Power network Point m

Fault point f Virtual terminal f Ia f Ib f Ic f Vc

f Va V f b

S R

R : arc-resistance (or may be tree-resistance) (a)

m Z1

E1′

I1′

Z′ mV1′ f 1

f Z1″

I1″

E1′′ f I1 f Z1

m Z2

I2′

Z′ m V2′ f 2

1 f V1

2 f Z2″

I2″

S f I2

3R f Z2

f V2

m Z0

I0′

m V0′

f Z0′

f Z0″

I0″ f I0 f Z0

Figure 7.2 Phase-a line-to-ground fault.

f V0

(b)

7.2 Line-to-Ground Fault (Phase-a to Ground Fault: 1ϕG)

The related equations are f V1

= E1 − f Z1 I1 = E1 − fZ1 I1

f V2

= − f Z 2 I2 = − f Z 2 I2

f V0

= − f Z 0 I0 = − f Z 0 I0

f I1

= I1 + I1

f I2

= I2 + I2

f I0

= I0 + I0

①

② 71

f Z1

=

f Z1

f Z1

=

f Z2

=

f Z2

f Z2

=

f Z0

=

f Z0

f Z0

=

f Z1f Z1 f Z1

+ f Z1

f Z2f f Z2

+ f Z2

f Z0f f Z0

Z2

③

Z0

+ f Z0

where Z1: positive-sequence impedance looking at the circuit at point f f Z1 : positive-sequence impedance looking at the left side at point f f Z : positive-sequence impedance looking at the right side at point f (the // symbol means parallel impedance values) 7.2.1

Condition Before the Fault

The outgoing currents f Ia, f Ib, f Ic on the virtual abc-terminals at point f are zero before the fault, so the corresponding symmetrical sequence currents f I1, f I2, f I0 are also zero: f Ia

= f Ib = f Ic = 0

f I0

= f I1 = f I2 = 0

72

= f V2 = 0

f V0

In the negative- and zero-sequence circuits, because no power source exists and the virtual terminals are open, all quantities including f I2, fV2, f I0, fV0 at point f are zero before the fault. In the positive-sequence circuit, we have f I1 = I1 + I1 = 0. Accordingly, for the balanced three-phase load current flowing through point f before the fault, E1 − E1 = I1 = − I1 Z f 1 + f Z1

73

Therefore, the voltage at point f before fault fE1 is f E1

7.2.2

= E1 − f Z1 I1 =

f Z1 f Z1

+ f Z1

E1 +

f Z1 f Z1

+ f Z1

E1

74

Phase-a to Ground Fault

Now, the phase-a conductor to ground short-circuit fault at point f means the phase-a virtual terminal is earth grounded (switch S is closed through arc resistance R) at point f, while the phase-b and phase-c virtual terminals remain open in Figure 7.2. Therefore, = R f Ia f Ib = f Ic = 0

f Va

75

129

130

7 Fault Analysis Based on Symmetrical Components

Transforming this equation from the abc-domain into the 012-domain, + f V1 + f V2 = R f I0 + f I1 + f I2 2 2 I + f 0 a f I1 + a f I2 = f I0 + a f I1 + a f V0

f I2

76

=0

Utilizing the relation a2 + a = −1, f I0

= f I1 = f I2

f V0

77

+ f V1 + f V2 = 3R f I0

This is the equation in the 012-domain transformed from Eq. (7.5). The condition in Eq. (7.7) can be expressed as the drawing circuit shown on the right outside the dashed line in Figure 7.2b. Figure 7.2b is the equivalent circuit for phase-a to ground fault (phase-a 1ϕG) by symmetrical components. 7.2.3

Voltages and Currents at Virtual Terminal Point f in the 012-domain

Now, phase-a to ground faults are realized by switching on the virtual switch S in Figure 7.2b: in other words, connecting the outside impedance f Z2 + f Z0 + 3R to virtual terminals ① ② of the positive-sequence circuit. The current flowing through terminals ① ② and the voltage can be easily found by applying Thévenin’s theorem. The current through terminals ① ② at point f is given by f E1 f I1 = f Z1 + f Z2 + f Z0 + 3R 1 78 f I1 = f I2 = f I0 = f E1 Z f total f Ztotal = f Z1 + f Z2 + f Z0 + 3R and the voltage at point f by = − f Z0 f I0 = − f Z0 f I1 f V2 = − f Z2 f I2 = − f Z2 f I1 f V1 = − f V0 + f V2 + 3R f I1 = f Z0 + f Z2 + 3R f I1

f V0

79

The voltage fE1 in Eqs. (7.4) and (7.9) is the voltage between terminals ① ② and is given by Eq. (7.4) as a known initial quantity before switch S closes, where E1 and E1 are known quantities. Finally, this solution for symmetrical voltages and currents on virtual terminals at point f are inverse-transformed into phase-abc quantities at point f: f Ia

1

1

1

f Ib

= 1 1

a2

a

= f I1 f I1

a

a2

f I2 = f I1

f Ic

1

f Va f Vb f Vc

1

= 1 1

f I0

1

a2

a

a

2

a

3f I1 = 0 = 0

− f Z 0 f I1 Z + f 0 f Z2 + 3R f I1 = − f Z 2 f I1

3 f E1 Z f total 0 0

①

3R a −1 f Z0 + a2 −a f Z2 + a2 3R 2

a −1 f Z0 + a− a

2

3R f1 fE Z f total =

a −1 f Z0 + a− a2 f Z2 + a 3R f Ztotal

f E1

= f

②

f E1

where f Ztotal = f Z1 + f Z2 + f Z0 + 3R f E1

+ a 3R 7 10

a2 −1 f Z0 + a2 −a f Z2 + a2 3R f Ztotal

f Z2

f I1

2

f Z1 f Z1 E + E Z1 + f Z1 1 f Z1 + f Z1 1

③ ④

7.2 Line-to-Ground Fault (Phase-a to Ground Fault: 1ϕG)

All the solutions f Ia, f Ib, f Ic, fVa, fVb, fVc in the abc-domain were found. Incidentally, f E1 is the positive-sequence voltage (i.e. the phase-a voltage) at point f before the fault. If the load-flow current on the line at point f before the fault is zero, the voltage at point f is the same value as that of the generator source voltage: E = E = fE1.

7.2.4

Voltages and Currents at an Arbitrary Point under Fault Conditions

Let’s examine the voltages and currents at point m under the phase-a 1ϕG fault condition at point f shown in Figure 7.2. Figure 7.2b is the mathematical representation at any point of the system connection diagram in Figure 7.2a. Therefore, the voltages and currents at points m(mV1, mV2, mV0, mI1, mI2, mI0) and (mVa, mVb, mVc, mIa, mIb, mIc) correspond to each other by the symmetrical transformation: f V1 f I1

∴

= E1 − f Z1 I1 = E1 − f Z1 I1

= I1 + I1

Z1 I1 Z1 + f Z1 f f

I1 =

f

+ f

The fault current supplied from the left side through point m to point f

7 11a

E1 − E1 Z1 + f Z1

C1 f I1 + Iload ①

The load current before fault

The load current is not included in the negative-and zero-sequence circuits, so I2 = I0 =

C2 × f I2

f

Z2 I Z2 + f Z2 f 2

②

C0 f I0

f

f Z0 I0 Z0 + f Z0 f

③

f

Z1 Z1 + f Z1 f

where C1 = f

7 11b

C2 , C0 are defined by the same equation forms

C1 is the coefficient of the branched current mI1/f I1 from the left side through point m from the total current f I1 and is the vector value of 0–1.0∠δ. C2, C0 are also defined in the same way. As we know already, the value of f I0 = f I1 = f I2, the currents I1, I2, I0 at point m are calculated with Eq. (7.11). Finally, the currents in the abc-domain at point m are Ia 1 Ib = 1 Ic 1

1 a2 a

1 a a2

C0 C1 C2

1 1 f I1 + 1

Fault current term

1 a2 a

1 a a2

0 Iload 0

7 12

Load current term

The second term on the right side is the load-current components that existed before the fault, and the first term is the fault-current components caused by the fault at point f. This equation explains the fact that the fault-current component at any point in the system is not affected by the load-current component just before the fault. In other words, we can calculate any fault under the condition of zero load current and then use vectors to superpose the load current if necessary. The voltages at point m can be calculated from the voltages and current quantities already found at point f, by utilizing the following equations: m V0 m V1 m V2

f V0

=

f V1 f V2

m Z0

+

0 0

0 m Z1 0

0 0 m Z2

I0 I1 I2

Finally, the voltages can be inverse-transformed into the abc-domain.

7 13

131

132

7 Fault Analysis Based on Symmetrical Components

Voltage at point f before fault f V1 = E1′ = E2′

– f Z1″

f Z1′

S

1

+ f Z1

f V1

f Z1

3R f Z2 f Z0

2

Figure 7.3 No-load fault calculation.

7.2.5

Fault Under No-Load Conditions

A fault under no-load conditions is a special case of E1 = E1 = f E1 in Figure 7.2(b) and Eq. (7.10)④. The power system looking from point f under this condition can be regarded as a black box with an internal power source, whose voltage across the terminals ① ② is fV1 = E1 = E1 and internal impedance is f Z1 = f Z1

f Z1

, where the symbol // means a

parallel circuit of f Z1 and f Z1 . On the other hand, as shown in Figure 7.3, the equivalent circuit of the phase-a to ground fault at point f connects the outer impedance f Z2 + f Z0 + 3R to the terminals ① ② of the black box. Accordingly, the flow current at terminals ① ② is easily found by Thévenin’s theorem. That is, the current is f I1 = f E1/{ f Z1 + ( f Z2 + f Z0 + 3R)}. This is of course in accordance with Eq. (7.8). In conclusion, we can apply Figure 7.3 as the equivalent circuit, instead of Figure 7.2b, whenever we need fault-current components (without load currents). Then we can superpose load currents if necessary.

7.3

Fault Analysis at Various Fault Modes

Voltage and current equations for cases with different mode faults are summarized in Tables 7.1a and 7.1b, and equations in the 012-sequence domain and phase-abc-domain, as well as the equivalent circuits, are indicated. Case 7 is that of the phase-a to ground fault, which we have already examined in detail. The voltage and current equations and equivalent circuits for different fault modes can be derived by the same procedure. Note, incidentally, that the generator impedances are jxg1 jxg2 in a strict sense, while transmission-line impedances are exactly Z1 = Z2. Therefore, when a line fault is further from the generator, the condition jxg1, jxg2 Z1, Z2 and the approximation jxg1 + Z1 ≒ jxg2 + Z2 are justified, so the calculation’s accuracy will be improved by the dominant line impedances. Currents at point m in the case of a different mode fault at point f can be found by the following procedure: m I1

= C1

where C1 = f Z1 m Ia m Ib m Ic

f I1 , m I2 f Z1

= C2

f I2 , m I0

= C0

f I0

7 14

+ f Z1 , etc. The inverse transformed currents are

+ C1 f I1 + C2 f I2 2 = C0 f I0 + a C1 f I1 + aC 2 f I2 C0 f I0 + aC 1 f I1 + a2 C2 f I2 C0

f I0

Voltage equations are derived analogously.

7 15

7.3 Fault Analysis at Various Fault Modes

Table 7.1a Equations and equivalent circuits for various fault modes. Fault conditions (abc-domain)

#1

3ϕS three-phase line-to-line fault

Point f

f Ia

f Ia

+ f Ib + f Ic = 0

(1A)

f Va

= f Vb = f Vc

fVa

= fVb = fVc = 0 (2A)

f Ia

f Ib

= f Ic = 0

f Ib

f Va

f Ib f Ic f Vb

f Vc

#2

3ϕG three-phase line-to-ground fault

f Va

f Ia f Ib

Metallic fault

f Ic

f Va f Vb

f Vc

#3

1ϕG phase-a line-to-ground fault

f Vb

f Vc

2ϕS phase-b to -c line-to-line fault

(3A)

f Va

f Ic

#4

=0

f Ia

f Ia

=0

f Ib

f Ib

+ f Ic = 0

f Ic

f Vb = f Vc

(4A)

V V f Vc f b f a

#5

2ϕ G phase-b, -c double line-to-ground fault

f Ia

f Ia

f Ib

f Va

=0

f Vb

= f Vc = 0

(5A)

Arc fault arc resistance R

f Ic f Vc

#6

3ϕG three-phase line-to-ground fault

f Vb

fIa

r r

fIc fVc

#7

1ϕG phase-a line-to-ground fault

fVb

f Ib

f Va

fVa fVb

f Vc

= f Ic = 0 = R f Ia

f Ia

f Vb

f Va

(7A)

=0

f Vb − r

r r

f Ic

f Ic

R

f Ia f Ib

f Ib

= f Vc −r

R

fIb

fVc

2ϕG phase-b, -c line-to-ground fault

fVa

= f Vb −r

= R f Ia + f Ib + f Ic

fIa fIc

#8

f Va − r f Ia

r

fIb

f Ib

= f Vc −r

R

f Ic

= R f Ib + f Ic

(8A)

(6A)

133

134

7 Fault Analysis Based on Symmetrical Components

Table 7.1b Equations and equivalent circuit for various fault modes. Fault condition at point f and the equivalent circuits: Metallic fault

#1

3ϕS phase-abc

p

#2

#3

#4

V

I

n

I

V

0

I

V

3ϕG phase-abc

=0

f V1

p n

f I2

f V2

0

f I0

f V0

f V1

= f V2 = 0

f I0

1B

f I2

= f V1 = f V2 = 0

= f I0 = 0

f I0

2B

f I1

n

f I2

f V2

0

f I0

f V0

2ϕS phase-b, -c f I1

f V1

n

f I2

f V2

0

f I0

f V0

f I0

= f I1 = f I2

f V0

+ f V1 + f V2 = 0

f I0

=0

f I1

= − f I2

f V1

= f V2

f V0

=0

4B

f V1

2ϕG phase-b, -c

p

f I1

f V1

n

f I2

f V2

0

f I0

f V0

= f I1 = f I2 =

f V0

= f V1 = f V2

3C

f Ea f Z1

= f V2 = − f Z 2

f I2

+ f Z2

+

=

=

f Z1

f Z2

f Z2

+ f Z2

f Ea

f Z0

f Z0

− f Z0 f I2 = f Z 2 + f V2 f V0

f Z2

f Ea

=

where 5B

f Ea

4C

f Z1

+ f I1 + f I2 = 0

f Ea

=0

f I1

f I0

2C

= f V1 = f V2 = 0

= 0, I1 = − f I2 =

f V0

f Ea f Z1

Δ − f Z0 f V0 = − f Z0 f I0 = f Ea Δ f Z0 + f Z2 f V1 = − f V0 + f V2 = Δ − f Z2 f Ea f V2 = − f Z2 f I2 = Δ where Δ = f Z0 + f Z1 + f Z2

3B

f I0

1C

= f I2 = 0, f I1 =

f V0

f I0

f V1

f Z1

= f V1 = f V2 = 0

1ϕG phase-a

p

f Ea

= f I2 = 0, f I1 =

f V0

f I0 = 0, f I2 = 0

f V0

f I1

p

#5

f I0

=

f Z2 f Z0 f Z2

f I1 , f I0

=

= f V1 = f V2 = − f Z2

f Z2 f Z0 f Z2

+ f Z0

f I1

=

f Z2

+ f Z0 − f Z2 + f Z0

f Z2 f I2

f Z0

f I1

5C f I1

7.3 Fault Analysis at Various Fault Modes

Table 7.1b (Continued) Fault condition at point f and the equivalent circuits: Metallic fault

#6

#7

f V0 − r f I0

3ϕG phase-abc

p

f I1

f V1

r

n

f I2

f V2

r

0

f I0

f V0

r 3R

f V1 = r

f I1

=r

f I2

f V2

= 3R f I0

f I0

6B

f I2 = f I0 = 0 f V2

= f V0 = 0

= f I2 = 0, f I1 =

f V0

= f V2 = 0

f V1

=r

1ϕG phase-a

f I0

p

f I1

f V1

n

f I2

f V2

0

f I0

f V0

f I0

= f I1 = f I2 + f V1 + f V2 = 3R f I0

=

+r 6C

r f Z1

+r

= f I1 = f I2 =

f Ea

f Ea

Δ − f Z0 f V0 = − f Z0 f I0 = f Ea Δ f V1 = − f V0 + f V2 + 3R f I1

3R f V0

f I1

f Ea f Z1

7B

=

f Z0

+ f Z2 + 3R Δ

7C

f Ea

− f Z2 f Ea Δ Δ = f Z0 + f Z1 + f Z2 + 3R

f V2

#8

2ϕG phase-b, -c

f I1

f I1

f V1

n

f I2

f V2

0

f I0

r r 3R f V0

f Ea

Δ2 Δ0 Δ1 + Δ2 + Δ0 − Δ0 f I2 = f I1 Δ2 + Δ0

r

p

=

= − f Z2

f I0

= f I1 = f I1 = 0

f V0 −

r + 3R f I0

= f V1 −r

f I1

= f V2 −r

f I2

8B

f V0

= − f Z0

f V1

= r+

f V2

= − f Z2

f I0

=

f I2

, f I0 =

−Δ2 Δ2 + Δ0

f Z0 Δ2 Δ2 + Δ0

Δ2 Δ0 Δ2 + Δ0 f I2

=

=−

f I1

f I1

8C

f I1 f Z0 Δ2 Δ2 + Δ0

f I1

where Δ1 = f Z1 + r Δ2 = f Z2 + r Δ0 = f Z0 + r + 3R Notes: All the quantities for the negative- and zero-sequence circuits become zero for 1(3ϕS), 2(3ϕG), and 6(3ϕG), because power sources do not exist in these circuits. f Ea is the voltage at point f before the fault.

135

7 Fault Analysis Based on Symmetrical Components

Table 7.1c Equations and equivalent circuits for various fault modes. Phase voltages and currents

3ϕS phase-abc #1

f Ia

= f I1 , f Ib = a2 f I1 , f Ic = af I1

f I1

= f Ea f Z1

1D

f Va = f Vb = f Vc = 0

#2

3ϕG phase-abc

Same as above (2D)

#3

1ϕG phase-a

f Ia

= 3f I0 = 3f Ea Δ where Δ = f Z0 + f Z1 + f Z2

Metallic fault

f Ib = f Ic = 0, f Vb

#4

2ϕS phase-b, -c

f Ia

#5

2ϕG phase-b, -c

3ϕG phase-abc

f Ic

=

f Ia

1ϕG phase-a

f Ia

f E a , f Vc

a −1 f Z0 + a− a2 f Z2

=

Δ

f Ea Z f 1 + f Z2

= 2f V1 , f Vb = f Vc = − f V1 , where f V1 =

= 0, f Ib =

=

a2 −a f Z0 + a2 − 1 f Z2 f Z0

+ f Z2

a −a2 f Z0 + a− 1 f Z2 f Z0

3f Z2 f Z0 f Z2 + f Z0

+ f Z2 f I 1 , f Vb

+ f Z2

5D

= f Vc = 0, where f I1 =

f Ea Z + f Z2 f 1

f Z0

f Ea f Z1

+r

= f V1 , f Vb = a f V1 , f Vc = af V2 , where f V1 = f Ib

f Ea

f I1

2

= 3f I1 = 3f Ea Δ,

f Ea

4D

f Z2 f Z1

3D

f I1

= f I1 , f Ib = a2 f I1 , f Ic = af I2 , where f I1 =

f Va

#7

Δ

f Ia

f Va

#6

=

a2 − 1 f Z0 + a2 − a f Z2

= 0, Ib = − f Ic = a2 − a f I1 = a2 −a

f Va

Arc fault arc resistance R

136

6D

r f Z1

+r

f Ea

= f Ic = 0

a2 −1 f Z0 + a2 −a f Z2 + a2 3R 3R f Ea f Ea, f Vb = Δ Δ 2 a− 1 f Z0 + a −a f Z2 + a 3R f Vc = f Ea , where Δ = f Z0 + f Z1 + f Z2 + 3R Δ

f Va

#8

2ϕG phase-b, -c

= 3R f I1 =

a2 −a Δ0 + a2 −1 Δ2 f I1f Δ0 + Δ2 f Z0 Δ2 + Δ0 Δ2 + f Z2 Δ0 + r I1 f Va = Δ0 + Δ2 f f Ia

= 0, f Ib =

f Vb

=

f Z0

Δ2 + a2 Δ0 Δ2 + af Z2 Δ0 + a2 r Δ0 + Δ2

2 f Z0 Δ2 + aΔ0 Δ2 + a f Z2 Δ0 + ar f Vc = Δ0 + Δ2

where a−a2 = j 3, a2 −1 = j 3a, 1 −a = j 3a2

Ic =

a− a2 Δ0 + a −1 Δ2 Δ0 + Δ2 where f I1 = f Ea

f I1

Δ1 +

Δ1 = f Z1 + r f I1

f I1

7D

Δ2 = f Z2 + r Δ0 = f Z0 + r + 3R

Δ2 Δ0 Δ2 + Δ0 8D

7.4 Conductor Opening

7.4

Conductor Opening

This section examines one- and two-phase conductor openings. Conductor openings (or cut-offs) of one or two phases seldom happen by accident in actual power systems. However, a single-phase breaker tripping as a procedure of singlephase reclosing is a kind of one-phase conductor opening. Moreover, in the case of three-phase tripping by a circuit breaker, current tripping by breaker–pole opening of each phase occurs sequentially, and the timing of each phase tripping is different. In other words, a three-phase circuit is opened by the breaker through the transient states of trip-start 1ϕ opening 2ϕ opening and 3ϕ opening. Furthermore, breaker-tripping failure may occur and give rise to severe outcomes. Phase-imbalanced opening occurs often and at various places in practical engineering.

7.4.1

Single-Phase (Phase a) Conductor Opening

Referring to Table 7.2 (1A), the phase-a conductor is opened between points p and q; va, vb, vc, are the voltages across points p and q of each phase, and ia, ib, ic are the phase currents at points p and q. This condition can be described by the following equations: vb = v c = 0

7 16

ia = 0 The transformed equation in the 012-domain is v 0 = v 1 = v2

7 17

i0 + i1 + i2 = 0

This equation can be described with the equivalent circuits in Table 7.2 (1B). In the figure, the negative- and zero-sequence circuits are connected in parallel to the positive-sequence circuit. pZ1, pZ2, pZ0 are the impedances of the left-side circuit at point p, and qZ1, qZ2, qZ0 are the impedances of the right-side circuit at point q. Then, from the equivalent circuit, i1 =

G Ea − g Ea

Z2 Z0 Z1 + Z2 + Z0

,

i2 =

− Z0 −Z2 i1 , i0 = i1 Z2 + Z0 Z2 + Z0

Z2 Z0 v0 = v 1 = v2 = i1 Z2 + Z0

7 18

Z1 = p Z1 + q Z1 , Z2 = p Z2 + q Z2 , Z0 = p Z0 + q Z0 This equation is written again in Table 7.2 (1C), and the inverse-transformed equation for abc-phases is shown in Table 7.2 (1D).

137

138

7 Fault Analysis Based on Symmetrical Components

Table 7.2 Phase-opening modes (equations and equivalent circuits). Phase-b, -c opening

Phase-a opening

[1A]

[2A]

Point p Point q ia = 0 v a ia = 0 ib ib vb ic

ia

va

ib = 0

vb

ia ib = 0

ic = 0

vc

ic = 0

ic

vc

ib = ic = 0 ia = 0

1A

vb = vc = 0 [1B]

[2B]

Point p Point q i1 i1 v1 p GEa

pZ1

pv1

i2 n

pZ2

pv2

pZ0

pZ1

gEa

i2

n

pZ2

o

pZ0

qZ0

v0 = v1 = v2 i1 =

v0 + v1 + v2 = 0

1B

qZ2

q v2

pv2

i0

v0 p v0

gEa

qZ0

q v0

2B

G Ea − g Ea Z1 + Z2 + Z0 v1 = Z2 + Z0 i1

G Ea − gE a

Z2 Z0 Z1 + Z2 + Z0 − Z0 i2 = i1 Z2 + Z0 − Z2 i0 = i1 Z2 + Z0 Z2 Z0 v0 = v1 = v2 = i1 Z2 + Z0 Z1 = p Z1 + q Z1

qZ1

i2

v2

i0

i0 = i1 = i2 i0 + i1 + i2 = 0

v pv1 q 1

i0 qv0

i1

v1

i2

qZ2

qv2

v0 pv0

i1 p GEa

qZ1

qv1

v2

i0 o

2A

va = 0

i0 = i1 = i2 =

v2 = − Z2 i1 v0 = − Z0 i1

2C

where

1C

Z1 = p Z1 + q Z1 Z2 = p Z2 + q Z2 Z0 = p Z0 + q Z0

Z2 = p Z2 + q Z2 Z0 = p Z0 + q Z0 3 G Ea − g Ea Z1 + Z2 + Z0 ib = ic = 0, va = 0

ia = 0

ia = 3i1 =

a2 − a Z0 + a2 − 1 Z2 i1 ib = Z2 + Z0 a− a2 Z0 + a −1 Z2 i1 Z2 + Z0 3Z2 Z0 va = i1 Z2 + Z0 G Ea − g Ea i1 = Z2 Z0 Z1 + Z2 + Z0

vb =

ic =

1D

a2 −1 Z0 + a2 −a Z2 i1

a −1 Z0 + a− a2 Z2 i1 G Ea − g Ea i1 = Z1 + Z2 + Z0 vc =

2D

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

7.4.2

Two-Phase (Phase-b, -c) Conductor Opening

Voltages and currents in this case are found in a similar way, and the resulting equations as well as the equivalent circuits are shown in Table 7.2 (2A, 2B, 2C, 2D).

7.5

Visual Vector Diagrams of Voltages and Currents under Fault Conditions

This section introduces a diagrammatic solution for voltages and currents under various fault conditions.

7.5.1 Three-Phase Fault: 3ϕS, 3ϕG (Solidly Neutral Grounding System, High-Resistance Neutral Grounding System) The equivalent circuit of a three-phase line-to-line fault (3ϕS, 3ϕG) and the voltage distribution by distance between the generator and the faulting point f are shown in Figures 7.4a and b, where x1 is the positive sequence reactance of the total line length. The voltage at an arbitrary midpoint m can be derived as a function of k1 (k1 = 0 ~ 1) by the equation q x1

k1 =

x1

,

x1 = p x 1 + q x 1 = 1 −k1 x1 + k1 x1 ,

I1 =

E E = −j , Δ x1

f V1

= 0, f V2 = f V0 = 0

Δ = jx1

① ②

I2 = I 0 = 0

③ 7 19

I1 q x 1 = jq x1 = = k1 , E E x1

m V1

f Va

mV 0 = =0 E E

= k1 ,

m Vb

E

= a2 k1 ,

④ ⑤

= f Vb = f Vc = 0

m Va

E

m V2

m Vc

E

= ak 1

⑥

Accordingly, the voltage vectors at point m can be drawn as shown in Figure 7.4c with the parameter k1. Finally, we can draw the three-dimensional vector diagram, Figure 7.4d, where the current vectors are the same at any point because leakage current is neglected, and the phase angle is approximately 90 (say 85 , considering line resistances) lagged from the voltages.

139

7 Fault Analysis Based on Symmetrical Components

The figure for k1 = 0.4

Point f I1

Point m

E

qx1

px1 mV1

x1

mV1 = k1E

E

px1

qx1

(1–k1)x1

k1x1

fV1

x1 (b)

(a)

Point m Ic

E fVa = fVb = fVc =

0

mVa mVc

Ib

Ia 0.2 k = 0.4 0.6 0.8 1.0

mVb

(c)

Source voltage Ic Ia

E

Point m Ic

aE

140

2 aE

mVa

Point f

Ia Ic

Ia

Ib mVc

l–k

1

mV b

Ib fVa = fVb = fVc =0

k1 Ib (d)

Figure 7.4 Three-phase fault: 3ϕS, 3ϕG (solidly neutral grounding system, high-resistance neutral grounding system).

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

7.5.2 Phase b to c Fault: 2ϕS (for Solidly Neutral Grounding System, High-Resistance Neutral Grounding System) The equivalent circuit and the related voltage distribution through the series circuit of positive- and negative-sequence reactance are shown in Figures 7.5a and b.

The figure for : k1 = 0.4 Point m

E

px1

qx1 x1

mV1

f V1

qx1 mV2 x1

1 (1 + k1)E V = V = 1 2 E f 1 f 2 2 1 (1 – k1)E V = m 2 2

mV2 fV1

I2 px1

mV1 =

E

Point f I1

px 1

(1–k1)x1

fV2

fV2

qx1

qx1

k1x1

k1x1

px1

(1–k1)x1

x1

x1

(a)

(b)

Point m Ic

mVa =

Point f

E

fVa =

Ic fV2

m Vc

0

m Vb

k1 = 1 Remote fault k1 = 0.8 k1 = 0.6 k1 = 0.4 k1 = 0.2 k1 = 0 Near fault k1 = 0.2 k1 = 0.4 k1 = 0.6 k1 = 0.8 k1 = 1 Remote fault

fV 2

fVc

= 1E 2

1 E fV1 = 2 a 2· fV

a· fV 1

a 3·

E

1

fVb

V2 a· f

Ib

Ib

(c)

Source voltage

Ic

Point m aE

E

Ic

a2E

Point f mVa

Ib

mVc

Ic

mVb

l – k1

Ib k1

fVa fVc

fVb

Ib

(d) Figure 7.5 Phase-b to -c fault: 2ϕS (for solidly neutral grounding system, high-resistance neutral grounding system).

141

142

7 Fault Analysis Based on Symmetrical Components

The related equations are q x1

k1 =

x1

x1 = p x 1 + q x 1 = 1 − k1 x1 + k1 x1

,

Δ = j x1 + x1 = j2x1 , I1 = −I2 =

E E = −j , Δ 2x1

① I0 = 0

②

Ia = 0 + I1 + I 2 = 0 Ib = 0 + a2 I1 + aI 2 = a2 − a I1 = −

3 E 2 x1

Ic = 0 + aI 1 + a2 I2 = a2 −a I2 = +

3 E = − Ib 2 x1

f V0

=0

E f V1

=E−

E f V2

=−

E m V0

jx1 I1 1 = 2 E

m V1

=

E m V2

=

E f Va

jx1 I2 1 = 2 E

f V1

E f V2

E

=0+

E

+ jq x1

I1 1 = 1 + k1 E 2

+ jq x1

I2 1 = 1 − k1 E 2

1 1 1 = 0 + a2 + a = − 2 2 2 E

=0+

E m Vb

=0+a

m Vbc

E

⑥

1 1 1 + k1 + 1 − k1 = 1 2 2

= 0 + a2

E

E

7 20

1 1 1 = 0 + a + a2 = − 2 2 2

m Va

m Vc

⑤

1 1 + =1 2 2

f Vb

E

④

=0

E

f Vc

③

3 1 1 1 1 + k1 + a 1 − k1 = − −j k1 2 2 2 2

⑦

3 1 1 1 1 + k1 + a2 1 − k1 = − + j k1 2 2 2 2

= − j 3k1

The vector diagram in Figure 7.5c for an arbitrary point m is derived from Eq. (7.20)⑦, and the diagram in Figure 7.5d for fault point f is a special case with k = 0.

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

The fault currents in the case of 2ϕS become 3 E 2 x1

Ib = I c =

which is 0.87 times the current in the case of 3ϕS

E x1

The zero-sequence voltage and current are zero in this case, so the neutral voltage is zero potential at any point. Accordingly, the equation E=mVa=fVa (where Ia = 0) is found, and the phase-a to ground voltage mVa at an arbitrary point m is almost unaffected by the distance from point f.

7.5.3

Phase-a to Ground Fault: 1ϕG (Solidly Neutral Grounding System)

The equivalent circuit in this case is given by Figure 7.6a, from which the related Eq. (7.3) as well as the diagrams in Figures 7.6 (b1) and (b2) are derived in a way similar to that in Section 7.2. Figure 7.6 (b1) is drawn to satisfy the precise circuit condition in Figure 7.6a: the straight line ad is divided by points b and c, which are determined to satisfy the ratio ab bc cd = x1 x1 x0 = 1 1 v fV1,fV2 are drawn at point b and fV0 at point c. The arrows for fV2,fV0 are in the opposite direction from fV1 because fV1 = − (fV2+fV0). p q mV1 is drawn at the point b’, which divides ab by ab b b = p x 1 q x 1 m V2 , m V0 are derived analogously.

Point m px1

E

Point f I1 q x1 mV1 x1 fV1 I2

p x1

q x1 mV2 x1

p x0

q x0 mV0 x0

fV2

D

I0 fV0

(a) Equivalent circuit

fV2

E

mV1

a

b´

px1

(l–k1)x1

x1

qx1

k1x1

fV1

b

qx1

k1x1

The figure derived from x v = 0 = 1.5, k1=0.4, k 0 =0.2 x1 n + k1 + 1 Point E mV1 = n+2 m Point mV2 – (1 – k1) mV1 f E E mV2 = n+2 mV2 – (1 – k0) v E mV0 = mV0 n+2 fV0 mV0 m f c d px0 px1 qx0 p x1 qx1 (l–k1)x1 (l–k0)nx1 (l–k1) x1 k1x1 k nx

x1

0

1

x0=nx1

k0

l–k0

(b1) Figure 7.6 Phase-a to ground fault: 1ϕG (solidly neutral grounding system).

(b2)

v+1 v + 2E –1 E fV2 = v+2 –n fV0 = v+2E fV1

=

143

7 Fault Analysis Based on Symmetrical Components

E f Vc

3 E∠90° 2

=

fVb

fVa

v=0

3 E∠ 270° 2

=

Ia

=0

v=0

60° 30°

30°

v = 0.5 v = 1.0 v = 1.5 v=2 v=3 v=4 v=5 v = 10

3 E fV ∠1 c 50°

b

= 3 21

=

E∠

v=∞

0°

High-reactance neutral grounding system v = 5–∞

b

fV

n = 0 – 2.5

60° f V

fV c

Solidly neutral grounding system

fVc

v = 0.5 v = 1.0 v = 1.5 v=2 v=3 v=4 v=5 v = 10 v=∞

= –j 3 E

(c) Voltages at point f for parameter v = x0/x1

Point m mV1

E

mV0

Point f Ib = Ic = 0

mVa

a ·2 mV

a

V1 a· m Vc m mV0

mV b

fVa = 0 Ib = Ic = 0

fV1

E

fV0

mV2

I1 2 ·m V 1

fV2

= I2 = I0

Ia

m V0

2 a·mV2 a ·f V

V1 a· f Vf c fV0

fV b

I1 = I2 = a ·2 fV

j

3E

fV0

j

I0

Ia

1

2

2

a·fV2

3E

(d) Voltages, currents at points f, m

Source voltage Point m

Ia

E aE

144

Point f 2 aE

Ia mVa fVa =

l – k1

m Vc

k1

Ia fVb

fVc

Ib = Ic = 0

(e) Voltages and currents overview Figure 7.6 (Continued)

0

mVb

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

Figure 7.6 (b2) is obtained by folding Figure 7.6 (b1). The related equations are v=

x0 x0 = , x1 x2 q x1

k1 =

x1

x1 = x2 =

x0 v

x1 = p x 1 + q x 1 = 1 − k1 x1 + k1 x1

,

q x0 k0 = , x0

①

x0 = p x 0 + q x 0 = 1− k0 x0 + k0 x0

Δ = j x0 + x1 + x2 = j v + 2 x1 E 1 E = −j Δ v + 2 x1 I0 −v f V0 = − jx0 = E E v+2

②

I1 = I2 = I0 =

f V1

f V0

=−

E

+

E

f V2

E

v+1 v+2

=

③

I2 −1 = E E v+2 I0 f V0 k0 v −v k0 v f V0 m V0 = + = + jq x0 = + v+2 v+2 v+2 E E E E f V2

= − jx1

m V1

E m V2

E f V2

=

E f Va

f V1

=

E

+ jq x1

I1 f V1 k1 v+1 k1 + = + = E E v+2 v+2 v+2

+ jq x1

I2 f V2 k1 −1 k1 + = + = E E v+2 v+2 v+2

④

7 21

=0

f Vb

E f Vc

E

f V0

=

=

f Vbc

+ a2

E f V0

E

+a

f Vb

E

f V1

E

+a

+ a2

f V2

E f V2

E

=

3 a2 − 1 v + a2 − a − 3v = −j 2 v+2 2 v+2

=

3 a −1 v + a− a2 −3v = +j 2 v+2 2 v+2

⑤

f Vc

= −j 3 E E E k0 v + k1 + k1 k0 v + 2k1 f Va m Va = + = E E v+2 v+2 m Vb

E

=

=

f Vb

E

−

f V1

+

k0 v + a2 k1 + ak 1 f Vb k0 v− k1 = + v+2 E v+2

k1 v + ak 1 + a2 k1 f Vc k0 v −k1 = + = + E E v+2 E v+2

m Vc

m Vbc

E

f Vc

=

m Vb

E

−

m Vc

E

=

f Vbc

E

⑥

= −j 3

Accordingly, fV1, fV2, fV0 and mV1, mV2, mV0 are easily calculated by referring to Eq. (7.21)①–④ or Figure 7.6 (b2). Then, referring to Eq. (7.21), the phase voltages at fault point f, fVa, fVb, fVc can be drawn with the parameter v = x0/x1, as shown in Figure 7.6c. The parameter v is v = x0/x1 = 0 to 2 or 3 at most for the solidly neutral grounding system, so that the magnitudes of unfaulted phase voltages fVb, fVc are around 0.8–1.1. In other words, jumping overvoltage phenomena of unfaulted phases do not occur for 1 ϕG in the solidly neutral grounding system, which is obviously a great advantage from the viewpoint of reducing the level of insulation coordination. This is the essential reason why modern extra high voltage (EHV) and ultra high voltage (UHV) systems over 200 or 300 kV utilize the solidly neutral grounding system without exception (see Chapters 16 and 17). Figure 7.6d shows the vector diagrams at points f and m, and Figure 7.6e is the three-dimensional overview of the vector diagram of the total system.

145

7 Fault Analysis Based on Symmetrical Components

If a high-reactance neutral grounded system of large V (say, v ≧ 8 or 10) is assumed, the unfaulted phase voltages become 3 times E, as shown in Figures 7.6(c–e).

fVb, fVc

7.5.4

Double Line-to-Ground (Phases b and c) Fault: 2ϕG (Solidly Neutral Grounding System)

Equation (7.22) and Figure 7.7 can be derived in the same manner as in the previous section. The related equations can be found in the section that follows.

Point f

Point m

The figure for x v = x0 = 1.5, 1 k1 = 0.4, k0 = 0.2

p x1

E

m V1

qx 1

E

I2 px 1

qx 1 m V2 x1

px 0

qx 0 m V0 x 0

f V2

∆l

I0 fV0

(b) 2

(a)

a 2·

mV 2

Point m m V2 mVa m V0 m V1

Vc

m

V1 a· m

mV b

a 2·

mV 1

Ib

Ic

Ia = 0

I0 aI1 1

I0 aI2 2 V a· m mV0

Point f fV2 fVa

I1 I2

a 2I

Ic

mV0

(E – fV1)

v (1 – k1) fV2 = (l – k1) 2v + 1 E v fV0 = fV1 = fV2 = 2v + 1 E mV0 = (1 – k0) fV0 v = (1 – k0) E 2v +1 xx k0 1 0 x1 + x0 px1 qx1 xx (1 – k0) 1 0 (1 – k1)x1 k1x1 x1 + x0 vx1 x1 (1 – k0) x x v v 1 0 v +1 k0 x x = v + 1 1 x1 + x0 v + 1 1

fV1

x1

mV1 = fV1 + k1 mV2 =

I1

a 2I

146

I0

fV0

Ib

a 2·

f V2

fV1

a· fV2 fV0

fV0

a· fV1 a2· fV1 fVb = fVc = 0 Ia = 0

Ia = 0

(c) Source voltage

Ic

Point m aE

E a 2E

Ic

Point f

m Vc

mVa mVb

Ic fVa

Ib 1 – k1

Ib k1

Ib

fVb = fVc = 0

Ia = 0

(d) Figure 7.7 Double line-to-ground (phases b and c) fault: 2ϕG (solidly neutral grounding system).

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

v=

x0 x0 = , x1 x2 q x1

k1 =

x1 q x0

k0 =

x0

x1 = x2 =

x0 v

x1 = p x 1 + q x 1 = 1 − k1 x1 + k1 x1

,

① ,

x0 = p x 0 + q x 0 = 1− k0 x0 + k0 x0

x1 x0 2v + 1 x1 =j v+1 x1 + x0 E v+1 E I1 = = − j Δ 2v + 1 x1 Δ = j x1 +

I2 =

− x0 −v v E I1 = j I1 = v+1 2v + 1 x1 x1 + x0

②

− x1 −1 1 E I1 = j I1 = v+1 2v + 1 x1 x1 + x0 x1 x0 I1 vx1 I1 v f V0 f V1 f V2 = = =j =j = E E E x1 + x0 E v + 1 E 2v + 1 V V I k0 v v f 0 f V0 f V0 m 0 0 = 1 − k0 = + jq x0 = − = 1 − k0 2v + 1 E E E E 2v + 1 E I0 =

m V1

=

E m V2

=

f V1

E f V2

E Ia = 0

E

+ jq x1

I1 f V1 v + 1 f V1 v + k1 v + 1 f V1 = = = + k1 + k1 1 − 2v + 1 E E E E 2v + 1

+ jq x1

I2 f V2 k1 v v f V2 = 1 − k1 = − = 1 − k1 2v + 1 E E 2v + 1 E 3 a−v 2v + 1

E x1

1 −a v + 1 + a2 v E 3 − a2 + v = 2v + 1 x1 2v + 1 V V V V V 3v f a f 0 f 1 f 2 f 1 = + + =3 = 2v + 1 E E E E E

E x1

Ib = j1 −a2 v + 1 +

av E = 2v + 1 x1

Ic = j

f Vb

E

③

④ 7 22

⑤

⑥

=0

f Vc

=0 E −k0 v + k1 v + 1 + − k1 v k1 − k0 v f Va f Va m Va = + + = 2v + 1 E E E 2v + 1 =

3v k1 −k0 v + 2v + 1 2v + 1

a2 − j 3v k1 − k0 v −k0 v + a2 k1 v + 1 + a − k1 v = + = E E 2v + 1 2v + 1 f Vb

m Vb

m Vc

E

=

f Vc

E

+

a2 + j 3v k1 −k0 v − k0 v + ak 1 v + 1 + a2 − k1 v = 2v + 1 2v + 1

⑦

147

7 Fault Analysis Based on Symmetrical Components

7.5.5

Phase-a Line-to-Ground Fault: 1ϕG (High-Resistance Neutral Grounding System)

Equation (7.23) and Figure 7.8 can be derived in the same way. The related equations are

Point f

Point m

mV2

The figure for 3Rare 1 k1 = 0.4, kare = = 3RN 3

f V2 ⇋ 0

fVare

mV1

mV0

fV1 V f 0

q x1 q x1 (1–k1)x1 k1x1 x1

E

fVare – fV0

fV0

x1

(b)

Point m Point f

fV0

150 3E

(The figure for kare =

n

0°

I0

px 0 px 0 3RN NV0 mV0 Z0

–1 E 1 + kare

21

∆l

3Rare

Vf a

fV2 fVare

3E

qx 1 m V2 x1

fV0 =

°

fV1

I2 px 1

E

O

fV c

q x1

E

q x1 m V1 x 1

a

fVa = kare = / (1 + kare)

I1

3Rare

a 2E

aE

(c)

(a) Source voltage Neutral grounded point aE

148

E O a2 E

NVa NV0 NVc

Ia O n

Point m

NVb

mVa

Ia

Point f

O mVb

mVc mV0

fVa

Ia fV0 =

O

fVc

fVb

–1 E 1 + kare

n

(d)

3RN>>q x0,

NV0 ⇋ mV0

Figure 7.8 Phase-a to ground fault: 1ϕG (high-resistance neutral grounding system).

1 ) 3

7.5 Visual Vector Diagrams of Voltages and Currents under Fault Conditions

Z0 = 3RN + j

px0

+ q x 0 ≒ 3RN ,

3RN

px0

+ q x 0 = x0

p q Z0 3RN + j p x 0 + q x 0 3RN v= = ≒ jx1 jx1 jx1 q x1

k1 =

,

x1 = p x1 + q x1 = 1 − k1 x1 + k1 x1

3Rarc , 3RN

Δ = 3RN + j x0 + x1 + x2 ≒ 3RN

x1

karc =

E E E 1 ≒ = Δ + 3Rarc 3RN + 3Rarc 3RN 1 + karc

I1 = I2 = I0 = f Varc

= 3Rarc

E f V0

f V1

f V2

= − jx1

E m V0

=

E m V1

=

E m V2

=

E f Va

f Vb

E

=

m Va

E m Vb

E m Vc

E

f V1

E f V2

E

f V0

E f V0

= = =

E

E

=

E

f V0

f V0

=

E

②

1 1 = 1 RN 1+ 1+ karc Rarc

I0 −RN ≒ = E 3RN + 3Rarc

−1 −1 = Rarc 1 + karc 1+ RN

③

E − jx1 I1 jx1 ≒ 1− ≒1 E 3RN + 3Rarc

=

E

f Vc

I0 3Rarc ≒ = E 3RN + 3Rarc

= − 3RN + jx0

E

①

E

jq x0 I0 f V0 −1 f V0 = + = ≒ 1 + karc E E 3RN + 3Rarc E

+ jq x1

jq x1 I1 f V1 f V1 = + ≒1 ≒ E E 3RN + 3Rarc E

+ jq x1

jq x1 I2 f V2 f V2 = + ≒0 ≒ E E 3RN + 3Rarc E

f V1

+a

m V0

E m V0

+

E

f V1

+a

E

f V1

+ a2

E m V1

E

+ a2 +a

f V2

+

E

+ a2

E

7 23

+ jq x0

+

m V0

E

I2 −jx1 ≒ ≒0 E 3RN + 3Rarc

+

m V1

E

m V1

E

≒

−1 +1 1 + karc

f V2

E f V2

E

m V2

E +a

+ a2

④

≒

−1 + a2 1 + karc

≒

−1 +a 1 + karc

≒

f V0

m V2

E m V2

E

E ≒ ≒

+1≒ f V0

E f V0

E

⑤

f Va

E

+ a2 ≒ +a≒

f Vb

E

f Vc

E

where

f V0

E

=

−1 1 + karc

⑥

149

150

7 Fault Analysis Based on Symmetrical Components

7.5.6

Double Line-to-Ground (Phases b and c) Fault: 2ϕG (High-Resistance Neutral Grounding System)

Equation (7.24) and Figure 7.9 are derived in a similar fashion:

1+ k1 E 2 1– k1 mV2⇋ (1– k1) fV1 = 2 E 1 fV1 = fV2 = fV0 ⇋ 2 E 1 mV0 ⇋ fV0 ⇋ 2 E

The figure for k1 = 0.4

mV1⇋

I1 E

p x1

q x1 m V1 x 1

p x1

q x1 mV2 x1

px 0

qx 0

fV1

E

I2

∆

f V2

I0 3RN mV0 Z 0

px1 (1–k1)x1

fV0

(b)

mVa

Ic

fVa

mV2

O mVc

fV1

n a2 ·m V 1 mV0 mVb

fV2

Ic

mV1

V1 a· m mV 2

qx1 k1x1

x1

(a)

a 2·

(1+ k1) fV1 =

n a2 ·f V

V1 a· m

1

fV0

a 2·

f V2

V2 a· m

O

V2 a· f

fVb = fVc =

0

f Ia = 0

Ib

Ib

(c)

Source voltage

E O E a a2 E

Point m Ic Ic

Point f

m Va

O mVc

O

mVb

Ib

Ic

fVa

O Ib (d)

Ib

Figure 7.9 Double line-to-ground (phases b and c) fault: 2ϕG (high-resistance neutral grounding system).

7.6 Three-Phase-Order Misconnections

Z0 = 3RN + j v=

px

p

0 + q xq 0 ≒ 3RN

jx1

Z0 3RN ≒ jx1 jx1

k1 =

q x1

x1

x1 = p x1 + q x 1 = 1 − k1 x1 + k1 x1

,

Δ = jx1 +

①

Z0 jx1 ≒ jx1 + jx1 = j2x1 Z0 + jx1

I1 =

E 1 E ≒ −j Δ 2 x1

I2 =

− Z0 1 E I1 ≒ − I1 = j 2 x1 Z0 + jx1

I0 = −

jx1 I1 ≒ 0 Z0 + jx1

f V1

f V2

=

=

f V0

E E E 1 f V0 m V0 ≒ ≒ E E 2 m V1

E

=

f V1

E

= −jx1

+ jq x1

②

I2 1 ≒ E 2

③

I1 f V1 1 1 = + k1 ≒ 1 + k1 2 2 E E

④

I2 f V2 1 1 f V2 = + jq x1 = − k1 ≒ 1 − k 1 2 2 E E E E Ia = 0 m V2

Ib = 0 + a2 I1 + aI 2 = − j 3I1 = − Ic = 0 + aI 1 + a2 I2 = − Ib = f Va

E f Vb

E

=3

f V1

E

=

3 E 2 x1

⑤

3 E 2 x1

3 2

=0

⑥

f Vc

=0 E 1 1 + k 1 1 − k1 3 m Va ≒ + + = 2 2 E 2 2 m Vb

E m Vc

E

7.6

≒

3 1 1 + k1 1 −k1 +a = −j k1 + a2 2 2 2 2

≒

3 1 1 + k1 1 −k1 +a k1 + a2 =j 2 2 2 2

⑦

Three-Phase-Order Misconnections

As the final section of this chapter, we will try to calculate the circuit with a phase misconnection.

7 24

151

152

7 Fault Analysis Based on Symmetrical Components

Phase-abc to acb Misconnection

7.6.1

This problem is shown in Figure 7.10a. If the two balanced circuits are connected normally, the voltages and currents at the connecting point are given by the equations E1 − E1 Z1 + Z1 i2 = i0 = 0

i1 =

E1 − E1 Z1 + Z1

ia =

①

②

7 25

2

ib = a ia , ic = aia

Now we will examine the case of a wrong connection. The related equations are va vb vc

va vc , vb

ia ib ic

vabc

vacb

iabc

iacb

va vc vb

1 0 0

0 1 0

va vb , vc

ia ic ib

vabc

iacb

=

ia ic ib

=

7 26a

and =

0 0 1 k1

vacb

1 0 0

=

0 0 1

0 1 0

ia ib ic

k1

7 26b

iabc

Then, from Eqs. (7.26a, b), vacb = k 1 vabc

iacb = k 1 iabc

7 27

Transforming into symmetrical components, v012 = a vabc = a k 1 a −1 v012 , i012 = a k 1 a −1 i012 a k1 a

−1

1 1 1 = 3 1

E1

Z1, Z2, Z0

1 a a2

1 a2 a

1 0 0

ia

va va′

ia′

ib

vb vb′

ib′ ic′

ic vc

vc′

0 0 1

0 1 0

1 1 1

1 a2 a

1 1 a = 0 a2 0

1-circuit

2-circuit

i2′ v2′

v2

Z2

a

E′1 Z1′

Z 2′

E1 Z1 + Z1′

a2

ic

Z1 + Z1′

ic

a i2

E1

a

v1′

v1

Z1

7 28

Z1′, Z2′, Z0′

i1′

E1

0 1 0

E1′

(a) i1

0 0 1

E1′ Z1 + Z1′ E1 a2 Z1 + Z1′ E1

Z0

v′0

Z 0′

(b) Figure 7.10 Phase-abc to acb misconnection.

E1 Z1 + Z1′ ia

i′0 v0

Z1 + Z1′

Z1 + Z1′

ib 0-circuit

E1′

Z1 + Z1′

E1′

ia i0

a

a2

E1 Z1 + Z1′

E1′ Z1 + Z1′

(c1) normal connection

a2

ib

E1′

E1

Z1 + Z1′

Z1 + Z1′ ib (c2) misconnection

7.6 Three-Phase-Order Misconnections

Now we have found the following symmetrical equation in the case of an acb misconnection: v0 v0 v1 = v2 , v2 v1

i0 i0 i1 = i2 i2 i1

7 29

The equivalent circuit in Figure 7.10b is derived from this equation. The voltages and currents at the connecting point are derived from the equivalent circuit, and i1 =

E1 − E1 i2 = i0 = 0 Z1 + Z2 Z1 + Z2

7 30

v2 = − i 2 Z 2 v0 = 0

v1 = i1 Z2

0 1 ia ib = 1 ic 1

1

1

2

a

a

2

a

0

E1 Z1 + Z2

a

−

1 va vb = 1 vc 1

,

E1 Z1 + Z2

1

1

2

a

a

2

a

a

E1

Z2 Z1 + Z2

7 31

E1 Z2 Z1 + Z2

These are the voltages and currents at the connecting point. Accordingly, Normal connection E1 − E1 ia = Z1 + Z1 ib =

a2 E1 − E1 Z1 + Z1

ia =

a E1 − E1 Z1 + Z1

Misconnection E1 − E1 ia = Z1 + Z1 ①

ib =

a2 E1 −a E1 Z1 + Z1

ia =

aE 1 − a2 E1 Z1 + Z1

②

7 32

where Z1 = Z2, Z1 = Z2 . As can be seen in the vector diagram in Figure 7.10c2, a large negative-sequence current appears, and extremely large unbalanced phase currents ib, ic arise. Of course, such a situation would seriously affect generators, most other station equipment, and every kind of load.

7.6.2

Phase-abc to bca Misconnection

The next problem is shown in Figure 7.11a. The related equations are va vb vc vabc

1 1

= 1 k2

va ia vb , i b vc ic vabc

1 1

= 1

iabc

k2

ia ib ic

7 33

iabc

Transforming into symmetrical components, v012 = a k 2 a − 1 v012 , i012 = a k 2 a − 1 i012 1 where a k 2 a − 1 = 0 0

0 a2 0

0 0 a

7 34a

153

154

7 Fault Analysis Based on Symmetrical Components

E1

ia

va

ib

vb

v′a i′a v′ i′b

E′1

b

ic

i′c vc

(ic)

ic

v′c

ia =

E1 – a2E′1

(a) i1

ai1

a2E′1

i′1

E1 1-circuit

Z1 + Z′1

E′1 v1

a

av′1 a2i2

i2

Z1 + Z′1

v′1

(ia)

i′2

E1

(ib) a2

v2

2-circuit

a2v2

i0

Z1 + Z′1

v′2 i′0

ib

E1 Z1 + Z′1

v′0

v0

0-circuit

(b)

ia ib ic : current by misconnection (ia) (ib) (ic) : current by normal connection (c)

Figure 7.11 Phase-abc to bca misconnection.

So, the symmetrical boundary equations are v0 v0 v1 = a 2 v2 , v2 av1

i0 i0 i1 = a2 i2 i2 ai1

7 34b

The equations of the right- and left-side circuits are v0 = −Z0 i0

v0 = Z0 i0

v1 = E1 − Z1 i1

v1 = E1 + Z1 i1

v2 = −Z2 i2

v2 = Z2 i2

7 35

From Eqs. (7.34b) and (7.35), the currents on the misconnected point are derived as follows: E1 −a2 E1 Z1 + Z1 i2 = i0 = 0

i1 =

ia =

E1 − a2 E1 Z1 + Z1

ib = a2 ia

7 36

7 37

ic = aia The equivalent circuit is shown in Figure 7.11b, where the vector operators a and a2 are inserted. In this particular case, extremely large phase-balanced currents flow to all parts of the network, although negative- and zero-sequence currents do not exist. Needless to say, misconnections should be avoided.

155

8 Fault Analysis with the αβ0-Method 8.1

αβ0-Method (Clarke-Components)

The αβ0-coordinate method (αβ0-components or Clarke components) is another tool for fault analysis that can often compensate for weak points of symmetrical analysis. There are some circuit conditions for which αβ0-components can give solutions via easier calculation processes rather than the 012-method; further, some circuit conditions can be solved only with αβ0-components and cannot be solved with symmetrical components. A typical example is the calculation of transient phenomena caused at the time of current-zero tripping of the breaker’s second pole, a few milliseconds after the breaker’s first pole trips. This will be introduced in Chapter 15, Section 15.11. The αβ0-coordinate method is also mathematically a kind of variable transformation of the 3 × 3 matrix operators α, α−1; the important characteristic of αβ0-components is that the transformation operators α, α−1 contain only real-number matrix elements, while the symmetrical components method uses matrix operators a, a−1 based on complex numbers a, a−1. For example, in a circuit with badly distorted waveform currents and voltages, we may be able to observe the real-time waveforms of voltages and currents with various harmonic components as real-number time-series values in an oscillograph, but we do not know the equations for the source voltages and circuit elements. In this case, we need to find the circuit conditions from the observed real-number electrical quantities; the observed real-time abc-values can be transformed into αβ0-components, but they cannot be easily transformed into a 012-complex number expression. Complex number expressions including symmetrical components may not be useful tools when the circuit equations are unknown. We will study first the definition and concepts of the αβ0-coordinate method, and then the mutual relationship of abcphase quantities, αβ0-quantities, and 012-quantities. Finally, we will examine system modeling and fault analysis with αβ0-components. 8.1.1

Definition of the αβ0-Coordinate Method (αβ0-Components)

The αβ0-voltage and current quantities in the αβ0-coordinate method (αβ0-components) are defined with the following equations. The transformation is Vα Vβ V0

2 1 0 = 3 1

−1

−1

3 1

− 3 1

V αβ0 =

Iα Iβ I0 I αβ0 =

=

1 0 3 1

−1

−1

3 1

− 3 1

α

∴ V αβ0 = α V abc

V abc

α

2

Va Vb Vc

Ia Ib Ic

81 I αβ0 = α I abc

I abc

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

156

8 Fault Analysis with the αβ0-Method

and the inverse transformation is Va Vb Vc

1

0

1 2 1 − 2

3 2 3 − 2

−

=

V αβ0 =

1 Vα Vβ V0

1 1

V abc = α −1 V αβ0

V αβ0

α

Ia Ib Ic

1

0

1 2 1 − 2

3 2 3 − 2

−

=

I αβ0 =

8 2a

1 Iα Iβ I0

1 1

∴ I abc = α −1 I αβ0

I αβ0

α−1

where α × α −1 = α1 α = 1 α, α

−1

1 is the unit matrix

inverse matrices of each other

The transform operator matrices α, α−1 are real-number matrices in the αβ0-coordinate method. Therefore if Va, Vb, Vc are given as real-number quantities, Vα, Vβ, V0 are also real-number quantities, while if Va, Vb, Vc are given as complex-number quantities, Vα, Vβ, V0 are also complex-number quantities. The relation abc-phase quantities αβ0-quantities is shown in Figure 8.1a and b, where both quantities are demonstrated as vector values. Note: The operational matrix α−1 of the αβ0-method can be written as follows.

α −1 =

1

0

1 2 1 − 2

3 2 3 − 2

−

1 1

cos 0 =

1

2π 3 4π cos 3 cos

sin 0 2π 3 4π sin 3 sin

1 1

8 2b

1

The mathematical form of α−1 is similar to Eq. (10.12) D−1 of the dq0-method but under the condition θa = 0. Therefore, the αβ0-method may be said to be almost a special case of the dq0-method from a mathematical viewpoint.

8.1.2

The Transformation of Arbitrary Waveform Quantities

Suppose voltages have an arbitrary waveform that may include DC components, harmonic components, and powerfrequency components. Then n

Va t =

Vak

e j kωt + θak

Vbk

e j kωt + θbk

Vck

e j kωt + θck

k =0 n

Vb t = k =0 n

Vc t = k =0

Complex-number expression

8 3a

8.1 αβ0-Method (Clarke-Components)

Ia= Iα

b phase c phase

3 I +I 2 β 0 Iα – 3 Iβ +I0 2

Ib = –

1 I + 2 α

–

1 I 2 α

Ic = –

1 2

–

1 I 2 α

Ia+Ib+Ic = 3I0

Iα

2

Ia

–

I0

1 I 2 α

1 I 2 α

I0

1 I + 2 α

β circuit

0 circuit

Iα + Iβ + I0 = Ia +

I0 I0 I0

(a)

Phase-b current

2I 3 a 0

=

1I 3 a

3 I – 3 I 3 b 3 c

3I 0

3I0 earth

Iβ

3 I +I 2 β 0

Iα = 2 Ia – 1 Ib – 1 Ic 3 3 3 3 Iβ = 0 + I – 3 Ic 3 b 3 1 1 1 I0 = Ia + Ib + Ic 3 3 3

I0

Ia

Phase-a current α circuit

I0

+

3 2 Iβ

1

–

Iα +

0-circuit current I0

0

– 3 2 Iβ +I 0

–

–

Ic =

Iα

2

3 2 Iβ

1 2 Ia

– 3 2 Iβ 1

–

Ia =

Ib = –

+

0

I0

I0

β-circuit current 0 3 I 2 β 3 – I 2 β

– 3 2 Iβ

a phase

α-circuit current Iα

+I0

– 1 Ib 3 3 I 3 b 1I 3 b

+

Phase-c current

+

3 I 3 b

Ia

– 1 Ic 3 – 3 Ic 3 1I 3 c – 3 Ic 3 Ic

Ic

I0

2 3 Ia

Iβ

– 1 Ib 3

I Iα – 1 I a 3 c I 3 I b 3 b

2I 3 α 1I 3 α

3 I 3 b Ia

– 3 Ic 3 (b)

– 1 Ib 3

Ib

1I 3 b

1I 3 c – 1 Ic 3

– 3 Ic 3

α-quantity: The currents Iα , –1/2 Iα, –1/2 Iα flow out on the phase a, b, c circuits, respectively, in the same direction. In other words, the current Iα goes out from phase a and the half current 1/2 Iα comes back from phase b and another half current 1/2 Iα from phase c. As a result, the current through the ground pass is zero. β-quantity: The currents +

3/2 Iβ, – 3/2Iβ, flow in phase b, c, respectively, in the same direction. In other words, current

3/2 Iβ goes out from phase b and comes back from phase c. As a result, the currents through phase a and the ground pass are zero. 0-quantity: The currents of the same value I0 flow out from phase a, phase b, phase c, respectively, in the same direction and the current 3I0 comes back from the ground pass. In other words, the definition of 0-quantitiy is exactly the same as the zero-sequence quantity in symmetrical components.

Figure 8.1 Correlation of vectors Ia, Ib, Ic and Iα, Iβ, I0.

157

158

8 Fault Analysis with the αβ0-Method n

Vak cos kωt + θak

va t = k =0 n

Vbk cos kωt + θbk

vb t =

Real-number expression

8 3b

k =0 n

Vck cos kωt + θck

vc t = k =0

These quantities in the abc-domain are transformed into two different domains as follows. For symmetrical coordinates, referring to Eqs. (3.18) and (3.21) and recalling that a = e j120 , a2 = e− j120 , then the complex-number expression is Va t + Vb t + Vc t 1 Va t + aV b t + a2 Vc t = 3 Va t + a2 Vb t + aV c t

V0 t V1 t V2 t

n

Vak =

1 3

k =0 n

Vak k =0 n

Vak

n

e j kωt + θak +

Vbk k =0 n

e j kωt + θak +

k =0 n

e j kωt + θak +

Vck

e j kωt + θck

k =0

Vbk Vbk

k =0

n

e j kωt + θbk +

n

e j kωt + θbk + 120 +

e j kωt + θck −120

Vck

e j kωt + θck + 120

k =0 n

e j kωt + θbk −120 +

k =0

Vck

8 4a

k =0

For the real-number expression, taking the real part of the previous equation, Re V0 t = Re V1 t Re V2 t

v0 t v1 t v2 t n

n

n

Vak cos kωt + θak + =

1 3

k =0 n

Vbk

cos kωt + θbk +

k =0 n

Vak cos kωt + θak + k =0 n

Vck cos kωt + θck k =0

Vbk

cos kωt + θbk + 120 +

k =0 n

Vak cos kωt + θak + k =0

8 4b

n

Vck cos kωt + θck − 120 k =0 n

Vbk

cos kωt + θbk −120 +

k =0

Vck cos kωt + θck + 120 k =0

The αβ0-components are defined for complex-number and real-number expressions with Eqs. (8.1) and (8.2a). Then, for the complex-number expression, Vα t Vβ t V0 t

1 = 3

2Va t − Vb t −Vc t 3 Vb t −Vc t Va t + Vb t + Vc t n

2 k =0

=

1 3

n

Vak e j kωt + θak −

k =0 n

3

Vbk

k =0

Vak e j kωt + θak +

n

Vck e j kωt + θck

k =0

e j kωt + θbk −

k =0 n

Vbk e j kωt + θbk − n

Vck e j kωt + θck

k =0 n

Vbk k =0

e j kωt + θbk +

n k =0

Vck e j kωt + θck

8 5a

8.1 αβ0-Method (Clarke-Components)

and for the real-number expression vα t vβ t v0 t

1 = 3

2va t − vb t − vc t 3 vb t −vc t va t + v b t + v c t n

n

k =0

=

Vbk

Vck cos kωt + θck k =0

n

1 3

cos kωt + θbk −

k =0

8 5b

n

3

Vbk

cos kωt + θbk −

k =0

Vck cos kωt + θck k =0

n

n

n

Vak cos kωt + θak + k =0

8.1.3

n

Vak cos kωt + θak −

2

Vbk cos kωt + θbk + k =0

Vck cos kωt + θck k =0

Interrelation Between αβ0-Components and Symmetrical Components

The symmetrical-components method is a kind of one-to-one transformation between the domains abc 012. The αβ0-components method is also a one-to-one transformation between the domains abc αβ0. Therefore, symmetrical components and αβ0-components should be the same kind of one-to-one transformation between the domains 012 αβ0 for each other. Let’s examine the relation between 012 quantities and αβ0 quantities in detail. We will examine the relation using voltage symbols here, where a lowercase letter v means instantaneous real-number voltages, and an uppercase letter V means instantaneous complex-number voltages. We need to examine the mutual relation of both domains. Applying the complex-number expression, The equations of symmetrical components from Eqs 3 18 and 3 21 V 012 = a V abc ① V abc = a − 1 V 012 ②

86

The equations of αβ0 − components from Eqs 8 1 and 8 2 V αβ0 = α V abc ③ V abc = α − 1 V αβ0 ④ Accordingly, V αβ0 = α V abc = α a −1 V 012 = α a − 1 V 012

87

V 012 = a V abc = a α − 1 V αβ0 = a α −1 V αβ0 where α a−1 and a α−1 are calculated as follows:

α a

−1

2 1 0 = 3 1

−1 3 1

0 1 0 = 3 3

2 − a2 + a

2 − a + a2 2

1 a α

−1

1 1 = 3 1

−1 − 3 1

1 0 1

1 a2 a

2

3 a −a

3 a− a

2

1 + a + a2

1+a +a

1

1

a

a2

a2

a

1

0

1 2 1 − 2

3 2 3 − 2

−

1 a a2 0 = 0 1

1 −j 0

1 j 0

88

1 1 1

0 1 1 = 2 1

0 j −j

2 0 0

159

160

8 Fault Analysis with the αβ0-Method

Therefore Vα Vβ V0

=

0 0 1

V αβ0

1 −j 0

1 j 0

V0 V1 V2 V 012

α a −1

or Vα t = V1 t + V2 t

89

Vβ t = −j V1 t −V2 t V0 t = V0 t or

V0 V1 V2 V 012

=

0 1 1

0 j −j

2 0 0

a α−1

Vα Vβ V0 V αβ0

V0 t = V0 t 1 Vα t + jV β t V1 t = 2 1 Vα t − jV β t V2 t = 2

8 10

Equations (8.9) and (8.10) show the relation between αβ0-components and symmetrical components that are written in complex-number quantities. In words, Vα is the vector sum of positive-sequence voltage V1 and negative-sequence voltage V2: V1 + V2. Vβ is the product of (−j) and (V1 – V2), or the vector obtained with a 90 clockwise rotation of subtracted vector (V1 – V2). For the relation for power-frequency components, the symmetrical components are With complex-number expression

With real-number expression

V0 t = V0 e

j ωt + θ0

v0 t = V0 cos ωt + θ0

V1 t = V1 e

j ωt + θ1

v1 t = V1 cos ωt + θ1

V2 t = V2 e j ωt + θ2

v2 t = V2 cos ωt + θ2

8 11

and the αβ0-components are, for the complex-number expression, Vα t = V1 t + V2 t = V1 e j ωt + θ1 + V2 e j ωt + θ2 Vβ t = − j V1 t − V2 t

= e −j90

V0 t = V0 e j ωt + θ0 and for the real-number expression vα t = V1 cos ωt + θ1 + V2

V1 e j ωt + θ1 − V2 e j ωt + θ2

①

8 12

cosωt + θ2

vβ t = V1 cos ωt + θ1 − 90 − V2 cos ωt + θ2 − 90 = V1 sin ωt + θ1 − V2 sin ωt + θ2

②

v0 t = V0 cos ωt + θ0 Figure 8.2 is a summary of these concepts showing mutual relations of the abc-, 012-, and αβ0-domains.

8.1.4

Circuit Equation and Impedance with the αβ0-Coordinate Method

The general equation for a three-phase circuit is expressed with E abc − V abc = Z abc I abc

8 13

The equation is transformed into the 012-domain as follows: E abc = a −1 E 012 ,

V abc = a −1 V 012 ,

∴ E 012 −V 012 = a Z abc a − 1 I 012

I abc = a − 1 I 012 Z 012 I 012

where Z 012 = a Z abc a

−1

8 14

8.1 αβ0-Method (Clarke-Components)

V012 α –1 1 1 1 V0 1 a2 a V1 1 a a2 V2

Vabc Va = Vb Vc

a-b-c domain V Vabc= a Vb Vc

V012 Vabc α V0 = 1 1 1 1 Va 3 V1 1 a a2 Vb V2 1 a2 a Vc

0-1-2 domain V012=

V0 V1 V2

Vabc α –1 Va = 1 0 1 1 3 – Vb 2 2 1 1 – – 3 1 Vc 2 2 Vαβ0 α Vα = 1 2 –1 –1 Vβ 3 0 3 – 3 V0 1 1 1

Vαβ0 Vα Vβ V0 Vabc Va Vb Vc

2

V 01

V0 V1 1 V2 α 1 j 0 j 0 0 0 – β V Vα = 0 0 1 Vα – 1 Vβ Vα ·a 2 V0 α β 0 0 V 0 j 0 V0 1 1 –j 2 V 01 0 = 2 1 V V1 V2 –1

·a

- -0 domain

V

0=

Vα Vβ V0

Figure 8.2 Correlation of the abc-, 012-, and αβ0-domains.

Equation (8.13) can be transformed into the αβ0-domain in a similar way. The definition of αβ0-components is E αβ0 = α E abc ,

Eabc = α − 1 E αβ0

V αβ0 = α V abc ,

V abc = α −1 V αβ0

I αβ0 = α I abc ,

I abc = α

−1

8 15

I αβ0

Substituting this equation into Eq. (8.13), α − 1 E αβ0 −α −1 V αβ0 ,

= Z abc α −1 I αβ0

Left-multiplying with α and recalling that α α−1 = 1, the equation in the αβ0-domain is E αβ0 − V αβ0 = α Z abc α −1 I αβ0

Z αβ0 I αβ0

where Z αβ0 = α Z abc α −1

8 16

Extracting the equations for the impedance from Eqs. (8.14) and (8.16), Z 012 = a Z abc a − 1 Z αβ0 = α Z abc α −1

or

Z abc = a −1 Z 012 a Z abc = α − 1 Z αβ0 α

∴ Z αβ0 = α Z abc α −1 = α a − 1 Z 012 a α − 1 = α a −1 Z 012 a α −1

8 17a 8 17b

Accordingly, the circuit equation and impedance in the αβ0-domain are E αβ0 − V αβ0 = Z αβ0 I αβ0 where Z αβ0 = α Z abc α −1 = α a − 1 Z 012 a α − 1 α a−1 and a α−1 in this equation were derived in Eq. (8.8).

8 18

161

162

8 Fault Analysis with the αβ0-Method

Now we can draw the conclusion that the impedances in αβ0-domain circuits are given with Eq. (8.18). Further, Eq. (8.13) for the abc-domain, Eq. (8.14) for the 012-domain, and Eq. (8.18) for the αβ0-domain are in one-to-one correspondence with each other. In the next section, we will investigate Zαβ0 for lines and other equipment. 8.1.5

Single-Circuit Transmission Lines

A well-balanced three-phase single-circuit transmission line between points m and n, as shown in Figure 1.1, has its impedance matrix Z012 given with Eq. (3.34) and Figure 3.11, and is again shown here: 0 Z1 0

Z0 Z 012 = 0 0

0 0 Z2

Z1 = Z2 = Zs − Zm

8 19

Z0 = Zs + 2Zm

Accordingly, Zαα Z αβ0 = Zβα Z0α 0 = 0 1

Zαβ Zββ Z0β 1 −j 0

Zα0 Zβ0 Z00

1 j 0

Z0 0 0

1 Z1 + Z2 2 1 − j Z1 − Z2 2 0

=

= α a − 1 Z 012 a α − 1

0 Z1 0

1 j Z1 − Z2 2 1 Z1 + Z2 2 0

0 0 Z2

0 1 1 2 1

0 j −j

2 0 0

8 20

0 0 Z0

As Z1 = Z2 is always correct for transmission lines, then m Vα m Vβ

n Vα

−

n Vβ

m V0

n V0

m V αβ0

n V αβ0

=

Z1 0 0

0 Z1 0

0 0 Z0

Z αβ0

Iα Iβ I0

Z1 = Z2 = Zs − Zm Z0 = Zs + 2Zm

Iαβ0

I𝛼 Z1 = Z2 = Zs – Zm 𝛼-circuit

mV𝛼

nV𝛼

I𝛽 Z1 = Z2 = Zs – Zm 𝛽-circuit

mV𝛽

nV𝛽

I0 Z0 = Zs + 2Zm 0-circuit

mV0

nV0

Figure 8.3 The equivalent circuit for a transmission line in αβ0-components (single-circuit line).

8 21

8.1 αβ0-Method (Clarke-Components)

This is the fundamental equation for the transmission line in αβ0-components; the equivalent circuit of the equation is given in Figure 8.3. The impedance matrix Zαβ0 is the same as Z012: Zαβ0 = Z 012. In other words, the major feature of αβ0-components is that a well-balanced transmission line can be expressed with the same impedance matrix and the same equivalent circuits in the αβ0-domain as well as in the 012-domain. The strange coefficient ± 3 2 in the definition of αβ0-components was a device used to obtain this advantage.

8.1.6

Double-Circuit Transmission Lines

A well-balanced double-circuit line is shown in Eqs. (3.39a, b) in the 012-domain, and is again shown here: 1 m V 012 2 m V 012

−

1 n V 012 2 n V 012

Z 012 Z 0M

=

Z 0M Z 012

1 2

I 012 I 012

or 1 m

V 012 −

1 n

V 012 = Z 012 1 I 012 + Z 0M 2 I 012

2 m

V 012 −

2 n

V 012 = Z 0M 1 I 012 + Z 012 2 I 012 0 Z1 0

Z0 Z 012 = 0 0

where

0 0 Z1

8 22

Z0M Z 0M = 0 0

0 0 0

0 0 0

Equation (8.22) can be transformed into equations in the αβ0-domain in the same way as in Eqs. (8.14)–(8.18): 1 m

V αβ0 −

V αβ0 = α a −1 Z 012 a α −1

1 n

+ α a −1 Z 0M a α −1 2 m

V αβ0 −

+ α a −1 Z 012 a α −1 1 m V αβ0 2 m V αβ0

or

=

−

−1

Z αβ0 Z αβ0

I αβ0

I αβ0

1 2

I αβ0

I αβ0

1 n V αβ0

8 23

2 n V αβ0

α a − 1 Z 012 a α − 1 αa

2

V αβ0 = α a −1 Z 0M a α −1

2 n

1

Z 0M a α

−1

Z αβ0 Z αβ0

I αβ0 I αβ0

1 2

α a −1 Z 0M a α −1 αa

−1

Z 012 a α

Zαβ0 is in the same form as Eq. (8.15),

Z αβ0 = α a

−1

Z 012 a α

−1

Z1 = 0 0

0 Z1 0

0 0 Z0

−1

1

I αβ0

2

I αβ0

163

164

8 Fault Analysis with the αβ0-Method

and Z αβ0 is Z αβ0 = α a −1 Z 0M a α −1 0 0 1

=

1 −j 0

1 j 0

Z0M 0 0

0 0 0

0 0 0

0 1 1 2 1

0 j −j

2 0 0 = 0 0 0

0 0 0

0 0 Z0M

Therefore, the fundamental equation for the double-circuit line in αβ0-components is 1 m Vα 1 m Vβ 1 m V0 2 m Vα 2 m Vβ

−

1 n Vα 1 n Vβ 1 n V0 2 n Vα 2 n Vβ

2 m V0

2 n V0

α-circuit

1 m Vα 2 m Vα

β-circuit

1 m Vβ 2 m Vβ

0-circuit

1 m V0 2 m V0

Z1 0 0 = 0 0 0

0 Z1 0 0 0 0

0 0 Z0 0 0 Z0M

0 0 0 Z1 0 0

0 0 0 0 Z1 0

0 0 Z0M 0 0 Z0

1

Iα Iβ 1 I0 2 Iα 2 Iβ 2 I0 1

8 24a

or Z1 = 0

0 Z2

1

−

1 n Vα 2 n Vα 1 n Vβ 2 n Vβ

=

Z1 0

0 Z2

1

−

−

1 n V0 2 n V0

Z0 = Z0M

2

2

Iα Iα

Iβ Iβ

8 24b 1

Z0M Z2

2

I0 I0

where, referring to Eq. 8.19, Z1 = Z2 = Zs − Zm Z0 = Zs + 2Zm

Z0M = 3Zm

The equivalent circuit corresponding to this equation is given in Figure 8.4. Now we can conclude that the α-circuit and β-circuit of the double-circuit transmission line can be expressed with the positive-sequence equivalent circuit, and the 0-circuit of course with the zero-sequence equivalent circuit. 8.1.7

Generators

The generator circuit is described in Eq. 3.45b and Figure 3.16 in the 012-domain, and is again shown here:

0 Ea 0

−

V0 V1 V2

=

E 012

−

V 012

=

Z0 0 0

0 Z1 0 Z 012

0 0 Z1

I0 I1 I2

+

3Zn I0 0 0

I 012

+

3Z n I 0

8 25

8.1 αβ0-Method (Clarke-Components)

Z1 = Z2 = Zs – Zm

1I 𝛼

Z1

2I 𝛼 1V m 𝛼

𝛼-circuit

1V n 𝛼 2V n 𝛼

2V m 𝛼

Z1 = Z2 = Zs – Zm

1I 𝛽

Z1 = Z2

2I 𝛽 1V m 𝛽

𝛽-circuit

1V n 𝛽 2V n 𝛽

2V m 𝛽

1I

0

2I

0

Z0 = Zs + 2Zm Z0 Z0M = 3Z′m

1V m 0

0-circuit

1V n 0 2V n 0

2V m 0

Figure 8.4 The equivalent circuit for a transmission line in αβ0-components (double-circuit line).

Left-multiplying by α a−1, ∴ E αβ0 − V αβ0 = Z αβ0 I αβ0 + α a −1 3Z n I 0 where Zαβ0 is in the same form as Eq. (8.20), and

E αβ0 = α a

αa

−1

−1

0 E 012 = 0 1

0 3Zn I 0 = 0 1

1 −j 0

1 −j 0

1 j 0

1 j 0

0 Ea = 0

Ea − jE a 0

3Zn I0 0 0 0 = 0 3Zn I0

Therefore the generator equation in the αβ0-domain is 0 Vα −jE a − Vβ = 0 V0 where

Z1 = jx1 ,

1 2 Z1 + Z2 1 − j Z1 −Z2 2

1 j Z1 − Z2 2 1 Z1 + Z2 2

0

0

Z2 = jx2

Z0 = jx0

0 0 Z0

Iα 0 Iβ + 0 I0 3Zn I0

8 26

165

166

8 Fault Analysis with the αβ0-Method

Furthermore, the equation can be simplified as follows, if the assumption Z1≒Z2(jx1≒jx2) is justified for fault analysis distant from the generator terminal. For example:

jx1 I𝛼 𝛼-circuit

V𝛼

Ea

Ea Va Z1 −jE a − Vβ = 0 0 V0 0

jx1 = jx2

0 Z1 0

0 0 Z0

Ia 0 Iβ + 0 I0 3Zn I0

8 27a

I𝛽

𝛽-circuit –jEa

V𝛽

Ea −Vα = Z1 Iα

jx0 I0 V0

0-circuit

or

− jE a − Vβ = Z1 Iβ

3Zn

8 27b

−V0 = Z0 + 3Zn I0

Figure 8.5 is the equivalent circuit for Eqs. (8.27a) and (8.27b). As they are based on the assumption jx1 = jx2, some errors may appear if they are adopted for analysis of phenomena around the generator terminal. However, Eq. (8.26) is the equation for the generator with the αβ0-method where the circuit is described with the known symmetrical reactance values, although the equation cannot be replaced with a simple equivalent circuit. In addition, remember that the generator source voltages Ea, − jEa exist on the α-, and β-circuits, respectively, in the αβ0-domain. Figure 8.5 Equivalent circuit for a generator with the assumption jx1≒jx2.

8.1.8

Transformer Impedances and Load Impedances in the αβ0-Domain

Transformers do not include mutual impedances in the 012-domain, as shown in Table 6.1, so Eqs. (8.19) and (8.20) can be applied. Moreover, Z1 = Z2 is always correct, so Eq. (8.21) and Figure 8.3 can be applied for the transformer. In other words, positive-sequence impedance Z1 = jX1 is applied for the α- and β-circuits, and Z0 = jX0 is applied for the 0-circuit. The load-circuit equations assumed in Eqs. (3.46) and (3.47) are in the same form as Eqs. (3.34) and (8.19) for a single-circuit line. Therefore, Eq. (8.21) shows the load equations in the αβ0-domain under the approximation Z1≒Z2.

8.2

Fault Analysis with αβ0-Components

Now that the transformed equations and the equivalent circuit of three-phase circuits in the αβ0-domain have been completed, we can begin fault analysis with the αβ0-method using the process from Figure 3.1 or the similar one from Figure 7.1.

8.2.1

Line-to-Ground Fault (Phase a to Ground Fault: 1G)

We will try to solve phase-a to ground fault from Figure 8.6a with the αβ0-method. This is the same fault mode calculation we saw in Section 7.2, so the solution using the αβ0-method should be consistent with Eq. (7.10), which is the solution using the 012-method (a2–1 = a j 3, etc.) and is shown again here:

8.2 Fault Analysis with αβ0-Components

f Z1 Ea= Ea

I𝛼

f Z1

f I𝛽

–jEa = E𝛽

f

f Z0/2

f Ib =

f Va f Vb f Vc

f Z1

= f Z2 f Z0

×

f V𝛽 ×

f Ia

0 f Ic = 0

f V𝛼

f Va = 0

f I b = f Ic = 0

2f I0

f V0

f V𝛼 + f V0 = 0 f I𝛽 =

0

f I𝛼 = 2f I0

(b) Equivalent circuit in the 𝛼𝛽 0 domain

(a) Fault condition Figure 8.6 Phase-a to ground fault.

3 f Ea , f Ib = f Ic = 0 + f Z2 + f Z0

f Ia

=

f Vb

=

j 3 af Z0 − f Z2 f Z1 + f Z2 + f Z0

f Vc

=

j 3 −a2 f Z0 + f Z2 f Z1 + f Z2 + f Z0 f Ea

f Z1

f Ea

8 28

Referring to Figure 8.6a, the fault condition is f Va

= 0, f Ib = f Ic = 0

8 29

With the αβ0-domain, 1 1 3 3 f Iα + f Iβ + f I0 = 0, − f Iα − f Iβ + f I0 = 0 2 2 2 2 ∴ f Vα = − f V0 , f Iα = 2f I0 , f Iβ = 0 f Vα

+ f V0 = 0, −

The equivalent circuit in Figure 8.6b is derived, where the equation –f V0 = f Z0 (2f I0) in the zero-sequence circuit. Then f Iα

= 2 f I0 =

f Iβ

=0

f I0

is expressed with f V0 = −(f Z0/2)

2 f Ea = f Ea 2 Z + 1 2 Z Z f 1 f 1 + f Z0 f 0

2 f Vα = − f V0 = f Ea − Z1 f Iα = f Ea 2f Z1 + f Z0 f Vβ

8 30

= Eβ = − jf Ea

This is the solution in the αβ0-domain. Then, with inverse transformation,

8 31

167

168

8 Fault Analysis with the αβ0-Method

f Iα

= f Iα + f I0 = 1 5f Iα =

f Ib

=0

f Ic

=0

f Vα f Vb

f Vc

3 f Ea 2Z1 + Z0

= f Vα + f V0 = 0 3 1 − 3j −jf Ea + − f Va = f Vα + f Ea − 3jf Va 2 2 2 − 3j Z0 j 3 aZ 0 − Z1 = f Ea − 3j f Ea = f Ea 2 2Z1 + Z0 2Z1 + Z0 =−

8 32

3 3j 1 −jf Ea + − f Va = f Vα − f Ea + 3jf Va 2 2 2 Z0 j 3 − a2 Z0 + Z1 3j = f Ea + 3j f Ea = f Ea 2 2Z1 + Z0 2Z1 + Z0 =−

Equations (8.32) and (8.28) are the same under the condition f Z1 = f Z2.

8.2.2

The Phase-bc Line to Ground Fault

Suppose the phase-bc l–g fault at point f is as shown in Figure 8.7. The fault condition at f is f Vb

= f Vc = 0, f Ia = 0

8 33

Equation (8.33) can be rewritten as the αβ0-domain equation by using Eq. (8.2a): −

1 1 3 3 f Vα + f Vβ + f V0 = − f V0 − f Vβ + f V0 = 0 2 2 2 2

f Iα

+ f I0 = 0

8 34

∴ f Vα = 2 f V0 f Iα

①, f Vβ = 0 ②

+ f I0 = 0 ③

The network equations are Ea − f Vα = f Z1 f Iα

④

− jE a − f Vβ = f Z1 f Iβ

⑤ ⑥

− f V 0 = f Z 0 f I0 or − 2f V0

= 2f Z0

8 35

⑥

f I0

The equivalent circuit that satisfies both Eqs. (8.34) and (8.35) is shown in Table 8.1 #1A. In order to satisfy Eqs. ① and ⑥ together, the zero-sequence circuit is expressed with ⑥ (instead of ⑥) with terminal voltage 2f V0 and impedance 2f Z0.

Fault point f System

f Ia f Ib

f Va

f Ic

= f Z2 f Z0

f Z1

f Vb f Vc

Figure 8.7 Phase-bc l-g fault.

8.2 Fault Analysis with αβ0-Components

Table 8.1 The equivalent circuits for various faults in the αβ0-coordinates domain. System equations in the αβ0-domain

Fault point f Network

f Ia f Ic

f Z1 ⱌ f Z2 f Z0

Ea − f Vα = f Z1 f Iα

fVa

f Ib

fVb

fVc

f Ea: the voltage at point f before the fault f Z1 ≒ f Z2, f Z0: the system impedances at point f

−jE a − f Vβ = f Z1 f Iβ − f V0 = f Z 0 f I 0 Eq. (8.27b)

#1 Phase-bc line-to-ground fault #1A f Vb

#1B

#1C

= f Vc = 0

f Vβ f Iα

=0

+ f I0 = 0

𝛼-circuit 𝛽-circuit 0-circuit

Ea=E𝛼

f Z1

f Z1

–jEa=E𝛽

f I𝛼 f V𝛼 f I𝛽

Zero-sequence circuit is from the modified equation –2fV0 = (2Z0) f I0

a c b

nZ1 = nZ2 nZ0

nZ1 mZ1 fVa fIa nEa mEa m I𝛽

n I𝛼

mZ1 I –jmEa f bfVb

f Va f Ib

f Vα

=0

= f Ic = 0

+ f V0 = 0

m I0

2mZ0

fI0

= f Vc

f Vβ

=0

f Iβ

=0

f Ia

=0

f Iα

=0

f Iα

= 2f I0

f Ib

+ f Ic = 0

f I0

=0

2Z0′2 V

f Va f Ia

= f Vb = f Vc = 0

= f Ib = f Ic

–jEa=E𝛽 f

f Z0

2Z ′′ n 0 2Z′′ 0 2ZM

Zero-sequence circuit is from the modified f Z0 equation − f V0 = 2f I 0 2 Inverse transform equation to domain-abc f Va = f Vα + f V0 1 3 f Vα + f Vβ + f V0 2 2 1 3 f Vα − f Vβ + f V0 f Vc = − 2 2

f Z1

f I𝛼 fV𝛼

f Z1

I𝛽 V f 𝛽

f Z0

f I0 fV0

Eq 8 2a

f Vα

=0

f Vβ

=0

f Ib

f I 𝛽 f V𝛽

–jEa=E𝛽

2Z ′

#4 Three-phase fault

f Z1

f I𝛽 fV𝛽

nZ 1

f I0

f 0

Ea=E𝛼

fZ1

=−

2ZM′

2Z0′

f I𝛼 f V 𝛼

Ea=E𝛼

f Vb

fV𝛽

f Z1

I𝛼 V f 𝛼

Z0/2 2f I0 V f 0

Z1′′

Z1′′

Z1′ f I𝛽

2nZ0

#3 Phase-b to -c line-toline fault f Vb

nZ1

Z1′′

fV𝛼

Z1′

–jmEa

nI0

Z1 Ea=E𝛼 –jEa=E𝛽

nZ1 –jnEa

Z1′′

Z1′ f I𝛼

mEa

mZ 1

f Ic

Z1′

mZ1

f I𝛼

m I𝛼

2mZ0 2fV0 #2 Phase-a line-to-ground fault

f Ib

fIc

mZ1 = mZ2 mZ0

f V𝛽

2f Z 0 I 0 2f V0

f

f

a b c fIb

f Va = 0 f V α = 2f V 0

I0

f V0

=0

nEa

–jnEn

169

170

8 Fault Analysis with the αβ0-Method

From the equivalent circuit, the following equations are obtained: f Iα

= − f I0 = f Iβ

Ea Z + f 1 2f Z0

=−

jE a f Z1

①

f Vα

= 2f V0 = Ea

f Vβ

=0

2f Z0 Z f 1 + 2f Z0

②

8 36

These are the solutions (fIα, fIβ, fI0), (fVα, fVβ, fV0) in the αβ0-domain. Then these solutions can be inverse-transformed into the abc-domain with Eq. (8.1), or into the 012-domain with Eq. (8.10), as our final solution. This final solution coincides with Eqs. (5C) and (5D) in Table 7.1. Figures 1B and 1C in Table 8.1 show the same equivalent circuit, but for a circuit with a double-source line and double-circuit line.

8.2.3

Other Mode Short-Circuit Faults

The equations and equivalent circuits for the other mode faults are shown in Table 8.1.

8.2.4

Open-Conductor Mode Faults

The equivalent circuits of open-conductor mode faults in αβ0-components can be obtained analogously. Table 8.2 shows the equations and the equivalent circuits.

Table 8.2 Equivalent circuit of a conductor opening in the αβ0-domain. #6 Phase-b, -c conductors opening

#5 Phase-a conductor opening

ia = 0 ib ic

ia ib ic = 0

va vb vc

vα = 2v0

vb = vc = 0

iβ = 0 i𝛼

v𝛼 nZ1

mZ1

mEa

iα = 2i0

ib = ic = 0

Iα + I0 = 0

i𝛼

vα + v0 = 0

va = 0

vβ = 0

ia = 0

mE a

mEa

mZ1

nZ1

2v0

nZ1

nEa

nZ1

–jnEa

i𝛽

–jnEa

–jmEa

mZ1

i0 2mZ0

v𝛼

mZ1

i𝛽 v𝛽 –jmEa

va vb vc

2nZ0

Zero-sequence circuit is from modified equation 2vG = (2Z0)i0 instead of v0 = Z0i0

v𝛽 2i0

mZ0 /2

v0

nZ0 /2

Zero-sequence circuit is from modified equation v0 = instead of v0 = Z0i0

Z0 2

2i0 ,

8.4 Fault-Transient Analysis with Symmetrical Components and the αβ0-Method

8.3

Advantages of the αβ0-Method

Let’s try here to compare the equivalent circuits in αβ0-components shown in Tables 8.1 and 8.2 with the symmetrical components shown in Tables 7.1 and 7.2. The following conclusion may be derived. In the cases shown in Tables 7.1 and 7.2 with the symmetrical method, there is one complicated equivalent circuit in which positive-, negative-, and zero-sequence impedances are connected in series and/or parallel. On the other hand, in Tables 8.1 and 8.2, there are two or three simple, independent circuits. To solve three simple equations is generally easier than solving one complicated equation, regardless of whether a computer is used. This is one of the reasons the αβ0method is a useful approach as a complement to the symmetrical method. Again, it must be stressed that the αβ0-method is as precise as the symmetrical method, at least for Eq. (8.26), instead of Eq. (8.27a), which is adopted for generator equations. Figure 8.8 shows the current flow under the condition of a phase-a l–g fault. The phase currents are represented by arrows indicating the relative magnitudes of currents in each circuit. This is the original figure for αβ0-components first developed by W. W. Lewis. In the figure, the currents through points p1, p2, p3 are the β-, α-, and 0-currents, respectively. Physical current flow based on Kirchhoff’s law and the cancelation law of ampere-turns can be imagined from this figure as either real-number currents or complex-number currents.

8.4

Fault-Transient Analysis with Symmetrical Components and the αβ0-Method

8.4.1

Transmission-Line Equations for Transient Analysis

Now, let’s examine the equations of a three-phase circuit for transient phenomena. Equations (3.3) and (3.4) in Chapter 3 are the steady-state equations for the transmission line shown in Figure 3.1b. The original equation with regard to a phase-a conductor covering transient phenomena is a differential equation m Va

t − n Va t =

ra + rg + Laag + Lg

d d Ia t + rg + Labg + Lg Ib t dt dt

8 37

d Ic t + rg + Lacg + Lg dt

where V(t), I(t) are complex-number expressions. Equation (8.37) and Eqs. (3.3) and (3.4) are in the same form except for the displacement of jω ⟺ d/dt. In exact terms, the general form of the equation was originally a differential equation in d/dt, which can be replaced with jω for limited applications of steady-state analysis. Equation (8.38) using d/dt (instead of jω) can be transformed into the symmetrical domain with the same procedure explained in Chapter 3, resulting in the equation

Generator

p1 A

Tr1

Transmission line p2 p3

Tr2 0

C B

𝛽-current

0 0

b c 𝛼-current

A B C

b 0-current

Figure 8.8 The αβ0-component currents through transformers under l-g fault.

0

171

172

8 Fault Analysis with the αβ0-Method

r0 + L0 m V0

t m V1 t m V2 t

n V0

t − n V1 t n V2 t

=

where

d dt

0

0

d r1 + L1 dt

0

0

0 0 r1 + L1

r0 = rs + 2rm ,

L0 = Ls + 2Lm

r1 = rs −rm ,

L1 = Ls + Lm

d dt

I0 t I1 t I2 t 8 38

and V t , I t are complex-number quantities Extracting the real part of this equation,

r 0 + L0 m v0

t m v1 t m v2 t

n v0

t − n v1 t n v2 t

=

d dt

0

0

d r 1 + L1 dt

0

0

0 0 r 1 + L1

d dt

i0 t i1 t i2 t

8 39

where v(t), i(t) are real-number quantities. Equations (8.38) and (8.39) correspond to Eq. (3.34). Now we can recognize by the same analogy that all the equations and transformation procedures described in the previous chapters are perfectly preserved for transient phenomena with symbolic replacement of jω ⟺ d/dt (or the symbol s = d/dt of the Laplace transform). Of course, correlations among the abc-, 012-, and αβ0-domains are also preserved by the same transform/inverse transform equations.

8.4.2

Comparison of Transient Analysis with Symmetrical Components and the αβ0-Method

The transient analysis for a phase-bc l–l short-circuit fault (2ϕS) is demonstrated in Table 8.3, in which the following four approaches are compared: Case A1: Symmetrical component method (symbolic method with complex numbers) Case A2: Symmetrical component method (real-number expressions) Case α1: αβ0-component method (symbolic method with complex numbers) Case α1: αβ0-component method (real-number expressions) These different approaches are equivalent from a mathematical viewpoint, but their ease of use is quite different. In the demonstrated-fault analysis, the αβ0-component method provides the solution with or without the symbolic method. In contrast, the symmetrical-components method only provides the solution together with the symbolic method. In practical engineering, good sense is required to select the most appropriate method for the occasion from the approaches indicated in Table 8.3. Reviewing Tables 8.2 and 8.3, the following comments may be made. The symmetrical-coordinates method and the αβ0-coordinates method are vital basic analytical methods that enable three-phase circuit analysis as a practical, effective approach, whether for steady-state or for transient phenomena. However, the powerful analytical capability of these methods is displayed in particular when they are used in combination with the symbolic method of complex-number variables.

Table 8.3 Transient fault analysis with symmetrical components and the αβ0-method. Calculation with complex-number variables (symbolic method)

Three-phase circuit

Calculation with real-number variables

ia(t)

Ia(t) Ea t = Ea1

e

j ωt + θ

Ea1 e j ωt + θ −120

Eb t = a2 Ea t =

Ea(t) Eb(t)

Ec(t)

ea t = Ea1 cos ω + θ

Ib(t)

eb t = Ea1 cos ωt + θ −120

Ec t = a Ea t = Ea1 e j ωt + θ − 120

Calculation with symmetrical coordinates

ec t = Ea1 cos ωt + θ − 120

Ic(t)

RN

ec(t) RN

ic(t)

Calculation A2: ea(t) = Ea1 cos(ωt + θ)

Calculation A1: Ea(t) = Ea1 ej(ωt + θ)

R1 L1

Positive seq.

Ea(t) = |Ea1

Negative seq.

Rl

I1(t)

Positive seq.

i1(t) ea(t) R1

|e j(𝜔t + 𝜃) Negative seq.

Ll I2(t)

Rl

I0 t = 0

L1 Ll i2(t)

i0 t = 0 Ea1

I1 t = −I2 t = 2R1

2

+ ω2 2L1

2

e j ωt + θ − φ

2R1 t −e 2L1 e j θ −φ

2R1

①

=

Ib t = I0 t + a2 I1 t + aI 2 t = a2 −a I1 t = 3 e − j90 I1 t Ea1 R21 + ω2 L21

e

j ωt + θ − φ −90

Ea1 R21 + ω2 L21

+ ω2 2L1

2

②

R1 − t e j ωt + θ − φ −90 −e L1 e j θ − φ −90

Ea1 2

R21 + ω2 L21

③

cos ωt + θ − φ

R1 − − e L1 cos θ −φ

R1 t −e L1 e j θ − φ −90 −

Ic t = I0 t + aI 1 t + a2 I2 t = − a2 − a I1 t = − Ib t 3 2

2

2R1 − t cos ωt + θ −φ −e 2L1 cos θ − φ

2R1 L1 = tan −1 2L1 R1 Transforming into phase-a, -b, -c current Ia t = I0 t + I1 t + I2 t = 0

3 = 2

Ea1

i1 t = − i2 t =

−

where φ = tan − 1

=−

ea(t) eb(t) ib(t)

Solution ③ cannot be inverse-transformed into the abc-domain. Symmetrical components can be inverse-transformed into this domain only when they are given as solutions to complex-number equations. (Continued)

Table 8.3 (Continued) Calculation with complex-number variables (symbolic method)

Calculation with real-number variables

The real-number part extracted from Eq. ② is the same with Eqs. ⑥ and ⑧: Calculation with αβ0coordinates

Calculation α2: β-Circuit

Calculation α1: β-Circuit Ea t =

Ea1 e j ωt + θ

R1 L1

− jE a t = −j Ea1 e j ωt + θ

–jEa(t) = |Ea1|e j(

Ea1 e j ωt + θ90

=

I𝛽(t)

−jea t =

t + – 90°)

=

Iα t = 0, I0 t = 0 Iβ t =

Ea1 R21

+ ω2

L21

e j ωt + θ −φ − 90

Ea1 R21 + ω2 L21

Ea1 cos ωt + θ −φ − 90 R21 + ω2 L21 R1 − t − e L1 cos θ −φ −90

ia t = iα t + i0 t = 0

e j ωt + θ −φ −90

1 3 3 iβ t + i0 t = iβ t ib t = − iα t + 2 2 2

R1 − t −e L1 e j θ −φ− 90 ⑤

=

e

j ωt + θ −φ− 90

R1 t − e L1 e j θ − φ−90 −

Ea1 R21 + ω2 L21

cos ωt + θ − φ− 90

−

3 1 iβ t + i0 t ic t = − iα t − 2 2 =

taking the real-number part

3 2

R1 −e L1 cos θ −φ −90

3 3 1 Ic t = − Iα t − Iβ t = I 0 t = − Iβ t = −Ib t 2 2 2 3 Ea1 =− 2 R21 + ω2 L21

3 iβ t = −ib t 2

where the source voltage is

ia t = 0 3 2

Ea1 R21 + ω2 L21

ic t = − ib t Note: Equations ⑥ and ⑧ are the same.

ea(t) = |Ea1| cos (ωt + θ) eb(t) = |Ea1| cos (ωt + θ – 120 ) ec(t) = |Ea1| cos (ωt + θ + 120 )

cos ωt + θ − φ−90

R1 − t −e L1 cos θ − φ−90

⑦

Transforming into phase-a, -b, -c current

1 3 3 Iβ t + I0 t = Iβ t Ib t = − Iα t + 2 2 2

ib t =

i𝛽(t)

|Ea1|cos(𝜔t + 𝜃 – 90°)

Ea1 sin ωt + θ

iβ t =

④

Iα t = I α t + I0 t = 0

3 2

Ea1 cos ωt + θ −90

iα t = 0, i0 t = 0 R1 − t − e L1 e j θ −φ − 90

Transforming into phase-a, -b, -c current

=

R1 L1

Ea1 cos ωt + θ

ea t =

the solution ⑥

the solution⑧

175

9 Power Cables 9.1

Structural Features of Power Cables

Around the year 1910, 10–60 kV power cable began to prevail. Today, power-cable networks have been widely adopted as essential power-circuit lines for distribution networks and for high-voltage transmission networks of up to 400 and 500 kV as well as in industrial factories. Understanding the structures, physical features, and theoretical background of power cable is important for two reasons. First, the power cable is one of the most delicate power system components and is characterized by unique features such as its extreme longitudinal structure, non-self-restoring insulation (in particular, polyethylene (PE) for CV cables), a variety of layout environments and stresses, and so on. Second, the large stray capacitance C often causes fluctuations in power frequency or higher frequency, in combination with the inductances L of other equipment within a circuit. So, we will quickly review the structures and some physical background of power cables, and then examine the electrical features. 9.1.1

Structures of CV(XLPE) Cable and OF Cable

High-voltage power cables of 30 kV or higher voltages can be classified into two types based on their insulation materials and are called simply CV cables and oil-filled (OF) cables. The CV cable (cross-linked PE insulated with vinyl-sheathed cable) has another name, cross-linked polyethylene (XLPE). The main insulation is composed of the solid organic material XLPE. In continuous-production processes, colloidal high-viscosity PE in a temperature range of 120–130 C is extruded to cover the moving core conductor, and then chemical molecular cross-linking is reacted through the hardening process with water cooling; electric radiation is also processed with chemical molecular cross-linking (or polymerization) of the polyethylene insulation material. For the OF cable, the main insulation is oil-impregnated insulation paper. The conductor is wrapped in several layers of the paper in a continuous process and dried in a vacuum chamber to remove humidity, and then is impregnated with insulating oil. Under operating conditions, the cable is filled with pressurized oil, which is supplied from a pressure tank installed at the end or at jointing terminals. Typical structures and roles of each laminated layer of CV and OF cables are shown in Figures 9.1a–c. Also, representative standards of the required testing voltages for the main insulation and outer covering are given in Tables 9.1 and 9.2. Referring to the CV cable in Figure 9.1a, ① Conductor: This consists of stranded, compacted copper or aluminum wires. For large conductor sizes (typically 1000–2500 mm2), a Milliken construction or a compact segmental stranded conductor is used to reduce skin and proximity effects and AC conductor resistance. ② Conductor screen: This is employed to reduce excessive electrostatic stress or to smooth electric fields in voids between conductor and insulation. PE or other polymers with carbon black are extruded directly under the insulation. ③ Cross-linked PE insulation: XLPE is the insulation material for CV cables, whose molecular structure is cross-linking PE. XLPE is a chemically stable material and has outstanding properties of electrical insulation strength, mechanical strength (tension, compression, twisting, bending–flexibility, etc.), as well as stable thermal characteristics. PE’s characteristics change at 105–110 C, and it becomes fluid with high viscosity over 120 C. Therefore, in the manufacturing

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

176

9 Power Cables

1 2 3 4 5 6 7

Courtesy of Exsym corporation

(a) 1 2 3 4 Courtesy of Showa Wire and Cable Co.

5 6 7 8

(b) Figure 9.1 (a) CV cable. (b) OF cable. (c) Various CV and OF cables.

process, the conductor can be continuously covered by XLPE by extrusion at a temperature around 120–125 C. The three layers of conductor screen (②), insulation (③), and insulation screen (④) are generally extruded simultaneously in one process so that irregularities (gaps, crevices, voids, particles, etc.) can be avoided between the layers. Chemical cross-linking is treated in the cooling process after extrusion. Conversely, the allowable continuous operating temperature of CV cable is 90 C maximum. ④ Insulation screen: This is employed to reduce excessive stresses in voids between the insulation and metallic screen. PE or other polymers with carbon black are extruded directly on the insulation. ⑤ Cushion layer: For cables with a metallic sheath, cushion layers are applied between the insulation screen and metallic sheath to absorb the expansion and contraction of the core due to heat cycles. Semiconducting tape and/or fabric tape woven with copper wire are commonly used. ⑥ Metallic screen: This is used to confine the electric field within the cable. Copper tape is used for low- or mediumvoltage cables (33 kV and under), and copper wires or metallic sheaths are used for high- and ultrahigh-voltage cables (66 kV and over).

9.1 Structural Features of Power Cables

Courtesy of Showa Wire and Cable Co.

(c) Figure 9.1 (Continued)

⑦ Outer covering: This protects the cable from mechanical damage or chemical deteriorating factors. Polyvinyl chloride (PVC) or PE is used because of its weather-resistant, abrasion-resistant, and chemical-resistant properties. When a fire-retardant property is required, PVC is employed. In the case of a cable without a metallic sheath, plastic tape laminated with metal foil is often used underneath the outer covering to prevent ingress of water. Now, referring to the OF cable shown in Figure 9.1b: ① Oil duct: Insulation oil fills the oil duct, fed from the oil pressure tank installed at the cable end terminal (or jointing point, for long lines). The oil pressure is always kept at a constant value regardless of the expansion or contraction of the insulating oil caused by cable temperature fluctuations. A steel strip open spiral is typically used as the oil duct. ② Conductor: This consists of stranded, compacted copper or aluminum wires. For large conductor sizes (typically 1000–2500 mm2), a Milliken construction or a compact segmental stranded conductor is used to reduce skin and proximity effects and AC conductor resistance. ③ Conductor screen: This is employed to reduce excessive electrostatic stress or to smooth electric fields in voids between the conductor and insulation. Carbon paper or metallized carbon paper is typically adopted. ④ Oil-impregnated insulation paper: OF cables are insulated with a complex insulator composed of paper and oil. Several layers of insulation paper are wrapped around the conductor, dried in a vacuum chamber to remove humidity, and impregnated with mineral or synthetic insulating oil. For ultrahigh-voltage cable, a synthetic blend of alkyl benzenes may be used, due to its superior electrical properties. ⑤ Metallic screen: Nonmagnetic metal tape or metallized carbon paper is generally adopted as the metallic screen, which is employed mainly for the purpose of electrical shielding.

177

178

9 Power Cables

Table 9.1 Test voltages of CV cable.

Rated voltage

Highest voltage for equipment

Value of U0 for determination of test voltages

Voltage test (routine test)

Impulse voltage test

Switching impulse voltage test

(type test) U kV

IEC 60840

IEC 62067

Um kV

U0 kV

2.5U0 kV

Duration (min)

kV

kV

45 to 47

52

26

65

30

250

–

60 to 89

72.5

36

90

30

325

–

110 to 115

123

64

160

30

550

–

132 to 138

145

76

190

30

650

–

150 to 162

170

87

218

30

750

–

220 to 230

245

127

318

30

1050

–

275 to 287

300

160

400

30

1050

850

330 to 345

362

190

420

60

1175

950

380 to 400

420

220

440

60

1425

1050

500

550

290

580

60

1550

1175

U0: the rated power frequency voltage between conductor and core screen for which the cable and its accessories are designed (rms). Um: the highest line-to-line system voltage that can be sustained under normal operating conditions (rms). In addition, a partial discharge test at 1.5 U0 to confirm nondetectable discharge should be carried out. 1) U0 is the same as U/ 3 or a value a little larger than U/ 3. 2) Impulse and switching voltage tests are performed using a sample cable at least 10 m in length. 3) The applied waveform. Lightning impulse test: time to peak Tf: 1–5 μ s, time to half value Tt: 40–60 μs. Switching impulse test: time to peak Tf = 250 μs ± 20%, time to half-value Tt = 2500 μs ± 60%. 4) The test voltages for OF cables can be considered the same as in the table, although they are slightly different for historical reasons.

⑥ Binding tape: This is applied to protect the metallic screen from external damage in the manufacturing process. It also electrically conducts charging current to the metallic sheath. ⑦ Metallic sheath: This is made of corrugated aluminum or of lead and has the function of maintaining oil pressure, to protect the cable structure from external damage, and, furthermore, of acting as a conductor for the electrical earth– fault return pass. ⑧ Outer covering: This protects the cable from mechanical damage or chemical deteriorating factors. PVC or PE is used due to its excellent weather-resistant, abrasion-resistant, and chemical-resistant properties. When a fire-retardant property is required, PVC is employed. The practical application of OF cables for voltages of 30 kV or higher began to spread in the era around 1910. On the other hand, the first applications of CV cables were seen around 1970. CV cables are a modern development of the organic material XLPE realized by advanced chemical materials production technology. It has various outstanding properties, such as excellent electrical insulation characteristics, mechanical strength, bending flexibility, chemical stability, the ability to withstand heat, etc.; and outstanding production technology, including continuous extrusion and cross-linking reaction processes. Today, CV cables have been mostly adopted for lower voltage classes of 60/70 kV and distribution networks, while CV and OF are used together for higher voltages of 100–400 or 500 kV, according to individual users’ requirements. OF cable is based on the traditional insulation method of “paper and oil,” so it is generally considered a very reliable product; but it has some disadvantages from a practical application viewpoint that mainly concern maintenance work around the oil pressure tank and fear of fire. CV cable is an advanced large-size product that can be as much as 2000–2500 m in length, made entirely of solid insulation. Because it is based on solid insulation without oil, maintenance work is simple; but advanced, careful production technology/management is required in order to achieve chemically/mechanically stable and homogeneous insulation, and to avoid small particles, voids, irregular shapes, and so on. Advanced technology is also required for cable jointing.

9.1 Structural Features of Power Cables L 9

9

L = 1545 ± 100mm 8

5

4

7

1

2

3

6

10

ϕD = 355 ± 30mm (Courtesy of Showa Wire and Cable Co.)

① Conductor sleeve coupling ② Sleeve cover ③ Gum block insulator ④ Outer semiconducting layer ⑤ Protective tape ⑥ Waterproof compound ⑦ Protective tube ⑧ Waterproof cover ⑨ Grounding terminal ➉ Insulation tube cover Figure 9.2 Power cable jointing structure (400 kV, 2000 mm2, insulation gum block type for trench layout).

Prior to 1980–1985, CV cable was considered less reliable in comparison with OF cable, especially for higher voltage classes. Today, however, the idea has taken root that CV cables are as reliable as OF cables even for EHV classes. Thanks to advances in CV production technology and accumulated successful applications, CV cables have become the principal cables even for 200–500 kV classes. UHV AC 500 kV CV cable lines as well as ±500 kV DC OF submarine cable lines have been put into practice in Japan and are the highest-voltage applications. Figure 9.1c shows various types of CV cables and OF cables. Figure 9.2 shows the typical structure of a CV cable-joint (gum block insulation type) for the 400 kV class. The insulation gum block is pressure-connected to the cable insulation material, and the well-shaped electrode and semiconducting materials (stress-corn and outer semiconducting layer) mitigate dielectric stresses. 9.1.2

Features of Power Cables

Power cables may be distinguished from other equipment or overhead transmission lines as described next. 9.1.2.1

Insulation

Let’s compare power cables and overhead transmission lines. Overhead transmission line is based on naked conductors with self-restoring air insulation from natural circulation; therefore, appropriate insulation can be secured only by maintaining the necessary insulator distances and, as a result, by maintaining the appropriate distances between conductors, overhead grounding wires (OGWs), and tower structures. Breakdown by lightning is also allowable. On the other hand, power cable insulation is much more delicate due to its non-self-restoring insulation, and breakdown should be entirely avoided with regard to insulation coordination. CV cable is also much more delicate because of the native characteristics of organic compound insulation. As shown in Figure 9.3, PE (polyethylene) begins to soften at around 90 C and quickly loses insulation strength as well as mechanical strength. Furthermore, the dielectric loss tangent tan δ (= G/2πfC, where C, G are the capacitance and conductance of a cable per unit length. In Figure 9.4, tan δ is probably a value within 0.001–0.005, but it quickly increases as the temperature increases and deterioration of physical characteristics accelerates. These changes cannot be reversed by returning to a lower temperature. The material becomes fluid with high viscosity at 120 C. All these things have to be taken into account at any stage of power cable engineering: manufacturing, layout design, construction, operation, and maintenance. 9.1.2.2

Production Processes

Production sizes of EHV class power cables are typically 600–2500 mm2 and up to 2000 m in length (see Table 9.4) per lot, limited by the drum size for production and transportation. The production processes for CV cable consist of several cascading subprocesses: segment conductor (twisting of elemental copper wires) segmental stranded conductor simultaneous extrusion of three layers (inner conductor screen/PE insulation/outer conductor screen) chemical

179

180

9 Power Cables

Relative value Tensile strength Thermal deformation ratio

tanδ

0

20

60

100

120

Temperature [°C]

Figure 9.3 Physical characteristics of cross-linked polyethylene (XLPE).

ileak = iR + iC iC v

iR

Cp

ileak = iR + iC 𝛿

Rp

tan𝛿 =

Table 9.2 Withstanding voltages of outer-covering layers.

iC = j𝜔Cp · v v

iR =

v Rp

iC = 2𝜋f · CpRp iR

Parallel CR equivalent model Figure 9.4 Dielectric loss tangent tan δ and dielectric loss W.

Rated voltages (kV)]

Impulse withstanding voltages (kV)

~33 66–74 110–187 220–275

– 50 65 75

Note: The table is quoted from JEC 3402 (1990) in order to show the sensitive values for the withstanding voltages of the outer covering. The standard IEC 229 (1982) “Test on oversheath” indicates the routine tests of “Ad h C voltage of 8 kV/mm of specified nominal thickness of the extruded oversheath for 1 min.”

cross-linking cooling by water for hardening metallic screening outer covering. Winding and rewinding with drums is repeated at every subprocess. This production process with its many subprocesses looks like that in a chemical factory and is quite different from that of generation/substation equipment. 9.1.2.3 Environmental Layout Conditions and Stresses

The installation environments for power cables are varied, as illustrated by the following keywords picked randomly from this field:

•• •• •• •

Ground surface, soil-buried, duct, trench, tunnel, overhead, sea-bottom, river-bottom, sea-bottom-buried, hanging Flat, slope, vertical Hard, soft, muddy basis Hot, cold, chemical-dirty, contamination Tensile-tension, press, bending, mechanical beat, expansion, contraction Soil vibration, water current, sea tide, wind power Unequal-sink, earthquake, flood, fire

Cable installation involves large civil construction projects covering large, varied environmental areas. Furthermore, pre-detecting faults and repairing or retrying installations generally are not easy due to the specific nature of cables and the installed environment. Given all of these conditions, the stresses that power cables must be able to withstand may be classified as follows. i) Mechanical stress: Winding and rewinding are repeated throughout the production process (typical weight of EHV class is 40–48 kg/m), so withstanding gravitational weight and the mechanical stresses of compression, tension, and bending is essential. Further, the final layout environment may vary, as has already been explained. In addition, mechanical attraction and/or repulsion forces are inevitably caused among phase conductors due to continuous load currents and instantaneous fault currents. Expansion and contraction stresses are also caused by thermal conditions.

9.1 Structural Features of Power Cables

ii) Electrical stress: Power cables with thinner insulation layers (typically 23–27 mm for 275–500 kV CV cables) must withstand the electrical stresses specified in Table 9.1. The entire length of the cable must be produced in such a way that the insulation material is homogeneous and has no voids or particles. iii) Thermal stress: As explained earlier, abnormal temperature increases that exceed the specified critical limits should be avoided in particular for CV cable. OF cable may have a larger redundancy than CV cable. The operating temperature of a power cable is determined by the thermal balance between heat generation (caused by the resistive current losses of conductors, the metallic sheath, and dielectric losses of insulators) and heat discharge from the outer-covering surface to the installed environment. Heat generation is determined only by the cable structure and electric condition (load current, sheath connection), whereas heat discharge is widely affected by the environmental conditions of the cable installation. Accordingly, the thermal balance and temperature increase must be calculated section by section as an important part of the installation design, in order to maintain the cable operating temperature below an allowable upper limit (typically 90 C for continuous current, 105 C for temporary overcurrent, and 250 C for fault current within a few seconds). This is also why power cables do not have a specific rated capacity or rated current. The allowable current limit must be determined based on the cable conductor size and the individual conditions of the cable’s installation. iv) Chemical stress: Power cables may need to be laid under dirty mud or in poor trenches in a desert area or sea bottom. Weather resistant, abrasion-resistant, and chemical-resistant as well as ultraviolent resistant capabilities are essential requirements 9.1.2.4

Metallic Sheath Circuits and Outer-Covering Insulation

The metallic sheath plays an important role as a mechanical container to protect the delicate insulation and electrical earth–fault current return pass and reduce any induced current/voltage/noise interference from the outer environment. The outer covering also protects the cable from various mechanical, chemical, thermal, and electrical stresses from the environment – and may do so for a long time, perhaps as long as 30 years – and has allowable electrical and thermal upper limits. Table 9.2 shows the typical insulation level of the outer covering. Accordingly, we need to investigate not only the voltage/current behavior on the conductors and main insulation, but also the induced voltage/current on the metallic sheath and the outer covering. Note: The power cable’s insulation layer is accompanied by resistive loss leakage (R) as well as capacitive leakage (C), so it can be shown as the parallel CR equivalent model or as series CR. tan δ and dielectric loss W are given by the following equations, as shown in Figure 9.4. iC = 2πf Cp Rp iR Dielectric loss W = v i = 2πfC v2 tan δ Watt km

tan δ =

tan δ is a very important parameter to check for sound insulation, as one of the certification test items during commercial tests. The typical measured values for sound cables are 0.001–0.005. 9.1.2.5

Electrical Specifications and Factory Testing Levels

The outward appearances and sizes of CV cables and OF cables are almost the same, in spite of the structures being quite different. Their electrical specifications, such as voltage and current classification, as well as standardized insulation levels and testing methods are also almost the same. Table 9.1 shows the testing voltages of CV cables by IEC standards; the same table is commonly used for OF cables. The switching impulse test is used for power cables over 245 kV and is based on the fundamental concept of insulation coordination (explained in Chapter 17). Table 9.2 shows typical withstanding impulse voltages for the outer covering layer. Table 9.3 shows the typical thickness of the main insulation layer for CV and OF cables based on individual rated voltages. Impulse and switching-impulse tests are performed using a sample cable at least 10 m in length, because specified impulse voltages cannot be established by an impulse generator on a long cable with large capacitance C . l (where typically C = 0.25–0.7 μF/km; see Table 9.4 for typical values).

Table 9.3 Typical values of insulation thickness (D − r).

Rated voltages (kV)

6.6 22 33 66 77 110 154 220 275 500

OF cable (oilimpregnated, insulation paper) ε = 3.7)(mm)

CV cable (crosslinking polyethylene) ε = 2.3)(mm)

– – – 7~8 8~9 9.5~10.5 13~14 18 19.5

2.7~3.6 6~7 8~9 9~11 11~13 17 15~19 20~23 23 27

181

9 Power Cables

Table 9.4 Typical line constants for power cables.

500 275

154

66

33

6.6

Resistance

Working inductance

Working capacitance

Reactance

Leakage-current

Surge-impedance

(kV)

Cable diameter

2r 2

Sheath diameter

Insulation-thickness

Sectional area

Conductor thickness

CV cables

Line voltages

182

D

S

R

Ls – Lm

C

jX

Ic

L C

(A/km/ϕ)

(Ω)

(mm )

(mm)

(mm)

(mm)

(mm)

(Ω/km)

(mH/km)

(μ F/km)

(Ω/km)

2500

61.2

27

142

163

0.00746

0.383

0.25

0.112

22.7

39.1

2000

53.8

27

134

155

0.00933

0.400

0.23

0.116

20.9

41.7

2500

61.2

23

133

160

0.00746

0.381

0.28

0.108

14.0

36.9

2000

53.8

23

125

149

0.00933

0.392

0.25

0.112

12.5

39.6

1200

41.7

23

112

134

0.01560

0.422

0.21

0.122

10.5

44.8

2000

53.8

17

108

122

0.00933

0.352

0.26

0.103

7.3

36.8

1200

41.7

17

96

110

0.01560

0.382

0.22

0.112

8.7

41.7

800

34.0

17

88

100

0.02310

0.404

0.19

0.119

5.3

46.1

2000

53.8

10

95

95

0.00933

0.302

0.53

0.086

6.3

23.9

1200

41.7

10

82

82

0.0156

0.324

0.43

0.092

5.1

29.6

800

34.0

10

73

73

0.0231

0.340

0.37

0.097

4.4

30.3

1200

41.7

8

73

73

0.0156

0.301

0.46

0.086

2.8

25.6

600

29.5

8

58

58

0.0308

0.324

0.38

0.092

2.3

29.2

200

17.0

8

45

45

0.0915

0.383

0.26

0.108

1.6

38.4

600

29.5

5

47

47

0.0308

0.282

0.71

0.089

0.8

19.9

200

17.0

4

32

32

0.0915

0.315

0.51

0.102

0.6

24.9

OF cables (kV)

500 275 154

66

2

(mm )

(mm)

(mm)

(mm)

(mm)

(Ω/km)

(mH/km)

(μ F/km)

(Ω/km)

(A/km/ϕ)

(Ω)

2500

68.0

25.0

132

153

0.00732

0.305

0.37

0.101

33.5

28.7

2000

59.1

33.0

139

160

0.00915

0.388

0.27

0.113

24.5

37.9

2000

57.5

19.5

107

137

0.00915

0.363

0.41

0.098

20.4

29.8

1200

45.7

19.5

94

124

0.01510

0.389

0.34

0.105

17.0

33.8

2000

57.5

13.5

94

119

0.00915

0.333

0.57

0.09

15.9

24.2

1200

45.7

13.5

81

106

0.01510

0.367

0.45

0.095

12.6

28.6

800

40.6

12.5

74

96

0.02260

0.361

0.44

0.097

12.3

28.6

2000

57.0

8.0

82

106

0.00910

0.312

0.96

0.082

11.5

18.0

1200

45.2

8.0

69

92

0.01510

0.331

0.80

0.086

9.6

20.3

800

39.6

7.0

61

82

0.02230

0.334

0.79

0.087

9.5

20.6

Notes: The working inductance is calculated under the three-phase allocation of touching equilateral triangles. Accordingly,

Ls − Lm = 0 4605 log D r + 0 05 mH km where (Sab Sbc Sca)1/3 = (D D D)1/3 = D. If the average phase-to-phase distance S is larger, the inductance will be slightly larger. The reactance is calculated from jX = j2π 50(Ls – Lm) based on 50 Hz. Then the values should be multiplied by 1.2 for the 60 Hz system. The leakage current is calculated with Ic = 2π 50 C(1/ 3)V. The surge impedance is calculated with Ls −Lm C.

9.2 Circuit Constants of Power Cables

9.2

Circuit Constants of Power Cables

Now, let’s examine the constants of power cables; see Table 9.4.

9.2.1

Cable Inductance

Electrically, a power cable consists of concentric cylinders with three layers: metal conductor, insulation material, and metallic sheath; see Figure 9.5a. When current flows through the conductor, sheath current is induced by the mutual inductance between the conductor and the sheath circuit. The induced flux from this conductor current appears partly within the cable sheath as ϕin and partly beyond the cable sheath as ϕout. Furthermore, ϕout partly interlinks the cable conductors/metallic sheaths of two other parallel-installed phases so that mutual inductances exists between the conductors/metallic sheaths of different phase cables. In other words, the cable line consists of three cables of single-phase type as well as a cable of three-phase type and can be expressed as a six-parallel-wire inductive circuit consisting of conductors a, b, c and metallic sheaths a , b , c and installed in parallel underground–earth along with the earth surface, as shown in Figure 9.5b. Then, it is easy to obtain Figure 9.5c for a symmetrical equivalent circuit of phase conductors with metallic screen terminals (sheath circuit), and Figure 9.5d for a symmetrical equivalent circuit of phase conductors whose screen-end terminals are earth connected. Incidentally, three-phase cables or triplex type (the three twisted cables are bound) are naturally well-balanced threephase circuits. In the case of a three single-phase-type cable circuit, the installed layout would obviously affect it somewhat as a phase-unbalancing factor. However, a three single-phase-type cable circuit with an equilateral triangular installed layout (see Figure 9.5e) can be considered a well-balanced three-phase circuit. Also, long-line cables with cross-bond connections (see Section 9.3.4) can be considered almost balanced three-phase. We assume here a balanced three-phase circuit for simplification. Referring to the equations for inductances in Chapter 3, Eq. (3.3)–(3.6), the related equations for the balanced threephase cable inductances can be written as follows, where the earth effect is already taken into account: Conductor voltages m Va m Vb m Vc

n Va

−

m V abc

n Vb n Vc

=

Zs Zm Zm

n V abc

Zm Zs Zm

Zm Zm Zs

Z abc

Ia Ib Ic

Zs Zm Zm

+

Zm Zs Zm

I abc

Zm Zm Zs

Ia Ib Ic

Zabc

①

Iabc

Metallic sheath voltages m Va m Vb m Vc m V abc

n Va

−

n Vb n Vc n V abc

=

Zs Zm Zm

Zm Zs Zm Z abc

Zm Zm Zs

Ia Ib Ic

+

Zs Zm Zm

I abc

Zm Zs Zm

Zm Zm Zs

Zabc

Ia Ib Ic Iabc

mV abc , nV abc , Iabc voltages and currents of metal conductors mVabc , nVabc , Iabc voltages and currents of metal sheaths earth connected at n where Zabc = jωLabc impendance matrix of three −phase conductors Zabc = jωLabc impendance matrix of three −phase sheaths Zabc = jωLabc mutual impendance matrix between conductors and sheaths

②

91

183

Flux ϕout Flux ϕin

Eddy current in metallic sheath Load current Metallic sheath Insulation Core conductor

Thickness of the outer cover is exaggerated

rrent

Sheath cu

(a) Cable structure Earth surface

Point m

Point n

Ia

nVa

mVa

Phase-a

mVa′

Ib

mVb

Phase-b

Lab´

mVb′

nVb nV b′

Ic

mVc

Phase-c

nVa′

nVc

mVc′

nVc′

L′′s

Ls

Lm

L′′m

L′s

L′m

(b) Inductances circuit p-seq, n-seq m V1, m V2

I 1, I 2 I 1′, I 2′ rs′

Zero-seq mV0

rs

rs I0 I 0′

rs′

p-seq, n-seq Ls – Lm L s′ – Lm′

nV1, nV2

m V1, m V2

L s′′ – Lm′′

Ls′ + 2Lm′

nV0

r0 L0 = Ls + 2Lm – 𝛿0 nV0

m V0

I0

L s′′ + 2Lm′′

(c) Symmetrical equivalent circuit for Fig. (b)

(d) Symmetrical equivalent circuit under both screen terminals earth connected Equilateral triangular allocation (or triplex cable)

General allocation Sab

Sbc D

13

S = (Sab × Sbc × Sca)

D

nV1, nV2

I 1, I 2

Zero-seq Ls + 2Lm

r1 L1, L2 = Ls – Lm– 𝛿1

Sca

D : outer diameter of sheath S : equivalent core distance

Sab = Sbc = Sca = D S

(e) Three-phase cable allocation Figure 9.5 Inductance of power cables.

13

then, S = (D×D×D)

=D

9.2 Circuit Constants of Power Cables

The equations can be transformed into the symmetrical-components equations: Conductor voltages mV 0 mV 1 mV 2

Ls + 2Lm

nV 0

−

m V 012

nV 1 nV 2

Ls − Lm

= jω

Ls −Lm

n V 012

L012

Ls + 2Lm

I0 I1 I2

Ls −Lm

+

Ls − Lm

I012

L012

I0 I1 I2

①

I012

Metallic sheath voltages mV 0 mV mV

1

Ls + 2Lm

nV 0

−

nV nV

2

Ls − Lm

= jω

1

Ls − Lm

2

m V 012

n V 012

m V 012 , n V 012 ,

I012 voltages and currents of metal conductors

m V 012 , n V 012 ,

I

012

L012

I0 I1 I2

Ls + 2Lm +

I012

Ls −Lm Ls − Lm L012

I0 I1 I2

②

I012

voltages and currents of metal sheaths

Ls ,Lm self and mutual inductances among three conductors Ls ,Lm self and mutual inductances among three sheaths Ls mutual inductance between the conductor and the sheath of the same phase Lm mutual inductance between the conductor and one sheath of different phases

92 Eq. (9.2) ① shows that the positive-sequence inductance is given by (Ls − Lm) given the condition I1 = 0 (sheath current zero), whereas some correction is necessary under the condition I1 0. The situation is the same for the zero-sequence inductance (Ls + 2 Lm). Also, in Eq. (9.1) ①, we can imagine a special case in which the current I goes out through the phase-a conductor and comes back through the phase-b conductor so that no other currents exist on the phase-c conductor, or on sheaths of three-phase cables, as well as on the earth circuit. The equations are Ia = −Ib = I, Ic = 0, Ia = Ib = Ic = 0 ∴ m v a − n v a = jω Ls −Lm I This is the same case as in Figure 3.2b in Chapter 3; and the inductance (Ls − Lm) is of course the working inductance, which is given by Eq. (4.9) or (4.13). Accordingly, the working inductance of cable phase conductors is Ls −Lm = 0 4605 log10

S r

9 3a

where S is the average phase-to-phase distance of three parallel conductors (center to center) S = Sab Sbc Sca

1 3

refer to Figure 9 5e

r radius of each conductor = d 2; where d is the diameter

9 3b

In addition, (Ls + 2Lm) is the inductance, which is the same as that measured by the method in Figure 3.2b. The sheath terminals are generally earth-grounded at both end terminals as well as at the middle jointing terminals for long-distance cable lines, whereas it is true that in rare cases, only one end terminal may be earth-grounded for a shortdistance line (typically for factory or in-house cable lines). Thus we need to check both cases.

185

186

9 Power Cables

Case 1 When the metallic sheath is earth-grounded at one terminal point n (for short-distance lines), n n V abc = 0, Iabc = 0, so that nV 012 = 0 and I012 = 0 in Eqs. (9.1) and (9.2). Accordingly, Equationof main circuit mV 1 − nV 1

= jωL1 I1

mV 2 − nV 2

= jωL1 I2

mV 0 − nV 0

= jωL0 I0

①

Positive− negative − sequence inductance L1 = L2 = Ls −Lm Zero − sequence inductance L0 = Ls + 2Lm Induced sheath voltages at open sheath end point m

②

94

mV1 = jω Ls − Lm I1 ③

mV2 = jω Ls −Lm I2 mV0 = jω Ls + 2Ls I0

Case 2 When the metallic sheath is earth-grounded at both terminal points m and n (general case), mV abc = 0, = 0, and then mV 012 = 0, nV 012 = 0 in Eqs. (9.1) and (9.2). Accordingly, currents I1 ,I2 ,I0 can be eliminated from both equations:

nV abc

Equationof main circuit mV 1 − nV 1

= jωL1 I1

mV 2 − nV 2

= jωL1 I2

mV 0 − nV 0

= jωL0 I0

①

Positive− negative −sequence inductance L1 = L2 = LS − Lm − δ1 where δ1 =

Ls + Lm Ls + 2 Lm

2

②

Zero-sequence inductance

95 L0 = LS + 2Lm

L + 2 Lm − δ0 where δ0 = s Ls + 2 Lm

2

δ0 ,δ1 are correction factors

Induced sheath currents through the metallic sheath between points m and n I1 = −

Ls − Lm I1 Ls − Lm

I2 = −

Ls − Lm I2 Ls − Lm

I0 = −

•

③

Ls + 2 Lm I0 Ls + 2 Lm

To summarize cable inductance: The inductance of a cable line is affected by the dimensions of the cables, the mutual allocation of three-phase conductors, and the terminal condition of the metallic sheath.

9.2 Circuit Constants of Power Cables

• • •• •

The working inductance (Ls − Lm) is derived from Eq. (9.3a), whose variable parameters are the average distance S between the conductors and the radius r of the conductors. If the metallic sheath is open at one terminal, positive-sequence impedance is given by the working inductance, L1 = Ls − Lm. However, L1 is a little smaller than the working inductance under the metallic sheath current-flowing condition, because of the correction factors δ1, δ0 from Eq. (9.5) ②. As a numerical check for working inductance Ls − Lm from Eq. (9.3a): Case 1: r = 25 mm, S = 75 mm, Ls − Lm = 0.4605 log10(75/25) + 0.05 = 0.269 mH/km Case 2: r = 25 mm, S = 100 mm, Ls − Lm = 0.327 mH/km Case 3: r = 25 mm, S = 150 mm, Ls − Lm = 0.408 mH/km

The working inductances of cables are typically 0.15–0.4 mH/km. Positive-sequence inductance L1 would become much smaller than working inductance because most cable lines are metallic sheath earth-grounded at multiple terminal points. Accordingly, the typical value is L1 = 0.1–0.3 mH/km, which is one-fifth to one-tenth smaller than that of overhead lines. (Refer Table 9.4.) 9.2.2

Cable Capacitance and Surge Impedance

The capacitance of a three-phase cable line can be expressed as in Figure 9.6a, so it can be written as in the three-phase π-circuits from Figure 9.6b. Assuming the circuit constants are phase balanced for simplicity, the symmetrical equivalent circuit in Figure 9.6c can be derived from Figure 9.6b in the same way as that for Figure 3.13. The cable conductor is contained by the metallic sheath of an electrically concentric cylinder, and the gap is filled with insulation material of dielectric constant ε. If electric charge +q is given to the conductor and −q is given to the metallic sheath, all the electric lines of force start from the conductor and reach the metallic sheath through a radial path, so they are not affected by the electrical conditions on the outer side of the sheath. Therefore, the capacitance C across the conductor and the metallic sheath is the typical case of a concentric cylinder, which is quoted in most textbooks on electromagnetism. The equation is from Eq. (4.30b): C=

2πε ε 0 02413ε F m = μF km 9 D= D loge r 2 × 9 × 10 loge r log10 Dr

96

where r: radius of the conductor D: radius of the sheath (inner radius) ε = εs ε0: dielectric constant of insulation material OF cable: ε = 3.7 CV cable: ε = 2.3 The equation corresponds with Eq. (4.33b), whose matrix size is 1 × 1: va = paa × qa. The circuit equation in Figure 9.6(a–c) can be written as follows. For the abc-domain Leakage current Ia = jωC Va −Va Ib = jωC Vb −Vb Ic = jωC Vc −Vc

① 97

From conductor to sheath Ia = jωCa Va

Iab = jωCab Va − Vb

Ib = jωCb Vb

Ibc = jωCbc Vb −Vc

Ic = jωCc Vc

Ica = jωCca Vc − Va

From sheath to earth and for the 012-domain

From sheath to sheath

②

187

188

9 Power Cables

Ground surface C ′b

I ′a

I ′b

C ′a

C ′c

I ′bc I ′ab

C ′ab C ′bc

C Ia Va

I ′c

Metallic sheath (inner radius: D) C ′ca

V ′a

Conductor (radius: r)

I ′ca

(a) Main conductors Va Vb Vc

Metallic sheath

Ia Ib Ic

1 C′ 2 m

C

C

C

1 C′ 2 s

V ′a V ′b V ′c

Inductances 1 C′ 2 m

Sheath terminals

(b) Positive/negativesequence circuit m V1, V2

Main conductor C

n 1 L ′1 2 Metallic sheath 1 (C ′ + 3C ′ ) 1 C′ = s m 2 2 1

V ′1 , V ′2

Zero-sequence circuit V0

C

V ′0

1 1 C′ = C′ 2 0 2 s (c) Figure 9.6 Capacitance of cables.

1 L′ 2 0

9.2 Circuit Constants of Power Cables

I1 = jωC V1 −V1 = jωC1 V1 I2 = jωC V2 −V2 = jωC1 V2 I3 = jωC V0 −V0 = jωC0 V0

①

Induced sheath voltage C V1 V1 = C + C1 V2 =

C V2 C + C1

V0 =

C V0 C + C0

where

98

②

C1 = Cs + 3Cm , C0 = Cs Note that the metallic sheath end terminals are generally earth-grounded at both ends or at least at one end (for shortdistance lines), so Va = Vb = Vc = 0 and V0 = V1 = V2 = 0. Consequently, Eq. (9.7)① and Eq. (9.8)① are simplified as follows in the abc-domain Ia Va Ib = jωC Vb Ic Vc

99

and in the 012-domain I0 V0 I1 = jωC V1 I2 V2

9 10

That is, the capacitance of a cable line is C1 = C2 = C0 = C, where C is given by Eq. (9.6) in the symmetrical-components domain. The stray capacitance C is determined by the specific physical size D and r of the cable and the dielectric constant ε of the insulation material layer, so it is not affected by the cable-installation layout; inductance L is, however, affected by the layout, so the value of C is given in the specification supplied by the cable manufacturer. As a result, the surge impedance and the traveling-wave velocity of the cable are Surge impedance Surge velocity Positive-sequence circuit

Z1surge =

L1 C

u1 =

1 L1 C

Zero-sequence circuit

Z0surge =

L0 C

u0 =

1 L0 C

9 11

Typical constants for power cables are shown in Table 9.4. Further, in addition to the R, L, C constants, the leakage resistance G exists as the fourth line constant. This is typically the creepage resistance of insulators for transmission lines or station equipment, which is a parallel resistance with stray admittance jωC, and usually has an extraordinarily large ohmic value. G is an important constant that is significantly

189

190

9 Power Cables

affected by the insulation characteristics of individual high-voltage insulators, the attenuation ratio of surge phenomena, and so on. However, G can be ignored for most ordinary circuit analysis (except for surge analysis), because it has a large resistance on the order of, say, greater than a megaohms.

9.3

Metallic Sheaths and Outer Coverings

9.3.1

Role of Metallic Sheaths and Outer Coverings

High-voltage power cables are produced and transported as coiled cables on a drum; the length of cable per drum is typically 500–2000 m. These cable-production units are connected in series through cable joints when the route’s span is longer than the production unit length. Sheath terminals are prepared at each end of this production unit. There are various types of power cables, due to the variety of electrical requirements as well as requirements due to the mechanical and installed surroundings (including submarine cable). Accordingly, the metallic sheaths as well as the outer coverings have various structures. The metallic sheath for high-voltage CV cables and OF cables over 60 kV generally consists of corrugated aluminum or lead (reinforced with steel or copper wires if necessary) so that the cable can be electrically considered a metal cylinder containing concentric allocated conductor and insulation materials. The major roles of the metallic sheath are:

•• •• • • • •

Containing and protecting insulation materials (cross-linked PE or oil-immersed paper layers) Isolating from the outer environment (air, water, contaminants) Sharing mechanical strength Return pass of fault currents, continuous unbalanced sheath currents Acting as an electrical shield (electrostatic shield, electromagnetic shield) The major roles of the outer covering are as follows: Protecting the cable from mechanical damage or chemical deterioration, and isolating the metallic sheath in order to secure against electric shock from touching. The outer covering is a kind of insulation material made typically of PVC or PE that has some voltage-withstanding capability, which can protect humans from electrical shock that may be caused by induced sheath voltages. Further, the outer covering has temperature-withstanding capability, so all the heat loss caused in the conductor and the sheath is radiated to earth through the outer covering. Possessing outstanding physical characteristics including weather-resistant, abrasion-resistant, chemical-resistant, and fire-retardant properties.

9.3.2

Double Sheath-End Terminal Grounding Method (Solid-Sheath-Bonding Method)

A power cable is equipped with earth-connecting metallic sheath terminals at each end of the production unit. The sheath terminals are typically earth-grounded at each end, as well as at mid-joining points. A typical earth-grounding method earth-connects multiple (or all) sheath terminals; this is called the solid-sheath-bonding method. The features may be summarized as follows:

• •

Stationary sheath voltage is maintained by the earth potential (0 volt) at any point along the cable length. Accordingly, the outer covering is released from sheath-voltage stresses. (The situation is different for surge phenomena.) Sheath currents induced by unbalanced continuous loads or by temporary induced current flows cause thermal losses on the metallic sheath. Accordingly, the outer covering as well as the insulation layer are thermal heated by the main conductor current plus the sheath current. The thermal heating of the metallic sheath by the sheath current could be a disturbing factor in thermal heat diffusion from the main conductor to the earth.

Sheath current caused by electromagnetic induction will flow with this practice, and the positive-/negative-/zerosequence sheath currents I1 , I2 , I0 are calculated with Eq. (9.5) ③. As a closed circuit is formed through the metallic sheath and the earth ground, positive-sequence sheath current I1 is always induced by positive-sequence conductor current I1. Negative-/zero-sequence sheath currents I2, I0 are induced if negative-/zero-sequence conductor currents exist.

9.3 Metallic Sheaths and Outer Coverings

The temperature increase in the metallic sheath caused by sheath current as well as by conductor current must be considered from the viewpoint of heat balance in the cable, in that primary insulation materials as well as outer-covering materials have to be kept within allowable temperatures. The sheath potential is kept at earth voltages at both end terminals, so that serious electrostatic voltage will not be caused in most cases. 9.3.3

Single Sheath-End Terminal Grounding Method (Single-Sheath-Bonding Method)

With this method, the sheath terminal is grounded at only one end terminal and is not grounded at the other end terminal. This practice is adopted for short-length cable lines, typically less than 500 m. The features are as follows:

• •

Sheath voltage on the open-sheath terminal or in the surrounding zone occurs due to electrostatic induction, and is calculated with Eq. (9.4)③. The outer covering must be protected against the induced voltage with regard to electrical insulation as well as corrosion. Therefore this practice can be adopted only for short-length cable lines (typically, in-house lines). Sheath currents do not flow through the sheath, so no thermal heat generation occurs on the metallic sheath.

Referring to Eq. (9.4)③, the induced power frequency voltage on the sheath can be calculated with the following equation for the sheath voltage under normal conditions (power frequency) E = jω ls −lm I

V km

9 12

where ls −lm : mutual inductance per km between the conductor and the sheath (H/km) I: conductor current (A) As a numerical check, the nominal induced sheath voltage with the single-sheath-bonding method is, for the conditions f = 50 HZ, ls −lm = 2 mH/km, E = j2π 502 × 10 −3 I = 0 628 I V km ∴ E = 62 8 V km for I = 100 A , or E = 314 V km for 500 A This is a large voltage from the viewpoint of human security as well as the capability of the outer covering insulation, so the single-sheath-bonding method can be adopted only for short-distance lines of less than a few hundred meters. 9.3.4

Cross-Bonding Metallic-Shielding Method

One drum length of an EHV-class high-voltage power cable is typically 500 m. Then, for example, for a 30 km length of cable line, 60 cables are series-jointed per phase. That means 2 terminal ends and 59 mid-jointing points. With regard to long-distance high-voltage cable lines with single-conductor-type power cables and a number of cablejointing points, the constants for both the main conductors Labc, Cabc and the sheath circuits Labc , Cabc are forced to become unbalanced. In particular, the imbalance in Labc , Cabc causes unbalanced continuous sheath currents, leading to undesirable temperature increases in the cables even if the main conductor’s three-phase currents are balanced. Moreover, surge voltages transmitted to the main conductors for any reason will induce serious unbalanced surge voltages on the sheath metals of each section, which can damage or break the insulation of the outer-covering layer. The widely adopted practical countermeasure to overcome this problem is the cross-bonding metal-shielding method. The straightforward cross-bonding connection is a kind of transposition of three-phase sheath circuits, as shown in Figure 9.7a: the cable sheath ends are earth-connected every three sections, and the sheath connection is transposed within the three sections. Figure 9.7b shows the details of the jointing boxes of the three coupled sections, with four jointing points I, II, III, IV. The sheath terminal ends are earth-grounded at jointing boxes I and IV, while at jointing boxes II and III, the sheath terminal ends are cross-connected without earth grounding. In other words, the cable sheath terminals are earthgrounded at the jointing boxes every three spans, and they are cross-connected twice at the cross-bonding jointing boxes in the three spans. With this practice, the inductance Labc and the capacitance Cabc of the sheath circuits can be approximately balanced three-phase over the total length of the three spans, although they may not necessarily be balanced at

191

Sheath-metal a1

a2

a3

b1

b2

b3

c1

c2

c3

P

Q

(a) Outer-cover protection (arrester)

Sheath-insulated jointing-box Ordinal jointing-box

e(t)

(I)

Phase-a

(II)

Ordinal jointing-box (IV)

(III)

a´ Phase-b b´ Phase-c c´ Single-core cable Metallic-sheath-current limiter (inductance) (b) Cross-bonding metallic-shielding method 2e

Phase-a

a

Phase-b Phase-c

b a, b, c: conductor jointing points

c Zc

Zc

Zc

a′, b′, c′: metallic sheath jointing points

Zc

Zc

Zc: surge impedance across the conductor and metallic sheath

Zc

a′ b′ c′

Zs: self-surge impedance across the metallic sheath and earth

Zs – Zm

Zm: mutual surge impedance across the sheaths of different phases

Zs – Zm g1

g2 Zm

(c) The equivalent circuit-1 of insulation box II

Zm

2e

2e

(– e) a ec

eb i1

Zc Zc

Zc

c′

Zc

a′

a′ b′

c

(Zs – Zm)

(Zs – Zm)

i2

Zc b Zc

g2

Th

Zm

es

am

ote

nti

al

Figure 9.7 Cross-bonding metallic shielding method.

c′

ec Zc c Zc

b′

1 (Z – Zm) 2 s 1 (Z – Zm) 2 s

ep

(d) The equivalent circuit-2

Zc eb

(Zs – Zm)

g1 Zm

Zc

Zc

Zc

(e)

a

1 Z 2 m

(e) The equivalent circuit-3

9.3 Metallic Sheaths and Outer Coverings

each longitudinal section. The major expected effects are obviously (i) reducing induced sheath voltages, (ii) restraining insulator temperature increases, and (iii) restricting interference with the environment. One disadvantage of this practice is that the sheath terminals are earth-grounded longitudinally every three spans instead of every span. Therefore, protecting the sheath circuit from surge voltage and current is vitally important, because the interval of the earth connection is three times longer.

9.3.5

Surge Protection at Jointing Boxes

Now we will examine surge phenomena arising on the cable conductors and sheath circuits. Surge phenomena arising on the sheath circuit should be carefully investigated, because electromagnetic and electrostatic coupling densities between the cable conductor and the metal sheath are high for all frequency zones. Let’s undertake a surge-voltage analysis of the sheath ends at intermediate jointing box II with a cross-bond connection that would be caused by the incident surge voltage e(t) from the main conductors. In Figure 9.7b, incident surge voltage e traveling through the phase-a conductor arrives at insulated jointing box II. The surge impedance circuit of the jointing box can be written as in Figure 9.7c, where surge source voltage 2e is inserted at point a. Incidentally, the reason source voltage 2e, instead of e, is inserted in this case will be explained in Section 15.5 and Figure 15.7. Figure 9.7c is vertically symmetrical, so it can be modified to Figure 9.7d. That figure is also symmetrical, so points g1 and g2 should have the same potential. Accordingly, this figure can be simplified as in Figure 9.7e. The impedance looking down into the circuit from points a and b is Zb a

4Zc

Zs − Zm

=

4Zc Zs − Zm 4Zc + Zs − Zm

9 13

The voltage between points a and b is Zb a Zs −Zm 4e 2e = 2Zc + Zb a 4Zc + 3 Zs − Zm Zs − Zm where m = 4Zc + 3 Zs − Zm

eb −ea =

m 4e 9 14

This equation means an initial surge voltage of magnitude 4me appears between points a and b . This voltage is shared by the four Zc in Figure 9.7b. Accordingly, the surge potentials at each point are 2me for point b , me for point c, zero for point c , and –me for points b. To sum up, we have the solution as in the following equations. The initial induced surge arising on the conductors and the sheath of each phase at jointing box II is, for transmitted wave voltages, The conductor voltage

The sheath voltage

ea = e

ea = −2me

eb = − me ①

eb = 2me

ec = me

ec = 0

9 15a

②

Furthermore, we need to consider the longitudinal induced voltages across the right and left terminals of insulated jointing box II in Figure 9.7b, as follows: Sheath box phase-a ea = −2me

eb = 2me

ea −eb = − 4me

Sheath box phase-b eb = 2me

ec = 0

eb − ec = + 2me

Sheath box phase-c ec = 0

ea = −2me

ec − ea = + 2me

Left-side terminals Zs − Zm m= 4Zs + 3 Zs −Zm

Right-side terminals

Across voltage 9 15b

193

194

9 Power Cables

In the figure, the voltage 4me will appear across the left and right terminals of the phase-a insulating box. Arresters as well as current-restraining devices (inductances) must be adopted, as shown in this figure. As a numerical check, although the surge impedances of cables are affected by the design structure of the cable and the installed conditions, the individual differences are small. Let’s assume Zc = 15 Ω, Zs = 25 Ω, Zm = 13 Ω. In this case, m can be derived from Eq. (9.14) as m = 0.125, and, accordingly:

•• •

The surge voltage of the conductors (transmitted wave) is ea = e, eb = −0:125e, ec = +0 : 125e. The surge voltages of the sheaths are ea = −0.25e, eb = +0.25e, ec = 0. The surge voltage across the insulated joint box is ea – eb = −0.5e, eb – ec = +0.25e, ec – ea = +0.25e.

These values for sheath voltage of 0.25e as well as the voltage across the jointing box of 0.5e are relatively large, so without appropriate protection by the arresters, the insulation of the cable outer-covering layer or across the cable joint sheath will be damaged by conductor surge e. Figure 9.7b also shows the practice of surge protection by arresters. Finally, surge phenomena that would be caused on a power cable under various conditions are explained in Sections 15.3–15.16.

195

10 Synchronous Generators, Part 1 Circuit Theory Any large or small power system has unique dynamic characteristics and a wide variety of behaviors. Most of these characteristics are due to the nature of generators, because a power system is an aggregate of parallel connected generators and loads. Therefore, it is not too much to say that the dynamic features of a generator or rotating machine are the dominant features of a power system. Given this viewpoint, studying generator theory is the same as studying powersystem theory.

10.1

Generator Model in a Phase abc-Domain

The generator abc-domain model given by Eq. (3.44) and Figure 3.15, and the 012-domain model given by Eq. (3.45) and Figure 3.16, are not physically correct. We need to find another model that is physically reasonable and that can be utilized to calculate circuits for a power system as in the 012-domain symmetrical-coordinate model. Park’s generator model and equations and the resulting equations in the dq0-domain give us a satisfactory answer for this requirement. Due to these theoretical processes, generators can be connected with other network facilities in the symmetrical-coordinate domain as a circuit for the purpose of network calculations. In this chapter, starting from the basic concept of a three-phase rotating machine, we will introduce Park’s theory for generators based on dq0 transformation first, and then examine the generator’s equations and equivalent circuits in relation to the 012-domain as well as the dq0-domain; furthermore, we will demonstrate a transient analysis of the generator for short-circuit faults. A generator is an electromechanical machine composed of a static part (the stator) and a rotating part (the rotor) whose relative position is changed periodically by rotating angle θ = ωt. In other words, a generator is a three-phase electromagnetic machine composed of l(t) and resistance r of the stator and rotor windings (we ignore leakage capacitances of the stator and of the rotor), while the inductance l(t) should periodically change depending on the relative angular position θ = ωt between the stator and rotor windings. The voltages v(t) [V], currents i(t) [A], flux linkage ψ(t) [weber-turn], and magnetic reluctances of the related magnetic passes should also be functions of ωt. A generator’s fundamental electrical structure can be expressed as shown in Figure 10.1. Although a two-pole machine is shown, a multipole machine with any number of pairs of poles can be treated as a two-pole machine electrically, because armature (stator) windings are identically arranged with respect to each pair of poles. We will investigate Figure 10.1 in detail as the starting point for developing the mathematical model. 10.1.1

Stator and Rotor Windings Structure

The rotor has two axes of mechanically rectangular symmetry, and we will call them the d-axis and the q-axis. In other words, the rotor is designed symmetrically with a d-axis and q-axis.:

••

d-axis or direct-axis: The axis from the axial center point o in the pole direction q-axis or quadrature-axis: The axis from the axial center point o in the direction 90 ahead (leading) of the d-axis

Field windings: The field winding (called the d-axis field coil) is a closed circuit connected to a source of DC voltage Efd, with an inductance to produce flux only in the direction of the d-axis.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

10 Synchronous Generators, Part 1

d-a xis

196

Phase-b eb(t)

θa + 90°

ib(t)

θb = θc θ =

θa

= 0° 24 ° + 120 θ+ θ =

Ψb(t) is

rfd

q

rk

d

+

–

ifd

Ro tor

ikq

rk

or

d

t Sta

Ef

120° θ –

q-ax

ia(t)

Phase-a

Ψa(t) ea(t)

ik

d

Ψc(t) ic(t)

Neutral point ia(t) + ib(t) + ic(t)

ec(t) Phase-c

Ro

tati

ng

dir

ect

ion

Figure 10.1 Electrical concept of a synchronous generator.

The flux may partly flow into the left side (i.e. the +q-axis component) and partly into the right side (i.e. the-q-axis component) of the d-axis, but such ±q-axis components will be balanced because of the design symmetry and must be canceled in total. Therefore, we can assume that the field windings produce flux only in the d-axis direction and not in the q-axis direction. So, we can imagine that the d-axis field coil exists in our rotor circuit model but the q-axis field coil does not exist. Damper windings: Typical hydraulic-turbine-driven generators (vertical type with salient poles) have amortisseur or squirrel-cage windings in the pole face (damper windings), which consist of copper bars through the pole connected at their ends as a closed circuit. Thermal or nuclear-turbine-driven generators (horizontal type with cylindrical nonsalient poles) have only field windings and do not have damper windings. However, eddy currents are forced to flow into the pass of the solid steel rotor for the duration of transient or current-unbalanced conditions. Therefore, we have to assume that steam-driven generators also have damper windings. The currents in these damper windings may be assumed to flow in only two closed circuits – one in the d-axis and the other in the q-axis – as an approximate electrical model, again because of the rotor’s symmetrical design for the d-axis and q-axis. With all these bases, the rotor-circuit model consists of one field coil and one damper coil in the d-axis and one damper coil in the q-axis, as shown in Figure 10.1. Stator (armature) windings: The stator has three stator windings for phases a, b, c, and one end of each of these three coils is connected to the neutral terminal. The three windings are arranged electrically by 120 symmetrically to each other. We can justify the assumption that the stator windings are sinusoidally distributed along with the air gap, as far as the mutual effects of the rotor are concerned, because the generator windings are distributed so as to minimize harmonics in the design. On these bases, the stator-circuit model consists of three star-connected phase a, b, c coils, each of which has its own self-inductance and resistance as well as mutual inductances among all other stator coils and rotor coils, as shown in Figure 10.1.

10.1 Generator Model in a Phase abc-Domain

10.1.2

Relative Angular Position Between Rotor and Stator

The stator is immovable, and the rotor rotates counterclockwise at an angular speed of ω = dθ/dt; therefore, the relative position between the rotor and the stator is measured by the rotating angle of the rotor d-axis. That is, referring to Figure 10.1, the rotating position of each coil in time can be written as follows on a d-axis basis: Stator phase-a coil θa = θ = ωt = 2πft phase-b coil θb = θ + 240 = θ −120 = ωt − 120 phase-c coil θc = θ + 120 = ωt + 120 Rotor field coil 0

10 1

damper d-axis coil 0 damper q-axis coil + 90 The position of each stator coil is a function of time t on the d-axis basis. Now, we have a physically reasonable abc-domain generator circuit in Figure 10.1 that can be utilized as the original model for a generator circuit. 10.1.3

Three-Phase abc-Domain Circuit Equations and Inductance Matrices

We define the quantities of each coil in Figure 10.1 as follows: ψa t , ψb t , ψc t

total flux linkage of phase a, b, c coil, respectively Wb-turn

ea t , eb t , ec t

terminal voltage of phase a,b, c coil, respectively V

ia t , ib t , ic t

terminal current of phase a,b, c coil, respectively A

ψ fd t

total flux linkage of d-axis field coil Wb-turn

ψ kd t

total flux linkage of d-axis damper coil Wb-turn

ψ kq t

total flux linkage of q-axis damper coil Wb-turn field excitation voltage V

Efd ifd t

current in d-axis field coil A

ikd t

current in d-axis damper coil A

ikq t

current in q-axis damper coil A

With these definitions, the following equations can be derived. The equation of the stator (armature) coils’ voltage is

ea t eb t ec t eabc t

=

d ψ t dt a d ψ t dt b d ψ t dt c d ψ dt abc

ia t − r ib t ic t iabc t t

10 2

197

198

10 Synchronous Generators, Part 1

The equation of the rotor coils’ voltage is

Efd 0 0

=

EF t

d ψ t dt fd d ψ t dt kd d ψ t dt kq

+

rfd ifd t rkd ikd t rkq ikq t

10 3

Voltage drop

d ψF t dt The equation of the stator (armature) coils’ flux linkage is ψa t ψb t ψc t

t =

− laa t lba t lca t

ψ abc

lab t − lbb t lcb t

lac t lbc t − lcc t

l abc t

ia t ib t ic t

+

lafd t lbfd t lcfd t

lakd t lbkd t lckd t

iabc t

lakq t lbkq t lckq t

ifd t ikd t ikq t

labc −F t

10 4

iF t

And the equation of the rotor coils’ flux linkage is ψ fd t ψ kd t ψ kq t

lfad t l = − kad t lkaq t

ψF t

lfbd t lkbd t lkbq t

lfcd t lkcd t lkcq t

l F − abc t

ia t ib t ic t

+

Lffd Lfkd 0

iabc t

Lfkd Lkkd 0 LF

0 0 Lkkq

ifd t ikd t ikq t

10 5

iF t

where r rfd ,rkd ,rkq

resistance of each stator coil Ω resistance of field d-axis, damper d-axis coil ,damper q-axis coil

laa t , lbb t , lcc t

self-inductances of stator coils H

lab t ,lbc t ,etc

mutual inductances among stator coils H

labc t l abc − F t

mutual inductance matrix between stator coils and rotor coils H

l F − abc t

same as the above H

LF

mutual inductance matrix among the three rotor coils H

The matrices labc–F(t) and lF-abc(t) are obviously dependent on time t, because all the mutual inductances between the stator phase coils and rotor coils (lafd(t), lakd(t), lakq(t)) are affected by the changing relative position over time between the stator and the rotor. Therefore, lafd(t), lakd(t), lakq(t), etc. include the symbol (t) to emphasize their time dependency with rotation. Matrices labc–F(t) and lF-abc(t) also include (t) for the same reason. The relative position between the phase-a stator coil and phase-b stator coil does not change all the time. However, the self-inductance laa(t) of the phase-a coil and the mutual inductance lab(t) between phase-a and -b coils may be affected by the changing rotor position over time, because the flux passes of ψ aa as well as ψ ab may flow partly through the periodically changing air gap and rotor structure. Therefore, the self- and mutual-inductance laa(t), lab(t) must be time dependent. By such reasoning, the matrix labc(t) also has time-dependent inductances and thus uses the symbol (t). The self-inductance of each rotor coil Lffd, Lkkd, Lkkq in the incident matrix lF is not affected by ωt. In other words, they are time independent, because the field d-axis, damper d-axis, and damper q-axis coils are fixed on the d-axis or q-axis, and all the linking fluxes of these coils in the rotor are not affected by the relative position of the stator phase coils from the rotor. The mutual inductance Lfkd = Lkfd between the rotor d-axis coil and the damper d-axis coil also exists as timeindependent mutual inductance. On the other hand, mutual inductance between the rotor d-axis coil and the rotor q-axis does not exist physically. Therefore, the matrix LF is time independent and includes some zero elements, as shown in Eq. (10.5).

10.1 Generator Model in a Phase abc-Domain

Now, we will examine how labc(t), labc–F(t), lF–abc(t) can be written as time-dependent (ωt) inductance matrices. The conclusive equations are shown first by the following equations, followed by our justifications for these equations. The inductance of the matrix of the stator coils is − laa t lba t lca t

l abc t =

lab t − lbb t lcb t

lac t lbc t −lcc t

− Laa0 + Laa2 cos 2θa = Lab0 −Laa2 cos θa + θb Lab0 − Laa2 cos θa + θc

Lab0 − Laa2 cos θa + θb − Laa0 + Laa2 cos 2θb Lab0 −Laa2 cos θb + θc

− Laa0 + Laa2 cos 2θ = Lab0 −Laa2 cos 2θ − 120 Lab0 − Laa2 cos 2θ + 120

Lab0 − Laa2 cos θa + θc Lab0 − Laa2 cos θb + θc − Laa0 + Laa2 cos2θc

Lab0 − Laa2 cos 2θ −120 − Laa0 + Laa2 cos 2θ + 120 Lab0 − Laa2 cos 2θ

Lab0 −Laa2 cos 2θ + 120 Lab0 − Laa2 cos2θ − Laa0 + Laa2 cos 2θ − 120

θa = θ, θb = θ −120 , θc = θ + 120 10 6 And the inductance matrices between the stator and six rotor coils are

l abc − F

lafd t t = lbfd t lcfd t

lakd t lbkd t lckd t

Lafd cos θa = Lafd cosθb Lafd cos θc

lakq t lbkq t lckq t

Lakd cos θa Lakd cosθb Lakd cos θc

Lafd cos θ = Lafd cos θ −120 Lafd cos θ + 120

l F − abc

lfad t t = lkad t lkaq t

lfbd t lkbd t lkbq t

=

Lafd cos θa Lakd cos θa − Lakq sin θa

=

Lafd cos θ Lakd cos θ − Lakq sin θ

Lafd cos θa = Lafd cos θb Lafd cosθc

Lakd cosθa Lakd cos θb Lakd cosθc

Lakq cos θa + 90 Lakq cos θb + 90 Lakq cos θc + 90

− Lakq sin θa − Lakq sin θb −Lakq sin θc

Lakd cos θ Lakd cos θ − 120 Lakd cos θ + 120

10 7a

− Lakq sin θ − Lakq sin θ − 120 −Lakq sin θ + 120

lfcd t lkcd t lkcq t Lafd cosθb Lakd cosθb − Lakq sin θb

Lafd cosθc Lakd cosθc − Lakq sin θc

Lafd cos θ −120 Lakd cos θ −120 −Lakq sin θ − 120

Lafd cos θ + 120 Lakd cos θ + 120 −Lakq sin θ + 120

10 7b

= labc − F

t t

where the matrix []t is the transposed matrix of [] – that is, the rows and columns are interchanged. Now let’s examine how Eqs. (10.6) and (10.7a and b) are derived.

199

200

10 Synchronous Generators, Part 1

The total flux linkage of the phase-a armature coil can be described as follows: ψ a t = −ψ aa t + ψ ab t + ψ ac t + ψ afd t + ψ akd t + ψ akq t = −laa t ia t + lab t ib t + lac t ic t

10 8

+ lafd t ifd t + lakd t ikd t + lakq t ikq t Here laa(t) = ψ aa(t)/ia(t) is the self-inductance of the phase-a stator coil (the flux linkage of the phase-a coil induced by the unit current of the phase-a coil). lab(t) = ψ ab(t)/ib(t) is the mutual inductance between the phase-a stator coil and the phase-b coil (the flux linkage of the phase-a coil induced by the unit current of the phase-b coil). lafd(t) = ψ afd(t)/ifd(t) is the mutual inductance between the phase-a stator coil and the rotor d-axis coil (the flux linkage of the phase-a coil induced by the unit current of the rotor d-axis coil), and so on. 10.1.4

Introduction of Stator Inductance Matrix: labc(t)

The surface of any rotor is uneven, and the air-gap length between the stator and rotor varies depending on the relative position of the rotor to the stator. In other words, no rotor is a uniform cylinder from the viewpoint of magnetic passes. This means the magnetic reluctance labc(t) varies depending on the relative angular position from the d-axis. Therefore, the self-inductance of any armature winding varies periodically as a function of ωt, and it must be at its maximum when the pole (d-axis) is in line with the phase axis, and at its minimum when the interpole (q-axis) is in line with the phase axis. That is, laa(t) must be a periodic function with electrical angle 180 and an even function with θa. In other words, laa(t) can be written as an equation in the Fourier series expansion: laa t = Laa0 + Laa2 cos 2θa + Laa4 cos 4θa + Laa6 cos 6θa + … ∴ laa t = Laa0 + Laa2 cos 2θa

10 9a

θa = ωt The armature coil is designed as sinusoidal distribution windings, so the third term and other smaller terms on the right side can be ignored. Figure 10.2 shows the state. The equation lab(t) = lba(t) is presumed as follows in the same way: lab t = lba t = Lab0 −Lab2 cos θa + θb

10 9b

here θa = ωt, θb = ωt − 120

However, the reasoning behind Eqs. (10.9a) and (10.9b) is still not clear, and the physical images and values of Laa0, Laa2, Lab0, Lab2 cannot be obtained using this explanation. Let’s examine this further from the viewpoint of electromagnetism. The equation of a magnetic circuit is Magnetic motive force mmf = flux linkage ψ × magnetic reluctance ᑬ laa (t) = Laa0 + Laa2 cos2θa

Laa0

Laa2

0°

90°

180°

270° θa = ωt

360°

Figure 10.2 Self-inductance of a phase-a coil.

Using symbols, ammf, aψ: Magnetic motive force and flux linkage of the phase-a coil to produce the phase-a current ia(t). aψ d(t), aψ q(t): d-axis and q-axis component of the flux linkage aψ(t). ᑬd, ᑬq: Magnetic reluctance of the flux pass for aψ d(t) in the d-axis direction and for aψ q(t) in the q-axis direction, where the reluctances ᑬd for aψ d(t) and ᑬq for aψ q(t) are time-independent constants because aψ d(t), aψ q(t) are in synchronization with the rotating d-axis and q-axis.

10.1 Generator Model in a Phase abc-Domain

Then d-axis component of magnetic motive force = a mmf cosθa = a ψ d t ᑬd q-axis component of magnetic motive force = a mmf cos θa + 90 = a ψ q t ᑬq or aψ d

a mmf

cos θa ᑬd a mmf a mmf cos θa + 90 = − sin θa aψ q t = ᑬq ᑬq t =

10 9c

Now we will examine the behavior of the flux linkage ψ aa(t) [Weber turn] caused by current ia(t)[A] and the resulting flux linkages caused on the phase-a, -b, -c coils and the three rotor coils described by the symbols ψ ba(t), ψ ca(t) and ψ fad(t), ψ fkd(t) ψ fkq(t), respectively. As shown in Figure 10.3, the flux linkage caused by current ia(t) and interlinked with ia(t) itself (self-linkage) can be decomposed into a d-axis component aψ d(t) and a q-axis component aψ q(t). While the stator phase-a coil links to aψ d(t) at angle θa and links to aψ q(t) at angle (θa + 90 ), the total flux linkage on the phase-a coil ψ aa(t) = laa(t) ia(t) can be calculated as follows: ψ aa t = laa t ia t = a ψ d t cos θa + a ψ q t cos 90 + θa = =

a mmf

cos2 θa +

ᑬd

a mmf

ᑬq

sin2 θa =

1 1 a mmf + + ᑬd ᑬq 2

a mmf

2

a mmf

ᑬd

1+ cos 2θa a mmf 1− cos2θa + ᑬq 2 2

1 1 cos 2θa − ᑬd ᑬq

10 9d

= A + Bcos2θa = A + Bcos 2θ Where A=

a mmf

2

1 1 a mmf + , B= ᑬd ᑬq 2

1 1 − ᑬd ᑬq

10 9e

ax

a𝜓q (t) cos (𝜃a + 90º)

d-

q-

(t)

is

a𝜓

ax

ia (t)

Phase-a coil

The same linking flux aψ of current ia(t) links with the phase-b coil and causes flux linkage on the phase-b coil. While the stator phase-b coil links to aψ d(t) at angle θb and links to aψ q (t) at angle (θb + 90 ), the total flux linkage on the phaseb coil ψ ba(t) = lba(t) ia(t) can be calculated as follows:

a𝜓 q

a𝜓 d

(t)

is

(t)

a𝜓d (t)

cos𝜃a =

a mmf

𝜃a 𝜃a + 90º Figure 10.3 Linking flux of a phase-a winding coil.

d

cos2𝜃a

201

202

10 Synchronous Generators, Part 1

ψ ba t = lba t ia t = a ψ d t cos θb + a ψ q t cos θb + 90 = = =

a mmf

ᑬd a mmf

ᑬd a mmf

2

cos θa cosθb +

a mmf

ᑬq

sin θa sin θb

cos θa −θb + cos θa + θb a mmf cos θa − θb −cos θa + θb + ᑬq 2 2 1 1 a mmf + cos θa −θb + ᑬd ᑬq 2

1 1 − cos θa + θb ᑬd ᑬq

10 9f

= Acos θa −θb + Bcos θa + θb = Acos 120 + Bcos θa + θb =−

1 A− Bcos θa + θb 2

The inductance is defined as the flux linkage per one ampere, which means l(t) = ψ(t) under the condition i(t) = 1[A] (in other words, [H] = [Wb]/[A] defines the units). Then the equations for laa(t), lba(t) are obtained as follows: laa t = ψ aa t ia t = Laa0 + Laa2 cos 2θa

①

lba t = ψ ba t ia t = Lab0 − Laa2 cos θa + θb

②

where 10 9g

Laa0 =

A 1 1 a mmf 1 = 2Lab0 = + ia t i a t 2 ᑬd ᑬ q

③

Laa2 =

B mmf 1 1 1 =a − ia t i a t 2 ᑬd ᑬq

④

Equation (10.9g) proves that the intuitively written Eqs. (10.9a) and (10.9b) are correct. Also, Lab2 in Eq. (10.9b) is the same as Laa2 in Eq. (10.9g)②, so it can be replaced by Laa2. Furthermore, Eq. (10.9g)③④ indicates Laa0 > Laa2. Accordingly, laa(t) of Eq. (10.9a) can be represented by the curve drawn in Figure 10.2. The time-dependent inductances lbb(t), lcc(t), lac(t), lbc(t), etc. can be derived analogously. Now, arranging all these inductances, the inductance matrix of the armature coil, Eq. (10.6), is obtained. 10.1.4.1 Introduction of the Stator and Rotor Mutual Inductance Matrix labc–F(t), lF–abc(t)

For lafd = lfad(t), lafd(t) is the mutual inductance for the linkage of field flux ψ fd (induced by ifd(t)) to the stator phase-a coil. From Figure 10.1, lafd (t) reaches its peak value when the field d-axis coil and the stator phase-a coil are in line, and it has a negative peak value at the 180 reverse position. Accordingly, lafd t = lfad t = Lafd cos θa

10 9h

The mutual inductance is obtained as a term of the fundamental sinusoidal form cosθa without including higher harmonic components, because generators are mechanically designed so that the stator mmf is sinusoidally distributed. For lakd(t) = lkad(t), the damper d-axis coil is in line with the field d-axis coil, so lafd(t) = lfad(t) is in the same form as Eq. (10.9h), namely lakd t = lkad t = Lakd cos θa

10 9i

For lakq(t) = lkaq(t), the damper q-axis coil is in a position 90 ahead of the d-axis. Therefore lakq t = lkaq t = Lakq cos θa + 90 = − Lakq sin θa

10 9j

In the same way, the mutual inductances lbfd(t), lcfd(t), etc., in regard to the phase-b coil and phase-c coil can be obtained by replacing θa in Eqs. (10.8)–(10.10) with θb and θc. From all of these explanations, the mutual inductance matrices between the stator and the rotor, Eqs. (10.7a, b), are obtained. In conclusion, Eqs. (10.1)–(10.9j) are the equations for the generator in the abc-domain.

10.2 dq0 Method (dq0 Components)

10.2

dq0 Method (dq0 Components)

The derived equations are of no use as they are, because inductances are periodically time-dependent variables, and the equations cannot be connected with transmission lines and other equipment. Consequently, we now need to introduce the dq0 method. 10.2.1

Definition of the dq0 Method and Its Physical Meaning

The dq0 method is a transformation from three variables in the abc-domain to another three variables in the dq0domain from a mathematical viewpoint. The dq0 method is defined by the following equations, including ωt in its transformation matrix and eabc(t) as a real number and crest value expression: ed t eq t e0 t

cos θa 2 − sin θa = 3 1 2

edq0 t

id t iq t i0 t

cos θa 2 − sin θa = 3 1 2

cos θb − sin θb 1 2

=

=

cos θa cos θb cosθc

1 1 1

− sin θa − sin θb − sin θc

=

ψ abc t

cosθa cos θb cos θc

−sin θa − sin θb −sin θc

10 10b

ψa t ψb t ψc t

10 10c

ed t eq t e0 t

10 11a

edq0 t

1 1 1

D−1 t

iabc t

ψa t ψb t ψc t

− sin θa − sin θb −sin θc

ia t ib t ic t

ψ abc t

D −1 t

eabc t

ia t ib t ic t

cos θc −sin θc 1 2

Dt

cos θa cos θb cosθc

10 10a

iabc t

Dt

ψ dq0 t

ea t eb t ec t

cos θc − sin θc 1 2

cosθb −sin θb 1 2

cos θa 2 − sin θa = 1 3 2

ea t eb t ec t eabc t

Dt

idq0 t

ψd t ψq t ψ0 t

cos θc − sin θc 1 2

cosθb −sin θb 1 2

id t iq t i0 t

10 11b

idq0 t

1 1 1

D −1 t

ψd t ψq t ψ0 t

10 11c

ψ dq0 t

where θa = θ = ωt,

θb = θ − 120 = ωt −120 ,

θc = θ + 120 = ωt + 120

10 11d

203

204

10 Synchronous Generators, Part 1

and cos θa 2 − sin θa Dt = 1 3 2

cos θb − sin θb 1 2

cos θc − sin θc 1 2 ①

cosθa 2 cos θa + 90 = 3 1 2

D

−1

cosθa t = cos θb cos θc

cosθb cos θb + 90 1 2

−sin θa − sin θb − sin θc

1 cos θa 1 = cos θb 1 cosθc

cos θc cos θc + 90 1 2 cos θa + 90 cos θb + 90 cos θc + 90

10 12

1 1 1

②

The transformation matrices D(t), D−1(t) are functions of time t because they include θa, θb, θc. Now we will examine the physical meaning of the dq0 transformation by using the stator flux linkages ψ a(t), ψ b(t), ψ c(t) and their transformed flux linkages ψ d(t), ψ q(t), ψ 0(t). The positions of the stator phase-a, -b, -c coils to the d-axis are θa, θb, θc, respectively, so that the components in the d-axis direction of ψ a(t), ψ b(t), ψ c(t) are ψ a(t) cos θa, ψ b(t) cos θb, ψ c(t) θc, respectively. Then the definition of ψ d(t) from Eq. (10.10c) is 2 ψ t cos θa + ψ b t cos θb + ψ c t cos θc 3 a 2 d-axis component of ψ a t + d-axis component of ψ b t = 3

ψd t =

10 13a

+ d-axis component of ψ c t The components in the q-axis direction of ψ a(t), ψ b(t), ψ c(t) are ψ a(t) cos(θa + 90 ) = −ψ a(t) sin θa, ψ b(t) cos(θb + 90 ) = −ψ b(t) sin θb, ψ c(t) cos(θc + 90 ) = −ψ c(t) sin θc, respectively. Then, the equation for ψ q(t) from Eq. (10.10c) is 2 − ψ a t sin θa − ψ b t sin θb −ψ c t sin θc 3 2 q-axis component of ψ a t + q-axis component of ψ b t + q-axis component of ψ c t = 3

ψq t =

10 13b In other words, the physical meaning of ψ d(t) is two-thirds the value of the total sum of the d-axis components of ψ a(t), ψ b(t), ψ c(t). The meaning of ψ q(t) is the same, but for the q-axis. ψ 0(t) from Eq. (10.10c) is the same as that of the zero-sequence symmetrical component: ψ0 t =

1 ψ t + ψb t + ψc t 3 a

10 13c

The voltages and currents in the dq0-domain are defined analogously. 10.2.2

Mutual Relation of the dq0-, abc-, and 012-Domains

The theory of the double-axes armature reaction, which appears in books on synchronous machine design, can be said to be the same as the dq0 transformation but eliminating zero-sequence quantities in principle. Although the theory is appropriate for generator design, it may be imperfect or useless as a theoretical tool because it cannot explain unbalanced or transient phenomena; neither can it provide a means to connect the generator to other network equipment.

10.2 dq0 Method (dq0 Components)

On the other hand, the dq0 method is a mathematical transformation from the abc-domain to the dq0-domain using three variables, covering any phenomenon, and it furthermore provides a “generator circuit” to be connected to other network equipment. Let’s examine the mutual relations of the abc-, dq0-, and 012 domains, put together. The relations of quantities in the three different domains are defined by the following equations. For the 012 abc domain: e012 t = a eabc t

10 14

eabc t = a − 1 e012 t For the dq0

abc domain:

edq0 t = D t Re eabc t

10 15

Re eabc t = D −1 t edq0 t For the dq0

012 domain:

edq0 t = D t Re a − 1 e012 t

10 16

Re e012 t = Re a D − 1 t edq0 t

It is obvious that the transformation between the dq0- and 012-domains is defined by Eq. (10.16), which is derived from Eqs. (10.14) and (10.15). However, remember that ed(t) and eq(t) from Eqs. (10.10a) and (10.11a) are defined as real-number quantities, because they are DC values under a balanced three-phase condition, as is soon mathematically confirmed. Therefore, the threephase quantities eabc(t), iabc(t), ψ abc(t) in Eqs. (10.10a, b, c) and (10.11a, b, c) should be real-number expressions. Characteristics of dq0-Domain Quantities

10.2.3

Now we will examine how arbitrary power-frequency quantities having positive-, negative-, and zero-sequence components are transferred into dq0-domain quantities: ed t eq t e0 t

cos ωt 2 − sin ωt = 3 1 2

cos ωt − 120 − sin ωt −120 1 2

edq0 t

cos ωt + 120 −sin ωt + 120 1 2

Dt

×

Ea1 cos ωt + α1 Ea1 cos ωt + α1 − 120 Ea1 cos ωt + α1 + 120

+

Ea2 cos ωt + α2 Ea2 cos ωt + α2 −120 Ea2 cos ωt + α2 + 120

positive-seq

=

Ea1 cos α1 Ea1 sin α1 0

+

negative-seq

Ea2 cos 2ωt + α2 − Ea2 sin 2ωt + α2 0

p-seq

n-seq

+

+

Ea0 cos ωt + α0 Ea0 cos ωt + α0 Ea0 cos ωt + α0

10 17

zero-seq

0 0 Ea0 cos ωt + α0 zero-seq

namely ed t = Ea1 cos α1 + Ea2 cos 2ωt + α2 eq t = Ea1 sin α1 −Ea2 sin 2ωt + α2 e0 t = Ea0 cos ωt + α0

10 18

205

206

10 Synchronous Generators, Part 1

Equations (10.17) and (10.18) explain that the positive-sequence component appears as a DC component in the dq0-domain, which is time independent, while the negative-sequence component appears as a double-frequency component. Eq. (10.18) can be recast as ed t + jeq t

= Ea1 e jα1 + Ea2 e −j 2ωt + α2 jωt = Ea1 e

a 10 19

j ωt + α1

+ Ea2 e

−j ωt + α2

ed t + jeq t

e

ed t − jeq t

= Ea1 e −jα1 + Ea2 e j 2ωt + α2

b

or

ed t − jeq t

e

− jωt = Ea1 e −j

ωt + α1

+ Ea2 e

a 10 20

j ωt + α2

b

Then Eq. (10.19b) can be decomposed into real and imaginary parts as follows: ed cosωt − eq sin ωt = Ea1 cos ωt + α1 −Ea2 cos ωt + α2

10 21

ed sin ωt + eq cos ωt = Ea1 sin ωt + α1 − Ea2 sin ωt + α2

10 22

or

These are the equations including positive-sequence and negative-sequence components. Then, when only a positive-sequence component Ea1 cos(ωt + α1) exists, the following relation is preserved: e1 t = Ea1 cos ωt + α1 = ed cos ωt − eq sin ωt

10 23

Or, if only the positive-sequence component Ea1 sin(ωt + α1) exists, e1 t = Ea1 sin ωt + α1 = ed sin ωt + eq cosωt

• • •

10 24

Now, we have found that: Positive-sequence voltage rotates at angular velocity ω in synchronization with the rotor or d-axis and q-axis, so that it is at a standstill from the d- and q-axes viewpoints. Therefore, positive-sequence voltage appears as DC variables Ea1 ejα1 in ed(t) and eq(t): ed t = Ea1 cos α1 , eq t = Ea1 sin α1

10 25a

Negative-sequence voltage rotates in the abc-domain with inverse rotation (angular velocity – ω) to the rotor or d- and q-axes, so that it appears as the voltage components of angular velocity 2ω in ed(t), eq(t): ed t = Ea2 cos 2ωt + α2 , eq t = Ea2 sin 2ωt + α2

10 25b

Zero-sequence voltage is the same as that of the symmetrical components: e0 t = Ea0 cos ωt + α0

10 26

Equations (10.19a) and (10.20a) are the symmetrical components’ voltages looking from the stator, and Eqs. (10.19b) and (10.20b) may be said to be the same, but looking from the rotor.

10.3

Transformation of Generator Equations from the abc-Domain to the dq0-Domain

10.3.1

Transformation of Stator Voltage Equations to the dq0-Domain

We will examine the transformation of a generator’s equations, Eqs. (10.2)–(10.7), to equations in the dq0-domain using the definitions in Eqs. (10.10) and (10.11). Again, Eq. (10.2) is eabc t =

d ψ t − riabc t dt abc

10 27

10.3 Transformation of Generator Equations from the abc-Domain to the dq0-Domain

Recall that D(t), D−1(t) are the matrices of real-number elements. Left-multiplying both sides by D(t), d ψ t dt abc

edq0 t = Re D t eabc t = Re D t =D t

d D −1 t ψ dq 0 t dt d −1 D t dt

=D t

− D t riabc t

−D t rD − 1 t idq0 t

ψ dq0 t + D t D − 1 t

d ψ t − ridq0 t dt dq0

D−1(t) and ψ abc(t) are functions of time t, so the first term on the right side in this equation should be expressed as follows by applying a differential equation formula (Appendix A.5b): d D −1 t ψ dq0 t dt

=

d −1 D t dt

d −1 D t dt

∴ edq0 t = D t

ψ dq0 t + D −1 t

ψ dq0

d ψ t dt dq0

10 28

d t + ψ dq0 t −ridq0 t dt

where θa = ωt,

d −1 D dt

θb = ωt − 120 ,

cosθa d cos θb t = dt cos θc

ωt + 120 dθa dt 1 dθb 1 = −sin θb dt 1 dθc −sin θc dt − sin θa

−sin θa − sin θb −sin θc

dθa dt dθb − cosθb dt dθc −cos θc dt

− cos θa

0 − sin θa 0 = −sin θb − sin θc 0

− cos θa −cos θb − cosθc

0 0 0

dθ dt

dθa dθb dθc dθ =ω = = = dt dt dt dt Then

Dt

d −1 D t dt

cosθa 2 −sin θa = 3 1 2

cosθb −sin θb

cos θc − sin θc

1 2

1 2

− dθ

−sin θa −sin θb − sin θc

−cos θa − cosθb − cos θc

0 0 0

0 dθ dθ = dt dt 0

dt

0

0 0

0 0

d −1 D t dt

Dt

That is, the dq0-domain equation transformed from Eq. (10.2) is

ed t eq t e0 t

0 = dθ dt 0

where dθ = ω = 2πf dt

− dθ

dt

0

0

0

0

0

ψd t ψq t ψ0 t

d ψ t dt d d ψ t + dt q d ψ t dt 0

id t −r iq t i0 t

Park’s equation

10 29

207

208

10 Synchronous Generators, Part 1

10.3.2

Transformation of the Rotor Voltage Equation

The equation includes only dq0-domain quantities, so the transformation is not required. The equation is again provided here: d ψ t dt fd

Efd 0 = d ψ kd t dt 0 d ψ t dt kq

10.3.3

rfd ifd t r + kd ikd t rkq ikq t

10 30

Transformation of the Stator Flux Linkage Equation

Now ψ abc t = l abc t iabc t + labc− F t iF t

10 31

Then ψ dq0 t = D t labc t D − 1 t idq0 t + D t l abc − F t iF

10 32

D(t)labc(t)D−1(t)} and {D(t)labc–F(t)} in this equation can be calculated from labc(t) and labc–F(t) in Eqs. (10.6) and (10.7a), and the following equations are derived. The result can be proved manually, although the process is rather time-consuming and the demonstration is omitted from this book: 3 Laa0 + Lab0 + Laa2 2 D t labc t D − 1 t = −

3 Laa0 + Lab0 − Laa2 2 0

0 0

Ld − 0 0

0 Lq 0

0

0 0 Laa0 − 2Lab0

0 0 L0

This calculation can be proved rather easily by applying the hyperbolic sinusoidal formula described here. cosθa = {(ejθ + e−jθ)/2}, cosθb = {(ejθ a2 + e−jθ a)/2}, sinθa = {(ejθ − e−jθ)/2j}, sinθb = {(ejθ a2 − e−jθ a)/2j}, and so on (see Appendix A3):

D t labc − F

Lafd t = 0 0

Lakd 0 0

0 Lakq 0

Accordingly, the following equations in the dq0-domain are derived as the transformation of the original Eq. (10.4) from the abc- to the dq0-domain:

10.3 Transformation of Generator Equations from the abc-Domain to the dq0-Domain

ψd t ψq t ψ0 t

0 Lq 0

Ld =− 0 0

0 0 L0

id t iq t i0 t

Lafd + 0 0

0

Lakd 0 0

ifd t ikd t ikq t

Lakq 0

where

10 33

3 Ld = Laa0 + Lab0 + Laa2 2 3 Self-inductance of stator q-axis coil Lq = Laa0 + Lab0 − Laa2 2 Self-inductance of stator zero-sequence coil L0 = Laa0 −2Lab0 Self-inductance of stator d-axis coil

10.3.4

Transformation of the Rotor Flux Linkage Equation

Equation (10.5) is again ψ F t = − lF − abc t iabc t + LF iF t

10 34

Accordingly, ψ F t = − lF − abc t D −1 t idq0 t + LF iF t

10 35

and lF − abc t D −1 t =

Lafd cos θb Lakd cosθb −Lakq sin θb

Lafd cosθa Lakd cos θa − Lakq sin θa

Lafd cos θc Lakd cosθc −Lakq sin θc

cos θa cosθb cos θc

− sin θa −sin θb − sin θc

1 Lafd 3 1 = Lakd 2 1 0

0 0 Lakq

0 0 0

This equation is the same as the transposed matrix of D(t) labc–F(t) multiplied by 3/2. Then l F −abc t D − 1 t

=

3 D t labc −F t 2

t

That is, the dq0-domain equation transformed from Eq. (10.5) is ψ fd t ψ kd t ψ kq t

Lafd 3 Lakd =− 2 0

0 0 Lakq

0 0 0

id t iq t i0 t

Lffd + Lfkd 0

Lfkd Lkkd 0

0 0 Lkkq

ifd t ikd t ikq t

10 36

In conclusion, a generator’s equations in the dq0-domain are shown by Eqs. (10.29), (10.30), (10.33), and (10.36). Note that all of these equations are described using only fixed inductances L (independent of ωt), and all the ωt-dependent inductances have disappeared. Equation (10.29) is Park’s equation, named after R. H. Park, which is simply written as d dθ ψ t −rid t − ψ q t dt d dt d dθ eq t = ψ q t −riq t − ψ d t dt dt d e0 t = ψ 0 t − ri0 t dt

ed t =

Park’s equation

10 37a

209

210

10 Synchronous Generators, Part 1

Or by combined form of ed t + jeq t edq t = jωψ dq t +

d ψ t − ridq t dt dq

edq t = ed t + jeq t

Park’s equation phasor expression

10 37b

Where idq t = id t + jiq t ψ dq t = ψ d t + jψ q t

10.4

Physical Meanings of Generator Equations in the dq0-Domain

Now that we have derived the fundamental Eqs. (10.29)–(10.37) for a generator in the dq0-domain, let’s examine what the equations mean physically. The conceptual explanation of linking flux described in Supplement 1 at the end of this chapter may be a good reference to provide a better understanding of generators. 10.4.1

Main Fluxes and Leakage Fluxes

Figure 10.4a shows the flux paths of a cylindrical-type generator. Figure 10.4b is the conceptual image of the generator inductances in the dq0-domain. An observer is riding on the rotor. The stator coils of the d- and q-axes are the imaginary coils in the dq0-domain, which are derived as the transformed coils of the stator a-, b-, c-coils, so that they are stationary on the d-, q-axes. Then, all the coils of the stator d-, q-coils as well as the rotor fd-, kd-, kq-coils are stationary from the observer’s viewpoint. In other words, all of these coils are at a standstill on the d- and q-axes, so their inductances are of fixed values independent of time t. The physical meaning and value of the transformation from the abc-domain to the dq0-domain is indeed to realize this. Next, we will examine the physical meaning of each inductance in Eqs. (10.29)–(10.37). Figure 10.4b shows six coils electromagnetically coupled through a flux-core pass (the intersection of the d-, q-axes at a right angle is not shown in the image). The six coils are energy-coupled to each other through the linking flux ϕlink; therefore, the total ampereturns of the six coils are zero. Now we will focus on the excitation coil. When current ifd flows through the field coil, most of the flux induced by the electromotive force of ifd is effectively coupled with the three stator coils as the linking flux ϕlink (it is also called the main flux by engineers). However, a small amount of the induced flux from ifd may not be coupled with the stator coils. This is the leakage flux ϕleak of the excitation coil. Figure 10.5 shows the leakage flux in various parts of the generator. As shown in Figures 10.4 and 10.5, the flux starts from the S-pole of the rotor exciter coil, and most of it interlinks with the stator coils and then returns to the N-pole. However, a small amount of flux returns to the N-pole without reaching the stator coils. This is the leakage flux ϕleak. The leakage flux can be classified into three ϕair, ϕslot, ϕend by the location of the flux passes. The leakage flux that returns to the N-pole by way of the rotor to the air-gap (without reaching the stator surface) is the air-gap leakage flux ϕair. The leakage flux that reaches the surface of the stator (the slot-piece, wedges, etc.) but returns to the N-pole without reaching the armature coils is the slot-leakage flux ϕslot. The leakage flux that starts from the S-pole in the rotor-end sections of both sides and returns to the N-pole without coupling with the stator coils is the end-coil leakage flux ϕend. Regarding the inner sections of the generator, most of the flux produced by ifd reaches the armature coils, as shown in Figure 10.4; therefore, the leakage flux ϕair, ϕslot is quite small. Application of a silicon-steel plate with a high flux density and a smaller air-gap distance design will mitigate it effectively. While referring to Figures 11.4 and 11.5, we will examine the rotor end-coil sections of both sides. In these sections, most of the flux passes through the coil-end steel structures (such as parts of the core, yoke, coil-support, shield plate, cooling pipe, etc.) and incredibly large air gap, so that the flux cannot be coupled with the armature coils. In other words, the end-coil leakage ϕend is inevitably larger. Finally, the situation can be written as follows: ϕleak = ϕair + ϕslot + ϕend ϕair , ϕslot 0) requires strong excitation, while operation in the leading zone (q < 0) requires weak excitation. The special condition of Efd = 0 corresponds to the special point ⑤ (0, −1/xd). For a generator whose xd is smaller (short-circuit ratio 1/xd is larger), point ⑤ moves down to the lower point in the p–q coordinate plane, so larger excitation capacity is obviously required. Stability-Limit Curve (Curve ⑤−⑦)

11.5.3

The steady-state stability limit under the condition of δ = 90 is given by Eq. (12.18) in Chapter 12 and by the circle in Figure 12.4a. The circle can be transferred to Figure 11.11 as the curve of the steady-state stability limit ⑤−⑦. This curve also shares the same special point ⑤. It is interesting that this point is not only the center of the excitation-limit circle, but also one end of the diameter of the circle of steady-state stability. Limit Curve against Extraordinary Local Heating on the Stator Coil End (Curve ③−④)

11.5.4

A generator has the special problem of extraordinary local heating under operation in the leading power-factor (q < 0) zone (in other words, under weak field operation), which tends to be caused mainly around the structure of the stator coil end. Curve ③−④ in the figure indicates the critical curve for this problem, for the reason discussed in Section 11.9.2. This curve should be indicated by the generator manufacturer. Figure 11.12 shows a capability curve and voltage-versus-excitation-current characteristic curve of a typical largecapacity thermal generator, where the explanation of the latter curve (b) is shown in Figure 10.11. In this case, the capability characteristics are given by four curves depending on the pressure rates of hydrogen-cooling.

1.2

1000

1.1

0P

F

F 0P

F

5PF

0.9

400 200 0

200

400

600

800

1000

1200

LOAD (MW) Load (MW)

200 0.95P

F

400 F 0P

0.6

600

0.8

0.9

0PF

0P

0.8 F

5P

F

Terminal voltage [per unit]

F

0P

0.9

0.20 MPa.g

Lagging

1.0

5P

0.8

0.30 MPa.g

0.8

600

0.50 MPa.g 0.40 MPa.g

0.6

800

nNoo--lo loaaddcca hrarcatecrte isrtiisctiscs thhr T reee p-phha asese shsho ortrtc ciricrcu uititc chhaar aracct eterirs itsitci css

Q [MVA]

1200

Leading

274

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

800

0.0 P [MW]

(a) Capability characteristic curve (P-Q coordinate : MVA) curve

0

1000 2000 3000 4000 5000

ifd [A]

(courtesy of Toshiba)

(b) Voltage-excitation current characteristic curve

Figure 11.12 Characteristic curve of a large-capacity thermal turbine generator (1120 MVA).

11.6 Generator’s Locus in the pq-Coordinate Plane under Various Operating Conditions

11.6 Generator’s Locus in the pq-Coordinate Plane under Various Operating Conditions The generator’s specific capabilities were discussed with the parameter of generator terminal voltage eG in the previous section. Now let’s examine the generator’s loci with the parameter eB (usually eB = 1.0) instead of eG, under various operating conditions.

11.6.1

The Locus under Fixed Excitation Efd

Rewriting the left and right sides of Eq. (11.52)④, p2 + q + x1d

xd Efd = eB xl

p2

q − x1l

+

2

11 56

2

This is a second-order equation in p and q, and it can be modified into the equation of a circle as shown next. (See the Supplement for the calculation process.) For the circle, 2

b the circle equation p + q + 1 − a2 2

=

2

2

+c

Efd xl eB xd

where a = b=

b 1 − a2

Efd 1 a2 + = xd xl eB xd

2

xl +

1 xd

11 57

Efd 2 1 a2 1 − 2 − 2 2 x x xd eB xd c = l 2d = 2 1−a 1−a center 0, −

b , 1 −a2

2

b 1 − a2

radius

+c

The operational locus (p, q) of the generator is drawn as a circle under the conditions of eB = 1.0, fixed Efd, and constant xd, xl. This circle is shown as loci a1, a2 in Figure 11.13. Assuming generator operation under fixed Efd (i.e. AVR is out of service), if p is increased by the prime mover, the generator locus (p, q) moves down immediately in the direction of leading power-factor operation: that is, to the prohibited operating zone.

11.6.2

Locus under Fixed Terminal Voltage eG

The terminal voltage eG is given by Eq. (11.52) ①, which can be modified as the circle equation center

p2 + q −

1 xl

2

=

1 eB eG xl

2

1 1 eB 0, , radius xl eG xl

This is also drawn as a circle, which is shown as loci b1, b2, b3 in Figure 11.13.

11 58

275

276

11 Synchronous Generators, Part 2

Efd

eB = 1.0

eG xd xq

xl

The case eB = 1.0 xl = 0.33 xd = xq = 1.7 q 1.0

Locus of Efd fixed. Eq. (11.57)

P: fixed

a1: Efd = 1.4, center (0, –0.87), radius 1.06

c2

a1: Efd = 1.2, center (0, –0.79), radius 0.89

c1

P = 0.95

Locus of eG fixed. Eq. (11.58)

1.0 eG = 1.1

Efd : fixed a1 Efd = 1.4 a2 1.2

b1

1.0 p

0.5

1.0

eG : fixed

b1: eG = 1.1, center (0, + 3.0), radius 2.72 b2: eG = 1.0, center (0, + 3.0), radius 3.00 b3: eG = 0.9, center (0, + 3.0), radius 3.33

b2

Locus of P fixed. Eq. (11.60) c1: P = 1.0, center (4.5, + 3.0), radius 4.3

b3

c1: P = 0.95, center (4.7, + 3.0), radius 4.7

0.9

Locus of iG fixed. Eq. (11.61) d1: iG = 1.0, center (0, –0.375), radius 1.12

– 0.5

d2: iG = 0.9, center (0, –0.29), radius 0.99

iG = 0.8

d3 0.9

d3: iG = 0.8, center (0, –0.22), radius 0.86

d2

d1 1.0

iG : fixed Figure 11.13 Operational locus in the pq coordinate plane.

11.6.3

Locus under Fixed Effective Power P

From the real part of Eq. (11.52) ③, eB xl

P = e2G p =

2

p p2 + q − x1l

11 59

2

Then the circle equation

p−

with

e2B 2 Px2l

2

+ q−

1 xl

2

=

e2B 2 Px2l

2

11 60

e2B 1 e2B 2 , x , radius 2 Pxl l 2 Px2l

center

The circle is shown as loci c1, c2 in Figure 11.13. 11.6.4

Locus under Fixed Terminal Current iG

The equation for the fixed terminal current iG is derived from Eq. (11.52)②, which can be modified as the circle equation where A = center

p2 + q −

1 xl 1 − A2

2

eB xl iG

0,

1 xl 1 −A2

=

A xl 1 − A2

2

11 61 , radius

A xl 1 − A2

11.7 Leading Power-Factor (Under-Excitation Domain) Operation, and UEL Function by AVR

The circle is shown as loci d1, d2, d3 in Figure 11.13. The curve in which iG is replaced by iGmax is obviously the current limit curve. Finally, the vector diagram for the four typical operating points in the pq-coordinate plane are shown in Figure 11.13. The magnitudes of eB and i are drawn in a similar size for each diagram.

11.7 Leading Power-Factor (Under-Excitation Domain) Operation, and UEL Function by AVR 11.7.1

Generator as a Reactive Power Generator

The load of a power system {Pload + j(Ql − Qc)} (where P, Q consumed on the transmission lines are included) is incidental by nature and always changing with time. On the other hand, the simultaneity and equality of the demanding power and the supplying power are also essentials of the power system. Therefore, with regard to effective power, the total amount of generation Pgen is controlled to meet the capricious effective load Pload and to maintain the frequency within 50/60 ± α Hz (α is typically ±0.05 Hz, although it may be different for utilities) over time by means of generating power dispatching control and AFC based at the control center. Reactive and capacitive power also must be supplied and controlled over time to meet the load requirement j(Ql−Qc) or, in other words, to maintain the voltages of each local area within allowable levels (typically, 1.0 ± 0.05). However, reactive power control (VAR Q-control) must be used not only on a total system basis but also on a partial network basis because voltages must be controlled within the range of 1.0 ± 0.1 at any part of the grid. The generating sources of reactive power to meet reactive/capacitive load demand are as follows:

•• •

Reactors (for jQl) and capacitors (for jQc) installed at receiving substations. Generators (for jQl and jQc). Synchronous phase modifiers (rotary condensers, for jQl and jQc) installed at receiving substations. (Synchronous phase modifiers are uncommon today because they are expensive as compared to reactors and/or capacitors.)

Generators also play an important role as reactive/capacitive power generators and have a very large capacity for supplying reactive power (Var, MVar), but poor capacity for supplying capacitive power. Typically, in the daytime, each generator supplies reactive power and effective power to the load. At night, however, capacitive load is required, which can be partly provided in parallel by reactor banks installed at the receiving stations. Some generators are required to share the role to supply capacitive power with leading power-factor operation. Figure 11.14 shows four (p, q) operating points and the related vector diagrams. The infinite bus voltage eB is drawn with the same size and direction for each diagrams. In addition to scheduled leading power-factor operation, generators may be suddenly forced into leading power-factor operation as a result of the following conditions:

•• •• •

Sudden Sudden Sudden Sudden Sudden

tripping of another generator under leading power-factor operation. increase of excitation (increasing lagging reactive power) by another generator operating in parallel. tripping of reactors at receiving substations, or sudden increase in capacitive loads. increase of effective power caused by the prime mover. change of the AVR setting voltage (Vset) to a lower value (by mistake).

Now we need to study the reason why leading power-factor operation may cause severe conditions for generators. That is the reason for the under-excitation limit ③−③ in Figure 11.11. 11.7.2 Under-Excitation (Leading Power-Factor Operation) and the Problem of Overheating at the Stator Core End

Generators have a problem with abnormal overheating of the stator core end, caused during leading power-factor operation. Here we will study the limit curve ③−④ shown in the leading power-factor zone in Figure 11.11. Figures 11.15 and 11.16 show thermal and hydro-generators in which the structure of the coil-end section can be seen. A generator rotor is a rotating electromagnet with N- and S-poles. All the flux ϕ generated by excitation current if flows out from the N-pole and returns to the S-pole, as shown in Figure 11.15a, b. Most of the flux from the N-pole reaches the stator winding zone and returns to the S-pole after effectively interlinking twice with the armature coil. However, some

277

278

11 Synchronous Generators, Part 2

q

a b p

c d

Efd

eG (a) Lagging 80º

eB

𝜑

Efd

(c) Leading 10º

i

eG

i

eB Efd

(b) Lagging 10º

(d) Leading 80º eG eB

Efd

i

i eG

eB

Figure 11.14 Vector diagrams for various pq operation modes.

flux does not reach the stator winding zone and returns to the S-pole without interlinking with the armature coil. That is, the total flux produced by the rotor magnet can be categorized as follows: ϕtotal = ϕeff linking flux + ϕleak leakage flux ≒ ϕeff + ϕend where ψ total flux linkage = ϕeff N N coil turns ϕleak = ϕair + ϕslot + ϕend ≒ ϕend ϕend

• • •

ϕair,

11 62

ϕslot

ϕair (air-gap leakage flux): Flux starting from the N-pole and flowing down through the air gap to the S-pole without interlinking with the armature coil. ϕslot (slot leakage flux): Flux starting from the N-pole and flowing down around the stator surface to the S-pole without interlinking with the armature coil. ϕend (end-coil leakage flux): Flux starting from the N-pole around the coil end of the rotor cylinder and flowing down to the S-pole without interlinking with the armature coil.

The structure of a generator-rotor and stator-flux pass is made up of silicon steel-laminated plates with outstanding permeability and small hysteresis loss. Further, the air-gap distance between the stator and rotor is designed to be as small as possible (say, 100 mm for a 1000 MVA class), so magnetic resistance is small. In addition, the stator and rotor coil structures are arranged to minimize their leakage flux. Then, imagining round slices of a generator’s inner cylinder along the axial length, most of the flux generated by the rotor coil interlinks effectively with the armature coil (see Figures 11.15a and 10.4.) In other words, the effective flux linkage ψ total = ϕeff N is relatively large (where N is the number of turns of the armature coil) and leakage flux ϕair and ϕslot seldom exist.

11.7 Leading Power-Factor (Under-Excitation Domain) Operation, and UEL Function by AVR

Stator-frame Bearing-bracket Setting-plate Shield-plate

(c) Flux pass of inner round slices

Stator-core N Stator

coil-en

d

Rotor

Bearing

Rotor-windings (a) Structure of stator/rotor-end S

Casing Cooling-system

Laminated silicon steel plate armature Armature-coil Rotor Rotor-shaft

N-pole S-pole

𝜙eff

𝜙air 𝜙slot

𝜙end

To exciter

To turbine Bearing

Rotor coil (b) Concept of flux pass Figure 11.15 Flux passes of a thermal generator.

However, at the coil end, for a round slice of the cylinder end, most of the flux flowing out from the N-pole turns down to the S-pole as ϕend without interlinking with the stator coil, through various routes in the stator end’s metal structure (such as parts of the core, yoke, coil support, shield plate, cooling pipe, etc.). Accordingly, eddy currents appear in the highly resistive metal structure and tend to raise the temperature. This temperature increase is not too serious under operation at a power factor of 1.0, but it is extremely serious under operation at the leading power factor. The reason is explained next. Figure 11.17 shows vector diagrams for operation under three different power factors: (a) for a lagging power factor, (b) for a power factor of 1.0, and (c) for a leading power factor. The vector magnitudes of terminal voltage eG and current i are drawn equally in the three cases. If we change the operating power-factor angle δ slowly from 90 (lagging) 0 (cos δ = 1.0) −90 (leading) under the condition of fixed values of eG and i, the excitation voltage jEfd becomes smaller, as already discussed. Next, the effective linkage flux ϕeff is 90 leading from eG and obviously proportional to Efd. Accordingly, ϕeff is greatest in case (a).

279

280

11 Synchronous Generators, Part 2

Courtesy of Toshiba

Cooling method : Stator: water cooling Rotor: hydrogen gas cooling Stator: inside diameter Ω = 1432 mm Rotor: outside diameter Ω = 1432 mm Air-gap length = 138 mm Slot deep length = 184 mm Core length = 7700 mm, 79 ton

Rotor (hydrogen cooled) Direct water-colled coil Collector room Stator core Stator windings

Wedge Creepage block Inlet region Slot armour

Shaft

Bearing Terminal box Bearing bracket

Stator frame

Sealing ring

Figure 11.16 Thermal generator.

High-voltage bushing terminal (hydrogen gas cooled)

Rotation Flush inlet Scoop Turn insulation Outlet region Extruded copper channel

11.7 Leading Power-Factor (Under-Excitation Domain) Operation, and UEL Function by AVR

ϕeff : total linking flux ϕend: total end-coil leakage flux ϕtotal = ϕeff + (ϕleak)

ϕeff + (ϕend)

where (ϕleak ) = (ϕair + ϕslot) + ϕend) ϕeff + ϕend

(ϕend)

ϕair ϕslot

0

ϕeff ϕeff + ϕend

ϕeff

jEfd

jEfd

ϕend i

eleak

eG

ϕend

(a) Lagging power-factor

i

eleak = jxl i eG

(b) Power-factor 1.0

Leakage voltage (inner voltage drop)

ϕeff + ϕend

ϕeff i

eleak = jxleak i jEfd eG

ϕend (c) Leading power-factor (low excitation) Figure 11.17 Flux vector diagrams around the coil-end structure.

The counter-electromotive force eleak = (jxleak∙i) has the same value in each case because the stator current i has the same magnitude. The leakage flux ϕend is produced by eleak and is 90 leading from eleak. However, ϕend does not have the same magnitude for each case, even though the magnitudes of eleak are the same. In case (a) (lagging power-factor operation), ϕeff is large, and, in addition, the composed flux (ϕeff + ϕend) becomes larger. However, the flux density of ϕeff is so high that superposed flux ϕend cannot have a large magnitude due to magnetic saturation of the flux pass. The situation in case (b) (power factor of 1.0) is similar to case (a). On the other hand, in case (c) (leading power-factor operation), ϕeff is very small (because Efd is small), and, in addition, the composed flux (ϕeff + ϕend) becomes smaller. That is, the flux density of ϕeff is very low and becomes weaker with the addition of ϕend. Accordingly, superposed ϕend would have a very large magnitude, because magnetic saturation would not occur. Now we will focus on the leakage flux ϕend of a round slice of the coil end. Even though the magnitude of eleak is the same in all three cases, the flux produced at the round slice of the coil end ϕend becomes extremely large only in case (c). Therefore, a relatively large current is forced to flow through the metal structure of the cylinder end, and large eddycurrent loss and magnetic loss occur around the coil-end structure and cause serious local heating of the structure. If a generator’s operating mode is suddenly changed from the lagging power-factor zone to the leading power-factor zone, leakage flux around the coil end is suddenly increased: the temperature will increase rapidly in a very short time and cause decay of mechanical strength due to annealing, melting, and/or damage to insulation. It should be stressed that wedges (for example) must have the strength to withstand the huge centrifugal force of approximately Mach 1 peripheral speed occurring on the rotor coils. This is the reason for the generator’s leading power-factor operational limit (low-excitation problem). The capability curve for the low-excitation limit ③−④ in Figure 11.11 is to protect generators from such an effect. Again, consider a general view of Figure 11.11. With regard to operation in the lagging power-factor zone, a generator can withstand such operation for some time even if the excitation limit curve ①−② is suddenly exceeded: (i) the exciter can generate current exceeding its rating, and (ii) there is no stability problem. Therefore, exceeding the curve ①−② is not necessarily a major problem for the excitation limit or the generator’s lagging-mode operation.

281

282

11 Synchronous Generators, Part 2

On the other hand, during operation in the leading power-factor zone, time becomes very important if operation exceeds the coil end overheat limit curve ③−④ and the stability limit curve ⑤−⑦. Such operation should be prevented or avoided within a few or tens of seconds. 11.7.3

Under-Excitation Limit Protection by AVR

There are several reasons a generator may be forced to rush into the low-excitation zone and exceed the limit curves ③−④ and ⑤−⑦ without countermeasures being taken within a few seconds, because the time constants Tavr + Tf + T are less than a few seconds. The countermeasure to prevent this from happening is the under-excitation limit (UEL) function of AVR. This function is supplied by AVR as one of its essential functions. As shown in Figure 11.11, the UEL-setting zone is usually within some reserved margin of a straight line from (0, −q0) to the upper right. (The setting characteristics can be written as the equation q = ap − q0, where a, q0 are positive PU values.) AVR changes excitation freely within the setting area of Figure 11.11 in order to maintain the terminal voltage at the set value Vset. However, if the operating point in the p–q-domain reaches the UEL line, AVR will no longer weaken the excitation, and the voltage may consequently differ from the value of Vset. The UEL function of AVR is an essential protection for a generator. We will discuss loss of excitation protection further in Chapter 14, Section 14.6.

11.8

Operation at Over-Excitation (Lagging Power-Factor Operation)

In Figure 11.11, the distance between an arbitrary point (p, q) and point ⑤ is proportional to Efd, as already explained. Then, the operational limit curve ①−② means the rated capacity of the excitation circuit (generator rotor and exciter current). Although the generator should not be operated in the exceeded zone of the limit curve, the exceeded zone may have time redundancy because the excitation circuit has thermal redundancy. Then, engineering practices to prevent the generator from exceeding the over-excitation limit curve ①−② are rather simple, and may just involve time-delayed overcurrent relays for the armature winding current iG and/or field current ifd and alarming instead of automatic tripping.

11.9 Thermal Generators’ Weak Points (Negative-Sequence Current, Higher Harmonic Current, Shaft-Torsional Distortion) In addition to the capability curves, we need to examine some weak points of thermal generators that are common in synchronous motors/condensers as well as in large induction motors. As the starting premise, we need to recognize some basic features of modern advanced generators. 11.9.1

Generator Volume Size and Unit Capacity

The unit capacity MVA of a generator is determined by the total magnitude of the effective flux linkage, whose practical relation with the generator volume can be explained by the following equation: Rated MVA ∞ ψ a + ψ b + ψ c total flux linkage ∞ C Nturn 2

Nturn = stator coil volume ∞ π Dout 2 −π Din 2

2

B 2

L ∞D L

RPM ①

B flux density at rotor surface or air gap

where

C current density of armature Nturn number of turns in armature coil 11 63

Dout , Din outer and inner diameters of stator L effective axial length of stator and rotor 2

D L generator volume structure RPM rotating speeding per minute ∴ Rated MVA = B flux density

C current density

D2 L generator volume

RPM

②

11.9 Thermal Generators’ Weak Points (Negative-Sequence Current, Higher Harmonic Current, Shaft-Torsional Distortion)

The unit capacity of the largest thermal generators today is 1000–1300 MVA, while that of generators in 1950 was only 50–100 MVA. The capacity of the largest generators has increased 20 times in half a century; however, they have a relatively smaller physical size. In other words, a dramatic capacity enlargement has been realized not only by enlarging the generator’s physical size (D and L), but also by enlarging the flux density (B) and current density (C). Such enlargements have been based on advanced technology over the last half-century, which makes the following quantitative explanation possible. For C (stator current density):

•• •• •

Direct stator coil cooling by water or hydrogen gas coolant flowing through the hollow piped coil High-temperature-withstanding stator coil insulation For B (flux-linkage density): Direct rotor coil cooling by hydrogen gas coolant flowing through the hollow pipe coil Advanced silicon–steel plate with high flux density and heat-withstanding capability High-temperature-withstanding rotor-winding insulation

A rotor in a thermal generator is a horizontal cylinder that rotates at 50/60 turns per second (the peripheral rotating velocity is close to the velocity of sound, Mach 1). Furthermore, the rotor requires a very large dynamic metal strength and balancing accuracy in order to withstand mechanical effects such as bending caused by its weight, centrifugal forces, vibrations, eccentricity, mechanical resonance, heat expansion, torque, electromagnetic motive force, etc. Naturally, there are manufacturing upper limits on the sizes of D and L, especially in forging technology for large-scale metal bodies with sufficient homogeneity and strength.

11.9.2

Critical I2-Withstanding Capability

Whenever unbalanced faults occur in the network, the negative-sequence current I2 is forced to flow into a generator from outside the network. I2 immediately induces large eddy currents in the rotor surface metal and consequently causes a rapid, serious temperature increase around it. These phenomena can be explained by the d- and q-axis equivalent circuits in Figure 10.6. The currents id, iq of the d- and q-axis circuit are DC components for positive-sequence current I1 (under normal operating conditions), yet they become second-harmonic (2ωt) currents for negative-sequence current I2, because the rotating direction of the negative-sequence current is the opposite of the rotation of the rotor (see Eq. (10.93)). Now, in the d- and q-axis circuits in Figure 10.6, the second-harmonic currents are suddenly forced to flow into the circuits. In the d-axis circuit, recall that Ldi/dt + ri is dominated by ri for DC and by Ldi/dt for second-order harmonics. The second-harmonic current id seldom flows into the Lad branch or field branch and mostly is forced to flow into the damper branch. Accordingly, the damper branch quickly becomes overheated due to the Joule heat losses at high resistances, rkd. Incidentally, in the case of a generator with a cylindrical rotor, the damper circuit branch in the equivalent generator circuit corresponds to the rotor surface metal materials, particularly the metal wedges. The metal wedge pieces are quickly softened by the eddy current due to the annealing effect. In other words, the negative-sequence fault current I2 flowing into the stator winding causes extreme eddy currents on the rotor surface so that the wedge pieces as well as the laminated coil slot metal experiences a large increase in temperature. An advanced winding cooling system is only effective for conductor cooling and is useless against such a temperature increase on the rotor surface. The weakening of the mechanical strength of the wedges by annealing is the most serious effect, because they have the important role of withstanding centrifugal force on the rotor windings in the slots. This is why the ability of thermal generators to withstand negative-sequence current I2 is generally poor. Table 11.4 shows the Japanese Electrical Standards (JEC) for a generator’s withstanding capacity against negativesequence current indicated by typical standards. These values are decided primarily based on the critical metal strength of the rotor wedges, whose mechanical strength is rapidly weakened around 200 C due to metal annealing. The negative-sequence current I2 forced to flow into the generator is classified as follows: a) Continuous negative-sequence current caused by three-phase imbalance of the network and/or loads b) Intermittent negative-sequence current caused by special loads (rail traffic loads, electric furnace, etc.)

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Table 11.4 Withstanding limit of I2 (negative-sequence current) (JEC 2130 [2000]).

Salient-pole machine

Maximum limit of I2/Irate for continuous I2 current

Maximum limit of (I2/Irate)2 t for I2 fault current

0.1

20

1 Direct cooling Generator

0.08

20

Synchronous Phase modifier

0.1

20

0.08

15

2 Indirect cooling

Air cooling type

0.1

15

Hydrogen-gascooling type

0.1

10

0.08

8

Cylindrical machine

2 Direct cooling rotor

Motor Generator

0.05

Synchronous

15

0.08

15

Phase modifier

Note : ∗1

Maximum limit of (I2/Irate)2 t for fault current

1 Indirect cooling rotor

Motor

Time t: sec.

Maximum limit of I2/Irate for continuous I2 current

≦350 MVA ≦900 MVA

∗

( 1) ∗

(∗2)

≦1250 MVA

( 1)

5

≦1600 MVA

0.05

5

I2 SN −350 = 0 08− Irate 3 × 104

∗2

I2 Irate

2

t = 8 −0 00545 Srate −350

SN rated apparent power MVA

c) Unbalanced fault relay-tripping d) Breakers that fail to trip (open-mode failure), or breaker single-phase reclosing With (a) or (b), the generator’s life time is typically shortened due to insulation deterioration or wedge metal fatigue. With (c) or (d), I2/Irate must be large, and a generator may be in serious danger. As a check, for a 1000 MVA class cylindrical generator, I2 Irate

2

t≦5

from Table 11 4

Then for for for for

I2 I2 I2 I2

Irate = 0 1 the allowable time is t = 500sec Irate = 0 5 the allowable time is t = 20 sec Irate = 1 the allowable time is t = 5 sec Irate = 5 the allowable time is t = 0 2 sec 10− 12 cycles

The allowable duration for a large fault current is obviously small. Although this critical condition may have some redundancy, repetition of such a fault may cause serious metal fatigue and may badly damage the generator in the worst case. Now we can recognize that high-speed fault tripping by protective relays + breakers (typically 2 + 2 cycles time) is vitally important not only for maintaining stability or overcurrent protection of transmission lines and equipment, but also for protecting generators and motors against negative-sequence current. Further, high speed reclosing is often adopted as a useful practice: the non-eternal fault (arc-horn short-circuit fault) tripped line is put into service immediately after the trip. However, such reclosing, particularly in the area of generating plants’ close neighborhood lines, must be carefully investigated from the viewpoint of the individual generator’s ability to withstand I2. In particular, single-phase breaker reclosing on a transmission line near a thermal generating plant must be avoided. In the case of zero-sequence current I0 flowing into a generator as described here, I0 caused by a high-voltage line will not flow into the generator because I0 cannot flow through the delta winding of the generator-connected main transformer. In the case of a grounding fault between the generator and the step-up transformer, I0 will flow into the

11.9 Thermal Generators’ Weak Points (Negative-Sequence Current, Higher Harmonic Current, Shaft-Torsional Distortion)

generator. However, the magnitude of I0 is restricted to a small value (say, 200 A) because generators are mostly neutralresistive grounded so serious situations will not occur. Hydro-generators have relatively large ability to withstand negative-sequence current, due to the large size of the rotor, natural cooling, and smaller centrifugal force.

11.9.3

Rotor Overheating Caused by DC and Higher Harmonic Currents

We will now study the phenomena that arise whenever DC and higher harmonic currents flow into a generator. 11.9.3.1 nth-Order Harmonic Current Flowing into a Phase-a Coil of a Generator

This case can be described by the following equation, where the real-number symbolic method is applied (n = 0 means DC): ia = Icos nωt ib = ic = 0

11 64a

In the 012-domain, i0 = i1 = i2 =

1 1 Ia = Icos nωt 3 3

11 64b

Substituting Eq. (11.64a) into Eq. (10.10b), 2 1 id = cos ωt Icos nωt = + I cos n + 1 ωt+ cos n− 1 ωt 3 3 2 1 iq = − sin ωt Icos nωt = − I sin n + 1 ωt− sin n− 1 ωt 3 3 1 i0 = Icos nωt 3 ∴ or

11 64c

1 1 2 id + jiq = I e −j n + 1 ωt + e j n −1 ωt = I e −jnωt + ejnωt e − jωt = Icos nωt e −jωt 3 3 3 2 id + jiq e jωt = Icos nωt 3

Eq. (11.64b) indicates that i2 and i0 appear regardless of the value of n for the phase current. Accordingly, eddy currents are induced in the cores and wedges of the rotor surface and cause local overheating for the same reason explained in the previous section. Furthermore, Eq. (11.64c) indicates that id iq cannot be DC regardless of the value of n. In other words, harmonic current is forced to flow into the damper branch of the d- and q-axis equivalent circuits in Figure 10.6, and cause the rotor to overheat. Note that the special case of this equation with n = 1 corresponds to Eqs. (10.93b, c). Thus, current with higher harmonic distortion cause the generator rotor surface’s temperature to increase due to eddy currents. A similar situation is caused in the auxiliary driving motors of the station and any motor loads. 11.9.3.2 DC Current Flow

This is the case of n = 0 in the previous equations, with i2. The transient DC components included in the short-circuit fault usually have a duration within 1–3 cycles or 15–60 ms (see Ta in Section 10.10.4 and Eq. (10.131)), so it is not a serious matter for the generator, although it is a concern from the viewpoint of breaker zero-miss phenomena and the protective relay high-speed tripping capability. Transformer inrush current is inevitably caused when a transformer is energized by the associated breaker. The current is extremely unbalanced, including DC components as well as higher harmonics, and it lasts a relatively long time because of the transformer circuit’s long time-constant. The magnitude of the inrush current can exceed five times the

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rated current, and it may continue for a few cycles or even 1 to 10 seconds (see Figure 11.18), so special attention is required to prevent trouble in rotating machines, breakers, and protective relays.

Phase currents ia ib

11.9.3.3 Three-Phase nth-Order Current Flow

ic Excitation voltage

15 sec

Figure 11.18 Magnetizing inrush current caused by transformer charging.

In recent years, various types of load equipment and electric power conditioners are widely prevalent. These pieces of equipment have common characteristics for generating phase-balanced or -unbalanced higher harmonic currents. AC/DC converters and power-electronic motor-driving systems and power conditioners of small generating units (wind-generation, solar generation, etc.) are typical examples. So, balanced three-phase harmonic current should be carefully examined from the viewpoint of the withstanding capability of rotating machines.

This case can be written as the following equation: ia = Icos n ωt ib = Icos n ωt − 120

11 65a

ic = Icosn ωt + 120 ia = Iejnωt ib = Ie jn ωt − 120 = a2n Iejnωt

11 65b

ic = Ie jn ωt + 120 = a n Iejnωt a2 = e −j120

a = e j120

Then, the symmetrical components are 3i0 = 1 + a2n + a n Iejnωt 3i1 = 1 + a2n + 1 + a n + 2 Iejnωt

11 65c

3i2 = 1 + a2n + 2 + a n + 1 Iejnωt id and iq are derived by substituting Eq. (11.65a) into Eq. (10.10b): id = 2 3 I cos ωt cos n ωt + cos ωt − 120

cos n ωt −120 + cos ωt + 120

cosn ωt + 120

= 1 3 cos n + 1 ωt + cos n + 1 ωt −120 + cos n + 1 ωt + 120 + cos n −1 ωt + cos n− 1 ωt − 120 + cos n −1 ωt + 120 iq = − 2 3 I sin ωt cos n ωt + sin ωt −120

cos n ωt − 120 + sin ωt + 120

cosn ωt + 120

11 66

= − 1 3 I sin n + 1 ωt + sin n + 1 ωt − 120 + sin n + 1 ωt + 120 + sin n − 1 ωt + sin n− 1 ωt − 120 + sin n −1 ωt + 120 Then, when n = 3 m: DC, third-, sixth-, and ninth-order harmonic quantities: i1 = 0 i2 = 0 i0 = Icos nωt

id = 0 iq = 0

id + jiq = 0

11 67a

11.10 Transient Torsional Twisting Torque of a TG Coupled Shaft

When n = 3 m + 1: first-, fourth-, and seventh-order balanced harmonic quantities: i1 = Icos nωt i2 = 0 i0 = 0

id = Icos n + 1 ωt iq = − Isin n + 1 ωt

id + jiq = Iej n−1 ωt

11 67b

When n = 3 m + 2: second-, fifth-, and eighth-order balanced harmonic quantities: i1 = 0 i2 = Icos nωt i0 = 0

id = Icos n + 1 ωt iq = −Isin n + 1 ωt

id + jiq = Ie j n + 1 ωt

11 67c

When n = 0 (DC), third-, sixth-, and ninth-order harmonic quantities, only zero-sequence quantities of the nth order exist, so i1, i2, and id, iq are zero. When n = first-, fourth-, and seventh-order balanced harmonic quantities, only positivesequence quantities of the nth order exist, while id and iq of (n−1)th order harmonics appear on the dq-axes circuit. In the case of n = 2nd, fifth- and eighth-order balanced harmonic quantities, only negative-sequence quantities of the nth order exist, while id and iq of (n + 1)th order harmonics appear on the dq-axes circuit. The cases n = second, fifth, eighth, and eleventh are of nth-order negative-sequence currents and id + j iq is of (n + 1)th order AC. Recall that id, iq are DC components under normal conditions in the d- and q-axis equivalent circuit in Figure 10.6, so the rotor surface is protected from abnormal overheating. On the other hand, the rotor surface cannot avoid serious overheating if id, iq include AC components. Accordingly, higher harmonic currents cause abnormal conditions on the generator regardless of whether the current has positive-sequence components (n = 3 m + 1) or negative-sequence components (n = 3 m + 2).

11.10

Transient Torsional Twisting Torque of a TG Coupled Shaft

11.10.1

Transient Torsional Torque Caused by a Sudden Network Disturbance

A generator and a turbine are mechanically coupled as a rigidly connected TG unit and are stationary operated at speed ωm under the balanced condition of mechanical input Tm and electrical output Te. We note the relation Pm = Tm ωm from Eq. (11.17). If a sudden disturbance occurs on either the connected network side or the prime-mover side, the generator or the turbine is immediately accelerated or decelerated by the torque ΔT = Tm − Te, so that transient torsional torque stress is caused on the rigidly coupled TG shaft train. Typical examples are sudden short-circuit faults and/ or tripping of the on-load operating generator. Immediately after the electrical output is lost (Te = 0), all the mechanical input power from the prime mover becomes excess power (ΔT = Tm), and the turbine is suddenly accelerated by the torque ΔT on both the shafts and the coupling device. This is the TG torsional shaft-twisting phenomenon. Repetition of such phenomena may cause mechanical fatigue. It may also cause mechanical vibration or pulsation. Typical sudden disturbances causing shaft twisting are:

•• •• ••

Sudden tripping of the generator Sudden tripping of another generator operating in parallel Short-circuit fault at a relatively close point in the network Large-capacity load tripping Reclosing after fault tripping Emergency shutdown of the boiler–turbine (BT) system.

Figure 11.19 is a diagram of a typical thermal TG unit that consists of three turbine shafts and one generator shaft (system with four mechanical points), mutually and rigidly coupled as one rotating unit. Thermal TG units are generally very long and slender cylindrical structures with high-speed rotation (a 1300 MVA class tandem compound unit is approximately 100 m in length, and the diameter of the shaft is only 1–1.5 m), so a mechanical system cannot tolerate shaft-twisting phenomena. Hydro TG systems are tougher.

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11 Synchronous Generators, Part 2

Steam Turbine H H

M L

L

Generator

L

Reheater

Generator

Mt

Mg k Mt, Mg: inertia constant k: spring constants of coupling

H: high pressure turbine M: middle pressure turbine L: low pressure turbine (a) T-G unit

(b) Two material points system

Figure 11.19 Concept of TG unit.

• • •

The transient torsional torque modes can be classified as follows: DC offset-mode torsional torque (positive-sequence mode) ΔT1: Balanced three-phase disturbances (three-phase short-circuit fault, three-phase reclosing, load tripping, sudden shutdown of the prime mover, etc.) cause sudden stepping-mode changes ΔPe or ΔPm; the resulting shaft torque must also be of stepping mode (ΔT1). Double-harmonic mode torsional torque (negative-sequence mode) ΔT2: Phase-unbalanced disturbances (phaseunbalanced short-circuit fault, single-phase reclosing, etc.) cause negative-sequence current. The resulting shaft twisting must be of double-harmonic frequency pulsation mode (ΔT2). Fundamental-frequency mode torsional torque (transient DC mode) ΔT3: Whenever faults occur, unbalanced DC offset currents appear in each phase as superposed transient components of short-circuit currents (the duration is a few cycles, say 0.1 second), and the resulting torsional torque must be of fundamental-frequency pulsation mode (ΔT3). All of these torques cause severe stresses in the TG shafts, especially in the coupling mechanism.

11.10.2

Amplification of Torsional Torque

We will examine transient torsional torque phenomena in the case of a three-phase fault and the reclosing and final tripping as shown in Figure 11.20. We assume that the TG unit is a simplified system with two material points (one

Circuit #2 3𝜙S

Circuit #1

Te Dead voltage time

1.0 0 𝛥T = Tm – Te (pu) 5 4 3 2 1 0 –1 –2 –3 –4

1

t0

2

t1

3

t2

t3

3𝜙S Tripping Final tripping Reclosing fault

1

arcing time (breaker + relay time; 2–6 cycles) 2 dead-voltage time (tentative value) ~ 200 kV : 0.2–0.4 sec 250–500 kV : 0.5–1 sec 3

Figure 11.20 Transient torsional torque of TG coupled shaft under reclosing process.

11.10 Transient Torsional Twisting Torque of a TG Coupled Shaft

point for the turbine and one point for the generator), as shown in Figure 11.19b, and the mechanical input of the prime mover is not changed for the duration of the process. Time interval t0−t1: As shown in Figure 11.20, a three-phase short-circuit fault occurs at time t0 at a point very close to the generator terminal. The generator output power Pe is lost, and the corresponding electrical torque Te becomes almost zero (Te ≒ 1 0) for the duration t0−t1, because the terminal three-phase voltages are almost zero, so the continued transmission of power through the second circuit line is very small. Accordingly, accelerating torque of the stepping mode (ΔT = 0 1.0) occurs on the generator at time t0, in that torsional torque oscillation is caused by the mechanical natural oscillating frequency ω0 of the TG system, and continues oscillation with very slow damping by an amplitude of ±1.0 (between +2 − 0). Time interval t1−t2: The three-phase fault on circuit #1 is cleared at t1 (perhaps 50–100 ms after occurrence of the fault), and the power transmission Pe is recovered through the second circuit line for interval t1−t2. Te begins to recover (Te = 0 1) at time t1. The instantaneous value of the oscillating ΔT at time t1 is a value between +2 and 0 and is ΔT = ±2 randomly. Then ΔT begins to oscillate with an amplitude of ±2.0 (between ΔT = +2 and −2) for the new interval t1−t2. Time interval t2−t3: Three-phase reclosing of the circuit #1 occurs at time t2 (perhaps 0.5 second after t1). The sending power is lost again (Pe ≒ 1 0) because the metallic fault is still on circuit #1. If ΔT = −2 at t2 randomly, ΔT begins to oscillate between ΔT = +4 and − 2 with an amplitude of ±3 for the interval t2−t3. Time interval t3−t4: The circuit #1 line is finally tripped at time t3. If ΔT = +4 randomly at t3, ΔT begins to oscillate between ΔT = +4 and − 4 with an amplitude of ±4 for the interval after t3. In this example, four large disturbances (fault, fault tripping, reclosing, final tripping) occurred in succession. As a result, the amplitude of the shaft mechanical torsional torque oscillation could randomly become extremely large. Furthermore, in the intervals t0−t1 and t2−t3, the shaft-torsional torque of the fundamental-frequency pulsation mode (ΔT3) should be superposed, because unbalanced DC offset currents appear in each phase. In the case of phase-unbalanced faults, the shaft-torsional torque of the double-frequency pulsation mode (ΔT2) is caused by negative-sequence current. The damping of these mechanical oscillations is very slow (the time constants of the mechanical oscillation are long) in comparison with electrical phenomena, so rapid oscillation reduction cannot be expected. Figure 11.21 shows a simulation chart of a transient torsional oscillation in a four-mass model for fault tripping and reclosing. Mechanical resonance of the rotating TG unit is another matter to be considered. Any hardware has its own specific natural resonant frequency. Figure 11.22 shows example “danger curves” for generators. STs have similar curves whose first-order resonant zone is generally below the rated rotating speed and whose secondorder resonant zone is over the rated speed. Whenever the turbine is started up from zero speed to the rated 3000/3600 rpm, a quick pass through the first-order resonant area is required. Overspeed rotation should be strictly avoided regardless of the existing area of second-order resonance. Special attention must be paid to loads in a typical electrical furnace, because it has an intermittent repeated rushing g pin load and a large unbalanced load with negative-sequence ng g i p n t s i i tr ul clo ipp Fa components, as well as rippled harmonic components. alRe Tr Fin Generators and motors installed close to such loads require ΔTe special attention. ΔT1

rpm 4000 3600

ΔT2

3000

ΔT3

2000

LP-B Gen

ΔT1

Generator

LP-A ΔT2

2nd order

IP HP ΔT3

Turbines

Figure 11.21 Transient torsional torque caused on shaft coupling (simulation).

1000

1st order 100

500

1000

MVA

Figure 11.22 Speed–danger curve by rotational resonance.

289

290

11 Synchronous Generators, Part 2

es

Protective gap

jx

i

er

11.10.3

Subsynchronous Resonance

Subsynchronous resonance (SSR) is a unique phenomenon that is caused on a power system when hydro/thermal generator units are connected to loads through a series capacitor-compensated long transmission line. Series resonance with the frequency below is caused on this circuit:

– jxc

Figure 11.23 Transmission line with series capacitance.

f=

1 2π Ltotal CSC

where Ltotal = Lgen + Ltr + Lline

11 68

A power system where large hydro stations are connected through a 500–1000 km long series capacitor-compensated transmission line is a typical example. In such a circuit, a subsynchronous resonant current component with a lower frequency (say, 20–40 Hz) may appear continuously and flow into the connected generators over time, because the system is an LC series-connected circuit (see Figure 11.23). The subresonant frequency is very close to the natural mechanical frequency of the TG shaft. Therefore continuous tortional stress appears on the TG shaft, and the shaft is slowly damaged by mechanical fatigue. We will discuss series capacitors further in Chapter 21, Section 21.11.2.

11.11

General Description of Modern Thermal/Nuclear TG Units

Most electrical engineers are required to know something about modern TG units in thermal or nuclear generating stations in the general sense described in this section, although a detailed description is beyond this book.

11.11.1

ST Unit for Thermal Generation

The elements hydrogen (H) and carbon (C) can be burned to produce heat energy. The process is expressed by the following chemical equations: Hydrogen

1 H2 + O2 = H2 O + 14 2 × 107 J kg or 142 000 kW sec kg 2

Carbon

C + O2 = CO2 + 3 4 × 107 J kg or 34 000 kW sec kg

Sulfur

S + O2 = SO2 + 0 92 × 107 J kg or 9200 kW sec kg

This is why oil and coal, composed primarily of C and H, can be utilized as thermal energy sources (fuels). Typical examples of their composition are as follows: Oil

C 86

, H 12

Coal

C 62

, H5

,S 2 , S2

, N1

Incidentally, the minor elements sulfur (S) and nitrogen (N) are also burned together in steam boilers, so SOx, NOx, and CO2 are inevitably produced as undesirable by-products. However, with the recent advanced technology of lowNOx combustion, and NOx neutralization with ammonia into N2 and H2O, NOx emissions have been reduced to low levels. Practical SOx reduction is also being widely achieved. Figures 11.24a and b show the general layout of a TG arrangement for a 1000 MW class ST and generator. As shown in the figure, the TG unit based on oil or coal combustion typically consists of three sections of turbines (HP-T/IP-T/LP-T, with different steam pressure and temperature ratings) and a generator with two poles that have 3000/3600 rpm ratings, whose rotary shafts are rigidly coupled like a single train shaft (tandem type). Superheated steam (typically 550–600 C) is fed from the oil- or coal-fueled steam boiler to the TG area and flows through the HP-T boiler reheater IP-T LP-T main condenser. Exhaust steam at the outlet of the LP-T is fed into the main condenser and then cooled by water (from a river water reservoir with a cooling tower or from the sea, at 0–32 C). In the condenser, most of the exhaust steam is condensed or liquidized, resulting in low pressure (almost vacuum gauge pressure). The steam flow is then naturally extracted through the LP-T outlet to the main condenser. Typical examples of the dry-steam pressures and temperatures at each stage are shown in Figure 11.24b.

11.11 General Description of Modern Thermal/Nuclear TG Units

(a) TG-yard Total length 54 m T:1,540 ton G:560 ton

Generator LP-T (low-pressure turbine) LP-T rotor Shaft diameter 1727mm Diameter (the last stage) 3760 mm Peripheral speed 708 m/s

LP-T

IP-T (intermediatepressure turbine)

HP-T (highpressure turbine)

Courtesy of Toshiba

HP-T rotor (double-flow type) 240 atm

8.7 atm

24.1 MPa, 566ºC

0.88 MPa, 354ºC

Generator 1000 MW

4.1 atm Reheater HP=310MW

4.15MPa, 593ºC IP=306MW

722mmHg, 32ºC LP=206.5 2=413MW Total = 1029 MW

(b) Steam-pressure/temperature flow diagram

External Loss Exhaust Loss = 19MW Gen. Loss = 8MW Mechanical Loss = 2MW Loss Total = 29MW

Generator 1000 MW Noste : 1 atm = 1.013×105 Pa = 760 mm Hg Figure 11.24 Large-capacity TG unit (1000 MW).

291

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11 Synchronous Generators, Part 2

The thermal efficiency of the total BTG system with an ST unit is derived using the Rankine-cycle model in thermodynamics and is limited by the temperature difference Δt = t1 − t2 between the superheated steam temperature t1 and the cooling water temperature t2 in the main condenser (0–32 C). Because t2 is technically almost uncontrollable, temperature t1 is the primary factor in determining the thermal efficiency η. The total efficiency η of a typical modern generating unit with t1 = 550–600 C is within the range η = 38 – 43%, which is the theoretical upper limit with a steam temperature of 600 C or less. Now, let’s quickly run through today’s advanced thermal ST technology, which is the result of 100 years of significant technical advances. STs were used first in the early 1900s as a new prime-mover technology, replacing the reciprocal steam engines designed by Watt; however, the unit capacity was only 500 kW or less. In the 1950s, the most advanced unit was probably 75 MW with 420 or 480 C and 6 Mpa(60 atm. Then, in 1960, supercritical) (SC) turbines with 566 C and 246 atm were achieved; the thermal efficiency of the turbine room was approximately 35%. This was a significant milestone. Water (H2O) as the power-transmitting medium becomes a SC condition in conditions over 374 C and over 22.1 Mpa (218 atm), in that border between gaseous and liquidated condition disappears. For 25 years after the 1960s, turbines with 566 C, 25.0 Mpa (246 atm) were the most common advanced units, and plant thermal efficiency of 39–40% was achieved. In 1989, an ultra-supercritical (USC) turbine unit with 566 C and 32.1 Mpa (316 atm) was developed and put into service. Since then, more-advanced technology with over 600 C and total efficiency of 42% has been gradually achieved. In the near future, an advanced ultra-supercritical (A-USC) turbine with class 700 C will be developed. This is a longawaited key technology that will help protect against global temperature increase. Figure 11.12 shows a capability curve and voltage-excitation current characteristic curve of a typical large-capacity thermal generator; the explanation of the latter curve is shown in Figure 10.11. 11.11.2

Combined Cycle System with Gas and Steam Turbines

Liquefied natural gas (LNG) has recently become a very important fuel for power generation in parallel with coal and oil, supported by remarkably advanced excavation and liquefaction technology for easy transportation and storage, and combustion technology. The essential quality of natural gas (NG) is due to hydrocarbon cyclic compounds whose chemical formation is generally given by CnH2n + 2 (methane [CH4], about 90% ethane [C2H6], propane [C3H8], butane [C4H10], etc.). NG can be burned efficiently in the presence of oxygen (O2): in other words, under well-mixed conditions with plenty of compressed air supplied by the air compressor to the combustor (CB). The chemical equation of burning hydrocarbon gas is written as follows: Cn H2n + 2 +

3n + 1 O2 = nCO2 + n + 1 H2 O + Q J kg 2

For methane (n = 1), CH4 + 2O2 = CO2 + H2 O + Q J kg or 55, 700 kW sec kg NG liquefies into LNG at a very low temperature: −162 C (111 K). Liquefaction of gas obviously enables effective transportation by ship, and storage. Due to the advances in NG excavation and LNG ship transportation/storage technology achieved since about 1970, NG resources and large-scale, large-capacity LNG thermal power generation have become familiar practices, particularly since the 1990s. Figure 11.25a shows the configuration of (advanced) combined cycle power generation (ACC or CC) with LNG fuel combustion, based on state-of-the-art technology. As shown in the figure, the first section of the power system is the gas turbine (GT) unit, which consists of an air compressor, CB, and GT. LNG is vaporized in the vaporizer (LNG can be easily vaporized to gas through a warming process with water) and enters the CB. LNG is then mixed with highly compressed air and burned in the CB. The resulting combustible gas, at a very high temperature (say, 1500 C), enters the GT chamber and drives the GT’s rotary shaft. The principle and basic structure of the GT unit are similar to a jet engine for an aircraft. The exhaust gas at the outlet of the GT has a temperature of 600–700 C and enters the heat recovery steam generator (HRSG). This is a kind of large-scale, high-temperature heat exchanger, in which the thermal energy of the inlet exhaust gas is used to convert circulatory steam–water into superheated steam at 580–600 C. The steam then enters the high-pressure steam–turbine (HPST) and drives the shaft. The principles of the processes after the HPST are the same as in conventional ST generating stations.

11.11 General Description of Modern Thermal/Nuclear TG Units

Vaporized natural gas (fuel) HP steam (100–170 atm, 500–580ºC)

Gathering chimney stack

Exhaust gas (600ºC)

HRSG

LP steam (5–10 atm, 260–300ºC) AC

CB

LP

HP

Generator Condenser pump

GT (1400–1600ºC) Cold reheat steam

Air

Hot reheat steam

Water from sea or cooling tower

Water (15–20 atm, 30ºC)

Circulating water pump (a) Combined cycle system

CB: combustor AC: air compressor GT: gas turbine (15–20 atm, 1400–1600ºC) HP: high-pressure steam turbine (100–170 atm, 500–580ºC) LP: low-pressure steam turbin (5–10 atm, 260–300ºC) HRSG: heat recovery steam generator (b) Single-shaft combined cycle power train

Gene

rator AC Gasturbi CB ne un it

Steam

tur (HP/ bine unit IP/LP )

Typical power ratio

GT : ST

GT

2:1 Courtesy of TEPCO/GE/Toshiba

Figure 11.25 Combined cycle (CC) system.

In brief, the CC consists of a GT and ST unit. Figure 11.25b shows the configuration of a typical ACC system of 150–500 MW capacity with the GT, ST, and generator (G) mechanically directly coupled as a single-shaft power train. Another ACC system configuration has the GT/G train and ST/G train arranged separately without mechanical shaft coupling. In contrast to the CC system, a simple cycle system (SC, with only GT and without ST) is also available, mainly to generate a spinning reserve and meet the peak power demands of individual power systems; the virtue of this system is its quick-start capability.

293

294

11 Synchronous Generators, Part 2

HP-turbine LP-turbine Feed-in:280ºC-68atm Feed-in:265ºC-13atm Feed-out:190ºC-14atm Feed-out:25–37ºC-0.02–0.1atm Wet steam separator

Chimney stack

Main transformer

Main steam pipe To/from boiler (nuclear reactor)

HP-T

LP-T

LP-T

Generator

LP-T

Cooling water Water pipe

Dust collector (cinder catcher) Grand steam condenser

Main condenser Air ejector LPC-P Feed water pump

Condensate demineralizer

Water filter

HPC-P Feed water heater

Water supplement

Generator

(a) Configuration LP-T

LP-T

LP-T

HP-T

(b) TG area

Courtesy of Toshiba Figure 11.26 TG unit for nuclear power generating station.

The rated firing temperature at the CB/GT using recent technology is 1400–1500 C, compared to 1100–1300 C in the 1980s. The firing temperature at the CB/ GT is far higher than the inlet steam temperature for a conventional ST, so the heat efficiency η of an ACC or CC generating system is quite high compared to that of a conventional ST system. The efficiency of ACC at 1500 C, for example, is 50–54%, while that of conventional steam generation is 40–43%.

11.11 General Description of Modern Thermal/Nuclear TG Units

An extremely high temperature for the combustible gas could be close to the annealing/melting critical temperature of the alloys and metals used to construct the CB and GT (a few top stages) nozzles and buckets, so prudent thermal design is required in order to cool the hot-gas parts without relying on film cooling and to avoid causing abnormal hotspots. For example, in an advanced single-shaft power-train model with a class 1400–1600 C GT, the GT’s first-stage rotational and stationary aerofoils are made of single-crystal alloy and are thermal-barrier coated. Furthermore, a threedimensional geometry with closed-loop steam cooling (using high-temperature steam of 400 C extracted from the HRSG as the cooling fluid) is employed. Using today’s advanced technology, GT classes over 1600 C may be achieved in the near future. Note also that NG contains such impurities as dust, sulfur, and water, but almost all of these can be removed during the liquefaction process. Therefore, the emission levels of substances responsible for air pollution (NOx, SOx, CO2) are reduced to low levels, which is why thermal power generation by LNG has been recognized as “clean thermal energy” in comparison with coal and oil. 11.11.3

ST Unit for Nuclear Generation

Figure 11.26 shows a TG unit for nuclear generating stations, including typical turbine steam pressures and temperatures. The principles of the steam/water circulation system, including turbines and condensers, are the same as for conventional thermal units. However, the superheated steam fed from the nuclear reactor is typically at 280–300 C, which is much lower than a thermal steam boiler. Therefore, the TG unit of a nuclear plant consists of two sections of turbines, HP-T/LP-T (without IP-T), and a generator with four poles that have 1500/1800 rpm ratings. In other words, the TG-coupled rotor for the nuclear unit is half the speed, and the generator rotor has a relatively larger mechanical volume compared to a thermal unit.

Supplement Derivation of Eq. (11.57) from Eq. (11.52)④ Modifying Eq. (11.52)④ and making it equivalent to a, Efd xl = xd eB

p2 + q + xd−1

2

p2 + q − xl−1

2

1

a

Modifying the middle and right side of this equation, p2 + q2 + 2q

xd−1 + a2 xl−1 xd−2 + a2 xl−2 = − 1 −a2 1− a2

2

Putting the numerator of the third term on the left equivalent to b and the right to c, then b =c 1 −a2 b p2 + q + 1 −a2

p2 + q2 + 2q ∴

b=

1 a2 + xd xl

a2 1 − x2 x2 c = l 2d 1 −a Thus Eq. (11.57) is obtained.

2

=

b 1 −a2

2

+c 3

295

297

12 Steady-State, Transient, and Dynamic Stability We studied generator characteristics and capabilities in Chapters 10 and 11, mainly from the viewpoint of the electromechanical equipment. In this chapter, we will begin by examining the generator’s fundamental characteristics in combination with the connected network. Incidentally, various characteristics of a power grid are dominated by the generator’s characteristics, because the power grid is an aggregate of generators. We will study power-system behaviors as simple power-grid models in this chapter, but such behaviors are common for any large or small power system when it comes to practical engineering.

12.1

P-δ Curves and Q-δ Curves

A generator operates under balanced three-phase conditions and rotates at constant speed. This coincides with Eqs. (10.55a)–(10.56) and the vector diagram in Figure 10.7. ed + jeq = E 1 e jα

12 1a

id + jiq = I 1 e jβ ed + jeq e jt = E 1 ej t + α = e1 t

12 1b

id + jiq e jt = I 1 e j t + β = i1 t ed = E 1 cos α = E 1 sin δ

12 1c

eq = E 1 sin α = E 1 cosδ where δ = π 2 −α

12 1d

The symbol (t) in ed ,eq , id ,iq has been omitted because all the quantities are for time-independent DC components under balanced three-phase conditions. t is the unitized value [rad] based on Eq. (10.42). Referring to Figure 10.7, δ is the jxad ifd , proportional to the field excitation) and the angular difference between the generator’s induced voltage jE f jα terminal voltage E 1 e (the generator’s inner angular difference) and δ = (π/2) − α. The apparent power is, referring to Eqs. (12.1a, b), S 3ϕ = S 1ϕ = e1 t i1 t ∗ = E 1 e jα

I 1 e jβ

= ed + jeq e jt

∗

id − jiq e − jt

= ed id + eq iq + j eq id − ed iq ∴

12 2

P3ϕ = P 1ϕ = ed id + eq iq Q3ϕ = Q1ϕ = eq id − ed iq

This is the apparent power of the generator under balanced three-phase conditions, and all the d- and q-domain quantities are for DC components.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

298

12 Steady-State, Transient, and Dynamic Stability

Next, Eq. (10.59) is utilized under balanced three-phase conditions. Rewriting the equation as E f − eq − riq xd ed + rid iq = xq id =

12 3

omitting terms in r (because xd ,xq P gen

P 3ϕ = P1ϕ =

r) and substituting Eq. 12.3 into Eq. (12.2),

E f ed 1 1 + ed eq − xd xq xd

2 E f eq e2d eq Qgen = Q3ϕ = Q1ϕ = − + xd xq xd

① 12 4 ②

When we substitute Eq. (12.1c) into (12.4), the following very important equations are derived. For the P–δ and Q–δ curve characteristics: P −δ curve P gen = P3ϕ =

Ef E1 E2 1 1 sin 2δ sin δ + 1 − xd 2 xq xd

① 12 5

Q − δ curve Qgen = Q3ϕ =

Ef E1 E2 1 1 cos δ− 1 + xd 2 xq xd

+

E 21 1 1 cos 2δ − 2 xq xd

②

where the second terms on the right sides in Eqs. ① and ② are called the saliency effect terms. The equations explain the generator’s essential characteristics in terms of the P–δ curve and Q–δ curve. Figure 12.1a shows the vector diagram of the generator quantities under balanced three-phase operation. Figures 12.1b and c are the sets of the P–δ curve and Q–δ curve derived from Eq. (12.5), which show the effective power P and reactive power Q as parameters of δ. The saliency effect appears under the condition xd xq Equations (12.5) ① ② as well as the resulting P–δ curve and Q – δ curve are twins derived from a common process, so they cannot be treated independently. So, discussion of stability based on either the P–δ curve or the Q – δ curve rather than both is nonsense or a serious mistake. Equations (12.5) ① ② can be written as the combined equation S gen = Pgen + jQgen =

E f E 1 −jδ E 21 1 E 21 1 1 1 − e −j2δ + e − − 2 jxq jxd jxd 2 jxq jxd

① 12 6

for the non-salient-pole machine xd = xq S gen = Pgen + jQgen = E 1

E1 − Ef e jxd

−jδ

②

Equations (12.5) and (12.6) show the generator’s basic characteristics in combination with the connected outer network. In the equations overall, only xd ,xq are specified values for the generator machine; all the other variables are for electrical quantities at the generator terminal as part of the connected network. Therefore, the equations can be written as an implicit functional equation as follows: function P, Q, v, i, δ, ω = 2πf = 0

12 7

All the variables P, Q, v, i, δ, ω = 2πf are linked to each other, while P and f as well as Q and V are closely correlated. The steady-state stability limit has often explained as follows by using the P–δ curve: δ = 90 gives the steady-state stability limit, and the system can be stable within the range δ = 0 – 90 . This explanation is not correct from the viewpoint of practical engineering. Figure 12.1 obviously indicates that large var power Q is required in order to maintain nominal voltage E 1 1 0 in the range δ = 40 – 90 and to transmit the maximum power E f E 1 xd . Var generating capacity is limited in the range δ = 0 − 40 in Figure 12.1c, so E 1 decreases in the range 40 − 90 . In other words, power system operation with the maximum value P = E f E 1 xd sin δ at around δ = 90 cannot be realistic, because of limited

12.2 Power Transfer Limits of Grid-Connected Generators (Steady-State Stability)

P

Curve 2 (salient-pole generator) Curve 1 (non-salient-pole generator)

Pgen

jEf

.

j (xd – xq) id

E f E1 sin𝛿 xd

.

𝛿 · 𝛼

𝛿 90°

q-axis

180° E12 1 – 1 xq xd 2

(

id

jxq·i

·

egen

·

i

·

iq

) sin 2𝛿

d-axis

(b) P–𝛿 curve

(a)

Q EfE1 cos𝛿 xd

0

E12 2

90°

( x1

1 xd

q

) cos 2𝛿

𝛿

180°

E12 2

( x1 + x1 ) q

d

Curve 2 (salient-pole generator)

Qgen Steady-state stability zone

Curve 1 (non-salient-pole generator)

(The case of : x =E1.1,= 1.0x = 0.7, E d

q

f

)

= 1.5,

1

(c) Q–𝛿 curve Figure 12.1 P–δ and Q–δ curves of a generator.

reactive power capacity. Actual operation is limited to the conditions where E 1 can be kept within 1.0 ± 0.1, and the practical operating zone of δ is limited in the range 0 –40 . This explanation is discussed again in Section 12.7. The P–δ and Q–δ curves always exist as an inseparable couple, and they should always be treated together.

12.2

Power Transfer Limits of Grid-Connected Generators (Steady-State Stability)

12.2.1

Apparent Power of Generators

Figure 12.2a shows a system model with two machines, with generator 1 connected to generator 2 through a network reactance; and Figure 12.3a shows a system model with one machine connected to an infinite bus, with generator 1 is connected to the infinite bus. The vector diagrams for each case are also shown.

299

300

12 Steady-State, Transient, and Dynamic Stability

Figure 12.2 System model with two machines.

Ef ∠δ

δ #1

xd xq

. egen

. ebus = Ebus∠0°

÷ i

Ef∠δ

÷ i

xl

. jxq ⋅ i

. egen . ebus

q-axis j (xd – xq) ⋅id =⋅ ⋅0

. jxl ⋅ i

d-axis

(a)

(b)

Figure 12.3 System model with one machine to infinite bus.

First, let’s examine the equivalence of these figures. The electrical condition of generator 1 becomes equivalent in both figures if the condition in the following equation is satisfied: B xl

+ B xd ≒ B xl + B xq

xl

12 8

Equation (12.8) means the d-axis circuit and the q-axis circuit of the outer system can be considered the same in Figure 12.2a. (The difference for the zero-sequence circuit is ignored because we are studying balanced three-phase normal operating conditions.) The assumption in Eq. (12.8) can be justified if generator 2 is a cylindrical type xd = xq , or salient type B xd B xq Equation (12.8) is approximately justified by inserting some line reactance B xl between the two generators. In other words, the behavior of generator 1 can be examined using either a two-machine system or a one-machine to infinite-bus system. In an actual power system with a total number of generators n, we can imagine that generator 1 is connected to another equivalent generator that actually consists of n – 1 parallel connected generators through the network.

12.2 Power Transfer Limits of Grid-Connected Generators (Steady-State Stability)

We assume that Eq. (12.8) is justified and that generator 1 is a thermal-driven generator xd = xq in Figures 12.2 and 12.3. Then ebus = E bus ∠ 0 = E bus

①

egen = E bus + jxl i

②

E f = E f ∠ δ = E bus + j xq + xl i, j xd − xq id ≒ 0 Sbus = Pbus + jQbus = E bus i

③

∗

12 9

④ ∗

⑤

Sgen = Pgen + jQgen = E bus + jxl i i

where the suffix gen refers to quantities of generator 1. From Eq. (12.9) ④ ⑤, P gen + jQgen = P bus + jQbus + jxl i

2

12 10

From Eq. (12.9) ③, i=

E f e jδ − E bus −jE f e jδ + jE bus ∗ − jE f e −jδ − jE bus , i = = xq + xl xq + xl j xq + xl ∗

2

i =i i =

=

E 2f + E 2bus − E f E bus e jδ + e −jδ xq + xl

12 11

2

E 2f + E 2bus − 2E f E bus cos δ xq + xl

2

Accordingly, the apparent power of a generator is ∗

S bus = P bus + jQbus = E bus i = E bus =

E bus E f cosδ −E bus E f E bus sin δ + j xq + xl xq + xl

Sgen = Pgen + jQgen = Pgen = Qgen =

jE f e −jδ − jE bus xq + xl

E f E bus sin δ + j xq + xl

E f E bus sin δ xq + xl

①

E 2f xl − E 2bus xq + E f E bus xq −xl cos δ xq + xl

2

P − δ curve

E 2f xl − E 2bus xq + E f E bus xq − xl cosδ xq + xl

2

②

12 12

③

Q −δ curve

P gen ,Qgen are derived as P–δ and Q–δ characteristics with the parameters E f , E bus , where usually E bus = 1 0∠ 0 . The equation shows that Qgen has positive (+) magnitude (lagging operation) for large E f , and it has negative (−) magnitude (leading operation) for smaller E f (weak excitation). Also, P gen = P bus is recognized because line resistance is ignored, while Qgen Qbus because there is reactive power consumption on the transmission line. The special case xl 0, E bus E 1 ,xd = xq in Eq. (12.12) coincides with Eq. (12.6) ②.

12.2.2

Power-Transfer Limits of Generators (Steady-state Stability)

A generator normally operates in synchronization with the connected power system. This means E f ∠δ and E bus ∠0 are running at the same speed and the angular ∠δ displacement is always within some upper limit (δ < 90 ).

301

302

12 Steady-State, Transient, and Dynamic Stability

Suppose generators 1 and 2 are operating in synchronization, and generator 2 accelerates a little for some reason, leading to δ δ + Δδ. Then generator 1 automatically tries to recover the delay Δδ by releasing the kinetic energy stored in its rotor, so the electrical output of generator 1 immediately increases by P P + ΔP. These inherent recovery characteristics of the generator are called synchronizing power. The synchronizing power of the generator is effective as long as the recovering power ΔP has a positive sign for some disturbance Δδ. In other words, the critical condition in which the generator can be operated with the power system in synchronization is Synchronizing power

∂P gen ΔP ≧0 = Δδ ∂δ

12 13

Applying Eq. (12.12) ③ to Eq. (12.13), ∂Pgen Ef E bus = cosδ ≧ 0 xg + xl ∂δ

∴ − 90 ≦ δ ≦ 90

12 14

The generator reaches the power-transfer upper limit at large δ of approximately 90 . Such an upper-limit condition is called the steady-state stability limit. Using the suffix max for such conditions, S gmax = P gmax + jQgmax

①

E f E bus xq + xl

②

P gmax = Qgmax =

12 15

E 2f xl − E 2bus xq xq + xl

③

2

where δ = 90 with the condition xd = xq In Figure 12.1b, the steady-state stability limit is at δ = 90 for a generator with non-salient poles xd = xq , curve 1 and is at an angle of less than 90 (say, 70 ) for a generator with salient poles xd xq , curve 2 . Note that in order to transfer the maximum power Pgmax through the line, the corresponding large value of Qgen in Figure 12.1c must be supplied to the system so the terminal voltage E1 is kept as the normal voltage. Otherwise, E1 decreases, and the P–δ curve shrinks. The necessary value for Qgen becomes quite large when operating with δ exceeding 50 , as shown in Figure 12.1c. This is why we always need to treat P and Q together as coupled quantities.

12.2.3

Power Cycle Diagram

Pgmax ,Qgmax are functions of E f ,E bus and so cannot be independent of each other. Therefore, we want to find an equation for Pgmax and Qgmax by elimination of E f and E bus For this purpose, we need one more equation in addition to Eqs. (12.15) ② ③, which can be obtained from Figure 12.2c. Under the condition of δ = 90 ,AO CO = AB GB = xl + xq xl , Pythagoras’s theorem can be applied: e2gen =

xl xq + xl

2

E 2f +

xq xq + xl

2

E 2bus

12 16

Eliminating E f ,E bus from Eqs. (12.15) ② ③ and (12.16), we obtain the following equation (refer to Supplement 1 at the end of the chapter for the process): P2gmax + Qgmax −

1 1 1 2 e − 2 xl xq gen

2

=

1 1 1 2 e + 2 xl xq gen

2

12 17

12.2 Power Transfer Limits of Grid-Connected Generators (Steady-State Stability)

Unitizing P max, Qmax by e2gen and writing them as p, q, p2 + q −

2

1 1 1 − 2 xl xq

where p =

=

1 1 1 + 2 xl xq

2

Qgmax Pgmax , q= 2 e2gen egen

The circular locus in p −q coordinates Center

0,

12 18

1 1 1 − 2 xl xq

Radius

1 1 1 + 2 xl xq

Diameter

the straight line connecting the points 0,

1 1 and 0, − xl xq

Figure 12.4a can be drawn from Eq. (12.18). The circle in PQ coordinates gives the steady-state stability limit, and the generator cannot be operated outside of the circle. Equation (12.18) can be modified to the following equation: p xq

2

q xq −

+

1 xq −1 2 xl

2

=

1 xq +1 2 xl

2

For the circular locus in p xq − q xq coordinates Center

0,

1 xq −1 2 xl

12 19

1 xq +1 2 xl

Radius Diameter

the straight line connecting the points 0,

xq and 0, − 1 xl

Figure 12.4b can be drawn from the equation: circles based on the parameter xq xl .

Lagging

q=

Qgmax egen2

1 xl 1 2

( x1

l

1 xq

) 1.0

–⋅1 xq –1.0

qxq = Radius 1 2

1 1 + xl xq

0.5 0.3 xl = 0.2 0.6 0.4 1.00.8 Stable 1.0 Pgmax p= egen2 Non-salient pole generator (x–d = x–q = 1.7)

Qgmax .x egen2 q

2.0

Lagging

xl = 0.5 1.0

(0, xx )

0.75 1.0

q l

2.0 Stable 1.0

Leading –1.0

pxq =

Pgmax egen2

2.0 Non-salient Pole generator (x– = x– ) d

Leading

p – q coordinates (a)

Figure 12.4 Steady-state stability limit of a generator.

pxq–qxq coordinates (b)

q

.x

q

303

12 Steady-State, Transient, and Dynamic Stability

If the power system’s capacity is relatively small xl large , the generator’s stable operating zone shrinks. In other words, if the generator’s output power exceeds the limit, the generator has to be stepped out in deceleration mode. 12.2.4

Mechanical Analogy to Steady-State Stability

Power-system characteristics with a generator and a motor load, as shown in Figure 12.5a, can be explained as a mechanical analogy to Figures 12.5b and c. Figure 12.5b, called the NODA model, gives a perfect analogy for the stability between the electrical system and the mechanical model. In Figure 12.5b, if we turn the rotary handle slowly in the direction of the arrow, the force power P is transmitted to the hoist through the three rubber elastic cords, and the weight W is hoisted up. This is the same analogy as the generator power P being transmitted to the load motor through the line impedance. Figure 12.5c shows an electrical vector diagram of the power system, and Figure 12.5d shows the mechanical force vectors between two disks. If we observe the mechanical vectors on two disks from the right to the left, the diagram will be similar to Figure 12.5c: i) With this model, the radius VG, VM of both disks corresponds to the terminal voltages at the generator terminal G and the load terminal M. ii) If we increase the weight W (electrical load Pload), mechanical angular displacement δ (electrical angular difference δ) is gradually increased from 0 , and the three elastic cords (the voltage drop of the three-phase conductors) expands gradually. Finally, if δ exceeds the critical condition 90 , the elastic cords are completely twisted and could Line distance (L) Line reactance

1’

5’

G

Turbine

M Synchronous Mechanical load W motor (VM)

Generator VG

(a) Electrical two-machine model Distance FP •

F

VG

Hoist

•

VK

Fq

O

Rubber elastic cords

G

•

VM

O’

M

K

Rotary handle Disk G

Disk M

(b) Mechanical two-disk model Driving terminal

Load terminal

l0

VL = j •

PG

IX

•

VK

QG •

Fp •

VM motor voltage

𝛼

F

VG

l

𝛽

FQ

V0

sin 𝛿

𝛼 •

VM

304

𝛿

𝛿

L G

Generator voltage VG Angular 𝛿 displacement O

(c) Electrical vector diagram Figure 12.5 Mechanical analogy of power-system stability.

VM

(d) Mechanical vector diagram

M

12.2 Power Transfer Limits of Grid-Connected Generators (Steady-State Stability)

be broken (loss of synchronization), and the disk load VM can no longer follow the generator disk VG. The critical maximum load Wmax (Pload max) is proportional to the radius VG, VM of the disks (the operating voltage VG, VM) and is inverse proportional to the length l between the two disks (inductance L between points G and M). iii) Further, the radius VK of the midpoint K corresponds to the voltage at point K, which decreases gradually as δ increases and it reaches a value near zero (crash, loss of synchronization) at the critical condition δ = 90 . At δ = 180 after the crash, the situation is the same as a three-phase fault at the midpoint. iv) The tension force F of the elastic cord can be decomposed into Fp (circumference component) and FQ (centripetal component), which correspond to the effective power P and var power Q. If Fp(P) is to be increased, FQ(Q) should also be increased, or voltage will be lost. v) If a spacer is installed at the midpoint K in order to maintain the original distance of the three elastic cords, the region of stable power transmission expands beyond δ = 90 . The spacer works as a var generator at point K in order to maintain voltage VG ≈ 0 (SVG). Referring to the mechanical disk model shown in Figure 12.5c and d, we see complete mathematical similarity with the electrical two-machine power-system model, as follows L2 + VG − VM

l0 =

2

original length of the three elastic cords

L2 + VG − VM cos δ 2 + VM sin δ

l=

2

expanded length of the cords at

a b

angular displacement δ cos α =

VM sin δ α is the angle between F and FP l

VG − VM cosδ cos β = l

c 12 20

β is the angle between F and FQ

FP = Fcos α circumference component of the tension F

d

FQ = Fcos β centripetal component of the tension F l0 l = F0 F

e

X

where F0, F are the tension of the cords at angular displacement 0 and δ with the length l0 and l, respectively. If L is long enough, l l0. Then, F

F0

l −Δl l0 − Δl

l 1 − Δl l X 1 −Δl l0

l X

a

where l0 is the length of the cord with tension 0. Then, FP =

l VM sin δ VM sin δ = X l X

FQ =

l VG −VM cos δ VG −VM cos δ = X l X

b

P = FP VG 12 21 Q = FQ VG P=

c

VG VM sin δ X

VG2 − VG VM cos δ d X Equation (12.21) shows complete equivalency with Eqs. (12.12) and (12.15) for a two-machine system. Q=

305

306

12 Steady-State, Transient, and Dynamic Stability

12.3

Transient Stability

12.3.1

Definitions of Steady-State Stability, Transient Stability, and Dynamic Stability

The dynamic characteristics of a power system including all different stability modes that are deeply dependent on the generators’ characteristics. A power system is said to be stable when it remains at operating equilibrium in both normal and abnormal conditions. In order for interconnected synchronous machines to be stable, they must maintain synchronization in both normal and abnormal conditions. Stability can be classified into three categories, which we will introduce next. 12.3.1.1 Steady-State Stability

Steady-state stability is defined as an operating state of a power system characterized by gradual change. The steadystate stability limit is explained by Eqs. (12.5), (12.12), and (12.17) and Figures 12.1 and 12.4. 12.3.1.2 Transient Stability

The transient state of a power system is characterized by a sudden change in load or circuit conditions. Transient stability is defined as stability in such transient states. Short-circuit faults and fault tripping/reclosing, switching of circuits, abrupt significant load changes, sudden tripping of generators, etc. are typical disturbances. Other types of disturbances include sudden changes in generator excitation caused by irregular conditions in the automatic voltage regulator (AVR) (a sudden change in the AVR’s set value, for example) or the mechanical power of a prime mover, or a change of power flow in the network caused by changes in power distribution among generators or changes in network connections. Hunting phenomena among multiple generators is another kind of disturbance. 12.3.1.3 Dynamic Stability

The power-system stability limit can be improved far beyond the steady-state stability limit by using appropriately designed AVR equipment. In addition, automatically controlling the speed governor of each generator (automatically controlling the mechanical power of the prime movers) by detecting sudden significant frequency changes under fault or no-fault conditions may improve the stability limit. Dynamic stability can be defined as improving stability by applying appropriately quick excitation control (jEf control by an AVR) as well as appropriately quick speed-governor control (Pm control by frequency detection) at each generating station. Dynamic stability was named originally in contrast to steady-state stability. However, it is obvious that appropriate AVR control and speed-governor control at each generating plant improve not only steady-state stability but also transient-state stability caused by various cascades of sudden changes in power-system conditions. The time constants of the AVR + field excitation circuit are very small (say, 0.1–0.5 seconds), while those of the speed governor + prime mover must be larger (say, a few seconds). Accordingly, AVR is more effective for an initial rapid response to serious disturbances. 12.3.2 Mechanical Acceleration Equation for a Two-Generator System Generator B The generator’s mechanical acceleration equation was derived in

Generator G i

Ef

EB

Eq. (11.36) or (11.37). Now we will examine the power system shown in Figure 12.6, which contains generators G and B connected by a double-circuit line:

f PGe 𝜃G, 𝜔G MG

Figure 12.6 Two-generator system.

PBe 𝜃B, 𝜔B MB

Generator G d 2 θG t ωG t = PGm − PGe 2 MG dt

Generator B d 2 θB t ωB t = PBm −PBe 2 MB dt 12 22

12.3 Transient Stability

There are mutual relations between generators G and B, as follows: PGe = − PBe

Ignores line resistance

δ t = θG t −θB t =

ωG t −ωB t dt < 90∘

ωG t ≒ = ωB t ≒ 2π f0

ω0 ,

Angular difference of induced voltages of both generators

12 23

f0 power frequency = 50 60Hz

δ(t) is the angular difference between the two generators, which must be within ±90 under synchronization. From Eq. (12.23), d2 δ PGm − PGe PBm −PBe − 2 = ω0 MG MB dt

12 24

Now we will assume a non-salient-pole machine for simplicity and take the effective power from Eq. (12.12) PGe = −PBe =

Ef EB sin δ xq + xl

12 25

where Ef, EB: internal induced voltages of generators G and B, respectively. The mechanical input PGm, PBm cannot be changed for 0–3 seconds from the magnitudes just before the system disturbance (because of the inertia of the prime-mover system). Then PGm = PGe = − PBe = − PBm

12 26

From Eqs. (12.24) and (12.26), the mechanical-acceleration equation of a generator G is d2 δ 1 1 PGm − PGe = ω0 + MG MB dt 2 Ef EB ω0 MG MB PGm − sin δ where M0 = = M0 xq + xl MG + MB

12 27

where δ t phase angular difference between the induced voltages of both generators − 90 < δ t < 90 under normal conditions If generator B has a large capacity in comparison with generator G, this means Mb ∞ and M0 MG, which correspond to the one-machine to infinite bus. In Eq. (12.27), ω0 = 2π f0 (where f0 = 50/60 Hz) has a fixed value. Mg, Mb, are the specific machine constants. Reactance xl is the network reactance connected to the generator terminal, which will suddenly take on a large value if a fault occurs in the network. Therefore, Eq. (12.27) can be written as δ(t) = function(PGm, Ef), from the viewpoint of controllable quantities. In other words, this indicates that we can control at the generating station only the excitation of the generator and the mechanical input power from the prime mover. Incidentally, the equivalent inertia constant M0 for two machines is written as the weighted average value of each generator’s inertia constants. Analogously, the equivalent inertia constant M0 for multiple-machine systems can be written as the weighted average value of each generator’s inertia constants. That is, 1 = M0

n k

1 Mk

12 28

Equation (12.27) also explains the situation as follows: ① Whenever a fault disturbance is caused at an outer circuit of a generator, xl (the operational reactance of the outer circuit at the generator terminal, explained later in detail) increases, while PGm and Ef cannot be changed suddenly, so ∂δ/∂t = ω inevitably increases. Then if δ(t) increases over 90 , the generator is stepped out by

307

308

12 Steady-State, Transient, and Dynamic Stability

acceleration mode. However, if Ef is quickly increased by the appropriate operation of the AVR, or if PGm is quickly decreased by the appropriate operation of the speed governor, increasing δ(t) can be resisted effectively. In other words, the steady-state stability limit can be improved to the dynamic stability limit using the AVR and the speed governor. When fault reclosing occurs (a typical outer disturbance), the time sequence is fault (t = 0 +) fault-tripping (t = 0.1 second) reclosing (t = 0.5 seconds) final tripping (t = 0.6 seconds). So, a cascade disturbance (sudden change of xl) is caused four times, and then δ(t) swings in a complicated manner within a short time. More specifically, the breaker’s fault trip consists of two different times: the first-pole tripping at current-zero time, and the second and third poles trip after a few milliseconds. ② If the inertia constant M is larger, the swinging angle of δ(t) is smaller. Comparing generators with the same rated capacity, a larger physical volume contributes to power-system stability. Although GD2 for modern large-scale generators tends to be somewhat smaller, an AVR and speed governor that are highly responsive can cover reasonable synchronization.

12.3.3

Transient Stability under Fault Condition (Equal-Quadrant Method)

Let’s assume a power system with a parallel-circuit transmission line, as shown in Figure 12.6, and assume that a shortcircuit fault occurs at point f of the first circuit. A proper protective relay will detect the fault immediately, and the associated circuit breakers will trip to remove the fault successfully within 3 to 6 cycles so the rest of the system can continue ordinary operation without instability. We will study the concepts of transient stability and dynamic stability for the typical cascade disturbances just described. The system’s behavior in the face of such disturbances is illustrated step by step in Figure 12.7, which is a P–δ curve that explains stability using the equal-quadrant method. Figure 12.7 is the P–δ curve for generator G. The condition before the fault is shown by curve 0, whose peak value is specified by Eq. (12.5) ① or Eq. (12.12) ③. When the fault occurs at point f, reactance xl suddenly becomes large (see the next section), so the generator condition must be stepped down from curve 0 to curve 1. Next, immediately after the fault trip, the generator condition jumps from curve 1 to curve 2, because reactance x1 returns to a smaller value that is very close to the original value (curve 2 is a little lower than curve 0 because of the removal of the faulted line during this time). 12.3.3.1 Case Study 1: Transient Stability Is Successfully Maintained

In this case, transient stability is successfully maintained, and the system continues stable operation after removing the fault. Generator G operates with power output P0 (δ0) at point ① on curve 0 before the fault (P0 is the power from the prime mover). When the fault occurs at point f, the operating point moves suddenly from point ① P0(δ0) to ② P1(δ0) on curve 1. As a result, the generator causes excess power of P0(δ0) – P1(δ1) (say, accelerating mode), so δ begins to increase on curve Effective power 11

P

12

5 10

P0

6

1

Curve 0 (before fault) 4

8

13 7

9

Curve 2 (after fault tripping) Curve 1 (during fault)

3

Angular difference

2

𝛿0

𝛿0´

𝛿1

90°𝛿2 𝛿3 𝛿4

𝛿5

Figure 12.7 Transient stability (equal-area method).

180°

𝛿

12.4 Dynamic Stability

1 from ② P1(δ0) to ③ P1(δ1). Then the fault is removed by the associated breakers tripping at the time of point ③ P1(δ1), and the generator condition jumps from ③ to ⑤ P2(δ1). In this condition, the generator suddenly causes a power shortage of P2(δ1) – P0(δ1) (say, decelerating mode). However, δ continues to increase up to ⑥ P2(δ4) because of the rotor inertia, and then in turn decreases toward the new stable point ⑧ P0 δ0 , although δ may repeatedly over-swing a little across the new stable angular difference δ0 . In this process, the maximum angle δ4 at point ⑥ satisfies the following relation: Accelerating energy δ1 δ0

P0 δ0 − P1 δ dδ =

Decelerating energy δ4 δ1

P2 δ −P0 δ dδ

or

12 29a

the area ①②③ ④ = the area ④ ⑤⑥ ⑦ Note that δ exceeds 90 at point ⑥ for a short period. 12.3.3.2 Case Study 2: The System Condition Exceeds the Transient Stability Limit

In this case, the system condition unfortunately exceeds the transient stability limit, and the system fails to continue stable operation after the fault. After reaching point ⑤ P2(δ1) by the same process as in case 1, if δ continues to increase beyond ⑥, in spite of the decelerating mode, and exceeds point ⑨ P2(δ5), the generator again enters accelerating mode, and the synchronizing force is lost entirely. As a result, the generator loses synchronization. Point ⑨ P2(δ5) is the critical point of the synchronization, where critical angle δ5 has a value exceeding 90 . The critical condition of the transient stability limit is area ①②③ ④ ≦ area ④ ⑤⑥ ⑨

12 29b

It is obvious that δ1 should be small enough (this means fast fault tripping) to satisfy this condition. 12.3.3.3 Case Study 3: High-Speed Reclosing Is Conducted

After reaching point ⑤ P2(δ1) by the same process as in case 1, δ continues to increase along curve 2 by inertia. Next, reclosing of the fault line is executed at the time of point P2(δ2). If a faulted arc is extinguished and the insulation at the faulted point is recovered, the generator condition jumps from 10 P2(δ2) to 11 P0(δ2) due to the successful reclosing, and δ increases to point 12 P0(δ3) and then decreases. Point 12 P0(δ3) satisfies the following equation: area ①②③④ = area ④⑤ 10 11 12 13

12 29c

Returning to our main theme, let’s now examine how transient stability can be improved. First, the height of curve 1 is determined by 1/xl in fault mode, so it is out of control (we will discuss xl in the next section). Accordingly, the most effective countermeasure to improve transient stability is to shorten the fault-tripping time by decreasing the acceleration energy {area ①②③④}. Clearly, delaying fault tripping for any reason means point ③ in the figure is shifted to the right and acceleration energy increases dramatically. It must be stressed that stable power-system operation can be secured by high-speed fault tripping. Today, due to the advanced technology of protective relays and circuit-breakers, fast fault tripping for high-voltage trunk lines has been realized, where typical operating times are {fault-detecting time by the relays 1–3 cycles} + {tripping time by the breakers 1–3 cycles} = {total tripping time 2–6 cycles}.

12.4

Dynamic Stability

In addition to high-speed fault tripping, two other effective countermeasures to improve stability can be realized by decreasing the acceleration energy {area ①②③④} or increasing the deceleration energy {area ④⑤⑥⑦}. 12.4.1

Quick Excitation Control with an AVR

This countermeasure increases the decelerating energy {area ④⑤⑥⑦} by enlarging curve 2 immediately after the fault. The peak value of curve 2 is EfEB/(Xq + Xl), so it can be enlarged by increasing the generator’s excitation Ef. An AVR

309

310

12 Steady-State, Transient, and Dynamic Stability

increases excitation jEf very quickly immediately after detecting a voltage drop –ΔV caused by a fault, so curve 2 is enlarged and the decelerating energy increases. Today, due to the advanced technology of AVR and excitation equipment, rapid excitation control (time constant of, say, 0.1 second) can be achieved. (This is discussed further in Section 12.8.) 12.4.2

Quick Driving-Power Adjustment with a Speed-Governor Control

This countermeasure decreases the accelerating energy {area ①②③④} by depressing the mechanical input from the prime mover immediately after the fault. When the speed governor detects a sudden acceleration of the rotor speed, it reduces the mechanical input from the prime mover by decreasing the water/steam flow (i.e. P0 (P0 ΔP0 in Figure 12.7). However, the amount of input power the speed governor can decrease in a short time is limited (say, ΔP0/P0 = 3–10%). In addition, especially in the case of a hydro unit, water flow cannot be changed quickly due to the time-delay characteristics of the water system (time constant, say, 1–5 seconds). Accordingly, the contribution of a speed governor as a countermeasure to improve dynamic stability is limited, especially for the initial short time (0–1 second) just after a disturbance. Controlling power with a speed governor is important in reducing frequency fluctuations in the power system; on the other hand, it is a supplementary countermeasure to improve dynamic stability. As described earlier, dynamic stability means stability that is greatly improved beyond the steady-state stability limit, which can be realized by quick excitation (AVR) control and supplementary speed-governor control of the generators.

12.5

Four-Terminal Circuit and the P − δ Curve under Fault Conditions

12.5.1

Four-Terminal Circuits

In Figure 12.7, the peak value of the P–δ curve is given by Ef EB/(Xq + Xl). We conducted our study with the understanding that curve 0 before the fault has a large peak value, while curve 1 during the fault has a very small peak value, because the reactance xl (the equivalent reactance) included in the denominator becomes large under fault conditions in comparison with that under normal conditions before the fault (say, 10 or 20 times). Why would xl become so large under the fault condition in comparison with the value before the fault? What is the reactance xl that is included in the denominator of Eq. (12.27) or Eq. (12.12), in particular under fault conditions? We need to clear up these simple questions. Let’s examine Figure 12.8, which is a single-phase, four-terminal circuit between sending point s and receiving point r. The equation for the circuit is vs

=

is

A

B

vr

C

D

ir

12 30a

where vs = Vs ∠ δ = Vs e jδ vr = Vr ∠ 0 = Vr Eliminating ir , is =

D BC − AD vs + vr B B

12 30b

The apparent power at sending point s is vs=Vs ∠ 𝛿 A

xb

vr =Vr ∠ 0° B C D

is Figure 12.8 Four-terminal circuit.

S s = Ps + jQs = vs i∗s =

ir

=

D

∗

∗

∗ ∗ vs v s

B ∗ D ∗

B

+

Vs2 +

∗

∗

∗

B C −A D ∗

B ∗ ∗ ∗ ∗ B C −A D ∗

B

vs v∗r Vs Vr e jδ

12 30c

12.5 Four-Terminal Circuit and the P − δ Curve under Fault Conditions

Using this equation, we compare circuits 1 and 2 in Figure 12.9. Zs ≡ vs/is and Zr ≡ vr/ir are called the characteristic impedance of terminals r and s, respectively. Assuming the circuit between points s and r is symmetrical, the constants A,B,C,D satisfy the following condition:

𝜐s

𝜐r

Circuit 1 jx

xb

ir

is 𝜐s

A = D and Z s = Z r =

C B

s

Circuit 2

12 30d

jx'

a jx

xb is

𝜐r

b

r

jx' jxf

ir

Figure 12.9 Comparison of sending power.

12.5.2 Four-Terminal Circuit in a Transmission Line Before Fault This case corresponds to the Circuit 1 of Figure 12.9 and x + x’. The equation is Eq. (12.12) ② ③ given the relation Xl, vs =

is

1

j x+x

vr

0

1

ir

=

A

B

vr

C

D

ir

12 31

A = D = 1, B = j x + x , C = 0 ∗

∗

∗

∗

Substituting the conjugate A , B , C , D values from Eq. (12.31) into Eq. (12.30c), Ss = Ps + jQs =

Vs Vr Vs2 − Vs Vr cosδ sin δ + j x+x x+x

12 32

The first and second terms on the right side give the P–δ curve and the Q–δ curve at point s, and the denominator is x + x (the reactance between points s and r). 12.5.3

Four-Terminal Circuit in a Transmission Line under Fault

This is the case of Circuit 2 in Figure 12.9 where reactance xf is added in parallel at the midpoint of the line. The fourterminal circuit equations can be calculated by multiplying the impedance matrices of each section s–a, a–b, b–r, in the following order: Matrix s−a

vr = ir

1 0

x xf 1 −j xf

1+ =

jx 1

a−b

1 1 jxf

0 1

b −r

1 0

x x xf x 1+ xf

j x+x +

vr ir

jx 1

vr

A

B

vr

ir

C

D

ir

∗

∗

∗

12 33

∗

A,B,C,D are found, and their conjugates A , B , C , D are substituted into Eq. (12.30c): Ss = Ps + jQs Vs Vr = sin δ + j x x x+x + xf

Vs2

1+

x x xf

− Vs Vr cos δ

x x x+x + xf

12 34

311

312

12 Steady-State, Transient, and Dynamic Stability

Table 12.1 The P–δ curve during a fault and positive-sequence equivalent impedance. V sVr × sin δ Dx

P−δ

Ps =

Before fault D(x) = x + x

Impedance Zf = (rf + jxf) to be inserted in the positive-sequence circuit

Short-circuit modes D x = x + x +

xx xf

Phase opening modes D(x) = x + x + xf

3ϕS Zf = 0 2ϕS Zf = f Z2 1ϕG Zf = f Z0 + f Z2 2ϕG Zf = f Z0 f Z2 /(f Z0 + f Z2) 3ϕOp Zf = ∞ 2ϕOp Zf = f Z0 + f Z2 1ϕOp Zf = f Z0 f Z2 /(f Z0 + f Z2)

In this case, the denominator reactance, which corresponds to xl from Eq. (12.12) ②③, becomes x+x + x x

xf

Now, comparing the results of both circuits, the equivalent reactance of circuit 2 becomes larger than that of circuit 1: Circuit 1 x+x

Circuit 2 x x x+x + xf

12 35

That is, Eq. (12.32) under normal conditions is replaced by Eq. (12.34) under short-circuit fault conditions with the reactance jxf (or Zf) inserted at point f. This result can be applied to the positive-sequence circuit in Figure 7.5, for example, in which faults occur at the midpoint f of the line. For 1ϕG (phase a to ground fault), the reactance f x2 + f x0 is inserted at point f, which corresponds to the inserted reactance fx2 + fx0 of circuit 2. In the same way, this result can be applied to all other fault modes by referring to Tables 7.1 and 7.2, which are summarized in Table 12.1. With short-circuit mode faults, the P–δ curve obviously becomes smaller (lower) for smaller Zf = jxf . Then, referring 1ϕG 2ϕG 2ϕS 3ϕS. The to Table 12.1, the P–δ curve becomes smaller and on the order of before fault extreme case is 3ϕS, with xf 0, D(x) ∞, Ps (δ) 0, which means power cannot be transmitted at all during the three-phase fault. Conversely, for phase-opening modes, the P–δ curve becomes smaller (lower) for larger Zf = jXf. Then, referring to 1ϕOp 2ϕOp 3ϕOp. Table 12.1, the P–δ curve becomes smaller and on the order of before fault

12.6

P-δ Curve under Various Fault-Mode Conditions

12.6.1

Three-Phase Fault Mode (3ϕS)

This case corresponds to the condition xf 0, D(x) ∞, so the height of the P–δ curve is almost zero. In other words, power can seldom be transmitted to the outer system during the fault, and the generator will immediately begin to accelerate quickly. 12.6.2

Double-Phase Fault Mode (2ϕS)

Referring to Table 12.1 and Table 7.1, xx x+x

xf = f x2 = x

x =

D x =x+x +

xx =2 x+x xf

The height of the P–δ curve under 2ϕS is half of that before the fault.

12 36a

12.7 PQV Characteristics and Voltage Instability (Voltage Avalanche)

12.6.3

Single-Phase Fault Mode (1ϕG)

Referring to Tables 12.1 and Table 7.1, xf = f x2 + f x0 = x D x =x+x +

x + x0

xx x0 x0 + x+x x0 + x0

x0 =

xx xf

12 36b

If a solidly neutral grounded system with k = x0 x = x0 x is assumed, then xf = 1 + k xx

x+x

Dx =

1+k

2+k

12 36c

x+x

Accordingly, the height of the P–δ curve under the 1ϕG fault is (1 + k)/(2 + k) times that before the fault (4/5 times for k = 3). In the case of a high-resistance neutral grounding system, (x0⫽x’0) ∞, xf D(x) x + x’ so the height of the P–δ curve is not changed much from before the fault (see Table 16.2 [10]). 12.6.4

Double-Phase Opening Mode (2ϕOp)

Referring to Table 12.1 and Table 7.2 [2B], xf = f x2 + f x0 = x + x + x0 + x0 = 1 + α x + x D x = x + x + xf = 2 + α x + x

12 36d

where x0 + x0 = α x + x is assumed Then, the height of the P–δ curve under 2ϕOp becomes 1/(2 + α) times that before the fault. Then, in the case of two-phase open-mode, the peak value of the P-δ curve is 1/(2 + α) times, so if α = 3, the possible power transmission limit is only 20% of the normal condition. 12.6.5

Single Phase Opening Mode(1ϕOp)

Referring to Table 12.1 and Table 7.2 [1B], xf = f x2

f x0

= x+x

D x = x + x + xf = 1 +

x0 + x0 =

x + x x0 + x0 x + x + x0 + x0

x0 + x0 x + x + x0 + x0

12 36e

x+x

Then, the height of the P–δ curve under 2ϕOp becomes 1 1 + x0 + x0 x + x + x0 + x0 times that before the fault. If (x + x): (X0 + x’0) = 1 : 3, then xf = 3/4 and D(x) = 1.75(x + x’), so the peak value of the P–δ curve is 57% of that before the fault.

12.7

PQV Characteristics and Voltage Instability (Voltage Avalanche)

12.7.1

Apparent Power at the Sending Terminal and Receiving Terminal

Referring to the four-terminal circuit in Figure 12.8, the voltages and currents at both terminals are given by Eq. (12.30a), and the apparent power at the sending terminal is given by Eq. (12.32). We will calculate the apparent power at the receiving terminal. From Eq. (12.30a), 1 A i r = vs − vr B B

12 37a

313

314

12 Steady-State, Transient, and Dynamic Stability

𝜈s=Vs ∠𝛿

𝜈r=Vr ∠0°

Sr = Pr + jQr = vr i∗r

r + jx

s

r

(a)

ΔVr ΔPr

Gradient

Vr

The apparent power at the receiving terminal is given by

ΔVr ΔQr

Gradient

=

A

B

B

∗ ∗ vs vr −

∗

∗ ∗ vr vr =

1

A

B

B

− jδ − ∗ Vs Vr e

∗

2 ∗ Vr

12 37b

Next, we will derive the equations for apparent power in Figure 12.10a, where the line impedance between points s and r is z = r + jx. The four-terminal circuit equation is vs A = is C

Pr , Qr

(b)

1

B D

vr 1 = 0 ir

Z 1

vr ir

12 38

Z = r + jx

Figure 12.10 Voltage sensitivity characteristics ΔV/ΔQ, ΔV/ΔP.

Substituting the conjugates of A = D = 1, B = Z, C = 0 into Eqs. (12.30c) and (12.37b), the apparent power at the sending and receiving terminals, respectively, is derived as follows. The apparent power of the sending terminal: S s = Ps + jQs =

1 V 2 − Vs Vr e jδ r − jx s

Vs xV r sin δ + r Vs − Vr cosδ + j x Vs − Vr cos δ −rV r sin δ r 2 + x2 Apparent power of the receiving terminal 1 Vs Vr e −jδ − Vr2 S s = Pr + jQr = r − jx Vr = 2 2 xV s sin δ + r Vs cos δ− Vr + j x Vs cosδ − Vr − rV s sin δ r +x =

12.7.2

① 12 39

②

Voltage Sensitivity Characteristics with a Small Disturbance ΔP, ΔQ

We will examine the voltage sensitivity caused by a small disturbance of P, Q. From Eq. (12.39) ②, Pr + jQr r − jx = Vr Vs e −jδ −Vr

12 40a

∴ rP r + xQr + Vr2 + j rQr − xP r = Vs Vr e −jδ

We will find two equations by separating the real and imaginary parts of this equation and then eliminating δ from both equations: rP r + xQr + Vr2

2

+ rQr − xP r

2

= Vs2 Vr2

12 40b

Using Pr Pr + ΔPr, Vr Vr + ΔVr, then ΔVr/ΔPr can be calculated. ΔVr/ΔQr can be calculated similarly. (Refer to Supplement 2 at the end of the chapter for the derivation process.) Thus ∂Vr ΔVr x2 + r 2 Pr + rV 2r = = 2 ∂Pr ΔPr Vr Vs − 2 Vr2 −2 rPr + xQr

①

∂Vr ΔVr x2 + r 2 Pr + xV 2r = = ∂Qr ΔQr Vr Vs2 − 2 Vr2 −2 rP r + xQr

②

12 41

12.7 PQV Characteristics and Voltage Instability (Voltage Avalanche)

In this equation, Vr, Vs are real numbers of approximate value 1.0. Accordingly, both equations have negative values. In addition, we know that x r. Then 0>

∂Vr ∂Pr

∂Vr ∂Qr

12 42

This equation tells us that V is forced to decrease whenever P or Q increases, and the voltage sensitivity to Q is much larger than it is to P. Figure 12.10b shows the voltage-sensitivity characteristics given by the equation as the nature of the power system. Voltage V is more sensitive to Q than to P, and this can be explained by the fact that x r for the power system. 12.7.3

Circle Diagram for Apparent Power

We will examine the system shown in Figure 12.11, ignoring line resistance for simplicity. Using r = 0 in Eq. (12.39), Sr = Pr + jQr =

Vs Vr Vr Vs cosδ −Vr sin δ + j ① x x

Pr =

Vs Vr sin δ x

②

Qr =

Vr Vs cosδ − Vr x

③

12 43

The equation can be modified as follows, for the equation of the circle diagram in coordinates (Pr, Qr): Pr + j Qr + or Pr2 + Qr + or Pr2

Vr2 Vs Vr − jδ =j e ① x x 2

Vr2 x

Vs Vr x

= 2

k 2 Vs2 + Qr + x

kV 2s x

−

Center of the circle 0, − where k =

② 2

12 44

Vr2 Vs Vr , radius x x

③

Vr Vs Qr

Ps Qs

Pr Qr

s Vs ∠

r

jx

V k = r = 0.8 Vs 0.9

Load (power-factor cos 𝜑)

1.0 1.1 1.2

Rl

Vr ∠0

jXl Vs

jxI 𝛿

𝜑

Vr

Vr2 x Vs2

𝜑

Vs Vr x

x

I

(a)

(b)

Figure 12.11 Power circle diagram at the receiving terminal (line resistance ignored).

Pr

315

316

12 Steady-State, Transient, and Dynamic Stability

12.7.4

PQV Characteristics, and PV and QV Curves

Here we will introduce the load equation at the receiving terminal r, whose power factor is cos φ: S r = Pr + jQr = Sr cos φ + j Sr sin φ ① Qr = Pr tan φ, cos φ = Pr

12 45

Pr2 + Q2r ②

Substituting Pr, Qr from Eq. (12.43) into Eq. (12.45) ② and modifying it using formulae for trigonometric functions, Vr = Vs cos δ−tan φsin δ = Vs

cos φ + δ cosφ

12 46

Applying this equation to Eq. (12.43), either Vs or Vr can be eliminated, and we have the following equations. For the equation of Pr, Qr by Vs, Pr −Vs curve Vs2 cos φ + δ sin δ Vs2 = cos φ x 2x Qr − Vs curve

Pr =

Qr = Pr tan φ =

Vs2 2x

sin φ + 2δ −sin φ ① cos φ

sin φ + 2δ − sin φ tan φ cos φ

12 47

②

For the equation of Pr, Qr by Vr, Pr −Vr curve Vr2 cos φsin δ V2 = r x cos φ + δ 2x Qr − Vr curve

Pr =

Qr = Pr tan φ =

Vr2 2x

sin φ + δ − sin φ − δ cos φ + δ

①

sin φ + δ −sin φ −δ tan φ cos φ + δ

②

12 48

The receiving-point quantities Vr, Pr, Qr from Eqs. (12.46) and (12.47) are expressed as functions of variables Vs, δ, φ, where Vs = 1.0. Accordingly, Vr, Pr, Qr can be written with the parameters δ (the angular difference between the sending and receiving terminals) and φ (the angle of load power factor). In other words, the curved surface of PQV in threedimensional coordinates can be written with the parameters δ and φ. Also, the PV, QV, and PQ curves can be written as projections in two-dimensional coordinates. Figure 12.12 shows the curves of the PQV characteristics. On the PV curve in Figure 12.12a, if P gradually increases from zero under the load of the leading power factor (φ is negative), V tends to increase slowly. However, it then begins to decrease at roughly δ = 40 , and P reaches the critical point at the upper limit at roughly δ = 70 . If P gradually increases from zero under the load of the lagging power factor (φ is positive), V tends to decrease rather quickly, and P reaches the critical point at the upper limit at roughly δ = 45–70 . The QV curve in Figure 12.12b shows characteristics similar to Figure 12.12a. Incidentally, Figure 12.12a indicates that Eq. (12.48) gives two voltage solutions for one value of P: higher and lower solutions. In other words, the upper half of the PV curve includes the actual operating zone, and the lower half is meaningless. Also, Figure 12.12b indicates two voltage solutions for one value of Q: the actual stable voltage is given by the higher solution.

12.7.5

PQV Characteristics of Load

Now that we have discussed the PQV characteristics of the power system, we will examine the PQV characteristics of the load system. The load can be expressed using the parallel circuit of resistance Rl and reactance jXl (negative for the capacitive load), as shown in Figure 12.11a

12.7 PQV Characteristics and Voltage Instability (Voltage Avalanche)

Voltage at the receiving terminal cos (𝜑 + 𝛿) (Leading) Vr = Vs· cos 𝜑 30° 𝛿 = 7.5° 𝜑15° = – 45° 𝛿 = 0° – 30°

Angular difference 30° 𝛿 = 15°

45° 60° Operating𝛿 = 90° zone

Vs

Power-factor angle 𝜑 = – 45° – 30° (Leading power-factor)

Vs

15° 30°

– 15° 15° 30°

45°

Angular 60° difference 𝛿 = 75°

– 15° 0°

Operating zone

60°

𝜑 = 60° Qr leading Lagging Reactive power at the receiving terminal Qr = Prtan𝛿

Vs2 Pr 2x Effective power at the receiving terminal V 2 {sin (𝜑 + 2𝛿) – sin 𝜑} Pr = s · cos 𝜑 2x

(b) Q–V curve

(a) P–V curve Vr Vr

Light-load (Rl large) 1 2

2

Heavy-load (Rl small)

4

3

1

Leading

4

3

5 P-V curve of load

5

Lagging

P-V curve of power system 30°

Pr

15°

Power-factor angle 𝜑 = – 45° – 30° –15° 0 Pr

Qr Lagging

(d)

(c) P-Q-V curved-surface

P-V curve

Leading Ql

Ql small Vr ( –Xl large)

Ql small (Xl large) Lagging Ql Ql large (Xl small)

Ql large ( – Xl small)

Q-V curve of load (including reactors at the receiving station) Q-V curve of power system Qr

Qr

(e) Q-V curve Figure 12.12 PQV characteristics (PQV steady-state stability).

Pl + jQl = Vr Il∗ = Vr

Vr∗ 1 1 = Vr2 + ∗ Rl jX l Zl

Power factor cos φ =

①

Pl , Ql = Pl tanφ ② Pl + jQl

12 49

317

318

12 Steady-State, Transient, and Dynamic Stability

where the suffix l means load, and Pl = Ql =

Vr2 Rl Vr2

①

V r = Rl

Pl

V r = Xl

Ql

②

12 50

Xl

The voltage Vr is a function of the square root of Pl or Ql. This equation gives the PQV characteristics of the load system. The set of magnitudes Pl(t), Ql(t), Vl(t) at the intersecting point l and at time t gives a vector in PQV coordinates, and the arrowhead of the vector must be the intersecting point of the transmission-side PQV characteristics and the load-side PQV characteristics. In order to visualize the total PQV characteristics, let’s examine the phenomena in two-dimensional coordinates. 12.7.6

PV Mode Voltage Collapse (PV Avalanche)

The PV characteristics at the receiving point (Eq. (12.48)① ) and those of the load (Eq. (12.50)) are shown together in Figure 12.12d. Pr, Vr at the intersecting point are the actual operating power and voltage, respectively. If the load gradually increases ④ ⑤. Around point ④, the (under nearly constant Qr), the operating point moves from the load point ① to ② ③ voltage begins to decrease suddenly and rapidly, and before long Pr, Vr are forced to decrease or miss the critical point ⑤ (the maximum load point). In other words, if the system operates at point ④ and the load increases for some reason, exceeding the critical point ⑤, the stable operating point no longer exists, and the voltage and the power of the network are entirely lost. This phenomenon is usually called voltage collapse or PV mode voltage avalanche. Further, for Eq. (12.47), there are higher and lower solutions of V for a given value of P, and the real solution is obviously (P, V) for the higher solution V. The (P, V) zone of the lower solution V is the voltage-collapse zone and is also stable. So, the operational point should be kept within the critical point of the higher solution V. Suppose our power system is under operation at point ④ of Figure 12.7d; if the load increases a little (load curve 3 5), the operational point moves and exceeds point ⑤, and voltage collapse will inevitably result. Note that δ = 50–70 is a realistic stable operational limit as a result of this explanation. On the other hand, the operational limit of steady-state stability by definition is δ = 90 , as explained in Section 12.2. The reason for these two different explanations should be clearly understood. The explanation for the steady-state stability limit δ = 90 is true under the condition that the voltages Vs and Vr are kept within a realistic voltage band (say, 1 ± 0.05pu), which means enough Qs max, Qr max corresponding with Ps max, Pr max is put into system operation. Now, using δ = 90 in Eq. (12.43), we have Vs Vr x V2 Qrmax = r x Pr max =

12 51

The maximum power Pr max under δ = 90 can be transmitted only when this Qr max is supplied, where Qr max Pr max, because voltage Vs = Vr 1.0 should always be kept in this equation. However, installing such a huge amount of reactive power source equipment (typically reactors at large power-receiving substations) Q is economically unrealistic. In Figure 12.12e, the reasonable way to maintain maximum Pr under the condition of Vs = 1.0 ± 0.05 and δ ≤ 30–40 is to maintain the angle φ of the total power factor cos φ within 0 ± 10 at the receiving terminals. A reasonable number of reactive power-generation facilities (reactors and capacitors) are often installed at power-receiving key stations for this reason. 12.7.7

QV Mode Voltage Collapse (QV Avalanche)

The QV curve of the line at the receiving point (Eq. (12.48)②) and that of the load (Eq. (12.50)) are shown together in Figure 12.12e, where the reactive power of the reactors and capacitors for power-factor improvement purposes is included in the load curve. Qr, Vr at the intersecting point are the actual operating reactive power and voltage, respectively.

12.8 Generator Characteristics with an AVR

In Figure 12.12e, if the reactive power (lagging [+] or leading [−]) of the load gradually increases under constant P, for example, the intersecting point on the QV curve is lost in the very low-power-factor zone. This phenomenon is usually called voltage collapse or QV mode voltage avalanche. It must be stressed that not only the magnitude of the load (or the power flow, then δ), but also the power factor of the load cos φ affect the voltage stability limit. Also, this explanation in the two-dimensional plane is only to make understanding easier. The actual operating condition should be understood as the intersecting point of the threedimensional characteristics. With this in mind, the characteristics explained in Figure 12.12 should be called PQV steady-state stability characteristics and PQV collapse or PQV avalanche, respectively, because these are more exact expressions. Note also that the power system contains the smooth curved surface of the PQV characteristics in the three-dimensional coordinates of the V-axis, P-axis, and Q-axis, as shown in Figure 12.12c. On the other hand, the load at each receiving terminal also contains the PQV characteristics, but there may not be a smooth surface in the same three-dimensional coordinates because the load is the total composite characteristics of many individual unpredictable loads.

12.7.8

PQV Steady-State Stability

Again looking at Figure 12.12a, the gradient of the curve ∂Vr/∂Pr changes from plus (+) to minus (−) around δ = 40 . On the other hand, recall the explanation that the steady-state stability limit is given at δ = 90 in Figure 12.1 (for a non-salient-pole machine). We must clarify the difference. Let’s look again at Eq. (12.39)②, which shows the apparent power at the receiving point before omitting line resistance r. We can separate the real and imaginary parts and then eliminate δ to derive the following equation (refer to Supplement 3 at the end of the chapter for the process). For the power circle diagram (line resistance r is considered), 2

rV 2r xV 2 + Qr + 2 r 2 2 2 r +x r +x where Vs ≒ Vr ≒ 1 0, x r Pr +

2

=

Vs2 Vr2 r 2 + x2

①

Or omitting r2 in the denominators, Pr +

rV 2r x2

2

+ Qr +

xV 2r x2

rV 2 xV 2 with center − 2r , − 2 r x x

2

=

Vs Vr x

2

Vs Vr and radius x

②

12 52

The power circle diagram of the receiving terminal (Pr, Qr) considering line resistance is derived as shown in Figure 12.13, in which the diagram of the sending terminal is also shown. We can conclude the following from Eq. (12.51), Figure 12.13, and Figure 12.11:

• •

In order to increase Pr gradually without changing the voltage (k = Vr/Vs: fixed, say 1.0), the angular difference δ and Qr become larger, and before long much larger ΔQr is required (around δ = 30–35 ) for only a small increase in ΔPr. If Pr is gradually increased under fixed Qr, k = Vr/Vs has to become smaller; in other words, Vr must become smaller quickly under fixed Vs, which means voltage collapse.

This is yet another explanation of the characteristics given by Figure 12.1. In actual power system practice, operation under a condition of δ exceeding 40 is unrealistic, because the capacity of Qr is limited.

12.8

Generator Characteristics with an AVR

12.8.1

VQ Control (Voltage and Reactive Power Control) of Power Systems

The load characteristics of each load area are quite different, because they are related to the social and geographical environments of the area. Load naturally varies, so the actual load behavior may not be easy to express in simple

319

320

12 Steady-State, Transient, and Dynamic Stability

Q

Sending terminal

s 𝛿

90°

Ps + jQs 60°

O

P 𝛿 r

60° Pr + jQr 90°

Receiving terminal

Receiving terminal circle

(

Center point r : –

Radius : rO =

r Vr2 x

2

,–

x Vr2 x2

)

Vs Vr x

Figure 12.13 Power circle diagram (line resistance considered).

equational form. In addition, as a typical example, the load characteristics of an ordinary industrialized area may have the following features:

• •

At night, the magnitude of real power P consumption decreases dramatically, and the ratio of this load to the peak load from the same day may decrease to a value of one-half or even one-third. In addition, the power factor cos φ tends to be lower at night. This is because many facilities that have capacitive MVA (capacitors for power-factor improvement, cable stray capacities, etc.) are apt to be operated continuously without large changes despite the fact that facilities with real power MW decrease during this time.

In other words, the load quantities P, Q, cos φ (or angle φ) change widely over 24 hours, as a condition of the total network or local area networks. Furthermore, severe disturbances (faults, etc.) may occur unpredictably at any location. Under all these conditions, power systems must operate constantly, maintaining voltages within 1.0 ± 0.1 (typically) at any point and ensuring system stability. To control voltage V and reactive power Q appropriately over time and at all the locations of the power system is a very difficult engineering matter and is as important as total real power (Σ Pi) dispatching control (ALD) and frequency (f ) control (AFC). A generator’s stable operation is closely related to its AVR characteristics and the load condition. In this section, the operational characteristics of a generator with an AVR and its visualization in PQ coordinates are examined. Then, we will investigate the critical conditions of the generator’s stable operation. 12.8.2

eG Load AVR

Zload = R + jX

Figure 12.14 Model power system with one generator with an automatic voltage regulator (AVR) and load.

Generator Transfer Function

Our generator is running in combination with an AVR and supplying power to load Z, as shown in Figure 12.14. Whenever the generator terminal voltage e decreases a little, e e – Δe, the AVR detects the small deviation Δe and immediately increases the generator excitation Ef Ef + ΔEf so the terminal voltage is recovered quickly. This behavior must be examined as a problem of response characteristics with regard to the total system

12.8 Generator Characteristics with an AVR

including the generator, AVR, transmission line, and load. This will be investigated by applying the concept of transfer functions, which is a familiar tool in automation engineering theory based on the Laplace transform. In the Laplace transformation, s = d/dt is treated like an algebraic equation. We will start with the generator’s transfer function. We begin with the results of Eqs. (10.43)–(10.46) and Figure 10.6: Park’s theory for a generator, in Chapter 10. The equations are again written here as Laplace-transformed equations, while d/dt is replaced by symbol note s:

• •

ed s = − ψ q s sθ + sψ d s −rid s

①

eq s = + ψ d s sθ + sψ q s − riq s

②

Efd s = sψ fd s + rfd ifd s

③

ψ d s = − xd id s + xad ifd s + xad ikd s

④

ψ q s = − xq iq s + xaq ikq s

⑤

ψ fd s = − xad id s + xffd ifd s + xfkd ikd s

⑥

These are the general equations for a generator, which can be simplified as follows for practical reasons: The damper currents ikd(s), ikq(s) vanish within 0–3 cycles after the disturbance, so they can be ignored for the phenomena 3 cycles after the power-system disturbance. The generator operates in synchronization at a speed of 50/60 ± 0.05 Hz. Accordingly, the unitized angular velocity is 1.0: sθ = ω =

•

12 53

2π 50 ± 0 05 = 1 0 ± 0 001 ≒ 1 0 2π × 50

12 54

Flux linkages ψ d, ψ q are DC quantities at t = 0–, as explained in detail in Section 10.6 and Eq. (10.75), so the derivatives sψ d, sψ q (corresponding to transient DC current) can be ignored for the purposes of this chapter. Therefore, Eq. (12.53) is as simplified here: ed s = − ψ q s − rid s

①

eq s = + ψ d s − riq s

②

Efd s = sψ fd s + rfd ifd s

③

ψ d s = − xd id s + xad ifd s

④

ψ q s = − xq iq s

⑤

ψ fd s = − xad id s + xffd ifd s

⑥

12 55

Because we are studying phenomena of excitation control, Efd (s) must be treated as a function of s. Eliminating variables of flux linkage ψ, from ① ⑤ ed s = xq iq s −rid s

①

from ② ④ eq s = − xd id s + xad ifd s −riq s

②

from ③ ⑥ Efd s = −xad sid s + xffd s + rfd ifd s

③

12 56

321

322

12 Steady-State, Transient, and Dynamic Stability

Substituting ifd from Eq. (12.56) ③ into ②, eq s = =

xad x2ad s Efd s − xd − id s −riq s xffd s + rfd xffd s + rfd 1 xad 1 x2 xd + xd Td0 − ad s id s − riq s s − 1 + Td0 s rfd 1 + Td0 s rfd

x2 1 1 ef s − xd + xd − ad Td0 s id s − riq s ∴eq s = 1 + Td0 s 1 + Td0 s xffd xad Efd s , rfd

where ef s

Td0

① 12 57 ②

xffd rfd

The equation can be modified further. From Eq. (10.70), xd = xl +

1 1 1 + xad xfd

①

From Eq 10 72 , ②

xd = xl + xad

12 58

From Eq 10 50 , xfd = xffd −xad

③

∴ xd = xd − xad +

x2ad

xad xfd = xd − xad + xfd xffd

④

Then, replacing ( ) on the right side of Eq. (12.57) ② with xd , we obtain the transfer function of the generator as follows: ed s = xq iq s − rid s eq s =

xd + xd Td0 s 1 id s −r iq s ef s − 1 + Td0 1 + Td0 s

① ②

where xffd Td0 = time constant of the excitation circuit of the d-axis equivalent circuit rfd

12 59

see Figure 10 10 and Eq 10 107a xd = xd − ef s =

x2ad xffd

transient reactance of the d-axis circuit

xad Efd s excitation voltage xfd

The first term on the right side of Eq. (12.59) ② is the term that is changed proportionally by the excitation ef, and is a first-order time delay s function with time constant Td0 The second term on the right side of Eq. (12.59) ② is the term related to the armature current, and corresponds to the voltage drop caused by armature reactance. The second term obviously becomes xd id when t 0 + (s ∞), ∞ (s 0+). The result coincides with the explanations in Sections 10.6 and and finally becomes xdid when t Section 10.7.

12.8.3

Transfer Function of a Generator Plus Load

Next, the generator is connected to the load impedance Z = R + jX, as shown in Figure 12.14, which already includes line impedance. Also referring to Eq. (10.55b) or Figure 10.7,

12.8 Generator Characteristics with an AVR

①

e2d s + e2q s

eG s =

ed s + jeq s = id s + jiq s

②

R + jX

12 60

∴ id s =

R X ed s + 2 eq s R2 + X 2 R + X2

③

iq s =

R X eq s − 2 ed s 2 R + X2 +X

④

R2

By substituting id(s), iq(s) into Eq. (12.59) ①②, we obtain simultaneous equations of the first order and two variables ed(s), eq(s), and the following solution (see Supplement 4 at the end of the chapter for details): X + xq R − X R + r

ed s =

X + xd X + xq + R + r

2

+

X + xd X + xq + R + r

2

Td0 s

X X + xq + R R + r

eq s =

X + xd X + xq + R + r

2

+

X + xd X + xq + R + r

2

Td0 s

ef s

① 12 61

ef s

②

Substituting the results into Eq. (12.60) ① (see Supplement 5 for the solution), eG s =

A ef s 1+T s

①

eG s A = the transfer function of the greater terminal voltage 1+T s ef s

GG s

②

under load condition where A=

T=

X 2 + R2

X + xq

2

+ R+r

X + xd X + xq + R + r X + xd X + xq + R + r

2

X + xd X + xq + R + r

2

ef s =

xad Efq s rfd

12 62

2

2

the gain of generator + load

Td0 the time constant of generator + load

excitation

③

④ ⑤

Equations (12.61) and (12.62) explain that the total system of a generator plus load has the following features. The value of the time constant T is affected by the load condition (R, jX) and is smaller than the generator’s specific time constant Td0 , T < Td0 , because xd > xd in Eq. (12.62)④. Incidentally, the armature resistance r can be ignored in these equations, as we will now examine further. 12.8.4

Transfer Function of a Generator under Special Load Conditions

Case 1 No-Load Condition Using R = ∞, X = ∞, then A = 1, T = Td0 ∴ GG s =

1 1 + Td0 s

12 63

323

324

12 Steady-State, Transient, and Dynamic Stability

Case 2 Load with Power Factor 1.0 (cosφ = 1.0) Using X = 0, r = 0, R ∴GG s =

xd xq + R2 x xq + R2 1+ d T s xd xq + R2 d0

For a light load R GG s =

x2q + R2 12 64

xd, xq, then

1 1 + Td0 s

12 65

Case 3 Inductive Load with Power Factor Zero: (cosφ = 0) Using R = 0, X X + xd ∴ GG s = X + xd 1+ T s X + xd d0 For an inductive light load, X

12 66

xd , xd , so GG(S) is the same as in Eq. (12.65).

Case 4 Capacitive Load with Power Factor Zero (cosφ = 0) In this case, GG(s) is equal to Eq. (12.66), but X has a negative magnitude X = Xc. The typical case for this mode is a transmission line charged by a generator with no-load condition. If the absolute value of Xc is close to the value of xd, an abnormal solution of (X + xd) = (−Xc + xd) 0, A ∞, T ∞ occurs. Such an extraordinary large gain A means an unstable system condition from the viewpoint of automation theory. Of course, this condition physically means very unstable series resonance. This phenomenon is discussed further in Section 12.9.3. Figure 12.15 shows the transfer function of a generator and the behavior when a step-function signal ef (t) = 1(t) (ef(s) = 1/s) is input. 12.8.5

Duties of an AVR

We have found the transfer function of the generator plus load impedance. Now we will find the total transfer function of the AVR plus exciter plus generator plus load impedance, by which we can investigate the dynamic behavior of the generator system with an AVR. One-order s-function ef (s)

A 1 + Ts

G(s) =

eG(s) eG (s) =

A 1 + Ts

. 1s

ef (t) = 1(t) A T

t eG(t) = A(1 – e–(t/T ))

A 0

t

Figure 12.15 Transfer function of first-order time delay.

The inclination of the auxiliary line (tangential line at t = 0) is d e (t) dt G

t=0

=

A T

e–(t/T )

t=0

Accordingly, one side of the derived triangle becomes T

=

A T

12.8 Generator Characteristics with an AVR

• •• •• •• •• ••

The duties of an AVR are outlined as follows: To maintain generator terminal voltage within a permissible band around the rated voltage and to prevent generator operation in the unstable zone or in the prohibited operating zone. To generate appropriate var(Q) in relation to effective power (P) generation (VQ control and PQ control). To control voltage and reactive power automatically and maintain stability during large disturbances. To enlarge the generator’s stability limit, especially at small GD2 or short-circuit ratio. To improve transient and dynamic stability. To prevent hunting or unnecessary swinging of voltage and reactive power among generators. To make the most of the generator’s capabilities including all of these functions. The major requirements for an AVR are outlined as follows: High sensitivity, small offset (small dead-band) High response (small time constant) Fine controllability Wide control area, etc.

AVRs installed at every generator terminal of a power system play leading roles in power-system operation, in order to protect individual generators from prohibited operation, or to maintain stable power-system operation over time by avoiding unstable conditions among other generators, such as loss of synchronization or unnecessary hunting phenomena of voltages, cross-currents, and reactive power. If an AVR were categorized as a generator accessory, serving only to maintain its terminal voltage at a certain value, this would be a serious underestimation. 12.8.6

Transfer Function of a Generator Plus an AVR

We will study the response characteristics of the AVR and the exciter first, and then examine the dynamic characteristics of a generator with an AVR. Incidentally, the AVR is treated in this book only from a functional viewpoint, not a hardware one. For the transfer function of the exciter, Gf (s), the excitation system of a generator is typically classified as follows:

•• •

DC generator type (self-excitation method, separate excitation method) AC generator type with a rectifier circuit Solid-state type (by semiconductors for power use)

Regardless of the type of excitation circuit, the transfer function of the exciter can be assumed as a typical form of firstorder delay for our purposes of power-system behavior, although the hardware of individual excitation circuits may be different. The time constant Tf is on the order of, say, 0.1 second, and is usually smaller than the time constant Td0 of the paired generator. For the transfer function of the AVR Gavr(s), the AVR is a closed-loop control system with a negative feedback pass in which the generator terminal voltage eG is stabilized to a certain value. Figure 12.16 shows the block diagram of a total system including a generator, an exciter, and AVR equipment written as a combination of transfer functions. In the AVR

Voltage setting value

+

Vset

–

+ –

GeneratorAVR signal output Excitation voltage terminal voltage eG (s) (s) e (s) e f avr 𝜇a 𝜇f A Gavr(s) = GG(s) = Gf (s) = 1 + Tavr s 1 + Ts 1 + Tf s

Gk (s) =

ks 1 + Tk s

For hunting prevention (the derivative characteristics)

Figure 12.16 Transfer function of a generator with an AVR connected to the outer load.

325

326

12 Steady-State, Transient, and Dynamic Stability

figure, Vset is a value set by the AVR for the generator terminal voltage. The generator voltage eG (s) detected by the potential transformer (PT) is fed directly back to the AVR. The transfer function of the forward element Gavr (s) of the AVR can also be written in the form of a first-order time delay with gain μa and time constant Tavr. The AVR also has a duty to prevent hunting, so the supplementary negative feedback function Gk (s) is equipped with the Laplace derivative s, with which quick follow-up is possible for small fluctuations of voltages and current. Further, another supplementary function to minimize cross-currents from neighboring generators will be examined in Section 12.10.1. Advanced AVR equipment in recent years is mostly solid-state (digital) and equipped with outstanding characteristics such as high sensitivity and quick response times, where the time constant Tavr of the AVR itself is very small. Accordingly, the actual values of each time constant can be estimated as T, Tf Tavr. From Figure 12.16, eG s = GG s ef s

①

ef s = Gavr s Gf s Vset − eG s −Gk s ef s

② 12 67

From ① ② ef s =

Gavr s Gf s Vset 1 + Gavr s Gf Gk s + GG s

③

Then the total transfer function is eG s = GG s ef s =

Gavr s Gf s GG s Vset 1 + Gavr s Gf s Gk s + GG s

①

where A transfer function of generator in relation to load 1+T s μf ≒ μf transfer function of exciter Gf s = 1 + Tf s μa ≒ μa transfer function of AVR forward element Gavr s = 1 + Tavr s ks Gk s = transfer function of feedback circuit to prevent hunting ≒ ks 1 + Tk s GG s =

② ③

12 68

④ ⑤

Simplification of Gavr(s), Gf (s), Gk(s) in Eq. (12.68) is justified by the difference in time constants, by T, Tf Tavr. In addition, we can use k = 0 because the gain of Gk(s) is set by a very small value. Accordingly, for the total transfer function, eG s ≒

=

μa μf GG s μa μf GG s Vset ≒ Vset 1 + μa μf ks + GG s 1 + μa μf GG s A

A + 1 μa μf Atotal

1 T 1+ s 1 + μa μf A

Vset

Atotal

1 Vset 1 + Ttotal

12 69

Ttotal

In conclusion, the dynamic characteristics of a generator with an AVR can be written as the transfer function of a generator with AVR given by Eq. (12.69), which is in the form of a first-order delay mode with gain Atotal and time constant Ttotal. Now, the new gain Atotal has replaced generator-specific gain A, and new time constant Ttotal has replaced generatorspecific time constant T by using AVT. If the AVR gain μa is set larger, Ttotal can be set smaller, while Atotal remains almost the same value as A. If we use s 0 in the equation (i.e. t ∞ in the time domain), then we find eG Vset, which means the generator terminal voltage eG (t) is controlled to maintain the AVR’s set value Vset. Recall that, in Eq. (12.69), only the value of μa is controllable; all other symbols have the uncontrollable specific values of the actual machines.

12.8 Generator Characteristics with an AVR

We can introduce the following conclusion from the observation of Atotal and Ttotal in Eq. (12.69): if μa is set to a large value, the total system time constant Ttotal can be set to a much smaller value (almost to zero) in comparison with the generator’s specific time constant T, while the total system gain Atotal will not differ greatly from the generator’s specific gain A. In other words, due to the AVR, the time constant T can be minimized so quick recovery of generator terminal voltages eG(t) against ordinary fluctuations as well as against large system disturbances can be realized, maintaining eG(t) at its set value Vset of the AVR over time. Note that generator operation without an AVR (in other words, manual operation of excitation) is impossible. Without an AVR, the generator would be forced to rush into the specifically prohibited voltage operating zone, or (p + jq) coordinate zone; or serious voltages and reactive power hunting phenomena would be caused among other generators, regardless of their physical distance from our generator, which might lead to mechanical breakdown or damage to the generators. Furthermore, various modes of power-system instability might arise. These problems are discussed in detail in Section 12.7. In Figure 12.16, the special case where the gain of the AVR is set to zero (μa = 0) corresponds to the case when the AVR is out of service and the total transfer function consists of only Gf (s) and GG(s), although such operation is not possible. Incidentally, if we do not ignore the gain k of Gk(s) to prevent hunting, the equation is written as a second-order function, instead of Eq. (12.69). 12.8.7

Transfer Function of the Total Power System Including an AVR and Load

Our next task is to find conditions in which a generator with an AVR can be stably operated. Note, however, that the load condition must be taken into account for this purpose, because we know that the operating characteristics of a generator are closely affected by the load condition. Therefore, we will examine the behavior of a generator with an AVR in the power system shown in Figure 12.17a, where the load is expressed as a parallel circuit of Rl, jXl (Xl is negative for capacitive loads). The relation between the generator terminal eG, iG and the load impedance Z(Rl//jXl) is ① iG = Z −1 eG 1 1 = Rl−1 −jX l− 1 ② Z −1 = + Rl jX l

12 70

where the impedance of the transmission line is already contained within the load impedance. As the system is operated under balanced three-phase conditions, the following equation can be derived from Eqs. (10.55) and (12.69): id + jiq e jt = Z −1 ed + jeq e jt ∴

12 71

id + jiq = Z −1 ed + jeq

During a system disturbance, the quantities will suffer some deviations, such as ΔiG = Rl− 1 −jX l− 1 ΔeG

ΔeG =

∴Δid + jΔiq = Rl− 1 −jX l− 1

1 Rl−1 −

jX l− 1

Δed + jΔeq

ΔiG ① ②

Δid = Rl−1 Δed + Xl− 1 Δeq Δiq = −Xl− 1 Δed + Rl− 1 Δeq

12 72

③ ④

On the other hand, we can derive the following equations from Eq. (12.59) ①② and Eq. (12.60) ① (using armature resistance r = 0): Δed = xq Δiq Δeq = ΔeG =

①

1 xd + xd Td0 s Δef − Δid 1 + Td0 s 1 + Td0 s Δed

2

+ Δeq

2

② ③

12 73

327

328

12 Steady-State, Transient, and Dynamic Stability

Figure 12.17 Operating limit of the generator under weak excitation (leading power-factor operation).

Next, referring to Figure 12.16, the following equation is derived as the s function of the system including an AVR, here the subsidiary feedback function Gk (s) is ignored for simplicity, to prevent hunting (k = 0): ef s = Vset −eG s Gavr s Gf s ∴ Δef s = − Gavr s Gf s ΔeG s μ

① − μΔeG s

②

12 74

μa μf ≒ Gavr s Gf s total gain of AVR and exciter ③

Eliminating Δiq from Eqs. (12.72)④ and (12.73) ①, xq−1 + Xl− 1 Δed = Rl−1 Δeq

∴ Δed = xq−1 + Xl− 1

−1

Rl−1 Δeq

12 75

Substituting this equation into Eqs. (12.72) ③④ and (12.73) ④, Δid = Rl− 2 xq− 1 + Xl−1 + Xl−1 Δeq

①

Δiq = − Xl− 1 Rl−1 xq− 1 + Xl− 1 + Rl−1 Δeq

②

ΔeG =

xq−1 + Xl− 1

= xq− 1 + Xl−1

Rl− 1 Rl−1

2

2

12 76

+ 1 Δeq

+ xq−1 + Xl−1

2

Δeq

③

12.8 Generator Characteristics with an AVR

Eliminating Δef by substituting Eq. (12.74) ② into (12.73) ②, 1 + Td0 s Δeq + Gavr s Gf s ΔeG + xd + xd Td0 s Δid = 0

12 77

Substituting Δid, ΔeG from Eqs. (12.76) ① ③ into Eq. (12.77), then all terms including Δeq vanish, and the following equation is obtained: 1 + Td0 s + Gavr s Gf s xq−1 + Xl− 1

2 Rl− 1

2

Rl−1 −1

Rl− 2 xq−1 + Xl−1

+ xd + xd Td0 s ∴Gavr s Gf s

−1

2

+ xq− 1 + Xl− 1

+ xq−1 + Xl− 1

+ Xl−1 = 0

12 78

+ Xq− 1 + Xl−1 + xd Xl− 1 xq− 1 + Xl−1 + Rl−2

xq−1 + Xl−1 + xd Xl−1 xq−1 + Xl−1 + Rl− 2

+ Td0 s

2

=0

That is, 1 Rl

Gavr s Gf s

2

1 1 + x q Xl

+

2

+

1 1 xd 1 1 xd + + 2 + + xq Xl Xl x q Xl Rl

12 79

1 1 x 1 1 x + d + d2 = 0 + + xq Xl Xl x q Xl Rl

+ Td0 s

On the other hand, the relation between the generator’s apparent power and the load impedance Z (parallel impedance Rl//jXl) is P + jQ = eG i∗G = eG

eG eG + Rl jX l

∗

=

e2G e2 +j G Rl Xl

12 80

Unitizing P, Q by the base value e2G , p + jq =

P Q 1 1 +j 2 = +j e2G eG Rl Xl

Now, Eq. (12.79) can be expressed as follows by replacing the symbols 1/Rl p, 1/Xl q. For the s function of the total system, generator + exciter + AVR + load, under the load condition (p + jq), p2 + q +

Gavr s Gf s + Td0 s xd

1 xq

2

+ xd p2 + q +

1 p2 + q + xd

1 q+ xq

1 xd

q+

1 xq

12 81

=0

This is the Laplace-transformed equation (in PQ coordinates) for the operating condition for the total system, which consists of a generator, an exciter, and an AVR plus load, as shown in Figure 12.17a. The necessary condition for this equation to be stable can be found with the help of the method for determining system stability within the essential theory of automation. Here we ignore the time lag of AVR and the exciter, using Tavr = 0, Tf = 0 (which has no consequence for stability analysis): Gavr s Gf ≒ μa μf

μ the total gain of AVR and exciter

12 82

Accordingly, Eq. (12.81) can be modified in the form of a first-order s function: A + BTd0 s = 0 with one root only and s = − A BTd0 here A = μ

p2 + q +

1 xq

B = Td0 xd p2 + q +

2

+ x d p2 + q + 1 xd

q+

1 xq

1 xd

① q+

1 xq

② ③

12 83

329

330

12 Steady-State, Transient, and Dynamic Stability

12.9

Generator Operation Limit With and Without an AVR in PQ Coordinates

Our purpose here is to find the critical condition from Eq. (12.83) where the generator stable operation become critical. The method for determining system stability as in the theory of automation indicates that the first-order s function in the form of Eq. (12.83)① can be stable only when the root of s is a negative real number (Nyquist stability criterion). Accordingly, Condition of system stability A B≧0 Boundary condition of the stable condition

12 84

A = 0, B = 0 Consequently, the critical borderline for stable operation can be given by using A = 0, B = 0 in Eq. (12.83) ②③. 12.9.1

Generator Operation without an AVR

If the AVR is out of service, so that μa = 0, μ = μa μf = 0 in Eq. (12.74), then the equation of the critical stability limit is A = p2 + q +

1 xd

q+

1 =0 xq

①

B = p2 + q +

1 xd

q+

1 =0 xq

②

12 85

The stable condition A B≧0

③

Equation (12.85)①② can be written as in the pq coordinates diagram shown in Figure (12.17b). Equation (12.85)① is the circle whose diameter is given by the straight line connecting points (0, − 1/xd) and (0, − 1/xq), and the outer area of the circle is A≧ 0 On the other hand, Eq. (12.85)② is the circle whose diameter is given by the straight line connecting 0, −1 xd , 0 − 1 xq The outer area of the two circles satisfies the stable condition AB ≧ 0, while the area of − 1 xd ≧ q is unrealistic. Accordingly, for the realistic stable area, q > −1 xd

12 86a

That is, the upper area from the small circle in Figure 12.17b. 12.9.2

Generator Operation with an AVR

Eq. (12.83) ② with the condition A = 0 and Eq. (12.83) ③ with the condition B = 0 give the critical stability limit. These conditions can also be written in PQ coordinates as shown in Figure 12.17c. The larger circle at the bottom is given by B = 0, which is the same circle as that in Figure 12.17b. The smaller distorted circle at the top is given by the equation A = 0 and can be written in PQ coordinates with the parameter μ(>0). The case with μ = 0 coincides with case 1: generator operation without an AVR. If the parameter μ (the total gain) is made larger, the distorted circle shrinks and finally converges to the special point (0, − 1/xq). The value of μ at this special converged point can be calculated using the conditions p = 0, A = 0, B = 0, and is given by the equation xd μ = −1 12 86b xq Now, comparing Figures 12.17b and c, if the AVR gain μa is adjusted to larger values (μ = μaμf also becomes larger), the upper distorted circle shrinks toward the converged point, and as a result the stable operating zone in the capacitive area (q is negative) is enlarged to the point (0, − 1/xq). In other words, an AVR can enlarge the leading power-factor zone of the generator from the critical point (0, − 1/xd) to (0, − 1/xq). From an actual engineering viewpoint, the operating zone of hydro-generators can be greatly improved (perhaps by 40–50%) by virtue of xd > xq, but that of thermal generators can be only moderately improved because xd ≒ xq (refer to Table 10.1). The problem of the generator’s leading power-factor operation was discussed in Chapter 11.

12.9 Generator Operation Limit With and Without an AVR in PQ Coordinates

12.9.3

Charging Transmission Lines with a Generator with and without an AVR

We will examine the special case of p = 0, which means a generator operating with pure inductive (L) or capacitive (C) load, with power-factor zero. A typical example is a no-load transmission line being charged by a generator: one terminal of the line is opened, and the other terminal is charged by the generator, so the stray capacitance of the line becomes the pure capacitive load of the generator. Equation (12.81) becomes very simple by using p = 0: μ + xd q +

1 xd

1 s=0 xd

+ Td0 xd q +

12 87

For the system to be stable, the root of s should be a negative real number, for the stability condition of line charging: 1 xd ≦0 1 q+ xd

μ + xd q + s= − Td0 xd

∴ μ + xd q +

1 xd

① 12 88

q+

1 ≧0 xd

②

12.9.3.1 Charging Lines with a Generator without an AVR

This is the case where μ = 0, so the stable condition is 1 realistic stable zone xd 1 q≦ − unrealistic zone xd

q≧ −

① 12 89 ②

Equation ① is the actual stable condition, while ② has physically no meaning. Therefore, The case of capacitive load q < 0

critical point of stable operation is q = − 1 xd pu

The case of inductive load q > 0

stable for the entire inductive zone

12.9.3.2 Charging Lines with a Generator with an AVR

This is the case where μ 0. Equation (12.88) ② is a second-order inequality of q, so it can be solved. The enlarged stable zone of line charging is q≧

1 1 1 − 2 xd xd

2

−μ

1 1 1 1 + − xd 2 xd xd

12 90

The absolute value of the left side of Eq. (12.90) is larger than 1/xd of Eq. (12.89), and the special case of μ = 0 in the former equation gives the latter. Thus, due to the effect of the term μ/xd with the AVR, the stable limit of leading powerfactor operation in PQ coordinates is enlarged in the downward direction in Figure 12.17b. Trial calculation: For the case of generator reactance xd = 1.8 pu, xd = 0 4 pu Then, Operation without an AVR using Eq. (12.89) is q ≥ −0 556 and operation with an AVR using Eq. (12.90) is q ≥ − 0 863

for gain μ = 0 5

q ≥ − 0 904

for gain μ = 1 0

Obviously, the stable zone of the generator’s capacitive load operation is dramatically improved with the AVR.

331

332

12 Steady-State, Transient, and Dynamic Stability

The theme of charging transmission lines is again discussed in Chapter 15, Section 15.12 as a problem of overvoltage phenomena and insulation coordination.

12.10

VQ (Voltage and Reactive Power) Control with an AVR

12.10.1

Reactive Power Distribution for Multiple Generators and Cross-Current Control

Two generators are operating in parallel, as shown in Figure 12.18, and with the same operating output (P1 + jQ1) – (P2 + jQ2). Balance of the effective power distribution P1 = P2 is to be maintained by controlling the mechanical input volume from the prime mover. Balance of the reactive power distribution Q1 = Q2 is to be maintained by controlling the generator excitation, and each AVR plays a role. Assuming that the reactive power of generator 1 is suddenly increased from Q1 Q1 + ΔQ for some reason, generator 2 will then decrease from Q2 Q2 – ΔQ. This represents unnecessary hunting phenomena. Assuming a very small error difference in the AVR set values Vset1 and Vset2, one generator may run up to the lagging power-factor zone, while the other is forced to run down to the leading power-factor zone. The countermeasure to solve these problems is also provided by an AVR, the function of cross-current compensation.

Figure 12.18 An AVR and the principle of voltage detection.

12.10 VQ (Voltage and Reactive Power) Control with an AVR

In Figure 12.18a, the two generators are operating under the condition that the total power is constant, (P1 + jQ1) + (P2 + jQ2) = constant. The currents of generators 1 and 2 are i1(t), i2(t), and the averaged current is iav(t). Thus eG = Efd1 − i1 jxd = Efd2 − i2 jxd i1 + i2 iav the averaged current of i1 , i2 iav = 2 i1 −i2 Δi i1 −iav = 2 i2 −i1 −Δi i2 −iav = 2 Δi the difference between i1 , iav or between i2 , iav

12 91a

12 91b

Accordingly, i1 = iav + Δi i2 = iav − Δi i1 − i2 1 Efd1 −Efd2 1 ΔE = = Δi = 2 2 jxd 2 jxd

12 91c

Δi is called the cross-current between generators 1 and 2. The cross-current Δi can be explained as the current induced by the difference in the excitation voltages of 1 and 2, ΔE = Efd1 − Efd2. The currents i1, i2 have measurable values, so the averaged current iav is also measurable. Then, the excitation of each generator can be continuously controlled, i1 iav, i2 iav (i.e. to minimize the cross-current Δi 0), with high sensitivity and rapid response time, and as a result equal reactive power generation Q1(t) = Q2(t) is realized, and hunting can be prevented. This is the principle of cross-current compensation control with an AVR. The hunting-prevention circuit in the block diagram in Figure 12.16 is the supplementary feedback circuit of the AVR to realize a rapid response. Figure 12.18 shows the principle of cross-current control with an AVR, including a typical signal-detection method. Each AVR produces positive-sequence quantities vr(t) – jkir(t) at its signal-detecting circuit (as shown in (b)) and controls the excitation Efd using the AVR’s dropping characteristics (as shown in (c)). Accordingly, cross-current can always be minimized to zero. This principle can also be applied to a system with two parallel operating generators with different boiler–turbine– generator (BTG) units in the same plant, or for two generators belonging to a common boiler unit. In addition, this principle can be applied with a small modification to other operational modes, as follows. In the case of two generators with different reactive power-sharing ratios, operation by Q1:Q2 1:α is realized using the excitation control of each generator: Generator 1 Generator 2 where

1 iav α+1 α i2 iav α+1 i1 + i2 iav = 2

i1

In the case of n generators with equal reactive power generation, operation by Q1 : Q2 : generator k, ik

iav = i1 + i2 +

+ in n

12 92

: Qn = 1 : 1 :

: 1 is, for 12 93

There are cases where one thermal boiler unit is combined with two or more turbines and multiple generators that are electrically operated in parallel. Of course, appropriate excitation control by the same principle can be realized for a variety of BTG units, although the actual functional circuit of the AVR becomes rather complex. The application of various BTG structures has been realized due to recent trends toward large capacity and higher total efficiency as well as for various types of fuel. In fact, 7 to 10 generators form a single thermal unit in the case of a modern advanced combined cycle (ACC) thermal-generation unit, which is a combined system of gas turbines and steam turbines applied to the system and fueled with liquefied natural gas.

333

334

12 Steady-State, Transient, and Dynamic Stability

12.10.2

Pf Control and VQ Control

Each generator has its own dynamic equation, Eq. (11.36). By adding the equations of all the generators operating in synchronization, the following equation is derived for the total power system: Pmi − i

Pei = i

i

where ω1 = ω2 =

M1 Mi dωi d2π f i = ≒ dt ωi dt 2π f

Mi i

f

df dt

12 94

≒ ωaverage = 2πf

This equation indicates that system frequency can be maintained by controlling the total mechanical input of the prime movers i Pmi to meet the electrical load i Pei over time in the total power system. This is the duty of power dispatching, and automatic frequency control (AFC) with system frequency can be maintained within permissible limits. How to distribute total generating power to each generator (dispatching) does not matter in this case, although power generation dispatch (effectively distributing generating power among numbers of generators) is important from the viewpoint of generation economy, economic load dispatch (ELD), and the best fuel combination policy (fuel selection policy: hydro, coal, oil, natural gas, nuclear, various renewables). And further, current flow (A) of all the individual branched lines should always be strictly checked to maintain it within the current limit; otherwise, the overloaded line may be tripped by an overcurrent relay (OC-relay), and such a careless trip may cause a cascade trip of the other lines. Power-distribution control among multiple generating plants should be conducted within such conditions. The central dispatching center is responsible for all of them and commands the generation capacity to all the generation units through automatic load dispatch (ALD). On the other hand, V and Q must be controlled to maintain balance not only in the total power system, but also in individual local subsystems, because voltages have to be maintained within allowable bands at each part of the network. In other words, VQ control of the network must be individually locally dispersed. Accordingly, an automatic reactive/ capacitive power-control system (AQR) must be installed for major generating plants as well as for major substations, as the local area control system. In the generating plant, the AQR system controls the reactive power (or the terminal voltage) of each generating unit by controlling Vset (the set value) of the AVR. In the major receiving substations, the AQR system controls the operating capacity of reactor/capacitor banks as well as (on-load tap-changing transformers (LTC-Trs) to maintain the regional voltages and reactive power balance. An AQR installed at an individual generating unit is typically equipped with the function to select an operation mode as follows. Time-scheduled operation of Q(MVA) Constant power-factor operation Functional operation of P, Q = f(P) Var dispatching operation from the dispatching center An AQR installed at key substation also has the function of selecting similar modes.

•• ••

Supplement 1 Derivation of Circular Diagram Eq. (12.17) Using A and B as follows in Eq. (12.15), A=

Ef xq + xl

B=

E bus xq + xl

1

we obtain Pgmax = AB xq + xl

2

Qgmax = A2 xl −B2 xq

3

12.10 VQ (Voltage and Reactive Power) Control with an AVR

e2gen = A2 x2l + B2 x2q

4

From Eqs. (3) and (4), A2 =

e2gen + Qgmax xq xl xq + xl

B2 =

e2gen − Qgmax xl

5

xq xq + xl

Substituting Eq. (5) into Eq. (3),

P2gmax

=

e2gen −Qgmax xl

e2gen + Qgmax xq xq xl

= − Qgmax −

1 1 1 2 e − 2 xl xq gen

2

+

1 1 1 2 e + 2 xl xq gen

2

Thus Eq. (12.17) has been derived.

Supplement 2 Derivation of ΔV/ΔP, ΔV/ΔQ Sensitivity Equation (Eq. (12.41) from Eq. (12.40b)) In Eq. (12.40b), replacing Pr Pr + ΔPr, Vr Vr + ΔVr and Pr2 Pr2 + 2Pr ΔPr , Vr2 Vr2 + 2Vr ΔVr by omitting the secondorder terms (ΔPr)2, (ΔVr)2, we get the following two equations: ① The original equation is A2 + B2 = Vs2 Vr2 ② The derived equation is A + r ΔPr + 2Vr ΔVr 2 + B −x ΔPr 2 = Vs2 Vr2 + Vs2 2Vr ΔVr where A = rP r + xQr + Vr2 , B = rQr −xP r Substituting ① into ② and again omitting the second-order terms (ΔPr)2, (ΔVr)2, rA− xB ΔVr = Eq 12 41 ① = ΔPr Vr Vs2 −2A Equation (12.41)② can be derived analogously. Note that these calculations are mathematically a partial derivation of the implicit function.

Supplement 3 Derivation of Power Circle Diagram Eq. (12.52) From Eq. (12.39)②, rV 2r Vs Vr = x sin δ + r cos δ ① + x2 r 2 + x2 xV 2 Vs Vr Qr + 2 r 2 = 2 2 x cos δ −r sin δ ② r +x r +x

Pr +

r2

Then, the following equation is derived from ① 2 + ②2: Pr +

rV 2r 2 r + x2

2

+ Qr +

xV 2r 2 r + x2

2

=

Vs2 Vr2 r 2 + x2

12 52

335

336

12 Steady-State, Transient, and Dynamic Stability

Supplement 4 Derivation of ed(s), eq(s) as Functions of ef(s) Eq. (12.61) The calculation takes some time but can be solved. For the first step, substituting id(s), iq(s) from Eq. (12.60) into Eqs. (12.59) ① ②, we can derive simultaneous equations for ed(s), eq(s): R2 + X 2 + Xxq + Rr ed s + Xr −Rxq eq s = 0 Rxd −Xr + Rxd − Xr Td s ed s + 2

2

R2 + X 2 + Rr + Xxd

+ R + X + Xxd + Rr Td0 s eq s = ef s

2

R +X

1

2

The equations can be written as follows: m1 ed s + m2 eq s = 0 m3 + m4 Td0 s s ed s + m5 + m6 Td0 s s eq s = m7 ef s The solution is ed s =

m2 m7 ef s m2 m3 −m1 m5 + m2 m4 −m1 m6 Td0 s s

eq s = −

m1 ed s m2

where m1 = R2 + X 2 + Xxq + Rr = X X + xq + R R + r m2 = Xr −Rxq = X R + r −R X + xq m3 = Rxd −Xr = − X R + r + R X + xd m4 = Rxd −Xr = − X R + r + R X + xd m5 = R2 + X 2 + Rr + Xxd = R R + r + X X + xd m6 = R2 + X 2 + Rr + Xxd = R R + r + X X + xd m7 = R2 + X 2 m1 to m6 have been modified into the form shown on their right sides, leading to the following results: m2 m3 − m1 m5 = − X 2 + R2

X + xd X + xq + R + r 2

m2 m4 − m1 m6 = − X 2 + R2

X + xd X + xq + R + r 2

m2 m7 = − X 2 + R2 xq R− Xr Finally, we obtain Eq. (12.61).

Supplement 5 Derivation of eG(s) as a Function of ef(s) Eq. (12.62) Equation (12.62) is derived by substituting ed(s), eq(s) from Eq. (12.61) into Eq. (12.60)① . However, the following modification can be used in the process: xq R− Xr = −X R + r + R X + xq Accordingly, {numerator of Eq. (12.61)① }2 + {numerator of Eq. (12.61)②}2 = X 2 + R2

X + xq

2

+ R+r

which leads easily to Eq. (12.62).

2

337

13 Induction Generators and Motors (Induction Machines) 13.1

Introduction to Induction Motors and Generators

Motor drives are used in a very wide power range, from a few watts to many thousands of kilowatts, in various precise applications to control power, torque, position, direction, speed, acceleration, and so on. Typical examples are highperformance position-controlled drives in robotics; variable speed-, torque-, and position-control systems for industrial applications such as steel-mill driving, pumps in chemical plants and thermal generating plants; manufacturing processes in machinery production shops; and traffic applications including electric trains, ship engines, electric cars, elevators, wind-power generators of up to 5000–8000 kW ratings, and so on. Adjustable-speed pumped storage induction generator-motors of typically 5000 MW have the largest ratings. On the other hand, micro-motors are today essential built-in parts of home appliances. Induction motors are the workhorse of every kind of application as a means of converting electrical power to mechanical power. Traditionally, DC motor drives have been used for speed and torque control applications. However nowadays thanks to remarkable advance of power-electronics technologies, the use of AC-induction motors in these applications have become available for every kind of driving processes with precise power control in combination with advanced power-electronics technologies. There are two-phase servo-motors, typically used for position-follow-up control systems, and single-phase induction motors that are widely used in household appliances. However, these are not our focus in this book. We will discuss three-phase induction machines (IMs), which can be classified based on the type of rotor structure: three-phase coils (doubly fed induction machine) or squirrel-cage coil (squirrel cage motor). First we discuss the theory of three-phase IMs with Wye-connected rotor windings, which are common for induction generators and motors.

13.2

Doubly Fed Induction Generators and Motors

Figure 13.1 shows the coil winding structure of a three-phase induction generator motor (IM) equipped with threephase star-connected rotor coils as well as three-phase star-connected stator coils. This type of machine is called an electrical doubly fed induction machine. The stator coil is composed of three-phase star-connected windings (referred to here as, bs, cs) based on uniform distribution wave winding (see Chapter 11, Figures 11.4 and 11.5); the stator coils are equivalent to the coils in a synchronous generator motor. The rotor also has three-phase star-connected rotor coils (called ar, br, cr, for each phase). So, the rotor structure is quite different than that of a synchronous machine, which has a single DC-field coil. 13.2.1

abc-Domain Voltages and Currents Equations

Now, we will introduce the electrical circuit equations of IMs. Referring to Figure 13.1, the equations of the voltages and flux linkages can be written as follows, where the direction of the arrows for the stator coil current ias, ibs, ics, and the rotor coil current iar, ibr, icr point in the direction of flow into the stator as well as into the rotor.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

338

13 Induction Generators and Motors (Induction Machines)

𝜈cs

ics 𝜃m icr 𝜈ar i ar 𝜈cr 𝜈br ibr

𝜈as ias

+ 𝜈 rs cs ns

ics

Stator

Rotor ibs

icr

ias

𝜈bs

ibs

rs

ns

𝜈as

+ 𝜈cr rr nr

+

ns

𝜈bs

nr

𝜈br

rs

+

+ iar 𝜈ar

nr

ibr

rr

+ Rotor coil

Stator coil (a)

rr

(b)

Figure 13.1 Three-phase IM with Wye-connected rotor windings.

The equations for flux linkage ψ as and terminal voltage vas of the phase-a stator coil can be written as follows ψ as = ls ias + Ls ias + Ls cos120 ibs + Ls cos 240 ics + Lms cos θm iar + Lms cos θm +

2π 3

ibr + Lms cos θm −

2π 3

13 1a

icr

Vas = rs ias + sψ as

13 1b

where Lself = Ls + ls: self-inductance of the stator coil Ls: magnetizing inductance of the stator coil ls: leakage inductance of the stator coil Then, the equation of a Wye-connected symmetrical IM is given as follows (derived from the physical images of the machine structure): ls + Ls

Ls cos

ψ as ψ bs

− 2π 3

Ls cos

ψ cs

Ls cos

2π 3

Ls cos

ls + Ls

2π 3

Ls cos

− 2π 3

Ls cos

− 2π 3

2π 3

l s + Ls

Lms cos θm

2π 3

Lms cos θm +

Lms cos θm −

2π 3

Lms cos θm

Lms cos θm +

2π 3

Lms cos θm −

2π 3

Lms cos θm −

2π 3

Lms cos θm +

2π 3

Lms cos θm

ias ibs ics

= ψ ar ψ br ψ cr

Lmr cos θm

2π 3

2π Lmr cos θm + 3

Lmr cos θm

2π 3

Lmr cos θm +

Lmr cos θm −

where

Lmr cos θm −

Ls Lms Lr Lmr = , = ns nr nr ns

②

Lmr cos θm +

2π 3

2π Lmr cos θm − 3 2π 3

Lmr cos θm

lr + Lr − 2π Lr cos 3

Lr cos

2π 3

Lr cos

2π 3

Lr cos

− 2π 3

iar ibr

lr + Lr

Lr cos

− 2π 3

2π Lr cos 3

lr + Lr

icr

13.2 Doubly Fed Induction Generators and Motors

vas , vbs , vcs , ias , ibs , ics , ψ as , ψ bs , ψ cs voltages, currents, flux linkages of the stator coils var , vbr , vcr , iar , ibr , icr , ψ ar , ψ br , ψ cr voltages, currents, flux linkages of the rotor coils ns , nr number of turns in the stator coils and the rotor coils dθs θs , ωs = = 2πfs angular position and angular velocity of the stator phase-a outer circuit dt sinusoidal source power-grid side or driving-source side ; fs is the frequency of the stator source current dθm θm , ωm = angular position and speed of the rotor The rotor position is behind the stator dt by θm = θs − θr , and the speed is lower by ωm = ωs − ωr dθr = 2πfr angular position and angular velocity of the rotor θr = θs − θm , ωr = ωs − ωm = dt phase-a of an outer circuit sinusoidal source current fr = ωr 2π is the frequency of the rotor

13 2a

side source current or, with symbolic matrix equations, ψ abcs ψ abcr

=

Ls Lms

iabcs

Lmr Lr

iabcr

ψ as

13 3a

ψ ar

vas

ias

ψ abcs = ψ bs , ψ abcr = ψ br , vabcs = vbs , iabcs = ibs ψ cs

ψ cr

vcs

13 3b

ics

The magnetizing inductance matrix of the stator coil is ls + Ls Ls =

Ls cos

2π 3

− 2π 3 2π Ls cos 3

Ls cos

−2π ls + Ls 3 2π − 2π Ls cos Ls cos 3 3

Ls cos

=

ls + Ls

1 1 ls + Ls − Ls − Ls 2 2 1 1 − Ls ls + Ls − Ls 2 2 1 1 − Ls − Ls ls + Ls 2 2

13 3c

The magnetizing inductance matrix of the rotor coil is lr + Lr Lr =

Lr cos

2π 3

− 2π lr + Lr 3 2π − 2π Lr cos Lr cos 3 3

Lr cos

−2π 3 2π Lr cos 3

Lr cos

lr + L r

=

1 1 lr + Lr − Lr − Lr 2 2 1 1 − Lr lr + L r − Lr 2 2 1 1 − Lr − Lr lr + Lr 2 2

13 3d

The mutual-inductance matrix of the rotor coils toward the stator coils is cos θm Lms = Lms

cos θm +

2π 3

cos θm −

2π 3

cos θm

cos θm +

2π 3

cos θm −

2π 3

cos θm −

2π 3

cos θm +

2π 3

cos θm

=

Lms Lmr Lmr

t

13 3e

339

340

13 Induction Generators and Motors (Induction Machines)

And the mutual-inductance matrix of the stator coils toward the rotor coils is cos θm Lmr = Lmr

cos θm −

2π 3

cos θm +

2π 3

cos θm

cos θm −

2π 3

cos θm +

cos θm +

2π 3

cos θm −

2π 3

2π 3

=

Lmr Lms Lms

t

13 3f

cos θm

where 1 1 1 1 Lms t = Lmr , Lms = Lmr Lms Lmr Lms Lmr

t

Then, Eqs. (13.3a) and (13.3b) are again ψ abcs = Ls iabcs + Lms iabcr ①

13 3g

ψ abcr = Lmr iabcs + Lr iabcr ② where, due to the theory of electromagnetism, Ls =

2

μSn2s μSn2r ns , Lr = ∴ Ls = g g nr

13 3h

Lr

g : length of the air-gap loop pass through the stator coils and rotor coils S : sectional area of the flux linkage path μ : permeability of vacuum space (or the air-gap space) Note that when expressing the voltage equations in machine-variable form, it is convenient to replace all the rotor variables ψ abcr, iabcr, vabcr with the new variables ψ abcr , iabcr , vabcr . The rotor variables’ turn ratios are ns/nr times different in comparison with the real turn numbers ψ abcr , iabcr , vabcr , respectively. That is, we introduce new rotor variables ψ abcr , iabcr , vabcr instead of ψ abcr, iabcr, vabcr, preserving the physical relations from all the previous equations under the conditions of Eq. (13.4): ψ abcr

ns ψ ,i nr abcr abcr

nr iabcr , vabcr ns

ns vabcr nr

13 4

Eq. (13.3g) is modified as follows, preserving the original relation: ψ abcs = Ls iabcs +

∗

1

ns ns ∗ = 2 ψ Lmr nr abcr nr or ψ abcr =

∗

ns Lmr nr

2

ns Lms nr iabcs +

iabcs +

nr iabcr ns ∗3

∗

3

ns nr

2

ns nr

2

13 5a Lr

nr iabcr ns 13 5b

Lr

iabcr

and ∗

∗

∗

1=

ns Lms nr

2=

ns ns Lmr Lmr ns Lmr = Lms t = Lms nr nr Lms Lms nr

3=

ns nr

Lms

13 5c t

Lms

t

where

Lmr Lms t = Lmr Lms

13 5d

2

Lr

Lr

13 5e

13.2 Doubly Fed Induction Generators and Motors

2

ns nr

Lr

2

ns nr

Lr =

1 1 lr + Lr − Lr − Lr 2 2 1 1 − Lr lr + Lr − Lr 2 2 1 1 − Lr − Lr lr + Lr 2 2

1 1 lr + Ls − Ls − Ls 2 2 1 1 − Ls lr + Ls − Ls 2 2 1 1 − Ls − Ls lr + L s 2 2

13 5f

Equation (13.5f) is the new, rewritten version of Eq. (13.3f), with the condition of Eq. (13.3h). Therefore, comparing the inductance matrices Ls and Lr , the magnetizing inductance (main flux elements) Ls are common between the stator and the rotor, while the leakage inductance elements ls and lr are different. Then, substituting Eqs. (13.4 and 13.5c–13.5f) into Eqs. (13.5a) and (13.5b), the equation for the flux linkage in Eq. (13.3g) can be rewritten as follows, where Lms and Lmr are replaced by Lms and (Lms )t and all original physical relations are preserved: ψ abcs = Ls iabcs + Lms iabcr ψ abcr = Lms

t

ψ abcs

or

iabcs + Lr iabcr

ψ abcr

Ls =

Lms

Lms

iabcs

Lr

iabcr

t

13 6a

Remember that Ls, Lr are time independent, while Lms , (Lms )t are time dependent. The equation for the changing speed of the flux linkage is sψ abcs = Ls siabcs + sLms iabcr + Lms siabcr sψ abcr = sLms

t

13 6b

iabcs + Lms siabcs + Lr siabcr

The original Eq. (13.2a) has been rearranged as follows.

ψ as ψ bs ψ cs ψ ar

ls + L s

1 − Ls 2

1 − Ls 2

Lms cos θm

2π 3

1 − Ls 2

ls + L s

1 − Ls 2

Lms cos θm −

2π 3

Lms cos θm

1 − Ls 2

1 − Ls 2

ls + Ls

Lms cos θm +

2π 3

Lms cos θm −

Lms cos θm +

2π 3

Lms cos θm −

2π 3

Lms cos θm +

2π 3

ias ibs

Lms cos θm

ics

= t

t

Lms cos θm

Lms cos θm −

ψ br t

2π 3

Lms cos θm

t

2π 3

Lms cos θm +

Lms cos θm +

ψ cr

2π 3

Lms cos θm −

t

t

t

2π 3

lr + Ls

1 − Ls 2

1 − Ls 2

t

2π 3

1 − Ls 2

lr + L s

1 − Ls 2

1 − Ls 2

1 − Ls 2

lr + L s

Lms cos θm + Lms cos θm −

2π 3

t

Lms cos θm

iar ibr icr

13 7 The original inductance matrices Ls, Lr, and Lms, Lmr in Eq. (13.2a) have been rewritten in symmetrical form except for the leakage inductances ls, lr. The equation suggests that all the electromagnetic quantities of the stator coils and rotor coils reserve reciprocal relations in mathematical form as well as from a physical. The voltage equation is the derivative of Eq. (13.7) plus the resistive drop component of the coils: rs iabcs

vabcs vabcr

=

rr iabcr

sψ abcs + sψ abcr

rs iabcs =

r1 iabcr

0 + s Lms

t

sLms

iabcs

0

iabcr

Ls + Lms

t

13 8

Lms

siabcs

Lr

siabcr

341

342

13 Induction Generators and Motors (Induction Machines)

We have obtained the fundamental equations of the IM using abc-domain variables. Remember that the inductance matrix Lms, (Lms)t are time-dependent because they include θm = ωmt, while Ls and Lr are obviously timeindependent.

13.2.2

dq0-Domain Transformed Equations

To take the study further, it is necessary to transform the variables associated with the symmetrical stator and rotor windings in the abc-domain into the dq0-domain as we did in Chapter 10 for synchronous machines. In the case of synchronous machine theory, we defined one set of dq0-axes that rotate in synchronism with the angular position of the stator voltages as well as the rotor velocity; they always rotate together in synchronism. However, in contrast, the IM rotor may slip from the stator rotating flux position so that the angular position of the stator voltage and the rotor velocity are always different. Therefore, we need to introduce two sets of dq0-axes that are ds qs 0s-axes for the stator and dr qr 0r-axes for the rotor. Figure 13.2 shows a physical image of an IM with double sets of the dq0transformed domain. So, we will now introduce the operational equations Ds(t), Ds(t)−1 for transformation of the stator coil quantities and Dr(t), Dr(t)−1 for the rotor coil quantities: 2π 3

cos θs +

2π 3

2 2π 3 − sin θs −sin θs − 3

− sin θs +

2π 3

cos θs Ds t =

cos θs −

1 2

1 2

13 9a

1 2

qr 𝜈ds

ids

b

ds 𝜃s

𝜃m

rs

dr

Ls

Ls

ls

𝜃r

Lr

a

lr rr

Lr

ls lr

qs

rs iqr

rr

iqs

𝜈qr 𝜈qs

idr 𝜈dr

c

Figure 13.2 Structural image of a three-phase double-fed IM in the dq0-domain.

13.2 Doubly Fed Induction Generators and Motors

cos θs Ds t

−1

=

− sin θs

cos θs −

2π 3

−sin θs −

2π 3

1

cos θs +

2π 3

− sin θs +

2π 3

1 cos θs − θm +

2π 3

2 2π 3 − sin θs − θm − sin θs − θm − 3

− sin θs − θm +

2π 3

cos θs − θm −

1 2

1 2

2 −sin θr 3

− sin θr −

1 2

−1

=

− sin θr +

13 10a

2π 3

1 2 −sin θs −θm

1

cos θs − θm −

2π 3

− sin θs − θm −

2π 3

1

cos θs −θm +

2π 3

− sin θs − θm +

2π 3

1 13 10b

− sin θr

cos θr

=

2π 3

2π cos θr + 3

1 2

cos θs − θm

Dr t

1 2

2π cos θr − 3

cos θr =

13 9b

2π 3

cos θs − θm Dr t =

1

1

cos θr −

2π 3

− sin θr −

2π 3

1

cos θr +

2π 3

−sin θr +

2π 3

1

where θr = θs − θm Our purpose is to transform Eqs. (13.6)–(13.8) into the dq0-domain. However, before performing the calculations, we will discuss the mutual relations of the equations for the voltage, current, and flux linkages of the machine in the dq0-domain. The stator voltage equation is transformed into the ds, qs, 0s-domain as follows: vabcs = r s iabcs + sψ abcs vdq0s = Ds vabcs = Ds r s iabcs + Ds s Ds−1 ψ dq0s = Ds r s Ds−1 idq0s + Ds sDs− 1 ψ dq0s + Ds Ds−1 sψ dq0s ∴ vdq0s = r s idq0s + Ds sDs−1 ψ dq0s + sψ dq0s

13 11 13 12

343

344

13 Induction Generators and Motors (Induction Machines)

−sin θs

cosθs sDs−1

sin θs −

2π 3

cos θs −

2π 3

0

sin θs +

2π 3

cos θs +

2π 3

0

− sin θs −

2π 3

1

2π 3

− sin θs +

2π 3

1

cos θs Ds t

sin θs

2π dθs d cos θs − 3 t = dt dθs cos θs +

sDs−1

1

cos θs −

2π 3

2 2π t = 3 − sin θs − sin θs − 3 1 2

1 2

= − ωs

cos θs

0

2π 3

sin θs

2π −sin θs + 3

sin θs −

2π 3

cos θs −

2π 3

0

sin θs +

2π 3

cos θs +

2π 3

0

cos θs +

1 2

−ωs

cosθs

0

0 −1 0 = ωs 1 0 0 0 0 0 13 13a Then, 0 −1 0 vdq0s = r s idq0s + ωs 1 0 0 ψ dq0s + sψ dq0s

13 13b

0 0 0 or vds = rs ids − ωs ψ qs + sψ ds vds = rs ids + ωs ψ qs + sψ ds

13 14

v0s = rs i0s + sψ 0s In a similar way, the rotor variables are transformed into the dr, qr, 0r-domain, where the transformed equation Dr(t)−1 for the rotor variables with angular axes position θr = θs – θm is adopted vdr = rr idr − ωs − ωm ψ qr + sψ dr vqr = rr iqr + ωs −ωm ψ dr + sψ qr

13 15

v0r = rr i0r + sψ 0r where θs, ωs = dθs/dt, fs = ωs/2π: angular position (phase-a), angular velocity, and frequency of the stator outer electrical source, respectively θr = θs – θm, ωr = dθr/dt, fr = ωr/2π: angular position (phase-a), angular velocity, and frequency of the rotor outer electrical source, respectively. θm = θs – θr, ωm = dθm/dt = ωs – ωr: rotor mechanical angular displacement and mechanical speed, respectively. Equations (13.14) and (13.15) show the relations of the stator variables (vs, is, ψ s, ωs) in the stator ds, qs, 0s-axes and those of the rotor variables (vr , ir , ψ r , ωr = ωs – ωm) in the rotor dr, qr, 0r-axes. The rotor mechanical speed is derived as the difference between the stator electrical speed and the rotor electrical speed. Now, we will proceed to the transformation of ψ abcs by Ds, Ds−1 and ψ abcr by Dr, Dr−1 into the ds, qs, 0 s-axes domain and the dr, qr, 0r-axes domain, respectively.

13.2 Doubly Fed Induction Generators and Motors

Again, ψ abcs = Ls iabcs + Lms iabcr ψ abcs = Lms

t

13 6a

iabcs + Lms iabcr

−1 Dr−1 idq0s ∴ ψ dq0s = Ds ψ abcs = Ds Ls Ds−1 idq0s + Ds Lms

ψ dq0r = Dr ψ abcr = Dr Lms ψ dq0s ψ dq0r

=

Ds− 1 idq0s + Dr Lr Dr−1 idq0r

t

Ds Ls Ds−1 Dr Lms

t

−1 Ds Lms D −1

3 ls + Ls 2 Ds−1 =

1 1 − Ls − Ls ls + Ls 2 2

=

ls + Ls

0

0

0

ls + Ls

0

0

0

ls + Ls

13 16c

idq0r

Ds−1 Dr Lr Dr−1

1 1 − Ls ls + Ls − Ls 2 2

13 16b

idq0s

1 1 ls + Ls − Ls − Ls 2 2 Ds Ls Ds− 1 = Ds

13 16a

0

0

0

3 ls + Ls 0 2

0

0

ls

13 17

where 3 L = Ls stator inductance in the dq0-domain 2 Note: See Supplement 1, for the calculation of Eqs. (13.17)–(13.19).

Ds Lms Dr− 1 =

2π 3

2π 3

cos θs +

2π 3

cos θm

2 2π − sin θs − sin θs − 3 3

−sin θs +

2π 3

Lms cos θm −

2π 3

cos θm

cos θm +

2π 3

cos θm −

cos θs

cos θs −

1 2

1 2 cos θs − θm

1 2 cos θs −θm −

2π 3

2 3 = Lms 2π 3 2 − sin θs −θm −sin θs − θm − 3 0 1 0 0

0

cos θs − θm + −sin θs −θm + 0

2π 3 2π 3

cos θm +

2π 3

cos θm −

2π 3

cos θm +

2π 3

Dr−1

cos θm

cos θs −θm

− sin θs −θm

1

cos θs − θm −

2π 3

−sin θs − θm −

2π 3

1

cos θs − θm +

2π 3

−sin θs − θm +

2π 3

1

1 0 0

3 = Lms 0 1 0 = M 0 1 0 2 0 0 0

0 0 0

13 18

345

346

13 Induction Generators and Motors (Induction Machines)

3 where M = Lms rotor inductance in the dq0-domain 2

Ds Lms

t

Ds−1 =

2π 3

2π 3

cos θs − θm +

2π 3

cos θm

2 2π −sin θs − θm − sin θs − θm − 3 3

−sin θs − θm +

2π 3

Lms cos θm +

2π 3

cos θm

cos θm −

2π 3

cos θm +

cos θs −θm

cos θs −θm −

1 2

1 2

cos θs = Lms

cos θs −

1 2

2π 3

cos θs +

2π − sin θs − sin θs − 3 0

2π − sin θs + 3

0

1 0 0

− sin θs

cos θs

2π 3

0

cos θm −

2π 3

cos θm +

2π 3

cos θm −

2π 3

Ds− 1

cos θm

1

cos θs −

2π 3

− sin θs −

2π 3

1

cos θs +

2π 3

− sin θs +

2π 3

1

1 0 0

3 = Lms 0 1 0 = M 0 1 0 2 0 0 0

0 0 0

13 19 1 1 lr + Ls − Ls − Ls 2 2 1 1 − Ls lr + Ls − Ls 2 2

Dr Lr Dr− 1 = Dr

3 lr + Ls 2 Dr− 1 =

0

1 1 − Ls − Ls lr + Ls 2 2 lr + L

0

0

=

0

0

0

3 lr + Ls 0 2 0

lr

13 20a

0

lr + L 0

0

0

lr

We also need to check the relation of L and M. From Eqs. (13.17), (13.18), (13.5b), and (13.2a), 3 3 3 ns Ls Lms L = Ls , M = Lms = Lms , = 2 2 2 nr ns nr

13 20b

∴L=M L is now replaced by M, and the equation for ψ becomes quite simple: ψ ds

ls + M

ψ qs =

0

ψ 0s

0

ψ dr

M 0

ψ qr ψ 0r

=

0

0

ls + M 0 0 0

ls ids

ids

M 0

0

idr

iqs +

0 M 0

iqr

i0s

0

0 M

i0r

0

0

idr

lr + M 0

iqr

lr + M

0 M 0

iqs +

0

0

i0s

0

0 M

0

lr

i0r

13 21a

13.2 Doubly Fed Induction Generators and Motors

or ψ ds = ls ids + M ids + idr

ψ dr = lr idr + M ids + idr

ψ qs = ls iqs + M iqs + iqr

ψ qr = lr iqr + M iqs + iqr

ψ 0s = ls i0s

ψ 0r = lr i0r

13 21b

where M: magnetizing inductance of the stator as well as of the rotor ls, lr : leakage inductances of the stator and the rotor This is the dq0-transformed version of the original Eqs. (13.6a and 13.6b), where all the elements of the 6 × 6 inductance matrix have become time-independent constants, while the original matrix includes time-dependent θm(t). The derivative sψ dq0 domain, which is the total voltage across the inductances M, ls, lr , is sψ ds = ls sids + sM ids + idr

sψ ds = lr sidr + sM ids + idr

sψ qs = ls siqs + sM iqs + iqr

sψ qr = lr siqr + sM iqs + iqr

sψ 0s = ls si0s

sψ 0r = lr si0r

13 22

The voltage equations with the expression sψ = v are vds = rs ids − ωs ψ ds + sψ ds vds = rr idr − ωs − ωm ψ qr + sψ dr vqs = rs iqs − ωs ψ ds + sψ qs vqr = rr iqr − ωs − ωm ψ dr + sψ qr v0s = rs i0s + sψ 0s

13 23

v0r = r0r = rr i0r + sψ 0r

The voltage equations with the expression of i are vds

s ls + M

rs ids

− ωs ls + M

0

ids

s ls + M

0

iqs + ωM

0

sls

i0s

vqs = rs iqs + ωs ls + M v0s

rs i0s

0

vdr

rr idr

sM

vqr

= rr iqr

v0r

rr i0r

+

+

sM −ωM 0 0

− ωs −ωm M 0

ids

ωs − ωm M

sM

0

iqs

0

0

0

i0s

idr

sM

0

iqr

0

0

i0r

13 24a

13 24b

s lr + M

− ωs − ωm lr + M

0

idr

ωs − ωm lr + M

s lr + M

0

iqr

0

0

slr

i0r

or vds = rs ids − ωs ls iqs + M iqs + iqr vqs = rs iqs − ωs ls ids + M ids + idr

+ sls ids + sM ids + idr + sls iqs + sM iqs + iqr

13 25a

v0s = rs i0s + sls i0s vdr = rr idr − ωs −ωm

lr iqr + M iqs + iqr

vqr = rr iqr − ωs − ωm

lr idr + M ids + idr

v0r = rr i0r + slr i0r

+ slr idr + sM ids + idr + slr iqr + sM iqs + iqr

13 25b

347

348

13 Induction Generators and Motors (Induction Machines)

The braces {} can be replaced by the flux linkages ψ ds, and so on, from Eq. (13.21b). Then, vds −rs ids −sls ids + ωs ψ qs = vdr −rr idr −slr idr + ωs − ωm ψ qr = sM ids + idr 13 26

vqs − rs iqs − sls iqs + ωs ψ ds = vqr − rr iqr − slr iqr − ωs − ωm ψ dr = sM iqs + iqr

The voltage in Eqs. (13.25a, b) and the flux linkage in Eq. (13.26) suggest the equivalent circuits in the dq0-domain as shown in Figure 13.3, where the rotor-side resistance rr includes the equivalent-load resistance (see Figure 13.11). Now the equations for ψ, sψ, v in the dq0-domain have been obtained. Equations (13.21)–(13.24a and b) may be compared with Eqs. (10.29) and (10.33) in Chapter 10 for the synchronous generator, where we concluded that all dq0 variables must be time-independent DC values under steady-state conditions. This is also true for the previous equations, so the terms including s = d/dt in Eqs. (13.23) and (13.24a and 13.24b) are transient. Of course, the transient terms in the dq0-domain are omitted for the steady-state analysis, if necessary. Further, comparing Eqs. (13.23) and (13.24), it is interesting that the voltages vds, vqs can be expressed quite simply using the flux linkages ψ ds, ψ qs, while the expressions with currents ids, iqs, idr , iqr are more complex. It suggests from the viewpoint of IM control technology that controlling flux linkages ψ is easier than controlling currents i; but we will discuss that later, in Chapters 20 and 21. All of the previous equations can be per-unitized as follows, where the zero-sequence equation is omitted for simplicity. In the derived equation, the per-unitized inductance M has been replaced by the per-unitized reactance X because M = M Mbase , and the corresponding X = X Xbase becomes the same values using per-unitization (see Chapter 10, Eq. (10.51)). All the per-unitized inductances are replaced with per-unitized reactances

rs

𝜔sψqs +

–

(𝜔s–𝜔m)ψ ′qr

l ′r

ls

–

+

r ′r

+

+ i ′dr

ids d axis

𝜈ds

𝜈′dr

M

–

–

rs

𝜔sψds –

+

(𝜔s–𝜔m)ψ ′dr

l ′r

ls

+

–

r ′r +

+ i ′qr

iqs q axis

𝜈qs

𝜈′qr

M

–

– rs

r ′r

i0s

i ′0r

+

+ 0 axis

𝜈0s

ls

l ′r

– Figure 13.3 Equivalent circuit of a three-phase double-fed IM in the dq0-domain.

𝜈 ′0r –

13.2 Doubly Fed Induction Generators and Motors

ψ ds = X ls ids + X M ids + idr ψ qs = X ls iqs + X M iqs + iqr

13 27

ψ dr = X lr idr + X M ids + idr ψ qr = X lr iqr + X M iqs + iqr ωs ωs X ls iqs − X M iqs + iqr ωbase ωbase ωs ωs X ls ids + X M ids + idr vqs = r s iqs + ωbase ωbase ωs −ωm ωs − ωm X lr iqr − XM vdr = r r ids − ωbase ωbase ωs − ωm ωs − ωm vqr = r r iqr − X lr idr − XM ωbase ωbase vds = r s ids −

s

X ls ids +

s

X M ids + idr ωbase ωbase s s + X ls iqs + X M iqs + iqr ωbase ωbase s s X i + X M ids + idr iqs + iqr + ωbase ls ds ωbase s s X i + X M iqs + iqr ids + iqr + ωbase lr dr ωbase +

13 28

where X ls = lls , X lr = llr , X M = M reactance in PU values. And the inverse matrix of Eq. (13.21a) is ids iqs idr

=

X lr + X M

0

−XM

0

ψ ds

0

X lr + X M

0

−XM

ψ qs

−X M

0

X ls + X M

0

ψ dr

0

−X M

0

X ls + X M

ψ qr

1 Kx

iqr

13 29

where Kx = X ls + X M X lr + X M − X 2M The voltage v in Eq. (13.28) can also be expressed as a function of ψ by substituting Eq. (13.29) into Eq. (13.28)

vds vqs vdr vqr

1 s ωs 1 r s X lr + X M + − − rs X M Kx ωbase Kx ωbase ωs 1 s r s X lr + X M + 0 Kx ωbase ωbase 1 1 s − rr X M 0 r r X ls + X M + Kx Kx ωbase 1 − rr X M Kx

0

ωs − ωm ωbase

0 1 rs X M Kx ωs −ωm − ωbase −

1 s r X ls + X M + Kx r ωbase

ψ ds ψ qs ψ dr ψ qr

13 30 ωs − ωm angular velocity in PU of rotor electrical source where ωr = ωbase 13.2.3

Phasor Expressions for dq0-domain Transformed Equations

The derived dq0-domain voltages from Eq. (13.28) can be written in as vectors or phasors in the form vdqs = vds + jvqs , idqs = ids + jiqs , vdqr = vdr + jvqr and idqr = idr + jiqr : jωs jωs s s + X ls i dqs + X M idqs + idqr X ls + X M idqs + idqr ωbase ωbase ωbase ωbase j ωs −ωm j ωs −ωm vdqr = r r idqr + X lr idqr + X M idqs + idqr ωbase ωbase

v dqs = r s i dqs +

+

s ωbase

X

s

13 31

X M idqs + idqr ωbase The first terms and the second terms of the right sides of these equations are the steady-state terms and transient terms, respectively. lr i dqr +

349

350

13 Induction Generators and Motors (Induction Machines)

From Eq. (13.27), and in the form ψ dqs = ψ ds + jψ qs , ψ dqr = ψ dr + jψ qr , ψ dqs = X ls idqs + X M idqs + idqr

sψ dqs = sX ls idqs + sX M idqs + idqr

ψ dqr = X ls idqr + X M idqs + idqr

sψ dqr = sX lr idqr + sX M idqs + idqr

13 32

Now, the original fundamental equations for the flux linkage ψ, voltage v, and current i in the abc-domain have been transformed into equations in the ds-, qs-, dr-, and qr-domain and further have been rewritten as phasor expressions. Throughout all the processes we have discussed in previous sections, the equational formations of the stator variables and rotor variables have been symmetrical. This is because the stator coils and rotor coils are on an equal footing and are symmetrical from a physical viewpoint. This is, of course, always true. Furthermore, we have found that the equations for ds and qs, as well as those for dr and qr, are symmetrical. This is because the structures of the stator coils and cores as well as those of the rotor coils are completely symmetrical in their radial direction in the longitudinal inner section. All of these equations are essential for basic knowledge of IM computational models and practical motor-driving control, which we will discuss later as a major focus. Now, it is worth reviewing these equations from a practical viewpoint. Our IM is a double-excited or double-fed machine, and the stator windings are connected to the outer circuit with the frequency fs = ωs/2π; the stator outer circuit consists of a power grid in the case of direct driving applications, or frequency-adjustable inverters in other applications. On the other hand, the rotor windings are connected to another outer circuit that is probably a power-electronic threephase circuit with very low frequency fr = ωr/2π (say, 0–5 Hz) or DC. Note that we have defined the angular position, angular velocity, and frequency of the stator coils as θs, ωs = dθs/dt, fs = ωs/2π, respectively, and the angular displacement of the rotor is θm, which is behind the stator position by θs − θm = θr. Then the rotor speed must be (ωm), which means the rotor windings are excited by the outer circuit, whose frequency is fr = ωr/2π = (ωs − ωm)/2π. In other words, if the frequency of the stator-coil side (connected to the power grid) is fs and that of the rotor side (connected to a power-electronic power-source circuit) is fr, then the rotor is driven by the angular speed ωm = 2π(fs − fr). This explanation intuitively suggests the relation P stator + P rotor + P stator = 0, which is of course true. From a electrical

electrical

mechanical

practical viewpoint, the IM is power reciprocal and is operated in generating mode or motoring mode. The cases explained next are typical examples of various operation modes. Case 1 Synchronized Operation In this case, the rotor-coil currents iar, ibr, icr are DC currents (that is, fr = fs – fm = 0) supplied by the electronic outer source, whereas the stator coils are connected to the power grid with frequency fs. This is electrically the same operating condition as a synchronous machine, because DC current is supplied to the rotor coil as like the field DC current ifd of a synchronous machine. Then the rotor is driven by ωm = 2πfs – 2πfr = 2πfm, which means operation in synchronism with the power-grid frequency. This operating condition is the same as that of a synchronous generator or motor. Case 2 Variable-Speed Motor Operation In a motor-operating application driven by a 50 Hz (or 60 Hz) power grid, if very slow-frequency sinusoidal currents iar, ibr, icr of fr = 0–5 Hz are supplied to the rotor coil (from the power-electronic outer circuit), the rotor is driven at speed ωm = 2π(fs – fr) = 2π(50–45 Hz) (or 60–55 Hz). This is the principle of adjusted motoring operation with variable speed. Further, in the case of speed-adjustable, pumped-storage hydro-generating units, the mechanical speed is controlled in the range of 100–90% rpm by changing the fed power frequency of the rotor coil in the range of fr = 0–5 Hz (see Section 21.6). Case 3 Variable-Speed Generating Operation In the case where a prime-moving turbine (a wind generating unit or of a small hydro-generating unit) is driven by randomly fluctuating speed ωm = 2πfm, if the outer power (current iar, ibr, icr) of frequency ωr = ωs – ωm (≤ ωs = 50/ 60 Hz) is provided to the rotor coil, the generator is operated in synchronism with the power grid of frequency fs = 50/60 Hz. This is the principle of typical power conditioners for small hydro-generation and/or wind generation (see Sections 21.7 and 21.8).

13.2 Doubly Fed Induction Generators and Motors

13.2.4

IM Driving Power and Torque

Next, we will discuss the power and torque equations. Equation (13.7) in the abc-domain is again used, although in a slightly modified form. That is, the 6 × 6 inductance matrix Labcdomain from Eq. (13.7) can be symbolically decoupled from the matrices of leakage inductance l and magnetizing inductance L. The symbolic expression of Eq. (13.7) is ψ abc = l iabc + L iabc

13 33

where Labcdomain = l + L. Then, we change the speed of the flux linkages ∴ sψ abc = l siabc + sL iabc + L siabc

13 34

and voltages vabc = r iabc + sψ abc = r iabc + l siabc + sL iabc + L siabc

13 35

∴ vabc = r iabc + sψ abc = r iabc + sL iabc + l + L siabc

where the 6 × 6 inductance matrix L is time-dependent, because the rotor displacement θm changes over time. ls, lr, Ls, Lr, Ms are time-independent. We assume next that the power is supplied by the outside circuit of the rotor coils as well as from the power grid through the stator coils. The total power flowing into the stator coils and rotor coils from both outside circuits is written as follows t t P in t = iabc vabc = iabc

r

P resist t

t iabc + iabc l siabc + it L siabc + it abc abc

sL iabc

13 36

effectively received power 2

2

2

t P resist t = iabc r iabc = rs i2as + i2bs + i2cs + rr i ar + i br + i cr

13 37

where Presist(t) is the total thermal power loss within the stator coils and rotor coils. The total electrical power given from the outer circuits of the stator coils and/or rotor coils is converted over time into electromagnetic power Pψ (t), mechanical power Pm(t), or thermal loss Presist(t). Then Pin t = Presist t + Pψ t + Pm t

13 38

Then, comparing Eqs. (13.37) and (13.38), Pψ (t) + Pm(t) can be extracted as follows. The effectively received power is t t t Pψ t + Pm t = iabc l siabc + iabc sL iabc + iabc L siabc

13 39

In addition, we know that the general form of the total inductive energy stored in a machine coil is given by Therefore, the total stored inductive energy in this machine is written as follows: Pψ t dt =

1 t 1 t iabc l iabc + iabc L iabc 2 2

1 2 k 2 Lk ik .

13 40

And the derivative form is Pψ t =

1 2

t t siabc l iabc + iabc l siabc

+

1 2

t siabc L iabc

sL

t iabc + iabc L siabc

13 41

t t L iabc = iabc L siabc ,. Further, the leakage inductance l is timewhere we know the mathematical formulae siabc independent, while the effective magnetic inductance L is time-dependent; then Pψ (t) is rewritten as follows.

1 t t t Pψ t = iabc l siabc + iabc sL iabc + iabc L siabc 2

13 42

Now, comparing Eqs. (13.39), (13.41), and (13.42), the mechanical power Pm(t) can be extracted as the subtraction of the both equations: 1 t sL iabc Pm t = iabc 2

13 43

351

352

13 Induction Generators and Motors (Induction Machines)

This equation gives the mechanical power produced by the electrical power fed by the double outer electrical circuits through the stator coils and rotor coils. Now, we will write this equation in a non-symbolic style. The matrix sL, which is the derivative form of L in Eq. (13.7), can be written as follows

sL =

dL dθm d = dt dθm dt

ls + Ls

1 − Ls 2

1 − Ls 2

Lms cosθm

1 − Ls 2

l s + Ls

1 − Ls 2

Lms cos θm −

2π 3

Lms cos θm

1 − Ls 2

1 − Ls 2

ls + Ls

Lms cos θm +

2π 3

Lms cos θm −

t

Lms cos θm −

t

2π 3

Lms cosθm

t

2π 3

Lms cos θm +

Lms cos θm + Lms cos θm −

2 = − ωm M 3

t

Lms cos θm

2π 3

2π 3

2π 3

1 − Ls 2

lr + Ls

1 − Ls 2

1 − Ls 2

1 − Ls 2

lr + Ls

t

Lms cos θm

0

0

0

sin θm −

2π 3

sin θm

0

0

0

sin θm +

2π 3

sin θm −

2π 3

sin θm

sin θm −

2π 3

sin θm +

2π 3

sin θm +

2π 3

2π 3

sin θm −

2π 3

sin θm +

2π 3

sinθm

sin θm +

2π 3

0

0

0

sin θm −

2π 3

0

0

0

0

0

0

sin θm

Lms cos θm

t

sin θm

sin θm +

2π 3

1 − Ls 2

0

2π 3

Lms cos θm +

1 − Ls 2

0

sin θm −

2π 3

lr + Ls

0

sin θm

2π 3

Lms cos θm −

2π 3

Lms cos θm −

t

2π 3

t

Lms cos θm +

t

Lms cos θm +

13 44 where L is the same as Labcdomain in Eq. (13.7), except that the leakage inductances ls, lr are omitted. Then, the voltage equations vabc are derived from Eq. (13.35), substituting L and sL from Eq. (13.44); the description is omitted here for brevity. Equation (13.43) of the mechanical power Pm(t) is rewritten as follows by referring to Eq. (13.44): 1 t sL iabc Pm t = iabc 2 sin θm 1 = − ωm M 3

ias ibs ics

sin θm −

2π 3

sin θm

sin θm +

2π 3

sin θm −

sin θm + iar ibr icr

sin θm +

sin θm −

2π sin θm + 3 sin θm −

2π 3

2π 3

sin θm −

2π 3

iar

sin θm +

2π 3

ibr

2π 3

sin θm +

2π 3

sin θm

icr

sin θm 2π 3

2π sin θm − 3

sin θm sin θm +

2π 3

13 45a

ias ibs ics

13.2 Doubly Fed Induction Generators and Motors

Then, Pm t = − ωm M ias iar sinθm + ias ibr sin θm +

2π 2π + ias icr sin θm − 3 3

+ ibs iar sin θm −

2π 2π + ibs ibr sin θm + ibs icr sin θm + 3 3

+ ics iar sin θm +

2π 2π + ics ibr sin θm − + ics icr sin θm 3 3

13 45b

The power equation in the abc-domain has been obtained. Now we will derive the equation for Pm in the d, q-domain. We are interested in balanced three-phase phenomena in most cases, so the zero-sequence component can be omitted, and the transformation matrices Ds, Dr from Eqs. (13.9a) and (13.10a) become a little simpler, as follows: 2 cos θs ids− sin θs iqs 3 2 2π 2π cos θs − ids− sin θs − ibs = 3 3 3

2 cos θs −θm idr− sin θs − θm iqr 3 2 2π 2π cos θs − θm − ibr = idr− sin θs − θm − 3 3 3

ias =

ics =

2 2π cos θs + 3 3

ids− sin θs +

iar =

iqs

2π 3

icr =

iqs

2 2π cos θs − θm + 3 3

idr− sin θs −θm +

2π 3

iqr iqr 13 46a b

The calculation of Pm can be conducted with a straightforward substitution of Eq. (13.46a,b) into Eq. (13.45b); the result is the same as the intuitively presumed forms, as follows. (The formulae for trigonometric functions in Appendix A are useful for the calculation.) Mechanical power Pm t = ωm M iqs idr −ids iqr Mechanical power Tm t =

13 47a

Pm = M iqs idr − ids iqr ωm

13 47b

where ωm(t) = dθdtm , Pm(t), Tm(t), ωm(t) are scalar values. Also, Pm(t) and Tm(t) can be expressed by various forms: Pm

Re ωm M ji∗dqs

idqr

Re ωm Mj ids Re jωm ψ ds

Tm = Re M ji∗dqs

idqr

= Re Mj ids + jiqs

= Re j ψ ds + jψ qs

∗

jiqs jψ qs

∗

∗

idr

jiqr

ωm M iqs idr − ids iqr 13 48a

idr

idr + jiqr

idr + jiqr

where ψ dr = M ids , ψ qr = M iqs

∗

jiqr

ωm ψ qs idr − ψ ds idr

= M iqs idr − ids iqr

13 48b

= ψ qs idr − ψ ds iqr 13 48c

And ψ ds, ψ qs, ψ dr , ψ qr are given by Equation13.21(b) by the functional form of currents. Now, the general equations for v(t), i(t), ψ(t) and the mechanical power Pm and torque Tm of the three-phase IM with Wye-connected rotor windings have been obtained. Through this calculation process, it is obvious that electrical power can be provided to or taken from the stator windings as well as the rotor windings simultaneously. So, this is an electrical doubly fed (or doubly excited) IM. Of course, the principle of power conservation is always preserved in the machine; it is written as follows as a total of mechanical power and electrical power P stator electrical

where P loss electrical

+ P rotor electrical

0.

+ P rotor mechanical

+ P loss electrical

=0

13 49

353

354

13 Induction Generators and Motors (Induction Machines)

13.2.5

Steady-State Operation

Now we will check the behavior of the IM under balanced three-phase steady-state operation. Our generator is running under the condition of fed-in power with frequency fs = ωs/2π Hz from the stator coils (connected to the power grid) and with frequency fr = ωr/2π = (ωs – ωr)/2π Hz from the rotor coils (connected to the power-electronic circuit), where either of the power sources can be a negative value in the case of fed-out power. Then the rotor is driven with the angular velocity ωm = ωs − ωr = 2π( fs − fr). We know that dq0 quantities are DC values under balanced three-phase steady-state operation, as shown later in Eqs. (13.52b, c). Then, referring to the voltage vdqs , vdqr from Eq. (13.31), the terms including s can be omitted for steady-state behavior, and the equations are simplified as shown here: vdqs = r s i dqs +

jωs jωs X ls i + X M idqs + idqr ωbase ωbase dqs

①

j ωs − ωm j ωs − ωm X lr idqr + X M i dqs + i dqr ωbase ωbase ② may be rewritten

②

vdqr = r r i dqr +

v dqr r r jωs jωs = X i + X M idqs + idqr i + slip slip dqr ωbase lr dqr ωbase ωs − ωm where ωr = slip definition of slip ωs

13 50 ③

From ①③, vdqs − r s +

vdqr r jωs jωs X ls idqs = − r + X ωbase slip slip ωbase lr

idqr =

jωs X M idqs + idqr ωbase

13 51

Equation (13.51) gives the equivalent circuit shown in Figure 13.4 for the IM under balanced three-phase steady-state condition. Next, let’s discuss the steady-state condition in a more commonly used form: Steady-state voltages and currents of the stator and rotor windings vas = Vs cos ωs t + αs

ias = Is cos ωs t + γ s

vbs = Vs cos ωs t + αs −

2π 3

ibs = Is cos ωs t + γ s −

2π 3

vcs = Vs cos ωs t + αs +

2π 3

ics = Is cos ωs t + γ s +

2π 3

var = Vs cos ωs −ωm t + αr

~ 𝜈–dqs

iar = Ir cos ωs − ωm t + γ r

vbr = Vr cos

ωs − ωm t + αr −

2π 3

ibr = Ir cos

ωs −ωm t + γ r −

2π 3

vcr = Vr cos

ωs −ωm t + αr +

2π 3

icr = Ir cos

ωs −ωm t + γ r +

2π 3

r–s +

13 52a

~ – idqs

j𝜔s – X 𝜔base ls

j𝜔s – X 𝜔base M

j𝜔s – X 𝜔base lr

– r'r slip ~ – i'dqr

– Figure 13.4 Equivalent circuit of an IM under steady-state condition.

+ ~ – 𝜈' dqr slip –

13.3 Squirrel-Cage Induction Motors

The dq0-transformed equations become DC values as follows (see Eq. (10.54)): vds = Vs cos αs

ids = Is cosγ s

vqs = Vs sin αs

iqs = Is sin γ s

v0s = 0

i0s = 0

vdr = Vr cos αr

idr = Ir cos γ r

vqr = Vr sin αr

iqr = Ir sin γ r

v0r = 0

i0r = 0

13 52b

The phasor expression is vdqs

vds + jvqs = Vs e jαs

v dqr

vdr + jvqr = Vr e jαr

i dqs

ids + jiqs = Is e jγs

i dqr

idr + jiqr = Ir e jγr

13 52c The phase-a voltage and current are written in a general phasor form: vas = Vs e j ias = Is e j

ωs t + γ s

var = Vr e j iar = Ir e j

ωs t + αs

= e jωs t vds + jvqs = e jωs t vdqs

= e jωs t

ωs −ωm t + αr ωs −ωm t + γ r

ids + jiqs

= ej

= ej

13 53a

= e jωs t ids

ωs −ωm t

ωs −ωm t

vdr + jvqr = e j idr + jiqr = e j

ωs − ωm t

ωs −ωm t

vdr

13 53b

idr

Substituting Eqs. (13.52b, c) into Eq. (13.50) ①, vdqs = Vs e jαs = r s Is e jγ s + ∴ vdqs e

jωs t

jωs jωs X ls Is e jγs + X M Is e jγ s + Ir e jγr ωbase ωbase

①

π j ωs t + γ s + ωs 2 = Vs e = r s Is e + X ls Is e ωbase π π j ωs t + γ s + j ωs t + γ r + ωs 2 + Ir e 2 ② X M Is e + ωbase j ω s t + αs

j ωs t + γ s

13 54

And the real part of ② becomes a phase-a voltage equation: vas = Vs cos ωs t + αs = r s Is cos ωs t + γ s −

ωs ωs X ls Is sin ωs t + γ s + X M Is sin ωs t + γ s + Ir sin ωs t + γ r ωbase ωbase

13 55

Note that the stator quantities are the values from the ds-qs-domain, and the rotor quantities are from the dr-qr-domain. Equations (13.50)–(13.55) and the equivalent circuit give the bases of the power-electronic DQ control method that we will discuss in Chapters 17 and 18.

13.3

Squirrel-Cage Induction Motors

13.3.1

Circuit Equations

The majority of IMs are not equipped with coil-wound rotor windings, and instead have different rotor structures. The current flows through copper or aluminum bars that are uniformly distributed and embedded in a ferromagnetic

355

356

13 Induction Generators and Motors (Induction Machines)

Stator

Fan

Cooling air flow

Rotor

Drive end

Non-drive end

Figure 13.5 Small squirrel-cage IM.

material, with all bars terminating in a common ring at each end of the rotor. This type of induction motor is called a squirrel-cage IM, which is obviously a singly fed or singly excited IM. Figure 13.5 shows the typical structure of small IM. Squirrel-cage motors are the most popular driving motors among the various types of motors in industrial applications, due to their low cost and tough rotor structures. Although single-phase IMs have been widely adopted for most residential applications, our interest is the three-phase type. When a three-phase IM is driven directly from line voltages of 50/60 Hz, the motor operates at a nearly constant speed. However, the operation speed can be varied using power-electronic converters. The major applications are classified into two categories: i) Adjustable speed drives: Typical applications are fans, compressors, pumps, and so on. ii) Servo drives: IMs are adopted for various sophisticated driving control applications such as servo driving equipment in production machine tools, robotics, computer peripherals, and the like. First of all, we need to look at the fundamental equations of the three-phase IM. A squirrel-cage motor (singly fed motor) is theoretically a special case of the doubly fed machine in that the rotor-side coils are short-circuited. So, all the equations in the previous section are simplified using var = vbr = vcr = 0, vdr = vqr = 0 and vdqr = 0. Then, the equivalent circuit of the motor is obtained from Figures 13.3 and 13.4, where the terminal on the right side is short-circuited to give the condition vdqr = 0. Under the condition of the vdqr terminal being short-circuited in Figure 13.4, the following equation is derived using vdqr = 0 in Eq. (13.51) jωs jωs XM − XM ωbase ωbase i dqs = i dqs idqr = jωs ' jωs jωs rr rr ' X lr + XM X lr + X M slip + ω slip + ω ωbase base base −

13 56

where idqs stator current idqr rotor current iM

idqs −idqr current flowing through the magnetizing reactance X M

The current iM and associated voltage eM are the current and electromotive force voltages that transfer power effectively between the stator and the rotor through the air gap, so they are called the air-gap current and air-gap voltage.

13.3 Squirrel-Cage Induction Motors

While referring to Eq. (13.48b), the per-unitized torque T m is expressed as follows as a phasor expression, with the magnetizing inductance M replaced by the reactance X M : T m = Re X M ji∗dqs idqr

13 57

Substituting Eq. (13.56) into Eq. (13.57), the torque equation is

T m = Re X M j idqs

∗

idqr = Re X M j idqs

= Re X M j

∴ Tm =

jωs XM ωbase idqs rr jωs + X + XM slip ωbase lr

jωs jωs r XM r − X + XM ωbase slip ωbase lr rr slip

ωs ωbase 2

rr slip

∗

+

2

+

X 2M ωbase

ωs X + XM ωbase lr

2 2

i dqs

ωs ωbase

= rr slip

2

+

X 2M ωbase

rr slip

idqs

2

ωs X + XM ωbase lr

2

rr slip

ωs X + XM ωbase lr

2

13 58 Next, the impedance Z s = vdqs idqs = vabc iabc can be derived from Eqs. (13.50) and (13.56): Zs =

v dqs

=

i dqs

v abc i dqs

= rs +

i dqr jωs jωs jωs X ls + XM + XM ωbase ωbase ωbase i dqs

jωs − XM jωs jωs ωbase = rs + X ls + X M + XM ωbase ωbase jωs rr + X + XM slip ωbase lr rs =

rr ωs − slip ωbase

2

jωs r r jωs X + XM + X ls + X M ωbase lr slip ωbase rr jωs + X + XM slip ωbase lr

X ls + X M X lr + X M + r s

+

ωs ωbase

2

X 2M

13 59a Then, utilizing Eq. (13.54), the impedance equation for a singly excited (squirrel cage) IM in the abc-domain is summarized as follows. This is called the positive-sequence impedance or balanced one-phase impedance of the squirrelcage IM:

ZS =

v abcs i abcs

=

e jωs t v dqs e jωs t i dqs

=

v dqs

rs =

ωs rr + slip ωbase

i dqs

where X SS = X lS + X M X rr = X lr + X M

X 2M − X ss X rr +

jωs r r s X rr + r X ss ωbase slip

jωS rr + X rr slip ωbase

13 59b

357

358

13 Induction Generators and Motors (Induction Machines)

~ ~ – – Ir = –i'ar ~ – ias

𝜔 – j 𝜔 s Xls

rs

base

+

base

+

~ 𝜈–as

𝜔 – ~ e–M = j 𝜔 s XM iM

~ ~ – – ~ – IM = ias – Ir

base

–

– r'r slip

𝜔 – j 𝜔 s X 'lr

~ –𝜈 as

lm

𝜔 – j 𝜔 s XM

~ – Ir

𝛿

base

– 𝜃r e~ M ~ 𝜔 – – rs + j 𝜔 s Xls · ias

~ – Ias

– (a) Equivalent circuit

base

(b) Phasor diagram

Figure 13.6 Equivalent circuit of an IM (squirrel cage) under steady-state conditions.

Figure 13.6 shows an equivalent circuit and phasor diagram for a squirrel-cage IM under balanced three-phase steadystate operation. The term r slip includes the resistance of the rotor coil plus the equivalent load resistance. 13.3.2

Torque-Speed Characteristics Equation for a Squirrel-Cage Induction Machine

The relation of mechanical rotating speed and slip is defined as follows in the standard theory for IMs: Angular velocity of the rotor ωm = ωS 1 − slip =

2πfs 1 − slip p

ωs − ωm fs −fm = = ωr ωs fs where pair numbers of poles p

Slip: slip =

13 60

slipangularvelocity: ωslip = ωS − ωm = ωS slip slipfrequency: fslip = fs − fm = fs slip Then slip = 0 means synchronous operation, and slip = 1 means zero speed. The torque equation is again shown in the following equation, where idqs is replaced by ias :

Tm =

ωS ωbase 2

rr slip

X 2M ωbase

rr slip

ωS + X + XM ωbase lr

2

ias

slip ωS X 2M r 2 ωbase ωbase r = 2 s ωs X lr + X M r r + lip ωbase

2

ias

2

13 61a

Now we have obtained the equation for torque T m versus slip (slip) or speed (ωm = [1 − slip] ωs) characteristics with the parameters of the motor load resistance r r (smaller r r means larger load), magnetizing inductance between the stator coil and rotor coil X M , and stator current ias . Equation (13.61a) gives the torque-speed characteristics and is the most fundamental equation for an IM. Numerical check: Equation (13.61a) is simplified as follows by using ωs = 1, ωbase = 1 slip X 2M r r

Tm = r

2

+ slip 2 X lr + X M

2

ias

2

13 61b

where ωm = 1 − slip Then, T m versus slip (or ωm = [1 − slip] ωs) curves can be derived with the parameters r r , X M , and X lr . Figure 13.7 shows the derived curves, which are calculated by Excel under the condition of r r = 0.5 (light load), 0.3, and 0.1 (heavy load) and with the parameter X M = 1.0, 0.75, 0.5.

Torque

13.3 Squirrel-Cage Induction Motors

–r ′ =0.5 0.3 0.1 0.4 X– =1.0; ~ M r

~ – XM = 0.75; r–′r =0.5 0.3 0.1

0.3

0.2

– ~ r ′r =0.5 0.3 0.1 XM =0.5; –

0.1

–1.0

–0.5

2.0

1.5

0

0.5

1.0

1.5

1.0

0.5

0

–0.5

2.0

ωm/ωs

–1.0 Slip

–0.1

Rated speed

Speed zero –0.2 –0.3

Figure 13.7 Torque-speed characteristics (numerical calculation of Eq. (13.61b)).

13.3.3

Reverse, Startup, and Ordinary Running and Overspeed Operating Ranges

Figure 13.8 shows the torque-speed characteristics of IMs, which can be drawn by plotting Eq. (13.61b). This is the functional curve of torque T m drawn with the parameter of angular speed (ωs/ωbase or the frequency fs/fbase) from the supplied power or slip under steady-state operation. The operating characteristics range can be divided into four classified regions. i) Operating region between points b and a: slip =

fs − fm >1 fs

ωm and T m > 0, so the motor is driven as a braking operation. The rotating direction is opposite that ωs of the rotating flux, and lengthy operation in braking mode can cause extreme overheating of the rotor-cage coils. This is the same phenomena as overheating of the synchronous generator rotor surface due to negative sequence current (see Section 11.9.2). fs −fm ii) Startup operating region up between points b and c: 1 ≥ slip = >0 fs This is the startup region. The rotor is at a standstill at point b and begins to rotate to the required speed between points c and d. We will discuss this mode in section 13.3.5. ωm ≥ 0 Tm > 0 iii) Operating region between points c and d: 1 > slip > 0 1 ≥ ωs This is the region for motor operation. The motor begins rotation at point b, and the speed is increased to the operating range between points c and d. fs − fm ωm iv) Operating region between points d and e: 0 ≥ slip = > 1 0 > Tm fs ωs This is the operation region of regeneration (or generator operation). The machine works as a generator if the rotor is mechanically driven. In this case, 0 >

In the case of a simple motor-driving operation, the IM begins to start rotation at point b and increases to the usual operating range between point c (maximum torque point) and point d. The maximum torque equation at point c is calculated using dT m d ωr ωs = 0

359

13 Induction Generators and Motors (Induction Machines)

Braking

Power running

𝜔m 𝜔s < 0

Regeneration

𝜔 0 < 𝜔m < 1 s

𝜔 1 < 𝜔m s

𝜔s

𝜔s

𝜔s

𝜔m

𝜔m

𝜔m

Torque a

Motor

Starting torque Te

c Maximum torque b d

–1.0 2.0

–0.6 1.6

Speed zero

–0.2 1.2

0.2 0.8 Generator

360

0.6 0.4

1.2 –0.2

1.6 –1.6

2.0 speed 𝜔m/𝜔s –1.0 slip = (𝜔s–𝜔m)/𝜔s

Synchronous speed e

Reverse operating range

Startup range

Standstill

Overspeed range Ordinary operating range

Figure 13.8 Steady-state torque-speed characteristics of IM (singly fed) or singly excited IM (squirrel cage).

Figure 13.9 shows Torque speed characteristics as shown by the parameter of power frequency fs as well as by voltages vs and rotor equivalent resistance (which includes equivalent load loss). As are seen in both figures, the output torque Tm of IM can be controlled by the power frequency fs and by voltages vs. Figure 13.10 shows the output mechanical power Pm and power efficiency as parameters of speed or slip. Maximum efficiency is obtained by operating with almost-synchronous speed. We will discuss IM control further in the following sections and in Chapters 17 and 18. 13.3.4

Torque, Air-Gap Flux, Speed, and Power as the Basis of Power-Electronic Control

We need to further discuss the torque, flux linkage, and rotor speed of a squirrel-cage IM in relation to the voltage and current of the motor, which is given by Eqs. (13.56)–(13.61b). Figure 13.11 shows the equivalent circuit of one phase under balanced three-phase normal operation, which is the same as Figure 13.6; the equivalent load resistance rr slip has been modified to the additional form rr (pure rotor resistance) plus rr 1 slip − 1 (equivalent resistance of the rotor mechanical load). (The upper bar and upper wave symbols showing PU values and phasors are omitted here.) rr ωs =r equivalent total resistance of the rotor circuit including mechanical load slip r ωs −ωm rr

1 ωs − 1 = rr equivalent resistance of the load slip ωs − ωm

13 62

13.3 Squirrel-Cage Induction Motors

(b) vs :large

4 poles f1 = 50 Hz Torque characteristics of the load

Initial torque

Torque

Operating point

(a)

r'r :large Torque

20 Hz 30 Hz

40 Hz

vs :large r'r :small

vs :middle r'r :small vs :middle r'r :large

} v :small r' :large}

Characteristics of the load

s r

500 1000 1500 [rpm] (s = 0 for (s = 0 for (s = 0 for Slip = 0 20 Hz) 30 Hz) 40 Hz) Rotating speed

0 Slip = 1

→ Slip

Slip = 1

Speed zero

(a) Characteristics with different frequency fs = ωs / 2𝜋

Slip = 0

Synchronous speed

(b) Characteristics by different voltage vs and rotor resistance r'r

Figure 13.9 Torque-speed characteristics.

Tmax

Torque Tm

40 Characteristics of the load

2

[N-m] 60

20 5A

0

1 (0 rpm)

0.5

→ Slip

sm

0

s0 0 (1500 rpm)

Current ias , torque Tm

Mechanical output Pm

4

Pm

Current ias

Power factor PF efficiency η

Operating zone [A]

[kW] 6

(a) Mechanical output, current, and torque

[%] 100 PF

80 60 40

η

PF = 11.3%

Efficiency

20

η = 0%

0 1

0.5

→ Slip

(0 rpm)

0 (1500 rpm for four poles)

(b) Power factor and efficiency

Figure 13.10 Operating characteristics of a squirrel-cage IM.

Referring to Figures 13.6(b) and 13.11, if a balanced set of three-phase sinusoidal voltages vas at frequency fs = ωs/2π is supplied to the stator, it results in a balanced set of currents, which establishes a flux-density distribution in the air gap. The rotor rotates with a constant synchronous speed ωs: ωs =

2π p 2 2 2 = 2πfS = ωS rad s 1 fs p p

ωS 120 fS = Ns = 60 × p 2π where fs , ωs stator-supplied source frequency and angular velocity p number of poles Ns synchronous speed of the rotor in rms revolutions per minute

13 63

361

362

13 Induction Generators and Motors (Induction Machines)

Note: 𝜔s = 2𝜋fs : stator electrical angular speed 𝜔m = 2𝜋fr : rotor mechanical angular speed rs

ias

Lls

iar

+

+

𝜈as at 𝜔s = 2𝜋fs

eM

slip =

Llr

iM

r'r

LM

𝜔s – 𝜔 m : slip 𝜔s

Rotor resistance

Total rotor load resistance r'r / Slip

Equivalent load resistance –

– r'r

𝜔m 1 –1 = r'r · 𝜔s – 𝜔 m Slip

Figure 13.11 Equivalent circuit of an IM (balanced three-phase steady-state condition).

The air-gap flux (or main flux) ϕag and the flux linkage ψ ag = ns ϕag (ns is equivalent to the number of turns in the stator phase coil) rotate at a synchronous speed relative to the stationary stator windings. Consequently, a counter-EMF eM (air-gap voltage) is induced in each of the stator phases at frequency fs. Elements rs and Lls are the resistance and leakage inductance of the stator coil, which are ignored under ordinary operating conditions (except under the initial startup process of, say, ωm ≈ 0–0.5). The stator coil current ias in Figure 13.11 is branched to iM of the main air-gap flux pass and to iar of the equivalent resistive pass of the rotor with resistance rr slip . The magnetizing component current iM establishes the air-gap flux where the following equation is satisfied: ψ ag = ns ϕag = LM iM eM =

13 64

dψ ag dϕag = ns dt dt

The air-gap flux that links the stator coil and rotor coil has a sinusoidal waveform. Then ϕag t = ϕag sin ωs t eM =

dψ ag = ns ϕag ωs cos ωs t dt

ϕag ωs

13 65

Obviously, eM is proportional to the flux magnitude ϕag and the rotating velocity ωs. Torque is produced by the interaction of the air-gap flux and the rotor current. If the rotor is rotating at a synchronous speed, there is no relative motion between the rotating air-gap flux ϕag and the rotor, so there are no induced rotor voltages, rotor currents, and torque. In contrast, if the rotor is rotating with ωm at the slip speed ωslip = ωs – ωm, voltage and current are induced on the rotor coil: ωslip = ωS − ωm slip speed , fslip = fS − fm = slip fS slip frequency Slip =

ωS − ωm fS − fm NS −Nm = = slip ωS fS NS

13 66

Faraday’s law indicates that the induced voltages in the rotor circuit are at the slip frequency Slip fs = fs – fm, which is proportional to the slip speed. Then, induced current and flux in the rotor circuit are also at the slip frequency slip fs. The slip-frequency current produces field flux that rotates at the slip speed ωslip with respect to the rotor speed ωm: that is, at speed ωs = ωslip + ωm with respect to the stator. The resistance rr /slip in Figure 13.11 is the equivalent total resistance of the rotor-circuit pass, which is decomposed into pure rotor resistance rr and the equivalent motor-load resistance rr (1/slip – 1) = rr (ωm/[ωs – ωm]). If the rotor is operated in synchronism with the stator-coil frequency (slip 0, ωs ωm), then rr /slip becomes quite large, so current does not flow into the rotor circuit branch: ias = iM + iar iM, iar 0. In contrast, with the condition slip 1 or ωm 0,

13.3 Squirrel-Cage Induction Motors

then rr /slip becomes smaller and the rotor current becomes larger, which suggests that the rotor current becomes quite large in the rotor startup region from the standstill condition. Now, referring to Figure 13.11 and the current and torque in Eqs. (13.56) and (13.61b), we will examine torque, flux linkage, and speed in relation to voltages and currents. The following equations can be derived: jωs LM rr slip + jωs Llr ias jωs LM + rr slip + jωs Llr

vas = rs + jωs Lls ias +

13 67

The equation can be modified as follows: vas rs + ωs A + jωs B

ias =

ias e − jδ

vas

ias =

rs + ωs A 2 + ωs B

2

where A=

rr slip ωs L2M rr slip

2

2

rr slip

B = Lls + LM δ = tan −1

+ ω2s LM + Llr rr slip

2

13 68

+ ω2s Llr Llr + LM

2

+ ω2s LM + Llr

2

ωs B rs + ωs A

and vas =

rs + ωs A + jωs B ias

∴vas i∗as = rs ias i∗as + ωs A ias i∗as + jωs Bias i∗as

13 69a

The power balance of one phase is ∴vas i∗as =

2

2

rs ias + ωs A ias + jωs B ias

∗ 1 2 Stator power Stator copper loss ∗

2

∗

∗ 3 4 Rotor Excitation mechanical power input

13 69b

Referring to Figure 13.11, the equivalent total rotor resistance rr slip can be divided into the rotor-coil resistance rr and the equivalent mechanical-load resistance rr 1 slip − 1 ; and the rotor-input power (air-gap power) can be written as follows: 2

Pm = ωs A ias = ∗

5 Rotor mechanical input

rr fm 2 2 ir ir = r r + r slip fs −fm r ∗ ∗ 6 7 Rotor Equivalent coil resistance of resistance the load

fm 2 = rr ir + r ir fs − fm ∗ ∗ 8 9 Rotor heat Load mechanical loss work power 2

13 70a

363

364

13 Induction Generators and Motors (Induction Machines)

fm r ir fs − fm r

Pload =

2

load mechanical power

13 70b

The rotor input power Pm is consumed as load mechanical work power (∗9), but partly consumed as rotor heat loss (∗8). The rotor mechanical input (∗5) is equal to the product of the mechanical angular velocity ωm and torque T. Then 2

2

3 ωs A ias = 3 nωm A ias = ωm 3 nA ias ∴ T = 3 nA ias

2

= ωm T

2

where ωs electrical angular velocity of the stator coil

13 71

ωm mechanical angular velocity of the rotor n paired number of poles ωsA (neglect rs) for simplicity. Then, referring to Eq. (13.68),

We assume rs 1

ias =

A2

+ B2

vas ωs

2

T = 3 nA ias = 3 nA vas , T ωs

∴ias

i 2as

vas 1 A2 + B2 ωs vas ωs

13 72a

2

2

13 72b

Next, from the equivalent circuit, vas = jωs LM iM = jωs ϕag = where

rr slip

∴ ir

vas slip

rr + jωs Llr slip

ir =

rr slip

2

+ ωs Llr

2

ir

rr ir slip

ωs Llr in ordinary operating condition

13 73a 13 73b

The air-gap flux ϕag = LM iM is ∴ϕag = LM iM

1 ir ωs slip

vas ωs

13 74

In the next equation, we look from another physical viewpoint. We know Eq. (13.65) for the stator coil is eM = ns ∴eM

dψ ag = ns ϕag ωs cos ωs t dt ϕag ωs or eM ϕag fs

ϕag ωs = ϕag 2πfs

13 75a 13 75b

This is because the air-gap flux rotates with speed ωs with respect to the stator coil. The rotor-coil equation should be written in the same form, but ωs should be replaced by ωslip = 2πfslip = 2πfs slip because the same air-gap flux rotates with the speed ωslip or fslip. Then dψ ag = nr ϕag ωslip cos ωslip t dt r rr and er = r + jωr Llr iar iar slip slip er = nr

∴er

ϕag ωslip or er

ϕag fslip

ϕag ωslip = ϕag 2πfslip 13 76a

13 76b

13.3 Squirrel-Cage Induction Motors

We also need to discuss torque T: ias = iM + iar T = kψ ag iar sin δ

13 77a

where δ = ∠iM , iar ψ ag = ns ϕag = nr ϕag δ torque angle The magnetizing current iM, which produces flux linkage ψ ag, lags the air-gap voltage eM by torque angle 90 . The equivalent rotor current iar lags eM by θ, where θ is the power factor of the equivalent total rotor load rr slip + jωs Llr . And θ = tan −1

ωs Llr ≈ 0 because usually ωs Llr rr slip

rr slip

then

13 77b

δ = ∠iM , iar ≈ 90 Therefore T = kψ ag iar sin δ ≈ kψ ag iar α ϕag iar

13.3.5

13 78

Startup Operation

We will discuss a small startup region of an IM: point b to c, in Figure 13.8. Referring to Figure 13.12a, the ordinary operating region is very narrow within a band of, say, s = 0–0.1; and the accelerating torque ΔT must be provided by the stator electrical power in order to start up the motor from a standstill condition and to bring it to ordinary speed. However, as shown in Figure 13.12b, a large rotor current ir (probably five or seven times the ordinary operating current) flows during the speed-up process in the lower-speed region. This is because rr slip in Figure 13.11 is very small for its small speed of slip ≈ 1. Bearing this in mind, we intend to realize constant torque (T) startup control, as shown in Figure 13.13a. Referring to Figure 13.9a, and due to power-electronic inverter control, frequency fs = ωs/2π can be changed to be continuously higher: 10 20 50/60 Hz. On the other hand, we know the linear relationship between T and slip at any fs from Eq. (13.80), which will be given later as the last equation in this chapter. Therefore, the torque can be kept constant by maintaining slip at a constant value at any fs of the starting process. Figure 13.13 shows such an operating condition, assuming ϕag is kept constant. The torque-speed characteristics shift horizontally in parallel. Comparing two points f1 and f2, an equal torque can be delivered at both frequencies by maintaining ωs slip1 = ωs slip2. In this case, appropriate torque constant startup control can be realized by changing the frequency fs continuously, while maintaining ϕag at a constant value (to obtain the same gradient torque–speed curve) and keeping fs slip constant. 13.3.6

Rated-Speed Operation

Referring again to Figure 13.13, after reaching the ordinary operating region, operation should be near the rated speed, occasionally requiring power control, while the torque can be reduced by reducing ϕag, because the speed need not be accelerated further. If constant power Pload operation is intended in this region, it is realized typically by maintaining vas at the rated value and keeping the current ias at a reasonable required value, or by keeping the flux ϕag at a reasonable value, while also maintaining fs control. 13.3.7

Overspeed Operation and Braking Operation

By increasing the stator frequency above its normal value, it is possible to increase the motor speed beyond the rated speed. The region 1.8 ≥ ωm ≥ 1.0 in Figure 13.13 shows such operating conditions. In most adjustable-speed drive applications, the motor voltage does not exceed its rated value. Therefore, by keeping vas at its rated value and by increasing the frequency fs, vas/fs as well as ϕag are reduced. Further, for higher-speed operation with ωm ≥ 1.8, the speed is controlled by vas fs2 . Note that in the case of high-speed operation with ωm ≥ 1.0, ϕag decreases so much that the motor approaches the pull-out torque.

365

366

13 Induction Generators and Motors (Induction Machines)

Tr Trated Pull-out torque

2.0

1.5

Ordinary operating limit

Accelerating torque 𝛥T = Tem–Tload

Electromechanical torque Tem 1.0 (rated)

Load torque Tload

0.5

Rated 0.0

0 1.0fslip 1.0

0.2 0.8fslip 0.8

0.4 0.6 0.6fslip 0.4fslip 0.6 0.4 Rotating speed

0.8 0.2fslip 0.2

𝜔m 𝜔s 1.0 0 fsl 0 S

Ordinary operating region (a) Torque versus speed ir (ir)rated 6.0

Large current at slow speed

5.0

4.0

3.0

2.0 1.0 (rated) Rated

0.0 0 1.0fslip 1.0

0.2 0.8fslip 0.8

0.4 0.6 0.6fslip 0.4fslip 0.6 0.4 Rotating speed

0.8 0.2fslip 0.2

1.0 0 fsl 0 Slip

𝜔m 𝜔s

Ordinary operating region (b) Rotor current versus speed Figure 13.12 Startup process ordinal operating region.

In many applications, repeated braking operation is required in order to quickly reduce the motor speed or to bring it to a halt. Effective power can be collected by operating in generation mode (regeneration), if necessary. A typical example is in applications for railway car driving systems, where electrical braking has become a major practice instead of mechanical braking. The operation is performed by utilizing symmetrical torque-speed characteristics of the motor (see Figure 13.8) and power-electronic ignition switching control, which we will discuss in Chapter 17.

13.4 Proportional Relations of Mechanical Quantities and Electrical Quantities as a Basis of Power-Electronic Control

Constant-torque region T ≈ constant

Tm

f1

f2

f3

f4

Constant-speed High-speed region region T ∝ 12 𝜔m 1 T∝ 𝜔 as

Torque

T

0

fs . slip1

fs . slip2

fs . slip3 fs . slip4 → 𝜔r (a) Torque-speed characteristics

T, Vas, Pm, fslip

Torque T Rotor output Pm

Stator voltage 𝜈as

Slip angular velocity fslip = fs – fm = fs . slip

0 Stator resistive drop compensation

→ Speed 𝜔r (b) Torque, speed, and power control

Figure 13.13 IM characteristics.

13.4 Proportional Relations of Mechanical Quantities and Electrical Quantities as a Basis of Power-Electronic Control Rotating machines transform electrical power and mechanical power, either one-way or in reciprocal ways. The most popular applications are IM driving systems consisting of a combination of power-electronic control systems; mechanical measures (typically mechanical power Pm, torque T, and physical position, speed, and acceleration) are controlled using voltages v, currents i, or flux linkages φ. So, the equations to correlate mechanical quantities and electrical quantities are an important basis of motor-driving control. From this viewpoint, we present the following equations as the conclusion of our studies from the previous sections: From Eq 13 72b , ias

vas ; then i∞ v f or i ∞v ω ωs

From Eq 13 72 or Eq 13 79a , T

i2s

vas ωs

13 79a

2

; then T ∞ i2 ∞ v f

vas slip ; then i ∞v slip 1 vas ir ; then φ ∞i∞ v ω From Eq 13 74 , ϕag = LM iM ωs slip ωs

From Eq 13 73 , ir

From Eq 13 75b , eM

ϕag ωs or eM

ϕag fs ; then eM

φf

2

13 79b 13 79c 13 79d 13 79e

367

368

13 Induction Generators and Motors (Induction Machines)

vas ωs

From Eq 13 76b , T

2

2

vas fs

ϕ2ag ; then T

i2r ; then P Slip

From Eq 13 70b , Pload

fm 2 i fs − fm r

From Eq 13 71 , ωm T

i2as , ωm = ωs 1− slip

ias =

i 2M + i 2r i 2r = i 2as − i 2M =

v f

2

φ2

13 79f

i Slip

13 79g

i2as T

13 79h 13 79i

Pload slip

Further, referring to Figure 13.8, the ordinary operating region is within points c and e, and the torque T and the slip slip have an almost-linear relationship at any value of T. Then T = k slip within smaller region of slip ≈ −0 1 to + 0 1

13 80

Now we have found the basis of various motor-driving control systems, as follows: Torque T constant operation To keep vas/fs constant while changing fs Torque T constant operation To keep ϕ2ag constant Torque T constant and rotary speed ωm constant operation To keep vas/fs constant while changing fs, and to keep i2as T constant while changing ias Torque T constant and rotary speed ωm constant operation To keep vas/fs constant while changing fs, and to change i2as T while changing ias We will discuss these control practices in detail in Chapters 17 and 18.

Supplement 1 Calculation of Eqs. (13.17)–(13.19) cos θs −

cos θs Ds Ls Ds− 1 =

2 3

=

2 3

cos θs −

2 3

3 l s + Ls 2

1 1 ls + Ls − Ls − Lr 2 2

2π 3

− sin θs +

2π 3

1 1 − Ls ls + Ls − Ls 2 2

cos θs +

2π 3

3 l s + Ls 2

0

0

2π 3

− sin θs +

2π 3

0

3 ls + Ls 2

0

0

0

3 ls + L r 2

1 2

− sin θs 1 2

1 2 2π cos θs − 3 2π − sin θs − 3 1 2

Ds− 1

1 1 − Ls − Ls ls + Ls 2 2

1 2

2π 3

cos θs =

2π 3

1 2

− sin θs − sin θs − 1 2

cos θs +

− sin θs − sin θs − 1 2

cos θs

2π 3

cos θs +

2π 3

2π − sin θs + 3 1 2

0 0 0 1 − Ls 2

1 1 1 − Ls − Ls − Ls 2 2 2 +

1 1 1 − Ls − Ls − Ls 2 2 2

Ds− 1

1 1 1 − Ls − Ls − Ls 2 2 2 − sin θs

cos θs

1

0 0 0

cos θs −

2π 3

−sin θs −

2π 3

1

3 3 3 2 2 2

cos θs +

2π 3

−sin θs +

2π 3

1

13.4 Proportional Relations of Mechanical Quantities and Electrical Quantities as a Basis of Power-Electronic Control

3 0 0 2 =

2 3 ls + Ls 3 2

0

3 0 2

0 0

Ds M s Dr−1 =

2 3

cos θs −

0 0 0

2π 3

cos θs +

2π 3

2π 3

− sin θs +

2π 3

cos θm LM

1 2 2π 3

cos θs − θm

cos θs −θm −

− sin θs − θm

2π −sin θs −θm − 3 0

=

cos θs − θm +

0

2π 3

cos θm −

2π 3

cos θm

cos θm +

2π 3

cos θm −

2π 3

2π −sin θs − θm + 3 0

0

cos θm +

0

cos θs −θm

ls

cos θm −

2π 3

cos θm +

2π 3

2π 3

Dr−1

cos θm −sin θs −θm

1

cos θs − θm −

2π 3

− sin θs − θm −

2π 3

1

cos θs − θm +

2π 3

− sin θs − θm +

2π 3

1

Ds M s 1 0 0 3 = LM 2

0

3 ls + Lms 0 2

0

3 3 l s + Ls − L s 2 2

1 2

0

3 ls + Lms 2 3 l s + Ls 2

=

9 0 0 2

− sin θs − sin θs − 1 2

2 3 = LM 3 2

1 − Ls 3

3 2

cos θs

3 ls + Ls 2

0 0 0

0 1 0 0 0 0

The trigonometric formulae in Appendix A are useful for calculating Ds Ms and Ds Ms Dr−1 .

Dr−1

369

371

14 Directional Distance Relays and R–X Diagrams 14.1

Overview of Protective Relays

14.1.1

The Missions and Duties of Protective Relays

Rapid fault tripping by high-speed protective relays is essential for the stable operation of modern power systems. Among various types of relays, the two principal types are directional distance relays and differential relays (including carrier differential relay equipment for transmission-line protection). This chapter mostly concerns directional distance relays, because they are the most widely used as essential relays for various protective purposes, and their theoretical background and setting operation are complicated. Wherever a fault occurs in a network, a minimum section, including the faulted point, must be removed immediately so that the power system maintains sound operation. Operating times for protective relays are typically 1.0–3 cycles, and the tripping times of breakers are 1.5–3 cycles, giving a total fault-tripping time of 2.5–6 cycles (50–120 ms at 50 Hz, 42–100 ms at 60 Hz). Malfunction or a slight delay in fault tripping by the associated relays can, in rare cases, trigger serious blackouts as the result of a domino effect, meaning disruption of a power system caused by the following system behaviors:

•• • ••

Cascade trips caused by over/undervoltages, over-currents, or backup tripping at the adjacent substations. Power-stability collapse (step-out) or voltage-stability collapse (instability caused by the imbalance of real power and/or voltage loss). Cascade trips of generators caused by abnormal frequencies exceeding each machine’s over- or under-frequency limits. Tearing (cutting) of a network caused by simultaneous lightning faults. Tearing (cutting) due to extended cascade trippings (multiple faults, extensive damage to equipment, faulty operation, backup trips of protective relays, etc.) and so on.

Protective relays shoulder the vitally important duty of protecting the power system from such threats or at least minimizing the failed tripping zone. The duties of protective relays can be classified as follows:

• • ••

To detect the faulted section immediately and to order tripping commands to the associated breakers of the section that includes the fault: – Primary protection detects the minimum section, including the faulted point. – Backup protection detects the faults of adjacent sections and orders tripping commands to adjacent breakers whenever primary protection fails to trip the faulted section. Continuous monitoring of various electrical quantities (voltages v(t); currents i(t); power P, Q, or S; frequency f; phase angle δ; etc.). Reclosing (in the case of overhead transmission lines). Instability preventive control (a combination of several protective relays over a wide region).

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

372

14 Directional Distance Relays and R–X Diagrams

Protective relays have the responsibility to conduct instantaneous fault tripping or network connection changes. This last item is a relatively new application for protective relay engineering. For example, a few sections may be tripped almost simultaneously by plural faults, or by backup relays tripping (which may be caused when fault clearing by the associated primary relays or breakers fails). In such cases, a real-power imbalance condition (over- or underfrequency phenomena) or critical-stability condition (PQV critical condition) is apt to cause cascade behaviors. Control equipment to prevent instabilities continuously monitors the operation of relays installed at multiple stations and dispatches special commands (such as generator tripping, load shading, inter-tie tripping, etc.) whenever these conditions occur, in order to prevent power-system collapse due to the domino effect. 14.1.2

Classification of Relays

Protective relays may be classified based on the protective principles shown next: i) Directional distance relays (DZ relays) The appearance of high-speed directional distance relays (mechanical type) around 1950 should be remembered not only as part of the history of modern protective relay equipment, but also as the realization of modern powersystem operating practices. In the first half of the twentieth century, relays commonly had simple functions and slow operating speeds, typically like the induction disc-type OC (over-current) or UV (undervoltage) relays. Technical improvements led to the appearance of greatly advanced relays that could judge the direction and modes of faults as well as measure the distance from the relay to the fault points. Furthermore, faults were detected within 2–3 cycles. These high-speed directional distance relays became the state of the art in complicated, modern, high-voltage transmission-line networks, used as essential relays for primary protection (carrier-relaying equipment) as well as for backup protection and for detecting loss of excitation, step-outs, etc. Today, directional distance relays still have these important roles, although the hardware has been changed from mechanical to electrostatic (solid-state analogue or digital). ii) Differential relays and pilot-wire relays Differential relays including pilot-wire relays for line protection have a long history and have mainly been used for generator and transformer protection. The inflow current iin(t) and outflow current iout(t) of the equipment (generator or transformer) are sent to the relay through current transformers, and the relay operates whenever the differential current Δi(t) = iin(t) – iout(t) exceeds a specified threshold. Today, advanced digital communication facilities through space communication networks or overhead ground wire (OGW) with optical fiber (OPGW; see Chapter 3) enable fault-current waveforms to be sent from one terminal station to another. Due to the widespread adoption of such facilities in recent years, differential relay practice has begun to be widely applied not only for equipment protection at generating plants/substations, but also for the primary differential protection of overhead and power cable transmission lines. iii) Other relays:

• •

Relays with single input quantities: Voltage relays (OV-Ry, UV-Ry), current relays (OC-Ry), frequency relays (OF-Ry, UF-Ry) Relays with plural input quantities: Power relay (P-Ry, Q-Ry, S-Ry), angular difference-detecting relays (Δθ-Ry) Differentiated value-detecting relays (dQ/dt-Ry, etc.)

The principles of these various relays, except distance relays, are rather simple. However, the fault-detecting principles, as well as coordination of setting the operating zone, for the many distance relays installed in neighboring substations is very complicated from the viewpoint of relay application engineering.

14.2

Directional Distance Relays (DZ-Ry) and R–X Coordinate Plane

This chapter concentrates on the application theory of directional distance relays, because they are the most essential relays and are widely applied for purposes such as detecting primary and backup faults in lines and equipment, detecting generator loss of excitation, detecting step-outs, etc. Further, the fault-detecting principles, drawing the characteristics of the R-X coordinates, and setting work are complicated.

14.2 Directional Distance Relays (DZ-Ry) and R–X Coordinate Plane

14.2.1

Fundamental Algorithms of Directional Distance Relays

High-speed directional distance relays (DZ-Rys) have the fundamental function of detecting the direction and distance of faulted points on a transmission line from the associated substation. The relays are installed on each feeder at a substation, to which the secondary voltages and currents of the potential transformers (PTs; another name for voltage transformers) and current transformers (CTs) at the associated feeder line are provided as shown in Figure 14.1a. The fundamental principles of the relays will be briefly described from an application viewpoint, although the actual hardware/software practices of analogue and digital relays are much more sophisticated. Six DZ-Rys are utilized as a set for feeder protection at one station terminal:

••

Phase-to-phase fault-detection relays (DZ-S) for phase-ab, phase-bc, and phase-ca Phase-to-ground fault-detection relays (DZ-G) for phase-ag, phase-bg, and phase-cg. Each relay produces new quantities, as described here: Va −Vb Ia − Ib

DZ-S for phase-ab

RY Z ab

DZ-G for phase-ag

Va RY Z a = Ia

=

14 1

(Relays for the other phases are described analogously.) The composed quantities V a − V b , I a −I b are called delta-voltages and delta-currents. The composed quantities from Eq. (14.1) have the dimensions of impedances Z t = V t I t , so the characteristics of the relays are usually Phase-comparator

Polar-voltage

V (t), v′(t)  90° V (t) I (t)

Composite signal v (t) = k⬔𝛼 · I(t)

Setting of k,⬔𝛼

v′ (t) = v (t) – V (t)

(a) Principle of fault detection

·

·

k⬔𝛼 – V· I

·

v·′= k⬔𝛼·I – V

X

𝛽

P

𝛽 k⬔𝛼

·

·

v· = k⬔𝛼·I

I

V

Light load

𝛼

𝛼

·

I

· v· = k⬔𝛼 · I ·v′ = k 𝛼 · I· – V·

⬔ · · 𝛽 = V, v´  90°

(b) Vector diagram Figure 14.1 Mho-relay (M).

Load zone

· V ·

O

R

Maximum load curve

⬔𝛼: designed value (80°, 85°, etc.) k : setting · distance

(c) Relay characteristics

373

374

14 Directional Distance Relays and R–X Diagrams

drawn as phenomena in R–X coordinates. Therefore we first need to investigate the definition of the R–X coordinate plane. 14.2.2

R–X Coordinates (R–X Diagram) and P–Q Coordinates

The characteristics of distance relays are usually given by R–X coordinates (R–X diagram), so let’s examine what the R–X diagram is. An arbitrary impedance V I = Z = R + jX can be plotted as a point (R, X) on an R–X coordinate plane, where Z, R, X are mutually related to apparent power by the following equations: S = P + jQ = V I ∗ 14 2

Z = R + jX =

V I

P + jQ = V

V R + jX

Z ∗ = R −jX =

V2 1 = P + jQ p + jq

Z = R + jX =

V2 1 = P − jQ p− jq

∗

2

∗

=

V R −jX 14 3

where p = P/V2, q = Q/V2 (the same definitions as in Eq. (12.18)). R ± jX is the inverse complex number of P jQ or p q. In other words, the R–X coordinate plane is defined as the inverse plane of the p–q coordinate plane, and a point (R, X) on the R–X coordinate plane is in one-to-one correspondence with a point (p, q) in the p–q coordinate plane. The behavior of power systems before and after unbalanced fault conditions and their relay-detecting capabilities are usually discussed using the R–X diagram, whereas a p–q diagram is used to show generator- or motor-capability characteristics under balanced three-phase conditions in most cases. 14.2.3

R–X Characteristics and Algorithms of Distance Relays

Distance relays are usually explained using the characteristics of the R–X diagram. This is the generic name for relays with various characteristics as shown in Figure 14.2. A relay with particular characteristics is called an impedance relay, reactance relay, mho relay, offset-mho relay, etc. for historical reasons. A directional distance relay is a representative relay that typically includes two reactance units for zone-one and zone-two distant fault detection and a forward directional unit (mho-unit) for detecting in the forward direction and zone-three distant fault detection. X Forward offset mho characteristics Mho characteristics Backward offset mho characteristics Reactance characteristics Ohm characteristics O

R

Impedance characteristics Figure 14.2 Directional-distance relays, varying based on operational characteristics.

14.3 R–X Diagram Locus under Fault Conditions

Figure 14.1a shows the fundamental structural design diagram of an electrostatic Mho-Ry. The signal voltage V(t) and current I(t) from the secondary terminals of the associated PT and CT are applied to the relay, and the new quantities v(t), v (t) compose the detected part of the relay circuit, as shown in the figure. Then, the phase angular difference between v(t) and v (t) is continuously compared in the phase comparator part of the relay circuit. As shown in Figure 14.1b, if the condition β ≦ 90 is satisfied, the vector relation between V(t) and I(t) must be within the circle. In other words, a relay that is designed not to operate at β ≧ 90 but at β ≦ 90 gives V–I characteristics of the circle, as shown in the figure. The relay would be operated whenever the vectors V, I satisfy the condition β ≦ 90 in this figure. Incidentally, all the equations of the V–I coordinate plane in Figure 14.1b can be divided by the current I t , and the derived equations contain the variables V I, or Z = R + jX In other words, the V–I characteristics of the figure can be replaced with the R–X characteristics given in the R–X coordinate plane, and the resulting operating characteristics are also a circle with the dimensions of Ω, as shown in Figure 14.1c. The relay installed at point O in the R–X coordinate plane will operate if the vector V I enters the inner zone of the circle. This means the relay operates only when V I is in one direction and within a fixed-length value (distance). This explanation is the essence of DZ-Ry. The design details of these relays, manufactured using advanced technology as mentioned, are omitted in this book. Nevertheless, the relays must detect faults exactly and within 15–40 ms, and incorrect operation cannot be allowed under serious transient voltage and current conditions accompanied by bad DC or harmonic superposed waveform distortions. The relays must detect the direction and distance of the fault even if the voltage V of V/I is almost zero during the fault at a close point, or even if there are errors in the PT or CT, for example. Details of such countermeasures are beyond the scope of this book.

14.3

R–X Diagram Locus under Fault Conditions

14.3.1

Directional Distance Relay(44S-1, 2, 3 Relays) for Line-to-Line Fault Detection

Referring to Figure 14.3a, the behavior of the directional distance relay installed for line 1 at terminal m is examined, where a phase-bc line-to-line fault occurs at point f on line 1. The equivalent circuit in this case is shown in Figure 14.3a. The current at point f is supplied from both sides of the terminal at m and n. The following equations are derived from the equivalent circuit, where we assume that the positive impedance and the negative impedance are the same (this assumption can be justified in most cases, because the difference between the positive- and negative-sequence reactances of the generator is diluted by the line reactances Xline1 = Xline2): f V1 − f V2

= R f If

f V2

= f Z 1 If

f V1

=

m V1 − z1 m I1

f V2

=

m V2 − z1 m I2

m I1

= C1 I f ,

m I2

14 4

= − C 2 If

The load current is zero. Then m I1

= − m I2 = C 1 If , C 1 = C 2

where z1: positive-sequence impedance from relay point m to fault point f f Z1: positive- (and negative-) sequence impedance looking into the circuit from point f Rf: arc resistance C1: the ratio of mI1 to If in the positive-sequence circuit C2: the ratio of mI2 to If in the negative-sequence circuit (C1, C2 are the vector coefficients of (0 − 1) ∠ α, where α is almost 0 .) Since the positive- and negative-sequence reactances are equal, and the no-load condition is assumed, then C1 = C2 and mI1 = − mI2 in this case.

375

376

14 Directional Distance Relays and R–X Diagrams

m

Line #2

zb1

E

n E´

Z

f

Line #1

(a) System m

n

zb1 z1

mI1

E mV1

f V1

E´

If

f z1

Rf zb1 z1 m I2

mV2

f V2

f Z´ 1

f Z˝1

(b) Equivalent circuit 7

–a2Rf

jX

Ry Zca

C1

Phase b-c relay

j 3 f Z1 C1

6

Rf

(Vb – Vc)/(Ib – Ic)

2C1

2

z1

Load zone before fault

–j 3 f Z1 C1

3

4

RyZbc

R

60°

0 f Z1

zb1

60°

–aRf Ry Zab

60°

1

(c) The relay impedances when three relays

Ry Zbc, Ry Zca, Ry Zab

5

C1

observe phase b-to-c fault

Figure 14.3 DZ(S)-Ry operational characteristics.

Now, the delta quantities ΔV = Va − Vb, ΔI = Ia − Ib are applied to the relay at point m, and the relays look for the fault by measuring ΔV/ΔI = (Va − Vb)/(Ia − Ib). The measured impedances of each delta-phase RyZab, RyZbc, RyZca are calculated as follows: i) 44S-2 relay impedance R y Zbc

=

m Vb − m Vc m Ib −

= z1 +

m Ic

=

RyZbc

m V1 − m V2

f V1 − f V2

2

m I1

for phase bc line-to-line detection:

m I1 −

m I2

=

f V1

+ z1

R f If Rf = z1 + = z1 + 2C1 If 2C1

− f V2 + z 1 I − m 1 m I2

m I1

m I2

14 5

14.3 R–X Diagram Locus under Fault Conditions

ii) 44S-1relay impedance Ry Zab

=

m Va − m Vb m Ia −

= z1 + = z1 + = z1 +

m Ib

RyZab

=

=

m V1 − a m V2 m I1 − a

f V1 −a f V2 m I1 −a m I2

= z1 +

1 −a f Z1 If + Rf If − a 2 C 1 If

m I2

=

f V1

+ z1

− a f V2 + z 1 I − m 1 a m I2

m V1

m I2

1 −a f V2 + Rf If 1+a =

m I1

14 6

z1 + a2 − a f Z 1 − a Rf C1

− j 3f Z1 aRf − C1 C1

iii) 44S-31relay impedance

Ry Zca

for phase-a-b line-to-line detection:

m Vc − m Va m Ic − m Ia

for phase ca line-to-line detection:

RyZca

= z1 +

a −a2 f Z 1 −a2 Rf C1

14 7

2

j 3f Z1 a Rf = z1 + − C1 C1 Figure 14.3c shows the vector diagrams of these equations and explained here. i) Relay

RyZbc:

Eq. (14.5): Relay

RyZbc

RyZab,

RyZbc,

RyZca

that are derived in the R–X coordinate plane using

at point O sees the impedance as given by Eq. (14.5) and Figure 14.3c:

0 ② z1 + ② ③ Rf 2C1 = 0 ③ ② ③ is not long because Rf has a small value by nature, although C1 varies with the value C1 = 0 – 1. ii) Relay RyZab: Eq. (14.6): Referring to the equivalent circuit in Figure 14.3b, straight line 0 ② (z1: line impedance between f and m) and ① 0 (zb1: the back impedance at m) can be written first in the R–X coordinate plane as shown in Figure 14.3c. The straight line ① ② is of course z1 + zb1. Now, we will examine the special case of C1 = 1.0, which means all the current If at fault point f is supplied from point m through line 1 and is not supplied from point n. This means f Z1 = ∞ in the positive-sequence equivalent circuit. Accordingly, the positive-sequence impedance looking into the circuit at point f (fZ1) is equal to z1 + zb1. In other words, fZ1 = z1 + zb1 under the condition C1 = 1.0. Next, the straight line ① ② can be multiplied by − j 3, and − j 3f Z1 is obtained as the new straight line ② ④ This new line ② ④ corresponds to the second term on the right in Eq. (14.6), where C1 = 1. The straight line ② ③ corresponds to Rf/2C1 with C1 = 1.0, which can be multiplied by –2a (turning 120 clockwise and doubling the length) to obtain the new line ④ ⑤ The straight line ④ ⑤ is −aRf/C1, and it corresponds to the third term on the right in Eq. (14.6). As a result, the derived straight line 0 ⑤ satisfies the following relation under the condition C1 = 1, whose right side is the same as that of Eq. (14.6) with C1 = 1: 0⑤

Ry Zab

= ① ② z1 + ② ④ − j 3f Z1 C1 + ④ ⑤ − aRf C1

In other words, the straight line 0⑤ in Figure 14.3c gives the RyZab of Eq. (14.6) under the condition C1 = 1. The ratio C1 can be varied between 0 and 1 depending on the fault location and the power source condition of both terminals m and n. The magnitudes of fZ1/C1 and Rf/C1 therefore must be modified so that the length of straight lines ② ④ (then ① 0) and ④ ⑤ may be expanded. iii) Relay RyZca: Eq. (14.7): In the same way, the relay impedance RyZca corresponding to Eq. (14.7) can be drawn as shown in Figure 14.3c. Now, if the phase-bc line-to-line fault occurs at point f on line 1 (where the line impedance between m and f is z1), the DZ-Rly of each phase at the installed point m measures its own impedance zone within each parallelogram in

377

378

14 Directional Distance Relays and R–X Diagrams

Figure 14.3c. Note that relay RyZbc measures the length of z1 0 ② (the distance between m and f ) exactly, regardless of the magnitudes of Rf and C1 (where the length of ② ③ is small in comparison), and, accordingly, the relay will operate if the measured length z1 is shorter than the diameter of the circle (the preset value). The relay does not detect faults beyond the previous set distance in the forward direction, or faults in the backward direction. Figure 14.3 also explains why relays RyZca as well as RyZab do not detect the phase-bc line-to-line fault. Incidentally, the magnitude of P + jQ before the fault is within the capacity limit, while the relation between (P, Q) and (R, X) is defined by Eq. (14.3). Then, the operating point (R, X) before the fault must exist in the furthest area from point 0 (0, 0). In other words, before the fault, the three relays are observing the load (R, X) in the farthest-right area for the load flow in the point m-to-n direction (or farthest-left area for the load flow in the opposite direction). Whenever a phase-bc line-to-line fault occurs, the locus point (R, X) for each relay moves from the load area to the fault area of the parallelogram in Figure 14.3c, and relay RyZbc detects the fault as a result, although RyZca and RyZab may not. This theoretical visualization based on the R–X diagram was first presented by A. R. van C. Warrington in his famous AIEE Transactions paper “Performance of Distance Relays” in 1949. Following this seminal work, analytical methods of faulting phenomena and relay operation against various complicated phase-unbalanced faults were established, and the technology of high-speed protection for large networks progressed remarkably. The faulting phenomena as well as the behavior of protective equipment based on directional distance relays cannot be appropriately described by any other method today, except Warrington’s analytical method.

14.3.2

Directional-Distance Relay(44G-Relays) for Line-to-Ground Fault Detection The equivalent circuit for a phase-a line-to-ground fault (1 ϕ G) is given in Figure 14.4. In this case, a 44G-relay for line-to-ground fault detection cannot be realized by the principle of measuring Va/Ia by analogy with a 44S-relay in which (Vb − Vc)/(Ib − Ic) is measured. Special countermeasures are required for 44G-relays, because the zerosequence circuit has very different impedance constants than the pos3Rf itive-sequence circuit, and, furthermore, mutual reactance exists between parallel circuits of double-circuit transmission lines (see Chapter 3, Figure 3.12). From the equivalent circuit for Figure 14.4,

z1 mV1

mI1

mV2

m I2

f V1

z1

f V2

f Z1

f Z1

m I0´

mV0

m I0

If z0

f V0

f Z0

Figure 14.4 Equivalent circuit of a phase-a line-toground fault.

m V1 − f V1

= z1

m I1

m V2 − f V2

= z1

m I2

m V0 − f V0

= z0

m I0

+ Z0M

f V1

+ f V2 + f V0 = 3Rf If

m I1

=

m I0

= C 0 If

f V2

= − f Z 1 If ,

m I2

m I0

14 8

= C1 If assuming load flow is zero f V0

= − f Z0 If

where z0M : zero-sequence mutual reactance between circuit 1 and 2 (the same as Z0M in Figure 3.12 and Eq. (3.39a)) fZ1, fZ0 : impedance at the fault point f looking into the circuit Relay

RyZa:

m Va

=

Applying these equations, m V1

+ m V2 +

m V0

= z1

m I1

+ z1

m I2

m I0

+ z0M

= z1

m I1

+ m I2 + m I0 + z0 − z1

= z1

m Ia

+ z0 −z1

m I0

+ z0M

m I0

+ z0

+ 3Rf If

m I0

m I0

+ z0M

+ 3Rf If

m I0

+ 3Rf If 14 9a

14.3 R–X Diagram Locus under Fault Conditions

This equation can be modified as follows: m Va −

Ry Z a

z0 − z1

m I 0 − z0M m I 0

m Ia

where m Ia

2C1

3Rf

z1

If m Ia

14 9b

C0 If

Equation (14.9b) shows that the relay can measure the line-to-ground fault exactly by applying the composed voltages of the numerator on the left side of Eq. (14.9b) to the relay instead of Va. This is possible because the zero-sequence currents m I0 , m I0 of lines 1 and 2 are available in the same substation through the composition of the three-phase currents Ia, Ib, Ic and Ia , Ib , Ic For DZ(G)-Ry for phase w (where w = a,b,c), the measuring equation of the phase-to-ground relay is Ry Zw

m Vw −

=

z0 −z1

m I 0 − z0M m I 0

m Iw

Voltage quantity

m Vw −

Current quantity

m Iw

z0 − z1

m I0 −z0M m I0

where w = a, b, c = 1 3

m I0

− z0 −z1 − z0M

m Ia

m I0

= 1 3

m Ia

+ mI b + mI

14 10 c

zero-sequence compensation term of line 1

mI 0

zero-sequence compensation term of the parallel circuit line 2

m I0

m I0 , m I0

+ mI b + mI c ,

can actually be derived from the residual circuit of three CT secondary circuits

of composed values 1 3 Ia + Ib + Ic and 1 3 Ia + Ib + Ic i) Relay

RyZa:

Ry Za

Utilizing these equations,

= z1 + 3Rf

If 3Rf = z1 + 2C1 + C0 m Ia

14 11

Relay RyZa can measure (detect) phase-a to grounding faults accurately by the countermeasure of zero-sequence current compensation. ii) Relay RyZb: The voltage and current quantities for this relay are calculated here under the conditions from Eq. (14.8): m Vb −

Voltage quantity

f V1

+ z1

= z1

m Ib

+ f Vb

+a Current quantity

z0 − z1

2

m I 0 − z0M m I0

mI 1

= f V0 + z0

+ a f V2 + z1

mI 2

mI 0

− z0 − z1

+ z0M

m I0

m I 0 −z0M m I0

14 12a

m Ib

Then, the impedance for the phase-b relay observing the phase-a line-to-ground fault is Ry Zb

= z1 +

f Vb

14 12b

m Ib

Although this equation looks to be in good order, we need to examine the second term fVb/mIb on the right: f Vb

= f V0 + a2 f V1 + a f V2 = f V0 + a2 3Rf If − f V2 − f V0 + a = a −a =

m Ib

2

f V2

+ 1−a

2

f V0

2

+ 3a

R f If

a2 − a f Z1 + a2 −1 f Z0 + 3a2 Rf If

= m I 0 + a2

mI 1

+a

mI 2

f V2

= m I 0 − m I 1 = − C 1 − C 0 If

14 13a

379

380

14 Directional Distance Relays and R–X Diagrams

a −a2 f Z1 + 1 −a2 f Z0 −3a2 Rf

∴ Ry Zb = z1 +

C1 −C0

14 13b

j 3f Z1 − j 3a f Z0 − 3a2 Rf = z1 + C1 −C0 Relay RyZc The following equation is derived in the same way: Ry Zc

= z1 +

a2 − a f Z1 + 1 −a f Z0 − 3a Rf C1 − C0

14 14

−j 3f Z1 + j 3a2 f Z0 −3a Rf = z1 + C1 − C0

Equations (14.11), (14.13), and (14.14) specify the behavior of relays RyZa, RyZb, RyZc for the phase-a line-to-ground fault. Equation (14.11) tells us that the phase-a relay RyZa measures z1 (the direction and distance of the fault) appropriately. The operating zone of RyZa can be drawn in the R–X coordinate plane by using C1, C0 and Rf as parameters. On the other hand, RyZb, RyZc include fZ0 as well as C1, C2, Rf, so they can be drawn as a point in the R–X coordinate plane for the case when fZ0 is given as the individual fault condition. For the special case of C1 ≒ C0, the relays encounter the following impedances: Rf C1

Ry Za

= z1 +

Ry Zb

= z1 + ∞

Ry Zc

= z1 + ∞

14 15

The phase-a line-to-ground fault can be accurately detected by RyZa, while RyZb, RyZc do not operate in this case. Table 14.1 shows the impedances that the six relays see for line-to-line faults and line-to-ground faults under various fault modes. The table is based on that given by A. R. van C. Warrington in the previously mentioned AIEE paper. Each equation in the table can be derived, although this takes time.

Table 14.1 Impedances observed by directional distance relays for various short circuits. Relays 3ϕS, 3ϕG 2ϕS (phase-b to c) RyZab

RyZbc

RyZca

RyZa

RyZb

RyZc

z1 +

Rf C1

z1 +

a2 − a f Z1 −aRf C1

Rf z1 + C1 Rf z1 + C1

Rf z1 + 2C1

Rf C Rf z1 + C1

∞

z1 +

Rf z1 + C1

z1 +

1ϕG (phase-a to ground)

z1 +

a− a2 f Z1 + 1 − a2

2ϕG (phase-bc to ground) f Z0

+ 3Rf

3C1

∞

a −a2 f Z1 −a2 Rf C1

z1 + z1 +

z1 +

f Z1 − a a −a2

z1 +

f Z1 − aRf a 2 − a C1

z

Rf C1

z1 + z1 +

a2 −a f Z1 + 1 − a

f Z0

+ 3Rf

3C1 3Rf 2C1 + C0 a− a2 f Z1 + 1 − a2 f Z0 −3a2 Rf C1 − C0 a2 −a f Z1 + 1 − a f Z0 −3aRf

C1 −C0

Note: z1: positive-sequence line reactance from the relay to the fault point. fZ1, fZ0: positive/zero-sequence impedances looking into the network at the fault point C1 (C0): the ratio of mI1 to If (complex number) C1 = mI1/If, C0 = mI0/If Rf, rf: arc resistance (Rf, r in the case of 2ϕG corresponds to R, r in Table 7.1b)

rf + C1 C1 rf z1 + C1 z1 +

3 f Z1 + rf f Z0 + rf + 3Rf 2 f Z + rf + a− a f Z0 + rf + 3Rf

1 −a2

3 f Z1 + rf f Z0 + rf + 3Rf 2 f Z1 + rf + a −a f Z0 + rf + 3Rf

z1 +

rf + C1 C1

z1 +

3f Z0 + rf + 2Rf C1 −C0

z1 +

rf a2 f Z1 + rf + a2 −a f Z0 + rf + 3Rf − f Z1 + rf rf + 3Rf C1 a2 f Z1 + rf + a2 −a f Z0 + rf + 3Rf − C0 f Z1 + rf

z1 +

rf a f Z1 + rf + a−a2 f Z0 + rf + 3Rf − f Z1 + rf rf + 3Rf C1 a f Z1 + rf + a−a2 f Z0 + rf + 3Rf − C0 f Z1 + rf

a −1

14.4 Impedance Locus under Ordinary Load Conditions and Step-Out Conditions

X Ry Zc

∠(+30º) ∠(+ 90º)

Rf c

c′

30º

∠(+ 90º)

k

3C1

f Z1

3C1

∠(+30º)

b

f Z1

z1

3C1

30º Ry Zb

0

b′ 30º R

30º 30º

Rf 3C1

The relay Ry Zb looks at the impedance 0b or 0b′ with the condition Rf = 0 or Rf ≠ 0, respectively. The relay Ry Zc looks at 0c or 0c′, respectively, in a similar way.

f Z1

C1

h

Figure 14.5 Relay loci of line-to-ground relays

14.3.3

RyZa, RyZb, RyZc

in the case of a phase-bc to ground short-circuit fault.

Behavior of 44G-Relays against Line-to-Line Short-Circuit Faults

Again we return to the case of phase-bc line-to-line faults (Figure 14.3a and b) and examine the behavior of phase-toground directional distance relays (44G-1,2,3) against the 2ϕS fault. Referring to Table 14.1, or quoting Eq. (14.4), the measured distances for relays RyZa, RyZb, RyZc against the phase-bc fault (2ϕS) can be calculated as follows: Ry Za

=

Ry Zb

=

Ry Zc

=

m Va m Ia m Vb m Ib m Vc m Ic

=

m V1 m I1

+ +

m V2 m I2

=∞

=

2 a2 m V1 + am V2 f Z1 − a Rf = z1 + = z1 + ∠ −90 2 2 a m I1 + a m I2 a− a C1

f Z1

=

am V1 + a2 m V2 f Z1 −aRf = z1 + 2 = z1 + ∠ + 90 a m I1 + a 2 m I2 a −a C1

f Z1

3C1 3C1

+ ∠ − 30

Rf 3C1

+ ∠ + 30

Rf 3C1

14 16

Figure 14.5 shows the impedance loci of relays RyZa, RyZb, RyZc that are drawn from these equations. The figure obviously indicates that the phase-a relay RyZa never operates against the b-c line-to-line fault, while phase-b relay RyZb and phase-c relay RyZc operate if the setting zone is large. The straight line h0k kb = 1 1 3 is satisfied, and the relation of ∠khb = ∠ khc = 30 is also satisfied.

14.3.4

Directional-Grounding Relay (67G-Relays) for a High-Impedance Neutral Grounded System

Directional distance relays (44G-relays) cannot be used for the line-to-ground fault detection in a high-impedance neutral grounded system, although 44S-relays can be used for line-to-line fault detection. This is because the ratio of C0 = mI0/If in Eq. (14.8) would become very small, and the angle might differ by almost 90 from that of C1. Accordingly, in Eq. (14.11), the second term on the right rather than the first term z1 becomes dominant. However, in case of a high-impedance neutral grounded system, directional grounding relays (67G-relays) can detect the fault direction correctly, although the fault distance cannot be measured (the explanation of 67G-relays is omitted from this book).

14.4

Impedance Locus under Ordinary Load Conditions and Step-Out Conditions

14.4.1

Impedance Locus under Ordinary Load Conditions

Figure 14.6a shows the power system we will examine here, which operates under balanced three-phase conditions. The induced voltages at points s and r are given by es and er , and the electrical angular displacement is δ (es lags er by δ ).

381

382

14 Directional Distance Relays and R–X Diagrams

(a) s

i·

m

z· s

n

z· l

z·r

r

e·s = Es e j𝜔t

e·r = Ere j(𝜔t+ 𝛿)

(b) I

jX

·

rZ

r z· r n

·

nZ

z·l

·

mZ

·

sZ

m

R

z· s s

(c) jX

𝛿= 0

e

0°

36

k: ·constant (k = 1.3) 𝛿

𝛿 = 90° c(k = 1.1, 𝛿 = 20°)

k = 1.3

18

0°

d

1.2

𝛿 =180° r

0° 1.1 18 R

0

O s

k >1

40°

z·s + z· l + z· r

30° 20°

k 1.0. When k 1.0, point o becomes more and more distant from a; and when k = 1.0, it at last converges to the straight line crossing the midpoint of a and b at right angles. In the case of k < 1.0, k can be replaced by 1/k in these equations, and the locus moves to a symmetrical position on the straight line crossing the midpoint of a and b at right angles.

389

390

14 Directional Distance Relays and R–X Diagrams

Supplement 3 Drawing Method for Z = 1 1 A + 1 B from Eq. (14.24) The drawing method for the vector Z from the given vectors A and B is examined here in reference to Figure 14.11: Z=

1 1 1 + A B

1

Accordingly, A = A 1 + ke −jδ 1+ B Z + Z ke − jδ = A Z=

A

2

A = B ke −jδ Z=

B

=

B 1 1 + e jδ k

B A 1 Z + Z e jδ = B k 1+

3

B=A

1 jδ e k

From Eq. (2), vectors A, Z, Z ke − jδ make a closed triangle Δoac, where the length of vector Z ke −jδ is obtained by expanding the Z vector k times and turning it δ degrees clockwise. Accordingly, if δ is constant, the vertical angle ∠aco = π − δ is constant, so point c is on the circular arc whose chord is A, and the vertical angle is (π − δ). In the same way, vectors B, Z, Z 1 k e jδ make a closed triangle Δobc, where the vector Z 1 k e −jδ is obtained by expanding Z (1/k) times and turning it δ degrees counterclockwise. Accordingly, if δ is constant, the vertical angle ∠bco = π − δ is also constant, so point c is on the circular arc whose chord is B, and the vertical angle is (π − δ). As a result, it is proved that vector Z (straight line oc) is given by the crossing point c of two arcs whose chords are A and B, and the vertical angles are (π − δ) for both arcs. Incidentally, ac co = oc cb = k 1

Δoac ∞Δboc

and

4

In conclusion, Z total in Eq. (21.24) can be drawn from Z load and Z fault by putting A method.

· Z·

b

·

𝛿

B

𝜋–𝛿 c 𝛿 · Z o

1 j𝛿 e k

· Z · ke –j𝛿

𝛿

c´

𝜋–𝛿

·

·

·

·

·

·

A = Z + Z · ke –j𝛿

· A

a

B =Z+Z·

Figure 14.11 Drawing method for Z = 1 1 A + 1 B .

1 j𝛿 ke

Z load , B

Z fault , Z

Z total in this

391

15 Lightning and Switching Surge Phenomena and Breaker Switching We have focused in the previous chapters mainly on the behavior of power frequency or transient phenomena in the lower frequency zone, but this is only a partial view of the entire image of power-system dynamic behavior. Powersystem networks have profound dynamic characteristics including behavior in the very wide frequency zone from DC to surge phenomena, so it is vital to understand the fundamentals of power-system networks including such behaviors. Theories of surge phenomena include lightning surges and switching surges, and further surge-protection theories including insulation coordination are also essential background for power-system engineering. We will study surge phenomena in this chapter, highlighting traveling wave theories and switching surge phenomena; theories of surge protection and of insulation coordination are almost entirely omitted because of limited pages. Table 15.1 shows a frequency-diagram map of the phenomena of a power system classified according to three frequency categories for the sake of convenience: ① The lower-order harmonic zone of zero to a few kilohertz ② The higher-order harmonic zone of a few kilohertz to hundreds of kilohertz ③ The surge-order zone of megahertz and gigahertz. The voltage and current characteristics of a single-phase transmission line for an electromagnetic wave are expressed by the distributed-constants circuit shown in Figure 15.1, and the approximated circuit is the concentrated-constants circuit (L-circuit, T-circuit, π-circuit, etc.). Analytical treatment is far easier using a concentrated-constants circuit, although the error is larger for higher-frequency phenomena. Therefore, evaluation of errors by such an approximation is always important, whenever we study higher harmonic phenomena using concentrated-constants circuits.

15.1

Traveling Wave on a Transmission Line, and Equations

15.1.1

Traveling-Wave Equations

Electricity propagates through a vacuum with a constant velocity c0, always accompanying the electric field and magnetic field as is the nature of electromagnetic waves. In the case of an overhead transmission line, an electromagnetic wave travels through air along with the transmission line, so the velocity is almost the same as that for a vacuum. Now, we need to examine propagation phenomena of electricity in terms of the physics of voltage and current. For this purpose, the transmission line should be modeled as a distributed-constants circuit, as shown in Figure 15.1. The circuit consists of four constants: inductance L[H/km], resistance r[Ω/km], leakage (stray) capacitance C[F/km], and leakage conductance G[ /km] (inverse value of leakage resistance [Ω/km]). These are the values per unit length of the longitudinal line, or, in other words, the specific values at one point on the circuit line. The leakage current is given by Ω

ileak t = ileak t + ileak t = GV t + C

dV t dt

or, for commercial frequency,

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

Table 15.1 Frequency map of power-system phenomena (examples). Equivalent circuit of transmission lines

DC

Application

Phenomena or analysis

Obstacle

DC transmission electrochemical industry

DC component of faultcurrent inrush current (transformer)

Current-zero-missing (breaker) saturation (CT, relay)

0–50/60 Hz

Very low harmonic waveform distortion (relay)

Power frequency 50/60 Hz

R–X circuit

Power utilization

Steady-state analysis: PQVI analysis, fault analysis Dynamic analysis: PQV stability, temporary overvoltage

Overvoltage/overcurrent tripping Voltage instability and power instability

Ripple-control (distribution line) refining industry Fault locator (pulse-radar type) Power line carrier

Lower harmonicresonance waveform distortion AVR response analysis Higher harmonic resonance Corona noise transfer voltage (transformer) Lightning-surge switching surge (very fast) chopped wave

Voltage flickering (various loads, various controls) waveform distortion (relay) Local heat vibration (generator/motor) capacitor overload Vibration, local heat (transformer, generator) Noise protection Surge protection

Z = R+jX

100 Hz

L-circuit

500 Hz

Z

1 kHz

C

10 kHz T-circuit p-circuit

100 kHz

Z/2

1 MHz

Z

Z/2

C/2

C/2

C

10 MHz distribution circuit

100 MHz

1 GHz

Note: Equivalent circuits should be appropriately selected considering the length of transmission lines, frequency, and acceptable error. Concentrated circuits rather than distributed circuits are realistic approaches for most practical engineering applications, so error-rate evaluation for concentrated equivalent circuits is important.

Equipotential surface

𝜈

𝜈 – ∂𝜈 dx ∂x ∂i i– dx ∂x

i Rdx

Ldx

Rdx

Cdx

Gdx Cdx

i

Ldx

Rdx

Gdx Cdx

x – dx x x + dx

–𝜈

𝜈 + ∂𝜈 dx ∂x i + ∂i dx ∂x

𝜈

–i Images of voltage and current surge

Figure 15.1 Distribution circuit of a transmission line.

Ldx Gdx

Rdx

Electromagnetic field

Electric lines of force

15.1 Traveling Wave on a Transmission Line, and Equations

ileak t = ileak t + ileak t = G + jωC V t = Y V t

15 1

Y = G + jωC where Y, G are called admittance and conductance, respectively. G is taken into account only in cases of high-frequency phenomena (typically surge analysis) because it usually has large parallel resistance. In Figure 15.1, v(x, t), i(x, t) are the voltage and current at point x and time t, which is located at a distance x from starting point s. The following equation can be derived for the very small section between points x and x + Δx: v x, t − v x, t +

∂v x, t ∂i x, t dx = Ldx + Rdx i x,t ① ∂x ∂t

∂i x, t ∂v x, t dx = Cdx + Gdx v x, t ② − i x,t + i x, t − ∂x ∂t ∂v x, t ∂i x, t =L + Ri x, t ① ∂x ∂t ∂i x, t ∂v x, t − =C + Gv x, t ② ∂x ∂t

15 2a

∴ −

15 2b

Partial differentiation of both sides of Eq. (15.2) ① by ∂/∂x yields a new equation. Then, substituting Eq. (15.2) ② into the new equation, the variable i can be eliminated, and an equation in only variable v is derived. In the same way, another equation can be derived in only variable i. The resulting equations are ∂ 2 v x,t ∂ 2 v x, t ∂v x,t + RG v x, t ① = LC + LG + CR 2 2 ∂x ∂t ∂t ∂ 2 i x, t ∂ 2 i x, t ∂i x, t + RG i x, t ② = LC + LG + CR 2 2 ∂x ∂t ∂t

15 3

Each of these differential equations is called a traveling-wave equation and is essential as a theoretical starting point for surge phenomena of transmission lines. The equation also has other names such as telegraph equation, wave equation, and hyperbolic differential equation, because it also becomes the theoretical starting point for other electrical, physical, and mechanical phenomena based on wave phenomena such as theories of signal metal cables, coaxial cables, waveguides, fiber-optical cables, mechanical dynamics of rods/pipes/towers (e.g. the dynamics of penstock), and so on. Holding our solution of Eq. (15.3) until later, we will investigate some special cases in order to find the physical image of this equation. Incidentally, the original equation before Eq. (15.3) was derived by William Thomson (Lord Kelvin), but in the form of an equation in which constants L and G were missing. Heaviside derived Eq. (15.3) as the telegram equation of a communication line; it was first applied to a transmission line by Blakesley.

15.1.2

The Ideal (No-Loss) Line

The first case is the ideal (no-loss) line. If the losses are negligible, namely R = 0, G = 0, Eqs. (15.2b) and (15.3) become the following equations: −

∂v x, t ∂i x, t = L ∂x ∂t

∂i x, t ∂v x, t = C − ∂x ∂t ∂ 2 v x,t ∂ 2 v x, t = LC ∂x2 ∂t 2 ∂ 2 i x, t ∂ 2 i x, t = LC ∂x2 ∂t 2

15 4

15 5

393

394

15 Lightning and Switching Surge Phenomena and Breaker Switching

The general solution of Eq. (15.5) is v x,t = v1 x− ut + v2 x + ut 1 v1 x− ut − v2 x + ut i x, t = Z0 where Z0 =

L C

surge impedance , u =

15 6 1 LC

velocity

The relation of v(t)/i(t) = Z0 is maintained over time for the surge impedance Z0 = L C. That Eq. (15.6) is the solution of Eq. (15.5) can be easily proven by substituting Eq. (15.6) into Eq. (15.5) and by applying some theorems of differentiation, but deriving Eq. (15.6) (the solution by d’Alembert) from Eq. (15.5) is the work of mathematicians. Now, although Eq. (15.6) was found as the solution of the traveling-wave equation, we cannot say anything at this moment about v1, v2 for its characteristics or its relation with v. However, the dimension of (x–ut) is distance and of t is time, so u should be velocity. Therefore, if we observe v1(x–ut) by moving along the transmission line with a velocity u, we should observe the same v1(x–ut) unchanged. In other words, v1(x–ut) runs along the line for a distance x at velocity u. v2(x + ut) also runs along the line for a distance (−x) at velocity u, in the opposite direction. Thus, v1(x–ut) is called the forward wave and v2(x + ut) the backward wave (or reverse wave). Eq. (15.6) can be understood as the voltage at point x and time t, v(x, t) expressed as the summation (or composite) of the forward wave and the backward wave. Eq. (15.6) also tells us by analogy that the current i(x, t) is expressed as the summation (or composite) of the forward wave i1(x–ut) = (1 /Z0) ∙ v1 (x–ut) and the backward wave –i2(x + ut) = (−1/Z0) ∙ v2 (x + ut), while the polarity of the backward wave has a minus sign. Figure 15.2 shows the physical image of the actual voltage v, forward voltage v1, and backward voltage v2, and the actual current i, forward current i1, and backward current i2 at point x and time t. The phenomena at position x and time t are the same. We need to recall some points about the physical meaning of the equations:

• • • •

We stated that voltage and current exist at point x and time t and called them expressions of v(x, t) and i(x, t) (see Figure 15.1); however, we did not explain the reason v(x, t) and i(x, t) exist. The equation was derived without any explanation of the location and kind of power source (such as the switching-in of a generator or lightning surges). The equation is the specific characteristics of the transmission line, which are not affected by outer circuit conditions or adding a power source. The actual existing voltage and current that we can measure on the transmission line are v(x, t), i(x, t), and the equation only says that the actual quantities v(x, t) as well as i(x, t) can be understood as the addition of the forward wave and the backward wave. In other words, v(x, t) can be decomposed into v1(x, t) and v2(x, t). The forward wave and the backward wave cannot be measured, and as such they are the conceptual voltage and current. Accordingly, it is nonsense to discuss why and from where the forward wave or the backward wave originates. The forward wave v1(x–ut), i1(x–ut) and the backward wave v2(x + ut), i2(x + ut) can be specified by the condition of the power source (generator, lightning surge, or switching surge) connected at the line’s starting point s, the condition of the transmission line from point s to x, and the condition of the line after passing point x, as explained later. In the case of the ideal (no-loss) line, the forward wave, the backward wave, and the original v(x, t) and i(x, t) will not be changed by waveform or velocity.

Incidentally, if a power source (a generator or lightning strike) is suddenly connected (or if any voltage or current is injected) at point x, the actual voltage and current surge begin to travel from the point to both sides (the forward direction and the backward direction) of the line. All of these quantities are measurable existing quantities and should not be confused with the conceptual v1, i1, v2, i2. 15.1.3

The Distortion-Less Line

The distortion-less line satisfies the following relation regarding the constants: α=

R G = L C

15 7

15.1 Traveling Wave on a Transmission Line, and Equations

s

x

x t

x′

Δx Δx/u

𝜈 (x′, t′)

𝜈 (x, t) 𝜈1 (x – ut)

𝜈2 (x + ut)

t0 = 0 t1

𝜈1 (x′ – ut′)

t2

𝜈2 (x′ + ut′)

t0 + Δx u t1 + Δx u t2 + Δx u

𝜈 (x, t) = 𝜈1 (x – ut) + 𝜈2 (x + ut)

𝜈 (x′, t′) = 𝜈1 (x′ – ut′) + 𝜈2 (x′ + ut′)

t t

Voltage t0 = 0 t1

t2

t0 = 0

t

t2 + Δx u t1 + Δx u t0 + Δx u

i1 (x – ut) = 1 𝜈1 (x – ut) Z0

t0 = 0

i2 (x + ut) = t1

–1 𝜈 (x + ut) Z0 2

t2 i1 (x′ – ut′) =

t0 + Δx u

1 𝜈 (x′– ut′) Z0 1 –1 𝜈 (x′ + ut′) Z0 2

i2 (x′ + ut′) =

t1 + Δx u t2 + Δx u t

i (x, t) = i1 (x – ut) + i2 (x + ut) i

i (x′, t′) = i1 (x′ – ut′) + i2 (x′ + ut′)

i t

Current t0 = 0 t1

t2

t t0 = 0

t2 + Δx u Δx t1 + u t0 + Δx u

Figure 15.2 Traveling waves on a transmission line.

395

396

15 Lightning and Switching Surge Phenomena and Breaker Switching

In this case, Eqs. (15.2b) and (15.3) are simplified as ∂v x, t ∂i x, t =L + αi x, t ∂x ∂t

①

∂i x, t ∂v x, t =C + αv x, t − ∂x ∂t

②

−

15 8

∂ 2 v x,t ∂ 2 v x, t ∂v x, t + α2 v x,t = LC + 2α ∂x2 ∂t 2 ∂t

①

∂ 2 i x, t ∂ 2 i x, t ∂i x, t + α2 i x,t = LC + 2α 2 2 ∂x ∂t ∂t

②

15 9

The general solution of the equation is given next (see Supplement 1 at the end of the chapter for the derivation): v x,t = e −αt v1 x− ut + v2 x + ut

①

e − αt v1 x− ut − v2 x + ut Z0

②

i x,t =

15 10

This solution includes the attenuation term e−αt in comparison with Eq. (15.6). In this case, the forward wave and the backward wave of the voltage and current attenuate as similar waveforms along the line. The relation of v(t)/i(t) = Z0 is always preserved while v(t), i(t) is attenuating.

15.1.4

Laplace Transformed Solution of Voltage and Current

Let’s go back to Eq. (15.2) and apply Laplace transforms. The equation written in the s domain (d/dt −

s) is

dV x,s = Ls + R I x, s −L i x,0 dx

15 11

dI x, s = Cs + G V x, s − C v x,0 − dx where v(x, 0), i(x, 0) are the initial value of the voltage and current at point x and time t = 0, respectively. V(x, s) or I(x, s) can be eliminated from the previous two equations: d 2 V x, s = γ 2 s V x, s + φv x dx2

①

2

d I x, s = γ 2 s I x, s + φi x dx2 15 12

where γ s =

Ls + R Cs + G

the propagation constant ②

di x, 0 φv x = L − C Ls + R v x,0 dx dv x,0 φi x = C − L Cs + G i x,0 dx

③

15.1 Traveling Wave on a Transmission Line, and Equations

γ(s) can be modified as follows: γ s =

Ls + R Cs + G = LC

s+

R L

G C

s + α 2 − β2

s + α + β s + α −β = LC

= LC

s+

④

where

R =α+β L G = α −β C

attenuation constant

α=

1 R G + 2 L C

wave length constant

β=

1 R G − 2 L C

velocity of propagation u =

1 LC

φv(x), φi(x) are the initial values at x at time t = 0, respectively, so if the line is not charged before t = 0, φv x = 0, φi x = 0

15 13

Then Eq. (15.12) is d 2 V x, s = γ 2 s V x, s ① dx2

15 14

d 2 I x, s = γ 2 s I x, s ② dx2 The general solution of the equation is V x, s = A s e −γ s x + B s e γ s x I x, s =

Z0 s =

1 Z s

A s e −γ s x − B s e γ s x

Ls + R 1 = Cs + G Y0 s

① ②

15 15

③

where Z0(s): the characteristic impedance (the surge-impedance operator) Y0(s): the characteristic admittance (the surge-admittance operator) That Eq. (15.15) ① is the solution of Eq. (15.14) ① can be easily confirmed. That is, differentiating Eq. (15.15) ① V(x, s) twice by x gives the equation that satisfies Eq. (15.14) ①. Equation (15.15) is the general solution of voltage and current at arbitrary point x, and we need to add some initial conditions to find a concrete answer. The initial voltage v(0, t) is injected at point x = 0 as a given value. Then V 0, s =

v 0,t

is the Laplace transform

15 16

In conclusion, V(x, s), I(x, s) by Eq. (15.15) give the voltage and current when initial voltage V(0, s) is given at the starting point s but in the s-domain.

397

398

15 Lightning and Switching Surge Phenomena and Breaker Switching

15.2

Four-Terminal Network Equations between Two Arbitrary Points

We will now examine the voltage and current relations between an arbitrarily selected sending point s (distance x = 0) and a receiving point r (distance x). Using x = 0 in Eq. (15.15), V 0, s = A s + B s

I 0, s =

1 Z0 s

① A s −B s 15 17

1 ∴ A s = V 0, s + Z0 s I 0, s 2 ② Bs =

1 V 0, s −Z0 s I 0,s 2

Substituting Eq. (15.17) ② again into Eq. (15.15), V x, s = I x,s =

e γ s x + e −γ s x e γ s x −e −γ s x V 0, s − Z0 s I 0, s 2 2 1

Z0 s

e γ s x − e −γ s x eγ s x + e − γ s x − V 0,s + Z0 s I 0,s 2 2

① 15 18a ②

or as a matrix equation V x, s I x, s

cosh γ s x = −1 sinh γ s x Z0 s

− Z0 s sinhγ s x

V 0,s I 0,s

cosh γ s x

15 18b

Also, the inverse of Eq. (15.18b) is (see Supplement 2 at the end of this chapter for the procedure) V 0,s I 0,s

cosh γ s x 1 = sinh γ s x z0 s

Z0 s sinh γ s x cosh γ s x

V x,s I x, s

15 19a

Equations (15.18b) and (15.19a) are the equations of a four-terminal network for the distributed-constants circuit. Equation (15.19a) can be rewritten as follows by changing the symbols of the quantities with suffix s or r (meaning the sending point and the receiving point), where the distance between points s and r is l: Vs s Is s

cosh γ s l 1 = sinh γ s l Z0 s

Z0 s sinh γ s l cosh γ s l

Vr s Ir s

15 19b

where V(x, s) = [v(x, t)], I(x, s) = [i(x, t)] γ(s) is given by Eq. (15.12) ④, and Z0(S) is given by Eq. (15.15) ③. This is the four-terminal equation between points s and r in its common form. These equations show the exact relation of voltage and current quantities between the two arbitrary points, and so they are important equations, because we can begin most of the analysis from them instead of starting from the differential equations already introduced.

15.3 Examination of Line Constants

For the distortion-less line, this is a special case of Eqs. (15.12) and (15.19): α= Z0 =

R G = β = 0, then L C 1 = Y0

L C

γ s = LC s + α = u=

s+α u

1 LC

15 20

where Z0 surge impedance, Y0 surge admittance γ s propagation constant, u velocity of propagation For the ideal (no-loss) line, α = β = 0, then 1 = Y0

L C 15 21 s γ s = LC s = u 1 u= LC We have now obtained the exact transmission-line equations in the Laplace transformation domain, which can be applied to any transient analysis of every frequency zone (ignoring non-linear losses like corona losses). By application of Laplace transforms, the symbol s = d/dt is used, and Ld/dt, Cd/dt are replaced by Ls, Cs respectively, and then algebraic manipulation can be applied. If steady-state phenomena of angular velocity ω = 2π f (f does not necessarily mean commercial frequency) are to be treated, we can replace s j with that for the four-terminal network for steady-state phenomena. Thus, for the distortion-less line Z0 =

γ = LC jω + α =

jω + α ① u

15 22

and for the ideal (no-loss) line γ = LC jω =

15.3

jω ② u

Examination of Line Constants

We will now examine typical values of constants such as γ(s), Z, u with regard to overhead transmission lines and power cables.

15.3.1

Overhead Transmission-Line Constants

The working inductance L and the working capacitance C of a transmission line are given by Eqs. (4.9a, b) and (4.25b), and typical magnitudes are shown in Tables 4.1 and 4.2. These constants are determined by the physical allocation of the line conductors (in other words, by the structure of the towers) and do not depend on induced voltages or frequency.

399

400

15 Lightning and Switching Surge Phenomena and Breaker Switching

Again ignoring corona loss and eddy-current loss and rewriting Eqs. (4.9) and (4.25b), for the positive-sequence inductance L = 0 4605 log10

sll sll + 0 05 mH km ≒ 0 4605 log10 × 10 −3 H km ① r r

and for the positive-sequence capacitance C=

0 02413 0 02413 −6 sll μF km = sll × 10 F km ② log10 log10 r r

15 23

Applying this equation, γ(s), Z, u of the ideal (no-loss) overhead transmission line (R = G = 0) are calculated next. For the velocity of propagation, 1 = LC

u=

1 0 4605 × 10

−3

× 0 02413 × 10 −6

= 300000 km s = 300 m μs

km s 15 24

velocity of light c in a vacuum

For the surge impedance, 1 0 4605 × 10 − 3 Sll Sll L = uL = − 6 log10 r = 138log10 r Ω 0 02413 × 10 LC L = 1mH km, C = 0 01 μF km

L = C

Z= if

15 25

then Z = uL = 300 000 × 1 × 10 − 3 = 300 Ω And for the constant of propagation, γ s = LC s =

s u

15 26

The logarithmic terms of L and C are canceled by calculation of LC, so the velocity of propagation u becomes the velocity of light c unconditionally (u = c). We learned in Section 2.5 and Tables 4.1 and 4.2 that the typical constants of overhead lines are L = 1 mH/km and C = 0.01 μF/km regardless of the rated voltage classes. As a result, typical surge impedance Z is around 300 Ω (the surge impedance of an overhead transmission line is practically 250–500 Ω).

15.3.2

Power-Cable Constants

Referring to Table 9.2, we will calculate the constants for typical CV and OF cable lines. For CV cable at 275 kV ∙ 2000 square class (single core), L = 0 392 × 10 −3 H km, C = 0 25 × 10 −6 F km Accordingly, u = 101 000 km s = 101 m μs, Z = 39 6 Ω

15 27

For CV cable at 154 kV ∙ 800 square class (single core), L = 0 404 × 10 −3 H km, C = 0 19 × 10 −6 F km Accordingly, u = 109 000 km s = 109 m μs, Z = 46 1 Ω

15 28

The surge impedance of the power cable is generally 15~50 Ω, which means an order of magnitude less than that of the overhead transmission line; and the velocity of propagation u is approximately 100–140 m/μs, which is about half or less of that in an overhead transmission line (see Chapter 4).

15.4 Behavior of Traveling Waves at Transition Points

15.3.3

Approximation of Distributed-Constants Circuits and Accuracy of Concentrated-Constants Circuits

We return to Eqs. (15.18) and (15.19) and continue our examination of the no-loss line for simplicity. Using R = 0, G = 0 in Eqs. (15.12) and (15.19), the four-terminal network equation becomes V 0, s I 0,s

=

cosh γl 1 sinh γl Z0

Z0 sinh γl

s where γ = LC s = , Z0 = u

V l,s I l, s

cosh γl

15 29

L , R = 0, G = 0 C

The hyperbolic cosine and sine functions can be expanded as follows using Maclaurin’s expansion theorem: γl 2 γl 4 γl 6 + + +… 2 4 6 γl 3 γl 5 γl 7 + + +… sinh γl = γl + 3 5 7 cosh γl = 1 +

15 30

If we ignore the second term and other smaller terms on the right side under the condition of smaller γl, for (γl)2/2 2), or γl V 0, s I 0,s

=

1 γl Z0

Z0 γl 1

cosh γl ≒ 1, sinh γl ≒ γl =

V l, s I l, s

≒

1 Cls

Lls 1

V l,s I l, s

15 31

l s u

This is the equation of the type-L, concentrated-constants circuit in Table 15.1. Incidentally, when we study the single-frequency characteristics of a line, we can let s a transmission line with length l and single frequency f, γl =

1

jω or j2πf. Then if we examine

jωl j2π fl = u u

15 32

Accordingly, the accuracy-evaluation curves with line distance l and frequency f with a given smaller parameter of γl can be obtained. Figure 15.3 is the graphic diagram of Eq. (15.32) under the condition u = 300 m/μs (overhead 2 line) and u = 150 m/μs (power cable), and under the three conditions (γl)2/2 = 0.05,0.1,0.15. For example, an overhead line can be expressed as a type-L concentrated circuit within about 10% error ((γl)2/2 ≦ 0.1) for the phenomena within 200 Hz for 110 km or 400 Hz for 55 km, and for a power cable of 200 Hz for 55 km or 400 Hz for 27 km.

15.4

Behavior of Traveling Waves at Transition Points

15.4.1

Incident Wave, Transmitted Wave, and Reflected Wave at a Transition Point

A point on a line at which there is an abrupt change of circuit constants (and, accordingly, an abrupt change of surge impedance) is called a transition point. Points such as an open- or short-circuited terminal, a junction with another line, a connecting terminal of a machine winding or capacitor, and a short-circuit fault point are typical transition points. When a traveling wave on a line reaches a transition point, part of the wave is reflected back along the line, and part passes on to other sections of the circuit. The impinging wave is called the incident wave, and the two waves arising at the transition point are called the reflected wave and transmitted wave, respectively. We will now examine the behavior of traveling-wave (surge) phenomena at the transition point, as shown in Figure 15.4. The surge impedance Z1 on the left-side line is abruptly changed to Z2 at the transition point a. The behavior when the incident wave from the left reaches transition point a is explained by the following equations:

401

15 Lightning and Switching Surge Phenomena and Breaker Switching

Hz 2000

}

}

402

Calculating condition 1000 5

Overhead line 10 15%

Overhead line

u = 300 m/μs

Cable line

u = 150 m/μs

Error level 5% : (𝛾l)2 / 2 = 0.05 500 Cable line 5 10 15%

300

(𝛾 l = 0.32)

10%:

= 0.10

(

15%

= 0.15

(

= 0.45) = 0.55)

200

300 km

100 5

10

20

30

50

100

200

Figure 15.3 Accuracy evaluation of a concentrated circuit by line length and frequency.

e

et it

i a

Z1 er –ir

Z2 Transition point

e, i : incident wave et, it : transmitted wave er, –ir : reflected wave

Figure 15.4 Behavior of voltage and current traveling waves at a transition point.

Incident wave e, i Transmitted wave et , it Reflected wave er , ir

e = Z1 i et = Z2 it er = Z1 ir

Continuity of voltages at both sides of point a

e + er = et

Continuity of currents at both sides of point a

i− ir = it

15 33

The following equation is derived from Eq. (15.33): er =

Z2 − Z1 e Z2 + Z1

2Z2 et = e Z1 + Z2

ir =

Z2 − Z1 i Z2 + Z1

2Z1 it = i Z1 + Z2

ρ=

Z2 −Z1 the reflection operator Z2 + Z1

2Z2 μ= the refraction operator Z1 + Z2

15 34

15.4 Behavior of Traveling Waves at Transition Points

The surge impedance is defined by Eq. (15.15), where ωL Z1 =

L1 , Z2 = C1

R, ωC

G for no-loss line. Then

L2 for the ideal no-loss line C2

15 35

Simplification by ignoring losses of R, G is justified for most analyses of surge phenomena. For the continuity of power at a transition point, the power at the impinging side (left side) is PA = e + er i −ir =

2Z2 e Z1 + Z2

2Z1 i Z1 + Z2

15 36

and the power at the transmitted side (right side) is PB = et it =

2Z2 e Z1 + Z2

2Z1 i Z1 + Z2

∴ PA = PB Thus the continuity of power at the transition point is secured. Note that there is an explanation where the current-reflecting wave ir is defined with an opposite polarity sign, so the equation is given by er/ir = −Z1, i + ir = it. The results are the same for either explanation.

15.4.2

Voltage and Current Traveling Waves at Typical Transition Points

Figure 15.5 demonstrates the behavior over time of impinging voltage and current traveling waves at four typical types of transition points:

• • • •

Case 1 (the transition point is opened), Z2 = ∞: When the current traveling wave i reaches the transition point, the reflecting wave ir(=i) arrives at the transition point from the opposite direction (right side) simultaneously, so the current at the point becomes i – ir = 0. The reflecting voltage is er = e (because er has the same polarity as ir), so the resulting voltage becomes e + er = 2e. Case 2 (the transition point is earth grounded), Z2 = 0: When the voltage traveling wave e reaches the transition point, the reflecting wave er(= − e) arrives at the transition point from the opposite direction (right side) simultaneously, so the voltage at the point becomes e + er = 0. The reflecting current is ir = −i (because ir has the same polarity as er), so the resulting current becomes i – ir = 2i. Case 3 (the transition point is grounded through a capacitor), Z2 = 1/sC: The grounding through capacitance acts as if the terminal is directly grounded for the voltage incident wavefront (because C ∙ de/dt is large), which acts as if the terminal is open for the voltage incident wave-tail (because C ∙ de/dt 0). Accordingly, the behavior of this case can be drawn analogously to cases 1 and 2. Case 4 (the transition point is grounded through a reactor), Z2 = sL: The grounding through inductance acts as if the terminal is open for the current incident wavefront (because L ∙ di/dt is large), which acts as if the terminal is grounded for the current incident wave-tail (because L ∙ di/dt 0). Accordingly, the behavior of this case can be drawn analogously to cases 1 and 2.

Figure 15.6 demonstrates the behavior of traveling waves at a transition point that is the junction of three lines with the same surge impedance (300 Ω). The behavior of traveling waves with a waveform of approximately a step function, and at typical transition points, can be drawn as was shown in Figures 15.5 and 15.6. The previous description is very helpful in the investigation of every kind of surge phenomena.

403

404

15 Lightning and Switching Surge Phenomena and Breaker Switching

Case 1: Z2 = ∞ (opening)

Reflected wave

Total voltage Incident wave Reflected wave

Total current

e

0

e

er

0

er e + er

e + er

0

i

i

0

–ir

0

0

–ir i – ir

i – ir

Case 3 (the transition point is grounded through a capacitor): Z2 = 1/sC

Case 4 (the transition point is grounded through a reactor): Z2 = sL

C

Incident wave

0

Reflected wave

0

Total voltage

0

Incident wave Reflected wave

L e

e er er e + er

e + er i

0 0

Z2 = 0

Z1

Z2 = ∞

Z1 Incident wave

Case 2: Z2 = 0 (short circuit)

–ir

i

–ir

i – ir Total current

0

Z2 =1/sC behaves like Z2 = 0 (metallic earth) for the initial time t = 0+, and like Z2 = ∞ (opening) for the transient ending time t = ∞

i – ir

Z2 = sL behaves like Z2 = ∞ (opening) for the initial time t = 0+, and like Z2 = 0 for the transient ending time t = ∞

Figure 15.5 Behavior of traveling waves at typical terminal conditions.

15.5

Surge Overvoltages and Their Three Different, Confusing Notations

Surge overvoltage and overcurrent phenomena can be understood or analyzed using a surge-impedance circuit. In such surge-impedance circuits, various differently defined surge voltages and currents are often handled with the same notation e, i or e(t), i(t), so engineers may be confused. Three different typical cases described in Figure 15.7 are compared as follows, to clear up such confusion.

nA Sectio 00 Ω a Z1 = 3 v

50 Ω

Z2 = 1

t line)

-circui

nB

Sectio

osite (comp

f value o

ouble 00 Ω d

3

l u 2𝜏u 3𝜏 0 𝜏u e + er = er = 2 e 3

e 0

er = 150 – 300 e = – 1 e 150 + 300 3

𝜏

et = 2 × 150 e = 2 e 300 + 150 3

𝜏u

2𝜏

et

e+er 3𝜏

2𝜏u

er

t i

u 2𝜏u 3𝜏 0 𝜏u

i

l i – ir = it = 4 i 3 ir = 150 – 300 i = – 1 i 150 + 300 3

𝜏 2𝜏

it = 2 × 300 i = 4 i 300 + 150 3

it i–ir

nB

Sectio

ir nA Sectio

3𝜏 t

Figure 15.6 Behavior of voltage and current traveling waves at a typical transition point.

Case 1: Calculation of appearing voltage ea at the transition point a when surge voltage e is injected at the source point ea Z1 a Line 1

e

Z2 •e Z1 + Z2

ea = Z2 Line 2

Z1

e

(1a)

a

Z2

(1b)

Case 2: Calculation of appearing voltage ea at the transition point a when impinging surge voltage e arrives at point a. ea = µ • e =

e

2Z2 e Z1 + Z2

ea = 2e

Z1 a Line 1

Z2 Line 2

2e

Z1

(2a)

a

Z2 • (2e) Z1 + Z2

Z2

(2b)

Case 3: Calculation of the case when transmitted voltage e begins to travel to the right Impinging wave

e Z1 + Z2 • Transmitted e µ = 2Z2 wave

ea = e

Transition point a Z –Z Reflected wave e – e = 1 2 µ 2Z2 (3) Figure 15.7 Three confusing cases of surge overvoltage phenomena.

406

15 Lightning and Switching Surge Phenomena and Breaker Switching

Case 1 Figure 15.7(1a): Surge voltage e is injected at the source point, and surge voltage ea appears at transition point a as a result. In this case, voltage e may be of an ideal generator or potential voltage of a thundercloud. In Figure 15.7(1a), two lines of surge impedance Z1, Z2 are connected at transition point a. Now, the left-side source voltage e is switched on, and surge voltage/current ea,i are caused at transition point a. The equations in this case are in the Laplace domain: e s − ea s = i s Z1 s

①

ea s = i s Z2 s

②

∴ ea s =

Z2 s es Z1 s + Z2 s

15 37

③

Equation ③ means ea(s) can be simply calculated as voltage e(s) divided by the ratio of Z1(s) and Z1(s). Therefore, ea(s) can be calculated using the equivalent circuit shown in Figure 15.7(1b). Case 2 Figure 15.7(2a): Surge voltage e arrives at transition point a as an impinging surge voltage at a. In Figure 15.7(2a), the impinging surge is traveling on line Z1 and will arrive at transition point a as impinging voltage e. Immediately after arriving at time t = 0+, transmitted voltage ea = μ e is caused at point a and begins to travel on Z2 in the right direction. As the refraction operator is given by μ = 2Z2/(Z1 + Z2), caused voltage ea is given by the following equation. ea = μ e =

2Z2 Z2 e= 2e Z1 + Z2 Z1 + Z2

15 38

The equation means ea(s) can be calculated as the voltage of 2e divided by the ratio of Z1(s) and Z2(s). Therefore, ea(s) can be calculated using the equivalent circuit Figure 15.7(2b), where source voltage 2e is given. Case 3 Figure 15.7(3): A surge voltage/current arrives at transition point a, and surge voltage e appears (or is caused) at transition point a. In this case, voltage e is defined as the resulting voltage at transition point a, which can be measured. Then, the impinged voltage at point a must be e/μ, where μ is the refraction operator. Voltage e is the transmitted voltage, which is equivalent to μ e in Case 2. The voltage arriving at point a is obviously e/μ in this case. Figure 15.7(3) shows the notation of this case. Now, comparing Case 1 and Case 2, voltage e is divided into the ratio of Z1 and Z2 in Case 1,while voltage 2e is divided into the ratio of Z1 and Z2 in Case 2. These two cases should be strictly distinguished without confusion. Comparing Case 2 and Case 3, the difference is the naming of e for the arriving voltage or the transmitted voltage. This may be another confusing matter.

15.6

Behavior of Traveling Waves at a Lightning-Strike Point

Now, we will investigate the aspects of a lightning-strike point, as shown in Figure 15.8a. Point a in the figure is the transition point of surge impedance Z1, Z2, and we will examine a transmitted surge voltage E(t) or surge current I(t) that is eventually injected at the transition point. This is considered a traveling wave from a cloud (where Z0 is the surge impedance of the lightning pass) that reaches transition point a whose surge impedance is Ztotal = 1 Z1−1 + Z2− 1 . We need to distinguish the following three expressions, which are apt to be confused with each other:

15.6 Behavior of Traveling Waves at a Lightning-Strike Point

E (t) v2 (t)

I (t) =

v1 (t) a

Z2 i2 (t)

E (t) i2 (t) Z1

i1 (t)

va (t)

i1 (t)

Z1 + Z2 Z2

E (t)

· I (t)

E (t) I (t)

a Z1

(b) Lightning current I(t) injected at point a

(a)

Z1 · I (t) Z2

a Z2

Z0

E (t) Ztotal

Z0

Z2

Z1· E (t)

Z1

(c) Surge current I(t) travels from point a to the right

Figure 15.8 Behavior of traveling waves at a lightning-strike point.

•

Case 1: Lightning voltage E(t) is transmitted at transition point a: This case corresponds to that in Figure 15.8b, where source voltage E(t) is suddenly injected at point a by closing a switch: Injected voltage at point a Injected current at point a I t =

E t

E t 1 1 = + Z1 Z2 Ztotal t

Traveling waves to the right Traveling waves to the left v1 t = E t

i1 t =

•

15 39

v2 t = E t

1 E t Z1

i2 t =

1 E t Z2

Case 2: Lightning current I(t) is transmitted at transition point a: This case is the same as Case 1, where source voltage E(t) = Ztotal I(t) is injected at point a by closing a switch: Injected voltage at point a E t = Ztotal t I t =

v1 t = E t = Ztotal I t

i1 t =

Injected current at point a

Z1 Z2 I t Z1 + Z2

Traveling waves to the right

•

E t

Z2 I t Z1 + Z2

I t

Traveling waves to the left

15 40

v2 t = E t = Ztotal I t

i2 t =

Z1 I t Z1 + Z2

Case 3: Traveling waves E(t) and I(t) caused at point a to the right: This case corresponds to that in Figure 15.8c, where source voltage E(t) = Z1 I(t) and source current E(t)Z1 + E(t)/Z2 are injected at point a.

407

408

15 Lightning and Switching Surge Phenomena and Breaker Switching

Injected voltage at point a Injected current at point a E t = v1 t = v2 t = Z 1 I t

i1 t + i2 t =

Z1 + Z2 I t Z2

Traveling waves to the right Traveling waves to the left

15 41

v2 t = E t = Z1 I t

v1 t = E t = Z1 I t i1 t = I t

i2 t =

1 Z1 v2 t = I t Z2 Z2

If lightning surge current I(t) is injected at point a of the non-transition point of the transmission line, the previous equations can be simplified by Z = Z1 = Z2. Accordingly, the same surge currents of (1/2)I(t) start traveling in both the right and left directions, and the induced voltage at point a, v(t) = {(1/2)I(t)} Z = I(t) {(1/2)Z}, also starts traveling in both directions. Incidentally, if line constants per unit length are written using the parameters L, R, C, G, then the constants per length l are L l, R l, C l, G l. So, we need to recognize that L, R, C, G and L l, R l, C l, G l may be called by the same names although they are different. In contrast, surge impedance L C is a specific value that always belongs to each arbitrary section, so a concept of L C l is nonsense. In other words, surge impedance is always a concept of per-unit length. We studied Figures 15.7 and 15.8 by imagining a lightning strike as the power source. However, the explanation in this section actually applies to switching phenomena.

15.7

Traveling Wave Phenomena of Three-Phase Transmission Lines

15.7.1

Surge Impedance of Three-Phase Lines

We will consider an arbitrary point p of a three-phase transmission line, where self- and mutual-inductances (Laa, Lab, Lac, etc.) and self- and mutual-capacitances (Caa, Cab, Cac, etc.) exist for a very small section. Now, if a current traveling wave ia(t) is running through the phase-a conductor in a left-to-right direction, voltage traveling waves va(t) = Zaa ia(t), vb(t) = Zab ia(t), vc(t) = Zac ia(t) ought to be running through the phase-a, b, c conductors, respectively, in the same direction, where Zaa, Zab, Zac are the surge impedances and are given by Zaa =

Laa , Zab = Caa

Lab , Zac = Cab

Lac Cac

15 42a

This means the matrix equation with regard to the accompanying voltage and current traveling waves and the surge impedance matrix can be derived. Assuming a balanced three-phase transmission line, the equation is va t vb t vc t

=

vabc t

where Zs =

Ls , Zm = Cs

Zs Zm Zm

Zm Zs Zm Z abc

Zm Zm Zs

ia t ib t ic t

or vabc t = Z abc iabc t

①

iabc t

Lm self-surge impedance, mutual surge impedance ② Cm vabc t

traveling-wave voltages

iabc t

traveling-wave currents

Z abc t

surge-impedance matrix

15 42b

15.7 Traveling Wave Phenomena of Three-Phase Transmission Lines

The matrix equation can be of course transformed into symmetrical components: v0 t v1 t v2 t

Zs + 2Zm 0 = 0

0 Zs − Zm 0

0 0 Zs − Zm

i0 t i1 t i2 t

Z0 =

Z1 Z2 Z 012

v012 t

i0 t i1 t i2 t i012 t

or v012 t = Z 012 i012 t where Z1 = Z2 = Zs − Zm =

① 15 43 Ls − Cs

Lm positive- negative- sequence surge impedance ② Cm

v012 t

Ls Lm +2 zero-sequence surge impedance Cs Cm positive-, negative-, zero-sequence traveling-wave voltages

i012 t

positive-, negative-,zero-sequence traveling-wave currents

③

Z0 = Zs + 2Zm =

Now it is clear that the traveling-wave (surge) theory can also be treated using symmetrical components. Analogously, the αβ0-method can be a useful analytical tool for surge phenomena.

15.7.2

Symmetrical Coordinate Analysis of Lightning Strikes

We will examine the case where lightning from a thundercloud directly strikes the phase-a conductor of a transmission line at point p, and the strike current I(t) of an approximate step function is transmitted into the phasea conductor, as shown in Figure 15.9a. Here, ideal insulation of the line is assumed, so the line structure withstands the induced surge voltages at each phase. The surge current I(t)/2 begins to travel to the right and to the left. To the right, traveling-wave voltages ZsI(t)/2, ZmI(t)/2, ZmI(t)/2 are induced on the phase-a, b, c conductors, respectively, by the phase-a current I(t)/ 2, and a voltage of (Zs − Zm)I(t)/2 is induced between the phase-a and -b conductors. In other words, these surge voltages begin to travel on each phase conductor. This aspect can be calculated by symmetrical coordinates as follows.

I1 (t) =

Point p I (t)

Phase-a Phase-b Phase-c

Positivesequence

𝜈a (t)

𝜈a (t) =

Zm 𝜈b (t)

𝜈b (t) =

Zm 𝜈c (t)

1 Z I (t) 2 s

1 Z I (t) 2 m 1 𝜈c (t) = Zm I (t) 2

Negativesequence

Zerosequence

(a) Three-phase domain Figure 15.9 Lightning striking a phase-a conductor.

1 I (t) 3 𝜈1 (t) =

1 Z · I (t) 2 1 1

1 I (t) 2 1 1 I2 (t) = I (t) 1 3 𝜈2 (t) = Z · I (t) 2 1 2 1 I 2 2 1 I0 (t) = I (t) 3 1 𝜈0 (t) = Z · I (t) 2 0 0 1 I 2 0

(b) Symmetrical coordinate domain

409

410

15 Lightning and Switching Surge Phenomena and Breaker Switching

Surge current I(t) is transmitted into the phase-a conductor at point p. Then the surge currents are Ia = I, Ib = 0, Ic = 0 1 ∴ I0 = I1 = I2 = I t 3

15 44

This means the lightning current (1/3)I is equally injected into point p of positive-, negative- and zero-sequence circuits, as shown in Figure 15.8b. To the right, the traveling-wave current is one-half of I(t)/3 (i.e. I(t)/6) at each sequence circuit. The traveling-wave voltages are v1 t = Z1

1 I1 t 2

=

1 Z1 I t 6

v2 t = Z1

1 I2 t 2

1 = Z1 I t 6

v0 t = Z0

1 I0 t 2

1 = Z0 I t 6

15 45

Then the inverse-transformed voltages are 1 2Z1 + Z0 I t 6 1 vb t = v0 t + a2 v1 t + av2 t = Z0 −Z1 I t 6 1 vc t = v0 t + av1 t + a2 v2 t = Z0 −Z1 I t 6 v a t = v0 t + v 1 t + v 2 t =

15 46

or 1 va t = Zs I t 2

Zs = where

1 vb t = vc t = Z m I t 2

1 2Z1 + Z0 3

15 47

1 Zm = Z0 − Z1 3

Equations (15.46) and (15.47) gives the magnitudes of induced traveling-wave voltages at point p on each phase caused by the injection of transmitted current I(t) to the phase-a conductor (lightning hits the phase-a conductor). In actual practice, the insulation of the line probably could not withstand the induced surge voltages, and faults would occur at some structural parts of the transmission line. However, this is another subject: whether the induced overvoltage exceeded the designed insulation level of the line. The topics of surge protection and insulation are omitted from this book.

15.7.3

Line-to-Ground and Line-to-Line Traveling Waves

Line-to-ground traveling waves and line-to-line traveling waves are sometimes useful concepts in practical engineering regarding surge phenomena. This is a kind of transformation of variables from a mathematical viewpoint. (Let’s call them l–g and l–l transforms.) Figure 15.10 illustrates the concept of such waves. Traveling waves va, vb, vc are transformed into newly defined traveling waves v0 (line-to-ground traveling waves), v12 (phase-a to phase-b traveling waves), and v13 (phase-a to phase-c traveling waves). The current waves are of course defined in a similar way.

Phase-a

𝜈0

Phase-b

𝜈0

Li ne -to

𝜈13

wa ve g

Li ne -to

-li ne

-li ne

tra ve lli n

d un ro -g Li ne -to

𝜈12

𝜈0

tra ve lli n

g

tra ve lli n

g

wa ve

wa ve

15.7 Traveling Wave Phenomena of Three-Phase Transmission Lines

– 𝜈12

Phase-c

– 𝜈13

Ground – 3𝜈0 Figure 15.10 Line-to-ground and line-to-line traveling waves.

The transformed equation is va t vb t vc t

=

1 1 1

1 −1 0

1 0 −1

D

vabc t

v0 t v12 t v13 t

or vabc t = D vl −g t

15 48

the same form for currents

vl − g t

The inverse matrix equation is v0 t v12 t v13 t

1 1 1 = 3 1

vl −g t

1 −2 1

1 1 −2

va t vb t vc t

or vl − g t = D −1 vabc t the same form for currents

15 49

vabc t

D−1

Substituting Eq. (15.40) ① and the current equation from Eq. (15.46) into Eq. (15.47), we have vl − g t = D − 1 Z abc iabc t = D −1 Z abc D il − g t

①

where D − 1 Z abc D = Z 012

②

v0 t v12 t v13 t vl − g t

15 50 =

Zs + 2Zm 0 0

0 Zs −Zm 0

0 0 Zs − Zm

D −1 Z abc D = Z 012

i0 t i12 t i13 t

or vl − g t = Z 012 il − g t

③

il − g t

Equation ② is proved by a physical calculation of D−1 Zabc D. The resulting equation ③ means the impedance Z012 in symmetrical components can be commonly used for that of l–g and l–l transformation. In the balanced three-phase transmission line, the traveling waves va, vb, vc and ia, ib, ic can be transformed into traveling waves v0, v12, v13 and i0, i12, i13, where v0, i0 are called phase-to-ground traveling waves, and v12, i12 and v13, i13 are

411

412

15 Lightning and Switching Surge Phenomena and Breaker Switching

called phase-to-phase (or line-to-line) traveling waves. This is our l–g and l–l transformation, where the impedances of symmetrical components Z012 are commonly used. For the phase-to-phase (or line-to-line) traveling waves v12, i12 and v13, i13, Ls − Cs and for the line-to-ground traveling waves v0 ,i0

Lm velocity ull = Cm

l − l wave surge impedance Zll = Zs −Zm =

① 15 51

Ls + Cs

l − g wave surge impedance Zlg = Zs + 2Zm =

1 Ls Cs

1 ② Lm Cm

Lm velocity ulg = Cm

Incidentally, Zll < Zlg , ull > ulg

15 52

These inequalities can be predicted from our knowledge of the characteristics of positive- and zero-sequence circuits. It is true that the line-to-ground traveling waves v0, i0 are a little slower in velocity and attenuate a little faster compared to those in phase-to-phase (or line-to-line) traveling waves v12, i12. The 1–g and 1–1 transform is obviously similar to the αβ0-transformation, where the line constants (L, C, R, G) and the surge impedances of symmetrical components are commonly applied. Analytical examination of traveling waves using 1–g and 1–1 transformation is very useful.

p0

Line #0 Surge-impedance

#1

p1

#2

p2

Z0

p3

Z1 𝜌1′

𝜇1′ 𝜌1

Z2

𝜌2

𝜌1

Z3

𝜌 2′

𝜇2′

𝜌3′

𝜇3′ 𝜌3

𝜇2

p1

#3

𝜇3

p2

p3

1

0

𝛼

1T

2T 𝜇′ 1 𝛼2 𝜌 2

𝛼 𝜌2

𝜌2

𝛼2

2

𝛼𝛽

𝛼 2𝜌

′1 𝜌

2

𝛼3

𝜌′1 𝜌

2

3T

𝛼 3𝜌

′1 𝜌

2

2 4 𝜌′ 1 𝜌 2

𝛼

4T

5T

𝛼𝜇

𝛼 4𝜌′ 2 2 2 m′ 1 1 𝜌2 𝜌 2 ′1

𝛼4 𝜌

6T Time Figure 15.11 The reflection lattice.

2

𝜇2

𝜇2 3 ′ 𝜌2 2 𝛼𝜌1 𝛽 r3 a ′ 𝜇2 2 𝜌 3𝜇 2 𝛽 𝛼 2 𝛼 𝛼5 2 𝜌′1 𝜌 2 𝛽 𝜌′ 𝜌 2 3𝜇 2 2

𝛼𝛽

𝛼 3𝛽

𝜌′ 𝜌 1

2

𝜇2 𝜌 3𝜇 2 𝛼 𝛽 𝜇 2𝜇 3

𝜇2

15.8 Reflection Lattices and Transient Behavior Modes

15.8

Reflection Lattices and Transient Behavior Modes

15.8.1

Reflection Lattices

We will examine the surge phenomena shown in Figure 15.11, in which the lines 0, 1, 2, 3 (the surge impedances Z0, Z1, Z2, Z3, respectively) are connected in series. The incident surge voltage is traveling along line 0 from left to right and arrives at transition point p1 at time t = 0. Then the transmitted wave e = 1 (step waveform of value 1) begins to travel along line 1 from left to right. The surge arrives at point p2 with value α (α is the attenuation ratio from traveling one way along line 1, α = 0–1) at time t = l1/u1 (≡T, the unit time of line 1); then the reflected wave (value αρ2) and the transmitted wave (value αμ2) appear and begin traveling. Therefore, the voltage at point p2 is zero for t < T, while it becomes (α + αρ2) or (αμ2) for time T ≦ t. The traveling wave repeats such reflection/refraction at each transition point p1, p2, p3 until it disappears by attenuation; thus the reflection lattice can be written as shown in Figure 15.11. Assuming lines 0 and 2 are far longer than line 1, the surge voltages at transition points p1, p2 can be derived as follows: Point p1 0 −2 T

4−6 T

2− 4 T

v1 t = 1 + α ρ2 + α ρ1 ρ2 + α 2

=

2

4

ρ1 ρ22

+α ρ 4

6−8 T 2 2 1 ρ2

+ αρ 6

2

1 + α2 1 + ρ1 ρ2 1 + α2 ρ1 ρ2 + α2 ρ1 ρ2 converging to

2 3 1 ρ2

8 −10 T

+α ρ 6

+ α2 ρ1 ρ2

3

3 3 1 ρ2

+ αρ 8

3 4 1 ρ2

+ α8 ρ 41 ρ42 +

4

+ α2 ρ1 ρ2 +

1 + α2 ρ2 1 − α2 ρ1 ρ2

①

Point p2 0−1 T 1−3 T

0 + α + αρ2 + α ρ1 ρ2 + α

v2 t =

5−7 T

3− 5 T 3

3

ρ1 ρ22

+ α ρ 5

0 + α 1 + ρ2

1 + α2 ρ1 ρ2 + α2 ρ1 ρ2

converging to

α 1 + ρ2 αμ2 = 2 1− α ρ1 ρ2 1 − α2 ρ1 ρ2

=

2

2 2 1 ρ2

+α ρ 5

+ α2 ρ1 ρ2

7−9 T 2 3 1 ρ2 3

+ α ρ 7

3 3 1 ρ2

+ α7 ρ 31 ρ41 +

4

+ α2 ρ1 ρ2 + ②

The above v2 obviously coincides with the equation below Point p3 0 −1 T 1 − 3 T

3−5 T

5−7 T

7−9 T

0 + αμ2 + αμ2 α2 ρ1 ρ2 + αμ2 α2 ρ1 ρ2

v2 t =

converging to

2

3

+ αμ2 α2 ρ1 ρ2 +

αμ2 1− α2 ρ1 ρ2

③

where Z0 , Z1 , Z2 , Z3 surge impedances of lines 0, 1,2, 3 respectively Reflection operators ρ, ρ and refraction operators μ, μ are ρ1 = Z1 −Z0

Z1 + Z0 , μ1 = 2Z1 Z1 + Z0

ρ1 = Z0 −Z1

Z0 + Z1 , μ1 = 2Z0 Z0 + Z1

ρ2 = Z2 −Z1

Z2 + Z1 , μ2 = 2Z2 Z2 + Z1

T = l1 u1

the unit time of line 1 length

Note that 1 + x + x2 + x3 + x4 +

etc

l1 , velocity

u1 = 1

L1 C1

= 1 1 −x 15 53

413

414

15 Lightning and Switching Surge Phenomena and Breaker Switching

(a)

=

+

(b)

=

+

(c)

=

+

(d)

=

+

(e)

=

+

(f)

=

+

(g)

=

+

Figure 15.12 Composition of waveforms.

The voltages v1(t), v2(t) for all times t = (0 − n)T can be calculated using these equations. The converged values of v1(t), v2(t) mean the final values of transient voltages as t = ∞ (steady-state voltage); v2(t) can be derived using either equation ② or ③. Note that the terms of the reflected waves from point p0 or p3 are not yet included in the equation. Accordingly, whenever the reflected wave from point p3 (or p0) arrives at point p2 (or p1), the related transmitted wave terms should be added to these lattice equations according to the rules of lattice analysis. Needless to say, if the length of the analytical line section Z1 is relatively short compared to the length of the next sections Z0, Z2, the transient behavior early on can be examined using Eq. (15.53) without any additional reflected voltage terms. Furthermore, the transmitted voltage with step function v(t) = 1(t) (which means the tail length is infinitely long and the value never attenuates) was assumed at point p1, although it is unrealistic. Any surge voltage attenuates and disappears eventually, so the transmitted voltage with waveform 1(t) in Eq. (15.53) must be replaced by that with a more realistic waveform. Figure 15.12 shows that a realistic waveform with finite wave length can be realized by combining two waveforms with infinite wave length. As a typical example, the waveform of the standard impulse voltage shown in Table 17.2 can be divided into two exponential attenuation curves, as shown in Figure 15.12c. 15.8.2

Oscillatory and Non-oscillatory Convergence

Examining Eq. (15.53) ①②, the equations for v1(t), v2(t) are simple increasing arithmetic series under the condition ρ1 ρ2 > 0, and the oscillatory series under the condition ρ1 ρ2 < 0 over time. This can be summarized as follows. For case 1a Z0, Z2 > Z1, ρ1 , ρ2 > 0; or for case 1b Z0, Z2 < Z1, ρ1 , ρ2 < 0. That is, v1(t), v2(t) are simple increasing nonoscillatory series, or in other words dv1(t)/dt > 0, dv2(t)/dt > 0. The voltage v(t) at any arbitrary point within the section Z1 is also non-oscillatory.

15.9 Switching Surge Phenomena Caused by Breakers Tripping

I0 zI0 R

E0 =0

E E0 I0 = 0

0

E0

e′ = –E0

0 T –E0

2T 3T 4T 5T

–E0

E0

–E 0

–2E0 +2E0

E0

4T

azI 0 a 2zI

5T

a2 zI 0 a 3zI

–E

0

E0 E0 –E 0

-2E0

a = –0.6 z E zI 0 = 0 R T azI0

7T

–E

0

E0

4E0

0

0

a3 zI 0 a 4zI 0 a4 zI 0

0

9T

(a) Grounding a charged line

(b) Removal of a short circuit

Figure 15.13 Line grounding and removal of a short circuit.

For case 2a Z0 > Z1 > Z2 (ρ1 > 0, ρ2 < 0), or for case 2b Z0 < Z1 < Z2 (ρ1 < 0, ρ2 > 0). That is, v1(t), v2(t) are oscillatory series over time; in other words, the derivative dv(t)/dt alternates plus and minus values. In Eq. (15.53), if α = 1 : 0 and ρ1 = 1.0, ρ2 = 1.0 are assumed, the voltage finally diverges to an infinite value. Of course, this is unrealistic, because the original transmitted traveling wave voltage is of limited tail length instead of a step waveform with infinitive tail length; furthermore, the attenuation factor is actually the value of 0 < α < 1.0, so v(t) converges eventually to a finite steady-state value. However, the converged value obviously becomes very large under the conditions α ≒ 0 and Z0, Z2 Z1. This case may be compared to a landscape where a narrow lowland Z1 is surrounded by high ground Z0, Z2 and water floods in from high ground Z0 to lowland Z1. A violent wave will thus be caused at Z1. Of course, this violent behavior would be reduced considerably if Z2 were also lowland at a similar level or even lower. Figures 15.13a and b show the voltage behavior when a line terminal end is switched on and off. The figure is reproduced from the great classical work Travelling Waves on Transmission Systems (1933), by L.V. Bewley, who originated the lattice method.

15.9

Switching Surge Phenomena Caused by Breakers Tripping

Power-systems engineering cannot be discussed without mentioning the switching operation, which is closely related to overvoltages and high-frequency transient phenomena. Switching overvoltages caused by tripping/closing operations of circuit breakers or line switches (LS) are inevitable phenomena that can be overcome by a combination of various specialized practical engineering countermeasures. In this section, we will study switching surges first through a mathematical treatment; explanations of insulation and breaker design engineering are omitted. Manual calculation of transient phenomena is not easy, and a simple circuit with even a few elements L, C, R cannot generally be solved without proper approximations. However, engineers need to find transient solutions in practical engineering for more complicated circuits. Accordingly, deriving an accurate solution through a reasonable approximation is an important part of practical engineering. Furthermore, engineers can generally discover the essentials of

415

416

15 Lightning and Switching Surge Phenomena and Breaker Switching

engineering practice through properly simplified model themes rather than through a more complicated large-scale theme. Calculating transient behavior by a breaker switching operation, as introduced in this section, is useful not only for a better understanding of switching phenomena but also as a good exercise in transient calculation techniques.

15.9.1

Calculating Fault-Current Tripping (Single-Phase, Single-Source Circuit)

To begin, we will carry out a transient calculation of the single-phase circuit shown in Figure 15.14a, in which a shortcircuit fault has occurred at point f and the breaker is going to be opened. Our objective is to find the transient voltage (switching surge voltage) across the two contacts, vBr(t) and the phase-to-ground voltage vb(t), which are integral to the required breaker’s tripping duties and the required insulation levels of the network. The circuit for the calculation is given in Figure 15.14b, where an L-type concentrated circuit is used. The transient calculation of this circuit can be solved as the superposition of the calculations in Figures 15.14c and d, using Thévenin’s theorem.

t=0 Br

Line #1

Line #2

a

b 𝜈Br(t)

f

e(t) = Ee j t or e(t) = E cos 𝜔t

(a)

Br L

r

i(t)

a

L

L r

b 𝜈Br(t)

C

Load

𝜈a (t)

Power source

=

𝜈Br(t)

+

C

(b)

i(t) 1(t)

r

C

(c)

Z(s) (d)

𝜈Br (t) ( = 𝜈a (t)) 2E

Natural frequency f=

1 2𝜋 LC

E Recovery voltage

5 ms 4.15 ms

t 10 ms (50 Hz system) 8.3 ms (60 Hz system)

–E i(t) 0

t

(e)

Fault current i(t) Figure 15.14 Transient calculation for single-phase breaker tripping (single-source circuit).

15.9 Switching Surge Phenomena Caused by Breakers Tripping

Step 1: Calculating steady-state term Whenever a short-circuit fault occurs, a protective relay detects the fault within 1–3 cycles and dispatches a tripping signal to the associated circuit breaker. The breaker begins to open the contacts immediately after receiving the signal and completes fault-current tripping at the time of current zero and within 2–3 cycles. We assume that the transient term of the fault current is already attenuated at the time of tripping completion, although actually the DC current component may not yet have disappeared (because the DC time constant could be of order 0.05–0.1 second). The source voltage is e t = Re Ee jωt = E cos ωt

15 54

The fault current i(t) flowing through the breaker is seldom affected by C in the figure. Then i t = Re

E e jωt = jωL + r

E r2 ωL 1 + 2 2 ω L

Here we can utilize the condition ωL it =

sin ωt +

r cos ωt ωL

15 55a

r (the reason is explained later). Accordingly,

E sin ωt ωL

15 55b

This is the initial current before breaker tripping. Step 2: Calculating the transient term The transient term can be calculated by applying Thévenin’s law to Figure 15.14d: that is, inserting the current source i (t) across the breaker’s fixed contactor and moving contactor, so the flowing current just before the switch opens is canceled at t = 0. Current breaking finishes at the time of current zero according to the nature of circuit breakers E (the reason is explained later), so the initial current is given in the form i t = sin ωt, which becomes zero at time ωL 2 2 t = 0. Also, the Laplace transform of sin ωt is ω/(s + ω ) for t ≧ 0. Then, E ω ωL s2 + ω2 r s+ 1 1 L = Z s = 1 1 C 2 r s + s+ + Cs Ls + r L LC E ωL

is =

∴ vBr

sin ωt =

E s =i s Z s = LC

① ② 15 56

s+ s2 + ω2

r L

r 1 s2 + s L LC

E F s LC

③

Equation ② for Z(s) is the impedance looking into the circuit across the contacts from the current source in the s-domain, as in Figure 15.14b. Accordingly, equation ③ is the voltage solution vBr(s), but in the s-domain. F(s) defined in Equation ③ is s+ F s = s2 + ω2 =

s2 + ω2

where u =

r L

r 1 s2 + s + L LC s + 2α

=

s + 2α s2 + ω2 s2 + 2αs + u2 15 57

2

s + α + u2 − α2 r 1 , α= 2L LC

F(s) has a denominator with a fourth-order s function, so the inverse transformation to the t domain is not necessarily easy.

417

418

15 Lightning and Switching Surge Phenomena and Breaker Switching

Here we will investigate the magnitudes of constants in preparation for our justifiable approximation. Assuming L = 1 mH/km, = 0.01 μF/km, r = 0.01 Ω/km, f = 50 Hz, then ω = 2πf = 314, ω2 ≒ 105 ωL = 314 × 10 −3 Ω km ≒ 0 3 Ω km 1 1 Ω km ≒ 3 × 105 Ω km = ωC 314 × 10 − 8 1 ≒ 10 −6 3 × 105

ωL ωC ≒ 0 3 × u= 2α =

1 = LC

1 10

−3

× 10 −8

①

15 58

≒ 300 000 km s

r 0 01 = = 10, α = 5 L 10 − 3

Accordingly, to reasonable accuracy, 1 ωC

ωL

α②

r, u

Then we can adopt the approximation (u2 − α2) s + 2α

F s =

2

s + a 2 + u2 − α

s2 + ω2 =

=

u2, and the equation for F(s) is

s + 2α s + jω s− jω s + α + ju s + α −ju

↖

15 59

k1 k2 k3 k4 + + + s + jω s − jω s + α + ju s + α − ju

The arrow ↖ in this equation indicates the part that is ignored for simplicity. (These arrows are used as indicators of omission for reasonable simplifications.) From the formula for the inverse Laplace transform, −1

1 =e s±a

at

for t ≧ 0

15 60

Accordingly, F t = k1 e −jωt + k2 e jωt + k3 e − =

α + ju t

+ k4 e −

α−ju t

k1 + k2 cos ωt − j k1 −k2 sin ωt +e

−αt

15 61

k2 + k4 cos ut − j k3 − k4 sin ut

We still have work to do to find the coefficients k1, k2, k3, k4. The calculation procedure and the results are given in Supplement 3 at the end of this chapter. During the calculation, a total of 15 arrows of omission appear, which is unexpected for such a simple circuit. The general method to find these coefficients was explained in Chapter 10, Supplement 1. Consequently, the result obtained is k1 = k2 = − k3 = −k4 =

1 LC = 2 2u 2

15 62

15.9 Switching Surge Phenomena Caused by Breakers Tripping

Then F t = LC cos ωt −e −αt cos ut ∴ vBr t = va t =

①

E F t = E cos ωt − e − αt cos ut LC

=E

r − t cos ωt −e 2L cos steady -state term

1 t LC

②

15 63

transient term

natural oscillation frequency f0 =

where

1 2π LC

attenuation time constant T = 2L r This is the solution for this circuit, and the transient waveform of the equation is indicated in Figure 15.14e. The solution indicates the following conclusions:

• • • •

Current trip finishes at t = 0, at the time i(t) = 0 (current-zero tripping), and the source voltage at t = 0 is the peak value E (because the power factor of the fault current is almost 90 during the fault). The steady-state term of the voltage between contacts vBr(t) (the recovery voltage across the breaker contacts after breaker opening) is Ecos ωt. The transient oscillation term appears just after tripping, which has natural frequency f0 and attenuation time constant T = 2 L/r, and it oscillates by ± E within the band 0–2E. As a result, the voltage before attenuation vBr(t) (the transient recovery voltage (TRV) of the breaker just after breaker opening) reaches a maximum magnitude of 2E. In this particular case, vBr(t) = va(t), because the voltage at point b is vb(t) = 0. In other words, Eq. (15.63)② is also the phase overvoltage appearing on the left side of the breaker terminal.

The resulting oscillation is the total of the repeated reflections and refractions of the traveling waves at the transition points.

15.9.2

Calculating Fault-Current Tripping (Double Power Source Circuit)

The next circuit is shown in Figure 15.15a, where double power sources exist at both terminal sides of the breaker. The calculation of this case is more complicated compared to the previous case, because the denominator of F(s) here becomes a sixth-order s function. As a matter of fact, readers can count 48 arrows of omission throughout the calculation process of this problem. This is a typical example showing that proper approximation is a very important technique in practical engineering. Step 1: Calculating the steady-state term The generators are operating in synchronism with angular difference δ: e1 t = E1 e jωt

15 64

e1 t = E1 e j ωt − δ Then Δe t = e1 t − e1 t = 1 − k δ = 1−

E1 −jδ e E1

E1 − jδ e1 t = k δ E1 e jωt e E1

15 65

The current i(t) flowing through the breaker before tripping is not affected by the existence of C1, and C1 , so the current equation is

419

420

15 Lightning and Switching Surge Phenomena and Breaker Switching

I sin𝜔t 1(t)

𝜈Br (t)

+

i = I sin𝜔t

L1′r1′

L1 r1

= e1

C1′

C1

+

e1′ e1

e1′

(a)

(b)

𝜈Br (t)

(c)

𝜈Br (t) Note: Oscillation frequency of the transient terms is exaggerated extremely slowly.

k (𝛿) 2E1

(d)

Voltage across the breaker contacts 𝜈Br (t) [(19.22a)]

k (𝛿) E

Transient recovery voltage

t

Recovery voltage

(e)

t Oscillatory components r1 L1 t ·e 2L1 cos t k (𝛿) E1 – L1+ L1′ L1C1 k (𝛿) E1 –

L1′ ·e L1+ L1′

r1′ t 2L1′

cos

t L′1C′1

Figure 15.15 Transient calculation for single-phase breaker tripping (double-source circuit).

it =

Δe t − jΔe t − jE 1 ≒ =k δ e jωt ω L1 + L1 jωL1 + r1 + jω L1 + r1 ω L1 + L1 ↖

↖

15 66

E1 e j ωt −90 =k δ ω L1 + L1 Taking the real parts of all these equations, for the power sources e1 t = E1 cos ωt

15 67

e1 t = E1 cos ωt − δ and for the fault current before tripping (steady-state current), it = k δ

E1 cos ωt −90 = k δ ω L1 + L1

I sin ωt where I = k δ

E1 ω L1 + L1

k δ = 1−

i t is “Current Zero condition” at t = 0

E1 −jδ e E1

E1 sin ωt ω L1 + L1 15 68

15.9 Switching Surge Phenomena Caused by Breakers Tripping

Step 2: Calculating the transient term As shown in Figure 15.15c, the current i(t) = I sin ωt is inserted at the position of the breaker terminals at time t = 0. In the Laplace transform domain, Initial current ω i s =I 2 s + ω2

①

Circuit impedances ②

ΣZ s = Z1 s + Z1 s

Z1 s =

1 1 + C1 s L1 s + r1

1 = C1

s+

r1 L1

r1 1 s2 + s + L1 C1 L1 s+

Z1 s =

=

1 s + 2α1 C1 s + α1 2 + u21

r1' L'1

1 1 1 s + 2α2 = = 1 1 C1 2 r1 C1 s + α2 2 + u22 + C1 s s + s+ L1 s + r1 L1 C1 L1

③ 15 69

④

Transient recovery voltage across the contacts vBr s = i s ΣZ s = Iω

1 Z1 s + Z1 s s2 + ω2

Iω F s

⑤

Z1 s , Z1 s are of the same form as Z(s) in Eq. (15.56), so F(s) defined in Eq. (15.69)⑤ can be modified as

F s =

1 s2 + ω2

↙

↙

1 s + 2α1 1 s + 2α2 + C1 s + α1 2 + u21 C1 s + α2 2 + u22 2

2 C1 s + α↙ + u21 + C1 s + α↙ 1 2 + u2 s = 2 s + ω2 C1 C1 s + α1 2 + u21 s + α2 2 + u22

=

s C1 s2 + u21 + C1 s2 + u22 C1 C1 s + jω s − jω s + α1 −ju1 s + α2 + ju2 s + α2 − ju2

=

k1 k2 k3 k4 k5 k6 + + + + + s + jω s −jω s + α1 + ju1 s + α1 −ju1 s + α2 + ju2 s + d2 − ju2

15 70

F(s) is the addition of the two terms whose denominators contain the fourth-order s function. Accordingly, F(s) has a sixth-order denominator. As shown in Supplement 4, coefficients k1 − k6 are derived after applying 40 arrows of omission, and the results are k1 = k 2 =

1 L1 + L1 2

k3 = k 4 = −

L1 2

k5 = k 6 = −

L1 2

15 71

421

422

15 Lightning and Switching Surge Phenomena and Breaker Switching

Then F s =

1 L1 + L1 2 −

1 1 L1 1 1 + + − s + jω s −jω 2 s + α1 + ju1 s + α1 − ju1

L1 1 1 + 2 s + α2 + ju2 s + α2 − ju2 ↙

15 72

↙

1 s + α1 s + α1 −L1 −L1 = L1 + L1 2 2 2 2 s +ω s + α1 + u1 s + α2 2 + u22 1 s s = L1 + L1 2 −L1 −L1 2 2 s + ω2 s + α1 + u1 s + α2 2 + u22 From the formula of the Laplace inverse transform, −1

s2

s = cos ωt + ω2 2

s + α1

+ u21

s + α2 −1

α1 sin u t = e − α1 t cos u t = e −α1 t cos u1 t − u 1 1 1

15 73

↙

s

−1

∴

↙

s

−1

2

+ u22

α2 sin u t = e −α2 t cos u t = e −α2 t cos u2 t − u 2 2 2

F s = L1 + L1 cos ωt −L1 e −α1 t cos u1 t − L1 e − α2 t cos u2 t

15 74

Accordingly, the TRV of the breaker is vBr t = Iω

vBr t = k δ

−1

E L1 + L1

F s = k δ

E1

F s

r1 − t L1 2L 1 cos cos ωt − e L1 + L1

steady-state term

where k δ = 1 −

−1

r1 − t 1 L1 t− e 2 L1 cos L1 C1 L1 + L1

transient term of the right circuit

E1 − j e δ E1

1 L1 C1

transient term of the left circuit 15 75a

The equation can also be written as

vBr t = k δ

+

E1

L1 L1 + L1

L1 L1 + L1

r1 t cos ωt − e 2L1 cos

− r1 t cosωt −e 2 L1 cos

−

1 t L1 C1 15 75b

1 t L1 C1

This derived equation is very accurate because all the omissions in the processes were done after careful checks. Figure 15.15d shows the transient voltages of Eqs. (15.745a, b), where the oscillation frequency of the transient terms is exaggerated extremely slowly.

15.9 Switching Surge Phenomena Caused by Breakers Tripping

15.9.3

Breaker TRV and RRRV

Now, let’s further examine the calculated Eqs. (15.75a, b). The calculated transient voltage across the breaker contacts vBr(t) is called TRV, and the time derivative dvBr(t)/dt is called the rate of rise of recovery voltage (RRRV). TRV and RRRV are very important parameters that affect the breaker-tripping capability and the switching-surge magnitude that is inevitably caused by the breaker switching on and off. For the magnitude vBr(t), the transient oscillation terms for the right- and left-side circuits appear just after tripping. The magnitude inside the braces ( ) in the equations reaches a maximum of 2 when the oscillation angles of the three terms coincide. Accordingly, vBr(t) has a maximum of |k(δ)| 2E1. Incidentally, assuming E1 = E1 = 1, k δ = 2 for δ = 90 (tripping around the steady-state stability limit condition), the theoretical maximum value of the TRV is 2 2E. On the other hand, |k(δ)| = 0 for the case of δ = 0 (no-load tripping). We have Left-side circuit Right-side circuit Maximum peak value of Transient voltages line-to-ground voltages Natural oscillation frequency

L1 L1 2k δ E1 L1 + L1 L1 + L1 1 1 f1 = f1 = 2π L1 C1 2π L1 C1

2k δ E1

15 76

The total peak value of the transient terms of vBr(t) is a maximum of 2|k(δ)| E1. Assuming L1 = L1 = 1 mH/km, C1 = C1 = 0 01 μF/km as typical values, f1 = f 1 =

1 2π 10 −3 × 10 − 8

≒ 50 kHz one-wavelength time = 20 μs

15 77a

Therefore, the transient terms increase from zero to the peak value 2 k(δ) E1 in about 5 μs (quarter-wavelength time). The approximate value of the RRRV is 2 k δ

E1 ×

1 kV μs 5

15 77b

The exact value of the RRRV can be calculated by differentiating Eq. (15.75a) with respect to time t = 0+: d vBr t = k δ dt

E1 − ω sin ωt

L1 − L1 + L1

r1 r1 − 2L t 1 cos − e 2L1

L1 − L1 + L1

r1 r1 − 2 L t 1 cos − e 2 L1

r1 − t 1 2L 1 t −e L1 C1 r1 − t 1 2 t −e L1 L1 C1

1 sin L1 C1

1 t L1 C1

1 sin L1 C1

15 78a

1 t L1 C1

In this equation, the first transient term in parentheses ( ) may be ignored, because r1 and r1 are small. Also, at time t = 0 + (initial time just after tripping), the following replacements are possible: sin ωt

0, e

−

r1 t 2L1

1 0, e

−

r1 t 2 L1

1 0, at t = 0 +

15 78b

Accordingly, at t = 0+, d vBr t = k δ dt

E1

1 L1 + L1

L1 sin C1

1 1 t+ L1 + L1 L1 C1

L1 sin C1

1 t L1 C1

15 79

423

424

15 Lightning and Switching Surge Phenomena and Breaker Switching

The maximum value of the RRRV is Max

d vBr t = k δ dt

E1 L1 + L1

L1 + C1

L1 C1

15 80

Sum of surge impedances of both sides

The RRRV (usually expressed in kV/μs) is a very important concept that significantly affects the tripping duty of breakers. Equation (15.80) indicates that the RRRV is directly proportional to the addition of surge impedances L1 C1 , L1 C1 on both sides of lines.

15.10

Breaker Phase Voltages and Recovery Voltages after Fault Tripping

The study of switching phenomena in three-phase circuits is vitally important not only for engineers who are involved with breakers, but also for engineers who are engaged with system overvoltages or large currents as well as insulation, regardless of each engineer’s different interests or viewpoints. Considering the importance of such study, it may seem strange that a detailed description of the calculation of switching phenomena in three-phase circuits can seldom be found. In this section, we will introduce the equations for transient voltages appearing at the first, second, and third pole (phase) tripping. Readers can become acquainted with switching surge phenomena through the study of the mathematical treatments. Transient voltages and steady-state voltages appearing across the contacts (poles) just after breaker tripping are called transient recovering voltages and recovering voltages, respectively. Before studying these voltages in detail, the recovering voltage from first phase tripping can be observed. The methods and the results are shown in Figure 15.16. The figure illustrates six different fault cases for effective and non-effective neutral grounding systems. The steadystate voltage (the recovering voltage) appearing across the first tripping pole (the first tripping phase) is derived intuitively. Case 1 Three-Phase Fault Tripping The voltages at the time just after the first phase-a trip are, for the solidly neutral grounding system, va = 0, vb = vb = 0, vc = vc = 0, vn = 0 ∴ va = E, vaa = E and for the neutral-opening system, va = 0, vb = vb = 0, vc = vc = 0, vn = indefinite Then the voltage at the midpoint of b and c is zero: vab = va = 1 5E, ∴ vaa = 1 5E Case 3 Phase-a and Phase-b Grounding Fault Tripping This is the case with grounding at both sides of the breaker. For the solidly neutral grounding system, va = 0, vb = vb = 0, vn = 0 ∴ va = 1E, vaa = 1E and for the neutral opening system, va = 0, vb = vb = 0, vn = indefinite ∴ vab = 3E, vaa =

3E

Case 4 Step-Out Tripping In the worst case, the voltages va and va are in opposite directions at the time of the first phase-a trip. For the solidly neutral grounding system, va = 1E, va = − 1E, vn = 0 ∴ va = 1E, v1a = − 1E, vaa = 2E

15.10 Breaker Phase Voltages and Recovery Voltages after Fault Tripping

a

a′

a

1.5 E

E

n

c

b

0

b

b′

c

c′

Non-effective neutral grounded system a

Solidly neutral grounded system

Case-1 Three-phase to ground fault tripping

a E

0 0

c = c′

0

1.5 E

n

b = b′ c = c′ = 0

0

b = b′ = 0

0 Case-2 Phase-a to ground fault tripping a

a

0

E

b 0

a′

b

b′

c

c′

a

a

0

n

c

0

a′

E

n

c′

b′

0

c

a′ = 0

n

b

c = c′

b = b′

Case-3 Phase-a′ & phase-b to ground fault tripping a E

a

a

b

b′

0

c c

c′

Case-4 Step-out tripping a a

a′ 3E

b

3E b

a b′

c′

b′

c

c′

b = a′ = 0

a c′

b′

2E

b

c 3E

n

c

E

n b

E

a

3E

n

c

a′

0

0 c

b

a′

b 3E c′

c

a′

b′

a′ Case-5 Step-out with phase c, c′ to ground fault tripping a′ a a b′ c′ 2 3E E n 2E b′ b b c c 0 a′ c c′

a

a b′

c′ b′

c′

0

0

c 0

b

b 2 3E

a′ a′

Case-6 Step-out with phase b′, c to ground fault tripping

E c

a′

a

a

a

b′

b′

2 3E

n

2 3E

2E a′ b

b

b′

c′

a

c′

c

0

b′

b a′

c

b

0 c

0 potential

c′ a′

Figure 15.16 Recovery voltages appearing with the first pole (phase-a) tripping, vaa .

c′

425

426

15 Lightning and Switching Surge Phenomena and Breaker Switching

and for the neutral opening system, vb = vb , vc = vc , vn = indefinite, vbc = vb c ∴ vaa = 3E The phase voltages and recovery voltages derived using this method give a useful rough idea of the voltage behaviors. The magnitudes of these voltages directly affect the tripping capabilities of the breakers as well as coordination of the insulation engineering in a total power-system network. Taking a general view, the phase overvoltages appearing as well as the breaker’s recovery voltages are obviously smaller in solidly neutral grounding systems compared to non-effective neutral grounding systems. Also, breaker tripping under step-out conditions is very heavy-duty for the breakers.

15.11

Three-Phase Breaker TRVs across Independent Poles

We will introduce the equations for the transient voltages appearing across the first, second, and third poles when the breaker trips three-phase short-circuit fault currents, as shown in Figure 15.17a. In the figure, a three-phase fault occurs at point f on line l’, and the first-phase tripping by the breaker phase-a pole is going to be executed (of course, at the time of the phase-a current zero). The transient calculation for Figure 15.17b can be solved as the superposition of the steady-state term in (c) and the transient term in (d).

15.11.1

First Pole Tripping

Step 1: Calculating Steady-State Current before Phase-a Tripping We will calculate the steady-states currents of the three-phase short-circuit fault before tripping. The current can be calculated using the positive-sequence circuit. Accordingly, this case is equivalent to the case of Section 15.9.2 and Figure 15.15 under the additional condition e1 = 0. Therefore, we can appropriate Eqs. (15.64)–(15.68) for use here. That is, for the phase-a (positive-sequence) source voltage, ea t = e1 t = Ee jωt

15 81

The current flowing through the breaker is given by Eq. (15.66), where, using Δė(t) = ė(t), k(δ) = 1, Ee jωt

it = jωL1 + r1

↖

≒

+ jω L1 + r1

E e j ωt −90 ω L1 + L1

15 82

↖

Taking the real part of the equation, e1 t = Ecos ωt it =

E cos ωt − 90 = I sin ωt ω L1 + L1

where I =

15 83

E ω L1 + L1

The equations are for the initial conditions corresponding to Figure 15.17c, where the current i(t) is current zero at t = 0 and the voltage is at its peak value at t = 0, because the fault current is almost 90 lagging to the voltage. Step 2: Calculating TRV Just After First-Pole Tripping Now we begin the calculation of transient terms for Figure 15.17d, where the current I sin ωt 1(t) is suddenly inserted at t = 0. The equivalent circuit is given in Figure 15.17e, which corresponds to the equivalent circuit of symmetrical components in the case of phase-a opening given in Table 7.2 [1B].

15.11 Three-Phase Breaker TRVs across Independent Poles

Generator

Point f (3𝜙S)

Br

Line l

Line l′

a

a′

b

b′

c

c′

Line ·l′′

(a)

+

ia (t)1(t)

+

–

ia (t)

=

+

(b)

(d)

(c)

ia= i1 + i2 + i0

+

+

–

veq

i1

v1 L1′ r1′

L1 r1

Positivesequence

C1

Σ Z1(s)

C1′

ea

1.5

1.0

i2 v2 Negativesequence

C1

ec

eb Vector diagram after the first pole tripping

L1′ r1′

L1 r1 Σ Z1(s)

C1′

i0

(f) v0

Zerosequence Rn

L0′ r0′

L 0 r0 C0

Σ Z0(s)

Rn′

C0′

(e) Figure 15.17 Calculating transient recovery voltage (TRV) (first pole tripping).

The related equations in the Laplace domain are derived as follows. The circuit equations are ΣZ1 s = Z1 s + Z1 s

①

ΣZ0 s = Z0 s + Z0 s

②

where 1 + C1 s L1 s + r1

1 = C1

s+

r1 L1

1 s + 2α1 = ③ r 1 1 C1 s + α1 2 + u21 s2 + s + L1 C1 L1 r s+ 1 1 1 1 s + 2α2 L1 Z1 s = = ④ = 1 1 C1 2 r1 C1 s + α2 2 + u22 + C1 s s + s+ L1 s + r1 L1 C1 L1 Z1 s =

1

15 84

427

428

15 Lightning and Switching Surge Phenomena and Breaker Switching

where Z0(s), Z0 s are of the same form as Z1(s), Z1 s . The voltage and current equations are, for t ≧ 0, ω ① + ω2 veq s v1 s = v2 s = v0 s = i s ΣZtotal ② 1 ③ where ΣZtotal = 1 1 1 + + ΣZ1 s ΣZ2 s ΣZ0 s I sin ωt = I

is =

s2

15 85

∴ vaa s = v1 s + v2 s + v0 s = 3 i s ΣZtotal = 3I where

s2

ω ΣZtotal + ω2

15 86a

Ztotal is For solidly neutral grounding system ΣZ1 s ≒ ΣZ2 s ≒ ΣZ0 s

①

1 ΣZ1 s 3 For high-impedance neutral grounding system ∴ ΣZtotal =

ΣZ1 s ≒ ΣZ2 s ΣZ0 s 1 ∴ ΣZtotal s = ΣZ1 s 2

15 86b ②

Accordingly, the voltage across points a and a (the transient voltage) vaa (s) is vaa s = k i s ΣZ1 s = kI

s2

ω 1 s + 2α1 1 s + 2α2 + 2 2 2 + ω C1 s + α1 + u1 C1 s + α2 2 + u22

15 87

kIω F s where k ≒ 1 for solidly neutral grounding system k ≒ 1 5 for high-impedance neutral grounding system

This is the answer for the TRV of the first tripped pole in the Laplace domain. F(s) defined in the equation is fortunately in the same form as F(s) in Eq. (15.70), so we already know that the inverse transformed equation is given by Eq. (15.74). Accordingly, for the TRV of the first tripping pole, vaa =

−1

vaa s = kIω

−1

F s =k

−r1 t L = kE cos ωt − e 2L1 cos L1 + L1

E L1 + L1

−1

F s

− r1 1 L1 2 L t t− e 1 cos L1 C1 L1 + L1

15 88a 1 t L1 C1

The term in braces { } in this equation has a maximum of 2, so the maximum value of the TRV vaa t (or the overvoltage va(t) of terminal a, because va = 0 in this case) is Max vaa t = 2k ≒ 2E for solidly neutral grounding system ≒ 3E for ineffective neutral grounding system

① ②

15 88b

The equation for the steady-state voltage after the transient terms attenuate is vaa t = k E cos ωt, which coincides with Case 1 in Figure 15.16.

15.11 Three-Phase Breaker TRVs across Independent Poles

Calculating the RRRV (the maximum value) of vaa t This is derived by differentiating Eq. (15.88a) with respect to t and then following the same treatment as for Eqs. (15.78a)–(15.80). The result is Max

d E1 vBr t = k dt L1 + L1

L1 + C1

L1 C1

15 89

sum of surge impedances of both sides

Note that this equation is in the same form as Eq. (15.80) except for the coefficient k and k(δ), although Eq. (15.89) is the case for fault-current tripping, while Eq. (15.80) is the case for load-flow tripping. The equation again indicates that the RRRV (the maximum value), which is an important indication of a breaker’s fault-current tripping duty, is closely related to the surge impedances of the transmission lines. 15.11.2

Second- and Third-Pole Tripping

Our problem here is to solve the transient voltages across the second and third poles of the circuit shown in Figure 15.18a. The voltages appearing on the second and third poles are more severe in a non-effective neutral grounding system rather than a solidly neutral grounding system. Thus we will try to calculate the second/third-pole fault-tripping phenomena in the non-effective neutral grounding system. Step 1: Calculating the Steady-State Current Before Phase-b and -c Tripping (Phase-a Open) We will calculate the steady-state currents of the circuit in Figure 15.18b, referring to the equivalent circuit (for phase-a conductor opening) for the symmetrical components given in Table 7.2 [1B], [1C], [1D], and under the condition Z1 = Z2 Z0. The equations are ①

e t = Ee jωt et

i1 t = Z1 +

1 + Z2

et

=

1

Z1 +

1 Z0

Z1 Z1 Z0

≒

1 et 2 Z1

②

+1

↖

i2 t =

− Z0 i1 t ≒ −i1 t Z2 + Z0

③

i0 t =

− Z2 i1 t ≒ 0 Z2 + Z0

④

where Z1 =

Z2

Z0 for high impedance neutral grounding system

ib (t) + ib (t) a b c

=

(a)

a b c

+

(b)

a b c (c)

Figure 15.18 Calculating transient recovery voltage (TRV) (second and third poles tripping).

⑤

15 90

429

430

15 Lightning and Switching Surge Phenomena and Breaker Switching

The steady-state solution is ia t = 0 ib t = −ic t = a2 − a i1 t = − 3j ×

1 2

E jωL1 + r1 + jω L1 + r1 ↖

=−

e jωt ↖

15 91

3 E e jωt 2 ω L1 + L1

Taking the real part, ea t = Ecos ωt ia t = 0

15 92

E 3 ib t = −ic t = − cos ωt 2 ω L1 + L1 The current ib(= − ic) is in inverse phase with the source voltage ea, and accordingly leads the voltage vbc by 90 . These are the initial currents of the second and third poles before tripping. Step 2: Calculating the TRV for Second- and Third-Pole (Phase-b, c) Tripping The next calculation is to insert the previously derived current ib in the opposite direction (i.e. −ib in the forward direction) in Figure 15.18c. Then we will change the polarity of the current ib and ic. Furthermore, we will shift the timescale of Eq. (15.92) by 90 (by replacing ωt ωt −90 ) in order to get the condition current zero at new t = 0. Accordingly, ea t = Ecos ωt −90 = Esin ωt ia t = 0 −ib t = + ic t =

E E 3 3 cos ωt −90 = sin ωt 2 ω L1 + L1 2 ω L1 + L1

15 93

where ωt = ωt −90 This is the initial value of the calculation after phase-a is tripped. The currents ib = −ic are for current zero at t = 0 with the new timescale (which lags the former timescale t by 90 ). Now the transient phenomena of the circuit in Figure 15.18c can be solved by the αβ0-method, whereas solution by symmetrical components is impossible. Referring to Table 8.2 #5, the equivalent circuit to be solved is obtained as shown in Figure 15.19, in which the α0-circuit and the β-circuit can be calculated separately. The initial value can be derived by transforming Eq. (15.93) into the αβ0-domain: −1 ib t + ic t = 0 3 3 E ib t − ic t = − sin ωt iβ t = ω L1 + L1 3 1 i0 t = ib t + ic t = 0 3 iα t =

15 94

For the transient calculation of the α0-circuit, the initial values of iα and i0 to be injected are zero because of the Z0, so ia(t ) = i0(t ) = 0. In other words, all the quantities are zero: vα = v0 = 0 in the α0assumption Z1 = Z2 circuit.

15.11 Three-Phase Breaker TRVs across Independent Poles

i𝛽(0) + i𝛽

𝜈𝛽 L1′ r1′ C1′

L1 r1 C1

i𝛼

𝛽-circuit

𝜈𝛼 L1′ r1′ C1′

L1 r1 C1

𝛼-circuit

2i0 L0 r0 Rn

L0′ r0′

2𝜈0

C0

Rn′

C0′

0-circuit

Figure 15.19 Calculating the second and third poles tripping with the αβ0 method.

For the transient calculation of the β-circuit, the current iβ(t ) is to be injected at time t = 0 but with an inverse sign to cancel the initial current, and the equation in the Laplace domain is for t ≧ 0 iβ s =

E ω L1 + L1

vβ s = iβ s

sin ωt =

Z1 s =

E ω 2 ω L1 + L1 s + ω2

15 95

E F s L1 + L1

Z1(s), F(s) are again the same as in Eqs. (15.69) and (15.70) or (15.84). Then Eq. (15.74) can be utilized for the inverse Laplace transform: vβ t =

E L1 + L1

=

E L1 + L1

−1

= E cos ωt −

F s

L1 + L1 cos ωt − L1 e r1 − t L1 e 2L1 cos L1 + L1

−

r1 t 2L1

cos

r1 − t 1 t − L1 e 2L1 cos L1 C1

r1 − t 1 L1 t− e 2L1 cos L1 + L1 L1 C1

1 L1 C1

1 L1 C1

t

15 96

t

For the inverse transform from the αβ0-domain to the abc-domain, recalling that vx(t) = v0(t) = 0, vbb t

= −vcc t = =

3 vβ t 2

r1 − t 3 L1 E cosωt − e 2L1 cos 2 L1 + L1

r1 − t 1 L1 t− e 2L1 cos L1 + L1 L1 C1

1 L1 C1

15 97 t

In conclusion, the TRV with first-pole tripping is given by Eq. (15.88a), and that with second- and third-pole tripping is given by Eq. (15.97). Figure 15.20 shows the waveform aspects, which are explained by the equations.

431

432

15 Lightning and Switching Surge Phenomena and Breaker Switching

ia

ea

Transient recovery voltage

1.5 ea

𝜈aa′

Recovery voltage

Phase-a 𝜈aa′ 𝜈bb′

ib

eb

Phase-b ec

Phase-c

ic

Phase-a tripping

90º

𝜈cc′

Phase-b,c tripping

Figure 15.20 Transient recovery voltage (TRV) caused by three-phase fault tripping.

Non-effective neutral grounding system

Transient recovery voltage

Recovery voltage

First-pole tripping

3E

1.5E

Second- and third-pole tripping

3E

3 E 2

The TRV is larger from the first-pole tripping than the second- and third-pole tripping and is a maximum of 3E in the case of the non-effective neutral grounding system. For the solidly neutral grounding system, the TRV and the recovery voltage with the first-pole tripping are 2E and 1E, respectively, and for the second and third poles have smaller values (the calculation is omitted; they can be calculated by the same method if Z1, Z0 are given). It should also be noted that the RRRV (the maximum value) in three-phase circuits using the 3φS fault-current tripping is given by Eq. (15.89), which indicates that the surge impedances of the transmission line directly affect the breaker’s duty. Z0 was assumed, so ib(t) = − ic(t) was obtained; then the time of current zero Note: In this explanation, Z1 = Z2 for ib (t) and ic (t) are the same. This is why the second and third poles were tripped at the same time. For the solidly neutral grounding system Z1 = Z2 Z0), the timings of current zero for ib(t) and ic(t) may not be the same, so the tripping time for the second and third poles may not be the same. This case can be calculated, although it is omitted in this book.

15.12

Circuit Breakers and Switching Practices

The switching-off phenomena described in the previous sections are by ideal circuit breakers, which can trip fault current instantly at the time of current zero without any accompanying arcs. Actual breaker tripping is a little different from the ideal case, because a current arc appears transiently across the leaving contacts.

15.12.1

Fundamentals of Breakers

A high-voltage circuit breaker has a pair of contacts: typically, one is fixed and one is movable. Just after receiving the trip signal (from the relay or manually), the movable contact begins to slide off and leave the fixed contact. In the case of typical SF6-type breakers, the total stroke distance (wipe distance + departing distance) is 150–250 mm, and the necessary moving time is around 10 ms (less than 1 cycle). At the instant when the movable contact leaves the fixed contact, a current arc (plasma of approximately 5000–10 000 C by thermal ionization) appears across the leaving contacts. The breaker contrives to blow off the thermal ionized gas within a short time and complete tripping by around 20–60 ms (1–3 cycles). The technology of breakers is designed to disperse and remove high-density ionized plasma gas within the small chamber space in a very short time – in contrast to the technology of nuclear fusion, which is to contain such

15.12 Circuit Breakers and Switching Practices

high-density plasma gas within a small space. Furthermore, circuit breakers must be capable of tripping a fault current of 50 or 63 kA (RMS) repeatedly without damaging (melting) the contacts. Although the technology of high-voltage circuit breakers today is advanced, the long history of repeated failure and success is worthy of note among the other equipment in a power-system network. There are two different interpretations with regard to the principle of arc extinction: a) The theory of insulation balance across the contacts: The keen struggle of the TRV and RRRV appearing versus insulation recovery by sudden enlargement of the breaker’s stroke distance and by forcing blow-off of the ionized arc. b) The theory of energy balance: The keen struggle of the supplied energy (the product of TRV and leak current) versus the forced dispersing energy. Regardless of the academic interpretations, successful current breaking means rapid insulation enlargement across the contacts to overcome the supplied arc energy from the power system in a short time. Accordingly, it is understandable that the magnitude of the TRV and its initial increase speed (RRRV) have become the two essential parameters for the breaker’s tripping capability. Breakers today can be classified as types of SF6 gas, air, oil, and vacuum, based on the extinction media. Each type has its own specific characteristics. Oil-filled circuit breakers (OCBs) and air circuit breakers (ACBs) were the two leading types of breakers in the 1930s– 1960s, and then SF6-gas filled circuit breakers (GCBs) began to spread in the 1970s. The basic concept and the fundamental electrical characteristics of these three types of breakers are common because they consist of a pair of fixed or moving-sliding contacts, although the arc-extinction media (oil, air, SF6-gas) varies. The principal theories and practices explained later in this chapter are common for all of these breakers. Of course, they have different features, in particular from the viewpoint of arrangement and maintenance. With regard to high-voltage classes (EHV and UHV), SF6-gas type breakers have been predominant; however, such a comparison is beyond the scope of this book. Vacuum circuit breakers (VCBs) became pervasive in the 1960s mainly for lower voltages of 6–60 kV. VCBs consists of a glass or ceramic vacuum bottle in which a pair of bat-contacts are contained. VCBs can repeat load currents that trip often (typically 50 000 times or more) because they can break current quickly without causing a large amount of arc. With this unique characteristic, VCBs have been widely adopted mainly in the area of industrial factory applications, including auxiliary circuits of power substations. Vacuum bottle load switches have similar characteristics, although they don’t include short-circuit current-tripping duty. Figure 15.21 shows some fundamentals of a typical configuration of 362–420 kV GIS and a GCB. SF6 gas is a thermodynamically stable gas with outstanding insulation and arc-extinction characteristics. SF6 gas at typically 4–10 atm fills the extinction chamber, whose structure is a plunger-type puffer chamber. The movable contact and the plunger rod form one body: the gas is puffed out just after the contact begins to slide, in order to extinguish ionized current arcs and to recover insulation across the opening contacts. Figure 15.21e shows the typical gas-blowing characteristics of a puffer chamber. Breakers must break the current in a single stroke within an effective puffer time. In other words, if a breaker cannot break the current within a specified time (say, 3 cycles), the breaker will be damaged by the thermal energy, which causes a tripping failure and a new phase fault at the broken breaker (short-circuit or open-conductor mode). The application limit described here is common for all types of breakers. In the case of OCBs, for example, oil gasification due to the arc temperature occurs in the chamber, so the effective tripping time is limited. Note that a breaker fault due to tripping failure will severely impact the power system in most cases. Several breakers are directly connected to the same bus, and the associated breakers at adjacent substations must be immediately tripped (by various backup relays) to remove the original fault and the faulted breaker, possibly leading to other cascade faults or serious system disturbances such as line overcurrent, generator tripping from the I2 limit or frequency limit, system instability, and so on.

15.12.2

Terminology of Switching Phenomena

The duty of circuit breakers is not limited to the role of tripping fault/load currents as switches in a network. Another important duty is to reduce switching overvoltages (surge) within a certain limit, which is an essential part of insulation

433

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15 Lightning and Switching Surge Phenomena and Breaker Switching

Current transformer

Interrupting chamber

Motor charged spring mechanism

(a) GIS design (362–420 kV)

(b) Gas circuit breaker (one break for 63kA) Disconnector (DS)

High-speed earthing switch (H-ES)

Maintenance earthing switch (ES) (c) Disconnector and earthing switch Voltage transformer (VT)

Feeder side disconnector (DS) Current transformer (CT) Busbar side disconnector (DS) Busbar

Cable sealing end (CH) Maintenance earthing switch (ES) Busbar

SF6 gas Insulators Live parts

Gas circuit breaker (GCB) High-speed earthing switch (H-ES)

Busbar side disconnector (DS)

(d) 362–420 kV GIS (typical arrangement) Figure 15.21 GIS and gas circuit breaker (362–420 kV GIS). Source: Courtesy of Toshiba.

CT/VT Earthed parts

15.12 Circuit Breakers and Switching Practices

Initial stage of current interruption

Arcing contact

Nozzle First stage of heavy current interruption

Final stage of small current interruption

(e) Current interruption process of gas circuit breaker (self-blast puffer type) Figure 15.21 (Continued)

coordination. Bearing this in mind, we will continue our study of the phenomena arising from breakers switching (tripping and closing). The major terms related to breaker tripping are as follows: ① Transient recovery voltage: The initial transient voltage appearing across the paired contacts just after the tripping action. ② Ideal TRV: The theoretical TRV for an ideal breaker without accompanying arcs. ③ Recovery voltage: The steady-state voltage of power frequency appearing across the paired contacts just after tripping. ④ Peak value of TRV: The largest peak value among a few peak values of the oscillating TRV. ⑤ Initial peak value of TRV (the first peak value): The first peak value of the TRV. ⑥ Frequency of TRV: The oscillatory frequency (plural frequencies may be possible). ⑦ Amplitude ratio of TRV: The ratio of the TRV to the peak value of the recovery voltage. ⑧ RRRV: The velocity of the voltage arising at the initial part of the TRV (kV/μs). ⑨ Reignition of arcs: The reignition of arc currents within a quarter cycle (5 ms for 50 Hz, 4.1 ms for 60 Hz) of the initial current breaking. 10 Restriking of arcs: The reignition of arc currents after a quarter cycle (5 ms for 50 Hz, 4.1 ms for 60 Hz) of the initial current breaking. (The meaning of a quarter cycle will be explained later.) 11 Low-frequency extinction: Arc-current extinction at the time of low-frequency current zero. 12 High-frequency extinction: Arc-current extinction at the time of high oscillatory frequency current zero. All of these terms are important for a proper understanding of switching overvoltage phenomena. Readers should understand the meanings of ①–⑧ through a study of the calculation method for TRVs in the previous sections. There are various voltage and current conditions for which circuit breakers are good or bad at breaking, or the breakers may or may not cause severe switching overvoltages. We will examine various different current-switching phenomena under different voltage and current conditions from the viewpoint of practical breaker switching duties. 15.12.3

Short-Circuit Current (Lagging Power-Factor Current) Tripping

We studied the current-tripping phenomena of the circuit shown in Figure 15.14 for the ideal breaker. However, actual fault-current tripping by a breaker is a little different, because a current arc appears transiently across the contacts. Figure 15.22a is the same circuit as in Figure 15.14, with the actual voltage and current waveforms indicated. In this case, the breaker completes current tripping at the time of the second current zero. Detailed comments for the waveform are also written in Figure 15.22. The magnitude of the arc-voltage drop would be, say, 5% or less of the peak value of the phase voltage.

435

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15 Lightning and Switching Surge Phenomena and Breaker Switching

L

r b

a C (a)

Transient recovery voltage Recovery voltage

Arc voltage

Voltage across contacts 𝜈ab (t) E t

Successful trip

i t3

t5 Extinction

Fault current (90° lagging from voltage) Contact sliding stroke

Open Close

t1 t2

t3 t4

t5

t6

Arcing time (b) i

At t < t1 (before time t1), fault current of 90° lagging power factor is flowing through the breaker, while the phase voltage is almost zero under the condition of the fault.

ii

The movable contact begins to slide and leave the fixed contact at t1, and simultaneously a current arc with smaller voltage drop 𝜈ab(t) appears across the contacts. The current arc will continue to flow until the time of first or second current zero (t3 or t5).

iii The current arc is not extinguished at t3 (the first current zero) by chance, but is extinguished at t5 (the second current zero). The polarity of the arc drop voltage is changed at the time of every current zero. iv The current arc is extinguished at t5 (the second current zero), and simultaneously the transient recovery voltage 𝜈ab(t) as well as the surge overvoltage 𝜈a(t) at terminal a appear. v

The breaker completes tripping at t5 if reignition of arc does not occur.

Figure 15.22 Fault-current tripping.

Figure 15.23a shows a typical waveform of a TRV. Figure 15.23b is the explanation of the TRV versus the conceptual insulation withstanding voltage across the contacts, although the latter conceptual characteristics cannot be measured. Incidentally, it is not easy to indicate the breaker’s tripping capabilities with simple illustrations. Figure 15.23c is an attempt to illustrate the relation of the breaker’s capability versus the expected duty required by the power-system condition related to fault-current tripping. The figure shows the coordinates of rated short-circuit capacity (MVA) of the breaker and the expected RRRV (kV/μs) arising in the system. Two breakers of different capabilities are shown in the diagram. If the required system duty is given by the curve P2, breaker 1 with the larger capability should be used. Recall that breakers must always be selected to leave some margin to meet all possible conditions, including a future expanded power system.

15.12.4

Leading Power-Factor Small-Current Tripping

This is a very important case for understanding unique switching phenomena and the delicate tripping characteristics of breakers. Above all, the phenomena of the reignition and restriking of arcs during the tripping procedure produce quite different results.

15.12 Circuit Breakers and Switching Practices

Recovery voltage

Time Transient recovery voltage Arc voltage

Time Recovery voltage Arc current Transient recovery voltage (a) Arc

Movable contact

Fixed contact

Voltage (kV)

oltage tion v

Voltage (kV)

Insula

oltage tion v

Insula

Transient recovery voltage

Transient recovery voltage

Trip failure (broken) arc voltage Time

Time

(b1) Successful trip

(b2) Tripping failure (b)

RRRV(kV/μs)

Tripping failure

Su

cce

ss

Breaker 1 Breaker 2

Power system 1

P1

P2

Tripping capacity (MVA) Power system 2

(c) Figure 15.23 Breaker capability (concept).

437

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15 Lightning and Switching Surge Phenomena and Breaker Switching

In Figure 15.24a, line section 2 is charged from the generator side while the opposite terminal end is open. Accordingly, a very small line-charging current of leading power factor flows through the breaker. Breaker Br will be tripped to break the small leading power-factor current under this condition. We have to investigate two typical, different switching phenomena for the breaker. Incidentally, referring to Table 4.1, the typical line-charging current is 0.6 A/km for the 275 kV line, so a small current may mean 100 A or less per phase for a line of 100 km or less. Case 1 Successful Tripping Without Reignition or Restriking of Arcs This case is explained with Figure 15.24b. A small line-charging current i(t) flows before time t1 (i.e. t < t1) with 90 leading angular phase to voltage E. Immediately after a breaker receives a tripping signal, the movable contact of the breaker begins to slide off and leave the fixed contact at t1, while the small current continues to appear as an arc current. Next, the breaker breaks the current at the time of the first current zero t2 (this is called the low-frequency extinction), because the breaker can easily break such a small current. After time t2(t2 < t), the system voltage va(t) at the terminal point a sinusoidally changes (va(t) = E cos ωt) from E to −E. On the other hand, the charging DC voltage E remains on line section 2 as a residual potential voltage, because the current is 90 leading, so the voltage value at the time of current zero t2 is just the crest value E. In other words, the voltage of the breaker terminal point b remains unchanged as vb(t) = E. Accordingly, the voltage across the contacts vab(t) after t2 increases from zero (at t2) to 2E and soon decreases to zero. This behavior after t2 can be written symbolically as follows: va t = Ecos ωt = E, − E

or va t ≦ E

vb t = E where E, −E means betwee E and −E ∴ vab t

≦ 2E and 2E maximum

The breaker tripping is successfully completed because the breaker has enough capability to withstand the voltages within 2E. Case 2 Successful Tripping Including Reignition of Arcs In the previous process, let’s assume the occurrence of reignition at t3, which is a quarter cycle or less after t2 as shown in Figure 15.24b. When the reignition is caused at t = t3, the potential voltage vb(t) at point b just before reignition is between E and 0, so vab(t) across the contacts is also between E and 0 for t < t3. That is, before reignition, for t < t3 va t = E cos ωt = E, 0 vb t = d c potential charging voltage vab t ≦ E,and E maximum Then, the reignition voltage (high-frequency free-oscillatory transient voltage) caused just after t3 ≤ t is of magnitude between [E, −E] by amplitude ±E: For t3 < t < t4 va t = vb t = E, − E ∴ vab ≦ 2E and 2E maximum As the movable contact is still enlarging the stroke, the breaker easily breaks the small oscillatory current at t4, which is the time of oscillatory current zero (this is called high-frequency [current-zero] extinction). After arc extinction at t = t4(t4 ≦ t), the potential voltage vb(t) and source voltage va(t) are kept within ordinal magnitudes E. In conclusion, even if one time of reignition occurs within a quarter cycle after the initial arc extinction, the breaker’s recovery voltage vab(t) is 2E maximum. In other words, reignition does not cause severe transient voltages across the breaker contacts, so the breaker tripping is successfully completed in this case. Reignition within one time through the type test is usually permitted in the representative international standards for breakers.

15.12 Circuit Breakers and Switching Practices Open

Br

Line #1

Line #2 a

b

(a) Recovery voltage 𝜈ab (2E maximum) Voltage of contact a 𝜈a

Reignition

Voltage of contact b 𝜈b

Source voltage E E t Maximum E Low-frequency extinction

E

Reignition current High-frequency extinction t 3 t4

t2

Successful trip Leading small current (charging current) Open Close t1

t2

t3 t4 t5

Sliding start Moving stroke finished

Low-frequency extinction

High-frequency extinction

Reignition

(b) Successful tripping with reignition Maximum 5E 𝜈b

𝜈ab

𝜈ab (Maximum 6E)

(Maximum 4E)

E E t 𝜈a

–E

Maximum Restriking

t3

t2

i

𝜈b

– 3E t4

open close t1

t2

t3 t4 t5

t 6 t7 Second highfrequency extinction Second restriking

Sliding start Charging current tripping (low-frequency extinction) First restriking First high-frequency extinction

(c) Tripping failure with restriking

Figure 15.24 Leading power-factor small-current (charging current) tripping.

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15 Lightning and Switching Surge Phenomena and Breaker Switching

Case 3 Tripping Phenomenon Including Restriking of Arcs (Tripping Failure) This case is explained with Figure 15.24c. If reignition (which we call restriking of arcs) is caused at t3, which is at least a quarter cycle after t2, then before restriking, for t < t3 va t = E cos ωt = 0, − E vb t = E d c charging voltages ∴ vab t ≦ 2E maximum Then the restriking voltage (high-frequency free-oscillatory voltage) after t3 ≤ t is of magnitude between [E, −3E] by amplitude ±2E: for t3 < t < t4 va t = vb t = E, − 3E

high-frequency oscillatory voltage with amplitude ± 2E

Next, as the movable contact is still enlarging the stroke, the breaker breaks the oscillatory current at the time of oscillatory current zero (at t4; this is again high-frequency extinction). After arc extinction at t = t4(t4 ≦ t), for t < t4 va t = E cos ωt = 0, − E vb t = E, −E

d c potential charging voltages

∴ E ≦ vab t ≦ 4E and 4E maximum In other words, one time of restriking may cause a large restriking voltage of up to 4E across the contacts, and further −3E of phase voltages arise. Then, a second restriking is inevitably caused at t6: for t6 < t < t7 va t = vb t = 5E, − 3E transient restriking voltage of amplitude ± 4E maximum Further, if the arc is again extinguished at t7, the phase voltage vb(t) at terminal b becomes 5E maximum, and the voltage vab(t) across the contacts is 6E maximum. for t7 < t vb t ≤ 5 E and 5 E maximum vab t ≤ 6 E and 6 E maximum If we assume the third restriking, vb(t) and vab(t) could theoretically become a magnitude of even −7E and 8E maximum, respectively, using a similar analogy. Of course, the breaker cannot withstand such severe restriking voltages, and the breaking chamber is broken probably around the time of the first or second restriking. In addition, earth-grounding faults are newly induced on the breaker or on other closely installed equipment. Case 3 explains why the one-time occurrence of restriking has a high probability of leading to a large TRV vab(t) and transient phase voltage (switching surge) vb(t), by which a tripping failure and/or a grounding fault is caused and the breaker is immediately broken, or by which successive restrikings are caused and further extreme voltages vab(t), vb(t) are induced, so tripping failure and breakdown of the breaker occurs sooner or later. Furthermore, a new grounding failure from phase overvoltage on line sections or substation equipment is also caused. The results of restriking (Case 3) and reignition (Case 2) are significantly different. Obviously, the occurrence of restriking may cause severe effects, so this means the breaker’s tripping capability is lacking, whereas reignition does not necessarily mean a lack of capability. Note that restriking is caused not only by tripping of a small leading current, but also by tripping of an ordinal fault current if the tripping capability is critical to the circuit condition, although the previous explanations applied to the processes of small leading-current tripping. In practical engineering, the necessary tripping duties of each type of breaker are confirmed by checking all the items in a commercial test against the specified authorized standards. The occurrence of restriking for any reason in such a test is recognized as unsuccessful tripping (rejection due to lack of tripping capability), although one-time reignition may be granted.

15.12 Circuit Breakers and Switching Practices

15.12.5

Short-Distance Line Fault Tripping (SLF)

We assume three fault points f1, f2, f3, as shown in Figure 15.25a. In the early 1950s, it was found that f2 fault tripping (f2 is a few kilometers distant from the breaker) could be more difficult than f1 fault tripping, in spite of the smaller fault current through the breaker. This is because a higher RRRV appeared with the fault at f2 as a result of repeated reflection of the traveling-wave voltages caused by the fault between point b and f2. It was recognized that such aspects often appeared in cases of faults up to 10 km from the breaker point.

Line #1

Line #2

Br

L1, C1

L2, C2 a

L3, C3

b f1

f2

Load

f3

Case 1: The fault at point f1 (very close to the breaker terminal b) The voltage at terminal b is actually zero just after the fault, namely vb(t) = 0, accordingly va(t) = vab(t) and the RRRV will not become relatively large in comparison with case 2 in spite of the fault current being larger than that of case 2. Case 2: The fault at point f2 (up to 10 km distant from the breaker terminal b) The free oscillatory overvoltage vb(t) appears at point b as the result of repeated travelling-wave voltage reflection between f2 and b. In addition, the caused frequency is higher and accordingly the RRRV appearing for the breaker’s vab(t) could also be larger, because the distance between f2 and b is only a few to 10 km. (Assuming distance bf2 = 3 km and the velocity is 300 m/μs, one-way travelling time is 10 μs so that the caused natural frequency would be around 50 kHz.) Therefore, in total, this case could be more severe for the breaker tripping in comparison with case 1 in spite of the smaller fault current. Case 3: The fault at point f3 (rather distant from the breaker terminal b) In comparison with case 2, the fault current is smaller. In addition, the similar oscillatory overvoltage also appears in this case; however, the frequency is lower because of the distance between f3 and b. Therefore this case is easier for breaker tripping.

(a) Concept of short-line fault (SLF) tripping A B Xs Model circuit Cs Um = Un

Voltage distribution at the instant of breaker extinction t = 0+

2 3

X2 C2

U2 = Um

X2 X2 + X5

Um

Voltage distribution after trip

Um

Transient wave form of phase to earth voltage

UA U2 0.5

UB du/dt

Transient recovery voltage across the breaker contacts

1

t (ms)

1

t (ms)

alue rest v

rst c

The fi

U2 0.5 (b) Concept of voltage distribution Figure 15.25 Short line fault (SLF) tripping.

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15 Lightning and Switching Surge Phenomena and Breaker Switching

SLF fault current Current trip time Arc voltage

50 t (μs)

40 kV

40 kV/12 μs SLF-TRV 40 kV/3 ms

(c) Waveform of TRV in case of SLF TRV Figure 15.25 (Continued)

Such a fault was named a short line fault (SLF). In the past, only TRV and RV were recognized as closely affected parameters for breaker fault-current tripping. But later, engineers noticed that RRRV should be the third important parameter that affects the breaker SLF fault-current tripping capability. Figure 15.25b shows a concept of voltage distribution for a simple model, and Figure 15.25c shows a measured waveform at t = 0 +. In the 1950s, it was believed that the absolute value of the TRV and recovery voltage (both of which are proportional to the fault current) were the essential factors for breaker tripping duty. However, after repeated accidents caused by breaker-tripping failures, it was recognized that very severe RRRV is caused by repeated reflection of the surge voltage between the breaker and fault points a few kilometers ahead. Figure 15.25b shows a measured waveform of the TRV caused by a SLF. The RRRV of the first wave front recorded 40 kV by 3 μs, which means the surge voltage traveled a one-way distance by 3 μs (about 900 m with velocity 300 m/μs). This RRRV value is four times that of the recorded increasing rate of TRV 40 kv/12 μs, so it could be a serious cause of a breaker tripping.

15.12.6

Current-Chopping Phenomena with Small Current Tripping and Lagging Power Factor

Breakers are specifically designed to break large current arcs (of even 50 or 63 kA RMS maximum), but then some smaller currents may be cut off before the time of current zero. This phenomenon is called current chopping. This current chopping causes large TRVs physically, and the reason for the resulting high surge voltage can be explained simply as follows while i(t) is small: it

small,

v t =L

di t dt

di t dt large,

large, dv t d2 v t and dt dt 2

quite large because of repeated ignition extinction

A typical case is the tripping of transformer excitation current as shown in Figure 15.26a, in which a small lagging current of the transformer excitation current is going to be tripped. Figure 15.26b shows the typical waveforms of current chopping, where the small arc current is forcibly cut off just before its current zero (the voltage is almost the peak value in this timing), so free oscillatory overvoltages appear. The resulting oscillatory frequency f = 1 2π LC is also quite high because the leakage capacitance C (from the insulation structures of bus and feeders, etc.) has a small value (say, 100–1000 pF). Figure 15.26c shows the simple equivalent circuit, in which the following equations are found from the law of energy conservation:

15.12 Circuit Breakers and Switching Practices

1 1 1 C e2 0 − + L i2L 0 − = C v2b 0 + 2 2 2 Total stored energy at t = 0−

where e 0 − = E cos ωt

Max stored energy by C at t = 0 + t = 0−

①

=E

then L 2 i 0 − + E2 C L

vb 0 + = ±

L iL 0 − ,E C

>

15 98

②

L 2 1 iL 0 − + E 2 , f = C 2π LC 0 − , i 0 − the voltages and current at the time of current chopping

vab 0 + = E ± where vb 0 − , vab vb 0 +

the peak value of resulting oscillatory voltage at point b

vab 0 +

the peak value of resulting transient recovery voltage of the breaker

f oscillatory voltage frequency

Substation bus Note: at t = 0, when current chopping is initialized,

Open

Br a

Tr

Open

b i

iC

e(0 –) E b(0–), –) = arc drop voltage

ab(0

Load or generator

Power source e = Ecos t

a

Transformer or reactor

Stray capacitance C

(a) Transformer excitation-current tripping

iL

i C

L

(c) Simple equivalent circuit

Repeated current chopping

Current

𝜈b

b 𝜈ab

0

i

Breaker voltage across the contact 𝜈ab(t) Arc drop voltage

Recovery voltage

Transient recovery voltage

Repeated reignition voltage

(b) Current chopping Figure 15.26 Current-chopping phenomena (small lagging current tripping).

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15 Lightning and Switching Surge Phenomena and Breaker Switching

The voltage at the time of chopping is almost the peak value of the source voltage e(t), vb(0−) ≒ E. Accordingly, vb(0+) is at least as large as the crest value of the source voltage E. Furthermore, the current-chopping phenomena of small inductive current tripping may cause repeated reignition and high-frequency extinction due to large RRRV and the powerful breaker capability for forced arc extinction, so extremely large switching surge overvoltage vb(t) and TRV vab(t) could be caused. Serious current chopping is inevitably caused by repeated ignition and extinction, so it must be carefully examined from the viewpoints of breaker-tripping capability (a breaker duty against vab(0+)) and the caused switching surge level (caused phase-over voltage va(0+) and vb(0+)). Current chopping is caused by smaller current tripping of typically i(t) ≤ 10–20 A for GCB, and the situation is similar for OCB and ACB. Typical cases of current chopping are as follows: i) Tripping of a no-load transformer (excitation current tripping) Excitation current i(t) of a transformer under no-load charging operation is typically 1–5 A (see Section 6.9.5), so a breaker can easily trip, so there is probably a small current with current chopping mode. ii) Line-charging current tripping of a short transmission line The charging current of a 10–20 km transmission line that flows through the line stray capacitance C is 6–10A (assuming 0.6 A/km), which causes the charging voltage. This small leading current is broken by a breaker with chopping mode. In addition, the L and C constants of substation equipment (reactor, PT, and blocking coil, for example) are coupled with the line capacitance C, creating a free oscillatory tank-circuit. Then, very quick oscillatory transient phenomena are caused by chopping-mode current tripping, closely affected by all the L and C constants of the line and substation. iii) Reactor tripping Tripping of a smaller-capacity reactor is another example of smaller current chopping.

15.12.7

Step-Out Tripping

As we saw in Figure 15.16, Cases 4, 5, and 6, breaker tripping under the step-out procedure may cause large recovery voltages (3E maximum), so it is one of the most severe tripping duties for the breaker. High-voltage breakers are usually assigned to have the capability to trip under step-out conditions, although most breakers are blocked from step-out tripping by protective relay equipment, as explained in Section 14.5.

15.12.8

Current-Zero Missing

Finally, current-zero missing phenomena of breaker tripping is a vital subject of concern for engineers. Our breaker is an AC tripping breaker capable of breaking the current flowing at the time of current zero within a time duration of 2 or 3 cycles when the breaker current-tripping ability continues. Breakers can trip AC fault current only under such conditions (see Section 15.12.3). In other words, breakers cannot trip DC offset current. With this in mind, the following two cases should be considered. We know experimentally that tripping failure of breakers by current-zero missing seldom occurs. However, we cannot dismiss this subject as an unrealistic matter that would never occur. Let’s examine this problem in a little more detail. Recall that in Chapter 10, Figure 10.13 shows a waveform of a typical short-circuit current for a three-phase fault at a generator terminal, where phase currents ia, ib, ic include DC components idc − a, idc − b, icd − c. Obviously, the idc component of the three phases becomes larger or smaller by chance, primarily affected by the angular timing of current tripping. The idc component of one phase becomes the largest offset value, and the disappearing time duration becomes longer. We have also seen that the time constant of current DCcomponent attenuation is given by Ta (see Section 10.10.4) and the typical values are 0.2–0.4 seconds (see Table 10.1). Referring to Ta from Table 10.1, DC-component idc attenuates to a half value by 0.4–0.8 seconds. Further, Eqs. (10.109) and (10.111a) indicate that the time constant is modified to larger values when line faults occur, because line and transformer impedances are inserted between the generator terminal and the fault point. Moreover, an essential point to remember is that a breaker must have a current-zero time within 1 or 2 cycles (in the case of a 2 cycle tripping breaker) immediately after receiving a tripping signal.

15.12 Circuit Breakers and Switching Practices

Taking all these things into account, current zero-missing may not be unrealistic subject in an EHV/UHV system, where the time constants of individual transmission lines are apt to become larger due to multibundled conductors and for other reasons. However, breaker-tripping operation under current-zero conditions should be entirely avoided in any case, so current-zero missing is always an important subject from the viewpoint of practical breaker-switching engineering. Magnetizing inrush currents including extremely distorted DC components may arise 1–10 seconds after initial excitation of transformers (see Section 6.9). Therefore, an attempt to trip inrush current forcibly may cause a tripping failure of the breaker due to current-zero missing, or severe overvoltages due to current chopping (= L di/dt). Usually, a trip-lock function is included in transformer protective equipment as a typical practice, with which transformer inrush current tripping is blocked for the initial few seconds just after the transformer is charged. This is not only to prevent mis-operation of transformer differential relays but also to prevent breaker failure. 15.12.9

Overvoltages Caused by Breaker Closing (Close-Switching Surge)

We will now examine overvoltages caused by breaker closing. In Figure 15.27, breaker Br is going to be closed in order to charge the line. Just after the breaker receives the closing dispatch signal, the movable contact begins sliding to approach the fixed contact and completes mechanical closing within around 0.1 second. As the stroke approaches, and just before the two contacts mechanically touch, preignition is started, and the current arc begins to flow. The current arc will never be extinguished until the two contacts complete mechanical closing, because the stroke gap across the contacts is rapidly decreasing. Accordingly, breaker closing is electrically achieved at the time of ignition start. The time of arc ignition may be accidental, unlike the time of current zero for tripping. Also, so-called re-extinction is never caused during the closing process, though reignition or restriking may be caused during the tripping process. In Figure 15.27, vqr(0−) is the preignition voltage across the contacts q and r in the initial stage of the closing procedure. Now the transient phenomena caused by the breaker closing can be calculated by inserting the voltage with the opposite polarity –vqr(0−) across points q and r as the sudden forced voltage source. At the instant of insertion (of arc ignition), the surge traveling waves vq(t), iq(t) and vr(t), ir(t) begin to travel in the right and left directions, respectively. The surge equations can be written as follows: vq t −vr t = vqr 0 − , iq t = − ir t vq t = Z2 , iq t ∴ surge in left direction Z2 vqr 0 − vq t = Z1 + Z2 iq t =

15 99a

vr t = Z1 ir t

①

1 vqr 0 − Z1 + Z2

∴surge in right direction Z1 vqr 0 − vr t = − Z1 + Z2 ir t = −

1 vqr 0 − Z1 + Z2

②

These equations give the initial traveling surge voltages and currents in both directions.

𝜈qr (t = 0–) · 1(t)

Lin

e #4

Z4

Br

Z3 p Line #3 BusLine #2 Z2 = L´ C´

q

S r Line #1 L Z1 = C

s

Figure 15.27 Transient overvoltage caused by breaker closing (breaker-closing surge).

15 99b

445

446

15 Lightning and Switching Surge Phenomena and Breaker Switching

For the initial voltage across the breaker contacts just before preignition by closing, vqr(0−): Case 1 Points q and r are loop connected through another route or a parallel circuit At t < 0 vqr t = E1 e jωt − E2 e j ωt + δ = E1 1 −

E2 jδ jωt e e ≒ E1 1 − e jδ e jωt ≦ 2E1 E1

∴ vqr 0 − = 1 4E maximum Case 2 The switch S at point s is opened The residual voltage of 0 to ± E may exist at line 1 unless the line is forcibly grounded On the other hand , the phase angle of the generator source voltage is within 0 to 360

15 100

vq t = E cos ωt + α , vr t = − E, E for t ≤ 0, then vqr 0− = 2E maximum ② Case 3 Closing of inverse angular polarity by operational mistake vqr 0− = 2E maximum

③

Needless to say, breaker closing without synchronization has to be strictly avoided not only to prevent instability or generator shaft distortion, but also to prevent switching overvoltages and breaker failure

• ••

Numerical check of surge voltage caused by breaker closing: In Figure 15.27, the following line conditions are assumed, and attenuation is ignored: Overhead line 1: length 15 km, surge impedance Z1 = 300 Ω, velocity u1 = 300 m/μs; switch S at the opposite terminal s is opened. Cable line 2: length 5 km, surge impedance Z2 = 30 Ω, u2 = 150 m/μs. Overhead lines 3, 4: length infinite, surge impedance Z3 = Z4 = 300 Ω, u3 = u4 = 300 m/μs. For the initial surge in the left direction (from point q to p): Z2 30 vqr 0 − = vqr 0 − = 0 09 vqr 0− 300 + 30 Z1 + Z2 1 1 vqr 0 − = vqr 0 − = 0 03 vqr 0 − iq t = Z1 + Z2 300 + 30

vq t =

•• •

The coefficient of reflection at point p = (150–30)/(150 + 30) = 0.67. Total time going to and returning from points q, p = 2 × 5000/150 μs = 66.7 μs ∴ f = 15.0 kHz. The voltage at point q at the time of arrival of the first reflected wave: 1 + 0 67 vqr 0 − = 1 67 vqr 0 −

For the initial surge in the right direction (between points r and s): Z1 vqr 0− = 300 300 + 30 vqr 0 − = 0 91 vqr 0 − Z1 + Z2 1 vqr 0 − = 1 300 + 30 vqr 0 − = 0 003 vqr t ir t = Z1 + Z2

vr t =

•• •

The coefficient of reflection at open point s = (∞ − 300)/(∞ + 300) = 1.0. Total time for going to and returning from points r, s = 2 × 15 000/300 μs = 100 μs ∴ f = 10 kHz. The voltage at point r at the time when the first reflected wave arrives (1 + 1) vqr(0−) = 2.0 vqr(0−).

15.12 Circuit Breakers and Switching Practices

Higher frequency caused by short-distance reciprocating surge travel means greater steepness generally, while the attenuation caused by eddy-current losses and corona losses is quick. Recently developed gas-insulated switchgear (GIS) is generally of compact length, so switching surges with extraordinarily higher frequency phenomena (say 1 GHz) could appear within a limited distance. We will discuss this problem in the next chapter.

15.12.10

Resistive Tripping and Resistive Closing by Circuit Breakers

Resistive tripping and/or resistive closing may be used in a practical design of EHV/UHV class breakers in order to reduce switching surge voltages within specified standard voltage levels (see section 17.10.4), or to improve the tripping/closing capability of the breakers. Diagrams of two typical practices are given in Figure 15.28. As shown in the figure, a breaker has one main contact frame s1 (sliding contacts design) and one auxiliary contact frame s2 (resistive contacts: (probably bat contacts design) and a resistor element R (usually 300–500 Ω, short-duration time). For the tripping sequence of a breaker with resistive tripping design, in the original breaker-closing condition, the main contact s1 and the auxiliary contact s2 are in the closed position, so all the current i flows through the main contact s1. Just after the tripping signal is received, the main contact s1 is opened first at t1, then the current pass through the main contact is switched to the auxiliary pass through the resistive element and the auxiliary contact s2, and the resistive restrained current iaux flows through the auxiliary pass until s2 is opened at t2 (the time difference is, say, ΔT1 = t1 – t2 = 10 ms or less), thus completing the breaker tripping. Although the duration ΔT1 of the current iaux flowing through the resistive element is only 10 ms, the resistive element has to withstand the temperature rise caused by the extremely large resistive loss i2aux R dt. For the closing sequence of a breaker with resistive closing design, in the original breaker-opening condition, the main contact s1 and the auxiliary contact s2 are in the open position. Just after the closing signal is received, the auxiliary contact s2 is closed first at t1, and then the resistive restrained current iaux begins to flow through the auxiliary pass until s1 is closed at t2 (the time difference is, say, ΔT2 = t1 – t2 = 10 – 20 ms), completing the breaker closing. Although the duration ΔT2 of the current iaux flowing through the resistive element is only 10 ms or less, the resistive element has to withstand the temperature rise caused by the extremely large resistive loss i2aux Rdt

15.12.11

Standardized Switching Surge Level Requested by EHV/UHV Breakers

EHV/UHV breakers are requested to restrain switching surge voltages within specified voltage levels of associated standards or recommendations for insulation coordination, the details of which are examined in the next chapter. (Example 1) s2

s1: main contact frame

R

s2: auxiliary (resistive) contact frame s1 Tripping sequence Tripping signal → s1 opening → s2 opening (Example 2)

R Closing sequence s2

s1

Closing signal → s2 closing → s1 closing

Figure 15.28 Resistive-tripping method and resistive-closing method.

447

448

15 Lightning and Switching Surge Phenomena and Breaker Switching

Design practices for resistive tripping and/or resistive closing may be useful to reduce switching voltage, while breaker manufacturers are free to adopt such design practices if necessary. Tables 17.2B and 2C show the IEC and IEEE standards for power systems over 245 kV, which indicate the permissible switching surge level. The permissible switching surge level based on these standards can be written as the multiple PU value (α) based on the crest value of the rated phase voltage. For example, in the IEC standard (Table 17.2B) for a rated voltage 420 kV system, for the highest system voltage Vl-l = 420 kV (line-to-line, RMS value), the crest value of the line-to-ground phase voltage is 2 3 420 kV = 343 kV. The specified standard switching impulse-withstand voltage from the IEC for this operating voltage system is 850 or 950 or 1050 kV, corresponding to α = 2.48, 2.77 or 3.06 times 343 kV, respectively. A resistive element of typically 300–500 Ω is used to satisfy the requirements for both these reasons. Assuming R = 500 Ω for a 500 kV breaker, the current flow iaux of the auxiliary contact frame is 500 3 500 ≒ 0 6 kA rms. Then the resistive element has to withstand thermal energy of i2aux Rdt ≒ 600 2 ≒ 500 watt 10−2 sec = 1800 kJ, although the duration ΔT1 is only about 10 ms. Ceramic resistors have been used recently because of their outstanding heat-resisting stable characteristics. Further, resistive tripping and/or resistive closing design may be useful to realize smart design of breakers by reducing the requirements of the main contacts frame, although adoption of such a design is an individual manufacturer’s choice. A typical design of a GCB for 500 kV or higher voltages uses resistive closing with or without resistive tripping.

15.12.12

Overvoltage Reduction with Resistive Tripping

We will examine the tripping phenomenon for the circuit in Figure 15.29a, in which a grounding fault has occurred and the fault current (90 lagging) i(t) = I sin ωt is flowing through the breaker. The main contact frame s1 starts moving and breaks the through current at the time of current zero (t = 0). A little time ΔT later, the auxiliary contact frame s2 breaks the current iaux (t) at the time of current zero. The initial voltage and current (steady-state value) in Figure 15.29a are e t = Ecos ωt i t = i1 t = Isin ωt =

15 101

E sin ωt current zero at t = 0 ω L1 + L2

The transient phenomenon when the main contact frame starts moving can be calculated as in Figure 15.29b, where the total impedance looking into the circuit from points a and b is composed as the parallel circuit of the surge impedance (Z1 + Z2) and R. The equation for the Laplace transform is E E ω sin ωt = 2 ω L1 + L2 ω L1 + L 2 s + ω 2 R ii s α s i1 s is = R + Z1 s + Z1 s

for t7 ≥ 0 i1 s =

Z1 s + Z2 s i1 s R + Z1 s + Z2 s R where α s = R + Z1 s + Z2 s iaux s =

15 102

va s = Z1 s i s = α s Z1 s i1 s vb s = − Z2 s i s = −α s Z2 s i1 s vab s = Z1 s + Z2 s

i s = α s Z1 s + Z2 s

i1 s

The special case of a non-resistive-tripping breaker is obtained by letting R

∞. Then α(s)

1.

i s = i1 s va s = Z1 s i1 s

15 103

vb s = − Z2 s i1 s vb s = Z1 s Z2 s

i1 s

15.12 Circuit Breakers and Switching Practices

s2 i(t)

c

R

s1 j 𝜔L1

e = E cos 𝜔t

i1

a

b

j𝜔L2

(a) Before s1 opening

iaux(s)

R

i(s)

i(s) a

+

b

i1(s)

Z1(s)

Z2(s)

(b) s1 opening

s2

iaux R

c

a

sL1

b

sL2

(c) Before s2 opening

+

c

R

a

b

Z1(s)

iaux(s)

Z2(s)

(d) s2 opening Figure 15.29 Calculating breaker resistive-tripping surge voltages.

Compared to Eqs. (15.102) and (15.103), all the transient quantities i(s), va(s), vb(s), vab(s) can be reduced by adopting resistive tripping with resistance R, and the rate of switching surge reduction is given by α(s): αs =

R R + Z1 s + Z2 s

15 104a

For the initial time interval before the reflected waves return from the transition point at the opposite terminal, simL1 C1 , Z2 s L2 C2 is possible, and plification of Z1 s α= R+

R L1 + C1

L2 C2

15 104b

Accordingly, the initial levels of the switching surges va, vb, vab, i can be reduced by the rate of α in Eq. (15.104b). If the resistance R is selected to be smaller than (Z1 + Z2), R < (Z1 + Z2), the transient voltages can be reduced by half. If smaller R is selected, the tripping duty of the main contact frame and the induced switching surge voltages can be reduced, although a relatively larger tripping duty is required for the auxiliary contact frame.

449

450

15 Lightning and Switching Surge Phenomena and Breaker Switching

Next, the auxiliary contact frame s2 is opened with a time delay of about 10 ms from the opening of s1, and the transient phenomenon is calculated as in Figures 15.29c and d. The transient term of iaux at the opening of s1 disappears quickly (because of the insertion of large R), and the initial iaux at the time of s2 opening is given by iaux t =

E E sin ωt ≒ sin ωt R + jω L1 + L2 R

where R = 300 − 500Ω

current zero at t = 0

ω L1 + L2

The equation of the Laplace transforms is E R + s L1 + L2 va s = Z1 s iaux s

for t≧0 iaux s =

ω E ω ≒ s2 + ω2 R s2 + ω2 15 105

vb s = − Z2 s iaux s vc s = − R + Z2 s

iaux s

vac s = s = R + Z1 s + Z2 s

iaux s

These equations indicate the approximation of the induced surge voltages: va =

Z1 − Z2 Z1 + Z2 Z2 E, vb = E, vab = E, vc = − 1 + E ① R R R R

the current through the auxilary contact frame E iaux t ≒ sin ωt R

15 106 ②

Accordingly, tripping of the current iaux by the auxiliary contact frame is relatively easy, because the magnitude of iaux is small; and iaux is almost in phase with the voltage, so the TRV at the time of current zero of iaux is also small. Then iaux can be tripped by the relatively simple, small auxiliary contact frame. Meanwhile, the heat energy that causes a temperature rise in the resistive element can be calculated by the equation iaux t

2

Rdt ≒ i2aux R ΔT

15 107

where ΔT ≒ 0 01 sec The resistive element has to withstand this heat energy and maintain a relatively stable ohm value through the duration of ΔT. 15.12.13

Overvoltage Reduction with Resistive Closing

In Figure 15.30a, the main contact frame s1 and the auxiliary contact frame s2 are open when the breaker is initially opened. Breaker closing is started by s2 first and is completed by s1 closing with time delay ΔT. The initial voltage vab(0−) across the breaker terminals a–b is affected by the condition of the terminal d on the opposite side:

•

When d is in the open condition,

•

When d is under a grounding fault or manually grounded,

•

For the breaker closing under synchronization,

vd = −e to 0 to + e, then vab 0 − = 2e maximum

vd = 0, then vab 0 − = e maximum

va , vb are in phase, then vab 0 − ≒ 0

15.12 Circuit Breakers and Switching Practices

s2 Line #1 e = E cos (𝜔t + 𝛿)

R

a a

j𝜔L1

Line #2

c s1 b

j𝜔L2

d

(a) Before s2 closing i

𝜈ac(s)

i

R

c

a Z 1(s)

b

Z 2(s)

iaux (s)

(b) s2 closing R

i(s)

i1(s) a

Z 1(s)

d b

𝜈ab

Z 2(s)

(c) Figure 15.30 Calculating breaker resistive-closing surge voltages.

Accordingly, the initial voltage can be written as vab = k e (where k = 0–2) in our calculation, which is to be inserted between points a and b in Figure 15.30b. After the sliding action of s2 starts, preignition across s2 is caused before mechanical contact is complete, which is actually the time of electrical closing because ignition continues without extinction until the mechanical contact of s2 is finished. The timing of electrical closing t = 0 is a matter of chance. Accordingly, the initial voltage across the breaker terminals at t = 0 is vac t = k Ecos ωt + δ

15 108

where k = 0 − 2 decided by the condition of line 2 The equation just after the electrical closing of s2 at t = 0 is vac s = k E iaux =

cos ωt + δ = kE

scos δ− ωsin δ ① s2 + ω2

1 vac s R + Z1 s + Z2 s

②

va s = Z1 s iaux s =

Z1 s vac s R + Z1 s + Z2 s

③

vb s = − Z2 s iaux s =

− Z2 s vac s R + Z1 s + Z2 s

④

vc s = − R + Z2 s

iaux s =

15 109

− R + Z2 s vac s ⑤ R + Z1 s + Z2 s

The special case of a non-resistive-closing breaker is obtained by letting R 0 in these equations; then the denominator is changed to {R + Z1(s) + Z2(s)} {Z1(s) + Z2(s)}. Therefore, the resulting transient switching surge

451

452

15 Lightning and Switching Surge Phenomena and Breaker Switching

quantities i(s), va(s), vb(s), vc(s), vac(s) can be reduced by adopting the resistive-closing method, and the rate of switching surge reduction is given by β (s) as β s =

Z1 s + Z2 s R + Z1 s + Z2 s

15 110a

For the initial time interval before the reflected waves return from the transition points on the opposite sides, simplification of Z1 s L1 C1 , Z2 s L2 C2 is possible. Then, the initial surge-reduction rate with the use of a resistive closing breaker is L1 L2 + C1 C2 β= L1 L2 R+ + C1 C2

15 110b

If R is selected as R = Z1 + Z2, then the surge level is reduced to approximately one-half. The resulting transient voltages and currents can be calculated by solving Eq. (15.109). Generally, in the case of breaker closing, the behavior of the transient voltages va, vb, vab is not so severe, but the transient current iaux (the inrush current) is an important matter because the contacts have to withstand a large inrush current without melting; furthermore, the breakers under the resistive-closing method must withstand the thermal energy caused on the resistive element. If the breaker is closed for the faulted line (called breaker closing at fault), the short-circuit current i(t) begins to flow immediately, and the initial current i(0+) may be twice the value of the AC component because the DC component may be superposed. Then the maximum inrush current is as follows: Theoretical largest value 2 2 Irate = 2 83 Irate Practical largest value 2 5 Irate various system losses are taken into account

15 111

where Irate is the rated fault-tripping current Returning to the breaker with resistive closing, Eq. (15.109) shows that the initial current iaux is reduced by β, so the duty of the auxiliary contact frame to the inrush current is also reduced by β, while the accumulated thermal energy of the resistive element i2aux R ΔT must be treated as a breaker design factor. Next, the voltage vab(t) = R iaux(t) just before closing the main contact frame is almost at steady state, because iaux is already dominated by R. Accordingly, severe transient behavior is not generally caused, whose phenomena can be calculated as in Figure 15.30c, if necessary. Note that resistive closing can obviously reduce inrush current caused by transformers or capacitor bank closing operations.

15.13

Switching Surge Caused by Line Switches (Disconnecting Switches)

A line switch (LS, also called a disconnecting switch) is a switching device that does not have the duty to break or to close load current or fault current. Its duty is to change the network connection by switching on/off under no-load circuit conditions. As shown in Figure 15.31 as an example, LSs are important members of the bus system and carry out their duties in combination with breakers. LSs can be operated only under the condition that all the series-connected breakers are open, and no current flows through the LS before and after switching is assured. (Therefore, the sequential interlocking facilities in combination with breakers at the same substation are vitally important.) The LS is not required to operate quickly, so opening and closing are rather slow. The operating stroke time of a conventional open-air LS may be, say, 2–3 seconds, while that of an LS set in GIS is, say, 0.5–1 second. However, the LS has to trip leakage current that flows through a small leakage capacitance C in a neighboring section, by which unique switching surge phenomena by are caused by the LS. Now we will examine the mechanism of LS switching surge, as shown in Figure 15.31a. The substation here consists of a double-bus structure (buses A and B), and feeder line 1 is connected to bus A (i.e. LS11 closed, LS12 open). Now the connection of feeder line 1 is going to be changed from bus B to bus A:

15.13 Switching Surge Caused by Line Switches (Disconnecting Switches)

Bus-tie breaker LS21

t1

LS11

t2

a

b

Br1

Br2 Power source

LS22

Line #1 C

LS12

t3 Bus-A

Bus-B

(a)

Contact b 𝜈b (t)

Contact a 𝜈a (t)

(b) Opening operation of LS11 Figure 15.31 Switching surge caused by line switch (LS) (disconnecting switch) operation.

•• •

Time t1: The breaker Br1 is tripped (load current tripped). Time t2: LS11 is opened. At this time, the line switch LS11 has to break the leakage current that is flowing across the leakage capacitance C of the small energized section connecting Br1/LS11/LS12. Time t3: LS12 is closed, so the bus connection changing processes of feeder line 1 from bus B to bus A are completed.

•

Physical surge level: The magnitude of lightning surges arriving at a substation is repressed within a certain level by the arrester installed at each feeder terminal. Compared to these repressed lightning surges, LS surges can be even larger

Concerning the behavior of LS11 at time t2 when LS11 is going to break the leakage current, ic = jωC E, the waveform of the voltage switching surge caused by LS11 opening is shown in Figure 15.31b. The voltage at point b (one terminal of LS11) is obviously AC system voltage vb = E cos ωt with a sinusoidal waveform, while the waveform for the voltage va at point a (another terminal of LS11) is a repeated stepping form with quite high oscillatory transient terms. Just after contact a leaves contact b, free oscillatory overvoltage va appears in the small section including point a, whose natural frequency is extraordinarily high because the traveling distance l is quite short (say, l = 10 m or less for a conventional substation, and a few meters or less for GIS). If the traveling length l = 6 m is assumed, the velocity u ≒ 300 m/μs and the natural frequency caused at the small section is approximately 25 MHz. The oscillatory transient term of va is soon attenuated, and va remains on the small section as the DC charging voltage of the stray capacitance C. However, the voltage vb = E cos ωt still changes continuously, so the voltage vab across points a and b increases. On the other hand, the stroke speed of the LS is so slow that a second reignition is inevitably caused. The amplitude of the second oscillation may be larger than that of the first oscillation by chance. These reignitions are repeated after some cycles, and the amplitude of the transient terms could become larger by repetition. However, the LS can finally break the charging current because the stroke distance of the LS is long enough to recover the insulation across the contacts. The problem we need to keep in mind is severe LS switching surge attacks on all the energized parts of the substation (buses A and B and all other connected equipment). The LS switching surge could be even more severe than a lightning surge in practical engineering, especially for EHV and UHV substations, The reasons are:

453

454

15 Lightning and Switching Surge Phenomena and Breaker Switching

• •

in magnitude, higher in frequency, and of a much longer duration, often in daily operation. This tendency is conspicuous for the EHV/UHV system, because it is generally designed with a relatively lower total insulation level. Protection hardness: LS switching surges are caused inside the substation, so surge protection by arrestors or any other devices may not be easy. The LS switching surges appearing may affect nearby equipment directly without attenuation. Meanwhile, closely installed arresters (installed typically at the line-feeder terminal point for lightning surge protection) have to withstand the thermal energy caused by long LS surges. Chance: LS switching surges appear quite often whenever any one of several LSs is operated. We will discuss this matter again in the next chapter.

Again, as in Figure 15.31a, capable LSs that cause fewer switching surges are important. Further, if a leakage resistance exists in parallel with the stray capacitance C, it could be an effective countermeasure to reduce the surge level. Adopting a gap-less arrester at the small section may also be useful. The gap-less arrester could work as a kind of nonlinear highresistive device connected in parallel with the stray capacitance C, while the duty of the arrester might be severe. Incidentally, the LS is usually equipped with the earth-grounding function at its terminals, which should be grounded during maintenance work periods on the associated transmission line or any other associated part of the substation. The LS is usually equipped with a manual earth-grounding function as typical practice.

15.14

Surge Phenomena Caused on Power Cable Systems

We will study surge phenomena caused on power cable lines under a few typical conditions. We already discussed cable cross-bonding connection and the surge phenomena caused on the sheath-circuit in Sections 9.3.4 and 9.3.5. Figures 15.7a and b for Case 2 are again shown in Figure 15.32, where lines Z1 and Z2 are connected at transition point m, and surge voltage e is traveling on line Z1 (surge impedance Z1 may be an overhead line) from left to right and arrives at m. Immediately the transmitted wave voltage μe begins to travel on line Z2 (may be a cable line), where μ = 2Z2/ (Z1+Z2) is the transmitted wave operator from Z1 to Z2. Thus μ e=

2Z2 Z2 e= 2e Z1 + Z2 Z1 + Z2

15 112

This equation shows that the transmitted wave voltage μe in Figure 15.32a is given by the same equation for the voltage at point m in Figure 15.32b in which source voltage 2e is going to be switched to the circuit. In other words, the transmitted wave μe in Figure 15.32a can be calculated as the voltage of the source voltage 2e divided by Z1 and Z2. Bearing this explanation in mind, we will examine the surge phenomena arising in the cable conductors and metallic sheaths, as shown in Figure 15.33a. Surge Voltages at the Cable Infeed Terminal Point m Figure 15.33a is the diagram we will examine, where incident surge e is coming from the overhead transmission line Z0 to the cable line. For this condition, we need to calculate surge voltages e1 (conductor voltage across the cable insulation

𝜇e =

2Z2 Z1 + Z2

.e=

Z2 . (2e) Z1 + Z2

𝜇e

e

m

2e Z1

m

Z2 Cable

Overhead line

(a)

Z1

Z2 Z2

2e

Z1 + Z2 (b)

Figure 15.32 The transition point of overhead line and cable.

. 2e

15.14 Surge Phenomena Caused on Power Cable Systems The surge impedances Overhead line e m

Z0 : of overhead line

Terminal end n

Cable

Z12 : of across cable insulation layer (from conductor to sheath metal)

Z12

Z0

Z2 : of across outer layer (from sheath metal to earth)

Z2

Zs1

Z

Zs2

Zs1, Zs2 : of sheath wire (to earth) at points m and n Z : of the circuit connected to point n (other lines or load)

(a)

2e e

Z0

Z0

e1

e2

Z12

Z12

e1 Zs1

Z

Z2

e2 Zs1

(b1) The surge impedance at point m

Z2

(b2) The surge impedance and voltage at point m

e1 : induced voltage across the conductor and the sheath Conductor a

e1 Z12 e2

Z Z2

2e1 e´1 Z 12

Z12 Z

Zs2

Z2

Zs2

2e2 Z2

Z e´2

Zs2

sheath metal b (c1) The surge impedance around point n (the cable end)

(c2) The surge impedance and voltage distribution across the insulation layer at point n (coaxial mode)

(c3) The surge impedance and voltage distribution across the outer layer at point n (line-to-ground mode)

Figure 15.33 Surge voltages arising on the cable conductors and the metallic sheath.

layer) and e2 (sheath voltage across the outer layer) at the jointed cable terminal point m. The voltages can be calculated analogously to Eq. (15.112) as follows. The situation of surge impedances around point m and the equivalent circuit are as in diagrams b1 and b2. The voltage appearing across the cable insulation layer (coaxial mode wave) is e1 =

2Z12 Z2 Zs1 Z0 + Z12 + Z2 + Zs1

e

where

Z2 Zs1 = Z2 Z2 + Zs1

Zs1

15 113a

The voltage across the outer covering (sheath to ground mode wave) is Z2 Zs1 Z2 + Zs1 e e2 = Z2 Zs1 Z0 + Z12 Z2 + Zs1 2

15 113b

or e2 =

Z2 Zs1 e1 Z12

15 113c

These surge voltages e1 and e2 begin to travel along the cable conductor and the sheath metal from left to right. Surge voltages at the Cable Outfeed Terminal Point n The induced surge voltages e1 and e2 at point m travel on the cable toward the other terminal end point n (diagram c1). The traveling velocity of e1 may be 130–150 m/μs and that of e2 a little slower (due to the line-to-ground mode). The surge e1 arrives at the conductor (point a) at point n first, and the equivalent circuit of this timing is given in diagram c2.

455

456

15 Lightning and Switching Surge Phenomena and Breaker Switching

Next, e2 arrives at the sheath metal (point b), and the equivalent circuit of this timing is given in diagram c3. Then, the following equations are derived, where we ignore attenuation. The voltage across the cable insulation layer at point n is e1 =

2Z12 e1 = Z12 + Z2 Zs2 + Z

2Z12 e1 Z2 Zs2 Z12 + +Z Z2 + Zs2

15 114a

The voltage across the cable outer-covering layer (between the sheath and the earth) at point n is e2 = where

2Zr e2 Z2 + Zr

Zr = Zs2

Z12 + Z

15 114b

Zs2 Z12 + Z = Zs2 + Z12 + Z

The cable insulation layer has to withstand the surges e1, e1 and their following reflected waves. Of course, these equations give only the initial transferred wave voltages before reflection, so the actual voltage at arbitrary time t after arrival should be calculated by the lattice method. If our concern is the stress on the insulation layer, we can assume that the sheath is ideal earth grounded; in other words, the surge impedance of the sheath is zero (Zs1 0, Zs2 0), ignoring the earth-mode traveling surge e2, e2 on the outer-covering layer. In this case, only the traveling surges of the coaxial mode e1, e1 exist, and e1, e1 can be calculated as pessimistic (large) values: 2Z12 e Z0 + Z12 2Z12 e1 = e Z12 + Z where Zs1 0, Zs2 e1 =

15 115 0

Z surge impedance of outer circuit

15.15

Lightning Surge Caused on Cable Lines

Now we will examine the overvoltage behavior of the cable system when a lightning surge (incidental surge) is injected from the connected overhead line, as shown in Figure 15.34. Referring to Figure 15.11 and Eq. (15.53), the overvoltages appearing at junction points m and n can be calculated as follows: et = μ e =

2Z1 e Z0 + Z1

①

the transmitted wave voltage at point m

Point m 0 −2 T

2−4 T

4 −6 T

6−8 T

vm = et 1 + α2 1 + ρ1 ρ2 1 + α2 ρ1 ρ2 + α2 ρ1 ρ2

8 − 10 T 2

+ α2 ρ 1 ρ 2

3

1 + α2 ρ2 et 1 − α2 ρ1 ρ2

②

15 116

Point n 0−1 T

1−3 T

vn = et 0 + α 1 + ρ2

3 −5 T

5 −7 T

1 + α2 ρ1 ρ2 + α2 ρ1 ρ2

7 −9 T 2

+ α2 ρ 1 ρ 2

3

α 1 + ρ2 et 1 − α2 ρ1 ρ2

③

It is obvious from these equations that vm, vn would become larger if ρ1 ρ2 were positive large values (namely, Z0, Z2 > Z1) and would become largest under the conditions Z0 Z1, Z 2 Z1. We need then to investigate Figure 15.34 as a typical severe case with cable Z1 = 30 Ω, overhead line Z0 = 300 Ω (sufficiently long), and Z2 = ∞ (open end).

15.15 Lightning Surge Caused on Cable Lines

Lightning

Direct-stroke or Back-flashover e

𝜌′1

𝜌2

𝜇2

𝜇e m

#0 overhead line

n

#1 cable line

#2 overhead line

Z1

Z0

Z2

Figure 15.34 Lightning surge on the cable line.

Accordingly Point m ρ1 = Z0 − Z1

Z0 + Z1 = 1 300− 30

Point n Z2 = ∞ open end ρ2 = Z2 − Z1

300 + 30 = + 0 82

Z2 + Z1 = 1 ①

et = 2 × 30 300 + 30 e = 0 182e 0−2 T

2−4 T

4 −6 T

6 −8 T

8 −10 T 2

vm = et 1 + 1 82α2 1 + 0 82α2 + 0 82α2 + 0 82α2

3

1 + α2

1 −0 82α2 et

②

15 117

Point n 0−1 T

1 − 3 T 3− 5 T

5 −7 T 2 2

vn = et 0 + 2α 1 + 0 82α2 + 0 82α

7−9 T + 0 82α2

3

2α 1 −0 82α2 et

③

The equation means vm and vn repeat reflection at both terminals, and the values increase over time toward the final converged value. Assuming α = 1.0 (attenuation zero), the converged value of vm and vn is 11.1et, which is a large, severe value. Fortunately, however, these severe values are unrealistic for the following reasons: a) This is a calculation in which the incidental surge voltage e of the step-waveform continues infinitely. Lightning surges do not have infinite tail length, so they disappear within 100–200 μs. Therefore, the step voltage in the calculation should be replaced with the realistic impinging waveform of a typically standard impulse wave of 1.2 × 50 μs (see Table 17.2C). b) Z0 > Z1. Accordingly, the transmitted operator μ at point m from the overhead line (Z0) to the cable (Z1) is less than 1.0 (μ = 0.182 in the calculation). In other words, the original surge voltage e diminishes to the first transmitted wave voltage μ e. c) The diminishing effect from the attenuation operator α (0 < α < 1) may be significant. The converged values calculated by Eq. (15.116) are 3.45, 1.70, and 1.33 for α = 0.8, 0.6, and 0.4, respectively. The resistive constants R and G of the cable act as attenuating factors of the cable. d) In normal operation of the network, smaller Z2, Z0 are expected, which would significantly reduce the voltages. If the cable terminal point n is connected, for example, to five parallel circuits of overhead lines, Z2 is around 300/n = 60 Ω; accordingly ρ2 = 0.33, so vm, vn are significantly reduced. If the double-circuit cable line is connected to point n, then Z2 ≒ 30/2 = 15 Ω, and ρ2 has a further negative value between 0 and − 1. If ρ1 or ρ2 is negative (i.e. if Z0 < Z1 or Z2 < Z1), vm, vn would diminish very quickly under oscillatory modes. Regardless, this investigation indicates that transient surge-voltage phenomena on a cable line caused by an incidental lightning surge is more severe in value and duration when the surge impedances of the connected adjacent lines (Z0, Z2) are larger than those of the cable line Z1 (Z0, Z2 Z1). As cable insulation does not have self-restoring characteristics, detailed surge analysis and the necessary protection (typically by arresters) against incidental surges are essential in practical engineering.

457

458

15 Lightning and Switching Surge Phenomena and Breaker Switching

15.16

Switching Surge Caused on Cable Lines

Now we will examine switching surge phenomena arising on a cable line as shown in Figure 15.35a, where cable 1 may be installed within the power station yard or may be part of the outside network. When the breaker is closed, a transient switching surge appears on the adjacent cable line, and the transmitted voltage appearing at point m is {2Z1/(Z0 – Z1)} e0 (e0 is the voltage [crest value] across the breaker contacts at t = 0–). As the station bus probably has a number of parallel circuits, Z0 (the resulting total surge impedance of the station side) is small; accordingly, the voltage vm at point m becomes almost 2e0. Furthermore, a large, long duration for the transient surge is anticipated for larger Z2 (surge impedance of another connected line). Figure 15.35b shows the calculated result for vn at point n under the conditions Z0 = 0, Z1 = 30 Ω, Z2 = 400 Ω. The calculation indicates that an oscillatory switching surge of maximum 1.86 times the nominal voltage (crest value) is induced on the cable line. The voltage level would become more severe if terminal n were opened. The transmitted wave voltage at terminal point m is vm = 2e under the condition Z0 = 0. Actual switching surges appearing on the cable may be more complicated and probably include multifrequency components because the surge-impedance circuits of the station/outer lines are generally also complicated, although the mechanism illustrated by this investigation is unchanged. Of course, we need to confirm in practical engineering that the insulation of the cables (insulation layer and outer covering) is protected against switching surges. The cable should have an insulation level to withstand the maximum switching surge level specified by authorized standards or recommendations, as shown in Table 9.1a.

Z1 ·e Z0 + Z1

e

𝜌2

vm

o

l

Z0 = 0 W

𝜌′1

𝜇′1

𝜇2

n Cable line #1 Z1 = 30 Ω

e

𝜌2 = (400 – 30) / (400 + 30) = +0.860 𝜇2 = 2 × 400/(400 + 30) = +1.86

vn

m

𝜌′1 = (0 – 30) / (0 + 30) = –1

Surge impedance looking from the switch

Overhead line #2 Z2 = 400 Ω

(a) m

𝜈n

0

Voltage 2e

T

1.86e

1.64e

1.47e

2T

1.35e

3T

e 0.45e

0.26e 1T

4T

0.59e

5T

10T

(b)

5T time

6T

T: unit time of cable line #1

7T

T = l1/u1

8T 9T

n

𝜌2 = 0.86

𝜇′1 = 0

𝜇2 = 1.86

𝜌′1 = –1 1 1 0.86

–0.86 2 – 0.86 0.86 2 3 0.86 –0.86 3 4 –0.86 0.86 4

1.86

0.86 –0.86 – 0.86 2 0.86 3

0.86 –0.86 3

5

Time

Typical waveform of switching surge (10 kHz to 1 MHz) calculation by Eq. 18 · 51:

Figure 15.35 Switching surge on the cable and overhead line.

6

0.86 · 1.8

6

–0.86 · 1.8 6 4 6 –0.8 4 0.86 0.86 · 1.8 6 0.86

vn = (1+𝜌2) · {1+(𝜌1′𝜌2) + (𝜌1′ 𝜌2)2 + (𝜌1′ 𝜌2)3+ (𝜌1′ 𝜌2)4 + · · · } · e

–0.86 · 1.8

2

15.17 Surge Voltages Caused on Cables and GIS Jointed Points

In particular, in the case of a long cable line with a cross-bonding jointing connection, the insulation of the outer covering should be carefully investigated, because the metallic sheath terminals are earth grounded only every three jointed spans.

15.17

Surge Voltages Caused on Cables and GIS Jointed Points

A gas-insulated substation (GIS) contains all the functional equipment (bus conductors, breakers, LSs, arrester, PT, CT, bushings, etc.) in a hermetically sealed metal container filled with SF6 gas at a few atmospheres pressure (typically 4 atm) as an insulating material (see Figure 17.9). GISs are of extremely compact design compared to conventional open-air substations; and, furthermore, a number of (phase-unbalanced) feeders are connected through the bus conductors. In other words, the GIS is a complicated surge-impedance circuit containing various transition points in a narrow threedimensional space. Therefore, the behavior of switching surges or transferred lightning surges in GIS containers includes high-frequency components (even at 10 MHz) caused by short travel distances between the transition points. This is why very fast-front voltages have been designated in the standards for insulation coordination (see Section 17.2.5). Switching surge voltages with very fast fronts caused by a breaker in the GIS (time of wavefront, Tf = 0.1–1 μs) may cause particular problems to the GIS-connected cable lines. A typical phenomenon is indicated in the system in Figure 15.36a. Referring to Figure 15.36b, the incidental switching surge voltage e from the GIS is divided by the various surge impedances at the cable jointing point, where abnormal surge overvoltages eg1 and eg2 appear. The voltage eg1 may damage the outer covering, or the voltage (eg1 – eg2) may damage the longitudinal insulation cylinders in the jointing box. These voltages (the first wave before reflection) can be calculated as follows. GIS Cable line e Zc

ZGIS

g1 Zg1

g2

Zg2

(a) Junction point of cable and GIS 2e The surge impedances Zc : of the cable across the cable conductor and the metallic-sheath

ec ZGIS

Zc g1 Za

eg1 eg2 Z ′c

Zg1

Z ′c : of the cable across the metallic-sheath and the grounded earth

g2

Zg2

Zb Z ′GIS

ZGIS : of the GIS across the conductor and the earth terminal of the tank Z ′GIS : of the GIS across the tank and the earth Zg1, Zg2 : of the earth-wire across the earth-terminals to earth Za = (Z ′c //Zg1)

Zb = (Z ′GIS //Zg2)

(b) Equivalent circuit Figure 15.36 Surge voltages arising on the cable sheath by the switching operation of GIS.

459

460

15 Lightning and Switching Surge Phenomena and Breaker Switching

The surge voltages across the cable conductor to the sheath are ec =

Zc 2e Zc + Za + ZGIS + Zb

and across the cable sheath are eg1 =

Za 2e Zc + Za + ZGIS + Zb

and across the GIS conductor to the grounding terminal are eGIS =

ZGIS 2e Zc + Za + ZGIS + Zb

and across the GIS grounding terminal to earth are eg2 =

− Zb 2e Zc + Za + ZGIS + Zb

15 118

and across the insulated cylinder of the cable terminal box are eg1 −g2 =

Za + Zb 2e Zc + Za + ZGIS + Zb

where Za =

Zc Zg1 surge impedance across the sheath terminal to earth Zc + Zg1

Zb =

ZGIS Zg2 surge impedance across the GIS ground terminal to earth ZGIS + Zg2

Detailed computer analysis of such phenomena is possible by the lattice method if necessary. However, manual calculation of the duration of the first wave using these equations gives us a general concept of the behavior, because the attenuation of reflected waves is quite fast for such high-frequency phenomena, so the first wave contains dominant components at least for the absolute values. Finally, detailed analysis of surge phenomena in practical engineering must be conducted using specialized surgeanalyzing software (typically, Electromagnetic Transient Program [EMTP]), where a surge-impedance map including detailed switch models is prepared. However, surge phenomena can be pre-estimated with manual calculations as shown in this chapter.

Supplement 1 General Solution Eq. (15.10) for Differential Eq. (15.9) We introduce new variables v’(x, t) for convenience as follows: v x,t = e −αt v x,t

1

Differentiating Eq. (1) twice with respect to x, 2 ∂ 2 v x,t −αt ∂ v x,t = e ∂x2 ∂x2

2

Partial differentiation of Eq. (1) twice with respect to t gives ∂v x, t ∂v x, t = − αe −αt v x, t + e −αt ∂t ∂t =e

− αt

∂v x, t − αv x,t + ∂t

3

15.17 Surge Voltages Caused on Cables and GIS Jointed Points

∂2 v x, t ∂v x, t = − αe − αt − αv x,t + ∂t 2 ∂t = e −αt

+ e −αt − α

∂v x, t ∂ 2 v x,t + ∂t ∂t 2

∂v x,t ∂ 2 v x,t α2 v x,t −2α + ∂t ∂t 2

4

Substituting Eqs. (1), (3), and (4) into the right side of Eq. (15.9) ①, The right side of Eq 15 9 ① = LCe − αt

∂ 2 v x, t ∂t 2

5

On the other hand, Eq. (2) is the same as the left side of Eq. (15.9) ①. Accordingly, Eq. (15.9) ① for v(x, t) is replaced by the following equation for v’(x, t): ∂2 v x,t ∂2 v x,t = LC 2 ∂x ∂t 2

6

This equation is in the same form as Eq. (15.5), so the general solution is in the form of Eq. (15.6): v x,t = v1 x− ut + v2 x + ut

7

Therefore, the general solution of Eq. (15.9) ① is v x,t = e −αt v1 x−ut + v2 x + ut

8

The equation for the current can be derived analogously.

Supplement 2 Derivation of Eq. (15.19) from Eq. (15.18) Utilizing Eq. (15.18) ① ②, we derive ① ± ② Z0(s). Then V x, s + Z0 s I x,s = e −γ V x, s − Z0 s I x, s = e

sx

γ sx

V 0, s + Z0 s I 0,s

V 0, s −Z0 s I 0,s

1 2

Transferring the exponential terms of both equations from the right to the left, and adding both equations, gives Eq. (15.19).

461

Supplement 3 Calculating the Coefficients k1 – k4 of Eq. (15.59) Coefficient k1 can be calculated by using s = −jω in the equation F(s) = Chapter 10: k1 = F s

s + jω

s = −jω

=

− jω + 2α − j2ω − jω + α + ju −jω + α − ju

↙

− jω + 2α

j = 2ω

≒

u2 − ω2 + α2 − j2ωα ↖

(s + jω), as explained in Section 10.5 of

↖

j −jω 1 = 2 2ω u2 2u

↖

In the same way k2 = F s

s− jω

s = jω

=

jω + 2α↙

−j 2ω

u2 − ω2 + α2 + j2ωα ↖

s + α + ju

k3 = F s

s = − α−ju

s + α− ju

k4 = F s

≒

=

s = −α + ju

↙

−u2 +

≒

j −ju 1 ≒− 2 2u −u2 2u

ω 2 + α2 ↖

+

↖

≒

−j ju 1 ≒− 2 2u −u2 2u

j2uα ↖

↙

−j 2u

ju + α −u2 +

1

↖

− ju + α

j 2u

=

↖

j −jω 1 = 2 2ω u2 2u

ω 2 + α2 ↖

− j 2uα ↖

↖

The arrows ↖ are the “arrows of omission” indicating parts that are ignored for reasons of simplicity.

Supplement 4 Calculating the Coefficients k1 – k6 of Eq. (15.70) Here, k1 = F s

s + jω

s = −jω

−jω C1 −ω↙ + u21 + C1 −ω↙ + u22

=

2

− jω + α1

C1 C1

=

↖

↖

2

+

−jω + α2

u21

↖

C1 u21 + C1 u22 1 1 1 = + = L1 + L1 2 2 2 2 2C1 C1 u1 u2 2C1 u1 2 C1 u2 2

↖

+ u22

1

In the same way, 1 L 1 + L 1 = k1 2 k3 = F s s + α1 + ju1

k2 =

=

α↙ 1

−

s = − a1 + ju1

+ ju1 C1 α1 + ju1

C1 C1 =−

2

↖

α↙ 1 + ju1

2

2

−j2u1

+ ω2 ↖

α↙ 1 + ju1

2

+ u22

− α1 − ju1 + α2

2

+ u22

+ u21 + C1 ↖

1 L1 =− 2 2 2C1 u1

k4 = F s

s + α1 − ju1

k5 = k6 = −

↖

3 s = − α1 −ju1

=−

L1 2

These are the results given in Eq. (15.71).

1 L1 =− 2 2C1 u12

4 5

463

16 Overvoltage Phenomena The electrical insulation withstand levels of individual pieces of equipment, and power-system insulation coordination, are essential engineering subjects. However, before we discuss these topics, we need to examine all types of overvoltage phenomena, including steady-state and temporary overvoltages that occur under nominal or temporary conditions, as well as overvoltages due to lightning or switching. This is the object of the chapter.

16.1

Neutral-Grounding Methods

Neutral-grounding methods can be classified as effective neutral grounding (or solidly neutral grounding) and noneffective neutral grounding. The difference between the two practices is the difference in the zero-sequence circuit from the viewpoint of power network theory. Therefore, all power-system behavior characterized by the neutral-grounding method can be explained as phenomena related to the characteristics of the zero-sequence circuit. Neutral-grounding methods have a wide effect on the practices of various engineering fields, such as planning or operational engineering of short-circuit capacity, insulation coordination, surge protection, structure of transmission lines and towers, transformer insulation, breaker capability, protective relaying, noise interference, etc. In this section, some typical features of neutral-grounding methods are presented and explained. The neutral-grounding methods of power systems can be classified as follows: a) Effective neutral-grounded system: Solidly grounded system

• •• •

b) Non-effective neutral-grounded system: Resistive neutral-grounded system Arc-suppression coil (Peterson coil) neutral-grounded system Neutral ungrounded system (also called a neutral minute-grounded system); only used for distribution systems. Table 16.1 explains these neutral-grounding methods in detail. The features of each method are based on the zero-sequence circuit. By using a plain expression for the non-effective grounded system, grounding fault currents can be reduced considerably, but on the other hand, higher temporary overvoltages (TOV) are caused during faults. The effective neutral-grounded systems (solidly grounded system) has the opposite features. Table 16.2 presents typical features of the two grounding methods. AC power transmission systems were first constructed in around 1900 with lower voltages of 6–30 kV; today’s power systems are the result of continuous network growth since then. Individual power systems have their own history, which has led to the applied practices of the neutral-grounding method and today’s applied power-frequency and nominal voltages. Therefore, the applied neutral-grounding methods differ somewhat, in particular for older lower-voltage classes. Younger EHV (say, over 200 kV) and UHV (say, 500 kV or higher) trunk-line networks have been unified by solidly grounded systems all over the world, mainly to realize EHV/UHV networks with reduced insulation levels. However, this does not mean the high-impedance neutral-grounded method is inferior to the solidly grounded method, in particular for lower-operational voltage systems.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

464

16 Overvoltage Phenomena

Table 16.1 Various neutral-grounded systems. A

Solidly neutral-grounded system (effectively neutral-grounded system)

All the transformers installed at substations belonging to the same rated voltage are solidly neutral-grounded.

B

Resistive neutral-grounded system (impedance neutral-grounded system)

All or some selected key transformers installed at substations belonging to the same rated voltage section are neutral-grounded through a neutral grounding resistor (NGR). The resistive value (Ω) of NGR is determined so the grounding current through the NGR in a one-phase-to-ground fault is limited to 100 A or within 1000 A.

RN C

Arc-suppression coil neutral-grounded system (resonant neutral-grounded system)

XPC

D

R

Co

Resistive neutral-grounded system with neutral compensation reactor

R E

R′N

Some key transformers are neutral grounded through tap-changeable reactors (inductance Lpc), whose taps are selectively controlled so the inductive reactances ( j2πf Lpc) are well tuned with the capacitive reactances (−j/2πf C0) of transmission lines over time. The zero-sequence circuit is kept under parallel quasi-resonant conditions, and the zero-sequence impedance f Z0 of the systems have large values; therefore, effective arc extinction can be expected during one-phase-to-ground lightning faults: 1 1 = f Z0 = 1 1 + jωCs j − + ωC0 j3ωLPC 3ωLPC where 1 ωLPC ≒ 3ωC0 ∴ f Z0 ∞ f I 0 ≒ 0 This is essentially the same system as B, except the neutral compensation reactors are equipped to compensate for stray capacitances C0 of the transmission line, in particular for long transmission lines or cable lines.

Cs

jXN

Neutral ungrounded (isolated neutral ungrounded) systems

R

This practice is typically used only for distribution networks. In this system, intentional neutral-grounding connections do not exist, except through potential-indicating or measuring devices or other very high-impedance devices. The grounding current caused by one-phase-to-ground faults is limited to values of 10 mA to 1 A by a large neutral impedance fZ0 (the order of a few thousand ohms or more; f Z0 ≒ ∞ from an analytical viewpoint).

In contrast, non-effective (high-impedance) neutral-grounding methods are still widely used for lower-voltage lines and distribution networks in several countries, for reasons having to do with tradition and engineering. The latter are summarized as follows:

• • •

The largest feature of the non-effective neutral-grounding method is that the continuous/temporary earth-ground flowing current (3I0) is considerably reduced under normal or fault conditions. Its greatest advantages for distribution systems concern human security and suppression of noise interference. These are important matters, especially in residential areas covered by distribution networks. Reducing the system insulation level or cost by using the solidly grounded method cannot be expected in lowervoltage or distribution networks. For example, for the lowest-voltage network of 100–400 V, the insulation level cannot be reduced by adopting a solidly neutral-grounded method instead of the high-resistive neutral-grounded method, because toughness – including mechanical strength, thermal strength, and durability – and human safety are priorities for these voltage classes, rather than lower insulation levels. Similar situations are seen in distribution systems of 30 kV or lower. Changing the neutral-grounding method of existing networks is almost impossible, because major modifications to existing engineering practices, including basic insulation coordination of the power system, would be required. For example, in a substation, the basic design of the substation’s earth grounding (grounding varied mats, counterpoise, etc.) would have to be revised. Most arresters, breakers, protective relays, and other substation equipment would have to be replaced, and so on.

Table 16.2 Comparison of neutral-grounding methods.

(1) Temporary overvoltages (TOV) of the sound phases during LG fault (1ϕG)

Solidly neutral-grounding method (S-ngm)

High-resistive (impedance) neutral-grounding method (R-ngm)

(1S) ○ The TOV Vb, Vc is 0.8–1.3 E. The calculation check from Eq. (7.10) is j 3 af Z0 − f Z2 f Vb = f Ea . f Z0 + f Z1 + f Z2

(1R) ▼ The TOV Vb, Vc is 1.5–1.9 E. The calculation check from Eq. (7.10) is f Z2 j 3 a− j 3 af Z0 − f Z2 f Z0 f Vb = f Ea = f Ea f Z1 f Z2 f Z0 + f Z1 + f Z2 1+ + f Z0 f Z0 If the following equation is assumed: f Z0 f Z1 , f Z2

If the following typical impedance is substituted,f Z1 = f Z2 = jX 1 ≒ j namely jX0 ≒ j3X1, then f Vb ≒

j 3 3a− 1 3+1+1

f Ea

X0 f X0 =j , 3 3

= 1 25 f Ea

∴

f Vb

≒ j 3a f Ea = 3 f Ea

(2) The headway from simple faults to double or triple faults (multiphases, or multipoint faults)

(2S) ♦ The TOV of the sound phase under a one-phase-to-ground fault is approximately 1.25E, so the probability that the sound phases are forced to be sequentially faulted is relatively low.

(2R) ▼ The TOV of the sound phase under a one-phase-to-ground fault is approximately 3E, so the probability that the sound phases are forced to be sequentially faulted is relatively high.

(3) Required insulation level of station equipment and transmission lines

(3S) ○ The AC insulation level can be reduced. This is an essential advantage of S-ngm, particularly for higher-voltage systems, typically for EHV, UHV.

(3R) ▼ Relatively higher insulation levels are required for R-gnm, because TOV is higher. However, this is not necessarily disadvantageous for lower-voltage systems (especially distribution systems), because insulation levels are not the most sensitive factor for cost reduction (see Chapter 17).

(4) Arresters

(4S) ♦ Arresters with relatively low Ur ratings (PU values of the duty-cycle voltage rating) for lower maximum continuous operating voltage (MCOV) ratings are used because insulation levels are relatively low (see Chapter 17).

(4R) ▼ Arresters with relatively high Ur ratings for MCOV are used because insulation levels are relatively high.

(5) Transformers

(5S) ○ Stepping-reduced insulation design of the windings can be used. Autotransformers can be used.

(5R) ▼ Full insulation design of the windings is required. Autotransformers cannot be applied.

(6) The magnitudes of fault current passing through the earth–ground and OGW (3f10)

(6S) ▼ The earth-grounding currents during the 1ϕG or 2ϕG faults are large (say, 5–50 kA). The earth resistance R of substations must be designed to fall in order to reduce V = Ig ∙ Rd. The sheath current as well as the main-conductor current cause a temperature increase in the power cable. The calculation check for the system in the figure is 3ϕG.

(6R) ○ The earth-grounding currents during 1ϕG and 2ϕG faults are limited to within 100–400 A. The calculation check for the system in the figure, assuming a 154 kV system with 200 A NGR, is 154 3 3 10 = 445 Ω RNGR = 200 = j Z X + 3R f 0 f 0 NGR f Z1 = jf X1 ≒ f Z2 = jf X2 1ϕG f Ea = 200 A RNGR The grounding-current capacity in the stations can be reduced. Communication interference can be limited.

3f I0 = f Ia = 3f Em Δ ≒ 3 ×

f Ea

f Z0

≒

(Continued)

Table 16.2 (Continued) Solidly neutral-grounding method (S-ngm)

= f Ib = f Ic =

f Ia

275 3 = 15 9 kA j10

3f I0 = 0

High-resistive (impedance) neutral-grounding method (R-ngm) f Z1

= f Z2 = j10(Ω) f Z0

= j4(Ω) 275 kV

1ϕG 3f I0 = f Ia =

3 275 3 = 19 9 kA j 10 + 10 + 4

2ϕG f I1

=

275 3 = −j12 3 kA j10 + j10 j4

3f I 0 = 3 × f Ib

=

− j10 −j12 3 = 26 3 kA j10 + j4

a2 −a j4 + a2 −1 j10 −j12 3 j4 + j10

= 19 0 kA For 1ϕG, 2ϕG, the phase fault current is larger than that for 3ϕG and the earth current 3fI0 is larger than the phase fault current. (7) Continuous zero-sequence current through the ground

(7S) ▼ Continuous zero-sequence current flows through the ground via the transformer neutral point due to phase imbalance in the transmission lines. Continuous sheath current flows through cable sheaths due to the phaseunbalanced allocation.

(7R) ♦ The continuous zero-sequence current is zero. This is a very important advantage of R-ngm.

(8) Radio noise or communication interference

(8S) ▼ Special countermeasures are required against disturbances caused by zero-sequence flow-through current (including third, sixth, …, 3nth harmonics, especially in highly populated areas

(8R) ○ Serious problems do not arise, because zero-sequence flow-through current is limited.

(9) High-speed protection against LG faults

(9S) ○ High-speed selective protection schemes using differential relays and directional distance relays can be used to detect all fault modes.

(9R) ▼ High-speed selective protection for LG fault detection is relatively difficult due to limited grounding fault current I0. In particular, directional distance relays and differential relays cannot be used to detect LG faults in transmission lines.

(10) Power flow limit (P–δ curve characteristics) during 1ϕG or 2ϕG faults

(10R) ♦ (10S) ▼ Referring to Figure 12.7, Table 12.1, Eq. (12.33). and Table 7.1, the parallel-inserted reactance jxf = Z2 + Z0 is larger in R-ngm than S-ngm. Accordingly, the P–δ curve for the case stability limit during the fault is lower in S-ngm than R-ngm.

(11) Power flow limit (P–δ curve characteristics) during phaseopening modes

(11S) ♦ (11R) ▼ Referring to Figure 12.7, Table 12.1, Eq. (12.33), and Table 7.2, the P–δ curve for the stability limit during the one-phase opening mode (no-voltage time of singlephase reclosing) is lower in R-ngm. Furthermore, the power-flow limit during the two-phase opening mode is zero in R-ngm.

(12) Breaker’s required breaking capacity (kA)

(12S) ▼ The fault current for 1ϕG is larger than that for 3ϕG, as is demonstrated in (S6).

(12R) ♦ The fault current for 3ϕG is larger than that for any other fault modes.

(13) Voltage-resonance phenomena

(13S) ▼ Refer to Section 16.5. Resonance phenomena should be studied carefully, although they seldom occur.

(13R) ♦ Unrealistic.

○: Very advantageous; ♦: Advantageous;

▼:

Disadvantageous.

16.3 Overvoltages Caused by a Line-to-Ground Fault

16.2

Arc-Suppression Coil (Petersen Coil) Neutral-Grounded Method

The principle of the arc-suppression coil (Petersen coil, PC coil) neutral-grounded method is shown in Table 16.1C. The transmission line has stray capacitances C1, C2, C0, so the zero-sequence circuit is a parallel circuit with the impedance at neutral ground Zpc and zero-sequence line capacitance C0. Recall Eq. (7.10) and the equivalent circuit in Figure 7.2 of the fault 1ϕG. If the zero-sequence impedance of the neutral point Zpc is tuned with –jXco = 1/jωC, this means f Z0 ∞, ∞ and f Ia = 3I0 0 in Eq. (7.10), so we can expect easy extinction of the grounding current whenever the 1ϕG fZtotal fault occurs. This practice was adopted in Germany around 1918 and spread to several countries as a good example of resistive neutral-grounding or solidly grounding. However, the practice had some weak points:

• •• •

Tuning Zpc to –jXco in the zero-sequence circuit is easy for smaller power systems with radial feeder connections. However, it is not as easy for large power systems that include several substations that must be neutral grounded and/or for loop-connected power systems. High-speed detection of the 1ϕG fault by protective relay is not necessarily easy, because f Ia = 3f I0 0. The practice is useless against double-phase faults. If a transformer with a suppression coil is tripped for any reason, the best case is that the tuning condition of the system is broken, and the worst case is that the power-system loses its neutral grounding point and suffers unstable overvoltages.

Consequently, today, most power systems that used a PC coil have been switched to a resistive-grounded system or solidly neutral-grounded system. The PC coil neutral-grounding method has become a historical artifact that is seen only in smaller power systems with primarily radial connections.

16.3

Overvoltages Caused by a Line-to-Ground Fault

Power systems typically operate with a voltage within ± 5–10% of the rated voltage. Such systems are often exposed to TOVs and lightning and switching surges. Understanding these overvoltage is the basis of insulation coordination. TOVs with commercial frequency (including low-frequency components of 1000 Hz or lower) occur for various reasons. However, it is well known that overvoltages caused on unfaulted phases whenever a line-to-ground (LG) fault occurs at a transmission line must be the highest among all kinds of low-frequency overvoltage phenomena in most cases. So we need to examine this type of overvoltage phenomena in detail. If one phase-to-ground fault (LG fault 1ϕG) occurs at a phase-a conductor, the power-frequency voltages on the unfaulted phase-b and -c Vb, Vc are given by Eq. (7.10). Accordingly, the phase-c power-frequency voltage Vc becomes the value of the following equation during the phase-a fault: a− 1 f Z0 + a −a Vc k= = E f a f Z0 + 2f Z1

2

f Z1

a −1 =

f Z0 f Z1 f Z0 f Z1

+ a− a2 +2

δ + jv +j 3 σ+j = δ + jv +2 σ+j where f Z0 = f R0 + jf X0 , f Z1 = f R1 + jf X1 − a2 j 3

δ= f Z1

f R0 f X1

, v=

f X0 f X1

, σ=

f R1 f X1

,

f Z0 f Z1

16 1

=

δ + jv σ+j

= f Z2

In this equation, voltage Vc is expressed as a ratio of normal line-to neutral voltage (operating voltage) f Ea. The constant k = Vc/f Ea is the ratio of TOV with power frequency caused on the unfaulted phase-c conductor during the phase-a to ground fault. The absolute values of k for unfaulted phase voltages Vb, Vc are the same.

467

468

16 Overvoltage Phenomena

δ = R0 / X1 = 1

2.5 0

δ=

Where f R1 = 0 is assumed

7

f R0

2.0

=0

f X1

50

6

3 5

5

k

1.5

1

4 1

R 0 / X1 = 0

3

1.0

5

5 10

2 8

8

3 1 1 0

0.5

–10

–8

–6

–4

–2

0

2

4

6 v=

(a)

8

10

f X0 f X1

0 –10

(b)

–5

0 v = X0 /X1

5

10

Figure 16.1 Overvoltage ratio k of unfaulted sound phases during a line-to-ground (LG) fault.

Eq. (16.1) for the ratio can be expressed as curves with parameters δ, v, and σ on a coordinated graph. Figure 16.1a is a typical example under the parametric conditions δ = f R0/f X1 = 0 + ∞, v = −10 to +10, whereσ = f R1/f X1 ≒ 0 is assumed. Figure 16.1b is the local detail of the same curve. The term v = f X0/f X1 should have a positive value of 0–4, so the zone v < 0 is of course unrealistic under the practical conditions of a power system. The condition δ ≒ 0 to + 1 corresponds to a solidly grounded system and δ ≒ 5 to + ∞ to a non-effective neutral-grounded system. Figure 16.1 indicates the following.

• •

For the non-effective neutral-grounded system (δ ≒ 5 to +∞, v = f X0/f X1 = 0 to +4), whenever the phase-a LG fault occurs, the unfaulted (sound) phase voltages Vb, Vc increase, and the temporary phase voltages become approximately k = 3 times nominal voltages. For the solidly neutral-grounded system, (δ ≒ 0 to +1, v = f X0/f X1 = 0 to + 4), assuming δ ≒ 1.0, and the TOV caused by the same fault would increase by k = 1.2–1.3 times; furthermore, k is around 1.0–0.8 for the range 0 < δ < 1.

The TOV caused by this mode (1ϕG overvoltage) is one of the essential factors affecting insulation coordination for individual power-system networks. The required insulation level against continuous/temporary power-frequency overvoltages in individual power systems is significantly affected by the overvoltage coefficient k. This will be discussed later. Incidentally, the overvoltage caused by a double LG fault (2ϕG) is generally lower than that caused by 1ϕG. However, the overvoltage ratio for 2ϕG and 1ϕG should be investigated the same way. In contrast, three-phase faults (3ϕS, 3ϕG) and line-line faults (2ϕS) are of no interest. In conclusion, overvoltage for unfaulted phases is 0.8–1.3 E for solidly neutral grounding systems (δ = 0–1) and 1.5–1.9 E for non-effective neutral grounding systems (δ ≒ 5–∞). Figure 16.1 indicates the overvoltage ratio of the power-frequency term, and the transient overvoltage ratio is a little higher. (The transient overvoltage ratio can be derived using a curve format similar to that in Figure 16.1.) The TOV phenomenon and the overvoltage coefficient k are important factors for fundamental design based on insulation coordination, which we will discuss later.

16.4 Other Low-Frequency Overvoltage Phenomena (Non-resonant Phenomena)

Table 16.3 Overvoltage phenomena. 1) Fundamental (power) frequency overvoltages (temporary overvoltages [TOV]) 1a. Ferranti effect 1b. Self-excitation of generator 1c. Overvoltages of unfaulted phases by one LG fault 1d. Sudden load tripping or load failure 2) Lower-frequency harmonic resonant overvoltages 2a. Broad area resonant overvoltages (lower-order frequency resonance) 2b. Local area resonant overvoltages 3) Switching surge caused by breaker or LS switching 3a. Breaker-closing overvoltages 3b. Breaker-tripping overvoltages 3c. Switching surges by line switches 4) Lightning surge 4a. Direct stroke to phase conductors (direct flashover) 4b. Direct stroke to overhead grounding wire or tower structure (inverse flashover) 4c. Induced stroke (electrostatic-induced stroke, electromagnetic-induced stroke) 5) Overvoltage caused by abnormal conditions 5a. Interrupted ground fault of cable (in high-impedance neutral-grounding system) 5b. Overvoltage induced on cable sheath (see Section 9.3, Section 16.5 and Section 16.6) 5c. Touching of different kilovolt lines, etc.

16.4

Other Low-Frequency Overvoltage Phenomena (Non-resonant Phenomena)

In addition to LG fault overvoltages, other types of overvoltage phenomena are caused by various mechanisms. Table 16.3 lists our classification of overvoltages, and we will examine several of them. The phenomena listed in item 2a, for example, is not often discussed, because they seldom occur. However, engineers need to study the mechanism and ensure the absence of this sort of problem. Phenomena for four different power-frequency overvoltages are examined next.

16.4.1

Ferranti Effect

For the power system in Figure 16.2a, the vector diagrams for lagging and leading power factors are shown in Figures 16.2b and c. During leading power-factor operation, the receiving terminal voltage vr becomes larger than the sending terminal voltage vs (i.e. vs < vr). The phenomenon in which vr becomes larger than vs is called the Ferranti effect. We discussed much of this with regard to leading power-factor load operation in Chapter 11, Section 11.7.

16.4.2

Overvoltage Due to Transmission Line Charging

The ultimate case of pure capacitive load is transmission line charging from one terminal; the vector diagram is shown in Figure 16.2d. We will calculate the overvoltage when the line is charged from the sending terminal point s (the receiving terminal r is open). The four-terminal circuit equation between points s and r is given by Eq. (15.19b). Now, with Ir = 0, we have Vr x 1 = ratio of overvoltages at the receiving terminal cosh γ s x Vs x where γ s =

Ls + R Cs + G

x is the distance from point s to r

16 2

469

470

16 Overvoltage Phenomena

e

jx

jxd vs

Load

vr

i

(a) Power system e jxd ·i

vs φr

i

i e

jxd ·i

vs

jx·i

vr

jx·i vr (c) Load with leading power factor (–1.0 < cosϕ < 0)

(b) Load with lagging power factor (0 < cosϕ < 1.0) i

jxd ·i

e

jx·i vr

vs

(d) Pure capacitive load (cosϕ = 0) Figure 16.2 Vector diagrams under different power-factor loads.

Ignoring R and G, and letting s

jω (because fundamental frequency phenomena are investigated here),

coshγ s x = cosh jω LC x = = cos ωx

1 jω e 2

LCx

+ e − jω

LCx

LC

16 3

Vr 1 then, ∴ = ≧1 0 Vs cos ωx LC The equation is the rate of overvoltage caused at point r in comparison with the voltage at point s. Figure 16.3 shows the calculated result of the equation under the condition of typical line constants L = 1 mH/km, C = 0.01 μF/km, 50 Hz. The rate of overvoltage at point r is 1.05 for line length l = 312 km and 1.10 for l = 432 km. In the case of n parallel circuits, the constants are replaced with L 1/n L, C n C, so LC and the rate Vr/Vs are not affected. Overvoltages caused by the Ferranti effect under relatively light load conditions can be severe, especially at a receiving substation in a metropolitan area where many cable feeders are connected. An effective countermeasure to prevent excess overvoltage is to install reactor banks at the receiving stations; this is a normal countermeasure for substations in heavy-load areas of large power systems. V–Q control equipment based on the control of reactor banks and on-load tap-changing transformers is often used at key substations as another effective countermeasure to prevent excess overvoltages or to maintain stable voltages around the clock. Vr

Vs / Vr

Vs

=

1 cos(2π f

LC )

where L = 1 mH/km, C = 0.01 μF/km, f = 50 Hz

1.1 1.05 1.0 0

100

200

300

400

500 km

Figure 16.3 Ferranti effect, ratio of voltage at receiving end.

16.4 Other Low-Frequency Overvoltage Phenomena (Non-resonant Phenomena)

eG (a)

e

i

xd

C

v eG

Load characteristics v = Small C 1

Large C 2

a

i j𝜔C

Generator’s v–i characteristics by changing C under zero excitation and rated speed

1.0 (b) jEf

i i Armature leading current

By residual flux

jxd·i eG (c)

Induced voltage by residual magnetism Figure 16.4 Self-excitation of a generator.

16.4.3

Self-Excitation of Generators

If a capacitive load is connected to a generator as shown in Figure 16.4a, overvoltage will be caused on the generator terminal even if Efd is kept at zero. This phenomenon is called the self-excitation phenomenon of a generator. We will refer to Figure 16.2 to discuss this phenomenon. Under leading power-factor operation, a large terminal voltage eG is apt to arise in spite of the small excitation jEfd, as seen in Figure 16.2c. Now we will look at Figure 16.4b and examine the behavior of a generator under the leading power-factor condition. The generator is driven by the prime mover under the open-terminal condition and is running at constant speed with excitation current zero (jxad ifd = Ef = 0, in Eq. (10.60) and Figure 10.7). However, a small amount of residual magnetism remains as a magnetic field in the rotor since termination of the last operation, even if the excitation current is zero. A small terminal voltage eG also remains. If a very small capacitance C is connected to the generator under this condition, the same terminal voltage eG will be maintained. Furthermore, if the connected capacitance C increases, eG and the leading current i in the armature coil will increase; their characteristics can be written as curve a in Figure 16.4b. This is a specific v-i characteristic of a generator under capacitive load with excitation current zero. As shown in the vector diagram in Figure 16.4c, current i leads the voltage by 90 . Referring to curve (a) in Figure 16.4b, the generator characteristics are gradually saturated at around the rated voltage range. On the other hand, specific v–i characteristics of the capacitance C are given by v = i/(ωC), as shown in Figure 16.4b. The intersecting point of both characteristics is the operating point of the C load. In conclusion, the self-excitation phenomenon of a generator is caused by connection of a pure capacitive load, which causes extreme, sustained overvoltage (possibly the ceiling voltages of the saturated zone). This condition also means critical overheating of the generator caused by the rapid increase in iron loss as a result of core saturation. Obviously, we need to avoid the generator self-excitation. 16.4.4

Sudden Load Tripping or Load Failure

If load tripping or load failure occurs at a receiving substation, the voltage is boosted for the following three reasons:

•

The generator terminal voltage at the sending station is transiently boosted, so the voltage at the receiving point also must be boosted. Referring to Figure 10.7 or Eq. (10.63), if the generator current ia1 suddenly decreases, the terminal voltage vector ea1 = Ea1 e jα1 must change to a large vector closer to jEf in magnitude and angle. In other words, the

471

472

16 Overvoltage Phenomena

••

generator terminal voltage is transiently boosted by the sudden decrease of load current, until the AVR decreases the excitation jEf. The impedance drop voltage of the transmission line is decreased. The total power factor of the load is shifted a little in the leading power-factor direction, so the Ferranti effect only partially occurs.

16.5

Lower-Frequency Resonant Overvoltages

If one inductance L and one capacitance C exist as elements of a circuit, series resonant frequency and parallel resonant frequency exist. Our concern from the viewpoint of overvoltages and waveform distortion is series resonant phenomena of nth order resonant voltages and currents in the power system, whose condition is given by Z = j2πn fL + 1/(j2πnfL 0) (then f0 = 1 2π LC is the resonant frequency). Figure 16.1 indicates that the area v=f X0/f X1 − 2 is the dangerous series resonant zone. Obviously, we need to be careful in order to prevent ν from becoming the negative value. In Figure 7.2b, which shows a single-phase-to-ground fault, and in Table 7.2, which shows the conductor opening, we can imagine that for the cases where C1, C2, and C0 exist in positive-, negative- and zero-sequence circuits, LC series resonant local loops arise in the circuits. If j f X0 or j f X1 becomes negative (capacitive value), regardless of the time interval, extreme overvoltages are caused. Although such resonance conditions seldom occur, engineers must still examine irregular conditions including unbalanced shortcircuit modes and open-conductor modes under various network connections. We know that the stray network capacitance C is increasing in today’s networks, in particular in big cities. Trunk lines and low-voltage distribution lines in metropolitan areas are based on cable lines whose line constants are one-fifth smaller L and 20 times larger C per kilometer compared to overhead lines. Moreover, the system is a meshed network with several routes and a number of parallel circuits per route to meet large-load capacity. Such networks have very “overcrowded” L and C constants. In addition to the 50/60 Hz phenomena, lower-order harmonic phenomena in particular (under, say, 1 kHz) may cause problems over a broad area and continue for a long period without attenuation. We need to ensure the absence of these sorts of problems. 16.5.1

Positive-Sequence Series Resonance

Figure 16.5a is a simplified model connecting a generating plant and urban area with cable lines. If we represent the overhead line by L and the cable lines by C, the circuit can be simply written as an LC series-connected circuit as shown in Figure 16.5b, although the load circuit is usually connected. We have already studied the Ferranti effect of these circuits as phenomena caused by power frequency. Nevertheless, we need to check the possibility of series resonance in other frequency zones under different conditions, such as nominal load condition and irregular conditions (phase fault, grounding fault, phase opening, reclosing time, etc.). At the very least, we need to prove that such resonant phenomena do not occur during system operation. We will try to find a series resonance condition that may exist by chance in our power system. Let’s assume the following line constants for the positive-sequence circuit in Figure 16.5b: Overhead line 0 01 mH km , and for l1 km

L = 10 −5 l1 H

Cable lines 0 33 μF km, and for l2 km ,

16 4

C = 0 33 × 10 −6 l2 m F

numbers of circuits m ,

Assuming a no-load condition, the condition for LC series resonance for the nth-order higher-harmonic components is j2πn f0 L −j

2πn f0 2 LC = 314n 2 LC = 1 0

Then ∴ where

1 =0 2πn f0 C

10 n LC ≒ 1 0 5 2

f0 = 50Hz

16 5a

16.5 Lower-Frequency Resonant Overvoltages

Cable lines Overhead line e

L

L

e

C (a)

v C Total capacitance of cable lines

(b)

Overhead line (km)

n = 4 harmonic order 5

500 13 10

6

100 100

500 50

100 50

Average length of cable section (km)

800 200 100 50

1 circuit (m = 1) 4 circuits 8 circuits 16 circuits

(c) Figure 16.5 Series resonance of a power system with cable lines (positive-sequence resonance).

Substituting Eq. (16.4) into (16.5a) the resonant condition of the positive-sequence circuit is n2 l1 l2 m ≒ 3 × 106

16 5b

Figure 16.5c shows the series resonant condition derived from Eq. (16.5b). This figure gives us a rough idea of the existing zone of series resonance for the model system. Assuming that the length of the overhead line l1 = 200 km and the total length of cable l2 = 400 km or less, for example, resonant conditions do not exist, fortunately, for the lower frequency n = 2–6. Although the curves were derived under a somewhat unrealistic no-load condition, a similar condition is caused by a sudden bus fault (all the connected breakers must be tripped) in the key substation. Recall that only the existing zone of resonant frequency in the system was checked in this study, from which we concluded that natural resonance would seldom be caused in the system. However, we need to recognize that power-system networks include not only generators of almost ideal sinusoidal source voltages, but also various loads and power conditioners that generate large rippled currents. If the frequency of the harmonic components of such a “dirty forced current” matches the network natural frequency, abnormal overvoltages or waveform distortion can be significantly amplified in some local areas of the network. We must recognize that harmonic currents of various frequency are apt to flow forcibly into the power system, so resonance phenomena must be checked from the viewpoint of waveform distortion and overvoltage phenomena. (Waveform distortion is examined again in Chapter 18.) 16.5.2

Series Resonance under Temporary Conditions (Faults, Phase Opening, Reclosing Time, etc.)

We will investigate the possibility of resonance under irregular conditions including line-to-line (LL) fault modes, LG fault modes, and phase opening modes (including dead-voltage time for automatic faulting phase reclosing). Referring to Tables 7.1, 7.2, 8.1, and 8.2, we need to check whether these conditions exist: that is, whether the denominators of the equations in the tables tend to values close to zero.

473

474

16 Overvoltage Phenomena

For the modes of series resonance, Normal condition

Z1

2ϕG

Z1 + Z2 Z0

2ϕS

Z1

1ϕG

Z1 + Z2 + Z0

1ϕ opening

Z1

2ϕ opening

Z1 + Z2 + Z0

0 Z2 Z2

0

Z0

Z1

Z2 Z0 Z2

Z0

0

Z1

Z2 Z0 Z2

Z0

0

0

Z0

16 6

0

Perhaps we do not need to worry about the possibility of resonance caused by LL and LG faults, because the resonance condition is unlikely to occur by analogy to Figure 16.5c. Furthermore, these faults are cleared within a few cycles, generally by the associated protective relays. However, there may be little phase-opening mode trouble (typically caused by a breaker operation failure), because the protective relays may not be able to detect the phase open-mode faults and continue without noticing. In conclusion, it is worth assuring the absence of these sorts of resonant conditions at several different points in the network.

Br1

Br2 Cable line

16.5.3 Transformer Winding Resonant Oscillation Triggered by Switching Surges

In Figure 16.6, the transformer is energized through the cable line. This circuit can be simplified to the parallel circuit with C for the cable and L for the transformer’s excitation reactance, so overvoltage is not caused when breaker Br1 is closed. However, the resulting oscillatory inrush Figure 16.6 Resonance caused by transformer voltage and current could trigger a unique voltage oscillation in the excitation. transformer windings. Let’s assume a cable line of 275 kV, 2000 sq. (given in Table 9.4: 0.392 mH/km, C = 0.25 μF/km, velocity u = 1/ LC = 1.01 × 105 sec = 101 m/μs) and length 10 km. The traveling time of the surge from breaker Br to the transformer is 100 μs, so the oscillatory frequency of the switching surge is 5 kHz. The transient oscillation of this frequency zone (10th or lower order) is not attenuated quickly. Therefore, a reasonable countermeasure is required at the transformer engineering side to avoid winding-coil resonance with a frequency of the order described. This scheme is discussed again in Section 17.11. Note that the in-rush current of the transformer is, regardless of the condition of the feeding line, an extremely waveform-distorted offset current that continues for more than 10 seconds, as shown in Section 6.9, Figure 6.15. Because an inrush current contains DC and lower-order harmonics with slow attenuation, special consideration is generally required in the practical engineering of transformers, breakers, generators, protective relays, etc. Open

16.5.4

Ferroresonance Caused by Core Saturation

Figure 16.7 illustrates the ferroresonance phenomenon. As shown in Figure 16.7c, if the operating voltage exceeds 1.1 pu of the rated voltage, it begins to saturate rapidly, and the excitation current i greatly increases. In other words, the excitation impedance Zl (= jXl) of the transformer has nonlinear characteristics that rapidly decrease in value in the saturated zone. Figures 16.7a and b show the well-known series ferroresonance circuit and parallel ferroresonance circuit, respectively. Both circuits have unique behavior (although the explanation is omitted in this book), and we need to prevent their occurrence. Parallel ferroresonance may arise particularly in a circuit with a large cable capacitance, while series ferroresonance may arise especially in irregular cases with unbalanced phase opening or lines with a series capacitor. Ferroresonance causes unstable, jumping overvoltage phenomena and extreme flux saturation, leading to a severe temperature rise during the flux pass – all of which need to be avoided.

16.7 Switching Surge Overvoltages

vl

il

C

b a

1.0 Z

vl

L

0

(a) Series ferroresonance circuit

Z

C

vl

L

ZL =

il

vl il

il

0 (b) Parallel ferroresonance circuit

(c)

Figure 16.7 Ferroresonance phenomena.

16.6

Interrupted Ground Fault of a Cable Line in a Neutral-Ungrounded System

This is a unique feature that is limited to the distribution cable lines of neutral-ungrounded system (see Table 8.1). This system is widely used for distribution networks of 3–20 kV, whose advantage is that the grounding fault current for onephase-to-ground faults (1ϕG) can be greatly reduced to a value of only, say, 10–100 mA, which is zero amps. With this system, noise interference on the communication lines is also reduced, because the zero-sequence circuit is actually open. However, this practice has one disadvantage: interrupted ground faults in the cable lines. The solid insulation of a distribution cable may slowly deteriorate over a long period of operation. We will assume that a pinhole crack appears in the cable insulation. A very small 1ϕG current flows through the pinhole across the insulation layer (called minute grounding of perhaps 10 mA). However, this grounding current is halted immediately, and the insulation of the pinhole is soon recovered, because the pinhole pass is not adversely affected by such a small current through the pinhole. The pinhole grounding will probably occur at a time around the peak value of sinusoidal AC voltage and is always halted at the time of voltage zero. Incidentally, the minute grounding fault current is almost 90 of the lagging power factor, so halting the pinhole grounding current at the time of voltage zero gives rise to current-chopping phenomena, which cause large, needle-shaped, high-frequency transient overvoltages by the same mechanism explained in Section 15.12. When this happens at an arbitrary point in the cable system, the same process is intermittently repeated at the same point over time, so deterioration of the pinhole slowly accelerates and finally leads to a permanent breakdown fault. Moreover, this interrupted ground fault at one pinhole point intermittently generates needle-like high-frequency overvoltages, which deteriorate the insulation of other parts of the cable. Finally, the entire cable line within a limited area is frequently stressed by the high-frequency overvoltages caused at many pinhole points, after which cable deterioration accelerates quickly and broadly. Due to the excellent insulation technology of recent CV cables for distribution lines, interrupted ground faults seldom occur today. However, such phenomena can occur on a neutral-ungrounded system at any time and in any location, and are initialized without any indication or notice. Continuous observation or predetection of such overvoltages on a commercially operated system is almost impossible, so such phenomena, if they occur, are generally noticed only after the cable lines have been damaged or have seriously deteriorated.

16.7

Switching Surge Overvoltages

Opening (tripping) and closing circuit breakers and line switches can cause severe switching surges. It is fair to say that such phenomena can be even more difficult to manage than lightning surges. In order to understand switching surges, we will begin our discussion by comparing them to lightning surges.

475

476

16 Overvoltage Phenomena

Lightning surges are severe in nature. However, the following characteristics may mitigate the severity to some extent, as compared to switching surges:

•• •• •• •

Rare. Occur mostly on overhead transmission lines whose structures are made of metal (conductors, tower structures, arcing horns, etc.) and porcelain insulators and that have inherent fault self-recovery (except for a direct strike on a substation). Very short surge duration for a typical waveform 1.2 × 50 μs. Single-action phenomena (mostly). Attenuation through the line traveling to substations, in most cases. Upper limit of incidental voltage level (can be controlled by arcing horns). Relatively easy protection using advanced arresters installed at substations.

The characteristics of switching surges (SSs) are the opposite of these. SSs appear whenever any one of several breakers or line switches at the same substation are operated. SSs occur in the vicinity of apparatus with no self-recovery insulation (generators, transformers, breakers, cables, etc.) in the same substation. They are oscillatory, transient, extremely high-frequency surges with sharp waveforms, and they continue for a long time before attenuation or extinction. SSs are often caused during switching. They seldom attenuate before reaching weak insulation points at the substation, because of the short traveling distance. Arresters may not be able to protect weak insulation points at the substation, due to the relative distance of installed arresters from the surge-generating breakers or line switches and other equipment to be protected. Furthermore, SSs cause thermal heating of arrester–non-linear resistive elements, and the switching energy can break installed arresters.

16.7.1

Overvoltages Caused by Breakers Closing (Breaker-Closing Surge)

We studied breaker-closing surges in Chapter 15. Figure 16.8 shows the transient phenomenon caused by the breaker closing, which means the sudden insertion of initial voltage {e1(0) – e2(0)}1(0) with opposite polarity across the breaker contacts. The resulting voltage and current traveling waves are proportional to the initial voltage {e1(0) – e2(0)}, as shown by the equations in Figure 16.8b. Successive phenomena are affected by the circuit conditions, especially the included transition points.

16.7.2

Overvoltages Caused by Breakers Tripping (Breaker-Tripping Surge)

We studied breaker tripping in Chapter 15 in detail and will mention it only briefly here. A typical case of fault-current tripping is shown in Figure 16.9, where the initial surge voltage caused by breaker tripping can be calculated by the equations in the figure. The breaker’s reignition surges also must be taken into account. We know that restriking

e2(0)

e1(0)

Z1 · e1(0)–e2(0) Z 1 + Z2 –Z2 v2(t) = Z2 · i2(t) = · e1(0)–e2(0) Z1 + Z 2 1 i1(t) = · e1(0) – e2(0) Z1 + Z2 –1 · i2(t) = e1(0) – e2(0) Z1 + Z2 v v where 1 = Z1, 2 = Z2 i1 i2 v1(t) = Z1 · i1(t) =

(a) e1(0)–e2(0)

v1(t) i1(t) Z1

v2(t) i2(t) Z2

(b) Figure 16.8 Breaker-closing surge.

16.8 Overvoltage Phenomena Caused by Lightning Strikes

v1(t) = Z1·i(t) i(0)

i(0) Z1

Z2 = 0

e1(t)

e2(t) ⇋ 0

(b)

(a) Figure 16.9 Breaker-tripping surge.

the breaker should be avoided, not only to prevent breaker failure, but also to avoid extremely serious overvoltages that may cause other insulation failures in the network. 16.7.3

Switching Surges Caused by Line Switches

LS switching surge voltages are unique phenomena, as mentioned in Section 15.12 and shown in Figure 15.31, and substation insulation coordination is consequently very important. We will discuss this in detail later.

16.8

Overvoltage Phenomena Caused by Lightning Strikes

The mechanism of a lightning stroke is typically explained as follows. When the electric field gradient at some point in a charged concentration of a cloud exceeds the breakdown value for moisture-laden, ionized air, an electric streamer darts out toward the ground and halts after progressing about 100 m. After a short interval, the streamer again darts out and stops. The streamer progressing with a series of jumps is called a stepped leader. Just after the stepped leader reaches the ground, a heavy current (the main stroke of 1000–150 000 A) flows up the path blazed by the stepped leader with a velocity about one-tenth that of light. Our concern in this book is lightning that directly strikes electrical networks or causes surge voltages and current phenomena on them. Lightning surge phenomena can be classified into three stroke modes, which we will discuss in turn. 16.8.1

Direct Strike on Phase Conductors (Direct Flashover)

This is the case where a lightning stroke directly strikes one or more phase conductors. Figure 16.10 shows the main stroke directly hitting the phase-a conductor; current I(t) is injected into it. The successive surge phenomena were discussed in detail in Sections 15.5 and 15.6. Surge voltages va = Zaa (I/2), vg = Zag (I/2), etc. in the figure are induced due to the associated surge impedances. The induced surge voltage vag = (va – vg) exceeds the insulation strength across the phase-a conductor and point g (the overhead grounding wire [OGW], the top or arm of a tower, or arcing horn), flashover is caused, and a phase-a LG fault occurs. Furthermore, if the induced vab (= Zab (I/2)) exceeds the insulation strength across the phase-a and -b conductors, flashover also occurs between the two conductors, causing a LL fault. Of course, whenever a direct stroke occurs, line faults are caused in most cases. I(t)

Overhead grounding wire : g Phase a b c

Zag·I/ 2

p

Zag·I/ 2

Zaa·I/ 2

Zaa·I/ 2

Zab·I/ 2

Zab·I/ 2

Zac·I/ 2

Zac·I/ 2

Figure 16.10 Direct lightning stroke to a phase-a conductor.

477

478

16 Overvoltage Phenomena

I(t) Point p e

Zgg·I/2 Zgg·I/2

i Z0

Zag·I/2 Reff

g a b

Effective total surge impedance at the transition point p Zn overhead grounding wires (n)

v

c

R Tower (a)

(b)

Figure 16.11 Direct stroke to an overhead grounding wire or tower structure.

16.8.2

Direct Strike on an OGW or Tower Structure (Inverse Flashover)

This is the case where a lightning stroke directly strikes the OGW or tower structure, as shown in Figure 16.11a. The induced surge voltage vga = (vg – va) exceeds the insulation strength across point g and the phase-a conductor. Flashover is caused between point g (OGW, the top or arm of a tower, or arcing horn) and the phase-a conductor. Assuming lightning e(t) = i(t) Z0 strikes the transmission tower top directly or a nearby point on the OGW, the point is a transition point from surge impedance Z0 to parallel composed surge impedance Zn ⫽ Zn ⫽ R, and surge voltages and currents at the point can be calculated as follows v = e + erefl = itrans Zg = Zn in = R iR

①

i −irefl = itrans = 2in + iR

②

i = e Z0 ,irefl = erefl Z0 , in = v Zn , iR = v R ③ 1 Zg = ④ 2 1 + Zn R

16 7a

where Z0: surge impedance of lightning pass from the cloud to the striking point Zgw, Zn = Zgw/n: surge impedance of one and n stripes of OGW, respectively R: surge impedance of the tower structure Zg: total surge impedance of the grounding circuit at the struck point composed of a tower and both sides of the OGW (n stripes). Then, substituting ③ into ②① and eliminating e by rewriting, v=

2Zg 1 2e 1 e= = 1 1 Z0 1 2 1 Z0 + Zg + + + Z0 Zg Z0 Zn R

2e Z0

16 7b

Eq. (16.7b) can be also derived directly by applying the concept of Eq. (15.34). The equation v gives the induced voltage at the struck point, and the injected surge current i to the OGW and the tower. Numerical check: Assuming Z0 = 400 Ω, R = 350 Ω, Zn = Zgn = 300 Ω (where n = 1), then the effective surge impedance of the OGW and the tower is Zg = 1/(2/300 + 1/350) = 105 Ω.

16.8 Overvoltage Phenomena Caused by Lightning Strikes

If I(t) = 20 kA is assumed, the induced initial surge voltage v at the tower top is v = 20 kA × 105 Ω = 2100 kV. This is high-enough voltage to cause inverse flashover. The power-frequency voltage is of course superimposed on the injected surge voltage. If the rated line voltage is 400 kV, the power-frequency voltages va, vb, vc have values of ±400 × 2 3 = ± 327 kV (peak value). Accordingly, the superposed voltage vga = 1773~2427 kV(5.4–7.4E), so flashover is inevitably caused. If the number of OGWs is increased from one to n, the surge impedance Zn is decreased 1/n times, so the total surge impedance Reff can also be effectively decreased. Assuming n = 3 in this calculation, then Zn = Zg/3 = 100 Ω, and accordingly Zg = 43 Ω. That is, we can reduce the equivalent surge impedance by about half. 16.8.3

Induced Strokes (Electrostatic-Induced Strokes and Electromagnetic-Induced Strokes)

These are called induced lightning surges, and their mechanism is shown in Figure 16.12. Lightning may strike the earth a short distance from a transmission line. This phenomenon can be considered a virtual wire connecting a cloud and a point on the ground, with a surge current I(t) suddenly flowing along the virtual wire. The virtual wire has mutual capacitances and mutual inductances across the line conductors and the OGWs of the neighboring transmission line, so capacitive induced voltage and inductive induced voltage appear on the phase conductors and on the OGWs.

16.8.4

Capacitive Induced Lightning Surges

Figure 16.12 shows virtual conductor a and conductor b. If the potential voltage of the cloud is e(t), the capacitive induced voltage on conductor b is v(t) = {C/(C + C )}e(t). The insulation is broken down by the induced voltage, and the resulting flashover is called the capacitive induced lightning stroke. Now, we will examine the function and effect of the OGWs through a thought experiment. In Figure 16.12, virtual wire a (from the cloud to the earth) and line conductor b already exist. If one grounding wire g is installed in parallel close to conductor b, the equation for the capacitive induced voltage is modified as follows. The capacitive induced voltage on conductor b is C et C +C C With grounding wire g v ≒ e t , where C ≒ Cg C + C + Cg

①

Without grounding wire v =

Inductive induced voltage v = L

C < Cg ②

d it dt

16 8

③

The new equation ② includes an additional Cg in the denominator for the additional grounding wire. Conductor b and grounding wire g are closely linked, so the relation of the capacitance sizes Cg ≒ C < C < Cg is obviously justified. Accordingly, the capacitive induced voltage v(t) can be reduced by adding the grounding wire g (the shielding effect). This explanation of the shielding effect shown in Eq. (16.8) is a basic principle of electric field control that is widely used in designs for transmission lines, high-voltage equipment, and high-voltage testing facilities. Arcing horns for

Cloud

C

i(t)

Lightning pass: virtual line a

Lab

b

Figure 16.12 Induced lightning stroke.

e(t)

Cg⇋C

e(t)

Conductor b Grounding wire g

v C′

Cg′

Cg(⇋C)

C b C′

v

g Cg′

479

480

16 Overvoltage Phenomena

transmission lines (Figure 17.2), shielding rings for arresters (Figure 17.5), shielding plates for transformers (Figure 17.16), and shielding fences/shielding wire nets in high-voltage testing facilities are other examples of this principle. We will discuss this again later. 16.8.5

Inductive Induced Lightning Surges

As shown in Figure 16.12, mutual inductance Lab exists between virtual conductor a and the real conductor b, so inductive surge voltage v(t) = Lab di(t)/dt is caused on conductor b by the lightning inrush current i(t). The resulting flashover by the voltage is called the inductive induced lightning surge. If we also install grounding wire g, Lab can be reduced the same way we saw in Chapter 4. Therefore, the grounding wire can also reduce the electromagnetic induced voltage. Through this discussion, with regard to the grounding wire, installing OGWs can reduce L and increase C of the conductors so the surge impedance L C is also effectively reduced. This is discussed further in the next chapter.

481

17 Insulation Coordination We have studied various kinds of overvoltage phenomena in the previous chapters. Now we need to discuss the total process of insulation design for transmission lines, stations, and all the installed equipment including protection schemes for overvoltages: the power-system insulation coordination and overvoltage protection scheme as essential practical engineering for power systems. In the history of engineering development and commercial application of power systems in the twentieth century, the most dramatic turning point was the 1920s and 1930s: the fundamental concept of today’s power systems was established in the 1920s, and commercial construction expanded rapidly and widely in the 1930s. The key elements of modern power systems is summarized as follows: (i) parallel operation of three-phase 50/60 Hz generators, (ii) long power transmission lines with high voltages, (iii) interconnected power stations/substations, and (iv) one power system covering an individual area. Most of today’s modern power systems were first invented in these years. The success of insulation coordination was gradually recognized, and the first technical paper was published in 1928 in the United States, arguing for the need to establish the concept of insulation coordination. In addition, effective concepts and theories needed to be calculated, but the measures to observe steady-state, transient, and surge phenomena in three-phase power systems did not yet existin the 1920s. Practical theories and engineering knowledge advanced dramatically in the 1930s, with parallel rapid industrial applications. We can imagine the problems our predecessors had when struggling with repeated failures caused by casual, irregular, over- or under-insulation design in the early twentieth century, when they had a poor theoretical background, poor materials and application technology, poor experimental data, and no commonly shared guidelines to coordinate insulation strength, like today’s standards or recommendations. Insulation coordination is a combination of physical phenomena, practical engineering and technology, and economy, based on the experience accumulated over 100 years. The photos in Figure 17.1 show a 1000 kV transmission line with eight bundled conductors, and an invaluable photo that shows the instant when lightning struck the line.

17.1

Overvoltages as Insulation Stresses

Typical conduction materials are copper, aluminum, and other metals or alloys, and typical insulation materials are air, oil, SF6 gas, porcelain, fiberglass, paper, plastic, cross-linked polyethylene, etc. Every part of a power system is composed of a combination of conduction materials and insulation materials whose characteristics are very different. To obtain and control electricity means to prepare an extremely large container called a power system, every part of which is made of a skillful combination of different materials. In all materials, conduction is caused by the migration of charged particles. Conductors have large numbers of relatively free electrons, which drift in an applied electric field. On the other hand, insulation materials (insulators) have very few free electrons. When electric stress in an insulation material is increased to a sufficiently high level, the resistivity along a path through the material changes from a high value to a value comparable to that of conductors. This change is the breakdown. In order to achieve our purpose, we need to know the following from a practical engineering viewpoint:

•• •

Magnitudes and characteristics of resulting overvoltages and over-currents Required insulation strength of all the members that form the power system Countermeasures to reduce overvoltages and protect the insulation of lines and station equipment

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

482

17 Insulation Coordination

Double-circuit lines, nominal voltage: 1000 kV, maximum voltage: 1100 kV Conductor wire ACSR 810 mm2 (diameter 38.4 mm) 1258 A (continuous, in summer) eight-bundled conductors (w = 400 mm) maximum temperature 90°C OPGW: double circuit 500 mm2 × 2 Suspension insulator (320 mm type) double size 2 × 40 series connection horn gap: 6.3 m Towers (average weight: 310 tons) average height: 111 m, height of lowest arm: 65 m phase-to-phase within same circuit: 20 m phase-to-phase between different circuits 33 m conductor height >32 m span average 632 m, maximum 1056 m TOV (temporary overvoltage) 1 LG: 1.1 pu, LR (load rejection): 1.5 pu LIWV (lightning impulse withstand voltage) GIS 2250 kV, Tr 1950 kV SIWV (switching impulse withstand voltage) GIS 1550 kV, Tr 1425 kV

Direct strike to the upper conductor (1998)

Eight bundled conductors

Figure 17.1 1000 kv design double-circuit transmission line. Source: Courtesy of TEPCO.

Insulation coordination is the total process and practical guidelines for system insulation design, combining these three areas, which are based on accumulated theoretical, experimental, technical, and economic data. Possible overvoltage levels and the required insulation to withstand those levels must be estimated for each part of a power system, to realize continuous long-life equipment operation. Accordingly, we need to know the insulation strength as another aspect of overvoltage; this is a key theme of practical high-voltage engineering, although a detailed description of individual insulation strength is far beyond the scope of this book. The essential philosophy and concrete recommended practices based on practical data for insulation design have been issued as international and/or national standards (IEC, IEEE, ANSI, JEC, etc.), and include the characteristics of impinging overvoltages and methods to protect insulation; these are important guidelines for practical insulation design engineering. Today, insulation coordination is an established process throughout the world, and the standards issued by different organizations are substantially identical. Today, all parts of the power system’s primary circuit (transmission lines, power cable lines, power station/substation equipment, and related configurations) are planned, specified, designed, manufactured, tested, installed, and operated

17.2 Classification of Overvoltages

based on insulation coordination standards and related individual standards. All engineering activities with regard to power systems are deeply correlated with this process of insulation coordination.

17.2

Classification of Overvoltages

In previous chapters we have looked at overvoltage mechanisms. Lightning is an overvoltage generated by external events, while overvoltages generated by switching operations, fault occurrences, ferroresonance, load rejection, loss of ground, etc. are generated by internal events. Now we need to classify overvoltages again from the viewpoint of impinging overvoltages on individual insulation structures. The insulation strength of individual parts of the system or equipment is affected by the following factors of impinging overvoltages:

•• •• •

Magnitude, shape, duration, and polarity of the applied voltages Insulation design of the electric field distribution in the insulation Type of insulation (gaseous, liquid, solid, or a combination) Physical state of the insulation (temperature, pressure, mechanical stress, etc.) Operation and maintenance history of individual insulation materials

Impinging overvoltages may exceed permissible insulation levels of individual parts of the system, so these overvoltages should be reduced to within permissible levels, or the insulated equipment should be safely protected against such overvoltages. This is essential to avoid equipment insulation damage or prevent undesirable system performance. The magnitudes, wave shapes (steepness of the voltages), and duration of overvoltages are important factors with regard to stress on the insulation. Taking such important factors into account, overvoltages are generally classified into the following categories by authorities like the IEC and national standards bodies: a) Maximum continuous (power-frequency) overvoltage (MCOV) Us: This can originate from the system under normal operating conditions. b) Temporary overvoltage (TOV): This can originate from faults, switching operations such as load rejection, resonance conditions, nonlinearity (ferroresonance), or a combination of these. c) Slow-front overvoltages: These can originate from switching operations and direct lightning strikes to conductors of overhead lines. d) Fast-front overvoltages. These can originate from switching operations, lightning strikes, or faults. e) Very fast-front overvoltages. These can originate from faults or switching operations in gas-insulated switchgear (GIS). Overvoltages (c)–(e) are transient overvoltages. We will examine then next. 17.2.1

Maximum Continuous (Power-Frequency) Overvoltages: Us

Under normal operation, power-frequency voltages differ from one point of the system to another and vary in magnitude over time. For the purposes of insulation design and coordination, continuous overvoltages should be considered the highest possible operating voltages under normal operation, which is usually defined as the highest system voltage (symbol Us (kV). In usual practice, Us is presumed to have a value of 1.04–1.1 times the system normal voltage. For example, Highest system voltage Us

Representative TOV (Urp = k

230 kV (phase-to-phase)

245 kV (phase-to-phase)

212 kV (phase to earth, k = 1.5)

500 kV

550/525 kV

476/455 kV (k = 1.5)

735 kV

765 kV

662 kV (k = 1.5)

1000 kV

1100 kV

953 kV (k = 1.5)

Normal voltage

1

Phase-to-phase

3 Us)

483

484

17 Insulation Coordination

17.2.2

TOVs and Representative TOVs: Urp

Typical situations that may give rise to TOVs include:

•• •• •• •

Single line-to-ground (LG) faults Ferroresonance Load rejection Loss of neutral grounding Long unloaded transmission lines (Ferranti rise) Coupled-line resonance Transformer-line inrush

TOVs caused by all these situations are the result of internal events caused by circuit connection changes, so voltages tend to show unstable, slow, transient behaviors and often include somewhat lower-order harmonics or DC components, usually lasting hundreds of milliseconds or longer, up to a few minutes. The highest TOVs for each of these situations are considered the representative TOV Urp, considering current and future system configuration and operating practices. However, at minimum, overvoltages due to LG faults should be addressed, because they are typically the most significant and the highest TOVs in most cases. Single LG faults cause the largest power-frequency overvoltages among fault modes.

17.2.2.1 Single Line-to-Ground Faults

We examined TOVs caused by phase-grounding faults in Section 16.3 and Figure 16.1. The TOV caused by this mode is provably the largest type. The magnitudes of TOVs appearing in unfaulted phases during a LG fault are related closely to the system grounding conditions (δ = f R0/f X1, v = f X0/f X1) and can be estimated by earth–fault factor k (or the coefficient of grounding [COG]). Factor k is typically 1.5 for a solidly neutral-grounding system, or 3 or higher for a resistance/ reactance neutral-grounding system. 17.2.2.2 Load Rejection

If a large load is suddenly separated (by the bus protective relay tripping in a large substation, for example), overvoltage due to load rejection is caused and continues for 10–200 seconds until the voltage is regulated by AVR systems in power stations/substations. Although the amplitudes of overvoltages depend on the rejected load size and network configuration, the amplitudes are 1.2 pu or less in moderately extended systems due to the quick response characteristics of AVRs. However, overvoltages are apt to be larger in power systems with extended configurations. When full-load rejection is caused at the receiving substation of a long transmission line end, the voltage may increase by a magnitude close to 1.5, because the Ferranti effect at the open receiving end is added. 17.2.2.3 Loss of Neutral Grounding

In high-resistive/reactive neutral-grounded systems, special consideration is required. If one phase-to-ground fault occurs in the system, and the neutral-grounded transformer is tripped for some reason (by a backup relay tripping, for example), this means loss of neutral grounding (assuming another neutral-grounding transformer does not exist in the same system). As the potential of the system neutral point becomes free of the earth potential, unfaulted phase voltages become at least 3 pu during one-phase-to-earth fault (1φG). Furthermore, even after the one-phase-fault condition is removed, three-phase voltages and neutral-point voltages fluctuate wildly. In these conditions, installed arresters are broken, which means new faults at arrester points, and/or earth faults at various locations. Above all, loss of ground should be avoided. In high-resistive/reactive neutral-grounding systems, neutral groundings by as many transformers as possible (at least two transformers installed at different stations in one service area) are preferable. A TOV caused by one-phase-to-ground faults is generally considered the representative largest power-frequency overvoltage Urp (usually expressed by RMS kV value, phase-to-earth). Then representative TOV = Urp = k

1

3 Us

17.2 Classification of Overvoltages

As a typical example,

17.2.3

Nominal voltage 230 kV k = 1 5 TOV = 1 5

1

3

245 = 212 kV

400 kV k = 1 5 TOV = 1 5

1

3

400 = 346 kV

735 kV k = 1 5 TOV = 1 5

1

3

735 = 662 kV

Slow-Front Overvoltages

Slow-front overvoltages have durations of tens to thousands of microseconds and tail durations of the same order of magnitude, and are oscillatory by nature. They arise generally in the following occasions:

•• •• •

Line energization/re-energization (breaker closing/opening) Fault occurring or clearing Switching of capacitive or inductive current Load rejection Distant lightning strike (lightning strike wave front flattened by traveling)

•• ••

Switching of unloaded cables or capacitor banks Inductive current tripping (transformer magnetizing current tripping, for example) Arc-furnace load switching Current interruptions caused by high-voltage fuses

The voltage mechanisms of all of these have been studied in previous chapters. The representative voltage stress is characterized by the representative voltage amplitude and wave shape. Therefore, the representative switching impulse voltage of 250/2500 μs (time to peak 250 μs, and time to half-value on the tail 2500 μs (see Table 17.2c) have been standardized as a general concept throughout the world to represent standardized slow-front overvoltages. For the switching of capacitive and inductive currents, we have looked at overvoltages caused by current chopping in Section 15.12.6, caused when the current power factor is almost zero. In particular, the following switching operations require special attention:

The typical practice to limit slow-front overvoltages is to use surge arresters, which will be discussed later. Another useful method is to use resistive tripping/closing breakers (resistive tripping/closing method), which we studied in Section 15.12.10. The resistive tripping and/or closing method is utilized typically for UHV breakers over 500 kV, as we discuss later. 17.2.4

Fast-Front Overvoltages

Typical fast-front overvoltages are of course lightning strikes, although they can also originate from switching operations. We have already seen that lightning overvoltages can be classified s direct strikes, back-flashovers, and induced lightning strikes. Induced lightning surges generally occur below 400 kV and so are important only for lower-voltage systems of 100 kV. Back-flashover voltages are less probable on UHV systems of 500 kV or more, due to high insulation withstanding values. The representative wave shape of fast-front overvoltages is the well-known 1.2/50 μs wave (see Table 17.2c). 17.2.5

Very Fast-Front Overvoltages

Very fast-front overvoltages can originate from switching operations or from faults within GIS due to the fast breakdown of the gas gap and nearly undamped surge propagation within GIS, where the average distance between two adjacent transition points in the same GIS is very short. (If 7.5 m distance and u = 300 m/μs is assumed, the resulting traveling surges repeat 20 times every 1 μs, so the natural frequency is 20 MHz, and the first-front wavelength [quarter

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17 Insulation Coordination

cycle] is 0.0125 μs.) However, the amplitudes of the surges are rapidly dampened and flattened on leaving the GIS, so they are relieved to a limited extent at the external circuit of the GIS bushings. The overvoltage shape is characterized by a very fast increase of voltage to nearly its peak value, resulting in a front time less than 0.1 μs. For switching operations, this front is typically followed by an oscillation with frequencies of 1–20 MHz. The duration of very fast-front overvoltages is less than 2–3 ms; however, 20 MHz and 3 ms means 60 000 times of beating stresses. Furthermore, they may occur several times. The magnitudes of overvoltage amplitude depend on the structure of the disconnector and on the adjacent structure of station equipment. Very fast-front overvoltages can and must be dampened/flattened to some extent, and the typical countermeasure is application of gapless arresters, which means inserting a nonlinear high resistance in parallel across the phases and earth. Maximum amplitudes of 2.5 pu can be assumed to be achievable. Due to faults within GIS, the connected equipment, in particular a transformer, is stressed by overvoltages, which contain frequencies up to 20 MHz, and the amplitude may exceed the breakdown voltages of the transformer without effective countermeasures. This will be discussed in Sections 17.11 and 17.12.

17.3

Fundamental Process of Insulation Coordination

17.3.1

Insulation Coordination

Insulation coordination usually means the coordination or correlation of the transmission-line insulation with that of the station apparatus, and perhaps the correlation of insulation of pieces of apparatus and parts of the substation. Coordinating substation and equipment insulation protects of service and apparatus from overvoltages in excess of specified insulation withstanding values with optimum economy and reliability. It is obvious that various kinds of knowledge and accurate data based on experience and advanced technology are required in order to establish a reasonable process for insulation coordination, as follows:

•• •• •

Investigating mechanisms of different overvoltages Estimating possible overvoltages on the line or in the substation Countermeasures to reduce overvoltages, including development of protective devices Specifying insulation withstanding values for a transmission line Specifying insulation withstanding values of the station apparatus, and so on

The descriptions that IEC, IEEE, and other national standards use for insulation coordination have converged on the same criteria. This result has been achieved through a long process of evolution over the past 100 years. We study such this worldwide common process in this chapter. 17.3.2

Specific Principles of Insulation Strength and Breakdown

The basic process of coordination or correlation of transmission-line insulation and that of the station apparatus can be summarized as follows. 17.3.2.1 Insulation Design Criteria for Overhead Transmission Lines

The basic criteria are as follows:

• • •

Flashovers caused by lightning strikes are considered fatal phenomena, and damage (damage to conductors, cracks in insulators, etc.) to the transmission line should be avoided. Technically and economically balanced insulation distance (clearance) is to be ensured in the fundamental design, allowing some extent of failure rate caused by lightning strikes. Also, countermeasures should be used as much as possible to reduce the influence and frequency of effects on a substation. Flashover should not be caused by switching surges or sustained lower-frequency overvoltages.

This process is based on the characteristics in atmospheric air of insulation and cooling materials with infinite natural circulation, so that, once broken, insulation is restored (recovered) whenever the surge source disappears (self-restoring insulation characteristics).

17.4 Countermeasures on Transmission Lines to Reduce Overvoltages and Flashover

• • •• ••

The principal countermeasures are: Reducing the probability of lightning strikes, and limiting faulted circuits (in the case of multiple-circuit transmission lines) and faulted phases as much as possible (using overhead grounding wires (OGWs) and any other effective countermeasure) Reducing the probability of back-flashover caused by lightning strikes on the OGWs or on the towers (surge impedance reduction of towers and OGWs) Countermeasures to relieve the traveling waves to some extent before reaching the substation terminal point Insulation-level withstanding against switching surges from the substation Reducing the probability of simultaneous faults on plural circuits of the same route Using reclosing.

17.3.2.2 Insulation Design Criteria for Substation and Substation Apparatus

The basic criteria are that insulation of the station and station apparatus should be protected to withstand lightning surges and switching surges, to avoid insulation failure of the apparatus and loss of station services over a long time. The principal countermeasures are:

•• •

Countermeasures to reduce direct lightning strikes to the station as much as possible. Protecting the substation and station apparatus against transmitted lightning surge voltages from the overhead transmission lines without damage (arresters). Reducing switching surge levels and protecting the station apparatus against switching surges without damage. That is, flashover or insulation failure of equipment should not be caused by switching surges or by any sustained lowerfrequency overvoltages in the substation.

17.3.2.3 Insulation Design Criteria for Power Cable Line

The basic criteria are to protect the power cable against lightning surges, switching surges, and fundamental frequency overvoltages. The power cable does not have self-restoring insulation, so it should be protected entirely from overvoltages in the same way as transformers or other substation apparatus. All of these criteria are substantial aspects of the fundamentals of power systems, and thus are the objectives of insulation coordination.

17.4

Countermeasures on Transmission Lines to Reduce Overvoltages and Flashover

In this section, we will list the major countermeasures that are usually used in the concrete design for high-voltage transmission lines and substations in order to meet the design criteria we have described. We saw in previous chapters the reasons why using each countermeasure is technically effective and can be justified. We recommend that readers refer to the literature for details exceeding the scope of this book. 17.4.1

Using Plural OGWs and OPGWs

The following effects are expected from using OGWs:

• • • •

By locating the OGW at the top of a tower, the probability that lightning will directly strike the phase conductors can be reduced (shielding effect; see Section 16.8). A lightning strike directly on the OGW may occur, but most of the energy can be bypassed as strike current though the OGW and towers to earth. The surge impedance L C of the line conductors can be reduced. (L is decreased significantly and C is increased significantly by installing or increasing the number of OGWs; see Chapters 4, 6, and 16). The absolute magnitude of induced surge voltages appearing on the phase conductors can be reduced so the probability of phase faults due to back-flashover or induced strikes is improved (Section 16.8). Also, alleviation of steep wave-front (surge-front flattening) from the lightning surge can be expected (Chapter 15). The time constants T = 2 L/r of the line can also be reduced so the attenuation of traveling–wave or transient oscillation terms can be accelerated (Chapters 15). Positive-, negative-, and zero-sequence inductance L1 = L2, L0 can be reduced as an important feature.

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••

Power-frequency voltage drop due to line reactance can be reduced (Chapter 4). The stability limit and power circle diagram can be improved (Chapter 12).

Today, OPGW (grounding wire with optical fiber) is often used instead of OGW, and the fibers are utilized for communication channel media.

17.4.2

Reasonable Allocation and Air Clearances for Conductors and Grounding Wires

Phase conductors and OGWs must have sufficient clearance from each other so the necessary insulation withstanding strength is maintained within a margin against the predicted largest short-duration overvoltages and switching surges from neighboring substations. Of course, this is a trade-off between the probability of flashover failure and the construction cost of larger towers. The allocation of conductors and grounding wires should result in a reasonably phasebalanced smaller L and smaller L C within the necessary margin against physical movement (caused by wind, heat expansion, galloping, sleet jumping, etc.). Details of standard lightning impulse-withstanding voltages are examined in Section 17.10.

17.4.3

Reduction of Tower Surge Impedance

The surge impedance of towers must be reduced as much as possible, in that induced surge voltages on the transmission lines can be reduced, or the probability of striking can be reduced. In particular, back-flashover caused by direct strikes on the top of towers or grounding wires is effectively reduced (Section 16.8). The magnitude of tower surge impedance is related to the height of the tower and resistivity of the ground, and is typically 20–100 Ω for EHV class lines. 17.4.4

Using Arcing Horns (Arcing Rings)

Arcing horns are a kind of air gap with self-restoring insulation. They are arranged in parallel with each insulator as a single body on every tower. Figure 17.2 shows an example for a single-conductor line. The duty and purpose of arcing horns can be summarized as follows:

•

Conductor Figure 17.2 Arcing horn.

•

• •

Flashing overvoltages can be controlled by selecting the shape and air-gap length of the arcing horns so the magnitudes of traveling surge voltages caused by the lightning can be limited by the flashover voltage of the arcing horn. The arcing horn is said to be the intentionally arranged weak point of a conductor’s insulation (controllable limitation of surge-voltage magnitude). An arcing horn assembled together with an insulator can improve the potential gradients of the insulator by unifying the voltage distribution by series-connected individual porcelain pieces (improving withstand voltage). Flashover by arcing horn can avoid flashover along the surface of insulators, so the insulators can be protected from damage against thermal shock (protection of porcelain insulators).

On the other hand, arcing horns have this limitation: Arcing horns and any other part of the overhead transmission lines should not experience flashover by switching surges coming from adjacent substations.

The air-gap lengths of each arcing horn on series-connected towers are generally arranged with equal lengths, while those on the first to third towers of outgoing feeder lines from the station may be arranged with a smaller gap length. Lightning overvoltage caused by striking the transmission line point within one or two spans from the station appears as impinging traveling surges to the station without attenuation or shape flattening. Accordingly, in order to alleviate such severe surges caused by a close-point direct strike, the arcing horns at the first few spans are arranged with smaller gap lengths.

17.5 Tower-Mounted Arrester Devices

17.5

Tower-Mounted Arrester Devices

Tower-mounted arrester devices have been developed in recent years and tried by some utilities. The purpose of the equipment is to improve the function of arc horns and, by doing that, to reduce the probability of a line fault caused by a lightning surge. Figures 17.3a and b show the concept, and Figure 17.3c shows an application of a 154 kV transmission line. This is explained next in comparison with an arcing horn.

Lightning OGW

Arrester

Arrester device Arc-horn Series-gap

Line conductor

(Courtesy of Chubu Electric co.) (a) Structure

(c) An example of application for 154 kV system

Discharge characteristics of the arcing horn Discharge voltage

Discharge characteristics of the series gap Voltage

Surge voltage

v –i characteristics of ZnO disc element Conductor remaining voltage that is smaller than discharge voltage of arcing horn

Residual voltage System voltage Time

Current Current Surge current Cut-off of residual current (recovering insulation)

Time Residual current (smaller than 1A peak) (Courtesy of Chubu Electric Co.)

(b) Response characteristics during gap flash-over Figure 17.3 Tower-mounted arrester device.

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17 Insulation Coordination

Flash-over of an arcing-horn means a phase-to-ground fault of the transmission line, so fault tripping of the line by associated relays and breakers is unavoidably caused within a few cycles after flash-over, although automatic reclosing may be successfully performed. As seen in Figure 17.3c, the device consists of an arrester device in series with an air gap and is arranged in parallel with the insulator per phase. When impinging surge voltage occurs, the air gap is flashed-over. However, owing to the seriesconnected arrester device, the voltage based on v–i characteristics of the arrester disc-element remains between the phase-conductor and earthed tower arm, and discharge current is also limited. Therefore, the discharge current is cut off within 1/2 cycle at the time of current zero when ± polarity of the operational voltage of 50/60 Hz is changed. As a result, the fault current disappears within 1/2 cycle, so fault-tripping by relays is avoided if possible. Furthermore, instantaneous voltage drop can be avoided. Instantaneous voltage drop caused by a fault is unavoidable whenever faults occur on lines with conventional arc horns and continue for a few cycles until the fault is cleared by the associated breakers tripping. These phenomena often have a negative influence on the load side because the resulting voltage drop is 0–100% within a few cycles, depending on the distance between the fault point and the load. A tower-mounted arrester device is a new technology with advanced v–i characteristics and reduced weight. These devices will be used in the future as a key technology to reduce line faults.

17.6

Using Unequal Circuit Insulation (Double-Circuit Lines)

Line failures caused by lightning striking the transmission lines cannot be avoided; however, we need to reduce the likelihood by avoiding simultaneous double-circuit faults. The number of faulted phases of the same circuit should also be reduced as much as possible. For this purpose, unequal circuit insulation or unequal phase insulation can be applied, with horn-gap lengths of a single circuit or phase intentionally arranged within a short distance.

17.7

Using High-Speed Reclosing

Automatic high-speed reclosing is an important practice to reduce the influence of lightning failure. Its essence is presented here, although it is not directly related to insulation coordination. In single-phase reclosing for a phase-a-to-ground fault (aφG) (for example), immediately after the phase-a conductorto-earth fault occurs, the phase-a pole contactors of both line terminal breakers are tripped within 2–6 cycles (operating time of the relay + the breaker): this is single-phase tripping. Then, although the phase-a conductor is already separated from the station buses, the voltage va on the phase-a conductor remains, so the arc also continues for a small duration (arcing time; 0.2–0.5 seconds); in addition to the initial trapped potential charge (DC), the electrostatic (C-coupled) voltage is induced from unfaulted parallel phase voltages vb, vc. However, due to the outstanding self-restoring characteristics of natural air, the arc is soon extinguished. Accordingly, if the phase-a poles of the breakers of both terminals are reclosed after arc extinction, the faulted line will continue three-phase operation successfully. The time from the one-pole tripping to reclosing is called dead-voltage time, which is a set value in the line’s primary protective relay equipment. The arcing time is apt to be longer for higher-voltage systems, in particular UHV systems of 500 kV or more, so dead-voltage time as a relay setting value must be set longer. There is a presumption that in 1000 kV class power systems, self-arc extinction cannot be expected within a short time, so automatic reclosing is not available without using forced grounding switches. For the classification of high-speed reclosing, we will assume a double circuit with line 1 (a,b,c) and line 2 (A,B,C):

•• •

Single-phase reclosing: This occurs on one LG fault (phase-a reclosing against a phase-a fault). Three-phase reclosing: When a fault occurs on line 1, the three-phase-abc poles are tripped and reclosed regardless of the fault modes on the line. This practice can be applied only for double-circuit lines, or at least assured loopconnected lines. Multiphase reclosing: In the case of a fault of line 1 phase-a and line 2 phase-B, for example, reclosing is conducted for these two poles, because phases A, b, C are still soundly connected even though this is double-circuit fault. This is an effective method in comparison with three-phase reclosing, because it can minimize lines out of service or relieve system instability. In addition, very accurate fault phase-detection by protective relays (preferably by phasedifferential protection) is required.

17.8 Overvoltage Protection with Arresters at Substations

Incidentally, reclosing is allowed only when both terminal buses across the faulted line are operating in synchronization. Furthermore, any power sent during dead voltage time must be noted. Single-phase reclosing on a single-circuit line, for example, means a one-phase-open condition, as shown in Table 7.2. Then reclosing should be allowed only when the power flow during the dead-voltage time is within the stability limit under the transmission line condition where the faulted phases are cut off (see Section 12.5). We also need to pay attention to the undesirable electrical and mechanical effects on the thermal generators caused by negative- and zero-sequence currents during the reclosing dead-time.

17.8

Overvoltage Protection with Arresters at Substations

Now we will examine aspects of overvoltages at substations and countermeasures to protect against overvoltages and reduce stress. 17.8.1

Surge Protection Using Metal–Oxide Surge Arresters

Arresters are key devices to protect substations and station devices against lightning surges. Their surge-protective capability determines the required insulation levels in the power-system network. Gapless arresters may also be able to reduce switching surges. Typical high-voltage arresters are metal–oxide surge arresters that have metal–oxide resistive disc elements with excellent nonlinear v–i characteristics and thermal energy withstanding capability. The metal–oxide resistive elements are composed of a number of overlapped disc elements, each of which is made from zinc oxide (ZnO) powdered material with specially mixed inclusions; similar to pottery or porcelain, they are produced through a high-temperature baking procedure. Figure 17.4 shows the fundamental principles of arresters. In Figure 17.4a, the arrester is installed at point a. Arrester point a is a transition point because arrester resistance is connected to the conductor at this point. Now, incident overvoltage wave E comes from the left and passes through point a from the left-side line Z1 to the right-side circuit Z2 toward the station bus. The behavior of the traveling waves at point a can be written by the following equations: E + vr = var = vt i− ir = it + iar

where E incidental surge voltage from the left-side line var ; iar the terminal voltage and the current of the arrester at arrester point a Zar nonlinear resistance of the arrester

E = Z1 i vr = Z1 ir

vr , ir the reflected voltage and current at point a

vt = Z 2 i t

vt , it the transmitted voltage and current at point a to the substation

var = Zar iar

17 1a

Z1 the surge impedance of the transmission line Z2 the surge impedance of the substation gateway at point a

The relation between the arrester voltage var and current iar is shown as the nonlinear curve ① in Figure 17.4b, which represents the v–i characteristics of the arrester. Eliminating vr, ir in this equation, and by modification, var = vt =

Z2 2E −Z1 iar Z1 + Z2

17 1b

This equation is written as the straight line ② in Figure 17.4b. The voltage var at point a and the arrester current iar are given as those at the intersection of curve ① and the straight line ②. If the arrester did not exist, the voltage var at point a would be {Z2/(Z1 + Z2)} 2E (then, E under the condition Z1 = Z2), or 2E maximum under the condition Z2 = ∞ (i.e. the case when the feeding terminal is opened). However, if an arrester with appropriate nonlinear v–i characteristics is installed at point a, surge voltage var can be reduced to a value smaller than the original impinging surge value E and, of course, smaller than 2E. But to realize this condition, the arrester is required to withstand the resulting extremely large thermal energy ( var iardt) without losing the original v–i characteristics of the resistive elements and without breaking. It is furthermore required to immediately restore the original electrical operating condition before the impinging surge.

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17 Insulation Coordination

νr , ir

νt , it

E Line

a

i

Substation

Z1

Z2 νar

Arrester iar

(a) Voltage (kV) νa Z2 ·2E Z 1 + Z2 1 ν – i characteristics of arrester

Z2 ·E Z1 + Z2 νar Maximum operating voltage 1.0

2

iar

Z2 (2E – Z1 · iar) Z1 + Z2

2E Z1

Discharging current (kA) (b) Figure 17.4 Principle of surge protection by arrester.

As a natural consequence of this, the arrester must have the following characteristics as its inherent duty requirements:

•

Under power-frequency operation with voltage Us (MCOV) and Urp (representative TOV), the arrester must have high resistivity so its resistive elements can withstand the thermal stress caused by the small leakage current (continuous current of arrester, usually within a few milliamps, 1 mA or less).

For example, for a power system of nominal voltage 230 kV, Us = 245 kV; Urp = k 1 3 Us = 1.5 × 1 3 245 = 212 kV. So, assuming the arrester continuous current iar ≦ 1 mA at Urp = 212 kV, then var = 212 kV; iar ≦ 1 mA; Zar ≧ 212 MΩ is obtained and the arrester thermal loss (stress) is var iar = 212 kV 1 mA = 212 W.

• • • •

Whenever surge voltage and current arrive at the arrester point a, the arrester must discharge the surge current immediately; voltage var that arises should be limited within a specified upper limit (residual voltage, or discharge voltage, Ures) during the discharge. Thus the station equipment can be protected against the overvoltage. The arrester should withstand the thermal energy caused by the surge current through the arrester elements (arrester discharge current, say 10–100 kA), with thermally stable v–i characteristics. Immediately after the discharge current through the arrester disappears (within 50–100 μs), current iar from continuing power-frequency voltage var should return to the original small leakage current within a few milliamps. In other words, immediately after the surge voltage and current disappear, all electrical characteristics must be restored. The arrester must have switching surge-discharging capability within the specified levels. (The switching surge duty on metal–oxide arresters increases for higher system voltages.)

At the substation where the arrester is installed at the junction point with the transmission line, the surge voltage (caused by the impinging lightning surge from the same line) can be limited to the arrester-discharge voltage so station

17.8 Overvoltage Protection with Arresters at Substations

equipment with an insulation level that exceeds the arrester-discharge voltage (or residual voltage) is protected by the arrester. This is the principle of station equipment protection with a surge arrester. 17.8.2

Metal–Oxide Arresters

The fundamental configuration of these arresters can be classified as follows:

•• •

Gapless arresters Series-gapped arresters Shunt-gapped arresters

Figure 17.5 shows a typical example of a porcelain-type metal-oxide gapless arrester for high-voltage station use. Before the appearance of gapless arresters, around 1980, all high-voltage arresters were the series-gap type, arranged in series with nonlinear resistive elements. The thermal duty of resistive elements at the time was small, so the elements could not withstand the thermal energy of continuous flowing leakage current caused by power-frequency terminal voltages without the series gap. Accordingly, series-gap arresters were indispensable to avoid thermal damage to resistive elements caused by continuous leakage current flowing through the elements. Today, most arresters for high-voltage systems are gapless: the nonlinear resistive element block is always directly charged by power-frequency phase voltages so minute leakage currents flow through the arrester’s resistive elements continuously. Of course, the resistive elements need to withstand the thermal stress caused by continuous thermal energy ( var iardt) from the leakage current and maintain thermally stable v–i characteristics. In the case of series-gapped arresters, the arcing ignition across the series gap is initiated by the surge voltage. Therefore, arcing extinction across the gap immediately after the surge current disappears under charging conditions from the continuing power-frequency voltage is another important requirement for this arrester (following current tripping duty). Now, let’s briefly recap the technological history of power-system insulation coordination. The concept of insulation coordination was born in the late 1920s, and the first AIEE technical paper was presented in 1928: “Coordination of Transmission Line and Apparatus Insulation.” The first cathode-ray oscillograph was adopted in 1929 and observed voltages due to direct lightning strokes in 1930. Rapid progress was achieved in the 1930s, including the accumulation of experimental data on overvoltages and insulation practices, the development of the theory of surge-overvoltage phenomena, the final push to establish a process and associated standards for insulation design and testing in order to coordinate insulation for individual transmission lines and apparatus. The first forms of insulation coordination were established in 1935, which was said to be the turning point of power-system technology from infancy to youth. The period from the 1940s to the 1980s was the era of drastic expansion of power systems based on continuous development of higher voltages and larger capacities for all equipment, including EHV and UHV. Development of higherperformance arresters (with series gap) was an essential key technology to realize higher voltage ratings with reasonable insulation coordination. All arresters were equipped with a series gap because the resistive disc elements could not

Rupture diaphragm

Insulation spacer

Line–terminal

Pressurerelief aperture Shield ring to improve potential gradient

Porcelain housing

Metal–oxide (ZnO) resistive elements

ZnO disc element SF6 gas

Insulator (moulded on ground terminal)

Earth–terminal Mounting base

(a)

(b)

Figure 17.5 Arrester for station use (a) Arrester for GIS; (b) Porcelain arrester (Courtesy of Toshiba).

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d t2

e

t4

f

c 1pu

Residual voltage curve (upper half)

t3

Vmax

Arrester residual voltage: Vmax at the time t3: peak value appears between the terminals of an arrester during the passage of discharge current

t5

t1 a

Imax 0 i0

b

1A

10 A

100 A

1 kA

10 kA 100 kA

–1pu

Figure 17.6 Arrester characteristics (v-i-t loci of resistive element).

withstand the required continuous thermal stress v iar = v2/Rar caused by continuous leakage currents iar. A technological breakthrough was achieved in the 1980s: gapless-arrestors with outstanding thermal withstanding capacity in the resistive disc elements. The v–i characteristics of the elements (flatten v–i characteristic with larger surge current) enabled drastic reduction of insulation levels, in particular for EHV/UHV. Today, most newly installed arresters are gapless. Figure 17.5 shows examples of gapless arresters for station use. Now, we will examine arresters by looking at the v–i characteristics in Figure 17.6. The point where an arrester is installed is charged continuously by power-frequency voltage v (= 1.0 ± 0.1), so the arrester leakage current iar (of 1 mA or less and typically 100–500 μA) is flowing through the arrester disc element. In Figure 17.6 with this condition, (var, iar) is tracing points a and b once per cycle. iar never exceeds the assigned current value of the manufacturer (the reference current of the arrester, which is typically 1 mA). Now, transmitted surge voltage E appears at time t1. Immediately after the surge arrival time t1, the voltage exceeds point c and traces up to the voltage peak point d with Vmax (residual voltage or discharge voltage [kVcrest]) at time t3, and then reaches the maximum current point e with Imax (the maximum value of discharge current [kAcrest]) at time t4. Soon (200–500 μs after t4), the surge voltage and current disappear, so the tracing point goes back through point f and finally returns to the original zone between points a and b. The return path is a little shifted from the outward path, because the original v–i characteristics are temporally reduced by the thermal effect of the disc element. The peak voltage point c is called the reference point, and the voltage is the reference voltage. The resulting maximum voltage Vmax is the residual voltage ([kVcrest]). Throughout this process, the voltage at the arrester installed point is restrained within the residual voltage, so appropriate coordination can be achieved by using the insulation design criteria of withstanding voltages in all substation equipment that are higher than the residual voltages of the installed arresters. This is the fundamental process of insulation coordination. Of course, the arresters have to withstand the thermal stress caused by 50/60 Hz MCOV, TOV, and switching surges. Figure 17.7 shows a typical example of residual voltage characteristics based on international standards. The vertical axis shows the maximum voltage Vmax (by [kVcrest]), which corresponds to the peak voltage point d in the traced curve in Figure 17.6. These characteristics show the protective level of the arrester, which is the basis for deciding the insulation design criteria for the individual transmission lines/power cables and station apparatus. The resulting voltage at the arrester point is limited to the residual voltage, while the value is influenced by the surge current iar flowing through the arrester. The surge current through iar is usually 5–20–50 kA, while the presumed largest discharge-current values are 150 kA for the 400–500 kV, 100 kA for the 275–300 kV, 80 kA for the 160–230 kV, 60 kA for the 110–160 kV, and 30 kA for the 60–90 kV class. Figure 17.7a is an example of advanced characteristics by which the residual voltage is distributed within a voltage range of 1.5 times the power-frequency operating voltage in most cases; a larger discharge voltage of more than two times (2–3.5 times) the operating voltage can also appear if a large impulse discharge current flows through the arrester. The lower, flattened v–i characteristics of today’s arresters have enabled the reduction of the insulation level for EHV/UHV class power systems.

17.8 Overvoltage Protection with Arresters at Substations

Discharge voltage pu value of Ur(kVrms)

The guarantee zone of switching surge voltage (kVcrest)

The guaranteed discharge voltage against impulse (kVcrest)

2.0

1.5

The reference voltage for the specification of operating characteristics)

1.0

point b point c

point a

0.5 Discharge current (kAcrest) 0 10μA 100μA 1mA

10mA

100mA

1A

10A

100A

1kA

10kA

iar 100kA

Standard nominal discharge current (kAcrest) (selected from 1.5kA, 2.5kA, 5kA, 10kA, 20kA, 40kA) (guaranteed typically by 8/20 ms waveform current) Duty cycle voltage rating Ur(kVrms) The designated maximum permissible voltage between the line and ground terminals at which an arrester is designed to perform its duty cycle. Maximum continuous operating voltage rating MCOV (kVrms) The maximum designated power frequency voltage that may be applied continuously between the terminals of the arrester. Ordinary operating voltage V (line-to-ground kVrms) The 𝜈 – i characteristics should exceed the point a (the guaranted reference voltage) The 𝜈 – i characteristics should be below the point b (the guaranted discharge (residual) voltage)

(a) Arrester 𝜈 – i characteristic curve TOV (pu value of the duty cycle voltage rating) 1.25 1.20 1.15 1.10 1.05 1.00

ZnO elements Courtesy of Toshiba

0.95 0.1

1

10

100

1000

Permissible duration (sec) (b) TOV capability

Figure 17.7 v–i characteristics of arrester.

10000

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17 Insulation Coordination

Taking all this into consideration, it is concluded that, for insulation coordination of a substation and its equipment:

• •

To protect again a lightning surge to the station, it is necessary: – To limit the station surge voltage within the presumed largest residual voltages of the adopted arresters – To realize the withstanding insulation level of the station equipment against the largest discharge voltage (residual voltage) for the largest surge current For power-frequency voltages, the arrester should withstand the thermal stress of continuous leakage current caused by MCOV, TOV, and switching surge energy.

Figure 17.7b shows the typical TOV capability of arresters. Of course, the arresters have to withstand the thermal stress caused by 50/60 Hz MCOV, TOV, and switching surges. Note that the IEEE term discharge voltage and the IEC term residual voltage are synonyms, although the definitions are slightly different. The definition for arrester discharge voltage in the IEEE standard is “the voltage that appears across the terminals of an arrester during the passage of discharge current.” The definition for the residual voltage of an arrester in the IEC standard is “peak value of voltage that appears between the terminals of an arrester during the passage of discharge current.”

17.8.3

Arrester Ratings, Classification, and Selection

Now that we have looked at the fundamentals of arresters, we will examine the practices of arrester ratings and selecting arresters based on the IEC and IEEE standards. First, we will list the major defined terms for arrester voltages and currents according to the IEC and IEEE standards:

•

•

IEC 60099–4 standard for surge arresters: – Rated voltage: Maximum permissible RMS value of power-frequency voltage between its terminals at which it is designed to operate correctly under TOV conditions as established in the operating duty tests (Ur) – Continuous operating voltage: Designed permissible RMS value of power-frequency voltage that is applied continuously between the arrester terminals in accordance with 8.5 operating duty tests (Uc) – Reference voltage: Peak value of power-frequency voltage divided by 2 that is applied to the arrester to obtain the reference current (Uref) – Reference current: Peak value (the higher peak value of the two polarities if the current is asymmetrical) of the resistive component of a power-frequency current used to determine the reference voltage of the arrester (Iref) – Residual voltage: Peak value of voltage that appears between the terminals of an arrester during the passage of discharge current. (Ures) – Discharge current: Impulse current which flows through the arrester. IEEE C62.11 standard for surge arresters: – Duty-cycle voltage rating: The designated maximum permissible power-frequency voltage (RMS value) between the line and ground terminals at which an arrester is designed to perform its duty cycle. The duty-cycle voltage is used as a reference parameter for the specification of operating characteristics. – Maximum continuous operating voltage rating (MCOV rating): The maximum designed RMS value for powerfrequency voltage that is applied continuously between the terminals of the arrester. – TOV rating: The voltage the arrester is capable of operating for limited periods of time at power-frequency voltages above their MCOV rating. – Discharge voltage: The voltage that appears across the terminals of an arrester during the passage of discharge current. – Reference voltage (Uref ) and reference current (Iref ): The IEEE and IEC definitions are the same.

The terms are slightly different for the standards but the essential concepts are almost the same, as we will explain next. The IEEE duty-cycle voltage rating is the rated voltage of the arrester with RMS kV values, selected to exceed k times MCOV rating (duty-cycle voltage ≥k MCOV where k = 1.24 is the reserved coefficient and is selected from a series of integers in multiples of 3). The IEC rated voltages for an arrester are the RMS kV values that exceed all possible TOV and

17.8 Overvoltage Protection with Arresters at Substations

MCOV. These terms have almost the same meaning: an appropriately rated arrester voltage should be selected to exceed TOC and MCOV with a safety margin. The meaning of the arrester’s rated voltages is obviously different from that of other equipment such as generators, transformers, breakers, and system voltages. Needless to say, all voltages and currents for an arrester are based on lineto-ground phase quantities. Now let’s examine Figure 17.7a, which shows the essential v–i characteristics for an arrester including two guaranteed points a and b from the supplier. Point a is the reference voltage and current (va (kVrms), ia (mArms)), indicated by the supplier. It should be guaranteed that the v–i characteristics of an individual arrester at the reference current ia (typically 1 mA) exceed the guaranteed reference voltage va, and the arrester must withstand the thermal energy of the guaranteed power-frequency continuous current ia. Point b is the standard nominal discharge voltage and current (vb (kVcrest), ib(kAcrest)). The standard nominal current ib is specified with standard values such as 1.5, 2.5, 5, 10, 20, 40 kA for the purpose of arrester classification and to guarantee the discharge voltage characteristics. The discharge voltage of an individual arrester should be smaller than the guaranteed standard nominal discharge voltage vb at the standard nominal discharging current ib (tested typically with the 8/20 μs standard impulse wave current). By the way, the Discharge voltage by IEEE and the Residual voltage by IEC have the same meaning. The discharge voltage of an arrester should be type tested using a 8 × 20 μs standard test wave voltage, and it should be guaranteed that the measured voltage always indicates values less than the nominal discharge voltage Ures. In practical engineering, individual arresters should be selected so the guaranteed reference voltage (or the duty-cycle voltage) exceeds the MCOV or the maximum TOV (Urp). High-voltage arresters of EHV/UHV classes are also assigned a switching surge durability by the standards for the arresters: generally, the thermal energy absorbing capability (kJ) at the specified discharge current is type tested. The guaranteed value can be additionally written as point c in Figure 17.7a. With regard to switching surges, point c can be written as the guaranteed standard nominal discharge voltage and current (vc (kVcrest), ic (kAcrest)) for the switching surges. The standard nominal currents for the switching surges are specified as standard values such as 0.5, 1, 2 kAcrest. (A detailed description of the arrester’s switching surge durability is omitted.)

17.8.4

Separation Effects of Station Arresters

With regard to surge phenomena, the induced time-changing overvoltage of an arbitrary point is different from that of any other point in the same substation. Accordingly, the voltage at the transformer terminal (or at any other equipment) is different from that of the station arrester terminal. Generally, we need to consider that the voltage at the protected insulation may possibly be higher than that at the arrester terminals due to the traveling distance on connecting leads and the conductor circuit. This increase in voltage is called the arrester separation effect. This effect obviously lessens the surge-protection performance of arresters. Referring to Figure 17.8a, the separation effect can be explained as the behavior of traveling waves and is linked to (i) the increasing rate of rise of the incoming surge μ (kV/μs), (ii) the distance l between the arrester and under-protective equipment (a transformer), and (iii) the reflection factor ρtr of the equipment. The phenomena can be roughly calculated by the trial equation below, referring to Figures 17.8a, b, and c. Imagine that the transmitted lightning surge voltage at the arrester point with time front Tfront ≒ 1.2 μs and the initial steepness of α (kV/μs typically 200–500 kV/μs) appears at t = 0, so the voltage during the initial small time interval of 0 < t < Tfront (≒ 1.2 μ s) can be written Var(t) = α t(kv). The surge voltage begins to travel to the transformer terminal (distance l) at t = 0 and arrives at t = l/u ≡ T, so the surge voltage at the transformer terminal (reflection factor ρtr) appears at t = l/u (that is, t = t − l/u = 0) as the voltage form Vtr(t ) = (1 + ρtr) α t (kV), where attenuation is ignored. Numerical check: Assuming l = 60 m, u = 300 m/μs, T] = 60/300 = 0.2 μs for one-way travel. Z1 = 300 Ω (for station conductors), and Z2 = 5000 Ω (for a transformer): ρtr = 5000 − 300 Var t = α t kV

5000 + 300 = 0 9 reflection factor

for 0 < t < 2T = 0 4 μs

vtr t' = 1 + ρtr α t' kV = 1 9α t' kV

17 2 for 0 < t' < 2T = 0 4 μs

497

498

17 Insulation Coordination

Ztr

Surge impedance

Zl

Z

ρtr

0.9 – 1.0

ρar

𝜈tr

G

𝜈o

𝜈ar

tr ar l

Arrester

(a) [kV] 1000 𝜈o : original surge voltage

800 600 𝜈tr

400

𝜈ar : arrester terminal voltage 𝜈tr : transformer terminal voltage

𝜈ar

200 0 0

1

2

3

4 [μs]

5

6

7

8

(b) Surge voltages at arrester point and transformer terminals (simulation) (c) Typical values of surge impedances Overhead lines Cable lines Transformers Rotating machinery

: : : :

300–500Ω 20–60Ω 1000–10000Ω 500–1500Ω

Figure 17.8 Separation effects of arresters.

The equation shows that the transformer terminal voltage Vtr(t ) built up by 1 + ρtr = 1.9 times the steepness of the arrester terminal voltage Var(tr) during the initial time up to 2 T = 0.4 μs. Furthermore, Var continues to increase in magnitude until the interval t = Tfront = 1.2 μs and reaches the maximum residual voltage, so Vtr (t ) also continues to increase in magnitude for the interval of the wave front 1.2 μs, while the waveform becomes oscillatory after t > 2 T = 0.4 μs due to the negative reflection waves that return from the arrester point. As a result, the transformer terminal voltage Vtr(t) is a larger magnitude than the arrester terminal voltage Var and is oscillatory and almost twice as steep. These results indicate that the overvoltage stress to the transformer is more severe than the arrester’s protective level due to the separation effect from large distance l or large traveling time T = l/u. On the other hand, if distance l is small, such severe stress does not appear at the transformer terminal, because the negative reflected waves return quickly from the arrester point, and Vtr and Var are almost the same.

17.9 Station Protection Using OGWs and Reduced Grounding Resistance

In practical engineering, arresters installed very close to the junction tower of a transmission line are important the gateway barriers to protect entire substations against lightning surges from the transmission lines. However, these arresters cannot protect the transformers or other facilities properly because of the separation effects. This is why important transformers and other equipment (including cables) at large stations should be protected by bespoke arresters installed nearby. Such arresters for individual transformers are also very effective at reducing overvoltage stresses caused by repeated switching surges or rare cases of direct lightning strikes on the station. Figure 17.9 shows a typical example of GIS for a 500 kV outdoor substation with a double-bus system, with onepoint-breaking breakers. Arresters are installed at each main transformer terminal and at transmission-line feeding points.

17.9

Station Protection Using OGWs and Reduced Grounding Resistance

17.9.1

Direct Lightning Strikes on Substations

Surge arresters are generally installed at the gateway point of the substation, very close to the first tower of each feeding transmission line, and protect the substation against impinging lightning surges from transmission lines. However, direct lightning strikes on the substation are possible. We cannot prevent such occasions, but we should minimize the probability of a strike on the station, or at least protect against internal insulation failure (inner insulation failure) as much as possible. A direct lightning strike on a substation causes more severe results than a lightning strike on a transmission line, but the probability of occurrence is smaller. When lightning strikes a transmission line, first, arcing horns limit the magnitude of the surge voltage. Second, attenuation of the traveling waves before arrival at the substation is expected in most cases. The arresters installed at the tower junction point and other points in the substation will appropriately protect the substation and station apparatus. A direct strike on the substation is different:

• •• •

It is impossible to protect against external (air) insulation failure. Furthermore, the striking point cannot be anticipated or limited, because the physical configuration and electric circuits (distribution of surge impedance, for example) are complicated. Attenuation through long-line travel cannot be expected. Transition points and surge impedances exist within a narrow area, so unexpected voltage enlargement such as the arrester separation effect may occur. The surge energy may exceed the arrester’s duty, which means cascade failure of the arrester or insulation parts of other equipment.

17.9.2

OGWs in the Station Area

The major purpose of OGWs in the area of the substation is to reduce the probability of a direct lightning strike on the substation as much as possible, although they cannot be a perfect countermeasure to prevent a direct strike on the conductors. The OGWs can reduce the probability of a direct strike on the conductors by providing a shielding effect. Ample OGWs can also reduce the probability of inverse flashover, because the surge impedance is also reduced or the stuck current is bypassed.

17.9.3

Reduction of Station Grounding Resistance and Surge Impedance

The reduction of the surge impedance at the substation is a vitally important countermeasure to prevent or reduce every kind of surge stress, in particular direct lightning strikes on the substation and switching surges and traveling surges from the lines. The reduction of station surge impedance is the most important step to reduce the surge impedance Z2 in Figure 17.4 against incidental traveling surges, for example.

499

500

17 Insulation Coordination

Courtesy of TEPCO

Arr

DS for feeder line (straight type) Gas-bushing

CT PD(capacitive-divider type) SF6 gas breaker 500 kV, 8000 A, 50 kA (one-point breaking)

Detachable structure

Grounding switch High-duty arrester

spacer

Breakerdriving structure

Courtesy of Toshiba

Bus A

Bus B

DS (Disconnecting Switch) for bus

Detachable structure Figure 17.9 GIS substation (500 kV, 8000 A, 63 kA, double-bus system).

Numerical check: Assuming Z1 = 300 Ω as the line surge impedance, then the arrester terminal voltages are 2Z2 E = 0 5 E for Z2 = 100Ω Z1 + Z2 = 0 28 E for Z2 = 50Ω = 0 18 E for Z2 = 30Ω

17 3a

17.10 Insulation Coordination Details

This check clearly indicates the significant effect of reducing surge impedance at the station. In the case of a direct strike on the OGW in the substation, we can apply Eq. (16.7), referring to Figure 16.11 Reff =

1 1 1 1 + + Z0 ZOGW Z2

17 3b

where Z0 : surge impedance of the lightning surge (Z0 ≒ 400 Ω) ZOGW : total equivalent surge impedance of all the OGW wires Z2 : surge impedance of the substation Therefore, Reff can be reduced by a reduction in the station surge impedance Z2, and accordingly back-flashover is also reduced. The resistive ohm value of the substation ground system should also be kept within a specified value for human safety. If ground resistance R = 1 Ω and Ig = 1000 A are assumed, the induced voltage on the earth conductor could be Vg = 1 kV, which is too large for the human body to withstand. Typical practices for the ground system in substations involve ground pilings or pipes/rods, ground conductors, ground meshes, ground mats, or a combination of these, and the resistive ohm values are designed to stay within specified values, say 0.5–1 Ω, by which the surge impedance of the station is also reduced. Some apparatus is also used in practical engineering where the external insulation (mostly the bushings) is designed to be a little weaker than the internal insulation in order to avoid internal faults.

17.10

Insulation Coordination Details

We have reached the stage where we can study the details of insulation coordination, taking the previous discussion into account, and including material from previous chapters. Insulation is the essential technology to enclose electricity or electrical power within a huge container called a powersystem network. The process of insulation coordination is the result of well-balanced viewpoints including electrical and material physics, practical engineering, technical history, engineering economy, environmental adaptation, and social needs. It is now concentrated in the recognized international standards of the IEC and IEEE. Both standards are essentially the same, with slightly different terminologies and table values. 17.10.1

Definitions and Principal Topics in the Standards

The definitions of insulation coordination in the IEEE and IEC standards are as follows:

• •

Selecting insulation strength consistent with expected overvoltages to obtain an acceptable risk of failure (IEEE 1313.1-1996, standard for insulation coordination). Selecting the dielectric strength of equipment in relation to the voltages which can appear on the system for which the equipment is intended and taking into account the service environment and the characteristics of the available protective devices (IEC 71-1, 1993, insulation coordination).

The meanings are the same, although the expressions are different. Needless to say, any other national standards will have definitions with the same meanings. Table 17.1 summarizes the criteria with regard to the principal goals of insulation coordination that are recognized worldwide, and is a more concrete expression of these definitions. Standards are a kind of engineering consensus or practical policy guidelines for industrial applications, substantially based on expert detailed theories and technical facts obtained from a great deal of engineering experience with practical application in the field. This is why all the standards with regard to power-system insulation are almost identical. Readers should appreciate our intention to introduce the worldwide consensus and its technical background instead of quoting from the IEEE or IEC standards, although the terminology and the figures come primarily from these two representative standards.

501

Table 17.1 The basic aspects of insulation design criteria (general consensus). Overhead transmission lines

Switching overvoltages

Lightning overvoltages

Matters of policy

Major countermeasures

Principles for withstanding voltages

• •

•• •

Insulation failures are unavoidable but should be within a reasonable failure rate. Flashover passes should be limited through arcing horns so damage to conductors or insulators is avoided.

OGW Arcing horns Tower impedance reduction

Substation and substation equipment Temporary overvoltages (TOV)

Insulation failure should be entirely avoided without exception.

• •

Mainly dependent on the countermeasures at the substation. The gap lengths of arcing horns should be set to avoid discharge by the switching surges or by any TOVs.

The discharge probability function of self–restoring insulation based on a Gaussian cumulative frequency distribution is used. In typical practice, voltages with a normalized deviation 3σ value based on the 50% discharge voltages are applied.

Lightning overvoltages

• •

•• •

Insulation failures caused by lightning strikes to the overhead lines should be entirely avoided. Lightning strikes directly to stations are unavoidable, but countermeasures to minimize the probability of insulation failure, and in particular to protect internal insulation, are required as much as possible.

Switching overvoltages

Insulation failures should be entirely avoided.

Power cables must be entirely protected. OGW Arresters Station neutralgrounding system (meshes, counterpoise)

TOVs

•• •

Arresters (gapless) Breakers (resistive tripping/closing) Station neutralgrounding system (meshes, counterpoise)

• • • • •

Shunt reactors and capacitors Tap-changing transformers AVR, AQR, V–Q control (local) Load dispatching (total system) V–Q control (total system)

Station equipment must be guaranteed by the withstanding insulation levels specified by the authorized standards such as in Tables 17.2a, b and c.

17.10 Insulation Coordination Details

17.10.2

Insulation Configuration

Some important definitions for insulation configuration from the IEEE and IEC standards are as follows:

• • • • • • • • • •

The complete geometric configuration of the insulation, including all elements (insulating and conducting) that influence its dielectric behavior. Examples of insulation configurations are phase-to-ground insulation, phase-to-phase insulation and longitudinal insulation (IEEE 1313, similarly IEC 71-1). Longitudinal insulation: An insulation configuration between terminals belonging to the same phase, but which are temporarily separated into two independently energized parts (open-switch device) (IEEE). An overvoltage that appears between the open contact of a switch (IEC). External insulation: The air insulation and the exposed surfaces of solid insulation of equipment, which are both subject to dielectric stresses of atmospheric and other external conditions such as contamination, humidity, vermin, etc. (IEEE). The distances in atmospheric air, and the surfaces in contact with atmospheric air of solid insulation of the equipment which are subject to dielectric stresses and to the effects of atmospheric and other external conditions, such as pollution, humidity, vermin, etc. (IEC). Internal insulation: Internal insulation comprises the internal solid, liquid, or gaseous elements of the insulation of equipment, which are protected from the effects of atmospheric and other external conditions such as contamination, humidity, and vermin (IEEE, IEC similarly). Self-restoring insulation: Insulation that completely recovers its insulating properties after a disruptive discharge caused by the application of a test voltage; insulation of this kind is generally, but necessarily, external insulation (IEEE). Insulation which completely recovers its insulating properties after a disruptive discharge (IEC). Non-self-restoring insulation: An insulation that loses its insulating properties or does not recover them completely after disruptive discharge caused by the application of a test voltage; insulation of this kind is generally, but not necessarily, internal insulation (IEEE). Insulation which loses its insulating properties, or does not recover them completely, after a disruptive discharge (IEC).

17.10.3

Insulation Withstanding Level and BIL, BSL

We have studied in the previous sections kinds of overvoltages and the countermeasures to mitigate them or to protect insulation, in particular to protect internal insulation. Taking all this into consideration, we will introduce insulation strength and insulation-withstanding level, and, furthermore, provide clear guidelines for the latter: standard withstand voltages and standard insulation levels. The guideline criteria must be able to offer the important measures for insulation strength and withstanding level at least for the following four important engineering procedures: a) Measures to determine or to select entire levels of insulation strength for transmission lines and substations belonging to the same operating voltages of the individual power system. b) Measures to mitigate possible overvoltage stresses within the selected insulation levels. c) Measures to specify the required insulation strength of individual equipment or facilities. d) Measures to prove the specified insulation-withstanding strength by which required insulation levels of equipment or facilities can be tested and guaranteed. The insulation strength or insulation withstanding voltage levels are expressed in terms of three representative categories of overvoltages: basic lightning impulse insulation level (BIL), basic switching impulse insulation level (BSL), and the highest power-frequency voltages, as principal aspects of insulation coordination. The IEEE definitions of BIL and BSL are as follows:

503

504

17 Insulation Coordination

• • • •

BIL: The electrical strength of insulation expressed in terms of the crest value of a standard lightning impulse under standard atmospheric conditions. BIL is expressed as either statistical or conventional. BSL: The electrical strength of insulation expressed in terms of the crest value of a standard switching impulse. BSL is expressed as either statistical or conventional. Conventional BIL: The crest value of a standard lightning impulse for which the insulation shall not exhibit disruptive discharge when subjected to a specific number of applications of this impulse under specified conditions, applicable specifically to non-self-restoring insulations. Conventional BSL: The crest value of a standard switching impulse for which the insulation does not exhibit disruptive discharge when subjected to a specific number of impulses under specified conditions, applicable to non-selfrestoring insulations.

Let’s comment parenthetically about the meaning of conventional. Disruptive discharge or breakdown phenomena are probabilistic phenomena with regard to the measured breakdown voltages of a specified insulation being considered to be distributed by the Gaussian distribution curve, but equipment with non-self-restoring insulation cannot be tested many times. Therefore, if equipment successfully clear the overvoltage test a few times, it is recognized that the equipment has the capability of withstanding the specified test voltage. This is the meaning of conventional. We will also comment on the withstanding voltage test of an arc horn shown in Table 17.1. An arc horn has selfrestoring insulation, and the insulation level should be the weakest among the overhead transmission lines. Therefore the 3σ low-side voltage value of 50% flash-over voltage is typically assigned as the withstanding voltage value. 17.10.4

Standard Insulation Levels and Principles

Tables 17.2a–c show the standard withstand voltages for power systems specified by IEEE 1313 and IEC 71-1. The IEEE and IEC standards are divided into the following parts: a) IEEE, for the specified standard highest voltages: Class I: Medium (1–72.5 kV) and high (72.5–242 kV) voltages Class II: EHV and UHV: Above 242 kV b) IEC, for specified highest voltages for equipment: Range I: Above 1–245 kV Range II: Above 245 kV

•• ••

The contents and background reasoning of both standards are basically the same, although the terminologies are slightly different. Therefore, we have combined IEEE Class I and IEC Range I in Table 17.2a. Tables 17.2b and c are the standards for voltages above 245 kV. The standard withstanding BIL and BSL for equipment must also be proved by overvoltage tests with the same specified wave shapes. The IEC and IEEE definitions for the standard wave shapes for BIL and BSL tests are the same, and are given in Table 17.2c:

• •

Standard lightning impulse: 1.2/50 μs, with the wave shape having a time to peak of 1.2 μs and a time to half-value of 50 μs. Standard switching impulse: 250/2500 μs, with the wave shape having a time to peak of 250 μs and a time to half-value of 2500 μs.

In Table 17.2a, for systems under 245 kV, standard short-duration power-frequency withstand voltages and standard lightning impulse withstand voltages are assigned. On the other hand, in Tables 17.2b and c, for systems over 245 kV, standard switching impulse withstand voltages are assigned instead of short-duration power-frequency withstand voltages. The reason is explained next. 17.10.5

Insulation Levels for Power Systems under 245 kV (Table 17.2)

There is a tendency for TOVs to be high. In particular, many lower-voltage systems are non-effective neutral-grounding systems, so the ground–fault factor k (the ratio of the highest power-frequency voltage on an unfaulted phase during a LG fault to the phase-to-ground power-frequency voltage without the fault) is at least 1.7 or higher. Therefore, a lowfrequency (=power-frequency), short-duration withstand voltage test with large voltage is essential. Furthermore,

17.10 Insulation Coordination Details

Table 17.2a IEC 71-1 and IEEE 1313: Standard withstand voltages for power systems of up to 245 kV.

IEC 71-1, 1993-12 Insulation coordination Part I: Definitions, principles and rules, Range 1: 1 kV < Um ≤ 245 kV Highest voltage for equipment Um

Standard short-duration power frequency withstand voltage

Standard lightning impulse withstand voltage

IEEE 1313,1-1996 Insulation Coordination – Definitions, principles and rules, Class 1: 15 kV < Vm ≤ 242kV Maximum system voltage (phase-to-phase) Vm

Low-frequency, shortduration withstand voltage (phase-toground)

Basic lightning impulse insulation level (phaseto-phase) BIL

kV, rms value

kV, crest value

kV, rms value IEC

IEEE

IEC

IEEE

IEC

3.6

10

20, 40

7.2

20

40, 60

12

28

60, 75, 95

IEEE

17.5

15

38

34

75, 95

95, 110

24

26.2

50

50

95, 125, 145

150

36

36.2

70

70

145, 170

200

52

48.3

95

95

250

250

72.5

72.5

140

95

325

250

123

121

140

350

140

350

(185)

185

450

450

230

230

550

550

(185) 145

145

(450)

230

230

550

450

275

275

650

550

325

170

245

169

242

650

(230)

230

(550)

550

275

275

650

650

325

325

750

750

(275)

275

(650)

650

(325)

325

(750)

750

360

360

850

825

395

395

950

900

460

480

1050

950, 1050

505

Table 17.2b IEC 71-1: Standard withstand voltages for power systems, range 2: Um > 245 kV; standard insulation levels for range 2 (60071-1 Amend. 1 IEC 2010). Standard switching impulse withstand voltage Highest voltage for equipment Um kV (RMS value)

Longitudinal insulationa kV (peak value)

Phase-to-earth kV (peak value)

Phase-to-phase (ratio to the phase-to-earth peak value)

Standard lightning impulse withstand voltageb kV (peak value)

850 750

750

1.50 950

300c 950 750

850

1.50 1050 950

850

850

1.50 1050

362 1050 850

950

1.50 1175 1050

850

850

1.60 1175 1175

420

950

950

1.50 1300 1300

950

1050

1.50 1425 1175

950

950

1.70 1300 1300

550

950

1050

1.60 1425

950

1425 1050

1.50

1050

1550 1675

1175

1300

1.70 1800 1800

800

1175

1425d

1.70 1950

1175

1950 1550

1.60

1300

2100 1950

–

1425d

– 2100 2100

1100

1425

1550

1.70 2250 2250

1550

1875

1.65 2400

Table 17.2b (Continued) 2400 1675

1800

1.60 2550 2100

1550

1675

1.70 2250 2250

1675

1200

1800

1.65 2400 2550

1800

1950

1.60 2700

a

Value of the impulse voltage component of the relevant combined test while the peak value of the power-frequency component of opposite polarity is Um × 2 3. b These values apply for phase-to-earth and phase-to-phase insulation as well; for longitudinal insulation, they apply as the standard rated lightning impulse component of the combined standard rate withstand voltage, while the peak value of the power-frequency component of opposite polarity is 0 7 × Um × 2 3. c This Um is a non-preferred value in IEC60038. d This value is only applicable to the phase-to-earth insulation of single-phase equipment not exposed to air.

Table 17.2c IEEE 1313: Standard withstand voltages for power systems of over 242 kV, Class 2: Vm > 242 kV. Maximum system voltage (phase-tophase) Vm kV, RMS

Basic switching impulse insulation level (phase-to-ground) BSL kV, peak

Basic lightning impulse insulation level (phase-to-ground) BIL kV, peak

362

650 750 825 900 975 1050

900 975 1050 1175 1300

550

1300 1425 1550 1675 1800

1175 1300 1425 1550

800

1300 1425 1550 1675 1800

1800 1925 2050

Tf : the time-to-crest value (virtual time) Tt : the time-to-half value (virtual time) Voltage[%]

Voltage [%] Tf = 1.2 [μs] Tt = 50 [μs]

100

50

0

P

100

Tf = 250 [μs] Tt = 2500 [μs]

50

Tf

0 Tt

Standard lightning impulse test voltage 1.2/50 impulse

Time [μs]

Tf

Tt

Standard switching impulse test voltage 250/2500 impulse

Time [μs]

508

17 Insulation Coordination

especially in systems under 36 kV, temporary and irregular electrical conditions (e.g. neutral terminal opening, conductor crossing from higher-voltage systems) and mechanical damage to network facilities are apt to occur, so a reasonable margin for AC voltages is required for social and security purposes. These are the reasons the standard short-duration power-frequency withstand levels are assigned relatively large values. Switching impulse withstand levels (BSL) have no meaning in this class because they can be covered by BIL. Next, the ratio of BIL/{ 2 3 Um} for this voltage class is larger than that for the higher-voltage class in Tables 17.2b and 17.2c. Numerical check of BIL/{ 2 3 Um}: Table 17 2a

Tables 17 2b, c

200

2

3 36 = 6 80, 325

850

2

3 245 = 4 25

850

2

3 300 = 3 47, 1300

1800

2

2

3 72 5 = 5 49,

2

3 525 = 3 03,

3 800 = 2 76

In other words, BILs of Class 1 systems are relatively high, so BSLs must be relatively high. On the other hand, switching overvoltages are obviously proportional to the system operating voltages, so switching surges are lower for systems in this class. These are the reasons BSL is omitted for the lower-voltage systems in Class 1.

17.10.6

Insulation Levels for Power Systems over 245 kV (Tables 17.2b and 17.2c)

BSL is strictly assigned to this class. The reason can be explained by the inverse of the second item in the previous section. That is, first, BIL has been considerably reduced (due to the advance of arrester properties in particular) in EHV and UHV. Second, switching surges are proportional to operating voltages and so are relatively large in comparison with insulation levels. These are the reasons BSL must be assigned important items. In IEC 71–1, phase-to-phase and longitudinal BSL and phase-to-earth BSL are also assigned. Low-frequency, short-duration withstand voltage is omitted, because, nowadays, all EHV and UHV class systems are treated under the solidly neutral-grounding method (without exception), so the ground fault factor is small. Then, equipment that covers BIL and BSL withstand levels can withstand temporary overvoltage of about k = 1.5. This is the reason the low-frequency, short-duration withstand voltage test is omitted. Numerical check: BSL phase-to-earth 2 3 Um for IEC values: i BIL

2

3 Um

ii BSL phase-to-earth

2

3 Um

300 kV

850 −1050

550 kV

1175 − 1550

2

3 550 = 2 61 −3 45,

950− 1175

800 kV

1670 − 2100

2

3 800 = 2 56 −3 21,

1300− 1550

2

3 800 = 1 99 − 2 37,

1550− 1875

2

3 1100 = 1 73 −2 09,

1100 kV

1950 − 2400

2

3 300 = 3 47− 4 29,

2

3 1100 = 2 17 − 2 67,

750− 850

3 300 = 3 06 −3 48,

2 2

3 525 = 2 22 − 2 74,

This leads to the following:

• • •

The BIL ratio to the operating voltage is remarkably low for EHV and UHV systems, which means the reduction of the insulation level has been achieved; this owes much to advanced arrester technology to lower the protective levels. The BSL ratio is very close to the BIL ratio. Accordingly, the following points have to be carefully confirmed as essential for EHV and UHV systems: – EHV and UHV arresters are required to limit the lightning impulse level to lower values, while withstanding the large switching surge (the capability to absorb large switching surge energy). – EHV and UHV breakers are required to limit the switching surges to within specified levels. For example, UHV breakers of 500 kV class are required to limit BSL to within typically 2.5, and for this purpose breakers with resistive closing and/or resistive tripping mechanism are used. Of course, all the other EHV and UHV station equipment must guarantee the insulation levels specified in Tables 17.2b and 17.2c, including BSL.

17.11 Transfer Surge Voltages through Transformers, and Generator Protection

17.10.7

Evaluating the Degree of Insulation Coordination

The degree of coordination can be measured or evaluated by the protective ratio (PR) (IEEE PC62.22). That is, PR = insulation withstanding level

voltage at protected equipment

or the ratio of the insulation strength of the protected equipment to the overvoltages appearing across the insulation. The voltage at protected equipment is equal to arrester protective level, if the separation effect is insignificant: PRl = BIL LPL ratio for BIL acceptable level 1 15 where LPL is the lightning impulse protective level, and PRs = BSL SPL ratio for BIL acceptable level 1 2 where SPL is the switching impulse protective level The denominators of these ratios are the expected largest possible voltages. The insulation coordination of a power system can be evaluated by deriving the PR of individual lines and equipment using detailed overvoltage analysis. Note that, in addition to BIL and BSL, a chopped-wave overvoltage test is additionally introduced as an optional requirement for special conditions. The standard chopped-wave impulse voltage shape is a lightning impulse that is intentionally interrupted at the tail by sparkover of a gap or other chopping equipment. Usually the time to chop is 2–3 μs. (Further details are omitted because the matter is beyond the scope of this book.) All transmission lines, station facilities, and equipment have to be designed to satisfy the applied standard insulation levels. For this purpose, breakdown voltage characteristics under conditions are required as essential data regarding insulation characteristics. The breakdown characteristics of any insulation materials (air, gas, oil, paper, porcelain, etc.) are affected by the shapes of electrodes and the atmosphere (pressure, humidity, temperature, etc.), so a huge amount of valuable experimental data has been accumulated over time. Figure 17.10 is an example showing the breakdown characteristics of air gaps. 17.10.8

Insulation of Power Cables

Power cables contain non-self-restoring insulation, so any breakdown of cable insulation requires extensive outage time for repairs, at a high cost. Therefore, insulation failure of power cables for network lines and for power station use should be avoided. Cable circuits have a low surge impedance of typically 40–50 Ω, so surges from overhead lines are reduced significantly at the line–cable junction. However, switching surges originating in the vicinity of cable lines at a substation are significantly reflected at the junction point to the transmission lines (see Section 15.7.3). Switching-surge phenomena in the substation area are generally complicated. Metal–oxide gapless arresters can provide excellent cable protection, but the arrester is required to absorb the relatively large thermal energy 12 CV 2 caused by the high frequency (HF) oscillatory overvoltages. Regardless, insulation coordination of the power-system network, including cable lines of significant capacitance, have to be carefully examined. The overvoltage behavior of cable lines is also discussed in Sections 15.14–15.17.

17.11

Transfer Surge Voltages through Transformers, and Generator Protection

If lightning surges or switching surges flow into the high-voltage (HV) bushing terminal of a transformer, they put serious stress on the transformer windings. Furthermore, an induced transfer overvoltage appears at the low-voltage (LV) bushing terminal and threatens the insulation of all the lower-voltage-side equipment. This is a serious problem in lowtension-side insulation coordination. In particular, in the case of a generating plant, the rated voltage of the LV side of the generator terminals are relatively low (say, 10–30 kV) (in other words, a large transformation ratio), so the insulation level is low in comparison with that of the HV side. Appropriate countermeasures to protect generators and other lowvoltage equipment are therefore required. Problems of this sort are usually treated as matters of individual insulation for the plant’s internal equipment. In other words, well-arranged insulation among transmission lines and substations does not automatically guarantee the insulation coordination of the low-voltage plant-side circuit of the in-house factory loads or of the generators and the auxiliary circuit, for example. Insulation coordination of the in-house network is separate from the transmission network side. Given this viewpoint, it is essential to carefully study impinging transfer surge voltages that come through main-transformers and prepare the necessary protective countermeasures.

509

152 cm 125 cm 107 cm 88 cm 70 cm 52 cm 34.5 cm

4.1 cm

26 cm 18 cm 11.5 cm 8.4 cm 5.6 cm

+ (1.5 x40) µs wave 760 mmHg atmospheric pressure 25ºC temperature 15g/m3 humidity

0 2 4 6 8 10 12 14 16 18 20 22 24 26 Volt-time characteristic [μs] 0 20 40 60 80 100 120 140 160 180 200 220 240 260

cm

0.5 1 2 3 4 6 8 16 volt-time characteristic [µs]

179 cm 152 cm

Volt-time characteristics [μ s]

269

m

179 cm

Flashover voltage [Mv]

cm

0.5 1 2 3 4 6 8 16 volt-time characteristic [µs]

2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

c 224

9 26

cm 224

2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Volt-time characteristics [μ s]

17 Insulation Coordination

Flashover voltage [Mv]

510

125 cm 107 cm 88 cm 70 cm 52 cm 34.5 cm 26 cm

4.1 cm

18 cm 11.5 cm 8.4 cm 5.6 cm

+ (1.5 x40) µs wave 760 mmHg atmospheric pressure 25ºC temperature 15g/m3 humidity

0 2 4 6 8 10 12 14 16 18 20 22 24 26 Volt-time characteristic [μs] 0 20 40 60 80 100 120 140 160 180 200 220 240 260

Gap length [cm]

Gap length [cm]

(a) Positive polarity

(b) Negative polarity

(c)

Note: If the time to crest is shorter, the sparkover voltage increases as the wave front is made steeper. This variation of the sparkover voltage of gaps with the time to crest of the applied voltage or steepness of the front of the wave is the volt-time characteristics. Figure 17.10 Discharge characteristics: rod-to-rod gap in atmosphere. Source: Courtesy of Dr. K. Takuma.

17.11.1

Electrostatic Transfer Surge Voltage (Single-Phase Transformers)

The transformer we studied in Chapter 6 had a power frequency based on inductance L of the coil windings, ignoring stray capacitances C. We need to consider that the behavior of the transformer in the surge-frequency zone is dominated by stray capacitance C instead of inductance L of the coil windings. Accordingly, we need to establish an equivalent circuit for the transformer based on capacitance C.

17.11 Transfer Surge Voltages through Transformers, and Generator Protection

E

Point n : non-grounded

C2

S1 n

a

b k

k x

C12 Emax 𝛼E E 𝛼E

n

n

Point n : earth-grounded

m

Coil length x

LT-coil

m

HT-coil m

S2 (a)

(b)

(c)

Figure 17.11 Equivalent circuit of a transformer for high-frequency phenomena.

Figure 17.11a shows a typical simplified circuit model of a single-phase transformer for high-frequency phenomena. This is a distributed circuit with elemental capacitances from the high-tension (HT) coil to the low-tension (LT) coil and from the LT coil to grounding earth, where the capacitances from the HT coil to grounding earth are not written because they can be treated as part of the connected HT outer circuit. As we are not applying powerful computers in this book, we need to simplify Figures 17.11a and b where the distributed capacitances are concentrated by C12 (total capacitance from the HT to LT coil) and C2 (total capacitance from the LT coil to grounded steel-core). Then we examine two different cases in Figure 17.11b:

• •

Circuit 1 (switch S1 closed, S2 open): Terminals m and n are connected, while the earth terminal is open. In the figure, scale x shows the distributed coil position being measured from point n to point m. When the surge voltage traveling from the HT line arrives at the connected points m and n with magnitude E, the surges then rush into the primary coil simultaneously from m and n and frontally meet point k so the voltage at k obtains the largest value Emax (= 2E if attenuation is ignored). The initial voltage distribution in the HT coil is shown as curve a in Figure 17.11c. The averaged value of the distributed voltage can be written E = αE, where α = 1.4–1.6 > 1.0. Circuit 2 (switch S1 open, S2 closed): Terminal n is connected only to the earth, and surge E is injected at point m. Surge voltage E travels through the HT coil from point m to n, so the initial voltage distribution is shown as curve b in Figure 17.11c. The averaged value of the distributed voltage can be written E = αE, where α = 0.5–0.7 < 1.0. As a result, we can presume that the averaged voltage αE is injected at point k of circuits 1 and 2 as follows: For circuit 1 α = 1 4− 1 6

17 4

For circuit 2 α = 0 5− 0 7

Our problem is shown in Figure 17.11a, where surge voltages Em(t) and En(t) arrive at terminals m and n simultaneously, and the equation for deriving the electrostatically induced voltage at the LT coil terminal is as follows: 1 vm

= Em

1 vn

= En

17 5a

Now we can transform the set voltages Em and En into the LG traveling wave E (t) and line-to-line traveling wave E (t), and Figure 17.12a can be divided into Figures 17.12b and c: Em = E + E En = E −E

line-to-ground traveling wave E = 1 Em + En 2 ① 1 E = Em −En line-to-line traveling wave 2

②

17 5b

For Figure 17.12b, we can quote the result of circuit 1: that is, voltage αE is charged at point k: 2v

=

C12 α C12 αE = Em + En 2 C12 + C2 C12 + C2

17 6a

For Figure 17.12c, voltages E and −E are injected at points m and n, respectively, so the frontal voltage at k becomes zero. In other words, this is a special case of α = 0 on circuit 2:

511

512

17 Insulation Coordination

Phase-to-ground travelling wave

En

αE′

E′

Em

Phase-to-phase travelling wave E′′ C12

C12

1𝜈m

2𝜈

k 1𝜈n

E′

2𝜈′

k

C12

0V

2𝜈′′

k

C2

C2

C2 –E′′

(a)

(b)

(c)

Figure 17.12 Transfer voltage from HT coil to LT coil.

2v

=

C12 0=0 C12 + C2

17 6b

Thus the solution of Figure 17.12a is derived as the addition of the results from Figures 17.12b and c: 2v = 2v

17.11.2

+ 2v =

C12 α C12 αE = Em + En 2 C12 + C2 C12 + C2

17 6c

Electrostatic Transfer Surge Voltage (Three-Phase Transformers)

Figure 17.13 is a typical connection diagram of a main transformer for thermal or hydro-generating plants. Surge voltages Ea(t), Eb(t), Ec(t) arrive simultaneously at the HT terminal bushings. Our problem is to calculate the transfer surge voltages induced by the generator-side LT terminal bushings. We can quote the result of Eq. (17.6) for the initial transfer surge voltage induced at the LT side: Phase-a 2 va =

α C12 E a + 1 vn 2 C12 + C2

Phase-b 2 vb =

α C12 Eb + 1 vn 2 C12 + C2

Phase-c 2 vc =

α C12 Ec + 1 vn 2 C12 + C2

17 7

where1vn : surge voltage at the neutral point n. Or, by symmetrical components,

C12

1va

2vc 2vb 1 vb 1vb

2 va

a′

n

1vc

b′ (Stray capacitances of phase-a only are indicated)

c′ C2

1 vn

Figure 17.13 Stray capacitance between HT- and LT-windings and steel core.

17.11 Transfer Surge Voltages through Transformers, and Generator Protection

2 v0

=

1 2 va + 2 vb + 2 vc 3

where α C12 E 0 + 1 vn 2 C12 + C2 E0 1 1 E 1 1 = α C12 3 E1 2 v1 = E2 1 2 C12 + C2 =

2 v2

=

1 a a2

1 a2 a

Ea Eb Ec

17 8

α C12 E2 2 C12 + C2

Equations (17.7) and (17.8) are the resulting general equations giving the transfer surge voltages appearing at the bushings of the LT side, while the LT bushing terminals are in the open condition. By applying these derived equations, the equations for calculating the transfer voltages under different terminal conditions can be derived, as summarized in Table 17.3. Regarding the derivation processes shown in Table 17.3, the neutral point n should be carefully treated as a surge transition point. The transmittal wave coefficient at point n is obviously different for each case. For example, the coefficient of the voltage vn at the neutral point n is 2 for case 5, 2/3 for case 6, 4/3 for case 7, and so on, as explained in the table. We can find the equation for calculating the transfer surge voltages appearing on the LT side under different conditions on the HT voltage side using this general equation and Table 17.3. The resulting equations for the seven cases can be summarized as the following general equation. The transfer surge voltage from the HT to the LT side (the LT side is in open mode) is 2 v = kα

C12 E C12 + C2

17 9

where α : given by Eq. (17.4) k : transfer coefficient given in Table 17.3 For example, for case 6 phase-a, k = 5/6, and for phase-b, k = 2/6, with kα

C12 transfer voltage ratio C12 + C2

Incidentally, the transfer voltages for each case with C12 = 4000 pF, C2 = 8000 pF, α = 0.5 (for the solidly neutralgrounding system), and α = 1.5 (for the neutral-ungrounded system) are shown in the table as supplemental references. In the case of the high-resistive neutral-grounding system, α is initially between 1.5 and 0.5, and soon (after the wavefront passes) becomes very close to 0.5. Let’s consider a trial calculation for a 275 kV class power station, with main transformer 275 kV/24 kV, y–Δ windings, and a solidly neutral-grounding system. Then C12 = 4000pF, C2 = 8000pF, α = 0 7, k =

1 2

From Eq. (17.9), α 4000 1 E = αE = 0 12E 17 10 2 4000 + 8000 6 Now, assuming a 1050 kV impulse surge appearing at the HT side of the transformer, the induced transfer voltage at the LT side terminal is 2 va = 16 × 0 7 × 1050 = 122 5kV , which is 6.25 times the power-frequency normal voltage 2 3 19.6 kV and almost exceeds the BIL value 95 or 125 kV by IEC. Therefore, special countermeasures to protect all LT-side equipment are required. We also need to recognize that each case in the table is often caused by real surge modes. In the case of lightning striking the connected transmission line, the incidental surges Ea, Eb, Ec come together at the station regardless of the fault phases, so they include equal components E that correspond to case 3 or 4 or 5, for example. Another example is switching surges caused by a breaker’s first pole closing, which corresponds to case 1. All the cases shown in Table 17.3 are realistic phenomena. 2 va

=

513

Table 17.3 Electrostatic transfer voltage (calculation formula). C12 = 4000pF

Surge voltage (conditions)

Case 1

Primary

Secondary

E

1va 1vn

a c

n b

b′ a

a′

c

= E, 1vb = 1vc = 0 =0

b

Transfer voltage from HT side to LT side (phase voltages, neutral voltage) v = kα × C 12C 12 + C2 E

α C12 E, 2 C12 + C2 2 vb = 2 vc = 0 α C12 ∴ 2 v0 = E 6 C12 + C2 2 va

Transfer voltage

α=0 5 05 = 2 05 2 v0 = 6

=

2 va

C2 = 8000pF C12 2 = C12 + C2 3

1 1 E= E 3 12 1 1 E= E 3 36

c′ k =1 2

Case 2 E

1va 1vn

= 1vc = E, 1vb = 0 =0

n

2 va

= 2 vc =

2 vb

=0

∴ 2 v0 =

α C12 E, 2 C12 + C2

α=0 5

α C12 E 3 C12 + C2

2 va

= 2 vc =

2 v0

=

1 E 12

1 E 18

k = 1 2, 1 3

Case 3

E

1va 1vn

= 1vb = 1vc = E =0

2 va

= 2 vb = 2 vc =

α C12 E, 2 C12 + C2

α=0 5

α C12 E ∴ 2 v0 = 2 C12 + C2

n

2 va

= 2 vb = 2 vc =

2 v0

=

1 E 12

1 E 12

k =1 2

Case 4 E

1vn

n

= 1vb = 1vc = E voltage at point n (En 0)

1va

1vn:

Early duration: the same as in case 5 Later duration: the same as in case 2

2 va ; = 2 vb

= 2 vc α C12 = E + 1 vn 2 C12 + C2 α C12 E + 1 vn ∴ 2 v0 = 2 C12 + C2 k =1 2

Case 5 E

= 1vb = 1vc = E = 2E Reflection factor at point n is 2 1va

1vn

n

2 va

= 2 vb = 2 vc =

α=1 5

3α C12 E 2 C12 + C2

2 va

3α C12 E ∴ 2 v0 = 2 C12 + C2

= 2 vb = 2 vc = = 0 75E

k =3 2

2 v0 = 0 75E

3×1 5 1 E 2 3

Case 6 E

= E, 1vb = 1vc = 0 Z1: Z2 = 2 : 1 then 2Z2 2 E = E Z1, Z2: surge 1 vn = 3 Z1 + Z2 impedance before and after point n 1va

n

2va

2vb

α C12 2 E+ E 2 C12 + C2 3 5α C12 E = 6 C12 + C2

=

α C12 2 0+ E 2 C12 + C2 3

2α C12 E 6 C12 + C2 α C12 ∴ 2v0 = E 2 C12 + C2

2 va

=

5 E 12

1 = 2 vc = E 6 1 2 v0 = E 4

2 vb

= 2vc =

α=1 5

=

Case 7 E

= 1vc = E, 1vb = 0 Z1: Z2 = 1 : 2 then 2Z2 4 E= E 1 vn = 3 Z1 + Z2 1va

n

k = 1 2, 5 6, 1 3 α=1 5

2 va = 2 vc α C12 4 = E+ E 2 C12 + C2 3 7α C12 E 6 C12 + C2 α C12 4 0+ E 2vb = 2 C12 + C2 3 =

4α C12 E 6 C12 + C2 C12 E 2v0 = α C12 + C2 =

C12 Note: k: coefficient of transfer voltage kα ; ratio of transfer voltage α: refer to Eq. (17.3). C12 + C2

k = 7 6, 4 6, 1

1 = 2 vc = E 6 1 2 vb = E 3 1 2 v0 = E 3 2 va

516

17 Insulation Coordination

17.11.3

Transfer Voltage Caused at the Generator Terminal Side

Next, we need to calculate the transfer voltages arriving at the generator terminals when a generator (surge impedance Zg) is connected to the transformer. The circuit is shown in Figure 17.14a, where the incidental voltage is kα e(t) from Eq. (17.9). The circuit equations in the Laplace domain are kα e s − 2 v s sC 12 = i s = i2 s + ig s 17 11 i2 s = ig s Zg Zg surge impedance on the generator 2v s = sC 2 C12 s s C12 kα e s kα e s = ∴ 2v s = 1 C12 + C2 s + δ s C12 + C2 + Zg 17 12 1 where δ = C12 + C2 Zg Next, the incidental surge voltage coming from the HT side e(t) may have a waveform similar to the virtual standard waveform, which is expressed by e(t) = E(e−at − e−bt) as shown in Figure 17.14b. The incident surge is for t ≧ 0 e t = E e −at − e −bt e s =E

17 13

1 1 − s+a s+b

We calculate the LT side voltage 2v(s), 2v(t): 2v

C 12 C 12 + C2

1 s 1 s − s+a s+δ s+b s+δ

C 12 C 12 + C2

1 δ a 1 δ b − − − δ− a s + δ s + a δ− b s + δ s + b

s = kα E

= kα E

2𝜈(t)

C12

k𝛼e (t) i (t)

C2

i2 (t)

ig (t)

Generator Zg Surge impedance

(a) +E

Ee–at e(t) =

E(e –at –e –bt)

The standard waveform of 1.2 × 50 µs can be written as the equation e(t)  E{ e–0.015 × 106t –e–5.0 × 106t } using

0

t

a  0.015 × 106, b  5.0 × 106 then e(t)  0.983E at t = 1.2 × 10–6

–Ee–bt –E (b) Figure 17.14 Equivalent circuit of an LT coil with the generator side for high-frequency phenomena.

17.11 Transfer Surge Voltages through Transformers, and Generator Protection

= kα E

a −b δ δ − a δ− b

C 12 C 12 + C2

1 a 1 b 1 − + s + δ δ −a s + a δ− b s + b

17 14

The transfer surge voltage at the generator bushing terminals is 2v

t = kαE

a− b δ a − at b − bt e − δt − e + e δ −a δ −b δ− a δ− b

17 15

b (this is the case of a 1.2 × 50 μs standard wave or similar) and b

In the case of a 2v

C12 C12 + C2

t = kαE

δ in order,

C12 b C12 e − δt − e − bt = kαE δt e −δt C12 + C2 b− δ C12 + C2

1 C12 + C2 Zg

where δ =

α = 0 5 for solidly neutral-grounded system 17 16a

= 1 5 for neutral-ungrounded system = between 0 5 and 1 5 for high-resistive neutral-grounded system k = 1 3 ,1 2, 2 3, 5 6,1, 7 6 depending on the transformer neutral connection see Table 17 3 E kVcrest

surge transmitted voltage from HV bushing terminal

x2 x3 x4 − + − ≒ 1 − x for smaller x, then (e−δt − e−bt) = e−δt (1 − e[δ − b]t) −e−δt(δ − b)t 2 3 4 As a trial calculation, assuming a steep front wave b = 5 × 106 and t = 10−6 sec = (1 μs), then δt = 5.0, and C12 = 4000 pF, C2 = 8000 pF, then Note: e −x = 1 − x +

2v

t = kαE

4,000 × 5 0 e − δt = 1 67 e −δt kα E 4, 000 + 8, 000

17 16b

The equation shows that the LT-side induced voltage could be large enough to threaten the insulation. Appropriate countermeasures to protect the LT-side insulation against the transfer-surge voltages are essential, as described next. 17.11.4

Transfer-Surge Protection

Appropriate countermeasures are required to protect generators against transfer-surge voltage coming from the HT side, because the rated voltages of the generators are perhaps 10–35 kV, so the insulation level of the LT side against surges is relatively low. The typical countermeasure is to install a surge absorber at the LT bushing terminal of the transformer for each phase, To the generator which is a parallel circuit with a capacitor and arrester, as LT-coil of transformer shown in Figure 17.15. By adding capacitance C as the surge absorber, Eq. (17.16a) is modified as C2 C2 + C, so C12 C12 + C2

C12 C12 + C2 + C

17 17

Accordingly, the transfer voltages can be considerably reduced by adding C of value larger than C12, C2 (C C12, C2). Typically, C = 0.1–0.5 μF is a reasonable range. The arrester in parallel is to relieve the steep wave front.

Arrester Capacitor Surge absorber Figure 17.15 Surge absorber installed at an LT bushing terminal.

517

518

17 Insulation Coordination

17.11.5

Electromagnetic Transfer Voltage

A transfer voltage to the LT side by electromagnetic coupling also arises. Voltages caused by magnetic coupling can be written simply as Mdi/dt, so the transfer voltage is roughly proportional to the transformation ratio, T-ratio = nHT/nLT, the turns ratio of the HT and LT coils of the transformers. Considering a virtual standard waveform of 1.2 × 50 μs as the incident voltage E at the HT side and a current surge with a similar waveform, the transfer voltage to the LT side in an initial duration of t = 0–1.2 μs is roughly (T-ratio) E. In the case of a 275 kV/24 kV transformer (T-ratio 11.5) as an example, the incidental surge voltage E on the HT-side line is transferred to the LT side by (1/11.5)E = 0.087E. A less serious threat arises from the absolute value and the relatively gradual wave shape in comparison with the electrostatic transfer voltages. Electrostatic, rather than electromagnetic, coupling plays the lead role in transfer-voltage phenomena.

17.12

Transformer Internal High-Frequency Voltage Oscillation Phenomena

A transformer is a very compact piece of apparatus: HT, medium-tension (MT), and LT coils are concentrically and tightly arranged surrounding the laminated silicon-steel core, and each coil is composed of a number of winding sections. Therefore, when we examine transient voltage and current phenomena on the internal coils, we need to establish the equivalent circuit as a distributed circuit with a number of L and C elements. Internal transient oscillations are obviously caused by the impact of an external surge voltage wave, so effective suppression of such transient oscillatory voltages and effective insulation design for each sectional part of the coils are vital. This is a matter that engineers at transformer manufacturing companies have to take into account, especially during the structure design stage, because a transformer with enough withstanding capability to clear the individual specified standard insulation voltages is rarely broken by internal transient voltage oscillations that originate from by lightning or switching surges.

17.12.1

Equivalent Circuit of a Transformer in the HF Domain

Figure 17.16 shows sketches of a typical large-capacity transformer. As seen in Figures 17.16a and b, the LT, MT, and HT coils are concentrically arranged in order as cylinder coils. Each coil is assembled from a number of series-connected sectional windings. These figures also show typical coil-winding structures called multilayer cylindrical windings and disc windings. Now we will examine the voltage behavior of a transformer’s internal coil when an incident surge voltage E arrives at the HT bushing terminal. Figure 17.17a is the equivalent circuit of the HT coil against surge or high oscillatory frequency phenomena, although resistance and susceptance are ignored and the mutual couplings with MT and LT coils are not written. In the case of a disc-winding coil, each section of the ladder circuit means each disc winding is connected in series.

17.12.2

Transient Oscillatory Voltages Caused by Incident Surges

The initial conditions of the circuit just after the surge arrives can be very accurately expressed as in Figure 17.17b, which is equivalent to Figure 17.17a with regard to the C distribution, although all inductances are ignored. On the other hand, the final conditions after the transient terms disappear is expressed as circuit (a) but ignoring all the C elements, as shown in Figure 17.17c. Figure 17.17a is a very accurate equivalent circuit of the HT coil for extra high frequency (EHF) phenomena. Now we will calculate the surge voltage distribution of this circuit, where C : the total capacitance from the HT coil to earth-ground Cdx : the capacitance of one disc (between x and x + dx) to earth-ground K : the total series capacitance of the HT coil from the HT terminal bushing to the neutral terminal K/dx : the series capacitance across discs (between x and x + dx) ic : the current flowing through Cdx ik : the current flowing through K/dx

17.12 Transformer Internal High-Frequency Voltage Oscillation Phenomena

Center of core

LT-coil Barrier (paper press-board) HT-coil

Center of core

(a) Multilayer cylindrical winding method

Barrier (paper press-board)

LT-coil

HT-coil

(b) Disc-winding method

Bus duct (LT-side) Steel-core clamp plate Upper yoke Conservator Neutral bushing Transformer tank (inside wall magnetically shielded)

Cable head (HT-side)

Cooler units

Binding tape HT-coil LT-coil Side core Main core (laminated silicon steel plate) Core support plate (c) Structure Figure 17.16 Large-capacity transformer for power station use (y–Δ connection).

519

17 Insulation Coordination

E Ldx K/dx

E Cdx

1

2

M1

N th

3

(a) Equivalent circuit of transformer winding (HT-coil) E

𝜈–

∂𝜈 · dx K/dx ∂x 𝜈 ik

𝜈+

iC

M2

MN

(c) Final voltage distribution

∂𝜈 · dx ∂x

Cdx

ΔC´1

ΔKk

Shield plate ΔC´k Vk

dx

x

E

ΔC´N

ΔCk

l (d) Non-oscillatory windings by the parallel compensation method

(b) Equivalent circuit at initial time of inrush surge Voltage potential (pu) 𝜈/E 1.0 Cdx

0.8

𝛼=

0.6

𝛼

sinh 𝛼 (l–x) sinh 𝛼 l

K/dx

=

C K

C: the total capacitance from the HT-coil to earth-ground (Cdx for the length dx)

0 1

𝜈/E =

520

2

0.4

3

K: the total series capacitance of HT-coil from the HT terminal bushing to neutral terminal (K/dx for the length dx)

4 6

0.2

10 20 30 0 x=0 Line terminal

→x (e1) initial surge voltage distribution

Figure 17.17 Internal surge behavior of a transformer.

x=l Earth terminal

17.12 Transformer Internal High-Frequency Voltage Oscillation Phenomena

Voltage potential (pu) Envelope curve of the largest potential 𝜈/E

1.0

0.8 (ii) 0.6 s

10

5μ

(i) Initial distribution (ii) Quasi-steady-state voltage distribution

μs

0.4

4μ

(i)

s

3μ s

0 μs 1μ s

0.2

0 x=0

x=l

→x

Line terminal

Earth terminal

(e2) Surge voltage oscillation Figure 17.17 (Continued)

The equation at the winding section x to x + dx is

ik =

k ∂ dx ∂t

v

−

v+

voltage at x

∂v ∂x

∴ ik = −K

∂2 v ∂t∂x

voltage at x + dx

∂ik − dx ∂x

dv Cdx dt

=

17 18 ∂ik ∂v =C ∴− ∂t ∂x

Leakage current through Cdx Reduced current of ik through dx

Eliminating ik from both equations, K

∂ ∂2 v ∂v =C 2 ∂t ∂x ∂t

K

∂2 v = Cv ∂x2

17 19

then

∂2 v = α2 v where α = ∂x2

C K

17 20

The general solution of this partial differential equation is v = A cosh αx + B sinh αx where A and B are determined by the terminal conditions (refer to the Supplement for the proof ).

17 21

521

522

17 Insulation Coordination

The terminal conditions are ∴0 = A cosh αl + B sinh αl A + B αl A− B −αl = e + e 2 2 v = E at x = 0 ∴ E = A cosh 0 + B sinh 0 = A

v = 0 at x = l

17 22

∴A = E B = −E

e αl + e −αl cosh αl = −E cosh αl e αl −e −αl

17 23

Substituting Eq. (17.23) into (17.21) and modifying, v x,0 = E α=

e α l −x − e −α l −x sinh α l −x =E αl −αl e −e sinh αl

C K

17 24

This equation gives the initial voltage distribution of the HT coil at t = 0, where v(x, t) means the transient voltage on coil position x and at time t. Figure 17.17(e1) shows the curve of the initial voltage distribution v(x, 0) at t = 0 along with the coil position x and with the parameter α, which was derived from Eq. (17.24). The figure indicates that at initial time t = 0, surge voltage E cannot be uniformly charged to each coil section except in the exceptional case under the condition α = 0 (unrealistic case of C = 0). If α is larger (larger C and smaller K), most of the surge voltage is charged unequally on the coil sections close to the HT bushing. If α = 10, for example, a large voltage stress of approximately 0.8E is placed on the first 20% of coil sections close to the line terminal bushing. The potential gradient is ∂v αcosh α l −x = −E ∂x sinh αl

17 25

∂v ∂x

17 26

and = −E x=0

αcosh αl sinh αl

The equation shows that the potential gradient of the initial distribution is largest at the high-voltage terminal. Let’s investigate this behavior a little more. Just after the incident surge of a triangular waveform with a short wavefront and long wave-tail (1.0 × 50 to 200 μs, for example) is charged on the HT coil, the voltage at each position v(x, t) begins to oscillate from the initial distribution v(x, 0). Figure 17.17e2 shows the oscillatory behavior of the voltage distribution just after the initial condition. The surge voltage at each coil section repeats an over-swing oscillation across the straight line (ii). Soon, the transient voltage oscillation decreases over time and disappears, so the voltages converge to the final distribution of the straight line (ii), which is called the quasi-steady-state voltage distribution. The incident voltage E can be treated as the DC component in the duration of the long wave-tail, so dv/dt 0 at this time, although v is still large. Accordingly, at the end of the long voltage wave-tail, Cdv/dt 0 and K/(dv/dt) ∞, or in other words the capacitive elements act like open-circuit elements. Therefore, all the C and K branched circuits in Figure 17.17a can be ignored during this time, and the final voltage distribution can be derived from Figure 17.17c. The voltage distribution during this time converges to a uniform distribution: the distribution of (ii) in Figure 17.17e2. The final distribution in Figure 17.17c is expressed by the following equation, for disc coil number 1–N N

Δvk

E=

①

k =1

17 27

N

Δvk =

Lkj j=1

di dt

di ② Mk dt

17.12 Transformer Internal High-Frequency Voltage Oscillation Phenomena

where k = 1–N: the number of disc windings i: the current flowing through each winding (the same current) Δvk: shared voltage by the kth disc winding Lkj: mutual inductance between the kth and jth disc windings Mk = N j = 1 Lkj : total summation of self- and mutual-inductances of the kth disc winding (the specific value of each disc winding) At the time of the final distribution condition (quasi-steady-state voltage distribution), the voltage oscillation is terminated so the shared voltage of each disc Δvk has a constant value, and accordingly di/dt is constant from Eq. (17.27)②. In other words, at the time of the quasi-steady-state voltage distribution, the surge voltage oscillation is terminated, whereas the surge current i is still increasing at constant speed. Although this surge current continues to increase a little, sooner or later it will stop and disappear because it is resistively attenuated. It can be concluded from Figure 17.17(e2) that the surge voltage distribution v(x, t) of the coil sections initiates internal oscillation from the initial distribution of Figure 17.17(e1) and repeats oscillatory over-swing across the quasisteady-state voltage distribution line (ii) and soon terminates voltage oscillation, while the surge current continues to increase and then disappears. The envelope curve of the voltage oscillatory distribution is also shown in Figure 17.17(e2). In all events, transformer design with excess non-uniform initial voltage distribution should be avoided, to avoid excess concentrated stress on a few coil sections around the HT bushing side and to reduce oscillatory surge voltage.

17.12.3

Reducing Internal Oscillatory Voltages (Non-oscillatory Windings)

We need to make the initial distribution curve as coincident as possible with the final distribution curve (the straight line of α = 0) by reducing α: in other words, by decreasing C or by increasing K. However, decreasing C is impossible because enlarging the distance from the winding to the core/tank/other windings cannot be done realistically. K also cannot be increased because the winding discs are already very closely arranged. The widely applied effective countermeasure to reduce oscillation is known as the parallel compensation method of stray capacitances. In Figure 17.17d, the shield ring plate with the voltage potential of the HT bushing terminal (often called the rib shield) is arranged as a typical high-voltage transformer design practice. The distance between each winding disc and the rib shield is closer to the HT terminal and far apart around the neutral terminal (i.e. ΔC1 > > ΔCk > > ΔCN ) (parallel capacitance compensation method). At the initial time when surge E is charged, the current through ΔCk is supplied directly through ΔCk , so the current through direct capacitance ΔKk for each disc is uniform, which means the initial shared voltage for each disc winding is kept almost equal. We can write this symbolically as follows. At the initial time, for the parallel charging currents for each winding disc, ∂ E − vk ∂vk = ΔCk ∂t ∂t ∂vk 1 ∂E ∂E = ∴ δk ΔCk ∂t ∂t ∂t +1 ΔCk ΔCk

where k = 1 2,

N 17 28

≒ δk ≒ ), the initial voltageThen, if the δk for the winding discs are designed to be almost equal (δ1 ≒ δ2 ≒ increase velocity at each winding disc (δvk/δt) can be kept in almost equal, so the initial surge voltage is uniformly distributed. The oscillatory voltage behavior on the transformer winding is closely affected by the required insulation of the transformer. However, the phenomenon is not usually included in insulation coordination because it is not necessarily related to the insulation coordination of the power system. In other words, this is considered an insulation matter that the transformer engineers or suppliers have to solve for individual transformer products.

523

524

17 Insulation Coordination

17.13

Oil-Filled Transformers Versus Gas-Filled Transformers

Power transformers with large capacity (say, over 60 MVA) are usually oil filled because they have been the standard type for many years, although small-capacity transformers (typically 0.5–40 MVA) are dry (air or SF6 gas insulation/ coolant) or oil-filled. This is because oil is the one material that has superior electrical insulation and superior thermal loss discharge (coolant). Accordingly, it was believed for a long time that oil could not be technically replaced by SF6 gas for large-transformer applications, because SF6 gas has poor thermal capacity (or thermal conductivity), in spite of its outstanding insulation characteristics. This fact is in stark contrast to the engineering history of circuit breakers, because SF6-gas-type breakers have been widely used in place of oil-filled type/air-blast-type breakers over the last half century. However, SF6-gas-filled transformers with a large capacity of 300 MVA 275 kV were first utilized in the mid-1990s in the underground EHV substations of high buildings in Tokyo: their fundamental structures were the familiar disc windings, but they were filled with gas instead of oil. These achievements led to a breakthrough by changing the conservative concept of large-capacity/EHV transformers as only oil filled. Needless to say, oil-filled transformers have one major weakness: the severe damage to the transformer if a breakdown fault occurs in the internal coil, which can also affect the installed surroundings. Whenever a short circuit occurs in the coil insulation of oil-insulation-type transformers, liquid oil around the fault is immediately gasified by the arcing temperature, so the internal gauge pressure of the tank rapidly increases. The oil–gas pressure increase continues until fault tripping by the related breakers is completed, so the accumulated pressure may reach a very high level, even exceeding the mechanical withstanding strength of the tank within just a few cycles (50–100 ms), in particular the critical strength of the tank-cover cramping. As a result, a hot-oil-blasting overflow or even fire from the oil burning is caused in the worst cases. Of course, the shape of the coils is deeply distorted. On the other hand, in the case of a short-circuit fault in an SF6-gas-insulated transformer, the physical damage caused by the internal short-circuit fault is limited to a narrow spot on the coil where a breakdown arcing pass is produced, so the concentrically arranged HT/MT/LT coils may not be badly deformed, although carbonization of insulation tapes/ press-board barriers and the copper conductor melting in a limited area is caused. In other words, SF6 gas transformers do not produce serious blasting or fire even in the case of an internal short-circuit fault. This is obviously a major advantage from the safety point of view, in particular for receiving substations located in city areas, whether the substations are outdoors or in-house, or under high-rise buildings. The reasons this breakthrough was achieved can be summarized by the following three points: a) Extensive study of thermal discharge (conductivity) and insulation/breakdown characteristics of SF6 gas with various shapes of coils, gas-flowing passes, and gas-flow speed based on a detailed mathematical simulation approach and experimental model tests, and finally well-investigated smart coil structure and gas-flow pass design. b) Using 0.4 MPa (4 atm) SF6 gas pressure. c) Application of class F insulation materials (maximum temperature 130 C) instead of class A (105 C) for coil taping based on polyethylene teraphthalate (PET) films. Oil flow depends on its liquid viscosity characteristics, so oil may not flow easily through very narrow passes, while SF6 gas can flow easily through even narrow passes because its gas-flow distribution does not depend on viscosity. This characteristic can be said to be an advantage of SF6 gas in comparison with oil. However, SF6 gas may flow through passes in an unbalanced way, in contrast to a joule-loss distribution that should be cooled at a continuous withstanding temperature (130 C). This is obviously a disadvantage of SF6 gas in comparison with oil as a coil coolant material. Furthermore, the thermal capacity of SF6 gas under 1 atm pressure is only 1/200 = 0.5%, and that under 4 atm pressure is still only 2.4/200 = 1.2% in comparison with oil (see Table 17.4). These are the reasons accurate analysis and careful gas-flow pass design are required. Using 4 atm pressure gas is also a valuable improving factor to achieve effective cooling by SF6 gas, although its thermal capacity is still much smaller than that of oil. Figure 17.18 shows 400 MVA/330 kV/132 kV SF6 gas transformers installed in Australia. Large-capacity SF6 gas insulation transformers can be realized only by fully involving these three countermeasures. As can be seen in the figure, cylindrical tanks, instead of the traditional box type, are generally advantageous for gas transformers because the tank needs to withstand the high pressures. Gas insulation transformers also have a smaller volume and can save on installation space in comparison with oil insulation transformers of the same capacity; in particular their height is less, because the oil conservator, oil-absorbable saucer, fireproof barrier, etc. can be removed. Large-capacity SF6 gas transformers should rapidly prevail in the near future, in particular as large transformers installed in urban areas.

17.13 Oil-Filled Transformers Versus Gas-Filled Transformers

Table 17.4 Comparison of thermal capacity. Ratio of thermal capacity

Oil

200

SF6 gas: 0.125 MPa-g

1

SF6 gas: 0.40 MPa-g)

24

Note: 0.1 MPa-g (megapascal to gravity) = 1 atm = 1000 mb approximately.

Courtesy of Transgrid (Australia)/Toshiba

IIaymarket substation (Australia)

Primary: 330 kV ± 10% (21 taps) 400 MVA Secondary: 138.6 kV 400 MVA Tertiary: 11 kV 20 MVA Rated gas pressure: 0.43 MPa-g (20°) Heat exchanges: gas-to-water cooling

Figure 17.18 A 400 MVA SF6-gas-insulated transformer (with on-load-tap-changer).

Supplement Proof that Eq. (17.21) is the Solution of Eq. (17.20) Differentiating v of Eq. (17.21) twice, v = Acosh αx + Bsinh αx = ∴ ∴

A αx − αx B αx −αx + e −e e +e 2 2

∂v A + B αx A − B − αx = αe − αe ∂x 2 2

∂ 2 v A + B 2 αx A −B 2 −αx = α e + α e = α2 v ∂x2 2 2

Therefore, Eq. (17.21) satisfies Eq. (17.20).

525

527

18 Harmonics and Waveform Distortion Phenomena Continuous and temporary waveform distortions are slowly but steadily increasing for most power systems. It is a problem in the same category as air pollution or water contamination, because an effective simple solution does not exist. For this reason, electrical engineers should understand the true nature of waveform distortion.

18.1

Classification of Harmonics and Waveform Distortion

Nonlinear loads of any kind in industrial plants and commercial buildings generate harmonics that cause waveform distortion of voltages and currents. Linear loads with unbalanced three-phase circuit and single-phase loads can become harmonics; in other words, negative phase voltages and current are often the reason for increased harmonics. Further, power electronic equipment is a harmonic power source. A DC transmission system is another example, and countermeasures to reduce harmonic components, such as multiphase (12- or 24-phase) power conversion, power filters, etc., are generally adopted. Power that includes harmonics has a significant impact on power quality and can damage neighboring load operation and/or damage facilities where currents and voltages are waveform-distorted. All power electronic equipment has more or less the characteristics to generate harmonic voltages and/or currents, but ironically such equipment also is apt to go wrong due to harmonics. Excessive waveform distortion adversely affects most control equipment and measuring equipment. Capacitor banks for improving the power factor are thermally overheated, because I = j2π fC ν is increased by harmonic components of larger f. Electrical appliances driven by single-phase motors with supplementary capacitors may pulsate due to harmonics, or the capacitors may thermally overheat. Power network facilities are also badly affected by harmonic components or waveform distortion. If the ratio of harmonic components in the voltage or current exceeds 5%–10%, serious effects can result. Table 18.1 classifies waveform distortions based on active causes and passive causes. Table 18.2 summarizes the effects that are caused by harmonics and waveform distortion. Figure 18.1 is a typical example of continuous voltage and current harmonics.

18.2

Impact of Harmonics

In this section, we will discuss the effects of waveform distortion on power-network facilities, listed in Table 18.2b.

18.2.1

Generators and Industrial Motors

As discussed in Chapters 10 and 16, generators are fundamental frequency positive-sequence voltage and current generators, so they have weaknesses related to negative- and zero-sequence voltages/currents and DC or harmonic components:

••

If a negative-sequence current flows into a generator, the third, fifth, seventh, etc. odd-number harmonics are caused. If a DC (offset) component current flows into a generator, the second, fourth, sixth, etc. even-number harmonics are caused.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

528

18 Harmonics and Waveform Distortion Phenomena

Table 18.1 The background of waveform distortion (occurrence of low-order harmonics). (a) Active causes (active generation of harmonic voltages and currents) (a1) Causes by loads (continuous or periodic) Intermittent rushing loads (trains, rolling machines) High-frequency loads (induction furnaces) Discharging loads (welder machines, electric furnaces) Rectification loads (rectifiers, power conditioners) Powered semiconductor application loads (speed-control loads; elevators) Uninterruptable power supply (UPS); power sources for infrastructure such as computers, data-processing servers, telecommunications and broadcasting equipment, etc.) Battery chargers

•• •• •• • ••

(a2) Causes by power network (continuous) Converters for DC transmission lines, frequency changers Power conditioners for distributed small generating systems (for solar, wind, fuel cell, small gas turbine, secondary battery, etc.) (b) Passive causes (resonant phenomena under special conditions) (b1) Negative- and zero-sequence or DC currents flowing through rotating machines Causes by loads (continuous) Single-phase distribution loads Unbalanced three-phase loads (electric furnaces, trains)

•• •• •• •• ••

Causes by power network Negative- and zero-sequence voltages and currents caused by network imbalance (continuous) Inrush current of transformers (temporary) Dead-voltage time of single-phase reclosing (temporary) Breaker-tripping failure (temporary)

(b2) Saturation Saturable nonlinear loads (continuous) Ferroresonance phenomena (temporary) (b3) LC resonance under special conditions in a power network Local resonance under normal operation (continuous) Local resonance under fault conditions (temporary)

Table 18.2 Influences of waveform distortion.

•• •• • •• •• •• •

(a) Influences on loads Shunt capacitors: Increase in thermal losses or damage due to thermal overheating Industrial motors: Problems caused by pulsating torque Single-phase AC motors using capacitors for rotating field production: Problems with rotating field mechanisms using capacitors Control equipment: Increased errors Illumination: Flickering (b) Influences on power networks Generators and motors: Thermal overheating of rotor surface, etc.; pulsating torque Capacitors for power-factor control: Thermal overheating Transformers: Local heat; mechanical vibration of cores and windings PT, CT: Degradation of accuracy Protective relays: Malfunctions; inaccurate operation Control equipment (AVR, etc.): Irregular operation caused by voltage signal-detection errors Energy (watt-hour) meters: Measuring errors (c) Interference with communication networks

•

(d) Interference with power line carrier systems (lower-frequency carriers) Load-management system

18.3 Harmonic Phenomena Caused by Power Cable Line Faults

(a)

(b) Figure 18.1 Waveform distortion of voltage and current (oscillogram recorded experimentally on a local system).

We must pay special attention to reducing continuous negative- and zero-sequence currents. The following items are essential:

•• •

Keeping the phase-imbalance rate of three-phase overhead transmission lines within a reasonable limit. Maintaining three-phase load balance as much as possible. This is essential particularly in a distribution system that includes various single-phase loads. Using special countermeasures for large imbalanced industrial loads, DC transmission lines, etc. (three-phase load balancing over time, local power filters, multiphase [12 or 24] conversion, etc.).

18.2.2

Nonlinear Loads

Harmonics and waveform distortion cannot be discussed separately from the nonlinear and saturation characteristics of equipment. Transformers and rotating machines have nonlinear characteristics and are harmonic sources. Further, L and C coupled resonance phenomena can generate harmonics or cause unexpected abnormal phenomena including overvoltages and overcurrents, because an L and C coupled circuit always has the natural frequency f = 1 2π LC. A circuit that includes motors and capacitors may cause ferro-resonance over voltage phenomena with typically periodic harmonics (refer Section 16.5.4).

18.3

Harmonic Phenomena Caused by Power Cable Line Faults

Unique phenomena due to fault-current waveform distortion have been recognized on urban network systems with cable lines. This is a kind of fault-current distortion caused by free-energy oscillation between large capacitances C of cable lines and inductances L of overhead transmission lines, leading to serious technical problems with high-speed protective relays (in particular, current differential relays and directional distance relays). Cable lines are increasing dramatically in modern power system networks, and such phenomena cause serious problems with the line main-protective differential relay operation. We will next investigate these phenomena using equations. 18.3.1

Transient Current Equation

Figure 18.2a shows a power system in which overhead transmission line 1 and cable line 2 are connected at power receiving substation q, and a three-phase fault occurs at point f, which is length x distant from point q. The waveform of the transient fault current at points p, q, r during the fault is investigated here. 18.3.1.1 Step 1

As this is a study of transient phenomena, we need to treat the problem by Laplace transforms. Referring to Figure 15.3, an overhead transmission line within 100 km and a cable line within 50 km can be treated as a concentrated circuit within an error of 10% or less for a phenomenon of 500 Hz or lower frequency. Therefore, we adopt the concentrated circuit shown in Figure 18.2b as the equivalent circuit. Incidentally, we assume for simplicity that terminal r is open, although similar phenomena can be observed even if point r is connected to the load.

529

530

18 Harmonics and Waveform Distortion Phenomena

r + sL = (r1 + sL1)x + (r2 + sL2)l2

sLb + (r1 + sL1) · (l1 – x) p

q

f

iq

ip

(c)

r ir sC2l2

𝜈f (t) = 𝜈e j𝜔t

(a)

Section #1 (overhead line) x l1 – x q p f

𝜈f (s) = 𝜈 ·

Section #2 (cable line) l2

1 s – j𝜔

r

f

i(s) r

(r1 + sL1) l1 q

(r2 + sL2) l2

r

sC

(d) Section #1

sLb p

sL

p

Open

Section #2

Section #3

f iq

sC2l2

(e)

(b)

q

r = r1x + r2l2 L = L1x + L2l2 C = C2l2

CT1 CT2

m

Figure 18.2 Transient fault calculation.

18.3.1.2 Step 2

Now, a line-to-ground fault occurs at time t = 0 at point f between p and q. The transient phenomenon of the fault can be calculated as in Figure 18.2c, where initial source voltage νf(t) is switched in at point f using Thévenin’s theorem. The transient current at point q can be calculated as in Figure 18.2d, which is the right half of Figure 18.2c. That is, L = L1 x + L2 l2 C = C 2 l2

18 1

r = r1 x + r2 l2 We assume the fault occurs at the time of peak voltage, so the initial voltage is 1 s −jω and the circuit impedance and current equations are 1 Z s = r + sL + sC 1 1 s 1 s ∴ = = 2 1 Z s L 2 r L s + 2αs + u2 s + s+ L LC 1 s 1 s = ≒ L s + α 2 + u2 −α2↖ L s + α + ju s + α− ju

for t ≧ 0 vf t = Vf e jωt , vf s = Vf

where α = is =

r , u= 2L

vf s Vf = L Z s

1 LC s s − jω s + α + ju s + α− ju

①

②

18 2 ③ ④

Vf F s L

⑤

18.3 Harmonic Phenomena Caused by Power Cable Line Faults

Assuming the surge impedance Z2 = 20 Ω, and surge velocity u = (1/2)c = 150 000 km/s for the power cable, then u2 =

1 = 150 000 km s = 150 × 106 m s L2 C2

Z2 =

L2 = 20 Ω C2

18 3

r2 = 0 03 Ω km = 0 3 × 10 −4 Ω m Accordingly, Z2 20 = = 0 133 × 10 −6 H m = 0 133 mH km u2 150 × 106 1 1 = = 0 33 × 10 − 9 F m = 0 33 μF km C2 = Z2 u2 20 × 150 × 106 r2 0 3 × 10 − 4 α2 = = = 1 13 × 102 2L2 2 × 0 133 × 10 − 6 L2 =

18 4

Also, ω = 2π × 50 = 314. Then u2

α2 , ω

u

α, ω

18 5

This is why we can ignore α2 in the denominator of Eq. (18.2b) ③. For the calculation of F(s) defined in Eq. (18.2b) ⑤, F s =

s k1 k2 k3 + + = s −jω s + α + ju s + α −ju s − jω s + α + ju s + α −ju

18 6

The coefficients k1, k2, k3 can be calculated using the same method with Eqs. (15.59) and (15.71) and the associated supplement. The result is k1 = F s

s −jω

s = jω

jω

=

=

2

jω + α ↖

↖

+ u2

jω = jωLC u2

↙

k2 = F s

s + α + ju

s = − α + ju

− α + ju

=

=

j LC =j 2u 2

=

−j LC =j 2u 2

2ju α + j u + ω ↖

↖

↙

k3 = F s

s + α −ju

s = − α −ju

− α + ju

=

2ju α − j u− ω ↖

18 7

↖

The arrow ↖ shows the negligible part based on Eq. (18.5). Thus j 1 1 1 − = 2u s + α + ju s + α − ju s + α 2 + u2 ∴F s =

jω 1 1 u + u2 s −jω u s + α 2 + u2

18 8a

18 8b

531

532

18 Harmonics and Waveform Distortion Phenomena

1 = e jωt s − jω

−1

u

−1

= e −αt sin ut

s + α 2 + u2 u=

18 9

1 LC

Therefore, Eq. (18.2) ⑤ is it =

Vf L

jω jωt 1 − αt e + e sin ut = Vf ωCej ωt + 90 + u2 u

∴ i t = Re i t = Vf

ωCsin ωt Steady-state term

+

1 L C

r t − e 2L sin

1 L C

e −αt sin ut

t LC

Transient term

The second term on the right side is the transient term, which is of amplitude Vf f = 1 2π LC , and attenuation time constant T = 2 L/r. 18.3.2

18 10

L C, oscillatory frequency

Transient Fault Current

The derived equation i(t) is the current iq in Figure 18.2c, whose line constants are given by Eq. (18.1). Now, let’s examine this result from the viewpoint of practical engineering. Case 1: Single-Cable Circuit Line (Fundamental Case) Figure 18.2e shows the fundamental case where a cable section is a single-circuit line. The waveform-distorted transient current (the transient term) is magnitude iq =

Vf L C

=

Vf Zsurge 1

oscillatory frequency f =

18 11

2π LC attenuation time constant T = 2L r where L = L1 x + L2 l2 C = C 2 l2 r = r1 x + r2 l2 The magnitude of the current distortion can be seen as a function of l1 (the length of the overhead line), l2 (the length of the cable section), and x (the distance from point q to the fault point f on the overhead line). Case 2: Line with Multiple (n) Parallel Cable Circuits In this case, modifications of the cable sections are L2 (1/n) L2, C2 nC2. The surge impedance Zsurge increases 1/n times, but the natural frequency of the specific cable section is not changed. With regard to the total system, the constants are replaced as follows in Eq. (18.11): L = L1 x + L2 l2 C = C2 l2

L = L1 x +

L 2 l2 n

C = nC 2 l2

18 12

18.3 Harmonic Phenomena Caused by Power Cable Line Faults

In the case of a fault at point f very close to point q (x = 0), One circuit

L2 1 Zsurge = n C2

Zsurge = Vf

iq = f=

n circuits

iq = n

L2 C2

L2 C2

Vf 18 13

L2 C2

1 1 f= 2π L2 C2 2π L2 C2

T=

2L2 r2

T=

2L2 r2

That is, under the condition of a fault at x = 0, the transient fault current iq is n times larger proportional to the number of parallel cable circuits, while the oscillatory frequency and the time constant do not change. Case 3: Line with Multiple (n) Parallel Cable Circuits In this case, cables in a third section are connected, as in Figure 18.2e. The total capacitance of the third section C3 is added to C2, so the transient current iq becomes larger and of lower frequency. The situation does not change much even if there are transformers between the second and third sections. Numerical check: Replacing Eq. (18.12) in Eq. (18.11), the magnitude and frequency of the distorted transient fault current iq can be calculated as a function of circuit number n and line length x. Figure 18.3 shows the calculated result with the condition of n = 1 to 4 circuits and x = 0 to 100 km, where typical line constants are assumed. Overhead line p

Current (transient term) 275 · 3

1

q cable lines

x

[kA]

L /C

x = 0 km 4 Circuits

30

Rp1

Rq1 Rq2

x=1

3

20

Rr2

[Calculating condition] Overhead line section :

2

x=5 10

r

10

275/ 3kV one circuit l1 = 100 km L1 = 1 mH/km Cable section :

20

1 1–4 Circuits

100

100

500 Frequency [Hz]

1000 f=

1 2𝜋 LC

l2 = 20 km

Zsurge = L2 /C2 = 20 Ω 1 u2 = = 150 000 km/s L 2 C2 L2 = 0.133 mH/km C2 = 0.33 μF/km

Figure 18.3 Fault calculation in a power system with cables and overhead lines.

533

534

18 Harmonics and Waveform Distortion Phenomena

Phase voltage Va

Va Vb

Vb Vc

Vc Current Ia Ib Ic

Ia

Ib Ic

Overhead line

Cable

2ϕS Measured point Figure 18.4 Waveform distortion of fault current (simulation).

The result indicates that a distorted transient fault current of large magnitude and lower-order frequency is caused in urban networks with many cable routes and lines. The voltage at point q is also badly distorted, although the calculation is omitted. Figure 18.4 shows voltage and current waveforms for a line-to-line (phase-b to -c) fault that was conducted as an artificial fault test on a real power network under conditions similar to those in our calculation.

18.3.3

Waveform Distortion and the Impact on Protective Relays

This numerical demonstration shows the appearance of fault current with excessive waveform distortion. Referring to Figure 18.2, we need to remember that the fault current from the generator side measured at p is not distorted, although the current from the cable side is badly distorted. As we hinted in Chapter 14, the technology of protective relays is based on the principle of detection of the voltages and currents at power frequency. Therefore, the phenomena discussed here could significantly affect high-speed protective relays (in particular, current differential relays and directional distance relays) in sections 1 and 2 of the network. Assuming a fault at point f in Figure 18.2e, for example, the relays in section 2 can see only badly distorted current. This is an important matter in practical engineering of protective relays, and engineers must investigate it carefully.

535

19 Power Electronic Applications, Part 1 Devices The following three chapters highlight fundamentals of power electronics and their industrial applications, focusing in particular on the basics of devices, of conversion circuits, and of power electronics control technology.

19.1

Fundamental Concepts of Power Electronics

This chapter briefly examines fundamentals of power electronics and power electronics control technologies for powersupplying and power-consuming industries. Following the initial application of electric lamps and motors in the 1890s, the invention of the mercury rectifier by Cooper Hewitt in 1900 and the cathode ray tube by John A. Fleming in 1904 were the main accelerators of the industrial application of electricity in heavy industry and in radio communication. Then, after half a century, the first transistor was invented by W.B. Shockley, J. Berdeen, and W.H. Brattain in 1947, and the silicon controlled rectifier (SCR) was developed by GE in 1956. These were the second major technical turning points of the application of electricity. Semiconductors have revolutionarily changed every aspect of the modern era. Power electronic devices large integrated circuits (LSI) devices are the two primary branches that cover power application industries as well as all other applications. The definition of power electronics was advocated as follows by IEC in 1975: Power Electronics is a technology which is interstitial to all three of the major disciplines of electrical engineering: electronics, power, and control. Not only does power electronics involve a combination of the technologies of electronics, power, and control, … but it also requires peculiar fusion of the viewpoints which characterize these different disciplines. (IEEE Trans. IA-10 by W.E. Newell) The basic concept of power electronics technology is conversion and controlled application of electric power using the time-dependent instantaneous electrical quantities P, Q, v, i, ψ, ω = 2πf, which can be achieved by the fusion of power, electronics, and control, as shown in Figure 19.1. Consequently, today’s advanced power electronics technologies enable the conversion of power and energy from one mode to another as desired. Table 19.1 shows a lineup of power conversion equipment.

19.2

Power Switching with Power Devices

Let’s discuss first the switching function of the mechanical switch shown in Figure 19.2a, where the voltage between contacts a and b is vab and the flowing current is i, and the closing state of the switch may be written (vab, i) = (0, i) and the opening state (vab, i) = (0, i) . Figure 19.2b shows the process of switching on and off as (vab, i), and the closing and opening conditions are given by points A and B, respectively. In the figure, the closing/opening switching transient process can be traced with the v(t) & i(t) locus. If switching phenomena of a high-voltage breaker are applied to the figure, arc current and transient recovery voltage can be expected (see Chapter 15), and points A and B show steady-state closing/opening conditions. Assuming a switch-opening process from A to B, a locus through point c1 shows soft switching, while a locus through point c2 shows hard switching with large transient voltage and current. Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

19 Power Electronic Applications, Part 1

Cir cui ts

tic t Sta men ip equ

Electronics

Power and Electronics

Continuous

Power

ng tati nt Ro ipme equ

De vic es

536

Sampled data

Control

Figure 19.1 Technical concept of power electronics.

Table 19.1 Classification of power electronics conversion equipment. Name

Function

Rectifier

AC to DC conversion

Inverter

DC to AC conversion

Reversible converter

Reversible conversion of power between AC and DC

(Electronic) frequency converter

Power-frequency conversion

DC converter

Voltage and current control of DC power

Electronic AC power controller

Var power control and power-factor control

Electric (power) switch

Circuit switching

Active filter

Removing higher harmonics

Var generator

Var power generation

Of course, hard switching can cause switching failure, as with the restriking of a high-voltage breaker, or a significant loss of power. Obviously, a soft-switching locus passing through a point close to O(0,0) is better. Now let’s examine switching phenomena by semiconductor devices. The same expression from Figure 19.2 can be used for semiconductor device switches. However, in the case of a device switch, very small leakage current ileak( 0) continues to flow at the turn-on state. Also there is a very small leakage voltage vab( 0) at the turn-off state. Figure 19.3a shows these conditions. An ideal switch can be summarized as follows: i) ii) iii) iv) v)

Leakage current ileak at turn-off state is minimal. Voltage between the two terminals at the turn-on state vab is minimal. Turn-on switching time ton and turn-off switching time toff are minimal. Repeated switching can be performed almost infinitely. Thermal losses at on- and off-states and during the switching process are minimal.

Mechanical switches can do i, ii, and v despite their size, but they can’t do iii and iv. On the other hand, recent advanced power electronic devices can cover all of these points and create nearly ideal switches using appropriate circuit

19.2 Power Switching with Power Devices

is υs

On-state

Resistive load

S Switch

Power source

E

is A Hard switching C2

C1 Soft switching O

B Off-state

νs

(b) ν – i characteristics of a switching process

(a) Switching circuit

Figure 19.2 Switching v − i characteristics.

νoff ≅ Es

νon

Voltage ν

Current i

ion ≅ Es R

ioff

pon

poff

Power p tS2

ton

tS1

tS2

toff

T Figure 19.3 Switching loss.

design. Table 19.2 shows typical waveform conversions of ideal power switching by an ideal switch. The switching application for every frequency between DC and 1 MHz has been achieved. Now, we will examine thermal losses. Figure 19.3 shows voltage and current waves under repeated on and off states and switching transient processes. Power loss v(t) ∙ i(t) is caused when v(t) and i(t) simultaneously exist and thermal energy v(t) ∙ i(t)dt is consumed in the device as thermal loss. Power loss von ∙ ion (von is leakage voltage) and voff ∙ ioff (ioff is leakage current) caused under on and off states are relatively small because device voltage von and device current ioff are small. Figure 19.4a–d shows switching stages with three different transient loci. In pattern (a), v and i are changing in a straightforward way, so the switching trajectory locus corresponds with ① in Figure 19.4d. In pattern (b), current begins to decrease after voltage is established, so the locus corresponds with ②. In pattern (c), voltage begins to increase after the current disappears, so the locus corresponds to ③ and the switching loss is minimal. The space enclosed by the trajectory locus in Figure 19.4d causes the thermal loss. Reducing v(t) ∙ i(t) and switching time T is the key factor to achieve soft switching. von and ioff are determined by selecting a device, but soft trajectory design requires adequate circuit design. Power loss caused by pattern (a), where v and i change in a straightforward way, can be calculated as follows: ΔT

JSW =

ΔT

v t i t dt = 0

0 2

= EI

E 1−

t t − 2ΔT 3 ΔT

=

2 0

I

t dt = EI ΔT

ΔT 0

t t2 − ΔT ΔT

2

dt 19 1

ΔT

3

t ΔT

EI ΔT 6

Joule

537

19 Power Electronic Applications, Part 1

Table 19.2 Basic switching circuits and ideal switching waveforms. Pattern

Circuit

Output voltage waveform

1

S1 S4 S2 S3 S1 S4 S3 t

Load

S1

S2

2

t

DC voltages can be produced from input AC voltages using the switching combination S1–S4. DC voltages can be controlled by adjusting the switching duration.

S4

S1 t Load

S2

t

DC voltages can be produced from input AC voltages using the switching combination S1–S2. AC voltages can be controlled by adjusting the switching duration.

3

Averaged voltage

S

Load

538

S on

off

on

t

t

DC voltages can be controlled by adjusting the switching duration. 4

S1

S3

Load S2

S4

t

t

AC voltages can be produced from input DC voltages using the switching combination S1–S4. AC voltages can be controlled by adjusting the switching duration.

On-state Es

υ

Es

Es

0 i

2

B

IL 1

IL

IL

IL

i

A

0 Es IL Es I L 4

ps ts (a)

3

ts (b)

Figure 19.4 Switching loss and switching loci.

ts (c)

0

Off-state υ

Es

(d) v – i loci

19.3 Snubber Circuit

Switching is conducted twice during 1 cycle with period T, so the resulting thermal power P is P=

EI EI ΔT ΔT × 2f = 6 3 T

19 2

watt

where f = 1/T The switching loss is proportional to E, I, and ΔT, so a realistic design to reduce them is required.

19.3

Snubber Circuit

Figure 19.5a and b shows v, i waveforms and the corresponding trajectory loci of turn-on and -off switching at times t0–t5. With such hard-switching conditions, the device will soon be damaged by thermal heat or excessive contact with overvoltages. The attached auxiliary circuit to achieve soft switching is a snubber circuit. The roles of a snubber circuit can be summarized as follows. i) ii) iii) iv)

To reduce transient v and i in order to keep trajectory loci within adequate soft-switching area To shut out mis-switching caused by excess dv/dt or thermal breakdown caused by excess di/dt heat spots To reduce simultaneous v and i appearing at intervals, to minimize heat loss To achieve uniform voltage distribution in the case of high-voltage applications where multiple devices are cascade connected Turn-on process

Turn-off process

υCE

iC I0

E

t0

t1

t2

E

t3

t4

t5

t

(a) Switching waveform ic

t1

The case ignoring stray capacitance

Without snubber circuit t2

I0

t3

t4

With snubber circuit RBSOA border line Soft switching loci zone ZVS, ZCS 0

0

t0

t5

E (b) Switching trajectory loci

Figure 19.5 Switching wave-form and switching loci.

υCE RBSOA border line : (Reverse Bias Safe Operating Area) ZVS : Zero voltage switching ZCS : Zero current switching

539

Turn-on snubber to minimize large over currents

19 Power Electronic Applications, Part 1

Turn-off snubber to minimize large over voltages

540

is S υs

is υs S

Ic

Rs′

E

E

Ic

Ls′

D′

OFF

S

Ic

is Rs

S

ON

E

υs ZVS (zero voltage switch)

(a) Typical snubber circuit

OFF Ic

is

D

Cs

ON

E

υs

ZVS (zero current switch)

(b) Resonant snubber circuits

Figure 19.6 Typical snubber circuits.

Figure 19.6a shows typical turn-on snubber and turn-off snubber circuits. A turn-off snubber is typically composed of R and C elements (voltage snubber or parallel-connected snubber), and its function is to absorb stored inductive energy in the circuit caused by the surge voltage dv/dt and reduce v at turn-off. A turn-on snubber (current snubber or series-connected snubber) is designed to reduce di/dt. Figure 19.5b shows typical examples of zero-voltage switches (ZVS) and zero-current switches (ZCS), also known as resonant snubbers. The function of ZVS (or ZCS) is to make the voltage zero (or the current zero) by using the additional auxiliary switch S for a time duration Toff. Resonant snubbers help to reduce electromagnetic interference (EMI) and minimize switching loss.

19.4

Voltage Conversion with Switching

Referring to Figure 19.7a, where DC source voltage E, ideal switch S, and resistive load R are series connected, if on–off switching is repeated by S, voltage with a repetitive square waveform is obtained across load R, as shown in Figure 19.7b. The average voltage Vave across the load is smaller than E and can be calculated as follows: vave = d=

1 T

T

v t dt = 0

1 T

Ton

Toff

Edt + E 0

0dt = E Ton

Ton Ton =E =d E T Ton + Toff

19 3

Ton duty factor Ton + Toff

The average voltage value Vave can be controlled by changing Ton, Toff, under the fixed cycle period T. This control is known as duty factor control. Table 19.3 shows switching circuits and the corresponding waveforms for voltage conversion. υ υ Vave

E υ E

R

t Ton

Toff

Vave T

(a) Ideal switching circuit

(b) Voltage waveform

Figure 19.7 Generation of DC repetitive voltage by an ideal switch.

Table 19.3 Various power electronic devices. Uncontrollable Device name

On-controllable

Diode

Symbol

p n

On-off controllable

Thyristor

On-off controllable

GTO

On-off controllable

Power transistor

P1 J1 n1 J2 P2 J3 n2

Cathode : K

Anode : A

P1 n1

Gate : G

P2 n2

Cathode : K

Anode : A

P

Base : B

n

Cathode : K

P n

Emitter : E

Switching operation

Typical large devices

P n

Collector : C

P n

n

G

V-I waveform

C

Drain : D

n

Gate : G

IGBT

D

Collector : C

Anode : A

On-off controllable

Power MOSFET

Gate : G

P

P

n

Source : S

S

G

t Anode

t

Anode voltage

t

Collector voltage

t

Drain voltage

t voltage

Anode current

Anode current

t

Anode current

t

Collector current

t

Drain current

t current

Gate current Igate

t

Gate current Igate

t

Base current Ibase

t

Gate • source voltage Vgate

t

Emitter : E

E

Anode voltage

voltage

Base : B

n

Collector

t

Gate • emitter voltage Vgate

Turn-on

Keep anode voltage va positive

Give trigger current to the gate

Give base current Ibase

Give gate voltage Vgate

Give gate voltage Vgate

Holding turn-on state

To keep anode voltage va positive

The gate current is not required if the anode current exceeds the threshold value

Continue to flow Ibase

Maintain Vgate

Maintain Vgate

Turn-off

Stop anode current, or give inverse voltage to the anode terminal

Give reverse current to the gate

Base current Ibase to make zero

Gate voltage Vgate to make zero

Gate voltage Vgate to make zero

Rated voltage

6000 V

4500 V

1200 V

1000 V

1200 V

Rated current Turn-on voltage

2500 A

3000 A

800 A

8A

600 A

3.0 V/2500 A

4.0 V/3000 A

2.5 V/800 A

1.30 hm

4.0 V/600 A

Turn-on time

10 μs

10 μs

3 μs

0.2 μs

0.8 μs

Turn-off time Switching frequency (approximate order)

100 Hz

t

Collector

400 μs

20 μs

15 μs

0.8 μs

0.4 μs

100 Hz

1 kHz

10 kHz

1 MHz

100 kHz

t t

19 Power Electronic Applications, Part 1

19.5

Power Electronics Devices

19.5.1

Classification and Features of Power Semiconductors

A semiconductor is a power electronics device that is neither an insulator nor a conductor, so its properties are quite different from ordinary devices such as resistors and capacitors. Pure silicon (Si) and germanium (Ge) are elements belonging to group IV of the periodic table: they have some free electrons whose conductivities are larger than those of ordinal insulators, but smaller than those of ordinal conductors. If a small amount of energy (typically an electric field or light) is given to the semiconductor, free electrons or positive holes (or carriers) appear. Small amounts of substances from group III or V group are added to the Si or Ge in an n-type (electron carrier) or p-type (positive hole carrier) semiconductor. The pnjunction can cause a current to flow if enough voltage is given to the junction by ± polarity (forward-biased), but cannot be caused by polarity (reverse-biased). This type of current flow in one direction is called rectification. Table 19.3 shows a list of representative devices with a brief explanation of basic features and symbols, typical ratings, and so on. Figure 19.8 shows a typical application area by rated power capacity and applicable switching frequency. In addition to voltage and current ratings, switching frequency or on-off switching time is an important factor for practical applications. Next, we will begin to discuss individual devices. First is the diode, which is one of the most fundamental devices. 19.5.2

Diodes

A diode is a device composed of a simple p-n junction; it can conduct only when it is forward biased. Figure 19.9 shows the structure, elemental symbol, and v–i characteristics of a diode. It has terminals A (anode) and K (cathode), and current flows only from A to K as the forward direction. If forward polarity voltage is given to terminals A and K, a current flows (point a in Figure 19.9). On the other hand, if reverse polarity voltage is given, only negligibly small leakage current flows; current cannot be made to flow (point b). A diode should be operated with a source voltage lower than its reverse breakdown voltage (point c), to avoid breakdown or failure. A diode is used alone or in combination with other devices for most applications, as discussed in Chapters 20 and 21. 19.5.3

Thyristors

There are two kinds of electronic components: positive devices and passive devices. Passive devices cannot generate an increase in power. Diodes and other ordinal devices (resistors, capacitors, and so on) are examples of passive devices. In contrast, transistors or thyristors are positive devices that have low-power input (small current) and convert this to a high-power output (large current). In other words, the main current flow can be controlled (or modified) by a small 108

Thyristor GTO

107

IGBT Capacity [V · A]

542

106 Power transistor

105

104

MOSFET

103 102

103

104 105 106 Switching frequency [Hz]

107

Figure 19.8 Typical available range of ratings for various devices.

19.5 Power Electronics Devices

Current id a Anode (A)

A Conducting (forward-bias) region

Reverse breakdown voltage

p n

0

b

Voltage νd

Reverse blocking region Cathode (K) (a) Structure

K (b) Symbol

(c) ν – i characteristics

Figure 19.9 Diode.

signal current. The energy for this activity comes from the electrical supply to the circuit. A thyristor can be triggered into the on-state by applying a pulse of positive gate current for a short duration, provided the device is in its forwardblocking state. The forward voltage drop in the on-state is typically a few volts or less. In reverse bias at voltages below the reverse breakdown voltage, only a negligibly small leakage current flows in the thyristor. Thyristors belong to a group of components known as silicon controlled switches (SCSs) or SCRs. Their name suggests that their main use is for turning a current on and off. The name rectifier indicates that the current can flow only in one direction. As shown in Figure 19.10, a thyristor has three terminals: anode (A), cathode (K), and gate (G); current flows from A to K, but the third terminal G controls the flow of the current through the thyristor. Most often, a thyristor is used for switching a large current. With the source voltage condition vAK = vA–vK > 0 to a thyristor, whenever a gate signal current (pulse current) is given to G, the thyristor is shifted to the turn-on state; however, it cannot then be shifted to the turn-off state by removing the gate signal. (Note that the most popular gate signal is an optical pulse.) The current can be turned off only by removing the source voltage (reducing the source voltage to less than the holding level) or by changing the source voltage polarity. Such a function is classified as external commutation because the current can be turned off only by an external circuit condition. A thyristor can be considered a combination of a PNP transistor and an NPN transistor, as shown by the equivalent circuit in Figure 19.10c. If a gate signal is given to the gate terminal G, the two elemental transistors work together as a positive-feedback circuit by which the thyristor is turned on. This situation can be written using the following equations: IC1 = α1 IA + IC01 IC2 = α2 IK + IC02 IB1 = IA −IC1 = IC2 IK = IA + IG

19 4

α2 IG + IC01 + IC02 ∴ IA = 1 −α1 − α2 IC01 ,IC02 leakage current α1 ,α2 current supplying rate zero under turn-off state Under turn-off state, IG = 0, and α1 + α2 is also negligible. If the trigger signal is given to the G terminal, α1 + α2 is changed 0 1, and then IA( + IK) ∞, which means turn-on switching.

543

19 Power Electronic Applications, Part 1

A Anode A

IA A

Gate G

p n p n C

G I G

G Cathode

Tr1

p n p

Tr1 n p Tr2 n

C

α1

IA

IB1 IC2

IC1 G

IG

K

α2 Tr2

IK

K (a) Structure

(b) Symbol

(c) Equivalent sturucture model Current i

544

Holding current

On-state

IG large

Break over voltage

Reverse breakdown voltage Reverse blocking state

IG = 0

Voltage v Forward blocking state

(d) v – i characteristics Figure 19.10 Thyristor.

19.5.4

Gate Turn-Off (GTO) Thyristor

GTOs are on-off controllable devices, which can be turned on by a positive gate signal and turned off by a negative gate signal. A GTO is composed of a PNPN junction and can be considered two combined transistor circuits like a thyristor. However, unlike a thyristor, a GTO can be turned off using a short-duration gate current pulse signal: positive-feedback work is forced to stop by sufficient negative gate signal. This is the function of self-commutation. By applying negative gate-cathode voltage, sufficiently large negative gate current flows. GTO features and symbols are summarized in Table 19.3. GTO switching speeds are in the range of 3–25 μs, with voltage ratings up to 5 kV and current ratings of a few kilo amperes, so they are mainly used for applications with a switching frequency range of 100 Hz–10 kHz. 19.5.5

Bipolar Junction Transistor (BJT) or Power Transistor

Bipolar junction transistors (BJTs) are also controllable devices and are equipped with three terminals collector (C), emitter (E), and bias (B), as shown in Figure 19.11. The word bipolar comes physics: both electrons and positive holes work together as electricity carriers in this device. A BJT also has the function of linear amplification, but it is used as a switch for most heavy industrial applications. Let’s review what we know about power loss, regardless of application, for power amplification and switching. Referring to Figure 19.12a, the circuit can be written using the following equations. iE = iB + iC, where iE, iC, iB are the base currents of the emitter, collector, and base, respectively, and α = iC/iB is the DC current amplification rate (ordinarily α = 0.9–0 : 999): hFE =

iC iC α = 10− 1000 = = iB iE −iC 1 − α

So, assuming iB = 0.02 mA, then hFE = 100.

19 5

19.5 Power Electronics Devices Saturation zone (100A.600V)

Collector C

B

C B

E Emitter (a) Structure

E

Collector current ic [A]

n p n

Base

0.8 A

100 0.6 A

80 0.4 A

60 0.2 A

40

0.1 A

20

(b) Symbol

0

Turn-off zone IB = 0

8 10 0 2 4 6 Voltage between C and E vCE [V] (c) VCE – Ic characteristics Figure 19.11 Bipolar junction transistor (BJT).

IC I

Transistor characteristics

IC3

IB3

IC2

IB2

IC1

IB1 Load characteristics

Rload Vload

0

IC

(b) VCE – Ic characteristics

C IB

B

E

VCE E

eB VB

VCE

E

VCE3 VCE2 VCE1

PT c

E2 4RL

IE b 0

(a) Power amplification circuit

a E/2

E

VCE

(c) Vce – Ploss characteristics

Figure 19.12 Power amplification circuit with a transistor.

In addition to the device characteristics in Figure 19.12b, the straight line shows load characteristics, and the point of the intersection gives the operating point. The associated power loss can be calculated as follows. Ploss = VCE Ic = VCE

Vload E − VCE 1 E = VCE =− VCE − Rload Rload Rload 2

2

+

E2 4Rload

19 6

From this equation, Figure 19.11c can be derived as voltage VCE and power loss Ploss. The power loss Ploss is at its maximum at point c and minimum at point a (turn-off zone) and b (saturation zone). Therefore, device cooling is required regardless of application for amplification or switching. Further, from the viewpoint of switching, the circuit design must achieve operation points with on-off states very close to points a and b. BJTs are available in voltage ratings up to 1500 V and current ratings of a few hundred amperes.

545

19 Power Electronic Applications, Part 1

19.5.6

Power Metal Oxide Semiconductor Field Effect Transistors (MOSFET)

A metal oxide semiconductor field effect transistor (MOSFET) is a kind of MOS power transistor whose gate electrode is made of a metal oxide semiconductor. A MOSFET can achieve high-speed switching because it has so many carriers. It has a vertically oriented four-layer structure with alternating p and n doping (see Table 19.3). The v–i characteristics are shown in Figure 19.13. The device is fully on and approximates switching off when the gate source voltage is below the threshold value vgate. It requires the continuous application of a gate source voltage of appropriate magnitude in order to be in the on-state. However, it is a voltage-controlled device with a high-impedance gate, so no gate current flows except for a very small-duration on and off transition time, which results in low driving power. Further, the switching times are very short, typically in the range of 10–500 ns depending on the device type. So, high-speed switching of even 1 MHz can be achieved by 10 kVA class devices. MOSFETs are available in voltage ratings in excess of 1000 V but with small current ratings, typically 100 A. However, because their on-state resistance has a positive temperature coefficient, MOSFETs can be easily paralleled. With all these advantages, they have become widely used since the 1980s and are replacing BJTs in many applications, especially those where high-speed switching is important. 19.5.7

Insulated Gate Bipolar Transistors (IGBT)

The insulated gate bipolar transistor (IGBT) is a device with features of a BJT (high-voltage withstanding), a MOSFET (high-speed switching), and a GTO (self-commutation), as shown in Table 19.3. Figure 19.14 shows the symbol and v–i characteristics. Similar to a MOSFET, an IGBT has a high-impedance gate, which requires a small amount of energy to switch the device. Further, like a BJT, an IGBT has a small on-state voltage (say, von = 2–3 V) even in devices with large blocking voltage ratings of 1000 V. Furthermore, similar to a GTO, an IGBT can be designed to block negative voltages. IGBT turn-on and turn-off times are on the order of 1 μs, and typically, an available large module rating is 4.5 kV, 1500 A with a switching frequency of 20 kHz. IGBTs began to appear in the 1980s, and since then more advanced characteristics with larger VA and higher switching capabilities have been achieved. Using IGBTs, high voltage and large power quantities with any frequency and any waveform are available today.

100 10 V 6.0 V

(500V, 50A)

80 Drain current ID [A]

546

Drain D

5.5 V 60

40

5.0 V

20

4.5 V

G Gate S Source

VGS = 4.0 V 0

0

10

20

30

Source voltage VDS [V] Figure 19.13 MOSFET symbol and v–i characteristics.

40

19.6 Mathematical Background for Analyzing Power Electronics Applications

(1200V, 300A)

15 V 500

20 V

C

Collector current I C [A]

12 V 11 V

400 300

10 V

200 G

9V 100

VGE = 7 V

0 0

2

6

4

8V 8

10 E

Source voltage VCE [V]

(b) Equivalent circuit

(a) v – i characteristics Figure 19.14 IGBT symbol and v–i characteristics.

19.5.8

Intelligent Power Modules (IPM)

There are other kinds of devices, such as IEGT and SI-transistors, that we are not going to examine in this book. In addition, some modular components, based on devices already described and with all the necessary auxiliary elements, are now available and can help to achieve compact designs for power electronic applications.

19.6

Mathematical Background for Analyzing Power Electronics Applications

Most of the voltage and current quantities we have to handle in power electronics application engineering are cyclical repetitive quantities with distorted waveforms. We need to examine the mathematical expressions for these arbitrary waveform quantities. 19.6.1

Fourier Series Expansion

Every periodical function can be written using a Fourier series expansion as follows: ∞

∞

v t = a0 +

ak cos kωt + bk sin kωt = a0 + k∞ =1

k =1

2Vk sin kωt + φk

= V0 +

a2k + b2k sin kωt + φk 19 7a

k =1

1 HZ ω = 2πf = dθ dt rad sec θ = ωt radian T 1 T 1 T v t dt = v θ dθ DC component a0 = V 0 = T 0 2π 0

where f =

ak = bk = Vk =

2 T 2 T

T

v t coskωtdt = 0 T

v t sin kωtdt = 0

a2k

+

1 π 1 π

2π 0 2π

v θ coskθdθ 19 7b v θ sin kθdθ

0

b2k

2 a k φk = tan −1 bk

And with the fundamental component for k = 1, and kth order harmonics for k = 2, 3, 4….

547

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19 Power Electronic Applications, Part 1

19.6.2

Averaged Value and Effective Value of Arbitrary Waveform Quantities Averaged value Vave =

1 T

1 Iave = T

v t dt = 0 T 0

1 T

Effective value Veff =

1 T

Ieff =

19.6.3

T

1 2π

1 i t dt = 2π

2π 0 2π

v θ dθ where dt =

1 dθ ω

19 8a

i θ dθ

0

T

v2 t dt = 0 T

i2 t dt = 0

2π

1 2π

v2 θ dθ

0 2π

1 2π

19 8b i2 θ dθ

0

Power, Power Factor, and Distortion Factor of Arbitrary Waveforms

Assuming quantities with an arbitrary periodic function using the following equations, ∞

2Vk sin kωt + φt

vt = k =1

19 9a

∞

2Ik sin kωt + φt − θk

it = k =1

then effective power W[watt] can be derived by the following equation: W=

T

1 T

p t dt = 0

1 T

T

v t i t dt = 0

2π

1 2π

p ωt d ωt

0

= V1 I1 cos θ1 + V2 I2 cos θ2 + V3 I3 cos θ3 + ∞

19 9b

Vk Ik cos θk

= k =1

Averaged power is given by the summation of the products of the voltage and current components with the same order frequency (refer to Chapter 11, Section 11.1.2). And Veff = Ieff =

1 T 1 T

T

v2 t dt = 0

V12 + V22 + V32 +

effective voltage

T

i2 t dt = 0

I12 + I22 + I32 +

effective current

19 9c

The total power factor λ is λ=

effective power = apparent power

W V12

+

V22

+

V32 +

×

I12 + I22 + I32 +

averaged value of power = effective value of v effective value of i effective power of fundamental frequency Displacement power factor = apparent power of fundamental frequency V1 I1 cosθ1 = cos θ1 = V 1 I1

19 9d

19 9e

19.6 Mathematical Background for Analyzing Power Electronics Applications

T υ V t Ton

Toff

Figure 19.15 Repeat interrupted DC quantity.

The total harmonic distortion (THD) factor or distortion factor is THD =

V22 + V32 + V42 + V1

effective voltage of total harmonic frequency VH = = effective voltage of fundamental frequency V1

19 9f effective current of total frequency IH = = THD = effective current of fundamental frequency I1 19.6.4

I22 + I32 + I42 + I1

Repetitive On-Off Switching of DC-Quantities

Figure 19.15 shows repeated on and off switching of a DC quantity, whose averaged value and effective value can be calculated as follows: vavr = veff =

1 T

T

v t dt = 0

1 T

T

1 T

Ton

Vdt = 0

1 T

v2 θ dθ =

0

Ton

Ton V T V 2 dθ =

0

1 2 V Ton = T

19 10

Ton V T

where T = 2π 19.6.5

Alternate Rectangular Waveform

Figure 19.16a shows an alternate waveform quantity with repeated frequency 2π:

2π υ

V 0

π

2π

ωt

(a) 1.5 V V

υk = υ – υ1

0.5 V 0

1.5 V υ +υ +υ 1 3 5

υ1 υ

2π ωt

π

υ1 υ

V

υ3 υ 5

0.5 V 0

–0.5 V

–0.5 V

–V

–V

π

–1.5 V

–1.5 V (b)

Figure 19.16 Rectangular waveform AC quantity with periodic time 2π.

(c)

2π ωt

549

550

19 Power Electronic Applications, Part 1

for 0 ≤ θ ≤ π

V

vθ =

−V

19 11a

for π ≤ θ ≤ 2π where θ = ωt

v(θ) can be Fourier series expanded as follows. Substituting Eq. (19.11a) into (19.7b), 2π

1 2π

a0 =

π

1 2π

v θ dθ =

0

2π

Vdθ + 0

− V dθ

0

V π V 1 θ0+ − θ 2π π −2π + π = 0 0 = 2π 2π 2π

=

2π

2 2π

ak =

v θ cos kθdθ =

0

V 1 sin kθ π k

=

2π

2 bk = 2π

0

π

−

0

π

2π

V coskθdθ +

π

+ 0

π

0 2π

V 1 sin kθ π k

1 v θ sin kθdθ = π

V 1 − cos kθ π k

=

1 π

= π

−V coskθdθ

V sin kπ − sin 2kπ+ sin kπ = 0 kπ

π

2π

V sin kθdθ + 0

19 11b

−V sin kθdθ

0 2π

V 1 cos kθ π k

= 0

V − cos kπ + 1+ cos 2kπ − coskπ kπ

2V 1− coskπ = 0 for k = 2,4, 6, kπ 4 for k = 1, 3, 5, = kπ Then, substituting this into (19.7a), =

vt = 2

2 2 1 1 V sin ωt + sin 3ωt + sin 5ωt+ 3 5 π 19 11c

2 2 ∞ 1 sin 2k − 1 ωt V = 2 2k − 1 π k =1 or v t = v1 t + vharmonics 2 2 V sin ωt fundamental frequency component π 2 2 ∞ 1 sin 2k − 1 ωt total harmonic frequency components V t = 2 2k −1 π k =2

v1 t = 2 vharmonics Also,

19 11d

vave = 0 veff = v1eff

t

=

2π

1 2π

v2 θ dθ =

0

1 2π

π

2π

V 2 dθ +

0

π

− V 2 dθ = V

1 2 2 V sin ωt effective value of fundamental frequency component V1 t = π 2 ∞

2 − V 2 −V 2 Veff 1 0

Vk2 =

vHeff =

19 11e

total effective value of harmonics

k =2

Then for the waveform in Figure 19.16a, using V0 = 0, Veff = V, 2

VH =

V2−

2 2 V2 = π

2 2 1− π

2

V = 0 435V

19 11f

19.6 Mathematical Background for Analyzing Power Electronics Applications

2π 2π 3

v

V

π

0 π 6

𝜔t

2π

5π 6

Figure 19.17 Rectangular AC waveform.

∞

THD =

VH = V1

k =2

1−

Vk2

V1

=

2 2 π

2 2 V π

2

=

π2 −1 = 0 483 8

19 11g

As shown in Figure 19.16b, the addition of the first, second, and third components is similar to the original rectangular waveform.

19.6.6

Alternate Rectangular Waveform with Switch-On at α and Switch-Off at β

In Figure 19.17, if the quantity is switched on at α and switched off at β, then

vt =

0

for π + β ≤ ωt ≤ 2π + α and

V

for α ≤ ωt ≤ β

for β ≤ ωt ≤ π + α

−V for π + α ≤ ωt ≤ π + β Vave = 0 Veff = =

1 2π

2π

v θ dθ =

0

1 2π

β α

V 2 dθ +

1 V 2 β −α + V 2 β − α = 2π

π+β π+α

− V 2 dθ

19 12a

β−α V π

π 5π Assuming α = , β = , then 6 6 Veff =

2 V 3

19 12b

The Fourier series expansion of Figure 19.17 is given by vt = 2

6 1 1 1 1 V sin ωt − sin 5ωt − sin 7ωt + sin 11ωt + sin 13ωt− 5 7 11 13 π

∞ 6 −1 k V sin ωt + = 2 sin 6k ± 1 ωt π 6k ± 1 k =1

Then the effective value is V1 =

6 V. π

19 12c

551

552

19 Power Electronic Applications, Part 1

19.6.7

Power of Waveform Distorted Voltage and Current

We discussed this topic in Chapter 11, but it is important to mention it again: ∞

2Vk sin kωt + φk

v t = V0 + k =1 ∞

i t = I0 +

19 13a 2Ik sin kωt + φk − θk

k =1

The averaged power is given by the total products of voltage and current components of the same frequency: W=

1 T

T

p t dt = 0

1 T

T

v t i t dt = 0

1 2π

2π

p ωt d ωt

0

= V0 I0 + V1 I1 cosθ1 + V2 I2 cosθ2 + V3 I3 cosθ3 +

19 13b

∞

= V 0 I0 +

Vk Ik cosθk k =1

Total power factor λ is λ=

effective power = apparent power

W V12 + V22 + V32 +

×

I12 + I22 + I32 +

19 13c

averaged value of power = effective value of v effective value of i The displacement power factor is Power factor of fundamental frequency =

V1 I1 cos θ1 = cos θ1 V 1 I1

19 13d

553

20 Power Electronics Applications, Part 2 Circuit Theory In this chapter,we will discuss typical circuits for power converters with different input/output frequency. We will first look at rectifiers (AC-to-DC converters) and the basic applications of diodes and thyristors. Readers can also become familiar with reading power electronics circuits through the study of AC-DC converters.

20.1

AC-to-DC Conversion: A Rectifier with a Diode

20.1.1

Single-Phase Rectifier with Pure Resistive Load R

Figure 20.1a shows the fundamental circuit of a rectifier: a single-phase circuit with one diode D between the AC sinusoidal power source and the DC side load R. Load R is a resistive load or powered load, such as a driving motor load. Diode D is in the turn-on state (conductive state with a negligibly small voltage drop of, say, 0.8–0.5 V) whenever the AC source voltage v has forward directional polarity, while it is in the turn-off state (an almost infinitely high resistive state with a small leakage current) whenever the source voltage has reverse directional polarity. In other words, as our power source is a sinusoidal waveform voltage, current id flows through the diode every half cycle during the time the diode is forward biased (Figure 20.1b). Therefore, voltage vd as well as id on load R are pulsing and DC biased, which is the equivalent of DC. The suffix d indicates DC quantities. As the load is purely resistive, the equation id = vd/R is always satisfied, so vd and id are in-phase, as shown in Figure 20.1b. Next, the power P that is provided from the source to load R (average power per cycle) can be calculated as follows. The AC sinusoidal source voltage and current are given by the following equations v t = 2Vrms sin ωt = 2Vrms sin θ i t = 2Irms sin ωt = 2Irms sin θ = 2

20 1

Vrms sin θ R

where Vrms = Irms R. Then the average voltage and current of the load are calculated as follows. vavr =

1 T

T

v t dt = 0

π

1 2π

2π

2Vrms sin θdθ +

π

0

0dθ =

1 2Vrms −cos θ π0 2π

20 2a

2 V = 0 45V rms rms = π iavr =

1 T

T

i t dt = 0

1 2π

π 0

2Irms sin θdθ +

2π π

0dθ =

1 2Irms − cos θ π0 2π

20 2b

2 I = 0 45I rms rms = π

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

554

20 Power Electronics Applications, Part 2

Source voltage vs = √2Vrmssin ωt

0

id

D

t

DC output voltage vd These two areas are equal. vd

νavr

R 0

υs

(a) Circuit

π ω

2π ω

t

(b) Voltage wave forms

Figure 20.1 Half-wave rectifier with pure resistive load.

The two shaded areas between the average voltage lines are the same size as the nature of the average value (Figure 20.1b): P=

1 2π

2π

v θ i θ dθ =

0

1 2πR

2π

v2 θ dθ =

0

1 2πR

π

2Vrms sin θ

2

2π

dθ +

0

π

0 2 dθ 20 3

2 Vrms

1 = Vrms Irms = 2R 2 The time duration of the current flowing through load R is a half cycle, so P becomes a half value of Vrms Irms, which is an ordinary value in the case of an AC load circuit. Also, 1 vavr iavr = 0 45 Vrms 0 45 Irms < Vrms Irms = P 2

20 4

vavr iavr is smaller than P = (1/2) Vrms Irms for a pulsing waveform.

20.1.2

Inductive Load and the Role of Series-Connected Inductance L

Next, in Figure 20.2, inductance L is series-connected with load R. We will examine how DC-side voltage and current quantities are modified by this additional inductance. di θ The DC-side voltage vd is always divided by i(θ) R and vL = ωL through the time interval t = (0, π): dθ v θ = i θ R + ωL

di θ dθ

20 5

Then the current id(t) is forced to be delayed by phase angle φ = tan−1 ωL/R toward vd(t), where cosφ is the power factor of the total impedance of R and ωL. Figure 20.2b shows the current waveform. Note that the moment θm when the following equation is satisfied exists once per cycle: vd θ − id θ R = ωL

did θ =0 dθ

at θ = θm

20 6

20.1 AC-to-DC Conversion: A Rectifier with a Diode

υd

A id

D

Rid

RIm νd

υs

A'

𝜃m π

0

L R

√2Vrms

2π

3π

υd id =υsid Rid2

𝜃

B'

υs

δ D on-state

0 (b) νd and R·id

(a) Circuit

B

𝜃m 𝜋

δ

2π

3π

𝜃

(c) Instantaneous power νs ˙ id and R ˙ id2

Figure 20.2 Single-phase diode rectifier with inductive LR load.

did become zero, so that vd(θm) = dt R id(θm) is satisfied. The voltage condition within the time 0 ≤ θ ≤ θm (or the interval [0, θm]) is vs > vd, and the current did builds up, so the inductor’s stored energy increases. Gradually, after time θm, vL = ωL becomes a negative value dθ because the current id begins to decrease, so that the diode forward directional voltage continue not only through (0, π) but also through (0, δ). Then after 0 = π, current id continues to flow until θ = δ (δ; extinction angle) in spite of the fact that the polarity of vd (t) is negative during the interval [π, δ] because of the inductor’s stored energy. The current waveform i(θ) can be calculated by solving Eq. (20.6) under the initial condition of i(θ)|θ = 0 = 0, and the solution is given as follows: Because did (θm)/dθ = 0 at this moment, i(θm) takes the peak value and ωL

iθ =

=

2Vrms R2 + ωL

sin θ − φ + sin φ e − ωLθ R

2

2Vrms cos φ sin θ − φ + sin φ e − θ cot φ R

where tan φ =

ωL , cos φ = R

20 7

R R2 + ωL

2

Namely δ can be calculated by Eq. (20.8a). e δ cot φ sin δ− φ = −sin φ

20 8a

Next, we compare area A and area A . area A − area A =

θm

ωL

0

= ωL

did dθ + dθ

θm

0

Im

ωL

did dθ dθ 20 8b

did = 0

did + 0

δ

Im

So, in the graphical interpretation, δ is the time when the voltage-second areas A and A become the same. Next, the power through interval [0,2π] is illustrated in Figure 20.2c. During the interval [0, θm], the supplied source power is partly consumed by R and is partly stored by inductance L (the shaded area B shows energy stored by L). Then L discharges the energy through the interval [θm, δ] as power. The power is provided from L to load R during the interval [θm, π] and is consumed as i2d R. Furthermore, during the interval [π, θm], the polarities of vd and id are different,

555

556

20 Power Electronics Applications, Part 2

which means the power supplied by L is partly consumed by R but is partly returned to the power source. Of course, the total stored energy (area B) and discharged energy (area B ) of L are the same. The interval [δ, 2π] is the turn-off state. However, if inductance L is designed with a larger value, θm moves close to π, and δ moves close to 2π (symbolically, if L ∞, then θm π, δ 2π); in this way, the current waveform becomes flattened with a wider base. In other words, the turn-off state interval [δ, 2π] is reduced. If L is large enough (or, more exactly, if the time constant T = ωL/R in Eq. (20.7) is large enough), then the turn-off state interval disappears, so the load current id flows continuously. 20.1.3

Roles of Freewheeling Diodes and Current-Smoothing Reactors

Referring to Figure 20.3, diode D2 is inserted in parallel with the RL series-connected load branch. In the interval of positive polarity of AC source voltage vs(θ), diode D1 is in the on state with forward voltage, while diode D2 is in the off state with reverse voltage. Such an operating state is symbolized here by [D1, D2] = [on, off ]. The circuit condition of this state in Figure 20.3 is the same as in Figure 20.2, because D2 is off, so the source current is(θ) and load current id(θ) are the same: is(θ) = id(θ). In the second half-cycle interval π ≤ 0 < 2π with negative source polarity, the circuit becomes [D1, D2] = [off, on], so the load-current pass is switched from vs – D1 − L − R to D2 – L − R. Such current pas- switching is called current commutation. In this state, vs(θ) = vd(θ) is satisfied because D2 is on (Figure 20.3b). Also in this state, the energy stored by L is provided to load R through the D2 pass. Furthermore, the average value of the load voltage is calculated using Eq. (20.2a): Vavr = 0.45 Vrms. Due to diode D2, steady-state load current flows continuously through the interval with the waveform in Figure 20.3c. If L (or ωL/R) is larger, the load current id(θ) increases and flattens and approaches the same waveform as pure flattened DC current. Diode D2 is called a freewheeling diode, and inductance L is called a smoothing reactor. Owing to their roles, pure flattened DC current can be obtained with the circuit in Figure 20.3. Furthermore, we have the following equations: vd θ = R id t + L

did t did θ = R id θ + ωL dt dθ

20 9

vd id

D1 is

D2

L

i2

R

vd υs

π

0

D2 on-state

√2Vrms 3π

D1 on-state

D1 on-state

S

vs (b) Load voltage vd

(a) Circuit

id = is+i2=Id

id = is+ i2 is 0

π

θ

2π

i2

i2

is 2π

3π

θ

is 0

(c) is , i2 and load current id L small R Figure 20.3 Single-phase rectifier with a freewheeling diode.

π

i2

is 2π

i2 3π

(d) Pure flattened current id L quite large R

θ

Id

20.1 AC-to-DC Conversion: A Rectifier with a Diode

By integration, 1 2π 1 vd θ dθ = R 2π 0 2π ∴ vavr = R iavr

2π

vavr =

id θ dθ +

0

ωL 2π

did θ dθ

2π 0

dθ = R iavr

20 10a

Here, the first term of the middle side is equal to R iavr, and the second term is zero under steady state, so the relation vavr = R iavr is obtained. The average current is determined by vavr and R, so iavr is not affected by the inductance value L. Note that voltage source vs(θ) is sinusoidal, while current is(θ) is a distorted waveform with a DC component. So, if L/R is larger, id can be approximated by pure DC, while the source current is(θ) is Fourier-series expanded as follows: is θ =

iavr 2iavr 2iavr 1 1 sin 3θ + sin 5θ+ + sin θ + 5 2 π π 3

DC component AC component

20 10b

Harmonic component

is(θ) includes a DC component and a higher harmonic component in addition to an AC component. However, the DC component and higher harmonic components of is(θ) cannot contribute effective power to the load, because power source vs(θ) is composed of only fundamental frequency. In other words, effective power supplied to the load is given by vs(θ) × (fundamental component of is(θ)), while the DC component and harmonic component are unfortunately included in is(θ). The current DC component may cause flux bias on transformers (core flux saturation; see Section 6.9).

20.1.4

Single-Phase Diode Bridge Full-Wave Rectifier

Next, we will study the single-phase bridge circuit shown in Figure 20.4, which is a typical full-wave rectification circuit. The current pass in the interval θ = [0, π] is the source Gen D1 the load LR D2 . That in the interval θ = [π, 2π] is Gen D2 load LR D1 . Thus, commutation of the current loop pass is repeated every cycle. Figure 20.4b shows the waveforms for the load voltage and current where vd(θ) is always sinusoidal but with positive polarity through [0, π] and [π, 2π], and the current id(θ) is pure flattened DC under the condition of L/R ∞. The situation can be written using the following equations: Source side vs θ = 2Vrms sin θ

20 11a

is θ = 2Irms sin θ − φ

νd

id = Id √2Vrms

π

2π

3π

θ

υs

id is

D1

D2 D2, D'1 on

D1, D'2 on

L

D1, D'2 on

νd

νs

R D'1

D'2

is

0

(a) Figure 20.4 Single-phase full-wave bridge rectifier.

π

Id –Id (b)

2π

3π

θ

557

558

20 Power Electronics Applications, Part 2

On the load side, For θ = 0, π vd θ = 2Vrms sin θ id θ = =

1 2π

π

2Irms sin θ − φ dθ =

0

2 Irms − cos θ −φ 2π

π 0

2 2 Irms −cos π − φ + cos − φ = Irms cos φ 2π π

For θ = π,2π vd θ = 2Vrms sin θ 1 id θ = 2π

2π

2Irms

π

2 Irms cos θ −φ −1 sin θ −φ dθ = 2π

2π π

2 Irms cos φ = π

20 11b

Then vd θ = 2Vrms sin θ id θ =

2 Irms cos φ flattened DC magnitude because φ is a constant π

where tan φ =

20.1.5

ωL R

0

Roles of Voltage-Smoothing Capacitors

Figure 20.5 shows another single-phase bridge circuit where a capacitor element C (smoothing capacitor) is connected in parallel with the load, instead of a series-connected smoothing reactor. In other words, the total load is the parallel circuit of R and 1/ωC from the converter’s viewpoint. The capacitor C is charged by pulsing current from the AC power source only under the condition of |vs| ≥ vd once every half cycle by the states D1, D2 = [on, on] and D1 D2, = [on, on]. The charged energy is discharged to the load during the interval |v5| ≤ vd because all the diodes are in an off state. Figure 20.5b shows the waveforms. The transient voltage during the interval of all the diodes in an off state can be calculated as the solution of the following equation with the condition ω CR: id = C where

dvd vd + dt R

id t = 2Irms sin ωt −α , ω

where time-constant T = CR The solution is 1 1 vd t = 2Irms cos α e − CR ωC where time-constant T = CR

CR

20 12a

20 12b

Therefore, if C (then T = CR) is designed to be large enough, the load-voltage waveform is flattened into pure flattened DC voltage. In conclusion, the smoothing capacitor C can flatten the load voltage to pure DC.

20.1 AC-to-DC Conversion: A Rectifier with a Diode

νd υs

id ls

is

D1

D1'

iR

𝜋

2𝜋

D1, D'2 on

D2, D'1 on

0

𝜃

D2 + C

υs

√2Vrms

vd iC

rs

All the diodes All the diodes off-state off-state

R

νd

D2'

is

is

Ip

I1M Fundamental component (a)

(b)

Figure 20.5 Single-phase full-wave bridge rectifier with a voltage-smoothing capacitor.

20.1.6

Three-Phase Half-Bridge Rectifiers

Figure 20.6 shows a three-phase half-wave rectification circuit where balanced three-phase source voltage v1(θ), v2(θ), v3(θ) is provided to the anode side of diodes D1, D2, D3, and the DC load R and the smoothing reactor L (L/R is a large value) are series-connected between the cathode-side common terminal and the source neutral point N. Note that we will use the suffixes 1, 2, 3 for the phase-abc elements and variables in this chapter. We will discuss specific circuit characteristics in most cases instead of the condition of the connected outer circuit, so it is better to distinguish the circuital symbols 1, 2, 3 and the phase symbols a, b, c of the connected outer circuit. Diode D1 is in the on state during the interval when v1(θ) is of positive polarity and is larger than v2(θ), v3(θ), and the situation is the same for D2, D3. Then the diodes can be on alternatively for the interval 2π/3. Figure 20.6b shows the waveforms. υ1

υ2

υ3

0 i1

υ3

υ1 N

𝜃

D1

i2

2𝜋 3

D2

2𝜋 3

2𝜋 3

υa υ2

D3

i3

υR

υR (L=∞)

υR M id

υd R

υa

L

0 Diodes current 0 Diode on-state

(a) Circuit Figure 20.6 Three-phase half-bridge rectifier.

i1

i2

i3

i1

D1

D2

D3

D1

(b) Voltage waveform

559

560

20 Power Electronics Applications, Part 2

Diode D1 is in the on state for

v1avr =

1 2π

5π 6 π 6

2Vrms sin θdθ =

Also D2 is on for D3 is on for

π 5π , with source voltage v1(θ), and the average load voltage during this interval is 6 6 6 Vrms = 0 39 Vrms 2π

5π 9π , and v2avr = 0 39 Vrms 6 6

20 13a

9π π , and v3avr = 0 39 Vrms 6 6

Then, the average load voltage for the total interval [0, 2π] is vavr =

3 6 Vrms = 1 17Vrms 2π

20 13b π 5π , is 6 6

The average load current through D1 for 1 iθ = 4π 6

5π 6 π 6

2Irms sin θ −φ dθ =

3 6 Irms cos φ 2π

20 14a

And the situations are the same for other intervals through D2, D3. The average load current for the total interval [0, 2π] is 3 6 Irms cos φ = 1 17Irms cos φ 2π wL 0 where tan φ = R iavr =

20.1.7

20 14b

Current Overlapping

Referring to Figure 20.7, the bridge currents i1, i2 i3 overlap during the small commutating interval u in an actual circuit, because commutation of i1, i2 i3 is delayed a little by the inductances Xl, respectively. For a small interval u for the commutation of i1, i2, vu = v1 −Xl

di1 di2 = v2 − X l dθ dθ

① ②

i1 + i2 = id

20 15

where v1 = 2Vrms sin θ +

5π 6

, v2 = 2 Vrms sin θ +

π 6

③

∴v1 − v2 = 2Vrms − 3sin θ vu = Ll

2 v 1 + v2 Vrms cos θ = 2 2

di2 t di2 θ di2 θ v1 − v2 = ωLl = Xl = 2 dt dθ dθ

This is because the loop circuit of voltage (v1 − v2 ) and inductance 2L1 exists through the time u.

20 16

20.1 AC-to-DC Conversion: A Rectifier with a Diode

υ0

υ1

υ2

Voltage drop caused by current overlap

υ3

2𝜋 3 i1

D1 X

υ1 υ3

i2

N

D2 X

υ2

On-state element

D3

𝜃 D1

D2

X υd

id

i3

Id

υou

i3 = 0

i2

i1

D3

Current lap time u

id

𝜃 =𝛼

(a)

(b)

Figure 20.7 Overlapping bridge currents.

And i2 = 0 at θ = 0; then i2 θ =

1 Xl

θ

v1 − v2 dθ = 2 0

i1 θ = Id − i2 θ = Id −

6 Vrms − cos θ θ0 = 2Xl

6 Vrms 1− cos θ 2Xl

6 Vrms 1− cos θ 2Xl

20 17

Further, i1(θ = u) = 0 or i2(θ = u) = Id. Then u = cos − 1 1 −

2Xl Id 6V

20 18a

The average voltage drop caused by current overlapping is Ex =

3 2π

v1 −v2 3 dθ = Xl Id 2π 2 0 u

20 18b

From Eqs. (20.18b) and (20.13b), vavr =

3 6 3 Vrms − Ex = 1 17Vrms − Xl Id 2π 2π

20 19

This is the modified version of Eq. (20.13b) with the current lapping time u. The equation shows that the average voltage vavr is smaller for larger leakage inductance Xl.

20.1.8

Three-Phase Full-Bridge Rectifiers

Figure 20.8 shows a three-phase full-bridge rectifier. In the A side arm circuit, the diode with the highest voltage among D1, D2, D3 is switched-on, whereas the other two are switched-off, so each diode successively becomes switched-on for the time duration of 120 degrees. Therefore, the potential voltage vAN of point A from neutral point N is the same as that

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20 Power Electronics Applications, Part 2

v1

v2

v3

vAN vAB θ

N vBN

On-devices

D1

D3

D6

D5 A i1 v3

v1

D1

D2

vAB id

D3

vAB

vd

D3 D4

D5

π /3

vd – π /6

i2 v2

L=∞

D2

π /6

R

i3

vAB = vAN – vBN D'1

D'2

D'3 B

(a) Circuit

0

θ=0

(b) Voltage waveforms

Figure 20.8 Three-phase full-bridge rectifier.

in Figure 20.6. The situation for vBN of point B is similar, but with opposite polarity, because the diode with the lowest voltage among D1 , D2 , D3 is switched-on, successively. Now the potential voltages vAN and vBN of the arm points A and B become the waveforms, respectively, as shown Figure 19.8b. The voltage vAB = vd between A and B is given as a pulsing waveform, as shown in Figure 20.8b.

20.2

AC-to-DC Controlled Conversion: Rectifier with a Thyristor

In the previous section, we discussed diode rectifiers, which are widely used at the front end of power electronics systems to convert AC input to an uncontrolled DC output voltage. However, it is necessary for the DC voltage to be controllable in some applications, such as DC-motor drives, battery chargers, and so on. The AC to controlled DC conversion is accomplished in phase-controlled converters by means of thyristors. Now we will discuss controlled rectification circuits with thyristors in which diodes are replaced by thyristors in the same circuits we examined in the previous section.

20.2.1

Single-Phase Half-Bridge Rectifier with a Thyristor

Figure 20.9 is the same circuit as Figure 20.3, except that diode D1 has been replaced by thyristor T1. Again, ωL of the ωL 0 smoothing reactor L is large enough against R, and tan φ = R A thyristor does not conduct until it has been triggered by a positive pulse applied to its gate, even with forward voltage between A-K terminals. From then on, it conducts indefinitely until forward voltage vAK is cut off or the polarity is alternated. The thyristor vAK is under forward-charged voltage from the source during the interval [0, π], but it does not conduct until θ = α, so the load current flows through the loop circuit D L R during the interval [0, α]. When a gate pulse signal is given at θ = α, the thyristor conducts until vAK with positive polarity is lost or reversed at θ = π, and current flows through the loop T L R. When the thyristor conducts in [α, π], the diode current becomes zero (if = 0), so the thyristor current i becomes equal to the load current (i = id).

20.2 AC-to-DC Controlled Conversion: Rectifier with a Thyristor

iex (excitation current) i1

L=∞

i2

v1

vd

v2 n 1 : n2

Id R vd

(a) Circuit

v1 (𝜃) = n1 ·

2π θ(

= wt)

α

v1,v2

i2 Id i2

=

v2(𝜃) n2

d𝜑1 (𝜃)

for [𝜋, 2𝜋 + 𝛼]

– n2 · i''dc = – Φ''dc for [𝛼, 𝜋] for [𝜋, 2𝜋 + 𝛼]

for [𝜋, 2𝜋 + 𝛼] 𝜋–𝛼 2𝜋

· (Id – i'dc ) =

𝜋+𝛼 · i''dc = i1avr– 2𝜋

0.45 V

i1avr+ i'dc i'1avr+

A i1avr – Actual current i1 + iex

VRAV

0

n2 n1

0

d𝜃

for [𝛼, 𝜋]

where i1avr + =

0 i 1=

Id

i2 (𝜃) =

d𝜑1 (𝜃)

= √2 Φ rms sin𝜃 d𝜃 n2 · (Id –i'dc ) = n2 Id – Φ'dc for [𝛼, 𝜋]

n2 · (Id –i'dc ) n1 n2 – · i''dc n1

i1 (𝜃) =

π

=

n1

𝜑1 (𝜃) = n1 · i1 (𝜃) =

i0

α

d𝜃

v1 (t) = √2V rms sin𝜃

where

vd 0

v1(𝜃)

d𝜑1(𝜃)

B

0 0

0 (b) Voltage and current waveforms

α

π

(c) vd -α curve

Figure 20.9 Single-phase half-bridge rectifier with a thyristor.

Next, the thyristor is cut off in the interval [π, 2π] because of reversed voltage vAK, and the cut-off state continue in [2π, 2π + α] while vAK changes to positive polarity. Throughout the cut-off interval [π, 2π + α], flat current flows through the loop D L R, and then the load current is commutated to the loop T L R at the next gate signal time θ = 2π + α. The thyristor conducts again in [2π + α, 3π]. Now, the average load voltage is calculated as follows. vavr =

1 2π

π α

2 Vrms sin θdθ =

2 1+ cos α Vrms − cos θ πα = 0 45 Vrms 2 2π

20 20

vavr can be controlled using the ignition angular timing α; Figure 20.9c shows the vavr – α curve. In the Figure 20.9, the following equations describe the nature of the transformer: v1 θ = r i 1 θ + n 1

dφ θ dθ

v1 θ dφ θ v2 θ = = n1 dt n2

20 21a

20 21b

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20 Power Electronics Applications, Part 2

These equations indicate that v1(θ), v2(θ), dφ(θ)/dθ always have the same waveform. As the source-side voltage v1 is sinusoidal, v2 and the changing speed of the flux dφ1/dθ are also sinusoidal v1 θ = 2 V1rms sin θ

v2 θ = 2 V2rms sin θ

20 21c

dφ θ = 2 Φrms sin θ dθ

20 21d

φ θ = − 2 Φrms cos θ + Φdc

20 21e

where Φdc is a constant decided based on the initial condition. v1(θ), v2(θ), dφ(θ)/dθ satisfy the alternate condition because their integration through the interval [π, 2π] is zero, so they are alternate values. The equations also indicate that dφ(θ)/dθ depends only on v1(θ) and is not affected by the secondary current i2(θ). Further, the core flux φ(θ) is somewhat biased by the DC offset value Φdc. The secondary current can flow only in the interval [α, π]. Then i2 θ =

Id

for α, π

0

for π,2π + α

20 21f

Now we try to integrate v1(θ) by one cycle [0, 2π] 1 2π

2π

1 2π

v1 θ dθ = 0 =

0

1 = 2π

2π 0 2π 0

r i1 θ dθ + n1

1 φθ 2π

2π 0

20 21g

1 φ 2θ −φ 0 r i1 θ dθ + n1 2π

where φ(2θ) = φ(0) ∴

1 2π

2π

r i1 θ dθ = 0 and

0

2π

i1 θ dθ = 0

20 21h

0

Then i1(θ) is an alternate value, but it is not sinusoidal. Indeed, i1(θ) is a rectangular waveform, as explained next. We know n1 i1(θ) = n2 i2(θ), given the nature of the transformer ampere-turn cancelation. Then n2 Id n1 0

i1 θ =

for α, π

20 21i

for π,2π + α

The i1(θ) from this equation does not satisfy the alternate condition. Therefore, i1(θ) should include another term so the alternate condition is satisfied: i1 θ −idc =

i1 θ =

n2 Id − idc n1

for α, π

20 21j

for π, 2π + α

0 + idc = idc

Because i1(θ) satisfies the alternate condition (one-cycle integration becomes zero), then

0=

1 2π

∴ i1avr +

2π + α α

i1 θ dθ =

π − α n2 = 2π n1

1 2π

Id −idc

π α

n2 n1

Id − idc dθ +

1 2π

2π + α π

π+α i = − i1avr − =− 2π dc

idc dθ =

π − α n2 2π n1

Id −idc +

π+α i 2π dc 20 21k

Average current in [α, π] Average current in [π, 2π + α] Area A

Area B

In conclusion, i1(θ) becomes a asymmetrical rectangular wave with Eq. (20.21j), and it satisfies the alternate condition using Eq. (20.21k). The actual current is a little modified because the excitation current iex is superposed.

20.2 AC-to-DC Controlled Conversion: Rectifier with a Thyristor

The DC-biased flux causes transformer core saturation, and severe wave-distorted excitation current flows out to the primary side; furthermore, the transformer core is overheated due to greatly increased hysteresis loss from the core. So, the circuit in Figure 20.9a should not be used for important applications. We will study transformer saturation caused by DC-linking flux bias of the source transformer further in Chapter 6, Section 6.9.2.

20.2.2

Single-Phase Full-Bridge Rectifier with a Thyristor

Figure 20.10 is a full-bridge rectifier with four thyristors: is the same as in Figure 20.4, except that the diodes are replaced by thyristors. The load voltage and current waveforms are given in Figure 20.10b. The current id is a flattened DC waveform through the conductive interval [α, π] because ωL/R 0 is assumed. If α = 0 is selected, the situation is equal to that in Figure 20.4, which indicates that the diode rectifiers from Section 20.1 are a subset of the controllable thyristor rectifiers. Now we will discuss this circuit. The load current id is a continuous waveform or discontinuous waveform based on φ = tan−1 ωL/R and α. Figure 20.11a–d shows the load vd, id waveforms for four different cases. Let’s examine the critical condition between the continuous waveform current and the discontinuous waveform current. The on state of thyristors T1 and T2 is written using the following equation: ωL

did θ + Rid θ = vs θ dθ

20 22

where dθ = ωdt tan φ = ωL R The critical condition is found by using id = 0 at θ = α, and the solution is 2 Vrms cos φ sin θ −φ − sin α −φ e − θ −α cos φ R

id θ =

20 23

ed

vd ( = Rid ) vs

0

2π

π

Gate signal for T1,T2′

id is

vd T2′

T2,T1′on

T1,T2′on T2,T1′on

ed T1′

π+α

L

T1 T2

vs

α

Gate signal for T2,T1′

R

i1

EDC 0

(a) Circuit Figure 20.10 Single-phase full-bridge rectifier with thyristors.

Id

(b) Voltage and current waveforms

θ

565

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20 Power Electronics Applications, Part 2

vs

vs

id

ed δ π

0

α of T1,T2′

θ

2π

α of T2,T1′ T1,T2′ on

π

0

π

0

θ

2π

α of T1,T2′ α of T2,T1′ T1,T2′ on

T2,T1′ on

θ

2π

π

0

δ

is (a) id discontinuous (α >ϕ) vs

id

ed

θ

2π

is

(b) Limit of continuous id (α = ϕ) vs

id

ed

T2,T1′ on

ed

id = Id 2Vrms

2Vrms π

0

α of T1,T2′

2π

θ

α of T2,T1′ T1,T2′ on

π

0

π

θ

α of T1,T2′ α of T2,T1′ T1,T2′ on

T2,T1′ on

is 0

2π

T2,T1′ on

is 2π

(c) id continuous L : large R

θ

Id 0

π

– Id 2π

θ

(d) id continuous L : quite large R

Figure 20.11 Load current modes with continuous and discontinuous waveforms.

The first term in the brace [ ] is a steady-state term, and the second is a transient term. The transient term obviously becomes zero under the condition of α = φ, so id becomes sinusoidal (Figure 20.11b). In conclusion, id becomes a discontinuous wave current if α > φ (Figure 20.11a), a sinusoidal wave current if α = φ (Figure 20.11b), a continuous wave current if α < φ (Figure 20.11c), and finally a DC wave if ωL/R 0 (Figure 20.11d). On the other hand, the source-side current is becomes sinusoidal if α = φ but becomes a rectangular wave with delayed angle α. Furthermore, let’s examine the commutation of the current as shown in Figure 20.11d, where the DC current wave id is a continuous wave. The id pass changes from T1(T2 ) to T2(T1 ) at θ = π + α. At the moment when T2(T1 ) becomes the turn-on state, the source voltage vs with opposite polarity appears at the load terminal, so T1(T2 ) becomes the turn-off state simultaneously. In other words, this commutation is conducted by the change of the external source voltage instead of by a gate signal. Such commutation is called external commutation. Needless to say, the thyristor circuit has an externally commutated converter at its core. Such external commutation by thyristors can theoretically be conducted only within 0 ≤ α ≤ π, and cannot be done within −π ≤ α ≤ 0, which is called the self-commutation zone.

20.2 AC-to-DC Controlled Conversion: Rectifier with a Thyristor

α of T2,T1′

α of T1,T2′ vs

– vs 2Vrms

θ 2π

π

ed 0

β id

is

id – Id

β

0

π

0 i = –I d d

π

id = Id

T1,T2′ on

T2,T1′ on

Ed = – Eo

2π

θ

2π

θ

T2,T1′ on

Figure 20.12 Reverse conversion or externally commutated inverter operation mode.

The average voltage vd avr of the load voltage vd in the case shown in Figure 20.11d) is calculated as follows, where the id waveform is pure DC current: vavr =

1 π

π +α α

vs dθ =

2 2Vrms cos α = R Id π

20 24

π The average voltage vavr is proportional to cos α, so vave ≥ 0 in 0 ≤ α ≤ . In this zone, effective power P is provided from 2 the AC source side to the load as P = {average value of vs is = vd id} ≥ 0. So, this operation mode is called forward conversion or rectification mode. In contrast, vavr ≤ 0 is obtained in the zone π/2 ≤ α ≤ π, which means effective power is provided from the load side to the AC side. Of course, as shown in Figure 20.10a, a DC power source EDC(= −vavr) is required instead of load R at the same terminal positions in this operation mode. This operation mode is called reverse conversion or external inverter operation mode, and the waveforms are shown in Figure 20.12 (where L/R is larger). The angle defined by β = π − α is called the angle of advance. The reverse-conversion operation mode is the case where power is transmitted from DC to AC; however, AC-side sinusoidal voltage is always required in order to achieve turn-off commutation at the angle of advance β. Using a highvoltage DC transmission line system is general practice when a power-sending terminal station is controlled by forward conversion mode and the receiving terminal station is controlled by reverse conversion mode. 20.2.3

Three-Phase Full-Bridge Rectifier with Thyristors

Figure 20.13a–e shows the circuit and waveforms of a three-phase full-bridge rectifier with thyristors. The source voltage v1, v2, v3 is balanced three-phase; the lagging ignition angle is α for the top-group thyristors T1, T2, T3 with positive polarity and is also α for the bottom-group thyristors T1 ,T2 ,T3 with negative polarity. Also L/R is large enough because of large inductance L. Each thyristor conducts for 2π/3 in turn per cycle. The cathode-side voltage ed1 at the top side arm point is also v1 from T1 on, v2 from T2 on, and v3 from T3 on. The anode-side voltage ed2 at the bottom side arm point is also v1 from T1 on, v2 from T2 on, v3 from T3 on, and so on in the same manner (see Figure 20.13c). The load voltage is given by ed = ed1 − ed2, and the waveform in Figure 20.13b is obtained. This is the case with six pulses (number of commutations per cycle p = 6).

567

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20 Power Electronics Applications, Part 2

α of T1

α of T2

α of T3

v1

v2

v3

ed1 2Vrms

2π π

0 id

α of T1′

α of T3′

α of T2′

ed2

(c) DC voltage ed1, ed2

ed1

0

θ

v1

i1

v2

i2

v3

i3

T1

T2

T3

L

T3 on T1 on T3 on T1 on T2 on T1′ on T2′ on T3′ on T2′ on

ed

(d) On-states duration of the thyristors vdc R

ed2 T1′

T2′

EDC

(a) Circuit v12

v13

v23

v21

v31

v32

v12

6Vrms = 2Vl-lrms π

0

α

π 6

T3′

2π

θ

Id

i1

0

i2

0

i3

0

–Id Id

(b) DC voltage ed = ed1 – ed2

2π π

π

θ

–Id Id

2π θ 2π

π –Id

θ

(e) Source current

Figure 20.13 Three-phase full-bridge rectifier with thyristors.

The average voltage ed avr is calculated using the following equation: 3 ed avr = π

5π 6 +α π 6+α

2Vrms sin θdθ =

3 6Vrms 3 2Vl −lrms cos α = cos α = 1 35Vl −lrms cos α π π

P = ed avr Id

20 25

π π ed avr is proportional to cos α, so 0 ≤ α ≤ is the range for forward conversion if ed avr ≥ 0, and ≤ α ≤ π is the range of the 2 2 reverse-conversion zone if ed avr ≤ 0. Figure 20.13b shows the output waveforms of the DC-side voltage (where the condition L/R is larger because of smoothing reactor L) where six different controlled ignition angles α for every π/6 pitch are demonstrated as a function of α. These figures and Eq. (20.25) indicate that the average DC voltage ed avr can be expressed by K cos α, as shown in Figure 20.14a. As a result, the average DC voltage ed avr can be controlled by α. Figure 20.14b shows waveforms with different α. Furthermore, if DC power source EDC is provided, as shown in Figure 20.13a, the DC power can be converted into the AC mode with arbitrary frequency. Such DC-to-AC reverse operation is called the inverter operating range, and we will discuss it again in Section 20.6 as a function of synchro converters.

20.2.4

Higher Harmonics and the Ripple Ratio

The waveforms of phase currents i1, i2, i3 are shown in Figure 20.13e given the design condition of a larger L/R. Now we will calculate the Fourier series expansion of the phase-a current (refer to Figure 19.17 and Eq. (19.12c) for the process): i1 =

2 3Id 1 1 1 1 sin θ −α − sin 5 θ − α − sin 7 θ −α + sin 11 θ − α + sin 13 θ − α − 5 7 11 13 π

∞ 6Id −1 k sin θ −α + sin 6k ± 1 θ −α = 2 π 6k ± 1 k =1

20 26

20.2 AC-to-DC Controlled Conversion: Rectifier with a Thyristor

Ed 1 0.5 0

π 2

π a

–0.5 –1

(a) Ed -α curve Vp Vd 𝛼 = 0˚ VN

Forward conversion

𝛼 = 60˚

𝛼 = 90˚

Zero power factor

𝛼 = 120˚

Reverse conversion

𝛼 = 150˚

(b) Output DC voltage waveforms Ed = kcos α with ignition angle α Figure 20.14 Output DC voltage of Three-phase full-bridge rectifier by thyristor.

The phase-a current i1 includes harmonics of (6 k ± 1) with order k = 1, 2, 3;…. The fundamental component of i1 is delayed from v1 by α, so the power factor of the fundamental component (displacement power factor) is cosα: The effective magnitude of i1 Ieff =

2 3Id

The effective magnitude of the fundamental component I1 = 6 Ripple ratio THD refer to Eq 19 9f

THD =

Ie I1

2

−1 =

Id π

π2 − 1 = 0 311 9

Effective power provided from the source P = 3Vrms I1 cos α = ed avr Id =

3 6 Vrms Id cos α π

20 27

569

570

20 Power Electronics Applications, Part 2

In conclusion, the effective power P that is provided from the AC source is determined by a fundamental frequency effective component and is of the same magnitude as the power consumed by the load. The power caused by higher harmonic components of i1 remains non-effective. The total power factor λ of this bridge circuit is given by the following equation, where the waveform of id is pure flattened DC current.: λ=

E d Id 3 = cos α = 0 955 cos α 3Vrms × effective value of ia π

20 28

The total power factor λ is a little smaller than the power factor for fundamental component cos α. λ is also proportional to Ed and λ = 0 at α = π/2. 20.2.5

Commutating Reactance: Effects of Source-Side Reactance

In the previous explanations, we have ignored the AC source side impedance Zs = jωLs + rs. Referring to Figure 20.15, we will check the influence of Zs = jωLs for DC voltage vd, while rs is again ignored because ωLs rs. At the moment of v1 exceeding v3, T1 begins to conduct. However, T3 continues to flow current until all the stored energy (1/2)L3 I2d from reactance ωL3 is discharged. Therefore, the overlap angle of T1 and T3 conduction is caused as a transient phenomenon of the commutation. For the transient interval, ed = v1 − ωLs

di1 di3 = v3 − ωLs ① dθ dθ ②

i 1 + i 3 = i d = Id from ② di1 di3 + =0 dθ dθ

③

20 29

Substituting ③ into ① di1 v1 − v3 = dθ 2 v1 + v 3 v2 =− ed = 2 2

④

ωLs

⑤

α of T1

α of T2

u v1

v2 0 v3

u

𝜔ls

X

T1

ed

i2

𝜔ls

Y

T2

0

i3

𝜔ls

T3

ed

Load

L

id

i1 i2 i3

Figure 20.15 Effect of commutation reactance.

2

i3 0

(a) Three-phase half wave rectification circuit

v2

–

ed

v3

π –

Z

u

v2

v1 i1

α of T3

2π

v3

–

2

i1

v1 2

i2 π

θ

i3 2π

(b) Voltage and current waveforms (L is larger)

i1

Id = id θ

20.3 DC-to-DC Converters (DC-to-DC Choppers)

Figure 20.15b shows the waveform of ed, which includes a DC voltage-drop effect caused during the overlap time u. As a result, the average DC voltage decreases, as indicated by the shaded area in comparison with the ideal case with u = 0. Such overlap is caused three times per cycle (or p times for a circuit with p pulses), and the modified Ed with such overlap is given by the following equation. Ed, the average value of ed, can be calculated as twice the (average value of ed1): 3 Ed = 2π

α + 5π 6 α + π6

v1 dθ − ωLs

Id

di1 = 0

3 6Vrms 3ωLs cos α− Id 2π 2π

20 30

The second term on the right side is the voltage drop caused by the AC-side reactance. This phenomenon is called commutation reactance voltage drop. In other words, the second term reduces ed a little as if there is an internal resistive voltage drop. The second term is modified to (pωLs/2π)Id if the pulse number is p. Note that the voltage drop does not increase power loss from the overlap, because this is the voltage-drop phenomenon at point X, Y, Z in Figure 20.15a.

20.3

DC-to-DC Converters (DC-to-DC Choppers)

DC-to-DC converters are widely used in regulated switch mode–DC power supplies as well as in large and small DC motor-driving applications. In most cases, the input of the converters is unregulated DC voltage, which is obtained by rectifying the utility line voltage, so it will fluctuate due to fluctuation of the line AC voltages. Therefore, DC-to-DC converters are often used to convert such unregulated DC input voltage into a controlled DC output voltage at a desired voltage level. Figure 20.16 shows a typical example of such applications. DC-to-DC converters are indispensable as voltage source stabilizers or as reciprocal DC-to-DC transformers of primary and secondary DC voltages. Note that DC-to-DC converters are often called by the familiar name choppers, which originates from the traditional technical term current chop.

20.3.1

Voltage Step-Down Converters (Buck Choppers)

This is a one-directional step-down converter; Figure 20.17a shows the fundamental circuit in which transistor or thyristor devices are used. In DC-DC converters, the average output voltage must be controlled to equal a desired level, although the input voltage and output load fluctuate in most cases. The average output DC voltage can be adjusted by controlling the turn-on interval per cycle of the switching device. In Figure 20.17a, the smoothing reactor L (where ωL/R 0) is series-connected to load R; one freewheeling diode Df is parallel-connected, and as a result, flattened DC output voltage is obtained. Assuming that the transistor and the diode are ideal switching devices, the output voltage E2 is given by the following equation. Average output voltage: E2 =

ton E1 = d E1 ton + toff

d; duty factor or duty ratio

20 31

Battery AC line voltage (1-phase or 3-phase)

Uncontrolled Diode Rectifier

DC (unregulated)

Filter Capacitor

DC (unregulated)

DC-DC Converter

vcontrol Figure 20.16 Typical diagram of a DC-DC converter system.

DC (regulated)

Load

571

572

20 Power Electronics Applications, Part 2

L

S i1

i2

E1

vt

Df

C

E2

R

(a) Fundamental circuit L

S

i2

i1 Ed

vt

Df

M

Em

(b) Motor load application vt

vt

Ed

Ed Em 0

0

t

t i1

i2

i

i I20

I10

I10

0 ton

0 t

toff

i1

i2 I20

ton

ts toff

t

T

T

(c) Continuous and discontinuous load current Figure 20.17 Voltage step-down converter (buck chopper).

The energy stored by L during the interval ton is discharged to the load in the interval toff, so the average voltage of inductance L per cycle is zero; then the average output voltage E2 for load R is the same with E1 per cycle. The output voltage E2 can be controlled in the range of 0–100% of input voltage E1. Power consumption is caused only in load R, so the following equation can be justified by the law of energy conservation: E1 I1 = E2 I2 , I2 =

E1 1 I1 = I1 d E2

20 32

The form of the equation is almost the same as that for AC transformers. Figure 20.17b shows a typical example of a motor load. The ignition control α will be discussed later. We will now examine this circuit in more detail. The circuit condition can be written using the following equation, where Em is the counter-electromotive force of the motor. i) The turn-on state of S (Df is the turn-off state) is di1 + Ri1 + Em = Ed dt Ed − Em 1 −e − t τ ∴ i1 = I10 e − t τ + R where τ = L R and I10 is the initial current at the interval L

20 33a

20.3 DC-to-DC Converters (DC-to-DC Choppers)

ii) The turn-off state of S (Df is the turn-on state) is L

di2 + Ri2 + Em = 0 dt

∴ i2 = I20 e − t τ −

Em 1 − e−t R

20 33b

τ

where τ = L R and I20 is the initial current at the interval So, either continuous or discontinuous waveforms can be obtained based on design selection. Now, let’s calculate the pulse current with a continuous flattened waveform. Using i1 = I20 at t = ton from Eq. (20.33a), and i2 = I10 at t = toff from Eq. (20.33b). Ed − Em 1− e −ton R τ Em 1 −e −toff τ − R

I20 = I10 e −ton τ + I10 = I20 e −toff

τ

Then 1 − e − ton τ Ed Em 1− e −dρ Ed − = −ξ I10 = −ρ −T τ R R 1 −e R 1 −e I20 =

20 33c

e ton τ −1 Ed Em e dρ −1 Ed − = −ξ ρ T τ e −1 R e −1 R R

where τ = L R ξ = Em Ed ton ton T T ton = dρ, ρ = , d = = τ τ T τ T I10 and I20 give the largest value and the smallest value of pulse waves, respectively, in Figure 20.17. Then, using L ∞ and applying the first approximation of the Taylor expansion (see Appendix, A.4), I10 = I20

d −ξ

Ed d E d − Em = R R

ton Ed = d Ed T The equation shows that perfectly flattened load current can be obtained by making L equivalent to stepping down the source voltage to the desired magnitude level d Ed. Eavr =

20.3.2

20 33d 20 33e ∞. A step-down chopper is

Step-Up (Boost) Converters (Boost Choppers)

Figure 20.18a shows a step-up converter that boosts DC load voltage. If the smoothing reactor L is fairly large, then flattened DC current I1 flows through L during switch S turn-on state, and as a result energy E1I1ton is stored by L. Then all the stored energy is discharged during S turn-off state. Capacitor C is also large enough that the output voltage is flattened to pure DC E2. Then the discharged energy from L to the load during the toff state of S should be (E2 − E1)I1toff. Both energies should be equal under a steady state. Then E1 I1 ton = E2 − E1 I1 toff ton + toff T ∴ E2 = E1 = E1 toff toff 1 1 or E2 = ton E1 = 1 −α E1 1− T where T toff 1

20 34a

573

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20 Power Electronics Applications, Part 2

L

R

D i2

i1

Ed

S

Em

L

vs

i1 E1

(c) Equivalent circuit of switch S on and off states

D i2 vs

S

E2

R

C

Ed 0

t

(a) Circuit D

L i1 G

i

i2

Em S

Ed

vs

i1

i2

I20

I10

I10

0 ton

toff

t

T (b) Regeneration mode operation of DC motor

(d) Voltage and current waveforms

Figure 20.18 Step-up (boost) converters (boost chopper).

The output E2 is boosted to a value larger than E1 because T/toff The input and output power is E1 I1 = E2 I2

1. 20 34b

In other words, boost-up converters perform the function of what we call DC-DC transformers. Of course, boost converters can produce continuous current as well as discontinuous current. They can also be operated in generation mode and regeneration mode. Figure 20.18b shows the operating condition with a DC motor in regeneration braking mode. The equivalent circuit in the switch S on and off states is shown in Figure 20.18c. Then, For S on-state di1 + Ri1 = Em L dt ∴ i1 = I10 e −t τ +

20 35a Em 1 −e −t R

τ

For S off-state di2 L + Ri2 = Em − Ed dt Ed − Em 1 −e −t ∴ i2 = I20 e −t τ + R

20 35b τ

The equations i1 and i2 give the continuous and discontinuous waveform of the current, depending on the time constant τ, and Figure 20.18d shows the waveform of the continuous current.

20.3 DC-to-DC Converters (DC-to-DC Choppers)

20.3.3

Buck-Boost Converters (Step-Down/Step-Up Converters)

The main application of a buck boost (step-down/step-up) converter is regulated power supplies, where positive and negative polarity output are desired with respect to the common terminal of the input voltage, and the output voltage can be either higher or lower than the input voltage. Figures 20.19a and b show a typical circuit of a buck-boost converter and the voltage and current waveforms, respectively. When switch S is in the on state, vL become the same voltage with E1 and current i1 flows through the E1 – S − L loop, and energy is charged to L. When S is switched off, current i2 begins to flow in the reverse direction of i1, so the polarity of the output voltage E2 changes. Under steady-state operation, the average voltage vL per cycle is obviously zero. Then 2π

vL dt = 0

20 36a

0

Therefore, under the condition of large L, E1 ton = E2 toff ton ton d E1 E1 = E1 = 1−d toff T −ton ton turn-on ratio where d = T ∴ E2 =

20 36b

Therefore, E2 can be stepped up or stepped down by changing d: in other words, by changing the ignition time duration ton. As i1, i2 are flattened to I1, I2, I1 ton = I2 toff

20 36c

toff T − ton 1−d I1 ∴ I2 = I1 = I1 = d ton ton Therefore, ignoring the switching loss of S, E1 I1 = E2 I2

20 36d

The equation shows that the power on both sides is the same, so the buck-boost converter also works as a DC transformer. Figure 20.20 shows the voltage-changing ratio E2/E1 to d characteristics of these three different types of converters.

ton 0

D

S

iS 0

vL E1

i1

L

i2

C

R

E2

toff

iD

iR

Figure 20.19 Buck-boost converter (step-down/step-up converter).

E2

t

t t

0 iC 0

(a) Circuit

E1

t (b) Current waveforms

575

20 Power Electronics Applications, Part 2

5 Voltage changing rate E2/E1

576

4 Step-up (boost) converter 3

Step-down/step-up converter

2

Voltage step-down converter

1

0

0.2

0.6 0.4 Duty factor d

0.8

1

Figure 20.20 E2/E1 to d characteristics of the three different converters.

20.3.4

Two- and Four-Quadrant Converters (Composite Choppers)

In most DC motor applications, the source side and motor load side are fixed; however, the motors are often required to be in reverse rotation (regenerative operation). A converter for such an operation mode performs a two-quadrant operating function in that the current polarity is reversible while the voltage polarity is not changed. Further, if interchangeable function of the source side and load side is required, the converter has to provide a fourquadrant function because the polarities of the voltage and current are reversible. These are called composite choppers. Figures 20.21a and b show the two quadrants with voltage and current polarities. In the circuit, S1 and D1 work as a stepdown chopper from the source side, and S2 and D2 work as step-up converter from the load side. In the reverse mode, mechanical rotating motion energy is returned to the source side as electrical energy (regenerating function). In the circuit in Figure 20.21b, the load current flows continuously from S1 D1 S2 D2 in this order, so it is an alternate-pulse current. Then the average load current is biased to positive (forward operation) or to negative (regeneration), depending on the controlled switching interval. A typical application of such a converter is regenerating braking of electric cars. Note that S1 and S2 should not be turned on simultaneously, because doing so results in a short-circuit fault in the source.

vt Ed

0 S1

T L

im

Ed

ia S2

D1

t

ton

D2

vt

Em

M

iD1

iS1

0

t iD2

(a) Current reversible chopper Figure 20.21 Two-quadrant DC-DC converter.

iS2

(b) Voltage and current waveforms

20.3 DC-to-DC Converters (DC-to-DC Choppers)

Tm

vt

Second quadrant

First quadrant

Reverse braking (inverter mode)

Forward motoring (rectifier mode)

Em > 0, ia < 0

Em > 0, ia > 0

Tm > 0, 𝜔 < 0

Tm > 0, 𝜔 > 0 𝜔

ia Third quadrant

Reverse motoring (rectifier mode) Tm < 0, 𝜔 < 0

Fourth quadrant

Em < 0, ia < 0

Em < 0, ia > 0

(a) Operation modes by voltage and current polarities

(b) Operation modes by torque and rotating direction

S3

S1 D2 im

Ed S2

Forward braking (inverter mode) Tm < 0, 𝜔 > 0

L

D4

Em M S4

D1

D3

(c) Circuit Figure 20.22 Four-quadrant DC-DC converter.

Figures 20.22a–c show a typical circuit and operation modes with a bridge-type reversible chopper (four-quadrant converter), which is two sets of two-quadrant converter circuits. With this arrangement, the source side and load side are interchangeable. 20.3.5

Pulse-Width Modulation Control (PWM) of a DC-DC Converter

In Figure 20.17a, the average value E2 of the output DC voltage is given by Eq. (20.31), shown again here: E2 =

ton E1 = d E1 ton + toff

20 37

ton is the switch duty ratio. ton + toff Figure 20.23 shows the ignition-control method to obtain any desired output voltage value vdesired ≡ E2 regardless of fluctuated input voltage E1. Figure 20.23a shows a block diagram to generate a control signal. A control signal vcontrol(t) given by the following equation is generated in the amplifier over time and is introduced to the comparator: where d =

vcontrol t = vdesired t −vactual t

20 38

In the comparator, vcontrol(t) and a repetitive sharp saw-tooth voltage vst(t) are compared over time; then, the generated switch control signal is in the on state during the interval vcontrol(t) ≥ vst(t) and in the off state during the interval vcontrol(t) < vst(t). This method is called pulse-width modulation (PWM) control. These functions of signal generation and ignition control are made up of microprocessor-based digital control technology. Note that with this practice, any constant output voltage E2 can be obtained, regardless of fluctuating input voltage E1.

577

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20 Power Electronics Applications, Part 2

+

vdesired (t)

vcontrol (t)

Amplifier −

vactual (t)

Switch control signal

Comparator

Repetitive waveform (a) Block diagram

vst = Sawtooth voltage

vcontrol (amplified error)

V^st t

0 vcontrol > vst on

Switch control signal

ton

on

off

off toff

vcontrol < vst

Ts (b) Comparator singals Figure 20.23 PWM control of a converter.

Further, this explanation suggests that any waveform output can be obtained by changing vdesired to trace the desired waveform. It is possible, and we will discuss PWM control in more detail later. Finally, needless to say, these explanations are applicable for Figures 20.17a, 20.18a, 20.19 and similar circuits explained in later sections.

20.3.6

Multiphase Converters

Figure 20.24 shows a multiple-phase converter. It has pulse number m on the source side. This is m parallel circuits of the single-phase chopper in Figure 20.17. With this practice of m large parallel circuits, smooth operation of DC motors with smaller ripples can be achieved, and the reliability of the circuit is greater, whereas the circuit size is larger.

S1

S2

Sn

•••••••

L1

i1

in

•••••

i2

Ed

L2

i F

Ln D1

D2

••••••••

Figure 20.24 Multiphase converter.

Dn

M

20.4 DC-to-AC Inverters

20.4

DC-to-AC Inverters

20.4.1

Overview of Inverters

A device that converts power from DC power to sinusoidal AC power is called an inverter. Inverters are widely used in AC motor drives and uninterruptible AC power supplies where the objective is to produce sinusoidal AC output power whose magnitude and frequency can be controlled. Figure 20.25a shows the typical arrangement of an inverter for a motor-driving load. In an AC motor load, the voltage at its terminals should ideally be sinusoidal and adjustable in magnitude and frequency in spite of the fact that in most cases, the inverter’s input DC power is obtained from a rectifier circuit, which will probably fluctuate and include harmonics. All of the functions just described can be accomplished by inverters. A typical inverter is a converter through which the power flow is reversible. However, during most of the operating time, the power flow is from the DC side to the AC motor side, requiring inverter mode operation; it is rarely reversed from AC to DC. Therefore, these switch-mode converters are called switch-mode inverters in order to emphasize such reversible characteristics (Figure 20.25a). However, in applications where a braking operation is performed frequently (typically, railroad trains), a better alternative is regenerative braking, where the energy recovered from the motor-load inertia is fed back to the utility power grid. Figure 20.25b shows such an arrangement in which the switch-mode converter is used at the input side instead of at the rectifier. We are interested in the function of an inverter that controls both sinusoidal AC voltage magnitude and frequency. Controllable (self-commutation mode) switching devices (IGBT, MOSFET, and so on; see Sections 19.5.6 and 19.5.7) are used as the key elements. Further, very high-speed switching control-signal generation using a digital processor becomes another key technology. In addition, externally commutated inverters equipped with thyristors are frequently used when the primary DC voltage is supplied by voltage-controllable switch-mode converters instead of diode rectifiers. 20.4.2

Single-Phase Inverters

Figure 20.26 shows a voltage source inverter and a current source inverter. (Most major applications are based on the former model.) In Figures 20.26a and b, the ideal switches S1, S4 and S2, S3 repeat on/off switching alternately without time-lapping. Figure 20.26c shows the waveforms of the output voltage v on the node branch in Figure 20.26a (or the

+ Vdc -

50/60Hz ac

Dioderectifier

Filter capacitor

AC motor

Switch mode inverter

(a) DC to sinusoidal AC inverter

+

50/60Hz ac

Vdc -

Switch mode converter

Filter capacitor

Switch mode inverter

(b) Reversible converter Figure 20.25 Inverter and reversible converter.

AC motor

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20 Power Electronics Applications, Part 2

id

S1, S4 on S2, S3 off S3

S1 L

i

Vs

R 𝜈

S2

S4

S

Vs Output voltage 0 𝜈 –Vs

IS Output current t i –IS

Output current 0 i

t

Output voltage 𝜈

Output power 0 p = 𝜈i

t

Output power p = 𝜈i

Input current id

t

Input voltage 𝜈d

(a) Voltage type inverter

S1 IS

𝜈d

C

S3

i

S2

R 𝜈

S4

0

Current of S1, S4 0 iS1, iS4

t Voltage of S2, S3 𝜈S2, 𝜈S3

Current of S2, S3 0 iS2, iS3

t

For circuit(a)

(b) Current type inverter

T1

T2

Voltage of S1, S4 𝜈S1, 𝜈S4 For circuit(b)

(c) Waveforms

Figure 20.26 Inverter.

output current i in Figure 20.26b), the waveform of the output power pout, and the input current id (or input voltage vd). As switches S1, S4 and S2, S3 repeat on/off switching alternately, the load voltage v in Figure 20.26a (or the load current i in Figure 20.26b) alternates every T/2 with a rectangular waveform. The load current i (load voltage v) exponentially lags v, because the load (typically motor load) is inductive. Also, the output power pout and the input power pin can be written using the following equations: pout = v i pin = vs id =

20 39a Vs i for the interval T1 − Vs i for the interval T2

= DC voltage vs input current id

20 39b

pout = v i as the product of v and i has a saw-tooth waveform. Therefore, the input power Pin (= Pout) and the input current id should also have saw-tooth waveforms because of the relation in Figure 20.26c. The state during the small beginning interval of each cycle is vs > 0, id < 0, and pin = pout < 0, so the power flow is in a regenerative direction from the AC side to the DC side during this interval. This fact indicates that the inverter is reversible. The situation is the same for the more generic inverter model shown in Figure 20.27, where sinusoidal output voltage vac and current iac are assumed due to the filtering function of the inverter. Within one cycle, four quadrant modes of vac and iac polarities are included. In conclusion, the inverters have four operational quadrant mode. Again referring to Figures 20.26a and b, obviously the output current (or the output voltage) waveforms are affected by the load characteristics of inductive or capacitive loads and the power factor. We have studied Figure 20.26, with rectangular output voltage v versus output current i on the R and L seriesconnected inductive load. Figure 20.28 shows the waveforms of the output voltage v versus the output current i with three different load modes of R, L, C. In case (a), with a pure resistive load R, i is of the same form as v. In case (b), R and L are the same as in Figure 20.26a or Figure 20.21. In case (c), with R and C, i behaves almost the same way as in case (a) at the initial moment of change, because C dv/dt ∞. In case (d), with of RCL load, the waveform of i becomes oscillatory due to Qfactor = ω0 (L∕R) = L C/R ≥ 1; and the waveform can be made closer to sinusoidal if the output frequency f is close to f0 (= ω0/2π), where f0 is the natural frequency of L and C. With this condition, the load current i is switched off at current zero, which means switching loss is decreased almost to zero.

20.4 DC-to-AC Inverters

iac 2 Rectifier id

1 Inverter

iac

1–phase switch– mode inverter + filter

+ Vd

–

vac

0

+ 3 Inverter

vac –

(a)

4 Rectifier

(c) vac

iac t

0

4

1

2

3

(b) Figure 20.27 Four-quadrant function of an inverter.

Load

v

0

t

i

0

t

(c)

i

0

t

(d)

v R

(a)

(b)

L

R

C

L

R

C R

i

0

t

i

0

t f=f0 (Q, Large)

Figure 20.28 v–i waveform characteristics of an inverter with four different load modes.

Figures 20.29a and b show actual circuits of a single-phase type inverter. In Figure 20.29a, the ideal switches S1, S2, S3, S4 have been replaced with sets of a transistor and a diode in parallel connection. Because each switch has to flow current both ways, a parallel-diode attached to a transistor (a feedback diode) is indispensable in the actual circuit. Note that alternate voltage of repetitive rectangular waveforms is written using the following equation as a Fourier series expansion: vt = 2

6 1 1 1 1 V sin ωt − sin 5ωt − sin 7ωt + sin 11ωt + sin 13ωt− π 5 7 11 13

∞ 6 −1 k = 2 V sin ωt + sin 6k ± 1 ωt π 6k ± 1 k =1

20 40a 6 V sin ωt + vharmonic = 2 π

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20 Power Electronics Applications, Part 2

P id

P

Tr3

Tr1

Vs 2 Vs

D1 L i

Cd A

Vs 2

D2

Tr2 N

D3 B

Load

v

D4

Tr4

N

t

v 0

Output voltage

Reactor

Tr1, Tr4 Tr2, Tr3 on on i 0

Output current

Tr1, Tr4 Tr2, Tr3 Current flowing 0 through the switch

D1 A D2

vd

Vs t

D1, D4 Current flowing through the feedback diode

Tr1

Is

t

t

0

i Load

Tr2

D2, D3

(a) Voltage source type inverter

Current control

C

v

Tr3 D3 B D4 Tr4

Is Required current

(b) Current source type inverter

Figure 20.29 Single-phase bridge inverter.

6 V sin ωt π 6 1 1 1 1 V sin ωt − sin 5ωt − sin 7ωt + sin 11 ωt + sin 13ωt− t = 2 5 7 11 13 π

vfundamental t = 2 vharmonic

6 V = 2 π

20.4.3

∞

−1 k sin 6k ± 1 ωt 6k ± 1 k =1

20 40b

20 40c

Three-Phase Inverters

In the previous section, we studied rectangular voltage and showed that it can be obtained using a single-phase inverter. Then we can obtain three-phase alternate rectangular wave voltages using an expanded three-phase inverter. Figure 20.30 shows the actual circuit of a three-phase inverter (source voltage) and the related waveforms; it is equipped with six switching arms, with a turn-on angle of 60 for each arm. With this circuit, the stepped-form alternate voltage can be obtained at the output terminal for a load whose waveform is closer to a sinusoidal wave. Referring to the switching mode of six transistors in Figure 20.30b, Tr1 and Tr2 repeat switching every 180 alternately. Tr3 and Tr4, and Tr5 and Tr6, also repeat switching in a similar way, but at 120 and 240 different times. As a result, vu0, vv0, vw0 become rectangular waves with 180 width, and vuv, vvw, vwu become nearer to 120 . With such switching control, the output load voltage vun, vvn, vwn is obtained, each a stepping waveform but similar to balanced three-phase sinusoidal waves and shifted 0 , 120 , 240 . The voltage at neutral point nvn0 = (1/3) (vu0 + vv0 + vw0) becomes an alternate stepping voltage. Then, the inverter-phase voltages vu0, vv0, vw0 and the load-phase voltages vun, vvn, vwn are different than each other. Next, let’s check the input current id of the DC source side. Referring to interval mode ② in Figure 20.30b, Tr1, Tr4, Tr6 are turn-on states, and the DC-side input current id becomes equal to the load u-phase current iu (id = iu). Assuming a balanced three-phase load, the DC input-side circuit is switched six times per cycle; the situation is the same for all the

20.5 PWM Control of Inverters

id Tr1

Vs 2

υug

Tr3

Tr5

D1

u

D5 iu iv iw

D3 v

0 Vs 2

D2

D4

n L load

D6

Tr6

Tr4

Tr2

w

υun

(a) Circuit T 1

Output phase voltage Load neutral Output line to voltage line voltage Load voltage Output phase current Input current

υuv υuw

3

4

Tr1

On device

υuo υvo υwo

2

Vs 2 0 V – s 2 0 0 Vs 0 – Vs

6

1

Tr2

Tr4 Tr5

5

3

Tr1

Tr3 Tr6

2

Tr4 Tr5

Tr3 Tr6 t t t t

0

t

υwu

0

t

υno

0

υun

0

υun

0

υwu

0

t

iu

0

t

0

t

0

t

0

t

iv

t

Vs 6 Vs 3 2 3 Vs

t t

iw

id (b) Waveforms (three-phase balanced LR series load)

Figure 20.30 Three-phase inverter (voltage source type).

mode intervals ①–⑥, and id becomes six saw-tooth offset waveforms, or id includes sixth-harmonics in the output frequency.

20.5

PWM Control of Inverters

It is time to discuss microprocessor-based digital high-speed switching control schemes to produce sinusoidal output with magnitude and frequency control. Any waveforms can be obtained today, due to advanced control technology based on digital processers. We will begin by discussing inverter control technology.

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20 Power Electronics Applications, Part 2

20.5.1

Principles of PWM Control (Triangle Modulation)

We discussed PWM control of DC-DC converters in Section 20.3.5, where control signal vcontrol was compared with a repetitive switching frequency triangular waveform in order to generate a switching signal. In inverter applications, PWM control is much more complex, because we need to generate control signals by which an inverter can produce sinusoidal output with magnitude and frequency control. In order to produce a sinusoidal output voltage at a desired arbitrary frequency f1, a sinusoidal control signal at the desired frequency f1 is to be compared with a high-switchingfrequency repetitive triangular waveform voltage vtri at switching frequency fsw, as shown in Figure 20.31. We need to define a few basic terms related to PWM control. A triangular waveform voltage vtri at the switching frequency fsw is called the carrier signal, and frequency fsw is called the carrier frequency, which establishes the frequency with which the inverter switching devices are switched. On the other hand, the control signal voltage vcontrol is used to modulate the switch duty ratio and has a frequency f1 (f1 is called the modulation frequency), which is the desired frequency of the output voltage. Simply put, PWM control is when the carrier frequency triangular signal vtri with frequency fsw is modulated by the sinusoidal modulation signal vcontrol with frequency f1. The basic concepts of carrier waves and modulation waves are the same as the theory of carrier waves and modulation waves that is familiar from broadcasting and wireless communications, and so on. The carrier frequency fsw should be quite a bit higher than modulation frequency f1.

S1

Ed 2

S5

S3 D1

D3

D5

0 S2

Ed 2

S6

S4 D2

D4

U V W

D6

(a) Circuit

v*uo

v*vo

ec

v*wo

Tc

S1 Ed 2 vu0 0

2

Ed 2 vv0 0

vr1

vav1 vr2

vav2 t

vr2

Ed 2 0

vw0 vuv

P

Ed

vtri

var

Q

T2

T1

Ed 0

vbr (= –var)

R

v

Ed vvw 0

vwu

Ed 0

Pulse 1 (b) PWM control of a three-phase inverter

t

Pulse 2

(c) Details of triangle modulation

Figure 20.31 Pulse-width modulation (PWM) control (triangle modulation).

20.5 PWM Control of Inverters

PWM control is used in almost every modern power electronics application area for basic switching-signal control. Typical examples are driving industrial motors, railroads, broadcasting, audio amplifiers, computer power supplies, lamp dimmers, electric ovens, cooling systems (air or water), and of course power utilities, as we will discuss later. The PWM switching frequency (the carrier frequency) has to be much faster than the required load-control speed: the typical applied switching frequency is 10–120 Hz for home appliances, 1–20 kHz for driving motors, and 10–100 kHz for audio applications and computer power supplies. Of course, in power electronics, MOSFETs and IGBTs with high-speed switching capabilities are ideal components, because any arbitrary waveforms for voltages, currents, and fluxes can be obtained using PWM control; as a result, the power supply can be flexibly adjusted. Further, another primary advantage of PWM is that power loss in a switching device is very low. When a switch is on, there is almost no voltage drop across the switch; and when it is off, there is practically no current. Power loss, being the product v i of voltage v and current i, is thus, in both cases, close to zero (1–2% or less). Now we will discuss the principle of PWM control. Figure 20.31a is the same circuit as Figure 20.30a, so the output voltage and current are non-sinusoidal, as shown in Figure 20.30b. PWM control can generate switching signals to produce balanced three-phase sinusoidal voltages. Figure 20.31b shows the principle of PWM triangle modulation, which is the most popular PWM control scheme. When the value of the reference signal (modulation signal, the sinusoidal wave in Figure 20.31b) is more than the modulation waveform, the PWM signal is in the high state (1); otherwise, it is in the low state (zero). With this practice, each generated pulse alternates at two constant time values of Ed/2 and –Ed/2 over time, while in contrast, the pulse width is proportional to the instantaneous sinusoidal signal vcontrol(t). Figure 20.31c shows a section of Figure 20.31b: one cycle of the triangular carrier frequency wave is shown, including an enlarged triangular carrier wave and a small part of a sinusoidal waveform vcontrol(t). The modulated pulse widths of pulse1 and pulse 2 are T1 and T2, respectively. Based on this figure, the sharp slope of the triangular wave can be written in two ways: Ed/(Tc/2) and (vr1 + vr2)/T1. Then the following equation is derived: PQ vr1 + vr2 Ed = = T1 Tc 2 QR Tc vr1 + vr2 1 , Tc = vav1 where vav1 = ∴ T1 = fc Ed 2

20 41a 20 41b

PWM signal

Source signal

Therefore, the pulse width of the first pulse T1 is proportional to var1. In the same way, the pulse widths T1, T2, T3, in series, respectively. Then, if the modulation signal vav1, vav2, vav3, satisfies a are proportional to vav1, vav2, vav3, sinusoidal wave, T1, T2, T3, is the PWM signal of the sinusoidal wave. Note that the center point of each pulse width is always coincidental with the zero point of the triangular carrier wave. gives a sinusoidal wave voltage because the instanSwitching control using the PWM signal pulse series T1, T2, T3, taneous output voltage is proportional to the switching duration with duty ratio d. Triangle wave modulation can be achieved using a simple electronic circuit with a triangular-waveform-generating oscillator and a comparator. The triangular carrier waveform of vtri is replaced with a saw-tooth carrier waveform, as shown in Figure 20.32 or Figure 20.23. 1 We have discussed the principle of PWM control for driving a motor: the phase voltage (volts) of a load motor is modulated as a series of pulses that result in a sine-like flux density (teslas) waveform in the motor’s magnetic circuit. The smoothness of the resulting waveform can be controlled with the width and number of modulated impulses per given 0 cycle. Note that the same principles of PWM control can be used as con1 trol methods for any other variable quantities y(t). The variable y(t) can be v[volt], i[amp], ψ[tesla], P[MW], Q[MVA], or mechanical quantities such as torque[N m], revolving speed ω[rad/s], thermal 0 temperature [degree], flowing water volume [m3/sec], and so on. Such instantaneous analogue values y(t) can be expressed using pulse Time widths if higher-frequency on/off switching is available. This explanation for the small one-cycle time interval from Figure 20.32 PWM (saw-tooth carrier waveform modulation). Eq. (20.41) is expressed, in other words, as follows.

585

586

20 Power Electronics Applications, Part 2

During the one-cycle time interval 2π of high-speed signal frequency (carrier frequency), a switch may be or may not be switched once, so a rectangular pulse (height ypeak, width Ton, and 2π = Ton + Toff) can be created per cycle. The average value yavr per cycle is given by the following equation: yavr =

1 2π

2π

y t dt = = 0

where d =

Ton

1 2π

ypeak dt + 0

1 2π

2π

0dt = Ton

Ton ypeak Ton = ypeak = d ypeak 2π Ton + Toff 20 42

Ton duty factor Ton + Toff

2π = Ton + Toff The equation shows that the average value per cycle yavr of the signal y(t) is directly proportional to the duty factor d = Ton/(Ton + Toff) per cycle. 20.5.2

Another PWM Control Scheme (Tolerance Band Control)

A PWM control scheme using the intersective method can produce accurate output voltage to the control signal; however, it may require too much switching for control purposes. So, to reduce switching times, it is helpful to allow a reasonable error tolerance. Tolerance band control (delta modulation) is such a PWM control method. Figures 20.33a and b illustrate the principle and the control block diagram. An actual analog voltage va(t) (or any quantity) and the control (reference) signal v∗a with some tolerant ± error Δvlimit are introduced. The actual output voltage (which can be a phase voltage) va(t) is compared with the tolerance band around the reference voltage associated with that phase. If the actual voltage tries to go beyond the upper tolerance band, the associated switch T1 is turned off (that is, T1 is turned on). As a practical comparison, Δerror(t) = va(t) – v∗a (t) is added, and the result is compared with the limits. Every time the integral of the output reaches one of the limits, the PWM signal changes the state. Tolerance band control schemes are widely used not only for sinusoidal wave output inverters, but also for other applications, because appropriate control with allowable tolerances and reduced switching times can be achieved from a practical viewpoint. We will study another example later. Reference current i*A Actual current iA t

0 + SA+

DA+ iA

SA–

A DA–

Vd

υAN

– N 0 SA–: on

t SA+: on

(a) Comparator tolerance band i*A +

ie

Σ

Switchmode inverter

ie

– iA

(b) Figure 20.33 PWM tolerance band (current) control scheme.

A B C

20.6 AC-to-AC Converters (Cycloconverters)

20.6

AC-to-AC Converters (Cycloconverters)

In low-speed, very large power applications, cycloconverters for AC (with frequency finput = 50/60 Hz) to AC (with frequency foutput) power conversion can be used to control the speed of synchronous motors and induction motors. Typical applications of cycloconverters are rolling mill drives, paper-rolling mill drives, ball mills for ore processing, cement kilns, and ship propulsion in the 1–50 MW class. In typical applications, 50/60 Hz three-phase input power is introduced from a utility line through a transformer and the cycloconverter produce balanced three-phase sinusoidal AC power with a very low controllable frequency of foutput ≈ 0–20 Hz. Figure 20.34 shows a typical three-phase cycloconverter and waveforms. The cycloconverter output is derived directly from the power grid frequency input without intermittent DC links. The maximum output frequency is limited to about one-third of the input AC frequency, to maintain an acceptable waveform. Thyristor devices are used for most cycloconverters, and commutation can be executed externally either by primary source voltage or by secondary source voltage, so it has four quadrature conversion characteristics. We saw in Section 20.2.3 (Figure 20.14 and Figure 20.15) that average output voltage is controlled by changing ignition angle α, and that arbitrary sinusoidal waveform voltage can be obtained by adopting cos α-control. The same principle is used for cycloconverters. Figure 20.34a shows a three-phase cycloconverter where the P-converter and N-converter are symmetrically connected and both sides are controlled by the same cos α control signal. Figure 20.34b illustrates controlling α, and Negative converter

Positive converter

va ia

Output voltage 3-ϕ variable Output frequency current output

3-ϕ line–frequency input

ϕ 𝜋

P-CONV Rec. Inv.

P-CONV Rec. Inv.

𝛼

𝜋/2 Ignition angle 0

Inversion

Rectification Filtered output voltage

Inversion

(b) COS α ignition control

(a) Circuit Rectification

N-CONV Rec. Inv.

α < 𝜋 (Fundamental component α > 𝜋 of output voltage) 2 2

Output voltage 0 νa α> 𝜋 2 N-CONV 0 Output current ia

Figure 20.34 Three-phase cycloconverter.

θ

P-CONV

α 0, ifi > 0 or vfi < 0, ifi < 0, and the N-converter is in the on state under the condition vfi > 0, ifi < 0 or vfi < 0, ifi > 0. They are never on simultaneously. The fundamental component of the saw-tooth output voltage with frequency foutput is sinusoidal. So, the output voltage from the load viewpoint is sinusoidal with some higher harmonic distortion.

589

21 Power Electronics Applications, Part 3 Control Theory

21.1

Introduction

The first high-power electronic switching devices were mercury arc valve rectifiers, which were utilized in the 1900s. Cooper Hewitt invented the mercury rectifier in 1900, and C.P. Steinmetz achieved the next steps when he utilized the idea for practical industrial purposes. He explained his idea for creating DC current by using a mercury power switch to change batteries. Then, after half a century, the first transistors and the first silicon controlled rectifiers (SCRs) appeared in 1947 and 1956, respectively. And today, power electronics converters can be found wherever there is a need to modify a form of electrical power (that is, to create new voltages, currents, frequency, waveforms, and so on). The power range of these converters is from milliwatts to hundreds of megawatts. And this wide variety of applications may even be found in a single power station or substation, or in one factory. Advanced power electronics technology has also had a significant impact on conventional power equipment and has led to the creation of new power equipment. Speed-adjustable pumped-storage generator-motors, brushless generators, and var compensators without capacitors/inductors are typical examples. In addition, the most important common keyword for smart grids or flexible AC transmission systems) (FACTS) is modern power electronics technology. Modern factory automation and railway traffic systems are other typical examples. In this chapter, we will look at power electronics applications in the area of utility power systems, primarily from the viewpoint of control.

21.2

Driving Motors

21.2.1

Induction Motor (IM) Driving Control

Motor driving is used in a very wide power range, from a few watts to many thousands of kilowatts. Among DC motors, synchronous motors, and IMs, DC motor drives have traditionally been used to control speed and position. However, the use of AC IMs for these applications is increasing dramatically as a workhouse of industry. This is due to their low cost and rugged construction, in particular IMs with squirrel-cage rotors. Another reason is that precise power and speed control of IMs has become available due to advanced high-speed switching control technologies based on power electronic circuits. The discussion here assumes a three-phase squirrel-cage type IM. Figure 21.1a and b shows the general concept of a motor-driving system including load, IM, coupling mechanism, and control scheme. However, prior to discussing IM driving practices, we must determine the load requirements’ parameters, such as load inertia, torque control range, speed control range, and direction of motion. Then the motion profile as a function of time should be specified. Figure 21.2 shows some example motion profiles. Speed ωm, position θm, and load torque Tm under steady-state and transient state conditions should be specified as functions of time t. IM drives can be classified into two broad categories based on their use: i) Adjustable-speed drives: A typical application is process control with a feedback scheme or simple speed control without feedback, where the driving speeds and/or torques are controlled. ii) Servo drives: Typical applications are precise position control in machine tools, robotics, and computer peripherals. IMs operate at a nearly constant speed with some slip rate when connected directly from the power grid, with almost constant frequency (50/60 Hz) and voltages. However, today, IMs can be precisely controlled using speed and torque in Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

21 Power Electronics Applications, Part 3

Mechanical load

Power source

Power conversion

Coupling mechanism

Motor

Mechanism

Load

Measured value Control equipment

Energy flow

Signal flow

Control signal

Mechanical load

Position sensor

Speed sensor

Torque sensor

i*

Motor

T* + – T

Power conversion Current

ω* + – ω

Torque control

θ* + – θ

Speed control

(a) Configuration of motor driving system Position control

590

ω (b) Feedback scheme Figure 21.1 General concept of induction motor (IM) driving control.

Speed ωm

0

Time t

Position q

0

Time t

Tarque Tm

0

Time t

Figure 21.2 Load characteristics profile (an example).

combination with a power electronic control scheme, because they are driven by adjusted voltages, frequency, or other electrical quantities. Adjustable-speed and torque control methods of IMs are classified as follows. For IMs with squirrel-cage rotors, i) Primary voltage (Vl) control: Primary voltage is controlled by an AC power regulator. ii) Volts-per-hertz (V1/f1) control: Primary voltage and frequency are controlled by an inverter where the primary voltage is controlled as a scalar value.

21.2 Driving Motors

iii) Vector control (V1/f1): The primary voltage and frequency are controlled by an inverter where the voltage is controlled as a rotating vector. For IMs with Wye-connected rotor windings (double-fed induction machine), iv) Secondary voltage (V2) control: Secondary voltage of the Wye-connected rotor windings is controlled. v) Secondary power frequency (f2) control: Power frequency of the Wye-connected rotor windings is controlled. Typical examples include a speed-adjustable pumped-storage generator motor unit and wind generators.

21.2.2

Volts per Hertz (V/f) Control (or AVAF Inverter Control)

A volts per hertz (V/f ) motor-driving control or adjustable voltage adjustable frequency (AVAF) inverter control is perhaps the most popular and simplest IM speed-control practice. Recall the IM characteristics derived in Chapter 13, in Figure 13.10 and Eqs. (13.35) and (13.59b); these are repeated in Figure 21.3a and the following equations. The phase voltage from Eq. (13.35) is vabc = r iabc + sψ abc = r iabc + sL iabc + l + L siabc

21 1

And the phase voltage from Eq. (13.59b) is

Z=

v abcs i abcs

=

e

jωs t

e jωs t

v dqs i dqs

=

v dqs

rs =

ωs rr + slip ωbase

i dqs

2

X 2M −X ss X rr +

jωs rr r s X rr + X ss ωbase slip

jωs rr + X rr slip ωbase

21 2

where X ss = X ls + X M X rr = X lr + X M The principle of V/f control is based on two observations that use these equations for the IM characteristics. First, the torque-speed characteristics in Figure 21.3 are normally steep in the neighborhood of synchronous speed, so the rotor speed is close to the electrical frequency. The situation is illustrated in Eq. (21.2) and Figure 21.3. If we omit the terms including rs and rr in Eq. (21.2), the phase voltage equation under steady-state conditions is simplified as follows. vabc = − j

ωs X 2M −X ss X rr i abc ωbase X rr

21 3

Therefore, the speed of the motor can be approximately controlled by adjusting the frequency fs = ωs/2π of the supplied power to the stator coil from the inverter. The second observation is based on Eq. (21.1), vabc = r iabc + sψ abc for the phase voltages. For steady-state conditions at intermediate to high speeds, sψ abc dominates r iabc; replacing s = d/dt with ω = 2πf, the approximate relation of vabc ≈ 2πfs ψ abc is suggested. Therefore, in order to maintain constant flux linkage ψ abc, vabc/fs should also be maintained at constant magnitudes. In other words, the stator voltage magnitude should be proportional to the frequency. AVAF control usually means constant v/f (or ψ) control of an IM. Driving control to keep the constant flux linkage ψ enables smooth control of the motor. Figure 21.3a shows T–f characteristics of an IM, derived by keeping v/f at a constant value with valuable parameter f. The right side of the maximum torque point gives a downward approximately straight line, which indicates that v/f is almost a constant value. Then constant v/f operation can be achieved by operating the motor in the straight characteristic region. The intersecting points of the motor characteristics and the load characteristics a, b0, b1, c show operating points under different frequencies f. If the frequency is changed from f6 to fx, the operating point shifts from point a to b0. On the other hand, if the frequency is changed from f5 to fx, the operating point shifts from point b1 to b0. Smooth motor-driving control is achieved this way. However, with this practice, the motor rotating speed (or the slip Slip of the horizontal axis) is affected. Of course, the motor should be operated in the allowable operating range of v and f, as shown in Figure 21.3b. If v becomes too large, the insulation may be broken on the motor’s stator coils; and if v becomes too low,

591

Torque

21 Power Electronics Applications, Part 3

f3

f4

b΄0 a Torque c b1 b0 characteristics of the load Angular velocity a΄

Vrate

f5 fx f6 f2

Voltage

f1

Starting torque

592

Frequency

frate (b) Operation limit

(a) Torque-speed (or slip) characteristics

Mechanical load ωn

+ Power source

Inverter

Rectifier

IM

– f0, V0 (constant)

f1 , V1 (adjustable)

DC link (c) Typical arrangement

Frequency signal

Speed adjustment

f1*

ʃ(2πf1)dt

θ

sinθ

sin θ – 2π 3

sin θ – 4π 3

Va*

Vb* PWM pattern

Inverter

Vc* IM

Boosting signal

Voltage pattern generator (d) V/f control scheme

Figure 21.3 Volt per hertz (V/f) motor-driving control.

the motor cannot continue stable operation because resistive loss cannot be ignored. Thus, smooth motor driving with ψ≈ v/f constant operation can be accomplished. Figure 21.3c and d shows a typical arrangement and the diagram of v-f control subsystem based on PWM control. Figure 21.4 is an example of a single-phase inverter circuit (voltage type) with an illustration of PWM-controlled waveforms; this is the same circuit as in Figure 20.29.

21.2 Driving Motors

υA0 Q1

E/2

A

0 Q2

E/2

Q3

D1 i

Load υ

D2

υB0

D3

D4

–E/2 E/2 –E/2 E

B Q4

E/2

υ = υA0 –υB0 –E

(a) Circuit

(b) Output voltage waveform

Figure 21.4 Single-phase inverter with PWM sinusoidal control.

21.2.3

Constant Torque and Constant Speed Control

The torque equation for an IM is given by Eq. (13.58) in terms of slip frequency. Here is that equation again: ωs ωbase

Tm = rr slip

2

+

X 2M ωbase ωs ωbase

rr slip 2

X lr + X M

idqs

2

21 4a

then rr slip

idqs =

2

+

ωs ωbase

ωs ωbase

X lr + X M

2

Tm 21 4b

X 2M ωbase

rr slip

Therefore, if torque Tm and slip frequency ωs are used, the RMS value of the stator current idqs should be set in accordance with this equation. This current source-control-based operation is possible using phase current feedback in spite of the fact that the three-phase bridge inverter is fundamentally a voltage source device.

21.2.4

Space Vector PWM Control of IMs (Sinusoidal Control Method)

Now we will discuss space vector PWM control, which began to appear in recent years as an advanced PWM control technology. We discussed PWM control of IMs in Section 20.5 and showed sinusoidal modulation PWM control in Figure 20.31. However, we need to see how to generate the necessary sinusoidal modulation signals for PWM control. Straightforward methods of generating modulation signals for different motor-control purposes may require complicated control design. Space vector PWM control can obtain ideal PWM modulation signals using simple sequential practices. Looking at Figure 21.5a, our purpose is to create ideal modulation signals for switches S1–6 by ensuring that balanced three-phase sinusoidal voltage waves ea(t), eb(t), ea(t) are obtained as the ideal inverter output: ea t = Ee jωt , eb t = Ee j ωt − 2π

3

= Ee jωt e −j2π 3 , ec t = Ee j ωt − 4π3 = Ee −jωt e − j4π

3

21 5

593

594

21 Power Electronics Applications, Part 3

E (010)

E

Sa+

Sb+

Sc+

a b c

N E

Sa–

Sb–

E (110)

E (000) E (111)

E (011)

E (110)

e3 e2 e1

E (100)

e (110) e

E (111) E (000)

Sc– E (001)

(a) Circuit

e (100)

E (101) (b) Eight space vector

Notching pattern

a

0

b

0

0

1

1

1

0

0

0

0

1

E (000)

E (100)

E (110)

E (010)

E (011)

E (001)

c Space vector

1

1

0

0

0

1

1

0

0

1

1

1

1

E (101)

E (111)

State1 :+side switch on and – side switch off Switching condition

State0 : +side switch off and – side switch on

(c) Base voltage vectors and the switching conditions

E E (000) E (100) (110) Phase-a notching Phase-b notching Phase-c notching

E (111)

E (110) E (100) E (000)

E (000)

1 0 1 0 1 0 t0

T000

t1

T100

Starting point of a cycle

t2 T/2

t

T110

t3

T111

T111

T110 T100 t110 t5 T/2

Middle point of a cycle

T000

Starting point of the next cycle

(d) Switching patterns based on the base vectors Figure 21.5 Space-vector modulation.

E (100)

21.2 Driving Motors

Now we define eight vectors, as shown in Figure 21.5b and the Eq. (21.6a), which can be obtained by controlling the on/off patterns of three notches for phase-abc, as shown in Figure 21.5c: Esv

E 000 =

2π 2π 2 +j −j 0e j0 + 0e 3 + 0e 3 = 0 3

E 100 =

2π 2π 2 +j −j Ee j0 + 0e 3 + 0e 3 = 3

2 j0 Ee 3

E 110 =

2π 2π 2 +j −j Ee j0 + Ee 3 + 0e 3 = 3

2 jπ Ee 3 3

E 010 =

2π 2π 2 +j −j 0e j0 + Ee 3 + 0e 3 = 3

2 j 2π Ee 3 3

E 011 =

2π 2π 2 +j −j 0e j0 + Ee 3 + Ee 3 = 3

2 j 3π Ee 3 3

E 001 =

2π 2π 2 +j −j 0e j0 + 0e 3 + Ee 3 = 3

2 j 4π Ee 3 3

E 101 =

2π 2π 2 +j −j Ee j0 + 0e 3 + Ee 3 = 3

2 j 5π Ee 3 3

E 111 =

2π 2π 2 +j −j Ee j0 + Ee 3 + Ee 3 = 0 3

21 6

The output voltage of the phase-a notch, for example, becomes E with switch mode [Sa+, Sa−] = [on, off ] and 0 with switch mode [Sa+, Sa−] = [off, on]. So for notch pattern E(110), for example, where the selected notching patterns of phase-abc are [on, off], [on, off], [off, on], respectively, the output voltages become ea E, eb E, ec 0, and the defined π vector E 110 = 2 3Ee j3 is obtained. Now we can obtain any of the eight defined voltage outputs by changing the on/off patterns of three notches. For example, if the operation is conducted with mode E(000) for time interval T000 and then switched with mode E(100) for T100 within one cycle, the average voltage from the following equation is obtained. In other words, an arbitrary average vector voltage e ∠ θ within 60 between E(000) and E(100) can be produced by E(000) and E(100): e ∠θ = E 000 T000 + E 100 T100

T000 + T100

Now we will examine a typical operation mode, as shown in Figure 21.5d. The notch mode is changed by E(000) E(100) E(110) E(111) at time t0, t1, t2, t3 and for the interval T000, T100, T110, T111, respectively, in a half cycle T/2. Then we use this equation e∠ θ = E 000 T000 + E 100 T100 + E 110 T111 + E 111 T111 e ∠ θ = 0 T000 +

2 j0 Ee T100 + 3

2 jπ Ee 3 T111 + 0 T111 3

3 1 T111 T100 + T110 + j 2 2 T T000 + T100 + T110 + T111 = 2

∴e ∠ θ =

2 E 3

T 2

T 2 21 7

T 2

2 j0 2 jπ Ee , E 110 = Ee 3 , E 111 = 0 3 3 Now we can produce any arbitrary vector average voltage e ∠ θ (absolute value e and angle ∠θ) within 0–60 by utilizing E(000), E(100), E(110), E(111). Therefore, we can also produce e0, e1, e2, e3, as shown in Figure 21.5b, by controlling time t0, t1, t2, t3. where E(000) = 0, E 100 =

595

21 Power Electronics Applications, Part 3

With this method, we can produce 24 vectors whose lengths are the same, arranged in a time-series every 15 , as representative vectors per cycle. Then, a set of the 24 time-series vectors e1(t1), e2(t2), , e24(t24) per cycle is approximately one rotation of the vector e(t) = Eejωt. The approximate rotating vector e(t) of arbitrary length within the value E is obtained as follows. One cycle time interval T = 2π is divided into 24 intervals of (2π/24) = 15 . The first switching mode is given by mode 1 from the table Figure 21.5c for the 0th, first, second, and third intervals e0, e1, e2, e3, and the ignition interval is controlled time sequentially in the order e0 e1 e2 e3, as was explained by Eq. (21.7); then the switch mode is changed to mode 2 for the fourth, fifth, sixth, and seventh intervals e4, e5, e6, e7. Thus, one vector rotation is completed in six time-switching mode changes, and the obtained signal wave can be used as the modulation signal waves for balanced three-phase sinusoidal voltages ea(t), eb(t), ec(t), as shown in Figure 20.31.

21.2.5

Space Vector PWM Control (Rotor-Flux Oriented Control)

Most uses of AVAF control by inverters are for driving motors, so producing a constant rotation signal for the linking flux ψ(t) = ψejωt is desirable. Note that we know from Eq. (21.1) that the derivative of linking flux of the stator coil sψ 1(t) is equal to the terminal voltage e1(t). Then the following equation is satisfied: ψ t = ψe jωt

21 8

ψ t + Δt = ψe j ωt + Δt = ψe jωt e jΔt

Then, a rough approximate rotating vector of ψ(t) is obtained using switch-mode rotation in the order of the states E(100) E(110) E(010) . However, a much smoother rotating vector can be obtained using the following principle. Assuming linking flux ψ 1(t) and terminal voltage e1(t) = E(110) of the stator coil at time t, the flux ψ(t + Δt) at time t + Δt (where Δt is a small value) can be written with the following equation (see Figure 21.6a): ψ t + Δt = ψ t + E 110 Δt

21 9

Therefore, a very smooth approximation of ideal constant rotating ψ 1(t) can be achieved using the control as shown in Figure 21.6b. Note the switching operation in the time interval t1–t5 in Figure 21.5d in relation to Figure 21.6b. Operating mode E(000) is maintained during T000 from t0 to t1, so the flux value ψ(t0) is maintained. Then the operation mode is switched to E(100) at t1, so the linking flux gradually increases and reaches the value ψ(t2) = ψ(t0) + E(100) T100 at t2. Then the operation mode is switched to E(110) from t2 until t3, so the linking flux value increases gradually and becomes ψ(t3) = ψ(t2) + E(110) T110 at t3. At t3, the mode is switched to E(111) until t4, which means the value ψ(t3) is maintained until t4. With such time-sequential switching control, a zigzag approximate rotating linking flux signal is obtained in spite of dramatically reduced number of switches. The interval E(000) is the duration of the speed adjustment without switching. This is composed of regular multi-polygons based on the six base vectors E(100) T100,E(110) T110, … Because the frequency with which the switching mode changes depends on the PWM carrier frequency, the resulting regular polygon is an accurate, regular circle.

Δ

Ψ

(t+

110

t)

)∙Δ

t

Ψ (t)

E(

596

E (110) ∙ T110 E (100) ∙ T100 t3~t4 [E (111)]

t)

Ψ( (a)

t2 t0~t1 [E (000)] Ideal locus of the rotating flux Actual controlling signal (b) locus of the rotating flux

Figure 21.6 Phase-vector PWM control (rotor flux-oriented control).

21.3 Static Var Compensators (SVC: A Thyristor-Based Approach))

i1q*

* Current control i1q (q-axis)

+ –

i1d*

+ –

Current control (d-axis)

* i1d

iu* dq i* conver- v sion iw*

PWM control

PWM inverter

θ i1q i1d

ia dq conversion

ib ic Load

Figure 21.7 dq-sequence current PWM control (sinusoidal control).

Now we have obtained the PWM modulation signals for an ideal constant linking flux rotating vector: in other words, a sinusoidal changing linking flux modulation signal. 21.2.6

dq-Sequence Currents PWM Control (Sinusoidal Control)

Figure 21.7 shows an alternative method for generating sinusoidal current-modulation signals. We know that balanced three-phase sinusoidal currents ia, ib, ic can be transformed into DC values id, iq using Eq. (13.50). Then, modulating signals i∗a , i∗b , i∗c , can be generated from the DC current signals i∗d , i∗q . All of these signals can be generated through a digital process using a microprocessor.

21.3

Static Var Compensators (SVC: A Thyristor-Based Approach)

We have discussed power P[MW] conversion using converters, inverters, choppers, cycloconverters, and so on. However, this power electronics equipment can also handle non-effective power ± Q[MVar]. In this section, we will discuss var generation/compensation technology based on static power electronic applications. Power system voltages should be regulated within allowable deviation of +5% to −10% of normal values at any part of the system and over time, except a short fault-recovery time, within 50–100 ms. To get the correct grid voltage, we need the right amount of reactive power in the system. If there is not enough reactive power, the voltage will sag. If there is too much, the voltage will be too high. Maintaining the correct amounts at all times, and in the right part of the grid, is a task that can be performed using reactive power generation or compensation. 21.3.1

SVCs

SVC is the term generally used for combined systems of thyristor switched reactors (TSRs), thyristor controlled reactors (TCRs), thyristor switched capacitors (TSCs), thyristor controlled capacitors (TCCs), and/or fixed capacitors (FCs), and filters, which are based on thyristor-controlled (in other words, external commutation type) equipment. Of course, the individual configuration varies depending on individual requirements; Figure 21.8 shows some typical configurations. In most cases, SVC is connected to 6–35 kV stepped-down low-voltage buses, in spite of the fact that conductor sizes must be large to handle high currents, because the sizes and costs can be reduced. TSRs and TSCs are the same as conventional reactor and/or capacitor banks, except that the mechanical breakers are replaced by thyristor switches where the current-value-adjusting function is not used. The nature of SVCs lies in the use of thyristors connected in series and inverse-parallel. While TSRs and TSCs only control step-mode on- and off-states of the reactor and/or capacitor banks, very fast repeated switching is possible. Fast response of the var compensators is always crucial for grid voltage stability. Further, a fast response to help recover from voltage sags is a particular concern in regions with an abundance of high-tech

597

21 Power Electronics Applications, Part 3

TCR

Filter

TCR

(a) TCR / FC configuration

TSC

Filter

(b) TCR / TSC configuration

Figure 21.8 Static var compensator (SVC).

sensitive loads: it is often a matter of milliseconds for reactive compensators. Regions with heavy electricity-consuming industries such as steel works, ship building, petrochemical loads, and so on, are other examples. Typically, an SVC comprises one or more banks of fixed or switched capacitors with a TCR. With such a configuration, the thyristor for the reactor is ignition controlled within the range 0 ≤ α ≤ π, so smoothly adjusted inductive var is obtained. Then, under total control in combination with the continuous-mode adjusted inductive var and the stepmode capacitive var, ±linear adjusted var values (±Q) within the rated range are instantaneously obtained. Now, we will discuss actual circuits using TCR and TCC. 21.3.2

TCR and TCC

Figure 21.9 shows a TCR circuit consisting of a thyristor bridge and a DC reactor (air-core or iron-core). The circuit is similar to that in Figure 20.8, except load resistor R is replaced by reactor L. Referring to Eq. (20.25), P and Q of the AC source grid side are given by the following equations: P = ed ave Id =

3 2 Vl − lrms Id cos α ① π

3 2 Vl − lrms Id sin α ② π where 0 < α < π

Q=

21 10a

Id DC reactor current Vl − lrms grid-side voltage line-to-line voltage vd avr =

3 π

5π 6 +α π 6+α

2 Vl − lrms sin θdθ =

3 2 Vl −lrms cos α π

Id

Vs

Ed

Control α Figure 21.9 Thyristor-controlled reactor (TCR).

Ldc

Reactor

598

21 10b

21.3 Static Var Compensators (SVC: A Thyristor-Based Approach))

The ignition angular signal α is controllable within 0 ≤ α ≤ π, so Q can be controlled by α within the range 0 ≤ Q ≤ 3 2 π Vl − lrms Id. The equations show that if α is maintained close to the value π/2, large inductive power compensation Q 3 2 π Vl − lrms Id is possible; on the other hand, very small power P (thermal loss of the SVC) is required to build up Id. Obviously, this circuit can generate continuous value-adjustable reactive power (+Q) as a TCR. However, it cannot generate capacitive power (−Q), because the thyristors can be operated with 0 ≤ α ≤ π, so Q from Eq. (21.10) can take only positive values. If some switched capacitor banks are installed at the AC terminals, the equipment can be used as a TCC, or as a reactive/capacitive-controllable SVC. Another alternative to obtain ±Q is to replace the thyristors with GTOs. One disadvantage of this circuit is the difficult task of stable control. That is, α should be kept close to π/2 as much as possible in order to obtain an operating condition of large Q and small P; and Id should be controlled for smooth voltage build-up and adjustment of +Q. However, we cannot simultaneously keep sinα 1 and change Id with α control. We can overcome this dilemma using the asymmetrical control method, which we will discuss in the next section. Note that Figure 21.10a is a single-phase inductive load circuit with series and inverse parallel thyristor switch, which is a simplified single-phase circuit from Figure 21.9, Figures 21.10b, c, and d show the waveforms of the current iL flowing through the inductance and power source, where the ignition angle of the thyristor is α = 90 , 120 , 135 , respectively. The figure also shows the fundamental term iL1 of the Fourier series expanded iL: i) In the case of 0 ≤ α ≤ 90 , the thyristor cannot ignite. ii) In the case of Figure 21.10b for α = 90 , the thyristor can ignite, and the current iL is 90 delayed sinusoidal current whose RMS values equals Vs 21 11 iL = iL1 = ωL where v(t) = Vs sin ωt. iii) If α increases beyond 90 , iL can be controlled as shown in Figure 21.10c and d, corresponding to α values 120 and 135 . As α is increased, IL1 is decreased, thus allowing control over the effective value of inductance connected to the utility grid voltage. The effective inductance value is as follows: Vs Leff = ωIL1 Vs 21 12 2π − 2α+ sin 2α fundamental term of the Fourier series expanded iL IL1 = πωL π where ≤ α ≤ π 2 υs α = 90°

iL = iL1 ωt

0

90°

(b) iL Th1 iL

1

iL1

α = 120° 0

ωt

120°

(c) Th2

vs

L

1′ (a) Inductive load circuit Figure 21.10 TCR: principle of per-phase.

iL α = 135°

iL1 ωt

0 135°

(d)

599

600

21 Power Electronics Applications, Part 3

Therefore, the reactive power drawn by the per-phase TCR at the fundamental frequency is Q = Vs IL1 =

Vs2 ωLeff

21 13

The inductor current is not a pure sinusoidal wave at α > 90 , as shown in Figure 21.10c and d. Fourier analysis of iL consists of odd harmonics in the order 3, 5, 7, 9, 11…. However, it is possible to prevent thirdorder and multiples of third-order harmonics flowing out to the grid by using a delta-connected transformer, which is common practice in TCRs. 21.3.3

Asymmetrical Control Method with PWM Control for SVC

Figure 21.11 shows another circuit, where two sets of the same six-bridge converters (Figure 21.9) are arranged. Now Eq. (21.13) is replaced by the following equations: 3 2 Vl − l rms Id cos α1 + cosα2 ① π 3 2 Vl − l rms Id sin α1 + sin α2 ② Q= π where 0 ≤ α1 ≤ π, 0 ≤ α2 ≤ π, Id DC reactor current

P=

21 14

In this case, α1 and α2 are controllable independently. Then we can control (cos α1 + cos α2) in order to build up a stable value of Id, and control (sin α1 + sin α2) to adjust Q. Small effective power P is required, to establish stable current Id, which is consumed mostly as thermal switching loss. With this approach, Q and Id can be simultaneously controlled over time, so stable initial starting control and smooth adjustable var compensation control are possible. The SVC can generate var power of zero to 2 3 2 π Vl − lrms Id. maximum under the operating condition α1 α2 π/2. In addition to static var generators (SVGs), let’s discuss the asymmetrical control method. Eq. (21.14) can be modified as follows: 3 2 Vl −lrms Id Kd cos φ π 3 2 Vl − lrms Id Kd sin φ ① Q= π where Kd cos φ = cos α1 + cos α2 P=

21 15

Kd cos φ = sin α1 + sin α2 ② In these equations, α1 and α2 are controllable independently, so Kd and φ are also controllable independently, because (Kd, φ) and (α1, α2) are one-to-one correspondent and interchangeable using simple calculations. Therefore, from Figure 21.11, it is possible to control Id with φ and to control Q with Kd. This is called asymmetrical control because α1 and α2 are asymmetrically controlled. In addition, Q can be controlled by k using PWM control; fast, smooth ±Q control can be achieved, preventing annoying voltage flickers caused by industrial loads (arc furnaces, for example) and resulting in fast voltage recovery after Id faults or unstable voltages. Vs The asymmetrical control method has been widely adopted Converter 1 by power utilities and other industrial and residential power electronics applications. A typical example is a motor-driving Control angle α1 Ed L Reactor system, where driving torque T and rotating speed ω are simuldc taneously controlled. Vs

Converter 2 Control angle α2

Figure 21.11 Asymmetrical PWM control method for a TCR.

21.3.4

Statcom or SVG

A statcom (static synchronous compensator) or SVG is also a type of var power-control equipment, but it uses selfcommutation based on gate turn-off devices (typically GTOs

21.3 Static Var Compensators (SVC: A Thyristor-Based Approach))

or IGBTs). The circuit structures are similar to those of an SVC, except the thyristors are replaced with self-turn-off devices. However, the functionality of a statcom is quite different. Due to the self-turn-off devices, control is possible in the ranges 0 ≤ α1 ≤ 2π and 0 ≤ α2 ≤ 2π. Of course, asymmetrical control and PWM control are also commonly used for statcoms. As a result, statcoms can generate adjustable leading/lagging reactive power (±Q) without conventional reactors or capacitors. The statcom is one of the key elements in flexible AC transmission systems (FACTS), whose function is the same as SVG. Now we will discuss the principle of statcoms in detail. Referring to Figure 21.12a, arbitrary sending point s and receiving point r are connected by a line whose impedance is z = r + jx; a statcom is going to be installed at the receiving station bus r. The circuit conditions can be written with the following equations. The transmission line equations between point s and r (line resistance r is ignored) are vs = Vs ∠δ, vr = Vr ∠ 0 , iline = vs − vr jx

①

and vr = Vr sin ωt, iline = iload

②

Sr = Pr + jQr = vr i

∗ line

= vr

v∗s − v∗r Vs Vr ∠ −δ + 90 V2 = −j r ③ − jx x x

Vs Vr sin δ Pr = x V Vs cosδ −Vr Qr = x

21 16

④

Note: Eq. (21.16) ③ ④ is repeated from Chapter 12, Eq. (12.43). The load equations are ∗

Pload + jQload = vr i = Vr Iload cosφ + JV r Iload sin φ vr = Vr sin ωt

21 17

iline = iload = Iload sin ωt − φ = Iload cos φ sin ωt − Iload sin φ cos ωt + ihigh where cos φ is the power factor of the load. The first and second terms of iload are called the effective and non-effective current components, respectively. + is injected as follows: Now we assume that the statcom is installed at point r and current istatcom ①

+ iload = iline + istatcom + = kcosωt istatcom

+

21 18

+ + ihigh the control signal of Statcom ②

Then the current situation at the point r is modified as follows: + = Iload cosφ sin ωt − Iload sin φ cos ωt − kcosωt iline = iload − istatcom

+

+ + ihigh −ihigh

21 19

Due to self-commutation and PWM high-speed switching control of statcoms, we can generate any current waveform. Therefore we will try to control the statcom, so we make the second and third terms of Eq. (21.19) zero. Statcom control is as follows: kcos ωt + ihigh

+

ihigh or

Iload sin φ cos ωt or k + kcosωt

+

Iload sin φ for over time ①

Iload sin φ cos ωt for over time

②

21 20

Referring to Figure 21.12c and d, if ideal adjustments are made to the statcom so Eq. (21.20) is satisfied, the circuit condition is improved as follows: + iline = iload −istatcom

Pline = Pload + Pstatcom

Iload cos φ sin ωt Vr Iload cos φ and Pstatcom

① ②

0

Qline = Qload + Qstatcom = Vr Iload sin φ cosωt −Vr kcos ωt

+

21 21

0 ③

Therefore, we can conclude that the statcom generates QStatcom (positive or negative value) and cancels the reactive power Qload of the load, so the reactive power Qload + Qstatcom becomes zero and the power factor at point r is improved

601

602

21 Power Electronics Applications, Part 3

Vstatcom v⋅ s = Vs 𝛿

x

v⋅r = Vr 0 i⋅+statcom E Xstatcom

Pload , Qload (a) Power grid

ic L Ed vr

vc

(b) Var compensation

v· s v· r 𝛿

iload sin 𝜑

v· s

v· s – v· r = jx · iline

𝜑

iline = iload

i⋅+statcom

iload

iload sin 𝜑

(c1) Before compensation

v· s – v· r = jx · iline

· iline vr

𝛿

iload cos 𝜑

𝜑

Auxiliary small capacitor only for building up and maintaining circuit voltage (never for Var compensation)

(c2) Under compensation by Statcom

(c) Operation mode of better power factor (cos 𝜑 ≅ 20° lagging) v· s

v· s

v· s – v· r = jx · iline

i 𝜑 line

iload cos 𝜑

iload sin 𝜑

iline = iload

v· r

d

v· r

v· s – v· r = jx · iline

iload sin 𝜑

(d1) Before compensation

iload

i⋅+statcom

(d2) Under compensation by Statcom

(d) Operation mode of poor power factor (cos 𝜑 ≅ 70° lagging) Figure 21.12 Var compensation with a statcom (SVG).

to 1.0. Of course, the grid line current iline is minimized because the power factor at the receiving point of the line is also improved to 1.0. We also need to check the effective power of the statcom Pstatcom which is supplied from the line bus terminal and injected into the load. Most statcoms are high-speed switching devices, and very small capacitors are used to build up the terminal voltage; no large resistive load is included in the equipment. In other words, most of the power loss caused in a statcom Pstatcom is switching loss from the turn-off devices, and it is of very small magnitude. Although instantaneous power P statcom t is large, it is taken and given every cycle between the grid and the equipment, so power Pstatcom

21.4 Active Filters

(the average power of P statcom t according to the definition) must be negligibly small. The function of statcom is to generate appropriate reactive (inductive or capacitive) power and to cancel the load reactive power without consuming power other than very small switching loss. In conclusion, a statcom generates a current waveform that cancels the reactive component of the load current. Or, in + , which cancels the var power of the load ±Qload, theoretically other words, it can generate reactive (var) power ± Qload making it zero, in spite of the fact that it is not equipped with capacitors and reactors. Referring to Figures 21.12c and d, the line current iline becomes in-phase with vr, and as a result iline is minimized (the line resistive loss r i2line is minimized) and Qline becomes zero (the receiving power becomes Pline ± jQline Pline ± j0). The improvement is drastic in the case of a poor power factor. Inherently, statcoms (without fixed reactors or capacitors) have symmetrical ratings with respect to inductive rated power (+Qrate) and capacitive rated power (−Qrate). Note that if statcoms are installed at all the receiving points of a grid system, it will theoretically result in all loads operating with power factor 1.0. Then the power grid can be completely operated with a power factor close to 1.0 except for the var requirement for inductive/capacitive MVA of the inherent transmission lines and cables. By discussing the principle of statcoms, we have studied its advantage from the viewpoint of var power flow. However, the fairly fast response characteristics of statcoms should be emphasized as another important advantage. Their voltage recovery sensitivity is outstanding, not only for recovery of slow voltage sag or fluctuation, but also for recovery of large voltage drops caused by faults; inrush current when switching capacitors, reactors, or transformers; ordinal repeated voltage drops in electric furnaces; and so on. + ihigh shows that a A statcom can also cancel the higher harmonic components of the load current. Eq. (21.20) ② ihigh statcom can cancel the harmonic components of the current; we will discuss this in the next section as a technology for active filters. We will discuss the circuit of a statcom in Figure 21.15. Additional note: We discussed the circle diagram of a transmission line in Chapter 12, Figure 12.11 and Eqs. (12.43)–(12.51). If we assume a statcom is installed at point r, these equations are modified as follows: Pr = Pload 2 Qr = Qload + XSTAT ISTAT = Qload + ISTAT Vr Vs Vr sin δ Pr = Pload = x V 2 − VS Vr cos δ Qr = Qload + ISTAT Vr = r x

21 22

2

Pload x 2 + Vr2 −Qload x− ISTAT Vr x = VS Vr 2 Pload + Qload −

Vr2 − ISTAT Vr x

2

=

VS Vr x

2

2

Electrical effects caused by installing a statcom are evaluated with these equations. Obviously, the center point 0, Vr2 x of the original circle diagram can be modified to almost (0,0) using an appropriately adjusted statcom operation, with current generation ISTAT = Vr/x.

21.4

Active Filters

21.4.1

Basic Concepts of Active Filters

Voltages and current in power systems and all the connected loads may include harmonic components as part of their makeup, most of which originates from by nonlinear characteristics of the other members of power systems. Further, most power electronic switching equipment causes higher harmonics. Power electronic equipment has the characteristics to generate harmonics. Of course, large harmonics often cause negative effects on equipment, such as overheating motors, transformers, and capacitors; mal-operation of control schemes; communication interference; and so on. Conventional (passive) filters are often used as countermeasures to remove harmonics, but they can remove harmonic components that are included within the design specified frequency-tuning zones. In contrast, due to the high-speed switching characteristics of turn-on devices, power electronic active filters can remove every frequency component almost perfectly without any matching predesign.

603

604

21 Power Electronics Applications, Part 3

𝜐s =√ 2 Vs sin 𝜔 t

√ 2 Vs System voltage vs

𝜋

0

2𝜋

𝜔t

𝛼

√ 2 Vs Original load I d current (including 0 harmonics) –I iload = isign + ihigh d

𝜋 𝛼

Id Sinusoidal current 0 signal i+sign (t) –Id

2𝜋

𝜔t

if = 4 Id sin (𝜔 t – 𝛼) 𝜋

System System current Load current is = isign is + i+high Original load voltage vs Generated harmonic current i+high (t)

2𝜋 𝜋

Current generation 2𝜋 𝜋

iload = isign + ihigh

𝜔t

𝛼

Id Generated harmonic current i+high (t) 0 –Id

current (including harmonics)

Harmonic current detection Active filter ic∗ = ih Signal

𝜔t

(b) Harmonic wave

(a)

Figure 21.13 Active filter.

Referring to Figure 21.13a, vs is a sinusoidal source voltage of a power system, and load current iload with harmonic components is flowing at load point r. Now we will discuss an active filter installed at the point r. Any periodical waveform of distorted current iload can be Fourier expanded. Referring to Figure 21.13b for the alternate rectangular waveform current as an example, it =

4 1 1 sin ωt − α1 + sin 3ωt − α3 + sin 5ωt − α5 + π 3 5 21 23a

4 ∞ sin 2k −1 ωt − αm = π k =1 or ∴ i t = isign + ihigh 4 where isign = Id sin ωt −α1 π 4 ihigh = Id π

fundamental component

21 23b

∞

1 sin 2k − 1 ωt −αk 2k −1 k =2

higher harmonics

+ Then, if equipment generates current ihigh t whose waveform is the same with these total harmonic components + t can cancel the harmonic components ihigh(t). An active filter’s function is to genihigh(t), this generated current ihigh + t and to cancel the existing harmonic component ihigh(t). Figure 21.13b explains erate a current with the waveform ihigh this principle in the case of rectangular waveform current. Next, we will discuss the principle of generating a PWM modification signal i+(t) as the key function of an active filter. Conventional filtering practice is to use (analogue or digital) band-pass filtering for a switching signal. However, our interest is in advanced digital control technology, such as the dq-method and PQ method. Active filters can also remove unbalanced three-phase current components, or negative-sequence and zero-sequence current components.

21.4 Active Filters

21.4.2

Active Filtering with the dq Method

We discussed the dq0-method in Chapters 10, 11, and 13 as an essential theoretical tool for generators. However, our subject in this section is balanced three-phase steady-state phenomena, so zero-sequence quantities can be eliminated. Then the dq conversion equations are simplified as follows: id = iq

2 3

cos ωt

cos ωt −

2π 3

2π − sin ωt −sin ωt − 3

or idq = C iabc

=

ic t

cos ωt −

2π 3

2π cos ωt + 3

2π 3

ia t ib t

2π − sin ωt + 3

21 24a

ic t

−sin ωt

cosωt ia t ib t

cos ωt +

− sin ωt −

2π 3

id t 21 24b

iq t

2π − sin ωt + 3

iabc = C − 1 idq And P + jQ = va i∗a + vb i∗b + vc i∗c = vd id + vq iq

21 25

If ia(t), ib(t), ic(t) are for a fundamental sinusoidal wave, id(t), iq(t) should be DC quantities. The equations are va t = 2Vrms cos ωt + α1

ia t = 2Irms cos ωt + α2

vb t = 2Vrms cos ωt + α1 − 2π 3

a

vc t = 2Vrms cos ωt + α1 + 2π 3

ib t = 2Irms cos ωt + α2 − 2π 3

b

ic t = 2Irms cos ωt + α2 + 2π 3

21 26

Then vd = 2Vrms cos α1

id = 2Irms cosα2 c

d

vq = 2Vrms sin α1 3 vd id + vq iq = 3Vrms Irms cos α1 − α2 2 3 Q3ϕ = 3Qq = vq id − vd iq = 3Vrms Irms sin α1 −α2 2 ∴ P3ϕd − q domain = 3Vrms Irms cos φ P3ϕ = 3Pd =

Q3ϕd − q domain = 3Vrms Irms sin φ

iq = 2Irms sin α2 a b c

21 27

d

where cos φ power factor φ = α1 −α2 Then, id, iq, as well as P3φ d−q, Q3φ d−q become DC components. That is, if harmonic components are included in id(t), iq(t), those components should correspond to the components ihigh, included in ia(t), ib(t), ic(t). Therefore, all the harmonic components idqhigh in the dq-domain can be excluded from id(t), iq(t) using the digital low-pass filter (LPF), and the output signal id and iq corresponds to the fundamental sinusoidal components of ia, ib, ic. Figure 21.14 shows the control-flow diagram of an active filter using the dq method. In the figure, the current waveform signals ia, ib, ic are converted by C to the signals id, iq, and they are treated by the digital LPF. The extracted signals id ,iq are inversely converted to ia , ib , ic with the C−1 calculation. The signals id , iq are DC values, so ia , ib , ic must be the fundamental sinusoidal components of ia, ib, ic. Finally, the harmonic cancelling signals i∗a , i∗b , i∗c can be obtained, where i∗a is the difference between ia and ia: i∗a = ia −ia .

605

21 Power Electronics Applications, Part 3

Reference voltage va

ia C

ib ic

id iq

LPF

LPF

– id

sin 𝜔 t cos 𝜔 t

C –1

– iq

– ia – ib – ic

+ –

+ –

Compensation signals ia* + –

ib* i*c

Power grid

Figure 21.14 Active harmonic filter: dq method.

PWM inverter Transformer

+

icontrol

Breaker

isystem

DC voltage sensor

606

CT-2 esystem

Ripple-filter PT On-off switching singal Triangle wave – +

PWM control (Var harmonics control)

PWM control (auxiliary capacitor voltage control)

DC voltage present value

icontrol IL

CT-1

Modulation signal generation iload

Load

Figure 21.15 Active filter and/or statcom.

Note that only the load-side current is used as input signals for the equipment, and all the included harmonic components are compensated for over time, so voltages and currents at any other grid point may never be badly affected. Furthermore, if a negative-sequence component is included in ia(t), ib(t), ic(t), then id(t), iq(t) should theoretically include sinusoidal AC components with angular velocity 2ω1. So, the negative-sequence component can also be removed by extracting 2ω1 components from id(t), iq(t) with the active filter. In conclusion, using the dq method based on high-speed PWM switching control, active filters can remove higher harmonic components and negative-sequence (unbalance) components. All of these processes are performed automatically in real time without any predictive tuning process. Obviously, the active filter does not consume any power except switching loss caused in the circuit; the circuit structure is almost the same as that of a statcom. Or, more precisely, var compensation and harmonic removal can be achieved if necessary by generating appropriate modification signals. Figure 21.15 shows a typical circuit of an active filter based on a bridge circuit with six arms of self-commutation devices. PWM asymmetrical control is used to control the var power and maintain the anode-bus voltage. Note that the auxiliary capacitor C in Figure 21.15 can store energy (1/2)C ν2 every cycle. With a statcom (SVG), the capacitor C is has a very small capacity (negligibly small) whose purpose is only to supply power every cycle, which is thermally consumed in the equipment (primarily as switching loss), so the anode bus voltage is built up and maintained. In other words, a statcom, if its purpose is only to compensate for the reactive (var) power of the power system, does not require a var-compensating capacitor. It can compensate for large var power without a capacitor. Also, it does not consume much effective power, except for small switching-loss power caused in the device modules. Further,

21.4 Active Filters

if the purpose is to compensate for negative-sequence power or higher-harmonic power as an active filter, the equipment requires the corresponding power. Negative-sequence current compensation will be discussed again in Section 21.12. Vector PWM Control Based on the dq Method

21.4.3

Vector PWM control is also possible for var compensation as an expanded application of the dq method. Because we are handling balanced three-phase quantities from Eqs. (21.26a, b), the power P3φd − q domain, Q3φd − q domain in the dq-domain is the DC value shown in Eq. (21.27). Therefore, we can use the vector PWM control scheme based on the following equations: P3ϕd −q domain = 3Vrms Ka Irms cos φ Q3ϕd − q domain = 3Vrms Ka Irms sin φ

21 28

where 0 ≤ Ka ≤ 1 modification factor PWM control −π ≤ φ ≤ π ignition angle

Then if Q3ϕd − q domain is controlled to maintain a DC value over time, it results in constant var compensation. For this purpose, asymmetrical control of Qdq by Ka and control of Irms sinφ by φ( 90 ) is possible. Converter Modeling as a dq Coordinates Laplace Transfer Function

21.4.4

The DQ method is a useful approach that can be applied to a wide range of power electronics applications. A three-phase inverter circuit can be written as a functional model of a Laplace transfer function in the dq-domain, so, for example, dynamic analysis and appropriate power electronics control of motor-driving systems are possible. The three-phase converter is connected to the power grid as shown in Figure 21.16a, and the following equations are satisfied: ia va d e b = R ib + L ib − vb dt ec ic ic vc va ia + vb ib + vc ic = vd id + vq iq = vdc idc ia

ea

① ②

1 idc −iL dt ③ C ea ,eb , ec source voltage of the grid system

21 29

vdc =

va , vb ,vc phase voltage of the converter vdc ,idc DC voltage and current iL load current

idc

iL

𝜐d

AC grid source ia

ea

𝜐a

ib

eb ec

ic

RL

+ – ed +

1 + R + Ls

id

×

𝜔L 𝜐b 𝜐c

(a) Three-phase circuit

C

𝜐dc

Load

𝜔L 𝜐q

eq – +

+

–

iL +

idc – ÷ + +

1 𝜐dc Cs

1 × R + Ls iq

(b) Laplace transformed d-q-domain model

Figure 21.16 dq-domain modeling of a three-phase inverter as a Laplace transfer function.

607

608

21 Power Electronics Applications, Part 3

The equations can be transformed into the dq-domain using Eqs. (21.24a, b), where the matrices C and C −1are defined by Eqs. (21.24a,b). Then the following equation is derived: 1 0

C C −1 =

0 1

and C

0 −1 d −1 C =ω dt 1 0

21 30

Then from Eq. (21.29) ①, ed eq

=L

R −ωL d id + dt iq ωL R

d id = dt iq

R ω L R −ω − L

−

id

+

iq

id iq

+

vd

21 31a

vq

1 e d − vd L e q − vq

21 31b

From Eq. (21.29) ② ③, d 1 iL vdc = vd id + vq iq − dt Cvdc C Therefore, using

d dt

21 31c

s with Laplace transformation,

R + sL id = ωLiq + ed −vd

21 32a

R + sL iq = −ωLid + eq −vq

21 32b

vdc =

1 vd i d + v q i q 1 P − iL = − iL sC sC vdc vdc

21 32c

Figure 21.16b is derived from Eqs. (21.32a, b, c) as a Laplace transfer function diagram, which is a feedback control scheme. We have real-time values of ed, eq and vd, vq, so real-time vdc is obtained with this diagram. Now, we will revisit Eq. (11.25) from Chapter 11: P = ed id + eq iq

21 33a

Q = ed iq + eq id

21 33b

Of course, if we write the source voltage and current as follows cos ωt + α ea eb = 2E ec

cos ωt + β

cos ωt + α −

2π 3

ia

cos ωt + α +

2π 3

ic

ib = 2I

cos ωt + β −

2π 3

cos ωt + β +

2π 3

21 34a

then calculating the transformed equation edq = C eabc, ed eq

=

3Ecos α

id

3Esin α

iq

=

3Icosβ 3Isin β

21 34b

and P = 3EIcos α −β , Q = 3EIsin α − β

21 34c

So, it can be concluded that id corresponds to effective power, and iq corresponds to reactive power. Figure 21.16b can be applied as a functional block of the converter for analytical purposes or for power electronic P, Q, v, i control purposes.

21.5 Generator Excitation Systems

i𝛼

ia ib

C𝛼𝛽

ic

𝜐a 𝜐b 𝜐c

C𝛼𝛽

×

i𝛽

×

𝜐𝛼

×

𝜐𝛽

×

Instantaneous effective power 𝜐𝛼i𝛼 p

+

∑

– LPF (low pass filter)

𝜐𝛽i𝛽

Instantaneous reactive power 𝜐𝛼i𝛽 q

C𝛼𝛽 =

+

∑

–1

p

–

2 1 0 3 1

–1 –1 3– 3 1

1

q

LPF (low pass filter)

– 𝜐𝛽i𝛼

Figure 21.17 Signal generation of instantaneous AC power.

21.4.5

Active Filter Using the PQ Method or αβ Method

The PQ method or αβ method is another practice whose control algorithm is shown in Figure 21.17.We have discussed the relation between symmetrical components and αβ0 components (see Chapter 8 and Figure 8.2). Now we can omit the zero-sequence component, so the following relation is obtained: iα t iβ t

=

1 1

i1

−j j

i2

2 − 1 −1 1 0 3 − 3 = 3 1 1 1

2ia rms t 21 35

2ib rms t 2ic rms t

If the circuit currents are three-phase balanced, we know that i2(t) = 0 and iα(t) = i1(t), iβ(t) = −j i1(t) Instantaneous effective power p(t) and reactive power q(t) are given by the following equation in the αβ-domain: pt q t

=

vα t iα t + vβ t iβ t − vβ t iα t + vα t iβ t

=

vα t

vβ t

iα t

−vβ t

vα t

iβ t

21 36a

In the inverse form, iα t iβ t

=

v2α

1 t + v2β t

vα t

−vβ t

pt

vβ t

vα t

q t

21 36b

We know that instantaneous power p(t),q(t) are of alternate values with DC offset power components (see Chapter 11, Figure 11.1), and the average values of p(t), q(t) are defined as power p, q (see Chapter 11). Also, the power p, q is obtained by multiplying the voltage and current components with the same frequency (Chapter 11). Then, if we decompose p, q into the DC component p, q and the AC component p,q, the former is the power based on fundamental frequency voltage and current and the latter is the power corresponding to the harmonic voltage and current components. Therefore, the power p,q corresponding to harmonic power components can be extracted in the αβ-domain and inverse transformed into harmonic power components in the abc-domain. Figure 21.17 shows the signal-calculating algorithm for the instantaneous power components p,q. One disadvantage of the PQ method is that some harmonic current is included in the AC source side, because instantaneous power from source-side p(t),q(t) is used as the original input signals. From this viewpoint, the DQ method is better than the PQ method, at least for the purpose of active filtering, because it can generate sinusoidal current without depending on the waveform of the grid voltage.

21.5

Generator Excitation Systems

In modern hydro, thermal, and nuclear power-generating stations, most of the generators are equipped with solid-state generator excitation systems, which are of course composed of power electronic DC excitation current-generating systems. In earlier days, the exciters were small DC generators directly coupled with the generator rotor.

609

610

21 Power Electronics Applications, Part 3

Excitation transformer

PT

AVR

CT 41

G

Field circuit breaker

Surge absorber

Thyristor rectifier bridge Surge absorber Battery Field flashing

Figure 21.18 Static excitation system with a thyristor bridge circuit (ordinary scheme).

Figure 21.18 shows a configuration diagram for an ordinary static state excitation system; three-phase thyristor bridge excitation systems are generally used. Three-phase power is introduced from the generator terminal point to the thyristor bridge rectifying circuit, where an almost-flattened DC excitation current ifield is produced and is fed to the generator field circuit. The thyristor bridge circuit is cos α ignition controlled by the automatic voltage regulator (AVR) digital signals, resulting in a change to the excitation current ifd. Due to the high-speed feedback response from the solid-state excitation system, generators can be operated smoothly while operating conditions P + jQ and v of an individual generator unit are fluctuating on the P-Q coordinates capability curve. The rated capacity of an excitation system is typically 10% or less of the generator rated capacity. Another approach is a brushless excitation system, shown in Figures 21.19a and b. Here, the armature of the AC exciter is directly coupled with the primary turbine-generator shaft while the field coil is at a standstill. Further, a diode circuit is also arranged inside the rotating cylinder. As the first stage, the standstill thyristor bridge rectifier (A-AMP) produces DC current and feeds it to the standstill field coil of the AC exciter. Next, the armature of the AC exciter rotates while the field coil is at a standstill, so AC current is produced and then converted to DC current ifd by the rotating diode circuit. In conclusion, the generator field current power is transmitted from the standstill stage to the rotary stage through the air gap of the AC exciter whose armature is rotating while the field is at a standstill. The brushless excitation system can be used for thermal generators or hydro-generators and can eliminate the need for slip rings. Hence it reduces maintenance and improves reliability. Refer Section 11.8 for generator excitation system. In addition, in the event of a severe, widespread power outage that causes isolation of power stations, a station may need to perform a black start to excite the fields of its large generators. Such additional circuits based on battery and power electronics equipment are installed depending on individual requirements.

21.6

Adjustable-Speed Pumped-Storage Generator-Motor Units

Pumped-storage hydropower systems have been used for almost 100 years to effectively store large amounts of energy, and today they are one countermeasure of several large-scale commercial applications of energy storage. The basic concept of pumped-storage systems requires two (upper and lower) reservoirs and a reversible pumped turbine and generator-motors with a grid-connected electrical input/output feeder. The machine operates as a motor in pumping mode and as a generator by changing the direction of rotation when the system is operated in turbine mode. Such systems have been constructed in the range of 10 m to almost 1000 m water-pumping head and can be constructed in almost any power range.

21.6 Adjustable-Speed Pumped-Storage Generator-Motor Units

Plant control system D-AVR unit

Isolation unit

Analog input

PT CT

Digital input

Control circuit Keyboard Phase angle control

EXT

Digital output

52E

Analog output

Power supply DC power source

A-AMP CLR G

Indication

41E

Power supply for initial excitation

AC EX DR Rotating parts

64E 71ER 64ER

A A

(a) Thyristor and field coil of AC exciter AC sourse AC exciter Diodes

Standstill

Generator rotor

Rotating (b)

Figure 21.19 Static excitation system with a thyristor bridge circuit (brushless excitation scheme).

Generator stator

Standstill

611

612

21 Power Electronics Applications, Part 3

For the purposes of this book, most of the electrical practices of a pumped-storage unit are the same as those of a conventional hydro-generating unit, except that the mode change is conducted by changing the connection of the three-phase order (from a-b-c to a-c-b) to the grid. Although the technology of turbines and generator-motors has matured, the conventional approach has some weak points:

• •

The pumping-up efficiency η is widely affected by the water head H, the water volume V, and the driving speed ω0 = 2πf0/(n/2) (where n is the number of poles in the synchronous generator motor) of the mechanical nature of the turbine, and the highly efficient operating range is relatively narrow. As a result, the round-trip efficiency is in the range 70–80% or even lower, depending on the circumstances. The driving speed ω0 in motor mode and generator mode is always fixed at a synchronized speed value with the utility power frequency according to the nature of the synchronous generator motor. This is a severe limitation in view of highly efficient pumping-up. Further, the unit can operate only as a large fixed load in pumping-up mode, so loadleveling operation is impossible. This may cause problems during power system operation, due to the restrictions of frequency control (AFC or LFC [load frequency control]), voltage control, and the system stability limit.

Adjustable-speed pumped-storage hydrogenation is a major technical breakthrough for overcoming these weak points. An adjustable-speed pumped-storage system with a unit capacity of 350 MW was first constructed on a commercial basis in Japan in around 1990; today, units over 500 MW are available. The primary purpose is to achieve optimum efficiency pumping-up and load leveling (LFC). This approach has come into wide use, in particular in the last decade. That is, to connect a large speed-adjustable induction generator motor, instead of a synchronous generator motor, to a power system is possible due to power electronics MWP torque control technology. We discussed operation characteristics of IG motors with Wye-connected rotor windings in Chapter 13, in particular in Section 13.2.4; now we will discuss the variable-speed pumped-storage generator motor in more detail. Figure 21.20b shows an adjustable-speed generator motor in comparison with the conventional synchronous generator motor unit in Figure 21.20a.

Power grid

Power grid

Excitation system 0-5 Hz (GTO and PWM control)

Excitation system (thyristor rectifier) DC

AC Slip ring

Salient pole rotor

Slip ring Cylindrical rotor

Stator

Stator Reversible water turbine

(a) Synchronous generator motor

Reversible water turbine

(b) Double-fed adjustable induction generator-motor

Figure 21.20 Synchronous generator motor and double-fed adjustable-speed machine and ordinary machine.

21.6 Adjustable-Speed Pumped-Storage Generator-Motor Units

Cavitation coefficient (𝜎 %)

Water-head H (%)

Efficiency 𝜂 (%)

Mechanical rotor input power P (%)

Referring to the variable-speed unit in Figure 21.20b, the stator 90 90% coils are connected to the outer power grid, and the voltages vas, 80% vbs, vcs and currents ias, ibs, ics with power system frequency 70% Guide vane fs = ωs/2π = 50/60 Hz (where the suffix s means stator) are opening position 120 85 charged to the stator terminals. Therefore, the circuit condition of the stator side is the same as that of a conventional synchro100 nous generator motor unit. 80 In contrast, the rotor side is different. The Wye-connected 80 60 three-phase coils are in the rotor. Three-phase rotor-coil term120 inals are connected to a three-phase cycloconverter (fs fr conHP max version) or IGBT (or GTO) converter-inverter ( fs DC fr conversion), which supplies balanced three-phase low-frequency 110 current iar, ibr, icr of typically fr = 0–5 Hz to the rotor coils. So, the HP min rotor-winding coils receive very low-frequency balanced threephase voltages var, vbr, vcr and currents iar, ibr, icr with a typical 100 frequency range fr = 0–5 Hz. 0.90 The state of frequency fr = 0 means DC current is supplied to 0.80 𝜎c the rotor coils, which is the same as the operating condition of a 90 cylindrical synchronous generator, from an electrical viewpoint. 0.70 With this condition, the flux produced by the DC current rotates 70 80 90 100 110 120 with angular velocity ωm = 2πfs in synchronization with the Pumping-up water volume power system grid. (n sp = 30m m 3/s) When currents iar, ibr, icr with low-frequency fr = 0–5 Hz flow into the rotor coils, the flux produced by the current rotates with Figure 21.21 Francis water turbine. angular velocity ωm = ωs − ωr = 2π(fs − fr), so the rotor speed ωm is mechanically slowed down by slip = (ωs − ωr)/ωs (slip velocity by PU). Assuming fr = 0–5 Hz and fs = 50 Hz as a typical design for an adjustable-speed generator motor, the rotor speed is adjustable by 100–90% of the rated rotating speed. Although the adjustable speed of the rotor ωm is only 10% maximum in this case, the motor-driving power can be adjusted by 30–40%, because the driving power P of a motor is proportional to ω3m (see Section 21.7 for a similar explanation.). Figure 21.21 shows an example of a typical Francis pumping-up water wheel. As is seen in the figure, the pumping-up mechanical power Pm[%](= ωm T) and efficiency η[%] are more dependent on the pumping-up water volume Q[%] and water head H[m]. The situation is written here as an implicit nonlinear function: •

, P rotor

Function P stator electrical

Therefore, if P rotor electrical

, ωm , Pm = ωm T H

=0

electrical

and T = Pm/ωm are independently controlled, the operating condition Function (P stator

, ωm,

electrical

Pm: H) = 0 can be achieved with high pumping-up efficiency ηm under water-head conditions. To achieve such control, vector PWM control methods are used in recent advanced adjustable-speed pumped-storage generator motor units. Note that the functional difference between a cycloconverter and a IGBT (or GTO) converter-inverter is summarized as follows (the primary purpose, to produce low-frequency current to the rotor coil in this case remains the same). The GTO converter-inverter method is usually advantageous compared to cycloconverter methods, because the former enables better operation due to its self-commutating function:

• • •

The power factors at the output side and input side of the converter-inverter can be controlled to a value around 1.0, so the MVA capacity of the converter-inverter can be reduced. Smooth, stable operational control of high-efficiency speed and torque is possible using advanced the PWM vector control method. Higher-harmonic components can be removed using the same approach with an active filter, if necessary (see Section 21.4).

Figures 21.22 and 21.23 show a cylindrical rotor of a 345 MVA adjustable pumped-storage unit and the GTO inverterconverter control scheme, respectively.

613

Figure 21.22 Adjustable-speed pumped-storage induction generator motor (cylindrical rotor with Wye-connected windings: 345 MVA 500 rpm 0–5% GTO inverter control).

Main transformer Reactor Inverter I DC condenser

89G

Converter IA

89M Excitation transformer

Converter IB

52 89S Chopper circuit

Generator-motor

Inverter 2

Converter 2A Inverter N GTO inverter

Converter 2B NGR GTO converter

Figure 21.23 Double-fed adjustable-speed pumped-storage hydro generator motor.

21.7 Wind Generation

Note that the rotor structure (slots/wedges, coils, and so on) of the IG-motor with Wye-connected windings is cylindrical and is mechanically similar to a thermal generator in spite of the difference in horizontal or vertical structures. Keep in mind that IG motors share similar weaknesses of poor withstanding capabilities against negative-sequence current i22 t and higher-harmonic current. We discussed this in Section 11.9 while examining thermal generators.

21.7

Wind Generation

Wind generation has become widespread in recent years, because it is clean energy and can be obtained worldwide with low ongoing costs. It requires a moderate capital cost, most of which is the construction cost of the turbine/generators and transmission facilities, including power-conditioning facilities in order to connect with power systems. The typical unit capacity of large wind generators is 3500–5000 kW; the diameters of wind blades are around 80 m, and total height is close to 100 m. Figure 21.24 shows the statistics of worldwide wind power installed capacity from The World Wind Energy Association 2017 report. Worldwide wind power capacity reached 486 661 MW by the end of 2016, of which 54 846 MW were added in 2016. This is a growth rate of 11.8% (17.2% in 2015). All the wind turbines installed worldwide by the end of 2016 generated around 5% of the world’s electricity demand. Wind power and energy can be calculated using the following equation: 1 1 Kinetic energy E = MV 2 = AρV 3 2 2

21 37a

The output energy of a wind turbine is 1 AρV 3 2 where air density ρ kg m3 wind velocity V m sec wind power receiving area A = πr 2 m2 wind mass per sec M = AρV kg sec wind turbine efficiency η Eout = η

21 37b

The output energy is proportional to V3, where V is wind speed. The theoretical efficiency η of a wind turbine cannot exceed ηmax = 0.593 according to Betz’s law (explained shortly). Wind velocity V and available turbine rotating speed ωwind = 2πfwind are different because the parallel individual units fluctuate constantly and unpredictably, so wellorganized grid-connection management based on power electronic conversion technology is required to connect to the grid with angular velocity ω0 = 2πf0. Albert Betz derived his law in 1919, which is as follows. If we write V1 as the wind speed just in front of the turbine and V2 for the speed behind the turbine, the average wind speed is 12(V1 + V2), and the following equations are obtained (see Figure 21.25a). 540’000* 486’661 435’258 371’317 11,0%

318’914 11,8%

282’866 17,2%

16,4% 12,7% 2012

2013

2014

2015

2016

2017

Figure 21.24 Total Installed Wind Capacity 2012–2017. Source: World Wind Energy Association World Wind Energy Report, 2017.

615

21 Power Electronics Applications, Part 3

Theoretical maximum point P2 —— P1

0.7

vavg

v2

0.5 0.4 0.3 0.2 0.1 0 –0.1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

v1

max at v2/v1 = 0.33 where P/P0 is 0.593

0.6

Power efficiency 𝜂 =

S

(a) Wind stream

(b) Betz’s curve

V2 V1

0.5 American farm model

0.4 0.3 0.2 0.1 0

Darrieus model

Windmill model

Savonius model

Three-blade propeller

(c) Power efficiency of different wind turbine models

Two-blade propeller model

P2 —— P1

0.6

Power efficiency 𝜂 =

616

V Vwind

peripheral k = —————

Figure 21.25 Betz’s efficiency curve for a wind turbine.

The kinetic energy that the turbine blades receive from the upstream wind is 1 1 1 1 E = mV 21 − mV 22 = m V12 − V22 = Aρ V1 + V2 2 2 2 2

1 2 V − V22 2 1

21 38a

The original kinetic energy of the upstream wind is 1 1 E0 = mV 21 = AρV13 2 2

21 38b

The theoretical turbine efficiency (power coefficient or Betz’s coefficient) is ∴ η=

E V2 = 1+ E0 V1

1−

V2 V1

2

where V1 ,V2 wind speed per sec just in front of and behind the turbine rotating space 1 V1 + V2 average wind speed working to rotate the turbine 2 1 V1 + V2 wind mass per sec that works to rotate the turbine m = Aρ 2 A wind power receiving area

21 38c

ρ air density Figure 21.25b shows the derived V2/V1 to η curve, which shows ηmax = 0.593 at V2/V1 = 0.33. Betz’s efficiency η is given by Eq. (21.38c), and if the curve is substituted into Eq. (21.37b), then the fundamental equation for Eout based on the wind turbine size and wind velocity is derived. The wind turbine should be controlled to operate at the sweet spot given varying wind velocity. Figure 21.26 shows the typical structure of a large-capacity wind-generating unit and the interconnection to the grid.

21.7 Wind Generation Pitch system Converter Gear box

Airspeed meter

Blade

Main frame

Yaw system

Generator

Nacelle

Tower

(a) Hub structure (3,000-3,500 kW class)

60–Hz transformer

+ Vd –

AC generator

AC system

Filters

(b) Power conditioner circuit (converter DC inverter method) Lightning pass : Direct strike to the blades : Impinged surge voltages through transmission lines : Induced surge voltages through communication lines Air-speed meter Lightning rod Aero-lamp

Aero-lamp

Substation

Grounding earth

(c) Circuit structure and surge voltage passes Figure 21.26 Wind generation.

The power available from a wind wheel varies with the cube of the wind velocity V, so it is desirable to let the turbine speed vary over a wide range to an optimum value in order to extract the maximum amount of power, depending on the operating conditions. This would not be possible using a synchronous generator, which dictates constant angular speed in response to the power system frequency. Therefore, to allow the generator turbine speed to vary, in order to optimize

617

618

21 Power Electronics Applications, Part 3

efficiency of power generation, three-phase IGs with speed-controllable power electronic interconnection are generally used for large wind-generation units. Referring to Figure 21.26, a wind turbine is designed to produce maximum power at a wide spectrum of wind speeds. All wind turbines are designed for a maximum wind speed called the survival speed, which is 50–60 m/s. A control system involves three basic elements: sensors to measure process variables, actuators to manipulate energy capture and component loading, and control algorithms to coordinate the actuators based on information gathered by the sensors. Stalling works by increasing the angle at which the wind strikes the blades, and it reduces the induced drag when the wind speed increases. Furling works by pitch control of the blades, which reduces the induced drag from the lift of the rotor, and the cross-section. A fully furled turbine blade has the edge of the blade facing into the wind when stopped. Many turbines use hydraulic systems, so if hydraulic power fails, the blades automatically furl by spring action. Turbines can also use a servomotor for every rotor blade; they have a small battery reserve in case of a power-grid breakdown. Further, a mechanical drum brake or disk brake is used to hold the turbine at rest for maintenance. A smaller turbine may have an electrical dumping resistor bank. 21.7.1

Wind Generators

For large commercial-size horizontal-axis wind turbines, the generator is mounted in a nacelle at the top of a tower, behind the hub of the turbine rotor. As is the nature of wind generation, asynchronous generators (IGs) are used in most cases that are mechanically (direct or geared) coupled with wind turbines. Usually, the rotational speed of the wind turbine is 5–20 rpm, far slower than the electrical grid frequency speed between 700 and 3600 rpm. Therefore, a gearbox is inserted between the rotor hub and the generator. 21.7.2

Gearless Wind Turbine with Double-Fed IGs

The generator stator coils must be connected to the utility grid with frequency fs = ωs/2π = 50/60 Hz. So, if the generator is a double-fed IM with Wye-connected windings and is to be running with velocity ωm, it can be interconnected directly by supplying electrical power with frequency fr = ωr/2π = ωs − ωm to the rotor windings. This is obviously the same principle as the adjustable pumped-storage generating unit, except that wind generation need not be a reciprocal operation, while pumped-storage generation must be a reciprocal function with generating mode and motoring mode. Of course, this approach gets rid of the gearbox completely, and the generator spins at the same speed as the blade. Most of the large, commercial units are classified as follows:

• • •

Model 1: Adjustable-speed controlled scheme with geared coupling, connected with the power grid using the DC-link method (fwind DC fs based on a set of a converter and an inverter). Figure 21.26b shows this model. Model 2: Adjustable-speed controlled scheme with gearless coupling, where low-frequency AC power is fed to the rotor coils of the double-fed generator. Model 3: Non-adjustable-speed scheme for smaller units, which generally use a cycloconverter (fwind fs).

IGs require reactive power for excitation in order to avoid loss of excitation or leading power-factor operation (see Section 11.7.2 for the reason), so substations used in wind-power collection systems include substantial capacitor banks for power-factor correction. Further, IGs cannot support the system voltage during faults, unlike steam or hydro turbine-driven synchronous generators. Different wind turbine generators behave differently during transmission grid disturbances, so extensive modeling of the dynamic electromechanical characteristics of a new wind power installation is required for grid management. Since wind speed is not constant, a wind farm’s annual energy production is never as much as the sum of the generator nameplate ratings multiplied by the total number of hours in a year. The ratio of actual productivity in a year to this theoretical maximum is called the capacity factor. Typical capacity factors are 20–40%. The capacity factor is affected by the variability of the wind at the site and by the generator size. A smaller generator is cheaper and achieves a higher capacity factor, but it makes less electricity in high winds. Conversely, a larger generator costs more and generates little extra power and may stall out at low wind speed.

21.8

Small Hydro Generation

Small hydropower has various degrees of “smallness”: there is no agreed-on definition of “small” hydro, but a range of 500 KW–10 MW is the most widely accepted value. Even smaller machine units are called micro-hydro(0

1 s 1 sn 1 s±α ω s2 + ω2 s s2 + ω2 ω

t n −1 e

αt

1 t

αt

sinωt cos ωt

= e − αt cos ωt 1 t

e − αt sinωt sinhωt

=e

− αt

sinωt 1 t

s + α− jω s + α + jω = s + α 2 + ω2 ∴

F(s)

n −1

1 =e s±a

−1

s + α 2 + ω2 −1

f t e −st dt

f(t)

A∠θ A∠− θ + s + α −jω s + α + jω

= 2Ae − αt cos ωt + θ

1t

cosh ωt e − αt cos ωt ± θ e − αt sin ωt ± θ 2Ae − αt cos ωt ± θ

= [f(t)]

s + α 2 + ω2 ω s2 −ω2 s s2 −ω2 s + α cos θ

ωsinθ

s + α 2 + ω2 ± s + α sin θ + ωcos θ s + α 2 + ω2 A∠θ A∠−θ + s + α− jω s + α + jω

633

635

Appendix B Matrix Equation Formulae Matrix equation analysis is an essential approach for many kinds of analysis including three or more variables, not only as a computational tool but also to provide logical steps. Here we will explain some essential points with regard to matrix analysis from a practical viewpoint; for further mathematical details, refer to specialized works of mathematics. a) Matrix l × m with l rows and m columns: column m

A11 A21

A12 A22

row l Al1

Al2

A=

A1m A2m …

B1

Alm

b) Multiplication of two matrices A (l rows and m columns) and B (m rows and n columns). For example, multiplication of A (3 × 2) and B (3 × 3): A11 A21 A31

A12 A22 A32

B11 B21

B12 B22 B

A

B13 B23

=

A11 B11 + A12 B21 A21 B11 + A22 B21 A31 B11 + A32 B21

A11 B12 + A12 B22 A21 B12 + A22 B22 A31 B12 + A32 B22 A

A11 B13 + A12 B23 A21 B13 + A22 B23 A31 B13 + A32 B23

B2

B

c) If and only if the number of columns in matrix A and the number of rows in matrix B are equal, the multiplied new matrix A B with l rows and n(l × n) columns can exist as follows: colum m

n

row l A m B = l

n A B

B3

d) A matrix with the same number of rows and columns is called a square matrix. e) A square matrix in which the diagonal elements are 1 and all other elements are 0 is called a unit matrix. The unit matrix is usually written using the symbol 1. For example, the 3 × 3 unit matrix is 1 1= 0 0

0 1 0

0 0 1

B4

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

636

Appendix B Matrix Equation Formulae

When a square matrix and a unit matrix are multiplied, the resulting matrix is unchanged: C =C 1=1 C

B5

f) If a multiplied matrix C D of two square matrices C and D become a unit matrix 1, the square matrices C and D are called inverse matrices of each other: C D=1

B 6a

D may be written C−1, and C may be written D−1: C C − 1 = 1,

D D −1 = 1

B 6b

Typical examples of inverse matrices are the operational matrices a, a α – β – 0 method, and D(t), D−1 (t) of the d–q–0 method. Another example: 1 2 1

0 4 3

3 1 0

×

C

−1 1 3 2 3

−4 5 3 4 3

3 −1 −1

=

1 0 0

0 1 0

−1

−1

of symmetrical components α, α

of the

0 0 1

1

C −1

g) If two square matrices C and E are the same size (have the same number of rows and columns), then generally, the commutative law cannot be satisfied between C and E: C E

E C

B7

If either C or E is a unit matrix, or if C and E are inverse matrices of each other, the commutative law is satisfied as an exceptional case: C 1=1 C =C C C

−1

=C

−1

B8

C =1

B9

h) The following equation is always satisfied when the matrices A, B, C satisfy condition (c): A B C = A B C =A B C A B+A C =A B+C

B 11

A B+D B= A+D B

B 12

i) j)

B 10

Z1 = A11 y1 + A12 y2 Z2 = A21 y1 + A22 y2

B 13

Z3 = A31 y1 + A23 y2 Equation B.13 can be written symbolically as follows: Z1 Z2 Z3

A11 y1 + A12 y2 A21 y1 + A22 y2 A31 y1 + A32 y2

=

Z

A y

2 =

3

Z

3

A × 2

A12 A22 A32 A

where 1

=

A11 A21 A31

1 y

y1 or symbolically y2 Z=A y y

B 14

Appendix B Matrix Equation Formulae

y1 y2

B11 x1 + B12 x2 + B13 x3 B21 x1 + B22 x2 + B23 x3

=

y

=

B11 B21

B12 B22

B x

B13 B23

B

x1 or x2 x3 y = B x

B 15

x

The variables y1, y2 can be deleted from Eqs. (B.14) and (B.15) using the matrix procedure Z=A y=A B x = A B x

C x

B 16a

C =A B

where or

Z1 C11 Z2 = C21 Z3 C31

C12 C22 C32

C13 C23 C33

x1 x2 x3

Z=C x

B 16b

where C is 3

2

3 = 3

C

3

A

2

B

and C

C11 C21 C31

C12 C22 C32

C13 A11 C23 = A B = A21 C33 A31

A12 A22 A32

A11 B11 + A12 B21 = A21 B11 + A22 B21 A31 B11 + A32 B21

B11 B21

B12 B22

A11 B12 + A12 B22 A21 B12 + A22 B22 A31 B13 + A32 B22

B13 B23 A11 B13 + A12 B23 A21 B13 + A22 B23 A31 B13 + A32 B23

B 17

where C11: A11 B11 + A12 B12, C12 : A11 B12 + A12B22, etc. k) Given the following matrix equation C =A B

B 18 −1

then to derive the matrix B, we left-multiply by A : A − 1 C = A −1 A B = A − 1 A B = B ∴ B = A −1 C

B 19a

To derive matrix A from Eq. B.19a, we right-multiply by B−1: C B −1 = A B B −1 = A B B −1 = A ∴ A = C B −1 Generally, A

B − 1 C, B

C A −1

B 19b

637

638

Appendix B Matrix Equation Formulae

l) Given the matrix equation P=L M N

B 20

where P, L, M, N are square matrices, then the following equations can be derived by applying the process from (l) twice: L = P N −1 M −1 M = L − 1 P N −1

B 21

N = M −1 L −1 P F +G=G+F

m)

B 22

n) If F +G Q=K

B 23

then F + G + K Q −1 ∴ F = K Q −1 −G, G = K Q −1 −F l

o)

n1

B 24

l

n2

l

l

n2

l

m1 W1 = m1 R 11 × n1 V1 + m1 R 12 × n2 V2 l

n1

B 25

m2 W2 = m1 R 21 × n1 V1 + m2 R 22 × n2 V2 These two matrix equations can be combined into one matrix equation: m1 W 1 l m2 W l2

=

n1

n2

n1

n2

m1 R11 m1 R12 m2 R21 m2 R22

l

n1 V 1

B 26

l

n2 V 2

Needless to say, Eq. (B.26) can be divided into the two equations from Eq. (B.25). p) The new matrix in which the row and column elements of the original matrix A are interchanged is called the transposed matrix of A. It is written with the symbol tA. If C =A B

B 27

then t

C = tB

t

A

B 28

With regard to matrix C from Eq. (B.17), the transposed matrix is

t

C

C11 C12 C13

C21 C22 C23

C31 C32 = C33

B11 B12 B13

B21 B22 B23 tA

A11 A12

A21 A22 tB

A31 A32

= tB tA

B 29

639

Part B Digital Computation Theories

641

22 Digital Computation Basics 22.1

Introduction

The primary purpose of this section is to explain various circuit properties and useful algebraic results that can be expanded to study large power networks. We will start from the algebraic manipulation of a system of linear equations to be solved for the unknown values. To model and study an electrical power system, which is an intricate ensemble of equipment of various kinds, a high degree of abstraction is required in order to represent the electrical power system as a whole. To conduct this task, we start by describing the physical electrical system with the use of a reduced set of idealized elements with defined properties between variables through them. That is, by Ohm’s law we relate the electric current that flows when a voltage difference is applied at its terminals. We write a consistent set of equations for the network, using Kirchhoff’s current law (KCL) at each node, and work out a solution for state variables, which are those electric quantities known as node voltages that allow us to calculate currents and voltage drops on all elements in the network. The analysis of the system enables us to predict its behavior under a range of stimuli with practical applications. From a general network model and its numerical solution, we derive Norton’s equivalent, which is a numerical model that relates nodal voltages to nodal currents (those currents injected or extracted from a node by sources/loads) through admittance values or short-circuit coefficients. The numerical process is conducted by Gaussian elimination or its numerical equivalent, partial inversion, in a matrix sense. If we continue the partial-inversion process, we eventually get Thévenin’s equivalent. Thévenin’s circuit relates nodal voltages to nodal currents through impedance values known as open-circuit coefficients. The importance of network equivalents is that an extensive network can be reduced to a small equivalent circuit arrangement without losing the effects of connected elements as lines, voltage, or current sources. Using equivalent circuits, we can solve for state variables with less effort, given that we are only interested in a small portion of the original network. The solution to the equivalent can be fed into the complete network equations, and the effect of the equivalent’s solution can be reflected on the whole network. This approach will be used in fault current calculations and contingency analysis. A basic electrical power circuit is an interconnection of passive and active elements. Resistors, inductors, and capacitors are passive elements. Other elements are designed to convert energy from the primary thermal of mechanical sources to electric DC or AC. These current and voltage sources are active components of the power system. Other types of sources with nonlinear characteristics that are based on power electronics are also active elements; through switching devices, they accomplish the useful handling of electric energy for various industrial and largescale power applications. The electric elements may be series or shunt connected, forming a delta or a star configuration in the case of three-phase AC systems. In more elaborate circuits, the voltage and current sources may be of an independent or a controlled type. For passive elements, the voltage at their terminals with respect to the current flowing in the resistor, inductor, or capacitor may be through a linear or nonlinear relation, depending on the nature of the element. Furthermore, the response can vary linearly with frequency or in a nonlinear fashion. Among various system elements, the ideal transformer with a real or a complex transformer with a tap and phase-shifting angle is a handy device to represent voltage and current increases and drops at crucial sections of the circuit. Depending on the electrical length (with respect to the actual size of equipment), its circuit representation may be required to be described by distributed parameters or by lumped parameters – the electrical length is dictated by the actual physical dimension of the device and the voltage frequency at which the device is going to work. Ohm’s law determines the amount of current that flows through a given element (R, L, and C), and it depends on the voltage at terminals for a passive element and the properties that oppose the current to flow. Kirchhoff’s voltage and Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

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22 Digital Computation Basics

current laws govern the interaction between the voltages and currents in all passive and active elements within a circuit. We can deduce various voltage–current formulations to study electric circuits; it all depends on the way in which we apply the fundamental laws. Popular outcomes are nodal equations and mesh equations, but other formulations are possible. When dealing with how to find the solution of practical electric power networks – which generally have a substantial number of nodes and electric elements – the computational effort may be quite high, even for a digital computer. This is one of the reasons we have to carefully plan all computational steps in order to keep the numerical burden well under control. It is fortunate that most large physical systems (as in the case of electrical power circuits) are loosely interconnected in a structural sense; not all electrical nodes are connected through elements (transformers, lines, and cables). This characteristic makes the systems equations sparse and prone to an efficient numerical solution. When we recognize that, for high-voltage power systems, up to 80% or more of the numerical coefficients in the nodal model have zero values – and that such coefficients should neither be stored nor processed – then the nodal model is very compact in terms of storage requirements, even for extensive power networks. Quite elaborate algorithms called sparse techniques will store and process the nonzero essential information. The solution of the sparse system of equations is mainly based on carefully thought-out Gaussian elimination and are well-developed; these techniques and more are included in ETAP power-system simulation software that solves electric power networks.

22.2

Network Types

Network elements may be classified into various types based on some parameters:

•• •

Active elements and passive elements Linear elements and nonlinear elements Bilateral elements and unilateral elements

22.2.1

Active Elements and Passive Elements

Network elements may be classified as either active or passive, based on their ability to deliver power. Active elements deliver power to other elements present in an electric circuit. Sometimes, they may absorb power like passive elements. That means active elements can both deliver and absorb power: for example, voltage sources and current sources. Passive elements can’t deliver power (energy) to other elements. However, they can absorb power. That means these elements either dissipate power in the form of heat or store energy in the form of either magnetic fields or electric fields: for example, resistors, inductors, and capacitors. Active elements are the network elements that deliver power to other elements present in an electric circuit. So, active elements are also called sources of voltage or current. They are classified as follows. 22.2.1.1 Independent Sources

As the name suggests, independent sources produce fixed values of voltage or current and are not dependent on any other parameter. Independent sources can be further divided into the following two categories:

••

Independent voltage sources Independent current sources

22.2.1.1.1

Independent Voltage Sources

An independent voltage source produces a constant voltage across its two terminals. This voltage is independent of the amount of current that is flowing through the two terminals of the voltage source. An independent ideal voltage source and its V–I characteristics are shown in Figure 22.1. The V–I characteristics of an independent ideal voltage source is a constant line, which is always equal to the source voltage (Vs) regardless of the current value (I). Therefore, the internal resistance of an independent ideal voltage source is zero ohms. Hence, independent ideal voltage sources do not exist practically, because there will be some internal resistance. An independent practical voltage source and its V–I characteristics are shown in Figure 22.2. There is a deviation in the V–I characteristics of an independent practical voltage source from the V–I characteristics of an

22.2 Network Types

independent ideal voltage source, due to the voltage drop across the internal resistance (RS) of an independent practical voltage source. 22.2.1.1.2

Independent Current Sources

I

V

+ Vs

+ –

V = Vs Vs

An independent current source produces a constant current. – This current is independent of the voltage across its two term0 inals. An independent ideal current source and its V–I characteristics are shown in Figure 22.3. The V–I characteristics Figure 22.1 Ideal voltage source V–I characteristics. of an independent ideal current source is a constant line, which is always equal to the source current (IS) regardless of the voltage value (V). Therefore, the internal resistance Rs I of an independent ideal current source is infinite ohms. V + Ideal Independent ideal current sources do not exist practically, because there will be some internal resistance. An independent practical current source and its V–I characteristics are Vs + V = Vs – IRs Vs – shown in Figure 22.4. There is a deviation in the V–I characteristics of an independent practical current source from Practical – the V–I characteristics of an independent ideal current 0 source, due to the amount of current flowing through the internal shunt resistance (RS) of the independent practical Figure 22.2 Practical voltage source V–I characteristics. current source. I = Is

22.2.1.2 Dependent Sources

As the name suggests, dependent sources produce an amount of voltage or current that is dependent on some other voltage or current. Dependent sources are also called controlled sources. Dependent sources can be further divided into the following two categories. 22.2.1.2.1

Voltage-dependent voltage source (VDVS) Current-dependent voltage source (CDVS)

Dependent voltage sources are represented with the signs + and − inside a diamond shape. The magnitude of the voltage source can be represented outside the diamond.

22.2.1.2.2

V

Is

Dependent Current Sources

I

I Is

–

V

0

Dependent Voltage Sources

A dependent voltage source produces a voltage across its two terminals. The amount of this voltage is dependent on some other voltage or current. Hence, dependent voltage sources can be further classified into the following two categories:

••

+

I

Figure 22.3 Ideal current source V–I characteristics.

I = Is – V/ Rs

Is

I

+

V/ Rs Rs

V

Ideal Is Practical

– 0

V

Figure 22.4 Practical current source V–I characteristics.

A dependent current source produces a current. The amount of this current is dependent on some other voltage or current. Hence, dependent current sources can be further classified into the following two categories:

••

Voltage-dependent current source (VDCS) Current-dependent current source (CDCS)

Dependent current sources are represented with an arrow inside a diamond shape. The magnitude of the current source can be represented outside the diamond. We can observe these dependent or controlled sources in equivalent models of transistors.

643

644

22 Digital Computation Basics

Rs

Vs

+ –

Is

Rs

Figure 22.5 Practical voltage source to practical current source transformation.

22.2.1.3 Source Transformation

We know that there are two practical sources: voltage source and current source. We can transform (convert) one source into the other based on the requirement while solving network problems. The technique of transforming one source into the other is called a source transformation technique. Following are the two possible source transformations. 22.2.1.3.1

Practical Voltage Source into a Practical Current Source

A practical voltage source consists of a voltage source (VS) in series with a resistor (RS). This can be converted into a practical current source. It consists of a current source (IS) in parallel with a resistor (RS). The transformation of a practical voltage source into a practical current source is shown in Figure 22.5. The value of IS is equal to the ratio of VS and RS. Mathematically, it can be represented as IS =

22.2.1.3.2

VS RS

22 1

Practical Current Source into a Practical Voltage Source

A practical current source consists of a current source (IS) in parallel with a resistor (RS). This can be converted into a practical voltage source. The transformation of a practical current source into a practical voltage source is shown in Figure 22.6. It consists of a voltage source (VS) in series with a resistor (RS). The value of VS is equal to the product of IS and RS. Mathematically, it can be represented as VS = IS × RS.

22.2.2

Linear Elements and Nonlinear Elements

Network elements can be classified as linear or nonlinear based on how their characteristic follow the property of linearity. Linear elements show a linear relationship between voltage and current: for example, resistors, inductors, and capacitors. Nonlinear elements do not show a linear relationship between voltage and current: for example, voltage sources and current sources. Rs

Is

Rs

Vs

+ –

Figure 22.6 Practical current source to practical voltage source transformation.

22.3 Circuit Elements

22.2.3

Bilateral Elements and Unilateral Elements

Network elements can also be classified as either bilateral or unilateral based on the direction of current flows through the network elements. Bilateral elements allow current in both directions and offer the same impedance in either direction of current flow: for example, resistors, inductors, and capacitors. Figure 22.7 shows current (I) flowing from terminal A to B through a passive element having impedance of Z Ω. It is the ratio of voltage (V) across that element between terminals A and B and current (I). Figure 22.8 shows current (I) flowing from terminal B to A through a passive element having impedance of Z Ω. That means current (–I) is flowing from terminal A to B. Since both the current and voltage have negative signs with respect to terminals A and B, the calculated impedance values are the same. Unilateral elements allow current in only one direction. Hence, they offer different impedances in both directions.

22.3

Circuit Elements

22.3.1

Resistors

I

Z B

A +

V

–

Figure 22.7 Current from terminal A to B.

I

Z

B

A –

V

+

Figure 22.8 Current from terminal B to A.

The electrical resistance of an electrical conductor is a measure of the difficulty of passing an electric current through that conductor. The inverse quantity is electrical conductance and is the ease with which an electric current passes. Electrical resistance shares some conceptual parallels with the notion of mechanical friction. The SI unit of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S). Example How much current flows through a 5 Ω resistor if a voltage of 120 V is applied across it? Solution I=

V 120 = = 24 Amps R 5

I A

+

Ammeter (Z = 0)

120 V

V Voltmeter (Z = ∞)

R

–

Resistances in series add linearly, Rseries = R1 + R2 + R3. a

R2

R1

R3

a Rseries

b

b

Resistances in parallel add inversely, a

Rparallel

=

1 1 1 + + . R1 R2 R3

a R1

b

1

R2

R3

Rparallel b

645

646

22 Digital Computation Basics

22.3.1.1 Current-Divider Rule

Current division refers to the splitting or distribution of current between the branches of the divider. In a simple linear circuit, therefore, an output current is produced that is a fraction of its input current. Example Find the currents I1, I2, and I in the following circuit, if R1 = 2 Ω and R2 = 4 Ω. I1

I

I2

10 V

R1

R2

Solution I1 =

10 10 = 5 A, I2 = = 2 5 A, therefore I = I1 + I2 = 7 5 A 2 4

Another solution is to use the current-divider rule:

I 10 V

Req

Req =

1 R1 × R2 4 10 = 7 5A = = Ω and I = 1 1 R1 + R2 3 4 + R1 R2 3

Therefore, I1 =

R2 4 R1 2 × 7 5 = 5 A and I2 = × 7 5 = 2 5A ×I = ×I = 2+4 2+4 R1 + R2 R1 + R2

22.3.1.2 Voltage-Divider Rule

The voltage-divider rule finds the voltage over a load in a DC or AC circuit, when the loads are connected in series. Example Find the voltages V1 and V2 if R1 = 1 Ω and R2 = 3 Ω. I V1

R1

V2

R2

V = 40 volts

22.3 Circuit Elements

Table 22.1 Resistivity and temperature coefficients of metals. Material

ρ20

α20

Copper

1.673 × 10−8

4.05 × 10−3

Aluminum

2.655 × 10−8

4.03 × 10−3

Iron

9.71 × 10

−8

5.76 × 10−3

Silver

1.59 × 10−8

3.75 × 10−3

Gold

2.44 × 10

−8

3.4 × 10−3

Solution I=

V 40 = 10 A = R1 + R 2 1 + 3

V1 = I × R1 = 10 × 1 = 10 V and V2 = V – V1 = 40 – 10 = 30 V Using the voltage-divider rule, V1 =

R1 1 R2 3 × 40 = 10 V and V2 = × 40 = 30 V ×V = ×V = 1+3 1+3 R1 + R2 R1 + R2

22.3.1.3 Resistivity

The resistance of a conductor of length L, cross-section area A, and resistivity ρ is R =

ρ×L , A

where ρ is in Ω-m, L is in m, and A is in m2. The resistivity if a conductor is a function of temperature and varies approximately linearly with temperature: ρT = ρ20 × 1 + α20 × T − 20 where ρ20 = resistivity at 20 C α20 = temperature coefficient at 20 C In a general form, ρT = ρt × (1 + αt × (T − t)) Resistivity and temperature coefficients of some metals in pure form are given in Table 22.1. Example Determine the dc resistance of 600 ft. of #12 cable at 40 C ambient temperature. Diameter (D) of #12 cable is 0.0808 in. Solution D 2 0 00513 = 0.00513 in.2 = = 3.31 × 10−6 m2 2 3 281 × 12 2 Note: circular mils = D2 = (0.0808 × 1000)2 = 6530 circular mils = 6.53 Kcmil Resistivity at 40 C can be calculated as: Area = A = πr2 = π

ρ40 = ρ20 × 1 +

20

× 40 − 20 = 1 673 × 10 − 8 1 + 4 05 × 10 −3 × 20 = 1 81 × 10 − 8 Ω−m

Therefore DC resistance at 40 C = Rdc , 40 C = =

5 47 × 0 6 = 1 0Ω 3 281

ρ × L 1 81 × 10 −8 × 1000 = = 5 47 Ω km A 3 31 × 10 − 6

647

22 Digital Computation Basics

22.3.2 Inductors Inductance is the ability to store magnetic energy (a property that resists changes in current). The symbol of the inductor is L, while its l value is called inductance. The unit of inductance is the henry (H), i named after the famous American physicist Joseph Henry. As inductance increases, so does the inductor’s opposition to the flow of AC N v current, as shown in Figure 22.9. It can be shown that the AC voltage across an inductor leads the current by a quarter of a period. Viewed as phasors, the voltage is 90 A ahead (in a counterclockwise direction) of the current. In the complex plane, the voltage phasor is perpendicular to the current phasor, in the positive direction (with respect to the reference direction, counterFigure 22.9 Factors affecting inductance. clockwise). We can express this with complex numbers using an imaginary factor j as a multiplier. N2 × A Consider: L = μ × i(t) l where ϕ

μ = μ0 × μr = permeability N = number of turns A = core cross-section area l = average flux path length

L

e(t)

–S = P + jQ

Flux linkage = λ = L × i t

Figure 22.10 Inductive circuit.

dλ di dL 1 1 =L× +i× and i t = i 0 + v t × dλ or w t = × Li2 Therefore, v = dt dt dt L 2 0

For the inductive circuit in Figure 22.10, t

di t 1 e t =L and i t = L dt

−∞

t

1 e t dt = i 0 + e t dt L 0

An inductive circuit waveform can be represented as shown in Figure 22.11, where voltage e t = EM cosωt and current i t = IM sinωt = IM cos ωt − 90 EM = ωLIM may be represented in vector form as E = jX L I = XL I ∠90 or as shown in Figure 22.11.

Voltage

Current

V (rms)

I (rms)

V

1

0.5 0 0.005

0.01

–0.5

0.015

Imaginary

1

0.5 0 0.5

1

–0.5 –1

–1 –1.5

I

1.5

1.5 pu - Volatage and current

648

–1.5 Time (seconds)

Figure 22.11 Inductive circuit waveform and vectors.

Real

1.5

22.3 Circuit Elements

649

2 IM XL sin 2ωt = I 2 XL sin2ωt. 2 1 2 XL = I 2 XL (vars). Therefore, P = Pavg = 0 (watts) and Q = Qavg = IM 2 Hence, apparent power S = EI ∗ = jX L I × I ∗ = jX L I 2

Power p t = e t i t =

t

t

1 1 w t = p t dt = I XL sin 2ωt × dt = I 2 L 1− cos 2ωt = Li2 t 2 2 2

o

0

1 w t = I 2 L 1− cos 2ωt implies oscillations are at twice the angular frequency of e(t) and i(t). 2 Maximum power is Wmax = I2L joules, and reactive power is Q = I2(ωL) = ω × Wmax vars. Therefore, the reactive power (Q) is directly related to the stored energy in the reactive element L. Energy is temporarily stored in the magnetic field of an inductor during parts of each cycle, i.e. the energy is increased when both the current and voltage have the same sign. Later in the cycle, the field collapses when the voltage and current have different signs, and the energy is returned to the rest of the system, i.e. the energy decreases. 22.3.3

Capacitors

Capacitance is the ability to store electrical charges, as shown in Figure 22.12. Capacitive charge q = cv A Capacitance c = ϵ × l Permittivity ε = ϵ0 × ϵr 1 Permittivity of free space ϵ0 = 36π × 109 Dielectric constant εr = varies between 1 to 10 dq dv t Current i t = =C × dt dt Voltage v t = C1

t

−∞

t

i t × dt = C1 i t × d t + v 0

Power p t = v t i t = Cv t

i(t) +q v (t)

–q

Figure 22.12 Capacitance parameters.

0

dv t dt

t

t

Δw t = p λ dλ energy from time t0 to t = C v λ dv λ = t0

C 2 v t 2

t0

C w t − w t0 = v2 t − v2 t0 w t0 and v t0 = 0 2 1 2 w t = Cv 2 v = Vm cos wt and i = − CV m sin wt = CV m cos wt + 90

i(t)

Consider a capacitive circuit, as shown in Figure 22.13, where t

d 1 i t = C × e t and e t = dt C

−∞

t

1 i t dt = e 0 + i t dt C

C

e(t)

0

e t = EM cosωt and i t = − IM sinωt = IM cos ωt + 90 1 EM = IM = XC IM or E = −jX C I ωC IM EM sin 2ωt = − I 2 XC sin 2ωt watts Power p t = e t i t = − 2

–S = P + JQ Figure 22.13 Capacitive circuit.

l

650

22 Digital Computation Basics

Q = − I 2 XC vars t

1 1 W t = p t dt = Ce2 t = CE 2 1+ cos 2wt 2 2 0

22.3.4

R–L–C Networks

Resistance

i(t)

R

e(t)

I

Capacitance

e t = Eϵ jwt = 2Ecos wt i t = Iϵ jwt = 2Icos wt E E et = = R= it I I

E

E = RI

i(t)

e(t)

e t = Eϵ jwt d i t = C × e t = jwCEϵ jwt = Iϵ ji dt

C

I

–jXc

E

Xc =

1 ωC

I

E = −jX c I E

Inductance

i(t)

e(t)

L

i t = Iϵ jwt di t e t = L dt

–jXL

XL = ωL

I

E

E E = jX L I

I

= jwLIϵ jwt = Eϵ jwt

22.3 Circuit Elements

22.3.4.1 Parallel Connection of a Resistor and Inductor (RL Network) I IR

IL

R

V

L

IR =

V R

IL =

V XL

V

θ

IR

IL

I I cos θ

I = IR + IL ∠ − 90 = I∠− θ = Icosθ − jIsinθ

θ I sin θ

I

Power factor PF = cos θ Reactive factor RF = sinθ where θ is the angle between the voltage and current.

22.3.4.2 Parallel Connection of a Resistor and Capacitor (RC Network) I IR

IC

R

V

I R = VR

C

IC = j

V XC

I

IC

I = IR + IC ∠ − 900

θ

IR

V

I θ

I Sin θ

I ∠ θ = Icosθ + jsinθ

I Cos θ

Example Determine the input current and power factor for the circuit shown in Figure 22.14.

I IR

Solution V = 277 Volts

277 277 277 = 27 7 A, IL = = 69 3 A, IC = = 92 3 A, 10 4 3 IR = 27 7 ∠ 0 A, IL = 69 3 ∠ − 90 A and IC = 92 3 ∠90 A

10 Ω

IR =

Figure 22.14 Example circuit.

IL

IC

4Ω

3Ω

651

652

22 Digital Computation Basics

IC

3/4

j2

j1

I

IC - IL

+

θ

IR

IR

10√2 30°

j1

2

2

–j2

–E

–

IL

–Z

–Z

3

I = I R + I L + I C = 27 7 + j 92 3 −69 3 = 27 7 + j23A = 36 ∠40 A θ = 40

–Z

2

1

Figure 22.15 Example circuit.

Therefore, PF = cos 40 = 0.77 (leading)

E 30 –15

Example Find Z 1 ,Z 2 ,Z 3 , & I from the circuit shown in Figure 22.15.

I Figure 22.16 Vectorial representation of the example circuit.

Solution Z 1 = 2 − j2 + j1 = 2 − j1 Z 2 = 2 + j1 Z3 = I=

2 −j1 =

2 + j1 2− j1 5 = 4 2 + j1 + 2 − j1

3 5 + j2 + = 2 + j2 = 2 2 ∠ 45 4 4

E 10 2 ∠ 30 = = 5 ∠ − 15 , which can be visualized using vectors as shown in Figure 22.16. Z 3 2 2 ∠ 45

Example Given the circuit in Figure 22.17, find the indicated impedance Zin if ɷ = 2 rad/sec 2Ω

Solution 2 + j5 I 1 − jI 2 − j4I 3 = E

1/2 h

−jI 1 + I 2 − I 3 = 0

+

− j4I 1 − I 2 + 1 + j2 I 3 = 0

–

2 + j5 − j Z =

– E

−j

1

− j4

− 1 1 + j2

2h

−1

–Z Z in =

E = 3 −j7 5 = 8 08 ∠ −68 2 I1

1/2 f

¯I 3

1/4 f

1

¯I 1

−j4

¯I 2

in

Figure 22.17 Example circuit.

22.4 Ohm’s Law

22.3.5

Circuit with Lumped Elements

k i(t)

+ v(t) _

Reference

m

0

As a starting point to form a network, we interconnect lumped elements. Let’s start with two terminal elements: in this case, terminals are electrical nodes. We write the primary variables of interest: voltage v(t) at terminals and current i(t) that flows through the element. We use Ohm’s law; once we solve for voltages, other variables can be calculated as power flows and losses. We define the voltage v(t) at any time t across terminals k and m, in terms of voltages with respect to reference 0, and call the voltages Vk0 and Vm0 nodal voltages, at any time t. The following equation shows the voltage– current variables in a power-system device: v t = Vk0 t – Vm0 t Therefore, the product v(t) × i(t) is the instantaneous power p(t) delivered to the load element.

22.4

Ohm’s Law

In a series of experiments in 1825, Ohm demonstrated that there are no “perfect” electrical conductors. Every conductor he tested offered some level of resistance. Ohm’s law states that if temperature remains constant, the current flowing through a conductor is proportional to the potential difference (voltage) across it. In modern words, current equals voltage divided by resistance. In most equations and diagrams, I represents the current, V the voltage, and R the resistance. Ohm’s law is a crucial rule for analyzing electrical circuits and describing the relationship between three vital physical quantities: voltage, current, and resistance. It represents that the current is proportional to the voltage across two points, with the constant of proportionality being the resistance. The relationship defined by Ohm’s law is generally expressed in three equivalent forms:

• • •

V = I × R, where V represents the voltage measured across the conductor in volts V I = , where I represents the electrical current in units of amperes R V R = , where R represents the resistance of the conductor in ohms I In other words, if we increase the voltage, then the current will increase. However, if we increase the resistance, then the current will decrease. High values of resistivity imply that the material making up the wire is very resistant to the flow of electricity. Low values of resistivity imply that the material making up the wire transmits electrical current very quickly. Another way of representing Ohm’s law was developed by Gustav Kirchhoff and takes the following form: J =σ×E where

• • •

J = current density (or electrical current per unit area of cross-section) of the material. This is a vector quantity representing a value in a vector field, meaning it contains both a magnitude and a direction. σ = conductivity of the material, which is dependent upon the physical properties of the individual material. The conductivity is the reciprocal of the resistivity of the material. E = electric field at that location. It is also a vector field. Work done raising a charge Q through a potential V is given by W = QV. W = Power, Using the fact that Q = i × t and t Power = I × V or Watts = Amps × Volts

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22 Digital Computation Basics

I˙

V2 R The power is the rate at which energy is dissipated in a (resistive) current element, usually in the form of heat. Figure 22.18 shows a sinusoidal AC circuit. According to Ohm’s law, we obtain Using the definition of resistance, we find Power = I × V = I 2 R =

˙ U

Z = R + jX

Figure 22.18 Power through a resistive element.

V or U = Z × I U = UR + jU I , UR = Ucosϕu ,UI = Usinϕu 1 Z = R + jX, X = ωL or X = − ωC I = IR + jI I , IR = Icosϕi ,II = Isinϕi

Here, V or U and I are amplitudes of the voltage and current, respectively; ϕu and ϕi are phases of the voltage and current, respectively; and ω is the line angular frequency. Ohm’s law can be rewritten as UR + jU I = R + jX IR + jI I This is the phasor form of Ohm’s law for the impedance circuit, and the corresponding phasor form of Ohm’s law for the admittance circuit is written as

A + DC variable voltage source

V

v –

Figure 22.19 Ideal DC variable voltage source.

v

Gradient Dv Di

Figure 22.20 VI linear relationship.

R + jI I = G + jB UR + jU I

R

where G and B are conductance and susceptance, respectively. Further, consider an ideally controllable DC voltage source that feeds an ideal resistive load, as shown in Figure 22.19, below where current i through the load and voltage v across its terminals are measured. If the voltage applied by the source is increased gradually and the current through R is plotted for a range of values, Figure 22.20 reveals a linear relationship between the applied voltage v and the current response i. The gradient R in ohms is the resistance of the resistive element and acts as proportionality constant between the applied voltage and the current as a response. For a relation between the applied voltage and the output current, as shown in Figure 22.21, the behavior is not continuously linear. In this case, we should apply Ohm’s law in a piecewise form: v t = R1 i t

0 ≤ v t ≤ V1

v t = R2 i t

V1 ≤ v t ≤ V2

V2

Example A voltage U = 10 V is applied to a copper conductor with a length l = 200 m and a cross-section S = 50 mm2. Calculate the resistance R and the current I at 60 C. The resistivity of copper is ρ20 C = 17.5 nΩ-m.

V1

Solution

Gradient 2

Gradient 1

R20

C

=

ρ20

R60 C = R20

17 5 × 10 − 9 × 200 = 0 07Ω S 50 × 10 −6 C × 1 + α × T − T0 = 0 07 x 1 + 0 004 x 60 − 20 C

×l

=

= 0 0812 Ω Figure 22.21 Non-continuous VI relationship.

V 10 = 123 1 A I= = R 0 0812

22.5 Kirchhoff’s Circuit Laws

22.5

Kirchhoff’s Circuit Laws

Kirchhoff’s circuit laws, including Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL) deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff. This generalized the work of Georg Ohm and preceded the work of James Clerk Maxwell. Widely used in electrical engineering, they are also called Kirchhoff’s rules or simply Kirchhoff’s laws. Both of Kirchhoff’s laws can be understood as corollaries of Maxwell’s equations in the low-frequency limit. They are accurate for DC circuits, and for AC circuits at frequencies where the wavelengths of electromagnetic radiation are very large compared to the circuits.

22.5.1

Kirchhoff’s Current Law

This is also known as Kirchhoff’s first law, Kirchhoff’s point rule, and Kirchhoff’s junction rule or nodal rule. It states that the sum of all the currents flowing into a node is equal to the sum of all the currents flowing away from that node: i.e. the algebraic sum of currents in a network of conductors meeting at a point is zero, as shown in Figure 22.22. Charge (coulombs) is therefore conserved and is the product of current (amps) and time (sec). Or Ientering =

Ileaving

For basic circuit analysis, this can be summarized as n

Ik = 0 k =1

where n is the number of branches with currents flowing toward and away from the junction or node. For complex current, the equation can be rewritten as n

Ik = 0 k =1

KCL is typically used in conjunction with Ohm’s law for nodal analysis of power systems. KCL is applicable to any lumped network regardless of the nature of the network, whether unilateral or bilateral, active or passive, linear or nonlinear. 22.5.2

Kirchhoff’s Voltage Law

I1

This is also known as Kirchhoff’s second law, Kirchhoff’s second rule, and Kirchhoff’s loop rule or mesh rule. It states that the sum of all voltages in a closed loop is zero, as shown in Figure 22.23. Alternatively, the algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total EMF available in that loop:

I4

I2

I5

I3

Vs = V1 + V2 + V3 + V4

I1 + I2 + I3 = I4 + I5

Similar to KCL, a simplified form of KVL can be written for basic circuit analysis as well as complex voltage

Figure 22.22 Kirchhoff’s junction rule.

n

Vk = 0

V1

k =1

where n is the total number of voltages measured. For complex voltage, VS

V2

V3

+ –

n

Vk =0 k =1

Figure 22.23 Kirchhoff’s voltage law (KVL) or mesh rule.

V4

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22 Digital Computation Basics

22.5.2.1 Limitations

KCL and KVL both depend on the lumped element model. When the model is not applicable, the laws do not apply. KCL is dependent on the assumption that current flows only in conductors, and whenever current flows into one end of a conductor, it immediately flows out the other end. This is not a safe assumption for high-frequency AC circuits, where the lumped element model is no longer applicable. KCL violations can occur in high-voltage cables even at 60 Hz. KVL is based on the assumption that there is no fluctuating magnetic field linking the closed loop. This is not a safe assumption for high-frequency (short-wavelength) AC circuits. In the presence of a changing magnetic field, the line integral of the electric field around the loop is not zero, directly contradicting KVL.

22.6

Electrical Division Principle

There are two division principles for electrical quantities.

22.6.1

Current Division

When two or more passive elements are connected in parallel, the amount of current that flows through each element is divided (shared) among them from the current that is entering the node. Consider the circuit diagram shown in Figure 22.24, consisting of an input current source IS in parallel with two resistors R1 and R2. The voltage across each element is VS. The currents flowing through the resistors R1 and R2 are I1 and I2 respectively. The KCL equation at node P will be I S = I1 + I2 Substituting I1 = IS =

Vs Vs and I2 = in this equation, R1 R2

VS VS R2 + R1 + = VS R1 R2 R1 R2

V S = IS

Substituting the value of VS in I1 = I1 =

Is R1 × R2 R1 R1 + R2

I1 = IS

Is R1 × R2 R2 R1 + R2

VS , R1

R2 R1 + R2

Substituting the value of VS in I2 = I2 =

R1 R2 R1 + R2

I2 = IS ×

VS , R2

R1 R1 + R2

From the equations for I1 and I2, we can generalize that the current flowing through any passive element can be found by using the following formul IN = Is ×

P

Is

+

I1

Vs

R1

– Figure 22.24 Current division.

I2 R2

Z1 Z2 … ZN −1 Z1 + Z2 + … + ZN

where IN : current flowing through the passive element of the Nth branch IS : input current that enters the node Z1, Z2, … , ZN : impedances of the first branch, second branch, …, Nth branch, respectively This is known as the current-division principle, and it is applicable when two or more passive elements are connected in parallel and only one current enters the node.

22.7 Instantaneous, Average, and RMS Values

22.6.2

Voltage Division

Is

When two or more passive elements are connected in series, the amount of voltage present across each element is divided (shared) among them from the voltage that is available across that entire combination. Consider the circuit diagram shown in Figure 22.25, consisting of a voltage source VS in series with two resistors R1 and R2. The current flowing through these elements is IS. The voltage drops across the resistors R1 and R2 are V1 and V2, respectively. The KVL equation around the loop is Vs = V1 + V2. Substituting V1 = IS × R1 and V2 = IS × R2 in this equation,

+ Vs

R1 V1

–

+ –

+ V2

R2

–

Figure 22.25 Voltage division.

VS = IS × R1 + IS × R2 = IS × R1 + R2 VS IS = R1 + R2 Substituting the value of IS in V1 = IS × R1, V1 =

VS × R1 R1 + R2

V1 = VS ×

R1 R1 + R2

Substituting the value of IS in V2 = IS × R2, V2 =

VS × R2 R1 + R2

V2 = VS ×

R2 R1 + R2

From the equations of V1 and V2, we can generalize that the voltage across any passive element can be found by using the following formula ZN VN = VS × Z1 + Z2 + … + ZN where VN : voltage across the Nth passive element VS : input voltage, which is present across the entire combination of series passive elements Z1, Z2, …, Z3 : impedances of the first passive element, second passive element, …, Nth passive element, respectively This is known as the voltage-division principle, and it is applicable when two or more passive elements are connected in series and only one voltage is available across the entire combination.

22.7

Instantaneous, Average, and RMS Values

Instantaneous power is the energy absorbed in dt seconds by a circuit element or a network. For a network N, as shown d in Figure 22.26, instantaneous power is given by p t = w t = v t × i t watts (W). dt t w t = p t dt = w 0 + v t × i t dt 0

T

1 p t dt watts (P = Pavg). Average power is defined as P = T 0

p t = p t + T , which is a periodic function with a period = T Implies v t and i t both have to be periodic If v t = Vm cos ωt + θv and i t = Im cos ωt + θi p t = Vm × Im cos ωt + θv cos ωt + θi 1 2π where T = = f ω Vavg = 0 and Iavg = 0 1 Therefore, P = Pavg = Vm Im cosθ 2

i(t) Network v(t)

N

Figure 22.26 Instantaneous power circuit.

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22 Digital Computation Basics

22.7.1

a

The root mean square (RMS) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load:

0.707a VPK

Root Mean Square

VPP

VRMS y 0

T

Vrms =

v2

t

avg

=

1 2 v dt T 0

If i(t) = Im cos(ωt + θ), which is a sinusoidal with ω = 2πf ,

–a 0

180 t°

90

270

360 T

Irms =

Figure 22.27 Continuous periodic wave.

1 2 i t dt = T 0

T

1 2 1 1 Im + cos 2ωt + 2θ dt = Im T 2 2 2 0

If the waveform is a pure sine wave, the relationships between amplitudes (peak-to-peak, peak) and RMS are fixed and known, as they are for any continuous periodic wave, as shown in Figure 22.27. However, this is not true for an arbitrary waveform, which may or may not be periodic or continuous. For a zero-mean sine wave, the relationship between RMS and peak-to-peak amplitude (VPK) is Peak-to-peak = 2 2 × RMS ≈ 2 8 × RMS Note that Irms

Im for a non-sinusoidal i(t): 2 T

P = Pavg =

1 2 p t dt = T 0

22.8

T

1 2 2 v t i t dt = Vrms Irms cosθ = VIcosθ T 0

Nodal Formulation

Based on the previous sections, solving a circuit means finding a value for all node potentials and conductor currents. For circuits including branches, this means finding the values of all branch voltages and currents. The most natural and most common techniques are as follows:

• • •

Superposition allows the rank of the system to be limited based on the number of Kirchhoff equations, since constitutive equations are directly substituted into Kirchhoff’s. Nodal analysis allows further reduction of equations, giving rise to a system whose rank is equal to the number of circuit nodes minus one. Mesh analysis is another method that allows reduction of the number of equations up to the number of KVL equations.

22.8.1

Superposition Theorem

The superposition theorem is based on the concept of linearity between the response and excitation of an electrical circuit. It states that the response to several independent sources is the sum of the response to each of them with the remaining independent sources dead. In this method, we will consider only one independent source at a time. So, we have to eliminate the remaining independent sources from the circuit. We can eliminate the voltage sources by shorting their two terminals, and, similarly, eliminate the current sources by opening their two terminals. Therefore, we need to find the response in a particular branch n times if there are n independent sources. The response in a particular branch could be either current flowing through that branch or voltage across that branch: i t = i1 t + i2 t + … = 2 I1 cos ω1 t + θ1 + 2 I2 cos ω2 t + θ2 + …

22.8 Nodal Formulation

Total effective voltage V and effective current I of arbitrary voltage and current including various frequency components are defined as follows, where the variables with upper bar means “effective voltages and currents” (refer the Eq. (19.9c)).

5Ω

20 V

+ –

10 Ω

10 Ω

20 Ω

4A

V = V1 + V2 + … I = I1 + I2 + … where ω1

ω2

…

Figure 22.28 Example circuit.

including a possible DC component P = I 2 R = I12 R + I22 R+ … = P1 + P2 + … by superposition

5Ω

General procedure: 1) Find the response in a particular branch by considering one independent source and eliminating the remaining independent sources present in the network. Short-circuit a current source, and open-circuit a voltage source. 2) Repeat step 1 for all independent sources present in the network. 3) Add all the responses to get the overall response in a particular branch when all independent sources are present in the network.

20 V

+ –

10 Ω

10 Ω

20 Ω

Figure 22.29 Modified circuit.

Example Find the current flowing through a 20 Ω resistor of the circuit shown in Figure 22.28 using the superposition theorem. Solution Step 1 – Find the current flowing through the 20 Ω resistor by considering only the 20 V voltage source. In this case, eliminate the 4 A current source by assuming it is an open circuit. The modified circuit diagram is shown in Figure 22.29. There is only one primary node other than ground in this circuit. So, we can use the nodal-analysis method. The node voltage V1 is labeled in the following figure. Here, V1 is the voltage from node 1 with respect to ground. 5Ω

10 Ω

V1 1

20 V

+ –

10 Ω

20 Ω

The nodal equation at node 1 is V1 − 20 V1 V1 6V1 −120 + 3V1 + V1 + + =0 = 0 10V1 = 120 V1 = 12V 5 10 10 + 20 30 The current flowing through the 20 Ω resistor can be found by doing the following simplification: V1 10 + 20 Substituting the value of V1 in the previous equation, I1 =

I1 =

12 12 = 0 4A = 10 + 20 30

Therefore, the current flowing through the 20 Ω resistor is 0.4 A, when only the 20 V voltage source is considered. Step 2 – Find the current flowing through the 20 Ω resistor by considering only the 4 A current source. In this case, eliminate the 20 V voltage source by assuming it is a short circuit. The modified circuit diagram is shown in Figure 22.30.

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22 Digital Computation Basics

5Ω

10 Ω

A

10 Ω

20 Ω

In this circuit, there are three resistors on the left of terminals A and B. We can replace these resistors with a single equivalent resistor. Here, 5 and 10 Ω resistors are connected in parallel, and the entire combination is in series with the 10 Ω resistor. The 4A equivalent resistance on the left of terminals A and B will be: RAB =

5 × 10 10 40 + 10 = + 10 = Ω 5 + 10 3 3

The simplified circuit diagram is shown in Figure 22.31. We can find the current flowing through the 20 Ω resistor by using current-division principle:

B Figure 22.30 Modified circuit.

I2 = IS ×

A

R1 R1 + R2

Substituting Is = 4A, R1 = 40 Ω 3

20 Ω

4A

I2 = 4 × B

40 3

40 + 20 3

40 Ω, and R2 = 20 Ω in this equation, 3

=4

40 = 1 6A 100

Therefore, the current flowing through the 20 Ω resistor is 1.6 A, when only the 4 A current source is considered. Step 3 – We will get the current flowing through the 20 Ω resistor of the given circuit by adding the two currents we got in step 1 and step 2. Mathematically, it can be written as I = I1 + I2 Substituting the values of I1 and I2 in this equation, I = 0.4 + 1.6 = 2A Therefore, the current flowing through the 20 Ω resistor of the circuit is 2 A. Note that the superposition theorem cannot be applied directly in order to find the amount of power delivered to any resistor in a linear circuit, just by adding the power delivered to that resistor by each independent source. Instead, we can calculate either total current flowing through or voltage across that resistor by using the superposition theorem and, V2 . from that, calculate the amount of power delivered to that resistor using I2R or R Figure 22.31 Simplified circuit.

22.8.2

Nodal Analysis

Follow these steps while solving any electrical network or circuit using nodal analysis: 1) Identify the principal nodes, and choose one of them as a reference node. We will treat that reference node as the ground. 2) Label the node voltages with respect to ground from all the principal nodes except the reference node. 3) Write nodal equations at all the principal nodes except the reference node. The nodal equation is obtained by applying KCL first and then Ohm’s law. 4) Solve the nodal equations obtained in step 3 in order to get the node voltages. Now, we can find the current flowing through an element and the voltage across an element that is present in the given network by using node voltages. Example Find the current flowing through the 20 Ω resistor of the circuit shown in Figure 22.32 using nodal analysis. Solution Step 1 – There are three principle nodes in the Figure 22.32 circuit. They are labeled 1, 2, and 3 in Figure 22.33. Consider node 3 as the reference node (ground).

22.8 Nodal Formulation

Step 2 – The node voltages, V1 and V2, are labeled in Figure 22.34. V1 is the voltage from node 1 with respect to ground, and V2 is the voltage from node 2 with respect to ground. Step 3 – In this case, we will get two nodal equations, since there are two principal nodes, 1 and 2, other than ground. When we write the nodal equations at a node, assume all the currents are leaving from the node for which the direction of current is not mentioned and that the node’s voltage is greater than other node voltages in the circuit. The nodal equation at node 1 is V1 − 20 V1 V1 −V2 + + =0 5 10 10 2V1 − 40 + V1 + V1 − V2 =0 10 4V1 − 40 −V2 = 0 V2 = 4V1 − 40

5Ω

20 V

10 Ω

+ –

10 Ω

5Ω

10 Ω 1

20 V

+ –

2 10 Ω

20 Ω

4A

22 2

− 80 + V2 + 2V2 − 2V2 =0 20

3 Figure 22.33 Example circuit with nodes labeled.

22 3 Step 4 – Find the node voltages V1 and V2 by solving Eqs. (22.2) and (22.3). Substituting Eqs. (22.2) in (22.3), 3 4V1 −40 − 2V1 = 80

4A

Figure 22.32 Example circuit.

The nodal equation at node 2 is V2 V2 −V1 −4 + + =0 20 10 3V2 − 2V1 = 80

20 Ω

12V1 −120 −2V1 = 80 10V1 = 200 V1 = 20V

5Ω

10 Ω

V1 1

20 V

+ –

V2 2

10 Ω

20 Ω

4A

3

Substituting V1 = 20 V in Eq. (22.2), V2 = 4 20 −40

V2 = 40V

Figure 22.34 Example circuit with voltages labeled.

So, we get the node voltages V1 and V2 as 20 V and 40 V, respectively. Step 5 – The voltage across the 20 Ω resistor is nothing but the node voltage V2, and it is equal to 40 V. Now, we can find the current flowing through the 20 Ω resistor by using Ohm’s law: V2 I20Ω = R Substituting the values of V2 and R in this equation, 40 I20Ω = I20Ω = 2A 20 Therefore, the current flowing through the 20 Ω resistor of the given circuit is 2 A. Note: From this example, we can conclude that we have to solve n nodal equations if the electric circuit has n principal nodes (except the reference node). Therefore, we can choose nodal analysis when the number of principal nodes (except the reference node) is less than the number of meshes of any electrical circuit. 22.8.3

Mesh Analysis

In mesh analysis, we will consider the currents flowing through each mesh. Hence, mesh analysis is also called the meshcurrent method. A branch is a path that joins two nodes, and it contains a circuit element. If a branch belongs to only one mesh, then the branch current is equal to mesh current. If a branch is common to two meshes, then the branch current is equal to the sum (or difference) of two mesh currents, when they are in the same (or opposite) direction.

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22 Digital Computation Basics

22.9

30 Ω

5Ω

Procedure for Mesh Analysis

Follow these steps while solving any electrical network or circuit using mesh analysis:

20 V

+ –

10 Ω

1) Identify the meshes, and label the mesh currents in either a clockwise or counterclockwise direction. 2) Observe the amount of current that flows through each element in terms of mesh currents. Figure 22.35 Example circuit with voltages labeled. 3) Write mesh equations for all meshes. A mesh equation is obtained by applying KVL first and then Ohm’s law. 30 Ω 5Ω 4) Solve the mesh equations obtained in step 3 in order to get the mesh currents. Now, we can find the current flowing through any element + and the voltage across any element that is present in the given 20 V – network by using mesh currents.

I1

10 Ω

I2

– +

80 V

– +

80 V

Example Figure 22.36 Mesh current in the example circuit. Find the voltage across the 30 Ω resistor using mesh analysis for Figure 22.35. Solution Step 1 – There are two meshes in this circuit. The mesh currents I1 and I2 are considered in a clockwise direction. These mesh currents are shown in Figure 22.36. Step 2 – The mesh current I1 flows through the 20 V voltage source and the 5 Ω resistor. Similarly, the mesh current I2 flows through the 30 Ω resistor and the −80 V voltage source. But the difference of the two mesh currents, I1 and I2, flows through the 10 Ω resistor, since it is the common branch of the two meshes. Step 3 – In this case, we will get two mesh equations since there are two meshes in the given circuit. When we write the mesh equations, assume the mesh current of that particular mesh is greater than all other mesh currents in the circuit. The mesh equation of first mesh is 20 − 5I1 − 10 I1 −I2 = 0

20 − 15I1 + 10I2 = 0

10I2 = 15I1 − 20

Divide this equation by 5: 2I2 = 3I1 − 4 Multiply this equation by 2 4I2 = 6I1 − 8

22 4

The mesh equation of the second mesh is −10(I2 − I1) − 30I2 + 80 = 0 Divide this equation by 10: −(I2 − I1) − 3I2 + 8 = 0 − 4I2 + I1 + 8 = 0

4I2 = I1 + 8

22 5

Step 4 – Find mesh currents I1 and I2 by solving Eqs. (22.4) and (22.5). The left-side terms of Eqs. (22.4) and (22.5) are the same. Hence, equate the right-side terms of Eqs. (22.4) and (22.5) in order find the value of I1: 6I1 − 8 = I1 + 8

5I1 = 16

I1 =

16 A 5

Substituting the I1 value in Eq. (22.5), 4I2 =

16 +8 5

4I2 =

56 5

I2 =

14 A 5

So, we have the mesh currents I1 and I2 as

16 14 A and A, respectively. 5 5

22.9 Procedure for Mesh Analysis

Step 5 – The current flowing through the 30 Ω resistor is nothing but the mesh current I2, and it is equal to 145 A. Now, we can find the voltage across the 30 Ω resistor by using Ohm’s law: V30Ω = I2 R Substituting the values of I2 and R in this equation, V30Ω =

• •

14 30 5

V30Ω = 84V

Therefore, the voltage across the 30 Ω resistor of the given circuit is 84 V. Note: From this example, we can conclude that we have to solve m mesh equations if the electric circuit has m meshes. That’s why we can choose mesh analysis when the number of meshes is less than the number of principal nodes (except the reference node) in any electrical circuit. We can choose either nodal analysis or mesh analysis when the number of meshes is equal to the number of principal nodes (except the reference node) in any electric circuit.

22.9.1

Equivalent Circuits

If a circuit consists of two or more similar passive elements connected exclusively in series or in parallel, then we can replace them with a single equivalent passive element. Hence, this circuit is called an equivalent circuit. There are two types. 22.9.1.1 Series Equivalent Circuits

If similar passive elements are connected in series, then the same current flows through all of these elements. But the voltage is divided across each element. Consider the circuit diagram shown in Figure 22.37. It has a single voltage source (VS) and three resistors with resistances of R1, R2, and R3. All of these elements are connected in series. The current IS flows through all of these elements. This circuit has only one mesh. The KVL equation around this mesh is VS = V1 + V2 + V3 Substituting, V 1 = IS R 1 V 2 = IS R 2 and V3 = ISR3 in the previous equation. VS = IS R1 + IS R2 + IS R3

IS

R1

R2

VS = IS R1 + R2 + R3

This equation is in the form of VS = ISREq

VS

+ –

R3

where REq = R1 + R2 + R3 The equivalent circuit diagram of the given circuit is shown in Figure 22.38. Figure 22.37 Series equivalent circuit. That means if multiple resistors are connected in series, we can replace them with an equivalent resistor. The resistance of this equivalent resistor is equal to the sum of the resistances of all those multiple resistors. Note 1 – If N inductors having inductances of L1, L2, … , LN are connected in series, then the equivalent inductance is LEq = L1 + L2 + … + LN

VS

+ –

IS

REq

Note 2 – If N capacitors having capacitances of C1, C2, … , CN are connected in series, then the equivalent capacitance is 1 1 1 1 = + +…+ CEq C1 C2 CN

Figure 22.38 Equivalent circuit diagram.

663

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22 Digital Computation Basics

22.9.1.2 Parallel Equivalent Circuits

P +

IS

VS

I1

I2

I3

R1

R2

R3

– Figure 22.39 Parallel circuit diagram.

+ IS

VS

If similar passive elements are connected in parallel, then the same voltage will be maintained across each element. But the current flowing through each element is divided. Consider the circuit diagram shown in Figure 22.39. It has a single current source (IS) and three resistors having resistances of R1, R2, and R3. All of these elements are connected in parallel. The voltage (VS) is available across all of these elements. This circuit has only one principal node (P) except the ground node. The KCL equation at this principal node (P) is IS = I1 + I2 + I3. VS VS VS in the previous Substituting I1 = , I2 = , and I3 = R1 R2 R3 equation,

REq

IS =

VS VS VS + + R1 R2 R3

– Figure 22.40 Equivalent parallel circuit diagram.

IS = V S

1 1 1 + + R1 R 2 R3

VS = IS ×

1 1 1 1 + + R 1 R2 R 3

This equation is in the form of VS = ISREq, where REq =

1 1 1 1 1 or = + + 1 1 1 REq R1 R2 R3 + + R1 R2 R3

The equivalent circuit diagram of the given circuit is shown in Figure 22.40. That means if multiple resistors are connected in parallel, then we can replace them with an equivalent resistor. The resistance of this equivalent resistor is equal to the reciprocal of the sum of the reciprocal of each resistance of all those multiple resistors. Note 1 – If N inductors having inductances of L1, L2, … , LN are connected in parallel, then the equivalent inductance is 1 1 1 1 1 = + + +…+ LEq L1 L2 L3 LN Note 2 – If N capacitors having capacitances of C1, C2, … , CN are connected in parallel, then the equivalent capacitance is CEq = C1 + C2 + … + CN

22.10

Norton’s and Thévenin’s Equivalents

22.10.1

Thévenin’s Theorem

Any two-terminal network may be replaced by the open-circuit voltage in series with the dead network. A dead network implies the voltage source is short-circuited and current sources are open-circuited, as shown in Figure 22.41: Voc = Req Isc Voc =

R2 R1 R2 V1 and Req = R1 + R2 R1 + R2

22.10 Norton’s and Thévenin’s Equivalents

Req a R1

+ V1 –

R1

–

≡

a

Voc

R2

R1

+ a

≡ Voc

R2

R2 b

b

b Req

Figure 22.41 Open circuit voltages.

Example Find V0 by using Thévenin’s theorem. Solution

4

1

+

1

a

6V _

b

V0 2

2

Find Req: 1 1 a

4

a

Req =

b

2

2

2/3 Ω

8/6 Ω

b 2

4

2 8 + = 2Ω 3 6

2

Find Voc: 1 × 6 = 2V 1+2 4 V2 = × 6 = 4V 4+2 Voc = V2 − V1 = 2V V1 =

+

V1

6V

1 a

V2

Voc

_ 2

Req = 2 Ω + 2 V Voc

4 b

2

Vo =

a

Vo

– b

1Ω

1 2 ×2= V 1+2 3

665

666

22 Digital Computation Basics

a) Find Isc: 6 volts 30 = amps. 4 2×2 9 + 1+4 2+2 4 30 24 I2 = × = amps 4+1 9 9 2 30 15 × = amps I4 = 2+2 9 9 Isc = I2 − I4 = 1 amps

I1

I1 =

I2

I3

1

4

+

ISC

a 6V

b I5

I4

_

2

2 I1

Req =

Voc 2 = = 2Ω Isc 1

Example Replace the following network with its Thévenin equivalent. Solution R1 a – kV2 +

(R2 + R1)Isc + kV2 = 0

V2

VOC

+

(R2 + R1)Isc + k(V1 + R2Isc) = 0

R2

V1 –

b R1 – kV2 +

Isc = −

ISC

V2

k V1 R1 + 1 + k R2

+ R2

V1 –

Voc = − kV2 = − kV1

a + Voc –

Req = b

VS

Z

Figure 22.42 Norton circuit.

U

− kV 1 − kV 1 R1 + 1 + k R2

22.10.2

I

ZN

Voc = Isc

Zload

= R1 + 1 + k R2

Norton’s Theorem

Similar to Thévenin’s theorem for AC circuits, we can calculate the current equivalent source (Norton’s current source JN) with an impedance (Norton’s impedance ZN) connected parallel to the source, as shown in Figure 22.42. Therefore, any two-terminal network may be replaced with the short-circuit current in parallel with the dead network, as shown in Figure 22.43:

22.10 Norton’s and Thévenin’s Equivalents

Therefore Req =

Voc Isc

Req a

Example Find the current and voltage for the circuit shown in Figure 22.44.

+ ≡

Voc

ISC

Req

–

Solution

b

Isc = 1A Voc Req = = 2Ω Isc I0 =

a

b

Figure 22.43 Norton circuit equivalent. Io a

2 2 ×1= A 1+2 3

V0 =

1A

2 2 × 1 = V = 10 67 V 3 3

2

1 Vo

b

Figure 22.44 Example circuit.

Example Determine the value of current I1 by using Norton’s theorem for the circuit shown in Figure 22.45.

2

a +

I1

~ 8V 3A

6

Solution

–

1Ω 2 b

a) Find Isc:

Figure 22.45 Example circuit.

3

(3-I) 2 + 8

6

3

ISC

– 2

(3-I-ISC)

I

3 I= A 4 8 + 2(3 − I − Isc) + 2(3 − I) = 6I 3 20 = 10I + 2Isc = 10 × + 2Isc 4 Isc = 6.25A 6I = 2 3 − I

a

b

b) Find Voc: 3

(3-I) 2

a

I = 2A

+ 3A

8

6

–

VOC

2

(3-I-)

I

8 + (2 + 2)(3 − I) = 6I Voc = 8 + 2(3 − I) Voc = 10V

b

c) Req =

Voc 10 = = 1 6Ω Isc 6 25

(Req =

2× 2+6 = 1 6Ω) 2+ 2+6

a

d) ISC

Req

1

I1 = I1

b

Req 16 × 6 25 = 3 85A Isc = 1+1 6 1 + Req

667

668

22 Digital Computation Basics

22.11

Maximum Power Transfer Theorem

The amount of power received by a load is an important parameter in electrical and electronic applications. In DC circuits, we can represent the load with a resistor having resistance of RL ohms. Similarly, in AC circuits, we can represent it with a complex load having an impedance of ZL ohms. The maximum power transfer theorem states that the DC voltage source will deliver maximum power to the variableload resistor only when the load resistance is equal to the source resistance. Similarly, this theorem states that the AC voltage source will deliver maximum power to the variable complex load only when the load impedance is equal to the complex conjugate of source impedance. This section discusses the maximum power transfer theorem for DC circuits. 22.11.1

Proof of the Maximum Power Transfer Theorem

Replace any two terminal linear networks or circuits on the left side of a variable-load resistor having resistance of RL ohms with a Thévenin’s equivalent circuit. We know that Thévenin’s equivalent circuit resembles a practical voltage source. This concept is illustrated in Figure 22.46. The amount of power dissipated across the load resistor is PL = I2RL. VTh in this equation, Substituting I = RTh + RL PL =

VTh RTh + RL

2

RL RTh + RL

2 PL = VTh

RL

2

22.11.1.1 Condition for Maximum Power Transfer 2 For the maximum or minimum, the first derivative is zero. So, differentiate PL = VTh

RL RTh + RL

2

with respect to RL

and make it equal to zero: dP L 2 = VTh dRL

RTh + RL

2

× 1 − RL × 2 RTh + RL RTh + RL

RTh + RL 2 − 2RL RTh + RL = 0 RTh − RL = 0

=0

4

RTh + RL RTh + RL − 2RL = 0

RTh = RL

Therefore, the condition for maximum power dissipation across the load is RTh = RL. That means if the value of load resistance is equal to the value of source resistance, i.e. Thévenin’s resistance, then the power dissipated across the load will be of maximum value. 22.11.1.2 Maximum Power Transfer Value 2 Substituting RL = RTh & PL = PL, Max in PL = VTh

2 PL, Max = VTh

RTh RTh + RTh

2

RL RTh + RL

2 PL, Max = VTh

RTh 4 R2Th RTh

A Two-terminal linear circuit

2

RL B

Figure 22.46 Maximum power transfer theorem.

,

2 PL, Max = VTh

A I RL

VTh + – B

2 VTh 4RTh

22.11 Maximum Power Transfer Theorem

Since RTh = RL

PL, Max =

2 VTh 4RTh

Therefore, the maximum amount of power transferred to the load is PL, Max =

2 VTh V2 = Th 4RL 4RTh

22.11.1.3 Efficiency of Maximum Power Transfer

The efficiency of maximum power transfer ηmax can be calculated using the formula ηmax =

PL, Max PS

where PL, Max : maximum amount of power transferred to the load PS : amount of power generated by the source The amount of power generated by the source is PS = 2

VTh 2RTh

2

RTh

PS = 2

Substituting the values of PL, VTh 2 4RTh ηmax = 2 VTh 2RTh

ηmax =

Max

2 VTh RTh 4 R2Th

PS =

and PS in ηmax =

2 VTh 2RTh

PL, Max , PS

1 2

We can represent the efficiency of the maximum power transfer as a percentage as follows: ηmax = ηmax × 100 1 ηmax = × 100 2

ηmax = 50

Example Find the maximum power that can be delivered to the load resistor RL of the circuit shown in Figure 22.47.

10 Ω

5Ω

Therefore, the efficiency of the maximum power transfer is 50%. 20 V

+ –

10 Ω

A

RL

4A

Solution B Step 1 – Thévenin’s equivalent circuit on the left side of terminals A and B is shown in Figure 22.48. Figure 22.47 Maximum power transfer example circuit. 200 Here, Thévenin’s voltage VTh = V and Thévenin’s 3 40 Ω 40 3 A resistance RTh = Ω. 3 Step 2 – Replace the part of the circuit, which is left side of terminals A and B of the given circuit with this Thévenin’s equivalent circuit. The resulting cir200 V + cuit diagram is shown in Figure 22.49. – 3 Step 3 – We can find the maximum power that is delivered to the load resistor RL by using the following formula: PL, Max =

2 VTh 4RTh

B Figure 22.48 Thévenin equivalent circuit.

669

22 Digital Computation Basics

40 Ω 3

+ –

2

200 3 PL, Max = 40 4 3

I

200 V 3

200 40 V and RTh = Ω in this formula, 3 3

Substituting VTh =

A

RL

or PL, Max =

250 W 3

Therefore, the maximum power that is delivered to the load resistor RL of the given circuit is 83.33 W.

B Figure 22.49 Thévenin equivalent circuit diagram.

a12

…

a1n

a21

a22

…

a2n

…

…

[A]=

a11

…

22.12 Linear System Mathematics

…

670

am1

am2

…

amn

22.12.1

Matrix Algebra

In mathematics, a matrix (plural matrices) is an array of numbers, symbols, or expressions, arranged in rows and columns. The dimensions of matrix (1) are 2 × 3 (read “two by three”) because there are two rows and three columns.

where aij can be real or complex

Figure 22.50 m-by-n matrix.

1 9 − 13 20 5 − 6 The m rows are horizontal, and the n columns are vertical. Each element of a matrix is often denoted by a variable with two subscripts. For example, in Figure 22.50, a21 represents the element at the second row and first column of a matrix.

A =

22.12.2

a11 a12

a1n

a21 a22

a2n

am1 am2

amn

where aij can be real or complex

Matrix Types

There are several types of matrices, but the most commonly used are as follows: Matrix type

General form (m × n)

Conjugate matrix A =

Transpose matrix A

T

a1n

a21 a22

a2n

am1 am2

amn

a11 a12

a21 a22

am1

a1n a2n

amn

=

Rectangular matrix

m

Square matrix

m=n

Symmetrical matrix

a11 a12

n

A = A

T

(m = n)

am2

22.12 Linear System Mathematics

General form (m × n)

Matrix type

a1 0

Diagonal matrix

0

0 a2 0 0 Identity matrix

0 a3

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 a11 a12 a13 a14

Upper triangular matrix

0

Lower triangular matrix

a22 a23 a24

0

0

0

0

a11

0

a21 a22

a33 a34 0

a44

0

0

0

0

a31 a32 a33

0

a41 a42 a43 a44

22.12.3

Matrix Addition and Subtraction

Two matrices must have an equal number of rows and columns to be added. The sum of two matrices A and B is a matrix that has the same number of rows and columns as A and B. The sum of A and B, denoted A + B, is computed by adding corresponding elements of A and B: A ± B = C cij = aij ± bij

i = 1, 2,…, m j = 1, 2,…, n

22.12.4

Matrix Multiplication

if A is an n × m matrix and B is an m × p matrix, their matrix product AB is an n × p matrix, in which the m entries across a row of A are multiplied with the m entries down a column of B and summed to produce an entry of AB. When two linear maps are represented by matrices, then the matrix product represents the composition of the two maps: n

k

k

m A n B =m C n

Cij =

ail blj l=1

Multiplication rules include: A B C = A B A + B A B

T

C

C = A C + B C = BTCT

Example 2 0 −1 3 4 2

1 −1 5 0 0 3

−2 6 =

4 1

0 1

−2 − 3 10 −1 11 11 7 26

671

672

22 Digital Computation Basics

22.12.5

Matrix Determinant

A determinant of a matrix represents a single number. We obtain this value by multiplying and adding its elements in a special way. We can use the determinant of a matrix to solve a system of simultaneous equations. The determinant of the 2 × 2 matrix A a b

A=

c d

is a scalar given by det A =

a b c d

; det A =

a b c d

= a∗ d − c∗ b

The determinant of the 3 × 3 matrix A a b c A= d e f g h i is a scalar given by a b c det A = d e f = a ei − hf − d bi− hc + g bf − ec g h i A =

A =

a11 a12

a11 a12 a21 a22

−1 h a1i1 a2i2 …anin

A =

a21 a22

= a11 a22 − a12 a21

a11 a12 a13 A = a21 a22 a23 = a11 × a11 minor + a12 × a12 minor + a13 × a13 minor a31 a32 a33 = a11

a22 a23 a32 a33

− a12

a21 a23 a31 a33

+ a13

a21 a22 a31 a32

Signs associated with positions of minors (cofactors): + − + − + − + − + − + − + − + Some other common determinant operations include: AB = A B , A −1 =

22.12.6

1 A

Triangle Matrix – Lower (L) and Upper (U)

A matrix is called an upper triangular matrix if, in a square matrix, all the values below the diagonal are equal to zero. Similarly, a matrix is called a lower triangular matrix if, in a square matrix, all the values above the diagonal are equal to zero. Note that in a square matrix, the number of rows equals the number of columns. The LU decomposition method, which we will discussed later, is used to solve linear equations representing a certain system and is fundamental to most computer-based software. The matrix of coefficients is written as a product of the lower (L) and upper (U) matrix.

22.12 Linear System Mathematics

Consider this system of equations that can be converted or written in matrix form: x1 + x2 −x3 = 3 x1 −2x2 + 4x3 = −5 2x1 + 3x2 + x3 = 8 1 1

−1

1 −2 4 2 3

x1 × x2 =

1

x3

3 −5 which is of the form AX = B 8

If we let A = LU, then we can substitute into AX = B. Then we can solve LUX = B to derive X in order to solve this system. Let UX = Y; then LY = B and UX = Y. Thereafter, we can first solve for LY = B and then solve UX = Y for X.

22.12.7

Matrix Inversion

If [A] is a square matrix and nonsingular, (|A| a b

A =

A

−1

0) implies there exists a unique matrix [A]−1 such that [A][A]−1 = [I]:

c d =

d −b adj A 1 = ad − bc −c a A

Example 1 0 1 Find the inverse of A = 2 3 1 . 1 3 2 Solution The matrix can be written as 1 0 1 1 0 0 2 3 1 0 1 0 1 3 2 0 0 1 And add −2 times the first row to the second row, and add −1 times the first row to the third row, obtaining 1 0 1

1 0 0

0 3 −1 −2 1 0 0 3 1

−1 0 1

Multiply the second row by one-third to get the following: 1 1 1 2 0 1 − − 3 3 0 3 1 −1 1 0

0 0 1 0 3 0 1

673

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22 Digital Computation Basics

Add −3 times the second row to the third row to get the following: 1 1 1 2 0 1 − − 3 3 0 0 2 1 1 0

0 0 1 0 3 −1 1

Multiply the third row by one-half to get the following: 1 0 0 1 0 1 2 1 1 − 0 0 1 − 3 3 3 1 1 0 0 1 1 − 2 2 2 Add −1 times the third row to the first two, and add one-third times the third two to the second row, to get the following: 1 1 1 − 2 2 2 1 0 0 1 1 1 0 1 0 − 2 6 6 0 0 1 1 1 1 − 2 2 2 Hence 1 1 1 − 2 2 2 1 1 1 −1 A = − 2 6 6 1 1 1 − 2 2 2 To check, calculate [A][A]−1 and [A]−1[A] to see whether they are [I]. Note: If [A] has an inverse, a system of n linear equations in n unknowns x1, … , xn, AX = Y can be solve by calculating [A]−1 and multiplying by it, for [A]−1[A]X = [A]−1Y or X = [A]−1Y. 22.12.8

Adjugate and Cofactor of a Matrix

The matrix cofactor is the determinant obtained by deleting the row and column of a given element of a matrix or determinant. The adjugate or adjunct of a square matrix is the transpose of its cofactor matrix. If A is a square matrix, then the minor of entry aij is denoted by Mij and is defined to be the determinant of the submatrix that remains after the ith row and jth column are deleted from A. The number (−1)i + jMij is denoted called the cofactor of entry aij: i.e. cof aij is the cofactor of the element aij and is (−1)i + j times the determinant of the matrix formed by deleting the ith row and jth column of [A]. a11 a12 a13 Consider A = a21 a22 a23 a31 a32 a33 A

−1

=

adj A A cof a11 cof a12 cof a13

adj A = cof aij

t

= cof a21 cof a22 cof a23 cof a31 cof a32 cof a33

t

22.13 Network Topology

Example 2 −1 5 A= 1 3 1 2 0 1 A = 2 3 − 0 − − 1 1 − 2 + 5 0 − 6 = − 25 cof a11 =

3 1 0 1

cof a12 = −

=3

1 1 2 1

= − 1−2 = 1

cof a13 = − 6 cof a21 = − 1 −1 − 0 = 1 cof a22 = 2 − 10 = −8 cof a23 = − 1 + 2 = − 2 cof a31 = − 1− 15 = − 16 cof a32 = − 1 2 −5 = 3 cof a33 = 6 + 1 = 7 A

22.13

−1

1 = −25

3

1

− 16

1

−8

3

−6 − 2

7

Network Topology

Network topology is a graphical representation of electric circuits. It is useful for analyzing complex electric circuits by converting them into network graphs. Network topology is also called graph theory. 22.13.1

Basic Terminology of Network Topology

Some terms commonly used in graph theory are defined as follows:

• •• •• •• •• ••

Element: An individual component such as a generator, transformer, transmission line, or load, which, regardless of its characteristics, is represented in the single line diagram of a power system by single line segment. Node: The terminals of an element. Graph: Shows the physical interconnections of the elements of a network. Subgraph: Any subset of elements of the graph. Path: A subgraph of connected elements with no more than two elements connected to any one node. Connected graph: A graph where there is a path between every pair of nodes. Oriented graph: A connected graph where all the elements are assigned directions. Tree: A connected subgraph constituting all the nodes but containing no closed loops. Branch: An element of a tree. Link: An element of the connected graph that are is included in the tree. Co-tree: A subgraph that is constituted of links. A co-tree is the complement of a tree.

22.13.1.1 Graphs

A network graph is simply called a graph. It consists of a set of nodes connected by branches. In graphs, a node is a common point of two or more branches. Sometimes, only a single branch connects to a node.

675

676

22 Digital Computation Basics

30 Ω

20 V

10 Ω

5Ω

1

2

3

+ –

10 Ω

20 Ω

4A

4 Figure 22.51 Example circuit.

A branch is a line segment that connects two nodes. Any electric circuit or network can be converted into its equivalent graph by replacing the passive elements and voltage sources with short circuits and the current sources with open circuits. That means the line segments in the graph represent the branches corresponding b c 1 3 to either passive elements or voltage sources of the electric circuit. 2 Consider the electric circuit shown in Figure 22.51. In this circuit, there are four principal nodes labeled 1, 2, 3, and 4. There are seven branches in this circuit, d e f among which one branch contains a 20 V voltage source, another branch contains a 4 A current source, and the remaining five branches contain resistors having resistances of 30, 5, 10, 10, and 20 Ω, respectively. An equivalent graph corresponding to this electric circuit is shown in Figure 22.52. 4 In this graph, there are four nodes labeled 1, 2, 3, and 4, respectively. These are Figure 22.52 Equivalent graph. the same as the principal nodes in the electric circuit. There are six branches in this graph, labeled a, b, c, d, e, and f, respectively. In this case, the graph has one less branch because the 4 A current source is made an open circuit while converting the electric circuit into its equivalent graph. From this example, we can conclude the following: a

••

The number of nodes present in a graph is equal to the number of principal nodes present in an electric circuit. The number of branches present in a graph is less than or equal to the number of branches present in an electric circuit.

22.13.2

Types of Graphs

Following are the types of graphs:

•• ••

Connected graph Unconnected graph Directed graph Undirected graph

22.13.2.1 Connected Graphs

If there exists at least one branch between any two nodes of a graph, then it is called a connected graph. That means each node in the connected graph has one or more branches connected to it. So, no node is isolated or separated. The graph shown in the previous example is a connected graph. In that case, all the nodes are connected by three branches. 22.13.2.2 Unconnected Graphs

If there exists at least one node in the graph that remains unconnected by even a single branch, then it is called an unconnected graph. So, there will be one or more isolated nodes in an unconnected graph. Consider the graph shown in Figure 22.53.

22.13 Network Topology

In this graph, nodes 2, 3, and 4 are connected by two branches each. But no branch has been connected to node 1. So, node 1 becomes an isolated node. Hence, this graph is an unconnected graph.

1

c

2

e

22.13.2.3 Directed Graph

If all the branches of a graph are represented with arrows, then that graph is called a directed graph. These arrows indicate the direction of current flow in each branch. Hence, this type is also called an oriented graph. Consider the graph shown in Figure 22.54. In this graph, the direction of current flow is represented with an arrow in each branch. Hence, it is a directed graph.

3

f

4 Figure 22.53 Unconnected graph.

a

22.13.2.4 Undirected Graphs

If the branches of a graph are not represented with arrows, then that graph is called an undirected graph. Since there are no directions for current flow, this type is also called an unoriented graph.

b

1

c

3

2

22.13.2.5 Subgraphs and Their Types

A part of a graph is called a subgraph. We get subgraphs by removing some nodes and/or branches of a given graph. So, the number of branches and/or nodes of a subgraph will be less than in the original graph. Hence, we can conclude that a subgraph is a subset of a graph. Following are the two types of subgraphs:

••

d

e

f

4

Tree Co-tree

22.13.3

Figure 22.54 Directed graph.

1

Tree

A tree is a connected subgraph of a given graph, which contains all the nodes of a graph. But there should not be any loops in the subgraph. The branches of a tree are called twigs. Consider the connected subgraph shown in Figure 22.55. This connected subgraph contains all four nodes of the given graph, and no loops. Hence, it is a tree. This tree has only three branches out of six branches of the given graph. If we consider a single branch of the remaining branches of the graph, then there is a loop in this connected subgraph. So, the resulting connected subgraph is not a tree. From this tree, we can conclude that the number of branches present in a tree should be equal to n − 1, where n is the number of nodes in the graph. 22.13.4

2

3

d

e

f

4 Figure 22.55 Tree subgraph.

Co-Trees

A co-tree is a subgraph that is formed from the branches that are removed while forming a tree. Hence, it is the complement of a tree. For every tree, there is a corresponding co-tree, and its branches are called links or chords. In general, the links are represented with dotted lines. The co-tree corresponding to the previous tree is shown in Figure 22.56. This co-tree has only three nodes instead of four nodes from the graph, because node 4 is isolated from the co-tree. Therefore, the co-tree need a not be a connected subgraph. This co-tree has three branches, and they form a loop. The number of branches in a co-tree is equal to the difference between the number of branches of the graph and the number of twigs. Mathematically, it can be written as l = b − n −1 l = b −n + 1

1

b

c

2

Figure 22.56 Co-tree subgraph.

3

677

678

22 Digital Computation Basics

where l : number of links b : number of branches in the graph n : number of nodes in the graph If we combine a tree and its corresponding co-tree, then we get the original graph, as shown in Figure 22.57. The tree branches d, e, and f are represented with solid lines. The co-tree branches a, b, and c are represented with dashed lines. A single-line representation of a four-bus electrical power-system network is shown in Figure 22.58. The corresponding oriented connected graph is given in Figure 22.59. Figure 22.60 shows a tree and the corresponding co-tree of the connected graph, where branches and links are shown in solid and dashed lines, respectively.

a

b

1

c

3

2

d

e

f

4 Figure 22.57 Combined tree and co-tree.

22.13.5 2

3

4

Injected currents, network topology, and branch impedances govern the voltages at each node of a circuit. In many cases, the relationship between the known, or input, quantities and the unknown, or output, states is a linear relationship. Therefore, a linear system may be generically modeled as

~

1 ~

~

Figure 22.58 Single-line representation of a four-bus electrical system.

1

4

2

5

3

6

Ax = b where b is the n × 1 vector of known quantities, x is the n × 1 unknown state vector, and A is the n × n matrix that relates x to b. For the time being, it is assumed that the matrix A is invertible, or nonsingular; thus, each vector b will yield a unique corresponding vector x. Thus the matrix A−1 exists, and x∗ = A−1b is the unique solution for Ax = b. The natural approach to solving the equation Ax = b is to directly calculate the inverse of A and multiply it by the vector b. One method to calculate A−1 is to use Cramer’s rule.

4

7 3 1

2

0

Matrix Operations

Figure 22.59 Four-bus system oriented, connected graph.

22.13.5.1 Cramer’s Rule

1 T Aij for det A i = 1, … , n and j = 1, … , n where A−1(i, j) is the ijth entry of A−1 and Aij is the cofactor of each entry aij of A. This method requires the calculation of (n + 1) determinants, which results in 2(n + 1)! multiplications to find A−1!. For large values of n, the calculation requirement grows too rapidly for computational tractability; thus, alternative approaches have been developed to solving the equation Ax = b: Cramer’s rule calculates A−1 as A − 1 i,j =

1

4

2

5

3

6

4

7 3 1

0

Branch of tree

2

Link of co-tree

•

•

Direct methods, or elimination methods, find the exact solution (within the accuracy of the computer) through a finite number of arithmetic operations. The solution x of a direct method would be completely accurate were it not for computer round-off errors. Iterative methods, on the other hand, generate a sequence of progressively improving approximations to the solution based on the application of the same computational procedure at each step. The iteration is terminated when an approximate solution is obtained having some pre-specified accuracy or when it is determined that the iterates are not improving.

Figure 22.60 Co-tree of a connected graph.

22.13 Network Topology

The choice of solution methodology usually relies on the structure of the system under consideration. Certain systems lend themselves more amenably to one type of solution method versus the other. In general, direct methods are best for full matrices, whereas iterative methods are better for matrices that are large and sparse. As with most generalizations, there are notable exceptions to this rule of thumb 22.13.5.2 Gaussian Elimination

Gaussian elimination, or the row-reduction algorithm, is usually used to solve systems of linear equations. The algorithm consists of a sequence of operations that are applied to the matrix representing the set of equations. We first write the set of equations in a matrix form (A|xi=|yi). Then we make the elements of the lower triangle of matrix A zero using appropriate row transformations that are also applied to y. These row operations are performed in N steps; then we can find each value in x by continuous subtraction and division. Here, N represents the number of rows and columns in A. Gaussian elimination is usually carried out using matrices. This method reduces the effort in finding the solutions by eliminating the need to explicitly write the variables at each step. Example Solve this system: x+y=3 3x − 2y = 4 Solution Multiplying the first equation by −3 and adding the result to the second equation eliminates the variable x: −3x−3y = − 9 3x −2y = 4 −5y = − 5 This final equation, −5y = − 5, immediately implies y = 1. Back-substitution of y = 1 into the original first equation, x +y = 3, yields x = 2. Back-substitution of y = 1 into the original second equation, 3x − 2y = 4, would also yield x = 2. The solution of this system is, therefore(x, y) = (2, 1). Example Solve this system: x+y=3 3x − 2y = 4 Solution The first step is to write the coefficients of the unknowns in a matrix:

1 1

3 −2 This is called the coefficient matrix of the system. Next, the coefficient matrix is augmented by writing the constants that appear on the right sides of the equations as an additional column: 1 1 3 3 −2 4 This is called the augmented matrix, and each row corresponds to an equation in the given system. The first row, r1 = (1, 1, 3), corresponds to the first equation, 1x + 1y = 3, and the second row, r2 = (3, −2, 4), corresponds to the second equation, 3x − 2y = 4. We may choose to include a vertical line – as shown above – to separate the coefficients of the unknowns from the extra column representing the constants. The counterpart of eliminating a variable from an equation in the system is changing one of the entries in the coefficient matrix to zero. Likewise, the counterpart of adding a multiple of one equation to another is adding a multiple of one row to another row. Adding −3 times the first row of the augmented matrix to the second row yields

679

680

22 Digital Computation Basics

1 1 3

−3r1 added to r2

3 −2 4

1 1

3

0 −5 −5

The new second row translates into −5y = − 5, which means y = 1. Back-substitution into the first row (that is, into the equation that represents the first row) yields x = 2 and, therefore, the solution to the system: (x, y) = (2, 1). Gaussian elimination can be summarized as follows: given a linear system expressed in matrix form, Ax = b, first write down the corresponding augmented matrix: A b

•• •

Then, perform a sequence of elementary row operations, which are any of the following: Type 1: Interchange any two rows. Type 2: Multiply a row by a nonzero constant. Type 3: Add a multiple of one row to another row.

The goal of these operations is to transform – or reduce – the original augmented matrix into one of the form [A b ], where A is upper triangular (aij = 0 for i > j), any zero rows appear at the bottom of the matrix, and the first nonzero entry in any row is to the right of the first nonzero entry in any higher row; such a matrix is said to be in echelon form. The solutions of the system represented by the simpler augmented matrix, [A b ], can be found by inspection of the bottom rows and back-substitution into the higher rows. Since elementary row operations do not change the solutions of the system, the vectors x that satisfy the simpler system A x = b are precisely those that satisfy the original system, A x = b. 22.13.6

Gauss–Jordan Elimination

Gaussian elimination proceeds by performing elementary row operations to produce zeros below the diagonal of the coefficient matrix and reduce it to echelon form. (Recall that a matrix A = [aij ] is in echelon form when aij = 0 for i > j, any zero rows appear at the bottom of the matrix, and the first nonzero entry in any row is to the right of the first nonzero entry in any higher row.) Once this is done, inspection of the bottom row(s) and back-substitution into the upper rows determines the values of the unknowns. However, it is possible to reduce (or eliminate entirely) the computations involved in back-substitution by performing additional row operations to transform the matrix from echelon form to reduced echelon form. A matrix is in reduced echelon form when, in addition to being in echelon form, each column that contains a nonzero entry (usually made to be 1) has zeros not just below that entry but also above that entry. Loosely speaking, Gaussian elimination works from the top down, to produce a matrix in echelon form, whereas Gauss–Jordan elimination continues where Gaussian left off by then working from the bottom up to produce a matrix in reduced echelon form.

22.13.7

LU Factorization

LU factorization means factorizing a matrix into two matrices L and U, where if A is a matrix, then A = LU such that L= lower triangular matrix and U= upper triangular matrix. This method is used to solve an equation in matrix form. Consider this example: AX = B (X is unknown), or (LU)X = B [since A = LU], or L(UX) = B, or LY = B, where Y = UX. We first find L and U from the equation A = LU and then find Y from the equation LY = B. Finally, we calculate matrix X from the equation Y = UX. The main idea of LU decomposition is to record the steps used in Gaussian elimination on A in the places where the zero is produced. Consider this matrix: 1 −2

3

A = 2 −5 12 0 2

− 10

The first step of Gaussian elimination is to subtract 2 times the first row from the second row. In order to record what has been done, put the multiplier, 2, into the place it was used to make a zero: i.e. the second row, first column. In order

22.14 Power System Matrices

to make it clear that it is a record of the step and not an element of A, the following result is put into parentheses. This leads to 1 −2

3

2 −1

6

0

2 −10

There is already a zero in the lower-left corner, so there is no need to eliminate anything there. This fact is recorded with a (0). To eliminate the third row, second column, subtract −2 times the second row from the third row. Recording the −2 in the spot it was used, we have: 1

−2

3

2

−1

6

0

−2

2

Let U be the upper triangular matrix produced, and let L be the lower triangular matrix with the records and ones on the diagonal, i.e. 1 0 0

1 −2 3

L = 2 1 0 and U = 0 −1 6 0 −2 1

0 0 2

Then we have the following property: 1 0 0 LU = 2 1 0 0 −2 1

1 −2 3

1 −2

3

0 − 1 6 = 2 − 5 12 0 0 2

0 2

=A

− 10

Thus we see that A is actually the product of L and U. Here, L is lower triangular, and U is upper triangular. When a matrix can be written as a product of simpler matrices, we call that a decomposition of A, and this one we call the LU decomposition. 22.13.8

LU Factorization with Partial Inversion or Pivoting

Proper permutation in rows (or columns) is sufficient for LU factorization. LU factorization with partial pivoting (LUP) refers to LU factorization with row permutations only PA = LU where L and U are again lower and upper triangular matrices, and P is a permutation matrix, which, when left-multiplied by A, reorders the rows of A. All square matrices can be factorized in this form, and the factorization is numerically stable. This makes LUP decomposition a useful technique in practice. 22.13.9

LU Factorization with Complete Inversion

An LU factorization with full pivoting involves both row and column permutations: PAQ = LU L, U, and P are defined as before, and Q is a permutation matrix that reorders the columns of A.

22.14

Power System Matrices

The physical structure of a connected graph can be represented with the help of incidence matrices. Following are the three matrices that are used in graph theory.

681

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22 Digital Computation Basics

22.14.1

Incidence Matrices

An incidence matrix represents the graph of a given electric circuit or network. Hence, it is possible to draw the graph of that same electric circuit or network from the incidence matrix. We know that a graph consists of a set of nodes, and those are connected by branches. So, the connecting of branches to a node is called incidence. An incidence matrix is represented with the letter A. It is also called a node-to-branch incidence matrix or node incidence matrix. If there are n nodes and b branches in a directed graph, then the incidence matrix will have n rows and b columns. Here, rows and columns correspond to the nodes and branches of a directed graph. Hence, the order of the incidence matrix is n × b. The elements of the incidence matrix have one of the three values: +1, −1, and 0:

•• •

If the branch current leaves from a selected node, then the value of the element is +1. If the branch current enters toward a selected node, then the value of the element is −1. If the branch current neither enters at a selected node nor leaves from a selected node, then the value of element is 0.

22.14.1.1 Procedure to Find the Incidence Matrix

Follow these steps in order to find the incidence matrix of a directed graph: 1) Select a node at a time of the given directed graph, and fill in the values of the elements of the incidence matrix corresponding to that node in a row. 2) Repeat this step for all the nodes of the graph. Example Consider the following directed graph, shown in Figure 22.61. The incidence matrix corresponding to this directed graph is −1 1 A=

0

−1 0

0

0

1

0 1

0

−1 1

1

0

−1 0

0

0

0

0

−1 −1

1

The rows and columns of this matrix represents the nodes and branches of the graph. The order of this incidence matrix is 4 × 6. By observing this incidence matrix, we can conclude that the summation of its column elements is zero. That means a branch current leaves from one node and enters at another single node. Note: If the given graph is undirected, then we convert it into a directed graph by adding the arrows on each branch. We can consider the current flow in each branch to be arbitrary. a

22.14.2 b

1

c

3

2

d

e

f

4 Figure 22.61 Directed graph.

Fundamental Loop Matrices

A fundamental loop or f-loop is a loop that contains only one link and one or more twigs. So, the number of f-loops is equal to the number of links. A fundamental loop matrix is represented by the letter B. It is also called a fundamental circuit matrix or tie-set matrix. This matrix gives the relation between branch currents and link currents. If there are n nodes and b branches in a directed graph, then the number of links in a co-tree, which corresponds to the selected tree of the graph, is b − n + 1. So, the fundamental loop matrix will have b − n + 1 rows and b columns. Here, rows and columns correspond to the links of the co-tree and branches of the graph. Hence, the order of the fundamental loop matrix is (b − n + 1) × b.

22.14 Power System Matrices

•• • •

The elements of a fundamental loop matrix have one of the three values, +1, −1, and 0: The value of the element is +1 for a link of the selected f-loop. The value of the element is 0 for the remaining links and twigs, which are not part of the selected f-loop. If the direction of the twig current of the selected f-loop is the same as that of the f-loop link current, then the value of the element is +1. If the direction of the twig current of the selected f-loop is the opposite of the f-loop link current, then the value of the element is −1.

22.14.2.1 Procedure to Find the Fundamental Loop Matrix

Follow these steps in order to find the fundamental loop matrix of a directed graph: 1) Select a tree of the directed graph. 2) By including one link at a time, we get one f-loop. Fill the values of the elements corresponding to this f-loop in a row of the fundamental loop matrix. 3) Repeat this step for all links. Consider the tree of a directed graph shown in Figure 22.62, which is considered an incidence matrix. This tree contains three branches d, e, and f. Hence, branches a, b, and c are the links of the co-tree corresponding to the tree. By including one link at a time in the tree, we get one f-loop. So, there are three f-loops, since there are three links. These three f-loops are shown in Figure 22.63. In the figure, the branches, which are represented with colored lines, form f-loops. We can get the row-wise element values of the tie-set matrix from each f-loop. So, the tie-set matrix of this tree is 1 0 0 −1 0 B= 0 1 0 1 0 0 1 0

1

−1 0

−1 1

The rows and columns of this matrix represent the links and branches of the directed graph. The order of this incidence matrix is 3 × 6. The number of fundamental loop matrices of a directed graph is equal to the number of trees of that graph, because every tree has one fundamental loop matrix. 22.14.3

1

Fundamental Cut-Set Matrices

A fundamental cut set or f-cut set is the minimum number of branches that must be removed from a graph such that the original graph becomes two isolated subgraphs. The f-cut set contains only one twig and one or more links. So, the number of f-cut sets is equal to the number of twigs. A fundamental cut set matrix is represented with the letter C. It gives the relation between branch voltages and twig voltages. If there are n nodes and b branches in a directed graph, then the number of twigs present in a selected tree of the graph is n − 1. So, the fundamental cut set matrix

2

d

3

e

f

b

1

2

d

3

e

f

2

1

d

c

e

4

4

4

f-loop 1

f-loop 2

f-loop 3

Figure 22.63 F-loops.

e

f

4 Figure 22.62 Four-bus system oriented, connected graph.

3

f

3

d

a

1

2

683

684

22 Digital Computation Basics

will have n − 1 rows and b columns. Here, rows and columns correspond to the twigs of the selected tree and branches of the graph. Hence, the order of a fundamental cut set matrix is (n − 1) × b. The elements of the fundamental cut set matrix have one of the three values +1, −1, and 0:

•• • •

The value of the element is +1 for a twig of the selected f-cut set. The value of the element is 0 for the remaining twigs and links, which are not part of the selected f-cut set. If the direction of the link current of the selected f-cut set is same as that of the f-cut-set twig current, then the value of the element is +1. If the direction of the link current of the selected f-cut set is opposite that of the f-cut set twig current, then the value of the element is −1.

22.14.3.1 Procedure to Find the Fundamental Cut-Set Matrix

Follow these steps in order to find the fundamental cut set matrix of a directed graph: 1) Select a tree of the directed graph, and represent the links with dotted lines. 2) By removing one twig and necessary links at a time, we get one f-cut set. Fill the values of the elements corresponding to this f-cut set in a row of the fundamental cut set matrix. 3) Repeat this step for all twigs. Example Consider the same directed graph discussed in Section 22.14.1.1. Select branches d, e, and f of the graph as twigs. So, the remaining branches a, b, and c are the links. The twigs d, e, and f are represented with solid lines, and links a, b, and c are represented with dotted lines in Figure 22.64. By removing one twig and necessary links at a time, we get one f-cut set. So, there are three f-cut sets, since there are three twigs. These three f-cut sets are shown in Figure 22.65. We can get three f-cut sets by removing a set consisting of a twig and links C1, C2, and C3. We get the row-wise element values of the fundamental cut set matrix from each f-cut set. So, the fundamental cut set matrix of this tree is 1 −1 0 1 0 0 C = 0 −1 1 0 1 0

a

−1 0 0 1

1 0 b

1

c

3

2

d

e

The rows and columns of this matrix represent the twigs and branches of the directed graph. The order of this fundamental cut set matrix is 3 × 6. The number of fundamental cut set matrices of a directed graph is equal to the number of trees of the graph, because every tree has one fundamental cut set matrix.

f

22.14.4

Algorithms for Formation of Network Matrices

A power network is essentially an interconnection of two-terminal components. When a set of components is not connected, it constitutes a primitive network. The two-terminal components can be represented in impedance form and in admittance form, respectively, as shown in Figure 22.66.

4 Figure 22.64 Directed graph example.

a

a

a

C1 b 1

2

d

b

c

e

3

1

f

4 Figure 22.65 Directed graph f-cut sets.

C3

C2 c

2

d

e

4

f

3

1

b

c

2

d

e

4

3 f

22.14 Power System Matrices

ekm Vk

ikm

+

–

Zkm

+

–

Vm

m

k vkm jkm

Vk

+

ykm

ikm

k

ikm + jkm

–

Vm

m

vkm Figure 22.66 Two-terminal components.

The performance equation of a primitive network can be expressed with the help of voltage and current variables using either impedance or admittance form. Generally, the equation in impedance form is written vkm + ekm = zkm ikm where vkm : voltage across element k − m ekm : source voltage connected in series with the element k − m ikm : current through the element k − m zkm : self-impedance of the element Similarly, the performance equations in admittance form are ikm + jkm = ykm vkm where jkm : source current connected in parallel with the element k − m ykm : self-admittance of the element The performance equations for the primitive impedance and admittance networks, respectively, are written v + e = zi i + j = yv where e and j are specified voltage and current vectors, respectively, and v and i are the element voltage and current vectors, respectively. The matrices z and y are diagonal matrices of the primitive networks whose diagonal elements represent the self-impedance and self-admittance, respectively, when there is no mutual coupling between the elements. A nonzero, off-diagonal element in either the z matrix or y matrix gives the mutual impedance or mutual admittance, as the case may be, between the two elements. The primitive admittance matrix y is the inverse of primitive impedance matrix z. A network is made up of an interconnected set of primitive elements. The general form of the network performance equation in admittance and impedance form is given by the performance equation in admittance form Ibus = Ybus Vbus Vbus = Zbus Ibus where Ibus : a current vector of injected currents Vbus : a voltage vector of bus voltages The bus voltages are specified with respect to a reference node, usually, but not necessarily the ground. Ybus is the bus admittance matrix, and Zbus is the bus impedance matrix for the network. Network equations can be formulated systematically in a variety of forms.

685

686

22 Digital Computation Basics

The formulation of a network in the form of a bus admittance matrix is used extensively in the analyses of power systems. The bus admittance matrix of the interconnected network can be obtained using a singular transformation of the primitive admittance matrix using bus incidence matrix A. An alternate, simple method of forming the bus admittance matrix by inspection is widely used. This method uses the node voltage equations of the power-system network.

22.14.5

Singular Transformation

The bus admittance matrix of a interconnected network can be obtained by using the bus incidence matrix A to relate the variables and parameters of the primitive network to the bus quantities of the interconnected network. The performance equation of the primitive network in admittance form, given by i + j = yv when pre-multiplied by the transpose of the bus incidence matrix A, gives A t + A t j = A t yv By KCL, the term Ati = 0 since the product Ati represents a vector in which each element gives the sum of currents flowing through the elements terminating at a bus. A t + A t j = A t yv simplifies to A t j = A t yv The term Atj represents a vector in which each element is the algebraic sum of source currents injected into each bus and thus equals the vector of the injected bus currents. Therefore, the bus current vector Ibus is given by Ibus = A t j = A t yv

22 6

If the bus voltage vector is represented by Vbus, the complex power S in the network is given by ∗ S = Ibus Vbus t

Since power is invariant, the total power in the primitive network is given by (j∗)tv: ∗ Vbus = j ∗ t v S = Ibus t

22 7 t

∗ ∗ = j ∗ t A ∗ . Substitution of Ibus From Eq. (22.6), taking the conjugate on both sides gives Ibus

j ∗ t AV bus = j ∗ t v

t

in Eq. (22.7) yields 22 8

Since A is real, A∗ = A. Hence Eq. (22.8) may be rewritten as j ∗ t AV bus = j ∗ t v

22 9

Equation (22.9) holds for all values of source current j. Therefore, it is observed that AV bus = v

22 10

Substituting for v from Eq. (22.10) in Eq. (22.6) gives Ibus = A t yAV bus

22 11

The network performance equation is given by Ibus = Ybus Vbus

22 12

Comparing Eqs. (22.11) and (22.12), we get the following: Ybus = A t yA As bus incidence matrix A is singular, therefore, AtyA is a singular transformation of the primitive matrix y. The general form of the network performance equation in impedance form is given by Vbus = Zbus Ibus The Zbus matrix for the network can be obtained from −1 = A t yA Zbus = Ybus

−1

22.14 Power System Matrices

Example For the power system network shown in Figure 22.67, the primitive element data is as follows: Bus number Element number

From

To

Primitive impedance

1

1

0

0.05

2

3

0

0.1

3

1

2

0.5

4

2

3

0.4

5

1

3

0.25

••

Compute the Ybus matrix, assuming zero mutual coupling between the elements. Compute the Ybus matrix by taking a mutual coupling of 0.2 between elements 4 and 5.

Solution The oriented connected graph of the power-system network graph is shown in Figure 22.68. For the assumed orientation in the connected graph, the bus incidence matrix [A] is of order 5 × 3 and is given by −1 0

0

0

0

−1

1

1

0

0

−1 1

A=

−1 0

1

The transpose of the power system bus incidence matrix [A] is as follows: t

A =

−1 0

1

0

0

−1 − 1 0

0

−1 0

0 1

−1

z=

0

0

01 0

0

0

~

0

0 05 0

0

0

0

0 04

0

0

0

Figure 22.67 Example of a power-system network.

2 4 3

0

3

0 0 25

The Ybus matrix can be formed from the primitive impedance matrix: 20 0 0 0 0 0 10 0 0 0 Y = z−1 =

3

1

0

0

2

~

The primitive impedance matrix [z] can be formed as shown here: 0 05 0

1

0

0 2 0 0 or Ybus = A t yA =

0

0 0 25 0

0

0 0 0 4

5

1

2 1

24 5

−2

25

−2

45

−2 5

− 2 5 −2 5

15

0

Figure 22.68 Directed graph of a power-system network.

687

688

22 Digital Computation Basics

0 05 0 0 z=

0

0

0

01 0

0

0 0

−1

=

20 0 0

0

0

0 10 0

0

0

0

0 2

0

0

0

0 05 0

0

0

0 04 02

0

0 0 4 17 −3 33

0

0

0 0 2 0 25

0

0 0 3 33 6 67

and y = z

28 6667 − 5 3333 −3 3333 t

Ybus = A yA =

22.14.6

− 5 3333 6 1677

−0 8333

− 3 3333 0 8333

14 1667

0 0448 0 0405 0 0129 therefore

−1 Zbus = Ybus

= 0 0414 0 2009 0 0216 0 0193 0 0271 0 0767

Bus Admittance Matrix Using Nodal Analysis

Consider the simple power network shown in Figure 22.69a. The equivalent circuit of the system is drawn in by transforming voltage sources to current sources. The admittances of the elements are indicated on the diagram shown in Figure 22.69b. The ground is chosen as the reference. Applying KCL to all the nodes gives I1 = y10 V1 + y12 V1 − V2 + y13 V1 −V3 + y14 V1 − V4 I2 = y20 V1 + y12 V2 − V1 + y23 V2 −V3 0 = y23 V3 − V2 + y13 V3 −V1 + y34 V3 − V4 0 = y14 V4 − V1 + y34 V4 −V3 By rearranging the equations, we get I1 = y10 + y12 + y13 + y14 V1 − y12 V2 − y13 V3 − y14 V4 I2 = −y12 V1 + y20 + y12 + y23 V2 −y23 V3 0 = − y13 V1 −y23 V2 + y13 + y23 + y34 V3 − y34 V4 0 = − y14 V1 −y34 V3 + y14 + y34 V4

4

4

3

3

y34 y13 y14

y23 y12

1

2 1

~

y10

~ m

(a) Figure 22.69 Admittances of elements.

I2 2

y20

I1

m

m

(b)

m

22.14 Power System Matrices

In matrix form, this set of equations may be written as y10 + y12 + y13 + y14

− y12

− y13

−y14

V1

− y12

y20 + y12 + y23

− y23

0

V2

0

− y13

− y23

y13 + y23 + y34

−y34

V3

0

− y14

0

− y34

y14 + y34

V4

I1 I2 =

Y11 Y12 Y13 Y14

V1

Y21 Y22 Y23 Y24

V2

Y31 Y32 Y33 Y34

V3

Y41 Y42 Y43 Y44

V4

22 13

=

or Ibus = Ybus Vbus

22 14

where Y11 = y10 + y12 + y13 + y14

Y22 = y20 + y12 + y23

Y33 = y13 + y23 + y34

Y44 = y14 + y34

Y12 = Y21 = − y12

Y13 = Y31 = − y13

Y14 = Y41 = − y14

Y23 = Y32 = − y23

Y24 = Y42 = 0

Y34 = Y43 = − y34

Note that the diagonal element of each node is the sum of the admittances connected to the node. This is known as the self-admittance or driving-point admittance. The off-diagonal element is equal to the negative of the admittance connected between the nodes. It is known as the mutual admittance or transfer admittance. Extending the relation to an n-bus system, the Eq. (22.13) node voltage equations in matrix form are I1

Y11 Y12

Y1i

Y1n

V1

I2

Y21 Y22

Y2i

Y2n

V2

Yi1 Yi2

Yii

Yin

Vi

Yn1 Yn2

Yni

Ynn

Vn

=

Ii

In

22 15

Equation (22.15) can also be written mathematically as j=n

Yij Vj for i = 1, 2,3, …, n

Ii = j=1

Equation (22.15) is the network performance equation and is the same as Ibus = YbusVbus. The diagonal term Yii and off-diagonal term Yij of Ybus are given by n

Yii =

yij where j

i

j=0

Yij = Yji = − yij Ybus can also be obtained using the singular transformation of the primitive admittance matrix y; the result is the same.

689

690

22 Digital Computation Basics

22.14.6.1 Building the YBUS Matrix

Since power-system networks are composed of linear bilateral components, Ybus is a square symmetric matrix. If the elements connected to a bus are not mutually coupled to other elements, then the formulation of a Ybus matrix of a network is a two-step procedure as follows: 1) To obtain the diagonal term Yij, add the admittances of all the branches (including those connected between the node and ground) to the ith node. 2) To obtain the off-diagonal term connected between node i and node j, take the negative of the sum of all the primitive admittances of the branches connected between nodes i and j. Since the matrix is symmetric, Yij = Yji.

1

I12

2

y12

I1 V1

V2 I14

y14

I24

I23

y34

y23

V4

I3 y34 4

V3

I34 3

Figure 22.70 Power-system network.

Example For the power system shown in Figure 22.70, build the Ybus matrix. The branch impedances of the lines are as follows:

•• •• •

Line 1–2: Line 1–4: Line 2–3: Line 2–4: Line 3–4:

(10 + j40)Ω (15 + j50)Ω (5 + j25)Ω (15 + j20)Ω (10 + j30)Ω

Assume that an impedance of (20 + j40) is connected between node 4 and the ground.

Solution 1 = 0 0059 − j0 0235 Mho 10 + j40 1 = 0 0055 − j0 0183 Mho y14 = 15 + j50 1 = 0 0077 −j0 0385 Mho y23 = 5 + j25 1 = 0 024 −j0 032 Mho y24 = 15 + j20 1 y34 = = 0 01 − j0 03 Mho 10 + j30 1 = 0 01 − j0 02 Mho y40 = 20 + j40 Y11 = y12 + y14 = 0 0059− j0 0235 + 0 0055 − j0 0183 = 0 104− j0 0418 Mho y12 =

Y22 = y12 + y23 + y24 = 0 0059 −j0 0235 + 0 0077− j0 0385 + 0 024− j0 032 = 0 0376 −j0 094 Mho Y33 = y23 + y34 = 0 0077− j0 0385 + 0 01 − j0 03 = 0 0177 − j0 0685 Mho Y44 = y14 + y24 + y34 + y40 = 0 0055 − j0 0183 + 0 024− j0 032 + 0 01 − j0 03 + 0 01 − j0 02 = 0 0495 − j0 1003 Mho Y12 = Y21 = − 0 0059 − j0 0235 Mho Y14 = Y41 = − 0 0055 − j0 0183 Mho Y23 = Y32 = − 0 0077 − j0 0385 Mho Y24 = Y42 = − 0 024 −j0 032 Mho Y34 = Y43 = − 0 01 − j0 03 Mho Y13 = Y31 = 0

22.14 Power System Matrices

Having computed the elements of Ybus, the admittance matrix can be written

22.14.7

0 104− j0 0418

− 0 0059 + j0 0235

0

− 0 0055 + j0 0183

− 0 0059 + j0 0235

0 0376 −j0 094

−0 0077 + j0 0385

− 0 024 + j0 032

0

− 0 0077 + j0 0385

0 0177 −j0 0685

−0 01 + j0 03

− 0 0055 + j0 0183

−0 024 + j0 032

− 0 01 + j0 03

0 0495 + j0 1003

Sparse Matrix Techniques

A matrix is said to be sparsely filled (or simply sparse) if the number of zero elements in the matrix is greater than the number of nonzero elements. If the number of nonzero elements is less than 15%, the matrix is called a sparse matrix. In electrical power networks, the bus admittance matrix has a large number of zero elements compared to the nonzero elements. This is because the bus admittance matrix reflects only direct connections between buses. Also, the bus admittance matrix is symmetrical about the major diagonal. An example of a six-bus admittance matrix of a hypothetical network is given next. Crosses (X) are the only nonzero elements in the network: 1

2

X

X

X

X

3

4

5

X

X

6

X X

X

X

X

X

X

X

X

X

The number of elements required to be stored (along with the diagonal elements) for a symmetric square matrix of n n+1 order n is equal to . The sparsity in bus admittance matrices in electrical power networks can be used to reduce 2 storage requirements and increase speed of computation with the help of sparsity techniques. The nonzero elements of a sparse matrix are stored in a single-dimensional array with two tables: one gives the column number, and the other gives the starting position of the next row. The storage of only nonzero elements can be best explained by an example. The following example shows a sparse matrix of a seven-bus hypothetical network. Only the nonzero elements are indicated in the matrix: y11

y31

y13 y22

y23

y26

y32

y33

y34

y43

y44 y55

y57

y62

y66 y75

y77

The storage of the Ybus matrix, along with the two single-dimension tables, is as follows: Pos Elem Col

1

3

6

8

9

11

12

y11

y13

y22

y23

y26

y33

y34

y44

y55

y57

y66

y77

1

3

2

3

6

3

4

4

5

7

6

7

691

692

22 Digital Computation Basics

The nonzero elements of the Ybus matrix are stored in a single-dimensional table elem, and the position of the diagonal element of each row is shown in the pos table. Thus, the diagonal element of row two is in the third position, while the third row starts from the sixth position. Also note that the difference between any two values in the pos table gives the number of elements in a row. For example, the difference between the third and fourth positions in the pos table [pos (4) − pos (3)] = 8 − 6 = 21 shows that the number of elements in the third row is two. In addition, the length of the elem and col tables is equal to the number of nonzero elements of Ybus when only the upper triangle along with the diagonal elements are stored.

22.15

Transformer Modeling

Transformers are a vital constituent of an electrical power system since they step up the voltage at the generator end to a level suitable for transmission of bulk power over long distances. Similarly, at the receiving end, transformers step down the voltage to levels desirable for utilization of energy. Between the generating point and the consumers’ terminals, the voltage levels of electrical energy may undergo several (four to five) transformations. 22.15.1

An Ideal Transformer

An ideal transformer means that the permeability μ of the core is infinite, all the magnetic flux created by the primary winding links with the secondary winding core loss is zero, and the windings have zero resistance. The schematic representation of an ideal two-winding transformer is shown in Figure 22.71. The primary winding when connected to an AC source v1 induces EMF e1 in the primary winding and is given by v1 = e 1 = N 1

dϕ V dt

The whole of the flux ϕ links the secondary winding as well. Hence, the EMF e2 induced in the secondary winding is in phase with e1 and with the secondary open-circuited, as shown in Figure 22.72. The following equation holds: v2 = e 2 = N 2

dϕ V dt

The secondary voltage and secondary current can be referred to the primary side 1 V1 = V2 = aV 2 ; I1 = I2 = I2 a where V2 and I2 , respectively, denote secondary voltage and current referred to the primary. Similarly, the primary voltage and current referred to the secondary side are 1 V2 = V1 = V1 ; I2 = I1 = aI 1 a Figure 22.73 shows an ideal transformer, the secondary of which is connected across a load of impedance ZL and draws a current of I2 amps: ZL =

V2 I2

i1 +

Substituting the values of V2 and I2,

v1

V1 V1 = a2 ZL = ZL ZL = a or aI 1 I1 I1 =

+ e1

–

–

1 I a 2

N2

e2

+ v2

–

I2 = aI1 + N1

N1

–

Figure 22.71 Representation of an ideal two-winding transformer.

+ V1 = aV2

i2 +

N2

_

Figure 22.72 Ideal two-winding transformer open circuit.

V2 = _

1 a

V1

22.15 Transformer Modeling

I1

I1

+

+ N1

V1

N2

V2

–

ZL′ = a2ZL

I1

I2

+

+ IL

ZL V1 –

–

ZL′ = a2ZL

V1 –

(a)

(b)

(c)

Figure 22.73 Equivalent circuit for an ideal transformer with load connected on secondary.

r1

x1 I1

+ V1 _

l Gc = Rc

x2 I0

I'2 l Bm = jXm

+ E _ 1

E2

+ _

r2 I2

+ _ V2

N1 N2

Figure 22.74 Complete equivalent circuit of an ideal transformer.

ZL is the equivalent load impedance transformed to the primary side. Figure 22.73b shows the load impedance connected after transformation to the primary of the ideal transformer. Figure 22.73c is the final reduced network as looked at from the source of supply. In a practical two-winding transformer, core permeability is finite, and some of the magnetic flux linking the primary winding does not link the secondary winding. This flux is proportional to primary current and causes a voltage drop that is accounted for by an inductive reactance XL1 called leakage reactance, which is added in series with the primary winding of an ideal transformer. For similar reasons, a leakage reactance XL2 is added in series with the secondary winding. The winding resistances R1 and R2 are also accounted for by connecting them in series with their respective windings. In a real transformer, losses occur in the core due to hysteresis and eddy-current losses. A component of current IcA in phase with E1 accounts for the energy loss in the core and is represented in the equivalent circuit by the resistance Rc, as shown in the complete equivalent circuit of a transformer shown in Figure 22.74. Very often, the magnetizing branch is ignored because the magnetizing current is too small compared to the usual load currents, in which case the transformer is represented by the equivalent series impedance Zeq = Req + jXeq, as shown in Figure 22.75, where Req = r1 + r2 = r1 +

r2 x2 and Xeq = x1 + x2 = x1 + 2 a2 a

For large power transformers, the equivalent resistance is very small compared to the equivalent reactance and can be ignored. The equivalent circuit reduces to Figure 22.76. 22.15.2

Three-Phase Transformer Model

A three-phase transformer is made up of a bank of three singleXeq Req phase transformers. Alternatively, it may be made of three primary and three secondary windings on a common threelimbed magnetic core and contained in a single tank. The bank I2 of three transformers or the three windings can be connected I1 = a = I2' + + on the primary and secondary sides in four ways: delta–delta V1 V2' (Δ − Δ), star-star (Y − Y), delta-star (Δ − Y), and star-delta – – (Y−Δ). Note that for fixed line-to-line voltages, the total kVA rating of each transformer is one-third the kVA rating of the bank, regardless of the connections used, but the voltage and current ratings of the individual transformers depend on Figure 22.75 Transformer circuit, ignoring magnetizing branch. the connections:

693

22 Digital Computation Basics

•

Xeq

I1 = I'2 + V – 1

• •

+ V' – 2

Figure 22.76 Transformer circuit, ignoring magnetizing and equivalent resistance.

VA

I1

IA

+ √3N1/2

N2 YL1

V1

T phase – IB

VB

I2 + N1/2

N2

S

YL2

V2

N1/2

•

A star-delta (Y − Δ) connection is used for stepping down from high voltage to medium and low voltages with the star point connected to the ground. Since the star side carries the high voltage, insulation requirements are also less severe, and the neutral point is earthed. Similarly, a delta-star (Δ − Y) connection is used for stepping up to a high voltage. Both Y − Δ and Δ − Y provide stable connections under unbalanced load conditions. A delta–delta (Δ − Δ) connection has the advantage that even when one phase is out of service for maintenance and repair, the other two phases continue to function with a reduced rating of 58% of the original capacity. This type of a connection is called an open-delta or V-connection. A delta–delta (Δ − Δ) connection provides a path for the third harmonics, thereby eliminating the associated problems. However, since no neutral point is available, the winding insulation has to be designed to withstand full line-to-line voltage. A star-star (Y − Y) connection is rarely used due to the problems of third harmonics and unbalanced operation. To eliminate the problem of third harmonics, another set of delta-connected windings (called the tertiary winding) is employed. The tertiary winding is placed on the transformer core and provides a path for the third harmonic current. A transformer with a tertiary winding is also called a threewinding transformer. A star-star connection provides the facility of neutral grounding on both sides and decreased insulation level, and hence lowers the cost.

M phase

IC

VC

–

22.15.3

Figure 22.77 Scott-T connected transformer.

A V1

VAB VBC

V2

Voltage phasor in primary A

IAB S

Voltage phasor in secondary +

IAC

2 1 1 β VA − VB − VC YL1 2 2 3 I2 = YL2 V2 = β VB − VC YL2 I1 = YL1 V1 =

C

V1

Y2

– B

C ICB

Primary current flow

I2 Y2 Secondary current flow

Figure 22.78 Voltage and current diagrams for a Scott-T transformer.

22 17

3 3 3 αIAC + αIAB = αIA 2 2 2 2 ∴ IA = βI1 22 18 3 1 1 1 1 I2 = − αICB + αIAC − αIAB = −αICB + αIAC − αIAC − αIAB 2 2 2 2 1 1 = −αIC − αIA = − αIC − I1 2 3 1 ∴ IC = −βI2 − βI1 22 19 3 I1 =

V2

22 16

IA = IAC + IAB ; IB = − ICB − IAB ; IC = ICB −IAC

I1

+

S

B

Scott-T Connected Transformer

A feeding transformer with a Scott connection in substations is shown in Figure 22.77. Both the voltage-phasor and current-flow diagrams are shown in Figure 22.78. YL1 and YL2 denote, respectively, the loads in the two sides of the center-feed system. Using transformer flux coupling, KCL, and KVL, the following relations are found:

–

694

22.15 Transformer Modeling

Because the system primary is ungrounded, 1 βI1 3 Substituting Eqs. (22.16) and (22.18) into (22.20) yields IB = − IA − IC = βI2 −

22 20

2 1 1 2 β YL1 −β2 YL2 VC IB = − β2 YL1 VA + β2 YL1 + β2 YL2 VB + 3 3 3 IC = −β2 YL1 VA +

1 2 1 β YL1 −β2 YL2 VB + β2 YL1 + β2 YL2 VC 3 3

Given β2YL1 = Y1 and β2YL2 = Y2, the relation between VABC and IABC can be expressed as 4 2 Y1 − Y1 3 3 2 1 IB = − Y1 Y1 + Y2 3 3 IC 2 1 Y1 − Y2 − Y1 3 3 By transforming Eq. (22.21) into IA

0

I0

0

VA 22 21

VB VC

components, the relation between V012 and I012 can be obtained:

V0

I1 = 0 Y 1 + Y 2 Y 1 − Y 2

V1

0 Y1 −Y2 Y1 + Y2

V2

I2

22.15.4

0

2 − Y1 3 1 Y1 −Y2 3 1 Y1 + Y2 3 symmetrical

Le-Blanc Connected Transformer

A feeding transformer with a Le-Blanc connection in substations is shown in Figure 22.79. Both the voltage-phasor and current-flow diagrams are shown in Figure 22.80. Using transformer flux coupling, KCL, KVL, and Ohm’s law, 1 3

V1 =

N2 N1

VCA −VAB =

1 β VB + VC − 2VA 3

1 V2 = β 2VBC − VAB − VCA = β VB − VC 3

VA IA

I1

IAB N1

VB IB

+ N2/√3

YL1

N2 /3

V1 _

IBC N1

2 N2 /3 I2

VC IC ICA N1

N2/√3

YL2

+ V2 _

N2 /3 Figure 22.79 Le-Blanc connected transformer.

695

696

22 Digital Computation Basics

A

N2 VCA N1 √3

VAB

VCA

B

N2 VAB

V1

N1

N1

3

N2 VAB

C

VBC

N2 VCA

N1 √3

Voltage phasor in primary

3

N2 2 N1 3

VBC

V2

Voltage phasor in secondary

Figure 22.80 Voltage and current diagram for a Le-Blanc transformer.

I1 = YL1 V1 ; I2 = YL2 V2 IA = − ICA + IAB ; IB = −IAB + IBC ; IC = −IBC + ICA 1 1 2 1 1 I1 − I2 ; IBC = βI2 ; IAB = − β I2 + I1 3 3 3 3 3

ICA = β

We can determine the following: IA = −

2 1 1 βI1 ; IB = βI1 + βI2 ; IC = βI1 − βI2 3 3 3

By substitution, we can derive the following: 4 2 2 IA = β2 YL1 VA − β2 YL1 VB − β2 YL1 VC 3 3 3 2 1 1 IB = − β2 YL1 VA + β2 YL1 + β2 YL2 VB + β2 YL1 −β2 YL2 VC 3 3 3 2 1 1 IC = − β2 YL1 VA + β2 YL1 − β2 YL2 VB + β2 YL1 + β2 YL2 VC 3 3 3 Given β2YL1 = Y1 and β2YL2 = Y2, the relation between VABC and IABC can be expressed as IA IB IC

=

4 Y1 3 2 − Y1 3 2 − Y1 3

2 − Y1 3

2 − Y1 3

1 Y1 + Y2 3 1 Y1 − Y2 3

1 Y1 −Y2 3 1 Y1 + Y2 3

VA VB

22 22

VC

By transforming Eq. (22.22) into symmetrical components, the relation between V012 and I012 can be obtained: I0

0

0

V0

I1 = 0 Y1 + Y2 Y1 − Y2

V1

0 Y1 −Y2 Y1 + Y2

V2

I2

22.16

0

Transmission Line Modeling

The transmission line parameters – resistance, inductance, and capacitance – were derived as per-unit length values in Part A of this book. These parameters are not lumped but are uniformly distributed along the length of the line, and the general equations relating voltage and current of a transmission line are formulated accordingly. However, the use of lumped parameters gives good accuracy for short lines of lengths less than 80 km and medium lines of lengths roughly between 80 and 240 km.

22.16 Transmission Line Modeling

For short length lines, the value of shunt capacitance is very small and can be ignored without loss of accuracy. For short lines, only the series resistance and series inductance of the total length of the line are considered. A medium-length line can be represented well by the series resistance and series inductance of the total length of the line as lumped parameters and half the total capacitance of the line lumped at each end of the line. Normally, transmission lines are operated with balanced three-phase loads. Even when the lines are not transposed, the dissymmetry in line parameters is very small, and the phases may be considered balanced. As the transmission line is considered to be in a sinusoidal steady state, the phasors and impedances are used in the analysis. In order to differentiate the series impedance/shunt admittance per unit length and the total series impedance/shunt admittance of the line, the following terminology is used: z = r + jmL = series impedance per unit length per phase y = G + jmC = shunt admittance per unit length per phase to neutral l = length of line Z = zl = total series impedance per phase Y = yl = total shunt admittance per phase to neutral For power transmission lines, the shunt conductance G is usually ignored. 22.16.1

Modeling Long Transmission Lines

Figure 22.81 shows one phase and the neutral connection of a three-phase line: an elemental strip, of length Dx, of a transmission line, at a distance x from the receiving end. The length of the transmission line is l m, and z and y are, respectively, the series impedance and shunt admittance per unit length of the line. If I(x) is the current flowing through the elemental length Dx, then by Ohm’s law voltage at (x + Dx) from the sending end may be written as V x + Dx = V x + zDxI x

22 23

Similarly, applying Kirchhoff’s current law to the junction of the elemental strip, the current at (x + Dx) from the receiving end may be obtained: I x + Δx = I x + yΔxV x + Δx

22 24

Equations (22.23) and (22.24) may be rewritten as V x + Δx − V x = zI x Δx I x + Δx − I x = zV x Δx Taking the limit Δx

22 25 22 26

0, (22.25) and (22.26) can be modified as

dV x = zI x dx dI x = yV x dx IS

+

VS

I(x + Δx )

V(x + Δx )

22 27 22 28 I(x )

zΔx

yΔx

+

yΔx

V(x)

VR

–

– Δx l

Figure 22.81 Long transmission line model.

x

697

698

22 Digital Computation Basics

Equations (22.27) and (22.28) are two linear first-order homogenous differential equations with two unknowns V(x) dI x from Eq. (22.28), I(x) may be eliminated as and I(x). Differentiating (22.27) with respect to x and substituting for dx follows d 2 V x zdI x d2 V x d2 V x = yzV x or = −yzV x = 0 or − γ2V x = 0 dx dx2 dx2 dx2

22 29

where γ = yz = propagation constant Equation (22.29) is a linear, second-order homogeneous differential equation with one unknown, V(x). Similarly difdV x ferentiating Eq. (22.28) with respect to x and substituting for from Eq. (22.27), V(x) may be eliminated as follows: dx d2 I x ydV x d2 I x d2 I x = yzI x or = − yzI x = 0 or − γ2I x = 0 2 2 d dx dx dx2

22.16.2

Lumped Parametric Π Equivalent Circuits of Transmission Lines

Power-system analysis uses equivalent circuits of different components such as generators, transformers, and transmission lines. It is possible to find an equivalent circuit model of a transmission line that can replace the ABCD parameters of the line. In this section, the Π equivalent circuit of a long transmission line is summarized. Z'

IS

+

Y' 2

VS

IR

Y' 2

+

VR

–

–

sinh γl sinh γl =Z gives the exact total series impedance of a transmission line. γl γl However, for medium-length lines and short lines, the equivalent circuit and the equations are greatly simplified. Depending upon the length and operating voltage, further approximations can be made in the simulation of a transmission line, without significantly affecting the accuracy of the results. Z = zl

22.16.3

Short-Length Line

In transmission lines that are less than 80 km in length and operate at voltages below 66 kV, the shunt admittance can be ignored. The equivalent circuit of such a line is shown in Figure 22.82. Inspection of the figure yields the following equations: VS = VR + ZI R = VR + R + jX IR IS

Z = R+jX

IS = IR = 0 × VR + IR

IR

+

+

In matrix form, VS –

Figure 22.82 Short-line equivalent circuit

VR –

VS IS

=

1 Z

VR

0 1

IR

22.16 Transmission Line Modeling

At no-load, IR = 0, and therefore the no-load voltage is VR NL

VS

22.16.4

Z = (r + jωCl)l

IS

+

VS = A

jωCl Y = 2 2

jωCl Y = 2 2

IR

+

VR

Medium-Length Line

_ For lines of length greater than 80 km and less than 250 km, _ the shunt capacitance must be considered because the lineFigure 22.83 Nominal TC circuit. charging current becomes appreciable with increase in line length. Such lines are categorized as medium-length lines. It is common to lump the total shunt capacitance and locate half at each end of the line. Such a circuit, called a nominal TC circuit, is shown in Figure 22.83, where Z is the total series impedance of the line and Y is the total shunt admittance given by

Y = G + jωC lS The sending end voltage and current are V S = V R + IR + IS = Y

Y 1 + YZ VR Z = VR + ZI R 2 2

1 + YZ 1 + YZ VR + IR 4 2

In matrix form, VS IS

22.16.5

1+ =

YZ 2

Y 1+

YZ 4

Z 1+

VR

YZ 2

IR

Load Modeling

The variation of real and reactive power taken by a load with the variation of voltage is important when representing the load for power flow and stability studies. For system studies, usually the load on a substation is considered, and this load is of a composite nature consisting of industrial, commercial, and domestic consumers. At a typical substation bus, load may consist of the following:

•• ••

Induction motors 50–70% Heating and lighting 20–30% Synchronous motors 5–10% Transmission losses 10–12%

• •

Constant power implies that both the specified active power P and reactive power Q are assumed constant. This representation is used in load-flow studies. Constant current has current I, computed as

Although it would be accurate to consider the P–V and Q–V characteristics of each of these loads for simulation, the analytic treatment would be very complicated. For analytic purposes, there are primarily three ways to represent the load:

I=

P − jQ = I ∠ θ −φ V∗

where V = |V| ∠ θ and ϕ = tan −1

Q P

is the power-factor angle.

699

700

22 Digital Computation Basics

•

The magnitude of I is held constant. Constant impedance is mostly used in stability studies. If P and Q at a bus are known and assumed to remain constant, then the constant impedance Z connected between the bus and the ground represents the load and is computed as Z=

V V 2 = I P − jQ

or the load admittance is given as Y=

I P − jQ = V V 2

Combining all three representations gives P=a+b V +c V

2

Q=d+e V +f V

2

where a, b, c, d, e, and f are constants that separate the load into constant power, constant current, and constant impedance components. The constants a and d represent a part of the load as a combination of constant real and reactive power, constants b and e represent a part of the load as constant current load at a constant power factor, and constants c and f represent a part of the load as static impedance. There is no restriction on the magnitudes of the constants; they may be zero, positive, or negative. While emphasizing the importance of accurate representation of system loads, keep in mind that determining the load characteristics of an electric power utility is quite involved and requires accurate data for computing the constants.

701

23 Power-Flow Methods 23.1

Newton–Raphson Method

The approach to Newton–Raphson load flow is similar to that of solving a system of nonlinear equations using the Newton–Raphson method. The Newton–Raphson method formulates and solves iteratively the following load-flow equation J1 J2 ΔP = ΔQ J3 J4

Δδ ΔV

where ΔP and ΔQ: bus real-power and reactive-power mismatch vectors between the specified value and calculated value, respectively ΔV and Δδ: bus voltage magnitude and angle vectors in an incremental form J1 through J4: Jacobian matrices At each iteration, we have to form a Jacobian matrix and solve for the corrections from the following equation. For the load-flow problem, this equation is of the form Δδ2

J

Δδn Δ V2 V2 Δ V 1 + n0 V1 + n0

=

ΔP2

ΔP2

ΔPn

ΔPn

ΔQ2

ΔQ2

ΔQ1 + n0 ΔQ1 + n0

where the Jacobian matrix is divided into submatrices as J=

J11 J12 J21 J22

The size of the Jacobian matrix is (n + np − 1) × (n + np − 1). For example, for the example five-bus system, this matrix is size (7 × 7). The dimensions of the submatrices are as follows: J11

n −1 × n −1 , J12

n −1 × np, J21 np × n− 1 , and J22 np × np

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

702

23 Power-Flow Methods

The submatrices are ∂P2 ∂δ2

∂P2 ∂δn

∂Pn ∂δ2

∂Pn ∂δn

J11 =

V2

∂P2 ∂ V2

V 1 + n0

∂P2 ∂ V 1 + n0

V2

∂Pn ∂ V2

V 1 + n0

∂Pn ∂ V 1 + n0

J12 =

∂Q2 ∂δ2

∂Q2 ∂δn

J21 = ∂Q1 + n0 ∂δ2 ∂Q2 V2 ∂ V2

∂Q1 + n0 ∂δn V 1 + n0

∂Q2 ∂ V1 + n0

V 1 + n0

∂Q1 + n0 ∂ V1 + n0

J22 = V2

∂Q1 + n0 ∂ V2

The Newton–Raphson method possesses a unique quadratic convergence characteristic. It usually converges very quickly compared to other load-flow calculation methods. It also has the advantage that the convergence criteria are specified to ensure convergence for bus real-power and reactive-power mismatches. This criterion gives us direct control over the accuracy we want to specify for the load-flow solution. The convergence criteria for the Newton– Raphson method are typically set to 0.001 MW and Mvar. The Newton–Raphson method is highly dependent on the initial bus-voltage values. Careful selection of these values is strongly recommended. Before running load flow using the Newton–Raphson method, ETAP makes a few Gauss– Seidel iterations to establish a set of sound initial values for the bus voltages. The Newton–Raphson method is recommended for use with any system as a first choice.

23.2

Gauss–Seidel Method

From the system nodal voltage equation [I] = [YBUS][V], the accelerated Gauss–Seidel method derives the following load-flow equation and solves it iteratively ∗ P + jQ = V T YBUS V∗

where P and Q : specified bus’s real and reactive power vectors V : bus voltage vector YBUS : system admittance matrix Y∗BUS and V∗ are the conjugates of YBUS and V, respectively. VT is the transpose of V. The accelerated Gauss–Seidel method has relatively low requirements for the initial bus voltage values compared to the Newton–Raphson method and the fast-decoupled method. Instead of using the bus real-power and reactive-power mismatch as convergence criteria, the accelerated Gauss–Seidel method checks the bus-voltage magnitude tolerance

23.4 Fast-Decoupled Method

between two consecutive iterations to control the solution precision. The typical value for bus voltage magnitude precision is 0.000001 pu. The accelerated Gauss–Seidel method has a slower convergence speed. When we apply appropriate acceleration factors, a significant increase in the rate of convergence can be obtained. The range for the acceleration factor is between 1.2 and 1.7, and is typically set to 1.45.

23.3

Adaptive Newton–Raphson Method

This improved Newton–Raphson Method introduces a set of smaller steps for iterations where a potential divergence condition is encountered. The smaller increments may help reach a load-flow solution for some systems where the regular Newton–Raphson method fails to reach one. The Newton–Raphson method is based on the Taylor series approximation. For simplicity and incremental steps, a linear interpolation/extrapolation of the additional time-step increments is performed to improve the solution: f xk + αk × Δxk < f xk The incremental steps are controlled by adjusting the value of αk to find a possible solution for the following solution step. Test results prove that the adaptive load-flow method can improve the convergence for distribution and transmission systems with significant series capacitance effects (i.e. negative series reactance). It is also possible to improve convergence for systems with very small impedance values, but that is not guaranteed. One side effect of using this method is reduced calculation speed because of the incremental steps in the solution.

23.4

Fast-Decoupled Method

The fast-decoupled method is derived from the Newton–Raphson method. The basic premise is that a small change in the magnitude of the bus voltage does not vary the real power at the bus appreciably; and, likewise, for a small change in the phase angle of the bus voltage, the reactive power does not change appreciably. Thus, the load-flow equation from the Newton–Raphson method can be simplified into two decoupled sets of load-flow equations, which can be solved iteratively: ΔP = J1 Δδ ΔQ = J4 ΔV The fast-decoupled method reduces computer memory storage by approximately half, compared to the Newton– Raphson method. It also solves the load-flow equations using significantly less computer time than required by the Newton–Raphson method, since the Jacobian matrices are constant. As with the Newton–Raphson method, the convergence criteria of the fast-decoupled method are based on real-power and reactive-power mismatches, which are typically set to 0.001 in the order of MW and Mvar. Although for a fixed number of iterations, it is not as accurate as the Newton–Raphson method, the savings in computer time and the more favorable convergence criteria make for very good overall performance. In general, the fast-decoupled method can be used as an alternative to the Newton–Raphson method. It is commonly utilized if Newton–Raphson method has failed when dealing with long radial systems or systems that have long transmission lines or cables. For such systems, note that both the Gauss–Seidel and Newton–Raphson methods yield similar results. However, the Newton–Raphson method converges faster than the Gauss–Seidel method.

703

705

24 Short-Circuit Methods 24.1

ANSI/IEEE Calculation Methods

In ANSI/IEEE short-circuit calculations, an equivalent voltage source at the fault location, which equals the prefault voltage at the location, replaces all external voltage sources and machine internal voltage sources. All machines are represented by their internal impedances. Line capacitances and static loads are neglected. Transformer taps can be set at either the nominal position or at the tapped position, and different schemes are available to correct transformer impedance and system voltages if off-nominal tap settings exist. It is assumed that for a three-phase fault, the fault is bolted. Therefore, arc resistances are not considered. We can specify fault impedance in the ETAP Short-Circuit Study Case for a single-phase to ground fault. System impedances are assumed to be balanced threephase, and the symmetrical-components method is used for unbalanced fault calculations. Three different impedance networks are formed to calculate momentary, interrupting, and steady-state short-circuit currents, and corresponding duties for various protective devices. These networks are: ½ cycle network (subtransient network), 1.5–4 cycle network (transient network), and 30 cycle network (steady-state network). ANSI/IEEE standards recommend the use of separate R and X networks to calculate X/R values. X/R ratios are obtained for each individual faulted bus and short-circuit current. This X/R ratio is then used to determine the multiplying factor to account for the system DC offset. Using the ½ cycle and 1.5–4 cycle networks, the symmetrical root mean square (RMS) value of the momentary and interrupting short-circuit currents are solved first. These values are then multiplied by appropriate multiplying factors to finally obtain the asymmetrical value of the momentary and interrupting short-circuit currents. The following terms are helpful in understanding short-circuit calculations using ANSI/IEEE standards.

24.1.1

½ Cycle Network

This network is used to calculate momentary short-circuit current and protective device duties ½ cycle after the fault. Table 24.1 shows the type of device and its associated duties using the ½ cycle network. This network is also referred to as the subtransient network, primarily because all rotating machines are represented by their subtransient reactance. Table 24.2 shows ½ cycle machine impedances for the ½ cycle network.

24.1.2

1.5–4 Cycle Network

This network is used to calculate the interrupting short-circuit current and protective device duties 1.5–4 cycles after the fault. Table 24.3 shows the type of device and its associated duties using the 1.5–4 cycle network. This network is also referred to as the transient network. The type of rotating machine and its representation are shown in the Table 24.4.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

706

24 Short-Circuit Methods

Table 24.3 1½ – 4 cycle network duty.

Table 24.1 ½ cycle network duty. Type of device

½ cycle

Type of device

High-voltage circuit breaker

Closing and latching capability

High-voltage circuit breaker

Interrupting capability

Low-voltage circuit breaker

Interrupting capability

Low-voltage circuit breaker

N/A

Fuse

Interrupting capability

Fuse

N/A

Switchgear and MCC

Bus bracing

Switchgear and MCC

N/A

Relay

Instantaneous settings

Relay

N/A

Table 24.2 ½ cycle machine impedances.

Table 24.4 1½ – 4 cycle machine impedances.

Type of machine

Xsc

Utility

X

Turbo generator

Xd

Hydrogenerator with amortisseur winding

Xd

Hydrogenerator without amortisseur winding

0.75 Xd

Condenser

Xd

Synchronous motor

Xd

Induction machine >1000 hp. at 1800 rpm or less

Xd

>250 hp. at 3600 rpm

Xd

All other ≥50 hp

1.2 Xd

1000 hp. at 1800 rpm or less

1.5 Xd

>250 hp. at 3600 rpm

1.5 Xd

All other ≥50 hp

3.0 Xd

230 kV

1.10

1.00 Voltage factor c (±10% voltage tol)

Nominal voltage Un

For maximum short-circuit current calculation cmax

For minimum short-circuit current calculation cmin

Others 1–230 kV

1.10

1.00

High voltage: >230 kV

1.10

1.00

For more detailed information, refer to IEC 60909-2016, Table 1. 24.2.3

Initial Symmetrical Short-Circuit Current Calculation

The initial symmetrical short-circuit current (I k) is calculated using the following formula Ik =

cUn 3Zk

where Zk is the equivalent impedance at the fault location. 24.2.4

Peak Short-Circuit Current Calculation

The peak short-circuit current (ip) is calculated using the following formula ip = 2kIk where k is a function of the system R/X ratio at the fault location. IEC standards provide three methods for calculating the k factor:

• • •

Method A – Uniform ratio R/X or X/R. The value of the k factor is determined by considering the smallest ratio of R/X or the largest ratio of all the branches in the network. It is only necessary to choose the branches that carry partial short-circuit currents at the nominal voltage corresponding to the short-circuit location and branches with transformers adjacent to the short-circuit location. Any branch may be a series combination of several impedances. Method B – R/X ratio at the short-circuit location. The value of the k factor is determined by multiplying the k factor by a safety factor of 1.15, which covers inaccuracies caused after obtaining the R/X ratio from a network reduction with complex impedances. Method C – Equivalent frequency. The value of the k factor is calculated using a frequency-altered R/X. R/X is calculated at a lower frequency and then multiplied by a frequency-dependent multiplying factor.

721

722

24 Short-Circuit Methods

24.2.5

Symmetrical Short-Circuit Breaking Current Calculation

For a far-from-generator fault, the symmetrical short-circuit breaking current (Ib) is equal to the initial symmetrical short-circuit current: Ib = Ik For a near-to-generator fault, Ib is obtained by combining contributions from each individual machine. Ib for different types of machines is calculated using the following formula Ib =

μIk for synchronous machines μqIk for asynchronous machines

where μ and q are factors that account for AC decay. They are functions of the ratio of the minimum time delay and the ratio of the machine’s initial short-circuit current to its rated current, as well as real power per pair of poles of asynchronous machines. IEC standards allow us to include or exclude the AC decay effect from asynchronous machines in the calculation. 24.2.6

DC Component of Short-Circuit Current Calculation

The DC component of the short-circuit current for the minimum-delay time of a protective device is calculated based on the initial symmetrical short-circuit current and the system X/R ratio −

Idc = Ik

2e

2πf tmin X R

where f : system frequency tmin : minimum delay time of the protective device under concern X/R : system value at the faulted bus ETAP plots the DC component of the fault current vsersus time. The Idc component is printed in the “Breaking and DC Fault Current (kA)” section of the short-circuit report for each fault location. The currents in this report are always based on the total bus fault current. 24.2.7

Asymmetrical Short-Circuit Breaking Current Calculation

The asymmetrical short-circuit breaking current for comparison with the circuit-breaker rating is calculated as the RMS value of the symmetrical and DC components of the short-circuit current. ETAP plots the asymmetrical breaking current at every bus starting from 0.01 until 0.3 seconds. This information can be used in the selection of the circuit-breaker breaking current depending on the tmin value of the device. Ibasym is printed in the “Breaking and DC Fault Current (kA)” section of the short-circuit report for each fault location. The currents in this report are always based on the total bus fault current. 24.2.8

Steady-State Short-Circuit Current Calculation

The steady-state short-circuit current Ik is a combination of contributions from synchronous generators and the power grid. Ik for each synchronous generator is calculated using the following formula Ikmax = λmax IrG Ikmin = λmin IrG where λ : function of a generator’s excitation voltage, a ratio between its initial symmetrical short-circuit current and rated current, and other generator parameters IrG : generator’s rated current.

24.2 IEC Calculation Methods

The steady-state short-circuit current calculated depends on the option selected for the short-circuit current in the case study. If the Max and User-Defined c factor is selected, the maximum steady-state current short-circuit is reported. If the Min option is selected, the minimum steady-state short-circuit current is reported. This maximum steady-state short-circuit current is used to determine minimum device ratings. The minimum steady-state short-circuit value is used for relay-coordination purposes, to prevent the occurrence of nuisance trips and loading deviations. 24.2.9

Meshed and Non-Meshed Networks

According to IEC 60909-0, short-circuit contributions from meshed and non-meshed sources are calculated differently when it comes to various factors and the R/X ratio. In ETAP calculations, the short-circuit contribution in the following cases is assumed to be from a non-meshed network:

••

A contributing machine is connected directly to the faulted bus. A contributing machine is connected to the faulted bus through a radial network in which the machine is the only source making short-circuit contributions to the faulted bus. In all other cases, the short-circuit contributions assumed to be from a meshed network.

24.2.10

Adjustment of Ib

According to IEC 60909-0, to improve accuracy of Ib calculation for a near-to-generator three-phase short-circuit in a meshed network, the breaking current can be adjusted for decay in Ib from synchronous and induction machines based on the standard’s Eq. (77). This adjustment reduces Ib slightly from Ik. In ETAP, this adjustment is implemented according to Eq. (77) for each subnetwork that has near-to-generator short-circuit contributions to the faulted bus. A subnetwork with respect to a given faulted bus includes all elements that are connected together, except through the faulted bus. When a subnetwork has multiple contributions to a faulted bus, the total Ib adjustment (a phase value) is distributed among all contributions from the subnetwork based on the phase ratio of individual Ik contribution over total Ik of all the contributions from the subnetwork. 24.2.11

Modeling Power-Station Units

According to IEC 60909-0, modeling the impedance of a power-station unit involves special considerations. Depending on where the fault location is in the system and whether the unit transformer has an on-load tap-changer, the impedance values of the generator and unit transformer are adjusted by different factors. In short-circuit calculation, the generator and the transformer specified as a pair for a power unit are modeled as a power unit only when both the generator and the transformer are energized. If the transformer is not energized, the generator is modeled as a regular generator. If the generator is not energized, the transformer is modeled as a network transformer. The generator and the transformer specified as a pair for a power unit must also be connected either directly or through branches other than transformers; otherwise, they are modeled as a regular generator and a network transformer. 24.2.12

Network Bus, Connecting Bus, and Auxiliary System Bus for a Power-Station Unit

According to IEC 60909-0, the generator and transformer in a power-station unit are modeled differently depending on the fault location. In ETAP, a faulted bus can be classified as one of three types with respect to a power-station unit: a network bus, a connecting bus, or an auxiliary system bus. A connecting bus for a power-station unit is the bus on the shortest connecting path between the unit generator and the unit transformer. ETAP automatically determines the connecting path and connecting buses for a power-station unit. An auxiliary bus is a bus that is in the auxiliary system of a power-station unit, but not a connecting bus. The auxiliary system includes all the elements that are connected to the connecting buses without going across the unit transformer. Network buses are all the rest of the buses that are neither connecting buses nor auxiliary buses. 24.2.13

Wind Power-Station Units

Wind power-station units can be modeled per section 6.8 of IEC 60909-2016 by using the options shown in Figure 24.9.

723

724

24 Short-Circuit Methods

Figure 24.9 Wind power-station units.

LV

G 3–

T

HV

k3

L Un F

UrG S, SO

Figure 24.10 Wind power-station units from IEC 60909.

LV

G

T

HV

k3

L

3~

Un F AC

DC DC

AC

WD Figure 24.11 Type 3 – DFIG, wind power-station units from IEC 60909 with a crowbar.

The SC Model page of the wind turbine can be used to specify doubly fed and full-size converter (constant-current injection) wind-turbine technology as shown in Figure 24.10. For wind turbines using asynchronous generators (induction generators), the WTG editor Imp/Model page determines their short-circuit contribution. For wind turbines with doubly fed asynchronous generators (with crowbar), the IEC Short-Circuit section of the SC Model determines their short-circuit current, as shown in Figure 24.11. The values of these impedances are used by ETAP to find an equivalent impedance used to determine the contribution of the power station generator toward a fault on the primary or secondary side of its unit transformer. As shown in Figure 24.12, iWDmax, kWD, RWD/XWD, and μWD are used to find the initial symmetrical current. IkWDmax and IkWDmin are used to determine the steady-state contribution from the WTG power station. Note that the IEC short-circuit contribution of the wind power-station generator is determined based on the magnitudes of current specified in the IEC short-circuit section. IskPF is for three-phase faults, I(1)sk1PF is for LG and LLG faults, and I(1)sk2PF is for LL faults. IkPFmax is for steady-state current and, finally, IkPFmin is for minimum steady-state currents.

Figure 24.12 Wind Power Station Editor in ETAP.

24.2.14 Power-Station Units with Full-Size Converters To model power-station units with full-size converters in ETAP, as shown in Figure 24.13, make the selection in the Tap page of the two-winding transformer. A full-size inverter or WTG can be selected there to make a generator

G 3~

AC

LV

DC DC

AC

T

HV

k3

L Un F

WF Figure 24.13 Type 4 – wind power-station units from IEC 60909.

24.2 IEC Calculation Methods

/ unit transformer pair. For full-size, fully decoupled WTG terminals from AC systems, the SC model and FRT pages denote the behavior of the power station during faults.

24.2.15

IEC Short-Circuit Mesh Determination Method

IEC standards have published different benchmark sample calculation results based on IEC 60909-0 2001. These calculation examples are published in IEC 60909–4 2000 and are described in detail in Sections 3.1, 4.1, 5.1 and 6.1 of IEC 60909-4. These examples were created mostly for hand calculations (except for Example 4), and one problem that arises because of these multiple solutions is that they do not establish a single calculation method that produces consistent results for all four examples for a computer-based solution. Because of these inconsistencies in the standard, certain calculation assumptions have been added to ETAP’s short-circuit program in order for the results to match those published in these examples. These calculation preferences affect the selection of calculation methods for Idc, Ib, and Ik. One of the most important causes of the inconsistencies in the standard is the method used to determine the meshed or non-meshed (radial) parts of the systems. This determination is very important since the results are affected considerably once this determination is made. 24.2.15.1 Meshed/Non-Meshed Systems

A meshed system can be considered a looped system or one that has multiple source contributions meshed together through the same contributing branch. A non-meshed system is defined as a radial system or one that has only one contribution passing through a branch toward the faulted bus. Figure 24.14 illustrates the concept of meshed and non-meshed as described by IEC 60909-0 2016. The areas enclosed in red represent the meshed contributions in this system toward the faulted buses. In Figure 24.15, areas enclosed in green represent the non-meshed (radial) contributions to their connected bus. In other words, the contributions of G2, G1, G3, M3, and M2 are considered to be non-meshed as long as the fault is placed at the bus to which they are connected. These same contributions could be handled as meshed contributions to faults in other parts of the system. ETAP has a number of short-circuit mesh-determination methods. Refer to the ETAP user guide for more information about these options.

G2 G 3∼

8 LV T3

1 Q1

HV

2

MV

T2

3

G1 G 3∼

T1

R1 L2 10 km

L1 20 km Un = 110 kv

T4

5

L4 10 km

L5 15 km

L3(a) 5 km UnQ1= 380 kv L3(b) 5 km T5 Q2

6

G3 G 3∼ 7

T6 L6 1 km UnQ2 = 110 kv Figure 24.14 Meshed example from IEC 60909-4.

R6

Un = 10 kv

M 3 ∼ M3 M M2 3∼ (2x)

725

726

24 Short-Circuit Methods

G 3 ∼

I"kS

I"kT

IpS

IpT

IbS

IbT

IkS

IkT

M 3 ∼ I"kM IpM IbM

I"k Ip Ib k3

Ik F

IEC 1283/2000 Figure 24.15 Non-meshed example from IEC 60909-0.

Impact of the Meshed/Non-Meshed Determination on Ib, Ik, and Idc Once the program has determined the meshed and non-meshed parts of the system, it makes decisions based on this for calculating the values of Ib, Idc, and Ik as described throughout IEC 60909-0 (2001):

• • •

Idc: If the branch contribution is considered to be coming from a non-meshed source, then the R/X of the individual branch is used to determine the value of the Idc coming into the faulted bus. The equivalent R/X value of the meshed network is used to determine the value of Idc for the meshed contributions. See Sections 4.3.1.1, 4.3.1.2, and 4.4 of IEC 60909. Ib: If a contribution is non-meshed, then the program will use the method described in Section 4.5.2.2, Eqs. (71) and (72) of IEC 60909 to determine the contributions of Ib from different non-meshed components. If the system is meshed, then the program uses a very different approach to determine Ib: it uses Section 4.5.2.3, Eqs. (74) and (75). Ik: If the contribution or system is considered to be non-meshed, then the program uses the method described in Section 4.6.2, Eqs. (82) and (83) of IEC 60909. If the contribution of the system is considered to be meshed, then the program uses the method described in Section 4.6.3, Eqs. (84) and (85) to determine Ik. Note that the use of these equations may result in the value of Ik being higher than Ib, as can be observed in the results published in IEC 60909-4 2000 for Example 4.

From the previous description, it becomes apparent that determining meshed and non-meshed parts of the system can have a drastic effect on the results. The options that ETAP provides are designed to provide choices for how the analysis should be performed. 24.2.16

Comparison of Device Rating and Short-Circuit Duty

In the three-phase device duty calculation, ETAP compares the protective device rating against the bus short-current duty for the devices that are checked for compliance with IEC standards and also have a device rating entered. If the short-circuit duty is greater than the device duty, ETAP flags the device as underrated in both the one-line diagram and output reports. Table 24.9 lists the device ratings and short-circuit duties used for the comparison for MVCB, LVCB, and fuses. 24.2.17

Calculating IEC Device Capability

As shown in Table 24.9, some of the device capability values are calculated by ETAP based on values provided by users and default parameters given in IEC standards:

24.2 IEC Calculation Methods

Table 24.9 Comparison of device ratings and short circuit duties. Device type

Device capability

Short-circuit current duty

MVCB

Making

ip

AC breaking

Ib,symm

Ib,asymma

Ib,asymm

Idc

a

Ithr LVCB

Fuse

a

Ith

Making

Ip

Breaking

Ib,symm

Ib,asymma

Ib,asymm

Ithr

Ith

Breaking

Ib,symm

Ib,asymma

Ib,asymm

Device capability calculated by ETAP.

•

Medium-voltage circuit breaker (MVCB): The asymmetrical breaking and DC current ratings for MVCB are calculated as follows Ib, asymm = Ib, symm Idc = Ib, symm

•

1 + 2e − 2

2 e−

Tmin τ

Tmin τ

where tmin : minimum delay time Ib,symm : AC breaking current provided by the user Following IEC 62271-100, τ is equal to 45 ms. Low-voltage circuit breaker (LVCB): The asymmetrical breaking current rating for LVCB is calculated as follows −

Ib, asymm = Ib, symm

•

1+2×e

4πft min X R

where f : system frequency tmin : minimum delay time Ib,symm : breaking current provided by the user X/R is calculated based on a testing PF given in IEC 60947-2, Table 11. Fuse: The asymmetrical breaking current rating for a fuse is calculated as follows −

Ib, asymm = Ib, symm

1+2×e

4πft min X R

where f : system frequency tmin : assumed to be a half cycle Ib,symm : breaking current provided by the user X/R is calculated based on the default testing PF of 15%.

727

728

24 Short-Circuit Methods

•

Ith: The thermal equivalent short-circuit current through LVCB and MVCB is calculated based on specifications on IEC 60909-0 2001 Annex A, as follows: Tk 2 i2 dt = Ik 2 m + n TK = Ith TK and Ith = IK m + n 0

1+ n=

1 I K IK

2

Td 1 − e −20TK 20TK

Td 1− e −10TK 5TK

m=

Td

2

IK IK − IK IK 2

IK IK − IK IK

Td

Td 1 − e −11TK 5, 5TK

Td

+

+

Td 1 −e −20TK 2TK

2 Td 1 −e − TK TK

IK IK − IK IK

Td

Td

IK −1 IK

IK −1 IK

2

+

2

+

IK −1 IK

1 ∗ e4 fT K ln k −1 − 1 2fT K ln k − 1

These equations represent the Joule integral and the equivalent short-circuit current Ith that ETAP determines in order to compare against the value of Ithr specified on the circuit-breaker rating page (LV or MV). ETAP compares the circuit-breaker (CB) rated short-time withstand thermal energy in megajoules (calculated as current (Ithr)2 times the rated short-time in seconds (Tkr)) with the calculated thermal equivalent short-circuit energy in MJ (calculated as current (Ith)2 times, either the rated short-time in seconds (Tkr) or the user-defined short-time in seconds (Tk)). The comparison of thermal energy values is provided in the summary report. An example of the report is provided in the following image. As can be observed, the rated thermal energy for CB11 is 192.0 MJ, while the short-circuit thermal energy was determined to be 126.0 MJ. Short-Circuit Summary Report Device capacity Bus ID Sub2A

Device ID CB11

Ithr (kA)

Tkr (sec.)

8.000

Ithr = Rated short-time withstand current (Icw for low voltage circuit breaker) Tkr = Rated short-time Ith = Thermal equivalent short-time current * Indicates a device with calculated duty exceeding the device capability.

3.00

3-Phase Short-Circuit Duty Results Rated Thermal Energy (MJ) 192.00

Ith (kA) 6.481

Tkr (sec.)

Thermal Energy (MJ)

3.00

126.01

729

25 Harmonics 25.1

Problem Formulation

The majority of nonlinear loads consist of equipment that utilizes power semiconductor devices for power conversion (e.g. rectifiers). They include, for example, computer switched-mode power systems (SMPS) for converting AC to DC. Figure 25.1 illustrates a typical current waveform of a computer switched-mode power supply unit. SMPS uses capacitors to smooth the rectified DC voltage and current prior to it being supplied to other internal subsystems and components. The semiconductor diode rectifiers are unidirectional devices (i.e. they conduct in one direction only). The additional function of the capacitor is to store energy, which is drawn by the load as necessary. When the input voltage (Vi) is higher in value than the capacitor voltage (Vc), the appropriate diode will conduct and non-sinusoidal, “pulsed” current will be drawn from the supply. This non-sinusoidal current contains harmonic currents in addition to the sinusoidal fundamental (50 or 60 Hz) current, as shown in Figure 25.2. Harmonic voltages and currents are integer multiples of the fundamental frequency. On a 60 Hz power supply, the fifth harmonic is 300 Hz; the seventh harmonic is 420 Hz, and so forth. When all harmonic voltages and currents are added to the fundamental, a waveform known as a complex wave is formed. Any periodic waveform can be expressed as a sum of sinusoids. The sum of the sinusoids is referred to as a Fourier series. It is usually more convenient to interpret a complex wave by means of a Fourier series and associated analysis methods. Joseph Fourier, a nineteenth century French physicist, introduced a theory that any periodic function in an interval of time could be expressed by the sum of the fundamental component and a series of higher-order harmonic frequencies that are integral multipliers of the fundamental component, as shown in Figure 25.2. In a balanced three-phase system under non-sinusoidal conditions, the hth order harmonic voltage (or current) can be expressed as follows: Vh hw0 t − θh

Vah =

25 1

h 1

Vbh =

Vh hw0 t −

hπ θh 3

25 2

Vh hw0 t −

2hπ θh 3

25 3

h 1

Vch = h 1

Based on these equations and counterclockwise rotation of the fundamental phasors, we can write Va = V1 sinωt + V2 sin2ωt + V3 sin 3ωt + V4 sin 4ωt + V5 sin 5ωt+ … Vb = V1 sin ωt − 120

0

+ V2 sin 2ωt −240

0

+ V3 sin 3ωt − 360

0

+ V4 sin 4ωt − 480

25 4 0

+ V5 sin 5ωt −600 + … 0

= V1 sin ωt − 1200 + V2 sin 2ωt + 1200 + V3 sin3ωt + V4 sin 4ωt −1200 + V5 sin 5ωt + 1200 + … Vc = V1 sin ωt + 1200 + V2 sin 2ωt + 2400 + V3 sin 3ωt + 3600 + V4 sin 4ωt + 4800 + V5 sin 5ωt + 6000 + … = V1 sin ωt + 1200 + V2 sin 2ωt − 1200 + V3 sin 3ωt + V4 sin 4ωt + 1200 + V5 sin 5ωt −1200 + …

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

25 Harmonics

Vc

Vi

+ Voltage

Vdc

Nonlinear load current

AC –

Figure 25.1 Typical current waveform for SMPS. 70

% Current distortion magnitude

730

60 60 Hz fundamental 180 Hz (3rd harmonic)

50 40 30 20 10 0

1 2 3 4 5 6 7 8 9 11 12 13 14 15 17 19 21 23 25 27 29 31 33

Complex wave

Harmonic order

Figure 25.2 Harmonic distortion current and spectrum.

Table 25.1 Harmonic order and network sequence. Harmonic order

Sequence

1

+

2

−

3

0

4

+

5

−

6

0

7

+

8

−

9

−

10

+

11

−

12

0

The nth or hth harmonic (frequency of harmonic = n × fundamental frequency) of phase B lags n times 120 behind that of the same harmonic in phase A. The nth harmonic of phase C lags n times 240 behind that of the same harmonic in phase A. In the case of triplen harmonics, shifting the phase angles by three times 120 or three times 240 results in vectors that are in the same phase. The harmonic order may be written as positive-negative-zero, as shown in Table 25.1. All triplen harmonics generated by nonlinear loads are zero-sequence phasors. These add up in the neutral. In a three-phase four-wire system, with perfectly balanced single-phase loads between the phase and neutral, all positive- and negative-sequence harmonics cancel out, leaving only the zero-sequence harmonics. In an unbalanced single-phase load, the neutral carries zero-sequence and the residual unbalance of positive- and negativesequence currents. Generally, a time domain voltage or current waveform can be expressed by a Fourier series: ∞

at = 2

An sin 2πfn + φn

25 5

n=1

where a(t) : time domain voltage or current waveform An : rms coefficients of Fourier series fn : harmonic frequency of order n φn : phase angle of fn The Fourier series for a typical six-pulse converter is as follows: Iac =

2 3 1 1 1 1 Id cos ωt − cos 5ωt + cos 7ωt − cost11ωt + cos 13ωt 5 7 11 13 π

∞ h=1

Ih cos hωt + Φh

25.1 Problem Formulation

25.1.1

Characteristic Harmonic Currents

A typical nonlinear load (controlled rectifier) connected to a power-distribution system can be simply represented as a p-pulse controller that produces harmonic currents. A p-pulse rectifier produces nth order characteristic harmonic currents by the following relationship n = kp ± 1 where n : harmonic order, as a multiplication of fundamental frequency k : 1, 2, 3 … p : pulse number of rectifier All harmonic currents can be approximately represented as In =

I1 n

where I1 is the fundamental current Note that in the “ideal” harmonic theory, the following hypotheses are assumed for all rectifiers:

•• •• ••

The impedance of the AC supply network is zero. The DC component of the rectifier configuration is uniform. An AC line or commutating reactor is not used in front of the rectifier. The AC supply network is symmetrical (i.e. balanced). The AC supply is sinusoidal, free from harmonics. There are no overlap or delay angles for the devices.

Any divergence from any of these hypotheses will introduce non-characteristic harmonics, including possibly DC, into the harmonic series. In practical terms, note that supply networks or connected equipment are never idealized (i.e. based on these hypotheses), and, therefore, any actual harmonic currents measured will not be exactly as calculated using these simplified formulas. Therefore, in theory, the fifth harmonic current and seventh harmonic current, for example, should represent 20% and 14% of the total RMS current, respectively. However, again, this is never transferred into practical reality, as the magnitudes of the various harmonic currents are determined by the per-phase inductance of the AC supply connected, the rectifier, and the impedance of the rectifier as seen by the AC supply. In rectifiers without added inductance (e.g. AC line reactors), it is not uncommon to measure fifth harmonic current up to ~80% on single-phase rectifiers and 65% on three-phase rectifiers. On the basis of the current-injection method, the harmonic currents produced by the nonlinear load can be seen as a harmonic current source connected to the power system. From circuit theory, the harmonic currents injecting into the power system and their corresponding harmonic voltages can be easily calculated. Possible resonant conditions between capacitors and system impedance can also be predicted. The generated harmonic frequencies depend on the type of nonlinear load or device. Most nonlinear loads and devices produce odd harmonics with small even harmonics. However, loads such as arc furnaces produce the entire spectrum of harmonics: odd, even, and non-integer harmonics in between (non-integer harmonics are also referred as interharmonics). Generally, the amplitude of the harmonics decreases as the frequency (or the harmonic order) increases. When a nonlinear load draws distorted (non-sinusoidal) current from the supply, that distorted current passes through all of the impedance between the load and power source. The associated harmonic currents passing through the system impedance cause voltage drops for each harmonic frequency based on Ohm’s law (Vh = Ih × Zh), as shown in Figure 25.3. The vector sum of all the individual voltage drops results in the total voltage distortion, the magnitude of which depends on the system impedance and the levels of harmonic currents at each harmonic frequency. Figure 25.4 shows in detail the effect that individual harmonic currents have on the impedances within the power system and the associated voltage drops for each. Note that the total harmonic voltage distortion Vthd (based on the vector sum of all individual harmonics) is reduced as more impedance is introduced between the nonlinear load and the source.

731

732

25 Harmonics

Impedance Distorted voltage

Distorted current

No voltage distortion

Nonlinear load

Source

Figure 25.3 Distorted current through a nonlinear load.

Source ZSh

ZTh

ZCh

Ih Nonlinear load

ZSh

Transf. ZTh Sinusoidal voltage source

Cable ZCh

Vh

Vh

Vh

@ Source

@ Transf.

@ Load

Nonlinear load

Harmonic current source

Figure 25.4 Harmonic effect on impedances.

V = Ih × Zh (Ohm’s law) At the load: Vh = Ih × (ZCh + ZTh + ZSh) At the transformer: Vh = Ih × (ZTh + ZSh) At the source: Vh = Ih × (ZSh) where Z : impedance at the frequency of the harmonic (e.g. 250 Hz) Vh : harmonic voltage at the hth harmonic (e.g. fifth) Ih : harmonic voltage at the hth harmonic (e.g. fifth) Vthd : total harmonic voltage distortion 25.1.2

Interharmonics

Interharmonics are defined as follows: “Between the harmonics of the power frequency voltage and current, further frequencies can be observed which are not an integer of the fundamental. They appear as discrete frequencies or as a wide-band spectrum.” Interharmonics can therefore be considered the intermodulation of the fundamental and harmonic components of the system with any other frequency components. These can be observed increasingly with nonlinear loads, including large AC frequency converter drives (especially under unbalanced conditions and/or levels of preexisting voltage distortion), cycloconverters, and static slip-recovery drives. 25.1.3

Subharmonics

Subharmonics is an unofficial but common name given to interharmonics whose frequency is less than that of the fundamental (i.e. f > 0 Hz and f < f1). The more technically correct term, sub-synchronous frequency component, is also used for subharmonics. Therefore, the calculations provided for interharmonics are also valid for subharmonics.

25.2 Methodology and Standards

Both harmonics and interharmonics can be defined in a quasi-steady state in terms of their spectral components over a range of frequencies:

•• ••

Harmonics: f = h × f1, where h is an integer > 0 DC: f = 0 Hz (f = h × f1, where h = 0) Interharmonics: f is not equal to h × f1, where h is an integer > 0 Subharmonics: f > 0 and f < f1, where f1 is the fundamental frequency

25.2

Methodology and Standards

− Y12

Y22

… − Y2n

V2

…

−Y1n − Y2n … Ynm

VN

I1 =

I2 …

V1 …

−Y12 … − Y1n …

Y11 …

An ideal current source provides a constant current regardless of the system impedance seen by the source. In most studies for industrial applications, the nonlinear load or the harmonic source is considered an ideal current source without a Norton’s impedance across the source (i.e. Norton impedance is assumed to be infinite). This approximation is generally reasonable and yields satisfactory results. When the nonlinear device acts like a voltage source, meaning it has relative fixed voltage waveform distortion rather than fixed current waveform distortion (e.g. a pulse-width-modulated [PWM] inverter, an arc furnace, or a utility connection), a Thévenin equivalent voltage source model may be utilized. During harmonic analysis, the system is subjected to harmonic current injections and/or harmonic voltage affection at multiple frequencies, and the network is solved for voltage and current at each frequency separately. The total voltage or current in an element is then found either by a root mean square (RMS) sum or arithmetic sum, using the principle of superposition. The effect of harmonic current propagation through the network, including the power source, produces distortion of the voltage waveform depending upon harmonic voltage drops in various series elements of the network. Therefore, the voltage distortion at a given bus is dependent on the equivalent source impedance; the smaller the impedance, the better the voltage quality. Note that the harmonic sources, which are nonlinear loads, are not the sources of power, but are the cause of additional active and reactive power losses in the system. There are two basic modeling and simulation methods to represent harmonic generation from nonlinear devices in power systems. The first method is the harmonic load-flow method. This method models harmonic generation from nonlinear power system devices by injecting harmonic voltages and/or currents from these nonlinear devices into the power network. To determine the harmonic voltages and harmonic currents generated by the nonlinear devices, techniques such as fast Fourier transformation (FFT) can be employed to extract data from the total voltage and current waveforms. The harmonic voltage source and current source are expressed in a series of harmonics with both magnitude and initial phase angle specified for each harmonic component. The magnitude of the harmonic component is usually a percentage of the fundamental quantity and the phase angle. The power transmission or distribution system is modeled as an electrical impedance network in nodal voltage equations similar to the load-flow studies. Each network element is represented by a set of linear equations corresponding to its previously described circuit model. The reactance and admittance of the element circuit model must be adjusted for the harmonic order or frequency when harmonic sources are applied to the network at the same harmonic order of frequency:

IN

Depending on the network components’ frequency characteristics, sometimes a frequency-saturation effect needs to be taken into consideration when adjusting their impedance. Due to the sequence characteristics of harmonics, in general all three sequence impedance networks – the positive-sequence network, negative-sequence network, and zerosequence network – need to be formulated and used with harmonic sources of different orders. One important part of formulating a zero-sequence network is to make sure the data for the machine and transformer winding connection types and grounding configurations are included, because they will affect the zero-sequence network. With harmonic sources determined and modeled, and the network selected and adjusted to the harmonic orders, harmonic currents are injected into the network from harmonic current sources, and harmonic voltages are applied to the buses where harmonic voltage sources exist. By solving the previous equation at each frequency, we obtain

733

734

25 Harmonics

the nodal voltages. This computation is performed for each harmonic frequency of interest. From the harmonic voltages, we can compute the harmonic currents in each branch: Iij = Vi – Vj × Yij Overall bus voltages and branch currents are obtained by combining their fundamental values plus harmonic components. Bus voltages and branch currents can be expressed in different forms, such as total RMS value, maximum or peak value, etc. The inverse of the nodal admittance matrix is called the nodal impedance matrix. This matrix is rich in quantitative information. The diagonal entry on the ith row is the Thévenin impedance of the network seen from bus i. By computing values of this matrix over a range of frequencies, we obtain the frequency response of the network seen from each bus. Exact resonance frequencies can be determined from this computation. The off-diagonal values in the matrix show the effect of a harmonic current injection on the bus voltages. Consider a single harmonic source connected at bus i, forcing 1.0 A of current into the network. The harmonic voltage at bus j is simply Zij, the value found in the ith row and jth column of the nodal impedance matrix. The harmonic voltage at bus j due to numerous sources can be solved by superposition. There are a number of methodologies for calculation of harmonics and effects of nonlinear loads. Direct measurements can be carried out using suitable instrumentation. Analytical analysis can be carried out in the frequency and time domains. Another method is to model the system using a state-space approach. The differential equations relating current and voltages to system parameters are found through basic circuit analysis. For calculations in the frequency domain, the harmonic spectrum of the load is ascertained, and the current injection is represented by a Norton’s equivalent circuit. Harmonic current flow is calculated throughout the system for each of the harmonics. The system impedance data is modified to account for higher frequency and reduced to the Thévenin equivalent. The principal of superposition is applied. In Figure 25.5, Isource is the harmonic current from the pure current source, Iha is the harmonic current specified in the harmonic spectrum for the source, Zsource is the internal impedance of the harmonic source, and Zsys is the equivalent system impedance.

25.2.1

Ideal Current Source

In this model, the current source injects harmonic current as specified by its harmonic spectrum, independent of system impedance and source internal impedance, as shown in Figure 25.5 Model A.

25.2.2

Thévenin/Norton Equivalent Sources

A non-electronic harmonic current source is represented by an equivalent Thévenin/Norton model, as shown in Figure 25.5 Model B or Model C. The harmonic current source is represented by a pure current source connected in parallel with source internal impedance. For a harmonic order at which the source has zero current injection, it is represented by its internal impedance only. In Model B, the harmonic current (Iha, i.e. 5% for fifth order harmonic) from the current source (Isource) follows the harmonic spectrum specified for the source. Due to the current sharing of the source internal impedance, the amount of harmonic current injected into the system is smaller. The difference between the two depends on the values of source and system impedance. In Model C, the harmonic current from the pure current source (Isource) is adjusted so that the actual harmonic current injection into the system (Iha, i.e. 5% for fifth order harmonic) follows the harmonic spectrum specified for the source. Note that utility, generator, and static Iha

Iha

Iha

Zsource

Zsys Isource

Isource Model A

Figure 25.5 Norton equivalent circuit.

Model B

Zsource

Zsys Isource

Model C

Zsys

25.3 Harmonic Indices

load are considered non-electronic harmonic sources. When this option is selected, the electronic sources and transformer sources are modeled as an ideal current source. In order to conduct a harmonic study, knowledge of the harmonic currents or voltages generated by nonlinear loads or nonlinear sources is critical. Incorrect harmonic-injection information will yield to inaccurate results. There are several options to acquire data for harmonic analysis:

• • •

Harmonic distortion injection of the source from the equipment manufacturer. This is particularly important in cases where new technology or nonstandard equipment is applied. Calculate the generated harmonics by analytical methods where possible, such as at converters or static var compensators. Apply typical values based on similar applications or published data.

Harmonic measurements may not yield the worst-case harmonic spectrum, depending upon the equipment operating modes at the time the measurements are made. Calculation of harmonics by analytical methods can be challenging, as it requires determining an equation for the subject waveform and performing a Fourier analysis on that equation. Since the system configuration and load continually change, the harmonics also change, and it would be a formidable task to study all such conditions. Usually, the worst operating condition is determined, and the design is based on the “worst-generated” harmonics ratio. However, it needs to be recognized that even with the worst-generated harmonic case, the harmonic flows within different elements of the network can be different depending upon the number of transformers or tie breakers in service. This means that for the worst-generated case, the worst operating cases(s) must be analyzed. One other difficulty in the analysis arises from the fact that when multiple harmonic sources are connected to the same bus (or different buses), the phase angles between harmonics of the same order usually are not known. Generally speaking, arithmetic addition of harmonic magnitudes is reasonable if the harmonic sources are similar and have similar operating load points. However, this approach can lead to a more conservative filter design and distortion calculations, if the sources are different or operate at different load points. Determination of phase angles of harmonics and vectorial addition can be quite a complex and expensive approach for general industrial applications. This is often resolved by simplifying assumptions based on experience or by field measurements. More advanced techniques are used in highvoltage DC transmission and other utility applications where precision is important. Industrial harmonic studies are usually represented on a single-phase basis, i.e. based on the assumption that the system is balanced and positive-sequence analysis applies. A three-phase study is warranted only if the system or the load is severely unbalanced or a four-wire system with single-phase loads exists. However, with the development and operation of electrical railway systems in some countries, more and more unbalanced three-phase systems have been observed. In such a situation, it is desirable to determine the harmonics generated in all three phases. The cost of a three-phase study can be higher than a single-phase study and should be used only when the expense and purpose can be justified.

25.3

Harmonic Indices

Harmonic indices are used to determine the quality of power being delivered to the load. They are also used to measure the quality of current and voltage at the utility point of common coupling.

25.3.1

Harmonic Factor

The harmonic factor or harmonic distortion factor is the ratio of the RMS of the harmonic content to the RMS of the fundamental quantity, expressed as a percentage of the fundamental: DF =

of squares of amplitudes of all harmonics × 100 Square of the amplitude of the fundamental

The most commonly used index is total harmonic distortion (THD), which in common use is the same as DF. RMS voltage in the presence of harmonics can be written

735

736

25 Harmonics h= ∞

Vh2, rms

Vrms = h=1

RMS current in the presence of harmonics can be written Irms =

h= ∞ 2 h = 1 Ih, rms

The THD factor for the voltage is h= ∞ h=2

THDV =

Vh2, rms

Vf , rms

where Vf,rms is the fundamental frequency voltage. This can be written Vrms Vf , rms

THDV =

2

− 1 or Vrms = Vf , rms

1 + THD2V

Total harmonic distortion for current can be written as: h= ∞ 2 h = 2 Ih, rms

THDI =

If , rms

=

Irms If , rms

2

− 1 or Irms = If , rms

1 + THD2I

where If,rms is the fundamental frequency current. Total demand distortion (TDD) is defined as TDD =

h= ∞ 2 h = 2 Ih

, where IL is the load demand current. IL The partial weighted harmonic distortion (PWHD) of current is expressed as h = 40 h = 14

PWHDI =

hI 2h

If , rms

PWHD evaluates the influence of current or voltage harmonics of higher orders. An expression can be written for voltage that is similar to the expression for current. The sum parameters are calculated with single-harmonic current components Ih. 25.3.2

Individual Harmonic Distortion (IHD)

Individual harmonic distortion (IHD) simply calculates the ratio of a given harmonic component to the fundamental component. This value is sometimes used to track the effect of each individual harmonic and examine its magnitude. IHD is determined using IHD =

25.3.3

Fi F1

Arithmetic Summation (ASUM)

Arithmetic summation (ASUM) adds the magnitudes of the fundamental and all harmonic components directly to provide a conservative estimation of the crest value of voltage and current. It is useful for evaluating the maximum withstanding ratings of a device: ∞

ASUM =

Fi 1

25.3.4

Telephone Influence Factor

Harmonics influence telephone circuits through inductive coupling, and an index for measuring this effect is the telephone influence factor (TIF). TIF for a voltage or current wave in an electrical supply circuit is the ratio of the square

25.3 Harmonic Indices

root of the sum of the squares of the weighted RMS values of all the sine wave components (including AC waves both fundamental and harmonic) to the RMS value (unweighted) of the entire wave Wf2 If2

TIF =

Irms

where If : single-frequency RMS current at frequency f Wf : single-frequency TIF weighting at frequency f The voltage can be substituted for current. 25.3.5

IT Product

The I ∗T product is the inductive influence expressed in terms of the product of its RMS magnitude I in amperes times its TIF: IT = TIF × Irms =

W f If

2

The balanced I ∗T product (I ∗TB) is calculated the same as I ∗T but takes only the summation of positive- and negativesequence harmonics. Similarly, the residual I ∗T product (I ∗TR) takes only the summation of zero-sequence harmonics. 25.3.6

kVT Product

The kVT product is the inductive influence expressed in terms of the product of its RMS magnitude in kV times its TIF: kVT = TIF × kV rms =

25.3.7

Wf Vf

2

C Message

The C message weighted index is very similar to the TIF except that the weights ci are used in place of wi ∞ 2 h = 2 c h Ih

Ci =

Irms

Where ci factors are related to the TIF weights by wi = 5 (i) ( fo) ci. This index can also be applied to bus voltage and is used in the United States and Canada. The telephone weighting factor that reflects the present C message weighting and the coupling normalized to 1 kHz is given by Wf = 5Pf f where Pf is the C message weighting at frequency f under consideration. 25.3.8

Telephone Form Factor (TFF)

The International Consultation Commission on Telephone and Telegraph System (CCITT) used in Europe describes the level of harmonic interference in terms of TFF TFF =

1 V1

∞

Kh Ph V h

2

h=1

where h : coupling factor 800 Ph : harmonic weight divided by 1000 Kh =

737

738

25 Harmonics

25.3.9

Distortion Index (DIN)

This is basically the Fourier-based equivalent of the TIF without a weighting factor included ininside the summation in the numerator. The concept of the distortion index (DIN) is very close to that of the THD. The difference is that whereas the THD is the ratio of the square root of the power content ratio of the harmonics to the fundamental frequency, the DIN is simply the ratio of the harmonic power to the total power in the waveform: DIN =

∞ i=2

Vi

∞ i=1

Vi 2

2

=

THD THD 2 + 1

For low levels of harmonics, 1 DIN ≈ THD 1 − THD 2

25.3.10

Total Interharmonic Distortion (TIHD)

Total interharmonic distortion (TIHD) is the ratio of the RMS of all interharmonics to the fundamental component TIHD =

all interharmonics i

Fi2

F1

where Fi : interharmonic components F1 : fundamental component

25.3.11

Total Subharmonic Distortion (TSHD)

Total subharmonic distortion (TSHD) is the ratio of the RMS of all subharmonics to the fundamental component all subharmonics s

TSHD =

Fs2

F1

where Fs : subharmonic components F1 : fundamental component

25.3.12

Group Total Harmonic Distortion (THDG)

This is the ratio of the RMS value of the harmonic groups (g) to the RMS value of the group associated with the fundamental H

THDG = n=1

Ggn Gg1

2

where G : voltage or current Gn : harmonic group calculated based on IEC 61000-4-7, Eq. (8) n : harmonic order (n = 1 for fundamental) H : maximum harmonic order to account for

25.3 Harmonic Indices

25.3.13

Subgroup Total Harmonic Distortion (THDS)

This is the ratio of the RMS value of the harmonic subgroups (g) to the RMS value of the subgroup associated with the fundamental H

THDS = n=1

Gsgn Gsg1

2

where G : voltage or current Gsgn : subharmonic group calculated based on IEC 61000-4-7, Eq. (9) n : harmonic order (n = 1 for fundamental) H : maximum harmonic order to account for

25.3.14

Harmonic Power Factor

For sinusoidal voltages and currents, the power factor (PF) is defined as ϕ = cos −1

kW , and the power factor angle ϕ is kVA

kW kvar = tan −1 kVA kW

The PF in the presence of harmonics comprises two components: displacement and distortion. The effect of the two is combined in the total PF. The displacement component is the ratio of active power of the fundamental wave in watts to the apparent power of the fundamental wave in volt-amperes. This is the PF as seen by the watt-hour and var-hour meters. The distortion component is the part associated with harmonic voltages and currents: PF t = PF f × PF distortion The displacement PF is equal to the total PF, as the displacement PF does not include kVA due to harmonics, while the total PF does include it. This is true at the fundamental frequency. For harmonic loads, the total PF is always less than the displacement PF. The PF of a converter with a DC-link reactor is given by this expression from IEEE 519 q π Total PF = sin π q where q : number of converter pulses π∕q : angle in radians This ignores commutation overlap and no-phase overlap, and ignores transformer magnetizing current. For a sixpulse converter, the maximum PF is 3 π = 0.955. A 12-pulse converter has a theoretical maximum PF of 0.988. The PF drops drastically with an increase in firing angle. Note that the PF is a function of the drive topology: for example, with pulse-width modulation, the input PF is dependent on the type of converter, and the motor PF is compensated by a capacitor in the DC link. In the case of a nonlinear load or when the source has a non-sinusoidal waveform, the active power P and reactive power Q can be written in terms of voltage and current RMS values: h= ∞

P=

Vh Ih cos θh − δh

h=1 h= ∞

Q=

Vh Ih sin θh − δh

h=1

The apparent power (S) in MVA can be written with the distortion power (D) S=

P 2 + Q2 + D2

739

740

25 Harmonics

where D2 up to the third harmonic order is D2 = V02 + V12 + V22 + V32 I02 + I12 + I22 + I32 − V0 I0 + V1 I1 cosθ1 + V2 I2 cosθ2 + V3 I3 cosθ3 − V1 I1 sinθ1 + V2 I2 sinθ2 + V3 I3 sinθ3

2

2

The total PF in terms of current and voltage distortion can be written as P

PF tot = V f If

1+

2

THDV 100

1+

THDI 100

2

Ignoring the power contributed by harmonics and also voltage distortion, as it is generally small, i.e. THDV = 0, 1

PF tot = cos θf − δf ×

THDI 100

1+

2

= PF displacement × PF distortion That is, total PF is the product of displacement PF (which is the same as the fundamental PF) and is multiplied by the distortion PF. Example. Consider the following: RMS

60 Hz

180 Hz

420 Hz

Voltage (pu)

1 n NT

MAIFI: The momentary average interruption frequency index tracks the average frequency of momentary interruptions: MAIFI =

•

Li LT

Total number of customer momentary interruptions = Total number of customers served

IDi Ni NT

MAIFIE: The momentary average interruption event frequency index tracks the average frequency of momentary interruption events: MAIFI =

Total number of customer momentary interruption events = Total number of customers served

ΣIDE Ni NT

Ni is the number of customers experiencing momentary interruption events. This index does not include the events immediately preceding a lockout.

•

•

CEMSMI: The customers experiencing multiple sustained interruptions and momentary interruptions events index tracks the number n of both sustained interruptions and momentary interruption events experienced by a set of specific customers. Its purpose is to help identify customer trouble that cannot be seen by using averages. This index accounts for both momentary interruption events and sustained interruptions: CEMI n =

Total No of customers that experienced more than n interruptions Total number of customers served

CEMI n =

CNT k > n NT

Average failure rate at load point i, λi(f/yr): λi =

λe, j j Ne

where e, j : average failure rate of element j (or element combination j, such as double contingency) Ne : total number of the elements whose faults will interrupt load point i

753

754

26 Reliability

•

Annual outage duration at load point i, Ui (h/yr): λe, j rij

Ui = j Ne

where rij : failure duration at load point i due to a failed element j

•

Average outage duration at load point, ri (h): ri =

•

Ui λi

Expected energy not supplied index at load point, EENSi (MWhr/yr): EENS i = Pi Ui where Pi : average load of load point i

• •

System expected energy not supplied index, EENS (MWhr/yr): EENS = Total energy not supplied by the system = EENS i Expected interruption cost index at load point, ECOSTi (k$/yr): f rij λe, j

ECOST i = Pi j Ne

where f(rij) : SCDF

• •

System expected interruption cost index, ECOST(k$/yr): ECOST = ECOST i Interrupted energy assessment rate index at load point, IEARi ($/kWh): IEARi =

• •

Average service unavailability index, ASUI (pu): ASUI = 1 − ASAI Average energy not supplied index, AENS (MWhr/customer.yr): AENS =

•

ECOST i EENS i

Total energy not supplied by the system = Total number of customer served

EENS i Ni

System interrupted energy assessment rate index, IEAR ($/kWh): ECOST IEAR = EENS

755

27 Numerical Integration Methods Fundamental to the computer modeling of power-system transients is the numerical integration of the set of differential equations involved. This chapter identifies and defines the properties required from the numerical integration method in the context of power-system analysis.

27.1

Accuracy

This property is limited by two leading causes: round-off and truncation errors. Round-off error occurs while performing arithmetic operations and is due to the inability of the computer to represent numbers accurately. A word length of 48 bits usually is sufficient for scientific work and is acceptable for transient stability analysis. When stability studies are carried out on computers with a 32-bit word length, it is necessary to use double precision on some regions of the storage to maintain adequate accuracy. The difference between the exact and calculated results is mainly determined by the truncation error, which is due to the numerical method chosen. The exact solution at any one point can be expressed as a Taylor series based on some initial point, and by substituting these into the formulae, the order of accuracy can be found from the lowest power of step length h, which has a non-zero coefficient. In general terms, the truncation error T(h) of a method using a step length h is given by T h = O hp + 1 where superscript p represents the order of accuracy of the method. The true solution y(tn) at tn is thus y tn = yn + O h p + 1 + εn where yn : value of y calculated by the method after n steps εn : other possible errors

27.2

Stability

Two types of instability occur in the solution of ordinary differential equations: inherent and induced instability. Inherent instability occurs when, during a numerical step-by-step solution, errors generated by any means (truncation or round-off ) are magnified until the exact solution is swamped. Fortunately, transient stability studies are formulated in such a manner than inherent instability is not a problem. Induced instability is related to the method used in the numerical solution of the ordinary differential equation. The numerical method gives a sequence of approximations to the true solution, and the stability of the method is basically a measure of the difference between the approximate and exact solutions as the number of steps becomes large.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

756

27 Numerical Integration Methods

Consider the ordinary differential equation py = λy with the initial conditions y(0) = yo, which has the solution y(t) = y0eλt. Note that λ is the eigenvalue of the single variable system given by the ordinary differential equation: py = λy This may be solved by a finite difference equation of the general multistep form k

k

αi yn− i + 1 − h i=0

βi pyn −i + 1 = 0 i=0

where αi and βi are constants. Letting k

k

αi z i and σ z =

mz = i=0

βi z

i

i=0

and constraining the difference scheme to be stable when λ = 0, then the remaining part of the equation k

k

αi yn− i + 1 − h i=0

βi pyn −i + 1 = 0 i=0

is linear, and the solutions are given by the roots zi (for i = 1, 2,. …, k) of m(z) = 0. If the roots are all different, then yn = A1 z1

2

+ A2 z2 n …Ak zk

n

and the exact solution, in this case (λ = 0), is given by y tn = A1 z1

n

+ 0 h p + 1 = y0

where superscript p is the order of accuracy. The principal root z1, in this case, is unity, and instability occurs when |zi| ≥ 1 (for i = 2, 3,. …, k, i 1), and the exact solution will eventually be swamped by this root as n increases. If a method satisfies these criteria, then it is said to be stable, but the degree of stability requires further consideration. Weak stability occurs where a method can be defined as being stable, but because of the nature of the differential equation, the derivative part of equation gives one or more roots that are greater than or equal to unity: k

k

αi yn− i + 1 − h i=0

βi pyn −i + 1 = 0 i=0

A stable method that has the maximum order of accuracy is always weakly stable. The maximum order or accuracy of a method is either k + 1 or k + 2 depending on whether k is odd or even, respectively. Partial stability occurs when the step length (h) is critical to the solution and is particularly relevant when considering Runge–Kutta methods. In general, the roots zi of equation yn = A1(z1)n + A2(z2)n … Ak(zk)n are dependent on the product hλ and also on the equation k

αi z

mz =

i

i=0

The stability boundary is the value of hλ for which |zi| = 1, and any method that has this boundary is termed conditionally stable. A method with an infinite stability boundary is known as A-stable (unconditionally stable). A linear multistep method is A-stable if all solutions of the equation k

k

αi yn− i + 1 − h i=0

βi pyn −i + 1 = 0 i=0

tend to zero as n ➔ ∞ when the method is applied with fixed h > 0 to equation py = λy where λ is a complex constant with Reλ < 0.

27.4 Predictor–Corrector

For a multistep method to be A-stable, the order of accuracy cannot exceed p = 2, and hence the maximum k is unity, i.e. a single-step method. Backward Euler and the trapezoidal method are A-stable, single-step methods. Other methods not based upon the multistep principle may be A-stable and also have high orders of accuracy. In this category are implicit Runge–Kutta methods in which p < 2r, where r is the number of stages in the method. A further definition of stability has been introduced recently using Z-stability, which is the multivariable version of A-stability. The two are equivalent when the method is linear but may not be equivalent otherwise. Backward Euler and the trapezoidal method are C-stable single-step methods. The study of scalar ordinary differential equations of the form py = λy is sufficient for the assessment of stability in coupled equations, provided that λ are the eigenvalues of the ordinary differential equations. Unfortunately, not all equations used in transient stability analysis are of this type.

27.3

Stiffness

A system of ordinary differential equations in which the ratio of the largest to the smallest eigenvalue is very much greater than unity is usually referred to as being stiff. Only during the initial period of the solution of the problem are the largest negative eigenvalues significant, yet they must be accounted for during the whole solution. For methods that are conditionally stable, a very small step length must be chosen to maintain stability. This makes the method very expensive in computing time. The advantage of Σ-stability thus becomes apparent for this type of problem, as the step length need not be adjusted to accommodate the smallest eigenvalues. In an electrical power system, the differential equations that describe its behavior in a transient state have greatly varying eigenvalues. The largest negative eigenvalues are related to the network and the machine stators, but these are ignored by establishing algebraic equations to replace the differential equations. The associated variables are then permitted to vary instantaneously. However, the time constants of the remaining ordinary differential equations are still sufficiently varied to give a large range of eigenvalues. It is, therefore, important that if the fastest remaining transients are to be considered and not ignored, as so often done in the past, a method must be adopted that keeps the computation to a minimum.

27.4

Predictor–Corrector

These methods for the solution of the differential equation pY = F(Y, X) with Y(0) = Y0 and X(0) = X0 have all developed from the general k-step finite difference equation. Basically, the methods consist of a pair of equations, one being explicit (β0 = 0) to give a prediction of the solution at tn + 1 and the other being implicit (β0 0), which corrects the predicted value. There are a great variety of methods available, the choice being made based on the requirements of the solution. It is usual for simplicity to maintain a constant step length with these methods if k > 2. Each application of a corrector method improves the accuracy of the method by one order, up to a maximum given by the order of accuracy of the corrector. Therefore, if the corrector is not to be iterated, it is usual to use a predictor with an order of accuracy one less than that of the corrector. The predictor is thus not essential, as the value at the previous step may be used as a first crude estimate, but the number of iterations of the corrector may be large. While, for accuracy, there is a fixed number of relevant iterations, it is desirable for stability purposes to iterate to some predetermined level of convergence. The characteristic root (z1) of a predictor or corrector when applied to the single variable problem py = λy and y 0 = y0

27 1

may be found from k

αi −hλβi z k −i = 0

i=0

When applying a corrector to the problem defined by Eq. (27.1) and rearranging Eq. (27.2) to give yn + 1 =

−

k i=1

αi −hλβi yn− i + 1 α0 −hλβ0

27 2

757

758

27 Numerical Integration Methods

the solution to the problem becomes direct. The predictor is now not necessary as the solution only requires information about y at the previous steps, is at yn = i + 1 for i = 1, 2, …. k. Where the problem contains two variables, one non-integrable, such that py = λy + μx with y 0 = y0 and x 0 = y0 0 = g y,x then yn + 1 = cn + 1 + mn + 1 xn + 1 where Cn + 1 =

−

k i=1

αi −hλβi yn− i + 1 −hμβi xn− i + 1 α0 − hλβ0

and mn + 1 =

−hμβ0 α0 − hλβ0

The solution is iterative, although cn + 1 and mn + 1 are constant at a particular step. The convergence of this method is now a function of the nonlinearity of the system. Provided that the step length is sufficiently small, a simple Jacobi form of iteration gives convergence in only a few iterations. It is equally possible to form a Jacobian matrix and obtain a solution by a Newton iterative process, although the storage necessary is much larger and, as before, the step length must be sufficiently small to ensure convergence. For a multivariable system, equation pY = F(Y, X) with Y(0) = Y0 and X(0) = X0 is coupled with 0 = G(Y, X), and the solution of the integrable variables is given by the matrix equation Yn + 1 = Cn + 1 + Mn + 1[Yn + 1, Xn + 1]t. The elements of the vector Cn + 1 are given in the equation Cn + 1 =

−

k i=1

αi −hλβi yn− i + 1 −hμβi xn− i + 1 α0 − hλβ0

and the elements of the sparse matrix Mn + 1 are given in the equation mn + 1 =

−hμβ0 α0 − hλβ0

The iterative solution may be started at any point in the loop if Jacobi iteration is used. Because the number of algebraic variables (X) associated with equations pY = F(Y, X) with Y(0) = Y0 and X(0) = X0 or 0 = G(Y, X) are small, it is most advantageous to extrapolate these algebraic variables and commence with a solution using the equation 0 = G (Y, X). The disadvantage of any multistep method (k > 2) is that it is not self-starting. After a discontinuity, (k – 1) steps must be performed by some other self-starting method. Unfortunately, it is the period immediately after a step that is most critical as the largest negative eigenvalues are significant. As (k – 1) is usually small, it is not essential to use an A-stable starting method. Accuracy over this period is of more importance.

27.5

Runge–Kutta

Runge–Kutta methods are able to achieve high accuracy while remaining single-step methods. This is obtained by making further evaluation of the functions within the step. The general form of the equation is v

Yn + 1 = yn +

w i ki i=1

27.5 Runge–Kutta

where v

ki = hf tn + ci h, yn +

aij kj j=1

for i = 1, 2, … , v, and vi= 1 wi = 1 These methods are self-starting, and the step length need not be constant. If j is restricted so that j < i, then the method is explicit and ci must be zero. When j is permitted to exceed i, then the method is implicit, and an iterative solution is necessary. Also of interest are the forms developed by Merson and Scraton. These are fourth-order methods (p = 4) but use five stages (v = 5). The extra degree of freedom obtained is used to give an estimate of the local truncation error at that step. This can be used to control the step length automatically. Although they are accurate, the explicit Runge– Kutta methods are not A-stable. Stability is achieved by ensuring that the step length does not become large compared with any time constant. For a pth order explicit method, the characteristic root is p

z1 = 1 +

v 1 i i ai i i hλ hλ + i i i=1 i=p+1

where the second summation term exists only when v > p and where ai is constant and dependent on the method. For some implicit methods, the characteristic root is equivalent to a Pade approximant to ehλ. The Pade approximant of a function f(t) is given by M j j = 1 aj t N j j = 1 bj t

PMN f t = If ∞

f t =

cj , t j

j=0

then f t −PMN f t =

∞

M j=0 N j=0

N

cj t

bj t −

j

j=0

j

j=0

aj t j bj t j

If the approximant is to have accuracy of order M + N, and if f(0) = PMN(f(0)), then ∞ j=0

N j=0

∞

M

bj t j −

cj t j

aj t j j=0

=

dj t j

j=M+N +1

It has been demonstrated that for approximations of hλ where M = N, M = N + 1 and M = N + 2, the modulus is less than unity, and thus a method with a characteristic root equivalent to these approximants is A-stable as well as having an order of M + N.

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28 Optimization Kirchhoff’s current law requires that the sum of the currents injected and withdrawn at bus n equal zero. If we define current to ground to be yn0(vn − v0) and v0 = 0, we have in =

kynmk vn − vm + yn0 vn

r j r j inmk = ynmk vn − vm = gnmk vnr − vm −bnmk vnj − vm + j bnmk vnr − vm + gnmk vnj − vm r r j inmk = gnmk vnr − vm −bnmk vnj − vm j

r j + gnmk vnj − vm inmk = bnmk vnr −vm

Since current is a linear function of voltage, we can rearrange as in = vn yn0 +

kynmk −

kynmk vm

The current and voltage flow equations (linear) in matrix form are I = YV = G + jB V r + jV j = GV r − BV j + j BV r + GV j where Ir = GV r − BV j and I j = BV r + GVj Another way to rearrange is I = I r ,I j = Y V r , V j

T

or I = I r , I j =

G −B B G

Vr V

j

where Y =

G −B B G

If a and φ are constant, the I = YV equations are linear. If not, the linearity is lost since some elements of the Y matrix will be functions of V.

28.1

Power-Flow Injections

Power-flow equations (quadratic) defined in terms of real power, reactive power, and voltage are S = P + jQ = diag V I ∗ = diag v YV ∗ = diag V Y ∗ V ∗ The power injections (quadratic) are S = V I ∗ = V r + jV j

I r − jI j = V r I r + V j I j + j V j I r − V r I j

where P = Vr Ir + Vj Ij Q = V j Ir + V r I j

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

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28 Optimization

28.2

Voltage Magnitude Constraints

Two constraints limit voltage magnitude in rectangular coordinates at each bus m: r vm

2

min vm

j + vm 2

2

r ≤ vm

max ≤ vm

2

2

2

j + vm

28 1 28 2

Each nonlinear inequality involves voltage magnitudes at bus m only. In terms of matrix notation, voltage magnitude constraints are V r V r + V j V j ≤ V max V

min 2

min

≤V V +V V r

r

j

2

28 3

j

28 4

max

V and V are determined by power-system studies. The voltage magnitude bounds are generally between 95% and 105% of nominal voltage of each bus. High voltages are often constrained by capabilities of the circuit breakers. Low-voltage-magnitude constraints can be due to operational requirements of motors or generators.

28.3

Line-Flow Thermal Constraints

Skmax is a thermal transmission limit on k based on the temperature sensitivity of the conductor and supporting material in the transmission line and transmission elements. Transmission assets generally have three thermal ratings: steady-state, ½ hour, and 4 hours. These ratings vary with ambient temperature. The apparent power at bus n on transmission element k to bus m is snmk = vn i∗nmk = vn y∗nmk vn − vm

∗

r The thermal limit on snmk is Snmk

= vn y∗nmk v∗n − vn y∗nmk v∗m 2

2

j

+ Snmk

28 5 2

= Snmk 2 ≤ Skmax .

j

r r j These constraints are quadratic in Snmk and Snmk quartic in vnr , vnj , vm ,vm . j r Since vn = vn + jvn and ynmk = gnmk + jbnmk, j r + jvm vn y∗nmk v∗m = vnr + jvnj gnmk + jbnmk vm

Expanding, vn y∗nmk v∗m = vnr + jvnj

r j j r −bnmk vm −j gnmk vm + bnmk vm gnmk vm

In matrix notation, Re vn y∗nmk v∗m = vnr , vnj Im vn y∗nmk v∗m = vnr , vnj

gnmk − bnmk

r vm

bnmk

j vm

gnmk

−bnmk −gnmk

r vm

− bnmk

j vm

gnmk

Similarly, vn y∗nmk v∗n = vnr + jvnj gnn − jbnn vnr − jvnj Expanding, we obtain = vnr + jvnj

gnn vnr −bnn vnj −j gnn vnj + bnn vnr

= gnn vnr vnr + vnj vnj −jbnn vnr vnr + vnj vnj

28 6

28.5 Line-Flow Constraints as Voltage Angle Constraints

In matrix notation, Re vn y∗nmk v∗n = vnr , vnj Im vn y∗nmk v∗n = vnr , vnj

28.4

gnn

vnr

0

28 7

vnj

0 gnn − bnn

0

vnr

0

− bnn

vnj

28 8

Line-Flow Constraints as Current Limitations

As current increases, lines sag and equipment may be damaged by overheating. The constraints that limit the current magnitude in rectangular coordinates at each bus n or k are r inmk

2

j

+ inmk

2

max ≤ inmk

2

28 9

Again, the nonlinearities are convex quadratic and isolated to the complex current at the bus. Generally, the max maximum currents inmk are determined by material science properties of conductors and transmission equipment, or as a result of system stability studies.

28.5

Line-Flow Constraints as Voltage Angle Constraints

The power flowing over a transmission line is approximately proportional to the sine of the voltage phase angle difference at the receiving and transmitting ends. For stability reasons, the voltage angle difference for terminal buses n and m connected by transmission element k can be constrained as follows: min max θnm ≤ θn − θm ≤ θnm

In the rectangular formulation, the arctan function appears in some constraints: min ≤ arctan θnm

j vnj vm max − arctan ≤ θnm r vnr vm

The theoretical steady-state stability limit for power transfer between two buses across a lossless line is 90 . If this limit were exceeded, synchronous machines at one end of the line would lose synchronism with the other end of the line. In addition, transient stability and relay limits on reclosing lines constrain voltage-angle differences. The angle constraints used in optimal power flow (OPF) should be the smallest of these angle constraints, which depend on the equipment installed and configuration. In general, system engineers design and operate the system comfortably below the voltage angle limit to allow time to respond if the voltage angle difference across any line approaches its limit. Readers are encouraged to refer to Part A of this book for detailed theory of power-system devices, analytics, and power electronics as well as Part C for power-system analysis applications and examples for a more complete understanding of power-system theory and applications.

763

765

Part C Analytical Practices and Examples using ETAP

767

29 Introduction to Power System Analysis Electrical engineers, globally, are concerned with the complete process of generation, transmission, distribution, and utilization of power and energy. The electric/energy industry is the largest and most complex industry in the world where electrical engineers encounter challenging problems. These problems include designing future power systems to deliver increasing amounts of electrical energy in a safe, clean, and economical manner. For example, when electricity stops flowing in a manufacturing plant, the plant’s production stops. Plant owners typically invest in the most modern production machines, have an ample inventory of raw material, product design, highly trained and efficient labor, and everything else that is required to produce manufactured goods quickly and at low cost. However, if electric power is not available when and where needed in the plant, then the owner’s investment in the plant and inventory becomes idle capital. Electric power distribution is a vital link that carries electric power from the utility supply point to the production machine. Electric power distribution in a plant generally costs about 5% of the total plant cost, including process machinery. Since we get so much for so little from the investment in the plant power system, it pays to use only the best practices and analysis equipment to assure maximum profit from the overall plant investment.

29.1

Planning Studies

To obtain a power system that is adequate to meet the service reliability requirements of a plant and yet that is lowest in cost requires that the power-system engineer plan the system on an overall inclusive basis. System planning is split into two parts: conceptual and detailed design. System studies are carried out during the planning stage to analyze and evaluate that the design is optimum and it will be practical to build and operate. Electrical power-system conceptual design is the synchronized, balanced integration of various discrete but competing system-design requirements to meet the operational objectives economically. Best practice to achieve overall, credible performance of a power system is to develop a virtual system model via an intelligent one-line diagram and introduce the means to keep it up to date regularly. Thus, in the event of any failure of the system, quick problem identification and resolution may be done, resulting in minimal system downtime and increased productivity. The electrical system one-line diagram must be updated to illustrate the power system accurately. The one-line diagram is a vital maintenance document in any electrical system. When any modifications/retrofits or additions are made to a power system, the one-line diagram should be updated immediately to show that change. All persons concerned with the maintenance and operation of the electrical system should have access to the latest revised copies on a regular basis. These diagrams should be reviewed and updated periodically. Safety is an important aspect that must be maintained by utilizing adequate circuit protective equipment. Shortcircuit and load-flow analysis are typically performed during initial power-system design and operation whenever system changes are made, to ensure that device duty and continuous operational ratings are adequate for safe and reliable operation of the system. Designing the electrical system so that working on energized conductors is not necessary and determining the appropriate personal protective equipment (PPE) required to be worn in the event work must be safely performed on energized equipment.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

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29 Introduction to Power System Analysis

Multiple “what if” scenarios are created to ensure that an alternate power supply is available to minimize downtime. Under alternate supply conditions, the electrical system is analyzed to ensure adequate device duty and overload handling capacity are available. With an up-to-date one-line diagram, a power-system engineer can quickly analyze multiple conditions and make cost comparisons to determine the reasonable trade-off between economics and safe operation of the electrical network. For example, using simulation tools, it is critical while designing or retrofitting existing systems to determine whether a new transformer is required and then analyze the trade-offs of investing in a new transformer or using other methods like load-tap changers, capacitor banks, or synchronous condensers to improve overall system voltage.

29.2

Need for Power-System Analysis

Load-flow, short-circuit, and device coordination analyses must be continuously performed during the operational lifetime of an electrical system plant since each year the power demand increases by at least 5%. These analyses are always done in the following situations:

•• •

New electrical system design Retrofits to existing electrical system Electrical network failure

These power-system analyses are essential to any electrical system design and evaluation. However, there are multiple specialized, purpose-served analyses covered in this book, which must be performed as the need arises. Considering retrofits or load growth as one case, for example, if a new motor control center (MCC) or substation is to be added to an existing electrical train, the loading will increase, raising the stress on existing electrical devices. Old equipment cannot be retired all at once; hence, proper analysis has to be done to make sure existing equipment can still be used under new operating load.

29.3

Computers in Power Engineering

As electrical systems, especially power utilities, have grown in size, and the number of interconnections has increased, planning for future expansion has become increasingly complex. The increasing cost of additions and modifications has made it vital that electric utilities consider a range of design options and perform detailed studies of the effects on the system of each option, based on some assumptions: standard and abnormal operating conditions, peak and off-peak loadings, and present and future years of operation. A large volume of network data must also be collected and accurately included in a virtual system model. Highly developed computer programs have been developed in order to assist power-system planning and operation tasks. Such programs include load-flow, transient-stability, short-circuit, and fast-transient simulations.

29.4

Study Approach

The 10 significant steps in a typical electric system study are as follows: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)

Define the purpose and scope. State assumptions and methodology. Establish performance or evaluation criteria. Perform technical analysis, and establish a baseline/benchmark system. Develop alternative or “what if” scenario cases. Conduct a screening study to eliminate redundant or low-impact “what if” cases. Perform a detailed technical analysis of alternatives. Perform an economic analysis. Perform a sensitivity analysis. Select or recommend a cost-effect alternative for implementation.

29.4 Study Approach

The steps follow a logical sequence, but in practice, it may be necessary to bypass some of them. Further, these steps may also be iterative. For example, all the factors in the evaluation criteria may not have been defined up front. Though these steps may be performed formally or informally, nonetheless it is always recommended to document the steps, especially when solving complex systems. The thoroughness of following these steps and documenting the details is dependent upon the importance and cost magnitude of the system being analyzed. Every study has to start with an initial system established as a baseline or benchmark. For the majority of planning studies, a partial or complete existing system is used as a starting point. For such systems, the network topology, device types, engineering properties, measurements, historical events, and network connectivity are readily available, though most of the time they may be out of date depending upon the record-keeping practice. For new electrical systems, an initial computer-based virtual model of the system has to be put together based on the proposed design, experience, and similar concepts used in other facilities. In the screening stage, only major performance requirements are evaluated, and “what if” cases not meeting the requirements are eliminated or deferred for future consideration. Also, widely varying cost alternatives are discarded. An alternative that is many times the cost of another alternative need not be evaluated in detail or deferred until the lessexpensive alternative is found to be unacceptable. One of the critical points for an electrical engineer is to evaluate the performance of the electrical system and equipment, whether existing or new. It is, of course, desirable due to economic reasons to utilize every component up to its maximum potential in a safe manner. For this reason, new equipment may also be designed and constructed with smaller margins or reserves from physical loading and performance limits. It is accomplished by performing technical analysis or power-system analysis. The calculations required for this type of analysis can become tedious and complicated. These are calculations that cannot be accurately performed using calculators, hand calculations, or rule-of-thumb lookup tables. Access to and familiarity with digital computers has, therefore, become prevalent over the last 30 to 40 years to perform power-system analysis. The engineer working with digital computers for power-systems analysis should become familiar with system modeling assumptions, software modeling techniques, methods of analysis, data-entry guidelines, interpretation of results, and more. The types of studies commonly performed as the part of power-system studies include but not limited to load flow, short circuit, motor acceleration, system stability, protection and coordination, arc flash, transients and switching, reliability, harmonics, subsynchronous resonance, etc. Based on the sector and power-system design, specialized power-system studies have evolved to better understand system performance, limits, distinct cases, conditions, etc. For example, in addition to the studies mentioned, it is common to perform transmission-related studies like contingency, voltage stability, small signal stability, optimal power flow, insulation coordination, available transfer capacity, etc. For distribution systems, in addition to the core set of studies, it is common to perform additional studies including time series power flow, volt/var optimization, feeder balancing and reconfiguration, load allocation, feeder-hosting capacity for distributed energy resources (DER), and more. These studies primarily address the performance of the electrical power system during the period under consideration. For example, lightning is an impulse with a time-period of microseconds to milliseconds. A switching transient created by turning on/off a capacitor bank or submarine cable has a time-period of milliseconds. The thermal loading of electric equipment has a time-period of minutes to hours. Therefore, it is not necessary to consider lightning surges when determining the heating from load carried by a cable. Table 29.1 summarizes the electrical phenomena and study tools for various periods. Generally, an important concept to remember is that the severity of the applied quantity (voltage or current) is inversely related to the time duration of the application. Some of the critical features of these studies are summarized in Table 29.2. The electrical engineer should have specialized knowledge in performing these studies. The studies described in the remaining sections will assume the reader understands basic electrical engineering concepts and is familiar with linear and nonlinear circuit analysis, equivalent circuits, the per-unit system, and network theorems including but not limited to Thévenin, Norton, and Superposition. DC system analysis based on Ohms law and Kirchhoff’s law is much more straightforward than AC network analysis. Most of the examples use 50 or 60 Hz power-system frequency for sinusoidal waveforms. Phasor representation of these AC voltages and currents for three-phase and single-phase systems is also assumed. Knowledge of symmetrical components is essential for unbalanced system analysis. For more sophisticated analysis such as transient stability, knowledge of differential equations is essential; for calculating system response, knowledge of Laplace transformations, time and frequency domain concepts is needed; and for calculating harmonics in a waveform, knowledge of Fourier

769

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29 Introduction to Power System Analysis

Table 29.1 Power-system phenomena, studies, and suggested tools. Time scale

Phenomena

Power-system study

Suggested tools

Microseconds to milliseconds

Lightning impulse Switching surges Transients Relaying Protection Power swing

Electromagnetic transient (EMT)

PSCAD-EMTDC

Short-circuit Relay coordination Transient stability Load shedding Arc flash Impact start or power flow based Transient stability Automatic generation control Economic dispatch Microgrid control Energy storage dispatch Demand management Harmonic analysis

ETAP

Milliseconds to cycles

Seconds

Motor starting

Seconds to minutes

Operations

Minutes

Wave distortion

Hours Days Weeks

Operations planning

Hours days weeks years

Operations

System Planning

Online EMS/PMS or ADMS coupled online analysis Load flow Short circuit Stability Switch plan and optimization Unit commitment Load forecasting Load growth Load forecasting Economic analysis

ETAP ETAP

ETAP ETAP

ETAP

ETAP

Table 29.2 Power-system studies and significant features. Type of study

Significant features

Load flow or power flow

Steady-state analysis Uses network analysis methods Used to determine MVA or current rating of lines, cables, and transformers Voltage regulation Voltage drop Load tap changer, range, step size, control Shunt capacitor, location, size Losses Easy and fast to run on a PC Can run many cases to determine system performance for different circuit configurations and contingencies Output in both graphical and tabular format Used as the basis for stability and motor-starting studies Transformer and conductor sizing

•• •• ••

29.4 Study Approach

Table 29.2 (Continued) Type of study

Significant features

Short-circuit or fault studies

Steady-state analysis with fault conditions Uses network analysis techniques Used to determine Fault currents for three-phase, line-line, line-ground, and line-line-ground events Circuit-breaker and protective device duty Relay settings Protective device coordination Transformer and conductor sizing

Motor starting

Dynamics and stability

Protection and coordination

Transient and switching

Reliability

Harmonic

•• •• • •• •• ••

Quasi-steady state Used to determine Voltage drop during starting Load connected starting characteristics Acceleration time Coordination and protection Special starting methods Transformer and conductor sizing Dynamic analysis with disturbances Uses differential equations and network analysis Used to determine Synchronous stability Induction motor stability Corrective actions for unstable cases

•• • •• •• •

Used to determine Relay settings, types, and numbers Time-overcurrent and distance protection Fuse ratings Low-voltage circuit breaker trip unit settings CT sizing Transient analysis Uses differential equations or analog simulation Used to determine Transient overvoltages Breaker interrupting capability Surge arrester ratings Equipment failure investigation

•• ••

Steady-state (long-term) analysis Uses deterministic or probabilistic methods Used to determine Risk (qualitative/quantitative) Effect of outages/failures Redundancy needs

•• • •• ••

Uses Fourier transformation and network-solution techniques Used to determine Distortion, shunt filter design Resonance/overvoltage Flicker Telephone Interference

771

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29 Introduction to Power System Analysis

transformation and more. The reader is referred to many excellent introductory textbooks for network and controlsystem topics.

29.5

Operator Training

Many total system shutdowns have resulted from complicated systems and unreliable or incomplete electrical information available in one place. Operators do not get emergency switching practice every day and are bound to make mistakes more often in emergency conditions. The growing difficulty of obtaining adequately trained personnel justifies maintaining an electrical system model in a power simulator like Electrical Transient Analyzer Program (ETAP).

29.6

System Reliability and Maintenance

The power-system reliability or maintenance engineer should verify the availability of all major equipment regularly. The system design engineer can help by including suggested maintenance schedules in the power-system assetmanagement database and providing the information to the reliability engineer. For example, relay settings and operation should be tested/verified periodically (at least once a year). Integrated electrical engineering software solutions should be utilized that integrate a graphical power-system model with an advanced suite of protective device engineering libraries and management applications. Such systems include a relay-setting database that stores all protective device settings in a redundant central database and allows for modeling of all relevant workflows and sequences. Such integrated systems allow for viewing and modifying as-designed, as-built, as-tested, and as-operated power-system networks.

29.7

Electrical Transient Analyzer Program (ETAP)

ETAP is a nuclear-grade power-system analysis and system operation solution that accompanies this text. The purposes of integrating ETAP with the text are to provide computer solutions to examples in the text, to extend the examples, to demonstrate topics covered in the text, to provide a software tool for more realistic design projects, and to provide readers with experience using the foremost-integrated database for electrical systems. ETAP demo software may be downloaded from https://etap.com/demo-download. The remainder of this section provides the details for getting started with ETAP. ETAP allows the user to work directly with graphical one-line diagrams, underground cable raceway systems, threedimensional cable systems, innovative time-current coordination and selectivity plots, geographic information system schematics (GIS), as well as three-dimensional ground grid systems. ETAP design includes three key concepts. ETAP demo software may be downloaded from https://etap.com/demo-download. 29.7.1

Virtual Reality Operation

The ETAP virtual system model resembles a real electrical system operation as closely as possible. For example, when a circuit breaker is opened or closed, an element is placed out of service, or operating status of motors are changed, deenergized elements and subsystems are indicated on the one-line diagram in gray. ETAP incorporates new concepts for determining protective device coordination, contingency railway traction, distance protection, and more, directly from the one-line diagram. 29.7.2

Total Integration of Data

ETAP combines the electrical, logical, mechanical, and physical attributes of system elements in the same database. For example, a cable contains not only data representing its electrical properties and physical dimensions but also information indicating the raceways through which it is routed. Thus, the data for a single cable as shown in Figure 29.1 can be used for load-flow or short-circuit analyses (which require electrical parameters and connections) as well as cable-capacity calculations (which require physical routing data). This integration of the data provides consistency throughout the system and eliminates multiple data entry for the same element.

29.7 Electrical Transient Analyzer Program (ETAP)

Figure 29.1 Same cable represented in underground thermal analysis and power flow.

29.7.3

One-Line Diagrams

ETAP provides an easy-to-use, fully graphical user interface (GUI), as shown in Figure 29.2, for constructing one-line diagrams. It is a one-line representation of a three-phase system with multiple wires. This one-line diagram is the starting point for most studies, constructed by connecting the buses, branches, motors, generators, and protective devices. Elements may be connected graphically (by dragging lines from the device element) or by using the Info page of the Device Property Editor (double-click the element, and its property editor will open). Engineering properties of the element, such as its ratings, settings, loading, and connection may be defined. 29.7.4

Simplicity in Data Entry

ETAP keeps track of the detailed data for each electrical apparatus. Data editors can speed up the data-entry process by requiring the minimum data for a particular study. The ETAP one-line diagram supports many features to save time during modeling networks of varying complexities. 29.7.5

Multidimensional Database

ETAP organizes an electrical system into a single project. Within this project, ETAP creates significant system components, as shown in Figure 29.3:

• • • •

Presentations: Unlimited, independent graphical presentations of the one-line diagram that represent design data for any purpose (such as impedance diagrams, study results, or plot plans). Configuration: Unlimited, independent system configurations identify the status of switching devices (open and closed), motors and loads, and generator operating modes. Revision data: Base and unlimited revision data keep track of the changes and modifications to the engineering properties (e.g. nameplate or settings) of elements. Study case: Solution parameters stored for various “what if” cases including load and generation levels, system threshold limits, solution methods, and more.

These components are organized orthogonally to provide exceptional power and flexibility in constructing and manipulating power-system analysis scenarios.

773

774

29 Introduction to Power System Analysis

Figure 29.2 Electrical Transient Analyzer Program (ETAP) graphical user interface.

Base & rev. data Con

figu

on

ati ent

s

Pre

Loading

Generation

Study solutions Figure 29.3 ETAP multidimensional database.

rati

on

29.7 Electrical Transient Analyzer Program (ETAP)

29.7.6

Other ETAP Analysis Modules

ETAP includes over 65 power-system analysis modules. Some of the major ones are listed for the reader’s reference:

•• •• •• •• •• •• •• •• •• •• •• •

Unified AC and DC Power Flow Unified Time Series Load Flow Unbalanced Short Circuit Analysis Arc Flash Analysis Motor Acceleration Analysis Harmonic Analysis Transient and Voltage Stability Analysis Protective Device Coordination – Overcurrent and Distance/Impedance DC Load Flow, DC Short-Circuit, and DC Arc Flash Analysis Battery Sizing and Discharge Optimal Power Flow Analysis Reliability and Contingency Analysis Feeder Reconfiguration and Switching Optimization Fault Location, Isolation, and Service Restoration Volt/Var Optimization and Capacitor Placement Switching Sequence Management Railway Traction Power Analysis Underground Cable Thermal Analysis Control Circuit Diagram Analysis Ground Grid/Earth Mat Design Feeder Hosting Capacity Transfer Capacity Analysis Grid Code Evaluation

775

777

30 One-Line Diagrams 30.1

Introduction

Modern electric systems are customarily three-phase systems. The design of power distribution systems is such that equal loading on all the three phases of the network is ensured by allocating, as far as possible, similar domestic loads to each phase of three-phase low-voltage distribution feeders; while industrial loads are connected directly to threephase power supply. Thus, normal operation of the power system with three phases is almost balanced. Balanced three-phase networks are always solved as per-phase equivalent circuits. They are typically represented by one of the three phases and the neutral. Power-system devices such as transmission lines, generators, transformers, and system loads are visualized by standard symbols. The diagram may be further simplified by ignoring the neutral wire. Such a simplified diagram of an electric system is called a single-line or one-line diagram. A single-line diagram is a blueprint created for the purpose of visualization and electrical-system analysis. Modeling an electrical system via a single-line diagram is the first step in preparing a critical-response plan. Using a single-line diagram, a power-system engineer can become thoroughly familiar with the electrical-system layout and design. The single-line diagram also becomes a lifeline of information when updating or responding to an emergency. An accurate diagram ensures optimum system performance and coordination for all future testing and can highlight potential risks before a problem occurs. The one-line diagram is used in conjunction with other notational simplifications, such as the per-unit system and equipment ratings. Another advantage of utilizing a one-line diagram is that the simpler diagram may allow for nonelectrical information (economic data) to be included. Separate one-line diagrams are made for each of the positive-, negative-, and zero-sequence systems when modeling sequence networks. This simplifies the analysis of unbalanced conditions of a polyphase system. Devices that have dissimilar impedances for the different phase sequences are identified on the diagrams.

30.2

Engineering Parameters

Typical system parameters are indicated on the single-line diagram for each device type, as shown in the following tables. Device type

Rating

Generator

kW or MW and Subtransient reactance Xd”

Power grid (utility)

Short-circuit MVA (MVAsc) and Positive sequence impedance in % on 100 MVA base (R + jX)

Motor

HP or kW and % locked rotor current

Load / Panel

kW / MW, kVA or MVA and connection type (# of phases – # of wires)

Bus / Node

Bus bracing (kA)

Circuit breaker

Rated interrupting (kA)

Fuse

Interrupting (kA) (Continued)

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

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30 One-Line Diagrams

Device type

Rating

PT and CT

Transformer turns ratio

Relay

Relay function and settings

Branch, impedance

Resistance, reactance, and susceptance

Branch, reactor

Continuous amps

Transformer

kVA / MVA and Positive sequence impedance PS (%Z) for two-winding transformers and PS/PT/ST (%Z) for three-winding transformers

Lines and cables (size)

# of cables – # of conductor / cable – size

Busway and bus duct

# of conductors / phase

Device type

Rating

Impedance and reactor

Impedance in % (R + jX) or ohms

Voltage regulator

Impedance (%Z) (if included)

Line, cable, and bus duct

Positive sequence impedance (R + j X in ohms or per-unit length)

Charger

AC kVA and DC kW (or MVA / MW)

Inverter

DC kW and AC kVA (or MW / MVA)

UPS

kVA

VFD

HP / kW

Battery

Ampere-hour

Control circuit diagram

W / kW

Converter

kW / MW

Cable (Size)

# of cables – # of conductor / cable – size

Switch

Rated bracing current (kA)

The information to be provided on the single-line diagram depends on the type of analysis to be undertaken on the system. For example, to perform a load-flow analysis, information regarding power generation, load demands, line data, and transformer data are essential. However, to perform a transient-stability study of the system, additional details about machine parameters, the location of circuit breakers and relays, and the speed with which relays and circuit breakers operate to isolate the faulted part of the system are essential. The one-line diagram summarizes the relevant information about the system for the particular problem studied. For example, relays and circuit breakers are not necessary when dealing with a typical state problem. However, when fault conditions are considered, the location of relays and circuit breakers is essential and is thus included in the single-line diagram. Sometimes one-line diagrams include information on the current and potential transformers that are installed for connection to the relays and metering purposes. The formulation of power-system problems for load-flow, shortcircuit, and transient-stability analysis involves the formulation of suitable mathematical representations of all the components of the power system. For such computations, a one-line diagram of the system is first drawn, and then single-phase or per-phase equivalent circuits for the various electrical components are used to form the per-phase impedance diagram of the system. The perphase impedance diagram is sometimes called the per-phase positive-sequence diagram since it shows impedances as seen by positive sequence currents in one phase of a symmetrical three-phase system.

30.3

One-Line Diagram Symbols

For the sake understanding power-system single-line diagrams, certain conventions for representing various components are used. The International Electrotechnical Commission (IEC), the American National Standards Institute (ANSI), and the Institute of Electrical and Electronics Engineers (IEEE) have published a set of standard symbols for electrical diagrams. Graphical symbols must ideally conform to IEEE 315 or ISO 7000 or IEC 62648 and IEC 60417 (see Appendix C). A list of common graphic symbols is shown in Table 30.1.

30.3 One-Line Diagram Symbols

Table 30.1 Common graphic symbols based on IEC standards. Power grid

Busway (with impedance) and bus duct

Bus way 800 A Plug-in AL 256 ft

0.15987+j3.9968 % 220 kV Power grid 2500 MVAsc

Bus duct 400 A Isolated phase CU 300 ft

6.6 kA Power transformers

Dyn11

Two winding transformer 25 MVA 220/66 kV 10 %Z

Three winding transformer 200/150/50 MVA 220/66/33 kV 16/15/25 %Z

Scott-T transformer 20/20 MVA 138/27.5/27.5 kV 12.5/11/0 %Z

Open delta transformer 500 kVA 11/0.693 kV 5.2 %Z

Zig-zag grounding 6.6 kV

Autotransformers

Voltage regulator 1.5 MVA 11 kV Impedances

Resistor 0.1+j0.05% 100 MVAb

Inductor/reactor 0.056 ohms 300 A

AL

OH line 1 mile 0.09101+j0.48758 ohms/1mile WOLF

Note - The lines across the symbol designate phase information. In this example, it is a three-phase line.

Capacitor Switched capacitor 2×5 Mvar 2×5 Mvar 11 kV 11 kV 524.9 A 524.9 A

CU

Power cable 356 ft 6-1/C 14 3.1+j0.073 ohms/1000ft

Loads

Y Induction motor

1500 kW 4 kV 248.7 A 687 %LRC

Y Synchronous motor

5 MW 11 kV 297.9 A 650 %LRC

Const. impedance

Lumped load

560 kVA 0.4 kV 808.3 A

1.7 MVA 4 kV 245.4 A

Panel board 1Ph-3W 0.1 kV 100 A

(Continued)

779

780

30 One-Line Diagrams

Table 30.1 (Continued) Switching and protective devices

Power fuse 15 kV 2.5 kA 4.5 A SPST 27 kV 20.0 kA 600 A

Gnding switch

HVCB 1 31.5 kA 12 kV 2500 A

Open

ATS/SPDT 23 kV 400 A 8 kA Overload heater 49

HVCB 2 31.5 kA 12 kV 2500 A

Voltmeter

+ –

Overcurrent relay

l>

LVCB 0.22 kV 70 A 85 kA Contactor 400 A 4.8 kV

27/59 Voltage relay SST MV solid-state relay

Overload relay Multimeter

Ammeter

Current transformer +

Voltage transformer

Differential relay

d l>

Distance relay

21

R

Multi-function relay

Miscellaneous

Harmonic filter 3rd order damped

Y M/G

Rotary UPS 1500 kW 4 kV Static var comp. 251.5 A 230 kV 650 %LRC 125.51/87.86 A

HVDC 550/475/500 MVA

A standard symbol for three-phase Wye with the neutral solidly grounded is shown in Figure 30.1a. If a resistor or reactor is inserted between the neutral of the Wye point and ground, the appropriate symbol for resistance or inductance may be added to the standard symbol for the grounded Wye as shown in Figure 30.1b. If distribution transformer grounding is inserted between the neutral of the Wye point and ground, the appropriate symbol for resistance or inductance may be added to the standard symbol for the grounded Wye, as shown in Figure 30.1c. Most transformer neutrals in transmission systems are solidly grounded. Generator neutrals are usually grounded through current-limiting resistances or inductance coils during unbalanced short-circuits involving ground, as shown in Figure 30.1d. Power-system grounding is required information in order to calculate the ground fault current. For a load-flow study, the impedance diagram does not include the current-limiting impedances connected between the neutrals and the ground because the neutrals of the generators are at the same potential as the neutral of the system under balanced conditions. A typical single-line diagram for a simple power system is shown in Figure 30.2. The one-line diagram may serve as the basis for a circuit representation that includes the equivalent circuits of the components of the power system. Such a representation is called an impedance diagram, or a reactance diagram if resistances are ignored. A corresponding impedance diagram with resistance ignored and single-phase connections is shown in Figure 30.3. The corresponding reactance diagram with resistance ignored and single-phase connections are shown in Figure 30.4.

30.4

Power-System Configurations

Essential components of an electric power system are shown in Figure 30.5. The transmission system consists of a network of transmission lines and transmission substations, also called bulk power substations. The subtransmission system consists of step-down transformers, substations, and transmission lines that connect bulk power substations to distribution substations. In some cases, a line may be tapped, usually through a circuit breaker, to supply a single-customer distribution load such as a large industrial plant. Distribution substations include step-down transformers that lower subtransmission voltages to primary distribution voltages for local distribution. These transformers connect through the associated circuit breaker to substation buses,

30.4 Power-System Configurations

Generator

(a)

(b)

(c)

(d)

Figure 30.1 Grounding symbols.

G1

G4

T1

G2

Bus1

G5

T2

OH line Bus2

G3

Sub A

Sub B Load 2

Figure 30.2 Typical one-line diagram.

G1

G2

G3

G4

G5

G4

G5

Figure 30.3 Typical impedance diagram.

G1

G2

Transformer

Transformer

T1

T2

G3

Figure 30.4 Typical reactance diagram.

781

Power plant

Generator Step-up transformer

Circuit breaker

Transmission

Bulk power substation

Subtransmission

Step-down transformer

Distribution substation

Step-down transformer Distribution substation Primary distribution feeder

Sectionalizing fuse Primary lateral

Distribution transformer

Secondary main Utilization Figure 30.5 Components in a typical power system.

Distribution transformer Service conductor

30.4 Power-System Configurations

which in turn connect through circuit breakers to three-phase primary distribution lines called distribution circuits or feeders. Each substation bus usually supplies many feeders. Distribution substations may also include equipment for regulating the primary voltage, such as load-tap changers (LTCs) on the distribution substation transformers or separate voltage regulators. Feeders are usually separated into several three-phase sections connected through sectionalizing fuses or switches. Each feeder section may have several single-phase laterals connected to it through fuses. Three-phase laterals may also be connected to the feeders through fuses or reclosers. Separate, dedicated primary feeders supply industrial or large commercial loads. Feeders and laterals run along streets, as either overhead lines or underground cables, and supply distribution transformers that step the voltage down to the secondary distribution level. Distribution transformers are installed on utility poles for overhead lines, on pads at ground level, or in vaults for underground cables. From these transformers, energy flows through secondary mains and service conductors to supply single- or three-phase power directly to customer loads (residential, commercial, and light industrial).

30.4.1

Transmission and Distribution Substation Configurations

One of the most common or typical configurations used to connect a set of elements is that for substations and switching stations. Common transmission and distribution substation configurations are shown in Figure 30.6. Substations are the points in the power network where transmission lines and distribution feeders are connected through circuit breakers or switches via bus bars and transformers. Therefore, it is helpful to be familiar with some of the common substation layouts and their corresponding names. Substation types include transmission, distribution, collector, converter, switching, railways, and mobile.

L11

L12

L13

(a) Double bus single breaker Figure 30.6 Common transmission and distribution substation configurations.

L14

783

784

30 One-Line Diagrams Line13

Line14

L23

L24

+ –

T32

T33

R

R

R

R

Line18 Line19 Line20

R

R

Line15 Line16 Line17

L25

(b) Distribution substation

(c) Double bus double breaker

L1

L2

XFMR2

XFMR1

(d) Ring bus

L26

Incomer1

FDR1

Incomer2

FDR2

Incomer3

FDR3

(e) Breaker-and-half

Figure 30.6 (Continued)

30.4.2

Primary Distribution Configurations

Primary distribution systems include three basic systems: radial, loop, and primary meshed systems, as shown in Figure 30.7.

Primary feeder

Substation bus Recloser

Feeder

Fuse

Substation bus

Primary fuse or CB

Breaker

Lateral

Breaker

Sectionalizing fuse

Secondary circuit breaker Shunt capacitor bank

Service Service

Normally open tie switch

Service

Adjacent feeder Primary radial Figure 30.7 Common primary distribution configurations.

Primary selective – Dual fused switch

Individual primary feeder – Secondary unit substation

Substation bus

Service

Circuit breaker Substation bus

Substation bus

Primary feeder

Service Primary Recloser feeder 1 R

R Primary feeder

Breaker

Breaker Service Normally open tie switch (or tie recloser) Normally open switch

Primary feeder

Primary feeder 2

Fuse Distribution transformer Service

Recloser R

R

Breaker

Service

Transformer

Distribution transformer

Service Overhead primary loop

Figure 30.7 (Continued)

Underground primary loop

Primary meshed

30.5 Network Topology Processing

30.4.3

Secondary Distribution Configurations

Secondary distribution systems include four basic systems: per-customer transformer, common secondary main, secondary meshed, and spot network, as shown in Figure 30.8. Primary feeder

Primary feeder

Primary fuse or CB

Fuse or circuit breaker

Transformer Transformer Secondary cirucit breaker Service Service

Per customer transformer

Common secondary main

Substation bus Substation bus

Breaker Primary feeder

Breaker

Network transformer Network protector

Disconnect switch Network transformer Network protector Fuse Spot network bus

Service

Secondary meshed

Spot network

Figure 30.8 Common secondary distribution configurations.

30.5

Network Topology Processing

Once a single-line diagram has been created, it is possible to include advanced network topology processing and intelligent color coding, as shown in the examples that follow: 1) Phase-based coloring Phase color based on phase type (a, b, c, ab, bc, ca, abc) and a number of wires (1, 2, 3, 4, 5) can be automatically colorcoded as shown in Figure 30.9. 2-Phase 2 Wire

2-Phase 3 Wire

Phase AB

Phase AB

Phase BC

Phase BC

Phase CA

Phase CA

Figure 30.9 Phase-based color coding.

2) Area-based coloring In this case, two areas are automatically color-coded based on the fact that Sub A is Area 1 and Sub B is Area 2. Devices connected to Sub A and Sub B are colored differently based on device connectivity. 3) Voltage-based coloring Voltages less than the values specified in the table are automatically color-coded as shown in Figure 30.10.

787

Voltage ( 1 kV), the focus of grounding-system design is less on safety and more on the reliability of supply, the reliability of protection, and impact on the equipment in the presence of a short circuit. Preference of reliability over safety is chosen because these systems are not easily accessible to the general public. The magnitude of a line-ground short circuit, which is the most common, is significantly impacted by the choice of the grounding system. There are five types of neutral grounding, which can be color-coded as shown in Figure 30.11: Solid Ungrounded Resistance grounded Low-resistance (Z) High-resistance (Z) Reactance grounded Grounding transformers (such as the zigzag transformer)

•• •• •••

5) Earthing-based coloring for low-voltage systems In low-voltage networks, which distribute electric power to the broadest class of end users, the main concern for the design of earth systems is the safety of consumers who use electric appliances and protecting them against electric shocks. The earthing system, in combination with protective devices such as fuses and residual current devices, must ultimately ensure that a person does not come into contact with a metallic object whose potential relative to the person’s potential exceeds a safe threshold, typically set at about 50 V. International standard IEC 60364 distinguishes three families of earthing arrangements, using the two-letter codes TN, TT, and IT. The first letter indicates the connection between earth and the power-supply equipment (generator or transformer):

•• •• •

T: Direct connection of a point with earth (Latin: terra). I: No point is connected with earth (isolation), except perhaps via a high impedance. The second letter indicates the connection between earth or network and the electrical device being supplied: T: Earth connection is by a local direct connection to earth, usually via a ground rod. N: Earth connection is supplied by the electricity supply network, either as a separate protective earth (PE) conductor or combined with the neutral conductor.

In a TN earthing system, one of the points in the generator or transformer is connected with the earth, usually the star point in a three-phase system. The body of the electrical device is connected with earth via this earth connection at the transformer. This arrangement is a current standard for residential and industrial electric systems, particularly in Europe. The conductor that connects the exposed metallic parts of the consumer’s electrical installation is called PE. The conductor that connects to the star point in a three-phase system, or that carries the return current in a single-phase system, is called neutral (N). Three variants of TN systems are distinguished:

•• •

TN − S: PE and N are separate conductors that are connected only near the power source. TN − C: Combined PEN conductor fulfills the functions of both a PE and an N conductor. (on 230/400 V systems; commonly only used for distribution networks). TN − C − S: Part of the system uses a combined PEN conductor, which is at some point split up into separate PE and N lines. The combined PEN conductor typically occurs between the substation and the entry point into the building, and earth and neutral are separated in the service head. In the UK, this system is also known as protective multiple earthing (PME), because of the practice of connecting the combined neutral-and-earth conductor via the shortest practicable route to a local earth-rod at the source and each premises, to provide both system earthing and

789

790

30 One-Line Diagrams

equipment earthing at each of these locations. Similar systems in Australia and New Zealand are designated as multiple earthed neutral (MEN) and, in North America, as multigrounded neutral (MGN). In the United States National Electrical Code and Canadian Electrical Code, the feed from the distribution transformer uses a combined neutral and grounding conductor, but within the structure separate neutral and PE conductors are used (TN-C-S). The neutral must be connected to earth only on the supply side of the customer’s disconnecting switch. 6) Circuit Continuity-based coloring In this case, switching devices like circuit breakers are shown with their status (open or closed), and any part of the circuit not energized is indicated with gray, as shown in Figure 30.12. G1

G4 T5 Open

T4

Bus4 OH Line1

G2

G5

Bus3 Open T1

G3

T2

Bus1 OH Line Bus2

Open

Open Sub B

Sub A

Load 2

Figure 30.12 Dynamic circuit continuity.

30.6

13.5 kV U1 0 MVAsc

Illustrative Example – Per-Unit and Single-Line Diagram

Using the per-unit method theory discussed earlier, consider the one-line diagram shown in Figure 30.13, for which we need to calculate the corresponding impedance diagram. Here are the steps: 1) Choose 100 MVA to be the base MVA. 2) n stands for new values, and O stands for old values. For Transformer: T6, turn ratio: N1/N2 = 3.31, and X/R = 12.14:

Bus8 13.8 kV T6 5 MVA 13.8/4.16 kV

Xpu =

0 065 12 14 1 + 12 14

n o = Zpu Zpu

Bus9 4.16 kV

VBo VBn

2

2

= 0 06478 Rpu =

0 06478 = 0 005336 12 14

SBn = 5 33 × 10 − 3 + j0 06478 SBo

13 8 13 5

2

30 1 100 = 0 1115 + j1 3538 5 30 2

Z1 0.1+j1 ohms Bus10 4.16 kV

Z = 100 × Zpu = 11 15 + j135 38 3) For Impedance: Z1, the base voltage is determined with the transformer turn ratio: kV utility 13 5 VB2 4 0695 2 VB = = = 0 165608 = = 4 0695 ZB = 30 3 N1 3 31 MVA 100 N2 Zactual 0 1 + j1 = 0 6038 + j6 0382 = 0 1656 ZB Z = 100 × Zpu = 60 38 + j603 8

Figure 30.13 Single-line diagram for per-unit calculation.

a) Compare with the ETAP output. The ETAP branch connections report from the loadflow study is shown in Figure 30.14.

CKT/Branch ID

30 4

Zpu =

Load4 3 MVA 4.16 kV

Connected Bus ID Type

From Bus

% Impedance, Pos. Seq., 100 MVA Base To Bus

R

X

Z

T6

2W XFMR

Bus8

Bus9

11.15

135.38

135.84

Z1

Impedance

Bus9

Bus10

60.38

603.82

606.83

Figure 30.14 ETAP branch connections report.

Y

791

31 Load Flow 31.1

Introduction

The base study without completion of which no other studies can proceed is the power-flow, load-flow, or voltage-drop study. A load-flow study is a numerical analysis of the flow of electric power in an interconnected system regardless of the voltage level, frequency, or complexity. A power-flow study uses flow measurements such as MW, Mvar, power factor, amps, etc. denoted on a single-line or one-line diagram. Power-flow programs may use percent or per-unit values to perform computations, regardless of which the results should be very close to one another. Power-flow or load-flow studies are vital for planning the future expansion of power systems as well as determining the optimal operation of existing systems. A power-flow study provides the magnitude and phase angle of the voltage at each bus and the real and reactive power flow in each branch. Note that a branch may be a cable, transformer, overhead line, or current limiting reactor – any component or device that has specified impedance (Z) in ohms, per unit, or percent. Commercial power systems are usually too complex to allow for fast and accurate hand solution of the power flow. Specialized network analyzers were built between 1929 and the early 1960s to provide laboratory-scale physical models of power systems. The computer revolution in the 1980s resulted in these analyzers being available to the general public, and they are the backbone of all power-system studies performed today. A load-flow study is especially valuable for systems with multiple load centers, such as substations, motor control centers, switching stations, etc. Power-flow analysis gives an engineering insight into the system’s capability to adequately supply connected load and available margins. Total system losses, as well as individual line losses, also are tabulated. Voltage-correction equipment is selected and optimized to ensure adequate voltage at critical locations. Performing a load-flow study on an existing system provides insight and recommendations as to the system operation and optimization of control settings to obtain maximum capacity while minimizing operating costs. The results of such an analysis are in terms of active power, reactive power, magnitude, and phase angle. Furthermore, power-flow computations are crucial for optimal operations of groups of generating units. Therefore, the minimum output of any load-flow program is the following information:

•• •

Voltage magnitude and phase angle at each connection bus/node Real and reactive power flowing in each element/device Reactive power loading on each generator/source

31.2

Study Objectives

Load-flow analysis objectives tend to vary based on industrial, transmission, or distribution power systems; however, the core objectives tend to remain the same regardless of the system under analysis: 1) The most important information obtained from the load-flow analysis is the voltage profile of the system. Under unacceptably low voltage, the usual practice is to: a) Install capacitor banks in order to provide reactive compensation to the load. b) Install on-load-tap changers (OLTCs) or adjust fixed transformer tap settings. c) Conduct load-flow studies to determine how much reactive compensation is required, to bring the local voltage up to an appropriate level. Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

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31 Load Flow

2) Load-flow studies determine whether system voltages remain within rated insulation limits under normal or emergency operating conditions. 3) Determine equipment overloading such as transformers and conductors. 4) Optimize component or circuit loading. 5) Identify real and reactive power flow. 6) Minimize MW and Mvar losses. Load flow will also show how much reduction in losses will result from the new line (important for economic assessment). 7) Evaluate equipment rating and size equipment. This includes de-rating and/or re-rating the equipment based on manufacturer design specifications. 8) Determine the voltage and current imbalance between the phases that can cause unpredictable behavior, increasing the risk of equipment damage and unplanned outages. 9) Determine steady-state stability limits. 10) Select appropriate generator operating modes (isochronous/droop). 11) Identify circulating reactive power that results in heating and reduces system efficiency.

31.3

Problem Formulation

The load-flow problem is nonlinear because the power flow into load impedances is a function of the square of the applied voltages. Due to nonlinearity, in many cases, the analysis of large networks via an AC power-flow model is not feasible, and a linear (but less accurate) DC power-flow model is used instead. However, a DC power-flow model should not be used as the primary means for a power-flow study. DC power flow is used only where approximate solutions are needed in order to perform additional analysis using the DC solution as a base. This technique is mainly used in transmission system planning and, due to the elimination of reactive power, does not provide any accurate means to calculate and evaluate MVA limits. Due to the nonlinear nature of the load-flow problem, numerical methods are employed to obtain a solution that is within an acceptable tolerance iteratively. The power-flow solution begins with identifying the known and unknown variables in the system, as shown in Figure 31.1. Load-flow iterative solutions begin by assuming load voltage or initial conditions (VR). The typical initial condition value is 100% voltage and voltage angle δ = 0 . Main steps: 1) 2) 3) 4)

Assume VR Calculate: I = Sload / VR Calculate: Vd = I ∗ Z Recalculate: VR = VS −Vd VS = Voltage across source Vd = Voltage drop across system impedances (Z) VR = Voltage across load (S)

Current flowing through the load is calculated using the connected MVA and the assumed load voltage. This is the same current that flows through the impedance (Z). Therefore, the voltage drop can be calculated. Since the source voltage is known, the new voltage across the load can be calculated by subtracting the voltage drop (Vd) from the source voltage (VS). The resulting value from subtracting (VS − Vd) is VR again. This new value of VR is used again as the initial condition until the results are within the specified precision. Most load-flow programs include the number of iterations to solve and the solution precision as part of their case study or solution parameters. For time-invariant load-flow studies, loading is constant and defined by real and reactive power consumption. Generator terminal voltages are tightly regulated and therefore constant. The primary objective of the load flow study is to find the voltage magnitude of each bus and its angle when the powers generated and loads are prespecified. Vd

VS

VR

Figure 31.1 Iterative load-flow problem.

31.3.1

Generation and Load Bus Modeling

Consider a typical load-flow problem in which all the load demands are known. Even if the generation matches the sum total of these demands exactly, the mismatch

31.3 Problem Formulation

Table 31.1 Load-flow bus types. MVA (S)

Voltage (V)

Generator

Load-flow bus

P

Q

V

δ

Governor

Excitation

Swing or slack

Unknown

Unknown

Known

Known

Isochronous

AVRa

Voltage or PV

Known

Unknown

Known

Unknown

Droop

AVR

Load or PQ

Known

Known

Unknown

Unknown

–

No AVR

a

AVR = automatic voltage regulator.

between generation and load will persist because of the branch I2R losses. Since the I2R loss of a branch depends on the current, which, in turn, depends on the magnitudes and angles of voltages of the two buses connected to the branch, it is difficult to estimate the loss without calculating the voltages and angles. Therefore, a generator bus is usually chosen as the swing or slack bus without specifying its real power. It is assumed that the generator connected to this bus will supply the balance of the real power required and the line losses. To facilitate this, we classify the different buses of the power system as shown in Table 31.1. A slack or swing bus sets the angular reference for all the other buses. Since it is the angle difference between two voltage sources that dictates the real and reactive power flow between them, the particular angle of the slack bus is not essential. However, it sets the reference against which angles of all the other bus voltages are measured. Thus, the angle of this bus is usually chosen as 0 . Furthermore, it is assumed that the magnitude of the voltage of this bus is known. A voltage controlled bus has generator connections. Therefore, the power generation in such cases is controlled via a prime mover, while the terminal voltage is controlled through a closed-loop generator excitation or automatic voltage regulator (AVR). Keeping the input power constant through turbine-governor control and keeping the bus voltage constant using an AVR, we can specify constants P and V for these buses. This is why such buses are also referred to as PV buses. Note that the reactive power supplied by the generator Q depends on the system configuration and cannot be specified in advance. Furthermore, we have to find the unknown angle δ of the bus voltage. A load bus does not have any generator connections, and hence the generated real power P and reactive power Q are taken as zero. The load drawn by these buses is defined by real power -P and reactive power Q. This is why these buses are sometimes referred to as PQ buses. The objective of the load-flow study is to find the bus voltage magnitude V and its angle δ. 31.3.2

Modeling of Loads (ZIP)

I

31.3.2.1 Constant Power (P) Load

Constant-power loads include induction motors, synchronous motors, conventional loads, and unbalanced lumped loads (motor load portion), UPSs, and chargers. The power output remains constant for all changes in input voltage. Figure 31.2 shows I–V and P–V curves for a constant power load.

P

Motor load current

Running motor power output

V Vmin

Vmax

Vmin

Vmax

V

Figure 31.2 Constant kVA load.

31.3.2.2 Constant Impedance (Z) Load

Constant-impedance loads include static loads, capacitors, harmonic filters, motor-operated valves (MOVs), and conventional and unbalanced lumped loads (static load portion). The input power increases proportionally to the square of the input voltage. Figure 31.3 shows I–V and P–V curves for a constant impedance load.

I

P Load current

Load input power

31.3.2.3 Constant Current (I) Loads

Constant-current loads include unbalanced lumped loads (constant current load portion). The current remains constant for all changes in voltage. Figure 31.4 shows I–V and P–V curves for a constant current load.

V

Figure 31.3 Constant Z load.

V

793

794

31 Load Flow

Constant current

I

P

V

V

Figure 31.4 Constant I load.

31.4

3

2

1

4

5

Calculation Methodology

Consider the system shown in Figure 31.5, which has two generators and three load buses. Bus 1 is selected as the swing bus, and Bus 5 is the PV bus. Buses 2, 3, and 4 are PQ buses.

Figure 31.5 Simple five-bus one-line diagram.

Table 31.2 Bus loading and generation. Bus voltage

Power generated

Load

Bus

Mag. (%)

Angle ( )

P (MW)

Q (Mvar)

1

105

0

–

–

P (MW)

Q (Mvar)

0

0

2

100

0

0

0

96

62

3

100

0

0

0

35

14

4

100

0

0

0

16

8

5

102

0

48

–

24

11

Table 31.3 Line impedances and charging admittances.

Zs Line (bus to bus)

Impedance

Line charging (Y/2)

1–2

0.02 + j 0.10

j 0.030

1–5

0.05 + j 0.25

j 0.020

2–3

0.04 + j 0.20

j 0.025

2–5

0.05 + j 0.25

j 0.020

3–4

0.05 + j 0.25

j 0.020

3–5

0.08 + j 0.40

j 0.010

4–5

0.10 + j 0.50

j 0.075

Initial data including bus loading and generation is given in Tables 31.2 and 31.3. Let’s assume that an n-bus power system contains a total np number of PQ buses, while the number of PV (generator) buses is ng such that n = np + ng + 1. Consider the voltage source VS with a source (series) impedance of ZS, as shown in Figure 31.6. Using Norton’s theorem, this circuit can be replaced by a current source IS with a parallel admittance of YS, as shown in Figure 31.7.

+ Vs –

Figure 31.6 Voltage source behind impedance.

Is

Ys

Figure 31.7 Current source.

31.4 Calculation Methodology

The relations between the original system and the Norton equivalent are Is = Vs/Zs and Ys = 1/Zs. Based on this data, the Ybus matrix for the sample system is as follows: 1

2

3

4

5

1

2.6923 − j 13.4115

−1.9231 + j 9.6154

0

0

−0.7692 + j 3.8462

2

−1.9231 + j 9.6154

3.6538 − j 18.1942

−0.9615 + j 4.8077

0

−0.7692 + j 3.8462

3

0

−0.9615 + j 4.8077

2.2115 − j 11.0027

−0.7692 + j 3.8462

−0.4808 + j 2.4038

4

0

0

−0.7692 + j 3.8462

1.1538 − j 5.6742

−0.3846 + j 1.9231

5

−0.7692 + j 3.8462

−0.7692 + j 3.8462

−0.4808 + j 2.4038

−0.3846 + j 1.9231

2.4038 − j 11.8942

Note that sources and their internal impedances are not considered while forming the Ybus matrix for load-flow studies, which deal only with the bus voltages. Approximate load-flow results for each bus based on hand calculation are shown in Table 31.4, along with accurate results from ETAP, shown in Figure 31.8. Table 31.4 Approximate hand-calculation results for the system in Figure 31.5. Bus voltage Bus

Mag. (%)

1

105

Power generated Angle (0)

0

P (MW)

Q (Mvar)

126.60

57.11

Load P (MW)

Q (Mvar)

0

0

2

98.26

−5.0124

0

0

96

62

3

97.77

−7.1322

0

0

35

14

4 5

98.76 102

−7.3705

0

−3.2014

48

0 15.59

16

8

24

11

Figure 31.8 ETAP simulation results for the system in Figure 31.5.

31.4.1

Load-Flow Convergence

As in any iterative solution method, the convergence of the load-flow solution is affected by a number of factors specific to power systems.

795

796

31 Load Flow

31.4.1.1 Negative Impedance

Load-flow calculations may not converge if a significant value of negative impedance is used. As an example, the traditional method of modeling three-winding transformers by a Y equivalent model, using one impedance and two 2-winding transformers, sometimes results in a negative impedance value for one of the impedance branches. In this case, the negative impedance should be combined with other series circuit elements so that the result is a positive impedance value. 31.4.1.2 Negative Reactance

Series transmission line capacitance can create overall negative reactance on the branch element. 31.4.1.3 Zero or Very Small Impedance

A zero or minimal impedance value for any branch is not allowed since this will result in infinity or a substantial number in the system admittance matrix. Such impedance should be represented by a tie circuit breaker. 31.4.1.4 Widely Different Branch Impedance Values

Widely different branch impedance values on the same per-unit base may result in slow convergence. To avoid this situation, various techniques, such as combining series branches with low impedance values, ignoring short-length transmission lines and/or cables, or modeling a small impedance branch with tie circuit breakers, can be employed. 31.4.1.5 Long Radial System Configurations

Long radial system configurations usually take a longer time to converge than loop configurations. In general, the fast-decoupled method works faster than Newton–Raphson for radial system. 31.4.1.6 Bad Bus Voltage Initial Values

Solution convergence speed and computing time are functions of the initial voltages for load-type buses. The closer the initial voltages are to their final profile, the faster the solution converges. The solution may not converge if the initial voltages are too far from the final profile.

31.5

Required Data for ETAP

Table 31.5 is a summary of the required data for some of the standard device types. Note that some of the data requirements vary, based on the mode of operation. Refer to the ETAP user guide for a detailed list of required data. Table 31.5 Required data for the load-flow analysis. Device

Buses

Nominal kV

x

x

Operating kV

x

x

Angle

x

x

Impedance R, X Diversity

Line/Cables

x

Transformers

x

Sources

x

Tap changer

x

Tolerance

x

x x

x

x

Operating mode

x

Mvar limits

x

kW/HP

x

kvar

x

Power factor

x

Xd

x

Rated kV

Motors

x

Rating kVA

Temperature

Impedances

x

x

x x x

Efficiency

x

Loading levels

x

Generating levels

x

31.7 Model Validation

31.6

Data Collection and Preparation

Results from any power-system model are only as good as the data used to create it. If the data is not correct, then the model will not produce accurate results. Collecting the data for the first time is a very labor-intensive undertaking and can account for 50% of the cost of completing the initial study. If the facility has good single-line diagrams, manuals, and data in other forms, this will speed up the data-collection process. However, it is always a good idea to verify all this data in the field, especially for larger, existing facilities where changes are made on a frequent basis. Once completed, the model can be used as a change-management tool. ETAP includes a multidimensional database where facility ratings and operating conditions can be easily recorded to be used for “what-if” scenarios. ETAP also provides a mobile application that further speeds up the data-collection process: etapAPP™, which includes innovative, smart ways to collect data and synchronize the information with the desktop. Based on past experience, the gathering stage of the project will take about two hours for every substation, switchgear, and motor control center (MCC). The estimate for load data collection is roughly 0.1 hours per piece of equipment, including motors, panels, etc. If the facility is geographically dispersed, such as a utility distribution system, or has a lot of unique equipment (arc furnace, switched capacitors, pole-mounted fuses, transformers, etc.), these numbers may need to be increased.

31.7

Model Validation

Model construction and validation are essential tasks that form the foundation of all power-system studies. Periodic system model validation is necessary to ensure that the power-system models are accurate and up to date. These tasks need to be performed regularly in order to keep up with ongoing changes and additions to the power system. The best and most accurate method to validate electrical models is utilizing real-time data to compare with the results from the offline or planning model. The process of comparing real-time data with offline models starts with accounting for the system topology conditions, including open breakers, abnormal switching configurations, etc. Substations are required to be modeled using a node-breaker topology from the typical power-flow bus-line configuration. That process is often labor intensive (much of this is still done manually), prone to errors, and very time-consuming. Power-flow simulation results are compared with time-synchronized recordings of nodal voltages, angles, and key flows during steady-state conditions to ensure the adequacy of system-wide power-flow models. Adjusting the conditions in the power-flow model to match the observed conditions of the actual power system is a significant task but essential for model validation. The validation process may be performed using a state estimator, which is also included with ETAP software; however, the traditional approach is to measure critical locations such as sources, main feeders, and critical loads and compare with the load-flow simulation results. Any deviations are evaluated, and the model is calibrated or tuned such that the load-flow results match the measured quantities at the same point in the model as shown in the procedure given by the North American Electric Reliability Corporation (NERC) in Figure 31.9. Figure 31.10 shows measured values in blue; the state estimated values that may be used for tuning are shown in red. Using these values and the electrical model, the state estimator examines the data for apparent data errors, determines those portions of the network that have sufficient telemetry to be observable, generates artificial measurements (“virtual” measurements) at locations where they are required for observability, complements those with pseudo-measurements, and then computes the estimate of the system voltages and electrical power flows for the entire system. ETAP includes a state estimator comparator that provides an automated comparison for load flow versus measured, estimated versus measured, and estimated versus load flow values. Various aspects of a system power-flow model need to be matched to a specific time instance, including network topology (service status, general network connections, and impedances), generation dispatch, transformer tap position, the status of switched reactive devices, and loading (load profile). The degree of success in matching a power-flow model to observed system conditions is indicative of the validity of the power-flow model. Large unresolved errors in parts of the network (that cannot be attributed to measurement errors) strongly suggest that the system power-flow model itself has errors at those points.

797

798

31 Load Flow

Line impedances, transformers, etc.

Generation dispatch & load

Collect basic system data

Network topology

Build power flow case

Collect Steady-State measurements of key flows, voltages and angles

Compare Power Flow Model with Measured System Steady-State Condition: Model Validation

Comparison is good?

No (iterate)

Expand on power flow database to include other studies

Figure 31.9 Power-system model validation. Source: NERC (https://www.nerc.com/docs/pc/mvwg/MV%20White%20Paper_Final.pdf).

Figure 31.10 ETAP real-time state estimation.

31.8 Study Scenarios

31.8

Study Scenarios

In order to evaluate the system ratings, voltage drops, acceptable limits, etc. we must adjust loading and generation such that combinations or minimum, normal, and maximum operating conditions are simulated, which may be dependent on weather or production processes. The ETAP Unbalanced Load Flow Study Case editor shown in Figure 31.11 contains solution control variables, loading conditions, and a variety of options for output reports. ETAP allows you to create and save an unlimited number of study cases. It is possible to switch between study cases without having to reset the study case options for each load-flow simulation. The following set of combinations may be considered for load-flow simulation scenarios:

•• ••

Loading: Max, Normal, Min Generation: Max, Normal, Min Topology: Various combinations Controllable Devices: Taps, Switched Capacitors, SVC, etc.; operating or locked

Figure 31.11 ETAP Load Flow Study Case editor.

Load-flow simulation results from various combinations need to be analyzed in order to make an engineering determination. This can be done quickly in ETAP using the Load Flow Analyzer shown in Figure 31.12. The ETAP Load Flow Analyzer tabulates results of various load-flow studies in one spreadsheet view, allowing for convenient comparison and analysis of various results. We can compare the results of general information about the project or more specific information such as the results contained from buses, branches, loads, or sources in a load-flow study. The Load Flow Analyzer is a time-saving tool that allows us to compare and analyze different reports coming from different projects, within the same directory, in a single display. Alternatively, instead of using fixed combinations of generation and loading, a time-varying profile may be used for generation and loading based on the time step required to be analyzed.

Figure 31.12 ETAP Load Flow Analyzer.

799

800

31 Load Flow

31.9

Contingency Analysis

A contingency-analysis study uses a single method for load-flow calculation: fast-decoupled. AC load flow is first to run for the base case. Then for each outage, the elements with outages are kept out, and another load flow is run, called the contingency load flow. Depending on the outage, the bus voltage and branch loading can be affected significantly. Contingency load flow may determine undervoltage buses and overloading branches. The fast-decoupled method is derived from the Newton–Raphson method. It is based on the fact that a small change in the magnitude of bus voltage does not vary the real power at the bus appreciably; and likewise, for a small change in the phase angle of the bus voltage, the reactive power does not change appreciably. Thus the load-flow equation from the Newton–Raphson method can be simplified into two separate decoupled sets of load-flow equations, which can be solved iteratively. A performance index is used for outage ranking, to indicate the severity of the outage. The performance index is a number, and it has no unit. Therefore, we can add different indices as well as different weights. 31.9.1

Bus Voltage Security Index (PIv/vsp)

The bus voltage security index is calculated based on the bus voltage difference before and after the contingency. A smaller value for this index implies smaller differences in bus voltage: N

PI v

vsp

= n=1

Vn_PostCon −Vn_Spec Vlimit

2

N

31 1

where, for PIv/vsp based on equipment rating, Vn_PostCon : post-contingency bus voltage for Bus n in percent Vn_Spec : specified bus voltage base for Bus n in percent, 100% in this case Vlimit : limit for the bus voltage critical alert; Vlimit = |Vcritical − 100%| N : number of buses in the system and for PIv/vsp based on a base case, Vn_PostCon : post-contingency bus voltage for Bus n in percent Vn_Spec : specified bus voltage base for Bus n in percent; here, the base case Vlimit : limit for the bus voltage critical alert; Vlimit = VCrit_Dev, which is the critical voltage deviation alert in percent; VCrit_Dev is collected from the Alert page 31.9.2

Real Power Flow Change Index (PIΔP)

The real power flow change index is calculated based on the real power-flow difference before and after the contingency. A smaller value for this index implies smaller differences in real power flow: N

PI ΔP = n=1

Pn_PostCon −Pn_Base Pn_Base

2

N

31 2

where Pn_PostCon : post-contingency real power flow for Branch n Pn_Base : base-case real power flow for Branch n N : number of branches with real power-flow changes in the system 31.9.3

Reactive Power Flow Change Index (PIΔQ)

The reactive power flow change index is calculated based on the reactive power-flow difference before and after the contingency. A smaller value for this index implies smaller differences in reactive power flow: N

PI ΔQ = n=1

Qn_PostCon − Qn_Base Qn_Base

2

N

31 3

31.10 Optimal or Optimum Power Flow

where Qn_PostCon : post-contingency reactive power flow for Branch n Qn_Base : base-case reactive power flow for Branch n N : number of branches with reactive power-flow changes in the system

31.9.4

Branch Overloading Security Index (PIS/Ssp)

The branch overloading security index is calculated based on the branch-overloading condition after the contingency. Smaller values represent fewer overloading violations: N

PI S

Ssp

= n=1

Sn_PosCon SLimit

2

N

31 4

where, for PIS/Ssp based on equipment rating, Sn_PostCon : post-contingency branch load flow for Branch n SLimit : branch-overloading limit, in this case SLimit = Capacity × (Scrit_dev) N : number of branches in the system and for PIS/Ssp based on the base case, Sn_PostCon : post-contingency branch load flow for Branch n SLimit : branch-overloading limit, in this case SLimit = SBase × (1 + Scrit_dev) N : number of branches in the system

31.10

Optimal or Optimum Power Flow

Three types of problems are commonly referred to in power system literature: power flow (load flow), economic dispatch, and optimal power flow (OPF). Three other classes of power-system optimization – unit commitment, optimal topology, and long-term planning – involve binary and integer variables and are not discussed in this section. Unlike load-flow studies, where the computer solves for bus voltages and branch flows, OPF solves for bus voltages, branch flows, and control settings. OPF is a direct solution that requires the specification of bus-voltage and branch-flow constraints and control-setting ranges, and the selection of optimization objectives. Control variables may include:

•• •• •• •• •• •• •• •• ••

Load-tap changer (LTC) settings Generator AVR settings Generator MW generation Series or shunt VAR compensator settings Phase-shift transformer tap positions Switched capacitor settings Load shedding DC line flow Objective functions can include: Minimize real power losses in the system. Minimize reactive power losses in the system. Minimize real power generation at the swing bus(es). Minimize var generation from available shunt var control devices. Minimize total generation fuel cost. Minimize var generation from available series var control devices. Minimize load to be shed from the available bus-load-shed schedule. Minimize total number of controls. Minimize overall adjustment from all controls. Maximize the voltage security index

801

31 Load Flow AllBuses i

Vi − Vi, avg dV i

2n

where Vi, max + Vi, min 2 Vi, max −Vi, min dV i = 2 Maximize the line-flow security index Vi, avg =

•

AllBranches i

Si Si

2n

where dS i line rating

•

Voltage magnitude difference between all buses is minimum

OPF employs an optimization technique to automatically adjust the power-system control settings while it solves the load-flow equation at the same time. Moreover, we can specify a wide range of optimization criteria for the system and enforce limits on system quantities (bus voltage, line flow, etc.) during the optimization process. These optimization criteria are called objectives (usually the system performance indexes), and the limits are called constraints. Mathematically, the OPF study can be expressed as Min = f x, u

31 5

subject to the equality constraints P x, u = 0

31 6

Q x, u = 0

31 7

and the inequality constraints umin ≤ u ≤ umax y x, u

min

≤ y x, u ≤ y x, u

31 8 31 9

max

where x : bus voltage vector, called the state variable set u : system control vector, called the control variable set f : objective functions, expressed in terms of x and u y : system output vector, a variable set typically including line flows, etc. as a function of x and u P : real power, expressed in terms of x and u Q : reactive power, expressed in terms of x and u Equation (31.5) indicates the specified objective function to be minimized or optimized. Equations (31.6) and (31.7) show the system power balance equation (loadflow equation) to be solved. Equation (31.8) specifies the control upper and lower limits, and Eq. (31.9) sets the upper and lower limits for output variables. Consider the following objective function to be minimized: y(x) = 20x2 − 80x − 10. If the control variable x range is 0 ≤ x ≤ 10, then a solution can be found as shown in Figure 31.13.

Non-constrained 1400 1200 1000 800 Y

802

600 400 200 0 –200

0

5

10 X

Figure 31.13 OPF solution found.

15

31.11 Illustrative Examples

(a)

(b)

Constrained (Solution Found)

1400 1200

1200

1000

1000 800 Y

800 Y

Constrained (No Solution) 1400

600

400

400

200

200

0 –200 0

0 –200

600

0

5

10

15

5

10

15

–400

X

X

Figure 31.14 (a) OPF solution found (constrained condition); (b) no solution.

Assuming the same objective function and control variable range, if a constraint or balance equation is applied where y = 60x − 100, then a solution is also possible, as shown in Figure 31.14a. If the constraint-of-balance equation is changed to y = 20x − 300, then the problem is infeasible or no solution is found, as shown in Figure 31.14b. ETAP optimal power flow analysis uses the state-of-the-art interior point-optimization technique with the logarithm barrier function and the prime-dual direction searching method. The algorithm is very efficient and robust, suitable for large systems with both equality and inequality constraints. On the power-system modeling side, an accurate AC model is used, which makes it possible for this program to achieve the ultimate accuracy while solving power system OPF problems of any size under any feasible conditions.

31.11

Illustrative Examples

31.11.1

Example – Load-Flow Study for a Transmission System

31.11.1.1 System Modeling 31.11.1.1.1 One-Line Diagram

The one-line diagram for the Example Model for Transmission System is shown in Figure 31.15.

Figure 31.15 Example model for a transmission system.

803

804

31 Load Flow

31.11.1.1.2 Required Data for the Load-Flow Study

The required data for the load-flow study is summarized in Section 31.5. Data utilized for Example Model for Transmission System are as follows: Power grid Operating mode

Swing

Nominal kV

245 kV

Short-circuit capacity

1500 MVA

X/R ratio

20

Buses (nodes) Nominal kV

245 kV, 72.5 kV, 24 kV

Transmission lines Nominal kV

250 kV

Conductor

Manufacturer: Pirelli Code: 37175HDCU, Copper Size: 89 mm2 (strands = 37)

Impedance and ampacity

See Section 31.11.4

Configuration

(Length: Line1 = 25 km, Line 2 = 50 km) Line 1 and Line 2

(Length: Line 3 = 50 km) (distance: Line 3A and 3B = 7 m) Line 3

Transformers Rated kV ID

Primary

Secondary

MVA

%Z

X/R

Connection

Grounding

Tap

T-1

22

220

150

14

45

YNd11

Solid

As per required

T-2

220

66

100

12.5

45

YNd11

Solid

T-3

220

66

100

12.5

45

YNd11

Solid

31.11 Illustrative Examples

Synchronous generator Operating mode

Swing (for configuration UT-Stop) PF Control (for other configurations)

Rating

150 MW, 22 kV, 2 poles, %PF = 90, Eff. = 99%

Generation

90 MW, PF = 90%

Subtransient impedance

Xd" = 19%

Transient reactance

Xd’ = 28%

Synchronous reactance

Xd = 155%

X/R ratio

X"d/Ra = 19

Loads (Industrial Plant-1 and -2) Rating

Nominal kV = 66 kV, Rated Load = 100 MVA, PF = 95%, %LRC = 650%, X/R = 10

%Loading

80% (Constant kVA = 80%, Constant Z = 20%)

31.11.1.2 Load-Flow Study

The load-flow study was executed on the Example Model for Transmission System shown in Figure 31.15. ETAP provides four load-flow calculation methods: Adaptive Newton–Raphson, Newton–Raphson, Fast-Decoupled, and Accelerated Gauss–Seidel. 31.11.1.2.1

Analysis Objectives

The load-flow analysis objectives are summarized in Section 31.2. 31.11.1.2.2

Study Scenarios

Load flow was examined for the seven system configurations shown in Table 31.6. Table 31.6 Study scenarios. Scenario

System Configuration

Utility

Generation Plant

Line-1

Line-2

Line-3

Industrial Plant

LF-1

Normal (Run All System)

Run

Run

Run

Run

Run

Run

LF-2

Utility Stop

(Stop)

Run

Run

Run

Run

Run

LF-3

Generation Plant Stop

Run

(Stop)

Run

Run

Run

Run

LF-4

Line1 Stop (Open)

Run

Run

(Stop)

Run

Run

Run

LF-5

Line2 Stop (Open)

Run

Run

Run

(Stop)

Run

Run

LF-6

Line3 Stop (Open)

Run

Run

Run

Run

(Stop)

Run

LF-7

Industrial Plant-2 Stop

Run

Run

Run

Run

Run

(Stop)

31.11.1.2.3

Study Conditions

Calculation method

Adaptive Newton–Raphson

Alert (% of calculation results)

Bus voltage: marginal over ±2%, critical over ±5% Loading – marginal over 95%, critical over 100%

31.11.1.2.4

Result of the Load-flow Study

The result of the load-flow study on scenario LF-1 “Normal (run all system)” is shown in the one-line diagram in Figure 31.16. Transformer taps were set to maintain the bus voltage within the alert limit (marginal over ±2%, critical over ±5%) and adjusted as shown in Table 31.7.

805

806

31 Load Flow

Figure 31.16 Result of the load-flow study for the “Normal (run all system)” case.

Table 31.7 Transformer tap settings. Rated kV ID

Primary

Secondary

MVA

%Z

X/R

Connection

Grounding

Tap

T-1

22

220

150

14

45

YNd11

Solid

−4%

T-2

220

66

100

12.5

45

YNd11

Solid

−4%

T-3

220

66

100

12.5

45

YNd11

Solid

−4%

The results of bus voltages for all study scenarios, when transformer taps were set as mentioned in Table 31.7, are summarized in Table 31.8. Cells where the bus voltages exceeded the marginal limit (± 2%) or the critical limit (± 5%) are highlighted in light gray. The results of power flow (MW, Mvar, MVA, PF, and kA) for all study scenarios are shown in Table 31.9. Cells with a negative Mvar flow are highlighted in light gray. Table 31.8 Bus voltages for all study scenarios. Bus voltage (% of bus nominal kV)

Bus ID

Bus nominal kV

Scenario LF-1

Scenario LF-2

Scenario LF-3

Scenario LF-4

Scenario LF-5

Scenario LF-6

Scenario LF-7

Bus-1

245

100.00

88.92

100.00

100.00

100.00

100.00

100.00

Bus-2

245

99.84

88.92

99.44

99.08

99.42

100.00

100.20

Bus-3

245

99.33

88.73

98.18

98.94

98.14

98.61

100.30

Bus-4

24

100.40

91.67

96.22

100.00

99.28

99.72

101.40

Bus-5

72.5

101.7

90.26

100.50

101.30

100.50

101.00

102.80

Bus-6

72.5

101.7

90.26

100.50

101.30

100.50

101.00

—

Table 31.9 Power flow (MW, Mvar, MVA, PF, and kA) for all study scenarios. Power flow Branches

Bus nom. kV

MW

Mvar

Scenario

MVA

Power flow PF (%)

kA

MW

Mvar

LF-1

MVA

Power flow PF (%)

kA

MW

LF-2

70.4

1.9

70.4

99.96

0.166

—

—

24

90.0

43.6

100.0

90.00

2.396

152.4

66.7

166.3

91.61

4.365

—

Line-1

245

29.8

−2.5

29.9

−99.64

0.071

0.2

−2.1

2.1

−7.66

0.006

68.3

Line-2

245

40.5

4.4

40.8

99.41

0.096

0.2

−2.1

10.7

−1.57

0.028

Line-3

245

14.9

1.4

15.0

99.57

0.035

0.1

1.1

1.1

7.44

72.5

79.8

26.2

84.0

95.00

0.657

75.7

24.9

79.7

72.5

79.8

26.2

84.0

95.00

0.657

75.7

24.9

79.7

Generation

Industrial 1 Industrial 2

Scenario

Utility

LF-4

—

MVA

PF (%)

kA

LF-3

245

Utility

Mvar

—

—

160.8

166.2

96.74

—

—

13.5

69.6

98.09

0.164

92.5

28.5

96.8

95.55

0.228

0.003

34.0

0.0

35.2

96.56

0.084

95.00

0.703

79.3

26.1

83.5

95.00

0.662

95.00

0.703

83.5

95.00

0.662

79.3

42.1 —

26.1

LF-5

0.392 —

LF-6

245

70.3

8.1

70.7

99.35

0.167

69.9

14.1

71.3

98.04

0.168

70.0

15.5

71.7

97.65

0.169

24

90.0

43.6

100.0

90.00

2.405

90.0

43.6

100.0

90.00

2.423

90.0

43.6

100.0

90.00

2.412

Line-1

245

—

—

—

—

69.9

14.1

71.3

98.04

0.168

0.0

−5.4

5.4

−0.01

0.013

Line-2

245

70.3

8.1

70.7

99.35

0.167

—

—

—

—

70.0

20.9

73.1

95.82

0.172

Line-3

245

0.0

−6.3

6.3

−0.03

0.015

34.8

9.4

36.1

96.55

0.086

—

—

—

—

72.5

79.6

26.2

83.8

95.00

0.659

79.3

26.1

83.5

95.00

0.662

79.5

26.1

83.7

95.00

72.5

79.6

26.2

83.8

95.00

0.659

79.3

26.1

83.5

95.00

79.5

26.1

83.7

Generation

Industrial 1 Industrial 2

Scenario

Utility

—

LF-7

245

9.5

32.5

33.9

27.97

0.080

24

90.0

43.5

100.0

90.00

2.373

Line-1

245

4.1

10.8

11.6

35.29

0.027

Line-2

245

5.4

5.4

7.6

70.93

0.018

Line-3

245

2.1

−1.1

2.3

−88.56

0.005 0.654

Generation

Industrial 1

72.5

80.3

33.3

84.4

95.00

Industrial 2

72.5

—

—

—

—

—

—

0.662

95.00

— 0.660 0.660

808

31 Load Flow

Based on the results of these scenarios, we must consider the following with reference to the analysis objectives described in Section 31.2:

•• •

Validation of equipment rating Adjustment of bus voltages (setting of the transformer taps) Correction of reactive power flow in branches, and so on

31.11.2

Example – Load-Flow Study for an Industrial System

31.11.2.1 System Modeling 31.11.2.1.1 One-Line Diagram

The one-line diagram for the Example Model for Industrial System is shown in Figure 31.17.

Figure 31.17 Example model for an industrial system.

31.11 Illustrative Examples

31.11.2.1.2

Required Data for the Load-Flow Study

The required data for the load-flow study is summarized in Section 31.5. Applied data for the Example Model for Industrial System are as follows: Power grid Operating mode

Swing

Nominal kV

66 kV

Short-circuit capacity

500 MVA

X/R ratio

20

Buses (nodes) Nominal kV

66 kV, 11 kV, 6.6 kV, 0.65 kV

Transformers Two-winding transformer

•

Rated kV

ID

Primary

Secondary

MVA

T-Gen

11

66

T-1

66

11

T-LV-1

11

•

%Z

X/R

Connection

Grounding

Tap

50

12.5

45

YNd11

Solid

As per required

70

12.5

45

Dyn11

Solid

6

Dyn11

Solid

0.65

2.5

6.25

Three-winding transformer ID

Winding

Rated kV

MVA

%Z

X/R

Connection

Grounding

T-2

Primary

66

70

14a

40

Wye

Solid

b

40

Delta

40

Delta

Secondary

11

50

12

Tertiary

6.6

20

12c

Tap

As per required

a

Primary-secondary. Primary-tertiary. c Secondary-tertiary. b

Synchronous generator Operating mode

Voltage control

Rating

50 MW, 11 kV, 2 poles, %PF = 90, %Eff. = 98%

Generation

40 MW, %V = 100%

Subtransient impedance

Xd" = 25%

Transient reactance

Xd’ = 35%

Synchronous reactance

Xd = 99%

X/R ratio

X"d/Ra = 25

809

810

31 Load Flow

Loads

•

Induction motor

Motor ID

MW

Pole

kV

% Load

LRC (%)

IM-1

10

4

11

80

650

7.59

35.29

220.8

1.5

93.44

94.92

591.8

1

IM-2

2

4

11

80

650.08

15.05

99.65

220.2

1.5

93.2

95.45

118

5

IM-3

5.45

4

6.6

80

549.9

20.01

65.02

165

1.5

92.06

95.55

542

1

IM-4

300

4

0.65

100

600

20.06

35

210

0

91.99

93.11

311.1

1

•

PF lr (%)

LRTrq (%)

MaxTrq (%)

Slip (%)

PF (%)

Eff (%)

Current (A)

Q’ty.

Synchronous motor

Motor ID

MW

Pole

kV

% Load

LRC (%)

PF lr (%)

PF (%)

Eff (%)

Current (A)

Quantity

SM-1

5

2

11

80

650

9.2

93.15

96.3

292.6

1

Lumped Load ID

MVA

kV

PF (%)

%Motor Load

%Static Load

LRC (%)

X/R

% Load

L-HV-1

10

11

90

80

20

600

10

80

L-HV-2

5

6.6

90

80

20

600

10

80

L-LV

1.5

0.65

90

80

20

600

2.38

100

L-T1

30

11

90

80

20

650

10

100

•

•

Lumped loads

Power capacitors

Nominal kV

11 kV

Capacity

2Mvar × 5 banks

•

Feeder cables

Cable ID

Voltage (kV)

Type

Core

Size (mm2)

Q’ty

Length (m)

C-T1

66

XLPE

3

240

3

100

C-T2

66

XLPE

3

240

3

100

C-SM1

11

XLPE

3

120

2

200

C-HV1

11

XLPE

3

400

2

200

C-IM3

6.6

XLPE

3

185

3

200

C-HV2

6.6

XLPE

3

240

2

200

31.11.2.1.3 Lumped Motor

A lumped motor is an imaginary element created from more than one induction motors and used for dynamic simulation of electrical systems: dynamic motor-starting studies, transient stability analysis, and so on. When there are hundreds or more than a thousand motors in the system to be analyzed, convergence of the dynamic calculations can be

31.11 Illustrative Examples

difficult or impossible. Replacing a group of motors with a single lumped motor, as shown here, reduces the dimensions of the electrical system matrix and makes it easier to perform the calculations with an acceptable level of accuracy. Bus

Bus

M1

M2

M3

M4

M0 Lumped motor

Actual motors Capacity

kW1

kW2

kW3

kW4

kW0

No. of Poles

Pole1

Pole2

Pole3

Pole4

Pole0

Inertia Constant

H1

Loading Factor

%Load1

H2

H3

%Load2

%Load3

H4 %Load4

H0 %Load0

The lumped-motor method makes it possible to calculate or determine the following parameters as a single lumped motor: kW0

Capacity (kW)

Pole0

Number of poles

H0

Inertia constant (second)

%Load0

Loading factor (%)

Calculation Formulas When there are n sets of motors to be lumped into one motor and their capacities are represented as kW1, kW2, kW3, … kWn kVA1, kVA2, kVA3, … kVAn, the inertia constants are represented as H1, H2, H3, … Hn, and the loading factors are represented as %Load1, %Load2, %Load3, … %Loadn. The capacities kW0 and kVA0 of the lumped motor are calculated as a total of the individual motor capacities: n

kW 0 =

kW i i=1 n

kVA0 =

kVAi i=1

The inertia constant H0 and loading factor %Load0 of lumped motor are calculated as a weighted average using the individual capacities kVAi and kWi, respectively: n

Hi × kVAi H0 = i = 1

kVA0

sec

n

Load i × kW i Load 0 =

i=1

kW 0

The number of poles Pole0 of the lumped motor is determined with the following steps: 1) Calculate the sum kW of motors with the same number of poles. 2) Select the number of poles for which the summed kW is the maximum.

811

812

31 Load Flow

The following key parameters of the lumped motor are retrieved from the ETAP motor library and used for dynamic modeling of this lumped motor – that is, estimating the parameters of the motor equivalent circuit to get the motor torque, current, and power factor characteristic curves: LRC (%)

Locked rotor current

PFlr (%)

Locked rotor power factor

Tlr (%)

Locked rotor torque

Tmax (%)

Breakdown (Max) torque

S (%)

Motor slip

PF100 (%)

Power factor at full load

EFF100 (%)

Efficiency at full load

31.11.2.2 Load-Flow Study

The load-flow study was executed on the Example Model for Industrial System shown in Figure 31.17. 31.11.2.2.1 Analysis Objectives

The load-flow analysis objectives are summarized in Section 31.2. 31.11.2.2.2 Study Scenarios

Load flow was examined for the four system configurations shown in Table 31.10. Table 31.10 Study scenarios. Scenario

System Configuration

CB-tie

CB-66 kV-2

SW-T1

CB-66 kV-3

SW-T2

LF-1

11 kV CB-Tie Open

Open

Close

Close

Close

Close

LF-2

11 kV CB-Tie Close

Close

Close

Close

Close

Close

LF-3

Source: From T-1 Line

Close

Close

Close

Open

Open

LF-4

Source: From T-2 Line

Close

Open

Open

Close

Close

31.11.2.2.3 Study Conditions Calculation method

Adapted Newton–Raphson

Alert (% of calculation results)

Bus voltage: marginal over ±2%, critical over ±5% Loading: marginal over 95%, critical over 100%

31.11.2.2.4 Results of the Load-Flow Study

The result of the load-flow study on scenario “11 kV CB-Tie Open” is shown in the one-line diagram in Figure 31.18. Transformer taps were set to maintain the bus voltage within the alert limit (marginal over ±2%, critical over ±5%) and adjusted as shown in Table 31.11. The results for the bus voltages in all the study scenarios, when the transformer taps were set as mentioned in Table 31.11, are summarized in Table 31.12. Cells where the bus voltages exceed the marginal limit (± 2%) or the critical limit (± 5%) are highlighted in light gray. The results for power flow (MW, Mvar, MVA, PF, and kA) for all study scenarios are shown in Table 31.13. In this analysis, all Mvar flows were positive. Just as with the load-flow study for a transmission system, based on the results of the scenarios summarized here, we must consider the followings with reference to the analysis objectives described in Section 31.2:

•• •

Validation of equipment rating Adjustment of bus voltages (setting of the transformer taps) Correction of reactive power flow in branches, and so on

Figure 31.18 Results of the load-flow study for the “CB-Tie Open” case.

Table 31.11 Transformer tap settings. Two-winding transformer Rated kV %Z

X/R

Connection

Grounding

50

12.5

45

YNd11

Solid

−2.5%

70

12.5

45

Dyn11

Solid

−2.5%

6

Dyn11

Solid

−2.5%

ID

Primary

Secondary

MVA

T-Gen

11

66

T-1

66

11

T-LV-1

11

0.65

2.5

6.25

Tap

Three-winding transformer ID

T-2

Winding

Rated kV

MVA

%Z

X/R

Connection

Primary

66

70

14

40

Wye

Secondary

11

50

12

40

Delta

20

12

40

Delta

Tertiary

6.6

Grounding

Solid

Tap

−2.5%

814

31 Load Flow

Table 31.12 Bus voltages for all study scenarios. Bus voltage (% of bus nominal kV)

Bus ID

Bus nominal kV

Scenario LF-1

Scenario LF-2

Scenario LF-3

Scenario LF-4

Bus-66 kV

245

100.00

100.00

100.00

100.00

Bus-Gen

245

100.00

100.00

100.00

100.00

Bus-11 kV-T1

245

100.30

100.70

98.15

98.03

Bus-11 kV-T2

24

101.10

100.70

98.15

98.03

Bus-6.6 kV-T2

72.5

101.40

101.10

97.50

99.66

Bus-LV1

72.5

100.40

100.90

98.26

98.14

Table 31.13 Power flow (MW, Mvar, MVA, PF, and kA) for all study scenarios.

Branches

Bus nom. kV

Power flow MW

Mvar

Scenario

MVA

Power flow PF (%)

kA

MW

Mvar

LF-1

MVA

PF (%)

kA

LF-2

CB-66 kV

66

25.5

15.6

29.9

85.25

0.261

25.5

14.8

29.5

86.42

0.258

T-Gen Primary

11

40.0

11.6

41.6

96.07

2.185

40.0

11.6

41.6

96.07

2.185

T-1 Primary

66

45.8

13.9

47.9

95.67

0.418

32.4

10.8

34.1

94.85

0.299

T-2 Primary

66

19.6

9.1

21.6

90.63

0.189

33.0

11.5

35.0

94.46

0.306

T-1 Secondary

11

45.8

13.9

47.9

95.67

0.419

32.4

10.8

34.1

94.85

0.299

T-2 Secondary

11

11.4

5.1

12.5

91.22

0.648

24.8

6.3

25.6

96.97

1.333

8.2

3.5

8.9

92.09

0.766

8.2

3.5

8.9

92.09

0.767

1.4

0.7

1.5

88.56

0.081

1.4

0.7

1.5

88.57

0.080

1.4

0.7

1.5

88.56

0.081

1.4

0.7

1.5

88.57

0.080

T-2 Tertiary T-LV-1 Primary T-LV-1 Secondary

6.6 11 0.65

Scenario

LF-3

LF-4

CB-66 kV

66

25.2

20.0

32.2

78.26

0.281

25.2

19.8

32.1

78.59

0.281

T-Gen Primary

11

40.0

11.6

41.6

96.07

2.185

40.0

11.6

41.6

96.07

2.185

T-1 Primary

66

65.1

27.5

70.7

92.13

0.618

—

—

—

—

T-2 Primary

66

—

—

—

—

—

65.1

27.3

70.6

92.23

T-1 Secondary

11

65.1

27.5

70.7

92.13

0.618

—

—

—

—

T-2 Secondary

11

8.1

3.6

8.9

91.53

0.474

56.8

15.5

58.9

96.46

3.151

8.1

3.4

8.8

92.10

0.791

8.1

3.5

8.8

92.09

0.776

1.4

0.7

1.5

88.51

0.081

1.4

0.7

1.5

88.51

0.081

1.4

0.7

1.5

88.51

0.081

1.4

0.7

1.5

88.51

0.081

T-2 Tertiary T-LV-1 Primary T-LV-1 Secondary

6.6 11 0.65

— 0.618 —

(a) Wind power (WTG) plant

Figure 31.19 Example model for a renewable energy system.

(b)

Solar power (PV) plant

Figure 31.19 (Continued)

31.11 Illustrative Examples

31.11.3

Example – Load-Flow Study for a Renewable Energy System

31.11.3.1 System Modeling 31.11.3.1.1 One-line Diagram

The one-line diagram for the Example Model for Renewable Energy System is shown in Figure 31.19 (pages 815 and 816). 31.11.3.1.2

Required Data for the Load-Flow Study

The required data for the load-flow study is summarized in Section 31.5. Applied data for the Example Model for Renewable Energy System are as follows: Power grid Operating mode

Swing

Nominal kV

33 kV

Short-circuit capacity

500 MVA

X/R ratio

20

Buses (nodes) Nominal kV

33 kV, 1.8 kV (WTG Plant), 0.48 kV (PV Plant)

Transformers Rated kV ID

Primary

Secondary

MVA

%Z

X/R

Connection

Grounding

Tap

T-WTG1 to 6

33

1.8

6

11.2

8.5

Dd0

None

Normal

T-PV1 to 6

33

0.48

3

6.25

6

Dy11

None

Normal

Feeder cables Cable ID

Voltage (kV)

Type

Core

Size (mm2)

Quantity

Length (m)

C-WTG12, 23, 34, 45, 56

33

XLPE

3

150

1

1000

C-PV12, 23, 34, 45, 56

33

XLPE

3

60

1

200

31.11.3.1.3

Required Data for the Wind Power Plant

The required data for the wind turbine generator (WTG) are as follows: WTG type Type 1 is fixed-speed, conventional induction generator; Type 2 is variable-slip, induction generators with variable rotor resistance; Type 3 is variable speed, doubly-fed asynchronous generators with rotor-side converter; and Type 4 is variable speed, asynchronous generators with full converter interface. Control WECC, Generic, and UDM Controls are available when Type 3 is selected. For all other types, WECC and UDM are available.

Operation mode Operation modes for different WTG types and controls are as shown in the following list.

817

818

31 Load Flow

Type

Control

Operation Mode

Type 1

WECC/UDM

Induction Generator

Type 2

WECC/UDM

Induction Generator

Type 3

WECC/UDM

Voltage Control

Type 3

Generic / Existing Model (ETAP 5.0 onwards)

Mvar Control

Type 4

WECC/UDM

Voltage Control

WTG Rating for this example project Mvar limits and nominal operation voltages. Average of wind speed can entered on Wind page. For this example project, specified 5 types of Generation Category (100%, 75%, 50%, 25%, and 0%). Once WTG Type 4 is selected, the operation mode is preset to Voltage Control mode.

Generator’s Locked Rotor Current and Power Factor (for IEC short circuit current and motor starting study)

Generator model including converter model (Need for dynamic simulation)

Source of Short-Circuit Current: Max and Min Fault Current in percent of FLA (full load current) The image shows the Current Limiting Curve for the current magnitude, when an inverter operates in its current limiting mode. Control Adjustment Angle applied when PF cannot be maintained under SC conditions.

31.11.3.1.4

Required Data for the Solar Power Plant

The required data for the photovoltaic (PV) array is as follows: PV Array Manufacturer, Model, Type, and Size of PV panels Rating and Performance adjustment coefficients of PV panels

P-V Curve and I-V Curve registered to library by user

PV Array Manufacturer, Model, Type, and Size of PV panels

Arrangement and Total Quantities of PV panels PV Array-Total Rated Setting of Generation Category. For this example project, set the 5category as 100%, 75%, 50%, 25%, and 0% generation related to the irradiances.

PV Array Manufacturer, Model, Type and Size of PV panels

Rating of Inverter

PV Array -Total Rated

Inverter (PCS) DC Rating

Efficiency

AC Rating

AC Grounding

Inverter (PCS) Generation Categories For this example project, specified 5 types of generation category (100%, 75%, 50%, 25%, and 0%).

31.11 Illustrative Examples

31.11.3.2 Load-Flow Study

The load-flow study was executed on the Example Model for Renewable Energy System shown in Figure 31.19. 31.11.3.2.1

Analysis Objectives

The load-flow analysis objectives are summarized in Section 31.2. 31.11.3.2.2

Study Scenarios

The load flow was examined for the five scenarios shown in Table 31.14.

Table 31.14 Study scenarios. Operating conditions

Scenario

System

Presentation

Revision data

Configuration

Study mode

Study type

LF-1-100

LF-1-025

Network analysis

WTG plant PV plant

Base

All run

Load flow

Load flow

LF-1-000

31.11.3.2.3

Generation

100%

LF-1-075

75%

LF-1-050

50%

100%

LF-1-025

25%

LF-1-000

0%

Study Conditions

Calculation method

Adapted Newton–Raphson

Alert (% of calculation results)

Bus voltage: marginal over ±2%, critical over ±5% Loading: marginal over 95%, critical over 100%

31.11.3.2.4

Utility kV

LF-1-100

LF-1-075 LF-1-050

Study case

Results of the Load-Flow Study

The result of the load-flow study on scenario “S-LF-1-100” is shown in the one-line diagram in Figure 31.20. Transformer taps were set to maintain the bus voltage within the alert limit (marginal over ±2%, critical over ±5%) and adjusted as shown in Table 31.15. The results for the bus voltages and power flow (MW, Mvar, MVA, and PF) in all study scenarios, when the transformer taps were set, as mentioned in Table 31.15, are summarized in Table 31.16. Cells where the bus voltages exceed the marginal limit (± 2%) or the critical limit (± 5%) are highlighted in light gray. Also, cells with a negative Mvar flow are highlighted in light gray. Branch losses for all study scenarios are summarized in Table 31.17. Cells with a negative Mvar flow are highlighted in light gray. Just as for the load-flow study for a transmission system, based on the results of scenarios summarized above, we must consider the followings with reference to the analysis objectives described in Section 31.2:

•• •

Validation of equipment rating Adjustment of bus voltages (setting of the transformer taps) Correction of reactive power flow in branches, etc.

821

(a)

Wind power (WTG) plant

Figure 31.20 Results of the load-flow study for study scenario S-LF-1-100 (Utility kV 100%, and 100% generation case).

(b) Solar power (PV) plant

Figure 31.20 (Continued)

824

31 Load Flow

Table 31.15 Transformer tap settings. Rated kV ID

Primary

Secondary

MVA

%Z

X/R

Connection

T-WTG1 to 6

33

1.8

6

11.2

8.5

T-PV1 to 6

33

0.48

3

6

31.11.4

6.25

Grounding

Tap

Dd0

None

0%

Dy11

None

0%

Example – Impedance and Ampacity of Transmission Lines

This example illustrates calculation of impedance and ampacity of the transmission line Line-1 shown in Figure 31.15. 31.11.4.1 Impedance of Transmission Lines

ETAP automatically calculates the following impedances (per phase):

• •

Positive and zero-sequence resistances and reactances (R and X) in ohms or ohms per unit length, at base temperatures T1 and T2, according to the specified configuration and grounding information of a transmission line. Positive and zero-sequence susceptances (Y) in microsiemens or microsiemens per unit length, according to the specified configuration and grounding information of a transmission line. If the value is Y > 0, the transmission line is treated as a model, with one-half of the charging susceptance connected to neutral at each end of the line. If Y = 0, the transmission line is treated as an external impedance.

31.11.4.2 Results of Impedance Calculation

The result of the impedance calculation of transmission Line-1 are shown in Figure 31.21 (page 829) and the R, X, Y matrix in Table 31.18 (page 829). 31.11.4.3 Ampacity of Transmission Lines

ETAP determines the current-temperature relationship for transmission lines on the Transmission Line Editor page. The calculation is based on IEEE Standard 738–1993, “IEEE Standard for Calculating the Current-Temperature Relationship of Bare Overhead Conductors.” Conductor surface temperatures are a function of

•• •• •

Conductor material Conductor outside diameter (OD) Conductor surface conditions Ambient weather conditions Conductor electrical current

Based on the steady-state heat balance equation of a bare overhead conductor, the conductor current and temperature relationship can be given as the following equation I=

qc + qr + qs R Tc

where I : conductor current qc : converted heat loss qr : radiated heat loss qs : heat gain from the sun R : conductor AC resistance at conductor temperature Tc

31 10

Table 31.16 Bus voltages and power flow (MW, Mvar, MVA, and PF). For wind power (WTG) plant: Branches From

Power flow

To

Devices

Bus nom. kV

Bus kV (%)

MW

Scenario

Mvar

MVA

Power flow PF (%)

Bus kV (%)

MW

S-LF1-100

Bus-WTG6b

Bus-WTG5b

101.3

5.44

0.71

Bus-WTG5b

Bus-WTG4b

101.2

10.9

1.55

Bus-WTG4b

Bus-WTG3b

101.0

16.3

Bus-WTG3b

Bus-WTG2b

100.8

Bus-WTG2b

Bus-WTG1b

Cables

33

Bus-WTG1b

Utility-WTG

CB

33

Mvar

MVA

PF (%)

S-LF1-075

5.49

99.16

101.0

4.09

0.795

4.16

98.16

11.0

99.00

100.9

8.18

1.71

8.35

97.88

2.47

16.5

98.87

100.8

12.3

2.69

12.5

97.67

21.7

3.51

22.0

98.72

100.6

16.3

3.77

16.7

97.44

100.4

27.1

4.7

27.5

98.52

100.3

20.4

4.97

21.0

97.15

100.0

32.4

3.78

32.6

99.33

100.0

24.4

4.37

24.8

98.44

WTG-1

Bus-WTG1a

100.0

5.5

−0.362

5.51

−99.78

100.0

4.13

−0.324

4.14

−99.69

WTG-2

Bus-WTG2a

104.2

5.5

1.72

5.76

95.43

103.6

4.13

1.48

4.38

94.14

WTG-3

Bus-WTG3a

104.2

5.5

1.53

5.71

96.32

103.6

4.13

1.33

4.33

95.18

Transformers

1.8 / 33

WTG-4

Bus-WTG4a

104.2

5.5

1.39

5.67

96.93

103.6

4.13

1.22

4.30

95.9

WTG-5

Bus-WTG5a

104.2

5.5

1.3

5.65

97.31

103.6

4.13

1.15

4.28

96.35

WTG-6

Bus-WTG6a

104.2

5.51

1.25

5.65

97.50

103.6

4.13

1.11

4.27

96.57

Scenario

S-LF1-050

S-LF1-025

Bus-WTG6b

Bus-WTG5b

100.5

2.73

−0.654

2.81

−97.25

100.2

1.37

−0.488

1.45

−94.2

Bus-WTG5b

Bus-WTG4b

100.4

5.46

−1.21

5.6

−97.65

100.2

2.74

−0.89

2.88

−95.11

Bus-WTG4b

Bus-WTG3b

100.4

8.19

−1.73

8.37

−97.85

100.2

4.11

−1.29

4.31

−95.39

Cables

33

Bus-WTG3b

Bus-WTG2b

100.3

10.9

−2.2

11.1

−98.04

100.1

5.48

−1.7

5.73

−95.53

Bus-WTG2b

Bus-WTG1b

100.2

13.6

−2.6

13.9

−98.22

100.1

6.84

−2.1

7.16

−95.59

Bus-WTG1b

Utility-WTG

100.0

16.3

−2.93

16.6

−98.43

100.0

8.21

−2.2

8.4

−96.59

WTG-1

Bus-WTG1b

100.0

2.75

−0.252

2.76

−99.58

100.0

1.38

−0.144

1.38

−99.46

WTG-2

Bus-WTG2b

100.0

2.75

−0.339

2.77

−99.25

99.5

1.38

−0.449

1.45

−95.06

WTG-3

Bus-WTG3b

100.0

2.75

−0.407

2.78

−98.92

99.56

1.38

−0.449

1.45

−95.06

100.0

2.75

−0.459

2.79

−98.64

99.60

1.38

−0.449

1.45

−95.06

100.0

2.75

−0.493

2.79

−98.43

99.63

1.38

−0.449

1.45

−95.06

100.0

2.75

−0.509

2.8

−98.33

99.65

1.38

−0.449

1.45

−95.06

WTG-4

Bus-WTG4b

WTG-5

Bus-WTG5b

WTG-6

Bus-WTG6b

CB

Transformers

33

1.8 / 33

Scenario

S-LF1-000

Bus-WTG6b

Bus-WTG5b

100.0

0

−0.086

0.086

0

Bus-WTG5b

Bus-WTG4b

100.0

0

−0.172

0.172

0

Bus-WTG4b

Bus-WTG3b

100.0

0

−0.258

0.258

0

Bus-WTG3b

Bus-WTG2b

100.0

0

−0.345

0.345

−0.01

Bus-WTG2b

Bus-WTG1b

100.0

0

−0.431

0.431

−0.01

Bus-WTG1b

Utility-WTG

100.0

0

−0.431

0.431

−0.01

WTG-1

Bus-WTG1b

100.0

0

0

0

0

WTG-2

Bus-WTG2b

100.0

0

0

0

0

WTG-3

Bus-WTG3b

100.0

0

0

0

0

100.0

0

0

0

0

Cables

33

CB

33

Transformers

1.8 / 33

WTG-4

Bus-WTG4b

WTG-5

Bus-WTG5b

100.0

0

0

0

0

WTG-6

Bus-WTG6b

100.0

0

0

0

0

For solar power (PV) plants: Branches From

Power flow

To

Devices

Bus nom. kV

Bus kV (%)

Power flow Bus kV

MW

Scenario

Mvar

MVA

PF (%)

(%)

MW

S-LF1-100

Mvar

MVA

PF (%)

S-LF1-075

Bus-PV6b

Bus-PV5b

100.2

2.25

−0.104

2.25

−99.89

100.2

1.7

−0.06

1.7

−99.94

Bus-PV5b

Bus-PV4b

100.2

4.5

−0.193

4.5

−99.91

100.2

3.41

−0.105

3.41

−99.95

Bus-PV4b

Bus-PV3b

100.2

6.75

−0.284

6.75

−99.91

100.1

5.11

−0.151

5.11

−99.96

Bus-PV3b

Bus-PV2b

100.1

8.99

−0.375

9.0

−99.91

100.1

6.81

−0.197

5.82

−99.96

Bus-PV2b

Bus-PV1b

100.1

11.2

−0.467

11.2

−99.91

100.1

8.51

−0.244

8.52

−99.96

Bus-PV1b

Utility-PV

100.0

13.5

−0.561

13.5

−99.91

100.0

PV-1

Bus-PV1b

100.7

2.27

0

2.27

100.00

100.5

1.71

0

1.71

−100.00

PV-2

Bus-PV2b

100.7

2.27

0

2.27

100.00

100.6

1.71

0

1.71

−100.00

PV-3

Bus-PV3b

100.8

2.27

0

2.27

100.00

100.6

1.71

0

1.71

−100.00

100.8

2.27

0

2.27

100.00

100.7

1.71

0

1.71

−100.00

100.9

2.27

0

2.27

100.00

100.7

1.71

0

1.71

−100.00

100.9

2.27

0

2.27

100.00

100.7

1.71

0

1.71

−100.00

PV-4

Bus-PV4b

PV-5

Bus-PV5b

PV-6

Bus-PV6b

Cables

CB

Transformers

33

33

1.8/33

10.2

−0.292

10.2

−99.96

Scenario

S-LF1-050

S-LF1-025

Bus-PV6b

Bus-PV5b

100.1

1.14

−0.027

1.14

−99.97

100.1

0.566

−0.007

Bus-PV5b

Bus-PV4b

100.1

2.28

−0.039

2.28

−99.99

100.1

1.13

Bus-PV4b

Bus-PV3b

100.1

3.42

−0.052

3.42

−99.99

100.0

Bus-PV3b

Bus-PV2b

100.1

4.56

−0.065

4.56

−99.99

Bus-PV2b

Bus-PV1b

100.0

5.7

−0.078

5.7

Bus-PV1b

Utility-PV

100.0

6.84

−0.092

PV-1

Bus-PV1b

100.4

1.15

PV-2

Bus-PV2b

100.4

PV-3

Bus-PV3b

100.4

Cables

CB

Transformers

33

33

1.8/33

0.566

−99.99

0.001

1.13

100.00

1.7

0.009

1.7

100.00

100.0

2.26

0.016

2.26

100.00

−99.99

100.0

2.83

0.024

2.83

100.00

6.84

−99.99

100.0

3.39

0.031

3.39

100.00

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

1.15

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

1.15

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

PV-4

Bus-PV4b

100.5

1.15

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

PV-5

Bus-PV5b

100.5

1.15

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

PV-6

Bus-PV6b

100.5

1.15

0

1.15

−100.00

100.2

0.567

0

0.567

100.00

Scenario

S-LF1-000

Bus-PV6b

Bus-PV5b

100.0

0

−0.014

0.014

0

Bus-PV5b

Bus-PV4b

100.0

0

−0.028

0.028

0

Bus-PV4b

Bus-PV3b

100.0

0

−0.043

0.043

0

Bus-PV3b

Bus-PV2b

100.0

0

−0.057

0.057

0

Bus-PV2b

Bus-PV1b

100.0

0

−0.071

0.071

0

Bus-PV1b

Utility-PV

100.0

0

−0.071

0.071

0

PV-1

Bus-PV1b

100.0

0

0

0

0

PV-2

Bus-PV2b

100.0

0

0

0

0

PV-3

Bus-PV3b

100.0

0

0

0

0

100.0

0

0

0

0

Cables

CB

Transformers

33

33

1.8/33

PV-4

Bus-PV4b

PV-5

Bus-PV5b

100.0

0

0

0

0

PV-6

Bus-PV6b

100.0

0

0

0

0

Table 31.17 Summary of branch losses. For wind power (WTG) plants: Losses

Branches From

Bus nom. kV

To

kW

Losses

kvar

S-LF1-100

Scenario Transformers

Cables

T-WTG1

kvar

S-LF1-075

kW

Losses

kvar

S-LF1-050

kW

Losses kvar

S-LF1-025

kW

kvar

S-LF1-000

66.3

563.2

37.3

317.4

16.6

141.4

4.17

35.4

0

0

T-WTG2

66.7

567.2

39.0

331.6

16.7

142.3

4.61

39.2

0

0

T-WTG3

65.5

556.7

38.2

324.4

16.9

143.3

4.6

39.1

0

0

T-WTG4

64.7

549.7

37.6

319.6

17.0

144.1

4.6

39.1

0

0

T-WTG5

64.2

545.4

37.2

316.6

17.0

144.7

4.6

39.1

0

0

T-WTG6

64.0

544.4

37.1

315.2

17.1

145.0

4.6

39.1

0

0

Total

391.4

3326.6

226.4

1924.8

101.3

860.8

27.18

231

0

0

104.2

−5.44

60.8

−39.1

26.7

−65.5

7.1

−80.7

0.021

−86.1

C-WTG12

1.8/33

kW

Losses

33

C-WTG23

66.3

−35.6

38.6

−56.9

17.1

−73.2

4.55

−82.8

0.013

−86.1

C-WTG34

37.1

−58.8

21.6

−70.6

9.67

−79.2

2.56

−84.4

0.006

−86.1

C-WTG45

16.4

−75.3

9.55

−80.2

4.31

−83.5

1.14

−85.6

0.002

−86.2

C-WTG56

4.1

−85.1

2.38

−86.0

1.08

−86.1

0.288

−86.3

0.0

−86.2

Total

228.1

−260.24

132.93

−332.8

58.86

−387.5

15.638

−419.8

0.042

−430.7

For solar power (PV) plants: Losses

Branches From

To

Bus Nom. kV

Cables

kvar

S-LF1-100

Scenario Transformers

kW

Losses

T-PV1

kvar

S-LF1-075

kW

kvar

S-LF1-050

Losses kW

kvar

S-LF1-025

Losses kW

kvar

S-LF1-000

17.4

104.2

9.95

59.7

4.46

26.8

1.1

6.58

0

0

T-PV2

17.3

104.0

9.94

59.7

4.46

26.7

1.1

6.57

0

0

T-PV3

17.3

103.9

9.93

59.6

4.45

26.7

1.1

6.57

0

0

T-PV4

17.3

103.8

9.93

59.6

4.45

26.7

1.09

6.57

0

0

T-PV5

17.3

103.7

9.92

59.5

4.45

26.7

1.09

6.57

0

0

T-PV6

17.3

103.7

9.92

59.5

4.45

26.7

1.09

6.57

0

0

Total

103.9

623.3

59.59

357.6

26.72

160.3

6.57

39.43

0

0

8.78

−10.2

5.04

−11.9

2.26

−13.2

0.556

−13.9

0

−14.2

C-PV23

5.62

−11.7

3.22

−12.7

1.45

−13.5

0.356

−14.0

0

−14.2

C-PV34

3.16

−12.8

1.81

−13.4

0.813

−13.8

0.2

−14.1

0

−14.2

C-PV45

1.4

−13.6

0.805

−13.9

0.361

−14.0

0.089

−14.2

0

−14.2

C-PV56

0.351

−14.1

0.201

−14.1

0.09

−14.2

0.022

−14.2

0

−14.2

Total

19.311

−62.4

11.076

−66

4.974

−68.7

1.223

−70.4

0

−71

C-PV12

1.8/33

kW

Losses

33

31.11 Illustrative Examples

Target conductors

Calculated impedances

Calculated R, X, Y Matrix

Temperature condition

Figure 31.21 Results of the impedance calculation for transmission Line-1 in the ETAP editor.

Table 31.18 R, X, Y Matrix of transmission Line-1. R, X, Y matrix: phase domain Resistance matrix (ohms)

A

Reactance matrix (ohms)

A

B

C

A

B

3.701

1.168

1.175

15.551

7.737

Susceptance matrix (microsiemens)

C

A

7.743

74.765

B

C

−16.768

−15.780

B

1.168

3.686

1.168

7.737

15.571

7.737

−16.768

72.671

−16.768

C

1.175

1.168

3.701

7.743

7.737

15.551

−15.780

−16.768

74.765

R, X, Y matrix: sequence domain Resistance matrix (ohms) 0

1

2

Reactance matrix (ohms) 0

1

Susceptance matrix (microsiemens)

2

0

1

2

0

6.036

0.008

−0.000

31.036

0.004

−0.009

41.189

0.514

0.514

1

−0.000

2.525

0.009

−0.009

7.819

−0.005

0.514

90.505

0.020

2

0.008

−0.009

2.525

0.004

−0.005

7.819

0.514

0.020

90.505

For a bare-stranded conductor, if the conductor temperature (Tc) and the steady-state weather parameters are known, the heat losses due to convection and radiation, solar heat gain, and conductor resistance can be calculated. While the calculation given in IEEE Std. 738 can be performed for any conductor temperature and any weather condition, a maximum allowable conductor temperature and conservative weather conditions are often used to calculate the conductor’s steady-state thermal rating. ETAP calculates the operating temperature corresponding to the user-entered operation current for the specified installation and environment conditions, so it is possible to determine the maximum operating temperature for given

829

830

31 Load Flow

transmission line loading conditions. ETAP also calculates the derated ampacity for the user-defined conductor temperature limit, so maximum loading current may be determined for selected transmission lines. 31.11.4.4 Results of Ampacity Study

The results of the ampacity study for transmission Line-1 are shown in Figure 31.22.

Target conductors

Condition of atmosphere & installation

Calculated ampacity

Allowable ampacity

Figure 31.22 Results of the ampacity calculation for transmission Line-1.

31.11.5

Example – Cable Ampacity and Cable Sizing for an Industrial System

A cable ampacity and sizing study was executed on the Example Model for Industrial System shown in Figure 31.17. The target cables for the study were six cables listed in Section 31.11.2.1.2 in the Feeder Cables table.

31.11.5.1 Standards

There are many standards to determine cable ampacity. ETAP can apply all of the following standards for a cable ampacity study:

•• •• •• ••

IEEE 399 ICEA P-54-440 NEC (NFPA 70) BS 7671 IEC 60364 IEC 60502 IEC 60092 NF C 15-100

31.11 Illustrative Examples

31.11.5.2 Ampacity of Feeder Cables

Correction factors vary according to the related standards and regulations. In general, the following correction factors are used for cable ampacity studies:

a

Ca: Correction factor for ambient temperature Tc −T where, Ca = Tc− Ta (T is operating temperature) Cg: Correction factor for grouping

Ta = 15/35 ( C)a Tc = 90/90 ( C)a

Cr: Correction factor for soil thermal resistivity

RHO = 120/90 ( C-cm/W)a

Cu: User-defined correction factor

Cu = 0.9

Rows = 2, columns = 3

Base/Operating ( C).

Here, in accordance with IEEE 399, we determined the cable ampacity of the six cables. Figure 31.23 shows the applied standard, correction factors, and results of the ampacity study for the target cable C-HV1.

Target cable Standard

Correction factors

Results

Figure 31.23 Cable Ampacity Editor for cable ID C-HV1.

31.11.5.3 Results of the Ampacity Study

The results of the ampacity study for the six cables are summarized in Table 31.19. Derated (A) in the Cable Ampacity column were calculated based on the full-load amperage (FLA) of loads, not on their operating current. That is, Derated (A) should be greater than Load FLA (A), considering the correction factors Ca, Cg, Cr, and Cu. Derated (%) is calculated as (= Derated (A)/Base (A) × 100%). 31.11.5.4 Cable-Sizing Study

Parameters of Cable ID C-HV1 for cable sizing are as follows. Loading (FLA base)

524.9 A

Maximum voltage drop (at normal operation)

2%

Apply power factor

84.33%

Short-circuit current (kA and duration)

36.33kA, 0.2 seconds

Multiplication factor (MF)

1.25

831

832

31 Load Flow

Table 31.19 Results of cable-ampacity study. Applied cable (standard BS6622)

Load

Cable ampacity (FLA base) Derated

Cable ID

ID

Capacity (MVA or MW)

Voltage (kV)

FLA (A)

Voltage (kV)

C-T1

T-1

70

MVA

66

612.3

C-T2

T-2

70

MVA

66

5

MW

11

10

MVA

11

C-SM1 SM-1 C-HV1 L-HV-1

Cores

Size (mm2)

66

3

612.3

66

292.6

11

Q’ty

Base (A)

(A)

240

3

1440

742.7

52

3

240

3

1440

742.7

52

3

120

2

580

357.3

62

(%)

524.9

11

3

400

2

1040

607.5

58

C-IM3 IM-3

5.45

MW

6.6

542

6.6

3

185

3

1110

668.8

60

C-HV2 L-HV-2

5

MVA

6.6

437.4

6.6

3

240

2

850

508.1

60

Here, in accordance with IEEE 399, we determined the size of the six feeder cables. Figure 31.24 shows the constraint options; Multiplication Factor, Growth Factor, and Service Factor; and sizing result for the target cable C-HV1.

Target cables

Results of sizing

Constraints

MF, GF, & SF as Options

Figure 31.24 Cable Sizing (Phase) Editor for cable ID C-HV1.

31.11.5.5 Results of Cable Sizing Study

The results of the cable-sizing study for the six cables are summarized in Table 31.20.

31.11 Illustrative Examples

833

Table 31.20 Results of cable-sizing study.

Load

Constraints

ID

Capacity (MVA or MW)

Voltage (kV)

FLA (A)

FLA (A)

Max. VD (%)

PF (%)

C-T1

T-1

70

MVA

66

612.3

612.3

2

95.62

C-T2

T-2

70

MVA

66

612.3

612.3

2

90.46

C-SM1

SM-1

5

MW

11

292.6

292.6

2

93.15

C-HV1

L-HV-1

10

MVA

11

524.9

524.9

2

90.00

C-IM3

IM-3

5.45

C-HV2

L-HV-2

5

Cable ID

Cable sizing FLA base

Applied cable

SC Current

Voltage (kV)

Cores

Size (mm2)

Q’ty

Size (mm2)

Q’ty

9.699

66

3

240

3

240

3

9.699

66

3

240

3

240

3

20.82

11

3

120

2

120

2

20.82

11

3

400

2

400

2

MW

6.6

542

542

2

93.62

36.33

6.6

3

185

3

185

3

MVA

6.6

437.4

437.4

2

90.00

36.33

6.6

3

240

2

240

2

The formulas for voltage drop (%) during normal operation and when starting a three-phase motor are as follows: Normal operation

At motor start

Formula 1 3 × I × rl cosθ + xl sinθ ε= 10000 × Vs Formula 2 3 × Vm × I × rl cosθ + xl sinθ ε= 10000 × Vs2

Formula 1 3 × Is × rl cosθS + xl sinθS εs = 10000 × Vs Formula 2 3 × Vm × Is × rl cosθs + xl sinθs εs = 10000 × Vs2

where ε, εs : Calculated voltage drop (%) based on the selected cable size l : Actual cable length (m) of the load Vs : System (bus) voltage (kV, line-line) Vm : Motor rated voltage (kV) I, Is : Motor full-load current or starting current (A) r : Resistance (ohm/km) of the selected cable size x : Reactance (ohm/km) of the selected cable size cos θ, cos θs : Motor power factor at full load, or at start sin θ, sin θs : Motor reactance factor at full load, or at start The formulas to determine the optimal cross-sectional area A (mm2) of a cable based on a calculated short-circuit current (I) and duration (t): Copper conductor I = 224

A t

Aluminum conductor I = 150

In A t

234 + θ1 234 + θ0 In

230 + θ1 230 + θ0

where I : short-circuit current A A : Cross-sectional area (mm2) t : short-circuit duration (sec) θ0 : Max. permissible continuous operating temp (90 C) for XLPE insulated cable θ1 : Max. permissible temperature at short circuit (250 C) for XLPE insulated cable

834

31 Load Flow

31.11.6

Example – Underground Cables for an Industrial System

Cable-derating analysis is an important part of power-system design and analysis. When designing a new system, this determines the proper size of cables to carry the specified loads. An analysis of an existing system examines cable temperatures and determines their ampacities. ETAP provides five types of calculations for cable-derating analysis: steady-state temperature, uniform-ampacity cable ampacity, uniform-temperature cable ampacity, cable sizing, and transient temperature. The steady-state temperature calculation is based on IEC 60287 or the NEC accepted Neher-McGrath method. The IEC 60287 steady-state temperature calculation fully complies with the latest standards as listed here: Standard

Title

IEC 60287-1-1 Ed. 1.2 b:2001

Electric cables – Calculation of the current rating – Part 1–1: Current rating equations (100% load factor) and calculation of losses – General

IEC 60287-2-1 Ed. 1.1 b:2001

Electric cables – Calculation of the current rating – Part 2–1: Thermal resistance – Calculation of thermal resistance

IEC 60287-2-1 Amd.1 Ed. 1.0 b:2001

Amendment 1

IEC 60287-2-1 Amd.2, 2006-03

Amendment 2

IEC 60287-3-1 Ed. 1.1 b:1999

Electric cables – Calculation of the current rating – Part 3–1: Sections on operating conditions – Reference operating conditions and selection of cable type

IEC 60287-3-1 Amd.1 Ed. 1.0 b:1999

Amendment 1

31.11.6.1 Steady-State Temperature Calculation

The steady-state cable temperature calculation determines the temperature of all the cable conductors involved in the raceway system under a specified loading condition. The calculation is based on the IEC 60287 standard or the NEC accepted Neher-McGrath approach, which employs a thermal circuit model to represent heat-flow situations. It is assumed that the cables have been carrying the specified load long enough that the heat flow has reached its steady state and no more changes of temperature will occur throughout the raceway system. The cable temperature calculated is dependent on the raceway system configuration, cable loading, and location of each particular cable. Here, in accordance with Neher-McGrath method, we determined the steady-state temperature of the six cables listed in Section 31.11.2.1.2 in the Feeder Cables table. 31.11.6.2 Configuration of the Raceway (Duct Bank)

The dimensions of the duct bank and the cable layout of cables are described in Figure 31.25.

Figure 31.25 Dimension of duct bank, and layout of cables.

31.11 Illustrative Examples

31.11.6.3 Required Data

In addition to the cable data mentioned in Section 31.11.2.1.2, the following data are required for this study: Type of raceway (duct bank or direct buried)

This example uses a duct bank.

Dimension of duct bank and layout of cables

See Figure 31.25

Soil data (material type, RHO, temperature):

Raceway information:

Conduit and location information:

31.11.6.4 Results of the Steady-State Temperature Study

The results of the steady-state cable temperature study are shown in Figure 31.25 (page 834) and also summarized in the ETAP output reports shown in Figure 31.26 (page 836). In this study, the steady-state cable temperatures for all cables were in the allowable range (below 90 C).

31.11.7

Example – OPF for a Transmission System

OPF solves power-system load flows, but at the same time can optimize system operating conditions and automatically adjust control variable settings, taking into account the following conditions:

• •

Constraints for bus voltage, branch flow of various types (MVA, MW, Mvar, and amp), and control variable adjustable bounds Any practical controls in a power system, such as transformer LTC, generator AVR controls, shunts and series compensations, and load shed

835

Analysis Reults (RW-1)

No.

Cable ID

Conduit/Location ID

Conductor per Cable

Energized Conductor per Cable

Dielectric Rdc @ Losses Final Temp. μOhm/m

Watt/m

Conductor Yc

Ys

Losses

Current

Watt/m

Amp

Temp. °C

1

C-Tl-1

Cond-11

3

3

90.10

1.170

0.022

0.079

11.504

204.10

69.62

2

C-Tl-2

Cond-11

3

3

90.10

1.170

0.022

0.079

11.504

204.10

69.62

3

C-Tl-3

Cond-11

3

3

90.10

1.170

0.022

0.079

11.504

204.10

69.62

4

C-Tl-1

Cond-12

3

3

91.39

1.170

0.021

0.078

11.661

204.10

73.96

5

C-Tl-2

Cond-12

3

3

91.39

1.170

0.021

0.078

11.661

204.10

73.96

6

C-Tl-3

Cond-12

3

3

91.39

1.170

0.021

0.078

11.661

204.10

73.96

Losses

Current

Temp.

Watt/m

Amp

Analysis Reults (RW-2)

No.

Cable ID

Conduit/Location ID

Conductor per Cable

Energized Conductor per Cable

Dielectric Rdc @ Losses Final Temp. μOhm/m

Watt/m

Conductor Yc

Ys

0.080

0.103

13.113

262.45

83.52

°C

1

C-HV1-2

Cond-22

3

3

58.73

0.080

2

C-HV1-1

Cond-22

3

3

58.73

0.080

0.080

0.103

13.113

262.45

83.52

3

C-HV2-2

Cond-24

3

3

94.01

0.023

0.030

0.061

13.891

218.70

82.80

4

C-HV2-1

Cond-24

3

3

94.01

0.023

0.030

0.061

13.891

218.70

82.80

5

C-IM3-1

Cond-23

3

3

121.84

0.021

0.017

0.031

12.133

180.67

78.37

6

C-IM3-2

Cond-23

3

3

121.84

0.021

0.017

0.031

12.133

180.67

78.37

7

C-IM3-3

Cond-23

3

3

121.84

0.021

0.017

0.031

12.133

180.67

78.37

8

C-SM1-1

Cond-21

3

3

186.57

0.048

0.017

0.019

12.059

146.30

75.82

9

C-SM1-2

Cond-21

3

3

186.57

0.048

0.007

0.019

12.059

146.30

75.82

Figure 31.26 ETAP output report for the steady-state temperature study.

31.11 Illustrative Examples

Optimization techniques used in ETAP are summarized in Section 3.10. The optimized system will reduce installation and/or operating costs, improve overall system performance, and increase the system’s reliability and security. The example OPF study was executed on the Example Model for Transmission System shown in Figure 31.15. 31.11.7.1 Objective Conditions and Constraints for Optimization

In addition to the required data mentioned in Table 31.5, the objective conditions and constraints for optimization shown in Figure 31.27 are required. The objective conditions for this study are based on Scenario OPF-1 Minimize Real Power Loss and OPF-2 Minimize Swing Bus Power. Objective Conditions and Constraint

Objective Conditions Scenario OPF-1 = Minimize Real Power Losses OPF-2 = Minimize Swing Bus Power

• Bus voltage constraints

• Branch power constraints

• Generator AVR constraints

• Generator MW constraints

Bus ID Bus-2

kV

% Min

% Max

Weight

245

95

105

100

Bus-3

245

95

105

100

Bus-4

24

95

105

100

Bus-5

72.5

95

105

100

Bus-6

72.5

95

105

100

Branch ID

Type

Constr

Base

Max

Line-1

Line

Amp

871

871

0

Line-3A Line-2

Line

Amp

436

436

0

Line

Amp

871

871

0

Line-3B T-1

Line

Amp

436

436

0

Xfmr

MVA

150

150

0

T-2

Xfmr

MVA

100

100

0

T-3

Xfmr

MVA

100

100

0

Min

Gen ID

MVA

% Vmax

% Vmin

Weight

Generation Plant

167

105

95

100

Utility

1500

105

95

100

Gen ID

MVA

Max MW

167

0.15

Min MW 0

Weight

Generation Plant Utility

1500

0

0

100

Figure 31.27 Objective conditions and constraints for optimization.

100

837

838

31 Load Flow

31.11.7.2 Results of the OPF Study

The study scenarios were as follows: Scenario OPF-1

Minimize Real Power loss

Scenario OPF-2

Minimize Swing Bus Power

Scenario LF-1

Traditional load flow

We ran these three study scenarios with the system configuration “Normal (Run All System) case” shown in Table 31.6. The results are summarized in Figure 31.28. Comparisons of the results of the three scenarios are summarized here: Minimize Real Power Loss

Minimize Swing Bus Power

Traditional Load Flow

Scenario OPF-1

0.780 MW

Scenario OPF-2

0.830 MW

Scenario OPF-1

72.822 MW

Scenario OPF-2

66.986 MW

Scenario LF-1

70.367 MW

(a) Minimize real power losses (Results of scenario OPF-1)

MW Source (Swing Buses):

Mvar

MVA

72.822

–1.673

72.841

0.000

0.000

0.000

162.042

20.721

163.362

Total Line Charging:

0.000

–32.540

Apparent Losses:

0.780

21.195

System Mismatch:

0.000

0.000

Source (Non-Swing Buses): Total Demand:

Number of Iterations: 10

Figure 31.28 Results of the optimal and traditional load-flow study.

%PF 99.97 PF Lagging

99.19 PF Lagging

(Minimum)

(Minimum)

31.11 Illustrative Examples

(b) Minimize swing bus power (Results of scenario OPF-2)

MW Source (Swing Buses):

Mvar

MVA

66.986

3.090

67.057

0.000

0.000

0.000

156.156

23.271

157.881

Total Line charging:

0.000

–28.055

Apparent Losses:

0.830

23.408

System Mismatch:

0.000

0.000

Source (Non-Swing Buses): Total Demand:

Number of Iterations: 13

Figure 31.28 (Continued)

%PF 99.89 PF Lagging

98.91 PF Lagging

839

840

31 Load Flow

(c) Traditional load flow (Results of scenario LF-1)

MW

Mvar

MVA

%PF

Source (Swing Buses):

70.367

1.879

70.392

99.96 Lagging

Source (Non-Swing Buses):

90.000

43.589

100.000

90.00 Lagging

Total Demand:

160.367

45.468

166.688

96.21 Lagging

Total Motor Load:

121.600

39.968

128.000

95.00 Lagging

Total Static Load:

37.968

12.479

39.966

95.00 Lagging

Total Constant I Load:

0.000

0.000

0.000

Total Generic Load:

0.000

0.000

0.000

Apparent Losses:

0.799

–6.980

System Mismatch:

0.000

0.000

Number of Iterations: 4

Figure 31.28 (Continued)

841

32 Short-Circuit/Fault Analysis 32.1

Introduction

Faults usually occur in a power system due to physical damage, human error, insulation failure, flashover, and more. Faults may comprise all three phases in a symmetrical manner or may be asymmetrical, where only one or two phases are involved. Faults may also be caused by short circuits to ground/earth or between live conductors, or by broken conductors in one or more phases. Sometimes simultaneous faults occur involving both short-circuit (SC and broken-conductor faults (also known as open-circuit or series faults). SC analysis is carried out in electrical power utility transmission and distribution systems, industrial power systems, commercial power systems, and power station auxiliary systems. Other specialized applications are in concentrated power-system installations on board ships and aircraft.

32.2

Analysis Objectives

SC analysis is usually performed using per-unit quantities (similar to percentage quantities), as the solutions are relatively consistent over different voltage and power ratings, and operate on values on the order of unity. Fault-analysis calculations are performed for a number of reasons, listed here in no particular order or importance:

• •

•

System design and sizing: SC current calculations are made at the system-design stage to determine the SC ratings of new switchgear and substation equipment to be commissioned. System reinforcements may be triggered by network expansion and/or the connection of new generating plant to the power system. Routine calculations are also made to check the continued adequacy of existing equipment as system operating configurations are modified. Personnel safety: SC fault analysis is carried out to ensure the safety of workers as well as the general public (especially equipment 1000 Volts)

Momentary asymmetrical. RMS kA

Bracing asymmetrical

Momentary asymmetrical. crest kA

Bracing crest

Momentary symmetrical. RMS kA

Bracing symmetrical

Momentary asymmetrical. RMS kA

Bracing asymmetrical

LV bus ( = 50 HP Small group: 975 HP total with each motor 730.5 V (outside tolerable limits) Step potential = 658.1 V < 2429.7 V

Table 32.17 summarize the results of the “Optimized Number of Conductor” study (number of rods = 6, as in the original), where the number of conductors needed is 19 (X = 7, Y = 12). Based on these conductors and rods, both touch potential and step potential are within tolerance. Table 32.18 summarize the results of the “Optimized Number of Conductor and Rods” study, where the number of conductors needed is 19 (X = 7, Y = 12) and number of rods required is 4. Based on these conductors and rods, both touch potential and step potential are within tolerance.

Figure 32.26 Grid size and arrangement.

877

Figure 32.27 Conductor, rod, and soil data.

Figure 32.28 Study case (study conditions).

Table 32.16 Results for the originally designed system. Rg

GPR

Touch Potential

Step Potential

Ground Resistance

Ground Potential Rise

Tolerable

Calculated

Tolerable

Calculated

ohm

Volts

Volts

Volts

%

Volts

Volts

%

1.263

5320.8

730.5

738.3

101.1

2429.7

658.1

27.1

32.6 Illustrative Examples

As a result of optimal design, when we select a Conductors value of 17, X direction = 7, Y direction = 12, and a Rods value of 4,” all of following indexes are within tolerance:

•• ••

Step voltage Touch voltage GPR Grounding resistance

Table 32.17 Results for the study of the optimized number of conductors. Rg

GPR

Touch Potential

Step Potential

Ground Resistance

Ground Potential Rise

Tolerable

Calculated

Tolerable

Calculated

ohm

Volts

Volts

Volts

%

Volts

Volts

%

1.257

5297.2

730.5

709.9

97.2

2429.7

658.7

27.1

Table 32.18 Results for the study of the optimized number of conductors and rods. Rg

GPR

Touch Potential

Step Potential

Ground Resistance

Ground Potential Rise

Tolerable

Calculated

Tolerable

Calculated

ohm

Volts

Volts

Volts

%

Volts

Volts

%

1.259

5306.4

730.5

728.9

99.8

2429.7

670.8

27.6

879

881

33 Motor Starting 33.1

Methods

33.1.1

Motor Fundamentals

A motor is a device that converts electrical energy into mechanical energy. It is also known as a torque-producing device. Torque is defined as a turning or twisting force supplied by a drive to the load. Two significant assemblies make up an AC motor in order to produce this motor or machine torque: the rotor and the stator. The basic operation of an induction motor is similar to transformer, where the stator acts as the primary side of the transformer and the rotor as the secondary element of the transformer. The rotor is made up of the shaft, the rotor core (a stack of steel laminations that form slots, and aluminum conductors and end rings formed by either a die-cast or fabrication process), and sometimes a fan. This is the rotating part of the electromagnetic circuit. The other major part is the stator. The stator is formed from thin steel laminations stacked and fastened together, so the notches (called slots) form a continuous lengthwise slot on the inside diameter. Insulation is inserted to line the slots, and then coils wound with many turns of wire are inserted into the slots to form a circuit. Each grouping of coils, together with the steel core it surrounds, forms an electromagnet. Electromagnetism is the principle behind motor operation. The stator windings are connected directly to the power source. There are primarily two types of motors: induction motors and synchronous motors. 33.1.1.1 Induction Motors

The induction motor derives its name from the fact that the rotor is not connected electrically to the source of power supply. The currents that circulate in the rotor conductors are not produced directly by the voltage of the power supply; they result from the voltage being induced in the rotor by the magnetic field of the stator. For its operation, the induction motor depends on a rotating magnetic field. This field is set up by the current flowing in the stator windings. The magnetic field, rotating around the outer surface of the stator, cuts across the conductors embedded in the rotor and induces voltages in the conductors. The voltages induced in the rotor windings cause currents to flow in the rotor conductors. The rotor takes the form of an iron cylinder threaded with a symmetrical arrangement of short-circuited conductors (usually called a squirrel-cage rotor because of its appearance), and the rotating magnetic field generated by the stator induces large currents in the short-circuited conductors. The current-carrying conductors are then exposed to a magnetic field. This causes a repelling force between the conductors and the field and produces a torque that causes the rotor to turn. The strength of the current induced in the rotor is determined by the rate at which the field is changing within the rotor and is large when the rotor is stationary, falling to zero when the rotor speed equals the rotating field speed. This means the standard squirrel-cage rotor machine can never reach synchronous speed – the rotor speed must always be sufficiently lower than the rotating field speed (the slip speed or slip frequency) to induce sufficient current in the rotor conductors.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

882

33 Motor Starting Winding 1

Winding 2 0°

90°

180°

270°

0°

360°

N

90°

N

180°

270°

N

N

360°

N

N

N

Single-phase stator Current and magnetic Field vectors

Two-phase stator Current and magnetic Field vectors

Winding 3

0°

90°

180°

270°

360°

Winding 1

Winding 2

N N

N

N

N

N

Three-phase stator Current and magnetic Field vectors Figure 33.1 Current change in winding for one cycle.

Figure 33.1 shows how current in each winding changes with time through one complete cycle of the supply frequency for each of the three cases together with the direction of the resulting magnetic field at the different points in the supply cycle.

33.1.1.2 Synchronous Motors

Synchronous motors are AC machines that have a field circuit supplied by an external DC source. The word synchronous originates from Greek. The prefix syn means with, and chronos means time. A synchronous motor literally operates “in time with” or “in sync with” the power supply system. In a synchronous generator, a DC current is applied to the rotor winding, producing a rotor magnetic field. The rotor is then turned by external means, producing a rotating magnetic field, which induces a three-phase voltage within the stator winding. In a synchronous motor, a three-phase set of stator currents produces a rotating magnetic field, causing the rotor magnetic field to align with it. The rotor magnetic field is produced by a DC current applied to the rotor winding. Field windings are the windings producing the primary magnetic field (rotor windings for synchronous machines); armature windings are the windings where the main voltage is induced (stator windings for synchronous machines). The rotor of a synchronous machine is a large electromagnet, laminated to reduce eddy current losses. The magnetic poles can be either salient (sticking out of rotor surface) or non-salient construction.

33.1 Methods

33.1.1.3 Difference Between Synchronous and Induction Motors

Synchronous and induction motors differ in the following manners:

• •• • • • • •• •

Synchronous motors operate at synchronous speed (rpm = 120f/p), while induction motors operate at less-thansynchronous speed (rpm = 120f/p – slip). Slip is nearly zero at zero load torque and increases as load torque increases. Synchronous motors require DC excitation for the rotor windings, while induction motors do not. Synchronous motors require a DC power source for the rotor excitation. Synchronous motors require slip rings and brushes to supply rotor excitation. Induction motors do not require slip rings, but some induction motors have them for soft starting or speed control. Synchronous motors require rotor windings, while induction motors are most often constructed with conduction bars in the rotor that are shorted together at the ends to form a squirrel cage. Synchronous motors require a starting mechanism in addition to the mode of operation that is in effect once they reach synchronous hp speed. Three-phase induction motors can start by directly applying power, but single-phase motors require an additional starting circuit. The power factor of a synchronous motor can be adjusted to be lagSynchronous ging, unity, or leading, while induction motors must always operate motor 1000 with a lagging power factor. Synchronous motors are generally more efficient than induction motors. Induction Synchronous motors can be constructed with permanent magnets in Synchronous motor or the rotor, eliminating the slip rings, rotor windings, DC excitation 500 induction system, and power factor adjustability. motors Synchronous motors are usually built only in sizes larger than about 1000 HP (750 kW) because of their cost and complexity. However, permanent magnet synchronous motors and electronically controlled rpm permanent synchronous motors called brushless DC motors are avail0 1000 2000 able in smaller sizes. Figure 33.2 shows typical application criteria Figure 33.2 Application of induction versus based on power and rotational speed. synchronous motors.

33.1.2

Motor Rated Power

AC motors used in North America are generally rated in horsepower. Equipment manufactured in Europe is generally rated in kilowatts (kW). Horsepower can be converted to kilowatts with the following formula: kW = 0 746 × HP or HP = 1 341 × kW Torque is the turning or twisting force supplied by a drive to the load. Units of measure are inch-pounds or footpounds. Torque and horsepower are related to each other by a basic formula that states Power kW =

T ×S 9 5488

where T : torque in N-m S : speed in rpm HP =

Torque × Speed T ×S or HP = Constant 5252

where T : torque in lb.ft. S : speed in rpm 33.1.3

Torque – Speed

The majority of induction motors are fitted with low-resistance rotors (i.e. large area of cross-section conductors), as this permits large induced currents to flow in the rotor conductors when the rotor speed is only slightly less than synchronous speed. At this low slip frequency, the rotor inductance has little effect, and almost all the circulating current

883

33 Motor Starting

250 200 150 100 50

100

80

60

40

20

0

Machine %full load torque

884

0

Slip (%) Figure 33.3 Typical motor torque-speed curve.

produces useful torque at the output shaft. However, if for any reason the rotor speed is reduced by a large amount, the slip frequency increases, and at this higher frequency the rotor inductance becomes a major part of the rotor impedance; this both limits the induced current in the rotor conductors and moves the effective magnetic field pattern, so only part of the induced current produces useful output torque. This results in the motor characteristic torque versus slip curve shown in Figure 33.3. At synchronous speed, there is no relative movement between the rotor conductors and the rotating field generated by the stator, so the output torque is zero. As the rotor speed drops (increasing the slip frequency and rotor current), the torque generated by the induced rotor currents rapidly increases. With about 1% slip, enough torque is generated to overcome the motor no-load mechanical losses, giving a no-load speed of close to 99% of synchronous. The motor rated full load torque is typically reached at 95% of synchronous, giving an almost-constant speed characteristic – only 4% change from no load to full load. If the load on the motor output shaft is increased further, as the speed drops, the output torque increases to a peak of two or three times the full load torque at somewhere between half and two-thirds synchronous speed. At lower speeds, although the circulating current in the rotor is still increasing, the effect of the rotor inductance becomes dominant because of the high slip frequency, and the available torque reduces, dropping to about twice the rated full load torque at zero speed. This is the starting torque (sometimes called locked rotor torque) and is the maximum load torque that the motor can overcome and accelerate toward its rated full load speed. Although the motor can deliver somewhat more torque at its peak torque speed, the speed range between this and zero speed is potentially unstable. If any attempt is made to load the motor in excess of this peak torque capability, the motor speed will drop and reduce the available torque. This is a cumulative effect: each drop in speed further reduces the available torque until the motor abruptly stalls. The motor can only operate stably in this part of the speed-torque curve if it is driving a load whose torque requirements reduce rapidly as speed drops. Fan loads are typical of this class. 33.1.4

Motor-Connected Load Types

There are three basic load types, and these types are classified by the relationship of horsepower and speed. Constant-torque applications are those that have the same torque at all operating speeds, and horsepower varies directly with the speed. About 90% of all applications, other than pumps, are constant-torque loads. Examples include conveyors, hoisting loads, surface winding machines, positive displacement pumps, and piston and screw compressors, as shown in Figure 33.4. Constant-horsepower applications have higher values of torque at lower speeds and lower values of torque at higher speeds. Examples include lathes, milling machines, drill presses, and center winders. Variable torque implies that the torque required varies as the square of its speed, and horsepower requirements increase as the cube of the speed. Examples include centrifugal pumps, turbine pumps, centrifugal blowers, fans, and centrifugal compressors. When the motor torque versus slip curve is superimposed on the load torque versus slip curve, the plot in Figure 33.5 is generated. This curve provides immediate insight into the load demand versus available torque as a function of speed. At rated voltage, if the motor-torque curve is less than the load-torque curve at any point, then the motor will fail to start. The speed at which the motor will stall can be approximately identified by analyzing the intersection of the two curves.

33.1 Methods

Constant horsepower 500

100

80

80

400

80

60 40

60 40

20

20

0

0

20 40 60 80 100 % Speed

% Horsepower

100

% Torque

100 % Horsepower

% Torque

Constant torque

300 200 100 0

20 40 60 80 100 % Speed

60 40 20 0

20 40 60 80 100 % Speed

20 40 60 80 100 % Speed

Variable torque

80

80

% Horsepower

100

% Torque

100

60 40 20

60 40 20 0

0 20 40 60 80 100 % Speed

20 40 60 80 100 % Speed

Figure 33.4 Typical load torque-speed curve.

% Torque = A0 + A1w + A2w2 + A3w3

% Torque = A0 + A1w + A2w2 + A3w3

225.00

200

168.75

150

Accelerating torque

100

112.50

50

56.25 Load torque required

0

0 0

25

50

75

100

% Speed

Motor will start at rated voltage

0

25

50

75

100

% Speed

Motor will fail to start and stall approximately 50% speed

Figure 33.5 Typical motor and load torque-speed curves superimposed.

The brown curve is the machine-torque curve, while the green curve is the load-torque curve. The difference between the regions is known as the acceleration torque. Greater acceleration torque means higher inertia that can be handled by the motor without approaching thermal limits. Typical motor and load torque-speed curves superimposed are shown in Figure 33.5.

885

886

33 Motor Starting

33.1.5

Motor Inrush

When AC motors are started with full voltage applied, they draw line currents substantially more significant than their full load running current rating. The magnitude of this inrush current is a function of motor horsepower and the design characteristics of the motor. In order to define the inrush characteristics and present them in a simplified form, a series of code letters group motors depending on the range of inrush in terms of kVA. Table 33.1 lists the code letter designations. Generally, standard motors of 15 HP or larger have code letters of G or lower; 10 HP and smaller motors have code letters of H or higher.

Table 33.1 National Electric Manufacturers Association (NEMA) motor code letters. Code

a

kVA/HPa

Approx. mid-range value

A

0.00–3.14

1.6

B

3.15–3.54

3.3

C

3.55–3.99

3.8

D

4.00–4.49

4.3

E

4.50–4.99

4.7

F

5.00–5.59

5.3

G

5.60–6.29

5.9

H

6.30–7.09

6.7

J

7.10–7.99

7.5

K

8.00–8.99

8.5

L

9.00–9.99

9.5

M

10.00–11.99

10.6

N

11.20–12.49

11.8

P

12.50–13.99

13.2

R

14.00–15.99

15.0

Locked-rotor kVA per horsepower.

33.1.6

Motor-Starting Methods

Methods of starting AC induction motors can be broken down into four basic categories: full-voltage (across-the-line) starting, electromechanical reduced-voltage starting, solid-state reduced-voltage starting, and variable-frequency drive (VFD) starting. Electromechanical reduced-voltage starting has been in existence nearly as long as the induction motor itself. This starting method encompasses autotransformer starting, Wye-delta (star–delta) starting, and resistor/reactor starting. Each of these methods requires the use of some type of mechanical switch or contact. Electromechanical starting is the most common method of reduced-voltage starting used in industry today. Full-voltage starting produces the most significant amount of starting torque. High starting torque is generally desired when trying to start a high-inertia load, to limit the acceleration time. However, in some instances, this high starting torque may damage the mechanical system. Gears or chains might be broken or damaged. Strain or slippage may reduce belt life. Gearboxes are also put under a more significant stress and are subject to more abuse. Voltage drop on the system must be carefully studied, and the breakers and relays need to be coordinated with upstream devices to prevent nuisance tripping of these devices during startup. If the voltage-drop limitation for the system is exceeded, other methods of starting should be considered. Beyond the initial shock of inrush current and torque, this type of starting does result in a smooth acceleration characteristic with the shortest acceleration time, which offers an advantage over some of the other available methods of starting.

33.1 Methods

33.1.6.1 Direct Across-the-Line Start or Direct Online Starting (DOL)

This is the most simple and inexpensive method of starting a squirrel-cage induction motor. The motor is switched on directly to full supply voltage. The initial starting current is large, typically about 5–7 times the rated current, but the starting torque is likely to be 0.75–2 times the full load torque. To avoid excessive supply-voltage drops due to large starting currents, this method is restricted to small motors only. To decrease the starting current, cage motors of medium and larger sizes are started at a reduced supply voltage. 33.1.6.2 Star-Delta Starting

This is applicable to motors designed for delta connection in normal running conditions. Both ends of each phase of the stator winding are brought out and connected to a three-phase change-over switch. For starting, the stator windings are connected in star; and when the machine is running, the switch is thrown quickly to the running position, thus connecting the motor in delta for normal operation. The phase voltages and the phase currents of the motor in star connection are reduced to 1/√3 of the direct-on-line values in delta. The line current is one-third of the value in delta. A disadvantage of this method is that the starting torque (which is proportional to the square of the applied voltage) is also reduced to one-third of its delta value. 33.1.6.3 Rotor Resistance Starting

This type of starter is typically used with wound rotor induction motors where the motor has a stator like a squirrel-cage motor but a rotor with insulated windings is brought out via slip rings or brushes. The sole purpose of the slip rings is to allow resistance to be placed in series with the rotor windings, as shown here: ϕ1 ϕ2 M1 ϕ3 Stator

Rotor

Start resistance

These types of motor starters are used for a high inertia load or a load that requires a starting torque across a full speed range. A wound rotor induction motor may also act as a generator when driven over synchronous speed. The external resistance introduced in the rotor circuit is able to develop high torque at low speeds and is shorted out once the motor is started to make the rotor electrically equivalent to its squirrel-cage counterpart. By adding external resistance to the rotor circuit, the magnitude of the starting torque can be controlled using resistance values where the number of starting steps varies depending on the peak torque requirements of the application. High torque can be maintained throughout the starting period by gradually cutting out the resistance. The added resistance also reduces the starting current, so that a starting torque in the range of 2–2.5 times the full load torque can be obtained at a starting current of 1–1.5 times the full load current, as shown in Figure 33.6. RL

IST

XL

RLR

50% Tap

VMCC

0.5VMCC

VM XLR

Figure 33.6 Rotor-resistance motor starting.

The starting current in the rotor winding is Ir =

Er Rr + Rext

2

+ Xr

2

887

888

33 Motor Starting

where Rext : additional resistance per phase in the rotor circuit Er : voltage applied to the rotor Rr : rotor resistance Xr : rotor reactance 33.1.6.4 Reactor Starting

A reactor is inserted during motor starting that reduces start current and the voltage drop in the network. The starting current reduces linearly with the voltage across the motor. The reactor may be inserted in the line to the motor or between the motor and its neutral point and is furnished with 50, 65, and 80% taps. The reactor can obviously be left permanently in circuit after starting the motor. One advantage of the reactor method is that the distribution of the voltage drop over the reactor and the motor will vary during acceleration. The power factor of the motor increases with its speed and consequently so does the motor voltage. This has a boost effect on the motor torque, specifically for synchronous motors where the motor torque drops off at the end of the acceleration. The starting current (Ist) in amps is Ist =

En × 3 Xm + Xr + Xn

Where Xm : motor reactance En : nominal supply voltage (bus voltage) Xn : network reactance Xr : rotor reactance 33.1.6.5 Auto-Transformer Starting

An auto transformer is inserted in the line to the motor, reducing the voltage to the motor terminals, depending on the tap selected. After starting the motor, the transformer is bypassed. This method also reduces the initial voltage applied to the motor and therefore the starting current and torque. The motor, which can be connected permanently in delta or in star, is switched first on reduced voltage from a three-phase tapped autotransformer; when it has accelerated sufficiently, it is switched to the running (full voltage) position, as shown in Figure 33.7. The principle is similar to star/delta starting and has similar limitations. The advantage of the method is that the current and torque can be adjusted to the required value by taking the correct tapping on the autotransformer. Auto transformers are very similar to reactors and in some cases are readily interchangeable. The standard voltage taps provided are usually 80, 65, and 50%. However, this starter multiplies current by the transformer ratio, making it more effective than a reactor. This method is more expensive because of the additional autotransformer. The starting current (Ist) in amps is Ist = Is × K 2 ×

Xm Xm + Xr + Xn

2

+ Itrm

Where Xn : network reactance Xr : rotor reactance Xm : motor reactance Itrm : magnetizing current of the auto transformer K : voltage tap Example: 50% Tap

line

MCC

RL

0.3 IST

VMCC M

XL

IST VMCC

Autotransfer starter Figure 33.7 Auto-transformer motor starting.

0.5VMCC 50% Tap

RLR VM XLR

33.1 Methods

33.1.6.6 Comparison Table

Table 33.2 provides a comparison of typical locked-rotor current and acceleration time for a sample motor using various starting methods. Table 33.2 Comparison of motor-starting methods. Line voltage = motor-rated voltage

Type of starter

Motor voltage line voltage

Starting torque full-voltage starting torque

Line current full-voltage starting current

Full-voltage starter

1.0

1.0

1.0

0.80

0.64

0.68

a

Autotransformer: 80 per cent tap………………… 65 per cent tap…………………

0.65

0.42

0.46

50 per cent tap…………………

0.50

0.25

0.30

Resistor starter, single step (adjusted for motor voltage to be 80 per cent of line voltage)

0.80

0.64

0.80

0.50

0.25

0.50

Reactor: 50 per cent tap………………… 45 per cent tap…………………

0.45

0.20

0.45

37.5 per cent tap………………

0.375

0.14

0.375

Part-winding starter (low-speed motors only):

a

75 per cent winding………………

1.0

0.75

0.75

50 per cent winding………………

1.0

0.50

0.50

The settings given are the more common for each type.

33.1.7

Motor Inertia

Inertia is defined as a body’s resistance to a change in velocity. An object’s moment of inertia (commonly referred to as WR2 or Wk2) is the product of the weight of the object and the square of the object’s radius of gyration. The radius of gyration is a measure of how the object’s mass is distributed about the center of rotation and is commonly expressed in units of feet. Inertia of the given load is a major factor when determining both the acceleration time and the motor heating. Motor acceleration time may be calculated using the following formula: ta =

Wk 2 × ΔN ; 308 × Ta

where ta = acceleration time (seconds) Wk2 = total connected moment of inertia (lb-ft2) ΔN = speed change during time (rpm) Ta = average accelerating torque (lb-ft) Note: 1 × WR2 2 2) Stored Kinetic Energy in kW-sec = 2.31 × (Total WK2) × RPM2 × 10−7 Stored Kinetic Energy in kW Seconds 3) Inertia Constant H = HP × 0 746 4) Wk2 = 5.93 × GD2 = 23.72 × Gr2. The units of GD2 are expressed in kg-m2, while the units of Wk2 are lb-ft2.

1) WK 2 =

889

890

33 Motor Starting

GD2 is used in European countries to represent a body’s resistance to changes in rotational speed. Wk2 is more widely used in North America, but it expresses the same physical property as GD2. The primary difference between GD2 and Wk2 is that GD2 is calculated using the diameter of gyration instead of the radius of gyration. When the diameter of gyration is squared, the value will change the number by a factor of 4 relative to Wk2. GD2 is also a mass-based quantity, while Wk2 is a weight-based quantity. If the curve of the load torque is complex and the motor torque is not constant, it is advantageous to divide the computation into individual zones. Acceleration time for non-constant torques shows summation of all zones with multiplication of moment of inertia and change in speed. ta =

j × ΔN ; 9 55 × Ma

where ta = acceleration time (seconds) J = total connected moment of inertia (kg-m2) ΔN = speed change during time (rpm) Ma = accelerating torque (N-m) 33.1.8

Motor Nameplate

A stainless steel rating plate is attached permanently to the machine frame, and it must not be removed, as shown in Figure 33.8. These ratings are defined based on electric motor standards. The electric motor standards can be grouped into two major categories: National Electric Manufacturers Association (NEMA) and International Electrotechnical Commission (IEC) (and its derivatives). In North America, NEMA sets motor standards, including what should go on the nameplate (NEMA Standard MG 1). The IEC sets the standards for many other countries, or other countries base their standards very closely on the IEC standards. For example, Germany’s VDE 0530 standard and United Kingdom BS 2613 standard are close to IEC with minor exceptions. The motor nameplate indicates manufacturing, identification, electrical, and mechanical information, as indicated here: Type designation Manufacturing year Duty Type of connection Insulation class Machine weight (kg) or (lbs) Degree of protection (IP class) Type of cooling (IC code) Mounting arrangement (IM code) (IEC) Additional information Manufacturer Serial number

•• •• •• •• •• ••

EFF1

AC MOTOR IEC 60034 TYP

SER. NO.

YEAR

KW

r/min

V

A

HZ

KW

r/min

V

A

HZ

DUTY

INSUL

AMB

COS Ø

CODE

IP

°C RISE IC

GREASE DIAG

K DESIGN

SERVICE FACTOR DE BRG

IA/IN

MA/MN

Figure 33.8 Typical IEC motor nameplate.

3 PHASE

NDE BRG kg MOTOR WT

33.1 Methods

•• •• •• •• •• •

Output (kW) or (HP) Stator voltage (V) Frequency (Hz) Rotating speed (rpm) Stator current (A) Power factor (cos θ) Standard marking Standard Designation for locked-rotor kVA/HP (NEMA) Ambient temperature ( C) (NEMA) Service factor (NEMA)

33.1.9

Electric Motor Standard Comparison

IEC is the typical motor standard used in Europe and other parts of the world, while its equivalent in North America is NEMA. The main differences between the two standards are highlighted next and will be discussed in further detail:

• • • •

Frame relationships: Both IEC and NEMA motors use a letter code to specify the physical frame dimensions. The NEMA and IEC standards use different terms, but they are essentially analogous in ratings and, for most common applications, are mostly interchangeable as shown in Figure 33.9. IEC Design N motors are similar to NEMA Design B motors, the most common motors for industrial applications. IEC Design H motors are nearly identical to NEMA Design C motors. There is no specific IEC equivalent to the NEMA Design D motor. Enclosure designations: Like NEMA, IEC has designations indicating the protection provided by a motor’s enclosure. However, where NEMA designations are in words, such as Open Drip Proof or Totally Enclosed Fan Cooled, the IEC uses a two-digit Index of Protection code to describe how well the enclosure protects the motor from the environment. Cooling designations: IEC uses a letter and number index of cooling (IC) code to designate how a motor is cooled. There is an individual code for just about every type of cooling method, from small fan-cooled motors to large liquidcooled motors. Duty cycles: NEMA motors refer to the duty cycle in one of two or three terms: continuous, intermittent, or special duty (typically expressed in minutes). IEC breaks it into eight ratings instead (S1–S8).

300

Torque (% of full-load torque)

Design D 250 200

Design C or H

150 Design A, B or N 100 50 0

0

20 40 60 80 Speed (% of synchronous speed)

Figure 33.9 Frame relationships.

100

891

892

33 Motor Starting

• • • •

Design types: The IEC design rating code describes a motor’s speed versus torque characteristics. The IEC design codes nearly mirror NEMA design types, but with different letters. For example, the most common industrial motor is an IEC Design N motor, which is very similar to a NEMA Design B motor, the most common type of motor for industrial applications. By the same token, the characteristics of IEC Design H are nearly identical to those of NEMA Design C. There is no specific IEC equivalent to NEMA Design D. Logically, the IEC suffix has a meaning. Comparing torques of IEC Design N (think of it as “normal” torque) motors in general mirror those of NEMA Design B motors. The torque of IEC Design H (think of it as “high” torque) is nearly identical to that of NEMA Design C. Insulation designations: IEC and NEMA use the same classification system for winding insulation. It is based on the highest temperature the material can withstand continuously without degrading or reducing motor life. Most industrial-duty motors use Class B or Class F insulation, depending on the application. IEC and NEMA 1.0 service factor ratings are nearly identical; NEMA 1.15 ratings are higher. IEC motors do not have a service factor rating definition. Instead, the temperature rise, ambient temperature, and altitude ratings are defined via the kW output rating. If an increased service factor is required, use the next size larger motor. kW and HP: This is the rated shaft power output at the rated voltage, current, and frequency. IEC uses kilowatts (kW), and NEMA uses horsepower (HP). The conversion between the two is 1 HP = 745.7 W = 0.7457 kW. Rated voltage: IEC standard 34-1 requires that motors be able to provide their rated output at their rated efficiency for a voltage range of 95–105% of the rated voltage.

33.1.10

Motor and System Frequency Impact

Ideally, we should supply power to AC motors only at the frequency designated on the nameplate. In actual practice, it is often satisfactory to apply 50 Hz power to a 60 Hz motor or vice versa. Some AC motors are rated for use at either 50 or 60 Hz. On the other hand, motors designed for operation at either of these frequencies would be almost certain to malperform at 16.67 or at 400 Hz. There are a number of variables that enter the picture, including operational duty-cycle, ventilation, starting torque, rating conservatism, speed tolerance, allowable temperature rise, power line regulation, etc. Keep in mind that the ratedload speeds of induction and synchronous motors differ for 50 or 60 Hz lines. For example, four-pole synchronous motors, whether of 50 or 60 Hz nameplate rating, rotate at 1500 rpm on 50 Hz lines and at 1800 rpm when powered from 60 Hz lines. The disparity between the speeds of induction motors operated from 50 and 60 Hz lines span approximately the same percentage spread. When AC motors are operated at lower than their nameplate designated frequency, it is possible to obtain improvements in starting torque and in full-load torque. It may not always be wise to exploit such performance enhancements because of the inevitable temperature rise that are sure to accompany them. Another way of looking at this is to decide whether reduced life or increased maintenance would be acceptable. Operating a 50 Hz motor at 60 Hz tends to be a more benign situation and tends to yield a worthwhile improvement in breakdown torque. It is as if the motor is slightly over-designed. Finally, the use of motors at other than their nameplate frequencies can make them more vulnerable to the effects of low and high line voltage. This, too, translates into higher temperature rise. Providing these matters have been given due consideration, the resolution of possible problems with 50/60 Hz motors and power lines should prove fairly straightforward in most practical situations. Suppose a 50 Hz induction motor is to be used on a 60 Hz line of the same voltage. Inasmuch as horsepower is directly proportional to speed, it is anticipated that the motor will deliver about 60/50, or 20% more horsepower. The notion of a 20% increase in output horsepower derives from the basic equation HP =

Torque × N 9549

where T : torque in N-m N : speed in rpm 9549 : represents the number of N-m in 1 horsepower

33.2 Analysis Objectives

The probable weak point in the expectation of a 20% gain in output power is the reduction in torque from 60 Hz operation. This stems from the fact that the applied voltage must be increased by 20% to compensate for the 20% increase in inductive reactance for 60 Hz operation. This, of course, is an approximate approach, which neglects other factors, such as increased friction, windage, hysteresis, and eddy-current losses. It is, however, representative of adjustable-speed systems, which endeavor to obtain a speed range from an induction motor without also varying the applied voltage from a variable-frequency power supply.

33.2

Analysis Objectives

Motor-starting studies are performed on industrial and large commercial systems where a large motor can have an undesirable impact on its own performance and surrounding equipment. During the motor-starting period, the starting motor appears to the system as small impedance connected to a bus. It draws a large current from the system, about six times the motor rated current, which therefore results in voltage drops in the system and imposes disturbances to the normal operation of other system loads. Since the motor acceleration torque is dependent on motor terminal voltage, in some cases the starting motor may not be able to reach its rated speed due to extremely low terminal voltage: Torque

Voltage2

This makes it necessary to perform a motor-starting analysis. The purposes of performing a motor-starting study are multifold:

•• •• •• •

Whether the starting motor can be successfully started under the operating conditions Whether starting the motor will seriously impede normal operation of other loads The largest motor size that can be started with a given generator rating The rating of the generator(s) for a given load and largest motor size Voltage drop during starting conditions for given motor Starting type and current-reduction methods Cable sizing

33.2.1

Voltage Drop

During motor starting, the voltage level at the motor terminals should be maintained, as a minimum, at approximately 80% of rated voltage or above for a standard NEMA Design B motor (as specified in NEMA MG 1) having a standard 150% starting torque and with a constant-torque load applied. This value results from examination of speed-torque characteristics of this type motor (150% starting torque at full voltage) and the desire to successfully accelerate a fully loaded motor at reduced voltage (that is, torque varies with the square of the voltage T = 0.82 × 150% ≈ 100%). Assuming reduced voltage permits adequate accelerating torque, it should also be verified that the longer starting interval required at reduced torque caused by a voltage dip does not result in the I2t thermal damage limit of the motor being exceeded. In general, if the motors on the system are standard NEMA Design B, the speed-torque characteristics (200% breakdown torque at full voltage) should prevent a stall, provided the motor terminal voltage does not drop below about 71% of motor nameplate voltage. This is a valid guideline to follow anytime the shaft load does not exceed 100% rated, since the developed starting torque is again proportional to the terminal voltage squared (V2), and the available torque at 71% voltage would thus be slightly above 100%. If motors other than NEMA Design B motors are used on the system, a similar criterion can be established to evaluate reacceleration following a motor-starting voltage dip based on the exact speed-torque characteristics of each particular motor. By industry standards (see NEMA ICS 1, NEMA ICS 2, NEMA ICS 3, NEMA ICS 4, and NEMA ICS 6), AC control devices are not required to pick-up at voltages below 85% of rated nameplate voltage, whereas DC control devices must operate dependably (i.e. pick-up) at voltages above 80% of their rating. Critical control operations may, therefore, encounter difficulty during motor-starting periods where voltage dips are excessive. A motor-starting study might be required to determine if this is a problem with using devices rated at 110 V rather than the typical 115 V nominal devices. Contactors are required to hold-in with line voltage as low as 80% of their rating. The actual dropout voltages of contactors used in industrial applications commonly range between 60% and 70% of rated voltage, depending on the manufacturer. Voltages in this range, therefore, may be appropriate and are sometimes used as the criteria for the lower

893

894

33 Motor Starting

Table 33.3 Minimum allowable voltage drops – Table 51 of IEEE Std. 242-1986 and Table 1 of IEEE Std 3002.7TM-2018.

Location

Minimum allowable voltage (% rated)

Grid Point of Interconnection

95 to 105%

Motor terminal (starting motor)

80%a

All terminals of other motors that must reaccelerate

71%a

AC contactor pick-up (see 9.8, NEMA Std.)

85%

DC contactor pick-up (see 9.8, NEMA Std.)

80%

Contactor hold-in (average of those in use)

60–70%b

Solid-state control devices

90%c

Noticeable light flicker

3% change

a

Typical for NEMA Design B motors only. Value may be higher (or lower) depending on actual motor and load characteristics. b Value may be as high as 80% for certain conditions during prolonged starting intervals. c May typically vary by ±5% depending on available tap settings of power supply transformer when provided.

limit that contactors can tolerate. Depending on where lighting buses are located, with respect to large starting motors, this may be a factor requiring a motor-starting study. Table 33.3 summarizes some critical system voltage levels of interest when performing a motor-starting study for the purpose of evaluating the effects of voltage dips.

33.3

Methodology and Standards

Motor-starting analysis is recommended during initial design or retrofits of existing facilities under the conditions where the starting motor size is comparatively large with respect to the power source or the starting motor may potentially have a negative impact on the surrounding/sensitive critical loads. There are some basic rule of thumb practiced in the industry, a few of which are listed here:

•• •• •

Motor-starting kVA > 150% of the transformer bank rating. Fault level at the motor bus < six times motor-starting inrush. Motor rating exceeds 10–15% of the generator rating. Voltage drop due to motor starting exceeds 12–15%. Consider analyzing this impact on the contactors of other loads connected nearby. Acceleration time is approximately equal to or exceeds the motor hot-stall time. Consider analyzing this condition to ensure that the motor is not damaged under repeated operating duty cycles of start-stop.

Motor-starting studies using ETAP can be carried out using four main modeling techniques to determine the voltage drop and acceleration time of single or multiple motors. 33.3.1

Static Motor Starting or Nondynamic Motor Model

Starting motors are modeled by the locked-rotor impedance during starting time, simulating the worst impact on standard operating loads. This method is suitable for checking the effect of motor starting on the system when the dynamic model is not available for starting motors. The starting motor can always be started in this method as the objective is not to determine the acceleration time but rather the impact of voltage start on the immediate and surrounding electrical buses. Therefore, motors analyzed with this method are referred to as starting motors, not accelerating motors. During the starting period, the motor is represented by its locked-rotor impedance and draws the maximum possible current from the system. This is a conservative way to determine the impact on bus voltages and effects on other loads in the system. After the user-defined starting time is completed, the starting motor is changed to a constant kVA load. This is done so the user can simulate a sequence of events of multiple motors starting at various time steps. A static or impact start method is a recommended approach under any of the following conditions:

33.3 Methodology and Standards

• • ••

New system or conceptual design where the objective is to determine the voltage impact on buses, neighboring loads, protection settings, equipment/feeder sizing, etc. Motor and connected load dynamics are unavailable and cannot be estimated due to lack of torque information, or the starting motors are low voltage (LV) and impose no significant impact on the rest of the system. Starting times of the motors need not be calculated as there is no immediate problem associated with start-up times. Motor loads are grid connected (without a generator) or connected to a system fed by generator(s) only, but the size of the starting motor is less 10% of the generator kVA rating or one-sixth the size of the generator.

In static motor starting, the starting motors are modeled by the locked-rotor impedance during acceleration time, simulating the worst impact on standard operating loads. The static motor starting calculation method involves a time domain static model where starting motors are modeled as ZLR during starting and constant kVA load after starting. Intermediate load flow is run whenever there is any change in the system operating configuration, prior to starting any other motor. For the simple system shown in Figure 33.10: Rated kVA of the motor: kVAR =

0 746 × HP Eff × PF FL

Rated kV of the motor: kVR = 3 kVRIR Starting PF of the motor: PFST = cos(θ)ST, where the starting PF range is 0.15–0.3 (0.2 typical) Full-load PF of the motor: PFFL = cos(θ)FL Locked rotor current at kVR: ILR

RS(pu) XS(pu)

Inf bus

M

VS

kVST

Locked rotor kVA at kVR: kVALR = 3 × kV R × ILR Figure 33.10 Motor starting with infinite bus. kVALR Code-letter factor from NEC Table 430.7b: η = HP Starting kVA at kVST: kVAST = 3 × kV ST × IST Locked-rotor impedance of the motor: Z LR = RLR + jX LR , where RLR = ZLR cos(θ)ST and XLR = ZLR sin(θ)ST kVB: base kV at motor terminal kVAB: base kVA 33.3.1.1 Illustrative Example – Motor Starting

Consider the system shown in Figure 33.11, with a single induction motor: U1 200 MVAsc

U1 200 MVAsc

Bus1 13.8 kV

Bus1 13.8 kV

213 A 6% 97.4

T1 5 MVA 11%Z Bus2 4.16 kV

T1 5 MVA Bus2 4.16 kV

Mtr1 1000 HP Figure 33.11 Input data and simulation results.

7%

86.2 706.5 A

Mtr1 1000 HP

kVAR kVR PFST, PFFL ILR, kVALR, η

895

896

33 Motor Starting

base MVA 100 = 0.5 pu = MVASC 200 base MVA × transformer Z 100 × 0 11 = = 2.2 pu T1: transformer impedance = transformer MVA 5 base MVA × Xd 100 × 0 17 Mtr1: motor impedance = = = 18.7 pu motor MVA 4 16 × 0 126 × 3 transformer Z + motor Z 2 2 + 18 7 = = 0.9766 pu or 97.66% U1: system voltage = system Z + transformer Z + motor Z 0 5 + 2 2 + 18 7 motor Z 18 7 = = 0.8738 pu or 87.38% Mtr1: motor voltage = system Z + transformer Z + motor Z 0 5 + 2 2 + 18 7 U1: system impedance =

33.3.1.2 Illustrative Example – Motor Sizing

A 1000 kVA (4.16 – 0.48 kV) transformer with 5.75% impedance and X/R = 10 is connected to a 0.48 kV motor control center (MCC). The voltage flicker must be limited to 3%. Assume the motor efficiency and power factor at full load are 95% and 90%, respectively. The starting power factor is 20%, and the locked-rotor current is 650% of full-load amps. Ignore the source impedance. Using the approximate method: kV ST pu kV ST = ZLR pu kV B × ZLR kVAR kV ST × kV B kVAR 0 48 × 0 48 × × IST = 6 5 × =6 5× = 0 00774 kVAR pu 2 kVAB 1000 0 442 kV R

IST =

Vd = IST R cos θ

ST

+ X sin θ

ST

0 03 = 0 00774 kVAR 0 00575 × 0 2 + 0 0575 × 0 98

kVAR = 67 4

and kVA × PF × Eff 67 4 × 0 9 × 0 95 = = 77 3 HP 0 746 0 746 Using the exact method: HP =

1 kVAB 6 5 kVAR

kV R kV B 25 9 RLR = 0 2 × ZLR = kVAR

ZLR =

2

=

1 1000 6 5 kVAR

0 44 0 48

2

= 129 3 kVAR

and 126 7 kVAR kV ST kV B

XLR = 0 98 × ZLR = IST =

RT + RLR

2

+ XT + XLR

2

and Vm = IST × ZLR Vm = 0 97 =

129 3 0 48 × kVAR 0 48 25 9 0 00575 + kVAR

2

126 7 + 0 0575 + kVAR

2

= 0 00334 kVAR

2

+ 14 88 kVAR − 1047 5 = 0

kVAR = 69 3 and HP = 79 4 33.3.1.3 Illustrative Example – Static Starting

Consider another example, as shown in Figure 33.12, where base kV at Bus 2 is 13.8 kV and base kVA is 1000. kVST = 13.7 kV at Bus 2 before starting = 13.7/13.8 = 0.993 pu.

33.3 Methodology and Standards

Ignoring line impedances, we can write:

132 – 13.8 kV Inf. bus

Transformer: X = 0.06 pu and R = 0.006 pu Motor: kVR = 13.2 kVAR =

0 746 × HP ≈ HP = 1000 Eff × PF FL

132 kV (Bus 1)

kVALR = 5 5 × kVAR = 5500 ZLR =

1000 kVA 6%Ƶ X/R = 10

13.8 kV (Bus 2)

1000 HP 13.2 kV ILR = 550% PFST = 20% RFST = 98%

Figure 33.12 Motor starting with infinite bus.

2

1000 13 2 × 5500 13 8

M

= 0 1664 pu

RLR = 0 2 × 0 1664 = 0 0333 pu and XLR = 0 98 × 0 1664 = 0 163 pu

The starting current is IST =

0 993 0 006 + 0 0333 2 + 0 06 + 0 163

2

= 4 385 pu

The motor voltage is VM = 4 385 × 0 1664 = 0 73 pu system base 10 07 = 0 76 pu motor base = 0 73 × 13 8 = 10 07 kV = 13 2 Figure 33.13 shows the output from ETAP, which indicates that the motor voltage on system base is 10.1 kV vs. 10.07 kV. U2

Bus4 132 kV

U2

0A

132

kV

19.3 A

Bus4 132 kV

T2 1 MVA 13.7

Bus3 13.8 kV

kV

Mtr2 1000 HP t = 0– seconds (prior to motor start)

132

kV

T2 1 MVA Bus3 13.8 kV

10.1 185.5 A

kV

Mtr2 1000 HP t = 0+ seconds (after motor start)

Figure 33.13 Motor starting with ETAP simulation results.

33.3.1.4 Illustrative Example – Static Starting with an Autotransformer

Consider the example shown in Figure 33.12, now using the autotransformer starting method as shown in Figure 33.14. The impedance diagram for this system can be redrawn as shown in Figure 33.15. The motor voltage is VM = 0 8 × V2 From Table 33.2, IST = 0 68 × IST = Ideal autotransformer + Magnetizing current = 0 64 × IST + 0 04 × IST

897

898

33 Motor Starting

V1

V2 132 – 13.8 kV

Autotrans

Inf. bus

1000 HP M

13.2 kV

Vm 80% Tap

ILR = 550%

1000 kVA 6%Ƶ X/R = 10

PFST = 20% 13.8 kV (Bus 2)

132 kV (Bus 1)

Figure 33.14 Motor starting with autotransformer.

0.006

V1

IʹST

IST

j0.06

0.006 0.0333

VM

V2

V1 j0.163

j0.06

IʹST

RʹLR

a.0333 = 0.052 0.82

XʹLR

j0.163 = j0.255 0.82

XM V2

80% Figure 33.15 Impedance diagram of motor starting with autotransformer.

U2 100000 MVAsc

Note that Magnetizing current = 6 25 of IST 1 0625 × V1 1 0625 × 0 993 IST = = 2 2 0 006 + 0 052 2 + 0 06 + 0 255 R + RLR + X + XLR

2

= 3 294 pu

The voltage drop is Vd = I Rcos θ + Xsin θ = 3 294 × 0 006 × 0 2 + 0 06 × 0 98 = 0 198 pu

Bus4 132 kV

13.6 A 132

T2 1 MVA

V2 = 0 993− 0 198 = 0 795 pu = 10 98 kV = 79 5 The motor voltage is VM = 0 795 × 0 8 = 0 636 pu = 8 78 kV Results from ETAP simulation are shown in Figure 33.16, where the terminal voltage or motor voltage is 8.92 kV. Clearly, using numerical hand calculation with approximate methods is not advisable when solving large, complex systems due to numerical differences (simplifications) inherent in these methods and those used by computer-based software like ETAP.

kV

Bus3 13.8 kV

5 kV

11.1

131.2 A 8.92

kV

Mtr2 1000 HP

33.3.2

Dynamic Motor Starting

Figure 33.16 ETAP simulation results.

For a dynamic motor acceleration calculation, the starting motors are represented by dynamic models, and the Motor Acceleration Module simulates the entire process of motor acceleration assuming that the generator internal voltage remains constant throughout the entire simulation. This method is used to determine whether a motor can be started and how much time is needed for the motor to reach its rated speed, as well as to determine the effect of voltage dips on the system without considering the effects of generator excitation. It is a standard method used where the system generation is significantly larger than the starting motor, i.e. the size of the starting motor is less than one-sixth the size of the generator. This yields a conservative result guaranteeing that if generator excitation is not varied and the motor starts successfully, then in the practical system with generation-excitation control, there will be minimal likelihood of this motor failing to start due to voltage considerations.

33.3 Methodology and Standards

33.3.2.1 Dynamic Motor Acceleration with Motor Circuit Models

In the dynamic motor-starting method using motor circuit models, the entire dynamic model for the motor and connected load is used to simulate the acceleration behavior and voltage impact on the entire network. This method does not consider the generator or network dynamics. The motor circuit model can be defined in multiple ways:

•

An equivalent (Thévenin) circuit model with constant rotor resistance and reactance is shown in Figure 33.17. This modeling technique should be avoided since it yields the most inaccuracy in simulation. Thévenin equivalent circuits are shown before starting and after starting with their R and X in pu values, as shown in Figure 33.18. The locked-rotor impedance is 1000 Ω kVALR 1000 ZB = kV 2B × Ω kVAB

ZLR = kV 2R ×

Therefore, ZLR =

kVAB kV R × kVALR kV B

2

pu

where RLR = ZLR × PF ST and XLR = ZLR × PF ST The starting current is VS

IST = RS + RLR

2

+ XS + XLR

2

pu

The starting motor voltage is VM = IST × ZLR pu

Figure 33.17 Thévenin circuit motor for motor starting.

RS

puVS

jXS

RS

puVST

RLR

puVS

jXS

IST

jXLR Figure 33.18 Thévenin equivalent circuit before and after motor starting.

puVM

RLR jXLR

899

900

33 Motor Starting

Figure 33.19 shows a circuit model with deep-bar effect; rotor resistance and reactance change with speed. Figure 33.20 shows a double-cage circuit model with integrated rotor cages (DBL1). And Figure 33.21 shows a double-cage circuit model with independent rotor cages (DBL2). Dynamic motor starting using a circuit model is a recommended approach under any of the following conditions:

•• •• •

Existing system design change or expansion. Motor and connected load dynamics are available and/or can be estimated. Acceleration times of the motors are required to be calculated. Accelerating motors are primarily medium-voltage motors. Motors are connected to a system fed by generator(s), but the size of the starting motor is greater than 10% of the generator kVA rating.

Figure 33.19 Circuit model with deep-bar effect.

Figure 33.20 Double-cage with integrated rotor cages.

Figure 33.21 Double-cage with independent rotor cages.

33.3 Methodology and Standards

33.3.2.2 Torque Slip Characteristic Curve Model

In this dynamic motor-starting method, torque slip characteristics for motor and load are modeled as point-based or equation based. The entire dynamic model for the motor and connected load is used to simulate the acceleration behavior and voltage impact on the entire network. This method does not consider the generator or network dynamics. The motor circuit model can be defined as shown in Figure 33.22.

Figure 33.22 Torque slip characteristic curve model.

33.3.3

Control of Motor Characteristic Curves

Considering a motor circuit model as shown in the previous sections, the reactance X2 represents the motor-leakage reactance due to the rotor flux linkages that do not cross the stator windings. Therefore, the further away a rotor bar is from the stator, the greater its leakage reactance. In a squirrel-cage motor, if the rotor bars are placed closer to the surface of the rotor, then the leakage flux will be smaller and, correspondingly, the leakage reactance will be smaller. A motor manufacturer can achieve these characteristics by locating the rotor bars deeper in the iron rotor cylinder or by closing the slot at the air gap. An increase in rotor-bar reactance tends to lower starting torque, starting current, and breakdown torque, while not having any effect on full-load conditions. Further, if the size of the rotor bar is larger (greater area of cross-section) then the corresponding resistance is lower. Due to this low resistance, very little air-gap power is lost in the rotor resistance, the pull-out torque is closer to synchronous speed, and the motor is significantly efficient. A motor manufacturer can decrease the cross-sectional area of the rotor bars or use a rotor bar material having greater resistance. Increasing the rotor-bar resistance tends to affect motor performance by increasing starting torque and lowering starting current, full-load speed, and efficiency, while not having any effect on breakdown torque. As with most product engineering, the final design is usually a compromise. Therefore, a designer might lower full-load speed by increasing rotor-bar resistance and at the same time lower starting current by increasing rotor bar reactance.

33.3.4

Transient Stability or Full Network Dynamics

In transient stability motor acceleration, the starting motors and network model including generators are modeled dynamically, and all control actions such as switched capacitors, SVC, generator governors, and excitation controls are considered in order to calculate voltages and motor-acceleration times. The dynamic model for the motor and connected load is used to simulate the acceleration behavior, voltage, and frequency impact on the entire network. This method does consider generator and frequency dependent network dynamics such as generator dynamic model, excitation, governor response, FACTS devices, etc. Full network dynamics is a recommended approach under any of the following conditions:

•• •• •

Existing system design change or expansion. Motor and connected load dynamics are available and/or can be estimated. Acceleration times of the motors are required to be calculated. Accelerating motors are primarily medium-voltage motors. Motors are connected to a system fed by generator(s), and the size of the starting motor is greater than 10% of the generator kVA rating. A summary of differences in these calculation methods is shown in Table 33.4.

901

902

33 Motor Starting

Table 33.4 Summary of motor-starting modeling. Element

Load-flow model

Full network dynamics

Circuit model / torque slip

Static or impact start

Generators

Infinite bus

Dynamically modeled

Constant voltage behind Xd’

Constant voltage behind Xd’

Exciter/ Governors

Not applicable

Dynamically modeled

Not modeled

Not modeled

Utility ties

Infinite bus

Constant voltage behind X"

Constant voltage behind X"

Constant voltage behind X"

Operating motors

Constant kVA

Modeled dynamically or constant kVA

Constant kVA

Constant kVA

Starting motors

Not applicable

Single1, Single2, DBL1, and DBL2 models

Single1, Single2, DBL1, DBL2, and TSC models

Locked-rotor Z and power factor

Starters

Not applicable

Modeled

Modeled

Modeled

33.4

Required Data

Table 33.5 is a summary of the required data for some of the standard device types. Note that some of the data requirements vary based upon the mode of operation. Refer to the ETAP user guide for detailed list of required data.

Table 33.5 Summary of motor-starting required data. Device

Bus

Nominal kV

x

x

Operating kV

x

x

Angle

x

Impedance R, X Diversity

Line/Cable

Impedance

Source

x

x

x

x x

Tap changer

x

Tolerance

x x

x

x

Operating mode

x

Mvar limits

x

kW/HP

x

Kvar

x

Power factor

x

Xd Rated kV

Motors

x x

Rating kVA

Temperature

Transformer

x x

x x

x

x

Efficiency

x

Loading levels

x

Generating levels

x

MVAsc

x

Xd ’

x

Starting type

x

Starting settings

x

33.5 Illustrative Examples

33.4.1

Additional Data for Starting Motors

For static motor-starting studies, the additional data includes:

•• •• •

Motor locked-rotor impedance and power factor Motor acceleration time at no load and full load Start and final percent loading and begin and end of load change time Starting device data when needed No load and full load accelerated time (for static motor starting)

•• ••

Dynamic motor model for induction motors LR model for synchronous motors Load torque model Motor inertia

For dynamic motor-acceleration studies, the additional data includes:

33.5

Illustrative Examples

33.5.1

Static Motor Starting with Load Change

Consider a case with impact starting for a sequence of motors with load change after the motors are started, as shown in Figure 33.23. During the acceleration period, the motor is represented by its locked-rotor impedance, which draws the maximum possible current from the system and has the most severe effect on other loads in the system. Once the acceleration period has passed, the starting motor is changed to a constant kVA load, and ETAP simulates the load-ramping process according to the starting and final loads specified in the motor editor. The motor-starting condition is specified in Table 33.6. Consider the sequence of events given in Table 33.7, where static motor starting is performed using load change. Alerts are shown after running static motor starting, as shown in Figure 33.24. Critical alerts (typically 90%) are shown in magenta.

Loading

Loading

Ilr

Ilr If

If

IS tacc

IS time

tS

tst

tf

tf

Motor acceleration (dynamic model) Ilr = Locked rotor current (%) IS = Starting load (%) If = Final load (%) Figure 33.23 Motor starting with load change.

time

tS Motor starting (static model)

tacc tSt tS tf

= Acceleration time (dynamically calculated) = Starting time (fixed) = Beginning of load change after acceleration = End time for motor load change

903

904

33 Motor Starting

Table 33.6 Condition for motor starting with load change. Starting category Motor ID

Cat ID

% Loading

Utilize

Start

Time of load change Final

Begin

End

Mtr1

Normal

Yes

0

100

2

4

Mtr2

Backup

No

0

100

1

2

Mtr4

Normal

Yes

80

100

1

2

Mtr5

Normal

No

60

100

1

4

Syn1

Normal

Yes

25

90

2

4

Table 33.7 Motor-starting sequence of events. Event ID

Time (sec)

Action type

Parameter setting

Event 1

0.2

By element

Start Mtr5 using normal starting category

By element

Switch on static load 2, Loading Category: Design

By starting category

Start all motors in the system belonging to Normal Category (Mtr1, Mtr4, and Syn1)

Event 2

3

Figure 33.24 Motor-starting alerts.

33.5.2

Dynamic Motor Starting

This example helps to illustrate the difference between static and dynamic motor-starting analysis using the same example as the previous section. Dynamic acceleration requires a dynamic circuit model representation of the electrical behavior of the motor. The mechanical parameters and the inertia of the rotor, shaft, coupling, and load are also needed, as listed in Table 33.8.

33.5 Illustrative Examples

Table 33.8 Dynamic circuit model selection and mechanical information. Circuit model

Mechanical load

Motor inertia

Motor ID

Type

Library

Type

Library

H

Mtr1

CKT Single 2

HV-HS-LT, MV500HP2P

Polynomial

FAN

0.4

Mtr2

Charac.

LV-HS-HT, LV200HP2P

Polynomial

a k∗∗∗3

0.2

Syn1

CKT Single 2

HV-HS-HT, MV1000HP2P

Polynomial

Recip. Comp

0.5

When dynamic motor starting is performed, Mtr4 fails to start, as shown in the alert view (Figure 33.25). The key is lack of acceleration torque, as can be seen from the load and motor torque curves at rated voltage in Figure 33.26. Table 33.9 is a summary of the methods that can be used to solve the starting issue. The possible outcomes for this case are also discussed along with their pass/fail reasons. Changing the starting load on a motor is possible in cases like starting a high-pressure centrifugal pump. In order to unload the pump and achieve successful and faster acceleration time, the inlet valve is kept slightly open and the discharge valve closed at the time of motor start. Due to limited fluid in the motor chamber, the impeller does not require significant mechanical torque on the shaft. After the motor has come up to full speed, the inlet and the discharge valves are opened, allowing the fluid to flow through the pump and thereby increasing the fluid load and correspondingly the electrical load on the impeller. If this transition is not timed correctly, resulting cavitation can damage the equipment. Similarly, a large fan may be started with reduced load by keeping the exhaust vanes closed, encouraging the motor to start with the fan moving around within the enclosure only, i.e. no cross-flow. The exhaust vanes can be opened after the fan has come up to full speed. ETAP allows for simulation of such events using the Load Change option. Load-change simulation allows the motor to start with a user-defined initial load (< 100%). ETAP automatically determines whether the motor has reached rated speed or low slip before automatically ramping/stepping up the load.

Figure 33.25 Motor starting results showing Mtr4 failed to start.

905

906

33 Motor Starting

Figure 33.26 Insufficient acceleration torque.

Table 33.9 Motor failed to start – mitigation methods. Mitigation

Possible outcome

Reason

Change starting sequence.

Motor will fail to start.

This does not change available acceleration torque.

Transformer load-tap changer.

Motor will fail to start.

Though this will change available acceleration torque, the voltage rise is not sufficient to overcome the load torque demand. Motor will stall at a higher speed.

Starting capacitor.

Motor will fail to start.

This does not change available acceleration torque and only provides local var support and indirectly increases voltage close to rated voltage.

Larger motor.

Motor will start.

Not a practical approach in all cases.

Change the load.

Motor will start.

Not a practical approach in all cases.

Change Mtr4 starting load from 80% to 60% or 40%.

Motor will start.

This is the optimal solution because it requires changing the load controller such that the initial loading on the motor is reduced. Once the motor reaches operational slip, the load on the motor shaft can be increased.

33.5.3

Motor Starting with VFD

The example in Figure 33.27 illustrates the transient behavior of VFD for motor (induction motor) starting purposes.

33.5 Illustrative Examples

Gen1 5 MW Syn. Generator Data 5 MW, 34.5 kV, 85% PF Bus20 34.5 kV T1 5 MVA 7%Z

Transformer Data MVA Ratings = 5 MVA Prim/Sec Voltage = 34.5/4.16 kV

Bus1 4.16 kV

VFD Data HP Ratings = 1800 HP KVA = 1342.3 kVA Rated kV = 4.16 kV FLA = 183.3 Frequency = 60 to 150(Max) PF = 100%

CB9

CB5

CB6

VFD1 Bus3 4.16 kV

Lump2 2.5 MVA

CB16 Motor Data 1500 HP, 4 kV, MTR1 1500 HP 4 kV Figure 33.27 Motor starting with variable-speed drives (VFD).

Case 1 Stepped frequency during VFD start with 20% increase per step from 0 to 100% frequency with V/Hz kept constant. The stepped-frequency control scheme cannot start the Mtr1 as there is high current drawn from the system due the 20% frequency step, which dips the motor voltage and flux and results in VFD shut-down as per the transient graphical results shown in Figure 33.28. In Figure 33.29, as the motor reference frequency is increased in steps, motor slip increases substantially. This is because the motor frequency being changed at each step is always greater than the speed increase on motor and driven load. Thus it causes high motor slip that is dependent on the step increase in frequency magnitude. The motor and VFD input current increases, causing large voltage drop and eventual shutdown of the VFD.

907

33 Motor Starting

120.00 100.00

Slip (%)

80.00 60.00 40.00 20.00 0.00

0

2

4

6

8

10

Time (sec) Figure 33.28 Case 1 – Motor with VFD failed to start.

Case 2 Ramped frequency during VFD start from 0 to 100% in 8 seconds with V/Hz kept constant. The transient results of the case are given in Figure 33.29. As the motor reference frequency is gradually increased as a ramp, motor slip is maintained close to zero right from the starting time. This is because the motor frequency being changed now closely matches the motor and driven load speed increase. The motor and VFD input current can thus be controlled, resulting in a successful start and tolerable system voltage profile. The results show that a percentage step increase in frequency causes an increase in motor and VFD current, resulting in reduced voltage and flux and VFD shut off. A smooth frequency ramp is essential for VFD-based starting since the motor is able to accelerate close to the motor frequency (synchronous speed). 120 100 80 Slip (%)

908

60 40 20 0

0

2

4

6

8

10

Time (sec) Figure 33.29 Case 2 – Motor with VFD started successfully.

33.5.4

Synchronous Motor Starting

The purpose of this example, as shown in Figure 33.30, is to model a synchronous motor and study its effects during motor start. The synchronous motor nameplate data, torque-slip curve, rotor type, inertia (salient pole), and subtransient impedance model information are shown in Figure 33.31. The motor-excitation model is IEEE AC8B. During start-up, discharge resistance is 0.2 Ω, excitation is applied at 98% speed, and the discharge resistor is removed. ETAP simulation results for synchronous motor start are shown in Figure 33.32. Synchronous motor tabulated results are shown as plots versus time in Figures 33.33–33.36.

33.5 Illustrative Examples U1 1500 MVAsc Grid Data Rated Voltage 69 kV Fault MVA = 1500, X/R = 20

69 kV Bus5

Transformer Data MVA Rating = 100 MVA Prim/Sec Voltage = 69/13.8 kv Typical Z & X/R

T1 100 MVA Bus1 13.8 kV

Cable Data Length 200 ft, 1-3/C 250mmˆ2

Cable1

Bus2 13.8 kV

Mtr-1 17393 HP

Syn. Motor data: 17393 HP, 13.2 kV, –99.98%pf, Typical subtransient impedance model LR Torque = 104.9% LR pf =19% Exciter AC8B UDM model with AVR Mode

Figure 33.30 ETAP one-line diagram for synchronous motor starting.

Figure 33.31 Synchronous motor information.

909

U1 1500 MVAsc Grid Data Rated Voltage 69 kV Fault MVA = 1500, X/R = 20

69 kV Bus5

552.4 A 4.74% PF

Transformer Data MVA Rating = 100MVA Prim/Sec Voltage = 69/13.8 kv Typical Z & X/R

T1 100 MVA %

Bus1 90.32 13.8 kV 60

Cable Data Length 200 ft, 1-3/C 250mmˆ2

Hz

Cable1

Bus2 13.8 kV

4% Syn. Motor data: 90.0 6 2762 A 0 Hz 17393 HP, 13.2 kV, –99.98%pf, Typical subtransient impedance model 4.44% PF LR Torque = 104.9% LR pf =19% Mtr-1 Exciter AC8B UDM model with AVR Mode 17393 HP –55.17 Deg 626.5 RPM

0

Figure 33.32 ETAP simulation results for synchronous motor start.

25

MW (Electrical)

20 15 10 5 0

0

1

2

3

4

5 6 Time (sec)

7

8

9

10

7

8

9

10

Figure 33.33 Synchronous motor electrical power (MW).

14 13.8

Bus Voltage (kV)

13.6 13.4 13.2 13 12.8 12.6 12.4 12.2

0

1

2

3

4

5 6 Time (sec)

Figure 33.34 Synchronous motor bus voltage (kV).

33.5 Illustrative Examples

2000 1800

Motor Speed (RPM)

1600 1400 1200 1000 800 600 400 200 0 0

1

2

3

4

5 6 Time (sec)

7

8

9

10

7

8

9

10

Figure 33.35 Synchronous motor speed (rpm).

Motor Relative Power Angle (degree)

50 0 0

1

2

3

4

5

6

–50 –100 –150 –200 –250 Time (sec)

Figure 33.36 Synchronous motor relative power angle (degrees).

33.5.5

Example – Motor-Starting Study for an Industrial System

A motor-starting study was executed on the one-line diagram “Example Model for Industrial System” shown in Figure 33.37. 33.5.5.1 Analysis Objectives

Objectives for the motor-starting study are described in Section 33.2. In this example, we analyze whether the largestcapacity motor (IM-3, 5.45 MW) can start in a reasonable amount of time when the plant is running on the system configuration “11kV CB-Tie Open” case. 33.5.5.2 Methodology and Standards

The methodology and standards for the study are explained in Section 33.3.

911

33 Motor Starting

Starting Motor Data Output 5.45 MW Poles 4 Load Fan Motor Torque, Current, Power Factor Curve (Estimate based on the motor parameters below.) • LRC (%) 550 PFlr (%) 20 • Tlr (%) 65 • Tmax (%) 165 • Slip (%) 1.5 • PF100 (%) • 92.1 • EFF100 (%) 95.54 Inertia

WR2 = 500 kg-m2 (motor and fun)

Figure 33.37 Motor-starting study for industrial system.

33.5.5.3 Required Data

The required data for the study are summarized in Section 33.4. In this example, in addition to the required data for a load-flow study, the following motor and load data is utilized:

•• •

Motor torque and current curves, as shown in Figure 33.38 Load torque curve Inertia of motor and fan WR2 = 500 kg/m2 (motor and fan)

300

600 % Torque

Locked rotor current (100%) Locked rotor current (85%)

200

500

200

400

150

300

150 Torque curve (100%) 100

Torque curve (85%)

50

Current [%]

250

Torque [%]

912

100

200

50

100

0 0

Load torque curve 0

0 1

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Slip (pu)

0

Figure 33.38 Motor torque, current, and load torque curve given by manufacturer.

25

50 % Speed

75

100

33.6 Motor-Starting Plots and Results

33.5.5.4 Dynamic Modeling of Motor Starting (IM-3, 5.45 MW, 4-Pole)

Motor data for the dynamic modeling of IM-3 (5.45 MW, 4-pole) is shown in Figure 33.39.

Parameters for motor (Locked rotor current/power factor/ torque, Maximum torque, Full load slip/current/torque/power and factor/efficiency)

Estimated equivalent circuit

Rs

Xs

1.72

17.8

Kr Xrfl = 14.01 Xrlr = 0.01

Xm

Rc

1090

8379

Rr/s Rrfl = 1.41 Rrlr = 1.92

% Torque 200 150 100 50 0 100

75

50 % Slip

25

0

75

50 % Slip

25

0

%1 600 450 300 150 0 100 % PF

% PF

100 75 50 25 0 100

Estimated parameters for motor equivalent circuit

75

50 % Slip

25

0

Estimated motor torque, current and power factor curve Figure 33.39 Motor dynamic modeling by ETAP.

33.5.5.5 Results of the Motor-Starting Study

Results of the IM-3 starting study on the system configuration “11kV CB-Tie Open” case are summarized in Table 33.10.

33.6

Motor-Starting Plots and Results

Typical motor-starting plots available in ETAP are shown in Table 33.11. In addition to reports and plots, the software can display the calculation results on the one-line diagram. Once a motor-starting calculation is finished, a time slider starts from time = 0 seconds to the final simulation time. As the pointer is moved along the ruler, the displayed results change accordingly. In Figure 33.40, we can see results on the ETAP one-line diagram at t = 0.1 seconds prior to motor start and results at t = 0.12 seconds when the motor starts.

913

Table 33.10 Resultsof the largest-capacity motor IM-3 starting study. Largest motor starting

Event1: Start the largest-capacity motor IM-3 (5.45 MW) = > Action: start IM-3 at 0.1 sec Results: Starting time is approx. 9 sec, from induction machine slip curve. Bus voltage at starting the motor down to 90% at Bus-6.6 kV-T2, and 95% at Bus-66 kV The largest-capacity motor (IM-3, 5.45 MW) can start in a reasonable amount of time when the plant is running on the system configuration “11 kV CB-Tie Open” case. Induction Machine Acceleration Torque

Induction Machine Slip 120

80 MW

Slip (%)

100

60 40 20 0

0

1

2

MS-1

3

4

5

6

7 8 9 10 11 12 13 14 15 Time (Sec)

80 70 60 50 40 30 20 10 0

0

1

2

3

600

100

500

80

400

60 40

0

1

2

3

4

5

7 8 9 10 11 12 13 14 15 Time (Sec)

300 200

0

6 7 8 9 10 11 12 13 14 15 Time (Sec)

0

1

2

3

4

6 7 8 9 10 11 12 13 14 15 Time (Sec)

% of Bus Nominal kV

110

105 100 95 90 85

5

Bus-6.6kV-T2 Voltage

Bus-66kV Voltage 110 % of Bus Nominal kV

6

100

20 0

5

Induction Machine Teminal Current

120

I (Amps)

% of Bus Nominal kV

Induction Machine Terminal Voltage

4

0

1

2

3

4

5

6 7 8 9 10 11 12 13 14 15 Time (Sec)

105 100 95 90 85

0

1

2

3

4

5

6 7 8 9 10 11 12 13 14 15 Time (Sec)

Table 33.11 ETAP motor-starting plots by device type. Generator and power grid

Starting motor

Starting motor operated valve (MOV)

Static load and capacitor

Bus

Current

Slip

Current

Current

Voltage

Vt (machine base)

Current

Vt (MOV base)

Vt (load base)

MW

Vt (motor base)

Vt (Bus base)

Vt (Bus base)

Mvar

Vt (bus base)

Vbus

Vbus

MVA

Vbus

kW

kW

PF

Accel. torque

kVar

kVar

Motor torque Load torque kW, electrical kvar kW, mechanical

kVA

33.6 Motor-Starting Plots and Results

Figure 33.40 ETAP motor-starting time slider at t = 0.1 second (prior to motor start) and t = 0.12 seconds (motor start).

915

916

33 Motor Starting

33.7

Motor-Starting Alerts

It is possible to specify the limits for ETAP to raise critical and marginal alerts for a motor-starting simulation. These limits may be set globally for all machines or at individual motor/bus levels. A motor-starting alerts includes three categories: starting motors, generator rating, and bus voltages, as shown in Figure 33.41. Results of motor starting alerts are shown in a tabular manner, including critical and marginal alerts, in Figure 33.42.

Critical

Marginal

Starting Motors/MOV MOV Terminal Voltage Motor Terminal Voltage Failed to Start, Slip Kept

≤ 80.00 ≤ 80.00 ≥ 5.00

Generator/Engine/Exciter Rating Generator Rating Engine Continuous Rating Engine Peak Rating Exciter Peak Rating Bus Voltage Group Starting Motor Bus Grid/Generator Bus HV Bus, kV ≥ 10.00 MV Bus, 10.00 > kV >1.00 LV Bus, kV ≤ 1.00

100.00 100.00 100.00 100.00 VBus ≤ 80.00 VBus ≤ 92.00 VBus ≤ 90.00 VBus ≤ 90.00 VBus ≤ 90.00

Figure 33.41 Motor-starting alert categories.

Figure 33.42 Motor-starting alert view.

90.00 (Vmtr, rate) 90.00 (Vmtr, rate)

95.00 95.00 95.00 95.00

Min. Span (Sec) 2.00 2.00 2.00 2.00

90.00 95.00 95.00 95.00 95.00

Min. Dip Width (Sec) 2.00 2.00 2.00 2.00 2.00

917

34 Harmonics 34.1

Introduction

Nonlinear loads in commercial buildings and industrial plants are in the range of 30% to 50% of the total load. Their impact on power quality, especially harmonics within the system, and their effect on the utility and neighboring loads needs to be examined and understood in order to avoid equipment damage, production loss, or customer complaints. Harmonics are voltages or currents in the electrical system at some multiple of fundamental frequency (50 or 60 Hz). Harmonics are one of the main concerns related to power-quality problems. Major power-quality issues are summarized in the Table 34.1 in order for the reader to gain a broad understanding. As compared to linear time-invariant loads, the application of a sinusoidal voltage does not result in a sinusoidal flow of current due to the nonlinearity in impedance. Nonlinearity should not be confused with frequency dependency, as the load current is noncontinuous or pulsed. Nonlinear load examples include:

•• •• •• •• •• ••

Variable frequency drives (VFDs) Adjustable-speed drives (ASDs) Cycloconverters Arc furnaces Fluorescent lighting and electronic ballasts Switching mode power supplies (SMPSs) Static var compensators (SVCs) High-voltage DC (HVDC) transmission links Railway electric traction Wind and solar power generation Energy storage devices (ESDs), battery charging, and fuel cells Slip recovery schemes of induction motors

Nonlinear loads such as thyristors precisely “chop” the sinusoidal waveform to supply power to the load device. This basic chopping effect of the current generates a distorted voltage waveform I(t), shown in Figure 34.1. Therefore, power system harmonics are typically generated by nonlinear loads when a sinusoidal voltage is applied to a simple nonlinear resistor. Increasing the voltage by a few percent may cause the current to double. The presence of harmonics in a power system can give rise to a variety of problems including equipment overheating, reduced power factors, deteriorating performance of electrical equipment, incorrect operation of protective relays, interference with communication devices, and, in some cases, circuit resonance that causes electric apparatus dielectric failure and other types of severe damage. Even worse, harmonic currents generated in one area can penetrate into the power grid and propagate into other areas, resulting in voltage and current distortions for the entire system. This phenomenon has become a major concern for power quality due to the ever-increasing usage of electronic devices and equipment in power systems. The following describes some of the typical effects of harmonics on various power system components:

•

Motors and generators – Increased heating due to iron and copper losses – Reduced efficiency and torque

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

918

34 Harmonics

Table 34.1 Power-quality problems. Category

Spectral content

Typical duration

Typical voltage magnitude

0–5000 Hz

Steady state

0–0.25%

Steady state

Pst and Plt

Waveform distortion DC offset Flicker Harmonics

0–6 kHz

Steady state

0–30%

lnterharmonics

Broadband

Steady state

0–5%

Noise: common mode and normal mode

Steady state

Notching

Steady state

Voltage fluctuations

Intermittent

Voltage unbalance

0.1–7% 0.5–4%

Transients—impulsive Nanosecond

5 ns rise

1 ms

Low frequency

< 5 kHz

0.3–50 ms

0–4 pu

Medium frequency

5–500 kHz

20 μs

0–8 pu

High frequency

0.5% MHz

5 sec

0–4 pu

Interruptions

0.5–30 cycle

< 0.1 pu

Sags (dips)

0.5–30 cycle

0.1–0.9 pu

Swell

0.5–30 cycle

1.1–1.8 pu

Interruptions

30 cycles–3 sec

< 0.1 pu

Sags (dips)

30 cycles–3 sec

0.1–0.9 pu

Swell

30 cycles–3 sec

1.1–l.4 pu

Interruptions

3 sec–1 min

< 0.1 pu

Sags (dips)

3 sec–1 min

0.1–0.9 pu

Swell

3 sec–1 min

l.1–1.2 pu

Sustained interruption

1 min

0.0 pu

Undervoltage

1 min

0.8–0.9 pu

1 min

1.1–1.2 pu

Transients—oscillatory

Short-duration variations a. Instantaneous

b. Momentary

c. Temporary

Long-duration variations

Overvoltage Frequency variations

•

– Increased audible noise – Pulsating and reduced torque in rotating equipment Transformers – Parasitic heating – Increased copper, stray flux, and iron losses

< 10 sec

34.2 Analysis Objectives

I(t)

Nonlinear resistor

V(t)

V I

Figure 34.1 Nonlinear load distortion.

• • • • • •

Capacitors (var compensators) – Possibility of system resonance – Increased heating and voltage stress – Shortened capacitor life Power cables – Involved in system resonance – Voltage stress and corona leading to dielectric failure – Heating and derating Neutrals of four-wire systems – Overheating Instrumentation devices – Erroneous fuse operation – Possibility of relay misoperation – Affected meter readings Switchgear – Increased heating and losses – Reduced steady-state current carrying capability – Shortened insulation life General – Premature aging due to increased stress in the equipment insulation – Overvoltages due to resonance and damage to equipment – Communication interference due to inductive coupling between power and communication circuits.

34.2

Analysis Objectives

Harmonic-analysis studies are conducted for new or existing system for multiple reasons. Some of the primary reasons include:

•

Simulate and evaluate the impact of harmonic sources on a point of common coupling (PCC) with the utility, and compare harmonic indices with relevant standards or codes. Based on the comparison, it is possible to ascertain

919

920

34 Harmonics

•• • • •

whether any harmonic voltage, current, or telephone interference factor violations exist in the system that are not in compliance with the utility grid code. Voltage distortion is typically calculated to determine whether the return current is close to a sinusoidal waveform. There is a strong likelihood that if voltage distortion limits at a PCC are met, then the current distortion may also be within limits. Identify the location, type, and magnitude of harmonic sources in the system. Simulate and understand the impact of harmonic sources under various conditions of power grid negative-sequence impedance. Determine individual voltage and current total harmonic distortion (THD) in order to determine the health of power quality within the facility, to avoid unexpected critical failures on key power system equipment. Use a frequency sweep to identify parallel or series resonance conditions that can cause failures on capacitor tanks, incorrect protection operation (such as fuses), etc. Simulate and compare harmonic mitigation methods, and choose the most relevant technology based on economics, safety, and performance such as transformer phase shifting, passive and active harmonic filters, etc.

A harmonic study should be performed whenever:

•• •• • •

Capacitor banks are to be installed. ASDs and/or other nonlinear loads are installed. The power network or equipment is experiencing harmonic-related problems. A generator is added in the plant as an alternate standby power source. A new facility or power system is being designed, where the load-flow, power factor compensation, and harmonic analyses are considered one integrated study to determine how to meet the reactive power demands and harmonic performance limits. Harmonic current injection needs to be limited at the PCC or grid interconnection point. This is accomplished by compliance with relevant standards and best practices, as summarized shortly.

A harmonic study consists of six significant steps: 1) 2) 3) 4) 5) 6)

Calculating or measuring harmonic source currents Modeling the electrical system using a one-line diagram Performing an electrical power-flow analysis at each harmonic frequency Determining the harmonic performance indices for the system and equipment Comparing harmonic performance indices with applicable industry standards and best practices Evaluating harmonic mitigation techniques and alternatives as needed

The factors that influence the effect of harmonics on the system and equipment include electric supply short-circuit level, location and size of capacitors and filters, loading, equipment grounding, and balanced/unbalanced conditions. The following approach is considered in the study: 1) Model a one-line diagram, and include relevant harmonic voltage and current sources. If the facility is new, then consider worst-case distortion based on minimum loading and typical sources of harmonic distortion. It is possible to use ETAP to determine the harmonic spectrum injection based on converter pulse numbers, commutation, and firing angles. 2) Set up the voltage and current harmonic distortion limits per component, paying special attention to the component type, such as the PCC or other type of bus. These limits can be set up globally in ETAP using the harmonic rulebook for IEEE or IEC standards. 3) Run a harmonic load flow and obtain the harmonic indices and voltage and current distortions. 4) ETAP automatically compares the harmonic distortion limits with selected standards and provides an alert view of the violations. 5) If there are capacitances in the network, such as line or cable susceptances, capacitor banks, or any other adjustable var source, run a harmonic frequency scan in addition to the harmonic load flow. ETAP automatically evaluates the frequency scan and determines whether a parallel resonance exists in the system. A parallel resonance can be potentially catastrophic for equipment if large voltage or current distortion is determined at a harmonic order that has parallel resonance. Theoretically, the voltage amplification for that order can become infinite, but it is practically limited by the resistance in the circuit. 6) If resonance and/or harmonic distortion limits are found during the simulation, then select appropriate harmonic mitigation methods and rerun the simulation with the same network conditions. If harmonic filters are sized for

34.3 Required Data

harmonic mitigation, then the filter design should take into account the power factor requirements of the system. It is possible to utilize harmonic filters for power factor compensation and gain dual benefits: minimize harmonic injection, and improve the overall system power factor. 7) Note that for an existing facility, all possible operating conditions, existing harmonic sources, and configurations should be used to evaluate harmonic distortion and resonance. For a new facility or system expansion, all potential or future harmonic sources should be included as well.

34.3

Required Data

Table 34.2 is a summary of the required data for harmonic analysis. Note that some of the data requirements vary based on the mode of operation. Refer to the ETAP user guide for a detailed list of required data.

Table 34.2 Harmonic analysis required data. Device

Bus

Line/Cable

Transformer

Impedance

Source

Nominal kV

x

x

Operating kV

x

x

Angle

x

x

Impedance R, X (pos)

X

x

x

Impedance R, X (zero)

X

x

x

Impedance R, X (neg) Diversity

Motors

x

x

x

x

x

x

Rating kVA

x

Tap changer

x

Tolerance

x

Temperature

x

x

Winding connection

x

x

x

Grounding

x

x

x

Phase shift

x

Operating mode

x

Mvar limits

x

kW/HP

x

kvar

x

Power factor

x

Xd

x

Rated kV

x

x

x x x

Efficiency

x

Loading levels

x

Generating levels

x

MVAsc

x

Harmonic injection data

x

x

Harmonic limit category

x

x

VTHD limits

x

X

VIHD limits

x

921

922

34 Harmonics

34.3.1

Harmonic Spectrum Data

Voltage and current harmonic spectrum data should be provided by the manufacturer or based on any field measurements made directly at the terminal of the harmonic source: Harmonic measurements can be made at selected locations – for example, the PCC and the locations where nonlinear loads are connected. In customer-owned transformer locations, the PCC is the point where the utility will meter the customer, generally the high-voltage side of the transformer. If the utility meters the low-voltage side, then this becomes the PCC. The decision about the optimal period to conduct harmonic measurements is somewhat complicated. Industrial installations are usually composed of a mix of loads having a diversity of spectral contents, which may require longterm measurements to characterize harmonic content. This need may become more evident if cyclic loads exist, because measurements to characterize harmonics at the PCC would need to encompass all, or at least the most significant, duty cycles. Under steady-state operation, and where no loading variations occur, a few minutes of recording are sufficient, and averaging over a few seconds should meet the requirements. However, due to the changing nature of loads in most situations, measurements over a few days are needed to assure that load-variation patterns and their effects on harmonic distortion are considered. Note that any measurement made other than at the terminal of the harmonic source is a composite spectrum. In order to use such as composite spectrum, utilize harmonic state estimation to distribute the composite spectrum into each individual harmonic source downstream of the measurement. This is a very complex task and not a preferred approach for most studies. The alternative is to create an equivalent load downstream of the measurement, if the intention is to understand the impact of this source on the rest of the system, particularly upstream.

34.3.2

Effect of kVA and Source Impedance

Power sources in relation to harmonics are often characterized by the terms stiff source and soft source, and both have a significant effect on both the nonlinear currents drawn by the load and the resulting voltage distortion. Stiff sources are often associated with transformers, whether utility-owned or customer-owned within a system. Their source impedance is often on the order of 5–6% (Z). Generators, however, are considered soft sources whose source impedance is actually the subtransient reactance (Xd ), often in the range of approximately 0.1–0.18 pu. Stiff sources, having a higher short-circuit capability, permit a given nonlinear load(s) to draw higher magnitudes of harmonic current for a given kW value of nonlinear load, as it is not as limited by the leakage reactance of the source. The higher harmonic current does not usually significantly distort the voltage. The stiffer the source, the higher the harmonic current that will be drawn by the nonlinear load(s) and the less the subsequent voltage distortion for a given load. Soft sources, such as generators, tend to have reduced short-circuit capability and limit the magnitude of the harmonic currents drawn by the given nonlinear load(s). However, that lower value of harmonic current will produce significantly higher levels of voltage distortion. Generators for low harmonic distortion need a low value of subtransient reactance, Xd , such as that achieved by oversizing the kVA rating, which results in an increase in short-circuit current. The generator rotor damper cage, designed for linear loads, is subject to higher levels of current when nonlinear loads are present. The machine subtransient reactance, Xd , should be relatively low to maintain the voltage distortion within the necessary limits. The damper winding must be designed such that the sinusoidal voltage waveform is maintained. The solution will depend on the type of nonlinear load, the magnitude of the harmonic currents produced, and the subsequent voltage distortion permissible in the power system. The kVA rating also has an impact on the magnitude of harmonic currents and subsequent voltage distortion for a given load as the short-circuit capability and associated Isc / IL (load current to short-circuit current ratio) vary. The higher the kVA (or MVA), the higher the harmonic current distortion, ITHD, and the lower the resulting harmonic voltage distortion, VTHD, assuming the source impedance is unchanged. Transformer kVA rating and impedance (Z) or (subtransient reactance Xd for generators) have an important effect on the magnitude of the harmonic currents drawn and the resulting voltage distortion. It can be seen that the resulting voltage distortion for a given nonlinear load varies in proportion to the source impedance (i.e. for generators, the lower the Xd , the less the resulting voltage distortion) and installed kVA. Achieving a low level of Xd usually involves a special winding design with a high level of excitation and therefore high magnetic flux or the use of derated generators (i.e. of a larger kVA rating than one based on the kW load.). The value of Xd varies from generator to generator and is inversely proportional to the square of the generator working voltage.

34.4 Harmonic Load Flow and Frequency Scan

At reduced generator loading, the effective Xd is reduced, thus decreasing the voltage distortion for a given nonlinear load. Modern generators have excitation systems that can cope with voltage and current distortion, provided they are correctly designed for the nonlinear load(s) in terms of thermal rating and have an appropriate level of winding insulation.

34.4

Harmonic Load Flow and Frequency Scan

The Harmonic Load Flow Study first carries out a load-flow calculation at the fundamental frequency. The results of the fundamental load flow sets the base for the fundamental bus voltages and branch currents, which are used later to calculate different harmonic indices. Then, for each harmonic frequency at which any harmonic source exists in the system, a direct load-flow solution is found by using the Current Injection Method. The frequencies considered are all from the 2nd to 250th order, including both harmonics and interharmonics. Impedance of components is adjusted based on the harmonic frequencies and the types of components. For a triplen harmonic frequency, zero-sequence impedance is adjusted to the actual frequency, and the zero-sequence network is used. From the harmonic load-flow calculation, the harmonic components for bus voltages and branch currents are found, and then all harmonic indices are computed accordingly. The computed bus voltage THD and individual harmonic distortion (IHD) are compared with their limits as specified by the user in the Bus Editor. If any violations are detected, they are shown in Harmonic Load Flow Analysis Alert View, and flags are placed in the text report next to the associated bus in the Harmonic Information section. The Harmonic Load Flow Study generates output reports showing the system input data, fundamental load-flow results, system harmonic information, and tabulation of bus voltages and branch currents with all harmonic contents. These results can also be viewed directly from the one-line diagram using the Harmonic Load Flow Slider and the Harmonic Display Options Editor. Along with the text report and one-line display, bus voltage and branch current plots are also available to show both voltage and current waveforms in the time domain and the harmonic spectrums in a bar chart. Frequency Scan is used to investigate a potential system resonance problem. It calculates and plots the magnitudes and phase angles of bus-driving point impedance over a frequency range specified by the user; thus, any parallel resonance condition and its resonance frequency can be clearly identified. The harmonic frequency scan study also allows users to tune their harmonic filter parameters and test the final results. Frequency Scan uses the system positive sequence as base impedance. The frequency range for scanning is defined by the user, and the results from the study are provided in reports that include the system input data, the fundamental load-flow results, and a tabulation listing bus-driving point impedances. The same tabulated information is also given on the one-line diagram, as well as in a plot format. Some of the main features, results, and reports for ETAP harmonic load flow and frequency scan are summarized here:

•• •• •• •• •• •• •• •• •

Same system and component modeling capabilities for harmonic load flow and fundamental load flow Automatic load-tap changer (LTC) settings for fundamental load flow Frequency dependency of rotary machine impedance Nonlinearity and frequency dependency of cable/line and transformer impedance Frequency dependency of other power system components and loads Transformer phase shift Machine and transformer winding connections and grounding types Short-line and long-line model for cable and transmission lines Interharmonic, harmonic voltage source, and current source models Harmonic source defined by either spectrum or device parameters Calculation of various harmonic indices based on IEEE and IEC standards THD for both bus voltages and branch currents Total RMS value for both bus voltages and branch currents Total arithmetic summation value (ASUM) for both bus voltages and branch currents Telephone influence factors (TIF) for both bus voltages and branch currents I×T product for branch currents I×TB (balanced component of I×T) product for branch currents

923

924

34 Harmonics

•• •• •• •• ••

I×TR (residual component of I×T) product for branch currents Total interharmonic distortion (TIHD) for both bus voltages and branch currents Total subharmonic distortion (TSHD) for both bus voltages and branch currents Group total harmonic distortion (THDG) for both bus voltages and branch currents Subgroup total harmonic distortion (THDS) for both bus voltages and branch currents Built-in harmonic filters in different types and automatic filter sizing Graphical one-line display of study results Slider bar to display fundamental load flow and total and individual harmonic distortion Graphical plots of voltage and current waveforms and a spectrum for viewing and printing Harmonic rulebooks for ANSI and IEC distortion-limit evaluation, with violation flagging based on current and voltage distortion limits and total and individual harmonic distortion limits.

34.5

Illustrative Examples

34.5.1

Example – Harmonic Study for an Industrial System

A harmonic study was executed on the one-line diagram “Example Model for Industrial System” shown in Figure 34.2. 34.5.1.1 Analysis Objectives

Analysis objectives for the harmonic study are described in Section 34.2. In this example, we analyzed the current and voltage distortions at 66 kV PCC when the plant is running on the system configuration “11kV CB-Tie Open” case, and evaluated the results compared with the requirements of IEEE Std. 519.

34.5.1.2 Required Data

The required data for the study are summarized in Section 34.3. In this example, in addition to the applied required data for a load-flow study, the following harmonic source data is required

•

Harmonic source => IEEE 6Pulse1 model for VFD1 (11 kV, 10MVA)

34.5.1.3 Results of the Harmonic Study

Table 34.3 summarizes the current and voltage distortion and frequency scanning by plots at 66 kV PCC as a result of harmonic studies on scenario “11 kV CB-Tie Open.” Table 34.4 summarizes the evaluation results compared with the criteria as per IEEE Std. 519, highlighted in light gray cells where the results exceed the criteria:

•• ••

I-THD: 7.52% > 5.0% I-IHD: 6.24% at 5th, 4.11% at 7th > 4.0% In order to maintain the I-THD and I-IHD within the criteria, some improvements are required, for example: Replace the harmonic source from the 6-pulse model to 12-pulse model. Add 5th and/or 7th harmonic filters.

34.5 Illustrative Examples

Harmonic distortion current and voltage at 66kV power incoming

Harmonic source

Figure 34.2 Harmonic study on the “11kV CB-Tie Open” case.

925

926

34 Harmonics

Table 34.3 Results of harmonic study on “11kV CB-Tie Open” case. Scenario

Event/Action/Results/Countermeasures

Harmonic load flow HA-1

Current distortion 100

6 5

Current (%)

Current Spectrum (%)

7

4 3 2

50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 –50

1 –100

0 0

10

20 30 Harmonic Order

40

Time (Cycle)

50

(Waveform)

(Spectrum)

Voltage distortion 150 100 1.5

Voltage (%)

Voltage Spectrum (%)

2.0

1.0

50 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

–50

0.5 –100 0

–150 0

10

40

20 30 Harmonic Order

Time (Cycle)

50

(Waveform)

(Spectrum)

Frequency Scan Z angle and magnitude 1.6

120

1.4

100

Impedance (ohm)

Impedance Angle (rad)

HA-1

1.2 1.0 0.8 0.6 0.4

80 60 40 20

0

5

10

15 20 Hermonic Order (Z Angle)

25

30

35

0

0

5

10

15

20

Harmonic Order (Z Magnitude)

25

30

35

34.5 Illustrative Examples

Table 34.4 Evaluation as per IEEE Std. 519. Criteria (IEEE Std 519)

Calculation results

Harmonic index

Order

Limit (%)

Order

Calculated (%)

Conditions for criteria (IEEE Std. 519)

I-THD

–

5.0

–

7.52

I-IHD

Q

Load 3

Figure 35.60 System P and Q after fault clearance.

Load 3

P=0

P+jQ

P’ + j Q’ Gen

Load 2

Gen

Load 3

Figure 35.58 Single-generator system with Load 1 – Load 3 energized.

J QL

Gen

Load 2

Load 3

Figure 35.61 Slow load shedding: loss of Load 1 + Load 2.

After fault clearance, the system is faced with partially collapsed flux energy in rotating machines (generators and motors) and has P +jQ to balance its generation and load levels while rebuilding its magLoad 2 Gen netic energy. During this time, depending on the motor residual back EMF, the system also has an additional reactive power demand from the motor loads under reacceleration conditions, as shown in Load 3 Figure 35.60. The longer it takes to shed MW or P load and return the voltage Figure 35.62 Fast load shedding: loss of Load 1. and frequency to acceptable levels, the more load-shedding MW is required, as shown in Figure 35.61 and 35.62. Generator turbine governors and the type of the prime mover also have a dramatic impact on the performance of the power system during major system disturbances. The frequency conditions of the overall system directly depend on the amount of active power that the generator prime movers can deliver to the system. Also, the mechanical energy available to help the generator prime movers ride through a fault or other disturbances plays an important role in system behavior. This stored energy varies dramatically between a gas turbine, steam turbine, and hydro unit. As a consequence, the performance of power systems supplied by different types of prime movers, governors, and voltage regulators is very different. For quick evaluation, the total system inertia may be determined as follows: Load 2

Htotal =

N i = 1 Hi × MVAi N i = 1 MVAi

The time taken to reach a decayed frequency is f −f0 df pu f0 × 2H × 2H = t f = Pgen − Pload dP pu Pgen In addition to system upsets caused by faults, there are disturbances caused by switching surges or lightning strikes. As an example, some switching disturbances can result in a loss of generation or cause a system to separate from the utility grid (system-islanding condition). This condition can cause the power system to collapse and be adversely impacted by

35.11 System Simulation

inappropriate load reduction caused by a faulty load-shedding scheme. For some switching disturbances (that result in a loss of generation or system islanding), cascading effects may be the primary concern if the load-shedding action is not set correctly and/or timed properly. Moreover, the type of disturbance impacts the dynamic response of the prime mover. For instance, a short circuit at the power station bus bar results in acceleration of the generator prime mover (no more brake power). For a gas turbine generator, the speed regulator then initiates closing of the fuel or gas-inlet valve. After the fault has been cleared, the turbines face the impact of the load still connected. At this time, their fuel or gas-inlet valves are closed, resulting in difficult reacceleration conditions. Gas turbines are very sensitive to critical speeds affecting their low-pressure blades. These critical speeds may be close to the rated operating speed, leaving a small margin in the allowed frequency range before a protective trip. For example, on a 50 Hz system, the protective instantaneous low-speed trip on gas turbines is set for 48 Hz, i.e. at −4% of the nominal system frequency. Various load-shedding schemes are utilized to protect the system from low frequency due to an imbalance in load and generation:

•• ••

Breaker interlock scheme Under-frequency relay (81) scheme Programmable logic controller-based load shedding Fast intelligent load shedding (ILS)

• • ••

Each study scenario should include the system configuration to define system connections and topology, data revisions to specify electrical parameters for the network and machines, types, and parameters. Study scenarios should include study cases covering all possible system configurations and operating conditions. Study scenarios should also investigate oscillation conditions under all possible system configurations and operating conditions. Additional study scenarios can be added to the scenario base when necessary. Study scenarios and solutions should be saved for future retrieval or modification.

Following are some recommendations for this task:

35.10

Stability Improvement

Depending on the causes of instability problems in a particular system, a number of enhancements can be made to improve system stability. Typical enhancements include:

•• • •• •

Improve the configuration and system design. Increase synchronizing power. Design and selection of rotating equipment – use induction motors – Increase moment of inertia – Reduce transient reactance – Improve voltage regulator and exciter characteristics Apply Power System Stabilizer (PSS). Add system protection – fast fault clearance, system separation, etc. Add a load-shedding scheme.

However, note that each of these remedies requires careful consideration, and we recommend re-running all system studies, because changes brought about by the listed enhancements will probably impact system transient-stability, short-circuit, and motor-starting results.

35.11

System Simulation

35.11.1

Events

Table 35.1 includes typical transient-stability action and disturbance types that can be included in ETAP as a sequence of events or executed in parallel at the same time.

977

978

35 Transient Stability

Table 35.1 Transient stability actions in ETAP. Device type

Actions

Setting 1

Setting 2

Bus

—

—

Cable

3 Phase Fault / Clear Fault/ LG Fault1 LL, LLG Fault Fault / Clear Fault

% of total length

—

Line

Fault / Clear Fault

% of total length

—

Impedance

Fault / Clear Fault

% of total length

—

Circuit breaker

Open / Close

—

—

SPST switch

Open / Close

—

—

Contactor

Open / Close

—

—

Fuse

Trip

—

—

Generator

Make reference machine

—

—

Switch to Droop mode

—

—

Switch to Isochronous mode

—

—

Start

—

—

Loss Excitation

—

—

Generation Impact

% change in electrical power

—

Generation Ramp

% change in electrical power

Time (sec) for % change

Voltage Impact

% change in reference voltage

—

Voltage Ramp

% change in reference voltage

Time (sec) for % change

Delete

—

—

Make reference machine

—

—

Voltage Impact

% change in reference voltage

—

Voltage Ramp

% change in reference voltage

Time (sec) for % change

Delete

—

—

Accelerate

—

—

Load Impact

% change in loading

—

Load Ramp

% change in loading

Time (sec) for % change

Delete

—

—

Load Impact

% change in loading

—

Load Ramp

% change in loading

Time (sec) for % change

Utility

Syn. & ind. motor

Lumped load

Delete

—

—

MOV

Start

—

—

Wind turbine

Wind Disturbance Wind Gust Wind Ramp

Wind turbine (zone)

Wind Disturbance Wind Gust Wind Ramp

VFD

Frequency Change

% change in frequency setting

None

Load Flow (no action, print load flow at the event time)

—

—

35.12 Illustrative Examples

Table 35.2 Partial list of transient-stability tabulated data and graphical plots. Plot type

Synch. generator

Synch. motor

Induction machine

Buses

Lumped loads

Branches

Power angle (relative)

X

X

Power angle (absolute)

X

X

Speed

X

X

MW electrical

X

MW mechanical

X

X

X

X

X

X

X

X

Mvar Current

X

X

X

X

X

X

X

X

X

X

X

Field voltage

X

X

Field current

X

X

Impedance

X

X

X

Bus voltage

X

X

X

Terminal voltage

X

X

X

V/Hz

X

Slip

X

Accel. torque

X

X X

Voltage angle

X

Frequency

X

Voltage

X

MVA

35.11.2

X

X

Results, Reports, and Plots

Transient-stability calculation results are available in ETAP in the form of graphical plots and tabulated data. Typical results for a few devices are shown in Table 35.2.

35.12

Illustrative Examples

35.12.1

Determine Critical Fault-Clearing Time (IEEE 9-Bus System)

Consider an IEEE nine-bus system with the input data shown in Figures 35.63 and 35.64: 1) Create a study case called FaultLine3. 2) Add new event: FltLine3End. This event is a sustained fault and is added to Line3 at 0% length or between CB 11 and Line3, as shown in Figure 35.65. Note that the event is added at t = 1 second because it is good practice to observe and report the system operating condition prior to applying disturbances. If the system is observed to be unstable prior to application of an event, then the simulation thereafter does not have much importance. 3) Run the Transient Stability Study with the FaultLine3 study case, and check the generator rotor angle plots. Use the time slide to navigate through the results. The color contouring displayed in Figure 35.66 is based on bus voltage. Green is normal voltage, blue is overvoltage, and yellow to red indicates undervoltage. These results are at t = 0 seconds (initial condition). Figure 35.67 shows the results at t = 1 second (fault event). button to open the event list. 4) Use the button to navigate to the next event; can be used to navigate to the next time step. The following are 5) Use the the results at the next event (t = 1 second) when the fault is applied on Line3. Note that since the fault is closer to

979

980

35 Transient Stability

Figure 35.63 ETAP one-line diagram for an IEEE nine-bus system.

CB11, i.e. Bus7, the voltage is nearly 0%, while other buses have progressively increasing voltage gradients based on their electrical distance (impedance) from the fault location. 6) The G2 rotor relative rotor angle reaches 180 at approximately 1.352 seconds at the first swing or 0.352 seconds, as shown in Figure 35.68. 7) In order to determine the CFCT, add a new action to the study case to clear the fault on Line3 at 1.3 seconds. 35.12.1.1 Second Run

Rerun the case, and check the generator rotor angle plots. The G2 rotor relative angle reaches 180 at approximately 1.358 seconds (>1.352 seconds) at the first swing, as shown in Figure 35.69. This implies that although we are clearing the fault, the FCT is not fast enough, allowing the generator to lose synchronism. Therefore, the direction we must continue to reduce the clearing time is 180

0.358

First swing unstable

0.200

>180

0.461.

First swing

0.140

>180

2.911.

Second swing

0.138 (CFCT)

Action: open CB-U1 at 0.1 sec Results: Generator speed goes down, and mechanical power reaches up to 150 MW (generator rated power) => Generator trips, and the system will collapse.

Generator Mechanical Power

Generator Speed 3200 3000 2600

MW

RPM

2800 2400 2200 2000 1800

0

1

2

3

4

5

6

7

8

9

10

160 150 140 130 120 110 100 90 80 70 60

0

1

2

3

Time (Sec)

5

6

7

8

9

10

8

9

10

Time (Sec)

Countermeasure Load shed (trip Industrial Plant-1 (100 MVA, 80% Loading) after 0.2 sec from Event 1) Event 1: Islanding the system from power grid => Action: open CB-U1 at 0.1 sec Event 2: Load shedding after 0.2 sec from Event 1 (trip Industrial Plant-1, 100 MVA, 80% loading) => Action: open CB-T2A at 0.3 sec Results: Both generator speed and mechanical power return to stable conditions.

Generator Speed

Generator Mechanical Power

3200 3000 2800 2600

MW

RPM

TS-1A

4

2400 2200 2000 1800

0

1

2

3

4

5

6

Time (Sec)

7

8

9

10

160 150 140 130 120 110 100 90 80 70 60

0

1

2

3

4

5

6

7

Time (Sec)

(Continued)

Table 35.3 (Continued) Scenario

Event/Action/Results/Countermeasures

Fault at 245kV bus (Bus-1) Event 1: Short-circuit fault at 245kV bus (Bus-1) => Action: fault at Bus-1 at 0.1 sec Event 2: Clear fault after 0.2 sec from Event 1 => Action: clear fault at 0.3 sec Results: Voltage of the 245 kV bus (Bus-1) falls to 0% for 0.2 sec Although the power angle, speed, and mechanical power of the generator fluctuate somewhat, they return to a stable state within 5 seconds. Generator Relative Power Angle

120 100 80

Degree

% of Bus Nominal kV

Bus-1 Voltage

60 40 20 0

0

1

2

3

4

80 70 60 50 40 30 20 10 0 –10 –20

5

0

2

4

6

8

12

14

16

18

20

16

18

20

Generator Mechanical Power

Generator Speed 3300

120

3200

110

MW

3100 3000 2900

100 90 80

2800 2700

10

Time (Sec)

Time (Sec)

RPM

TS-2

0

2

4

6

8

10

12

Time (Sec)

14

16

18

20

70

0

2

4

6

8

10

12

Time (Sec)

14

Fault in the middle of 245kV transmission lines (Line-2) Event 1: Short-circuit fault in the middle of Line-2 => Action: fault in the middle of Line-2 at 0.1 sec Event 2: Clear fault after 0.5 sec from Event 1 => Action: open CB-L2A and CB-L2B at 0.6 sec Results: Bus voltages at 245kV Bus-1, Bus-2, and Bus-3 fall to approx. 0 to 20% at 0.6 sec Although the speed, power angle, and mechanical power of the generator fluctuate somewhat, they return to a stable state within 6 seconds.

Generator Speed

120

3300

100

3200

80

3100

RPM

% of Bus Nominal kV

Bus-1 Voltage

60

3000

40

2900

20

2800

0

2700

0

1

2

3

4

5

0

2

4

6

80

Degree

% of Bus Nominal kV

100

60 40 20 0

1

2

3

4

200 150 100 50 0 –50 –100 –150 –200

5

0

2

4

6

8

Bus-3 Voltage 100

120

80

100 MW

140

60

60

20

40 2

3 Time (Sec)

16

18

20

10

12

14

16

18

20

80

40

1

14

Generator Mechanical Power

120

0

12

Time (Sec)

Time (Sec)

0

10

Generator Relative Power Angle

Bus-2 Voltage 120

0

8

Time (Sec)

Time (Sec)

% of Bus Nominal kV

TS-3

4

5

20

0

2

4

6

8

10

12

14

16

18

20

Time (Sec)

(Continued)

Table 35.3 (Continued) Scenario

Event/Action/Results/Countermeasures

Fault in the middle of 245kV transmission lines (Line-3A) Event 1: Short-circuit fault in the middle of Line-3A => Action: fault in the middle of Line-3A at 0.1 sec Event 2: Clear fault after 0.5 sec from Event1 => Action: open CB-L3A1 and CB-L3A2 at 0.6 sec Results: Bus voltages at 245kV Bus-1, Bus-2, and Bus-3 fall to approx. 10 to 25% at 0.6 sec Although the speed, power angle, and mechanical power of the generator fluctuate somewhat, they return to a stable state within 6 seconds. Generator Speed

120

3300

100

3200

80

3100 RPM

% of Bus Nominal kV

Bus-1 Voltage

60

3000

40

2900

20

2800

0

2700 0

1

2

3

4

5

0

2

4

6

Time (Sec)

80

Degree

% of Bus Nominal kV

100

60 40 20 0

0

1

2

3

8 10 12 Time (Sec)

4

200 150 100 50 0 –50 –100 –150 –200

5

0

2

4

6

8

100

120

80

100 MW

140

60

60

20

40 2

3 Time (Sec)

20

10

12

14

16

18

20

18

20

80

40

1

18

Generator Mechanical Power

Bus-3 Voltage 120

0

16

Time (Sec)

Time (Sec)

0

14

Generator Relative Power Angle

Bus-2 Voltage 120

% of Bus Nominal kV

TS-4

4

5

20

0

2

4

6

8

10

12

Time (Sec)

14

16

35.12 Illustrative Examples

35.12.4

Example – Transient-Stability Analysis for an Industrial System

Transient-stability analysis was executed on the one-line diagram “Example Model for Industrial System” shown in Figure 35.79.

Scenario TS -2 Voltage Sag/Dip from Power Grid

Scenario TS-1 & TS-1A Islanding from Power Grid

Scenario TS-3 Fault inside facility at 11kV feeder Scenario TS-4 Largest Motor Starting

Figure 35.79 Transient-stability study for an industrial system.

35.12.4.1 Analysis Objectives

In this example, we analyze the stability of the system when various system disturbances occur at the locations highlighted in the one-line diagram.

•

Study scenario (TS-1 and TS-1A) – industrial plant islanding from power grid: The purpose of this study is to verify the effect of an industrial power system islanding from its utility connection. The industrial plant may be inadvertently islanded due to malfunction of a protection relay or a forced outage caused by the trip of a main utility feeder circuit breaker at 66 kV. When the industrial plant is islanded, its onsite generation or cogeneration is unable to supply the power demand. If the situation is not corrected via load shedding, there is a possibility of a complete blackout of the facility. In this type of study, it is important to not only calculate the required load-shed MW but also calculate the load-shedding time delay (time from

995

996

35 Transient Stability

• • •

disturbance to mitigative action). If the load-shedding time delay is too long, the frequency and voltage will fall to the point of no return/collapse. This example shows how to determine the load-shedding MW and load shedding time. Study scenario (TS-2) – voltage sag/dip from the power grid: The purpose of this study is to simulate and understand the impact of a voltage sag/dip experienced by an industrial power plant due to a fault deep in the external power grid network or near the industrial plant. This type of fault is a temporary fault that is cleared by the utility protective devices. An industrial plant should be able to sustain and ride through such temporary faults without loss of critical equipment/motors. This example simulates a voltage dip experienced by the industrial plant for 200 ms; the critical synchronous machines (motors and generator) are shown to survive the event without going out of step. Study scenario (TS-3) – fault inside a plant at an 11 kV feeder: The purpose of this study is to simulate and understand the impact of a short-circuit fault on an 11 kV feeder inside the industrial facility power system network. This type of fault is more likely than the fault examined in the previous case. The intention is to observe the impact of such a fault on critical machines and generators. Typically, each 11 kV and other voltage levels must be analyzed; however, for illustrative purposes, one 11 kV feeder is selected for this analysis. Study Scenario (TS-4) – largest motor starting: In this example, another type of sudden, large change is introduced by means of starting the largest induction motor in the industrial plant. The purpose is to analyze whether this motor can start in a reasonable amount of time. This example also illustrates the relationship that motor torque is dependent on terminal voltage (T V2), which directly affects the acceleration time under various starting conditions.

35.12.4.2 Required Data

In this example, dynamic models and parameters for the generator and motor are required, in addition to the required data for the load-flow study. Further details for dynamic models are described in Section 35.4.

•

Generator parameters:

35.12 Illustrative Examples

•

AVR model => IEEE Type 1: SE = f(Efd)

Vref VT IT

VC1= VF+(Rc+jXc)IT

VC1

Vc – + 1 𝛴 1+sTR –

VRmax KA 1+sTA

+

𝛴

–

1 KE + sTE

Efd

VRmin sKA 1+sTF

•

Governor model => GT model

Wref W

+

𝛴

–

P Kw 1 + sTsr

–

𝛴

P = Pe for Isoch Pref for droop +

Pmax 1 1 + sTc

1 1 + sTt

Pm Pmin

• •

Inertia of generator and prime mover: H = 5 MW-s/MVA (WR2 = 5626 kg/m2) total for the generator and prime mover. Dynamic model of induction motor (IM-3, 5.45 MW, four-pole): see Section 35.4.3.

35.12.4.3 Results of Transient-Stability Analysis

Study scenarios (events and actions that occur on this network) and the results of the analysis are summarized in Table 35.4. The countermeasures necessary to maintain/restore the stability of the system when the system remains unstable are also listed in Table 35.4. This study is executed on the system configuration “11 kV CB-Tie Open” case. More analysis of other system configurations caused the severe results is also required.

997

Table 35.4 Summary of transient-stability analysis for an industrial system. Scenario

Event/Action/Results/Countermeasures

Industrial plant islanding from power grid TS-1

Event 1: Islanding the system from power grid = > Action: open CB-66 kV at 0.1 sec Results: Generator speed falls, and mechanical power reaches up to 50 MW (generator rated power) = > Generator trips, and the system will collapse.

Generator Speed

Generator Mechanical Power 55

3200 3000

50

2600

MW

RPM

2800 2400 2200

45 40

2000 1800

0

1

2

3

4

5

6

7

8

9

35

10

0

1

2

3

Time (Sec)

5

6

7

8

9

10

Time (Sec)

Countermeasure Load shed (trip L-T3 [30MVA] after 0.2 sec from Event 1) Event 1: Islanding the system from power grid = > Action: open CB-66 kV at 0.1 sec Event 2: Load shedding after 0.2 sec from Event 1 (trip L-T3, 30 MVA) = > Action: open CB-T1–3 at 0.3 sec Results: Both generator speed and mechanical power returned to stable conditions. Generator Speed

Generator Mechanical Power

3200

55

3000 50

2800 2600

MW

RPM

TS-1A

4

2400 2200

40

2000 1800

45

0

1

2

3

4 5 6 Time (Sec)

7

8

9

10

35

0

1

2

3

4

5 Time (Sec)

6

7

8

9

10

Voltage sag/dip from power grid Event 1: Voltage sag/dip (down to 20% for 0.2 sec) occurred in power grid, and the voltage at Bus-66 kV fell to 20% = > Action: fault at Fault-Bus at 0.1 sec Event 2: Voltage at Bus-66 kV returns to 100% after 0.2 sec from Event 1 = > Action: clear fault at 0.3 sec Results: Voltage of the 66 kV bus (Bus-66 kV) falls to 20% for 0.2 sec Voltage of the 11 kV bus (Bus-11 kV-T1) also falls to approx. 20% for 0.2 sec Although the speed and power angle of the generator fluctuate somewhat, they return to a stable state within approx. 20 seconds. The synchronous motor SM-1 (5 MW) also returns to a stable state within approx. 2 seconds. The industrial plant will continue operation even with the voltage sag/dip at the grid (falls to 20% for 0.2 sec).

Bus-11kV-T1 Voltage

Bus-66kV Voltage

% of Bus Nominal kV

% of Bus Nominal kV

120 100 80 60 40 20 0

0

1

2

3

4

120 100 80 60 40 20 0

5

0

1

2

RPM

Degree 0

2

4

6

8

10

12

14

16

18

20

80 60 40 20 0 –20 –40 –60 –80 –100 –120 –140

0

2

4

6

Time (Sec) Synchronous Machine Speed

0

2

4

6

8

10

12

Time (Sec)

5

14

8 10 12 Time (Sec)

14

16

18

20

18

20

Synchronous Machine Relative Power Angle

Degree

3200 3150 3100 3050 3000 2950 2900 2850 2800

4

Generator Relative Power Angle

Generator Speed 3200 3150 3100 3050 3000 2950 2900 2850 2800

3

Time (Sec)

Time (Sec)

RPM

TS-2

16

18

20

80 60 40 20 0 –20 –40 –60 –80 –100 –120 –140 0

2

4

6

8

10

12

14

16

Time (Sec)

(Continued)

Table 35.4 (Continued) Scenario

Event/Action/Results/Countermeasures

Fault inside plant at 11 kV feeder Event 1: Short-circuit fault at 11 kV bus (Bus-CB-T1–3) = > Action: fault at Bus-CB-T1–3 at 0.1 sec Event 2: Clear fault after 0.2 sec from Event 1 = > Action: clear fault at 0.3 sec Results: Voltage of the 66 kV bus (Bus-66 kV) falls to approx. 50% for 0.2 sec Voltage of the 11 kV bus (Bus-11 kV-T1) falls to 0% for 0.2 sec due to bolted (zero-impedance) fault near this bus Although the speed and power angle of the generator fluctuate somewhat, it returns to a stable state within approx. 20 seconds. The synchronous motor SM-1 (5 MW) also returns to a stable state within approx. 2 seconds. The industrial plant will continue operation if it is cleared within 0.2 sec against a short-circuit fault occurring on the 11 kV bus (Bus-CB-T1–3). Bus-66kV Voltage

Bus-11kV-T1 Voltage 120 % of Bus Nominal kV

% of Bus Nominal kV

120 100 80 60 40 20 0

0

1

2

3

4

100 80 60 40 20 0

5

0

1

2

Time (Sec)

0

2

4

6

8

10

12

14

16

18

20

80 60 40 20 0 –20 –40 –60 –80 –100 –120 –140

0

2

4

6

Time (Sec)

Degree 0

2

4

6

8

10

5

12

Time (Sec)

14

8 10 12 Time (Sec)

14

16

18

20

Synchronous Machine Relative Power Angle

Synchronous Machine Speed 3200 3150 3100 3050 3000 2950 2900 2850 2800

4

Generator Relative Power Angle

Degree

RPM

3200 3150 3100 3050 3000 2950 2900 2850 2800

3 Time (Sec)

Generator Speed

RPM

TS-3

16

18

20

80 60 40 20 0 –20 –40 –60 –80 –100 –120 –140

0

2

4

6

8

10

12

Time (Sec)

14

16

18

20

Largest motor starting Event 1: Start the largest-capacity motor IM-3 (5.45 MW) = > Action: start IM-3 at 0.1 sec Results: Starting time is approx. 0.9 sec from induction machine slip curve. Bus voltage at starting the motor falls to 90% at Bus-6.6 kV-T2 and 95% at Bus-66 kV. The largest-capacity motor (IM-3, 5.45 MW) can start in a reasonable amount of time when the plant is running on the system configuration “11 kV CB-Tie Open” case. Induction Machine Acceleration Torque

120

3

100

2

80

1 MW

Slip (%)

Induction Machine Slip

60

0

40

–1

20

–2

0

0

1

2

3

4

5

6

7

8

–3

9 10 11 12 13 14 15

0

1

2

3

4

5

6

Time (Sec)

100

2500

80

2000

I (Amps)

% of Bus Nominal kV

3000

60 40

9 10 11 12 13 14 15

1500 1000 500

20

0 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

0

1

2

3

4

5

Time (Sec) Bus-66kV Voltage

6 7 8 9 10 11 12 13 14 15 Time (Sec)

Bus-6.6kV-T2 Voltage

110 % of Bus Nominal kV

110

105 100 95 90 85

8

Induction Machine Teminal Current

120

0

7

Time (Sec)

Induction Machine Terminal Voltage

% of Bus Nominal kV

TS-4

105 100 95 90 85

0

1

2

3

4

5

6

7

8

Time (Sec)

9 10 11 12 13 14 15

0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

Time (Sec)

1003

36 Reliability Assessment 36.1

Introduction

An electric power system has a primary function to satisfy load requirements as economically as possible and with a reasonable assurance of stability and quality. Power system planners and operators have utilized a wide range of criteria to achieve the required degree of reliability while maintaining safety and economy. In the past, new construction was planned to meet load growth and justified by the additional revenue generated or by the needs of the service area. Sometimes, new facilities were constructed solely to improve reliability, even though little effort was made to quantify reliability benefits. Today, the question is asked by the ratepayer and the regulators –“What is the cost benefit to the customer based on the resulting change in reliability?” Initially, all of these reliability criteria were deterministic or qualitative, with many of these criteria and associated methods still in use today. Reliability assessment using qualitative techniques implies dependence solely upon engineering experience and judgment. The main weakness of deterministic criteria is that they do not respond to nor do they reflect the probabilistic or stochastic nature of system behavior, of customer demands, or of component failures. The stochastic or random nature can be incorporated in a probabilistic approach to electric power system reliability assessment with a wide range of techniques and criteria. Probabilistic or quantitative reliability studies may, therefore, be defined as the likelihood of a system performing its function adequately for the period and intended operating conditions. Quantitative methodologies use statistical approaches to reinforce engineering judgments. These techniques can describe historical performance of existing systems and therefore predict the effects of changing conditions on system performance. The concept of power system reliability is expansive and encompasses many aspects of the power system to satisfy customer requirements. There is a reasonable subdivision of the concern designated as system reliability, as shown in Figure 36.1. System adequacy relates to the existence of sufficient capability within the power system to satisfy the load demand. This includes the summation of generation plants necessary to produce sufficient energy and the interconnected transmission and distribution infrastructure required to deliver the energy to the consumer load points. Security relates to the ability of the system to respond to disturbances arising within that system. Security is therefore associated with the response of the system to perturbations. Most of the probabilistic techniques presently available for power system reliability evaluation are in the domain of adequacy assessment.

36.2

Analysis Objectives

System planners can apply probability assessment techniques to a variety of tasks, such as:

•• •• ••

Determination of reliability trends Comparison of alternative system plans Development of reliability criteria and design standards System performance assessment against reliability criteria Selection of appropriate substation schemes Identification and ranking of system weakness and sensitivity

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

1004

36 Reliability Assessment

•• •• •

Reliability assessment is typically carried out for: Generation station and capacity Composite generation and transmission system Distribution system Substation and switching stations Protection system

System reliability

System adequacy System security

Figure 36.1 Concept of system reliability.

Various indices are calculated in order to rank and compare the reliability of one electrical system versus another, including but not limited to:

•• •• •• •• •• •• •• •

System Bus Failed component Frequency and duration Probability of transfer limits exceeded Unserved energy Number of customers affected Customer interruption frequency Customer interruption duration Customer curtailment or power/energy not served Reliability cost and consumer interruption cost Seasonal indices Running costs Incremental line-loss costs Incremental fuel costs

36.3

Problem Formulation

36.3.1

Generation Reliability Assessment

The standard definition of failure while evaluating generation capacity adequacy is “loss of load.” Loss of load is an outage due to lack of capacity. Reliability may be defined in terms of the loss of load probability in a given time interval, typically a year, i.e. the loss of load expectation (LOLE) given in days per year. For a loss of load to occur, the system capacity must fall to a level due to scheduled maintenance and/or forced outages of other generating units by a margin exceeding the spinning reserve to meet the system peak load. When the generation capacity cannot meet the actual load, the load-duration curve has to considered in order to calculate the amount of outage time. The most commonly used generation reliability index of LOLE can be calculated if all parameters, namely, forced outage rates of different generating units, the load forecast, the load duration curve, the spinning reserve, and the other refinements deemed necessary (e.g. the reliability of the transmission system), are known. Electric utilities routinely perform probabilistic assessments of generation reserve margin requirements using sophisticated tools based on Monte Carlo simulations and a contingency enumeration approach. 36.3.2

Transmission Reliability Assessment

The probability of system capacity outage levels calculated based on the forced-outage rates of the generators alone will lead to overly optimistic results. The average forced-failure rate and outage duration of each component, such as overhead lines, transformers, and circuit breakers of the transmission system, can be calculated, along with the reliability of a load point using an appropriate reliability model. Load-point reliability depends on individual components and other factors such as system configuration and environment. Since the transmission system is meshed with multiple network loops, failure of one component does not necessarily result in system failure. Failure of outdoor components is caused by lightning, snow, high winds, and so on. Failure of one component may increase the chance of failure of another, as assumed in statistical calculations. This

36.5 Illustrative Examples

is commonly known as common-mode failure, where more than one component fails due to the same cause. Therefore, different failure rates are allocated based on weather conditions and the dependency of failures.

36.3.3

Distribution Reliability Assessment

The application of reliability concepts to distribution systems differs from generation and transmission applications in that it is more customer load-point oriented instead of being system oriented, and the local distribution system is considered rather than the whole integrated system involving the generation and transmission facilities. Generation and transmission reliability also emphasizes capacity and loss-of-load probability, with some attention paid to components, whereas distribution reliability looks at all facets of engineering: design, planning, and operations. Because the distribution system is less complex than the integrated generation and transmission system, the probability mathematics involved are much simpler than those required for generation and transmission reliability assessments. In summary, a detailed analysis can be very complex, and it gets more complicated when the composite generation, transmission, and distribution system is taken together. The use of powerful computer software is almost mandatory for any system reliability analysis.

36.4

Required Data

To run a distribution system reliability analysis study, an engineer must obtain reliability-related data, such as failure rates, repair times, and switching times of network elements. In addition to network connectivity and a one-line diagram, a summary of reliability data for different types of elements is given in Table 36.1. Refer to the ETAP user guide for detailed information. Table 36.1 Required data for the reliability assessment. Device

Buses

Active failure rate

x

Lines/Cables

Passive failure rate

Transformers

Impedances

Sources

Loads

x

x

x

x

x

x x

Repair time

x

x

x

x

Switching time

x

x

x

x

Replacement time

x

x

x

x

Length

x

x

x

Load sector

x

Quantity (no. of loads)

x

36.5

Illustrative Examples

36.5.1

Simple Radial System

Consider the simple radial system shown in Figure 36.2, with the reliability data shown in Tables 36.2–36.6. The tables are the input data for the simple radial system including a bus, branch-switching device, load, and source data. Solution. The main bus is de-energized if:

•• ••

CB1 fails actively or passively CB2 and CB3 fail actively Utility fails The main bus itself fails

1005

1006

36 Reliability Assessment

U1 1500 MVAsc

Main bus 34.5 kV

CB1

CB2

CB3 T1 10 MVA

Bus2 4.16 kV

CB4

T2 10 MVA

Bus3 4.16 kV

CB6

CB5

CB7

Mtr1 25 HP

Mtr2 25 HP

Figure 36.2 ETAP one-line diagram for the simple radial system.

Table 36.2 Bus input data. Replacement λA

Momentary rate

MTTR

Switch time

f/yr

λA%

hour

4.160

0.0010

80.00

4.150

0.0010

34.500

0.0010

Bus ID

kV

Bus2 Bus3 Main bus

Avail.

Time

hour

Yes/No

hour

2.00

2.00

No

10.00

80.00

2.00

2.00

No

10.00

80.00

2.00

2.00

No

10.00

Table 36.3 Branch input data. Replacement Transformer

λA

λP

Momentary rate

MTTR

Switch time

Avail.

Time

ID

Type

f/yr

f/yr

λA%

hour

hour

Yes/No

hour

T1

2 W XFMR

0.0250

0.0150

80.00

200.00

200.00

No

10.00

T2

2 W XFMR

0.0150

0.0100

80.00

200.00

200.00

No

10.00

36.5 Illustrative Examples

Table 36.4 Switching device input data. Replacement λA

Switching device

λP

Momentary rate

MTTR

Switch time

Avail

Time

f/yr

f/yr

λA%

hour

hour

Yes/No

hour

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB4

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB3

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB5

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB6

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB7

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

ID

Type

CB1 CB2

Status

Table 36.5 Load input data. Replacement Load ID

λA

No. of Type

Connected bus ID

kW

Sector

Loads

f/yr

Momentary Rare

MTTR

Avail.

Time

λA%

hour

Yes/No

hour

Mtr1

Ind. Mtr.

Bus2

24.04

Industrial

1

0.0200

80.00

50.00

No

10.00

Mtr2

Ind. Mtr.

Bus3

24.04

Commercial

1

0.0200

80.00

50.00

No

10.00

Table 36.6 Source input data. Replacement Source ID

U1

λA

Momentary rate

MTTR

Snitch Time

Avail

Time

Type

Connected bus ID

f/yr

λA%

hour

hour

Yes/No

hour

Power Grid

Main Bus

0.0010

80.00

2.00

2.00

No

10.00

The failure rate for the main bus is given as λLP1 = λa + λp

CB1

+ λa

CB2

+ λa

CB3

+ λUtility + λMain Bus

= 0 005 + 0 003 + 0 003 + 0 001 + 0 001 = 0 013 failure year Annual unavailability for the main bus is ULP1 = MTTRCB1 × λa + λp + MTTRCB3 × λa

CB1

+ MTTRCB2 × λa

CB2

CB3 + MTTRUtility × λUtility

+ MTTRMain Bus × λMain Bus = 30 × 0 005 + 30 × 0 003 + 30 × 0 003 + 2 × 0 001 + 2 × 0 001 = 0 334 hour year

1007

1008

36 Reliability Assessment

Time to replace the main bus is rLP 1 =

U 0 334 = = 25 692 hours λ 0 013

Simulation results using the Reliability Assessment module from ETAP are shown in Figure 36.3. U1 1500 MVAsc 23 hr 0.004 f/yr Main Bus 34.5 kV

f/yr 0.013 25.7 hr

CB1 CB3

CB2

T1 10 MVA

Bus2 4.16 kV

CB4

CB6

0.064

T2 10 MVA

f/yr Bus3 4.16 kV

134.

f/yr

0.071

Average Average interrupting outage rate duration

Bus Load sector

Connected Bus ID

f/yr

f/yr

Mtr2 25 HP 94.3 hr

Mtr1 25 HP 112.7 hr

ID

0.049 115 hr

CB7

9 hr

0.086

CB5

f/yr

Annual outage duration

hour

hr/yr

EENS

ECOST

MW hr/yr

$/yr

IEAR $/kW hr

Bus2

N/A

0.0640

134.94

8.6360

0.2076

1582.49

7.624

Bus3

N/A

0.0490

115.02

5.6360

0.1355

1727.38

12.751

Main Bus

N/A

0.0130

25.69

0.3340

0.0000

0.00

0.000

Mtr1

Industrial

Bus2

0.0860

112.74

9.6960

0.2331

1774.82

7.615

Mtr2

Commercial

Bus3

0.0710

94.31

6.6960

0.1609

2045.86

12.711

U1

None

Main Bus

0.0040

23.00

0.0920

0.0000

0.00

0.000

Figure 36.3 ETAP results for the simple radial system.

36.5.2

Single and Double Contingency

Consider the simple system shown in Figure 36.4, which illustrates the effect of single and double contingency. Tables 36.7–36.11 show the required data for the system. Solution. We simplify the hand calculations by assuming failure rates for the breakers connecting the transformers to the buses are zero. Failure rates of the two transformers are taken to be 1, and MTTR = 200 hour. Failure rate at Bus2 due to double contingency is λ = λ1 λ2 r1 + r2 = 1×1×

8760

200 + 200 8760

1 + λ1 r1 + λ2 r2 8760 1+

1 × 200 + 1 × 200 = 0 0436681 failures year 8760

U1 1500 MVAsc

Main Bus 34.5 kV

CB1

CB2

CB3

T1 10 MVA

T2 10 MVA

CB4

Bus2 4.16 kV

CB5 CB6

Mtr1 25 HP Figure 36.4 ETAP one-line diagram for a parallel system.

Table 36.7 Bus input data. Replacement Bus

λA

Momentary rate

MTTR

Switch time

Avail

Time

ID

kV

f/yr

λA%

hour

hour

Yes/No

hour

Bus2

4.160

0.0010

80.00

2.00

2.00

No

10.00

Main Bus

34.500

0.0010

80.00

2.00

2.00

No

10.00

Table 36.8 Branch input data. Replacement Transformer

λA

λP

Momentary rate

MTTR

Switch time

Avail.

Time

ID

Type

f/yr

f/yr

λA%

hour

hour

Yes/No

Hour

T1

2 W XFMR

1.0000

0.0000

80.00

200.00

200.00

No

10.00

T2

2 W XFMR

1.0000

0.0000

80.00

200.00

200.00

No

10 00

Table 36.9 Source input data. Replacement Source

λA

Momentary rate

MTTR

Switch time

Avail

Time

ID

Type

Connected bus ID

f/yr

λA%

hour

hour

Yes/No

hour

U1

Power Grid

Main Bus

0.0010

80.00

2.00

2.00

No

10.00

1010

36 Reliability Assessment

Table 36.10 Load input data. Replacement Load

No. of

λA

Momentary rate

MTTR

Avail.

Time

ID

Type

Connected bus ID

kW

Sector

loads

f/yr

λA%

hour

Yes/No

hour

Mtr1

Ind. Mtr.

Bus2

24.04

Agricultural

1

0.0200

80.00

50.00

No

10.00

Table 36.11 Switching device input data. Replacement Switching device ID

Type

Status

λA

λP

Momentary rate

MTTR

Switch time

Avail.

Time

f/yr

f/yr

λA%

hour

hour

Yes/No

hour

CB1

HVBreaker

0.0030

0.0020

80.00

30.00

30.00

No

10.00

CB2

HVBreaker

0.0000

0.0000

80.00

30.00

30.00

No

10.00

CB4

HVBreaker

0.0000

0.0000

80.00

30.00

30.00

No

10.00

CB3

HVBreaker

0.0000

0.0000

30.00

30.00

30.00

No

10.00

CB5

HVBreaker

0.0000

0.0000

80.00

30.00

30.00

No

10.00

CB6

HVBreaker

0.0030

0.0020

30.00

30.00

30.00

No

10.00

The failure rate for the single-contingency case is P A + λMainBus + λBus2 + λCB6 λsingle = λU1 + λACB1 + λCB1

= 0 011failures year The total failure rate at Bus2 is λBus2 = λdouble + λsingle = 0 0436681 + 0 011 = 0 0546681failures year Simulation results from ETAP for the single-contingency case are given in Table 36.12. Simulation results from ETAP are shown in Figure 36.5 for the double-contingency case and tabulated in Table 36.13.

Table 36.12 ETAP results for single-contingency case.

Bus

ID

Load sector

Connected bus ID

Average interrupting rate

Average outage duration

Annual outage duration

f/yr

hour

hr/yr

EENS

MW hr./yr

ECOST

$/yr

IEAR

$/kW hr

Bus2

N/A

0.0110

22.36

0.2460

0.0059

3 05

0.516

Main BUS

N/A

0.0070

22.00

0.1540

0.0000

0.00

0.000

Mtr1

Agricultural

Bus2

0.0330

39.58

1.3060

0.0314

16.15

0.514

U1

None

Main Bus

0.0040

23.00

0.0920

0.0000

0.00

0.000

36.5 Illustrative Examples

U1 1500 MVAsc

Main Bus 34.5 kV

CB1

CB2

0.007

22 h

CB3

r

T2 10 MVA

T1 10 MVA

CB4

Bus2 4.16 kV

f/yr

CB5 CB6

Mtr1 25 HP

f/yr 0.011 22.4 hr

f/yr 0.033 39.6 hr

Figure 36.5 ETAP results for single and double contingency.

Table 36.13 ETAP results for double-contingency case.

Bus

ID

Load sector

Bus2

N/A

Connected bus ID

Average interrupting rate

Average outage duration

Annual outage duration

f/yr

hour

hr/yr

MW hr./yr

$/yr

0.0547

84.38

4.6128

0.1109

57.01

0.514

EENS

ECOST

IEAR

$/kW hr

Main Bus

N/A

0.0070

22.00

0.1540

0.0000

0.00

0.000

Mtr1

Agricultural

Bus2

0.0768

73.93

5.6755

0.1364

70.14

0.514

U1

None

Main Bus

0.0040

23.00

0.0920

0.0000

0.00

0.000

36.5.3

Reliability Index Calculation

Consider the simple system shown in Figure 36.6, with input data obtained from IEEE Trans. on Power Systems, vol. 6, no. 2 (May 1991): 813–820. SAIFI = total number of customer interruptions / total number of customers served =

λ i Ni

Ni

0 2403 × 210 + 0 2533 × 210 + 0 2533 × 210 + 0 2403 × 1 + 0 2533 × 1 + 0 25 × 10 + 0 2533 × 10 + 0 1407 × 1 + 0 1407 × 1 210 + 210 + 210 + 1 + 1 + 10 + 10 + 1 + 1 = 162 657 654 =

= 0 2487 f customer yr

1011

U1 110 MVAsc Bus0 11 kV CB3

CB1 Line3 Bus2 11 kV

Bus1 11 kV

Bus4 11 kV

Bus5 11 kV

Line2

Bus3 11 kV

Fuse5 LP2 535 kVA

Cable12

T5 0 MVA

Bus10 11 kV

Fuse8

SW3 Cable10

Fuse9 LP4 0.566 MVA

Fuse6

Cable13

T6 0 MVA

Bus7 11 kV

Bus6 11 kV

Bus9 11 kV

SW1

T1 0 MVA

Cable8

Cable9

Bus8 11 kV

Line5

Bus15 11 kV

Bus14 11 kV

Bus11 11 kV

Bus12 11 kV

Fuse10

LP1 535 kVA

Bus18 11 kV

LP7 454 kVA

Cable14

T7 0 MVA Bus17 11 kV

T8 0 MVA

Fuse14 Line6 Bus16 11 kV

Cable15

Fuse11

Figure 36.6 ETAP one-line diagram for reliability index calculation.

Cable16

Bus22 11 kV Fuse2

LP3 535 kVA

Cable4 LP9 1.15 MVA

Open Fuse12

Bus13 11 kV

T9 0 MVA

LP8 1 MVA

Cable1 Bus21 11 kV

SW4 LP5 0.566 MVA

Fuse1 Cable2

T3 0 MVA

SW2

Bus20 11 kV

Bus19 11 kV

LP6 454 kVA

36.5 Illustrative Examples

SAIDI = total number of customer interruptions / total number of customers served =

Ui Ni

Ni

3 5773 × 210 + 3 6423 × 210 + 3 6423 × 210 + 3 5773 × 1 + 3 6423 × 1 + 3 626 × 10 + 3 6033 × 10 + 0 5447 × 1 + 0 5058 × 1 210 + 210 + 210 + 1 + 1 + 10 + 10 + 1 + 1 = 2361 56 654

=

= 3 6109 hr customer yr CAIDI: ΣUiNi/ΣλiNi = 2361.56/162.657 = 14.5186 hr/customer interruption ASAI: ( 8760Ni − UiNi)/ 8760Ni = 0.9996 ASUI: 1 − ASAI = 0.0004 EENS: EENSi = 14.3 MWhr/yr ECOST: ECOSTi = 78.865 k$/yr AENS: UiPi/ Ni = 0.022 MWhr/customer.yr IEAR: ECOST/EENS = 5.513 $/kWhr Simulation results from ETAP are shown in Figure 36.7.

36.5.4

Example – Reliability Assessment for a Transmission System

A reliability assessment was executed on the one-line diagram “Example Model for Transmission System” shown in Figure 36.8. 36.5.4.1 Analysis Objectives

Distribution system reliability assessment deals with the availability and quality of the power supply at each customer’s service entrance. Analysis of customer failure statistics shows that, compared to other portions of electrical power systems, distribution system failures contribute as much as 90% toward the unavailability of supply to a load. These statistics show how important the reliability evaluation of distribution systems can be. The basic reliability indices normally used to predict or assess the reliability of a distribution system are as follows:

•• •

Load-point average failure rate λ Average outage duration r Annual unavailability U

•• •• ••

System average interruption frequency index (SAIFI) System average interruption duration index (SAIDI) Customer average interruption duration index (CAIDI) Average energy not supplied (AENS) Average service availability index (ASAI) Average service unavailability index (ASUI)

•• •

Expected energy not supply (EENS) Expected interruption cost (ECOST) Interrupted energy assessment rate (IEAR)

In order to evaluate the severity or significance of a system outage, using these three basic indices, two expanded sets of indices must also be calculated. The first set is the system reliability index:

These additional indices can be used to assess the overall behavior of the distribution system. The second set includes the reliability cost/worth indices:

The indices EENS, ECOST, and IEAR can be specifically for each load point or for the overall system. All of these indices can be used to evaluate the reliability of an existing distribution system and to provide useful planning information regarding improvements to existing systems and the design of new distribution systems.

1013

1014

36 Reliability Assessment U1 110 MVAsc Bus0 11 kV

0.002

0.002 hr/yr 0.001 f/yr

CB3 r 53 f/y Bus5 0.2 3 .6 4 hr/y 11 kV r

Bus4 11 kV

f/yr LP2 0.253 T5 535 kVA 0 MVA 3.64 hr/yr Bus10 /yr 11 kV 0.24 f 3.58 h r/yr

Bus9 11 kV

LP4 T6 0.566 MVA f/yr 0 MVA 0.24 3.58 hr/yr f/yr 0.253 Bus15 3.64 h 11 kV r/yr 0.253

Bus14 11 kV

f/yr

LP5 T7 0.566 MVA 0 MVA f/yr 0.253 Bus18 hr/yr 11 kV 3.64 3.64 hr/yr LP7 454 kVA f/yr 0.253

Bus1 11 kV

0.186 0.382 hr/yr

f/yr

Bus2 11 kV

Fuse14

T8 0 MVA

/yr

1 hr

0.00

f/yr f/yr 0.24 Bus3 0.141 Bus20 0.545 11 kV Bus19 r hr/yr f/y 3.58 h 11 kV r/yr 0.243.58 11 kV hr/yr 0.28859 f/yr r Fuse5 0.0 hr/yr 1 f/y 4 r .1 0 f/y Fuse1 SW3 0.24SW1 Fuse8 T1 0.545 hr/yr 0 MVA LP8 f/yr f/yr LP1 1 MVA 3 5 .186 r 0 .2 535 kVA Bus7 0 f/y Bus6 f/yr 0.253 f/yr 11 kV 11 kV 0.24 f/yr Bus21 Bus22 11 kV 0.141 0.382 0.089 3 3 3.58 h 11 kV 11 kV .6 .6 h 0 r/ .506 h 4 hr/ 4 hr/ yr r/yr 0.246 r/yr yr yr hr/yr Fuse9 3.58 hr/yr Fuse6 T3 SW2 Fuse2 f/yr 0 MVA LP3 0.253 LP9 535 kVA 1.15 MVA f/yr 3.64 hr/yr 0.141 r Open 0.506 hr/yr /y f r 5 /y f Fuse12 0.2 f/yr Bus11 0.186 Bus12 Bus13 0.253 f/yr 11 kV 11 kV 11 kV 0.25 3.64 h 3 0 .3 .6 3 .6 8 3 2 hr/y 3 hr/y r/yr Fuse10 hr/yr r r SW4 f/yr 0.25

f/yr 0.253 3.64 h r/yr

T9 0 MVA

r

Bus17 11 kV

CB1

hr/yr

f/y 0.253

Bus16 11 kV

3.6 hr/

yr

f/yr 0.186 0.343 h

r/yr

LP6 454 kVA 3.63 hr/yr

Fuse11

Sytem Indexes ACCI

1.85 kVA/ customer

AENS

0.0219 MW hr / customer.yr

ALII

0.21 pu (kVA)

ASAI

0.9996 pu

ASUI

0.00041 pu

CAIDI

14.521 hr / customer interruption

CTAIDI

3.611 hr / customer.yr

ECOST

78,862.86 S / yr

EENS

14.306 MW hr / yr

IEAR

5.513 S / kW hr

SAIDI

3.6109 hr / customer.yr

SAIFI

0.2487 f / customer.yr

Figure 36.7 ETAP results for the reliability index calculation.

36.5.4.2 Required Data

Required data for reliability assessment are summarized in Table 36.1. In this assessment, the reliability parameters listed in the Table 36.14 are utilized.

36.5.4.3 Study Scenarios

Reliability assessments were examined on system configurations RA-1 to RA-7, listed in Table 36.15.

Utility 245 kV 1500 MVAsc

CB-U1 250 kV 1250 A 31.5 kA

0.019 f/yr 0.794 hr/yr Bus-1 245 kV

CB-L1A 250 kV 1250 A 31.5 kA

Line-1 25 km

CB-L2A 250 kV 1250 A 31.5 kA

0.648 f/yr 128.8 hr/yr

CB-L1B 250 kV 1250 A 31.5 kA

0.02 f/yr 0.797 hr/yr Bus-2 245 kV

CB-L3A1 250 kV 1250 A 31.5 kA

CB-L3B1 250 kV 1250 A 31.5 kA

Line-3A 50 km

Line-3B 50 km

Transmission Lines

Distribution Lines Line-2 50 km

CB-L3A2 250 kV 1250 A 31.5 kA

CB-L2B 250 kV 1250 A 31.5 kA

0.037 f/yr 7.65 hr/yr

CB-G 24 kV 3150 A 25 kA

Generation Plant 22 kV 150 MW

0.012 f/yr 0.448 hr/yr Bus-4 24 kV

CB-T1B 24 kV 3150 A 25 kA

T-1 22/220 kV 150 MVA

YNd11 14 %Z –4% TapP

CB-L3B2 250 kV 1250 A 31.5 kA CB-T2A T-2 250 kV 220/66 kV 1250 A 100 MVA 31.5 kA

CB-T2B 72.5 kV 1250 A 31.5 kA

0.103 f/yr 12.5 hr/yr Bus-5 72.5 kV

CB-T3B 72.5 kV 1250 A 31.5 kA

0.103 f/yr 12.5 hr/yr Bus-6 72.5 kV

YNd11 12.5 %Z –4% TapP

CB-T1A 250 kV 1250 A 31.5 kA Bus-3 245 kV 0.035 f/yr 1.25 hr/yr

CB-T3A T-3 250 kV 220/66 kV 1250 A 100 MVA 31.5 kA

Industrial Plant-1 66 kV 100 MVA 0.103 f/yr 12.5 hr/yr

Industrial Plant-2 66 kV 100 MVA 0.103 f/yr 12.5 hr/yr

YNd11 12.5 %Z –4% TapP

Figure 36.8 Result of study scenario RA-1, “Normal (Run All System)” case. Average failure rate (failures/year) and annual outage duration (hours/year) are indicated in the diagram.

1016

36 Reliability Assessment

Table 36.14 Reliability parameters.

Parameters

Library

Reliability parameter

Utility

Generator

Bus

Transformer

Circuit breaker

Transmission line

Loads

Source

IEEE Std 493–1997

ETAP typical

Type

Power supply

Steam turbine

Typical

LqdFill >15 kV

Fixed

Open wire

Class

All

All kV

0.0-33 W

>15 kV

All kV

>15 kV

0.032

0.001

0.013

0.0052

0.0246

0.013

0.0078

0.0246

23.8692

276.3407

1095

175.2

λA

(Failure/yr)

0.643

λP

(Failure/yr)

–

μ

(Repair/yr)

43.8

37.4359

4380

1.4468E-2

8.54E-4

2.2831E-7 1E-3

4.7E-5

4.4929E-5

1.1414E-4

FOR

0.02

MTTF

(yr)

1.6

31.2

1000

38.5

76.9

203

50

MTTR

(hr)

200

234

2

367

31.7

8

50

Replacement available

rΡ

(hr)

200

201

1

71.5

4.5

8

10

Swiching time

(hr)

200

234

1

367

31.7

8

–

Where

λA (Failure/yr): active failure rate in number of failures per year per unit length λP (Failure/yr): passive failure rate in number of failures per year μ (Repair/yr): mean repair rate in number of repairs per year, calculated automatically based on MTTR (μ = 8760/MTTR). FOR: forced MTTF (yr): mean time to repair in hours MTTR (hr): mean time to failure in years, calculated automatically based on λΑ and λΡ (MTTF = 1.0/[λΑ + λΡ]) rp (hr): replacement time in hours for replacing a failed element with a spare one

Table 36.15 Study scenario. Study scenario

RA-1

RA-2

RA-3

RA-4

RA-5

RA-6

RA-7

System Operation

Normal (Run All System)

Utility Stop

Genertion Plant Stop

Line 1 Stop

Line 2 Stop

Line 3 Stop

Industrial Plant-2 Stop

36.5.4.4 Results of Reliability Assessment

Results of the reliability assessment for the study scenario RA-1 “Normal (Run All System)” case are shown in the ETAP one-line diagram in Figure 36.8. Average failure rate (failures/year) and annual outage duration (hours/year) of the power sources, buses, and loads are shown in this diagram. The average failure rate (failures/year) and annual outage duration (hours/year) of the power sources, buses, and loads for all study scenarios are shown in Tables 36.16 and 36.17.

36.5 Illustrative Examples

Table 36.16 Summary of reliability analysis (average failure rate and outage duration). Study scenario

RA-1

RA-2

RA-3

RA-4

RA-5

RA-6

RA-7

System Operation

Normal (Run All Syatem)

Utility Stop

Genertion Plant Stop

Line 1 Stop

Line 2 Stop

Line 3 Stop

Industrial Plant-2 Stop

Utility

0.6482

(Stop)

0.6482

0.6482

0.6482

0.6482

0.6482

Generation Plant

0.0372

0.0372

(Stop)

0.0372

0.0372

0.0372

0.0372

Bus-1

0.019

0.142

0.667

0.051

0.033

0.019

0.019

Bus-2

0.02

0.142

0.69

0.063

0.022

0.02

0.02

Bus-3

0.035

0.125

0.765

0.04

0.032

0.029

0.029

Bus-4

0.012

0.051

0.754

0.015

0.014

0.012

0.012

Bus-5

0.103

0.193

0.774

0.108

0.1

0.098

0.098

Bus-6

0.103

0.193

0.774

0.108

0.1

0.098

(Stop)

Industrial Plant − 1

0.103

0.193

0.774

0.108

0.1

0.098

0.098

Industrial Plant − 2

0.103

0.193

0.774

0.108

0.1

0.098

(Stop)

Utility

128.8

(Stop)

128.8

128.8

128.8

128.8

128.8

Generation Plant

7.65

7.65

(Stop)

7.65

7.65

7.65

7.65

Bus-1

0.794

19.5

129.3

0.92

0.78

0.792

0.777

Bus-2

0.797

19.9

129.9

1.66

0.814

0.654

0.778

Bus-3

1.25

19.1

120.4

1.29

1.1

1.08

1.08

Bus-4

0.448

8.07

140.6

0.465

0.456

0.448

0.448

Bus-5

12.5

30.3

141.6

12.5

12.3

12.3

14.3

Bus-6

12.5

30.3

141.6

12.5

12.3

12.3

(Stop)

Industrial Plant − 1

12.5

30.3

141.6

12.5

12.3

12.3

14.3

Industrial Plant −2

12.5

30.3

141.6

12.5

12.3

12.3

(Stop)

Average failure rate (f/yr)

Annual outage duration (hr/yr)

Table 36.17 Summary of reliability analysis (system index). Scenario RA-2

Scenario RA-1 AENS

985.7620 MW hr. / customer yr

AENS

2397.9690 MW hr. / customer.yr

ASAI

09986 pu

ASAI

09965 pu

ASUI

0.00142 pu

ASUI

0.00346 pu

CAIDI

121.085 hr. / customer interruption

CAIDI

156.909 hr. / customer interruption

ECOST

0.00$/yr

ECOST

0.00 $/yr

EENS

1971.524 MWhr/yr

EENS

4795.937 MW hr./yr

IEAR

0000 $ / kW hr

IEAR

0.000 $ / kW hr

SAIDI

12.4557 hr. / customer.yr

SAIDI

30.2998 hr. / customer.yr

SAIFI

0.1029 f/ customer.yr

SAIFI

0.1931 f / customer.yr

1017

1018

36 Reliability Assessment Table 36.17 (Continued) Scenario RA-3

Scenario RA-4

AENS

11 203.6100 MW hr. / customer AT

AENS

988.8796 MW hr. / customer.yr

ASAI

0.9838 pu

ASAI

09986 pu

ASUI

001616 pu

ASUI

0.00143 pu

CATDI

182 995 hr. / customer interruption

CAIDI

115.739 hr. / customer interruption

ECOST

0.00 $ / yr

ECOST

0.00 $ / yr

EENS

22 407.220 MW hr./yr

EENS

1977.759 MW hr./yr

IEAR

0.000 $ / kW hr

IEAR

0.000 $ kW hr

SAIDI

1 415 646 hr. / customer.yr

SAIDI

12.4951 hr. / customer.yr

SAIFI

07736 f / customer.yr

SAIFI

01080 f / customer.yr

Scenario RA-6

Scenario RA-5 AENS

974.2431 MW hr. / customer.yr

AENS

9 726 724 MW hr. / customer.yr

ASAI

09986 pu

ASAI

09986 pu

ASUI

000141 pu

ASUI

000140 pu

CAIDI

122.917 hr. / customer interruption

CAIDI

125.950 hr. / customer interruption

ECOST

0.00 $ / yr

ECOST

0.00 $ / yr

EENS

1948 486 MW hr./yr

EENS

1945.345 MW hr./yr

IEAR

0000 $ / kW hr

IEAR

0000 $ / kW hr

SAIDI

12.3102 hr. / customer.yr

SAIDI

122 903 hr. / customer.yr

0.1002 f/ customer.yr

SAIFI

00976 f / customer.yr

SAIFI

Scenario RA-7 AENS

972.7164 MW hr. / customer.yr

ASAI

09986 pu

ASUI

000140 pu

CAIDI

125 844 hr. / customer interruption

ECOST

0.00 $ / yr

EENS

972.716 MW hr. / yr

IEAR

0000 $ / kW hr

SAIDI

122 909 hr. / customer.yr

SAIFI

00977 f / customer.yr

1019

37 Protective Device Coordination 37.1

Introduction

Electrical power systems must be designed to serve a variety of loads safely and reliably. Effective control of short-circuit current, or fault current as it is commonly called, is a major consideration. The three major objectives of electric power system protection are:

•• •

Personnel safety Prevention of equipment damage Improved coordination and selectivity

These requirements are necessary, first for early detection and localization of faults, and second for prompt removal of faulty equipment from service. In order to carry out these duties, protection must have the following qualities:

•• • • ••

Selectivity: To detect and isolate the faulty item as shown in Figure 37.1. Stability: To leave all healthy circuits intact to ensure continuity of supply. Sensitivity: To detect even the smallest fault, current, or system abnormalities and operate correctly at its setting before the fault causes irreparable damage. Protection should be as sensitive as possible to detect faults at the lowest possible current level. At the same time, however, it should remain stable under all permissible load, overload, and through-fault conditions. Speed: To operate speedily when it is called upon to do so, thereby minimizing damage to the surroundings and ensuring safety to personnel. To meet all of these requirements, protection must be reliable, which means it must be: Dependable: It must trip when called upon to do so. Secure: It must not trip when it is not supposed to.

Tremendous work has gone into personnel safety as part of IEEE Std. 1584 (IEEE Guide for Performing Arc Flash Hazard Calculations) and will be covered in a later revision of this book. During the time of writing of this book, IEEE 1584-2018 was in the draft process. The purpose of this chapter is to discuss protective device coordination, equipment protection, and selectivity. One of the primary types of studies performed for equipment protection, selective coordination is overcurrent coordination. Overcurrent coordination is a systematic application (study) of current responsive (actuated) devices in an electrical power system, which in response to faults or overloads will remove only the minimum amount of electrical equipment from service. The objective of an overcurrent coordination study is to determine the ratings and settings of fuses, breakers, relays, etc. to ensure that minimum load is interrupted when the protective device isolates the fault or overloads in an affected region.

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

1020

37 Protective Device Coordination

Without selective coordination

With selective coordination

Opens Opens Not affected Unnecessary power loss

Not affected

Figure 37.1 Selective coordination vs. lack of selective coordination.

37.1.1

Study Criteria

A protection and coordination study is typically performed when:

•• • •• •• •

A new electrical system is being designed There is an electrical system expansion or retrofit Coordination failure results in misoperation of protective devices and loss of selectivity in equipment interruption Some of the critical criteria while performing the study are: Open the protective device nearest (upstream) of the fault or overload. leaving the rest of the system operational. Provide adequate protection for overloads (less than 250% of full-load current). Interrupt short circuit (5–20 times full-load current) as rapidly (instantaneously) as possible. Comply with all standards, including IEC, IEEE/ANSI, and national codes. Demonstrate by plotting the time/current characteristics of the different protective devices (fuses, breakers, relays) on log–log digital graph paper.

37.1.2

Study Objectives

Disconnection of equipment and system circuits in order to prevent loss of equipment life (deterioration) when subjected to:

•• •• •• •• •• •

Overload Overtemperature Under- and overvoltage Unbalanced current Under- and overfrequency Loss of field Loss of synchronism Reverse power (anti-motoring) Reverse current Phase reversal and open phase Phase and ground faults.

37.1.3

Low-Voltage Circuit Breakers

Low-voltage circuit breakers (LVCBs) are classified as power circuit breakers (PCBs) or air circuit breakers (ACBs), insulated case circuit breakers (ICCBs), molded case circuit breakers (MCCBs), and miniature circuit breakers (MCBs). The differences between various PCBs is described in Table 37.1. ETAP has Verified & Validated (V&V) libraries containing thousands of LVCB frames and sizes, as shown in Figure 37.2. LVCB trip units are classified as:

37.1 Introduction

Table 37.1 Differences between various circuit breakers. Insulated case circuit breaker (ICCB)

Molded case circuit breaker (MCCB)

5-cycle for electrically operating devices

5-cycle for electrically operating devices

>5-cycle for electrically operating devices

Interrupting rating

Interrupting duty at 635 V : 42–100 kA and current-limiting with or without fuses up to 200 kA

Interrupting duty at 508 V : 35–150 kA

Interrupting duty at 480 V : 22–100 kA without fuses and up to 200 kA with integral fuses

Enclosed rating

100% continuous current

80% continuous current, unless 100% stated in enclosure

80% continuous current, unless 100% stated in enclosure

Series rating

Not available

Not available

Available

Enclosure types

MCCs, switchboards, and switchgear

MCCs and switchboards

Panelboards, switchboards, MCCs, and control panels

Category

Power circuit breaker (PCB)

Closing speed

Maintenance

Extensive

Limited

Very limited

Available frame sizes

Typical 800-6000A

Typical 800-6000A

Typical 100-2500A

Relative cost

High

Medium

Low

Currentlimiting

Current-limiting with or without fuses up to 200 kA

Current-limiting with or without fuses up to 150 kA

Current-limiting with or without fuses up to 200 kA

Short-time rating selective trip

Over full range of fault current up to interrupting rating

Partial range of fault currents within interrupting rating. Typically up to 35 kA.

Smaller range of fault currents within interrupting rating. Typically 10–14 times the frame size.

Figure 37.2 ETAP low-voltage circuit breaker (LVCB) library.

1021

1022

37 Protective Device Coordination

Figure 37.3 ETAP low-voltage circuit breaker (LVCB) trip unit libraries.

•• ••

Thermal magnetic Motor circuit protector (MCP) Solid-state trip (SST) or microprocessor based Electromechanical

•• •

LT (ANSI; I > IEC) ST (ANSI; I >> IEC) IT (ANSI; I >>> IEC)

ETAP libraries for various trip units such as SST are shown in Figure 37.3. Note that in each case, the trip unit is classified based on manufacturer and model. Each low-voltage circuit breaker comes premapped with an associated trip unit that works with that breaker frame size. The ETAP SST library includes settings for long-time (LT), short-time (ST), instantaneous (IT), override, ground, and maintenance mode functions, as shown in Figure 37.4. Trip segment designations based on ANSI and IEC are:

The ETAP Thermal Magnetic library includes settings for thermal, magnetic, and ground fault interrupter / residual current device GFI/RCD functions, as shown in Figure 37.5. The ETAP MCP library includes settings for instantaneous functions, as shown in Figure 37.6. And the ETAP Electro-Mechanical library includes settings for LT, ST, and IT functions, as shown in Figure 37.7.

37.2

Relays

37.2.1

Electromechanical Relays

These were the earliest forms of relay used for the protection of power systems, and they date back nearly 100 years. They work on the principle of a mechanical force causing operation of a relay contact in response to a stimulus. The mechanical force is generated through current flow in one or more windings on a magnetic core or cores; hence the term electromechanical relay. The principal advantage of such relays is that they provide galvanic isolation between the inputs and outputs in a simple, cheap, and reliable form – therefore for simple on/off switching functions where the output

37.2 Relays

Figure 37.4 ETAP solid-state trip (SST) trip unit library.

Figure 37.5 ETAP thermal magnetic (TM) trip unit library.

1023

1024

37 Protective Device Coordination

Figure 37.6 ETAP motor circuit protector (MCP) trip unit library.

Figure 37.7 ETAP electro-mechanical (EM) trip unit library.

37.2 Relays

contacts have to carry strong currents, they are still used. Electromechanical relays can be classified into several different types, as follows:

•• •• ••

Attracted armature Moving coil Induction Thermal Motor operated Mechanical

37.2.2

Static Relays

The term static implies that the relay has no moving parts. This is not strictly the case for a static relay, as the output contacts are still generally attracted armature relays. In a protection relay, the term static refers to the absence of moving parts to create the relay characteristic. A number of design problems had to be solved with static relays. In particular, the relays generally require a reliable source of DC power, and measures to prevent damage to vulnerable electronic circuits had to be devised. Substation environments are particularly hostile to electronic circuits due to common forms of electrical interference (e.g. switching operations and the effect of faults). While it is possible to arrange for DC supply to be generated from the measured quantities of the relay, this has the disadvantage of increasing the burden on the current transformers (CTs) or voltage transformers (VTs), and there will be a minimum primary current or voltage below which the relay will not operate. This directly affects the possible sensitivity of the relay. So provision of an independent, highly reliable, and secure source of relay power supply was an important consideration. To prevent incorrect operation or destruction of electronic devices during faults or switching operations, sensing circuitry is housed in a shielded case to exclude common mode and radiated interference. The devices may also be sensitive to static charge, requiring special precautions during handling, as damage from this cause may not be immediately apparent but become apparent later in the form of premature failure of the relay. Therefore, radically different relay manufacturing facilities are required compared to those for electromechanical relays. Calibration and repair are no longer tasks performed in the field without specialized equipment. 37.2.3

Digital Relays

Digital protection relays introduced a step change in technology. Microprocessors and microcontrollers replaced analog circuits used in static relays to implement relay functions. Early examples began to be introduced into service around 1980, and, with improvements in processing capacity, can still be regarded as current technology for many relay applications. However, such technology will be superseded entirely within the next five years by numerical relays. Compared to static relays, digital relays introduce A/D conversion of all measured analog quantities and use a microprocessor to implement the protection algorithm. The microprocessor may use some kind of counting technique or use the discrete Fourier transform (DFT) to implement the algorithm. However, the typical microprocessors used have limited processing capacity and memory compared to that provided in numerical relays. The functionality tends, therefore, to be limited and restricted mostly to the protection function itself. Additional functionality compared to that provided by an electromechanical or static relay is usually available, typically taking the form of a broader range of settings, and greater accuracy. A communications link to a remote computer may also be provided. The limited power of the microprocessors used in digital relays restricts the number of samples of the waveform that can be measured per cycle. This, in turn, limits the speed of operation of the relay in specific applications. Therefore, a digital relay for a particular protection function may have a longer operation time than the static relay equivalent. However, the extra time is not significant in terms of overall tripping time and possible effects of power system stability. 37.2.4

Numerical Relays

The distinction between digital and numerical relays rests on points of fine technical detail and is rarely found in areas other than protection. They can be viewed as natural developments of digital relays as a result of advances in technology.

1025

1026

37 Protective Device Coordination

Typically, they use a specialized digital signal processor (DSP) as the computational hardware, together with the associated software tools. The input analog signals are converted into a digital representation and processed according to the appropriate mathematical algorithm. Processing is carried out using a specialized microprocessor that is optimized for signal processing applications: the DSP. Digital processing of signals in real time requires a very high-power microprocessor. In addition, the continuing reduction in the cost of microprocessors and related digital devices (memory, I/O, etc.) naturally leads to an approach where a single item of hardware is used to provide a range of functions (a “one-box solution” approach). By using multiple microprocessors to provide the necessary computational performance, a large number of functions previously implemented in separate items of hardware can now be included within a single item. There are two methods for indicating protection relay functions in everyday use. One is given in ANSI standard C37.2 and uses a numbering system for various functions. The functions are supplemented by letters where amplification of the function is required. The other is given in IEC 60617 and uses graphical symbols. The comparison between the ANSI and IEC symbols is shown in Table 37.2.

Table 37.2 Relay symbols and their representation in ANSI and IEC standards. Description

ANSI

IEC 60617

Description

ANSI

IEC 60617

Overspeed relay

12

ω>

Inverse time earth fault overcurrent relay

51G

Underspeed relay

14

ω


Power factor relay

55

Undervoltage relay

27

U


Directional overpower relay

32

59 N

Ursd >

Underpower relay

37

Undercurrent relay

37

Negative sequence relay

46

Z
P> P< I< I2 >

Negative sequence voltage relay

47

Thermal relay

49

Instantaneous overcurrent (IOC) relay

50

I >>

Inverse time overcurrent (TOC) relay

51

I>

U2 >

Neutral point displacement relay

Earth-fault relay

64

Directional overcurrent relay

67

Directional earth fault relay

67 N

I

>

I

>

U

I>

cosφ >

I

> I> > I

Phase angle relay

78

Autoreclose relay

79

Underfrequency relay

81 U

Overfrequency relay

810

Differential relay

87

>

>

φ > O

I f< f>

Id >

37.2 Relays

37.2.5

Relay Types/Functions

37.2.5.1 Phase Fault Relays

The pick-up values of phase overcurrent relays are normally set 30% above the maximum load current, provided that sufficient short-circuit current is available. This practice is recommended in particular for mechanical relays with reset ratios of 0.8–0.85. Numerical relays have high reset ratios near 0.95 and therefore allow about a 10% lower setting. Feeders with large transformer and/or motor load require special consideration. The energizing of transformers causes inrush currents that may last for seconds, depending on their size. Selection of the pickup current and assigned time delay have to be coordinated so that the rush current decreases below the relay overcurrent reset value before the set operating time has lapsed. The inrush current typically contains about 50% of the fundamental frequency component. Numerical relays that filter out harmonics and the DC component of the rush current can, therefore, be set more sensitively. 37.2.5.2 Instantaneous Overcurrent Protection (50)

This is typically applied on the final supply load or on any protective device with sufficient circuit impedance between itself and the next downstream protective device. The setting at transformers, for example, must be chosen about 20–30% higher than the maximum through-fault current. 37.2.5.3 Inverse-Time Relays (51)

For the time grading of inverse-time relays, the same rules apply in principle as for the definite time relays. The time grading is first calculated for the maximum fault level and then checked for lower current levels. If the same characteristic is used for all relays, or when the upstream relay has a steeper characteristic (e.g. very much over normal inverse), then selectivity is automatically fulfilled at lower currents. The time-current characteristics of time overcurrent (TOC), instantaneous overcurrent (IOC) and TOC + IOC in combination are shown in Figure 37.8. 37.2.5.4 Ground Fault Relays

Residual-current relays enable a much more sensitive setting, as load currents do not have to be considered (except fourwire circuits with single-phase load). In solidly and low-resistance grounded systems, a setting of 10–20% rated load current is generally applied. High-resistance grounding requires much more sensitive settings, on the order of some Device 51

Device 50

t

t 50

51 CT

CT

i Device 51/50

t 51 50 CT

i Figure 37.8 Inverse time characteristics.

i

1027

1028

37 Protective Device Coordination

t

t 59

27 VT

VT

v

v

Figure 37.9 Undervoltage and overvoltage relay characteristics.

t 67 CT

i Figure 37.10 Directional relay characteristics.

amperes primary. The ground-fault current of motors and generators, for example, should be limited to values below 10 A in order to avoid iron burning. Residual-current relays in the star-point connection of CTs can in this case not be used, in particular with rated CT primary currents higher than 200 A. The Pickup value of the zero-sequence relay would, in this case, be on the order of the error currents of the CTs. A particular zero-sequence CT is therefore used in this case as a ground current sensor. An even more sensitive setting is applied in isolated or Peterson-coil-grounded networks where very small ground currents occur with single-phase-to-ground faults. 37.2.5.5 Differential Relays (87)

Transformer differential relays are generally set to pick-up values between 20% and 30% rated current. The higher value has to be chosen when the transformer is fitted with a tap changer. Restricted ground-fault relays and high-resistance motor/generator differential relays are, as a rule, set to about 10% rated current. 37.2.5.6 Undervoltage and Overvoltage Relays

Undervoltage and overvoltage relays are typically applied on buses and other equipment such as generators and medium voltage (MV) motors. Typical characteristics are shown in Figure 37.9. 37.2.5.7 Directional Relays

Directional relays are typically applied on equipment such as cables and lines to either allow or prevent flow of power in the opposite direction and standard setup. Typical characteristics are shown in Figure 37.10.

37.3

Methodology

Correct overcurrent relay application requires knowledge of the fault current that can flow in each part of the network. Since large-scale tests usually are impracticable, system analysis must be used. The data required for a relay-setting study are:

37.3 Methodology

• •• •• • •

A one-line diagram of the power system involved, showing the type and rating of the protective devices and their associated current transformers The impedances in ohms, percent, or per unit, of all power transformers, rotating machine, and feeder circuits The maximum and minimum values of short-circuit currents that are expected to flow through each protective device The maximum load current through protective devices The starting current requirements of motors and the starting and locked rotor/stalling times of induction motors The transformer inrush, thermal withstand, and damage characteristics decrement curves showing the rate of decay of the fault current supplied by the generators Performance curves of the current transformers

Relay settings are first determined to give the shortest operating times at maximum fault levels and then checked to see whether operation will also be satisfactory at the minimum fault current expected. It is always advisable to plot on a standard scale the curves of relays and other protective devices, such as fuses, that are to operate in series. It is usually more convenient to use a scale corresponding to the current expected at the lowest voltage base or to use the predominant voltage base. The alternatives are a common MVA base or a separate current scale for each system voltage. The basic rules for correct relay coordination can generally be stated as follows:

••

Whenever possible, use relays with the same operating characteristic in series with each other. Make sure the relay farthest from the source has current settings equal to or less than the relays behind it – that is, that the primary current required to operate the relay in front is always equal to or less than the primary current required to operate the relay behind it.

Among the various possible methods used to achieve correct relay coordination are those using either time or overcurrent, or a combination of both. The common aim of all three methods is to give correct discrimination. That is to say, each one must isolate the faulted section of the power system network, leaving the rest of the system undisturbed. 37.3.1

Discrimination by Time

In this method, an appropriate time setting is given to each of the relays controlling the circuit breakers in a power system to ensure that the breaker nearest to the fault opens first. A simple radial distribution system is shown in Figure 37.11, to illustrate the principle. Overcurrent protection is provided at B, C, D, and E: that is, at the infeed end of each section of the power system. Each protection unit comprises a definite-time delay overcurrent relay in which the operation of the current-sensing element initiates the time-delay element. Provided the setting of the current element is less than the fault current value, this element plays no part in the achievement of discrimination. For this reason, the relay is sometimes described as an “independent definite-time delay relay,” since its operating time is for practical purposes independent of the level of overcurrent. It is the time-delay element, therefore, that provides the means of discrimination. The relay at B is set at the shortest time delay possible to allow the fuse to blow for a fault at A on the secondary side of the transformer. After the time delay has expired, the relay output contact closes to trip the circuit breaker. The relay at C has a time-delay setting equal to t1 seconds, and similarly for the relays at D and E. If a fault occurs at F, the relay at B will operate in t seconds, and the subsequent operation of the circuit breaker at B will clear the fault before the relays at C, D, and E have time to operate. The time interval t1 between each relay time setting must be long enough to ensure that the upstream relays do not operate before the circuit breaker at the fault location has tripped and cleared the fault. The main disadvantage of this method of discrimination is that the longest fault-clearance time occurs for faults in the section closest to the power source, where the fault level (MVA) is highest. E

D

t1

C

t1

B

A

t1 F

Figure 37.11 Simple radial distribution system.

1029

1030

37 Protective Device Coordination

37.3.2

Discrimination by Current

Discrimination by current relies on the fact that the fault current varies with the position of the fault because of the difference in impedance values between the source and the fault. Hence, typically, the relays controlling the various circuit breakers are set to operate at suitably tapered values of current such that only the relay nearest to the fault trips its breaker.

37.3.3

Discrimination by Current and Time

Each of the two methods described so far has a fundamental disadvantage. In the case of discrimination by time alone, the disadvantage is due to the fact that the more severe faults are cleared in the longest operating time. On the other hand, discrimination by current can only be applied where there is appreciable impedance between the two circuit breakers concerned. It is because of the limitations imposed by the independent use of either time or current coordination that the inverse TOC relay characteristic has evolved. With this characteristic, the time of operation is inversely proportional to the fault current level, and the actual characteristic is a function of both time and current settings. Figure 37.12 illustrates the

Figure 37.12 Characteristics of two relays with different current/time settings.

37.3 Methodology

characteristics of two relays given different current/time settings. For a significant variation in fault current between the two ends of the feeder, faster operating times can be achieved by the relays nearest to the source, where the fault level is the highest. The disadvantages of grading by time or current alone are overcome. The selection of overcurrent relay characteristics generally starts with selection of the correct characteristic to be used for each relay, followed by choice of the relay current settings. Finally, the grading margins and hence time settings of the relays are determined. An iterative procedure is often required to resolve conflicts and may involve use of non-optimal characteristics, current, or time-grading settings. The current/time-tripping characteristics of inverse definite minimum time (IDMT) relays may need to be varied according to the tripping time required and the characteristics of other protective devices used in the network. The following equation demonstrates that three parameters determine the operating characteristics of the TOC relay: the pickup current (IPU), the time dial setting, TDS, and the degree of inverseness: Trip time = TDS ×

Kd τ s I Ip

= TDS ×

2

−1

A M 2 −1

where M : multiples of pickup current,

I IPU

A = Kd/τs In practice, there are actually three parameters that determine the degree of inverseness: tr = TDS

C 1− M 2

where tr represents the reset time. Time to trip (tt) is valid only if the relay is the reset condition. It is accurate only if the relay starts when the timeovercurrent function has timed out: tt = TDS

A + B for M ≥ 1 M p −1

Disk inertia is considered by using the parameter B. The result is that relay manufacturers design relays with particular characteristics using various factors for A, B, and p. Figure 37.13 illustrates the family of curves for very inverse characteristics and a TDS of 1 through 10. This set of curves demonstrates that the time to operate is linearly dependent on the TDS for a given characteristic and multiple of pickup current. IEC 60255 defines a number of standard characteristics, as shown in Table 37.3, and typical curves based on this standard are shown in Figure 37.14. North American inverse relay characteristics are shown in Figure 37.13, and the complete set of equations are given in Table 37.4. Table 37.5 is a summary of degrees of inverseness as a function of A, B, and p for US and IEC standard curves. I Ir = , Is where Is : relay setting current TMS : time multiplier setting TD : time dial setting

37.3.4

Scale Selection

The traditional approach for scale selection is to choose the smallest device on the left side of the time-current characteristic curve (TCC) (log–log) plot. The choice of scale should require the least amount of manipulation (calculations). A voltage level should be selected that has the most devices. Once a scale is selected, other (different voltage levels) scaling factors must be calculated. The current values of one device at the selected voltage level must be converted

1031

1032

37 Protective Device Coordination

Figure 37.13 Family of curves for U.S. inverse U2 characteristics.

Table 37.3 Relay characteristics according to IEC 60255. Relay characteristic

Standard inverse (SI) Very inverse (VI) Extremely inverse (EI) Long-time standard earth fault

Equation

t = TMS ×

0 14 Ir0 02 − 1

13 5 Ir − 1 80 t = TMS × 2 Ir −1

t = TMS ×

t = TMS ×

120 Ir − 1

37.3 Methodology

Figure 37.14 Typical relay curves based on IEC characteristics.

Table 37.4 North American relay characteristics and equations according to IEC 60255. Relay characteristics

IEEE Moderately Inverse IEEE Very Inverse Extremely Inverse (EI) US CO8 Inverse US CO2 Short Time Inverse

Equation

t=

TD 0 0515 + 0 114 7 Ir0 02 − 1

t=

TD 19 61 + 0 491 7 Ir2 −1

t=

TD 28 2 + 0 1217 7 Ir2 −1

t=

TD 5 95 + 0 18 7 Ir2 −1

t=

TD 0 02394 + 0 01694 7 Ir0 02 − 1

1033

37 Protective Device Coordination

Table 37.5 Degrees of Inverseness as function of A, B, and p for North American (United States) and IEC. Curve

A

B

C

P

US Moderately Inverse (U1)

0.0104

0.2256

1.08

0.02

US Inverse (U2)

5.95

0.18

5.95

2.00

US Very Inverse (U3)

3.88

0.963

3.88

2.00

US Extremely Inverse (U4)

5.67

0.352

5.67

2.00

US Short-time Inverse (U5)

0.00342

0.00262

0.323

0.02

IEC Class A – Standard Inverse (C1)

0.14

IEC Class B – Very Inverse (C2) IEC Class C – Extremely Inverse (C3) IEC Long-time Inverse (C4) IEC Short-time Inverse (C5)

0.0

13.5

0.02

13.5

0.0

47.3

2.00

80.0

0.0

80.0

2.00

120.0

0.0

120.0

2.00

0.05

0.0

4.85

0.04

to the equivalent current value at the voltage level of the other devices. Scale selection is accomplished automatically in ETAP. The power engineer has the option to customize the scale as needed.

37.3.5

Log–Log Plot

A log–log plot is typically used to show TCC, because the log–log scale compresses values to a more-manageable range, and thermal damage curves I2t can be plotted as straight lines, as shown in Figure 37.15.

1000 Effectively steady state 100 l2t withstand curves plot as straight lines

1 minute

Time in seconds

1034

10 Typical motor acceleration

1 Typical fault clearing 0.1

1 cycle (momentary) 0.01

5 cycles (interrupting) FLC = 1 pu

0.5

1

Fp = 577 pu

Fs = 13.9 pu

10

100

Current in amperes Figure 37.15 Typical scale selection on a log–log plot.

1000

10000

37.4 Required Data

37.4

Required Data

The following is a list of required data that must be collected prior to beginning a protection and coordination study:

• • • • • • • • • •

Power grid – Ratings, types, and settings of the first upstream power-grid protective device Transformers – kVA ratings (self-cooled/forced air [OA/FA]) – Primary and secondary voltages – Connections (Wye-delta, delta-Wye, etc.) – Percent impedance – Liquid-filled or dry type – Overload capacities (capability) – Damage curves Motors – Horsepower or kW – Full load and locked rotor amperes – Service factor (1.0 or 1.15) – Starting and locked rotor (stall) times – Starting type (full voltage, reduced, etc.) – Thermal damage curve Other load types – Maximum loads – Special considerations Cables – Type and insulation – Type of conduit – Ampacity – Number per phase – Short-circuit withstand curves from ICEA or similar Short-circuit currents (calculated by ETAP or manually defined) – Maximum and minimum available symmetrical RMS fault current at each protective device location. – X/R ratio and asymmetrical fault currents. Busway, panelboards, switchboards, or trunking busbar – Ampacities – Fault current ratings LV circuit breakers – Type and manufacturer – Frame sizes – Ampere ratings, sensor ratings, plug ratings – Long-time, short-time, instantaneous, I2t, and ground-fault pickup and adjustment ranges – Time/current characteristic curves MV circuit breakers – Type and manufacturer – Operating (trip) times (3 cycles, 5 cycles, etc.) Fuses – Type and manufacturer – Continuous current ratings – Time/current characteristic curves, minimum melt, and total clearing or average melting

1035

37 Protective Device Coordination

•

Relays – – – –

Type and manufacturer Ampere tap (AT), time dial (TD), and instantaneous pickup (IT) and adjustment ranges CT ratios Time/current characteristic curves

37.5

Principle of Protection

When we prepare a protective device TCC, the protective device must be set to the left and below the equipment damage curve. Protective devices and protected equipment represent the “protection zone.” The motor protection example in Figure 37.16 shows the motor-starting curve, thermal protection, locked rotor protection, and fault protection. .5 1K

1

3

5

10

30

500

50

100

300 500

1K

3K 5K

10K 1K 500

Stator damage curve

300

300

Cold start stall time

100 Full load amps (FLA)

50

100 50

Hot start stall time

30

Seconds

1036

30

10

10

5

5

3

3

Estimated starting time

1

1 Locked Rotor Amps (LRA)

.5

.5

.3

.3

.1

.1

.05

Inrush Current (Asymmetrical LRC)

.03

.01 .5

1

3

5

10

30 50 100 300 500 Amps X 10 (Plot Ref. kV=4.16)

Figure 37.16 Typical motor protection plot.

1K

3K 5K

.05 .03

.01 10K

37.6 Principle of Selectivity/Coordination

Table 37.6 Protected power systems and parameters measured.

Rotating

Generators

Devices Branches

V

I

F

ϕ

T

M

M

M

M

M

M

M

M

Motors

M

X

Synch. condensers

M

M

P

t

M

Transmission

M

M

D

D

M

Distribution

M

M

D

D

M

M

Cable

M

M

D

D

Series

M

M

M

Shunt reactor

M

M

M

M

Breaker

M

M

M

M

Bus fault

M

M

M

Transformer

M

M

M

Shunt

M

M

M

M

Capacitors Station

Failure

Capacitor System Stability

Load shedding

M

M

D

D

Load frequency control

M

M

D

D

Reclosing

M

M

D

D

Sync check

M

M

D

D

Similarly, protection rules are established for cables, generators, overhead bare conductors, transformers, etc., as discussed in Section 37.7. Table 37.6 shows typical protected power systems and parameters. The parameter symbols are as follows:

•• •• •• •• •

V – volts M – measured I – amps D – derived F – frequency t – time φ – phase T – temperature P – pressure

37.6

Principle of Selectivity/Coordination

The downstream device curve is located to the left and below the upstream device curve for the range of applicable currents, as shown in Figure 37.17. A sufficient time margin must be maintained for operation of the downstream device before the upstream device, as shown in Figure 37.18. Ground-fault-based selective coordination requires device ground-fault overcurrent coordination with other devices with ground detection, other devices with phase-overcurrent detection, and a combination of phase and ground-fault detection. The minimum and maximum faults are typically used for checking the range of coordination. Phase- and

1037

1038

37 Protective Device Coordination

T (sec)

T (sec)

10

10

1

I1

1 T2

T2

T1

0.1

T1 I2

0.1 I1

I2 I (kA)

0.01

I (kA)

0.01 1 10 100 Inverse time overcurrent selectivity

1 10 100 Definite time current selectivity Figure 37.17 Downstream device vs. upstream device selectivity.

.5 1K

1

3

5

10

30

50

100

300 500

1K

3K 5K

Transformer FLA

500

10K 1K 500

300

300 Pickup

100

100

50

50

30

30 Transformer

10

10 Moderately inverse

5

5

3

3 Inverse

1

Time margin

.5

.5

Very inverse

.3

.3

Extremely inverse

.1

.1

Transformer inrush

.05

.05 Max fault current

.03

.01

1

.5

1

3

5

10

30

50

Figure 37.18 Time margin between devices.

100

300 500

1K

3K 5K

.03

.01 10K

37.6 Principle of Selectivity/Coordination

.5 1K

1

3

5

10

30

50

100

300 500

1K

3K 5K

10K 1K

500

500

300

300 LVCB-1-Phase

100

100 LVCB-1 Ground

50

LVCB-2-Phase 50

30

30

Seconds

10

10

LVCB-2 Ground

5

Relay-Phase

5 3

3 Relay-Ground 1

1

.5

.5

.3

.3

.1

.1

.05

.05

.03

.03

.01

.5

1

3

5

10

30

50

100

300 500

1K

3K 5K

.01 10K

Amps X 100 (Plot Ref. kV=0.48) Figure 37.19 Individual phase and ground curves.

single-line-to-ground fault coordination is also typically performed. Figure 37.19 shows individual phase and ground curves in ETAP. Phase and ground curves may be combined on the same TCC in order to determine operation of phase or ground function based on a phase fault, as shown in Figure 37.20.

37.6.1

Selectivity Time Margins

Typical selectivity or coordination time margins are given in Table 37.7.

1039

37 Protective Device Coordination

1K

.5

10K 1K

500

500

300

300

1

3

5

10

30 50

100

300 500

1K

3K 5K

LVCB-1 Phase and Ground Combination

100

100

50

50 LVCB-2 Phase and Ground Combination

30

30

10

10 Seconds

1040

5

Relay Phase and Ground Combination

3

5 3

1

1

.5

.5

.3

.3

.1

.1 Ground Fault

.05 .03

.01

.5

1

3

5

10

30 50 100 300 500 Amps X 100 (Plot Ref. kV=0.48)

1K

3K 5K

.05 .03

.01 10K

Figure 37.20 Combined phase and ground curves.

Table 37.7 Typical time margins between devices.

a b

Device between

Recommended minimum time margin

Relay to Relaya

0.12 to 0.22 seconds + downstream breaker opening time

Relay to Fuse / LVCB

0.12 to 0.22 seconds

Fuse / LVCB to Relay

0.12 seconds + downstream breaker opening time

Fuse / LVCB to Fuse / LVCB

Clear space between curvesb

Calibrated protective devices. Possible adjustment for upstream fuse preloading and safety factors.

37.7

Art of Protection and Coordination >600 V

This section is intended to provide rule-of-thumb or typical settings for equipment more significant than 600 V. It is the power engineer’s responsibility to verify application on an individual basis using the ETAP software. Care must be taken when coordinating a microprocessor TOC element with an electromechanical relay downstream. The electromechanical relay may respond to a fundamental phasor magnitude, true root mean square (RMS), or rectified magnitude. Note

37.7 Art of Protection and Coordination >600 V

Figure 37.21 ETAP Star Rule Book for automated protection and coordination evaluation.

that majority of these rules and additional coordination rules can be obtained from the ETAP Protection & Coordination rule book, as shown in Figure 37.21. 37.7.1

Bus Relays

Pickup is set between 100% and 125% of full-load amperage (FLA) (150% FLA maximum) and should be set to coordinate with transformer primary protective relaying. The IOC element on main breaker relays should be defeated or disabled. 37.7.2

Feeder Relays

Set Pickup to comply with NEC 240-100 (limited to 600% of rated ampacity of conductor) or similar local electrical code. Actually, the pickup permitted by NEC is slightly higher. Keep it down in the neighborhood of 200%; the intent is not to provide overload protection, but rather to provide short-circuit protection. Set TD as required to coordinate with downstream devices while protecting the conductor against damage. Enable instantaneous elements only if the load has a notable impedance (i.e. transformer, motor, capacitor, etc.) or if the load is the end of a radial circuit. 37.7.3

Induction Motors

Set Pickup at 101–120% of the nameplate rating, depending on the service factor and average load:

•• •• •

Motor 1500 hp: set just above FLA × S.F If using thermal O/L (Device 49), Pickup is set accordingly: Motors with marked service factor ≥ 1.15: pickup = 125% of FLA Motors with temperature rise not over 40 C: pickup = 125% of FLA All other motors: 115% of FLA

For Zero Sequence CT (BYZ), set Ground Trip at 10A primary and Alarm at 5A primary. Set for Instantaneous if using electromechanical or a 20 ms delay (minimum) if using digital relays. For solidly grounded systems, ensure that the

1041

1042

37 Protective Device Coordination

Ground Trip setting will not cause a motor starter to attempt interrupting a fault beyond its rating. Set Mechanical Jam to 150% FLA at 2 seconds unless certain applications do not allow this. Set Instantaneous Trip at 200% locked rotor current (LRC). A higher pickup may be used depending on system available short circuit; however, do not lower it below 160% LRC unless the relay filters/removes the DC component. Ensure that the instantaneous trip setting will not cause a motor starter to attempt interrupting a fault beyond its rating. Locked rotor current may be calculated as follows: ILR = 1 XS + Xd For relay-based protection, the recommended instantaneous setting is given as Relay Pickup = Ipickup/ILR × 1.6 to 2 If the recommended setting criteria cannot be met, or where more-sensitive protection is desired, the instantaneous relay (or a second relay) can be set more sensitively if delayed by a timer. This permits the asymmetrical starting component to decay out. A typical setting for this is Relay Pickup = Ipickup/ILR × 1.2 with a time delay of 0.10 s (six cycles at 60 Hz) 37.7.4

Motor Protection

The energizing of motors causes a starting current of initially 5 to 6 times rated current (locked rotor current). In the first 100 ms, a fast-decaying asymmetrical inrush current appears. With conventional relays, it was standard practice to set the instantaneous O/C step for short-circuit protection 20–30% above the locked-rotor current, with a short-time delay of 50 to 100 ms to override the asymmetrical inrush period. Numerical relays are able to filter out the asymmetrical current component very quickly, so setting an additional time delay is no longer applicable. The overload protection characteristic should follow the thermal motor characteristic as closely as possible. The adaption is to be made by setting of the pickup value and the thermal time constant, using the data supplied by the motor manufacturer. Further, the locked-rotor protection timer has to be set according to the characteristic motor value. Motor relays also monitor local and remote status contacts controlled by relays and switches. Figure 37.22 shows motor protection for LV and MV motors, respectively. 37.7.5

Cables

The actual temperature rise of a cable when exposed to a short-circuit current for a known time is calculated by A=

I2 t T2 + 234 0 0297log T1 + 234

where A : conductor area (circular-mils) I : short-circuit current (amps) t : time of short circuit (seconds) T1 : initial operation temperature (75 C) T2 : maximum short circuit temperature (150 C)

37.7.6

Capacitors

For individual protection, the fuse protecting the capacitor is chosen such that its continuous current capability is greater than or equal to 135% of rated capacitor current. The feeder cable should be sized as such for continuous operation. This overrating is due to 10% for allowable overvoltage conditions, 15% for capacitor kvar rating tolerance (this correlates to a 15% percent deviation from nominal capacitance), and 10% for overcurrent due to harmonics. For unbalance, set Alarm for the loss of one capacitor, and set Trip for an overvoltage of 110% rated (nameplate). For feeder protection, set Pickup at 135% of FLA, set Time Dial at 1.0, set the 50P element above maximum inrush, and include a slight time delay to coordinate with the individual fuse clear time. Plot TOC to protect the capacitor case rupture curve.

.5 1K

1

10

Amps X 10 (Plot Ref. kV=0.48) 100 1K

.5

10K 1K

1

3

5

Amps X 10 (Plot Ref. kV=4.16) 10 30 50 100 300 500 1K

3K 5K 10K 1K

1K 500

500

300

300 MV-IM

OL1 100

100

MCP1

100

100 RL–IM

50 IM1-Cold IM1-Hot

Seconds MCP1

1

IM1-100%

.1

1

10

5

5

3

3 CT – IM RL-IM

1

OL1-3P MCP1-3P

10

100 1K Amps X 10 (Plot Ref. kV=0.48)

1 .5

MV-IM-100%

.3

.1

.01 .5

10

.5

OL1

IM1

30

MV-IM-Cold

.01 10K

Figure 37.22 Typical motor-protection plot for LV and MV motors respectively.

.3

.1

.1

.05

.05 MV – IM

.03

RL-IM - 3P

.01 .5

1

3

5

10 30 50 100 300 500 1K Amps X 10 (Plot Ref. kV=4.16)

.03

.01 3K 5K 10K

Seconds

CONT1 1

CB – IM

MV-IM-Hot 10 Seconds

Seconds

10

50

OCR

30

37 Protective Device Coordination

37.7.7

Power Transformers

Transformer I2t curves based on ANSI are shown in Figure 37.23 based on frequent and infrequent fault types as defined in Figure 37.24.

FLA 200

Thermal mechanical font change gray I2t = 1250 (Δ–Δ LL) 0.87 Infrequent fault

t (sec)

1044

(Δ–Y LG) 0.58 Frequent fault 2 Mechanical

K=(1/Z)2t Inrush

2.5

Isc

25

I (pu)

Figure 37.23 Transformer damage curve based on frequent and infrequent faults.

Source Transformer primary-side protective device (fuses, relayed circuit breakers, etc.) may be selected by reference to the infrequent-faultincidence protection curve Infrequent-fault Incidence zone*

Category II or III Transformer Fault will be cleared by transformer primary-side protective device Optional main secondary–side protective device. may be selected by reference to the infrequent-faultincidence protection curve Fault will be cleared by transformer primary-side protective device or by optional main secondaryside protection device Feeder protective device

Frequent-fault Incidence zone*

Fault will be cleared by feeder protective device Feeders

* Should be selected by reference to the frequent-fault-incidence protection curve or for transformers serving industrial, commercial, and institutional power systems with secondary-side conductors enclosed in conduit, bus duct, etc., the feeder protective device may be selected by reference to the infrequent-fault-incidence protection curve. (Frequent Fault = More than 10 through faults (lifetime) for category II and 5 faults for category III)

Figure 37.24 Transformer fault zone definition. Source: IEEE C57.

37.7 Art of Protection and Coordination >600 V

•• • • •

37.7.7.1 Transformer Primary-Phase Fault

CT Ratio: 200% FLA Set Pickup to comply with NEC 450-3, but as a rule of thumb, it should be less than 300% of transformer self-cooled rating or 150% of transformer maximum rating. Try to set the TD such that pickup time for maximum through fault is in the neighborhood of 1.0 seconds or less. If higher, ensure that ANSI damage points are not exceeded. Set Instantaneous at between 160% and 200% of maximum through fault (assume infinite bus). Ensure that the available system short circuit allows this. Set TD at 1.0–1.5 seconds at maximum fault. Do not exceed 2.0 seconds, which is the mechanical damage point.

37.7.7.2 Transformer Ground Fault

Ground-fault protection requires measurement of Residual (IR) or Zero Sequence current (3I0), as shown in Figure 37.25. IR = 3I0 = Ia + Ib + Ic vector summation Therefore, for a balanced fault, Ia = Ib = Ic and IR = 3I0 = 0; and for an unbalanced system, Ia Ib Ic and IR = 3I0 > 0 Grounded-phase (3I0) current is detected directly with a current transformer installed in the grounded-neutral conductor, as shown in Figure 37.26. Grounded-phase current (IR) is directly detected by a doughnut-type current transformer installed around the three-phase conductors. This CT is also known as core balance CT or zero-sequence CT; see Figure 37.27. Grounded-phase current is detected as the unbalance in the current produced by the phase-current transformers, as shown in Figure 37.28. Therefore, Relay Ground Function (50/51G) externally measures residual current (two inputs) and indicates the relay has a Ground CT source. Relay Neutral Function (50/51N) implies relay internally measured residual current (six inputs) and indicates that the relay internally calculates residual (3Io) current from Phase CTs. Figure 37.29 shows the modeling of residual connections in ETAP. Source

Figure 37.26 Grounded phase current (3I0) measurement using CT.

Figure 37.27 Grounded phase current (IR) measurement using CT.

I¯a I¯b I¯c I¯R A Figure 37.25 Residual current.

1045

1046

37 Protective Device Coordination

Internal residual

External residual Phase CTs

Ground inputs 50G / 51G

Phase CTs A

A

B

B

C

C

Phase inputs for 50 / 51 and 50N / 51N

Figure 37.28 Residual connections.

Figure 37.29 Residual connections in ETAP.

•• • •

37.7.7.3 Transformer Primary Ground

Set 50G if the primary winding is delta connected. Provide a time delay (approximately 20 ms) when setting digital relays with zero-sequence CTs. Do not use a time delay when using electromechanical relays with zero-sequence CTs. Use a residually connected neutral. CT mismatch and residual magnetization will not allow the most sensitive setting. It is recommended to delay above inrush. For zero-sequence CT, care must be taken to ensure that cables are appropriately placed and cable shields are appropriately terminated.

37.7.7.4 Transformer Primary Fuse Phase

135% FLA < fuse < 250% FLA. Try to stay in the range of 150%. The primary fuse rating of the power transformer should be approximately 200% FLA if the transformer has a secondary main. Generally, use E-rated fuses. Note that the TOC characteristics of fuses are not all the same.

37.7 Art of Protection and Coordination >600 V

37.7.7.5 Transformer Secondary, Low-Resistance Grounded

Set Pickup for 20–50% of maximum ground fault. Note that ground resistors typically have a continuous rating of 25–50% of nominal. This value can be specified when purchasing the equipment. For example, for a 2000 A main breaker (2000 : 5 CTs), it may make sense to specify a 400 A ground resistor with a continuous rating of 50% (200 A) such that a 2000 : 5 residually connected CT input can be used with a minimum pickup (0.1 × CT = 0.5 A secondary, 200 A primary). Set the TD such that at the time to trip is 2.0 seconds at maximum ground fault. Protect the resistor curve for 10 seconds using the I2t curve. A typical resistor is rated in seconds at nominal current (to be specified at time of order). 37.7.7.6 Transformer Secondary, Solid Grounded

If the secondary is solidly grounded and a neutral relay is available (using CT on X0 bushing), set Pickup at approximately 50% of phase element and ensure the transformer two-second damage point is protected. Coordinate TOC with the main breaker (or partial differential) ground relay. Decrease the primary phase element by 58% (to account for transformer damage curve shift). This is the equivalent current seen on the primary (delta) for a secondary ground fault. It is certainly preferable to rely on the transformer primary-phase overcurrent relay for backup of transformer secondary ground faults. However, downstream coordination does not always afford us that luxury (shifting the transformer damage curve and associated transformer primary relay 58%). For solidly grounded transformer secondary installations, an argument can be made that the 87T is the primary protection and 51NT is the 51 main backup protection for a transformer secondary ground fault. This allows us to set the 50T/51T relay without consideration of the 58% shift, as shown in Figure 37.30.

87T

50T 51T

51 NT Phase-Gnd

51 Main

51 N

Figure 37.30 Transformer solidly grounded connection.

1047

37 Protective Device Coordination

50T 51T

50T 51T

51M

51M

67

N.C.

67

Trip direction

51 Tie

Trip direction

1048

Figure 37.31 Fault between transformer and main breaker.

37.7.7.7 Primary Neutral (YD)

If the primary is solidly grounded and a neutral relay is available, set Pickup at approximately 50% of the phase element. This must coordinate with upstream line-protective devices (i.e. 21P, 21G, 67, 67G). If it is at the utility level, they will review and provide settings. For generator step-up transformers (GSU), the HV 51NT should typically be the last device to trip for upstream ground faults. Ensure that the GSU damage curve and the H0 grounding conductor are protected. 37.7.7.8 Directional Overcurrent

For a fault between the transformer and main breaker, the partial differential bus relays 51B will not detect current (other than motor contribution), as shown in Figure 37.31. Both transformer primary overcurrent relays will detect the same current. A directional overcurrent relay is required to prevent tripping of both transformers via 50T/51T. Therefore, set 67 pickup at 40% of transformer FLA. Coordinate with the time curve with 51Tie. For a fault between the transformer and main breaker, the main and tiebreaker relays will all see the same current (other than motor contribution). The tiebreaker will trip, followed by the respective transformer primary overcurrent. A directional overcurrent relay is required to prevent loss of one bus.

37.8

Illustrative Examples

37.8.1

Basic Operation and Phase Protection

Consider the system shown in Figure 37.32, where the protective devices are to be coordinated for reliable operation of the power system. For Mtr1, go to the Nameplate page, and select a 75 kW motor with voltage rating of 0.415 kV using the Library button. Go to the Protection page, and enter the data shown in Figure 37.33. Click the Points button to view the stator thermal current versus time curve, and use the values shown in Figure 37.34. The stator curve defines the thermal-limit curve for the stator. This information is typically provided by the motor manufacturer. Hot or cold stator curves can be shown on the TCC by selecting the check box next to Stator-Hot and/or Stator-Cold. Click the overload relay (OL1), and go to the Setting page; select the manufacturer and model from the library, as shown in Figure 37.35.

37.8 Illustrative Examples

Figure 37.32 ETAP one-line diagram for protection and coordination analysis.

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37 Protective Device Coordination

Figure 37.33 Motor protection – starting, locked rotor current, and thermal limit curves.

Figure 37.34 Thermal limit curve data.

37.8 Illustrative Examples

Figure 37.35 Overload heater library.

Figure 37.36 Contactor information.

Double click the contactor (Cont1), and enter the data shown in Figure 37.36. Double click the fuse (Fuse1), and select the manufacturer data from the ETAP Fuse Library, as shown in Figure 37.37. Similarly, update all the relay data from the library on the OCR page, as listed in Table 37.8. Update all HV and LV circuit-breaker operating times. Double-click the circuit breaker, go to the Rating page, and update the minimum time delay as shown:

••

LV circuit breaker: 0.04 seconds HV circuit breaker: 0.06 seconds Fault at all buses from the ETAP Star study case, and click the Run/Update Short-Circuit kA button; see Figure 37.38.

1051

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37 Protective Device Coordination

Figure 37.37 ETAP fuse library.

Table 37.8 Relay equipment manufacturer and model. Equipment ID

Location

Equipment

Make and model from ETAP library

Relay1

Incomer MCC

Multifunction relay

OCR, AREVA P122

Relay2

Incomer PMCC

Multifunction relay

OCR, AREVA P122

Relay3

Transformer neutral

Multifunction relay

OCR, AREVA P122

Relay4

Transformer primary

Multifunction relay

OCR, AREVA P122

Relay5

Grid relay

Multifunction relay

OCR, AREVA P122

Short-circuit through fault current is automatically updated in all protective devices. Graphically highlight the area as shown with its protective devices (i.e. overload relay [OLR], contactor, fuse, and combustor) using a left-mouse click, and click Create Star View to plot the TCC curve, as shown in Figure 37.39. Set the motor protection as shown in Figure 37.40 for Thermal and Jam settings:

• •

Thermal – Current pickup = 1.1 times FLC of motor (123.5 × 1.1 = 135.9 A). – Time delay should be above the motor start and below the stator thermal damage curve. Jam – Current pickup (Trip) = 150% of motor FLA = 137% of thermal overload pickup. – Time delay should be above the motor start and below the stator thermal damage curve.

37.8 Illustrative Examples

Figure 37.38 ETAP star short-circuit results.

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37 Protective Device Coordination

Figure 37.39 Create Star View to plot time-current characteristic curve (TCC) curves.

Figure 37.40 Motor-protection settings.

Adjust the time delay using the time handle on the OLR jam protection curve. Set it above the motor starting curve and below the motor stator and rotor thermal damage curve. Create a star view to plot the TCC curve. Rename it MCC Incomer. Double click Relay 1, go to the OCR page, enter the curve type IEC – Extremely Inverse, and enter the Pickup value 0.94 (i.e. 800 × 0.94 = 752 A) on the Phase page; see Figure 37.41. Adjust the time delay by using the time handle on the plot such that it is above the characteristics of the fuse and below the MCC incomer cable thermal withstand characteristics. Now select the MCC incomer relay and PMCC incomer relay, as shown in Figure 37.42, create a new star view, and rename it 03. PMCC to MCC circuit. Double-click Relay 2, go to the OCR page, enter the curve type IEC – Standard Inverse, and enter the Pickup value 0.98 (i.e. 1.1 × transformer secondary full load current/CT ratio) on the Phase page, as shown in Figure 37.43. Adjust the time delay by using the time handle on the plot such that it is above the characteristics of Relay1 and below the transformer thermal withstand characteristics, as shown in Figure 37.44.

37.8 Illustrative Examples

Figure 37.41 MCC incomer time-current characteristic curve (TCC).

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37 Protective Device Coordination

Figure 37.42 PMCC to MCC circuit time-current characteristic curve (TCC).

37.8 Illustrative Examples

Figure 37.43 Relay OCR settings.

Check the time delay of Relay2 by using time handle such that it is above the Relay1 curve with a time difference of 300 ms, using the Time Difference tool on the ETAP Star View toolbar shown in Figure 37.45. The time difference is displayed on the ETAP Star View; see Figure 37.46. Now graphically select the one-line diagram and create a star view; see Figure 37.47. Double-click Relay4, go to the OCR page, enter the curve type IEC – Standard Inverse, and enter the Pickup value 0.92 (i.e. 1.05 times of transformer primary full load current/CT ratio) in the Phase page, as shown in Figure 37.48. Adjust the time delay by using the time handle on the plot such that it is above the characteristics of Relay2, transformer inrush current, and below the transformer thermal withstand characteristics. Also, check the time difference between Relay2 and Relay4 by using Time Difference tool, as shown in Figure 37.49. Current seen by the transformer incomer relay is shown in Table 37.9 for faults on the primary and secondary side. The table shows that fault at the primary side is 42.86 kA while the fault current seen at the primary for a secondary fault is 2.73 kA. Therefore the transformer incomer relay protection is enabled for fault current above 140% of the transformer secondary fault level So the transformer incomer relay IOC protection is enabled for fault current greater than 140% above the transformer secondary fault level. Set Instantaneous Current Protection to 18.77 (i.e. 1.4 times greater

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37 Protective Device Coordination

Figure 37.44 PMCC to MCC circuit.

Figure 37.45 ETAP Star View toolbar.

37.8 Illustrative Examples

Figure 37.46 ETAP star view showing time difference.

than the transformer secondary fault level/CT ratio) and Time Delay at a minimum time, say 0.01 seconds. Use the Curve Integration option in the Relay Editor. The relay curve integration of the two types of curves of the same relay is shown in Figure 37.50. Double-click Relay5, go to the OCR page, enter the curve type IEC – Very Inverse, and enter the Pickup value 1 (i.e. 100% CT ratio) on the Phase page. Adjust the time delay by using the time handle on the plot such that it is above the characteristics of Relay4. Also, check the time difference between Relay4 and Relay5 by using the Time Difference tool; see Figure 37.51.

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37 Protective Device Coordination

Figure 37.47 6.6 kV switchgear time-current characteristic curve (TCC).

37.8 Illustrative Examples

Figure 37.48 Relay4 OCR page.

Figure 37.49 Time difference between Relay2 and Relay4.

1061

37 Protective Device Coordination

Table 37.9 Transformer incomer relay fault current summary. Fault current seen by transformer incomer relay at 6.6 kV in kA

Fault location

Fault at transformer secondary (0.415 kV side)

2.73

Fault at transformer primary (6.6 kV side)

42.86

II

After integration

Time

Before integration

Time

1062

I Figure 37.50 Curve integration.

Figure 37.51 Curve integration in star view.

I

37.8 Illustrative Examples

37.8.2

Illustrative Example – Ground Fault Protection

Using the system from the previous example, open the “02. MCC Incomer” star view TCC plot, and click Mode to view the earth fault, as shown in Figure 37.52. Double-click Relay1, go to the OCR page, and enter the curve type IEC – Standard Inverse, and Pickup: 0.2 on the Ground Settings page. Adjust the TD by using the graphical adjustment handle, and set TD: 0.05. Open the “03.PMCC to MCC Circuit” TCC plot, and click Mode to view the earth fault. Double-click Relay2, go to the OCR page, and enter the curve type IEC – Standard Inverse and Pickup value 0.2 on the Ground page. Similarly, doubleclick Relay3, go to the OCR page, and enter the curve type IEC – Standard Inverse, and Pickup value 0.2 on the Ground page. Adjust the time delay by using the time handle on the plot such that the minimum discrimination between the relay curves is 300 ms, as shown in Figure 37.53. Open the “04. 6.6KV SWGR IC” TCC plot, and click Mode to view the ground/earth fault. Hide the Relay2 and Relay3 curves from the TCC plot using the plot options. Double-click Relay4, go to the OCR page, and enter the curve type IEC – Standard Inverse, and Pickup value 0.5 on the ground page. Similarly, double-click Relay5 and go to the Phase page. Enter

Figure 37.52 Ground-fault view on ETAP star time-current characteristic curve (TCC).

1063

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37 Protective Device Coordination

Figure 37.53 Relay discrimination between Relay1, Relay2, and Relay3.

the curve type IEC – Standard Inverse, and Pickup value 0.5 on the Ground page. Adjust the time delay by using time handle on the plot such that the minimum discrimination between the the relay curves is 300 ms, as shown in Figure 37.54.

37.8.3

Evaluating Phase and Ground Settings Using a Sequence of Operation

Double-click Relay1, go to the Output page, and then click Add to Interlock Relay1 with CB2; see Figure 37.55. Similarly, update all relay interlocks as given in Table 37.10. Go to the study case, and set the Seq of Op. solution parameters as shown in Figure 37.56.

37.8 Illustrative Examples

Figure 37.54 Relay discrimination between Relay4 and Relay5.

Figure 37.55 Relay interlock with a circuit breaker.

1065

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37 Protective Device Coordination

Table 37.10 Relay interlocks. Interlock Editor page settings Relay ID

Relay element

Level/Zone

Device

ID

Action

Relay1

Any

Any

LVCB

CB2

Open

Relay2

Any

Any

LVCB

CB3

Open

Relay3

Any

Any

LVCB

CB3

Open

Relay4

Any

Any

HVCB

CB4

Open

Relay5

Any

Any

HVCB

CB5

Open

Figure 37.56 ETAP star study case for sequence of operations.

Click Fault Insertion, and apply a fault at the motor terminals. Check for the flashing devices to determine the sequence of operations, as shown in Figure 37.57. To check the normalized star view, create one more star view with all protective devices, and click Normalized TCC on the Star View toolbar. Normalized (Shifted) TCC mode provides a graphical view (TCC plot) of the operation times of protective devices based on their corresponding settings and characteristics for specified a fault location and type; i.e. curves are shifted by a factor calculated based on the ratio of the through fault current seen by a PD and the total fault current at the point of the fault. The effected TCC curves are then shifted according to the total fault current.

37.8 Illustrative Examples

Figure 37.57 ETAP star sequence of operations showing the location of faults and protective device operating sequence based on through-fault current in each device.

1067

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37 Protective Device Coordination

Figure 37.58 ETAP star sequence of operations viewer.

Click Sequence Viewer to view a complete list of the sequence of operation events with time, through fault current, operating device, and reason for the trip as shown in Figure 37.58. The following table shows the time difference summarized from the ETAP Sequence-of-Operation Events viewer for a three-phase fault applied on the motor terminal. Three-phase fault applied on motor terminal: current of 15.342 kA flows at 0.415 kV level

Location

Fuse 1

Motor feeder

0.010

0.010

Minimum operating time OK

Relay1

MCC incomer

0.135

0.125

Fuse-Relay >150 ms OK

Relay2

PMCC incomer

0.962

0.827

Relay-Relay >300 ms OK

Relay4

Transformer incomer

1.512

0.55

Relay-Relay >300 ms OK

Relay5

6.6 kV SWGR incomer

20.769

19.25

Relay-Relay >300 ms OK

37.8.4

Operating time (s)

Time diff with downstream element (s)

Element

Remark

Illustrative Example – Star Auto-Evaluation

The Star Automated Protection and Coordination Evaluation software provides automatic detection and evaluation of system protection and coordination/selectivity based on customized design criteria and industry guidelines. Protection rules are applied to evaluate the settings of the protective devices for equipment protection against possible damage. Coordination rules help identify time-overcurrent conflicts between downstream and upstream protective devices, to assist with choosing optimal settings. The ETAP Star Rule Book offers a set of default rules based on industry standards and guidelines. The Rule Book can be customized to apply different evaluation criteria rules based on company or industry practices. For the Star Auto-Evaluation, go to the relay coordination study case, and choose the IEC Rule Book. Go to the Rating page, and update the type and continuous ampere rating of each bus:

•• •

Bus 1 = 600 A Bus 2 = 3000 A Bus 3 = 800 A

37.8 Illustrative Examples

Figure 37.59 ETAP auto-protection and coordination evaluation interactive result viewer.

Graphically select the complete one-line diagram, and run the Star Auto-Evaluation. Automatic evaluation results are shown in a tabular manner, with pass, fail, warnings, and error messages clearly color-coded as shown in Figure 37.59.

1069

1071

Appendix C Standards, Regulations, and Best Practice C.1

Standards for One-Line Diagram Symbols

Category

Standard

Title

General

IEC 61082-1

Preparation of documents used in electrotechnology – Part 1: Rules

Graphical symbols

IEC 60417

Graphical symbols for use on equipment

ISO 7000

Graphical symbols for use on equipment – Registered symbols

IEC 62648

Graphical symbols for use on equipment – Guidelines for the inclusion of graphical symbols in IEC publications

IEC 60617

Graphical symbols for diagrams

IEC 80416-1

Basic principles for graphical symbols for use on equipment – Part 1: Creation of graphical symbols for registration

ISO/IEC 81714-1

Design of graphical symbols for use in the technical documentation of products – Part 1: Basic rules

Line types

ISO 128-20

Technical drawings – General principles of presentation – Part 20: Basic conventions for lines

Dimensioning

ISO 129

Technical drawings – Indication of dimensions and tolerances

Dimensional and geometrical product specifications (GPS)

ISO 1101

GPS – Geometrical tolerancing – Tolerances of form, orientation, location and run-out

Projection

ISO 128-30

Technical drawings – General principles of presentation – Part 30: Basic conventions for views

Flowcharts or Organigrams

ISO 5807

Information processing – Documentation symbols and conventions for data, program and system flowcharts, program network charts and system resources charts

C.2

IEC Standards for Short-Circuit Calculation and Analysis

Standard

Pub. year

61660-1

1997

Short-circuit currents in DC auxiliary installations in power plants and substations – Part 1: Calculation of short-circuit currents

61660-2

1997

Short-circuit currents in DC auxiliary installations in power plants and substations - Part 2: Calculation of effects

61660-3

2000

Short-circuit currents in DC auxiliary installations in power plants and substations – Part 3: Examples of calculations

Title

60865-1

2011

Short-circuit currents – Calculation of effects – Part 1: Definitions and calculation methods

60865-2

2015

Short-circuit currents – Calculation of effects – Part 2: Examples of calculation (Continued)

Power System Dynamics with Computer-Based Modeling and Analysis, First Edition. Yoshihide Hase, Tanuj Khandelwal and Kazuyuki Kameda. © 2020 John Wiley & Sons Ltd. Published 2020 by John Wiley & Sons Ltd.

1072

Appendix C Standards, Regulations, and Best Practice

Standard

Pub. year

Title

60909-0

2016

Short-circuit currents in three-phase AC systems – Part 0: Calculation of currents

60909-1

2002

Short-circuit currents in three-phase AC systems – Part 1: Factors for the calculation of short-circuit currents according to IEC 60909-0

60909-2

2008

Short-circuit currents in three-phase AC systems – Part 2: Data of electrical equipment for short-circuit current calculations

60909-3

2009

Short-circuit currents in three-phase AC systems – Part 3: Currents during two separate simultaneous line-toearth short circuits and partial short-circuit currents flowing through earth

60909-4

2000

Short-circuit currents in three-phase AC systems – Part 4: Examples for the calculation of short-circuit currents

C.3

Short-Circuit Calculations per ANSI/IEEE and IEC Standards

Standard

Pub. year

Title

IEEE IEEE IEEE IEEE IEEE IEEE IEEE

C37.04 C37.04f C37.04g C37.04h C37.04i C37.04 C37.010

Standard Rating Structure for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current Basis and Supplements

IEEE IEEE IEEE IEEE

C37.010b C37.010e C37.010 C37.13

1979 (1988) 1990 1986 1990 1991 1999 1979, 1988, 1999 1985 1985 1999 1990

IEEE Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Basis, and supplements

Standard for Low-Voltage AC Power Circuit Breakers Used in Enclosures

IEEE C37.013

1997

Standard for AC High-Voltage Generator Circuit Breakers Rated on a Symmetrical Current Basis

IEEE C37.20.1

2002

Standard for Metal Enclosed Low-Voltage Power Circuit Breaker Switchgear

IEEE 399

1990 & 1997

Recommended Practice for Power System Analysis (IEEE Brown Book)

IEEE 141

1986 & 1993

Recommended Practice for the Electric Power Distribution for Industrial Plants(IEEE Red Book)

IEEE 242

1986 & 2001

Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems (IEEE Buff Book)

UL 489_9

1996, 2000, 2002

Standard for Safety for Molded-Case Circuit Breakers, Molded-Case Switches, and Circuit-Breaker Enclosures

IEC 62271-100

2003

High-voltage switchgear and controlgear – Part 100: High-voltage alternating-current circuit breakers

IEC 62271-200

2003

High-voltage switchgear and controlgear – Part 200: AC metal-enclosed switchgear and controlgear for rated voltages above 1 kV and up to and including 52 kV

IEC 62271-203

2003

High-voltage switchgear and controlgear – Part 203: Gas-insulated metal-enclosed switchgear for rated voltages above 52 kV

IEC 60282-2

1997

High-voltage fuses – Part2: Expulsion fuses

IEC 61363-1

1998

Electrical installations of ships and mobile and fixed offshore units – Part 1: Procedures for calculating short-circuit currents in three-phase AC

IEC 60909-0

2016

Short-circuit currents in three-phase AC systems – Part 0: Calculation of currents

IEC 60909-1

2002

Short-circuit currents in three-phase AC systems – Part 1: Factors for the calculation of short-circuit currents according to IEC-60909-0

IEC 60909-2

1992

Electrical equipment – Data for short-circuit current calculations in accordance with IEC 60909 (1988)

Appendix C Standards, Regulations, and Best Practice

Standard

Pub. year

Title

IEC 60909-4

2000

Short-circuit currents in three-phase AC systems Part 4: Examples for the calculation of short-circuit currents

IEC 60947-1

2004

Low voltage switchgear and control gear, Part 1: General rules

IEC 60947-2

2003

Low voltage switchgear and control gear, Part 2: Circuit breakers

C.4

Major Standards Related to Harmonic Analysis

Standard

Title

IEEE C37.99-2012

Guide for Protection of Shunt Capacitor Banks

IEEE 519-2014

Recommended Practice and Requirements for Harmonic Control in Electric Power Systems

IEEE 1036-2010

Guide for the Application of Shunt Power Capacitors

IEEE C57.110-2008

Recommended Practice for Establishing Liquid-Filled and Dry-Type Power and Distribution Transformer Capability When Supplying Nonsinusoidal Load Currents

IEEE 18-2012

Standard for Shunt Power Capacitors

IEEE 1531-2003

Guide for Application and Specification of Harmonic Filters

IEEE 3002.8-2018

Recommended Practice for Conducting Harmonic Studies and Analysis of Industrial and Commercial Power Systems

IEEE 141-1993

Recommended Practice for Electric Power Distribution for Industrial Plants

IEC 61000-4-7 : 2009

Electromagnetic compatibility (EMC) – Part 4-7: Testing and measurement techniques – General guide on harmonics and interharmonics measurements and instrumentation, for power supply systems and equipment connected thereto

IEC 61000-3-6 : 2009

EMC – Part 3-6: Limits – Assessment of emission limits for the connection of distorting installations to MV, HV and EHV power systems

IEC 61000-3-6 : 2018

EMC – Part 3-2: Limits – Limits for harmonic current emissions (equipment input current ≤16 A per phase)

C.5

Harmonic Limits Based on IEEE 519–2014

Voltage distortion limits: Bus voltage V at PCC

Individual harmonic (%)

THD (%)

V ≤ 1 kV

5.0

8.0

1 kV < V ≤ 69 kV

3.0

5.0

69 kV < V ≤ 161 kV

1.5

2.5

161 kV < V

1.0

1.5

1073

1074

Appendix C Standards, Regulations, and Best Practice

Current distortion limits for systems 120 V–69 kV and above 69 kV: Bus Nominal Voltage (kV)

Isc / IL for kV ≤ 1

h

Hz

THD

60

5

8

12

2

120

1

1.75

3

180

4

7

4

240

1

1.75

5

300

4

6

360

1

7

420

4

8

480

1

Isc/IL for kV ≤ 69

kV ≤ 161

(Isc/IL)