Myatt / Hammond Symmetrical Components

The Commonwealth and International Library: Applied Electricity and Electronics Division
1. Auflage 2014
ISBN: 978-1-4831-8120-2
Verlag: Elsevier Science & Techn.
Format: EPUB
Kopierschutz: Adobe DRM (»Systemvoraussetzungen)

E-Book, Englisch, 184 Seiten

Symmetrical Components introduces the fundamental concepts involved in the method of symmetrical components. The book also demonstrate the method for analyzing simple power networks that are subjected to unbalanced fault conditions. The text first discusses the h operator, and then proceeds to detailing symmetrical components. The next two chapters cover the properties and measurement of symmetrical components. Chapter 5 tackles the short-circuit faults on an alternating current generator, while Chapter 6 discusses the equivalents circuits for unbalanced faults. The seventh chapter talks about the sequence networks and faults on three-phase systems, while the last chapter deals with unbalanced loads. The text will be of great use to students of electrical engineering. Professional electrical engineers and technicians will also benefit from the book.

Autoren/Hrsg.

Myatt, L. J.

Hammond, P.

Inhaltsverzeichnis

1;Front Cover;1
2;Symmetrical Components;4
3;Copyright Page;5
4;Table of Contents;6
5;Preface;8
6;Acknowledgements;9
7;CHAPTER 1. The h Operator;10
7.1;1. Representation of sinusoidal quantity by a vector;11
7.2;2. The h operator;13
7.3;3. Functions of h;16
7.4;4. Simplification of hn;17
7.5;5· Simplification of (h + k)-1 and (h2 + k)-1;18
7.6;6. Useful functions of h;20
7.7;7. Symmetrical three-phase voltages;21
7.8;8. Positive-phase sequence;21
7.9;9 . Negative phase sequence;22
7.10;10. Zero-phase sequence;23
7.11;11. Star-delta impedance transformations;24
7.12;12. Percentage and per unit values;26
7.13;Exercises;32
8;CHAPTER 2. Symmetrical Components;33
8.1;14. Resolution into symmetrical components;33
8.2;15. Solution of symmetrical components;35
8.3;16. Sum and difference equations;38
8.4;17. Zero-sequence currents;43
8.5;18. Line-to-line voltages;46
8.6;19. Graphical determination of phase voltages from line voltages when zero-sequence component is absent;48
8.7;20. Alternative graphical method to determine zero-phase sequence voltage;49
8.8;21. Alternative graphical method to determine positive- and negative-phase sequence voltages from phase voltages;51
8.9;22. Graphical determination of positive- and negative-phase sequence voltages from line voltages;53
8.10;23. Unbalance factor;55
8.11;Exercises;56
9;CHAPTER 3. The Properties of Symmetrical Components;57
9.1;24. Phase sequence currents and voltages in a symmetrical network;57
9.2;25. Generated e.m.f. of a three-phase generator;65
9.3;26. Neutral current;66
9.4;27. Generator with neutral earthed through an impedance;67
9.5;28. Kirchhoff's first law;69
9.6;29. Kirchhofes second law;71
9.7;30. Star–delta transformations;72
9.8;31. Star–delta transformers;75
9.9;32. Effects of unbalanced loading on synchronous machines;82
9.10;33. Power due to unbalanced three-phase loads;83
9.11;Exercises;86
10;CHAPTER 4. The Measurement of Symmetrical Components;88
10.1;34. Zero-sequence voltage;88
10.2;35. Zero-sequence current;89
10.3;36. Positive- and negative-sequence voltage;91
10.4;37. Simultaneous measurement of positive- and negative-sequence voltage;94
10.5;38. Positive- and negative-sequence current without residue;95
10.6;39. Simultaneous measurement of positive- and negative-sequence currents without residue;97
10.7;40. Elimination of zero-sequence component;99
10.8;41. Negative-sequence bridge;100
10.9;42. Positive- and negative-sequence filter;102
10.10;43· Zero-sequence power;106
10.11;44. Positive-sequence power;107
10.12;45. Negative-sequence power;110
10.13;Exercises;110
11;CHAPTER 5. Short-circuit Faults on an Alternating current Generator;113
11.1;46. Line-to-Iine fault;114
11.2;47. Line-to-line fault through an impedance;116
11.3;48. Single line-to-earth fault;119
11.4;49. Single line-to-earth fault through an impedance;121
11.5;50. Double line-to-earth fault;126
11.6;51. Double line-to-earth fault through an impedance;129
11.7;52. Three-phase fault;134
11.8;53. Three-phase fault through an impedance;136
11.9;Exercises;138
12;CHAPTER 6. Equivalent Circuits for Unbalanced Faults;139
12.1;54. Line-to-line fault;139
12.2;55. Single line-to-earth fault;141
12.3;56. Double line-to-earth fault;144
12.4;57. Three-phase fault;145
12.5;Exercise;148
13;CHAPTER 7. Sequence Networks and Faults on Three-phase Systems;149
13.1;58. Unbalanced fault on a three-phase system;149
13.2;59. Positive-sequence network;152
13.3;60. Negative-sequence network;152
13.4;61. Zero-sequence network;152
13.5;62. Zero-sequence impedances;153
13.6;63. Method of fault calculation for a network;160
13.7;64. Use of analogues;166
13.8;Exercises;168
14;CHAPTER 8. Unbalanced Loads;169
14.1;65. Unbalanced star-connected load;169
14.2;66. Currents in unbalanced load;172
14.3;67. Voltages across the unbalanced load;174
14.4;68. Use of unbalanced load to determine phase sequence;176
14.5;Exercise;178
15;APPENDIX: Use of Matrix Notation;179
16;Bibliography;182
17;Answers to Exercises;183
18;Index;184

Leseproben

CHAPTER 1

The h Operator

Publisher Summary

This chapter provides an overview of operator. When performing calculations on single-phase circuits an operator was used, which, when multiplied by a vector, had the effect of rotating the vector through an angle of 90°, and for this purpose the operator was used. In dealing with three-phase quantities, and in particular when applying the method of symmetrical components, there are definite advantages to be gained by having an operator that is capable of rotating a vector through an angle of 120°, because of the angular displacement of balanced three-phase vectors. The chapter further discusses symmetrical three-phase voltages, positive-phase sequence, negative phase sequence, zero-phase sequence, and Star-delta impedance transformations.

In a three-phase system, the three-phase voltages or currents may be said to be balanced if they are sinusoidal and, when represented by vectors, are of equal magnitude and displaced from each other by equal phase angles of 120°. A three-phase circuit is also balanced, or symmetrical, when each of the three phases contain equal impedances. Furthermore, if a balanced system of three-phase voltages is applied across the terminals of a balanced three-phase network, then the currents flowing in each of the three phases will also be balanced.

The solution of a problem involving a balanced three-phase system may be found by considering only one phase in which the voltage, or current, is taken as the reference vector and solving for that phase alone as in the case of a single-phase problem. The magnitude of the currents and voltages in the other two phases will then be the same as in the reference phase but there will be a corresponding phase displacement of ±120° as appropriate.

If now the applied voltage is unbalanced, or the impedances in each phase are no longer identical, the three-phase currents will also become unbalanced and the problem cannot be solved by considering a single phase as in the above method. It would be possible in a simple problem of unbalance to find a solution by applying Kirchhoff’s laws, but in the more involved type of problem, such as that involving a power network, a less laborious method is necessary.

Considerable thought was given to the problem of unbalance, at the beginning of the century, with particular reference to the unbalanced loading of three-phase machines, and L. G. Stokvis showed that, under such conditions, the armature m.m.f. of a three-phase generator may be treated as two separate components, equivalent to the effect of two balanced loads. It remained for C. L. Fortesque to develop the method, now known as the method of symmetrical components, after working on problems associated with the use of phase balancers in the single-phase electrification of railways. Following on from this, he studied the problem of unbalance in general when, in 1918, he published his very important paper “The method of symmetrical coordinates applied to the solution of polyphase networks”. In this paper he showed how it was possible for a set of unbalanced polyphase currents to be resolved into a number of component systems of balanced currents equal to the number of phases. A further development of the theory provides considerable simplification in solving problems involving unbalanced faults on power networks.

1 Representation of sinusoidal quantity by a vector

The instantaneous value of sinusoidal voltage of frequency Hz is given by

?=Vsin?t, (1)

(1)

where = maximum value and ? = 2 p .

This may be represented by a rotating vector of magnitude having a constant angular velocity ? rad/s as shown in Fig. 1. The vector rotates anticlockwise about starting from when = 0. The angle turned through, at any time , is then given by ? and the instantaneous value at any time is given by sin ? which is the projection of on the -axis, .

FIG. 1 Vector representation of a sinusoidal quantity.

If the vector is now shown on an Argand diagram as in Fig. 2 where ? = ?, it may be represented by ( + ).

FIG. 2 Representation of a vector on an Argand diagram.

a=Vcos?,

where

b=Vsin?,a+jb=V(cos?+jsin?). (2)

(2)

Another way of expressing the vector is as , where is known as the and would correspond to the peak value of the alternating voltage wave of Fig. 1, and ? is known as the and represents the angular displacement of the vector from the real axis .

Now

cos?+jsin?=ej?, (3)

(3)

and

V(cos?+jsin?)=Vej?. (4)

(4)

Thus the expression may be used to represent a vector of modulus and argument ?.

2 The operator

When performing calculations on single-phase circuits use was made of an operator, which, when multiplied by a vector, had the effect of rotating the vector through an angle of 90°, and for this purpose the operator was used. In dealing with three-phase quantities, and in particular when applying the method of symmetrical components, there are definite advantages to be gained by having an operator which is capable of rotating a vector through an angle of 120°, due to the angular displacement of balanced three-phase vectors.

First let us consider the multiplication of two complex quantities which may be represented by and , then we may write

r1/?1_r2/?2_=r1(cos?1+jsin?1)r2(cos?2+jsin?2)=r1r2(cos?1cos?2+jsin?1cos?2+jcos?1sin?2-sin?1sin?2)=r1r2(cos?1cos?2-sin?1sin?2)+j(sin?1cos?2+cos?1sin?2)=r1r2[cos(?1+?2)+jsin(?1+?2)]=r1r2/?1+?2_. (5)

(5)

Thus it may be observed that the modulus of the product of two complex quantities is equal to the product of their moduli, and that the argument of their product is equal to the sum of their arguments.

Suppose now that is replaced by a voltage vector , and is used to denote an operator such that it will rotate the voltage vector through an angle of 120° to give a new vector /? + 120°, then

r1/??1_,V/?_=V/?+120°_.

Comparing the moduli and arguments with the results of eqn. (5) gives

r1V=V.

Therefore

r1=1;

also

?+?1=?+120°

Therefore

?1=120°.

The operator 1 /?1 is then given by

r1/?1_=1/120°_=ej2p/3. (6)

(6)

Various symbols, including , ?, and , have been assigned to this operator by different writers on the subject, but the latter is now used in accordance with standard practice, as recommended by the British Standards Institution. Hence,

h=ej2p/3,=cos2p/3+jsin2p/3,=-12+j12v3. (7)

(7)

Considering again the results of eqns. (5) and (6) it may be stated that, in general, a vector multiplied by will cause an angular rotation of 2p/ in the vector, and so it is a simple matter to produce an operator capable of rotating a vector through any desired angle. For example, the operator which rotates a vector through an angle of 90° would be represented by /2, where

ejp/2=(cosp/2+jsinp/p)=j. (8)

(8)

A further significance may be attached to the operator if we consider a vector represented by as shown in Fig. 3 and which is shown drawn on the reference axis . If this vector is multiplied by , giving , it will be rotated through 120°. A second multiplication by , giving 2, will give a further rotation of 120° and a third multiplication by , giving 3, will cause a further rotation of 120° bringing us back to the original position of vector on the reference axis. It is now apparent that

FIG. 3 The operator.

h3V=V,h3=1. (9)

(9)

In other words, ...