Lecture 19

Symmetrical Components

Read: Chapter 8.1 - 8.6 ; 8.8

Homework 6 – will be posted on the course website

Dr. Lei Wu

Department of Electrical and Computer Engineering

EE 333

POWER SYSTEMS ENGINEERING

Outline

Symmetric Components

Zero

Positive

Negative

Sequence Diagram of Systems Components

System Analysis Using Sequence Diagram

2

0 0 0a b cI I I= =2

1 1 1 1b a c aI I I Iα α= =2

2 2 2 2b a c aI I I Iα α= =

02

1

22

1 1 1

1

1

a a

p b a s

c a

I I

I I

I I

α α

α α

= = =

I A I

0 0

1 1

2 2

3 0 0

0 0

0 0

y n

s s s y

y

Z ZV I

V Z I

V IZ

+ = = =

V Z I

The system is decoupled

0 0( 3 )y nV Z Z I= + 1 1yV Z I= 2 2yV Z I=

Sequence Diagrams of Impedance Loads, cont’d

3

Each sequence network is uncoupled with the other two

This separation underlines the advantage of symmetrical

components

Neutral impedance does not appear in positive and negative

sequence networks

Positive and negative currents do not flow in the neutral

impedance

Sequence Diagrams of Impedance Loads

4

0 0( 3 )y nV Z Z I= + 1 1yV Z I= 2 2yV Z I=

Sequence Diagrams of Impedance Loads, cont’d

For a delta-connected balanced 3Φ load

Delta-connected balanced 3Φ load is equivalent to wye-connected

balanced 3Φ load with infinite neutral to ground impedance

3YZ

Z ∆=

5

01 2

1

22

1 1 1 01

13

34.111

ag

bg

cg

VV

V V

V V

α α

α α

−

100∠ ° 18.63∠63.43° = = 75∠180° = 67.32∠ − 26.45° 50∠90° ∠22.99°

A

00 2.446

3y n

VI

Z Z= = ∠ − 3.37°

+

11 13.46

y

VI

Z= = ∠ − 79.58° 2

2 6.822y

VI

Z= = ∠ − 30.14°

Sequence Diagrams of Impedance Loads Example

6

Sequence Diagrams of Impedance Loads Example

02

1

22

1 1 1 2.446

1 13.46

6.822 8.531

a

b

c

I I

I I

II

α α

α α

∠ − 3.37° 19.97∠ − 57.34° = = ∠ − 79.58° = 15.16∠132.42° ∠ − 30.14° ∠37.45°

A

7

For a symmetrical series impedance, which means

, we can obtain

Sequence Diagrams for a General 3Φ Load

ag aaa ab ac

bg ab bb bc b

ac bc cc ccg

V IZ Z Z

V Z Z Z I

Z Z Z IV

=

1

2 0 0

0 0

0 0

aa ab

s p aa ab

aa ab

Z Z

Z Z

Z Z

−+

= = −

−

Z A Z A

aa bb cc ab ac bcZ Z Z and Z Z Z= = = =

8

However, if the loads are not symmetrical, then

is not diagonal. The sequence networks are coupled and the

voltage drop across any one sequence network depends on all

three sequence currents.

The sequence diagrams can be directly applied for series

impedances and transmission lines, which usually have the same

impedance value in three phases.

sZ

Sequence Diagrams for a General 3Φ Load

9

Sequence diagrams for generators

Balanced 3Φ positive sequence generators only produce positive

sequence voltages; therefore only the positive sequence network

has a voltage source.

01 2

1

22

1 1 1 01

13

01

ag

bg

cg

VV V

V V V V

VV V

α α

α α

−

∠ ° 0 = = ∠ −120° = ∠0° ∠120°

A

10

Sequence diagrams for generators

Balanced 3Φ negative sequence generators only produce negative

sequence voltages; therefore only the negative sequence network

has a voltage source.

01 2

1

22

1 1 1 01

1 03

1

ag

bg

cg

VV V

V V V

V VV V

α α

α α

−

∠ ° 0 = = ∠120° = ∠ −120° ∠0°

A

11

Sequence diagrams for generators, cont’d

During a fault Z1 ≈ Z2 ≈ Xd”. The zero sequence impedance is

usually substantially smaller. The value of Zn depends on whether the generator is grounded.

Synchronous motors have the same sequence networks as synchronous generators, except the sequence currents are referenced into the networks.

12

Sequence diagrams for generators, example

0

1

2

277 0 5

260 120 15

10295 115

10 40 30 40

, , ?

ag

bg

cg

Y

a b c

V Z j

V Z j

Z jV

Z Z

I I I∆

∠ = Ω

= ∠ − = Ω = Ω∠

= ∠ Ω = ∠ Ω

13

Get the sequence networks

Sequence diagrams for generators, example

( )0

11

2

15.91 62.11

277.10 1.77 // 3 5 40

9.22 216.59

ag

bg Y Y

cg

VV

V A V Z Z Z

V V

−∆

∠ ′= = ∠ − = = ∠ Ω ∠

14

Calculate sequence currents

Calculate phasor currents

Sequence diagrams for generators, example

0

1

2

46.94 50.05

49.68 70.49

47.98 166.94

a

b

c

I I

I A I

II

∠ = = ∠ − ∠

0 0I = ( )1

1 // 3

48.19 48.83

l Y

VI

Z Z Z∆=

+

= ∠ −

( )2

2 // 3

1.60 169.53

l Y

VI

Z Z Z∆=

+

= ∠

15