SECTION 7: FAULT ANALYSIS ESE 470 Energy Distribution Systems
2 Introduction
Power System Faults 3 Faults in three-phase power systems are short circuits Line-to-ground Line-to-line Result in the flow of excessive current Damage to equipment Heat burning/melting Structural damage due to large magnetic forces Bolted short circuits True short circuits i.e., zero impedance In general, fault impedance may be non-zero Faults may be opens as well We ll focus on short circuits
Types of Faults 4 Type of faults from most to least common: Single line-to-ground faults Line-to-line faults Double line-to-ground faults Balanced three-phase (symmetrical) faults We ll look first at the least common type of fault the symmetrical fault due to its simplicity
5 Symmetrical Faults
Symmetrical Faults 6 Faults occur nearly instantaneously Lightening, tree fall, arcing over insulation, etc. Step change from steady-state behavior Like throwing a switch to create the fault at tt = 0 Equivalent circuit model: R: generator resistance L: generator inductance ii tt = 0 for tt < 0 vv tt = 2VV GG sin ωωωω + αα Source phase, αα, determines voltage at tt = 0 Short circuit can occur at any point in a 60 Hz cycle
Symmetrical Faults 7 The governing differential equation for tt > 0 is dddd dddd + ii tt RR LL = 2VV GG LL The solution gives the fault current ii tt = 2VV GG ZZ sin ωωωω + αα where ZZ = RR 2 + ωωωω 2 1 ωωωω and θθ = tan RR sin ωωωω + αα θθ sin αα θθ ee ttrr LL This total fault current is referred to as the asymmetrical fault current It has a steady-state component ii aaaa tt = 2VV GG ZZ And a transient component sin ωωωω + αα θθ ii dddd tt = 2VV GG ZZ sin αα θθ ee tt RR LL
Symmetrical Faults 8 Magnitude of the transient fault current, ii dddd, depends on αα ii dddd 0 = 0 for αα = θθ ii dddd 0 = 2II aaaa for αα = θθ 90 II aaaa = VV GG /ZZ is the rms value of the steady-state fault current Worst-case fault current occurs for αα = θθ 90 ii tt = 2VV GG sin ωωωω ππ ZZ 2 + ee ttrr LL
Symmetrical Faults 9 Important points here: Total fault current has both steady-state and transient components asymmetrical Magnitude of the asymmetry (transient component) depends on the phase of the generator voltage at the time of the fault
Generator Reactance 10 The reactance of the generator was assumed constant in the previous example Physical characteristics of real generators result in a time-varying reactance following a fault Time-dependence modeled with three reactance values XX dd : subtransient reactance XX dd : transient reactance XX dd : synchronous reactance Reactance increases with time, such that XX dd < XX dd < XX dd
Sub-Transient Fault Current 11 Transition rates between reactance values are dictated by two time constants: ττ dd : short-circuit subtransient time constant ττ dd : short-circuit transient time constant Neglecting generator resistance and assuming θθ = ππ, the synchronous 2 portion of the fault current is ii aaaa tt = 2VV GG 1 XX dd 1 XX dd ee tt ττ 1 dd + XX 1 dd XX dd ee tt ττ 1 dd + sin ωωωω + αα ππ XX dd 2 At the instant of the fault, tt = 0, the rms synchronous fault current is II FF = VV GG XX dd This is the rms subtransient fault current, II FF This will be our primary metric for assessing fault current
12 Symmetrical Three-Phase Short Circuits
Symmetrical 3-φφ Short Circuits 13 Next, we ll calculate the subtransient fault current resulting from a balanced three-phase fault We ll make the following simplifying assumptions: Transformers modeled with leakage reactance only Neglect winding resistance and shunt admittances Neglect Δ-YY phase shifts Transmission lines modeled with series reactance only Synchronous machines modeled as constant voltage sources in series with subtransient reactances Generators and motors Induction motors are neglected or modeled as synchronous motors Non-rotating loads are neglected
Symmetrical 3-φφ Short Circuits 14 We ll apply superposition to determine three-phase subtransient fault current Consider the following power system: Assume there is a balanced three-phase short of bus 1 to ground at tt = 0
Symmetrical 3-φφ Short Circuits 15 The instant of the fault can be modeled by the switch closing in the following line-to-neutral schematic The short circuit (closed switch) can be represented by two back-to-back voltage sources, each equal to VV FF
Symmetrical 3-φφ Short Circuits 16 Applying superposition, we can represent this circuit as the sum of two separate circuits: Circuit 1 Circuit 2
Symmetrical 3-φφ Short Circuits 17 Assume that the value of the fault-location source, VV FF, is the pre-fault voltage at that location Circuit 1, then, represents the pre-fault circuit, so II FF1 = 0 The VV FF source can therefore be removed from circuit 1 Circuit 1 Circuit 2
Symmetrical 3-φφ Short Circuits 18 The current in circuit 1, II LL, is the pre-fault line current Superposition gives the fault current II FF = II FFF + II FFF = II FFF The generator fault current is II GG = II GGG + II GGG II GG = II LL + II GGG The motor fault current is II MM = II MMM + II MMM II MM = II LL + II MMM
Symmetrical 3-φφ Fault Example 19 For the simple power system above: Generator is supplying rated power Generator voltage is 5% above rated voltage Generator power factor is 0.95 lagging A bolted three-phase fault occurs at bus 1 Determine: Subtransient fault current Subtransient generator current Subtransient motor current
Symmetrical 3-φφ Fault Example 20 First convert to per-unit Use SS bb = 100 MMMMMM Base voltage in the transmission line zone is VV bb,tttt = 138 kkkk Base impedance in the transmission line zone is ZZ bb,tttt = VV 2 bb,tttt = SS bb 138 kkkk 2 100 MMMMMM = 190.4 Ω The per-unit transmission line reactance is XX tttt = 20 Ω 190.4 Ω = 0.105 pp. uu.
Symmetrical 3-φφ Fault Example 21 The two per-unit circuits are These can be simplified by combining impedances
Symmetrical 3-φφ Fault Example 22 Using circuit 2, we can calculate the subtransient fault current II FF = 1.05 0 jjj.116 = 9.079 90 pp. uu. To convert to ka, first determine the current base in the generator zone II bb,gg = SS bb 3VV bb,gg = The subtransient fault current is 100 MMMMMM 3 13.8 kkkk II FF = 9.079 90 4.18 kkkk II FF = 37.98 90 kkkk = 4.18 kkkk
Symmetrical 3-φφ Fault Example 23 The pre-fault line current can be calculated from the pre-fault generator voltage and power II LL = SS GG 3VV GG = 100 MMMMMM 3 1.05 13.8 kkkk cos 1 0.95 Or, in per-unit: II LL = 3.98 18.19 kkkk II LL = 3.98 18.19 kkkk 4.18 ka = 0.952 18.19 pp. uu. This will be used to find the generator and motor fault currents
Symmetrical 3-φφ Fault Example 24 The generator s contribution to the fault current is found by applying current division 0.505 II GGG = II FF 0.505 + 0.15 = 7.0 90 pp. uu. Adding the pre-fault line current, we have the subtransient generator fault current Converting to ka II GG = II LL + II GGG II GG = 0.952 18.19 + 7.0 90 II GG = 7.35 82.9 pp. uu. II GG = 7.35 82.9 4.18 kkkk II GG = 30.74 82.9 kkkk
Symmetrical 3-φφ Fault Example 25 Similarly, for the motor 0.15 II MMM = II FF 0.505 + 0.15 = 2.08 90 pp. uu. Subtracting the pre-fault line current gives the subtransient motor fault current Converting to kkkk II MM = II LL + II MMM II MM = 0.952 18.19 + 2.08 90 II MM = 2.0 116.9 II MM = 2.0 116.9 4.18 kkkk II MM = 8.36 116.9 kkkk
26 Symmetrical Components
Symmetrical Components 27 In the previous section, we saw how to calculate subtransient fault current for balanced three-phase faults Unsymmetrical faults are much more common Analysis is more complicated We ll now learn a tool that will simplify the analysis of unsymmetrical faults The method of symmetrical components
Symmetrical Components 28 The method of symmetrical components: Represent an asymmetrical set of NN phasors as a sum of NN sets of symmetrical component phasors These NN sets of phasors are called sequence components Analogous to: Decomposition of electrical signals into differential and common-mode components Decomposition of forces into orthogonal components For a three-phase system (NN = 3), sequence components are: Zero sequence components Positive sequence components Negative sequence components
Sequence Components 29 Zero sequence components Three phasors with equal magnitude and equal phase VV aaa, VV bbb, VV ccc Positive sequence components Three phasors with equal magnitude and ± 120, positive-sequence phase VV aa1, VV bb1, VV cc1 Negative sequence components Three phasors with equal magnitude and ± 120, negative-sequence phase VV aa2, VV bb2, VV cc2
Sequence Components 30 Note that the absolute phase and the magnitudes of the sequence components is not specified Magnitude and phase define a unique set of sequence components Any set of phasors balanced or unbalanced can be represented as a sum of sequence components VV aa VV bb VV cc = VV aa0 VV bb0 VV cc0 + VV aa1 VV bb1 VV cc1 + VV aa2 VV bb2 VV cc2 (1)
Sequence Components 31 The phasors of each sequence component have a fixed phase relationship If we know one, we know the other two Assume we know phase aa use that as the reference For the zero sequence components, we have VV 0 = VV aaa = VV bbb = VV ccc (2) For the positive sequence components, VV 1 = VV aaa = 1 120 VV bbb = 1 240 VV ccc (3) And, for the negative sequence components, VV 2 = VV aaa = 1 240 VV bbb = 1 120 VV ccc (4) Note that we re using phase aa as our reference, so VV 0 = VV aaa, VV 1 = VV aaa, VV 2 = VV aaa
Sequence Components 32 Next, we define a complex number, aa, that has unit magnitude and phase of 120 aa = 1 120 (5) Multiplication by aa results in a rotation (a phase shift) of 120 Multiplication by aa 2 yields a rotation of 240 = 120 Using (5) to rewrite (3) and (4) VV 1 = VV aaa = aavv bbb = aa 2 VV ccc (6) VV 2 = VV aaa = aa 2 VV bbb = aavv ccc (7)
Sequence Components 33 Using (2), (6), and (7), we can rewrite (1) in a simplified form VV aa VV bb VV cc = 1 1 1 1 aa 2 aa 1 aa aa 2 VV 0 VV 1 VV 2 (8) The vector on the left is the vector of phase voltages, VV pp The vector on the right is the vector of (phase aa) sequence components, VV ss We ll call the 3 3 transformation matrix AA We can rewrite (8) as VV pp = AAVV ss (9)
Sequence Components 34 We can express the sequence voltages as a function of the phase voltages by inverting the transformation matrix where So VV ss = AA 1 VV pp (10) AA 1 = 1 3 VV 0 VV 1 VV 2 = 1 3 1 1 1 1 aa aa 2 1 aa 2 aa 1 1 1 1 aa aa 2 1 aa 2 aa (11) VV aa VV bb VV cc (12)
Sequence Components 35 The same relationships hold for three-phase currents The phase currents are II pp = II aa IIbb IIcc And, the sequence currents are II ss = II 0 II 1 II 2
Sequence Components 36 The transformation matrix, AA, relates the phase currents to the sequence currents II pp = AAII ss And vice versa II aa IIbb IIcc = 1 1 1 1 aa 2 aa 1 aa aa 2 II 0 II 1 II 2 (13) II ss = AA 1 II pp II 0 II 1 II 2 = 1 3 1 1 1 1 aa aa 2 1 aa 2 aa II aa IIbb (14) IIcc
Sequence Components Balanced System 37 Before applying sequence components to unbalanced systems, let s first look at the sequence components for a balanced, positive-sequence, three-phase system For a balanced system, we have VV bb = VV aa 1 120 = aa 2 VV aa VV cc = VV aa 1 120 = aavv aa The sequence voltages are given by (12) The zero sequence voltage is VV 0 = 1 3 VV aa + VV bb + VV cc = 1 3 VV aa + aa 2 VV aa + aavv aa VV 0 = 1 3 VV aa 1 + aa 2 + aa Applying the identity 1 + aa 2 + aa = 0, we have VV 0 = 0
Sequence Components Balanced System 38 The positive sequence component is given by VV 1 = 1 3 VV aa + aavv bb + aa 2 VV cc VV 1 = 1 3 VV aa + aa aa 2 VV aa + aa 2 aavv aa VV 1 = 1 3 VV aa + aa 3 VV aa + aa 3 VV aa Since aa 3 = 1 0, we have VV 1 = 1 3 3VV aa VV 1 = VV aa
Sequence Components Balanced System 39 The negative sequence component is given by VV 2 = 1 3 VV aa + aa 2 VV bb + aavv cc VV 2 = 1 3 VV aa + aa 2 aa 2 VV aa + aa aavv aa VV 2 = 1 3 VV aa + aa 4 VV aa + aa 2 VV aa Again, using the identity 1 + aa 2 + aa = 0, along with the fact that aa 4 = aa, we have VV 2 = 0
Sequence Components Balanced System 40 So, for a positive-sequence, balanced, three-phase system, the sequence voltages are VV 0 = 0, VV 1 = VV aa, VV 2 = 0 Similarly, the sequence currents are II 0 = 0, II 1 = II aa, II 2 = 0 This is as we would expect No zero- or negative-sequence components for a positive-sequence balanced system Zero- and negative-sequence components are only used to account for imbalance
Sequence Components 41 We have just introduced the concept of symmetric components Allows for decomposition of, possibly unbalanced, three-phase phasors into sequence components We ll now apply this concept to power system networks to develop sequence networks Decoupled networks for each of the sequence components Sequence networks become coupled only at the point of imbalance Simplifies the analysis of unbalanced systems
42 Sequence Networks
Sequence Networks 43 Power system components each have their own set of sequence networks Non-rotating loads Transmission lines Rotating machines generators and motors Transformers Sequence networks for overall systems are interconnections of the individual sequence network Sequence networks become coupled in a particular way at the fault location depending on type of fault Line-to-line Single line-to-ground Double line-to-ground Fault current can be determined through simple analysis of the coupled sequence networks
Sequence Networks Non-Rotating Loads 44 Consider a balanced Y-load with the neutral grounded through some non-zero impedance Applying KVL gives the phase-aa-to-ground voltage VV aaaa = ZZ yy II aa + ZZ nn II nn For phase bb: VV aaaa = ZZ yy II aa + ZZ nn II aa + II bb + II cc VV aaaa = ZZ yy + ZZ nn II aa + ZZ nn II bb + ZZ nn II cc (15) VV bbgg = ZZ yy II bb + ZZ nn II nn = ZZ yy II bb + ZZ nn II aa + II bb + II cc VV bbgg = ZZ nn II aa + ZZ yy + ZZ nn II bb + ZZ nn II cc (16) Similarly, for phase cc: VV aaaa = ZZ nn II aa + ZZ nn II bb + ZZ yy + ZZ nn II cc (17)
Sequence Networks Non-Rotating Loads 45 Putting (15) (17) in matrix form or VV aagg VV bbbb VV cccc = VV pp = ZZ pp II pp ZZ yy + ZZ nn ZZ nn ZZ nn ZZ nn ZZ yy + ZZ nn ZZ nn ZZ nn ZZ nn ZZ yy + ZZ nn II aa IIbb IIcc (18) where VV pp and II pp are the phase voltages and currents, respectively, and ZZ pp is the phase impedance matrix We can use (9) and (13) to rewrite (18) as AAVV ss = ZZ pp AAII ss Solving for VV ss VV ss = AA 1 ZZ pp AAII ss or VV ss = ZZ ss II ss (19)
Sequence Networks Non-Rotating Loads 46 VV ss = ZZ ss II ss (19) where ZZ ss is the sequence impedance matrix ZZ ss = AA 1 ZZ pp AA = ZZ yy + 3ZZ nn 0 0 0 ZZ yy 0 (20) 0 0 ZZ yy Equation (19) then becomes a set of three uncoupled equations VV 0 = ZZ yy + 3ZZ nn II 0 = ZZ 0 II 0 (21) VV 1 = ZZ yy II 1 = ZZ 1 II 1 (22) VV 2 = ZZ yy II 2 = ZZ 2 II 2 (23)
Sequence Networks Non-Rotating Loads 47 Equations (21) (23) describe the uncoupled sequence networks Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks Non-Rotating Loads 48 We can develop similar sequence networks for a balanced Δ- connected load ZZ yy = ZZ Δ /3 There is no neutral point for the Δ-network, so ZZ nn = - an open circuit Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks 3-φφ Lines 49 Balanced, three-phase lines can be modeled as The voltage drops across the lines are given by the following system of equations VV aaaa VV bbbb VV cccc = ZZ aaaa ZZ aaaa ZZ aaaa ZZ bbbb ZZ bbbb ZZ bbbb ZZ cccc ZZ cccc ZZ cccc II aa IIbb IIcc = VV aann VV aa nn VV bbbb VV bb nn VV cccc VV cc nn (24)
Sequence Networks 3-φφ Lines 50 Writing (24) in compact form VV pp VV pp = ZZ pp II pp (25) ZZ pp is the phase impedance matrix Self impedances along the diagonal Mutual impedances elsewhere Symmetric Diagonal, if we neglect mutual impedances
Sequence Networks 3-φφ Lines 51 We can rewrite (25) in terms of sequence components AAVV ss AAVV ss = ZZ pp AAII ss VV ss VV ss = AA 1 ZZ pp AAII ss VV ss VV ss = ZZ ss II ss (26) where ZZ ss is the sequence impedance matrix ZZ ss = AA 1 ZZ pp AA (27) ZZ ss is diagonal so long as the system impedances are balanced, i.e. Self impedances are equal: ZZ aaaa = ZZ bbbb = ZZ cccc Mutual impedances are equal: ZZ aaaa = ZZ aaaa = ZZ bbbb
Sequence Networks 3-φφ Lines 52 For balanced lines, ZZ ss is diagonal ZZ ss = ZZ aaaa + 2ZZ aabb 0 0 0 ZZ aaaa ZZ aaaa 0 = 0 0 ZZ aaaa ZZ aaaa ZZ 0 0 0 0 ZZ 1 0 0 0 ZZ 2 Because ZZ ss is diagonal, (26) represents three uncoupled equations VV 0 VV 0 = ZZ 0 II 0 (28) VV 1 VV 1 = ZZ 1 II 1 (29) VV 2 VV 2 = ZZ 2 II 2 (30)
Sequence Networks 3-φφ Lines 53 Equations (28) (30) describe the voltage drop across three uncoupled sequence networks Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks Rotating Machines 54 Consider the following model for a synchronous generator Similar to the Y-connected load Generator includes voltage sources on each phase Voltage sources are positive sequence Sources will appear only in the positive-sequence network
Sequence Networks Synchronous Generator 55 Sequence networks for Y-connected synchronous generator Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks Motors 56 Synchronous motors Sequence networks identical to those for synchronous generators Reference current directions are reversed Induction motors Similar sequence networks to synchronous motors, except source in the positive sequence network set to zero
Sequence Networks - Y-Y Transformers 57 Per-unit sequence networks for transformers Simplify by neglecting transformer shunt admittances Consider a Y-Y transformer Similar to the Y-connected load, the voltage drops across the neutral impedances are 3II 0 ZZ NN and 3II 0 ZZ nn 3ZZ NN and 3ZZ nn each appear in the zero-sequence network Can be combined in the per-unit circuit as long as shunt impedances are neglected
Sequence Networks Y-Y Transformers 58 Impedance accounting for leakage flux and winding resistance for each winding can be referred to the primary Add together into a single impedance, ZZ ss, in the per-unit model Y-Y transformer sequence networks Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks Y-ΔTransformers 59 Y-Δ transformers differ in a couple of ways Must account for phase shift from primary to secondary For positive-sequence network, Y-side voltage and current lead Δ- side voltage and current For negative-sequence network, Y-side voltage and current lag Δ- side voltage and current No neutral connection on the Δ side Zero-sequence current cannot enter or leave the Δ winding
Sequence Networks Y-ΔTransformers 60 Sequence networks for Y-ΔTransformers Zero-sequence network: Positive-sequence network: Negative-sequence network:
Sequence Networks Δ-ΔTransformers 61 Δ-Δ transformers Like Y-Y transformers, no phase shift No neutral connections Zero-sequence current cannot flow into or out of either winding
Sequence Networks Δ-ΔTransformers 62 Sequence networks for Δ-ΔTransformers Zero-sequence network: Positive-sequence network: Negative-sequence network:
Power in Sequence Networks 63 We can relate the power delivered to a system s sequence networks to the three-phase power delivered to that system We know that the complex power delivered to a threephase system is the sum of the power at each phase SS pp = VV aaaa II aa + VV bbbb II bb + VV cccc II cc In matrix form, this looks like SS pp = VV aann VV bbbb VV cccc II bb II aa SS pp = VV pp TT II pp II cc (28)
Power in Sequence Networks 64 Recall the following relationships VV pp = AAVV ss (9) II pp = AAII ss (13) Using (9) and (13) in (28), we have SS pp = AAVV ss TT AAII ss SS pp = VV ss TT AA TT AA II ss (29) Computing the product in the middle of the right-hand side of (29), we find AA TT AA = 3 0 0 0 3 0 0 0 3 where II 3 is the 3 3 identity matrix = 3II 3
Power in Sequence Networks 65 Equation (29) then becomes SS pp = VV ss TT 3II 3 II ss SS pp = 3VV TT ss II ss SS pp = 3 VV 0 VV 1 VV 2 II 1 II 0 II 2 SS pp = 3 VV 0 II 0 + VV 1 II 1 + VV 2 II 2 The total power delivered to a three-phase network is three times the sum of the power delivered to the three sequence networks The three sequence networks represent only one of the three phases recall, we chose to consider only phase aa
66 Unsymmetrical Faults
Unsymmetrical Faults 67 The majority of faults that occur in three-phase power systems are unsymmetrical Not balanced Fault current and voltage differ for each phase The method of symmetrical components and sequence networks provide us with a tool to analyze these unsymmetrical faults We ll examine three types of unsymmetrical faults Single line-to-ground (SLG) faults Line-to-line (LL) faults Double line-to-ground (DLG) faults
Unsymmetrical Fault Analysis - Procedure 68 Basic procedure for fault analysis: 1. Generate sequence networks for the system 2. Interconnect sequence networks appropriately at the fault location 3. Perform circuit analysis on the interconnected sequence networks
Unsymmetrical Fault Analysis 69 To simplify our analysis, we ll make the following assumptions 1. System is balanced before the instant of the fault 2. Neglect pre-fault load current All pre-fault machine terminal voltages and bus voltages are equal to VV FF 3. Transmission lines are modeled as series reactances only 4. Transformers are modeled with leakage reactances only 5. Non-rotating loads are neglected 6. Induction motors are either neglected or modeled as synchronous motors
Unsymmetrical Fault Analysis 70 Each sequence network includes all interconnected powersystem components Generators, motors, lines, and transformers Analysis will be simplified if we represent each sequence network as its Thévenin equivalent From the perspective of the fault location For example, consider the following power system:
Unsymmetrical Fault Analysis Sequence Networks 71 The sequence networks for the system are generated by interconnecting the sequence networks for each of the components The zero-sequence network: The positive-sequence network: Assuming the generator is operating at the rated voltage at the time of the fault
Unsymmetrical Fault Analysis Sequence Networks 72 The negative sequence network: Now, let s assume there is some sort of fault at bus 1 Determine the Thévenin equivalent for each sequence network from the perspective of bus 1 Simplifying the zero-sequence network to its Thévenin equivalent
Unsymmetrical Fault Analysis Sequence Networks 73 The positive-sequence network simplifies to the following circuit with the following Thévenin equivalent Similarly, for the negative-sequence network, we have Next, we ll see how to interconnect these networks to analyze different types of faults
Unsymmetrical Fault Analysis SLG Fault 74 The following represents a generic three-phase network with terminals at the fault location: If we have a single-line-toground fault, where phase aa is shorted through ZZ ff to ground, the model becomes:
Unsymmetrical Fault Analysis SLG Fault 75 The phase-domain fault conditions: II aa = VV aaaa ZZ ff (1) II bb = II cc = 0 (2) Transforming these phase-domain currents to the sequence domain II 0 II 1 II 2 = 1 3 1 1 1 1 aa aa 2 1 aa 2 aa VV aagg /ZZ ff 0 0 = 1 3 VV aagg /ZZ ff VV aagg /ZZ ff VV aagg /ZZ ff (3) This gives one of our sequence-domain fault conditions II 0 = II 1 = II 2 (4)
Unsymmetrical Fault Analysis SLG Fault 76 We know that and Using (5) and (6) in (1), we get II aa = VV aaaa ZZ ff = II 0 + II 1 + II 2 (5) VV aaaa = VV 0 + VV 1 + VV 2 (6) II 0 + II 1 + II 2 = 1 ZZ ff VV 0 + VV 1 + VV 2 Using (4), this gives our second sequence-domain fault condition II 0 = II 1 = II 2 = 1 3ZZ ff VV 0 + VV 1 + VV 2 (7)
Unsymmetrical Fault Analysis SLG Fault 77 The sequence-domain fault conditions are satisfied by connecting the sequence networks in series along with three times the fault impedance We want to find the phase domain fault current, II FF II FF = II aa = II 0 + II 1 + II 2 = 3II 1 II 1 = VV FF ZZ 0 +ZZ 1 +ZZ 2 +3ZZ FF II FF = 3VV FF ZZ 0 +ZZ 1 +ZZ 2 +3ZZ FF (8)
SLG Fault - Example 78 Returning to our example power system The interconnected sequence networks for a bolted fault at bus 1:
SLG Fault - Example 79 The fault current is II FF = 3VV FF ZZ 0 +ZZ 1 +ZZ 2 +3ZZ FF II FF = 3.0 30 jjj.487 The current base at bus 1 is II bb = SS bb VV bbb = = 6.16 60 pp. uu. 150 MMMMMM So the fault current in ka is 3 230 kkkk = 376.5 AA II FF = 2.32 60 376.5 AA II FF = 2.32 60 kkkk
Unsymmetrical Fault Analysis LL Fault 80 Now consider a line-to-line fault between phase bb and phase cc through impedance ZZ FF Phase-domain fault conditions: II aa = 0 (9) II bb = II cc = VV bbbb VV cccc ZZ FF (10) Transforming to the sequence domain II 0 II1 II 2 = 1 3 1 1 1 1 aa aa 2 1 aa 2 aa 0 II bb II bb = 1 3 So, the first two sequence-domain fault conditions are 0 aa aa 2 II bb aa 2 aa 2 II bb (11) II 0 = 0 II 2 = II 1 (12) (13)
Unsymmetrical Fault Analysis LL Fault 81 To derive the remaining sequence-domain fault condition, rearrange (10) and transform to the sequence domain VV bbbb VV cccc = II bb ZZ FF VV 0 + aa 2 VV 1 + aavv 2 VV 0 + aavv 1 + aa 2 VV 2 = II 0 + aa 2 II 1 + aaii 2 ZZ FF aa 2 VV 1 + aavv 2 aavv 1 aa 2 VV 2 = aa 2 aa II 1 ZZ FF aa 2 aa VV 1 aa 2 aa VV 2 = aa 2 aa II 1 ZZ FF The last sequence-domain fault condition is VV 1 VV 2 = II 1 ZZ FF (14)
Unsymmetrical Fault Analysis LL Fault 82 Sequence-domain fault conditions II 0 = 0 II 2 = II 1 (12) (13) VV 1 VV 2 = II 1 ZZ FF (14) These can be satisfied by: Leaving the zero-sequence network open Connecting the terminals of the positive- and negative-sequence networks together through ZZ FF
Unsymmetrical Fault Analysis LL Fault 83 The fault current is the phase bb current, which is given by II FF = II bb = II 0 + aa 2 II 1 + aaii 2 II FF = aa 2 II 1 aaii 1 II FF = jj 3 II 1 = jj 3 VV FF ZZ 1 +ZZ 2 +ZZ FF II FF = 3 VV FF 90 ZZ 1 +ZZ 2 +ZZ FF (15)
LL Fault - Example 84 Now consider the same system with a bolted line-to-line fault at bus 1 The sequence network:
LL Fault - Example 85 II 1 = 1 30 jjj.363 = 2.75 60 The subtransient fault current is given by (15) as II FF = 3 2.75 60 90 II FF = 4.76 150 pp. uu. Using the previously-determined current base, we can convert the fault current to ka II FF = II bbb 4.76 150 II FF = 4.76 150 376.5AA II FF = 1.79 150 kkkk
Unsymmetrical Fault Analysis DLG Fault 86 Now consider a double line-to-ground fault Assume phases bb and cc are shorted to ground through ZZ FF Phase-domain fault conditions: II aa = 0 (16) II bb + II cc = VV bbbb ZZ FF = VV ccgg ZZ FF (17) It can be shown that (16) and (17) transform to the following sequence-domain fault conditions (analysis skipped here) II 0 + II 1 + II 2 = 0 (18) VV 1 = VV 2 (19) II 0 = 1 3ZZ FF VV 0 VV 1 (20)
Unsymmetrical Fault Analysis DLG Fault 87 Sequence-domain fault conditions II 0 + II 1 + II 2 = 0 (18) VV 1 = VV 2 (19) II 0 = 1 3ZZ FF VV 0 VV 1 (20) To satisfy these fault conditions Connect the positive- and negative-sequence networks together directly Connect the zero- and positive-sequence networks together through ZZ FF
Unsymmetrical Fault Analysis DLG Fault 88 The fault current is the sum of the phase bb and phase cc currents, as given by (17) In the sequence domain the fault current is II FF = II bb + II cc = 3II 0 II FF = 3II 0 (21) II 0 can be determined by a simple analysis (e.g. nodal) of the interconnected sequence networks
DLG Fault - Example 89 Now determine the subtransient fault current for a bolted double line-to-ground fault at bus 1 The sequence network: Here, because ZZ FF = 0, VV 0 = VV 1 = VV 2
DLG Fault - Example 90 To find II FF, we must determine II 0 We can first find VV 0 by applying voltage division VV 0 = VV FF ZZ 2 ZZ 0 ZZ 1 + ZZ 2 ZZ 0 jjj.194 jjj.124 VV 0 = 1.0 30 jjj.169 + jjj.194 jjj.124 VV 0 = 0.309 30
DLG Fault - Example 91 Next, calculate II 0 II 0 = VV 0 ZZ 0 = 0.309 30 jjj.124 = 2.49 120 pp. uu. The per-unit fault current is II FF = 3II 0 = 7.48 120 pp. uu. Using the current base to convert to ka, gives the subtransient DLG fault current II FF = 7.48 120 376.5 AA II FF = 2.82 120 kkkk