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SHORT CIRCUIT CALCULATIONS REVISITED. pptx

This document discusses short circuit calculations and fault analysis in power systems. It begins by explaining that while fault-proof systems are not practical, determining currents, voltages, and impedances during faults is important for several applications including equipment ratings, protective device coordination, and stability studies. The document then covers per-unit calculations, symmetrical components analysis using positive, negative, and zero sequence networks, and modeling components like generators, transformers and transmission lines in these networks. Various types of faults are classified including balanced, unbalanced, shunt and series faults. The derivation of connecting these sequence networks for fault analysis is also presented.

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SHORT CIRCUIT CALCULATIONS REVISITED Short Circuit Calculations IIEE Presentation FORTUNATO C. LEYNES Chairman Professional Regulatory Board of Electrical Engineering Professional Regulation Commission
Short Circuit (Fault) Analysis FAULT-PROOF SYSTEM  not practical  neither economical  faults or failures occur in any powersystem In the various parts of the electrical network under short circuit or unbalanced condition, the determination of the magnitudes and phase angles  Currents  Voltages  Impedances Short Circuit Calculations IIEE Presentation
Application of Fault Analysis Short Circuit Calculations IIEE Presentation 1. The determination of the required mechanical strength of electrical equipment to withstand the stresses brought about by the flow of high short circuit currents 2. The selection of circuit breakers and switch ratings 3. The selection of protective relay and fuse ratings
4. The setting and coordination of protective devices 5. The selection of surge arresters and insulation ratings of electrical equipment 6. The determination of fault impedances for use in stability studies 7. The calculation of voltage sags caused resulting from short circuits Short Circuit Calculations IIEE Presentation Application of Fault Analysis
8. The sizing of series reactors to limit the short circuit current to a desired value 9. To determine the short circuit capability of series capacitors used in series compensation of long transmission lines 10. To determine the size of grounding transformers, resistances, or reactors Short Circuit Calculations IIEE Presentation Application of Fault Analysis
Per Unit Calculations Short Circuit Calculations IIEE Presentation
Three-phase Systems Short Circuit Calculations IIEE Presentation )2 base MVA3 (base voltage, kV base kVA3 (base voltage, kV )2 1000 ZB LL ZB LL
Per Unit Quantities B Short Circuit Calculations IIEE Presentation pu B pu B pu Base impedance (Z ) actual impedance BaseVoltage (kV ) V Base Current (I ) actual current Z   actual voltage (kV) I 
Changing the Base of Per Unit Quantities pu[new] B[new] Short Circuit Calculations IIEE Presentation [new] [new] B[new] [old] pu[old] [old] pu[old] Z Z base kVA basekV Z base kVA[old] base kVA[old] Z  Z() 2 1000  Z base kV 2 1000 Z()  base kV 2 1000  actual impedance, Z()
Changing the Base of Per Unit Quantities B[new] Short Circuit Calculations IIEE Presentation pu[new] [new] [new] B[new] [old] pu[old ] [old ] pu[old] Z Z base kVA basekV Z base kVA[old] base kVA[old] Z  Z() 2 1000  Z base kV 2 1000 Z()  base kV 2 1000  actual impedance, Z()
kVA Base for Motors Short Circuit Calculations IIEE Presentation kVA/hp hp rating 1.00 Induction < 100hp 1.00 Synchronous 0.8 pf 0.95 Induction 100 < 999hp 0.90 Induction > 1000hp 0.80 Synchronous 1.0 pf
SYMMETRICA L COMPONENT S Short Circuit Calculations IIEE Presentation
Vc1 Va1 Va2 V Vb2 Unbalanced Phasors Zero Sequence Vb1 Positive Sequence Va0  Vb0 Vc0 c Short Circuit Calculations IIEE Presentation V Vb Vc0 c1 V Vc2 Vb0 Vb2 Vb1 Va1 a2 V Negative Sequence c2 Va 0 Va
Symmetrical Components of Unbalanced Three-phase Phasor a Short Circuit Calculations IIEE Presentation b c a a V   V Vb Vc   a2 V  aV  b c a2 2 a1 a0 3 1 3 V  1 V  aV  a V  3 V  1 V a2 a1 a0 c a2 a1 a0 b a2 a1 V V  aV  a2 V V V  a2 V  aV V V Va Va0
In matrix form: Symmetrical Components of Unbalanced Three-phase Phasor     2 1 1 1 Vc Vb   1 a a  Short Circuit Calculations IIEE Presentation aVa1    V 2  a2 Va2  Va2  a1  Va  1Va0  Va0 1 1Va  1 1 3 1 a a Vb  a2 aVc    1 
Sequence Networks Short Circuit Calculations IIEE Presentation Power System Short Circuit Calculations
Fault Point Short Circuit Calculations IIEE Presentation The fault point of a system is that point to which the unbalanced connection is attached to an otherwise balanced system.
Definition of Sequence Networks Positive-sequence Network Ea1 = Thevenin’s equivalent voltage as seen at the fault Z1 = point Thevenin’s equivalent impedance as seen from the fault point Va1  Ea1 Ia1Z1 Z1 Ea1 Ia1 Va1 + Short Circuit Calculations IIEE Presentation -
Negative-sequence Network Z2 = Thevenin’s equivalent negative-sequence impedance as seen at the fault point Definition of Sequence Networks Va2 Ia 2Z2 Z2 Ia2 Va 2 + - Short Circuit Calculations IIEE Presentation
Zero-sequence Network Z0 = Thevenin’s equivalent zero-sequence impedance as seen at the fault point Definition of Sequence Networks a0 0 a0 V  I Z Z0 Ia0 Va 0 + - Short Circuit Calculations IIEE Presentation
Sequence Network Models of Power System Components Short Circuit Calculations IIEE Presentation Power System Short Circuit Calculations
Synchronous Machines (Positive Sequence Network) Z1  jx''d Short Circuit Calculations IIEE Presentation
Synchronous Machines (Negative Sequence Network) 2 2 x''d x''q Z  j Z2 Ia2 Va2 + - Where: x”d = direct-axis sub-transient reactance x”q = quadrature-axis sub-transient reactance Short Circuit Calculations IIEE Presentation
Synchronous Machines (Zero Sequence Network) Solidly-Grounded Neutral jX0 n I0 ground Short Circuit Calculations IIEE Presentation
Synchronous Machines (Zero Sequence Network) I0 Impedance-Grounded Neutral jX0 n 3Zg ground Short Circuit Calculations IIEE Presentation
Synchronous Machines (Zero Sequence Network) Ungrounded-Wye or Delta Connected Generators jX0 n I0 ground Short Circuit Calculations IIEE Presentation
Two-Winding Transformers (Positive Sequence Network) Standard Symbol P S ZPS Equivalent Positive-seq. Network P S Short Circuit Calculations IIEE Presentation
Three-Winding Transformers (Positive Sequence Network) t Zst  Zs Z Z ps  Zp  Zs Z pt  Z p  Zt   pt st ps s pt st p ps Zpt  Zst  Z ps Zt   Z Z 2 1  Z  1 Z 2 Z  1 Z  Z  Z  P T Standard Symbol Zt T Equivalent Positive-Sequence Network ZS Zp S S P 2 Short Circuit Calculations IIEE Presentation
Transformers (Negative Sequence Network) Short Circuit Calculations IIEE Presentation The negative-sequence network of two- winding and three-winding transformers are modeled in the same way as the positive-sequence network since the positive-sequence and negative-sequence impedances of transformers are equal.
Simplified Derivation of Transformer Zero- Sequence Circuit Modeling Grounded wye S1 = 1 and S3 = 0 or S2 = 1 or S4 = 0 Delta S1 = 0 and S3 = 1 or S2 = 0 and S4 = 1 Ungrounded wye S1 = 0 and S3 = 0 or S2 = 0 and S4 = 0 cuit Calculations (Thanks to Engr. Antonio C. Coronel, Retired VP, Meralco, and former member, Board of Electrical Engineering) P Q Short Cir IIEEPresentation
Simplified Derivation of Transformer Zero-Sequence Circuit Modeling P Q S1 Short Circuit Calculations IIEE Presentation S3 S4 S2 Grounded wye – Grounded wye S1 = 1 S2 = 1 S3 = 0 S4 = 0 Z
Simplified Derivation of Transformer Zero-Sequence Circuit Modeling P Q Grounded wye – Ungrounded wye S1 = 1 S2 = 0 S3 = 0 S4 = 0 Short Circuit Calculations IIEE Presentation
Simplified Derivation of Transformer Zero-Sequence Circuit Modeling S1 = 1 S2 = 0 S3 = 0 S4 = 1 P Q Grounded wye – Delta Short Circuit Calculations IIEE Presentation
Simplified Derivation of Transformer Zero-Sequence Circuit Modeling S1 = 0 S2 = 0 S3 = 0 S4 = 0 P Q Delta – Delta Short Circuit Calculations IIEE Presentation
Transformers (Zero-Sequence Circuit Model) Transformer Connection Zero-Sequence Circuit Equivalent P Q ZPQ P Q P Q ZPQ P Short Circuit Calculations IIEE Presentation Q
Transformer Connection Q P ZPQ P Q P Q ZPQ P Short Circuit Calculations IIEE Presentation Q Transformers (Zero-Sequence Circuit Model) Zero-Sequence Circuit Equivalent
Transformer Connection P Q R ZP ZQ ZR P R Q P Q ZPQ Zero-Sequence Circuit Equivalent P Q Short Circuit Calculations IIEE Presentation Transformers (Zero-Sequence Circuit Model)
Transformer Connection Zero-Sequence Circuit Equivalent ZP ZQ ZR P R Q P Q R Q P R ZP ZQ ZR P R Q Transformers (Zero-Sequence Circuit Model) Short Circuit Calculations IIEE Presentation
Transmission Lines (Positive Sequence Network) ra jXL Z1 Short Circuit Calculations IIEE Presentation
Transmission Lines (Negative Sequence Network) Short Circuit Calculations IIEE Presentation The same model as the positive-sequence network is used for transmission lines inasmuch as the positive-sequence and negative-sequence impedances of transmission lines are the same
Transmission Lines (Zero Sequence Network) Short Circuit Calculations IIEE Presentation The zero-sequence network model for a transmission line is the same as that of the positive- and negative-sequence networks. The sequence impedance of the model is of course the zero-sequence impedance of the line. This is normally higher than the positive- and negative-sequence impedances because of the influence of the earth’s resistivity and the ground wire/s.
Power System Short Circuit Calculations Short Circuit Calculations IIEE Presentation Classification of Power System Short Circuits
Shunt Faults Short Circuit Calculations IIEE Presentation Single line-to-ground faults Double line-to-ground faults Line-to-line faults Three-phase faults
Series Faults Short Circuit Calculations IIEE Presentation One-line open faults Two-line open faults
Combination of Shunt and Series Faults Short Circuit Calculations IIEE Presentation Single line-to-ground and one-line open Double line-to-ground and one-line open faults Line-to-line and one-line open faults Three-phase and one-line open faults
Single line-to-ground and two-line open faults Double line-to-ground and two-line open faults Line-to-line and two-line open faults Three-phase and two-line open faults Short Circuit Calculations IIEE Presentation Combination of Shunt and Series Faults
Balanced Faults Short Circuit Calculations IIEE Presentation Symmetrical or Three-Phase Faults
Derivation of Sequence Network Interconnections Vc  Ic Z f Boundary conditions: Ia  Ib  Ic 0 VF  Va  Ia Z f  Vb  Ib Z f Eq’n (1) Eq’n (2) Zf Zf Vf System A System B A B C Va Vb Vc Ia Ib Ic Zf Short Circuit Calculations IIEE Presentation
Short Circuit Calculations IIEE Presentation Z0 Ia0 Va0 + - Z1 Ea1 Ia1 Va1 + - Zf Z2 Ia2 Va2 + - Zf 1 1 Z a1 f f Ea1 1 f Ea1 a1  Ea1 I  I Z  Z I f  Ia1  Zf  0, I f  Ia 0  Ia1 Ia 2 Z  Z I 
Unbalanced Faults Short Circuit Calculations IIEE Presentation Single Line-to-Ground Faults
System A System B A B C Zf Va Vb Vc Ia Ib Ic Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ib  Ic  0 Va  Ia Z f  0
E Z0  Z1  Z2  3Zf Ia0  Ia1  Ia2 a1 Z1 Ea1 Ia1 Va1 + - Z2 Ia2 Va2 + - Z0 Ia0 Va0 + - 3Zf 1 Short Circuit Calculations IIEE Presentation 0 2 0 1 2 1 2 1 0 I 3Ea1 Ea1 Ea1 a2 a1 a0 Z 2Z I f  Ia  If Z f  0 and Z1  Z I f  Ia  Ia0  Ia1  Ia 2  3Ia1 3Ia0 Z  2Z Ia 0  Ia1  Ia 2  Z  Z Z  Z  Z  I   I If Z f  0
Unbalanced Faults Short Circuit Calculations IIEE Presentation Line-to-Line Faults
Vc  Ib Zf Zf System A System B A B C Va Vb Vc Ia Ib Ic Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ia  0 Ib  Ic Vb  IbZ f Vc; or Vb
2 2Z1 E 1 2 f Ea1 a1 Ia1 a1 If Z f  0 and Z1  Z Z  Z Z I  a1 a1 f a2 a1 a0 c The fault current   j 3I Ia1  Ia 2 I  a2  aI Iao  0;  I  a2 I  aI I f  Ib  I Z1 Ea1 Va1 + - Zf Ia1 Z2 Ia2 Va2 + - Z0 Ia0 Va0 + - Short Circuit Calculations IIEE Presentation
1 Short Circuit Calculations IIEE Presentation 1 1 1 2 f [3] f [3] f 3 I I Z I 2Z 3Ea1 Ea1   Ea1 3  Ea1       2 Z I f   j if Z f  0  f     2Z  Z I   j 3  thus, with Z1  Z2 f [LL]
Unbalanced Faults Short Circuit Calculations IIEE Presentation Double-to-line Ground Fault
 Ib  Ic Zg Vb  Ib Zf Vc  IcZ f  Ib  Ic Zg Eq’n BC-1 Eq’n BC-2 Eq’n BC-3 Zf SystemA System B A B C Va Vb Vc Ia Ib Ic Zf Vf Zg (Ib+Ic) Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ia  0
g f Ea1 a1 Z2  Z0  2Z 3Z  3Zg  Z  Z Z  Z 2 f 0 f Z1 Z f  I  Z0 Ia0 Va0 + - Z2 Ia2 Va2 + - Zf Zf Zf+3Zg Z1 Ea1 Ia1 Va1 + - Short Circuit Calculations IIEE Presentation
Negative-sequence Component:    g   Z0 2Z f 3Z 2  Z 0 Z f  3Zg Ia 2  Ia1  Z Zero-sequence Component: The fault current f a0 a1 a2 a2 a1 a0 c a0 a1 a2 f b I f  2Ia0 1Ia1  1Ia 2  2Ia0 Ia1  Ia2  but Ia0  Ia1  Ia2  0; or Ia0  Ia1  Ia2  thus, I f 3Ia 0  a2  aI  a  a2 I I  2I  aI  a2 I   a2 I  aI I I  I  I  I  Short Circuit Calculations IIEE Presentation       g Z  Z0  2Z f 3Z 2 Z2  Z f Ia0  Ia1 
1 1 0 Short Circuit Calculations IIEE Presentation 0 1 Z Z1Z0 E Z Z Z Z0 Ea1 a1 a1 a2 Ea1 a1 Z 2  2Z Z         Z1  Z0    1 0         Z  Z Z        0    Z1  Z0  I  I Z 2  2Z Z 1 1 0  Z1  Z0 Ea1 Z1 Z0 Z1 1 0 I  If Z f  Zg  0 and Z1  Z2
Short Circuit Calculations 0 1 1 0 1 1 1 1 1 1 Z2 Z1Z0 a0 f Z1Ea1 a1 a 0 a1 3Ea1 Z  2Z I  3I  Ea1 Z 2Z   2Z Z   0    Z  Z  1 0      E Z    Z  Z  Z    0     Z  Z  Z  I  I  1 1 0 1 0 IIEEPresentation 0 0 1 1 0 1 0 1 0 1 1 Z2 Z0 Ea1 a1 Z1Z0 a2 Ea1 Z1Z0 a1 Z 2  2Z Z       1 0  Z Z        E  Z     Z Z Z   a1      Z   Z1  Z0  I  I  2Z Z  Z1  Z0 Ea1 Z  Z Z  I  If Z f  Zg  0 and Z1  Z2
Voltage Rise Phenomenon Short Circuit Calculations IIEE Presentation Single-to-line Ground Fault
Short Circuit Calculations IIEE Presentation Unfaulted Phase B Voltage During Single Line-to-Ground Faults                                            1 1 1  1 2 0 1 2 1   1 0  1   1 0  2  0  1 0   2 1  neglecting resistances, R1 and R0 ;  2 0 1  1  X1 X0  X    X0  a2 V  E Z1  Z   Z    Z0  a2 V  E  Z   Z   Z0  Z1  2  Z0   Z1   E a   2Z1  Z0   Z  Z  Z 2Z  Z Z a 2Z  Z  a E Z  Z a1  b a1   b a1 Vb  Ea1 a   Ea1 Ea1 a1 Ea1 Vb   2Z a2 V V  a2 V aV b a0 a1
Short Circuit Calculations Unfaulted Phase B Voltage During Single Line-to-Ground Faults Neglecting resistances R0 &R1 50 45 40 35 30 25 20 15 10 5 0 -10 -6 -3 -2.8 -2.6 -2.4 -1 0 1 2 5 9 25 45 65 85 -2.2 -2 -1.8 -1.6 -1.4 -1.2 X0/X1 Phase B Voltage (p.u.) IIEEPresentation
Short Circuit Calculations Fault MVA 1 IIEEPresentation 3 for singe line - to - ground fault in p.u.: Z I F(SLG) Z0 2Z1 I   1 MVAF  IF  MVAbase where, for E a1  1.0p.u.; for three  phase fault in p.u.: F(3)
Fault MVA Short Circuit Calculations IIEE Presentation 1 0 0 1 1 I p.u. I p.u. I FSLG FSLG Z  3 2Z  3 2Z  Z Single line to  ground fault MVA : MVAFSLG IF SLG( p.u.) MVAbase Z  1 F 3 Three  phase fault MVA: MVAF3 IF 3( p.u.) MVAbase
Assumptions Made to Simplify Fault Calculations Short Circuit Calculations IIEE Presentation 1. Pre-fault load currents are neglected. 2. Pre-fault voltages are assumed equal to 1.0 per unit. 3. Resistances are neglected (only for 115kV & up). 4. Mutual impedances, when not appreciableare neglected. 5. Off-nominal transformer taps are equal to1.0 per unit. 6. Positive- and negative-sequence impedances are equal.
Outline of Procedures for Short Circuit Calculations Short Circuit Calculations IIEE Presentation 1 Setup the network impedances expressed in per unit on a common MVA base in the form of a single-line diagram 2 Determine the single equivalent (Thevenin’s) impedance of each sequence network. 3 Determine the distribution factor giving the current in the individual branches for unittotal sequence current.
Outline of Procedures for Short Circuit Calculations Short Circuit Calculations IIEE Presentation 4 Interconnect the three sequence networks for the type of fault under considerations and calculate the sequence currents at the faultpoint. 5 Determine the sequence current distribution by the application of the distribution factors to the sequence currents at the fault point 6 Synthesize the phase currents from the sequence currents.
Outline of Procedures for Short Circuit Calculations Short Circuit Calculations IIEE Presentation 7 Determine the sequence voltages throughout the networks from the sequence current distribution and branch impedances 8 Synthesize the phase voltages from the sequence voltage components 9 Convert the pre unit currents and voltages to actual physical units
CIRCUIT BREAKING SIZING (Asymmetrical Rating Factors) Momentary Rating Multiplying Factor = 1.6 Interrupting Rating Multiplying Factor Short Circuit Calculations IIEE Presentation 8 cycles 5 cycles 3 cycles 1 ½ cycles = 1.0 = 1.1 = 1.2 = 1.5
EXAMPLE PROBLEM In the power system shown, determine the momentary and interrupting ratings for primary and secondary circuit breakers of transformer T2. Short Circuit Calculations IIEE Presentation
Solution: 10 1 0 F(SLG) x  3  20.125  0.05pu x1  8  0.125pu 100  I  1000 10 pu 100 F(3) G1:   Short Circuit Calculations IIEE Presentation Equivalent 69kVsystem@100MVA: T1: 100 75  M1, M2 : kVAB  0.95000  4500 or MVAB 4.5 x  0.15 4.5  3.3333pu x  0.06100  0.80pu 100 100 x" 0.1025   0.40pu x0  0.0625   0.24 pu T2,T3, T3: 30  I  800  8pu x  0.08100  0.2667pu
Solution: Positive sequence network: Short Circuit Calculations IIEE Presentation
Solution: Zero sequence network: Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models Short Circuit Calculations IIEE Presentation
Improper Sequence Network Models Short Circuit Calculations IIEE Presentation
ions Incorrect Sequence Network Models Short Circuit Calculat IIEE Presentation
QUESTIONS ? Short Circuit Calculations IIEE Presentation
Short Circuit Calculations IIEE Presentation

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SHORT CIRCUIT CALCULATIONS REVISITED. pptx

  • 1.
    SHORT CIRCUIT CALCULATIONS REVISITED Short CircuitCalculations IIEE Presentation FORTUNATO C. LEYNES Chairman Professional Regulatory Board of Electrical Engineering Professional Regulation Commission
  • 2.
    Short Circuit (Fault)Analysis FAULT-PROOF SYSTEM  not practical  neither economical  faults or failures occur in any powersystem In the various parts of the electrical network under short circuit or unbalanced condition, the determination of the magnitudes and phase angles  Currents  Voltages  Impedances Short Circuit Calculations IIEE Presentation
  • 3.
    Application of FaultAnalysis Short Circuit Calculations IIEE Presentation 1. The determination of the required mechanical strength of electrical equipment to withstand the stresses brought about by the flow of high short circuit currents 2. The selection of circuit breakers and switch ratings 3. The selection of protective relay and fuse ratings
  • 4.
    4. The settingand coordination of protective devices 5. The selection of surge arresters and insulation ratings of electrical equipment 6. The determination of fault impedances for use in stability studies 7. The calculation of voltage sags caused resulting from short circuits Short Circuit Calculations IIEE Presentation Application of Fault Analysis
  • 5.
    8. The sizingof series reactors to limit the short circuit current to a desired value 9. To determine the short circuit capability of series capacitors used in series compensation of long transmission lines 10. To determine the size of grounding transformers, resistances, or reactors Short Circuit Calculations IIEE Presentation Application of Fault Analysis
  • 6.
    Per Unit Calculations ShortCircuit Calculations IIEE Presentation
  • 7.
    Three-phase Systems Short CircuitCalculations IIEE Presentation )2 base MVA3 (base voltage, kV base kVA3 (base voltage, kV )2 1000 ZB LL ZB LL
  • 8.
    Per Unit Quantities B ShortCircuit Calculations IIEE Presentation pu B pu B pu Base impedance (Z ) actual impedance BaseVoltage (kV ) V Base Current (I ) actual current Z   actual voltage (kV) I 
  • 9.
    Changing the Baseof Per Unit Quantities pu[new] B[new] Short Circuit Calculations IIEE Presentation [new] [new] B[new] [old] pu[old] [old] pu[old] Z Z base kVA basekV Z base kVA[old] base kVA[old] Z  Z() 2 1000  Z base kV 2 1000 Z()  base kV 2 1000  actual impedance, Z()
  • 10.
    Changing the Baseof Per Unit Quantities B[new] Short Circuit Calculations IIEE Presentation pu[new] [new] [new] B[new] [old] pu[old ] [old ] pu[old] Z Z base kVA basekV Z base kVA[old] base kVA[old] Z  Z() 2 1000  Z base kV 2 1000 Z()  base kV 2 1000  actual impedance, Z()
  • 11.
    kVA Base forMotors Short Circuit Calculations IIEE Presentation kVA/hp hp rating 1.00 Induction < 100hp 1.00 Synchronous 0.8 pf 0.95 Induction 100 < 999hp 0.90 Induction > 1000hp 0.80 Synchronous 1.0 pf
  • 12.
  • 13.
    Vc1 Va1 Va2 V Vb2 Unbalanced Phasors ZeroSequence Vb1 Positive Sequence Va0  Vb0 Vc0 c Short Circuit Calculations IIEE Presentation V Vb Vc0 c1 V Vc2 Vb0 Vb2 Vb1 Va1 a2 V Negative Sequence c2 Va 0 Va
  • 14.
    Symmetrical Components of UnbalancedThree-phase Phasor a Short Circuit Calculations IIEE Presentation b c a a V   V Vb Vc   a2 V  aV  b c a2 2 a1 a0 3 1 3 V  1 V  aV  a V  3 V  1 V a2 a1 a0 c a2 a1 a0 b a2 a1 V V  aV  a2 V V V  a2 V  aV V V Va Va0
  • 15.
    In matrix form: SymmetricalComponents of Unbalanced Three-phase Phasor     2 1 1 1 Vc Vb   1 a a  Short Circuit Calculations IIEE Presentation aVa1    V 2  a2 Va2  Va2  a1  Va  1Va0  Va0 1 1Va  1 1 3 1 a a Vb  a2 aVc    1 
  • 16.
    Sequence Networks Short CircuitCalculations IIEE Presentation Power System Short Circuit Calculations
  • 17.
    Fault Point Short CircuitCalculations IIEE Presentation The fault point of a system is that point to which the unbalanced connection is attached to an otherwise balanced system.
  • 18.
    Definition of Sequence Networks Positive-sequenceNetwork Ea1 = Thevenin’s equivalent voltage as seen at the fault Z1 = point Thevenin’s equivalent impedance as seen from the fault point Va1  Ea1 Ia1Z1 Z1 Ea1 Ia1 Va1 + Short Circuit Calculations IIEE Presentation -
  • 19.
    Negative-sequence Network Z2 =Thevenin’s equivalent negative-sequence impedance as seen at the fault point Definition of Sequence Networks Va2 Ia 2Z2 Z2 Ia2 Va 2 + - Short Circuit Calculations IIEE Presentation
  • 20.
    Zero-sequence Network Z0 =Thevenin’s equivalent zero-sequence impedance as seen at the fault point Definition of Sequence Networks a0 0 a0 V  I Z Z0 Ia0 Va 0 + - Short Circuit Calculations IIEE Presentation
  • 21.
    Sequence Network Modelsof Power System Components Short Circuit Calculations IIEE Presentation Power System Short Circuit Calculations
  • 22.
    Synchronous Machines (Positive SequenceNetwork) Z1  jx''d Short Circuit Calculations IIEE Presentation
  • 23.
    Synchronous Machines (Negative SequenceNetwork) 2 2 x''d x''q Z  j Z2 Ia2 Va2 + - Where: x”d = direct-axis sub-transient reactance x”q = quadrature-axis sub-transient reactance Short Circuit Calculations IIEE Presentation
  • 24.
    Synchronous Machines (Zero SequenceNetwork) Solidly-Grounded Neutral jX0 n I0 ground Short Circuit Calculations IIEE Presentation
  • 25.
    Synchronous Machines (Zero SequenceNetwork) I0 Impedance-Grounded Neutral jX0 n 3Zg ground Short Circuit Calculations IIEE Presentation
  • 26.
    Synchronous Machines (Zero SequenceNetwork) Ungrounded-Wye or Delta Connected Generators jX0 n I0 ground Short Circuit Calculations IIEE Presentation
  • 27.
    Two-Winding Transformers (Positive SequenceNetwork) Standard Symbol P S ZPS Equivalent Positive-seq. Network P S Short Circuit Calculations IIEE Presentation
  • 28.
    Three-Winding Transformers (Positive SequenceNetwork) t Zst  Zs Z Z ps  Zp  Zs Z pt  Z p  Zt   pt st ps s pt st p ps Zpt  Zst  Z ps Zt   Z Z 2 1  Z  1 Z 2 Z  1 Z  Z  Z  P T Standard Symbol Zt T Equivalent Positive-Sequence Network ZS Zp S S P 2 Short Circuit Calculations IIEE Presentation
  • 29.
    Transformers (Negative Sequence Network) ShortCircuit Calculations IIEE Presentation The negative-sequence network of two- winding and three-winding transformers are modeled in the same way as the positive-sequence network since the positive-sequence and negative-sequence impedances of transformers are equal.
  • 30.
    Simplified Derivation ofTransformer Zero- Sequence Circuit Modeling Grounded wye S1 = 1 and S3 = 0 or S2 = 1 or S4 = 0 Delta S1 = 0 and S3 = 1 or S2 = 0 and S4 = 1 Ungrounded wye S1 = 0 and S3 = 0 or S2 = 0 and S4 = 0 cuit Calculations (Thanks to Engr. Antonio C. Coronel, Retired VP, Meralco, and former member, Board of Electrical Engineering) P Q Short Cir IIEEPresentation
  • 31.
    Simplified Derivation ofTransformer Zero-Sequence Circuit Modeling P Q S1 Short Circuit Calculations IIEE Presentation S3 S4 S2 Grounded wye – Grounded wye S1 = 1 S2 = 1 S3 = 0 S4 = 0 Z
  • 32.
    Simplified Derivation ofTransformer Zero-Sequence Circuit Modeling P Q Grounded wye – Ungrounded wye S1 = 1 S2 = 0 S3 = 0 S4 = 0 Short Circuit Calculations IIEE Presentation
  • 33.
    Simplified Derivation ofTransformer Zero-Sequence Circuit Modeling S1 = 1 S2 = 0 S3 = 0 S4 = 1 P Q Grounded wye – Delta Short Circuit Calculations IIEE Presentation
  • 34.
    Simplified Derivation ofTransformer Zero-Sequence Circuit Modeling S1 = 0 S2 = 0 S3 = 0 S4 = 0 P Q Delta – Delta Short Circuit Calculations IIEE Presentation
  • 35.
    Transformers (Zero-Sequence Circuit Model) Transformer Connection Zero-Sequence CircuitEquivalent P Q ZPQ P Q P Q ZPQ P Short Circuit Calculations IIEE Presentation Q
  • 36.
    Transformer Connection Q P ZPQ P Q PQ ZPQ P Short Circuit Calculations IIEE Presentation Q Transformers (Zero-Sequence Circuit Model) Zero-Sequence Circuit Equivalent
  • 37.
    Transformer Connection P Q R ZP ZQ ZR P R Q PQ ZPQ Zero-Sequence Circuit Equivalent P Q Short Circuit Calculations IIEE Presentation Transformers (Zero-Sequence Circuit Model)
  • 38.
    Transformer Connection Zero-Sequence Circuit Equivalent ZP ZQ ZR P R Q PQ R Q P R ZP ZQ ZR P R Q Transformers (Zero-Sequence Circuit Model) Short Circuit Calculations IIEE Presentation
  • 39.
    Transmission Lines (Positive SequenceNetwork) ra jXL Z1 Short Circuit Calculations IIEE Presentation
  • 40.
    Transmission Lines (Negative SequenceNetwork) Short Circuit Calculations IIEE Presentation The same model as the positive-sequence network is used for transmission lines inasmuch as the positive-sequence and negative-sequence impedances of transmission lines are the same
  • 41.
    Transmission Lines (Zero SequenceNetwork) Short Circuit Calculations IIEE Presentation The zero-sequence network model for a transmission line is the same as that of the positive- and negative-sequence networks. The sequence impedance of the model is of course the zero-sequence impedance of the line. This is normally higher than the positive- and negative-sequence impedances because of the influence of the earth’s resistivity and the ground wire/s.
  • 42.
    Power System ShortCircuit Calculations Short Circuit Calculations IIEE Presentation Classification of Power System Short Circuits
  • 43.
    Shunt Faults Short CircuitCalculations IIEE Presentation Single line-to-ground faults Double line-to-ground faults Line-to-line faults Three-phase faults
  • 44.
    Series Faults Short CircuitCalculations IIEE Presentation One-line open faults Two-line open faults
  • 45.
    Combination of Shuntand Series Faults Short Circuit Calculations IIEE Presentation Single line-to-ground and one-line open Double line-to-ground and one-line open faults Line-to-line and one-line open faults Three-phase and one-line open faults
  • 46.
    Single line-to-ground andtwo-line open faults Double line-to-ground and two-line open faults Line-to-line and two-line open faults Three-phase and two-line open faults Short Circuit Calculations IIEE Presentation Combination of Shunt and Series Faults
  • 47.
    Balanced Faults Short CircuitCalculations IIEE Presentation Symmetrical or Three-Phase Faults
  • 48.
    Derivation of SequenceNetwork Interconnections Vc  Ic Z f Boundary conditions: Ia  Ib  Ic 0 VF  Va  Ia Z f  Vb  Ib Z f Eq’n (1) Eq’n (2) Zf Zf Vf System A System B A B C Va Vb Vc Ia Ib Ic Zf Short Circuit Calculations IIEE Presentation
  • 49.
    Short Circuit Calculations IIEEPresentation Z0 Ia0 Va0 + - Z1 Ea1 Ia1 Va1 + - Zf Z2 Ia2 Va2 + - Zf 1 1 Z a1 f f Ea1 1 f Ea1 a1  Ea1 I  I Z  Z I f  Ia1  Zf  0, I f  Ia 0  Ia1 Ia 2 Z  Z I 
  • 50.
    Unbalanced Faults Short CircuitCalculations IIEE Presentation Single Line-to-Ground Faults
  • 51.
    System A SystemB A B C Zf Va Vb Vc Ia Ib Ic Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ib  Ic  0 Va  Ia Z f  0
  • 52.
    E Z0  Z1 Z2  3Zf Ia0  Ia1  Ia2 a1 Z1 Ea1 Ia1 Va1 + - Z2 Ia2 Va2 + - Z0 Ia0 Va0 + - 3Zf 1 Short Circuit Calculations IIEE Presentation 0 2 0 1 2 1 2 1 0 I 3Ea1 Ea1 Ea1 a2 a1 a0 Z 2Z I f  Ia  If Z f  0 and Z1  Z I f  Ia  Ia0  Ia1  Ia 2  3Ia1 3Ia0 Z  2Z Ia 0  Ia1  Ia 2  Z  Z Z  Z  Z  I   I If Z f  0
  • 53.
    Unbalanced Faults Short CircuitCalculations IIEE Presentation Line-to-Line Faults
  • 54.
    Vc  IbZf Zf System A System B A B C Va Vb Vc Ia Ib Ic Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ia  0 Ib  Ic Vb  IbZ f Vc; or Vb
  • 55.
    2 2Z1 E 1 2 f Ea1 a1 Ia1a1 If Z f  0 and Z1  Z Z  Z Z I  a1 a1 f a2 a1 a0 c The fault current   j 3I Ia1  Ia 2 I  a2  aI Iao  0;  I  a2 I  aI I f  Ib  I Z1 Ea1 Va1 + - Zf Ia1 Z2 Ia2 Va2 + - Z0 Ia0 Va0 + - Short Circuit Calculations IIEE Presentation
  • 56.
    1 Short Circuit Calculations IIEEPresentation 1 1 1 2 f [3] f [3] f 3 I I Z I 2Z 3Ea1 Ea1   Ea1 3  Ea1       2 Z I f   j if Z f  0  f     2Z  Z I   j 3  thus, with Z1  Z2 f [LL]
  • 57.
    Unbalanced Faults Short CircuitCalculations IIEE Presentation Double-to-line Ground Fault
  • 58.
     Ib Ic Zg Vb  Ib Zf Vc  IcZ f  Ib  Ic Zg Eq’n BC-1 Eq’n BC-2 Eq’n BC-3 Zf SystemA System B A B C Va Vb Vc Ia Ib Ic Zf Vf Zg (Ib+Ic) Derivation of Sequence Network Interconnections Short Circuit Calculations IIEE Presentation Boundary conditions: Ia  0
  • 59.
    g f Ea1 a1 Z2  Z0 2Z 3Z  3Zg  Z  Z Z  Z 2 f 0 f Z1 Z f  I  Z0 Ia0 Va0 + - Z2 Ia2 Va2 + - Zf Zf Zf+3Zg Z1 Ea1 Ia1 Va1 + - Short Circuit Calculations IIEE Presentation
  • 60.
    Negative-sequence Component:    g  Z0 2Z f 3Z 2  Z 0 Z f  3Zg Ia 2  Ia1  Z Zero-sequence Component: The fault current f a0 a1 a2 a2 a1 a0 c a0 a1 a2 f b I f  2Ia0 1Ia1  1Ia 2  2Ia0 Ia1  Ia2  but Ia0  Ia1  Ia2  0; or Ia0  Ia1  Ia2  thus, I f 3Ia 0  a2  aI  a  a2 I I  2I  aI  a2 I   a2 I  aI I I  I  I  I  Short Circuit Calculations IIEE Presentation       g Z  Z0  2Z f 3Z 2 Z2  Z f Ia0  Ia1 
  • 61.
    1 1 0 ShortCircuit Calculations IIEE Presentation 0 1 Z Z1Z0 E Z Z Z Z0 Ea1 a1 a1 a2 Ea1 a1 Z 2  2Z Z         Z1  Z0    1 0         Z  Z Z        0    Z1  Z0  I  I Z 2  2Z Z 1 1 0  Z1  Z0 Ea1 Z1 Z0 Z1 1 0 I  If Z f  Zg  0 and Z1  Z2
  • 62.
    Short Circuit Calculations 0 1 10 1 1 1 1 1 1 Z2 Z1Z0 a0 f Z1Ea1 a1 a 0 a1 3Ea1 Z  2Z I  3I  Ea1 Z 2Z   2Z Z   0    Z  Z  1 0      E Z    Z  Z  Z    0     Z  Z  Z  I  I  1 1 0 1 0 IIEEPresentation 0 0 1 1 0 1 0 1 0 1 1 Z2 Z0 Ea1 a1 Z1Z0 a2 Ea1 Z1Z0 a1 Z 2  2Z Z       1 0  Z Z        E  Z     Z Z Z   a1      Z   Z1  Z0  I  I  2Z Z  Z1  Z0 Ea1 Z  Z Z  I  If Z f  Zg  0 and Z1  Z2
  • 63.
    Voltage Rise Phenomenon ShortCircuit Calculations IIEE Presentation Single-to-line Ground Fault
  • 64.
    Short Circuit Calculations IIEEPresentation Unfaulted Phase B Voltage During Single Line-to-Ground Faults                                            1 1 1  1 2 0 1 2 1   1 0  1   1 0  2  0  1 0   2 1  neglecting resistances, R1 and R0 ;  2 0 1  1  X1 X0  X    X0  a2 V  E Z1  Z   Z    Z0  a2 V  E  Z   Z   Z0  Z1  2  Z0   Z1   E a   2Z1  Z0   Z  Z  Z 2Z  Z Z a 2Z  Z  a E Z  Z a1  b a1   b a1 Vb  Ea1 a   Ea1 Ea1 a1 Ea1 Vb   2Z a2 V V  a2 V aV b a0 a1
  • 65.
    Short Circuit Calculations UnfaultedPhase B Voltage During Single Line-to-Ground Faults Neglecting resistances R0 &R1 50 45 40 35 30 25 20 15 10 5 0 -10 -6 -3 -2.8 -2.6 -2.4 -1 0 1 2 5 9 25 45 65 85 -2.2 -2 -1.8 -1.6 -1.4 -1.2 X0/X1 Phase B Voltage (p.u.) IIEEPresentation
  • 66.
    Short Circuit Calculations FaultMVA 1 IIEEPresentation 3 for singe line - to - ground fault in p.u.: Z I F(SLG) Z0 2Z1 I   1 MVAF  IF  MVAbase where, for E a1  1.0p.u.; for three  phase fault in p.u.: F(3)
  • 67.
    Fault MVA Short CircuitCalculations IIEE Presentation 1 0 0 1 1 I p.u. I p.u. I FSLG FSLG Z  3 2Z  3 2Z  Z Single line to  ground fault MVA : MVAFSLG IF SLG( p.u.) MVAbase Z  1 F 3 Three  phase fault MVA: MVAF3 IF 3( p.u.) MVAbase
  • 68.
    Assumptions Made toSimplify Fault Calculations Short Circuit Calculations IIEE Presentation 1. Pre-fault load currents are neglected. 2. Pre-fault voltages are assumed equal to 1.0 per unit. 3. Resistances are neglected (only for 115kV & up). 4. Mutual impedances, when not appreciableare neglected. 5. Off-nominal transformer taps are equal to1.0 per unit. 6. Positive- and negative-sequence impedances are equal.
  • 69.
    Outline of Proceduresfor Short Circuit Calculations Short Circuit Calculations IIEE Presentation 1 Setup the network impedances expressed in per unit on a common MVA base in the form of a single-line diagram 2 Determine the single equivalent (Thevenin’s) impedance of each sequence network. 3 Determine the distribution factor giving the current in the individual branches for unittotal sequence current.
  • 70.
    Outline of Proceduresfor Short Circuit Calculations Short Circuit Calculations IIEE Presentation 4 Interconnect the three sequence networks for the type of fault under considerations and calculate the sequence currents at the faultpoint. 5 Determine the sequence current distribution by the application of the distribution factors to the sequence currents at the fault point 6 Synthesize the phase currents from the sequence currents.
  • 71.
    Outline of Proceduresfor Short Circuit Calculations Short Circuit Calculations IIEE Presentation 7 Determine the sequence voltages throughout the networks from the sequence current distribution and branch impedances 8 Synthesize the phase voltages from the sequence voltage components 9 Convert the pre unit currents and voltages to actual physical units
  • 72.
    CIRCUIT BREAKING SIZING (Asymmetrical RatingFactors) Momentary Rating Multiplying Factor = 1.6 Interrupting Rating Multiplying Factor Short Circuit Calculations IIEE Presentation 8 cycles 5 cycles 3 cycles 1 ½ cycles = 1.0 = 1.1 = 1.2 = 1.5
  • 73.
    EXAMPLE PROBLEM In the powersystem shown, determine the momentary and interrupting ratings for primary and secondary circuit breakers of transformer T2. Short Circuit Calculations IIEE Presentation
  • 74.
    Solution: 10 1 0 F(SLG) x  3  20.125 0.05pu x1  8  0.125pu 100  I  1000 10 pu 100 F(3) G1:   Short Circuit Calculations IIEE Presentation Equivalent 69kVsystem@100MVA: T1: 100 75  M1, M2 : kVAB  0.95000  4500 or MVAB 4.5 x  0.15 4.5  3.3333pu x  0.06100  0.80pu 100 100 x" 0.1025   0.40pu x0  0.0625   0.24 pu T2,T3, T3: 30  I  800  8pu x  0.08100  0.2667pu
  • 75.
    Solution: Positive sequence network: ShortCircuit Calculations IIEE Presentation
  • 76.
    Solution: Zero sequence network: ShortCircuit Calculations IIEE Presentation
  • 77.
    Improper Sequence NetworkModels Short Circuit Calculations IIEE Presentation
  • 78.
    Improper Sequence NetworkModels Short Circuit Calculations IIEE Presentation
  • 79.
    Improper Sequence NetworkModels Short Circuit Calculations IIEE Presentation
  • 80.
    ions Incorrect Sequence NetworkModels Short Circuit Calculat IIEE Presentation
  • 81.
  • 82.