modern power system fault calculation

Hong Kong Polytechnic University Department of Electrical Engineering

Modern Power System Protection Chapter 2 – Fault Analysis

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Outcome on this chapter After you have read this chapter you should be able to:

i. Understand the basic terms used in power system fault calculation. ii. Carry out symmetrical and unsymmetrical fault calculations. iii. Represent 2 and 3 windings transformers with sequence networks. iv. Understand the merits of different earthing methods v. Model the zero Sequence network of a 3 phase auto-transformer. vi. Carry out open circuited fault calculations. vii. Carry out simultaneous fault calculations. 2.1 Per Unit Values

The per unit values of any quantity is defined as the ratio of the quantity expressed as a decimal to its base value. 2.1.1 Change of MVA base

bpu V

VV =

bpu MVA

MVAMVA =

bpu I

II =

ph 3 bphbb IVMVA =

( )2

233

where

lineb

b

b

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b

b

b

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newb

oldpu

newpu

lineb

oldboldpu

lineb

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ZZ

VZMVA

Z

VZMVA

Z

2

2

)(

)(

=

=

=



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2.1.2 Change of voltage base 2.1.3 Change of both MVA base and voltage base 2.1.4 Example: For the system shown in Fig. 2.1, calculate:

(i) the emf of G1 and G2 if the voltage magnitude at 11kV B/B is 11kV; (ii) the fault level of 11kV B/B; and (iii) the voltage of the 110kV B/B near T3 and T4 under fault conditions.

(1.375pu or 14.3kV, 43.7MVA, 0.219pu or 24.05kV)

2

2

2

2

)()(

)(

)(

newlb

oldlb

oldpu

newpu

oldlb

boldpu

newlb

bnewpu

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ZZ

VZMVA

Z

VZMVA

Z

=

=

=

2

2

)()(

newlb

oldlb

oldb

newb

oldpu

newpu

VV

MVAMVA

ZZ

=

12.5kV 30MW 45% 12.5/132kV 45MVA 20% 110kV X = j40Ω 110kV 110/11kV 20MVA 20% 11kV 30MW unity p.f.

Fig. 2.1



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2.2 Usefulness of Fault Calculation in the Study of Power System Protection To understand the relay performance, you must know how severe is the fault. Fault

calculation, mostly symmetrical fault, aided with the knowledge of unsymmetrical fault is required. This knowledge is required to find out the condition at the relaying point under fault conditions, for example, voltage at relaying point and current passing through relays. Based on this you can deduce what the relay will see. 2.3 Balanced Three Phase Faults 2.3.1 Behaviour of synchronous machine under fault conditions Immediately after the application of the short-circuit the armature current endeavours to create an armature reaction M.M.F., but the main air-gap flux cannot change to a new value immediately as it is linked with low-resistance circuits consisting of, (a) the rotor winding which is effectively a closed circuit, and (b) the damper bars, i.e., a winding which consists of short-circuited turns of copper strip set in the poles to dampen oscillatory tendencies. As the flux remains unchanged initially, the stator currents are large and can only flow through the medium of the creation of opposing currents in the rotor and damper windings by what is essentially transformer action. Owing to the higher resistance, the current induced in the damper winding decays rapidly and the armature current commences to fall. After this the currents in the rotor winding and body decay, the armature reaction M.M.F. is gradually established, and the generated e.m.f. and stator current fall until the steady-state condition on short-circuit is reached. Here the full armature reation effect is operational and the machine represented by the synchronous reactance Xs. The oscillograms of the currents in the three phases of a generator when a sudden short circuit is applied is shown in Fig. 2.2. To represent the initial short-circuit conditions, two additional reactances are needed to represent the machine, the very initial conditions requiring what is called the Subtransient Reactance (X") and the subsequent period the Transient Reactance (X') [Fig. 2.3]. It is assumed that the generator is on no-load prior to the application of the short circuit and is of the round-rotor type.

Fig. 2.2 The currents in the three phases of a generator

when a sudden short circuit is applied.



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2.3.2 Effect of load If the effect of load is taken into consideration, the actual fault current in any branch (by Superpositon Theorem) becomes the phasor sum of pre-fault load current in that branch and the fault current calculated from above. 2.3.3 Simplification by making assumption i. Load impedances and hence load currents neglected. ii. All voltages sources remain balanced and constant (usually at 1 p.u.) and they are all in

phase. Hence all Vpfn becomes equal and in phase with the source e.m.f. iii. All transformers are at normal taps. iv. Line capacitances are neglected. v. Transformer magnetising currents are negligible. 2.4 Unsymmetrical Fault Analysis In a 3 phase system, (2.1) where Zs = self impedance per phase. Zm = mutual impedance between any phase pair. or [ V ] = [ Z ] [ I ] (2.2)

Fig. 2.3 Subtransient reactance (X”) and the transient reactance (X’)

I Z + I Z + I Z = VI Z + I Z + I Z = VIZ+IZ+IZ=V

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=

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=

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smm

msm

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c

b

a

c

b

a



14

where a = 1 / 120° 1 + a + a2 = 0 [ V ] = [ Z ][ A ][ A ]-1[ I ] [ A ]-1[ V ] = [ A ]-1[ Z ][ A ] • [ A ]-1[ I ] (2.3) Examining Vo = 1/3 ( Va + Vb + Vc ) (2.4) V1 = 1/3 ( Va + a Vb + a2 Vc ) (2.5) V2 = 1/3 ( Va + a2 Vb + a Vc ) (2.6) Similarly Io = 1/3 ( Ia + Ib + Ic ) (2.7) I1 = 1/3 ( Ia + a Ib + a2 Ic ) (2.8) I2 = 1/3 ( Ia + a2 Ib + a Ic ) (2.9) Now, [ A ]-1 [ Z ] [ A ] = [ Zs ] (2.10)

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Z00

0Z0

00Z

=

Z - Z00

0Z - Z0

00Z2 + Z

=

aa1

aa1

111

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aa1

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31

2

1

o

ms

ms

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2

2

smm

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2



15

Therefore (2.11) Also [ I ] = [ A ] [ Is ] (2.12) Similarly (2.13) 2.5 Transformer Sequence Impedance 2.5.1 Positive and negative sequence equivalent circuit of two winding transformer A two winding power transformer may be represented by an equivalent circuit, representing the self-impedance and the mutual coupling between windings, as shown in Fig. 2.4. The total leakage impedance Zl is usually much smaller than the magnetization impedance Zm. Their magnitudes are of the order of 10 % and 2000 % respectively. Thus, in fault calculations, the positive sequence network of the transformer may be represented by the series leakage impedance only (Fig. 2.5). As the structure of the transformer is symmetrical, its equivalent circuit to negative sequence currents will be identical to the positive sequence network.

Fig. 2.4 Equivalent circuit of a two winding transformer

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=

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16

2.5.2 Zero sequence network of two winding transformer The transformer can still be considered as a three terminal network, but it is necessary to remember that the exciting branch is now the exciting impedance to zero sequence voltages or currents which are identical in all three phases. In three phase, three limb core type transformers this may be only of the order of 100 % to 400 % owing to the high reluctance of the flux path. Account may be taken of the method of interconnection of the windings and of the presence or absence of a neutral connection by considering that the transformer in the zero sequence representation is provided with sets of links 'a' and 'b' on both the primary and secondary sides (Fig. 2.6) and leaving these links open or close in accordance with the answers to the following questions: I. Does a physical circuit exist by means of which zero sequence currents can be passed into

the winding in question from the external circuit on that side (i.e. is there a neutral point on the transformer or elsewhere which is connected to earth or to a neutral wire, possibly through an impedance). If so, the link 'a' is closed; if not it is left open. This is done for each side of the transformer in turn.

II. Can zero sequence currents circulate in the winding in question, without flowing in the external circuit (i.e. does a delta connection exist). If so, the appropriate link 'b' is closed.

Fig. 2.5 Simplified equivalent circuit of a two winding transformer

Fig. 2.6 Zero sequence circuit connection of a two winding transformer



17

Fig. 2.7



18

2.5.3 Positive and negative sequence network of a three winding transformer Provided that the magnetising impedance is neglected, the positive and negative sequence network can be represented by an equivalent circuit consisting of three star impedances (Fig. 2.9).

Fig. 2.8

Fig. 2.9 Equivalent circuit of a three winding transformer



19



20

It should be pointed out that the star point of the equivalent circuit is a fictitious point and does not represent the system neutral and that loads or short circuits can be applied only to terminals. All impedances must be expressed in the same MVA and voltage base. One of the branches, usually the least important, may exhibit negative impedance. 2.5.4 Zero Sequence Networks of Three Winding Transformer The same rule as the two winding transformers applies. Typical examples are shown below: Fig. 2.10



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2.6 System Earthing The method of system earthing does not influence system operations. It only influence the system for faults involving earth. As the majority of faults in power systems are earth faults, system earthing becomes important. The purpose of earthing is to:

i) prevent damage to people, and ii) prevent or limit plant damage.

2.6.1 Earthing methods If a system is directly earthed or earthed through a low impedance, the fault current is high and the damage by fault current is considerable. If a high impedance is used to earth the system, the earth fault current is limited but it may cause transient overvoltage on unfaulted phases. The various earthing methods are:

i) Solidly earthed, ii) Reactance earthed, iii) Resistance earthed, iv) Resonant or Peterson Coil earthed, v) Earthing transformer, and vi) Insulated earthed.

2.6.2 Solidly earthed It is the simplest method of earthing. The zero sequence impedance is a minimum. The earth fault current is highest. However, overvoltage in unfaulted phases is a minimum. Earth fault protection is simple as the fault current is usually high. Overvoltage during earth faults is usually less than 0.8 times phase to phase voltage. 2.6.3 Resistance earthed There are two main types of earthing resistor, metallic grid and liquid types. The metallic grid is usually constructed using a cast or steel element. The temperature rise can be around 250°C. The liquid type earthing resistor consists of an electrode immersed in a tank containing a solution of sodium carbonate. This type of resistor has a negative temperature coefficient. Both types of earthing resistor have a short time thermal rating of 30 seconds. Earthing the system by means of a resistor reduces both the fault current and transient overvoltages. However, for high value of earthing resistance, overvoltage in unfaulted phases may be produced during earth faults approaching to phase to phase voltage. 2.6.4 Earthing reactor Reactors are usually smaller and less expensive than resistors, it can reduce fault current but this is associated with a increase in transient overvoltages during earth faults. Transient overvoltages are a maximum when the value of earthing reactance is approximately one third of the value required for resonant earthing. In order to reduce the transient overvoltages to an admissible level the earthing reactance has to be reduced so that the earth fault current approaches that for a solidly earthed system. Overvoltages during earth faults will be between 0.8 times and full phase to phase voltage.

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2.6.5 Resonant earthed – arc suppression coil (Peterson Coil) When the earthing reactance is equal to the total system capacitive reactance to earth, arcing earth faults will be self extinguishing. For short time rated reactors, if the fault has not extinguished after a certain time the reactor is shorted. The system capacitance must however remain fairly constant, otherwise the efficiency of this earthing method is greatly diminished. If the system is expanded, the value of the grounding reactance should be altered. The steady state voltages in the sound phases during a phase to earth fault is the full phase to phase voltage. 2.6.6 Insulated neutral The earth fault current is capacitive and if small may be self extinguished. Automatic segregation of faulty zones is extremely difficult. The system can be run with an earth fault for long periods. Overvoltages during earth faults may be greater than phase to phase voltage. Arcing earth faults are very likely and these can result in high transient overvoltages. 2.6.7 Earthing transformer In some cases where the neutral of the power transformer is not available, the system is earthed via a earthing transformer of zig-zag type (Fig. 2.11). A minimum impedance is offered to the flow of zero sequence currents. Under normal operating conditions the currents flowing through the windings are the magnetising currents of the earthing transformer. The earthing transformer is designed to carry the maximum fault current for 30 seconds. 2.7. Selection of Earthing Methods In unearthed distribution system, a break down between HV and LV winding may cause a severe overvoltage in the LV winding and it can cause a hazard for the plant and personnel safety in the LV distribution system. In general, high impedance earthing is used in medium voltage networks up to around 33 kV. At these voltages insulation of the power system components is not too costly, and the systems can be insulated to allow for overvoltages which

Fig. 2.11

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can occur during earth faults. For high and extra high voltage systems the cost of providing the necessary insulation can be very expensive, and these systems are normally solidly earthed or earthed via low impedance. Peterson coil earthing is used on overhead line systems up to around 132 kV, but requires high insulation and restricts the use of auto transformers. 2.7.1 System voltages up to 660 V These are normally solidly earthed for safety reasons. In some special cases, where the continuity of supply is very important, the system can be operated with insulated neutral. 2.7.2 System voltages from 660 V to 33 kV If the fault level is low the system may be solidly earthed. Generally, these systems are resistance or reactance earthed. When resistance earthing is used, the earth fault current is limited to a value close to the rated load current. Resonant earthing may be used where a system consists mainly of overhead lines and is particularly useful in areas having high iso-keraunic levels. These have a high incidence of lightning strokes and transient earth faults caused by lightning can be extinguished by the Peterson Coil without the need for line outages. Isolated systems are only used where long lines are involved and alternative supplies not easily available. 2.7.3 Systems above 33 kV The problems of overvoltages becomes more important than high fault currents. For this reason they are usually solidly earthed. 2.8 Variation of Healthy Phase Voltages for an Earth Fault 2.8.1 Single phase to earth fault Let Z2 = K2 Z1 and Zo = Ko Z1 For A-E fault,

(2.14)

(2.15)

(2.16)

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abb

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02

022

022

1

02

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1

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−=++=

++++

−=++=

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(2.17) Usually, K2 = 1, therefore 2.8.2 Phase to phase to earth fault For B-C-E fault,

(2.18)

(2.19)

A-E fault Fig. 2.12

Effectively earthed system

Non-effectively earthed system

( ) ares EKK

KV

02

0

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0

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a

a

a

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1 if 21

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25

The above equations show how the healthy phase voltages are affected by the method of earthing. K2 may be assumed constant and approximately equal to 1.0 0ºE. For solidly earthed system and with a earth fault close to generation, K0 may be approaching 0.5. In this case, the voltage in the sound phase is below rated phase to neutral voltage. For solidly earthed system and with a earth fault more distant from generation, K0 may be approaching 1.0. In this case, the voltage in the sound phase equals the rated phase to neutral voltage. Even on a solidly earthed system a earth fault far away from the source of generation, can result in a value of K0 approaching 2.5 to 4.5. The voltage in the sound phase will increase above the normal rated phase to neutral voltage. In general, effective earthing includes direct earthing and low reactance earthing. Non effective earthing is characterised by a high Z0 / Z1 ratio (large K0) and includes resistance earthing, distribution transformer earthing, resonant earthing and insulated systems. In BS 5311 – 1976, effective earthing is defined as “During phase to earth faults, the voltage to earth voltage of the sound phases does not exceed 80 % of the voltage between lines of the system”. The AIEE Standard No. 32 defines effectively earthed system as that in which X0 / X1 < 3 and R0 / X1 < 1 A resistance earthed system will result in a value of K0 having an associated angle. Due to this angle, the voltage of the lagging sound phase will be less than that of the leading sound phase, for a single phase to earth fault. The voltage between the sound phases will remain at the rated value provided K2 = 1 for the duration of the fault. For large values of earthing resistance where 3 RE >> Z1, the voltages on the sound phases will be equal to or slightly greater than rated phase to phase voltage. 2.9 Neutral Displacement and Residual Voltages The neutral displacement voltage is the voltage between the system neutral and ground during earth faults. Let us consider a general case as shown in Fig. 2.14, a L-E fault will cause a fault current IF flowing through the earthing impedance ZE. If all impedances (Z1, Z2, Z0, & ZE) have the same angle, G will be on the phasor VAN at some point determined by the relative magnitudes. This is equivalent to the position of G shown on Fig. 2.14 for K2 and K0 having zero angle.

B-C-E fault Fig. 2.13

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for the fault of BC see the variation of A ph voltage for variation of earthing z i.e K0 when KO=inf then Va is 1and 1/2 times the Ea voltage and is the Ko=1 then Ea=Va and Va=0 if K0=0

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At the source, Therefore the neutral displacement voltage equals the zero sequence voltage V0 and can be obtained if necessary from the average of the phase to ground voltages or by measuring the voltage across the earthing impedance. The residual voltage can be measured by using either a three phase five limb VT or three single phase VTs each with the secondary windings of the phases connected in series or open delta.

(2.20)

Fig. 2.14

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1 if

313

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021

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Z

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033 VVVVVV NGCGBGAGres ==++=

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Residual voltage measurement can be used to detect earth faults on a system and to provide a polarising quantity for directional earth fault relays. It can be seen that the residual voltage Vres depends mainly on the value of K0 and ZE. On systems with high earthing impedance or insulated neutral, the value of Vres will approach 3 times rated phase to neutral voltage. On solidly earthed systems the value of K0 may be small and there may be insufficient neutral displacement to provide the necessary polarising quantities. For this reason, where directional earth fault relays are used at generating stations where the system is solidly earthed, current polarisation is preferred to voltage polarisation. As the voltage between the neutral of power transformers and earth is small during earth faults on solidly earthed systems, graded insulation can be used with considerable reduction in cost. 2.10 Residual Currents during Earth Faults The residual current is the sum of the phase currents and is obtained from the sum of CT output placed in each phase. The residual current contains only zero sequence current. As this is only present during earth faults, Ires is a good method of earth fault detection. On solidly earthed systems Ires will normally be large and can be measured with normal ring type CTs. On resistance earthed systems Ires will normally be of the order of rated full load current and again can be measured using normal ring type CTs. On high impedance earthed systems Ires will be very small and should therefore be measured using core type CTs or very accurate ring type CTs. On insulated and Peterson Coil earthed system, residual current will be measured on healthy and faulty feeders and therefore very sensitive directional relays are used. These are polarised from the residual voltage and will discriminate between faulty and non-faulty outgoing feeders. For a phase to earth fault, it can be shown that

(2.21)

where I3φ equals to the 3 phase fault current and equals to Ea / Z1 For a double phase to earth fault, it can be shown that

(2.22) It can be shown that for K2 = 1, the residual current for a double phase to earth fault is greater than that for a single phase to earth fault for values of K0 less than 1.0.

03IIIII cbares =++=

( ) ϕ3021

3 IKK

I res ++=

( ) ϕ30202

23 IKKKK

KI res ++=

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28

2.11 Sequence circuit of 3 phase auto transformer The positive and negative sequence network of a 3 phase auto transformer is shown in Fig. 2.15. Basically it is identical to ordinary 3 winding transformer. The zero sequence network of a 3 phase auto transformer is shown in Fig. 2.16 With the tertiary winding open circuited, let us consider the flow of zero sequence currents as shown in Fig. 2.17.

Fig. 2.16

Fig. 2.17

Fig. 2.15



29

Now consider the HV side open circuited but with the tertiary winding closed as shown in Fig. 2.18.

( )

NNI

NIII

Nn

nnEE

nnN

NI

nnnII

nnInI

L

LHL

C

CS

L

H

C

SL

CS

CLH

CSHCL

+=

⎥⎦⎤

⎢⎣⎡

+−=−

+=+

=

=+

=

+=

+=

1

111

1

where 1

1

0

000

0

00

00

( )

( )side LV toreferring

13

3

3

:of consists be toseems side, HV on the Similarly,1

3

3:of consists be toseems network, sequence zero In the

2

'

0

00''

0

00

NNZZZ

NZZ

IIIZZZ

Z

ZN

NZ

IZIIZZ

Z

nHY

nH

L

HLnHY

'Y

nL

L

nHLLX

X

+−=

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

++=

−+=

Fig. 2.18

(2.23)

(2.24)



30

This is equivalent to a short circuit test between the common and tertiary winding. 2.12 Open circuit fault – open circuit of phase conductor Consider the less extreme case where an impedance is connected in series with one phase open as shown in Fig. 2.19. Then That is, the voltage drops in the three sequence networks are each equal to: These conditions are satisfied by the following sequence network interconnection as shown in Fig. 2.20.

NZZZ

NNZZZ

ZZZZZZ

nTZ

nLX

nTLZXTC

++=∴

++=

++=+=−

113 ,

13 As

3

(2.25)

Fig. 2.19

ycxc

ybxb

aayaxa

EE

EE

ZIEE

=

=

+=

aayx

aayx

ZIEE

ZIEE

3131 similarly and

22

11

+=

+=

( )02131

31 IIIZZI aaa ++=

( )[ ]

( )[ ]

aayx

aaycybya

ycybaayax

ZIEE

ZIEEE

EEZIEE

31

3131 and

00

0

+=

+++=

+++=

(2.26)



31

In the extreme case of one phase being open circuit then Za /3 becomes infinite and the above circuit with Za /3 removes becomes applicable. 2.13 Simultaneous fault –one phase open circuit on one side and phase to earth fault on

the other side in the same phase It is very rare to have two faults happen in different locations simultaneously. The only simultaneous fault that is most likely to happen is the breakage of the overhead line conductor. One side of the broken conductor will thus have an open circuited fault while the conductor on the other side falls to ground causing a line to earth fault. The physical line arrangement is shown in Fig. 2.21 and the positive sequence network of this fault is shown in Fig. 2.22. Solving the simultaneous fault cannot be carried out superposition theory as the sequence network are not physical networks. The solution of this problem can only be carried out by using sequence network analysis and the analysis must fulfill the rules of both open circuit fault and one phase to earth fault analysis.

Fig. 2.20



32

Ia

Ib

Ic

Ia’

Ib’

Ic’

P Q

GFig. 2.21

Fig. 2.22

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )'00

'22

'11

'00

'22

2'11

'

'00

'22

'11

2

'021

true,be toFor this0 ,

0

0 ,

0

)( ,0

side, Pat fault circuit open For the

aaaaaa

aaaaaa

cc

aaaaaa

bb

aaa

a

VVVVVV

VVVVaVVa

VV

VVVVaVVa

VV

IIII

−=−=−

=−+−+−∴

=−

=−+−+−∴

=−

+−=∴=

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )'00

'22

'11

'00

'22

2'11

'

'00

'22

'11

2

'

'0

'2

'1

'

true,be toFor this0 ,

0

0 ,

0

0 ,

0 side, Qat fault earth tophase 1 For the

aaaaaa

aaaaaa

cc

aaaaaa

bb

aaa

a

IIIIII

IIIIaIIa

II

IIIIaIIa

II

VVV

V

+=+=+

=+++++∴

=+

=+++++∴

=+

=++∴

=

(2.18) (2.19)



33

The sequence network connections are shown in Fig. 2.23. 2.14 Further Reading 1. Paul M. Anderson, ‘Analysis of Faulted Power Systems’, IEEE Press Power Systems

Engineering Series, IEEE, 1995. 2.15 Self Assessment i. Do you understand the physical meaning of positive and negative sequence components? ii. What is the relationship between mutual coupling impedance and sequence impedance in

the power system? iii. How do you transform between the actual system voltages and currents into their

corresponding sequence components? iv. Do you understand why the actual system network can be transformed into sequence

networks? v. How the sequence networks are connected under various unsymmetrical faults? vi. Do you know how to form the positive and negative sequence networks?

Fig. 2.23



34

Fig. Q3

R

Z1=Z2=j10Ω Z0=j40Ω

Z1=Z2 Z0=2Z1

400 kV •

G

E=1.0 pu

vii. Why the zero sequence network of a power transformer is different and do you know how to form the zero sequence network of a power transformer? Now do you understand how to form the zero sequence network of the power system?

viii. Do you know how to solve tutorial questions from 1 to 10? ix. Can you name the earthing systems that are used for power systems? x. Why the earthing methods will affect the voltage on unfaulted phases under earth fault

condition? xi. How do you connect the sequence networks for an open circuited fault? xii. How to represent the simultaneous fault by sequence networks? xiii. Are you able to solve the open circuit and simultaneous fault problems from Q11 to Q15? 2.16 Tutorials Q1. A 10 MVA, 13.8 KV, 50 Hz generator has a direct-axis sub-transient reactance of 0.2 p.u., a negative sequence reactance of 0.15 p.u. and a zero sequence reactance of 0.1 p.u. In order to reduce the short-circuit current in case of a fault to earth, a current limiting reactor is connected to the neutral of the generator. Determine the inductance required to limit the sub-transient line current for a single-line-to-earth fault current to that of a three-phase fault. If a resistor is used instead of the reactor, what value of resistance is required?

(3.03 mH, 2.52 Ω) Q2. A 20 MVA, 11 KV, 3 phase alternator which is connected in star and solidly earthed, is connected directly to a cable. Determine the current in each phase, the fault current, and the voltage between the unfaulted end of the cable and earth, when a B-C phase to earth fault occurs at the far end of the cable. (0 A, -2451 + j 1769A, 2451 + j 1769A, j3538A, 4.5kV) The p.u. phase sequence reactances are: Positive Negative Zero Generator 0.3 0.2 0.05 Cable 0.1 0.1 0.16 Q3. In the single source system as shown in Fig Q3, a 3 phase fault occurs at the end of the line. By using a MVA base of 600 MVA, calculate the fault current and the fault MVA if:

i) The 400 kV busbar has a short circuit level of 35 GVA; and ii) The 400 kV busbar has a short circuit level of 5 GVA. (-j15.85kA, 10.98GVA, -j5.5kA, 3.81GVA)



35

Z1=Z2= j20Ω Z0= j80Ω

(for the whole line)

G

G • A

•BZLA ZLB

50%

100%

Z1=Z2 Z0=2Z1

E=1.0p E=1.0p

Z1=Z2 Z0=2Z1

Q4. Repeat Q3 to calculate the fault current for: i) A single phase to earth fault on A phase; and (8.85kA, 3.68kA) ii) A phase to phase fault on B to C phase. (-13.74kA, 4.76kA)

Q5. In the two sources system as shown in Fig Q5, the short circuit level of the 400 kV busbars in side A and B is 5 GVA and 35 GVA respectively. An A-E fault occurs at the location shown in the figure. By using a MVA base of 600 MVA, calculate:

i) the total fault current; and (-j18.3, 0.686) ii) the fault current contributing from both ends of the line. (-j6.35, 0.238) What is the ratio of fault current contribution from both ends of the line? What causes that?

What effect can you think of in this situation? Q6. A 20 MVA 132/11 kV 3 phase transformer with winding connected in Dy11 is supplied from a source of negligible impedance. The leakage impedance of the transformer is 10 % and the neutral of the star connected winding is floating. If a B-C phase fault happens on the LV side near the transformer terminals, calculate the p.u. current in all three phases on the HV side of the transformer based on 20 MVA base. (-j5, -j5, j10) Q7. If the transformer in Q6 is supplied from a source of fault level 2000 MVA and the star point is now solidly earthed. Calculate the p.u. current in all three phases on the HV side of the transformer based on 20 MVA base if an A phase to earth fault occurs on the 11 kV side. (5.41, -5.41, 0)

Fig. Q5



36

Q8. Fig. Q8 shows a 275 KV, 3 phase overhead transmission line 80 km long which is supplied from both ends through identical delta/star connected transformers ‘A’ and ‘B’, each rated at 100 MVA, 11/275 kV. The star-point of both transformers are solidly earthed and the symmetrical short circuit level at the input terminals of ‘A’ is 1000 MVA, while that at the input terminals of ‘B’ is 2000 MVA. If an A phase to earth fault occurs on one overhead conductor at a position 20 km from ‘B’, find the fault current flowing into the fault F and the current seen by protective relays installed on each phase of the HV side of the transformer windings. The relevant reactance are given below: Transformer X1 = X2 = Xo = 0.15 p.u. based on rating Transmission line X1 = X2 = 0.8 Ω/km Xo = 1.6 Ω/km.

(1689 A, 696 A, 0 A, 0 A, 993 A, 0 A, 0 A.) Q9. For the same network as shown in Q8, if the source of Tx. B is disconnected at the L.V. side, calculate again, for an A phase to earth fault, the total fault current and the current seen by relays on the H.V. side of the transformers. (850 A, 683 A, 167 A, 167 A, 167 A, 167 A, 167 A.) Q10. In the two sources system as shown in Fig Q10, the short circuit level of the 400 kV busbars in side X and Y is 30 GVA and 10 GVA respectively. The neutral of the source on side X is solidly earthed while the neutral of the source on side Y is floating. A fault occurs at the location shown in the figure. By using a MVA base of 1000 MVA, calculate the fault current in each phase in side X and Y if

i) the fault is a three phase fault; and (11.36kA, 7.45kA) ii) the fault is a single phase to earth fault on A phase. (6.78kA, 1.22kA, 1.22kA; 2.43kA, 1.22kA, 1.22kA)

Suggest a method to limit the earth fault current in this case. Neglect the effect of load in the above calculations.

Fig. Q8



37

Fig. Q10

Z1=Z2= j30Ω Z0= j80Ω

(for the whole line)

G

G • X

•YZLA ZLB

50%

100%

Z1=Z2 Z0=2Z1

E=1.0p E=1.0p

Z1=Z2

Q11. In the distribution circuit shown in Fig. Q11, ‘A’ phase is open-circuited at load terminals, calculate:

i) magnitude of current in healthy phases; and ii) voltage across the open circuit.

(0.95∠222°, 0.95∠102°, 0.32∠-7.1°) Q12. An unloaded 3 phase star connected bank of single phase transformers having the star point isolated is fed over a cable having a capacitance of 3μF/phase from a large 66 kV, 50 Hz system having an earthed neutral point. The transformer takes a magnetizing circuit of 30A at 66 kV and the magnetizing reactance may be assumed constant. If one conductor breaks between the cable and the supply, estimate, neglecting reactance, the steady state voltage which would appear across the break. (14.2 kV)

kV 11 10MVA, toreferring pu 2.0pu 2.0pu 4.0

01

2

1

⎪⎭

⎪⎬

⎫

===

jZjZjZ

T

T

T

Fig. Q11



38

Q13. For the circuit shown in Fig. Q13, calculate the current distribution in the transformer windings and the neutral currents at the source and transformer earthing points.

Note: Voltages are phase voltage. All impedance referred to the 100 kV winding, i.e., √3 x 100 kV phase to phase.

Q14. If the source in Q13 is unearthed, calculate the new current distribution.

Fig. Q13



39

Q15. In the generator / generator transformer circuit shown in Fig. Q15, find the current distribution in the primary and secondary circuit for:

i) A L-E fault at P of If = 900A with delta opened, ii) A L-E fault at Q of If = 900A with delta opened, iii) A L-E fault at P of If = 900A with delta closed, and iv) A L-E fault at Q of If = 900A with delta closed.

Fig. Q15



40

Q16. (a) An EHV transmission system is shown in Fig. Q16. An ‘A’ phase to earth fault occurs at 70% of the line length from the relaying point X with a fault resistance of 100Ω. Neglect the effect of the load, calculate the current and phase to ground voltage of the faulted phase at locations X of the 400kV line. (1375A, 189.5 kV) (b) Distance relays are installed at locations X to protect the 400kV line. The zone 1 setting of the relay is set to cover 80% of the line length. Based on the above calculations, calculate the impedance seen by the phase-earth fault distance relays at the faulted phase and comment on the result obtained. (89.6 + j 33.9 Ω) The relay input currents are compensated by:

Where IR = Relay current Iph = phase current In = neutral current Z1 = line positive sequence impedance Z0 = line zero sequence impedance

(c) Comment on the accuracy of distance relays due to the effect of remote end infeed and the fault resistance in the above case.

Fig. Q16

1

10

3ZZZ

III nphR−

+=



41

Q17. (a) An e.h.v. transmission system is shown in Fig. Q17. During a typhoon, the A phase conductor at 70 % of the line length from the relaying point X was broken thus forming an open circuit fault. The conductor on the other side (Y) fell to the ground forming a line to earth fault. Nelect the effect of the load, calculate the current and phase to ground voltage of the faulted phase at locations X and Y on both sides of the 400 kV line. (0, 0.957-j0.0084, 0.158-j0.809, 0.085-j0.015)

(b)Distance relays are installed at locations X and Y protecting the 400 kV line. Each relay on both sides will set to cover 80 % of the line length. Based on the above calculations, calculate the impedance seen by both phase-earth fault distance relays at the faulted phase and comment on the result obtained. (42.6-j236Ω, 0.9+j10.8Ω)

The relay input current is compensated by:

Where IR = relay current Iph = phase current

In = neutral current Z1 = line positive sequence impedance Z0 = line zero sequence impedance (c) When a fault happens in the power system, what will be the transient voltage and current waveform that will appear at the relay location in the first few cycles. Does it affect the relay performance.

1

10

3ZZZ

III nphR−

+=

XH= XL= XT=0.05

Fig. Q17



42

Q18. (a) Name the sources of error in impedance measurement in distance relays? (a) An e.h.v. transmission System is shown in Fig. Q18 Distance relays are installed at locations X to protect the 400 kV line. The relay is set to cover 80 % of the line length. An A phase to earth fault with a fault resistance of 10Ω occurs at 70 % of the line length from the relaying point X. Neglect the effect of the load, calculate the impedance seen by the relay of the faulted phase at locations X when (i) the circuit breaker (C.B.) at the other side of the line is opened; and (ii) the circuit breaker (C.B.) at the other side of the line is closed. (7.2+j26.2Ω, 14.8+j27Ω)

Explain why the zero sequence impedance compensation does not give correct measurement in this case.



In = neutral current Z1 = line positive sequence impedance Z0 = line zero sequence impedance

1

10

3ZZZ

III nphR−

+=

Fig. Q18



43

Q19. A 275 kV transmission line is protected by two distance relays located at both ends of the line (X and Y) as shown in Fig.Q19. During a typhoon, the A phase conductor at 50% of the transmission line was broken. The conductor on side Y fell to ground thus forming a line to earth fault. The conductor of the other side (X) remain open circuited.

i) Sketch the sequence network and show how the fault current can be calculated. ii) Neglect the effect of the load, calculate the impedance seen by the distance

relay of the faulted phase at Y. (j12.5Ω)



In = neutral current Z1 = line positive sequence impedance Z0 = line zero sequence impedance Use 500 MVA as the MVA base in your calculation.

1

10

3ZZZ

III nphR−

+=

Fig. Q19

25/275kV 750MVA X1= X2= X0

= 0.15p.u.

275/33kV 500MVA X1= X2= X0 = 0.125p.u.

25kV 500MW X1= 0.15p.u. X2= 0.07p.u. X0= 0.05p.u.

Z1= Z2= j 25Ω Z0= j 42Ω

(for the whole length)

Y

50%

100%

275kV X

modern power system fault calculation

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