chapter 11b

Symmetrical Components and Sequence Networks

Chapter 11

Balanced sequences

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

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

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

Equivalent Y and Δ loads

Example 11.2 Textbook

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.

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

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

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

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

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

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

The previous set of equations can be written in matrix form as

𝑉𝒂𝟎

𝑉𝒂𝟏

𝑉𝒂𝟐

= 𝑍𝑌 A

𝐼𝒂𝟎

𝐼𝒂𝟏

𝐼𝒂𝟐

+ 3𝐼𝑎0𝑍𝑛 111

Where A =𝟏 𝟏 𝟏𝟏 𝒂𝟐 𝒂𝟏 𝒂 𝒂𝟐

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

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

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

The previous three equations results in three separate circuits for the Y connected load

𝑍0 is the zero sequence impedance

𝑍1 is the zero sequence impedance

𝑍2 is the negative sequence impedance

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

𝒁Δ

Sequence circuits of Δ impedance loads

𝒁Δ

𝟑 𝒁Δ

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 j10 j10

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

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)

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

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

The volt drops across the line can be calculated from the following equations

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′

𝐼𝑎

𝐼𝑏

𝐼𝑐

Solution

𝑉𝑎𝑎′

𝑉𝑏𝑏′

𝑉𝑐𝑐′

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

262.5 − 𝑗175262.5 − 𝑗175262.5 − 𝑗175

Solution

Using Symmetrical component

The sequence components of the line volt drops are

𝑉𝑎𝑎′0

𝑉𝑎𝑎′1

𝑉𝑎𝑎′2

1 1 11 𝑎 𝑎2

1 𝑎2 𝑎

𝑉𝑎𝑎′

𝑉𝑏𝑏′

𝑉𝑐𝑐′

Solution

𝑉𝑎𝑎′0

𝑉𝑎𝑎′1

𝑉𝑎𝑎′2

1 1 11 𝑎 𝑎2

1 𝑎2 𝑎

28 + 𝑗4228 + 𝑗4228 + 𝑗42

= 28 + 𝑗42

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

The sequence equations are:

𝐿𝑠 and 𝑀𝑠 are the self and mutual inductances of the windings

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

Case 1: Y-Y bank – both neutrals grounded Sequence equations: Positive sequence Negative sequence Zero sequence

Zero Sequence equivalent circuit:

𝑍0 = 3𝑍𝑁 + 3 𝑍𝑛

Zero Sequence equivalent circuit with leakage impedance Z:

𝑍0 = Z + 3𝑍𝑁 + 3 𝑍𝑛

Positive Sequence equivalent circuit with leakage impedance Z:

𝑍1 = Z

Negative Sequence equivalent circuit with leakage impedance Z:

𝑍2 = Z

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

Case 3: Δ-Δ bank

Case 3: Δ-Δ bank Sequence equations:

VAB0 = Vab

0 = 0 Zero sequence

VAB1 =

𝑁2Vab

1 Positive sequence

VAB1 =

𝑁2Vab

1 Negative sequence

Case 3: Δ-Δ bank

Case 3: Y-Δ bank – Y grounded

Case 3: Y-Δ bank – Y grounded Sequence equations:

VA0 - 3 𝑍𝑁 IA

0 = 𝑁1𝑁2

Vab0 = 0 Zero sequence

𝑁2Vab

1 Positive sequence

𝑁2Vab

2 Negative sequence

Case 4: Y-Δ bank – Y grounded Zero Sequence equivalent circuit with leakage impedance Z:

𝑍0 = Z + 3𝑍𝑁

Case 5: Y-Δ bank- Y ungrounded

Case 5: Y-Δ bank – Y ungrounded Sequence equations:

VAB0 =

𝑁1𝑁2

Vab0 = 0 Zero sequence

VAB1 =

𝑁2Vab

1 Positive sequence

VAB2 =

𝑁2Vab

2 Negative sequence

Case 5: Y-Δ bank – Y ungrounded Zero Sequence equivalent circuit with leakage impedance Z:

Y – Δ transformers

VAB1 = 3

𝑁2Vab

1 ∠300

VAB2 = 3

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

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

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

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:

XT1-1 Xline-1 XT2-1

Xm1-1 Xm2-1

Sequence Networks

Negative sequence network:

XT1-2 Xline-2 XT2-2

Xm1-1 Xm2-2 Xm1-2

Sequence Networks

Zero sequence network:

XT1-0 Xline-0 XT2-0

3Xn1-0

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%.

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.

Generator 𝑋𝑑” = j0.2

Transformer 𝑇1 𝑋1 = j0.0857

Transmission line 𝑋𝑙𝑖𝑛𝑒 = j0.182

Transformer 𝑇2 𝑋2 = j0.0915

Motor M1 𝑋𝑑” = j0.274

Draw the negative sequence network for the system. Assume the negative sequence reactance of each machine is equal to its subtransient reactance.

All negative sequence reactances are equal to positive sequence reactances.

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.

Generator 𝑋𝑑” = j0.05

Transformer 𝑇1 𝑋0 = j0.0857 (𝑍𝑛 = 0)

Transmission line 𝑋𝑙𝑖𝑛𝑒 = j0.545

Transformer 𝑇2 𝑋2 = j0.0915 ((𝑍𝑛 = 0)

Generator 3𝑍𝑛 = j0.902

Motor M1 3𝑍𝑛 = j1.89

chapter 11b

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