SECTION 7: FAULT ANALYSIS ESE 470 – Energy Distribution Systems 2 K. Webb Introduction ESE 470 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 K. Webb We’ll focus on short circuits ESE 470 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 K. Webb ESE 470 5 K. Webb Symmetrical Faults ESE 470 Symmetrical Faults 6 Faults occur nearly instantaneously Step change from steady-state behavior Lightening, tree fall, arcing over insulation, etc. Like throwing a switch to create the fault at 𝑡𝑡 = 0 Equivalent circuit model: R: generator resistance L: generator inductance 𝑣𝑣 𝑡𝑡 = 2𝑉𝑉 sin 𝜔𝜔𝜔𝜔 + 𝛼𝛼 𝑖𝑖 𝑡𝑡 = 0 for 𝑡𝑡 < 0 Source phase, 𝛼𝛼, determines voltage at 𝑡𝑡 = 0 𝐺𝐺 K. Webb Short circuit can occur at any point in a 60 Hz cycle ESE 470 Symmetrical Faults 7 The governing differential equation for 𝑡𝑡 > 0 is 𝑑𝑑𝑑𝑑 𝑅𝑅 2𝑉𝑉𝐺𝐺 sin 𝜔𝜔𝜔𝜔 + 𝛼𝛼 + 𝑖𝑖 𝑡𝑡 = 𝐿𝐿 𝑑𝑑𝑑𝑑 𝐿𝐿 The solution gives the fault current 𝑅𝑅 2𝑉𝑉𝐺𝐺 −𝑡𝑡 𝑖𝑖 𝑡𝑡 = sin 𝜔𝜔𝜔𝜔 + 𝛼𝛼 − 𝜃𝜃 − sin 𝛼𝛼 − 𝜃𝜃 𝑒𝑒 𝐿𝐿 𝑍𝑍 where 𝑍𝑍 = 𝑅𝑅2 + 𝜔𝜔𝜔𝜔 2 and 𝜃𝜃 = tan−1 𝜔𝜔𝜔𝜔 𝑅𝑅 This total fault current is referred to as the asymmetrical fault current K. Webb It has a steady-state component 2𝑉𝑉𝐺𝐺 sin 𝜔𝜔𝜔𝜔 + 𝛼𝛼 − 𝜃𝜃 𝑖𝑖𝑎𝑎𝑎𝑎 𝑡𝑡 = 𝑍𝑍 And a transient component 𝑖𝑖𝑑𝑑𝑑𝑑 𝑅𝑅 2𝑉𝑉𝐺𝐺 −𝑡𝑡 sin 𝛼𝛼 − 𝜃𝜃 𝑒𝑒 𝐿𝐿 𝑡𝑡 = − 𝑍𝑍 ESE 470 Symmetrical Faults 8 Magnitude of the transient fault current, 𝑖𝑖𝑑𝑑𝑑𝑑 , depends on 𝛼𝛼 𝑖𝑖𝑑𝑑𝑑𝑑 0 = 0 for 𝑖𝑖𝑑𝑑𝑑𝑑 0 = 2𝐼𝐼𝑎𝑎𝑎𝑎 for 𝛼𝛼 = 𝜃𝜃 𝛼𝛼 = 𝜃𝜃 − 90° 𝐼𝐼𝑎𝑎𝑎𝑎 = 𝑉𝑉𝐺𝐺 /𝑍𝑍 is the rms value of the steady-state fault current Worst-case fault current occurs for 𝛼𝛼 = 𝜃𝜃 − 90° K. Webb 𝑅𝑅 2𝑉𝑉𝐺𝐺 𝜋𝜋 −𝑡𝑡 𝑖𝑖 𝑡𝑡 = sin 𝜔𝜔𝜔𝜔 − + 𝑒𝑒 𝐿𝐿 𝑍𝑍 2 ESE 470 Symmetrical Faults 9 Important points here: fault current has both steady-state and transient components – asymmetrical Total Magnitude of the asymmetry (transient component) depends on the phase of the generator voltage at the time of the fault K. Webb ESE 470 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 ′′ 𝑋𝑋𝑑𝑑 : ′ 𝑋𝑋𝑑𝑑 : modeled with three reactance values subtransient reactance transient reactance 𝑋𝑋𝑑𝑑 : synchronous reactance Reactance K. Webb increases with time, such that 𝑋𝑋𝑑𝑑′′ < 𝑋𝑋𝑑𝑑′ < 𝑋𝑋𝑑𝑑 ESE 470 Sub-Transient Fault Current 11 Transition rates between reactance values are dictated by two time constants: 𝜏𝜏𝑑𝑑′′ : short-circuit subtransient time constant 𝜏𝜏𝑑𝑑′ : short-circuit transient time constant 𝜋𝜋 Neglecting generator resistance and assuming 𝜃𝜃 = − , the synchronous 2 portion of the fault current is 𝑖𝑖𝑎𝑎𝑎𝑎 𝑡𝑡 = 2𝑉𝑉𝐺𝐺 𝑡𝑡 𝑡𝑡 − ′′ − ′ 1 1 1 1 1 𝜋𝜋 𝜏𝜏 𝜏𝜏 𝑑𝑑 𝑑𝑑 − 𝑒𝑒 + ′− 𝑒𝑒 + sin 𝜔𝜔𝜔𝜔 + 𝛼𝛼 − 𝑋𝑋𝑑𝑑′′ 𝑋𝑋𝑑𝑑′ 𝑋𝑋𝑑𝑑 𝑋𝑋𝑑𝑑 𝑋𝑋𝑑𝑑 2 At the instant of the fault, 𝑡𝑡 = 0, the rms synchronous fault current is K. Webb 𝑰𝑰′′ 𝐹𝐹 𝑽𝑽𝐺𝐺 = ′′ 𝑋𝑋𝑑𝑑 This is the rms subtransient fault current, 𝑰𝑰′′ 𝐹𝐹 This will be our primary metric for assessing fault current ESE 470 12 K. Webb Symmetrical Three-Phase Short Circuits ESE 470 13 Symmetrical 3-𝜙𝜙 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-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 K. Webb ESE 470 14 Symmetrical 3-𝜙𝜙 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 𝑡𝑡 = 0 K. Webb ESE 470 15 Symmetrical 3-𝜙𝜙 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 𝑽𝑽𝐹𝐹 K. Webb ESE 470 16 Symmetrical 3-𝜙𝜙 Short Circuits Applying superposition, we can represent this circuit as the sum of two separate circuits: Circuit 1 K. Webb Circuit 2 ESE 470 17 Symmetrical 3-𝜙𝜙 Short Circuits Assume that the value of the fault-location source, 𝑽𝑽𝐹𝐹 , is the pre-fault voltage at that location Circuit The 𝑰𝑰′′ 𝐹𝐹1 = 0 𝑽𝑽𝐹𝐹 source can therefore be removed from circuit 1 Circuit 1 K. Webb 1, then, represents the pre-fault circuit, so Circuit 2 ESE 470 18 Symmetrical 3-𝜙𝜙 Short Circuits The current in circuit 1, 𝑰𝑰𝐿𝐿 , is the pre-fault line current Superposition gives the fault current ′′ ′′ ′′ 𝑰𝑰′′ 𝐹𝐹 = 𝑰𝑰𝐹𝐹𝐹 + 𝑰𝑰𝐹𝐹𝐹 = 𝑰𝑰𝐹𝐹𝐹 The generator fault current is ′′ ′′ 𝑰𝑰′′ = 𝑰𝑰 + 𝑰𝑰 𝐺𝐺 𝐺𝐺𝐺 𝐺𝐺𝐺 ′′ 𝑰𝑰′′ = 𝑰𝑰 + 𝑰𝑰 𝐿𝐿 𝐺𝐺 𝐺𝐺𝐺 The motor fault current is ′′ ′′ 𝑰𝑰′′ = 𝑰𝑰 + 𝑰𝑰 𝑀𝑀𝑀 𝑀𝑀𝑀 𝑴𝑴 K. Webb ′′ 𝑰𝑰′′ = −𝑰𝑰 + 𝑰𝑰 𝐿𝐿 𝑀𝑀 𝑀𝑀𝑀 ESE 470 19 Symmetrical 3-𝜙𝜙 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: K. Webb Subtransient fault current Subtransient generator current Subtransient motor current ESE 470 20 Symmetrical 3-𝜙𝜙 Fault – Example First convert to per-unit Use 𝑆𝑆𝑏𝑏 = 100 𝑀𝑀𝑀𝑀𝑀𝑀 Base voltage in the transmission line zone is 𝑉𝑉𝑏𝑏,𝑡𝑡𝑡𝑡 = 138 𝑘𝑘𝑘𝑘 Base impedance in the transmission line zone is 𝑍𝑍𝑏𝑏,𝑡𝑡𝑡𝑡 2 𝑉𝑉𝑏𝑏,𝑡𝑡𝑡𝑡 138 𝑘𝑘𝑘𝑘 2 = = = 190.4 Ω 𝑆𝑆𝑏𝑏 100 𝑀𝑀𝑀𝑀𝑀𝑀 The per-unit transmission line reactance is 𝑋𝑋𝑡𝑡𝑡𝑡 = K. Webb 20 Ω = 0.105 𝑝𝑝. 𝑢𝑢. 190.4 Ω ESE 470 21 Symmetrical 3-𝜙𝜙 Fault – Example The two per-unit circuits are These can be simplified by combining impedances K. Webb ESE 470 22 Symmetrical 3-𝜙𝜙 Fault – Example Using circuit 2, we can calculate the subtransient fault current 𝑰𝑰′′ 𝐹𝐹 1.05∠0° = 9.079∠ − 90° 𝑝𝑝. 𝑢𝑢. = 𝑗𝑗𝑗.116 To convert to kA, first determine the current base in the generator zone 𝐼𝐼𝑏𝑏,𝐺𝐺 = 𝑆𝑆𝑏𝑏 3𝑉𝑉𝑏𝑏,𝐺𝐺 = 100 𝑀𝑀𝑀𝑀𝑀𝑀 3 ⋅ 13.8 𝑘𝑘𝑘𝑘 The subtransient fault current is = 4.18 𝑘𝑘𝑘𝑘 𝑰𝑰′′ 𝐹𝐹 = 9.079∠ − 90° ⋅ 4.18 𝑘𝑘𝑘𝑘 𝑰𝑰′′ 𝐹𝐹 = 37.98∠ − 90° 𝑘𝑘𝑘𝑘 K. Webb ESE 470 23 Symmetrical 3-𝜙𝜙 Fault – Example The pre-fault line current can be calculated from the pre-fault generator voltage and power 𝑰𝑰𝐿𝐿 = Or, 𝑆𝑆𝐺𝐺 3𝑽𝑽′′ 𝐺𝐺 = 100 𝑀𝑀𝑀𝑀𝑀𝑀 3 ⋅ 1.05 ⋅ 13.8 𝑘𝑘𝑘𝑘 𝑰𝑰𝐿𝐿 = 3.98∠ − 18.19° 𝑘𝑘𝑘𝑘 ∠ − cos −1 0.95 in per-unit: This 3.98∠ − 18.19° 𝑘𝑘𝑘𝑘 = 0.952∠ − 18.19° 𝑝𝑝. 𝑢𝑢. 𝑰𝑰𝐿𝐿 = 4.18 kA will be used to find the generator and motor fault currents K. Webb ESE 470 24 Symmetrical 3-𝜙𝜙 Fault – Example The generator’s contribution to the fault current is found by applying current division 𝑰𝑰′′ 𝐺𝐺𝐺 = 𝑰𝑰′′ 𝐹𝐹 0.505 = 7.0∠ − 90° 𝑝𝑝. 𝑢𝑢. 0.505 + 0.15 Adding the pre-fault line current, we have the subtransient generator fault current ′′ 𝑰𝑰′′ 𝐺𝐺 = 𝑰𝑰𝐿𝐿 + 𝑰𝑰𝐺𝐺𝐺 𝑰𝑰′′ 𝐺𝐺 = 0.952∠ − 18.19° + 7.0∠ − 90° Converting to kA 𝑰𝑰′′ 𝐺𝐺 = 7.35∠ − 82.9° 𝑝𝑝. 𝑢𝑢. 𝑰𝑰′′ 𝐺𝐺 = 7.35∠ − 82.9° ⋅ 4.18 𝑘𝑘𝑘𝑘 K. Webb 𝑰𝑰′′ 𝐺𝐺 = 30.74∠ − 82.9° 𝑘𝑘𝑘𝑘 ESE 470 25 Symmetrical 3-𝜙𝜙 Fault – Example Similarly, for the motor 𝑰𝑰′′ 𝑀𝑀𝑀 = 𝑰𝑰′′ 𝐹𝐹 0.15 = 2.08∠ − 90° 𝑝𝑝. 𝑢𝑢. 0.505 + 0.15 Subtracting the pre-fault line current gives the subtransient motor fault current ′′ 𝑰𝑰′′ 𝑀𝑀 = −𝑰𝑰𝐿𝐿 + 𝑰𝑰𝑀𝑀𝑀 𝑰𝑰′′ 𝑀𝑀 = −0.952∠ − 18.19° + 2.08∠ − 90° 𝑰𝑰′′ 𝑀𝑀 = 2.0∠ − 116.9° Converting to 𝑘𝑘𝑘𝑘 𝑰𝑰′′ 𝑀𝑀 = 2.0∠ − 116.9° ⋅ 4.18 𝑘𝑘𝑘𝑘 K. Webb 𝑰𝑰′′ 𝑀𝑀 = 8.36∠ − 116.9° 𝑘𝑘𝑘𝑘 ESE 470 26 K. Webb Symmetrical Components ESE 470 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 K. Webb method of symmetrical components ESE 470 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 K. Webb ESE 470 Sequence Components 29 Zero sequence components Positive sequence components Three phasors with equal magnitude and equal phase 𝑽𝑽𝑎𝑎𝑎 , 𝑽𝑽𝑏𝑏𝑏 , 𝑽𝑽𝑐𝑐𝑐 Three phasors with equal magnitude and ± 120°, positive-sequence phase 𝑽𝑽𝑎𝑎1 , 𝑽𝑽𝑏𝑏1 , 𝑽𝑽𝑐𝑐1 Negative sequence components K. Webb Three phasors with equal magnitude and ± 120°, negative-sequence phase 𝑽𝑽𝑎𝑎2 , 𝑽𝑽𝑏𝑏2 , 𝑽𝑽𝑐𝑐2 ESE 470 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 K. Webb 𝑽𝑽𝑎𝑎 𝑽𝑽𝑎𝑎0 𝑽𝑽𝑎𝑎1 𝑽𝑽𝑎𝑎2 𝑽𝑽𝑏𝑏 = 𝑽𝑽𝑏𝑏0 + 𝑽𝑽𝑏𝑏1 + 𝑽𝑽𝑏𝑏2 𝑽𝑽𝑐𝑐 𝑽𝑽𝑐𝑐0 𝑽𝑽𝑐𝑐1 𝑽𝑽𝑐𝑐2 (1) ESE 470 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 𝑽𝑽0 = 𝑽𝑽𝑎𝑎𝑎 = 𝑽𝑽𝑏𝑏𝑏 = 𝑽𝑽𝑐𝑐𝑐 (2) 𝑽𝑽1 = 𝑽𝑽𝑎𝑎𝑎 = 1∠120° ⋅ 𝑽𝑽𝑏𝑏𝑏 = 1∠240° ⋅ 𝑽𝑽𝑐𝑐𝑐 (3) 𝑽𝑽2 = 𝑽𝑽𝑎𝑎𝑎 = 1∠240° ⋅ 𝑽𝑽𝑏𝑏𝑏 = 1∠120° ⋅ 𝑽𝑽𝑐𝑐𝑐 (4) For the positive sequence components, And, for the negative sequence components, Note that we’re using phase 𝑎𝑎 as our reference, so K. Webb 𝑽𝑽0 = 𝑽𝑽𝑎𝑎𝑎 , 𝑽𝑽1 = 𝑽𝑽𝑎𝑎𝑎 , 𝑽𝑽2 = 𝑽𝑽𝑎𝑎𝑎 ESE 470 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 𝑎𝑎2 yields a rotation of 240° = −120° Using (5) to rewrite (3) and (4) 𝑽𝑽1 = 𝑽𝑽𝑎𝑎𝑎 = 𝑎𝑎𝑽𝑽𝑏𝑏𝑏 = 𝑎𝑎2 𝑽𝑽𝑐𝑐𝑐 𝑽𝑽2 = 𝑽𝑽𝑎𝑎𝑎 = 𝑎𝑎2 𝑽𝑽𝑏𝑏𝑏 = 𝑎𝑎𝑽𝑽𝑐𝑐𝑐 K. Webb (6) (7) ESE 470 Sequence Components 33 Using (2), (6), and (7), we can rewrite (1) in a simplified form 1 𝑎𝑎 𝑎𝑎2 𝑽𝑽0 𝑽𝑽1 𝑽𝑽2 (8) The vector on the left is the vector of phase voltages, 𝑽𝑽𝑝𝑝 The vector on the right is the vector of (phase 𝑎𝑎) sequence components, 𝑽𝑽𝑠𝑠 We’ll call the 3 × 3 transformation matrix 𝑨𝑨 𝑽𝑽𝑎𝑎 1 1 𝑽𝑽𝑏𝑏 = 1 𝑎𝑎2 𝑽𝑽𝑐𝑐 1 𝑎𝑎 We can rewrite (8) as K. Webb 𝑽𝑽𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 (9) ESE 470 Sequence Components 34 We can express the sequence voltages as a function of the phase voltages by inverting the transformation matrix where So K. Webb 𝑽𝑽𝑠𝑠 = 𝑨𝑨−1 𝑽𝑽𝑝𝑝 𝑨𝑨−1 = 1 3 1 1 1 𝑎𝑎 1 𝑎𝑎2 𝑽𝑽0 1 1 1 𝑽𝑽1 = 1 𝑎𝑎 3 𝑽𝑽2 1 𝑎𝑎2 (10) 1 𝑎𝑎2 𝑎𝑎 1 𝑎𝑎2 𝑎𝑎 (11) 𝑽𝑽𝑎𝑎 𝑽𝑽𝑏𝑏 𝑽𝑽𝑐𝑐 (12) ESE 470 Sequence Components 35 The same relationships hold for three-phase currents The phase currents are 𝑰𝑰𝑎𝑎 𝑰𝑰𝑝𝑝 = 𝑰𝑰𝑏𝑏 𝑰𝑰𝑐𝑐 And, the sequence currents are K. Webb 𝑰𝑰0 𝑰𝑰𝑠𝑠 = 𝑰𝑰1 𝑰𝑰2 ESE 470 Sequence Components 36 The transformation matrix, 𝑨𝑨, relates the phase currents to the sequence currents 𝑰𝑰𝑝𝑝 = 𝑨𝑨𝑰𝑰𝑠𝑠 And vice versa 𝑰𝑰𝑎𝑎 1 1 𝑰𝑰𝑏𝑏 = 1 𝑎𝑎2 𝑰𝑰𝑐𝑐 1 𝑎𝑎 𝑰𝑰𝑠𝑠 = 𝑨𝑨−1 𝑰𝑰𝑝𝑝 K. Webb 𝑰𝑰0 1 1 1 𝑰𝑰1 = 1 𝑎𝑎 3 𝑰𝑰2 1 𝑎𝑎2 1 𝑎𝑎 𝑎𝑎2 1 𝑎𝑎2 𝑎𝑎 𝑰𝑰0 𝑰𝑰1 𝑰𝑰2 𝑰𝑰𝑎𝑎 𝑰𝑰𝑏𝑏 𝑰𝑰𝑐𝑐 (13) (14) ESE 470 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 𝑽𝑽𝑏𝑏 = 𝑽𝑽𝑎𝑎 ⋅ 1∠ − 120° = 𝑎𝑎2 𝑽𝑽𝑎𝑎 𝑽𝑽𝑐𝑐 = 𝑽𝑽𝑎𝑎 ⋅ 1∠120° = 𝑎𝑎𝑽𝑽𝑎𝑎 The sequence voltages are given by (12) The zero sequence voltage is 1 1 𝑽𝑽0 = 𝑽𝑽𝑎𝑎 + 𝑽𝑽𝑏𝑏 + 𝑽𝑽𝑐𝑐 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎 2 𝑽𝑽𝑎𝑎 + 𝑎𝑎𝑽𝑽𝑎𝑎 3 3 1 𝑽𝑽0 = 𝑽𝑽𝑎𝑎 1 + 𝑎𝑎 2 + 𝑎𝑎 3 Applying the identity 1 + 𝑎𝑎2 + 𝑎𝑎 = 0, we have K. Webb 𝑽𝑽0 = 0 ESE 470 Sequence Components – Balanced System 38 The positive sequence component is given by K. Webb 1 𝑽𝑽1 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎𝑽𝑽𝑏𝑏 + 𝑎𝑎2 𝑽𝑽𝑐𝑐 3 1 𝑽𝑽1 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎 ⋅ 𝑎𝑎2 𝑽𝑽𝑎𝑎 + 𝑎𝑎2 ⋅ 𝑎𝑎𝑽𝑽𝑎𝑎 3 1 𝑽𝑽1 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎3 𝑽𝑽𝑎𝑎 + 𝑎𝑎3 𝑽𝑽𝑎𝑎 3 Since 𝑎𝑎3 = 1∠0°, we have 1 𝑽𝑽1 = 3𝑽𝑽𝑎𝑎 3 𝑽𝑽1 = 𝑽𝑽𝑎𝑎 ESE 470 Sequence Components – Balanced System 39 The negative sequence component is given by 1 𝑽𝑽2 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎2 𝑽𝑽𝑏𝑏 + 𝑎𝑎𝑽𝑽𝑐𝑐 3 1 𝑽𝑽2 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎2 ⋅ 𝑎𝑎2 𝑽𝑽𝑎𝑎 + 𝑎𝑎 ⋅ 𝑎𝑎𝑽𝑽𝑎𝑎 3 1 𝑽𝑽2 = 𝑽𝑽𝑎𝑎 + 𝑎𝑎4 𝑽𝑽𝑎𝑎 + 𝑎𝑎2 𝑽𝑽𝑎𝑎 3 Again, using the identity 1 + 𝑎𝑎2 + 𝑎𝑎 = 0, along with the fact that 𝑎𝑎4 = 𝑎𝑎, we have K. Webb 𝑽𝑽2 = 0 ESE 470 Sequence Components – Balanced System 40 So, for a positive-sequence, balanced, three-phase system, the sequence voltages are 𝑽𝑽0 = 0, 𝑽𝑽1 = 𝑽𝑽𝑎𝑎 , 𝑽𝑽2 = 0 Similarly, the sequence currents are 𝑰𝑰0 = 0, 𝑰𝑰1 = 𝑰𝑰𝑎𝑎 , 𝑰𝑰2 = 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 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 K. Webb ESE 470 42 K. Webb Sequence Networks ESE 470 Sequence Networks 43 Power system components each have their own set of sequence networks 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 Non-rotating loads Transmission lines Rotating machines – generators and motors Transformers 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 Consider a balanced Y-load with the neutral grounded through some non-zero impedance Applying KVL gives the phase-𝑎𝑎-to-ground voltage 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑛𝑛 For phase 𝑏𝑏: 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑐𝑐 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑐𝑐 (15) 𝑽𝑽𝑏𝑏𝑔𝑔 = 𝑍𝑍𝑦𝑦 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑛𝑛 = 𝑍𝑍𝑦𝑦 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑐𝑐 𝑽𝑽𝑏𝑏𝑔𝑔 = 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑐𝑐 (16) Similarly, for phase 𝑐𝑐: K. Webb 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑏𝑏 + 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑐𝑐 (17) ESE 470 Sequence Networks – Non-Rotating Loads 45 Putting (15) – (17) in matrix form or 𝑽𝑽𝑎𝑎𝑔𝑔 𝑽𝑽𝑏𝑏𝑏𝑏 = 𝑽𝑽𝑐𝑐𝑐𝑐 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑽𝑽𝑝𝑝 = 𝒁𝒁𝑝𝑝 𝑰𝑰𝑝𝑝 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑛𝑛 𝑍𝑍𝑦𝑦 + 𝑍𝑍𝑛𝑛 𝑰𝑰𝑎𝑎 𝑰𝑰𝑏𝑏 𝑰𝑰𝑐𝑐 (18) where 𝑽𝑽𝑝𝑝 and 𝑰𝑰𝑝𝑝 are the phase voltages and currents, respectively, and 𝒁𝒁𝑝𝑝 is the phase impedance matrix We can use (9) and (13) to rewrite (18) as Solving for 𝑽𝑽𝑠𝑠 or K. Webb 𝑨𝑨𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑝𝑝 𝑨𝑨𝑰𝑰𝑠𝑠 𝑽𝑽𝑠𝑠 = 𝑨𝑨−1 𝒁𝒁𝑝𝑝 𝑨𝑨𝑰𝑰𝑠𝑠 𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑠𝑠 𝑰𝑰𝑠𝑠 (19) ESE 470 Sequence Networks – Non-Rotating Loads 46 𝑽𝑽𝑠𝑠 = 𝒁𝒁𝑠𝑠 𝑰𝑰𝑠𝑠 (19) where 𝑍𝑍𝑠𝑠 is the sequence impedance matrix 𝒁𝒁𝑠𝑠 = 𝑨𝑨−1 𝒁𝒁𝑝𝑝 𝑨𝑨 = 𝑍𝑍𝑦𝑦 + 3𝑍𝑍𝑛𝑛 0 0 0 𝑍𝑍𝑦𝑦 0 0 0 𝑍𝑍𝑦𝑦 (20) Equation (19) then becomes a set of three uncoupled equations 𝑽𝑽0 = 𝑍𝑍𝑦𝑦 + 3𝑍𝑍𝑛𝑛 𝑰𝑰0 = 𝑍𝑍0 𝑰𝑰0 𝑽𝑽1 = 𝑍𝑍𝑦𝑦 𝑰𝑰1 = 𝑍𝑍1 𝑰𝑰1 K. Webb 𝑽𝑽2 = 𝑍𝑍𝑦𝑦 𝑰𝑰2 = 𝑍𝑍2 𝑰𝑰2 (21) (22) (23) ESE 470 Sequence Networks – Non-Rotating Loads 47 Equations (21) – (23) describe the uncoupled sequence networks Zero-sequence network: Positive-sequence network: K. Webb Negative-sequence network: ESE 470 Sequence Networks – Non-Rotating Loads 48 We can develop similar sequence networks for a balanced Δconnected load K. Webb 𝑍𝑍𝑦𝑦 = 𝑍𝑍Δ /3 There is no neutral point for the Δ-network, so 𝑍𝑍𝑛𝑛 = ∞ - an open circuit Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 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 K. Webb 𝑽𝑽𝑎𝑎𝑎𝑎′ 𝑍𝑍𝑎𝑎𝑎𝑎 𝑽𝑽𝑏𝑏𝑏𝑏′ = 𝑍𝑍𝑏𝑏𝑏𝑏 𝑍𝑍𝑐𝑐𝑐𝑐 𝑽𝑽𝑐𝑐𝑐𝑐 ′ 𝑍𝑍𝑎𝑎𝑎𝑎 𝑍𝑍𝑏𝑏𝑏𝑏 𝑍𝑍𝑐𝑐𝑐𝑐 𝑍𝑍𝑎𝑎𝑎𝑎 𝑍𝑍𝑏𝑏𝑏𝑏 𝑍𝑍𝑐𝑐𝑐𝑐 𝑽𝑽𝑎𝑎𝑛𝑛 − 𝑽𝑽𝑎𝑎′𝑛𝑛 𝑰𝑰𝑎𝑎 𝑰𝑰𝑏𝑏 = 𝑽𝑽𝑏𝑏𝑏𝑏 − 𝑽𝑽𝑏𝑏′𝑛𝑛 𝑰𝑰𝑐𝑐 𝑽𝑽𝑐𝑐𝑐𝑐 − 𝑽𝑽𝑐𝑐 ′𝑛𝑛 (24) ESE 470 Sequence Networks – 3-𝜙𝜙 Lines 50 Writing (24) in compact form 𝑽𝑽𝑝𝑝 − 𝑽𝑽𝑝𝑝′ = 𝒁𝒁𝑝𝑝 𝑰𝑰𝑝𝑝 (25) 𝒁𝒁𝑝𝑝 is the phase impedance matrix Self impedances along the diagonal Mutual impedances elsewhere Symmetric Diagonal, if we neglect mutual impedances K. Webb ESE 470 Sequence Networks – 3-𝜙𝜙 Lines 51 We can rewrite (25) in terms of sequence components 𝑨𝑨𝑽𝑽𝑠𝑠 − 𝑨𝑨𝑽𝑽𝑠𝑠′ = 𝒁𝒁𝑝𝑝 𝑨𝑨𝑰𝑰𝑠𝑠 𝑽𝑽𝑠𝑠 − 𝑽𝑽𝑠𝑠′ = 𝑨𝑨−1 𝒁𝒁𝑝𝑝 𝑨𝑨𝑰𝑰𝑠𝑠 𝑽𝑽𝑠𝑠 − 𝑽𝑽𝑠𝑠′ = 𝒁𝒁𝑠𝑠 𝑰𝑰𝑠𝑠 (26) where 𝒁𝒁𝑠𝑠 is the sequence impedance matrix 𝒁𝒁𝑠𝑠 = 𝑨𝑨−1 𝒁𝒁𝑝𝑝 𝑨𝑨 (27) 𝒁𝒁𝑠𝑠 is diagonal so long as the system impedances are balanced, i.e. Self impedances are equal: 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑏𝑏𝑏𝑏 = 𝑍𝑍𝑐𝑐𝑐𝑐 Mutual impedances are equal: 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑎𝑎𝑎𝑎 = 𝑍𝑍𝑏𝑏𝑏𝑏 K. Webb ESE 470 Sequence Networks – 3-𝜙𝜙 Lines 52 For balanced lines, 𝒁𝒁𝑠𝑠 is diagonal 𝑍𝑍𝑎𝑎𝑎𝑎 + 2𝑍𝑍𝑎𝑎𝑏𝑏 0 𝒁𝒁𝑠𝑠 = 0 0 𝑍𝑍𝑎𝑎𝑎𝑎 − 𝑍𝑍𝑎𝑎𝑎𝑎 0 0 𝑍𝑍0 0 = 0 0 𝑍𝑍𝑎𝑎𝑎𝑎 − 𝑍𝑍𝑎𝑎𝑎𝑎 0 𝑍𝑍1 0 0 0 𝑍𝑍2 Because 𝑍𝑍𝑠𝑠 is diagonal, (26) represents three uncoupled equations 𝑽𝑽0 − 𝑽𝑽0′ = 𝑍𝑍0 𝑰𝑰0 (28) 𝑽𝑽2 − 𝑽𝑽2′ = 𝑍𝑍2 𝑰𝑰2 (30) 𝑽𝑽1 − 𝑽𝑽1′ = 𝑍𝑍1 𝑰𝑰1 K. Webb (29) ESE 470 Sequence Networks – 3-𝜙𝜙 Lines 53 Equations (28) – (30) describe the voltage drop across three uncoupled sequence networks K. Webb Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 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 K. Webb Sources will appear only in the positive-sequence network ESE 470 Sequence Networks – Synchronous Generator 55 Sequence networks for Y-connected synchronous generator K. Webb Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 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 K. Webb ESE 470 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 3𝐼𝐼0 𝑍𝑍𝑁𝑁 and 3𝐼𝐼0 𝑍𝑍𝑛𝑛 3𝑍𝑍𝑁𝑁 and 3𝑍𝑍𝑛𝑛 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 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, 𝑍𝑍𝑠𝑠 , in the per-unit model Y-Y transformer sequence networks K. Webb Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 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 Δside voltage and current For negative-sequence network, Y-side voltage and current lag Δside voltage and current No neutral connection on the Δ side K. Webb Zero-sequence current cannot enter or leave the Δ winding ESE 470 Sequence Networks – Y-ΔTransformers 60 Sequence networks for Y-ΔTransformers K. Webb Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 61 Sequence Networks – Δ-ΔTransformers Δ-Δ transformers Like Y-Y transformers, no phase shift No neutral connections Zero-sequence current winding K. Webb cannot flow into or out of either ESE 470 62 Sequence Networks – Δ-ΔTransformers Sequence networks for Δ-ΔTransformers K. Webb Zero-sequence network: Positive-sequence network: Negative-sequence network: ESE 470 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 𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑎𝑎𝑎𝑎 𝑰𝑰𝑎𝑎∗ + 𝑽𝑽𝑏𝑏𝑏𝑏 𝑰𝑰𝑏𝑏∗ + 𝑽𝑽𝑐𝑐𝑐𝑐 𝑰𝑰𝑐𝑐∗ In matrix form, this looks like 𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑎𝑎𝑛𝑛 K. Webb 𝑆𝑆𝑝𝑝 = 𝑽𝑽𝑇𝑇𝑝𝑝 𝑰𝑰𝑝𝑝∗ 𝑽𝑽𝑏𝑏𝑏𝑏 𝑽𝑽𝑐𝑐𝑐𝑐 𝑰𝑰𝑎𝑎∗ 𝑰𝑰∗𝑏𝑏 𝑰𝑰𝑐𝑐∗ (28) ESE 470 Power in Sequence Networks 64 Recall the following relationships 𝑽𝑽𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 𝑰𝑰𝑝𝑝 = 𝑨𝑨𝑰𝑰𝑠𝑠 (13) Using (9) and (13) in (28), we have 𝑺𝑺𝑝𝑝 = 𝑨𝑨𝑽𝑽𝑠𝑠 (9) 𝑇𝑇 𝑨𝑨𝑰𝑰𝑠𝑠 𝑺𝑺𝑝𝑝 = 𝑽𝑽𝑇𝑇𝑠𝑠 𝑨𝑨𝑇𝑇 𝑨𝑨∗ 𝑰𝑰∗𝑠𝑠 ∗ (29) Computing the product in the middle of the right-hand side of (29), we find 3 0 0 = 0 3 0 = 3𝑰𝑰3 0 0 3 where 𝑰𝑰3 is the 3×3 identity matrix 𝑨𝑨𝑇𝑇 𝑨𝑨∗ K. Webb ESE 470 Power in Sequence Networks 65 Equation (29) then becomes 𝑺𝑺𝑝𝑝 = 𝑽𝑽𝑇𝑇𝑠𝑠 3𝑰𝑰3 𝑰𝑰∗𝑠𝑠 𝑺𝑺𝑝𝑝 = 3𝑽𝑽𝑇𝑇𝑠𝑠 𝑰𝑰∗𝑠𝑠 𝑺𝑺𝑝𝑝 = 3 𝑽𝑽0 𝑽𝑽1 𝑽𝑽2 𝑰𝑰∗0 𝑰𝑰1∗ 𝑰𝑰∗2 𝑺𝑺𝑝𝑝 = 3 𝑽𝑽0 𝑰𝑰∗0 + 𝑽𝑽1 𝑰𝑰1∗ + 𝑽𝑽2 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 K. Webb The three sequence networks represent only one of the three phases – recall, we chose to consider only phase 𝑎𝑎 ESE 470 66 K. Webb Unsymmetrical Faults ESE 470 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 K. Webb ESE 470 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 K. Webb ESE 470 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 𝑉𝑉𝐹𝐹 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 Unsymmetrical Fault Analysis 70 Each sequence network includes all interconnected powersystem components Analysis will be simplified if we represent each sequence network as its Thévenin equivalent Generators, motors, lines, and transformers From the perspective of the fault location For example, consider the following power system: K. Webb ESE 470 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: K. Webb Assuming the generator is operating at the rated voltage at the time of the fault ESE 470 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 K. Webb ESE 470 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 K. Webb ESE 470 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 𝑍𝑍𝑓𝑓 to ground, the model becomes: K. Webb ESE 470 Unsymmetrical Fault Analysis – SLG Fault 75 The phase-domain fault conditions: 𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑎𝑎𝑎𝑎 (1) 𝑍𝑍𝑓𝑓 𝑰𝑰𝑏𝑏 = 𝑰𝑰𝑐𝑐 = 0 (2) Transforming these phase-domain currents to the sequence domain 𝑰𝑰0 1 1 1 𝑰𝑰1 = 1 𝑎𝑎 3 𝑰𝑰2 1 𝑎𝑎2 1 𝑎𝑎2 𝑎𝑎 𝑽𝑽𝑎𝑎𝑔𝑔 /𝑍𝑍𝑓𝑓 𝑽𝑽𝑎𝑎𝑔𝑔 /𝑍𝑍𝑓𝑓 1 = 𝑽𝑽𝑎𝑎𝑔𝑔 /𝑍𝑍𝑓𝑓 0 3 𝑽𝑽𝑎𝑎𝑔𝑔 /𝑍𝑍𝑓𝑓 0 (3) This gives one of our sequence-domain fault conditions K. Webb 𝑰𝑰0 = 𝑰𝑰1 = 𝑰𝑰2 (4) ESE 470 Unsymmetrical Fault Analysis – SLG Fault 76 We know that and 𝑰𝑰𝑎𝑎 = 𝑽𝑽𝑎𝑎𝑎𝑎 𝑍𝑍𝑓𝑓 = 𝑰𝑰0 + 𝑰𝑰1 + 𝑰𝑰2 𝑽𝑽𝑎𝑎𝑎𝑎 = 𝑽𝑽0 + 𝑽𝑽1 + 𝑽𝑽2 (5) (6) Using (5) and (6) in (1), we get 𝑰𝑰0 + 𝑰𝑰1 + 𝑰𝑰2 = 1 𝑍𝑍𝑓𝑓 𝑽𝑽0 + 𝑽𝑽1 + 𝑽𝑽2 Using (4), this gives our second sequence-domain fault condition 𝑰𝑰0 = 𝑰𝑰1 = 𝑰𝑰2 = K. Webb 1 3𝑍𝑍𝑓𝑓 𝑽𝑽0 + 𝑽𝑽1 + 𝑽𝑽2 (7) ESE 470 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, 𝑰𝑰𝐹𝐹 𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑎𝑎 = 𝑰𝑰0 + 𝑰𝑰1 + 𝑰𝑰2 = 3𝑰𝑰1 𝑰𝑰1 = K. Webb 𝑰𝑰𝐹𝐹 = 𝑽𝑽𝐹𝐹 𝑍𝑍0 +𝑍𝑍1 +𝑍𝑍2 +3𝑍𝑍𝐹𝐹 3𝑽𝑽𝐹𝐹 𝑍𝑍0 +𝑍𝑍1 +𝑍𝑍2 +3𝑍𝑍𝐹𝐹 (8) ESE 470 SLG Fault - Example 78 Returning to our example power system The interconnected sequence networks for a bolted fault at bus 1: K. Webb ESE 470 SLG Fault - Example 79 The fault current is 𝑰𝑰𝐹𝐹 = 𝑰𝑰𝐹𝐹 = 3𝑽𝑽𝐹𝐹 𝑍𝑍0 +𝑍𝑍1 +𝑍𝑍2 +3𝑍𝑍𝐹𝐹 3.0∠30° 𝑗𝑗𝑗.487 = 6.16∠ − 60° 𝑝𝑝. 𝑢𝑢. 𝑆𝑆𝑏𝑏 𝑉𝑉𝑏𝑏𝑏 150 𝑀𝑀𝑀𝑀𝑀𝑀 3 230 𝑘𝑘𝑘𝑘 The current base at bus 1 is 𝑰𝑰𝑏𝑏 = = So the fault current in kA is = 376.5 𝐴𝐴 𝑰𝑰𝐹𝐹 = 2.32∠ − 60° 376.5 𝐴𝐴 K. Webb 𝑰𝑰𝐹𝐹 = 2.32∠ − 60° 𝑘𝑘𝑘𝑘 ESE 470 Unsymmetrical Fault Analysis – LL Fault 80 Now consider a line-to-line fault between phase 𝑏𝑏 and phase 𝑐𝑐 through impedance 𝑍𝑍𝐹𝐹 Phase-domain fault conditions: 𝑰𝑰𝑎𝑎 = 0 (9) 𝑰𝑰𝑏𝑏 = −𝑰𝑰𝑐𝑐 = 𝑽𝑽𝑏𝑏𝑏𝑏 −𝑽𝑽𝑐𝑐𝑐𝑐 (10) 𝑍𝑍𝐹𝐹 Transforming to the sequence domain 𝑰𝑰0 1 1 𝑰𝑰1 = 1 3 𝑰𝑰2 1 1 𝑎𝑎 𝑎𝑎2 1 𝑎𝑎2 𝑎𝑎 0 0 1 𝑎𝑎 − 𝑎𝑎2 𝑰𝑰𝑏𝑏 𝑰𝑰𝑏𝑏 = 3 −𝑰𝑰𝑏𝑏 𝑎𝑎2 − 𝑎𝑎2 𝑰𝑰𝑏𝑏 (11) So, the first two sequence-domain fault conditions are 𝑰𝑰0 = 0 K. Webb 𝑰𝑰2 = −𝑰𝑰1 (12) (13) ESE 470 Unsymmetrical Fault Analysis – LL Fault 81 To derive the remaining sequence-domain fault condition, rearrange (10) and transform to the sequence domain 𝑽𝑽𝑏𝑏𝑏𝑏 − 𝑽𝑽𝑐𝑐𝑐𝑐 = 𝑰𝑰𝑏𝑏 𝑍𝑍𝐹𝐹 𝑽𝑽0 + 𝑎𝑎2 𝑽𝑽1 + 𝑎𝑎𝑽𝑽2 − 𝑽𝑽0 + 𝑎𝑎𝑽𝑽1 + 𝑎𝑎2 𝑽𝑽2 = 𝑰𝑰0 + 𝑎𝑎2 𝑰𝑰1 + 𝑎𝑎𝑰𝑰2 𝑍𝑍𝐹𝐹 𝑎𝑎2 𝑽𝑽1 + 𝑎𝑎𝑽𝑽2 − 𝑎𝑎𝑽𝑽1 − 𝑎𝑎2 𝑽𝑽2 = 𝑎𝑎2 − 𝑎𝑎 𝑰𝑰1 𝑍𝑍𝐹𝐹 𝑎𝑎2 − 𝑎𝑎 𝑽𝑽1 − 𝑎𝑎2 − 𝑎𝑎 𝑽𝑽2 = 𝑎𝑎2 − 𝑎𝑎 𝑰𝑰1 𝑍𝑍𝐹𝐹 The last sequence-domain fault condition is K. Webb 𝑽𝑽1 − 𝑽𝑽2 = 𝑰𝑰1 𝑍𝑍𝐹𝐹 (14) ESE 470 Unsymmetrical Fault Analysis – LL Fault 82 Sequence-domain fault conditions 𝑰𝑰0 = 0 𝑰𝑰2 = −𝑰𝑰1 𝑽𝑽1 − 𝑽𝑽2 = 𝑰𝑰1 𝑍𝑍𝐹𝐹 (12) (13) (14) These can be satisfied by: K. Webb Leaving the zero-sequence network open Connecting the terminals of the positive- and negative-sequence networks together through 𝑍𝑍𝐹𝐹 ESE 470 Unsymmetrical Fault Analysis – LL Fault 83 The fault current is the phase 𝑏𝑏 current, which is given by 𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑏𝑏 = 𝑰𝑰0 + 𝑎𝑎2 𝑰𝑰1 + 𝑎𝑎𝑰𝑰2 𝑰𝑰𝐹𝐹 = 𝑎𝑎2 𝑰𝑰1 − 𝑎𝑎𝑰𝑰1 𝑰𝑰𝐹𝐹 = −𝑗𝑗 3 𝑰𝑰1 = K. Webb 𝑰𝑰𝐹𝐹 = 3 𝑽𝑽𝐹𝐹 ∠−90° 𝑍𝑍1 +𝑍𝑍2 +𝑍𝑍𝐹𝐹 −𝑗𝑗 3 𝑽𝑽𝐹𝐹 𝑍𝑍1 +𝑍𝑍2 +𝑍𝑍𝐹𝐹 (15) ESE 470 LL Fault - Example 84 Now consider the same system with a bolted line-to-line fault at bus 1 The sequence network: K. Webb ESE 470 LL Fault - Example 85 𝑰𝑰1 = 1∠30° 𝑗𝑗𝑗.363 = 2.75∠ − 60° The subtransient fault current is given by (15) as 𝑰𝑰𝐹𝐹 = 3 2.75∠ − 60° ∠ − 90° 𝑰𝑰𝐹𝐹 = 4.76∠ − 150° 𝑝𝑝. 𝑢𝑢. Using the previously-determined current base, we can convert the fault current to kA 𝑰𝑰𝐹𝐹 = 𝐼𝐼𝑏𝑏𝑏 ⋅ 4.76∠ − 150° 𝑰𝑰𝐹𝐹 = 4.76∠ − 150° 376.5𝐴𝐴 K. Webb 𝑰𝑰𝐹𝐹 = 1.79∠ − 150° 𝑘𝑘𝑘𝑘 ESE 470 Unsymmetrical Fault Analysis – DLG Fault 86 Now consider a double line-to-ground fault Assume phases 𝑏𝑏 and 𝑐𝑐 are shorted to ground through 𝑍𝑍𝐹𝐹 Phase-domain fault conditions: 𝑰𝑰𝑎𝑎 = 0 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑐𝑐 = 𝑽𝑽𝑏𝑏𝑏𝑏 𝑍𝑍𝐹𝐹 = 𝑽𝑽𝑐𝑐𝑔𝑔 𝑍𝑍𝐹𝐹 (16) (17) It can be shown that (16) and (17) transform to the following sequence-domain fault conditions (analysis skipped here) 𝑰𝑰0 + 𝑰𝑰1 + 𝑰𝑰2 = 0 𝑽𝑽1 = 𝑽𝑽2 K. Webb 𝑰𝑰0 = 1 3𝑍𝑍𝐹𝐹 𝑽𝑽0 − 𝑽𝑽1 (18) (19) (20) ESE 470 Unsymmetrical Fault Analysis – DLG Fault 87 Sequence-domain fault conditions 𝑰𝑰0 + 𝑰𝑰1 + 𝑰𝑰2 = 0 𝑽𝑽1 = 𝑽𝑽2 𝑰𝑰0 = 1 3𝑍𝑍𝐹𝐹 𝑽𝑽0 − 𝑽𝑽1 (18) (19) (20) To satisfy these fault conditions K. Webb Connect the positive- and negative-sequence networks together directly Connect the zero- and positive-sequence networks together through 𝑍𝑍𝐹𝐹 ESE 470 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 𝑰𝑰𝐹𝐹 = 𝑰𝑰𝑏𝑏 + 𝑰𝑰𝑐𝑐 = 3𝑰𝑰0 𝑰𝑰𝐹𝐹 = 3𝑰𝑰0 (21) 𝐼𝐼0 can be determined by a simple analysis (e.g. nodal) of the interconnected sequence networks K. Webb ESE 470 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 𝑍𝑍𝐹𝐹 = 0, 𝑽𝑽0 = 𝑽𝑽1 = 𝑽𝑽2 K. Webb ESE 470 DLG Fault - Example 90 To find 𝑰𝑰𝐹𝐹 , we must determine 𝑰𝑰0 We can first find 𝑉𝑉0 by applying voltage division 𝑍𝑍2 ||𝑍𝑍0 𝑽𝑽0 = 𝑽𝑽𝐹𝐹 𝑍𝑍1 + 𝑍𝑍2 ||𝑍𝑍0 𝑗𝑗𝑗.194||𝑗𝑗𝑗.124 𝑽𝑽0 = 1.0∠30° 𝑗𝑗𝑗.169 + 𝑗𝑗𝑗.194||𝑗𝑗𝑗.124 K. Webb 𝑽𝑽0 = 0.309∠30° ESE 470 DLG Fault - Example 91 Next, calculate 𝑰𝑰0 −𝑽𝑽0 −0.309∠30° 𝑰𝑰0 = = = 2.49∠120° 𝑝𝑝. 𝑢𝑢. 𝑍𝑍0 𝑗𝑗𝑗.124 The per-unit fault current is 𝑰𝑰𝐹𝐹 = 3𝑰𝑰0 = 7.48∠120° 𝑝𝑝. 𝑢𝑢. Using the current base to convert to kA, gives the subtransient DLG fault current 𝑰𝑰𝐹𝐹 = 7.48∠120° 376.5 𝐴𝐴 K. Webb 𝑰𝑰𝐹𝐹 = 2.82∠120° 𝑘𝑘𝑘𝑘 ESE 470
