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Engineering

SYMMETRICAL COMPONENT OF ELECTRICAL POWER SYSTEMS

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- Symmetrical Components I An Introduction to Power System Fault Analysis Using Symmetrical Components Dave Angell Idaho Power 21st Annual Hands-On Relay School
- What Type of Fault? -25 0 25 -25 0 25 -25 0 25 -2500 0 2500 -2500 0 2500 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 V A V B V C I A I B I C Cycles VA VB VC IA IB IC
- What Type of Fault? -10000 0 10000 -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- What Type of Fault? -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C Cycles IA IB IC
- What Type of Fault? -5000 0 5000 -5000 0 5000 -5000 0 5000 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- What Type of Fault? -100 0 100 -100 0 100 -100 0 100 -200 -0 200 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- What Type of Fault? -200 0 200 -200 0 200 -200 0 200 -500 0 500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- What Type of Fault? -250 0 250 -250 0 250 -250 0 250 -100 0 100 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- Basic Course Topics Terminology Phasors Equations Fault Analysis Examples Calculations
- Unbalanced Fault Ia Ib Ic Ia Ib Ic Ia Ib Ic
- Symmetrical Component Phasors The unbalanced three phase system can be transformed into three balanced phasors. – Positive Sequence – Negative Sequence – Zero Sequence
- Positive Phase Sequence (ABC) -1.0 -0.5 0.0 0.5 1.0 0.000 0.017 0.033 0.050 Time Magnitude Va Vb Vc
- Positive Phase Sequence Each have the same magnitude. Each positive sequence voltage or current quantity is displaced 120° from one another. Va1 Vb1 Vc1
- Positive Phase Sequence The positive sequence quantities have a- b-c, counter clock- wise, phase rotation. Va1 Vb1 Vc1
- Reverse Phase Sequence (ACB) -1.0 -0.5 0.0 0.5 1.0 0.000 0.017 0.033 0.050 Time Magnitude Va Vb Vc
- Negative Phase Sequence Each have the same magnitude. Each negative sequence voltage or current quantity is displaced 120° from one another. Va2 Vc2 Vb2
- Negative Phase Sequence The negative sequence quantities have a- c-b, counter clock- wise, phase rotation. Va2 Vc2 Vb2
- Zero Phase Sequence -1.0 -0.5 0.0 0.5 1.0 0.000 0.017 0.033 0.050 Time Magnitude Va Vb Vc
- Zero Phase Sequence Each zero sequence quantity has the same magnitude. All three phasors with no angular displacement between them, all in phase. Va0 Vb0 Vc0
- Symmetrical Components Equations Each phase quantity is equal to the sum of its symmetrical phasors. Va = Va0 + Va1 +Va2 Vb = Vb0 + Vb1 +Vb2 Vc = Vc0 + Vc1 +Vc2 The common form of the equations are written in a-phase terms.
- The a Operator Used to shift the a-phase terms to coincide with the b and c-phase Shorthand to indicate 120° rotation. Similar to the j operator of 90°. Va
- Rotation of the a Operator 120° counter clock-wise rotation. A vector multiplied by 1 /120° results in the same magnitude rotated 120°. Va aVa
- Rotation of the a2 Operator 240° counter clock-wise rotation. A vector multiplied by 1 /240° results in the same magnitude rotated 240°. Va a2Va
- B-Phase Zero Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator. Vb0 = Va0 Va0 Vb0 Vc0
- B-Phase Positive Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator Vb1 = a2Va1 Va1 Vb1 Vc1
- B-Phase Negative Sequence We replace the Vb sequence terms by Va sequence terms shifted by the a operator Vb2 = aVa2 Va2 Vc2 Vb2
- C-Phase Zero Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator. Vc0 = Va0 Va0 Vb0 Vc0
- C-Phase Positive Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator Vc1 = aVa1 Va1 Vb1 Vc1
- C-Phase Negative Sequence We replace the Vc sequence terms by Va sequence terms shifted by the a operator Vc2 = a2Va2 Va2 Vc2 Vb2
- What have we produced? Va = Va0 + Va1 + Va2 Vb = Va0 + a2Va1 + aVa2 Vc = Va0 + aVa1 + a2Va2
- Symmetrical Components Equations Analysis – To find out of the amount of the components Synthesis – The combining of the component elements into a single, unified entity
- Symmetrical Components Synthesis Equations Va = Va0 + Va1 + Va2 Vb = Va0 + a2Va1 + aVa2 Vc = Va0 + aVa1 + a2Va2
- Symmetrical Components Analysis Equations Va0 = 1/3 ( Va + Vb + Vc) Va1= 1/3 (Va + aVb + a2Vc) Va2= 1/3 (Va + a2Vb + aVc)
- Symmetrical Components Analysis Equations - 1/3 ?? Where does the 1/3 come from? Va1= 1/3 (Va + aVb + a2Vc) Va = Va0 + Va1 + Va2 When balanced 0 0
- Symmetrical Components Analysis Equations - 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases Va
- Symmetrical Components Analysis Equations - 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases yields Va aVb Vc Va Vb
- Symmetrical Components Analysis Equations - 1/3 ?? Va1= 1/3 (Va + aVb + a2Vc) Adding the phases yields 3 Va. Divide by the 3 and now Va = Va1 a2 Vc Va aVb Vc Va Vb
- Example Vectors An Unbalanced Voltage Va Vc Vb Va = 13.4 /0° Vb = 59.6 /- 104° Vc = 59.6 /104°
- Analysis Results in These Sequence Quantities Va0 Vb0 Vc0 Va2 Vc2 Vb2 Va1 Vb1 Vc1 Va0 = -5.4 s Va1 = 42.9 s Va2 = -24.1
- Synthesize by Summing the Positive, Negative and … Va2 Vb2 Vc2 Va1 Vb1 Vc1
- Zero Sequences Va2 Vb2 Vc2 Va0 Vb0 Vc0 Va1 Vb1 Vc1
- The Synthesis Equation Results in the Original Unbalanced Voltage Va2 Vb2 Vc2 Va0 Vb0 Vc0 Va1 Vb1 Vc1 Va Vc Vb
- Symmetrical Components Present During Shunt Faults Three phase fault – Positive Phase to phase fault – Positive – Negative Phase to ground fault – Positive – Negative – Zero
- Symmetrical Component Review of Faults Types Let’s return to the example fault reports and view the sequence quantities present
- Three Phase Fault, Right? -25 0 25 -25 0 25 -25 0 25 -2500 0 2500 -2500 0 2500 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 V A V B V C I A I B I C Cycles VA VB VC IA IB IC
- A Symmetrical Component View of an Three-Phase Fault Component Magnitude Angle Ia0 7.6 175 Ia1 2790 -64 Ia2 110 75.8 Va0 0 0 Va1 18.8 0 Va2 0.7 337 0 45 90 135 180 225 270 315 I1 V1
- A to Ground Fault, Okay? -10000 0 10000 -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- A Symmetrical Component View of an A-Phase to Ground Fault Component Magnitude Angle Ia0 7340 -79 Ia1 6447 -79 Ia2 6539 -79 Va0 46 204 Va1 123 0 Va2 79 178 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- Single Line to Ground Fault Voltage – Negative and zero sequence 180 out of phase with positive sequence Current – All sequence are in phase
- A to B Fault, Easy? -10000 0 10000 -10000 0 10000 -10000 0 10000 1 2 3 4 5 6 7 8 9 10 11 I A I B I C Cycles IA IB IC
- A Phase Symmetrical Component View of an A to B Phase Fault Component Magnitude Angle Ia0 3 -102 Ia1 5993 -81 Ia2 5961 -16 Va0 1 45 Va1 99 0 Va2 95 -117 0 45 90 135 180 225 270 315 I1 I2 V1 V2
- C Phase Symmetrical Component View of an A to B Phase Fault Component Magnitude Angle Ic0 3 138 Ic1 5993 279 Ic2 5961 104 Vc0 1 -75 Vc1 99 0 Vc2 95 2.5 0 45 90 135 180 225 270 315 I1 I2 V1 V2
- Line to Line Fault Voltage – Negative in phase with positive sequence Current – Negative sequence 180 out of phase with positive sequence
- B to C to Ground -5000 0 5000 -5000 0 5000 -5000 0 5000 -2500 0 2500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- A Symmetrical Component View of a B to C to Ground Fault Component Magnitude Angle Ia0 748 97 Ia1 2925 -75 Ia2 1754 101 Va0 8 351 Va1 101 0 Va2 18 348 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- Line to Line to Ground Fault Voltage – Negative and zero in phase with positive sequence Current – Negative and zero sequence 180 out of phase with positive sequence
- Again, What Type of Fault? -100 0 100 -100 0 100 -100 0 100 -200 -0 200 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- C Symmetrical Component View of a C-Phase Open Fault Component Magnitude Angle Ic0 69 184 Ic1 101 4 Ic2 32 183 Vc0 0 162 Vc1 79 0 Vc2 5 90 0 45 90 135 180 225 270 315 I0 I1 I2 V1 V2
- One Phase Open (Series) Faults Voltage – No zero sequence voltage – Negative 90 out of phase with positive sequence Current – Negative and zero sequence 180 out of phase with positive sequence
- What About This One? -200 0 200 -200 0 200 -200 0 200 -500 0 500 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- Ground Fault with Reverse Load Ic0 164 -22 Ic1 89 -113 Ic2 41 -6 Vc0 4 -123 Vc1 38 0 Vc2 6 -130 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- Finally, The Last One! -250 0 250 -250 0 250 -250 0 250 -100 0 100 1 2 3 4 5 6 7 8 9 10 11 I A I B I C I R Cycles IA IB IC IR
- Component Magnitude Angle Ic0 45 40 Ic1 153 -4 Ic2 132 180 Vc0 0.5 331 Vc1 40 0 Vc2 0.5 93 Fault on Distribution System with Delta – Wye Transformer 0 45 90 135 180 225 270 315 I0 I1 I2 V0 V1 V2
- Use of Sequence Quantities in Relays Zero Sequence filters – Current – Voltage Relay operating quantity Relay polarizing quantity
- Zero Sequence Current Ia Ib Ic Direction of the protected line Ia+Ib+Ic 3I0
- Zero Sequence Voltage (Broken Delta) Va Vb Vc 3V0
- Zero Sequence Voltage Va Va Vc Vb Va Vb 3Vo Va Vb
- Sequence Operating Quantities Zero and negative sequence currents are not present during balanced conditions. Good indicators of unbalanced faults
- Sequence Polarizing Quantities Polarizing quantities are used to determine direction. The quantities used must provide a consistent phase relationship.
- Zero Sequence Voltage Polarizing 3Vo is out of phase with Va -3Vo is used to polarize for ground faults Va Vb 3Vo
- Learning Check Given three current sources How can zero sequence be produced to test a relay? How can negative sequence produced?
- How can zero sequence be produced to test a relay? A single source provides positive, negative and zero sequence – Note that each sequence quantity will be 1/3 of the total current Connect the three sources in parallel and set their amplitude and the phase angle equal to one another – The sequence quantities will be equal to each source output
- How can negative sequence produced? A single source provides positive, negative and zero sequence – Each sequence quantity will be 1/3 of the total current Set the three source’s amplitude equal to one another and the phase angles to produce a reverse phase sequence (Ia at /0o , Ib at /120o and Ic at /-120o ) – Only negative sequence will be produced
- Advanced Course Topics Sequence Networks Connection of Networks for Faults Per Unit System Power System Element Models
- References Symmetrical Components for Power Systems Engineering, J Lewis Blackburn Protective Relaying, J Lewis Blackburn Power System Analysis, Stevenson Analysis of Faulted Power System, Paul Anderson
- Conclusion Symmetrical components provide: – balanced analysis of an unbalanced system. – a measure of system unbalance – methods to detect faults – an ability to distinguish fault direction

- A Phase to Ground Fault
- A to B Phase Fault
- B to C to Ground Fault
- C Phase Open Fault
- A to B Phase Fault
- B to C to Ground Fault
- C Phase Open Fault