A step-by-step approach to prepare fault studies of electrical power systems

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The following are covered: the classification of faults, sources of fault currents, sequence impedance networks, the calculation of the fault currents for different types of shunt faults, the preparation of coordination studies and the inclusion of the different current time characteristics curves, damage curves/points and inrush (energization) currents.

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A step-by-step approach to prepare fault studies of electrical power systems

  1. 1. Fault calculations & protective relays co-ordination studies:Fault classifications: faults can be classified into symmetrical short circuit currents producing faultsand unsymmetrical ones, another method of classification is whether it is of the shunt or series types.Shunt types faults can be further classified into: single line to ground, line to line, double line toground, three phase grounded or ungrounded. The series types can be an open phase on one line or 2open phases. These faults can be bolted or faulty through a resistance (impedance).Va = Va1 + Va2 + Va0Vb = Vb1 + Vb2 + Vb0 = a2 Va1 + a Va2 + Va0Vc = Vc1 + Vc2 + Vc0 = a Va1 + a2 Va2 + Va0a = 1 120o = -.5+j .866 = cos 120 + j sin 120, a2 = 1 240o = -.5-j.866 , a2 + a + 1 = 0a3 = 1, a = ej2π/3.Va1 = (Va + a Vb + a2 Vc)/3, Va2 = (Va + a2 Vb + Vc)/3, Va0 = (Va + Vb + Vc)/3Where Va is the voltage of phase A, Vb of phase B & Vc of phase C (in 3-phase power systems); Va1 isthe positive sequence voltage (of the symmetrical components) of bus A, Vb1 is the positive sequencevoltage (of the symmetrical components) of bus B, Vc1 is the positive sequence voltage (of thesymmetrical components) of bus C, Va2 is the negative sequence voltage (of the symmetricalcomponents) of bus A, Vb2 is the negative sequence voltage (of the symmetrical components) of bus B,Vc2 is the negative sequence voltage (of the symmetrical components) of bus C, Va0 is the zerosequence voltage (of the symmetrical components) of bus A, Vb0 is the zero sequence voltage (of thesymmetrical components) of bus B & Vc0 is the zero sequence voltage (of the symmetrical components)of bus C. The same is applicable phase currents and symmetrical components of phase currents.The data required to perform fault calculations: they are the positive/negative/zero sequenceimpedance values of the system under study, to achieve this requirement the following data have to beavailable: size, length and methods of laying all cables in the system under study, the overheadconductors size, type and configuration, transformers KVA/impedance (with its base)/tap changerlimits and step size/impedance per tap step/primary and secondary voltage/method of neutral groundingand pertinent data, for the induction or synchronous motors: the rating/sub-transient reactance/transientreactance/speed/X to R ratio and the current limiting reactors ratings (if any).The sources of fault current: they are synchronous generators, synchronous motors & condensers,induction machines and electric utility systems. The fault current passes through two stages beforereaching steady-state, which are the sub-transient and transient states (of current), the first lasting about6 cycles, the second for approximately 30 cycles after which the steady state is reached. The shortcircuit currents are calculated for first cycle (post fault), after 5 or 6 cycles and after 30 cycles. Theimpedance of the circuit elements are established, referred to the base values (to be consistent witheach other for combination in series, parallel and to convert them from wye to delta & vice versa). Inthe per unit system there are four base quantities: base apparent power (MVAb), base voltage (KVb),base current (Ab) and base impedance (ohmb). Base MVA is taken as 100 MVA or any other suitableMVA rating of any equipment in the system under study and stays constant through out the study. KVbis taken at a starting point in the study (arbitrary point) and at every transformer, its value is adjustedaccording to the transformer ratio.Per unit (pu) quantity = actual value/ base value.Base current (Ab) = base MVA x 1000 / (3)0.5 base KV.Base impedance (ohmb) = (base KV)2 / base MVA.Per unit impedance = actual impedance (base MVA) / (base KV)2.Per unit impedance for transformers =percent impedance (base KVA) / [(KVA rating of transformer) (100)]Per unit reactance for motors = per unit reactance (base KVA) / motor KVA rating, assuming the perunit reactance is given based on the motor KVA rating.Shunt types fault calculations:
  2. 2. -The positive and negative sequence impedances are equal for a static device (transmission lines andtransformers), the zero sequence is different as it includes the impedance of the return path through theground. For rotating machines, the +ve, -ve and 0 sequence impedances are, generally, unequal. Zerosequence impedance is much smaller than +ve & -ve sequence ones. It is, always, assumed that the +vesequence current flows only in the +ve sequence network impedance producing +ve sequence voltagedrop across it; the –ve sequence current flows through the –ve sequence impedance producing a –vesequence voltage drop across it and, as well, the zero sequence current flows through the zero sequenceimpedance (network).-Three phase faults: the fault current = Pre-fault voltage divided by equivalent +ve sequence impedance(or reduced +ve sequence network equivalent). The prevailing conditions for this type of faults are: Va = Vb = Vc and Ia + Ib + Ic = 0.-Single line to ground: assuming the fault on phase A, the following relations are valid: Va = 0, Ib = Ic= 0 (assuming no-load conditions just prior to the fault instant). Ia1 = Ea/(Z1 + Z2 + Z0), the fault current, Ia = 3Ia1, Va1 = Ea - Ia1 Z1, Va2 = - Ia2 Z2, Va0 = - Ia0 Z0, Vb = Vb1 + Vb2 + Vb0, Vc = Vc1 + Vc2 + Vc0, Vab = - Vb, Vac = - Vc, Vbc = Vb - Vc.-Line to line: assuming the fault is between phases B & C, Ia = 0 (assuming no-load conditions justprior to the fault instant), Ib + Ic = 0, Vb = Vc, Va1 = Va2, Ia1 = Ia2 = Ea/(Z1 + Z2), Va = Va1 + Va2 +Va0, Vb = a2 Va1 + a Va2 + 0, the fault current (Ib) = Ib1 + Ib2 + Ib0 = a2 Ia1 + a Ia2 + 0 = 2(.866)Ia1.-Double line to ground fault: assuming the fault is between phases B & C and the ground, Ia = 0, Vb =Vc = 0, Va0 = Va1 = Va2, Ia1 + Ia2 = - Ia0, Ia1 = Ea / [Z1 + (Z0 Z2)/(Z0 + Z2)], Ia0 = - Va0 / Z0, Va= 3Va1, fault current (In) = Ib + Ic = a²Ia1 + aIa2 + Iao + a²Ia2 + aIa1 + Iao = - Ia1 - Ia2 + 2 Iao =3Ia0.-Faults with neutral impedance and fault impedance: For line to ground: Ia1 = Ea / [Z1 + Z2 + (Z0 + 3Zn) + 3Zf]; where Zn is the impedance in the neutral path & Zf the fault impedance. For line to line: Ia1 = Ea / [Z1 + (Z2 + Zf)] For double line to ground: Ia1 = Ea / { Z1 + [ Z2 (Zo + 3Zn + 3Zf) / (Z2 + Zo + 3Zf + 3Zn)]}.Co-ordinaion studies: for the coordination study the following data are essential: the relays current-time characteristics curves, fuses total clearing time-current & minimum melting time-currentcharacteristic curves, total available 3-phase short circuit current in the system, the 58% IEEE point fortransformers connected in delta - wye, damage curves for cables & transformers, inrush currents to
  3. 3. induction motors and their durations, inrush currents for transformers and durations, relays burden data,instrument transformers saturation & excitation curves, accuracies, ratios and taps number & ratiosdata.Coordination curves:Current Coordination curves: they are produced for protection against over-currents/ short circuitsand over-voltages. The co-ordination between the series protective devices, relays and fuses, isimportant to achieve selectivity and the operation of the device closest to the fault while leaving theupstream devices in the normal closed condition.Fuses are defined by two curves for each fuse size and type, the minimum melting time of the link vs.the current flowing through it and the total clearing time vs. the current. The o/c relays depending on itsdesign can have settings to choose the curve shape that would give the required relation between thecurrent flowing in the circuit and the operating time of the relay. A safety zone, usually, exists betweenthe two curves of two devices connected in series (to allow for the over travel, inaccuracies, etc. in theoperating time of the devices).Other points or curves that may appear on such composite curves are inrush currents when transformersare energized (12 times full load current for 0.1 second), inrush current for induction motors (for 10seconds, 4 times full load), damage curves for cables, damage curves for transformers and the ANSI58% point for delta - wye transformers. After obtaining all the curves for the relays and fuses used toprotect the system under study, construct the curves by starting with the most downstream load goingupstream and converting each curve to the base voltage level.For low voltage circuit breakers each of the following is an independent adjustment: long time setting,long time delay, short time pick-up setting, short time delay and instantaneous adjustment. Add theother curves previously mentioned as deemed applicable. Show the total short circuit available for athree-phase fault. Construct the 1-phase ground fault coordination curves for the system under study, ona different log-log sheet. Show on the same sheet having the set of curves the applicable portion of thesingle line diagram.The second method used (and is more suitable for studies run by computers) is the impedance matrixmethod that also provide for the voltage sensitivity studies (voltages at instant of fault appearing on the
  4. 4. unfaulty buses or nodes).Impedance network models: building the impedance network model for a power system will assist incalculating the fault studies including voltage sensitivity for radial systems, (the voltage on the busesupstream the faulty one at time of fault). The coverage here will cover only the representation ofbranches with no mutual coupling and will cover networks and radial systems (with or without theeffect of S.C.I.M. contribution). The 4 conditions that may exist in a system are: the presentation of animpedance connecting between the reference node (bus) which can be the neutral or ground of a systemand a new bus, the connection between an old bus and a new one, the connection between an old busand the reference node and finally the connection between 2 old buses. Lets assume a network withreference node 0, then equipment and lines go out radially from the reference node or bus 0 (which isthe neutral, the 3 phases are assumed to be balanced and a single phase out of the 3 phases isrepresented here) to bus 1, bus 1 is connected to bus 2, bus 2 to 3 and, finally, 1 to 3.Step 1: Zbus = [Z01]; bus 0 to 1Z44 = Z01+Z01+Z12+Z23 - 2 [Z01] + Z13Step 4a:the elimination of column and row 4 using Kron reduction method. The final Zbus matrix willhave the following elements:Z11 = Z01 - [(Z01-Z01)(Z01-Z01)/Z44]Z12 = Z01 - [(Z01-Z01)(Z01-Z01-Z12)/Z44]Z13 = Z01 - [(Z01-Z01)(Z01-Z01-Z12-Z23)/Z44]Z21 = Z01 - [(Z01-Z01-Z12) (Z01-Z01)/Z44]Z22 = Z01 + Z12 - [(Z01-Z01-Z12)(Z01-Z01-Z12)/Z44]Z23 = Z01 + Z12 - [(Z01-Z01-Z12) (Z01-Z01-Z12-Z23)/Z44]Z31 = Z01 - [(Z01-Z01) (Z01-Z01-Z12-Z23)/Z44]Z32 = Z01 +Z12 - [(Z01-Z01-Z12) (Z01-Z01-Z12-Z23)/Z44]Z33 = Z01+Z12+Z23 -[(Z01-Z01-Z12-Z23)²/Z44]The last possibility would be the presence of an equipment ( a generator or an induction motor that itseffect has to be taken while calculating fault currents) between bus 3 - for example - and node 0(reference). The step 4 above is replaced by the following step, and then Step 4 will be performed onthe resultant matrix and it will become Step 5.Step 4: from Step 3 above obtain the following matrix:
  5. 5. Z44 = Z01+Z12+Z23+Z03.Step 4a: the elimination of column and row 4 using Kron elimination method mentioned above under4a. Then Step 5 will become the addition of the impedance between buses 3 and 1 and this will addcolumns 4 and row 4 to the matrix which are then eliminated (using Kron method) to yield the finalnetwork impedance matrix with the final impedance elements that can be used in fault studies.Grid power systems modelling: the data required to perform fault studies on grid systems are more orless similar to those required for radial systems. The difference is in the degree of complexity of thegrid having more than one source, feeding the connected loads at the same time. The method bestsuited for grids is the impedance network. The neutral is taken as the reference node (node 0) and thenthe buses are numbered sequentially, each load bus, generator bus, motor bus (if applicable). For thesynchronous machines, the subtransient reactance is used to calculate the short circuit current duringthe first cycle (momentary, close and latch ratings of breaking devices), the transient reactance tocalculate the short circuit current during the 2nd or 3rd cycle up to the eighth cycle (breaking devicesinterrupting rating) and finally the synchronous reactance to calculate the short circuit current fordelayed tripping. For the induction motor and to calculate its contribution for the first few cycles, thetransient reactance is used in the calculation. For stationary components like overhead conductors,underground cables, transformers, the impedance or reactance of the component is used. Parallelcircuits, lines connecting, lets say, bus 1 to bus 3 and bus 1 to bus 2 can be presented in this model.Types of faults calculations: the purpose of fault calculations is to get the magnitude of the short circuitcurrent and the voltage on the buses, other than the faulty (in radial systems will be for those upstreamthe faulty bus, in grid systems, it will be for the upstream as well as the downstream faulty bus). Thevoltage sensitivity calculation in this section will only be limited to the two types of faults the mostprobably will happen (single phase to ground) and the most severe (3 phase faults).•• Radial systems:For line to ground faults: (examples: failed cable or splice, broken insulator or insulator flashover,reduced clearances, failed single phase transformers, etc.).The calculations here will be given for 4-bus radial system (for one feeder, as during faults the feederwith the fault is important and is part of the study the system, with the others either supplying the faultwith motor contribution or they are not part of the study - if there are no source of short circuit current).Step 1: construct the impedance network (that will be also used with other types of faults for +ve, -veand zero sequence). The fault is assumed to be on bus 4.X111 = XX011+XX121+XX231+XX341Y24 = XX011+XX121Y34 = XX011+XX121+XX231
  6. 6. where XX011 is the +ve sequence reactance between node 0 and bus 1, XX121 between bus 1 and 2,XX231 between bus 2 and 3, XX341 between bus 3 and 4 (even if the feeder had more buses, thecalculation would stop here - as the voltage on the downstream buses will be equal to that of the faultyas long as there is no motor contribution from the downstream buses), a similar negative sequenceimpedance model can be constructed.where XX112 = XX012 + XX122 + XX232 +XX342Y242 = XX012 + XX122, Y342 = Y242 + XX232For the zero sequence network, it’s a little bit tricky as the connections of the transformer and neutralgrounding method (if applicable) have to be taken into consideration when building the zero sequenceimpedance matrix. If the transformer connection (it is assumed in this example system that thetransformer is connected between node 0 and bus 1) is such that for a single phase to ground, the faultcurrent will have no path to ground eg. wye-wye (isolated ground), delta-delta, wye (grounded)-delta,wye (grounded)-wye then the short circuit (fault) current = 0. A zig-zag transformer connection canreplace a delta and it will provide the same phase angle between the primary and secondary windingsand will provide a neutral terminal. If the transformer is connected delta - wye (grounded) or wye(grounded) - wye (grounded) the zero sequence impedance network matrix has to be built for each caseincluding the source (system ahead of the transformer) of short circuit capacity or the generator (aheadof the transformer, if applicable). The relevant elements for the first transformer connection are: X110 =XX010 + XX120 + 3(d) + XX230 + XX340Y240 = XX010 + XX120, Y140 = XX010 + 3(d)Y340 = XX120 + XX230 + XX010 + 3(d)As the example, here, is a special condition that the supply of the short circuit current can supplyinfinity amperes, then the equations for the grounded wye-grounded wye transformer connection willbe identical to the above condition.XX010: is the zero sequence reactance between node 0 and bus 1.XX120: is the reactance between 1 and 2, and so on.d: is the reactance in the ground (fault) current path.Step 2: From above the short circuit current for a single phase to ground fault on bus 4 can becalculated and is equal to 3 (Vf)/(X111 + X112 + X110). For the voltage sensitivity calculations, onlyvoltage at bus 1 will be given here to correlate between the elements given in the above sequencenetworks and the equations:VS10 = -Y140 (Ifault)/3, VS11 = Vf - Y14 (If/3)VS12 = - (If) (Y142)/3, VSAf = VS10 + VS11 +VS12VSBr = VS10 + (-.5) (VS11) + (-.5) (VS12)VSBi = (-.866) (VS11) + (.866) (VS12)VSBf = [VSBr)² + (VSBi)²]VSCr = VS10 + (-.5) (VS11) + (-.5) (VS12)VSCi = .866 (VS11) + (-.866) (VS12); where VS10: zero sequence voltage on bus 1, VS11: +ve sequence
  7. 7. voltage, VS12 is the -ve sequence voltage, VSAf is the voltage on phase A of bus 1, VSBf is the voltageon phase B, VSCf is the voltage on phase C.For 3 phase faults:The S.C. fault current = Vf/X111The voltage on bus 1 = Vf (1-Y14/X111) in p.u.The voltage on bus 2 = Vf (1-Y24/X111) in p.u.The voltage on bus 3 = Vf (1-Y34/X111) in p.u.For line to line faults:The fault current = (.866)(2)(Vf/(X111+X112))For line to line to ground:Int = Intermediate value = Vf/[X111 + (X112) (X110)/(X112+X110)]The fault current = 3 (Int) (X112)/(X112 + X110)•• Grid or network systems:The analysis hereafter will be based on a hypothetical system having the following components: asynchronous generator between bus 1 and the reference node 0, other sources of fault currents arebetween buses 2 and reference node and another one between bus 3 node 0. The last 2 sources can besquirrel cage induction motors and the fault current is to be calculated for the first few cycles (i.e. ifthere are generators in the circuit, their reactance will be the subtransient for the calculation of the faultcurrent during the first few cycles and the transient value for the next period in the short circuitduration) with motor reactance equal the transient one (afterwhich the contribution of the SCIM isreduced to 0 - in a few cycles from fault inception). The other components are lines between thefollowing buses: 1& 2, 2 &3, 3 & 4, 1 & 4, 2 & 4. The full set of equations to solve this network isgiven in the “Example Programs” section.The impedance matrix is calculated based on the process given above “Impedance network models”and will have the following form:The same format can be used to build the negative and zero sequence networks:For line to ground fault on bus 3:If = 3 (Vf)/(XXC33 + XXC332 + XXC330)The voltage sensitivity calculations can proceed following the steps given under radial systems.For line-line-line fault on bus 3:If = 1/XXC33For line to line and double line to ground faults: the equations given in the radial system can beapplied here.

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