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Symmetrical Components and Sequence Networks
Chapter 11
Balanced sequences
Synthesis equations
Synthesis equations
Analysis equations
๐๐0 = 1
3 (๐๐+ ๐๐ + ๐๐)
๐๐1 = 1
3 (๐๐+๐ ๐๐+๐2๐๐)
๐๐2 = 1
3 (๐๐+๐2 ๐๐+๐๐๐)
Example: Balanced line to neutral voltages with positive sequence
Calculate the sequence components of the following line to neutral voltages with abc sequence
๐๐๐ = 277โ 00 V
๐๐๐ = 277โ โ1200 V
๐๐๐ = 277โ 1200 V
Solution
Example: Balanced line currents with a negative sequence
Calculate the sequence components of the current in line a of a balanced star connected load with acb sequence
๐ผ๐ = 10โ 00 A
๐ผ๐ = 10โ 1200 A
๐ผ๐ = 10โ โ1200 A
Solution
๐ผ๐0 = 0 A
๐ผ๐1 = 0 A
๐ผ๐2 = 10โ 00 A
This example shows that a balanced negative sequence network has only negative sequence components
Example 11.1 Textbook
Example 11.1 Solution
Balanced ฮ circuits
Only line voltages are applicable
Line currents and phase currents are applicable
Balanced ฮ circuits
๐ผ๐ = ๐ผ๐๐ โ ๐ผ๐๐
๐ผ๐ = ๐ผ๐๐ โ ๐ผ๐๐
๐ผ๐ = ๐ผ๐๐ โ ๐ผ๐๐
For positive phase sequence
๐ผ๐1 = 3โ (โ300 ) ๐ผ๐๐1
The magnitude of the line current is 3 the phase current and lags by 300
Balanced ฮ circuits
For negative phase sequence
๐ผ๐2 = 3โ (+300 ) ๐ผ๐๐2
The magnitude of the line current is 3 the phase current and leads by 300
Phasors for balanced ฮ circuits
Balanced Y circuits
Only line currents are applicable
Line voltages and phase voltages are applicable
Balanced Y circuits
๐๐๐ = ๐๐๐ โ ๐๐๐
๐๐๐ = ๐๐๐ โ ๐๐๐
๐๐๐ = ๐๐๐ โ ๐๐๐
For positive phase sequence
๐๐๐1 = 3โ (+300 ) ๐๐๐1
The magnitude of the line voltage is 3 the phase voltage and leads by 300
Balanced Y circuits
For negative phase sequence
๐๐๐2 = 3โ (โ300 ) ๐๐๐2
The magnitude of the line voltage is 3 the phase voltage and lags by 300
Phasors for balanced Y circuits
Equivalent Y and ฮ loads
Given a ฮ connected load of impedance ๐ฮ per phase, it can be shown that the equivalent Y connected load will have an impedance of ๐๐ = ๐ฮ
3 .
Equivalent Y and ฮ loads
Example 11.2 Textbook
Example 11.2 solution
Example 11.2 solution
Example 11.2 solution
Example 11.2 solution
Example 11.2 solution
Example 11.2 solution
Power in terms of symmetrical components
๐3ฮฆ = ๐๐๐ผ๐โ+๐๐๐ผ๐
โ+๐๐๐ผ๐โ
= 3๐๐0๐ผ๐0โ+ 3๐๐1๐ผ๐1
โ+ 3๐๐2๐ผ๐2โ
Example 11.3
Using symmetrical components calculate the power absorbed in the load of example 11.2 and check the answer.
Example 11.3 solution
๐3ฮฆ = ๐๐๐ผ๐โ+๐๐๐ผ๐
โ+๐๐๐ผ๐โ
= 3๐๐0๐ผ๐0โ+ 3๐๐1๐ผ๐1
โ+ 3๐๐2๐ผ๐2โ
When working in per unit the factor of 3 falls away. Therefore
๐3ฮฆ =๐๐0๐ผ๐0โ+ ๐๐1๐ผ๐1
โ+ ๐๐2๐ผ๐2โ
= 0 + (0.9857โ 43.60)(0.9857โ -43.60) +(0.2346โ 250.30)(0.2346โ -250.30)
= 1.023โ 00
Actual
= 1.023โ 00(500)
= 513.320KW
Example 11.3 solution
Verifying the answer
๐ผ๐1 = 0.9857โ 43.60
๐ผ๐1 = 0.9857โ โ76.40
๐ผ๐1 = 0.9857โ 163.60
๐ผ๐2 = 0.2346โ 250.30
๐ผ๐2 = 0.2346โ 370.30
๐ผ๐2 = 0.2346โ 130.30
๐ผ๐0 = ๐ผ๐0= ๐ผ๐0=0
Example 11.3 solution
๐ผ๐ = ๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2 = 0.7832โ 35.870 ๐ผ๐ = ๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2 = 1.026โ โ63.20 ๐ผ๐ = ๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2 = 1.189โ 157.40
Base current Ib = 500000
3 (2300) = 125.5 A
Actual currents ๐ผ๐= 98.3โ 35.870 A ๐ผ๐= 128.8โ โ63.20 A ๐ผ๐= 149.2โ 157.40 A
Example 11.3 solution
Power = ๐ผ๐2๐ ๐+๐ผ๐
2๐ ๐+๐ผ๐2๐ ๐
= 98.32(10.58)+128.82(10.58)+149.22(10.58)
= 513 kW
Sequence circuits of Y and ฮ impedance loads
Sequence circuits of Y and ฮ impedance loads
Neutral current
๐ผ๐ = ๐ผ๐+ ๐ผ๐+ ๐ผ๐
= (๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2) + (๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2)+(๐ผ๐0+ ๐ผ๐1+ ๐ผ๐2)
= (๐ผ๐0+ ๐ผ๐0+ ๐ผ๐0) + (๐ผ๐1+ ๐ผ๐1+ ๐ผ๐1)+(๐ผ๐2+ ๐ผ๐2+ ๐ผ๐2)
= (๐ผ๐0+ ๐ผ๐0+ ๐ผ๐0) + 0 +0
= 3๐ผ๐0
The neutral current consists only of the zero sequence current
Sequence circuits of Y and ฮ impedance loads
The volt drop across ๐๐ is ๐๐๐ = 3๐ผ๐0๐๐ This means that the voltages to neutral (๐๐๐, ๐๐๐, ๐๐๐) and the voltages to ground (๐๐, ๐๐ , ๐๐) are different under balanced conditions ๐๐ = ๐๐๐ + ๐๐๐ = ๐๐๐ผ๐ + 3๐ผ๐0๐๐ ๐๐= ๐๐๐ + ๐๐๐ = ๐๐๐ผ๐ + 3๐ผ๐0๐๐ ๐๐ = ๐๐๐ + ๐๐๐ = ๐๐๐ผ๐ + 3๐ผ๐0๐๐
Sequence circuits of Y and ฮ impedance loads
The previous set of equations can be written in matrix form as
A
๐๐๐
๐๐๐
๐๐๐
= ๐๐ A
๐ผ๐๐
๐ผ๐๐
๐ผ๐๐
+ 3๐ผ๐0๐๐ 111
Where A =๐ ๐ ๐๐ ๐๐ ๐๐ ๐ ๐๐
Sequence circuits of Y and ฮ impedance loads
Multiplying throughout by ๐จโ๐ gives
๐๐๐
๐๐๐
๐๐๐
= ๐๐
๐ผ๐๐
๐ผ๐๐
๐ผ๐๐
+ 3๐ผ๐0๐๐ ๐จโ๐ 111
Which reduces to
๐๐๐
๐๐๐
๐๐๐
= ๐๐
๐ผ๐๐
๐ผ๐๐
๐ผ๐๐
+ 3๐ผ๐0๐๐ 100
Sequence circuits of Y and ฮ impedance loads
Writing as previous equations as separate equations Multiplying throughout by ๐จโ๐ gives
๐๐0 = (๐๐ + 3๐๐) ๐ผ๐0 = ๐0 ๐ผ๐0
๐๐1 = ๐๐๐ผ๐1 = ๐1 ๐ผ๐1
๐๐2 = ๐๐๐ผ๐2 = ๐2 ๐ผ๐2
Where
๐0 is the impedance to zero sequence current
๐1 is the impedance to positive sequence current
๐2 is the impedance to negative sequence current
Sequence circuits of Y impedance loads
The previous three equations results in three separate networks:
Positive sequence network
Negative sequence network
Zero sequence network
Sequence circuits of Y impedance loads
The previous three equations results in three separate circuits for the Y connected load
Sequence circuits of Y impedance loads
๐0 is the zero sequence impedance
๐1 is the zero sequence impedance
๐2 is the negative sequence impedance
Sequence circuits of Y impedance loads
๐๐ can assume the following values:
0 (short circuit โ solidly bolted, solidly grounded)
Some positive value
โ
Solidly grounded neutral Open circuited neutral
Sequence circuits of ฮ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
๐ฮ
๐
๐ฮ
๐
Sequence circuits of ฮ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
๐ฮ
๐
๐ฮ
๐ ๐ฮ
๐
Example
Three equal impedances of j30 ฮฉ are connected in ฮ. Determine the sequence impedances and draw the sequence networks. Repeat the problem for the case where a mutual impedance of j5 ฮฉ exists between each branch of the load.
Solution
j30 j30
j30
j30 j10 j10
Zero sequence network
Positive sequence network
Negative sequence network
Solution with mutual impedance
Solution with mutual impedance
j40 J8.3 J8.3
Sequence circuits of a transmission line
๐๐๐ - self impedance of each phase conductor ๐๐๐ - self impedance of the neutral conductor ๐๐๐ - mutual impedance between phase conductors ๐๐๐ - mutual impedance between neutral and phase conductors
Sequence circuits of a transmission line
The presence of the neutral conductor changes the self impedance and mutual impedance of the phase conductors:
๐๐ = ๐๐๐+ ๐๐๐ - 2 ๐๐๐ (self impedance)
๐๐ = ๐๐๐+ ๐๐๐ - 2 ๐๐๐ (mutual impedance)
Sequence circuits of a transmission line
Using the self and mutual impedance of the line , the volt drops across the line can be calculated from the following set of equations:
๐๐๐โฒ
๐๐๐โฒ
๐๐๐โฒ
=
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
=
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐ผ๐
๐ผ๐
๐ผ๐
Sequence circuits of a transmission line
The sequence impedances of the transmission line are defined as:
๐0 = ๐๐ + 2๐๐ = ๐๐๐+ 2 ๐๐๐ + 3 ๐๐๐ - 6 ๐๐๐
๐1 = ๐๐ - ๐๐ = ๐๐๐- ๐๐๐
๐2 = ๐๐ - ๐๐ = ๐๐๐- ๐๐๐
Sequence circuits of a transmission line
The volt drops across the line can be calculated from the following equations
Sequence circuits of a transmission line
Zero sequence network
Positive sequence network
Negative sequence network
Example
A three phase transmission line has the following voltages at the sending and receiving ends
๐๐๐=182+j70 kV ๐๐โฒ๐โฒ=154+j28 kV
๐๐๐=72.24-j32.62 kV ๐๐โฒ๐โฒ=44.24-j74.62 kV
๐๐๐=-170.24+j88.62 kV ๐๐โฒ๐โฒ=-198.24+j46.62 kV
The line impedances are
๐๐๐=j60 ฮฉ ๐๐๐=j20 ฮฉ ๐๐๐=j80 ฮฉ ๐๐๐=0 ฮฉ
Determine the line currents ๐ผ๐, ๐ผ๐ and ๐ผ๐
Solution
๐๐ = ๐๐๐+ ๐๐๐ - 2 ๐๐๐ = j60 + j80 โ j60 = j80 ฮฉ
๐๐ = ๐๐๐+ ๐๐๐ - 2 ๐๐๐ = j20 + j80 โ j60 = j40 ฮฉ
Then using
๐๐๐โฒ
๐๐๐โฒ
๐๐๐โฒ
=
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
=
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐ผ๐
๐ผ๐
๐ผ๐
Solution
๐๐๐โฒ
๐๐๐โฒ
๐๐๐โฒ
=
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
๐๐๐ โ ๐๐โฒ๐โฒ
=
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐๐ ๐๐ ๐๐
๐ผ๐
๐ผ๐
๐ผ๐
28 + ๐4228 + ๐4228 + ๐42
x103 =
๐80 ๐40 ๐40๐40 ๐80 ๐40๐40 ๐40 ๐80
๐ผ๐
๐ผ๐
๐ผ๐
Solution
๐ผ๐
๐ผ๐
๐ผ๐
=
๐80 ๐40 ๐40๐40 ๐80 ๐40๐40 ๐40 ๐80
28 + ๐4228 + ๐4228 + ๐42
x103
=
262.5 โ ๐175262.5 โ ๐175262.5 โ ๐175
A
-1
Solution
Using Symmetrical component
The sequence components of the line volt drops are
๐๐๐โฒ0
๐๐๐โฒ1
๐๐๐โฒ2
= 1
3
1 1 11 ๐ ๐2
1 ๐2 ๐
๐๐๐โฒ
๐๐๐โฒ
๐๐๐โฒ
Solution
๐๐๐โฒ0
๐๐๐โฒ1
๐๐๐โฒ2
= 1
3
1 1 11 ๐ ๐2
1 ๐2 ๐
28 + ๐4228 + ๐4228 + ๐42
x103
= 28 + ๐42
00
x103 V
Solution
๐0 = ๐๐๐+ 2 ๐๐๐ + 3 ๐๐๐ - 6 ๐๐๐ = j60 + j40 + j240 โ j180 = j160 ๐1 = ๐2 = ๐๐๐- ๐๐๐ = j60 โ j20 = j40 So ๐๐๐โฒ
0 = ๐0 ๐ผ๐0
๐๐๐โฒ1 = ๐1 ๐ผ๐
1 ๐๐๐โฒ
2 = ๐2 ๐ผ๐2
Solution
(28 + ๐42) 103 = (j160) ๐ผ๐0 ๐ผ๐
0 = 262.5 โ j175 A 0 = j40 ๐ผ๐1 ๐ผ๐
1 = 0 0 = j40 ๐ผ๐2 ๐ผ๐
2 = 0 Therefore ๐ผ๐ = ๐ผ๐ = ๐ผc = 262.5 โ j175 A
Sequence circuits of a synchronous machine
Sequence circuits of a synchronous machine
The sequence equations are:
๐ฟ๐ and ๐๐ are the self and mutual inductances of the windings
Zero sequence network
Positive sequence network
Negative sequence network
Sequence circuits of a synchronous machine
Sequence circuits of a synchronous machine
Sequence circuits of a synchronous machine
Example
A generator rated at 20 MVA, 13.8 kV has a direct axis subtransient reactance of 0.25 per unit. The negative and zero sequence reactances are 0.35 and 0.10 per unit respectively. The neutral of the generator is solidly grounded. If the generator is unloaded at rated voltage ๐ธ๐๐= 1.0โ 00 per unit and a single line to ground fault occurs at the machine terminals, which results in the following terminal voltages to ground
Example
๐๐= 0
๐๐= 1.0.13โ โ102.250
๐๐= 1.0.13โ โ102.250
Determine the subtransient current in the generator and the line-to-line voltages for subtransient conditions due to the fault.
Sequence circuits of Y โ ฮ transformers
5 Cases will be considered
Case 1: Y-Y bank โ both neutrals grounded
Case 2: Y-Y bank โ one neutral grounded
Case 3: ฮ-ฮ bank
Case 4: Y-ฮ bank โ Y grounded
Case 5: Y-ฮ bank โ Y ungrounded
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded Sequence equations: Positive sequence Negative sequence Zero sequence
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Zero Sequence equivalent circuit:
๐0 = 3๐๐ + 3 ๐๐
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
๐0 = Z + 3๐๐ + 3 ๐๐
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
๐0 = Z + 3๐๐ + 3 ๐๐
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Positive Sequence equivalent circuit with leakage impedance Z:
๐1 = Z
๐1
Sequence circuits of Y โ ฮ transformers
Case 1: Y-Y bank โ both neutrals grounded
Negative Sequence equivalent circuit with leakage impedance Z:
๐2 = Z
๐2
Sequence circuits of Y โ ฮ transformers
Case 2 Y-Y bank โ one neutral grounded
Sequence circuits of Y โ ฮ transformers
Case 2 Y-Y bank โ one neutral grounded
Zero Sequence equivalent circuit
๐ผ0 cannot flow due to the absence of a path for current flow between the windings
Sequence circuits of Y โ ฮ transformers
Case 3: ฮ-ฮ bank
Sequence circuits of Y โ ฮ transformers
Case 3: ฮ-ฮ bank Sequence equations:
VAB0 = Vab
0 = 0 Zero sequence
VAB1 =
๐1
๐2Vab
1 Positive sequence
VAB1 =
๐1
๐2Vab
1 Negative sequence
Sequence circuits of Y โ ฮ transformers
Case 3: ฮ-ฮ bank
Zero Sequence equivalent circuit with leakage impedance Z:
Sequence circuits of Y โ ฮ transformers
Case 3: Y-ฮ bank โ Y grounded
Sequence circuits of Y โ ฮ transformers
Case 3: Y-ฮ bank โ Y grounded Sequence equations:
VA0 - 3 ๐๐ IA
0 = ๐1๐2
Vab0 = 0 Zero sequence
VA1 =
๐1
๐2Vab
1 Positive sequence
VA2 =
๐1
๐2Vab
2 Negative sequence
Sequence circuits of Y โ ฮ transformers
Case 4: Y-ฮ bank โ Y grounded Zero Sequence equivalent circuit with leakage impedance Z:
๐0 = Z + 3๐๐
Sequence circuits of Y โ ฮ transformers
Case 5: Y-ฮ bank- Y ungrounded
Sequence circuits of Y โ ฮ transformers
Case 5: Y-ฮ bank โ Y ungrounded Sequence equations:
VAB0 =
๐1๐2
Vab0 = 0 Zero sequence
VAB1 =
๐1
๐2Vab
1 Positive sequence
VAB2 =
๐1
๐2Vab
2 Negative sequence
Sequence circuits of Y โ ฮ transformers
Case 5: Y-ฮ bank โ Y ungrounded Zero Sequence equivalent circuit with leakage impedance Z:
Y โ ฮ transformers
VAB1 = 3
๐1
๐2Vab
1 โ 300
VAB2 = 3
๐1
๐2Vab
2 โ โ300
In per unit
VAB1 = Vab
1 โ 300
VAB2 = Vab
2 โ โ300
The same applies for currents in the per unit case
IA1 = Ia
1 โ 300
IA2 = Ia
2 โ โ300
Example
Three identical Y connected resistors form a load with a three phase rating of 2300 V and 500 kVA. If the load has applied voltages
๐๐๐ = 1840 V ๐๐๐ = 2760 V ๐๐๐ = 2300 V
Find the line and currents in per unit into the load. Assume that the neutral of the load is not connected to the neutral of the system and select a base of 2300 V, 500 KVA.
Example
If the resistive Y connected load bank is supplied from the low voltage Y side of a Y-ฮ transformer, find the line voltages and currents in per unit on the high voltage side of the transformer.
Solution
Voltages and currents on the load side has been calculated previously ๐๐๐ = 0.8โ 82.80 ๐๐๐ = 1.2โ โ41.40 ๐๐๐ = 1.0โ 1800 Vab
1 = 0.9857โ 73.60 Vab
2 = 0.2346โ 220.30 Van
1 = 0.9857โ 43.60 Van
2 = 0.2346โ 250.30 Ia
1 = 0.9857โ 43.60 Ia
2 = 0.2346โ 250.30
Solution
The calculated load voltages will be the voltages on the low voltage side of the transformer Vab
1 = 0.9857โ 73.60 Vab
2 = 0.2346โ 220.30 Therefore high voltage side voltages in per unit will be VAB
1 = Vab1 โ 300
VAB2 = Vab
2 โ โ300 So VAB
1 = 0.9857 โ 103.60 VAB
2 = 0.2346 โ 190.30 Next calculate the sequence components of the other lines VBC
1 = 0.9857 โ (103.6โ120)0 = 0.9857 โ โ16.40 VCA
1 = 0.9857 โ (103.6+120)0 = 0.9857 โ 223.60 VBC
2 = 0.2346 โ (190.3+120)0 = 0.2346 โ 310.30 VCA
2 = 0.2346 โ (190.3โ120)0 = 0.2346โ 70.30
Solution
The line-line voltages on the high voltage side are VAB = VAB
0 + VAB1 + VAB
2 = 0 + 0.9857 โ 103.60 + 0.2346 โ 190.30 = 1.026 โ 116.80 VBC = VBC
0 + VBC1 + VBC
2 = 0 + 0.9857 โ โ16.40 + 0.2346 โ 310.30 = 1.19 โ โ22.60 VCA = VCA
0 + VCA1 + VCA
2 = 0 + 0.9857 โ 223.60 + 0.2346 โ 70.30 = 0.783 โ โ144.10
Solution
The calculated load currents will be the line currents on the low voltage side of the transformer Ia
1 = 0.9857โ 43.60 Ia
2 = 0.2346โ 250.30 Therefore high voltage side voltages in per unit will be IA
1=Ia1โ 300
IA2 =Ia
2โ โ300
So IA
1= 0.9857 โ 73.60 IA
2= 0.2346 โ 220.30 Next calculate the sequence components of the other lines IB
1 = 0.9857 โ (73.6โ120)0 = 0.9857 โ โ46.40 IC
1 = 0.9857 โ (73.6+120)0 = 0.9857 โ 193.60 IB
2 = 0.2346 โ (220.3+120)0 = 0.2346 โ 340.30 IC
2 = 0.2346 โ (220.3โ120)0 = 0.2346โ 100.30
Solution
The line currents on the high voltage side are IA = IA
0 + IA1 + IA
2 = 0 + 0.9857 โ 73.60 + 0.2346 โ 220.30 = 0.8 โ 82.90 IB = IB
0 + IB1 + IB
2 = 0 + 0.9857 โ โ46.40 + 0.2346 โ 340.30 = 1.2 โ โ41.40 IC = IC
0 + IC1 + IC
2 = 0 + 0.9857 โ 193.60 + 0.2346 โ 100.30 = 1.0 โ 179.90
Sequence Networks
Sequence circuits have been developed for the following components: Loads (Y and ฮ) Synchronous machines Transmission lines Transformers The components when combined make up a network Thus combining sequence circuits together make up a sequence network
Sequence Networks
Recapping sequence circuits:
1. Separate volt drop equations for each sequence can be set up
2. Z1 and Z2 are equal for static components (loads, lines and transformers)
Z1 and Z2 are approximately equal for synchronous machines under subtransient conditions
Sequence Networks
Recapping sequence circuits:
3. Z0 is generally different different from and Z1 and Z2
4. Only the positive sequence of synchronous machines contains a voltage source (E)
5. The neutral is reference for positive and negative sequence circuits. Voltage to neutral and voltage to ground are the same
Sequence Networks
Recapping sequence circuits:
6. No positive or negative sequence currents flow between neutral and ground
7. The impedance Zn is not included in positive
and negative sequence circuits but is represented as 3 Zn
in the zero sequence.
Sequence Networks
Balanced 3 phase systems generally make up a positive sequence set.
In such cases, the per phase equivalent circuit is the positive sequence network.
Changing a positive sequence network to a negative sequence only involves changing the impedances of rotating machines
Sequence Networks
Consider the one line diagram shown below:
Sequence Networks
Positive sequence network:
Xg-1
XT1-1 Xline-1 XT2-1
Xm1-1 Xm2-1
Sequence Networks
Negative sequence network:
Xg-2
XT1-2 Xline-2 XT2-2
Xm1-1 Xm2-2 Xm1-2
Sequence Networks
Zero sequence network:
XT1-0 Xline-0 XT2-0
Xm2-0
Xm1-0
3Xn1-0
Xg-0
3Xgn-0
Sequence Networks - Example
A 300 MVA 20 kV three phase generator has a subtransient reactance of 20%. The generator supplies 2 synchronous motors over a of 64 km long transmission line having two transformers at both ends. The motors are rated at 13.2 kV. The neutral of motor M1 is grounded through a reactance of 0.4 ฮฉ. M2 is not grounded. Rated input for M1 is 200 MVA and M2 is 100 MVA. For both motors ๐๐
โ = 20%.
Sequence Networks - Example
Transformer T1 is rated at 350 MVA, 230/20 kV with a leakage of 10%. Transformer T2 is rated at 300 MVA, 220/13.2 kV with a leakage of 10%. Series reactance of the transmission line is 0.5 ฮฉ/km. Draw the positive sequence network. Use the generator values as base values.
Sequence Networks - Example
Sequence Networks - Example
Generator ๐๐โ = j0.2
Transformer ๐1 ๐1 = j0.0857
Transmission line ๐๐๐๐๐ = j0.182
Transformer ๐2 ๐2 = j0.0915
Motor M1 ๐๐โ = j0.274
Motor M2 ๐๐โ = j0.549
Sequence Networks - Example
Positive sequence network
Sequence Networks - Example
Draw the negative sequence network for the system. Assume the negative sequence reactance of each machine is equal to its subtransient reactance.
Sequence Networks - Example
All negative sequence reactances are equal to positive sequence reactances.
Sequence Networks - Example
Draw the zero sequence network. Assume the zero sequence reactance of each machine (generator and motors) is equal to 0.05 per unit. A current limiting reactor of 0.4 ฮฉ is in each of the neutrals of the generator and M1 . The zero sequence reactance of the transmission line is 1.5 ฮฉ/km.
Sequence Networks - Example
Generator ๐๐โ = j0.05
Transformer ๐1 ๐0 = j0.0857 (๐๐ = 0)
Transmission line ๐๐๐๐๐ = j0.545
Transformer ๐2 ๐2 = j0.0915 ((๐๐ = 0)
Motor M1 ๐๐โ = j0.0686
Motor M2 ๐๐โ = j0.1372
Generator 3๐๐ = j0.902
Motor M1 3๐๐ = j1.89
Sequence Networks - Example
Zero sequence network