3 Fault Calculation

8/6/2019 3 Fault Calculation

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33.1

A power system is normally rreared as a balancea tnree-phase network. In general, when a fault occurs, thesymmetry of a balanced network is upset, resulting inunbalanced currents and voltages appearing in the net-work. The exception to this rule is the three-phase fault,which, because it involves all three phases equally at thesame location, is described as a symmetrical fault. By usingsymmetrical component theory and employing the conceptof replacing the normal system sources by a sowce at thepoint of fault it is possible to analyze these fault conditions.

From the protective gear application point of view it is

essential to know the fault current distribution through thesystem and the voltages in different parts of the system dueto the fault. Further, boundary values of current at anyrelaying point must be known if the fault is to be clearedwith discrimination. The information normally required is:

i .

i i .

Maximum fault current for a fault at the relaying point.Minimum fault current for a fault at the relaying point.

i i i . Maximum through fault current at the relaying point.

To obtain the above information the limits of stable gener-ation and possible operating conditions including themethod of system earthing must be known, and the faultsare always assumed to be through zero fault impedance.3 .2

,( i,&)$ S!S._ . , *.,. ii ._ I. . .

It can be shown that, by applying the Principle of Super-position, any general three-phase system of vectors maybe replaced by three sets of balanced (symmetrical) vec-tors; two sets are three-phase but having opposite phaserotation and one set is co-phasal. These vector sets aredescribed as positive, negative and zero sequence setsrespectively.

The equations between phase and sequence quantities aregiven below:

F + E,,Fb = aF, + aE2+F

~ I0 .Equation 3 . 1

FL = ar, + a& +EoE, = :(r = +(ru + a& +aE


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where all quantities are referred to the reference phase A.A similar set of equations can be written for phase andsequence currents. Figure 3.1 illustrates the resolution of asystem of unbalanced vectors.

When a fault occurs in a power system, the phase im-pedances are no longer identical (except in the case ofthree-phase faults) and the resulting currents and voltagesare unbalanced, the point of greatest unbalance being atthe fault point. It has previously been shown that the faultmay be studied by short-circuiting all normal driving vol-tages in the system and replacing the fault connection bya source whose driving voltage is equal to the pre-fault

voltage at the fault point. Hence, the system impedancesremain symmetrical, viewed from the fault, and the faultpoint may now be regarded as the point of injection ofunbalanced voltages and currents into the system.

This is a most important approach in defining the faultconditions since it allows the system to be represented bysequence networkst using the method of symmetricalcomponents.

The networks are described as positive, negative and zerosequence networks and only the appropriate sequencecurrents and voltages appear in them, there being nomutual connections whatsoever between the networks.

32.1

Pos i t i ve sequence ne tw orkDuring normal system conditions only positive sequencecurrents and voltages can exist in the system, and thereforethe normal system impedance network is a positivesequence network.

When a fau_lt occurs the current in the fault branch changesfrom 0 to I, and the p_ositiye sequence voltage across thebranch changes from !I to V,: replacing the fault branch bya source equal to the change in voltage and short-circuiting all nprmal driving voltages in the system resultsin a current AI flowing into the system, and:

.Equation 3 . 3where7, is the positive sequence impedance of the system

viewed from the fault. As before the fault no current wasflowing from the fault into the system, it follows that r,, thefault current flowing from the system into the fault, mustequal ~ AI. Therefore:v, = v - I,Z, .Equation 3 . 4is the relationship between positive sequence currents andvoltages in the fault branch during a fault.

In Figure 3.2,-which represents a simple system, the vol-tage drops I;Z; and i;Z; are equal to (v -v,) where therents i; and ilenter the fault from the left and right respec-

z and .??Ypedances viewed from either side of the fault branch. The

voltage Vis usually equal to the open,circuit voltage in thesystem, and it has been shown that V-Te E (see Sec-tion 2.7). So the positive sequence voltages in the systemdue to the fault are greatest at the sourca, as shown in thegradient diagram, Figure 3.2(b).

3.2.2

Negat ive sequence network

On the premise that only positive sequence quantmes ap-pear in a power system under normal conditions, thennegative sequence quantities can exist only during anunbalanced fault.

K(b)GRADIENT DIAGRAMFigure 3.2 Positive sequence diagrams of a simplesysrem wth

a fault at F

If no negative sequence quantities are present in the faultbranch prior to the ault, then, when a fault occurs, thechange in voltage is V,, and the resulting current I2 flowingfrom the network into the fault is:

orv , = -1222 .Equation 3 . 5The impedances in the negative sequence network aregenerally the same 5s these in the positive sequence net-work. In machines Z, #Z2, but the difference is generallyignored, particularly on large systems.

The negative sequence diagrams, shown in Figure 3.3, aresimilar to the positive sequence diagrams, with two impor-tant differences; no driving voltage> exist before the faultand the negative sequence voltage V, is greatest at the fault ;point.

Method of Symmetrical Co-ordinates Applied to the Solution ofPolyphase Networks. C. L.Fortescue. A.I.E.E.Trans. Vol. 37. PartII1918.pp . i,027-l 140.

t Power Sysrem Analysis. J. R. Mortlock and M. W. Humphrey Davies, Chapman &Hall. c

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2.3ero sequence network

he same current and voltage relationships apply in thesequence network as in the negative sequence net-

work during a fault condition. Hence:

-Iz .Equation 3 . 6Also. the zero sequence diagram is as shown in Figure 3.3when IO is substituted for I,, and so on.

hecurrents and voltages in the zero sequence network arethat is, all the same phase. So for zero sequence

urrents to flow in a system there must be a return connec-on, through either a neutral conductor or the general

mass of earth. Note must be taken of this fact when deter-mining Nero-sequence equivalent circuits. Further, in

eneral Z,#Z,, and the value of Z,, varies according to theype of plant, the winding arrangement and the method ofarthing.

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The most important types of faults are as follows

. Single-phase-earth.

b. Double-phase.

Double-phase-earth.

. Three-phase (with or without earth).The above faults are described as single shunt faultsbecause they occur at one location and involve a connec-ion between one phase and another or to earth.

determining the currents and voltages at the fault pointt is possible to define the fault and connect the sequence

networks in such a manner as to represent the fault con-dition. From the initial equations and the network diagramt is then possible to determine the nature of the fault

currents and voltages in different branches of the system.

Neglecting load currant and assuming that the shunt faultsisted above are through zero impedance, the equations

defining each fault can be written down as follows:

a . Single-phase+earth ( A - E ) Ib = 0I, = 0

>, = 0.Equation 3 . 7

b . Double-Dhase ( B - C ) I, = 0

Ih = I,>, = V

.Equation 3 . 8

c. Double-phase-earth (B-C-E)

I, = 0

v , = 01, = 0

.Equation 3 . 9

d. Three-phase (A-B-C or A-B-C-E)I,, + I,> + I, = 0V = v ,

1.Eauatlon 3 . 1 0

Vb = v ,It should be noted from the above that for any type of faultthere are three equations which define the fault conditions.

All currents and voltages are phase to neutral values at thefault point and it is assumed that no load current is flowing.When there is a fault impedance, this must be taken intoaccount when writing down the equations. For example,with a single-phase earth fault through a fault impedanceZ, Equations 3.7 are rewritten:

.Equation 3.11

3.3 .1

Single-phase-earth faul t (A-E)

Consider a fault defined by Equations 3.7 and by Figure3.4(a). Converting Equations3.7 into sequence quantitiesby using Equations 3.1 and 3.2, then:

i, = i2 = i. = :i., .Equation 3 . 1 2 v, = (V? + Vo) .Equation 3 . 1 3

- -Substituting for V,, Vz and v,, in Equatron 3.13from Equa-tions 3.4, 3.5and 3.6:

but, from Equation 3.12. i, = iI = i,. therefore:V=I,(z ,+T ,+z ,) .Equation 3 . 1 4 The constraints imposed by Equation 3.14 indicate that theequivalent circuit for the fault is obtained by connectingthe sequence networks in series, as shown in Figure3.4(b).


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3.3.2Double-phase faul t (B-C)From Equation 3.8 and using Equations 3.1 and 3.2:

f , = -i2& = 0 .Equation 3.15v , = VI .Equation 3 . 1 6 From network Equations 3.4 and 3.5. Equation 3.16 can ben-written:

v - I,Z, = -12Z,and substituting for fz from Equation 3.15:v = I, (Z, +.&) .Equation 3 . 1 7 The constraints imposed by Equations 3.15 and 3.77 in-dicate that there is no zero sequence network connectionin the equivalent circuit and that the positive and negativesequence networks are connected in parallel. Figure 3.5shows the defining and equivalent circuits satisfying theabove equations.

3.3.3Double-phase-earth faul t (B-C-E)Again, from Equation 3.9and Equations 3.1 and 3.2:

i , = - (i , + io ) .Equation 3 . 1 8 and

v, = V? = i7 .Equation 3 . 1 9 Substituting for vz and V, using network Equations3.5 and3.6:

I2Z1 = I,Z,Thus, using Equation 3.18:

.Equation 3.20

.Equation 3.21Now, equating V, and i?? and using Equation 3.4gives:ii~ I,Z, = - I,Z>orv = I,Z,~I>Z,Substituting for f, from Equation3.27:

i , = V (Z +72)Zf + ZlZ +ZZ2 .Equation 3 . 2 2 From the above equations it follows that a double-phaseearth fault may be represented by connecting the threesequence networks in parallel as shown in Figure 3.6(b).

Figure 3.6 Doiihle-phase-earrh fmlr (S C E)3.3.4Three-phase faul t (A -BmC o r A B C-E)Assuming that the fault includes earth, then, from Equa-tions 3.10 and 3.1, 3.2, it follows that:

.Equation 3 . 2 3

and

i , = 0 .Equation 3 . 2 4 Substituting !iz = 0in Equation 3.5gives:i, = 0 .Equation 3 . 2 5 and substituting v, = 0in Equation 3.4:

- .~0 = v - I ,Z,or

v = I ,Z, .Equation 3 . 2 6 Fbrther. since froE Equation 3.24 1, = 0, it follows fromEquation3.6thatV,iszero when .?,,isfinite. Theequivalentsequence connections for a three-phase fault are shown inFigure 3.7.

lRENT ANDVOLTAGE DISTRIBUTION.__..__. .__.. __.__.l..l.. ..-..- ... -xamination of the

effect of a fault in branches of a network other than thefault branch, so that protection can be applied correctly toisolate the section of the system directly involved in thefault. It is therefore not enough to calculate the faultcurrent in the fault itself: the fault currant distribution mustalso be established. Further, abnormal voltage stresses mayappear in a system because of a fault, and these may affectthe operation of the protection. A knowledge of currentand voltage distribution in a system due to a fault is essen-tial for the application of protection.

The approach to system fault studies for assessing protec-five gear application may be summarized as follows:

a. From thesystem diagram and accompanying data assessthe limits of stable generation and possible operating con-ditions for the system.

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OTE: When full information is not available assumptionsay have to be made.

With faults assumed to occur at each relaying point inrn, maximum and minimum fault currents entering theult are calculated for each type of fault.

OTE: The fault is assumed to be through zero impedance.

By calculating the current distribution for faults at dif-rent points in the system (from (b) above) the maximum

hrough fault currents at each relaying point arestablished for each type of fault.

At this stage more or less definite ideas on the type ofrotection to be applied are formed. Further calculations

or establishing voltage variation at the relaying point, ore stability limit of the system with a fault on it, are now

arried out in order to determine the class of protectionecessary, such as high or low speed, unit or non-unit. ando forth.

4.1

urrent d is t r ibut ion

he phase current in any branch of a network is dererminedom the sequence current distribution in the equivalentrcuit for the fault. The sequence currents are expressed iner unit terms of the sequence current in the fault branch.

ince normally in power system calculations the positiveequence and negative sequence impedances are equal,he division of sequence currents in the two networks isentical.

he impedance values and configuration of the zeroequence network are usually different from those of theositive and negative sequence networks, so the zeroequence current distribution is calculated separately.

Ca and C, are described as zero and positive sequenceistribution factors then the actual current in a sequence

ranch is given by multiplying the actual current in theequence fault branch by-the appropriate distribution fac-or. For this reason, if ii,I; and 7; are sequence currents inn arbitrary branch of a network due to a fault at someoint in the network, then the phase currents in that branch

may be expressed in terms of the distribution constants andhe sequence currents in the fault. These are given belowor the various common shunt faults. using Equation 3.1nd the appropriate fault equations:

. Single-phase-earth (A-f)

= (2C, + &Jr,= (C, - C)i

I.Equation 3 .27

= -(C, -C)i. Double-phase (B-C)

.Equation 3 .28

c. Double-phase-earth (B-C-E)i: = -(C, -&)I

f: =r (2-.)c,$-ac , + coi.I Jd. Three-phase (A-B-C orA-&C-E)i:, =c,I,

,c, =aT,I,-- I

.Equation 3 . 3 0 i: = aC,I,As an example of current distribution technique, considerthe system in Figure 3.8(a) and the equivalent sequencenetworks in Figure 3.8(b) and (c). A fault is assumed at Aand it is desired to find the currents in branch 06 due to thefault. In each network the distribution factors are given foreach branch. with the current in the fault branch taken as1 .O p.u. From the diagram the zero sequence distributionfactor Co in branch OB is 0.112 and the positive sequencefactor C, is 0.373. For an earth fault at A the phase currentsin branch 05 from Equation 3.27are:i:, = (0.746 + 0.1 I 2)io

= 0.85810and

f6 = i: = (0.373 - 0.1 I2)i,= - 0.261 i,

I I

I POWER SYSTEM I

i7-5 * O-OS,04 n 10.4 n

A B,2+0 o iVS*

111--1.0 0.155 0~1120.755 i4-8*0.053 I0.192(b) ZERO SEQUENCE NETWORK.0.42

L

(c, POSITIVE AND NEGATIVE SEOUENCE NETWORKSFigure3.8 Typicalpower sysrem

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By using network reduction methods and.assumiEg that allimpedances are reactive, it can be shown that 2,=2, =j0.68 ohms.

Therefore, from Equation3.14. the current in fault branchII, 1= 1Vl/O.68.Assuming that !/ = 63.5 volts, then:II,,I= 4 I LI= 63.5

3 x 0.68 = 31.2A

If ii is taken as the reference vector, then:I: = 26.81-90Af;, = f: = 8.15/90AThe vector diagram for the above fault condition is shownin Figure 3.9.

/i;=2fi.s~090

!7px .5& 6.4.

branch OEdue TOs;ngle-phase f&Ita,busA3.4.2Vol tage d is t r ibu t ion

The voltage distribution in any branch of a network isdetermined from the sequence voltage distribution. Asshown by Equations 3.4, 3.5 and 3.6 and the gradientdiagrams, Figure 3.2(b) and 3.3(b), the positive sequencevoltage is a minimum at the fault, whereas the zero andnegative sequence voltages are a maximum. Thus, the

sequence voltages in any part of the system may be givengenerally as:

Iquation 3 . 3 1

Using the above equation the fault voltages at bus B in theprevious example can be found.From the positive sequence distribution diagram Figure3.8(c):

v; = Y- I,[Z, -j{(O.395x 0.75) + ( 0 . 3 7 3 x0.45)j]= ii - I, [Z, jO.4641

E =~I>[Z, ~jO.4641From the zero sequence distriburlon diagram Figure3.8(b):

%= In[ZO-j((O.165 x 2.6) + (0.112 x1.6))]= Io[ZO ~ jO.6081

For earth faults, at the fault i, = I, = i . = -j31.2A, whenIV: = 63_5 voJts and is taken as the reference vector.Further, Z, = Z, = j0.68 ohms.Hence:

Vi = 63.5 - (0.216 x 31.2)= 56.76m volts

i% = 6.74/1 voltsV;l = 2.25m voltsand using Equations 3. I:v=p+v+p// I 2

= 56.76 - (6.74 + 2.25)!? = 47.8,!!X voltsvi = aq + aP +P

= 56.76a*~(6.74a + 2.25)v;, = 61.5/- 116.4 voltsli: = av; + a2v; + v0

= 56.753 -(6.74a2 + 2.25)v: =61.5/116.4 voltsThese voltages are shown on vector diagram Figure 3.9.

3.5

It has been shown previously that zero sequence currents

flow in the earth path during earth faults. and it follows thatthe nature of these currents will be influenced by themethod of earthing. Because these quantities are unique intheir association with earth faults they can be utilized inprotection. provided their measurement and character areunderstood for all practical system conditions.

3.5 .1

Residual cur rent and vo l tage

Residual currents and voltages depend for their existenceon two factors:

a. A system connection to earth at two or more points.

b. A potential difference between the earthed pointsresulting in a current flow in the earth paths.

Under normal system operation there is capacitance bet-ween the phases and between phase and earth; thesecapacitances may be regarded as being symmetrical anddistributed uniformly through the system. So even when(a) above is satisfied, if the driving voltages ate symmetri-cal the vector sum of the currents will equate to zero andno current will flow between any two earth points in thesystem. When a fault to earth occurs in a system anunbalance results in condition (b) being satisfied. From thedefinitions given above it follows that residual currents andvoltages are the vector sum of phase currents and phasevoltages respectively.

Hence:

fn = i ,, f ih +I,and

v, = v,,, + Vb, +v,,.Equation 3 . 3 2

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lso, from Equations 3.2:

= si,= 3vo > .Equation 3 .33

should be further noted that:

= V,,! + VI,


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3.5.3Var ia t ion o f res idua l quant i t ies

The variation of residual quantities in a system due T Odifferent earth arrangements can be most readily under-stood by using vector diagrams. Three examples have beenchosen, namely solid fault-isolated neutral, solid fault-resistance neutral, and resistance fault-solid neutral. Theseare illustrated in Figures 3.12, 3.13 and 3.14 respectively.

a. Solid fault-isolated neutral.

From Figure 3.12 it can be seen that the capacitance toearth of the faulted phase is short circuited by the fault and

the resulting unbalance causes capacitance currents toflow into the fault, returning via sound phases throughsound phase capacitances to earth.

At fault:

Vd = 0and

v, = v,>, + v,*-= 3E , , ,

At source

since

Thus, with an isolated neutral system, the residual voltageis three times the normal phase neutral voltage of thefaulted phase and there is no variation between iiR atsource and vK at fault.In practice there is some leakage impedance betweenneutral and earth and a small residual current would bedetected at X if a very sensitive relay were employed.

b. Solid fault&resistance neutral.

Figure 3.13 shows that the capacitance of the faultedphase is short circuited by the fault and the neutral current

combines with the sound phase capacitive currents to giveI, in the faulted phase.

With a relay at X, residually connected as shown in Figure3.10, the residual current will be i,. that is, the neutral-earth loou current.

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(b) ECTCW DIAGRAM

(C) RESIDUAL VOLTAGE DIAGRAM

At fault:

VR = IT,>, + I/,, since ii , = 0At source:

From the residual voltage diagram it is clear that there islittle variation in the residual voltages at source and fault,as most residual voltage is dropped across the neutralresistor. The degree of variation in residual quantities istherefore dependent on the neutral resistor value.

c. Resistance fault-solid neutral.Capacitance can be neglected because, since thecapacitance of the faulted phase is not short-circuited, the 1circulating capacitance currents will be negligible. iAt fault:

At relaying point X:v, = v, + v,,, + V,From the residual voltage diagrams, shown in Figure 3.14,it is apparent that the residual voltage is greatest at the faultand reduces towards the source. If the resistance in the

&rlt approaches zero, that is, the fault becomes solid, thenV,, approaches zero and the voltage drops in & and 7,become greater. The ultimate value of i?Fe will depend uponthe effectiveness of the earthing, and this is a function ofthe system Z,/Z, ratio.


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