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7-Unsymmetrical_Faults.pptxiiiiiiiiiiiiii

Unsymmetrical faults occur more commonly on power systems than symmetrical faults. Unsymmetrical faults include single line-to-ground faults, line-to-line faults, and double line-to-ground faults. The symmetrical component method is used to analyze unsymmetrical faults by separating the system into positive, negative, and zero sequence networks. Currents flowing into a fault on the system are represented by their symmetrical components flowing out of each sequence network. Voltage changes on the networks due to a fault can be calculated using the bus impedance matrix for each sequence network.

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UNSYMMETRICAL FAULTS updated 11/11/13 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 1
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 2 Introductory Comments Most of the faults that occur on power systems are unsymmetrical faults, which may consist of unsymmetrical short circuits, unsymmetrical faults through impedances, or open conductors. Unsymmetrical faults occur as single l i n e -‐ to -‐g rou ndfaults, l i n e - ‐ t o - ‐ line faults, or double l i n e -‐ to -‐g roun dfaults. The path of the fault current from line to line or line to ground may or may not contain impedance. One or two open conductors result in unsymmetrical faults, through either the breaking of one or two conductors or the action of fuses and other devices that may not open the three phases simultaneously.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 3 Introductory Comments Since any unsymmetrical fault causes unbalanced currents to flow in the system, the method of symmetrical components is very used in the analysis to determine the currents and voltages in all parts of the system aLer the occurrence of the fault. We will consider faults on a power system by applying Thévenin's theorem, which allows us to find the current in the fault by replacing the entire system by a single generator and series impedance, and we will show how the bus impedance matrix is applied to the analysis of unsymmetrical faults.
Unsymmetrical Faults This shows the three lines a, b, and c of the three-‐phasesystem at the part of the network where the fault occurs. The flow of current from each line into the fault is indicated by arrows shown beside hypothetical stubs connected to each line at the fault location. In the derivation of equations for the symmetrical components of currents and voltages in a general network the currents flowing out of the original balanced system from phases a , b, and c at the fault point will be designated as Ia, lb, and lc, respectively. We can visualize these currents by as follows: a Ifa b Ifb I 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 4 fc c
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 5 Unsymmetrical Faults Appropriate connections of the stubs represent the various types of fault. For instance, direct connection of stubs b and c produces a l i n e - ‐ t o - ‐ l i n efault through zero impedance. The current in stub a is then zero, and lb equals - ‐lc. The l i n e -‐ to -‐g roun dvoltages at any bus j of the system during the fault will be designated Vja, Vjb and Vjc and we shall continue to use superscripts 1, 2, and 0, respectively, to denote positive-‐, negative-‐,and zero-‐sequencequantities. Thus, for example, V(1) ja, V(2) jb and V(0) jc will denote, respectively, the positive-‐,negative-‐,and zero-‐sequencecomponents of the l i n e - ‐ t o - ‐ ground voltage Vja at bus j during the fault.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 6 Unsymmetrical Faults The line-‐to-‐neutralvoltage of phase a at the fault point before the fault occurs will be designated simply by Vf, which is a positive-‐ sequence voltage since the system is balanced. We considered the prefault voltage Vf previously when calculating the currents in a power system with a symmetrical three-‐phase fault applied.
Unsymmetrical Faults Consider a single-‐linediagram of a power system containing two synchronous machines. This simple system is sufficiently general that the equations derived are applicable to any balanced system regardless of the complexity. The point where a fault is assumed to occur is marked P, and in this example it is called bus k on the single-‐linediagram and in the sequence networks. Single line diagram of a balanced three-‐phase system P k 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 7
Unsymmetrical Faults P k Referen ce fa I1 P  k Vf  Positive-‐Sequence Network   Vf P fa 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 8 I1 k kk Z1  Thévenin Equivalent of Positive-‐Sequence Network  ka V1
Unsymmetrical Faults Negative-‐Sequence Network P k Referen ce P k fa I2 P fa 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 9 I2 k kk Z2  Thévenin Equivalent of Negative-‐Sequence Network  ka V2
Unsymmetrical Faults Reference Zero-‐SequenceNetwork P k P k fa I0 P fa I0 k kk Z0  Thévenin Equivalent of Zero-‐Sequence Network  ka V0 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 10
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 11 Unsymmetrical Faults Machines are represented by their subtransient internal voltages in series with their subtransient reactances when subtransient fault conditions are being studied. Previously we used the bus impedance matrix composed of positive-‐sequenceimpedances to determine currents and voltages upon the occurrence of a symmetrical three-‐phase fault. The method can be easily extended to apply to unsymmetrical faults by realizing that the negative-‐and zero-‐sequencenetworks also can b e represented by bus impedance matrices. The bus impedance matrix will now be written symbolically for the positive-‐,negative-‐, and zero-‐sequencenetworks in the following form…
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 12 Unsymmetrical Faults bus 0,1,2    11 12 Z0,1,2 1k Z0,1,2 1N 21 Z0,1,2 Z0,1,2 Zk1 0,1,2   Z0,1,2 Z0,1,2 Z 0,1,2    Z0,1,2 N1 N 2 Z0,1,2 Nk Z0,1,2 22 2k 2N k 2 kk kN NN Z0,1,2     Z0,1,2        Z0,1,2   Z0,1,2        Z0,1,2 
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 13 Unsymmetrical Faults The Thévenin equivalent circuit between the fault point P and the reference node in each sequence network can be used for the analysis. As before, the voltage source in the positive-‐sequencenetwork and its Thévenin equivalent circuit is Vf, the prefault voltage to neutral at the fault point P, which happens to be bus k in this illustration. The Thévenin impedance measured between point P and the reference node of the positive-‐sequencenetwork is Z(1) kk, and its value depends on the values of the reactances used in the network. Recall that subtransient reactances of generators and 1.5 times the subtransient reactances (or else the transient reactances) of synchronous motors are the values used in
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 14 Unsymmetrical Faults There are no negative-‐or zero-‐sequencecurrents flowing before the fault occurs, and the prefault voltages are zero at all buses of the negative-‐and zero-‐sequencenetworks. Therefore, the prefault voltage between point P and the reference node is zero in the negative-‐and zero-‐sequencenetworks and no electromotive forces (emfs) appear in their Thévenin equivalents. The negative-‐and zero-‐sequenceimpedances between point P a t bus k and the reference node in the respective networks are represented by the the impedances Z(2) kk and Z(0) kk, the Diagonal elements of Z(2) bus and Z(0) bus, respectively.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 15 Unsymmetrical Faults Since Ifa is the current flowing from the system into the fault, its symmetrical components flow out of the respective sequence networks and their equivalent circuits at point P, as shown. Thus, the currents –I(1) fa, –I(2) g and –I(0) fc represent injected currents into the faulted bus k of the positive-‐, negative-‐, and zero-- ‐sequencenetworks due to the fault. These current injections cause voltage changes at the buses of the positive-‐,negative-‐,and zero-‐sequencenetworks, which can be calculated from the bus impedance matrices in the manner similare to what we have done before.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 16 Unsymmetrical Faults For instance, due to the injection –I(1) fa into bus k, the voltage changes in the positive-‐sequencenetwork of the N - ‐ b u ssystem are given in general terms by: 1a 1 2a 1   V Na 1     Vk a    1    11 12 1k Z1 1N Z1 21 22 1 1 Z1 Z1 Zkk N1 N 2 Z1     1 1  V Z Z   V        Z1      Nk Z1 2k 2N 1 Z1 kN NN Z1 Z1    Zk1 Zk 2      Z1                    0 0 1  I 0 Z I 1k fa 1 1 Z I 2k fa 1 1   Zkk I fa Z I Nk fa        fa                          1 1     1 1   
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 17 Unsymmetrical Faults Once again, it is industry practice to regard all prefault currents as being zero and to designate the voltage Vf as the positive-‐sequence voltage at all buses of the system before the fault occurs. Using superposition, the total positive-‐sequence voltage of phase a at each bus during the fault is: 1a 2a Na      Vka    V 1      V 1   1      V 1  f f f       V     V     V         Vf   V 1a 1  V 2a 1  V Na 1        Vk a             1       Z I f 1k fa 1 1 f  Z I 2k fa   f 1  Z I Nk fa     V       V  Z I f kk fa    V     V     1 1    1 1     1   
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 18 Unsymmetrical Faults This is the same equation as found for symmetrical faults, the only difference being the added superscripts and subscripts denoting the positive-‐sequencecomponents of the phase a quantities.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 19 Unsymmetrical Faults The negative-‐and zero-‐sequencevoltage changes due to the fault at bus k of the N - ‐ b u ssystem are similarly written with the superscripts changed accordingly. Because the prefault voltages are zero in the negative-‐and zero-‐sequencenetworks, the voltage changes express the total negative-‐and zero-‐sequence voltages during the fault, namely, 1a 2a Na      Vka    V 2     V 2   2      V 2  Z I 1k fa 2 2 Z I 2k fa 2 2    Zkk Ifa Z I Nk fa                          2 2     2 2     , 1a 2a Na      Vka    V 0     V 0   0      V 0  Z I 1k fa 0 0 Z I 2k fa 0 0    Zkk Ifa Z I Nk fa    0 0     0 0                          
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 20 Unsymmetrical Faults When the fault is at bus k, note that only the entries in columns k of Z(2) bus and Z(0) bus are involved in the calculations of negative-‐and zero-‐sequencevoltages. Thus, knowing the symmetrical components I(0) fa, I(1) fa and I(2) fa of the fault currents at bus k, we can determine the sequence voltages at any bus j of the system from the jth rows of 1a 2a Na      Vka    V 1     V 1   1      V 1   Z I f 1k fa 1 1  Z I f 2k fa  f  Z I Nk fa         V  Z I f kk fa    V     V        V 1 1   1 1     1 1     , 1a 2a Na      Vka    V2     V2   2      V2  Z I 1k fa 2 2 Z I 2k fa  Zkk Ifa Z I Nk fa                      2 2   2 2     2 2     , 1a 2a Na      Vka    V 0     V 0   0      V 0  Z I 1k fa 0 0 Z I 2k fa  Zkk Ifa Z I Nk fa  0 0                       0 0       0 0   
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 21 Unsymmetrical Faults That is, during the fault at bus k the voltages at any bus j are: If the prefault voltage at bus CD is not Vf, then simply replace Vf in by the actual value of the prefault (positive-‐sequence) voltage at that bus. Since Vf is by definition the actual prefault voltage at the faulted bus k, we always have at that bus: and these are the terminal voltage equations for the Thévenin equivalents of the sequence networks previously shown. V 0 ja jk fa 0 0       Z I , V 1 ja f jk fa 1 1     V  Z I , V 2 ja  Z I jk fa 2 2      Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1  V  Z I , ka 2    V  Z I kk fa 2 2   
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 22 Unsymmetrical Faults It is important to remember that the currents I(0) fa, I(1) fa and I(2) fa are symmetrical-‐componentcurrents in the stubs hypothetically attached to the system at the fault point. These currents take on values determined by the particular type of fault being studied, and once they have been calculated, they can be regarded as negative injections into the corresponding sequence networks.
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 23 Unsymmetrical Faults If the system has Δ - ‐ Ytransformers, some of the sequence voltages may have to be shiLed in phase angle before being combined with other components to form the new bus voltages of the faulted system. There are no phase shiLs involved in when the voltage Vf at the fault point is chosen as reference, which is customary. Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1  V  Z I , ka 2    V  Z I kk fa 2 2   
Unsymmetrical Faults Δ - ‐ Ytransformer with Positive-‐sequence Zero-- ‐sequence leakage impedance Z circuit circuit The negative-‐sequencecircuit is the same as the positive-‐sequence cir1 c1 / u1 1 i/ t1 .3 Unsymmetrical Faults (c) 2013 H. Zmuda 2 n m In a system with Δ - ‐ Ytransformers the open circuits encountered in the zero-‐sequence network requires some care in the Zbus building algorithm. Consider, for instance, the solidly grounded Δ - ‐ Y transformer connected between buses m and n as shown along with their positive and zero-‐sequencecircuits: Z m n Reference m n Reference Z
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 25 Unsymmetrical Faults It is straightforward using the circuit representations shown to generate the bus impedance matrices Zbus (0,1,2). This will be done subsequently. Suppose, however, that we wish to represent removal of the transformer connections from bus n in a computer algorithm which cannot make use of circuit (schematic) representations. We can easily undo the connections to bus n in the positive-‐and negative-‐sequencenetworks by applying the building algorithm the Zbus (1,2) matrices in the usual manner, i.e., by adding the negative of the leakage impedance Z between buses m and m in the positive-‐and negative-‐sequencenetworks. (Next slide)
Unsymmetrical Faults m n Reference Positive-‐sequence circuit: - ‐Z Z m n Reference 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 26  
Unsymmetrical Faults This strategy does not apply to the zero-‐sequencematrix Zbus (0) if it has been formed directly from the schematic representation shown. Adding - ‐Z between buses m and m does not remove the z e r o - ‐sequence connection from bus n. - ‐Z m n Reference Z 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 27
Unsymmetrical Faults To permit uniform procedures for all sequence networks, one strategy is to include an internal node p, as shown below. Note that the leakage impedance is now subdivided into two parts between node p and the other nodes as shown. Connecting –Z/2 between buses n and p in each of the sequence circuits will open the transformer connections to bus n. Z m n m n Reference Reference Z m n Reference Z/2 m n Reference Z/2 Z/2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 28 p p Z/2    
Unsymmetrical Faults Connecting –Z/2 between buses n and p in each of the sequence circuits will open the transformer connections to bus n. - ‐Z/2 - ‐Z/2 m n Reference Z/2 m n Reference Z/2 Z/2 p Z/2 p m n Reference Z/2 m n Reference Z/2 p 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 29 p    
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 30 Unsymmetrical Faults The faults to be discussed in succeeding sections may involve impedance Zf between lines and from one or two lines to ground. When Zf = 0, we have a direct short circuit, which is called a bolted fault . Although such direct short circuits result in the highest value of fault current and are therefore the most conservative values to use when determining the effects of anticipated faults, the fault impedance is seldom zero.
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 31 Unsymmetrical Faults Most faults are the result of insulator flashovers, where the impedance between the line and ground depends on the resistance of the arc, of the tower itself, and of the tower footing if ground wires are not used. T ower-‐footingresistances form the major part of the resistance between line and ground and depend on the soil conditions. The resistance of dry earth is 10 to 100 times the resistance of swampy ground.
Unsymmetrical Faults Three-‐PhaseFault Ifa Ifb Ifc Connections of the hypothetical stubs for faults through impedance Zf are as follows: a b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 32 Zf Zf
Unsymmetrical Faults b I fb c I fc Single L in e -‐ to -‐ Gro u n dFault Ifa Connections of the hypothetical stubs for faults through impedance Zf are as follows: a Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 33
Unsymmetrical Faults Ifb I fc L i n e - ‐ t o - ‐ L i n eFault Connections of the hypothetical stubs for faults through impedance Zf are as follows: a I fa b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 34
Unsymmetrical Faults Ifb I fc Double L in e -‐ to -‐ Gro u n dFault Connections of the hypothetical stubs for faults through impedance Zf are as follows: a I fa b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 35
Unsymmetrical Faults Other types of faults: a Ia c Ic Open-‐Conductor Faults b Ib 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 36
Unsymmetrical Faults Other types of faults: a Ia c Ic Open-‐Conductor Faults b Ib 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 37
Unsymmetrical Faults A balanced system remains symmetrical aLer the occurrence of a three-‐phasefault having the same impedance between each line and a common point. Only positive-‐sequencecurrents flow. With the fault impedance Zf equal in all phases, as in a three-‐phasefault, we simply add impedance Zf to the usual (positive-‐sequence) Thévenin equivalent circuit of the system at the fault bus k and calculate the fault current from the equation: For each of the other types of faults, formal derivations of the equations for the symmetrical-‐componentcurrents follow. In each case the fault point P is designated as bus k. fa 1  I  Vf 1  Z  Z kk f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 38
Exampl e Two synchronous machines are connected through three-- ‐phasetransformers to the transmission line as shown. The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8% On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15% and X0 = 50%. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 39
Exampl e Positive-‐and negative-‐sequence circuit: The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn =5% Transformers T1, T2: 100 MVA, 20Δ/345Y kV, X = 8% (split as 4% + 4%) Transmission line reactances are X1 = X2 = 15% and X0 = 50%. 1 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 40 3 4  V  f Vf   j0.20 j0.20 j0.04 j0.04 j0.15 j0.04 j0.04 Referen ce (1 ) (2) (3 ) (4 ) (5 )
Exampl e Add branch (1): Z 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 41 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) 1 1,2     j0.20   1,2 j0.20 j0.20 Add branch (2):  Z2     j0.20 j0.20  j0.08         j0.20  j0.20  j0.20 j0.28  
Exampl e 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 42 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) 3     Add branch (3):  j0.20 Z1,2   j0.20  j0.20 j0.20 j0.28 j0.28 j0.20 j0.28 j0.28  j0.15          j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.20 j0.28 j0.43     
Exampl e 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 43 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) Z4 1,2    j0.20 j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.28 Add branch (4):  j0.20 j0.28 j0.43 j0.43 j0.20 j0.28 j0.43 j0.43 j0.08                    j0.20 j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.28 j0.20  j0.20  j0.28 j0.28  j0.43 j0.43 j0.43   j0.51    
Exampl e Add branch (5):  1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 44 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) Z4 1,2    j0.20 j0.28 j0.28 j0.28 j0.20 j0.28 j0.43 j0.43 j0.20 j0.28 j0.43 j0.51       j0.20 j0.20 j0.20 j0.20        Znew ij  Zij Z Z  ik kj Z  Z kk b
Exampl e 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 45
Exampl e Z1,2: bus 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 46
Example Zero-‐sequence circuit: j0.04 1 3 j0.08 4 j0.04 Reference The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8% Transmission line reactances are X1 = X2 = 15% and X0 = 50%. j0.5 j0.04 3Xn  j0.15 (1 ) (2 ) (3 ) (4 ) (5 ) (6 ) 3Xn  j0.15 See Slide 199 of Symmetrical Components Transform er Node Bus Transform er Node Bus See Slide 196 of Symmetrical 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 47 Components j0.04 2
Exampl e 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 48 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 )
Exampl e Add branch (1): Z Add branch (2): 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 49 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 ) 0  1    j0.19  2 0   Z       j0.19 0 0 j0.08   3 0  Z  j0.19 0 0 j0.08 0 j0.08 Add branch (3):  0 j0.08        j0.08 j0.5  
Exampl e Add branch (5): 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 50 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 ) Z 0   j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus          
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 51 Exampl e We’ll use these in subsequent examples. Z 0      bus    j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19        Z 1,2       bus    j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437       
Single L in e -‐ to -‐ Gro u n dFaults The single l i n e -‐ to -‐g roundfault, the most common type, is caused b y lightning or by conductors making contact with grounded structures. A single l i n e -‐ to -‐g roundfault on phase a through impedance Zf is shown below: The relations to be developed will apply only when the fault is on phase a, but any phase can be designated as phase a. The conditions at the fault bus k are expressed by the following equations: I fa b I fb c I fc a Z f k 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 52 fb fc ka I  I  0, V  Z I f fa
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 53 Single L in e -‐ to -‐ Gro u n dFaults The symmetrical components of the current are, with I fb  I fc  0: But recall from Slide 21: Thus : Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1    V  Z I , Vka 2   Z I kk fa 2 2    Vka 0  kk fa 0 0     Z I , Vka 1  f kk fa 1 0  V  Z I , ka 2     V  Z I kk fa 2 0    fa fa I1 I  I0    2       fa     1 1 a a2 a2 a  1 3  1  1  1  0 0 I fa               I fa 0 fa 1  I  I 2 fa      I fa 3
Single L in e -‐ to -‐ Gro u n dFaults Summing, and noting that V This has the circuit representation shown on the next slide. V V ka ka ka 0 1 V Vka 2     f  V  Z 0 kk   Z 1 kk  Z 2 kk    I fa 0   3Z I f fa 0  ka  3Z I f fa 0  fa Solve for I 0 I fa 0 fa 1  I  I 2 fa      Vf Z 0 kk kk 1 kk 2      Z  Z 3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 54
Single L in e -‐ to -‐ Gro u n dFaults Vf   kk Z1 kk Z2 kk Z0 3Z f fa I1 fa I2 fa I0 I 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 55 fa 0 fa 1  I  I 2 fa     k k k  ka  V1  ka  V0  ka  V2
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 56 Single L in e -‐ to -‐ Gro u n dFaults The last result are the fault current equations particular to the single l i n e -‐ to -‐g roundfault through impedance Zf and they are used with the symmetrical-‐componentrelations to determine all the voltages and currents at the fault point P. If the Thévenin equivalent circuits of the three sequence networks of the system are connected in series, as shown on the previous slide, we see that the resulting currents and voltages satisfy the above equations – for the Thévenin impedances looking into the three sequence networks at fault bus k are then in series with the fault impedance 3Zf and the prefault voltage source Vf.
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 57 Single L in e -‐ to -‐ Gro u n dFaults With the equivalent circuits so connected, the voltage across each sequence network is the corresponding symmetrical component of the voltage Vka at the fault bus k, and the current injected into each sequence network at bus k is the negative of the corresponding sequence current in the fault. The series connection of the Thévenin equivalents of the sequence networks, as shown on Slide 55 is a convenient means of remembering the equations for the solution of the single l i n e - ‐ t o - ‐ ground fault, for all the necessary equations for the fault point can be determined from the sequence-‐network connection.
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 58 Single L in e -‐ to -‐ Gro u n dFaults Once the sequence components of the fault currents are known, the components of voltages at all other buses of the system can be determined from the bus impedance matrices of the sequence networks according to Slide 21, namely: V V V 0 ja  Z I jk fa 0 0      1 ja  f  V  Z I jk fa 1 1   2 ja   Z I jk fa 2 2   
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Y/345Y kV, X = 8% Two synchronous machines are connected through three-- ‐phasetransformers to the transmission line as shown. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 59
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 60 EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Both transformers are solidly grounded on on two sides. On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15 % and X0 = 50 %. The system is operating at nominal voltage without prefault currents when a bolted (Zf = 0) single l i n e -‐ to-‐ g ro u n dfault occurs o n phase A at bus 3. Using the bus impedance matrix for each of the three sequence networks, determine the subtransient current to ground at the fault, the l i n e -‐ to -‐groundvoltages at the terminals of machine 2, andthe subtransient current out of phase c of machine 2.
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults The system is the same as on Slide 39, except that the transformers are now Y - ‐ Yconnected. Therefore, we can continue to use Zbus (1,2) from Slide 51, however, because the transformers are solidly grounded on both sides, the zero-- ‐sequencenetwork is fully connected, as shown below, and has the bus impedance matrix 1 2 3 4 j0.04 j0.08 j0.15 j0.08 j0.04 j0.15 j0.15 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 61 Referen ce
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 62 3 4 j0.15 j0.15 j0.15 The bus impedance matrix is: 1 j0.04 j0.08 j0.08 j0.04 Referen ce Zbus 0   j0.1553 j0.1407 j0.0493 j0.0347 j0.1407 j0.1999 j0.0701 j0.0493 j0.0493 j0.0701 j0.1999 j0.1407 j0.0347 j0.0493 j0.1407 j0.1553              
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Since the l i n e -‐ to-‐ groundfault is at bus 3 : The total current in the fault is: f V 1.0 33 Z1 Z2 33 33 Z0 f 3Z  0 fA 1 I  fA I2 fA I0 IfA 3 k k k   V3a 1  3a  V0  3a  V2  0  Z     bus  Z  j0.1553 j0.1407 j0.0493 j0.0347 j0.1407 j0.1999 j0.0701 j0.0493 j0.0493 j0.0701 j0.1999 j0.1407 j0.0347 j0.0493 j0.1407 j0.1553 j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     1,2  bus                   I fA 0   I fA 1 fA 2    I  Vf kk 0 kk 1 kk 2     Z  Z  Z 3Z f  1 j0.1999  j1.696  j1.696  0   j1.8549 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 63 fA fA 0   I  3I   j5.5648
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Since the base current in the high-‐voltagetransmission line is: amp s Then: amp s The p h a se -‐asequence voltages at bus 4, the terminals of machine 2, are (from Slide 21 with k = 3 and j = 4): per unit base 3 345 I  100,000  167.35 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 64 I fA   j5.5648167.35 931270 V4a 43 fA 0 0 0       Z I     j0.1407  j1.8549  0.261 V4a 1  f 43 fA 1 1       V  Z I  1 j0.1211  j1.8549  0.7754 V4a 2   Z I 43 fA 2 2         j0.1211  j1.8549  0.2246
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 65 EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Note that the subscripts A and a denote voltages in the high-- ‐voltageand low-‐voltagecircuits, respectively, of the Y - ‐ Yconnected transformer. No phase shiL is involved. From the symmetrical components we can calculate the a - ‐ b - ‐ c l i n e - ‐ to- ‐groundvoltages at bus 4 as: per unit 4a 4b 4c   V  V    V     1  1 1 1 1 a2 a a a2 3 1   V4a 0 4a 4a               V 2     V 1   1  1 1 1 1 a2 a a a2 3 1          0.261 0.7754    0.2246   0.2898         0.5346  j0.8660   0.5346  j0.8660 
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults To express the l i n e -‐ to-‐g roundvoltages of machine 2 (in kV) multiply by 20/31/2: k V 4b 4c    V  4a     V     V  20 0.2898 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 66 0.5346  j0.8660    3       3.346 1.0187121.8       0.5346  j0.8660  1.0187 121.8  
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults To determine p h a s e-‐ccurrent out of machine 2 we must first calculate the symmetrical components of the p h a se -‐acurrent in the branches representing the machine in the sequence networks. From the zero-‐sequencecircuit, the zero-‐sequencecurrent out of the machine is: per unit 1 2 3 4 j0.15 j0.04 j0.15 j0.08 j0.15 j0.08 j0.04 Referen ce  4a V0  a I0 a 0  I   4a V0 0  jX j0.04 0.261   j6.525 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 67
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults per unit a 1 I  V V1  f 4a jX   1 0.7754 j0.2   j1.123  V  f Vf   j0.20 j0.20 j0.04 j0.04 j0.15 j0.04 j0.04 Referen ce  4a V1,2  a Similarly for the positive-‐and negative-‐sequence currents: 1 2 3 4 I1,2 a 2  I   4a V1 jX2  0.2246 j0.2   j1.123 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 68
EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults The p h a se -‐ ccurrent in machine 2 is per unit Since the base current in the machine circuit is: amp s c a 0 1    2 I  I  aI  a I a a 2    j6.525 a j1.123 a2 j1.123   j5.402 Ibase 3 20  100,000  2886.751 amps Ic 15,994 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 69
L i n e - ‐ t o - ‐ L i n eFaults To represent a l i n e - ‐ t o - ‐ l i n efault through impedance Zf the hypothetical stubs on the three lines at the fault are connected as shown. Bus k is again the fault point P, and without any loss of generality, the l i n e - ‐ t o - ‐ l i n efault is regarded as being on phases b and c. Clearly: Ifb 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 70 Ifc a b c Zf k I fa I  0 fa k k I fb  I fc Vkb Vkc  Z I f fb
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 71 L i n e - ‐ t o - ‐ L i n eFaults Substitutin g: The voltages throughout the zero-‐sequencenetwork must be zero current is not being injected into that network due to the fault. Hence, l i n e - ‐ t o - ‐ l i n efault calculations do not involve the z e r o - ‐ sequence network, which remains the same as before the fault - ‐adead network. since there are no zero-‐sequencesources, and because I fa fa I1      I0    2   I fa      1 1 1 1 a a2 a2 a   1 3   1  I fc          0  I  fb      I  I fa 0   0  I fa 1  2 fa  fa 0   0,
L i n e - ‐ t o - ‐ L i n eFaults equivalents of the positive-‐and negative-‐sequencenetworks i n parallel, as shown: To satisfy the requirement that I 1   I we connect the Thévenin fa fa 2  Vf   kk Z1 kk Z2 Z f fa I1 fa I2 k  ka V1  k 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 72  ka V2  Referen ce
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 73 L i n e - ‐ t o - ‐ L i n eFaults To show that this connection of the networks also satisfies the voltage equation: we now expand each side of that equation separately as follows: and : V V  Z I kb kc f fb kb kc V V  Vkb 1 kb 2     V  Vkc 1 Vkc 2      Vkb 1  Vkc 1   Vkb 2 Vkc 2        2 ka 1   a V  a  2     a Vka 2   a  a2   ka  a V Vka 1 2     f fb Z I  Z f fb 1 I  I 2 fb   f 2 fa 1  Z a I  aI fa 2      
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 74 L i n e - ‐ t o - ‐ L i n eFaults Equate both terms and set: But this is precisely the voltage drop across Zf in the figure on Slide 66. kb kc 2   V V  a  a Vka 1 ka 2     f fb f 2 V  Z I  Z a I fa 1  aI 2 fa     2    a  a Vka 1 ka 2     f 2   V  Z a  a I fa 1  ka 1 V Vka 2    Z I f fa 1  I fa 1   I 2 fa 
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 75 L i n e - ‐ t o - ‐ L i n eFaults Thus, all the fault conditions are satisfied by connecting the positive-‐and negative-‐sequencenetworks in parallel through impedance Zf as was shown. The zero-‐sequencenetwork is inactive and does not enter into the l i n e - ‐ t o - ‐ l i n ecalculations. The equation for the positive-‐sequence current in the fault can be determined directly from the circuit as:
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 76 Example – L i n e - ‐ t o - ‐ L i n eFaults The same system as before is operating at nominal system voltage without prefault currents when a bolted l i n e - ‐ t o - ‐ l i n efault occurs at bus 3. Determine the currents in the fault, the l i n e - ‐ t o - ‐ l i n e voltage a tthe fault bus, and the l i n e - ‐ t o - ‐ l i n evoltages at the terminals of machine 2.
Example – L i n e - ‐ t o - ‐ L i n eFaults Note that we will not need the zero-‐sequencebus impedance matrix since the fault is l i n e - ‐ t o - ‐ l i n e(see Slide 71). From the circuit: f 33 The Thévenin equivalent circuit for the positive-‐and n e g a t i v e - ‐ sequence is: Z1 33 Z2 f Z  0 IfA 1 fA 3  V3A 1  3  I2 V3A 2  Referen ce  V 1,2 Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437    bus            I fA fA 1 2    I  Vf 33 1 Z  Z33 2   2  j0.1696 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 77  1.0   j2.9481 per unit
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 78 Example – L i n e - ‐ t o - ‐ L i n eFaults Sinc e (Uppercase since the fault is in the high-- ‐voltage transmission line) Then I and since Then I fA 0 Also I per unit an d per unit Multiply these by Ibase = 167.35 A to get the actual currents. fA  I fA 1  I fA 2   I fA 0   0 I fA 1   I fA 2  1 fB fA fA 2 2    a I  aI  5.1061 I fC  I fB  5.1061
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 79 Example – L i n e - ‐ t o - ‐ L i n eFaults The symmetrical components of p h a s e -‐Avoltage to ground at bus 3are: per unit V3A 0   0 V3A 1 3A 2    V  1 Z I kk fA 1 1    1 j0.1696  j2.9481     0.5  j0
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 80 Example – L i n e - ‐ t o - ‐ L i n eFaults The l i n e -‐ to -‐g roun dvoltages at fault bus 3 are: V3C  V3B 0.5 all per unit V V 3A 3A 3A 0 1 V V3A 2      0  0.5 0.5  1.0 V V 3B 3B 0   a V3B 1 2   2V3B 2  2  0 a 0.5 a  0.5  0.5
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 81 Example – L i n e - ‐ t o - ‐ L i n eFaults The l i n e - ‐ t o - ‐ l i n evoltages at fault bus 3 are: V3,AB  V3A V3B  1 j0 0.5 j0 1.5 V3,BC  V3B V3C  0.5 j0 0.5 j0 0 all per unit. Multiply these by 345/31/2 to obtain the actual voltage. V3,CA V3C  V3 A  0.5  j0 1 j0 1.5
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 82 Example – L i n e - ‐ t o - ‐ L i n eFaults For the moment, let us avoid phase shiLs due to the Δ - ‐ Y transformer connected to machine 2 and calculate the sequence voltages of phase A at bus 4 using the the results from Slide 21 with k = 3 and j = 4: per unit Vja jk fa ja f jk fa  0  Z0I0, V 1V  Z1I1, V 2  Z2I2 ja jk fa 0 Z 0I0 V4a 43 fa 0 V4a 1  f jk fa 1 1    V  Z I 1 j0.1211 j2.9481 0.643 V4a 2   Z I 43 fa 2 2         j0.1211  j2.9481  0.357
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 83 Example – L i n e - ‐ t o - ‐ L i n eFaults To account for phase shiLs in stepping down from the high-- ‐voltagetransmission line to the low-‐voltageterminals of machine 2, we must retard the positive-‐sequencevoltage and advance the negative-‐sequencevoltage by 30°. At machine 2 terminals, indicated by lowercase a, the voltages are: V4a 43 fa 0 0 0       Z I  0 V4a 1 4a 1    V   30  0.643  30  0.5569  j0.3215 V4a 2  4a 2   V 30  0.35730  0.3092 j0.1785 V V 4a 4a 0  4a 1 V V4a 2    0.8861 j0.1430  0.8778 9.4
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 84 Example – L i n e - ‐ t o - ‐ L i n eFaults P h a se-‐bvoltages at the terminals of machine 2 are: V4b 0 4a 0     V  0 V4b 1  2  a V4a 1  1240 0.643  30  0.5569  j0.3215 V4b 2   aV4a 2   1120 0.35730  0.3092  j0.1785 V V 4b 4b 0  4b 1 V V4b 2    0.8861 j0.1430  0.8778170.6
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 85 Example – L i n e - ‐ t o - ‐ L i n eFaults P h a se-‐ cvoltages at the terminals of machine 2 are: V4c 0 4a 0     V  0 V4c 1   aV4a 1   1120 0.643  30  0.64390 V4c 2  2  a V4a 2   1240 0.35730  0.357  90 V V 4c 4c 0  4c 1 V V4c 2    j0.286
Example – L i n e - ‐ t o - ‐ L i n eFaults L i n e - ‐ t o - ‐ l i n evoltages at the terminals of machine 2 are: per unit For the voltage, multiply by 20 kV/31/2 k V V4,ab 4a 0 0 4b 0      V4,bc 0  V4b 0 V4c 0    0.8661 j0.429 j0.429 V 0   V V 4,ca 4c 4a 0 0     V V 1.7322    0.8661 V4,ab 0  20 3  1.7322   20 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 86 V4,bc 0  11.2 153.65  11.2153.65 V4,ca 0 
Ifa Ifb b c Zf I fa 0 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 87 Double L in e -‐ to -‐ Gro u n dFaults Again it is clear, with fault taken on phases b and c, that the relations at fault bus k are: k a Vkb  Vkc  Z f Ifb  I fc  k Ifc k
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 88 Double L in e -‐ to -‐ Gro u n dFaults Thus Vkb  Vkc  Z Substituting into: fa Since I  0 I fa 0   1 3 fa fb I  I  I fc   1 3 fb I  I fc   f fb fc   I  I  3Z I f fa 0  ka ka ka    V0       V2     V1    1  1 1 1 1 a a2 a2 a 3  1 ka kb kc     V  V     V      1  1 1 1 1 a a2 a2 a 3  1 ka kb kb  V     V  V       
Double L in e -‐ to -‐ Gro u n dFaults Expanding the second and third equations: ka ka ka    V0       V2   V1    1  1 1 1 1 a a2 a2 a 3  1   1 3 ka kb kb  V     V  V        Vka 1   2   a  a Vkb Vka 2 1 3  a2  aV 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 89 kb Vka 1  Vka 2 
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 90 Double L in e -‐ to -‐ Gro u n dFaults From the first equation: recall: Vkb  Vkc  Z f Ifb  Ifc  ka ka ka   V0      V2   V1    1  1 1 1 1 a a2 a2 a 3  1  3V ka kb kb     V  V     V     ka 0  ka kb    V  2V  Vka 0 ka 1    V Vka 2    2 3Z I f fa 0     3I fa 0  fb  I  I fc
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 91 Double L in e -‐ to -‐ Gro u n dFaults Collect zero-‐sequenceterms and recall that Now solve: Thus : and since Ifa = 0 These last two results characterize the double l i n e -‐ to -‐g roundfault. Vka 1 Vka 2   3Vka ka 0 0     V  6Z I f fa 0   2Vka 1  Vka ka 1 0    V  3Z I f fa 0  Vka 1  Vka 2 ka 0     V  3Z I f fa 0  I 0 fa fa 1  I  I 2 fa      0
Double L in e -‐ to -‐ Gro u n dFaults These can be realized by putting all three sequence networks in parallel as follows: Clearl y: kk Z2 kk Z0 3Zf fa I2 fa I0 k k  ka  V0  ka  V2 Vf   kk Z1 fa I1 k  ka  V1 I fa 1   Vf kk 1  Z  Z 0 kk  f    3Z Z 2 kk   Vf kk 1  Z  Z 0 kk  f    3Z Z 2 kk  Z 0 kk  Z 2 kk     3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 92
Double L in e -‐ to -‐ Gro u n dFaults By current divider: kk Z2 3Zf fa I2 fa I0 k k  kk ka Z0  V0  ka  V2 Vf   kk Z1 fa I1 k  ka  V1 I  I fa 2 1 fa    Z 0 kk   3Z f kk Z 0   Z 2 kk   3Z f , I fa 0   I fa 1  kk Z2 kk Z 0   Z 2 kk   3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 93
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 94 Double L in e -‐ to-‐ Gro u n dFaults For a bolted fault Zf is set equal to 0. When Zf = ∞, the z e r o - ‐ sequence circuit becomes an open circuit, no zero-‐sequence current an flow, and the equations revert back to those for the l i n e - ‐ t o - ‐ l i n efault. Again we observe that the sequence currents, once calculated, can be treated as negative injections into the sequence networks at the fault bus k and the sequence voltage changes at all buses of the system can then be calculated from the bus impedance matrices, as we have done in all along.
Example Double L in e -‐ to -‐ Gro u n dFaults Find the subtransient currents and the l i n e - ‐ t o - ‐ l i n e voltages at the fault under subtransient conditions when a double l ine -‐ to -‐g roun d fault with Zf = 0 occurs at the terminals of machine 2 in the system below. Assume that the system in unloaded and operating at rated voltage when the fault occurs. 1 T1 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 95 3 T2 4 Machine 1 Machine 2
Example Double L in e -‐ to -‐ Gro u n dFaults Model (the fault is a bus k = 4): kk Z2 kk Z0 3Zf fa I2 fa I0 k k  ka  V0  ka  V2 Vf   kk Z1 fa I1 k  11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 96 ka  V1 0  Z  j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus        , 1,2   Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     bus              
Example Double L in e -‐ to -‐ Gro u n dFaults Model (the fault is a bus k = 4): per unit j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 1  I  Vf Zkk 1  0  kk f   Z  3Z Z 2 kk  Zkk 0 2  Zkk  3Zf 44 1 Z  44 44 Z0Z2 44 0 Z  Z44 2    j0.1437  j0.19j0.1437 j0.19  j0.1437  1.0  1.0   j4.4342 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 97
Example Double L in e -‐ to -‐ Gro u n d Faults The sequence voltages at the fault are: j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 98 4a j0.1437  V1 V4a 1  V 2  0   V  V  Z I 4a 4a f 44 fa 1 1    1.0  jj0.1437 j4.4342 0.3628 per unit
Example Double L in e -‐ to -‐ Gro u n dFaults per unit j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 2  I  I fa 1  Z 0 kk   3Z f kk Z 0 kk 2     Z 3Z f  I fa 1  kk Z0 kk Z 0   Z 2 kk   I fa 1  44 Z0 Z 0  Z 44 44 2     j4.4342 j0.19 j0.19  j0.1437  j2.5247 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 99 fa Now that we have I 1
Example Double L in e -‐ to -‐ Gro u n dFaults per unit j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 0  I  I fa 1  kk Z2 kk Z 0 kk 2     Z 3Z f  I fa 1  44 Z2 44 Z 0  Z44 2     j4.4342 j0.1437 j0.19  j0.1437  j1.9095 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 100 fa Now that we have I 1
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 101 Example Double L in e -‐ to -‐ Gro u n dFaults The currents out of the system at the fault point are: The current If into ground is: all in per unit fa fa fa fa 0 1 2     I  I  I  I  j1.9095  j4.4342  j2.5247  0 fb fa 0  fa I  I  a I  aI fa 1 2 2     6.0266  j2.8642  6.0266  j2.8642 0 fc fa fa 1 I  I  aI  a I 2 fa    2  f fa fb fc fa 0  I  I  I  I  2I  j5.7285
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 102 Example Double L in e -‐ to -‐ Gro u n dFaults Calculating the a - ‐ b - ‐ cvoltages at the fault bus: V V 4a 4a 4a 0 1 V V4a 2     1.0884 V4b  V4c 0 V4,ab  V4a  V4b 1.0884 V4,bc  V4b  V4c 0 V4,ca  V4c V4 a  1.0884
Example Double L in e -‐ to -‐ Gro u n dFaults Recall the base current in the circuit of machine 2 is: amp s and the base line-‐to-‐neutralvoltage in machine 2 is: k V Ibase  100103 3 20  2887 Vbase  20 3 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 103
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 104 Phase Shias The previous two examples show that phase shiLs due to Δ - ‐ Y transformers do not enter into the calculations of sequence currents and voltages in that part of the system where the fault occurs provided Vf at the fault point is chosen as the reference voltage for the calculations. However, for those parts of the system which are separated by Δ - ‐ Y transformers from the fault point, the sequence currents, and voltages calculated by bus impedance matrix must be shiLed in phase before being combined to form the actual voltages. This is because the bus impedance matrices of the sequence networks are formed without consideration of phase shiLs, and so they consist of per-‐unitimpedances referred to the part of the network which includes the fault
Example Phase Shias Solve for the subtransient voltages to ground at bus 2, the end of the transmission line remote from the double l i n e -‐ to -- ‐ g ro u ndfault for the same system. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 105 0  Z  j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus        , 1,2   Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     bus              
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 106 Example Phase Shias We just solved for the values of the fault-‐current components. Neglecting the phase shiL of the Δ - ‐ Ytransformer for the moment we have, from Slide 21: At bus 2: V 0 ja jk fa 0 0       Z I , V 1 ja f jk fa 1 1     V  Z I , V 2 ja  Z I jk fa 2 2      V2a 0   Z I 24 fa 0 0      0    j1.9095  0 V2a 1  f 24 fa 1 1    V  Z I 1.0  j0.0789     j4.4342  0.6501 V2a 2   Z I 24 fa 2 2         j0.0789 j2.5247  0.1992
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 107 Example Phase Shias Accounting for phase shiL in stepping up to the transmission- ‐line circuit from the fault at bus 4 we have: Now the required voltages can be calculated: all per unit V2A 0   0 V2A 1 2A 1    V 30  0.650130 V 2  2 A 2A 2   V   30  0.1992  30 V V 4 A 4A 4A 0 1 V V4A 2      0.7355  j0.2255 4B V V4B 0  2 4B 1   aV4B 2   0.1275j0.5535  0.5656  j0.1274 4C V V4C 0   a V  aV4C 1  2  a V4C 2 
Example Phase Shias The per-‐unitvalues can be converted to volts by multiplying by the line-‐to-‐neutralbase voltage of: k V of the transmission line. Vbase  345 3 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 108
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 109 Open-‐Conductor Faults When one phase of a balanced three-‐phasesystem opens, an unbalance is created and asymmetrical currents flow. A similar type of unbalance occurs when any two of the three phases are opened while the third phase remains closed. These unbalanced conditions are caused, for example, when o n e - ‐or two-‐phaseconductors of a transmission line are physical broken by accident or storm. In other circuits, due to current overload, fuses or other switching devices may operate in one or two conductors and fail to operate in other conductors. Such open-‐conductorfaults can be analyzed by means of the bus impedance matrices of the sequence networks, as we now show.
Open-‐Conductor Faults Consider a section of a three-‐phase circuit in which the line currents in the respective phases are la, Ib, and Ic, with positive direction from bus m to bus n, with phase a open between points p and p’: a Ia b Ib c Ic m m m n n n p 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 110 p ’
Open-‐Conductor Faults Also consider the case where phases b and c are open between points p and p’: a Ia b Ib c Ic m m m n n n p p ’ 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 111
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 112 Open-‐Conductor Faults The same open-‐conductorfault conditions will result is all three phases are first opened between points p and p’ and short circuits are then applied in those phases which are shown to be closed in the preceding figures. The ensuing development follows this reasoning. Opening the three phases is the same a removing the line m - ‐ n altogether then adding appropriate impedances from buses m and n to the points p and p’.
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 113 Open-‐Conductor Faults If line m - ‐ nhas the sequence impedances Z0, Z1, and Z2,simulate the opening of the three phases by adding the negative impedances – Z0, – Z1, and – Z2 between buses m and n in the corresponding Thévenin equivalents of the three sequence networks of the intact system.
Open-‐Conductor Faults For example, consider the connection of – Z1 to the positive-‐sequence Thévenin equivalent between buses m and n: V mm Z  Zmn 1 1     m m n 0 Reference Voltages Vm and Vn are the normal (positive-‐sequence)voltages of phase a at buses m and n before the open-‐conductorfaults occur.   mn 1 Z  Znm 1   Vn nn 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 114 1 Z  Znm 1   Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1 
Vm mm Z  Zmn 1 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 115 Open-‐Conductor Faults The positive-‐sequenceimpedances kZ1 and (1 – k)Z1, 0<k<1, is added to represent the fractional lengths of the broken line m - ‐ n from bus m to point p and bus n to point p’. Z1
Open-‐Conductor Faults Simplify… Vm mm 1 Z  Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 kZ1 1 kZ1 p p’   Va 1  Add the do- ‐ nothing source 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 116
Open-‐Conductor Faults Simplify… Vm mm 1 Z  Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 kZ1 1 kZ1 p p’   a V1 Combine these 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 117
Open-‐Conductor Faults Simplify … Vm mm 1 Z  Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 Z 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 118 1 p p’   a V1 Perform a source conversion
Open-‐Conductor Faults Simplify … Vm mm 1 Z  Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 p p ’ a V1 Z1 Open circuit Z1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 119
Open-‐Conductor Faults Final result (positive-‐sequenceequivalent circuit): V mm Z  Zmn 1 1     m m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   th,mn Z1 V1 a Z1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 120
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 121 Open-‐Conductor Faults The above considerations for the positive-‐sequencenetwork also apply directly to the negative-‐and zero-‐sequencenetworks, but we must remember that the latter networks do not contain any internal sources of their own.
Open-‐Conductor Faults Negative-‐sequenceequivalent circuit: mm 2 Z  Zmn 2    m n 0 Referen ce 2 Z  Z mn nm 2    Znn  Znm 2 2 Z2 th,mn Z2 mm 2 nn 2     Z  Z  2Zmn 2  kZ2 1 kZ2 p p ’  a V2  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 122 Z2
Open-‐Conductor Faults Negative-‐sequenceequivalent circuit simplified: mm Z  Zmn 2 2    m n 0 Referen ce 2 Z  Z mn nm 2    nn 2 Z  Znm 2    th,mn Z2 V2 a Z2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 123
Open-‐Conductor Faults Zero-‐sequenceequivalent circuit: mm 0 Z  Zmn 0    m n 0 Referen ce 0 Z  Z mn nm 0    Znn  Znm 0 0 Z0 th,mn Z0 mm 0 nn 0     Z  Z  2Zmn 0  kZ0 1 kZ0 p p ’  a V0  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 124 Z0
Open-‐Conductor Faults Zero-‐sequenceequivalent circuit simplified: mm Z  Zmn 0 0    m n 0 Referen ce 0 Z  Z mn nm 0    nn 0 Z  Znm 0    th,mn Z0 V0 a Z0 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 125
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 126 Open-‐Conductor Faults component of the voltage drops Vpp’a, Vpp’b, and Vpp’c from p to p' in the phase conductors. , take on different values depending on which one of the open-‐conductorfauIts is being considered. a Let the voltage V 1denote the p h a s e -‐apositive-‐sequence a We will soon see that V 1and the corresponding negative-‐ and zero-‐sequence components Va and Va 2 0
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 127 Open-‐Conductor Faults In drawing the sequence equivalent circuits it is understood that the currents sources owe their origin to the open-‐conductor fault between points p and p' in the system. If there is no open conductor, the voltages: V0, V1, V2 a a a are all zero and the current sources disappear.
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 128 Open-‐Conductor Faults Note that each of the sequence currents sources: can be regarded in turn as a pair of injections into buses m and n of the corresponding sequence network of the intact system. Hence, we can use the bus impedance sequence matrices of the normal configuration of the system to determine the voltage changes due to the open-‐conductor faults. Z0 Z1 V0 V1 V2 a , a , a Z2
Open-‐Conductor Faults First we must find expressions for the symmetrical components of Va (i.e., of the voltage drops across the fault points p and p’ for each type of fault, (one or two open lines). These voltage drops can be regarded as giving rise to the following sets of injection currents into the sequence networks of the normal system configuration: POSITIVE NEGATIVE ZERO SEQUENCE SEQUENCE SEQUEN CE At bus m: At bus n: V1 Z a 1 a V2 Z2 a V0 Z0 Va 1  Z1 Va 2  Z2 Va 0  Z0 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 129
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 130 Open-‐Conductor Faults Recall Slide 87 from Network Calculations: 1   V j k   V N   V           V        Z1jI j  Z1k Ik jj j Z I  Z I jk k kj j Z I  Z I kk k I  Z I           Z             Nj j Nk k  
Open-‐Conductor Faults The changes in the symmetrical components of the p h a se -‐ a voltage of each bus i is:: Zero - ‐ s e quence: Positiv e - - ‐ sequence: Negative- ‐ s e q u e n c e :  V i 0   Zim 0  Z 0 in    Z0 a V0  V i 1   Zim 1  Z 1 in   Z1 a V1  V i 2   Zim 2  Z 2 in    Z2 a 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 131 V2
11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 132 Open-‐Conductor Faults Before developing the equations for the sequence components of the voltage for each type of o p e n - ‐conductor fault, let us derive expressions for the Thévenin equivalent impedances of the sequence networks as seen from fault points p and p’.
Open-‐Conductor Faults Looking into the positive-‐sequencenetwork between p and p’, we see the impedance: mm 1 Z  Zmn 1   m n 0 Referen ce 1 Z  Z mn nm 1   Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 133 Z1 Z 1 kZ  Z p p  1 th,mn 1  Z1  1 kZ1  Z1 Z Z1 th,mn Z1 1 th,mn 1  Z1 Z 2 th,mn Z1  Z1
Open-‐Conductor Faults The open-‐circuitvoltage from p to p’ is: Thus : Vm mm 1 Z  Zmn 1     m n   mn 1 Z  Znm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp Z1 pp 1  V  Z Vm Vn th,mn 1 Z 1  Z1 pp But: Z 1 Z1 2 th,mn Z1  Z1 V 0 11/11/13 Referen ce Unsymmetrical Faults (c) 2013 H. Zmuda 134 pp 1   p p  Z1 Z1 m n V  V 
Open-‐Conductor Faults Before any conductor opens, the current Imn in phase a of the line m - ‐ n is positive sequence and is given by: Thus : Where : Similarl y: and : pp 1 1   V  Z I p p  mn Imn  Vm  Vn Z1 Z 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 13 5 1 p p  1 Z 2 th,mn Z1  Z1 Z2 p p  Z 2 2 Z2  Z th,mn 2 Z0 p p  0 Z 2 Z0  Z th,mn 0 a Now we can find:V0, a V1, a V2
Open-‐Conductor Faults Thévenin equivalents: Positiv e - - ‐ Sequence Negative- ‐ S e q u e n c e Zero - ‐ S e quence:  Imn pp Z1  p p ’  Va 1  pp Z1 p p ’  Va 2  pp Z2 p p ’  a V0  pp Z0 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 136 I0 I2 a I1 a
Open-‐ConductorFaults – One Open Conductor Ia Ib Ic a b c m m m n n p 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 137 p ’ a I  0 I 0 a   I 1 a  I 2 a    0 n pp ,b V  0 Vpp,c 0
Open-‐ConductorFaults – One Open Conductor The open conductor in phase a causes equal voltage drops to appear from p to p' in each of the sequence networks. We can satisfy this requirement by connecting the Thévenin equivalents of the sequence networks in parallel at the points p and p’. a a a    V0     V2   V1   3    1  1 1 1  1 a a2 1 a2 a   V pp ,a 0 0           V      pp ,a pp ,a  1  3    V    V  pp,a   V  Va 0  a 1 a 2    V  V  pp ,a 3 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 138
Open-‐ConductorFaults – One Open Conductor From this circuit the expression for the positive-‐sequencecurrent is found as: Zpp 2 Zpp 0 a I2 a I0 p p   Va 0   Va 2  Imn pp Z1  pp Z1 a I1 p  Va  1 p ’ p ’ p ’ a 1 I  I  mn p p  Z 1 0  Z Z 2      Imn p p  Z1 Z1 pp p p  p p  p p  Z1 p p  p p  Z0Z2 pp pp Z0  Z2  I Zpp 1  Z 0  Z pp pp 2      pp pp pp pp pp pp mn Z0Z1 Z1Z2  Z0Z2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 139
Open-‐ConductorFaults – One Open Conductor The sequence voltage drops are: Zpp 2 Zpp 0 a I2 a I0 p p  Va  0   Va 2  Imn pp Z1  pp Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 140 I1 p   Va 1 p ’ p ’ p ’ Va 0 a 1 a 2  V  V  I 1 a      p p  p p  Z0Z2 pp pp Z0  Z2  I p p  p p  p p  Z0Z1Z2 mn Z0Z1 Z1Z2  Z0Z2 p p  p p  p p  p p  p p  p p  These terms are known from the impedance parameters of the sequence networks and the prefault current in phase a of the line m - ‐ n .
Open-‐ConductorFaults – One Open Conductor The currents: a , a , a V0 V1 V2 Z0 Z1 Z2 for injection into the sequence networks can now be determined. Zmm 0,1,2  Zmn 0,1,2     m n 0 Referen ce Zmn 0,1,2  Znm 0,1,2     Znn 0,1,2  Znm 0,1,2     th,mn Z0,1,2 V0,1,2 a Z0,1,2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 141
Open-‐ConductorFaults – Two Open Conductors Clearly: Ia Ib Ic a b c m m m n n p 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 142 p ’ b I  0 Ic  0 pp ,a V Va 0 a 1 a 2     V V  0 n
Open-‐ConductorFaults – Two Open Conductors Resolving the current into its symmetrical components: These can be satisfied by connecting the Thévenin equivalents of the sequence networks in series between the points p and p’. a a I1     Ia  I0   2    3   1  1 1 1  1 a a2 1 a2 a 0 0  Ia            a a 3   I    I   1    Ia      0 a    I  Along with: a 1 a 2     I  I  Ia 3 V V 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 143 0 pp ,a a a a 1 2     V V  0
Open-‐ConductorFaults – Two Open Conductors pp Z1 pp Z2 pp Z0 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 144 I1 a I2 a I0 I 0 a  I 1 a  I 2 a     p p p  1 a  V0  2 Va p’  Va p’  p’   ImnZpp 1 
Open-‐ConductorFaults – Two Open Conductors pp Z1 pp Z2 pp Z0 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 145 I1 a I2 a I0 I 0 a  I 1 a  I 2 a     p p p  1 a  V0   Va 2 Va p’  p ’ p’   Imn Zpp 1  I 0 a   I 1 a  I 2 a   I mn p p  Z1 pp pp pp Z0  Z1  Z2
Open-‐ConductorFaults – Two Open Conductors Zpp 1 Z2 p p  pp Z0 a I1 Ia 2 a I0 p p p a  V1 a  V0  2 p’ Va p’  p’   11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 146 Imn pp Z1  Va 1 1    I Z  Z I 1 1 mn p p  p p  a    Imn pp I Z 1  Z1mn p p  Z1 pp pp pp p p  Z0  Z1  Z2  Imn Z1p p  p p  Z0  Z2 pp pp pp p p  Z0  Z1  Z2 Va 2   Z I pp a 2 2      Imn p p  p p  Z1Z2 pp pp pp Z0  Z1  Z2 Va 0   Z I pp a 0 0      Imn pp p p  Z0Z1 pp pp pp Z0  Z1  Z2
11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 147 Open-‐ConductorFaults – Two Open Conductors In each of these equations the right-‐hand side quantities are all known before the fault occurs. The net effect of the open conductors on the positive-- ‐sequence network is to increase the transfer impedance across the line in which the open-‐conductorfault occurs. For one open conductor this increase in impedance equals the parallel combination of the negative-‐and zero-‐sequence networks between points p and p’. For two open conductors the increase in impedance equals the series combination of the negative-‐andzero-‐sequence networks between points p and p'.
Open-‐ConductorFaults – EXAMPLE Determine the change in voltage at bus 3 when the transmission line undergoes a. a one-‐open-‐conductorfault and b. a two-‐open-‐conductorfault along its span between buses 2 and 3 . Choose a base of 100 MVA, 345 kV in the transmission line. For the system of Slide 39, consider that Machine 2 is a motor drawing a load equivalent to 50 MVA at 0.8 power-‐factorlagging and nominal system voltage of 345 kV at bus 3. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 148
Open-‐ConductorFaults – EXAMPLE Prefault current (per unit) in line 2 - ‐ 3 : Sbase 100 S  50  50  0.5 * 3 23 23 S  V I  I  P  jQ 3 V*  S P  jQ S 1.0 1.0 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 149  0.5 0.8  j0.6  0.4 j0.3
Open-‐ConductorFaults – EXAMPLE Recal l: Also recall: From Slide 108: 0  Z     bus         j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    , 1,2  Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563   j0.0789 j0.1211   j0.1437      bus      Z1  Z2  j0.15, Z0  j0.5 Z1 p p  1 Z 2 th,mn Z1  Z1 Z0 p p  0 Z 2 Z0  Z th,mn 0 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  Z 1 Z2 p p  p p  Z1 2 22 1 33 1   1  Z  Z  2Z  Z 23 1  j0.152 j0.1696  j0.1696  2 j0.1104 j0.15  j0.712 pp Z 0 Z0 2 22 0 Z  33 0    23 0  Z  2Z  Z0  j0.52 j0.08  j0.58  2j0.08 j0.5   11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 150
Open-‐ConductorFaults – EXAMPLE Thus, if the line from bus 2 to bus 3 is opened, then an infinite impedance is seen looking into the zero-‐sequencenetwork between points p and p' of the opening. The zero-‐sequencecircuit confirms this fact since bus 3 would be isolated from the reference by opening the connection between bus 2 and bus 3. Slide 47: j0.04 1 j0.04 2 4 j0.04 Referen ce j0.5 3 j0.08 j0.04 3Xn  j0.15 (1 ) (2 ) (3 ) (4 ) (5 ) (6 ) 3Xn  j0.15 Transform er Node Bus 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 151 Transform er Node Bus
Open-‐ConductorFaults – EXAMPLE One open conductor: From Slide 118, From Slide 104: Va 0  Va 1 2    V  I a mn p p  p p  p p  Z0Z1Z2 Z Z pp pp pp pp 0 1 1 2  Z Z  Z Z pp pp 0 2          pp 0  Z   I p p  p p  Z1Z2 pp pp mn Z1 Z2  0.4  j0.3 j0.7122 2j0.712  0.1068  j1424  V i 0   Zim 0   Z 0 in  Z0 a V0,  V i 1   V i 2   im 1 Z  Z 1 in   Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 152 V1
Open-‐ConductorFaults – EXAMPLE One open conductor:  V 3 0   32 Z  Z33 0 0    Z0 a V0,  V 3 1   V 3 2   32 1 Z  Z 1   Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 153 33 V1  V 3 0   j0.08  j0.58 j0.5   0.1068 j1424  0.1068  j0.1424  V 3 1   V 3 2   j0.1104  j0.1696 j0.15   0.1068 j1424  0.0422 j0.0562 3 3 0  V  V   V3 3 1 2    V  0.1912  j0.24548 V3 new 3  V   3 V  1.0  0.1912  j0.24548  0.8088 j0.2548
Open-‐ConductorFaults – EXAMPLE Two open conductors: Inserting the infinite impedance of the z e r o - ‐ sequence network in series between points p and p' of the positive-‐sequence network causes an open circuit in the latter . No power transfer can occur in the system, since power cannot be transferred by only one phase conductor of the transmission line in this case since the zero-‐sequencenetwork offers no return path for current. 1 2 3 4  V  f Vf   j0.20 j0.04 j0.20 j0.04 j0.04 j0.04 Referen ce (1 ) 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 154 (2 ) (3) (4 ) Open (5 )

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7-Unsymmetrical_Faults.pptxiiiiiiiiiiiiii

  • 1.
  • 2.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 2 Introductory Comments Most of the faults that occur on power systems are unsymmetrical faults, which may consist of unsymmetrical short circuits, unsymmetrical faults through impedances, or open conductors. Unsymmetrical faults occur as single l i n e -‐ to -‐g rou ndfaults, l i n e - ‐ t o - ‐ line faults, or double l i n e -‐ to -‐g roun dfaults. The path of the fault current from line to line or line to ground may or may not contain impedance. One or two open conductors result in unsymmetrical faults, through either the breaking of one or two conductors or the action of fuses and other devices that may not open the three phases simultaneously.
  • 3.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 3 Introductory Comments Since any unsymmetrical fault causes unbalanced currents to flow in the system, the method of symmetrical components is very used in the analysis to determine the currents and voltages in all parts of the system aLer the occurrence of the fault. We will consider faults on a power system by applying Thévenin's theorem, which allows us to find the current in the fault by replacing the entire system by a single generator and series impedance, and we will show how the bus impedance matrix is applied to the analysis of unsymmetrical faults.
  • 4.
    Unsymmetrical Faults This shows thethree lines a, b, and c of the three-‐phasesystem at the part of the network where the fault occurs. The flow of current from each line into the fault is indicated by arrows shown beside hypothetical stubs connected to each line at the fault location. In the derivation of equations for the symmetrical components of currents and voltages in a general network the currents flowing out of the original balanced system from phases a , b, and c at the fault point will be designated as Ia, lb, and lc, respectively. We can visualize these currents by as follows: a Ifa b Ifb I 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 4 fc c
  • 5.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 5 Unsymmetrical Faults Appropriate connections of the stubs represent the various types of fault. For instance, direct connection of stubs b and c produces a l i n e - ‐ t o - ‐ l i n efault through zero impedance. The current in stub a is then zero, and lb equals - ‐lc. The l i n e -‐ to -‐g roun dvoltages at any bus j of the system during the fault will be designated Vja, Vjb and Vjc and we shall continue to use superscripts 1, 2, and 0, respectively, to denote positive-‐, negative-‐,and zero-‐sequencequantities. Thus, for example, V(1) ja, V(2) jb and V(0) jc will denote, respectively, the positive-‐,negative-‐,and zero-‐sequencecomponents of the l i n e - ‐ t o - ‐ ground voltage Vja at bus j during the fault.
  • 6.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 6 Unsymmetrical Faults The line-‐to-‐neutralvoltage of phase a at the fault point before the fault occurs will be designated simply by Vf, which is a positive-‐ sequence voltage since the system is balanced. We considered the prefault voltage Vf previously when calculating the currents in a power system with a symmetrical three-‐phase fault applied.
  • 7.
    Unsymmetrical Faults Consider a single-‐linediagramof a power system containing two synchronous machines. This simple system is sufficiently general that the equations derived are applicable to any balanced system regardless of the complexity. The point where a fault is assumed to occur is marked P, and in this example it is called bus k on the single-‐linediagram and in the sequence networks. Single line diagram of a balanced three-‐phase system P k 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 7
  • 8.
    Unsymmetrical Faults P k Referen ce fa I1 P  k Vf  Positive-‐Sequence Network  Vf P fa 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 8 I1 k kk Z1  Thévenin Equivalent of Positive-‐Sequence Network  ka V1
  • 9.
    Unsymmetrical Faults Negative-‐Sequence Network P k Referen ce P k fa I2 P fa 11/11/13Unsymmetrical Faults (c) 2013 H. Zmuda 9 I2 k kk Z2  Thévenin Equivalent of Negative-‐Sequence Network  ka V2
  • 10.
    Unsymmetrical Faults Reference Zero-‐SequenceNetwork P k P k fa I0 P fa I0 k kk Z0  Thévenin Equivalent ofZero-‐Sequence Network  ka V0 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 10
  • 11.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 11 Unsymmetrical Faults Machines are represented by their subtransient internal voltages in series with their subtransient reactances when subtransient fault conditions are being studied. Previously we used the bus impedance matrix composed of positive-‐sequenceimpedances to determine currents and voltages upon the occurrence of a symmetrical three-‐phase fault. The method can be easily extended to apply to unsymmetrical faults by realizing that the negative-‐and zero-‐sequencenetworks also can b e represented by bus impedance matrices. The bus impedance matrix will now be written symbolically for the positive-‐,negative-‐, and zero-‐sequencenetworks in the following form…
  • 12.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 12 Unsymmetrical Faults bus 0,1,2    11 12 Z0,1,2 1k Z0,1,2 1N 21 Z0,1,2 Z0,1,2 Zk1 0,1,2   Z0,1,2 Z0,1,2 Z 0,1,2    Z0,1,2 N1 N 2 Z0,1,2 Nk Z0,1,2 22 2k 2N k 2 kk kN NN Z0,1,2     Z0,1,2        Z0,1,2   Z0,1,2        Z0,1,2 
  • 13.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 13 Unsymmetrical Faults The Thévenin equivalent circuit between the fault point P and the reference node in each sequence network can be used for the analysis. As before, the voltage source in the positive-‐sequencenetwork and its Thévenin equivalent circuit is Vf, the prefault voltage to neutral at the fault point P, which happens to be bus k in this illustration. The Thévenin impedance measured between point P and the reference node of the positive-‐sequencenetwork is Z(1) kk, and its value depends on the values of the reactances used in the network. Recall that subtransient reactances of generators and 1.5 times the subtransient reactances (or else the transient reactances) of synchronous motors are the values used in
  • 14.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 14 Unsymmetrical Faults There are no negative-‐or zero-‐sequencecurrents flowing before the fault occurs, and the prefault voltages are zero at all buses of the negative-‐and zero-‐sequencenetworks. Therefore, the prefault voltage between point P and the reference node is zero in the negative-‐and zero-‐sequencenetworks and no electromotive forces (emfs) appear in their Thévenin equivalents. The negative-‐and zero-‐sequenceimpedances between point P a t bus k and the reference node in the respective networks are represented by the the impedances Z(2) kk and Z(0) kk, the Diagonal elements of Z(2) bus and Z(0) bus, respectively.
  • 15.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 15 Unsymmetrical Faults Since Ifa is the current flowing from the system into the fault, its symmetrical components flow out of the respective sequence networks and their equivalent circuits at point P, as shown. Thus, the currents –I(1) fa, –I(2) g and –I(0) fc represent injected currents into the faulted bus k of the positive-‐, negative-‐, and zero-- ‐sequencenetworks due to the fault. These current injections cause voltage changes at the buses of the positive-‐,negative-‐,and zero-‐sequencenetworks, which can be calculated from the bus impedance matrices in the manner similare to what we have done before.
  • 16.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 16 Unsymmetrical Faults For instance, due to the injection –I(1) fa into bus k, the voltage changes in the positive-‐sequencenetwork of the N - ‐ b u ssystem are given in general terms by: 1a 1 2a 1   V Na 1     Vk a    1    11 12 1k Z1 1N Z1 21 22 1 1 Z1 Z1 Zkk N1 N 2 Z1     1 1  V Z Z   V        Z1      Nk Z1 2k 2N 1 Z1 kN NN Z1 Z1    Zk1 Zk 2      Z1                    0 0 1  I 0 Z I 1k fa 1 1 Z I 2k fa 1 1   Zkk I fa Z I Nk fa        fa                          1 1     1 1   
  • 17.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 17 Unsymmetrical Faults Once again, it is industry practice to regard all prefault currents as being zero and to designate the voltage Vf as the positive-‐sequence voltage at all buses of the system before the fault occurs. Using superposition, the total positive-‐sequence voltage of phase a at each bus during the fault is: 1a 2a Na      Vka    V 1      V 1   1      V 1  f f f       V     V     V         Vf   V 1a 1  V 2a 1  V Na 1        Vk a             1       Z I f 1k fa 1 1 f  Z I 2k fa   f 1  Z I Nk fa     V       V  Z I f kk fa    V     V     1 1    1 1     1   
  • 18.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 18 Unsymmetrical Faults This is the same equation as found for symmetrical faults, the only difference being the added superscripts and subscripts denoting the positive-‐sequencecomponents of the phase a quantities.
  • 19.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 19 Unsymmetrical Faults The negative-‐and zero-‐sequencevoltage changes due to the fault at bus k of the N - ‐ b u ssystem are similarly written with the superscripts changed accordingly. Because the prefault voltages are zero in the negative-‐and zero-‐sequencenetworks, the voltage changes express the total negative-‐and zero-‐sequence voltages during the fault, namely, 1a 2a Na      Vka    V 2     V 2   2      V 2  Z I 1k fa 2 2 Z I 2k fa 2 2    Zkk Ifa Z I Nk fa                          2 2     2 2     , 1a 2a Na      Vka    V 0     V 0   0      V 0  Z I 1k fa 0 0 Z I 2k fa 0 0    Zkk Ifa Z I Nk fa    0 0     0 0                          
  • 20.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 20 Unsymmetrical Faults When the fault is at bus k, note that only the entries in columns k of Z(2) bus and Z(0) bus are involved in the calculations of negative-‐and zero-‐sequencevoltages. Thus, knowing the symmetrical components I(0) fa, I(1) fa and I(2) fa of the fault currents at bus k, we can determine the sequence voltages at any bus j of the system from the jth rows of 1a 2a Na      Vka    V 1     V 1   1      V 1   Z I f 1k fa 1 1  Z I f 2k fa  f  Z I Nk fa         V  Z I f kk fa    V     V        V 1 1   1 1     1 1     , 1a 2a Na      Vka    V2     V2   2      V2  Z I 1k fa 2 2 Z I 2k fa  Zkk Ifa Z I Nk fa                      2 2   2 2     2 2     , 1a 2a Na      Vka    V 0     V 0   0      V 0  Z I 1k fa 0 0 Z I 2k fa  Zkk Ifa Z I Nk fa  0 0                       0 0       0 0   
  • 21.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 21 Unsymmetrical Faults That is, during the fault at bus k the voltages at any bus j are: If the prefault voltage at bus CD is not Vf, then simply replace Vf in by the actual value of the prefault (positive-‐sequence) voltage at that bus. Since Vf is by definition the actual prefault voltage at the faulted bus k, we always have at that bus: and these are the terminal voltage equations for the Thévenin equivalents of the sequence networks previously shown. V 0 ja jk fa 0 0       Z I , V 1 ja f jk fa 1 1     V  Z I , V 2 ja  Z I jk fa 2 2      Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1  V  Z I , ka 2    V  Z I kk fa 2 2   
  • 22.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 22 Unsymmetrical Faults It is important to remember that the currents I(0) fa, I(1) fa and I(2) fa are symmetrical-‐componentcurrents in the stubs hypothetically attached to the system at the fault point. These currents take on values determined by the particular type of fault being studied, and once they have been calculated, they can be regarded as negative injections into the corresponding sequence networks.
  • 23.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 23 Unsymmetrical Faults If the system has Δ - ‐ Ytransformers, some of the sequence voltages may have to be shiLed in phase angle before being combined with other components to form the new bus voltages of the faulted system. There are no phase shiLs involved in when the voltage Vf at the fault point is chosen as reference, which is customary. Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1  V  Z I , ka 2    V  Z I kk fa 2 2   
  • 24.
    Unsymmetrical Faults Δ - ‐ Ytransformer with Positive-‐sequenceZero-- ‐sequence leakage impedance Z circuit circuit The negative-‐sequencecircuit is the same as the positive-‐sequence cir1 c1 / u1 1 i/ t1 .3 Unsymmetrical Faults (c) 2013 H. Zmuda 2 n m In a system with Δ - ‐ Ytransformers the open circuits encountered in the zero-‐sequence network requires some care in the Zbus building algorithm. Consider, for instance, the solidly grounded Δ - ‐ Y transformer connected between buses m and n as shown along with their positive and zero-‐sequencecircuits: Z m n Reference m n Reference Z
  • 25.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 25 Unsymmetrical Faults It is straightforward using the circuit representations shown to generate the bus impedance matrices Zbus (0,1,2). This will be done subsequently. Suppose, however, that we wish to represent removal of the transformer connections from bus n in a computer algorithm which cannot make use of circuit (schematic) representations. We can easily undo the connections to bus n in the positive-‐and negative-‐sequencenetworks by applying the building algorithm the Zbus (1,2) matrices in the usual manner, i.e., by adding the negative of the leakage impedance Z between buses m and m in the positive-‐and negative-‐sequencenetworks. (Next slide)
  • 26.
  • 27.
    Unsymmetrical Faults This strategy doesnot apply to the zero-‐sequencematrix Zbus (0) if it has been formed directly from the schematic representation shown. Adding - ‐Z between buses m and m does not remove the z e r o - ‐sequence connection from bus n. - ‐Z m n Reference Z 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 27
  • 28.
    Unsymmetrical Faults To permit uniformprocedures for all sequence networks, one strategy is to include an internal node p, as shown below. Note that the leakage impedance is now subdivided into two parts between node p and the other nodes as shown. Connecting –Z/2 between buses n and p in each of the sequence circuits will open the transformer connections to bus n. Z m n m n Reference Reference Z m n Reference Z/2 m n Reference Z/2 Z/2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 28 p p Z/2    
  • 29.
    Unsymmetrical Faults Connecting –Z/2 betweenbuses n and p in each of the sequence circuits will open the transformer connections to bus n. - ‐Z/2 - ‐Z/2 m n Reference Z/2 m n Reference Z/2 Z/2 p Z/2 p m n Reference Z/2 m n Reference Z/2 p 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 29 p    
  • 30.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 30 Unsymmetrical Faults The faults to be discussed in succeeding sections may involve impedance Zf between lines and from one or two lines to ground. When Zf = 0, we have a direct short circuit, which is called a bolted fault . Although such direct short circuits result in the highest value of fault current and are therefore the most conservative values to use when determining the effects of anticipated faults, the fault impedance is seldom zero.
  • 31.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 31 Unsymmetrical Faults Most faults are the result of insulator flashovers, where the impedance between the line and ground depends on the resistance of the arc, of the tower itself, and of the tower footing if ground wires are not used. T ower-‐footingresistances form the major part of the resistance between line and ground and depend on the soil conditions. The resistance of dry earth is 10 to 100 times the resistance of swampy ground.
  • 32.
    Unsymmetrical Faults Three-‐PhaseFault Ifa Ifb Ifc Connections of thehypothetical stubs for faults through impedance Zf are as follows: a b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 32 Zf Zf
  • 33.
    Unsymmetrical Faults b I fb c I fc SingleL in e -‐ to -‐ Gro u n dFault Ifa Connections of the hypothetical stubs for faults through impedance Zf are as follows: a Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 33
  • 34.
    Unsymmetrical Faults Ifb I fc L i n e - ‐ t o - ‐ L i n eFault Connections ofthe hypothetical stubs for faults through impedance Zf are as follows: a I fa b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 34
  • 35.
    Unsymmetrical Faults Ifb I fc Double L in e -‐ to -‐ Gro u n dFault Connectionsof the hypothetical stubs for faults through impedance Zf are as follows: a I fa b c Zf 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 35
  • 36.
    Unsymmetrical Faults Other types of faults: a Ia c Ic Open-‐ConductorFaults b Ib 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 36
  • 37.
    Unsymmetrical Faults Other types of faults: a Ia c Ic Open-‐ConductorFaults b Ib 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 37
  • 38.
    Unsymmetrical Faults A balanced systemremains symmetrical aLer the occurrence of a three-‐phasefault having the same impedance between each line and a common point. Only positive-‐sequencecurrents flow. With the fault impedance Zf equal in all phases, as in a three-‐phasefault, we simply add impedance Zf to the usual (positive-‐sequence) Thévenin equivalent circuit of the system at the fault bus k and calculate the fault current from the equation: For each of the other types of faults, formal derivations of the equations for the symmetrical-‐componentcurrents follow. In each case the fault point P is designated as bus k. fa 1  I  Vf 1  Z  Z kk f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 38
  • 39.
    Exampl e Two synchronous machinesare connected through three-- ‐phasetransformers to the transmission line as shown. The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8% On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15% and X0 = 50%. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 39
  • 40.
    Exampl e Positive-‐and negative-‐sequence circuit: Theratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn =5% Transformers T1, T2: 100 MVA, 20Δ/345Y kV, X = 8% (split as 4% + 4%) Transmission line reactances are X1 = X2 = 15% and X0 = 50%. 1 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 40 3 4  V  f Vf   j0.20 j0.20 j0.04 j0.04 j0.15 j0.04 j0.04 Referen ce (1 ) (2) (3 ) (4 ) (5 )
  • 41.
    Exampl e Add branch (1): Z 1 11/12/13Unsymmetrical Faults (c) 2013 H. Zmuda 41 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) 1 1,2     j0.20   1,2 j0.20 j0.20 Add branch (2):  Z2     j0.20 j0.20  j0.08         j0.20  j0.20  j0.20 j0.28  
  • 42.
    Exampl e 1 11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 42 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) 3     Add branch (3):  j0.20 Z1,2   j0.20  j0.20 j0.20 j0.28 j0.28 j0.20 j0.28 j0.28  j0.15          j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.20 j0.28 j0.43     
  • 43.
    Exampl e 1 11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 43 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) Z4 1,2    j0.20 j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.28 Add branch (4):  j0.20 j0.28 j0.43 j0.43 j0.20 j0.28 j0.43 j0.43 j0.08                    j0.20 j0.20 j0.20 j0.20 j0.20 j0.28 j0.28 j0.28 j0.20  j0.20  j0.28 j0.28  j0.43 j0.43 j0.43   j0.51    
  • 44.
    Exampl e Add branch (5):  1 11/12/13 UnsymmetricalFaults (c) 2013 H. Zmuda 44 2 (3) 3 4  V  f Vf   j0.20 j0.04 j0.15 j0.04 j0.04 j0.04 Referen ce (1 ) (5) j0.20 (2 ) (4 ) Z4 1,2    j0.20 j0.28 j0.28 j0.28 j0.20 j0.28 j0.43 j0.43 j0.20 j0.28 j0.43 j0.51       j0.20 j0.20 j0.20 j0.20        Znew ij  Zij Z Z  ik kj Z  Z kk b
  • 45.
  • 46.
  • 47.
    Example Zero-‐sequence circuit: j0.041 3 j0.08 4 j0.04 Reference The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8% Transmission line reactances are X1 = X2 = 15% and X0 = 50%. j0.5 j0.04 3Xn  j0.15 (1 ) (2 ) (3 ) (4 ) (5 ) (6 ) 3Xn  j0.15 See Slide 199 of Symmetrical Components Transform er Node Bus Transform er Node Bus See Slide 196 of Symmetrical 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 47 Components j0.04 2
  • 48.
    Exampl e 1 11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 48 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 )
  • 49.
    Exampl e Add branch (1): Z Addbranch (2): 1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 49 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 ) 0  1    j0.19  2 0   Z       j0.19 0 0 j0.08   3 0  Z  j0.19 0 0 j0.08 0 j0.08 Add branch (3):  0 j0.08        j0.08 j0.5  
  • 50.
    Exampl e Add branch (5): 1 11/12/13 UnsymmetricalFaults (c) 2013 H. Zmuda 50 2 3 j0.08 4 j0.08 Referen ce j0.5 j0.19 j0.19 (1 ) (2 ) (3 ) (4 ) (5 ) Z 0   j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus          
  • 51.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 51 Exampl e We’ll use these in subsequent examples. Z 0      bus    j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19        Z 1,2       bus    j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437       
  • 52.
    Single L in e -‐ to -‐ Gro u n dFaults The singlel i n e -‐ to -‐g roundfault, the most common type, is caused b y lightning or by conductors making contact with grounded structures. A single l i n e -‐ to -‐g roundfault on phase a through impedance Zf is shown below: The relations to be developed will apply only when the fault is on phase a, but any phase can be designated as phase a. The conditions at the fault bus k are expressed by the following equations: I fa b I fb c I fc a Z f k 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 52 fb fc ka I  I  0, V  Z I f fa
  • 53.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 53 Single L in e -‐ to -‐ Gro u n dFaults The symmetrical components of the current are, with I fb  I fc  0: But recall from Slide 21: Thus : Vka 0 0    kk fa 0   Z I , Vka 1  f kk fa 1 1    V  Z I , Vka 2   Z I kk fa 2 2    Vka 0  kk fa 0 0     Z I , Vka 1  f kk fa 1 0  V  Z I , ka 2     V  Z I kk fa 2 0    fa fa I1 I  I0    2       fa     1 1 a a2 a2 a  1 3  1  1  1  0 0 I fa               I fa 0 fa 1  I  I 2 fa      I fa 3
  • 54.
    Single L in e -‐ to -‐ Gro u n dFaults Summing, andnoting that V This has the circuit representation shown on the next slide. V V ka ka ka 0 1 V Vka 2     f  V  Z 0 kk   Z 1 kk  Z 2 kk    I fa 0   3Z I f fa 0  ka  3Z I f fa 0  fa Solve for I 0 I fa 0 fa 1  I  I 2 fa      Vf Z 0 kk kk 1 kk 2      Z  Z 3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 54
  • 55.
    Single L in e -‐ to -‐ Gro u n dFaults Vf   kk Z1 kk Z2 kk Z0 3Z f fa I1 fa I2 fa I0 I 11/12/13Unsymmetrical Faults (c) 2013 H. Zmuda 55 fa 0 fa 1  I  I 2 fa     k k k  ka  V1  ka  V0  ka  V2
  • 56.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 56 Single L in e -‐ to -‐ Gro u n dFaults The last result are the fault current equations particular to the single l i n e -‐ to -‐g roundfault through impedance Zf and they are used with the symmetrical-‐componentrelations to determine all the voltages and currents at the fault point P. If the Thévenin equivalent circuits of the three sequence networks of the system are connected in series, as shown on the previous slide, we see that the resulting currents and voltages satisfy the above equations – for the Thévenin impedances looking into the three sequence networks at fault bus k are then in series with the fault impedance 3Zf and the prefault voltage source Vf.
  • 57.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 57 Single L in e -‐ to -‐ Gro u n dFaults With the equivalent circuits so connected, the voltage across each sequence network is the corresponding symmetrical component of the voltage Vka at the fault bus k, and the current injected into each sequence network at bus k is the negative of the corresponding sequence current in the fault. The series connection of the Thévenin equivalents of the sequence networks, as shown on Slide 55 is a convenient means of remembering the equations for the solution of the single l i n e - ‐ t o - ‐ ground fault, for all the necessary equations for the fault point can be determined from the sequence-‐network connection.
  • 58.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 58 Single L in e -‐ to -‐ Gro u n dFaults Once the sequence components of the fault currents are known, the components of voltages at all other buses of the system can be determined from the bus impedance matrices of the sequence networks according to Slide 21, namely: V V V 0 ja  Z I jk fa 0 0      1 ja  f  V  Z I jk fa 1 1   2 ja   Z I jk fa 2 2   
  • 59.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults The ratings and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Y/345Y kV, X = 8% Two synchronous machines are connected through three-- ‐phasetransformers to the transmission line as shown. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 59
  • 60.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 60 EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Both transformers are solidly grounded on on two sides. On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15 % and X0 = 50 %. The system is operating at nominal voltage without prefault currents when a bolted (Zf = 0) single l i n e -‐ to-‐ g ro u n dfault occurs o n phase A at bus 3. Using the bus impedance matrix for each of the three sequence networks, determine the subtransient current to ground at the fault, the l i n e -‐ to -‐groundvoltages at the terminals of machine 2, andthe subtransient current out of phase c of machine 2.
  • 61.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults The system is the same as on Slide 39, except that the transformers are now Y - ‐ Yconnected. Therefore, we can continue to use Zbus (1,2) from Slide 51, however, because the transformers are solidly grounded on both sides, the zero-- ‐sequencenetwork is fully connected, as shown below, and has the bus impedance matrix 1 2 3 4 j0.04 j0.08 j0.15 j0.08 j0.04 j0.15 j0.15 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 61 Referen ce
  • 62.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 62 3 4 j0.15 j0.15 j0.15 The bus impedance matrix is: 1 j0.04 j0.08 j0.08 j0.04 Referen ce Zbus 0   j0.1553 j0.1407 j0.0493 j0.0347 j0.1407 j0.1999 j0.0701 j0.0493 j0.0493 j0.0701 j0.1999 j0.1407 j0.0347 j0.0493 j0.1407 j0.1553              
  • 63.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults Since the l i n e -‐ to-‐ groundfault is at bus 3 : The total current in the fault is: f V 1.0 33 Z1 Z2 33 33 Z0 f 3Z  0 fA 1 I  fA I2 fA I0 IfA 3 k k k   V3a 1  3a  V0  3a  V2  0  Z     bus  Z  j0.1553 j0.1407 j0.0493 j0.0347 j0.1407 j0.1999 j0.0701 j0.0493 j0.0493 j0.0701 j0.1999 j0.1407 j0.0347 j0.0493 j0.1407 j0.1553 j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     1,2  bus                   I fA 0   I fA 1 fA 2    I  Vf kk 0 kk 1 kk 2     Z  Z  Z 3Z f  1 j0.1999  j1.696  j1.696  0   j1.8549 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 63 fA fA 0   I  3I   j5.5648
  • 64.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults Since the base current in the high-‐voltagetransmission line is: amp s Then: amp s The p h a se -‐asequence voltages at bus 4, the terminals of machine 2, are (from Slide 21 with k = 3 and j = 4): per unit base 3 345 I  100,000  167.35 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 64 I fA   j5.5648167.35 931270 V4a 43 fA 0 0 0       Z I     j0.1407  j1.8549  0.261 V4a 1  f 43 fA 1 1       V  Z I  1 j0.1211  j1.8549  0.7754 V4a 2   Z I 43 fA 2 2         j0.1211  j1.8549  0.2246
  • 65.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 65 EXAMPLE – Single L in e -‐ to -‐ Gro u n dFaults Note that the subscripts A and a denote voltages in the high-- ‐voltageand low-‐voltagecircuits, respectively, of the Y - ‐ Yconnected transformer. No phase shiL is involved. From the symmetrical components we can calculate the a - ‐ b - ‐ c l i n e - ‐ to- ‐groundvoltages at bus 4 as: per unit 4a 4b 4c   V  V    V     1  1 1 1 1 a2 a a a2 3 1   V4a 0 4a 4a               V 2     V 1   1  1 1 1 1 a2 a a a2 3 1          0.261 0.7754    0.2246   0.2898         0.5346  j0.8660   0.5346  j0.8660 
  • 66.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults To express the l i n e -‐ to-‐g roundvoltages of machine 2 (in kV) multiply by 20/31/2: k V 4b 4c    V  4a     V     V  20 0.2898 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 66 0.5346  j0.8660    3       3.346 1.0187121.8       0.5346  j0.8660  1.0187 121.8  
  • 67.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults To determine p h a s e-‐ccurrent out of machine 2 we must first calculate the symmetrical components of the p h a se -‐acurrent in the branches representing the machine in the sequence networks. From the zero-‐sequencecircuit, the zero-‐sequencecurrent out of the machine is: per unit 1 2 3 4 j0.15 j0.04 j0.15 j0.08 j0.15 j0.08 j0.04 Referen ce  4a V0  a I0 a 0  I   4a V0 0  jX j0.04 0.261   j6.525 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 67
  • 68.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults per unit a 1 I  V V1  f 4a jX   1 0.7754 j0.2   j1.123  V  f Vf   j0.20 j0.20 j0.04 j0.04 j0.15 j0.04 j0.04 Referen ce  4a V1,2  a Similarly for the positive-‐and negative-‐sequence currents: 1 2 3 4 I1,2 a 2  I   4a V1 jX2  0.2246 j0.2   j1.123 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 68
  • 69.
    EXAMPLE – SingleL in e -‐ to -‐ Gro u n dFaults The p h a se -‐ ccurrent in machine 2 is per unit Since the base current in the machine circuit is: amp s c a 0 1    2 I  I  aI  a I a a 2    j6.525 a j1.123 a2 j1.123   j5.402 Ibase 3 20  100,000  2886.751 amps Ic 15,994 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 69
  • 70.
    L i n e - ‐ t o - ‐ L i n eFaults To represent al i n e - ‐ t o - ‐ l i n efault through impedance Zf the hypothetical stubs on the three lines at the fault are connected as shown. Bus k is again the fault point P, and without any loss of generality, the l i n e - ‐ t o - ‐ l i n efault is regarded as being on phases b and c. Clearly: Ifb 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 70 Ifc a b c Zf k I fa I  0 fa k k I fb  I fc Vkb Vkc  Z I f fb
  • 71.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 71 L i n e - ‐ t o - ‐ L i n eFaults Substitutin g: The voltages throughout the zero-‐sequencenetwork must be zero current is not being injected into that network due to the fault. Hence, l i n e - ‐ t o - ‐ l i n efault calculations do not involve the z e r o - ‐ sequence network, which remains the same as before the fault - ‐adead network. since there are no zero-‐sequencesources, and because I fa fa I1      I0    2   I fa      1 1 1 1 a a2 a2 a   1 3   1  I fc          0  I  fb      I  I fa 0   0  I fa 1  2 fa  fa 0   0,
  • 72.
    L i n e - ‐ t o - ‐ L i n eFaults equivalents of thepositive-‐and negative-‐sequencenetworks i n parallel, as shown: To satisfy the requirement that I 1   I we connect the Thévenin fa fa 2  Vf   kk Z1 kk Z2 Z f fa I1 fa I2 k  ka V1  k 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 72  ka V2  Referen ce
  • 73.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 73 L i n e - ‐ t o - ‐ L i n eFaults To show that this connection of the networks also satisfies the voltage equation: we now expand each side of that equation separately as follows: and : V V  Z I kb kc f fb kb kc V V  Vkb 1 kb 2     V  Vkc 1 Vkc 2      Vkb 1  Vkc 1   Vkb 2 Vkc 2        2 ka 1   a V  a  2     a Vka 2   a  a2   ka  a V Vka 1 2     f fb Z I  Z f fb 1 I  I 2 fb   f 2 fa 1  Z a I  aI fa 2      
  • 74.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 74 L i n e - ‐ t o - ‐ L i n eFaults Equate both terms and set: But this is precisely the voltage drop across Zf in the figure on Slide 66. kb kc 2   V V  a  a Vka 1 ka 2     f fb f 2 V  Z I  Z a I fa 1  aI 2 fa     2    a  a Vka 1 ka 2     f 2   V  Z a  a I fa 1  ka 1 V Vka 2    Z I f fa 1  I fa 1   I 2 fa 
  • 75.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 75 L i n e - ‐ t o - ‐ L i n eFaults Thus, all the fault conditions are satisfied by connecting the positive-‐and negative-‐sequencenetworks in parallel through impedance Zf as was shown. The zero-‐sequencenetwork is inactive and does not enter into the l i n e - ‐ t o - ‐ l i n ecalculations. The equation for the positive-‐sequence current in the fault can be determined directly from the circuit as:
  • 76.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 76 Example – L i n e - ‐ t o - ‐ L i n eFaults The same system as before is operating at nominal system voltage without prefault currents when a bolted l i n e - ‐ t o - ‐ l i n efault occurs at bus 3. Determine the currents in the fault, the l i n e - ‐ t o - ‐ l i n e voltage a tthe fault bus, and the l i n e - ‐ t o - ‐ l i n evoltages at the terminals of machine 2.
  • 77.
    Example – L i n e - ‐ t o - ‐ L i n eFaults Notethat we will not need the zero-‐sequencebus impedance matrix since the fault is l i n e - ‐ t o - ‐ l i n e(see Slide 71). From the circuit: f 33 The Thévenin equivalent circuit for the positive-‐and n e g a t i v e - ‐ sequence is: Z1 33 Z2 f Z  0 IfA 1 fA 3  V3A 1  3  I2 V3A 2  Referen ce  V 1,2 Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437    bus            I fA fA 1 2    I  Vf 33 1 Z  Z33 2   2  j0.1696 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 77  1.0   j2.9481 per unit
  • 78.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 78 Example – L i n e - ‐ t o - ‐ L i n eFaults Sinc e (Uppercase since the fault is in the high-- ‐voltage transmission line) Then I and since Then I fA 0 Also I per unit an d per unit Multiply these by Ibase = 167.35 A to get the actual currents. fA  I fA 1  I fA 2   I fA 0   0 I fA 1   I fA 2  1 fB fA fA 2 2    a I  aI  5.1061 I fC  I fB  5.1061
  • 79.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 79 Example – L i n e - ‐ t o - ‐ L i n eFaults The symmetrical components of p h a s e -‐Avoltage to ground at bus 3are: per unit V3A 0   0 V3A 1 3A 2    V  1 Z I kk fA 1 1    1 j0.1696  j2.9481     0.5  j0
  • 80.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 80 Example – L i n e - ‐ t o - ‐ L i n eFaults The l i n e -‐ to -‐g roun dvoltages at fault bus 3 are: V3C  V3B 0.5 all per unit V V 3A 3A 3A 0 1 V V3A 2      0  0.5 0.5  1.0 V V 3B 3B 0   a V3B 1 2   2V3B 2  2  0 a 0.5 a  0.5  0.5
  • 81.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 81 Example – L i n e - ‐ t o - ‐ L i n eFaults The l i n e - ‐ t o - ‐ l i n evoltages at fault bus 3 are: V3,AB  V3A V3B  1 j0 0.5 j0 1.5 V3,BC  V3B V3C  0.5 j0 0.5 j0 0 all per unit. Multiply these by 345/31/2 to obtain the actual voltage. V3,CA V3C  V3 A  0.5  j0 1 j0 1.5
  • 82.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 82 Example – L i n e - ‐ t o - ‐ L i n eFaults For the moment, let us avoid phase shiLs due to the Δ - ‐ Y transformer connected to machine 2 and calculate the sequence voltages of phase A at bus 4 using the the results from Slide 21 with k = 3 and j = 4: per unit Vja jk fa ja f jk fa  0  Z0I0, V 1V  Z1I1, V 2  Z2I2 ja jk fa 0 Z 0I0 V4a 43 fa 0 V4a 1  f jk fa 1 1    V  Z I 1 j0.1211 j2.9481 0.643 V4a 2   Z I 43 fa 2 2         j0.1211  j2.9481  0.357
  • 83.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 83 Example – L i n e - ‐ t o - ‐ L i n eFaults To account for phase shiLs in stepping down from the high-- ‐voltagetransmission line to the low-‐voltageterminals of machine 2, we must retard the positive-‐sequencevoltage and advance the negative-‐sequencevoltage by 30°. At machine 2 terminals, indicated by lowercase a, the voltages are: V4a 43 fa 0 0 0       Z I  0 V4a 1 4a 1    V   30  0.643  30  0.5569  j0.3215 V4a 2  4a 2   V 30  0.35730  0.3092 j0.1785 V V 4a 4a 0  4a 1 V V4a 2    0.8861 j0.1430  0.8778 9.4
  • 84.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 84 Example – L i n e - ‐ t o - ‐ L i n eFaults P h a se-‐bvoltages at the terminals of machine 2 are: V4b 0 4a 0     V  0 V4b 1  2  a V4a 1  1240 0.643  30  0.5569  j0.3215 V4b 2   aV4a 2   1120 0.35730  0.3092  j0.1785 V V 4b 4b 0  4b 1 V V4b 2    0.8861 j0.1430  0.8778170.6
  • 85.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 85 Example – L i n e - ‐ t o - ‐ L i n eFaults P h a se-‐ cvoltages at the terminals of machine 2 are: V4c 0 4a 0     V  0 V4c 1   aV4a 1   1120 0.643  30  0.64390 V4c 2  2  a V4a 2   1240 0.35730  0.357  90 V V 4c 4c 0  4c 1 V V4c 2    j0.286
  • 86.
    Example – L i n e - ‐ t o - ‐ L i n eFaults L i n e - ‐ t o - ‐ l i n evoltagesat the terminals of machine 2 are: per unit For the voltage, multiply by 20 kV/31/2 k V V4,ab 4a 0 0 4b 0      V4,bc 0  V4b 0 V4c 0    0.8661 j0.429 j0.429 V 0   V V 4,ca 4c 4a 0 0     V V 1.7322    0.8661 V4,ab 0  20 3  1.7322   20 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 86 V4,bc 0  11.2 153.65  11.2153.65 V4,ca 0 
  • 87.
    Ifa Ifb b c Zf I fa 0 11/12/13Unsymmetrical Faults (c) 2013 H. Zmuda 87 Double L in e -‐ to -‐ Gro u n dFaults Again it is clear, with fault taken on phases b and c, that the relations at fault bus k are: k a Vkb  Vkc  Z f Ifb  I fc  k Ifc k
  • 88.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 88 Double L in e -‐ to -‐ Gro u n dFaults Thus Vkb  Vkc  Z Substituting into: fa Since I  0 I fa 0   1 3 fa fb I  I  I fc   1 3 fb I  I fc   f fb fc   I  I  3Z I f fa 0  ka ka ka    V0       V2     V1    1  1 1 1 1 a a2 a2 a 3  1 ka kb kc     V  V     V      1  1 1 1 1 a a2 a2 a 3  1 ka kb kb  V     V  V       
  • 89.
    Double L in e -‐ to -‐ Gro u n dFaults Expanding thesecond and third equations: ka ka ka    V0       V2   V1    1  1 1 1 1 a a2 a2 a 3  1   1 3 ka kb kb  V     V  V        Vka 1   2   a  a Vkb Vka 2 1 3  a2  aV 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 89 kb Vka 1  Vka 2 
  • 90.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 90 Double L in e -‐ to -‐ Gro u n dFaults From the first equation: recall: Vkb  Vkc  Z f Ifb  Ifc  ka ka ka   V0      V2   V1    1  1 1 1 1 a a2 a2 a 3  1  3V ka kb kb     V  V     V     ka 0  ka kb    V  2V  Vka 0 ka 1    V Vka 2    2 3Z I f fa 0     3I fa 0  fb  I  I fc
  • 91.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 91 Double L in e -‐ to -‐ Gro u n dFaults Collect zero-‐sequenceterms and recall that Now solve: Thus : and since Ifa = 0 These last two results characterize the double l i n e -‐ to -‐g roundfault. Vka 1 Vka 2   3Vka ka 0 0     V  6Z I f fa 0   2Vka 1  Vka ka 1 0    V  3Z I f fa 0  Vka 1  Vka 2 ka 0     V  3Z I f fa 0  I 0 fa fa 1  I  I 2 fa      0
  • 92.
    Double L in e -‐ to -‐ Gro u n dFaults These canbe realized by putting all three sequence networks in parallel as follows: Clearl y: kk Z2 kk Z0 3Zf fa I2 fa I0 k k  ka  V0  ka  V2 Vf   kk Z1 fa I1 k  ka  V1 I fa 1   Vf kk 1  Z  Z 0 kk  f    3Z Z 2 kk   Vf kk 1  Z  Z 0 kk  f    3Z Z 2 kk  Z 0 kk  Z 2 kk     3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 92
  • 93.
    Double L in e -‐ to -‐ Gro u n dFaults By current divider: kk Z2 3Zf fa I2 fa I0 k k  kkka Z0  V0  ka  V2 Vf   kk Z1 fa I1 k  ka  V1 I  I fa 2 1 fa    Z 0 kk   3Z f kk Z 0   Z 2 kk   3Z f , I fa 0   I fa 1  kk Z2 kk Z 0   Z 2 kk   3Z f 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 93
  • 94.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 94 Double L in e -‐ to-‐ Gro u n dFaults For a bolted fault Zf is set equal to 0. When Zf = ∞, the z e r o - ‐ sequence circuit becomes an open circuit, no zero-‐sequence current an flow, and the equations revert back to those for the l i n e - ‐ t o - ‐ l i n efault. Again we observe that the sequence currents, once calculated, can be treated as negative injections into the sequence networks at the fault bus k and the sequence voltage changes at all buses of the system can then be calculated from the bus impedance matrices, as we have done in all along.
  • 95.
    Example Double L in e -‐ to -‐ Gro u n dFaults Findthe subtransient currents and the l i n e - ‐ t o - ‐ l i n e voltages at the fault under subtransient conditions when a double l ine -‐ to -‐g roun d fault with Zf = 0 occurs at the terminals of machine 2 in the system below. Assume that the system in unloaded and operating at rated voltage when the fault occurs. 1 T1 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 95 3 T2 4 Machine 1 Machine 2
  • 96.
    Example Double L in e -‐ to -‐ Gro u n dFaults Model(the fault is a bus k = 4): kk Z2 kk Z0 3Zf fa I2 fa I0 k k  ka  V0  ka  V2 Vf   kk Z1 fa I1 k  11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 96 ka  V1 0  Z  j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus        , 1,2   Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     bus              
  • 97.
    Example Double L in e -‐ to -‐ Gro u n dFaults Model(the fault is a bus k = 4): per unit j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 1  I  Vf Zkk 1  0  kk f   Z  3Z Z 2 kk  Zkk 0 2  Zkk  3Zf 44 1 Z  44 44 Z0Z2 44 0 Z  Z44 2    j0.1437  j0.19j0.1437 j0.19  j0.1437  1.0  1.0   j4.4342 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 97
  • 98.
    Example Double L in e -‐ to -‐ Gro u n dFaults The sequence voltages at the fault are: j0.1437 j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 98 4a j0.1437  V1 V4a 1  V 2  0   V  V  Z I 4a 4a f 44 fa 1 1    1.0  jj0.1437 j4.4342 0.3628 per unit
  • 99.
    Example Double L in e -‐ to -‐ Gro u n dFaults per unit j0.1437j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 2  I  I fa 1  Z 0 kk   3Z f kk Z 0 kk 2     Z 3Z f  I fa 1  kk Z0 kk Z 0   Z 2 kk   I fa 1  44 Z0 Z 0  Z 44 44 2     j4.4342 j0.19 j0.19  j0.1437  j2.5247 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 99 fa Now that we have I 1
  • 100.
    Example Double L in e -‐ to -‐ Gro u n dFaults per unit j0.1437j0.19 fa I2 fa I0 4 4  4a  V0  4a  V2 1.0   fa I1 4  4a j0.1437  V1 fa 0  I  I fa 1  kk Z2 kk Z 0 kk 2     Z 3Z f  I fa 1  44 Z2 44 Z 0  Z44 2     j4.4342 j0.1437 j0.19  j0.1437  j1.9095 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 100 fa Now that we have I 1
  • 101.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 101 Example Double L in e -‐ to -‐ Gro u n dFaults The currents out of the system at the fault point are: The current If into ground is: all in per unit fa fa fa fa 0 1 2     I  I  I  I  j1.9095  j4.4342  j2.5247  0 fb fa 0  fa I  I  a I  aI fa 1 2 2     6.0266  j2.8642  6.0266  j2.8642 0 fc fa fa 1 I  I  aI  a I 2 fa    2  f fa fb fc fa 0  I  I  I  I  2I  j5.7285
  • 102.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 102 Example Double L in e -‐ to -‐ Gro u n dFaults Calculating the a - ‐ b - ‐ cvoltages at the fault bus: V V 4a 4a 4a 0 1 V V4a 2     1.0884 V4b  V4c 0 V4,ab  V4a  V4b 1.0884 V4,bc  V4b  V4c 0 V4,ca  V4c V4 a  1.0884
  • 103.
    Example Double L in e -‐ to -‐ Gro u n dFaults Recallthe base current in the circuit of machine 2 is: amp s and the base line-‐to-‐neutralvoltage in machine 2 is: k V Ibase  100103 3 20  2887 Vbase  20 3 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 103
  • 104.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 104 Phase Shias The previous two examples show that phase shiLs due to Δ - ‐ Y transformers do not enter into the calculations of sequence currents and voltages in that part of the system where the fault occurs provided Vf at the fault point is chosen as the reference voltage for the calculations. However, for those parts of the system which are separated by Δ - ‐ Y transformers from the fault point, the sequence currents, and voltages calculated by bus impedance matrix must be shiLed in phase before being combined to form the actual voltages. This is because the bus impedance matrices of the sequence networks are formed without consideration of phase shiLs, and so they consist of per-‐unitimpedances referred to the part of the network which includes the fault
  • 105.
    Example Phase Shias Solve forthe subtransient voltages to ground at bus 2, the end of the transmission line remote from the double l i n e -‐ to -- ‐ g ro u ndfault for the same system. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 105 0  Z  j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    bus        , 1,2   Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563 j0.0789 j0.1211 j0.1437     bus              
  • 106.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 106 Example Phase Shias We just solved for the values of the fault-‐current components. Neglecting the phase shiL of the Δ - ‐ Ytransformer for the moment we have, from Slide 21: At bus 2: V 0 ja jk fa 0 0       Z I , V 1 ja f jk fa 1 1     V  Z I , V 2 ja  Z I jk fa 2 2      V2a 0   Z I 24 fa 0 0      0    j1.9095  0 V2a 1  f 24 fa 1 1    V  Z I 1.0  j0.0789     j4.4342  0.6501 V2a 2   Z I 24 fa 2 2         j0.0789 j2.5247  0.1992
  • 107.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 107 Example Phase Shias Accounting for phase shiL in stepping up to the transmission- ‐line circuit from the fault at bus 4 we have: Now the required voltages can be calculated: all per unit V2A 0   0 V2A 1 2A 1    V 30  0.650130 V 2  2 A 2A 2   V   30  0.1992  30 V V 4 A 4A 4A 0 1 V V4A 2      0.7355  j0.2255 4B V V4B 0  2 4B 1   aV4B 2   0.1275j0.5535  0.5656  j0.1274 4C V V4C 0   a V  aV4C 1  2  a V4C 2 
  • 108.
    Example Phase Shias The per-‐unitvaluescan be converted to volts by multiplying by the line-‐to-‐neutralbase voltage of: k V of the transmission line. Vbase  345 3 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 108
  • 109.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 109 Open-‐Conductor Faults When one phase of a balanced three-‐phasesystem opens, an unbalance is created and asymmetrical currents flow. A similar type of unbalance occurs when any two of the three phases are opened while the third phase remains closed. These unbalanced conditions are caused, for example, when o n e - ‐or two-‐phaseconductors of a transmission line are physical broken by accident or storm. In other circuits, due to current overload, fuses or other switching devices may operate in one or two conductors and fail to operate in other conductors. Such open-‐conductorfaults can be analyzed by means of the bus impedance matrices of the sequence networks, as we now show.
  • 110.
    Open-‐Conductor Faults Consider asection of a three-‐phase circuit in which the line currents in the respective phases are la, Ib, and Ic, with positive direction from bus m to bus n, with phase a open between points p and p’: a Ia b Ib c Ic m m m n n n p 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 110 p ’
  • 111.
    Open-‐Conductor Faults Also considerthe case where phases b and c are open between points p and p’: a Ia b Ib c Ic m m m n n n p p ’ 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 111
  • 112.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 112 Open-‐Conductor Faults The same open-‐conductorfault conditions will result is all three phases are first opened between points p and p’ and short circuits are then applied in those phases which are shown to be closed in the preceding figures. The ensuing development follows this reasoning. Opening the three phases is the same a removing the line m - ‐ n altogether then adding appropriate impedances from buses m and n to the points p and p’.
  • 113.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 113 Open-‐Conductor Faults If line m - ‐ nhas the sequence impedances Z0, Z1, and Z2,simulate the opening of the three phases by adding the negative impedances – Z0, – Z1, and – Z2 between buses m and n in the corresponding Thévenin equivalents of the three sequence networks of the intact system.
  • 114.
    Open-‐Conductor Faults For example,consider the connection of – Z1 to the positive-‐sequence Thévenin equivalent between buses m and n: V mm Z  Zmn 1 1     m m n 0 Reference Voltages Vm and Vn are the normal (positive-‐sequence)voltages of phase a at buses m and n before the open-‐conductorfaults occur.   mn 1 Z  Znm 1   Vn nn 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 114 1 Z  Znm 1   Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1 
  • 115.
    Vm mm Z  Zmn 11     m n 0 Referen ce   1 Z  Z mn nm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 115 Open-‐Conductor Faults The positive-‐sequenceimpedances kZ1 and (1 – k)Z1, 0<k<1, is added to represent the fractional lengths of the broken line m - ‐ n from bus m to point p and bus n to point p’. Z1
  • 116.
    Open-‐Conductor Faults Simplify… Vm mm 1 Z Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 kZ1 1 kZ1 p p’   Va 1  Add the do- ‐ nothing source 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 116
  • 117.
    Open-‐Conductor Faults Simplify… Vm mm 1 Z Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 kZ1 1 kZ1 p p’   a V1 Combine these 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 117
  • 118.
    Open-‐Conductor Faults Simplify … Vm mm 1 Z Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   Z1 th,mn Z1 Z 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 118 1 p p’   a V1 Perform a source conversion
  • 119.
    Open-‐Conductor Faults Simplify … Vm mm 1 Z Zmn 1     m n 0 Referen ce   1 Z  Z mn nm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 p p ’ a V1 Z1 Open circuit Z1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 119
  • 120.
    Open-‐Conductor Faults Final result(positive-‐sequenceequivalent circuit): V mm Z  Zmn 1 1     m m n 0 Referen ce   1 Z  Z mn nm 1   Vn nn 1 Z  Znm 1   th,mn Z1 V1 a Z1 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 120
  • 121.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 121 Open-‐Conductor Faults The above considerations for the positive-‐sequencenetwork also apply directly to the negative-‐and zero-‐sequencenetworks, but we must remember that the latter networks do not contain any internal sources of their own.
  • 122.
    Open-‐Conductor Faults Negative-‐sequenceequivalent circuit: mm 2 Z Zmn 2    m n 0 Referen ce 2 Z  Z mn nm 2    Znn  Znm 2 2 Z2 th,mn Z2 mm 2 nn 2     Z  Z  2Zmn 2  kZ2 1 kZ2 p p ’  a V2  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 122 Z2
  • 123.
    Open-‐Conductor Faults Negative-‐sequenceequivalent circuitsimplified: mm Z  Zmn 2 2    m n 0 Referen ce 2 Z  Z mn nm 2    nn 2 Z  Znm 2    th,mn Z2 V2 a Z2 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 123
  • 124.
    Open-‐Conductor Faults Zero-‐sequenceequivalent circuit: mm 0 Z Zmn 0    m n 0 Referen ce 0 Z  Z mn nm 0    Znn  Znm 0 0 Z0 th,mn Z0 mm 0 nn 0     Z  Z  2Zmn 0  kZ0 1 kZ0 p p ’  a V0  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 124 Z0
  • 125.
    Open-‐Conductor Faults Zero-‐sequenceequivalent circuitsimplified: mm Z  Zmn 0 0    m n 0 Referen ce 0 Z  Z mn nm 0    nn 0 Z  Znm 0    th,mn Z0 V0 a Z0 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 125
  • 126.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 126 Open-‐Conductor Faults component of the voltage drops Vpp’a, Vpp’b, and Vpp’c from p to p' in the phase conductors. , take on different values depending on which one of the open-‐conductorfauIts is being considered. a Let the voltage V 1denote the p h a s e -‐apositive-‐sequence a We will soon see that V 1and the corresponding negative-‐ and zero-‐sequence components Va and Va 2 0
  • 127.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 127 Open-‐Conductor Faults In drawing the sequence equivalent circuits it is understood that the currents sources owe their origin to the open-‐conductor fault between points p and p' in the system. If there is no open conductor, the voltages: V0, V1, V2 a a a are all zero and the current sources disappear.
  • 128.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 128 Open-‐Conductor Faults Note that each of the sequence currents sources: can be regarded in turn as a pair of injections into buses m and n of the corresponding sequence network of the intact system. Hence, we can use the bus impedance sequence matrices of the normal configuration of the system to determine the voltage changes due to the open-‐conductor faults. Z0 Z1 V0 V1 V2 a , a , a Z2
  • 129.
    Open-‐Conductor Faults First wemust find expressions for the symmetrical components of Va (i.e., of the voltage drops across the fault points p and p’ for each type of fault, (one or two open lines). These voltage drops can be regarded as giving rise to the following sets of injection currents into the sequence networks of the normal system configuration: POSITIVE NEGATIVE ZERO SEQUENCE SEQUENCE SEQUEN CE At bus m: At bus n: V1 Z a 1 a V2 Z2 a V0 Z0 Va 1  Z1 Va 2  Z2 Va 0  Z0 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 129
  • 130.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 130 Open-‐Conductor Faults Recall Slide 87 from Network Calculations: 1   V j k   V N   V           V        Z1jI j  Z1k Ik jj j Z I  Z I jk k kj j Z I  Z I kk k I  Z I           Z             Nj j Nk k  
  • 131.
    Open-‐Conductor Faults The changesin the symmetrical components of the p h a se -‐ a voltage of each bus i is:: Zero - ‐ s e quence: Positiv e - - ‐ sequence: Negative- ‐ s e q u e n c e :  V i 0   Zim 0  Z 0 in    Z0 a V0  V i 1   Zim 1  Z 1 in   Z1 a V1  V i 2   Zim 2  Z 2 in    Z2 a 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 131 V2
  • 132.
    11/12/13 Unsymmetrical Faults(c) 2013 H. Zmuda 132 Open-‐Conductor Faults Before developing the equations for the sequence components of the voltage for each type of o p e n - ‐conductor fault, let us derive expressions for the Thévenin equivalent impedances of the sequence networks as seen from fault points p and p’.
  • 133.
    Open-‐Conductor Faults Looking intothe positive-‐sequencenetwork between p and p’, we see the impedance: mm 1 Z  Zmn 1   m n 0 Referen ce 1 Z  Z mn nm 1   Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp 11/12/13 Unsymmetrical Faults (c) 2013 H. Zmuda 133 Z1 Z 1 kZ  Z p p  1 th,mn 1  Z1  1 kZ1  Z1 Z Z1 th,mn Z1 1 th,mn 1  Z1 Z 2 th,mn Z1  Z1
  • 134.
    Open-‐Conductor Faults The open-‐circuitvoltagefrom p to p’ is: Thus : Vm mm 1 Z  Zmn 1     m n   mn 1 Z  Znm 1   Vn Znn  Znm 1 1 Z1 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  kZ1 1 kZ1 p p ’  a V1  pp Z1 pp 1  V  Z Vm Vn th,mn 1 Z 1  Z1 pp But: Z 1 Z1 2 th,mn Z1  Z1 V 0 11/11/13 Referen ce Unsymmetrical Faults (c) 2013 H. Zmuda 134 pp 1   p p  Z1 Z1 m n V  V 
  • 135.
    Open-‐Conductor Faults Before anyconductor opens, the current Imn in phase a of the line m - ‐ n is positive sequence and is given by: Thus : Where : Similarl y: and : pp 1 1   V  Z I p p  mn Imn  Vm  Vn Z1 Z 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 13 5 1 p p  1 Z 2 th,mn Z1  Z1 Z2 p p  Z 2 2 Z2  Z th,mn 2 Z0 p p  0 Z 2 Z0  Z th,mn 0 a Now we can find:V0, a V1, a V2
  • 136.
  • 137.
    Open-‐ConductorFaults – OneOpen Conductor Ia Ib Ic a b c m m m n n p 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 137 p ’ a I  0 I 0 a   I 1 a  I 2 a    0 n pp ,b V  0 Vpp,c 0
  • 138.
    Open-‐ConductorFaults – OneOpen Conductor The open conductor in phase a causes equal voltage drops to appear from p to p' in each of the sequence networks. We can satisfy this requirement by connecting the Thévenin equivalents of the sequence networks in parallel at the points p and p’. a a a    V0     V2   V1   3    1  1 1 1  1 a a2 1 a2 a   V pp ,a 0 0           V      pp ,a pp ,a  1  3    V    V  pp,a   V  Va 0  a 1 a 2    V  V  pp ,a 3 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 138
  • 139.
    Open-‐ConductorFaults – OneOpen Conductor From this circuit the expression for the positive-‐sequencecurrent is found as: Zpp 2 Zpp 0 a I2 a I0 p p   Va 0   Va 2  Imn pp Z1  pp Z1 a I1 p  Va  1 p ’ p ’ p ’ a 1 I  I  mn p p  Z 1 0  Z Z 2      Imn p p  Z1 Z1 pp p p  p p  p p  Z1 p p  p p  Z0Z2 pp pp Z0  Z2  I Zpp 1  Z 0  Z pp pp 2      pp pp pp pp pp pp mn Z0Z1 Z1Z2  Z0Z2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 139
  • 140.
    Open-‐ConductorFaults – OneOpen Conductor The sequence voltage drops are: Zpp 2 Zpp 0 a I2 a I0 p p  Va  0   Va 2  Imn pp Z1  pp Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 140 I1 p   Va 1 p ’ p ’ p ’ Va 0 a 1 a 2  V  V  I 1 a      p p  p p  Z0Z2 pp pp Z0  Z2  I p p  p p  p p  Z0Z1Z2 mn Z0Z1 Z1Z2  Z0Z2 p p  p p  p p  p p  p p  p p  These terms are known from the impedance parameters of the sequence networks and the prefault current in phase a of the line m - ‐ n .
  • 141.
    Open-‐ConductorFaults – OneOpen Conductor The currents: a , a , a V0 V1 V2 Z0 Z1 Z2 for injection into the sequence networks can now be determined. Zmm 0,1,2  Zmn 0,1,2     m n 0 Referen ce Zmn 0,1,2  Znm 0,1,2     Znn 0,1,2  Znm 0,1,2     th,mn Z0,1,2 V0,1,2 a Z0,1,2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 141
  • 142.
    Open-‐ConductorFaults – TwoOpen Conductors Clearly: Ia Ib Ic a b c m m m n n p 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 142 p ’ b I  0 Ic  0 pp ,a V Va 0 a 1 a 2     V V  0 n
  • 143.
    Open-‐ConductorFaults – TwoOpen Conductors Resolving the current into its symmetrical components: These can be satisfied by connecting the Thévenin equivalents of the sequence networks in series between the points p and p’. a a I1     Ia  I0   2    3   1  1 1 1  1 a a2 1 a2 a 0 0  Ia            a a 3   I    I   1    Ia      0 a    I  Along with: a 1 a 2     I  I  Ia 3 V V 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 143 0 pp ,a a a a 1 2     V V  0
  • 144.
    Open-‐ConductorFaults – TwoOpen Conductors pp Z1 pp Z2 pp Z0 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 144 I1 a I2 a I0 I 0 a  I 1 a  I 2 a     p p p  1 a  V0  2 Va p’  Va p’  p’   ImnZpp 1 
  • 145.
    Open-‐ConductorFaults – TwoOpen Conductors pp Z1 pp Z2 pp Z0 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 145 I1 a I2 a I0 I 0 a  I 1 a  I 2 a     p p p  1 a  V0   Va 2 Va p’  p ’ p’   Imn Zpp 1  I 0 a   I 1 a  I 2 a   I mn p p  Z1 pp pp pp Z0  Z1  Z2
  • 146.
    Open-‐ConductorFaults – TwoOpen Conductors Zpp 1 Z2 p p  pp Z0 a I1 Ia 2 a I0 p p p a  V1 a  V0  2 p’ Va p’  p’   11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 146 Imn pp Z1  Va 1 1    I Z  Z I 1 1 mn p p  p p  a    Imn pp I Z 1  Z1mn p p  Z1 pp pp pp p p  Z0  Z1  Z2  Imn Z1p p  p p  Z0  Z2 pp pp pp p p  Z0  Z1  Z2 Va 2   Z I pp a 2 2      Imn p p  p p  Z1Z2 pp pp pp Z0  Z1  Z2 Va 0   Z I pp a 0 0      Imn pp p p  Z0Z1 pp pp pp Z0  Z1  Z2
  • 147.
    11/11/13 Unsymmetrical Faults(c) 2013 H. Zmuda 147 Open-‐ConductorFaults – Two Open Conductors In each of these equations the right-‐hand side quantities are all known before the fault occurs. The net effect of the open conductors on the positive-- ‐sequence network is to increase the transfer impedance across the line in which the open-‐conductorfault occurs. For one open conductor this increase in impedance equals the parallel combination of the negative-‐and zero-‐sequence networks between points p and p’. For two open conductors the increase in impedance equals the series combination of the negative-‐andzero-‐sequence networks between points p and p'.
  • 148.
    Open-‐ConductorFaults – EXAMPLE Determine thechange in voltage at bus 3 when the transmission line undergoes a. a one-‐open-‐conductorfault and b. a two-‐open-‐conductorfault along its span between buses 2 and 3 . Choose a base of 100 MVA, 345 kV in the transmission line. For the system of Slide 39, consider that Machine 2 is a motor drawing a load equivalent to 50 MVA at 0.8 power-‐factorlagging and nominal system voltage of 345 kV at bus 3. 1 T1 2 3 T2 4 Machine 1 Machine 2 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 148
  • 149.
    Open-‐ConductorFaults – EXAMPLE Prefault current(per unit) in line 2 - ‐ 3 : Sbase 100 S  50  50  0.5 * 3 23 23 S  V I  I  P  jQ 3 V*  S P  jQ S 1.0 1.0 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 149  0.5 0.8  j0.6  0.4 j0.3
  • 150.
    Open-‐ConductorFaults – EXAMPLE Recal l: Also recall: From Slide 108: 0  Z    bus         j0.19 0 0 0 0 j0.08 j0.08 0 0 j0.08 j0.58 0 0 0 0 j0.19    , 1,2  Z  j0.1437 j0.1211 j0.0789 j0.0563 j0.1211 j0.1696 j0.1104 j0.0789 j0.0789 j0.1104 j0.1696 j0.1211 j0.0563   j0.0789 j0.1211   j0.1437      bus      Z1  Z2  j0.15, Z0  j0.5 Z1 p p  1 Z 2 th,mn Z1  Z1 Z0 p p  0 Z 2 Z0  Z th,mn 0 th,mn Z1 mm 1 nn 1    Z  Z  2Zmn 1  Z 1 Z2 p p  p p  Z1 2 22 1 33 1   1  Z  Z  2Z  Z 23 1  j0.152 j0.1696  j0.1696  2 j0.1104 j0.15  j0.712 pp Z 0 Z0 2 22 0 Z  33 0    23 0  Z  2Z  Z0  j0.52 j0.08  j0.58  2j0.08 j0.5   11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 150
  • 151.
    Open-‐ConductorFaults – EXAMPLE Thus, ifthe line from bus 2 to bus 3 is opened, then an infinite impedance is seen looking into the zero-‐sequencenetwork between points p and p' of the opening. The zero-‐sequencecircuit confirms this fact since bus 3 would be isolated from the reference by opening the connection between bus 2 and bus 3. Slide 47: j0.04 1 j0.04 2 4 j0.04 Referen ce j0.5 3 j0.08 j0.04 3Xn  j0.15 (1 ) (2 ) (3 ) (4 ) (5 ) (6 ) 3Xn  j0.15 Transform er Node Bus 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 151 Transform er Node Bus
  • 152.
    Open-‐ConductorFaults – EXAMPLE One openconductor: From Slide 118, From Slide 104: Va 0  Va 1 2    V  I a mn p p  p p  p p  Z0Z1Z2 Z Z pp pp pp pp 0 1 1 2  Z Z  Z Z pp pp 0 2          pp 0  Z   I p p  p p  Z1Z2 pp pp mn Z1 Z2  0.4  j0.3 j0.7122 2j0.712  0.1068  j1424  V i 0   Zim 0   Z 0 in  Z0 a V0,  V i 1   V i 2   im 1 Z  Z 1 in   Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 152 V1
  • 153.
    Open-‐ConductorFaults – EXAMPLE One open conductor:  V 3 0  32 Z  Z33 0 0    Z0 a V0,  V 3 1   V 3 2   32 1 Z  Z 1   Z1 a 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 153 33 V1  V 3 0   j0.08  j0.58 j0.5   0.1068 j1424  0.1068  j0.1424  V 3 1   V 3 2   j0.1104  j0.1696 j0.15   0.1068 j1424  0.0422 j0.0562 3 3 0  V  V   V3 3 1 2    V  0.1912  j0.24548 V3 new 3  V   3 V  1.0  0.1912  j0.24548  0.8088 j0.2548
  • 154.
    Open-‐ConductorFaults – EXAMPLE Two openconductors: Inserting the infinite impedance of the z e r o - ‐ sequence network in series between points p and p' of the positive-‐sequence network causes an open circuit in the latter . No power transfer can occur in the system, since power cannot be transferred by only one phase conductor of the transmission line in this case since the zero-‐sequencenetwork offers no return path for current. 1 2 3 4  V  f Vf   j0.20 j0.04 j0.20 j0.04 j0.04 j0.04 Referen ce (1 ) 11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 154 (2 ) (3) (4 ) Open (5 )