new symmetrical components · 2020. 4. 8. · synthesis of unsymmetrical phases from their...

Symmetrical Components

Submitted By:Santosh Kumar GuptaAssistant ProfessorEE Departmet, SIT Sitamarhi


⚫ Symmetrical Component Analysis

⚫ Synthesis of Unsymmetrical

Phases from Their Symmetrical

Components

⚫ The Symmetrical Components of

Unsymmetrical Phasors

⚫ Phase Shift of Symmetrical

Components in Y− or −Y Transformer Banks

⚫ Power in Terms of Symmetrical

Components


⚫ Unsymmetrical Series

Impedance

⚫ Sequence Impedance and

Sequence Network

⚫ Sequence Networks of

Unloaded generators

⚫ Sequence Network

⚫ Zero-Sequence Network

Symmetrical Component Analysis

Goal :

Symmetrical component analysis is a very useful tool

for dealing with unbalanced three-phase faults.

Synthesis of Unsymmetrical Phases from TheirSymmetrical Components 1

“An unbalanced system of n related phasors can be resolved into

n systems of balanced phasors called the symmetrical components

of the original phasors. The n phasors of each set of components

are equal in lengths , and the angles between adjacent phasors of

the set are equal.”

by C.L Fortescue , 1918


(2) Negative-sequencecomponents

c2

n

b2

2. For negative-sequence

components

(1) Positive- sequence components

1a

1b

c1

1. For positive- sequence

components

a2n


(3) Zero-sequence components

Va0

Vb0

Vc0

0 For zero-sequence components


b 0

V a = V a 1 + V a 2 + V a 0

V + V= V b 1 + V b 2b

V c = V c 1 + V c 2 + V c 0

Va0

Va2

Va

a1V

Vb1

Vb0

Vb2

Vc0

Vc1

Vc

Vb

Vc2


Use a to designate the operator that causes a rotationof

counterclockwise direction ,

= 1 2 4 0 0 = − 0 . 5 − j 0 . 8 6 6

= 1 3 6 0 0 = 1

a 2

a 3

a = 1120 0 = − 0 . 5 + j 0 . 8 6 6

1200 in the

− a2

−1,−a3

a2− a

1,a3

a

1

a 2

V b 0

c 2

= V a 0 , V c 0 = V a 0

, V = a 2 VVb 2 = a V a 2

, Vc 1 = aV a 1V = a 2 V

b1 a1

a1b1

1c

a2

c2

b2

The Symmetrical Componentsof Unsymmetrical Phasors 2

= a V + a2V + Va1 a 2 a 0Vc = Vc1 + Vc 2 + Vc0

= a2V + aV + Va1 a 2 a 0Vb = Vb1 + Vb 2 + Vb0

Va = Va1 + Va 2 + Va 0

1 a 2

a 1

c

b

1 V a 0

a 2 V

a V

V

1 1V = 1 a 2

a

Va

3

1

a 2

a 1 1 1

a 2

a

A = 1 a

a 2

13

1 1 1

A− 1 =1 1 a

a 2

c

b a V V

a 2 V

a 2

a1

Va 0

1

1 Va 1 1

a

a 2 3 V =

1 1

* When three phase phasors are balanced , only the

positive-sequence component exists .

4

1.Sequence component representation of L-Lvoltage

c a

bc a V V

a 2 V

a b 2

ab1

Va b0

1

1 Va b 1 1

a

a 2 3 V =

1 1

2.Sequence component representation of current

c

b a I I

a 2 I

a 2

a 1

I a 0

1

1 I a 1 1

a

a 2 3 I =

1 1

5

3

3

cbaa 0

a b b c c aa b 0

a b ca 0

I

V

+ I + I )=1

( I3

=1

(V + V + V )

V =1

(V + V + V )

No zero-sequence components exist if the sum of the three

phasors is zero.


3a b ca 0

V =1

(V + V + V )

Va0 = 0

Va0 0

When 3 is balanced

When (Va +Vb +Vc ) 0

* If is unbalanced.Va0 0 then 3

* Unbalanced 3 does not guarantee Va0 0 .


3a b b c c aa b 0V =

1(V + V + V )

(Vab +Vb c +V c a ) is always zero (form closed loop)

Vab 0 is always zero whether the three phase system

is balanced or not.

b

a

c

8

Ia

bI

Ic

bI

Ic

I a

Ib

c

Ia

I nI

1Ia0 = (Ia + Ib + Ic ) = 0

3( ungrounded Y)

1Ia0 = 3

(Ia + Ib + Ic ) 0

(Ia + Ib + Ic ) = 3Ia0 = In

Y with a path to neutral

1Ia0 = 3

(Ia + Ib + Ic ) =0

connected

VECTOR

SOFTWARE

The Symmetrical Componentsof Unsymmetrical Phasors 9example : One conductor of a three-phase line is open. 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 the symmetrical components

of the line currents

Z Z

Z

a

b

c

aI = 1 0 0 0 a m p

b

0180 ampI =10

I c = 0

aI = 1 0 0 0 A

bI = 101800 A

I c = 0 A

The line currents are :

10

3+ 1 0 1 8 0 0 + 0 ) = 0

a 0I =

1( 1 0 0 0

a 1I =

1( 1 0 0 0 + 1 0 1 8 0 0 + 1 2 0 0 + 0 )

3

= 5 − j 2 . 8 9 = 5 . 7 8 − 3 0 0 A

a 2I =

1( 1 0 0 0 + 1 0 1 8 0 0 + 2 4 0 0 + 0 )

3

= 5 + j 2 . 8 9 = 5 . 7 8 3 0 0 A

I b 1 = 5 . 7 8 − 1 5 0 A0

= 5 . 7 8 1 5 0 0 A

= 0

I b 2

I b 0

I c 1 = 5 . 7 8 9 0 A0

= 5 .78 − 90 0 A

= 0

I c 2

I c 0

a 0Since there no neutral current involved , I

should be zero .

Phase Shift of SymmetricalComponents

Y −in or −Y Transformers Banks 1

to neutral leads the voltage drop from X3 to

.

drop from H3

neutral by 300

The American standard for designatingterminal H1 and X1 on Y − or −Y

transformer requires that the positive-sequence voltage drop from H1 to

neutral leads the positive-sequence voltage drop from X1 to neutral by 300 ,

regardless of whether the Y or winding is on the high tension side .

Similarly, the positive-sequence voltage drop from H2 to neutral leads the

voltage drop from X2 to neutral by 300 and the positive-sequence voltage


Transformers Banks 2in Y − or −Y

Example :

A

B

C a

b

c

H3

H1 X1

H2 X2

X3

A

B

C

a

b

c

H1

H3

H2

X1

X2

X3

VA1leads Vb1 by 300 leads Va1VA1 by 300

Phase Shift of SymmetricalComponentsin Y − or −Y Transformers Banks 3

,

The American standard for designating terminal H1and X1 on Y − or −Y

transformer requires that the negative-sequence voltage drop from H1 to

neutral lags the negative-sequence voltage drop from X1to neutral by 30

0

regardless of whether the Y or winding is on the high tension side .

Similarly, the negative-sequence voltage drop from H2 to neutral lags

the voltage drop from X2 to neutralby 300 and the negative-sequence

voltage drop from H3 to neutral lags the voltage drop from X3 to neutral

by 300 .


Transformers Banks 4in Y− or −Y

A

B

C a

b

c

H3

H1

H2

X1

X2

X3

A

B

C

a

b

c

H1

H3

H2

X1

X2

X3

lags Va2A2(b) V by 300

b2V(a) lags byA2V 030

Example :

Phase Shift of Symmetrical Components in Y

− or −Y Transformers Banks 5

A

B

C a

b

c

H3

H1

H2

X1

X2

X3

2B

A2

C2

2b

c2

a 2

1B

1A

C1

b1

a1

c1

Va1 leads VA1 by 900

A2Va2 lags V by 900

leads Vb1VA1 by 300

b2A2V lags V by 300

Y

Example 11.7. The resistive Y-connected load bank of Example 11.2 is supplied from the low-voltage

Y-side of a Y- transformer. The voltages at the load are the same as in that example. Find the line

voltages and currents in per unit on the high-voltage side of the transformer.

I (1)a = 0.9857 43.60 per unit

I ( 2)a = 0.2346 250.30 per unit

V (1)an = 0.985743.60 per unit(line− to − neutral voltage base)

V (2)an = 0.2346 250.30 per unit(line − to − neutral voltage base)

V (1) A = 0.9857 43.60 + 300 = 0.9857 73.60 = 0.2785 + j0.9456

V (2)A = 0.2346250.30 − 300 = 0.2346220.30 = −0.1789 − j0.1517

VA = V A +V A = 0.0994 + j0.7939 = 0.882.8 per unit(1) (2) 0

V (1)B = a2V (1)

A = 0.9857 − 46.40 = 0.6798 − j0.7138



V (2)B = a2V ( 2)

A = 0.2346 −19.70 = 0.2209 − j0.0791

VB +V (2)B = 0.9007 − j0.7929 = 1.20 − 41.40 per unit= V (1)

B

V (1)C = a2V (1)

A = 0.9857 193 .60 = −0.9581 − j0.2318

V (2)C = a2V ( 2)

A = 0.2346 100.30 = −0.0419 − j0.2318

VC +V (2)C = −1.0 + j0 = 1.0180 0 per unit= V (1)

C

VAB = VA −V B= 0.0994 + j0.7939 − 0.9007 + j0.7929 = −0.8013 + j1.5868

= 1.78116 .80 per unit(line − neutral voltagae base)

3=

1.78116 .80 per unit(line − to − line voltagae base)




3

VBC = VB −V C= 0.9007 − j0.7939 + 1.0 = 1.9007 − j0.7939

= 2.06 − 22.70 per unit(line − neutral voltagae base)

=2.06

− 22.70 = 1.19 − 22.70 per unit(line − to − line voltagae base)

3

VCA = VC −V A= −1.0 − 0.0994 − j0.7939 = 1.0994 − j0.97939

= 1.356 215 .80 per unit(line − neutral voltagae base)

=1.356

215 .80 = 0.783215 .80 per unit(line − to − line voltagae base)

IA = 0.8082.8 per unit0

IB = 1.20 − 41.4 per unit0

IC = 1.0180 per unit0

in Y− or Transformers Banks 8−Y

A

B

C a

b

c

H3

H1 X1

H2 X2

X3

A

B

C

a

b

cH3

H2

H1 X1

X2

X3

(b) by 300(a) V(1) A leads V(1)

a by 300 V(1) A leads V(1)

a

Figure 11.23

labeling of lines connected to a three-phase Y- transformer.



Power in terms of Symmetrical Components

S = P + j Q = V I *a + V I *

b + V I *c

a b c

*

0 1 20 1 2

IVa

T

a

= A V T A Ib

*

I

I c

b

Vc

S = V

0 1 20 1 2

*

0 1 20 1 2

T I *= 3 V= V T A T A * I

*

a 1

a 0

a 0 a 1 a 2

I a 2

I

= 3V V V I

a 1a 0 I *a 2 )I *

a 0I *

a 1 + Va 2

+ V= 3 ( V

, A T A *= 3 I

1

a

b

c

a '

b'

c'c

bZ

Z

Za

Ia

IbZcaZac

Zab

I c Z c Z c a

Z a c I a

Z b c I b

Z a b

= Z b a Z b

Z c b

Ic

Z a

Vb b '

V

c c '

V ' a a

2

Ia 2

I a 0

Z c Zca

Z b c A I a 1 V

c c ' 2

V ' Z a a a 0

A Vb b ' 1 = Z b a

−1 I a 0

Z c Ia 2

Z b c A I a 1

Z a c

V

c c ' 2

Vb b '1 = A

V ' aa 0

Z a b Z a c

Z b

Z c b

Z a Z a b

Z b a Z b

Zca Z c b

Z

3

Z 0 1 2

)

s1 M 1

(Z s 0 − 2ZM 0 )− 2ZM 2 )(Zs2

(Z s 2 + 2ZM 2= (Z − 2Z ) (Z − 2Z )

s 0 M 0

(Z s1 + 2ZM 1)

(Z s1 − 2ZM 1) (Z s 2 − 2ZM 2 )

= A − 1 ZA

(Z s 0 + 2ZM 0 )

Where

3

3

3

cbas2

cbas1

cbas0

+ aZ )+ a2 ZZ =1

(Z

+ aZ + a2 Z )Z =1

(Z

+ Z + Z )Z =1

(Z

3

abcabcM 2

abcabcM 1

abcabcM 0Z

Z =1

(Z + a 2Z + aZ ) 3

Z =1

(Z + aZ + a2 Z )

+ Z + Z )=1

(Z3

4

V ' = (Z + 2Z )I + (Z − 2Z )I + (Z − 2Z )Iaa 0 s 0 M 0 a0 s 2 M 2 a1 s1 M 1 a 2

V ' = (Z − 2Z )I + (Z − 2Z )I + (Z + 2Z )Iaa 1 s1 M 1 a0 s0 M 0 a1 s2 M 2 a2

V ' = (Z − 2Z )I + (Z + 2Z )I + (Z − 2Z )Iaa 2 s2 M 2 a0 s1 M 1 a1 s0 M 0 a2

5

= 0Case 1. If no coupling , Z i j ( i j )

V '

a a 0= Z s 0 I a 0 + Z s 2 I a 1 + Z s 1 I a 2

3 3 3a0 a b c a1 a b c

(Z +aZ + a2 Z )a2 a b c

=1

I (Z +Z + Z ) +1

I (Z +a2Z + aZ ) +1

I

V '

a a 2= Z s 2 I a 0 + Z s 1 I a 1 + Z s 0 I a 2

V '

a a 1= Z s 1 I a 0 + Z s 0 I a 1 + Z s 2 I a 2

3

1

3 3

2

a1 a b ca0 a b c(Z + a2 Z + Z )

a2 a b c(Z + Z + Z ) +

1I=

1I (Z + aZ + a Z ) + I

3

1

3

2

a1 a b ca 0 a b c a2 (Za + Zb + Zc )(Z + aZ + a Z ) + I=1

I (Z + a2 Z + aZ ) +1

I3

then = Z M 1 = Z M 2 = 0Z M 0

6

= 0Z i j ( i j )

Z a = Z b = Z c

V ' = I Zaa 1 a1 a

V ' = I Za a 2 a 2 a

V ' = I Za a 0 a 0 a

Symmetrical components of unbalanced currents flowing in a balanced- Y load

or in balanced series impedances produce voltage drops of the same sequence ,

provided no coupling exists between phases.

If the impedances are unequal, the voltage drop of any one sequence is dependent on the

current of all three sequences.

If coupling such as mutual inductance exists among the three impedances, then the

formula will become more complicated.

Case 2 . If 1.

2.

Complete transportation assumed

7

Assume:

1. No coupling

2.

Positive-sequence currents produce positive-sequence voltage drops.

Negative-sequence currents produce negative sequence voltage drops.

zero-sequence currents produce zero-sequence voltage drops.

Z a = Z b = Z c

Sequence Impedance and Sequence Network 1

The impedance of circuit when positive- sequence

current alone are flowing is called positive-sequence

impedance.

The impedance of circuit when negative-sequence

currents alone are flowing is called negative

sequence impedance.

When only zero-sequence currents are present, the

impedance is called zero sequence impedance.


The single-phase equivalent circuit composed of the impedance to

current of any one sequence only is called the sequence network.

Positive-sequence network contains positive sequence current and

positive sequence impedance only.

Negative-sequence network contains negative sequence current

and negative sequence impedance only.


Zero-sequence network contains zero sequence current and

zero sequence impedance only.

Sequence network carrying the individual currents Ia1 , Ia 2

and Ia 0 are interconnected to represent various

unbalanced fault condition.


Sequence Impedance of Various Devices

Positive Negative Zero

Line same same different

Transformer same same same

Machine different * different * different

* Usually they are assumed to be the same

Sequence Networks of Unloaded Generators 1

I a

I c

I n

Ea

b

Ib

c

a

+

-

- - Eb

+ Ec +

nZ

The generator voltage (Ea , Eb , Ec )

are of positive sequence only,

since the generator is designed

to supply balanced three-phase

voltage.


Ia 2

c2I

b2I

a

cb

2Z

Z 2

Z 2Va 2

a 2Ia

Z 2

+

-

Negative-sequence

network

a1V

a1I

a

E a

Z 1

+

-

Positive-sequence

network

Reference

Reference

Ia1

b1

bE

E

b

Ic

a

+

++Ec

- a

--

1Z

Z1Z1

aI

V a 1 = E a − I a 1 Z 1

a 2 2a 2V = − I Z


Ic1

Ia 0

Ia 0

Ia 0

a

Z g 0

Z g 0

Z g 0

b

V a 0

ag 0

Z+

-3Z n

Z 0

I a 0

Zero-sequence

network

Reference

Zn only appears in the zero-sequence

network

V a 0Zn

a 0

c

3I

= − I a 0 Z 0

= − I a 0 (Z g 0 + 3Zn )

na03I = I


Determine the subtransient current in the generator and the line-t0-line voltages for subtransient

conditions due to the fault.

Example 11.6. A salient-pole generator without dampers is rated 20 MVA, 13.8kV and has a

direct=axis subtransient reactance of 0.25 per unit. The negative-and-zero-sequence reactance

are, respectively, 0.35 and 0.10 per unit. The neutral of the generator is solidly grounded. With the

generator operating unloaded at rated voltage with Ean = 1.00 per unit , a singleline-to-0

ground fault occurs at the machine terminals, which then have per-unit voltages to ground,

V a = 0 V = 1 . 0 1 3 − 1 0 2 . 2 5 0 V = 1 . 0 1 3 1 0 2 . 2 5 0

b c

I c = 0

I a

I a = I n

b

I b = 0

c

a

E a n

E b n

++ E c n

+

-

- -

Z n

n

Figure 11.15


Figure 11.15 shows the line-to-ground fault of phase a of the machine.

Vb = −0.215 − j0.990 per unit

Vc = −0.215 + j0.990 per unit

V

V

c

a

=

b

− 0.143+ j0

a − 0.215 + j0.990 − 0.500 + j01

1

1 1 1 0

3 1

a 2

a a 2 − 0.215 − j0.990 = 0.643+ j0 per unit(0 )

V (0)

(0 )

Z g o j0.1 0

V ( 0 )a( 0 ) = − j1.43per un i t

( − 0 . 1 4 3 + j0)= −I a = −


EI

j0.25

(1 ) = − j1.43per unit(1.0 + j0) − (0 .643+ j0)− V (1 )

aa = − a n =

Z 1

Ij0.35Z 2

V ( 2 )a( 2 ) = − j1.43per unit

(−0.500 + j0)= −a = −

+ I ( 2 )a = 3I (0 )

a = − j4.29+ I (1)a

Ia = I a(0 )

There, the fault current into the ground is

The base current is 20,000 ( 3 13.8) = 837 A and so the subtransient current in line a is

Ia = − j4.29 837 = − j3,590 A


Line-to-line voltage during the fault are

ab a bV = V −V = 0.215 + j0.990 = 1.0177.70 per unit

bc b cV = V −V = 0 − j1.980 = 1.980270 0 per unit

ca c aV = V − V = −0.215 + j0.990 = 1.0177.70 per unit

ab3

V = 1.01 13.8

77.70 = 8.0577.70 kV

bc3

V = 1.980 13.8

270 0 = 15.78270 0 kV

ca3

V = 1.01 13.8

102.30 = 8.05102 .30 kV


Before the fault the line voltages were balanced and equal to 13.8kV. For comparison with the line

voltages after the fault occurs, the prefault voltages, with Van = Ean as reference, are given as

abV =13.8300 kV

bcV = 13.8270 0 kV

caV = 13.81500 kV

Figure 11.6 shows phasor diagrams of prefault and postfault voltages.

Figure 11.6

(a) Prefault

Van

Vca

bcVVbc

VabVab

a an

Vca

bb

c c

(b) Postfault


The positive-sequence diagram of a generator is

composed of an emf in series with the positive-sequence

impedance of the generator.

The negative and zero-sequence diagrams contain no

emfs but include the negative and zero-sequence

impedances of the generator respectively.

Sequence Networks 1

The matching reactance in positive-sequence network is the subtransient ,transient,

or synchronous reactance, depending of whether subtransient , transient, or

steady- state condition are being studied.

The reference bus for the positive and negative sequence networks is the neutral

of the generator. So for as positive and negative sequence components are

concerned , the neutral of the generator is at ground potential even if these is Zn

connection between neutral and ground.

The reference bus for the zero sequence network is the ground (not necessary

the neutral of the generator).

Sequence Networks 2

Convert a positive sequence network to a negative sequence

network by changing, if necessary, only the impedance

that represent rotating machine , and by omitting the emf.

The normal one-line impedance diagram plus the induced emf is the

positive sequence network.

Three-phase generators and motors have internal voltage of positive

sequence only.

Example of Positive and Negative-Sequence Network 1

Example: Draw the positive and negative-sequence networks

for the system described as below . Assume that the

negative-sequence reactance of each machine is

equal to its subtransient reactance .Omit resistance.

1T

M 1

T pm nk l

rM 2

Example of Positive and Negative-Sequence Network 2

+

--

j 0 .0 8 5 7 j 0 .0 8 1 5

j 0 .0 2

j 0 .2 7 4 5 j 0 . 5 4 9 0

+

E

- m 2

+

E m 1

gE

k l m

j 0 .0 9 1 5

n

p r

j 0 .0 8 5 7 j 0 .0 9 1 5j 0 .0 8 1 5

j 0 .0 2 j 0 .2 7 4 5

Reference bus

k l m

p

n

j 0 .5 4 9 0

q

(Positive)

(Negative)

Zero sequence Network 1

1 . Zero-sequence network currents will flow only if a return path exists.

2 . The reference bus of the zero-sequence network is the ground.

ZZ

Z

N

Z

Z

Z

N

Z N

R

N

Reference


Z

Z

Z

NZ

R

Nn

In = 3Ia 0

3Zn

NIa 0

Z Z

Z

R

Z


Zero-sequence equivalent circuit of three phase transform banks.

Symbols Connection DiagramsZero-Sequence

Equivalent Circuit

p Q

pQ p Q

Z0

Reference bus

p p Q pQ

Z0

Reference bus



Equivalent Circuit

p

pQ

p Z0

Reference bus

Q

p

p Q

p Z0

Reference bus

Q



Equivalent Circuit

p

p

pQ Q

Z0

Reference bus


3 Z n

g 0Z

Q

R

S

T

M N

P

(Zero-Sequence)

Example:

Z n

Q S

P

R T

M N


g 1Z

Q S

R T

MNP

E g 1

++

Z g 2

- E g 1-

Q

R TM

N

P

S

(Positive-Sequence)

(Negative-Sequence)

SEQUENCE

NETWORK

SOFTWARE


Example 11.9. Draw the zero-sequence network for the system described in Example 6.1. Assume

zero-sequence network for the generator and motors of 0.05 per unit. A current-limiting reactor of

is in each of the each of the neutrals of the generator and the large motor. The zero-sequence0.4

reactance of the transmission line is 1.5 km

Generator:

Motor 1:

Motor 2:

X 0 = 0.05 per uni t

1 3.2

2 0 0 1 3.8

2

0 ) = 0 .0686 per uni tX = 0.05(3 0 0

) (

1 3.2

1 0 0 1 3.8

2

0 ) = 0 .1 3 7 2 p er uni tX = 0.05(3 0 0

) (

= 1.333300

(20) 2

Base Z =

= 0.635300

(13.8)2

Base Z =


n1.333

3Z = 3(0.4

) = 0.900per unit

n0.635

0.4) = 1.890per unit3Z = 3(

176.30

Z =1.5 64

= 0.5445per unit

The zero-sequence network is shown in Fig. 11.28

j0.05

j0.900

j0.0857 j0.5445 j0.0915

l m

kn

p r

j0.0686

j1.890

j0.1372

reference

Thank You

new symmetrical components · 2020. 4. 8. · synthesis of unsymmetrical phases from their...

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