By Paul M. Anderson
This vintage textual content provides you with the foremost to figuring out brief circuits, open conductors and different difficulties with regards to electrical strength structures which are topic to unbalanced stipulations. utilizing the tactic of symmetrical parts, stated specialist Paul M. Anderson presents entire tips for either discovering suggestions for faulted energy platforms and preserving protecting process functions. you will discover ways to clear up complicated difficulties, whereas gaining an intensive heritage in common configurations.
Features you will positioned to rapid use:
- a number of examples and problems
- transparent, concise notation
- Analytical simplifications
- Matrix tools acceptable to electronic machine technology
- huge appendices
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Extra resources for Analysis of Faulted Power Systems
We apply a similar technique to investigate the possibility of finding symmetrical components of the unbalanced impedances. 1 1 We emphasize the word of because later we will define and use sequence impedances Zo, Zh and Z2 which will express impedance to the flow of laO, lalt and la'1. respectively. These are not the impedances under investigation here. 4, which we will call Vmn. 41) where V mn is actually V mn-abc. 45) = A -1 Z A The concept of the similarity transformation gives us mathematical insight into the operations being performed.
11. Sequence network connection for a LL fault. 5, where a fault impedance of 4 ohms is assumed between phases b and c. Let h = 1. 11. 0125L1L.. 9) with h Ib pu = 1320 A This system voltages may also be synthesized from a knowledge of the sequence currents and sequence network connections. 276 pu which is a good check, considering the algebra involved. - __ Ie I I / Ie2 Va REF (a ) (b) Fig. 12. Phasor diagrams for a LL fault: (a) currents (b) voltages. 12. Note that if Z, were zero the voltage "triangle" would collapse to zero on the b-c side and the Val - V a2 phasor would go to zero.
9 and note that the LL fault is placed between lines band c to retain symmetry with respect to phase a. - - . - - - - - - + 0 - - 0 + - - - - 4..... C Fig. 9. Diagram of a LL fault at F. 2. 8) fa Vb - 3. 9) lao = 0 and lal = - la2. 10) 4. 9), lao = 0, so the zero sequence network is open. 10. 5. 11. 11) Analysis of Unsymmetrical Faults: Three-Component Method 43 Fig. 10. Sequence network partial connection specified by the current equation. Zf Fig. 11. Sequence network connection for a LL fault.