symmetrical components explained by example
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Symmetrical Components Explained by Example
Symmetrical components is the name given to a methodology discovered by Charles Legeyt Fortescue in 1913.Fortescue demonstrated that any set of unbalanced three-phase quantities could be expressed as the sum of three symmetrical sets of balanced phasors. Using this method, unbalanced system conditions, like those caused by common fault types may be analyzed with a "Per Phase" approach.
According to Fortescueβs methodology, there are three sets of independent phasors (components) in a three-phase system: Positive Sequence: β’ Supplied by the "Generator " or "Source" and are always present. β’ Equal in Magnitude.β’ Rotate counter-clockwise with the "Sequence" A-B-C-A ....etc.β’ First A, then B, then C, then A again, etc.β’ In other words: B Lags A, C Lags B, A Lags C (by 120 Degrees.... or 5.56 sec in a 60Hz system).
Negative Sequence: β’ Negative Sequence "Generator " or "Source" Voltages will only exist if the "Source" is unbalanced. β’ Negative Sequence System Voltages and Currents will exist during unbalanced fault conditions.β’ Equal in magnitude.β’ Rotate counter-clockwise with the "Sequence" A-C-B-A ....etc.β’ First A, then C , then B, then A again, etc.β’ In other words: C Lags A, B Lags C, A Lags B (by 120 Degrees.... or 5.56 sec in a 60Hz system).
Zero Sequence: β’ Zero Sequence "Generator " or "Source" Voltages will not exist in a "Normal" 3-Phase Source. β’ Zero Sequence System Voltages and Currents will exist during unbalanced fault conditions when "Ground" Currents
flow.β’ Equal in magnitude.β’ Rotate counter-clockwise with no "Sequence" .β’ In other words: A and B and C then 360 Degrees later... A and B and C again ....etc.
Question Is .....How to "Decompose" an Unbalance set of 3-Phase Phasors into their Positive, Negative, and Zero Sequence ComponentsOr .....How to "Compose" a set of 3-Phase Vectors Given Phase "A" Positive, Negative, and Zero Sequence VectorsNote: The A, B, and C Phase Vectors could be Voltage or Current
Positive Sequence Phasors
Negative Sequence Phasors
Symmetrical Components Explained by Example (cont.)
π΄ = π΄+ + π΄β + π΄0
π΅ = π΅+ + π΅β + π΅0
πΆ = πΆ+ + πΆβ + πΆ0
A,B,C Phasors each have 3 components (positive, negative, zero)
where:
π΅+ = π΄+β 240π΅β = π΄ββ 120π΅0 = π΄0
πΆ+ = π΄+β 120πΆβ = π΄ββ 240πΆ0 = π΄0
πΌ = 1β 120Β°πΌ2 = 1β 240
let:
then:
π΄ = π΄0 + π΄+ + π΄β
π΅ = π΄0 + πΌ2π΄+ + πΌπ΄β
πΆ = π΄0 + πΌπ΄+ + πΌ2π΄β
solving for A+, A-, A0
π΄0 =1
3π΄ + π΅ + πΆ
π΄+ =1
3π΄ + πΌπ΅ + πΌ2πΆ
π΄β =1
3π΄ + πΌ2π΅ + πΌπΆ
in matrix form:
π΄0
π΄+
π΄β=1
3
π΄π΅πΆ
1 1 11 πΌ πΌ2
1 πΌ2 πΌ
π΄π΅πΆ
=1
3
π΄0
π΄+
π΄β
1 1 11 πΌ2 πΌ1 πΌ πΌ2
start:
Symmetrical Components Explained by Example (cont.)
Now ... How to Apply Symmetrical Components to Faults in Three-Phase Power Systems:β’ First .... important to realize that the source or generator voltage is balanced (in three-phase power systems)β’ Phase A source voltage is chosen for the reference angle (0)
|πβ | =|ππΏπΏ|
3ππ΄ = |πβ |β 0Β° ππ΅ = |πβ |β 240Β° ππΆ = |πβ |β 120Β°
Because the source or generator voltage is balanced ..... negative and zero sequence source voltages are zero (do not exist)
ππ΄β = ππ΅β = ππΆβ = 0β 0ππ΄0 = ππ΅π = ππΆπ = 0β 0ππ΄+ = |πβ |β 0ππ΅+ = |πβ |β 240Β°ππΆ+ = |πβ | β 120Β°
In other words .... β’ The source voltages are nothing new ... the same per-phase voltages used in normal circuit analysis.
Symmetrical Components Explained by Example (cont.)
Now that a per-phase fault analysis is possible ... How to model phase impedances?β’ First .... A,B,C phase impedances are assumed equal (in three-phase power systems)β’ Next... Draw the Circuit to be Analyzed .... Using only Phase A.
This circuit represents the source voltage and Thevenin impedance looking into some fault on the system⒠the Thevenin impedance consists of positive, negative, and zero sequence components⒠notice that loads are not considered in fault analysis.⒠important to realize that all values are phasor quantities⒠next⦠draw the three sequence networks
ππ΄+ = |πβ |β 0Β° Load = 0
Z
π = π + ππ
ππ΄+
Z+ Zo
ππ΄0 = 0
Z-
ππ΄β = 0
Symmetrical Components Explained by Example (cont.)
Next... The Trick is how to Connect the Sequence Networks for Different Fault Conditions.Although it may not be Intuitive ..... This is how it works:
Three-Phase to Ground Fault (3LG):⒠this is a balanced fault... no ground current will flow⒠since this is a balanced Fault..... no negative sequence current will flow⒠since there is no ground current... no zero sequence current will flow⒠the three sequence networks are not connected⒠this means that you only have to deal with the positive sequence network⒠the currents in the negative and zero sequence networks are zero⦠(no voltage to source them)⒠analyze the positive sequence network to obtain the A phase current or voltages⒠B and C phase current will have the same magnitude as A⦠but shifted by +/- 120
Single Line to Ground Fault (1LG):β’ this is an unbalanced fault.... ground Current will Flowβ’ since this is an unbalanced fault..... negative sequence current will flowβ’ since there is ground current..... zero sequence current will flowβ’ the positive, negative, and zero sequence networks are connected in seriesβ’ analyze this connection to obtain the A phase currents or voltagesβ’ you will find that the three sequence currents in phase A are equalβ’ use the phase A currents to calculate the B and C phases (using the transformations described above)
Line-to-Line Fault (LL):β’ this is an unbalanced fault.... but ground current will not Flowβ’ since this is an unbalanced fault..... negative sequence current will flowβ’ since there is no ground current..... zero sequence current will not flowβ’ the positive and negative sequence networks are connected in parallelβ’ analyze this connection to obtain the A phase currents or voltagesβ’ you will find that the positive and negative sequence currents are equal but shifted by 180β’ use the phase A currents to calculate the B and C phases (using the transformations described above)
Line-to-Line to Ground Fault (2LG):β’ this is an unbalanced fault.... ground current will flowβ’ Since this is an Unbalanced Fault..... Negative Sequence Current will Flowβ’ Since there is Ground Current..... Zero Sequence Current will Flowβ’ the positive, negative and zero sequence networks are connected in parallelβ’ analyze this connection to obtain the A phase currents or voltagesβ’ you will find that the positive sequence current is equal to the sum of the negative and zero sequence currents shifted by 180β’ use the phase A currents to calculate the B and C phases (using the transformations described above)
ExampleThree-Phase Fault (3L) -or- Three-Phase to Ground Fault (3LG):
Z+
ππ΄+πΌπ΄+
Z-
πΌπ΄β = 0
Zo
πΌπ΄π = 0
πΏππ‘:π+ = 2 + π10 = 10.2β 78.7Β° Ξ©πβ = πππ‘ πππππππ0 = πππ‘ ππππππππ¦π π‘ππ ππΏπΏ = 138ππ
|πβ | =ππΏπΏ
3= 79674
πΌπ΄+ =ππ΄+
π+=
79674β 0Β°
10.2β 78.7Β°= 7811β 281.3Β°
ππ πΆππ πππ€ πΆππππ’πππ‘π π‘βπ π΅+and πΆ+ Values:πΌπ΅+ = πΌπ΄+ 1β 240Β° = 7811β 281.3Β° 1β 240Β° = 7811β 161.3Β°πΌπΆ+ = πΌπ΄+ 1β 120Β° = 7811β 281.3Β° 1β 120Β° = 7811β 41.3Β°πππ€ πΆππππ’πππ‘π π‘βπ "Total" π΄, π΅ and πΆ Values:πΌπ΄ = πΌπ΄+ + πΌπ΄β + πΌπ΄π = πΌπ΄+ + 0 + 0 = πΌπ΄+ = 7811β 281.3Β° π΄πππ πΌπ΅ = πΌπ΅+ + πΌπ΅β + πΌπ΅π = πΌπ΅+ + 0 + 0 = πΌπ΅+ = 7811β 161.3Β° π΄πππ πΌπΆ = πΌπΆ+ + πΌπΆβ + πΌπΆπ = πΌπΆ+ + 0 + 0 = πΌπΆ+ = 7811β 41.3Β° π΄πππ
πΌπππ’ππ‘ = πΌπ΄+ + πΌπ΅+ + πΌπΆ+ = 0
3L or 3LG Fault
πΌπ΄β = πΌπ΅β = πΌπΆβ = 0 πΌπ΄0 = πΌπ΅0 = πΌπΆ0 = 0
1LG Fault
ExampleSingle Line-to-Ground Fault (1LG) on Phase "A":
Z+
ππ΄+πΌπ΄+
Z-
πΌπ΄β
Zo
πΌπ΄π
πΏππ‘:ππ¦π π‘ππ ππππ‘πππ = 69ππ
|πβ | =ππΏπΏ
3=69ππ
3= 39837 π
ππ΄+ = 39837β 0Β° = 39837 + π0 ππ+ = 9 + π20 = 21.93Ξ©β 65.77Β° Ξ©πβ = 9 + π20 = 21.93Ξ©β 65.77Β° Ξ©π0 = 16 + π39 = 42.15Ξ©β 67.69Β° Ξ©
πΉπππ π‘ πππ‘πππ: πΌπ΄+= πΌπ΄β = πΌπ΄π
ππΈπ = π+ + πβ + ππππΈπ = 9 + π20 + 9 + π20 + 16 + π39 = 34 + π79 = 86.0β 66.71Β°
πΌπ΄+ = πΌπ΄β = πΌπ΄π =ππ΄+
ππΈπ=
39837β 0Β°
86.0β 66.71Β°= 463.2β 293.3Β°
π πππππ πβππ‘:πΌπ΄π = πΌπ΅π = πΌπΆπ = 463.2β 293.3Β° = 183.2 β π425.4
πΆππππ’πππ‘π π‘βπ π΅+and πΆ+ Values:πΌπ΅+ = πΌπ΄+ 1β β 120Β° = 463.2β β 66.71 Β° 1β 240Β° = 463.2β 173.3Β° = β460.0 + π54.0πΌπΆ+ = πΌπ΄+ 1β 120Β° = 463.2β β 66.71 Β° 1β 120Β° = 463.2β 53.3Β° = 276.9 + π371.3
πΆππππ’πππ‘π π‘βπ π΅βand πΆβ Values:πΌπ΅β = πΌπ΄β 1β 120Β° = 463.2β β 66.71 Β° 1β 120Β° = 463.2β 53.3Β° = 276.9 + π371.3πΌπΆβ = πΌπ΄β 1β β 120Β° = 463.2β β 66.71 Β° 1β 240Β° = 463.2β 173.3Β° = β460.0 + π54.0
πΆππππ’πππ‘π π‘βπ "Total" π΄, π΅ and πΆ Values:πΌπ΄ = πΌπ΄+ + πΌπ΄β + πΌπ΄π = 3πΌπ΄π = 3 463.2β 293.3Β° = 1389.6β 293.3Β°πΌπ΅ = πΌπ΅+ + πΌπ΅β + πΌπ΅π = β460.0 + π54.0 + 276.9 + π371.3 + 183.1 β π425.5 = 0 + π0 = 0πΌπΆ = πΌπΆ+ + πΌπΆβ + πΌπΆπ = 276.9 + π371.3 β 460.0 + π54.0 + 183.1 β π425.5 = 0 + π0 = 0
πβππ π πππ π’ππ‘π π βππ€ π‘βππ‘ ππ πππ’ππ‘ ππ’πππππ‘ ππ ππππ‘πππ’π‘ππ ππππ π΅ πππ πΆ πβππ ππ β¦ .πππππ π πππ π
1LG Fault (cont.)
πΌπππ’ππ‘ = πΌπ΄ + πΌπ΅ + πΌπΆπΌπππ’ππ‘ = 1389.6β 293.3Β°
πππ‘:πΌπ΅ππ π = πΌπ΄+ = 463.2π‘βππ:
πΌπππ’ππ‘ππ’
=πΌπππ’ππ‘πΌπ΅ππ π
=1389.6β 293.3Β°
463.2πΌπππ’ππ‘ππ’
= 3.0β 293.3Β° = 3.0β β 66.7
LL Fault
ExampleLine-to-Line Fault (LL) on Phases B to C:
πΌπ΄+
Z+
ππ΄+
-Z
πΌπ΄β
Zo
πΌπ΄π = 0
πΏππ‘:ππ¦π π‘ππ ππππ‘πππ = 138πππ+ = 1Ξ© + π8Ξ© = 8.06Ξ©β 82.87Β°πβ = 1Ξ© + π8Ξ© = 8.06Ξ©β 82.87Β°π0 = πππ‘ ππππππ
|πβ | =ππΏπΏ
3=138ππ
3= 79674
ππ΄+ = 79674β 0Β° = 79674 + π0
πΉπππ π‘ πππ‘πππ: πΌπ΄+= βπΌπ΄β πππ πΌπ΄π = 0ππΈπ = π+ + πβ = 1 + π8 + 1 + π8 = 2 + π16 = 16.1245β 82.87Β°
πΌπ΄+ = βπΌπ΄β=ππ΄+
ππΈπ=
79674β 0Β°
16.1245β 82.87Β°= 4941.18 β 277.13Β°
πΌπ΄+ = 4941.18 β 277.13Β° = 613.30 β π4903.0πΌπ΄β = 4941.18 β 97.13Β° = β613.30 + π4903.0
ππππ ππππ’ππππ:πΌπ΄π = πΌπ΅π = πΌπΆπ = 0
πΆππππ’πππ‘π π‘βπ π΅+and πΆ+ Values:πΌπ΅+ = πΌπ΄+ 1β 240Β° = 4941.18β 277.13Β° 1β 240Β° = 4941.18β 157.13Β° = β4552.75 + π1920.35πΌπΆ+ = πΌπ΄+ 1β 120Β° = 4941.18β 277.13Β° 1β 120Β° = 4941.18β 37.13Β° = 3939.44 + π2982.62
LL Fault (cont.)
πΆππππ’πππ‘π π‘βπ π΅βand πΆβ Values:πΌπ΅β = πΌπ΄β 1β 120Β° = 4941.18β 97.13Β° 1β 120Β° = 4941.18β 217.13Β° = β3939.44 β π2982.62πΌπΆβ = πΌπ΄β 1β 240Β° = 4941.18β 97.13Β° 1β 240Β° = 4941.18β 337.13Β° = 4552.75 β π1920.35
πΆππππ’πππ‘π π‘βπ "Total" π΄, π΅ and πΆ Values:πΌπ΄ = πΌπ΄+ + πΌπ΄β + πΌπ΄π = 613.30 β π4903.0 β 613.30 + π4903.0 + 0 = 0πΌπ΅ = πΌπ΅+ + πΌπ΅β + πΌπ΅π = β4552.75 + π1920.35 β 3939.44 β π2982.6 + 0 = β8492.2 + π1062.25 = 8558.19β 187.1Β°πΌπΆ = πΌπΆ+ + πΌπΆβ + πΌπΆπ = 3939.44 + π2982.62 + 4552.75 β π1920.35 + 0 = 8492.2 β π1062.25 = 8558.19β 7.1
πβππ π πππ π’ππ‘π π βππ€ π‘βππ‘:
ππ πππ’ππ‘ ππ’πππππ‘ ππππ€π ππ π΄ πβππ πβ¦
ππ π ππππ π΄ πβππ π π€ππ πππ‘ πππ’ππ‘ππ.
π΅ πππ πΆ πβππ π πππ’ππ‘ ππ’πππππ‘π πππ πππ’ππ πππ πππππ ππ‘πβ¦
πππππ π πππ π πππ πππ π ππ£π π πππ ππππ£πππ‘πππ.
πππ‘ πππππ’πππ‘ππ βπππ, ππ’π‘β¦
π΅ π‘π πΆ πππ’ππ‘ π£πππ‘πππ ππ πππ£πππ’π ππ¦ π§πππ, βππ€ππ£ππβ¦
π‘βπ π£ππ‘πππ ππ‘ π‘βπ π΅πΆ "πππ’ππ‘ πππππ‘" π€ππ‘β πππ ππππ‘ π‘π πππ’π‘πππ
ππ 1/2 ππ π‘βπ πβππ π π£πππ‘πππ
πππ‘:πΌπ΅ππ π = πΌπ΄+ = 4941.18π‘βππ:
πΌπ΅ππ’
=πΌπ΅
πΌπ΅ππ π=8558.19β 187.1Β°
4941.18= 1.732β 187.1Β°
πΌπΆππ’
=πΌπΆ
πΌπ΅ππ π=8558.19β 7.1Β°
4941.18= 1.732β 7.1Β°
2LG Fault
ExampleLine-to-Line-to-Ground Fault (2LG) on Phases B to C to Ground:
πΌπ΄+
Z+
ππ΄+
πΏππ‘:π+ = 0.6 + π4 = 4.0447β 81.469Β° Ξ©πβ = 0.6 + π4 = 4.0447β 81.469Β° Ξ©π0 = 1.4 + π6 = 6.161β 76.866Β° Ξ©ππ¦π π‘ππ ππππ‘πππ = 69ππ
|πβ | =ππΏπΏ
3=69ππ
3= 39837.2
ππ΄+ = 39837β 0Β° = 39837 + π0
πΉπππ π‘ πππ‘πππ: πΌπ΄++ πΌπ΄β+ πΌπ΄π = 0ππΈπ = π+ + πβ β₯ π0
πβ β₯ π0 =πβπ0
πβ + π0=(4.0447β 81.47Β°)(6.161β 76.87Β° )
0.6 + π4 + (1.4 + π6)=24.919β 158.34Β°
10.198β 78.69Β°= 2.4436β 79.645Β° = 0.4392 + π2.4038
ππΈπ = 0.6 + π4 + 0.4392 + π2.4038 = 1.03923 + π6.40385 = 6.487623β 80.782Β°
πΌπ΄+ =ππ΄+
ππΈπ=
39837β 0Β°
6.487623β 80.782Β°= 6140.46 β 279.218Β° = 983.62 β π6061.19
πΌπ΄β =βπΌπ΄+π0πβ + π0
πΌπ΄0 =βπΌπ΄+πβπβ + π0
2LG Fault (cont.)
2LG Fault (cont.)
πΊπππ’ππ πππ’ππ‘ πΆπ’πππππ‘:πΌπΊ = πΌπ΄ + πΌπ΅ + πΌπΆ = 0 + π0 β 9194.4 + π2308.1 + 7675.7 + π4838.8 = β1518.3 + π7146.9 = 7306.4β 102.0Β°π‘βππ π πππ π’ππ‘π π βππ€ π‘βππ‘ ππ πππ’ππ‘ ππ’πππππ‘ πΉπππ€π ππ πβππ π π΄β¦ ππ π ππππ π΄ πhππ π π€ππ πππ‘ πππ’ππ‘ππ.πππ πβ¦ . π‘βπ π π’π ππ π΄, π΅, πππ πΆ πππ’ππ‘ ππ’πππππ‘π = ππππ’ππ πππ’ππ‘ ππ’πππππ‘ 3πΌπ΄0
πππ‘:πΌπ΅ππ π = πΌπ΄+ = 6140.46π‘βππ:
πΌπ΅ππ’
=πΌπ΅
πΌπ΅ππ π=9479.7β 165.9Β°
6140.46= 1.544β 165.9Β°
πΌπΆππ’
=πΌπΆ
πΌπ΅ππ π=9073.6β 32.23Β°
6140.46= 1.478β 32.23Β°
πΌπΊππ’
=πΌπΊ
πΌπ΅ππ π=7306.4β 102.0Β°
6140.46= 1.190β 102.0Β°