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Balanced faults

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Balanced faults

  1. 1. ELE B7 Power SystemsELE B7 Power Systems EngineeringEngineering Balanced (Symmetrical) FaultsBalanced (Symmetrical) Faults
  2. 2. Fault Analysis • Fault? When the insulation of the system breaks down or a conducting object comes in touch with a live point, a short circuit or a fault occurs. • Sources? • This breakdown can be due to a variety of different factors • lightning • wires blowing together in the wind • animals or plants coming in contact with the wires • salt spray or pollution on insulators
  3. 3. Fault Types • There are two classes of faults – Balanced faults (symmetric faults): A Fault involving all the three phases • system remains balanced; • these faults are relatively rare, but are the easiest to analyze so we’ll consider them first. – Unbalanced faults (unsymmetric faults): A fault involving only one or two phases • The majority of the faults are unsymmetrical. • system is no longer balanced; • very common, but more difficult to analyze • The most common type of fault on a three phase system is the single line-to-ground (SLG), followed by the line-to- line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults
  4. 4. Fault Types IFb Transformer IFa IFc IF Transformer IF Single Line to Ground Fault IF IF IF Transformer IF Line to Line Fault IF IFb Transformer IF IF IF IFc Double line to ground Fault Three Phases Fault (Balanced Fault)
  5. 5. Fault Analysis • Fault Analysis involve finding the voltage and current distribution throughout the system during fault condition. • Fault currents cause equipment damage due to both thermal and mechanical processes • Why do we need that? – To adjust and set the protective devices so we can detect any fault and isolate the faulty portion of the system. – To protect the human being and the equipment during the abnormal operating conditions. – need to determine the maximum current to insure devices can survive the fault
  6. 6. Symmetrical Faults (Balanced Faults) • It is the most severe fault but can be easily calculated • It is an important type of fault • The circuit breaker rated MVA breaking capacity is selected based on the three phase short circuit MVA. A Fault involving all the three phases. Transformer Three Phase Short Circuit
  7. 7. Symmetrical Faults (Balanced Faults) • It is the most severe fault but can be easily calculated • It is an important type of fault • The circuit breaker rated MVA breaking capacity is selected based on the three phase short circuit MVA. A Fault involving all the three phases. Transformer Three Phase Short Circuit
  8. 8. 1. Synchronous machines are represented by a constant voltage sources behind subtransient reactances. The EMF of all generators are 1 per unit making zero angle. o V 01 The following assumptions are made in three phase fault calculation : This means that the system voltage is at its nominal value and the system is operating at no-load conditions at the time of fault. Therefore, All the generators can be replaced by a single generator since all EMFs are equal and are in phase.
  9. 9. 2. Transformers are represented by their leakage reactances Winding resistances, shunt admittances and Δ-Y phase shifts are neglected 3. The shunt capacitances of the transmission Lines are neglected. 4. The system resistances are neglected and only the inductive reactance of different elements are taken into account.
  10. 10. Symmetrical Fault Calculation Transformer Three Phase Short Circuit Simple Circuits Load Ignored Use Thevenin's Equivalent Simple Circuits Load not Ignored Use Thevenin's Equivalent Find Pre-fault voltage Large Circuits Construct The Bus Impedance Matrix Three Phase Fault Calculation 2 3 1
  11. 11. 1. Simple Circuits and Load is ignored 1. Draw a single line diagram for the system. The Calculations for the three phase fault are easy because the circuit is completely symmetrical and calculations can be done for only one phase. T1 T2 G1 G2 1 2 3 phase fault e Steps For Calculating Symmetrical Faults: 2. Select a common base and find out the per unit reactances of all generators, transformers, transmission lines, etc.
  12. 12. 4. Reduce the single line reactance diagram by using series, parallel and Delta-Why transformations keeping the identity of the fault point intact. Find the total reactance of the system as seen from the fault point (Using Thevenin’s Theorem) 3. From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram. GG 15.0j 20.0j 16.0j 10.0j 23.0j GG 01 01 23.0j
  13. 13. 5. Find the fault current and the fault MVA in per unit. Convert the per unit values to actual values. 6. Retrace the steps to calculate the voltages and the currents throughout different parts of the power system
  14. 14. T1 T2 G1 G2 X=0.2 ohm/phase/km 1 2 50 MVA, 11 kV 10% Reactance 100 MVA, 11 kV 15% Reactance 100 MVA, 11/132 kV 10% Reactance 50 MVA, 11/132 kV 8% Reactance 200 km 3 phase fault e Example: A three phase fault occurs in the system as shown in the Figure. Find the total fault current, the fault level and fault current supplied by each generator.
  15. 15. Step 1: Draw a single line diagram for the system. The single line diagram for the system is given in the example as shown. Solution: T1 T2 G1 G2 X=0.2 ohm/phase/km 1 2 50 MVA, 11 kV 10% Reactance 100 MVA, 11 kV 15% Reactance 100 MVA, 11/132 kV 10% Reactance 50 MVA, 11/132 kV 8% Reactance 200 km 3 phase fault e
  16. 16. Step 2: Select a common base and find the per unit reactances of all generators, transformers, etc. Select the common base as: 100 MVA (100,000 kVA) 11 kV for Transformer low voltage side (LV) 132 kV for Transformer high voltage side (HV) T1 T2 G1 G2 X=0.2 ohm/phase/km 1 2 50 MVA, 11 kV 10% Reactance 100 MVA, 11 kV 15% Reactance 100 MVA, 11/132 kV 10% Reactance 50 MVA, 11/132 kV 8% Reactance 200 km 3 phase fault e kVVBase 11 kVVBase 132
  17. 17. T1 T2 G1 G2 X=0.2 ohm/phase/km 1 2 50 MVA, 11 kV 10% Reactance 100 MVA, 11 kV 15% Reactance 100 MVA, 11/132 kV 10% Reactance 50 MVA, 11/132 kV 8% Reactance 200 km 3 phase fault e kVVBase 11 kVVBase 132 G1 G2 15.01 jXG The per unit reactance 2.0 50 100 *1.02 jjXG The per unit reactance 1.01 jXT The per unit reactance 16.0 50 100 *08.02 jjXT The per unit reactance 23.0 )132( 100 )200*2.0( 2 j kV MVA j Z X X Base LINE LINE The per unit reactance T1 T2 TL
  18. 18. From the single line diagram of the system draw a single line reactance diagram showing one phase and neutral. Indicate all the reactances, etc. on the single line reactance diagram. T1 T2 G1 G2 1 2 3 phase fault e G 15.0j 20.0j 16.0j 10.0j 23.0j G 01 01 23.0j
  19. 19. G 25.0j 36.0j G 01 01 115.0j Find the total impedance (reactance) of the system as seen from the fault side. 25.0j 36.0j 115.0j pujX j jj jj X Total Total 2625.0 115.0 36.025.0 36.0*25.0    
  20. 20. puj j IF 8095.3 2625.0 01    AI sidekVBase 4.437 132*3 1000*100 )( 132  AjI o ActualF 9027.16664.437*8095.3)(  01 2625.0j LLBase Base Lbase V S I ||3 || || 3  BasepuFActualF III *)()( 
  21. 21. The Fault Level is: 100)( BaseMVA 95.380100*8095.3)( LevelActualMVA puMVA Level 8095.3)(  T1 T2 G1 G2 X=0.2 ohm/phase/km 1 2 50 MVA, 11 kV 10% Reactance 100 MVA, 11 kV 15% Reactance 100 MVA, 11/132 kV 10% Reactance 50 MVA, 11/132 kV 8% Reactance 200 km 3 phase fault e 1)( GFI 2)( GFI FI kVVBase 132kVVBase 11
  22. 22. The Fault Current at 11 kV side supplied by the two generators is: AI sidekVBase 8.5248 11*3 1000*100 )( 11  AjI o ActualF 90199958.5248*8095.3)(  At 11 kV Side: G 15.0j 20.0j 16.0j 10.0j 23.0j G 01 01 23.0j TotalFI )( 1)( GFI 2)( GFI kVVBase 132kVVBase 11
  23. 23. 111,2 )()()( GFkVTFGF III  A jj j I GF 903.11800 25.036.0 36.0 )9019995()( 1    AI GF 907.8194)( 2  G 25.0j 36.0j G 01 01 115.0j T1 T2 G1 G2 1 2 1)( GFI 2)( GFI 1)( GFI 2)( GFI TotalFI )( A jj j II kVTFGF 903.11800 25.036.0 36.0 )()( 11,1   
  24. 24. Example: Consider the single-line diagram of a power system shown below. The transient reactance of each part of the system is as shown and expressed in pu on a common 100 MVA base. G1 G22.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5 Assuming that all generators are working on the rated voltages, when a three-phase fault with impedance of j0.16 pu occurs at bus 5. Find: The fault currents and buses voltages.
  25. 25. G1 G22.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5 2.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5 16.0j 2.0j 4.0j 4.0j 8.0j 4.0j Bus 3 Bus 4 Bus 5 16.0j
  26. 26. 5.016.034.0 jjjZTH  pu0.2j 5.0j 1 )I( 0 5BusF   puj jj j II FGF 2.1 4.06.0 6.0 )( 1    puj jj j II FGF 8.0 4.06.0 4.0 )( 2    16.0j Bus 5 34.0j FI 16.0j Bus 5 34.0j FI 2.0j 4.0j 16.0j Bus 3 Bus 4 Bus 5 2.0j 2.0j 1.0j FI 1FGI 2FGI 16.0j Bus 5 6.0j 1.0j 4.0j FI 2FGI1FGI
  27. 27. 0V&01V :VoltagesPrefault Ground o BUS  pu VprefaultV VoltageFault o 76.024.001 : 33   pu jj jIV isFaultDuringVariationVoltage G 24.0 )2.1()2.0(0 )2.0(0 : 13    68.0 )0.2)(16.0(01 : 5   jjV FaultduringVariationVoltage o pu VprefaultV VoltageFault o 32.068.001 : 55   NOTE: If the fault impedance is zero. 0.1)I)(0.0j(01V :FaultduringVariationVoltage F o 5  pu0.00.101VprefaultV o 55  and 1GI 2GI FI o 01 2.0j 4.0j 16.0j Bus 3 Bus 4 Bus 5 2.0j 2.0j 1.0j +_    5V    3V
  28. 28. G1 G22.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5 Problem: For the network shown in the figure, find the current flowing between buses 3 and 4 during symmetrical three phase fault at bus 5 as in the previous example. 16.0j 2.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5
  29. 29. pu jjprefault VprefaultV pu VprefaultV VoltageFaultDuring o o 68.032.001 )8.0)(4.0(0 76.024.001 : 44 33      puj jj j II FGF 2.1 4.06.0 6.0 )( 1    puj jj j II FGF 8.0 4.06.0 4.0 )( 2    puj j VV IF 1.0 8.0 )( 43 34    4.0 )( 53 35 j VV IF   4.0 )( 54 45 j VV IF  and 2.0j 2.0j 4.0j 8.0j 4.0j 1.0j 1.0j Bus 1 Bus 3 Bus 4 Bus 2 Bus 5 16.0j 1GI 2GI 34F )I( 45F )I(35F )I(   4V   5V
  30. 30. Symmetrical Fault Calculation Transformer Three Phase Short Circuit Three Phase Fault Calculation Simple Circuits Load Ignored Use Thevenin's Equivalent Simple Circuits Load not Ignored Use Thevenin's Equivalent Find Pre-fault voltage Large Circuits Construct The Bus Impedance Matrix 2
  31. 31. 2. Symmetrical Faults Considering the load current Generally, the fault currents are much larger than the load currents. Therefore, the load current can be neglected during fault calculations. The pre-fault voltage and the ZTH are used in calculating the fault current. The fault current represents two components: 1. The Load Current 2. The Short Circuit Current There are some cases where considering the load current is an essential factor in fault calculations. Superposition technique is proposed for such cases to compute the fault current. For such case, it is necessary to compute the terminal voltage at fault location before fault takes place. This terminal voltage is known as the pre-fault voltage. Connecting the load to the system causes the current to flow in the network. Voltage drop due to system impedance cause the voltage magnitude at different buses to be deviated from 1.0 pu.
  32. 32. Consider a load (synchronous Motor) connected to a synchronous generator through a transmission line. The circuit model of the network could be represented as shown in the Figure MIGI jXG jXM G M jXL FI G M FMIFGI jXL jXG jXM Normal Operation Condition: LMG III  Fault condition at Load terminal: LGFG III  )( LMFM III  )( MFGFF III )()(  During fault condition in the system, the generator as well as the load (synchronous motor) will supply the faulted terminals with power from the energy stored in their windings.
  33. 33. The Motor is drawing 40 MW at 0.8 pf. Leading with terminal voltage of 10.95 kV. Calculate the total current in the generator and motor during 3 phase s.c. Example 100 MVA, 11 kV Generator 0.25 pu sub-transient reactance 50 MVA, Motor 0.2 pu sub-transient reactance 0.05 pu. Reactance On 100 MVA Base 3 phase fault G M
  34. 34. A kV MVA CurretBase kVkVBase MVAMVABase 88.5248 )11(3 100 11 100    Solution: The network could be represented as shown in the Fig. A4.2636 )8.0)(kV95.10(3 MW40 CurretLoadThe  pujI jI I L L L )301.0402.0( )6.08.0(502.0 )8.0(cos502.0 1     puI puL 502.0 88.5248 4.2636 )(  Under Normal Operation Condition (Prefault condition) leading0.8isfactorpowerThe 25.0j 4.0j G M 05.0j MIGI
  35. 35. Consider the motor terminal voltage as a reference. puvoltage)terminal(Motor pu 995.0 11 95.10  pu)0201.0j9799.0(V 995.0)050.0j)(301.0j402.0(V V)050.0j(IV G G MLG    pu)(V o puM 0995.0  Using KVL, calculate the pre-fault voltage at the generator terminal. Find the equivalent reactance as seen from the fault terminals. pujX jj jj X TH TH 1607.0 )45.0()25.0( )45.0)(25.0(    25.0j 4.0j 05.0j THX 25.0j 4.0j G M 05.0j MV
  36. 36. Calculate the fault current puj)(I j j )(I X Voltaegefaultpre )(I puF puF TH puF )09.6125.0( 1607.0 0201.09799.0      pu(I) puj(I) jj(I) I)(I(I) o G G G LGFG 4.8264.3 )609.3482.0( )301.0402.0()91.308.0(     Fault Current from Generator Side puj)(I pujj)(I )(I)(I)(I MF MF GFpuFMF )18.2045.0( )91.308.0()09.6125.0(    Fault Current from Motor Side Current from Generator Side: puj)(I j0.25j0.45 j0.45 )(I)(I GF puFGF )91.308.0(    pu(I) puj(I) jj(I) I)(I(I) o M M M LMFM 8.261505.2 )48.2357.0( )301.0402.0()18.2045.0(     Current from Motor Side: 25.0j 4.0j G M 05.0j FI G(I) M(I)
  37. 37. Consider the system shown in the Figure. The two generators are supplying the buses 1, 2, 3 and 4. Let the pre-fault voltage between bus 2 and ref. Bus be (Vf ). 3 - Symmetrical Fault Calculation Using Bus Impedance Matrix 3 T1 T2 G1 G2 1 2 4 fV
  38. 38. If a 3 phase short circuit occurs a bus 2. A huge current will run in the system during the fault condition. " fI 3 2 1 4 " fI 3 T1 T2 G1 G2 1 2 4 fV 3 phase fault e If the pre-fault voltage between bus 2 and ref. bus is Vf , then the voltage during fault is zero. Then, the system can be analyzed using the reactance circuit.
  39. 39. During the three-phase fault, the voltage between bus 2 and the reference becomes zero. Therefore, the 3 phase fault at bus 2 can be simulated by making the voltage between bus 2 and Ref. is equal to zero during the fault condition. 3 2 1 4 " fI fVfV This is simulated by inserting a source with a magnitude equal to Therefore, the voltage between bus 2 and reference is zero and fault current runs in the system. )( fV This means that, the Fault current during the three-phase short circuit results from the voltage source " fI )( fV
  40. 40. Therefore, to compute the short circuit current, we have to find the current entering bus 2 due to only )( fV             4 3 2 1 V V V V                  4 3 1 V V Vf V       BUSZ 44434241 34333231 24232221 14131211 ZZZZ ZZZZ ZZZZ ZZZZ             =              0 0 I 0 '' f= The Nodal Impedance Equation during the fault is: Note: is the voltage difference between node i and the reference node due to the current . iV " fI Ignoring all other currents entering other buses (ignore sources). 3 2 1 4 " fI fV
  41. 41. Or             4 3 2 1 V V V V                  4 3 1 V V Vf V                      '' f42 '' f32 '' f22 '' f12 IZ IZ IZ IZ = = And from the second row, the fault current is: 22 '' Z V I f f  Substituting in the previous equation:             4 3 2 1 V V V V                  4 3 1 V V Vf V                            f 22 42 f 22 32 f f 22 12 V Z Z V Z Z V V Z Z = =
  42. 42. Assuming no load is connected in the system, no current will flow before fault and all bus voltages are the same and equal to the pre- fault voltage )V( f Using Superposition, then during the fault:             4 3 2 1 V V V V     fV                        22 42 22 32 22 12 Z Z 1 Z Z 1 0 Z Z 1 = + i)voltagefaultPre(f)voltageiBus(i VVV   iV is the voltage change in bus i due to fault current in bus 2.             4 3 2 1 V V V V               f f f f V V V V =
  43. 43. Example: A three-phase fault occurs at bus 2 of the network shown in the Figure. Determine: 1. the sub-transient current during the fault, 2.the voltages at all the buses 3.The current flow during the fault from buses 3-2, 1-2 and 4-2. Consider the pre-fault voltage at bus 2 equal to 1 pu. and neglect the pre-fault currents. 3 T1 T2 G1 G2 1 2 4 fV 3 phase fault e
  44. 44. Solution: Construct the reactance circuit under normal operation condition. pu2.0j pu1.0j pu125.0j pu25.0j pu4.0j pu25.0j up2.0j 3 2 1 4 pu2.0j pu1.0j Construct the bus impedance Matrix (ZBUS)                1954.0j1046.0j1506.0j1456.0j 1046.0j1954.0j1494.0j1544.0j 1506.0j1494.0j2295.0j1938.0j 1456.0j1544.0j1938.0j2436.0j ZBUS NOTE: Constructing ZBus will be discussed later.
  45. 45. Since there are no-load currents, the pre-fault voltage at all the buses is equal to 1.0 pu. When the fault occurs at bus 2, pu357.4j 2295.0j 0.1 Z V I 22 f'' f  fV                        22 42 22 32 22 12 Z Z 1 Z Z 1 0 Z Z 1 =             4 3 2 1 V V V V fV                        22 42 22 32 22 12 Z Z 1 Z Z 1 0 Z Z 1 =             4 3 2 1 V V V V =                        2295.0j 1506.0j 1 2295.0j 1494.0j 1 0 2295.0j 1938.0j 1             343.0 349.0 0 155.0 =                1954.0j1046.0j1506.0j1456.0j 1046.0j1954.0j1494.0j1544.0j 1506.0j1494.0j2295.0j1938.0j 1456.0j1544.0j1938.0j2436.0j ZBUS 3 T1 T2 G1 G2 1 2 4 fV 3 phase fault e
  46. 46. The current flows from bus 1 to bus 2 is: 125.0j VV I 21 12   pu2448.1jI12  125.0j 1556.0 I12  pu2.0j pu1.0j pu125.0j pu25.0j pu4.0j pu25.0j pu2.0j 3 2 1 4 " fI pu2.0j pu1.0j The current flows from bus 3 to bus 2 is: 25.0j V3V I 2 32   pu396.1jI32  pu719.1jI42  The current flows from bus 4 to bus 2 is: The current flows from bus 3 to bus 1 is: 25.0j VV I 13 31   pu7736.0j
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  • TanvirHassan20

    Jul. 11, 2018
  • TasneemBanu5

    Dec. 14, 2017
  • bhaskarshriman

    Nov. 17, 2017

solving problems symmetrical faults balanced

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