Fault Analysis Using Z-bus 1.0 Introduction

Fault Analysis Using Z-bus1.0 IntroductionThe previous notes on Z-bus resulted insome useful knowledge: diagonal elementZkk of the Z-bus is the Thevenin impedanceseen looking into the network from bus k.Combining this knowledge with equation (4)from notes called “Symmetrical Faults 2”,which wasI f VfZ f ZThev(1)enables us to efficiently obtain the faultcurrent for any bus in the network.This is easy if we have Z-bus and Vf.However, one other thing that we willfrequently need is the line currents, becausethe circuit breakers are going to be in serieswith the lines, not with the faults.1

2.0 Fault calculations using ZbusRecall thatV ZI(2)Since (2) represents a set of linear equations,superposition holds, and we may write: V Z I(3)This says that the change in voltage at allbuses ΔV may be computed if the change ininjections at all buses ΔI are known. We canwrite eq. (3) in expanded form as: V1 Z11 Z1k Z1N I1 Vk Z k1 Z kk Z kN I k (4) VN Z N 1 Z Nk Z NN I N Now consider a fault at bus k, where the prefault voltage at bus k is Vf. Let the faultcurrent be I’’f, and assume that the faultimpedance Zf 0 (this is typically worst-casescenario).2

Since a fault is a short circuit, then ΔVk -Vf.Also, since the fault current is out of bus k,then ΔIk - I’’f. Substituting these into eq. (4)results in V1 Z11 Z1k Z1N 0 V f Z k1 Z kk Z kN I f row k VN Z N 1 Z Nk Z NN 0 (5)Note that the right-hand-side results in, foreach row j, only the Zjk being multiplied bya non-zero current. Therefore: V1 Z1k I f V f Z kk I f row k VN Z Nk I f (6)We observe from row k that: V f Zkk I f (7)Solving (7) for I’’f results inVfI f Z kk(8)3

Notice that eq. (8) is consistent with eq. (1)when Zf 0.Now substitute eq. (8) into eq. (6) to get: V1 Z1k / Z kk V f Vf V f row k VN Z Nk / Z kk V f (9)Now eq. (9) provides the change in the busvoltages due to the fault. Change from what?It is the change from the voltage without thefault, i.e., it is the pre-fault voltage.Consider any bus, let’s say bus j, with a prefault voltage of Vj. Then we can computethe bus j voltage under the faulted conditionasV jf V j V jFrom eq. (9), we know that4(10)

V j Z jkZ kkVf(11)Substitution of (11) into (10) results inV jf V j Z jkZ kkVf(12)Now eq. (12) is useful for computing faultcurrents in the circuits. Consider Fig. 1.Zbbus ibus jFig. 1We can use eq. (12) to write down thevoltages under the faulted condition forbuses i and j, asZ ikVif Vi VfZ kk5(13)

V jf V j Z jkZ kkVf(14)Now we can compute the subtransientcurrent flowing from bus i to bus j under thefault condition asI ij Vif V jfZb(15)Substituting eqs. (13) and (14) into (15)results inZ jkZ ikVi V f V j VfZ kkZ kkI ij Zb Vi V jZb V fZ ik Z jk(16)Z b Z kkWe can use eq. (16) to get the fault currentin the circuits. These values provide us withthe appropriate information for selecting thecircuit breakers in the lines.6

3.0 Some important comments Zbus should be developed usingsubtransient reactances in generator/motormodels. Because fault currents are typicallymuch larger than load currents, it may beassumed that there are no loads.o All pre-faults currents are 0.o All buses have voltage (pre-fault) equalto Vf.o Equation (16) becomes:I ij V f Z ik Z jkZ b Z kk(17)From (16) and (17), we see that only thekth column of the Z-bus is required toanalyze a fault at bus k.The lasteffectiveanalysis.computeobservation can be utilized in anfashion when performing faultLet’s assume that we want tothe short circuit currents for a fault7

at only one bus k. So we just want to get thekth column of Z-bus, but we do not need theentire Z-bus.There is an efficient way to get the kthcolumn of Z-bus. Let’s study it.Consider that the Z-bus and the Y-bus areinverses of each other, i.e.,Z Y 1(18)This means that their product gives theidentity matrix.YZ I(19)where I is given by a matrix of zeros exceptthe diagonal which contains all ones, i.e., 1 0 0 0 1 0 I 0 0 1 8(20)

Our approach will depend on two ideas:1. Column of Z: We can just consider asingle column of Z, instead of the entirematrix. Call it Zk; it is a column vector.The right hand side of (19) will just be acolumn of I. Call it Ik. It will also be justa vector and will contain zeros in everyrow except for row k. The resultingrelation is:YZk Ik(21)2. LU-Decomposition: If you have takenEE 456 or a linear algebra course inmath, then you are familiar with LUdecomposition.LUdecompositionprovides a way to solve for the vector xin the matrix relationAx b(22)where A is a n n square matrix, x is anunknown n 1 vector, and b is an knownn 1 vector. The advantage to LUdecomposition is it does not require9

inverting the matrix A. The basis of LUdecomposition is that we may factor Ainto a matrix product LU, i.e.,A LU(23)where L is a lower diagonal n n matrixof the form l11 0 0 L l21 l22 0 (24) l31 l32 l33 and U is an upper diagonal n n matrix ofthe form 1 u12 u13 U 0 1 u23 (25) 0 01 Substitution of eq. (23) in (22) yieldsLU x b(26)DefiningUx wprovides that eq. (25) becomes:10(27)

Lw b(28)If we have L and U, then (28) is easilysolved for w (without inverting L) usingforward substitution, and then (27) canbe easily solved for x (without invertingU) using backwards substitution. Moredetails on LU decomposition may ow observe that eq. (21) and (22) are in thesame form. Therefore we want to solve thefollowing equations in the order they aregiven:Lw I kUZk w(29)(30)Factorization of a matrix Y can be done efficiently and easily using the matlabcommand:[L,U] lu(Y)Then it is easy to find w by hand using forward substitution from:Lw I kAnd then it is easy to find Z2 by hand using backwards substitution from:UZ2 wAlternatively, the manual steps of LU decomposition can be performed per the notesfrom “LU Decomposition.”11

Homework #3: Due Tuesday, Feb 3.1. Work problem 9.8 in book.2. Consider the 4-bus system shownbelow. Both machines have subtransientreactances of 0.20 pu (you can combinethe machine subtransient reactance withthe transformer impedance to get asingle reactance connecting the machineinternal voltage with the network).Bus 2j0.25j0.125j0.25Bus 3j0.20j0.40Bus 4Bus 1j0.10j0.10a. Construct the Y-bus for this network (should bea 4 4 matrix).b. Consider that there is a three-phase(symmetrical) fault at bus 2.12

i. Use LU decomposition to obtain the 2ndcolumn of the Z-bus.ii. Compute the subtransient fault current.iii. Use eq. (12) to find the voltages during thefault.iv. Use eq. (17) to find the subtransient currentsin lines 3-2, 1-2, and 4-2.3. See notes called “Examples.”4. See notes called “Examples.”13

voltages due to the fault. Change from what? It is the change from the voltage without the fault, i.e., it is the pre-fault voltage. Consider any bus, let’s say bus j, with a pre-fault voltage of V j. Then we can compute the bus j voltage under the faulted co