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Solution:The base voltage at the generator is (220*13.5/220) 13.5 kV, and on the load side is(220*33/220) 33 kV. [Since we have selected the voltage base as that corresponding to thevoltage on that side of the transformer, we automatically get the voltage on the other sideof the transformer as the base on that side of the transformer and the above calculation is infact unnecessary.Generators GReactance of 0.85 pu corresponds 0.355 pu on 500 MVA, 13.5 kV base (see earlierexample)Generator voltage of 1.15 corresponds to 1.176 on 500 MVA, 13.5 kV baseTransformer T1Reactance of 8% (or 0.08 pu) remains unchanged as the given base is the same as the newchosen base.Transmission LineReactance of 7.8 Ω corresponds to 7.8 * 500/2202 0.081 puTransformer T2Reactance of 11% (0.11 pu) corresponds to 0.11 * 500/400 0.1375 pu(voltage base is unchanged and does not come into the calculations)LoadLoad of 250 MVA at a power factor of 0.85 corresponds to 250/500 0.5 pu at a powerfactor of 0.85 lag (power factor angle 31.79 ) resistance of load 0.5 * 0.85 0.425 puand reactance of load 0.5 * sin 31.79 0.263 puThe circuit may be expressed in per unit as shown in figure 2.4.1.176 puj0.3551.176 puj0.355j0.08j0.081j0.1380.425 j0.263Figure 2.4 - Circuit with per unit values2.3 Symmetrical Three Phase Fault AnalysisA three phase fault is a condition where either (a) all three phases of the system are shortcircuited to each other, or (b) all three phase of the system are earthed.3 φ faultSupplyside3 φ to earth faultabaSupplysidebcFigure 2.5a – Balanced three phase faultcFigure 2.5b – Balanced three phase fault toThis is in general a balanced condition, and we need to only know the positive-sequencenetwork to analyse faults. Further, the single line diagram can be used, as all three phasescarry equal currents displaced by 120o.EE 423 - Power System Analysis: Faults – J R Lucas – October 20056

Typically, only 5% of the initial faults in a power system, are three phase faults with orwithout earth. Of the unbalanced faults, 80 % are line-earth and 15% are double line faultswith or without earth and which can often deteriorate to 3 phase fault. Broken conductorfaults account for the rest.2.3.1 Fault Level CalculationsIn a power system, the maximum the fault current (or fault MVA) that can flow into a zeroimpedance fault is necessary to be known for switch gear solution. This can either be thebalanced three phase value or the value at an asymmetrical condition. The Fault Leveldefines the value for the symmetrical condition. The fault level is usually expressed inMVA (or corresponding per-unit value), with the maximum fault current value beingconverted using the nominal voltage rating.MVAbase 3 . Nominal Voltage(kV) . Ibase (kA)MVAFault 3 . Nominal Voltage(kV) . Isc (kA)whereMVAFault – Fault Level at a given point in MVAIbase– Rated or base line currentIsc– Short circuit line current flowing in to a faultThe per unit value of the Fault Level may thus be written asFault Level 3 . Nominal Voltage . I sc3 . Nominal Voltage . I base 3I sc3I base I sc , pu V No min al , puZ puThe per unit voltage for nominal value is unity, so thatFault Level ( pu ) 1,Z puFault MVA Fault Level ( pu ) MVAbase MVAbaseZ puThe Short circuit capacity (SCC) of a busbar is the fault level of the busbar. The strengthof a busbar (or the ability to maintain its voltage) is directly proportional to its SCC. Aninfinitely strong bus (or Infinite bus bar) has an infinite SCC, with a zero equivalentimpedance and will maintain its voltage under all conditions.Magnitude of short circuit current is time dependant due to synchronous generators. It isinitially at its largest value and decreasing to steady value. These higher fault levels taxCircuit Breakers adversely so that current limiting reactors are sometimes used.The Short circuit MVA is a better indicator of the stress on CBs than the short circuitcurrent as CB has to withstand recovery voltage across breaker following arc interruption.The currents flowing during a fault is determined by the internal emfs of machines in thenetwork, by the impedances of the machines, and by the impedances between themachines and the fault.Figure 2.6 shows a part of a power system, where the rest of the system at two points ofcoupling have been represented by their Thevenin’s equivalent circuit (or by a voltagesource of 1 pu together its fault level which corresponds to the per unit value of theeffective Thevenin’s impedance).EE 423 - Power System Analysis: Faults – J R Lucas – October 20057

Fault Level 8 puFault Level 5 pu12Z1 0.3 puZ2 0.3 puCB2CB13IFFaultFigure 2.6 – Circuit for Fault Level CalculationWith CB1 and CB2 open, short circuit capacities areSCC at bus 1 8 p.u. gives Zg1 1/8 0.125 puSCC at bus 2 5 p.u. gives Zg2 1/5 0.20 puEach of the lines are given to have a per unit impedance of 0.3 pu.Z1 Z2 0.3 p.u.With CB1 and CB2 closed, what would be the SCCs (or Fault Levels) of the busbars in thesystem ?o0.3 puZin0.2 pu0.125 pu1V3V 1 puV 1 pu30.3 pu 23ZfZfSystem Equivalent CircuitThevenin’s Equivalent at 3Figure 2.7a Determination of Short circuit capacitiesThis circuit can be reduced and analysed as in figure 2.7b.0.4250.425 3 0.5E 1 E 10.23 Bus3Bus3 0.51Figure 2.7b Determination of Short circuit capacity at Bus 3Thus, the equivalent input impedance is given by to give Zin as 0.23 pu at bus 3,so that the short circuit capacity at busbar 3 is given as SCC3 1/0.23 4.35 p.uThe network may also be reduced keeping the identity of Bus 1 as in figure 2.7c.0.125E 10.1251 0.8E 1 0.8Bus1 0.108 Bus11Figure 2.7c Determination of Short circuit capacity at Bus 1EE 423 - Power System Analysis: Faults – J R Lucas – October 20058

Thus, the equivalent input impedance is given by to give Zin as 0.108 pu at bus 1, so thatthe short circuit capacity at busbar 1 is given as SCC1 1/0.108 9.25 p.uThis is a 16% increase on the short circuit capacity of bus 1 with the circuit breakers open.The network may also be reduced keeping the identity of Bus 2. This would yield a valueof Zin as 0.157 pu, giving the short circuit capacity at busbar 2 as SCC2 1/0.157 6.37 p.uThis is a 28% increase on the short circuit capacity of bus 2 with the circuit breakers open.Typical maximum values of short circuit capacities at substations in Sri Lanka in 2000 areshown in table 2.1. Actual fault currents are lower than these values due to the presence offault impedance in the adampeMataraMatugamaNuwara 3435428628641763389697Puttalam 680572274463429183423509Average SCCStd deviationAverage Isc377 MVA166 MVA6.6 kATable 2.1 – Maximum 3φ Fault Levels at 33kV Substations in Sri Lanka in 20002.4 Fault Currents in synchronous machines14current(pu)12108Envelope of transient decayd.c. transient decaya.c. transient decay6420-2050100150200250time (ms)- - sub transient - - transient - - - - - - - - - - - - - - - - - - - steady state - -Figure 2.8 – Transient decay of current in synchronous generatorEE 423 - Power System Analysis: Faults – J R Lucas – October 20059

As mentioned earlier, the currents flowing in the power system network during a fault isdependant on the machines connected to the system. Due to the effect of armature currenton the flux that generates the voltage, the currents flowing in a synchronous machinediffers immediately after the occurrence of the fault, a few cycles later, and under sustainedor steady-state conditions.Further there is an exponentially decaying d.c. component caused by the instantaneousvalue at the instant of fault occurring. These are shown in figure 2.8.2.5current (pu)22.5current (pu)21.51.5110.50.500- 0.5-0.5-1-1- 1.5050100150-1.50200250time (ms)50100Figure 2.9a & b – Steady state and Transient current150200250time (ms)Figure 2.9a and 2.9b show the steady state current waveform, and the transient waveformof a simple R-L circuit, to show the decay in the d.c. component. In addition to this, in thesynchronous machine, the magnitude of the a.c. current peak also changes with time asshown in figure 2.9c, with the unidirection component of the transient waveform removed.currentcbao050100150200Figure 2.9c–Synchronous machine transient250Due to the initial low backemf at the instant of faultresulting in high current, theeffective impedance is verylow. Even when the d.c.transient component is notpresent, the initial currentcan be several times thesteady state value.Thusthree regions are identifiedfordeterminingthereactance. These are the subtransient reactance xd” for thefirst 10 to 20 ms of fault, thetransient reactance xd' for upto about 500 ms, and thesteady state reactance xd(synchronous reactance). The sub-transient mustusually be used in faultanalysis.time (ms)oa - peak value of steady state short-circuit currentob - peak value of transient short-circuit currentoc - peak value of sub-transient short-circuit currentEE 423 - Power System Analysis: Faults – J R Lucas – October 200510

The r.m.s. values of current are given bysubtransient transient steady-stateEoa I reactance reactance reactance2XdturboEob10-20 %15-25 % 150-230 % I′ generator2X d′salient-poleEoc15-25 %25-35 %70-120 % I ′′ generator2X d′′The typical generator reactance values are given above for reference.2.5 Revision of Symmetrical Component AnalysisUnbalanced three phase systems can be split into three balanced components, namelyPositive Sequence (balanced and having the same phase sequence as the unbalancedsupply), Negative Sequence (balanced and having the opposite phase sequence to theunbalanced supply) and Zero Sequence (balanced but having the same phase and hence nophase sequence). These are known as the Symmetrical Components or the SequenceComponents and are shown in figure 2.10.c1cb2abUnbalanced system3 unknown magnitudes3 unknown angles b1a2 a1Positive Sequence1 unknown magnitude1 unknown anglec2Negative Sequence c0a0b0Zero Sequence1 unknown magnitude 1 unknown magnitude1 unknown angle1 unknown angleFigure 2.10 – Symmetrical Components of unbalanced 3 phaseThe phase components are the addition of the symmetrical components and can be writtenas follows.a a1 a2 a0b b1 b2 b0c c1 c2 c0The unknown unbalanced system has three unknown magnitudes and three unknownangles with respect to the reference direction. Similarly, the combination of the 3 sequencecomponents will also have three unknown magnitudes and three unknown angles withrespect to the reference direction.Thus the original unbalanced system effectively has 3 complex unknown quantities a, band c (magnitude and phase angle of each is independent), and that each of the balancedcomponents have only one independent complex unknown each, as the others can bewritten by symmetry. Thus the three sets of symmetrical components also have effectively3 complex unknown quantities. These are usually selected as the components of the firstphase a (i.e. a0, a1 and a2) . One of the other phases could have been selected as well, butall 3 components should be selected for the same phase.Thus it should be possible to convert from either sequence components to phasecomponents or vice versa.EE 423 - Power System Analysis: Faults – J R Lucas – October 200511

2.5.1 Definition of the operator αWhen the balanced components are considered, we see that the most frequently occurringangle is 1200.In complex number theory, we defined j as the complex operator which is equal to -1and a magnitude of unity, and more importantly, when operated on any complex numberrotates it anti-clockwise by an angle of 900.i.e. j -1 1 900In like manner, we define a new complex operator α which has a magnitude of unity andwhen operated on any complex number rotates it anti-clockwise by an angle of 1200.i.e. α 1 1200 - 0.500 j 0.866Some Properties of αi.e.α 1 2π/3 or 1 1200α2 1 4π/3 or 1 2400 or 1 -1200α3 1 2πor 1 3600 or 1α3 - 1 ( α - 1)( α2 α 1) 0Since α is complex, it cannot be equal to 1, so that α - 1 cannot be zero.αα2 α 1 0 α2This also has the physical meaning that the three sides of anequilateral triangles must close as in figure 2.11.Also α 1 α2 andα 2 α1Figure 2.11 Phasor Addition2.5.2 Analysis of decomposition of phasorsLet us again examine the sequence components of the unbalanced quantity, with each ofthe components written in terms of phase a components, and the operator α, as in figure2.12.c1 αa1cb2 αa2a b1 α2a1bUnbalanced systema2 a1c2 α2a2Positive SequenceNegative SequenceFigure 2.12 – Expressing components in terms of phase a b0 c0 a0Zero SequenceWe can express all the sequence components in terms of the quantities for A phase usingthe properties of rotation of 00, 1200 or 2400.Thusa a0 a1 a2b a0 α2 a1 α a2c a0 α a1 α2 a2This can be written in matrix form.EE 423 - Power System Analysis: Faults – J R Lucas – October 200512

a 1 12 1 a0 α a1 b 1 α c 1 α α 2 a2 [Λ]PhSyThis gives the basic symmetrical component matrix equation, which shows the relationshipbetween the phase component vector Ph and the symmetrical component vector Sy usingthe symmetrical component matrix [Λ]. Both the phase component vector Ph and thesymmetrical component vector Sy can be either voltages or currents, but in a particularequation, they must of course all be of the same type. Since the matrix is a [3 3] matrix, itis possible to invert it and express Sy in terms of Ph.2.5.3 Decomposition of phasors into symmetrical componentsNow let us invert the symmetrical component matrix [Λ]. 1[Λ ] 1 α 4 α 2 (α 2 α ) α α 2 1 1 11 1 α α 2 1 (α 2 α ) 1 α 2 α α α 2 1 α α 2 1 αα 2 1 α α 21 α α 2 α α 2 α α 2 α α 2 α (1 α )α (1 α ) α (1 α )1 2 α 1 1 α α (1 α ) (1 α )(α 1)1 α α (1 α )α 2 1 1 α1 α (1 α )(α 1) and the discriminent 3(α - α2) 3α (1-α)Substituting, the matrix equation simplifies to give α1 α[Λ]-1 3α αSince α 1 α2, α 2 αα (α 1)1 1 (α 1) α and 1 α α2 0, the matrix equation further simplifies to1 1 11 1 α α 2 [Λ]-1 3 1 α 2 α It is seen that α is the complex conjugate of α2, and α2 is the complex conjugate of α.Thus the above matrix [ ]-1 is one-third of the complex conjugate of [ ].i.e.[Λ]-1 13[Λ]*This can now be written in the expanded form as1 A A0 1 1 A 1 1 α α 2 B 1 3 A2 1 α 2 α C [Λ ]SyPhEE 423 - Power System Analysis: Faults – J R Lucas – October 200513

2.5.4 Sequence ImpedancesConsider how the impedance appears in sequence components.To do this we must first look at the impedance matrix in phase components.Vp [Zp].IpSubstituting for Vp and Ip in terms of the symmetrical components we have[Λ] Vs [Zp]. [Λ] Is-1pre-multiplying equation by [Λ] we have [Λ]-1.[Zp]. [Λ] IsVsThis gives the relationship between the symmetrical component voltage Vs and thesymmetrical component current Is, and hence defines the symmetrical componentimpedance matrix or Sequence Impedance matrix.Thus [Λ]-1.[Zp]. [Λ][Zs]13 [Λ]*.[Zp]. [Λ]In a similar manner, we could express the phase component impedance matrix in terms ofthe symmetrical component impedance matrix as follows.[Zp] [Λ].[Zs]. [Λ]-113 [Λ].[Zs]. [Λ]*The form of the sequence impedance matrix for practical problems gives one of the mainreasons for use of symmetrical components in practical power system analysis.If we consider the simple arrangement of a 3 phase transmission line (figure 2.13), wewould have the equivalent circuit asRaLaRbRcMabMbcLbMcaLcFigure 2.13 – 3 phase transmission lineIf we think of an actual line such as from Victoria to Kotmale, we would realise that all 3phase wires would have approximately the same length (other than due to differences insagging) and hence we can assume the self impedance components to be equal for eachphase.i.e. Ra Rb Rc and La Lb LcWhen a current passes in one phase conductor, there would be induced voltages in theother two phase conductors. In practice all three phase conductors behave similarly, sothat we could consider the mutual coupling between phases also to be equal.i.e.Mab Mbc McaIn such a practical situation as above, the phase component impedance matrix would befully symmetrical, and we could express them using a self impedance term zs and amutual impedance term zm.EE 423 - Power System Analysis: Faults – J R Lucas – October 200514

Thus we may write the phase component impedance matrix as[Z ]p zs z m z mzmzszmzm z m z s We may now write the symmetrical component impedance matrix asz1 1 1 12 s*1[Z s ] 3 [Λ] . Z p .[Λ] 1 α2 α zm3 1 αα zm [ ]zmzszmz m 1 1 1 z m 1 α 2 α 2z s 1 α α 22 1 1 1 zs 2zm zs (α α )zm zs (α α )zm 1 1 α α 2 zs 2zm α 2 zs (1 α)zm α zs (1 α 2 )zm 3 1 α 2 α z s 2zm α zs (1 α 2 )zm α 2 zs (1 α)zm This can be simplified using the property 1 α α2 0 as followszs zmzs zm 1 1 1 z s 2z m1 2 2[Z s ] 1 α α z s 2z m α ( z s z m ) α ( z s z m ) 3 1 α 2 α z s 2z m α ( z s z m ) α 2 ( z s z m ) 00 3( z s 2z m )1 0(1 α 3 α 3 )(z s z m )0 3 00(1 α 3 α 3 )(z s z m ) i.e.00 ( z s 2 z m ) Z 0 0[Z s ] 0(zs zm )0 000( z s z m ) 0Z100 0 Z 2 We see an important result here. While the phase component impedance matrix was a fullmatrix, although it had completely symmetry, the sequence component impedance matrixis diagonal. The advantage of a diagonal matrix is that it allows decoupling for ease ofanalysis.2.6 Power associated with Sequence ComponentsWith phase components, power in a single phase is expressed asPphase V I cos φThus in three phase, we may either write P 3 VL IL cos φ 3 Vp Ip cos φ for a balancedthree phase system. However, with an

1. Power Flow Analysis: (8 hrs) Analogue methods of power flow analysis: dc and ac network analysers Digital methods of analysis: Power Flow algorithms and flow charts, analysis using iterative techniques. 2. Power system faults (8 hrs) Causes and effects of faults. Review of per unit system and symmetrical components. Symmetrical three-phase .