Electric Power Systems
`
`Third Edition
`
`B. M. WEEDY
`
`Department cj'Electrical Engineering,
`The Universig/ (j Southampton
`
`
`
`JOHN WILEY 8; SONS
`
`GE2014
`CHICHESTER NEW YORK BRISBANE TORONTO Vestasv.GE
`|PR2018-01015
`
`GE 2014
`Vestas v. GE
`IPR2018-01015
`
`i
`
`

`

`Copyright © 1967, 1972, 1979 by John Wiley & Sons Ltd.
`
`All rights reserved.
`
`No part of this book may be reproduced by any means, nor
`transmitted, nor translated into a machine language without the
`written permission of the publisher.
`
`Library of Congress Cataloging in Publication Data:
`
`Weedy, Birron Mathew
`Electric power systems—w3rd ed.
`1. Electric power transmission
`I. Title
`
`621.319’1
`
`TK3001
`
`79-40081
`
`ISBN 0 471 27584 0
`
`Printed in Great Britain by J. W. Arrowsmith Ltd.,
`Winterstoke Road, Bristol B83 ZNT
`
`ii
`
`

`

`246
`
`7
`
`Fault Analysis
`
`7.1
`
`Introduction
`
`An essential part of the design of a power supply network is the calculation of
`the currents which flow in the components when faults of various types occur.
`In a fault survey, faults are applied at various points in the network and the
`resulting currents obtained by hand calculation, or, more likely now on large
`networks, by analogue or digital processes. The magnitude of the fault currents
`give the engineer the current settings for the protection to be used and the
`ratings of the circuit breakers.
`if” The types of fault commonly occurring in practice are illustrated1n Figure
`57.1 and the most common of theseis the short circuit of a single conductor to
`earth. Often the path to earth contains resistance in the form of an are as shown
`;
`in Figure 7.1(f). Although the single line to earth fault is the most common,
`:1
`calculations are frequently performed with the three-line, balanced short
`l
`circuit (Figure 7.1(d) and (e)). This is the most severe fault and also the most
`2“agamenable to calculation. The causes of faults are summarized in Table 7.1
`
`(a; j: (b)
`
`T
`
`E
`
`I
`
`1:;9"
`
`Figure 7.1 Common types of fault.
`
`246
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`

`

`247
`
`Table 7.1 Causes of overhead~line faults, British system
`66 kV and above
`
`Faults/160 km
`of line
`
`
`Lightning
`Dew, Fog, Frost
`Snow, ice
`Gales
`
`Salt spray
`
`Total
`
`1.59
`0.15
`0.01
`0.24
`
`0.01
`
`2 faults per 160 km giving a
`total of 232 faults on system
`
`which gives the distribution of faults due to various causes on the British
`Central Electricity Generating Board system in a typical year. Table 7.2 shows
`the components affected. In tropical countries the incidence of lightning is
`much greater than in Britain, resulting in larger numbers of faults.
`
`Table 7.2 Distribution of faults, British system
`
`Type
`
`Number of faults
`
`Overhead lines
`Cables
`
`Switchgear
`Transformers
`
`Total
`
`289
`67
`
`56
`59
`
`471
`
`As well as fault current, fault MVA is frequently considered; this is obtained
`from the expression \/(3) VLIF X 10~6 fault, where VL is the nominal line voltage
`of the faulted part. The fault MVA is often referred to as the fault level; The
`calculation of fault currents can be divided into the following two main types.
`
`three phases when the network remains
`(a) Faults short-circuiting all
`these calculations normal
`single-phase
`balanced electrically. For
`equivalent circuits may be used as in ordinary load-flow calculations.
`(b) Faults other than three~phase short circuits when the network is elec—
`trically unbalanced. To facilitate these calculations a Special method for
`dealing with unbalanced networks is used known as the method of
`symmetrical components.
`
`The main objects of fault analysis may be enumerated as follows:
`
`1. To determine maximum and minimum three-phase short—circuit cur-
`rents.
`
`2. To determine the unsymmetrical fault current for single and double
`Iine-to-earth, Iine—to—line faults, and open—circuit faults.
`3. Investigation of the operation of protective relays.
`
`

`

`248
`
`4. Determination of rated rupturing capacity of breakers.
`5. To determine fault-current distribution and busbar—voltage levels during
`faults.
`
`7.2 Calculation of Three-phase Balanced Fault Currents
`
`The action of synchronous generators on three—phase short circuits has, been
`described in Chapter 3. There it was seen that dependent on the time elapsing
`from the incidence of the fault, either the transient or the subtransient
`
`reactance should be used to represent the generator. For specifying switchgear
`the value of the current flowing at the instant at which the circuit breakers open
`is required, It has been seen, however, that the initially high fault current
`associated with the subtransient reactance decays with the passage of time.
`Modern air-blast circuit breakers usually operate in 2.5 cycles of 60 Hz
`alternating current and are associated with extremely fast protection. Older
`circuit breakers and those on lower voltage networks usually associated with
`relatively cruder protection can take in the order of 8 cycles or more to operate.
`In calculations it is usual to use the subtransient reactance of generators and to
`ignore the effects of induction motors. The calculation of fault currents ignores
`the direct-current component, the magnitude Of which depends on the instant -
`in the cycle that the short circuit occurs. If the circuit breaker opens a
`reasonable time after the incidence of the fault the direct-current component
`will have decayed considerably. With fast—acting circuit breakers the actual
`current to be interrupted is increased by the direct-current component and it
`must be taken into account. To allow for the direct-current component of the ' 2'7;
`fault current the symmetrical r.m.s. value is modified by the use of multiplying 533-5".
`factors such as the following:
`Mi
`8 cycle circuit breaker opening time, multiply by 1
`3 cycle circuit breaker opening time, multiply by 1.2
`2 cycle circuit breaker opening time, multiply by 1.4
`
`
`
`
`
`
`
`
`
`g
`5M...
`
`ff:
`"-3-:-
`
`_
`
`'_
`
`
`
`Consider an initially unloaded generator with a short circuit across the three.
`terminals as in Figure 7.2. The generated voltage per phase is E and therefore
`the short-circuit current is [E/Z(Q)]A, where Z is either the transient "or;
`
`Figure 7.2 VoItage source with short circuit and equivalent circuit.
`
`