lecture 2 pp

EET 331Symmetrical Components

Lectures

Tim SkvareninaSchool of Technology

Purdue University

Introduction to symmetrical components

Objectives

Following this section, you should be able to :

a. Define the terms positive, negative, and zero sequenceand draw phasor diagrams for each sequence

b. Define the operator "a"

c. Convert an unbalanced set of phase quantities intosequence components or vice-versa

d. Calculate the power in an unbalanced three phase circuitusing sequence voltages and currents

Figure 1 shows three sets of rotating phasors. These could represent voltagesor currents.

The set on the left consists of a balanced set (same magnitude and 120o apart). Since phasors rotate in the CCW direction, this set of phasors has an A-B-Crotation sequence, called the positive sequence.

The set in the middle is also a balanced set. Its phase rotation is A-C-B ornegative sequence.

The set of phasors to the right all have the same magnitude, but they are inphase with each other. This set of phasors is called the zero sequence.

An example of zero sequence would be third harmonics.

These three sets of phasors are given the name symmetrical components. They are useful because any unbalanced set of voltages or currents can bebroken down into symmetrical components. The theory of symmetricalcomponents was developed by Charles Fortesque who presented it at ameeting of the American Institute of Electrical Engineers (a predecessor of theIEEE) in 1918. Fortesque's paper was over 110 pages long. Symmetricalcomponents allow us to break down the problem of solving unbalanced faultconditions in the power system.

a

b

c

a

cb

a, b, c

Positive Sequence

Negative Sequence

Zero Sequence

Figure 1

As an example of the process consider figure 2.

In figure 2, the like phases from each of the three sequences havebeen added together.

That is the "a" phasors from the positive, negative, and zerosequences have been added. The same has been done for "b"and "c".

The resultant set of phasors is clearly very unbalanced. The threephases all have different magnitudes and they are not 120o apart.

Obviously, by changing the relative magnitudes of the sequencesand the phase angles between them, we could create an infinitenumber of unbalanced sets. The question is how can we find thesequences if we have a set of unbalanced voltages or currents. To do that we must first look at some mathematical principles anddefinitions.

Figure 2

a

b

c

ac

b

a

b

c

a

b

c

Mathematical Operators

Given an equation:

we know the equation must have "n" roots, or solutions

Thus for

there are two solutions, namely:

and for

we could rewrite it as

so

xn = k

x2 = 1

x = ± 1

x2 = -1

x2 = 1p180o

x = ± 1p90o = 1p90o and 1p-90o


Since we use the roots of -1 a lot we assign them a specialsymbol:

x = ± j

"j" is a mathematical operator

Multiplying a phasor by "j"doesn't change its magnitude,but shifts its phase angleforward by 90o I

j I

Multiplying by "-j" shifts thephase angle backward by 90o.

-j I

I


Now consider the equation:

There should be three solutions.

The other two are in the complex plane

Rewrite the equation as

we can see the other solutions are:

x3 = 1

x = 1 is a solution

x3 = 1p360o or x3 = 1p720o

x = 1p120o and x = 1p240o


W e can define another mathematical operator:

Squaring "a":

a = 1p120o

a2 = 1p240o

"a" rotates phasors by 120oV

aV

"a2 " rotates phasors by 240o VaV

Sequences

Using V to represent our phasors and subscripts 0, 1, 2 for the 0, +, and - sequences, respectively we can write for the positivesequence:

Va1

Vc1 = aVa1

Vb1 = a2Va1

Sequences

For the negative sequence:

Va2

Vc2 = aVa2

Vb2 = a2Va2

For the zero sequence: Vao = Vbo = Vco

Va

Vb

Vc

'

1 1 1

1 a 2 a

1 a a 2

V0

V1

V2

V0

V1

V2

'13

1 1 1

1 a a 2

1 a 2 a

Va

Vb

Vc

Va

Vb

Vc

Expressing Unbalanced Sets in Terms of Sequences

Va = Va0 + Va1 + Va2 = V0 + V1 + V2

In Matrix Form

These are called the synthesis equations

Vb = Vb0 + Vb1 + Vb2 = V0 + aV1 + a2V2

Vc = Vc0 + Vc1 + Vc2 = V0 + a2V1 + aV2

by inverting the A Matrix we can solve for the sequence quantitiesin terms of a set of known unbalanced phase quantities

which can be found to be:

In shorthand notation:

[Vabc] = [A] [V012]

[V012] = [A]-1 [Vabc]

Solving for Sequences in Terms of Phase Quantities

These are called the analysis equations

Consider the unbalanced Ysystem shown in figure 3. The+ and -sequence componentsadd to zero at the center of theY. The sum of the abc currentsflows on the neutral.

In = Ia + Ib + Ic = 3I0

thus: I0 = In / 3

Zero Sequence

Ia Ib

Ic

In

Example

Va = 120p15o , Vb = 30p-90o , and Vc = 90p165o

Find the sequence voltages

V0 = a (Va + Vb + Vc) = 12.62 40.04O

V1 = a (Va + aVc + a2Vb) = 77.66 28.09O

V2 = a (Va + a2Vb + aVc) = 40.13 -19.85O

Power in Symmetrical Components

S3N = P3N + jQ3N = Sa + Sb + Sc

S3N = VaIa* + VbIb

* + VcIc*

It can be shown: S3N = 3 ( Vo Io* + V1I1

* + V2I2* )

This means there is no coupling between the sequences.

If we can develop a circuit diagram for each sequence, we cancombine them to find phase conditions under unbalancedconditions (such as faults).

Vag

Vbg

Vcg

'

(Z%Zn) Zn Zn

Zn (Z%Zn) Zn

Zn Zn (Z%Zn)

Ia

Ib

Ic

Note: When installing the program, you may get amessage “cannot copy DDEML.DL_ because it is in use” Just click OK, and continue installation.

Sequence networks for symmetrical components

Objectives


a. Draw positive, negative, and zero sequence circuits for athree-phase system including transmission lines,transformers, generators, and Y or delta loads

b. Calculate the currents in a three-phase system whenunbalanced voltages are applied

Sequence Networks

Consider a Y system as shown. The voltage from "a" to groundis:

Similarly for the other twophases:

g

a

b

c

Z Z

Z

Zn

Vag = Z Ia + ZnIn

Vag = Z Ia + Zn (Ia + Ib + Ic )

Vag = (Z+Zn )Ia + Zn Ib + Zn Ic

Vbg = ZnIa + (Z+Zn ) Ib + Zn Ic

Vcg = ZnIa + Zn Ib + (Z+Zn )Ic

Sequence Networks

In matrix form:

in shorthand: Vp = Zp Ip

- subscript "p" indicates phase quantities- phase quantities are coupled by the neutral - replacing phase quantities by sequence quantities:

Vp = A Vs and Ip = A IsA Vs = ZpA Is

Vs = ( A-1ZpA ) Is = Zs Is

Zs '

(Z%3 Zn) 0 0

0 Z 0

0 0 Z

'

Z0 0 0

0 Z1 0

0 0 Z2

Sequence Impedances

If the phase impedances are all the same, it can be shown:

The sequence impedances for this load are uncoupled and thematrix is diagonal. Thus the sequences are also uncoupled.

V0

I0 Z0

3Zf

V1

I1Z1 V2

I2Z2

If Z11 = Z22 = Z33 and the off diagonal elements of the phase Zmatrix are equal, the sequence networks will be decoupled.

For a delta load, we can convertto an equivalent Y load with noneutral connection as shown.

Also shown are the sequencenetworks for the delta load.

Because of the conversion to aY, the sequence currents onlyrepresent the delta from itsterminals and not the internalload characteristics.

Delta Load

Z

Z

Z

Z/3

Z/3

Z/3

0Z/3 Z/3

Z/3

Sequence Networks of Series Impedances

If the impedances are equal in all three phases and there is nomutual impedance:

Zo = Z1 = Z2 = Z

When we calculate the inductance and capacitance of atransmission line, we accounted for the mutual reactance so wecan use this for a transmission line.

Z

Z

Z

a

b

c

Vas

Vbs

Vcs

Var

Vbr

Vcr

However, the zerosequenceimpedance of atransmission lineis typically asmuch as 3 timesthe + and -sequences

Sequence Networks of Generators

Below is a simple model for the synchronous machine that showsall three phases. A generator is designed to generate a positivesequence of voltages.

Thus there is no source in the negative and zero sequencenetworks.

E

E

E

c

a

bZn

For balanced + sequence currents from thegenerator, the flux due to the armaturecurrent penetrates the rotor and Z1 . Zs (thesynchronous impedance).

Z0

3Zn

Z1

For balanced negative sequence currents,there is a flux wave rotating in the oppositedirection of the rotor motion. Relatively littleflux penetrates the rotor so:

Z2 « ZS Z2 . Z" (subtransientimpedance)

Z2Zero sequence currents would producefluxes that essentially cancel, so the zerosequence impedance is the smallest of all.

Generator Sequence Circuits

Typical Per-Unit Values ofMachine Impedances(Round Rotor)

Xs . 1.1 pu

Xd' . .23 pu

Xd" . .12 pu

X2 . .13 pu

X0 . .05 pu

These sequence networksneglect:

Saliency of rotorSaturation effectsTransient effects

but they are accurate enough formany types of studies.

Recall there are three values ofsynchronous impedance:

- Steady-state (Zs or Zd )- Transient (Zd') used for stability studies- Subtransient (Zd") used for fault studies

Generator Impedances

Z)

Zn

Z

Z

Z Z)

Z)

Example: Balanced 480 V generator

T-line: Z = 1p85o S Load: Z)= 30p40o S

Generator: Zn = j10 S Zgo = j1 S Zg1 = j15 SZg2 = j3 S

3Zn =j30

Zg0 =j1

Zline =1p85O

10p40O

open) load

Zline =1p85O

10p40OZg1 =j15

Zline =1p85O

10p40OZg2 =j3

For balanced voltages, V0 = V2 = 0, so there are no sources in thezero and negative sequence circuits.

If voltages are unbalanced, then there will be sources in all threesequence circuits

Asymmetrical faults

Objectives


a. Combine the sequence networks of a three-phasesystem to represent unbalanced fault conditions

b. Calculate the sequence currents in an unbalancedfault and convert them back to phase currents

c. Calculate the phase voltages during an unbalancedfault condition

Asymmetrical Faults

The figure shows a three-phase power system at some point in thesystem (the rest of the system being represented by theamorphous shape).

a

b

c

I

I

I

a

b

c

a

b

c

I

I

I

a

b

c

line-line fault

a

b

c

I

I

I

a

b

c

line-line-groundfault

line-ground fault

By connecting lines to ground or each other, we can applyunbalanced faults. We will have to use symmetrical componentsto find the fault current.

W e can represent a three-phase system by sequencenetworks, each of which canbe reduced to a Theveninequivalent at the point of thefault, as shown.

Connecting the sequencestogether will allows calculationof the fault current

0 seqnetwork

Zo

+ seqnetwork

Z1

- seqnetwork

Z2

Sequence Networks Assumptions

- Balanced steady-state operation before the fault. Thus,the +, -, and 0 sequence networks are uncoupled beforethe fault.

- During the fault, the only interconnection of the sequencenetworks will be at the point of the fault

- Prefault load current is neglected. Thus voltageeverywhere is the same.

- Neglect resistances and shunt admittances forcalculations by hand. Neglect induction motors unlessthey are very large

I0

I1

I2

'13

1 1 1

1 a a 2

1 a 2 a

Ia

Ib

Ic

I0

I1

I2

'13

1 1 1

1 a a 2

1 a 2 a

0

Ib

&Ib

a

b

c

I

I

I

a

b

c Zf

Single Line to Ground Fault

Vag = V0 + V1 + V2 = Ia Zf = 3I1 Zf

Ia = 3 I0 = 3 I1 = 3 I2

After fault, Ib = Ic = 0

Single Line to Ground Fault

Vag = V0 + V1 + V2

Vag = Ia Zf = 3I1 Zf

Ia = 3 I0 = 3 I1 = 3 I2 Zo

Z

Z

1

2

3ZfV1

V0

V2

Since the current is thesame in all three sequencecircuits, the circuits mustbe in series

The sequence circuits areconnected together by animpedance = 3Zf

E1

a

b

c

I

I

I

a

b

c

Zf

Line to Line Fault

I1 = -I2

I1 = a(a - a2)Ib

but (a - a2) = j %3

I1 = a(j %3 )Ib º Ib = -j %3 I1

a

-aa2

After fault, Ib = -Ic and Ia = 0

from synthesis equations:

Vbg = V0 + a2V1 + aV2 Vcg = V0 + aV1 + a2V2

Vbc = Vbg - Vcg = Ib Zf = -j %3 I1 Zf = (a2 - a)(V1 - V2)

I1 Zf =(V1 - V2)

Ib = -j %3 I1

These results indicate that the + and - sequence circuitsmust be in parallel with the fault impedance betweenthem

Line to Line FaultZf

V1 V2

Z1 Z2

Line to Line to Ground Fault

a

b

c

II

I

a

b

c Zf

a2V1 - aV1 = a2V2 - aV2

Ia = 0 = I0 + I1 + I2

Thus sequence circuits musthave a common node

Vbg = Vcg

V0 + a2V1 + aV2 = V0 + aV1 + a2V2

ˆ V1 = V2

Thus + and - sequence circuits must be in parallel


Vbg = V0 + a2V1 + aV2 = Zf [( I0 + a2I 1 + aI2 ) + (I0 + aI1 + a2I2 )]

ˆ V0 - V1 = Zf ( 2I0 - I 1 - I2 ) = 3 Zf I0

Zf

V1 V2

Z1 Z2 Z0

Connecting the sequences asshown in figure 13 will satisfythe conditions

I1'

Vf

Z1 % [Z2 ** (3Zf % Z0)]

I1 '

Vf

Z1 %Z2 (3 Zf % Z0)

Z0 % Z2 % 3 Zf

&I1

3Zf % Z0

Z0 % Z2 % 3 Zf

I0 ' &I1

Z2

Z0 % Z2 % 3Zf


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lecture 2 pp

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