Chapter 11b
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Transcript of Chapter 11b
Symmetrical Components and Sequence Networks
Chapter 11
Balanced sequences
Synthesis equations
Synthesis equations
Analysis equations
ππ0 = 1
3 (ππ+ ππ + ππ)
ππ1 = 1
3 (ππ+π ππ+π2ππ)
ππ2 = 1
3 (ππ+π2 ππ+πππ)
Example: Balanced line to neutral voltages with positive sequence
Calculate the sequence components of the following line to neutral voltages with abc sequence
πππ = 277β 00 V
πππ = 277β β1200 V
πππ = 277β 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
πΌπ = 10β 00 A
πΌπ = 10β 1200 A
πΌπ = 10β β1200 A
Solution
πΌπ0 = 0 A
πΌπ1 = 0 A
πΌπ2 = 10β 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 Ξ circuits
Only line voltages are applicable
Line currents and phase currents are applicable
Balanced Ξ circuits
πΌπ = πΌππ β πΌππ
πΌπ = πΌππ β πΌππ
πΌπ = πΌππ β πΌππ
For positive phase sequence
πΌπ1 = 3β (β300 ) πΌππ1
The magnitude of the line current is 3 the phase current and lags by 300
Balanced Ξ circuits
For negative phase sequence
πΌπ2 = 3β (+300 ) πΌππ2
The magnitude of the line current is 3 the phase current and leads by 300
Phasors for balanced Ξ circuits
Balanced Y circuits
Only line currents are applicable
Line voltages and phase voltages are applicable
Balanced Y circuits
πππ = πππ β πππ
πππ = πππ β πππ
πππ = πππ β πππ
For positive phase sequence
πππ1 = 3β (+300 ) πππ1
The magnitude of the line voltage is 3 the phase voltage and leads by 300
Balanced Y circuits
For negative phase sequence
πππ2 = 3β (β300 ) πππ2
The magnitude of the line voltage is 3 the phase voltage and lags by 300
Phasors for balanced Y circuits
Equivalent Y and Ξ loads
Given a Ξ connected load of impedance πΞ per phase, it can be shown that the equivalent Y connected load will have an impedance of ππ = πΞ
3 .
Equivalent Y and Ξ 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
π3Ξ¦ = πππΌπβ+πππΌπ
β+πππΌπβ
= 3ππ0πΌπ0β+ 3ππ1πΌπ1
β+ 3ππ2πΌπ2β
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
π3Ξ¦ = πππΌπβ+πππΌπ
β+πππΌπβ
= 3ππ0πΌπ0β+ 3ππ1πΌπ1
β+ 3ππ2πΌπ2β
When working in per unit the factor of 3 falls away. Therefore
π3Ξ¦ =ππ0πΌπ0β+ ππ1πΌπ1
β+ ππ2πΌπ2β
= 0 + (0.9857β 43.60)(0.9857β -43.60) +(0.2346β 250.30)(0.2346β -250.30)
= 1.023β 00
Actual
= 1.023β 00(500)
= 513.320KW
Example 11.3 solution
Verifying the answer
πΌπ1 = 0.9857β 43.60
πΌπ1 = 0.9857β β76.40
πΌπ1 = 0.9857β 163.60
πΌπ2 = 0.2346β 250.30
πΌπ2 = 0.2346β 370.30
πΌπ2 = 0.2346β 130.30
πΌπ0 = πΌπ0= πΌπ0=0
Example 11.3 solution
πΌπ = πΌπ0+ πΌπ1+ πΌπ2 = 0.7832β 35.870 πΌπ = πΌπ0+ πΌπ1+ πΌπ2 = 1.026β β63.20 πΌπ = πΌπ0+ πΌπ1+ πΌπ2 = 1.189β 157.40
Base current Ib = 500000
3 (2300) = 125.5 A
Actual currents πΌπ= 98.3β 35.870 A πΌπ= 128.8β β63.20 A πΌπ= 149.2β 157.40 A
Example 11.3 solution
Power = πΌπ2π π+πΌπ
2π π+πΌπ2π π
= 98.32(10.58)+128.82(10.58)+149.22(10.58)
= 513 kW
Sequence circuits of Y and Ξ impedance loads
Sequence circuits of Y and Ξ impedance loads
Neutral current
πΌπ = πΌπ+ πΌπ+ πΌπ
= (πΌπ0+ πΌπ1+ πΌπ2) + (πΌπ0+ πΌπ1+ πΌπ2)+(πΌπ0+ πΌπ1+ πΌπ2)
= (πΌπ0+ πΌπ0+ πΌπ0) + (πΌπ1+ πΌπ1+ πΌπ1)+(πΌπ2+ πΌπ2+ πΌπ2)
= (πΌπ0+ πΌπ0+ πΌπ0) + 0 +0
= 3πΌπ0
The neutral current consists only of the zero sequence current
Sequence circuits of Y and Ξ impedance loads
The volt drop across ππ is πππ = 3πΌπ0ππ This means that the voltages to neutral (πππ, πππ, πππ) and the voltages to ground (ππ, ππ , ππ) are different under balanced conditions ππ = πππ + πππ = πππΌπ + 3πΌπ0ππ ππ= πππ + πππ = πππΌπ + 3πΌπ0ππ ππ = πππ + πππ = πππΌπ + 3πΌπ0ππ
Sequence circuits of Y and Ξ impedance loads
The previous set of equations can be written in matrix form as
A
πππ
πππ
πππ
= ππ A
πΌππ
πΌππ
πΌππ
+ 3πΌπ0ππ 111
Where A =π π ππ ππ ππ π ππ
Sequence circuits of Y and Ξ impedance loads
Multiplying throughout by π¨βπ gives
πππ
πππ
πππ
= ππ
πΌππ
πΌππ
πΌππ
+ 3πΌπ0ππ π¨βπ 111
Which reduces to
πππ
πππ
πππ
= ππ
πΌππ
πΌππ
πΌππ
+ 3πΌπ0ππ 100
Sequence circuits of Y and Ξ impedance loads
Writing as previous equations as separate equations Multiplying throughout by π¨βπ gives
ππ0 = (ππ + 3ππ) πΌπ0 = π0 πΌπ0
ππ1 = πππΌπ1 = π1 πΌπ1
ππ2 = πππΌπ2 = π2 πΌπ2
Where
π0 is the impedance to zero sequence current
π1 is the impedance to positive sequence current
π2 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
π0 is the zero sequence impedance
π1 is the zero sequence impedance
π2 is the negative sequence impedance
Sequence circuits of Y impedance loads
ππ can assume the following values:
0 (short circuit β solidly bolted, solidly grounded)
Some positive value
β
Solidly grounded neutral Open circuited neutral
Sequence circuits of Ξ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
πΞ
π
πΞ
π
Sequence circuits of Ξ impedance loads
Zero sequence network
Positive sequence network
Negative sequence network
πΞ
π
πΞ
π πΞ
π
Example
Three equal impedances of j30 Ξ© are connected in Ξ. Determine the sequence impedances and draw the sequence networks. Repeat the problem for the case where a mutual impedance of j5 Ξ© 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
πππ - self impedance of each phase conductor πππ - self impedance of the neutral conductor πππ - mutual impedance between phase conductors πππ - 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:
ππ = πππ+ πππ - 2 πππ (self impedance)
ππ = πππ+ πππ - 2 πππ (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:
πππβ²
πππβ²
πππβ²
=
πππ β ππβ²πβ²
πππ β ππβ²πβ²
πππ β ππβ²πβ²
=
ππ ππ ππ
ππ ππ ππ
ππ ππ ππ
πΌπ
πΌπ
πΌπ
Sequence circuits of a transmission line
The sequence impedances of the transmission line are defined as:
π0 = ππ + 2ππ = πππ+ 2 πππ + 3 πππ - 6 πππ
π1 = ππ - ππ = πππ- πππ
π2 = ππ - ππ = πππ- πππ
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
πππ=182+j70 kV ππβ²πβ²=154+j28 kV
πππ=72.24-j32.62 kV ππβ²πβ²=44.24-j74.62 kV
πππ=-170.24+j88.62 kV ππβ²πβ²=-198.24+j46.62 kV
The line impedances are
πππ=j60 Ξ© πππ=j20 Ξ© πππ=j80 Ξ© πππ=0 Ξ©
Determine the line currents πΌπ, πΌπ and πΌπ
Solution
ππ = πππ+ πππ - 2 πππ = j60 + j80 β j60 = j80 Ξ©
ππ = πππ+ πππ - 2 πππ = j20 + j80 β j60 = j40 Ξ©
Then using
πππβ²
πππβ²
πππβ²
=
πππ β ππβ²πβ²
πππ β ππβ²πβ²
πππ β ππβ²πβ²
=
ππ ππ ππ
ππ ππ ππ
ππ ππ ππ
πΌπ
πΌπ
πΌπ
Solution
πππβ²
πππβ²
πππβ²
=
πππ β ππβ²πβ²
πππ β ππβ²πβ²
πππ β ππβ²πβ²
=
ππ ππ ππ
ππ ππ ππ
ππ ππ ππ
πΌπ
πΌπ
πΌπ
28 + π4228 + π4228 + π42
x103 =
π80 π40 π40π40 π80 π40π40 π40 π80
πΌπ
πΌπ
πΌπ
Solution
πΌπ
πΌπ
πΌπ
=
π80 π40 π40π40 π80 π40π40 π40 π80
28 + π4228 + π4228 + π42
x103
=
262.5 β π175262.5 β π175262.5 β π175
A
-1
Solution
Using Symmetrical component
The sequence components of the line volt drops are
πππβ²0
πππβ²1
πππβ²2
= 1
3
1 1 11 π π2
1 π2 π
πππβ²
πππβ²
πππβ²
Solution
πππβ²0
πππβ²1
πππβ²2
= 1
3
1 1 11 π π2
1 π2 π
28 + π4228 + π4228 + π42
x103
= 28 + π42
00
x103 V
Solution
π0 = πππ+ 2 πππ + 3 πππ - 6 πππ = j60 + j40 + j240 β j180 = j160 π1 = π2 = πππ- πππ = j60 β j20 = j40 So πππβ²
0 = π0 πΌπ0
πππβ²1 = π1 πΌπ
1 πππβ²
2 = π2 πΌπ2
Solution
(28 + π42) 103 = (j160) πΌπ0 πΌπ
0 = 262.5 β j175 A 0 = j40 πΌπ1 πΌπ
1 = 0 0 = j40 πΌπ2 πΌπ
2 = 0 Therefore πΌπ = πΌπ = πΌc = 262.5 β j175 A
Sequence circuits of a synchronous machine
Sequence circuits of a synchronous machine
The sequence equations are:
πΏπ and ππ 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 πΈππ= 1.0β 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
ππ= 0
ππ= 1.0.13β β102.250
ππ= 1.0.13β β102.250
Determine the subtransient current in the generator and the line-to-line voltages for subtransient conditions due to the fault.
Sequence circuits of Y β Ξ transformers
5 Cases will be considered
Case 1: Y-Y bank β both neutrals grounded
Case 2: Y-Y bank β one neutral grounded
Case 3: Ξ-Ξ bank
Case 4: Y-Ξ bank β Y grounded
Case 5: Y-Ξ bank β Y ungrounded
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded Sequence equations: Positive sequence Negative sequence Zero sequence
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Zero Sequence equivalent circuit:
π0 = 3ππ + 3 ππ
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
π0 = Z + 3ππ + 3 ππ
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Zero Sequence equivalent circuit with leakage impedance Z:
π0 = Z + 3ππ + 3 ππ
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Positive Sequence equivalent circuit with leakage impedance Z:
π1 = Z
π1
Sequence circuits of Y β Ξ transformers
Case 1: Y-Y bank β both neutrals grounded
Negative Sequence equivalent circuit with leakage impedance Z:
π2 = Z
π2
Sequence circuits of Y β Ξ transformers
Case 2 Y-Y bank β one neutral grounded
Sequence circuits of Y β Ξ transformers
Case 2 Y-Y bank β one neutral grounded
Zero Sequence equivalent circuit
πΌ0 cannot flow due to the absence of a path for current flow between the windings
Sequence circuits of Y β Ξ transformers
Case 3: Ξ-Ξ bank
Sequence circuits of Y β Ξ transformers
Case 3: Ξ-Ξ bank Sequence equations:
VAB0 = Vab
0 = 0 Zero sequence
VAB1 =
π1
π2Vab
1 Positive sequence
VAB1 =
π1
π2Vab
1 Negative sequence
Sequence circuits of Y β Ξ transformers
Case 3: Ξ-Ξ bank
Zero Sequence equivalent circuit with leakage impedance Z:
Sequence circuits of Y β Ξ transformers
Case 3: Y-Ξ bank β Y grounded
Sequence circuits of Y β Ξ transformers
Case 3: Y-Ξ bank β Y grounded Sequence equations:
VA0 - 3 ππ IA
0 = π1π2
Vab0 = 0 Zero sequence
VA1 =
π1
π2Vab
1 Positive sequence
VA2 =
π1
π2Vab
2 Negative sequence
Sequence circuits of Y β Ξ transformers
Case 4: Y-Ξ bank β Y grounded Zero Sequence equivalent circuit with leakage impedance Z:
π0 = Z + 3ππ
Sequence circuits of Y β Ξ transformers
Case 5: Y-Ξ bank- Y ungrounded
Sequence circuits of Y β Ξ transformers
Case 5: Y-Ξ bank β Y ungrounded Sequence equations:
VAB0 =
π1π2
Vab0 = 0 Zero sequence
VAB1 =
π1
π2Vab
1 Positive sequence
VAB2 =
π1
π2Vab
2 Negative sequence
Sequence circuits of Y β Ξ transformers
Case 5: Y-Ξ bank β Y ungrounded Zero Sequence equivalent circuit with leakage impedance Z:
Y β Ξ transformers
VAB1 = 3
π1
π2Vab
1 β 300
VAB2 = 3
π1
π2Vab
2 β β300
In per unit
VAB1 = Vab
1 β 300
VAB2 = Vab
2 β β300
The same applies for currents in the per unit case
IA1 = Ia
1 β 300
IA2 = Ia
2 β β300
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
πππ = 1840 V πππ = 2760 V πππ = 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-Ξ 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 πππ = 0.8β 82.80 πππ = 1.2β β41.40 πππ = 1.0β 1800 Vab
1 = 0.9857β 73.60 Vab
2 = 0.2346β 220.30 Van
1 = 0.9857β 43.60 Van
2 = 0.2346β 250.30 Ia
1 = 0.9857β 43.60 Ia
2 = 0.2346β 250.30
Solution
The calculated load voltages will be the voltages on the low voltage side of the transformer Vab
1 = 0.9857β 73.60 Vab
2 = 0.2346β 220.30 Therefore high voltage side voltages in per unit will be VAB
1 = Vab1 β 300
VAB2 = Vab
2 β β300 So VAB
1 = 0.9857 β 103.60 VAB
2 = 0.2346 β 190.30 Next calculate the sequence components of the other lines VBC
1 = 0.9857 β (103.6β120)0 = 0.9857 β β16.40 VCA
1 = 0.9857 β (103.6+120)0 = 0.9857 β 223.60 VBC
2 = 0.2346 β (190.3+120)0 = 0.2346 β 310.30 VCA
2 = 0.2346 β (190.3β120)0 = 0.2346β 70.30
Solution
The line-line voltages on the high voltage side are VAB = VAB
0 + VAB1 + VAB
2 = 0 + 0.9857 β 103.60 + 0.2346 β 190.30 = 1.026 β 116.80 VBC = VBC
0 + VBC1 + VBC
2 = 0 + 0.9857 β β16.40 + 0.2346 β 310.30 = 1.19 β β22.60 VCA = VCA
0 + VCA1 + VCA
2 = 0 + 0.9857 β 223.60 + 0.2346 β 70.30 = 0.783 β β144.10
Solution
The calculated load currents will be the line currents on the low voltage side of the transformer Ia
1 = 0.9857β 43.60 Ia
2 = 0.2346β 250.30 Therefore high voltage side voltages in per unit will be IA
1=Ia1β 300
IA2 =Ia
2β β300
So IA
1= 0.9857 β 73.60 IA
2= 0.2346 β 220.30 Next calculate the sequence components of the other lines IB
1 = 0.9857 β (73.6β120)0 = 0.9857 β β46.40 IC
1 = 0.9857 β (73.6+120)0 = 0.9857 β 193.60 IB
2 = 0.2346 β (220.3+120)0 = 0.2346 β 340.30 IC
2 = 0.2346 β (220.3β120)0 = 0.2346β 100.30
Solution
The line currents on the high voltage side are IA = IA
0 + IA1 + IA
2 = 0 + 0.9857 β 73.60 + 0.2346 β 220.30 = 0.8 β 82.90 IB = IB
0 + IB1 + IB
2 = 0 + 0.9857 β β46.40 + 0.2346 β 340.30 = 1.2 β β41.40 IC = IC
0 + IC1 + IC
2 = 0 + 0.9857 β 193.60 + 0.2346 β 100.30 = 1.0 β 179.90
Sequence Networks
Sequence circuits have been developed for the following components: Loads (Y and Ξ) 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 Ξ©. M2 is not grounded. Rated input for M1 is 200 MVA and M2 is 100 MVA. For both motors ππ
β = 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 Ξ©/km. Draw the positive sequence network. Use the generator values as base values.
Sequence Networks - Example
Sequence Networks - Example
Generator ππβ = j0.2
Transformer π1 π1 = j0.0857
Transmission line πππππ = j0.182
Transformer π2 π2 = j0.0915
Motor M1 ππβ = j0.274
Motor M2 ππβ = 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 Ξ© is in each of the neutrals of the generator and M1 . The zero sequence reactance of the transmission line is 1.5 Ξ©/km.
Sequence Networks - Example
Generator ππβ = j0.05
Transformer π1 π0 = j0.0857 (ππ = 0)
Transmission line πππππ = j0.545
Transformer π2 π2 = j0.0915 ((ππ = 0)
Motor M1 ππβ = j0.0686
Motor M2 ππβ = j0.1372
Generator 3ππ = j0.902
Motor M1 3ππ = j1.89
Sequence Networks - Example
Zero sequence network