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Symmetrical Components and Sequence Networks
Chapter 11
Balanced sequences
Synthesis equations
Synthesis equations
Analysis equations
𝑉𝑎0 = 1
3 (𝑉𝑎+ 𝑉𝑏 + 𝑉𝑐)
𝑉𝑎1 = 1
3 (𝑉𝑎+𝑎 𝑉𝑏+𝑎2𝑉𝑐)
𝑉𝑎2 = 1
3 (𝑉𝑎+𝑎2 𝑉𝑏+𝑎𝑉𝑐)
Example: Balanced line to neutral voltages with positive sequence
Calculate the sequence components of the following line to neutral voltages with abc sequence
𝑉𝑎𝑛 = 277∠00 V
𝑉𝑏𝑛 = 277∠−1200 V
𝑉𝑐𝑛 = 277∠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
𝐼𝑎 = 10∠00 A
𝐼𝑏 = 10∠1200 A
𝐼𝑐 = 10∠−1200 A
Solution
𝐼𝑎0 = 0 A
𝐼𝑎1 = 0 A
𝐼𝑎2 = 10∠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 Δ circuits
Only line voltages are applicable
Line currents and phase currents are applicable
Balanced Δ circuits
𝐼𝑎 = 𝐼𝑎𝑏 − 𝐼𝑐𝑎
𝐼𝑏 = 𝐼𝑏𝑐 − 𝐼𝑎𝑏
𝐼𝑐 = 𝐼𝑐𝑎 − 𝐼𝑏𝑐
For positive phase sequence
𝐼𝑎1 = 3∠(−300 ) 𝐼𝑎𝑏1
The magnitude of the line current is 3 the phase current and lags by 300
Balanced Δ circuits
For negative phase sequence
𝐼𝑎2 = 3∠(+300 ) 𝐼𝑎𝑏2
The magnitude of the line current is 3 the phase current and leads by 300
Phasors for balanced Δ circuits
Balanced Y circuits
Only line currents are applicable
Line voltages and phase voltages are applicable
Balanced Y circuits
𝑉𝑎𝑏 = 𝑉𝑎𝑛 − 𝑉𝑏𝑛
𝑉𝑏𝑐 = 𝑉𝑏𝑛 − 𝑉𝑐𝑛
𝑉𝑐𝑎 = 𝑉𝑐𝑛 − 𝑉𝑎𝑛
For positive phase sequence
𝑉𝑎𝑏1 = 3∠(+300 ) 𝑉𝑎𝑛1
The magnitude of the line voltage is 3 the phase voltage and leads by 300
Balanced Y circuits
For negative phase sequence
𝑉𝑎𝑏2 = 3∠(−300 ) 𝑉𝑎𝑛2
The magnitude of the line voltage is 3 the phase voltage and lags by 300
Phasors for balanced Y circuits
Equivalent Y and Δ loads
Given a Δ connected load of impedance 𝑍Δ per phase, it can be shown that the equivalent Y connected load will have an impedance of 𝑍𝑌 = 𝑍Δ
3 .
Equivalent Y and Δ 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
𝑆3Φ = 𝑉𝑎𝐼𝑎∗+𝑉𝑏𝐼𝑏
∗+𝑉𝑐𝐼𝑐∗
= 3𝑉𝑎0𝐼𝑎0∗+ 3𝑉𝑎1𝐼𝑎1
∗+ 3𝑉𝑎2𝐼𝑎2∗
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
𝑆3Φ = 𝑉𝑎𝐼𝑎∗+𝑉𝑏𝐼𝑏
∗+𝑉𝑐𝐼𝑐∗
= 3𝑉𝑎0𝐼𝑎0∗+ 3𝑉𝑎1𝐼𝑎1
∗+ 3𝑉𝑎2𝐼𝑎2∗
When working in per unit the factor of 3 falls away. Therefore
𝑆3Φ =𝑉𝑎0𝐼𝑎0∗+ 𝑉𝑎1𝐼𝑎1
∗+ 𝑉𝑎2𝐼𝑎2∗
= 0 + (0.9857∠43.60)(0.9857∠-43.60) +(0.2346∠250.30)(0.2346∠-250.30)
= 1.023∠00
Actual
= 1.023∠00(500)
= 513.320KW
Example 11.3 solution
Verifying the answer
𝐼𝑎1 = 0.9857∠43.60
𝐼𝑏1 = 0.9857∠−76.40
𝐼𝑐1 = 0.9857∠163.60
𝐼𝑎2 = 0.2346∠250.30
𝐼𝑏2 = 0.2346∠370.30
𝐼𝑐2 = 0.2346∠130.30
𝐼𝑎0 = 𝐼𝑏0= 𝐼𝑐0=0
Example 11.3 solution
𝐼𝑎 = 𝐼𝑎0+ 𝐼𝑎1+ 𝐼𝑎2 = 0.7832∠35.870 𝐼𝑏 = 𝐼𝑏0+ 𝐼𝑏1+ 𝐼𝑏2 = 1.026∠−63.20 𝐼𝑐 = 𝐼𝑐0+ 𝐼𝑐1+ 𝐼𝑐2 = 1.189∠157.40
Base current Ib = 500000
3 (2300) = 125.5 A
Actual currents 𝐼𝑎= 98.3∠35.870 A 𝐼𝑏= 128.8∠−63.20 A 𝐼𝑐= 149.2∠157.40 A
Example 11.3 solution
Power = 𝐼𝑎2𝑅𝑌+𝐼𝑏
2𝑅𝑌+𝐼𝑐2𝑅𝑌
= 98.32(10.58)+128.82(10.58)+149.22(10.58)
= 513 kW
Sequence circuits of Y and Δ impedance loads
Sequence circuits of Y and Δ impedance loads
Neutral current
𝐼𝑛 = 𝐼𝑎+ 𝐼𝑏+ 𝐼𝑐
= (𝐼𝑎0+ 𝐼𝑎1+ 𝐼𝑎2) + (𝐼𝑏0+ 𝐼𝑏1+ 𝐼𝑏2)+(𝐼𝑐0+ 𝐼𝑐1+ 𝐼𝑐2)
= (𝐼𝑎0+ 𝐼𝑏0+ 𝐼𝑐0) + (𝐼𝑎1+ 𝐼𝑏1+ 𝐼𝑐1)+(𝐼𝑎2+ 𝐼𝑏2+ 𝐼𝑐2)
= (𝐼𝑎0+ 𝐼𝑎0+ 𝐼𝑎0) + 0 +0
= 3𝐼𝑎0
The neutral current consists only of the zero sequence current
Sequence circuits of Y and Δ impedance loads
The volt drop across 𝑍𝑛 is 𝑉𝑍𝑛 = 3𝐼𝑎0𝑍𝑛 This means that the voltages to neutral (𝑉𝑎𝑛, 𝑉𝑏𝑛, 𝑉𝑐𝑛) and the voltages to ground (𝑉𝑎, 𝑉𝑏 , 𝑉𝑐) are different under balanced conditions 𝑉𝑎 = 𝑉𝑎𝑛 + 𝑉𝑍𝑛 = 𝑍𝑌𝐼𝑎 + 3𝐼𝑎0𝑍𝑛 𝑉𝑏= 𝑉𝑏𝑛 + 𝑉𝑍𝑛 = 𝑍𝑌𝐼𝑏 + 3𝐼𝑎0𝑍𝑛 𝑉𝑐 = 𝑉𝑐𝑛 + 𝑉𝑍𝑛 = 𝑍𝑌𝐼𝑐 + 3𝐼𝑎0𝑍𝑛
Sequence circuits of Y and Δ impedance loads
The previous set of equations can be written in matrix form as
A
𝑉𝒂𝟎
𝑉𝒂𝟏
𝑉𝒂𝟐
= 𝑍𝑌 A
𝐼𝒂𝟎
𝐼𝒂𝟏
𝐼𝒂𝟐
+ 3𝐼𝑎0𝑍𝑛 111
Where A =𝟏 𝟏 𝟏𝟏 𝒂𝟐 𝒂𝟏 𝒂 𝒂𝟐
Sequence circuits of Y and Δ impedance loads
Multiplying throughout by 𝑨−𝟏 gives
𝑉𝒂𝟎
𝑉𝒂𝟏
𝑉𝒂𝟐
= 𝑍𝑌
𝐼𝒂𝟎
𝐼𝒂𝟏
𝐼𝒂𝟐
+ 3𝐼𝑎0𝑍𝑛 𝑨−𝟏 111
Which reduces to
𝑉𝒂𝟎
𝑉𝒂𝟏
𝑉𝒂𝟐
= 𝑍𝑌
𝐼𝒂𝟎
𝐼𝒂𝟏
𝐼𝒂𝟐
+ 3𝐼𝑎0𝑍𝑛 100
Sequence circuits of Y and Δ impedance loads
Writing as previous equations as separate equations Multiplying throughout by 𝑨−𝟏 gives
𝑉𝑎0 = (𝑍𝑌 + 3𝑍𝑛) 𝐼𝑎0 = 𝑍0 𝐼𝑎0
𝑉𝑎1 = 𝑍𝑌𝐼𝑎1 = 𝑍1 𝐼𝑎1
𝑉𝑎2 = 𝑍𝑌𝐼𝑎2 = 𝑍2 𝐼𝑎2
Where
𝑍0 is the impedance to zero sequence current
𝑍1 is the impedance to positive sequence current
𝑍2 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
𝑍0 is the zero sequence impedance
𝑍1 is the zero sequence impedance
𝑍2 is the negative sequence impedance
Sequence circuits of Y impedance loads
𝑍𝑛 can assume the following values:
0 (short circuit – solidly bolted, solidly grounded)
Some positive value
∞
Solidly grounded neutral Open circuited neutral
Sequence circuits of Δ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
𝒁Δ
𝟑
𝒁Δ
𝟑
Sequence circuits of Δ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
𝒁Δ
𝟑
𝒁Δ
𝟑 𝒁Δ
𝟑
Example
Three equal impedances of j30 Ω are connected in Δ. Determine the sequence impedances and draw the sequence networks. Repeat the problem for the case where a mutual impedance of j5 Ω 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
𝑍𝑎𝑎 - self impedance of each phase conductor 𝑍𝑛𝑛 - self impedance of the neutral conductor 𝑍𝑎𝑏 - mutual impedance between phase conductors 𝑍𝑎𝑛 - 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:
𝑍𝑠 = 𝑍𝑎𝑎+ 𝑍𝑛𝑛 - 2 𝑍𝑎𝑛 (self impedance)
𝑍𝑚 = 𝑍𝑎𝑏+ 𝑍𝑛𝑛 - 2 𝑍𝑎𝑛 (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:
𝑉𝑎𝑎′
𝑉𝑏𝑏′
𝑉𝑐𝑐′
=
𝑉𝑎𝑛 − 𝑉𝑎′𝑛′
𝑉𝑏𝑛 − 𝑉𝑏′𝑛′
𝑉𝑐𝑛 − 𝑉𝑐′𝑛′
=
𝑍𝑠 𝑍𝑚 𝑍𝑚
𝑍𝑚 𝑍𝑠 𝑍𝑚
𝑍𝑚 𝑍𝑚 𝑍𝑠
𝐼𝑎
𝐼𝑏
𝐼𝑐
Sequence circuits of a transmission line
The sequence impedances of the transmission line are defined as:
𝑍0 = 𝑍𝑠+ 2𝑍𝑚 = 𝑍𝑎𝑎+ 2 𝑍𝑎𝑏 + 3 𝑍𝑛𝑛 - 6 𝑍𝑎𝑛
𝑍1 = 𝑍𝑠- 𝑍𝑚 = 𝑍𝑎𝑎- 𝑍𝑎𝑏
𝑍2 = 𝑍𝑠- 𝑍𝑚 = 𝑍𝑎𝑎- 𝑍𝑎𝑏
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
𝑉𝑎𝑛=182+j70 kV 𝑉𝑎′𝑛′=154+j28 kV
𝑉𝑏𝑛=72.24-j32.62 kV 𝑉𝑏′𝑛′=44.24-j74.62 kV
𝑉𝑐𝑛=-170.24+j88.62 kV 𝑉𝑐′𝑛′=-198.24+j46.62 kV
The line impedances are
𝑍𝑎𝑎=j60 Ω 𝑍𝑎𝑏=j20 Ω 𝑍𝑛𝑛=j80 Ω 𝑍𝑎𝑛=0 Ω
Determine the line currents 𝐼𝑎, 𝐼𝑏 and 𝐼𝑐
Solution
𝑍𝑠 = 𝑍𝑎𝑎+ 𝑍𝑛𝑛 - 2 𝑍𝑎𝑛 = j60 + j80 – j60 = j80 Ω
𝑍𝑚 = 𝑍𝑎𝑏+ 𝑍𝑛𝑛 - 2 𝑍𝑎𝑛 = j20 + j80 – j60 = j40 Ω
Then using
𝑉𝑎𝑎′
𝑉𝑏𝑏′
𝑉𝑐𝑐′
=
𝑉𝑎𝑛 − 𝑉𝑎′𝑛′
𝑉𝑏𝑛 − 𝑉𝑏′𝑛′
𝑉𝑐𝑛 − 𝑉𝑐′𝑛′
=
𝑍𝑠 𝑍𝑚 𝑍𝑚
𝑍𝑚 𝑍𝑠 𝑍𝑚
𝑍𝑚 𝑍𝑚 𝑍𝑠
𝐼𝑎
𝐼𝑏
𝐼𝑐
Solution
𝑉𝑎𝑎′
𝑉𝑏𝑏′
𝑉𝑐𝑐′
=
𝑉𝑎𝑛 − 𝑉𝑎′𝑛′
𝑉𝑏𝑛 − 𝑉𝑏′𝑛′
𝑉𝑐𝑛 − 𝑉𝑐′𝑛′
=
𝑍𝑠 𝑍𝑚 𝑍𝑚
𝑍𝑚 𝑍𝑠 𝑍𝑚
𝑍𝑚 𝑍𝑚 𝑍𝑠
𝐼𝑎
𝐼𝑏
𝐼𝑐
28 + 𝑗4228 + 𝑗4228 + 𝑗42
x103 =
𝑗80 𝑗40 𝑗40𝑗40 𝑗80 𝑗40𝑗40 𝑗40 𝑗80
𝐼𝑎
𝐼𝑏
𝐼𝑐
Solution
𝐼𝑎
𝐼𝑏
𝐼𝑐
=
𝑗80 𝑗40 𝑗40𝑗40 𝑗80 𝑗40𝑗40 𝑗40 𝑗80
28 + 𝑗4228 + 𝑗4228 + 𝑗42
x103
=
262.5 − 𝑗175262.5 − 𝑗175262.5 − 𝑗175
A
-1
Solution
Using Symmetrical component
The sequence components of the line volt drops are
𝑉𝑎𝑎′0
𝑉𝑎𝑎′1
𝑉𝑎𝑎′2
= 1
3
1 1 11 𝑎 𝑎2
1 𝑎2 𝑎
𝑉𝑎𝑎′
𝑉𝑏𝑏′
𝑉𝑐𝑐′
Solution
𝑉𝑎𝑎′0
𝑉𝑎𝑎′1
𝑉𝑎𝑎′2
= 1
3
1 1 11 𝑎 𝑎2
1 𝑎2 𝑎
28 + 𝑗4228 + 𝑗4228 + 𝑗42
x103
= 28 + 𝑗42
00
x103 V
Solution
𝑍0 = 𝑍𝑎𝑎+ 2 𝑍𝑎𝑏 + 3 𝑍𝑛𝑛 - 6 𝑍𝑎𝑛 = j60 + j40 + j240 – j180 = j160 𝑍1 = 𝑍2 = 𝑍𝑎𝑎- 𝑍𝑎𝑏 = j60 – j20 = j40 So 𝑉𝑎𝑎′
0 = 𝑍0 𝐼𝑎0
𝑉𝑎𝑎′1 = 𝑍1 𝐼𝑎
1 𝑉𝑎𝑎′
2 = 𝑍2 𝐼𝑎2
Solution
(28 + 𝑗42) 103 = (j160) 𝐼𝑎0 𝐼𝑎
0 = 262.5 – j175 A 0 = j40 𝐼𝑎1 𝐼𝑎
1 = 0 0 = j40 𝐼𝑎2 𝐼𝑎
2 = 0 Therefore 𝐼𝑎 = 𝐼𝑏 = 𝐼c = 262.5 – j175 A
Sequence circuits of a synchronous machine
Sequence circuits of a synchronous machine
The sequence equations are:
𝐿𝑠 and 𝑀𝑠 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 𝐸𝑎𝑛= 1.0∠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
𝑉𝑎= 0
𝑉𝑏= 1.0.13∠−102.250
𝑉𝑐= 1.0.13∠−102.250
Determine the subtransient current in the generator and the line-to-line voltages for subtransient conditions due to the fault.
Sequence circuits of Y – Δ transformers
5 Cases will be considered
Case 1: Y-Y bank – both neutrals grounded
Case 2: Y-Y bank – one neutral grounded
Case 3: Δ-Δ bank
Case 4: Y-Δ bank – Y grounded
Case 5: Y-Δ bank – Y ungrounded
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded Sequence equations: Positive sequence Negative sequence Zero sequence
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Zero Sequence equivalent circuit:
𝑍0 = 3𝑍𝑁 + 3 𝑍𝑛
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
𝑍0 = Z + 3𝑍𝑁 + 3 𝑍𝑛
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
𝑍0 = Z + 3𝑍𝑁 + 3 𝑍𝑛
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Positive Sequence equivalent circuit with leakage impedance Z:
𝑍1 = Z
𝑍1
Sequence circuits of Y – Δ transformers
Case 1: Y-Y bank – both neutrals grounded
Negative Sequence equivalent circuit with leakage impedance Z:
𝑍2 = Z
𝑍2
Sequence circuits of Y – Δ transformers
Case 2 Y-Y bank – one neutral grounded
Sequence circuits of Y – Δ transformers
Case 2 Y-Y bank – one neutral grounded
Zero Sequence equivalent circuit
𝐼0 cannot flow due to the absence of a path for current flow between the windings
Sequence circuits of Y – Δ transformers
Case 3: Δ-Δ bank
Sequence circuits of Y – Δ transformers
Case 3: Δ-Δ bank Sequence equations:
VAB0 = Vab
0 = 0 Zero sequence
VAB1 =
𝑁1
𝑁2Vab
1 Positive sequence
VAB1 =
𝑁1
𝑁2Vab
1 Negative sequence
Sequence circuits of Y – Δ transformers
Case 3: Δ-Δ bank
Zero Sequence equivalent circuit with leakage impedance Z:
Sequence circuits of Y – Δ transformers
Case 3: Y-Δ bank – Y grounded
Sequence circuits of Y – Δ transformers
Case 3: Y-Δ bank – Y grounded Sequence equations:
VA0 - 3 𝑍𝑁 IA
0 = 𝑁1𝑁2
Vab0 = 0 Zero sequence
VA1 =
𝑁1
𝑁2Vab
1 Positive sequence
VA2 =
𝑁1
𝑁2Vab
2 Negative sequence
Sequence circuits of Y – Δ transformers
Case 4: Y-Δ bank – Y grounded Zero Sequence equivalent circuit with leakage impedance Z:
𝑍0 = Z + 3𝑍𝑁
Sequence circuits of Y – Δ transformers
Case 5: Y-Δ bank- Y ungrounded
Sequence circuits of Y – Δ transformers
Case 5: Y-Δ bank – Y ungrounded Sequence equations:
VAB0 =
𝑁1𝑁2
Vab0 = 0 Zero sequence
VAB1 =
𝑁1
𝑁2Vab
1 Positive sequence
VAB2 =
𝑁1
𝑁2Vab
2 Negative sequence
Sequence circuits of Y – Δ transformers
Case 5: Y-Δ bank – Y ungrounded Zero Sequence equivalent circuit with leakage impedance Z:
Y – Δ transformers
VAB1 = 3
𝑁1
𝑁2Vab
1 ∠300
VAB2 = 3
𝑁1
𝑁2Vab
2 ∠−300
In per unit
VAB1 = Vab
1 ∠300
VAB2 = Vab
2 ∠−300
The same applies for currents in the per unit case
IA1 = Ia
1 ∠300
IA2 = Ia
2 ∠−300
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
𝑉𝑎𝑏 = 1840 V 𝑉𝑏𝑐 = 2760 V 𝑉𝑐𝑎 = 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-Δ 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 𝑉𝑎𝑏 = 0.8∠82.80 𝑉𝑏𝑐 = 1.2∠−41.40 𝑉𝑐𝑎 = 1.0∠1800 Vab
1 = 0.9857∠73.60 Vab
2 = 0.2346∠220.30 Van
1 = 0.9857∠43.60 Van
2 = 0.2346∠250.30 Ia
1 = 0.9857∠43.60 Ia
2 = 0.2346∠250.30
Solution
The calculated load voltages will be the voltages on the low voltage side of the transformer Vab
1 = 0.9857∠73.60 Vab
2 = 0.2346∠220.30 Therefore high voltage side voltages in per unit will be VAB
1 = Vab1 ∠300
VAB2 = Vab
2 ∠−300 So VAB
1 = 0.9857 ∠103.60 VAB
2 = 0.2346 ∠190.30 Next calculate the sequence components of the other lines VBC
1 = 0.9857 ∠(103.6−120)0 = 0.9857 ∠−16.40 VCA
1 = 0.9857 ∠(103.6+120)0 = 0.9857 ∠223.60 VBC
2 = 0.2346 ∠(190.3+120)0 = 0.2346 ∠310.30 VCA
2 = 0.2346 ∠(190.3−120)0 = 0.2346∠70.30
Solution
The line-line voltages on the high voltage side are VAB = VAB
0 + VAB1 + VAB
2 = 0 + 0.9857 ∠103.60 + 0.2346 ∠190.30 = 1.026 ∠116.80 VBC = VBC
0 + VBC1 + VBC
2 = 0 + 0.9857 ∠−16.40 + 0.2346 ∠310.30 = 1.19 ∠−22.60 VCA = VCA
0 + VCA1 + VCA
2 = 0 + 0.9857 ∠223.60 + 0.2346 ∠70.30 = 0.783 ∠−144.10
Solution
The calculated load currents will be the line currents on the low voltage side of the transformer Ia
1 = 0.9857∠43.60 Ia
2 = 0.2346∠250.30 Therefore high voltage side voltages in per unit will be IA
1=Ia1∠300
IA2 =Ia
2∠−300
So IA
1= 0.9857 ∠73.60 IA
2= 0.2346 ∠220.30 Next calculate the sequence components of the other lines IB
1 = 0.9857 ∠(73.6−120)0 = 0.9857 ∠−46.40 IC
1 = 0.9857 ∠(73.6+120)0 = 0.9857 ∠193.60 IB
2 = 0.2346 ∠(220.3+120)0 = 0.2346 ∠340.30 IC
2 = 0.2346 ∠(220.3−120)0 = 0.2346∠100.30
Solution
The line currents on the high voltage side are IA = IA
0 + IA1 + IA
2 = 0 + 0.9857 ∠73.60 + 0.2346 ∠220.30 = 0.8 ∠82.90 IB = IB
0 + IB1 + IB
2 = 0 + 0.9857 ∠−46.40 + 0.2346 ∠340.30 = 1.2 ∠−41.40 IC = IC
0 + IC1 + IC
2 = 0 + 0.9857 ∠193.60 + 0.2346 ∠100.30 = 1.0 ∠179.90
Sequence Networks
Sequence circuits have been developed for the following components: Loads (Y and Δ) 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 Ω. M2 is not grounded. Rated input for M1 is 200 MVA and M2 is 100 MVA. For both motors 𝑋𝑑
” = 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 Ω/km. Draw the positive sequence network. Use the generator values as base values.
Sequence Networks - Example
Sequence Networks - Example
Generator 𝑋𝑑” = j0.2
Transformer 𝑇1 𝑋1 = j0.0857
Transmission line 𝑋𝑙𝑖𝑛𝑒 = j0.182
Transformer 𝑇2 𝑋2 = j0.0915
Motor M1 𝑋𝑑” = j0.274
Motor M2 𝑋𝑑” = 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 Ω is in each of the neutrals of the generator and M1 . The zero sequence reactance of the transmission line is 1.5 Ω/km.
Sequence Networks - Example
Generator 𝑋𝑑” = j0.05
Transformer 𝑇1 𝑋0 = j0.0857 (𝑍𝑛 = 0)
Transmission line 𝑋𝑙𝑖𝑛𝑒 = j0.545
Transformer 𝑇2 𝑋2 = j0.0915 ((𝑍𝑛 = 0)
Motor M1 𝑋𝑑” = j0.0686
Motor M2 𝑋𝑑” = j0.1372
Generator 3𝑍𝑛 = j0.902
Motor M1 3𝑍𝑛 = j1.89
Sequence Networks - Example
Zero sequence network