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Symmetrical Components & Fault Analysis in Power Systems | Engineering
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9 same original system sequence. The second set of phasors has an opposite sequence which is called the negative sequence. The zero sequence has three components in phase with each other. The symmetrical components theory will be discuss into more detail in chapter 3 of this project. 10 Chapter 3 THE MATHEMATICAL MODEL 3.1 Introduction This chapter describes the mathematical model that is used in the analysis of faulted power systems and the assumptions that are used in this project's analysis. 3.2 Fortescue’s Theory A three-phase balanced fault can be defined as a short circuit with fault impedance called Zf between the ground and each phase. The short circuit will be called a solid fault when Zf is equal to zero. This type of fault is considered the most sever short circuit which can affect any electrical system. Fortunately, it is rarely taking place in reality. Fortescue segregated asymmetrical three-phase voltages and currents into three sets of symmetrical components in 1918 [8]. Analyzing any symmetrical fault can be achieved using impedance matrix method or Thevenin’s method. Fortescue’s theorem suggests that any unbalanced fault can be solved into three independent symmetrical components which differ in the phase sequence. These components consist of a positive sequence, negative sequence and a zero sequence. 3.2.1 Positive Sequence Components The positive sequence components are equal in magnitude and displayed from each other by 120o with the same sequence as the original phases. The positive sequence currents and voltages follow the same cycle order of the original source. In the case of 11 typical counter clockwise rotation electrical system, the positive sequence phasor are shown in Fig 3.1. The same case applies for the positive current phasors. This sequence is also called the “abc” sequence and usually denoted by the symbol “+” or “1” [9]. Vc1 Positive Sequence Components Va1 Vb1 Figure 3.1 Positive sequence components 3.2.2 Negative Sequence Components This sequence has components that are also equal in magnitude and displayed from each other by 120o similar to the positive sequence components. However, it has an opposite phase sequence from the original system. The negative sequence is identified as the “acb” sequence and usually denoted by the symbol “-” or “2” [9].The phasors of this sequence are shown in Fig 3.2 where the phasors rotate anti- clockwise. This sequence occurs only in case of an unsymmetrical fault in addition to the positive sequence components, 12 Vb2 Negative Sequence Components Va2 Vc2 Figure 3.2: Negative sequence components 3.2.3 Zero Sequence Components In this sequence, its components consist of three phasors which are equal in magnitude as before but with a zero displacement. The phasor components are in phase with each other. This is illustrated in Fig 3.3. Under an asymmetrical fault condition, this sequence symbolizes the residual electricity in the system in terms of voltages and currents where a ground or a fourth wire exists. It happens when ground currents return to the power system through any grounding point in the electrical system. In this type of faults, the positive and the negative components are also present. This sequence is known by the symbol “0” [9]. Zero Sequence Components Vc0 Vb0 Va0 Figure 3.3 Zero sequence components 13 The following are three sets of components to represent three-phase system voltages as positive, negative and zero components: Positive Va1 Vb1 Vc1 Negative Va2 Vb2 Vc2 Zero Va0 Vb0 Vc0 The addition of all symmetrical components will present the original system phase components Va, Vb and Vc as seen below: Va Va 0 Va1 Va 2 Vb Vb 0 Vb1 Vb 2 (3.1) Vc Vc 0 Vc1 Vc 2 The “a” operator is defined below: a 10 (3.2) The following relations can be driven from 3.2: a 2 1 120 a3 10 From the above definition and using the “a” operator, it can be translated into a set of equations to represents each sequence: a) Zero sequence components: Va 0 Vb0 Vc 0 b) Positive sequence components: (3.3) 14 Vb1 a 2Va1 (3.4) Vc1 aVa1 c) Negative sequence components: Vb 2 aVa 2 (3.5) Vc 2 a 2Va 2 Now, the original system phasors Va, Vb and Vc can be expressed in terms of phase “a” components only. Equation 3.1 can be written as follows: Va Va 0 Va1 Va 2 Vb Va 0 a 2Va1 aVa 2 (3.6) Vc Va 0 aVa1 a Va 2 2 Writing the above equations can be accomplished in a matrix form: Va 1 1 V 1 a 2 b Vc 1 a 1 Va 0 a Va1 a 2 Va 2 (3.7) Defining A as: 1 1 A 1 a 2 1 a 1 a a 2 (3.8) Equation 3.7 can be written as: Va 0 Va V A V a1 b Va 2 Vc (3.9) This equation can be reversed in order to obtain the positive, negative and zero sequences from the system phasors: 15 Va 0 Va 1 Va1 A Vb Va 2 Vc (3.10) Where A1 is equal to the following: 1 1 1 A 1 a 3 1 a 2 1 1 a 2 a (3.11) These equations can be applied for the phase voltages and currents. In addition, it can express the line currents and the line-to-line voltages of any power system under fault conditions. 3.3 Sequence Impedances and Sequence Networks Knowing specific data of the sequence-impedances of synchronous machine, transmission lines and transformers is a need when a numerical analysis of a power system under fault conditions is required. Due to the extensive information about sequence impedances and sequence network theory a general introduction of the comprehensive concepts of the sequence impedances and sequence networks is here explained [3]. 3.3.1 Synchronous Machines The positive, negative and zero sequence current impedances in synchronous machines and other rotating machines have generally different values. For a synchronous machine the selection of its positive-sequence impedance depends on the time that is assume to elapse from the instant the fault initiates to the instant at which values are 16 desired (e.g., for relay response, breaker opening, or sustained fault conditions), therefore the positive-sequence impedance of a synchronous machine can be selected to be its subtransient ( X d'' ), transient ( X d' ), or synchronous ( X d ) reactance. However, it is the subtransnient reactance of the synchronous machine the one usually taken for fault analysis purposes studies [1]. The subtransient and negative-sequence reactances are the same in the case of a cylindrical-rotor synchronous machine. In general, the negative-sequence impedance of a synchronous machine is usually determined from X d'' X q'' Z 2 jX 2 j 2 (3.11) The determination of the zero-sequence impedance of a synchronous machine is a little more complicated since it varies widely and depends on the pitch of the armature coils. The zero-sequence impedance is much smaller than the corresponding positive, negative-sequence impedances. An easy way to measure the zero-sequence impedance is by connecting the three armature windings in series and applying a single-phase voltage [1]. 3.3.2 Transmission Line A transmission line is considered a passive device since no voltage or current sources are present in its equivalent model. It is also considered a bilateral device which means the line behavior remains the same way no matter the direction of the current. It is important to note that although a single transmission line is bilateral, an interconnected 17 transmission network is not this is due to the dispersion of active components (generators) throughout the network. Due to these characteristics the phase sequence of the applied voltage makes no difference, this means that the a-b-c (positive-sequence) voltages produce the same voltage drops as a-c-b (negative-sequence) voltages. Therefore, the impedances of a transmission line for its positive- and negative-sequence are the same, provided that the line is transposed. A transmission line is transposed when the phase conductors of the line physically exchange positions along the length of the line [7]. Figure 3.4 shows a representation of an untransposed transmission line with unequal self-impedances and unequal mutual impedances [1] a Ia I b b Zbb Va c Ic a’ Zaa Zcc Zab Zbc b’ Va' Zca c’ Vb Vb' Vc n Ia + Ib + Ic= In n’ Vc' Figure 3.4 Transmission line diagram with unequal impedances From Figure 3.4 it is observed Vabc Zabc I abc Where, (3.12) 18 Z aa Z abc Zba Z ca Z ab Z bb Z cb Z ac Z bc Z cc (3.13) In general the self impedances are Z aa Zbb Zcc (3.14) And the mutual impedances Z ab Zbc Zca (3.15) If both sides of Equation 3.12 are multiplying by A 1 and knowing that I abc A I012 (3.16) then, A Vabc A Zabc A I 012 (3.17) Z012 A Zabc A (3.18) 1 1 where, 1 Then sequence impedance matrix of an untransposed transmission line cam be obtained from Equation 3.18 and is expressed as Z 00 Z012 Z10 Z 20 Z 01 Z11 Z 21 Z 02 Z12 Z 22 (3.19) it can be also expressed as Z s 0 2 Z m 0 Z012 Z s1 Z m1 Z s 2 Z m 2 Z s 2 Z m 2 Z s1 Z m1 Z s 0 Z m0 Z s 2 2Z m 2 Z s1 2Z m1 Z s 0 Z m0 (3.20) 19 it is known by definition that, Z s 0 zero-sequence self impedance 1 Z aa Zbb Zcc , 3 (3.21) Z s1 positive-sequence self impedance 1 Z aa aZbb a 2 Z cc , 3 (3.22) Z s 2 negative-sequence self impedance 1 Z aa a 2 Z bb aZ cc , 3 (3.23) Z m 0 zero-sequence mutual impedance 1 Zbc Zca Z ab , 3 (3.24) Z m1 positive-sequence mutual impedance 1 Zbc aZ ca a 2 Z ab , 3 (3.25) Z m 2 negative-sequence mutual impedance 1 Zbc a 2 Z ca aZ ab , 3 (3.26) Therefore, V012 Z012 I012 (3.27) The resultant matrix from Equation 3.20 is not symmetrical this means that the result obtained from Equation 3.27 will show a non desirable result such as mutual coupling among the three sequence. There are two solutions to this problem, one is to completely transpose the line and the other one is to obtained equal mutual impedances by placing the conductors with 20 equilateral spacing among them. Figure 3.5 shows a representation of a completely transposed line with equal series and equal mutual impedances. [1] a Ia I b b a’ Zs Zm Zs Va c Ic Zm Zs b’ Va' Zm c’ Vb Vb' Vc n Ia + Ib + Ic= In n’ Vc' Figure 3.5 Transmission line diagram with equal series and equal mutual impedances For a transposed line such as the one represented in Figure 3.5 the mutual impedances are Zab Zbc Zca Z m (3.28) And the self-impedances, Zaa Zbb Zcc Z s (3.29) With Equations 3.28 and 3.29, Equation 3.13 can be expressed as Zs Z abc Z m Z m Zm Zs Z 21 Zm Z m Z s (3.30) Where the self and mutual impedance are given by D Z s ra re j 0.1213ln e l , Ds (3.31) 21 D Z m re j 0.1213ln e l, Deq (3.32) Where re is the resistance of Carson’s, Ds is the geometric mean radius (GMR) and De is a function of both the earth resistivity and the frequency. And, Deq Dm Dab Dbc Dca 3 , 1 (3.33) When Equation 3.18 is applied Z s 2 Z m Z012 0 0 0 Zs Zm 0 0 0 , Z s Z m (3.34) Equation 3.34 can also be expressed as Z0 Z 012 0 0 0 Z1 0 0 0 , Z 2 (3.35) As it can be seen from equations 3.34 and 3.35, the result is desirable since there is no mutual coupling among the three sequences. This means that each sequence network only produce voltage drop in their respective sequence network. [1] 3.3.3 Transformers Three identical single-phase transformers can be connected to make up a three- phase transformer. This is known as a three-phase transform bank. The positive and negative sequence currents in a transformer are the same. However, the zero-sequence series impedances of three-units are a little different than the positive-, and negative- 22 sequence series impedance, but in practice all the sequences impedances are assumed to be the same regardless the transformer type. Z0 Z1 Z 2 Ztrf (3.36) Z0 is infinite when the flow of zero-sequence current is prevented by the transformer connection. The zero-sequence networks diagrams of three phase transformer banks are shown in Figure 3.6. If the zero sequence current path is not indicated in the diagram this means the transformer connection prevents the flow of it. Some points are important to highlight, zero sequence current flows inside the delta windings for a delta-delta bank but prevents it to flow outside the windings by not providing a path for it. [1] Also, it is important to notice that there is no existing path that allows the flow of the zero-sequence current in a wye-grounded-wye-connected three-phase transformer bank. The reason is no zero-sequence current is present in any given winding on the wye side of the transformer bank since it has an ungrounded wye connection. [1] 23 Symbols Transfomer connection diagram Ia0 Zero-sequence network equivalent Ia0 P P Z0 S n S Ia0 3Ia0 3Ia0 Ia0 Ia0 Ia0 N0 Z0 S P P S n n N0 Z0 Ia0 S P n P S Ia0 3Ia0 Ia0 N0 Z0 S P P S N0 Z0 S P P S N0 Z0 S P P S n n N0 Figure 3.6 Zero-sequence network equivalents of three-phase banks The equivalent networks for the zero-sequence of three-phase transformer banks made of three indentical single-phase transformers with three windings are shown in figure 3.7. It is important to note that for a three-phase transformer bank made with three 24 identical single-phase transformers with three windings the wye-wye connection with delta tertiary is the only one that allows the flow of zero-sequence current from either wye line (as long as the are grounded). [1] Tranformer connection diagram P Zero-sequence network equivalent S ZP ZS P S ZT T N0 T P ZP S ZS P S ZT T N0 T P ZP S ZS P S ZT T N0 T ZP S P ZS P S ZT T N0 T Figure 3.7 Zero-sequence network equivalents of three windings transformer banks 25 3.4 Fault Analysis in Power Systems In general, a fault is any event, unbalanced situation or any asymmetrical situation that interferes with the normal current flow in a power system and forces voltages and currents to differ from each other. It is important to distinguish between series and shunt faults in order to make an accurate fault analysis of an asymmetrical three-phase system. When the fault is caused by an unbalance in the line impedance and does not involve a ground, or any type of inter-connection between phase conductors it is known as a series fault. On the other hand, when the fault occurs and there is an inter-connection between phase-conductors or between conductor(s) and ground and/or neutral it is known as a shunt fault. [3] Statistically, series faults do not occur as often as shunt faults does. Because of this fact only the shunt faults are explained here in detail since the emphasis in this project is on analysis of a power system under shunt faults. 3.4.1 Three-Phase Fault By definition a three-phase fault is a symmetrical fault. Even though it is the least frequent fault, it is the most dangerous. Some of the characteristics of a three-phase fault are a very large fault current and usually a voltage level equals to zero at the site where the fault takes place. [3] A general representation of a balanced three-phase fault is shown in Figure 3.8 where F is the fault point with impedances Zf and Zg . Figure 3.9 shows the sequences networks interconnection diagram. 26 F a b c Iaf Iaf Iaf Zf Zf Zg Zf Iaf +Ibf +Icf = 3Iaf N n Figure 3.8 General representation of a balanced three-phase fault Ia0 + Va0 - Zf +3Zg F0 Z0 N0 Ia1 + Va1 - Zf F1 + Z1 Va2 + o N1 1.0 0 - Ia2 Zf F2 Z2 N2 Figure 3.9 Sequence network diagram of a balanced three-phase fault From Figure 3.9 it can be noticed that the only one that has an internal voltage source is the positive-sequence network. Therefore, the corresponding currents for each of the sequences can be expressed as Ia0 0 Ia2 0 I a1 1.00 Z1 Z f (3.37) 27 If the fault impedance Zf is zero, I a1 1.00 Z1 ( 3.38) If equation is substituted into equation I af 1 1 2 I bf 1 a I cf 1 a 1 0 a I a1 a 2 0 (3.39) Solving Equation 3.39 I af I a1 1.00 , Z1 Z f I bf a 2 I a1 I cf aI a1 1.0240 , Z1 Z f (3.40) 1.0120 Z1 Z f Since the sequence networks are short-circuited over their own fault impedance Va 0 0 Va1 Z f I a1 (3.41) Va 2 0 If Equation is substituted into Equation Vaf 1 1 2 Vbf 1 a Vcf 1 a Therefore, 1 0 a Va1 a 2 0 (3.42) 28 Vaf Va1 Z f I a1 Vbf a 2Va1 Z f I a1240 (3.43) Vcf aVa1 Z f I a1120 The line-to-line voltages are Vab Vaf Vbf Va1 1 a 2 3Z f I a130 Vbc Vbf Vcf Va1 a 2 a 3Z f I a1 90 (3.44) Vca Vcf Vaf Va1 a 1 3Z f I a1150 If Zf equals to zero, I af 1.00 Z1 I bf 1.0240 , Z1 I cf 1.0120 Z1 (3.45) The phase voltages becomes, Vaf 0 Vbf 0 (3.46) Vcf 0 And the line voltages, Va 0 0 Va1 0 (3.47) Va 2 0 3.4.2 Single Line-to-Ground Fault The single line-to-ground fault is usually referred as “short circuit” fault and occurs when one conductor falls to ground or makes contact with the neutral wire. The 29 general representation of a single line-to-ground fault is shown in Figure 3.10 where F is the fault point with impedances Zf. Figure 3.11 shows the sequences network diagram. Phase a is usually assumed to be the faulted phase, this is for simplicity in the fault analysis calculations. [1] F a b c + Vaf Iaf Ibf = 0 Icf = 0 Zf - n Figure 3.10 General representation of a single line-to-ground fault. Ia0 F0 + Va0 Iaf Z0 N0 - Ia1 3Zf + Va1 - N1 F1 Z1 + 1.0 - Ia2 F2 + Va2 - Z2 N2 Figure 3.11 Sequence network diagram of a single line-to-ground fault. 30 Since the zero-, positive-, and negative-sequence currents are equals as it can be observed in Figure 3.11. Therefore, I a 0 I a1 I a 2 1.00 Z 0 Z1 Z 2 3Z f (3.48) I af 1 1 2 I bf 1 a I cf 1 a 1 Ia0 a I a1 a 2 I a 2 (3.49) Since Solving Equation the fault current for phase a is I af I a 0 I a1 I a 2 (3.50) it can also be I af 3I a 0 3I a1 3I a 2 (3.51) From Figure 3.10 it can be observed that, Vaf Z f I af (3.52) The voltage at faulted phase a can be obtained by substituting Equation 3.49 into Equation 3.52. Therefore, Vaf 3Z f I a1 (3.53) Vaf Va 0 Va1 Va 2 (3.54) but, therefore, Va 0 Va1 Va 2 3Z f I a1 (3.55) 31 With the results obtained for sequence currents, the sequence voltages can be obtained from Va 0 0 1 1 V 1.00 1 a 2 b1 Vc 2 0 1 a 1 Ia0 a I a1 a 2 I a 2 (3.56) By solving Equation Va 0 Z 0 I a 0 Va1 1.0 Z1 I a1 (3.57) Va 2 Z 2 I a 2 If the single line-to-ground fault occurs on phase b or c, the voltages can be found by the relation that exists to the known phase voltage components, Vaf 1 1 2 Vbf 1 a Vcf 1 a 1 Va 0 a Va1 a 2 Va 2 (3.58) as Vbf Va 0 a 2Va1 aVa 2 Vcf Va 0 aVa1 a 2Va 2 3.4.3 (3.59) Line-to-Line Fault A line-to-line fault may take place either on an overhead and/or underground transmission system and occurs when two conductors are short-circuited. One of the characteristic of this type of fault is that its fault impedance magnitude could vary over a wide range making very hard to predict its upper and lower limits. It is when the fault impedance is zero that the highest asymmetry at the line-to-line fault occurs [3]. 32 The general representation of a line-to-line fault is shown in Figure 3.12 where F is the fault point with impedances Zf. Figure 3.13 shows the sequences network diagram. Phase b and c are usually assumed to be the faulted phases; this is for simplicity in the fault analysis calculations [1], F a b c Iaf = 0 Ibf Icf = -Ibf Zf Figure 3.12 Sequence network diagram of a line-to-line fault. Zf Ia0 = 0 + Va0 = 0 - Ia1 F0 Z0 N0 + Va1 - N1 Ia2 F1 + Z1 Va2 + 1.0 0o - F2 Z2 N2 Figure 3.13 Sequence network diagram of a single line-to-ground fault. From Figure 3.13 it can be noticed that I af 0 Ibf I cf Vbc Z f I bf (3.60) 33 And the sequence currents can be obtained as Ia0 0 (3.61) 1.00 Z1 Z 2 Z f (3.62) 1.00 Z1 Z 2 (3.63) I a1 I a 2 If Zf = 0, I a1 I a 2 The fault currents for phase b and c can be obtained by substituting Equations 3.61 and 3.62 into Equation 3.49 Ibf I cf 3 I a1 90 (3.64) The sequence voltages can be found similarly by substituting Equations 3.61 and 3,62 into Equation 3.56 Va 0 0 Va1 1.0 - Z1 I a1 (3.65) Va 2 Z 2 I a 2 Z 2 I a1 Also substituting Equation 3.65 into Equation 3.58 Vaf Va1 Va 2 1.0 Ia1( Z 2 - Z 1) Vbf a 2Va1 aVa 2 a 2 Ia1(aZ 2 - a 2 Z 1) Vcf aVa1 a 2Va 2 a Ia1(a 2 Z 2 - aZ 1) (3.66) Finally, the line-to-line voltages for a line-to-line fault can be expressed as Vab Vaf Vbf Vbc Vbf Vcf Vca Vcf Vaf (3.67) 34 3.4.4 Double Line-to-Ground Fault A double line-to-ground fault represents a serious event that causes a significant asymmetry in a three-phase symmetrical system and it may spread into a three-phase fault when not clear in appropriate time. The major problem when analyzing this type of fault is the assumption of the fault impedance Zf , and the value of the impedance towards the ground Zg. [3] The general representation of a double line-to-ground fault is shown in Figure 3.14 where F is the fault point with impedances Zf and the impedance from line to ground Zg . Figure 3.15 shows the sequences network diagram. Phase b and c are assumed to be the faulted phases, this is for simplicity in the fault analysis calculations. [1] a b c F Iaf = 0 Ibf Zf Icf Zf Zg n Ibf +Icf N Figure 3.14 General representation of a double line-to-ground fault. 35 Ia0 + Va0 - Zf +3Zg F0 Z0 N0 Ia1 Zf N1 F1 + Z1 Va2 + 1.0 0o - + Va1 - Ia2 - Zf F2 Z2 N2 Figure 3.15 Sequence network diagram of a double line-to-ground fault. From Figure 3.15 it can be observed that I af 0 Vbf ( Zf Zg ) Ibf ZgIcf Vcf ( Zf Zg ) Icf ZgIbf (3.68) Based on Figure 3.15, the positive-sequence currents can be found as Ia1 1.00 ( Z 2 Zf )( Z 0 Zf 3Zg ) ( Z 1 Zf ) ( Z 2 Zf ) ( Z 0 Zf 3Zg ) Ia 2 [ ( Z 0 Zf 3Zg ) ]Ia1 ( Z 2 Zf ) ( Z 0 Zf 3Zg ) Ia 0 [ ( Z 2 Zf ) ]Ia1 ( Z 2 Zf ) ( Z 0 Zf 3Zg ) (3.69) An alternative method is, Iaf 0 Ia 0 Ia1 Ia 2 Ia 0 ( Ia1 Ia 2) (3.70) If Zf and Zg are both equal to zero, then the positive-, negative-, and zerosequences can be obtained from 36 Ia1 1.00 ( Z 2)( Z 0) ( Z 1) ( Z 2 Z 0) Ia 2 [ ( Z 0) ]Ia1 ( Z 2 Z 0) Ia 0 [ ( Z 2) ]Ia1 ( Z 2 Z 0) (3.71) From Figure 3.14 the current for phase a is I af 0 (3.72) Now, substituting Equations 3.71 into Equation 3.49 to obtain phase b and c fault currents I bf I a 0 a 2 I a1 aI a 2 I cf I a 0 aI a1 a 2 I a 2 (3.73) The total fault current flowing into the neutral is I n 3I a 0 Ibf I cf (3.74) And the sequences voltages can be obtained by using Equation 3.51 V 0 a Z 0 Ia 0 Va1 1.0 Z 1Ia1 Va 2 Z 2 Ia 2 (3.75) The phase voltages are equal to Vaf Va 0 Va1 Va 2 Vbf Va 0 a 2Va1 aVa 2 Vcf Va 0 aVa1 a Va 2 2 The line-to-line voltages can be obtained from (3.76) 37 Vab Vaf Vbf Vbc Vbf Vcf Vca Vcf Vaf (3.77) If Zf = 0 and Zg = 0 then the sequence voltages become, and the positive-sequence current is found by using Equation 3.71. Va 0 Va1 Va 2 1.0 Z 1Ia1 (3.78) Now the negative- and zero-sequence currents can be obtained from Va 2 Z2 Va 0 Ia 0 Z0 Ia 2 (3.79) The resultant phase voltages from the relationship given in Equation 3.78 can be expressed as Vaf Va 0 Va1 Va 2 3Va1 Vbf Vcf 0 (3.80) And the line-to-line voltages are Vabf Vaf Vbf Vaf Vbcf Vbf Vcf 0 Vcaf Vcf Vaf Vaf (3.81) 38 Chapter 4 APPLICATION OF MATHEMATICAL MODEL In order to do the analysis a 6-node network well-known circuit introduced by Ward and Hale is used in the following section [2]. The mathematical model previously explained in chapter 3 is applied by doing hand calculations as well as a Matlab code to confirm the results obtained by hand. Figure 4.1 shows the 6-network system. The parameters of the system are given in Appendix A and were taken from Faulted Power System by Paul M. Anderson. 4 1 3 Y 1 2 9 6 7 8 3 6 2 5 2 Y 10 5 Figure 4.1 Ward and Hale 6-node network [2]. Y 39 4.1 Hand Calculations Hand calculation was done based on the mathematical model and Equations explained in Chapter 3 and shows the computations for total fault current, buses voltages, fault currents for each line, sequence currents and phase currents for each shunt fault at bus 4. 4.1.1 Three-Phase Fault The total fault current at bus 4 for a three-phase fault is given by Ik (F ) Vk (0) (Z Z f ) 1 kk (4.1) where k is the number of the bus where the fault takes place, I4 (F ) V4 (0) (Z Z f ) 1 44 10 [(0.13269 j 0.57694) 0] (4.2) 1.689183 77.04782 Voltages at all buses Vi ( F ) Vi (0) Zik I k Vi (0) Zi 4 I 4 (4.3) For buses 1, 3, 4 and 6, then i = 1,3,4,6 their voltages respectively are V1 ( F ) V1 (0) Z14 I 4 1.050 (0.02254 j 0.17266)(1.689183 77.04782 ) 0.75776 2.137 (4.4) 40 V3 ( F ) V3 (0) Z34 I 4 10 (0.14333 j 0.53368)(1.689183 77.04782 ) (4.5) 0.0752526.769 V4 ( F ) V4 (0) Z 44 I 4 10 (0.13269 j 0.57694)(1.689183 77.04782 ) (4.6) 00 V6 ( F ) V6 (0) Z 64 I 4 10 (0.06881 j 0.34726)(1.689183 77.04782 ) (4.7) 0.40269 2.59 the line fault currents are given by Il 4 ( F ) Vl ( F ) V4 ( F ) V (F ) 0 Vl ( F ) l ( Zl 4 )Actual ( Zl 4 ) Actual ( Zl 4 ) Actual (4.8) For line between bus 1 and bus 4 and applying Equation (4.8) with l = 1 I14 ( F ) V1 ( F ) ( Z14 ) Actual (4.9) 0.75776 2.137 1 79.936 (0.16 j 0.74) For line between bus 3 and bus 4 and applying Equation (4.8) with l = 3 I 34 ( F ) V3 ( F ) ( Z 34 ) Actual (4.10) 0.0752526.769 0.2894 63.231 (0 j 0.26) For line between bus 6 and bus 4 and applying Equation (4.8) l = 4 41 I 64 ( F ) V6 ( F ) ( Z 64 ) Actual (4.11) 4.1.2 0.40269 2.59 0.4812 79.1848 (0.194 j 0.814) Single Line-to-Ground Fault The voltage at bus 4 before the fault is V4 (0) 10 (4.12) The sequence currents at bus 4 for a single line-to-ground fault are given by I 40 I 41 I 42 V4 (0) ( Z Z Z 442 3Z f ) 0 44 1 44 10 (0.00756 j 0.24138) 2(0.13269 j 0.57694) 0 (4.13) 0.70338 78.931 From Equation 3.49 1 1 abc 012 I 4 ( F ) AI 4 1 a 2 1 a 1 1 1 1240 1 1120 1 I 40 a I 41 a 2 I 42 1 0.70338 78.931 1120 0.70338 78.931 1240 0.70338 78.931 2.1101 78.931 0 0 (4.14) 42 The phase currents are I 4a ( F ) 2.1101 78.931 I 4b ( F ) I 4c ( F ) 0 (4.15) The total fault current at bus 4 is I 4 ( F ) 3I 40 3(0.70338 78.931 ) 2.1101 78.931 (4.16) From Equation 3.57 the sequence voltages are Vi 0 ( F ) 0 Z i04 I 40 Vi1 ( F ) Vi (0) Z i14 I 41 (4.17) Vi 2 ( F ) 0 Z i24 I 42 For Bus 1 then i = 1 V10 ( F ) 0 Z140 I 40 0 (0.0112 j 0.11474)(0.70338 78.931 ) (4.18) 0.08108 163.355 V11 ( F ) V1 (0) Z141 I 41 1.050 (0.02254 j 0.17266)(0.70338 78.931 ) 0.9278 0.479 (4.19) V12 ( F ) 0 Z142 I 42 0 (0.02254 j 0.17266)(0.70338 78.931 ) (4.20) 0.1224 176.368 For Bus 3 then i = 3 V30 ( F ) 0 Z 340 I 40 0 (0)(0.70338 78.931 ) 00 (4.21) 43 1 1 V31 ( F ) V3 (0) Z 34 I4 10 (0.14333 j 0.53368)(0.70338 78.931 ) 0.61282.513 (4.22) V32 ( F ) 0 Z342 I 42 0 (0.14333 j 0.53368)(0.70338 78.931 ) (4.23) 0.3889176.035 For Bus 4 then i = 4 0 0 V40 ( F ) 0 Z 44 I4 0 (0.00756 j 0.24138)(0.70338 78.931 ) (4.24) 0.1698 170.72 1 1 V41 ( F ) V4 (0) Z 44 I4 10 (0.13269 j 0.57694)(0.70338 78.931 ) 0.58391.3426 (4.25) V42 ( F ) 0 Z 442 I 42 0 (0.13269 j 0.57694)(0.70338 78.931 ) (4.26) 0.416178.11 For Bus 6 then i = 6 V60 ( F ) 0 Z 640 I 40 0 (0.01706 j 0.05554)(0.70338 78.931 ) 0.0408 151.85 (4.27) 1 1 V61 ( F ) V6 (0) Z 64 I4 10 (0.06881 j 0.34726)(0.70338 78.931 ) 0.75090.0461 (4.28) V62 ( F ) 0 Z 642 I 42 0 (0.06881 j 0.34726)(0.70338 78.931 ) 0.249179.86 (4.29) 44 the line fault currents are given by I l04 ( F ) Vl 0 ( F ) V40 ( F ) Vl 0 ( F ) 0.1698 170.72 ( Zl04 )line ( Zl04 )line (4.30) I l14 ( F ) Vl1 ( F ) V41 ( F ) Vl1 ( F ) 0.58391.3426 ( Zl14 )line ( Zl14 )line (4.31) Vl 2 ( F ) V42 ( F ) Vl 2 ( F ) 0.416178.11 ( Zl24 )line ( Zl24 )line (4.32) I 2 l 4 (F ) For line between bus 1 and bus 4 and applying Equations 4.30 through 4.32, with l = 1 I140 ( F ) V10 ( F ) V40 ( F ) ( Z140 )line 0.08108 163.355 0.1698 170.72 (0.8 j1.85) (4.33) 0.0446 63.96 I141 ( F ) V11 ( F ) V41 ( F ) ( Z141 )line 0.9278 0.479 0.58391.3426 (0.16 j 0.74) (4.34) 0.455 81.365 I142 ( F ) V12 ( F ) V42 ( F ) ( Z142 )line 0.1224 176.368 0.416178.11 (0.16 j 0.74) 0.3888 81.98 From Equation 3.49 (4.35) 45 1 1 abc 012 I14 ( F ) AI14 ( F ) 1 a 2 1 a 1 1 1 1240 1 1120 1 I140 ( F ) a I141 ( F ) a 2 I142 ( F ) (4.36) 1 0.0446 63.96 1120 0.455 81.365 1240 0.3888 81.98 0.0446 63.96 0.455 81.365 0.3888 81.98 The phase currents for line between Bus 1 and 4 are I14a ( F ) 0.0446 63.96 I14b ( F ) 0.455 81.365 c 14 ( F ) I (4.37) 0.388 81.98 For line between Bus 3 and Bus 4 and applying Equations 4.29 through 4.31 with l = 3 I 340 ( F ) V30 ( F ) V40 ( F ) ( Z 340 )line 00 0.1698 170.72 00 (0) 1 I 34 (F ) V31 ( F ) V41 ( F ) 1 ( Z 34 )line 0.61282.513 0.58391.3426 0.1179 65.15 ( j 0.266) I 342 ( F ) V32 ( F ) V42 ( F ) ( Z342 )line (4.38) (4.39) (4.40) 46 0.3889176.035 0.416178.11 0.1156 64.65 ( j 0.266) From Equation 3.49 I 34abc ( F ) AI 34012 ( F ) 1 1 1 a 2 1 a 1 I 340 ( F ) 1 a I 34 (F ) 2 2 a I 34 ( F ) 1 1 1 1240 1 1120 (4.41) 1 00 1120 0.1179 65.15 1240 0.1156 64.65 The sequence currents for line between Bus 3 and 4 are I 34a ( F ) 0.2335 63.90 I 34b ( F ) 0.1176116.07 I c 34 ( F ) (4.42) 0.1159114.11 For line between Bus 6 and Bus 4 and applying Equations 4.30 through 4.32 with l = 6 I 0 64 ( F ) V60 ( F ) V40 ( F ) ( Z 640 )line 0.0408 151.85 0.1698 170.72 (0.9 j 2.06) (4.43) 0.0586 62.86 1 I 64 (F ) V61 ( F ) V41 ( F ) 1 ( Z 64 )line 0.75090.0461 0.58391.3426 (0.194 j 0.814) 0.2003 81.068 (4.44) 47 I 642 ( F ) V62 ( F ) V42 ( F ) ( Z 642 )line 0.249179.86 0.416178.11 (0.194 j 0.814) (4.45) 0.1999 81.09 From Equation 3.49 I 34abc ( F ) AI 34012 ( F ) 1 1 1 a 2 1 a 1 I 340 ( F ) 1 a I 34 (F ) 2 2 a I 34 ( F ) 1 1 1 1240 1 1120 (4.46) 1 0.0586 62.86 1120 0.2003 81.068 1240 0.1999 81.09 The sequence currents for line between Bus 6 and 4 are I 64a ( F ) 0.4562 78.78 b I 64 ( F ) 0.145591.82 (4.47) I 64c ( F ) 0.145791.56 4.1.3 Line-to-Line Fault The voltage at bus 4 before the fault is V4 (0) 10 (4.48) The sequence currents at bus 4 for a single line-to-ground fault are given by I 40 0 I 41 (4.49) V4 (0) ( Z Z 442 Z f ) 1 44 (4.50) 48 10 0.8446 77.048 2*(0.13269 j 0.57694) 0 I 42 I 41 0.8446102.95 (4.51) The phase currents are I 4abc ( F ) AI 4012 1 1 1 a 2 1 a (4.52) 1 I 40 a I 41 a 2 I 42 1 1 1 1240 1 1120 1 0 1120 0.8446 77.048 1240 0.8446102.95 0 1.4628 167.05 1.462812.95 The total fault current at Bus 4 is I 4b ( F ) I 4c ( F ) j 3I 41 j 3(0.8446 77.048 ) (4.53) 1.4628 167.05 I 4c ( F ) I 4b ( F ) 1.462812.952 (4.54) The sequence voltages are given by Vi 0 ( F ) 0 Zi04 I 40 Vi1 ( F ) Vi (0) Zi14 I 41 Vi ( F ) 0 Z I 2 For Bus 1 then i = 1 2 2 i4 4 (4.55) 49 V10 ( F ) 0 Z140 I 40 0 (0.0112 j 0.11474)(00 ) (4.56) 00 V11 ( F ) V1 (0) Z141 I 41 1.050 (0.02254 j 0.17266)(0.8446 77.048 ) (4.57) 0.9037 0.895 V12 ( F ) 0 Z142 I 42 0 (0.02254 j 0.17266)(0.8446102.95 ) (4.58) 0.1475.5123 For Bus 3 then i = 3 V30 ( F ) 0 Z 340 I 40 0 (0)(00 ) (4.59) 00 1 1 V31 ( F ) V3 (0) Z 34 I4 10 (0.14333 j 0.53368)(0.8446 77.048 ) 0.5331.82 (4.60) V32 ( F ) 0 Z342 I 42 0 (0.14333 j 0.53368)(0.8446102.95 ) (4.61) 0.4667 2.083 For Bus 4 then i = 4 0 0 V40 ( F ) 0 Z 44 I4 0 (0.00756 j 0.24138)(00 ) (4.62) 00 1 1 V41 ( F ) V4 (0) Z 44 I4 10 (0.13269 j 0.57694)(0.8446 77.048 ) 0.4990.002 (4.63)
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