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Section 7 Fault Analysis.pdf

This document section discusses fault analysis in power systems. It begins by defining different types of faults like line-to-ground and line-to-line faults that can result in excessive current. It then discusses fault current calculations for a symmetrical three-phase fault on a simple system using superposition and per-unit calculations. As an example, it calculates the subtransient fault current, generator current, and motor current for a bolted three-phase fault at a bus.

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ESE 470 – Energy Distribution Systems SECTION 7: FAULT ANALYSIS
K. Webb ESE 470 Introduction 2
K. Webb ESE 470 3 Power System Faults  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
K. Webb ESE 470 4 Types of Faults  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
K. Webb ESE 470 Subtransient Fault Current 5
K. Webb ESE 470 6 Fault Current  R: generator resistance  L: generator inductance  𝑖 𝑡 = 0 for 𝑡 < 0  Source phase, 𝛼, determines voltage at 𝑡 = 0  Short circuit can occur at any point in a 60 Hz cycle  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 𝑡 = 0  Consider an unloaded synchronous generator  Equivalent circuit model: 𝑣 𝑡 = 2𝑉𝐺 sin 𝜔𝑡 + 𝛼
K. Webb ESE 470 7 Fault Current  The governing differential equation for 𝑡 > 0 is 𝑑𝑖 𝑑𝑡 + 𝑖 𝑡 𝑅 𝐿 = 2𝑉𝐺 𝐿 sin 𝜔𝑡 + 𝛼  The solution gives the fault current 𝑖 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 + 𝛼 − 𝜃 − sin 𝛼 − 𝜃 𝑒−𝑡 𝑅 𝐿 where 𝑍 = 𝑅2 + 𝜔𝐿 2 and 𝜃 = tan−1 𝜔𝐿 𝑅  This total fault current is referred to as the asymmetrical fault current  It has a steady-state component 𝑖𝑎𝑐 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 + 𝛼 − 𝜃  And a transient component 𝑖𝑑𝑐 𝑡 = − 2𝑉𝐺 𝑍 sin 𝛼 − 𝜃 𝑒−𝑡 𝑅 𝐿
K. Webb ESE 470 8 Fault Current  Magnitude of the transient fault current, 𝑖𝑑𝑐, depends on 𝛼  𝑖𝑑𝑐 0 = 0 for 𝛼 = 𝜃  𝑖𝑑𝑐 0 = 2𝐼𝑎𝑐 for 𝛼 = 𝜃 − 90°  𝐼𝑎𝑐 = 𝑉𝐺/𝑍 is the rms value of the steady-state fault current  Worst-case fault current occurs for 𝛼 = 𝜃 − 90° 𝑖 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 − 𝜋 2 + 𝑒−𝑡 𝑅 𝐿
K. Webb ESE 470 9 Fault Current  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  In this class, we will use the steady-state current component, 𝐈𝑎𝑐, as our primary fault current metric
K. Webb ESE 470 10 Generator Reactance  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  𝑋𝑑 ′′ : subtransient reactance  𝑋𝑑 ′ : transient reactance  𝑋𝑑: synchronous reactance  Reactance increases with time, such that 𝑋𝑑 ′′ < 𝑋𝑑 ′ < 𝑋𝑑
K. Webb ESE 470 11 Sub-Transient Fault Current  Transition rates between reactance values are dictated by two time constants:  𝜏𝑑 ′′ : short-circuit subtransient time constant  𝜏𝑑 ′ : short-circuit transient time constant  Neglecting generator resistance, i.e. assuming 𝜃 = 90°, the synchronous portion of the fault current is 𝑖𝑎𝑐 𝑡 = 2𝑉𝐺 1 𝑋𝑑 ′′ − 1 𝑋𝑑 ′ 𝑒 − 𝑡 𝜏𝑑 ′′ + 1 𝑋𝑑 ′ − 1 𝑋𝑑 𝑒 − 𝑡 𝜏𝑑 ′ + 1 𝑋𝑑 sin 𝜔𝑡 + 𝛼 − 𝜋 2  At the instant of the fault, 𝑡 = 0, the rms synchronous fault current is 𝐼𝐹 ′′ = 𝑉𝐺 𝑋𝑑 ′′  This is the rms subtransient fault current, 𝐼𝐹 ′′  This will be our primary metric for assessing fault current
K. Webb ESE 470 Symmetrical Three-Phase Short Circuits 12
K. Webb ESE 470 13 Symmetrical 3-𝜙 Short Circuits  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-𝑌 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
K. Webb ESE 470 14 Symmetrical 3-𝜙 Short Circuits  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 𝑡 = 0
K. Webb ESE 470 15 Symmetrical 3-𝜙 Short Circuits  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 𝑽𝐹
K. Webb ESE 470 16 Symmetrical 3-𝜙 Short Circuits  Applying superposition, we can represent this circuit as the sum of two separate circuits: Circuit 1 Circuit 2
K. Webb ESE 470 17 Symmetrical 3-𝜙 Short Circuits  Assume that the value of the fault-location source, 𝑽𝐹, is the pre-fault voltage at that location  Circuit 1, then, represents the pre-fault circuit, so 𝑰𝐹1 ′′ = 0  The 𝑽𝐹 source can therefore be removed from circuit 1 Circuit 1 Circuit 2
K. Webb ESE 470 18 Symmetrical 3-𝜙 Short Circuits  The current in circuit 1, 𝑰𝐿, is the pre-fault line current  Superposition gives the fault current 𝑰𝐹 ′′ = 𝑰𝐹1 ′′ + 𝑰𝐹2 ′′ = 𝑰𝐹2 ′′  The generator fault current is 𝑰𝐺 ′′ = 𝑰𝐺1 ′′ + 𝑰𝐺2 ′′ 𝑰𝐺 ′′ = 𝑰𝐿 + 𝑰𝐺2 ′′  The motor fault current is 𝑰𝑴 ′′ = 𝑰𝑀1 ′′ + 𝑰𝑀2 ′′ 𝑰𝑀 ′′ = −𝑰𝐿 + 𝑰𝑀2 ′′
K. Webb ESE 470 19 Symmetrical 3-𝜙 Fault – Example  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
K. Webb ESE 470 20 Symmetrical 3-𝜙 Fault – Example  First convert to per-unit  Use 𝑆𝑏 = 100 𝑀𝑉𝐴  Base voltage in the transmission line zone is 𝑉𝑏,𝑡𝑙 = 138 𝑘𝑉  Base impedance in the transmission line zone is 𝑍𝑏,𝑡𝑙 = 𝑉𝑏,𝑡𝑙 2 𝑆𝑏 = 138 𝑘𝑉 2 100 𝑀𝑉𝐴 = 190.4 Ω  The per-unit transmission line reactance is 𝑋𝑡𝑙 = 20 Ω 190.4 Ω = 0.105 𝑝. 𝑢.
K. Webb ESE 470 21 Symmetrical 3-𝜙 Fault – Example  The two per-unit circuits are  These can be simplified by combining impedances
K. Webb ESE 470 22 Symmetrical 3-𝜙 Fault – Example  Using circuit 2, we can calculate the subtransient fault current 𝑰𝐹 ′′ = 1.05∠0° 𝑗0.116 = 9.079∠ − 90° 𝑝. 𝑢.  To convert to kA, first determine the current base in the generator zone 𝐼𝑏,𝐺 = 𝑆𝑏 3𝑉𝑏,𝐺 = 100 𝑀𝑉𝐴 3 ⋅ 13.8 𝑘𝑉 = 4.18 𝑘𝐴  The subtransient fault current is 𝑰𝐹 ′′ = 9.079∠ − 90° ⋅ 4.18 𝑘𝐴 𝑰𝐹 ′′ = 37.98∠ − 90° 𝑘𝐴
K. Webb ESE 470 23 Symmetrical 3-𝜙 Fault – Example  The pre-fault line current can be calculated from the pre-fault generator voltage and power 𝑰𝐿 = ൗ 𝑺𝐺 3 ൘ 𝑽𝐺 ′′ 3 ∗ = 𝑺𝐺 3𝑽𝐺 ′′ ∗ = 100∠ cos−1 0.95 𝑀𝑉𝐴 ∗ 3 ⋅ 1.05 ⋅ 13.8∠0° 𝑘𝑉 ∗ 𝑰𝐿 = 100∠ − 18.19° 𝑀𝑉𝐴 3 ⋅ 1.05 ⋅ 13.8∠0° 𝑘𝑉 𝑰𝐿 = 3.98∠ − 18.19° 𝑘𝐴  Or, in per-unit: 𝑰𝐿 = 3.98∠ − 18.19° 𝑘𝐴 4.18 kA = 0.952∠ − 18.19° 𝑝. 𝑢.  This will be used to find the generator and motor fault currents
K. Webb ESE 470 24 Symmetrical 3-𝜙 Fault – Example  The generator’s contribution to the fault current is found by applying current division 𝑰𝐺2 ′′ = 𝑰𝐹 ′′ 0.505 0.505 + 0.15 = 7.0∠ − 90° 𝑝. 𝑢.  Adding the pre-fault line current, we have the subtransient generator fault current 𝑰𝐺 ′′ = 𝑰𝐿 + 𝑰𝐺2 ′′ 𝑰𝐺 ′′ = 0.952∠ − 18.19° + 7.0∠ − 90° 𝑰𝐺 ′′ = 7.35∠ − 82.9° 𝑝. 𝑢.  Converting to kA 𝑰𝐺 ′′ = 7.35∠ − 82.9° ⋅ 4.18 𝑘𝐴 𝑰𝐺 ′′ = 30.74∠ − 82.9° 𝑘𝐴
K. Webb ESE 470 25 Symmetrical 3-𝜙 Fault – Example  Similarly, for the motor 𝑰𝑀2 ′′ = 𝑰𝐹 ′′ 0.15 0.505 + 0.15 = 2.08∠ − 90° 𝑝. 𝑢.  Subtracting the pre-fault line current gives the subtransient motor fault current 𝑰𝑀 ′′ = −𝑰𝐿 + 𝑰𝑀2 ′′ 𝑰𝑀 ′′ = −0.952∠ − 18.19° + 2.08∠ − 90° 𝑰𝑀 ′′ = 2.0∠ − 116.9°  Converting to 𝑘𝐴 𝑰𝑀 ′′ = 2.0∠ − 116.9° ⋅ 4.18 𝑘𝐴 𝑰𝑀 ′′ = 8.36∠ − 116.9° 𝑘𝐴
K. Webb ESE 470 Symmetrical Components 26
K. Webb ESE 470 27 Symmetrical Components  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
K. Webb ESE 470 28 Symmetrical Components  The method of symmetrical components:  Represent an asymmetrical set of 𝑁 phasors as a sum of 𝑁 sets of symmetrical component phasors  These 𝑁 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 (𝑁 = 3), sequence components are:  Zero sequence components  Positive sequence components  Negative sequence components
K. Webb ESE 470 29 Sequence Components  Zero sequence components  Three phasors with equal magnitude and equal phase  𝑽𝑎0, 𝑽𝑏0, 𝑽𝑐0  Positive sequence components  Three phasors with equal magnitude and ± 120°, positive-sequence phase  𝑽𝑎1, 𝑽𝑏1, 𝑽𝑐1  Negative sequence components  Three phasors with equal magnitude and ± 120°, negative-sequence phase  𝑽𝑎2, 𝑽𝑏2, 𝑽𝑐2
K. Webb ESE 470 30 Sequence Components  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 𝑽𝑎 𝑽𝑏 𝑽𝑐 = 𝑽𝑎0 𝑽𝑏0 𝑽𝑐0 + 𝑽𝑎1 𝑽𝑏1 𝑽𝑐1 + 𝑽𝑎2 𝑽𝑏2 𝑽𝑐2 (1)
K. Webb ESE 470 31 Sequence Components  The phasors of each sequence component have a fixed phase relationship  If we know one, we know the other two  Assume we know phase 𝑎 – use that as the reference  For the zero sequence components, we have 𝑽0 = 𝑽𝑎0 = 𝑽𝑏0 = 𝑽𝑐0 (2)  For the positive sequence components, 𝑽1 = 𝑽𝑎1 = 1∠120° ⋅ 𝑽𝑏1 = 1∠240° ⋅ 𝑽𝑐1 (3)  And, for the negative sequence components, 𝑽2 = 𝑽𝑎2 = 1∠240° ⋅ 𝑽𝑏2 = 1∠120° ⋅ 𝑽𝑐2 (4)  Note that we’re using phase 𝑎 as our reference, so 𝑽0 = 𝑽𝑎0, 𝑽1 = 𝑽𝑎1, 𝑽2 = 𝑽𝑎2
K. Webb ESE 470 32 Sequence Components  Next, we define a complex number, 𝑎, that has unit magnitude and phase of 120° 𝑎 = 1∠120° (5)  Multiplication by 𝑎 results in a rotation (a phase shift) of 120°  Multiplication by 𝑎2 yields a rotation of 240° = −120°  Using (5) to rewrite (3) and (4) 𝑽1 = 𝑽𝑎1 = 𝑎𝑽𝑏1 = 𝑎2𝑽𝑐1 (6) 𝑽2 = 𝑽𝑎2 = 𝑎2 𝑽𝑏2 = 𝑎𝑽𝑐2 (7)
K. Webb ESE 470 33 Sequence Components  Using (2), (6), and (7), we can rewrite (1) in a simplified form 𝑽𝑎 𝑽𝑏 𝑽𝑐 = 1 1 1 1 𝑎2 𝑎 1 𝑎 𝑎2 𝑽0 𝑽1 𝑽2 (8)  The vector on the left is the vector of phase voltages, 𝑽𝑝  The vector on the right is the vector of (phase 𝑎) sequence components, 𝑽𝑠  We’ll call the 3 × 3 transformation matrix 𝑨  We can rewrite (8) as 𝑽𝑝 = 𝑨𝑽𝑠 (9)
K. Webb ESE 470 34 Sequence Components  We can express the sequence voltages as a function of the phase voltages by inverting the transformation matrix 𝑽𝑠 = 𝑨−1 𝑽𝑝 (10) where 𝑨−1 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 (11) So 𝑽0 𝑽1 𝑽2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑽𝑎 𝑽𝑏 𝑽𝑐 (12)
K. Webb ESE 470 35 Sequence Components  The same relationships hold for three-phase currents  The phase currents are 𝑰𝑝 = 𝑰𝑎 𝑰𝑏 𝑰𝑐  And, the sequence currents are 𝑰𝑠 = 𝑰0 𝑰1 𝑰2
K. Webb ESE 470 36 Sequence Components  The transformation matrix, 𝑨, relates the phase currents to the sequence currents 𝑰𝑝 = 𝑨𝑰𝑠 𝑰𝑎 𝑰𝑏 𝑰𝑐 = 1 1 1 1 𝑎2 𝑎 1 𝑎 𝑎2 𝑰0 𝑰1 𝑰2 (13)  And vice versa 𝑰𝑠 = 𝑨−1𝑰𝑝 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑰𝑎 𝑰𝑏 𝑰𝑐 (14)
K. Webb ESE 470 37 Sequence Components – Balanced System  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 𝑽𝑏 = 𝑽𝑎 ⋅ 1∠ − 120° = 𝑎2 𝑽𝑎 𝑽𝑐 = 𝑽𝑎 ⋅ 1∠120° = 𝑎𝑽𝑎  The sequence voltages are given by (12)  The zero sequence voltage is 𝑽0 = 1 3 𝑽𝑎 + 𝑽𝑏 + 𝑽𝑐 = 1 3 𝑽𝑎 + 𝑎2 𝑽𝑎 + 𝑎𝑽𝑎 𝑽0 = 1 3 𝑽𝑎 1 + 𝑎2 + 𝑎  Applying the identity 1 + 𝑎2 + 𝑎 = 0, we have 𝑽0 = 0
K. Webb ESE 470 38 Sequence Components – Balanced System  The positive sequence component is given by 𝑽1 = 1 3 𝑽𝑎 + 𝑎𝑽𝑏 + 𝑎2𝑽𝑐 𝑽1 = 1 3 𝑽𝑎 + 𝑎 ⋅ 𝑎2𝑽𝑎 + 𝑎2 ⋅ 𝑎𝑽𝑎 𝑽1 = 1 3 𝑽𝑎 + 𝑎3𝑽𝑎 + 𝑎3𝑽𝑎  Since 𝑎3 = 1∠0°, we have 𝑽1 = 1 3 3𝑽𝑎 𝑽1 = 𝑽𝑎
K. Webb ESE 470 39 Sequence Components – Balanced System  The negative sequence component is given by 𝑽2 = 1 3 𝑽𝑎 + 𝑎2𝑽𝑏 + 𝑎𝑽𝑐 𝑽2 = 1 3 𝑽𝑎 + 𝑎2 ⋅ 𝑎2𝑽𝑎 + 𝑎 ⋅ 𝑎𝑽𝑎 𝑽2 = 1 3 𝑽𝑎 + 𝑎4 𝑽𝑎 + 𝑎2 𝑽𝑎  Again, using the identity 1 + 𝑎2 + 𝑎 = 0, along with the fact that 𝑎4 = 𝑎, we have 𝑽2 = 0
K. Webb ESE 470 40 Sequence Components – Balanced System  So, for a positive-sequence, balanced, three-phase system, the sequence voltages are 𝑽0 = 0, 𝑽1 = 𝑽𝑎, 𝑽2 = 0  Similarly, the sequence currents are 𝑰0 = 0, 𝑰1 = 𝑰𝑎, 𝑰2 = 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
K. Webb ESE 470 41 Sequence Components  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
K. Webb ESE 470 Sequence Networks 42
K. Webb ESE 470 43 Sequence Networks  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
K. Webb ESE 470 Sequence Networks – Non-Rotating Loads 44
K. Webb ESE 470 45 Sequence Networks – Non-Rotating Loads  Consider a balanced Y-load with the neutral grounded through some non-zero impedance  Applying KVL gives the phase-𝑎-to-ground voltage 𝑽𝑎𝑔 = 𝑍𝑦𝑰𝑎 + 𝑍𝑛𝑰𝑛 𝑽𝑎𝑔 = 𝑍𝑦𝑰𝑎 + 𝑍𝑛 𝑰𝑎 + 𝑰𝑏 + 𝑰𝑐 𝑽𝑎𝑔 = 𝑍𝑦 + 𝑍𝑛 𝑰𝑎 + 𝑍𝑛𝑰𝑏 + 𝑍𝑛𝑰𝑐 (15)  For phase 𝑏: 𝑽𝑏𝑔 = 𝑍𝑦𝑰𝑏 + 𝑍𝑛𝑰𝑛 = 𝑍𝑦𝑰𝑏 + 𝑍𝑛 𝑰𝑎 + 𝑰𝑏 + 𝑰𝑐 𝑽𝑏𝑔 = 𝑍𝑛𝑰𝑎 + 𝑍𝑦 + 𝑍𝑛 𝑰𝑏 + 𝑍𝑛𝑰𝑐 (16)  Similarly, for phase 𝑐: 𝑽𝑎𝑔 = 𝑍𝑛𝑰𝑎 + 𝑍𝑛𝑰𝑏 + 𝑍𝑦 + 𝑍𝑛 𝑰𝑐 (17)
K. Webb ESE 470 46 Sequence Networks – Non-Rotating Loads  Putting (15) – (17) in matrix form 𝑽𝑎𝑔 𝑽𝑏𝑔 𝑽𝑐𝑔 = 𝑍𝑦 + 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑦 + 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑦 + 𝑍𝑛 𝑰𝑎 𝑰𝑏 𝑰𝑐 or 𝑽𝑝 = 𝒁𝑝𝑰𝑝 (18) where 𝑽𝑝 and 𝑰𝑝 are the phase voltages and currents, respectively, and 𝒁𝑝 is the phase impedance matrix  We can use (9) and (13) to rewrite (18) as 𝑨𝑽𝑠 = 𝒁𝑝𝑨𝑰𝑠  Solving for 𝑽𝑠 𝑽𝑠 = 𝑨−1𝒁𝑝𝑨𝑰𝑠 or 𝑽𝑠 = 𝒁𝑠𝑰𝑠 (19)
K. Webb ESE 470 47 Sequence Networks – Non-Rotating Loads 𝑽𝑠 = 𝒁𝑠𝑰𝑠 (19) where 𝑍𝑠 is the sequence impedance matrix 𝒁𝑠 = 𝑨−1𝒁𝑝𝑨 = 𝑍𝑦 + 3𝑍𝑛 0 0 0 𝑍𝑦 0 0 0 𝑍𝑦 (20)  Equation (19) then becomes a set of three uncoupled equations 𝑽0 = 𝑍𝑦 + 3𝑍𝑛 𝑰0 = 𝑍0𝑰0 (21) 𝑽1 = 𝑍𝑦𝑰1 = 𝑍1𝑰1 (22) 𝑽2 = 𝑍𝑦𝑰2 = 𝑍2𝑰2 (23)
K. Webb ESE 470 48 Sequence Networks – Non-Rotating Loads  Equations (21) – (23) describe the uncoupled sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 49 Sequence Networks – Non-Rotating Loads  We can develop similar sequence networks for a balanced Δ- connected load  𝑍𝑦 = 𝑍Δ/3  There is no neutral point for the Δ-network, so 𝑍𝑛 = ∞ - an open circuit  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 Sequence Networks – 3-𝜙 Lines 50
K. Webb ESE 470 51 Sequence Networks – 3-𝜙 Lines  Balanced, three-phase lines can be modeled as  The voltage drops across the lines are given by the following system of equations 𝑽𝑎𝑎′ 𝑽𝑏𝑏′ 𝑽𝑐𝑐′ = 𝑍𝑎𝑎 𝑍𝑎𝑏 𝑍𝑎𝑐 𝑍𝑏𝑎 𝑍𝑏𝑏 𝑍𝑏𝑐 𝑍𝑐𝑎 𝑍𝑐𝑏 𝑍𝑐𝑐 𝑰𝑎 𝑰𝑏 𝑰𝑐 = 𝑽𝑎𝑛 − 𝑽𝑎′𝑛 𝑽𝑏𝑛 − 𝑽𝑏′𝑛 𝑽𝑐𝑛 − 𝑽𝑐′𝑛 (24)
K. Webb ESE 470 52 Sequence Networks – 3-𝜙 Lines  Writing (24) in compact form 𝑽𝑝 − 𝑽𝑝′ = 𝒁𝑝𝑰𝑝 (25)  𝒁𝑝 is the phase impedance matrix  Self impedances along the diagonal  Mutual impedances elsewhere  Symmetric  Diagonal, if we neglect mutual impedances
K. Webb ESE 470 53 Sequence Networks – 3-𝜙 Lines  We can rewrite (25) in terms of sequence components 𝑨𝑽𝑠 − 𝑨𝑽𝑠′ = 𝒁𝑝𝑨𝑰𝑠 𝑽𝑠 − 𝑽𝑠′ = 𝑨−1𝒁𝑝𝑨𝑰𝑠 𝑽𝑠 − 𝑽𝑠′ = 𝒁𝑠𝑰𝑠 (26) where 𝒁𝑠 is the sequence impedance matrix 𝒁𝑠 = 𝑨−1𝒁𝑝𝑨 (27)  𝒁𝑠 is diagonal so long as the system impedances are balanced, i.e.  Self impedances are equal: 𝑍𝑎𝑎 = 𝑍𝑏𝑏 = 𝑍𝑐𝑐  Mutual impedances are equal: 𝑍𝑎𝑏 = 𝑍𝑎𝑐 = 𝑍𝑏𝑐
K. Webb ESE 470 54 Sequence Networks – 3-𝜙 Lines  For balanced lines, 𝒁𝑠 is diagonal 𝒁𝑠 = 𝑍𝑎𝑎 + 2𝑍𝑎𝑏 0 0 0 𝑍𝑎𝑎 − 𝑍𝑎𝑏 0 0 0 𝑍𝑎𝑎 − 𝑍𝑎𝑏 = 𝑍0 0 0 0 𝑍1 0 0 0 𝑍2  Because 𝑍𝑠 is diagonal, (26) represents three uncoupled equations 𝑽0 − 𝑽0′ = 𝑍0𝑰0 (28) 𝑽1 − 𝑽1′ = 𝑍1𝑰1 (29) 𝑽2 − 𝑽2′ = 𝑍2𝑰2 (30)
K. Webb ESE 470 55 Sequence Networks – 3-𝜙 Lines  Equations (28) – (30) describe the voltage drop across three uncoupled sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 Sequence Networks –Rotating Machines 56
K. Webb ESE 470 57 Sequence Networks – Rotating Machines  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
K. Webb ESE 470 58 Sequence Networks – Synchronous Generator  Sequence networks for Y-connected synchronous generator  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 59 Sequence Networks – Motors  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
K. Webb ESE 470 Sequence Networks –Transformers 60
K. Webb ESE 470 61 Sequence Networks - Y-Y Transformers  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 3𝐼0𝑍𝑁 and 3𝐼0𝑍𝑛  3𝑍𝑁 and 3𝑍𝑛 each appear in the zero-sequence network  Can be combined in the per-unit circuit as long as shunt impedances are neglected
K. Webb ESE 470 62 Sequence Networks – Y-Y Transformers  Impedance accounting for leakage flux and winding resistance for each winding can be referred to the primary  Add together into a single impedance, 𝑍𝑠, in the per-unit model  Y-Y transformer sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 63 Sequence Networks – Y-ΔTransformers  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
K. Webb ESE 470 64 Sequence Networks – Y-ΔTransformers  Sequence networks for Y-ΔTransformers  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 65 Sequence Networks – Δ-ΔTransformers  Δ-Δ transformers  Like Y-Y transformers, no phase shift  No neutral connections  Zero-sequence current cannot flow into or out of either winding
K. Webb ESE 470 66 Sequence Networks – Δ-ΔTransformers  Sequence networks for Δ-ΔTransformers  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
K. Webb ESE 470 Power in Sequence Networks 67
K. Webb ESE 470 68 Power in Sequence Networks  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 three- phase system is the sum of the power at each phase 𝑆𝑝 = 𝑽𝑎𝑛𝑰𝑎 ∗ + 𝑽𝑏𝑛𝑰𝑏 ∗ + 𝑽𝑐𝑛𝑰𝑐 ∗  In matrix form, this looks like 𝑆𝑝 = 𝑽𝑎𝑛 𝑽𝑏𝑛 𝑽𝑐𝑛 𝑰𝑎 ∗ 𝑰𝑏 ∗ 𝑰𝑐 ∗ 𝑆𝑝 = 𝑽𝑝 𝑇 𝑰𝑝 ∗ (28)
K. Webb ESE 470 69 Power in Sequence Networks  Recall the following relationships 𝑽𝑝 = 𝑨𝑽𝑠 (9) 𝑰𝑝 = 𝑨𝑰𝑠 (13)  Using (9) and (13) in (28), we have 𝑺𝑝 = 𝑨𝑽𝑠 𝑇 𝑨𝑰𝑠 ∗ 𝑺𝑝 = 𝑽𝑠 𝑇 𝑨𝑇 𝑨∗ 𝑰𝑠 ∗ (29)  Computing the product in the middle of the right-hand side of (29), we find 𝑨𝑇𝑨∗ = 3 0 0 0 3 0 0 0 3 = 3𝑰3 where 𝑰3 is the 3×3 identity matrix
K. Webb ESE 470 70 Power in Sequence Networks  Equation (29) then becomes 𝑺𝑝 = 𝑽𝑠 𝑇3𝑰3𝑰𝑠 ∗ 𝑺𝑝 = 3𝑽𝑠 𝑇 𝑰𝑠 ∗ 𝑺𝑝 = 3 𝑽0 𝑽1 𝑽2 𝑰0 ∗ 𝑰1 ∗ 𝑰2 ∗ 𝑺𝑝 = 3 𝑽0𝑰0 ∗ + 𝑽1𝑰1 ∗ + 𝑽2𝑰2 ∗  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 𝑎
K. Webb ESE 470 Example Problems 71
K. Webb ESE 470 A bolted, symmetric, three-phase fault occurs 60% of the way from bus 1 to bus 2. Determine the subtransient fault current in per-unit and in amperes. The load is consuming rated power at rated voltage and unity power factor.
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K. Webb ESE 470 Determine the sequence components for the following unbalanced set of three-phase voltage phasors: 𝐕𝑎 = 1∠0° 𝑝. 𝑢. 𝐕𝑏 = 0.5∠ − 60° 𝑝. 𝑢. 𝐕𝑐 = 2∠200° 𝑝. 𝑢.
K. Webb ESE 470
K. Webb ESE 470 Determine the phase components for the following set of sequence components: 𝐕0 = 1∠60° 𝑝. 𝑢. 𝐕1 = 1∠0° 𝑝. 𝑢. 𝐕2 = 0 𝑝. 𝑢.
K. Webb ESE 470
K. Webb ESE 470 Unsymmetrical Faults 80
K. Webb ESE 470 81 Unsymmetrical Faults  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
K. Webb ESE 470 82 Unsymmetrical Fault Analysis - Procedure  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
K. Webb ESE 470 83 Unsymmetrical Fault Analysis  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 𝑉𝐹 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
K. Webb ESE 470 84 Unsymmetrical Fault Analysis  Each sequence network includes all interconnected power- system 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:
K. Webb ESE 470 85 Unsymmetrical Fault Analysis – Sequence Networks  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
K. Webb ESE 470 86 Unsymmetrical Fault Analysis – Sequence Networks  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
K. Webb ESE 470 87 Unsymmetrical Fault Analysis – Sequence Networks  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
K. Webb ESE 470 Single-Line-to-Ground Fault 88
K. Webb ESE 470 89 Unsymmetrical Fault Analysis – SLG Fault  The following represents a generic three-phase network with terminals at the fault location:  If we have a single-line-to- ground fault, where phase 𝑎 is shorted through 𝑍𝑓 to ground, the model becomes:
K. Webb ESE 470 90 Unsymmetrical Fault Analysis – SLG Fault  The phase-domain fault conditions: 𝑰𝑎 = 𝑽𝑎𝑔 𝑍𝑓 (1) 𝑰𝑏 = 𝑰𝑐 = 0 (2)  Transforming these phase-domain currents to the sequence domain 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑽𝑎𝑔/𝑍𝑓 0 0 = 1 3 𝑽𝑎𝑔/𝑍𝑓 𝑽𝑎𝑔/𝑍𝑓 𝑽𝑎𝑔/𝑍𝑓 (3)  This gives one of our sequence-domain fault conditions 𝑰0 = 𝑰1 = 𝑰2 (4)
K. Webb ESE 470 91 Unsymmetrical Fault Analysis – SLG Fault  We know that 𝑰𝑎 = 𝑽𝑎𝑔 𝑍𝑓 = 𝑰0 + 𝑰1 + 𝑰2 (5) and 𝑽𝑎𝑔 = 𝑽0 + 𝑽1 + 𝑽2 (6)  Using (5) and (6) in (1), we get 𝑰0 + 𝑰1 + 𝑰2 = 1 𝑍𝑓 𝑽0 + 𝑽1 + 𝑽2  Using (4), this gives our second sequence-domain fault condition 𝑰0 = 𝑰1 = 𝑰2 = 1 3𝑍𝑓 𝑽0 + 𝑽1 + 𝑽2 (7)
K. Webb ESE 470 92 Unsymmetrical Fault Analysis – SLG Fault  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, 𝑰𝐹 𝑰𝐹 = 𝑰𝑎 = 𝑰0 + 𝑰1 + 𝑰2 = 3𝑰1 𝑰1 = 𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 𝑰𝐹 = 3𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 (8)
K. Webb ESE 470 93 SLG Fault - Example  Returning to our example power system  The interconnected sequence networks for a bolted fault at bus 1:
K. Webb ESE 470 94 SLG Fault - Example  The fault current is 𝑰𝐹 = 3𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 𝑰𝐹 = 3.0∠30° 𝑗0.473 = 6.34∠ − 60° 𝑝. 𝑢.  The current base at bus 1 is 𝑰𝑏 = 𝑆𝑏 3𝑉𝑏1 = 150 𝑀𝑉𝐴 3 230 𝑘𝑉 = 376.5 𝐴  So the fault current in kA is 𝑰𝐹 = 6.34∠ − 60° 376.5 𝐴 𝑰𝐹 = 2.39∠ − 60° 𝑘𝐴
K. Webb ESE 470 Line-to-Line Fault 95
K. Webb ESE 470 96 Unsymmetrical Fault Analysis – LL Fault  Now consider a line-to-line fault between phase 𝑏 and phase 𝑐 through impedance 𝑍𝐹  Phase-domain fault conditions: 𝑰𝑎 = 0 (9) 𝑰𝑏 = −𝑰𝑐 = 𝑽𝑏𝑔−𝑽𝑐𝑔 𝑍𝐹 (10)  Transforming to the sequence domain 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 0 𝑰𝑏 −𝑰𝑏 = 1 3 0 𝑎 − 𝑎2 𝑰𝑏 𝑎2 − 𝑎 𝑰𝑏 (11)  So, the first two sequence-domain fault conditions are 𝑰0 = 0 (12) 𝑰2 = −𝑰1 (13)
K. Webb ESE 470 97 Unsymmetrical Fault Analysis – LL Fault  To derive the remaining sequence-domain fault condition, rearrange (10) and transform to the sequence domain 𝑽𝑏𝑔 − 𝑽𝑐𝑔 = 𝑰𝑏𝑍𝐹 𝑽0 + 𝑎2 𝑽1 + 𝑎𝑽2 − 𝑽0 + 𝑎𝑽1 + 𝑎2 𝑽2 = 𝑰0 + 𝑎2 𝑰1 + 𝑎𝑰2 𝑍𝐹 𝑎2 𝑽1 + 𝑎𝑽2 − 𝑎𝑽1 − 𝑎2 𝑽2 = 𝑎2 − 𝑎 𝑰1𝑍𝐹 𝑎2 − 𝑎 𝑽1 − 𝑎2 − 𝑎 𝑽2 = 𝑎2 − 𝑎 𝑰1𝑍𝐹  The last sequence-domain fault condition is 𝑽1 − 𝑽2 = 𝑰1𝑍𝐹 (14)
K. Webb ESE 470 98 Unsymmetrical Fault Analysis – LL Fault  Sequence-domain fault conditions 𝑰0 = 0 (12) 𝑰2 = −𝑰1 (13) 𝑽1 − 𝑽2 = 𝑰1𝑍𝐹 (14)  These can be satisfied by:  Leaving the zero-sequence network open  Connecting the terminals of the positive- and negative-sequence networks together through 𝑍𝐹
K. Webb ESE 470 99 Unsymmetrical Fault Analysis – LL Fault  The fault current is the phase 𝑏 current, which is given by 𝑰𝐹 = 𝑰𝑏 = 𝑰0 + 𝑎2 𝑰1 + 𝑎𝑰2 𝑰𝐹 = 𝑎2𝑰1 − 𝑎𝑰1 𝑰𝐹 = −𝑗 3 𝑰1 = −𝑗 3 𝑽𝐹 𝑍1+𝑍2+𝑍𝐹 𝑰𝐹 = 3 𝑽𝐹∠−90° 𝑍1+𝑍2+𝑍𝐹 (15)
K. Webb ESE 470 100 LL Fault - Example  Now consider the same system with a bolted line-to-line fault at bus 1  The sequence network:
K. Webb ESE 470 101 LL Fault - Example 𝑰1 = 1∠30° 𝑗0.349 = 2.87∠ − 60°  The subtransient fault current is given by (15) as 𝑰𝐹 = 3∠ − 90° 2.87∠ − 60° 𝑰𝐹 = 4.96∠ − 150° 𝑝. 𝑢.  Using the previously-determined current base, we can convert the fault current to kA 𝑰𝐹 = 𝐼𝑏1 ⋅ 4.96∠ − 150° 𝑰𝐹 = 4.96∠ − 150° 376.5𝐴 𝑰𝐹 = 1.87∠ − 150° 𝑘𝐴
K. Webb ESE 470 Double-Line-to-Ground Fault 102
K. Webb ESE 470 103 Unsymmetrical Fault Analysis – DLG Fault  Now consider a double line-to-ground fault  Assume phases 𝑏 and 𝑐 are shorted to ground through 𝑍𝐹  Phase-domain fault conditions: 𝑰𝑎 = 0 (16) 𝑰𝑏 + 𝑰𝑐 = 𝑽𝑏𝑔 𝑍𝐹 = 𝑽𝑐𝑔 𝑍𝐹 (17)  It can be shown that (16) and (17) transform to the following sequence-domain fault conditions (analysis skipped here) 𝑰0 + 𝑰1 + 𝑰2 = 0 (18) 𝑽1 = 𝑽2 (19) 𝑰0 = 1 3𝑍𝐹 𝑽0 − 𝑽1 (20)
K. Webb ESE 470 104 Unsymmetrical Fault Analysis – DLG Fault  Sequence-domain fault conditions 𝑰0 + 𝑰1 + 𝑰2 = 0 (18) 𝑽1 = 𝑽2 (19) 𝑰0 = 1 3𝑍𝐹 𝑽0 − 𝑽1 (20)  To satisfy these fault conditions  Connect the positive- and negative-sequence networks together directly  Connect the zero- and positive-sequence networks together through 3𝑍𝐹
K. Webb ESE 470 105 Unsymmetrical Fault Analysis – DLG Fault  The fault current is the sum of the phase 𝑏 and phase 𝑐 currents, as given by (17)  In the sequence domain the fault current is 𝑰𝐹 = 𝑰𝑏 + 𝑰𝑐 = 3𝑰0 𝑰𝐹 = 3𝑰0 (21)  𝐼0 can be determined by a simple analysis (e.g. nodal) of the interconnected sequence networks
K. Webb ESE 470 106 DLG Fault - Example  Now determine the subtransient fault current for a bolted double line-to-ground fault at bus 1  The sequence network:  Here, because 𝑍𝐹 = 0, 𝑽0 = 𝑽1 = 𝑽2
K. Webb ESE 470 107 DLG Fault - Example  To find 𝑰𝐹, we must determine 𝑰0  We can first find 𝑉0 by applying voltage division 𝑽0 = 𝑽𝐹 𝑍2||𝑍0 𝑍1 + 𝑍2||𝑍0 𝑽0 = 1.0∠30° 𝑗0.18||𝑗0.124 𝑗0.169 + 𝑗0.18||𝑗0.124 𝑽0 = 0.303∠30°
K. Webb ESE 470 108 DLG Fault - Example  Next, calculate 𝑰0 𝑰0 = −𝑽0 𝑍0 = −0.303∠30° 𝑗0.124 = 2.44∠120° 𝑝. 𝑢.  The per-unit fault current is 𝑰𝐹 = 3𝑰0 = 7.33∠120° 𝑝. 𝑢.  Using the current base to convert to kA, gives the subtransient DLG fault current 𝑰𝐹 = 7.33∠120° 376.5 𝐴 𝑰𝐹 = 2.76∠120° 𝑘𝐴
K. Webb ESE 470 Example Problems 109
K. Webb ESE 470 Draw the sequence networks for the following power system. Assume the generator is operating at rated voltage.
K. Webb ESE 470
K. Webb ESE 470 Reduce the sequence networks to their Thévenin equivalents for a fault occurring half of the way along the transmission line.
K. Webb ESE 470
K. Webb ESE 470
K. Webb ESE 470 Determine the subtransient fault current resulting from a DLG fault, half way along the transmission line, through an impedance of j0.2 p.u.
K. Webb ESE 470

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Section 7 Fault Analysis.pdf

  • 1.
    ESE 470 –Energy Distribution Systems SECTION 7: FAULT ANALYSIS
  • 2.
    K. Webb ESE470 Introduction 2
  • 3.
    K. Webb ESE470 3 Power System Faults  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
  • 4.
    K. Webb ESE470 4 Types of Faults  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.
    K. Webb ESE470 Subtransient Fault Current 5
  • 6.
    K. Webb ESE470 6 Fault Current  R: generator resistance  L: generator inductance  𝑖 𝑡 = 0 for 𝑡 < 0  Source phase, 𝛼, determines voltage at 𝑡 = 0  Short circuit can occur at any point in a 60 Hz cycle  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 𝑡 = 0  Consider an unloaded synchronous generator  Equivalent circuit model: 𝑣 𝑡 = 2𝑉𝐺 sin 𝜔𝑡 + 𝛼
  • 7.
    K. Webb ESE470 7 Fault Current  The governing differential equation for 𝑡 > 0 is 𝑑𝑖 𝑑𝑡 + 𝑖 𝑡 𝑅 𝐿 = 2𝑉𝐺 𝐿 sin 𝜔𝑡 + 𝛼  The solution gives the fault current 𝑖 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 + 𝛼 − 𝜃 − sin 𝛼 − 𝜃 𝑒−𝑡 𝑅 𝐿 where 𝑍 = 𝑅2 + 𝜔𝐿 2 and 𝜃 = tan−1 𝜔𝐿 𝑅  This total fault current is referred to as the asymmetrical fault current  It has a steady-state component 𝑖𝑎𝑐 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 + 𝛼 − 𝜃  And a transient component 𝑖𝑑𝑐 𝑡 = − 2𝑉𝐺 𝑍 sin 𝛼 − 𝜃 𝑒−𝑡 𝑅 𝐿
  • 8.
    K. Webb ESE470 8 Fault Current  Magnitude of the transient fault current, 𝑖𝑑𝑐, depends on 𝛼  𝑖𝑑𝑐 0 = 0 for 𝛼 = 𝜃  𝑖𝑑𝑐 0 = 2𝐼𝑎𝑐 for 𝛼 = 𝜃 − 90°  𝐼𝑎𝑐 = 𝑉𝐺/𝑍 is the rms value of the steady-state fault current  Worst-case fault current occurs for 𝛼 = 𝜃 − 90° 𝑖 𝑡 = 2𝑉𝐺 𝑍 sin 𝜔𝑡 − 𝜋 2 + 𝑒−𝑡 𝑅 𝐿
  • 9.
    K. Webb ESE470 9 Fault Current  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  In this class, we will use the steady-state current component, 𝐈𝑎𝑐, as our primary fault current metric
  • 10.
    K. Webb ESE470 10 Generator Reactance  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  𝑋𝑑 ′′ : subtransient reactance  𝑋𝑑 ′ : transient reactance  𝑋𝑑: synchronous reactance  Reactance increases with time, such that 𝑋𝑑 ′′ < 𝑋𝑑 ′ < 𝑋𝑑
  • 11.
    K. Webb ESE470 11 Sub-Transient Fault Current  Transition rates between reactance values are dictated by two time constants:  𝜏𝑑 ′′ : short-circuit subtransient time constant  𝜏𝑑 ′ : short-circuit transient time constant  Neglecting generator resistance, i.e. assuming 𝜃 = 90°, the synchronous portion of the fault current is 𝑖𝑎𝑐 𝑡 = 2𝑉𝐺 1 𝑋𝑑 ′′ − 1 𝑋𝑑 ′ 𝑒 − 𝑡 𝜏𝑑 ′′ + 1 𝑋𝑑 ′ − 1 𝑋𝑑 𝑒 − 𝑡 𝜏𝑑 ′ + 1 𝑋𝑑 sin 𝜔𝑡 + 𝛼 − 𝜋 2  At the instant of the fault, 𝑡 = 0, the rms synchronous fault current is 𝐼𝐹 ′′ = 𝑉𝐺 𝑋𝑑 ′′  This is the rms subtransient fault current, 𝐼𝐹 ′′  This will be our primary metric for assessing fault current
  • 12.
    K. Webb ESE470 Symmetrical Three-Phase Short Circuits 12
  • 13.
    K. Webb ESE470 13 Symmetrical 3-𝜙 Short Circuits  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-𝑌 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
  • 14.
    K. Webb ESE470 14 Symmetrical 3-𝜙 Short Circuits  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 𝑡 = 0
  • 15.
    K. Webb ESE470 15 Symmetrical 3-𝜙 Short Circuits  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 𝑽𝐹
  • 16.
    K. Webb ESE470 16 Symmetrical 3-𝜙 Short Circuits  Applying superposition, we can represent this circuit as the sum of two separate circuits: Circuit 1 Circuit 2
  • 17.
    K. Webb ESE470 17 Symmetrical 3-𝜙 Short Circuits  Assume that the value of the fault-location source, 𝑽𝐹, is the pre-fault voltage at that location  Circuit 1, then, represents the pre-fault circuit, so 𝑰𝐹1 ′′ = 0  The 𝑽𝐹 source can therefore be removed from circuit 1 Circuit 1 Circuit 2
  • 18.
    K. Webb ESE470 18 Symmetrical 3-𝜙 Short Circuits  The current in circuit 1, 𝑰𝐿, is the pre-fault line current  Superposition gives the fault current 𝑰𝐹 ′′ = 𝑰𝐹1 ′′ + 𝑰𝐹2 ′′ = 𝑰𝐹2 ′′  The generator fault current is 𝑰𝐺 ′′ = 𝑰𝐺1 ′′ + 𝑰𝐺2 ′′ 𝑰𝐺 ′′ = 𝑰𝐿 + 𝑰𝐺2 ′′  The motor fault current is 𝑰𝑴 ′′ = 𝑰𝑀1 ′′ + 𝑰𝑀2 ′′ 𝑰𝑀 ′′ = −𝑰𝐿 + 𝑰𝑀2 ′′
  • 19.
    K. Webb ESE470 19 Symmetrical 3-𝜙 Fault – Example  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
  • 20.
    K. Webb ESE470 20 Symmetrical 3-𝜙 Fault – Example  First convert to per-unit  Use 𝑆𝑏 = 100 𝑀𝑉𝐴  Base voltage in the transmission line zone is 𝑉𝑏,𝑡𝑙 = 138 𝑘𝑉  Base impedance in the transmission line zone is 𝑍𝑏,𝑡𝑙 = 𝑉𝑏,𝑡𝑙 2 𝑆𝑏 = 138 𝑘𝑉 2 100 𝑀𝑉𝐴 = 190.4 Ω  The per-unit transmission line reactance is 𝑋𝑡𝑙 = 20 Ω 190.4 Ω = 0.105 𝑝. 𝑢.
  • 21.
    K. Webb ESE470 21 Symmetrical 3-𝜙 Fault – Example  The two per-unit circuits are  These can be simplified by combining impedances
  • 22.
    K. Webb ESE470 22 Symmetrical 3-𝜙 Fault – Example  Using circuit 2, we can calculate the subtransient fault current 𝑰𝐹 ′′ = 1.05∠0° 𝑗0.116 = 9.079∠ − 90° 𝑝. 𝑢.  To convert to kA, first determine the current base in the generator zone 𝐼𝑏,𝐺 = 𝑆𝑏 3𝑉𝑏,𝐺 = 100 𝑀𝑉𝐴 3 ⋅ 13.8 𝑘𝑉 = 4.18 𝑘𝐴  The subtransient fault current is 𝑰𝐹 ′′ = 9.079∠ − 90° ⋅ 4.18 𝑘𝐴 𝑰𝐹 ′′ = 37.98∠ − 90° 𝑘𝐴
  • 23.
    K. Webb ESE470 23 Symmetrical 3-𝜙 Fault – Example  The pre-fault line current can be calculated from the pre-fault generator voltage and power 𝑰𝐿 = ൗ 𝑺𝐺 3 ൘ 𝑽𝐺 ′′ 3 ∗ = 𝑺𝐺 3𝑽𝐺 ′′ ∗ = 100∠ cos−1 0.95 𝑀𝑉𝐴 ∗ 3 ⋅ 1.05 ⋅ 13.8∠0° 𝑘𝑉 ∗ 𝑰𝐿 = 100∠ − 18.19° 𝑀𝑉𝐴 3 ⋅ 1.05 ⋅ 13.8∠0° 𝑘𝑉 𝑰𝐿 = 3.98∠ − 18.19° 𝑘𝐴  Or, in per-unit: 𝑰𝐿 = 3.98∠ − 18.19° 𝑘𝐴 4.18 kA = 0.952∠ − 18.19° 𝑝. 𝑢.  This will be used to find the generator and motor fault currents
  • 24.
    K. Webb ESE470 24 Symmetrical 3-𝜙 Fault – Example  The generator’s contribution to the fault current is found by applying current division 𝑰𝐺2 ′′ = 𝑰𝐹 ′′ 0.505 0.505 + 0.15 = 7.0∠ − 90° 𝑝. 𝑢.  Adding the pre-fault line current, we have the subtransient generator fault current 𝑰𝐺 ′′ = 𝑰𝐿 + 𝑰𝐺2 ′′ 𝑰𝐺 ′′ = 0.952∠ − 18.19° + 7.0∠ − 90° 𝑰𝐺 ′′ = 7.35∠ − 82.9° 𝑝. 𝑢.  Converting to kA 𝑰𝐺 ′′ = 7.35∠ − 82.9° ⋅ 4.18 𝑘𝐴 𝑰𝐺 ′′ = 30.74∠ − 82.9° 𝑘𝐴
  • 25.
    K. Webb ESE470 25 Symmetrical 3-𝜙 Fault – Example  Similarly, for the motor 𝑰𝑀2 ′′ = 𝑰𝐹 ′′ 0.15 0.505 + 0.15 = 2.08∠ − 90° 𝑝. 𝑢.  Subtracting the pre-fault line current gives the subtransient motor fault current 𝑰𝑀 ′′ = −𝑰𝐿 + 𝑰𝑀2 ′′ 𝑰𝑀 ′′ = −0.952∠ − 18.19° + 2.08∠ − 90° 𝑰𝑀 ′′ = 2.0∠ − 116.9°  Converting to 𝑘𝐴 𝑰𝑀 ′′ = 2.0∠ − 116.9° ⋅ 4.18 𝑘𝐴 𝑰𝑀 ′′ = 8.36∠ − 116.9° 𝑘𝐴
  • 26.
    K. Webb ESE470 Symmetrical Components 26
  • 27.
    K. Webb ESE470 27 Symmetrical Components  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
  • 28.
    K. Webb ESE470 28 Symmetrical Components  The method of symmetrical components:  Represent an asymmetrical set of 𝑁 phasors as a sum of 𝑁 sets of symmetrical component phasors  These 𝑁 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 (𝑁 = 3), sequence components are:  Zero sequence components  Positive sequence components  Negative sequence components
  • 29.
    K. Webb ESE470 29 Sequence Components  Zero sequence components  Three phasors with equal magnitude and equal phase  𝑽𝑎0, 𝑽𝑏0, 𝑽𝑐0  Positive sequence components  Three phasors with equal magnitude and ± 120°, positive-sequence phase  𝑽𝑎1, 𝑽𝑏1, 𝑽𝑐1  Negative sequence components  Three phasors with equal magnitude and ± 120°, negative-sequence phase  𝑽𝑎2, 𝑽𝑏2, 𝑽𝑐2
  • 30.
    K. Webb ESE470 30 Sequence Components  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 𝑽𝑎 𝑽𝑏 𝑽𝑐 = 𝑽𝑎0 𝑽𝑏0 𝑽𝑐0 + 𝑽𝑎1 𝑽𝑏1 𝑽𝑐1 + 𝑽𝑎2 𝑽𝑏2 𝑽𝑐2 (1)
  • 31.
    K. Webb ESE470 31 Sequence Components  The phasors of each sequence component have a fixed phase relationship  If we know one, we know the other two  Assume we know phase 𝑎 – use that as the reference  For the zero sequence components, we have 𝑽0 = 𝑽𝑎0 = 𝑽𝑏0 = 𝑽𝑐0 (2)  For the positive sequence components, 𝑽1 = 𝑽𝑎1 = 1∠120° ⋅ 𝑽𝑏1 = 1∠240° ⋅ 𝑽𝑐1 (3)  And, for the negative sequence components, 𝑽2 = 𝑽𝑎2 = 1∠240° ⋅ 𝑽𝑏2 = 1∠120° ⋅ 𝑽𝑐2 (4)  Note that we’re using phase 𝑎 as our reference, so 𝑽0 = 𝑽𝑎0, 𝑽1 = 𝑽𝑎1, 𝑽2 = 𝑽𝑎2
  • 32.
    K. Webb ESE470 32 Sequence Components  Next, we define a complex number, 𝑎, that has unit magnitude and phase of 120° 𝑎 = 1∠120° (5)  Multiplication by 𝑎 results in a rotation (a phase shift) of 120°  Multiplication by 𝑎2 yields a rotation of 240° = −120°  Using (5) to rewrite (3) and (4) 𝑽1 = 𝑽𝑎1 = 𝑎𝑽𝑏1 = 𝑎2𝑽𝑐1 (6) 𝑽2 = 𝑽𝑎2 = 𝑎2 𝑽𝑏2 = 𝑎𝑽𝑐2 (7)
  • 33.
    K. Webb ESE470 33 Sequence Components  Using (2), (6), and (7), we can rewrite (1) in a simplified form 𝑽𝑎 𝑽𝑏 𝑽𝑐 = 1 1 1 1 𝑎2 𝑎 1 𝑎 𝑎2 𝑽0 𝑽1 𝑽2 (8)  The vector on the left is the vector of phase voltages, 𝑽𝑝  The vector on the right is the vector of (phase 𝑎) sequence components, 𝑽𝑠  We’ll call the 3 × 3 transformation matrix 𝑨  We can rewrite (8) as 𝑽𝑝 = 𝑨𝑽𝑠 (9)
  • 34.
    K. Webb ESE470 34 Sequence Components  We can express the sequence voltages as a function of the phase voltages by inverting the transformation matrix 𝑽𝑠 = 𝑨−1 𝑽𝑝 (10) where 𝑨−1 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 (11) So 𝑽0 𝑽1 𝑽2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑽𝑎 𝑽𝑏 𝑽𝑐 (12)
  • 35.
    K. Webb ESE470 35 Sequence Components  The same relationships hold for three-phase currents  The phase currents are 𝑰𝑝 = 𝑰𝑎 𝑰𝑏 𝑰𝑐  And, the sequence currents are 𝑰𝑠 = 𝑰0 𝑰1 𝑰2
  • 36.
    K. Webb ESE470 36 Sequence Components  The transformation matrix, 𝑨, relates the phase currents to the sequence currents 𝑰𝑝 = 𝑨𝑰𝑠 𝑰𝑎 𝑰𝑏 𝑰𝑐 = 1 1 1 1 𝑎2 𝑎 1 𝑎 𝑎2 𝑰0 𝑰1 𝑰2 (13)  And vice versa 𝑰𝑠 = 𝑨−1𝑰𝑝 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑰𝑎 𝑰𝑏 𝑰𝑐 (14)
  • 37.
    K. Webb ESE470 37 Sequence Components – Balanced System  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 𝑽𝑏 = 𝑽𝑎 ⋅ 1∠ − 120° = 𝑎2 𝑽𝑎 𝑽𝑐 = 𝑽𝑎 ⋅ 1∠120° = 𝑎𝑽𝑎  The sequence voltages are given by (12)  The zero sequence voltage is 𝑽0 = 1 3 𝑽𝑎 + 𝑽𝑏 + 𝑽𝑐 = 1 3 𝑽𝑎 + 𝑎2 𝑽𝑎 + 𝑎𝑽𝑎 𝑽0 = 1 3 𝑽𝑎 1 + 𝑎2 + 𝑎  Applying the identity 1 + 𝑎2 + 𝑎 = 0, we have 𝑽0 = 0
  • 38.
    K. Webb ESE470 38 Sequence Components – Balanced System  The positive sequence component is given by 𝑽1 = 1 3 𝑽𝑎 + 𝑎𝑽𝑏 + 𝑎2𝑽𝑐 𝑽1 = 1 3 𝑽𝑎 + 𝑎 ⋅ 𝑎2𝑽𝑎 + 𝑎2 ⋅ 𝑎𝑽𝑎 𝑽1 = 1 3 𝑽𝑎 + 𝑎3𝑽𝑎 + 𝑎3𝑽𝑎  Since 𝑎3 = 1∠0°, we have 𝑽1 = 1 3 3𝑽𝑎 𝑽1 = 𝑽𝑎
  • 39.
    K. Webb ESE470 39 Sequence Components – Balanced System  The negative sequence component is given by 𝑽2 = 1 3 𝑽𝑎 + 𝑎2𝑽𝑏 + 𝑎𝑽𝑐 𝑽2 = 1 3 𝑽𝑎 + 𝑎2 ⋅ 𝑎2𝑽𝑎 + 𝑎 ⋅ 𝑎𝑽𝑎 𝑽2 = 1 3 𝑽𝑎 + 𝑎4 𝑽𝑎 + 𝑎2 𝑽𝑎  Again, using the identity 1 + 𝑎2 + 𝑎 = 0, along with the fact that 𝑎4 = 𝑎, we have 𝑽2 = 0
  • 40.
    K. Webb ESE470 40 Sequence Components – Balanced System  So, for a positive-sequence, balanced, three-phase system, the sequence voltages are 𝑽0 = 0, 𝑽1 = 𝑽𝑎, 𝑽2 = 0  Similarly, the sequence currents are 𝑰0 = 0, 𝑰1 = 𝑰𝑎, 𝑰2 = 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
  • 41.
    K. Webb ESE470 41 Sequence Components  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.
    K. Webb ESE470 Sequence Networks 42
  • 43.
    K. Webb ESE470 43 Sequence Networks  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
  • 44.
    K. Webb ESE470 Sequence Networks – Non-Rotating Loads 44
  • 45.
    K. Webb ESE470 45 Sequence Networks – Non-Rotating Loads  Consider a balanced Y-load with the neutral grounded through some non-zero impedance  Applying KVL gives the phase-𝑎-to-ground voltage 𝑽𝑎𝑔 = 𝑍𝑦𝑰𝑎 + 𝑍𝑛𝑰𝑛 𝑽𝑎𝑔 = 𝑍𝑦𝑰𝑎 + 𝑍𝑛 𝑰𝑎 + 𝑰𝑏 + 𝑰𝑐 𝑽𝑎𝑔 = 𝑍𝑦 + 𝑍𝑛 𝑰𝑎 + 𝑍𝑛𝑰𝑏 + 𝑍𝑛𝑰𝑐 (15)  For phase 𝑏: 𝑽𝑏𝑔 = 𝑍𝑦𝑰𝑏 + 𝑍𝑛𝑰𝑛 = 𝑍𝑦𝑰𝑏 + 𝑍𝑛 𝑰𝑎 + 𝑰𝑏 + 𝑰𝑐 𝑽𝑏𝑔 = 𝑍𝑛𝑰𝑎 + 𝑍𝑦 + 𝑍𝑛 𝑰𝑏 + 𝑍𝑛𝑰𝑐 (16)  Similarly, for phase 𝑐: 𝑽𝑎𝑔 = 𝑍𝑛𝑰𝑎 + 𝑍𝑛𝑰𝑏 + 𝑍𝑦 + 𝑍𝑛 𝑰𝑐 (17)
  • 46.
    K. Webb ESE470 46 Sequence Networks – Non-Rotating Loads  Putting (15) – (17) in matrix form 𝑽𝑎𝑔 𝑽𝑏𝑔 𝑽𝑐𝑔 = 𝑍𝑦 + 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑦 + 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑛 𝑍𝑦 + 𝑍𝑛 𝑰𝑎 𝑰𝑏 𝑰𝑐 or 𝑽𝑝 = 𝒁𝑝𝑰𝑝 (18) where 𝑽𝑝 and 𝑰𝑝 are the phase voltages and currents, respectively, and 𝒁𝑝 is the phase impedance matrix  We can use (9) and (13) to rewrite (18) as 𝑨𝑽𝑠 = 𝒁𝑝𝑨𝑰𝑠  Solving for 𝑽𝑠 𝑽𝑠 = 𝑨−1𝒁𝑝𝑨𝑰𝑠 or 𝑽𝑠 = 𝒁𝑠𝑰𝑠 (19)
  • 47.
    K. Webb ESE470 47 Sequence Networks – Non-Rotating Loads 𝑽𝑠 = 𝒁𝑠𝑰𝑠 (19) where 𝑍𝑠 is the sequence impedance matrix 𝒁𝑠 = 𝑨−1𝒁𝑝𝑨 = 𝑍𝑦 + 3𝑍𝑛 0 0 0 𝑍𝑦 0 0 0 𝑍𝑦 (20)  Equation (19) then becomes a set of three uncoupled equations 𝑽0 = 𝑍𝑦 + 3𝑍𝑛 𝑰0 = 𝑍0𝑰0 (21) 𝑽1 = 𝑍𝑦𝑰1 = 𝑍1𝑰1 (22) 𝑽2 = 𝑍𝑦𝑰2 = 𝑍2𝑰2 (23)
  • 48.
    K. Webb ESE470 48 Sequence Networks – Non-Rotating Loads  Equations (21) – (23) describe the uncoupled sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 49.
    K. Webb ESE470 49 Sequence Networks – Non-Rotating Loads  We can develop similar sequence networks for a balanced Δ- connected load  𝑍𝑦 = 𝑍Δ/3  There is no neutral point for the Δ-network, so 𝑍𝑛 = ∞ - an open circuit  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 50.
    K. Webb ESE470 Sequence Networks – 3-𝜙 Lines 50
  • 51.
    K. Webb ESE470 51 Sequence Networks – 3-𝜙 Lines  Balanced, three-phase lines can be modeled as  The voltage drops across the lines are given by the following system of equations 𝑽𝑎𝑎′ 𝑽𝑏𝑏′ 𝑽𝑐𝑐′ = 𝑍𝑎𝑎 𝑍𝑎𝑏 𝑍𝑎𝑐 𝑍𝑏𝑎 𝑍𝑏𝑏 𝑍𝑏𝑐 𝑍𝑐𝑎 𝑍𝑐𝑏 𝑍𝑐𝑐 𝑰𝑎 𝑰𝑏 𝑰𝑐 = 𝑽𝑎𝑛 − 𝑽𝑎′𝑛 𝑽𝑏𝑛 − 𝑽𝑏′𝑛 𝑽𝑐𝑛 − 𝑽𝑐′𝑛 (24)
  • 52.
    K. Webb ESE470 52 Sequence Networks – 3-𝜙 Lines  Writing (24) in compact form 𝑽𝑝 − 𝑽𝑝′ = 𝒁𝑝𝑰𝑝 (25)  𝒁𝑝 is the phase impedance matrix  Self impedances along the diagonal  Mutual impedances elsewhere  Symmetric  Diagonal, if we neglect mutual impedances
  • 53.
    K. Webb ESE470 53 Sequence Networks – 3-𝜙 Lines  We can rewrite (25) in terms of sequence components 𝑨𝑽𝑠 − 𝑨𝑽𝑠′ = 𝒁𝑝𝑨𝑰𝑠 𝑽𝑠 − 𝑽𝑠′ = 𝑨−1𝒁𝑝𝑨𝑰𝑠 𝑽𝑠 − 𝑽𝑠′ = 𝒁𝑠𝑰𝑠 (26) where 𝒁𝑠 is the sequence impedance matrix 𝒁𝑠 = 𝑨−1𝒁𝑝𝑨 (27)  𝒁𝑠 is diagonal so long as the system impedances are balanced, i.e.  Self impedances are equal: 𝑍𝑎𝑎 = 𝑍𝑏𝑏 = 𝑍𝑐𝑐  Mutual impedances are equal: 𝑍𝑎𝑏 = 𝑍𝑎𝑐 = 𝑍𝑏𝑐
  • 54.
    K. Webb ESE470 54 Sequence Networks – 3-𝜙 Lines  For balanced lines, 𝒁𝑠 is diagonal 𝒁𝑠 = 𝑍𝑎𝑎 + 2𝑍𝑎𝑏 0 0 0 𝑍𝑎𝑎 − 𝑍𝑎𝑏 0 0 0 𝑍𝑎𝑎 − 𝑍𝑎𝑏 = 𝑍0 0 0 0 𝑍1 0 0 0 𝑍2  Because 𝑍𝑠 is diagonal, (26) represents three uncoupled equations 𝑽0 − 𝑽0′ = 𝑍0𝑰0 (28) 𝑽1 − 𝑽1′ = 𝑍1𝑰1 (29) 𝑽2 − 𝑽2′ = 𝑍2𝑰2 (30)
  • 55.
    K. Webb ESE470 55 Sequence Networks – 3-𝜙 Lines  Equations (28) – (30) describe the voltage drop across three uncoupled sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 56.
    K. Webb ESE470 Sequence Networks –Rotating Machines 56
  • 57.
    K. Webb ESE470 57 Sequence Networks – Rotating Machines  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
  • 58.
    K. Webb ESE470 58 Sequence Networks – Synchronous Generator  Sequence networks for Y-connected synchronous generator  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 59.
    K. Webb ESE470 59 Sequence Networks – Motors  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
  • 60.
    K. Webb ESE470 Sequence Networks –Transformers 60
  • 61.
    K. Webb ESE470 61 Sequence Networks - Y-Y Transformers  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 3𝐼0𝑍𝑁 and 3𝐼0𝑍𝑛  3𝑍𝑁 and 3𝑍𝑛 each appear in the zero-sequence network  Can be combined in the per-unit circuit as long as shunt impedances are neglected
  • 62.
    K. Webb ESE470 62 Sequence Networks – Y-Y Transformers  Impedance accounting for leakage flux and winding resistance for each winding can be referred to the primary  Add together into a single impedance, 𝑍𝑠, in the per-unit model  Y-Y transformer sequence networks  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 63.
    K. Webb ESE470 63 Sequence Networks – Y-ΔTransformers  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
  • 64.
    K. Webb ESE470 64 Sequence Networks – Y-ΔTransformers  Sequence networks for Y-ΔTransformers  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 65.
    K. Webb ESE470 65 Sequence Networks – Δ-ΔTransformers  Δ-Δ transformers  Like Y-Y transformers, no phase shift  No neutral connections  Zero-sequence current cannot flow into or out of either winding
  • 66.
    K. Webb ESE470 66 Sequence Networks – Δ-ΔTransformers  Sequence networks for Δ-ΔTransformers  Positive-sequence network:  Negative-sequence network:  Zero-sequence network:
  • 67.
    K. Webb ESE470 Power in Sequence Networks 67
  • 68.
    K. Webb ESE470 68 Power in Sequence Networks  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 three- phase system is the sum of the power at each phase 𝑆𝑝 = 𝑽𝑎𝑛𝑰𝑎 ∗ + 𝑽𝑏𝑛𝑰𝑏 ∗ + 𝑽𝑐𝑛𝑰𝑐 ∗  In matrix form, this looks like 𝑆𝑝 = 𝑽𝑎𝑛 𝑽𝑏𝑛 𝑽𝑐𝑛 𝑰𝑎 ∗ 𝑰𝑏 ∗ 𝑰𝑐 ∗ 𝑆𝑝 = 𝑽𝑝 𝑇 𝑰𝑝 ∗ (28)
  • 69.
    K. Webb ESE470 69 Power in Sequence Networks  Recall the following relationships 𝑽𝑝 = 𝑨𝑽𝑠 (9) 𝑰𝑝 = 𝑨𝑰𝑠 (13)  Using (9) and (13) in (28), we have 𝑺𝑝 = 𝑨𝑽𝑠 𝑇 𝑨𝑰𝑠 ∗ 𝑺𝑝 = 𝑽𝑠 𝑇 𝑨𝑇 𝑨∗ 𝑰𝑠 ∗ (29)  Computing the product in the middle of the right-hand side of (29), we find 𝑨𝑇𝑨∗ = 3 0 0 0 3 0 0 0 3 = 3𝑰3 where 𝑰3 is the 3×3 identity matrix
  • 70.
    K. Webb ESE470 70 Power in Sequence Networks  Equation (29) then becomes 𝑺𝑝 = 𝑽𝑠 𝑇3𝑰3𝑰𝑠 ∗ 𝑺𝑝 = 3𝑽𝑠 𝑇 𝑰𝑠 ∗ 𝑺𝑝 = 3 𝑽0 𝑽1 𝑽2 𝑰0 ∗ 𝑰1 ∗ 𝑰2 ∗ 𝑺𝑝 = 3 𝑽0𝑰0 ∗ + 𝑽1𝑰1 ∗ + 𝑽2𝑰2 ∗  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 𝑎
  • 71.
    K. Webb ESE470 Example Problems 71
  • 72.
    K. Webb ESE470 A bolted, symmetric, three-phase fault occurs 60% of the way from bus 1 to bus 2. Determine the subtransient fault current in per-unit and in amperes. The load is consuming rated power at rated voltage and unity power factor.
  • 73.
  • 74.
  • 75.
  • 76.
    K. Webb ESE470 Determine the sequence components for the following unbalanced set of three-phase voltage phasors: 𝐕𝑎 = 1∠0° 𝑝. 𝑢. 𝐕𝑏 = 0.5∠ − 60° 𝑝. 𝑢. 𝐕𝑐 = 2∠200° 𝑝. 𝑢.
  • 77.
  • 78.
    K. Webb ESE470 Determine the phase components for the following set of sequence components: 𝐕0 = 1∠60° 𝑝. 𝑢. 𝐕1 = 1∠0° 𝑝. 𝑢. 𝐕2 = 0 𝑝. 𝑢.
  • 79.
  • 80.
    K. Webb ESE470 Unsymmetrical Faults 80
  • 81.
    K. Webb ESE470 81 Unsymmetrical Faults  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
  • 82.
    K. Webb ESE470 82 Unsymmetrical Fault Analysis - Procedure  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
  • 83.
    K. Webb ESE470 83 Unsymmetrical Fault Analysis  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 𝑉𝐹 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
  • 84.
    K. Webb ESE470 84 Unsymmetrical Fault Analysis  Each sequence network includes all interconnected power- system 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:
  • 85.
    K. Webb ESE470 85 Unsymmetrical Fault Analysis – Sequence Networks  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
  • 86.
    K. Webb ESE470 86 Unsymmetrical Fault Analysis – Sequence Networks  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
  • 87.
    K. Webb ESE470 87 Unsymmetrical Fault Analysis – Sequence Networks  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
  • 88.
    K. Webb ESE470 Single-Line-to-Ground Fault 88
  • 89.
    K. Webb ESE470 89 Unsymmetrical Fault Analysis – SLG Fault  The following represents a generic three-phase network with terminals at the fault location:  If we have a single-line-to- ground fault, where phase 𝑎 is shorted through 𝑍𝑓 to ground, the model becomes:
  • 90.
    K. Webb ESE470 90 Unsymmetrical Fault Analysis – SLG Fault  The phase-domain fault conditions: 𝑰𝑎 = 𝑽𝑎𝑔 𝑍𝑓 (1) 𝑰𝑏 = 𝑰𝑐 = 0 (2)  Transforming these phase-domain currents to the sequence domain 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 𝑽𝑎𝑔/𝑍𝑓 0 0 = 1 3 𝑽𝑎𝑔/𝑍𝑓 𝑽𝑎𝑔/𝑍𝑓 𝑽𝑎𝑔/𝑍𝑓 (3)  This gives one of our sequence-domain fault conditions 𝑰0 = 𝑰1 = 𝑰2 (4)
  • 91.
    K. Webb ESE470 91 Unsymmetrical Fault Analysis – SLG Fault  We know that 𝑰𝑎 = 𝑽𝑎𝑔 𝑍𝑓 = 𝑰0 + 𝑰1 + 𝑰2 (5) and 𝑽𝑎𝑔 = 𝑽0 + 𝑽1 + 𝑽2 (6)  Using (5) and (6) in (1), we get 𝑰0 + 𝑰1 + 𝑰2 = 1 𝑍𝑓 𝑽0 + 𝑽1 + 𝑽2  Using (4), this gives our second sequence-domain fault condition 𝑰0 = 𝑰1 = 𝑰2 = 1 3𝑍𝑓 𝑽0 + 𝑽1 + 𝑽2 (7)
  • 92.
    K. Webb ESE470 92 Unsymmetrical Fault Analysis – SLG Fault  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, 𝑰𝐹 𝑰𝐹 = 𝑰𝑎 = 𝑰0 + 𝑰1 + 𝑰2 = 3𝑰1 𝑰1 = 𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 𝑰𝐹 = 3𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 (8)
  • 93.
    K. Webb ESE470 93 SLG Fault - Example  Returning to our example power system  The interconnected sequence networks for a bolted fault at bus 1:
  • 94.
    K. Webb ESE470 94 SLG Fault - Example  The fault current is 𝑰𝐹 = 3𝑽𝐹 𝑍0+𝑍1+𝑍2+3𝑍𝐹 𝑰𝐹 = 3.0∠30° 𝑗0.473 = 6.34∠ − 60° 𝑝. 𝑢.  The current base at bus 1 is 𝑰𝑏 = 𝑆𝑏 3𝑉𝑏1 = 150 𝑀𝑉𝐴 3 230 𝑘𝑉 = 376.5 𝐴  So the fault current in kA is 𝑰𝐹 = 6.34∠ − 60° 376.5 𝐴 𝑰𝐹 = 2.39∠ − 60° 𝑘𝐴
  • 95.
    K. Webb ESE470 Line-to-Line Fault 95
  • 96.
    K. Webb ESE470 96 Unsymmetrical Fault Analysis – LL Fault  Now consider a line-to-line fault between phase 𝑏 and phase 𝑐 through impedance 𝑍𝐹  Phase-domain fault conditions: 𝑰𝑎 = 0 (9) 𝑰𝑏 = −𝑰𝑐 = 𝑽𝑏𝑔−𝑽𝑐𝑔 𝑍𝐹 (10)  Transforming to the sequence domain 𝑰0 𝑰1 𝑰2 = 1 3 1 1 1 1 𝑎 𝑎2 1 𝑎2 𝑎 0 𝑰𝑏 −𝑰𝑏 = 1 3 0 𝑎 − 𝑎2 𝑰𝑏 𝑎2 − 𝑎 𝑰𝑏 (11)  So, the first two sequence-domain fault conditions are 𝑰0 = 0 (12) 𝑰2 = −𝑰1 (13)
  • 97.
    K. Webb ESE470 97 Unsymmetrical Fault Analysis – LL Fault  To derive the remaining sequence-domain fault condition, rearrange (10) and transform to the sequence domain 𝑽𝑏𝑔 − 𝑽𝑐𝑔 = 𝑰𝑏𝑍𝐹 𝑽0 + 𝑎2 𝑽1 + 𝑎𝑽2 − 𝑽0 + 𝑎𝑽1 + 𝑎2 𝑽2 = 𝑰0 + 𝑎2 𝑰1 + 𝑎𝑰2 𝑍𝐹 𝑎2 𝑽1 + 𝑎𝑽2 − 𝑎𝑽1 − 𝑎2 𝑽2 = 𝑎2 − 𝑎 𝑰1𝑍𝐹 𝑎2 − 𝑎 𝑽1 − 𝑎2 − 𝑎 𝑽2 = 𝑎2 − 𝑎 𝑰1𝑍𝐹  The last sequence-domain fault condition is 𝑽1 − 𝑽2 = 𝑰1𝑍𝐹 (14)
  • 98.
    K. Webb ESE470 98 Unsymmetrical Fault Analysis – LL Fault  Sequence-domain fault conditions 𝑰0 = 0 (12) 𝑰2 = −𝑰1 (13) 𝑽1 − 𝑽2 = 𝑰1𝑍𝐹 (14)  These can be satisfied by:  Leaving the zero-sequence network open  Connecting the terminals of the positive- and negative-sequence networks together through 𝑍𝐹
  • 99.
    K. Webb ESE470 99 Unsymmetrical Fault Analysis – LL Fault  The fault current is the phase 𝑏 current, which is given by 𝑰𝐹 = 𝑰𝑏 = 𝑰0 + 𝑎2 𝑰1 + 𝑎𝑰2 𝑰𝐹 = 𝑎2𝑰1 − 𝑎𝑰1 𝑰𝐹 = −𝑗 3 𝑰1 = −𝑗 3 𝑽𝐹 𝑍1+𝑍2+𝑍𝐹 𝑰𝐹 = 3 𝑽𝐹∠−90° 𝑍1+𝑍2+𝑍𝐹 (15)
  • 100.
    K. Webb ESE470 100 LL Fault - Example  Now consider the same system with a bolted line-to-line fault at bus 1  The sequence network:
  • 101.
    K. Webb ESE470 101 LL Fault - Example 𝑰1 = 1∠30° 𝑗0.349 = 2.87∠ − 60°  The subtransient fault current is given by (15) as 𝑰𝐹 = 3∠ − 90° 2.87∠ − 60° 𝑰𝐹 = 4.96∠ − 150° 𝑝. 𝑢.  Using the previously-determined current base, we can convert the fault current to kA 𝑰𝐹 = 𝐼𝑏1 ⋅ 4.96∠ − 150° 𝑰𝐹 = 4.96∠ − 150° 376.5𝐴 𝑰𝐹 = 1.87∠ − 150° 𝑘𝐴
  • 102.
    K. Webb ESE470 Double-Line-to-Ground Fault 102
  • 103.
    K. Webb ESE470 103 Unsymmetrical Fault Analysis – DLG Fault  Now consider a double line-to-ground fault  Assume phases 𝑏 and 𝑐 are shorted to ground through 𝑍𝐹  Phase-domain fault conditions: 𝑰𝑎 = 0 (16) 𝑰𝑏 + 𝑰𝑐 = 𝑽𝑏𝑔 𝑍𝐹 = 𝑽𝑐𝑔 𝑍𝐹 (17)  It can be shown that (16) and (17) transform to the following sequence-domain fault conditions (analysis skipped here) 𝑰0 + 𝑰1 + 𝑰2 = 0 (18) 𝑽1 = 𝑽2 (19) 𝑰0 = 1 3𝑍𝐹 𝑽0 − 𝑽1 (20)
  • 104.
    K. Webb ESE470 104 Unsymmetrical Fault Analysis – DLG Fault  Sequence-domain fault conditions 𝑰0 + 𝑰1 + 𝑰2 = 0 (18) 𝑽1 = 𝑽2 (19) 𝑰0 = 1 3𝑍𝐹 𝑽0 − 𝑽1 (20)  To satisfy these fault conditions  Connect the positive- and negative-sequence networks together directly  Connect the zero- and positive-sequence networks together through 3𝑍𝐹
  • 105.
    K. Webb ESE470 105 Unsymmetrical Fault Analysis – DLG Fault  The fault current is the sum of the phase 𝑏 and phase 𝑐 currents, as given by (17)  In the sequence domain the fault current is 𝑰𝐹 = 𝑰𝑏 + 𝑰𝑐 = 3𝑰0 𝑰𝐹 = 3𝑰0 (21)  𝐼0 can be determined by a simple analysis (e.g. nodal) of the interconnected sequence networks
  • 106.
    K. Webb ESE470 106 DLG Fault - Example  Now determine the subtransient fault current for a bolted double line-to-ground fault at bus 1  The sequence network:  Here, because 𝑍𝐹 = 0, 𝑽0 = 𝑽1 = 𝑽2
  • 107.
    K. Webb ESE470 107 DLG Fault - Example  To find 𝑰𝐹, we must determine 𝑰0  We can first find 𝑉0 by applying voltage division 𝑽0 = 𝑽𝐹 𝑍2||𝑍0 𝑍1 + 𝑍2||𝑍0 𝑽0 = 1.0∠30° 𝑗0.18||𝑗0.124 𝑗0.169 + 𝑗0.18||𝑗0.124 𝑽0 = 0.303∠30°
  • 108.
    K. Webb ESE470 108 DLG Fault - Example  Next, calculate 𝑰0 𝑰0 = −𝑽0 𝑍0 = −0.303∠30° 𝑗0.124 = 2.44∠120° 𝑝. 𝑢.  The per-unit fault current is 𝑰𝐹 = 3𝑰0 = 7.33∠120° 𝑝. 𝑢.  Using the current base to convert to kA, gives the subtransient DLG fault current 𝑰𝐹 = 7.33∠120° 376.5 𝐴 𝑰𝐹 = 2.76∠120° 𝑘𝐴
  • 109.
    K. Webb ESE470 Example Problems 109
  • 110.
    K. Webb ESE470 Draw the sequence networks for the following power system. Assume the generator is operating at rated voltage.
  • 111.
  • 112.
    K. Webb ESE470 Reduce the sequence networks to their Thévenin equivalents for a fault occurring half of the way along the transmission line.
  • 113.
  • 114.
  • 115.
    K. Webb ESE470 Determine the subtransient fault current resulting from a DLG fault, half way along the transmission line, through an impedance of j0.2 p.u.
  • 116.