symmetrical components.pdf

Lecture # Symmetrical Components

EE 308 Power Systems 1 18-Apr-13

Introduction

Under the normal (or healthy ) operation condition, balanced 3-Ø system can be performed using per-phase analysis.

Fault types:

This lecture deals to analyze the abnormal system behavior under condition of unsymmetrical short circuits

Unsymmetrical fault analysis requires new tools

– Symmetrical Components

– augmented component models

18-Apr-13 EE 308 Power Systems 2

L-L-L Faults 5 %

L-L-G Faults 10 %

L-L Faults 15 %

L-G Faults 70 %

Unsymmetrical faults

Symmetrical Components Allow unbalanced three-phase phasor quantities

decomposed into three separate but balanced symmetrical components (+ve , -ve and zero sequences)

– applicable to current and voltages

– permits modeling of unbalanced systems and networks

symmetrical components


Positive Sequence

– balanced and having the same phase sequence as the unbalanced supply

Negative Sequence

– balanced and having the opposite phase sequence to the unbalanced supply

Zero Sequence

– balanced but having the same phase and hence no phase sequence

Symmetrical Components


Representative symmetrical components

The phase components are the addition of the symmetrical components

The unknown unbalanced system has three unknown magnitudes and three unknown angles with respect to the reference direction. Similarly, the combination of the 3 sequence components will also have three unknown magnitudes and three unknown angles with respect to the reference direction.


For example, phase voltages,


210

210

210

cccc

bbbb

aaaa

VVVVVVVVVVVV

++=

++=

++=

• A so called “ α “ operator is commonly used in symmetrical component representation.

• It is defined as a vector having 1 as its magnitude and 120 degree as its phase angle.

866.05.01201 j+−=∠= α

( ) ( ) 12012401120112012 −∠=∠=∠×∠=α

Some properties of “α “ operator

012 =++αα Since α is complex, it can not be equal to 1, so that α -1 can not be zero

2α

α

1


Analysis of decomposition of phasors


symmetrical component matrix equation



Decomposition of phasors into symmetrical components

⇒

[ ]( )

( )( )

−−−−−−−−−−−

−=

=Λ −

1111

131

11

111

22

22

2224

2

21

αααααααααααααα

αααααα

αααα == −− 221 ,Since [ ]

=Λ −

αααα

2

21

11

111

31

PhSym

cba

aaa

=

αααα

2

2

2

1

0

11

111

31

Power in Terms of Sequence Components


***3 ccbbaa IVIVIVjQPS ++=+=−φ

[ ] [ ] [ ] *

012012*

*

*

*

3 IVIVIII

VVVST

abcTabc

c

b

a

cba ΛΛ==

=−φ

[ ] [ ] *012

*0120120123

*

IVIV ΛΛ=ΛΛ=−TTTS φ

=

=ΛΛ

111111111

311

111

11

111

2

2

2

2*

αααα

ααααT

( )*22

*11

*00

*0120123 33 aaaaaa

T IVIVIVS ++==− IVφ

( )2*

221*110

*003 coscoscos3 φφφφ aaaaaa IVIVIVP ++=−

Example

One conductor of a 3-Ø line is open as shown in Fig. The current flowing to the ∆-connected load through line a is 10 A. With the current in line a as reference and assuming that line c is open, find symmetrical components of the line currents.


Ex: 11.1 Grainger and Stevenson

a

b

c

Z

Z

Z

AIa010∠=

AIb18010∠=

AIc 0=

=

c

b

a

a

a

a

III

III

αααα

2

2

2

1

0

11

111

31

( )

( )( )

( )( ) AI

AI

AI

a

a

a

3078.502401801001031

3078.501201801001031

001801001031

2

1

0

∠=++∠+∠=

−∠=++∠+∠=

=+∠+∠=

Example … Cont’d


=

2

1

0

2

2

11

111

a

a

a

c

b

a

III

III

αααα

AIIAII

AII

ab

ab

ab

15078.515078.5

0

22

12

1

00

∠==

−∠==

==

αα⇒ ;

AIIAII

AII

ac

ac

ac

9078.59078.5

0

22

2

11

00

−∠==

∠==

==

αα

The result holds for any 3-wire system (absence of neutral connection).

AIII cba 0000 ===

Components have non-zero values although line c is open and can carry no net current.

21 cc IandI

Sequence Impedances

The impedance offered to the flow of a sequence current creating sequence voltages

– positive, negative, and zero sequence impedances

Augmented network models

– Y-connected load

– Transmission line

– 3-Ø transformers

– Synchronous generators


Network components

Balanced Load


3-Ø balanced load with self and mutual elements

Model and governing equations

nncsbmamc

nncmbsamb

nncmbmasa

IZIZIZIZVIZIZIZIZVIZIZIZIZV

+++=+++=+++=

cban IIII ++=

+++++++++

=

a

a

a

nsnmnm

nmnsnm

nmnmns

c

b

a

III

ZZZZZZZZZZZZZZZZZZ

VVV

abcabcabc IZV =

aV

sZ

bV

cV

sZ

sZ

aI

bI

cI

nI

nZ

mZ

mZmZ

Balanced Load


[ ] [ ]ΛΛ==⇒ΛΛ=

Λ=Λ⇒=−−

abcabc

abcabcabcabc

where ZZIZVIZVIZVIZV

1012012012012012

1012

012012

+++++++++

=

2

2

2

2012

11

111

11

111

31

αααα

αααα

nsnmnm

nmnsnm

nmnmns

ZZZZZZZZZZZZZZZZZZ

Z

−−

++=

ms

ms

mns

ZZZZ

ZZZ

00000023

012Z

Transmission Lines


sZ

sZ

sZ

aI

bI

cI

mZ

mZmZ

b

a

c

'b

'a

'c

−−

+=

ms

ms

ms

ZZZZ

ZZ

0000002

012Z

Fully transposed transmission line (i.e., symmetrical) has:

The conductors of a transmission line, being passive and stationary, do not have an inherent direction. Equal +ve and –ve sequence impedances However, as the zero sequence path also involves the earth

wire and or the earth return path, Zero sequence impedance much larger than the +ve (or –ve ) sequence impedance (approx. 2.5 times)

In practice all three phase conductors behave similarly, so that we could consider the mutual coupling between phases also to be equal

Transmission Lines: Summary

While the phase component impedance matrix was a full matrix, although it had completely symmetry

the sequence component impedance matrix is diagonal.

The advantage of a diagonal matrix is that it allows decoupling for ease of analysis.

Ratio of zero sequence impedance to +ve sequence impedance – Order of 2 (single circuit x’mission with earth wire) – About 3.5 (single circuit x’mission with no-earth wire or double circuit)


=

smm

msm

mms

ZZZZZZZZZ

Z

−−

+=

ms

ms

ms

ZZZZ

ZZ

0000002

012Z

Phase component impedance matrix symmetrical component impedance matrix

Synchronous Generator


aI

bI

cI

nI

b

a

c

n

anE

bnE

cnE

nZ

( )[ ]( )[ ]

( )[ ] 2222

1111

0000 2

ZIMLjRIVZIEMLjRIEV

ZIMLjRIV

assaan

aanssaanan

gassaan

−=++−=−=++−=

−=−+−=

ωω

ω

The generator (or a synchronous machine) has a inherent direction of rotation,

and the sequence considered may either have the same direction (no relative motion) or the opposite direction (relative motion at twice the speed).

Thus, the rotational emf developed for the positive sequence and the negative sequence would also be different.

Thus the generator has different values of positive sequence, negative sequence and zero sequence impedance.



1aI

1bI

1cIb

a

c

n

anE

bnE

cnE1Z

2aI

2bI

2cIb

a

c

n

1Z

1Z

2Z

2Z2Z

0aI

0bI

0cIb

a

c

n

0gZ 0gZ

0gZ

03 aInZ

1aI

anE

1Z

Reference for +ve

2aI

2Z

Reference for -ve

0aI

0gZ

nZ3

Reference for zero

positive sequence negative sequence zero sequence


Typical value of sequence impedances of turbo-generator rated 5 MVA, 6.6 kV, 3000 RPM are:


( )( )( )

%5%12%110

%20%12

0

2

1

1

1

=====

ZZ

ssynchronouZtransientZ

transientsubZ

Transformers

Series Leakage Impedance

– the magnetization current and core losses represented by the shunt branch are neglected (they represent only 1% of the total load current)

– the transformer is modeled with the equivalent series leakage impedance


lZZZ == 21

Wye-delta transformers create a phase shifting pattern for the various sequences

• the positive sequence line voltage on HV side leads the corresponding line voltage on LV side by 30 degrees

• For negative sequence line voltage the corresponding phase shift is -30 degrees

Two-winding transformers

Two winding (primary and secondary), three phase transformers may be categorized into

1) star-star

2) earthed star – star

3) earthed star – earthed star

4) delta – star

5) delta – earthed star

6) delta – delta


Two-winding transformers


Reference bus

0tZa b

c d

P S

1) star-star

Keep all a,b,c, and d switches open

2) earthed star – star

Close the switch a and keep open rest of the switches

3) earthed star – earthed star

Close the switches a and b. keep open rest of the switches

4) delta – star (Special case)

Close the switch b and keep open rest of the switches

5) delta – earthed star

Close the switches c and b. keep open rest of the switches

6) delta – delta

Close the switches c and d. keep open rest of the switches

Transformers


single-line networks for zero sequence of two-winding transformers

Transformer : Summary

An unearthed star winding does not permit any zero sequence current to flow so that it could be represented in the single line diagram by a 'break' between the line terminal and the winding.

If the star point is solidly earthed, it could be represented by a solid connection across the break and for an earth connection through an impedance, by 3 times the earthing impedance across the break.

In the case of a delta winding, no current would flow from the line, but a current is possible in the winding depending on the secondary winding connections. This could be represented by a break in connection with the line but with the winding impedance being connected to the reference.


Example For the power system of Fig. draw the positive, negative and zero sequence networks. The generators and transformers are rated as follows:

– Generator1: 25 MVA, 11 kV, X”=0.2, X2 =0.15, X0 = 0.03 pu

– Generator 2: 15 MVA, 11 kV, X”=0.2, X2 =0.15, X0 = 0.05 pu

– Synchronous Motor : 25 MVA, 11 kV, X”=0.2, X2 =0.2, X0 = 0.1 pu

– Transformer 1: 25 MVA, 11 ∆ / 120 Y kV, X = 10 %

– 2: 12.5 MVA, 11 Y / 120 ∆ kV, X = 10 %

– 3: 10 MVA, 120 Y / 11 ∆ kV, X = 10 %

Choose a base of 50 MVA, 11 kV in the circuit of generator 1.

Note: zero sequence reactance of each line is 250 % of its positive reactance.


symmetrical components.pdf

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