lecture13

lecture13 - power flow - evaluate normal operating...

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Unformatted text preview: power flow - evaluate normal operating conditions fault analysis - evaluate abnormal operating conditions three-phase ￿ ￿ single-line to ground and double-line to ground line-to-line faults unbalanced faults ￿ balanced faults ￿ ￿ ￿ specifying ratings for circuit breakers and fuses protective relay settings specifying the impedance of transformers and generators Results used for: ￿ ￿ F a u l t ty p e s : ￿ ￿ Analysis types Power Systems I ￿ ￿ ￿ Fault Analysis the impedance of the network the internal impedances of the generators the resistance of the fault (arc resistance) transmission line impedances transformer connections and impedances grounding connections and resistances ￿ ￿ ￿ sub-transient period, lasting for the first few cycles transient period, covering a relatively longer time steady state period Generator behavior is divided into three periods ￿ ￿ ￿ Network impedances are governed by ￿ ￿ ￿ Magnitude of fault currents depend on: Power Systems I ￿ ￿ ￿ Fault Analysis determine the interrupting capacity of HV circuit breakers determine the operation timing of the protective relay system for high-voltage networks ￿ ￿ ￿ determine the interrupting capacity of MV circuit breakers determine the operation timing of the protective relay system for medium-voltage networks transient stability studies Transient period, XG = Xd’ ￿ ￿ Sub-transient period, XG = Xd" Power Systems I ￿ ￿ Fault Studies equivalent to the addition of an impedance at the place of the fault if the fault impedance is zero, the fault is referred to as a bolted fault or solid fault ￿ ￿ ￿ ￿ ￿ the faulted network can be solved conveniently by Thévenin’s method network resistances are neglected generators are modeled as an emf behind the sub-transient or transient reactance shunt capacitances are neglected system is considered as having no-load First order method ￿ ￿ A fault represents a structural network change Power Systems I ￿ ￿ Fault Representation 1 3 2 1 3 j 0.4 j0.2 j0.8 2 j0.4 j0.4 The fault is simulated by switching a fault impedance at th e f a u l t e d b u s The change in the network voltages is equivalent to adding the prefault bus voltage with all other sources short curcuited Power Systems I ￿ ￿ Thévenin’s Method 1 j 0.4 j0.2 3 j0.8 Xf = 0.16 2 j0.4 j0.4 = 1 3-phase fault with Zf = j0.16 on bus 3 Power Systems I ￿ Thévenin’s Method If j 0.4 j0.2 3 j0.8 -+ V th 2 Xf = 0.16 j0.4 j0.4 1 3 j 0.2 If j0.1 j0.2 j0.4 Xf = 0.16 2 V th Power Systems I j0.2 V3[ 0] = Z 33 + Z f j 0.4 + j 0.6 Z 33 = j 0.34 Z 33 ( j 0.4)( j 0.6) + j 0.1 = Z 1s = Z 2 s = ( j 0.4)( j 0.8) = ( j1.6) ( j 0.4)( j 0.4) = j 0.1 Z 3s = ( j1.6) V1[ 0 ] = V2[ 0] = V3[ 0] = 1.0 [ I3 f ] Thévenin’s Method j 0 .2 If 3 j0.16 V th j0.1 j0.24 j0.16 V th Power Systems I If 3 Z33 = j0.34 I [f] 3 V3[ 0] 1.0 = = = − j 2. 0 Z 33 + Z f j 0.34 + j 0.16 Z 33 = j 0.34 Thévenin’s Method ￿ ￿ ￿ ￿ ￿ use the pre-fault bus voltages which can be obtained from the results of a power flow solution include loads - to preserve linearity, convert loads to constant impedance model Thevenin’s theorem allows the changes in the bus voltages to be obtained bus voltages are obtained by superposition of the pre-fault voltages and the changes in the bus voltages current in each branch can be solved For more accurate solutions Power Systems I ￿ Thévenin’s Method ￿ in per unit: SB X kk Vk[ pre−f ] = X kk SCC = I [f ] k SCC = 3 VL − L ,k I k Measures the electrical strength of the bus Stated in MVA Determines the dimension of bus bars and the interrupting capacity of circuit breakers Definition: [ pre − f ] [ f ] Power Systems I ￿ ￿ ￿ ￿ Short Circuit Capacity (SCC) 1 3 j 0.4 j0.2 S base 100 MVA SCC3 = = = 294 MVA Z 33 0.34 S base = 100 MVA Z 33 = j 0.34 Find the SCC for bus #3 Power Systems I ￿ Short Circuit Capacity (SCC) j0.8 2 j0.4 j0.4 difficult to apply to large networks ￿ ￿ ￿ ￿ ￿ operating under balanced conditions each generator represented by a constant emf behind a proper reactance (Xd, Xd+ , or Xd + ) lines represented by their equivalent π model G j i Si Sk k Zf seek a matrix where the diagonal elements represent the source impedance for the buses consider the following system Matrix algebra formation ￿ Network reduction by Thévenin’s method is not efficient Power Systems I ￿ ￿ Fault Analysis Using Impedance Matrix Z i −load = The change in the network voltage caused by the fault is equivalent to placing a fault voltage at the faulted bus with all the other sources short-circuited Replace the loads by a constant impedance model using the prefault bus voltages j Si*−load Vi [pre − f] 2 Place the prefault voltages into a vector Power Systems I ￿ ￿ ￿ i Zi [pre Vbus −f] Zk k Zf -Vk ȺV1[pre −f] Ⱥ Ⱥ Ⱥ ￿ Ⱥ Ⱥ = ȺVk[pre −f] Ⱥ Ⱥ Ⱥ Ⱥ ￿ Ⱥ ȺV [pre −f] Ⱥ Ⱥ n Ⱥ Fault Analysis Using Impedance Matrix [ I kf ] The change in bus voltages can be calculated from the network matrix ΔVbus Ⱥ 0 Ⱥ Ⱥ ￿ Ⱥ Ⱥ Ⱥ [Fault] [Fault] I bus = Ⱥ− I k Ⱥ Ⱥ Ⱥ Ⱥ ￿ Ⱥ Ⱥ 0 Ⱥ Ⱥ Ⱥ I[Fault] = Ybus ΔVbus bus I bus = Ybus Vbus [f [ pre Vbus] = Vbus −f ] + ΔVbus Using superpositioning, the fault voltages are calculated from the prefault voltages by adding the change in bus voltages due to the fault Power Systems I ￿ ￿ ȺΔV1 Ⱥ Ⱥ ￿ = ȺΔVk Ⱥ Ⱥ ￿ ȺΔV Ⱥ n Fault Analysis Using Impedance Matrix Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ ￿ ￿ ￿ ￿ ￿ y1k ￿￿ ￿ ykk ￿￿ ￿ ynk ￿ Power Systems I Fault] ΔVbus = Z bus I[bus −1 Z bus = Ybus −1 ΔVbus = Ybus I[Fault] bus Ⱥ 0 Ⱥ Ⱥ y11 Ⱥ ￿ Ⱥ Ⱥ ￿ Ⱥ Ⱥ Ⱥ [f] Ⱥ− I k Ⱥ = Ⱥ yk 1 Ⱥ Ⱥ Ⱥ Ⱥ ￿ Ⱥ Ⱥ ￿ Ⱥ 0 Ⱥ Ⱥ yn1 Ⱥ Ⱥ Ⱥ I[Fault] = Ybus ΔVbus bus y1n Ⱥ ￿ Ⱥ Ⱥ ykn Ⱥ Ⱥ ￿ Ⱥ ynn Ⱥ Ⱥ ȺΔV1 Ⱥ Ⱥ ￿ ȺΔVk Ⱥ Ⱥ ￿ ȺΔV Ⱥ n Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Fault Analysis Using Impedance Matrix Power Systems I [ I kf ] Vk[ pre−f ] = Z kk + Z f Ⱥ 0 Ⱥ Ⱥ ￿ Ⱥ Ⱥ Ⱥ [f] Ⱥ− I k Ⱥ Ⱥ Ⱥ ￿ Ⱥ Ⱥ Ⱥ 0 Ⱥ Ⱥ Ⱥ [ & Vk[ f ] = Z f I kf ] Ⱥ Ⱥ z11 ￿ z1k ￿ z1n Ⱥ Ⱥ Ⱥ ￿ ￿ ￿ ￿ ￿ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ = Ⱥ z k1 ￿ z kk ￿ z kn Ⱥ Ⱥ Ⱥ Ⱥ ￿ ￿ ￿ ￿ ￿ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ z n1 ￿ z nk ￿ z nn Ⱥ Ⱥ Ⱥ Ⱥ [f [ pre Vbus] = Vbus −f ] + ΔVbus [f [ pre f] Vbus] = Vbus −f ] + Z bus I[bus [ Vk[ f ] = Vk[ pre−f ] + Z kk I kf ] ȺΔV1 Ⱥ Ⱥ ￿ ȺΔVk Ⱥ Ⱥ ￿ ȺΔV Ⱥ n Fault Analysis Using Impedance Matrix ￿ 1 bus 3 j 0.4 j0.2 3 j0.8 Xf = 0.16 2 j0.4 j0.4 3 - p h a s e fa u l t w i t h Z f = j 0 . 1 6 o n Power Systems I ￿ Example 1 If j 0.4 j0.2 3 j0.8 V th 2 Xf = 0.16 j0.4 j0.4 j 0.16 j 0.08 j 0.24 j 0.12Ⱥ j 0.16Ⱥ Ⱥ j 0.34Ⱥ Ⱥ j 2.50 Ⱥ − j 6.25 j 2.50 Ⱥ Ⱥ j 2.50 − j 5.00Ⱥ Ⱥ j1.25 V3[ pre − f ] 1.0 pu = = = − j 2.0 pu Z 33 + Z f j 0.34 + j 0.16 Power Systems I I [f] 3 Ⱥ− j8.75 Ybus = Ⱥ j1.25 Ⱥ Ⱥ j 2.50 Ⱥ Ⱥ j 0.16 Z bus = Ⱥ j 0.08 Ⱥ Ⱥ j 0.12 Ⱥ Example Power Systems I [ V3[ f ] = V3[ pre − f ] − Z 33 I 3 f ] = 1.0 pu − ( j 0.34 )(− j 2.0 ) = 0.32 pu [ V2[ f ] = V2[ pre − f ] − Z 23 I 3 f ] = 1.0 pu − ( j 0.16 )(− j 2.0 ) = 0.68 pu [ V1[ f ] = V1[ pre − f ] − Z13 I 3 f ] = 1.0 pu − ( j 0.12 )(− j 2.0 ) = 0.76 pu Example ...
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