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Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich 8. Short Circuit Analysis 8-1 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.1 Comparison of a short circuit in RL circuit and a generator a. DC component of the short-circuit current Solution of the differential equation: 2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ i + L di → dt Homogenous solution: di =0 dt di = λ ⋅ e λt i = e λt ; dt R ⋅i + L i = C ⋅e − t τ → R ⋅ e λt + L λ ⋅ e λt = 0 → λ = − ( C :int egration R 1 =− L τ cons tan t ) Non-homogenous solution: i = A. cos( ωt + α ) + B sin( ωt + α ) → di = − A.ω ⋅ sin( ωt + α ) + B ⋅ ω ⋅ cos( ωt + α ) → dt di → dt 2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ ( A. cos( ωt + α ) + B sin( ωt + α )) + L ⋅ (− A.ω ⋅ sin( ωt + α ) + B ⋅ ω ⋅ cos( ωt + α )) → 2 ⋅ U ⋅ sin( ωt + α ) = R ⋅ i + L 2 ⋅U = R ⋅ B − ω ⋅ L ⋅ A → B = 0 = R ⋅ A + X ⋅ B → B = −A ⋅ 2 ⋅ U + A ⋅ ωL R R = X R 2 ⋅ U + A ⋅ ωL = −A ⋅ → R X 2 ⋅U ⋅ X A=− with Z = R 2 + X 2 2 Z 2 ⋅U ⋅ R B= → Z2 ( i=− 2 ⋅U ⋅ X ) 2 ⋅U ⋅ R ⋅ sin( ωt + α ) = Z Z2 = 2 ⋅ I ⋅ (− sin θ . cos( ωt + α ) + cos θ ⋅ sin( ωt + α )) 2 . cos( ωt + α ) + ( ωL = X ) X R and cos θ = with sin θ = Z Z 8-2 Exercises Electric Power Systems I - III Trigonometric identitiy : cosu ⋅ sin v = UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich 1 ⋅ (sin( u + v ) − sin( u − v )) → 2 1 i = 2 ⋅ I ⋅ − ((sin( ωt + α + θ ) − sin( ωt + α − θ ) + (sin( ωt + α + θ ) + sin( ωt + α − θ )) 2 = 2 ⋅ I ⋅ sin( ωt + α − θ ) Overall solution: i( t ) = C ⋅ e − t τ + 2 ⋅ I ⋅ sin( ωt + α − θ ) Determination of the integration constant C and the final solution: i( 0 ) = 0 = C + 2 ⋅ I ⋅ sin( α − θ ) → C = − 2 ⋅ I ⋅ sin( α − θ ) ⇒ t − τ i( t ) = 2 ⋅ I ⋅ sin( ωt + α − θ ) − sin( α − θ ) ⋅ e 8-3 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich 8-4 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III t − i( t ) = 2 ⋅ I ⋅ sin( ωt + α − θ ) − sin( α − θ ).e τ with , U = 215V L = 0.27 H R = 15 Ω I= U R 2 + ( ωL )2 = 2 .5 A ωL 0 θ = tan −1 = 80 R L τ = = 18 ms R The DC component: u( t ) = 2 ⋅ 215 V ⋅ sin α = 100 V → α = 19.2 0 i d.c. = − 2 ⋅ I ⋅ sin( α − θ) ⋅ e − t τ = 3.08 A b. Instantaneous value of the voltage that produces maximum DC 8-5 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III sin(α − θ) = −1 → α − θ = −90 0 α = −10 0 → i d.c. = 3.54 A c. Instantaneous value of the voltage that results in the absence of any DC component sin(α − θ) = 0 → α − θ = 0 0 α = 80 0 → i d.c. = 0 d. Current values for switching at instantaneous voltage value of zero α=0 i( t ) = 2 ⋅ I ⋅ (sin( ωt − θ) + sin(θ).e − t τ 0 ) = 3.5 A ⋅ (sin( ωt − 80 ) + 0.985.e − t 0.018 ) t = 0.5 ⋅ 0.02 = 0.01s → i = 5.42 A t = 1.5 ⋅ 0.02 = 0.03 s → i = 4.1A t = 5.5 ⋅ 0.02 = 0.11s → i = 3.5 A e. Difference between short circuit currents in an R - L circuit and a synchronous generator While the short-circuit current in an R – L circuit maintains a constant amplitude for the entire duration of the short-circuit (assuming that the DC component is neglected), the short-circuit current in a synchronous generator does not. The flux across the air gap of a synchronous machine is determined by the combined action of the field, the armature, and the damper windings (or iron parts of the cylindrical rotor). Depending on the level of the transient processes in these windings and their interaction with one another, the post-fault period is categorized into sub-transient, transient and steady-state phases, each with its own characteristic reactance (Xd”, Xd’ and Xd) and the corresponding components of the shortcircuit current have decreasing magnitudes, i.e. Ik” > Ik’ > Ik. 8-6 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich 8.2 Symmetrical short-circuit in a generator – transformer unit Single-line diagram (with the fault on the low voltage side of the transformer): ∼ a. the sustained short-circuit current in the breaker " E =1 " Ik = E 1 S 100 = = −0.769 p.u. ⇒ Ik = 0.769 ⋅ = 0.769 ⋅ = 2.467 kA jx d j1.3 3 ⋅U 3 ⋅ 18 b. sub-transient, transient and the initial symmetrical RMS currents Sub-transient short-circuit current: " Ik = E " jx "d = 1 S 100 " = − j5.263 p.u. ⇒ Ik = 5.263 ⋅ = 5.263 ⋅ = 16.88 kA j0.19 3 ⋅U 3 ⋅ 18 Transient short-circuit current ' Ik = E " jx 'd = 1 100 ' = − j3.846 p.u. → Ik = 3.846 ⋅ = 12.34 kA j0.26 3 ⋅ 18 The initial symmetrical RMS current is the same as the sub-transient short-circuit current. c. the maximum possible dc component in the breaker i d.c. = 2 ⋅ Ik" = 23.874 kA Single-line diagram (with the fault on the high voltage side): ∼ d. the initial symmetrical RMS current on the high voltage side 8-7 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III x ps = 0.1 ⋅ 10.5 2 E " Ik = j( " + x ps ) x"d S " I k = 4.464 ⋅ = 4.464 ⋅ = 0.034 18 2 3 ⋅U 100 = 1 = − j 4.464 p .u . ⇒ j ( 0.19 + 0.034 ) = = 0.683kA 18 3 ⋅ 220 ⋅ 10.5 e. the initial symmetrical RMS current on the low voltage side. " I k = − j 0.683 kA ⋅ 220 = − j14.32kA 10.5 f. the breaker interrupting current. For the time delay specified (t = 0.1 s): " μ = 0.62 + 0.72 ⋅ e −0.32⋅IkG / IrG (DIN EN 60909-0) where: I "kG = 14.32kA IrG = 100 MVA 3 ⋅ 18 kV (initial symmetrical RMS current) = 3.2075 kA → (rated generator current) μ = 0.62 + 0.72 ⋅ e −0.32⋅14.32 / 3.2075 = 0.62 Breaker interrupting current: I b = μ ⋅ I "k = 0.62 ⋅ 14.32kA = 8.9 kA 8-8 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.3 Short-circuit current computation: internal emf method vs. Thévenin theorem I a. Short-circuit current computation using the internal emf method I=− 400 MW 3 ⋅ 20 kV X "d = x "d ⋅ " E = Un2 20 2 = 0. 2 ⋅ Ω = 0.16 Ω 500 Sn 20 kV 3 = −11.547 kA ∠0 0 − I ⋅ jX "d = 11.547 + j1.8475 kV = 11.69 kV∠9.09 0 " E 11.69 kV∠9.09 0 = 73.06 kA∠ − 80.910 Ik = " = j0.16 Ω jX d " b. Short-circuit current computation using the Thévenin theorem. Xd“ ZL Pre-fault voltage at the fault location: U th = 20 3 kV = 11.547 kV Equivalent impedance at the fault location: ZL = 20 kV 2 = 1Ω 400 MW Z th = jX "d // Z L = j0.16 // 1 = 0.158 Ω∠80.9 0 8-9 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Fault current: " Ik = U th 11.547 kV = = 73.08 kA∠ − 80.9 0 0 Z th 0.158 Ω∠80.9 Current flowing through the generator: (part of the current flowing through the generator + pre-fault current ) " " IkG = Ik ⋅ 1 " + j11.547 kA = − j72.162 + 11.547 kA = 73.08 kA∠ − 80.8 0 = Ik 1 + j0.16 8-10 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.4 Short-circuit current computation: internal emf method vs. Thévenin theorem II a. The sub-transient short-circuit currents using the internal EMF method Internal voltage of the generator: IM = 20 MW 0.8 ⋅ 12.8 kV ⋅ 3 2 ( 13.2kV ) = 0. 1⋅ X line 30 MVA X "d = X M" = 0.2 ⋅ E "G = I " kG 13.2 3 30 MVA j( X + X line ) 13.2 3 = 1.1616 Ω kV∠0 0 + IM ⋅ ( jX line + jX "d ) = 6.41kV∠14.2 0 " d E M" = = 0.58 Ω (13.2kV )2 E "g = ∠36.9 0 = 1127.64 A ∠36.9 0 = 3.679 kA∠ − 75.8 0 kV∠0 0 − IM ⋅ jX M" = 8.24 kV∠ − 7.3 0 " M " M E = 7.1kA∠ − 97.3 0 jX " = IkM Ik" = I"fG + I"fM = 3.679 kA∠ − 75.8 0 + 7.1kA∠ − 97.3 0 = − j10.6 kA b. Solution using the Thévenin theorem. Pr efualt voltage : U th = 12.8 kV∠0 0 3 Thevenin Equivalent impedance : X th = ( X "d + X line ) // X M" = Ik" = 1.74 ⋅ 1.1616 Ω = 0 .7 Ω 2 .9 U th 12.8 / 3 = kA = −10.6 kA jX th j0.7 Contribution of the motor and the generator to the short-circuit current: 8-11 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Pr e − fault currents : IM = S 3 ⋅U IG = −IM = 20 MW ∠36.9 0 = 1127.64 A∠36.9 0 0.8 ⋅ 12.8 kV Thevénin current (as a result of fault): IGth X M" 1.16 = Ik ⋅ " = −10.6 kA ⋅ = − j4.24 kA " 2 .9 X M + X d + X line IMth = Ik ⋅ X "d + X line 1.7424 = −10.6 kA ⋅ = −6.37 kA " " 2 .9 X M + X d + X line " IkG = 1127.64∠36.9 0 A " IkM = IMth − IM = − j6.37 kA − 1127.64 A∠36.9 0 = −0.9 − j7.047 kA = 7.1kA∠ − 97.3 0 " IkG = IGth − IG = − j4.24 kA + 1127.64 A∠36.9 0 = 0.9 − j3.563 kA = 3.675 kA∠ − 75.82 0 How does the pre-fault load current affect the current flowing into the fault location? The current flowing into the fault location is the same whether the pre-fault load current is considered or neglected. But in the equipments (generator, transformer, lines, etc.) the current during fault is the resulting current when the pre-fault current is superimposed on the fault current. Additionally, the pre-fault current indirectly affects the fault current in the sense that the Thevénin voltage at the fault location is determined by the load flow situation in the network. 8-12 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.5 Short-circuit in a generator connected to motors I a. The impedance diagram xd“ XM“ x "d = 0.2 p.u. x M" = 0.2 ⋅ 625 = 0.67 p.u. 187.5 b. Symmetrical short-circuit current (interrupted by breakers A and B) for a fault at P. Brea ker A : " IkA = 1 1 = = − j5 p.u. " jx d j0.2 Brea ker B : " IkB = 1 1 1 1 1 1 + " + " = + + = − j8 p.u " jx d jx M jx M j0.2 j0.67 j0.67 c. For a fault at point Q. Brea ker A : " IkA = 1 1 = = − j5 p.u. " jx d j0.2 Brea ker B : " IkB = 1 1 = = − j1.5 p.u " jx M j0.67 8-13 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III d. For a fault at point R. Brea ker A : 1 1 1 1 1 1 " IkA = " + " + " = + + = − j4.5 p.u jx M jx M jx M j0.67 j0.67 j0.67 Brea ker B : " IkB = 1 = − j1.5 p.u jx M" R A Q P B 8-14 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.6 Short-circuit in a generator connected to motors II a. the sub-transient current into the fault location All reactances on a base of 5 MVA: Generator : x "d = 0.2 p.u. Transforme r : x ps = 0.1p.u. Each Motor x "d = 0.2 ⋅ 25 MVA = 1p.u. 5 MVA 1 X th = 1//( // (0.1 + 0.15 ) = 0.125 3 1 = − j8.0 p.u. I"k = j0.125 b. the sub-transient current in breaker A " = IkA 1 1 j //(0.1 + 0.15) 3 = − j7.0 p.u. 8-15 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.7 Symmetrical short-circuit current computation 110kV G1 T3 220kV T6 110kV 220kV T1 T4 T5 T2 G2 T7 Q 2X240/40-Al/St T9 G9 T8 50km, double line SS2 SS3 240/40-Al/St 50km, double line SS1 SS4 Data of the network components: Xb’=0.298 Ω/km (220-kV overhead line; for the bundle in each system) Xb’=0.393 Ω/km (110-kV overhead line; for each system) Overhead line 220 kV: R/X = 0.26 Overhead line 110 kV: R/X = 0.3 G1, G2: 10.5 kV; 100 MVA; xd“ = 0.16; RsG/Xd“ = 0.05; G9: 21 kV; 225 MVA; xd“ = 0.19; RsG/Xd“ = 0.05; T1, T2: 120 MVA; uk =10%; ür = 115 kV/10.5 kV T3 …T8: 200 MVA; uk =12%; ür = 240 kV/110 kV T9: 250 MVA; uk =10%; ür = 112 kV/21 kV; R/X = 0.03 Q: SkQ“ = 20 GVA Calculation of reactances (all referred to the 110-kV side): 10.5 2 100 2 115 ⋅ Ω = 21.16 Ω 10.5 G1, G2: X "d = 0.16 ⋅ G9: 212 112 ⋅ X = 0.19 ⋅ Ω = 10.6 Ω 225 21 2 " d 8-16 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 115 2 = 0. 1⋅ Ω = 11.02 Ω 120 T1, T2: X ps T3…T8: X ps = 0.12 ⋅ T9: X ps = 0.1⋅ Q: 1.1⋅ 220 2 110 ⋅ XN = Ω = 0.559 Ω 20000 240 Xb220: 1 110 X b 220 = 0.298 ⋅ 50 ⋅ ⋅ Ω = 1.565 Ω 2 240 Xb110: 1 X b110 = 0.393 ⋅ 50 ⋅ Ω = 9.825 Ω 2 110 2 Ω = 7.26 Ω 200 112 2 Ω = 5.02 Ω 250 2 2 Calculation of impedances: G1, G2: Z G1 = 21.16 ⋅ (0.05 + j)Ω = 1.058 + j21.16 Ω = 21.186 Ω∠87.10 G9: Z G 9 = 10.6 ⋅ (0.05 + j)Ω = 0.53 + j1.06 Ω = 10.61Ω∠87.10 T1, T2: Z psT1 = j11.02 Ω T3…T8: Z psT 3 = j7.26 Ω T9: Z psT 9 = 5.02 ⋅ (0.03 + j)Ω = 0.1506 + j5.02 Ω = 5.022 Ω∠88.28 0 Q: Z N = j0.559 Ω " " 220-kVLine: Z L 220 = 1.565 ⋅ (0.26 + j)Ω = 0.4069 + j1.565 Ω = 1.617 Ω∠75.4 0 110-kVLine: Z L110 = 9.825 ⋅ (0.3 + j)Ω = 2.9475 + j9.825 = 10.26 Ω∠73.3 0 Calculation of equivalent impedance at the fault location: The impedance diagram of the network is given below. 8-17 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III SS3 SS2 SS1 Z3 Z2 SS4 Z1 After Delta → Star conversion: Z 1 = 0.4688 + j2.4349 Ω Z 2 = 0.1311 + j1.4921Ω Z 3 = −0.0596 + j0.5819 Ω Z eq1 " Z G1 + Z psT1 = Z1 + Z 2 + // (Z 3 + Z Q ) = 0.4147 + j3.5065 Ω 2 " Z eq2 = Z psT 9 + Z G 9 = 0.6806 + j15.62 Ω Z eq = Z eq 1 // Z eq 2 = 0.2993 + j2.866 Ω = 2.882 Ω∠84.04 0 a. The initial short-circuit current ( Ik" ) for a fault at SS4 " Ik = 1.1⋅ 110 kV 3 ⋅ Z eq = 24.24 kA∠ − 84.04 0 b. The first amplitude of the fault current (peak short-circuit current) κ = 1.02 + 0.98 ⋅ e −3R / X = 1.7364 ip = κ ⋅ 2 ⋅ Ik" = 59.525 kA c. The fault interrupting current for the maximum interrupting time delay of 0.25 s Distribution of the fault current: 8-18 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III " Q: IQ = 2.1151 − j18.4589 kA = 18.58 kA∠ − 83.46 0 Generator 9 : IG9 = 0.1945 − j4.464 kA = 4.468 kA∠ − 87.5 0 " Generators 1& 2 : I"G = 0.2087 − j1.189 kA = 1.2072 kA∠ − 80 0 Ik" = 2.5183 − j24.112 kA = 24.24 kA∠ − 84.04 0 Determination of µ: µ=1 Q: (Distance fault) IbQ = 18.58kA 2 IrG G1, G2: 2 ⋅ 100 MVA 10.5 = ⋅ = 91.68 A 3 ⋅ 10.5 kV 115 I"G 1207.2 = = 13.17 IrG 91.68 μ = 0.56 + 0.94 ⋅ e −0.38⋅13.17 = 0.566 IbG = 0.566 ⋅ 1207.2 A = 683.65 A 2 IrG9 225 MVA 21 = ⋅ = 217.47 A 3 ⋅ 21kV 112 I"G 4468 = = 20.55 IrG 217.47 G9: μ = 0.56 + 0.94 ⋅ e −0.38⋅20.55 = 0.56 IbG9 = 0.56 ⋅ 4.468 kA = 2.5 kA Total interrupting current at the fault location: Ib = 18.58 kA+0.68365 kA + 2.5 kA = 21.76 kA d. The maximum dc component of the short-circuit current i d.c. = 2 ⋅ Ik" = 34.28 kA e. The influence of the resistive components of the impedances on the fault current Neglecting all the resistive components results in the following values: After Delta → Star conversion: Z1 = j2.41Ω Z 2 = j1.469 Ω Z 3 = j0.594 Ω 8-19 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Z eq1 " Z G + Z psT1 = Z1 + Z 2 + 1 // (Z 3 + Z Q ) = j3.492 Ω 2 " Z eq2 = Z psT 9 + Z G 9 = j15.62 Ω Z eq = Z eq1 // Z eq 2 = j2.854 Ω " Ik = 1.1⋅ 110 kV 3 ⋅ Z eq = 24.48 kA∠ − 90 0 i d.c. = 2 ⋅ Ik" = 34.62 kA κ = 1.02 + 0.98 ⋅ e −3R / X = 2 ip = κ ⋅ 2 ⋅ Ik" = 69.23 kA By neglecting the resistive components in the impedances of the system elements, the computational effort can be immensely reduced without any significant loss of accuracy. 8-20 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.8 Mechanical and thermal stress caused by the short-circuit current 21 kV G 110 kV T G: UrG = 21 kV; SrG = 225 MVA; xd“ = 0.18; xd = 2; R/X= 0.05 T: SrT = 250 MVA; uk = 10% X "d = 0.18 ⋅ 21kV 2 = 0.3528 Ω 225 MVA Ik" = R = 0.05 → κ = 1.02 + 0.98 ⋅ e −3⋅0.05 = 1.86 X → 1.1⋅ 21kV 3 ⋅ 0.3528 Ω = 37.8 kA ip = κ ⋅ 2 ⋅ Ik" = 99.62 kA Arrangement of the Al – bars: Phase A 1 2 Phase C Phase B 3 1 2 1 3 2 3 b am am am = 350mm b = 200 mm as = 15mm d = 15 mm d as a11 a12 a13 8-21 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III a. The type of fault that causes maximum stress The symmetrical fault calls for the maximum short-circuit current and therefore causes the maximum mechanical (as well as thermal) stress on the aluminium bars. The bars of the middle phase experience the largest force. b. The total force on the middle phase bars by the fault current flowing in the other two phases General: Force on conductor 1 due to a current flowing in conductor 2 (placed at a distance a): ( ) F = l 1 X B 2 ⋅ I1 = μ0 ⋅ I1 ⋅ I2 ⋅ l 2π ⋅ a Three-phase system: In a three-phase system, the force on the middle conductor (phase b) assuming phases a and c are placed at a distance a on either side of b: F= μ0 ⋅ Ib ⋅ (Ia − Ic ) ⋅ l 2π ⋅ a The maximum force (Fmax) for the following current relation: i a = im ⋅ cos θ F( t ) = ib = im ⋅ cos(θ − 2π / 3) ic = im ⋅ cos(θ + 2π / 3) ( θ = ωt ) → μ0 ⋅ im2 ⋅ (cos(θ − 2π / 3) ⋅ (cos θ − cos(θ + 2π / 3))) ⋅l 2π ⋅ a dF( t ) d μ0 ⋅ im2 ⋅ (cos(θ − 2π / 3) ⋅ (cos θ − cos(θ + 2π / 3))) = ⋅ l = 0 dt dt 2π ⋅ a − sin(θ − 2π / 3) ⋅ ((cos θ − cos(θ + 2π / 3))) + cos(θ − 2π / 3) ⋅ (− sin θ + sin(θ + 2π / 3)) = 0 ⇒ − sin( 2θ − 2π / 3) + sin 2θ = 0 [Follows from the trigonomet ric identity ] sin( α + β) = sin α ⋅ cos β + cos α ⋅ sin β) The solution of the trigonometric equation: − sin( 2θ − 2π / 3) + sin 2θ = 0 → sin 2θ = sin( 2θ − 2π / 3) → 2θ − 2π / 3 = π − 2θ θ= [Follows from sin δ = sin( π − δ) ] 5 ⋅ π = 75 0 ⇒ 12 The peak force, therefore, is: Fmax = ( ( )) μ0 ⋅ im2 ⋅ cos(75 0 − 120 0 ) ⋅ cos 75 0 − cos(75 0 + 120 ) 3 μ0 ⋅ im2 ⋅l = ⋅ ⋅l 2π ⋅ a 2 2π ⋅ a 8-22 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III This relationship, however, is only valid under the assumption that the current is concentrated at the centre of the conductor and the mutual distance between the conductors is much larger than the conductor dimensions. To account for the deviation from this idealised assumption, the mutual distance should be replaced by an effective distance using a correction factor in the form: a' = a , where a = the actual distance and k = the correction factor. k The above peak force (including the correction factors) thus becomes: Fmax 3 μ0 ⋅ im2 ⋅ l k 11 k 12 k 13 k 21 k 22 k 23 k 31 k 32 k 33 = ⋅ ⋅ + + + + + + + + 2 2π a11 a12 a13 a 21 a 22 a 23 a 31 a 32 a 33 where kij are the correction factors. The mutual distances aij are shown in the figure above. The correction factors for rectangular bars can be obtained from Figure 1 (given in the problem) as follows: height b 200 = = = 13.33 → identify (or estimate) the applicable curve in Figure 1. breadth d 15 Then, a11 a m 350 = = = 23.33 → from curve : k 11 = 0.95 = k 22 = k 33 d d 15 a12 a m + 2d 380 = = = 25.33 → from curve : k 12 = 0.96 = k 23 d d 15 a13 a m + 4d 410 = = = 27.33 → from curve : k 13 = 0.97 d d 15 a 21 a m − 2d 320 = = = 21.33 → from curve : k 21 = 0.94 = k 32 d d 15 a 31 a m − 2d 290 = = = 19.33 → from curve : k 31 = 0.93 d d 15 ' max F Fmax 3 μ0 ⋅ im2 0.95 0.96 0.97 0.94 0.95 0.96 0.93 0.94 0.95 = = ⋅ ⋅ + + + + + + + + 2 2π 350 380 410 320 350 380 290 320 350 l = 3 μ0 ⋅ im2 ⋅ ⋅ 24.64 m −1 2 2π According to DIN VDE 0103, the peak current is to be used for the calculation of the maximum stress. Thus, im = ip 3 = 99.62 kA = 33.2 kA 3 (Division by 3 is necessary since the total current divides between the three bars in each of the phases) 8-23 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 3 4π ⋅ 10 −7 H ⋅ m −1 ⋅ (33.2 kA ) F = ⋅ ⋅ 24.64 m −1 = 4.7 kN / m 2 2π V ⋅ s ⋅ A ⋅ A V ⋅ s ⋅ A W ⋅ s N⋅m Unit check : = = = = N/m A ⋅m⋅m m⋅m m⋅m m⋅m 2 ' m c. The force exerted on the outer bar of the middle phase due to the currents flowing in the middle phase itself In part (a) it was stated that the middle phase experiences the biggest mechanical stress. Within the conductors of the middle phase, however, the middle bar experiences the least force (due to the currents flowing in the other two bars of the same phase) since the two outer bars apply antipodal forces on it with the resultant being zero). The force on one of the outer bars: Fs' = μ0 ⋅ im2 k 11 k 12 ⋅ + 2π 2d 4d Please note that the factor 3 is not necessary here since the two currents are in phase 2 and the force has its peak value when the current has its maximum value. 2d = 2 → k 11 = 0.35 d 4d = 4 → k 12 = 0.55 d Fs' = μ0 ⋅ im2 k 11 k 12 0.35 0.55 2 = 4.59 kN / m ⋅ + + = 2 ⋅ (33.2 kA ) ⋅ 10 −7 ⋅ 2π 2d 4d 30 mm 60 mm d. The combined force exerted on the mounting and its direction Although the two forces (one arising from the currents in the other two phases and the other from currents of the same phase) obviously have their maxima at two different instants. In spite of this they should be added directly (as an additional safety margin) in the dimensioning of the setup (DIN VDE 0103). Thus, the combined force: F = (Fs' + Fm' ) ⋅ l = ( 4.59 + 4.71) kN = 9.3 kN Each of the mountings absorbs half of this force. The force is horizontally directed. e. Capability of the Al bars to withstand the thermal stress during fault 8-24 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Determination of m: Fault current: Ik" = 37.8 kA (Calculated at the beginning). RG = 0.05 → κ = 1.86 X "d Tk = 0.2 s → Tk ⋅ f = 10 Using these values, we obtain from the curve → m = 0.35 Determination of n: Rated generator current: IrG = 225 MVA 3 ⋅ 21kV = 6185 .9 A Ik = 1.76 ⋅ IrG = 1.76 ⋅ 6185.9 A = 10887 .18 A Ik" 37.8 kA = = 3.47 Ik 10887 .18 A With Tk = 0.2 s from curve → n = 0.69 The thermal equivalent of the short circuit current: I th = Ik" ⋅ m + n = 38.55 kA Short-time current density: S th = Ith 38.55 kA = = 4.28 A / mm 2 A 9000 mm 2 ( A = 3 ⋅ 200 mm ⋅ 15 mm = 9000 mm 2 ) The Al – bar is designed for the allowable short-time current density of: S th _ max = 87 A / mm 2 (given in the problem), which leads to a temperature rise of 650C (operating temperature) to 2000C (maximum allowable temperature) in 1 s. The fault duration in this case is Tk = 0.2 s. This means, even a larger current density is tolerable, i.e. ( S 2th _ max ⋅ Tk = S 'th _ max ) 2 ⋅ 1s → S 'th _ max = S th _ max ⋅ 1s = 194.5 A / mm 2 Tk While the Al – bars are designed to withstand a short-time short-circuit current density of up to 194.5 A/mm2 for the duration of the short circuit considered here, the current density occurring here is only a tiny proportion of that, namely 4.28 A/mm2 → Capable of withstanding the thermal stress caused by the short circuit. 8-25 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.9 Calculation of short circuit currents in a meshed network Reactances (referred to the 110-kV level): 1.1⋅ Un2 1.1⋅ (110kV ) 2 XN = = = 3.33Ω 4GVA S k'' X G1 = X G2 = X T1 = X T 2 = X T 3 = X T 4 X T5 = X T6 XM = 0.12 ⋅ Un2 0.12 ⋅ (110kV ) 2 = = = 46.1Ω Sn 31.5MVA 0.11⋅ Un2 0.11⋅ (110kV ) 2 = = = 83.19Ω Sn 16MVA x M' ⋅ Un2 Pn / cos = n 0.3 ⋅ (110kV ) 2 = 242Ω 12MW / 0.8 X L1 = X b' ⋅ l = 0.4Ω / km ⋅ 15km = 6Ω X L 2 = X b' ⋅ l ⋅ 110 2 = 0.4Ω / km ⋅ 2.5km ⋅ 121 = 121Ω 10 2 Currents at fault locations: Fault location F1: U" = '' IkN = 1.1 ⋅ 110kV 3 = 69.859 kV U" = 20.98kA XN '' '' IkG 1 + IkG2 = U" = 0.93kA ( X G1 + X T1 + X T 3 ) / 2 + X L1 '' IkM ≈0 '' '' '' '' Ik'' = IkN + IkG 1 + IkG2 + IkM = 21.91kA Fault location F2: '' IkN = U" = 7.49kA X N + X L1 '' '' IkG 1 + IkG2 = U" ( X G1 + X T1 + X T3 ) / 2 = 1.01kA '' IkM ≈0 '' '' '' '' Ik'' = IkN + IkG 1 + IkG 2 + IkM = 8.5kA 8-26 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Fault location F3: U" = 2.16kA X N + X L1 + X T 3 // X T 4 '' = IkN '' '' IkG 1 = IkG 2 = U" = 0.76kA X G1 + X T1 U" = 0.215kA XM + X T5 '' = IkM '' '' '' Ik'' = IkN + 2 ⋅ IkG 1 + IkM = 3.895kA Referred to the 30-kV level: Ik'' 30 = Ik'' ⋅ 110 = 14.28kA 30 Fault location F4: '' '' + IkG = IkN U" = 683.5 A ( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 '' '' '' = (IkN + IkG IkN )⋅ ( X G1 + X T1 ) / 2 = 401.5 A X N + X L1 + X T 3 / 2 + ( X G1 + X T1 ) / 2 '' '' '' '' '' IkG 1 = IkG 2 = 0.5 ⋅ (IkN + IkG ) − IkN = 141A 1.1⋅ 110kV '' = IkM 3 I =I '' k '' kN ⋅ 1 = 289 A XM '' '' + 2 ⋅ IkG 1 + IkM = 972 A Referred to the 10-kV level: Ik'' 10 = Ik'' ⋅ 110 = 10.7kA 10 Fault location F5: Ik'' = U" = 0.3622kA [( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] // X M + X L 2 '' '' (IkN ) = Ik'' ⋅ + IkG XM = 254.6 A [( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] + X M '' '' (IkN ) = Ik'' ⋅ + IkG XM = 254.6 A [( X N + X L1 + X T 3 / 2) //( X G1 + X T1 ) / 2 + X T 5 ] + X M '' '' '' IkM = Ik'' − (IkN + IkG ) = 107.6 A 8-27 Exercises Electric Power Systems I - III '' '' '' IkN = (IkN + IkG )⋅ UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich ( X G1 + X T1 ) / 2 = 149.6 A X N + X L1 + X T 3 / 2 + ( X G1 + X T1 ) / 2 '' '' '' '' '' IkG 1 = IkG2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 52.5 A Referred to the 10-kV side: Ik'' 10 = Ik'' ⋅ 110 = 3.984kA 10 8-28 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich 8.10 Short Circuit Current Limitation a. Possibilities for short circuit current limitation ◊ Measure 1: opening the bus bar coupling A – B (for fault location F3) ◊ Measure 2: installation of an inductance in series with L2 (for fault location F5) b. Effectiveness of the alternative options Measure 1: No effect for fault locations F1 and F2. Fault location F3: '' '' + IkG IkN 2 = U" = 1.274kA ( X N + X L1 ) //( X G2 + X T 2 + X T 4 ) + X T 3 '' '' '' IkG 2 = (IkN + IkG 2 ) ⋅ X N + X L1 = 0.0805kA X N + X L1 + X G2 + X T 2 + X T 4 '' '' '' '' = (IkN + IkG IkN 2 ) − IkG 2 = 1.1935kA I '' kG1 U" = = 0.76kA X G1 + X T1 I '' kM U" = = 0.215kA XM + X T5 '' '' '' '' Ik'' = IkN + IkG 2 + IkG1 + IkM = 2.249kA F3 (30 kV): Ik'' 30 = Ik'' ⋅ 110 = 8.246kA 30 → Only 57.7 % of the short-circuit current without the countermeasure flows. Fault location F4: 8-29 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III '' '' + IkG = IkN [( X N + X L1 ) //( X G2 + X T 2 '' '' + IkG IkN + IkG 2 = (IkN )⋅ I = (I '' kG1 '' kN + I ) − (I '' kG '' kN '' '' '' IkG 2 = (IkN + IkG 2 ) ⋅ U" = 594.2A + X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5 ( X N + X L1 ) //( X G2 X G1 + X T1 = 372.6 A + X T 2 + X T 4 ) + X T 3 + X G1 + X T1 '' + IkG 2 ) = 221 .6 A X N + X L1 = 23.5 A X N + X L1 + X G2 + X T 2 + X T 4 '' '' '' '' = (IkN + IkG IkN 2 ) − IkG 2 = 349 .1A I '' kM U" = = 288.7 A XM '' '' '' '' Ik'' = IkN + IkG 2 + IkG1 + IkM = 882 .9 A F4 (10 kV): Ik'' 30 = Ik'' ⋅ 110 = 9.71kA 10 → The fault current now is reduced to 90% of the value without the reactance. Fault location F5: I = '' k ({[( X N + X L1 ) //( X G2 + X T 2 '' '' IkN + IkG = Ik'' ⋅ U" = 349.1A + X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5 } // X M ) + X L 2 ({[( X N + X L1 ) //( X G2 + X T 2 XM = 234.9 A + X T 4 ) + X T 3 ] //[ X G1 + X T1 ] + X T 5 } + X M ) '' '' '' IkM = Ik'' − (IkN + IkG ) = 114.2A '' '' '' '' (IkN + IkG 2 ) = (IkN + IkG ) ⋅ ( X N + X L1 ) //( X G2 X G1 + X T1 = 147.3 A + X T 2 + X T 4 ) + X T 3 + X G1 + X T1 '' '' '' '' '' IkG 1 = (IkN + IkG ) − (IkN + IkG 2 ) = 87.6 A '' '' '' IkG 2 = (IkN + IkG 2 ) ⋅ X N + X L1 = 9 .3 A X N + X L1 + X G2 + X T 2 + X T 4 '' '' '' '' IkN = (IkN + IkG 2 ) − IkG 2 = 138 A F5 (10 kV): Ik'' 30 = Ik'' ⋅ 110 = 3.84kA 10 Measure 2: XD = x D ⋅ Un2 3 ⋅ U n ⋅ In = 36.963Ω (referred to 110-kV level) Calculations similar to Measure 1, only transformer reactances XT3 and XT4 need to be increased by 36.936 Ω each Fault currents: Fault location F1: 8-30 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III I '' kN U" = = 20.98kA XN '' '' IkG 1 + IkG 2 = U" = 0.746kA ( X G1 + X T1 + X T 3 + X D ) / 2 + X L1 '' IkM ≈0 '' '' '' '' Ik'' = IkN + IkG 1 + IkG 2 + IkM = 21 .726kA Fault location F2: U" = 7.49kA X N + X L1 '' = IkN '' '' IkG 1 = IkG 2 = 1.1⋅ 110kV 3 ⋅ 1 = 0.399kA X G1 + X T1 + X T 3 + X D '' IkM ≈0 '' '' '' Ik'' = IkN + 2 ⋅ IkG 1 + IkM = 8.29kA Fault location F3: '' = IkN X N + X L1 + ( X T 3 '' '' IkG 1 = IkG 2 = U" = 1.374kA + X D ) || ( X T 4 + X D ) 1.1⋅ 110kV 3 ⋅ 1 = 0.76kA X G1 + X T1 " U = 0.215kA XM + X T5 '' = IkM '' '' '' Ik'' = IkN + 2 ⋅ IkG 1 + IkM = 3.109kA F3 (30 kV): Ik'' 30 = Ik'' ⋅ 110 = 11.4kA 30 Fault location F4: I '' kN +I '' kG U" = = 650.6 A ( X N + X L1 + X T 3 / 2 + X D / 2) || ( X G1 + X T1 ) / 2 + X T 5 '' '' '' = (IkN + IkG IkN )⋅ ( X G1 + X T1 ) / 2 = 309.3 A X N + X L1 + ( X T 3 + X D ) / 2 + ( X G1 + X T1 ) / 2 '' '' '' '' '' IkG 1 = IkG 2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 171 .15 A '' = IkM U" = 288.7 A XM '' '' '' Ik'' = IkN + 2 ⋅ IkG 1 + IkM = 939 .3 A F4 (10 kV): Ik'' 10 = Ik'' ⋅ 110 = 10.3kA 10 8-31 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Fault location F5: Ik'' = [( X N + X L1 + X D / 2 + X T 3 '' '' (IkN ) = Ik'' ⋅ + IkG U" = 357.6 A / 2) || ( X G1 + X T1 ) / 2 + X T5 ] || X M + X L 2 [( X N + X L1 + X D / 2 + X T 3 XM = 247.7 A / 2) || ( X G1 + X T1 ) / 2 + X T 5 ] + X M '' '' '' IkM = Ik'' − (IkN + IkG ) = 109.9 A '' '' '' IkN = (IkN + IkG )⋅ ( X G1 + X T1 ) / 2 = 117.8 A X N + X L1 + X D / 2 + X T 3 / 2 + ( X G1 + X T1 ) / 2 '' '' '' '' '' IkG 1 = IkG2 = 0.5 ⋅ [(IkN + IkG ) − IkN ] = 65 A F5 (10 kV) : Ik'' 10 = Ik'' ⋅ 110 = 3.934kA 10 8-32 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.11 Analysis of unsymmetrical faults a. Fault conditions (a) I A + IB + IC = 0 U A =U B =UC (b) IB = IC = 0 UA =0 (c) IA =0 (d ) I A = 0 ; I B = −I C U B =UC = 0 U B =UC b. Fault conditions in symmetrical components and sequence networks (a) I A + I B + I C = 0 → I A0 = 0 U A = U B = U C → U A0 = U A1 = U A2 = 0 UG“ UA1 UA2 IA0 = 0 (b) UA0 I B = I C = 0 → I A0 = I A1 = I A2 U A = 0→ U A0 + U A1 + U A2 = 0 8-33 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III IA1 UG“ UA1 IA2 UA2 IA0 UA0 (c) I A = 0 → I A0 + I A1 + I A2 = 0 U B = U C = 0 → U A0 = U A1 = U A 2 IA1 UG“ UA1 IA2 UA2 IA0 UA0 (d ) I A = 0 → I A0 + I A1 + I A2 = 0 ; I B = − I C → I A0 = 0 U B = U C → U A1 = U A 2 8-34 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III IA1 UG“ UA1 IA2 UA2 Positive sequence network: 01 U“ U“ Xd“ Xps 2 1 XN XL F 3 Negative sequence network: 01 Xd“ Xps 2 1 XN XL F 3 Zero sequence network: 00 XG0+3Xn Xps 1 XG0→∞ XL 2 F 3 Reactances: Generators G1 external network: 8-35 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III X "d = X 1 = X 2 = 0.19 ⋅ X G 0 = 0.067 ⋅ 380 2 Ω = 144.07Ω 13.8 2 380 2 Ω = 50.8Ω 13.8 2 380 2 X n = 0.095 ⋅ Ω = 72.03Ω. 13.8 2 Transformers T1: X ps = 115.5Ω Transmission line: X L1 = X L 2 = 21.66Ω X L 0 = 72.2Ω X N = 1.1 ⋅ U " = 1. 1 (380kV )2 10000MVA 380 3 = 15.88Ω kV = 241.33kV c. Single line-to-ground fault at bus 3 Z 1 = Z 2 = j ( X d" + X ps + X L1 ) // jX N = j (144.07 + 115.5 + 21.66) // 15.88Ω = j15.03Ω Z 0 = j ( X L 0 + X ps ) = j187.7Ω i. " The sub-transient fault current for a single line to ground fault I k (1) = 3 ⋅ U" 241.33kV = 3⋅ = − j 3.325kA Z 0 + Z1 + Z 2 j (15.03 + 15.03 + 187.7) ii. voltage at the fault location and at the terminals of G1 • voltage at the fault location 8-36 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III " − j 3.325kA = − j1.108kA 3 3 = U "− I a1−G1 ⋅ jX 1 = 241.33kV − (− j1.108kA ⋅ j15.03Ω ) = 224.68kV I a 0 = I a1 = I a 2 = U a1 I k (1) = U a 2 = − I a 2−G 2 ⋅ jX 2 = −(− j1.108kA ⋅ j15.03Ω ) = −16.65kV U a 0 = − I a 0−G1 ⋅ jX 0 = − − j1.108kA ⋅ j187.7Ω = −207.97kV U a = U a 0 + U a1 + U a 2 = −207.97kV + 224.68kV − 16.65kV ≈ 0 U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = −207.97kV + 224.68kV∠240 0 − 16.65kV∠120 0 = 2 − 311.99 − j 209 = 375.52∠213.82 0 U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = −207.97kV + 224.68kV∠120 0 − 16.65kV∠240 0 = 2 − 311.99 + j 209 = 375.52∠146.18 0 • voltage at the terminals of generator 1 Current out of generator 1 : I a1−G1 = I a 2−G1 = − j1.108kA ⋅ 15.88 = − j 0.0592kA 15.88 + 21.66 + 115.5 + 144.07 I a 0−G1 = − j1.108kA • Voltages Ua, Ub, Uc: U a1 = U "− I a1−G1 ⋅ jX d" = 241.33kV − (− j 0.0592kA ⋅ j144.07Ω ) = 232.8kV U a 2 = − I a 2−G 2 ⋅ jX d" = −(− j 0.0592kA ⋅ j144.07Ω ) = −8.53kV U a0 = 0 U a = U a 0 + U a1 + U a 2 = −8.53kV + 232.8kV = 224.27kV 2 U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 232.8kV∠240 0 − 8.53kV∠120 0 = → U b = −112.14 − j 209 = 237.18kV∠241.78 0 U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 232.8kV∠120 0 − 8.53kV∠240 0 = 2 → U c = −112.14 + j 209 = 237.18kV∠92.910 iii. Phase c current out of generator 1 ( ) I c = a ⋅ I a1−G1 + a ⋅ I a 2−G1 = − j 0.0592kA ⋅ a ⋅ + a = j 0.0592kA 2 2 d. Line-to-line fault at bus 3 i. fault current I a1 = − I a 2 = 1 241.33kV = = − j8.03kA Z1 + Z 2 j (15.03 + 15.03) I a0 = 0 2 I b = − I c = (a − a ) ⋅ (− j8.03kA) = − j 3 (− j8.03kA) = −13.9kA 8-37 Exercises Electric Power Systems I - III ii. UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Line-to-line voltage at the fault location U a1 = 241.33kV − I a1 ⋅ Z 1 = 241.33kV − (− j8.03kA) ⋅ j15.03Ω = 120.64kV U a 2 = −( j8.03kA) ⋅ j15.03Ω = 120.69kV U a = 120.64kV + 120.69kV = 241.33kV 2 U b = a ⋅ U a1 + a ⋅ U a 2 ≈ −U a1 = −120.64kV 2 U c = a ⋅ U a1 + a ⋅ U a 2 = U b U ab = 362kV U bc = 0 e. Double line-to-ground fault at bus 3 Fault current I a1 = 241.33kV 241.33kV = = − j8.34 kA Z 1 + Z 2 // Z 0 j (15.03 + 15.03 // 187.7)Ω I a 2 = − I a1 ⋅ Z0 = j 7.69kA Z0 + Z2 I a 0 = − I a1 ⋅ Z2 = j 0.62kA Z0 + Z2 " I k ( 2) = 3 ⋅I a 0 = j1.85kA Voltage at the fault location U a1 = 241.33kV − I a1 ⋅ Z 1 = 115.98kV U a 2 = − I a 2 ⋅ Z 2 = 115.58kV U a 0 = − I a 0 ⋅ Z 2 = 116.37 kV U a 0 ≈ U a1 ≈ U a 2 U a = U a 0 + U a1 + U a 2 = 348kV 2 U b = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 0 2 U c = U a 0 + a ⋅ U a1 + a ⋅ U a 2 = 0 8-38 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.12 Effect of transformer neutral point connection on system performance during fault Network parameters: Overhead line B – C: Positive and negative sequence: j0.32 ⋅ 100 = j16 Ω 2 Zero-sequence: j1.12 ⋅ 100 = j56 Ω 2 Overhead line B – A: Positive and negative sequence: j0.32 ⋅ 50 = j8 Ω 2 Zero-sequence: j1.12 ⋅ 50 = j28 Ω 2 Overhead line C – A: Positive and negative sequence: j0.32 ⋅ 50 = j8 Ω 2 Zero-sequence: j1.12 ⋅ 50 = j28 Ω 2 220-kV Cable Positive and negative sequence: j 0.16 ⋅ 30 = j 2 .4 Ω 2 Zero-sequence: j 0.56 ⋅ 30 = j 8 .4 Ω 2 110-kV Network: Substation A: Positive and negative sequence: Z N = 1 .1 ⋅ Zero-sequence: Z0 → 220 2 = 17.75 Ω 3000 ∞ 8-39 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Substation B Positive and negative sequence: Z N = 1 .1 ⋅ Zero-sequence: Z0 → 220 2 = 11.83 Ω 4500 ∞ 220-kV Network: Positive and negative sequence: Z N = 1.1 ⋅ Zero-sequence: Z0 → 220 2 = 21.3 Ω 2500 ∞ 380-kV Network: 220 2 = 5.32 Ω 10000 Positive and negative sequence: Z N = 1 .1 ⋅ Zero-sequence: 231 Z 0 = 50 ⋅ = 16.68 Ω 400 2 Transformer A Positive and negative sequence: Zero-sequence: Xp = 2312 2312 2312 1 + 0.148 ⋅ − 0.09 ⋅ 0.155 ⋅ 83 83 250 4 = 17.59Ω Xs = 2312 2312 2312 1 − 0.148 ⋅ + 0.09 ⋅ 0.155 ⋅ 83 83 250 4 = −1.05 Ω Xt = 2312 2312 2312 1 − 0.155 ⋅ + 0.148 ⋅ + 0.09 ⋅ 83 83 250 4 = 29.98 Ω X 0 = X pt = 47.57 Ω 8-40 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Transformer B Positive and negative sequence: X p = 17.59 / 2 = 8.8 Ω X s = - 1.05 / 2 = -0.53Ω X t = 29.98 / 2 = 15.0 Ω X 0 = X pt = 23.79 Ω Zero-sequence Transformer C 1 0.145 ⋅ 2312 ⋅ = 8.23 Ω 4 235 Positive and negative sequence: X ps = Zero-sequence X 0 = X ps = 8.23 Ω Transformer D Positive and negative sequence: 2312 2312 2312 1 + 0.166 ⋅ − 0.113 ⋅ X p = j ⋅ 0.1 ⋅ 210 210 630 2 = j10.97 Ω 2312 2312 2312 1 − 0.166 ⋅ + 0.113 ⋅ X s = j ⋅ 0.1 ⋅ 210 210 630 2 = -j 2.5 Ω 2312 2312 2312 1 + 0.166 ⋅ + 0.113 ⋅ X t = j ⋅ − 0.1 ⋅ 210 210 630 2 = j31.21Ω Generator C Positive and negative sequence X "d = 1 0.19 ⋅ 2312 ⋅ = 10.79 Ω 4 235 X 2 = 10.79 Ω 8-41 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Positive-sequence network: 01 B110 C220 B220 C015 A220 A110 D220 Branch B110 - G A110 - G B220 - G C015 - G D380 - G A220 - A110 D220 - D380 A220 - D220 B220 - B110 C220 - C015 B220 - A220 B220 - C220 A220 - C220 D380 Reactance [Ω] 11.83 17.75 21.3 10.79 5.32 16.54 8.47 2.40 8.27 8.23 8.00 16.00 8.00 8-42 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Negative sequence: The negative-sequence network is similar to the positive sequence network without the voltage sources. For generators: X 2G = X "d + X "q 2 ≈ X1G (during the sub-transient phase); otherwise X1G ≠X2G Zero-sequence network: 00 B110 A110 C220 B220 A220 Branch B220 - 00 (Xpt) (4 trafos in parallel) A220 - 00(Xpt) (2 trafos in parallel) C220 - 00 (Xps) (4 trafos in parallel) A220 - D220 (cable) D220 - D0 (Xs) D0 - 00 (Xt) D0 - D380 – 00 (ZN0+Xp) D0 - D380 - 00 (380-kV side isolated) A220 - C220 (line) A220 - B220 (line) C220 - B220 (line) D220 D380 D0 Reactance [Ω] 23.79 47.57 8.23 8.40 -2.50 31.21 27.65 →∋ 28.00 28.00 56.00 8-43 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich The fault location is assumed to be the bus bar A 220. a. The initial single-phase (I‘‘k(1)) and three-phase (I‘‘k(3)) short-circuit currents Delta → Star conversion of the positive-sequence line reactance: 8⋅8 = 2Ω 32 16 ⋅ 8 C220 − 0 : X = = 4Ω 32 16 ⋅ 8 B220 − 0 : X = = 4Ω 32 A 220 − 0 : X = U" = 1.1 ⋅ 220 3 = 139.72kV Equivalent reactance of the positive (negative)-sequence network: X B220 = ((11.83 + 8.27 ) // 21.3) + 4 = 14.34 Ω X C220 = ( 4 + 8.23 + 10.79) = 23.02 Ω X A220 = (2.4 + 8.47 + 5.32) // (16.54 + 17.75) = 11.0 Ω X1 = X 2 = X A220 //(2 + X B220 //X C220 ) = 5.46Ω I"k(3) = U" = − j25.59kA jX1 Delta → Star conversion of the zero-sequence line reactance: 28 ⋅ 28 = 7Ω 112 28 ⋅ 56 C220 − 0 : X = = 14 Ω 112 28 ⋅ 56 B220 − 0 : X = = 14 Ω 112 A 220 − 0 : X = X 0 = ((23.79 + 14 ) //(14 + 8.23 ) + 7) // 47.57 //(8.4 − 2.50 + 31.21 // 27.65 ) = 8.527 Ω I"k(1) = 3 ⋅ U" = − j21.55kA 2 ⋅ X1 + X 0 b. Earth fault factor U a0 = − Ia0 ⋅ jX 0 " Ua1 = U − Ia1 ⋅ jX1 Ua2 = − Ia 2 ⋅ jX 2 8-44 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Ua 1 1 2 Ub = 1 a U 1 a c 1 U a0 a ⋅ Ua1 2 a Ua2 Ua = 0 " Ub = −Ia0 ⋅ jX 0 − a 2 ⋅ (U − Ia1 ⋅ jX1 ) − a ⋅ Ia 2 ⋅ jX 2 = U" ⋅ X 0 ⋅ (a 2 − 1) + X1 ⋅ (2a 2 + 1) X 0 + 2 X1 − " =U ⋅ 3 X0 3 X0 ⋅ −j ⋅ + 2 2 X1 2 X1 X0 +2 X1 " Uc = −Ia0 ⋅ jX 0 − a ⋅ (U − Ia1 ⋅ jX1 ) − a 2 ⋅ Ia 2 ⋅ jX 2 " =U ⋅ X 0 ⋅ (a − 1) + X1 ⋅ (2a + 1) X 0 + 2 X1 − " =U ⋅ 3 X0 3 X0 ⋅ +j ⋅ + 2 2 X1 2 X1 X0 +2 X1 2 X0 X + 0 + 1 X1 X1 Ub = Uc = U" ⋅ 3 ⋅ X0 +2 X1 = UN 3 ⋅ cf 2 X0 X + 0 + 1 X1 X1 c f (Earth fault factor ) = 1.1 ⋅ 3 ⋅ X0 +2 X1 c. Voltage of the phases that are not directly affected by the fault i. For the given case 8-45 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III X 0 8.53 = = 1.56 → c f = 1.196 X1 5.46 Ub = Uc = 1.196 ⋅ 220 3 = 151.91 kV ii. only one 220-kV neutral point grounded X1 = 5.46 Ω X 0 = (( 4 ⋅ 23.79 + 14 ) //(14 + 4 ⋅ 8.23 ) + 7) //(2 ⋅ 47.57 ) // (8.4 − 2.50 + 31.21 // 27.65 ) = 11.51Ω X 0 11.51 = = 2.11 → c f = 1.275 X1 5.46 Ub = Uc = 1.275 ⋅ 220 3 = 161.95 kV iii. transformer with isolated neutral point in substation D X1 = 5.46 Ω X 0 = (( 4 ⋅ 23.79 + 14 ) //(14 + 4 ⋅ 8.23) + 7) //( 2 ⋅ 47.57) //(8.4 − 2.50 + 31.21) = 15.98 Ω X 0 15.98 = = 2.93 → c f = 1.367 X1 5.46 Ub = Uc = 1.367 ⋅ 220 3 = 173.63 kV d. The neutral point to ground voltage (UEM) of the isolated transformers Iao = U" 139.72 kV = = − j5.194 kA j(2 ⋅ X1 + X 0 ) j(2 ⋅ 5.46 + 15.98 ) Ω Distribution of the zero-sequence current: Substation Zero-sequence current A -j0.872kA B -j0.62678kA C -j1.458kA D 2.237kA 8-46 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 0 UEM C220 A110 A220 D220 D0 D380 The neutral to ground voltage (UEM) for substation A is illustrated in the figure above. As can be seen in the figure, this voltage is equal to the zero-sequence voltage at the corresponding bus bar (in this case A). Thus, A UEM = 0.872 kA ⋅ 95.14Ω = 83kV UBEM = 0.62678 kA ⋅ 95.16Ω = 59.64kV C UEM = 1.458 kA ⋅ 32.92Ω = 48kV 8-47 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.13 Neutral point grounding in medium voltage networks a. sequence network Upa1 XN Xps XN Xps 3ZME Xst CE X ps = 1.26 Ω X st = 0.51Ω S k" = 2500 MVA → X N = 1.1 ⋅ 20 2 Ω = 0.176 Ω 2500 IC = 0.02 ⋅ 100 + 1.47 ⋅ 30 = 46.1 A IC = X CE 20 kV ⋅ ωCE → CE = 12.708 μF 3 ≥ 3Z ME ≥ 0 → X CE = 1 = 250.48 Ω ωCE 8-48 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III b. Single line-to-ground fault current for isolated neutral point (3ZME→∞) Upa1 " Ik (1) = 3 ⋅ 2 ⋅ j( X ps + X N ) + j( X st + 3Z ME ) // Z ME → ∞ and ( X ps + X N ) << 1 jωC E 1 ⇒ ωC E " Ik (1) ≈ 3 ⋅ ωC E ⋅ Upa1 = 3 ⋅ IC = 3 ⋅ 46.1A = 138.3 A This same result can be obtained directly from the network without using the symmetrical components, as illustrated below. Uac.jωCE c Uab.jωCE b a CE ICE Ik" (1) ≈ ICE = Uab ⋅ jωC E + Uac ⋅ jωC E = jωCE ⋅ (Ua − Ub + Ua − Uc ) = 3 ⋅ jωC E ⋅ Ua = 3 ⋅ jωCE ⋅ Upa1 = 3 ⋅ IC with: I“k(1) = single line-to-ground fault current; ICE = capacitive earth leakage current IC = charging current And the single line-to-ground fault current considering the positive and negative sequence reactances is: Ik" (1) = 3. Upa1 2 jX1 + ( − jX C ) = 3 ⋅ 20 kV = j139.90 A j(2 ⋅ (0.176 + 1.26 ) − 250.48 ) Ω c. Resonant grounding (3ZME=3XME=XCE) i. reactance for resonant tuning 8-49 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III For resonance tuning: IL (coil current) = ICE (capacitive earth leakage current) = 138.3 A→ ω ⋅ L ⋅ Upa1 = 138.3 A → ω ⋅ L = X L = 20 kV 3 ⋅ 138.3 A = 83.5 Ω = X ME ii. Equivalent circuit of the zero-sequence network for a damping factor of 4% and a detuning (off-resonance factor) of -3% (XME ≠ XL) Y0 = 1 1 1 1 + + = j3 ⋅ X ME − jX CE R D X CE 3 ⋅ X ME − X CE X CE ⋅ j + 3 ⋅ X ME RD X st << 3 X ME → 3 X ME + X st ≈ 3 X ME Definition: v (off - resonance factor) = d (damping factor) = Y0 = 1 X CE 3 ⋅ X ME − X CE ICE − IL = 3 ⋅ X ME ICE X CE RD ⋅ (d + jv ) X CE = 0.04 → R D = 6.262kΩ RD 3 ⋅ X ME − X CE = −0.03 → 3 ⋅ X ME = 243.18Ω 3 ⋅ X ME iii. Single line-to-ground fault current 1 " Ik(1) = Ir = 3 ⋅ Upa1 ⋅ Y0 = 3 ⋅ Upa1 ⋅ X CE ⋅ (d + jv ) = ICE ⋅ (d + jv ) = 0.05 ⋅ ICE ∠ − 36,87 0 = 6.915 A∠ − 36,87 0 Ir = residual current Irw = ICE ⋅ d= 138.3 ⋅0.04 = 5.532 A (real part of the residual current ) Irb = ICE ⋅ v = 138.3 ⋅ 0.03 = 4.149 A (imaginary part of the residual current ) d. Impedance grounding (0<3XME<3XCE) i. Grounding reactance for a maximum fault current of 1000A 8-50 Exercises Electric Power Systems I - III 1 = 1000 A → 3 2 ⋅ X1 + X 0 2 ⋅ X1 + X 0 = 36.37 Ω Ik" (1) = 3 ⋅ 1.05 ⋅ 20 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich ⋅ X1 = X 2 = X ps + X N = 1.26 Ω + 0.176 = 1.436 Ω X st = 0.51Ω X 0 = 3 ⋅ X ME + X st = 36.37Ω − 2.52 Ω = 33.5 Ω → X ME = X 0 − X st = 11 Ω 3 ii. The effect of neglecting the zero-sequence capacitance X 0 = ( j3 ⋅ X ME + jX st ) // − jX CE = ( j33 + j0.51) // − j250.48 = j38.69Ω Ik" (1) = 3 ⋅ 1.05 ⋅ 20 1 = − j875.15 A 3 2 ⋅ X1 + X 0 ⋅ The earth capacitance compensates part of the grounding reactance, and, as a result, the actual fault current is significantly smaller. Neglecting the zero-sequence capacitance, therefore, would lead to a significant error (the error in this example = 1000 A – 875.15 A= 124.85 A). e. Transformer neutral point solidly grounded (3ZME = 0) i. Single line-to-ground fault Ik" (1) = 3 ⋅ 1.1 ⋅ 20 1 = 11.267kA 3 2 ⋅ X 1 + X st ⋅ ii. Effect of the zero-sequence capacitance on the fault current X 0 = j051 // − 250.58 = j0.511 Ik" (1) = 3 ⋅ 1.1⋅ ◊ 20 1 = 11.264kA → 3 2 ⋅ X1 + X0 ⋅ The effect of the zero-sequence capacitance on the fault current is negligible. As a result of the small zero-sequence transformer reactance (Xst), the capacitance is practically parallel short-circuited, and can thus be neglected. f. Discussion of the results Basically, all methods of neutral point grounding can be viewed as alternative options. However, for this network practical considerations reduce the options to only resonant and impedance grounding because of the following reasons: ◊ Isolating the neutral point is not an option since the capacitive earth leakage current of ICE ≈ 138A does not guarantee a self-extinguishing arc. 8-51 Exercises Electric Power Systems I - III ◊ UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich In case of solid grounding, the high short-circuit current (I“k(1) ≈ 11.3kA) would cause an unacceptably high stress on equipment along the short-circuit path. Of the two remaining options, the resonant grounding offers a clear advantage over the (low) impedance grounding in this network. Because, although the large overhead line contributes only a small amount to the capacitive earth leakage current, it gives rise to a much larger probability of earth fault. In case of impedance grounding, this would lead to tripping of the protective relays to isolate the faulted section of the distribution network, and, therefore, to service interruption. In the event of resonant grounding, a fault does not necessarily lead to immediate tripping, resulting in improved supply continuity. 8-52 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.14 Problem of voltage un-symmetry in three phase transmission systems c M b a Note: only one of the 3 impedances YaE=GaE+jωCaE is drawn for clarity. ZEM E Applying Kirchoff’s current law: I EM = I aE + I bE + I cE → I EM = U aE ⋅ Y aE + U bE ⋅ Y bE + U cE ⋅ Y cE U aE = U aM − U EM because (Z LE >> Z L ) 2 U bM = a ⋅ U aM U cM = a ⋅ U aM → ZL: line impedance; ZLE: line – earth impedance U EM (Y = U EM = = = d= ) + a 2 ⋅ Y bE + a ⋅ Y cE ⋅ U aM 3 ⋅ Y 01 = ⋅ U aM (Y aE + Y bE + Y cE ) + Y EM 3 ⋅ Y 00 + Y EM aE G EM G EM 3Y 01 ⋅ U aM + jB EM + 3(G 00 + jB00 ) 3Y 01 ⋅ U aM + 3G 00 + j (B EM + 3B00 ) 3Y 01 G + 3G 00 + 3B00 B 3B00 EM + j EM 3B00 3B00 G EM + 3G 00 I = wr 3B00 I CE v= ⋅ U aM = B EM + 3B00 = 3B00 Y 01 ⋅ U aM B00 (d + jv ) 3ωC 00 − 1 X EM 3ωC 00 = I CE − I L I CE 8-53 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III I wr = (G ME + 3G 00 ) ⋅ U EM I CE = 3ωC 00 ⋅ U EM IL = 1 X ME ⋅ U EM a. Ohmic symmetry, capacitive un-symmetry G aE = GbE = G cE C aE ≠ C bE = C cE → C aE + ∆C = C bE = C cE ( for example if C aE > C bE ) UEM = j (Y 2 aE ) + a ⋅ Y bE + a ⋅ Y cE ⋅ UaM 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) = (C 2 aE ) + ∆C + a ⋅ C bE + a ⋅ C cE ⋅ jω ⋅ UaM = 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) ∆C ⋅ ( v + jd) ∆C ⋅ UaM = ⋅ UaM 3 ⋅ C 00 ⋅ (d + jv ) 3 ⋅ C 00 ⋅ d 2 + v 2 v = 0 → UEM = j ∆C ⋅U 3 ⋅ C 00 ⋅ d aM v = ±∞ → UEM = ∆C ⋅ UaM 3 ⋅ C 00 b. Capacitive symmetry, Ohmic un-symmetry G aE ≠ GbE = G cE C aE = C bE = C cE → G aE + ∆G = GbE = G cE ( for example if G aE > G bE ) UEM (Y = 2 aE ) + a ⋅ Y bE + a ⋅ Y cE ⋅ UaM 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) (G = 2 aE ) + ∆C + a ⋅ GbE + a ⋅ G cE ⋅ ω ⋅ UaM = 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) ∆G ⋅ (d − jv ) ∆G ⋅ UaM = ⋅ UaM 3 ⋅ C 00 ⋅ (d + jv ) 3 ⋅ C 00 ⋅ d 2 + v 2 v = 0 → UEM = ∆G ⋅U 3 ⋅ C 00 ⋅ d aM v = +∞ → UEM = − j v = −∞ → UEM = j ∆G ⋅ UaM 3 ⋅ C 00 ∆G ⋅ UaM 3 ⋅ C 00 c. Ohmic symmetry, capacitive un-symmetry (general) 8-54 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III G aE = GbE = G cE C aE ≠ C bE ≠ C cE → C aE = C aE C bE = C aE + ∆C bE C cE = C aE + ∆C cE UEM = (Y 2 aE ) + a ⋅ Y bE + a ⋅ Y cE ⋅ UaM 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) (a ) 2 ⋅ ∆C bE + a ⋅ ∆C cE ⋅ jω ⋅ UaM 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) = with : Δc bE = ΔC bE 3 ⋅ ω ⋅ C 00 Δc cE = ΔC cE → 3 ⋅ ω ⋅ C 00 ( per unit ) ( ) ( 2 2 U EM a ⋅ ΔC bE + a ⋅ ΔC cE j a ⋅ Δc bE + a ⋅ Δc cE = ⋅ jω = 3 ⋅ ω ⋅ C 00 ⋅ (d + jv ) U aM (d + jv ) ) d. Conductor-to-earth capacitances UEM UaM = ( ) 2 j a ⋅ ∆c bE + a ⋅ ∆c cE → (d + jv ) ( 2 ) j a ⋅ ∆c bE + a ⋅ ∆c cE 11.6 kV∠245 0 = → 110 0.025 kV 3 4.5663 ⋅ 10 −3 ∠245 0 = ∆c bE (cos( −30 0 ) − j sin 30 0 ) + ∆c cE (cos 210 0 + j sin 210 0 ) → ∆c bE = 3.0242 ⋅ 10 −3 ∆c cE = 5.2528 ⋅ 10 −3 ICE = 3 ⋅ ω ⋅ C 00 ⋅ Un 3 = 3 ⋅ C 00 ⋅ 100 ⋅ π ⋅ 110 3 kV = 395 A → 3 ⋅ C 00 = 19.8µF ∆C bE = ∆c bE ⋅ 3 ⋅ C 00 = 3.0242 ⋅ 10 −3 ⋅ 19.8 µF = 0.059879 µF ∆C cE = ∆c cE ⋅ 3 ⋅ C 00 = 5.2528 ⋅ 10 −3 ⋅ 19.8 µF = 0.104 µF C aE + C bE + C cE = 3 ⋅ C aE + ∆C bE + ∆C cE = 3 ⋅ C 00 → CaE = 6.5454 µF CbE = 6.5454 µF + 0.059879 µF = 6.6 µF CcE = 6.5454 µF + 0.104 µF = 6.65 µF 8-55 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.15 Low impedance grounding in cable networks Sequence networks: 01 1.05 UN 3 j0.176Ω A110 j1.26Ω A20-2 0.45+j0.3 Ω B2 j1.26Ω A20-1 0.45+j0.3Ω B1 0.45+j0.3Ω 0.6+j0.4Ω C2 0.45+j0.3Ω D2 C1 0.3+j0.2Ω D1 Negative sequence equivalent circuit is identical to the positive sequence network without the voltage sources. 8-56 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Zero-sequence network: 00 3ZME 3ZME j0.51Ω j0.51 XCE A20-2 0.3+j1.92Ω B2 0.3+j1.92Ω C2 0.3+j1.92Ω D2 X CE = 3 ⋅ Un = ICE A20-1 0.3+j1.92Ω B1 0.4+j2.56Ω C1 0.2+j1.28Ω D1 3 ⋅ 20 ⋅ 10 3 Ω; ICE [ A ] X CE = 679.24Ω (ICE = 51A ) X CE = 407.54Ω (ICE = 85 A ) X ME = R ME = 12 Ω Upa1 = 1.05 ⋅ 20 3 = 12.124kV XN = 0.176 Ω Xps = 1.26 Ω a. Single line-to-ground fault at bus A i. Coupling switch in bus bar A open, loop opened at bus D, transformer neutral points not interconnected 8-57 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Z 0 = ( j3 ⋅ X ME + jX st ) // − jX CE = j38.58Ω reac tan ce grounding Z 0 = (R ME + jX st ) // − jX CE = 36Ω∠ − 2.23 0 resis tan ce grounding Z1 = Z 2 = 0.176 Ω + 1.26 Ω = 1.436 Ω Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 1011 .1∠ − 2.2 0 A ⇔ for grounding through R ME Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 877.47∠ − 90 0 A ⇔ for grounding through X ME ii. Coupling switch in bus bar A open, loop opened at bus D, transformer neutral points not interconnected Z 0 = − jX CE //( jX st + (3Z ME // 3Z ME //( jX st − jX CE ))) = j19.55Ω ( X ME ) = 17.97Ω∠ − 1.5 0 (R ME ) Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 2007.2∠ − 7.7 0 A ⇔ for grounding through R ME Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 1620.8∠ − 90 0 A ⇔ for grounding through X ME iii. Coupling switch in bus bar A closed, loop opened at bus D, transformer neutral points not interconnected Z0 = − jX CE //( jX st + 3Z ME ) − j679.24 //( j0.51Ω + 3Z ME ) = 2 2 = j19.29Ω ( Z ME = j3 X ME = j36 Ω) = 17.99Ω∠ − 2.23 0 ( Z ME = 3R ME = 36 Ω) Z1 = Z 2 = j1.436Ω Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 2007.73∠ − 6.88 0 A ⇔ for grounding through R ME Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 1640.65∠ − 90 0 A ⇔ for grounding through X ME iv. Coupling switch in bus bar A closed, loop opened at bus D, transformer neutral points interconnected - same as (iii) - b. Fault at Bus B2, CB to the left of B2 open 8-58 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Z 0 = − jX CE //( jX st + 3Z ME ) + (1.5 + j9.6) Ω X CE = 407.54 Ω Z 0 = 1.5 + j49.7 = 49.73Ω∠88.27 0 ( Z ME = jX ME ) Z 0 = 37.95Ω∠10.54 0 ( Z ME = R ME ) Z1 = Z 2 = 2.25 + j2.936 Ω = 3.7∠52.5 Ω Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 36.372kV 2 ⋅ 3.7Ω∠52.5 0 + 37.95Ω∠10.54 0 = 831.74 A∠ − 17.03 0 ⇔ (grounding through R ME ) Ik" (1) = 3 ⋅ Upa1 2 ⋅ Z1 + Z 0 = 36.372kV 2 ⋅ 3.7Ω∠52.5 0 + 49.72Ω∠88.27 0 = 650.78∠ − 83.84 0 A ⇔ ( for grounding through X ME ) c. Discussion of the results • In parts (a) and (b) the network is to be considered as not effectively grounded. • The cable impedances have significant bearing on the magnitude of the fault current. This fact needs to be taken into consideration during the setting of protective relays. • As the grounding impedance becomes smaller, the effect of the earth capacitance on the fault current becomes also smaller. • As a result of the compensating effect of the grounding reactance, grounding of the neutral point through a resistance will always cause a larger fault current during a single line-to-ground fault. 8-59 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.16 Transient stability analysis I An elementary principle of dynamics states that: d 2 δm J ⋅ 2 = Tm − Te = Ta dt (1) where: J = the total moment of inertia of the rotating masses in kg.m2 δm = the angular displacement of the rotor with respect to a synchronously rotating reference in mechanical radians Tm = the mechanical torque in N.m. T2 = the electrical torque in N.m. Ta = accelerating torque in N.m. Multiplying both sides of the equation by ωm ωm ⋅ J ⋅ d 2 δm = ωm ⋅ (Tm − Te ) = ( Pm − Pe ) dt 2 (2) Definition: 1 ⋅ J ⋅ ωm2 H= 2 Sr where: H = inertia constant in seconds Sr = machine rated power in MVA ωm = angular speed in mechanical radians/second Dividing both sides of equation (2) by Sr/2ωm and re-arranging, we have: d 2δ m dt 2 = ωm 2H ⋅ ( Pm − Pe ) , Pm, Pe: in per unit. Introducing δm = δ/p (δ = angle in electrical radians; p = number of pole pairs), we have the swing equation in final form: d 2δ π ⋅ f = ⋅ (Pm − Pe ) H dt 2 (3) where: δ = the angular displacement of the rotor (in electrical radians) with respect to a reference frame rotating at synchronous speed f = the frequency Pm = shaft mechanical power in per unit Pe = electrical output power in per unit 8-60 Exercises Electric Power Systems I - III 8.17 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Transient stability analysis I a. Kinetic energy, inertia constant S = P / cos φ = 150 / 0.85 MVA = 176.47 MVA Un = 15 kV, n = 1500 rpm, J =75 ·103 kgm2 i. the stored kinetic energy of the rotating masses at nominal speed 1 ⋅ J ⋅ ωm2 2 2⋅π ⋅n ωm = = 157.08 mech.rad / s → 60 75 ⋅ 10 3 ⋅ 157.08 2 Wk = kW ⋅ s = 925.279 MW ⋅ s 2 Wk = ii. the inertia constant H (in seconds) H= 1 J ⋅ ωm2 Wk ⋅ = = 5.243 s 2 S S Mechanical time constant Tm is defined as follows: Tm = J ⋅ ωm2 = 2H S iii. the inertia constant M (in joule-seconds/mechanical radian). M = J ⋅ ωm = 11.777 MJ / rad J ⋅ ωm 1 J ⋅ ωm2 H= ⋅ → 2⋅ H = = M [ p .u ] S 2 S / ωm b. Stored kinetic energy of a 100-MVA generator having H = 8.0 s. i. Kinetic energy at synchronous speed H= 1 1 J ⋅ ωm2 ⋅ = 8s → ⋅ J ⋅ ωm2 = H ⋅ S = 800 MWs 2 2 100 MVA ii. Speed of a 40-tonne lorry to have the same amount of kinetic energy 1600⋅ 10 6 Ws 1 ⋅ m ⋅ v 2 = 800MWs → v = = 200m / s 2 40 ⋅ 10 3 kg Unit check :1W ⋅ s = 1N ⋅ m;1 N = 1kg ⋅ m / s 2 → W ⋅ s / kg = (m / s )2 iii. Total mass of a train travelling at 100 km/h to have the same kinetic energy 2 10 5 1 ⋅ m ⋅ m / s = 800 ⋅ 10 6 Ws → m = 2073.6 tonne 2 3600 c. Angular acceleration, angular speed, displacement angle i. Angular acceleration for a mechanical power increase from 50 MW to 80 MW 8-61 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III π ⋅ 50 (80 − 50) d 2δ π ⋅ f = ⋅ (Pm − Pe ) = ⋅ = 5.89 ele. rad / s 2 2 8 100 H dt ii. The rotor angle and rotor speed for a 10-cycle constant acceleration d 2δ 1 dδ = k = 5.89 rad / s 2 → = k ⋅ t + ω0 = Δω + ω0 → Δω = 5.89 ⋅ 10 ⋅ ⋅ rad / s 2 50 dt dt = 1.178rad / s ( all in ele. radians ) δ( t ) = 1 ⋅ k ⋅ t 2 + ω0 ⋅ t + δ0 → 2 2 1 1 1 Δδ = δ( t ) − ω0 ⋅ t = ⋅ k ⋅ t 2 + δ0 = ⋅ 5.89 s − 2 ⋅ 10 ⋅ s = 0.1178 el . rad . = 6.75 0 el . 50 2 2 (In the last expression, subtraction of ω0 ⋅ t is necessary since the angle is measured with respect to a reference frame rotating at the constant speed of ω0 .) Brief explanatory note on the relationship between mechanical and electrical radians: Assume the generator is running at the speed n rpm→n/60 revolutions per second (rps). That means the machine is traversing in 1 s : rotational speed of ωm = n ⋅ 2π 60 ( mech.) rad leading to the n ⋅ 2π ( mech.) rad / s . Whereas one complete revolution is 60 always 2π rad mechanically, electrically it is 2π.p (p = number of pole pairs). Thus, the speed in electrical radians/s is: ωe = p⋅ n ⋅ 2π ( el .)rad / s = p ⋅ ωm . Generally: 60 ωe δ = e = p. ωm δ m d. The critical fault clearing angle 2,5 2 1,5 P [p.u.] pre-fault during fault 1 post fault mech. power 0,5 0 0 0,5 1 1,5 2 2,5 3 3,5 -0,5 delta [rad.] 8-62 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III The accelerating and decelerating areas are indicated above. Assuming the following general values: Pm = Mechanical power P0max = Pre-fault amplitude of the power vs. power angle curve Pfmax = Amplitude during fault Ppfmax = Post-fault amplitude The critical fault clearing angle can be obtained by equating the accelerating and decelerating areas: δc ∫ (Pm − δ0 f Pmax ) ⋅ sin δ ⋅ dδ = δ max ∫ (Pmax ⋅ sin δ − Pm )⋅ dδ → pf δc0 f pf Pm ⋅ (δc − δ0 ) + Pmax ⋅ (cos δc − cos δ0 ) = Pmax ⋅ (cos δc − cos δ max ) + Pm ⋅ (δc − δmax ) → A general formula for calculating critical fault clearing angle: pf f P ⋅ (δ − δ ) + Pmax ⋅ cos δmax − Pmax ⋅ cos δ0 δc = cos −1 m max 0 pf f Pmax − Pmax With: δ0 = sin −1 ( 1 / 2 ) = 0.523 rad = 30 0 δmax = π − δ0 = 2.618rad = 150 0 We have for the critical fault clearing angle: δc = cos −1 (δmax − δ0 + 1.5 ⋅ cos δmax − 0.5 ⋅ cos δ0 + ) = 68.710 = 1.2rad 8-63 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.18 Transient stability analysis III a. Assuming a pre-fault generator load of 1 p.u, calculate the critical fault clearing angle for a solid three-phase fault occurring at point P in the system given below. j0.15 j0.1 j0.15 j0.15 j0.28 UN=1.0 j0.15 ∼ P E=1.2 j0.15 j0.14 j0.14 j0.15 Infinite bus Before fault: P= E ⋅U N 0 ⋅ sin δ = Pmax ⋅ sin δ XΣ X Σ = 0.15 + 0.1 + 0.15 + 0.28 + 0.15 + 0.15 = 0.69 → 2 1 .2 = 1.74 0.69 E ⋅U N 1= P = ⋅ sin δ0 → δ0 = 35.10 XΣ 0 = Pmax The equivalent circuit of the network during fault: j0.15 j0.1 j0.15 j0.15 j0.28 UN=1.0 j0.15 ∼ P E=1.2 j0.15 j0.14 j0.14 j0.15 Infinite bus j0.58 j0.15 E=1.2 ∼ Generator - Trafo j0.15 j0.1 j0.29 j0.29 ∼ External network The Thevenin equivalent circuit of the external network: 8-64 UN=1 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III X Th = 0.29 // (0.58 + 0.15 // 0.29 ) = 0.203 U Th = 0.29 ⋅ 1 0.29 ⋅ = 0.197 0.15 + (0.58 + 0.29 ) // 0.29 0.58 + 0.29 + 0.29 The equivalent circuit during fault: j0.15 E=1.2 j0.1 j0.203 ∼ ∼ UTh=0.197 The power equation for the duration of the fault: Pf = 1.2 ⋅ 0.197 ⋅ sin δ = 0.522 ⋅ sin δ 0.25 + 0.203 The power equation after the fault is cleared: pf = Pmax 1 .2 ⋅ 1 ⋅ sin δ = 1.22 ⋅ sin δ 0.98 With Pfmax = 0.522, δ0 = 35.10 =0.61rad δmax = 124.950 = 2.18rad and using the formula derived in 8.17 (d), we have for the critical clearing angle. pf f P ⋅ (δ − δ0 ) + Pmax ⋅ cos δmax − Pmax ⋅ cos δ0 δc = cos −1 m max pf f − Pmax Pmax → 2.18 − 0.61 − 1.22 ⋅ 0.573 − 0.522 ⋅ 0.818 0 δc = cos −1 = 50.5 0.698 b. A solid three- phase fault is assumed to occur at point P in the system given below. All the necessary data are given in the figure and the system frequency is 50 Hz. The fault is cleared by disconnecting the faulted line in both ends. What are the critical fault clearing angle and critical fault clearing time for a pre-fault generator load of 1 p.u. 8-65 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III j0.15 j0.1 UN=1.0 j0.5 j0.05 ∼ E =1.2 p.u H=4s j0.4 P Infinite bus The approach as the same as (a). Pre-fault: 0 = Pmax 1 .2 = 2 .3 0.15 + 0.1 + 0.4 // 0.5 + 0.05 During fault: The external network incorporates the rest of the network except generator and transformer. Parameters of the Thevenin equivalent for this part of the network are: f Pmax =0 After fault: pf Pmax = 1 .2 = 1 .5 0.15 + 0.1 + 0.5 + 0.05 With δmax = 138.190 = 2.41rad δ0 = 25.770 =0.45rad and again using the formula derived in 8.17 (d), we have for the critical clearing angle. pf f P ⋅ (δ − δ0 ) + Pmax ⋅ cos δmax − Pmax ⋅ cos δ0 δc = cos −1 m max pf f Pmax − Pmax −1 δc = cos (2.41 − 0.45) + 1.5 ⋅ cos 138.19 0 = 55.85 0 1.5 → = 0.974rad c. The generator in the figure below has an inertia constant H = 4 s and feeds 1 p.u. power into the infinite bus. Data for the system are: UN = Ut = 1.0 p.u., xd’ = 0.25 p.u., xt = 0.10 p.u., xl = 0.5 p.u., f = 50 Hz. xd’ xt xl UN ∼ Up’ Ut xl Infinite bus 8-66 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III i. What is the value of the emf (Up’) behind the transient reactance? The magnitude of the generator terminal voltage (Ut) is given, but its phase angle is unknown. It can be determined from the power relationship: P =1= U t ⋅U N 1 ⋅ sin δG = ⋅ sin δG → δG = 20.48 0 → U t = 1∠20.48 0 xt + xl / 2 0.1 + 0.25 The current flowing out of the generator: IG = Ut − U N 1∠20.48 0 − 1 − 0.063 + j 0.35 = = = 1 + j 0.18 = 1.016∠10.2 0 j ( xt + xl / 2 ) j 0.35 j 0.35 The transient voltage of the generator can now be calculated using the relationship: ' U p = U t + jx'd ⋅ I G = 1∠20.48 0 + j 0.25 ⋅ 1.016∠10.2 0 = 0.89 + j 0.6 = 1.07∠34.0 0 ii. Calculate the maximum power that can be transmitted if: − the topology of the system is as shown in the figure. 0 Pmax = U 'p ⋅ U N x'd + xt + xl / 2 = 1.07 = 1.78 0.25 + 0.1 + 0.25 − When one of the lines experiences a three-phase fault midway along the line. X Th = xl // xl / 2 = 0.167 U Th = f Pmax = 1 3 1.07 ⋅ 1 / 3 = 0.69 0.25 + 0.1 + 0.167 iii. What is the accelerating power immediately after the fault described in (ii)? d 2δ π ⋅ f = ⋅ ΔP H dt 2 At the instant of fault: δ = δ0 and the electrical output f ⋅ sin δ0 = 0.69 ⋅ sin( 34 0 ) = 0.3858 → ΔP = 1 − 0.3858 = 0.61 → power: P = Pmax π ⋅ 50 d 2δ π ⋅ f = ⋅ ΔP = ⋅ 0.61 = 23.95 rad / s 2 2 4 H dt iv. What is the value of the rotor angle if the acceleration is maintained constant at the value calculated in (iii) for 0.05 s? Δδ = 1 π ⋅ f ⋅ ΔP ⋅ ⋅ 0.05 2 = 1.72 0 2 4 v. What is the accelerating power at the rotor angle calculated in (iv)? f δ = δ0 + Δδ = 34 0 + 1.72 0 = 35.72 0 → P = Pmax ⋅ sin 35.72 0 = 0.4 → Pa = Pm − P = 1 − 0.4 = 0.6 vi. Discuss the difference between the values obtained in (iii) and (v)! 8-67 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich The value obtained in (v) is smaller since the output electrical power increases as the power angle increases. 8-68 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III 8.19 Transient stability analysis Derivation of the Equal Area Criterion: Equation of motion: J 1 dωL = MT − M E = ⋅ ( PT − P ) ωL dt dδGN → dt d 2δGN 1 J = MT − M E = ⋅ ( PT − PE ) dt ωL ωL = J= Tm ⋅ S n → from the definition of Tm ωL2 0 d 2δGN = k ⋅ ( PT − PE ) dt with k= ω0 S nTm Equal area criterion: d 2δGN dδ dδ ⋅ 2 GN = k ⋅ ( PT − PE ) ⋅ 2 GN → dt dt dt d 2δGN d δ ⋅ 2 GN = dt dt dt dδ ⋅ GN dt 2 2 dδGN d dδGN d dx dδGN d 2δGN 2 (Check: x = → ⋅ =2 ⋅ ) = ⋅ x = 2x ⋅ dt dt dt dt dt dt dt Thus, 2 dδ d GN = k ⋅ ( PT − PE ) ⋅ 2 ⋅ dδGN → dt δ dδ 2 GN ( max ) = GN dt δGN (0) δGN ( max ) ∫ 2k ⋅ ( PT − PE ) ⋅dδGN δGN ( 0 ) dδGN = 0 at δGN ( 0 ) and δGN (max) dt 8-69 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III δGN ( max ) ∫ ( PT − PE )⋅dδGN = 0 δGN ( 0 ) Prefault power output of the generator: Us ⋅ U∞ ⋅ sin δ = 1.4 x 0.1 + 0.2 + 0.1 x= = 0.2 p.u. 2 P= δ s∞ = 16.26 0 I= Us − U∞ E "G = jx 1∠16.26 0 − 1 = 1.414∠8.13 0 j0.2 = U s + I ⋅ jx "d = 1∠16.26 0 + j0.12 ⋅ 1.414∠8.13 0 = 0.936 + j0.448 = 1.04∠25.58 0 → δ 0 = 25.58 0 Pre-fault power vs. power angle curve: Pe = E "G ⋅ U ∞ x+ x "d ⋅ sin δ = 1.04 ⋅ 1 ⋅ sin δ = 3.25 ⋅ sin δ 0.32 a. Symmetrical fault Equivalent circuit during the symmetrical fault: xps xd xL xps “ EG“ U∞ Generator External system Output power during fault: E th = x ps x ps + x L + x ps 0.1 = = 0. 2 0. 5 xTh ETh x Th = x ps //( x ps + x L + x ps ) = 0.1// 0.4 = 0.08 P= E "G ⋅ E Th x "d + x Th ⋅ sin δ = 1.04 ⋅ 0.2 ⋅ sin δ = 1.04 ⋅ sin δ 0 .2 8-70 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Output power, when one of the parallel lines is disconnected: P= E "G ⋅ U ∞ x "d + x ps + x L + x ps ⋅ sin δ = 1.04 ⋅ 1 ⋅ sin δ = 2 ⋅ sin δ 0.12 + 0.1 + 0.2 + 0.1 Power vs. power angle curve 3,5 3 power [p.u.] 2,5 2 1,5 1 0,5 0 0 0,5 1 1,5 2 2,5 3 3,5 angle [rad] ∆t = 0.05 s k= π⋅f (∆t )2 = 0.0785398 H δ n = δ n−1 + ∆δ n (in electrical radians ) ∆δ n = ∆δ n−1 + k ⋅ Pa,n−1 Pa, n = Pm, n − Pel, n (Pm,n, Pel,n: mechanical input and output electrical power, respectively) t[s] δ Pa ∆δ t[s] δ Pa 0 0.05 0.4464 0.4755 0.0373 0.1 0.4838 0.9162 0.1093 0.15 0.5931 0.8187 0.1736 0.2 0.7667 0.6785 0.2269 0.25 0.9936 0.5285 0.2684 1.2620 -0.04811 0.2646 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 2.528 1.5266 1.7443 1.9172 2.0523 2.1581 2.2431 2.3152 2.3817 2.45 -0.5980 -0.5700 -0.4812 -0.3726 -0.2649 -0.1648 -0.071 0.0223 0.12437 0.2483 8-71 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III ∆δ 0.2177 0.1729 0.1351 0.1058 0.0850 0.0721 0.0665 0.0683 0.078 0.0975 The angles corresponding to the fault clearing and switching times can be obtained from the table: t =0 s → δ =0.446 rad t =0.25 s → δ =1.262 rad t =0.75 s → δ = 2.528 rad Application of the equal area criterion and determining whether or not the generator remains in synchronism with the rest of the network: Area corresponding to 0 ≤ t ≤0.25 s: 1.262 ∫ (1.4 − 1.04 ⋅ sin δ) ⋅ dδ = 0.52 A1 = 0.446 Area corresponding to 0.25 < t ≤0.75 s: 2.528 ∫ (1.4 − 2 ⋅ sin δ) ⋅ dδ = −0.47 A2 = 1.262 Accelerating area: A1 + A2 = 0.52 – 0.47 = 0.05 Reserve area (t>0.75 s): 2.6958 ∫ (3.25 ⋅ sin δ − 1.4) ⋅ dδ = 0.04 A3 = 2.528 Accelerating area (0.05) > Reserve area (0.04) → the generator will go out of synchronism! This is demonstrated by the following plot of the power angle against time. Swing curve 4,5 4 Delta [rad] 3,5 3 2,5 2 1,5 1 0,5 0 0 0,2 0,4 0,6 0,8 1 1,2 1,4 t [s] 8-72 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III In this example, the generator loses synchronism just marginally. If the fault is cleared, for example, at t = 0.2 s (instead of t=0.25) the generator will retain synchronism, as shown in the following curve. Swing curve 2 Delta [rad] 1,5 1 0,5 0 0 0,2 0,4 0,6 0,8 1 1,2 1,4 -0,5 t [s] b. Double line-to-ground fault Equivalent circuits of the sequence networks: 01 EG“ xd“ xps xL xps U∞ 02 xd“ xps xL xps x2=0.1172 00 xps xL xps x0=0.1 8-73 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Double line-to-ground fault calls for the parallel connection of the three networks, with the fault location as the terminal. 01 EG“ xd“ xps xL xps U∞ xeq= x2//x0=0.054 External network xd“ xTh=0.1066 ETh=0.3776 EG“ Generator Output power during the double line-to-ground fault: P= E "G ⋅ E Th x "d + x Th ⋅ sin δ = 1.73 ⋅ sin δ 8-74 UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich Exercises Electric Power Systems I - III Power vs. power angle curve Double line-to-ground fault 2 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 0 0,5 1 1,5 2 2,5 3 3,5 Accelerating area: A+ = 0.943 ∫ (1.4 − 1.73 ⋅ sin δ) ⋅ dδ = 0.151 0.446 Reserve (decelerating) area: A− = π−0.943 ∫ (1.73 ⋅ sin δ − 1.4 ) ⋅ dδ = 0.2744 0.943 A- > A+→ Even for a sustained fault (fault not cleared), the generator would remain in synchronism, as demonstrated by the following swing curve. Sustained double line-to-ground fault 1,8 1,6 1,4 Delta [rad] 1,2 1 0,8 0,6 0,4 0,2 0 0 0,2 0,4 0,6 0,8 1 1,2 1,4 t [s] Maximum swing angle: 8-75 Exercises Electric Power Systems I - III UNIVERSITÄT DUISBURG - ESSEN FG Elektrische Anlagen und Netze Univ. Prof. Dr.-Ing. habil. I. Erlich δx ∫ (1.73 ⋅ sin δ − 1.4) ⋅ dδ = 1.4 ⋅ (0.943 − δ x ) + 1.73 ⋅ (cos 0.943 − cos δ x ) = 0.151 → 0.943 cos δ x + 0.809 ⋅ δ x = 1.2632 → δ x = 1.61 = 92.25 0 → Note: This value corresponds to the maximum angle in the curve above. 8-76