Power Systems Modelling and Fault Analysis: Theory and Practice [2 ed.] 9780128151174

Table of contents :
Cover......Page 1
Power Systems Modelling and Fault Analysis: Theory and Practice
......Page 3
Copyright......Page 4
Dedication......Page 5
List of electrical symbols......Page 6
Foreword to second edition......Page 9
Foreword to first edition......Page 12
Preface to the second edition......Page 14
Preface to first edition......Page 16
Biography......Page 20
1.1 General......Page 21
1.2 Structure of modern power systems......Page 22
1.3.2 Health and safety considerations......Page 23
1.3.4 Design of power system equipment......Page 29
1.4.2 Types of faults......Page 30
1.4.3 Causes of faults......Page 31
1.4.4 Characterisation of faults......Page 32
Three-phase short-circuit currents......Page 34
One-phase short-circuit current......Page 37
1.5.2 Terminology of short-circuit current interruption......Page 39
1.6.2 Mechanical effects......Page 42
1.7.2 Single-phase systems......Page 45
1.7.3 Change of base quantities......Page 48
1.7.4 Three-phase systems......Page 49
1.7.5 Mutually coupled systems having different operating voltages......Page 50
Base and per-unit values of mutual inductive impedance......Page 51
Base and per-unit values of mutual capacitive admittance......Page 53
1.7.6 Examples......Page 56
Further reading......Page 59
Chapter Outline......Page 60
2.1 General......Page 61
2.2.1 Symmetrical components of balanced three-phase voltage and current phasors......Page 62
2.2.2 Symmetrical components of unbalanced voltage and current phasors......Page 64
2.2.4 Definition of phase sequence component networks of a three-phase power system containing balanced voltage sources......Page 67
2.2.5 Sequence impedances of coupled unbalanced three-phase impedances......Page 70
2.2.6 Sequence impedances of coupled balanced three-phase impedances......Page 73
2.2.8 Examples......Page 74
2.3.2 Balanced three-phase to earth short-circuit faults......Page 77
2.3.3 Balanced three-phase clear of earth short-circuit faults......Page 79
2.3.4 Unbalanced one-phase to earth short-circuit faults......Page 81
2.3.5 Unbalanced phase-to-phase or two-phase short-circuit faults......Page 83
2.3.6 Unbalanced two-phase to earth short-circuit faults......Page 85
2.3.7 Unbalanced one-phase open-circuit faults......Page 88
2.3.8 Unbalanced two-phase open-circuit faults......Page 90
2.3.9 Example......Page 92
2.4.2 One-phase to earth short-circuit faults......Page 94
2.4.3 Two-phase to earth short-circuit faults......Page 95
2.5.2 Simultaneous short-circuit faults at the same location......Page 97
2.5.3 Cross-country faults or simultaneous faults at different locations......Page 100
2.5.4 Simultaneous open-circuit and short-circuit faults at the same location......Page 101
2.5.5 Simultaneous faults caused by broken and fallen to earth conductors......Page 102
2.5.6 Simultaneous short-circuit and open-circuit faults on distribution transformers......Page 103
2.6.1 General......Page 107
2.6.2 Symmetrical components of voltage and current sources......Page 108
2.6.3 Balanced three-phase to earth short-circuit faults......Page 109
Current sources produce positive-sequence current only......Page 110
Current sources produce positive- and negative-sequence currents......Page 112
Current sources produce positive-sequence current only......Page 114
Current sources produce positive- and negative-sequence currents......Page 116
Books......Page 117
Paper......Page 118
Chapter Outline......Page 119
3.2.1 Background......Page 120
General......Page 122
Potential coefficients, shunt capacitances and susceptances......Page 124
Self impedance......Page 125
Earth return path impedances......Page 128
Summary of self and mutual impedances......Page 130
Stranded conductors......Page 132
Bundled phase conductors......Page 134
Average height of conductor above earth......Page 136
Phase and sequence series impedance matrices......Page 137
Phase and sequence shunt susceptance matrices......Page 142
3.2.4 Transposition of single-circuit three-phase overhead lines......Page 145
Phase and sequence series impedance matrices......Page 151
Phase and sequence shunt susceptance matrices......Page 154
3.2.6 Transposition of double-circuit overhead lines......Page 156
3.2.7 Phase and sequence parameter matrices of untransposed and transposed multiple-circuit lines......Page 172
Lines with three coupled circuits......Page 173
Lines with four coupled circuits......Page 174
3.2.8 Examples......Page 176
Potential coefficients, capacitance, phase and sequence susceptance matrices......Page 179
Phase and sequence impedance matrices......Page 182
Potential coefficients, capacitance, phase and sequence susceptance matrices......Page 185
Phase and sequence impedance matrices......Page 187
3.3.1 Background......Page 190
Single-point bonded cables......Page 192
Cross-bonded cables......Page 193
General......Page 194
Shunt capacitances and susceptances......Page 195
Single-core underground cables......Page 198
Single-core submarine cables......Page 202
Three-core armoured cables and pipe-type cables......Page 203
Power frequency impedance equations......Page 205
ac resistance at power frequency......Page 207
General......Page 209
Three-core cables, three single-core cables in touching trefoil and three single-core cables in equilateral layout......Page 210
Single-core cables in a flat layout......Page 212
Single-core armoured cables......Page 215
Three-core armoured cables and pipe-type cables......Page 218
Screened cables with no armour......Page 219
Unscreened or belted cables......Page 221
3.3.6 Impedance matrix of three-phase double-circuit cables......Page 223
3.3.7 Examples......Page 225
Cable impedances......Page 226
3.4.1 Background......Page 228
Core dc resistance is available......Page 229
Insulation, semiconductive screens and bedding tape......Page 230
Metallic sheath made of copper wire screen only......Page 232
Stranded copper core......Page 233
Composite sheath......Page 234
3.5.1 Background......Page 235
Overhead lines......Page 237
Cables......Page 238
3.5.3 Sequence π models of double-circuit overhead lines......Page 239
3.5.4 Sequence π models of double-circuit cables......Page 242
3.6 Sequence π models of three-circuit overhead lines......Page 243
3.7.2 Single-circuit overhead lines and cables......Page 245
3.7.3 Double-circuit overhead lines and cables......Page 246
General......Page 249
Single-circuit overhead lines with one earth wire......Page 250
Errors between model lumped parameters and measured distributed parameters......Page 253
Double-circuit overhead lines......Page 254
Measurement of cable dc resistances......Page 256
Measurement of cable positive-sequence/negative-sequence impedance......Page 257
Measurement of positive-/negative-/zero-sequence susceptance of a three-core belted cable......Page 259
3.9.1 Overhead lines......Page 260
Papers......Page 261
Chapter Outline......Page 263
4.2.1 Background......Page 264
Transformer core construction and general equivalent circuit......Page 267
Transformer equivalent circuit in actual physical units......Page 269
Transformer equivalent circuit in per unit......Page 271
Transformer equivalent circuit in per unit based on nominal impedance......Page 274
Interpreting transformer equivalent circuit and off-nominal tap ratio......Page 275
π equivalent circuit model......Page 278
4.2.3 Three-phase two-winding transformers......Page 280
Positive-phase sequence and negative-phase sequence equivalent circuits......Page 281
Zero-sequence equivalent circuits......Page 283
Effect of winding connection phase shifts on sequence voltages and currents......Page 287
4.2.4 Three-phase three-winding transformers......Page 292
Positive-phase sequence and negative-phase sequence equivalent circuits in actual physical units......Page 293
Positive-phase sequence and negative-phase sequence equivalent circuits in per unit......Page 296
Transformer equivalent circuit in per unit based on nominal impedance......Page 298
4.2.5 Three-phase autotransformers with and without tertiary windings......Page 301
Positive-sequence and negative-sequence equivalent circuits in physical units......Page 303
Positive-sequence and negative-sequence equivalent circuits in per unit......Page 308
Zero-sequence equivalent circuit in physical units and in per unit......Page 311
Zero-sequence equivalent circuit of an autotransformer with an isolated neutral......Page 314
General......Page 317
Positive-sequence equivalent circuit of a four-winding transformer......Page 318
4.2.7 Three-phase earthing or zigzag transformers......Page 320
4.2.8 Single-phase traction transformers connected to three-phase systems......Page 322
4.2.9 Variation of transformer’s positive-sequence leakage impedance with tap position......Page 323
4.2.10 Practical aspects of zero-sequence impedances of three-phase transformers and effect of core construction......Page 324
Three-phase transformers made up of three single-phase banks......Page 325
Three-phase transformers of five-limb core-form construction and shell-type core construction including seven-limb shell-form......Page 326
Three-phase transformers of three-limb core-form construction......Page 328
4.2.11 Measurement of sequence impedances of three-phase transformers......Page 329
Positive-sequence and zero-sequence impedance tests on two-winding transformers......Page 332
Positive-sequence and zero-sequence impedance tests on three-winding transformers......Page 333
4.2.12 Examples......Page 335
4.3.1 Background......Page 343
Positive-sequence equivalent circuit model......Page 345
Negative-sequence equivalent circuit model......Page 348
4.3.3 Measurement of sequence impedances of QB and PS transformer......Page 350
4.3.4 Examples......Page 352
4.4.2 Modelling of series reactors......Page 355
4.4.3 Modelling of shunt reactors and capacitors......Page 358
General modelling aspects of series capacitors......Page 361
Modelling of series capacitors for short-circuit analysis......Page 363
4.5.1 Background......Page 367
4.6.1 Background......Page 368
4.7.2 Three-phase modelling of reactors and capacitors......Page 370
Single-phase two-winding transformers......Page 371
Three-phase banks two-winding Ynd connected transformers with no interphase mutual couplings......Page 376
Three-phase common-core two-winding Ynd connected transformers with interphase mutual coupling......Page 379
4.7.4 Three-phase modelling of QB and PS transformers......Page 384
Books......Page 386
Papers......Page 387
Chapter Outline......Page 388
5.2 Overview of synchronous machine modelling in the phase frame of reference......Page 389
5.3.1 Transformation from ryb reference to dq0 reference......Page 392
5.3.2 Machine dq0 equations in per unit......Page 394
q-Axis operator reactance......Page 396
d-Axis operator reactance......Page 397
q-Axis parameters......Page 398
d-Axis parameters......Page 401
5.4.1 Synchronous machine sequence equivalent circuits......Page 403
Short-circuit currents......Page 404
Positive-sequence reactance and resistance......Page 408
Short-circuit fault through an external impedance......Page 412
Simplified machine short-circuit current equations......Page 414
5.4.3 Unbalanced two-phase (phase-to-phase) short-circuit faults......Page 415
Negative-sequence reactance and resistance......Page 418
Simplified machine short-circuit current equations......Page 419
5.4.4 Unbalanced single-phase to earth short-circuit faults......Page 420
Zero-sequence reactance and resistance......Page 422
Simplified machine short-circuit current equations......Page 423
5.4.5 Unbalanced two-phase to earth short-circuit faults......Page 424
Machine internal voltages......Page 429
One-phase to earth short circuit......Page 430
5.4.7 Effect of automatic voltage regulators on short-circuit currents......Page 431
5.4.8 Modelling of synchronous motors/compensators/condensers......Page 434
Three-phase short-circuit fault......Page 435
One-phase short-circuit fault......Page 436
Three-phase short-circuit fault......Page 439
One-phase to earth short-circuit fault......Page 440
Measurement and separation of ac and dc current components......Page 441
Transient reactance and transient short-circuit time constant......Page 443
Subtransient reactance and subtransient short-circuit time constant......Page 444
5.5.2 Measurement of negative-sequence impedance......Page 445
5.5.3 Measurement of zero-sequence impedance......Page 446
5.5.4 Example......Page 448
Direct axis transient reactance and time constant......Page 449
Transient reactance at the end of the subtransient period......Page 450
Negative-sequence impedance......Page 451
5.6.1 General......Page 452
5.6.2 Overview of induction motor modelling in the phase frame of reference......Page 453
5.7.1 Transformation to Park dq axes......Page 457
5.7.3 Operator reactance and parameters of a single-winding rotor......Page 459
5.7.4 Operator reactance and parameters of double-cage or deep-bar rotor......Page 461
Total short-circuit current contribution from a motor on no load......Page 465
Simplified motor short-circuit current equations......Page 469
Positive-sequence reactance and resistance......Page 470
Effect of short-circuit fault through an external impedance......Page 472
Short-circuit current equations......Page 474
Negative-sequence reactance and resistance......Page 475
5.8.3 Modelling the effect of initial motor loading......Page 476
Steady-state equivalent circuit: stator resistance test......Page 477
Steady-state equivalent circuit: locked rotor test......Page 478
Steady-state equivalent circuit: no-load test......Page 480
Transient parameters from a sudden three-phase short-circuit immediately after motor disconnection......Page 481
5.8.5 Examples......Page 482
Papers......Page 485
Chapter Outline......Page 486
6.1 General......Page 488
6.2.1 Basic operation of voltage-source converters......Page 489
6.3 Types of wind turbine generator technologies......Page 494
6.4.1 Modelling and analysis of short-circuit current contribution......Page 495
6.4.2 Example......Page 496
6.5.1 Modelling and analysis of short-circuit current contribution......Page 497
6.5.2 Example......Page 500
6.6.1 Background......Page 501
6.6.2 Basic operation principle......Page 502
6.6.3 Rotor protection......Page 503
6.6.4 Passive and active crowbar protection......Page 504
6.6.5 dc chopper protection......Page 506
6.6.7 DFIG steady-state equivalent circuit......Page 507
6.6.8 DFIG natural stator and rotor short-circuit currents under constant ac excitation......Page 508
6.6.9 DFIG stator and rotor short-circuit currents under crowbar action......Page 514
6.6.10 DFIG short-circuit currents with dc chopper control......Page 522
6.6.11 DFIG short-circuit currents with rotor converter control......Page 523
6.6.12 Examples......Page 524
6.8 Type 5 variable-speed wind turbine synchronous generators......Page 528
6.9.2 Solar PV generator components......Page 529
6.9.3 Solar PV voltage-source inverters......Page 530
6.9.4 Grid connection of solar PV power plant......Page 531
6.10 Technologies interfaced to the ac grid through voltage-source inverters......Page 533
6.11.1 General inverter model......Page 534
6.11.2 Phase-locked loop......Page 537
6.11.3 Inverter inner current control loop......Page 538
6.11.4 Inverter outer control loops......Page 542
6.11.5 Short-circuit current contribution of voltage-source inverters with frozen controls......Page 544
6.12 Grid code requirements for dynamic reactive current injection from inverters......Page 547
6.13.1 Dynamic reactive current control......Page 550
6.13.2 Examples of inverter positive-sequence short-circuit current contribution......Page 551
6.14.1 Inverter model in positive- and negative-sequence synchronous reference frames......Page 554
6.14.2 Examples of inverter positive- and negative-sequence short-circuit current contributions......Page 556
6.15 Sequence network representation of voltage-source inverters during balanced and unbalanced short-circuit faults......Page 559
6.16.1 General......Page 560
6.16.2 Steady-state and transient characteristics of inverter short-circuit reactive currents......Page 561
6.16.3 Analysis of three-phase short-circuit fault currents......Page 562
Inverters do not contribute a fault current......Page 563
Inverter current source is open-circuited......Page 564
Inverters do not contribute a fault current......Page 565
ac grid voltage source is short-circuited......Page 566
Inverters contribute positive- and negative-sequence currents......Page 569
ac grid voltage source is short-circuited......Page 570
Inverters contribute positive-sequence current only......Page 571
ac grid voltage source is short-circuited......Page 572
ac grid voltage source is short-circuited......Page 574
6.16.6 Examples......Page 576
6.17.2 Emerging challenges in power systems dominated by grid-following voltage-source inverters......Page 585
6.17.3 Control structure of grid-forming inverters......Page 586
6.18.1 Possible VSM inverter features of real synchronous machines......Page 587
6.18.2 A model of grid-forming VSM inverters......Page 588
VSM inverter's virtual rotor model with inertia and damping......Page 590
6.19 Grid-forming voltage-source inverters using droop control......Page 592
6.20.1 Natural three-phase short-circuit current of grid-forming inverters......Page 596
6.20.2 Natural two-phase short-circuit current of grid-forming inverters......Page 597
6.21.1 General......Page 598
Strategy 1: Clipping PWM voltage reference......Page 599
Strategy 3: Static virtual resistor......Page 600
Strategy 4: Transient virtual resistor......Page 601
Strategy 1: Saturation of individual current references......Page 603
Strategy 2: Saturation of magnitude of current reference......Page 604
6.22 Symmetrical components sequence equivalent circuits of ‘grid-forming’ inverters......Page 606
6.23 Examples......Page 608
Papers......Page 612
Chapter Outline......Page 614
7.2.1 Simulation of short-circuit faults in networks containing active voltage sources only......Page 615
7.2.2 Simulation of short-circuit faults in networks containing mixed active voltage and current sources......Page 618
7.2.3 Simulation of open-circuit faults in networks containing active voltage sources only......Page 619
7.2.4 Simulation of open-circuit faults in networks containing active voltage and current sources......Page 621
7.3.1 Background......Page 622
7.3.3 The ac short-circuit analysis technique......Page 623
7.3.5 Estimation of ac short-circuit current component variation with time......Page 624
7.4 Time domain short-circuit analysis techniques in large-scale power systems......Page 625
7.5.1 Single voltage source connected by a radial network......Page 626
Three-phase short-circuit fault......Page 627
7.5.2 Parallel independent voltage sources connected by radial networks......Page 629
Network ac time constant......Page 630
Network dc time constant......Page 631
7.5.3 Multiple voltage sources in interconnected meshed networks......Page 633
Network ac time constant......Page 634
Network dc time constant......Page 637
Thévenin’s and Norton’s positive-sequence equivalent models of machines......Page 639
Positive-sequence admittance and impedance matrix equations......Page 640
General analysis of three-phase short-circuit faults......Page 643
Three-phase short-circuit fault at one location......Page 644
Simultaneous three-phase short-circuit faults at two different locations......Page 647
Three-phase short-circuit fault between two nodes......Page 649
Nodal sequence impedance matrices......Page 651
General mathematical analysis......Page 652
One-phase open-circuit faults......Page 656
General analysis of three-phase short-circuit faults......Page 660
Step 1: Fault current due to conventional machines in the network and all inverter current sources are open-circuited......Page 661
Step 2: Fault current due to inverter current sources in the network and all conventional machines' voltage sources are sho.........Page 664
7.7.1 Background......Page 665
Synchronous machines......Page 666
Induction machines......Page 669
7.7.3 Three-phase analysis of ac short-circuit current in the phase frame of reference......Page 670
Two-phase fault through a fault impedance ZF......Page 673
Two-phase to earth fault through a fault impedance ZF......Page 674
One-phase to earth short-circuit fault......Page 675
Three-phase to earth short-circuit fault......Page 677
7.7.5 Example......Page 678
Three-phase to earth fault......Page 679
Books......Page 680
Papers......Page 681
8.1 General......Page 682
8.2.2 Analysis technique and voltage source at the short-circuit location......Page 683
Directly connected synchronous generators and compensators (factor KG)......Page 685
Power station units with and without on-load tap-changers (factors KS; KG,S; KT,S; and KSO; KG,SO; KT,SO)......Page 686
Step-up transformer with an on-load tap-changer (factors KS; KG,S and KT,S)......Page 687
Asynchronous or induction motors......Page 688
Wind power station units with asynchronous generators......Page 690
Calculation of initial rms short-circuit current......Page 691
Calculation of X/R ratio......Page 692
Calculation of symmetrical rms breaking current......Page 693
Calculation of symmetrical rms steady-state current......Page 696
8.2.8 Example......Page 697
Asynchronous machines......Page 698
Small induction motors forming part of the general power system load......Page 699
General......Page 700
8.3.5 Implementation of ER G7/4 in the UK......Page 701
8.4.1 Background......Page 703
General......Page 704
E/X simplified method......Page 705
Closing and latching current......Page 706
8.4.5 Example......Page 707
8.5 Examples of calculations using IEC 60909, UK ER G7/4 and IEEE C37.010......Page 708
dc current rating......Page 715
Asymmetrical ratings......Page 716
8.6.2 Assessment of aymmetrical circuit-breaker short-circuit duty against rating......Page 718
Books......Page 720
Papers......Page 721
9.1 General......Page 722
9.2.1 Theory of static network reduction......Page 723
9.2.2 Need for power system equivalents......Page 724
Conventional bus impedance or admittance matrices......Page 726
Illustration of equivalents of one, two and three boundary nodes......Page 727
Direct derivation of admittance matrix of power system equivalents......Page 729
Generalised time-dependent power system equivalents......Page 733
9.3.1 Representation of power-generating stations......Page 734
9.3.2 Representation of transmission, distribution and industrial networks......Page 735
9.4.2 Effect of mutual coupling between overhead line circuits......Page 737
9.4.3 Severity of fault types and substation configuration......Page 739
9.5.1 Confidence in calculations......Page 742
9.5.2 Data, models and analysis techniques......Page 743
9.5.4 Quality control and precision versus accuracy......Page 744
9.6.2 Probabilistic analysis of an ac short-circuit current component......Page 745
Factors affecting dc short-circuit current magnitude......Page 747
Probabilistic analysis of dc short-circuit current magnitude......Page 750
9.6.4 Example......Page 754
9.7.1 Background......Page 755
9.7.3 Theory of quantified risk assessment......Page 756
9.7.4 Methodology of quantified risk assessment......Page 757
Papers......Page 758
Chapter Outline......Page 759
10.2.2 Recertification of existing plant short-circuit rating......Page 760
10.2.3 Substation splitting and use of circuit-breaker autoclosing......Page 761
10.2.4 Network splitting and reduced system parallelism......Page 762
10.2.5 Sequential disconnection of healthy then faulted equipment......Page 763
10.2.6 Increasing short-circuit fault clearance time......Page 764
10.2.9 Example......Page 765
10.3.2 Opening of unloaded delta-connected transformer tertiary windings......Page 767
10.3.4 Upgrading to higher nominal system voltage levels......Page 768
10.3.8 Examples......Page 769
10.4.2 Earthing resistor or reactor connected to transformer neutral......Page 771
10.4.3 Pyrotechnic-based fault current limiters......Page 772
10.4.4 Permanently inserted current-limiting series reactors......Page 773
10.4.6 Limiters using magnetically coupled circuits......Page 774
Flux-cancelling limiter with reverse parallel to series reconnection......Page 775
10.4.8 Passive damped resonant limiter......Page 776
Series resonant limiter using a thyristor protected series capacitor......Page 777
Solid-state limiter using normally conducting power electronics switches......Page 778
Background......Page 779
Saturated inductance superconducting limiter......Page 780
Air-gap superconducting limiter......Page 781
10.5.1 Operation principles and design of resistive superconducting fault current limiters......Page 782
10.5.2 Modelling of resistive superconducting fault current limiters for short-circuit analysis......Page 783
10.5.3 Short-circuit analysis behaviour of resistive superconducting fault current limiters......Page 785
10.5.4 Example......Page 788
10.6 Characteristics of the ideal fault current limiter......Page 789
10.7 Applications of fault current limiters......Page 790
10.8 Examples......Page 794
Papers......Page 796
Chapter Outline......Page 798
11.1 Background......Page 799
11.2.2 Electrical resistance of the human body......Page 800
11.2.3 Effects of ac current on the human body......Page 802
General......Page 804
One driven vertical rod......Page 805
Multiple driven vertical rods in a hollow square......Page 806
Buried horizontal strip or wire......Page 807
Buried vertical or horizontal flat plate......Page 808
Buried horizontal grid or mesh......Page 809
Combined horizontal mesh with driven vertical rods around periphery......Page 810
11.4.1 Earthing network of overhead line earth wire and towers......Page 811
11.4.2 Equivalent earthing network impedance of an infinite overhead line......Page 813
11.5 Analysis of earth fault zero-sequence current distribution in overhead line earth wire, towers and in earth......Page 814
11.6 Earthing system impedance of cables......Page 819
11.7 Overall substation earthing system and its equivalent impedance......Page 820
11.9 Screening factors for overhead lines......Page 821
11.10.2 Single-phase cable with metallic sheath......Page 824
11.10.3 Three-phase cable with metallic sheaths......Page 826
11.11 Analysis of earth return currents for short-circuits in substations......Page 829
11.12 Analysis of earth return currents for short circuits on overhead line towers......Page 830
11.13 Calculation of rise of earth potential in substations and at towers......Page 833
11.14 Examples......Page 834
11.15.2 Effects of environment on earth (soil) resistivity......Page 839
11.15.3 Need for measurement......Page 840
11.15.4 Wenner four-electrode method......Page 841
11.15.5 Schlumberger–Palmer four-electrode method......Page 842
11.15.6 Driven rod three-electrode fall of potential method......Page 843
11.15.7 Interpretation of apparent earth (soil) resistivity measurements and derivation of two-layer earth model......Page 844
11.15.9 Examples......Page 847
Books......Page 848
Papers......Page 849
12.1 Background......Page 850
12.2.1 General......Page 851
12.2.2 Calculation of pipeline voltage to earth and discharge currents through a person’s body......Page 852
Earthed pipelines......Page 853
Nonparallel exposure......Page 854
12.3.1 Electromagnetic coupling mechanism......Page 855
12.3.3 Mutual impedance between power line and pipeline......Page 856
12.3.4 Analysis of induced EMF on the pipeline during steady-state system conditions......Page 858
12.3.5 Analysis of induced EMF on the pipeline during a short-circuit fault in the power system......Page 859
12.4.1 Modelling and analysis of distributed parameter pipelines......Page 860
12.4.2 Calculation of pipeline voltages caused by inductive coupling......Page 863
Case 1: the pipeline continues for several kilometres beyond the parallelism at both ends A and B......Page 864
Case 2: the pipeline continues for several kilometres beyond the parallelism at end A but stops at end B where it is insula.........Page 865
Case 3: the pipeline is earthed at end A but continues for several kilometres beyond the parallelism at end B......Page 866
12.5.2 Pipeline shunt admittance......Page 867
12.6 Resistive or conductive coupling from power systems to pipelines......Page 868
12.7 Examples......Page 869
Papers......Page 876
Appendix A.1 Analysis of distributed multiconductor overhead lines and cables......Page 877
A.2.2 Induction machine model in an arbitrary reference frame......Page 881
A.2.2.2 Synchronously rotating reference frame......Page 884
A.2.3 Complex space vector representation of induction machines......Page 885
A.2.3.3 Rotor reference frame......Page 886
Appendix A.3 Root mean square value of an asymmetrical current waveform......Page 887
A.4.1.2 For transformers......Page 890
A.4.2 Data......Page 891
Index......Page 905
Back Cover......Page 915

Citation preview

Power Systems Modelling and Fault Analysis

Power Systems Modelling and Fault Analysis Theory and Practice Second Edition

Nasser Tleis, BSc (Hons), MSc, PhD, CEng, FIET, M-CIGRE Fellow of the Institution of Engineering and Technology (FIET), London, United Kingdom Member of the Engineering Council, London, United Kingdom Member of Conseil International des Grands Re´seaux E´lectriques (CIGRE), Paris, France

Academic Press is an imprint of Elsevier 125 London Wall, London EC2Y 5AS, United Kingdom 525 B Street, Suite 1650, San Diego, CA 92101, United States 50 Hampshire Street, 5th Floor, Cambridge, MA 02139, United States The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, United Kingdom Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/ permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN: 978-0-12-815117-4 For Information on all Academic Press publications visit our website at https://www.elsevier.com/books-and-journals

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Dedicated to the late Mr. Rafik Hariri former Prime Minister of Lebanon without whom this book would not have been written

List of electrical symbols

Resistor

Short-circuit fault

Resistor with nonlinear current/voltage characteristics Varistor or surge arrester with nonlinear current/voltage characteristics Inductor

Normally closed circuit breaker

Iron-cored inductor

Normally open disconnector

Capacitor

Fast closing switch

Impedance

Fast opening switch

Reactor

Fuse

Two-winding transformers of various winding connection arrangements

Normally open circuit breaker

Normally closed disconnector

Three-winding transformers

Autotransformers

xxii

List of electrical symbols

Interconnected star or zigzag earthing transformer Variable tap ratio ideal transformer

Earth

Autotransformers with a deltaconnected tertiary winding Neutral solidly earthed

Direct current (dc) voltage source Current source

Neutral earthing resistor

Voltage-dependent current source

RE

Neutral earthing reactor jXE

M

Cable sealing ends

Spark gap ac generator

Triggered spark gap

Induction motor

Substation earth mesh/mat or grid

Diode Solar PV field Thyristor

Insulated gate bipolar transistor (IGBT) Thyristor switched capacitor

dc/dc converter dc chopper

List of electrical symbols

Thyristor controlled reactor

xxiii

DC

AC

≡

VDC

Two-level voltage source inverter

R Y B

Foreword to second edition

In the past, one of the main issues in power system planning and operation that needed regular attention depending on the generation dispatch was to be sure that short circuit fault currents would be within the capability of equipment, particularly circuit breakers. Such a key dependency on safety of course means that the analysis of system faults is something every power engineer should know about. However, the ever-increasing need for breadth in a modern electrical engineer’s education, not least, at the time of writing this Foreword, the ‘smarter’ grid’s need for knowledge of communications, the latest monitoring and control techniques, power electronics, software engineering, optimisation and data analytics squeezes the depth with which ‘traditional’ topics can be treated. Thus, many of the critical details in accurate representation of real power system equipment and the practical assessment of system behaviour are often omitted from degree courses. The first edition of this book was therefore extremely welcome in filling that gap. The process had already started in 2007 when the first edition was published, but, since then, the pace of change on power systems around the world has accelerated. In my own home country, the United Kingdom, nearly 30% of electricity produced in 2017 came from renewable sources, something that would have been unthinkable just 10 years earlier when the figure was only 5%. This has meant a squeezing out of the market of fossil-fuelled generation, much of which has now closed, with a lot of the rest running only when wind and solar conditions are unfavourable. While the reduction in carbon emissions associated with electricity generation is essential, the change in energy sources has profound consequences for operation of the system with wind turbines and inverter-connected infeeds increasingly prevalent. Thus, just as, at certain times and in certain places, fault levels can still be too high, there are periods and locations when the worry is that they are too low, for example, the intended operation of protection or avoidance of commutation failure on line-commutated high voltage direct current (HVDC). However, the fault current contributions from equipment other than directly connected synchronous machines cannot be neglected. They must therefore be calculated accurately. Moreover, given both their increasing significance and the broad pallet available in respect of the control of power electronics, grid codes need to be clear in what performance is required under a variety of system conditions, including unbalanced faults. Indeed, in 2019, a subject of research is the cost and potential benefits of changes to short-term current ratings and control of voltage source converters to

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Foreword to second edition

provide more reliable operation under short-circuit conditions, lessen the depth of voltage dips and provide a ‘grid-forming’ capability so aiding system stability. Against the background outlined above, the second edition of Nasser’s book is extremely timely. He has enriched it to include analysis of both Type 3 and Type 4 wind turbines and all inverter-connected sources and has described their behaviour under short-circuit conditions and the nature of extant international grid codes. He has also addressed the emerging technology of grid-forming inverters, both virtual synchronous machines and droop-controlled inverters. Other enhancements include new material on superconducting fault current limiters and environmental influences on earth return currents and rise of earth potential. In the Preface to the first edition, Nasser noted that, since liberalisation of electricity supply industries around the world, ‘many new engineers entering industry are neither adequately equipped academically nor are they finding many experienced engineers to train them.’ A number of the companies with which I have contacts do now succeed in recruiting bright, motivated young people. When I’m interviewing prospective undergraduates and teaching them before they leave, I am encouraged by how many of them show an acute awareness of challenges faced by a globally warming planet and an active desire to do their bit to address them. However, it appears to me to be increasingly the case that senior managers in companies with a strong cost-reduction focus are losing sight of the need for both solid educational foundations and experienced, patient mentoring. I see little problem with the fundamental calibre of young people other than the ability of those of us established in the sector to attract them to dedicate their careers to electrical energy and to give them the knowledge they need. I worry that the sector, certainly where I live, is not taking seriously enough the need for technical excellence to keep ahead of the curve of change. The lights might not yet be going out much more often, but for how long will that continue? More broadly, how do we most costeffectively provide access to low-carbon electricity, not just for those already on the grid but also the millions around the world currently lacking it, and whose lives would be improved by it? Nasser’s career has been notable from when he was an outstanding PhD student at the University of Manchester at a time when there were many good students through to when he had senior positions with National Grid in England and, now, with the Dubai Electricity and Water Authority. He shows a rare combination of academic rigour, breadth, clarity and industrial know-how with a sense of the importance of detail and which details matter in practical analysis and decision making. He is also unfailingly polite and generous with his knowledge. In my own career, I have reached a position of responsibility with respect to education of the next generation of professional engineers and, potentially, sector leaders. I also find myself being regularly asked for advice by utilities, regulators and government officials, and yet I am all too conscious of the limits to my own knowledge and understanding. However, I know how much I learned from working alongside Nasser and others when we were in National Grid in England, and how much I could still learn

Foreword to second edition

xxvii

from him. It is, sadly, many years since I had a desk across the partition from his, but at least I, and many others, can benefit from this new edition of his excellent and important book. Keith Bell Professor of Smart Grids, University of Strathclyde, Glasgow, United Kingdom

January 2019

Foreword to first edition

This is a new era for the electricity sector. The challenges we face in the near future are greater than at any time since the major network development programmes of the mid-20th century. Thankfully, power transmission technology and its control and protection have made enormous leaps, enabling better utilisation of assets, greater efficiency and improved quality of supply. This will help us meet the challenges ahead. From a technology perspective we are now seeing the construction of new national networks, the formation or integration of regional networks and major network renewal programmes. There is also the need to develop and integrate newgeneration technologies and implement new control and power electronics solutions in more active and integrated transmission and distribution networks. The technology problem is therefore becoming richer and more complex and demanding of novel solutions; it also requires a greater understanding of the characteristics and performance of the systems we need to build. But technology is not the only development over the last few decades. We have also seen the development of a more competitive marketplace for electricity with second- and third-generation market models now being implemented, the unbundling of utility companies all providing benefits for consumers. And crucially we are now understanding the impact of human activity on the environment and seeking to reduce emissions and develop more sustainable networks. This creates new pressures to incorporate the new greener technologies and meet planning and amenity constraints. From a social perspective we know that electricity has entwined itself into the very heart and veins of society and all the services we now take for granted. We have learnt this lesson very keenly in the opening of the 21st century with rude reminders on what can happen when electricity supplies are lost. Academia and the industry need to help the next generations of engineers to rise to these challenges. I believe that now is the time for the resurgence in engineering and electrical engineering disciplines; in particular the power generation, transmission and distribution sectors. It is vital that we develop and equip engineers with the verve, excitement, knowledge and talent they require to serve society’s needs. This book fills a major gap in providing the tools for this generation of engineers. It carefully targets the knowledge required by practitioners as well as academics in understanding power systems and their characteristics and how this can be modelled and incorporated into the development of the networks of the future. Nasser Tleis is distinguished as an academic and a senior manager in the industry. Nasser has been at the forefront in developing academic capability as well as

xxx

Foreword to first edition

building generations of engineers capable of taking forward this knowledge and experience in the practical application of the techniques. This book captures Nasser’s unique blend of the theoretical and applied knowledge to become a reference text and work book for our academics and engineers. It gives me great pleasure to write the Foreword to this book. It comes at a time where its contents are most relevant and I am confident it will bridge a gap between academic treatments and the very real need for application to power systems for the future by a new breed of practitioners. Nick Winser Director of Transmission, National Grid Company, Warwick, United Kingdom

Preface to the second edition

The first edition was published in 2008 and has become popular among practicing power engineers, universities and postgraduate students. The adoption of the book as a main reference text in some postgraduate teaching programmes worldwide and the use by patent offices is gratifying. Feedback from practicing engineers and university academics has been valuable and significant new material has been identified over the years. These were the drivers behind this second edition. The first edition material has been almost entirely preserved and the new edition adds new and substantial material in terms of volume, depth and the latest developments in equipment modelling and analysis. The following is a summary of the new material added in this new edition: 1. Application of the symmetrical components theory and use of The´venin’s and superposition theorems in the analysis of faulted power systems containing mixed voltage and current sources, for example, mixed conventional synchronous generators and modern inverter-based generators and systems such as type 4 wind-turbine generators and solar photovoltaic (PV) generators. The techniques are applicable to all inverter-based systems such as high-voltage direct current links, battery energy storage systems, fuel cells, etc. 2. Extending the modelling and analysis of actual as-built underground power cables to include semiconducting screens. 3. Inclusion of four-winding transformers and new seven-limb shell core transformer designs and significant expansions in the modelling of power transformers operation with off-nominal tap ratios. 4. Extensive modelling and analysis of all types of wind-turbine generators including extensive and original analysis of type 3 doubly fed induction generators and type 4 inverterbased wind-turbine generators. 5. New and general fault current analysis techniques in large-scale power systems containing mixed voltage and current sources. 6. Voltage-source inverters covering their applications, operation, modelling, control strategies and short-circuit current contribution to meet international grid code connection rules for dynamic reactive current injection. 7. Voltage-source inverter-based solar PV generator technology, applications, components and grid connection. 8. Emerging technology of grid-forming inverters, both virtual synchronous machines and droop-controlled inverters, system needs, their principles, control strategies, short-circuit current contribution and evolving fault current limitation approaches. 9. IEC 60909-0 treatment of wind and solar PV inverter-interfaced power stations in the calculation of short-circuit currents. 10. Design, modelling and analysis of resistive super-conducting fault current limiters with a practical example of their design and short-circuit current-limiting performance.

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Preface to the second edition

11. The chapter on the analysis of earth return currents and rise of earth potential, and electromagnetic interference from overhead power lines to metal pipelines, has been split into two dedicated chapters. 12. The dedicated chapter on earth return currents and rise of earth potential has been expanded to include treatment of earth (soil) resistivity, how it is affected by environmental factors, single- and two-layer models and practical earth (soil) resistivity measurement techniques. 13. The dedicated chapter on electromagnetic interference has been expanded with the inclusion of new material on the modelling and analysis of distributed parameter metal pipelines. 14. A new appendix has been added on the calculation of the root mean square value of an asymmetric current waveform containing AC and DC current components. 15. A new appendix has been added on the reference frames and transformations used in the analysis of AC induction machines. 16. New data have been added to the appendix on power system equipment data. 17. The subject index has been revised and expanded to facilitate reference to the new material.

In most chapters, many new, practical, worked examples have been added. Some typographical errors in various chapters in the first edition have been corrected. I am grateful for all the valuable comments I received and I hope the second edition will continue to be of great interest and benefit to practicing engineers, university academics and students. As in my first edition, I welcome any comments and suggestions for improvements on any aspect of this book. Finally, since the first edition was issued, I have had two new additions to my household; Tarek and Lynn. My wife Hanadi, daughters Serene, Lara and Lynn, and son Tarek, who accompanied me during this journey, have been my joy and strength and I would like to thank them for their tremendous support and unlimited love throughout. Finally, I would be grateful to receive comments of any errors, typographical or otherwise, as well as suggestions for improvements on any aspect of this book on [email protected] Nasser Tleis, BSc (Hons), MSc, PhD, CEng, FIET, M-CIGRE Dubai, United Arab Emirates January 2019

Preface to first edition

The objective of this book is to present a practical treatment of modelling of electrical power systems, and the theory and practice of power system fault analysis. The treatment is designed to be sufficiently in-depth and generally adequate to serve the needs of practising electrical power engineers. Practical knowledge of power systems modelling and analysis techniques is essential for power system engineers working in the planning, design, operation, protection and incident analysis of generation, transmission, distribution and industrial power systems. In many universities, undergraduate levels cover very little electrical engineering and, even at postgraduate levels, course contents have become more fundamental, theoretical and basic. Nowadays, many undergraduate and postgraduate university teachers have no or very little practical industrial experience. This book is intended to provide a practical source of material for postgraduate students, researchers and university teachers in electrical power engineering. Further, over the last twenty years or so, the ongoing liberalisation and restructuring of electricity supply industries have been accompanied by significant loss of experienced electrical power engineers, mostly to retirement. New engineers entering industry are neither adequately equipped academically nor are they finding many experienced engineers to train them! Technical learned society papers are necessarily concise and specialised. Though not necessarily brief, books on power system modelling and fault analysis generally tend to follow a highly theoretical treatment and lack sufficient practical information and knowledge that leaves the reader with inadequate understanding. In writing this book, one of my aims has been to attempt to bridge a gap between those theoretical books and the specialised technical papers. The aim of this book is to present practical power system modelling and analysis techniques as applied in modern industry practices. Therefore, strict academic and basic fundamental theories have largely been omitted to save valuable space. Basic knowledge is presumed in the following areas: analysis of three-phase alternating current electrical circuits; theory of electrostatic and electromagnetic fields; calculation of resistance, inductance and capacitance of lines; basic theories of electromagnetic transformers and ac rotating machines, complex phasor algebra, matrix algebra, linear differential equations and Laplace transforms. These basic topics are well covered in many power systems and mathematics textbooks. In support of the in-depth material I present in this book, I have included in every chapter a comprehensive and most relevant list of technical references.

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Preface to first edition

I have used SI units throughout and I hope this is not seen as a disadvantage where non-SI units are still in use. Chapter 1, Introduction to power system faults, discusses the nature, causes and effects of faults in power systems, presents fundamental concepts and definitions of short-circuit currents and circuit-breaker interruption as well as a practical treatment of per-unit system of analysis. Chapter 2, Symmetrical components of faulted threephase networks containing voltage and current sources, presents the theory of symmetrical components and a practical and detailed treatment of the connection of sequence networks under various fault conditions including simultaneous faults. Chapter 3, Modelling of multiconductor overhead power lines, and underground and submarine cables, is concerned with the advanced modelling and analysis of practical multiconductor overhead lines and cables in the phase coordinates and sequence reference frames. Chapter 4, Modelling of transformers, phase shifters, static power plant and static load, presents the modelling, in the phase coordinates and sequence reference frames, of transformers, quadrature boosters, phase shifters, series and shunt reactors, series and shunt capacitors, static variable compensators and power system load. Chapter 5, Modelling of rotating AC synchronous and induction machines, presents the modelling in the phase coordinates, dq0 and sequence reference frames of synchronous generators and induction motors. Modern wind-turbine generators such as squirrel-cage induction generators, woundrotor doubly fed induction generators and generators connected to the ac grid through power-electronics converters are also covered. In Chapters 3 5, practical measurement techniques of the electrical parameters of various power system equipment are presented. The models presented in these three chapters can be used in various power system analysis applications including positive phase sequence (PPS) load flow and PPS transient stability, multiphase unbalanced load flow and multiphase fault analysis, etc. Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators, presents methods for the simulation of short-circuit and open-circuit faults as well as static and dynamic shortcircuit analysis techniques in ac power systems. New and strong emphasis is given to the analysis of the time variation of the ac and dc components of short-circuit current. This emphasis is important in generation, transmission and industrial power systems. In addition, the expansion in the connection of small-scale distributed generation in distribution systems is exacerbating short-circuit current problems where switchgear are traditionally not designed either for make duties or for significant dc short-circuit current components. The emphasis on the analysis of the dc short-circuit current component also reflects the increase in X/R ratios of power system equipment due to the use of higher system voltages and/or more efficient and lower loss transformers. An introduction to modern short-circuit analysis in the phase coordinates frame of reference is given. Chapter 7, Short-circuit analysis techniques in ac power systems, describes and highlights the differences among the three international approaches to the analysis of short-circuit currents; the International Electro-technical Commission IEC 60909 Standard, the UK Engineering Recommendation ER G7/4 and the American IEEE C37.010 Standard. Chapter 8, International standards for short-circuit analysis

Preface to first edition

xxxv

in ac power systems, presents the formulation of power system equivalents by network reduction techniques. It discusses uncertainties present in short-circuit analysis, gives an introduction to probabilistic short-circuit analysis and to the theory of quantified risk assessment including safety considerations. Chapter 9, Network equivalents and practical short-circuit current assessments in large-scale ac power systems, presents practical methods for the control and limitation of high shortcircuit currents in power system design, operational planning and real-time operation. In addition, the various technologies of existing and some future short-circuit fault current limiters are described, including their applications. Chapter 10, Control of high short-circuit fault currents and fault current limiters, describes the effects of ac currents on the human body and its electrical resistance. It describes the components that make up extended substation earthing systems, and gives an introduction to the analysis of short-circuit earth or ground return current and rise of earth potential. The phenomenon of electrical interference from power lines is discussed and analysis techniques of induced voltages are presented with a particular focus on coupling interference from overhead power lines to metallic pipelines. Chapter 11, An introduction to the analysis of short-circuit earth return current and rise of earth potential, includes two appendices; the first presents the analysis of multiconductor lines and cables and the second presents typical data for power system equipment. I have used actual power system equipment data and solved practical examples representing some of the type of problems faced by practising power engineers. I have solved or shown how to solve the examples using hand calculations and electronic calculators. I believe in the ‘feel’ and unique insight that hand calculations provide which can serve as a good foundation for the power engineer to adequately specify, model, analyse and interpret complex and large power system analysis results. Many colleagues in National Grid have given me help and encouragement. Mr. Andy Stevenson who gave his and National Grid’s support at the start of this project, Mr. Tony Johnson, Mr. Tom Fairey and Dr. Andrew Dixon for reviewing various chapters and preparing their figures. I am indebted to Dr. Zia Emin for reviewing Chapter 3, Modelling of multiconductor overhead power lines, and underground and submarine cables, for the high-quality figures prepared and for our many useful discussions. Writing this book whilst full-time employed and raising a young family has been difficult. I would like to thank my wife Hanadi for her patience, unfailing support and encouragement throughout. I also want to thank my daughters Serene and Lara for never complaining why I ignored them for such a long time! Like most books that contain material of a reference nature, the published work is only the visible part of a huge iceberg. It is hoped that the nature of the material and list of references included may give a ‘taster’ of the size of the invisible iceberg!

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Preface to first edition

Finally, the book may contain errors of a typographical nature or otherwise. Should a second edition be required, I would be grateful to receive your comments and any suggestions for improvements on any aspect of this book on [email protected]. Nasser Tleis England, United Kingdom June 2007

Biography

Dr. Nasser Tleis obtained his PhD degree in Electrical Power Engineering from The University of Manchester Institute of Science and Technology in 1989. He then worked for the CEGB, then the National Grid UK, until 2010 when he joined the Dubai Electricity and Water Authority. During his career, he worked closely with UK universities and European research institutions engaged in research, development and demonstration projects in electrical power systems. He has extensive experience in several fields in electrical power engineering, such as: demand forecasting, generation capacity planning, power system operation, transmission network planning and design, integration of wind power generation, advanced modelling and analysis of power systems, detailed studies covering electromagnetic transients, insulation coordination, power plant specifications, voltage control strategies, reactive compensation planning, thermal and load flow analysis, voltage stability, frequency control and stability, transient and dynamic stability, faults and short-circuit analysis, multiphase power flow and unbalance analysis, measurement and analysis of harmonic distortion and voltage flicker. He has developed and validated new short-circuit analysis techniques, dynamic models of synchronous generators, excitation and governor control systems, voltage control systems and inverter control systems. He has been active in the integration of large-scale wind and solar PV generators in ac grids and in short-circuit and dynamic analysis of inverter-interfaced generators. He is registered with the Engineering Council, London, is a Chartered Engineer, a Fellow of the Institution of Engineering and Technology (formerly Institution of Electrical Engineers) and a member of CIGRE.

Introduction to power system faults

1

Chapter Outline 1.1 General 1 1.2 Structure of modern power systems 2 1.3 Need for power system fault analysis 3 1.3.1 1.3.2 1.3.3 1.3.4

General 3 Health and safety considerations 3 Design, operation and protection of power systems 9 Design of power system equipment 9

1.4 Characteristics of power system faults 10 1.4.1 1.4.2 1.4.3 1.4.4

Nature of faults 10 Types of faults 10 Causes of faults 11 Characterisation of faults 12

1.5 Terminology of short-circuit current waveform and current interruption

14

1.5.1 Short-circuit current waveforms 14 1.5.2 Terminology of short-circuit current interruption 19

1.6 Effects of short-circuit currents on equipment

22

1.6.1 Thermal effects 22 1.6.2 Mechanical effects 22

1.7 Per-unit analysis of power systems 1.7.1 1.7.2 1.7.3 1.7.4 1.7.5 1.7.6

Further reading

1.1

25

General 25 Single-phase systems 25 Change of base quantities 28 Three-phase systems 29 Mutually coupled systems having different operating voltages 30 Examples 36

39

General

In this introductory chapter, we present structures of modern power generation and network systems that include conventional power generation technologies and new and renewable generation technologies such as wind power, solar photovoltaic (PV) generation plant and distributed renewable energy resources. We then discuss the need for power system fault analysis and the characteristics of faults, introduce the important terminology of fault current waveform, and the thermal and mechanical effects of fault currents in power systems. Practical per-unit analysis of

Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00001-1 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

2

Power Systems Modelling and Fault Analysis

single-phase and three-phase power systems is presented including the base and per-unit equations of self and mutual impedances and admittances.

1.2

Structure of modern power systems

Electrical ac power systems consist of three-phase generation systems, transmission and distribution networks, and loads. The networks may supply large three-phase industrial loads at various distribution and transmission voltages, as well as singlephase residential and commercial loads. In some countries, for example, in North America, the term subtransmission is used to denote networks with voltage classes between transmission and distribution. Distribution voltages are typically up to 60 kV, subtransmission voltages are typically 66
138 kV and transmission voltages are typically above 138 kV. Generated voltages are up to 35 kV for generators used in large electrical power stations. Power station auxiliary supply systems and industrial power systems supply a significant amount of induction motor load. Residential and commercial loads include a significant amount of single-phase induction motor loads. In very warm and hot climates where air cooling is a necessity, three-phase induction motors are extensively used in air conditioning installations in commercial buildings and district cooling plant. For over a century, electric power systems used synchronous machines for the generation of electricity. However, in the 21st century, the generation of electricity from renewable energy sources such as wind and solar PV has begun to expand at a fast pace. Generally, wind generation systems use a variety of asynchronous machines or machines interfaced to the three-phase ac network through a lowvoltage direct current link. Solar PV generation systems and battery energy storage systems invariably connect to the ac power systems through power electronics inverters. Typical ratings of wind turbine generators are currently up to 7 MW and typical generated voltages range from 0.4 to 5 kV. Large solar PV generators currently operate at dc voltages of up to 1.5 kV and inverter output ac voltages up to 1 kV. The use of high-voltage direct current technologies for the transmission of power, either embedded within the ac power system or connecting large off-shore wind farms or interconnecting separate nonsynchronous power systems, is expected to increase. The mix of synchronous machines, asynchronous machines, converter-interfaced electrical generators, HVDC systems, battery energy storage systems and electric vehicles, etc., is changing the nature and behaviour of three-phase power systems following network disturbances. Distribution networks are changing from passive networks to active networks with the connection of large amounts of distributed resource generators at low and medium voltage levels. Fig. 1.1 illustrates a typical structure and components of a modern generation, transmission and active distribution power system and Fig. 1.2A
E illustrate in more detail the typical internal components of boxes A, B, C, D and E shown in Fig. 1.1.

Introduction to power system faults

3

External HVDC interconnector

15 kV

B1,B2

Coal-fired power station

Combined cycle gas power station

A

400 kV

E

275 kV Nuclear power station

35 kV

C1,C2 Wind power plant

132 kV

132 kV

D 11 kV 33 kV

Solar PV power plant

Distributed resource generator 415 V

Figure 1.1 Typical structure and components of a modern generation, transmission and distribution power system.

In practical power system short-circuit analysis, knowledge of substation configuration is important. Fig. 1.3A
E show typical substation configurations employed in practical generation, transmission and distribution systems.

1.3

Need for power system fault analysis

1.3.1 General Short-circuit analysis is carried out in electrical power utility systems, industrial power systems, commercial power systems and power station auxiliary systems. Other special applications are in concentrated power system installations on board military and commercial ships and aircraft. Short-circuit calculations are generally performed for a number of reasons. These are briefly described in the next sections.

1.3.2 Health and safety considerations Short-circuit fault analysis is carried out to ensure the safety of workers as well as the general public. Power system equipment, such as circuit-breakers, can fail catastrophically if they are subjected to fault duties that exceed their rating. Other

(A) 400 kV

15 kV 3.3 kV Auxiliaries

15 kV

250 MW

3.3 kV Auxiliaries

15 kV 225 MW

250 MW

GT1

GT2 ST

HRSG1

HRSG2 HPSH

Condenser

LPSH

GT = Gas Turbine ST = Steam Turbine HRSG = Heat Recovery Steam Generator HPSH/LPSH = High/Low Pressure Steam Header WDU = Water Desalination Unit

77,300 m3 /day 77,300 m3 /day WDU2 WDU1

(B1) +250 kV 500 MVA

DC line DC filter

1000 MW, 400 kV AC, 500 kV DC, HVDC link Current source converter technology

–250 kV

Smoothing reactor

AC filters

Figure 1.2 Typical structure and components of boxes A, B, C1/C2, D and E in Fig. 1.1. (A) Typical combined cycle gas turbine plant with water desalination plant (cogeneration plant) (B1) Typical HVDC converter station employing conventional current source converter technology (B2) Typical HVDC submarine transmission link employing modern voltage-source converter (VSC) technology (C1) Typical utility-scale on-shore or off-shore wind power plant (C2) Typical utility-scale off-shore wind power plant connected through HVDC link employing modern voltage source converter technology (D) Typical utility-scale solar photovoltaic (PV) power plant (E) Typical auxiliary supply system of a large power station.

Introduction to power system faults

5

(B2)

400 kV AC

400 kV AC

±1000 MW, 400 kV AC, 50 Hz ±320 kV DC, 160 km cable Phase reactor

DC capacitors (voltage sources)

AC harmonic filter (C1) 4 MVA 0.69/33 kV

WTG 3.6 MW 0.69 kV

33 kV 110 MVA 2 132 kV 33/132 kV 630 mm 120 MVA Grid

100.8 MW Wind Power Plant: 28 X 3.6MW Wind Turbine Generators (WTGs) (C2)

WTG

400 kV AC

155 kV ±320 kV, 140 km

4 MVA 0.69/33 kV

110 MVA 33/155 kV

5 MW 1 kV

On-shore HVDC converter platform 900 MW off-shore wind power plant

Off-shore HVDC converter platform Voltage source converters (VSC) technology

Figure 1.2 (Contiuned).

6

Power Systems Modelling and Fault Analysis

(D) 1.1 MVA DC/AC

1 MW 1000 VDC

2.2 MVA 0.6/33 kV

33 kV

Solar PV

110 MVA 630 mm2 132 kV 33/132 kV 120 MVA Grid

100 MW Solar Photovoltaic (PV) Plant : 100 X 1.1MVA Voltage-Source Inverters (E) 400 kV

Generator terminals 23 kV

132 kV GTG 11 kV

11 kV M

M

M

3.3 kV

M 3.3 kV

3.3 kV M

M

M

M

M

415 V

415 V M

M

M

M

M

M

3.3 kV M

M

GTG = Gas Turbine Generator

M = Induction Motor

Figure 1.2 (Contiuned).

equipment, such as busbars, transformers and cables, can fail thermally or mechanically if subjected to fault currents in excess of ratings. In addition, to ensure safety, short-circuit fault analysis is carried out and used in the calculation of the rise of earth potential at substations and overhead line towers. Other areas where fault analysis is carried out are for the calculation of induced voltages on adjacent communication circuits, pipelines, fences and other metallic objects.

Introduction to power system faults

7

(A)

M1

BS CB1

M2

BS CB2

BC CB1 R1

M3 BC CB3

BS CB3

R2

Key: BS CB = Bus Section Circuit-Breaker BC CB = Bus Coupler Circuit Breaker

BS CB4

R3

M = Main R = Reserve

(B)

Figure 1.3 Typical substation configurations used in generation, transmission and distribution systems (A) Typical generation/transmission substation configuration: double-busbar (B) Typical generation/transmission substation configuration: one-and-half breaker (C) Typical transmission substation configuration: double-bus (UK) or transfer-bus (America) (D) Typical transmission substation configuration: four-switch-mesh (UK) or ring-bus (America) (E) Typical MV distribution substation configuration: single-busbar arrangement.

8

Power Systems Modelling and Fault Analysis

(C)

Main 1

BS CB

Main 2 BC CB

BC CB

Reserve 2

Reserve 1

A second BS CB is also used by some utilities (D)

Figure 1.3 (Continued).

Feeder CB

Introduction to power system faults

9

(E)

M1

BS CB1

M2

BS CB2

M3

Figure 1.3 (Continued).

1.3.3 Design, operation and protection of power systems Short-circuit current calculations are made at the system design stage to determine the short-circuit ratings of new switchgear and substation infrastructure equipment to be procured and installed. 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. In addition, calculations of minimum short-circuit currents are made and these are used in the calculation of protection relay settings to ensure accurate and coordinated relay operations. In transmission systems, short-circuit currents must be quickly cleared to avoid loss of synchronism of generation plant and major power system blackouts. Maximum short-circuit current calculations are carried out for the design of substation earth electrode systems. Short-circuit analysis is also carried out as part of initial power quality assessments for the connection of disturbing loads to electrical power networks. These assessments include voltage flicker, harmonic analysis and voltage unbalance. Other areas where short-circuit analysis is carried out are in the modification of an existing system or at the design stage of new electrical power installations such as a new off-shore oil platform, new petrochemical process plant or the auxiliary electrical power system of a new power station. The aim is to determine the short-circuit ratings of new switchgear and other substation infrastructure equipment that will be procured and installed.

1.3.4 Design of power system equipment Switchgear manufacturers design, manufacture and test their circuit-breakers to ensure that they are capable of making, breaking and carrying, for a short time, the specified short-circuit current. Equipment with standardised short-circuit ratings are

10

Power Systems Modelling and Fault Analysis

designed and produced by manufacturers. Also, manufacturers of substation infrastructure equipment and other power system plant, for example, transformers and cables, use the short-circuit current ratings specified by their customers to ensure that the equipment is designed to safely withstand the passage of these currents for the duration specified.

1.4

Characteristics of power system faults

1.4.1 Nature of faults A fault on a power system is an abnormal condition that involves an electrical failure of power system equipment operating at one of the primary voltages within the system. Generally, two types of failure can occur. The first is an insulation failure that results in a short-circuit fault and can occur as a result of overstressing and degradation of the insulation over time or due to a sudden overvoltage condition. The second is a failure that results in a cessation of current flow or an open-circuit fault.

1.4.2 Types of faults Short-circuit faults can occur between phases, or between phases and earth, or both. Short circuits may be one-phase to earth, phase to phase, two-phase to earth, threephase clear of earth and three-phase to earth. The three-phase fault that symmetrically affects the three phases of a three-phase circuit is the only balanced fault, whereas all the other faults are unbalanced. Simultaneous faults are a combination of two or more faults that occur at the same time. They may be of the same or different types and may occur at the same or at different locations. A broken overhead line conductor that falls to earth is a simultaneous one-phase open-circuit and onephase short-circuit fault at one location. A short-circuit fault occurring at the same time on each circuit of a double-circuit overhead line, where the two circuits are strung on the same tower, is a simultaneous fault condition. A one-phase to earth short-circuit fault in a high-impedance earthed distribution system may cause a sufficient voltage rise on a healthy phase elsewhere in the system that a flashover and short-circuit fault occurs. This is known as a cross-country fault. Most faults do not change in type during the fault period but some faults do change and evolve from say a one-phase to earth short circuit to engulf a second phase where it changes to a two-phase to earth short-circuit fault. This can occur on overhead lines or in air-insulated substations, for example, during a disconnector opening operation, where the flashover arc of the faulted phase spreads to other healthy phases. Internal short circuits to earth and open-circuit faults can also occur on windings of transformers, reactors and machines, as well as faults between a number of winding turns of the same phase.

Introduction to power system faults

11

1.4.3 Causes of faults Open-circuit faults may be caused by the failure of joints on cables or overhead lines or the failure of all the three phases of a circuit-breaker or disconnector to open or close. For example, during a closing operation, two phases of a circuitbreaker may close and latch successfully but not the third phase, or, during an opening operation, two phases may properly open but the third remains stuck in the closed position. Except on mainly underground systems, the vast majority of shortcircuit faults are weather-related, followed by equipment failure. The weather factors that usually cause short-circuit faults are: lightning strikes, accumulation of snow or ice, heavy rain, strong winds or gales, salt pollution depositing on insulators on overhead lines and in substations, floods and fires adjacent to electrical equipment, for example, beneath overhead lines. Vandalism may also be a cause of short-circuit faults as well as contact with or breach of minimum clearances between overhead lines and trees due to current overload. Lightning strikes discharge steep-fronted impulse currents with peak values in the range of a few kiloamps up to 100 or 200 kA for a duration of several microseconds. If the strike hits an overhead line or its earth wire, the voltage produced across the insulator may be so large that a back-flashover and short circuit occurs. This may involve one or all three phases of a three-phase electrical circuit and as a result a transient power frequency short-circuit current flows. For example, consider a 132-kV three-phase overhead transmission line of steel tower construction and one earth wire. The surge impedances of the tower, line phase conductors and earth wire are given as 220, 350 and 400 Ω, respectively, and the tower’s earthing or footing resistance is 50 Ω. The line’s rated lightning impulse withstand voltage to earth p isffiffiffi650 pffiffiffikV peak phase to earth (nominal peak phase to earth voltage is 132 kV 3 2= 3 5 107:8 kV). A lightning strike with a modest peak current of 10 kA hits a tower on the line and, to a reasonable approximation, ‘sees’ half the surge impedance of the earth wire in parallel with the surge impedance of the tower. The voltage produced across the line’s insulator is approximately equal to 10 kA 3 105 Ω 5 1050 kV. This significantly exceeds the line’s insulation strength, causes back-flashover on all three phases of the line and a three-phase to earth short-circuit fault. If the shielding of the earth wire fails and the lightning strike hits one of the phase conductors near a tower, then the voltage produced across the line’s insulator is approximately equal to 10 kA 3 175 Ω 5 1750 kV. In this case, a smaller lightning current in the order of 3.8 kA would be sufficient to cause a back-flashover and hence a short-circuit fault. On lower-voltage distribution lines, even ‘indirect’ lightning strikes, that is, those that hit nearby objects to the line may produce a sufficiently large voltage on the line to cause an insulator flashover and a short-circuit fault. Other causes of short-circuit faults are fires. The smoke of fires beneath overhead lines consists of small particles which encourage the breakdown of the air that is subjected to the intense electric field of a high-voltage power line. The hot air in the flames of a fire has a much lower insulation strength than air at ambient temperature. A flashover across an insulator to earth or from a phase conductor to a tree may occur.

12

Power Systems Modelling and Fault Analysis

Equipment failure of machines, transformers, reactors, cables, etc., causes many short-circuit faults. These may be caused by failure of internal insulation due to ageing and degradation, breakdown due to high switching or lightning overvoltages, by mechanical incidents or by inappropriate installation. An example is a breakdown of a cable’s polymer insulation due to ageing or the creation of voids within the insulation caused by an external mechanical force being accidentally applied on the cable. Short-circuit faults may also be caused by human error. A classical example is one where maintenance staff inadvertently leave isolated equipment connected through safety earth clamps when maintenance work is completed. A three-phase to earth short-circuit fault occurs when the equipment is re-energised to return it to service. On mainly overhead line systems, the majority of short-circuit faults, typically 80%
90%, tend to occur on overhead lines and the rest on cables, substation equipment and busbars combined. Typically, on a high-voltage transmission system with overhead line steel tower construction, such as the England and Wales transmission system, long-term average short-circuit fault statistics show that around 300 shortcircuit faults occur per annum. Of these, 67% are one-phase to earth, 25% are phase to phase, 5% are three-phase to earth and three-phase clear of earth, and 3% are two-phase to earth. About 77% of one-phase to earth faults are caused by lightning strikes followed by wind and gales then salt pollution on insulators. Although lightning can cause some phase-to-phase faults, by far the most common causes of these faults are snow/ice followed by wind/gales that cause two line conductors to clash. The majority of three-phase to earth and two-phase to earth faults in England and Wales are caused by lightning then wind and gales. On wood-pole overhead lines, for example, some 132 kV and lower voltages in England and Wales, between 50% and 67% of short-circuit faults are two-phase and three-phase faults.

1.4.4 Characterisation of faults Because they are unbalanced, one-phase open-circuit and two-phase open-circuit faults are characterised by the negative and zero-phase sequence voltages and currents they generate at the fault location and elsewhere in the power system particularly at substations where electrical machines are connected. Machines are vulnerable to damage by overheating due to excessive negative-phase sequence currents flowing into them. Short-circuit faults are characterised by the short-circuit current and its components. These are the ac or symmetrical root mean square (rms) short-circuit current, dc short-circuit current or dc time constant or X/R ratio, and the overall asymmetrical short-circuit current. These are described in the next section. To characterise the short-circuit current of one source, a group of sources or the short-circuit current of an entire system, the concept of short-circuit fault level is useful: Short
circuit fault levelðMVAÞ 5

pffiffiffi 3VPhase2PhaseðkVÞ Irms

ðkAÞ

(1.1)

Introduction to power system faults

13

where Irms(kA) is the rms short-circuit current at the point of fault and VPhase
Phase(kV) is the prefault phase-to-phase voltage at the point of fault. For example, for an rms short-circuit current of 54 kA and a 404-kV prefault voltage in a 400-kV system, the short-circuit fault level is equal to 37,786.4 MVA. The system short-circuit fault level or infeed gives a measure of the strength or weakness of the system at the point of fault. For a given system short-circuit fault level or MVAFault Level at a busbar, the equivalent system impedance seen at the busbar in per unit on MVABase and phase-to-phase VPrefaultðkVÞ is given by ZSðpuÞ 5

2 VPrefault MVABase ðkVÞ 3 2 MVAFault Level VBaseðkVÞ

(1.2)

where the definition of base quantities is presented in Section 1.7. Where the prefault and base voltages are equal, we have ZSðpuÞ 5

MVABase MVAFault Level

(1.3)

High system strength is characterised by a high short-circuit fault level and thus low system impedance, and vice versa. ZS is also equal to the The´venin’s impedance of the system. Sometimes, an MVA figure is used to describe the short-circuit rating of a circuit-breaker in MVA. This practice is discouraged as it can easily lead to confusion and errors. For example, consider a 5000-MVA, 132-kV circuit-breaker that is used in a 110-kV system. The 5000-MVA rating is not valid in the 110-kV system in which the 132-kV circuit-breaker is used. The correct figure should be pffiffiffi 5000 MVA 3 3 110 kV 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 4166:7 MVA 3 3 132 kV Another important characteristic of a short-circuit fault in the case of earth faults is its fault impedance. In general, in the event of a flashover, the fault impedance consists of the flashover arc resistance and the earthing impedance of the object to which the flashover occurs. For example, where this object is an overhead line’s steel tower, the earthing impedance is that of the tower footing resistance in parallel with the line’s earth wire impedance, where it exists (this is discussed in detail in Chapter 11: An introduction to the analysis of short-circuit earth return current and rise of earth potential). The fault impedance is generally neglected in higher-voltage systems when calculating maximum short-circuit currents to obtain conservative results. However, earthing impedances are taken into account when calculating the rise of earth potential for short-circuit faults in substations and on overhead lines’ towers. In lower-voltage distribution systems, many short-circuit faults occur due to a flashover from a line to a tree which may present a significant earthing impedance,

14

Power Systems Modelling and Fault Analysis

depending on earth resistivity. The resistance of the flashover arc can be estimated using the A. C. Van Warrington empirical formula: RArc 5 1:811 3

Lo 1 vt Ω 1:4 Irms

(1.4)

where Lo is the length of the arc in m in still air, v is wind speed in m/s, t is arc duration in s and Irms is the rms short-circuit current in kA. The formula is adjusted to include arc elongation due to wind speed. For example, for Lo 5 7.75 m for a typical 400-kV overhead line, v 5 10 m/s, t 5 0.05 seconds and Irms 5 20 kA, we obtain RArc 5 0.21 Ω in still air and 0.225 Ω at 0.05 seconds including the effect of wind speed.

1.5

Terminology of short-circuit current waveform and current interruption

1.5.1 Short-circuit current waveforms Three-phase short-circuit currents In order to calculate short-circuit current duties on power system equipment, it is important to define the terminology used in characterising the short-circuit current waveform. Consider the simple balanced three-phase electric circuit equivalent of a source or system shown in Fig. 1.4A. L and R are the equivalent inductance and resistance for each phase, and Le and Re are the earth return path inductance and resistance, respectively. The balanced three-phase voltage sources are given by υ i ðt Þ 5

pffiffiffi   2Vrms sin ωt 1 ϕi i 5 r; y; b

(1.5)

where Vrms is rms voltage magnitude, ω 5 2πf in rad/s, f is power frequency in Hz and ϕi is voltage phase angle in rad given by ϕy 5 ϕr 2 2π=3

ϕb 5 ϕr 1 2π=3

(1.6)

If a solid three-phase to earth connection or short-circuit fault is made simultaneously between phases r, y, b and earth e at t 5 0, as shown in Fig. 1.4A, we can write L

dii ðtÞ die ðtÞ 1 Rii ðtÞ 1 Le 1 Re ie ðtÞ 5 υi ðtÞ i 5 r; y; b dt dt

Substituting i 5 r, y, b in Eq. (1.7) and adding the three equations, we obtain

(1.7)

Introduction to power system faults

15

(A) r R

+ – Vb (t)

ir (t)

L

Vr (t) – Vy (t) +

–

+

iy (t) R

y

L ib (t)

R

L

Re

Le

b

e

ie (t)

(B) 80

80

Phase R

60

60

40

40

20

20

0

Phase Y

0 0

10

20

30

40

50

60

70

80

90 100 110 120

0

–20

–20

–40

–40

–60

ac

dc

80

asymmetric

10

–60

60

60

40

40

20

20

0

30

40

50

ac

80

Phase B

20

60

70

dc

80

90 100 110 120

asymmetric

Asymmetric 3-Phase Short-Circuit Currents

0 0

10

20

30

40

50

60

70

80

90 100 110 120

0

–20

–20

–40

–40

–60

10

20

30

40

50

60

70

80

90 100 110 120

–60

ac

dc

asymmetric

Phase Y

Phase B

Phase R

Figure 1.4 (A) Basic balanced three-phase electric circuit with earth return and a solid three-phase short-circuit fault. (B) Three-phase short-circuit current waveforms.

16

Power Systems Modelling and Fault Analysis

L

   d ir ðtÞ 1 iy ðtÞ 1 ib ðtÞ 1 R ir ðtÞ 1 iy ðtÞ 1 ib ðtÞ dt 1 3Le

die ðtÞ 1 3Re ie ðtÞ 5 υr ðr Þ 1 υy ðtÞ 1 υb ðtÞ dt

(1.8)

Since the three-phase voltage sources are balanced we have υr ðtÞ 1 υy ðtÞ 1 υb ðtÞ 5 0

(1.9)

Also, from Fig. 1.4A, we have ir ðtÞ 1 iy ðtÞ 1 ib ðtÞ 5 ie ðtÞ

(1.10)

Therefore, substituting Eqs. (1.9) and (1.10) into Eq. (1.8), we obtain ðL 1 3Le Þ

die ðtÞ 1 ðR 1 3Re Þie ðtÞ 5 0 dt

(1.11)

The solution of Eq. (1.11) is given by 2

3

ie ðtÞ 5 K 3 exp4

2t

L 1 3Le R 1 3Re

5

(1.12)

where K is a constant that satisfies the initial conditions. Since the three-phase system is symmetrical and balanced, ie (t 5 0) 5 0. Thus, Eq. (1.12) gives K 5 0 and ie (t) 5 0. That is, following a simultaneous three-phase short circuit, no current will flow in the earth return connection and the three fault currents ii (t) will flow independently as in single-phase circuits. Therefore, with ie (t) 5 0, the solution of Eq. (1.7) is given by ( "

# pffiffiffi ωL ii ðtÞ 5 2Irms sin ωt 1 ϕi 2 tan21 R " " #)

# 2t 21 ωL  2 sin ϕi 2 tan 3 exp  R L=R

(1.13)

where Vrms Irms 5 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R2 1 ðωLÞ2

(1.14)

Introduction to power system faults

17

Eq. (1.13) can be written as the sum of an ac component and a unidirectional dc component as follows: ii ðtÞ 5 iiðacÞ ðtÞ 1 iiðdcÞ ðtÞ

i 5 r; y; b

(1.15)

where iiðacÞ ðtÞ 5

 pffiffiffi ωL 2Irms sin ωt 1 ϕi 2 tan21 R

(1.16)

and   pffiffiffi 2t 21 ωL 3 exp iiðdcÞ ðtÞ 5 2 2Irms sin ϕi 2 tan R ðL=RÞ

(1.17)

Fig. 1.4B shows the short-circuit currents for each phase, including their ac and dc components for a fault in a typical 132-kV system where Irms 5 30 kA, ωL/R 5 20 and ϕr 2 tan21 ωL R 5 2 π=2 to give maximum dc current offset on phase r.

One-phase short-circuit current Fig. 1.5A shows the case of a solid one-phase to earth short-circuit fault from phase r to earth e at t 5 0, we can write L

dir ðtÞ die ðtÞ 1 Rir ðtÞ 1 Le 1 Re ie ðtÞ 5 υr ðtÞ dt dt

ir ðtÞ 5 ie ðtÞ and iy ðtÞ 5 ib ðtÞ 5 0

(1.18) (1.19)

Substituting Eq. (1.19) in Eq. (1.18), we obtain ðL 1 Le Þ

dir ðtÞ 1 ðR 1 Re Þir ðtÞ 5 υr ðtÞ dt

(1.20)

The solution of Eq. (1.20) is given by ( "

# pffiffiffi ωðL 1 L Þ e ir ðtÞ 5 2Irms sin ωt 1 ϕr 2 tan21 R 1 Re 3 2 " 7) 6

# 7 6 2 t 21 ωðL 1 Le Þ

7 2 sin ϕr 2 tan 3 exp6 7 6 R 1 Re 4 L 1 Le 5 R 1 Re

(1.21)

18

Power Systems Modelling and Fault Analysis

(A)

Vb (t) +

+ Vr (t) – –Vy (t) +

–

r

R

L

R

L

y

R

L

b

Re

Le

ir (t)

e ie (t)

(B) 100

One-phase short-circuit current

80 60 40 20 0 0

10

20

30

40

50

60

70

80

90

100

110

120

–20 –40 –60

ac

dc

asymmetric

Figure 1.5 (A) Basic balanced three-phase electric circuit with earth return and a solid onephase short-circuit fault. (B) One-phase short-circuit current waveforms.

where Vrms Irms 5 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðR1Re Þ2 1 ½ωðL1Le Þ2

(1.22)

Eq. (1.22) can be written as the sum of an ac component and a unidirectional dc component as follows: ir ðtÞ 5 irðacÞ ðtÞ 1 irðdcÞ ðtÞ

(1.23)

Introduction to power system faults

19

where irðacÞ ðtÞ 5



pffiffiffi ωðL 1 Le Þ 2Irms sin ωt 1 ϕr 2 tan21 R 1 Re

(1.24a)

and 2 3 

pffiffiffi ωðL 1 L Þ 2 t e 5 irðdcÞ ðtÞ 5 2 2Irms sin ϕr 2 tan21 3 exp4 L 1 Le R 1 Re

(1.24b)

R 1 Re

Fig. 1.5B shows the short-circuit current including the ac and dc components for a fault in a typical  132-kV system where Irms 5 36 kA, ω(L 1 Le)/(R 1 Re) 5 14 and and ϕr 2 tan21 ωL R 5 2 π=2 to give maximum dc current offset. The initial magnitude of the dc current component in any phase depends on the instant on the voltage waveform when the short circuit occurs, that is, on ϕr and on the magnitude of the ac current component Irms. The rate of decay of the dc current component in the three phases depends on the circuit time constant L/R or circuit X/ R ratio where X/R 5 ωL/R in the case of a three-phase fault, or (L 1 Le)/(R 1 Re) or equivalent circuit X/R ratio where X/R 5 ω(L 1 Le)/(R 1 Re) in the case of a onephase to earth fault. In the above analysis of three-phase and one-phase short-circuit faults, the magnitude of the ac current component is constant because we assumed that the source inductance L is constant or time independent. Also, this assumption results in a time-independent X/R ratio or constant rate of decay of dc current component. This assumption is only generally valid if the location of the short-circuit fault is electrically remote from electrical machines feeding short-circuit current into the fault. This aspect is discussed in detail later in this book where practical time-dependent source inductance is considered. In some power system analysis, the elements R, L, Re and Le are not immediately known and have to be derived from knowledge of the circuit or equivalent source positive sequence and zero sequence impedances ZP and ZZ. Using Eq. (2.51) with ZF 5 0, it can easily be shown that Z 5 R 1 jωL 5 Z P and Ze 5 Re 1 jωLe 5

ZZ 2 ZP 3

(1.25)

1.5.2 Terminology of short-circuit current interruption Short-circuit currents are detected by protection relays which initiate the interruption of these currents by circuit-breakers. Fig. 1.6 shows a general asymmetrical short-circuit current waveform and the terminology used to describe the various current components as well as the short-circuit current interruption.

20

Power Systems Modelling and Fault Analysis

(A) A"

B" 2 2I″k A' B' A t

B tB

tA ∆t2

∆t1 tF

∆t3 t

t1

t2

t3

Time sequence to circuit-breaker current interruption (B)

Contacts and arc-quenching mechanism

Figure 1.6 (A) Short-circuit current interruption process. (B) Modern 420-kV open-terminal SF6 circuit breaker. (C) Modern 132-kV three-phase metal enclosed gas insulated switchgear (GIS).

Introduction to power system faults

21

(C)

Figure 1.6 (Continued).

From Fig. 1.6, the following quantities and terms are defined: tF 5 Instant of short-circuit fault. Δt1 5 Protection relay time. tA 5 Instant of ‘initial peak’ of short-circuit current. t1 5 Instant of energisation of circuit-breaker trip circuit. Δt2 5 Circuit-breaker opening time. t2 5 Instant of circuit-breaker contact separation 5 instant of arc initiation. Δt3 5 Circuit-breaker current arcing time. t3 5 Instant of final arc extinction 5 instant of short-circuit current interruption. tBp5ffiffiffi Instant of peak of major current loop just before current interruption. 2 2Ik00 5 2:828Ik00 5 Theoretical current at the instant of short-circuit fault tF where Ik00 is the rms short-circuit current at t 5 tF.

22

Power Systems Modelling and Fault Analysis

AAv 5 ‘Initial peak’ short-circuit current at t 5 tA and is denoted ip. ip is also termed ‘1/2 cycle peak’ current, ‘peak make’ current or ‘making’ current. AAv 5 AA0 1 A0 Av where 0 AA0 5 magnitude of dc current pffiffiffi component at t 5 tA and A Av 5 peak ac current component at t 5 tA and is equal to 2 3 rms current at t 5 tA. BBv 5 Peak short-circuit current at t 5 tB and is denoted ib. BBv 5 BB0 1 B0 Bv where 0 BB0 5 magnitude of dc current pffiffiffi component at t 5 tB and B Bv 5 peak ac current component at t 5 tB and is equal to 2 3 Ib , where Ib is rms current at t 5 tB.

Percentage dc current component at t 5 tB is given by (BB0 3 100)/B0 Bv.

1.6

Effects of short-circuit currents on equipment

1.6.1 Thermal effects Short-circuit currents flowing through the conductors of various power system equipment create thermal effects on conductors and equipment due to heating and excess energy input over time as measured by I2T, where I is the short-circuit current magnitude and T is the short-circuit current duration. Because the short-circuit duration, including short circuits cleared by protection in back-up clearance times, is quite short, the heat loss from conductors during the short circuit is usually very low. Generally, both ac and dc components of the short-circuit current contribute to the thermal heating of conductors. Extreme values of the time constant of the dc short-circuit current component may be up to 200 ms, so that by 600 ms from the instant of short circuit, the dc component is nearly vanished and nearly all its generated heat will have dissipated by around 1 seconds. For the ac current component, the heat dissipation depends on the ratio of the initial subtransient rms current to steady-state rms current. For a typical ratio of about 2.5, the amount of conductor heat dissipation will be around 50% at 1 second and nearly 65% at 3 seconds. The three-phase short-circuit fault normally gives rise to the highest thermal effect on equipment. For cables of larger conductor sizes, the thermal limit is usually imposed by the sheath/armour as opposed to the core conductor under short-circuit earth fault conditions. During the making of high shortcircuit currents by inadequately rated circuit-breakers, their contacts may weld together encouraged by a prestriking flashover arc and possibly contact popping.

1.6.2 Mechanical effects Short-circuit currents flowing through the conductors of various power system equipment create electromagnetic forces and mechanical stresses on equipment such as overhead line conductors, busbars and their supports, cables and transformer windings. Mechanical forces on transformer windings are both radial and axial. The radial force is a repulsion force between the inner and outer windings and tends to crush the inner winding towards the transformer core and burst the outer winding outwards. The axial force tends to displace the windings, or part of the winding, with respect to each other. The transformer windings must be designed to withstand

Introduction to power system faults

23

L

a i2(t)

i1(t)

d

d i1(t)

F

F

Currents flowing in same direction

i2(t)

d i1(t) F

Currents flowing in opposite direction

i2(t) F

Figure 1.7 Electromagnetic forces on two parallel round current-carrying conductors.

the mechanical forces created by the short-circuit currents. The electromagnetic forces produced by short-circuit currents in three-core unarmoured cables tend to repel the cores from each other and could burst the cable altogether, leading to insulation damage if the cores are not adequately bound. The electromagnetic forces acting on two parallel, round conductors carrying currents i1 and i2, as shown in Fig. 1.7, are given by 2sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 μo L i1 ðtÞi2 ðtÞ4 F ðtÞ 5 1 1 2 15N d 2π

(1.26)

where d is the distance between the centres of the conductors in m and L is the conductor length between supports in m. μo 5 4π1027 H=m is the permeability of vacuum. Since i1(t) and i2(t) are instantaneous currents in amps as given in Eq. (1.13), the electromagnetic force F(t) is clearly time-dependent. In the general case, F(t) will contain a dc component, a power frequency component and a double-power frequency component, that is, a 100-Hz component in a 50-Hz system. In practical installations, L . . d and d . . a where a is the conductor diameter in m. Therefore, Eq. (1.26) reduces to

24

Power Systems Modelling and Fault Analysis

F ðtÞ 5

μo L i1 ðtÞi2 ðtÞ N d 2π

(1.27)

Where the currents flow in the two conductors in the same direction, the forces are compressive, that is, pulling the conductors together but if the currents are flowing in opposite directions, the forces would repel the conductors away from each other. This is illustrated in Fig. 1.7. To illustrate how this equation may be used, consider an overhead line with a twin conductor bundle per phase (overhead lines are discussed in detail in Chapter 3: Modelling of multiconductor overhead power lines, underground and submarine cables) and d 5 0.5 m. The short-circuit current satisfies Eq. (1.13) and has an rms value of 20 kA per phase and an X/R ratio equal to 15. The maximum attraction forces on each conductor per metre produced by the first peak of the asymmetrical short-circuit current are to be calculated. Individual conductor rms current is 20/2 5 10 kA. Using Eq. (1.13), the maximum value of the initial peak current is calculated assuming maximum dc current offset, which corresponds to   a short-circuit fault that occurs at a voltage phase angle ϕ 5 tan21 XR 2 π2. Therefore, using Eq. (1.13), the initial peak current is equal to ip1 5 ip2 5

h  2 πi pffiffiffi 2 3 10 3 1 1 exp 5 2:56 3 10 5 25:6 kA 15

The attraction forces on each conductor are calculated as F5

4π1027 ð25:6 3 103 Þ2 1 5 262 N=m 0:5 2π

The electromagnetic forces acting on three rigid, parallel and round conductors, shown in Fig. 1.8, under a balanced three-phase short-circuit fault condition, can be calculated approximately using the following equations:  μo ir ðtÞib ðtÞ L N ir ðtÞiy ðtÞ 1 Fr ðtÞ 5 2 d 2π

(1.28)

L

a ir (t)

ib(t)

iy(t) d

d

Figure 1.8 Electromagnetic forces on three-phase parallel round conductors.

Introduction to power system faults

L μo  N iy ðtÞib ðtÞ 2 iy ðtÞir ðtÞ d 2π  2 μo ib ðtÞir ðtÞ L Fb ðtÞ 5 1 ib ðtÞiy ðtÞ N 2 d 2π

Fy ðtÞ 5

25

(1.29)

(1.30)

In Eqs. (1.28)
(1.30), ir(t), iy(t) and ib(t) are the instantaneous currents given in Eq. (1.13) in amps. Substituting these currents into Eqs. (1.28)
(1.30), it can be shown that the maximum force occurs on the middle conductor and is given by pffiffiffi μo 3 2 L N Fy 5 i 2π 2 p d

(1.31)

where ip is the initial peak short-circuit current that corresponds to AAv in Fig. 1.6A. During a phase-to-phase short-circuit fault, it can be shown that the maximum electromagnetic force between the two faulted conductors is given by Fy 5

μo 2 L N i 2π p d

(1.32)

where ip is the initial peak short-circuit current. Where the conductors are not round, for example, rectangular, d in all the electromagnetic force equations is replaced by an effective distance deff equal to the effective geometric mean distance between the rectangular conductors.

1.7

Per-unit analysis of power systems

1.7.1 General Steady-state network analysis on initially balanced three-phase networks generally employs complex phasors and involves the calculation of active and reactive power flows, voltages and currents in the network. Usually, such calculations are carried out using per-unit values of actual physical quantities such as voltages, currents or impedances. In addition, some dynamic analysis that employ time domain rms quantities also use per-unit values. The advantages of per-unit systems of analysis are generally covered in most introductory power systems textbooks and will not be repeated here. However, as we will discuss later in this book, the use of actual physical units such as volts, amps, ohms and Siemens is more advantageous in multiphase network analysis.

1.7.2 Single-phase systems The per-unit system of analysis is based on the application of Ohm’s law to a single impedance or admittance as illustrated in Fig. 1.9A.

26

Power Systems Modelling and Fault Analysis

I

(A)

V

Z

I

Y V

(B) ILine

Iph

ILine Vph

Iph

Vph

Vph-ph

Vph-ph

Delta connection

Star connection

Figure 1.9 Per-unit analysis of (A) single-phase impedance and admittance and (B) threephase connections.

Using actual physical units of kilovolts, kiloamps, ohms and Siemens, the voltage drop across the impedance and injected current into the admittance are given by VActual;kV 5 ZActual;Ω 3 IActual;kA

(1.33a)

IActual;kA 5 YActual;S 3 VActual;kV

(1.33b)

where V, I, Z and Y are complex phasors representing actual physical quantities of voltage, current, impedance and admittance, respectively. To calculate a per-unit value for each of these quantities, a corresponding base quantity must be defined first. Let VBase,kV and IBase,kA be the base voltage and base current, respectively. Therefore, Eqs. (1.33a) and (1.33b) can be written as follows: VActual;kV ZActual;Ω I  3 Actual;kA 5 VBase;kV VBase;kV IBase;kA IBase;kA

Introduction to power system faults

27

and IActual;kA YActual;S V  3 Actual;kV 5 I Base;kA IBase;kA VBase;kV VBase;kV

or Vpu 5 Zpu 3 Ipu

(1.34a)

Ipu 5 Ypu 3 Vpu

(1.34b)

where Vpu 5

VActual;kV VBase;kV

Zpu 5

ZActual;Ω ZBase;Ω

ZBase;Ω 5

Ypu 5

YActual;S YBase;S

YBase;S 5

Ipu 5

IActual;kA IBase;kA

(1.35a)

and VBase;kV IBase;kA IBase;kA 1 5 ZBase;Ω VBase;kV

(1.35b)

(1.35c)

By defining VBase,kV and IBase,kA, ZBase,Ω and YBase,S are also defined as shown in Eqs. (1.35b) and (1.35c), and the per-unit values of the actual voltage and current are also defined in Eq. (1.35a). It should be noted that the base quantities VBase,kV and IBase,kA are defined as real numbers so that the phase angles of Vpu and Ipu remain unchanged from VActual,kV and IActual,kA, respectively. Using Eq. (1.35a), we have VActual;kV 3 IActual;kA 5 VBase;kV Vpu 3 IBase;kA Ipu or Vpu 3 Ipu 5

VActual;kV 3 IActual;kA VBase;kV 3 IBase;kA

or MVApu 5

MVAActual MVABase

(1.36a)

28

Power Systems Modelling and Fault Analysis

where MVAActual 5 VActual;kV 3 IActual;kA MVABase 5 VBase;kV 3 IBase;kA and MVApu 5 Vpu 3 Ipu

(1.36b)

We note that by defining VBase,kV and IBase,kA, MVABase is also defined according to Eq. (1.36b). In practical power system analysis, it is more convenient to define or choose MVABase and VBase,kV and calculate IBase,kV if required. Therefore, using Eqs. (1.35a) and (1.36b), we have IActual;kA V  and Vpu 5  Actual;kA Ipu 5  MVABase VBase;kV

MVABase IBase;kA

(1.37)

Also, using Eqs. (1.28) and (1.36b), we can write 2 VBase;kV VBase;kV VBase;kV 3 5 and IBase;kA VBase;kV VBase;kV IBase;kA 1 VBase;kV IBase;kA YBase;S 5 5 2 ZBase;Ω VBase;kV

ZBase;Ω 5

or using Eq. (1.35b), we have ZBase;Ω 5

2 VBase;kV MVABase and YBase;S 5 2 MVABase VBase;kV

(1.38)

Substituting Eq. (1.37) in Eqs. (1.35a) and (1.35b), we obtain ZActual;Ω YActual;S

 and Ypu 5 Zpu 5  V 2 Base;kV

MVABase

(1.39)

MVABase 2 VBase;kV

To convert per-unit values to per cent, the per-unit values are multiplied by 100.

1.7.3 Change of base quantities The per-unit values that are calculated by generator manufacturers, transformer and reactor manufacturers are usually based on the rated voltage and rated apparent power of the equipment. However, transmission and distribution system analysis using per-unit values is based on a single MVA base of typically 100 MVA. In industrial power systems and power station auxiliary supply systems, 10 MVA base is generally found to be more convenient. For transformers, it is often the case that the rated voltage of at least one of the windings is not equal to the base voltage of the network to which the winding is connected. This is discussed in detail in Chapter 4, Modelling of transformers, phase shifters, static power plant and static

Introduction to power system faults

29

load. To determine how, in general, per-unit values are calculated using new base quantities, let the two base quantities be MVABase-1, VBase-1,kV and MVABase-2, VBase-2,kV, and it is required to convert from Base-1 to the new Base-2. Using Actual;kV Actual;kV Eq. (1.35a), we can write Vpu
1 5 VVBase
1;kV and Vpu
2 5 VVBase
2;kV giving Vpu
2 5

VBase
1;kV 3 Vpu
1 VBase
2;kV

(1.40)

Using Eq. (1.37), we have Ipu
1 5 

IActual;kA

MVABase
1 VBase
1;kV

 and Ipu
2 5 

IActual;kA

MVABase
2 VBase
2;kV



giving Ipu
2 5

VBase
2;kV MVABase
1 3 3 Ipu
1 VBase
1;kV MVABase
2

(1.41a)

or using Eq. (1.36b), we have Ipu
2 5

IBase
1;kV 3 Ipu
1 IBase
2;kV

(1.41b)

From Eq. (1.39), we have ZActual;Ω ZActual;Ω  and Zpu
2 5  V 2  Zpu
1 5  V 2 Base
1;kV

Base
2;kV

MVABase
1

or 2 VBase
1;kV Zpu
2 5 2 VBase
2;kV

MVABase
2

!

MVABase
2 Zpu
1 MVABase
1

and similarly for the per-unit admittance !

2 VBase
2;kV MVABase
1 Ypu
2 5 Ypu
1 2 MVABase
2 VBase
1;kV

(1.42)

(1.43)

1.7.4 Three-phase systems The above analysis of single-phase systems can be easily extended to three-phase systems or equipment using the following basic relations that apply irrespective of whether the equipment is star or delta connected

30

Power Systems Modelling and Fault Analysis

MVA3
Phase 5 3 3 MVA1
Phase

(1.44a)

and MVA3
Phase 5

pffiffiffi 3 3 VPhase2Phase;kV 3 ILine;kA

(1.44b)

where MVA3-Phase is three-phase base apparent power, MVA1-Phase is one-phase base apparent power, VPhase
Phase,kV is phase-to-phase base voltage and ILine,kA is base line current. The relationships between base voltages and base currents for star- and delta-connected three-phase equipment is shown in Fig. 1.9B: Star : VPhase2Phase;kV 5

pffiffiffi 3 3 VPhase;kV and ILine;kA 5 IPhase;kA

Delta : VPhase2Phase;kV 5 VPhase;kV and ILine;kA 5

pffiffiffi 3 3 IPhase;kA

(1.45a) (1.45b)

where VPhase,kV and IPhase,kA correspond to the base voltage and current quantities as used in Section 1.7.2 for a single-phase system. Using Eqs. (1.44a) and (1.45a) in Eqs. (1.37), (1.39), (1.40), (1.41b), (1.42) and (1.43), we obtain Zpu 5 V 2

ZActual;Ω

ðPhase2PhaseÞBase;kV



Ypu 5

MVAð32PhaseÞBase

Vpu 5

VðPhase2PhaseÞActual;kV   MVAð3
PhaseÞBase pffiffi

Ipu 5 

3IðLineÞBase;kA

Vpu
2 5

YActual;S



(1.46a)

MVAð32PhaseÞBase 2 VðPhase2PhaseÞBase;kV

IðLineÞActual;kA

MVAð3
PhaseÞBase pffiffi 3VðPhase2PhaseÞBase;kV

VðPhase2PhaseÞBase
1;kV 3 Vpu
1 VðPhase2PhaseÞBase
2;kV

Ipu
2 5



(1.46b)

ILineðBase
1Þ;kA 3 Ipu
1 ILineðBaseÞ
2;kA (1.46c)

!

MVAð32PhaseÞBase
2 Zpu
2 5 Zpu
1 2 MVAð32PhaseÞBase
1 VðPhase2PhaseÞBase
2;kV 2 VðPhase2PhaseÞBase
1;kV

!

MVAð32PhaseÞBase
1 Ypu
2 5 Ypu
1 2 MVAð32PhaseÞBase
2 VðPhase2PhaseÞBase
1;kV 2 VðPhase2PhaseÞBase
2;kV

(1.46d)

(1.46e)

1.7.5 Mutually coupled systems having different operating voltages In many electrical power systems, double-circuit overhead lines are used for the transport of power. The modelling of these lines is presented in detail in Chapter 3,

Introduction to power system faults

31

ZS1

I1

Circuit 1 V1

V '1 ZM I2

Circuit 2

V2

ZS2

V '2

Figure 1.10 Per-unit analysis of two mutually coupled inductive circuits having different design and operating voltages.

Modelling of multiconductor overhead power lines, underground and submarine cables. For now, we state that these lines consist of two three-phase circuits strung on the same tower. In this section, we present the general case of the derivation of the per-unit values of the mutual impedance and mutual susceptance of two mutually coupled circuits operating at different voltages. The analysis is general and applies also to mutually coupled cable circuits and mutually coupled transformer windings.

Base and per-unit values of mutual inductive impedance Fig. 1.10 shows two mutually coupled inductive circuits having different design and operating voltages. The series voltage drop across each circuit is given by V1 2 V 01 5 ZS1 I1 1 ZM I2

(1.47a)

V2 2 V 02 5 ZS2 I2 1 ZM I1

(1.47b)

where V is in volts, I is in amps, ZS1 and ZS2 are the self-impedances of circuits 1 and 2, respectively, in ohms, and ZM is the mutual impedance between the two circuits in ohms. Using the same base apparent power for both circuits, let the base quantities of circuits 1 and 2 be defined as follows: V1ðpuÞ 5

V1

I1ðpuÞ 5

I1

V1ðBaseÞ ZS1 ZS1ðpuÞ 5 I1ðBaseÞ ZS1ðBaseÞ V2ðBaseÞ ZS2 ZS2ðBaseÞ 5 ZS2ðpuÞ 5 I2ðBaseÞ ZS2ðBaseÞ ZS1ðBaseÞ 5

V1ðBaseÞ I1ðBaseÞ V2 I2 V2ðpuÞ 5 I2ðpuÞ 5 V2ðBaseÞ I2ðBaseÞ SðBaseÞ 5 V1ðBaseÞ I1ðBaseÞ 5 V2ðBaseÞ I2ðBaseÞ

(1.48) We note that, for now, we have not yet defined the base or per-unit values for ZM. Using Eq. (1.48) and substituting the actual quantities, for example, V1 5 V1 (pu) 3 V1(Base), into Eqs. (1.47a) and (1.47b) we obtain

32

Power Systems Modelling and Fault Analysis

h i V1ðBaseÞ V1ðBaseÞ V1ðpuÞ 2 V 01ðpuÞ 5 ZS1ðpuÞ I1ðBaseÞ I1ðpuÞ 1 ZM I2ðBaseÞ I2ðpuÞ I1ðBaseÞ

(1.49a)

h i V2ðBaseÞ V2ðBaseÞ V2ðpuÞ 2 V 02ðpuÞ 5 ZS2ðpuÞ I2ðBaseÞ I2ðpuÞ 1 ZM I1ðBaseÞ I1ðpuÞ I2ðBaseÞ

(1.49b)

and

Dividing Eq. (1.49a) by V1(Base) and Eq. (1.49b) by V2(Base), we obtain V1ðpuÞ 2 V 01ðpuÞ 5 ZS1ðpuÞ I1ðpuÞ 1 ZM

I2ðBaseÞ I2ðpuÞ V1ðBaseÞ

(1.50a)

V2ðpuÞ 2 V 02ðpuÞ 5 ZS2ðpuÞ I2ðpuÞ 1 ZM

I1ðBaseÞ I1ðpuÞ V2ðBaseÞ

(1.50b)

and

or V1ðpuÞ 2 V 01ðpuÞ 5 ZS1ðpuÞ I1ðpuÞ 1 ZMðpuÞ I2ðpuÞ

(1.51a)

V2ðpuÞ 2 V 02ðpuÞ 5 ZS2ðpuÞ I2ðpuÞ 1 ZMðpuÞ I1ðpuÞ

(1.51b)

From Eqs. (1.50a), (1.50b) and (1.51a), (1.51b) the per-unit value of ZM is defined as ZMðpuÞ 5

ZM ZM Z 5 M  5 V1ðBaseÞ V2ðBaseÞ ZMðBaseÞ I2ðBaseÞ

(1.52a)

I1ðBaseÞ

And the base impedance of ZM is defined as ZMðBaseÞ 5

V1ðBaseÞ V2ðBaseÞ 5 I2ðBaseÞ I1ðBaseÞ

(1.52b)

From Eq. (1.48), S(Base) 5 V1(Base)I1(Base) 5 V2(Base)I2(Base) or I2ðBaseÞ 5

SðBaseÞ SðBaseÞ and I1ðBaseÞ 5 V2ðBaseÞ V1ðBaseÞ

(1.53)

Substituting Eq. (1.53) into Eq. (1.52b), and using S(Base) in MVA, V1(Base) and V2 (Base) in kV, we obtain

Introduction to power system faults

ZMðBaseÞ
Ω 5

33

V1ðBaseÞ
kV V2ðBaseÞ
kV SðBaseÞ
MVA

(1.54)

Substituting Eq. (1.54) into Eq. (1.52a), we have ZMðpuÞ 5 

ZMðΩÞ  V1ðBaseÞ
kV V2ðBaseÞ
kV SðBaseÞ
MVA

(1.55)

In the above equations, if the voltages used are phase to earth then the apparent power is single-phase power but if the voltages used are phase to phase, then the apparent power to be used is the three-phase value. An alternative and equivalent approach is to define a per-unit mutual impedance on the base of each circuit as follows: ZM1ðpuÞ 5 

ZMðΩÞ  V1ðBaseÞ
kV I1ðBaseÞ
kA

and ZM2ðpuÞ 5 

ZMðΩÞ  V2ðBaseÞ
kV I2ðBaseÞ
kA

(1.56)

now ZMðΩÞ  V1ðBaseÞ
kV I1ðBaseÞ
kA

ZM1ðpuÞ 3 ZM2ðpuÞ 5 

ZMðΩÞ  V2ðBaseÞ
kV I2ðBaseÞ
kA

3

ZMðΩÞ  V1ðBaseÞ
kV I2ðBaseÞ
kA

5

ZMðΩÞ  V2ðBaseÞ
kV I1ðBaseÞ
kA

3

2 and using Eq. (1.52b), we obtain ZM1ðpuÞ 3 ZM2ðpuÞ 5 ZMðpuÞ ZMðpuÞ 5 ZMðpuÞ or

ZMðpuÞ 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ZM1ðpuÞ 3 ZM2ðpuÞ

(1.57)

Eqs. (1.55) and (1.57) are identical.

Base and per-unit values of mutual capacitive admittance Fig. 1.11 shows two mutually coupled capacitive circuits having different design and operating voltages. Using self and mutual admittance terms to maintain generality, the injected current into each circuit at one end of the circuit is given by I1 5 YS1 V1 1 YM ðV1 2 V2 Þ 1 J1

(1.58a)

I2 5 YS2 V2 1 YM ðV2 2 V1 Þ 1 J2

(1.58b)

where V is in volts, I is in amps, YS1 and YS2 are the self-admittances of circuits 1 and 2 in Siemens, respectively, and YM is the mutual admittance between the two

34

Power Systems Modelling and Fault Analysis

I1

J1

Circuit 1 V1 YS1

YM

I2 J2

Circuit 2 V2 YS2

Figure 1.11 Per-unit analysis of two mutually coupled capacitive circuits having different design and operating voltages.

circuits in Siemens. Using the same base apparent power for both circuits, let the base quantities of circuits 1 and 2 be defined as follows: V1ðpuÞ 5

V1

I1ðpuÞ 5

I1

I1ðBaseÞ YS1 YS1ðpuÞ 5 V1ðBaseÞ YS1ðBaseÞ I2ðBaseÞ YS2 YS2ðBaseÞ 5 YS2ðpuÞ 5 V2ðBaseÞ YS2ðBaseÞ YS1ðBaseÞ 5

V1ðBaseÞ I1ðBaseÞ V2 I2 V2ðpuÞ 5 I2ðpuÞ 5 V2ðBaseÞ I2ðBaseÞ SðBaseÞ 5 V1ðBaseÞ I1ðBaseÞ 5 V2ðBaseÞ I2ðBaseÞ

(1.59) We note that, for now, we have not yet defined the base or per-unit values for YM. Using Eq. (1.59) in Eqs. (1.58a) and (1.58b), we obtain I1ðBaseÞ I1ðpuÞ 5 ½YS1ðpuÞ YS1ðBaseÞ 1 YM V1ðpuÞ V1ðBaseÞ 2 YM V2ðpuÞ V2ðBaseÞ 1 I1ðBaseÞ J1ðpuÞ

(1.60a)

I2ðBaseÞ I2ðpuÞ 5 ½YS2ðpuÞ YS2ðBaseÞ 1 YM V2ðpuÞ V2ðBaseÞ 2 YM V1ðpuÞ V1ðBaseÞ 1 I2ðBaseÞ J2ðpuÞ

(1.60b)

and

Dividing Eq. (1.60a) by I1(Base) and Eq. (1.60b) by I2(Base), we obtain 2

3 YM 5  V1ðpuÞ I1ðBaseÞ V1ðBaseÞ

I1ðpuÞ 5 4YS1ðpuÞ 1 

YM  V2ðpuÞ I1ðBaseÞ V2ðBaseÞ

2

1 J1ðpuÞ

(1.61a)

Introduction to power system faults

35

and 2

3

I2ðpuÞ 5 4YS2ðpuÞ 1 

YM

I2ðBaseÞ V2ðBaseÞ

YM  V1ðpuÞ I2ðBaseÞ V1ðBaseÞ

5V2ðpuÞ 2 

1 J2ðpuÞ

(1.61b)

or I1ðpuÞ 5 ½YS1ðpuÞ 1 YM1ðpuÞ V1ðpuÞ 2 YMðpuÞ V2ðpuÞ 1 J1ðpuÞ

(1.62a)

I2ðpuÞ 5 ½YS2ðpuÞ 1 YM2ðpuÞ V2ðpuÞ 2 YMðpuÞ V2ðpuÞ 1 J2ðpuÞ

(1.62b)

and

where YMðpuÞ 5 

YM  I1ðBaseÞ V2ðBaseÞ

5

YM  I2ðBaseÞ V1ðBaseÞ

5

YM YMðBaseÞ

(1.63a)

and the base admittance of YM is given by YMðBaseÞ 5

I1ðBaseÞ I2ðBaseÞ 5 V2ðBaseÞ V1ðBaseÞ

(1.63b)

also YM1ðpuÞ 5 

YM  I1ðBaseÞ V1ðBaseÞ

YM2ðpuÞ 5 

YM  I2ðBaseÞ V2ðBaseÞ

(1.64)

Using SBase 5 V1(Base)I1(Base) 5 V2(Base)I2(Base) in Eq. (1.63b), S(Base) in MVA, V1 (Base) and V2(Base) in kV, we obtain YMðBaseÞ
S 5

SðBaseÞ2MVA V1ðBaseÞ2kV V2ðBaseÞ2kV

(1.65)

Substituting Eq. (1.65) in Eq. (1.63a), we obtain YMðpuÞ 5 

YMðSÞ  SðBaseÞ2MVA V1ðBaseÞ2kV V2ðBaseÞ2kV

(1.66)

36

Power Systems Modelling and Fault Analysis

In the above equations, if the voltages are phase to earth then the apparent power is single-phase power, but if the voltages used are phase to phase, then the apparent power to be used is the three-phase value. From Eq. (1.64), we have YM  I1ðBaseÞ
kA V2ðBaseÞ
kV

YM1ðpuÞ 3 YM2ðpuÞ 5 

YM  I2ðBaseÞ
kV V1ðBaseÞ
kA

3

2 5 YMðpuÞ YMðpuÞ 5 YMðpuÞ

or YMðpuÞ 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi YM1ðpuÞ 3 YM2ðpuÞ

(1.67)

The admittance of an overhead transmission line or underground or submarine cable is Y 5 G 1 jB but in practical power system analysis at power frequency, for example, 50 or 60 Hz, the conductance G is negligible. The mutual admittance is therefore represented as YM(pu) 5 jBM(pu), YM1(pu) 5 jBM1(pu) and YM2(pu) 5 jBM2(pu) where B is shunt susceptance. Therefore, Eq. (1.67) reduces to jBMðpuÞ 5 j

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi BM1ðpuÞ 3 BM2ðpuÞ

(1.68)

1.7.6 Examples Example 1.1 The measured series leakage impedance of a three-phase, 1500-MVA, 400-kV current-limiting series reactor is Z 5 (0.08 1 j16) Ω/phase. Calculate the series impedance in pu and percent on 100 MVA base. Base impedance 5 (400 kV)2/100 MVA 5 1600 Ω. Zpu 5 (0.08 1 j16) Ω/1600 Ω 5 (0.00005 1 j0.01) pu or (0.005 1 j1) % on 100 MVA base.

Example 1.2 Calculate the susceptance of a three-phase, 132-kV, 60-Mvar shunt capacitor in pu and % on 100 MVA base. Base admittance 5 100 MVA/(132 kV)2. From the basic reactive power equation Q 5 BV2, Susceptance B 5 60 Mvar/(132 kV)2. Bpu 5 (60/1322)/(100/1322) 5 60/100 5 0.6 pu or 60% on 100 MVA base.

Example 1.3 Fig. 1.12 shows two inductively coupled circuits having operating (and design) voltages of 275 and 400 kV phase to phase, carrying currents of 1 and 2 kA, and with actual voltage

Introduction to power system faults

37

magnitudes at end A of 270 and 406 kV, respectively. The self and mutual impedances of the circuits are shown in the figure. Calculate all quantities shown in the figure in per unit using a 100-MVA base and 275- and 400-kV base voltages for the circuits. In addition, calculate the induced voltages in each circuit in actual kV and in per-unit values. 1 kA Circuit 1 275 kV nominal voltage

(6+j30)Ω

270 kV (2.5+j10)Ω

2 kA Circuit 2 400 kV nominal voltage

End A

(3.5+j25)Ω

3 406 kV 3

Figure 1.12 Two inductively coupled circuits having different operating (and design) voltages for use in Example 1.3.

The per-unit voltages at end A are 270 kV 5 0.9818 pu and 406 kV/400 kV 5 1.015 pffiffikV/275 ffi pu. Circuit 1 base current is 100 MVA=ð 3 3 275 kVÞ 5 0:21 kA and circuit 1 current in per unit is pffiffiffi 1 kA/0.21 kA 5 4.76 pu. Circuit 2 base current is 100 MVA=ð 3 3 400 kVÞ 5 0:1443 kA and circuit 2 current in per unit is 2 kA/ 0.1443 kA 5 13.86 pu. The base impedance of circuit 1 is (275 kV)2/100 MVA 5 756.25 Ω. The per-unit value of circuit 1 self-impedance is (6 1 j30)Ω/756.25 Ω 5 (0.00793 1 j0.0396)pu or (0.793 1 j3.96)%. The base impedance of circuit 2 is (400 kV)2/100 MVA 5 1600 Ω. The per-unit value of circuit 2 self-impedance is (3.5 1 j25)Ω/1600 Ω 5 (0.0022 1 j0.0156)pu or (0.22 1 j1.56)%. The base impedance of mutual impedance is (400 3 275 kV)/100 MVA 5 1100 Ω and the per-unit value of mutual impedance is (2.5 1 j10)Ω/1100 Ω 5 (0.00227 1 j0.00909)pu or (0.227 1 j0.909)%. Alternatively, ð2:5 1 j10Þ ð2:5 1 j10Þ  and ZM2ðpuÞ 5   ZM1ðpuÞ 5  275 kV pffiffi pffiffi 400 kV 3 3 0:21 kA

3 3 0:1443 kA

and ZMðpuÞ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 ZM1ðpuÞ 3 ZM2ðpuÞ 5 ð0:00227 1 j0:00909Þpu or ð0:227 1 j0:909Þ%

Induced voltage in circuit 2 due to current pffiffiffi in circuit 1 is (2.5 1 j10)Ω 3 1 kA 5 (2.5 1 j10) kV or in per unit ð2:5 1 j10ÞkV=ð400 kV= 3Þ 5 ð0:0108 1 j0:0433Þpu. Alternatively, it can be calculated as (0.00227 1 j0.00909)pu 3 4.76 pu 5 (0.0108 1 j0.0433)pu. The induced per-unit voltage in circuit 1 due to current in circuit 2 can be similarly calculated.

38

Power Systems Modelling and Fault Analysis

Example 1.4 Fig. 1.13 shows two 400 and 275 kV capacitively coupled circuits. The actual voltages at end A, the self susceptance of each circuit and the mutual susceptance between the two circuits are as shown in the figure. Calculate all quantities shown in the figure in per-unit using a 100-MVA base. In addition, calculate the charging current of each circuit at end A including the current flowing through the mutual susceptance. All calculations are to be carried out in physical units and in per unit. End A

Circuit 1 275 kV nominal voltage

283 kV 3

j150 μS

j40 μS Circuit 2 400 kV nominal voltage

404 kV 3

j200 μS

Figure 1.13 Two capacitively coupled circuits having different operating (and design) voltages for use in Example 1.4. The per-unit voltages at end A are 283 kV/275 kV 5 1.029 pu and 404 kV/400 kV 5 1.01 pu. Circuit 1 base current is 0.21 kA and circuit 2 base current is 0.1443 kA. The base admittance of circuit 1 is 100 MVA/(275 kV)2 5 0.00132 S and per-unit value of self-susceptance is j150 3 1026 S/0.00132 S 5 j0.1136 pu or j11.36%. The base admittance of circuit 2 is 100 MVA/(400 kV)2 5 6.25 3 1024 S and per-unit value of self-susceptance is j200 3 1026 S/(6.25 3 1024 S) 5 j0.32 pu or j32%. The base value of mutual susceptance is 100 MVA/(400 kV 3 275 kV) 5 9.09 3 1024 S and per-unit value of mutual susceptance is j40 3 1026 S/(9.09 3 1024 S) 5 j0.044 pu or j4.4%. Alternatively, YM1ðpuÞ 5

j40 3 1026 S j40 3 1026 S 5 j0:0303 pu and YM2ðpuÞ 5 5 j0:064 pu 0:00132 S 6:25 3 1024 S

thus YMðpuÞ 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi j0:0303 3 j0:064 5 j0:044 pu or j4:4%

26 3 Circuit pffiffiffi 1 charging current due to self susceptance is equal to j150 3 10 S 3 (283 3 10 V/ 3) 5 j24.51 A or j24.51 A/210 A 5 j0.1167 pu. Similarly, circuit 2 charging current due to self susceptance is equal to j46.65 A or j46.65 A/144.3 A 5 j0.3230 pu. 26 Current flowing through S3 pffiffiffi the mutual susceptance is equal to j40 3 10 3 3 (404 3 10
283 3 10 )V/ 3 5 j2.79 A.

Introduction to power system faults

39

Charging current of circuit 1 is equal to j24.51A
j2.79 A 5 j21.72 A or j21.72 A/210 A 5 j0.103pu. Charging current of circuit 2 is equal to j46.65 A 1 j2.79 A 5 j49.44 A or j49.44 A/144.3 A 5 j0.3426pu.

Further reading Tleis, N.D., Travelling wave calculations, Technical Development, System Technical Branch, National Grid Company, London, November 1990. Warrington, A., Reactance relays negligibly affected by arc impedance, Electrical World, 502
505.

Symmetrical components analysis of faulted three-phase networks containing voltage and current sources

2

Chapter Outline 2.1 General 42 2.2 Symmetrical components of a three-phase power system 2.2.1 2.2.2 2.2.3 2.2.4 2.2.5 2.2.6 2.2.7 2.2.8

43

Symmetrical components of balanced three-phase voltage and current phasors 43 Symmetrical components of unbalanced voltage and current phasors 45 Apparent power in symmetrical component terms 48 Definition of phase sequence component networks of a three-phase power system containing balanced voltage sources 48 Sequence impedances of coupled unbalanced three-phase impedances 51 Sequence impedances of coupled balanced three-phase impedances 54 Advantages of symmetrical components frame of reference 55 Examples 55

2.3 Symmetrical components analysis of faulted networks containing balanced voltage sources only 58 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5 2.3.6 2.3.7 2.3.8 2.3.9

General 58 Balanced three-phase to earth short-circuit faults 58 Balanced three-phase clear of earth short-circuit faults 60 Unbalanced one-phase to earth short-circuit faults 62 Unbalanced phase-to-phase or two-phase short-circuit faults 64 Unbalanced two-phase to earth short-circuit faults 66 Unbalanced one-phase open-circuit faults 69 Unbalanced two-phase open-circuit faults 71 Example 73

2.4 Fault analysis and choice of reference frame

75

2.4.1 General 75 2.4.2 One-phase to earth short-circuit faults 75 2.4.3 Two-phase to earth short-circuit faults 76

2.5 Analysis of simultaneous faults 78 2.5.1 2.5.2 2.5.3 2.5.4 2.5.5 2.5.6

General 78 Simultaneous short-circuit faults at the same location 78 Cross-country faults or simultaneous faults at different locations 81 Simultaneous open-circuit and short-circuit faults at the same location 82 Simultaneous faults caused by broken and fallen to earth conductors 83 Simultaneous short-circuit and open-circuit faults on distribution transformers 84

Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00002-3 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

42

Power Systems Modelling and Fault Analysis

2.6 Symmetrical components analysis of faulted networks containing mixed voltage and current sources 88 2.6.1 2.6.2 2.6.3 2.6.4 2.6.5

General 88 Symmetrical components of voltage and current sources 89 Balanced three-phase to earth short-circuit faults 90 Unbalanced one-phase to earth short-circuit faults 91 Unbalanced two-phase short-circuit faults 95

Further reading

2.1

98

General

The analysis of three-phase AC power systems is greatly simplified if the system is assumed to be perfectly balanced. The three-phase system can then be modelled and analysed on a per-phase basis because knowledge of voltages and currents in one-phase allows us to calculate those in the other two phases where they differ by 6 120
phase displacement from those of the known phase. However, in practical three-phase power systems, there are a number of sources of unbalance which can be divided into internal and external sources. Internal sources of unbalance are those due to the inherent small differences in the phase impedances and susceptances of three-phase power plant such as overhead lines and transformers. Three-phase generators, however, produce, by design, a balanced set of three-phase voltages. External sources of unbalance are those that impose an unbalanced condition on the power system network such as the connection of unbalanced three-phase loads, or single-phase loads, to a three-phase power system, the occurrence of unbalanced short-circuit and open-circuit faults. These external conditions create unbalanced voltages and currents in the three-phase network, even if this network is balanced in terms of impedance and susceptance elements. The modelling and analysis of external balanced and unbalanced faults are the core topic of this chapter. The analysis of unbalanced short-circuit and open-circuit faults in practical power systems makes extensive use of the theory of symmetrical components. Fortescue proposed this general theory in a famous paper in 1918. It applies to a system of N unbalanced phasors, including the case of three unbalanced phasors representing three-phase power systems. The theory enables the transformation of three unbalanced phasors into a three set of balanced phasors called the positivephase sequence (positive sequence), negative-phase sequence (negative sequence) and zero-phase sequence (zero sequence) phasors. This property presents an extremely powerful analysis tool that enables the formation of three separate and uncoupled equivalent networks called the positive-sequence, negative-sequence and zero-sequence networks provided that the three-phase power system network in physical phase terms is internally balanced. The types of external unbalanced conditions imposed on the actual three-phase network determine the methods of connection of these sequence networks.

Symmetrical components analysis of faulted three-phase networks

43

This chapter introduces the theory of symmetrical components, the definitions of sequence networks, the methods of connecting these networks for different fault types, and the analysis of sequence and phase currents and voltages for various conditions of external unbalances that may be imposed on the three-phase power system network. The analysis includes traditional power systems with voltage sources only and modern power systems with both voltage and current sources; the latter representing certain types of new and renewable power generation technologies such as type 4 wind turbine generators and inverter-interfaced solar photovoltaic generators.

2.2

Symmetrical components of a three-phase power system

2.2.1 Symmetrical components of balanced three-phase voltage and current phasors Mathematically, balanced three-phase voltages (or currents) can be defined as complex instantaneous, real instantaneous or complex phasor quantities. Denoting the three phases as R, Y and B, three-phase complex instantaneous voltages are defined as follows: VR ðtÞ 5

pffiffiffi pffiffiffi 2Vrms exp½jðω t 1 φÞ 5 ð 2Vrms ejφ Þejωt 5 VR ejωt

(2.1a)

  pffiffiffi pffiffiffi 2π V Y ðtÞ 5 2Vrms exp j ω t 1 φ 2 5 ð 2Vrms ejðφ22π=3Þ Þejωt 5 VY ejωt 3 (2.1b)   pffiffiffi pffiffiffi 2π VB ðtÞ 5 2Vrms exp j ω t 1 φ 1 5 ð 2Vrms ejðφ12π=3Þ Þejωt 5 VB ejωt 3 (2.1c) where Vrms is the root mean square (rms) value of any of the three-phase voltage waveforms and φ is the phase angle that determines the magnitude of the threephase voltages at t 5 0. From Eqs. (2.1a)(2.1c), the three-phase real instantaneous voltages are defined as follows: υR ðtÞ 5 Real ½VR ðtÞ 5

pffiffiffi 2Vrms cosðωt 1 φÞ

  pffiffiffi 2π υY ðtÞ 5 Real ½VY ðtÞ 5 2Vrms cos ωt 1 φ 2 3

(2.2a) (2.2b)

44

Power Systems Modelling and Fault Analysis

υB ðtÞ 5 Real ½VB ðtÞ 5

  pffiffiffi 2π 2Vrms cos ωt 1 φ 1 3

(2.2c)

and the three-phase complex phasor voltages are defined as follows: VR 5

pffiffiffi 2Vrms ejφ

(2.3a)

VY 5

pffiffiffi 2Vrms ejðφ22π=3Þ

(2.3b)

VB 5

pffiffiffi 2Vrms ejðφ12π=3Þ

(2.3c)

pffiffiffi and the peak phase voltages are given by V^ R 5 V^ Y 5 V^ B 5 2Vrms . Eqs. (2.3a)(2.3c) show that for a balanced three-phase set of voltage or current phasors R, Y and B, all three phasors are equal in magnitude and are displaced from each other by 120
in phase. Fig. 2.1 illustrates Eqs. (2.1a)(2.1c) to (2.3a)(2.3c) in their instantaneous and phasor forms. We now define a new complex number h that has a magnitude of unity and a phase angle of 120
such that the effect of multiplying any phasor by h is to advance or rotate the original phasor counterclockwise by 120
, whilst keeping its magnitude unchanged. h is thus known as an operator and is given by

(B)

(A)

Imaginary Vˆ R

Vˆ R cos φ

t=0

Real

(D)

(C)

1

Y

B

R

Y

B

Phase rotation ωt

VB

120

0 –1

VR

o

0.5 –0.5

ωt

φ

t

φ

R

Vˆ R

o

0

10

20

30

40

120

o

120

Time (ms) VY

Figure 2.1 Instantaneous and phasor definitions of balanced three-phase quantities: (A) instantaneous; (B) phasor at t 5 0; (C) balanced set of three-phase voltages or currents with phase rotation RYB, YBR, BRY, etc. and (D) balanced set of three phasors at t 5 0.

Symmetrical components analysis of faulted three-phase networks

h5e

j2π=3

pffiffiffi 3 1 5 2 1j 2 2

h3 5 ej2π 5 1

2j2π=3

h 5e 2

pffiffiffi 3 1 5 2 2j 2 2

45

j5

pffiffiffiffiffiffiffiffi 21

1 1 h 1 h2 5 0

(2.4a) (2.4b)

Using the operator h, and phase R as the reference phasor, Eqs. (2.3b) and (2.3c) can be written as VY 5 h2VR and VB 5 hVR. Therefore, we can write the balanced three-phase voltages in vector form as follows: 2

3 2 3 VR VR V RYB 5 4 VY 5 5 4 h2 VR 5 VB hVR

(2.5)

where VRYB denotes a column vector. The above equation equally applies to threephase balanced currents, i.e., 2

3 2 3 IR IR IRYB 5 4 IY 5 5 4 h2 IR 5 IB hIR

(2.6)

2.2.2 Symmetrical components of unbalanced voltage and current phasors Consider a set of three-phase unbalanced voltage phasors VR, VY and VB that are unbalanced in both magnitude and phase as shown in Fig. 2.2A. According to the symmetrical components theory, each of these phasors can be decomposed into three balanced phasors known as the positive-sequence, negative-sequence and zero-sequence sets as shown in Fig. 2.2BD. The three positive-sequence phasors, shown in Fig. 2.2B, are equal in magnitude, displaced from each other by an equal phase of 120
and have the same phase sequence or rotation as the original unbalanced set, i.e., RYB, YBR, BRY, etc. The three negative-sequence phasors, shown in Fig. 2.2C, are equal in magnitude, displaced from each other by an equal phase of 120
and have a phase sequence or rotation that is opposite to that of the original unbalanced set, i.e., RBY, BYR, YRB, etc. The three zero-sequence phasors, shown in Fig. 2.2D, are equal in magnitude and are all in phase with each other. We choose the letters P, N, Z, as opposed to 1, 2 and 0, as superscripts to denote positive-sequence, negative-sequence and zero-sequence quantities, respectively. Therefore, the three unbalanced phasors can be written in terms of their three positive-sequence, negative-sequence and zero-sequence symmetrical components as follows: VR 5 VRP 1 VRN 1 VRZ

(2.7a)

46

Power Systems Modelling and Fault Analysis

(B)

+

P

VB

(A) 1

R

120

R Y

0.5

P

VR

o

Y

B

o

B

0 –0.5 0

120

0 –0.5 0

10

20

30

–1

VY

Time (ms) N

N VB

VR

P

VB

P

VR

Z VB

(C)

P

VY

B

R

Y

B

10

20

30

40

Time (ms)

1

_

Z VR

0.5

N

VB

VR

VB

Y

–1

P

40

R

0.5 o

120

1

N

VY

B

R

Y

R

B

Y

0

N VR

–0.5 0

VY

10

20

30

40

Time (ms)

–1 N

VY Z

VY

1

(D) 0

0.5 Z

VR Z VY Z

VB

RYB

0 –0.5 0 –1

10

20

30

40

Time (ms)

Figure 2.2 Unbalanced three-phase voltages and their symmetrical components: (A) unbalanced instantaneous voltages and their phasors; (B) balanced positive-sequence phasors; (C) balanced negative-sequence phasors and (D) zero-sequence phasors.

VY 5 VYP 1 VYN 1 VYZ

(2.7b)

VB 5 VBP 1 VBN 1 VBZ

(2.7c)

Using phase R as the reference phase, the h operator and Fig. 2.2BD, the following equations can be written: For phase Y VYP 5 h2 VRP VYN 5 hVRN VYZ 5 VRZ

For phase B VBP 5 hVRP VBN 5 h2 VRN VBZ 5 VRZ

(2.8)

Substituting Eq. (2.8) into Eqs. (2.7b) and (2.7c), we obtain VR 5 VRP 1 VRN 1 VRZ

(2.9a)

VY 5 h2 VRP 1 hVRN 1 VRZ

(2.9b)

VB 5 hVRP 1 h2 VRN 1 VRZ

(2.9c)

Symmetrical components analysis of faulted three-phase networks

47

or in matrix form 2

3 2 VR 1 4 V Y 5 5 4 h2 h VB where

2

1 H 5 4 h2 h

1 h h2

2 3 32 P3 VRP VR 1 6 7 6 7 1 5 4 VRN 5 5 H 4 VRN 5 1 VRZ VRZ

3 1 15 1

1 h h2

(2.10a)

(2.10b)

or VRYB 5 HVPNZ R

(2.11)

H is the transformation matrix that transforms positive-sequence, negative-sequence and zero-sequence quantities into their corresponding phase quantities. Conversely, if the unbalanced phase quantities are known, the phase R sequence quantities can be calculated from VPNZ 5 H21 VRYB R where

2 1 1 H21 5 4 1 3 1

h h2 1

(2.12a) 3 h2 h5 1

(2.12b)

It is interesting to note that H21 5

1 T ðH Þ 3

(2.13)

where  and T represent conjugate and transpose, respectively. Expanding Eq. (2.12a) using Eq. (2.12b), we obtain VRP 5

1 ðVR 1 hVY 1 h2 VB Þ 3

(2.14a)

VRN 5

1 ðVR 1 h2 VY 1 hVB Þ 3

(2.14b)

VRZ 5

1 ðVR 1 VY 1 VB Þ 3

(2.14c)

48

Power Systems Modelling and Fault Analysis

The positive-sequence, negative-sequence and zero-sequence components of phases Y and B phasors can be calculated using Eq. (2.8). The above analysis applies to an arbitrary unbalanced three-phase set of phasors so that currents as well as voltage phasors could have been used. When current phasors are used, Eqs. (2.11) and (2.12a) can be written as IRYB 5 HIPNZ R

(2.15a)

5 H21 IRYB IPNZ R

(2.15b)

2.2.3 Apparent power in symmetrical component terms In a three-phase power system, the total apparent power can be expressed in terms of actual phase voltages and currents as follows: 

S3Phase 5 P3Phase 1 jQ3Phase 5 VR

VY

VB



2

3 IR 3 4 IY 5 5 VTRYB 3 IRYB IB (2.16)

Substituting Eqs. (2.11) and (2.15a) into Eq. (2.16), we obtain T PNZ  S3Phase 5 P3Phase 1 jQ3Phase 5 ðH 3 VPNZ R Þ 3 ðH 3 IR Þ T T PNZ  5 ðVPNZ R Þ H 3 H  ðIR Þ

(2.17)

It can readily be shown that HTH 5 3U, where U is the unit matrix, therefore, Eq. (2.17) can be written as T PNZ  S3Phase 5 P3Phase 1 jQ3Phase 5 3ðVPNZ R Þ ðIR Þ

5 3 VRP ðIRP Þ 1 3 VRN ðIRN Þ 1 3 VRZ ðIRZ Þ

(2.18)

Z 5 3ðSPR 1 SN R 1 SR Þ

The total apparent power in the three-phase unbalanced system can be calculated as three times the sum of the positive-sequence, negative-sequence and zero-sequence apparent powers. It is interesting to note that only like sequence terms are involved in each multiplication and that no cross-sequence terms appear.

2.2.4 Definition of phase sequence component networks of a three-phase power system containing balanced voltage sources We have already established that the symmetrical component theory allows us to replace a three-phase set of unbalanced voltages (or currents) with three separate

Symmetrical components analysis of faulted three-phase networks

49

sets of balanced voltages (or currents) defined as the positive-sequence, negativesequence and zero-sequence sets. When an external unbalanced condition, such as a single-phase short-circuit fault, is imposed on the network, positive-sequence, negative-sequence and zero-sequence voltages and currents appear on the network at the point of fault. A phase sequence network is one that carries currents and voltages of one particular phase sequence such as positive-sequence, negativesequence or zero-sequence networks. It should be remembered that because the actual three-phase network is assumed perfectly balanced, the positive-sequence, negative-sequence and zero-sequence networks are separate, i.e., there is no intersequence mutual coupling between them, and they are only connected at the point of unbalance in the system, as will be seen later. In addition, the assumption of a perfectly balanced three-phase network means that positive-sequence voltage sources and positive-sequence current sources exist in the positive-sequence network only. Although the negative-sequence and zero-sequence networks still exist and can still be artificially constructed, they are totally redundant because they carry no negative-sequence or zero-sequence voltages or currents. Each sequence network contains terms of one sequence only and hence each one can be considered as an independent single-phase network. That is the active positive-sequence network contains only positive-sequence voltages, positivesequence currents and positive-sequence impedances/susceptances. The passive negative-sequence network contains only negative-sequence voltages, negativesequence currents and negative-sequence impedances/susceptances. The passive zero-sequence network contains only zero-sequence voltages, zero-sequence currents and zero-sequence impedances/susceptances. The negative-sequence and zero-sequence voltages and currents in the negative-sequence and zero-sequence networks appear as a result of the unbalanced condition imposed on the actual three-phase network; they should not be confused with the voltage and current sources that exist in the positive-sequence network. Let us consider first the type of unbalanced conditions that can appear in the three-phase network at a point F relative to the network neutral point N, as shown in Fig. 2.3A. The three sequence networks can then be constructed from the actual three-phase network components and network topology, and illustrated as shown in Fig. 2.3B. It should be noted that the derivation of sequence models of power system components is the subject of later chapters. For now, it suffices to state that the entire positive-sequence, negative-sequence and zero-sequence networks can each be reduced using The´venin’s theorem. The reduction results in a single equivalent voltage source at point F and a single The´venin’s equivalent impedance seen looking back into the relevant network from point F, as illustrated in Fig. 2.3B. Remembering that only positive-sequence voltages existed in the network prior to the occurrence of the unbalanced condition, the resultant positive-sequence, negative-sequence and zero-sequence equivalent circuits are illustrated in Fig. 2.3C. Modern power system networks with both positive-sequence voltage and current sources are dealt with in Section 2.6. From Fig. 2.3C, sequence voltage and current relations at the point of unbalance F can be written for each sequence network. For the active positive-sequence network:

50

Power Systems Modelling and Fault Analysis

(A)

FR

Balanced three-phase power system (generators, motors, network, load)

FY FB V F N

(B)

F

Active singlephase network

Z

P

IR Z

VF

P

VR N

Passive singlephase network

P

VR

NP Thevénin PPS equivalent circuit

Z

NPS network N

IR

Z

N

N

F

N

P

FP

P

+ –

F

P

PPS network (C)

P

N

VR

N

Passive singlephase network

N

ZPS network Z

IR

FN N

VR

NN Thévenin NPS equivalent circuit

Z

Z

Z

Z

Z Z

VR N

Z

FZ Z

VR

NZ Thévenin ZPS equivalent circuit

Figure 2.3 Definition of The´venin’s positive-sequence, negative-sequence and zerosequence sequence networks. (A) Three-phase system as seen from point F, (B) positive, negative and zero sequence representations, and (C) positive, negative and zero sequence Thevenin’s equivalent circuits.

VRP 5 VF 2 Z P IRP

(2.19)

where VF is the positive-sequence voltage at the point of fault F immediately before the unbalance condition is applied, VRP is the resultant positive-sequence voltage at the point of fault, and IRP is the positive-sequence current flowing out of the positive-sequence network into the point of fault. For the passive negative-sequence network: VRN 5 2 Z N IRN

(2.20)

where VRN is the negative-sequence voltage at the point of fault and IRN is the negative-sequence current flowing out of the negative-sequence network into the point of fault. For the passive zero-sequence network: VRZ 5 2 Z Z IRZ

(2.21)

where VRZ is the zero-sequence voltage at the point of fault and IRZ is the zerosequence current flowing out of the zero-sequence network into the point of fault. As will be seen in later chapters, the positive-sequence and negative-sequence impedances are the same for static power equipment but are generally different for rotating machines. In a zero-sequence network, because zero-sequence voltages and

Symmetrical components analysis of faulted three-phase networks

51

zero-sequence currents are cophasal, the zero-sequence currents can only flow if there is a return connection, through either a neutral or earth wire or the general body of the earth. This is a very important consideration when determining the zero-sequence equivalent circuits for transformers and the analysis of earth return currents as will be seen in Chapter 5, Modelling of rotating AC synchronous and induction machines, and Chapter 11, An introduction to the analysis of short-circuit earth-return current and rise of earth potential respectively.

2.2.5 Sequence impedances of coupled unbalanced three-phase impedances Fig. 2.4 shows a system of static unbalanced three-phase mutually coupled impedance elements. The voltage drop equations across phases R, Y and B, in matrix form, are given by R 3 2 VR R ZRR 4 VY 5 5 Y 4 ZYR B ZBR VB 2

Y ZRY ZYY ZBY

B 32 3 ZRB IR ZYB 5 4 IY 5 ZBB IB

(2.22a)

or in concise matrix notation form: VRYB 5 ZRYB IRYB

(2.22b)

where ZRYB is defined as the phase impedance matrix of the unbalanced threephase mutually coupled system shown in Fig. 2.4. The diagonal elements are the self-impedances of each phase and the off-diagonal elements are the mutual impedances between the phases. It is important to note that this is the impedance matrix in the actual or physical phase frame of reference, RYB, where each impedance is an actual physical element. What are the positive-sequence, negative-sequence and zero-sequence impedances of this system and what light do they shed on the use of the symmetrical components’ theory in practice? To answer this question, we substitute Eqs. (2.11) and (2.15a) into Eq. (2.22b), and obtain R

IR

IB

B

Z RB VR

VB

Y IY

Z RY

Z YB

VY

Figure 2.4 Unbalanced, static, inductive, mutually coupled three-phase impedances.

52

Power Systems Modelling and Fault Analysis

HVPNZ 5 ZRYB HIPNZ R R and premultiplying by H21, we obtain VPNZ 5 H21 ZRYB HIPNZ 5 ZPNZ IPNZ R R R

(2.23)

where ZPNZ is defined as the phase sequence impedance matrix of the unbalanced three-phase system shown in Fig. 2.4 and is given by ZPNZ 5 H21 ZRYB H

(2.24)

It is important to note that this is the impedance matrix in the sequence frame of reference obtained by transforming the phase impedance matrix according to Eq. (2.24) using the transformation matrices shown. Assuming that the mutual phase impedances are bilateral or reciprocal, that is ZRY 5 ZYR, etc., Eq. (2.24) becomes P P Z PP PNZ 5 N 4 Z NP Z Z Z ZP

N Z PN Z NN Z ZN

2

Z 3 Z PZ Z NZ 5 Z ZZ

(2.25)

where Z PP 5 Z NN 5

1 ½ðZRR 1 ZYY 1 ZBB Þ 2 ðZRY 1 ZRB 1 ZYB Þ 3

(2.26a)

5 SPP 2 M PP Z PN 5

 1 ZRR 1 h2 ZYY 1 hZBB Þ 1 2ðh2 ZRB 1 hZRY 1 ZYB Þ 3

(2.26b)

5 SPN 1 2M PN Z PZ 5 Z ZN 5 5S Z NP 5

PZ

 1 ðZRR 1 hZYY 1 h2 ZBB Þ 2 ðh2 ZRY 1 hZRB 1 ZYB Þ 3

2M

 1 ðZRR 1 hZYY 1 h2 ZBB Þ 1 2ðh2 ZRY 1 hZRB 1 ZYB Þ 3

5S

PZ

1 2M

Z NZ 5 Z ZP 5 5S

PN

(2.26c)

PZ

 1 ðZRR 1 h2 ZYY 1 hZBB Þ 2 ðh2 ZRB 1 hZRY 1 ZYB Þ 3

2M

(2.26d)

PZ

PN

(2.26e)

Symmetrical components analysis of faulted three-phase networks

Z ZZ 5

1 ½ðZRR 1 ZYY 1 ZBB Þ 1 2ðZRY 1 ZRB 1 ZYB Þ 3

5 S 1 2M PP

53

(2.26f)

PP

Dropping the R notation from the current vector IPNZ for convenience and using R Eqs. (2.23), (2.24) and (2.25), we can write 2

3 2 PP VP S 2 M PP N 4 V 5 5 4 SPZ 1 2M PZ VZ SPN 2 M PN

SPN 1 2M PN SPP 2 M PP SPZ 2 M PZ

3 2 P3 I SPZ 2 M PZ SPN 2 M PN 5 4 I N 5 SPP 1 2M PP IZ

(2.27)

which can be written as 2

3 2 PP VP a 4 V N 5 5 4 d NP VZ eNZ

bPN aPP cPZ

3 2 P3 I cPZ eNZ 5 4 I N 5 f ZZ IZ

(2.28)

Eq. (2.28) is important and requires a physical explanation. The sequence impedance matrix is full, nondiagonal and nonsymmetric. A nonsymmetric matrix is exemplified by the different mutual coupling terms b and d (also c and e) in the matrix. A nondiagonal matrix means that mutual coupling between the sequence circuits exists. For example, taking the positive-sequence circuit, the total positivesequence voltage drop in this circuit consists of three components as follows: V P 5 aPP I P 1 bPN I N 1 cPZ I Z

(2.29)

The first term on the right is the positive-sequence voltage drop induced by the flow of positive-sequence current in the positive-sequence circuit. The second term is an additional positive-sequence voltage drop caused by the flow of negativesequence current in the negative-sequence circuit acting on the mutual impedance from the positive-sequence to the negative-sequence circuit. The third term is an additional positive-sequence voltage drop caused by the flow of zero-sequence current in the zero-sequence circuit acting on the mutual impedance from the positivesequence to the zero-sequence circuit. We have calculated the sequence impedance matrix of a general three-phase mutually coupled system and this turns out to be nondiagonal and nonsymmetric. What does this mean in practice? Let us remember that the basic principle of the symmetrical component theory rests on the fact that the positive-sequence, negative-sequence and zero-sequence circuits are separate, with no mutual coupling between them. In mathematical terms, this means that the sequence impedance matrix has to be diagonal so that a positive-sequence voltage drop is produced by the flow of positive-sequence current only, and likewise for the negative-sequence and zero-sequence circuits. The transformation of the phase impedance matrix of

54

Power Systems Modelling and Fault Analysis

Eq. (2.22a) from the phase frame of reference to the sequence frame of reference, Eq. (2.28), produced no advantages whatsoever. In such situations, the symmetrical component theory may not be used and analysis in the phase frame of reference is more advantageous.

2.2.6 Sequence impedances of coupled balanced three-phase impedances The application of the symmetrical component theory in practical power system short-circuit analysis requires the positive-sequence, negative-sequence and zerosequence circuits to be separate and uncoupled. This is achieved by the removal of the off-diagonal mutual elements in the sequence impedance matrix of Eq. (2.28) so that the matrix becomes diagonal. This can be accomplished if the physical phase impedances of the original three-phase system shown in Fig. 2.4 are assumed to be balanced, i.e., the self-impedances are assumed to be equal, and the mutual impedances are also assumed to be equal, i.e., ZRR 5 ZYY 5 ZBB 5 ZSelf

(2.30a)

ZRY 5 ZYB 5 ZBR 5 ZMutual

(2.30b)

Substituting Eqs. (2.30a) and (2.30b) into Eqs. (2.26a)(2.26f) and (2.27), we obtain 2

3 2 ZSelf 2 ZMutual VP 4 VN 5 5 4 0 0 VZ

0 ZSelf 2 ZMutual 0

3 2 P3 0 I 5 4 IN 5 0 ZSelf 1 2ZMutual IZ

(2.31)

where, PPS impedance 5 NPS impedance 5 Z P 5 Z N 5 ZSelf 2 ZMutual

(2.32a)

ZPS impedance 5 Z Z 5 ZSelf 1 2ZMutual

(2.32b)

Also, if the positive-sequence, negative-sequence and zero-sequence impedances are known, the balanced self and mutual phase impedances can be calculated as follows: ZSelf 5

1 Z ðZ 1 2Z P Þ 3

1 ZMutual 5 ðZ Z 2 Z P Þ 3

(2.33a) (2.33b)

Symmetrical components analysis of faulted three-phase networks

55

2.2.7 Advantages of symmetrical components frame of reference The significant advantage of the symmetrical component theory is that it provides a mathematical method that allows us to decompose the original complex mutually coupled three-phase system equations into three separate sets of equations. The positive-sequence, negative-sequence and zero-sequence sets of currents and voltages are each calculated from the positive-sequence, negative-sequence and zerosequence circuits, respectively, i.e., in the sequence frame of reference. The sequence currents and voltages are then easily transformed back into the phase frame of reference. However, the creation of three uncoupled single-phase sequence circuits is based on the assumption that the original three-phase system is perfectly balanced. In reality, this is not the case, as will be seen in later chapters. Nonetheless, reasonable assumptions of balance can be made to allow the use of the symmetrical component theory in practical applications. Nowadays, almost all large-scale short-circuit analysis in practical power systems is carried out in the sequence frame of reference with transformation to the phase frame of reference carried out as a final step in the analysis. There are, nonetheless, specialised shortcircuit applications where analysis in the phase frame of reference is used. This is discussed in Chapter 7, Short-circuit analysis techniques in large-scale AC power systems.

2.2.8 Examples Example 2.1 Prove that a balanced set of three-phase voltages or currents contains only a positive-sequence voltage or current component. Substituting Eqs. (2.5) and (2.6) into Eqs. (2.14a)(2.14c), we obtain 1 VRP 5 ðVR 1 hh2 VR 1 h2 hVR Þ 5 VR 3 1 VRN 5 ðVR 1 h2 h2 VR 1 hhVR Þ 5 0 3

since h3 5 1 since 1 1 h 1 h2 5 0

1 VRZ 5 ðVR 1 h2 VR 1 hVR Þ 5 0 3

Example 2.2 What is the phase sequence nature of MW and MVAr power flows in a balanced three-phase power system? In a balanced three-phase system, negative-sequence and zero-sequence voltages and currents are zero and only positive-sequence voltages and currents exist. Therefore, and according to Eq. (2.18), only positive-sequence apparent power exists and hence the MW and MVAr power flows in a balanced three-phase system are positive-sequence quantities.

56

Power Systems Modelling and Fault Analysis

Example 2.3 What is the relationship between the zero-sequence current and the neutral current in a threephase power system with a neutral wire or neutral connection to earth. From Eqs. (2.7a)(2.7c), rewritten for phase currents, we have IR 5 IPR 1 INR 1 IZR

IY 5 IPY 1 INY 1 IZY

IB 5 IPB 1 INB 1 IZB

The neutral current is the sum of the currents in phases R, Y and B, thus INeutral 5 IR 1 IY 1 IB 5 ðIPR 1 IPY 1 IPB Þ 1 ðINR 1 INY 1 INB Þ 1 ðIZR 1 IZY 1 IZB Þ or INeutral 5 0 1 0 1 ðIZR 1 IZY 1 IZB Þ 5 3IZR Since the positive-sequence (and negative-sequence) currents are balanced, by definition, their sum is zero and neither produces any neutral current. Since the zero-sequence currents are equal and in phase, the neutral current is the sum of the zero-sequence currents in each phase and this is equal to three times the zero-sequence current in phase R.

Example 2.4 The three voltages of a three-phase system are balanced in phase but not in magnitude and are VR 5 1+0 pu VY 5 0:97+ 2 120 pu

VB 5 1:01+120 pu

Calculate the positive-sequence, negative-sequence and zero-sequence voltages of phase R voltage. Express the negative-sequence and zero-sequence voltage magnitudes as percentages of the positive-sequence voltage magnitude. Using Eqs. (2.14a)(2.14c), we obtain 1 VRP 5 ð1 1 1 +120 3 0:97+ 2 120 1 1 + 2 120 3 1:01 + 120Þ 5 0:99 pu 3 1 VRN 5 ð1 1 1 + 2 120 3 0:97+ 2 120 1 1 +120 3 1:01 + 120Þ 3 5 0:012+ 2 73:89 pu 1 VRZ 5 ð1 1 0:97+ 2 120 1 1:01+120Þ 5 0:012+73:89 pu 3 N Z V V 0:012 R 5 R 5 5 0:01212% or 1:212% V P V P 0:99 R

R

Symmetrical components analysis of faulted three-phase networks

57

Example 2.5 The three voltages of a three-phase system have equal magnitudes of 1 pu but phase Y lags phase R by 115
and the phase displacement between phases Y and B is 120
. Calculate the positive-sequence, negative-sequence and zero-sequence voltages of phase R. Express the negative-sequence and zero-sequence voltage magnitudes as percentages of the positivesequence voltage magnitude. Taking phase R as an arbitrary reference, the phase voltages can be written as VR 5 1+0 pu VY 5 1+ 2 115 pu

VB 5 1+125 pu

Thus, the three voltages are balanced in magnitude but not in phase. Using Eqs. (2.14a) (2.14c), we obtain 1 VRP 5 ð1 1 1 +120 3 1+ 2 115 1 1 + 2 120 3 1 + 125Þ 5 0:9991+3:3 pu 3 1 VRN 5 ð1 1 1 + 2 120 3 1+ 2 115 1 1 +120 3 1 + 125Þ 5 0:029+ 2 87:5 pu 3 1 VRZ 5 ð1 1 1 + 2 115 1 1+ 125Þ 5 0:029+ 2 87:5 pu 3 N Z V V 0:029 R 5 R 5 V P V P 0:999 5 0:02903% or 2:9% R

R

Example 2.6 The three unbalanced voltages of a three-phase system under a very large unbalanced condition are VR 5 0+0 pu VY 5 1:1+ 2 140 pu VB 5 1:1+140 pu: Calculate the positive-sequence, negative-sequence and zero-sequence voltages of phase R. Express the negative-sequence and zero-sequence voltage magnitudes as percentages of the positive-sequence voltage magnitude. Using Eqs. (2.14a)(2.14c), we obtain 1 VRP 5 ð0 1 1 +120 3 1:1+ 2 140 1 1 + 2 120 3 1:1 + 140Þ 5 0:689+0 pu 3 1 VRN 5 ð0 1 1 + 2 120 3 1:1+ 2 140 1 1 +120 3 1:1 + 140Þ 5 2 0:1273+0 pu 3 1 VRZ 5 ð0 1 1:1 + 2 140 1 1:1+ 140Þ 5 2 0:5617+0 pu 3 N Z V 0:1273 V 0:5617 R 5 R 5 5 0:185% or 18:5% 5 0:815% or 81:5 % V P 0:689 0:689 V P R

R

58

2.3

Power Systems Modelling and Fault Analysis

Symmetrical components analysis of faulted networks containing balanced voltage sources only

2.3.1 General In general, there are three types of faults in three-phase power systems. These are short-circuit faults between one or more phases which may or may not involve earth, open-circuit faults on one or two phases and simultaneous faults where more than one fault occurs at the same time at the same or at different locations in the network. Short-circuit faults are sometimes referred to as shunt faults whereas open-circuit faults are series faults. When a short-circuit fault occurs at point F on the three-phase network, the conditions imposed by the fault at F must be observed between the relevant phase(s) and the neutral point in the sequence networks. However, when an open-circuit fault occurs at a point F in the network, the conditions imposed by the fault must be observed between the two sides of the open circuit, say points F and F0 , in the sequence networks. In this section, we will derive methods of connecting the three positive-sequence, negative-sequence and zerosequence networks for various fault types that can occur in power systems. The unbalanced fault condition applied in the three-phase network is arranged to be symmetrical with respect to phase R which is taken as the reference phase. This results in simpler mathematical derivation as will be shown later.

2.3.2 Balanced three-phase to earth short-circuit faults A three-phase to earth short-circuit fault at a point in a three-phase system is a balanced or symmetrical fault that can still be analysed using the symmetrical components theory. Fig. 2.5A shows the representation of this fault. FR, FY and FB are points in the three-phase system where the three-phase fault is assumed to occur through the balanced fault impedances ZF, and F0R ; F0Y and F0B are the true points of fault. From Fig. 2.5A, the voltages at point F are given by VR 5 ZF IR

VY 5 ZF IY

VB 5 ZF IB

(2.34)

Using Eqs. (2.6), (2.14a) and (2.19), it can be shown that the sequence voltages and currents at the fault point are given by VRP 5 ZF IRP 5 VFP 2 Z P IRP

VRN 5 0

VRZ 5 0

(2.35)

and IRP 5

VFP Z P 1 ZF

IRN 5 0 IRZ 5 0

(2.36)

Symmetrical components analysis of faulted three-phase networks

(A) R Y

ZF

B

Balanced three-phase power system

59

FR

ZF

FY FB

ZF ZF

ZF

ZF

(B)

F'N

F'P FP

ZF IP

IR IY IB

F'R F'Y F'B

IR + IY + IB

F'Z ZF

FN I N = 0

FZ

ZF IZ = 0

P VN = 0 VZ = 0 ZZ VP + Z P ZN – VF NN NZ NP Thévenin’s Thévenin’s Thévenin’s PPS network NPS network ZPS network

Figure 2.5 Balanced three-phase to earth short-circuit fault through a fault impedance ZF. (A) Fault representation and (B) positive, negative and zero sequence Thevenin’s equivalent circuits.

Therefore, the three-phase power system remains balanced and symmetrical after the occurrence of such a fault because the fault impedances are equal in the three phases. This means that only positive-sequence voltages exist and only positivesequence currents can flow. Since phase R is used as the reference, it is convenient from now on to drop the R notation in the positive-sequence, negative-sequence and zero-sequence voltage and current equations whilst always remembering that these sequence quantities are those of phase R. Using Eqs. (2.15a) and (2.10b), the phase fault currents are given by IR 5 I P 5

VFP Z P 1 ZF

(2.37a)

I Y 5 h2 I R 5

h2 VFP Z P 1 ZF

(2.37b)

IB 5 hIR 5

hVFP 1 ZF

(2.37c)

ZP

and, using Eq. (2.34), the phase fault voltages at the fault point F are given by VR 5 I P ZF 5

VFP ZF Z P 1 ZF

(2.38a)

60

Power Systems Modelling and Fault Analysis

V Y 5 h 2 VR 5 VB 5 hVR 5

h2 VFP ZF Z P 1 ZF

(2.38b)

hVFP ZF Z P 1 ZF

(2.38c)

As expected for a balanced and symmetrical short-circuit fault, the sum of the three-phase currents IR 1 IY 1 IB is equal to zero, hence the net fault current flowing into earth is zero. Similarly, the sum of the three-phase voltages VR 1 VY 1 VB is equal to zero. Fig. 2.5B shows the connections of the The´venin’s positivesequence, negative-sequence and zero-sequence equivalent circuits that satisfy Eqs. (2.35) and (2.36). The case of a solid or bolted three-phase to earth short-circuit fault is obtained by setting ZF 5 0.

2.3.3 Balanced three-phase clear of earth short-circuit faults A three-phase short-circuit fault clear of earth at a point F in a three-phase system is represented by connecting an equal fault impedance between each pair of phases as shown in Fig. 2.6A, i.e., as a delta connection. In order to calculate the equivalent fault impedance ‘seen’ in each phase, a transformation of delta to star with isolated neutral is needed as follows: ZR 5 ZY 5 ZB 5

ZF 3 ZF 1 5 ZF 3 ZF 1 ZF 1 ZF

(2.39)

The fault representation is illustrated in Fig. 2.6B. The phase fault currents and voltages are given by IR 1 IY 1 IB 5 0

(2.40a)

and V R 5 VY 5 VB 5 0

at point F0

(2.40b)

and VR 5

1 ZF IR 3

VY 5

1 ZF IY 3

VB 5

1 ZF IB 3

(2.40c)

Using Eqs. (2.14a)(2.14c), (2.19), (2.20) and (2.21), we obtain VP 5

1 ZF I P 5 VFP 2 Z P I P 3

VN 5 0

VZ 5 0

(2.41a)

Symmetrical components analysis of faulted three-phase networks

61

(A) R

ZF

ZF B (B)

ZF Y

R

ZF 3 ZF 3 Y 3 B ZF

(C)

R

Balanced three-phase power system

FR FY

ZF 3 IR Z F 3 IY IB

FB ZF 3

F'R F'Y F'B

F'Z

F'N Z F /3

Z F /3 FP I P

ZF ZF

B

Balanced three-phase power system

F'P

ZF

Y

FN I N = 0

Z F /3 FZ

IZ = 0

P VN = 0 V Z= 0 ZZ VP + Z N P Z V F – NZ NP NN Thévenin’s Thévenin’s Thévenin’s PPS network NPS network ZPS network

Figure 2.6 Balanced three-phase short-circuit fault clear of earth through a fault impedance ZF. (A) Fault representation with delta fault impedances, (B) Fault representation with star fault impedances and (C) positive, negative and zero sequence Thevenin’s equivalent circuits.

IP 5

VFP Z P 1 ZF =3

IN 5 0 IZ 5 0

(2.41b)

The phase fault currents are given by VFP 1 ZF =3

(2.42a)

I Y 5 h2 I R 5

h2 VFP Z P 1 ZF =3

(2.42b)

IB 5 hIR 5

hVFP 1 ZF =3

(2.42c)

IR 5 I P 5

ZP

ZP

The phase fault voltages at point F are given by 1 VFP VR 5 I P ZF 5 3 1 1 3Z P =ZF

(2.43a)

62

Power Systems Modelling and Fault Analysis

V Y 5 h 2 VR 5 VY 5 hVR 5

h2 VFP 1 1 3Z P =ZF

hVFP 1 1 3Z P =ZF

(2.43b)

(2.43c)

Fig. 2.6C shows the connections of the positive-sequence, negative-sequence and zero-sequence equivalent networks that satisfy the fault condition at the fault point F. The case of a solid or bolted three-phase clear of earth short-circuit fault is obtained by setting ZF 5 0.

2.3.4 Unbalanced one-phase to earth short-circuit faults Fig. 2.7A shows a representation of an unbalanced one-phase to earth fault on phase R through a fault impedance ZF. For ease of notation in the rest of this chapter, we will replace VFP with VF. The conditions at the point of fault are IY 5 IB 5 0

(2.44a)

VR 5 ZF IR

(2.44b)

and

Using Eqs. (2.14a)(2.14c) for currents instead of voltages, we obtain 1 I P 5 I N 5 I Z 5 IR 3

or IR 5 3 3 I Z

(2.45)

Substituting Eqs. (2.19), (2.20) and (2.21) into Eq. (2.9a) and using Eq. (2.45), we obtain VR 5 ZF 3 3I P 5 V P 1 V N 1 V Z 5 VF 2 ðZ P 1 Z N 1 Z Z ÞI P or VF 5 ðZ P 1 ZF ÞI P 1 ðZ N 1 ZF ÞI N 1 ðZ Z 1 ZF ÞI Z

(2.46)

Thus, the sequence fault currents are given by IP 5 IN 5 IZ

VF ½ðZ P 1 ZF Þ 1 ðZ N 1 ZF Þ 1 ðZ Z 1 ZF Þ VF 5 P ðZ 1 Z N 1 Z Z 1 3ZF Þ

5

(2.47)

Symmetrical components analysis of faulted three-phase networks

(A)

R B

Y

Balanced three-phase power system

ZF FN

F'N ZF IN

VN

ZN

ZF

(B)

F'P FP VP

IP ZP

+

FR FY FB

63

ZF ZF ZF

FZ VZ

IR

F'R

F'Z ZF IZ ZZ

P VF

NP Thévenin’s PPS network

NN Thévenin’s NPS network

NZ Thévenin’s ZPS network

3Z F

(C) FP

FN IN

IP VP

ZP +

VN

ZN

FZ IZ VZ

ZZ

P VF

NP Thévenin’s PPS network

NN Thévenin’s NPS network

NZ Thévenin’s ZPS network

Figure 2.7 Unbalanced one-phase to earth short-circuit fault through a fault impedance ZF. (A) Fault representation, (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits, and (C) as (B) but with alternative fault impedance representation.

Eqs. (2.45) and (2.46) show that the positive-sequence, negative-sequence and zero-sequence networks should be connected in series as shown in Fig. 2.7B with the fault impedance ZF appearing as an external impedance in series with each sequence network. The positive-sequence, negative-sequence and zero-sequence voltages at the fault point are calculated using Fig. 2.7B or C giving VP 5

ðZ N 1 Z Z 1 3ZF Þ VF ðZ P 1 Z N 1 Z Z 1 3ZF Þ

(2.48)

VN 5

2 ZN VF ðZ P 1 Z N 1 Z Z 1 3ZF Þ

(2.49)

2 ZZ VF 1 Z Z 1 3ZF Þ

(2.50)

VZ 5

ðZ P

1 ZN

The phase fault current is calculated using Eqs. (2.45) and (2.47) giving IR 5

ðZ P

1 ZN

3VF 1 Z Z 1 3ZF Þ

(2.51)

64

Power Systems Modelling and Fault Analysis

The phase fault voltage is calculated from Eqs. (2.44a) and (2.44b) giving VR 5

ðZ P

1 ZN

3ZF VF 1 Z Z 1 3ZF Þ

(2.52)

The phase voltages on the healthy or unfaulted phases at the point of fault can be calculated using Eqs. (2.9a), (2.9b) and (2.9c) and Eqs. (2.48), (2.49) and (2.50) giving VY 5 KE1 3 h2 VF

(2.53)

where KE1 5 1 2

ðZ P 1 h2 Z N 1 hZ Z Þ ðZ P 1 Z N 1 Z Z 1 3ZF Þ

(2.54)

and VB 5 KE2 3 hVF

(2.55)

where KE2 5 1 2

ðZ P 1 hZ N 1 h2 Z Z Þ ðZ P 1 Z N 1 Z Z 1 3ZF Þ

(2.56)

KE1 and KE2 are termed the earth fault factors and have a magnitude that typically ranges from 1 to 1.8 depending on the method of system earthing used. This factor is the ratio of the rms phase to earth voltage on a healthy phase at the fault point during the fault to the rms phase to earth voltage without a fault. This factor determines the extent of voltage rise on the healthy phases during the fault. Systems having an earth fault factor less than or equal to 1.4 are defined as ‘effectively earthed’. The case of a solid or bolted single-phase to earth short-circuit fault is obtained by setting ZF 5 0.

2.3.5 Unbalanced phase-to-phase or two-phase short-circuit faults Fig. 2.8A shows a representation of an unbalanced two-phase fault on phases Y and B through a fault impedance ZF. The conditions at the point of fault are IR 5 0 IY 5 2 IB

VY 2 VB 5 Z F I Y

(2.57a)

Symmetrical components analysis of faulted three-phase networks

(A) B

FR FY

Balanced three-phase power system

R Y

FB

ZF

(B)

F'N

F'P Z F /2 FP I P VP

ZP +

ZF /2 FN VN

IN ZN

65

Z F /2 I R Z F /2

IY

Z F /2

IB

F'R F'Y F'B

F'Z ZF /2 FZ I Z VZ

ZZ

P VF

NP Thévenin’s PPS network

NN Thévenin’s NPS network

NZ Thévenin’s ZPS network

Figure 2.8 Unbalanced two-phase short-circuit fault through a fault impedance ZF. (A) Fault representation and (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits.

Substituting the above phase currents into Eq. (2.15b), we obtain IZ 5 0

and

IP 5 2 IN

(2.57b)

Substituting IZ 5 0 into Eq. (2.21) gives VZ 5 0 and using Eqs. (2.7a)(2.7c) and (2.8), we can write VY 2 VB 5 ðh2 2 hÞðV P 2 V N Þ

(2.58a)

Also, using Eqs. (2.7b) and (2.8) for phase currents, we obtain ZF IY 5 ZF ðh2 I P 1 hI N Þ

(2.58b)

Equating VY 2 VB from Eqs. (2.57a) and (2.58a), using Eq. (2.58b) and IP 5 2 IN, we obtain V P 2 V N 5 ZF I P

(2.59a)

Substituting Eqs. (2.19) and (2.20) into Eq. (2.59a), and using IP 5 2 IN, we obtain     1 1 VF 2 Z P 1 ZF I P 5 2 Z N 1 ZF I N 2 2

(2.59b)

66

Power Systems Modelling and Fault Analysis

The sequence fault currents are calculated from Eq. (2.59b) using IP 5 2 IN giving VF VF 5 P IP 5 2 IN 5  P N ðZ 1 Z N 1 ZF Þ ðZ 1 ZF =2Þ 1 ðZ 1 ZF =2Þ

(2.60)

Eqs. (2.57b) and (2.59b) show that the positive-sequence and negative-sequence networks should be connected in parallel as shown in Fig. 2.8B with half the fault impedance, ð1=2ÞZF , appearing as an external impedance in series with the positivesequence and negative-sequence networks. The zero-sequence currents and voltages are zero. The phase fault currents are calculated using Eq. (2.15a) giving pffiffiffi I Y 5 2 I B 5 2 j 3I P 5

pffiffiffi 2 j 3 VF ðZ P 1 Z N 1 ZF Þ

(2.61)

The phase voltages on the faulted phases Y and B are calculated using Eqs. (2.9b), (2.9c), (2.19), (2.20), (2.21) and (2.60) giving 

h2 Z F 2 Z N P Z 1 Z N 1 ZF

VY 5 

hZF 2 Z N VB 5 ZP 1 Z N 1 ZF

 VF

(2.62a)

VF

(2.62b)



The phase voltage on the healthy or unfaulted phase R can be calculated using Eq. (2.9a) giving  VR 5

 2Z N 1 ZF VF ZP 1 Z N 1 ZF

(2.63a)

The case of a solid or bolted two-phase short-circuit fault is obtained by setting ZF 5 0. In this case, the voltage difference between the healthy phase R and one of the faulted phases, Y or B, is calculated using Eqs. (2.62a), (2.62b) and (2.63a) and is given by 

 VF VRðHealthyÞ 2 VYðFaultedÞ 5 3Z N ZP 1 Z N

(2.63b)

2.3.6 Unbalanced two-phase to earth short-circuit faults Fig. 2.9A shows a representation of an unbalanced two-phase to earth fault on phases Y and B through a fault impedance ZF.

Symmetrical components analysis of faulted three-phase networks

(A) R Y

B ZF

FN

ZP

VN

F'R

IY

F'Y F'B

IB IE = I Y + I B

ZF FZ IZ

ZF

ZF FP IP +

ZF

F'Z

F'N

F'P

VP

ZF

ZF

ZF

(B)

IN ZN

VZ

ZZ

P VF

NP Thévenin’s PPS network (C)

FR FY FB

Balanced three-phase power system

67

Balanced three-phase power system

NZ Thévenin’s ZPS network

NN Thévenin’s NPS network

R ZF ZF

IY IB

Y

B IE = I Y + I B ZE

ZF

ZF ZE

Figure 2.9 Unbalanced two-phase to earth short-circuit fault through a fault impedance ZF. (A) Fault representation, (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits, and (C) as (A) but with additional common fault impedance.

The conditions at the point of fault are given by IR 5 0 V Y 5 Z F IY

VB 5 ZF IB

(2.64)

From Eq. (2.7a) for currents, and IR 5 0, we obtain IP 1 IN 1 IZ 5 0

or

I Z 5 2 ðI P 1 I N Þ

(2.65)

Rewriting Eqs. (2.9b) and (2.9c), and using Eq. (2.64), we have VY 5 h2 V P 1 hV N 1 V Z 5 ZF IY 5 ZF ðh2 I P 1 hI N 1 I Z Þ

(2.66a)

VB 5 hV P 1 h2 V N 1 V Z 5 ZF IB 5 ZF ðhI P 1 h2 I N 1 I Z Þ

(2.66b)

Subtracting Eq. (2.66b) from Eq. (2.66a), we obtain VY 2 VB 5 ðh2 2 hÞV P 1 ðh 2 h2 ÞV N 5 ZF ½ðh2 2 hÞI P 1 ðh 2 h2 ÞI N 

68

Power Systems Modelling and Fault Analysis

pffiffiffi which, using h2 2 h 5 2 j 3, reduces to V P 2 ZF I P 5 V N 2 ZF I N

(2.67)

Now, substituting Eqs. (2.19) and (2.20) into Eq. (2.67), we obtain 2 ðZ N 1 ZF ÞI N 5 VF 2 ðZ P 1 ZF ÞI P

(2.68)

Adding Eqs. (2.66a) and (2.66b), we obtain VY 1 VB 5 ðh2 1 hÞV P 1 ðh2 1 hÞV N 1 2V Z 5 ZF ½ðh2 1 hÞI P 1 ðh2 1 hÞI N 1 2I Z  which, using h2 1 h 5 2 1 and Eqs. (2.20), (2.21) and (2.67), reduces to V N 2 ZF I N 5 V Z 2 ZF I Z

(2.69a)

2 ðZ Z 1 ZF ÞI Z 5 2 ðZ N 1 ZF ÞI N 5 VF 2 ðZ P 1 ZF ÞI P

(2.69b)

and

Eqs. (2.65), (2.69a) and (2.69b) show that the positive-sequence, negativesequence and zero-sequence networks should be connected in parallel, as shown in Fig. 2.9B, with the fault impedance ZF appearing as an external impedance in each sequence network. The sequence fault currents are calculated using Eqs. (2.65), (2.69a) and (2.69b), or alternatively from Fig. 2.9B. The use of the equations is illustrated for deriving the sequence currents. Using Eqs. (2.69a) and (2.69b) and expressing IN and IZ in terms of IP then obtain  substituting   into  Eq.  (2.65),  we I P 1 Z P 1 ZF I P 2 VF = Z N 1 ZF 1 Z P 1 ZF I P 2 VF = Z Z 1 ZF 5 0 which, after a little algebra, gives IP 5

VF ðZ P 1 ZF Þ 1 ðZ N 1 ZF ÞðZ Z 1 ZF Þ=½ðZ N 1 ZF Þ 1 ðZ Z 1 ZF Þ

(2.70a)

IN 5

2 VF ðZ P 1 ZF Þ 1 ðZ N 1 ZF Þ 1 ðZ P 1 ZF ÞðZ N 1 ZF Þ=ðZ Z 1 ZF Þ

(2.70b)

IZ 5

2 VF ðZ P 1 ZF Þ 1 ðZ Z 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ=ðZ N 1 ZF Þ

(2.70c)

The phase fault currents are calculated from the sequence currents using Eq. (2.15a) giving

Symmetrical components analysis of faulted three-phase networks

pffiffiffi 2 j 3VF ½ðZ Z 1 ZF Þ 2 hðZ N 1 ZF Þ IY 5 N ðZ 1 ZF ÞðZ Z 1 ZF Þ 1 ðZ N 1 ZF ÞðZ P 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ

69

(2.71a)

and pffiffiffi j 3VF ½ðZ Z 1 ZF Þ 2 h2 ðZ N 1 ZF Þ IB 5 N ðZ 1 ZF ÞðZ Z 1 ZF Þ 1 ðZ N 1 ZF ÞðZ P 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ

(2.71b)

The total fault current flowing into earth at the point of fault is the sum of the phase fault currents IY and IB giving IE 5 IY 1 IB 5 3I Z 5

ðZ P

1 ZF

Þ 1 ðZ Z

2 3VF 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ=ðZ N 1 ZF Þ

(2.72)

The phase voltage on the healthy or unfaulted phase R can be calculated using Eq. (2.9a) giving VR 5

3VF ½Z N Z Z 1 ZF ðZ N 1 Z Z 1 ZF Þ ðZ N 1 ZF ÞðZ Z 1 ZF Þ 1 ðZ N 1 ZF ÞðZ P 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ

(2.73)

The phase voltages on the faulted phases Y and B can be calculated using Eq. (2.64) giving pffiffiffi 2 j 3VF ZF ½ðZ Z 1 ZF Þ 2 hðZ N 1 ZF Þ VY 5 N ðZ 1 ZF ÞðZ Z 1 ZF Þ 1 ðZ N 1 ZF ÞðZ P 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ (2.74a) pffiffiffi j 3VF ZF ½ðZ Z 1 ZF Þ 2 h2 ðZ N 1 ZF Þ VB 5 N ðZ 1 ZF ÞðZ Z 1 ZF Þ 1 ðZ N 1 ZF ÞðZ P 1 ZF Þ 1 ðZ P 1 ZF ÞðZ Z 1 ZF Þ (2.74b) The case of a solid or bolted two-phase to earth short-circuit fault is obtained by setting ZF 5 0. Also, in the case where an earth impedance ZE is present in the common connection to earth, as shown in Fig. 2.9C, it can be shown that ZZ in Fig. 2.9B must be replaced by (ZZ 1 3ZE). The reader is encouraged to prove this statement.

2.3.7 Unbalanced one-phase open-circuit faults Fig. 2.10A shows the representation of an unbalanced one-phase open-circuit fault occurring on phase R and creating points F and F0 in a balanced three-phase power system.

70

Power Systems Modelling and Fault Analysis

Node i R YB

(A)

Balanced three-phase power system

IR = 0 I F F' Y

IB F F' F F'

FN

FP

(B)

VN

ZP +

FZ IN

IP VP

Node j R YB

ZN

IZ VZ

ZZ

P

IL Z

F'P Thévenin’s PPS network

F'N Thévenin’s NPS network

F'Z Thévenin’s ZPS network

Figure 2.10 Unbalanced one-phase open-circuit fault. (A) Fault representation and (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits.

The conditions imposed by the fault are IR 5 0 hence I P 1 I N 1 I Z 5 0

(2.75)

Eq. (2.75) represents the sequence currents of the phase current IR from points F to F0 . In addition, since these two points are still connected together on phases Y and B, we have 0

VYFF 5 0

and

0

VBFF 5 0

(2.76)

The sequence components of the voltages in Eq. (2.76) can be calculated using Eqs. (2.14a)(2.14c) giving VP 5 VN 5 VZ 5

1 FF0 V 3 R

(2.77)

Eqs. (2.75) and (2.77) are satisfied by connecting the positive-sequence, negative-sequence and zero-sequence equivalent networks in parallel at points F to F0 as shown in Fig. 2.10B. It should be noted that ZP, ZN and ZZ are the positivesequence, negative-sequence and zero-sequence equivalent The´venin impedances, respectively, as ‘seen’ looking back into the respective sequence network from between the points F and F0 . The voltage source in the positive-sequence equivalent network in Fig. 2.10B has not been derived yet. This voltage source is the opencircuit positive-sequence voltage appearing across F and F0 . This voltage is calculated as the multiplication of the prefault load current in phase R, i.e., the current flowing before the open circuit occurs (note that this is a positive-sequence current because the three-phase power system is balanced), and ZP, i.e.,

Symmetrical components analysis of faulted three-phase networks 0

VRFF 5 IL Z P

71

(2.78)

From Fig. 2.10B, the positive-sequence current is given by IP 5

IL Z P ZN 1 ZZ 5 IL Z P Z P 1 Z N Z Z =ðZ N 1 Z Z Þ Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.79)

The negative-sequence current is given by IN 5 2 IP

ZN

ZZ 1 ZZ

or IN 5

ZPZN

2 ZZ IL Z P 1 ZNZZ 1 ZZZP

(2.80)

The zero-sequence current is given by IZ 5 2 IP

ZN

ZN 1 ZZ

or IZ 5

2 ZN IL Z P ZPZN 1 ZNZZ 1 ZZZP

(2.81)

The sequence voltages are given by VP 5 VN 5 VZ 5

ZNZZ IL Z P Z PZ N 1 Z NZ Z 1 Z ZZ P

(2.82)

and the phase voltage across the open circuit is given by VR 5

 ZNZZ 3IL Z P P N N Z Z P Z Z 1Z Z 1Z Z

(2.83)

2.3.8 Unbalanced two-phase open-circuit faults Fig. 2.11A shows a representation of an unbalanced two-phase open-circuit fault occurring on phases Y and B and creating points F and F0 in a balanced three-phase power system.

72

Power Systems Modelling and Fault Analysis

Node i R YB I

(A)

Balanced three-phase power system

R

Node j R Y B

F F'

IY = 0 F F' IB = 0 F F'

(B)

FN

FP VP

ZP

FZ IN

IP VN

ZN

IZ VZ

ZZ

+ I ZP L F'P Thévenin‘s PPS network

F'N Thévenin‘s NPS network

F'Z Thévenin‘s ZPS network

Figure 2.11 Unbalanced two-phase open-circuit fault. (A) Fault representation and (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits.

The conditions imposed by the fault are given by IY 5 0

and

IB 5 0

(2.84a)

The sequence current components in Eq. (2.84a) are given by 1 I P 5 I N 5 I Z 5 IR 3

(2.84b)

In addition, since points F to F0 are still connected together on phase R 0

VRFF 5 V P 1 V N 1 V Z 5 0

(2.85)

Eqs. (2.84b) and (2.85) are satisfied by connecting the positive-sequence, negativesequence and zero-sequence equivalent networks in series at the points F to F0 as shown in Fig. 2.11B. From this, the sequence currents are given by IP 5 IN 5 IZ 5

ZP

IL Z P 1 ZN 1 ZZ

(2.86)

where IL is the prefault current flowing in phase R between F and F0 just before the open-circuit fault occurs on phases Y and B.

Symmetrical components analysis of faulted three-phase networks

73

The sequence voltages are given by

 V P 5 IP ZN 1 ZZ 5 V N 5 2 INZN 5 V Z 5 2 IZZZ 5

ZP

ZN 1 ZZ IL Z P 1 ZN 1 ZZ

(2.87a)

ZP

2 ZN IL Z P 1 ZN 1 ZZ

(2.87b)

ZP

2 ZZ IL Z P 1 ZN 1 ZZ

(2.87c)

The phase voltages on the faulted phases Y and B are calculated using Eqs. (2.9b) and (2.9c) giving pffiffiffi j 3ðhZ Z 2 Z N Þ IL Z P ZP 1 ZN 1 ZZ pffiffiffi 2 j 3ðh2 Z Z 2 Z N Þ 0 VBFF 5 IL Z P ZP 1 ZN 1 ZZ 0

VYFF 5

(2.88a)

(2.88b)

The case of a three-phase open-circuit fault does not require any consideration since this is not normally a fault but rather a normal switching operation in power systems such as the opening of the three phases of a circuit-breaker. Only positivesequence currents and voltages continue to exist in the resulting balanced power system. Such studies that involve the opening of the three phases of circuitbreakers, for example, to simulate the disconnection of a circuit and calculate the resultant currents, voltages active and reactive power flows, are known as positivesequence power flow studies.

2.3.9 Example Example 2.7 In this example, we assume ZF 5 0 and ZP 5 ZN: 1. Compare the relative magnitudes of a two-phase short-circuit fault and a three-phase shortcircuit fault. From Eqs. (2.36) and (2.61), we have VF I3φ 5 ZP

and

I2φ 5

pffiffiffi 3 VF 2 ZP

hence I2φ 5 0:866 3 I3φ

2. For both one-phase to earth and two-phase to earth faults, derive general expressions for the earth fault currents in terms of three-phase fault currents and expressions for residual voltages. Comment on the effect of zero-sequence to positive-sequence impedance ratio.

74

Power Systems Modelling and Fault Analysis

Let KZP 5 ZZ/ZP. The residual voltage is the sum of the three-phase voltage phasors. In actual three-phase power systems, both residual voltages and earth fault currents are measured using current and voltage transformers. These measurements are used for the detection of earth faults by protection relays. One-phase short-circuit fault The earth fault current is given by Eq. (2.48). Hence, I1φ 5

3 VF 3 5 I3φ 2Z P 1 Z Z Z P 2 1 KZP

I1φ 3 5 2 1 KZP I3φ

or

Using Eqs. (2.50) and (2.51), the residual voltage is equal to VEð1φÞ 5 VR 1 VY 1 VB 5

2 3KZP VF 2 1 KZP

VEð1φÞ 2 3KZP 5 VF 2 1 KZP

or

Two-phase to earth short-circuit fault The earth fault current is given by Eq. (2.72). Hence, IEð2φ2EÞ 5

2 3 VF 23 5 I3φ 1 1 2KZP Z P 1 1 2KZP

or

IEð2φ2EÞ 23 5 1 1 2KZP I3φ

Using Eqs. (2.73), (2.74a) and (2.74b), the residual voltage is equal to VEð2φ2EÞ 5 VR 1 VY 1 VB 5 KZP 5

3KZP VF 1 1 2KZP

or

VEð2φ2EÞ 3KZP 5 VF 1 1 2KZP

ZZ RZ 1 jX Z RZ 1 jX Z XZ RZ 5 P  5 P 2j P P P P Z R 1 jX jX X X

The complex ratio KZP has been simplified assuming that in high-voltage networks, the positivesequence XP/RP ratio is generally larger than 5. It is quite interesting to examine the variation of I1φ 3 5 I3φ 2 1 KZP VEð1φÞ 2 3KZP 5 VF 2 1 KZP

IEð2φ2EÞ 23 5 I3φ 1 1 2KZP VEð2φ2EÞ 3KZP 5 VF 1 1 2KZP

as functions of KZP where the latter varies from 0 to 5. For KZP 5 0, 1, 5 and N, I1φ/I3φ 5 1.5, 1, 0.43 and 0, respectively, and |IE(2φ E)/I3φ| 5 3, 1, 0.27 and 0, respectively. In other words, for KZP 5 (ZZ/ZP) , 1, I1φ . I3φ and IE(2φE) . I3φ, and IE(2φE) . I1φ, and vice versa. For KZP 5 0, 1, 5 and N, |VE(1φ)/VF| 5 0, 1, 2.14 and 3, respectively, and |VE(2φE)/VF| 5 0, 1, 1.36 and 1.5, respectively.

Symmetrical components analysis of faulted three-phase networks

2.4

75

Fault analysis and choice of reference frame

2.4.1 General The mathematical equations derived in previous sections for unbalanced shortcircuit and open-circuit faults were based on arranging the unbalance conditions to be symmetrical about phase R which was taken as the reference phase. In this section, we will show that whilst any phase could be chosen as the reference, the choice of phase R results in the simplest mathematical derivations.

2.4.2 One-phase to earth short-circuit faults In the case of a one-phase to earth short circuit on phase R with phase R chosen as the reference, the symmetrical components of the faulted phase are given by VRP 1 VRN 1 VRZ 5 0

IRP 5 IRN 5 IRZ

and

(2.89)

The above equations are represented in Fig. 2.7. However, for a one-phase to earth short circuit on phase Y with phase Y chosen as the reference, the symmetrical components of the faulted phase are given by VYP 1 VYN 1 VYZ 5 0

IYP 5 IYN 5 IYZ

and

(2.90)

Expressing the sequence currents and voltages of Eq. (2.90) in terms of phase R using Eq. (2.8), we obtain h2 VRP 1 hVRN 1 VRZ 5 0

and

h2 IRP 5 hIRN 5 IRZ

(2.91)

Similarly, for a one-phase to earth short circuit on phase B with phase B chosen as the reference, the symmetrical components of the faulted phase are given by VBP 1 VBN 1 VBZ 5 0

and

IBP 5 IBN 5 IBZ

(2.92)

Expressing the sequence currents and voltages of Eq. (2.92) in terms of phase R using Eq. (2.8), we obtain hVRP 1 h2 VRN 1 VRZ 5 0

and

hIRP 5 h2 IRN 5 IRZ

(2.93)

Eqs. (2.89), (2.91) and (2.93) can be represented by the connection of the positive-sequence, negative-sequence and zero-sequence networks as shown in Fig. 2.12. The sequence networks are still connected in series but through three complex multipliers kP, kN and kZ. These multipliers are applied to the sequence voltages and currents and are equal to either 1, h or h2 according to Eqs. (2.89), (2.91) and (2.93) and as shown in Case 1 in Fig. 2.12. The function of the multiplier is to apply the same phase shift to the relevant sequence current and voltage whilst

76

Power Systems Modelling and Fault Analysis

FP Z

I

P

V

P

1:k P

P

One-phase shortcircuit on phase: (R – E R is reference) (Y – E Y is reference) (B – E B is reference)

P P

k I P

k V

P

VF NP FN Z

N

I

N

V

N

I

Z

V

Z

N N

1:k N k I N

k V

N

h = e j2π/3

NN FZ Z

Z

1:k

Z

Z Z

k I Z

k V

Z

NZ

kP

kN

kZ

1

1

1

h2

h

1

h

h2

1

Case 1

One-phase shortcircuit on phase: (R – E R is reference) (Y – E Y is reference) (B – E B is reference)

kP

kN

kZ

1

1

1

1

2

h

h

1

h

h2

Case 2

Figure 2.12 One-phase short-circuit fault on either phase R, Y or B.

keeping their magnitudes unchanged. In describing this multiplier, we have deliberately avoided the use of the term phase-shifting transformer so as to avoid confusion with the property of physical transformers that transform voltages and currents by inverse ratios or multipliers. Fig. 2.12 shows that the phase shifts are applied to the positive-sequence and negative-sequence voltages and currents for faults on either phase Y or B. However, the zero-sequence current and voltage multiplier is always equal to unity. For convenience in practical analysis, it is normal to avoid applying phase shifts to the active positive-sequence network, which contains generating or voltage sources, and instead apply the phase shifts to the passive negative-sequence and zero-sequence networks. This is easily accomplished by dividing Eqs. (2.91) and (2.93) by h2 and h, respectively, and the resulting multipliers kP, kN and kZ are as shown in Case 2 in Fig. 2.12. In summary, in the case of a single-phase to earth short-circuit fault on any phase in a three-phase network, the assumption of the fault being on phase R results in the simplest mathematical equations, because the three multipliers kP, kN and kZ are all equal to unity.

2.4.3 Two-phase to earth short-circuit faults For a two-phase to earth short-circuit fault, we have three cases as follows: 1. Phase Y to B to earth short-circuit fault, phase R is the reference phase VRP 5 VRN 5 VRZ

and

IRP 1 IRN 1 IRZ 5 0

(2.94)

Symmetrical components analysis of faulted three-phase networks

77

2. Phase R to B to earth short-circuit fault, phase Y is the reference phase VYP 5 VYN 5 VYZ

IYP 1 IYN 1 IYZ 5 0

and

(2.95a)

or expressed in terms of phase R h2 VRP 5 hVRN 5 VRZ

h2 IRP 1 hIRN 1 IRZ 5 0

and

(2.95b)

3. Phase R to Y to earth short-circuit fault, phase B is the reference phase VBP 5 VBN 5 VBZ

IBP 1 IBN 1 IBZ 5 0

and

(2.96a)

or expressed in terms of phase R hVRP 5 h2 VRN 5 VRZ

and

hIRP 1 h2 IRN 1 IRZ 5 0

(2.96b)

Eqs. (2.94), (2.95b) and (2.96b) can be represented by the connection of the positive-sequence, negative-sequence and zero-sequence networks as shown in Fig. 2.13. The sequence networks are still connected in parallel but through three complex multipliers kP, kN and kZ that are equal to either 1, h or h2 according to Eqs. (2.95b) and (2.96b) as shown in Case 1 in Fig. 2.13. Again, applying phase shifts to the active positive-sequence network can be avoided by dividing Eqs. (2.95b) and (2.96b) by h2 and h, respectively, and the resulting multipliers kP, kN and kZ are as shown in Case 2 in Fig. 2.13. A comparison of the tables showing the complex multipliers kP, kN and kZ in Figs. 2.12 and 2.13 show that for the same reference phase, each multiplier, kP, kN or kZ, has the same value irrespective of the fault type. P

I P

Z

V

P

1:k

P

P P

k I P

k V

I

P

Z

N

N

V

N

1:k

N

N N

I

k I N

k V

N

Z

Z

Z

V

1:k

Z

Z

Z Z

k I Z

k V

Z

VF

h = e j2π/ 3

Two-phase to earth short-circuit on phases: (Y – B – E R is reference) (R – B – E Y is reference) (R – Y – E B is reference)

Case 1

kP

kN

kZ

1

1

1

h2

h

1

h

h2

1

Two-phase to earth short-circuit on phases: (Y – B – E R is reference) (R – B – E Y is reference) (R – Y – E B is reference)

kP

kN

kZ

1

1

1

1

h2

h

1

h

h2

Case 2

Figure 2.13 Two-phase to earth short circuit on Y 2 B 2 E, R 2 B 2 E or R 2 Y 2 E.

78

Power Systems Modelling and Fault Analysis

The methodology presented can be easily extended to any other unbalanced short-circuit or unbalanced open-circuit fault. The reader is encouraged to repeat the analysis for other fault conditions. These include: (a) a one-phase open-circuit fault on any phase, i.e., R or Y or B; (b) a two-phase open-circuit fault on any two phases, i.e., RY or YB or RB; and (c) a two-phase short-circuit fault on any two phases, i.e., RY or YB or RB.

2.5

Analysis of simultaneous faults

2.5.1 General Simultaneous faults are where more than one fault occur at the same time in a three-phase power system either at the same or at different locations. Because there is a very large theoretical combination of such faults, we will limit our attention to a representative number of cases that are of practical interest. These are three cases of two simultaneous faults and one case of three simultaneous faults. The analysis of simultaneous faults can be simplified by deriving the conditions at the fault locations with respect to the same reference phase, i.e., phase R. These conditions determine the method of connection of the positive-sequence, negative-sequence and zero-sequence networks. In this analysis, we assume that the full network has been reduced to an equivalent as seen from the faulted boundary locations and the latter, denoted as nodes J and L, have been retained. Network reduction is discussed in detail in Chapter 9, Network equivalents and practical short-circuit assessments in large scale AC power systems.

2.5.2 Simultaneous short-circuit faults at the same location The two simultaneous faults we consider are illustrated in Fig. 2.14A and consist of a solid one-phase to earth short-circuit fault and a solid two-phase short-circuit fault. Conceptually, since all three phases are faulted, the two faults can also be considered as a single three-phase unbalanced fault! Using phase R as the reference phase, the single-phase short circuit assumed to occur on phase R and the twophase short circuit assumed to occur on phases Y and B, the conditions at the fault location are given by VR 5 0

IY 1 IB 5 0

VY 5 VB

(2.97)

Using Eqs. (2.9b) and (2.9c) for IY and IB instead of voltages, we obtain I P 1 I N 5 2I Z

(2.98)

Symmetrical components analysis of faulted three-phase networks

(A)

Balanced three-phase power system

79

IR IY IB

Y

B R

P N I + I 1: 2

(B) P

I

I

VP

N

V

N

V

Z

I

Z

P

Z

Z

+

Z

N

Z

VF

NP Thévenin’s PPS network

NN Thévenin’s NPS network

Ideal transformer

NZ Thévenin’s ZPS network

Figure 2.14 Simultaneous one-phase to earth and two-phase short-circuits at the same location, or three-phase unbalanced short-circuit fault. (A) Fault representation and (B) Connection of positive, negative and zero sequence Thevenin equivalent circuits.

Substituting Eqs. (2.19), (2.20) and (2.21) for positive-sequence, negative-sequence and zero-sequence voltages at the fault point into Eqs. (2.9a), (2.9b) and (2.9c), we have VR 5 ðVF 2 Z P I P Þ 1 ð 2 Z N I N Þ 1 ð 2 Z Z I Z Þ 5 0

(2.99a)

VY 5 h2 ðVF 2 Z P I P Þ 1 hð 2 Z N I N Þ 1 ð 2 Z Z I Z Þ

(2.99b)

VB 5 hðVF 2 Z P I P Þ 1 h2 ð 2 Z N I N Þ 1 ð 2 Z Z I Z Þ

(2.99c)

and

Adding Eqs. (2.99a)(2.99c), we obtain VR 1 VY 1 VB 5 2 3ZZIZ 5 2VY 5 2VB or V Y 5 VB 5

2 3Z Z I Z 2

(2.100)

Also, using Eqs. (2.99b) and (2.99c) as well as VY 5 VB from Eq. (2.97), we have VY 2 hVB 5 (1 2 h)VY 5 (1 2 h)(ZNIN 2 ZZIZ) or VY 5 Z N I N 2 Z Z I Z

(2.101)

80

Power Systems Modelling and Fault Analysis

Equating Eqs. (2.100) and (2.101), we obtain 2 2Z N N I ZZ

IZ 5

(2.102)

Substituting Eq. (2.102) into Eq. (2.98), we obtain ZZ IP 1 ZZ

IN 5

4Z N

and

IZ 5

2Z N IP 1 ZZ

4Z N

(2.103)

Substituting Eq. (2.103) into Eq. (2.99a), the sequence fault currents are 4Z N 1 Z Z VF 1 ZNZZ 1 ZZZP

(2.104a)

IN 5

2 ZZ VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.104b)

IZ 5

2Z N VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.104c)

IP 5

4Z P Z N

Using Eqs. (2.9a), (2.9b) and (2.9c), the phase fault currents are given by 6Z N VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P pffiffiffi 2 j 3ð2Z N 1 Z Z Þ IY 5 IB 5 P N VF 4Z Z 1 Z N Z Z 1 Z Z Z P

IR 5

(2.105a)

(2.105b)

Subtracting Eq. (2.9c) from Eq. (2.9b) and using VY 5 VB, we obtain VP 5 VN. Using Eq. (2.9a) with VR 5 0 and VP 5 VN, we obtain VZ 5 2 2VP. Using Eq. (2.20), we obtain VP 5 VN 5

ZNZZ VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.106a)

and VZ 5

2 2Z N Z Z VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.106b)

Finally, using Eq. (2.100), the phase voltages on phases Y and B are given by V Y 5 VB 5

2 3Z N Z Z VF 4Z P Z N 1 Z N Z Z 1 Z Z Z P

(2.106c)

Symmetrical components analysis of faulted three-phase networks

81

Using Eq. (2.98) or IZ 5 (IP 1 IN)/2 and VZ 5 2 2VP 5 2 2VN, the equivalent circuit of this simultaneous fault condition is shown in Fig. 2.14B. It should be noted that in this case, the real transformation ratio acts on the voltage and current as a normal ideal transformer does.

2.5.3 Cross-country faults or simultaneous faults at different locations A cross-country fault is a condition where there are two simultaneous one-phase to earth short-circuit faults affecting the same circuit but at different locations and possibly involving different phases. Therefore, the fault conditions are a one-phase to earth, phase Rearth (E) short-circuit fault at location J and one-phase to earth short-circuit fault on either RE, YE or BE at location L. The fault conditions at both fault locations are given as follows: Location J: RE short-circuit fault VJP 1 VJN 1 VJZ 5 0

and

IJP 5 IJN 5 IJZ

(2.107)

Location L: RE short-circuit fault VLP 1 VLN 1 VLZ 5 0

and

ILP 5 ILN 5 ILZ

(2.108)

Location L: YE short-circuit fault h2 VLP 1 hVLN 1 VLZ 5 0

and

h2 ILP 5 hILN 5 ILZ

VLP 1 h2 VLN 1 hVLZ 5 0

and

ILP 5 h2 ILN 5 hILZ

or (2.109)

Location L: BE short-circuit fault hVLP 1 h2 VLN 1 VLZ 5 0

and

hILP 5 h2 ILN 5 ILZ

VLP 1 hVLN 1 h2 VLZ 5 0

and

ILP 5 hILN 5 h2 ILZ

or (2.110)

Fig. 2.15 shows the connection of the positive-sequence, negative-sequence and zerosequence networks at the two fault locations J and L using the three sets of Eqs. (2.107) and (2.108), (2.107) and (2.109) and (2.107) and (2.110).

Straightforward analysis using Kirchoff’s voltage and current laws can be carried out using Fig. 2.15.

82

Power Systems Modelling and Fault Analysis

Node J R Y B

Node L R Balanced Y three-phase power system B P

Type of fault at node L kP R-E 1 Y-E 1 B-E 1

JP

IJ

LP IPL

PPS network

P

VJ

1:k

P

P

kZ 1 h h2

h = e j2π/ 3

P P

k IL P

VL

kN 1 h2 h

P

k VL

NP

N

JN

IJ

LN

NPS network

N

VJ

N

1:k

IL

N

N

N N

k IL N

VL

N

k VL

NN

Z

JZ

IJ

LZ

ZPS network

Z

VJ

Z

IL

1:k

Z

Z

VL

Z Z

k IL Z

Z

k VL

NZ

Figure 2.15 Cross-country simultaneous one-phase to earth short-circuit faults.

2.5.4 Simultaneous open-circuit and short-circuit faults at the same location We now consider a one-phase open-circuit fault on phase R at location J and a onephase to earth short-circuit fault RE or YE or BE at the same location. The conditions created by the open-circuit fault on phase R at location J creating open-circuit JJ0 are given by VJJP 0 5 VJJN0 5 VJJZ 0

IJJP 0 1 IJJN0 1 IJJZ 0 5 0

and

(2.111)

The conditions created by an RE short-circuit fault at location J are given by VJP 1 VJN 1 VJZ 5 0

and

IJP 5 IJN 5 IJZ

(2.112)

The conditions created by a YE short-circuit fault at location J are given by h2 VJP 1 hVJN 1 VJZ 5 0

and

h2 IJP 5 hIJN 5 IJZ

VJP 1 h2 VJN 1 hVJZ 5 0

and

IJP 5 h2 IJN 5 hIJZ

or (2.113)

Symmetrical components analysis of faulted three-phase networks

R Y B

Nodes J

J'

R Y B

P P

P

k IJ P

Single-phase shortcircuit at node J R-E Y-E B-E JP IPJJ ′

P

k :1

IJ P

P

VJ

k VJ

83

kP 1 1 1 P

P

V JJ ′

IJJ ′

kN 1 h2 h

kZ 1 h = e j2π/ 3 h h2

J′P

PPS network NP

N

N N k IJ k N:1 IJ

k

N

N VJ

N

IJJ ′

JN

N VJ

N

N

V JJ ′

IJJ ′

J′N

NPS network NN

Z Z k IJ Z

Z

k :1 Z

k VJ

Z IJJ ′

Z IJ

JZ Z

VJ

Z

Z

V JJ ′

IJJ ′

J′Z

ZPS network NZ

Figure 2.16 Simultaneous one-phase open-circuit and one-phase short-circuit faults.

The conditions created by a BE short-circuit fault at location J are given by hVJP 1 h2 VJN 1 VJZ 5 0

and

hIJP 5 h2 IJN 5 IJZ

VJP 1 hVJN 1 h2 VJZ 5 0

and

IJP 5 hIJN 5 h2 IJZ

or (2.114)

Fig. 2.16 shows the connection of the positive-sequence, negative-sequence and zero-sequence networks using the three sets of Eqs. (2.111) and (2.112), (2.111) and (2.113) and (2.111) and (2.114). Straightforward analysis using Kirchoff’s voltage and current laws can be carried out using Fig. 2.16.

2.5.5 Simultaneous faults caused by broken and fallen to earth conductors Fig. 2.17 illustrates a one-phase open-circuit fault on phase R caused by a broken phase conductor. The conductors on both sides of the open circuit are assumed to fall to earth thus creating a one-phase to earth short-circuit fault on each side. The conditions created by the open-circuit fault on phase R creating an opencircuit JJ0 are given by

84

Power Systems Modelling and Fault Analysis

B Y R

B Y R

J'

J

P

IJ

JP IPJJ ′

P

IJ

1:1

P

P

VJ

VJ

P

P

V JJ ′

IJJ ′

J′P

P

IJ′

1:1 P

PPS network

VJ

P

IJ′ P

VJ

NP N

IJ

1:1 N

N

IJ

N

IJJ ′

JN

N

VJ

VJ

N

N

J′N IJ′

N

N

V JJ ′

IJJ ′

1:1 IJ′ N

VJ′

NPS network

N

VJ′

NN Z IJ

Z IJJ ′

Z IJ

1:1

JZ

Z

Z

VJ

VJ

Z

Z

Z

V JJ ′

IJJ ′

ZPS network

J′Z

Z IJ′ 1:1 IJ′ Z

VJ′

Z

VJ′

NZ

Figure 2.17 Three simultaneous faults caused by a broken and fallen to earth conductor.

VJJP 0 5 VJJN0 5 VJJZ 0

and

IJJP 0 1 IJJN0 1 IJJZ 0 5 0

(2.115)

Short-circuit fault on phase R at location J VJP 1 VJN 1 VJZ 5 0

and

IJP 5 IJN 5 IJZ

(2.116)

Short-circuit fault on phase R at location J0 VJP0 1 VJN0 1 VJZ0 5 0

and

IJP0 5 IJN0 5 IJZ0

(2.117)

Fig. 2.17 shows the connection of the positive-sequence, negative-sequence and zero-sequence networks using Eqs. (2.115), (2.116) and (2.117). Straightforward analysis using Kirchoff’s voltage and current laws can be carried out using Fig. 2.17.

2.5.6 Simultaneous short-circuit and open-circuit faults on distribution transformers We will now analyse in detail a simultaneous fault case that involves a twowinding distribution transformer that is extensively used in distribution substations. The high-voltage winding is delta-connected with fuses being used as incoming protection against high currents. The low-voltage winding is star-connected with the neutral solidly earthed and the transformer supplies a balanced three-phase load.

Symmetrical components analysis of faulted three-phase networks

85

The simultaneous faults may be created when a short-circuit fault occurs on the transformer low-voltage side which causes one fuse to blow and clear before the other fuses, or a circuit-breaker upstream, open and clear the fault. This can lead to a situation where the short-circuit currents are too low to operate any further protection so that the prolonged duration of such currents may overstress or even damage power plant. The transformer is assumed to supply a balanced star-connected three-phase static load having an impedance per phase of ZL and therefore the corresponding load positive-sequence, negative-sequence and zero-sequence impedances are all equal to ZL. The detailed sequence modelling of transformers will be covered in Chapter 4, Modelling of transformers, phase shifters, static power plant and static load, but we will present this case now because of its practical importance and relevance in this chapter. We will also make use of the transformer phase shifts introduced by the stardelta winding connections and presented in Chapter 4, Modelling of transformers, phase shifters, static power plant and static load, and we will consider the transformer vector group to be D11yn or in accordance with American ANSI standard. This means that when stepping up from the low-voltage to the high-voltage side, the transformer’s positive-sequence currents (and voltages) are advanced by 30
whereas the negative-sequence currents (and voltages) are retarded by 30
. The simultaneous faults are illustrated in Fig. 2.18A and the equivalent sequence network connections are shown in Fig. 2.18B. From Fig. 2.18B, the following relations can be written


P P j30 5 ILV e ISP 5 IHV




P P j30 VHV 5 VLV e

and


N N 2j30 ISN 5 IHV 5 ILV e

and

(2.118)


N N 2j30 VHV 5 VLV e

ISP 5 2 ISN

(2.119) (2.120)

kP IFP 5 kN IFN 5 kZ IFZ

and

kP VFP 1 kN VFN 1 kZ VFZ 5 0

(2.121)

Therefore, from the above current equations, we obtain


P j30 N 2j30 e 5 2 ILV e ILV







N P j60 ILV 5 2 ILV e

(2.122)

and IFP 5 kN IFN

IFZ 5

kN N I kZ F

kP 5 1

(2.123)

Applying Kirchoff’s voltage law to the high-voltage side of the positive-sequence P N network of Fig. 2.18B, we obtain VS 5 ISP ZSP 2 ISN ZSN 1 VHV 2 VHV or using Eqs. (2.118)(2.121)

86

Power Systems Modelling and Fault Analysis

(A)

High-voltage side open-circuit FHV F'HV

Balanced three-phase I Y(HV) power system IB(HV) Fuses

(B)

Low-voltage side one-phase short-circuit FLV IR(LV) Load ZL IY(LV) ZL ZL IB(LV)

D11

jπ/ 6 :1 P e P IS FHV F′HV I HV

ZHL

P

FLV

IF

VF

P

ZL

FLV

IF

P

ILV

1:k

P

P P

k IF

P

ZS + V – S

N

IS

P

VHV

P

P

VLV

N e− jπ/ 6 :1 F′HV IHV

ZHL

N

ILV

N

P

k VF

1:k

N

N N

k IF

FHV N

N

VHV

ZS

N

N

VLV

VF

Z

Z IS = 0

ZHL FHV

FLV

Z

ILV

N

Z

IF

1:k

F′HV Z

Z

VF

ZS

Type of fault on LV side R-E Y-E B-E

N

k VF

ZL

Z

ZL

Z

Z Z

k IF Z

Z

k VF

kP kN kZ 1 1 1 1 h2 h 1 h h2

h = e j2π / 3

Figure 2.18 Simultaneous one-phase short circuit on transformer low-voltage side and onephase open circuit on transformer high-voltage side. (A) Fault representation and (B) Connection of positive, negative and zero sequence equivalent circuits.








P j30 P j30 N 2j30 VS 5 ðZSP 1 ZSN ÞILV e 1 VLV e 2 VLV e

(2.124)

Applying Kirchoff’s voltage law to the low-voltage side of the positive-sequence network of Fig. 2.18B, and after a little algebra, we obtain P P VLV 5 ILV ðZHL 1 ZL Þ 2 IFP ZL

(2.125)

Applying Kirchoff’s voltage law to the low-voltage side of the negative-sequence network of Fig. 2.18B, and using Eq. (2.122), we obtain


N P j60 5 2 ðZHL 1 ZL ÞILV e 2 IFN ZL VLV

(2.126)

Symmetrical components analysis of faulted three-phase networks

87

Substituting Eqs. (2.125) and (2.126) into Eq. (2.124), and using Eq. (2.123), we obtain, after a little algebra








P j30 e 2 ZL IFN ðkN ej30 2 e2j30 Þ VS 5 ½ðZSP 1 ZSN Þ 1 2ðZHL 1 ZL ÞILV

(2.127)

Using Eqs. (2.122) and (2.123) as well as Fig. 2.18B, the positive-sequence, negativesequence and zero-sequence voltages at the short-circuit fault point are given by P VFP 5 ZL ðILV 2 kN IFN Þ

(2.128)




P j60 VFN 5 2 ZL ðILV e 1 IFN Þ

(2.129)

From the low-voltage side of the zero-sequence network of Fig. 2.18B, we have Z Z Z VFZ 5 ZLZ ðILV 2 IFZ Þ 5 2 ZHL ILV which, using Eq. (2.123) and Fig. 2.18B becomes VFZ 5 2

Z kN ZLZ ZHL IN Z Z F Z k ZL 1 ZHL

(2.130)

Now, by substituting Eqs. (2.128), (2.129) and (2.130) into the sequence voltages of Eq. (2.121), we can express INF in terms of IPLV as


IFN 5

ZL ð1 2 kN ej60 Þ  IP Z =ðZ Z 1 Z Z Þ LV kN 2ZL 1 ZLZ ZHL HL L 

(2.131)

P Substituting Eq. (2.131) into Eq. (2.127) and solving for ILV , we obtain


P ILV 5

VS e2j30 

  P  ZL2 kN ð2 2 kN ej60 Þ 2 e2j60 N

 ðZS 1 ZS Þ 1 2ðZHL 1 ZL Þ 2 N Z = ZZ 1 ZZ k 2ZL 1 ZLZ ZHL HL L

(2.132)

P The calculation of ILV enables us to calculate the required sequence currents and voltages in Fig. 2.18B using Eqs. (2.118)(2.131) by back substitution. We will present in Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators, that the voltage source VS is calculated from the initial load flow solution just before the occurrence of the short-circuit fault as follows:

VS 5 ðZSP 1 ZHL ÞILoad 1 VLoad

(2.133)

The short-circuit fault current on the faulted phase can be calculated from Eq. (2.123) as follows: IF 5 3kN IFN

(2.134)

88

Power Systems Modelling and Fault Analysis

The corresponding currents on the high-voltage side can be calculated using Eqs. (2.15a), (2.118), (2.119) and (2.122) to give IY 5

pffiffiffi P 2j60
3ILV e

(2.135a)

IB 5

pffiffiffi P j120
3ILV e 5 2 IY

(2.135b)

and

The reader is encouraged to repeat the above analysis for a solid two-phase to earth short-circuit fault on the transformer low-voltage side considering all three fault combinations of R 2 Y 2 E, R 2 B 2 E and Y 2 B 2 E.

2.6

Symmetrical components analysis of faulted networks containing mixed voltage and current sources

2.6.1 General The symmetrical components analysis of balanced and unbalanced faults presented in Section 2.3 considered networks where only voltage sources supply short-circuit current to the fault. The voltage sources are generally conventional rotating machines such as synchronous generators/motors and induction generators/motors directly connected to the AC network. The physical behaviour of such equipment is presented in detail in Chapter 5, Modelling of rotating AC synchronous and induction machines and Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators. In addition, modern renewable generation technologies, high-voltage direct-current technology employing a voltage-source converter and other equipment that are inverter-interfaced with the power network can behave as current sources rather than voltage sources during network faults. For example, many modern large-scale wind farms and solar photovoltaic parks are designed to behave as voltagedependent current sources thanks to their control strategies during network faults. The behaviour of such equipment is presented in detail in Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators. In this section, we present the method of symmetrical components analysis of faulted networks containing both voltage and current sources. The application of the superposition theorem in the analysis of faulted networks that contain voltage sources and voltage-controlled current sources is presented in Chapter 7, Shortcircuit analysis techniques in large-scale AC power systems. Here, we present the symmetrical component analysis of balanced and unbalanced short-circuit faults in such networks.

Symmetrical components analysis of faulted three-phase networks

89

2.6.2 Symmetrical components of voltage and current sources The sequence components of a balanced three-phase voltage source, ERYB, are calculated using Eqs. (2.12a) and (2.12b) and are given by 2 3 EP 1 4 EN 5 5 1 4 1 3 EZ 1 2

h h2 1

32 3 2 3 ER ER h2 h 54 EY 5 5 4 0 5 0 EB 1

(2.136a)

EZ 5 0

(2.136b)

or E P 5 ER ;

EN 5 0;

Similarly, for a balanced three-phase current source IRYB, we have I P 5 IR ;

I N 5 0;

IZ 5 0

(2.136c)

Eqs. (2.136a)(2.136c) show that balanced three-phase voltage/current sources produce only positive-sequence voltage/current components. Negative-sequence and zero-sequence components are not produced. This is a basic but important result. In Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators, we cover new types of unbalanced three-phase current sources that are controlled to produce a positive-sequence current component under balanced network faults, and both positive-sequence and negative-sequence current components, but not zero-sequence current components, under unbalanced network faults. The sequence components of such an unbalanced three-phase current source, IRYB, are calculated using Eqs. (2.12a) and (2.12b) and are given by 2

3 2 IP 1 1 4 IN 5 5 4 1 3 0 1

h h2 1

32 3 IR h2 h 54 IY 5 IB 1

(2.137a)

or I P 1 I N 5 IR

(2.137b)

  1 1 IR 1 j pffiffiffi ðIY 2 IB Þ I 5 2 3

(2.137c)

P

IN 5

 

 1 1 IR 2 j pffiffiffi ðIY 2 IB Þ 5 I P 2 3

(2.137d)

90

Power Systems Modelling and Fault Analysis

Conversely, the unbalanced three-phase current source, IRYB, is calculated from the positive- and negative-sequence current components using Eqs. (2.10a) and (2.10b) as follows 2

3 2 IR 1 4 IY 5 5 4 h 2 h IB

1 h h2

32 P 3 1 I 1 54 I N 5 1 0

(2.138a)

or IR 5 I P 1 I N

(2.138b)

IY 5 2

 pffiffiffi

i 1 h P I 1 IN 1 j 3 IP 2 IN 2

(2.138c)

IB 5 2

 pffiffiffi

i 1 h P I 1 I N 2 j 3 I P 2 I N 5 IY 2

(2.138d)

2.6.3 Balanced three-phase to earth short-circuit faults Fig. 2.19A shows a balanced three-phase short-circuit fault in a balanced power system network that contains balanced three-phase current sources as well as balanced voltage sources. Fig. 2.19B shows the equivalent positive-sequence circuit with both positive-sequence voltage and current sources active. The solution of the fault current is based on the superposition theorem. Fig. 2.19C shows the superposition of two circuits: a The´venin’s equivalent of the network and its voltage sources but with all current sources open-circuited, and an equivalent current source at the fault point calculated with all voltage sources short-circuited. A general method for the calculation of this current source is presented in Chapter 7, Short-circuit analysis techniques in large-scale AC power systems. From Fig. 2.19C, and using the superposition theorem, the positive-sequence fault current is given by IF 5

VF 1 Is ZP

(2.139a)

where Is may be an independent current source or a voltage-dependent current source. In Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators and Chapter 7, Short-circuit analysis techniques in large-scale AC power systems, we discuss the modelling of voltagedependent current sources that can be used to represent inverter-interfaced systems such as solar photovoltaic generators and type 4 wind turbine generators. For

Symmetrical components analysis of faulted three-phase networks

91

(A) IR

Balanced three-phase network with balanced three-phase voltage sources and three-phase current sources

(B)

F

IB

Voltagedependent

Independent

F

IY

IP IF

ZP

or

= kVC

+ P

VF

_

(C)

VoltageIndependent dependent F

F IP IF

ZP + P

VF

_

Thévenin’s PPS network with current sources open-circuited

IF

ZP

or

= kVC

+ _ Equivalent current source at fault location F with voltage sources short-circuited

Figure 2.19 Three-phase fault in a balanced power system network with balanced threephase voltage sources and three-phase current sources. (A) Fault representation, (B) Equivalent circuit with equivalent voltage and current sources at fault point, and (C) Superposition of separate voltage and current sources networks.

now, we give the equation of a voltage-dependent positive-sequence current source as Is 5 kVc

(2.139b)

where k is a factor or a constant multiplier synonymous with a conductance and VC is the voltage at the connection point of the current source.

2.6.4 Unbalanced one-phase to earth short-circuit faults Current sources produce positive-sequence current only Fig. 2.20A shows an unbalanced one-phase short-circuit fault in a balanced power system network that contains balanced three-phase voltage sources and balanced

92

Power Systems Modelling and Fault Analysis

(A)

IR

Balanced three-phase network with balanced three-phase voltage sources and balanced three-phase current sources

F

(B) P

+

_ VF

ZP F Voltage-dependent PPS current source

ZN F ZZ F

Thévenin’s sequence networks

Equivalent current sources at fault location F independent PPS current source

(C) ZP P

VF

ZZ + _

ZN

(D) ZP P

VF

ZZ + _

ZN

ZP

ZZ ZN

Figure 2.20 One-phase fault in a balanced power system network with balanced three-phase voltage sources and balanced three-phase current sources. (A) Fault representation, (B) Connection of positive, negative and zero sequence equivalent circuits, (C) Reduced equivalent circuit representation, and (D) Superposition of separate voltage and current sources networks.

three-phase current sources. The current sources produce positive-sequence currents only and their negative- and zero-sequence currents are represented as open switches in Fig. 2.20B. Fig. 2.20C shows the equivalent circuit with both voltage and current sources active. The solution of the fault current is based on the superposition theorem. Fig. 2.20D shows the superposition of two circuits; the The´venin’s

Symmetrical components analysis of faulted three-phase networks

93

equivalent circuit of the network and its voltage sources active but with all current sources open-circuited, and an equivalent circuit with an equivalent current source at the fault point with all voltage sources short-circuited. From Fig. 2.20D, and using the superposition theorem, the positive-sequence fault current with the current source open-circuited is given by P 5 IF1

VF ZP 1 ZN 1 ZZ

(2.140a)

Now, the positive-sequence fault current with the voltage source short-circuited is given by P 5 IF2

ZP IP ZP 1 ZN 1 ZZ S

(2.140b)

The total positive-sequence fault current is given by P P 1 IF2 5 IFP 5 IF1

VF 1 Z P ISP ZP 1 ZN 1 ZZ

(2.140c)

The phase fault current is given by IF 5 3IFP

(2.140d)

As presented in Section (2.6.3), ISP may also be a voltage-dependent positivesequence current source as in Eq. (2.139b).

Current sources produce positive- and negative-sequence currents Fig. 2.21A shows an unbalanced three-phase current source that produces both positive- and negative-sequence currents. Fig. 2.21B shows the interconnected positive-, negative- and zero-sequence circuits, and Fig. 2.21C shows a simplified equivalent circuit with both voltage and current sources active. Using the superposition theorem, Fig. 2.21D shows the superposition of three circuits; the The´venin’s equivalent of the network and its active voltage sources but with all current sources opencircuited, an equivalent circuit with equivalent positive-sequence current sources at the fault point active but with voltage sources short-circuited and negative-sequence current sources open-circuited, and an equivalent circuit with equivalent negativesequence current sources at the fault point active but with voltage sources shortcircuited and positive-sequence current sources open-circuited. Using the superposition theorem and Fig. 2.21D, the positive-sequence fault current with both current sources open-circuited is given by P 5 IF1

VF ZP 1 ZN 1 ZZ

(2.141a)

94

Power Systems Modelling and Fault Analysis

(A)

Balanced three-phase network with balanced three-phase voltage sources and unbalanced three-phase current sources

(B)

P

+

_ VF

IR

F

ZP F

Voltage-dependent PPS current source

ZN F Voltage-dependent NPS current source

ZZ F

Thévenin’s sequence networks

Equivalent current sources at fault location F independent PPS and NPS current sources

(C) ZP + P

VF

_

ZN

ZZ

(D)

ZP

ZP

ZN

ZN

ZN

ZZ

ZZ

ZZ

ZP + P

VF

_

Figure 2.21 One-phase fault in a balanced power system network with balanced three-phase voltage sources and unbalanced three-phase current sources. (A) Fault representation, (B) Connection of positive, negative and zero sequence equivalent circuits, (C) Reduced equivalent circuit representation, and (D) Superposition of separate voltage and current sources networks.

Now, from Fig. 2.21D, the positive-sequence fault current with the voltage source short-circuited and negative-sequence current source open-circuited is given by P 5 IF2

ZP IP ZP 1 ZN 1 ZZ S

(2.141b)

Symmetrical components analysis of faulted three-phase networks

95

Similarly, from Fig. 2.21D, the positive-sequence fault current with the voltage source short-circuited and positive-sequence current source open-circuited is given by P 5 IF3

ZP

ZN IN 1 ZN 1 ZZ S

(2.141c)

The total positive-sequence fault current is given by P P P 1 IF2 1 IF3 5 IFP 5 IF1

VF 1 Z P ISP 1 Z N ISN ZP 1 ZN 1 ZZ

(2.141d)

The phase fault current is given by IF 5 3IFP

(2.141e)

2.6.5 Unbalanced two-phase short-circuit faults Current sources produce positive-sequence current only Fig. 2.22A shows an unbalanced two-phase short-circuit fault in a balanced power system network that contains balanced three-phase voltage sources and balanced threephase current sources. Fig. 2.22B shows the interconnected sequence circuits and Fig. 2.22C shows the equivalent circuit with both voltage and current sources active. The solution of the fault current is based on the superposition theorem. Fig. 2.22D shows the superposition of two circuits; the The´venin’s equivalent circuit of the network and its voltage sources active but with all current sources open-circuited, and an equivalent circuit with an equivalent current source at the fault point with all voltage sources short-circuited. From Fig. 2.22D, and using the superposition theorem, the positive-sequence fault current with the current source open-circuited is given by P IF1 5

VF ZP 1 ZN

(2.142a)

Now, the positive-sequence fault current with the voltage source short-circuited is given by P IF2 5

ZP

ZP IP 1 ZN S

(2.142b)

The total positive-sequence fault current is given by P P 1 IF2 5 IFP 5 IF1

VF 1 Z P ISP ZP 1 ZN

(2.142c)

96

Power Systems Modelling and Fault Analysis

(A)

IR

Balanced three-phase network with balanced three-phase voltage sources and balanced three-phase current sources

(B)

P

+

_ VF

F

IY

ZP F Voltage-dependent PPS current source ZN

Thévenin’s sequence networks

F

Equivalent current sources at fault location F independent PPS current source

(C) ZP P VF

ZN

+ _

(D) ZP

ZP P VF

+ _

ZN

ZN

Figure 2.22 Two-phase fault in a balanced power system network with balanced three-phase voltage sources and balanced three-phase current sources. (A) Fault representation, (B) Connection of positive and negative sequence equivalent circuits, (C) Reduced equivalent circuit representation, and (D) Superposition of separate voltage and current sources networks.

The phase fault current flowing on phase B for a fault on phases Y and B is given by pffiffiffi

 pffiffiffi j 3 VF 1 Z P ISP IF 5 j 3IFP 5 ZP 1 ZN

(2.142d)

Symmetrical components analysis of faulted three-phase networks

97

As presented in Section (2.6.3), ISP may also be a voltage-dependent positivesequence current source as in Eq. (2.139b).

Current sources produce positive- and negative-sequence currents Fig. 2.23A shows an unbalanced three-phase current source that produces both positive- and negative-sequence currents. Fig. 2.23B shows the interconnected positive-, negative- and zero-sequence circuits and Fig. 2.23C shows a simplified equivalent circuit with both voltage and current sources active. Using the superposition (A)

IR

Balanced three-phase network with balanced three-phase voltage sources and unbalanced three-phase current sources

(B)

P

IY

ZP

+

_ VF

F

F

Voltage-dependent PPS current source

F

ZN

Voltage-dependent PPS current source

Thévenin’s sequence networks

Equivalent current sources at fault location F independent PPS and NPS current sources

(C) ZP +

ZN

P

VF

_

(D) + P VF

ZP

ZP

ZP ZN

ZN

ZN

_

Figure 2.23 Two-phase fault in a balanced power system network with balanced three-phase voltage sources and unbalanced three-phase current sources. (A) Fault representation, (B) Connection of positive and negative sequence equivalent circuits, (C) Reduced equivalent circuit representation, and (D) Superposition of separate voltage and current sources networks.

98

Power Systems Modelling and Fault Analysis

theorem, Fig. 2.23D shows the superposition of three circuits; the The´venin’s equivalent of the network and its active voltage sources but with all current sources open-circuited, an equivalent circuit with equivalent positive-sequence current sources at the fault point active but with voltage sources short-circuited and negative-sequence current sources open-circuited, and an equivalent circuit with equivalent negative-sequence current sources at the fault point active but with voltage sources short-circuited and positive-sequence current sources open-circuited. Using the superposition theorem and Fig. 2.23D, the positive-sequence fault current with both current sources open-circuited is given by P 5 IF1

VF ZP 1 ZN

(2.143a)

Now, from Fig. 2.23D, the positive-sequence fault current with the voltage source short-circuited and negative-sequence current source open-circuited is given by P IF2 5

ZP IP ZP 1 ZN S

(2.143b)

Similarly, from Fig. 2.23D, the positive-sequence fault current with the voltage source short-circuited and positive-sequence current source open-circuited is given by P IF3 5

Z N N I ZP 1 ZN S

(2.143c)

The total positive-sequence fault current is given by P P P 1 IF2 1 IF3 5 IFP 5 IF1

VF 1 Z P ISP  Z N ISN ZP 1 ZN

(2.143d)

The phase fault current flowing on phase B for a fault on phases Y and B is given by pffiffiffi

 pffiffiffi P j 3 VF 1 Z P ISP  Z N ISN IF 5 j 3IF 5 ZP 1 ZN

(2.143e)

Further reading Books Anderson, P.M., Analysis of Faulted Power Systems, Iowa State Press, Ames, IA, 1973. Blackburn, J.L., Symmetrical Components for Power Systems Engineering, 1993, ISBN 0824787676.

Symmetrical components analysis of faulted three-phase networks

99

Elgerd, O.I., Electric Energy Systems Theory, McGraw-Hill Int. Ed., 1983, ISBN 0-0766273-8. Grainger, J., Stevenson, W.D., Power System Analysis, 1994. ISBN 0701133380. Wagner, C.F., et al., Symmetrical Components, McGraw-Hill Book Company, Inc., 1933. Weedy, B.M., Electric Power Systems, John Wiley & Sons, 19670-471-92445-8.

Paper Fortescue, C.L., Method of symmetrical coordinates applied to the solution of polyphase networks, Transactions of AIEE, Vol. Vol. 37, 1918, 10271140.

3

Modelling of multiconductor overhead power lines, underground and submarine cables Chapter Outline 3.1 General 102 3.2 Phase and sequence modelling of three-phase overhead power lines

102

3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.2.6 3.2.7

Background 102 Formulation of overhead power line parameters: impedances and susceptances 104 Untransposed single-circuit three-phase lines with and without earth wires 119 Transposition of single-circuit three-phase overhead lines 127 Untransposed double-circuit overhead lines with earth wires 133 Transposition of double-circuit overhead lines 138 Phase and sequence parameter matrices of untransposed and transposed multiple-circuit lines 154 3.2.8 Examples 158

3.3 Phase and sequence modelling of three-phase cables 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.3.6 3.3.7

172

Background 172 Cable sheath bonding and earthing arrangements 174 Formulation of cable electrical parameters
loop and phase impedances, and susceptances 176 Sequence impedance matrices of single-circuit cables of different layouts 191 Shunt phase and sequence susceptance matrices of single-circuit cables 201 Impedance matrix of three-phase double-circuit cables 205 Examples 207

3.4 Modelling of actual cables with semiconducting screens 3.4.1 Background 210 3.4.2 Accounting for all layers of actual cables 3.4.3 Example 215

210

211

3.5 Sequence π models of single-circuit and double-circuit overhead lines and cables 217 3.5.1 3.5.2 3.5.3 3.5.4

Background 217 Sequence π models of single-circuit overhead lines and cables 219 Sequence π models of double-circuit overhead lines 221 Sequence π models of double-circuit cables 224

3.6 Sequence π models of three-circuit overhead lines 225 3.7 Three-phase modelling of overhead lines and cables (phase frame of reference) 227 3.7.1 Background 227 3.7.2 Single-circuit overhead lines and cables 227 3.7.3 Double-circuit overhead lines and cables 228 Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00003-5 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

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Power Systems Modelling and Fault Analysis

3.8 Computer calculations and measurements of overhead line and cable parameters 231 3.8.1 Computer calculations of overhead line and cable parameters 231 3.8.2 Measurement of overhead line parameters 231 3.8.3 Measurement of cable parameters 238

3.9 Practical aspects of phase and sequence parameters of overhead lines and cables 242 3.9.1 Overhead lines 242 3.9.2 Cables 243

Further reading

3.1

243

General

In this chapter, we present the modelling and calculation of electrical parameters of multiconductor overhead lines and cables, both in the RYB phase frame of reference as well as in the positive-, negative- and zero-phase sequence frame of reference. The correct calculation of these parameters is crucial for power system analysis and protection studies. Calculations and measurement techniques of the electrical parameters, or constants, of lines and cables based on material properties, geometrical and physical layouts are described. Modelling of cables with semiconducting screens is presented. Transposition analysis of single-circuit and multiplecircuit overhead lines, and sheaths and cores of cables, are presented as well as their π models in the sequence and phase frames of reference. Practical guidelines on computer calculations and measurements of line and cable parameters are given.

3.2

Phase and sequence modelling of three-phase overhead power lines

3.2.1 Background The transmission and distribution of three-phase electrical power on overhead lines requires the use of at least three-phase conductors. Most low-voltage lines use three-phase conductors forming a single three-phase circuit. Many higher-voltage lines consist of a single three-phase circuit or two three-phase circuits strung or suspended from the same tower structure and usually called a double-circuit line. The two circuits may be strung in a variety of configurations such as vertical, horizontal or triangular configurations. Infrequently, multi-circuit towers are used with all circuits having the same or different voltages are strung on one tower. Fig. 3.1 illustrates typical single-circuit and double-circuit overhead lines in horizontal, triangular and vertical-phase conductor arrangements. A line may also consist of two circuits running physically in parallel but on different towers.

Modelling of multiconductor overhead power lines, underground and submarine cables

103

Earth wire (A) 3-phase

3-phase

circuit

circuit

Near Vertical Horizontal

Triangular

Triangular

(i) Single-circuit lines

Counterpoise

(ii) Double-circuit lines

(B) Earth wires Quad Twin Insulators Towers

Suspension tower

Tension/angle tower

Figure 3.1 (A) Typical single-circuit and double-circuit overhead lines and (B) doublecircuit overhead lines with one earth wire: twin bundle 5 two conductors per phase and quad bundle 5 four conductors per phase.

In addition, a few lines have been built with three, four or even six three-phase circuits strung on the same tower structure in various horizontal and/or triangular formations. In England and Wales, almost 99% of the 400 and 275 kV overhead transmission system consists of vertical or near-vertical double-circuit line configurations. In addition to the phase conductors, earth wire conductors may be strung to the tower top and normally bonded to the top of the earthed tower. Earth wires perform two important functions: shielding the phase conductors from direct lightning strikes and providing a low-impedance path for the short-circuit fault current in the event of a back flashover from the phase conductors to the tower structure. The ground itself over which the line runs is an important additional lossy conductor having complex and distributed electrical characteristics. In the case of high resistivity or lossy earths, it is usual to use a counterpoise, that is, a wire buried underground beneath the tower base and connected to the footings of the towers. This serves to reduce the effective tower footing resistance. Where a metallic pipeline

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runs in close proximity to an overhead line, a counterpoise may also be used in parallel with the pipeline in order to reduce the induced voltage on the pipeline from the power line. Therefore, a practical overhead transmission line is a complex arrangement of conductors, all of which are mutually coupled not only to each other but also to earth. The mutual coupling is both electromagnetic (i.e., inductive) and electrostatic (i.e., capacitive). The asymmetrical positions of the phase conductors with respect to each other, the earth wire(s) and/or the surface of the earth cause some unbalance in the phase impedances, and hence currents and voltages. This is undesirable and in order to minimise the effect of line unbalance, it is possible to interchange the conductor positions at regular intervals along the line route, a practice known as transposition. The aim of this is to achieve some averaging of line parameters and hence balance for each phase. However, in practice, to avoid the inconvenience, costs and delays, most lines are not transposed along their routes but transposition is carried out where it is physically convenient at the line terminals, that is, at substations. Bundled phase conductors are usually used on transmission lines at 220 kV and above but are also used on lower HV voltages, for example, 132 kV where a higher line rating is required. Bundled phase conductors are constructed with more than one conductor per phase separated at regular intervals along the span length between two towers by metal spacers. Conductor bundles of two, three, four, six and eight are in use in various countries and, in Great Britain, two, three (triangle) and four (square or rectangle) conductor bundles are used at 275 and 400 kV. The purpose of bundled conductors is to reduce the voltage gradients at the surface of the conductors because the bundle appears as an equivalent conductor of larger diameter than that of the component conductors. This minimises active losses due to corona, reduces noise generation, for example, radio interference, reduces the inductive reactance and increases the capacitive susceptance or capacitance of the line. The reduction of inductive reactance improves the steady-state power transfer capability of the line. Fig. 3.1A(ii) shows a typical 400-kV double-circuit line of vertical-phase conductor arrangement having four bundled conductors per phase, one earth wire and one counterpoise wire. The total number of conductors in such a large multiconductor system is (4 3 3) 3 2 1 1 1 1 5 26 conductors, all of which are mutually coupled to each other and to earth.

3.2.2 Formulation of overhead power line parameters: impedances and susceptances General A line is a static power plant that has electrical parameters distributed along its length. The basic parameters of the line are conductor series impedance and shunt admittance. Each conductor has a self-impedance and there is a mutual impedance between any two conductors. The impedance generally consists of a resistance and a reactance. The shunt admittance consists of the conductor’s conductance to

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105

ground and the susceptance between conductors and between each conductor and earth. The conductance of the air path to earth represents the leakage current along the line insulators due to corona. This is negligibly small and is normally ignored in network models used in short circuit, power flow and transient stability analysis. Practical calculations of multiconductor line parameters with series impedance expressed in pu length (e.g., Ω/km) and shunt susceptance in μS/km are carried out using digital computer programs. These parameters are then used to form the line series impedance and shunt admittance matrices in the phase frame of reference as will be described later. The line capacitances or susceptances are calculated from the line potential coefficients which are essentially dependent on the line and tower physical dimensions and geometry. The calculations use the method of image conductors, and assume that the earth is a plane at a uniform zero potential and that the conductor radii are much smaller than the spacings among the conductors. The self and mutual impedances depend on the conductor material, construction, tower or line physical dimensions or geometry, and on the earth’s resistivity. Fig. 3.2 is a general illustration of overhead line tower physical dimensions and spacings of conductors above the earth’s surface as well as conductor images below earth used for the calculation of line electrical parameters. Practical digital computer-based calculations used in industry consider the effect of earth as a lossy conducting medium. The fundamental theories used in the calculations of resistance, inductance and capacitance parameters of overhead lines are extensively covered in most basic power system textbooks and will not be repeated here. The equations used in the calculation of line parameters are presented below.

y

Tower

Conductor bundle i

d ij

yi

Conductor bundle j

yj θij xi

Dij xj x earth

–yi Image of conductor bundle i

Image of conductor bundle j

–yj

Figure 3.2 A general illustration of an overhead line physical dimensions and conductor spacings relative to tower centre and earth.

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Power Systems Modelling and Fault Analysis

Potential coefficients, shunt capacitances and susceptances Using Fig. 3.2, given a set of N conductors, the potential V of conductor i due to conductor’s own charge and charges on all other conductors is given by Vi 5

N X

Pij Qj

V

i 5 1; 2; :::; N

(3.1)

j51

where Pij is the Maxwell’s potential coefficient expressed in km/F and Qj is the charge in C/km. The equations assume infinitely long, perfectly horizontal conductors above earth whose effect is included using the method of electrostatic images. This method is generally valid up to a frequency of about 1 MHz. The potential of a conductor i above earth due to its own charge and an equal but negative charge on its own image enables us to calculate the self-potential coefficient of conductor i as follows:   1 2yi Pii 5 3 ln 2πεo ri

km=μF

(3.2a)

where εo 5 8.854 187 8176 1023 μF/km and yi is the height of conductor i above earth in m and ri is the radius of conductor i in m. Clearly yi is much greater than ri. The potential of conductor i due to a charge on conductor j and an equal but negative charge on the image of conductor j enables us to calculate the mutual potential coefficient between conductor i and conductor j as follows: Pij 5

  1 Dij 3 ln 2πεo dij

km=μF

and

Pji 5 Pij

(3.2b)

where Dij is the distance between conductor i and the image beneath the earth’s surface of conductor j in m, and dij is the distance between conductor i and conductor j in m. Eq. (3.1) can be rewritten in matrix form as V 5 PQ V

(3.3a)

where P is a potential coefficient matrix of N 3 N dimension. Multiplying both sides of Eq. (3.3a) by P21, we obtain Q 5 CV

Coulomb

(3.3b)

μF=km

(3.3c)

where C 5 P21

Modelling of multiconductor overhead power lines, underground and submarine cables

107

C is the line’s shunt capacitance matrix and is equal to the inverse of the potential coefficient matrix P. Under steady-state conditions, the current and voltage vectors are phasors and are related by I 5 YV 5 jBV 5 jω CV

Amps

(3.4a)

since the line’s conductance is negligible at power frequency f. The line’s shunt susceptance matrix is given by B 5 ω C 5 2 πf C

μS=km

(3.4b)

Series self and mutual impedances Self impedance The equations assume infinitely long and perfectly horizontal conductors above a homogeneous conducting earth having a uniform resistivity ρe (Ωm) and a unit relative permeability. The proximity effect between conductors is neglected. Using Fig. 3.2, the series voltage drop in V of each conductor due to current flowing in the conductor itself and currents flowing in all other conductors in the same direction is given by Vi 5

N X

Zij Ij

V=km i 5 1; 2; :::; N

(3.5a)

j51

where Z is the impedance expressed in Ω/km and I is the current in amps. The selfimpedance of conductor i is given by     Zii 5 RiðcÞ 1 jXiðcÞ 1 jXiðgÞ 1 RiðeÞ 1 jXiðeÞ Ω=km

(3.5b)

where subscript c represents the contribution of conductor i resistance and internal reactance, g represents a reactance contribution to conductor i due to its geometry, that is, an external reactance contribution and e represents correction terms to conductor i resistance and reactance due to the contribution of the earth return path. If the skin effect is neglected, that is, assuming a direct current (dc) condition or zero frequency, the internal dc impedance of a solid magnetic round conductor, illustrated in Fig. 3.3A, is given by 1000ρc 1 j4π1024 f ZiðcÞ 5 RiðcÞ 1 jXiðcÞ 5 πrc2 ρc 5 ρc20 ½1 1 α20 ðTc 2 20Þ Ωm



μr 4

 Ω=km

(3.6)

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Power Systems Modelling and Fault Analysis

(A)

rc

ro

(B)

ri

(C) N

1

(D)

2 r i Dij j

ro

Figure 3.3 Illustration of some conductor types: cross-section of (A) a solid round conductor, (B) tubular round conductor and (C) stranded 30/7 ACSR conductor (30 aluminium, 7 steel strands) and (D) direction of layers of stranded 54/7 ACSR conductor (54 aluminium, 7 steel strands).

Where ρc is resistivity in Ωm at temperature Tc in  C, rc is conductor radius in m and μr is conductor relative permeability. ρc20 and α20 are resistivity and temperature coefficient at 20 C. With skin effect is taken into account, the following exact equation for the internal impedance of a solid round conductor can be used ZiðcÞ 5

1000ρc m I0 ðmrc Þ Ω=km 3 2πrc I1 ðmrc Þ

(3.7a)

where sffiffiffiffiffiffiffiffiffiffiffiffiffiffi jωμo μr pffiffiffi 21 jπ=4 1 1 m5 5 2δ e 5 1j δ δ ρc

m21

(3.7b)

is defined as the complex propagation constant and δ is skin depth or depth of penetration into the conductor and is given by δ 5 503:292 3

rffiffiffiffiffiffiffi ρc m f μr

(3.7c)

and Ii are modified Bessel functions of the first kind of order i. For calculation of line parameters close to power frequency, the following alternative equation,

Modelling of multiconductor overhead power lines, underground and submarine cables

109

suitable for hand calculations using electronic calculators, is found to be accurate up to about 200 Hz: ZiðcÞ 5

1000ρc πrc2

 11

   π4 rc4 f 2 μ2c π4 rc4 f 2 μ2c 24 μr 1 2 f 1 j4π10 Ω=km 4 3 3 1014 ρ2c 6 3 1014 ρ2c (3.8)

Eq. (3.8) shows that skin effect causes an increase in the conductor’s effective ac resistance and a decrease in its effective ac internal reactance. Also, for f 5 0, Eq. (3.8) reduces to Eq. (3.6). In the case of a tubular or hollow conductor, illustrated in Fig. 3.3B, the dc internal impedance is given by "  # 1000ρc 2ri2 4ri4 ro 24 μr Ω=km 12 2 ln ZiðcÞ 5 1 j4π10 f 1 2 2 2 2 2 2 4 ri πðro 2 ri Þ ðro 2 ri Þ ðro 2ri Þ (3.9) Mathematically, the case of a solid round conductor is a special case of the hollow conductor since by setting ri 5 0, Eq. (3.9) reduces to Eq. (3.6). Aluminium conductor steel-reinforced (ACSR) or modern gapped-type conductors can be represented as hollow conductors if the effect of steel saturation can be ignored. Saturation may be caused by the flow of current through the helix formed by each aluminium strand that produces a magnetic field within the steel. With skin effect taken into account, the exact equation for the internal impedance of a hollow conductor is given by ZiðcÞ 5

  1000ρc m I0 ðmro ÞK1 ðmri Þ 1 I1 ðmri ÞK0 ðmro Þ Ω=km 2πro I1 ðmro ÞK1 ðmri Þ 2 I1 ðmri ÞK1 ðmro Þ

(3.10)

where Ii and Ki are modified Bessel functions of the first and second kind of order i, respectively. Mathematical solutions suitable for digital computations that provide sufficient accuracy are available in standard handbooks of mathematical functions as well as in modern libraries of digital computer programs. Eq. (3.7a) that represents the case of a solid conductor can be obtained from Eq. (3.10) by substituting ri 5 0. The external reactance of conductor i due to its geometry is given by   2yi XiðgÞ 5 4π10 f ln Ω=km ri 24

(3.11)

In practice, the internal reactance of a conductor is much smaller than its external reactance except in the case of very large conductors at high frequencies.

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Power Systems Modelling and Fault Analysis

Generally, precision improvements in the former would have a very small effect on the total reactance. The contribution of the earth’s correction terms to the self-impedance of conductor i is presented after the next section.

Mutual impedance The mutual impedance between conductor i and conductor j is given by   Zij 5 jXijðgÞ 1 RijðeÞ 1 jXijðeÞ Ω=km

(3.12a)

and jXijðgÞ 5 j4π1024 f ln

  Dij Ω=km dij

(3.12b)

Earth return path impedances The contributions of correction terms to the self and mutual impedances, due to the earth return path, are generally given as infinite series. The general correction terms of resistance and reactance are calculated in terms of two parameters mij and θij, where θij is shown in Fig. 3.2, as follows " RijðeÞ 5 4π1024 f 3 2

π 2 b1 mij cos θij 1 b2 m2ij 8 8 0 1 9 < = C2 e 3 ln@ Acosð2θij Þ 1 θij sin ð2θij Þ 1 b3 m3ij cos ð3θij Þ : ; mij # 2 d4 m4ij cos ð4θij Þ 2 b5 m5ij cos ð5θij Þ 1 ? (3.13a)

0 1 1 1:85138 A 1 b1 mij cos θij 2 d2 m2 cos ð2θij Þ XijðeÞ 5 4π1024 f 3 2 ln@ ij 2 mij 0 1 ( eC4 3 4 1 b3 mij cos ð3θij Þ 2 b4 mij ln@ Acosð4θij Þ mij ) # "

1 θij sin ð4θij Þ 1 b5 m5ij cos ð5θij Þ 2 . . . (3.13b)

Modelling of multiconductor overhead power lines, underground and submarine cables

111

where 8 pffiffiffi 2yi > > 23 > < δ mij 5 pffiffiffi D ij > > > : 23 δ

for the self
impedance terms or i 5 j for the mutual impedance terms or i 6¼ j

8

> > >   2n N 6 < K ðmqÞ X 7= d 1 o k 24 6 7 Ω=km 12 5 j4πf μr 10 4 q

Kn21 ðmqÞ5> > mqK1 ðmqÞ > > n51 ; : n 1 1 μr 1 mq Kn ðmqÞ (3.106b)

where μr is the relative permeability of steel pipe/armour. The mutual earth return impedance of Eq. (3.102h) must also be replaced since the cables are now coupled through the pipe/armour instead of the earth. Therefore, the mutual impedance between cables i and k with pipe/armour return is given by 2 3 μ K ðmqÞ q o r 5 Zi2k 5 j4πf 1024 1 ln4pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mqK1 ðmqÞ di2 1 dk2 2 2di dk cosθik 2 3 ) N 6 X 2μr 17 6 7 Cn 6 2 7 Ω=km 1

Kn21 ðmqÞ n5 4 n51 n 1 1 μr 1 mq Kn ðmqÞ (

where

(3.107)

Modelling of multiconductor overhead power lines, underground and submarine cables

Cn 5

187

  di dk n cosðnθik Þ q2

θik is the angle subtended between the pipe/armour’s centre and the centres of cables i and k. The complete matrix that describes a three-core armoured cable or a three-phase pipe-type cable with an infinite pipe/armour thickness can be written as a 6 3 6 matrix involving the three cores and three sheaths. Where, at lower frequencies, the condition of armour/pipe infinite thickness does not hold, the electromagnetic coupling between the armour/pipe and the earth must be included through an additional armour/pipe earth loop where the armour/pipe is treated as an additional tubular conductor. For a land pipe-type cable and a land three-core cable, the earth return is included but for a three-core submarine cable, the sea return is included. The complete matrix that describes a three-core armoured cable or pipe-type cable with a finite pipe/armour thickness can be written as a 7 3 7 matrix involving the three cores, three sheaths and armour/pipe as follows Cable 1 3 2 ZCC1 ZCS1 VC1 6 VS1 7 6 ZCS1 ZSS1 7 6 6 6 ? 7 6 ? ? 7 6 6 6 VC2 7 6 Z12 Z 12 7 6 6 6 VS2 7 6 Z12 Z 12 7 6 6 6 ? 756 ? ? 7 6 6 6 VC3 7 6 Z13 Z13 7 6 6 6 VS3 7 6 Z13 Z13 7 6 6 4 ? 5 4 ? ? VA=P Z1A=P Z1A=P 2

Cable 2 ^ Z12 Z12 ^ Z12 Z12 ^ ? ? ^ ZCC2 ZCS2 ^ ZCS2 ZSS2 ^ ? ? ^ Z23 Z23 ^ Z23 Z23 ^ ? ? ^ Z2A=P Z2A=P

Cable 3 ^ Z13 Z13 ^ Z13 Z13 ^ ? ? ^ Z23 Z23 ^ Z23 Z23 ^ ? ? ^ ZCC3 ZCS3 ^ ZCS3 ZSS3 ^ ? ? ^ Z3A=P Z3A=P

Arm:=pipe 32 3 IC1 ^ Z1A=P 7 6 ^ Z1A=P 7 76 IS1 7 7 6 ^ ? 76 ? 7 7 7 6 ^ Z2A=P 7 76 IC2 7 6 IS2 7 ^ Z2A=P 7 76 7 7 6 ^ ? 7 76 ? 7 6 IC3 7 ^ Z3A=P 7 76 7 7 6 ^ Z3A=P 7 76 IS3 7 ^ ? 54 ? 5 IA=P ^ ZAA=PP (3.108)

Power frequency impedance equations For power flow, transient and dynamic studies as well as large-scale short-circuit fault studies in power networks, impedances calculated at power frequency, that is, 50 or 60 Hz, are used. In many cases, the frequency-dependent impedance equations presented in the preceding sections can be simplified to close form solutions. For single-core underground cables where the cable geometry is as shown in Fig. 3.23, the self-impedance of a tubular core conductor with earth return is given by Zcc 5 RcðacÞ 1 π2 1024 f 1 j4π1024 f

   μc Derc 3 f ðroc; ric Þ 1 ln Ω=km 4 roc (3.109a)

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Power Systems Modelling and Fault Analysis

where f ðroc ; ric Þ 5 1 2

  2ric2 4ric4 roc ln 1 2 2 r2 Þ 2 2r 2 Þ2 ric ðroc ðr ic oc ic

(3.109b)

and Derc is the depth of equivalent earth return conductor given in Eq. (3.15), μc and Rc(ac) are the relative permeability and ac resistance of the core conductor. The self-impedance of a sheath with earth return is given by Zss 5 RsðacÞ 1 π2 1024 f 1 j4π1024 f

   μs Derc 3 f ðros; ris Þ 1 ln Ω=km 4 ros (3.110a)

where   2ris2 4ris4 ros ln f ðros ; ris Þ 5 1 2 2 1 2 2 2 2 ris ðros 2 ris Þ ðros 2ris Þ

(3.110b)

and μs and Rs(ac) are the relative permeability and ac resistance of the sheath conductor. The self-impedance of an armour with earth return is   μa Derc 3 f ðroa; ria Þ 1 ln Zaa 5 RaðacÞ 1 π 10 f 1 j4π10 f Ω=km 4 roa 2

24

24



(3.111a) where   2ria2 4ria4 roa ln 1 f ðroa ; ria Þ 5 1 2 2 2 2r 2 Þ2 ria ðroa 2 ria2 Þ ðroa ia

(3.111b)

and μa and Ra(ac) are the relative permeability and ac resistance of the armour conductor. The mutual impedance between core or sheath or armour i conductor, and core or sheath or armour j conductor, with earth return, as illustrated in Fig. 3.25A, is given by   Derc Zij 5 π2 1024 f 1 j4π1024 f 3 ln Ω=km dij

(3.112)

where dij is the distance between the centres of cables i and j if the conductors belong to different cables. If the conductors belong to the same cable, dij is the geometric mean distance (GMD) between the two conductors. For example, the GMD

Modelling of multiconductor overhead power lines, underground and submarine cables

189

between the core and sheath of a cable is given by (ros 1 ris)/2 which is sufficiently accurate for practical cable dimensions.

ac resistance at power frequency The power frequency impedance equations include an ac resistance term. The ac resistance is derived from the dc resistance and takes into account the skin and proximity effects. The ac resistance can be calculated as follows   Rac 5 Rdc 1 1 kcore ðys 1 yp Þ Ω=km

(3.113a)

where kcore 5 1.5 for pipe-type cables. For single-core, two-core and three-core cables, kcore 5 1. yS and yP are skin effect and proximity effect factors, respectively. Rdc 5 Rdcð20Þ ½1 1 α20 ðT 2 20Þ Ω=km

(3.113b)

where Rdc/Rdc(20) is the dc resistance of the conductor at temperature T/20 C. The calculation of Rdc(20) for solid and stranded conductors is presented in Section 3.4. T is conductor temperature in  C and α20 is the conductor constant mass temperature coefficient in  C21 at 20 C. Table 3.5 illustrates typical values for α20 and conductor resistivity at 20 C (ρ20). The skin effect factor yS is the incremental resistance factor produced by ac current in an isolated conductor due to skin effect and is given by

ys 5

8 > > >
> 0:0563x2s 2 0:0177xs 2 0:136 > : 0:354xs 2 0:733

0 , xs # 2:8 2:8 , xs # 3:8 xs . 3:8

(3.114a)

Table 3.5 Typical values of conductor temperature coefficient of resistance at 20 C (α20) and electrical resistivity (ρ20) at 20 C Material

Temperature coefficient α20 ( C21) at 20 C

Resistivity ρ20 (Ωm) at 20 C

Copper (hard drawn) Copper (annealed) Aluminium 1350 (HD) Aluminium alloy 1120 Aluminium alloy 6201 Aluminium alloy 6101 Lead or lead alloy Bronze Galvanised steel Stainless steel 304

0.00381 0.00393 0.00403 0.00390 0.00360 0.00347 0.0040 0.0030 0.0044 0.00094

1.7770 3 1028 1.7240 3 1028 2.8264 3 1028 2.930 3 1028 3.380 3 1028 3.310 3 1028 21.4 3 1028 3.5 3 1028 17.0 3 1028 70 3 1028

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Power Systems Modelling and Fault Analysis

where xs 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 3 8πfks =Rdc 50

(3.114b)

The skin effect coefficient ks depends on the type of conductor and its insulation. For copper conductors, ks 5 1 for round solid, stranded and sector-shaped conductors of any insulation and ks 5 0:435 for round Milliken conductors with fluid/ paper/PPL inulation. For round Milliken copper conductors with extruded insulation, ks 5 0:35 for insulated wires, ks 5 0:62 for bare unidirectional wires and ks 5 0:8 for bare bidirectional wires. For aluminium wires of any insulation, ks 5 1 for solid and stranded conductors, and ks 5 0:25 for round Milliken wires. For tubular helically stranded copper and aluminium conductors of any insulation,    ro 2 ri ro 12ri 2 ks 5 , where ri and ro are the tubular conductor’s inner and ro 1 ri ro 1ri outer radii, respectively. Generally, xs is less than 2.8 in the majority of practical applications. Although based on a simplified approach, Eqs. (3.114a)and (3.114b) for the skin effect factor involve an error of less than 0.5% at power frequency. The proximity effect factor is the incremental resistance factor due to the proximity of other ac current-carrying conductors. For two-core cables and two singlecore cables, yp is given by  2 x4p dc yp 5 2:9 4 0:8xp 1 192 S

(3.115a)

For three-core and three single-core cables of circular conductors, yp is given by 2 yp 5

 2 6  2 x4p dc 6 dc 0:312 1 6 0:8x4p 1 192 S 4 S

3 1:18 x4p 0:8x4p 1 192

1 0:27

7 7 7 5

(3.115b)

where xp 5

1 3 50

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8πfkp =Rdc

(3.115c)

dc is conductor diameter and S is the axial spacing between conductors. The proximity effect coefficient kp depends on the type of conductor and its insulation. For copper conductors, kp 5 1 for round solid conductor of any insulation, for round stranded and sector-shaped with extruded insulation, and kp 5 0:8 for round stranded with fluid/paper/PPL insulation. For round Milliken conductors with fluid/paper/PPL insulation, kp 5 0:37. For round Milliken copper conductors

Modelling of multiconductor overhead power lines, underground and submarine cables

191

with extruded insulation, kp 5 0:20 for insulated wires, kp 5 0:37 for both bare unidirectional wires and bare bidirectional wires. For sector-shaped fluid/paper/PPL and extruded insulation, kp 5 0:8 and 1, respectively. For aluminium wires of any insulation, kp 5 1 for round solid conductor and 0.8 for round stranded conductors. For round Milliken, kp 5 0:15. For tubular helically stranded copper and aluminium conductors of any insulation, kp 5 0:8. Eqs. (3.115a
3.115c) are accurate for xp less than 2.8, which represents the majority of practical applications. For multicore cables with shaped conductors, the value of yp calculated by Eqs. (3.115a
3.115c) shall be multiplied by 2/3. The values of the various factors used in Eqs. (3.114a),(3.114b) and (3.115a
3.115c) are usually supplied by cable manufacturers.

3.3.4 Sequence impedance matrices of single-circuit cables of different layouts General Having calculated the basic phase parameters and series phase impedance matrices in Section 3.3.3, we will now present the sequence impedance matrices for specific cable installation layouts. Fig. 3.26 shows six practical cable layouts for three-core and single-core cables. Pipe-type cables were shown in Fig. 3.24. Recall we used suffixes C, S and A to denote core, sheath and armour conductors, respectively, for cable phases 1, 2 and 3.

s1 c1 s2 c2 c2

s1 c1

s1 c1

s2

s3

c3

s3

s2 d c2

c3

d

s3 c3

d Single-core in a touching trefoil

Three-core

s1 c1

Single-core equilateral

d c2 c1 s d c3 s s2 3 1 Single-core flat symmetrical

h2 + d2

s2

h c2 d

d

Single-core trefoil

s3 c3 c1

d2 d1 c3 s c2 s1 3 s2 Single-core flat asymmetrical

Figure 3.26 Typical practical single- and three-core cable layouts.

192

Power Systems Modelling and Fault Analysis

Sequence impedances of cables with no armour Eqs. (3.102a
3.102h) show the relationship between core and sheath voltages and currents for each cable arranged in the order of C1S1C2S2C3S3. Rearranging the conductors to C1C2C3S1S2S3, Eqs. (3.102a
3.102h) can be rewritten as 2 C1 C1 ZCC1 VC1 7 6 Cores 6 6 VC2 7 C2 6 Z12 6 VC3 7 C3 6 Z13 7 6 6 7 6 ? 6 6 ? 75 ?6 ? 6 VS1 7 S1 6 ZCS1 7 6 6 Sheaths 4 VS2 5 S2 4 Z12 S3 Z13 VS3 2

3

Cores C2 Z12 ZCC2 Z23 ? Z12 ZCS2 Z23

C3 Z13 Z23 ZCC3 ? Z13 Z23 ZCS3

S1 ^ ZCS1 ^ Z12 ^ Z13 ? ? ^ ZSS1 ^ Z12 ^ Z13

Sheaths S2 S3 32 3 IC1 Z12 Z13 7 6 ZCS2 Z23 7 76 IC2 7 6 IC3 7 Z23 ZCS3 7 76 7 7 6 ? ? 7 76 ? 7 6 IS1 7 Z12 Z13 7 76 7 ZSS2 Z23 54 IS2 5 Z23 ZSS3 IS3 (3.116)

Eq. (3.116) can be simplified depending on the cable layout and spacings between the phases as shown in Fig. 3.26. Fig. 3.27 shows a 3D view of two cable layouts for three-phase single-core unarmoured cable systems with core and sheath conductors for each cable.

Three-core cables, three single-core cables in touching trefoil and three singlecore cables in equilateral layout These three cable layouts, shown in Fig. 3.26, are fully symmetrical. From the configurations of these cables, a number of practical equalities can be deduced as follows: a 5 ZC1S1 5 ZC2S2 5 ZC3S3 ; b 5 Z12 5 Z13 5 Z23 ; e 5 ZC1C1 5 ZC2C2 5 ZC3C3 ; f 5 ZS1S1 5 ZS2S2 5 ZS3S3

(3.117)

Therefore, the cable full impedance matrix of Eqs. (3.113a)and (3.113b) can be written as

(B)

(A)

s1

c1

s2 c2 c1 s 1

c2

s2

c3 s 3

c3 s 3

Figure 3.27 A 3D view of three-phase unarmoured cables for (A) single-core flat symmetrical and (B) single-core touching trefoil.

Modelling of multiconductor overhead power lines, underground and submarine cables

193

Cores Sheaths C1 C2 C3 S1 S2 S3 32 2 3 3 2 IC1 C1 e b b ^ a b b VC1 6 7 7 6 e b ^ b a b 7 Cores 6 76 IC2 7 6 VC2 7 C2 6 b 7 6 VC3 7 C3 6 b 6 b e ^ b b a 76 IC3 7 6 7 7 6 6 7 7 6 6 ? 6 ? 75 ?6? ? ? ? ? ? ?7 76 ? 7 7 6 VS1 7 S1 6 a 6 b b ^ f b b 76 IS1 7 6 7 7 6 4 5 4 S2 b a b ^ b f b 54 IS2 5 Sheaths VS2 S3 b b a ^ b b f VS3 IS3 

  VC ZCC 5 VS ZCS

ZCS ZSS



IC IS

(3.118a)

 (3.118b)

where 2

VC1

3

6 7 VC 5 4 VC2 5

2

VS1

3

2

6 7 VS 5 4 VS2 5

b

e

3

2

6 7 IC 5 4 IC2 5 and

VC3 VS3 2 3 2 e b b a b 6 7 6 ZCC 5 4 b e b 5 ZCS 5 4 b a b

IC1

b b

IS1

3

6 7 IS 5 4 IS2 5

IC3 3 2 b f 7 6 b 5 ZSS 5 4 b

b f

7 b5

a

b

f

b

b

IS3 3

(3.118c)

It is interesting to observe that the matrices ZCC, ZCS and ZSS are all balanced matrices since all diagonal terms are equal, and all off-diagonal terms are equal. This is due to the symmetrical cable layouts. Where the cable sheaths are solidly bonded and earthed at both ends, the sheaths can be eliminated. Therefore, the phase impedance matrix involving the cores only can be calculated by setting VS 5 0 in Eq. (3.118b). It can be shown that the resultant core or phase impedance matrix is given by 2

ZCðSelfÞ 4 ZPhase 5 ZCC 2 ZCS Z21 SS ZCS 5 ZCðMutualÞ ZCðMutualÞ

ZCðMutualÞ ZCðSelfÞ ZCðMutualÞ

3 ZCðMutualÞ ZCðMutualÞ 5 ZCðSelfÞ

(3.119)

Because the individual impedance matrices ZCC, ZCS and ZSS are all balanced, it can be shown that the resultant phase impedance matrix ZPhase of Eq. (3.119) that includes the effect of the eliminated sheaths, will also be balanced. The reader is encouraged to prove this statement. Therefore, using ZPNZ 5 H21ZPhaseH, the sequence impedance matrix of the three-phase cable circuit is given by 2

ZP PNZ 4 Z 5 0 0

0 ZN 0

3 0 0 5 ZZ

(3.120a)

194

Power Systems Modelling and Fault Analysis

where Z P 5 Z N 5 ZCðSelf Þ 2 ZCðMutualÞ and Z Z 5 ZCðSelf Þ 1 2ZCðMutualÞ

(3.120b)

Single-core cables in a trefoil layout This general layout arrangement in Fig. 3.26 shows asymmetrical spacings between the cables. From the cable configuration, the impedance matrix of this cable layout cable can be written as Cores

Sheaths

S1 S2 S3 32 2 C1 C2 C3 3 IC1 e b b ^ a b b 76 6 7 7 6 e c ^ b a c 76 IC2 7 Cores 6 VC2 7 C2 6 b 76 6 7 7 6 7 6 VC3 7 C3 6 b 6 c e ^ b c a 7 76 IC3 7 6 7 6 76 6 7 7 6 ? 6 ? 7 5 ? 6 ? ? ? ? ? ? ? 76 ? 7 76 6 7 7 6 7 6 VS1 7 S1 6 a 6 b b ^ f b b 7 76 IS1 7 6 7 6 76 6 7 7 6 a c ^ b f c 54 IS2 5 Sheaths 4 VS2 5 S2 4 b S3 b c a ^ b c f VS3 IS3 2

VC1

3

C1

(3.121a) where a 5 ZC1S1 5 ZC2S2 5 ZC3S3 ; b 5 Z12 5 Z13 ; c 5 Z23 e 5 ZC1C1 5 ZC2C2 5 ZC3C3 ; f 5 ZS1S1 5 ZS2S2 5 ZS3S3

(3.121b)

The calculation of the phase impedance matrix involving the cores only and the corresponding sequence impedance matrix of the three-phase circuit follows the same steps as above. The sheaths are first eliminated then the resultant 3 3 3 core impedance matrix ZPhase is transformed to the sequence reference frame using ZPNZ 5 H21ZPhaseH . In this case, the phase impedance matrix ZPhase will not be balanced and, as a result, the sequence impedance matrix ZPNZ will contain offdiagonal intersequence mutual terms.

Single-core cables in a flat layout The general asymmetrical flat layout of single-core three-phase cables is shown in Fig. 3.26. From the cable configuration, the impedance matrix and equalities of this cable layout cable can be written as follows

Modelling of multiconductor overhead power lines, underground and submarine cables

Cores

195

Sheaths

S1 S2 S3 32 2 C1 C2 C3 3 IC1 e b d ^ a b d 76 6 7 7 6 e c ^ b a c 76 IC2 7 Cores 6 VC2 7 C2 6 b 76 6 7 7 6 7 6 VC3 7 C3 6 d 6 c e ^ d c a 7 76 IC3 7 6 7 6 76 6 7 7 6 ? 6 ? 7 5 ? 6 ? ? ? ? ? ? ? 76 ? 7 76 6 7 7 6 7 6 VS1 7 S1 6 a 6 b d ^ f b d 7 76 IS1 7 6 7 6 76 6 7 7 6 a c ^ b f c 54 IS2 5 Sheaths 4 VS2 5 S2 4 b S3 d c a ^ d c f VS3 IS3 2

VC1

3

C1

(3.122a) a 5 ZC1S1 5 ZC2S2 5 ZC3S3 ; b 5 Z12 5 ; c 5 Z23 ; d 5 Z13 ; e 5 ZC1C1 5 ZC2C2 5 ZC3C3 ; f 5 ZS1S1 5 ZS2S2 5 ZS3S3

(3.122b)

For the cross-bonded cable arrangement shown in Fig. 3.22, the cores of the minor sections are transposed so that each occupies the three positions, but the sheaths are not. This is illustrated in Fig. 3.28. Eq. (3.122a) is the full impedance matrix of the first minor section. The full impedance matrices of the second and third minor sections are given by 2

e

6c 6 6 6b ZSection22 5 6 6b 6 6 4a c

1 2 3

c

b b

a

e

d d

c

d d

e a

a f

b b

c a

b b d d

f c

c

3

a7 7 7 d7 7 d7 7 7 c5 f

2

e

6d 6 6 6c ZSection23 5 6 6d 6 6 4c a

d e b a b d

C1, S1

C3, S1

C2, S1

C2, S2

C1, S2

C3, S2

C3, S3

C2, S3

C1, S3

Section 1

Section 2

Section 3

Core

Sheath

Figure 3.28 Core transposition of cross-bonded cables.

c

d

c

b

a

b

a

3

d7 7 7 e b a c7 7 b f b d7 7 7 a b f c5 c d c f (3.122c)

196

Power Systems Modelling and Fault Analysis

Using Eq. (3.122a) and all impedances are in per unit length, that is, Ω/km, the full impedance matrix of a cross-bonded cable with transposed cores in each major section is given by Z5

3 1X ZSection2i 3 i51

2

3e 6b1c1d 6 6 16b1c1d Z5 6 36 6a1b1d 6 4a1b1c a1c1d

(3.122d) b1c1d 3e

b1c1d b1c1d

a1b1d a1b1d

a1b1c a1b1c

b1c1d a1b1d

3e a1b1d

a1b1d 3f

a1b1c 3b

a1b1c

a1b1c

3b

3f

a1c1d

a1c1d

3d

3c

3 a1c1d a1c1d 7 7 7 a1c1d 7 7 7 3d 7 7 5 3c 3f (3.123a)

or including core and sheath voltage vectors 

VC VS



 5

ZCC

ZCS

ZTCS

ZSS



IC

 (3.123b)

IS

As expected, the core matrix ZCC is balanced, due to the perfect core transpositions assumed, but the sheath matrix ZSS is not because the sheaths are untransposed. The mutual impedance matrix between the cores and the sheaths ZCS is also unbalanced. In both cases of a solidly bonded cable where the sheaths are bonded and earthed at the cable ends, and the case of a cross-bonded cable where the sheaths are bonded and earthed at the ends of major sections, the phase impedance matrix involving the cores only can be obtained by setting VS 5 0 in Eq. (3.123b). Therefore, the core or phase impedance matrix takes the following general form T Zphase 5 ZCC
ZCS Z21 SS ZCS 2 ZC1C12ðSheath=EarthÞ 6 56 4 ZC2C12ðSheath=EarthÞ ZC3C12ðSheath=EarthÞ

ZC1C22ðSheath=EarthÞ

ZC1C32ðSheath=EarthÞ

3

ZC2C22ðSheath=EarthÞ

7 ZC2C32ðSheath=EarthÞ 7 5

ZC3C22ðSheath=EarthÞ

ZC3C32ðSheath=EarthÞ (3.124)

The elements of the matrix ZPhase are the self-impedances of each phase (core) and the mutual impedances between phases (cores) with the effects of the sheaths and earth return included. Generally, for solidly bonded cables, the self or diagonal terms of ZPhase are not equal among each other nor are the off-diagonal mutual

Modelling of multiconductor overhead power lines, underground and submarine cables

197

terms. In other words, the matrix is generally not balanced. This might also be expected to be the case for cross-bonded cables but in practical cable spacings and layouts, the core transposition effectively eliminates the unbalance. The sequence impedance matrix is obtained by transforming ZPhase of Eq. (3.124) to the sequence frame using ZPNZ 5 H21ZPhase H and is given by 2

Z PP 6 NP PNZ Z 54Z Z

ZP

Z PN Z NN

3 Z PZ 7 Z NZ 5

Z

Z

ZN

(3.125)

ZZ

where ZPP, ZNN and ZZZ are the positive-sequence, negative-sequence and zerosequence impedances of the cable and ZPP 5 ZNN. The off-diagonal terms are intersequence mutual terms which are normally small in comparison with the diagonal terms for solidly bonded cables and practically negligible for cross-bonded cables. The above steps can be applied to the alternative cross-bonding cable arrangement where the sheaths of the minor sections are transposed but the cores are not. This is left for the reader. For an isolated cable with a transposed earth continuity conductor, two full impedance matrices would need to be built for each half of the cable including the earth continuity conductor then combined into a single matrix. The calculation of the phase impedance matrix of such a cable makes use of the fact that no current can flow in the sheath that is IS 5 0. The derivation of the phase impedance matrix for such a cable is left for the reader.

Cables with armour Some land-based cables and almost all submarine cables have metallic armour. The armour acts as a third conductor in addition to the core and the sheath, as already presented.

Single-core armoured cables Fig. 3.29 shows a 3D view of single-core and three-core armoured cables. Eq. (3.103) shows the relationship between core, sheath and armour voltages and currents for each cable arranged in the order of C1S1A1C2S2A2C3S3A3.

(B)

(A)

a s1 c1

s2 a1 s1 c1

c2 a 2 s2

a3 s3

c3

c2

c3 s3

Figure 3.29 A 3D view of three-phase armoured cables for (A) single-core flat symmetrical and (B) three-core.

198

Power Systems Modelling and Fault Analysis

Rearranging the conductors to C1C2C3S1S2S3A1A2A3, Eq. (3.103) can be rewritten as Cores 2

3

2

C1

C1 ZCC1 7 6 6 6 VC2 7 C2 6 Z12 7 6 6 7 6 6 6 V 7 C3 6 Z 6 C3 7 6 13 7 6 6 7 6 6 6 ? 7 ?6 ? 7 6 6 7 6 6 6 VS1 7 S1 6 ZCS1 7 6 6 7 6 6 7 6 6 6 VS2 7 5 S2 6 Z12 7 6 6 7 6 6 6 VS3 7 S3 6 Z13 7 6 6 7 6 6 6 ? 7 ?6 ? 7 6 6 7 6 6 7 6 6 6 VA1 7 A1 6 ZCA1 7 6 6 7 6 6 6 VA2 7 A2 6 Z12 5 4 4 VC1

VA3

A3

Z13

Sheaths

C2

C3

Z12

Z13

ZCC2 Z23

S1

^ ZCS1 Z12 ^

Z23 ZCC3 ^

Z13

? ? ?

Z12

Z13

Z23 ZCS3 ^

?

^

Z12

Z13

Z23 ZSS3 ^ ?

^ ZSA1 Z12

Z23 ZCA3 ^

? ?

Z12 ZSS2 Z23

? ? ?

^

^

^

Z13

? ? Z13

^

Z12 ZSA2 Z23

^

Z13

A2

^ ZCA1 Z12

Z13

?

ZCA2 Z23

Z13

A1

Z23 ZCS3 ^

^ ZSS1 Z12 ^

S3

Z12 ZCS2 Z23

?

ZCS2 Z23

S2

Armours

Z23 ZSA3 ^

A3

32

Z13

IC1

3

76 7 6 7 Z12 ZCA2 Z23 7 76 IC2 7 76 7 6 7 Z13 Z23 ZCA3 7 76 IC3 7 76 7 76 7 ? ? ? 76 ? 7 76 7 76 7 6 7 ZSA1 Z12 Z13 7 76 IS1 7 76 7 76 7 Z12 ZSA2 Z23 76 IS2 7 76 7 76 7 6 7 Z13 Z23 ZSA3 7 76 IS3 7 76 7 6 7 ? ? ? 7 76 ? 7 76 7 76 7 ZAA1 Z12 Z13 76 IA1 7 76 7 76 7 6 7 Z12 ZAA2 Z23 7 54 IA2 5 Z13

Z23 ZAA3

IA3 (3.126)

In order to control the voltages between the sheath and the armour, the sheath is usually bonded to the armour at a number of points along the route. The armour covering is not normally an electric insulation so that the armour, and the sheath bonded to it, are effectively earthed. Therefore, calculation of the cable’s phase (core) impedance matrix requires the elimination of both sheath and armour. Writing Eq. (3.126) in partitioned matrix form, we have 2

3 2 VC ZCC 6 7 6 5 V 4 S 5 4 ZCS VA

ZCA

ZCS ZSS

32 3 ZCA IC 76 7 ZSA 54 IS 5

ZSA

ZAA

(3.127)

IA

The sheaths and armours are eliminated by setting VS 5 VA 5 0 in Eq. (3.127) and combining their matrices as follows: 

VCð3 3 1Þ 0S;Að6 3 1Þ



 5

ZCCð3 3 3Þ

ZCS;CAð3 3 6Þ

ðZCS;CAð3 3 6Þ ÞT

ZSS;AAð6 3 6Þ



ICð3 3 1Þ IS;Að6 3 1Þ

 (3.128a)

Modelling of multiconductor overhead power lines, underground and submarine cables

199

Therefore, the phase (core) impedance matrix of the cable with sheaths, armours and earth return, or sea return in the case of a submarine cable, is obtained as follows: T ZPhaseð3 3 3Þ 5 ZCCð3 3 3Þ 2 ZCS;CAð3 3 6Þ Z21 SS;AAð6 3 6Þ ðZCS;CAð3 3 6Þ Þ

(3.128b)

The sequence impedance matrix is obtained by transforming ZPhase(3X3) of Eq. (3.128b) to the sequence frame using ZPNZ 5 H21ZPhase H. For single-core submarine cables where the individual phases are laid physically very far apart from each other, the mutual electromagnetic coupling between the phases is very weak and negligible. Therefore, from the configuration of such three-phase submarine cables, Eq. (3.126) can be rewritten as Cores

Sheaths

Armours

C1 C2 C3 S1 S2 S3 2 d 0 0 ^ a 0 0 7 6 6 6 VC2 7 C2 6 0 d 0 ^ 0 a 0 7 6 6 7 6 6 6 VC3 7 C3 6 0 0 d ^ 0 0 a 7 6 6 7 6 6 6 ? 7 ?6? ? ? ^ ? ? ? 7 6 6 7 6 6 6 VS1 7 S1 6 a 0 0 ^ e 0 0 7 6 6 7 6 6 6 VS2 7 5 S2 6 0 a 0 ^ 0 e 0 7 6 6 7 6 6 6 VS3 7 S3 6 0 0 a ^ 0 0 e 7 6 6 7 6 6 6 ? 7 ?6? ? ? ^ ? ? ? 7 6 6 7 6 6 6 VA1 7 A1 6 b 0 0 ^ c 0 0 7 6 6 7 6 6 6 VA2 7 A2 6 0 b 0 ^ 0 c 0 5 4 4 2

VC1

VA3

3

C1

A3

0

0

b

^

0

0

c

^

A1 b

^

0

^

0

^

?

^

c

^

0

^

0

^

?

^

f

^

0

^

0

A2 0

A3 3 32 0 IC1 7 76 7 6 b 0 7 76 IC2 7 7 76 7 6 0 b 7 76 IC3 7 7 76 7 6 ? ?7 76 ? 7 7 76 7 6 0 0 7 76 IS1 7 7 76 7 6 c 0 7 76 IS2 7 7 76 7 6 0 c 7 76 IS3 7 7 76 7 6 ? ?7 76 ? 7 7 76 7 6 0 0 7 76 IA1 7 7 76 7 6 f 0 7 54 IA2 5 0

f

IA3 (3.129a)

a 5 ZCS1 5 ZCS2 5 ZCS3 ; b 5 ZCA1 5 ZCA2 5 ZCA3 ; c 5 ZSA1 5 ZSA2 5 ZSA3 ; d 5 ZCC1 5 ZCC2 5 ZCC3 ; e 5 ZSS1 5 ZSS2 5 ZSS3 ; f 5 ZAA1 5 ZAA2 5 ZAA3 ; Zij 5 0 (3.129b) Using Eq. (3.128b), the phase (cores) impedance matrix is given by 2

Zα ZPhaseð3 3 3Þ 5 4 0 0

0 Zα 0

3 0 0 5 Zα

(3.130a)

200

Power Systems Modelling and Fault Analysis

where

Zα 5 d
a2 f 1 b2 e 2 2abc

(3.130b)

or using the equalities from Eq. (3.129b), 2 2 Zα 5 ZCC 2 ðZCS ZAA 1 ZCA ZSS 2 2ZCS ZCA ZSA Þ

(3.130c)

Zα is the impedance of each core or phase including the effect of the combined sheath, armour and seawater return. The corresponding positive-sequence/negativesequence and zero-sequence impedances for the cable are all equal to Zα which is a direct result of the absence of mutual coupling between the three phases.

Three-core armoured cables and pipe-type cables The complete matrix that describes a three-core armoured cable or pipe-type cable with a finite pipe/armour thickness was given in Eq. (3.108) and can be rewritten as Cores 2

VC1

3

6V 7 6 C2 7 7 6 6 VC3 7 7 6 7 6 6?7 7 6 6 VS1 7 5 7 6 7 6 6 VS2 7 7 6 6 VS3 7 7 6 7 6 4?5 VA

C1

2

C1 e

C2 b

Sheaths C3 b ^

S1 a

S2 b

^

b

a

e b C2 6 6 b 6 6 b e C3 6 b 6 ?6? ? ? 6 b b S1 6 6 a 6 a b S2 6 b 6 6 b a S3 6 b 6 ?4? ? ? A

c

c

c

^ ^ ^ ^ ^ ^ ^

b b ? ? f b

b f

b b ? ? c

c

Arm=Pipe S3 b

A=P 32 3 c IC1 7 6 b ^ c 7 76 IC2 7 7 76 7 6 a ^ c 7 76 IC3 7 7 76 ? ^ ? 76 ? 7 7 76 7 6 b ^ c 7 76 IS1 7 7 76 b ^ c 76 IS2 7 7 76 7 6 f ^ c 7 76 IS3 7 7 76 ? ^ ? 54 ? 5 c ^ d IA ^

(3.131)

where, besides the equalities given in Eq. (3.117), we have c 5 ZAC 5 ZAS and d 5 ZAA . As for single-core cables, where the sheaths and armours are bonded and earthed, they can be eliminated by combining the matrices that correspond to them and setting VS 5 VA 5 0. Writing Eq. (3.131) in partitioned matrix form, we have 

  VCð3 3 1Þ ZCCð3 3 3Þ 5 0S;Að4 3 1Þ ðZCS;CAð3 3 4Þ ÞT

ZCS;CAð3 3 4Þ ZSS;AAð4 3 4Þ



ICð3 3 1Þ IS;Að4 3 1Þ

 (3.132a)

The phase (core) impedance matrix of the cable with sheath, armour and earth, or sea water in the case of a submarine cable, is given by

Modelling of multiconductor overhead power lines, underground and submarine cables T ZPhaseð3 3 3Þ 5 ZCCð3 3 3Þ 2 ZCS;CAð3 3 4Þ Z21 SS;AAð4 3 4Þ ðZCS;CAð3 3 4Þ Þ

201

(3.132b)

The phase (core) impedance matrix will virtually be balanced. The corresponding sequence impedance matrix calculated using ZPNZ 5 H21ZPhaseH will be a diagonal matrix, that is, having virtually no intersequence terms. The diagonal terms are the required positive-sequence/negative-sequence and zero-sequence impedances.

3.3.5 Shunt phase and sequence susceptance matrices of singlecircuit cables In this section, we derive the cable phase and sequence susceptance matrices using the basic phase parameters calculated in Section 3.3.3.

Screened cables with no armour For screened three-core and screened single-core cables, we have already established that since the sheaths are earthed, there is no capacitance between one cable phase and another. From Fig. 3.25B, we note that there are two capacitances, or susceptances, corresponding to the core insulation and the sheath insulation layers and these are illustrated in Fig. 3.30A. For each cable phase, the core and sheath voltage and current equations can be written as IC 5 jBCS(VC
VS) and IC 1 IS 5 jBSEVS where BCS and BSE are the core-to-sheath and sheath-to-earth susceptances, respectively. Therefore, writing similar equations for the three-phase cable, the full shunt susceptance matrix is given by

(B)

(A) IC

Core

VC

Sheath IS

jB CS

Core

IC

VC

IS

jB CS Sheath

I C + IS

I C + IS

VS

jB SE

VS

Armour I A

jB SA I C + IS + I A

VA jB AE

Figure 3.30 Equivalent capacitances of screened cables: (A) no armour and (B) with armour.

202

Power Systems Modelling and Fault Analysis Cores

2 C1 2 3 IC1 C1 jBCS 6 I 7 C2 6 0 6 6 C2 7 6 6 7 6 IC3 7 C3 6 0 6 6 7 6 6 7 6?75 ?6 ? 6 6 7 6 IS1 7 S1 6 2 jBCS 6 6 7 6 6 7 4 IS2 5 S2 4 0 IS3 0 S3

C2 0

Sheaths C3 0

^

S1 2 jBCS

jBCS

0

^

0

0

jBCS

^

0

?

?

?

?

0

0

^

jðBCS 1 BSE Þ

2 jBCS 0

0 2 jBCS

^ ^

0 0

S2 0

S3 0

32

3 VC1 6 7 7 2 jBCS 0 76 VC2 7 76 7 6 7 0 2 jBCS 7 76 VC3 7 76 7 ? ? 76 ? 7 76 7 76 VS1 7 0 0 76 7 76 7 jðBCS 1 BSE Þ 0 54 VS2 5 0 jðBCS 1 BSE Þ VS3 (3.133)

For solidly bonded and cross-bonded cables, the sheaths can be eliminated using VS15 VS2 5 VS3 5 0. The resultant shunt phase (cores) susceptance matrix is given by 2

BCS BPhase 5 4 0 0

0 BCS 0

3 0 0 5 BCS

(3.134)

And the positive-sequence, negative-sequence and zero-sequence susceptances are all equal, and equal to BCS. For cables that are bonded and earthed at one end only or at the cable centre, there will also be a capacitance between the earth continuity conductor and earth with IECC 5 jBECC-E.VECC. This increases the dimension of the full admittance matrix given in Eq. (3.133) by one row and one column, that is, from 6 3 6 to 7 3 7. The 3 3 3 phase susceptance matrix for such cables is calculated with the constraint that no sheath current can flow, that is, IS1 5 IS2 5 IS3 5 0. The resultant phase (cores) susceptance matrix is given by 2

BCS BSE 6 BCS 1 BSE 6 6 6 6 0 BPhase 5 6 6 6 6 4 0

3 0

0

BCS BSE BCS 1 BSE

0

0

BCS BSE BCS 1 BSE

7 7 7 7 7 7 7 7 7 5

(3.135a)

This result shows that the cable-to-sheath capacitance and the sheath-to-earth capacitance of each core are effectively in series, as can be seen from Fig. 3.30A. The corresponding shunt sequence susceptance matrix of Eqs. (3.134) and (3.135a) is BPNZ 5 BPhase

(3.135b)

that is the positive-sequence, negative-sequence and zero-sequence susceptances are all equal to the relevant phase susceptances.

Modelling of multiconductor overhead power lines, underground and submarine cables

203

Screened cables with armour From Fig. 3.30B, we note that there are three capacitances, or susceptances, corresponding to the core insulation layer, the sheath insulation layer and the armour coating where this is made of insulation material. For each phase, the core, sheath and armour current/voltage equations can be written as IC 5 jBCS(VC
VS), IC 1 IS 5 jBSA (VS
VA) and IC 1 IS 1 IA 5 jBAEVA. Therefore, writing similar equations for the three cores, three sheaths and three armours, the full shunt susceptance matrix B of the three-phase cable is given by 2

3

2

C1 BCS 0 0 ? 2 BCS 0 0

IC1 6I 7 6 6 C2 7 6 6 7 6 6 IC3 7 6 6 7 6 6?7 6 6 7 6 6 7 6 6 7 6 6 IS1 7 6 6 7 6 6 IS2 7 5 j6 6 7 6 6 IS3 7 6 6 7 6 6 7 6 6?7 6 ? 6 7 6 6 IA1 7 6 0 6 7 6 6 7 6 4 IA2 5 4 0 0 IA3

Cores C2 0 BCS 0 ? 0 2 BCS 0

C3 0 0 BCS ? 0 0 2 BCS

? 0 0 0

? 0 0 0

^ ^ ^ ^ ^ ^ ^

S1 2 BCS 0 0 ? BCS 1 BSA 0 0

Sheaths S2 0 2 BCS 0 ? 0 BCS 1 BSA 0

S3 0 0 2 BCS ? 0 0 BCS 1 BSA

^ ^ ^ ^

? 2 BSA 0 0

? 0 2 BSA 0

? 0 0 2 BSA

^ ^ ^ ^ ^ ^ ^

A1 0 0 0 ? 2 BSA 0 0

^ ? ^ BSA 1 BAE ^ 0 ^ 0

Armours A2 0 0 0 ? 0 2 BSA 0

A3 0 0 0 ? 0 0 2 BSA

? 0 BSA 1 BAE 0

? 0 0 BSA 1 BAE

32

3 VC1 76 V 7 76 C2 7 76 7 76 VC3 7 76 7 76 ? 7 76 7 76 7 76 7 76 VS1 7 76 7 76 VS2 7 76 7 76 VS3 7 76 7 76 7 76 ? 7 76 7 76 VA1 7 76 7 76 7 54 VA2 5 VA3 (3.136a)

If the sheath and armour are bonded together and earthed at both ends, then circulating currents can flow through the sheath and armour in parallel with earth return, or seawater return in the case of a submarine cable. The phase (core) susceptance matrix is easily obtained by deleting the last six rows and columns that correspond to the sheath and armours. The earthing of the sheath and armour, in effect, short circuits the sheath-to-armour and armour-to-earth capacitances, as can be seen in Fig. 3.30B. The positive-sequence, negative-sequence and zero-sequence susceptances are equal to the phase susceptance BCS. If the sheath and armour are unearthed or earthed at one point only, then no sheath or armour current would flow and the equivalent core or phase capacitance is equal to the three capacitances of Fig. 3.30B in series, that is BPhase 5

1 1 1 1 1 1 BCS BSA BAE

(3.136b)

The positive-sequence, negative-sequence and zero-sequence susceptances are equal to BPhase.

Unscreened or belted cables Belted cables are three-core cables where each core has an insulation layer but no screen or metallic sheath. There is also a belt insulation layer around all three cores and a single metallic sheath cover, as illustrated in Fig. 3.31A.

204

Power Systems Modelling and Fault Analysis

(A)

(B)

Belt insulation

C1

C3

Sheath CCC

Insulation

CCC CCC

Filler Core

C2 CCS

CCS

(C) CCS

IC1 V C1

jB CC

Sheath

jB CC

V C2 IC2

CSE

jB CC V C3 IC3

jB CS

jB CS

jB CS

Figure 3.31 Three-core belted cable and capacitance equivalent circuits: (A) cross-section; (B) capacitance equivalent circuit; and (C) capacitance equivalent circuit with earthed sheath.

There are capacitances among the three cores, between each core and sheath and between the sheath and earth. Fig. 3.31B shows the equivalent capacitance circuit of the cable where CCC is the core-to-core capacitance, CCS is the core-to-sheath capacitance and CSE is the sheath-to-earth capacitance. Where the sheath is solidly earthed, as is usually the case, the equivalent capacitance circuit reduces to that shown in Fig. 3.30C. The injected currents into the three cores can be expressed as follows: IC1 5 jBCS VC1 1 jBCC ðVC1
VC2 Þ 1 jBCC ðVC1
VC3 Þ IC1 5 jðBCS 1 2BCC ÞVC1
jBCC VC2
jBCC VC3

(3.137)

and similarly for the two remaining cores. The nodal phase susceptance matrix is 2

3 2 IC1 BCS 1 2BCC 4 IC2 5 5 j 4 2 BCC IC3 2 BCC

2 BCC BCS 1 2BCC 2 BCC

32 3 VC1 2 BCC 2 BCC 54 VC2 5 BCS 1 2BCC VC3

(3.138a)

and the corresponding sequence susceptance matrix is given by 2

BP BPNZ 5 4 0 0

0 BN 0

3 0 0 5 BZ

(3.138b)

Modelling of multiconductor overhead power lines, underground and submarine cables

205

where BP 5 BN 5 BCS 1 3BCC

and BZ 5 BCS

(3.138c)

It is interesting to note that the zero-sequence susceptance of a belted cable is smaller than its positive-sequence/negative-sequence susceptance. This contrasts with the equal positive-sequence/negative-sequence and zero-sequence susceptances of screened three-core and single-core cables. It is useful to calculate the core-to-core cable susceptance in terms of the sequence susceptances giving BCC 5 (BP
BZ)/3.

3.3.6 Impedance matrix of three-phase double-circuit cables In some installations, double-circuit cables may be laid close to each other. These may be three-core circuits or single-core circuits laid in flat or trefoil arrangements. Near-vertical arrangements may be used in tunnels. We will describe the case of single-core unarmoured cables. As each circuit consists of six conductors, three cores and three sheaths, 12 conductors form this complex mutually coupled multiconductor arrangement. Let the two circuits be designated A and B and let C1, C2, C3, S1, S2 and S3 be the cores and sheaths of circuit A, and C4, C5, C6, S4, S5 and S6 be the cores and sheaths of circuit B. The full 12 3 12 impedance matrix is given by Circuit A Cores 2 C1 zC1C1 zC1C2 zC1C3 6 A C2 6 zC2C1 zC2C2 zC2C3 6 C3 6 6 zC3C1 zC3C2 zC3C3 6 C4 6 zC4C1 zC4C2 zC4C3 6 B C5 6 6 zC5C1 zC5C2 zC5C3 6 C6 6 zC6C1 zC6C2 zC6C3 6 Z5 S1 6 6 zS1C1 zS1C2 zS1C3 6 A S2 6 6 zS2C1 zS2C2 zS2C3 S3 6 6 zS3C1 zS3C2 zS3C3 6 S4 6 6 zS4C1 zS4C2 zS4C3 B S5 6 4 zS5C1 zS5C2 zS5C3 S6

zS6C1

zS6C2

zS6C3

Circuit B Cores

Circuit A Sheaths

Circuit B Sheaths

zC1C4

zC1C5

zC1C6

zC1S1

zC1S2

zC1S3

zC1S4

zC1S5

zC1S6

zC2C4 zC3C4

zC2C5 zC3C5

zC2C6 zC3C6

zC2S1 zC3S1

zC2S2 zC3S2

zC2S3 zC3S3

zC2S4 zC3S4

zC2S5 zC3S5

zC2S6 zC3S6

zC4C4

zC4C5

zC4C6

zC4S1

zC4S2

zC4S1

zC4S4

zC4S5

zC4S6

zC5C4

zC5C5

zC5C6

zC5S1

zC5S2

zC5S1

zC5S4

zC5S5

zC6S6

zC6C4

zC6C5

zC6C6

zC6S1

zC6S2

zC6S1

zC5S4

zC6S5

zC6S6

zS1C4

zS1C5

zS1C6

zS1S1

zS1S2

zS1S3

zS1S4

zS1S5

zS1S6

zS2C4

zS2C5

zS2C6

zS2S1

zS2S2

zS2S3

zS2S4

zS2S5

zS2S6

zS3C4

zS3C5

zS3C6

zS3S1

zS3S2

zS3S3

zS3S4

zS3S5

zS3S6

zS4C4 zS5C4

zS4C5 zS5C5

zS4C6 zS5C6

zS4S1 zS5S1

zS4S4 zS5S4

zS4S3 zS5S3

zS4S4 zS5S4

zS4S5 zS5S5

zS4S6 zS5S6

zS6C4

zS6C5

zS6C6

zS6S1

zS6S4

zS6S3

zS6S4

zS6S5

3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

zS6S6 (3.139a)

or in concise matrix form, including series voltage drop and current vectors 2

VCðAÞ

3

2

ZCCðAÞ

6 VCðBÞ 7 6 ZTCðAÞCðBÞ 6 7 6 6 756 6 4 VSðAÞ 5 4 ZTCðAÞCðBÞ VSðBÞ ZTCðAÞSðBÞ

ZCðAÞCðBÞ

ZCðAÞSðAÞ

ZCCðBÞ

ZTCðAÞSðBÞ

ZTCðBÞSðAÞ ZTCðBÞSðBÞ

ZSSðAÞ ZTSðAÞSðBÞ

3 ZCðAÞSðBÞ 2 ICðAÞ 3 ZCðBÞSðBÞ 7 76 ICðBÞ 7 7 76 6 7 7 ZSðAÞSðBÞ 54 ISðAÞ 5 ZSSðBÞ

ISðBÞ

(3.139b)

206

Power Systems Modelling and Fault Analysis

Circuit A

Circuit B

C1, S1

C3, S1

C2, S1

C2, S2

C1, S2

C3, S2

C3, S3

C2, S3

C1, S3

Section 1

Section 2

Section 3

C4, S4

C6, S4

C5, S4

C5, S5

C4, S5

C6, S5

C6, S6

C5, S6

C4, S6

Section 1

Section 2

Section 3

Core position 1

2 3

Core position 5

6 4

Figure 3.32 Double-circuit cross-bonded cable with similar core transposition arrangements.

In practical installations, various double-circuit cable earthing arrangements may be used. Some of these are: circuit A and circuit B are solidly bonded, circuit A and circuit B are cross-bonded, circuit A and circuit B are single-point bonded, circuit A is solidly bonded in a trefoil formation and circuit B is single-point bonded in a flat formation, etc. The matrix of Eqs. (3.139a)and (3.139b) can be modified for any of these arrangements to calculate the reduced 6 3 6 phase impedance matrix for the two circuits that include their cores only. Fig. 3.32 shows one arrangement where circuit A and circuit B are both cross-bonded with identical core transposition arrangements. It is instructive for the reader to use the methodology presented in the singlecircuit case to calculate the average 12 3 12 impedance matrix for the transposed double-circuit cable from the full 12 3 12 matrix of each minor section. Using Eq. (3.139b), this average 12 3 12 matrix can be written as follows: 

VCðABÞ6 VSðABÞ6

3 1 3 1





ZCCðABÞ6 5 ZTCSðABÞ6

3 6 3 6

ZCSðABÞ6 3 ZSSðABÞ6 3

 6 6

ICðABÞ6 ISðABÞ6

3 1



3 1

(3.140a)

The calculation of the 6 3 6 phase impedance matrix of the cores only, with both circuits loaded, makes use of VS(A) 5 VS(B) 5 0 or VS(AB) 5 0 in Eq. (3.140a) giving VCðABÞ6 3 1 5 ZPhase6 3 6 ICðABÞ6 3 1

(3.140b)

where ZPhase6

3 6

5 ZCCðABÞ6

3 6

2 ZCSðABÞ6 3 6 Z21 SSðABÞ6

T 3 6 ZCSðABÞ6 3 6

(3.140c)

Modelling of multiconductor overhead power lines, underground and submarine cables

207

Eq. (3.140b) can be rewritten as follows: 

  ZPhaseðAÞ3 3 3 VCðAÞ3 3 1 5 T Z VCðBÞ3 3 1 PhaseðABÞ3 3 3

ZPhaseðABÞ3 3 3 ZPhaseðBÞ3 3 3



ICðAÞ3 3 1 ICðBÞ3 3 1

 (3.141a)

If one circuit, for example, circuit A, is open-circuited, that is, the cores carry no currents, IC(A) 5 0 and the resultant 3 3 3 matrix for circuit B is directly obtained as ZPhase(B)3 3 3. If circuit A is open and earthed, then the resultant 3 3 3 impedance matrix of circuit B is derived using VC(A) 5 0 in Eq. (3.141a) giving ZEqðBÞ3 3 3 5 ZPhaseðBÞ3 3 3 2 ZTPhaseðABÞ3 3 3 Z21 PhaseðAÞ3 3 3 ZPhaseðABÞ3 3 3

(3.141b)

These are practical cases that can arise in normal power system design and operation for which the cable impedance parameters would be required.

3.3.7 Examples Example 3.5 A 275-kV underground self-contained three-phase cable laid out in a symmetrical flat arrangement as shown in Fig. 3.33. Air Earth 40.9 mm 43.9 mm 1000 mm

37.9 mm 21.9 mm 6.8 mm

127 mm

Core

127 mm

Insulation Sheath PVC oversheath

1 R

2 Y

3 B

Figure 3.33 Self-contained single-core cable layout used in Example 3.5. The cable’s geometrical and physical data are as follows:

Copper core: inner radius 5 oil duct radius 5 6.8 mm, outer radius 5 21.9 mm, relative permeability 5 1.0, ac resistance 5 0.01665 Ω/km; Core insulation: paper, relative permeability 5 1.0, relative permittivity 5 3.8;

208

Power Systems Modelling and Fault Analysis

Lead sheath: inner radius 5 37.9 mm, outer radius 5 40.9 mm, relative permeability 5 1.0, ac resistance 5 0.28865 Ω/km; PVC oversheath: thickness 5 3 mm, relative permittivity 5 3.5, relative permeability 5 1.0; Earth resistivity 5 20 Ωm. Nominal frequency f 5 50 Hz. εo 5 8.854 187 8176 1023 μF/km. Calculate the phase and sequence susceptance and impedance parameters for a solidly bonded cable, and a cross-bonded cable whose cores are perfectly transposed within each major section.

Cable susceptances The core insulation susceptance is equal to BCS 5 2π 3 50

2π 3 8:854187 3 1023 3 3:8   5 121:09 μS=km 37:9 ln 21:9

and the PVC oversheath susceptance is BCS 5 2π 3 50

2π 3 8:854187 3 1023 3 3:5   5 864:19 μS=km 43:9 ln 40:9

The full susceptance matrix of the cable is equal to 2 C1 121:091 C2 6 0 6 6 6 C3 0 6 B5 S1 6
121:091 6 6 S2 4 0 S3 0

0 121:091

0 0


121:091 0

0
121:091

0 0

121:091 0

0 985:282

0 0


121:091 0

0
121:091

0 0

985:282 0

0 0

3

7 7 7
121:091 7 7 7 0 7 7 5 0 985:282

The phase and sequence susceptance matrix for a solidly bonded or crossbonded cable is 2

121:091 6 PNZ BPhase 5 B 54 0 0

0 121:091

0 0

0

121:091

Cable impedances Depth of earth return conductor 5 658:87

3 7 5

pffiffiffiffiffiffiffiffiffiffiffiffiffi 20=50 5 416:7 m.

Modelling of multiconductor overhead power lines, underground and submarine cables

209

For the tubular core conductor f ðro ; ri Þ 5 1 2

  2 3 ð6:82 Þ 4 3 ð6:84 Þ 21:9 5 0:839863 1 ln 2 2 2 2 2 6:8 21:9 2 6:8 ð21:9 26:8 Þ

24

ZC1C1 5 0:01665 1 π 10 2

24

3 50 1 j4π10

2 0 13 1 416:7 A5 3 504 3 0:8398631 ln@ 4 21:9 3 1023

5 ZC2C2 5 ZC3C3 5 ð0:066 1 j0:632Þ Ω=km

For the sheath, f ðro ; ri Þ 5 1 2

  2 3 ð37:92 Þ 4 3 ð37:94 Þ 40:9 5 0:09774 1 ln 37:9 40:92 2 37:92 ð40:92 2 37:92 Þ2

24

ZS1S1 5 0:28865 1 π 10 2

24

3 50 1 j4π10

2 0 13 1 416:7 A5 4 @ 3 50 3 0:09774 1 ln 4 40:9 3 1023

5 ZS2S2 5 ZS3S3 5 ð0:338 1 j0:5814Þ Ω=Km 0

1

B C B C 416:7 C ZC1S1 5 π2 1024 3 50 1 j4π1024 3 50 3 lnB Bð40:9 1 37:9Þ C @ 3 1023 A 2 5 ZS1C1 5 ZC2S2 5 ZS2C2 5 ZC3S3 5 ZS3C3 5 ð0:0493 1 j0:5822Þ Ω=Km 0

24

ZC1C2 5 π 10 2

24

3 50 1 j4π10

1 416:7 A 3 50 3 ln@ 127 3 1023

5 ð0:0493 1 j0:5086Þ Ω=km 0

24

ZC1C3 5 π 10 2

24

3 50 1 j4π10

1 416:7 A 3 50 3 ln@ 2 3 127 3 1023

5 ð0:0493 1 j0:4651Þ Ω=km

All mutual impedances between cable phase 1 core and sheath conductors, and cable phase 2 core and sheath conductors are equal, and similarly for those impedances between cable phases 2 and 3. The full impedance matrix is equal to 2 C1 0:066 1 j0:632 C2 6 6 0:0493 1 j0:508 6 C3 6 0:0493 1 j0:465 Z5 6 S1 6 6 0:0493 1 j0:582 6 S2 4 0:0493 1 j0:508 S3 0:0493 1 j0:465

3 0:0493 1 j0:508 0:0493 1 j0:465 0:0493 1 j0:582 0:0493 1 j0:508 0:0493 1 j0:465 0:066 1 j0:632 0:0493 1 j0:508 0:0493 1 j0:508 0:0493 1 j0:582 0:0493 1 j0:508 7 7 7 0:0493 1 j0:508 0:066 1 j0:632 0:0493 1 j0:465 0:0493 1 j0:508 0:0493 1 j0:582 7 7 0:0493 1 j0:508 0:0493 1 j0:465 0:338 1 j0:5814 0:0493 1 j0:508 0:0493 1 j0:465 7 7 7 0:0493 1 j0:582 0:0493 1 j0:508 0:0493 1 j0:508 0:338 1 j0:5814 0:0493 1 j0:508 5 0:0493 1 j0:508 0:0493 1 j0:582 0:0493 1 j0:465 0:0493 1 j0:508 0:338 1 j0:5814

For a solidly bonded cable, the phase impedance matrix is equal to

210

Power Systems Modelling and Fault Analysis

ZPhase 5 ZCC 2 ZCS Z21 SS ZCS

2 C1 0:129 1 j0:126 6 5 C2 4 0:089 2 j0:003 C3 0:072 2 j0:025

0:089 2 j0:003

0:072 2 j0:025

3

7 0:089 2 j0:003 5 0:129 1 j0:126

0:12 1 j0:104 0:089 2 j0:003

and the corresponding sequence impedance matrix is equal to 2 0:043 1 j0:129 P 6 ZPNZ 5 H21 ZPhase H 5 N 4 0:026 2 j0:002 Z

2 0:012 1 j0:023 0:043 1 j0:129

2 0:001 2 j0:002

3 2 0:001 1 j0:002 7 2 0:001 2 j0:002 5

2 0:001 1 j0:002

0:292 1 j0:1

ZP 5 ZN 5 (0.043 1 j0.129)Ω/km and ZZ 5 (0.292 1 j0.1)Ω/km. For a cross-bonded cable, the full impedance matrix for a major cable section that includes three perfectly transposed minor sections is equal to 2 C1 0:066 1 j0:632 C2 6 6 0:0493 1 j0:494 6 C3 6 0:0493 1 j0:494 6 Z5 S1 6 6 0:0493 1 j0:518 6 S2 4 0:0493 1 j0:533 S3 0:0493 1 j0:518

0:0493 1 j0:494 0:066 1 j0:632

0:0493 1 j0:494 0:0493 1 j0:494

0:0493 1 j0:494 0:0493 1 j0:518

0:066 1 j0:632 0:0493 1 j0:518

0:0493 1 j0:533 0:0493 1 j0:518

0:0493 1 j0:533 0:0493 1 j0:518

3 0:0493 1 j0:518 0:0493 1 j0:518 7 7 7 0:0493 1 j0:518 0:0493 1 j0:533 0:0493 1 j0:518 7 7 0:338 1 j0:5814 0:0493 1 j0:508 0:0493 1 j0:465 7 7 7 0:0493 1 j0:508 0:338 1 j0:5814 0:0493 1 j0:508 5 0:0493 1 j0:465 0:0493 1 j0:508 0:338 1 j0:5814

0:0493 1 j0:518 0:0493 1 j0:518

0:0493 1 j0:533 0:0493 1 j0:533

The phase impedance matrix is equal to ZPhase 5 ZCC


2

0:108 1 j0:125 6 5 C2 4 0:092
j0:013 C3 0:092
j0:013 C1

ZCS Z
1 SS ZSC

0:092
j0:013

0:092
j0:013

3

7 0:092
j0:013 5 0:108 1 j0:125

0:108 1 j0:125 0:092
j0:013

and the corresponding sequence impedance matrix is equal to Z

PNZ

2 P 0:017 1 j0:138 6 0 5 H ZPhase H 5 N 4
1

Z

0

0 0:017 1 j0:138

0 0

0

0:292 1 j0:100

3 7 5

ZP 5 ZN 5 (0.017 1 j0.138)Ω/km and ZZ 5 (0.292 1 j0.1)Ω/km.

3.4

Modelling of actual cables with semiconducting screens

3.4.1 Background The calculation of cable parameters for both steady-state and electromagnetic transient analysis is based on cable models that assume a solid core, insulation, a single metal sheath and a plastic oversheath layer. However, in actual cables, copper cores are generally compacted stranded or compacted and segmented stranded for large core cross-sectional areas. Also, actual cables consist of an additional number of

Modelling of multiconductor overhead power lines, underground and submarine cables

(A) Model

ris ros

Core

211

(B) Actual

Semiconductor screen

rc

Insulation Semiconductor screen Bedding tape Cu wire screen Tape Lead sheath HDPE oversheath

Figure 3.34 Land cable: (A) model and (B) actual with stranded core and semiconducting layers.

semiconducting layers and semiconducting tapes, and some cables employ two metallic sheath layers of different material such as copper and lead. Accounting for these differences is presented in this section. Fig. 3.34 shows a comparison between a model cable and an actual 132-kV 630 mm2 single-core XLPE copper cable.

3.4.2 Accounting for all layers of actual cables Cable core Many cable aluminium cores are solid but some, and most copper cores, are generally compacted stranded. Although the process of compaction is quite efficient, it is nonetheless imperfect and will leave some spaces between the strands. Therefore, the calculation of core dc resistance from the core nominal resistivity and nominal cross-sectional area as obtained from nominal outer radius is not valid. For a given nominal outer radius, the effect of the air gap spaces between the strands can be accounted for by increasing the nominal core resistivity, whilst keeping the manufacturer-declared core dc resistance constant since this is a valid measured value. Two methods can be used depending on the availability of cable data as follows.

Core dc resistance is available The resistivity of the core can be modified as follows 0

ρc 5 πrc2 3 Rdc 3 1029 Ωm

(3.142a)

where Rdc 5 manufacturer-supplied core dc resistance at 20 C in Ω/km, and rc 5 manufacturer-supplied nominal radius of the core in mm.

212

Power Systems Modelling and Fault Analysis

Core dc resistance is not available The dc resistance can be expressed as Rdc 5 resistivity is 0

ρc 5

ρc AcðeÞ

5

0

ρc πrc2 .

Therefore, the modified

πrc2 ρ Ωm A cð e Þ c

(3.142b)

where ρc 5 nominal resistivity and AcðeÞ 5 effective cross-sectional area of the core.

Insulation, semiconductive screens and bedding tape Modern XLPE cables have a core screen, a screen over the XLPE insulation and a swelling bedding tape over the insulation screen. The core screen, insulation screen and bedding tape are made of semiconductor material. Electrically, each semiconductive layer can be represented as a conductance and capacitance in parallel and the equivalent shunt admittance between the core and the sheath is shown in Fig. 3.35 where G2 and C2: Conductance and capacitance of core screen layer; G3 and C3: Conductance and capacitance of XLPE insulation; G4 and C4: Conductance and capacitance of insulation screen layer; G5 and C5: Conductance and capacitance of bedding tape layer.

The admittance (conductance and capacitance) of each semiconductive layer depends on its geometric and material properties. For layer k, the admittance is given by 2πε0 εk ðf Þ  5 Gk 1 j2πfCk yk 5 jω  rk ln rk21

(3.143a)

Core G2

C2 Semiconductor screen

G3

C3 XLPE insulation

G4

C4 Semiconductor screen

G5

C5 Bedding tape Sheath

Figure 3.35 High-frequency representation of semiconducting screens and insulation.

Modelling of multiconductor overhead power lines, underground and submarine cables

213

where 0

00

εk ðf Þ 5 εk ðf Þ 2 jεk ðf Þ and 00

Gk 5

2πωε0 εk ðωÞ 2πσ   5  k rk rk ln ln rk21 rk21

(3.143b)

00

σk 5 ωε0 εk ðf Þ is the ac conductivity 5 1=ρk 0

Ck 5

2πε0 εk ðf Þ   rk ln rk21

(3.143c)

The semiconductive layers are made of polymer and contain 30
40% carbon. IEC 60840 requires that the resistivity of the core and insulation screens be lower than 1000 Ωm and 500 Ωm, respectively. However, measurements on sample cables show that manufacturers normally produce these layers with a resistivity as low as 0.1
10 Ωm. This is still much higher than the resistivity of copper of 1:7241 1028 Ωm, aluminium of2:8264 1028 Ωm and lead alloy of 21.4 21:4 1028 Ωm. As a result, the longitudinal current will not flow through these layers. Similarly, the high carbon content of the semiconductive layers results in a very high relative permittivity, typically around 5000 at low frequency and 1000 at 20 MHz. Also, the nominal thickness of a semiconductive layer ranges typically from 0.4 to 4 mm depending on the cable rated voltage. However, IEC 60840 and IEC 60502 limit the minimum thickness of each cable layer in relation to nominal thickness, but not maximum thickness. Therefore, cable manufacturers tend to choose thicker layers than nominal to account for the disparity in production and ageing effects. This practice is known to be prevalent for the semiconductive screens, main insulation and oversheath. The combination of very high permittivity and low thickness result in a very large capacitance for each semiconductive layer compared to the capacitance of the main XLPE insulation, whose permittivity is equal to 2.3. Thus, the capacitance of each semiconductive layer can be considered as a short circuit. Therefore, the manufacturer-declared core-to-metallic sheath capacitance will in effect be equal to that of the XLPE insulation. This means that the three semiconductive layers can be combined with the XLPE insulation by increasing its thickness and relative permittivity whilst keeping its capacitance constant and equal to the value declared by the manufacturer. The modified thickness of the insulation is given by: tequiv:2ins 5 tcore screen 1 tinsulation 1 tinsulation screen 1 tbedding tape

(3.144a)

214

Power Systems Modelling and Fault Analysis

And the modified inner radius of the metallic sheath is: ris 5 rc 1 tequiv:2ins

(3.144b)

Where tequiv:2ins 5 thickness of equivalent core to sheath insulation; tcore screen 5 thickness of core semiconductive screen layer; tinsulation 5 thickness of XLPE insulation; tinsulation screen 5 thickness of insulation semiconductive screen layer; tbedding tape 5 thickness of bedding tape (over insulation screen).

The modified relative permittivity can be calculated using:   ris ln 0 rc εr 5 C 2πεo

(3.145a)

where εo 5 8.854 187 8176 1023 μF/km and C is the manufacturer-declared core
sheath capacitance in μF=km. Or, where the capacitance is not known, the equivalent relative permittivity can be calculated as   ris ln 0 r εrðinsÞ 5 εrðinsÞ  c (3.145b) b ln a where a 5 rc 1tcore screen b 5 rc 1tcore screen 1 tinsulation

Metallic sheath Metallic sheath made of copper wire screen only Where a copper wire screen is used as the sole metallic sheath, the screen can be replaced by a tubular conductor having a cross-sectional area equal to that declared by the cable manufacturer. With an inner sheath radius ris , the outer sheath radius is given by rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ACu2screen 1 ris2 ros 5 (3.146) π where ACu2screen 5 Area of copper wire screen

Modelling of multiconductor overhead power lines, underground and submarine cables

215

Composite metallic sheath made of copper wire screen and lead sheath Some cables employ a copper wire screen and a lead sheath to achieve good electrical and mechanical performance. An equivalent composite metallic sheath can be derived to represent the two different sheaths characterized by an equivalent resistivity and an equivalent outer radius. These parameters must result in a dc resistance that is equal to the equivalent of the two sheaths. With an inner sheath radius ris , the outer sheath radius is given by: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Aequiv:2sheath 1 ris2 ros 5 π

(3.147a)

Aequiv:2sheath 5 ACu screen 1 ALead sheath

(3.147b)

is the cross-sectional area of the equivalent tubular sheath. Depending on the availability of data from the cable manufacturer, two methods can be used to calculate the equivalent resistivity of the equivalent sheath. Where the equivalent dc resistance of the copper screen and lead sheath is known, the equivalent resistivity of the equivalent sheath is ρequiv:2sheath 5 Aequiv:2sheath 3 Rdc2equiv:2sheath Ωm

(3.148a)

Alternatively, if the equivalent dc resistance of copper screen and lead sheath is not known, the equivalent resistivity can be calculated using: ρequiv:2sheath 5

ACu 1

A  Cu ρCu ρLead

ALead

ρCu 1

ALead  ρLead Ωm ALead 1 ρρLead ACu

(3.148b)

Cu

where ACu 5 nominal manufacturer-declared cross-section of copper wire screen; ALead 5 nominal manufacturer-declared cross-section of lead sheath; ρCu 5 nominal resistivity of copper; ρLead 5 nominal resistivity of lead alloy.

3.4.3 Example The following data are available from cable manufacturers:

Stranded copper core Area: Number of strands: Radius:

630 mm2 91 strands 15.2 mm

Thickness of tape
semiconductor core screen
tape: 0:25 mm
1 mm
0:25 mm

216

Power Systems Modelling and Fault Analysis

XLPE insulation Thickness: Capacitance: Relative permittivity:

16 mm 0:207μF=km 2.4

Thickness of insulation semiconductor core screen
tape nominal: 1:2 mm
0:5 mm

Composite sheath Copper wire screen area: 232.35 mm2 (100 wires, d 5 1.72 mm); Lead sheath nominal area: 710 mm2 (thickness 5 2.9 mm).

Recalculate the cable electrical parameters to take account of core stranding, semiconductive screens and composite sheath. The effective area of the core is 630 mm2 compared to π 3 15:22 5 725:83mm2 as calculated from the nominal outer radius. The modified core resistivity is calculated using Eq. (3.142b) 0

ρc 5

π 3 15:22 1:7241 3 1028 5 1:9864 3 1028 Ωm 630

The modified thickness of the insulation is calculated using Eq. (3.144a) tequiv:2ins 5 ð0:25 1 1 1 0:25Þ 1 16 1 ð1:2 1 0:5Þ 5 19:2 mm The modified inner radius of the metallic sheath is calculated using Eq. (3.144b) ris 5 15:2 1 19:2 5 34:4 mm The modified relative permittivity is calculated using Eq. (3.145a)   34:4 ln 0 15:2 εr 5 0:207 3 53 2π 3 8:854187 3 1023 The dc resistance of copper wire screen is calculated as Rdcð20CÞ 5 1:7241 3 1028 5 0:074315 Ω=km. 1000 232 3 1026 The dc resistance of the lead sheath is calculated as Rdcð20CÞ 5 21:4 3 1028 5 0:3014 Ω=km 1000 710 3 1026 1 Equivalent dc resistance of composite sheath 5 5 1 1 1 0:074315 0:3014 0:0596 Ω=km. The cross-sectional area of the equivalent sheath is Aequiv:2sheath 5 232 1 710 5 942mm2 .

Modelling of multiconductor overhead power lines, underground and submarine cables

217

The equivalent resistivity of the equivalent composite sheath is calculated as ρequiv:2sheath 5 942 3 1026 3 0:0596 3 1023 5 5:614 3 1028 Ωm Alternatively, the equivalent resistivity of the equivalent composite sheath is calculated using Eq. (3.148b) as 0

ρequiv:2sheath 5

232 1 @

232

1:7241A 3 710 21:4 0

1

3 1:7241 3 1028

1

710 1 @

710

1

3 21:4 3 1028 5 5:616 3 1028 Ωm

21:4 A 3 232 1:7241

The outer radius of the equivalent sheath is calculated using Eq. (3.147a) as rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 942 1 34:42 5 38:5 mm ros 5 π

3.5

Sequence π models of single-circuit and doublecircuit overhead lines and cables

3.5.1 Background In Sections 3.3 and 3.4, we presented calculations of the positive-sequence/negative-sequence and zero-sequence impedances and susceptances of overhead lines and cables per km of circuit length. The electrical parameters of lines and cables are distributed over their length. The derivation of T and π equivalent circuits using lumped parameters for the entire length of a line or a cable is covered in most basic power system textbooks and will not be repeated here. The π equivalent circuit is most extensively used in practical applications. Fig. 3.36A illustrates the distributed nature of the parameters of a line or cable. Fig. 3.36C is the nominal π lumped parameter equivalent circuit of the line or cable where the total impedance and susceptance are calculated by multiplying the per km parameters by the line/cable length. In Fig. 3.36B, the terms in brackets represent correction terms that allow for the distributed nature of the circuit parameters over the entire circuit length. These terms can be represented as infinite series as follows: pffiffiffiffiffiffiffiffiffi 0 0 2 0 0 3 0 0 sinh Z 0 Y 0 L Z LY L Z LY L Z LY L ffi pffiffiffiffiffiffiffiffi 1 511 1 1? 0 0 6 120 5040 ZYL

(3.149a)

218

Power Systems Modelling and Fault Analysis

tanh

pffiffiffiffiffiffi ffi 0 0 ZY L 2

pffiffiffiffiffiffi ffi Z0 Y0 L 2

0 0 2 0 0 3 0 0 Z LY L 17 Z LY L Z LY L 1 2 1? 512 12 120 20160

(3.149b)

The hyperbolic sine and tangent of a complex argument are approximately equal to the complex argument itself when this has small values. For a lossless line or cable, the complex argument is equal to (2π/λ)L where λ is the wavelength in km pffiffiffiffi and is given by λD3 3 105 =ðf 3 εr Þ, where εr is the relative permittivity of the dielectric. For overhead lines (εr 5 1 for air), the wavelength is approximately 6000 km at 50 Hz and 5000 km at 60 Hz. The definition of an ‘electrically short’ line or cable for which a nominal π circuit can be used depends on the acceptable error magnitude in the total series impedance and total shunt susceptance, and this determines the physical length in km for such a short line or cable. (A)

Z′ R

jB

G

jX G

Y′

jB

Self series impedance = Z ' = R + jX Ω/km L (km) Self shunt admittance = Y ' = jB mho/km, G ≈ 0 at 50/60 Hz L = line length (km) (B) Y

(C)

Figure 3.36 Representation of lines and cables: (A) distributed parameter circuit; (B) accurate π lumped parameter equivalent circuit; and (C) nominal π lumped parameter equivalent circuit.

Modelling of multiconductor overhead power lines, underground and submarine cables

(A)

P

P

P I t1

IS

V1 1

P

2 V2

P

It2

P

B

j

2

IS

V1 1

I2

P

Z

Z

Z

I t1

P

P

I1

j

(B)

Z

Z

Z

219

B

P

2

Z

2 V 2 IZ t2

Z

Z

I1

j

B

2

Z

I2

j

B

Z

2

Figure 3.37 Single-circuit overhead line or cable: (A) positive-sequence/negative-sequence and (B) zero-sequence π models.

For an error of less than 1% in these parameters, the physical length should be less than about 3% of the wavelength. For overhead lines, this corresponds to 180 km for 50-Hz and 150 km for 60-Hz systems. For underground cables, with εr being typically between 2.3 and 4.2, the corresponding cable lengths using εr 5 2.4 are 116 km for 50-Hz and 96.8 km for 60-Hz systems. In practice, these cable lengths are not reached above voltages of around 132 kV due to the adverse impact of cable charging current on the cable thermal rating. For such lines or cables, the first term in the infinite series expansion of Eqs. (3.149a)and (3.149b), which is equal to unity, is taken and all other terms are neglected. Therefore, the exact equivalent π circuit of Fig. 3.36B reduces to that shown in Fig. 3.36C.

3.5.2 Sequence π models of single-circuit overhead lines and cables Overhead lines The positive-sequence, negative-sequence and zero-sequence impedances and susceptances of perfectly transposed single-circuit overhead lines were derived in Section 3.2.4. Recalling the sequence impedance and susceptance matrices of Eqs. (3.50a) and (3.54a), denoting 1 and 2 as the sending and receiving ends of the line or cable circuit and connecting 1/2 the susceptance at each end, as shown in Fig. 3.37, we can write 3 2 P3 3 2 P 3 2 P3 2 P3 2 P IS I1 V1P 2V2P V1 Z 0 0 jB =2 0 0 76 7 6 N 7 6 76 7 6 7 6 4 V1 2V2N 5 5 4 0 Z N 0 5 4 ISN 5 and 4 I1N 5 5 4 0 jBN =2 0 5 4 V1N 5 ISZ V1Z 2V2Z I1Z V1Z 0 0 ZZ 0 0 jBZ =2 2

220

Power Systems Modelling and Fault Analysis

2

I2P

3 2

6 N7 6 4 I2 5 5 4 I2Z

jBP =2

0

0

jBN =2

0

0

0

32

V2P

3

7 6 N7 5 4 V2 5 Z V2Z jB =2

(3.150a)

0

For the positive-sequence, negative-sequence and zero-sequence circuits, the total injected current at ends 1 and 2 are given by It1 5 IS 1 I1

and

It2 5 2 IS 1 I2

(3.150b)

Therefore, using Eqs. (3.150a)and (3.150b), we obtain three separate positivesequence, negative-sequence and zero-sequence sequence π models for the line or cable as follows: 2

1 Bx 1 j  x  6 Zx 2 6 It1 56 6 x It2 4 2 1 Zx

3 1   Zx 7 7 V1x 7 x 5 P; N or Z x 1 Bx 7 5 V2 1 j Zx 2 2

(3.151)

As the positive-sequence and negative-sequence impedances and susceptances are equal, only two sequence π models are needed, as shown in Fig. 3.37. For convenience, we will only refer to the positive-sequence and zero-sequence sequence parameters and models in subsequent sections.

Cables The sequence impedance matrices of single-circuit cables of various sheath earthing arrangements were derived in Section 3.3.4. The sequence impedance matrix of Eq. (3.125), combined with the circuit sequence currents and voltages, gives 2

V1P 2 V2P

3

2

Z PP

6 N 7 6 4 V1 2 V2N 5 5 4 Z NP V1Z 2 V2Z Z ZP

Z PN

Z PZ

32

IP

3

Z NN

76 7 Z NZ 5 4 I N 5

Z

Z

ZN

ZZ

I

(3.152)

Z

The mutual intersequence terms are normally much smaller than the self-terms for solidly bonded cables and practically zero for cross-bonded cables. For the former, it is usual practice to ignore the mutual terms and use the self-terms only for the positive-sequence, negative-sequence and zero-sequence impedances of the cable. In the case of submarine cables where the three phases are laid a significant distance apart, the sequence impedance matrix is diagonal with no mutual intersequence terms. This is also the case for three-core cables, and single-core cables in touching trefoil or equilateral arrangements. The sequence susceptance matrix of all shielded and belted cables is diagonal and hence the positive-sequence,

Modelling of multiconductor overhead power lines, underground and submarine cables

221

negative-sequence and zero-sequence susceptances are equal to the phase terms unaffected by any intersequence terms. Therefore, as in the case of single-circuit overhead lines, positive-sequence and zero-sequence sequence π models can be derived as shown in Fig. 3.37.

3.5.3 Sequence π models of double-circuit overhead lines The positive-sequence, negative-sequence and zero-sequence impedances and susceptances of double-circuit overhead lines were derived in Section 3.2.6. We recall the sequence impedance and susceptance matrices of Eqs. (3.76a) and (3.78a) and illustrate their physical meaning as shown in Fig. 3.38A. Eq. (3.76a) suggests that each circuit has a self positive-sequence, negative-sequence and zero-sequence impedance. In addition, positive-sequence, negative-sequence and zero-sequence intercircuit mutual coupling exists between the positive-sequence, negative-sequence and zero-sequence impedances of each circuit. Importantly, we note that the three positive-sequence, negative-sequence and zero-sequence circuits are separate and independent. Eq. (3.78a) contains similar information for susceptances. Allocating half of the susceptances to each circuit end, Eqs. (3.76a) and (3.78a) of circuit 1 and circuit 2 are represented in Fig. 3.38B as a π model.

(A)

Circuit 1

Z1PNZ

1′

B1PNZ

1′′

PNZ PNZ B12 Z12

Circuit 2

2′

Z 2PNZ

B2PNZ

I1PNZ

Z1PNZ

(B) PNZ Circuit 1 V1′

I′t1PNZ PNZ I′1SH

j j Circuit 2

PNZ B12 2

V2′PNZ

B1PNZ 2 I 2PNZ

I′t2PNZ PNZ 2SH

I′

j

B2PNZ 2

2′′

′′PNZ I t1 ′′PNZ I1SH

PNZ Z12

j

PNZ

j

B12 2

Z 2PNZ

B1PNZ 2

′′PNZ I t2 I ′′

PNZ 2SH

j

V1′′PNZ

V2′′PNZ

B2PNZ 2

Figure 3.38 Double-circuit overhead line model in the sequence reference frame using sequence matrices: (A) schematic and (B) sequence π model.

222

Power Systems Modelling and Fault Analysis

Using Fig. 3.38B and Eqs. (3.76a) and (3.78a), we can write 2

V 01 P 2 V100 P

3

2

Z1P 6 V 0 N 2 V 00 N 7 6 0 6 1 6 1 7 7 6 6 0 6 V 1 Z 2 V100 Z 7 6 0 7 6 6 6 V 0 P 2 V 00 P 7 5 6 Z P 6 2 6 12 2 7 7 6 6 0 4 V 2 N 2 V200 N 5 4 0 V 02 Z 2 V200 Z 0 " ZPNZ 1 5 ZPNZ 12 3 2 P I 01;SH P B1 6 0 6 I0 N 7 6 6 1;SH 7 7 6 6 0 6 6 I 1;SH Z 7 75j16 0 6 6 I0 P 7 P 26 6 B12 6 2;SH 7 7 6 6 0 4 0 4 I 2;SH N 5 I 02;SH Z 0 " 1 BPNZ 1 5j 2 BPNZ 12 2

2

00 P I1;SH

3

2

BP1

6 0 6 I 00 N 7 6 6 1;SH 7 7 6 6 00 6 6 I1;SH Z 7 1 75j 6 0 6 6 BP 6 I 00 P 7 2 6 12 6 2;SH 7 7 6 6 00 4 0 4 I2;SH Z 5 00 I2;SH Z 0 " 1 BPNZ 1 5j 2 BPNZ 12

0 Z1P

0 0

P Z12 0

0 P Z12

0 0

Z1Z 0

0 Z2P

0 0

P Z12

0

0

Z2P

32 P 3 I1 0 6 IN 7 0 7 76 1 7 6 7 Z 7 76 I1Z 7 Z12 76 7 6 P7 0 7 76 I2 7 76 N 7 0 54 I2 5

Z Z12 0 0 #" # IPNZ ZPNZ 1 12 PNZ IPNZ Z2 2

Z2Z

0

I2Z

0 BP1

0 0

BP12 0

0 BP12

0

BZ1

0

0

0 BP12

0 0

BP2

0 BP2

32 0 3 V 1P 0 7 6 0 76 V 01 N 7 7 76 0 7 6 V 1Z 7 BZ12 7 7 76 6 0 7 0 7 76 V 2 P 7 7 76 0 54 V 02 N 5

0

BZ12 #"

0

0

BZ2

0

BPNZ 12 BPNZ 2

V0 PNZ 1 V0 PNZ 2

#

0

0

BP12

0

BP1 0

0 BZ1

0 0

BP12 0

0

0

BP2

0

BP12

0 BZ12 #"

0 0

BP2 0 #

0

BPNZ 12 BPNZ 2

V00 PNZ 1

BZ2

(3.153b)

V 02 Z

3 V100 P 6 00 7 0 7 7 6 V1 N 7 7 6 00 7 6V Z7 BZ12 7 76 1 7 7 00 7 0 76 6 V2 P 7 76 00 7 0 5 4 V2 N 5 0

(3.153a)

32

(3.153c)

V200 Z

V00 PNZ 2

The equivalent admittance matrix can be derived for each sequence circuit independently because the positive-sequence, negative-sequence and zero-sequence circuits are separate. The form of the admittance matrix is the same for the positivesequence, negative-sequence and zero-sequence circuits so the derivation of one would suffice. Collecting the positive-sequence terms from Eq. (3.153a), we have 

V 01 P 2 V100 P



"

Z1P 5 P V 02 P 2 V200 P Z12

P Z12 Z2P

#"

I1P I2P

#

Modelling of multiconductor overhead power lines, underground and submarine cables

"

I1P I2P

#

1 5 D

"

Z2P

P 2 Z12

P 2 Z12

Z1P

#"

V 01 P 2 V100 P

223

# (3.154a)

V 02 P 2 V200 P

P 2 Þ : Similarly, collecting the shunt positive-sequence terms where D 5 Z1P Z2P 2 ðZ12 from Eqs. (3.153b) and (3.153c), we have

"

#

" P 1 B1 5j 2 BP I 02;SH P 12

I 01;SH P

BP12

#"

V 01 P V 02 P

BP2

#

"

#

" P 1 B1 5j and 00 2 BP I2;SH P 12 00 P I1;SH

BP12

#"

V100 P

#

V200 P

BP2

(3.154b) The total injected currents into circuit 1 and circuit 2 at both ends are given by 2

I 01;t P

3

2

I 01;SH P

3

2

I1P

3

7 7 6 6 00 7 6 00 6 I1;t P 7 6 I1;SH P 7 6 2 I1P 7 7 7 6 7 6 6 7 716 6 0 756 0 P 7 7 6 6I P7 6I 4 2;t 5 4 2;SH P 5 4 I2 5 00 I2;t P

00 I2;SH P

(3.155a)

2 I2P

Substituting Eqs. (3.154a) and (3.154b) into Eq. (3.155a), we obtain 20

1 P P B Z 6 @j 1 1 2 A 6 2 D 6 6 6 2 0 3 6 6 I 1;t P 2Z2P 6 6 6 00 7 6 D 6 I1;t P 7 6 7 6 6 6 0 7560 1 6I P7 6 P 4 2;t 5 6 BP Z 6 @j 12 2 12 A 00 6 I2;t P 2 D 6 6 6 6 P 6 Z12 4 D

1 P P B Z @j 12 2 12 A 2 D

3

0 2Z2P D

P Z12 D

7 7 7 7 0 1 0 17 72 P P P P P 7 V0 P3 Z12 1 @j B1 1 Z2 A @j B12 2 Z12 A 7 7 7 2 D D 2 D 76 00 7 6 7 6 V1 P 7 7 7 0 1 76 V 02 P 7 76 P P P P 5 4 Z12 2Z1 7 @j B2 1 Z1 A 7 00 7 V2 P D 2 D D 7 7 0 1 0 17 7 P P P P 7 2Z1P @j B12 2 Z12 A @ j B 2 1 Z1 A 5 2 D D 2 D (3.155b)

The negative-sequence and zero-sequence admittance matrices of the doublecircuit line with negative-sequence and zero-sequence impedance and susceptance coupling between the two circuits are identical to that given in Eq. (3.155b) except that the zero-sequence parameter values are different.

224

Power Systems Modelling and Fault Analysis

The sequence model of a double-circuit overhead line derived above represents the case where the line is assumed to have had six transpositions. However, as we have shown in Eqs. (3.89a)and (3.89b), when the line is assumed to be ideally transposed with nine transpositions, there would only be zero-sequence coupling between the two circuits. In this case, the zero-sequence admittance matrix would be identical to that of Eq. (3.155b) but the positive-sequence/negative-sequence admittance matrices would have zero intercircuit mutual impedances and susceptances as follows: 3 1 P B 1 2 1 7 6 @j 1 1 A 0 0 7 6 2 Z1P Z1P 7 6 7 6 7 6 0 1 72 6 2 0 3 6 P 7 V0 P 3 I 1;t P 21 B 1 7 6 1 1 @ A 1 0 0 j 76 P P 7 6 00 7 6 2 Z Z 7 6 00 1 1 6 I1;t P 7 6 V1 P 7 76 7 6 7 6 6 76 6 0 756 0 1 76 0 7 6I P7 6 7 4 V 2P 7 P 5 4 2;t 5 6 B 1 2 1 7 2 @j 1 A 7 6 0 0 00 00 P P 7 V2 P 6 I2;t P 2 Z2 Z2 7 6 7 6 6 0 17 7 6 P 6 21 B2 1 A7 @ 4 0 0 j 1 P 5 2 Z2P Z2 20

(3.156)

3.5.4 Sequence π models of double-circuit cables As presented in Section 3.3.3, the series sequence impedance matrix of each cable circuit in a double-circuit cable will effectively be diagonal. There will also be intercircuit mutual inductive coupling between the two circuits but the intercircuit phase impedance mutual matrix is in general not balanced. For cross-bonded single-core double-circuit cables, there would be very small intersequence mutual coupling between the two circuits in the positive sequence and negative sequence but a nonnegligible coupling in the zero sequence depending on spacing between the two circuits. The same applies for a three-core double-circuit cable. As presented in Section 3.3.4, the shunt sequence susceptance matrix of a single-circuit cable is diagonal, that is, there is no intersequence mutual coupling. The presence of a second cable circuit has no effect other than introducing a second diagonal shunt sequence susceptance matrix for this circuit. Therefore, the positive-sequence and zero-sequence sequence π models shown in Fig. 3.37 for a single-circuit cable may also be used for double-circuit cables. Where zero-sequence impedance coupling between the two cable circuits is significant and needs to be represented, then the approach described in Section 3.5.3 for double-circuit overhead lines can be used.

Modelling of multiconductor overhead power lines, underground and submarine cables

225

Sequence π models of three-circuit overhead lines

3.6

Using a similar approach to that of Section 3.5.3, Fig. 3.39 illustrates three mutually coupled circuits and their self and mutual sequence impedances and susceptances. The sequence π model represents either the positive-sequence, negativesequence or zero-sequence equivalent circuits. Where ideal transposition is used with only zero-sequence intercircuit mutual coupling, the positive-sequence and negative-sequence π model of each circuit is completely independent from the other circuits. We will now derive the admittance matrix for the general case where intercircuit mutual coupling exists and we will use the zero-sequence π model parameters but drop the superscript Z to simplify the notation. From Fig. 3.39B, we can write 2

3 2 V 01 2 V100 Z1 4 V 02 2 V200 5 5 4 Z12 V 03 2 V300 Z13

Z12 Z2 Z23

32 3 I1 Z13 Z23 54 I2 5 Z3 I3

2

3 V 01 2 V100 00 5 4 0 I1;2;3 5 Z21 1;2;3 V 2 2 V2 0 V 3 2 V300

or

or

(A) Circuit 1

Z1PNZ

1′

B1PNZ

PNZ PNZ B12 Z12

Circuit 2

2′

PNZ

PNZ

PNZ

Z 2PNZ B2

1′′ PNZ Z13

2′′

PNZ B13

Z 23 B23 Circuit 3

3′

(B) V1′

Circuit 1

j

B12 2

V 2′

Circuit 2 B j 13 2

I 1′ I 1′ SH

I1

B 23 2

V3′

I 1′′

I2 B2 2

B j 12 V 2′′ 2

I ′2′

Z2

′ I ′2SH Z 23

V1′′

B j 1 2

Z12

I ′2 I′2SH

Z1

′′ I 1SH

B j 1 2

j j

Circuit 3

3′′

PNZ Z 3PNZ B3

Z13

j

B2 2

j j

I ′3 I ′3SH

I3 j

B3 2

I ′3′ V3′′

Z3

′ I ′3SH j

B13 2

B 23 2

B3 2

Figure 3.39 Three-circuit overhead line model in the sequence reference frame: (A) schematic and (B) zero-sequence π model.

226

Power Systems Modelling and Fault Analysis

I1;2;3 5

h

Z21 1;2;3

2 Z21 1;2;3

" # i V0 1;2;3

(3.157a)

V 001;2;3

where 2

Z1

6 Z1;2;3 5 4 Z12 Z13

Z12

3

Z13

7 Z23 5 Z3

Z2 Z23

(3.157b)

The total injected currents into each circuit at both ends are given by 2

jB1 6 2 6 6 6 jB12 6 2 03 6 6 2 I1 6 6 I0 7 6 jB13 6 27 6 6 07 6 6 6 I3 7 6 2 6 756 6 I 00 7 6 6 17 6 0 6 00 7 6 4 I2 5 6 6 6 I 003 6 0 6 6 6 6 4 0

3

jB12 2

jB13 2

0

0

jB2 2

jB23 2

0

0

jB23 2

jB3 2

0

0

0

0

jB1 2

jB12 2

0

0

jB12 2

jB2 2

0

0

jB13 2

jB23 2

0 7 7 7 7 0 7 72 3 2 7 V0 7 1 76 0 7 6 7 V 6 27 0 76 7 6 76 V 03 7 6 76 16 76 00 7 6 jB13 76 V1 7 7 6 76 7 6 7 00 2 74 V 5 6 4 2 7 00 jB23 7 7 V3 2 7 7 7 jB3 7 5 2

3 I1 I2 7 7 7 I3 7 7 2 I1 7 7 7 2 I2 5

(3.158)

2 I3

or written in concise matrix form using Eq. (3.157a), we have 3

2

jB1;2;3  0  6 2 6 I 1;2;3 56 6 I00 1;2;3 4 0

7 0  " 21 Z1;2;3 7 V 1;2;3 7 1 00 jB1;2;3 7 V 2 Z21 1;2;3 5 1;2;3 2 0

2 Z21 1;2;3 Z21 1;2;3

#

V0 1;2;3



V0 1;2;3

or 2

jB1;2;3 21  0  6 2 1 Z1;2;3 6 I 1;2;3 56 6 00 I 1;2;3 4 2 Z21 1;2;3

2 Z21 1;2;3 jB1;2;3 1 Z21 1;2;3 2

3 7 0  7 V 1;2;3 7 7 V00 1;2;3 5

(3.159a)

Modelling of multiconductor overhead power lines, underground and submarine cables

227

where 2

B1 B1;2;3 5 4 B12 B13

3.7

B12 B2 B23

3 B13 B23 5 B3

(3.159b)

Three-phase modelling of overhead lines and cables (phase frame of reference)

3.7.1 Background In advanced studies, such as multiphase loadflow and short-circuit fault analysis, where the natural unbalance of untransposed lines and cables is to be taken into account, the modelling and analysis may be carried out in the phase frame of reference. Such analysis allows the calculation of the current distribution on overhead line earth wires and cable sheaths under balanced and unbalanced power flow and short-circuit faults.

3.7.2 Single-circuit overhead lines and cables For single-circuit overhead lines, the series phase impedance matrices with earth wires present and eliminated were given in Eqs. (3.29) and (3.32c), respectively. The corresponding shunt phase susceptance matrices are obtained from the inverse of Eqs. (3.36a) and (3.36b) and from Eqs. (3.39a)and (3.39b). Considering the case of an overhead line with one earth wire, the line can be represented by a threephase equivalent nominal π model as shown in Fig. 3.40A. In Fig. 3.40, the line is represented by its series phase impedance matrix and half its shunt phase susceptance matrix connected at each end. We denote as 1 and 2 the two ends of the line, and we use ‘ph’ for ‘phase’ quantities, which also include the earth wires if these have not been eliminated. From Fig. 3.40B, we have ph ph ph Vph 1 2 V2 5 V IS

Iph SHð1Þ 5 j

Bph ph V 2 1

Iph SHð2Þ 5 j

Bph ph V 2 2

(3.160a)

Therefore, the nodal phase admittance matrix is given by 2

Bph ph 21 1 Z " ph # 6 j 6 2 I1 56 6 21 I2ph 4 2 Zph

3 7" ph # 7 V1 7 ph ph 21 7 5 V2 1 Z

21 2 Zph Bph j 2

(3.160b)

If the earth wire has already been eliminated, then it is simply removed from Fig. 3.40A and the dimension of the series phase impedance and shunt phase

228

Power Systems Modelling and Fault Analysis

(A) I1S

IS2

Earth wire S

S I B2

I1B

I1Y

B

B

Y

Y

I1R

I Y2 I R2

R

R

(B) I1RYB

I SRYB RYB I SH(1)

V1RYB

j

B RYB 2

I 2RYB

Z RYB RYB I SH(2)

j

V2RYB

B RYB 2

Figure 3.40 Three-phase model of a single-circuit line including earth wire(s): (A) π equivalent circuit and (B) π model (matrices).

susceptance matrices is reduced accordingly. The series phase impedance and shunt phase admittance matrices for cables, with their sheaths present or eliminated, can be represented using a similar approach taking into account the particular sheath earthing arrangement. Fig. 3.41 shows a three-phase model for a single-circuit underground cable with a solidly bonded sheath.

3.7.3 Double-circuit overhead lines and cables For double-circuit overhead lines, the series phase impedance matrices with earth wires present and eliminated are given in Eqs. (3.57a) and (3.58b), respectively. The corresponding shunt phase susceptance matrices are obtained from the inverse of Eqs. (3.60a) and (3.62c), respectively. Using the π matrix model shown in Fig. 3.40B together with the intercircuit mutual impedance and susceptance matrices, we obtain the π model representation of a double-circuit overhead line in the phase frame of reference as shown in Fig. 3.42A. We denote 1 and 10 , and 2 and 20 as the two ends of circuit 1 and circuit 2, respectively. We assume that the intercircuit mutual coupling is bilateral, that is, B21 5 BT12. From Fig. 3.42B, the series voltage drop across the two circuits is given by

Modelling of multiconductor overhead power lines, underground and submarine cables

229

C1 S1

C2 S2

C3 S3

Figure 3.41 Three-phase π equivalent of a single-circuit solidly bonded underground cable.

(A)

R Circuit 1

R Y B

Y B

Earth wire

R Circuit 2 Y B

R Y B

(B) I t1RYB

RYB I S1

Circuit 1

I

j j RYB

I t2 Circuit 2

RYB SH1 RYB 11

V1RYB

RYB 12

I′t1RYB RYB

Z11

B

Z12RYB

2

RYB

2 j B21 2

V2RYB

RYB I S2 RYB SH2 RYB 22

I

j

B

2

I′

j

2

I′t2RYB

RYB

Z 22

RYB SH2 RYB 22

I′

j

V1′RYB

B

B RYB j 12 2 B RYB j 21 2

RYB Z 21

B

RYB SH1 RYB 11

V2′RYB

B

2

Figure 3.42 Three-phase model of a double-circuit overhead line with one earth wire: (A) π equivalent circuit and (B) π model (matrices).

230

Power Systems Modelling and Fault Analysis

2 4

ph Vph 1 2 V10 ph Vph 2 2 V20

2 4

Iph Sð1Þ Iph Sð2Þ

3

3

2

554 2

5 54

ph Zph 11 Z12

32 54

ph Zph 21 Z22

ph Yph 11 Y12

32 54

ph Yph 21 Y22

Iph Sð1Þ

3 5

Iph Sð2Þ

ph Vph 1 2 V10

3 5

(3.161)

ph Vph 2 2 V20

The shunt susceptance currents are given by 2

2 32 3 ph Bph Vph j 11 B12 1 4 55 4 54 5 and ph ph ph ph 2 ISHð2Þ B21 B22 V2 Iph SHð1Þ

3

2

2 32 3 ph Bph Vph j 11 B12 10 4 55 4 54 5 ph ph ph ph 2 ISHð2Þ0 B21 B22 V20 Iph SHð10 Þ

3

(3.162) Since the total current injected into each circuit end is the sum of the series and shunt currents, it can be shown that the nodal phase admittance matrix equation of the mutually coupled double-circuit line is given by 2 Yph 11

Bph 1 j 11 2

6 6 6 6 2 ph 3 6 6 Itð1Þ 6 2 Yph 11 7 6 6 6 ph 7 6 6 Itð10 Þ 7 6 7 6 6  T 756 6 6 Iph 7 6 Bph 12 6 tð2Þ 7 6 ph 5 6 Y21 1 j 4 6 2 6 Iph 6 tð20 Þ 6 6 6 4 2 Yph 21

2 Yph 11 Yph 11 1 j

Yph 12

Bph 11 2

2 Yph 21  Yph 21

1j

Bph 1 j 12 2

2 Yph 12

Yph 22 1 j

Bph 12 2

Bph 22 2

T 2 Yph 22

3 2 Yph 12

7 7 7 7 ph 7 B12 7 7 Yph 12 1 j 2 7 7 7 7 7 7 7 ph 2 Y22 7 7 7 7 7 ph 7 7 B 22 5 Yph 22 1 j 2

2

Vph 1

6 6 Iph 6 10 6 6 ph 6 I2 4

3 7 7 7 7 7 7 5

Iph 20

(3.163) All quantities in Eq. (3.163) are matrices. Where the earth wires are eliminated, the dimensions of the impedances and susceptances are 3 3 3 and those of the current and voltage vectors are 3 3 1. Therefore, the dimensions of the nodal current and voltage vectors are 12 3 1 and the dimension of the admittance matrix of the double-circuit line is 12 3 12. The same approach presented above can be followed in the case of double-circuit cables although the dimensions of the matrices will be increased to a greater extent by the number of sheaths and armours as appropriate.

Modelling of multiconductor overhead power lines, underground and submarine cables

3.8

231

Computer calculations and measurements of overhead line and cable parameters

3.8.1 Computer calculations of overhead line and cable parameters The days of hand or analytical calculations of overhead line and cable phase and sequence impedances and susceptances have long gone together with, unfortunately, some of the great insights such methods provided. The calculations of overhead line and cable electrical parameters at power frequency, that is, 50 or 60 Hz, as well as higher frequencies, are nowadays efficiently carried out using digital computer programs. For overhead lines, the input data are phase conductor and earth wire material and resistivity, physical dimensions, dc or ac resistance, conductor bundle data, tower geometry, that is, spacing between conductors and conductor height above ground, average conductor sag and earth resistivity. The output information usually comprises the line’s ac resistance matrix, inductance matrix, series phase reactance matrix, series phase impedance matrix, potential coefficient matrix, shunt capacitance matrix and shunt susceptance matrix, all including and excluding the earth wires. Assumptions of phase transpositions can be easily implemented using efficient matrix analysis techniques. The sequence data calculated consist of the sequence series impedance and sequence shunt susceptance matrices. The dimension of these sequence matrices is 3N, where N is the number of three-phase circuits. The output information is usually calculated in physical units, for example, μF/km/phase or Ω/km/phase or per circuit and in pu or % on some defined voltage and MVA base. For cables, the input data are core, sheath and, where applicable, armour conductor or pipe material and resistivity, their physical dimensions, dc or ac resistance, the cable layout or spacing between phases, relative permittivity of core, sheath and, where applicable, armour insulation or pipe coating, as well as earth resistivity. The output information usually comprises the cable’s full and reduced series phase impedance and shunt susceptance matrices, that is, including and excluding the sheaths and armour/pipe. The sequence data calculated consist of the sequence series impedance matrix and sequence shunt susceptance matrix.

3.8.2 Measurement of overhead line parameters General Measurements of overhead line power frequency electrical parameters or constants can be made at the time of commissioning when the line construction has been completed so that the tests replicate the line in normal service operation. The parameters that may be measured are phase conductor self and mutual impedances with earth return, and self and mutual susceptances. The series phase impedance and shunt phase susceptance matrices are then constructed and the sequence impedances

232

Power Systems Modelling and Fault Analysis

and susceptances calculated. There are numerous designs of overhead lines used in practical installations but we will illustrate the techniques that can be used for a single-circuit line with one earth wire and a double-circuit line with one earth wire. The technique can be extended for application to any other line construction.

Single-circuit overhead lines with one earth wire Fig. 3.43A illustrates a technique for series phase impedance measurements where the three-phase conductors are earthed at the far end of the line but kept free from earth at the near measurement end. A test voltage is applied to a one-phase conductor and the current flowing through it is measured. The induced voltages to earth on the remaining conductors are also measured. In addition to the calculation of the impedances from the measured voltages and currents, a wattmeter or phase angle meter can be used to find the phase angle of these voltages with respect to the injected current. This allows the calculation of the resistance and reactance components of the ffiimpedance using pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 5 P/ 2 (P is measured phase power in Watts) and X 5 Z 2 2 R2 in the case of a wattmeter, but R 5 (V/I) 3 cos φ and X 5 (V/I) 3 sin φ in the case of a phase angle meter measurement where Z 5 V/I. The process is repeated for all other conductors. Where earth wire(s) are present, the measured impedances should represent values with the earth wire in parallel with the earth return path to represent normal service (A) Earth wire

2 V

1 3

Measuring end

(B)

Earth wire 2

V

Short circuited and earthed end

Open circuited end

1

Measuring end

3

Figure 3.43 Measurement of self and mutual electrical parameters of a single-circuit overhead line: (A) series phase impedances with earth return and (B) shunt phase susceptances.

Modelling of multiconductor overhead power lines, underground and submarine cables

233

operation. Using the single-circuit phase impedance matrix that includes the effect of earth wire that is with the earth wire mathematically eliminated, we have 2

3 2 V1 Z11 6 7 6 4 V2 5 5 4 Z21 V3

Z31

Z12 Z22

32 3 I1 Z13 76 7 Z23 54 I2 5

Z32

Z33

(3.164)

I3

With the voltage source applied to conductor 1, V1 is known and the measured values are I1, V2 and V3 with I2 5 I3 5 0. Using the measured values in Eq. (3.164), we obtain Z11 5

V1 I1

Z21 5

V2 I1

Z31 5

V3 I1

(3.165a)

Applying the voltage source to conductor 2 then conductor 3, and using the measured values in Eq. (3.164), we obtain Z12 5

V1 I2

Z22 5

V2 I2

Z13 5

V1 I3

Z23 5

V2 I3

Z32 5

V3 I2

(3.165b)

and Z33 5

V3 I3

(3.165c)

For sequence impedance calculations, and with the line assumed to be perfectly transposed, the self and mutual impedances per phase are calculated as the average of the measured self and mutual impedances as follows: ZS 5

Z11 1 Z22 1 Z33 3

(3.166a)

ZM 5

Z12 1 Z21 1 Z13 1 Z31 1 Z23 1 Z32 6

(3.166b)

and

Therefore, the positive-sequence and zero-sequence impedances are given by ZP 5

2ðZ11 1 Z22 1 Z33 Þ 2 ðZ12 1 Z21 1 Z13 1 Z31 1 Z23 1 Z32 Þ 6

(3.167a)

ZZ 5

ðZ11 1 Z22 1 Z33 Þ 1 ðZ12 1 Z21 1 Z13 1 Z31 1 Z23 1 Z32 Þ 3

(3.167b)

234

Power Systems Modelling and Fault Analysis

Fig. 3.43B illustrates a technique for phase susceptance measurements where the three-phase conductors are earthed through milliammeters at the near measurement end of the line but kept free from earth, that is, open-circuited at the far end. A test voltage is applied in series with a one-phase conductor and the currents flowing through all conductors are measured. The process is repeated for all other conductors. As in the case of impedance measurements, the measured susceptances include the effect of the earth wire. Using the single-circuit phase susceptance matrix that includes the effect of earth wire that is with the earth wire mathematically eliminated, we have 2

3 2 I1 B11 6 7 6 4 I2 5 5 j4 2B21 I3

2B31

2 B12 B22 2B32

32 3 V1 2B13 76 7 2B23 54 V2 5 B33

(3.168)

V3

With the test voltage source applied to conductor 1, V1 is known and the measured values are I1, I2 and I3 with V2 5 V3 5 0. Using the measured values in Eq. (3.168), we obtain jB11 5

I1 V1

jB21 5

2 I2 V1

jB31 5

2 I3 V1

(3.169a)

Applying the test voltage source to conductor 2 then conductor 3, and using the measured values in Eq. (3.168), we obtain jB12 5

2 I1 V2

jB22 5

I2 V2

jB32 5

2 I3 V2

(3.169b)

jB13 5

2 I1 V3

jB23 5

I2 V3

jB33 5

2 I3 V3

(3.169c)

and

For sequence susceptance calculations and with the line assumed to be perfectly transposed, the self and mutual susceptances per phase are calculated as the average of the measured self and mutual susceptances as follows: BS 5

B11 1 B22 1 B33 3

(3.170a)

BM 5

B12 1 B21 1 B13 1 B31 1 B23 1 B32 6

(3.170b)

and

Modelling of multiconductor overhead power lines, underground and submarine cables

235

Therefore, the positive-sequence and zero-sequence susceptances are given by BP 5

2ðB11 1 B22 1 B33 Þ 1 ðB12 1 B21 1 B13 1 B31 1 B23 1 B32 Þ 6

(3.171a)

BZ 5

ðB11 1 B22 1 B33 Þ 2 ðB12 1 B21 1 B13 1 B31 1 B23 1 B32 Þ 3

(3.171b)

Errors between model lumped parameters and measured distributed parameters The measured voltages and currents inherently take into account the distributed nature of the line’s impedances and susceptances. However, we have used lumped linear impedance and susceptance matrix models for the line which will introduce small errors in the calculated impedances and susceptances depending on line length. The calculated impedances and susceptances can be corrected for the line length effect if the equivalent π circuit of a distributed multiconductor line derived in Appendix A.1 is used. The impedance and susceptance constants now become impedance and susceptance matrices and the general nodal admittance equations for a distributed multiconductor line are given by IS 5 A Vs
B Vr

(3.172a)

IR 5
B Vs 1 A Vr

(3.172b)

where A5Y

pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 21 pffiffiffiffiffiffiffi 21 ZY coth ZYL andB 5 Y ZY cosech ZYL

(3.172c)

and Is and Ir are the line’s sending and receiving end current vectors, respectively; and Vs and Vr are the line’s sending and receiving end voltage vectors, respectively. In the impedance measurement technique, the voltage and current vectors Vs and Is are known from the measurements, Vr 5 0 but Ir is unknown. Therefore, from pffiffiffiffiffiffiffi  pffiffiffiffiffiffiffi  Eqs. (3.172a
3.172c), we can write VS 5 A21 IS 5 Y21 YZtanh YZL IS . Expanding the hyperbolic tangent into an infinite series, we obtain   1 2 VS 5 Z 2 ZYZL2 1 ZYZYZL4 2 . . . LIS 3 15

(3.173)

In the susceptance measurement technique, the voltage and current vectors Vs and Is are known from the measurements, Ir 5 0 but Vr is unknown. Therefore,

236

Power Systems Modelling and Fault Analysis

from Eqs. (3.172a
3.172c), we can write IS 5 ðA 2 BA21 BÞVS . Using the values of A and B from Eqs. (3.172a
3.172c) and expanding the hyperbolic tangent into an infinite series, we obtain   1 2 IS 5 Y 2 YZYL2 1 YZYZYL4 2 . . . LVS 3 15

(3.174)

We can use the voltages and currents obtained from the test measurements to calculate to a first approximation the elements of the impedance and susceptance matrices using the approximate lumped parameter equations as follows: VS  Z1 LIS and IS  Y1 LVS

(3.175)

Z1 and Y1 can now be equated to the series impedance and susceptance values of Eqs. (3.173) and (3.174) and used to obtain a second approximation as follows: Z2  Z1 1

1 2 Z1 Y1 Z1 L2 2 Z1 Y1 Z1 Y1 Z1 L4 1 . . . 3 15

(3.176a)

Y2  Y1 1

1 2 Y1 Z1 Y 1 L2 2 Y1 Z1 Y1 Z1 Y1 L4 1 . . . 3 15

(3.176b)

Similarly, the new values of the impedance and susceptance matrices can again be used to obtain an improved approximation and the process can be repeated until the desired level of accuracy is obtained. In general, we can write the following recursive formulae ZN11  ZN 1

1 2 ZN YN ZN L2 2 ZN YN ZN YN ZN L4 1 . . . 3 15

(3.177a)

YN11  YN 1

1 2 YN ZN YN L2 2 YN ZN YN ZN YN L4 1 . . . 3 15

(3.177b)

Double-circuit overhead lines The technique presented above for a single-circuit overhead line with one earth wire can be extended to double-circuit overhead lines with any number of earth wires. Fig. 3.44 shows the impedance and susceptance test arrangements. The measured phase impedance and susceptance matrices are given as

Modelling of multiconductor overhead power lines, underground and submarine cables

237

(A) Short circuited

Earth wire

V

1

4

and

Measuring 2

end

earthed end

5

3

6

(B) Measuring V

end

1

Open circuited

4

end

2 5 3 6

Figure 3.44 Measurement of self and mutual electrical parameters of a double-circuit overhead line: (A) series phase impedances with earth return and (B) shunt phase susceptances.

3 2 Z11 V1 6 V 7 Cct 1 6 Z 6 27 6 21 6 7 6 6 V3 7 6 Z31 6 75 6 6V 7 6Z 6 47 6 41 6 7 6 4 V5 5 Cct 2 4 Z51 2

V6

Z61

Circuit 1 Z12 Z13 Z22 Z23 Z32 Z42

Z33 Z43

Z52

Z53

Circuit 2 32 3 I1 Z14 Z15 Z16 7 6 Z24 Z25 Z26 76 I2 7 7 76 7 7 6 Z34 Z35 Z36 I 7 76 3 7 7 6 Z44 Z45 Z46 76 I4 7 7 76 7 Z54 Z55 Z56 54 I5 5

Z62

Z63

Z64

Z65

Z66

(3.178a)

I6

and 2

I1

3

2

jB11

6 I 7 Cct 1 6 2 jB 21 6 27 6 6 7 6 6 I3 7 6 2 jB31 6 75 6 6I 7 6 2 jB 41 6 47 6 6 7 6 4 I5 5 Cct 2 4 2 jB51 I6

2 jB61

Circuit 1 2 jB12

2 jB13

2 jB14

jB22 2 jB32

2 jB23 jB33

2 jB24 2 jB34

2 jB42 2 jB52

2 jB43 2 jB53

jB44 2 jB54

2 jB62

2 jB63

2 jB64

Circuit 2 3 2 jB15 2 jB16 2 jB25 2 jB26 7 7 7 2 jB35 2 jB36 7 7 2 jB45 2 jB46 7 7 7 jB55 2 jB56 5 2 jB65

jB66

2

V1

3

6V 7 6 27 6 7 6 V3 7 6 7 6V 7 6 47 6 7 4 V5 5

V6 (3.178b)

238

Power Systems Modelling and Fault Analysis

In Fig. 3.44A, we apply a voltage source V1 to conductor 1 and measure I1, V2, V3, V4, V5 and V6. In Fig. 3.44B, we apply a voltage source V1 to conductor 1 and measure I1, I2, I3, I4, I5 and I6. From these tests, and using Eqs. (3.178a)and (3.178b), the self and mutual phase impedances and susceptances are calculated as follows: Z11 5

V1 I1

Z21 5

V2 I1

Z31 5

V3 I1

Z41 5

V4 I1

Z51 5

V5 I1

Z61 5

V6 I1 (3.179a)

jB11 5

I1 V1

jB21 5

2 I2 V1

jB31 5

2 I3 V1

jB41 5

2 I4 V1

jB51 5

2 I5 V1

jB61 5

2 I6 V1

(3.179b) In both impedance and susceptance tests, the voltage source is applied to each phase conductor in turn in order to measure all the self and mutual impedances and susceptances of Eqs. (3.178a)and (3.178b). The balanced phase and resultant sequence impedances can be calculated assuming an appropriate number of phase transpositions such as three, six or ideal nine transpositions. The reader is encouraged to derive the balanced phase and sequence parameters. The sequence susceptance parameters can be similarly calculated.

3.8.3 Measurement of cable parameters General Measurements of power frequency electrical parameters or constants of cables can be made at the time of commissioning when the cable installation has been completed so that the tests replicate the final cable layout of normal service operation. The relevant parameters that are usually measured are the conductor dc resistances at prevailing ambient temperature, the positive-sequence/negative-sequence and zero-sequence impedances and sometimes the positive-sequence/negative-sequence/ zero-sequence susceptances. As there are numerous designs and layouts of cables in use in practical installations, we will illustrate the techniques that can be used for a typical single-circuit single-core three-phase coaxial cable with core and sheath conductors. An illustration of the positive-sequence/negative-sequence susceptance and zero-sequence susceptance measurement of a three-core belted cable is also presented.

Measurement of cable dc resistances The dc resistances of the cores and sheaths (screens) are measured using a dc voltage source as illustrated in Fig. 3.45A and B. In the first case, the dc voltage source V is applied between cores 1 and 2 with the third core floating and remote ends of

Modelling of multiconductor overhead power lines, underground and submarine cables

(A)

I

Core

Sheath

Single-core single circuit 3-phase cable

1

239

V 2

3

(B) I 1 V 2

Single-core single circuit 3-phase cable

Core Sheath

3

Figure 3.45 Measurement of a three-phase cable dc resistances: (A) core and (B) sheath/screen.

the three cores short-circuited. The dc voltage source V is next applied between cores 1 and 3, then 2 and 3. The measured core loop resistance in each case is calculated from RCxCy 5 V/I and the three measured loop resistances are given by RC1C2 5 RC1 1 RC2

RC1C3 5 RC1 1 RC3

RC2C3 5 RC2 1 RC3

(3.180a)

Therefore, the individual core resistances are given by RC1C2 1 RC1C3 2 RC2C3 2 RC1C3 1 RC2C3 2 RC1C2 RC3 5 2

RC1 5

RC2 5

RC1C2 1 RC2C3 2 RC1C3 2

(3.180b)

The test procedure for the sheath or screen dc resistances is similar to that for the core but the dc voltage source is now applied between two sheaths at a time as shown in Fig. 3.45B.

Measurement of cable positive-sequence/negative-sequence impedance The measurement of the cable’s positive-sequence/negative-sequence impedance requires a three-phase power frequency source. On each of the single cores,

240

Power Systems Modelling and Fault Analysis

IP

Core Single-core single circuit 3-phase cable

VP

h 2V P

hV P

(B)

VZ

I

Sheath

Core

IZ 3

IZ 3

Z

IZ 3

Single-core single circuit 3-phase cable

(A)

Sheath

Figure 3.46 Measurement of sequence impedances of a three-phase cable: (A) positivesequence/negative-sequence impedance and (B) zero-sequence impedance.

measurements of voltage, current and phase power or phase angle are made. The sheath’s bonding and earthing must correspond to that in actual circuit operation. Fig. 3.46A illustrates the positive-sequence/negative-sequence impedance measurement for a cable with a sheath that is solidly bonded at both ends. From the measurements, the following calculations for each phase are made. If a wattmeter is used to measure input power, we have ZP 5

VP IP

RP 5

PP ðI P Þ2

XP 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðZ P Þ2 2 ðRP Þ2

(3.181)

If a phase angle meter is used to measure the angle between the input voltage and current ZP 5

VP IP

RP 5 Z P 3 cosφ

X P 5 Z P 3 sinφ

(3.182)

where PP is the measured phase power in Watts. The three values obtained for each phase are then averaged and this is particularly important in the case of untransposed cable systems. The final positive-sequence/negative-sequence cable parameters are given as ZP 5

3 1X ZP Ω 3 i51 i

RP 5

3 1X RP Ω 3 i51 i

XP 5

3 1X XP Ω 3 i51 i

(3.183)

Modelling of multiconductor overhead power lines, underground and submarine cables

241

Measurement of cable zero-sequence impedance The measurement of the cable’s zero-sequence impedance requires a single-phase power frequency source. The three cores at the test end are bonded together, and at the remote end are bonded together and to the sheaths. Single-phase currents are injected into the cable cores to return via the sheath and earth. Fig. 3.46B illustrates the test circuit. The sheaths are solidly bonded at both ends as in actual cable operation. The calculations of the zero-sequence impedance, resistance and reactance use similar equations to those used for the positive-sequence/negative-sequence impedance except that the current should now be divided by 3, that is, the impedance equation becomes ZZ 5

3V Z IZ

(3.184)

Measurement of earth loop impedances As in the case of an overhead line, and if required, the earth loop impedances for each core and sheath may also be measured using a single-phase power frequency source. The technique is similar to that presented in Section 3.8.2.

Measurement of positive-/negative-/zero-sequence susceptance of single-core cable The measurement of a screened cable core-to-sheath, and if required, sheath-toearth susceptances uses a single-phase power frequency source. The three cores at the remote end are left floating, that is, open circuit. The applied voltage V, charging current I and phase angle φ between applied voltage and current, the latter measured using a low power factor meter, are measured. This is usually called the power factor and capacitance test of the dielectric. The calculations made are Y5

I V

G 5 Y 3 cosφ

B 5 Y 3 sinφ

(3.185)

B is the positive-sequence/negative-sequence/zero-sequence susceptance per phase. G is very small and is usually neglected in power frequency steady-state analysis.

Measurement of positive-/negative-/zero-sequence susceptance of a three-core belted cable For three-core belted cables, the core-to-core and core-to-sheath capacitances can be calculated from two measurements. The core to sheath, that is, zero-sequence susceptance can be measured by short-circuiting the three cores and applying a

242

Power Systems Modelling and Fault Analysis

voltage source V between them and the earthed sheath, see Fig. 3.31C. From the measured injected current I, we can write BCS 5

I 3V

(3.186)

The core-to-core capacitance can be obtained from a measurement where a voltage source V is applied between two cores with the third core short-circuited to the sheath. Let the measured susceptance be BMeas. Therefore, using Fig. 3.31C, we can write BMeas 5

I BCC 1 BCS 3BCC 1 BCS 5 BCC 1 5 V 2 2

Therefore, the core-to-core susceptance is calculated as BCC 5

2BMeas 2 BCS 3

(3.187)

Using Eq. (3.138c), the positive-sequence and zero-sequence susceptances of the cable are given by BP 5 2BMeas

3.9

and BZ 5 BCS

(3.188)

Practical aspects of phase and sequence parameters of overhead lines and cables

3.9.1 Overhead lines During the installation and commissioning of new overhead line designs, it was common practice to carry out field test measurements of the phase self and mutual impedances and susceptances. From these tests, the sequence impedances and susceptances are obtained. In addition, measurements of sequence parameters could also be directly made. Measurements served as benchmark values for the validation of the line’s power frequency models and calculations carried out on digital computers. The range of errors in the calculations of the sequence impedance parameters is usually very small. The assumption of conductor temperature used in the calculation is important primarily for resistance calculation and to a small extent reactance and susceptance calculations due to changes in conductor sag. Also, a reasonable value of uniform earth resistivity may be used and this may be an average value to represent the route over which the line runs. In England and Wales, and in America, average uniform earth resistivities of 20 and 100 Ωm are generally used.

Modelling of multiconductor overhead power lines, underground and submarine cables

243

3.9.2 Cables For cables, available models and computer calculations of the positive-sequence/ negative-sequence impedances and sequence susceptances usually provide results with good accuracy in comparison with field test measurements despite the large variety of cable designs and installations. However, this is not normally the case for the zero-sequence impedance, particularly for high-voltage cables. In addition to the cable design including its sheath or screen and cable layout, the zero-sequence impedance strongly depends on any other local and nearby metal objects that may form a return path in parallel with the cable sheath itself. These might be other parallel cables, nearby water pipes and pipelines, gas mains, subway structures, railway tracks and lines, steel or concrete tunnels and any earth electrodes that may be used, etc. Information about most of these additional factors is practically unknown and impossible to obtain so they cannot be appropriately included in the calculation. The only practical solution to obtain accurate results is to carry out field test measurement of the zero-sequence impedance with the cable installed as in actual service operation. In addition, such a measurement will only be accurate at the time it is taken but it may be different a few years later. This may be caused by underground construction or modifications of conducting material, for example, water and gas mains.

Further reading Books Abramowitz, M., Stegun, I., et al., Handbook of Mathematical Functions, Dover publications, Inc, 1964. Library of Congress catalog card number: 65-122253. Bickford, J.P., et al., Computation of Power-System Transients, Peter Peregrinus Ltd, 1980. ISBN 0-906048-35-4. IEC 60287-1-1, Electric Cables
Calculation of Current Rating, Edn 2.1 2014-11. McAllister, D., Electric Cables Handbook, Granada, 1982, ISBN 0-246-11467-3.

Papers Carson, J.R., Wave propagation in overhead wires with ground return, Bell System Technical Journal, Vol. 5, 1926, 539
554. Pollaczek, F., Ueber das feld einer unendlich langen wechsel stromdurchflossenen einfachleitung, Elektrische Nachrichlen Technik, Vol. 3, No. 5, 1926, 339
359. Schelkunoff, S.A., The electromagnetic theory of coaxial transmission lines and cylindrical shells, Bell System Technical Journal, Vol. XIII, 1934, 532
578. Haberland, G., Theorie der Leitung von Wechselstrom durch die Erde, Zeitschrifi fir Angewand te Mathematic and Mechanik, Vol. 5, No. 6, 1926, 366. Wang, Y.J., A review of methods for calculation of frequency-dependent impedance of overhead power transmission lines, Proceedings of the National Science Council Republic of China(A), Vol. 25, No. 6, 2001, 329
338.

244

Power Systems Modelling and Fault Analysis

Galloway, R.H., et al., Calculation of electrical parameters for short and long polyphase transmission lines, Proceedings IEE, Vol. 111, December 1964, 2051
2059. Dommel, H.W., Overhead line parameters from handbook formulas and computer programs, IEEE Transactions on PAS, Vol. PAS-104, No. 2, February 1985, 366
372. Nielsen, H., et al., Underground Cables and Overhead Lines Earth Return Path Impedance Calculations with Reference to Single Line to Ground Faults, Aalborg University, Denmark, 20041
11. Ametani, A., A general formulation of impedance and admittance of cables, IEEE Transactions on PAS, Vol. PAS-99, No. 3, May/June 1980, 902
910. Brown, G.W., et al., Surge propagation in three phase pipe-type cables, Part 1_unsaturated pipe, IEEE Transactions on PAS, Vol. PAS-95, No. 1, January/February 1976, 89
95. Bianchi, G., et al., Induced currents and losses in single-core submarine cables, IEEE Transactions on PAS, Vol. PAS-95, No. 1, January/February 1976, 49
58. Wedepohl, L.M., et al., Transient analysis of underground power-transmission systems, IEE Proceedings, Vol. 120, No. 2, February 1973, 253
260. Gustavsen, B., Panel session on data for modelling system transients insulated cables, SINTEF Energy Research, N-7465 Trondhein, Norway, 2001. Jensen, C.F., et al., Switching studies for the Horns Rev 2 wind farm main cable, Energinet, DK, 2009.

Modelling of transformers, phase shifters, static power plant and static load Chapter Outline 4.1 General 246 4.2 Sequence modelling of transformers

246

4.2.1 4.2.2 4.2.3 4.2.4 4.2.5 4.2.6 4.2.7 4.2.8 4.2.9

Background 246 Single-phase two-winding transformers 249 Three-phase two-winding transformers 262 Three-phase three-winding transformers 274 Three-phase autotransformers with and without tertiary windings 283 Three-phase four-winding transformers 299 Three-phase earthing or zigzag transformers 302 Single-phase traction transformers connected to three-phase systems 304 Variation of transformer’s positive-sequence leakage impedance with tap position 305 4.2.10 Practical aspects of zero-sequence impedances of three-phase transformers and effect of core construction 306 4.2.11 Measurement of sequence impedances of three-phase transformers 311 4.2.12 Examples 317

4.3 Sequence modelling of quadrature booster and phase-shifting transformers 325 4.3.1 4.3.2 4.3.3 4.3.4

Background 325 Positive-, negative- and zero-phase sequence modelling of QB and PS transformers 327 Measurement of sequence impedances of QB and PS transformer 332 Examples 334

4.4 Sequence modelling of series and shunt reactors and capacitors 4.4.1 4.4.2 4.4.3 4.4.4

337

Background 337 Modelling of series reactors 337 Modelling of shunt reactors and capacitors 340 Modelling of series capacitors 343

4.5 Sequence modelling of static variable compensators

349

4.5.1 Background 349 4.5.2 Positive-, negative- and zero-phase sequence modelling 350

4.6 Sequence modelling of static power system load 350 4.6.1 Background 350 4.6.2 Positive-, negative- and zero-phase sequence modelling 352

4.7 Three-phase modelling of static power plant and load in the phase frame of reference 352

Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00004-7 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

4

246

Power Systems Modelling and Fault Analysis

4.7.1 4.7.2 4.7.3 4.7.4 4.7.5

Background 352 Three-phase modelling of reactors and capacitors 352 Three-phase modelling of transformers 353 Three-phase modelling of QB and PS transformers 366 Three-phase modelling of static load 368

Further Reading

4.1

368

General

In this chapter, the modelling and analysis of various types of power plant and apparatus are presented. The correct modelling, calculation and measurements of parameters of such equipment are crucial for power system analysis and protection studies. Equipment covered includes: various types of transformers, quadrature boosters (QBs), phase shifters (PSs), series and shunt reactors, series and shunt capacitors, static variable compensators and static power system load. Positivephase sequence, negative-phase sequence and zero-phase sequence models, as well as three-phase models suitable for analysis in the phase frame reference, are presented. Sequence impedance measurement techniques of transformers, quadrature boosters and phase shifters are also described.

4.2

Sequence modelling of transformers

4.2.1 Background The invention of the power transformer towards the end of the 19th century having the basic function of stepping up or down the voltages (and currents but in opposite ratios) created the advantage of transmitting alternating current (ac) electric power over longer distances. This was not possible with direct current (dc) electric power and hence the locations of power generation plant and demand centres could not be a long distance apart. This is to ensure that power loss and voltage drops are kept to a reasonably low levels. The technology of power transformers has undergone significant developments during the 20th century and many types of transformers are nowadays used in various types and sizes in power systems. Three banks of single-phase transformers are generally used in power stations to connect very large generators to transmission networks with the low-voltage (LV) and high-voltage (HV) windings connected as required, for example, in delta or star. The main reasons for using separate banks are to reduce transport weight, improve overall transformer availability and reliability, and reduce the cost of spare holding. Single-phase transformers are also used in remote rural distribution networks to supply small single-phase demand consumers where three-phase supplies are uneconomical to use. There are also special applications where single-phase

Modelling of transformers, phase shifters, static power plant and static load

247

transformers are used, such as in the case of high-capacity traction transformers connected to three-phase power networks, usually connected to two phases of the three-phase transmission or distribution network. Other applications of single-phase transformers are those used to provide neutral earthing for generators. Three-phase power transformers are invariably used in transmission, subtransmission and distribution substations for essentially voltage transformation. In addition, they are also used for voltage or reactive power flow control and active power flow control. For the latter application, they are called phase shifters (PS) or quadrature booster (QB) transformers. Other applications of three-phase power transformers are for connections to converters in high-voltage direct-current (HVDC) links and inverters in wind turbine generators and solar photovoltaic generators. Other applications are providing supply to arc furnaces and artificial earthing of isolated power systems. Transformers used for voltage control are equipped with off-load or on-load tapchangers that vary the number of turns on the associated winding and hence the turns ratio of the transformer. In some countries, for example, the United Kingdom, generator transformers are generally equipped with on-load tap-changers on the HV winding with the HV side voltage being the controlled voltage. Typical voltage transformations are 11
142, 15
300 and 23.5
432 kV. Two-winding transformers having large transformation ratios may be equipped with on-load tap-changers on the HV winding but with the LV winding voltage being the controlled voltage. Typical transformations are 275
33, 132
11, 380
110 and 110
20 kV. Distribution transformers, for example, 11
0.433 or 6.6
0.433 kV usually have off-load tap-changers. Where the transformation ratio is generally less than about three, for example, 400/132 and 500/230 kV, autotransformers are generally more economical to use. Nowadays with increased trends towards automation, autotransformers that supply power to subtransmission systems tend to be equipped with on-load tap-changers. These may be located either at the HV winding line-end, or LV winding line-end or neutral-end, with the LV winding voltage being the controlled voltage. Autotransformers may also have a third delta-connected winding called the ‘tertiary’ winding with a typical MVA capacity around 25% of the main transformer MVA capacity. These may be used for the connection of reactive compensation plant, for example, capacitors, reactors or synchronous compensators (condensers) or simply left unloaded. They also provide a path for the circulation of triplen harmonic currents and hence prevent or reduce their flow on the power network as well as reduce network voltage unbalance. Figs. 4.1 and 4.2 show two large network autotransformers used in 400/132 and 400/275 kV transmission networks. The calculation of unbalanced currents and voltages due, for example, to shortcircuit faults, requires correct and practical modelling of power transformers using factory-measured data. In this section, we present the theory and models of singlephase transformers and three-phase power transformers having various numbers of windings and different types of winding connections for use in the analysis of power networks. Knowledge of basic electromagnetic transformer theory is assumed.

248

Power Systems Modelling and Fault Analysis

Figure 4.1 240-MVA 400-kV/132-kV autotransformer.

Figure 4.2 1000-MVA 400-kV/275-kV autotransformer.

Modelling of transformers, phase shifters, static power plant and static load

249

4.2.2 Single-phase two-winding transformers Transformer core construction and general equivalent circuit The power transformer is a complex static electromagnetic apparatus with windings and a nonlinear magnetic steel core. Fig. 4.3 illustrates practical core constructions of core-form and shell-form cores. Normally, the lower voltage winding is the inner winding. In the core form, half of each winding is placed around each limb. Fig. 4.3 illustrates the main linkage flux flowing in the core through the limbs and yokes, LV and HV windings leakage fluxes, the leakage flux between the windings, and a flux in the channel between the core limb and the innermost LV winding which tends to be overlooked in much of the transformer literature. Fig. 4.4A illustrates a general electrical circuit representation of the single-phase two-winding transformers shown in Fig. 4.3. The transformer is represented as two coupled windings around a magnetic steel core. Fig. 4.4B shows a low-frequency equivalent circuit of the transformer where all quantities are in physical units of volts, amps, ohms and Henries and VH and VL are the HV and LV terminal voltages and EH and EL are the induced winding voltages. NH and NL are the actual number of turns of HV and LV windings, respectively, and NH(nominal) and NL(nominal) are HV and LV windings number of turns at nominal tap positions, respectively. RH 5 HV winding resistance, LH 5 HV winding leakage inductance RL 5 LV winding resistance, LL 5 LV winding leakage inductance

LH and LL represent HV and LV winding leakage fluxes, respectively. RM 5 resistance representing core losses, LM 5 nonlinear magnetising inductance of core. Both RM and LM are referred to the LV side. The core includes all limbs and yokes. LAir is inductance representing the flux in the channel between the inner LV winding and core limb and represents an air component of the magnetisation branch as shown in Fig. 4.3. LAir is not a leakage inductance because this flux channel is enclosed by both LV and HV windings. Based on typical distances from the centre of the core limb to the LV winding, and to the HV winding, this inductance can be taken, to a good approximation, equal to half the leakage inductance between the HV and LV  1
windings, that is, LAir 5 LL 1 L0H , where L0H is equal to LH referred to the 2 LV side. It is noted that the magnetising branch is connected to the innermost winding, that is, the winding closest to the core which, in the case of our two-winding transformer, is normally the LV winding. The traditional connection of the magnetisation branch in the middle between LH and LL is generally adequate provided no core saturation occurs. However, in studies where core saturation occurs, such as under high quasi-steady state power frequency voltages that may occur under load rejection, energisation inrush currents and other low-frequency studies, this traditional connection becomes inaccurate. The connection shown in Fig. 4.4 is an accurate representation in such studies.

250

Power Systems Modelling and Fault Analysis

(A) Core-form Limb

LV winding HV winding

Yoke

Linkage flux

Flux between core limb and LV winding

Leakage flux between LV and HV windings

(B) Shell-form

(C) Core-form Limb

Yoke

Primary winding leakage flux

Flux between core limb and LV winding Secondary winding leakage flux

LV winding HV winding

Linkage flux

(D) Shell-form

Figure 4.3 Transformer leakage flux representation: integral flux (A) core-form and (B) shell-form; divided flux (C) core-form and (D) shell-form.

Modelling of transformers, phase shifters, static power plant and static load

(A)

NH : NL

IH

IL

H EH

VH

251

L

VL

EL

Magnetic core

(B)

IH

RH

LH

H

EH

VH

RL

LL

NH : NL

EL

IL IM L Air

RM

Ideal transformer L

=

1 2

L +

L VL

LM

N L N

=

1 L 2

(C)

ZL

ZH

IH

NH : NL

H RH

XL

XH

VH

EH

EL

IL RM X Air

XL

Ideal transformer X

RL

=

1 2

X +

L VL

XM

N X N

=

1 X 2

Figure 4.4 Single-phase transformer equivalent circuit in actual physical units: (A) representation of two iron-core mutually coupled windings, (B) basic divided leakage flux equivalent circuit and (C) as (B) but with reactances instead of inductances.

In the next sections, we first present the transformer equivalent circuit in actual physical units of volts, amps and ohms, then derive the per-unit equivalent circuit for the cases where the transformer has nominal and off-nominal turns ratios.

Transformer equivalent circuit in actual physical units Fig. 4.4C shows the power frequency equivalent circuit derived from Fig. 4.4B where the inductances are replaced by reactances using X 5 2πfL where f is the power frequency, that is, 50 or 60 Hz. From basic transformer theory and since the same core flux will link both HV and LV windings, the induced voltage per turn on EH EL 5 or the HV and LV windings are equal, that is, NH NL EL NL 1 5 5 5 NLH NHL EH NH

(4.1a)

252

Power Systems Modelling and Fault Analysis

For a core flux Ψ and finite core magnetic reluctance R, the ampere-turns or MMF balance is ΨR 5 NH IH 1 NL IL

(4.1b)

Now, the induced LV winding voltage is eL 5 NL LM 5

NL2 and NL IM 5 NL IL 1 NH IH R

dΨ diM 5 LM , we can wite dt dt (4.1c)

If the magnetising current IM is neglected, the HV and LV winding MMFs are related by NH IH 1 NL IL 5 0 or

IH NL 52 IL NH

(4.1d)

Also, the voltage drop equations for the HV and LV windings can be written as follows: V H 2 E H 5 ZH I H

(4.2a)

VL 2 EL 5 ZL IL

(4.2b)

Using Eqs. (4.1a)
(4.1d), (4.2a) and (4.2b), and after a little algebra, we obtain VL 2 ZLH IL 5 ZLH 5 ZL 1

VH NHL

ZH 2 NHL

(4.3a)

(4.3b)

ZLH is the equivalent transformer leakage impedance referred to the LV side and ZH is the HV winding leakage impedance referred to the LV side. Fig. 4.5A is 2 NHL drawn using Eq. (4.3a) and (4.3b) with the magnetising branch reinstated. The HV voltage and current VH and IH are referred to the LV side as VH/NHL and NHLIH, respectively. Fig. 4.5B shows an alternative equivalent circuit where the transformer leakage impedance is referred to the HV side. Generally, when the transformer core is not saturated, the magnetising impedance is much larger than the leakage impedance, typically by a factor of 400
1. Nevertheless, it is good practice to include the magnetising impedance branch in large-scale network models used in short-circuit, power flow and transient stability analyses.

Modelling of transformers, phase shifters, static power plant and static load (A)

I

N

H

N

:1

Z

I

253

I

L

I jX

V

V

V /N R Ideal transformer Z

= Z + Z /N

N

=

N N

jX

X

= X

/2

I

L

(B) Z

I

H

1: N I jX

V

V /N

V

R

Z

= Z + Z /N

Ideal transformer N X N = N

jX

= N

X

/2

Figure 4.5 Single-phase transformer equivalent circuit in actual physical units: (A) total leakage impedance referred to L side, (B) total leakage impedance referred to H side.

Transformer equivalent circuit in per unit In many types of steady-state analysis, we are interested in the transformer equivalent circuit where all quantities are in per unit on some defined base quantities. Let the base quantities be defined as SðBÞ 5 Transformer rating in VA VHðBÞ 5 HV network base voltage IHðBÞ 5 HV network base current VHðBÞ 5 ZHðBÞ IHðBÞ VHðBÞ IHðBÞ 5 VLðBÞ ILðBÞ 5 SðBÞ

VLðBÞ 5 LV network base voltage ILðBÞ 5 LV network base current VLðBÞ 5 ZLðBÞ ILðBÞ

VHðBÞ NHðnominal 5 V L ðB Þ NLðnominal

tap positionÞ tap positionÞ

(4.4a)

(4.4b)

In Eq. (4.4b), the HV and LV windings’ nominal tap positions are chosen to correspond to the HV and LV base voltages of the network sections to which the windings are connected. Let us now define the following per-unit quantities: ZHðpuÞ 5

ZH ZHðBÞ

ZLðpuÞ 5

ZL ZLðBÞ

(4.5a)

VHðpuÞ 5

VH V H ðB Þ

VLðpuÞ 5

VL V L ðB Þ

(4.5b)

IHðpuÞ 5

IH IHðBÞ

ILðpuÞ 5

IL I L ðB Þ

(4.5c)

254

Power Systems Modelling and Fault Analysis

Dividing Eq. (4.3a) by S(B) and using Eqs. (4.4a), (4.4b) and (4.5a)
(4.5c), we obtain 

1 NL NLðnominalÞ

 VL ZL IL 2 5 V L ðB Þ ZLðBÞ ILðBÞ

1 NH NHðnominalÞ



VH ZH IH 2 V H ðB Þ ZHðBÞ IHðBÞ



which can be rewritten as  1  1 VLðpuÞ 2 ZLðpuÞ ILðpuÞ 5 VHðpuÞ 2 ZHðpuÞ IHðpuÞ tL tH

(4.6a)

where the pu tap ratios of the LV winding tL and HV winding tH are given by NHðat given HV tap positionÞ NH 5 NHðnominalÞ NHðat nominal HV tap positionÞ Rated voltage of HV winding at given tap position 5 Rated voltage of HV winding at nominal tap position

(4.6b)

NLðat given LV tap positionÞ NLðnominalÞ NLðat nominal LV tap positionÞ Rated voltage of LV winding at given tap position 5 Rated voltage of LV winding at nominal tap position

(4.6c)

tH 5

NL

tL 5

5

Eq. (4.6a) is shown in Fig. 4.6A with, again, the magnetising branch reinstated and its parameters are also calculated in pu and referred to the LV side base. However, the individual leakage reactances of each winding cannot be measured in practice because of the presence of the magnetic core. Therefore, an equivalent circuit expressed in terms of the practical and measured leakage impedance between the two windings is required. This can be done once we have derived the relationship between the HV and LV pu currents from Eq. (4.1b) or NL IL 5 2 NH IH which, using Eqs. (4.4b), (4.6b) and (4.6c), can be written as tL ILðpuÞ 5 2 tH IHðpuÞ

(4.7a)

IHðpuÞ 5 2 ILðpuÞ =tHL

(4.7b)

and tHL 5

tH tL

tLH 5

1 tHL

(4.7c)

Now, substituting Eq. (4.7a)
(4.7c) in Eq. (4.6a), we obtain VHðpuÞ 5 tHL ½VLðpuÞ 2 ZLHðpuÞ ILðpuÞ 

(4.8a)

Modelling of transformers, phase shifters, static power plant and static load

(A)

I

(

H V

(

Z

)

(

Z

t :t

)

(

255

I

)

I ( jX

(

)

L

) (

)

)

V

R

(

)

jX

(

(

)

)

Ideal transformer

(B)

I

(

Z

:1

t

)

(

I

)

H V

(

I ( jX

t t

=

Z

(C)

I

(

(

Ideal transformer ( ) = Z ( ) + Z

Z

)

(

)

X

) /t

(

=

jX

(

)

(

)

= X

I

(

(

Ideal transformer Z ( )= Z (

t t

I

(

)

Z

(

= Z

)

)

+ Z

(

)

)

I I ( jX

(

)

(

Z

(

X

) (

= Z )

(

= X

) (

= Z

(

(

X

= t

)

L

)

) /2 = X

)

+ Z

(

(

jX (

)=

(

)

)

X

(pu )/2

) )

V (

) /2

)

jX

)

R

)

V (

) /t

(

(

L

)

)

H (

)

I ( jX

(D)

V

(

1: t

)

R

t

L V

(

H V

)

)

R

t

(

)

Z

(

(

)

)

(

,

)

) /2

Figure 4.6 Transformer equivalent circuits in pu: (A) general off-nominal-ratio transformer; (B) H side off-nominal ratio; (C) L side off-nominal ratio and (D) nominal-ratio transformer.

where 2 ZHðpuÞ 5 ZLðpuÞ 1 ZLHðpuÞ 5 ZLðpuÞ 1 tLH

ZHðpuÞ 2 tHL

(4.8b)

where ZLHðpuÞ is the transformer leakage impedance referred to the LV side. The equivalent circuit that corresponds to Eq. (4.8a) and (4.8b) is shown in Fig. 4.6B with the magnetisation branch reinstated. Fig. 4.6C shows an alternative equivalent circuit where the transformer leakage impedance is referred to the HV side.

256

Power Systems Modelling and Fault Analysis

Under nominal tap positions, tH 5 tL 5 1 and the equivalent circuit reduces to that shown in Fig. 4.6D.

Transformer equivalent circuit in per unit based on nominal impedance We will discuss later that the leakage impedance varies with the transformer tap position. However, for now, we will discuss the situation where the only impedance data available is the impedance at nominal tap position. A practical approach can still be used and is found generally acceptable in practical applications where we assume that the impedance of each winding is proportional to the square of the number of turns of the winding. We know from basic transformer theory that this is generally valid for the winding inductive reactance but not for the winding resistance. However, because the resistance is generally much smaller than the reactance, this approximation is found acceptable in practice. We define ZH(pu,nominal) as the pu HV winding impedance at nominal HV winding tap position and ZL(pu,nominal) as the pu LV winding impedance at nominal LV winding tap position. Therefore, for each winding, we can write ZHðpuÞ 5 tH2 ZHðpu;nominalÞ

(4.9a)

ZLðpuÞ 5 tL2 ZLðpu;nominalÞ

(4.9b)

Substituting Eq. (4.9a) and (4.9b) into Eq. (4.6a), we obtain  1  1 VLðpuÞ 2 tL2 ZLðpu;nominalÞ ILðpuÞ 5 VHðpuÞ 2 tH2 ZHðpu;nominalÞ IHðpuÞ tL tH

(4.10)

Eq. (4.10) is represented in Fig. 4.7A. Substituting Eq. (4.7a) into Eq. (4.10) and rearranging, we obtain VHðpuÞ 2 tL2 ZLHðpu;nominalÞ 3 tHL IHðpuÞ 5 VLðpuÞ tHL

(4.11a)

VLðpuÞ 2 tH2 ZHLðpu;nominalÞ 3 tLH ILðpuÞ 5 VHðpuÞ tLH

(4.11b)

and for tH 5 tL 5 1, VLðpuÞ 2 ZHLðpu;nominalÞ ILðpuÞ 5 VHðpuÞ

(4.11b)

where ZHLðpu;nominalÞ 5 ZLHðpu;nominalÞ 5 ZHðpu;nominalÞ 1 ZLðpu;nominalÞ

(4.11c)

Fig. 4.7B
D represent Eq. (4.11a)
(4.11c) with the magnetisation branch reinstated.

Modelling of transformers, phase shifters, static power plant and static load

257

(A) t Z

(

,

t

)

H

t Z

t

(

,

I

)

I ( jX R

X

(B)

(

Z t

H

:1

t [

(

= t [X

)

,

(

(

)

(

) (

)

jX )+

,

X

(

) ]/2

,

=

(

)

)+

Z

(

)]

,

I I ( jX

VH(pu)

R

X

(C)

H

(

(

)

)

jX

= t X

)

(

)

(

)

) /2

,

I I ( jX

V L(pu)

VH(pu)

(

) (

t :t 1: t

Z HL(pu) = t 2H Z HL(pu,nominal)

I

)

t LH

) (

)

V L(pu)

R

(

)

= t X

(

jX

(

)

H X

(D)

(

)

) /2

,

ZHL(pu) = ZLH(pu) = ZHL(pu,nominal)

I

I

H

I ( jX

t

=t

= 1

X

(

)

L

) (

)

V L(pu)

R

(

)

= X

(

) /2

jX =X

(

(

)

) /2

Figure 4.7 Transformer equivalent circuits in pu based on nominal impedance: (A) general off-nominal-ratio transformer; (B) H side off-nominal ratio; (C) L side off-nominal ratio and (D) nominal-ratio transformer.

Interpreting transformer equivalent circuit and off-nominal tap ratio We will use Fig. 4.7B and C to provide a practical interpretation of tap changing of an off-nominal ratio transformer. However, in practice, it is more convenient to use one or the other depending on the actual transformer characteristics. That is,

258

Power Systems Modelling and Fault Analysis

Fig. 4.7B would be used to represent a transformer with the following characteristics: 1. The HV winding side is equipped with an on-load tap-changer and the LV winding side is fixed nominal or fixed off-nominal ratio. 2. The HV winding side is fixed off-nominal ratio and the LV winding side is fixed nominal ratio or fixed off-nominal ratio.

Fig. 4.7C, however, would be used to represent a transformer with the following characteristics: 1. The LV winding side is equipped with an on-load tap-changer and the HV winding side is fixed nominal or fixed off-nominal ratio. 2. The LV winding side is fixed off-nominal ratio and the HV winding side is fixed nominal ratio or fixed off-nominal ratio.

We will make a number of important practical observations that apply to Fig. 4.7B and C. Considering Fig. 4.7B, and in the case of a tap-changer located on the HV winding with a fixed nominal or off-nominal ratio on the LV winding, then only tH changes but ZLH does not change with changes in HV tap position. Similar observation applies to Fig. 4.7C. In both circuits, the variable tap ratio of the ideal transformer, tHL or tLH, is located on the side equipped with the on-load tapchanger. That is, it is located away from the transformer impedance which has a value that corresponds to the fixed nominal or off-nominal ratio of the other side. It should be noted that our chosen location of the variable tap ratio is purely arbitrary and it could have been placed on the impedance side provided that Eq. (4.11a) and (4.11b) continue to be satisfied. A nominal-ratio transformer is one where the chosen base voltages of the network sections to which the transformer is connected are the same as the transformer turns ratio or rated voltages. For such a transformer, Eq. (4.6b) and (4.6c) give tH 5 tL 5 1, hence the ideal transformers shown in Fig. 4.7B and C can be eliminated as shown in Fig. 4.7D. However, for many transformers used in power systems, the transformer voltage ratio is not the same as the ratio of base or rated voltages of the associated network sections. Therefore, the choice of voltage base quantities for the network has to be made independent of the transformer’s actual voltage or turns ratio. The most common reason for this is that a very large number of transformers used in power systems are equipped with tap-changers, on the HV or LV winding, whose function is to vary the number of turns of their associated winding. In addition, some distribution transformers, in particular, can have one winding with a fixed off-nominal ratio. To ensure a proper understanding of the concepts introduced in deriving Fig. 4.7, we now present a number of practical cases. In the first case, consider a single-phase 11/0.25 kV distribution transformer interconnecting an 11-kV network and a 0.25-kV network. The transformation ratio or turns ratio of 11/0.25 kV corresponds exactly to the network base voltages. Therefore, Fig. 4.7D represents such a transformer. We also note that with the transformer on open circuit, VH(pu) 5 VL(pu).

Modelling of transformers, phase shifters, static power plant and static load

259

In the second case, we consider a 10/0.25 kV single-phase transformer interconnecting an 11-kV network and a 0.25-kV network. The HV winding is clearly a fixed off-nominal-ratio winding. The pu tap ratio tHL in Fig. 4.7B is calculated as tHL 5

tH 10=11 10 5 5 0:9091 5 0:25=0:25 11 tL

In the third case, we consider a generator transformer equipped with an on-load tap-changer on the HV winding side. Consider a 239 MVA, 231/21.5 kV singlephase transformer (this is actually one of a 717 MVA, 400/21.5 kV three singlephase banks) having a nominal-ratio leakage reactance of 15% on rated MVA (the resistance is ignored for the purposes of this example but not in practical calculations). The transformer is connected to a network with 231 kV and 21.5 kV base voltages. The transformer, therefore, has a fixed LV nominal-ratio, that is, tL 5 1. The tap-changer is located on the 231-kV winding and has a variable tapping range, hence tH is variable. We will calculate the transformer reactance and redraw the equivalent circuit of Fig. 4.7B to correspond to minimum, nominal and maximum transformation ratios. The minimum, nominal and maximum tap ratios correspond to the transformation ratios of 207.9/21.5, 231/21.5 and 254.1/21.5 kV, respectively. The corresponding tH values are 207.9/231 5 0.9, 1.0 and 254.1/231 5 1.1. The equivalent circuit of Fig. 4.8A shows that the transformer reactance value does not change from

(A)

(B) 0.9:1

H

0.9:1 j15%

L

H

= t = 0.9; t = 1; t

t = 0.9; t = 1; t

= 0.9

1:1

= 0.9

1:1 j15%

H

L

j12.15%

L

H

L

j15%

= t = t = 1; t

t = t = 1; t

=1

1.1:1 H

=1

1.1:1

j15%

L

H

L

j18.15%

= t = 1.1; t = 1; t

= 1.1

t = 1.1; t = 1; t

= 1.1

Figure 4.8 Example of effect of off-nominal tap ratio on transformer impedance.

260

Power Systems Modelling and Fault Analysis

nominal of 15%. It is instructive for the reader to demonstrate that if the variable tap ratio is placed on the impedance side, then, by making appropriate use of Fig. 4.7C, the reactance values that correspond to tH 5 0.9, 1.0 and 1.1 become 12.15%, 15% and 18.15%, respectively. This is illustrated in Fig. 4.8B. The reader should also demonstrate that the corresponding circuits of Fig. 4.8A and B are equivalent to each other. This can be done by calculating, for the corresponding equivalent circuits, the short-circuit reactances, that is, the reactances seen from one side, for example, H side with the other side, L side, short-circuited and vice versa. The calculated short-circuit reactances will be identical. The variation of the tap position of a transformer equipped with an on-load tapchanger results in a value of tHL, that for most transformers used in electric power systems, is generally within 0.8 and 1.2 pu.

π equivalent circuit model Although suitable for use on network analysers, the ideal transformer in the general pu equivalent circuits of an off-nominal-ratio transformer shown in Fig. 4.6B, C, 4.7b and C is not satisfactory for modern calculations that are almost entirely based on digital computation. Therefore, the ideal transformer must be replaced with a suitable equivalent circuit and this will be achieved by deriving an equivalent π circuit model for Fig. 4.6B. The reader is encouraged to do so for the remaining figures. For convenience, we will drop the explicit pu notation and
recall that all VL 2 VH =tHL quantities are in pu. From Fig. 4.6B, we can write IL 5 1 IM , ZLH VL giving 2 tHL IH 5 IM 2 IL and IM 5 ZM   21 1 1 IL 5 VH 1 1 (4.12a) VL tHL ZLH ZLH ZM IH 5

1 1 VH 2 VL 2 Z t tHL HL ZLH LH

(4.12b)

where  ZM 5 j XAir 1

R M XM RM 1 jXM



or in matrix form 2

1  6 t2 Z IH 6 HL LH 56 21 IL 4 tHL ZLH

3 21 7  tHL ZLH 7 VH 1 1 7 5 VL 1 ZLH ZM

(4.13a)

Modelling of transformers, phase shifters, static power plant and static load

(A)

VH

(B)

H

YC

Z A = t HL Z LH

IH

YC

t HL − 1 ⎛ − 1 ⎞ ⎜ ⎟ t HL Z LH ⎜⎝ t HL ⎟⎠

YC =

VH

YB

YB =

t LH − 1 t LH Z HL

VL

IL L

YC

YC =

VL

t HL − 1 + Y Mag t HL Z LH

Z A = t LH Z HL

IH

L

IL L

YB

YB =

H

IL

YB

VH

(C)

ZA

IH

H

261

VL

t LH − 1 ⎛ − 1 ⎞ ⎜ ⎟ + Y Mag t LH Z HL ⎜⎝ t LH ⎠⎟

Figure 4.9 π equivalent circuit of an off-nominal-ratio transformer: (A) general π equivalent circuit, (B) π equivalent circuit of Fig. 4.6B and π equivalent circuit of Fig. 4.6C.

The admittance matrix of Fig. 4.6C can be similarly derived and the result is 2 3 1 21  6Z 7  tLH ZHL IH 6 HL 7 VH 56 21 (4.13b) 1 1 7 IL 4 5 VL 1 2 tLH ZHL ZM tLH ZHL The equations of the general equivalent π circuit shown in Fig. 4.9A are   1 1 IH 5 Y B 1 VL (4.14a) VH 2 ZA ZA   1 1 IL 5 2 V H 1 YC 1 (4.14b) VL ZA ZA or in matrix form 2 1  6 YB 1 Z A IH 6 5 6
1 IL 4 ZA

3 7 7 VH 7 1 5 VL YC 1 ZA
1 ZA

(4.14c)

262

Power Systems Modelling and Fault Analysis

Equating the admittance terms in Eqs. (4.13a) and (4.14c), we have ZA 5 tHL ZLH and

(4.15a) 

21 tHL



YB 5

tHL 2 1 tHL ZLH

YC 5

tHL 2 1 1 1 tHL ZLH ZM

(4.15b)

(4.15c)

Similarly, equating the admittance terms in Eqs. (4.13b) and (4.14c), we have ZA 5 tLH ZHL

(4.16a)

and YB 5

tLH 2 1 tLH ZHL

tLH 2 1 YC 5 tLH ZHL

(4.16b) 

 21 1 1 tLH ZM

(4.16c)

The equivalent π circuit models of the off-nominal-ratio transformer of Figs. 4.6B, C, 4.7B and C are shown in Fig. 4.9B and 4.9C. The equivalent π circuit model parameters for the three conditions shown in Fig. 4.8A are calculated and shown in Fig. 4.10. Also shown in Fig. 4.10 are the short-circuit impedances calculated from side H with side L short-circuited, as well as from side L with side H short-circuited. It should not come as a surprise to the reader that these impedances are identical to those that the reader may have calculated, as recommended previously, for the equivalent circuits shown in Fig. 4.8A and B. As mentioned before, it is generally no longer the case in modern industry practice to neglect the shunt magnetising impedance branch (RM//jXM) that represents the core iron losses and magnetising losses. The air path reactance XLH connected in series with the magnetising branch is negligibly small compared to XM when the core is not saturated. The usual practice of inserting the exciting branch admittance midway through the series impedance, that is, splitting it in half and connecting each half at either end of the equivalent circuit, is not valid from a physical point of view although it would still be a good approximation provided the transformer core does not saturate.

4.2.3 Three-phase two-winding transformers The majority of power transformers used in power systems are three-phase transformers since these have significant economic benefits compared to three banks of

Modelling of transformers, phase shifters, static power plant and static load Z A = j13.5%

H

263

L ZH(L short -circuited) = j12.5% ZL(H short -circuited) = j15%

Y B = − j82.3% Y C = j74.07% t H = 0.9 t L = 1.0 t HL = 0.9 Z A = j15%

H

L Z H(L short -circuited) = j15% Z L(H short -circuited) = j15%

t H = 1.0

t L = 1.0

t HL = 1.0

Z A = j16.5%

H

L Z H(L short -circuited) = j18.15% Z L(H short -circuited) = j15%

Y B = j55.09%

Y C = − j60.6% tH = 1.1

t L = 1.0

tHL = 1.1

Figure 4.10 π equivalent circuit of example shown in Fig. 4.8A.

single-phase units that have the same total rating and perform similar duties. On the other hand, reliability considerations for very large three-phase transformers may result in the use of three banks of single-phase units with one single-phase unit held as spare that can be used in the event of failure. These are generally used in very large generator-transformers. In double-wound transformers, there is a low-voltage winding and an electrically separate high-voltage winding.

Winding connections In three-phase double-wound power transformers, the three phases of each winding are usually connected as star or delta. There are thus four possible connections of the primary and secondary windings, namely star
star, star
delta, delta
star and delta
delta connections. Interconnected star or zigzag windings are dealt with in Section 4.2.6.

Positive-phase sequence and negative-phase sequence equivalent circuits Like all static three-phase power system plants, the impedance such a plant presents to the flow of positive-sequence or negative-sequence currents is independent of the phase rotation or sequence of the three-phase applied voltages R, Y, B or R, B, Y. Therefore, the plant positive-sequence and negative-sequence impedances are equal.

264

Power Systems Modelling and Fault Analysis

It will be shown later in this section that three-phase transformers can introduce a phase shift between the primary and secondary winding currents, and the same phase shift between the primary and secondary voltages, depending on the winding connections. In addition, where the phase shift of positive-sequence currents and voltages is ϕ, it is
ϕ for negative-sequence currents and voltages. The effect of this phase shift is to change the ideal transformer off-nominal turns ratio from a real to a complex number. Therefore, the general positive-sequence and negative-sequence equivalent circuits of a three-phase two-winding transformer that correspond to the equivalent circuit of Fig. 4.6B are shown in Fig. 4.11A and B, respectively. Similarly, the general positive-sequence and negative-sequence equivalent circuits of a three-phase two-winding transformer that correspond to the equivalent circuit of Fig. 4.6C are shown in Fig. 4.12A and B, respectively. As discussed later in this section, the phase shift that may be introduced by a three-phase transformer is not required to be represented initially in short-circuit analysis (or positive-sequence based power flow or stability analysis). Due account of any phase shifts can be made from knowledge of the location of transformers that introduce phase shifts in the system once the sequence currents and voltages are calculated by initially ignoring the phase shifts. Therefore, the transformer positive-sequence and negative-sequence equivalent circuits, ignoring the phase shift, are shown in Figs. 4.11C and 4.12C, and the corresponding π equivalent circuit model is the same as that in Fig. 4.9A and B. It should be noted that for a three-phase transformer, the nominal turns ratio is equal to the ratio of the phase-to-phase base voltages on the HV and LV sides of the (A)

H IH

t HLe jφ : 1

Z LH

IL L

VH

(B) H I H

VL

t HLe– jφ : 1

Z LH

IL L

VH

(C) H I H

VH

VL

t HL : 1

Z LH

IL

L VL

Figure 4.11 Positive-sequence and negative-sequence equivalent circuits of Fig. 4.6B for an off-nominal ratio two-winding three-phase transformer: (A) positive-sequence equivalent circuit, (B) negative-sequence equivalent circuit and (C) positive-sequence and negativesequence equivalent circuit ignoring phase shift.

Modelling of transformers, phase shifters, static power plant and static load (A)

H

IH

Z HL

1 : t LHe jφ

IL

VH

(B)

H

H

L

VL

IH

Z HL

1 : t LHe– jφ

IL

L VL

VH

(C)

265

IH

Z HL

VH

1 : t LH

IL

L VL

Figure 4.12 Positive-sequence and negative-sequence equivalent circuits of Fig. 4.6C for an off-nominal ratio two-winding three-phase transformer: (A) positive-sequence equivalent circuit, (B) negative-sequence equivalent circuit and (C) positive-sequence and negativesequence equivalent circuit ignoring phase shift.

transformer irrespective of the primary and secondary winding connections. This means that the base voltage ratio and the nominal turns ratio are equal pffiffiffi for star
star and delta
delta winding connections but also includes the factor 3 for star
delta winding connection.

Zero-sequence equivalent circuits The zero-sequence equivalent circuit of a two-winding transformer is primarily dependent on the method of connection of the primary and secondary windings because the zero-sequence currents in each phase are equal in magnitude and are in phase. Also, the zero-sequence equivalent circuit and currents are affected by the winding earthing arrangements. In addition, a practical factor that can influence the magnitude of the zero-sequence impedances and make them appreciably different from the positive-sequence impedances is the type of construction of the transformer magnetic circuit. The effect of transformer core construction will be dealt with later in this chapter, but we will now focus on the effect of the winding connection and any neutral earthing that may be present. Before considering three-phase two-winding transformers of various winding connections, we will restate a basic understanding regarding zero-sequence currents in simple star-connected or delta-connected three sets of impedances. For zerosequence currents to flow in a star-connected set of impedances, the neutral point of the star must be earthed either directly or indirectly, for example, through an impedance. The zero-sequence current that flows to earth through this path is the sum of the zero-sequence currents in the three phases of the star impedances

266

Power Systems Modelling and Fault Analysis

(A) IZ

IZ = 0

IZ = 0

Z

Z

Z

IZ

Z Z

Z

Z

IZ

IZ = 0 IZ = 0

(B)

Z

3I Z

IZ

IZ Z

ZE

IZ = 0 IZ = 0

IZ

I Z 3Z E

Z VZ

Zero voltage node

IZ

Z

Z

Zero voltage node

Zero voltage node

Figure 4.13 Zero-sequence equivalent circuits for three-phase impedances connected in star or delta: (A) three-phase star- and delta-connected impedances and (B) zero-sequence equivalent circuits.

or three times the zero-sequence current of one phase, say phase R. For a delta-connected three-phase set of impedances, no zero-sequence currents will flow in the output terminals and hence into the delta connection. However, zerosequence currents can circulate inside the delta by mutual coupling but no zerosequence current can emerge out into the output terminals of the delta. Fig. 4.13A shows the three-phase connections of the impedances and zerosequence current flows for the above three cases and Fig. 4.13B shows the corresponding zero-sequence equivalent circuits. It is noted that because the voltage drop across the earthing impedance ZE is 3I Z ZE , the effective earthing impedance appearing in the zero-sequence equivalent circuit is 3ZE. We will now derive approximate zero-sequence equivalent circuits for twowinding transformers of various winding connection arrangements. We emphasise that these are approximate because for now we ignore the effect of transformer core or magnetic circuit construction on the magnetising impedance. This is an important practical aspect that should not be ignored in practical short-circuit analysis. This is discussed in detail in Section 4.2.8. The basic and fundamental principle that underpins the derivation of the approximate zero-sequence equivalent circuits is based on Eq. (4.1b). This states that there is always a magnetic circuit MMF or ampere-turn balance for the transformer primary and secondary windings. That is, no current can flow in one winding unless a corresponding current, allowing for the winding turns ratio, flows in the other winding. In deriving the zero-sequence equivalent circuits of two-winding transformers of various winding connections, the winding terminal is connected to the external circuit if zerosequence current can flow into and out of the winding. If zero-sequence current can circulate inside the winding without flowing in the external circuit, the winding

Modelling of transformers, phase shifters, static power plant and static load

(A) H VH

t HL : 1

I H 3Z E(H)

Z Z LH

3Z E(L) I L

Z E(H)

Z E(L)

267

L VL

Y MZ

Zero voltage node Key :

t HL : 1

t HL : 1

≡

Ideal transformer

(B) H VH

Z Z HL

I H 3Z E(H)

1 : t LH

Z E(H)

3Z E(L) I L Z E(L)

L VL

Y MZ

Zero voltage node Key :

1 : t LH

1 : t LH

≡

Ideal transformer

Figure 4.14 Generic zero-sequence equivalent circuit for two-winding transformers: (A) generic zero-sequence equivalent circuit of Fig. 4.11C, (B) generic zero-sequence equivalent circuit of Fig. 4.12C.

terminal is connected directly to the reference zero voltage node. Fig. 4.14 shows zero-sequence equivalent circuits for an off-nominal ratio transformer that correspond to Figs. 4.11C and 4.12C. These take account of any HV and LV winding connection arrangements, and any neutral impedance earthing that may be used. By making use of Fig. 4.14A and B, and noting that an arrow on a winding signifies the presence of a tap-changer on this winding, Fig. 4.15 can be derived. This shows approximate zero-sequence equivalent circuits for the most common threephase two-winding transformers used in power systems. It should be noted that ZE is in per unit on the voltage base of the winding to which it is connected. Returning to the example shown in Fig. 4.8, and assuming the transformer is a three-phase two-winding star (HV winding)
delta (LV winding) transformer, the zero-sequence equivalent circuits that correspond to the three tap ratios are shown in Fig. 4.16. The reader should find it instructive to obtain the three equivalent circuits shown in Fig. 4.16 from the corresponding equivalent π circuits shown in Fig. 4.10 by applying a short-circuit at the L side and calculating the equivalent impedance seen from the H side. The elements of the admittance matrices of the equivalent circuits shown in Fig. 4.15 can be easily derived using the general 2 3 2 admittance matrix given by 

 IH Y 5 11 IL Y21

Y12 Y22



VH VL

(4.17a)

268

Power Systems Modelling and Fault Analysis

(A) H

t HL : 1

H

L

Z Z LH

L Y MZ

Zero voltage node (B)

H

L

t HL : 1

H

Z Z LH

L Y MZ

Zero voltage node (C)

H

t HL : 1

H

L

Z Z LH

ZE

3Z E

L

Y MZ Zero voltage node

(D)

H

H

L

3Z E

1 : t LH

Z Z HL

ZE

L

Y MZ Zero voltage node

(E)

H

L

t HL : 1

H

Z Z LH

L

Zero voltage node (F) H

L

Z Z LH

H

L

Zero voltage node (G)

H

L

H

3Z E

t HL : 1

Z Z LH

L

ZE Zero voltage node

(H)

H

L

H

t HL : 1

Z Z LH

ZE

3Z E

L

Y MZ Zero voltage node

Figure 4.15 Approximate zero-sequence equivalent circuits for common three-phase twowinding transformers: (A) star
star with isolated neutrals; (B) star with solidly earthed neutral
star isolated neutral; (C) star solidly earthed
star neutral earthing impedance; (D) star neutral earthing impedance
star solidly earthed neutral; (E) star isolated neutral
delta; (F) delta
delta; (G) star neutral earthing impedance
delta and (H) delta
star neutral earthing impedance.

Modelling of transformers, phase shifters, static power plant and static load

269

j12.15%

H

L

t H = 0.9 t L = 1.0 t HL = 0.9

j15%

H

L

t H = 1.0 t L = 1.0 t HL = 1.0 j18.15% H

L

t H = 1.1 t L = 1.0 t HL = 1.1

Figure 4.16 Zero-sequence equivalent circuits of two-winding transformer of Fig. 4.8.

 IH  Y11 5 VH VL50

 IL  Y22 5 VL VH50

 IH  Y12 5 VL VH50

 IL  Y21 5 VH VL50

(4.17b)

The results are shown in Table 4.1. The reader is encouraged to derive these results.

Effect of winding connection phase shifts on sequence voltages and currents The effect of three-phase transformer phase shifts on the sequence currents and voltages will now be considered. The presence of a phase shift between the transformer primary and secondary voltages, and currents, depends on the transformer primary and secondary winding connections. The winding connections designated by British, IEC and North American Standards are: High voltage: Capital letters, Y Star, D Delta, Z Zigzag, A Auto, N Neutral brought out. Low voltage: Small letters, y star, d delta, z zigzag, a auto, n neutral brought out.

For transformers with a star
star or a delta
delta winding connection, the primary and secondary currents, and voltages, in each of the three phases are either in phase or out of phase, that is, the windings are connected so that the phase shifts are either 0 degree or 6 180 degrees. The former case is illustrated in Fig. 4.17A and B. British and IEC practices use ‘vector group reference’ number and symbol. In the symbol Yd1, the digit 1 indicates a phase shift of 2 30 degrees using a 12 3 30 degrees clock reference. For example, 0 degrees indicates 12 o’clock,

270

Power Systems Modelling and Fault Analysis

Table 4.1 Elements of zero-sequence admittance matrix of two-winding three-phase transformers Fig.

Two-winding threephase transformer winding connections

4.15A

H

L

Y11 5 Y12 5 Y21 5 Y22 5 0

4.15B

H

L

Y11 5 Y12 5 Y21 5 Y22 5 0

4.15C

H

L ZE

Elements of zero-sequence admittance matrix of Eq. (4.19b)

Y11 5

1
Z  2 Z ZLH 1 3ZE :ZM tHL

Y22 5

1 Z :Z Z 3ZE 1 ZLH M

Y12 5

21

 Z =Z Z Z Z tHL 1 1 ZLH M 3ZE 1 ZLH :ZM

Y21 5


Z 1 1 3ZE =ZM

21 
Z  Z ZLH 1 3ZE :ZM

Z Z 5 1=YM ZM

4.15D

H

L ZE

Straight forward derivation left for the interested reader

4.15E

H

L

Y11 5 Y22 5 Y12 5 Y21 5 0

4.15F

H

L

Y11 5 Y22 5 Y12 5 Y21 5 0

4.15G

H

L

Y11 5

ZE

1 2 Z Z 1 3Z tHL E LH

Y22 5 Y12 5 Y21 5 0 4.15H

H

L ZE

Y22 5

1 Z :Z Z ZLH M

1 3ZE

Y11 5 Y12 5 Y21 5 0

180 degrees indicates 6 o’clock, 2 30 degrees indicates 1 o’clock and 1 30 degrees indicates 11 o’clock. In Fig. 4.17, the 0 degrees phase shift is achieved by ensuring that the parallel windings, that is, same phase windings, are linked by the same magnetic flux. Fig. 4.15 also shows that the absence of phase shifts in the phase currents and voltages also translates into the positive-sequence and negative-sequence currents and voltages. Consequently, the presence of such transformers in the three-phase network requires no special treatment in the formed positive-sequence and

Modelling of transformers, phase shifters, static power plant and static load

(A)

271

r

R

Star-Star Yy0 n

N

y

Y b

B R

r

Phase to neutral voltages and PPS voltages n

N

y

Y b

B

r

R

NPS phase to neutral voltages n

N

Y

(B)

B

b

y

r

R

Dd0 y

Yb

B

r

R

Phase to phase and phase to neutral PPS voltages

y

Yb

B

r

R

Phase to neutral NPS voltages

Y

B

y

b

Figure 4.17 Positive-sequence (PPS) and negative-sequence (NPS) voltage phase shifts for (A) Yy0 and (B) Dd0 connected transformers.

negative-sequence networks under balanced or unbalanced conditions. It should be noted that for the delta winding, although a physical neutral point does not exist, a voltage from each phase terminal to neutral does still exist because the network to which the delta winding is connected would in practice contain a neutral point.

272

Power Systems Modelling and Fault Analysis

Star R

r

d11

d1

Ir

nIR

Winding Connections

b

R

B PPS phase-to-phase voltages B

Y

y

Ib

Iy

nI Y nIY

Iy

nIB

Ib r

r

r y

b

y

b Y

R

b

300 r

r

30

0

y

b

Y y

R

NPS phase-to-neutral voltages Y

nI R

nI B

y PPS phase-to-neutral voltages B

Ir

r

b 300 r

300 b

y

B y

b

Figure 4.18 Positive-sequence (PPS) and negative-sequence (NPS) voltage phase shifts for Yd1 and Yd11 transformers.

In the case of transformers with windings connected in star
delta (or delta
star), voltages and currents on the star winding side will be phase shifted by a 6 30 degrees angle with respect to those on the delta side (or vice versa depending on the chosen reference). According to British practice, Yd11 results in the positivesequence phase-to-neutral voltages on the star side lagging by 30 degrees the corresponding ones on the delta side. Also, Yd1 results in the positive-sequence phase-to-neutral voltages on the star side, leading by 30 degrees the corresponding ones on the delta side. The example vector diagrams shown in Fig. 4.18 for Yd1 and Yd11 illustrate this effect. For RBY/rby negative phase sequence or rotation, Fig. 4.18 also shows the effect of the Yd1 and Yd11 on the negative-sequence voltage phase shifts and shows that these are now reversed. These phase shifts also apply to the positive-sequence and negative-sequence currents in these windings because the phase angles of the currents with respect to their associated voltages are determined by the balanced load impedances only. In summary, where the positive-sequence voltages and currents are shifted by 1 30 degrees, the corresponding negative-sequence voltages and currents are shifted by 2 30 degrees and vice versa depending on the specified connection and phase shift, that is, Yd1 or Yd11. Mathematically, this is now derived for the Yd1 transformer shown in Fig. 4.18 where n is the turns ratio as follows. The red phase current in amps flowing out of the d winding r phase is equal to Ir 5 n(IR 2 IB). Using Eq. (2.9a) from Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, for phase

Modelling of transformers, phase shifters, static power plant and static load

273

currents and noting that IRZ 5 0 because the in-phase zero sequence currents cannot exit the d winding, we can write pffiffiffi pffiffiffi   Ir 5 n½ð1 2 hÞIRP 1 ð1 2 h2 ÞIRN  5 n 3IRP e
j30 1 n 3IRN ej30 or Ir 5 IrP 1 IrN where pffiffiffi  IrP 5 n 3IRP e2j30

and

pffiffiffi  IrN 5 n 3IRN ej30

(4.18a)

or  1 IRP 5 pffiffiffi IrP ej30 n 3

and

 1 IRN 5 pffiffiffi IrN e
j30 n 3

(4.18b)

1 or in pu where n 5 pffiffiffi, 3 

IRP 5 IrP ej30

and

IRN 5 IrN e
j30



(4.18c)

Similarly, from Fig. 4.18, the phase-to-neutral voltage in volts on the star winding R phase is VR 5 nðVr 2 Vy Þ and using Eq. (2.9b) for phase r and y voltages, we have pffiffiffi pffiffiffi   VR 5 n½ð1 2 h2 ÞVrP 1 ð1 2 hÞVrN  5 n 3VrP ej30 1 n 3VrN e2j30 or VR 5 VRP 1 VRN where

pffiffiffi  VRP 5 n 3VrP ej30

pffiffiffi  and VRN 5 n 3VrN e2j30

(4.19a)

or  1 VrP 5 pffiffiffi VRP e2j30 n 3 1 or in pu where n 5 pffiffiffi, 3

VRP 5 VrP ej30



and

 1 and VrN 5 pffiffiffi VRN ej30 n 3



VRN 5 VrN e2j30

(4.19b)

(4.19c)

The reader is encouraged to derive the equations for the Yd11 transformer.

274

Power Systems Modelling and Fault Analysis

The American standard for designating winding terminals on star
delta transformers requires that the positive-sequence (negative-sequence) phase-to-neutral voltages on the high-voltage winding to lead (lag) the corresponding positive-sequence (negative-sequence) phase-to-neutral voltages on the low-voltage winding. This is so regardless of whether the star or the delta winding is on the high-voltage side. In terms of sequence analysis, this means that when stepping up from the low-voltage to the high-voltage side of a star
delta or delta
star transformer, the positivesequence voltages and currents should be advanced by 30 degrees, whereas the negative-sequence voltages and currents should be retarded by 30 degrees. It is interesting to note the following observation on the British and American standards. In American practice, when the star winding in a star
delta transformer is the high-voltage winding, this would correspond, in terms of phase shifts, to the Yd1 in British practice. However, when, in American practice, the delta winding in a star
delta transformer is the high-voltage winding, this would correspond, in terms of phase shifts, to the Yd11 in British practice. In terms of fault analysis in power system networks using the positive-sequence and negative-sequence networks, it is common practice to initially ‘ignore’ the phase shifts introduced by star
delta transformers by assuming them as equivalent star
star transformers and to calculate the sequence voltages and currents on this basis. Then, having noted the locations in the network of such star
delta transformers, the appropriate phase shifts can be easily applied using the above equations as appropriate for the specified Yd transformer.

4.2.4 Three-phase three-winding transformers Three-phase three-winding transformers are widely used in power systems. When the VA rating of the third winding is appreciably lower than the primary or secondary winding ratings, the third winding is called a tertiary winding. These windings are usually used for the connection of reactive compensation plant, such as shunt rectors, shunt capacitors, static variable compensators or synchronous compensators or synchronous condensers. Tertiary windings may also be used to supply auxiliary load in substations and to generators. Delta-connected tertiary windings may also be used (and left unloaded) in order to provide a low-impedance path to zerosequence triplen harmonic currents. The B
H curve of the transformer magnetic circuit is nonlinear and, under normal conditions, transformers do operate on the nonlinear knee part of the curve. Thus, for a sinusoidal primary voltage, the magnetising current will be nonlinear and will contain harmonic components which are mainly third harmonics. Since in three-phase systems, the third-order harmonic currents in each phase are in phase, they can be considered as zero-sequence currents of three times the fundamental frequency. Consequently, as for the zero-sequence fundamental frequency currents, the tertiary winding allows the circulation of third-order harmonic currents. Other benefits of delta-connected tertiary windings include an improvement, that is, a reduction, in the unbalance of three-phase voltages. Sometimes, for economic benefits, a third winding is provided to form a transformer with double-secondary windings. These transformers tend to be used to

Modelling of transformers, phase shifters, static power plant and static load

275

supply high-density load in cities and this also provides additional benefits of fewer HV switchgear and also limiting LV system short-circuit currents where the two secondary LV terminals are not connected to the same busbar. Other uses are for the connection of two relatively small generators to a power network or for the connection of networks operating at different voltage levels.

Winding connections There are several winding connections for three-phase three-winding transformers which may be used in power systems. Some examples are YNdd, Ydd, YNynyn, YNynd, YNyd and Yyd.

Positive-phase sequence and negative-phase sequence equivalent circuits in actual physical units In a three-winding transformer with one HV and two LV windings, the three windings are generally mutually coupled although the degree of coupling between the two LV windings of a double secondary transformer can be chosen by design so that the LV windings are either closely or loosely coupled. Fig. 4.19 (A) shows a single-phase divided flux representation of a three-winding three-phase transformer iron-cored transformer. The shunt exciting impedance is connected to the side of the innermost winding. From Fig. 4.19A, the following equations in actual physical units can be written V 1 2 E 1 5 Z1 I 1

(4.20a)

V 2 2 E 2 5 Z2 I 2

(4.20b)

V 3 2 E 3 5 Z3 I 3

(4.20c)

E1 N1 5 E3 N3

(4.20d)

E2 N2 5 E3 N3

and N1 I1 1 N2 I2 1 N3 I3 5 N3 IM

(4.20e)

From Eq. (4.20a)
(4.20e), neglecting initially the shunt exciting or no-load current IM, we can write 1 1 ðV1 2 Z1 I1 Þ 5 ðV3 2 Z3 I3 Þ N1 N3

(4.21a)

1 1 ðV2 2 Z2 I2 Þ 5 ðV3 2 Z3 I3 Þ N2 N3

(4.21b)

Eq. (4.21a) and (4.21b) are illustrated in Fig. 4.19B and C which are identical.

(A)

Z1

I1

1

Z2 E1 N1

V1

I2

2

E2 N2

V2

I3 E3 N3

V3

1: N 2

(B)

1

I1

Z1

3

Z3

Z2

I2

2

V2

N1 :1 1: N 3

V1

Z3

I3

3

I jXAir

V3

RM

jXM

Side 3 is innermost winding (C)

2

Z2 /N2 1

I1

N1 :1

1: N 2

I2

2

2 Z1/N1

V2 2

Z3/N3

V1

1: N 3

I3

3 V3

IM jXAir RM

jXM

Side 3 is innermost winding (D)

2

Z2/N23 1

V1

I1

N 13 : 1

1 : N 23

I2

2 Z1/N13

Z3

2 V2

I3 3 V3

IM jXAir RM

jXM

Side 3 is innermost winding

Figure 4.19 Equivalent circuits for three-winding three-phase transformers: (A) single-phase divided flux representation of iron-cored three-winding three-phase transformer, (B) positivesequence and negative-sequence equivalent circuit in actual physical units with threeoff-nominal turns ratios; (C) as in (B) but with turns ratios placed outside the impedances, (D) positive-sequence and negative-sequence equivalent circuit with two off-nominal turns ratios.

Modelling of transformers, phase shifters, static power plant and static load

277

In Fig. 4.19B and C, we replaced the three-winding transformer with three twowinding transformers connected as three branches in a star or T configuration. Each branch corresponds to a winding and has its own general off-nominal tap ratio to represent a nominal, fixed off-nominal ratio or an on-load tap-changer, as required. Finally, the impedance of the magnetising branch ZM is now inserted at the side that corresponds to the innermost winding, that is, the winding closest to the core, which for the purpose of illustration in our analysis is assumed to be side 3. For example, if terminal 1 is HV, terminal 2 is MV and terminal 3 is LV, then ZM is connected to terminal 3. The ’air’ part reactance XAir of this magnetisation branch can be taken as half the leakage reactance between side 3 and side 2 where the latter corresponds to the middle winding around the core, that is, XAir 5 X23/2 and should be referred to side 3. For improved applications in practice, Fig. 4.19C can be further simplified to include only explicitly two off-nominal tap ratios instead of three as follows. Let N13 5

N1 N3

N23 5

N2 N3

N13 I1 1 N23 I2 1 I3  0

(4.22)

Using Eqs. (4.21a), (4.21b) and (4.22), the equivalent circuit of Fig. 4.19D can be obtained. As we will discuss later, the individual impedances of the T equivalent circuit are derived from measurements of the leakage impedances between the windings namely Z12, Z13, and Z23 where Z12 is measured from side 1 with side 2 shortcircuited and side 3 open-circuited, and similarly for Z13 and Z23. For example, Z12 is measured with V2 5 0 and I3 5 0. Therefore, to calculate the impedance of each branch of the T equivalent circuit in ohms with all impedances referred to side 1 voltage base, let us define the measured impedances Z12, Z13 and Z23 as follows: 0

Z12 5 Z1 1 Z2

(4.23a)

0

Z13 5 Z1 1 Z3 

N1 N2

2

(4.23b) 0

0

Z23 5 Z2 1 Z3

(4.23c)

where the prime indicates quantities referred to side 1 voltage base. 2

1

1

0

3

7 6 0 1 7 2 Z1 3 61 6 1 1 7 76 0 7 6 7 6 or in matrix form; 4 Z13 5 5 6 0 0 12 0 12 74 Z2 5 7 0 6 7 6 Z23 4 @N1 A @N1 A 5 Z3 N2 N2 2

Z12

3

(4.23d)

278

Power Systems Modelling and Fault Analysis

Solving Eq. (4.23d) for each branch impedance, we obtain "  2 # 1 N1 Z12 1 Z13 2 Z1 5 Z23 2 N2 "  2 # 1 N1 Z12 2 Z13 1 Z2 5 Z23 2 N2 0

(4.24b)

"  2 # 1 N1 2 Z12 1 Z13 1 Z3 5 Z23 2 N2 0

Or in matrix form 2

0

6 1 6 6 Z1 6 6 6 7 6 0 7 16 6 Z2 7 5 6 1 6 7 26 6 4 5 6 0 6 Z3 6 6 4 21 2

3

21

1

(4.24c)

12 3

N1 A N2 0 12 @N1 A N2 0 12 @N1 A N2

2@

1

(4.24a)

7 72 3 7 7 Z12 76 7 76 7 76 76 Z13 7 7 74 5 7 7 7 Z23 7 5

(4.24d)

Positive-phase sequence and negative-phase sequence equivalent circuits in per unit In most large-scale power system steady-state analysis, we are more interested in deriving the three-winding transformer equivalent circuit in pu rather than in actual physical units. The following base and pu quantities are defined V1ðBÞ 5 Z1ðBÞ I1ðBÞ V1ðpuÞ 5

I1ðpuÞ 5

Z1ðpuÞ 5

V1 V1ðBÞ I1 I1ðBÞ Z1 Z1ðBÞ

V2ðBÞ 5 Z2ðBÞ I2ðBÞ V2ðpuÞ 5

I2ðpuÞ 5

V2 V2ðBÞ

I2 I2ðBÞ

Z2ðpuÞ 5

Z2 Z2ðBÞ

V3ðBÞ 5 Z3ðBÞ I3ðBÞ V3 V 3 ðB Þ

V3ðpuÞ 5

I3ðpuÞ 5

I3 I 3 ðB Þ

Z3ðpuÞ 5

Z3 Z3ðBÞ

S1ðBÞ 5 S2ðBÞ 5 S3ðBÞ 5 V1ðBÞ I1ðBÞ 5 V2ðBÞ I2ðBÞ 5 V3ðBÞ I3ðBÞ

(4.25a) (4.25b)

(4.25c)

(4.26a) (4.26b)

Modelling of transformers, phase shifters, static power plant and static load

V 1 ðB Þ I3ðBÞ N1ðnominalÞ 5 5 V 3 ðB Þ I1ðBÞ N3ðnominalÞ

V2ðBÞ I 3 ðB Þ N2ðnominalÞ 5 5 V 3 ðB Þ I 2 ðB Þ N3ðnominalÞ

279

(4.26c)

Using Eqs. (4.25a)
(4.25c) and (4.26a)
(4.26c), Eq. (4.21a) and (4.21b) can be rewritten as  1  1 (4.27a) V1ðpuÞ 2 Z1ðpuÞ I1ðpuÞ 5 V3ðpuÞ 2 Z3ðpuÞ I3ðpuÞ t1 t3  1  1 V2ðpuÞ 2 Z2ðpuÞ I2ðpuÞ 5 V3ðpuÞ 2 Z3ðpuÞ I3ðpuÞ t2 t3

(4.27b)

where, the pu tap ratios t1, t2 and t3 are defined as t1 5 5 t2 5 5 t3 5 5

N1 N1ðnominalÞ

5

N1 ðat given winding 1 tap positionÞ N1 ðwinding 1 nominal tap positionÞ

Rated voltage of winding 1 at given tap position Rated voltage of winding 1 at nominal tap position N2 N2 ðat given winding 2 tap positionÞ 5 N2 ðwinding 2 nominal tap positionÞ N2ðnominalÞ Rated voltage of winding 2 at given tap position Rated voltage of winding 2 at nominal tap position N3 N3 ðat given winding 3 tap positionÞ 5 N3 ðwinding 3 nominal tap positionÞ N3ðnominalÞ Rated voltage of winding 3 at given tap position Rated voltage of winding 3 at nominal tap position

(4.28a)

(4.28b)

(4.28c)

The equivalent circuit of Eq. (4.27a) and (4.27b) is illustrated in Fig. 4.20A, where the exciting branch is reinstated and connected to the innermost winding terminal side 3. Now, as in the case of actual physical units and for improved applications in practice, Fig. 4.20A can be further simplified to include only two off-nominal tap ratios instead of three using t13 5

t1 t3

t23 5

t2 t3

t13 I1ðpuÞ 1 t23 I2ðpuÞ 1 I3ðpuÞ  0

(4.29)

Using Eq. (4.29) in Eq. (4.27a) and (4.27b), we obtain V1ðpuÞ Z1ðpuÞ 2 2 3 t13 I1ðpuÞ 5 V3ðpuÞ 2 Z3ðpuÞ I3ðpuÞ t13 t13

(4.30a)

280

Power Systems Modelling and Fault Analysis

(A)

1:t2

Z2(pu)/t22 1

I1(pu)

t1: 1

I 2(pu) 2 V 2(pu)

Z1(pu)/t12 Z3(pu)/t32

V1(pu)

1:t3 I 3(pu) 3 V3(pu)

IM(pu) jXAir(pu) RM(pu)

jXM(pu) Side 3 is innermost winding

(B)

V3(pu)

1:t23

2 Z2(pu)/t23

I1(pu)

t13: 1

V1(pu)

2 Z1(pu)/t13

I 2(pu) 2 V 2(pu) I3(pu) 3 V3(pu) IM(pu) jXAir(pu)

Z3(pu)

RM(pu)

jXM(pu)

Side 3 is innermost winding

Figure 4.20 Equivalent circuits for three-winding three-phase transformers: (A) positivesequence and negative-sequence equivalent circuit in per unit with three-off-nominal turns ratios; (B) positive-sequence and negative-sequence equivalent circuit in per unit with two off-nominal turns ratios.

V2ðpuÞ Z2ðpuÞ 2 2 3 t23 I2ðpuÞ 5 V3ðpuÞ 2 Z3ðpuÞ I3ðpuÞ t23 t23

(4.30b)

The equivalent circuit of Eq. (4.30a) and (4.30b) is illustrated in Fig. 4.20B.

Transformer equivalent circuit in per unit based on nominal impedance As in the case of two-winding transformers, we define Z1(pu,nominal), Z2(pu,nominal) and Z3(pu,nominal) as the pu leakage impedances of windings 1, 2 and 3 at nominal winding tap position, respectively. Since we assume that the impedance of each winding is proportional to the square of the number of turns of the winding, then for each winding, we can write Z1ðpuÞ 5 t12 Z1ðpu;nominalÞ

(4.31a)

Z2ðpuÞ 5 t22 Z2ðpu;nominalÞ

(4.31b)

Modelling of transformers, phase shifters, static power plant and static load

(A)

1:t2

Z2(pu,nominal)

1

I1(pu)

281

I 2(pu)

t1: 1 Z (pu,nominal) 1 1:t3

V1(pu)

IM(pu) jXAir(pu)

Z3(pu,nominal) RM(pu)

2 V 2(pu)

I3(pu) 3 V3(pu)

jXM(pu)

Side 3 is innermost winding

(B)

2

1:t23

t3Z2(pu,nominal)

1

I1(pu)

t13: 1 t2Z (pu,nominal) 3 1

V1(pu)

2

t3Z3(pu,nominal) IM(pu)

I 2(pu) 2 V 2(pu) I3(pu) 3 V3(pu)

jXAir(pu) RM(pu)

jXM(pu)

Side 3 is innermost winding

Figure 4.21 Equivalent circuits for three-winding three-phase transformers based on nominal tap impedance: (A) positive-sequence and negative-sequence equivalent circuit in per unit three-off-nominal turns ratios; (B) positive-sequence and negative-sequence equivalent circuit with two off-nominal turns ratios.

Z3ðpuÞ 5 t32 Z3ðpu;nominalÞ

(4.31c)

Substituting Eq. (4.31a)
(4.31c) into Eq. (4.27a) and (4.27b), we obtain V1ðpuÞ
Z11ðpu;nominalÞ 3 t1 I1ðpuÞ 5 t1

V3ðpuÞ
Z33ðpu;nominalÞ 3 t3 I3ðpuÞ t3

(4.32a)

V2ðpuÞ 2 Z22ðpu;nominalÞ 3 t2 I2ðpuÞ 5 t2

V3ðpuÞ 2 Z33ðpu;nominalÞ 3 t3 I3ðpuÞ t3

(4.32b)

The equivalent circuit of Eq. (4.32a) and (4.32b) is shown in Fig. 4.21A. Now, as previously, and for improved applications in practice, Fig. 4.21A can be further simplified to include only two off-nominal tap ratios instead of three. Using Eqs. (4.29) and (4.31a)
(4.31c) in Eq. (4.30a) and (4.30b), we obtain V1ðpuÞ 2 t32 Z1ðpu;nominalÞ 3 t13 I1ðpuÞ 5 V3ðpuÞ 2 t32 Z3ðpu;nominalÞ 3 I3ðpuÞ t13

(4.33a)

282

Power Systems Modelling and Fault Analysis

V2ðpuÞ 2 t32 Z2ðpu;nominalÞ 3 t23 I2ðpuÞ 5 V3ðpuÞ 2 t32 Z3ðpu;nominalÞ 3 I3ðpuÞ t23

(4.33b)

The equivalent circuit of Eq. (4.33a) and (4.33b) is illustrated in Fig. 4.21B. It should be noted that branch 3 now has an effective tap ratio of unity, t33 5 t3/ t3 5 1, as implied from Eq. (4.29), although t3 itself does not necessarily have to be equal to unity. The tap ratios included in the remaining two branches include the effect of branch 3 off-nominal tap ratio. This model needs to be used with care since, for example, the presence of a variable tap on one winding, say winding 3, must result in a coordinated and consistent change in the turns ratios between windings 1 and 3, and between windings 2 and 3. The equivalent circuits derived so far represent the three-winding transformer’s positive-sequence and negative-sequence equivalent circuit if we ignore the windings phase shift. Otherwise, the only difference between the positive-sequence and negative-sequence equivalent circuits would be introduced by the windings phase shift. This was described in Section 4.2.3 and shown to result in a complex transformation ratio instead of a real one. In practice, the impedances of the three-winding transformer required in the derived equivalent circuits are calculated from shortcircuit test data supplied by the manufacturer. This will be covered in detail in Section 4.2.11. As before, the equivalent circuits shown in Figs. 4.20A, B, 4.21A and B can all be converted to π equivalents suitable for digital computer computation. However, we will convert Fig. 4.20B only. We obtain three π equivalents, one for each transformer branch as shown in Fig. 4.22. It should be remembered that the π equivalents were derived with the convention that the currents are injected into them at both ends. Therefore, the currents flowing into the start point from the three branches would need to be reversed in sign.

Z2(pu)/t23 t23(t23 – 1) I1(pu) V2(pu)

Z1(pu)/t3

Z2(pu)

I3(pu) I 2(pu)

(1 – t23) Z2(pu)

V3(pu)

V1(pu)

(1 – t13) Z1(pu)

t13(t13 – 1) Z1(pu)

Z3(pu)

V1(pu) 1

IM(pu) jXAir(pu) RM(pu)

V3(pu)

jXM(pu)

Side 3 is innermost winding

Figure 4.22 Positive-sequence and negative-sequence π equivalent circuit for three-winding transformer.

Modelling of transformers, phase shifters, static power plant and static load

1:t23

Z2Z 1 I1 3Z E1 V1 ZE1

t13:1

283

3Z E2 ZE2

Z1Z

I2

2

V2

P

Z 3Z Y MZ

3Z E3 I3 ZE3

3 V3

Figure 4.23 Generic zero-sequence equivalent circuit of three-phase three-winding transformer.

Zero-phase sequence equivalent circuits The approximate zero-sequence equivalent circuits of three-winding transformers using various windings arrangements will be derived with the aid of a generic zerosequence equivalent circuit as already presented in the case of two-winding transformers. Remembering that we have converted the three-winding transformer to three two-winding transformers connected in T or star equivalent. Therefore, the generic zero sequence equivalent circuit, ignoring the shunt exciting branch for now is shown in Fig. 4.23 and is based on the following assumptions: 1. The secondary of each two-winding transformer equivalent is assumed to be starconnected with solidly earthed neutral, that is, directly connected to point P. 2. The primary of each two-winding transformer is assumed to have the same winding connection as one of the windings of the three-winding transformer.

The equivalent zero-sequence circuits for three-winding transformers having YNdd, Ydd, YNyd, YNynd, YNynyn and Yyd winding arrangements are shown in Fig. 4.24. The elements of the admittance matrix of Eq. (4.17a) and (4.17b) can be derived for each equivalent zero-sequence circuit shown in Fig. 4.24. This is similar to the derivation steps that resulted in Table 4.1. We will leave this exercise for the motivated reader!

4.2.5 Three-phase autotransformers with and without tertiary windings The autotransformer consists of a single continuous winding part of which is shared by the high- and low-voltage circuits. This is called the ‘common’ winding and is connected between the low-voltage terminals and the neutral. The rest of the total winding is called the ‘series’ winding and is the remaining part of the high-voltage circuits. The combination of the series-common windings produces the high-voltage winding. Fig. 4.25 shows a three-phase representation of a

1: t23

Z 2Z

(A)

1

3Z E

t13: 1

2

Z1Z

2

1

Z 3Z

ZE

3

3

1: t23

Z 2Z

(B)

t13: 1

1

Z1Z

2

1

2

Z 3Z

3

3

1: t23

Z 2Z

(C)

1

3Z E

t13: 1

2

Z1Z

2

1

Z 3Z

ZE

3

3

Y MZ

1: t23

Z 2Z

(D)

1

t13: 1

3Z E

Z1Z

2

1

Z E1

2

Z 3Z

3Z E3

Z E3 3

3

Y MZ

(E)

Z 2Z

3Z E1

1

t13: 1

1: t23

3Z E2 2

Z1Z

Z E2 2

1 Z E1

Z 3Z

Z E3

3Z E3

3

Z 2Z

(F)

1

1

3

Y MZ

t13: 1

1: t23

2

Z1Z

2 Z 3Z

3

3

Figure 4.24 Zero-sequence equivalent circuits of three-phase three-winding transformer: (A) star neutral earthing impedance
delta
delta; (B) star isolated neutral
delta
delta; (C) star neutral earthing impedance
delta
star isolated neutral; (D) star neutral earthing impedance
delta
star neutral earthing impedance; (E) all windings are star neutral earthing impedances and (F) star isolated neutral
star isolated neutral
delta.

Modelling of transformers, phase shifters, static power plant and static load

(A)

H R

285

I H(R)

series L

I L(r) Common

I H(R) + I L(r)

T

r

IT(y)

I L(y)

IE ZE

I H(B)

IT(r)

r

y

IT(b)

b

I L(b)

B

y

b

I H(Y)

Y (B)

H

IH EH

NS

IL NC E L ZC IH

IH + IL

IT

NT

ZS

VH

Z TT L

ET

T VT

VL IL

Figure 4.25 Star
star autotransformer with a delta tertiary winding: (A) three-phase representation and (B) single-phase representation.

star
star autotransformer with a delta tertiary winding. In the United Kingdom, autotransformers interconnect the 400 and 275 kV networks, and the 400/275 and 132 kV networks, whereas in North America, they interconnect the 345 and 138 kV networks, the 500 and 230 kV networks, etc.

Winding connections Three-phase autotransformers are most often star
star connected as depicted in Fig. 4.25A. The common neutral point of the star
star connection requires the two networks they interconnect to have the same earthing arrangement, for example, solid earthing in the United Kingdom, although impedance earthing may infrequently be used. Most autotransformers have a lower MVA rating tertiary winding connected in delta with a typical rating equal to 25% of the main rating.

Positive-sequence and negative-sequence equivalent circuits in physical units Autotransformers that interconnect extra high-voltage transmission systems are not generally equipped with tap-changers due to high costs. However, those that interconnect the transmission and subtransmission or distribution networks are usually equipped with on-load tap-changers in order to control or improve the quality of

286

Power Systems Modelling and Fault Analysis

their LV output voltage under heavy or light load system conditions. In autotransformers, tap-changers are connected on either the HV winding or the LV winding. In the latter case, the connection may be either at the LV winding line-end or the LV winding neutral-end. A single-phase representation of the general case of an autotransformer with a tertiary winding is shown in Fig. 4.25B. The tertiary winding is assumed, as is generally the practice, the innermost winding closest to the core. Using S, C and T to denote the series, common and tertiary windings, we can write in actual physical units VH 2 EH 5 ZS IH 1 ZC ðIH 1 IL Þ

(4.34a)

VL 2 EL 5 ZC ðIH 1 IL Þ

(4.34b)

VT 2 ET 5 ZTT IT

(4.34c)

The MMF balance equation is given by NS IH 1 NC ðIH 1 IL Þ 1 NT IT 5 NT IM

(4.35a)

NH 5 NS 1 NC

(4.35b)

EH NH 5 EL NL

and

NL 5 NC

EL NL 5 ET NT

EH NH 5 ET NT

(4.35c)

Neglecting initially the no-load exciting current IM and using Eqs. (4.35b), (4.35c) and (4.34a), Eq. (4.34b) and (4.34c) can be written as      VL 1 NH 2 NL VH 1 NH 2 NL 2 NL IL 2 2 NH IH 2 ZS 2 ZC 5 ZC NL NH NH NL NL NH (4.36a)      VT 1 NT2 VH 1 NH 2 NL 2 NT IT 2 ZTT 1 ZC 5 2 NH IH 2 ZS 2 ZC NT NL NH NH NL NT NH (4.36b) Eq. (4.36a) and (4.36b) can be represented by the star or T equivalent circuit shown in Fig. 4.26A, which includes three ideal transformers placed at the H, L and T terminals of the autotransformer. The measurement of the positive-sequence and zero-sequence impedances of an autotransformer with a tertiary winding using short-circuit tests between two winding terminals is dealt with later in this section. However, it is instructive to use

Modelling of transformers, phase shifters, static power plant and static load

(A)

287

1 NH–NL 2 NL ZC 1:NL NL

H VH IH

NH:1

NH–NL 1 ZC Z – NL NH2 S

IL

L VL

2

NT 1 2 ZTT + N N ZC NL H L

1:NT

IT

IM jXAir RM

T VT

jX M

Side T is innermost winding (B)

ZL′ ZH NH:1

H VH IH

NH2

1:NL

NH2

IL

L VL

Z T′ 1:NT I T

NH2 IM

T

VT

jXAir RM

jXM

Side T is innermost winding (C)

H

IH

NH:1

1 Z + ′ = 1 Z + Z NH – NL S C H ZL NL NH2 NH2

2

1:NL IM jXAir

VH RM

IL

L

VL

jX M

Side L is innermost winding

Figure 4.26 Positive-sequence/negative-sequence equivalent circuit of an autotransformer with a tertiary winding: (A) equivalent circuit in actual physical units in terms of series, common and tertiary impedances; (B) as (A) above but leakage impedances are referred to H side; (C) as (B) above but for an autotransformer without a tertiary winding; (D) as (C) above but with impedances referred to H side and one ideal transformer (E) as (C) above but with impedances referred to L side and one ideal transformer.

Eq. (4.36a) and (4.36b) to explain the results that are obtained from such tests in the construction of the autotransformer equivalent circuit. Using Eq. (4.36a) and (4.36b), the positive-sequence impedance measured from the H side with the L side VH j giving short-circuited and T side open-circuited is ZHL 5 IH VL 5 0; IT 5 0

288

Power Systems Modelling and Fault Analysis

(D)

H IH

ZHL = ZH + ZL′ 2 N – NL = ZS + ZC H NL

NH:NL IM 1:NL/NH jXAir

VH RM

IL L

VL

jX M Side L is innermost winding

(E)

ZHL =

H

NH:NL

IH

VH

=

NH2 NH2

NH2 NH2

ZS + ZC

(ZH + ZL′ ) NH – NL NL

2

IL IM jXAir

NH /NL:1 RM

L

VL

jX M

Side L is innermost winding

Figure 4.26 (Continued).

  NH 2NL 2 ZHL 5 ZS 1 ZC NL

(4.37a)

Also, the impedance measured from the H terminals with the T terminals shortVH circuited and L terminals open-circuited is ZHT 5 j giving IH VT 5 0; IL 5 0 ZHT 5 ZS 1 ZC 1

NH2 ZTT NT2

(4.37b)

Similarly, the impedance measured from the L terminals with the T terminals VL j giving short-circuited and H terminals open-circuited is ZLT 5 IL VT 5 0; IH 5 0 ZLT 5 ZC 1

NL2 ZTT NT2

(4.37c)

Now, given the three measured impedances ZHL, ZHT and ZLT in ohms, the impedance of each branch of the T equivalent circuit with all impedances referred to the H side voltage base can be calculated. Let 0

ZH 1 ZL 5 ZHL 0

ZH 1 ZT 5 ZHT

(4.38a) (4.38b)

Modelling of transformers, phase shifters, static power plant and static load

0

0

Z L 1 ZT 5

NH2 ZLT NL2

(4.38c)

Or in matrix form 2 1 1 2 3 6 61 0 ZHL 6 1 6 7 6 4 ZHT 5 5 6 0 0 1 6 2 6 ZLT @NH A 4 NL2

3

0

72 3 7 ZH 7 0 76 7 0 1 74 ZL 5 7 0 2 7 @NH A 5 ZT 2 NL 1 1

where the prime indicates impedances referred to the H side voltage base. Solving Eq. (4.38a)
(4.38d) for each branch impedance, we obtain   1 N2 ZHL 1 ZHT 2 H2 ZLT ZH 5 2 NL 0

ZL 5

  1 N2 ZHL 2 ZHT 1 H2 ZLT 2 NL

  1 NH2 ZT 5 2 ZHL 1 ZHT 1 2 ZLT 2 NL Or in matrix form 2 6 1 6 6 6 6 6 7 6 0 7 16 6Z 75 6 1 6 L 7 26 6 4 5 6 0 6 ZT 6 4 21 ZH

0 1

3

(4.38d)

(4.39a)

(4.39b)

0

2

289

(4.39c)

13

N2 2 @ H2 A 7 2 NL 7 7 Z

3 1 7 HL 76 7 2 76 7 @NH A 76 ZHT 7 7 6 7 2 NL 74 5 0 1 7 7 Z 2 7 LT @NH A 5 2 NL 0

21

1

(4.39d)

Now, substituting Eq. (4.37a) and (4.37b) into Eq. (4.39a)
(4.39d), we obtain ZH 5 ZS 2 ð 0

ZL 5 0

ZT 5

NH 2 NL ÞZC NL

(4.40a)

NH ðNH 2 NL Þ ZC NL NL

(4.40b)

NH N2 ZC 1 H2 ZTT NL NT

(4.40c)

290

Power Systems Modelling and Fault Analysis

Or in matrix form 2 61 6 6 6 7 6 6 0 7 6 6Z 756 0 6 L7 6 4 5 6 6 0 6 ZT 6 40 2

ZH

3

0

1 N 2 N H LA
@ NL 0 1 NH @NH 2 NL A NL NL NH NL

3 0 7 3 72 7 ZS 76 7 76 7 76 7 Z 7 0 76 C 7 5 74 7 2 7 ZTT NH 5 NT2

(4.40d)

Substituting Eq. (4.40a)
(4.40c) in Eq. (4.36a) and (4.36b), we obtain 0

VL Z VH ZH 2 NL IL L2 5 2 NH IH 2 NL NH NH NH

(4.41a)

0

VT Z VH ZH 2 NT IT T2 5 2 NH IH 2 NT NH NH NH

(4.41b)

Fig. 4.26B shows the equivalent circuit of Eq. (4.41a) and (4.41b). Where the autotransformer does not have a tertiary winding or where the tertiary winding is unloaded, the T terminal in Fig. 4.26B is unconnected to the power system network and the T branch impedance has no effect on the network currents and voltages. Thus, this branch can be disregarded and the effective autotransformer impedance would then be the sum of the H and L branch impedances. Fig. 4.26B can be simplified to the equivalent circuit shown in Fig. 4.26C which, in turn, can be further simplified to the equivalent circuits shown in Fig. 4.26D and E, where the equivalent leakage impedance is referred to the H side and L side, respectively.

Positive-sequence and negative-sequence equivalent circuits in per unit Now, we will convert Eq. (4.41a) and (4.41b) from actual physical units to pu values. To do so, we define the following pu quantities Vpu 5

V I ZH Z0 Z0 Ipu 5 ZHðpuÞ 5 ZLðpuÞ 5 L ZTðpuÞ 5 T V ðB Þ IðBÞ Z H ðB Þ Z H ðB Þ ZHðBÞ

(4.42a)

VHðBÞ 5ZHðBÞ IHðBÞ VLðBÞ 5ZLðBÞ ILðBÞ

(4.42b)

SHðBÞ 5SLðBÞ 5STðBÞ 5VHðBÞ IHðBÞ 5VLðBÞ ILðBÞ 5VTðBÞ ITðBÞ

(4.42c)

VHðBÞ ILðBÞ NHðnominalÞ VHðBÞ ITðBÞ NHðnominalÞ 5 5 5 5 VLðBÞ IHðBÞ NLðnominalÞ VTðBÞ IHðBÞ NTðnominalÞ

(4.42d)

Modelling of transformers, phase shifters, static power plant and static load

291

Using Eq. (4.42a)
(4.42d) in Eq. (4.41a) and (4.41b), and after a little algebra, we obtain VLðpuÞ ZLðpuÞ VHðpuÞ ZHðpuÞ 2 tL ILðpuÞ 2 5 2 tH IHðpuÞ 2 tL tH tH tH

(4.43a)

VTðpuÞ ZTðpuÞ VHðpuÞ ZHðpuÞ 2 tT ITðpuÞ 2 5 2 tH IHðpuÞ 2 tT tH tH tH

(4.43b)

where the pu tap ratios are defined as tH 5

NHðat given tap positionÞ NHðnominal tap positionÞ

tL 5

NLðat given tap positionÞ NLðnominal tap positionÞ

tT 5

NTðat given tap positionÞ NTðnominal tap positionÞ (4.43c)

Eq. (4.43a)
(4.43c) are represented by the pu equivalent circuit shown in Fig. 4.27A, which represents the autotransformer positive-sequence/negativesequence equivalent circuit ignoring the delta tertiary phase shift. The autotransformer is represented as three two-winding transformers that are star- or T-connected. This flexible model allows the representation of any off-nominal tap ratio on any winding, for example, a variable tap on-load tap-changer on one winding and the other two windings having nominal or fixed off-nominal tap ratios. In practice, it is found more convenient to represent the autotransformer with two ideal transformers instead of three by including the effect of the third ideal transformer in the tap ratios of the other two ideal transformers. It can be shown that Eq. (4.43a)
(4.43c) can be rewritten as VLðpuÞ ZLðpuÞ VHðpuÞ ZHðpuÞ 2 tLT ILðpuÞ 2 5 2 tHT IHðpuÞ 2 tLT tHT tHT tHT VTðpuÞ 2 ITðpuÞ

ZTðpuÞ VHðpuÞ ZHðpuÞ 5 2 tHT IHðpuÞ 2 2 tHT tHT tHT

(4.44a)

(4.44b)

where the new pu tap ratios are defined as tHT 5

tH tT

tLT 5

tL tT

(4.44c)

Fig. 4.27B shows the equivalent circuit of Eq. (4.44a)
(4.44c). Eq. (4.35c) shows that the variable tap ratios must be consistent and coordinated where an active tap-changer on only one winding can in effect change the effective turns ratio on another. For example, for a 400/132/13 kV autotransformer with a tertiary winding having an on-load tap-changer acting on the neutral end of the common winding, the variation of tLT caused by LV to TV turns ratio changes will also

292

Power Systems Modelling and Fault Analysis ZL(pu)/tH2

(A)

I H(pu)

H

1:tL

IL(pu) L

ZH(pu) /tH2

tH:1

VL(pu) Z T(pu)/tH2

VH(pu)

1:tT I T(pu)

T

I M(pu) jX Air(pu) RM(pu)

VT(pu)

jX M(pu) Side T is innermost winding

(B)

2 ZL(pu)/tHT

H

tHT:1

I H(pu)

2 ZH(pu) /tHT

1:tLT

IL(pu) L

VL(pu)

2 Z T(pu)/tHT

VH(pu)

IT(pu) IM jX Air RM

T

VT(pu)

jX M

Side T is innermost winding ZH(pu) + ZL(pu)

(C) H

I H(pu)

tH:1

1:tL

tH2

VH(pu)

IM jX Air RM

I H(pu)

1:tLH

ZHL(pu) = ZH(pu) + ZL(pu) IM jX Air

VH(pu) RM

(E) H

VH(pu)

I H(pu)

tHL:1

Z H(pu) =

tL2 tH2

L

VL(pu)

jX M

(D) H

IL(pu)

I L(pu)

L

VL(pu)

tH:tL

jX M

ZL(pu) + ZH(pu)

I L(pu) L IM jX Air

tH:tL RM

VL(pu)

jX M

Figure 4.27 Positive-sequence/negative-sequence equivalent circuit of an autotransformer with a tertiary winding: (A), (B), (C), (D) and (E) as in Fig. 4.26 but in per unit.

Modelling of transformers, phase shifters, static power plant and static load

293

cause corresponding changes in the HV to TV turns ratio and hence in tHT. Therefore, tHT is a function of tLT which is varied as part of the function to control the LV (132 kV) terminal voltage to a specified target value around a given deadband. In British practice, the tap-changer of such autotransformers is connected to the line-end or neutral-end of the ‘common’ winding. In North American and Asian practice, the tap-changer is normally connected to the line-end of the ‘series’ winding. Where the autotransformer does not have a tertiary winding or where the tertiary winding is unloaded, Eq. (4.44a)
(4.44c) can be reduced to VHðpuÞ ZHðpuÞ VLðpuÞ ZLðpuÞ 2 tH IHðpuÞ 2 5 2 tL ILðpuÞ 2 tH t tH tH L VHðpuÞ 2 ZHLðpuÞ IHðpuÞ 5

VLðpuÞ tLH

ZHLðpuÞ 5 ZHðpuÞ 1 ZLðpuÞ

(4.45a)

tLH 5

tL tH (4.45b)

VHðpuÞ 2 ZLH ðpuÞ tHL IHðpuÞ 5 VLðpuÞ tHL

ZLH ðpuÞ 5

 tL2  ZLðpuÞ 1 ZHðpuÞ 2 tH

tHL 5

tH tL

(4.45c) The equivalent circuits of Eq. (4.45a) and (4.45b) are shown in Fig. 4.27C
E. The latter two can be used to represent an autotransformer with ‘common’ winding or ‘series’ winding tap-changer, respectively.

Zero-sequence equivalent circuit in physical units and in per unit In addition to the primary, secondary and, where present, tertiary winding connections, the presence of an impedance in the neutral point of the star-connected ‘common’ winding must be taken into account in deriving the autotransformer’s zero-sequence equivalent circuit. We will ignore, as before, the shunt exciting impedance for the time being, and consider the general case of a star
star autotransformer with a tertiary winding with the common star neutral point earthed through an impedance ZE. The zero-sequence equivalent circuit, taking into account the presence of ZE in the neutral, can be derived in a similar way to the positive-sequence/negativesequence circuit. However, we will leave this as a minor challenge for the interested reader but we have drawn Fig. 4.28A to help the reader in such a derivation. For us, it suffices to say that in Eq. (4.36a) and (4.36b), ZC should be replaced by ZC 1 3ZE. Therefore, the zero-sequence equations are given by      VL 1 NH 2NL VH 1 NH 2NL 2NL IL 2 ðZC 13ZE Þ5 2NH IH 2 ZS 2 ðZC 13ZE Þ NL NH NH NL NL NH (4.46a)

294

Power Systems Modelling and Fault Analysis

(A)

H IH

R

IL 3IH

3(I H + I L )

IL

ZE

IH

I=0

r

IH + IL

VH

B

L

IT

I=0

IT

y VT = 0

IT

y

T r VT = 0

I=0

IL IH

b VT = 0

b

Y (B)

T H L

ZE H

IH(pu)

VH(pu)

3(NHL–1) 1 ZL(pu) + ZE(pu) NHL tH2

tH:1 3(NHL–1) 1 Z ZE(pu) H(pu) – 2 tH2 NHL

1 3 Z Z + E(pu) tH2 T(pu) NHL

1: tL

IT(pu) IM jXAir

RM

IL(pu) L VL(pu)

1: tT

T VT(pu)

jX M

Side T is innermost winding

Figure 4.28 Derivation of zero-sequence equivalent circuit of an autotransformer with a neutral-earthing impedance and a delta tertiary winding: (A) circuit under HV zero-sequence excitation; (B) zero-sequence equivalent circuit.

    1 NT2 VH 1 NH 2NL 02NT IT 2 ZTT 1 ðZC 13ZE Þ 5 2NH IH 2 ZS 2 ðZC 13ZE Þ NL NH NH NL NT NH (4.46b) Similar to the derivation of the positive-sequence impedances, the measured zero-sequence impedances between two terminals and the corresponding T equivalent branch impedances are ZHL 5 ZS 1

  NH 2NL 2 ðZC 1 3ZE Þ NL

(4.47a)

 2 NH ZTT NT

(4.47b)

ZHT 5 ZS 1 ZC 1 3ZE 1

Modelling of transformers, phase shifters, static power plant and static load

 ZLT 5 ZC 1 3ZE 1

NL NT

2

295

ZTT

(4.47c)

 NH 2 NL 3ZE NL

(4.48a)

  NH NH 2 NL 1 3ZE NL NL

(4.48b)

or  ZHðEÞ 5 ZH 2 Z 0LðEÞ

5 Z 0L

Z 0TðEÞ 5 Z 0T 1

NH 3ZE NL

(4.48c)

where ZH , Z’L and Z’T are as given in Eq. (4.40a)
(4.40c). Substituting Eq. (4.40a)
(4.40c) in Eq. (4.46a) and (4.46b), we obtain    VL 1 NH NH 2 NL 2 NL IL 2 Z 0L 1 3ZE NL NL NL NH    VH 1 NH 2 NL 2 NH IH 2 ZH 2 5 3ZE NH NL NH

(4.49a)

    1 NH VH 1 NH 2 NL 0 3ZE 5 2 NH IH 2 ZH 2 3ZE 0 2 NT IT 2 Z T 1 NL NH NL NH NH (4.49b) To convert Eq. (4.49a) and (4.49b) to pu, we will use the base quantities defined in Eq. (4.42a)
(4.42d) and define ZE(pu) 5 ZE/ZL(B), that is, we choose to use the L terminal as the base voltage for converting ZE to pu. It should be noted that the choice of the voltage base for ZE is arbitrary. Therefore, Eq. (4.49a) and (4.49b) become  VLðpuÞ 1 3ðNHL 2 1Þ 2 tL ILðpuÞ 2 ZLðpuÞ 1 ZEðpuÞ NHL tL tH  VHðpuÞ 1 3ðNHL 2 1Þ 5 2 tH IHðpuÞ 2 ZHðpuÞ 2 Z EðpuÞ 2 tH tH NHL

(4.50a)

 VTðpuÞ 1 3 2 tT ITðpuÞ 2 ZTðpuÞ 1 ZEðpuÞ NHL tT tH  VHðpuÞ 1 3ðNHL 2 1Þ 5 2 tH IHðpuÞ 2 ZHðpuÞ 2 ZEðpuÞ 2 tH tH NHL

(4.50b)

NHL 5

NH NL

(4.50c)

296

Power Systems Modelling and Fault Analysis

and VT(pu) 5 0. Eq. (4.50a)
(4.50c) are represented by the zero-sequence T equivalent circuit of Fig. 4.28B for a star
star autotransformer with a delta tertiary winding and the common neutral point earthed through an impedance ZE. Using Eq. (4.50a)
(4.50b), it is informative to observe how the neutral earthing impedance appears in the three H, L and T branches of the equivalent circuit ZEðpuÞ appearing in H branch terminal 5

2 3ðNHL 2 1Þ ZEðpuÞ 2 NHL

(4.51a)

ZEðpuÞ appearing in L branch terminal 5

3ðNHL 2 1Þ ZEðpuÞ NHL

(4.51b)

ZEðpuÞ appearing in T branch terminal 5

3 ZEðpuÞ NHL

(4.51c)

where ZE(pu) is in pu on L terminal voltage base. In the absence of a tertiary winding, the equivalent zero-sequence circuit cannot be obtained from Fig. 4.28 by removing the branch that corresponds to the tertiary winding. A shunt branch to the zero-voltage node may indeed still be required depending on the transformer core construction. This is covered in detail in Section 4.2.10. The case of a solidly earthed neutral is represented in Fig. 4.28 by setting ZE 5 0.

Zero-sequence equivalent circuit of an autotransformer with an isolated neutral The case of a star
star autotransformer with an isolated neutral, that is, ZE!N cannot be represented by Fig. 4.28 because the branch impedances of the equivalent circuit become infinite. This indicates no apparent paths for zero-sequence currents between the windings, although a physical circuit does exist. A mathematical solution is to convert the star or T circuit of Fig. 4.28 to delta or π and then setting ZE!N. However, we prefer the physical engineering approach to derive the equivalent zero-sequence circuit of such a transformer. Therefore, using the actual circuit of an unearthed star autotransformer with a delta tertiary winding, shown in Fig. 4.29A, we can write VH 2 VL 2 EH 5 ZS IH

(4.52a)

0 2 ET 5 ZTT IT

(4.52b)

NS IH 1 NT IT 5 NT IM  0

(4.52c)

EH NS 5 ET NT

(4.52d)

(A)

IH

NS ZS

EH

IL VL

ZC N C

3IH ZC VH IH

ZS

I=0 VT = 0

NT Z TT

ZC NC NC

VL

IL

ZS N S

NS

ET

IT

IL VL

IH

(B)

H

L

ZS

VH

T

IH

IL

VL

2

(C)

H

L

(D)

Z TT

NS : NT

ZS

VH

⎛ NS ⎞ ⎜⎜ N ⎟⎟ Z TT ⎝ T⎠

NS : NT

T

IH

VL

IL

Z Z HL = Z S- T =

VH IH H

IL

VL

L

2

⎛N ⎞ Z S + ⎜⎜ S ⎟⎟ Z TT ⎝ NT ⎠

T 2

(E)

V H(pu) H I H(pu)

⎛ N HL − 1 ⎞ ⎜⎜ ⎟⎟ Z S-T(pu) ⎝ N HL ⎠

1 : N HL

V L(pu) I L(pu)

L

T

Figure 4.29 Derivation of zero-sequence equivalent circuit of an autotransformer with an isolated neutral and a delta tertiary winding: (A) circuit under HV zero-sequence excitation; (B) single-phase representation; (C) tertiary impedance referred to series winding base turns; (D) H to L equivalent zero-sequence circuit and impedance in physical units; (E) H to L equivalent zero-sequence circuit in pu with impedance referred to H side; (F) H to L equivalent zero-sequence circuit in pu with impedance referred to L side; (G) π zero sequence-equivalent circuit of (E).

298

Power Systems Modelling and Fault Analysis

(F)

V H(pu) H I H(pu)

(N HL − 1)2 Z S-T(pu)

N LH : 1

V L(pu) I L(pu)

L

T Z = Z HL(pu–H side)

(G)

= VH IH H

N HL Z Z N HL − 1 HL(pu–H side) N HL − 1 Z S–T(pu) = N HL

N HL− 1 N HL

2

Z S–T(pu)

=

IL VL L 2 –N HL Z = Z N HL− 1 HL(pu–H side) = −(N HL− 1)ZS–T(pu)

T

Figure 4.29 (Continued).

Using Eq. (4.52b
d) in Eq. (4.52a), we obtain " VH 2 IH



NS ZS 1 NT

2

# ZTT 5 VL

(4.53a)

VH Z 5 j The measured zero-sequence leakage impedance in ohms ZHL IH VL50 V L Z or ZLH 5 j is given by IL VH50  Z Z 5 ZLH 5 ZS2T 5 ZS 1 ZHL

NS NT

2

ZTT 5 ZS 1 Z 0TT

(4.53b)

The zero-sequence leakage impedance measured from the H side or L side is the leakage impedance between the series and tertiary winding that is the sum of the series winding leakage impedance and the tertiary winding leakage impedance referred to the series side base turns. Fig. 4.29B shows a single-phase representation of the series and tertiary winding impedances and turns ratios. Fig. 4.29C shows the tertiary winding impedance referred to the series winding side and Fig. 4.29D represents the zero-sequence equivalent circuit and total zero-sequence equivalent impedance in actual physical units. The physical explanation for this result is as follows. The zero-sequence currents flow on the H side through the series windings only and, without transformation, flow out of the L side. This is possible because the delta tertiary winding circulates equal ampere-turns to balance that of the series winding but no zero-sequence currents flow in the common winding of the autotransformer. To convert Eq. (4.53a) to pu, we first calculate a pu value for ZS
T and note that this is the same whether referred to the series winding or tertiary winding because

Modelling of transformers, phase shifters, static power plant and static load

299

the base voltages in the windings are directly proportional to the number of turns in the windings. Using Eq. (4.42a)
(4.42d), Eq. (4.53a) can be rewritten as VHðpuÞ VHðBÞ 2 IHðpuÞ IHðBÞ ZS2TðpuÞ ZSðBÞ 5 VLðpuÞ VLðBÞ  2   ZSðBÞ NS NHL 21 2 Dividing by VH(B) and using 5 5 , we obtain NHL ZHðBÞ NS 1NC Z VHðpuÞ 2 ZHL ðpu referred to H sideÞ IHðpuÞ 5

1 VLðpuÞ NHL

(4.54a)

or Z VLðpuÞ 2 ZLH ðpu

referred to L sideÞ ILðpuÞ

5

1 VHðpuÞ NLH

(4.54b)

where the zero-sequence leakage impedance measured from the H side with the L side shorted is given by Z ZHL ðpu referred to H sideÞ 5

  NHL 21 2 ZS2TðpuÞ NHL

(4.54c)

and the zero-sequence leakage impedance measured from the L side with H side shorted is given by 2 Z ZLH ðpu referred to L sideÞ 5 ðNHL 21Þ ZS2TðpuÞ

(4.54d)

Eq. (4.54a) and (4.54b) are represented by the pu equivalent zero sequence circuits shown in Fig. 4.29E and F. The delta or π equivalent circuit can be derived using Fig. 4.6C. The result for Fig. 4.29E is shown in Fig. 4.29G using shunt impedances instead of admittances. Another similar test may be made but with the delta tertiary winding open to obtain the shunt zero-sequence magnetising impedance which may be low enough to be included depending on the core construction. This is dealt with in detail in Section 4.2.10.

4.2.6 Three-phase four-winding transformers General Four-winding transformers are not common in power transmission and distribution networks, but are found in special applications such as rectifier transformers in electric traction substations, converter transformers for variable-frequency drives in industrial power system applications, inverter transformers for connection of solar photovoltaic power inverters, static var compensators, modern static compensators (STATCOMS) and low-voltage electronic circuits.

300

Power Systems Modelling and Fault Analysis

Positive-sequence equivalent circuit of a four-winding transformer The principles used in Section 4.2.4 to derive the mathematical model and equivalent circuit of three-winding transformers can be extended to model four-winding transformers. However, we will only present the equivalent circuits here. Fig. 4.30A shows the electrical diagram representation of such a transformer. A four-winding transformer cannot be represented as a star or T equivalent circuit because the equivalent circuit should have NðN 2 1Þ=2 independent impedances when N . 3. Therefore, for N 5 4, the transformer should have six impedances. The equivalent circuit can be constructed from six short-circuit measurements of load losses and impedances between different pairs of windings. The measured impedances are denoted Z12, Z13, Z14, Z23, Z24, and Z34. For example, Z12 is measured with terminal 1 supplied, terminal 2 shorted, and terminals 3 and 4 open. Fig. 4.30B shows the equivalent circuit where winding 4 is assumed to be the innermost winding, the prime represents impedances referred to terminal 1 voltage base and ZM is the magnetising impedance. Fig. 4.30C shows the individual impedance equivalent circuit derived from the six impedance measurements. From Fig. 4.30C we can write Z12 5 Z1 1 Z2 1

Z5 ðZ5 1 2Z6 Þ ; 2ðZ5 1 Z6 Þ

Z14 5 Z1 1 Z4 1

Z6 ðZ6 1 2Z5 Þ ; 2ðZ5 1 Z6 Þ

Z 023 5 Z2 1 Z3 1

Z6 ðZ6 1 2Z5 Þ 2ðZ5 1 Z6 Þ

Z5 1 Z6 ; 2

Z 034 5 Z3 1 Z4 1

Z5 ðZ5 1 2Z6 Þ 2ðZ5 1 Z6 Þ

Z 024 5 Z2 1 Z4 1

Z13 5 Z1 1 Z3 1

Z 5 1 Z6 2 (4.55a)

The solution to the above system of six linear equations is Z1 5 Z2 5 Z3 5 Z4 5 Z5 5 Z6 5 ZC 5

 1
Z12 1 Z14 2 Z 024 2 ZC 2  1
Z12 2 Z13 1 Z 023 2 ZC 2  1
0 Z 2 Z 024 1 Z 034 2 ZC 2 23  1
2 Z13 1 Z14 1 Z 034 2 ZC 2  1
Z13 2 Z14 2 Z 023 1 Z 024 1 ZC 2  1
2 Z12 1 Z13 1 Z 024 2 Z 034 1 ZC 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðZ13 2 Z14 2 Z 0 23 1 Z 0 24 Þð 2 Z12 1 Z13 1 Z 0 24 2 Z 0 34 Þ

(4.55b)

Z2

I2

(A) E2 N2

V2

Z1

I1

1

2

I3 E1 N1

V1

3

Z3

E3 N3

V3

I4

4

Z4

E4 N4

V4

(B)

2

Z'23

3

Z'24

Z12

Z13

Z'34

Z14

1

4 ZM

(C)

Z6

Z2

2

Z5

Z3

Z5 Z6

Z1

3

Z4

4

1 ZM

(D)

1

Z1

Z2

2

Z5 Z3 3

Z4

4 ZM

Figure 4.30 Equivalent circuits for four-winding three-phase transformers: (A) divided flux representation of iron-cored four-winding transformer, (B) positive-sequence and negativesequence circuit representation of measured six pairs of leakage impedances; (C) positivesequence and negative-sequence equivalent circuit based on six derived leakage impedances, (D) positive-sequence and negative-sequence equivalent circuit based on five derived leakage impedances.

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Power Systems Modelling and Fault Analysis

For an N winding transformer, an alternative negative reactance model gives 2N 2 3 degrees of freedom, which is always less than the NðN 2 1Þ=2 possible short-circuit measurement combinations between pairs of windings for N . 3. This alternative model is sufficiently accurate and reproduces all measured leakage impedances except Z14, although the inaccuracy in this is very small and is of no practical significance, particularly for concentric windings. The alternative model is shown in Fig. 4.30D where Z1 5 Z2 5 Z3 5 Z4 5 Z5 5

 1
Z12 1 Z13 2 Z 023 2  1
Z12 2 Z13 1 Z 023 2  1
0 Z 23 2 Z 024 1 Z 034 2  1
2 Z 023 1 Z 024 1 Z 034 2  1
2 Z12 1 Z13 1 Z 024 2 Z 034 2

(4.56)

As discussed above, this model forces Z14 to be equal to Z14 5 Z13 1 Z 024 2 Z which should be very close to the actual measured value.

0

23

Zero-sequence equivalent circuit of a four-winding transformer As for two-winding and three-winding transformers, the zero-sequence equivalent circuit depends on the winding connections and neutral earthing arrangements and therefore will not be repeated here.

4.2.7 Three-phase earthing or zigzag transformers Three-phase earthing transformers are connected to unearthed networks in order to provide a neutral point for connection to earth. In the United Kingdom, this arises in delta-connected tertiary windings and 33-kV distribution networks supplied via star
delta transformers from solidly earthed 132- or 275-kV networks. Elsewhere, this can also arise on 11-kV distribution networks supplied via star
delta transformers from solidly earthed 132-kV networks. The purpose of using these transformers is to provide a low zero-sequence impedance path to the flow of zero-sequence short-circuit fault currents under unbalanced earthed faults. Such an aim can be achieved by the use of: 1. A star
delta transformer where the star winding terminals are connected to the network and the delta winding is left unloaded. The star neutral may be earthed through an impedance.

Modelling of transformers, phase shifters, static power plant and static load

303

2. An interconnected star or zigzag transformer which may also have a secondary star auxiliary winding to provide a neutral and substation supplies. The zigzag neutral may be earthed through an impedance. 3. An interconnected star or zigzag as (2) above but with a delta-connected auxiliary winding.

The star
delta transformer has already been covered. However, in this case, its positive-sequence and negative-sequence impedance is effectively the very large shunt exciting impedance. The zero-sequence impedance is the leakage impedance presented by the star
delta connection discussed previously. The interconnected star or zigzag transformer has each phase winding split into two halves and interconnected as shown in Fig. 4.31A and B. It should be obvious from the vector (A)

(B)

R

R

R 2

Y B

Y 2

B 2

B 2

R 2

B

B 2

ZE

Y 2

B 2

ZE

R 2

Y 2

R 2

Y

(C)

(D)

R 2

Z TZ

B 300 2 B 2

Y 2

Y 2

3Z E

R 2 Y 2

Figure 4.31 Interconnected-star or zigzag transformer: (A) connection to a three-phase network, (B) winding connections, (C) vector diagram and (D) zero-sequence equivalent impedance.

304

Power Systems Modelling and Fault Analysis

diagram (Fig. 4.31C) that this winding connection produces a phase shift of 30 degrees as if the zigzag winding had a delta-connected characteristic. Again, the positive-sequence/negative-sequence impedance presented by this transformer if positive-sequence/negative-sequence voltages are applied is the very large exciting impedance. However, under zero-sequence excitation, MMF balance or cancellation occurs between the winding halves wound on the same core due to equal and opposite currents flowing in each winding half. The zero-sequence impedance per phase is therefore the leakage impedance between the two winding’s halves. In practical installations, the impedance connected to the neutral is usually much greater than the transformer leakage impedance. The zigzag transformer zerosequence equivalent circuit is shown in Fig. 4.31D. The zigzag delta-connected transformer presents a similar zero-sequence equivalent circuit as the zigzag transformer shown in Fig. 4.31D. If the delta winding is loaded, the transformer positive-sequence/negative-sequence equivalent circuit includes a 90 degrees phase shift and is similar to that of a two-winding transformer with an impedance derived from the star equivalent of the three windings.

4.2.8 Single-phase traction transformers connected to threephase systems Single-phase traction transformers are used to provide electricity supplies to railway overhead catenary systems and are connected on the HV side to two phases of the transmission or distribution network. In the United Kingdom, the secondary nominal voltage to earth of these single-phase transformers is 25 kV and their HV voltage may be 132, 275 or 400 kV. The 132/25 kV are typically two-winding transformers, whereas modern large units derive their supplies from 400 kV and tend to be three-winding transformers, for example, 400/26.25
0
26.25 kV. Fig. 4.32 illustrates such connections. (A)

(B)

R

R

Y

Y

B

B HV

HV N N

LV

LV2

LV1

Figure 4.32 Single-phase traction transformer connected to a three-phase system: (A) two windings and (B) three windings.

Modelling of transformers, phase shifters, static power plant and static load

305

The model of two-winding single-phase traction transformers is the same as the single-phase model presented at the beginning of this chapter. Three-winding single-phase traction transformers can also be represented as a star equivalent of three leakage impedances. Under normal operating conditions, the traction load impedance referred to the primary side of the traction transformer appears in series with the transformer leakage impedance and is very large. However, under a shortcircuit fault on the LV side of the traction transformer, the transformer winding leakage impedance appears directly on the HV side as an impedance connected between two phases. This can be modelled as a phase-to-phase short-circuit fault through a fault impedance equal to the transformer leakage impedance. The proof of this statement is straightforward and is left for the interested reader.

4.2.9 Variation of transformer’s positive-sequence leakage impedance with tap position We have so far covered the modelling of the transformer with an off-nominal turns ratio due to the use of a variable tap-changer or fixed off-nominal ratio. We have shown that, for some models, the leakage impedance, both positive sequence and zero sequence, needs to be multiplied by the square of the off-nominal ratio. In addition, since the operation of the tap-changer involves the addition or cancellation of turns from a given winding, the variation of the number of turns causes variations in the leakage flux patterns and flux linkages. The magnitude of the leakage impedance variation and its direction in terms of whether it remains broadly unaffected, increases or decreases by the addition or removal of turns, depends on a number of factors. One factor is the tapped winding, for example, whether in the case of a two-winding transformer, the HV or LV winding is the tapped winding. For an autotransformer, it is whether the series or common winding and in the latter case, whether the common winding is tapped at the line, that is, output end or at the neutral end of the winding. Other factors are the range of variation of the number of turns and the location of the taps, for example, for a two-winding transformer, whether they are located in the body of the tapped winding itself, inside the LV winding or outside the HV winding. In summary, the addition of turns above nominal tap position or the removal of turns below nominal tap position may produce any of the following effects depending on specific transformer and tap-changer design factors: 1. The leakage impedance remains fairly constant and unaffected. 2. The leakage impedance may increase either side of the nominal tap position. 3. The leakage impedance may consistently decrease across the entire tap range as the number of turns is increased from minimum to maximum. 4. The leakage impedance may consistently increase across the entire tap range as the number of turns is increased from minimum to maximum.

From a network short-circuit analysis viewpoint, the leakage impedance variation across the tap range can be significant and this should be taken into account in the analysis. It is general practice in industry that the data of the impedance

306

Power Systems Modelling and Fault Analysis

Leakage impedance (% on rated 400 MVA)

(A) 36 35 34 33 32 31 30 29 28 27 26 25 24

400 kV Nominal Tap

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

460 kV

Leakage impedance (% on rated 50 MVA)

(B)

Tap position/number

340 kV

30 29 28 27 26 25 24

132 kV Nominal Tap

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

151.8 kV

Tap position/number

112.2 kV

Figure 4.33 Illustration of the variation of positive-sequence HV to LV leakage reactance with tap position for (A) a 500-MVA 400-kV/132-kV autotransformer and (B) a 50-MVA 132-kV/11-kV transformer.

variation across the entire tap range are usually requested by transformer purchasers and supplied by manufacturers in short-circuit test certificates. The measurement of transformer impedances is covered in Section 4.2.11. Fig. 4.33 illustrates the variations of positive-phase sequence HV to LV leakage reactance with tap position for a 500 MVA 400/132 kV autotransformer and a 50 MVA 132/11 kV transformer.

4.2.10 Practical aspects of zero-sequence impedances of three-phase transformers and effect of core construction In deriving the zero-phase sequence (zero-sequence) equivalent circuits for the various transformers we have presented so far, we have only considered the primary

Modelling of transformers, phase shifters, static power plant and static load

307

effects of the winding connections and neutral earthing impedances in the case of star-connected windings. We have temporarily neglected the effect of the transformer core construction and hence the characteristics of the zero-sequence flux paths on the zero-sequence leakage impedances. Contrary to what is generally published in most literature, we recommend that the effect of transformer core construction on the magnitude of the zero-sequence leakage impedance is taken into account in setting up transformer zero-sequence equivalent circuits in network models for use in short-circuit analysis. We consider this as international best practice that, in the author’s experience, can have a material effect on the assessment of short-circuit duties on circuit-breakers where margins are low. We will now consider this aspect in some detail and in order to aid our discussion, we will recall some of the basics of magnetic circuit theory. The relative permeability of transformer iron or steel core is hundreds of times greater than that of air. The reluctance of the magnetic core is its ability to oppose the flow of flux and is inversely proportional to its permeability. The reluctance and flux of a magnetic circuit are analogous to the resistance and current in an electric circuit. Therefore, transformer iron or steel cores present low-reluctance paths to the flow of flux in the core. Also, the core magnetising reactance is inversely proportional to its reluctance. Therefore, where the flux flows within the transformer core, the transformer magnetising reactance will have a very large value and will therefore not have a material effect on the transformer leakage impedance. However, where the flux is forced to flow out of the transformer core, for example, into the air, and complete its circuit through the tank and/or air/oil, then the effect of this external very high-reluctance path is to significantly lower the overall magnetising reactance. This will, in turn, substantially lower the leakage impedance of the transformer. It should be remembered that under positive-sequence/negative-sequence excitation, nearly all flux is confined to the iron or steel magnetic circuit, the magnetising current is very low (less than 1%) and hence the positive-sequence/negative-sequence magnetising reactance is very large (typically 10,000%) and has no practical effect on the positive-sequence/negative-sequence leakage impedance. We will now discuss the effect of various transformer core constructions on the zero-sequence leakage impedance.

Three-phase transformers made up of three single-phase banks Fig. 4.34 illustrates one of three single-phase banks making up the three-phase transformer. Both core-form type and shell-form type core constructions are illustrated. In both cases, the zero-sequence flux set up by zero-sequence voltage excitation can flow within the core in a similar way to positive-sequence flux. Consequently, the zero-sequence magnetising reactance will be very large and the zero-sequence leakage impedance of such transformers will be substantially equal to the positive-sequence leakage impedance.

308

Power Systems Modelling and Fault Analysis (A)

ΦZ

ΦP

φP = Positive sequence (PPS) flux φZ = Zero sequence (ZPS) flux (B) P ΦZ Φ

Figure 4.34 Zero sequence flux path in three single-phase banks three-phase transformer: (A) core-type with both limbs wound and (B) shell-type.

Three-phase transformers of five-limb core-form construction and shell-type core construction including seven-limb shell-form Figs. 4.35
4.37 illustrate three different core constructions; a five-limb core-form type, a standard and common shell-form type and a seven-limb shell-form type, respectively. The five-limb design is widely used in the United Kingdom and Europe, whereas the shell-type tends to be widely used in North America and parts of Asia. In all these designs, the zero-sequence flux set up by zero-sequence excitation can flow within the core and return in the outer limbs. Consequently, as in the case of three banks of single-phase transformers, the zero-sequence magnetising reactance will be very large and the zero-sequence leakage impedance of such transformers will be substantially equal to the positive-sequence leakage impedance. The measurement of sequence impedances will be dealt with in the next section. However, it is appropriate to explain now that positive-sequence leakage impedances are normally measured at nearly rated current, whereas zero-sequence impedances are measured, where this is done, at quite low current values, typically 10%
20% of rated current. However, under actual earth fault conditions in networks giving rise to sufficiently high zero-sequence currents and voltages on nearby transformers with currents similar to or exceeding those of the positive-sequence tests, the outer limbs and outer yokes of the five-limb core-form type construction shown in Fig. 4.35 may approach saturation. This is because these limbs and yokes carry 50% more flux and, in practical designs, have a cross-sectional area typically 40%
67% of the main core limbs and thus their flux density can reach around three times that in the main core limbs. This means that if the main limbs begin to

Modelling of transformers, phase shifters, static power plant and static load

V P/Z

V P/Z 0.5φZ

2 φY

309

V P/Z 0.5φZ

1 φR

3 φB

1.5φZ

1.5φZ

0.5φZ

0.5φZ

Under balanced positive sequence (PPS) voltage excitation: φR + φY + φB = 0 Under zero sequence (ZPS) voltage excitation: φR = φY = φB = φZ and ZPS fluxes remain inside the core

Figure 4.35 Zero sequence flux path in five-limb core-form three-phase transformer.

VZ 1

2 0.5φZ

φ

3

0.5φZ

Z

0.5φZ

φZ φZ

0.5φZ

φZ

φZ

φZ

φZ

0.5φZ

0.5φZ

Under zero sequence (ZPS) voltage excitation: ZPS fluxes remain inside the core

Figure 4.36 Zero sequence flux path in standard shell-type core three-phase transformer with centre limb wound.

saturate at 1 pu zero-sequence voltage, then the outer limbs and outer yokes would begin to saturate at around 0.3 pu zero-sequence voltage. Saturation causes flux to exit these limbs/yokes into the air/oil/tank, which lowers the zero-sequence magnetising reactance. The zero-sequence leakage impedance of such transformers that is measured in the factory at low current, as discussed above, may be somewhat higher than the actual value for core-form type design and substantially higher for common shell-form type design that exhibits much more variable core saturation than the

310

Power Systems Modelling and Fault Analysis

VZ φZ/2 φZ/2

φZ/2 φZ

φZ/2

φZ/2

φZ/2

φZ/2

φZ

φZ/2

φZ/2

φZ/2

Under zero sequence (ZPS) voltage excitation: ZPS fluxes remain inside the core

Figure 4.37 Zero sequence flux path in seven-limb shell-type core three-phase transformer.

limb-type core. Nonetheless, where the zero-sequence voltage does not exceed around 0.3 pu, these magnetising impedances remain relatively large in comparison with the leakage impedances and hence it is usually assumed that the positivesequence and zero-sequence leakage impedances of such transformers are equal.

Three-phase transformers of three-limb core-form construction Fig. 4.38 illustrates a three-phase transformer of three-limb core-form construction which is extensively used in the United Kingdom and worldwide. Under zerosequence voltage excitation, the zero-sequence flux must exit the core and its return path must be completed through the air with the dominant part being the tank then the oil and perhaps the core support framework. This zero-sequence flux will induce large zero-sequence currents through the central belt of the transformer tank and the overall effect of this very high reluctance path is to significantly lower the zerosequence magnetising reactance. This reactance is comparable to other plant values and may be 4
7 times the H to L positive-sequence leakage reactance (on rated MVA) for two- and three-winding transformers, and 6
10 times the H to L leakage reactance (on rated MVA) for autotransformers. These figures translate to typical values of around 70%
100% for two and threewinding transformers and 150% for autotransformers, for a typical H to L impedance on rating of 20%. These figures can be compared with 5000%
20,000% (which correspond to 2% and 0.5% no-load currents) for the corresponding positive-sequence magnetising reactance. Therefore, the effect of the tank of three-limb core-form three-phase transformer can be treated as if the transformer has a virtual magnetic delta-connected winding. In addition, the effect of the tank in the three-limb coreform construction is nonlinear with the reluctance increasing with increasing current that is the magnetising reactance decreasing with increasing current. This and the practice that the zero sequence leakage impedance measurements are made at low current value, mean that, the actual zero-sequence leakage impedance of three-limb transformers may be slightly lower than the measured values.

Modelling of transformers, phase shifters, static power plant and static load

311

(A) VR

VY φR

VB φY

φB

Under balanced positive phase sequence (PPS) voltage excitation: PPS flux remain inside the core and φR + φY + φB = 0 (B) VZ Air 3φZ φZ

φZ

φZ Oil

Tank

Tank wall

Under zero phase sequence (ZPS) voltage excitation, ZPS flux of each phase φZ must exit the core and return in air and tank/oil φR = φY = φB = φZ

Figure 4.38 Flux paths of three-limb core-form three-phase transformer: (A) positivesequence flux and (B) zero sequence flux path.

4.2.11 Measurement of sequence impedances of three-phase transformers Transformers are subjected to a variety of tests by their manufacturers to establish correct design parameters, quality and suitability for a 40
50-year service life. The tests from which the transformer impedances are calculated are iron loss test, noload current test, load loss test, short-circuit positive-sequence impedance and sometimes zero-sequence impedance tests. The iron loss and no-load current tests are carried out simultaneously. The rated voltage at rated frequency is applied to the LV winding, or tertiary winding where present, with the HV winding open-circuited. The shunt exciting impedance or admittance is calculated from the applied voltage and no-load current, and its resistive part is calculated from the iron loss. The magnetising reactance is then calculated from the resistance or conductance and impedance or admittance. The no-load

312

Power Systems Modelling and Fault Analysis

current is obtained from ammeter readings in each phase and usually an average value is taken. The iron loss would be the same if measured from the HV side but the current would obviously be different as it will be in inverse proportion to the turns ratio. Also, the application of rated voltage to the LV or tertiary is more easily obtainable and the current magnitude would be larger and more conveniently read. The measured no-load loss is usually considered equal to the iron loss by ignoring the dielectric loss and the copper loss due to the small exciting current. The load loss and short-circuit impedance tests are carried out simultaneously. A low voltage is applied to the HV winding with the LV winding being shortcircuited. The applied voltage is gradually increased until the HV current is equal to the full load or rated current. The applied voltage, measured current and measured load loss are read. As the current is at rated value, the short-circuit impedance voltage in pu is given by the applied voltage in pu of rated voltage of the HV winding. The load-loss would be the same if measured from the LV side. It is usual for transformers having normal impedance values to ignore the iron loss component of the load loss and the measured loss to be taken as the load loss. The reason being that the iron loss is negligible in comparison with the load loss at the reduced shortcircuit test voltage. Where high-impedance transformers are specified, however, the iron loss at the short-circuit test voltage may not be negligible. In this case, the load loss can be found after reading the load-loss from the short-circuit test by removing the shortcircuit (i.e., convert the test to an open-circuit test at the short-circuit applied voltage) reading the iron-loss and subtracting it from the load-loss. The resistance is calculated from the measured load loss and the reactance is calculated from the impedance voltage and the resistance. The measurements made in volts and amps allow the calculation of the impedances in ohms but for general power system analysis these need to be converted to per unit on some defined base. The equations presented below are straightforward to derive and we recommend that the reader does so. From the open-circuit test, the magnitude of the exciting admittance is defined as YMðpu on rated MVAÞ 5

No load MVA SRatedðMVAÞ

or

YMðpu on rated MVAÞ 5

INo load ðAÞ IRated ðAÞ (4.57a)

where No load MVA 5

pffiffiffi 3VLL rated ðkVÞ INo load ðkAÞ

(4.57b)

or on a new MVA base SBase YMðpu on SBase Þ 5

No load MVA SBase ðMVAÞ

or

YMðpu on SBase Þ 5

INo load ðAÞ IBase ðAÞ

(4.57c)

Modelling of transformers, phase shifters, static power plant and static load

313

Also YMðpuÞ 5 GMðpuÞ
jBMðpuÞ 5

1 1
j RMðpuÞ XMðpuÞ

(4.58a)

and the conductance/resistance part of the exciting admittance is calculated as Fe½32Phase loss ðMWÞ SRatedðMVAÞ 1 RMðpu on rated MVAÞ 5 GMðpu on rated MVAÞ GMðpu on rated MVAÞ 5

and (4.58b)

where Fe[3-phase loss (MW)] is the three-phase iron loss. The magnetising susceptance and reactance are calculated as BMðpu on rated MVAÞ 5

1 SRated ðMVAÞ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðNo load MVAÞ2 2 ðFe32Phase lossðMWÞ Þ2 (4.58c)

XMðpu on rated MVAÞ 5

1 BMðpu on rated MVAÞ

(4.58d)

or on a new MVA base SBase GMðpu on SBase Þ 5

Fe½32Phase loss ðMWÞ SBase ðMVAÞ

and BMðpu on SBase Þ 5

1 SBase ðMVAÞ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðNo load MVAÞ2 2 ðFe32Phase lossðMWÞ Þ2

From the short-circuit test, Z(pu) 5 R(pu) 1 jX(pu) is defined as Zðpu on rated MVAÞ 5

(4.59a)

the

VLL testðkVÞ VLL ratedðkVÞ

short-circuit

leakage

(4.59b) impedance

(4.60a)

This is equal to the applied test voltage in pu which is why the term short-circuit impedance voltage is used by transformer manufacturers. This applies at nominal and off-nominal tap position. The resistive part of the leakage impedance is calculated as Rðpu on rated MVAÞ 5

Load Loss½32PhaseðMWÞ SRated ðMVAÞ

(4.60b)

314

Power Systems Modelling and Fault Analysis

or on a new MVA base SBase Rðpu on SBase Þ 5

Load Loss½32Phase ðMWÞ 3 SBaseðMVAÞ ðSRatedðMVAÞ Þ2

(4.60c)

and the magnetising reactance is calculated as 5 Xðpu on rated MVA rÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi h i2  pffiffiffi 2 1 3VLL testðkVÞ IRatedðkAÞ 2 LoadLoss32PhaseðMWÞ SRatedðMVAÞ

(4.60d)

or on a new MVA base SBase Xðpu on SBase Þ 5 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi hpffiffiffi i2  2 SBaseðMVAÞ 3VLL test ðkVÞ IRated ðkAÞ 2 LoadLoss32PhaseðMWÞ  2 SRatedðMVAÞ (4.60e) Whereas positive-sequence impedance measurement tests for a number of tap positions such as minimum, maximum, nominal and mean have always been the norm, it is only recently becoming standard industry practice to similarly specify zerosequence impedance tests. In the case of zero-sequence impedance measurements, the three-phase terminals of the winding from which the measurement is made are joined together and a single-phase voltage source is applied between this point and neutral. Voltage, current and load loss may be measured, although the latter is normally not measured; the zero-sequence resistance is calculated from the load loss and input zero-sequence current. The zero-sequence impedance is calculated from the applied voltage and a third of the source current because the source zero-sequence current divides equally between the three phases. The basic principle of the positive-sequence and zerosequence leakage impedance tests is illustrated in Fig. 4.39 for a two-winding transformer with neutral-earthed star
delta winding connections. It should be noted that, in the zero-sequence test, if the LV winding is delta-connected, it must be closed but not necessarily short-circuited.

Positive-sequence and zero-sequence impedance tests on twowinding transformers The positive-sequence impedance test is HV
LV//N, that is, the HV winding is supplied with the LV winding phase terminals joined together, that is, shortcircuited and connected to neutral. The same test may be done for the zerosequence impedance if the secondary winding is delta connected. However, for a Ynyn transformer, the zero-sequence equivalent circuit that correctly represents a three-limb core transformer, is a star or T equivalent circuit since we represent the tank zero-sequence contribution as a magnetic delta tertiary winding. Therefore, at

Modelling of transformers, phase shifters, static power plant and static load

(A)

315

HV

IR

VR

Ir

LV

Iy P Z HL = V R I R

Ib

VB IB IY

VY (B)

Z = 3V Z HL

HV

Z

IZ 3

IZ

LV

IZ V

Z

IZ 3 IZ 3

Figure 4.39 Illustration of short-circuit impedance test circuits on a star-delta two winding transformer: (A) positive-sequence leakage impedance test, HV winding supplied, LV winding short circuited with all three phases measured and an average taken and (B) zerosequence leakage impedance test, HV winding terminals joined and supplied, LV winding closed.

least three measurements are required. It is the author’s practice that four zerosequence tests are specified to enable a better estimate of the impedance to neutral branch that represents the tank contribution. The zero-sequence tests are HV
N with LV phase terminals open-circuited, LV
N with HV phase terminals opencircuited, HV
LV//N with LV phase terminals short-circuited and LV
HV//N with HV phase terminals short-circuited.

Positive-sequence and zero-sequence impedance tests on threewinding transformers In order to derive the star or T equivalent positive-sequence circuit for such transformers from measurements, let the three windings be denoted as 1, 2 and 3, where: Z12 is the positive-sequence leakage impedance measured from winding 1 with winding 2 short-circuited and winding 3 open-circuited, and P12 is the measured load loss. Z13 is the positive-sequence leakage impedance measured from winding 1 with winding 3 short-circuited and winding 2 open-circuited, and P13 is the

316

Power Systems Modelling and Fault Analysis

measured load loss. Z23 is the positive-sequence leakage impedance measured from winding 2 with winding 3 short-circuited and winding 1 open-circuited, and P23 is the measured load loss. In three-winding transformers, at least one winding will have a different MVA rating but all the pu impedances must be expressed on the same MVA base. From the above tests, and with the impedances in ohms referred to the same voltage base, and load losses in MW, we have Z12 5 Z1 1 Z 02 Ω

R12 5

P12ðMWÞ kV12 Ω MVAðTestÞ MVAðTestÞ

R12 5 R1 1 R02 Ω (4.61a)

Z13 5 Z1 1 Z 03 Ω

R13 5

P13ðMWÞ kV12 Ω MVAðTestÞ MVAðTestÞ

R13 5 R1 1 R03 Ω (4.61b)

0

12 N 1 Z 023 5 @ A Z23 5 Z 02 1 Z 03 Ω N2

R023 5

P23ðMWÞ kV12 Ω MVAðTestÞ MVAðTestÞ

R023 5 R02 1 R03 Ω (4.61c)

Solving Eq. (4.61a)
(4.61c), we obtain Z1 5

 1
Z12 1 Z13 2 Z 023 Ω 2

R1 5

 1
R12 1 R13 2 R023 Ω 2

(4.62a)

Z 02 5

 1
Z12 1 Z 023 2 Z13 Ω 2

R02 5

 1
R12 1 R023 2 R13 Ω 2

(4.62b)

Z 03 5

 1
Z13 1 Z 023 2 Z12 Ω 2

R03 5

 1
R13 1 R023 2 R12 Ω 2

(4.62c)

X1 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z12 2 R21 Ω

and X 02 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi Z 02 2 2 R02 2 Ω

X 03 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi Z 03 2 2 R03 2 Ω (4.62d)

The conversion of the T equivalent circuit impedances R1 1 jX1, R0 2 1 jX0 2 and R0 3 1 jX0 3 from ohms to pu must be based on one common MVA base. Regarding the zero-sequence impedances, the zero-sequence T equivalent circuit that correctly represents a three-limb core three-winding transformer even when all

Modelling of transformers, phase shifters, static power plant and static load

317

windings are star-connected, is a star or T equivalent circuit since we represent the tank zero-sequence contribution as a magnetic delta tertiary winding. Therefore, as in the case of the two-winding transformer, the zero-sequence tests are HV
N with LV phase terminals open-circuited, LV
N with HV phase terminals open-circuited, HV
LV//N with LV phase terminals short-circuited and LV
HV//N with HV phase terminals short-circuited.

Autotransformers Generally, two cases are of most practical interest for autotransformers. The first is an autotransformer with a delta-connected tertiary winding, and the second is without a tertiary winding. With a tertiary winding, the positive-sequence tests are similar to those of a three-winding transformer and without a tertiary winding, the positive-sequence test is similar to that of a two-winding transformer. The zerosequence equivalent circuit that correctly represents a three-limb core autotransformer, with or without a tertiary winding, is a star or T equivalent circuit since we represent the tank zero-sequence contribution as a magnetic delta tertiary winding. Therefore, the zero sequence tests are: With delta tertiary

Without tertiary

Comments

HV
tertiary//N LV
tertiary//N HV
LV//tertiary//N LV
HV//tertiary//N

HV
N LV
N HV
LV//N LV
HV//N

LV phase terminals open-circuited HV phase terminals open-circuited LV phase terminals short-circuited HV phase terminals short-circuited

4.2.12 Examples Example 4.1 The rated and measured test data of a three-phase two-winding transformer are: Rated data: 120 MVA, Ynd1 or 275 kV star with neutral solidly earthed and 66 kV delta, 6 15% HV tapping range Test data: No-load current measured from LV terminals at rated voltage 5 7A Iron loss 5 103 kW, Full load loss 5 574 kW Short-circuit positive-sequence impedance measured from HV side with LV side shorted 5 122.6 Ω Short-circuit zero-sequence impedance measured from HV side with LV side shorted 5 106.7 Ω Calculate the positive-sequence and zero sequence equivalent circuit parameters in pu on 100 MVA base

YðMÞpu on 100 MVA 5

pffiffiffi 3 3 66 kV 3 0:007 kA 5 0:008 pu 5 0:8% 100 MVA

GðMÞpu on 100 MVA 5

0:103 5 0:00103 pu or 0:103% 100

318

Power Systems Modelling and Fault Analysis

and RðMÞpu on 100 MVA 5 BðMÞpu on 100 MVA 5

1 5 970:9 pu or 97; 087% 0:00103 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð0:008Þ2 2 ð0:00103Þ2 5 0:00793 pu or 0:793%

and XðMÞpu on 100 MVA 5

1 5 126:1 pu or 12; 610% 0:00793

ZHL PPS pu on 100 MVA 5

122:6 5 0:162 pu or 16:2% ð275Þ2 =100

RHL PPS pu on 100 MVA 5

0:574 3 100 5 0:00398 pu or 0:398% ð120Þ2

XHL PPS pu on 100 MVA 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð0:162Þ2 2 ð0:00398Þ2 5 0:16195 pu or 16:195%

It is worth noting that the positive-sequence X/R ratio is 16.195/0.398 5 40.7 ZHL ZPS pu on 100 MVA 5

106:7 5 0:141 pu or 14:1% ð275Þ2 =100

This is 87% of the positive-sequence impedance. In the absence of zero-sequence load loss measurement, the zero-sequence X/R ratio may be assumed equal to the positive-sequence X/R ratio of 40.7. Therefore, RHL ZPS pu on 100 MVA 5 0:346% XHL ZPS pu on 100 MVA 5 14:095% If the positive-sequence impedance were measured from the LV side, its ohmic value would be 122.6 3 (66 kV/275 kV)2 5 7.06 Ω and its pu value would also be equal to 16.2%.

Example 4.2 Consider a three-phase two-winding transformer with the following rated data: 45 MVA, 132/33 kV Ynd11 (star solidly earthed neutral
delta), HV winding on-load tapchanger with a tapping range of
20% to 1 10%. positive-sequence impedance tests were carried out at all 19 tap positions with the HV terminals supplied by a three-phase voltage

Modelling of transformers, phase shifters, static power plant and static load

319

source and the LV winding short-circuited. The applied test voltages and measured full load currents on the HV side at minimum, nominal and maximum tap positions are: Tap position (%) 1 ( 1 10%) 7 (nominal) 19 (
20%)

Phase
phase voltage (V) 17,800 15,560 11,130

Current (A) 179 197 246

Calculate the short-circuit impedance at each tap position both in ohms and in pu on MVA rating: Tap position 1 Rated tap voltage 5 132 3 1.1 5 144.2 kV. 17:8 3 100 5 12:3% and impedance in ohm Impedance in pu on 45 MVA 5 pffiffiffi 145:2 17; 800= 3 5 57:4 Ω: 5 179 57:4 Alternatively, impedance in pu on 45 MVA 5 3 100 5 12:3%: ð145:2Þ2 =45 Tap position 7 Rated tap voltage 5 132 3 1 5 132 kV, the impedance is 45.6 Ω, or 11.8% on 45 MVA. Tap position 19 Rated tap voltage 5 132 3 0.8 5 105.74 kV, the impedance is 26.1 Ω or 10.6% on 45 MVA. The reader should notice that the apparent small variation in impedance in percentage terms masks the significant change in the impedance when actual units of ohms are used. Also, the base impedance in each case is different because it is dependent on the rated tap voltage which is a function of tap position.

Example 4.3 The rated and measured test data for a three-phase three-winding transformer are: Primary HV winding 60.6 MVA, 132 kV Yn star solidly earthed, outer winding. Secondary LV1 winding 30.3 MVA, 11.11 kV yn star solidly earthed. Secondary LV2 winding 30.3 MVA, 11.11 kV yn star solidly earthed, inner winding. The construction is a three-limb core-form. The no load current measured from side LV2 at nominal voltage is 15 A. The short-circuit impedance and load loss measurement test data at nominal tap position is: Test Positive-phase sequence (positive-sequence) HV/LV1 shorted, LV2 open HV/LV2 shorted, LV1 open LV1/LV2 shorted, HV open Zero-phase sequence (zero-sequence) HV/LV1 and LV2 open

Volts

Amps

Full load loss (kW)

43,843 43,750 5,940

132.6 132.6 1,517

201.5 on 30.3 MVA 202.2 on 30.3 MVA 413 on 30.3 MVA

19,760

120




320

Power Systems Modelling and Fault Analysis

Derive the positive-sequence and zero-sequence impedances and corresponding transformer equivalent circuits and assume
10% turns HV winding tap position. The network base voltages are 132 and 11 kV. Use 60.6 MVA as a common base and neglect the core losses. 11;000 XM 5 pffiffiffi 5 423:39 Ω 3 3 15 pffiffiffi 43;843= 3 5 190:9 Ω ZHL1 5 132:6 pffiffiffi 43;750= 3 5 190:5 Ω ZHL2 5 132:6

ZL1L2 5

X 0M

pffiffiffi 5;940= 3 5 2:26 Ω 1;517

60:6 XM 5 pffiffiffi 5 212 pu 5 21; 200% 3 3 11 3 0:015 ZHL1 5

190:9 3 100 5 66:4% 1322 =60:6

ZHL2 5

ZL1L2 5

190:5 3 100 5 66:2% 1322 =60:6

2:26 100 5 113% 112 =60:6



 132 2 5 423:39 3 5 60;968 Ω 11

"  2 # 1 132 5 27:98 Ω 190:9 1 190:5 2 2:26 3 ZH ðΩÞ 5 2 11 0 12 2 3 1 132 Z 0L1 ðΩÞ 5 4190:9 1 2:26 3 @ A 2 190:55 5 162:92 Ω 2 11 0 12 2 3 1 132 Z 0L2 ðΩÞ 5 4190:5 1 2:26 3 @ A 2 190:95 5 162:52 Ω 2 11 The impedances in Ω in the star or T equivalent of the three-winding transformer have all been referred to the same voltage base. ZH ð%Þ 5

27:98 3 100 5 9:73% 1322 =60:6

Z 0L1 ð%Þ 5

162:92 3 100 5 56:66% 1322 =60:6

Z 0L2 ð%Þ 5

162:52 3 100 5 56:52% 1322 =60:6

Modelling of transformers, phase shifters, static power plant and static load

201:5 3 RHL1 5

103 3

202:2 3

103 3

321

413 3

5 3:82 Ω RHL2 5 5 3:83 Ω RL1L2 5 132:62 132:62 2 0 13 14 132 3:82 1 3:83 2 0:0598 3 @ A5 5 2 0:48 Ω RH ðΩÞ 5 2 11 0 12 2 3 14 132 RL1 ðΩÞ 5 3:82 1 0:0598 3 @ A 2 3:835 5 4:3 Ω 2 11 0 12 2 3 14 132 RL2 ðΩÞ 5 3:83 1 0:0598 3 @ A 2 3:825 5 4:31 Ω 2 11

103 3

15172

5 0:0598 Ω

Negative RH should not alarm the reader since the T equivalent is a fictitious mathematical model! RH 5

2 0:48 3 100 5 2 0:167% ð132Þ2 =60:6

RL2 5

4:31 3 100 5 1:5% ð132Þ2 =60:6

RL1 5

4:3 3 100 5 1:49% ð132Þ2 =60:6

The reactances of the T equivalent are given by XH 5 27:97 Ω or 9:7%

X 0L1 5 162:86 Ω or 56:6%

X0L2 5 162:46 Ω or 56:5%

XAir 5 ðX0L1 1 X0L2 Þ=2 5 162:66 Ω or 56:55% Remaining core part of XM 5 60; 968 Ω
162:66 Ω 5 60; 805:3 Ω 5 21; 200%
56:55% 5 21; 143:4% Z ZHN 5

19; 760 5 494 Ω 120=3

Z 5 ZHN

494 3 100 5 171:8% 1322 =60:6

0

Z Z ZHN 5 ZH 1 ZL2 1 jXAir 1 ZM Z ZM 5 494 2 27:98 2 162:52 2 94:21 5 208:3 Ω Z 5 171:8 2 9:73 2 56:52 2 33:1 5 72:45% ZM

Since impedance measurements are available at only nominal tap position, we will use Figs. 4.21A and 4.24E (with zero neutral-earthing impedances) for nominal tap and adjusted to three 0:9 3 132 5 0:9 pu pu tap ratios as in Fig. 4.21A. The HV, LV1 and LV2 pu tap ratios are t1 5 132 and t2 5 t3 5 11:11=11 5 1:01 pu. The positive-sequence and zero-sequence equivalent circuits are shown in Fig. 4.40.

322

Power Systems Modelling and Fault Analysis

(A)

(1.49 + j56.6)% (4.3 + j162.86)Ω

1:1.01

VL1

0.9:1 1:1.01 VH

(−0.167 + j9.7)% (−0.48 + j27.97)Ω

(1.59 + j56.5)% (4.31 + j162.46)Ω

VL2

j 33.1% j 94.21Ω j 21,200% j60,968Ω

Side L2 is innermost winding (B)

(1.49 + j 56.6)% (4.3 + j162.86)Ω

1:1.01

VL1

0.9:1 1:1.01

VH

(−0.167 + j9.7)% (−0.48 + j27.97)Ω

(1.59 + j56.5)% (4.31 + j162.46)Ω

VL2

j 33.1% j 94.21Ω j72.455% j 208.3Ω

Side L2 is innermost winding Figure 4.40 Example 4.3 three-winding transformer impedance tests: (A) positivesequence equivalent circuit and (B) zero-sequence equivalent circuit.

Example 4.4 A three-limb core autotransformer has the following rated data: 275/132 kV, 240 MVA, star
star solidly earthed neutral and a 60 MVA/13 kV tertiary stabilising delta-connected winding. The HV to LV positive-sequence impedance is 18.95% on a 240 MVA rating. Zero-sequence impedance tests were carried out with the delta winding closed and open to identify the change in zero-sequence impedances if the transformer had no tertiary winding and to identify the effect of the core construction. The measured zero-sequence impedances at nominal tap position in both cases are: Measured impedance (Ω/phase) Zero-sequence (zero-sequence test) Z ZH2N , H supplied, L open

Delta closed 107.3

Delta open 741.4

Z ZL2N , L supplied, H open

14.3

200.5

Z ZH2L==N , Z ZL2H==N ,

H supplied, L short-circuited

50.2

50.0

L supplied, H short-circuited

6.8

13.5

Modelling of transformers, phase shifters, static power plant and static load

323

Derive general equations for the zero-sequence T equivalent circuit impedances and calculate the impedances for the above two cases in percentage on 100 MVA: Z ZH2N ð%Þ 5

Z ZH2N ðΩ Þ 3 100 2 275 =100

Z ð%Þ 5 ZH2L==N

Z ZL2N ð%Þ 5

Z ZL2N ðΩÞ 3 100 1322 =100

Z ð%Þ 5 ZL2H==N

Z ZH2L==N ðΩÞ

2752 =100 Z ZL2H==N ðΩ Þ

1322 =100

3 100

3 100

Using these equations, we obtain Zero-sequence impedance (% on 100 MVA) Z ZH2N Z ZL2N Z ZH2L==N

Delta closed 14.2 8.2 6.6

Z ZL2H==N

Delta open 98.0 115.0 6.6

3.9

7.8

The zero-sequence T equivalent circuit has three unknowns and therefore only three tests are required, for example, the first three above. However, the fourth test, where available, can be used to improve the prediction of the shunt branch impedance. The four tests above are illustrated in Fig. 4.41A from which we have the following:

(A)

H

Z LZ

Z HZ

L

H

ZHZ

Z ZN Z HZ - N

N Z LZ

Z H ZH

L

Z ZN

(B)

H 6.35%

L

Z ZN

N

Z HZ - L //N

Z LZ

Z LZ

Z H ZH

N

7.84% N Delta tertiary closed

L

Z ZN

N 0.37% L

Z LZ- N

H – 4.52%

Z LZ- H//N 12.49% L 102.6%

N Delta tertiary open

Figure 4.41 Zero-sequence leakage impedance tests on an autotransformer: (A) zero-sequence leakage impedance tests; (B) zero-sequence equivalent circuits.

324

Power Systems Modelling and Fault Analysis

Z ZH2N 5 ZHZ 1 ZNZ Z ZH2L==N 5 ZHZ 1

Z ZL2N 5 ZLZ 1 ZNZ

ZLZ ZNZ 1 ZNZ

Z ZL2H==N 5 ZLZ 1

ZZL

ZHZ ZNZ ZHZ 1 ZNZ

Substituting ZHZ and ZLZ from the first two equations into the second two, we obtain ZNZ ð1Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Z Z Z 5 ZH2N ZL2N 2 ZL2H==N

ZNZ ð2Þ

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffi Z z Z 5 zL2H ZH2N 2 ZH2L==N

and a mean value is calculated as ZNZ ðmeanÞ 5

ZNZ ð1Þ 1 ZNZ ð2Þ 2

The remaining H and L branch impedances are Z 2 ZNZ ðmeanÞ ZHZ 5 ZH2N

Z ZLZ 5 ZL2N 2 ZNZ ðmeanÞ

The calculated ZHZ ; ZLZ and ZNZ for the closed and open delta tertiary are shown in Fig. 4.41B.

Example 4.5 The autotransformer with a tertiary winding used in Example 4.4 has a neutral earthing reactor of 10 Ω connected to its neutral. Calculate the earthing impedance values in percent on 100 MVA base that would appear in each branch of the autotransformer zero sequence T equivalent circuit. The transformer turns ratio is 275/132 5 2.0834. The percent impedance value of the earthing reactor is ZEZ ð%Þ 5

10 3 100 5 5:739% 1322 =100

The earthing impedances appearing in the three branches of the zero-sequence T equivalent circuit are ZEZ ðappearing in H terminalÞ 5

2 3 3 ð2:0834 2 1Þ 3 5:739 5 2 4:297% 2:08342

ZEZ ðappearing in L terminalÞ 5

3 3 ð2:0834 2 1Þ 3 5:739 5 8:953% 2:0834

ZEZ ðappearing in T terminalÞ 5

3 3 5:739 5 8:264% 2:0834

Modelling of transformers, phase shifters, static power plant and static load

4.3

325

Sequence modelling of quadrature booster and phase-shifting transformers

4.3.1 Background Quadrature booster (QB) and phase-shifting (PS) transformers are widely used in transmission and subtransmission power systems for controlling the magnitude and direction of active power flow mainly over parallel circuits in order to increase the power transfer capability across boundaries or interfaces. A single-core design is used for lower MVA rating and lower phase shift requirement and characterised by its simplicity and economy. However, it suffers from disadvantages that include the overvoltages that may be imposed on the tap-changer due to its position in series with the main line, as well as having a virtually zero impedance at nominal tap which may not be desirable if some short-circuit current limitation is additionally required. By far, the most commonly used design is the two-core design in either a single tank or two tanks, where a large MVA rating and phase angle shift are required. Fig. 4.42 shows a 400 kV QB rated at 2000 MVA and 6 11.3 degrees phase shift. Figs. 4.43A and 4.44A show QB and PS one-line diagrams, respectively. Simplified representations of the three-phase connections of typical large QBs and PSs are shown in Figs. 4.43B and 4.44B, respectively. The designs shown are those where both QB and PS consist of two transformers; a shunt transformer connected in star
star and a series transformer connected in delta
series star. The primary

Figure 4.42 2000-MVA 400-kV 6 11.3 degrees quadrature booster transformer consisting of a shunt transformer and a series transformer in separate tanks.

326

Power Systems Modelling and Fault Analysis (A)

Vi(RYB)

V o(RYB)

R 1Y1B1

R 2 Y2 B2

Shunt transformer

(B)

Series transformer

Series winding

R1

Input

Output R2

Y1

Y2

B1

B2

Exciting winding

Booster winding

(C)

VoB ΔViB

ΔViR = tViR ViR VoR φ φ

φ

ViB VoY

ViY ΔViY

Regulating winding

Figure 4.43 Quadrature booster transformer: (A) one-line diagram, (B) winding connections and (C) open-circuit vector diagram.

winding of the shunt transformer is called the exciting winding and the secondary is called the regulating winding. The primary delta of the series transformer is called the booster winding and the secondary is called the series winding because it is in series with the line. The principles of operation of the QB and PS are similar. The control of the magnitude and direction of active power flow on the line is achieved by varying the phase angle shift across the series winding. The phase shift and its variation are obtained by an on-load (under-load) tap-changer acting on the regulating winding and deriving a variable voltage component across two phases, for example, Y and B. This voltage is in quadrature with the input voltage. It is then injected by the booster transformer delta winding across the series winding third phase, for example, phase R, as shown in Figs. 4.43C and 4.44C. One important difference to notice between a QB and a PS is that for a QB, the input voltage of the shunt transformer is derived from one side of the series transformer so that the output voltage is slightly higher than the input voltage. However, for a PS, the shunt transformer input voltage is derived from the mid-point on the series winding of the series transformer so that the magnitudes of the input and output voltages remain equal.

Modelling of transformers, phase shifters, static power plant and static load

327

(A) Series transformer

Vi(RYB)

V o (RYB) R 2 Y2 B2

R1Y1B1

Shunt transformer

(B)

(C)

Y1

Output R2 Y2

B1

B2

R1

Input

Series winding

Booster Exciting

winding

winding

ViR

ΔVisR = tVisR VoR

φ VoB

φ

ViY

φ

ΔVisB ViB

VoY

ΔVisY

Regulating winding

Figure 4.44 Phase shifting transformer: (A) one-line diagram, (B) winding connections and (C) open-circuit vector diagram.

4.3.2 Positive-, negative- and zero-phase sequence modelling of QB and PS transformers Positive-sequence equivalent circuit model The derivation of QB and PS detailed equivalent circuits needs to take into account both the shunt and series transformers, their winding connections, tapchanger operation, complex turns ratio and leakage impedance variations with tap position. However, we will instead present a simplified treatment but one that is still sufficient for use in practice in positive-sequence power flow, stability and short-circuit analysis. The analysis below is based on the vector diagrams of Figs. 4.43C and 4.44C. These allow us to represent the QB or PS as an ideal transformer with a complex turns ratio in series with an appropriate impedance. This ratio is in series with a single effective leakage impedance representing both the series and shunt transformers leakage impedances including the effect of the tapchanger on the shunt transformer’s impedance. This representation is shown in Fig. 4.37.

328

Power Systems Modelling and Fault Analysis

From the QB vector diagram shown in Fig. 4.43C, the phase R open-circuit output voltage phasor is given by VoðRÞ 5 ViðRÞ ð1 1 jtÞ

(4.63a)

t 5 ΔViðRÞ =ViðRÞ

(4.63b)

t is the magnitude of the injected quadrature voltage in pu of the input voltage and ϕ is the corresponding no-load phase angle shift. Now, Eq. (4.63a) can be written as N5

VoðRÞ 5 1 1 jt 5 V i ðR Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 ðtan ϕÞ2 ejϕ

t 5 tan ϕ

(4.63c) (4.63d)

N is defined as the complex no-load turns ratio. Similar equations apply for the Y and B phases. In practice, the rated QB no-load phase angle shift does not exceed around 6 20 degrees. For example, at a 20 degrees angle, Eq. (4.63d) shows that the no-load output voltage of the QB reaches 1.064 pu of the input voltage. Similarly, from the PS vector diagram shown in Fig. 4.44C, and using t 5 ΔVi (R)/Vi(R) defined in Eq. (4.63b), we can write for phase R t VoðRÞ 5 ViSðRÞ 1 1 j 2

(4.64a)

t ViðRÞ 5 ViSðRÞ 1 2 j 2

(4.64b)

where ViS(R) is the input voltage at the mid-point of the series winding. The complex turns ratio of the PS, at no load, is given by

N5

VoðRÞ V i ðR Þ

t 2 5 ejϕ 5 t 12j 2

ϕ 5 2tan21

11j

t 2

(4.64c)

(4.64d)

In practice, the rated PS no-load phase angle shift can be up to around 6 60 degrees. Let Ze be the effective or equivalent QB or PS leakage impedance appearing in series with the line or the network where the device is connected. The QB and PS

Modelling of transformers, phase shifters, static power plant and static load

(A)

1: N

Ii

Ze

Io

V′

Vi

Vo

φ = tan−1(t)

For a Quadrature Booster: N = 1 + j tanφ For a Phase Shifter:

N=

329

φ = 2 tan−1(t/2)

e jφ

(B)

Vi

Y io = e − jΦ/Z e

Ii

Io

Vo

Y oi = e jΦ/Ze Yii = (1 − e − j Φ)/Z e

Yoo = (1 − e j Φ)/Ze

Asymmetrical π equivalent circuit representation for a Phase Shifter

Figure 4.45 Equivalent circuit for a quadrature booster or a phase shifter: (A) positivesequence equivalent circuit and (B) asymmetrical π equivalent circuit for a PS.

are both represented by the same positive-sequence equivalent circuit shown in Fig. 4.45. The positive-sequence admittance matrix for this circuit can be easily derived as in the case of a two-winding off-nominal ratio transformer but with an important difference that the turns ratio is a complex number. From Fig. 4.45, we can write V0 5N Vi

Vi Ii 5 V 0 ð 2Io Þ

Ii  5 2N Io

Vo 2 NVi 5 Ze Io

From these equations, the following admittance matrix is obtained 

  2   Ii 5 N =Ze Io 2 N=Ze



2 N =Ze 1=Ze



Vi Vo

(4.65a)

which, using Eq. (4.63c) and (4.63d), can be written for a quadrature booster as 

  Ii 1 1 ðtan φÞ2 =Ze 5 Io 2 ð1 1 j tan φÞ=Ze

2 ð1 2 j tan φÞ=Ze 1=Ze



or, using Eq. (4.64b), can be written for a phase shifter as

Vi Vo

(4.65b)

330

Power Systems Modelling and Fault Analysis



 Ii 1=Ze 5 Io 2 ejφ =Ze

2 e
jφ =Ze 1=Ze



Vi Vo

(4.65c)

An equivalent magnetising impedance branch that represents the magnetising impedances of both series and shunt transformers can be connected to the output side of the QB/PS if required. However, this is an approximation that is acceptable for power frequency studies with a normal system voltage range but not high overvoltage studies, inrush current or low-frequency studies. The above two admittance matrices for a QB and a PS cannot be represented by a physical π equivalent circuit because the complex turns ratio has resulted in matrices that are nonsymmetric, that is, the off-diagonal transfer terms are not equal. Nevertheless, noting that an asymmetrical nonphysical π equivalent circuit is simply a mathematical tool, such a circuit is shown in Fig. 4.45B for a PS. A similar π equivalent for a QB can be derived but this is left for the reader.

Negative-sequence equivalent circuit model Since the QB and PS are static devices, the negative-sequence impedance is the same as the positive-sequence impedance. However, we have shown that the positive-sequence model includes a complex turns ratio which is a mathematical operation that introduces a phase angle shift in the output voltage (and current) with respect to the input voltage (and current). The mathematical derivation of the phase shift in the QB or PS negative-sequence model requires a detailed representation of all windings of the shunt and series transformers, which, as outlined above, is not presented here. However, we will use vector diagrams to show that the phase shift in the negative-sequence circuit is equal to that in the positive-sequence circuit but reversed in sign. Fig. 4.46i shows the positive-sequence open-circuit vector diagrams of a QB and a PS and the resultant phase angle shift φ in each case. Fig. 4.46ii shows the negative-sequence vector diagrams of a QB and a PS and the resultant phase angle shift in each case. It is clear that with the reversed negative-sequence phase rotation RBY, the injected series quadrature voltage vector is also reversed, for both QB and PS. Therefore, if the positive-sequence phase angle shift is φ, then the negativesequence phase angle shift is
φ. Fig. 4.47A shows the QB or PS negativesequence equivalent circuit model where the complex ratio is given by

N5

8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > > 1
jtan φ 5 1 1 ðtan φÞ2 e2jφ > < > > e
jφ > :

φ 5 tan21 ðtÞ 0 1 t φ 5 2 tan21 @ A 2

for a QB for a PS

(4.66)

Using the positive-sequence admittance matrices for a QB and a PS, the negative-sequence admittance matrices are easily obtained by replacing φ with
φ in Eq. (4.65b) and (4.65c).

Modelling of transformers, phase shifters, static power plant and static load

(A)

331

(B)

ΔViRP ViRP

P VoR

Φ

(i) PPS

P ΔVisR

ViRP

P VoR

Φ

(i) PPS

P VisYB

P ViYB

ViBP

ΔViRN N VoR

–Φ

P VisB

ViYP

P VisY

ViRN

N ΔVisR

N VoR

(ii) NPS

–Φ

(ii) NPS

N ViYB

ViRN

N VisYB

ViYN

ViBN

N VisY

ViRZ ViYZ ViBZ

(iii) ZPS

N VisB Z VisR Z VisY Z VisB

(iii) ZPS

Figure 4.46 Positive-sequence (PPS), negative-sequence (NPS) and zero-sequence (ZPS) vector diagrams for a quadrature booster or a phase shifter: (A) QB vector diagram and (B) PS vector diagram.

(A)

1: N

Ii

Ze V′

Vi

Vo

For a Quadrature Booster: N = 1 − j tanφ For a Phase Shifter:

(B)

V i Ii

N=

Zi

Io

e−jφ

φ = tan−1(t) φ = 2 tan−1(t/2)

Z o Io Vo

ZN

Figure 4.47 Negative-sequence and zero-sequence equivalent circuits for a quadrature booster or a phase shifter: (A) negative-sequence equivalent circuit and (B) zero-sequence equivalent circuit.

332

Power Systems Modelling and Fault Analysis

Zero-sequence equivalent circuit model We have already stated that the basic principle of operation of the QB or PS is to inject a voltage across, say phase R, series transformer series winding that is in quadrature with the phase R input voltage. In the positive-sequence and negativesequence circuits, this is achieved by deriving the injected voltage so as to be in phase or out of phase with the voltage difference between phases Y and B, as shown in Fig. 4.46i and 4.46ii, respectively. However, since the three zero-sequence voltages are in phase with each other, a quadrature voltage component cannot be produced. Therefore, the phase shift between the input and open-circuit output voltages (and currents) in the zero-sequence circuit is zero. Hence, the complex turns ratio in the zero-sequence equivalent circuit for both a QB and a PS becomes unity. The QB or PS practical zero-sequence equivalent circuit, however, is different from the positive-sequence and negative-sequence equivalent circuits. The zerosequence equivalent circuit is dependent on the winding connections of the series and shunt transformers, their core construction and whether the shunt transformer has any delta-connected tertiary winding. In a similar way to a three-winding transformer, or a three-limb core autotransformer with or without a tertiary winding, the general zero-sequence equivalent circuit of a QB or a PS is a star or T equivalent as shown in Fig. 4.47B. The shunt branch in this equivalent represents either the impedance of a tertiary winding or the stray air path zero-sequence flux through the tank/oil. In addition, in some QBs and PSs, both series and shunt transformers are located in the same tank so that the zero-sequence flux can link both windings either directly or through the tank. In practice, the QB or PS zero-sequence impedances are derived from a series of manufacturer works-tests and this is described in the next section. The zero-sequence admittance matrix model of the QB or PS is given by 

 Ii Y 5 i Io 2 Yio

2 Yio Yo



Vi Vo

(4.67a)

where Yi 5

1 Zo ZN Zi 1 Zo 1 ZN

Yio 5

1

Zi Zo ðZi 1 Zo Þ 1 ZN

Yo 5

1 Zi ZN Zo 1 Zi 1 ZN (4.67b)

4.3.3 Measurement of sequence impedances of QB and PS transformer Like transformers, quadrature boosters and phase shifters are subjected to a variety of works-tests by their manufacturers. As most QBs and PSs are three-phase twotransformer devices, tests are usually carried out on each transformer alone and then on the combined units. No-load loss, load loss, short-circuit positive-sequence

Modelling of transformers, phase shifters, static power plant and static load

333

and zero-sequence impedance tests are carried out on the combined units. The positive-sequence impedance and load loss tests are carried out simultaneously with the output end of the series transformer supplied by three-phase voltage sources and the shunt transformer input terminals short-circuited. The ohmic value of the impedance is the applied voltage divided by the measured current. The pu value of the impedance must be expressed on the voltage base of the winding energised or supplied under the tests. Under such a test, the measured positive-sequence impedance is the series combination of the leakage impedances of the series and shunt transformers. This can be easily visualised by referring back to the QB and PS winding connections shown in Figs. 4.43 and 4.44. Therefore, this impedance is the QB or PS effective positive-sequence impedance Ze that appears in series with the line as seen from the QB or PS series transformer output end. The series transformer does not have a tap-changer so its leakage impedance, when measured from its output terminals, is constant. However, when measured from the same terminals, the shunt transformer leakage impedance varies with tap position as the tap-changer acts on the shunt transformer regulating winding. The variation of tap position varies the magnitude of the injected quadrature voltage and, to a reasonably good approximation, the positive-sequence/negative-sequence leakage impedance will vary with the square of the injected quadrature voltage t. Therefore,  ZeP ðtÞ 5 ZeN ðtÞ 5 ZSeries 1

t

tmax

2

3 ZShunt at end tap

(4.68)

which can be expressed in terms of the phase angle shift as follows: 8 0 12 > > tan φ > > A 3 ZShunt at end tap for a Quad Booster > ZSeries 1 @ > > tan φMax > > > > 0 1 32 2 > > < φ P N 6 tan@ A 7 Ze ðφÞ 5 Ze ðφÞ 5 7 6 > 2 > 7 6 > > 7 3 ZShunt at end tap for a Phase Shifter 6 > 0 1 Z 1 > Series 7 6 > > 7 6 > φ > Max > 4 A5 > tan@ > : 2 (4.69) where φMax is the QB or PS rated or maximum phase shift design value, φ the actual phase shift in operation and ZShunt at end tap is the shunt transformer leakage impedance measured at the end tap position that is at maximum quadrature voltage or maximum phase shift. It should be noted that Ze (φ 5 0) is equal to ZSeries because the booster delta winding of the series transformer is effectively shortcircuited by the tap-changer when a zero quadrature voltage is injected. Like transformers, the zero-sequence leakage impedance tests are carried out by applying a single-phase voltage source to the three-phase input or output terminals joined together. Usually three or four such tests may be carried out to derive the

334

Power Systems Modelling and Fault Analysis

star or T zero-sequence equivalent circuit. Some tests that may be carried out are described below where T represents a tertiary winding, where present in the shunt transformer, and N represents neutral. Reference is made to Figs. 4.43 and 4.44 to see the designated QB and PS terminals. Terminals supplied

Designation

Comments

R1Y1B1 terminals joined together R2Y2B2 terminals joined together R1Y1B1 terminals joined together R2Y2B2 terminals joined together

Zol
T//N

R2Y2B2 terminals open-circuit, tertiary closed, where present R1Y1B1 terminals open-circuit, tertiary closed, where present R2Y2B2 terminals short-circuited, tertiary closed, where present R1Y1B1 terminals short-circuited, tertiary closed, where present

Zo2
T//N Zol
2//T//N Zo2
1//T//N

4.3.4 Examples Example 4.6

16 15 14 13 12 11 10 9 8 7 6

Measured Nominal tap position

Calculated

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39

Effective positive sequence impedance (%)

A 400 kV 2000 MVA throughput QB consisting of a series and a shunt transformer and an onload tap-changer with 39 tap positions. The measured positive-sequence impedances of the series transformer, and shunt transformer at end tap position, are 7.01% and 7.04% respectively, both on rated 2000 MVA. The QB’s effective positive-sequence impedance has been measured in the factory at all tap positions. However, where only a few measurements are available, or even only nominal-tap impedance measurement, we will use Eq. (4.69) to provide a good approximation of the variation of the effective QB impedance with tap position. The result is shown in Fig. 4.48.

Figure 4.48 Example 4.6 comparing actual impedance measurements of quadrature booster transformer at all tap positions or phase shifts, and our model of Eq. (4.69).

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335

Example 4.7 Consider the QB of Example 4.6. The QB has a rated quadrature injected voltage of 6 0.2 pu of input voltage. The following positive-sequence and zero-sequence impedance tests were carried out. In the positive-sequence tests, the output, that is, series transformer terminals R2Y2B2 are supplied and the input of the shunt transformer terminals R1Y1B1 are short-circuited. Positive-sequence rest Tap 1 Tap 20 (nominal) Tap 39 Zero-sequence test Zo1
T//N Tap 1 Tap 20 (nominal) Tap 39 Zo2
T//N Tap 1 Tap 20 (nominal) Tap 39 Zo2
1//T//N Tap 1 Tap 20 (nominal) Tap 39

Phase
phase voltage (V) 56,800 28,049 56,920 Voltage (V)

Current (A) 2887 2886 2888 Current (A)

20,817 20,823 20,818

260.1 260.4 260.4

21,395 21,360 21,377

260.8 260.4 260.5

506 507 506

260.0 260.7 260.4

Calculate the open-circuit rated phase shift and output voltage of the QB for a 1 pu input voltage. Also, calculate the effective positive-sequence impedance of the QB and the zerosequence star equivalent impedances at nominal, minimum and maximum tap positions. The open-circuit rated phase shift is tan21( 6 0.2) 5 6 11.3 degrees. The QB range of rated injected voltage or phase shift is from 2 0.2 pu corresponding to 2 11.3 degrees phase shift to 1 0.2 pu corresponding to 1 11.3 degrees phase shift. The open-circuit output voltage at rated pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi phase shift is 1 1 0:22  1:02 pu: The positive-sequence impedance is calculated as follows: Tap 1: 56:8 3 100 5 14:2%: pffiffiffi Impedance in pu on 2000 MVA 5 400 ð56;800= 3Þ=2;887 Alternatively, impedance in % on 2000 MVA 5 3 100 5 14:2%: 4002 =2000 Tap 20: Impedance is 7.01% on 2000 MVA. Tap 39: Impedance is 14.23% on 2000 MVA. We note that for this QB, the effective positive-sequence impedance almost doubles between nominal tap position (tap 20) and maximum/minimum tap positions (tap 39 and tap 1). The results of the calculated zero-sequence impedances are given as: Zero-sequence impedance Tap 1 Tap 20 Tap 39

Zo1
T//N (Ω/phase)

Zo2
T//N (Ω/phase)

240.1 239.9 239.8

246.1 246.1 246.2

Zo2
1//T//N (Ω/phase) 4.84 4.83 4.83

336

Power Systems Modelling and Fault Analysis

We note that the zero-sequence impedances of the equivalent star are effectively unaffected by variation of tap position. The impedances of the individual T branch equivalent circuit at nominal tap position in percent on 2000 MVA and 400 kV are: Z1Z 5 7:582% where 1 corresponds to the shunt transformer’s input end R1Y1B1; Z2Z 5 2 0:294% where 2 corresponds to the series transformer’s output end R2Y2B2; ZNZ 5 300:0% where N corresponds to neutral or zero voltage reference in this case.

Example 4.8 A 132 kV, 300 MVA throughput PS has a rated phase shift of 6 30 degrees and the shunt transformer unit includes a tertiary winding. The positive-sequence impedance measured at nominal tap position is 2% on 100 MVA at 132 kV. The zero-sequence impedances of the PS are largely unaffected by variation of tap position and the zero-sequence impedance test data are as follows: Zo1
T//N (Ω/phase)

Zo2
T//N (Ω/phase)

Zo1
2//T//N (Ω/phase)

14.38

14.9

3.16

Zo2
1//T//N (Ω/phase) 3.16

Calculate the zero-sequence star equivalent circuit impedances in percent on 100 MVA. Like autotransformers, we have four tests but only three unknowns so we will calculate two values of the shunt impedance ZN in the star equivalent and obtain a mean value as follows: ZNZ ð1Þ 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 14:38 3 ð14:9 2 3:16Þ 5 13 Ω

ZNZ ð2Þ 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 14:9 3 ð14:38 2 3:16Þ

and 5 12:93 Ω

and the mean value is ZNZ ðmeanÞ 5 12:96 Ω Z Zð1Þ 5 14:38 2 12:96 5 1:42 Ω

Z Zð2Þ 5 14:9 2 12:96 5 1:94 Ω

The impedance values in percent on 100 MVA are ZðZ1Þ ð%Þ 5

1:42 3 100 5 0:81% 1322 =100

ZðZ2Þ ð%Þ 5

1:94 3 100 5 1:11% 1322 =100

ZNZ ðmeanÞ ð%Þ 5

12:96 3 100 5 7:44% 1322 =100

Modelling of transformers, phase shifters, static power plant and static load

4.4

337

Sequence modelling of series and shunt reactors and capacitors

4.4.1 Background Series reactors are widely used in power systems for power flow control but mostly as fault current limiters. The latter application will be covered in Chapter 10, Control of high short-circuit fault currents and fault current limiters. Series reactors are generally used in electric distribution, subtransmission and transmission systems as well as power stations and industrial power systems. Air core reactors are usually used at nominal system voltages up to and including 36 kV, but iron-cored reactors are generally used at higher voltages. The former is usually coreless and magnetically or electromagnetically shielded and the latter are usually oil-filled and gapped. Three-phase gapped iron core reactors look in appearance similar to a three-phase three-limb core transformer. However, reactors have only one winding on each limb. The function of the gaps in the core is to lower the flux density so that under high-current conditions, the core barely enters into saturation and the reactor’s impedance remains substantially constant. In practice a small decrease in the impedance from the rated current value is typical. Magnetically shielded coreless reactors use shields to surround the windings in order to provide a return path for the winding flux and if these saturate, impedance reduction occurs. Shunt reactors are widely used in power networks at various voltage levels for controlling and limiting transient and steady-state voltages. Shunt reactors are costeffective and robust reactive compensation devices used in absorbing surplus reactive power supply in cable systems where they may be switched in and out of service with the cable itself by the cable circuit-breakers. Many aspects of their design are similar to those of three-limb series reactors except that, at higher voltages, many shunt reactors are star-connected with their neutral point either solidly or impedance earthed. Shunt capacitors are usually mechanically switched, that is, by circuit-breakers and are very cost-effective reactive compensation devices that are extensively used in transmission, distribution and industrial power networks. They provide a source of reactive power supply and help to improve network voltage stability, voltage levels and power factors. Shunt capacitors can be star-connected with the neutral isolated or solidly earthed, or delta-connected, depending on the nominal system voltage. Series capacitors are also widely used in transmission and distribution networks. In distribution, they are mainly used to improve the voltage profile of heavily loaded feeders. In transmission, they are mainly used to increase network power transfer capabilities by improving generator and network transient stability, damping of power oscillations, network voltage stability or load sharing on parallel circuits.

4.4.2 Modelling of series reactors In a three-limb design of a three-phase series reactor, the winding around each limb represents one phase and the flux in any phase does not link that in the other two phases. Therefore, there is practically little or no mutual inductive coupling between

338

Power Systems Modelling and Fault Analysis

VR VY VB

IR

Z

IY

Z

IB

Z

V R' V Y' V B'

Figure 4.49 Equivalent circuit of a three-limb core series reactor.

the phases of the reactor. In addition, the phase windings are practically identical by design. A three-phase equivalent circuit is shown in Fig. 4.49 where Z 5 R 1 jX is the leakage impedance per phase. The series phase voltage drop is given by 2 3 2 3 2 32 3 VR 2 V 0R ΔVR Z 0 0 IR 4 VY 2 V 0 5 5 4 ΔVY 5 5 4 0 Z 0 54 IY 5 (4.70a) Y VB 2 V 0B 0 0 Z ΔVB IB and its inverse is given by 2 3 2 IR 1=Z 0 4 IY 5 5 4 0 1=Z 0 0 IB

32 3 ΔVR 0 0 54 ΔVY 5 1=Z ΔVB

(4.70b)

or in the sequence component reference frame expressed in terms of phase R, 32 2 P3 2 3 1=Z 0 0 ΔVRP IR 76 6 N7 6 7 1=Z 0 5 4 ΔVRN 5 (4.70c) 4 IR 5 5 4 0 IRZ

0

0

1=Z

ΔVRZ

Therefore, the positive-sequence, negative-sequence and zero-sequence admittances are all self-terms, as the phase admittances, and all are equal to 1/Z. In some designs, the three-phase windings of the reactor are vertically stacked on top of each other, as illustrated in Fig. 4.50. In such an arrangement, the flux in one phase can link with the other phases and mutual inductive coupling between the three phases can therefore exist with the middle phase seeing more linked flux than the outer phases. This means that whilst the selfimpedances are equal, the mutual impedances between the outer phases and that between the adjacent phases are not equal. Therefore, from Fig. 4.50, we define the following: ZRR 5 ZYY 5 ZBB 5 ZS

ZRY 5 ZYR 5 ZYB 5 ZBY 5 ZM1

Therefore, the series phase impedance matrix is written as 2 3 ZS ZM1 ZM2 ZRYB 5 4 ZM1 ZS ZM1 5 ZM2 ZM1 ZS

ZRB 5 ZBR 5 ZM2

(4.71a)

Modelling of transformers, phase shifters, static power plant and static load

339

R

R′ Z RY

Y Z RB

Y′ Z YB

B B′

Figure 4.50 Series reactor with a vertically stacked three-phase windings.

The series sequence impedance matrix can be calculated using ZPNZ 5 H21ZRYBH, where H is the transformation matrix. Thus, 2 3 3ZS 2 ðZM2 1 2ZM1 Þ 2h2 ðZM2 1 ZM1 Þ 2 hðZM2 2 ZM1 Þ 16 7 ZPNZ 5 4 2hðZM2 2 ZM1 Þ 3ZS 2 ðZM2 1 2ZM1 Þ 2 h2 ðZM2 2 ZM1 Þ 5 3 2 hðZM2 2 ZM1 Þ 3ZS 1 ðZM2 1 2ZM1 Þ 2 h2 ðZM2 1 ZM1 Þ (4.71b) This sequence impedance matrix is full and is not symmetric, that is, it shows unequal intersequence mutual coupling, which, as discussed in Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, eliminates the fundamental advantage of using the sequence component reference frame. However, in practice, for many vertical reactor designs, the vertical insulators between the phases are deliberately chosen to be sufficiently long that the mutual coupling between the adjacent phases is quite small. In such cases, two practical options exist. The simplest, which is the one mostly used, is to ignore the small mutual coupling and set ZM1 5 ZM2 5 0 so that the sequence impedance matrix reduces to 2 3 ZS 0 0 (4.71c) ZPNZ 5 ZRYB 5 4 0 ZS 0 5 0 0 ZS The second option, where the mutual impedances are known from factory measurements, is to set them all equal to a mean value ZM(mean) 5 (ZM1 1 ZM2)/2. In this case, the series phase impedance matrix of Eq. (4.71a) becomes 2 3 ZS ZMðmeanÞ ZMðmeanÞ 6 7 ZS ZMðmeanÞ 5 (4.71d) ZRYB 5 4 ZMðmeanÞ ZMðmeanÞ

ZMðmeanÞ

ZS

340

Power Systems Modelling and Fault Analysis

Reactance (pu)

1.05 1 0.95 0.9 0.85 0.8

0

1

2 3 Current (pu)

4

5

Figure 4.51 Relationship between reactance and current for a magnetically shielded 1500MVA, 400-kV series reactor.

The corresponding series sequence impedance matrix is given by 2 6 ZPNZ 5 4

ZS 2 ZMðmeanÞ

0

0

0

ZS 2 ZMðmeanÞ

0

0

0

ZS 1 2ZMðmeanÞ

3 7 5

(4.71e)

where ZP 5 ZN 5 ZS
ZM(mean) and ZZ 5 ZS 1 2ZM(mean). The corresponding series sequence admittance matrix is given by 3

2

1 6 ZS 2 ZMðmeanÞ 6 6 6 PNZ 0 Y 56 6 6 6 4 0

0

0

1 ZS 2 ZMðmeanÞ

0

0

1 ZS 1 2ZMðmeanÞ

7 7 7 7 7 7 7 7 5

(4.71f)

Fig. 4.51 shows the effect of saturation on the reactance of a magnetically shielded oil-immersed 1500-MVA, 400-kV series reactor. Load loss and impedance measurements are carried out in the factory and these are used to calculate the resistance and reactance of the reactor, both self and mutuals. By design, series reactors are highly efficient devices and have X/R ratios, depending on nominal system voltage, that typically range from 100 to 500, the latter is typical for a 400-kV series reactor.

4.4.3 Modelling of shunt reactors and capacitors In three-phase shunt reactors, there is practically negligible mutual coupling between the phases. Three-phase shunt reactors and capacitors are each represented

Modelling of transformers, phase shifters, static power plant and static load

(A)

R

S MVA V kV

Y

R

S MVA V kV

Y B

B Y = −jB

(B)

341

Y

Y

Y = jB Y

Y

Figure 4.52 Star-connected three-phase shunt reactors and capacitors: (A) three-phase shunt reactor and (B) three-phase shunt capacitor.

as shown in Fig. 4.52 as three identical shunt susceptances. The resistive part of the shunt reactor is normally neglected since, by design, these are required to have very high X/R ratios in the order of 100
400 to minimise active power losses. The internal series resistance of shunt capacitors is negligibly small. The phase and sequence admittance matrix of a shunt reactor or shunt capacitor is given by 2 3 Y 0 0 YRYB 5 YPNZ 5 4 0 Y 0 5 (4.72a) 0 0 Y where, Y is the shunt admittance, B is the shunt susceptance, and Y 5 jB for a shunt capacitor, Y 5 2 jB for a shunt reactor. However, under unbalanced conditions, and where the shield or steel tank of a shunt reactor saturate, the zero sequence impedance of the reactor can drop significantly. The phase impedance matrix is given in Eq. (4.71d) and the sequence admittance matrix is given in Eq. (4.71f). The positive-sequence, negative-sequence and zero-sequence susceptances are equal. The magnitude of the susceptance is derived from the three-phase rated MVAr data of the reactor or capacitor as follows: B ðS Þ 5

SRatingðMVArÞ 2 VLL ðkVÞ

(4.72b)

The pu susceptance is given by SRatingðMVArÞ 2 VLL ðkVÞ BðpuÞ 5 SBase ðMVAÞ 2 VLL ðkVÞ Bð%Þ 5

SRating ðMVArÞ 3 100 SBaseðMVAÞ

(4.72c)

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Power Systems Modelling and Fault Analysis

For example, the susceptance of a 150-MVAr shunt reactor or capacitor in percent on a 100-MVA base is equal to 150%. Fig. 4.53 shows two typical shunt capacitor banks used in 132 and 400-kV power systems.

Figure 4.53 Three-phase mechanically switched capacitor banks: (A) 60-Mvar, 132-kV bank and (B) 225-Mvar, 400-kV bank (with an RLC damping network).

Modelling of transformers, phase shifters, static power plant and static load

343

Example 4.9 Consider a 40 MVAr 132 kV oil-immersed shunt reactor. The following factory measurements were carried out. All voltages are phase to ground. 1. Three-phase source Applied voltage

Measured current

76,210 V

174.95 A

2. Single-phase source Applied voltage (R)

Current (R)

Voltage (Y)

Voltage (B)

62,900 V

166.0 A

10,616 V

10,546 V

3. Zero-sequence impedance measurement (single-phase source) Applied voltage

Measured current

14,724 V

180.1 A

76;210V 5 435:6 Ω 174:95A 10;616V 5 63:95 Ω The mutual impedance between phases R and Y is 166A 10;546V 5 63:53 Ω The mutual impedance between phases R and B is 166A Average value of mutual impedance 63:53 Ω 3 3 14; 724V 5 245:26 Ω The zero-sequence impedance is calculated as 180:1A All measurements on other phases yielded similar results. The self-impedance of the balanced phase impedance matrix is calculated as 435:6 1 ð 2 63:53Þ 5 372:07 Ω. We can check the zero-sequence impedance measurement using the measured positive-sequence impedance and mutual impedance giving 372:07 1 2ð 2 63:53Þ 5 245:01 Ω which is practically similar to the measured value of 245:26 Ω. The balanced phase impedance matrix and3 corresponding sequence impedance matrix 2 2 3 372:07 2 63:53 2 63:53 435:6 0 0 can be written as 4 2 63:53 372:07 2 63:53 5Ω, 4 0 435:6 0 5Ω. 2 63:53 2 63:53 372:07 0 0 245:26 The positive-sequence impedance is

4.4.4 Modelling of series capacitors Types of series capacitor schemes There are several three-phase series capacitor scheme designs in use in power systems, four of which are depicted in Fig. 4.54.

General modelling aspects of series capacitors The modelling of series capacitor schemes in power flow analysis is straightforward because the magnitude of current flowing through the capacitor would be within its appropriate rated design value. These rated values include not only the highest continuous current, but also the highest short-term overload value permitted and the

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Power Systems Modelling and Fault Analysis

(A)

(B) C

C

V

D D

G VC

TG CB CB CB (D)

(C) C

C

V

V

D D CB

T

Figure 4.54 Typical series capacitor schemes: (A) basic spark gap scheme, (B) modern scheme with varistor and triggered spark gap, (C) modern scheme with varistor but no spark gap and (D) modern thyristor controlled scheme with varistor.

highest permitted value under power system oscillation conditions. The capacitors are therefore represented as three identical series reactances and the ohmic value of the reactance is known from the rated data. The resistive part of the series capacitor is negligibly small. Therefore, the sequence and phase admittance matrix of a threephase series capacitor is given by 2

Y YPNZ 5 YRYB 5 4 0 0

0 Y 0

3 0 05 Y

(4.73a)

where YC 5 jBC 5 j

1 1 or ZC 5 2 jXC 5 2 j XC BC

(4.73b)

The positive-sequence, negative-sequence and zero-sequence admittances are equal. However, for short-circuit analysis, the modelling of series capacitors is not straightforward and requires knowledge of the capacitor protection scheme and capacitor design ratings. It is worth noting that IEC 60909-0:2016 short-circuit analysis standard states that the effect of series capacitors can be neglected in the calculation of short-circuit currents if they are equipped with voltage-limiting devices in parallel acting if a short-circuit occurs. However, as we will see later, this approach is generally inappropriate for varistor protected series capacitors and could result in significant underestimates in the magnitude of short-circuit currents.

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Modelling of series capacitors for short-circuit analysis Old series capacitor schemes were protected by spark gaps against a short-circuit fault that can cause a large capacitor current to flow and thus a substantial increase in the voltage across the capacitor. The flashover across the spark gap bypasses the series capacitor and the bypass circuit-breaker is immediately closed to extinguish the gap. In modern series capacitor installations, nonlinear resistors are used and these are called metal (usually zinc) oxide varistors (MOVs) known as ZnO. MOVs are used to provide the overvoltage protection mainly against external short-circuit faults, that is, external to the series capacitor and its line protection zone. For internal faults, within the series capacitor protected line zone, a forced triggered spark gap can operate typically within 1 ms to limit the energy absorption duty on the varistors. This is then followed by closure of the circuit-breaker to extinguish the gap and hence the capacitor is completely bypassed and removed from the circuit. The modelling and analysis of varistor protected series capacitors under external or remote short-circuit fault conditions is quite complex and generally requires three-phase time domain simulations. However, we will briefly and qualitatively describe the operation of the modern varistor protected capacitor schemes under external through-fault conditions that result in varistor operation only. As the instantaneous short-circuit current flowing through the capacitor increases, so will the voltage across the capacitor or varistor until this voltage reaches a preset threshold where the varistor starts conducting current in order to keep the voltage across the capacitor constant. This applies under both positive- and negative-polarity current conditions so that for the first part of each half cycle, the current flows through the capacitor only but for the other part of the half cycle, the current switches to the varistor. The behaviour of the varistor and series capacitor combination is repeated every half cycle during the fault period whenever the voltage across the capacitor/varistor exceeds the preset capacitor protection level. Using Fig. 4.55A, the behaviour of the series capacitor/varistor parallel combination under a through fault condition is illustrated using an electromagnetic transient analysis program. The voltage across the capacitor/varistor and the currents through the capacitor and varistor are shown in Fig. 4.55B
D, respectively. The fault current is shown in Fig. 4.55E curve (ii). Curve (i) of Fig. 4.55E shows the fault current that would flow through the capacitor if there were no varistor to protect it. Curve (iii) of Fig. 4.55E shows the fault current if the capacitor is assumed to be permanently bypassed. However, curve (ii) shows that the conduction of the MOV causes an effective reduction in the fault current magnitude from that of curve (i) but it remains higher than if the capacitor is assumed to be bypassed. This generic study illustrates that, in general, both ignoring the presence of the varistor or assuming the capacitor is permanently bypassed during the fault period can result in a large error in the fault current magnitude. The former can result in an overestimate, whereas the latter can result in an underestimate. The magnitude of the error depends on the electrical proximity of the fault location to the capacitor/varistor location among other factors.

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Power Systems Modelling and Fault Analysis

MOV

IV IF

– jX C

(A) VC

IC

(B) V C

IC (C)

(D) I V

(i) No MOV (ii) (iii) X Bypassed C

IF (E)

Figure 4.55 Behaviour of a series capacitor/varistor combination under external throughfault conditions: (A) generic circuit; (B) capacitor voltage; (C) capacitor current; (D) varistor current; and (E) fault current.

i

v Figure 4.56 Typical varistor nonlinear current-voltage characteristic.

The modelling and analysis of short-circuit faults in large-scale power systems is based on the assumption that all power system plant are linear or can be approximated as piece-wise linear. The varistor is, however, a highly nonlinear resistor that cannot be directly included in standard quasi-steady state short-circuit modelling and analysis simulations. Fig. 4.56 shows a typical current/voltage characteristic of a varistor which presents an almost infinite resistance below the threshold conduction voltage then a very low resistance above it. The impedance presented by the combined capacitor reactance and varistor resistance over a half cycle is dependent on the total current flowing through their parallel combination. The capacitor current or voltage protection threshold is known because the capacitor reactance is a known design parameter. Therefore, below this

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threshold, the equivalent impedance of the parallel capacitor/varistor combination is the capacitor reactance in parallel with a very high resistance or, alternatively, a very low resistance (practically zero) in series with the capacitor reactance. When a current higher than the threshold flows through the capacitor, the varistor will conduct current for part of the half cycle so that the combination will appear as a low resistance in parallel with the capacitor reactance or, alternatively, a higher resistance than zero in series with a much reduced capacitor reactance. The higher the current, the more resistive the equivalent impedance becomes reflecting the increased short-circuiting of the capacitor by the varistor. Therefore, the capacitor/ varistor parallel combination can be approximated as a resistance and a capacitance in series, with both being functions of the total current flowing through them. Fig. 4.57A shows a three-phase representation of a series capacitor protected by a varistor. Fig. 4.57B shows the corresponding equivalent circuit where the equivalent series impedance is a function of the total current flowing through the parallel capacitor/varistor combination. Denoting the capacitor current protective threshold as Ithr, the equivalent series impedance for each phase is given by  ZEq ðIÞ 5

2 jXC REq ðI Þ 2 jXEq ðI Þ

for I , Ithr for I $ Ithr

(4.74)

where XC is the capacitor reactance when all current flows through it, that is, its nominal design value, I is the total current flowing through the capacitor/varistor parallel combination and Ithr is a known capacitor design threshold parameter. The equations that describe REq(I) and XEq(I) as functions of I when I $ Ithr can be calculated from the known design data or parameters of the series capacitor and varistor. Fig. 4.58 illustrates typical equivalent impedance characteristics. (A)

(B)

R

IR

IR

IR

R REq. (I R )

− jX REq. (I R )

– jX C R

IY

IY

IY

R YEq. (I Y )

R IB

–jX C

IY

− jX YEq. (I Y)

– jX C IB

IR

IB

R BEq. (I B)

IB

− jX BEq. (I B)

Figure 4.57 Varistor-protected three-phase series capacitor and equivalent circuit: (A) threephase representation and (B) equivalent circuit model.

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Power Systems Modelling and Fault Analysis

XC X Eq. (I)

R Eq. (I)

I Figure 4.58 Typical relationship of equivalent resistance and reactance with current for a series capacitor/varistor combination under external through-fault conditions.

The series phase admittance matrix of the three-phase series capacitor/varistor device is 2

3 R ZEq ðIÞ 0 0 6 Y ðIÞ 0 7 ZEq ZRYB 5 4 0 5 B ðIÞ 0 0 ZEq

(4.75)

Prior to the short-circuit fault or when the current is less than Ithr, the equivalent phase admittances are balanced and equal. This will also be the case under balanced three-phase short-circuit faults. However, under unbalanced short-circuit faults, they are generally unequal depending on the fault type. Therefore, the sequence admittance matrix of Eq. (4.75) is given by 2 3 P ZS ZM1 ZM2 (4.76a) ZPNZ 5 N 4 ZM2 ZS ZM1 5 Z ZM1 ZM2 ZS where i 1h R Y B ZEq ðIÞ 1 ZEq ðIÞ 1 ZEq ðIÞ 3 i 1h R Y B ZM1 5 ZEq ðIÞ 1 h2 ZEq ðIÞ 1 hZEq ðIÞ 3 i 1h R Y B ZM2 5 ZEq ðIÞ 1 hZEq ðIÞ 1 h2 ZEq ðIÞ 3

ZS 5

(4.76b)

Eq. (4.76a) shows the presence of unequal intersequence mutual coupling which should be taken into account if significant errors are to be avoided. Because fixed-impedance short-circuit analysis techniques are linear, Eq. (4.74) may be implemented in an iterative linear calculation process. The fault currents

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349

flowing in the network and the series capacitors are initially calculated without varistor action, the capacitor current is then used to calculate a new value for the series capacitor equivalent impedance using Eq. (4.74). The new impedance is then used to calculate new fault currents in the network and the series capacitor. The final solution would be arrived at when there is a little change in the last two equivalent impedances or capacitor currents calculated. In power system networks with modern varistor protected series capacitor installations, the operation of the varistor when the capacitor current exceeds Ithr cannot be ignored if very large and unacceptable errors (overestimates) in the calculated short-circuit currents in the network are to be avoided. Also, if these series capacitors are not included in the network model at all, that is, assumed to be completely bypassed under short-circuit faults, as per IEC 60909, then this could result in significant underestimates in the magnitude of short-circuit currents.

4.5

Sequence modelling of static variable compensators

4.5.1 Background Static variable compensators, with one typical arrangement shown in Fig. 4.59A, are shunt connected devices used in renewable generation systems, transmission, subtransmission, distribution and industrial power systems to provide fast-acting dynamic reactive power and voltage control. The reactive power is usually provided by various combinations of fixed capacitors, thyristor switched capacitors and thyristor controlled reactors. These are usually connected at low voltage levels of typically 5
20 kV. Therefore, normally, a star
delta transformer is used to connect these reactive elements to the higher voltage power network. Fig. 4.60 shows a 400-kV static variable compensator (SVC) rated at
75 absorption to 150 MVAr generation. (A)

(B) RYB

(C)

I P

Z HL

Z

Z HL

V I P + jQ

Y

Y

Figure 4.59 Typical static variable compensator: (A) one-line diagram, (B) positivesequence/negative-sequence equivalent circuit and (C) zero-sequence equivalent circuit.

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Power Systems Modelling and Fault Analysis

Figure 4.60 400-kV, 2 75/ 1 150-Mvar static variable compensator (SVC).

4.5.2 Positive-, negative- and zero-phase sequence modelling The positive-sequence and negative-sequence equivalent circuits of a static variable compensator are identical and consist of the compensator transformer in series with the inductive and/or capacitive susceptance of the reactive elements. The admittance is calculated from Fig. 4.59B using S 5 P 1 jQ 5 VI , I 5 YV and P 5 0. Thus, the positive-sequence and negative-sequence admittance is given by YP 5 YN 5

2 jQ jV j2

(4.77)

where Q is the reactive power output and V is the voltage at the LV side of the compensator transformer. Both Q and V are known from an initial load flow study. In the zero-sequence network, the compensator should be represented by its transformer leakage impedance only, as shown in Fig. 4.59C.

4.6

Sequence modelling of static power system load

4.6.1 Background The term power system load does not have an agreed and clear meaning in academia or industry and, in practice, it can be interpreted to mean many different things by practicing engineers. Much research and development has been carried out to define the term load as well as its parameters and characteristics for use in shortterm power flow and transient stability studies. However, load representation has

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351

received little attention in large-scale network short-circuit analysis. A primary reason for this is that it has been a general practice to ignore the presence of load and calculate the short-circuit current changes in the network due to the short-circuit faults only. This is the case of IEC 60909 and American IEEE C 37.010 standards, as will be discussed in Chapter 8, International standards for short-circuit analysis in ac power systems. However, as we will see later, the pressures in recent years for increased accuracy and precision in short-circuit calculations resulted in the introduction of new techniques that require better and improved consideration of the load in the analysis. It is straightforward to include a model of load equipment in the analysis where this equipment is clearly identifiable because its electrical characteristics would generally be known. For example, a heating load, lighting load or induction motor load, the latter will be covered in detail in Chapter 5, Modelling of rotating AC synchronous and induction machines, can easily be modelled if one knew how much there is of each component. Nearly all the load components in an industrial power system or a power station auxiliary system would be known. However, in distribution, subtransmission and transmission system short-circuit analysis, the load in MW and MVAr at any given time is that supplied at a major grid supply substation, for example, a 132-kV substation, or a major bulk supply substation, for example, a 33 or 11-kV substation. This load therefore consists of thousands or tens of thousands of components of different types and characteristics and such composition is generally unknown to the network utility. The load will contain static components, that is, those that do not provide a short-circuit current contribution to a fault in the host network such as a fixed impedance load. Importantly, from a short-circuit analysis viewpoint, the load will contain a numerous number of single-phase induction motors, three-phase induction motors, embedded large- or small-scale three-phase synchronous generators or distributed renewable resource generators, both synchronous and nonsynchronous, which can feed a short-circuit current to a fault in the host network. In addition, these load components may be separated from the grid and bulk supply substations by distribution substations and LV networks. Therefore, the general loads seen at these substations for modelling in large-scale analysis will consist of distribution networks as well as passive and active components; the latter may be rotating machines or static renewable generators. Better precision short-circuit analysis should model the contribution of equipment that can feed short-circuit current and forming part of the general load at each substation in the network. Such equipment includes conventional rotating plant, that is, motors and generators, and static renewable generators, for example, solar photovoltaic generators. The modelling of rotating machines is covered in detail in Chapter 5, Modelling of rotating AC synchronous and induction machines, and that of renewable power generators in Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators but we will now restrict our attention to static or passive load as seen from major load supply substations.

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Power Systems Modelling and Fault Analysis

4.6.2 Positive-, negative- and zero-phase sequence modelling The positive-sequence and negative-sequence model of a passive load supplied radially at a substation that includes the load components and the intervening distribution network can be generally represented as a shunt admittance to earth as follows: YP 5 YN 5

P 2 jQ jV j2

(4.78)

where P and Q are the load supplied in pu MW and pu MVAr and V is the voltage at the substation just before the occurrence of the short-circuit fault, in pu. The zero-sequence model of a passive load can be substantially different from the positive-sequence/negative-sequence model. Generally, this model is determined by the presence of low zero-sequence impedance paths for the flow of zerosequence currents such as those provided by star
delta transformers. The zerosequence model is therefore the minimum zero-sequence impedance seen from the supplied substation looking down into the distribution network topology and equipment characteristics including transformer winding connections.

4.7

Three-phase modelling of static power plant and load in the phase frame of reference

4.7.1 Background In Chapter 7, Short-circuit analysis techniques in AC power systems, we describe a short-circuit analysis technique in large-scale power systems in the phase frame of reference. This type of analysis requires the use of three-phase models for power plant and load rather than positive-sequence, negative-sequence and zero-sequence models. Three-phase models are presented in this section.

4.7.2 Three-phase modelling of reactors and capacitors The three-phase series impedance and admittance matrix models for series reactors are given in Eq. (4.70a) and (4.70b), respectively. Where interphase mutual coupling exists, the series phase impedance matrices given in Eq. (4.71a) and (4.71d) apply for unequal and equal interphase couplings, respectively. The corresponding phase admittance matrices are given by 3 2 2 2 ZS2 2 ZM1 2 ðZS ZM2 2 ZM1 Þ 2 Z M1 7 6 ZS 2 ZM2 ZS 2 ZM2 7 6 1 7 6 2 Z Z 1 Z 2 Z YRYB 5 7 6 M1 S M2 M1 2 7 6 ZS ðZS 1 ZM2 Þ 2 2ZM1 2 ðZ Z 2 Z 2 Þ 2 ZS2 2 ZM1 S M2 5 4 M1 2 ZM1 ZS 2 ZM2 ZS 2 ZM2 (4.79a)

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353

in the case of unequal interphase mutual coupling, and 2 YRYB 5

ZS 1 ZM

1 6 2 ZM 2 4 ZS ðZS 1 ZM Þ 2 2ZM 2 ZM

2 ZM

2 ZM

7 2 ZM 5

ZS 1 ZM 2 ZM

3 (4.79b)

ZS 1 ZM

in the case of equal interphase mutual coupling, that is, ZM1 5 ZM2 5 ZM. The three-phase shunt admittance matrix of shunt reactors and capacitors was given in Eq. (4.72a). For modern varistor protected series capacitors, the currentdependent three-phase series impedance matrix was given in Eq. (4.75). The corresponding series phase admittance matrix is given by 2 6 YRYB 5 6 4

R ðIÞ 1=ZEq

0

0

0

Y 1=ZEq ðIÞ

0

0

B 1=ZEq ðIÞ

0

3 7 7 5

(4.80)

4.7.3 Three-phase modelling of transformers General The sequence modelling of single-phase and three-phase transformers of various winding connections was extensively covered in Section 4.2. In three-phase steadystate analysis in the phase frame of reference, three-phase models are required. For the purpose of steady-state fixed-impedance short-circuit analysis, it is sufficient to model the transformer as a set of mutually coupled windings with a linear magnetising reactance. We will not provide a similar extensive coverage to that in Section 4.2 but will instead present the basic modelling technique of a single-phase two-winding transformer, three-phase transformer banks and a three-limb core-form three-phase transformer. For the three-phase transformer, only a star
delta winding connection is considered. The principles provide the reader with the information and methodology required to develop a three-phase model for a transformer of any winding connection.

Single-phase two-winding transformers A single-phase two-winding transformer can be represented as a set of two mutually coupled windings as shown in Fig. 4.61A. The impedance matrix relating the terminal voltages and currents where Z21 5 Z12 can be written as 

 V1 Z 5 11 V2 Z12

Z12 Z22



I1 I2

(4.81a)

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Power Systems Modelling and Fault Analysis

(A)

Z11

V1

(B)

Z12

I1

H

Z H = Z11 − Z12

H

I2 Z 22

L

V2

Z L = Z 22 − Z12

I1

L I2

V1

Z12

V2

Figure 4.61 Single-phase two-winding transformer: (A) two iron-cored mutually coupled windings and (B) equivalent circuit in terms of self and mutual impedances.

where Z11 is winding 1 self-impedance, Z22 is winding 2 self-impedance and Z12 5 Z21 is mutual impedance between the two windings. Eq. (4.81a) can rewritten as V1 5 ðZ11 2 Z12 ÞI1 1 Z12 ðI1 1 I2 Þ

(4.81b)

V2 5 ðZ22 2 Z12 ÞI2 1 Z12 ðI1 1 I2 Þ

(4.81c)

V1 5 ZH I1 1 Z12 ðI1 1 I2 Þ and V2 5 ZL I2 1 Z12 ðI1 1 I2 Þ

(4.81d)

and

or

Eq. (4.81b)
(4.81d) are represented by the equivalent circuit shown in Fig. 4.61B where ZH and ZL are the leakage impedances of each winding, ZHL 5 ZH 1 ZL is the leakage impedance between the two windings and YHL 5 1/ ZHL. The mutual impedance Z12 consists of the series combination of the magnetising impedance and half ZHL as discussed previously, that is, Z12 5 ZM 1 0.5ZHL. From Eq. (4.81a), the self-impedance Z11 of winding 1 and the mutual impedance Z12 between the two windings are obtained from the results of an open-circuit test with winding 1 supplied and winding 2 open-circuited and are given by Z11 5

 V1  I 50 I1  2

and

Z12 5

V2  I 50 I1 2

(4.82a)

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355

From Eq. (4.82a), the mutual impedance can be expressed as Z12 5 Z11

V2 V1

(4.82b)

where V1/V2 is the transformer no-load voltage ratio. Similarly, the self-impedance Z22 of winding 2 and the mutual impedance Z12 between the two windings can be obtained from the results of an open-circuit test with winding 2 supplied and winding 1 open-circuited. Thus  V2  V1  Z22 5 I1 50 and Z12 5 I1 50 (4.83a) I2 I2 From Eq. (4.83a), the mutual impedance can be expressed as Z12 5 Z22

V1 V2

(4.83b)

From Eqs. (4.82b) and (4.83b), we obtain  Z11 5 Z22

V1 V2

2 (4.84)

The leakage impedance ZHL between the two windings is obtained from the results of a short-circuit test with winding 1 supplied and winding 2 short-circuited. Thus, from Eq. (4.81a), we obtain ZHL 5

 2 V1  Z12 V2 50 5 Z11 2  I1 Z22

(4.85a)

Alternatively, if winding 2 is supplied and winding 1 is short-circuited, the leakage impedance ZLH between the two windings is obtained as ZLH 5

 2 V2  Z12 V1 50 5 Z22 2  I2 Z11

(4.85b)

Having obtained all four elements of the impedance matrix of Eq. (4.81a), it is more convenient to use the corresponding admittance matrix given by "

I1 I2

#

" 5

Y11

Y12

Y12

Y22

#"

V1 V2

# (4.86a)

356

Power Systems Modelling and Fault Analysis

where Y11 5

Z22 2 Z11 Z22 2 Z12

Y12 5

2 Z12 2 Z11 Z22 2 Z12

Y22 5

Z11 2 Z11 Z22 2 Z12

(4.86b)

Using Eqs. (4.82b), (4.83b), (4.84) and (4.85a) in Eq. (4.86b), we obtain 0 Y11 5 YHL

V1 YHL V2

Y12 5 2

Y22 5 @

12

V1 A YHL V2

YHL 5

1 ZHL

(4.87a)

and 2 

YHL

6 6 I1 6 56 I2 6 V1 42 YHL V2

3 V1 YHL 7 V2 7 7 V1 0 12 7 7 V2 V @ 1 A YHL 5 V2 2

(4.87b)

or YLH 5

1 ZLH

Y11 5 0

1

Y12 5 2 0

12 YLH

1

1 YLH

@ V1 A V2

@ V1 A V2

Y22 5 YLH

(4.87c) and 2

1 Y 6 0 12 LH 6 V 1 6 @ A  6 V2 6 I1 56 6 I2 6 2 0 1 1 YLH 6 6 4 @ V1 A V2

20

1

1 YLH

@V 1 A V2 YLH

3 7 7 7 7 7 V1 7 7 V2 7 7 7 5

(4.87d)

In the derivation of the above impedance and admittance matrices, we note that, in practice, the only data most likely to be available are the measured short-circuit leakage impedance and no-load current. The derivation is illustrated using an example.

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Example 4.10 Consider a 239-MVA 231-kV/21.5-kV single-phase two-winding transformer having a leakage reactance of 15 %. The no load current measured from the HV side is 2.985 A and from the LV 239 side is 32.1 A. The HV and LV base currents are I1ðHÞðBaseÞ 5 pffiffiffi 3 1000 5 597:345A and 3 3 231 239 I2ðLÞðBaseÞ 5 pffiffiffi 3 1000 5 6; 417:987A. The pu no-load current measured from the HV 3 3 21:5 2:985 3 100 5 0:499775% and the pu no-load current measured from the side is IM2HVðpuÞ 5 597:345 32:1 LV side is IM2LVðpuÞ 5 3 100 5 0:5%. 6; 417:987 Therefore, from Fig. 4.61B, the measured no-load impedance from the HV side is 100 5 200:09pu and that from the LV side is Z11 5 ZH 1 Z12 5 0:499775 100 Z22 5 ZL 1 Z12 5 5 200pu. Now using ZH 1 ZL 5 0:15pu, we obtain Z12 5 199:97pu, 0:5 ZH 5 0:12pu and ZL 5 0:03pu and ZM 5 199:895pu. Where only one no-load current measurement is available, normally from the LV side, then two practical options can be considered. A 80/20 rule for the division of ZHL that is to assume ZH and ZL are 80% and 20% of ZHL, respectively. Alternatively, a simpler and more advantageous approach particularly in large-scale network analysis is to keep ZHL undivided and place Z12 at the L side. This is the approach used in Section 4.2 in this chapter. Ignoring the magnetising losses and the resistive part of the leakage impedance, the impedance and admittance matrices, in pu, are given by   200:09 199:97 6:6668666 2 6:66586657 Zpu 5 j and Ypu 5 2 j 199:97 200 2 6:66586657 6:66986669 The differences between the self and mutual reactances are the very small value ZH and ZL reactances. This results in very small differences between the self and mutual susceptances and necessitates calculations to at least seven or even eight decimal places if the small but important leakage reactance is not to be lost. The corresponding reactance and susceptance matrices in Ω and S are 2 3 2312 231 3 21:5 200:09 3 199:97 3 6 7 239 239 6 7 6 7 Z 5 j6 7Ω 6 7 21:52 4 199:97 3 231 3 21:5 5 200 3 239 239 " # 44; 673:6506 4; 155:4435 5j Ω 4; 155:4435 386:820 3 2 239 239 2 6:66586657 3 6:6668666 3 6 231 3 21:5 7 2312 7 6 7 6 Y 5 2 j6 7S 7 6 239 239 5 4 2 6:66586657 3 6:66986669 3 231 3 21:5 21:52 " # 0:0298604 2 0:3207776 5 2j S 2 0:3207776 3:4485627

358

Power Systems Modelling and Fault Analysis

The admittance matrix could also be calculated using Eq. (4.87b). The leakage impedance 2312 5 j33:49 Ω giving YHL 5 2 j0:02986 S referred to the HV side is equal to ZHL 5 j0:15 3 239 2 21:5 5 j0:290115 Ω giving YLH 5 2 j3:445908 S. The elements of the admitor ZLH 5 j0:15 3 239 tance matrix of Eq. (4.87b) are calculated using Eq. (4.87a). The above approach can easily be extended to single-phase three-winding transformers where the dimension of the admittance matrix of Eq. (4.86a) becomes 3 3 3. Obviously, we now have three self-admittances, three mutual admittances and three short-circuit leakage admittances. The derivation of the admittance equations following a similar method to the twowinding case is a simple exercise in algebra that is left for the interested reader.

Three-phase banks two-winding Ynd connected transformers with no interphase mutual couplings Consider a large two-winding Ynd star
delta transformer constructed as three single-phase banks as shown in Fig. 4.62. In such an arrangement, there is negligible or no interphase coupling between the three phases. Using the admittance matrix given in Eq. (4.86a) for each phase and Fig. 4.62, the admittance matrix equation that relates the winding currents and voltages for the three-phases is given by 2 3 I1 6 Y11 6 I2 7 6 0 6 7 6 6 I3 7 6 0 6 756 6 I4 7 6 Y14 6 7 6 4 I5 5 6 0 4 0 I6 2

0 Y22 0 0 Y25 0

0 0 Y33 0 0 Y36

Y14 0 0 Y44 0 0

3 2 3 0 7 V1 6 7 0 7 7 6 V2 7 6 7 Y36 7 7 6 V3 7 6 7 0 7 7 6 V4 7 4 5 0 7 5 V5 Y66 V6

0 Y25 0 0 Y55 0

VB IB

VY IY

IR I1

HV windings

V1

I2 Y11

V2

I4 V4

I3 Y 22

Y 44 I5

V5

Y 55 I6

Iy

Ib

Vr

Vy

Vb

Phase y

Y 33 Y 36

Ir

Phase r

V3

Y 25

Y14

LV windings

Phase B

Phase Y

Phase R VR

(4.88a)

V6

Phase b

Figure 4.62 Three single-phase banks two-winding star
delta transformer.

Y 66

Modelling of transformers, phase shifters, static power plant and static load

359

or IWinding 5 YWinding VWinding

(4.88b)

Eq. (4.88b) applies to a two set of three-phase windings irrespective of the type of winding connection. Thus, for our star
delta transformer, the three-phase nodal admittance matrix that relates the nodal, that is, phase RYB/ryb currents and voltages can be derived from the relationship between the winding currents and voltages, and the nodal currents and voltages. With all nodal voltages being with respect to the reference earth, the relationship between the winding and nodal voltages is found by inspection from Fig. 4.62 as follows: 3 2 3 2 3 2 1 0 0 0 0 0 V1 7 VR 6 7 6 V2 7 6 0 1 0 6 0 0 0 7 7 6 VY 7 6 7 6 7 6 VB 7 6 V3 7 6 0 0 1 0 0 0 76 7 6 756 (4.89a) 7 6 V4 7 6 0 0 0 6 1 21 0 7 7 6 Vr 7 6 7 6 5 4 V5 5 6 0 0 0 4 0 1 217 5 Vy 4 0 0 0 21 0 1 V6 Vb or Vwinding 5 CVNode

(4.89b)

where C is defined as a connection matrix. Similarly, the relationship between the nodal and winding currents is found by inspection from Fig. 4.62 and is given by 2

3

3

2

IR 61 6 IY 7 6 0 6 7 6 6 IB 7 6 0 6 756 6 Ir 7 6 0 6 7 6 4 Iy 5 6 0 4 0 Ib

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 21 0

0 0 0 0 1 21

2 3 0 7 I1 6 7 0 7 7 6 I2 7 7 7 0 76 6 I3 7 7 6 2 1 7 6 I4 7 7 4 5 0 7 5 I5 1 I6

(4.89c)

or INode 5 CT Iwinding

(4.89d)

Using Eq. (4.89a)
(4.89d) in Eq. (4.88b), we obtain INode 5 YNode VNode

(4.90a)

where YNode 5 CT YWinding C

(4.90b)

360

Power Systems Modelling and Fault Analysis

and, after a little algebra, it can be shown that the nodal admittance matrix YNode is given by 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 YNode 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

1

0

12 YLH14

0

20

0

1

1 YLH14

BV1 C @ A V4

BV1 C @ A V4 0

0

1

12 YLH25

0

1

0

0

0

0

1

12 YLH36

1 YLH25

20

1

1 YLH14

0

1

1 YLH25

0

0

1

1 YLH36

BV3 C @ A V6

0

20

1

1 YLH25

0

BV2 C @ A V5 1

1 YLH25

YLH14 1 YLH25

2 YLH25

2 YLH25

YLH25 1 YLH36

2 YLH14

2 YLH36

BV2 C @ A V5

0

0

1

BV2 C @ A V5

BV3 C @ A V6

BV1 C @ A V4

1 YLH14

1

1 YLH36

BV3 C @ A V6 20

0

BV1 C @ A V4

1

1 YLH36

1

1 YLH14

3

7 7 7 7 7 7 7 7 7 0 7 7 7 7 7 7 7 7 7 1 2 0 1 YLH36 7 7 7 7 BV3 C 7 @ A 7 V6 7 7 7 7 2 YLH14 7 7 7 7 7 7 7 7 7 7 2 YLH36 7 7 7 7 7 7 7 7 YLH14 1 YLH36 7 7 7 7 5 BV1 C @ A V4

BV2 C @ A V5

BV2 C @ A V5

20

0

0

BV3 C @ A V6

(4.90c)

Transformers are designed in practice so that the leakage impedances between the LV and HV windings are practically equal for all three phases and so are the V1 V2 V3 5 5 5 tHL . turns ratios. This means YLH14 5 YLH25 5 YLH36 5 YLH and V4 V5 V6 Using Eq. (4.90c) and denoting the star-connected winding as H and the deltaconnected winding as L, Eq. (4.90a) can be rewritten as follows 2

1

R

HV winding Y

YLH

6 t2 6 HL 6 6 6 0 6 2 3 6 6 IR 6 6 IY 7 6 0 6 7 6 6 IB 7 6 6 756 6 Ir 7 6 6 7 6 2 1 YLH 4 Iy 5 6 tHL 6 6 Ib 6 6 0 6 6 6 6 1 4 YLH tHL

B

LV winding r y 1 2 YLH tHL

b

0

0

1 YLH 2 tHL

0

1 YLH tHL

0

1 2 YLH tHL

0

1 YLH tHL

1 YLH tHL

0

2YLH

2 YLH

1 YLH tHL

2 YLH

2YLH

2 YLH

2 YLH

2

1 YLH tHL 0

2

1 YLH tHL

0

2

1 YLH tHL

3 YLH 7 tHL 7 7 7 7 0 7 72 3 7 VR 7 1 6 7 2 YLH 7 76 VY 7 tHL 76 VB 7 76 7 76 Vr 7 6 7 2 YLH 7 74 Vy 5 7 7 Vb 7 2 YLH 7 7 7 7 7 2YLH 5 1

(4.91)

Modelling of transformers, phase shifters, static power plant and static load

361

The reader is encouraged to show that the nodal admittance matrix of a twowinding Ynyn star
star transformer with both neutrals solidly earthed and no interphase mutual couplings has the form as the winding admittance given in Eq. (4.88a). Using these principles, the reader can easily obtain the nodal admittance matrix for any winding connections of three-phase transformers.

Three-phase common-core two-winding Ynd connected transformers with interphase mutual coupling In the above analysis, we considered mutual coupling between the HV and LV windings of the same phase Rr, Yy and Bb. However, in general, for common-core transformers, whether of the core-form or shell-form core design, some interphase mutual coupling does exist and this is generally weak and asymmetrical, as can be expected from Figs. 4.35
4.38, between the outer-limb phases and the phases connected to adjacent limbs. However, in practice, little or no information is available on these asymmetrical coupling affects although they can be measured in the factory. Fig. 4.63 illustrates a common-core Ynd star
delta transformer where each phase winding is coupled to all other phase windings and windings 1 and 4, 2 and 5, and 3 and 6, are wound on the same limb. With the assumption of full flux symmetry, we obtain a 6 3 6 winding admittance matrix that consists of four 3 3 3 symmetrical submatrices. The corresponding nodal admittance matrix can be obtained using Eq. (4.90b). However, standard transformer test certificates include only the normally measured parameters, that is, the positive-sequence and zero-sequence leakage impedances and the magnetising current from which the magnetising impedance is calculated. The general case where the positive-sequence and zero-sequence impedances are different implies that interphase mutual couplings do exist in the full

VY IY

IR I1 HV windings

V1

Y14 LV windings I4

Phase B VB IB

Phase Y

Phase R VR

V4

I2 Y11 Y12

I3

V2

V3

Y13

Y15

Y16

I5

Ir

Iy

Vr Phase r

Vy

V5

I6

V6

Ib Vb Phase y

Phase b

Figure 4.63 Common-core three-phase star
delta transformer with symmetrical interphase coupling.

362

Power Systems Modelling and Fault Analysis

three-phase admittance matrix. The asymmetrical interphase coupling effects are also indirectly averaged out by their short-circuit measurement procedures as described in Section 4.2.10. Since in nearly all practical situations only sequence and magnetising impedances are available, we will present a different technique for the formulation of the transformer three-phase model with interphase coupling. The technique is based on using the known positive-sequence and zero-sequence leakage impedances, and magnetising impedance measurements and therefore imply averagedout interphase couplings. To illustrate the technique, we consider the star
delta transformer of Fig. 4.63 with a tap-changer on the HV winding Yn star side. In Fig. 4.63, phase R of the star winding is assumed symmetrically coupled with phases Y and B of the star winding on different limbs, with phase ‘ry’ on the delta winding on the same limb, and with phases ‘rb’ and ‘by’ on the delta winding on different limbs. Similar symmetrical couplings are assumed for other phases. The positive-sequence and negative-sequence equivalent circuits are shown in Fig. 4.11A and B. The zero-sequence equivalent circuit is shown in Fig. 4.15G with ZE 5 0. Eq. (4.13a) corresponds to Fig. 4.11C in which the transformer winding phase shift of 30 degrees for Yd1 and 2 30 degrees for Yd11 connections was neglected. In the derivation to follow, we denote the star side as H and the delta side as L, include the winding phase shift ϕ and let all impedances be expressed in pu. It can be shown that the positive-sequence and negative-sequence nodal admittance matrices of Fig. 4.11A and B are given by 2

"

IHP

#

ILP

1 2 ZP 6 tHL LH 6 56 6 e2jϕ 4 2 P tHL ZLH 2

1 " # 6 2 P t N 6 HL ZLH IH 6 5 6 ILN ejϕ 4 2 P tHL ZLH

3 ejϕ P 7 " P# tHL ZLH 7 VH 7 1 1 7 5 VLP 1 P P ZLH ZM

2

3 e2jϕ P 7 " N# tHL ZLH 7 VH 7 1 1 7 5 VLN 1 P P ZLH ZM

(4.92a)

2

(4.92b)

and the zero-sequence nodal admittance matrix of Fig. 4.15G with ZE 5 0 is given by "

IHZ ILZ

#

2

1 6 t2 Z Z 5 4 HL LH 0

3 " # 0 7 VHZ 5 VLZ 0

(4.92c)

Modelling of transformers, phase shifters, static power plant and static load

363

Eq. (4.92a), (4.92b) and (4.92c) can be assembled to obtain the transformer sequence admittance matrix, including sequence current and voltage vectors, as follows: 2

1 6 t2 Z P 6 HL LH 6 6 2 P3 6 6 0 IH 6 6 N7 6 6 IH 7 6 6 7 6 6 IZ 7 6 0 6 H7 6 6 P 756 6 IL 7 6 6 7 6 e2jϕ 6 N7 6 4 IL 5 6 6 2 t ZP HL LH 6 6 ILZ 6 6 6 0 6 4 0

0 0

0

0

1 2 ZZ tHL LH

0

0

0

0

1 1 1 P P ZLH ZM

0

0

0

1 1 1 P P ZLH ZM

0

0

0

0 1 2 ZP tHL LH

2

3

ejϕ 2 P tHL ZLH

ejϕ P tHL ZLH 0

0 2

e2jϕ P tHL ZLH

07 7 7 7 72 P 3 07 7 VH 76 N7 7 6V 7 76 H7 07 76 VHZ 7 7 76 6 7 6 VP 7 76 L7 7 76 N7 07 7 4 VL 5 7 7 VLZ 7 7 07 7 5 0 (4.93a)

or 

 PNZ IPNZ YHH H 5 PNZ  IPNZ ðY L HL Þ

YPNZ HL YPNZ LL



VPNZ H VPNZ L

(4.93b)

The phase admittance matrix is obtained using the sequence to phase transformapffiffiffiffi tion matrix H noting that, in pu, the effective delta winding turns ratio is 3: Thus RYB RYB IRYB ðnodeÞ 5 YðnodeÞ VðnodeÞ

"

IRYB HðnodeÞ Iryb LðnodeÞ

#

" 5

(4.94a)

YRYB HH

Yphase HL

Yphase LH

Yryb LL

#"

VRYB HðnodeÞ

# (4.94b)

Vryb LðnodeÞ

where R IRYB HðnodeÞ 5 IH  R VRYB HðnodeÞ 5 VH

IY H VY H

PNZ 21 YRYB HH 5 HYHH H

IBH

T VBH

r Iryb LðnodeÞ 5 IL T

IyL

 r Vryb LðnodeÞ 5 VL

PNZ 21 Yryb LL 5 3HYLL H

IbL

T

VyL

(4.95a) VbL

T

(4.95b) (4.95c)

364

Power Systems Modelling and Fault Analysis

Yphase HL 5

pffiffiffi 21 3HYPNZ HL H

Yphase LH 5

pffiffiffi  21 3HðYPNZ HL Þ H

(4.95d)

Therefore, using Eq. (4.93a) and (4.93b) in Eq. (4.95c) and (4.95d), the self (H and L windings) and mutual (H
L and L
H windings) phase admittance matrices are given by 3 2 1 2 1 1 1 1 1 2 2 Z P Z P Z P 7 6 ZLH ZLH ZLH ZLH ZLH ZLH 7 6 7 6 1 1 1 2 1 1 7 6 1 6 2 1 2 (4.96a) YRYB 5 Z P Z P Z P 7 HH 2 6 Z Z Z Z Z Z LH LH LH LH LH 7 3tHL 6 LH 7 6 1 1 1 1 1 2 7 5 4 2 2 1 Z P Z P Z P ZLH ZLH ZLH ZLH ZLH ZLH 2 0

1

1 1 6 2@ P 1 P A 6 Z Z LH M 6 6 0 1 6 6 1A 6 @ 1 Yryb LL 5 6 2 P 1 ZP 6 ZLH M 6 6 0 1 6 6 4 2@ 1 1 1 A P P ZLH ZM

Yphase HL

pffiffiffi 5 3

0

1

1 1 1 PA P ZLH ZM 0 1 1 1 2@ P 1 P A ZLH ZM 0 1 1 1 2@ P 1 PA ZLH ZM

2@

2
cos ϕ 2 4
cosðϕ 2 2π=3Þ P 3tHL ZLH
cosðϕ 1 2π=3Þ

0

13

1 1 1 P A7 P ZLH ZM 7 7 0 17 7 7 1 1 7 2@ P 1 P A7 ZLH ZM 7 7 0 1 7 7 7 1 1 2@ P 1 P A 5 ZLH ZM 2@


cosðϕ 1 2π=3Þ
cos ϕ
cosðϕ 2 2π=3Þ

(4.96b) 3
cosðϕ 2 2π=3Þ
cosðϕ 1 2π=3Þ 5
cos ϕ (4.96c)

and phase T Yphase LH 5 ðYHL Þ

(4.96d)

It should be noted that for Ynyn star
star and Dd delta
delta connected twowinding transformers, φ 5 0 or φ 5 180 degrees. However, for our Yd1 transformer of Fig. 4.63, φ 5 30 degrees and the mutual phase admittance matrices become 2 3 21 1 0 1 6 7 21 1 5 Yphase 4 0 HL 5 tHL ZPLH 1 0 21 (4.96e) 2 3 21 0 1 1 6 7 phase T Yphase 1 21 0 5 LH 5 ðYHL Þ 5 P 4 tHL ZLH 0 1 21

Modelling of transformers, phase shifters, static power plant and static load

365

It is useful to present an overall picture of the full phase impedance matrix of 1 1 1 Z P YMP 5 P , Y P 5 Y Z 1 2Y P , 5 P YLH 5 Z Z the transformer. Letting, YLH LH LH SðHÞ ZM LH ZLH P Z P P P P YMðHÞ 5 YLH 2 YLH , YSðLÞ 5 YLH 1 YM and using Eq. (4.96a), (4.96b) and (4.96e) in Eq. (4.94b), Eq. (4.94a) can be rewritten as HV winding LV winding Y B r y b 2 3 2 2 3 2 33 P P IR YSðHÞ YMðHÞ YMðHÞ 2 YLH YLH 0 2 3 6 7 6 1 6 1 6 7 7 7 VR P P 6 IY 7 6 2 YLH YLH 5 76 7 6 7 6 3t2 4 YMðHÞ YSðHÞ YMðHÞ 5 tHL 4 0 76 V Y 7 6 7 6 HL P P 76 7 6 IB 7 6 YMðHÞ YMðHÞ YSðHÞ YLH 0 2 YLH 76 V B 7 6 756 76 7 2 3 2 3 6I 7 6 P P Vr 7 2 YSðLÞ 2 YSðLÞ 7 2YSðLÞ 2 YLH 0 YLH 6 r7 6 76 4 Vy 5 6 7 6 1 6 7 7 6 7 P P 6I 7 4 2 YSðLÞ 5 5 2 YLH 0 5 4 2 YSðLÞ 2YSðLÞ 4 YLH 4 y5 Vb tHL P P 2 YSðLÞ 2 YSðLÞ 2YSðLÞ 2 YLH 0 YLH Ib R

(4.97)

Example 4.11 Consider the single-phase two-winding transformer of Example 4.10. This is one phase of a three-phase two-winding Ynd generator-transformer having the following data: P 717 MVA, 400/21.5 kV, positive-sequence leakage reactance ZLH 5 15% on rating, zeroZ sequence leakage reactance ZLH 5 13:5% and positive-sequence reactance of the entire magneP tising branch and denoted ZM 5 199:97 pu. Assume the transformer is operating at nominal tap ratio and ignore the resistive parts of leakage impedances and magnetising impedance, calculate the full phase admittance matrix of the transformer. P P YLH 5 1=ðjXLH Þ 5 1=ðj0:15Þ 5 2 j6:6666667 pu; Z Z YLH 5 1=ðjXLH Þ 5 1=ðj0:135Þ 5 2 j7:4074074 pu;

YMP 5 1=ðjXMP Þ 5 1=ðj199:97Þ 5 2 j0:00500075 pu; YSðHÞ 5


 1 Z P 1=ðjXLH Þ 1 2=ðjXLH Þ 5 2 j 1=0:135 1 2=0:15 =3 5 2 j6:913580 pu 3

YMðHÞ 5


 1 Z P Þ 2 1=ðjXLH Þ 5 2 j 1=0:135 2 1=0:15 =3 5 2 j0:2469136 pu 1=ðjXLH 3

P YSðLÞ 5 1=ðjXLH Þ 1 1=ðjXMP Þ 5 2 jð1=0:15 1 1=199:97Þ 5 2 j6:671667pu

tHL 5 1

366

Power Systems Modelling and Fault Analysis

HV winding 2

IR

3

2

6 7 6 6 IY 7 6 6 7 6 6 IB 7 6 6 7 6 6 756 6 Ir 7 6 6 7 6 6 7 6 4 Iy 5 4

R

LV winding

Y

B

r

y

b

j6:666667

0

2 j6:666667

2 j13:34334

j6:671667

j6:671667

2 j6:666667

j6:666667

0

j6:671667

2 j13:34334

j6:671667

3 VR 76 7 76 VY 7 76 7 76 VB 7 76 7 76 7 76 Vr 7 76 7 76 7 54 Vy 5

0

2 j6:666667

j6:666667

j6:671667

j6:671667

2 j13:34334

Vb

2 j6:913580

2 j0:246913

2 j0:246913

2 j0:246913

2 j6:913580

2 j0:246913

2 j0:246913

Ib

j6:666667

2 j6:666667

0

2 j0:246913

0

j6:666667

2 j6:666667

2 j6:913580

2 j6:666667

0

j6:666667

32

(4.98)

Three-phase models for other transformer winding connections including three-winding transformers that result in a 9 3 9 nodal phase admittance matrix can be derived using a similar technique.

4.7.4 Three-phase modelling of QB and PS transformers In some special analysis, both the shunt and series transformers of QBs and PSs need to be explicitly modelled as three-phase devices. However, for most steadystate, for example, unbalanced short-circuit analysis on the external network to which this equipment is connected, the two transformers may be combined into one single model. Similar to a three-phase transformer with symmetrical interphase mutual couplings, we will make use of the positive-sequence and zero-sequence impedances and equivalent circuits of the combined transformers to derive the three-phase admittance matrix that relates the input and output phase currents and voltages. Therefore, using Fig. 4.45A, 4.47A and B as well as the sequence admittance Eqs. (4.65a), (4.66) and (4.67a), the nodal sequence admittance matrix including sequence current and voltage vectors is given by 2  2  P N =Z P 6  e Ii 6 6 IN 7 6 6 i 7 6 0 6 Z7 6 6 Ii 7 6 6 756 0 6 IP 7 6 6 o7 6 6 N 7 6 2 N P =ZeP 4 Io 5 6 6 4 0 IoZ 0 2

P3

3

0  2  N N  =ZeP

0

2 ðN Þ =ZeP

0

0

0

2 ðN Þ =ZeP

0

YZi

0

0

0

1=ZeP

0

2 N =ZeP

0

0

1=ZeP

0

2 YioZ

0

0

0 N

P

N

2 3 7 ViP 7 76 N7 0 7 6 Vi 7 76 Z7 7 6 Vi 7 6 7 2 YioZ 7 7 6 VP 7 76 o 7 6 7 0 7 7 4 VoN 5 7 0 5 VZ 0

YoZ

o

(4.99a)

or "

IPNZ i IPNZ o P

#

" 5

YPNZ ii

YPNZ io

YPNZ oi

YPNZ oo

N

#"

VPNZ i VPNZ o

#

" 5Y

PNZ

VPNZ i VPNZ o

# (4.99b)

where N and N are the positive-sequence and negative-sequence complex turns P N P ratios and ðN Þ and ðN Þ are their conjugates. N is given in Eqs. (4.63c) and (4.64c)

Modelling of transformers, phase shifters, static power plant and static load

367

N

for a QB and a PS, respectively, and N is given in Eq. (4.66) for both QB and PS. Therefore, using YRYB 5 HYPNZH21 it can be shown that the nodal phase admittance matrix is given by "

IRYB i

#

" 5

IRYB o

YRYB ii

YRYB io

YRYB oi

YRYB oo

#"

VRYB i

# (4.100a)

VRYB o

where 2 a b 1 4b a 5 YRYB ii 3 b b ( a5 ( b5

3 b b5 a

(4.100b)

YiZ 1 2ð1 1 tan2 φÞ=ZeP

for a Quadrature Booster

YiZ

for a Phase Shifter

1 2=ZeP

YiZ 2 ð1 1 tan2 φÞ=ZeP

for a Quadrature Booster

YiZ

for a Phase Shifter

2 ZeP

(4.100c)

(4.100d)

and 2 YRYB oo 5

YoZ 1 2=ZeP

16 Z 4 Yo 2 1=ZeP 3 YoZ 2 1=ZeP

YoZ 2 1=ZeP

YoZ 2 1=ZeP

3

YoZ 1 2=ZeP

7 YoZ 2 1=ZeP 5

YoZ

YoZ

2 1=ZeP

(4.101)

1 2=ZeP

for both quadrature boosters and phase shifters. And 2

x y 14 z x YRYB 5 io 3 y z

3 z y5 x

(4.102a)

where ( x5

2 ðYioZ (

y5

2 ðYioZ 1 2=ZeP Þ 1 2 cos φ

for a Quadrature Booster =ZeP Þ

for a Phase Shifter

pffiffiffi 2 ½YioZ 1 ð 3tan φ 2 1Þ=ZeP 

for a Quadrature Booster

2 ½YioZ

for a Phase Shifter

1 2cos

ðφ 2 2π=3Þ=ZeP 

(4.102b)

(4.102c)

368

Power Systems Modelling and Fault Analysis

( z5

2 ½YioZ 2 ð1 1

pffiffiffi 3 tan φÞ=ZeP 

for a Quadrature Booster

2 ½YioZ 1 2 cos ðφ 1 2π=3Þ=ZeP  for a Phase Shifter

(4.102d)

T 5 ðYRYB The reader should easily find that YRYB oi io Þ :

4.7.5 Three-phase modelling of static load A static three-phase load was represented by equal positive-sequence and negativesequence impedances or admittances as given by Eq. (4.78) and a zero-sequence impedance or admittance as described in Section 4.6.2. The sequence admittance matrix of the load is given by 2 P YLP YPNZ 5 N4 0 L Z 0

3 0 0 5 YLZ

0 YLP 0

(4.103a)

The corresponding three-phase admittance matrix is calculated using 21 YRYB 5 HYPNZ giving the following balanced three-phase admittance matrix L L H 2 YRYB 5 L

YLZ 1 2YLP

16 Z 4 YL 2 YLP 3 YLZ 2 YLP

YLZ 2 YLP

YLZ 2 YLP

3

YLZ 1 2YLP

7 YLZ 2 YLP 5

YLZ 2 YLP

YLZ 1 2YLP

(4.103b)

In the unusual case where the zero-sequence admittance YLZ is equal to the positive-sequence admittance YLP , then the load can be represented as three shunt admittances as follows: 2

YLP RYB 4 YL 5 0 0

0 YLP 0

3 0 0 5 YLP

(4.103c)

Further Reading Books The Electricity Council, Power system protection, Vol. 1, Principles and Components, 1981, ISBN 0-906048-47-8. Arrilaga, J., et al., Computer Modeling of Electrical Power Systems, John Wiley & Sons, 1983. ISBN 0-471-10406-X. Grainger, J.J., Power System Analysis, McGraw-Hill Int., 1994. Ed., ISBN 0-07-061293-4. Weedy, B.M., Electric Power Systems, John Wiley & Sons, 1967. ISBN 0-471-92445-8.

Modelling of transformers, phase shifters, static power plant and static load

369

Papers Bratton, E., et al., Tertiary windings in star/star connected power transformers, Electrical Power Engineer, 6
9. Zirka, S.E., et al., Topology-correct reversible transformer model, IEEE Trans. On Power Delivery, Vol. 27, No. 4, October 2012. Zirka, S.E., et al., Accounting for the influence of the tank walls in the zero-sequence topological model of a three-phase three-limb transformer, IEEE Trans. On Power Delivery, Vol. 29, No. 5, October 2014. Zirka, S.E., et al., Duality-derived transformer models for low-frequency electromagnetic transients
Part I: Topological models, IEEE Trans. On Power Delivery, Vol. 31, No. 5, October 2016. Bengtsson, C., et al, Design and testing of 300 Mvar shunt reactors, CIGRE, A2-305, 2008. Villa, A., Autotransformer and reactor models development for core and shell type windings for frequency analysis studies, CIGRE, A2-212, 2008. Henriksen, T., Transformer leakage flux modeling, SINTEF Enery Research, N-7465, Trondheim, Norway. Avendano, A., et al, Transformer short-circuit representation: Investigation of phase-to-phase coupling International Conference on Power System Transients (IPST), Lyon, France, June 2007. Riedel, P., Modeling of zigzag-transformers in the three-phase systems, Institute for Electrical Engineering D-3000 Hanover 1, FRG, pp. T3/1-T3/9. Jarman, P., Tleis, N., et al, Experience of specifying and using reactors on the National Grid UK transmission system, CIGRE, A2-302, 2008. Chen, M.-S., et al., Power system modeling, IEE Proceedings, Vol. 62, No. 7, July 1974, 901
914. Laugton, M.A., Analysis of unbalanced polyphase networks by the method of phase coordinates, IEE Proceedings, Vol. 115, No. 8, August 1968, 1163
1171. Bowman, W.I., et al., Development of equivalent Pi and T matrix circuits for long untransposed transmission lines, IEEE Summer General Meeting and Nuclear Radiation Effects Conference, Toronto, Ont., June 1964, pp. 625
631. Jimenez, R., et al., Protecting a 138 KV phase shifting transformer: EMTP modelling and model power system testing, 2002, 1
14. Mahseredjian, J., et al., Superposition technique for MOV-protected series capacitors in short-circuit calculations, IEEE Transactions On Power Delivery, Vol. 10, No. 3, July 1995, 1394
1400. Goldworthy, D.L., A linearised model for MOV-protected series capacitors, IEEE PES Transactions Paper 86 SM 357
8. Mexico City, 1986.

5

Modelling of rotating ac synchronous and induction machines Chapter Outline 5.1 General 372 5.2 Overview of synchronous machine modelling in the phase frame of reference 372 5.3 Synchronous machine modelling in the Park dq0 frame of reference 375 5.3.1 5.3.2 5.3.3 5.3.4

Transformation from ryb reference to dq0 reference 375 Machine dq0 equations in per unit 377 Machine operator reactance analysis 379 Machine parameters
subtransient and transient reactances and time constants 381

5.4 Synchronous machine behaviour under short-circuit faults and modelling in the sequence reference frame 386 5.4.1 5.4.2 5.4.3 5.4.4 5.4.5 5.4.6 5.4.7 5.4.8 5.4.9

Synchronous machine sequence equivalent circuits 386 Three-phase short-circuit faults 387 Unbalanced two-phase (phase-to-phase) short-circuit faults 398 Unbalanced single-phase to earth short-circuit faults 403 Unbalanced two-phase to earth short-circuit faults 407 Modelling the effect of initial machine loading 412 Effect of automatic voltage regulators on short-circuit currents 414 Modelling of synchronous motors/compensators/condensers 417 Examples 418

5.5 Determination of synchronous machine parameters from measurements

424

5.5.1 Measurement of positive-sequence reactance, positive-sequence resistance and d-axis short-circuit time constants 424 5.5.2 Measurement of negative-sequence impedance 428 5.5.3 Measurement of zero-sequence impedance 429 5.5.4 Example 431

5.6 Modelling of induction motors in the phase frame of reference

435

5.6.1 General 435 5.6.2 Overview of induction motor modelling in the phase frame of reference 436

5.7 Modelling of induction motors in the dq frame of reference 440 5.7.1 5.7.2 5.7.3 5.7.4

Transformation to Park dq axes 440 Complex form of induction motor equations 442 Operator reactance and parameters of a single-winding rotor 442 Operator reactance and parameters of double-cage or deep-bar rotor 444

5.8 Induction motor behaviour under short-circuit faults and modelling in the sequence reference frame 448 5.8.1 Three-phase short-circuit faults 448 5.8.2 Unbalanced single-phase to earth short-circuit faults 457 Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00005-9 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

372

Power Systems Modelling and Fault Analysis

5.8.3 Modelling the effect of initial motor loading 459 5.8.4 Determination of a motor’s electrical parameters from tests 5.8.5 Examples 465

Further reading

5.1

460

468

General

The short-circuit performance of rotating ac machines in power systems is of fundamental importance in the analysis of short-circuit currents and hence the safety of people and reliability of power systems. The modelling of rotating machines presents, in general, one of the most complex problems in the analysis of power systems. Fortunately, however, the required modelling detail of rotating machines depends on the type of study to be undertaken. In practice, different models are developed and used for load flow, short circuit, dynamic or electromechanical analysis and three-phase electromagnetic transient analysis, etc. The aim of this chapter is to present the modelling of rotating machines in the sequence frame of reference for use in short-circuit analysis of power systems. However, in order to obtain an insight into the behaviour of rotating machines during transient fault conditions, the well-known machine model in the Park dq0 axis reference frame will be briefly presented and the machine models in the sequence frame of reference will then be derived. The presentation of the machine model in the dq0 reference is important since it provides an understanding of the meaning and origin of machine parameters, such as the machine d and q axes reactances and time constants, that can directly affect the magnitude and decay rate of short-circuit fault current. The reader is expected to have a basic understanding of the theory of rotating ac machines. New types of synchronous and nonsynchronous generators used predominantly in renewable energy systems such as wind parks and solar parks are discussed in Chapter 6, Modelling of voltage-source inverters, wind-turbine and solar photovoltaic (PV) generators.

5.2

Overview of synchronous machine modelling in the phase frame of reference

A brief overview of synchronous machine modelling in the phase frame of reference is presented in this section. This model is a prerequisite to the derivation of the dq0 machine model as well as to the process of transformations from one reference to the other. The basic structure of a salient two-pole three-phase synchronous machine is shown in Fig. 5.1A and a schematic is shown in Fig. 5.1B illustrating stator and rotor circuits.

Modelling of rotating ac synchronous and induction machines

373

R phase

(A)

B' phase

Y' phase DC field winding

Direction of rotor rotation

Y phase

B phase

R' phase

r

(B)

ir Ra Stator windings rr

ry

br bb

Ra

yy

iy

Ra

y

yb

ib

rfd

b

θ

– Mutual inductance stator (r phase) to field – Mutual inductance stator (r phase) to d-axis damper winding kd

rkd rfd

d-axis

fd

rkq

efd ωr

rkd

kq

kd

ikq

Direction of rotor rotation

rkq –

Mutual inductance stator (r phase) to q-axis damper winding kq

ikd Rotor q-axis

Figure 5.1 Three-phase salient-pole synchronous machine: (A) cross-section and (B) illustration of stator and rotor circuits including mutual coupling.

The machine comprises three windings, r, y, b on the stator displaced by 120 degrees; a field winding f that carries the direct current (dc) excitation and a shortcircuited damper winding k on the rotor. The rotor can be cylindrical or salient pole in construction, with the former mainly used in steam power plant and confined to two- or four-pole turbo-machines. Salient-pole machines are mainly used in hydro plant and low-speed plant in general, and can have more than 100 poles. From the basic theory of magnetic coupling between circuits, the self and mutual inductances between stator and rotor circuits are defined as shown in Fig. 5.1B.

374

Power Systems Modelling and Fault Analysis

The damper winding k is represented as two short-circuited windings; one in the same magnetic axis as the field winding f on the d-axis and termed kd and the other is a winding in an q-axis chosen to be 90 degrees ahead of the field or d-axis and termed kq. The q-axis could have been chosen to lag the d-axis by 90 degrees since the choice is rather a convention and has no practical significance. We include one damper winding on the q-axis in our model. Fig. 5.1B illustrates the field and damper windings on the rotor with respect to the d and q axes. The flux linkages in all six windings, namely stator phases r, y and b, field winding f, damper windings kd and kq can be written in terms of self and mutual inductances. Using a generator convention i.e. stator currents flowing out of the stator, we have




ψryb


5

ψðfdÞðkdÞðkqÞ

Lss ðθÞ Lsr ðθÞ Lrs ðθÞ Lrr




2 iryb iðfdÞðkdÞðkqÞ

 (5.1)

where r

2

Lss 1 LM cos 2θ 6 6 6 r 6 2 Lms 6 Lss ðθÞ 5 y6 2 1 LM cosð2θ 2 2π=3Þ 6 b6 6 2 Lms 4 1 LM cosð2θ 1 2π=3Þ 2

y 2 Lms 1 LM cosð2θ 2 2π=3Þ 2 Lss 1 LM cosð2θ 1 2π=3Þ 2 Lms 1 LM cos 2θ 2

b 3 2 Lms 1 LM cosð2θ 1 2π=3Þ 7 2 7 7 7 2 Lms 7 1 LM cos 2θ 7 2 7 7 7 Lss 1 LM cosð2θ 2 2π=3Þ 5 (5.2a)

r

2

f Lrfd cosθ

kd Lrkd cosθ

kq Lrkq sinθ

3

6 7 Lsr ðθÞ 5 y 4 Lrfd cosðθ 2 2π=3Þ Lrkd cosðθ 2 2π=3Þ Lrkq sinðθ 2 2π=3Þ 5 b Lrfd cosðθ 1 2π=3Þ Lrkd cosðθ 1 2π=3Þ Lrkq cosðθ 1 2π=3Þ (5.2b) f f Lffd 6 Lrr 5 kd 4 Lkdf 0 kq 2

kd Lfkd Lkkd 0

kq 3 0 7 0 5 Lkkq

Lrs ðθÞ 5 ðLsr ðθÞÞT

Lss 5 Lσ 1 Lms (5.2c)

where Lσ is stator leakage inductance and Lms is stator magnetising inductance. The expressions LM cos[f(θ)] in Eq. (5.2a) represent the fluctuating part of the stator self and mutual inductances caused by the magnetically unsymmetrical structure of the rotor of a salient-pole machine. For a symmetrical round rotor machine, LM 5 0. The subscripts rfd, rkd and rkq in Eq. (5.2b) denote mutual inductances between stator winding phase r, and rotor field, d-axis damper and q-axis damper windings, respectively.

Modelling of rotating ac synchronous and induction machines

375

The zeros in the rotor inductance matrix Lrr represent the fact that there is no magnetic coupling between the d and q axes because they are orthogonal, that is, displaced by 90 degrees. If the machine were a static device such as a transformer, then all inductances in Eq. (5.1) would be constant. However, as can be seen in Fig. 5.1B, because of the rotation of the rotor, the self-inductances of the stator winding and the mutual inductances between the stator and the rotor windings vary with rotor angular position. This is defined as θ 5 ωrt, where ωr is rotor angular speed, and is the angle by which the d-axis leads the magnetic axis of phase r winding in the direction of rotation, as illustrated on Fig. 5.1B. The stator and rotor voltage relations for all six windings are given by 3 3 3 2 2 ψr 2 Ra ir er 6ψ 7 6 e 7 6 2 Ra iy 7 7 6 y 7 6 y 7 6 7 7 6 6 7 6 6 eb 7 6 2 Ra ib 7 d 6 ψb 7 71 6 7 6 756 6 e 7 6 R i 7 dt 6 ψ 7 6 fd 7 6 fd 7 6 fd fd 7 7 7 6 6 7 6 4 ψkd 5 4 0 5 4 Rkd ikd 5 2

0

Rkq ikq

(5.3)

ψkq

The zeros on the left-hand side of Eq. (5.3) represent the short-circuited damper windings. The inductances of Eqs. (5.2a) and (5.2b) show that the flux linkages of Eq. (5.1), and also the voltages of Eq. (5.3) are nonlinear functions of rotor angle position. Generally, with the exception of studies of three-phase electromagnetic transients and subsynchronous resonance analysis, this modelling detail is rather unwieldy and impractical for use in large-scale multimachine short circuit, load flow or electromechanical stability analysis.

5.3

Synchronous machine modelling in the Park dq0 frame of reference

5.3.1 Transformation from ryb reference to dq0 reference The inductances of Eqs. (5.2a) and (5.2b) are time-varying through their dependence on rotor angle position. Thus, the flux linkages are also functions of rotor angle position as given by Eq. (5.1). Using Eqs. (5.2a) and (5.2b) in Eq. (5.1), the rotor d and q axes damper winding flux linkages are given by ψkd 5 Lfkd ifd 1 Lkkd ikd 2 Lrkd ½ir cosθ 1 iy cosðθ 2 2π=3Þ 1 ib cosðθ 1 2π=3Þ (5.4a) ψkq 5 Lkkq ikq 1 Lrkq ½ir sinθ 1 iy sinðθ 2 2π=3Þ 1 ib sinðθ 1 2π=3Þ

(5.4b)

376

Power Systems Modelling and Fault Analysis

Eqs. (5.4a) and (5.4b) show a pattern involving the stator currents and rotor position. The outcome is new currents which can be expressed as follows: id 5

 2 ir cosθ 1 iy cosðθ 2 2π=3Þ 1 ib cosðθ 1 2π=3Þ 3

iq 5 2

(5.5a)

 2 ir sinθ 1 iy sinðθ 2 2π=3Þ 1 ib sinðθ 1 2π=3Þ 3

(5.5b)

It can be shown that the constant multiplier in Eqs. (5.5a) and (5.5b) is arbitrary and the choice of two-thirds results in the peak value of id being equal to the peak value of the stator phase current. A third variable can be conveniently defined as the zero-sequence current which is i0 5 13 ðir 1 iy 1 ib Þ. Therefore, the transformation from the stator ryb reference frame to the dq0 reference frame, written using stator currents, is given as iryb 5 TðθÞidq0

(5.6a)

idq0 5 T21 ðθÞiryb

(5.6b)

where 2

2 sinθ

cosθ

6 TðθÞ 5 4 cosðθ 2 2π=3Þ

T21 ðθÞ 5 2

3 ir 6 7 iryb 5 4 iy 5 ib

(5.7a)

2 sinðθ 1 2π=3Þ 1 cosðθ 2 2π=3Þ

cosθ

26 4 2 sinθ 3 1=2

3

7 2 sinðθ 2 2π=3Þ 1 5

cosðθ 1 2π=3Þ 2

1

2 sinðθ 2 2π=3Þ 1=2

3 id 6 7 idq0 5 4 iq 5

cosðθ 1 2π=3Þ

3

7 2 sinðθ 1 2π=3Þ 5 1=2

(5.7b)

2

(5.7c)

i0

The transformation matrix T21(θ) allows us to transform quantities from the ryb reference frame to the dq0 reference frame, and vice versa using matrix T(θ). This transformation method with the frame of reference fixed on the rotor applies to the stator fluxes, voltages and currents, and is generally known as Park’s transformation.

Modelling of rotating ac synchronous and induction machines

377

This transformation process produces a new set of variables associated with two fictitious d and q stator windings that rotate with the rotor such that the stator inductances become constants as seen from the rotor during steady-state operation. Therefore, applying the transformation matrices of Eqs. (5.6a) and (5.6b) to Eq. (5.1), and after much matrix and trigonometric analysis, we obtain 2

ψd 6ψ 6 q 6 6 ψ0 6 6ψ 6 fd 6 4 ψkd

3

2

Ld

0

0

Lrfd

Lrkd

Lq 0

0 L0

0 0

0 0

0 0

0 0

Lffd Lfkd

Lfkd Lkkd

1:5Lrkq

0

0

0

7 6 0 7 6 7 6 7 6 0 756 7 6 1:5L rfd 7 6 7 6 5 4 1:5Lrkd

ψkq

0

0

32

2 id

3

7 6 Lrkq 7 76 2 iq 7 76 7 0 76 2 i0 7 76 7 7 6 0 7 76 ifd 7 76 7 0 54 ikd 5 Lkkq

(5.8a)

ikq

where 3 Ld 5 Lσ 1 ðLms 1 LM Þ 2

Lq 5 Lσ 1

3 ðLms 2 LM Þ 2

L0 5 Lσ

(5.8b)

Similarly, we apply the transformation matrices of Eqs. (5.6a) and (5.6b) to the stator voltages of Eq. (5.3) and note that the matrix T(θ) is derivable because its elements are functions of time through the rotor angular position. Therefore, after much trigonometric analysis, we obtain 3 dθ 2 ψ 32 3 6 2 3 q dt 7 ψd id 0 7 6 7 6 d6 7 76 7 6 7 dθ 7 1 4 ψq 5 0 54 iq 5 1 6 6 ψd dt 7 dt Ra i0 ψ0 5 4 2

2

3

2

Ra 6 7 6 4 eq 5 5 2 4 0 0 e0 ed

0 Ra 0

(5.9)

0

5.3.2 Machine dq0 equations in per unit The equations derived so far are in physical units. However, the analysis is greatly simplified if they are converted into a per-unit form. Several per-unit systems have been proposed in the literature but we will use the equations that correspond to the system known as the Lad base reciprocal per-unit system. In this system, the pu mutual inductances among the three-stator field and damper windings are reciprocal and all mutual inductances between the stator and, field and damper windings, in the d and q axes are equal thus LrfdðpuÞ 5 LfdrðpuÞ 5 LrkdðpuÞ 5 LkdrðpuÞ 5 LadðpuÞ LrkqðpuÞ 5 Lkqr ðpuÞ 5 LaqðpuÞ

and

LfkdðpuÞ 5 Lkdf ðpuÞ

(5.10a)

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Power Systems Modelling and Fault Analysis

where LdðpuÞ 5 LσðpuÞ 1 LadðpuÞ LqðpuÞ 5 LσðpuÞ 1 LaqðpuÞ

(5.10b)

Dropping the explicit pu notation for convenience, it can be shown that Eqs. (5.8a) and (5.9) can be written in pu form as follows: 2

ψd

6ψ 6 q 6 6 ψ0 6 6ψ 6 fd 6 4 ψkd

3

2

Ld 7 6 0 7 6 7 6 7 6 0 756 7 6L 7 6 ad 7 6 5 4 Lad

ψkq

0

0 Lq

0 0

Lad 0

Lad 0

0

L0

0

0

0 0

0 0

Lffd Lfkd

Lfkd Lkkd

Laq

0

0

0

32 3 2 id 0 7 6 Laq 7 76 2 iq 7 76 7 0 76 2 i0 7 76 7 7 6 0 7 76 ifd 7 76 7 0 54 ikd 5 Lkkq

(5.11a)

ikq

and 3 2 3 3 3 2 2 ψd 2 Ra id ed 2 ω r ψq 6ψ 7 6 e 7 6 2 Ra iq 7 6 ω ψ 7 r d 7 7 6 6 q 7 6 q 7 6 7 6 7 7 6 6 7 6 6 e0 7 6 2 Ra i0 7 6 0 7 d 6 ψ0 7 716 7 71 6 6 756 7 6 6e 7 6R i 7 6 0 7 7 dt 6 ψfd 7 6 fd 7 6 fd fd 7 6 7 6 7 7 6 6 7 6 4 ψkd 5 4 0 5 4 Rkd ikd 5 4 0 5 ψkq Rkq ikq 0 0 2

(5.11b)

1 In this pu system, we note that time is also in pu where tBase 5 ωBase 5 1=ð2πfBase Þ with ωBase 5 314:159 rad/s for a 50 Hz system. Also, since any reactance, for example, Xd (Ω) 5 ωLd 5 2πf Ld and XBase ðΩÞ 5 ωBase LBase 5 2πfBase LBase , it follows that if the stator frequency is equal to the base frequency, that is, f 5fBase , then

Xd ðpuÞ 5

Xd ðΩÞ 2πfLd 5 5 Ld ðpuÞ XBase ðΩÞ 2πfBase LBase

that is the pu values of Xd and Ld are equal. It is interesting to note that pu Eqs. (5.11a) and (5.11b) retain the same form as the physical ones, but the factor 1.5 has been eliminated from the rotor flux equations. Eq. (5.11a) can be substituted in Eq. (5.11b) and the result can be visualised using the simplified d and q axes equivalent circuits shown in Fig. 5.2, where Xfd 5 Xffd 2 Xad Xfkd 5 Xad

Xkd 5 Xkkd 2 Xad Xd 5 Xσ 1 Xad

Xkq 5 Xkkq 2 Xaq Xq 5 Xσ 1 Xaq

(5.12)

Fig. 5.2A shows that the machine’s d-axis equivalent circuit consists of a field winding resistance and reactance in series with a voltage source efd, a damper

Modelling of rotating ac synchronous and induction machines

(A)

id ed

(B)

Xσ

Ra +

ifd

ikd + ifd –id

dψ d dt

379

Xad

Rfd efd

–

iq

Xkd Rkd

Xσ

Ra

ikd

Xfd

eq

+ dψ q dt

ikq –iq Xaq

ikq Xkq Rkq

–

Figure 5.2 Synchronous machine d and q axes equivalent circuits: (A) d-axis and (B) q-axis.

winding resistance and reactance, a mutual reactance, a stator leakage reactance and stator resistance. Fig. 5.2B shows the q-axis machine equivalent circuit. This is similar to the d-axis equivalent circuit except that there is no field winding on the q-axis.

5.3.3 Machine operator reactance analysis In order to gain an understanding of the origins of the various machine parameters, for example, subtransient and transient reactances and time constants, we will use the method of d-axis and q-axes operator reactances for its simplicity and the clear insight it provides.

q-Axis operator reactance We will take the Laplace transform of the  stator and rotor flux linkages of Eq. (5.11a) remembering that Laplace dtd f ðtÞ 5 sf ðsÞ 2 f ð0Þ and s 5 jω. Also, we replace the inductance symbol L by the symbol X for reactance since pu inductance and pu reactance are equal. It is also convenient to express the variables in terms of changes about the initial operating point in order to remove the terms that correspond to the initial condition that is Δf (s) 5 f (s)
f (0)/s. Thus Δψq ðsÞ 5 2 Xq Δiq ðsÞ 1 Xaq Δikq ðsÞ

(5.13a)

Δψkq ðsÞ 5 2 Xaq Δiq ðsÞ 1 Xkkq Δikq ðsÞ

(5.13b)

and

Taking the Laplace transform of the q-axis damper voltage equation, that is, that of the last row of Eq. (5.11b), and rearranging, we obtain Δψkq ðsÞ 5

2 Rkq Δikq ðsÞ s

(5.14a)

380

Power Systems Modelling and Fault Analysis

Substituting Eq. (5.14a) into Eq. (5.13b) and rearranging, we obtain Δikq ðsÞ 5

sXaq Δiq ðsÞ Rkq 1 sXkkq

(5.14b)

Substituting Eq. (5.14b) into Eq. (5.13a) and rearranging, we obtain ! 2 sXaq Δψq ðsÞ 5 2 Xq 2 Δiq ðsÞ Rkq 1 sXkkq or Δψq ðsÞ 5 2 Xq ðsÞΔiq ðsÞ

(5.15a)

where the q-axis operator reactance is defined as Xq(s) and is given by Xq ðsÞ 5 Xq 2

2 sXaq Rkq 1 sXkkq

(5.15b)

d-Axis operator reactance Taking the Laplace transform of the first, fourth and fifth rows of Eq. (5.11a), we can write 3 2 Δψd ðsÞ Xd 6 7 6 4 Δψfd ðsÞ 5 5 4 Xad 2

Δψkd ðsÞ

Xad

32 3 2 Δid ðsÞ Xad 76 7 Xad 54 Δifd ðsÞ 5 Δikd ðsÞ Xkkd

Xad Xffd Xad

(5.16a)

Also, taking the Laplace transform of the fourth and fifth rows of Eq. (5.11b),


Δefd ðsÞ 0




5

sΔψfd ðsÞ sΔψkd ðsÞ




 1

Rfd

0

0

Rkd




Δifd ðsÞ Δikd ðsÞ

 (5.16b)

By substituting Δψfd (s) and Δψkd (s) from Eq. (5.16a) into Eq. (5.16b), we can solve for Δifd (s) and Δikd (s) in terms of Δefd (s) and Δid (s). We then substitute this result back into Eq. (5.16a), and use Eq. (5.12), we obtain, after some algebra, Δψd ðsÞ 5 2 Xd ðsÞΔid ðsÞ 1 Gd ðsÞΔefd ðsÞ

(5.17a)

Δifd ðsÞ 5 2 sGd ðsÞΔid ðsÞ 1 Fd ðsÞΔefd ðsÞ

(5.17b)

Modelling of rotating ac synchronous and induction machines

381

where the d-axis operator reactance Xd (s) is given by Xd ðsÞ 5 Xd 2

2 2 Xad ðRfd 1 Rkd Þs 1 Xad ðXfd 1 Xkd Þs2 AðsÞ

(5.18a)

also Gd ðsÞ 5

Xad ðRkd 1 sXkd Þ AðsÞ

(5.18b)

Fd ðsÞ 5

Rkd 1 sXkd AðsÞ

(5.19a)

and 2 AðsÞ 5 Rfd Rkd 1 ðRfd Xkkd 1 Rkd Xffd Þs 1 ðXkkd Xffd 2 Xad Þs2

(5.19b)

5.3.4 Machine parameters
subtransient and transient reactances and time constants q-Axis parameters Using Eq. (5.12), we can rewrite the q-axis operator reactance of Eq. (5.15b) as follows: ! 2 2 Xaq Xkkq Xaq X q 1 Xq 2 s s 2 Xkkq Rkq sXaq Rkq 5 Xq 2 5 Xq ðsÞ 5 Xq 2 Xkkq Xkkq Rkq 1 sXkkq 11s 11s Rkq Rkq or 1 1 sTq00 00 1 1 sTqo

(5.20a)

Xkq 1 Xaq sec ωs Rkq

(5.20b)

Xq ðsÞ 5 Xq where 00 Tqo 5

and

0 Tq00 5

1

C 1 B 1 BXkq 1 Csec @ 1 1A ωs Rkq 1 Xaq Xσ

(5.20c)

382

Power Systems Modelling and Fault Analysis

are the q-axis open-circuit and short-circuit subtransient time constants. We note that we divided these time constants by ωs to convert from pu to seconds. These time constants could also be derived by inspection of Fig. 5.3A. In the case of the open-circuit time constant, the equivalent reactance seen looking from the q-axis damper winding with the machine terminals open-circuited is Xkq in series with Xaq. This is then divided by the q-axis damper winding resistance Rkq to obtain the time constant. In the case of the short-circuit time constant, the equivalent reactance seen looking from the q-axis damper winding with the machine (A)

Xσ Xkq

Opencircuit

Shortcircuit

Xaq

T" q q-axis subtransient T" qo time constants

Rkq

Xσ

Xσ Xkq

Xaq

X" q

Xaq

Xq

q-axis subtransient reactance

q-axis steady-state reactance

(B)

Xσ Opencircuit

Shortcircuit

Xkd Xad

Xfd Rkd

Xσ Shortcircuit

Opencircuit

Rfd

Xσ

X" d

Xfd Xad

T" d d-axis T" do subtransient time constants

T'd d-axis T'do transient time constants Xσ

Xσ

Xad

Xfd

Xkd

d-axis subtransient reactance

X'd

Xad

Xfd

d-axis transient reactance

Xd

Xad

d-axis steady state reactance

Figure 5.3 Machine parameters from its equivalent circuits: (A) q-axis: subtransient and steady-state equivalent circuits and (B) d-axis: subtransient, transient and steady-state equivalent circuits.

Modelling of rotating ac synchronous and induction machines

383

terminals short-circuited is Xkq in series with the parallel combination of Xaq and Xσ. This is then divided by the q-axis damper winding resistance Rkq to obtain the time constant. The effective machine q-axis reactance at the instant of an external disturbance is defined as the subtransient reactance Xq00 . Using Eq. (5.20a), this is given by Xq00 5 lim Xq ðsÞ 5 Xq s!N

Tq00 00 Tqo

and

Tq00 5

Xq00 00 T Xq qo

(5.21a)

Substituting Eqs. (5.20b) and (5.20c) into Eq. (5.21a) and using Eq. (5.12), we obtain after some algebra Xq00 5 Xσ 1

1 Xaq

1 1

1 Xkq

(5.21b)

Again, the above q-axis subtransient reactance could be derived by inspection from Fig. 5.3A with Rkq 5 0 because the rotor flux linkages cannot change instantly following a disturbance. This is equivalent to looking into the machine from its q-axis terminals. This shows Xσ is in series with the parallel combination of Xaq and Xkq. In the absence of a disturbance or under steady-state conditions, the effective machine q-axis reactance is defined as the synchronous reactance Xq. Using Eq. (5.20a) Xq ð0Þ 5 lim Xq ðsÞ 5 Xq 5 Xσ 1 Xaq s!0

(5.21c)

It should be noted that using the q-axis operator reactance of Eq. (5.20a), no q-axis transient time constant or transient reactance are defined and the latter can be assumed equal to the steady-state value. This is due to our use of one q-axis damper winding on the q-axis as shown in Fig. 5.2B. This representation is accurate for laminated salient-pole machines and still reasonably accurate for a round rotor machine which, sometimes, might be represented as two parallel damper winding circuits on the q-axis. In the latter case, the second winding may be used to represent the body of the solid rotor. The q-axis operator reactance of Eq. (5.20a) can be expressed in terms of partial fractions as follows: ! 00 1 1 sTqo sTq00 1 1 1 1   5 5 1 2 Xq ðsÞ Xq ð1 1 sTq00 Þ Xq Xq00 Xq 1 1 sTq00

(5.21d)

384

Power Systems Modelling and Fault Analysis

d-Axis parameters Following extensive algebraic manipulations, we can show that Eq. (5.18a) can be written in the following form: Xd ðsÞ 5 Xd

1 1 ðT4 1 T5 Þs 1 ðT4 T6 Þs2 1 1 ðT1 1 T2 Þs 1 ðT1 T3 Þs2

(5.22a)

where T1 5

T3 5

T5 5

Xad 1 Xfd ωs Rfd 0

T2 5 1

C 1 B 1 B C BXkd 1 1 1 C A ωs Rkd @ 1 Xad Xfd 0 1 C 1 B 1 B C BXkd 1 1 1C A ωs Rkd @ 1 Xad Xσ

Xad 1 Xkd ωs Rkd 0

T4 5

T6 5

1

C 1 B 1 B C BXfd 1 1 1C A ωs Rfd @ 1 Xad Xσ 0

1

C 1 B 1 B C BXkd 1 1 1 1 C A ωs Rkd @ 1 1 Xσ Xad Xfd (5.22b)

The numerator and denominator of Eq. (5.22a) can be expressed in terms of factors as follows: Xd ðsÞ 5 Xd

ð1 1 sT 0d Þð1 1 sTd00 Þ 00 Þ ð1 1 sT 0do Þð1 1 sTdo

(5.23a)

where 00

T 0do and Tdo are the transient and subtransient open-circuit time constants; 00 T 0d and Td are the transient and subtransient short-circuit time constants.

Accurate expressions for the machine open-circuit and short-circuit time constants can be derived by equating the numerator of Eq. (5.22a) with that of Eq. (5.23a), and similarly for the denominators. However, such a procedure would involve the solutions of quadratic equations and the values of the open-circuit and short-circuit time constants would result in rather involved expressions in terms of the time constants of Eq. (5.22b). Reasonable approximations can be made on the basis that the field winding resistance is much smaller than the damper winding resistance. This means that T2 (and T3) , , T1 and T5 (and T6) , , T4. Therefore, Eq. (5.22a) can be written as Xd ðsÞDXd

ð1 1 sT4 Þð1 1 sT6 Þ ð1 1 sT1 Þð1 1 sT3 Þ

(5.23b)

Modelling of rotating ac synchronous and induction machines

385

Therefore, from Eqs. (5.23a) and (5.23b), we have T 0do 5 T1

00 Tdo 5 T3

T 0d 5 T4

Td00 5 T6

(5.23c)

The effective machine d-axis reactance at the instant of an external disturbance is defined as the subtransient reactance Xd00 . Using Eq. (5.23a), this is given by Xd00 5 lim Xd ðsÞ 5 Xd s!N

T 0d Td00 00 T 0do Tdo

(5.24a)

This can also be expressed in terms of the internal machine d-axes parameters using Eqs. (5.22b) and (5.23c), hence Xd00 5 Xσ 1

1 1 1 1 1 1 Xad Xfd Xkd

(5.24b)

After the decay of the currents in the damper winding following an external disturbance, the d-axis operator reactance can be represented by setting Xkd 5 0 and Rkd!N in Eq. (5.23b) and making use of Eq. (5.23c). The result is given by Xd ðsÞ 5 Xd

1 1 sT 0d 1 1 sT 0do

(5.25a)

Eq. (5.25a) can in fact also represent the case of a machine with only a field winding on the rotor but no damper winding. The effective machine d-axis reactance at the beginning of a disturbance is now defined as the transient reactance X 0d . Using Eq. (5.25a), this is given by X 0d 5 lim Xd ðsÞ 5 Xd s!N

T 0d T 0do

(5.25b)

This can also be expressed in terms of the internal machine d-axis parameters using Eqs. (5.22b) and (5.23c), hence X 0d 5 Xσ 1

1 1 1 1 Xad Xfd

(5.25c)

From Eqs. (5.24a) and (5.25b), the following useful expression can be obtained Xd00 5 X 0d

Td00 00 Tdo

and

Td00 5

Xd00 00 T X 0d do

(5.26a)

386

Power Systems Modelling and Fault Analysis

In the absence of a disturbance or under steady-state conditions, the effective machine d-axis reactance is defined as the synchronous reactance Xd. Using Eqs. (5.23a) or (5.25a) Xd ð0Þ 5 lim Xd ðsÞ 5 Xd 5 Xσ 1 Xad s!0

(5.26b)

Again, the d-axis subtransient and transient time constants and reactances, and the steady-state reactance can be calculated by inspection from Fig. 5.3B. The d-axis operator reactance of Eq. (5.23a) can be rewritten as follows:  00 00 00 Þ 1 1 1 1 sT 0do ð1 1 sTdo T 0do Tdo ðs 1 1=T 0do Þðs 1 1=Tdo Þ  5 5 0 00 0 00 0 00 Xd ðsÞ Xd ð1 1 sT d Þð1 1 sTd Þ Xd T d Td ðs 1 1=T d Þðs 1 1=Td Þ This expression can be approximated to a sum of three partial fractions assuming 00 that Td00 and Tdo are small compared with T 0d and T 0do . The result is given by



 1 1 1 1 sT 0d 1 1 sTd00 5 1 1 2 2 Xd ðsÞ Xd X 0d Xd ð1 1 sT 0d Þ Xd00 X 0d ð1 1 sTd00 Þ

(5.27)

Eq. (5.27) describes the machine reactance variation in the s domain as given in both IEC and IEEE standards for synchronous machines.

5.4

Synchronous machine behaviour under short-circuit faults and modelling in the sequence reference frame

Having established the machine d and q axes reactances, we are now able to proceed with analysing the behaviour of the machine under sudden network changes such as the occurrence of balanced or unbalanced short-circuit faults. Our objective is to understand the nature of short-circuit currents of synchronous machines and, in order to do so, it is important to understand the meaning of positive-phase sequence, negative-phase sequence and zero-phase sequence machine impedances.

5.4.1 Synchronous machine sequence equivalent circuits In Section 5.3, we defined the machine reactances and time constants in the dq0 reference frame. In short-circuit analysis, the positive-sequence, negative-sequence and zero-sequence models or equivalent circuits and sequence reactances of the machine need to be defined. The three-phase synchronous generator is designed to produce a set of balanced three-phase voltages having the same magnitude and displaced by 120 degrees. As we have already seen in Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, the

Modelling of rotating ac synchronous and induction machines

(A)

I

(B)

P

+

+ –

V

P

I jX N

j XP

R

387

P

R

N

E

N

+ V

N

–

–

(C) (i)

I

(ii)

r

Stator windings

jXZ R

Z

Z

+ V

Z

3ZE y

–

:1

Ze

b

Figure 5.4 Synchronous machine: (A) positive-sequence, (B) negative-sequence and (C) zero-sequence equivalent circuits.

transformation of these balanced three-phase voltages into the sequence reference frame produces a positive-sequence voltage source in the positive-sequence network but zero voltages in the negative-sequence and zero-sequence networks. The stator windings of the machine are usually star-connected and the neutral is normally earthed through an earthing impedance, typically a resistance, that is, generally connected to the secondary of a distribution transformer which is connected to the neutral as illustrated in Fig. 5.4(C). The effective neutral impedance ZE appearing on the stator neutral side is N2 times the earthing impedance Ze. The earthing impedance is usually chosen to limit the machine current under a single-phase short-circuit terminal fault to the rated current of the machine. We designate the machine positive-sequence, negative-sequence and zerosequence reactances as XP, XN and XZ, respectively. The machine positivesequence, negative-sequence and zero-sequence equivalent circuits are shown in Fig. 5.4, which includes an illustration of the neutral earthing through a distribution transformer. The sequence quantities are those that correspond to phase r. As expected, the neutral earthing impedance appears only in the zero-sequence equivalent circuit and is multiplied by a factor of 3 since three times the zero-sequence current flows through it. The values of the sequence reactances and resistances are discussed in the following sections.

5.4.2 Three-phase short-circuit faults Short-circuit currents Before the occurrence of the short circuit, the machine is assumed to be in an opencircuit steady-state condition and the rotor speed is the synchronous speed ωs.

388

Power Systems Modelling and Fault Analysis

The field voltage and current are constant, the damper currents are zero and the armature phase voltages are balanced three-phase quantities. Let t 5 0 be the instant of the short circuit that occurs at the machine terminals and let θo be the angle between the axis of phase r and d-axis at t 5 0. Thus, θo defines the point in the voltage waveform at which the short-circuit occurs. The following initial conditions just before the short circuit, using Eqs. (5.11a) and (5.11b), are obtained pffiffiffi er ðtÞ 5 2Eo cosðωs t 1 θo Þ ido 5 iqo 5 ikdo 5 ikqo 5 ψqo 5 edo 5 0 pffiffiffi ψdo 5 Lad ifdo eqo 5 ψdo 5 2Eo

(5.28a)

We assume that, during the brief short-circuit period, the machine rotor speed and field voltage remain constant. The latter assumes no automatic voltage regulator (AVR) action. Therefore, immediately at the instant of the short circuit, we have Δefd 5 0

efd 5 efdo 5 Rfd ifdo

ed 5 eq 5 0

(5.28b)

Remembering that Δf(s) 5 f(s) 2 f (0)/s, the stator d- and q-axis voltages in Eq. (5.11b), transformed into the Laplace domain and using Eqs. (5.28a) and (5.28b), can be written as follows:


 ψdo 0 5 2 Ra id ðsÞ 1 s ψd ðsÞ 2 2 ω s ψ q ð sÞ s

(5.29a)

0 5 2 Ra iq ðsÞ 1 sψq ðsÞ 1 ωs ψd ðsÞ

(5.29b)

Using Eq. (5.28a), Eq. (5.15a) becomes ψq ðsÞ 5 2 Xq ðsÞiq ðsÞ

(5.30a)

Using Eqs. (5.28a) and (5.28b), Eq. (5.17a) becomes ψd ðsÞ 2

ψdo 5 2 Xd ðsÞid ðsÞ s

(5.30b)

Substituting Eqs. (5.30a) and (5.30b) into Eqs. (5.29a) and (5.29b), and rearranging, we obtain ½Ra 1 sXd ðsÞid ðsÞ 2 ωs Xq ðsÞiq ðsÞ 5 0 

pffiffiffi  ω s 2 Eo Ra 1 sXq ðsÞ iq ðsÞ 1 ωs Xd ðsÞid ðsÞ 5 s

(5.31a) (5.31b)

Modelling of rotating ac synchronous and induction machines

389

Solving Eqs. (5.31a) and (5.31b) for d and q axes stator currents, we obtain id ðsÞ 5

pffiffiffi Xq ðsÞ 2Eo ω2s sBðsÞ

and

iq ðsÞ 5

pffiffiffi Ra 1 sXd ðsÞ 2 Eo ω s sBðsÞ

(5.32a)

where BðsÞ 5 ½Ra 1 sXd ðsÞ½Ra 1 sXqðsÞ 1 ω2s Xd ðsÞXq ðsÞ

(5.32b)

We will make a number of simplifications that help us to obtain the closed-form time domain solution of the d and q axes stator currents of Eq. (5.32a). From Eq. (5.32b) 8 9 < = ½R 1 sX ðsÞ½R 1 sX ðsÞ a d a q BðsÞ 5 Xd ðsÞXq ðsÞ ω2s 1 : ; Xd ðsÞXq ðsÞ 8 9 0 1 < = 2 R 1 1 a A 1 s2 5 Xd ðsÞXq ðsÞ ω2s 1 1 sRa @ 1 : ; Xd ðsÞ Xq ðsÞ Xd ðsÞXq ðsÞ 8 9 0 1 < = 1 1  Xd ðsÞXq ðsÞ s2 1 sRa @ 00 1 00 A 1 ω2s : ; Xd Xq

  Xd ðsÞXq ðsÞ ðs11=Ta Þ2 1 ω2s

(5.33a)

where 2

Ta 5 ω s Ra

1 1 1 00 Xd00 Xq

!5

2Xd00 Xq00   sec ωs Ra Xd00 1 Xq00

(5.33b)

Ta is the stator or armature short-circuit time constant and is divided by ωs to convert it from pu to seconds and Ra is the stator resistance. In arriving at the final form of Eq. (5.33a), we made two approximations which, in practice, would have a negligible effect. First, we ignored R2a =½Xd ðsÞXq ðsÞ because it is too small in comparison with ω2s . We then added 1=ðTa Þ2 where 1=ðTa Þ2 , , ω2s . We also replaced Xd (s) with Xd00 , and Xq(s) with Xq00 in the term that multiplies sRa because we assumed that the rotor field and damper winding resistances are very small. This means that the factors (1 1 sT) in Eqs. (5.20a) and (5.23a) can be replaced with sT. Therefore, substituting Eq. (5.33a) into Eqs. (5.32a), we obtain pffiffiffi 2Eo ω2s id ðsÞ 5 s½ðs11=Ta Þ2 1 ω2s Xd ðsÞ

(5.34a)

390

Power Systems Modelling and Fault Analysis

and pffiffiffi 2 Eo ω s iq ðsÞ 5 ½ðs11=Ta Þ2 1 ω2s Xq ðsÞ

(5.34b)

Substituting Eqs. (5.21d) and (5.27) into Eqs. (5.34a) and (5.34b) and taking the inverse Laplace transform, it can be shown that the time domain d and q axes currents are given by id ðtÞ 5






 pffiffiffi 1 1 1 1 1 1 2t=Ta 2t=T 0d 2t=Td00 2Eo 1 2 1 2 2 e cosω t e e s Xd X 0d Xd Xd00 X 0d Xd00 (5.35a)

iq ðtÞ 5

pffiffiffi 1 2t=T a 2Eo 00 e sinωs t Xq

(5.35b)

The stator or armature phase r current can be calculated by transforming the d and q axes currents into the phase frame of reference using Eqs. (5.6a) and (5.7a) and noting that under balanced conditions io 5 0. Therefore, using θ 5 ωst 1 θo, ir(t) 5 id (t) cosθ
iq (t) sinθ, and after much trigonometric analysis, we obtain 2

0 1 0 1 3 pffiffiffi 00 1 1 1 1 1 0 ir ðtÞ  4 1 @ 0 2 Ae2t=T d 1 @ 00 2 0 Ae2t=Td 5 2Eo cosðωs t 1 θo 2 π=2Þ Xd Xd Xd Xd Xd |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Power frequency 0 ðacÞ current component: steady
state and two transient components 1 pffiffiffi 1 1 1 2 @ 00 1 00 Ae2t=Ta 2Eo cosðθo 2 π=2Þ 2 Xd Xq |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} unidirectional ðdcÞ current component 0 Transient 1 pffiffiffi 1 1 1 2 @ 00 2 00 Ae2t=Ta 2Eo cosð2ωs t 1 θo 2 π=2Þ 2 Xd Xq |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Transient double
frequency current component

(5.36) The currents in the two stator phases iy(t) and ib(t) are calculated with θo replaced by (θo
2π/3) and (θo 1 2π/3), respectively. Eq. (5.36) represents the total phase r current following a sudden three-phase short circuit fault at the machine terminals with the machine initially unloaded but running at rated speed. The short-circuit current contains three main components; a power frequency (e.g., 50 or 60 Hz) component, a transient unidirectional or dc component and a transient double-frequency (e.g., 100 or 120 Hz) component.

Modelling of rotating ac synchronous and induction machines

391

Both the dc component and the double-frequency component decay to zero with a time constant equal to Ta which is typically between 0.1 and 0.4 seconds. These current components are illustrated in Fig. 5.5. The initial magnitude of the dc component is dependent on the instant of time at which the short circuit occurs, that is, θo and therefore the initial magnitudes of the dc components in the three phases are different but they all decay at the same rate given by the time constant Ta. The dc component in each phase appears in order to satisfy the physical condition that the current cannot change instantaneously at the instant of fault. The initial value of the transient double-frequency component is small as it is due to the difference between the d-axis and q-axis subtransient reactances ðXq00 2 Xd00 Þ; an effect termed subtransient saliency. This current component disappears if the d and q subtransient reactances are equal. This component is neglected in network short-circuit analysis. The power frequency component itself consists of three subcomponents as shown in Eq. (5.36) and illustrated in Fig. 5.5B for a 165-MVA synchronous generator. These are termed the subtransient, transient and steady-state components. The subtransient component decays to zero with a time constant Td00 and typically lasts for up to 0.15 seconds. The transient component decays much more slowly to zero with a time constant T 0d and typically lasts for up to 5 seconds. The steady-state or sustained component is constant. The parameters of the machine that determine the magnitude and rate of decay of each current component are the various reactances and short-circuit time constants shown in Eq. (5.36). The sum of the dc and power frequency ac components produces an asymmetrical short-circuit current waveform. The phase r asymmetrical current is illustrated in Fig. 5.5C.

Positive-sequence reactance and resistance The three-phase short-circuit is a balanced condition resulting in balanced ac currents in the three phases of the machine. From Eq. (5.36), we can express the instantaneous power frequency component of the short-circuit current as ir(t) 5 Real[Ir(t)] where Ir(t) is phase r complex instantaneous current given by Ir ðtÞ 5

1 p Xð3φÞ ðtÞ

where Er ðtÞ 5

pffiffiffi jðω t1θ 2π=2Þ Er ðtÞ 2E o e s o 5 p jXð3φÞ ðtÞ

pffiffiffi jðω t1θ Þ 2Eo e s o is phase r complex instantaneous voltage and



 1 1 1 2t=T 0 1 1 2t=T 00 d d 1 2 1 2 0 e 5 e p X 0d Xd Xd00 Xd Xð3φÞ ðtÞ Xd 1

(5.37a)

(5.37b)

is the time-dependent positive-sequence reactance of the machine and is shown in Fig. 5.6. As presented in Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, the phase r complex voltage can also

392

Power Systems Modelling and Fault Analysis (A)

8 Short-circuit current components (pu) ac power frequency component

6 4

2 0 –2

0

100

200

300

400

500

600

700

800

Time (ms)

–4 –6 –8

(B)

double-frequency component

dc component

5 3

ac power frequency: subtransient current component

1 –1 0

100

200

300

400

500

600

700 800 Time (ms)

600

700

–3 –5 5

ac power frequency: transient current component

3 1 –1 0

100

200

300

400

500

800

Time (ms)

–3 –5 5

3

ac power frequency: steady state current component

1 –1 0

100

200

300

400

500

600

700 800 Time (ms)

500

600

700

–3 –5

(C)

Phase r asymmetrical current (pu)

3 0 –3

0

100

200

300

400

800

Time (ms) –6 –9 –12 –15

Figure 5.5 Three-phase short-circuit fault at synchronous machine terminals: (A) the three components of the short-circuit current, (B) the three subcomponents of the ac power frequency component and (C) phase r asymmetrical short-circuit current.

Modelling of rotating ac synchronous and induction machines

1/XP(t)

6

393

Xd = 2pu, X' d = 0.25pu, X''d = 0.18pu

5 4

XP(t)

1/XP(t)

3

2 XP(t) pu 1.5 1

2

0.5 1 0

0 0

0.5

1

1.5

2 2.5 Time (s)

3

3.5

4

Figure 5.6 Synchronous machine time-dependent positive-sequence reactance under a threephase short-circuit at machine terminals. (A) P

jX 3φ ( t )

Ra

IP (t )

equivalent circuit with time-dependent machine reactance

+ E r (t) –

(B)

P jX 3φ

Ra + Eo –

X ′′d t = 0, start of subtransient period I ′′ X ′d t = 0, transient reactance P X 3φ I′ = P X ′ Eq. (5.37c) I t = 5T ′′d , end of subtransient period P I′ Xd steady state, end of transient period I

Figure 5.7 Synchronous machine equivalent circuits for a three-phase short-circuit at machine terminals: (A) positive-sequence time-dependent equivalent circuit and (B) fixedimpedance positive-sequence equivalent circuits at various time instants.

pffiffiffi be written as Er ðtÞ 5 Er ejωs t where Er is a complex phasor given by Er 5 2Eo ejθo . As in the case of instantaneous currents, the complex phase y and phase b currents Iy (t) and Ib (t) are obtained by replacing θo of Ir (t) with (θo 2 2π/3) and (θo 1 2π/3), respectively. The complex instantaneous positive-sequence current is given by I P(t) 5 [Ir(t) 1 hIy(t) 1 h2Ib(t)]/3 5 Ir(t). Fig. 5.7A shows the machine positive-sequence equivalent circuit with a timedependent equivalent reactance. We define such an equivalent circuit as a transient

394

Power Systems Modelling and Fault Analysis

positive-sequence symmetrical component equivalent circuit. Fig. 5.7B shows the fixed-impedance approach of the machine positive-sequence reactance that is conventionally considered to consist of three components; subtransient, transient and steady-state components. If a time domain short-circuit analysis technique is used, then Eq. (5.37b) or Eq. (5.36) can be used directly to calculate the machine reactance or current at any time instant following the occurrence of the fault. However, the vast majority of large-scale short-circuit analysis computer programs used in practice use fixedimpedance analysis techniques and this will be discussed in Chapter 7, Short-circuit analysis techniques in large-scale ac power systems. Essentially, different values of reactances are used to calculate short-circuit currents at different time instants following the onset of the fault. At the instant of short-circuit fault, t 5 0, that is, the start of the subtransient period, Eq. (5.37b) gives X p 5 Xd00 . Also, neglecting the subtransient current component, the value of the transient reactance at t 5 0 is obtained by putting t 5 0 in Eq. (5.37b) giving X p 5 X 0d . The steady-state reactance XP 5 Xd applies from the end of the transient period when the transient current component has vanished, that is, t $ 5T 0d . Another machine positive-sequence reactance, found useful in practice, is the reactance that applies at the end of the subtransient period, that is, at t 5 5Td00 . Thus, using Eq. (5.37b), this is given by Xp 5

1

 1 1 1 25T 0 0 =T 0 1 e d d 0
Xd Xd Xd

(5.37c)

In summary, the four positive-sequence reactances are given by

X 5 P

8 00 Xd > > > > > > < Equationð5:37cÞ > > X 0d > > > > : Xd

t 5 0; start of subtransient period t 5 5T 00d ; end of subtransient period t 5 0; neglecting the subtransient current component t $ 5T 0d ; steady state; end of the transient period (5.37d)

For typical salient-pole and round rotor synchronous machines, Eq. (5.37c) shows that the positive-sequence reactance at the end of the subtransient period is typically equal to 1.1 X 0d to 1.5 X 0d . The stator or armature dc resistance Ra is very small. The positive-sequence power frequency stator resistance includes, in addition to stator losses, stray, hysteresis and eddy current losses and may be 1.5
2 times the stator dc resistance. The machine positive-sequence equivalent circuit, depending on the calculation time period of interest, is shown in Fig. 5.7B.

Modelling of rotating ac synchronous and induction machines

395

Short-circuit fault through an external impedance In many practical situations, the location of short-circuit fault will be on the network to which the machine is connected either directly or through a dedicated transformer. Therefore, an equivalent impedance will be present between the machine and the fault location. The effect of such an external impedance, denoted (Re 1 sXe), shown in Fig. 5.8, is now considered. The effect of the external impedance (Re 1 sXe) can be considered to be equivalent to modifying the stator leakage reactance and stator resistance as shown in Fig. 5.8. The analysis is in fact identical to the previous case without an external impedance except that the operator reactances will need to be modified to include the external impedance (Re 1 sXe). Using Eqs. (5.32a) and (5.32b), it can be shown that   pffiffiffi ω2s Xq ðsÞ 1 Xe id ðsÞ 5 2Eo sDðsÞ

iq ðsÞ 5

pffiffiffi ωs ½Ra 1 Re  1 s½Xd ðsÞ 1 Xe  2 Eo sDðsÞ

where

  DðsÞ 5 ðRa 1 Re Þ 1 s½Xd ðsÞ 1 Xe  ðRa 1 Re Þ 1 s½Xq ðsÞ 1 Xe  1 ω2s ½Xd ðsÞ 1 Xe ½Xq ðsÞ 1 Xe  The direct analytical solution of these equations is very tedious. An alternative approach is to calculate the short-circuit current components individually by

(A)

Machine terminals Re

Xe

Shortcircuit

Xσ Xad

Xfd

Xkd

Rfd efd+

Rkd

−

Machine terminals

(B)

Re Shortcircuit

Xe

Shortcircuit

Xσ

Xkq Xaq Rkq

Figure 5.8 Three-phase short-circuit seen by a synchronous machine through an external impedance: (A) d-axis and (B) q-axis.

396

Power Systems Modelling and Fault Analysis

substituting the relevant reactances. For example, to calculate the subtransient, transient and steady-state power frequency current components, we substitute Xd00 ; X 0d and Xd in place of ωs Xd ðsÞ; and Xq00 and Xq in place of ωsXq(s). Without going through the long mathematical analysis, it can be shown that the phase r shortcircuit current, ignoring the double-frequency component, is given by " 1 1 ir ðtÞ 5 X d 1 Xe



 1 1 0 2 e2t=T de 0 X d 1 Xe Xd 1 Xe

#  pffiffiffi 1 1 00 2t=Tde 2Eo cosðωs t 1 θo 2 π=2Þ 1 2 e Xd00 1 Xe X 0d 1 Xe

 pffiffiffi 2t=T 1 1 1 ae 2E o e 2 1 00 cosðθo 2 π=2Þ 2 Xd00 1 Xe Xq 1 Xe

(5.38a)

As before, the other two stator phases iy(t) and ib(t) are obtained with θo replaced by (θo 2 2π/3) and (θo 1 2π/3), respectively. Similar to the case with no external impedance, it can be shown that the effective short-circuit time constants in Eq. (5.38a) can be expressed in terms of the open-circuit time constants as follows: T 0de 5

ðRa 1Re Þ2 1 ðX 0d 1Xe Þ2 T 0do 2 0 ðRa 1Re Þ 1 ðX d 1 Xe ÞðXd 1 Xe Þ

ðRa 1Re Þ2 1 ðX 0d 1Xe Þ2 Xd 0 3 5 Td ðRa 1Re Þ2 1 ðX 0d 1 Xe ÞðXd 1 Xe Þ X 0d 00 Tde 5

ðRa 1Re Þ2 1 ðXd00 1Xe Þ2 00 Tdo ðRa 1Re Þ2 1 ðXd00 1 Xe ÞðX 0d 1 Xe Þ

ðRa 1Re Þ2 1 ðXd00 1Xe Þ2 X 0d 00 3 5 Td ðRa 1Re Þ2 1 ðXd00 1 Xe ÞðX 0d 1 Xe Þ Xd00 Tae 5

ωs ðRa 1 Re Þ

h

2

Xd00

1 1 Xe

1

Xq00

1 1 Xe

i

(5.38b)

(5.38c)

(5.38d)

Eq. (5.38d) can be approximated, with insignificant loss of accuracy, to Tae 5

1 00 2 ðXd

1 Xq00 Þ 1 Xe Xd00 1 Xe  ωs ðRa 1 Re Þ ωs ðRa 1 Re Þ

assuming Xd00 5 Xq00

(5.38e)

Modelling of rotating ac synchronous and induction machines

397

(A)

j Xe R e

P (t) j X 3φ

IP (t )

Ra + Er (t) -

equivalent circuit with time-dependent machine reactance

(B)

P j X 3φ

Ra + Ra -

j Xe

Re P

X 3φ = I

X ′′d t = 0, start of subtransient period P I ′′ X ′d t = 0, transient reactance P I′ Eq. (5.37c) t = 5T ′′d , end of subtransient period P I′ Xd steady state, end of transient period P I

Figure 5.9 Synchronous machine equivalent circuits for a three-phase short-circuit through an external impedance: (A) positive-sequence time-dependent equivalent circuit and (B) fixed-impedance positive-sequence equivalent circuits of various time instants.

Usually the stator and external resistances are much smaller than the reactances. Thus, the transient and subtransient time constants reduce to the following: T 0de 5

X 0d 1 Xe 0 X 0 1 Xe Xd T do 5 d 3 0 T 0d Xd 1 X e Xd 1 Xe Xd

(5.39a)

00 Tde 5

Td00 1 Xe 00 Xd00 1 Xe X 0d 00 T 5 3 T X 0d 1 Xe do X 0d 1 Xe Xd00 d

(5.39b)

The effect of the external resistance Re should not be neglected in calculating the armature or dc time constant given in Eq. (5.38e). The positive-sequence equivalent circuits of the machine, ‘seeing’ a short circuit through an external impedance, depending on the calculation time period of interest, are shown in Fig. 5.9.

Simplified machine short-circuit current equations In many practical short-circuit calculations, simplified short-circuit equations can be used assuming Xd00 5 Xq00 . Assuming maximum dc current offset and ignoring the

398

Power Systems Modelling and Fault Analysis

double-frequency component, the peak current envelope of Eq. (5.36), at any time instant, is given by i^r ðtÞ 5






  pffiffiffi 1 1 1 2t=T 0 1 1 2t=T 00 1 2t=Ta d 1 d 1 2E o 1 2 2 e e e Xd X 0d Xd Xd00 X 0d Xd00 (5.40a)

where Ta 5

Xd00 ω s Ra

(5.40b)

In the case of a fault through a predominantly inductive external impedance, the peak current envelope of Eq. (5.38a) at any time instant is given by i^r ðtÞ 5

"



 1 1 0 2 e2t=T de X 0d 1 Xe Xd 1 Xe #

 00 1 1 1 1 2 0 e2t=Tae e2t=Tde 1 00 Xd00 1 Xe X d 1 Xe X d 1 Xe

pffiffiffi 2Eo

1 1 Xd 1 X e

(5.40c)

5.4.3 Unbalanced two-phase (phase-to-phase) short-circuit faults Consider a two-phase or phase-to-phase short-circuit fault at the machine terminals involving phases y and b with the machine initially running at synchronous speed, rated terminal voltage and on open circuit. The initial conditions prior to the fault are given in Eqs. (5.28a) and (5.28b). The constraint equations that define the twophase short-circuit fault are given by ey 5 eb ir 5 0

or and

ey 2 eb 5 0 iy 1 ib 5 0 or

(5.41a) iy 5 2 ib

(5.41b)

Transforming Eq. (5.41a) and (5.41b) to the dq0 reference using Eq. (5.7b), we obtain ed sinðωs t 1 θo Þ 1 eq cosðωs t 1 θo Þ 5 0

(5.41c)

id cosðωs t 1 θo Þ 5 iq sinðωs t 1 θo Þ

(5.41d)

io 5 0

(5.41e)

Modelling of rotating ac synchronous and induction machines

399

Eqs. (5.41c)
(5.41e) and the machine Eqs. (5.11a) and (5.11b) that relate currents to fluxes and rate of change of fluxes to voltages provide a set of equations that are sufficient to obtain the solution of currents under a phase-to-phase short-circuit fault at the machine terminals. However, the resulting current equations are complex and nonlinear and a closed form solution is extremely tedious. Instead, a step-by-step simplification process, similar to that described for a three-phase fault, can be used. The short-circuit current can be shown to contain a power frequency component, a dc component and both even and odd harmonic orders of the power frequency, the latter due to subtransient saliency. However, for practical calculations of machine phase-to-phase short-circuit currents involving the power frequency and dc components only and neglecting the harmonic components, it can be shown that the phase y and b short-circuit currents are given by pffiffiffipffiffiffi iy ðtÞ 5 2 ib ðtÞ  3 2Eo

"

0

1

1 1 1 A 2t=T 0dð2φÞ e 1@ 0 2 N N Xd 1 X Xd 1 X Xd 1 X N 0 1 # pffiffiffipffiffiffi 00 1 1 1 Ae2t=Td ð2φÞ cosðωs t 1 θo Þ 2 3 2Eo cosθo 1 @ 00 2 e2t=Tað2φÞ Xd 1 XN X 0d 1 X N X 00d 1 X N (5.42a)

where T 0dð2φÞ 5

X 0d 1 X N 0 T Xd 1 X N do

(5.42b)

00 5 Tdð2φÞ

Xd00 1 X N 00 T X 0d 1 X N do

(5.42c)

Tað2φÞ 5

XN ω s Ra

(5.43a)

XN 5

qffiffiffiffiffiffiffiffiffiffiffi Xd00 Xq00

(5.43b)

and X N is the machine negative-sequence reactance. The maximum dc current component in Eq. (5.42a) occurs when θo 5 0. From Eq. (5.42a), we can express the instantaneous power frequency component of the short-circuit current as iy(t) 5 Real[Iy(t)] where Iy(t) is phase y complex instantaneous current and is given by pffiffiffipffiffiffi pffiffiffi 3 2Eo jðωs t1θo Þ 3Er ðtÞ e Iy ðtÞ 5 P 5 p 5 2 Ib ðtÞ Xð2φÞ ðtÞ Xð2φÞ ðtÞ

(5.44a)

400

Power Systems Modelling and Fault Analysis

and Er ðtÞ 5

pffiffiffi jðω t1θ Þ 2Eo e s o 5 Er ejωs t

and

Er 5

pffiffiffi jθ 2 Eo e o

Er(t) and Er are phase r complex instantaneous voltage and complex phasor voltage, respectively. And 0 1 1 1 1 Ae2t=T 0dð2φÞ 1@ 0 2 5 p N N N Xd 1 X Xd 1 X Xð2φÞ ðtÞ Xd 1 X 0 1 00 1 1 A 2t=Tdð2φÞ e 1 @ 00 2 0 N N Xd 1 X Xd 1 X 1

(5.44b)

is the time-dependent machine positive-sequence reactance. At the fault instant t 5 p 5 Xd00 1 X N . At t 5 0 and 0, that is, the start of the subtransient period, Xð2φÞ p 0 N neglecting the subtransient component, Xð2φÞ 5 X d 1 X . At t $ 5T 0dð2φÞ , that is, the p end of the transient period, Xð2φÞ 5 Xd 1 X N . The machine reactance at the end of 00 the subtransient period, that is, at t  5Tdð2φÞ is given by p Xð2φÞ 5

1

  00 1 1 1 25Tdð2φÞ =T 0dð2φÞ 1 2 e Xd 1 X N X 0d 1 X N Xd 1 X N

(5.44c)

In summary, the four fixed machine reactances are given by

p 5 Xð2φÞ

8 00 Xd 1 X N t 5 0; start of subtransient period > > > > < Equationð5 .44 cÞ t 5 5T 00 ; end of subtransient period dð2φÞ > X 0d 1 X N > > > : X 1 XN d

t 5 0; neglecting subtransient component t $ 5T 0dð2φÞ ; steady state (5.44d)

As we presented in Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, the transient positive-sequence and negative-sequence currents of the power frequency components of the complex instantaneous phase currents Iy(t) and Ib(t) of Eq. (5.44a) can be calculated using I P ðtÞ 5

  1 1 Ir ðtÞ 1 hIy ðtÞ 1 h2 Ib ðtÞ and I N ðtÞ 5 Ir ðtÞ 1 h2 Iy ðtÞ 1 hIb ðtÞ 3 3 (5.45a)

Modelling of rotating ac synchronous and induction machines

(A)

IP (t )

jX P2φ ( t )

equivalent circuit with time-dependent machine reactance

Ra + Er (t) –

(B) Xe

I jX P2φ

Ra + –

401

I

X ′′d P I ′′ X ′d

N

jX N

Ra

Eo

P

X 2φ = I

I′ Eq. (5.44c) P

I′ Xd P

I

P

t = 0, start of subtransient period t = 0, transient reactance t = 5T ′′d , end of subtransient period steady state, end of transient period

Figure 5.10 Synchronous machine equivalent circuits for a two-phase short-circuit fault at machine terminals: (A) positive-sequence time-dependent equivalent circuit and (B) fixedimpedance equivalent circuits of various time instants.

Substituting Eq. (5.44a) into Eq. (5.45a), we obtain I P ðtÞ 5

Er ðtÞ p jXð2φÞ ðtÞ

and

I N ðtÞ 5 2 I P ðtÞ

(5.45b)

Fig. 5.10A shows the resultant time-dependent symmetrical component equivalent circuit that satisfies Eq. (5.45b) with the reinstated machine stator resistance for completeness. Fig. 5.10B shows equivalent circuits using the fixed reactances of Eq. (5.44d). The time constants of Eq. (5.42b) and (5.42c), and the time-dependent machine reactance of Eq. (5.44b) show, from a positive-sequence current view point, that under a two-phase short-circuit fault, the machine behaves as if it has ‘seen’ a balanced three-phase short-circuit fault through an external reactance X N.

Negative-sequence reactance and resistance We return to the negative-sequence reactance X N given in Eq. (5.43b). This is the machine reactance that results from the flow of negative-sequence stator currents. Let us first recall that during a balanced three-phase short-circuit, only positivesequence currents flow into the short circuit. These currents set up an MMF wave that rotates at synchronous speed in the same direction of rotation of the rotor. Because of the very high X/R ratio of the machine, the MMF wave lines up almost exactly with the d-axis, hence d-axis currents only meet d-axis reactances. However, when negative-sequence currents flow in the stator of the machine, these currents produce an MMF wave that rotates at synchronous speed in an

402

Power Systems Modelling and Fault Analysis

opposite direction to the rotation of the rotor. It therefore rotates backward at twice synchronous speed with respect to the rotor. Thus, the currents induced in the rotor field and damper windings are double-frequency currents. At such a high frequency, the machine reactances are effectively the subtransient reactances and as the MMF wave sweeps rapidly over the d and q axes of the rotor, the equivalent negativesequence reactance alternates to d and q axes subtransient reactances. Therefore, the negative-sequence reactance is defined, by both IEC and IEEE standards, as the arithmetic mean of the d and q axes subtransient reactances or XN 5

 1  00 Xd 1 Xq00 2

(5.46)

We have stated previously that when subtransient saliency is present, the shortcircuit current will contain both even and odd harmonics. In this case, we have a different machine negative-sequence reactance, which is given by the geometric mean of the d and q axes subtransient reactances as shown in Eq. (5.43b). However, because Xd00 and Xq00 are nearly equal, the difference between the arithmetic mean and the geometric mean is very small. The difference can be illustrated for both a round rotor machine and a salient-pole machine as follows. For a typical round rotor synchronous generator with Xd00 5 0:2 pu and Xq00 5 0:24 pu, their arithmetic and geometric means are 0.22 pu and 0.2191 pu, respectively. For a salientpole machine with Xd00 5 0:18 pu and Xq00 5 0:275 pu, their arithmetic and geometric means are 0.2275 pu and 0.22248 pu, respectively. Given the uncertainty associated with the measurements of these quantities which can be up to 10%, differences in X N of less than 2.5% are generally acceptable. X N takes the same value under subtransient, transient and steady-state conditions as illustrated in Fig. 5.10B. The stator negative-sequence resistance is significantly higher (some 10
20 times) than the positive-sequence resistance because the second harmonic backward rotating MMF causes significant additional I2R losses in the rotor circuits due to double-frequency currents flowing in the rotor field and damper windings. For short-circuit analysis, the use of Ra instead of the negative-sequence resistance in the calculation of the fault point X/R ratio is highly conservative.

Simplified machine short-circuit current equations For maximum dc current offset, the peak current envelope of Eq. (5.42a) at any instant of time is given by pffiffiffipffiffiffi i^y ðtÞ 5 3 2Eo

"

0

1

1 1 1 A 2t=T 0dð2φÞ e 1@ 0 2 Xd 1 X N Xd 1 XN Xd 1 X N 0 1 # 00 1 1 1 2t=T 2t=T að2φÞ Ae dð2φÞ 1 1 @ 00 2 0 e Xd 1 X N Xd 1 XN Xd00 1 X N

(5.47)

Modelling of rotating ac synchronous and induction machines

403

5.4.4 Unbalanced single-phase to earth short-circuit faults Consider a single-phase to earth short-circuit fault at the machine terminals involving phase r with the machine initially running at synchronous speed, rated terminal voltage and on open circuit. The neutral of the stator winding, normally earthed through a high-impedance is assumed solidly earthed and the initial conditions prior to the fault are given in Eqs. (5.28a) and (5.28b). The constraint equations that define the single-phase short-circuit fault are given by er 5 0

(5.48a)

iy 5 ib 5 0

(5.48b)

Transforming Eq. (5.48a)
(5.48e) to the dq0 reference using Eq. (5.7b), we obtain ed cosðωs t 1 θo Þ 2 eq sinðωs t 1 θo Þ 1 eo 5 0

(5.48c)

id sinðωs t 1 θo Þ 5 2 iq cosðωs t 1 θo Þ

(5.48d)

io 5

id 2 cosðωs t 1 θo Þ

(5.48e)

As in the case of a two-phase short-circuit, Eqs. (5.48a)
(5.48e) and (5.11a) and (5.11b) are sufficient to obtain the solution of current under a single-phase short-circuit fault at the machine terminals. However, the resulting current equations are complex and nonlinear and a closed form solution is extremely tedious. Instead, a step-by-step simplification process, similar to that described for a three-phase fault can be used. The single-phase short-circuit current can be shown to contain, in general, a power frequency component, a dc component and both even and odd harmonic orders of the power frequency, the latter due to subtransient saliency. However, for practical calculations of machine single-phase short-circuit currents involving the power frequency and dc component only, neglecting the harmonic components, it can be shown that the phase r short-circuit current is given by pffiffiffi ir ðtÞ  3 2Eo

"

0

1

1 1 1 Ae2t=T 0dð1φÞ 1@ 0 2 N Z N Z N Z Xd 1 X 1 X Xd 1 X 1 X Xd 1 X 1 X 0 1 # 00 1 1 2t=T Ae dð1φÞ 1 @ 00 2 0 Xd 1 XN 1 XZ Xd 1 XN 1 XZ 3 cosðωs t 1 θo 2 π=2Þ pffiffiffi  2 3 2Eo cos θo 2 π=2

1 e2t=Tað1φÞ X 00d 1 X N 1 X Z (5.49a)

404

Power Systems Modelling and Fault Analysis

where T 0dð1φÞ 5

X 0d 1 X N 1 X Z 0 T Xd 1 X N 1 X Z do

(5.49b)

00 Tdð1φÞ 5

Xd00 1 X N 1 X Z 00 T X 0d 1 X N 1 X Z do

(5.49c)

Tað1φÞ 5

Xd00 1 X N 1 X Z ωs ð3Ra Þ

(5.49d)

Eq. (5.49a) shows that the magnitude of the dc component of short-circuit current is maximum when θo 5 π/2. In the time constant equations, XZ is the machine zero-sequence reactance. XN is the machine negative-sequence reactance applicable under a single-phase short-circuit fault condition and is given by s

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 

 XZ XZ XZ N 00 00 X 5 2 Xd 1 Xq 1 2 2 2

(5.49e)

Since, Xd00 and Xq00 are nearly equal, then if we set Xd00 5 Xq00 in Eq. (5.49e), we obtain X N 5 Xd00 . Although the expression of Eq. (5.49e) is different from the expressions given in Eqs. (5.43b) and (5.46), the differences in numerical values are practically insignificant. From Eq. (5.49a), we can express the instantaneous power frequency component of the short-circuit current as ir(t) 5 Real[Ir(t)] where Ir(t) is phase r complex instantaneous current given by pffiffiffi 3 2Eo jðωs t1θo 2π=2Þ 3Er ðtÞ Ir ðtÞ 5 p 5 p e jXð1φÞ ðtÞ Xð1φÞ ðtÞ

(5.50a)

pffiffiffi phase where Er ðtÞ 5 2Eo ejðωs t1θo Þ 5 Er ejωs t is p ffiffiffi r complex instantaneous voltage, Er is phase r complex phasor given by Er 5 2Eo ejθo and 0 1 p Xð1φÞ ðtÞ

5

1

1 1 1 Ae2t=T 0dð1φÞ 1@ 0 2 N Z N Z N Z Xd 1 X 1 X Xd 1 X 1 X Xd 1 X 1 X 0 1 00 1 1 Ae2t=Tdð1φÞ 1 @ 00 2 0 N Z N Z Xd 1 X 1 X Xd 1 X 1 X (5.50b)

is the equivalent time-dependent machine positive-sequence reactance applicable under a single-phase short-circuit fault condition. At the fault instant t 5 0, that is,

Modelling of rotating ac synchronous and induction machines

405

p start of subtransient period, Xð1φÞ 5 Xd00 1 X N 1 X Z . At t 5 0 and neglecting the subp p 0 transient component, Xð1φÞ 5 X d 1 X N 1 X Z . At t $ 5T 0dð1φÞ ; Xð1φÞ 5 Xd 1 X N 1 X Z . The machine reactance at the end of the subtransient period is given by p 5 Xð1φÞ

1

  1 1 1 25T 00dð1φÞ =T 0dð1φÞ 1
e Xd 1 X N 1 X Z X 0d 1 X N 1 X Z Xd 1 X N 1 X Z (5.50c)

In summary, the four fixed reactances are given by

p 5 Xð1φÞ

8 00 t 5 0; start of subtransient period Xd 1 XN 1 XZ > > > > < Equationð5 .50 cÞ t 5 5T 00 ; end of subtransient period dð1φÞ > X0d 1 X N 1 X Z > > > : X 1 XN 1 XZ d

t 5 0; neglecting subtransient component t $ 5T 0dð1φÞ ; steady state (5.50d)

The transient positive-sequence, negative-sequence and zero-sequence currents of the complex phase current Ir(t) of Eq. (5.50a) are calculated using I P ðtÞ 5 I N ðtÞ 5 I Z ðtÞ 5

1 Ir ðtÞ 3

(5.51a)

Substituting Eq. (5.50a) into Eq. (5.51a), we obtain I P ðt Þ 5 I N ðt Þ 5 I Z ðt Þ 5

Er ðtÞ p jXð1φÞ ðtÞ

(5.51b)

Fig. 5.11A shows the time-dependent positive-sequence equivalent circuit that satisfies Eq. (5.51b) with the reinstated machine stator resistance for completeness. Fig. 5.11B shows equivalent circuits using the fixed machine reactances of Eq. (5.50d). The time constants of Eqs. (5.49a)
(5.49e) and the time-dependent machine reactance of Eq. (5.50b) show, from a positive-sequence current view point, that under a single-phase to earth short-circuit fault, the machine behaves as if it has ‘seen’ a balanced three-phase short-circuit through an external reactance equal to Xe 5 X N 1 X Z. Fig. 5.11 shows the resultant symmetrical component equivalent circuits with the machine positive-sequence reactance connected in series with the machine negative-sequence and zero-sequence reactances to satisfy the single-phase shortcircuit fault condition.

Zero-sequence reactance and resistance The machine zero-sequence impedance to the flow of zero-sequence armature currents applies where the stator winding is star-connected and the neutral is not

406

Power Systems Modelling and Fault Analysis

(A)

IP (t ) = I N (t ) = IZ (t ) j X1Pφ ( t )

equivalent circuit with time-dependent machine reactance

Ra +

Er (t)

–

(B)

Xe

I

j X1Pφ

Ra + -

–

Eo

I

N

jXN

I

Ra

Z

jX Z

Ra

P

X 1φ = I

X ′′d t = 0, start of subtransient period P I ′′ X ′d t = 0, transient reactance P I′ Eq. (5.50c) t = 5T ′′d , end of subtransient period P I′ Xd steady state, end of transient period P I

Figure 5.11 Synchronous machine equivalent circuits for a one-phase to earth short-circuit fault at machine terminals: (A) positive-sequence time-dependent equivalent circuit and (B) fixed-impedance equivalent circuits of various time instants.

isolated. Recalling the spacial distribution of the stator phases and that zerosequence currents are equal in magnitude and phase, these currents produce no net power frequency MMF wave across the air gap. Therefore, the zero-sequence impedance to the flow of zero-sequence currents is only due to some stator winding slot leakage flux, which is greatly dependent on the machine coil pitch design. The zero-sequence reactance is always smaller than the stator leakage reactance Xσ and may be less than half of it depending on coil pitch. Since there is no power frequency stator MMF wave across the air gap, the zerosequence reactance is independent of the rotor’s rotation. It is therefore constant and has the same value under subtransient, transient and steady-state conditions as illustrated in Fig. 5.11B. The zero-sequence resistance can be assumed equal to the stator or armature ac resistance.

Simplified machine short-circuit current equations For maximum dc current offset, the peak current envelope of Eq. (5.49a) at any instant of time is given by 0 1 " pffiffiffi 1 1 1 Ae2t=T 0dð1φÞ i^r ðtÞ 5 3 2Eo 1@ 0 2 Xd 1 X N 1 X Z Xd 1 XN 1 XZ Xd 1 X N 1 X Z 0 1 # 00 1 1 1 2t=Tdð1φÞ 2t=Tað1φÞ @ A e 1 2 0 1 00 e Xd00 1 X N 1 X Z Xd 1 XN 1 XZ Xd 1 X N 1 X Z (5.52)

Modelling of rotating ac synchronous and induction machines

407

5.4.5 Unbalanced two-phase to earth short-circuit faults Consider a two-phase to earth short-circuit fault at the machine terminals involving phases y and b with the machine initially running at synchronous speed, rated terminal voltage and on open circuit. The machine neutral is assumed solidly earthed and the constraint equations that define the two-phase-to-earth short-circuit are given by ey 5 eb 5 0

(5.53a)

ir 5 0

(5.53b)

Transforming Eqs. (5.53a) and (5.53b) to the dq0 reference using Eq. (5.7b), we obtain id cosðωs t 1 θo Þ 2 iq sinðωs t 1 θo Þ 1 io 5 0

(5.54a)

ed sinðωs t 1 θo Þ 5 2 eq cosðωs t 1 θo Þ

(5.54b)

eo 5

ed 2cosðωs t 1 θo Þ

(5.54c)

As in the previous short-circuit cases, Eqs. (5.54a)
(5.54c) and (5.11a) and (5.11b) are theoretically sufficient to obtain the solution of currents under a twophase to earth short-circuit fault at the machine terminals. However, the resulting equations are very complex and nonlinear and a closed form solution is extremely unwieldy. Instead, a step-by-step simplification process can be used to derive the various components of the short-circuit currents. For practical calculations of machine short-circuit currents involving the power frequency and dc components only assuming that Xd00 5 Xq00 , it can be shown that the phase y and phase b shortcircuit currents are given by 3 pffiffiffi 2 3 2 Eo 4 21 ðX 00d 1 Xe Þ 1 iy ðt Þ  cosθo e2t=Ta1ð2φ2EÞ 5 cosðωs t 1 θo Þ 1 00 p ðX 00d 1 2X Z Þ Xð2φ2EÞ ðX d 1 2X Z Þ 2 ðtÞ 3 pffiffiffipffiffiffi 2 3 2Eo 4 1 ðX 00d 1 Xe Þ 1 1 sinðωs t 1 θo Þ 2 00 sinθo e2t=Ta2ð2φ2EÞ 5 p X 00d Xð2φ2EÞ Xd 2 ðtÞ (5.55a) 3 pffiffiffi 2 00 3 2 Eo 4 21 ðX d 1 Xe Þ 1 cosθo e2t=Ta1ð2φ2EÞ 5 ib ðt Þ  cosðωs t 1 θo Þ 1 00 p ðX 00d 1 2X Z Þ Xð2φ2EÞ ðX d 1 2X Z Þ 2 ðtÞ 3 pffiffiffipffiffiffi 2 3 2Eo 4 2 1 ðX 00d 1 Xe Þ 1 1 sinðωs t 1 θo Þ 1 00 sinθo e2t=Ta1ð2φ2EÞ 5 p X 00d Xð2φ2EÞ Xd 2 ðtÞ (5.55b)

408

Power Systems Modelling and Fault Analysis

The earth fault current is the sum of phase y and phase b currents, hence pffiffiffi
 3 2E o ðX 00d 1 Xe Þ 2t=Ta1ð2φ2EÞ 2 cosðωs t 1 θo Þ 1 cosθo e iE ðtÞ 5 00 Xð2φ2EÞ ðtÞ ðX d 1 2X Z Þ

(5.55c)

where 0

1

1 1 1 1 A 2t=T 0dð2φ2EÞ 5 e 1@ 0 2 Xð2φ2EÞ ðtÞ Xd 1 Xe X d 1 Xe Xd 1 Xe 0 1 00 1 1 Ae2t=Tdð2φ2EÞ 1 @ 00 2 0 Xd 1 Xe X d 1 Xe Xe 5

1 1 1 1 Z N X X

(5.56a)

(5.56b)

T 0dð2φ2EÞ 5

X 0d 1 Xe 0 T Xd 1 Xe do

T 00dð2φ2EÞ 5

Ta1ð2φ2EÞ 5

X N 1 2X Z ωs ð3Ra Þ

Ta2ð2φ2EÞ 5

X 00d 1 Xe 00 T X 0d 1 Xe do

(5.57a)

XN ω s Ra

(5.57b)

X N is the machine negative-sequence reactance applicable under a two-phase to earth short-circuit fault condition which, in the general case when Xd00 6¼ Xq00 , is given by XN 5

Xd00 Xq00 1

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Xd00 Xq00 ðXd00 1 2X Z ÞðXq00 1 2X Z Þ Xd00 1 Xq00 1 2X Z

(5.58a)

The limiting values, that is, the lower and upper bounds of Eq. (5.58a), can be found by letting X Z!0 and X Z!N, respectively. Thus, X N is bound as follows qffiffiffiffiffiffiffiffiffiffiffi 2Xd00 Xq00 , X N , Xd00 Xq00 00 00 X d 1 Xq

(5.58b)

The lower limit is the same as that found in the case of a three-phase fault given in Eq. (5.33b). Also, the upper limit is the same as that found in the case of a twophase fault given in Eq. (5.43b). In practice, the differences between the various expressions for X N, that depend on the short-circuit fault type, are negligible. The value of X N calculated from the arithmetic average of Xd00 and Xq00 given in Eq. (5.46) is sufficiently accurate for practical short-circuit analysis purposes. It should be noted that if we let Xd00 5 Xq00 in Eq. (5.58a), we obtain, as we should expect, X N 5 Xd00 .

Modelling of rotating ac synchronous and induction machines

409

Eqs. (5.55a) and (5.55b) of phase y and phase b short-circuit currents are quite different from those obtained for all short-circuit fault types considered previously. Each phase current now consists of two asymmetrical current components and each one of these consists of a power frequency ac component and a dc component. However, as expected, at t 5 0, the sum of the ac and dc components of each asymmetrical current is zero. The earth fault current, on the other hand, consists of one asymmetrical current with only one power frequency ac component and one dc component as the other terms cancel out when the phase currents are added together. It is important to note that the two dc components have different initial magnitudes and different armature time constants. The maximum dc current offset occurs at θo 5 0 for the first dc component but at θo 5 π/2 for the second dc current. It can be shown that the value of θo that results in maximum asymmetrical phase y and phase b currents is given by


 

  1 2X Z 1 1 2 θoðyÞ 5 2 θoðbÞ 5 2 tan21 pffiffiffi 1 1 00 exp 2 t Ta2ð2φ2EÞ Ta1ð2φ2EÞ Xd 3 (5.59a) Typically Xd00  2X Z hence at t 5 0, θo(y) 5 2 49.1 degrees and θo(b) 5 49.1 degrees. Also, for a typical 50-Hz synchronous machine with Xd00 5 0:2pu; X Z 5 0:1 pu and Ra 5 0.002 pu, the armature dc time constants calculated using Eq. (5.57b) are Ta1(2φ
E) 5 0.2122 seconds and Ta2(φ
E) 5 0.318 seconds. Thus, using Eq. (5.59a), we obtain θo(y) 5 2 49.87 degrees, (θo(b) 5 49.87 degrees) at t 5 10 ms and θo(y) 5 2 53.1 degrees (θo(b) 5 53.1 degrees) at t 5 50 ms. From Eq. (5.56a) of the machine equivalent reactance under a two-phase to earth short-circuit, we distinguish four machine fixed-impedance values. At t 5 0, that p is, the start of the subtransient period, Xð2φ2EÞ 5 Xd00 1 Xe . At t 5 0, and neglecting p 0 the subtransient component, Xð2φ2EÞ 5 X d 1 Xe . At t $ 5T 0dð2φ2EÞ , that is, end of tranp sient period, Xð2φ2EÞ 5 Xd 1 Xe . The machine reactance at the end of the subtransi00 ent period, that is, at t  5Tdð2φ2EÞ , is given by p Xð2φ2EÞ 5

1

  1 1 1 25T 00dð2φ2EÞ =T 0dð2φ2EÞ 1
e X d 1 Xe X 0d 1 Xe Xd 1 Xe

(5.59b)

In summary, the four fixed reactances are given by 8 00 X d 1 Xe > > > < Equationð5:59bÞ p Xð2φ2EÞ 5 > X 0d 1 Xe > > : Xd 1 X e

t 5 0; start of subtransient period t 5 5T 00dð2φ2EÞ ; end of subtransient period t 5 0; neglecting subtransient component t $ 5T 0dð2φ2EÞ ; steady state (5.59c)

410

Power Systems Modelling and Fault Analysis

From Eqs. (5.55a) and (5.55b), we can express the instantaneous power frequency component of the short-circuit currents as iy(t) 5 Real[Iy(t)] and ib(t) 5 Real[Ib(t)] where Iy(t) and Ib(t) are the phase y and phase b complex instantaneous currents given by Iy ðtÞ 5

pffiffiffi pffiffiffipffiffiffi
 3 2Eo jðωs t1θo 2π=2Þ ðXd00 1 Xe Þ 2 3 2Eo jðωs t1θo Þ e 1 e p Xd00 2Xð2φ2EÞ ðtÞ ðXd00 1 2X Z Þ

or pffiffiffi
 3Er ðtÞ ðXd00 1 Xe Þ 2 3Er ðtÞ 2 Iy ðtÞ 5 p Xd00 2Xð2φ2EÞ ðtÞ jðXd00 1 2X Z Þ

(5.60a)

pffiffiffi pffiffiffipffiffiffi
 3 2Eo jðωst1θo 2π=2Þ ðXd00 1 Xe Þ 2 3 2Eo jðωs t1θo Þ e 2 e Ib ðtÞ 5 p Xd00 2Xð2φ2EÞ ðtÞ ðXd00 1 2X Z Þ or Ib ðtÞ 5

pffiffiffi
 3Er ðtÞ ðXd00 1 Xe Þ 2 3Er ðtÞ 1 p Xd00 2Xð2φ2EÞ ðtÞ jðXd00 1 2X Z Þ

(5.60b)

pffiffiffi pffiffiffi where Er ðtÞ 5 2Eo ejðωs t1θo 1π=2Þ 5 Er ejðωs tÞ and Er 5 j 2Eo ejθo . The transient positive-sequence, negative-sequence and zero-sequence currents of the complex phase currents Iy(t) and Ib(t) of Eqs. (5.60a)
(5.60c) are calculated using I P ðtÞ 5

  1 1 hIy ðtÞ 1 h2 Ib ðtÞ I N ðtÞ 5 h2 Iy ðtÞ 1 hIb ðtÞ 3 3

I Z ðtÞ 5

 1 Iy ðtÞ 1 Ib ðtÞ 3 (5.60c)

Substituting Eqs. (5.60a) and (5.60b) in Eq. (5.60c), and using Eq. (5.56b) for Xe with X N 5 Xd00 , we obtain, after some algebraic manipulations, which the reader is encouraged to prove, the following: I P ðtÞ 5

I N ðtÞ 5

I Z ðtÞ 5

Er ðtÞ p jXð2φ2EÞ ðtÞ

(5.61a)

2 XZ Er ðtÞ 3 p 1 X Z Þ jXð2φ2EÞ ðtÞ

(5.61b)

2 XN Er ðtÞ 3 p 1 X Z Þ jXð2φ2EÞ ðtÞ

(5.61c)

ðX N

ðX N

Modelling of rotating ac synchronous and induction machines

IP (t )

(A)

j X 2P φ − E ( t )

equivalent circuit with time-dependent machine reactance

Ra + –

(B) I

E r (t)

j Xe N

j X P2 φ − E

Ra + –

411

Z

I jX N

I jX Z

Ra

Ra

Eo

P

X 2φ − E = I

X ′′d t = 0, start of subtransient period P I ′′ X ′d t = 0, transient reactance P I′ Eq. (5.59b) t = 5T ′′d , end of subtransient period P I′ Xd steady state, end of transient period P I

Figure 5.12 Synchronous machine equivalent circuits for a two-phase to earth short-circuit fault at machine terminals: (A) positive-sequence time-dependent equivalent circuit and (B) fixed-impedance equivalent circuits of various time instants.

Fig. 5.12A shows the time-dependent positive-sequence equivalent circuit that satisfies Eq. (5.61a) with the reinstated stator resistance for completeness. Fig. 5.12B shows equivalent circuits using the fixed machine reactances of Eq. (5.59c). From Fig. 5.12B and using complex phasor notation, the positivesequence phasor current is given by Eo  IP 5

XNXZ P j X 1 N X 1 XZ

with Ra neglected

(5.62a)

where the value of the positive-sequence reactance XP is as shown in Fig. 5.12B and depends on the calculation time instant. The negative-sequence and zerosequence phasor currents are given by IN 5

2 XZ 3 IP XN 1 XZ

(5.62b)

2 XN 3 IP 1 XZ

(5.62c)

and IZ 5

XN

The machine equivalent reactance of Eq. (5.56a) and the time constants of Eq. (5.57a) show that from a positive-sequence current view point, under a

412

Power Systems Modelling and Fault Analysis

two-phase to earth short-circuit fault, the machine behaves practically as if it has ‘seen’ a balanced three-phase short circuit through an external impedance Xe. This impedance is given by the parallel combination of the negative-sequence and zerosequence machine reactances as shown in Eq. (5.56b).

5.4.6 Modelling the effect of initial machine loading Machine internal voltages The short-circuit current equations already presented cover all cases of short-circuit currents at the machine terminals assuming the machine is initially unloaded, that is, the internal machine rms voltage is equal to Eo. However, the equations can be slightly modified to represent the situation where the machine is initially loaded by accounting for its prefault active and reactive power outputs as well as its terminal voltage. Because the machine is initially only producing a balanced set of threephase voltages into a balanced network, the negative-sequence and zero-sequence terminal voltages and output currents are all zero. Similarly, the initial machine active and reactive power outputs are positive-sequence MW and positive-sequence Mvar quantities. Assuming that the initial machine terminal voltage, active and reactive power outputs are Vt, P and Q, the machine stator current I can be calculated using the apparent power equation: P 1 jQ 5 Vt I . We have already established that the machine positive-sequence reactance XP can be equated to one of four reactances depending on the calculation time instant after the occurrence of the fault. Therefore, using Fig. 5.13, the internal machine voltages that correspond to these four reactances, for a given terminal voltage Vt, are Ev, E0 (at t 5 0), E0 (at t  5Td00 ) and E, respectively, and are given by

 P 2 jQ Vt

(5.63a)

 P 2 jQ E0 5 Vt 1 ðRa 1 jX 0d Þ Vt

(5.63b)

E

00

5 Vt 1 ðRa 1 jXd00 Þ

P + jQ

jXP

Ra + –

E

P

I

Vt

X E

P

=

X′′d t = 0, start of subtransient period E′′ X′ d t = 0, transient reactance E′ P X at t = 5T′′d end of subtransient period E′ at t = 5T′′d Xd steady state, end of transient period Eo

Figure 5.13 Positive-sequence equivalent circuits for an initially loaded synchronous machine.

Modelling of rotating ac synchronous and induction machines

0

E ðt 

5Td00 Þ 5 Vt



1 Ra 1 jX ðt  P

  P 2 jQ Vt

5Td00 Þ

 P 2 jQ E 5 Vt 1 ðRa 1 jXd Þ Vt

413

(5.63c)

(5.63d)

Clearly, when the machine is initially unloaded, I 5 0 and Ev 5 E0 5 E 5 Eo 5 Vt.

Machine short-circuit currents Using Eqs. (5.63a)
(5.63d) for Ev, E0 (at t 5 0) or E0 (at t  5Td00 ) and E, the equations that describe the machine peak short-circuit current envelope for each fault type are given below.

Three-phase short circuit Using Eq. (5.40a) and neglecting the double-frequency component, we have # "

0 

00  0 00 pffiffiffi E 0 00 E E E E E 2t=T 2t=T 2t=T að3φÞ dð3φÞ 1 dð3φÞ 1 i^r ðtÞ 5 2 1 2 2 0 e e e Xd Xd X 0d Xd00 Xd Xd00 (5.64)

Two-phase short circuit Using Eq. (5.47), we have 0 1 " 0 pffiffiffipffiffiffi E E E A 2t=T 0dð2φÞ i^y ðtÞ 5 3 2 e 1@ 0 2 Xd 1 X N Xd 1 X N Xd 1 XN 0 1 # 00 0 00 00 E E E 2t=T 2t=T að2φÞ Ae dð2φÞ 1 1 @ 00 2 0 e Xd 1 X N Xd 1 XN Xd00 1 X N

(5.65)

One-phase to earth short circuit Using Eq. (5.52), we have 0 1 " 0 pffiffiffi E E E Ae2t=T 0dð1φÞ 1@ 0
i^r ðtÞ 5 3 2 Xd 1 X N 1 X Z X d 1 X N 1 X Z Xd 1 X N 1 X Z 0 1 00 0 00 E E Ae2t=Tdð1φÞ 1 @ 00 2 Xd 1 X N 1 X Z X 0d 1 X N 1 X Z # E00
t=Tað1φÞ 1 00 e Xd 1 X N 1 X Z (5.66)

414

Power Systems Modelling and Fault Analysis

Two-phase to earth short circuit In the ac components of Eqs. (5.55a) and (5.55b), we simply replace the term Eo/X (2φ2E) (t), using Eq. (5.56a), with the following term 0 1 0 Eo E E Ae2t=T 0dð2φ2EÞ 1@ 0 2 Xd 1 X e Xd 1 Xe X d 1 Xe 0 1 (5.67) 00 0 00 E E 2t=T Ae dð2φ2EÞ 1 @ 00 2 0 Xd 1 Xe X d 1 Xe In the dc components of Eqs. (5.55a), (5.55b) and (5.55c), the term Eo should be replaced with Ev.

5.4.7 Effect of automatic voltage regulators on short-circuit currents Discussion of the function, design, analysis or tuning of AVRs is the subject of power system dynamics, stability and control. In our book, we simply mention that the AVR generally attempts to control the machine terminal voltage by sensing its variations from a given set point or target and causing an increase or decrease in the excitation or field voltage. In the case of short-circuit faults on the host network to which a synchronous machine is connected, the machine terminal voltage can see a significant drop during the fault period depending on the electrical distance, that is, impedance to the fault point. The question we are interested in is what effect can the AVR have, if any, on the machine short-circuit current for a three-phase fault at the machine terminals from an unloaded initial condition? We recall that we have already determined the machine short-circuit current assuming no AVR action with Δefd 5 0. To determine the machine current due to a change in field voltage Δefd, representing automatic AVR action, we recall Eq. (5.17a), Δψd(s) 5 2 Xd (s)Δid (s) 1 Gd (s)Δefd (s) from which we obtain   Δid ðsÞ

caused by Δefd

5

Gd ðsÞ Δefd ðsÞ Xd ðsÞ

(5.68a)

and from Eq. (5.15a) Δiq ðsÞ 5 0

(5.68b)

Eqs. (5.18a) for Xd (s) and (5.18b) for Gd (s) were derived for the general case of a machine with a damper winding. For a machine without a damper winding, a similar method can be used to derive these expressions and the reader is encouraged to do so. However, for us, we will directly derive the expressions for Xd (s) and Gd (s) from the general ones of Eqs. (5.18a), (5.18b) and (5.19a). To do so, we represent the absence of the damper winding by substituting Xkd 5 0 and letting Rkd!N. After a little algebra, the results can be easily shown as

Modelling of rotating ac synchronous and induction machines

Xd ðsÞ 5 Xd 2

415

2 sXad Rfd 1 sXffd

(5.69a)

and Gd ðsÞ 5

Xad Rfd 1 sXffd

(5.69b)

Substituting Eqs. (5.69a) and (5.69b) into Eq. (5.68a) and using Xffd 5 Xfd 1 Xad and Xd 5 Xad 1 Xσ from Eq. (5.12), we obtain Δid ðsÞ 5

Xad Δefd ðsÞ Rfd Xd

1 1 11 Rfd

Xfd 1

1 Xσ

1 1

! 1 Xad

s

or using ΔEfd ðsÞ 5

Xad Δefd ðsÞ Rfd

and Eqs. (5.22b) for T4 and (5.23c) for T0 d, we obtain   Δid ðsÞ

due to ΔEfd

5

1 ΔEfd ðsÞ 3 Xd 1 1 sT 0dð3φÞ

(5.70a)

For simplicity, we consider an instantaneous or a step change in field voltage ΔEfd where ΔEfd (s) 5 ΔEfd/s. Therefore, Eq. (5.70a) can be written as   Δid ðsÞ

ΔEfd 1 ΔEfd 5 5 0 Xd sð1 1 sT dð3φÞ Þ Xd due to ΔEfd

1 1 2 s s 1 1=T 0dð3φÞ

!

and taking the inverse Laplace transform, we obtain   Δid ðtÞ

due to ΔEfd

5

0 ΔEfd ð1 2 e
t=T dð3φÞ Þ Xd

(5.70b)

Also, from Eq. (5.68b), we have Δiq ðtÞ 5 0

(5.70c)

Eq. (5.70b) shows that an instantaneous increase in Efd causes an exponential increase in short-circuit current with a time constant equal to T 0dð3φÞ . That is, despite the very fast change in Efd, the resultant change in machine current is slowed down

416

Power Systems Modelling and Fault Analysis

or delayed by the d-axis short-circuit transient time constant T 0dð3φÞ , which falls in the range of 0.5 2 2 seconds. For example, for a typical value of T 0dð3φÞ 5 1 s, the 0 factor ð1 2 e
t=T dð3φÞ Þ reaches 0.11 at 120 ms. This indicates some change in current towards the end of the subtransient period at around 120 ms. However, in practice, even for modern fast excitation control systems such as static exciters, the change in field voltage will itself occur after a definite time delay. Thus, the effective change in short-circuit current will only begin to occur after the subtransient period. In practice, a static excitation control system will not be able to continue to operate by maintaining free thyristor firing if the terminal voltage, where it derives its supply from, drops below around 0.2
0.3 pu. In other words, the subtransient component of short-circuit current will have decayed to zero by the time the effect of the AVR is felt through the machine. Therefore, the effect of the AVR is to cause a possible increase in the transient and steady-state components of short-circuit current. It should be noted that the change in field voltage has no effect on the dc component of short-circuit current, as shown in Eqs. (5.70b) and (5.70c). We have now obtained the machine current changes due to both the short-circuit itself and the change in field voltage. Therefore, using the superposition theorem, the total machine short-circuit current change is obtained by the addition of the two values. Thus, examining the effect on the ac component of short-circuit current after the subtransient component has decayed to zero, Eq. (5.35a) can be rewritten as   Δid ðtÞ




  pffiffiffi 1 1 1 2t=T 0dð3φÞ 5 2 Eo 1 2 e Xd X 0d Xd due to short circuit

(5.71a)

Therefore, using Eqs. (5.70b) and (5.71a), the total change in machine current is given by   Δid ðtÞ

5 total

  i pffiffiffi
1 0 ΔEfd h 1 1 2t=T 0dð3φÞ 1 2 e2t=T dð3φÞ 1 2Eo 1 2 e Xd X 0d Xd Xd (5.71b)

A simplification can be made by using pffiffiffiΔEfd 5 Efd
Efdo and since the machine is initially unloaded, we have Efdo 5 2Eo . Therefore, Eq. (5.71b) simplifies to   Δid ðtÞ

i pffiffi2ffiE Efd h o 2t=T 0dð3φÞ 2t=T 0dð3φÞ 5 12e 1 0 e X X d d total

(5.72)

Eq. (5.72) shows two transient components; one is a rising exponential and one is a decaying exponential and both have the same time constant T 0dð3φÞ . The decaying exponential component represents, as expected, the decay in flux due to induced currents in the field winding, whereas the rising exponential component is due to field voltage change by AVR action. It should be noted that time t represents the beginning of the transient period. Clearly, whether the short-circuit current continues

Modelling of rotating ac synchronous and induction machines

417

to decrease, remains constant or starts to increase is dependent on the magnitude of Efd caused by AVR action. This depends on the extent of voltage drop at the machine terminals ‘seen’ by the AVR pffiffiffi and this in turn depends on the electrical distance to the fault point. If Efd =Xd 5 2Eo =X 0d , the transient current component due to the AVR effect cancels exponential and the short-circuit current remains conpffiffiffi out the decaying p ffiffiffi stant at 2Eo =X 0d . If Efd =Xd . 2Eo =X 0d , the current component due to the effect of the AVR is greater than the decaying exponential component and the short-circuit pffiffiffi current will increase. The converse is true for Efd =Xd , 2Eo =X 0d . For a typical turbo-generator running pffiffiffi at 0 no-load and rated voltage, Eo 5 1 pu, Xd 5 1.5 pu and X 0d 5 0:25 pu; 2Eo =X d 5 5:65 pu, and a high-gain static excitation control systems, a small drop in terminal voltage of only a few percent is sufficient to cause a field voltage change to maximum ceiling of typically 7
8 pu where 1 pu field voltage is that which produces 1 pu terminal voltage on open circuit, giving Efd/Xd 5 5.34 pu. The possible effects of the AVR on the machine short-circuit current are illustrated in Fig. 5.14. It is worth noting that the phase r current after the subtransient contribution has decayed to zero is calculated by substituting Eqs. (5.70c) and (5.72) into Δir ðtÞ 5 Δid ðtÞcosðωt 1 θo Þ 2 Δiq ðtÞsinðωt 1 θo Þ giving Δir ðtÞ 5

Efd Xd

pffiffiffi  2 Eo Efd 2t=T 0dð3ΦÞ 2 cosðωs t 1 θo Þ e X 0d Xd

(5.73)

5.4.8 Modelling of synchronous motors/ compensators/condensers Although in the preceding analysis we used the general term of synchronous machine, we nonetheless deliberately biased our attention towards the synchronous ac rms current (pu of rated)

6 5

(iv)

4

(iii)

3 2

(ii)

(i)

1 0 0

0.4

0.8 1.2 Time (s)

(i) no AVR, (ii) with AVR and (iii) with AVR and

1.6

2

E fd < 2E o Xd X′ d

E fd = 2E o , (iv) with AVR and E fd > 2E o X′ d Xd X′ d Xd

Figure 5.14 Effect of synchronous machine AVR on short-circuit current.

418

Power Systems Modelling and Fault Analysis

generator. However, synchronous motors are also used in industry essentially in two general applications; as motors driving large mechanical loads, for example, compressors, or as reactive power compensators and these are generally known as synchronous condensers in North America or synchronous compensators in the UK. The preceding analysis for synchronous generators applies equally to synchronous motors. The only difference being the initial loading conditions of the motor.

5.4.9 Examples Example 5.1 A 50-Hz three-phase synchronous generator has the following data: Rated apparent power 5 165 MVA, rated power 5 132 MW, rated voltage 5 15 kV, Xd 5 2:04 pu, X 0d 5 0:275 pu, Xd00 5 0:19 pu, Xq00 5 0:2 pu, X Z 5 0:095 pu, T 0o 5 8:16 s, To00 5 0:058 s, Ra 5 0:002 pu. All parameters are in pu on 165 MVA rating. Calculate the rms and dc short-circuit currents for a three-phase, two-phase and one-phase to earth short-circuit faults at the machine terminals at t 5 0, 10 and 100 ms. The machine is initially operating on open circuit and has a terminal voltage of 1 pu. Although in practice, the neutral point of the star-connected stator winding of such a generator would be earthed through a high resistance, assume in this example that the neutral is solidly earthed. Calculate the machine resistance and reactances in Ohm.

Three-phase short-circuit fault T 0d 5

0:275 0:19 3 8:16 5 1:1 s and Td00 5 3 0:058 5 0:04 s 2:04 0:275

Ta 5

0:19 5 0:302 s 2π50 3 0:002

From Eq. (5.40a), the rms fault current at time t is given by irms ðtÞ 5 1=2:04 1 ð1=0:275 2 1=2:04Þe2t=1:1 1 ð1=0:19 2 1=0:275Þe2t=0:04 pffiffi and the dc fault current is given by idc ðtÞ 5 0:192 e2t=0:302 The rms fault current at t 5 0, 10 ms and 100 ms is equal to 5.26, 4.87 and 3.5 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 7.44, 7.2 and 5.34 pu, respectively.

Modelling of rotating ac synchronous and induction machines

419

Two-phase short-circuit fault From either Eqs. (5.43b) (or (5.46)), we have X N 5 T 0d 5

0:275 1 0:195 3 8:16 5 1:716 s; 2:04 1 0:195

Td00 5

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:19 3 0:2 5 0:195 pu.

0:19 1 0:195 3 0:058 5 0:0475 s 0:275 1 0:195

and Ta 5

0:195 5 0:31 s 2π50 3 0:002

From Eq. (5.47), the rms and dc fault currents are 0 1 1 1 1 Ae2t=1:716 1@ 2 2:04 1 0:195 0:275 1 0:195 2:04 1 0:195 0 1 # 1 1 Ae2t=0:0475 2 1@ 0:19 1 0:195 0:275 1 0:195

pffiffiffi irms ðtÞ 5 3

idc ðtÞ 5

"

pffiffiffipffiffiffi 2t=0:31 3 2e =ð0:19 1 0:195Þ

The rms fault current at t 5 0, 10 and 100 ms is equal to 4.5, 4.33 and 3.62 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 6.36, 6.16 and 4.6 pu, respectively.

One-phase short-circuit fault

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi From either Eq. (5.49e), where X N 5 ð0:19 1 0:095=2Þð0:2 1 0:095=2Þ 2 0:095=2 5 0:1949 pu or a value of 0.195 pu as calculated above is acceptable from a practical viewpoint: 0:275 1 0:195 1 0:095 3 8:16 5 1:978 s 2:04 1 0:195 1 0:095 0:19 1 0:195 1 0:095 3 0:058 5 0:0493 s Td00 5 0:275 1 0:195 1 0:095 0:19 1 0:195 1 0:095 5 0:2546 s Ta 5 2π50 3 ð3 3 0:002Þ

T 0d 5

420

Power Systems Modelling and Fault Analysis

From Eq. (5.52), the rms and dc fault currents are "

1 2:04 1 0:195 1 0:095 0 1 1 1 Ae2t=1:978 2 1@ 0:275 1 0:195 1 0:095 2:04 1 0:195 1 0:095 0 1 # 1 1 Ae2t=0:0493 2 1@ 0:19 1 0:195 1 0:095 0:275 1 0:195 1 0:095 pffiffiffi 3 2 e2t=0:2546 idc ðtÞ 5 0:19 1 0:195 1 0:095

irms ðtÞ 5 3

The rms fault current at t 5 0, 10 and 100 ms is equal to 6.25, 6.06 and 5.24 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 8.84, 8.5 and 5.97 pu, respectively. We now convert the pu machine parameters to Ohms. Base impedance 5 (15 kV)2/165 MVA 5 1.3636 Ω Xd 5 2:04 3 1:3636 5 2:782 Ω; X 00d 5 0:19 3 1:3636 5 0:259 Ω

X 0d 5 0:275 3 1:3636 5 0:375 Ω;

X 00q 5 0:2 3 1:3636 5 0:2727 Ω; X Z 5 0:095 3 1:3636 5 0:1295 Ω; Ra 5 0:002 3 1:3636 5 0:002727 Ω

Example 5.2 Repeat the calculations for the three-phase and one-phase short-circuit faults in Example 5.1 assuming the machine is initially loaded and operating at rated power and power factor and a terminal voltage of 1 pu. Using Eqs. (5.63a)
(5.63d), let Vt 5 1 pu. The machine rated lagging power factor is equal to 132/165 5 0.8 and the rated lagging reactive power output is equal to 132 MW 3 tan (cos210.8) 5 99 MVAr. The machine’s real and reactive power output in pu on machine MVA rating are equal to 132/165 5 0.8 pu and 99/165 5 0.6 pu. Thus,

 0:8 2 j0:6 E 5 1 1 ð0:002 1 j0:19Þ 5 1:1257 pu+7:7 1 

0:8 2 j0:6 5 1:187 pu+10:7 E0t50 5 1 1 ð0:002 1 j0:275Þ 1

 0:8
j0:6 E 5 1 1 ð0:002 1 j2:04Þ 5 2:759 pu+36:2 1 00

Modelling of rotating ac synchronous and induction machines

421

Using Eq. (5.64) for a three-phase fault, we have

irms ðtÞ 5



 2:759 1:187 2:759
t=1:1 1:1257 1:187 2t=0:04 1 2 e 2 e 1 2:04 0:275 2:04 0:19 0:275

and the dc fault current is given by pffiffiffi 2 3 1:1257 2t=0:302 e idc ðtÞ 5 0:19 The rms fault current at t 5 0, 10 and 100 ms is equal to 5.93, 5.54 and 4.2 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 8.38, 8.1 and 6.0 pu. Using Eq. (5.66) for a one-phase to earth fault, we have "

2:759 2:04 1 0:195 1 0:095 0 1 1:187 2:759 Ae2t=1:978 2 1@ 0:275 1 0:195 1 0:095 2:04 1 0:195 1 0:095 0 1 # 1:1257 1:187 2t=0:0493 @ A 2 e 1 0:19 1 0:195 1 0:095 0:275 1 0:195 1 0:095 pffiffiffi 3 2 3 1:1257 e2t=0:2546 idc ðtÞ 5 0:19 1 0:195 1 0:095 irms ðtÞ 5 3

The rms fault current at t 5 0, 10 and 100 ms is equal to 7.17, 7.02 and 6.39 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 9.95, 9.57 and 6.72 pu, respectively.

Example 5.3 The generator of Example 5.1 is now assumed to be connected to a high-voltage 132-kV busbar through a star
delta transformer that has a positive-sequence leakage impedance equal to (0.00334 1 j0.2)pu on 165 MVA. The star-connected winding is the high-voltage winding. The diagram is shown in figure (A) below. The transformer’s zero-sequence reactance is equal to 95% of the positive-sequence reactance. Calculate the rms and dc short-circuit currents for three-phase and one-phase short-circuit faults on the 132-kV busbar assuming that the machine is initially operating on open circuit and the initial voltages at its terminals and on the 132-kV busbar are both 1 pu.

422

Power Systems Modelling and Fault Analysis

(A)

(B)

– (C)

–

Three-phase short-circuit fault The new short-circuit time constants can be calculated using Eqs. (5.38b) and (5.38c). The external impedance ‘seen’ by the machine up to the fault point on the transformer high-voltage side is simply that of the transformer. However, in this example, Eqs. (5.39a) and (5.39b) can be used because the armature resistance of the machine and the transformer resistance are very small in comparison with the machine and transformer reactances. Thus T 0de 5

0:275 1 0:2 3 8:16 5 1:73 s 2:04 1 0:2

00

Tde 5

0:19 1 0:2 3 0:058 5 0:0476 s 0:275 1 0:2

From Eq. (5.38e), we have Tae 5

ð0:19 1 0:2Þ=2 1 0:2 5 0:2328 s 2π50 3 ð0:002 1 0:0034Þ

The rms and dc fault currents are calculated using Eq. (5.40b), 0 1 1 1 1 Ae2t=1:73 1@ 2 irms ðtÞ 5 2:04 1 0:2 0:275 1 0:2 2:04 1 0:2 0 1 1 1 Ae2t=0:047 2 1@ 0:19 1 0:2 0:275 1 0:2

Modelling of rotating ac synchronous and induction machines

423

and pffiffiffi 2 e2t=0:2328 idc ðtÞ 5 0:19 1 0:2 The rms fault current at t 5 0, 10 and 100 ms is equal to 2.56, 2.46 and 2.06 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 3.63, 3.47 and 2.36 pu, respectively.

One-phase to earth short-circuit fault The first step is to calculate the external impedance ‘seen’ by the machine up to the fault point on the transformer high-voltage side. The positive-sequence, negativesequence and zero-sequence networks for this fault condition are connected in series as shown in figure (B) of Example 5.3. The equivalent external impedance seen by the machine is shown in figure (C) of Example 5.3 above. Therefore, the external impedance is given by Ze 5 (3Rt 1 Ra) 1 j(2.95Xt 1 XN) and the component resistance and reactance parts are Re 5 3 3 0.00334 1 0.002 5 0.01202 pu and Xe 5 2.95 3 0.2 1 0.195 5 0.785 pu. The short-circuit time constants can be calculated using Eqs. (5.38b) and (5.38c). However, we will use Eqs. (5.39a) and (5.39b) because the armature resistance and external resistance are very small in comparison with the machine’s reactances and external reactance. Thus T 0de 5

0:275 1 0:785 3 8:16 5 3:06 s 2:04 1 0:785

T 00de 5

0:19 1 0:785 3 0:058 5 0:0533 s 0:275 1 0:785

From Eq. (5.38e), we have Tae 5

ð0:19 1 0:2Þ=2 1 0:785 5 0:2225 s 2π50 3 ð0:002 1 0:01202Þ

The rms and dc fault currents are calculated using Eq. (5.40b), 0 1 1 1 1 Ae2t=3:06 1@ 2 irms ðtÞ 5 3 2:04 1 0:785 0:275 1 0:785 2:04 1 0:785 0 1 1 1 Ae2t=0:0533 2 1@ 0:19 1 0:785 0:275 1 0:785 "

and pffiffiffi 3 2 idc ðtÞ 5 e2t=0:2225 0:19 1 0:785

424

Power Systems Modelling and Fault Analysis

The rms fault current at t 5 0, 10 and 100 ms is equal to 3.07, 3.02 and 2.81 pu, respectively. The dc fault current at t 5 0, 10 and 100 ms is equal to 4.35, 4.16 and 2.77 pu, respectively.

5.5

Determination of synchronous machine parameters from measurements

Although calculations of machine parameters are made by machine manufacturers at the design stage, factory or field tests are generally carried out on the built machine. These are to identify the machine parameters and confirm that they are within the guaranteed or declared design values which are typically 6 10% or as agreed between the manufacturer and the customer. The reactances and time constants of the machine are determined from measurements as defined in IEC and IEEE standards.

5.5.1 Measurement of positive-sequence reactance, positivesequence resistance and d-axis short-circuit time constants Measurement and separation of ac and dc current components Several parameters can be calculated from measurements of the stator short-circuit currents during a sudden three-phase short circuit at the machine terminals. These are the positive-sequence or the d-axis subtransient, transient and steady-state reactances, and the d-axis short-circuit subtransient, transient and armature (dc) time constants. The machine is on open circuit and running at rated speed just before the application of the simultaneous three-phase short-circuit fault. The unsaturated reactances and time constants are determined by performing tests at a few low values of prefault stator voltage, for example, 0.1
0.4 pu. The saturated reactances are determined from tests at higher voltages typically at rated 1 pu prefault stator voltage. Oscillograms of the three-phase short-circuit currents are taken. Recalling Eq. (5.40a).




  00 pffiffiffi Eo Eo Eo 2t=T 0dð3φÞ Eo Eo 2t=Tdð3φÞ Eo 2t=Tað3φÞ 1 2 1 2 0 e 1 00 e i^r ðtÞ 5 2 e Xd X 0d Xd X 00d Xd Xd (5.74a) or i^r ðtÞ 5 I^ac ðtÞ 1 idc ðtÞ I^ac ðtÞ 5

00 pffiffiffi pffiffiffi pffiffiffi 0 2Id 1 2ðI 0d 2 Id Þe2t=T dð3φÞ 1 2ðId00 2 I 0d Þe2t=Tdð3φÞ

(5.74b) (5.75a)

Modelling of rotating ac synchronous and induction machines

425

and idc ðtÞ 5

pffiffiffi 00 2t=T að3φÞ 2I d e

(5.75b)

Eq. (5.75a) represents the envelope of the ac rms component of the short-circuit current where I 00d 5

Eo X 00d

I 0d 5

Eo X 0d

Id 5

Eo Xd

(5.75c)

To illustrate the process of determining the machine d-axis parameters, Fig. 5.15A illustrates the measured normalised phase r asymmetrical current. The first step is to separate the dc and ac components of the measured current. The dc component can be calculated as the algebraic half sum of the ordinates of the upper and lower envelopes of the current, whereas the ac components can be similarly determined from the algebraic half difference of the upper and lower envelopes. In practice, experience shows that this manual process is dependent on the engineer doing the analysis and can be subject to some error, whereas the use of numerical filters is both more accurate and consistent. The outcome of separating the ac and dc components of the short-circuit current is illustrated in Fig. 5.15B and the dc component alone is shown in Fig. 5.15C. (A)

2

Normalised short-circuit currents

1.5 1

Phase R asymmetrical current

0.5 0 –0.5 (B) 1

Normalised short-circuit currents dc component

0.5

(C) 1 Normalised dc short-circuit current 0.8 0.6

0

0.4 0.2

–0.5

ac power frequency component

0

Ta

–1

Figure 5.15 Measurement of short-circuit current and separation of ac and dc current components.

t

426

Power Systems Modelling and Fault Analysis

(A)

Peak ac current

)

)

t (B)

Logarithmic scale

) )

t

Figure 5.16 Measured ac current component plotted on linear and logarithmic scales: (A) peak ac current envelope and components and (B) time constants from current envelopes.

The dc (armature) time constant As a decaying exponential, the magnitude of the dc component drops from its initial value at t 5 0 to 1/e 5 0.36788 of the initial value at t 5 Ta(3φ) where Ta(3φ) is the dc or stator or armature time constant. Ta(3φ) is indicated in Fig. 5.15C.

Steady state d-axis reactance The peak envelope of the ac short-circuit current is shown in Fig. 5.16A. One method of determining the steady-state reactance is from the oscillogram’s current value after a sufficient time so that the current component has completely pffiffitransient ffi decayed or vanished. Therefore, with 2Id read from Fig. 5.16A, we have Xd 5

Eo Id

(5.76)

An alternative method that is usually used by machine manufacturers is to calculate Xd from the ratio of the field current at rated short-circuit stator current to the airgap field current at rated open-circuit stator voltage.

Transient reactance and transient short-circuit time constant Using Fig. 5.16A, the machine current attributable to the transient reactance is determined from the transient envelope of the ac current after subtracting the

Modelling of rotating ac synchronous and induction machines

427

steady-state current value then the transient current result is plotted on linearlogarithmic coordinates as shown in Fig. 5.16B. The envelope of the transient current component is given by ΔI 0d ðtÞ 5

pffiffiffi 0 0 2ðI d 2 Id Þe2t=T dð3φÞ

(5.77a)

When this component is extrapolated back to zero time, it cuts the ordinate axis at pffiffiffi ΔI 0d ðt 5 0Þ 5 2 I 0d 2 Id and this is read from the ordinate. Therefore, the transient current is calculated as follows: I 0d 5

ΔI 0d ðt 5 0Þ pffiffiffi 1 Id 2

(5.77b)

The transient reactance can now be calculated as follows X 0d 5

Eo I 0d

(5.77c)

The transient time constant is the time required for the transient current component to drop to 1/e 5 0.36788 of its initial value as illustrated in Fig. 5.16B.

Subtransient reactance and subtransient short-circuit time constant The machine current attributable to the subtransient reactance is determined from the upper envelope of the ac current after subtracting the transient current value. The subtransient current component is then plotted on linear-logarithmic coordinates as shown in Fig. 5.16B. The envelope of the subtransient current component is given by ΔI 00d ðtÞ 5

pffiffiffi 00 00 2ðI d 2 I 0d Þe2t=Td ð3φÞ

(5.78a)

When this component is extrapolated back to zero time, it cuts the ordinate axis at pffiffiffi ΔI 00d ðt 5 0Þ 5 2ðI 00d 2 I 0d Þ and this is read from the ordinate. Therefore, the subtransient current is calculated as follows: I 00d 5

ΔI 00d ðt 5 0Þ pffiffiffi 1 I 0d 2

(5.78b)

The subtransient reactance can now be calculated as Xd00 5

Eo Id00

(5.78c)

428

Power Systems Modelling and Fault Analysis

The subtransient time constant is the time required for the subtransient current component to drop to 1/e 5 0.36788 of its initial value as illustrated in Fig. 5.16B.

Transient reactance at the end of the subtransient period t 5 5T00 d 00 The transient current component at the end h i of the subtransient period at t 5 5Td p ffiffi ffi is given by ΔI 0dðt55T 00 Þ 5 2 I 0dðt55T 00 Þ
Id and this is read from the ordinate of d

d

Fig. 5.16B. Therefore, the transient current at t 5 5Td00 is given by I 0dðt55T 0 0 Þ 5 d

ΔI 0dðt55T 0 0 Þ pffiffiffi d 1 Id 2

and

X 0dðt55T 00 Þ 5 d

Eo I 0dðt55T 0 0 Þ

(5.78d)

d

Alternatively, Eq. (5.37b) can be used to calculate X 0dðt55T 0 0 Þ . d The measured reactances and short-circuit time constants are usually the unsaturated values. Also, since the currents in three phases are measured, the above calculations are made for each phase and an average value of each parameter is taken.

Positive-sequence resistance The total three-phase power during the three-phase short-circuit test is measured when the short-circuit current is equal to the rated machine current. The rated pffiffiffi machine current is calculated from IRated 5 MVARated =ð 3 3 VRated Þ. Therefore, the positive-sequence resistance is given by RP 5

Total losses under a three
phase short circuit 2 3 3 IRated

(5.79)

5.5.2 Measurement of negative-sequence impedance The negative-sequence reactance and resistance can be determined using one of two test methods. The first is by applying a solid two-phase sustained short circuit on any two phases when the machine is running at rated synchronous speed as illustrated in Fig. 5.17A. The following quantities are measured: the short-circuit current I(A), the voltage between the healthy phase and one of the short-circuited phases, say, Y and denoted VDiff (V) 5 [VR(Healthy) (V) 2 VY(Faulted)](V), and the electric input power P3φ (W). Using Eqs. (2.61) and (2.63) with ZF 5 0, it can be shown that the negative-sequence impedance, reactance and resistance are given by VDiff Ω Z 5 pffiffiffi 33I N

P3φ X 5 pffiffiffi Ω 3 3 I2 N

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R 5 ðZ N Þ2 2 ðX N Þ2 Ω N

(5.80)

Modelling of rotating ac synchronous and induction machines

429

(A)

Rotor

R

Stator

I

=

Y

I B (B)

Rotor

Stator

Stator

Rotor

Rotor direction

Rotor rotation

E Short-circuited field winding Machine under test

Machine as negative sequence source

Figure 5.17 Measurement of synchronous machine negative-sequence resistance and reactance: (A) sustained solid two-phase short-circuit and (B) three-phase negative-sequence voltage source test.

The second test method is illustrated in Fig. 5.17B. It involves applying a threephase voltage source having a negative-sequence phase rotation RBY to the machine being tested with the machine running at rated speed in a positivesequence rotation RYB and its field winding short circuited, that is, having zero internal EMF. The reason for short-circuiting the field winding is that X N, like X 0d and Xd00 , is due to induced currents in the rotor and these currents must be able to flow unhindered. The slip of the machine being tested is therefore 200% and large double-frequency currents are induced in the damper windings. Because the negative-sequence reactance is quite low, the applied test voltage must be very low, for example, 0.02
0.2 pu, so as to avoid overheating. The current I(A), the applied phase voltage E(V), and the electric input power P(W) are measured. The negativesequence reactance and resistance are given by E Ω Z N 5 pffiffiffi 33I

RN 5

P Ω 3 3 I2

XN 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðZ N Þ2 2 ðRN Þ2 Ω

(5.81)

The results obtained from calculations from the three phases are averaged.

5.5.3 Measurement of zero-sequence impedance The zero-sequence impedance can be determined using one of two test methods. In the first method, illustrated in Fig. 5.18A, a solid two-phase to earth sustained

430

Power Systems Modelling and Fault Analysis

(A)

Rotor

R

Stator

I

Y

I

=

B

(B)

Rotor

Stator

/3

Short-circuited field winding /3 /3

+

E – Single-phase source

Machine under test

Figure 5.18 Measurement of synchronous machine zero-sequence resistance and reactance.

short-circuit fault is applied with the machine initially running at rated speed and a very low stator voltage to avoid rotor overheating or vibration. The voltage to neutral (V) on the open healthy phase and the earth (neutral) current (A) are measured. It can be shown, using Eqs. (2.72) and (2.73) with ZF 5 0, that the zero-sequence impedance can simply be calculated by dividing the measured voltage by the earthfault current or ZZ 5

VRðHealthyÞ IE

(5.82a)

The second test method is illustrated in Fig. 5.18B. It involves applying a single-phase voltage source to the machine being tested at the point where the three stator windings terminals are joined together. As for the negative-sequence impedance test, the machine is running at rated speed and its field winding is short-circuited. The source current IZ(A), the applied source voltage E(V), and the total electric input power P(W) are measured. The zero-sequence reactance and resistance are given by ZZ 5

3E Ω IZ

RZ 5

P Ω 3 3 ðI Z Þ2

XZ 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðZ Z Þ2 2 ðRZ Þ2 Ω

(5.82b)

Modelling of rotating ac synchronous and induction machines

431

5.5.4 Example Example 5.4 Consider a 50-Hz three-phase synchronous generator having rated apparent power of 165 MVA, rated power of 132 MW and rated terminal voltage of 15 kV. The results below are obtained from various tests and measurements on the generator. The armature dc resistance is 0.2% on rated voltage and rated MVA: 1. Direct axis synchronous reactance Unsaturated or air-gap field current at rated open-circuit terminal voltage is 682 A. The field current at rated armature (stator) current on three-phase terminal short-circuit is 1418 A. 2. Sudden three-phase short-circuit test at rated synchronous speed Prefault stator phase-to-phase voltage is 3 kV Steady-state rms short-circuit current Id 5 623 A Three-phase load losses at rated current 5 477 kW Fig. 5.19 shows the results of phase R current envelopes (peaks) plotted on a semilogarithmic scale.

(A)

Peak current (A) 10000.0

)

1000.0 0

(B)

0.2 0.4 0.6 0.8 Time (s)

1

1.2 1.4

10000.0

) 1000.0

100.0 0

0.02

0.04 0.06 Time (s)

0.08

0.1

Figure 5.19 Measured short-circuit current envelopes plotted on semilogarithmic scale: (A) subtransient and transient components and (B) subtransient component.

432

Power Systems Modelling and Fault Analysis

3. Negative-sequence impedance test A phase-to-phase sustained short-circuit test at the generator terminals gave the following results:

VDiff (V) rms fault current I(A) Three-phase power P (kW)

453 1,006 455

626 1,407 879

806 1,802 1,451

4. Zero-sequence reactance test A two-phase to earth sustained short-circuit test at the generator terminals gave the following results: Voltage to neutral, healthy phase (V) rms earth fault current I(A)

149 1,146

194 1,490

232 1,780

Solution Direct axis synchronous reactance Unsaturated Xd 5

1; 418A 5 2:08 pu or 208% 682A

Alternatively, using Id 5 623 A from the sudden three-phase short-circuit test, we have pffiffiffi ð3000= 3ÞV 5 2:78Ω Xd 5 623A Base impedance 5

ð15 kVÞ2 5 1:3636Ω 165 MVA

thus Xd 5

2:78 5 2:04 pu or 204% 1:3636

Direct axis transient reactance and time constant From Fig. 5.19A, we have pffiffiffi 0 2 I d 2 Id 5 5; 655:4A

I 0d 5

5; 655:4 pffiffiffi 1 623 5 4; 622A 2

Modelling of rotating ac synchronous and induction machines

433

thus X 0d 5

pffiffiffi 3; 000= 3V 5 0:375Ω 4; 622A

X 0d 5

or

0:375Ω 5 0:275 pu or 27:5%: 1:3636Ω

At t 5 T0 d, the transient current component is 1/e times the initial value of 5,655.4 A, that is, 2,080.5 A. Thus, the transient time constant is found from Fig. 5.19 as T0 d 5 1.1 seconds.

Direct axis subtransient reactance and time constant From Fig. 5.19B, we have pffiffiffi 00 2 I d 2 I 0 d 5 3; 010:9A

I 00d 5

3; 010:9 pffiffiffi 1 4; 622 5 6; 751A 2

thus X 00d

pffiffiffi 3; 000= 3V 5 0:25656Ω 5 6; 751A

or

X 00d 5

0:25656Ω 5 0:188 pu or 18:8%: 1:3636Ω

At t 5 Td00 , the subtransient current component is 1/e times the initial value of 3010.9 A, that is, 1107.6 A. Thus, the subtransient time constant is found from Fig. 5.19B as Td00 5 39 ms.

Transient reactance at the end of the subtransient period The transient current component at the end the subtransient period, that i pffiffiffih of is, at t 5 5Td00 5 195 ms is given by 2 I 0dðt5195msÞ 2 Id 5 4; 709A giving I 0dðt5195msÞ 5 3; 953A. Therefore X 0dðt5195msÞ 5

pffiffiffi 3; 000= 3 V 5 0:438Ω 3; 953 A

or X 0d ðt 5 195 msÞ 5

0:438Ω 5 0:321 pu or 32:1% 1:3636Ω

Alternatively, using Eq. (5.37c), we obtain X 0dðt5195msÞ 5



1

 5 0:32 pu or 32% 1 1 1 1
e2195=1100 2:04 0:275 2:04

434

Power Systems Modelling and Fault Analysis

Usually, similar analysis is carried out for the other two phases and an average is taken. The above test results were made at 20% rated voltage. The same tests may also be carried out at other voltages.

Positive-sequence resistance Rp 5

Load losses during short 2 circuit at rated current 2 3 3 IRated

165 MVA 3 1; 000 5 6; 351A Rated armatureðstatorÞcurrent 5 pffiffiffi 3 3 15 kV

Rp 5

477 3 103 W 5 3:942 3 1023 Ω 3 3 ð6351AÞ2

or Rp 5

3:942 3 1023 Ω 5 0:00289 pu or 0:289% 1:3636Ω

Negative-sequence impedance Using the first set of results, the negative-sequence impedance, reactance and resistance are calculated as follows: 453 V 1 3 5 0:1906 pu or 19:06% Z N 5 pffiffiffi 3 3 1; 006A 1:3636Ω 455 3 103 W 1 5 0:1903 pu or 19:03% X N 5 pffiffiffi 3 2 1:3636Ω 3 3 ð1; 006AÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi RN 5 ð0:1906Þ2 2 ð0:1903Þ2 5 0:0107 pu or 1:07%% Similar calculations are carried out using the other test results and these will give broadly similar results. The negative-sequence impedance/reactance is then plotted pffiffiffi against the short-circuit current and extrapolated to 3IRated 5 10; 827A. The negative-sequence reactance value should be 19%.

Modelling of rotating ac synchronous and induction machines

435

Zero-sequence impedance The zero-sequence impedance is calculated as follows: ZZ 5

149 V 1 3 5 0:095 pu or 9:5% 1; 146A 1:3636Ω

Similar calculations are carried out using the other test results and these will give broadly similar results. The zero-sequence impedance is then plotted against the short-circuit current and extrapolated to 3 3 IRated 5 19,053 A. This value should be 9.5%.

Armature (dc) time constant Using the measured positive-sequence, negative-sequence and zero-sequence reactances, and the known armature dc resistance, the armature dc time constants for various short-circuit faults are calculated as follows: a. Three-phase short circuit Ta 5

18:8% 5 0:30 s 2π 3 50 Hz 3 0:2%

b. Two-phase short circuit Ta 5

19:03% 5 0:302 s 2π 3 50 Hz 3 0:2%

c. One-phase to earth short circuit Ta 5

ð18:8 1 19:03 1 9:5%Þ 5 0:251 s 2π 3 50 Hz 3 ð3 3 0:2%Þ

d. Two-phase to earth short circuit

5.6

Ta1 5

ð19:03 1 2 3 9:5Þ% 5 0:20 s; 2π 3 50 Hz 3 ð3 3 0:2Þ%

Ta2 5

19:03% 5 0:302 s 2π 3 50 Hz 3 0:2%

Modelling of induction motors in the phase frame of reference

5.6.1 General For over a century, the induction motor has been the workhorse of the electric energy industry. Induction motors are used to drive a variety of mechanical loads, for

436

Power Systems Modelling and Fault Analysis

example, fans, blowers, centrifugal pumps, hoists, conveyors, boring mills and textile machinery. The modelling of induction motors and the analysis of their short-circuit current contribution is fully taken into account in industrial power systems, for example, petrochemical plant, oil refineries, offshore oil platforms and power station auxiliary systems. The important characteristic in such installations is the concentrated large number of induction motors used which are clearly identifiable in terms of size, rating, location in the industrial network and, generally, electrical parameters. Three-phase and single-phase motors are also used in many commercial installations with the latter being utilised in a variety of domestic appliances. Single-phase motors can form a significant part of the general power system load ‘seen’ at distribution network substations, particularly in very warm and hot climates where the use of air conditioning is widespread. The importance of the short-circuit contribution from such a cumulative motor load has only recently been recognised but, unfortunately, there is no international consensus regarding its treatment. This aspect is discussed in Chapter 8, International standards for short-circuit analysis in ac power systems.

5.6.2 Overview of induction motor modelling in the phase frame of reference Similar to a synchronous machine, the stator of an induction machine consists of a set of three-phase stator windings which may be star- or delta-connected. The rotor may be of a wound rotor or a squirrel-cage rotor construction with conducting bars placed in slots on the rotor surface. In squirrel-cage motors, the bars are shortcircuited at both ends by end rings. Due to the absence of slip rings or brushes, squirrel-cage induction motors are most rugged and are virtually maintenance-free machines. For some applications where high starting torques are required, for example, boring mills, conveyor equipment, textile machinery, wood working equipment, etc., double squirrel-cage rotors are used. As the name implies, two layers of bars are used; a high-resistance one nearer the rotor surface and a low-resistance one located below the first. The highresistance bars are effective at starting in order to give higher starting torque, whereas the low-resistance bars are effective under normal running operation. The connection of a balanced three-phase power supply to the stator produces an MMF that rotates at synchronous speed. This MMF induces voltages in the rotor windings and because these are short-circuited, rotor currents flow. The interaction of the rotor currents and the stator MMF produces a torque that acts to accelerate the rotor in the same direction of rotation of the stator MMF. The frequency of the rotor currents is determined by the relative motion of the stator MMF and rotor speed and is equal to ns 2 nr where nr is the rotor speed. The relative speed of the rotor, called slip speed, in per unit of the synchronous speed, is given by s5

ns 2 nr ns

(5.83a)

Modelling of rotating ac synchronous and induction machines

437

where ns 5

60 3 fs NPP

(5.83b)

s is slip speed ns is synchronous speed in rpm f is supply frequency in Hz NPP is number of pole pairs nr is rotor speed in rpm

Clearly, when the rotor is stationary, for example, at starting, nr 5 0 and the slip s 5 1. Eq. (5.83a) can be written in terms of stator MMF and rotor angular speeds ωs and ωr, in electrical radians per second, as follows: s5

ωs 2 ωr ωs

(5.83c)

or ωr 5 ð1 2 sÞωs

(5.83d)

As for synchronous machines, one of our objectives is to gain an insight into the meaning of the induction motor parameters required for use in short-circuit analysis. Prior to the derivation of the motor equations in the ryb phase frame of reference, we note that even for a squirrel-cage rotor, the rotor can be represented as a set of three-phase windings. This is because induced currents in the rotor produce an MMF with the same number of poles as that produced by the stator windings. Also, the rotor is symmetrical as the air gap is uniform. This means that only the mutual inductances between stator and rotor windings are dependent on rotor angle position. Fig. 5.20 illustrates the stator and rotor circuits of an induction motor. Assuming that the axis of phase r rotor winding leads that of phase r stator winding by θr where θr 5 ω r t

(5.84a)

Using Eq. (5.83c), we can write θr 5 ð1 2 sÞωs t

(5.84b)

438

Power Systems Modelling and Fault Analysis

r

ir R s Stator windings 0.5L ms

iy

Rs

Rs

y θr

r

ωr ir Direction of rotation

Rr

d axis Axis of phase r

ib b

Lsr f (θr )

Rr

b

ib q axis

0.5L mr Rr

Rotor windings

y iy

Figure 5.20 Illustration of stator and rotor circuits of an induction machine.

Initially, we consider the rotor to have a single rotor winding to correspond to that of wound rotor or a single squirrel-cage rotor. The time domain stator and rotor voltage equations, with (s) and (r) denoting stator and rotor, respectively, can be written in matrix form as erybðsÞ 5 Rs irybðsÞ 1

d ψ dt rybðsÞ

(5.85a)

erybðrÞ 5 Rr irybðrÞ 1

d ψ dt rybðrÞ

(5.85b)

where e, i and ψ are 3 3 1 column matrices and Rs and Rr are 3 3 3 diagonal matrices of Rs and Rr elements, respectively. Rs and Rr are stator and rotor resistances, respectively. The time domain stator and rotor flux linkage equations for a magnetically coupled linear system are given by



ψrybðsÞ Lss 5 ψrybðrÞ Lrs ðθr Þ

Lsr ðθr Þ Lrr




irybðsÞ irybðrÞ

 (5.86a)

Modelling of rotating ac synchronous and induction machines

where rðsÞ

2

6 Lσs 1 Lms 6 rðsÞ 6 1 6 2 Lms yðsÞ 6 6 2 bðsÞ 6 6 4 2 1L ms 2

Lss 5

2

Lrr 5

yðsÞ 1 2 Lms 2 Lσs 1 Lms 2

rðrÞ

6 Lσr 1 Lmr 6 rðrÞ 6 1 6 2 Lmr yðrÞ 6 6 2 bðrÞ 6 6 4 2 1L mr 2

yðrÞ 1 2 Lmr 2 Lσr 1 Lmr 2

rðrÞ cosθr Lsr ðθr Þ 5 Lsr 4 cosðθr 2 2π=3Þ cosðθr 1 2π=3Þ 2

1 Lms 2

1 Lmr 2

bðsÞ 1 2 Lms 2 1 2 Lms 2 Lσs 1 Lms

439

3 7 7 7 7 7 7 7 7 5

bðrÞ 1 2 Lmr 2 1 2 Lmr 2 Lσr 1 Lmr

yðrÞ cosðθr 1 2π=3Þ cosθr cosðθr 2 2π=3Þ

3 7 7 7 7 7 7 7 7 5

bðrÞ 3 cosðθr 2 2π=3 cosðθr 1 2π=3Þ 5 cosθr

(5.86b)

rðsÞ yðsÞ bðsÞ (5.86c)

and Lrs ðθr Þ 5 LTsr ðθr Þ Lσs and Lσr are the leakage inductances of the stator and rotor windings, Lms and Lmr are the magnetising inductances of the stator and rotor, and Lsr is the amplitude of the mutual inductances between the stator and rotor windings. The magnetising and mutual inductances are associated with the same magnetic flux paths, hence Lms, Lmr and Lsr are related by the stator/rotor turns ratio Ns =Nr as follows: Ns Lms 5 Lsr Nr

and

2 Nr Lmr 5 Lms Ns

and

L0σr 5

Defining L0sr 5

Ns Lsr Nr

2 Ns Lσr Nr

(5.87a)

440

Power Systems Modelling and Fault Analysis

where the prime denotes rotor quantities that are referred to the stator. Thus 2

cosθr cosðθr 1 2π=3Þ 6 L0 sr ðθr Þ 5 Lms 4 cosðθr 2 2π=3Þ cosθr cosðθr 1 2π=3Þ cosðθr 2 2π=3Þ

3 cosðθr 2 2π=3Þ 7 cosðθr 1 2π=3Þ 5

(5.87b)

cosθr

and 2

0 6 Lσr 1 Lms 6 6 1 6 2 Lms L0 rr 5 6 6 2 6 6 4 2 1L ms 2

2

1 Lms 2

L0σr 1 Lms 2

3

1 Lms 2 1 2 Lms 2 2

1 Lms 2

L0σr 1 Lms

7 7 7 7 7 7 7 7 5

(5.87c)

thus "

#
ψrybðsÞ Lss 5 0 ψrybðrÞ L rs ðθr Þ

L0 sr ðθr Þ L0 rr




irybðsÞ irybðrÞ

 (5.87d)

Substituting Eq. (5.87d) into Eq. (5.85a) and (5.85b), we obtain


erybðsÞ erybðrÞ




5

Rs 1 pLss

pL0 sr ðθr Þ

pðL0 sr ðθr ÞÞT

R0 r 1 pL0 rr




irybðsÞ



irybðrÞ

(5.87e)

where R0 r 5

2 Ns Rr Nr

and

p5

d : dt

We have used p as the Laplace operator instead of s since the latter is now used to denote induction motor slip.

5.7

Modelling of induction motors in the dq frame of reference

5.7.1 Transformation to Park dq axes The next step is to transform the stator and rotor quantities to the d and q axes reference frame. However, because we are concerned in this book with short circuit

Modelling of rotating ac synchronous and induction machines

441

rather than electromechanical analysis, we will fix our reference frame to the rotor as in the case of synchronous machines. This introduces an advantage that will become apparent later. The d-axis is chosen to coincide with stator phase r axis at t 5 0 and the q-axis leads the d-axis by 90 degrees in the direction of rotation. Therefore, the stator and rotor quantities can be transformed using the transformation matrix given in Eq. (5.7b). It can be shown that the transformation of Eqs. (5.87d) and (5.87e) gives 2

3 2 ψds Lσs 1 Lm 6ψ 7 6 0 6 qs 7 6 6 756 4 ψdr 5 4 Lm ψqr

0

0 Lσs 1 Lm

Lm 0

0 Lm

0

L0σr 1 Lm

0

Lm

0

L0σr 1 Lm

32

3 ids 7 6i 7 7 6 qs 7 76 7 5 4 idr 5

(5.88a)

iqr

where Lm 5 1:5Lms 3 3 2 3 2 2 3 ψds eds Rs ids ωr ψqs 6e 7 6R i 7 6 2ω ψ 7 d 6ψ 7 r ds 7 6 qs 7 6 s qs 7 6 6 qs 7 71 6 6 756 0 716 7 5 dt 4 ψdr 5 4 edr 5 4 Rr idr 5 4 0 ψqr eqr R0r iqr 0 2

(5.88b)

Because the rotor windings are short-circuited, edr 5 eqr 5 0

(5.88c)

We note that the speed voltage terms in the rotor voltages of Eq. (5.88b) are zeros because of our choice of reference frame for the rotor quantities to coincide with the rotor rather than with the synchronously rotating MMF. The above equations are in physical units. However, it can be easily shown that they take the same form in per unit because the stator rated quantities are taken as base quantities. The d and q axes transient equivalent circuits are obtained by substituting Eq. (5.88a) into Eq. (5.88b). The result is the operational equivalent circuit shown in Fig. 5.21A that is identical in both the d and q axes because of the symmetrical structure of the rotor. The steady-state equivalent circuit, shown in Fig. 5.21B, is derived by converting the time variables into phasors and setting all derivatives to zero. We note that in order to simplify the notation, we have dropped the prime from the rotor resistance and inductance parameters.

442

Power Systems Modelling and Fault Analysis

(A)

Ids or Iqs

(B)

pLσ s

+

Is +

pLσ r

pψds

pLm

or pψqs

Rr

Rs

jX σs

jX m

Vs

Rr s

–

–

jX σr

Figure 5.21 Equivalent circuits of an induction motor with one rotor winding: (A) operational equivalent circuit and (B) steady-state equivalent circuit.

5.7.2 Complex form of induction motor equations To simplify the mathematics, we can replace all the voltage and flux linkage equations for the d and q axes of Eqs. (5.88a) and (5.88b) by half the number of equations if we define new complex variables Vs, Is, It, ψs, ψr such that Vs 5 eds 1 jeqs

Is 5 ids 1 jiqs

ψs 5 ψds 1 jψqs

ψr 5 ψdr 1 jψqr

(5.89)

Therefore, combining the fluxes and voltages from Eqs. (5.88a) and (5.88b) according to Eq. (5.89), we obtain Vs 5 pψs 1 jωr ψs 1 Rs Is

(5.90a)

0 5 Rr Ir 1 pψr

(5.90b)

ψs 5 ðLσs 1 Lm ÞIs 1 Lm Ir

(5.91a)

ψr 5 Lm Is 1 ðLm 1 Lσr ÞIr

(5.91b)

5.7.3 Operator reactance and parameters of a single-winding rotor In order to gain an understanding of the origin and meaning of the various motor parameters, for example, transient reactance and time constant, we will use the method of operator axis reactance analysis for the simplicity and clear insight it provides. Since the motor operational equivalent circuit of Fig. 5.21A is identical for both the d and q axes, the d and q axes operator reactances will be identical. By substituting Eq. (5.91b) for ψr into Eq. (5.90b), we obtain Ir 5

2 Lm p Is Rr 1 ðLm 1 Lσr Þp

Modelling of rotating ac synchronous and induction machines

443

which when substituted into Eq. (5.91a) for ψs and simplifying, we obtain
ψs 5 ðLσs 1 Lm Þ 2

 L2m p Is Rr 1 ðLσr 1 Lm Þp

(5.92a)

It can be shown that after further manipulation, Eq. (5.92a) can be written as ψs 5 ðLσs 1 Lm Þ

1 1 pT 0 Is 1 1 pT 0o

(5.92b)

where T 0o 5

Lσr 1 Lm Rr

(5.93a)

and 0 T0 5

1

C 1B 1 BLσr 1 C @ A 1 1 Rr 1 Lσs Lm

(5.93b)

T0 o and T0 are the motor’s open-circuit and short-circuit transient time constants, respectively. As for the synchronous machine, to obtain the time constants in seconds, the per-unit time constants of Eqs. (5.93a) and (5.93b) should be divided by ωs 5 2πfs where fs is the system power frequency. The motor operational reactance is defined as X(p) 5 ωsψs/Is, hence, from Eq. (5.92b), we obtain XðpÞ 5 ðXσs 1 Xm Þ

1 1 pT 0 1 1 pT 0o

(5.94a)

The effective motor reactance at the instant of an external disturbance is defined as the transient reactance X0 . Using Eq. (5.94a), this is given by X 0 5 lim XðpÞ 5 ðXσs 1 Xm Þ p!N

T0 T 0o

(5.94b)

This can also be expressed in terms of the electrical parameters of the stator and rotor circuits using Eqs. (5.93a) and (5.93b) as follows: X 0 5 Xσs 1

1 1 1 1 Xm Xσr

(5.94c)

444

Power Systems Modelling and Fault Analysis

5.7.4 Operator reactance and parameters of double-cage or deep-bar rotor Many induction motors are designed with a double squirrel-cage rotor. Also, many large MW size motors have deep-bar cage windings on the rotor designed to limit starting currents. For both double-cage and deep-bar rotors, we assume that the end-rings resistance and the part of the leakage flux which links the two rotor windings, but not the stator, are neglected. Therefore, we represent both rotor types by two parallel rotor circuits as shown in the operational equivalent circuit of Fig. 5.22A. The corresponding steady-state equivalent circuit is shown in Fig. 5.22B. Fig. 5.22C shows an alternative steady-state equivalent circuit where the two rotor branches are converted into a single equivalent branch whose resistance and reactance are functions of motor slip s as follows:

(A)

Ids or Iqs +

pLσ s

pψds

pLm

or pψqs

pLσ r1

pLσ r2

Rr1

Rr2

jXσr1

jXσr 2

R r1 s

R r2 s

– Is +

(B)

Rs

jX σs

jX m

Vs –

Is +

(C)

Rs

jX σs

jX m

Vs –

jX σ r (s)

R r (s) s

Figure 5.22 Equivalent circuits of an induction motor with a double squirrel-cage or deepbar rotor: (A) operational equivalent circuit, (B) steady-state equivalent circuit and (C) circuit (B) with a single equivalent rotor branch.

Modelling of rotating ac synchronous and induction machines

445

Rr ðsÞ 5

2 2 Rr1 Rr2 ðRr1 1 Rr2 Þ 1 s2 ðRr2 Xσr1 1 Rr1 Xσr2 Þ 2 2 ðRr1 1Rr2 Þ 1 s2 ðXσr1 1Xσr2 Þ

(5.95a)

Xr ðsÞ 5

Xσr2 R2r1 1 Xσr1 R2r2 1 s2 Xσr1 Xσr2 ðXσr1 1 Xσr2 Þ ðRr1 1Rr2 Þ2 1 s2 ðXσr1 1Xσr2 Þ2

(5.95b)

From the operational circuit of Fig. 5.22A, and as in the case of a single rotor winding, the complex form of the motor voltage and flux linkage equations can be written by inspection as follows: Vs 5 pψs 1 jωr ψs 1 Rs Is

(5.96a)

0 5 Rr1 Ir1 1 pψr1

(5.96b)

0 5 Rr2 Ir2 1 pψr2

(5.96c)

ψs 5 ðLσs 1 Lm ÞIs 1 Lm Ir1 1 Lm Ir2

(5.97a)

ψr1 5 Lm Is 1 ðLσr1 1 Lm ÞIr1 1 Lm Ir2

(5.97b)

ψr2 5 Lm Is 1 Lm Ir1 1 ðLσr2 1 Lm ÞIr2

(5.97c)

The derivation of the operational reactance is similar to that in the case of a single rotor winding except that more algebra is involved. The steps are summarised as follows: substitute Eq. (5.97b) in Eq. (5.96b), Eq. (5.97c) in Eq. (5.96c), eliminate Is and express Ir1 in terms of Ir2, express Ir1 and Ir2 in terms of Is and substitute these results back in Eq. (5.97a). It can be shown that the result is given by ψs 5 ðLσs 1 Lm Þ Is 2

L2m ðRr1 1 Rr2 Þp 1 L2m ðLσr1 1 Lσr2 Þp2 Rr1 Rr2 1 ½Rr1 ðLσr2 1 Lm Þ 1 Rr2 ðLσr1 1 Lm Þp 1 ½Lσr1 Lm 1 Lσr2 ðLσr1 1 Lm Þp2 (5.98a)

Using X(p) 5 ωsψs/Is, the motor operational reactance can be written as XðpÞ 5 ðXσs 1 Xm Þ 2

2 2 Xm ðRr1 1 Rr2 Þp 1 Xm ðXσr1 1 Xσr2 Þp2 Rr1 Rr2 1 ½Rr1 ðXσr2 1 Xm Þ 1 Rr2 ðXσr1 1 Xm Þp 1 ½Xσr1 Xm 1 Xσr2 ðXσr1 1 Xm Þp2

(5.98b) As in the case of synchronous machines, in order to derive the electrical parameters of the motor, it can be shown that Eq. (5.98b) can be rewritten as

446

Power Systems Modelling and Fault Analysis

XðpÞ 5 ðXσs 1 Xm Þ

1 1 ðT4 1 T5 Þp 1 ðT4 T6 Þp2 1 1 ðT1 1 T2 Þp 1 ðT1 T3 Þp2

(5.99)

where T1 5

Xσr1 1 Xm Rr1

T2 5

Xσr2 1 Xm Rr2

0 T3 5

1

C 1 B 1 BXσr2 1 C @ 1 1A Rr2 1 Xσr1 Xm 0

T5 5

0 T4 5

1

C 1 B 1 BXσr2 1 C 1 1A Rr2 @ 1 Xσs Xm

(5.100a)

C 1 B 1 BXσr1 1 C @ 1 1A Rr1 1 Xσs Xm 0

T6 5

1 (5.100b) 1

C 1 B 1 BXσr2 1 C 1 1 1 A Rr2 @ 1 1 Xσs Xm Xσr1 (5.100c)

Again, as in the case of synchronous machines, we make use of an approximation that reflects practical double squirrel-cage rotor or deep-bar rotor designs. The winding resistance of the lower cage winding Rr1 is much smaller than Rr2 of the second cage winding nearer the rotor surface. This means that T2 (and T3) is much smaller than T1, and T5 (and T6) is much smaller than T4. Therefore, Eq. (5.99) can be expressed in terms of factors as follows: ð1 1 pT 0 Þð1 1 pT 00 Þ  XðpÞ 5 ðXσs 1 Xm Þ  1 1 pT 0o 1 1 pTo00

(5.101)

where T 0o 5 T1

T 00o 5 T3

(5.102a)

are the open-circuit transient and subtransient time constants, and T 0 5 T4

T 00 5 T6

(5.102b)

are the short-circuit transient and subtransient time constants. The effective motor reactance at the instant of an external disturbance is defined as the subtransient reactance Xv. Using Eq. (5.101), this is given by X 00 5 lim XðpÞ 5 ðXσs 1 Xm Þ p!N

T 0 T 00 T 0o To00

(5.103a)

Modelling of rotating ac synchronous and induction machines

447

which, using Eqs. (5.100a)
(5.100c), (5.102a) and (5.102b), can be written in terms of the motor reactance parameters as follows: X 00 5 Xσs 1

1 1 1 1 1 1 Xm Xσr1 Xσr2

(5.103b)

As in the case of synchronous machines, the second rotor cage is no longer effective at the end of the subtransient period and the beginning of the transient period. Thus, using Eqs. (5.95a) and (5.95b) previously derived for the single rotor cage, we have X 0 5 ðXσs 1 Xm Þ

T0 1 5 Xσs 1 1 1 T 0o 1 Xm Xσr1

(5.104)

Using Eq. (5.104) for X0 in Eq. (5.103a) for Xv, we obtain the following useful relationship between the subtransient and transient reactances and time constants X 00 T 00 5 X0 To00

(5.105)

In order to determine the motor reactance under steady-state conditions, we recall that the rotor slip is equal to s and thus the angular frequency of the dq axes stator currents is sωs. Therefore, the steady-state motor impedance can be determined by putting p 5 jsωs in X(p) expression given in Eq. (5.101). We can express the induction machine operational reactance given in Eq. (5.101) as a sum of partial fractions as follows: 1 1 ð1 1 pT 0o Þð1 1 pTo00 Þ T 0o To00 ðp 1 1=T 0o Þðp 1 1=To00 Þ 5 5 0 00 0 00 XðpÞ ðXσs 1 Xm Þ ð1 1 pT Þð1 1 pT Þ ðXσs 1 Xm ÞT T ðp 1 1=T 0 Þðp 1 1=T 00 Þ which, using Eq. (5.103a) for Xv and Eq. (5.104) for X0 , can be written as



 1 1 1 1 p 1 1 p 5 1 1 2 2 0 XðpÞ Xσs 1 Xm X0 Xσs 1 Xm ðp 1 1=T 0 Þ X 00 X ðp 1 1=T 00 Þ (5.106)

448

5.8

Power Systems Modelling and Fault Analysis

Induction motor behaviour under short-circuit faults and modelling in the sequence reference frame

The most common types of short-circuit faults studied in power systems are the balanced three-phase and unbalanced single-phase to earth faults. Having derived the motor reactances and time constants, we are now able to proceed with analysing the behaviour of the motor under such fault conditions.

5.8.1 Three-phase short-circuit faults Armature (stator) short-circuit time constant Using X(p) 5 ωsψs/Is in Eq. (5.96a), the stator current is given by Is 5

ω s Vs h XðpÞ ðp 1 jωr Þ 1

ω s Rs XðpÞ

i

(5.107)

As for synchronous machines, we can make an assumption in the term ωsRs/X(p) to neglect the rotor resistances in the factors of X(p) given in Eq. (5.101). This is equivalent to setting infinite rotor time constants or neglecting the decay in the corresponding short-circuit currents. Mathematically speaking, this is equivalent to setting p!N in Eqs. (5.94a) and (5.101). Thus, using Eq. (5.94b) for a single rotor s Rs winding, we obtain X(p) 5 X0 giving ωXðpÞ 5 ωXs R0 s . Also, using Eq. (5.103a) for a s Rs double squirrel-cage or deep-bar rotor, we obtain X(p) 5 Xv giving ωXðpÞ 5 ωXs R00 s . Therefore, the armature or stator time constant Ta is defined as follows:

Ta 5

8 0 X > > >

> > :ω R s s

for a double squirrel
cage or deep
bar rotor

s s 00

(5.108)

Total short-circuit current contribution from a motor on no load Unlike a synchronous machine, since the induction motor has no rotor excitation, it draws its magnetising current from the supply network whatever the motor loading is, that is, from no load to full load operation. Therefore, the motor cannot be operated open-circuited in a steady-state condition! Prior to the short-circuit fault, the motor is assumed connected to a balanced three-phase supply and operating at rated terminal voltage. To simplify the mathematics, we assume that the motor is running at no load and hence the slip is very small and nearly equal to zero. Let the real instantaneous voltage, complex instantaneous voltage and complex phasor voltage values of the phase r stator voltage, in the rotor reference frame, just before the short-circuit, be given by vs ðtÞ 5

pffiffiffi 2Vrms cosðsωs t 1 θo Þ 5 Real½Vs ðtÞ

(5.109a)

Modelling of rotating ac synchronous and induction machines

Vs ðtÞ 5 Vs 5

pffiffiffi 2Vrms ejðsωs t1θo Þ 5 Vs ejsωs t

pffiffiffi 2Vrms ejθo

449

(5.109b) (5.109c)

where θo is the initial phase angle that defines the instant of fault on the phase r voltage waveform at t 5 0. Neglecting stator resistance, that is, letting Rs 5 0, using Eq. (5.83c) and putting p 5 0 in Eq. (5.107), we obtain Is 5

Vs jð1 2 sÞXðpÞðp50Þ

(5.110a)

Using Eq. (5.101), X(p)(p50) 5 Xσs 1 Xm and using Eq. (5.109c), the stator steady state current phasor is given by Is 5

pffiffiffi 2Irms ejðθo 2π=2Þ

(5.110b)

where Irms 5

Vrms ð1 2 sÞðXσs 1 Xm Þ

Is ðt Þ 5

pffiffiffi 2Irms e jðθo 2π=2Þ ejsωs t

(5.110c)

and (5.110d)

Eq. (5.110a) shows that the current lags the voltage by 90 degrees, as expected. Using the superposition principle, the total motor current after the application of a solid short circuit is the sum of the initial steady-state current and the change in motor current due to the short circuit. The latter is obtained by applying a voltage source at the point of fault equal to but out of phase with the prefault voltage, pffiffiffi that is, using Eq. (5.109b), ΔVs ðtÞ 5 2 2Vrms ejθo ejsωs t . The Laplace transform of pffiffi pffiffi jθo 2Vrms ejθo D 2 2Vprms e because, as already stated, the slip is ΔVs(t) is ΔVs ð pÞ 5 2 p
jsω s s Rs almost zero. Now, using ωXðpÞ 5 T1a , we obtain from Eq. (5.107)

pffiffiffi 2 ωs 2Vrms ejθo  ΔIs ð pÞ 5  p ðp 1 1=Ta Þ 1 jωr XðpÞ

(5.111a)

450

Power Systems Modelling and Fault Analysis

Substituting the partial fractions given by Eq. (5.106) for X(p), we obtain 0 1 " pffiffiffi 2 ωs 2Vrms ejθo 1 1 1 p A 1@ 0 2 ΔIs ð pÞ 5 X Xσs 1 Xm ðp 1 1=T 0 Þ pðp 1 1=Ta 1 jωr Þ Xσs 1 Xm 0 1 # 1 1 p 1 @ 00 2 0 A X X ðp 1 1=T 00 Þ (5.111b) The complex instantaneous current ΔIs(t) is obtained by taking the inverse Laplace transform of Eq. (5.111b). The total short-circuit current Is(total)(t) is the sum of the prefault current given in Eq. (5.110d) and ΔIs(t). The real instantaneous phase r stator short-circuit current is given by ir ðtÞ 5 Real½IsðtotalÞ ejωr t . For all practical purposes, 1/Ta, 1/T0 , and 1/Tv are negligible in comparison with ωs and ωr. Therefore, and after much mathematical operations, it can be shown that the total phase r short-circuit current is given by 20 1 0 1 3 pffiffiffi 1 1 1 1 0 00 Ae2t=T 1 @ 2 Ae2t=T 5 ir ðtÞ  2Vrms 4@ 0 2 X Xσs 1 Xm X 00 X0   (5.112a) 3 cos ð1 2 sÞωs t 1 θo 2 π=2 pffiffiffi  2Vrms cos θo 2 π=2 e2t=Ta 2 X 00 The first term is the ac component that consists of a subtransient and a transient component which decay with time constants T0 and Tv, respectively. The second term is the dc component that has an initial magnitude that depends on θo and it decays with a time constant Ta. Unlike a synchronous machine, there is no steadystate ac component term and hence the short-circuit current decays eventually to zero. The frequency of the short-circuit current is equal to the rotor frequency ð1 2 sÞfs and is slightly lower than the power frequency by the factor (1 2 s). For a single cage/winding rotor, it can be shown that the instantaneous current is given by

 pffiffiffi   1 1 0 ir ðtÞ  2Vrms 0 2 e2t=T cos ð1 2 sÞωs t 1 θo 2 π=2 X Xσs 1 Xm pffiffiffi  2Vrms cos θo 2 π=2 e2t=Ta (5.112b) 2 0 X In this case, the current consists of only one ac component and a dc component. The phase y and b currents are obtained by replacing θo with (θo 2 2π/3) and (θo 1 2π/3) respectively. Fig. 5.23 shows typical short-circuit current waveforms of a large 5.6-kV 4,700-hp induction motor with a deep-bar rotor having the following parameters: Xv 5 0.16 pu, X0 5 0.2 pu, X 5 Xσs 1 Xm 5 3.729 pu and Rs 5 0.0074 pu.

Modelling of rotating ac synchronous and induction machines

(A)

Subtransient current component (pu)

7

–5

0

50

100

3 –1

150 200 Time (ms)

Total ac current component (pu)

100

150 200 Time (ms)

dc current component (pu)

9

3

6

0

50

100

3

150 200 Time (ms)

16

Time (ms)

0 0

–9 (B)

50

–9

7

–5

0

–5

–9

–1

Transient current component (pu)

7

3 –1

451

50

100

150

200

Phase r asymmetric current (pu)

12 8 4

Time (ms)

0 –50

–4

0

50

100

150

200

0

50

100

150

200

4 0 –50

–4

Time (ms)

–8 –12

Phase y asymmetric current

–16 4 0 –50

–4

0

50

100

150

200 Time (ms)

–8 –12

Phase b asymmetric current

–16

Figure 5.23 Three-phase short-circuit fault at the terminals of a deep-bar rotor induction motor from no load: (A) components of phase r short-circuit current and (B) asymmetric currents of phases r, y and b.

452

Power Systems Modelling and Fault Analysis

Simplified motor short-circuit current equations For practical applications, the peak current envelope at any time instant, for a double-cage or deep-bar rotor, assuming maximum dc offset, is given by




  1 1 1 1 2t=T 00 1 2t=Ta 2t=T 0 2 1 2 0 e 1 00 e e X0 Xσs 1 Xm X 00 X X (5.113a)

i^r ðtÞ 5

pffiffiffi 2Vrms

i^r ðtÞ 5

pffiffiffi 0 pffiffiffi 0 00 2½ðI 2 IÞe2t=T 1 ðI 00 2 I 0 Þe2t=T  1 2I 00 e2t=Ta

(5.113b)

i^r ðtÞ 5

pffiffiffi 2Iac ðtÞ 1 Idc ðtÞ

(5.113c)

or

or

where I5

Vrms Xσs 1 Xm

I0 5

Vrms X0

I 00 5

Vrms 00 X

(5.114a)

and Idc ðtÞ 5

pffiffiffi 00 2t=T a 2I e

(5.114b) 0

Iac ðtÞ 5 ðI 0 2 IÞe2t=T 1 ðI 00 2 I 0 Þe2t=T

00

(5.114c)

For a single cage/winding rotor, the peak current envelope is given by


  1 1 1 2t=Ta 2t=T 0 2 1 e e X0 Xσs 1 Xm X0

i^r ðtÞ 5

pffiffiffi 2Vrms

i^r ðtÞ 5

pffiffiffi 0 pffiffiffi 0 2ðI 2 IÞe2t=T 1 2I 0 e2t=Ta

(5.115b)

i^r ðtÞ 5

pffiffiffi 2Iac ðtÞ 1 Idc ðtÞ

(5.115c)

(5.115a)

or

or

Modelling of rotating ac synchronous and induction machines

453

where Idc ðtÞ 5

pffiffiffi 0 2t=T a 2I e

(5.116a)

and Iac ðtÞ 5 ðI 0 2 IÞe2t=T

0

(5.116b)

Motor current change due to the short-circuit fault In the above analysis, we determined the total motor current under a three-phase short-circuit at the motor terminals when the motor is running on no load. This current consisted of the steady-state current just before the fault and the current change due to the short circuit. For some practical analysis, it is useful to calculate the current change only and account differently for the motor initial loading conditions. From Eqs. (5.113a) and (5.115a), we can show that the equations that give the current change for a double-cage and a single-cage/winding rotors are given by i^r ðtÞ 5




  pffiffiffi 1 2t=T 0 1 1 2t=T 00 1 2t=Ta 2Vrms 0 e 1 2 0 e 1 00 e X X 00 X X

i^r ðtÞ 5

pffiffiffi 0 2t=T 0 pffiffiffi 00 2½I e 1 ðI 00 2 I 0 Þe2t=T  1 2I 00 e2t=Ta

(5.117a)

or (5.117b)

and pffiffiffi 2Vrms 2t=T 0 i^r ðtÞ 5 ðe 1 e2t=Ta Þ X0

(5.118a)

or i^r ðtÞ 5

pffiffiffi 0 2t=T 0 2I ðe 1 e2t=Ta Þ

(5.118b)

Positive-sequence reactance and resistance From Eq. (5.112a) for a double-cage or deep-bar rotor, the instantaneous current changepdue ffiffiffi to the short-circuit fault is obtained by ignoring the prefault current Irms 5 2Vrms =ðXσs 1 Xm Þ. Thus with 1 2 s  1, we can express the instantaneous power frequency component of the short-circuit current in the stator reference frame as ir(t) 5 Real[Ir(t)], where Ir(t) is phase r complex instantaneous current given by

454

Power Systems Modelling and Fault Analysis

Ir ðt Þ 5

1 pffiffiffi Vr ðtÞ 2Vrms ejðωs t1θo 2π=2Þ 5 P jX ðtÞ

X P ðtÞ

(5.119a)

where Vr ðtÞ 5

pffiffiffi 2Vrms ejðωs t1θo Þ 5 Vr ejωs t

Vr 5

pffiffiffi 2Vrms ejθo

and

 1 1 2t=T 0 1 1 2t=T 00 5 e 1 2 0 e X P ðtÞ X 0 X 00 X

(5.119b)

is the equivalent time-dependent motor positive-sequence reactance. In the case of a single-winding rotor, using Eq. (5.112b), the time-dependent motor positivesequence reactance is given by 1 X P ðtÞ

5

1 2t=T 0 e X0

(5.119c)

From Eq. (5.119b) for a motor with a double-cage or deep-bar rotor, XP 5 Xv at t 5 0 or the instant of short-circuit, and XP 5 X0 at t 5 0 with the subtransient current component neglected. At the end of the subtransient period, that is, at t  5 Tv, then from Eq. (5.119b), we obtain X P ðtÞ 5 X 0 e5T

00

=T 0

(5.119d)

The effective motor steady-state positive-sequence reactance is infinite because the motor does not supply a steady-state short-circuit current. From Eq. (5.119c) for a single winding rotor, XP 5 X0 at the instant of short-circuit and is infinite in the steady state. The complex instantaneous positive-sequence current is given by I p ðtÞ 5

Ir ðtÞ 1 hIy ðtÞ 1 h2 Ib ðtÞ 5 Ir ðtÞ 3

Fig. 5.24 shows the positive-sequence motor equivalent circuits for a three-phase short-circuit at the motor terminals. The positive-sequence impedance, ZP 5 RP 1 jXP, at the instant of shortcircuit fault can also be derived from the motor steady-state equivalent circuit. For example, for a single winding rotor shown in Fig. 5.21B, ZP is the impedance that would be seen at the motor terminals at starting (i.e., when s 5 1). Thus, it can be shown that the positive-sequence resistance and positive-sequence reactance are given by

Modelling of rotating ac synchronous and induction machines

(A)

455

IP (t ) jXP (t )

Rs +

E r (t)

– (B)

P

I

jXP

=0

P

I

X=∞

Rs +

Vrms

–

Steady state, end of transient period

Figure 5.24 Induction motor positive-sequence equivalent circuits for a three-phase short circuit at its terminals: (A) positive-sequence time-dependent equivalent circuit and (B) positive-sequence fixed-impedance equivalent circuits at various time instants.

R p 5 Rs 1

R2r

2 Xm Rr 1 ðXm 1Xσr Þ2

X p 5 Xσs 1

Xm ½R2r 1 Xσr ðXm 1 Xσr Þ R2r 1 ðXm 1Xσr Þ2

(5.120a)

The term R2r is negligible in comparison with either of the two terms Xσr(Xm 1 Xσr) or (Xm 1 Xσr)2, thus, XP and RP reduce to X P 5 Xσs 1

Xm Xσr 5 X0 Xm 1 Xσr

(5.120b)

and R P 5 Rs 1

X m Rr Xm 1 2Xσr

(5.120c)

Although unnecessary, further simplification in RP and XP can be made if we consider that 2Xσr/Xm , , 1 to obtain RP 5 Rs 1 Rr and XP 5 Xs 1 Xσr.

Effect of short-circuit fault through an external impedance The effect of external impedance (Re 1 jXe) on the current change can be considered in the same manner as for synchronous machines. For the double-cage or deep-bar rotor, the peak current envelop and time constant equations are given by

456

Power Systems Modelling and Fault Analysis

pffiffiffi i^r ðtÞ 5 2Vrms 1

X 00

"

0 1  1 1 1 Aexp 2 t=T 00 expð 2 t=T 0e Þ 1 @ 00 2 0 e X 0 1 Xe X 1 Xe X 1 Xe #

1 expð 2 t=Tae Þ 1 Xe

(5.121a) where ðRs 1Re Þ2 1 ðX 00 1Xe Þ2 T 00 ðRs 1Re Þ2 1 ðX 00 1 Xe ÞðX 0 1 Xe Þ

Te00 5

5

ðRs 1Re Þ2 1 ðX 00 1Xe Þ2 X0 3 00 T 00 2 00 0 ðRs 1Re Þ 1 ðX 1 Xe ÞðX 1 Xe Þ X

(5.121b)

and T 0e 5 5

ðRs 1Re Þ2 1 ðX 0 1Xe Þ2 T 0o ðRs 1Re Þ2 1 ðX 0 1 Xe ÞðX 1 Xe Þ ðRs 1Re Þ2 1 ðX 0 1Xe Þ2 X 3 0 T0 2 0 ðRs 1Re Þ 1 ðX 1 Xe ÞðX 1 Xe Þ X X 00 1 Xe ωs ðRs 1 Re Þ

Tae 5

X 5 Xσs 1 Xm

(5.121c)

(5.121d)

Where the stator and external resistances are much smaller than the reactances, the transient and subtransient time constants reduce to T 0e 5

X 0 1 Xe X 3 0 T0 X X 1 Xe

T 00e 5

X 00 1 Xe X 0 00 3 T X 0 1 Xe X 00

(5.121e)

For a single-cage rotor, the corresponding equations are given by i^r ðtÞ 5

pffiffiffi  2Vrms  expð 2 t=T 0e Þ 1 expð 2 t=Tae Þ X 0 1 Xe

(5.121f)

where T 0e is as given in Eq. (5.121c) or Eq. (5.121e), and Tae 5

X 0 1 Xe ωs ðRs 1 Re Þ

(5.121g)

Modelling of rotating ac synchronous and induction machines

457

5.8.2 Unbalanced single-phase to earth short-circuit faults Short-circuit current equations The stator winding of induction motors may be star- or delta-connected. The neutral point of the star-connected winding is normally isolated. However, an earth fault in the first part or turns of the winding close to the neutral may go undetected until an earth fault external to the motor, that is, on the supply network, occurs. We will therefore present the short-circuit current equations of an induction motor with the neutral point of the star winding connected to earth. When an unbalanced single-phase short circuit occurs at the terminals of a loaded induction motor, the faulted phase will supply a short-circuit current. However, under such a fault condition, there is an interaction between the motor and other short-circuit current sources in the network, for example, synchronous machines through the zero-sequence network. The current contribution will therefore be affected by other sources in the network. Further, the phase current contribution would not decay to zero as in the case of a three-phase fault because the other two healthy phases remain supplied from the network and continue to provide some excitation which maintains some flux. Although the positivesequence current contribution will decay to zero, the negative-sequence current contribution will not so long as the single-phase fault condition continues to exist. In Section 5.8.4, we describe a sudden short-circuit test immediately after motor disconnection from the supply network. It can be shown that the change in motor current supplied under a single-phase fault at the motor terminals, in the case of a double-cage or a deep-bar rotor, expressed in terms of the peak current envelope, is given by pffiffiffi 2Vrms

"

1 expð 2 t=T 0ð1φÞ Þ X0 1 XN 1 XZ 0 1 1 1 Aexpð 2 t=T 00 Þ 1 @ 00 2 0 ð1φÞ X 1 XN 1 XZ X 1 XN 1 XZ # 1 1 00 expð 2 t=Tað1φÞ Þ X 1 XN 1 XZ

i^r ðtÞ 5

(5.122a)

where T 00ð1φÞ 5

X 00 1 X N 1 X Z 00 T X 1 XN 1 XZ o

Tað1φÞ 5

X 00 1 X N 1 X Z ωs ðRs 1 RN 1 RZ Þ

T 0ð1φÞ 5

X0 1 XN 1 XZ 0 T X 1 XN 1 XZ o

(5.122b)

458

Power Systems Modelling and Fault Analysis

and for a single-cage rotor i^r ðtÞ 5

pffiffiffi i 2Vrms h expð 2 t=T 0ð1φÞ Þ 1 expð 2 t=Tað1φÞ Þ 0 N Z X 1X 1X

(5.123a)

where T 0ð1φÞ is as given in Eq. (5.122b) and Tað1φÞ 5

X0 1 XN 1 XZ ωs ðRs 1 RN 1 RZ Þ

(5.123b)

Negative-sequence reactance and resistance As for a synchronous machine, the negative-sequence reactance of the induction motor X N does not vary with time. X N 5 Xv for a double-cage or deep-bar rotor and X N 5 X0 for a single-cage rotor. In addition, the negative-sequence impedance, ZN 5 RN 1 j XN, can be derived from the motor steady-state equivalent circuit shown in Fig. 5.21B for a single-cage rotor and Fig. 5.22B for a double-cage or deep-bar rotor. We recall that under a three-phase positive-sequence excitation, both the stator MMF and the rotor rotate in the same direction. However, under three-phase negative-sequence excitation, the negative-sequence stator currents will produce an MMF that rotates in the opposite direction to the rotor. Using Eq. (5.83a), the motor’s negative-sequence slip can be calculated as follows: Negative sequence slip 5

 Ns 2 ð 2 Nr Þ Ns 2 Nr 522 522s Ns Ns

(5.124a)

that is, in the motor negative-sequence equivalent circuit, s should be replaced by (2 2 s). For large motors, s is typically less than around 1% and even for small motors, s is less than around 4% or 5%. Therefore, if we replace (2 2 s) with 2 in Fig. 5.21B, we can calculate the motor negative-sequence resistance and negativesequence reactance, as follows: 2 2Xm Rr R 5 Rs 1 2 Rr 1 4ðXm 1Xσr Þ2 N

X 5 Xσs 1 N

  Xm R2r 1 4Xσr ðXm 1 Xσr Þ R2r 1 4ðXm 1Xσr Þ2 (5.124b)

As in the case of positive-sequence impedance, the term R2r is negligible and can be neglected. Therefore, Eq. (5.124b) reduces to R N 5 Rs 1

Xm Rr 2ðXm 1 2Xσr Þ

X N 5 Xσs 1

Xm Xσr 5 X0 5 XP Xm 1 Xσr

(5.124c)

Modelling of rotating ac synchronous and induction machines

459

Further approximation in RN can be made if we assume that 2Xσr/Xm , , 1 giving RN 5 Rs 1 ðRr =2Þ

(5.124d)

This produces an error in RN of typically a couple of per cent.

Zero-sequence reactance and resistance The zero-sequence impedance of an induction motor is infinite if the stator winding is either delta-connected, or star-connected with the neutral isolated. For the case above where the neutral of the star-connected winding is inadvertently connected to earth, then, as for a synchronous machine, the zero-sequence reactance X Z is finite, being smaller than the motor’s starting reactance (subtransient or transient as appropriate to the rotor construction), does not vary with time and is similar to the stator winding leakage reactance. The zero-sequence resistance is equal to the stator winding ac resistance.

5.8.3 Modelling the effect of initial motor loading When the motor is loaded, it draws active and reactive power from the supply network. Using the motor transient and subtransient reactances for a double-cage rotor or transient reactance for a single-cage rotor, the motor internal voltages behind 00 subtransient and transient reactances, E0m and Em , are given by

 P 2 jQ 5 Vrms 2 ðRs 1 jX Þ  Vrms

(5.125a)

 P 2 jQ E0m 5 Vrms 2 ðRs 1 jX 0 Þ  Vrms

(5.125b)

h i P 2 jQ E0mðt5T00 Þ 5 Vrms 2 Rs 1 jX 0ðt5T0 0 Þ  Vrms

(5.125c)

00 Em

00

Eqs. (5.125a)
(5.125c) are illustrated in Fig. 5.25A and the corresponding equivalent circuits during the short-circuit fault are illustrated in Fig. 5.25B. The peak short-circuit current changes in the case of a three-phase fault given in Eqs. (5.117a) and (5.118a) are modified as follows: i^r ðtÞ 5




00   00 pffiffiffi E0m 2t=T 0 Em E0m 2t=T 00 Em 2t=Ta e 2 0e 1 2 0 e 1 X X 00 X X}

(5.126a)

i^r ðtÞ 5

 pffiffiffi E0m  2t=T 0 2 0 e 1 e2t=Ta X

(5.126b)

460

Power Systems Modelling and Fault Analysis

(A)

IL =

(i)

P − jQ V ∗rms

(ii)

jX ′′

Equation (5.119d)

V rms

–

(B)

E ′′m

–

(i)

+

E ′m

–

I′′

P

jXP

jX

P

jX′

Equation (5.119d)

Rs + –

E′′m

E ′m

(iii)

(ii) jX′′

V rms

Rs

V rms

Rs +

IL jX ′

j XP

Rs +

(iii)

IL

Rs + –

E′m

I′

P

Rs + –

E′m

Figure 5.25 (A) Representation of initial motor loading just before the short-circuit fault: (i) motor internal voltage behind subtransient reactance, (ii) motor internal voltage behind transient reactance at t 5 5 Tv and (iii) motor internal voltage behind transient reactance at t 5 0; (B) representation of motor equivalent circuits during the short-circuit fault for (A) (i), (ii) and (iii) cases above.

From Eqs. (5.125a)
(5.125c) we observe that the motor internal voltage is equal to the terminal voltage less the voltage drop across the motor subtransient or transient impedance. For a 1 pu motor terminal voltage, typical values of motor subtransient and transient reactances, and various motor loading, the internal voltage magnitude can typically vary between 0.85 and 0.95 pu. Therefore, ignoring the initial motor loading and using 1 pu instead of the actual internal voltages represents a conservative assumption. This assumption is made by the American IEEE C37.010 Standard and UK ER G7/4 Guideline. This will be discussed in Chapter 8, International standards for short-circuit analysis in ac power systems.

5.8.4 Determination of a motor’s electrical parameters from tests Steady-state equivalent circuit: stator resistance test The stator winding dc resistance of a three-phase induction motor is measured by connecting a dc voltage source across two stator terminals. No induced rotor voltage can occur under dc stator excitation and hence the resistance ‘seen’ by the dc voltage source is the stator dc resistance. By applying a known dc voltage source and measuring the dc cource current, the stator dc resistance can be calculated. The calculation depends on whether the stator winding is star- or delta-connected as

Modelling of rotating ac synchronous and induction machines

(A)

(B)

Idc RS(dc)

Vdc RS(dc)

461

Idc

Vdc

RS(dc)

RS(dc)

RS(dc) RS(dc)

Figure 5.26 Measurement of induction motor stator dc resistance: (A) star-connected stator winding with isolated neutral and (B) delta-connected stator winding. (A)

Is

jX σs

Rs

+

~I

jX σr

(B)

I ≈0 R

Vs

Rm

–

jX m

Is +

Rs

jX σs

jX σ r R m

~I Rr

Vs Rm

jX m

s

–

Figure 5.27 Measurement of internal electrical parameters of an induction motor: (A) locked rotor test and (B) no-load test.

shown in Fig. 5.26A and B, where the star is assumed to have an isolated neutral. In physical units, the stator dc resistance Rs(dc) is given by

RsðdcÞ 5

8 1 Vdc > > >

> > : 2 Idc

for a delta
connected stator winding

dc

Ω=phase

(5.127)

If the test is repeated for the other two sets of terminal connections, then an average is taken. To obtain the ac resistance, Rs(dc) has to be corrected for skin effect by multiplying it by the skin effect factor, which is typically between 1.05 and 1.25.

Steady-state equivalent circuit: locked rotor test This test circuit is shown in Fig. 5.27A. The stator is connected to a balanced three-phase ac voltage source and the rotor shaft is locked mechanically to prevent it from turning. Since the slip s 5 1, the rotor impedance is several times smaller than that under rated running condition (because operating slip is typically between 0.001 and 0.05pu) and is typically 30
50 times smaller than the magnetising reactance. Thus, the locked rotor test is broadly analogous to a transformer short-circuit test. The test voltage applied should initially be very small and then gradually increase until rated or full load stator current is drawn. To avoid overheating, the applied voltage is significantly smaller than the rated voltage so that the motor current is limited to the rated value.

462

Power Systems Modelling and Fault Analysis

If rated voltage is applied, the locked rotor current drawn is very large and is typically four to seven times the rated current value. The combination of a low-test voltage and a low rotor impedance results in a very small exciting current and hence the iron loss and magnetising branches can be neglected with an insignificant loss of accuracy. As shown in Fig. 5.27A, the equivalent motor impedance ‘seen’ from the test terminals is given by ZEq 5 REq 1 jXEq  Rs 1 Rr 1 jðXσs 1 Xσr Þ

(5.128)

The measured stator quantities are line-to-line voltage VLL, stator current Is, three-phase input power P3φ and three-phase input reactive power Q3φ. For either a star- or a delta-connected stator winding, the equivalent resistance and reactance of Eq. (5.128) are calculated as follows: REq 5 Rs 1 Rr 5

P3φ Ω=phase 3Is2

XEq 5 Xσs 1 Xσr 5

(5.129a)

Q3φ Ω=phase 3Is2

(5.129b)

Alternatively, using the line-to-line stator voltage instead of the reactive power 8 VLL > > > pffiffiffi > < 3I s ZEq 5 pffiffiffi 3VLL > > > > : Is

for a star
connected stator winding Ω=phase

(5.130a)

for a delta
connected stator winding

and the equivalent reactance is calculated as follows: XEq 5 Xσs 1 Xσr 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 R2 ZEq Eq

Ω=phase

(5.130b)

Using the measured and corrected stator resistance, the rotor resistance can be calculated from Eq. (5.129a) as Rr 5 REq 2 Rs

Ω=phase

(5.130c)

In Eq. (5.130b), the sum of the stator and rotor leakage reactances is calculated. In the absence of manufacturer design data that provide the ratio of these reactances, the division shown in Table 5.1 for different types of motors as classified by the National Electrical Manufacturers Association (NEMA) may be used. If the motor design class is not known, then Xσs and Xσr may be assumed to be equal. The locked rotor test may also be carried out at reduced frequency to account for skin depth and a more accurate rotor resistance under rated operating conditions

Modelling of rotating ac synchronous and induction machines

463

Table 5.1 Estimation of induction motor stator and rotor leakage reactances using NEMA motor class in the absence of design data NEMA motor design class

Xσs in % of XEq

Xσr in % of XEq

Wound rotor Class A Class B Class C Class D

50 50 40 30 50

50 50 60 70 50

where the rotor current frequency may be between 0.5 and 2.5 Hz. IEEE Standard 112 recommends a frequency of one-quarter of the rated power frequency. In general, this method tends to provide better results for larger motors above 20 hp but for small motors, the results obtained from the power frequency test are generally satisfactory.

Steady-state equivalent circuit: no-load test This test circuit is shown in Fig. 5.27B. The stator winding is connected to a threephase source supplying rated voltage and frequency and the motor is run without shaft load. As a result, the rotor will rotate at almost synchronous speed, that is, slip speed s  0. Thus, Rr/s!N and the motor equivalent circuit is broadly similar to that of a transformer on open-circuit. Therefore, the rotor impedance branch is effectively open-circuit and the current drawn by the stator is fully attributed to the exciting current. The stator quantities measured are; line-to-line voltage VLL, current Is, three-phase input power P3φ and three-phase input reactive power Q3φ. The iron loss resistance associated with the magnetising impedance can be calculated, if required, by connecting this resistance in parallel with the magnetising reactance. Using the inequality Rm/Xm . . 1, the motor equivalent impedance as seen from the test terminals is approximately given by Z e 5 Rs 1

2 Xm 1 jðXσs 1 Xm Þ Rm

(5.131a)

The measured input power includes rotational losses (friction and windage, iron and stray load losses) as well as stator copper losses and is given by P3φ 5 3Is2 Rs 1 3Is2

2 Xm 1 Lossesðfriction=windage1strayÞ Rm

(5.131b)

Using the measured input reactive power, we have Xσs 1 Xm 5

Q3φ 3Is2

(5.131c)

464

Power Systems Modelling and Fault Analysis

Using the measured input reactive power and stator current, and the calculated value of Xσs, Xm is calculated using Eq. (5.131c). Also, Rm is calculated from Eq. (5.131b) using the measured input power and stator current, as well as the calculated quantities Rs and Xm. If the friction/windage and stray losses are not known, then they may be neglected with little error. If required, the very small stator copper losses in Eq. (5.131b) may also be neglected and the calculation of Rm and Xm proceeds similarly.

Transient parameters from a sudden three-phase short-circuit immediately after motor disconnection This method consists of two steps; the first is to disconnect the motor from the supply network and the second is to quickly apply a simultaneous three-phase short-circuit fault at the motor terminals. As for synchronous machines, this is usually a suitable test method for the determination of the motor subtransient and transient parameters. Immediately upon supply disconnection, the flux linking the closed rotor inductive circuits prevents the rotor current from changing instantaneously to zero. This flux will decay in a manner determined by the rotor open-circuit subtransient and transient time constants To00 and T 0o , respectively. It can be shown that the phase r open-circuit motor terminal voltage following motor disconnection, where the motor has a double squirrel-cage or deep-bar rotor, is given by υr ðtÞ 5

pffiffiffi

 0 00 2Irms ½ ðXσs 1 Xm Þ 2 X 0 e2t=T o 1 ðX 0 2 X 00 Þe2t=To cosðωs t 1 δÞ (5.132a)

where Irms is the initial stator current before motor disconnection and δ is its phase angle. The three-phase short circuit is applied a few cycles after the removal of the motor supply. The range of values of Tvo, even for larger induction motors, tends to be less than about 30 ms so that the voltage component associated with Tvo will be negligible by the time the short circuit is applied at time ts as illustrated in Fig. 5.28. The only relevant component is therefore the one associated with T0 o.

1

Peak open-circuit voltage (pu)

⁄

V′

0.75

)

0.5 0.25 0 –500 0

Time (ms)

0

ts

ts

500

1500

2500

3500

Figure 5.28 Upper envelope of open-circuit voltage of an induction motor upon supply disconnection.

Modelling of rotating ac synchronous and induction machines

465

It should be noted that this is the only component applicable in the case of a rotor with a single rotor winding, for example, a single squirrel-cage rotor or a wound rotor. Therefore υr ðtÞ 5

pffiffiffi 0 2Irms ½ðXσs 1 Xm Þ 2 X 0 e2t=T o cosðωs t 1 δÞ

t $ 5To00

(5.132b)

Fig. 5.28 illustrates the peak envelope of the motor phase r open-circuit voltage after supply disconnection. At the instant of short-circuit t 5 ts, following the decay of the subtransient voltage component, the rms value of the motor opencircuit voltage is given by 0

0

Voc 5 Irms ½ðXσs 1 Xm Þ 2 X 0 e2ts =T o 5 V 0 e2ts =T o

(5.133)

This voltage can be measured from the recorded voltage envelope. Since the motor is on open circuit, the equations that describe the motor short-circuit currents will be the same as those derived in Eqs. (5.117a) and (5.118a) or (5.122a) and (5.123a) but with Vrms replaced with the magnitude of Voc of Eq. (5.133). The envelope of the current can be used to determine the motor subtransient and transient reactances and time constants as described for synchronous machines.

5.8.5 Examples Example 5.5 A 50-Hz three-phase induction motor of a single squirrel-cage rotor construction drives a condensate pump in a power generating station and has the following data: rated phaseto-phase voltage 5 3.3 kV, rated three-phase apparent power 5 900 kVA, Rs 5 0.0968 Ω, Xσs 5 1.331 Ω, Xm 5 38.7 Ω, Rr 5 0.0726 Ω and Xσr 5 0.847 Ω at rated slip. The rotor parameters are referred to the stator voltage base. Calculate the motor parameters in pu on kVA rating, the motor transient reactance, open circuit, short circuit and armature time constants. Base voltage and base apparent power are 3.3 kV and 0.9 MVA, respectively. Base impedance is equal to (3.3)2/0.9 5 12.1 Ω. Therefore, Rs 5 0.0968/12.1 5 0.008 pu, Xσs 5 1.331/12.1 5 0.11 pu, Xm 5 38.7/12.1 5 3.2 pu, Rr 5 0.0726/12.1 5 0.006 pu, Xσr 5 0.847/12.1 5 0.07 pu. From Eq. (5.108) X 0 5 0:11 1

3:2 3 0:07 5 0:1785 pu 3:2 1 0:07

and Ta 5

0:1785 3 1000 5 71 ms 2π50 3 0:008

466

Power Systems Modelling and Fault Analysis

From Eqs. (5.93a), (5.93b) and (5.95b) T 0o 5

3:2 1 0:07 5 1:73 s 2π50 3 0:006

T0 5

0:07 1 ð3:2 3 0:11Þ=ð3:2 1 0:11Þ 3 1000 5 93:5 ms 2π50 3 0:006

and

In many situations, the internal motor parameters may not be known but T0 and X0 may be known from measurements. In this case, the following empirical formula for T0 that provides sufficient accuracy may be used T0 5 X0 /ωsRr. This is equivalent to replacing the numerator of Eq. (5.93b) by X0 . The numerator of Eq. (5.93b) is equal to 0.1763 pu and hence the use of T0 5 X0 /ωsRr to calculate the transient time constant gives a slightly higher and thus conservative time constant by about 1.2%.

Example 5.6 In the system shown in Fig. 5.29, the two induction motors are identical and have the data given in Example 5.5. Calculate the peak short-circuit current contribution of the motors at t 5 0, 10 and 100 ms for a three-phase short-circuit fault at (a) the 3.3-kV terminals of the motors, and (b) the 15-kV medium voltage busbar. Assume a 1 pu initial voltage at the 3.3- and 15-kV busbars and that the motors are initially unloaded. Consider only the change in motor current due to the fault. a. Three-phase short-circuit fault at the motor 3.3-kV busbar terminals Using Eq. (5.118a), the rms and dc short-circuit currents from each motor are equal to irms ðtÞ 5

1 e2t=93:5 0:1785

132kV

15kV Three-phase short-circuit faults

2 × 3MVA X = 10% X/R = 10 3.3kV

M M 2 × 0.9MVA Induction motors

Figure 5.29 Simple network for the calculation of induction motor short-circuit current contribution.

Modelling of rotating ac synchronous and induction machines

467

and idc ðtÞ 5

pffiffiffi 2 3 1 2t=71 e 0:1785

The rms current at t 5 0, 10 and 100 ms is equal to 5.6, 5.03 and 1.92 pu, respectively. The dc current at t 5 0, 10 and 100 ms is equal to 7.92, 5.88 and 1.94 pu, respectively. The peak current contribution at t 5 0, 10 and 100 ms is equal to 15.84, 13.76 and 3.88 pu respectively. The total current contribution from both motors at t 5 0, 10 and 100 ms is equal to 31.68, 27.52 and 7.76 pu, respectively. b. Three-phase short-circuit fault at the 15-kV medium-voltage busbar The equivalent transient reactance of the two motors is 0.1785/2 5 0.08925 pu. The external impedance is equal to the transformer’s impedance and is calculated on 0.9 MVA base as Ze 5 ð0:01 1 j0:1Þ 3

0:9 5 ð0:003 1 j0:03Þpu 3

The time constants of the equivalent motor representing the two parallel motors do not change but the stator resistance and transient reactance are halved so that Rs 5 0.004 pu and X0 5 0.08925 pu. The external impedance modifies the motor’s short-circuit time constant according to Eq. (5.121e), thus T 0e 5

0:08925 1 0:03 3 1:75 5 122:4 ms 1:655 1 0:03

where X 5 (3.2 1 0.11)/2 5 1.655 pu. The armature time constant is modified using Eq. (5.121g) to Tae 5

0:08925 1 0:03 5 54:2 ms 2π50 3 ð0:004 1 0:003Þ

Using Eq. (5.121f), the rms and dc short-circuit currents are given by irms ðtÞ 5

1 e2t=122:4 0:08925 1 0:03

and idc ðtÞ 5

pffiffiffi 2 e2t=54:2 0:08925 1 0:03

The rms current at t 5 0, 10 and 100 ms is equal to 8.386, 7.73 and 3.7 pu, respectively. The dc current at t 5 0, 10 and 100 ms is equal to 11.86, 9.86 and 1.87 pu, respectively. The peak current contribution at t 5 0, 10 and 100 ms is equal to 23.7, 19.72 and 3.74 pu, respectively.

468

Power Systems Modelling and Fault Analysis

Further reading Books Adkins, B., et al., The General Theory of Alternating Current Machines, Chapman and Hall, 19750-412-15560-5. Concordia, C., Synchronous Machines
Theory and Performance, John Wiley, New York, 1951. Krause, P.C., Analysis of Electric Machinery, McGraw-Hill, New York, 1986. Kundur, P., Power System Stability and Control, McGraw-Hill, Inc, 19940-07-035958-X.

Papers Canay, I.M., Determination of model parameters of synchronous machines, Proceedings IEE, Vol. 130, No. 2, March 1983, 86
94. Cooper, C.B., et al., Application of test results to the calculation of short-circuit levels in large industrial systems with concentrated induction-motor loads, Proceedings IEE, Vol. 116, No. 11, November 1969, 1900
1905. Huening, W.C., Calculating short-circuit currents with contributions from induction motors, IEEE Transactions on Industry Applications, Vol. IA-18, No. 2, March/April 1982, 85
92. Hwang, H.H., Mathematical analysis of double line to-ground short circuit of an alternator, IEEE Transactions on Power Apparatus and Systems, Vol. PAS-86, No. 10, October 1967, 1254
1257. Hwang, H.H., Transient Analysis of unbalanced short circuits of synchronous machines, IEEE Transactions on Power Apparatus and Systems, Vol. PAS-88, No. 1, January 1969, 67
71. Kai, T., et al., A simplified fault currents analysis method considering transient of synchronous machine, IEEE Transactions on Energy Conversion, Vol. 12, No. 3, September 1997, 225
231. Kalsi, S.S., et al., Calculation of system-fault currents due to induction motors, Proceedings IEE, Vol. 118, No. 1, January 1971, 201
215. Kamwa, I., et al., Experience with computer-aided graphical analysis of sudden-short-circuit oscillograms of large synchronous machines, IEEE Transactions on Energy Conversion, Vol. 10, No. 3, September 1995, 407
414. Kamwa, I., et al., Phenomenological models synchronous machines from short-circuit test during commissioning
a classical/modern approach, IEEE Transactions on Energy Conversion, Vol. 9, No. 1, March 1994, 85
97. Luke, Y.Y., et al., Motor contribution during three-phase short circuits, IEEE Transactions On Industry Applications, Vol. IA-18, No. 6, November/December 1982, 593
599.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators Chapter Outline 6.1 General 471 6.2 Three-phase voltage-source converters

472

6.2.1 Basic operation of voltage-source converters 472 6.2.2 Basic control of voltage-source converters 477 6.2.3 Current capability of converter insulated gate bipolar transistor (IGBT) switches 477

6.3 Types of wind turbine generator technologies 477 6.4 Type 1 fixed-speed wind turbine induction generators 478 6.4.1 Modelling and analysis of short-circuit current contribution 478 6.4.2 Example 479

6.5 Type 2 variable slip wind turbine wound rotor induction generators 480 6.5.1 Modelling and analysis of short-circuit current contribution 480 6.5.2 Example 483

6.6 Type 3 variable-speed wind turbine doubly fed induction generators (DFIG) 484 6.6.1 6.6.2 6.6.3 6.6.4 6.6.5 6.6.6 6.6.7 6.6.8 6.6.9 6.6.10 6.6.11 6.6.12

Background 484 Basic operation principle 485 Rotor protection 486 Passive and active crowbar protection 487 dc chopper protection 489 General model of doubly fed induction generators (DFIG) 490 DFIG steady-state equivalent circuit 490 DFIG natural stator and rotor short-circuit currents under constant ac excitation 491 DFIG stator and rotor short-circuit currents under crowbar action 497 DFIG short-circuit currents with dc chopper control 505 DFIG short-circuit currents with rotor converter control 506 Examples 507

6.7 Type 4 variable-speed inverter-interfaced wind turbine generators 511 6.8 Type 5 variable-speed wind turbine synchronous generators 6.9 Solar photovoltaic (PV) generators and connection to the ac grid 512 6.9.1 Solar PV applications 512

Power Systems Modelling and Fault Analysis. DOI: https://doi.org/10.1016/B978-0-12-815117-4.00006-0 Copyright © 2019 Dr. Abdul Nasser Dib Tleis. Published by Elsevier Ltd. All rights reserved.

511

6

470

Power Systems Modelling and Fault Analysis

6.9.2 6.9.3 6.9.4 6.9.5

Solar PV generator components 512 Solar PV voltage-source inverters 513 Grid connection of solar PV power plant 514 Short-circuit current contribution of solar PV inverters 516

6.10 Technologies interfaced to the ac grid through voltage-source inverters 516 6.11 Modelling and analysis of grid-connected voltage-source inverters 517 6.11.1 6.11.2 6.11.3 6.11.4 6.11.5

General inverter model 517 Phase-locked loop 520 Inverter inner current control loop 521 Inverter outer control loops 525 Short-circuit current contribution of voltage-source inverters with frozen controls 527

6.12 Grid code requirements for dynamic reactive current injection from inverters 530 6.13 Short-circuit current contribution of voltage-source inverters during balanced three-phase voltage dips 533 6.13.1 Dynamic reactive current control 533 6.13.2 Examples of inverter positive-sequence short-circuit current contribution 534

6.14 Short-circuit current contribution of voltage-source inverters during unbalanced voltage dips 537 6.14.1 Inverter model in positive- and negative-sequence synchronous reference frames 537 6.14.2 Examples of inverter positive- and negative-sequence short-circuit current contributions

539

6.15 Sequence network representation of voltage-source inverters during balanced and unbalanced short-circuit faults 542 6.16 Short-circuit currents in ac grids containing mixed synchronous generators and inverters 543 6.16.1 6.16.2 6.16.3 6.16.4 6.16.5 6.16.6

General 543 Steady-state and transient characteristics of inverter short-circuit reactive currents 544 Analysis of three-phase short-circuit fault currents 545 Analysis of single-phase short-circuit fault currents 548 Analysis of two-phase short-circuit fault currents 554 Examples 559

6.17 Grid-forming voltage-source inverters

568

6.17.1 Background to grid-following and grid-forming inverters 568 6.17.2 Emerging challenges in power systems dominated by grid-following voltage-source inverters 568 6.17.3 Control structure of grid-forming inverters 569

6.18 Grid-forming virtual synchronous machine (VSM) inverters

570

6.18.1 Possible VSM inverter features of real synchronous machines 570 6.18.2 A model of grid-forming VSM inverters 571

6.19 Grid-forming voltage-source inverters using droop control 575 6.20 Natural short-circuit current contribution of grid-forming inverters 6.20.1 Natural three-phase short-circuit current of grid-forming inverters 579 6.20.2 Natural two-phase short-circuit current of grid-forming inverters 580

579

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

6.21 Over-current limitation strategies of grid-forming inverters

471

581

6.21.1 General 581 6.21.2 Current limitation in inverters using single voltage controller structure 582 6.21.3 Current limitation strategies in inverters using cascaded control structure 586

6.22 Symmetrical components sequence equivalent circuits of ‘grid-forming’ inverters 589 6.23 Examples 591 Further reading 595

6.1

General

Until the end of the 20th century, almost all of the world’s electric power supply was produced by synchronous generators in thermal power and hydropower generation plant. With mounting concerns over climate change and the need for reduced green house gas such as CO2 emissions, renewable energy, particularly wind and solar photovoltaic (PV) generators, have now become a mainstream source of electric energy supply. Wind turbine generators utilising a range of nonsynchronous generators are used as a major new energy source in many countries around the world that have good wind resources. Medium- to large-scale onshore and offshore wind parks consist of tens to hundreds of wind turbine generators. At the time of writing, individual turbines have reached a rating of 4.8 MW onshore and 8 MW offshore. Wind turbine sizes of 10
12 MW offshore are nearing full-scale deployment. The most installed wind turbine generator technologies are the doubly fed and full-converter generators and the latter is likely to dominate most future installations. The largest wind turbine generators offshore are direct drive, that is, without a gearbox, employing permanent magnet synchronous generators which are interfaced with the ac grid through a full-size ac/dc/ac converter. Solar PV generators utilising full-size dc to ac voltage-source inverters are also used as a major new energy source in many countries around the world that have a good or even average solar irradiance resource. Solar PV generators range from the very small with a few kW in size installed on domestic rooftops to very large, utility-scale, ground-mounted, solar parks of hundreds of MW in size. An individual solar PV generator interfaced to the ac grid through a dc/ac voltage-source inverter can be as small as 1 or 2 kVA single-phase or as large as 4.4 MVA three-phase in a large solar park containing tens to hundreds of such individual generators. High-voltage direct-current (HVDC) transmission systems employing voltagesource converter technology represent a system that is interfaced to the ac grid through a dc/ac voltage-source converter. Typical uses are as a transmission link connecting a remote, large, offshore wind farm to the onshore ac grid, a long submarine cable HVDC link and an interconnection between two asynchronous ac systems.

472

Power Systems Modelling and Fault Analysis

Battery energy storage systems (BESS) are interfaced to the ac grid through a dc/ac voltage-source inverter technology. They are used to provide energy storage or ancillary services to the ac grid such as fast frequency response, reserves, reactive power and voltage control, and other ancillary services. Currently, installations in the few MW to a few tens of MW capacity are emerging but with further advances in battery technologies and reduction in installed costs, large BESS in the hundreds of MW capacity may become cost-effective and an indispensable solution to the integration of very large amounts of variable and nondispatchable wind and solar PV generation systems. Voltage-source inverters that use alternative control strategies to become gridforming inverters are an exciting alternative in grid-connected applications to conventional current-controlled grid-following converters. The latter follow the already available grid voltage and control their current output. However, grid-forming inverters can form their voltage and act as a true voltage source. Grid-forming inverters can either be implemented to replicate certain aspects of the behaviour of real synchronous machines known as virtual synchronous machines (VSM or VISMA) or be droop controlled. The latter have been used in small isolated grids, microgrids, and special applications such as marine and aeronautical power systems but connections to large-scale grids are also possible. Grid-forming inverters are expected to be adopted by inverter manufacturers in large-scale grid-connected applications.

6.2

Three-phase voltage-source converters

6.2.1 Basic operation of voltage-source converters There are many types and configurations of power electronics converters in use. However, within the scope of this book, we will briefly introduce some of the basic characteristics of three-phase voltage-source converters used in wind turbine generators, solar PV generators, battery energy storage, HVDC transmission and other applications. A three-phase voltage-source converter can be viewed as a three-phase voltage source whose magnitude, frequency and phase can be controlled simultaneously. A dc to ac converter is called an inverter and an ac to dc converter is called a rectifier. Fig. 6.1 shows a three-phase two-level six-switch voltage-source converter using insulated gate bipolar transistor (IGBT) switches predominantly used in low-voltage systems. Three-level converters and the more recent modular multilevel converters typically used in higher voltage systems are outside the scope of this book but the operation principle is similar. In this topology, a relatively large capacitor feeds the three-phase bridge inverter circuit with a dc voltage, hence the term voltage-source inverter. The bridge consists of six switches; each is a power transistor IGBT with a free-wheeling diode to provide bidirectional current flow and a unidirectional voltage-blocking capability. With proper control algorithms, this enables the converter to operate either as a

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

473

+C 2

–

,

( ) ( )

+ C 2 –

( )

Figure 6.1 Six-switch two-level voltage-source converter.

rectifier or an inverter without any change in topology. The dc capacitor is charged to a certain voltage level which ensures the basic function of the converter which is to control the ac current through the switching strategy. The LC, or in many installations LCL filter on the ac side, is used as a low-pass filter as well as to smooth the ripples in the ac current and dc voltage waveforms. The two-level refers to the number of voltage levels that can be produced at the output of each leg of the bridge. The converter switches are switched on and off with a fixed frequency but with a pulse width that is varied in order to control the output voltage. Fig. 6.2A shows one leg or one phase, phase r, of the converter with two IGBT switches S1 and S2, each with its antiparallel diode. When switch S1 is closed, the ac output voltage ero with respect to the zero-voltage node is Vdc/2 but when switch S2 is on, ero is 2 Vdc/2. The output voltage ero can be controlled between 2 Vdc/2 and Vdc/2. In a three-level inverter, the inverter is controlled so that in the positive half cycle of reference waveform, the output voltage is controlled between Vdc/2 and zero and in the negative half cycle between 2 Vdc/2 and zero, hence the three levels 2 Vdc/2, zero and Vdc/2. In the positive half cycle, switch S1 would be on and off but switch S2 would be off. The converse applies in the negative half cycle. The ac output voltage can be changed by modulating the width of the Vdc pulse. Fig. 6.2B illustrates a sinusoidal pulse-width modulated waveform and shows how the width of the pulse can be adjusted by comparing a modulating or reference waveform with a carrier waveform of a triangular shape. The reference waveform is the desired waveform at the output and the triangular waveform determines the frequency with which the IGBT switches are switched. When the reference waveform is higher than the triangular waveform, a switching pulse is generated and switch S1 is turned on. The carrier frequency should always be an odd multiple of 3 as this provides halfand quarter-wave symmetry which eliminates even harmonics and generates symmetrical three-phase voltages. The frequency modulation index fm is defined as fm 5

fsw fo

(6.1)

where fsw is the switching frequency in Hz and fo is the fundamental power frequency in Hz.

474

Power Systems Modelling and Fault Analysis

(A)

+

2

– 0

+

2

(B)

–

Reference waveform

Carrier waveform

0

2 0

− 2

ΔT1, on

ΔT2, on

Fundamental er0

Figure 6.2 (A) One leg of six-switch two-level voltage-source inverter and (B) generation of an ac waveform using carrier-based pulse width modulation.

The duty cycle or amplitude modulation index is given by mo 5

E^ out Vdc =2

(6.2a)

0 , mo # 1

(6.2b)

where E^ out is the peak of the fundamental component of the ac output voltage ero . Eq. (6.2a) can also be expressed as mo 5 ΔT1;on 3 fsw 5 Tsw 5

1 fsw

ΔT1;on Tsw

(6.3a) (6.3b)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

475

and ΔT1;on is the on-time of the switch S1 and Tsw is the period of the carrier waveform. mo lies between 0 (with S2 on continuously) and 1 (with S1 on continuously). However, mo can be varied in time and therefore any desired voltage and frequency can be generated for a given switching frequency and dc voltage. In Fig. 6.2B, the following parameters have been used: fo 5 50Hz; mo 5 0:9; fm 5 15 or fsw 5 750 Hz. A low value for fsw has been deliberately chosen to improve the clarity of the illustration although typical values in practice are much higher (a few kHz). Sinusoidal pulse-width modulation of two-level converters generates odd harmonics at the switching frequency fsw and its multiples as well as the sidebands around them. The high-frequency harmonics are filtered using passive tuned filters. An alternative to modelling every switching state of the converter is to use an average converter model which assumes that mo is constant during one switching cycle as implied in Eq. (6.2a). This is generally sufficiently accurate because the carrier signal switching frequency is at least 15
40 times higher than the frequency of the modulating signal. Therefore in the three-phase converter as shown in Fig. 6.1, there are three modulation indices mr ðtÞ, my ðtÞ and mb ðtÞ; and if these are designed to vary sinusoidally, the converter average output voltages with respect to the zero-voltage node of the dc capacitors, ignoring high-order harmonics, are given by er0 ðtÞ 5 mr ðtÞ

Vdc 2

(6.4a)

ey0 ðtÞ 5 my ðtÞ

Vdc 2

(6.4b)

eb0 ðtÞ 5 mb ðtÞ

Vdc 2

(6.4c)

where mr ðtÞ 5 mo sinðωo t 1 ϕÞ
 2π my ðtÞ 5 mo sin ωo t 1 ϕ 2 3
 2π mb ðtÞ 5 mo sin ωo t 1 ϕ 1 3

(6.5a) (6.5b) (6.5c)

and ωo 5 2πfo , fo is the power frequency. Therefore Eqs. (6.4a)
(6.4c) can be written as Vdc sinðωo t 1 ϕÞ 2
 Vdc 2π sin ωo t 1 ϕ 2 ey0 ðtÞ 5 mo 3 2 er0 ðtÞ 5 mo

(6.6a) (6.6b)

476

Power Systems Modelling and Fault Analysis


 Vdc 2π sin ωo t 1 ϕ 1 eb0 ðtÞ 5 mo 3 2

(6.6c)

Also, the converter output line-to-line voltages are given by ery 5 er0 2 ey0 5

pffiffiffi Vdc  π sin ωo t 1 ϕ 1 3mo 6 2

pffiffiffi Vdc  π sin ωo t 1 ϕ 2 3m o 2 2
 pffiffiffi Vdc 5π sin ωo t 1 ϕ 1 ebr 5 eb0 2 er0 5 3mo 6 2 eyb 5 ey0 2 eb0 5

(6.7a) (6.7b) (6.7c)

and the converter output rms line or phase-to-phase voltage is related to the dc voltage by pffiffiffi 3 Ery rms 5 Eyb rms 5 Ebrrms 5 pffiffiffi mo Vdc 2 2

(6.8)

Eq. (6.8) represents balanced three-phase output line voltages whose amplitude is controlled by mo and output frequency and phase are controlled by the frequency of the modulating or reference waveform. Fig. 6.3 shows the three-phase pulse-width modulated output waveforms of the three-phase two-level converter.

er0 ey0 eb0 ery eyb ebr Figure 6.3 Three-phase line output voltages of six-switch two-level voltage-source inverter.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

477

6.2.2 Basic control of voltage-source converters The currently dominant technology of grid-connected voltage-source converters operates as controlled current sources. This means that the pulse width modulation (PWM) switching of the converter is controlled so that its output current is forced to follow a reference value in its inner current control loop. Even during large changes in grid voltage, the converter output current continues to follow the reference values for both positive- and negative-sequence currents because, as ‘seen’ from the grid, the converter appears to have a very high impedance. In this method of control, the current-controlled voltage-source converter controls the output current indirectly. The converter uses the grid voltage feedback and the voltage drop across the output filter inductance to synthesise the converter voltage feedback that will produce the required converter output currents. This is presented in detail in Section 6.11.

6.2.3 Current capability of converter insulated gate bipolar transistor (IGBT) switches The transient output current of the converter that can be sustained for a short period of a few seconds is generally limited to no more than 1.1 pu of rated continuous current. This protects the IGBT switches from damage, avoid oversizing the converter and help to select an economic converter rating that is consistent with the rated active power of the feeding energy source, be it a solar field, a wind turbine generator or a HVDC transmission link, etc. Typically, therefore, the IGBT switches can carry up to 1.55 pu instantaneous current and up to 2 pu current for a maximum of 1 ms. Internal protection in the power stack IGBT unit samples voltages and currents at typically 1 μs and can block the IGBT current, that is, reduce it to zero in a few microseconds. External voltage dips cause an outrush current from the filter capacitor to flow out of the converter terminals. This is a high-frequency, oscillatory transient current with an initial peak of typically 2.5
5 pu and a frequency of several kHz. This damped current decays completely to zero within a few milliseconds. Obviously, this physical current does not flow through the converter IGBTs and hence cannot be controlled by the converter.

6.3

Types of wind turbine generator technologies

Several types of wind turbine generators are used in onshore and offshore wind parks. Wind turbine generators that are directly connected to the ac grid employ induction, rather than synchronous, generators. Synchronous generators are also widely used but, currently, these are not directly connected to the ac network. Rather, they are connected through a back-to-back power electronics voltage-source converter. Directly connected synchronous generators may be used in future if the development of variable-speed gearbox technology proves sufficiently reliable and

478

Power Systems Modelling and Fault Analysis

commercially attractive. The variable-speed gearbox has a variable input speed on the turbine side but produces constant output synchronous speed. At the time of writing, their large-scale commercial use is yet to materialise. Wind turbine generator technologies have been classified for convenience into five types based on their specific technology. These are discussed in the following sections.

6.4

Type 1 fixed-speed wind turbine induction generators

6.4.1 Modelling and analysis of short-circuit current contribution Type 1 fixed-speed induction generators are essentially similar to squirrel-cage induction motors. These are driven by a wind turbine prime mover at a speed just above synchronous speed, normally up to 1% rated slip for today’s large wind turbines. Because the speed variation from no load to full load is very small, the term ‘fixed’ speed is widely used. The generator is coupled to the wind turbine rotor via a gearbox as shown in Fig. 6.4. The design and construction of the stator and rotor of an induction generator are similar to those of a large induction motor with a squirrel-cage rotor. In Chapter 5, Modelling of rotating ac synchronous and induction machines, we presented the modelling of induction motors and analysed their short-circuit current contribution. The positive- and negative-sequence short-circuit contribution of a fixed-speed induction generator can be represented in a similar way to that of an induction motor. The equations of transient reactances and time constants derived for induction motors can also be used for induction generators. The stator windings of these generators are usually connected in delta or star with an isolated neutral. Thus their zero-sequence impedance to the flow of zero-sequence currents is infinite.

Wind turbine

Squirrel-cage Gearbox induction generator

Generator- MV transformer Grid

Power factor correction capacitors

Figure 6.4 Type 1 fixed-speed wind turbine induction generator.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

479

Using Eq. (5.112b), to represent a single-cage fixed-speed induction generator, the instantaneous short-circuit currents delivered by the machine for a three-phase solid fault at its terminals are given by ii ðtÞ 5

pffiffiffi 2Vrms




  1 1 2t=T 0 1 2t=Ta 2 sin ð ω t 1 θ Þ 2 e sinθ e o i i X0 X X0

(6.9)

where θr 5 θo

i 5 r; y; b

θy 5 θo 2 2π=3

θb 5 θo 1 2π=3

and θo is the instant of fault occurrence on machine stator phase r voltage waveform and the machine parameters are as given in Chapter 5, Modelling of rotating ac synchronous and induction machines.

6.4.2 Example Example 6.1 Consider a type 1 fixed-speed squirrel-cage wind turbine induction generator having the following parameters:

Machine voltages (pu)

Rated voltage 0.69 kV Stator resistance 0.0014 Ω Stator leakage reactance 0.018 Ω

Apparent power 2.3 MVA, 50 Hz Rotor resistance (referred to stator) 0.0023 Ω Rotor leakage reactance (referred to stator) 0.014 Ω, magnetising reactance 0.7 Ω

1.5

Phase R Phase Y Phase B

1 0.5

0 –40 –0.5

0

40

80

120

160

200 Time (ms)

40

80

120

160

200 Time (ms)

–1

Machine currents (pu)

–1.5 15

R

10

5 0 –40 –5 –10

0 Y

B

–15

Figure 6.5 Three-phase short-circuit currents of type 1 fixed-speed induction generator.

480

Power Systems Modelling and Fault Analysis

15

R

Components of phase r fault current (pu)

10 5 0 –40

0

40

80

120

–5

160

200 Time (ms)

–10

10 dc component of fault current

5 0 –40

0

–5

40

80

120

160 200 Time (ms)

ac component of fault current

–10

Figure 6.6 Components of phase R short-circuit current of type 1 fixed-speed induction generator.

Using the rated voltage and apparent power as base quantities, the machine parameters in per unit are: stator resistance 0.00676 pu, rotor resistance 0.01111 pu, magnetising reactance 3.3816 pu and stator leakage reactance 0.086957 pu and rotor leakage reactance 0.07198 pu. Using Eqs. (5.93b), (5.94c) and (5.108), we obtain: 0

0

X 5 0:1574 pu, T 5 44:9ms and Ta 5 74:1ms. We assume that θo 5 0 to obtain maximum dc current offset on phase r. The fault currents are shown in Fig. 6.5. Fig. 6.6 shows the ac and dc components of phase r which has the maximum dc current offset.

6.5

Type 2 variable slip wind turbine wound rotor induction generators

6.5.1 Modelling and analysis of short-circuit current contribution This type of induction generator historically uses a three-phase wound rotor winding that is accessible via brushes and slip rings. The rotor windings are connected to an external resistor circuit through an ac/dc power electronics converter that modifies the rotor circuit resistance by injecting a variable external resistance on the rotor circuit. The variation in the effective rotor resistance enables control of

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

481

the magnitude of rotor currents and hence generator electromagnetic torque. The rated slip of large fixed-speed squirrel-cage induction generators is typically around 2 1% but the speed of the wound rotor induction generator can be varied over a small range; typically up to 10% giving a variable slip range of up to 2 10%. Typical values of effective external resistance, as a multiple of rotor winding resistance, required to drive the machine at slips of 2 2%, 2 5% and 2 10% are 1, 4 and 8, respectively. Fig. 6.7A illustrates this type of wind turbine generator with a typical ac/dc power converter for external resistance control. The converter typically consists of a three-phase diode bridge rectifier, three-phase resistors and an IGBT chopper. The rectifier consists of six diodes and converts the three-phase ac voltages at the rotor terminals to a dc voltage and the IGBT is inserted on the dc side of the rectifier. When the IGBT is on, it short-circuits the rotor winding and this reduces the external resistance to zero. However, when the IGBT is off, the entire external resistance appears in series with the rotor circuit. The average value of the external resistor is controlled by the duty cycle mo of the IGBT switch using a classical PWM technique. Eq. (6.3a) gives the IGBT duty cycle and if ideal diodes and IGBT chopper are assumed, then the effective value of the external resistor appearing in series with the generator rotor winding resistor would be Rext ð1 2 mo Þ where 0 , mo # 1. An alternative arrangement is the use of the diode bridge with a single external resistor on the dc side in parallel with the IGBT chopper instead of the three resistances on the rotor winding ac side. This arrangement, shown in Fig. 6.7B,

(A) Wind turbine

Wound rotor Gearbox induction generator

Generatortransformer MV Grid Power factor correction capacitors

Rext Three-phase diode bridge and IGBT chopper (B)

To rotor Rext

IGBT

Three-phase diode bridge with single resistor and IGBT chopper

Figure 6.7 Type 2 variable slip wind turbine induction generator: (A) General diagram and (B) alternative variable external rotor resistor circuit.

482

Power Systems Modelling and Fault Analysis

requires only one resistor but tends to cause mechanical vibrations in the generator drive train which result from current rectification and hence harmonics in rotor currents. Modern designs have also been built that avoid slip rings and brushes, the socalled brushless designs. The most widely used approach is to mount the resistors, rectifier, current sensors and controller on the rotating rotor of the generator with communication signals transmitted over fibre optic cables. When a three-phase short-circuit fault occurs at the terminals of an induction generator, the generator will inherently supply a large stator short-circuit current. As we have already seen for synchronous generators and induction motors, due to the theorem of constant flux linkages, a corresponding increase in the generator’s rotor currents occurs. For a variable slip wound rotor induction generator, the rotor converter used to control the rotor currents is suddenly subjected to large overcurrents. However, the short-term over-current capability of these switches is extremely limited and if their junction temperature is exceeded, they would be damaged. When the instantaneous rotor current reaches the IGBT current threshold limit, the IGBT is immediately blocked and this effectively inserts the entire external resistor circuit in series with the rotor circuit. The blocking of the IGBT switch is extremely fast and typically occurs in less than a few microseconds for the higher rated IGBTs. This causes the rotor winding of the generator to become effectively similar to that of a conventional wound rotor induction motor but with an additional external resistance. Fig. 6.8 shows the induction generator’s equivalent circuit. Therefore the generator’s initial short-circuit current contribution to a fault on the ac grid can be calculated with an effective rotor resistance equal to the sum of the rotor winding resistance and the external resistance Rext . The effect of Rext on the time constant of the ac short-circuit current component is given as 0

Te 5

0

Xr ðRr 1 Rext Þ

Figure 6.8 Steady state equivalent circuit of type 2 variable slip induction generator.

(6.10a)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

483

! 0

Xr 5 Xσr 1

1 Xσs

1 1

(6.10b)

1 Xm

The additional rotor resistance Rext reduces the generator ac transient time constant and hence increases the rate of decay of the ac current component of the short-circuit current. To the best of the author’s knowledge, the effect of Rext on the dc time constant of the stator dc short-circuit current component of a type 2 induction generator has never been addressed in the published literature. The implicit assumption has been that Rext does not affect the dc time constant. However, for the first time, we show in Section 6.6.9 that this is only valid for small values of Rext but not moderate and large values. The effect can be calculated using Eq. 6.33b. As the stator short-circuit current decays, and the rotor current drops below the protection or the converter-controlled reference value, some generator manufacturers design their converters to quickly unblock the IGBT switch and regain control of the rotor currents back to some specified value. This means that the generator starts to supply a constant, predefined, low-value stator short-circuit current.

6.5.2 Example Example 6.2 Consider a type 2 wind turbine wound rotor induction generator having the following parameters: Rated voltage 0.69 kV Stator resistance 0.0054 pu Stator leakage reactance 0.102 pu

Apparent power 2.5 MVA, 60 Hz Rotor resistance (referred to stator) 0.00608 pu Rotor leakage reactance (referred to stator) 0.11 pu, magnetising reactance 4.36 pu

Calculate and plot the phase r fault current for a three-phase solid fault at the machine terminals assuming that θo 5 0 to obtain maximum dc current offset on phase r for the following cases: a. The external resistance is short-circuited. b. The external resistance is chosen by design to be equal to four times the rotor winding resistance. c. The external resistance is chosen by design to be equal to nine times the rotor winding resistance. Neglect the effect of the external resistance on the dc time constant Using Eqs. (5.93b), (5.94c) and (5.108), and Eqs. (6.10a) and (6.10b), we obtain 0 Xr 5 0:20929 pu, Ta 5 102:81ms and Case (a): Case (b): Case (c):

0

T 5 91:475 ms 0 Te 5 18:295ms 0 Te 5 9:147ms

Fig. 6.9 illustrates the effect of external rotor resistance on the stator ac rms short-circuit current and asymmetric current.

Power Systems Modelling and Fault Analysis

Machine currents (pu)

Machine currents (pu)

484

14

Asymmetric phase R fault currents

= 0 = 4 = 9

10 6 2 –2 –50

8

0

50

100

150

200

250 300 Time (ms)

=0 =4 =9

dc components of fault currents

4 0 –50

0

50

100

150

–4

200

250 300 Time (ms)

ac components of fault currents

–8

Figure 6.9 Effect of external resistor on short-circuit current of type 2 variable slip induction generator. Case (b) shows that a moderate value of external resistance is sufficient to cause a substantial reduction in the rotor transient time constant and a reduction in the first peak of the shortcircuit current which is caused by the reduction in the ac component of the short-circuit. Within four cycles, the ac current component is almost completely damped out. In case (c), the ac current is damped in around two cycles.

6.6

Type 3 variable-speed wind turbine doubly fed induction generators (DFIG)

6.6.1 Background These are variable-speed wound rotor induction generators with two bidirectional back-to-back voltage-source converters, as shown in Fig. 6.10. The three-phase stator winding is directly connected to the grid. The three-phase rotor windings are connected to a voltage-source inverter via slip rings. This rotor-side inverter injects three-phase rotor voltages of variable frequency, magnitude and phase. The ac terminals of the grid-connected converter are connected to the grid either at the generator stator terminals or the tertiary winding of a three-winding generator step-up transformer. The grid-connected converter controls the dc voltage and sometimes some reactive power output, if it was sized and specified to do so. The rotor-side converter injects a rotor voltage and controls the rotor currents almost instantaneously. This control provides two important functions. The first is variation of generator

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

Wind turbine

Wound rotor doubly-fed Gearbox induction generator

485

Generatortransformer Grid

ac-rotor frequency

PWM converters Grid frequency

Crowbar circuit

dc Rotor-side Grid-side converter chopper converter

Figure 6.10 Type 3 variable-speed doubly fed wind turbine induction generator.

electromagnetic torque and hence rotor speed. The second is usually stator terminal voltage control or, unusually, constant stator reactive power control or stator power factor control. The typical speed control range for a modern MW class wind turbine doubly fed induction generator (DFIG) may be between 70% and 130% of nominal synchronous speed, with the 130% usually referred to as the rated speed at which rated MW output is produced. The rating of the back-to-back converter corresponds to the speed variation so that, for a speed range of 30%, the rating of the converter is 30% of the generator rating. This reduces the cost of the converter compared to type 4 full-size converter-interfaced generators discussed in Section 6.7.

6.6.2 Basic operation principle In a synchronous machine, dc excitation produces a static magnetic field and this rotates at the speed of the rotor when the latter is driven by a prime mover. As a result, this rotating magnetic field produces ac voltages in the stator windings. For a conventional or singly fed induction generator, electrical power is produced when the generator’s rotor speed is driven above synchronous speed, resulting in a negative slip. The frequency of the rotor speed corresponds to the frequency of the stator plus the electrical frequency of the rotor current. This principle can be applied to DFIG with ac current excitation fed into the three-phase rotor windings. The speed of the rotating stator magnetic field is given by the rotational speed of the rotor and the frequency of the ac currents supplied to the rotor windings. In other words, we can write fstator 5

Nrotor 3 npp 6 frotor 60

(6.11)

486

Power Systems Modelling and Fault Analysis

where fstator is the frequency of the induced stator voltage in Hz; Nrotor is the mechanical speed of the rotor in rpm; npp is the number of pole pairs of the rotor; andfrotor is the frequency of rotor currents in Hz. The positive sign in Eq. (6.11) applies when the generator rotor rotates below synchronous speed and the rotor current excitation creates a magnetic field that rotates in the same direction as the generator rotor. However, the negative sign applies when the generator rotor rotates above synchronous speed and the ac current excitation applied to the rotor windings is negative sequence so that it creates a magnetic field that rotates in the opposite direction to the generator rotor. It is noted that due to the ac rotor current excitation, a laminated rotor design is used which results in short rotor flux time constants. The phase angle of the ac rotor current supplied to the rotor windings sets up the phase angle of the internal stator source voltage and hence affects the flow of active power. The magnitude of the ac rotor currents determines the magnitude of the internal stator source voltage and hence the flow of reactive power. Eq. (6.11) shows that in order to produce a constant stator frequency fstator of say, 50 or 60 Hz, the frequency of the rotor ac currents frotor supplied to the rotor must be continuously adjusted to counteract variations in rotor speed Nrotor caused by wind turbine mechanical power variations. Also, when the generator is connected to a network having a fixed power frequency equals to fnetwork , the adjustment of the rotor current frequency allows control of the rotor speed in such a way as to maintain a constant stator frequency equal to the network frequency. Example 6.3 Consider a wind turbine DFIG with four magnetic poles. The generator supplies power to a 50 Hz ac grid. The wind turbine drives the generator rotor at a speed of 1650 rpm. Calculate the frequency of the rotor ac currents that need to be fed into the rotor windings to produce a constant stator frequency of 50 Hz. The number of pole pairs is npp 5 2. The rotor synchronous speed is given by Nrotor 5 ð50 3 60Þ=2 5 1500 rpm. At 1650 rpm, the generator rotor is running above synchronous speed. Using Eq. (6.11), the frequency of the rotor ac currents is calculated as frotor 5 ð1650 3 2Þ=60 2 50 5 5Hz. If the rotor were running at a subsynchronous speed of 1410 rpm, then the frequency of the rotor ac currents is calculated as frotor 5 50 2 ð1410 3 2Þ=60 5 3Hz.

6.6.3 Rotor protection A simultaneous three-phase short-circuit fault in the network causes a symmetrical voltage dip at the generator terminals and large oscillatory currents in the stator windings and rotor windings connected to the rotor-side converter. Controlling large rotor currents requires a large and uneconomic rotor voltage rating. The large rotor currents flow through the converter switches and the dc link capacitor causing a steep rise in the dc link voltage. However, to protect the converter switches and dc capacitor from damage, a variety of methods and protection circuits have been used. These are generally called crowbar and dc chopper circuits.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

487

6.6.4 Passive and active crowbar protection Many DFIGs use a rotor converter protective circuit called a ‘crowbar’. The crowbar circuit is connected between the rotor winding and the rotor-side converter and is activated when either an instantaneous rotor current in any phase exceeds the allowable converter limit, or an overvoltage on the dc link capacitor exceeds its voltage limit. The rotor current limit is typically in the order of 1.1 pu of rated, since converter IGBTs can carry a very limited overcurrent for a very short time. The dc link voltage limit is typically set at 1.1
1.2 pu of rated dc link voltage. When these limits are exceeded, the converter switches are immediately blocked by blocking the firing pulses and the controllable switch of the crowbar circuit is simultaneously fired. This crowbar action effectively short-circuits the rotor winding, directly or through a small resistor. An earlier generation of DFIG used the so-called passive crowbar circuit where, following a large voltage dip on the network and a corresponding large dip on the generator terminals, the crowbar short-circuits the rotor winding then the main circuit breaker of the generator eventually disconnects the generator from the network. In these passive crowbars, the rotor winding remains short-circuited by the crowbar circuit until the generator stator is disconnected from the grid and the rotor currents decay to zero. Fig. 6.11 shows three typical passive crowbar circuits. The thyristor switches used in passive crowbar circuits cannot turn off the current until it decays to zero. However, following a grid voltage dip, the rotor current will invariably contain a dc current component that can typically take a few hundred milliseconds to decay. This precludes the ability to resume normal operation and rotor-side converter control during the voltage dip or disturbance. (A)

Thyristor crowbar

(B)

To rotor

Rotor-side converter

To rotor

Diode bridge thyristor crowbar

Rotor-side converter

To rotor

(C)

R Diode bridge thyristor crowbar

Rotor-side converter

Figure 6.11 DFIG passive crowbar circuits: (A) phase-to-phase antiparallel thyristor crowbar circuit, (B) diode bridge and a controlled thyristor crowbar circuit and (C) diode bridge and a controlled thyristor crowbar circuit with a crowbar resistor.

488

Power Systems Modelling and Fault Analysis

With increased penetration of wind turbine generation in ac power grids, network operators imposed a requirement called ‘low-voltage ride-through’ capability which generally dictates that the generator must not be disconnected and must quickly regain normal active and reactive power operation and control during or immediately following the low-voltage disturbance. In response, wind turbine generator manufacturers introduced the so-called active crowbar circuit with earlier gate turn-off (GTO) switches, but nowadays IGBT switches can switch off a current several times their rated value. Again, when rotor current or, mostly, dc link voltage limit is exceeded, the control system simultaneously disconnects the rotor-side converter by blocking IGBTs and connects the active crowbar circuit. This results in rerouting much of the high rotor currents through the external crowbar resistance which causes the high rotor currents to decay. Importantly, with an active crowbar, the crowbar resistor voltage and the dc link voltage are monitored during crowbar operation. When both voltages are low enough, the crowbar is turned off. After a short delay, to allow for the rotor currents decay, the rotor-side inverter is restarted, and the reactive current component of the generator is ramped up in order to support the grid. The crowbar may be reactivated after typically 40
60 ms of the onset of the disturbance. Fig. 6.12 shows typical active crowbar circuits. During the period when the crowbar is activated, the rotor winding of the generator appears effectively similar to that of a conventional singly fed wound rotor induction generator with an external rotor resistance Rcb or a rectified equivalent resistance of Rcb , and no control over active or reactive power. The generator goes from generating to absorbing reactive power from the network which depresses the grid voltage and may delay voltage recovery. This is controlled by minimising the operation time of the crowbar by increasing the value of the crowbar resistance and hence increasing the rate of decay of rotor currents. However, too high a value of crowbar resistance can cause high voltage at the rotor-side converter terminals which drives a higher current through the antiparallel diodes of the IGBT switches and charges the dc link capacitor. This increases the dc link voltage and, if the limit is exceeded, the crowbar circuit could be triggered again after its deactivation, leading to an entire generator shutdown. Therefore the selection of the optimum value of resistor in an active crowbar circuit is very important.

To rotor IGBT R Active diode bridge crowbar with IGBT chopper

Figure 6.12 An active crowbar circuit of a DFIG.

Rotor-side converter

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

489

6.6.5 dc chopper protection A recent alternative to the active crowbar circuit is the use of a chopper circuit in the intermediate dc bus of the ac/dc/ac converter, as shown in Fig. 6.13A. The chopper circuit is connected in parallel with the dc link capacitor and consists of a controlled IGBT switch and a chopper resistor sized to handle a specific amount of energy as dictated by the specified grid code low-voltage ride-through requirement. In Fig. 6.13A, the dc chopper is used alone without a crowbar and therefore the rotor winding itself is not short-circuited. In the event of a short-circuit, the rotorside converter IGBT switches are blocked, causing the rotor current to be diverted through the antiparallel diodes to the dc link capacitor. The dc chopper resistor is switched by the chopper control in order to regulate and maintain the dc link voltage within specified limits, typically using a hysteresis controller that is generally set to around 1.1
1.2 pu. The chopper resistor quickly decreases the rotor current and the converter is restarted when the rotor current and dc link voltage have dropped below certain thresholds. However, the disadvantage of this method is that it requires the antiparallel diodes of the converter switches to be oversized to carry and withstand the highest rotor transient currents during grid voltage dips which may be economically unattractive. An alternative operation in this configuration is that the converter remains in operation, maintains control of the generator and supplies reactive current during grid faults or dips as required by most grid codes. However, in this case, the rotor-side converter IGBTs have to be oversized which, again, is generally economically unattractive. In some recent designs, both dc chopper and crowbar circuits are used as shown in Fig. 6.13B. The active crowbar offers the overcurrent protection of the converter, whereas the dc chopper controls the dc bus voltage to within specified limits. (A)

To rotor Rch

Rotor-side converter dc chopper

(B)

To rotor Rch

IGBT R Active diode bridge crowbar with IGBT chopper

Rotor-side converter

dc chopper

Figure 6.13 (A) dc chopper in parallel with dc link capacitor and (B) dc chopper with an active crowbar.

490

Power Systems Modelling and Fault Analysis

Overall, the rotor protection philosophy, implementation method, sizing of converter components, and settings of limits of rotor-side converter in a type 3 wind turbine DFIG vary among manufacturers. Actual equipment data, control parameters and models for specific project installations need to be based on actual information provided by manufacturers.

6.6.6 General model of doubly fed induction generators (DFIG) In Appendix A.2, we present modelling of induction machines in various reference frames including reference frame transformations. Because of the symmetry of the magnetic system, the analysis of stator and rotor short-circuit currents of induction generators is simplified using a complex space vector induction machine model in a rotor reference frame. Using the generator convention for stator current and all rotor quantities referred to the stator, the DFIG equations in per unit are v s ðtÞ 5 2 Rs i s ðtÞ 1 jωr ψ s ðtÞ 1 v r ðtÞ 5 Rr i r ðtÞ 1

dψ s ðtÞ dt

dψ r ðtÞ dt

(6.12a)

(6.12b)

ψ s ðtÞ 5 2 Ls i s ðtÞ 1 Lm i r ðtÞ

(6.13a)

ψ r ðtÞ 5 2 Lm i s ðtÞ 1 Lr i r ðtÞ

(6.13b)

where Ls 5 Lσs 1 Lm

Lr 5 Lσr 1 Lm

ωr 5 ð1 2 sÞωs

(6.13c)

ωr and ωs are rotor and synchronous angular frequencies, respectively, and s is the rotor slip. The variables and parameters in Eqs. (6.12) and (6.13) are described in Appendix A.2.

6.6.7 DFIG steady-state equivalent circuit In the steady state, voltages and currents are represented as complex phasors and in the rotor reference frame, the equivalent circuit is derived by substituting d=dt 5 jsωs in Eq. (6.12). Using Eqs. (6.13a)
(6.13c), Eq. (6.12a) can be written as Vs 5 2 Rs Is 1 ½jð1 2 sÞωs 1 jsωs ð 2Ls Is 1 Lm Ir Þ 5 2 ðRs 1 jωs Ls ÞIs 1 jωs Lm Ir or Vs 5 2 ðRs 1 jXs ÞIs 1 jXm Ir

(6.14a)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

491

/

(A)

+ / −

(B)

/ + / −

Figure 6.14 (A) DFIG steady-state equivalent circuit and (B) DFIG equivalent circuit with crowbar activated.

where Xs 5 ωs Ls and Xm 5 ωs Lm . Now, using Xs 5 Xσs 1 Xm from Eq. (6.13c), we have Vs 5 2 ½Rs 1 jðXσs 1 Xm ÞIs 1 jXm Ir

(6.14b)

Similarly, Eq. (6.12b) can be written as Vr 5 Rr Ir 1 jsωs ð 2 Lm Is 1 Lr Ir Þ or Vr 5 ðRr 1 jsXr ÞIr 2 jsXm Is

(6.14c)

where Xr 5 ωs Lr . Dividing by s throughout and using Xr 5 Xσr 1 Xm from Eq. (6.13c), we have   Vr Rr 5 1 jðXσr 1 Xm Þ Ir 2 jXm Is s s

(6.14d)

Eqs. (6.14b)and (6.14d) are represented in the DFIG steady-state equivalent circuit shown in Fig. 6.14A.

6.6.8 DFIG natural stator and rotor short-circuit currents under constant ac excitation In this section, we derive the equations that model the natural responses of the DFIG stator and rotor currents to a solid three-phase short-circuit fault at the DFIG stator terminals or to a symmetrical voltage dip at the DFIG stator terminals which may be caused by remote three-phase short-circuit faults on the ac grid. The natural

492

Power Systems Modelling and Fault Analysis

stator and rotor current responses under a symmetrical voltage dip at the DFIG terminals assume constant ac rotor excitation and no crowbar action. In practice, the rotor currents can be too high under electrically close faults, that is, large voltage dips, and could destroy the rotor-side converter if allowed to flow freely as discussed in Section 6.6.3. However, our objective in this section is to provide an insight into the natural stator and rotor short-circuit current responses of the DFIG that would result if allowed to flow. The currents due to converter crowbar protection action are presented in Section 6.6.9. We assume that the rotor speed does not change during these disturbances and neglect saturation of leakage and magnetising inductances. The large rotor currents would flow unhindered because the rotor-side converter appears as a zero impedance to the flow of these currents. We maintain the use of s as the machine slip and use the complex frequency p 5 σ 1 jω as the Laplace operator. We choose a solution method that, although demanding, provides an insight into the various components of stator and rotor short-circuit currents. Therefore we take the Laplace transforms of Eqs. (6.12) and (6.13) then substitute Eq. (6.13b) into Eq. (6.12b) and, after some algebra, we obtain the following general expression that relates stator and rotor voltages to their corresponding currents 

  2 ½Rs 1 Ls ðp 1 jωr Þ V s ðpÞ 5 2 Lm p V r ðpÞ

Lm ðp 1 jωr Þ Rr 1 Lr p



  Ls I s ðpÞ 1 L I r ðpÞ m

2 Lm 2 Lr

   is  ir

(6.15a) 



where is and ir represent the initial stator and rotor currents, respectively. Dropping the complex frequency operator ðpÞ for convenience, the stator and rotor currents can be expressed as " #  " # 2 Lm ðp 1 jωr Þ Is Vs 21 Rr 1 Lr p 5 D Lm p 2 ½Rs 1 Ls ðp 1 jωr Þ V r Ir " #"  #

2 Lm ðRr 2 jωr Lr Þ is 1 Rr Ls 1 Lr Ls 2 L2m p 2 jωr L2m

1  2 D 2 Lm ðRs 1 jωr Ls Þ Lr Ls 2 Lm p 1 Lr ðRs 1 jωr Ls Þ ir (6.15b) where





D 5 Lr Ls 2 L2m p2 1 Rr Ls 1 Rs Lr 1 jωr Lr Ls 2 L2m p 1 Rr ðRs 1 jωr Ls Þ (6.15c)

Our research experience shows that the analytical solutions of the stator and rotor current equations can be greatly simplified by making use of a few useful equations that use the basic transient parameters of the machine as follows: L0s 5 Ls 2

L2m Lr

T dc 5

L0s Rs

(6.16a)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

L0r 5 Lr 2

L2m Ls

T 0r 5

L0r Rr

T 0o 5

493

Lr Rr

(6.16b)

Ls Lr T0 5 0 5 o0 0 Ls Lr Tr

Lr Ls 2 L2m 5 Lr L0s 5 Ls L0r

(6.16c)

where L0s is stator transient inductance; L0r is rotor transient inductance; Tdc is stator short-circuit dc time constant; T 0r is rotor transient short-circuit time constant; and T 0o is rotor transient open
circuit time constant Therefore using Eq. (6.16) in Eqs. (6.15b) and (6.15c), the stator and rotor currents are obtained as

Is 5

21 L0s



p1

1 T 0o

     L2 V s 1 LL0 mLr ðp 1 jωr ÞV r 1 p 1 T10 2 jωr L0 mLr is 2 s r s     1 1 1 Rs 2 p 1 T 0 1 Tdc 1 jωr p 1 T 0 Ls 1 jωr r

Lm L0s



1 T 0o

 2 jωr ir

r

(6.17a) Ir 5

2Lm L0s Lr

pV s 1

1 L0s Lr

  ½Rs 1 ðp 1 jωr ÞLs V r 2 LL0 mLr ðRs 1 jωr Ls Þis 1 p 1 s     p2 1 T10 1 T1dc 1 jωr p 1 T10 RLss 1 jωr r

1 Tdc

 1 jωr LLs0 ir s

r

(6.17b) The stator and rotor currents have the same denominator and therefore the same eigenfrequencies and damping time constants. The two roots can be determined from 2 3
 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
 1 1 1 1 1 4 Rs p1;2 5 4 2 1 1 jωr 6 1 1jωr 2 0 1 jωr 5 2 T 0r Tdc T 0r Tdc T r Ls (6.18) For practical DFIG parameters and rotor speed range of typically 0:7 pu # ωr # 1:3 pu, the small rotor resistance allows us to make simplifications that enable us to derive a closed form solution of the two roots with negligible errors in practical applications. As shown in Eqs. (6.17a) and (6.17b), the Laplace transforms of the initial stator and rotor voltages are required. Since the frequency of the original steady-state stator voltage and current in the rotor reference frame is sωs , the predisturbance instantaneous stator voltage can be written as  o pffiffiffi   vs ðtÞ 5 Real V s ðtÞ 5 2Vs cosðsωs t 1 θo Þ

(6.19a)

494

Power Systems Modelling and Fault Analysis o

where V s ðtÞ is the initial complex instantaneous stator voltage given by pffiffiffi o V s ðtÞ 5 2Vos ejsωs t 



Vs 5 Vs ejθo

(6.19b) (6.19c)

where Vos is the initial stator voltage phasor with an angle θo . Similarly, the DFIG original steady-state instantaneous rotor voltage is given by  o pffiffiffi   vr ðtÞ 5 Real V r ðtÞ 5 2Vr cosðsωs t 1 δo Þ (6.20a) o

where V r ðtÞ is the initial complex instantaneous rotor voltage given by 

V r ðtÞ 5 

pffiffiffi  jsω t 2V r e s



Vr 5 Vr ejδo

(6.20b) (6.20c)

where Vor is the initial rotor voltage phasor with an angle δo . When a symmetrical voltage dip occurs at the DFIG terminals, the retained complex instantaneous stator voltage at the dip instant can be expressed as pffiffiffi  V s ðtÞ 5 k 2Vs ejθo ejsωs t (6.21a) where 0 # k , 1 and the case of k 5 0 it represents a solid three-phase fault at the DFIG terminals. Taking the Laplace transform of Eq. (6.21a), we have pffiffiffi  k 2Vs ejθo V s ðpÞ 5 (6.21b) p 2 jsωs The natural DFIG stator and rotor currents are calculated with a constant rotor voltage. Therefore taking the Laplace transform of Eq. (6.20b), we have pffiffiffi  jδ 2 Vr e o V r ðpÞ 5 (6.21c) p 2 jsωs Substituting Eqs. (6.21b) and (6.21c) into Eqs. (6.18b) and (6.18c) and taking their inverse Laplace transforms, we obtain the complex instantaneous currents I s ðtÞ and I r ðtÞ. Since our equations are in the rotor reference frame, the real instantaneous stator current in the stationary reference frame is obtained as follows  (6.22a) is ðtÞ 5 Real I s ðtÞejωr t and the real instantaneous rotor current is obtained using  ir ðtÞ 5 Real I r ðtÞ

(6.22b)

Following extensive mathematical analysis, it can be shown that the stator shortcircuit fault current is given by

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

495

pffiffiffi pffiffiffi 

Xm 2Vr

2 k 2Vs Zr 0 0 is ðtÞ 5 cos ωs t 2 α01 1 α04 1 δo 0 cos ω s t 1 θo 2 α1 1 α2 1 0 Xs Zr Xs Zr ( 0 1 0 1 pffiffiffi  k 2 Vs @ 1 1 A Rr π 2 2 cos@ωr t 1 θo 2 α01 2 A Xs Zr0 2 1 2 s Xs0 pffiffiffi 

Xm 2 V r 1 cos ωr t 2 α01 1 α05 1 δo 0 Xs Zr ) t 20 1 3 2 0 Tr 1 1 X   m 1 4@ 0 2 AXs is 2 0 ir 5cosðωr tÞ e Xs Xs Xs

(

1

pffiffiffi pffiffiffi 

Xm 2Vr

2 k 2Vs 1 0 cos α04 1 δo cos θo 1 α3 1 0 0 Xs Xs Xr ½ð1 2 sÞωs Tdc 

)

1  

1 0 Xs i s 2 Xm i r Xs

t 2T

dc

e

(6.23a) The phase y and b currents are obtained by replacing θo by θo 2 2π=3 and θo 1 2π=3, respectively, and replacing δo by δo 2 2π=3 and δo 1 2π=3, respectively, The rotor short-circuit fault current is given by pffiffiffi 

2 Vr cos sωs t 2 α01 1 α04 1 δo 0 Zr ( 0 1 pffiffiffi  pffiffiffi 

2 V Xm k 2 Vs 1 1 r 1 2 cos 2 ωr t 1 θo 1 α03 1 Rs @ 0 2 A Xs Xs Xs Xr0 ð1 2 sÞXr0 ) t 2 0 1 3 2T

dc X 1 1   m 3 cos 2 ωr t 1 α04 1 δo 1 4 0 is 2 @ 0 2 AXr ir 5cosð 2 ωr tÞ e Xr Xr Xr

ir ðt Þ 5 2



Xm pffiffiffi  s k 2Vs 0 cos sωs t 1 θo 2 α01 1 α04 1 Zr Xs

(

2

pffiffiffi  pffiffiffi 



2V r Xm k 2 V s Rr 0 cos 2 α01 1 α05 1 δo 0 0 sin θo 2 α1 1 0 Xs ð1 2 sÞXr Zr Zr

)

2

1  

1 0 Xm is 2 Xr ir e Xr

t Tr0

(6.23b)

496

Power Systems Modelling and Fault Analysis

where Zr 5

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R2r 1 ðsXr Þ2

0 0

α1 5 tan21 sωs Tr 2

13 0 0 sT X α2 5 Argument4sωs Tr 1 j@ r 2 s A5 Tdc Xs 2 3 1 1 0 2 1 j5 α3 5 Argument4 ð1 2 sÞωs Tr0 ωs Tdc 0

0

0

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ffi R2r 1 sXr0 0 1 1 0 A α4 5 tan21 @ ωs Tdc 2 3 1 0 5 α5 5 tan21 4 ð1 2 sÞωs Tr0 0

Zr 5

0

0

Xs 5 ωs Ls

0

0

Xr 5 ωs Lr (6.23c)

Eq. (6.23a) shows that the stator current consists of three current components: 1. a steady-state current component due to the retained stator terminal voltage and constant rotor voltage and has a frequency equal to ωs ; 2. a transient ac current component that has a frequency equal to rotor speed ωr and decays with a time constant T 0r ; and 3. a transient dc or zero frequency current component that decays with a stator dc time constant Tdc .

Eq. (6.23b) shows that the rotor current consists of three current components: 1. a steady-state current component due to the retained stator terminal voltage and constant rotor voltage and has a frequency equal to slip frequency sωs ; 2. a transient ac current component that has a frequency equal to 2ωr and decays with a time constant Tdc ; and 3. a transient dc or zero frequency current component that decays with a time constant T 0r .

The predisturbance stator and rotor currents and rotor voltage are calculated from the DFIG’s known initial operating conditions of stator voltage, active power, reactive power and rotor slip. The initial rms stator current phasor is given by ðPs 1jQs Þ   Is 5  jθ  5 Is ejðθo 2ϕo Þ Vs e o

(6.24a)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

  where Is 5 ð P2s 1 Q2s Þ=Vs , ϕo 5 tan21 Qs =Ps and Ps and Qs are initial stator active and reactive power outputs, respectively. The initial, real, instantaneous stator current is given by 

is ðtÞ 5

pffiffiffi 

2Is cos ωs t 1 θo 2 ϕo

(6.24b)

The initial rms rotor current phasor is calculated from Eq. (6.14a) as follows

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators 



Ir 5

497



Vs 1 ðRs 1 jXs ÞIs  5 Ir ejβo jX m

(6.25a)

and the initial, real, instantaneous rotor current can be written as pffiffiffi ior ðtÞ 5 2Iro cosðsωs t 1 β o Þ

(6.25b)

The predisturbance rotor voltage is calculated using Eq. (6.14c) as follows 

Vr 5 ðRr 1 jsXr ÞIr 2 jsXm Is 5 Vr ejδo and the initial, real, instantaneous rotor voltage can be written as pffiffiffi   vr ðtÞ 5 2Vr cosðsωs t 1 δo Þ

(6.26a)

(6.26b)

The stator and rotor short-circuit currents are illustrated in the examples given in Section 6.6.12.

6.6.9 DFIG stator and rotor short-circuit currents under crowbar action As discussed in Sections 6.6.3 and 6.6.4, crowbar protection operates to protect the converter from the high and damaging rotor currents calculated in Section 6.6.8. In this section, we limit our attention to the derivation of the DFIG’s short-circuit currents during the period of crowbar operation. The short-circuit current in the period after the crowbar is removed and the converter regaining control of rotor currents is discussed in Section 6.6.11. As shown in Fig. 6.12, the blocking of the converter switches and firing of the crowbar inserts an effective crowbar resistance Rcb that appears in series with the rotor winding. Rcb is the effective resistance seen on the ac side of the threephase diode bridge and is therefore the ‘rectified’ value of R. The crowbar action is equivalent to short-circuiting the rotor winding through an equivalent resistance Rcb as illustrated in the steady-state equivalent circuit in Fig. 6.14B. Therefore the rotor voltage can be written as V r 5 2 Rcb I r

(6.27a)

The subscript cb is used throughout to denote crowbar. Substituting Eq. (6.27a) in Eq. (6.15a) and rearranging, we obtain " #  " # 2Lm ðp1jωr Þ Is Vs 21 Rr 1Rcb 1Lr p 5 D Lm p 2 ½Rs 1Ls ðp1jωr Þ V r Ir " #"  #

2Lm ½ðRr 1Rcb Þ2jωr Lr  is 1 ðRr 1Rcb ÞLs 1 Lr Ls 2L2m p2jωr L2m

1  2 D 2Lm ðRs 1jωr Ls Þ Lr Ls 2Lm p1Lr ðRs 1jωr Ls Þ ir (6.27b)

498

Power Systems Modelling and Fault Analysis

where



D 5 Lr Ls 2 L2m p2 1 ðRr 1 Rcb ÞLs 1 Rs Lr 1 jωr Lr Ls 2 L2m p 1 ðRr 1 Rcb ÞðRs 1 jωr Ls Þ

(6.27c)

The crowbar action is mathematically equivalent to setting the rotor voltage to zero and replacing the rotor winding resistance Rr with the total resistance Rr 1 Rcb . Similar steps to those used in Section 6.6.8 are followed making use of the following equations that greatly simplify the analysis 0

L2m Lr

0

Ls 5 Ls 2

T dc 5

Ls Rs

0

Lr 5 Lr 2

L2m Ls

(6.28a)

0

Lr Ls 2 L2m

0

0

5 Lr Ls 5 Ls Lr

Lr T cb 5 Rr 1 Rcb 0

0

T o;cb 5

Lr Rr 1 Rcb

(6.28b)

0

T Ls Lr 5 0 5 o;cb 0 L0s Lr T cb

(6.28c)

In defining the above equations, we note that we have intentionally reused L0s and Tdc that do not include the effect of crowbar resistance as defined in Eq. (6.16a) since this simplifies the mathematical solutions greatly but, as we show later, the effective stator transient reactance and stator dc time constant are affected and modified by the high rotor resistance. We also used L0r as defined in Eq. (6.16b) and, as we show later, this is not affected by the high rotor resistance. 0 0 Also, we included the crowbar resistance Rcb in the time constants Tcb and To;cb in Eq. (6.28b) which, again, facilitate the derivation of a closed form solution of stator and rotor currents and will be shown to be a valid practical representation of the DFIG short-circuit and open-circuit time constants. Now, substituting Eqs. (6.28a)
(6.28c) in Eqs. (6.27b) and (6.27c) and rearranging, it can be shown that the stator and rotor currents are given by
Is 5

21 L0s

p1


    L2 V s 1 p 1 T10 2 jωr L0 mLr is 2 LLm0 T 01 2 jωr ir s s cb o;cb     p2 1 T10 1 T1dc 1 jωr p 1 T10 RLss 1 jωr 1

0 To;cb

cb

cb

   2Lm Lm Lr  1 pV 2 ð R 1 jω L Þi 1 p 1 1 jω s s r s s r L0 i r Tdc L0 Lr L0 Lr s    r Ir 5 s p2 1 T10 1 T1dc 1 jωr p 1 T10 RLss 1 jωr cb

(6.29a)

(6.29b)

cb

The denominator of Eqs. (6.29a) and (6.29b) shows that the stator and rotor currents have the same frequencies and decay time constants as those of Eq. (6.17) and

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

499

0

that T 0r is now replaced by Tcb . However, since the crowbar resistance Rcb can be tens of times greater than the rotor winding resistance Rr , the roots of the denominator of Eq. (6.29) cannot be obtained using the simplification of negligibly small rotor resistance as was used in obtaining the roots of Eq. (6.18a). Physically, the high rotor resistance Rr 1 Rcb substantially increases the magnetic coupling between the stator and rotor fluxes. Fortunately, our extensive research has succeeded in obtaining an accurate closed-form solution for these roots and hence the formulation of a closed form solution for stator and rotor current transients. We start next by analysing the effect of the high rotor resistance on the rotor and stator transient inductances and stator time constant. In the stationary reference frame presented in Appendix A.2 and with a generator sign convention for stator current, substituting ψs of Eq. (A.2.19a) and V s 5 0 in Eq. (A.2.20a), we obtain I s 5 pLm =ðRs 1 pLs ÞI r . Now, substituting this into ψr of Eq. (A.2.19b), we obtain
ψ r 5 Lr 2

 pL2m I r 5 2 Lr ðpÞI r Rs 1 pLs

(6.30a)

Lr ðpÞ 5 Lr 2

pL2m Rs 1 pLs

(6.30b)

where Lr ðpÞ is the DFIG operator inductance, as seen from the rotor with the stator short-circuited. Since Rs is very small compared to Ls , Lr ðpÞ reduces to L2

Lr ðpÞ 5 Lr 2 Lms 5 L0r . This means that L0r is not affected by the rotor resistance. Also, in the rotor reference frame presented in Appendix A.2, substituting ψr of Eq. (A.2.19b) and Vr 50 in Eq. (A.2.22b), we obtain I r 5 pLm =½ðRr 1 Rcb Þ 1 pLr I s . Now, substituting this into ψs of Eq. (A.2.19a), we obtain  ψ s 5 2 Ls 2 Ls ðpÞ 5 Ls 2

 pL2m I s 5 2 Ls ðpÞI s ðRr 1 Rcb Þ 1 pLr

(6.31a)

pL2m ðRr 1 Rcb Þ 1 pLr

(6.31b)

where Ls ðpÞ is the DFIG operator inductance, as seen from the stator with the rotor short-circuited through a high crowbar resistance Rcb . With Rcb 5 0, we obtain the L2

familiar equation of stator transient inductance L0s 5 Ls 2 Lmr since Rr is very small compared to Lr which means that L0s is not affected by the small rotor resistance Rr . However, with practical values of crowbar resistance Rcb , Rr 1 Rcb can be comparable to or even greater than Lr and therefore cannot be neglected in the denominator of Eq. (6.31b).

500

Power Systems Modelling and Fault Analysis

In our chosen rotor reference frame, a stationary stator flux with respect to the stator, induces a voltage in the rotor whose angular frequency is ωr and, as implied in Eq. (6.22b) and shown in Eq. (6.23b), the frequency of the resultant rotor current is 2ωr . Therefore the effective stator transient inductance that includes the effect of the high crowbar resistance Rcb can be derived from the operator inductance given in Eq. (6.31b) by setting p 5 2 jωr . Therefore the resultant complex stator transient inductance is given by 0

L s;cb 5 Ls ð 2 jωr Þ 5 Ls 2

2 jωr L2m Rr 1 Rcb 2 jωr Lr

(6.32a)

Dividing by stator resistance Rs , the complex stator time constant is given by 0
 L s;cb 1 2 jωr L2m T dc;cb 5 5 Ls 2 (6.32b) Rs Rs Rr 1 Rcb 2 jωr Lr Following further extensive analysis, it can be shown that the effective stator transient inductance and stator dc time constant, in the presence of a high rotor resistance, are given by
 Rr 1Rcb 2 1 1 L10 2 r 0 0 1
s Ls;cb 5 Ls 3 (6.33a)
2 R r 1Rcb 1 1 1 L L0 r r 1
s


 Rr 1Rcb 2 1
s Tdc;cb 5 Tdc 3
 Rr 1Rcb 2 1 1 L 1L0 r r 1
s 11

1 L0r2

(6.33b)

The complex stator time constant given in Eq. (6.32b) means that the stator flux is no longer stationary with respect to the stator but it slowly rotates at a certain angular frequency. It can be shown that the angular frequency of the slowly rotating component of stator flux is given by

 1 1 Rr 1 Rcb 2 1 L0r Lr 1
s ωdc 5 3 (6.33c)
 Tdc 1 Rr 1Rcb 2 1 1 02 Lr 1
s Eqs. (6.33a) and (6.33b) show the effect of high rotor resistance as a multiplier applied to L0s and Tdc which apply in the case the machine has a negligibly small

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

501

rotor resistance. In fact, Eqs. (6.33a) and (6.33b) are general and apply for any 0 value of rotor resistance. Ls;cb and Tdc;cb show strong dependence on crowbar or total rotor resistance and rotor slip. To gain a feel of the effect of high rotor resistance, consider a DFIG machine having the following parameters in per unit: Rs 5 0:0069 pu, Xs 5 3:3158 pu, Rr 5 0:00867 pu, Xr 5 3:3239 pu, 0 0 Xs 5 0:2351 pu, Xr 5 0:2356 pu, Tdc 5 34:07 pu. Network frequency is 50Hz. With Rcb 5 0, the factor that multiplies L0s and Tdc in Eqs. (6.33a) and (6.33b) ranges 0 from 1.0007 to 1.0026 for 0:7 pu # ωr # 1:3 pu thus Ls;cb 5 L0s and Tdc;cb 5 Tdc . However, for a moderately high rotor resistance, for example, Rcb 5 20Rr , this factor increases significantly and now ranges from 1.32 to 2.04 for 0:7 pu # ωr # 1:3 pu. Also, Eq. (6.34) shows that with Rcb 5 0, the angular frequency of the rotating stator flux ranges from 0:038 to 0:07Hz for 0:7 pu # ωr # 1:3 pu and hence it is practically negligible. However, for Rcb 5 20Rr , this angular frequency ranges from 0:6 to 0:68Hz. pffiffi  k 2Vs ejθo Now, we return to Eqs. (6.29a) and (6.29b), use Eq. (6.21b) or V s ðpÞ 5 p 2 jsω , s take the inverse Laplace transforms and obtain I s ðtÞ and I r ðtÞ. It can be shown that following extensive mathematical analysis, the stator phase r short-circuit current response of the DFIG with crowbar action is given by pffiffiffi  2 k 2Vs Zr1 0 0

is ðtÞ 5 cos ωs t 1 θo 2 β 1 1 β 2 Xs Zr2 9 8
 0 pffiffiffi  1  1 Rr 1 Rcb Xr 0 0 > > > > > k 2 Vs cos ðωr 2 ωdc Þt 1 θo 2 β 1 1 β 3 > > > 0 2 > > X X Z Z > > s r3 r2 s > > > >
   > > 0 = t < 1 1 ð1 2 sÞXr  0 π 2 Xs is cos ðωr 2 ωdc Þt 1 β 3 2 e2Tac 1 0 2 Xs Xs 2 Zr3 > > > > > > > > > > Xm Zr4   0 0 > > > > > > 1 ir cos ðωr 2 ωdc Þt 1 β 3 1 β 5 > > ; : Xs Zr3 9 8 pffiffiffi  0  > 0 > 2 k 2Vs ð1 2 sÞXr > > > cos ðωr 2 ωac Þt 1 θo 1 β 4 > > > = < Xs0 Zr3 2 t
 e Tdc;cb 1  0 0 > > Z X   r4 m > > > > > ; : 1 Zr3 is 2 Xs ir cos ðωr 2 ωac Þt 1 β 3 1 β 5 > and the rotor phase r short-circuit fault current is given by

(6.34a)

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Power Systems Modelling and Fault Analysis

pffiffiffi  Xm k 2 Vs s 0

ir ðtÞ 5 2 cos sωs t 1 θo 2 β 1 Xs Zr2 9 pffiffiffi  0 Xm k 2Vs Rr 1 Rcb Xr 0 0

> > > cos 2 ωdc t 1 θo 2 β 1 1 β 3 0 > = t Xr X s Zr3 Zr2
 e2Tac 1 0 > 0 0 0 > X ð 1 2 s ÞX π Z   m r4 > > r > > > ; : 1 Xr Zr3 is cos 2 ωdc t 1 β 3 1 2 1 Zr3 ir cos 2 ωdc t 1 β 3 1 β 5 > 8 > > > >
0

Xm k 2Vs ð1 2 sÞXr > > cos 2 ωac t 1 θo 1 β 3 > 0 = Xr Xs Zr3 2 t  e Tdc;cb  
2 0 > > ð 1 2 s Þ X π 0  0  m > > > > X i 2 ðXr 2 Xr Þir cos 2 ωac t 1 β 3 1 > ; : 1 Zr3 2 > Xr r s 8 > > > >
1 Rr 1 Rcb Xr > > > > k 2 Vs 2 cos ωr t 1 θo 2 β 1 1 β 3 > > > > > Xs0 Xs Zr3 Zr2 > > > > > >

  > > 0 = t < 0 1 1 ð1 2 sÞXr  π 2 2 X i cos ω t 1 β 2 e2Tac 1 s s r 0 3 X X 2 Z > > s r3 s > > > > > > > > 0 0

Xm Zr4  > > > > > > 1 i cos ω t 1 β 1 β r > > r 3 5 ; : Xs Zr3 9 8 pffiffiffi  0 > 2 k 2Vs ð1 2 sÞXr 0 > > > > cos θo 1 β 4 > > > = < Xs0 Zr3 2 t
 e Tdc;cb 1 0 Zr4  Xm  0 > > > > > > > ; : 1 Zr3 is 2 Xs ir cos β 3 1 β 5 > and

(6.39a)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

505

pffiffiffi  Xm k 2 Vs s 0

i r ðt Þ 5 2 cos sωs t 1 θo 2 β 1 Xs Zr2 9 8 pffiffiffi  0 > Xm k 2Vs Rr 1 Rcb Xr 0 0 > > > > cos θo 2 β 1 1 β 3 > > > 0 > > X X Z Z > > r r3 r2 s > > > >
 > > 0 = t < 0 Xm ð1 2 sÞXr  π 1 is cos β 3 1 e2Tac 1 > > 2 Xr Zr3 > > > > > > > > Zr4  0 0

> > > > > > 1 i cos β 1 β > > r 3 5 ; : Zr3 9 pffiffiffi  0 > Xm k 2Vs ð1 2 sÞXr 0

> > cos 2 ωr t 1 θo 1 β 3 > = Xr Xs0 Zr3 2 t   
e Tdc;cb 2 0 > ð 1 2 s Þ Xm 0  π > 0  > > > > Xr is 2 ðXr 2 Xr Þir cos 2 ωr t 1 β 3 1 > ; : 1 Zr3 2 > Xr 8 > > > >
> kΔVc > < > > > : Imax

Imax k Imax ΔVc $ k ΔVc ,

(6.72)

where Ic is the steady-state short-circuit reactive current supplied by the inverter; k is the k factor given in Section 6.12; ΔVc is the voltage change at the inverter terminals; and Imax is the maximum transient current rating of the inverter. The above equation applies to both positive-sequence and negative-sequence reactive currents and, as presented in Section 6.11, represents the final steady-state current. This current is dependent on the inverter terminal voltage during the fault but this injected current also modifies the inverter terminal voltage. This means that the steady-state value of the injected reactive current can be determined by an iterative calculation process until the change in current magnitude is within a prespecified tolerance. Based on the grid code requirement shown in Fig. 6.35 with k 5 2, the injected current is linearly dependent on the inverter terminal voltage up to a voltage dip of 0:5 pu. For larger voltage dips, the injected current becomes constant and equal to the maximum transient current the inverter is designed to deliver. In Fig. 6.45A, the inverter transformer impedance is typically 6%
10% on rating and the 33/132 kV transformer impedance is typically 12%
24% on rating, giving a total impedance to the fault point of 18%
34%. With a typical value of maximum transient current of 1.1
1.2 pu on rating, the inverter terminal voltage could range from 0.2 to 0.41 pu. Therefore in such cases, the injected inverter steady state current would be constant and equal to the maximum inverter transient current rating. In Section 6.11, we showed that the actual value of inverter current from the onset of fault or voltage dip is dependent on the measurement and detection time of the voltage dip and the response of the current control system. In practical fault current analysis, calculations of initial make-and-break short-circuit currents are required. This is presented in detail in Chapter 8, International standards for

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

545

short-circuit analysis in ac power systems. For now, we simply state that the former current is required to be calculated at half cycle of power frequency after the onset of the fault, that is, 10 and 8.33 ms in 50 and 60 Hz power systems, respectively. The latter correspond to the minimum short-circuit current break time, which is typically in the range of 1.5
6 cycles of power frequency. The inverter inner current control loop is very fast and the injected reactive current variation with time following detection of the voltage dip, assuming a welldesigned fast-acting control system with no overshoot, can be approximated as follows: 8 0 t # Tm <  ic ðtÞ 5 Ic 1 2 e2ðt2Tm Þ=τ c Tm , t # Tm 1 4τ c : Ic t . Tm 1 4τ c

(6.73)

where ic ðtÞ is inverter current variation with time from onset of voltage dip; Tm is the measurement and detection time delay of the rms voltage dip during which the inverter injected reactive current is taken as zero. Tm is typically 10
20 ms; τ c is the effective time constant of the inner current control loop, typically 3
8 ms; and Ic is the maximum steady-state injected reactive current given in Eq. (6.72). To illustrate use of the above equation, consider the following: Tm 5 20ms, τ c 5 5ms and a minimum circuit-breaker opening time t 5 30ms. Therefore the inverter current contribution to the initial make fault current is obviously zero, and the contribution to the break fault current is 0:865Ic . For a minimum circuit-breaker opening time t 5 50ms, the contribution to the break fault current is Ic . For τ c 5 8ms, the contributions to the break fault current become 0:713Ic and 0:976Ic for t 5 30 and 50ms, respectively. In all phasor short-circuit current analysis to follow, the ac grid includes conventional rotating machines that are represented as voltage sources behind their subtransient or transient impedances. However, the technique is extended in Chapter 7, Short-circuit analysis techniques in large-scale ac power systems, and applied to the general analysis of large-scale networks containing mixed conventional machines and inverters. The inverter reactive current calculated in the analysis represents its injected current immediately upon fault occurrence, that is, at maximum voltage dip at the inverter terminals. However, the inverter injected current increases the inverter terminal voltage and hence reduces the voltage change. This causes a reduction in injected current according to Eq. (6.72), which, in turn, reduces the inverter voltage. This calculation can be carried out iteratively until the change in injected current is within a small tolerance.

6.16.3 Analysis of three-phase short-circuit fault currents The positive-sequence equivalent circuit of Fig. 6.45B is shown in Fig. 6.46A. The grid is represented as a voltage source behind an equivalent impedance and the inverter is represented as a voltage-dependent positive-sequence current source.

546

Power Systems Modelling and Fault Analysis

(A)

+

=

+

=

F

=

+

Three-phase solid short circuit fault

+

(B)

F ( )

(C)

(D) +

+

F

( )

+

F ( )

Figure 6.46 Analysis of three-phase faults: (A) system representation, (B) voltage-source inverters do not inject currents, (C) voltage-source inverters inject positive-sequence reactive current and (D) grid fault current.

Inverters do not contribute a fault current The inverter is represented as an open circuit in the positive-sequence equivalent circuit as shown in Fig. 6.46B. The positive-sequence fault current contribution of the ac grid is given by 1 IFðgÞ 5

Vg1 Zg1

(6.74a)

and the asymmetric fault current is given by " !# pffiffiffi 1 2 2πft  iF ðtÞ 5 2IFðgÞ 1 1 exp  X=R g

(6.74b)



where X=R g is the positive-sequence X/R ratio calculated from the positivesequence equivalent impedance Zg1 of the ac grid, that is,

X=R



5 g

Xg1 R1 g

(6.74c)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

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Inverters contribute a positive-sequence current As presented in Chapter 2, Symmetrical components of faulted three-phase networks containing voltage and current sources, the total short-circuit fault current at the 132 kV busbar is the superposition of the short-circuit current contributions of the voltage and current sources, that is, the inverter and grid contributions as shown in Fig. 6.46C and D.

ac grid voltage source is short-circuited The equivalent circuit is shown in Fig. 6.46C. The positive-sequence fault current supplied by the inverters with the ac grid voltage source short-circuited, that is, excluding the ac grid current contribution is given by 1 5 Ic1 IFðcÞ

(6.75a)

Inverter current source is open-circuited The equivalent circuit is shown in Fig. 6.46D. The positive-sequence fault current supplied by the ac grid with the inverter source current open-circuited, that is, excluding the inverter contributions, is given in Eq. (6.74a). The total positive-sequence fault current is given by 1 1 IF1ðcÞ IF1 5 IFðgÞ

(6.75b)

Assuming the short-circuit fault occurs at zero voltage on the voltage waveform, then the ac grid fault current will also include a dc current component that can be calculated from the X/R ratio ‘seen’ at the point of fault. The current injected by the inverter is a symmetrical ac current with no dc current component. Therefore the total asymmetric fault current is given by iF ðtÞ 5


 pffiffiffi 1 2 2πft pffiffiffi 2IF ðgÞ 1 1 exp 1 2IF1ðcÞ fX=Rgg |fflfflffl{zfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ac grid contribution

(6.75c)

Inverter contribution



where X=R g is as given in Eq. (6.74c). Combining the ac fault current components, we have " # i pffiffiffi pffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 IF ðgÞ 1 IFðcÞ 1 2IFðgÞ exp X=R g

(6.75d)

1 1 5 0, IF1 5 IFðgÞ and For calculation of initial make fault current, we have IFðcÞ

" ( )# pffiffiffi 1 2 2πft

iF ðtÞ 5 2IFðgÞ 1 1 exp X=R g

(6.76a)

548

Power Systems Modelling and Fault Analysis

For the calculation of break fault current, we have " # i pffiffiffi pffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 IF ðgÞ 1 IFðcÞ 1 2IFðgÞ exp X=R g

(6.76b)

For the next iteration in the calculation of the fault current, the required inverter terminal voltage is calculated using Fig. 6.46C as follows:

Vc1 5 Ic1 Zt1 1 ZT1

(6.76c)

6.16.4 Analysis of single-phase short-circuit fault currents In deriving the single-phase short-circuit fault current at the 132 kV busbar, we consider three practical cases and derive the total symmetrical and asymmetrical fault current for each case.

Inverters do not contribute a fault current The inverter is represented as an open-circuit current source in the positive-, negative- and zero-sequence circuits. The three sequence equivalent circuits and their interconnection are shown in Fig. 6.47.

= 0

+

F

( )

F

( )

F

( )

= 0

= 0

+

( )

Figure 6.47 Analysis of one-phase faults: voltage-source inverters do not inject currents.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

549

The ac component of the short-circuit fault current is calculated as follows: IF 5 3IF1 5 3IF2 5 3IF0 5

3Vg1 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo

(6.77a)

and the asymmetric fault current is given by " !# pffiffiffi 2 2πft  iF ðtÞ 5 2IF 1 1 exp  X=R F

(6.77b)

  The Thevenin’s fault point X=R F ratio is calculated from the series combination of the positive-, negative- and zero-sequence fault point equivalent impedances. From Fig. 6.47, we obtain

  ZoZo Imaginary Zg1 1 Zg2 1 Z o g1 TZ o g  T X=R F 5 Zgo ZTo 1 2 Real Zg 1 Zg 1 Z o 1 Z o

g

(6.77c)

T

Inverters contribute positive-sequence current only The inverter is represented as a positive-sequence current source in the positivesequence circuit but as an open-circuit in the negative- and zero-sequence circuits. The three sequence equivalent circuits and their interconnections are shown in Fig. 6.48A. The positive-sequence fault current can be calculated using the superposition theorem.

Inverter current source is open-circuited The resultant equivalent circuit is shown in Fig. 6.48B. The positive-sequence fault current supplied by the ac grid with inverter current source open-circuited, that is, excluding the solar PV converter contribution is given by 1 IFðgÞ 5

Vg1 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo

(6.78a)

ac grid voltage source is short-circuited The resultant equivalent circuit is shown in Fig. 6.48C. The positive-sequence fault current supplied by the solar PV inverter with ac grid voltage source shortcircuited, that is, excluding the ac grid contribution is given by

550

Power Systems Modelling and Fault Analysis

(A)

+ ( )

F

( )

F

( )

F

( )

= 0

= 0

+

( )

(B)

(C) +

( )

( )

( )

F

F + ( )

+ ( )

( )

+

Figure 6.48 Analysis of one-phase faults: (A) voltage-source inverters inject positivesequence reactive current and (B and C) illustration of the application of the superposition theorem.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

Zg1 Ic1

1 IFðcÞ 5

Zg1 1 Zg2 1

551

(6.78b)

Zgo ZTo Zgo 1 ZTo

By superposition, the total positive-sequence fault current is given by 1 1 1 IFðcÞ IF1 5 IFðgÞ

IF1 5

(6.78c)

Vg1 1 Zg1 Ic1 Zg1 1 Zg2 1

(6.78d)

Zgo ZTo Zgo 1 ZTo

The single-phase fault current including ac grid and inverter contributions is given by 1 1 1 3IFðcÞ 5 IF 5 3IF1 5 3IFðgÞ

  3 Vg1 1 Zg1 Ic1 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo

(6.79a)

The Thevenin’s fault point X/R ratio is as given in Eq. (6.77c). The asymmetric single-phase fault current is given by iF ðtÞ 5


 pffiffiffih 1 i 2 2πft pffiffiffih 1 i 2 3IFðgÞ 1 1 exp 1 2 3IFðcÞ fX=RgF |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} ac Grid Contribution

(6.79b)

Inverter Contribution

Or, combining the ac fault current components, " # i pffiffiffih i pffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 3IFðgÞ 1 3IFðcÞ 1 2 3IFðgÞ exp X=R F

(6.79c)

For the next iteration in the calculation of the fault current, it can be shown that the inverter positive- and negative-sequence voltages, calculated using Fig. 6.48B and C, are given by 1

1

Vc1 5 Ic Zt1 1 ZT 1

Vc2 5

2 Zg2 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo ZoZo Zg1 1 Zg2 1 Z o g1 TZ o g T

Zgo ZTo Zgo 1 ZTo

Zg2 1

  Vg1 1 Zg1 Ic1

  Vg1 1 Zg1 Ic1

(6.79d)

(6.79e)

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Power Systems Modelling and Fault Analysis

Inverters contribute positive- and negative-sequence currents The inverter is represented as a positive-sequence current source in the positivesequence circuit, a negative-sequence current source in the negative-sequence circuit and an open-circuit in the zero-sequence circuit. The three sequence equivalent circuits and their interconnections are shown in Fig. 6.49A. Again, the positivesequence fault current can be calculated using the superposition theorem.

(A) + ( )

F

( )

( )

F

( )

+

= 0

(C) ( )

F

+

( )

F ( )

+

(B) +

( )

( )

F

+

( )

( )

( )

F

( )

= 0

( ) ( )

F

( )

Figure 6.49 Analysis of one-phase faults: (A) voltage-source inverters inject positive- and negative-sequence reactive currents and (B and C) illustration of the application of the superposition theorem.

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Inverter positive- and negative-sequence current sources are open-circuited The resultant equivalent circuit is as previously shown in Fig. 6.47. The positivesequence fault current supplied by the ac grid with both inverter current sources open-circuited, that is, excluding the inverter contributions, is given in Eq. (6.78a).

ac grid voltage source is short-circuited The resultant equivalent circuit contains two current sources as shown in Fig. 6.49B. Various methods can be used to solve for the required currents and voltages in large-scale power networks and these methods are presented in Chapter 9, Network equivalents and practical short-circuit current assessments in large-scale ac power systems. Here, we choose to use the straightforward superposition method. With the negative-sequence current source Ic2 open-circuited, the positivesequence fault current is given in Eq. (6.78b). With the positive-sequence current source Ic1 open-circuited, as shown in Fig. 6.49C, the positive-sequence fault current is given by 1 IFðc2Þ 5

Zg2 Ic2 Zg1 1 Zg2 1

(6.80a)

Zgo ZTo Zgo 1 ZTo

The total inverter current contribution is given by 1 1 1 IFðcÞ 5 IFðc1Þ 1 IFðc2Þ 5

Zg1 Ic1 1 Zg2 Ic2 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo

(6.80b)

Finally, by superposition, the total positive-sequence fault current supplied by the ac grid and inverters is given by 1 1 1 IFðcÞ IF1 5 IFðgÞ

(6.81a)

Using Eqs. (6.79a) and (6.80b), we obtain IF1 5

Vg1 1 Zg1 Ic1 1 Zg2 Ic2 Zg1 1 Zg2 1

(6.81b)

Zgo ZTo Zgo 1 ZTo

The single-phase fault current supplied by the inverter and ac grid is given by IF 5 3IF1 5

  3 Vg1 1 Zg1 Ic1 1 Zg2 Ic2 Zg1 1 Zg2 1

Zgo ZTo Zgo 1 ZTo

The Thevenin’s fault point X/R ratio is as given in Eq. (6.78c).

(6.81c)

554

Power Systems Modelling and Fault Analysis

Where Zg1 5 Zg2 , Eq. (6.81c) becomes IF 5 3IF1 5

h

i 3 Vg1 1 Zg1 Ic1 1 Ic2 2Zg1 1

(6.81d)

Zgo ZTo Zgo 1 ZTo

The asymmetric single-phase fault current is given by 
 pffiffiffih 1 i 2 2πft pffiffiffih 1 i iF ðtÞ 5 2 3IFðgÞ 1 1 exp 1 2 3IFðcÞ fX=RgF |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} ac grid contribution

(6.82a)

Inverter contribution

or, combining the ac fault current components, " # i pffiffiffih i pffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 3IFðgÞ 1 3IFðcÞ 1 2 3IFðgÞ exp X=R F

(6.82b)

For the next iteration in the calculation of the fault current, it can be shown that the inverter positive- and negative-sequence voltages, calculated using Figs. 6.48B, C and 6.49C are given by Vc1 5 Ic1



Zt1 1 ZT1



1

Zgo ZTo Zgo 1 ZTo ZoZo Zg1 1 Zg2 1 Z o g1 TZ o g T

Zg2 1

2 4V 1 1 Z 1 I 1 2 g g c

Zg1 Zg2 Zg1 1 Zg2 1

3 Zgo ZTo Zgo 1 ZTo

Ic2 5

(6.82c) 2

1

Vc2 5 Ic Zt1 1 ZT 2

Zg2 Zg1 1 Zg2 1

" Zgo ZTo Zgo 1 ZTo

! # o o Z Z g T Vg1 1 Zg1 Ic1 2 Zg1 1 o I2 Zg 1 ZTo c (6.82d)

6.16.5 Analysis of two-phase short-circuit fault currents In deriving the two-phase short-circuit fault current at the 132 kV busbar, we consider two practical cases and derive the total symmetrical and asymmetrical fault current for each case.

Inverters contribute positive-sequence current only The inverter representation is as described in Section 6.16.3. The two sequence equivalent circuits and their interconnection are shown in Fig. 6.50. The positivesequence fault current can be calculated using the superposition theorem.

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(A)

F

F

+

+

F

= 0

(B)

(C) ( )

+

+

+

F

+

( )

F

Figure 6.50 Analysis of two-phase faults: (A) voltage-source inverters inject positivesequence reactive current and (B and C) illustration of the application of the superposition theorem.

Inverter current source is open-circuited The resultant equivalent circuit is shown in Fig. 6.50B. The positive-sequence fault current supplied by the ac grid with inverter current source open-circuited, that is, excluding the solar PV inverter contribution, is given by 1 IFðgÞ 5

Vg1 Zg1 1 Zg2

(6.83a)

ac grid voltage source is short-circuited The resultant equivalent circuit is shown in Fig. 6.50C. The positive-sequence fault current supplied by the solar PV inverter with ac grid voltage source shortcircuited, that is, excluding the ac grid contribution, is given by 1 IFðcÞ 5

Zg1 Ic1 Zg1 1 Zg2

(6.83b)

By superposition, the total positive-sequence fault current is given by 1 1 1 IFðcÞ IF1 5 IFðgÞ

IF1 5

Vg1 1 Zg1 Ic1 Zg1 1 Zg2

(6.83c) (6.83d)

556

Power Systems Modelling and Fault Analysis

The fault current flowing on phase B for a fault on phases Y and B including ac grid and inverter contributions is given by  pffiffiffi 1 1 i j 3 V1 pffiffiffi 1 pffiffiffih 1 g 1 Zg Ic I F 5 j 3I F 5 j 3 I FðgÞ 1 I 1 FðcÞ 5 2 Z1 g 1 Zg

(6.83e)

or, where Zg1 5 Zg2 , ! pffiffiffi 3 Vg1 1 1 Ic IF 5 j 2 Zg1

(6.83f)

The Thevenin’s fault point X/R ratio is calculated with current sources opencircuited and voltage sources short-circuited. From Fig. 6.50A, we obtain

  Imaginary Zg1 1 Zg2   X=R F 5 Real Zg1 1 Zg2

(6.83g)

The asymmetric fault current is given by iF ðtÞ 5


 pffiffiffihpffiffiffi 1 i 2 2πft pffiffiffihpffiffiffi 1 i 2 3IFðgÞ 1 1 exp 1 2 3IFðcÞ fX=RgF |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ac grid contribution

(6.84a)

Inverter contribution

Or, combining the ac fault current components, " # i pffiffiffihpffiffiffi i pffiffiffipffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 3 IFðgÞ 1 IFðcÞ 1 2 3IFðgÞ exp X=R F

(6.84b)

For the next iteration in the calculation of the fault current, it can be shown that the inverter positive- and negative-sequence voltages, calculated using Fig. 6.50B and C, are given by

Vc1 5 Ic1 Zt1 1 ZT1 1 Vc2 5

i Zg2 h 1 Vg 1 Zg1 Ic1 1 2 Zg 1 Zg

i Zg2 h 1 1 1 V 1 Z I g c Zg1 1 Zg2 g

(6.84c)

(6.84d)

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

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Inverters contribute positive- and negative-sequence currents The inverter representation is as described in Section 6.16.3. The two sequence equivalent circuits and their interconnection are shown in Fig. 6.51A. Again, the positive-sequence fault current can be calculated using the superposition theorem.

Inverter positive- and negative-sequence current sources are open-circuited The resultant equivalent circuit is shown in Fig. 6.51B. The positive-sequence fault current supplied by the ac grid with both inverter current sources open-circuited, that is, excluding the inverter contributions, is as given in Eq. (6.82a).

ac grid voltage source is short-circuited The equivalent circuit contains two current sources. We first open-circuit the negative-sequence current source Ic2 and calculate the positive-sequence fault current using Fig. 6.51C as follows: 1 IFðc1Þ 5

Zg1 Ic1 Zg1 1 Zg2

(6.85a)

Next, we open-circuit the positive-sequence current source Ic1 and calculate the positive-sequence fault current using Fig. 6.51D as follows 1 2 IFðc2Þ 5 2 IFðc3Þ 5

2 Zg2 Ic2 Zg1 1 Zg2

(6.85b)

By superposition, the positive-sequence fault current supplied by the inverters is given by 1 1 1 IFðcÞ 5 IFðc1Þ 1 IFðc2Þ 5

Zg1 Ic1 2 Zg2 Ic2 Zg1 1 Zg2

(6.85c)

Finally, by superposition, the total positive-sequence fault current is given by IF1 5

Vg1 1 Zg1 Ic1 2 Zg2 Ic2 Zg1 1 Zg2

(6.85d)

The two-phase fault current flowing on phase B for a fault on phases Y and B including ac grid and inverter contributions is given by   i pffiffiffi Vg1 1 Zg1 Ic1 2 Zg2 Ic2 pffiffiffi 1 pffiffiffih 1 1 IF 5 j 3IF 5 j 3 IFðgÞ 1 IFðcÞ 5j 3 Zg1 1 Zg2

(6.85e)

558

Power Systems Modelling and Fault Analysis

(A)

F

F

F

(B)

F

F

=0

=0

(C)

F

F

=0

(D)

F

F

=0

Figure 6.51 Analysis of two-phase faults: (A) voltage-source inverters inject positive- and negative-sequence reactive currents, (B
D) illustration of the application of the superposition theorem.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

559

or, where Zg1 5 Zg2 , # pffiffiffi " pffiffiffi 1 j 3 Vg1 1

2 1 Ic 2 Ic I F 5 j 3I F 5 2 Zg1

(6.85f)

The Thevenin’s fault point X/R ratio is given in Eq. (6.83c). The asymmetric fault current is given by iF ðtÞ 5


 pffiffiffihpffiffiffi 1 i 2 2πft pffiffiffihpffiffiffi 1 i 2 3IFðgÞ 1 1 exp 1 2 3IFðcÞ fX=RgF |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ac grid contribution

(6.86a)

Inverter contribution

or, combining the ac fault current components, " # i pffiffiffipffiffiffi pffiffiffipffiffiffih 1 2 2πft 1 1

iF ðtÞ 5 2 3 IFðgÞ 1 IFðcÞ 1 2 3IFðgÞ exp X=R F

(6.86b)

For the next iteration in the calculation of the fault current, it can be shown that the inverter positive- and negative-sequence voltages, calculated using Fig. 6.50B and C, are given by

Vc1 5 Ic1 Zt1 1 ZT1 1



i Zg2 h 1 1 1 2 V 1 Z I 1 I g c c Zg1 1 Zg2 g

(6.86c)



Vc2 5 Ic2 Zt1 1 ZT1 1



i Zg2 h 1 1 1 2 V 1 Z I 1 I g g c c Zg1 1 Zg2

(6.86d)

Similar steps can be followed to derive the expressions for the two-phase to earth short-fault current. However, we leave this to the interested researcher/reader.

6.16.6 Examples Example 6.11 Fig. 6.52 shows a typical grid code reactive current injection requirement curve for a currentcontrolled voltage-source inverter. The curve shows the implemented dynamic reactive current support response in the case of voltage dips at the inverter terminals. Discuss if the inverter can be represented as a voltage source behind a reactance, like a synchronous generator, as opposed to a current source representation. Referring to our discussion in Section 6.12, the grid code curve can be mathematically described as

560

Power Systems Modelling and Fault Analysis

2

Synchronous generator curve with X” = 1/ k

I (pu)

1.6

1.1 pu Grid code curve

1.2 0.8

Vc

0.4 0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

V (pu) Figure 6.52 To compare the representation of current-controlled voltage-source inverter and a synchronous generator, used in Example 6.11.

I5

Imax k ðV o 2 V Þ

V , Vc V $ Vc

(6.87a)

where V o is the inverter initial voltage in pu; V is the inverter voltage during the voltage dip in pu; k is the gain that defines the slope of the current
voltage (IV) relationship; and Vc is the inverter terminal voltage level below which the inverter reaches and maintains a constant current equal to its transient limit Imax . Given the following inverter parameters: V o 5 1 pu, Imax 5 1:1 pu, k 5 2:2 and Vc 5 0:5 pu, Eq. (6.87a) becomes

I5

1:1 2:2ð1 2 V Þ

V , 0:5 V $ 0:5

(6.87b)

Eq. (6.87b) is shown as the grid code curve on Fig. 6.52. The k factor has a physical unit of a susceptance and its inverse is therefore a reactance. If the inverter is represented as a synchronous generator, that is, a voltage source behind a sub1 transient reactance equal to X 00 5 1k 5 2:2 5 0:4545 pu, then the following equation for a synchronous generator applies I5

1 ð1 2 V Þ X00

V ,1

(6.87c)

The synchronous generator curve is shown on Fig. 6.52. We conclude that the inverter can be represented as a voltage source behind a reactance equal to 1=k for V $ 0:5 pu, that is, for more remote faults. However, for V , 0:5 pu, that is, faults that are electrically closer to the inverter, a synchronous generator representation of a voltage source behind a reactance equal to 1=k results in an overestimate in the short-circuit reactive current infeed of the inverter. The closer the electrical distance from the inverter to the fault, the larger the overestimate.

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Example 6.12 Consider the ac grid network shown in Fig. 6.53A. The grid consists of current-controlled voltage-source inverters that are designed to inject positive-sequence reactive currents. There are no conventional synchronous generators in this grid. Discuss the fault current characteristics for the balanced three-phase fault shown at F. For simplicity, the load current is neglected. Fig. 6.53B shows the three-phase equivalent circuit with balanced three-phase current sources producing positive-sequence currents only. Using the superposition theorem and Fig. 6.53C, the fault current is given by 1 1 I1 F 5 IFð1Þ 1 IFð2Þ 5

2 jXc1 I1 2 jXc2 I1 1 1 1 2 jXc1 1 Z1 2 jXc2 1 Z2

In practical networks, the capacitive reactances of lines and cables dominate compared to the series inductive impedances of lines, cables and transformers. Hence 1 1 I1 F DI1 1 I2

Therefore the currents injected by the inverters minus the negligibly small network capacitive charging currents are fed to the fault. (A)

VSC1

Fault F

VSC2

(B)

(C)

Figure 6.53 Three-phase fault in a power system containing voltage-source inverters only, used in Example 6.12: (A) System, (B) illustration of fault in a three-phase circuit and (C) positive sequence equivalent circuit.

562

Power Systems Modelling and Fault Analysis

Example 6.13 Using the example ac grid network shown in Fig. 6.53A, discuss the fault current characteristics for an unbalanced two-phase fault at F. Consider two cases where the inverters are designed to inject only positive-sequence currents, or both positive- and negative-sequence currents. Case 1: Inverters inject balanced phase currents, that is, positive-sequence currents The three-phase system representation with a two-phase fault is shown in Fig. 6.54A and the inverters are represented by balanced three-phase current sources. The sequence network representation is shown in Fig. 6.54B. The expressions of the positive- (and negative-) sequence 2 fault current I1 F (and IF ) can be easily calculated but we leave this simple exercise for the interested reader. Again, with the dominant capacitive reactances, we can easily write 2 I1 F 5 2 IF D

1 1 1

I 1 I2 2 1

and the phase fault current is given by pffiffiffi IF 5 j 3I1 F Dj

pffiffiffi 3 1 1

I 1 I2 2 1

However, it is important to note that these positive- and negative-sequence fault currents and hence phase fault current can only flow in the shunt capacitances of the network as

(A)

(B)

Positive-sequence network

Negative-sequence network

Figure 6.54 Two-phase fault in a power system containing voltage-source inverters only, which inject only positive-sequence currents, used in Example 6.13: (A) Illustration of the fault in a three-phase circuit and (B) connection of positive and negative sequence networks.

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

563

illustrated on Fig. 6.54B. If, theoretically, these capacitances are considered nonexistent, no shunt path would exist and hence no fault current could flow. Case 2: Inverters inject positive- and negative-sequence currents. The three-phase system representation of Fig. 6.55A shows three-phase inverter currents that are not balanced. The sequence network is shown in Fig. 6.55B. The expressions of the pos2 itive- (and negative-) sequence fault current I1 F (and IF ) can be easily calculated but, again, we leave this simple exercise to the interested reader. Again, with the dominant capacitive reactances, we can easily obtain 2 I1 F 5 2 IF D

1 1 1 1 2 2

I 1 I2 2 I1 1 I2 2 1 2

and the phase fault current is given by pffiffiffi pffiffiffi 1 3  1 1 2 2 I1 1 I2 2 I1 1 I2 IF 5 j 3IF Dj 2 However, as in Case 1, these positive- and negative-sequence fault currents and hence phase fault current can only flow in the shunt capacitances of the network as illustrated in Fig. 6.55B. If, theoretically, these capacitances are considered nonexistent, no shunt path would exist and hence no fault current could flow. In other words, the feeding of both positive- and

(A)

(B)

Positive-sequence network

Negative-sequence network

Figure 6.55 Two-phase fault in a power system containing voltage-source inverters only, which inject positive- and negative-sequence currents, used in Example 6.13: (A) Illustration of the fault in a three-phase circuit and (B) connection of positive and negative sequence networks.

564

Power Systems Modelling and Fault Analysis

(A)

(B)

Positive-sequence network

Negative-sequence network

Figure 6.56 Two-phase fault in a power system containing voltage-source inverters only which inject only positive-sequence currents and including induction motor loads, used in Example 6.13: (A) Illustration of the fault in a three-phase circuit and (B) connection of positive and negative sequence networks. negative-sequence currents by inverters in a system that consists of only inverters provides no additional benefits under two-phase fault conditions. In practical power system networks supplying mixed loads, the load mix includes directly connected induction motors mostly connected to the LV network and some connected to the MV network. As we presented in Chapter 5, Modelling of rotating ac synchronous and induction machines, these motors supply a positive-sequence current and present a low negative-phase sequence impedance under unbalanced faults. The motor’s initial positive-sequence and negative-sequence impedances are equal to the motor’s transient impedance. The positivesequence reactance increases with time whereas the negative-sequence reactance is constant. Therefore an equivalent impedance that represents the aggregate motors’ impedance in series with the intervening network impedances should be included in the positive- and negativesequence network representation. To illustrate, consider the network of Fig. 6.56A supplying a load at point F through stepdown transformers. The equivalent sequence network of Fig. 6.53B changes to that shown in Fig. 6.56B. This shows that the induction motor load will initially feed a small positive-sequence fault current and will present a shunt path in the negative-sequence circuit that allows the flow of some negative-sequence current. Overall, this effect is relatively small unless the induction motor composition within the load mix is very high over the whole system.

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Example 6.14 In Fig. 6.53A, replace the second inverter VSC2 with a conventional synchronous generator as shown in Fig. 6.57A. The generator is represented as a balanced three-phase voltage source. Discuss the fault current characteristics for the unbalanced two-phase fault shown at F. For simplicity, the load current is neglected but its effect is as discussed in Example 6.13. The three-phase circuit representation is shown in Fig. 6.57B and the sequence equivalent circuit is shown in Fig. 6.57C. The positive-sequence, negative-sequence and phase fault currents can easily be derived using the superposition theorem but we leave this to the interested reader. Fig. 6.57C shows that the negative-sequence circuit has a low impedance shunt path that corresponds to the generator negative-sequence impedance. In this circuit topology, the presence of the conventional voltage-source synchronous generator ensures that a large negative-sequence current and phase fault current flow even though the converter injects positive-sequence current only. In other words, in this example, we show that the feeding of negative-sequence current by the inverter provides no additional benefits.

(A)

VSC1

Two-phase fault F

Synchronous generator

(B)

(C)

Positive-sequence network

Negative-sequence network

Figure 6.57 Two-phase fault in a mixed power system of voltage-source inverters and synchronous generation, voltage-source inverters inject only positive-sequence currents, used in Example 6.14: (A) One-line diagram, (B) illustration of the fault in a three-phase circuit and (C) connection of positive and negative sequence networks.

566

Power Systems Modelling and Fault Analysis

Example 6.15 Consider the 100 MW solar PV power park shown in Fig. 6.45A. The data of this system is: Inverter: 2.2 MVA, 0.6 kV, reactor impedance 15% on 2.2 MVA and X/R 5 50, no. of inverters 5 50, plant rating 110 MVA, 100 MW, inverter transient current limit 1.1 pu. Inverter transformer: 0.6/33 kV, 2.2 MVA, impedance 10% on 2.2 MVA and X/R ratio 5 10. Plant transformer: 33/132 kV, 100 MVA, positive-sequence impedance 17.75% and X/R ratio 5 43, zerosequence impedance 16.5% and X/R ratio 5 30.7. 132 kV cable: Length 1.3 km, positive-sequence impedance (0.033 1 j0.1245) Ω/km, zero-sequence impedance (0.099 1 j0.06586) Ω/km. 132 kV ac grid fault infeeds: Source voltage 1 pu, three-phase fault current infeed: 20 kA, X/R 5 57; one-phase fault current infeed: 24 kA, X/R 5 48. The k factor of the inverters dynamic current injection is k 5 2. Calculate the short-circuit fault currents for a solid fault on the 132 kV busbar. The time of interest is 50 ms after fault inception and the inverters are assumed to have reached their steady-state reactive current magnitude in accordance with Eq. (6.75). The fault currents are to be calculated for threephase, two-phase and one-phase to earth faults. For the unbalanced faults, consider inverters that supply positive-sequence current only, and both positive- and negative-sequence currents with equal priority. The equivalent circuit of the 100 MW solar PV power plant of Fig. 6.45A is shown in Fig. 6.58. Solution We will use the equations developed in Sections 6.16.3
6.16.5 for the cases of three-phase fault, one-phase fault and two-phase fault, respectively. Base MVA and voltage are 100 MVA and 132 kV. 132 kV cable: positive-sequence and zero-sequence impedances are (0.0246 1 j0.0929)% and (0.0738 1 j0.0491)%. ac grid source equivalent: positive-sequence and negative-sequence impedance are (0.03836 1 j2.186)%, zero-sequence impedance is (0.03715 1 j1.093)%.

Fault 33 kV 132 kV Grid

110 MVA VSC 100 MW PV

Figure 6.58 Equivalent network used in Example 6.15.

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567

Three-phase fault  ac grid current infeed 5 45:73e2j89 pu. Immediately upon fault occurrence, inverters ‘see’ a  voltage change of 2100% and their steady-state current is 1:1e2j90 pu. The total ac fault cur   rent is 46:83e2j89 pu or 20:48e2j89 kA. The inverter current of 1:1e2j90 pu raises the inverter  positive-sequence terminal voltage to 0:3e2j2:8 pu and represent the final solution since the change in voltage remains greater than 0:55 pu. Two-phase fault The calculations of the short-circuit fault currents in the case of the inverters feeding positivesequence current only are left for the interested reader. The solution for the case where the inverters feed both positive- and negative-sequence currents of equal priority is given below.  The positive-sequence fault current from the ac grid is 22:86e2j89 pu. The change in positive- and negative-sequence voltages at the fault point and at the inverter terminal immediately upon fault occurrence are 2 0:5 and 0:5 pu; respectively. With equal priority and k 5 2, the injected positive- and negative-sequence inverter currents are 2 j0:55 and j0:55 pu, respectively. The positive- (negative-) sequence fault current due to injected positive- (negative-) sequence inverter current is 2 j0:275 pu ðj0:275 puÞ and the total positive-sequence fault current from the inverters is 2 j0:55 pu. The total positive-sequence fault current from the ac grid and inver   ters is 23:4e2j89 pu and the total phase B fault current is 40:55ej0:98 pu or 17:74ej0:98 kA. The injected positive-sequence inverter current of 2 j0:55 pu raises the inverter positivesequence terminal voltage from 0:5 to 0:65 pu. Conversely, the injected negative-sequence inverter current of j0:55 pu reduces the inverter negative-sequence terminal voltage from 0:5 to 0:35 pu. The inverter dynamic reactive current controller responds to these voltage changes and adjusts the injected positive- and negative-sequence currents. In this case, the full steady-state current magnitudes will still be supplied.  The phase voltages at the inverter terminal are calculated as vr 5 1ej0 pu,   vy 5 0:57e2j153:5 pu and vb 5 0:55ej152:4 pu. One-phase to earth fault We will provide the solution of the short-circuit fault currents for the case where inverters feed both positive- and negative-sequence currents.  The positive-sequence fault current from the ac grid is 18:56e2j88:8 pu. The positive- and negative-sequence voltages at the inverter terminal immediately upon fault   occurrence are calculated as 0:59e2j0:1 pu and 0:41e2j179:8 pu giving equal magnitude 2j179:8 changes of 0:41e . With k 5 2 and equal priority, the injected positive- and negativesequence currents are each equal to 2 j0:55 pu. The positive- and negative-sequence fault currents due to injected positive- and negative sequence inverter currents are each equal to 0:223e2j89:8 pu and the total positive-sequence 2j89:8 fault current from the inverters is 0:446e pu. The total positive-sequence current flowing  into the fault from the ac grid and inverters is 19:0e2j88:8 pu and the total phase R fault current   is 56:9e2j88:8 pu or 24:9e2j88:8 kA. The injected positive- and negative-sequence inverter currents of 2 j0:55 pu raise the inverter positive-sequence terminal voltage from 0:59 to 0:75 pu and reduce the inverter negative-sequence terminal voltage magnitude from 0:41 to 0:27 pu. The inverter dynamic reactive current controller responds to these voltage changes and adjusts the injected positive- and negative-sequence currents to 2 j0:56 and 2 j0:54 pu, respectively. The phase voltages at the inverter terminal immediately upon fault occurrence are    vr 5 0:19e2j0:76 pu, vy 5 0:87e2j96:27 pu, vb 5 0:872ej96:25 pu and after inverter current injec   tions vr 5 0:477e2j2:0 pu, vy 5 0:90e2j105:3 pu, vb 5 0:92ej105 pu.

568

6.17

Power Systems Modelling and Fault Analysis

Grid-forming voltage-source inverters

6.17.1 Background to grid-following and grid-forming inverters In Sections 6.10 and 6.11, we presented modelling and analysis of the currently dominant technology of grid-connected voltage-source inverters which are designed to operate as controlled, balanced or unbalanced three-phase current sources. These inverters are also known as grid-following since they follow the already-established voltage and frequency of the ac grid. In subsequent sections, we present the emerging technology of voltage-source inverters that are controlled in such a way that they operate as controlled, balanced, three-phase voltage sources behind an impedance during steady-state and transient conditions that involve relatively small over-currents. These inverters are also known as grid-forming inverters. Broadly speaking, grid-forming inverters can be controlled to mimic or replicate certain features of real synchronous generators and there are known as virtual synchronous machines (VSM or VISMA) inverters. Other grid-forming inverters employ droop control strategies. The latter are used in low-voltage microgrid applications and small island power systems. However, they can also be applied in large interconnected ac grids as an alternative to VSM inverters. Grid-forming inverters may have a flexible control system that enables seamless transition from grid-connected to islanded modes of operation. When operating as a balanced three-phase voltage-source, grid-forming inverters present their filter reactor impedance in the positive-sequence system and this same impedance as an inherent negative-sequence impedance in the negative-sequence system. This means that, just like real synchronous machines, the inverter supplies without delay inherent positive-sequence and negative-sequence currents under unbalanced network conditions. The latter reduces negative-sequence voltages and hence grid phase voltage unbalance.

6.17.2 Emerging challenges in power systems dominated by grid-following voltage-source inverters New challenges in the operation of power systems with high penetration of gridfollowing inverter-interfaced generators have recently emerged in a number of power systems and these issues are expected to appear in other power systems as the rate of penetration of current-controlled inverters increases. Some of the main challenges are presented next. Grid-following current-controlled voltage-source inverters do not contribute to the overall power system inertia. Only rotating machines make this contribution through the stored kinetic energy in their rotating masses. Reduced system inertia causes an increase in the initial rate of change of system frequency following sudden tripping of generation power infeed or load. This introduces a risk of further and cascade generation trippings that could lead to widespread customer demand disconnections. Grid-following inverters cannot supply an instantaneous or half cycle short-circuit current like rotating machines that have a stored electromagnetic energy. These

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569

inverters supply almost zero current at half cycle after the occurrence of a voltage dip or short-circuit fault on the ac grid due to the need to measure, process and detect the voltage dip, switch to reactive current injection mode then deliver the response of the inner current control system as discussed in Sections 6.11, 6.13 and 6.16.2. The delay in the supply of reactive current in the first 15
20 ms in inverter dominated power systems might slightly prolong the operation time of some types of protection relays. However, the practical significance of a short delay of one to two cycles will vary among power systems. The effect will be negligible in modern power systems that already employ very fast circuit breakers with opening times of around one cycle compared to three- or four-cycle circuit breakers. Another modern practice is the removal of trip relays and hence the reduction of overall protection time of 10
20 ms. At the time of writing, further research is required in this area. Both grid-following and grid-forming inverters make a small and limited contribution to short-circuit current levels on the power system. Their maximum steadystate short-circuit current contribution of typically 1.1
1.2 pu of rated current can be contrasted with 2.2
3.5 pu supplied by synchronous machines for faults on the higher voltage side of their generator-transformers. Grid-following inverters do not allow operation in very weak parts of networks because the extent of dq cross-coupling discussed in Section 6.10 becomes amplified besides the issues caused by network voltage unbalance and harmonic voltage distortion. Also, operation of such inverters alone, that is, without synchronous generation, in islanded networks is not possible. There are other technical challenges in the operation of power systems and particularly small- to medium-sized systems that result from high penetrations of gridfollowing inverter-interfaced systems. These include voltage stability, synchronising torques, supersynchronous control interactions, increased phase voltage unbalance and harmonic amplifications. However, these topics are outside the scope of this book. We will therefore restrict our attention to the short-circuit current contribution of inverter-interfaced generators.

6.17.3 Control structure of grid-forming inverters At the time of writing, the application of grid-forming VSM inverters and inverters based on droop control strategies in large utility-grid is still emerging although the latter are now being offered by some manufacturers. However, inverters based on droop control strategies have already been implemented in microgrids and small island power systems that comprise battery energy storage, wind and/or solar PV generators and/or conventional diesel generators. Importantly, and as discussed in Section 6.17.1, currently, there is no consensus on the preferred grid-forming inverter approach, VSM or droop-control based, in large-scale power systems. Moreover, for VSM inverters, there is no consensus on a unified set of features of a real synchronous machine a VSM inverter should emulate. The literature shows varying complexity of synchronous machine models used. Accordingly, there are significant differences in the implementation of control system structures and designs of VSM inverters.

570

Power Systems Modelling and Fault Analysis

Solar PV

Battery ES Wind

dc/dc

dc

dc/dc VSC ac/dc

Grid Gate signals

HVDC

Grid-forming algorithm and control

Fuel cell, etc.

Figure 6.59 Conceptual grid-forming voltage-source inverter structure.

Some research proposes a control system that essentially provides a delayed ‘synthetic’ inertia rather than the true instantaneous inertial response of a real synchronous machine. Some implementations use an inner current-control loop requiring current references while others do not. Some use a PLL in various forms for deriving the grid voltage angle and/or frequency and some have dispensed with the PLL apart from initial synchronisation to the grid, like a real synchronous machine. Some research dispenses completely with the PLL and proposes that initial synchronisation to the grid may be done by the inverter control system. At the time of writing, manufacturers and suppliers of grid-connected voltagesource inverters are beginning to embrace the utilities requirement for gridconnected, grid-forming inverters in large-scale power systems. As with other similar historical developments, it is inevitable that the preferred functional performance, structure and design of grid-forming inverters will emerge over the next few years. Grid-forming inverters can interface a range of technologies to the ac grid such as type 4 wind turbine generators, solar PV generators, battery energy storage, fuel cells, certain distributed energy resources, HVDC links, etc. Fig. 6.59 illustrates the conceptual structure of a grid-forming inverter and various potential applications.

6.18

Grid-forming virtual synchronous machine (VSM) inverters

6.18.1 Possible VSM inverter features of real synchronous machines The dynamic behaviour of a real synchronous machine is exhibited through the dependence of grid frequency on machine rotor speed, the machine having a

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571

rotating mass and stored kinetic energy, delivering real inertial response without delay, controlling active power/frequency and reactive power/voltage, providing synchronising and power oscillation damping torques, and enabling seamless parallel operation of many machines in a grid of any size. Under grid voltage dips, a real synchronous machine feeds a short-circuit current to the fault that is dependent on the external impedance to the fault point immediately and without any delay. Like a real synchronous machine, a grid-forming VSM inverter should have a virtual rotor with mass and damping and exhibits an inertia constant whose value can be set by the user. Such a VSM inverter should also feed without delay positive- and negative-sequence short-circuit currents under network voltage dips. The VSM should have an inherent negative-sequence impedance that allows the flow of negative-sequence current from the network. Moreover, and like a real synchronous machine, the grid-forming VSM inverter should contribute to the damping of harmonic voltage distortion on the network by presenting an inherent frequencydependent harmonic impedance. Therefore as seen from the grid, and apart from high-frequency harmonics in the IGBT bridge output voltage, the electrical behaviour of a grid-forming VSM inverter should replicate some of the essential characteristics of a real electromechanical synchronous machine. To achieve this objective, the control strategy of a grid-forming VSM inverter should control the amplitude, frequency and phase angle of the inverter output voltage. It may also be desirable to be able to synchronise and operate without the need for a PLL. The PLL can produce erroneous signals under large three-phase grid voltage dips at the inverter terminal and it can go unstable in weak power grids with a high grid impedance. Consequently, the removal of a PLL from a gridforming inverter may be desirable.

6.18.2 A model of grid-forming VSM inverters The common feature found in almost every grid-forming VSM inverter implementation is the representation of a real synchronous machine rotor swing dynamics and hence inertial response and sometimes electrical damping. The delivery of a predefined inertial response of typically H 5 2 MWs/MVA to H 5 8 MWs/MVA requires a suitable and sufficient amount of energy storage on the inverter’s dc busbar which may take the form of ‘power’ rather than ‘energy’ batteries or ultracapacitors that can store the required amount of energy over the delivery timescale required which may be in the order of 10
25 s in practical power systems. In this edition of our book, we have chosen a grid-forming VSM inverter control structure that is based on a third-order dynamic model of a synchronous machine and hence the following subset of functionalities: 1. a virtual governor-turbine model; 2. a virtual rotor model with inertia and damping; and 3. a virtual field or internal voltage model.

This grid-forming VSM inverter control structure is shown in Fig. 6.60.

572

Power Systems Modelling and Fault Analysis

PWM switching

Park transformation Virtual field voltage model

Inverse park

PQ measurements Virtual rotor model

Virtual governorturbine model

Figure 6.60 Possible virtual synchronous machine (VSM) inverter control structure.

+

+

−

− − −

+

Virtual rotor

Virtual rotor Virtual governorturbine

+

Virtual field voltage

Virtual governorturbine

Virtual field voltage

Figure 6.61 Representation of a three-phase grid-forming virtual synchronous machine (VSM) inverter.

Using Fig. 6.28 of a three-phase voltage-source inverter, a three-phase gridforming VSM inverter may be represented as shown in Fig. 6.61A and B, where the three-phase IGBT bridge is represented by a balanced three-phase voltage source behind a fixed impedance equal to the inverter actual reactor impedance.

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573

VSM inverter’s virtual governor-turbine model The virtual governor-turbine model of the VSM inverter can be arbitrarily chosen to represent the dynamics of a heavy-duty gas turbine but without any complexities such as ambient temperature dependence, etc. The virtual governor-turbine model uses a steady-state speed or frequency droop slope Df and low-pass filters representing a governor with a time constant Tg and a gas turbine with a time constant Tt . The dynamic equations can be written as ref

ref

Pm 5 Pe 1

 1  ref f 2f Df

(6.88a)

Tg

dP v ref 5 Pm 2 Pv dt

(6.88b)

Tt

dP m 5 Pv 2 Pm dt

(6.88c)

or using the Laplace operator S 5 d=dt, we have Pv 5

1 ref P 1 1 STg m

(6.89a)

Pm 5

1 Pv 1 1 STt

(6.89b) ref

where f is virtual frequency of the VSM inverter in pu; f is frequency set point in pu; Df is frequency droop slope in pu frequency deviation per pu power deviaref tion or in physical units of Hz/MW; P is electric power set point in pu; andP m is virtual mechanical power of VSM inverter in pu. Fig. 6.62A shows the transfer function block diagram of the virtual governorturbine of the VSM inverter.

VSM inverter’s virtual rotor model with inertia and damping The virtual rotor model of the VSM inverter is based on the electromechanical swing equation of a real synchronous generator and, with electrical damping, is given by

dΔω 1 5 P m 2 P e 2 kdm Δω dt 2H

(6.90a)

dδ 5 ωs Δω dt

(6.90b)

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Power Systems Modelling and Fault Analysis

(A) +

1

Σ

−

+

1

1

Σ +

̅ (B)

+

−

1

Σ

+

1

Σ

+

−

+

1

+

Σ

(C) − +

Σ

+ −

Σ

̅

1

̅

1

Figure 6.62 Possible virtual synchronous machine inverter controllers: (A) virtual governorturbine model, (B) virtual rotor model with electrical damping and (C) virtual AVR field voltage model.

or, using the Laplace operator S 5 dtd , Eq. (6.90) can be written as Δω 5 δ5

1 P m 2 P e 2 kdm Δω 2HS

ωs Δω S

(6.91a) (6.91b)

and δr 5 δ 1 kdd ωs Δω

(6.91c)

where t is time in seconds; ω is VSM virtual rotor angular speed in pu; H is VSM virtual inertia constant in MWs/MVA; P m is VSM virtual mechanical power in pu; P e is VSM measured electric power in pu; kdm is mechanical damping factor term in pu power per pu speed deviation; kdd is electrical damping factor term in pu angle per pu speed deviation; δ is VSM virtual rotor angle in electrical radian; and ωs 5 2πfs is rated angular speed in electrical rad/s. The beneficial damping of the physical damper windings of a real synchronous machine is replicated by the addition of a feed-forward term that superimposes a virtual rotor angle change proportional to rotor speed deviation through a virtual damping factor kdd in accordance with Eq. (6.91c). This factor can be set to a high value to achieve a targeted level of high damping that may even dispense with the need for additional power system stabilisers that are used as part of the excitation control systems of real synchronous machines.

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575

Eq. (6.91) can be represented as a transfer function block diagram as shown in Fig. 6.62B.

VSM inverter’s virtual field or internal voltage control model The virtual field or internal voltage regulator model of the VSM inverter is chosen to represent a high static gain or low-voltage droop, a fast-acting static exciter and the field winding of a real synchronous machine. In this model, the measured reactive power output is multiplied by the voltage droop slope Dv then subtracted from the voltage error which is the voltage reference minus the measured feedback voltage. The error signal is then taken directly to the exciter with a low-pass filter with a time constant Te then to the virtual field winding with a time constant Tf . The dynamic equations can be written as e ref rotor 5 V

ref

2 V 2 Dv Q

(6.92a)

Te

de x 5 e ref rotor 2 e x dt

(6.92b)

Tf

de rotor 5 e x 2 e rotor dt

(6.92c)

or using the Laplace operator S 5 d=dt, we have ex 5

1 e ref 1 1 STe rotor

(6.93a)

1 ex 1 1 STf

(6.93b)

e rotor 5

ref

where V is terminal voltage of VSM inverter in pu; V is voltage set point in pu; Dv is voltage droop slope in pu voltage deviation per pu reactive power deviation; Te is virtual VSM exciter time constant in seconds; and Tf is virtual VSM field time constant in seconds. Fig. 6.62C shows the transfer function block diagram of the virtual field voltage of the VSM inverter.

6.19

Grid-forming voltage-source inverters using droop control

As discussed in Section 6.17, grid-forming inverters employing a control strategy based on frequency and voltage droops or slopes schemes are an alternative to gridforming VSM inverters. The potential use of these inverters with sufficient dc

576

Power Systems Modelling and Fault Analysis

Grid-forming inverter

A

Figure 6.63 Power flow from a grid-forming inverter through an inductive impedance.

energy storage in large utility-scale wind and solar PV power generation systems presents an attractive application on technical grounds and perhaps economic grounds in the not too distant future. Since the inverter lacks physical inertia, the basis of droop control can be derived from the general relationship among active power, frequency, voltage and reactive power with powers flowing through an inductive impedance as shown in Fig. 6.63. Using phasor notation, the grid-forming inverter is represented as a voltage source E 5 Eejδ and supplies an apparent power S 5 P 1 jQ 5 VI to the ac bus through a complex inductive impedance Z 5 R 1 jX 5 Zejϕ . Using I 5 ðE 2 VÞ=Z and V 5 V+0 degrees, we can write after a little algebra 

  V R P 5 2 Q Z X

X 2R

Z 2 5 R2 1 X 2



Ecosδ 2 V Esinδ

 (6.94a) (6.94b)

Eq. (6.94a) shows that P and Q are coupled through R and X if neither is negligibly small. This means that both active and reactive powers are dependent on both voltage difference and power angle. This is the case in low-voltage microgrids where the network resistance can be larger than the reactance but both are significant. There are several droop control approaches proposed in the literature to resolve the power coupling challenges in microgrids that discuss conventional droop, reverse/opposite droop, indirect operation, virtual impedance and adaptive droop. However, such a discussion is outside the scope of this book. In our case, we focus on the potential application of droop control on inverters for use in higher voltage grids that are characterised by high X=R ratios so that the resistance R can be neglected. Substituting R 5 0 in Eq. (6.94), we obtain the following decoupled equations P5

EV sinδ X

(6.95a)

Q5

V ðEcosδ 2 V Þ X

(6.95b)

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and for a small power angle, sinδDδ and cosδD1, we can write δD

X P EV

E 2 VD

X Q V

(6.96a) (6.96b)

Eq. (6.96b) shows that by regulating reactive power flow Q, the voltage difference and hence ac bus voltage V can be controlled. Also, Eq. (6.96a) shows that by regulating active power flow P, the power angle δ can be controlled. Droop control mimics speed governor control on conventional synchronous power plant which provides speed or frequency regulation where frequency is ‘drooped’ against active power. At the same time, the voltage V is drooped against the measured reactive power Q. Using Eq. (6.96), the conventional f =P and V=Q droop equations for grid-forming inverters are given by

f 5 f ref 1 Df P ref 2 P 0 P0 5 δ5

1 P 1 1 STt

fs f S



V 5 V ref 1 Dv Q ref 2 Q 0 Q0 5

1 Q 1 1 STex

(6.97a) (6.97b)

(6.97c) (6.98a) (6.98b)

From Eqs. (6.97) and (6.98), we obtain e ref i ðtÞ 5 V sinð2πft 1 δ 1 αi Þ

(6.99)

where f ref , fs and f are set point, nominal and measured frequencies, respectively; P ref and P are set point and measured powers, respectively; Q ref and Q are set point and measured reactive powers, respectively;Df and Dv are frequency and voltage droop slopes as defined previously; and e ref is reference internal voltage of the i inverter generated by a voltage-controlled oscillator using V , f and δ, and αi 5 0; 232π and 2π 3 for phase r, y and b respectively.

Δf =fs ΔV=Vn Df 5 2ΔP=P where Pr is rated power, Dv 5 2ΔQ=Q where Vn is nominal voltage r r and Qr is rated reactive power. In large grids, typical droop slopes are 3%
4% for Df and 2%
4% for Dv . For example, in the context of droop controlled inverters,

578

Power Systems Modelling and Fault Analysis

Df 5 3%, means that a 100% change in power P causes a 3% change in frequency f , and Dv 5 4% means that a 100% change in reactive power Q causes a 4% change in voltage V. Eqs. (6.97) and (6.98a) are illustrated graphically in Fig. 6.64A and B, respectively. A negative Pr is used to illustrate a battery source for the grid-forming inverter. Similar to a VSM inverter, Tt and Tex time constants in Eqs. (6.97b) and (6.98b) can be selected to represent the mechanical time constants of a steam or gas turbine, and a modern fast-acting static exciter for a synchronous generator, respectively. Fig. 6.65 shows a simple control structure of a grid-forming inverter employing droop controls and the necessary over-current limitation strategy required similar to a VSM grid-forming inverter.

(A)

(B)

0

0

Figure 6.64 Illustration of droops of a droop-controlled grid-forming inverter: (A) f =P droop and (B) V=Q droop.

PWM switching

Control and fault current limitation

+

+

Σ + Σ + +

Σ

+

1 −

Σ −

1

Voltage/reactive power droop control

Figure 6.65 Possible grid-forming inverter droop controllers.

Instantaneous PQ measurements

Frequency/power droop control

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With droop controls, a voltage-source inverter becomes grid-forming, acts as a voltage source and provides inherent inertia, frequency and voltage control. Moreover, and like conventional synchronous generators, the use of similar droops enables parallel inverters to share their active and reactive power outputs in proportion to their ratings. This droop control is simple and reliable and requires no communication among inverters, and, arguably, simpler than the control structure of a VSM inverter. The supervisory controller of each inverter provides settings for the reference frequency, reference voltage and slopes.

6.20

Natural short-circuit current contribution of gridforming inverters

Since the inverter is controlled to behave as a true three-phase balanced voltage source, the fault current output under voltage dips on the ac grid is instantaneous and its phase angle with respect to the inverter initial voltage is determined by the network characteristics, namely the external inverter impedance to the fault point. This characteristic is inherently highly inductive in grid-connected inverters although it may be more resistive in low-voltage microgrids as well as marine and aeronautical power systems. The inverter can inherently deliver the desirable instantaneous short-circuit current during the fault detection period without the delay necessitated by measurements and controls required by grid-following inverters.

6.20.1 Natural three-phase short-circuit current of grid-forming inverters Fig. 6.66A shows a grid-forming inverter represented as a balanced three-phase voltage source behind a fixed impedance equal to its output filter impedance R 1 jωs L. The natural short-circuit current contribution of the inverter is the potential current that would be supplied without any current limitation. Earth faults are not considered due to the absence of a zero-phase sequence path. (A)

(B) −

+

−

+

−

+

−

+

−

+

−

+

Figure 6.66 Three-phase faults at grid-forming inverter (A) terminals and (B) through an external grid impedance.

580

Power Systems Modelling and Fault Analysis

The fundamental frequency component of the balanced three-phase voltage source of the inverter is given by pffiffiffi (6.100) ei ðtÞ 5 2Esinðωs t 1 δi Þ i 5 r; y; b where E is internal rms voltage of inverter; ωs 5 2πfs in rad/s, fs is power frequency in Hz; and δi is internal voltage phase angle of inverter in radian. δy 5 δr 2 2π 3 and δb 5 δ r 1 2π=3 For a solid three-phase short-circuit fault at the inverter terminals, the natural short-circuit current is given as 2 0 13 9 8 > > > > ω L > > > sin4ωs t 1 δi 2 tan21 @ s A5 > > > > > pffiffiffi > > R = <

2E
 (6.101) ii t 5 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 0 13 L > > 2t= R > R2 1 ðωs LÞ2 > > > ω s L A5 > > > > > > 2 sin4δi 2 tan21 @ e > > ; : R The natural fault current consists of a steady-state ac current component and a transient dc current component that decays exponentially with a dc time constant equal to L=R. In the case of a fault through an external grid impedance Re 1 jωs Le , shown in Fig. 6.66B, the fault current can be calculated using Eq. (6.101) by replacing R with ðR 1 Re Þ and L with ðL 1 Le Þ.

6.20.2 Natural two-phase short-circuit current of grid-forming inverters This case is shown in Fig. 6.67A. For a solid two-phase clear-of-earth short-circuit fault that occurs on phases y and b, we can write ey ðtÞ 2 eb ðtÞ 5 2Riy ðtÞ 1 2L (A)

diy ðtÞ dt

(6.102a) (B)

−

+

−

+

−

+

−

+

−

+

−

+

Figure 6.67 Two-phase faults at grid-forming inverter (A) terminals and (B) through an external grid impedance.

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581

and, using Eq. (6.100), we have ey ðt Þ 2 eb ðt Þ 5 2

pffiffiffipffiffiffi 3 2Ecosðωs t 1 δr Þ

(6.102b)

Substituting Eq. (6.102b) in Eq. (6.102a), the current solution is given by 2 0 13 9 8 > > > > ω L > 4ωs t 1 δr 2 tan21 @ s A5 > > > 2 cos > > > > pffiffiffi pffiffiffi > > R = < 3 2E
 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi iy ðt Þ 5 2 0 13 L > > 2 2t= R > R2 1 ðωs LÞ2 > > > > > 21 @ω s LA5 > > 4 > > 1 cos δr 2 tan e > > ; : R ib ðtÞ 5 2 iy ðtÞ

(6.103a)

(6.103b)

As in the case of a three-phase short-circuit fault, the fault current consists of a steady-state ac current component and a transient dc current component that decays exponentially with a dc time constant equal to L=R. In the case of a fault through an external grid impedance Re 1 jωs Le , shown in Fig. 6.67B, the fault current can be calculated using Eq. (6.103) by replacing R with ðR 1 Re Þ and L with ðL 1 Le Þ.

6.21

Over-current limitation strategies of grid-forming inverters

6.21.1 General For typical utility-scale grid-forming inverters, the inverter total output filter impedance is typically 0.05
0.15 pu, giving an unlimited ac rms fault current of 6.7
20 pu for a solid three-phase fault at the inverter output terminals. The combined impedance of the inverter output filter and inverter medium-voltage transformer is typically 0.13
0.25 pu, giving an unlimited ac rms fault current on the medium-voltage busbar of 4
7.7 pu. For an effective dc time constant (L 1 Le )/ðR 1 Re Þ of around 45 ms and maximum dc current offset, the prospective first peak instantaneous current can reach 7.2
13.9 pu. However, as discussed in Section 6.17.2, if allowed to flow, these current magnitudes would destroy the inverter unless the current is very quickly limited to the inverter transient current limit of typically 1.1
1.2 pu. The instantaneous fault current could exceed 1.5 pu even for remote grid faults that result in a voltage dip at the inverter terminals of more than 10%
14%. Therefore a fault, or more generally, overcurrent limiting strategy is an essential prerequisite for the application grid-forming inverters. The inverter output current limiting strategy needs to operate effectively and reliably under all types of balanced, unbalanced, close-up and remote short-circuit

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faults, as well as large phase angle jumps perhaps up to 60 degrees. The phase angle of the limited current should not be changed. The current limitation strategy should act to limit the instantaneous asymmetric current which may include a different magnitude of dc current component depending on the fault instant on the voltage waveform. The dc current magnitude could be zero for a fault close to voltage peak or maximum for a fault close to zero voltage. Different dc fault current magnitudes can occur in the case of multiphase faults. These objectives mean that the action to limit the converter over-current magnitude may have to be completed within typically a couple of milliseconds after the onset of the dip/fault. Integrated gate commutated thyristor (IGCT) switches and injection enhanced gate thyristor (IEGT) switches are used in medium-voltage high-power converters. IEGT switches can handle short-circuit fault current and enable the converter to deliver up to two or three times rated current for a short period of time. IGCT switches, however, cannot handle a fault current more than around 1.5 pu but this would still offer practical benefits. One application of IGCT- and IEGT-based highpower converters is in doubly fed induction machines and full-size converterinterfaced machines used in variable-speed pumped storage hydropower stations.

6.21.2 Current limitation in inverters using single voltage controller structure Several strategies for preventing inverter over-current are described below.

Strategy 1: Clipping PWM voltage reference This strategy can be used with either VSM or droop-controlled grid-forming inverters where the magnitude of the internal voltage is controlled or clipped in the time domain if the inverter current in any phase exceeds the maximum instantaneous current limit of the inverter. The strategy is illustrated in the block diagram in Fig. 6.68. This is a simple and effective method but the clipped current waveform is distorted and contains odd harmonic orders of decreasing amplitude, for example, third, fifth and seventh harmonics, as well as some high-frequency small-amplitude oscillations around the clipped current mean. The transient harmonic currents exist

Figure 6.68 Grid-forming inverter over-current limitation strategy 1: clipping PWM voltage reference.

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Figure 6.69 Grid-forming inverter over-current limitation strategy 2: limiting internal voltage in dq reference frame then clipping.

during the clipping period only, which should generally occur for a few cycles only until the current drops below the threshold. The effects of the current distortion on the operation of protective current transformers and protection relays need careful assessment.

Strategy 2: Limiting internal voltage in dq reference frame This strategy is based on the VSM inverter control structure shown in Fig. 6.60. The strategy is illustrated in the block diagram in Fig. 6.69. The magnitude of the inverter internal voltage is limited or reduced immediately upon current build-up to a maximum value given by   edq 

Limited

  5 vdq Unlimited 1 ωs LImax

(6.104)

  where edq Limited is the limited internal positive-sequence voltage of inverter;   vdq  is the unlimited positive-sequence output voltage of inverter; Imax is the Unlimited maximum allowed positive-sequence current of inverter; and X 5 ωs L is the reactance of the inverter filter reactor. In effect, this approach limits the difference between the inverter internal voltage and the inverter terminal voltage, that is, the voltage drop across the filter reactor to a maximum value given by the maximum inverter current limit and the reactance   of the filter reactor which is a constant given by ωs LImax . However, limiting edq  in accordance with Eq. (6.104) is not sufficient if the current contains dc current component.   Additional limitation by clipping the PWM voltage reference of the inverter eryb  in the ryb reference frame is required which, as in strategy 1, causes some harmonic distortion in the current waveform.

Strategy 3: Static virtual resistor Again, this strategy is based on the VSM structure shown in Fig. 6.60 and is illustrated in the block diagram in Fig. 6.70. It is based on the principle of adding a static virtual resistor in series with the inverter filter resistor in order to reduce or limit the internal voltage of the VSM inverter to a value given by

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Inverse park

Figure 6.70 Grid-forming inverter over-current limitation strategy 3 with static virtual resistance.



R edq Limited 5 edq Unlimited 2 vr 0

 0 i Rvr dq

(6.105)

where edq Limited is the limited internal positive-sequence voltage of inverter; edq Unlimited is the unlimited internal positive-sequence voltage of inverter; idq is the positive-sequence output current of inverter; and Rvr is virtual resistor in series with inverter filter resistor. The addition of a virtual series resistor to the output filter of the inverter maintains and replicates the inherent fault current waveform of a real synchronous machine. However, a sufficiently large value of the virtual resistor Rvr , for example, 20
40 times the resistance of the output filter may be required. This should also reduce the dc time constant and hence the dc fault current component as well as the ac component of the fault current but the latter may depend on the proximity of the fault location to the inverter terminals and the X=R ratio of the inverter external circuit. Being virtual, no physical active power losses occur in Rvr , so the greater the value of the virtual resistor Rvr , the greater the reduction in fault current magnitude. However, a disadvantage is that the static resistor Rvr is always present in the operation and control of the inverter. High values in weak ac grids have to be avoided to avoid inverter angle instability that may be caused by increased coupling between active and reactive power and hence reduced power swing damping. Therefore if used on its own, the magnitude of fault current reduction achievable may not be sufficient and additional reduction may be required.

Strategy 4: Transient virtual resistor Like strategy 1, this strategy can be applied to VSM and droop-controlled inverter control structures and is illustrated in the block diagram in Fig. 6.71A. However, the resistor added in series with the inverter filter resistor is transient and hence nonlinear. By definition, the virtual resistance should be zero for inverter currents up to the rated current then a resistance value that increases with inverter current magnitude up to inverter current limit may be effective. This strategy is particularly useful if the inverter is slightly oversized so that its current limit is greater than its

Modelling of voltage-source inverters, wind turbine and solar photovoltaic (PV) generators

(A)

0

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