Analysis of Faulted Power Systems (IEEE Press Series on by Paul M. Anderson

By Paul M. Anderson

This vintage textual content will give you the most important to realizing brief circuits, open conductors and different difficulties with regards to electrical energy platforms which are topic to unbalanced stipulations. utilizing the strategy of symmetrical parts, stated professional Paul M. Anderson presents complete counsel for either discovering options for faulted energy platforms and conserving protecting process functions. you are going to learn how to remedy complex difficulties, whereas gaining a radical heritage in undemanding configurations.

Features you are going to positioned to rapid use:

  • Numerous examples and problems
  • Clear, concise notation
  • Analytical simplifications
  • Matrix tools appropriate to electronic machine technology
  • Extensive appendices

Diskette records can now be chanced on via coming into in ISBN 978-0780311459 on booksupport.wiley.com.

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Additional info for Analysis of Faulted Power Systems (IEEE Press Series on Power Engineering)

Example text

3. ood A two-machine system. 20 MV A. Find the pu impedances of all components referred to these bases. 8 kV, x = 0. 8-161 kV , x = 0 . 8 pf lag Solution Using equation ( 1 . 1 5 ) , we proceed directly with a change in base for the apparatus. 2) 2 = 0 . 41 ) For the transmission line we must convert from ohmic values to pu values. 22 ). 33 VJ pu. 2 = 6 . 9 pf lead and the terminal voltage is 1 . 1 pu. What is the voltage at MVA 10 the generator terminals? Solution as First we compute the total load current.

4 . Sequence network connection for a SLG fault. 3. 4. With this connection we com­ pute h VF �O = � 1 = � 2 = Zo + Z 1 + Z2 + 3 Zf (3. 5) and knowing the sequence currents we easily find the sequence voltages from equation ( 2. 58). Example 3. 1 The simple power system shown in Figure 3 . 5, consists of a generator, trans­ former, transmission line, load transformer, and load. 5 kV ___ 2+ j4 ohm � SLG 0 6Y FAULT Fig. 5. Power system for Example 3 . 1 . LOAD P - IO MW Q a 5 M VAR Ana lysis of U nsym metrica l Fa u lts : Three -Component M ethod 39 C with a fault resistance of 4 ohms.

1y + J 0. OO- 1 0 , - 1 02 3Zf - 0. 6+ J O Fig. 3 . 7 . Sequence networks for SLG fault a t bus C. 7, where Z, is shown to be Z, + jO = --z;= 20 = 0 . 2 4 4 The total circuit impedance is Zt = Zo + Z l + Z2 + 3Z, From (3. :;/���� 05 = = 0. 4 A. We may also synthesize the phase voltages by first computing the sequence voltages. 1995 - jO. 3319/67. 1905)(0. 236 + jO. 7765 - jO. 33 19/67. 2348 - jO. 4591 = 0 . 6496 Analysis of Uns y m metrical Fa u lts : Three - Component M ethod 41 (0 ) b o (b) Fig.

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