lec6 power system

8/18/2019 lec6 power system

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POWER SYSTEMS ILecture 6

06-88-590-68

Electrical and Computer Engineering

University of Windsor

Dr. Ali Tahmasebi


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1

Analysis of Unsymmetric Systems

l Except for the balanced three-phase fault, faults

result in an unbalanced system.

l The most common types of faults are single line-

ground (SLG) and line-line (LL). Other types aredouble line-ground (DLG), open conductor, and

balanced three phase.

l System is only unbalanced at point of fault!

l The easiest method to analyze unbalanced system

operation due to faults is through the use of

symmetrical components


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Symmetric Components

l The key idea of symmetrical component analysis is

to decompose the system into three sequence

networks. The networks are then coupled only at

the point of the unbalance (i.e., the fault)

l The three sequence networks are known as the

– positive sequence (this is the one we’ve been using)

– negative sequence

– zero sequence


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Positive Sequence Sets

l The positive sequence sets have three phase

currents/voltages with equal magnitude, with phase

b lagging phase a by 120°, and phase c lagging

phase b by 120°.l We’ve been studying positive sequence sets (also

known as ‘abc’ sequence)

Positive sequencesets have zero

neutral current


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Negative Sequence Sets

l The negative sequence sets have three phase

currents/voltages with equal magnitude, with phase

b leading phase a by 120°, and phase c leading

phase b by 120°.l Negative sequence (also known as ‘acb’ sequence)

sets are similar to positive sequence, except the

phase order is reversed

Negative sequence

sets have zero

neutral current


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Sequence Set Representation

Any arbitrary set of three phasors (vectors), say Ia, I b, Iccan be represented as a sum of the three sequence sets

0

0

0

0 0 0

where

, , is the zero sequence set

, , is the positive sequence set

, , is the negative sequence set

a a a a

b b b b

c c c c

a b c

a b c

a b c

I I I I

I I I I

I I I I

I I I

I I I

I I I

+ -

+ -

+ -

+ + +

- - -

= + +

= + +

= + +

I a

I b

I c


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Conversion from Sequence to Phase

0a

2 3 3

0 0 0a b c

2

Only three of the sequence values are unique,

I , , ; the others are determined as follows:

1 120 0 1

I I I (since by definition they are all equal)

a a

b a c a b a c

I I

I I I I I I I

a a a a a

a a a

+ -

+ + + + - - +

Ð ° + + = =

= =

= = = =

2

0

0 + 2 2a a

2 2

1 1 1 11 1I 1 I 1

1 1

a

aa

b a a

c a

I

I I I I I

I I

a

a a a a

a a a a

-

- +

-

é ùé ù é ùé ùé ù é ù ê úê ú ê úê úê ú ê ú= + + = ê úê ú ê úê úê ú ê ú

ê úê ú ê úê úê ú ê úë û ë û ë ûë û ë ûê úû

=

-


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Conversion Sequence to Phase

2

2

0 0

Define the symmetrical components transformation

matrix

1 1 1

1

1

Then

aa

b a s

c a

I I I

I I I

I I I

a a

a a

+ +

- -

é ù

ê ú= ê úê úë û

é ù é ùé ù

ê ú ê úê ú= = = =ê ú ê úê úê ú ê úê úë û ê ú ê úû ë û

A

I A A A I

Is :

sequence

currents


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Conversion Phase to Sequence

1

1 2

2

By taking the inverse we can convert from the

phase values to the sequence values

1 1 11

with 13

1

Sequence sets can be used with voltages as well

as with currents

s

a a

a a

-

-

=

é ùê ú

= ê úê úë û

I A I

A


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Conversion Phase to Sequence

It can be easily derived that: =1

3 + +

In any Y-connected 3-phase system: I n = I a + I b + I c , so:

I n = 3 I 0

If this system is also balanced, then I n = I 0 = 0

(Also, any system with no neutral path will have I n = I 0 = 0)

NOTE: Some books use the notation 0, 1 and 2 instead of ±

signs. So for example instead of I 0, I 1 and I 2, we will have I 0, I 1and I 2, respectively.


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Symmetrical Component Example 1

1 2s

2

s

10 0

Let 10 Then

10

1 1 1 10 01

1 10 10 03

10 01

10 0 0

If 10 0

10 10 0

a

b

c

I

I

I

a a

a a

-

Ð °é ù é ùê ú ê ú= = Ð - 120°ê ú ê ú

Ð 120°ê ú ê úë û ë û

é ù Ð ° 0é ù é ùê úê ú ê ú= = Ð - 120° = Ð °ê úê ú ê úê ú Ð 120°ê ú ê úû ë ûë û

Ð °é ù é ùê ú ê ú= Ð +120° ® =ê ú ê ú

Ð - 120° Ð °ê ú ê úû ë û

I

I A I

I I

(‘abc’ or positive sequence), then:

(‘acb’ or negative sequence)


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1 2s

2

0

Let

Then

1 1 1 0 01

13 6.121

a

b

c

V

V

V

a a

a a

-

5Ð 9 °é ù é ùê ú ê ú= = 8Ð 150°ê ú ê ú

8Ð - 30°ê úê ú ë ûë û

é ù 5Ð 9 ° 1.67Ð 9 °é ù é ùê úê ú ê ú= = 8Ð 150° = 3.29Ð - 135°ê ú

ê ú ê úê ú 8Ð - 30° Ð 68°ê ú ê úû ë ûû

V

V A V


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0

2

2

10 0

Let 10

Then

1 1 1 10 0

1 10

1

s

s

I

I

I

a a

a a

+

-

é ù Ð °é ùê ú ê ú= = - Ð 0°ê ú ê úê ú 5Ð 0°ê ú

ë ûê úë û

é ù Ð ° 5.0Ð 0°é ù é ùê úê ú ê ú

= = - Ð 0° = 18.0Ð 46.1°ê úê ú ê úê ú 5Ð 0° 18.0Ð - 46.1°ê ú ê úû ë ûë û

I

I AI


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Phase b open:

=

=

10Ð 0°

0

10Ð 120°

(this is obviously an

unbalanced system)

=

=

3.33Ð 60°

6.66Ð 0°3.33Ð − 60°

I n = I a + I b + I c = 10Ð 0° + 0 + 10Ð 120° = 10Ð 60° = 3 I 0

(as we expected!)


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Use of Symmetrical Components

Consider this balanced, wye-connected impedance load:

( )

( )

( )

n a b c

ag a y n n

ag Y n a n b n c

bg n a Y n b n c

cg n a n b Y n c

I I I I

V I Z I Z

V Z Z I Z I Z I

V Z I Z Z I Z I

V Z I Z I Z Z I

= + +

= +

= + + += + + +

= + + +

ag y n n n a

bg n y n n b

ccg n n y n

V Z Z Z Z I

V Z Z Z Z I

I V Z Z Z Z

é ù é ù+ é ùê ú ê úê ú= +ê ú ê úê úê ú ê úê ú+ ûû ë û


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Use of Symmetrical Components

1

13 0 0

0 0

0 0

ag y n n n a

bg n y n n b

ccg n n y n

s s

s s s s

y n

y

y

V Z Z Z Z I

V Z Z Z Z I

I V Z Z Z Z

Z Z

Z

Z

-

-

é ù é ù+ é ùê ú ê úê ú= +ê ú ê úê úê ú ê úê ú+ ûë û ë û

= = =

= ® =

é ù+ê ú= ê ú

ê úû

V Z I V A V I A I

A V Z A I V A Z A I

A Z A

( Z is called the phase impedance matrix)

= Zs = sequence

impedance

matrix

® Vs = Zs Is


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Networks are Now Decoupled

0 0

0 0

3 0 0

0 00 0

Systems are decoupled

( 3 )

y n

y

y

y n y

y

V I Z Z

V Z I Z V I

V Z Z I V Z I

V Z I

+ +

- -

+ +

- -

é ù é ùé ù+ê ú ê úê ú

=ê ú ê úê úê ú ê úê úê ú ê úë ûë û ë û

= + ==

(because Zs is a diagonal matrix):


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Sequence Networks

® I0

+

-

V0Zy

3Zn

Z0 = Zy + 3ZnZero-sequence network:

® I+

+

-V+ Zy Z

+ = ZyPositive-sequence network:

®I

-

+

-V- Zy Negative-sequence network: Z- = Zy

Z0 = zero-sequence

impedance


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Grounding

l When studying unbalanced system operation how a

system is grounded can have a major impact on the

fault flows

l Ground current only impacts zero sequence system

l In previous example if load was ungrounded the

zero sequence network is (with Zn equal infinity):


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Grounding, cont’d

l Voltages are always defined as a voltage

difference. The ground is used to establish the

zero voltage reference point

– ground need not be the actual ground (e.g., an airplane)

l During balanced system operation we can ignore

the ground since there is no neutral current

l There are two primary reasons for grounding

electrical systems1. safety

2. protect equipment


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How good a conductor is dirt?

l There is nothing magical about an earth ground. All

the electrical laws, such as Ohm’s law, still apply.

l Therefore to determine the resistance of the ground

we can treat it like any other resistive material:

8

8

onductor lengthResistance

cross sectional area

2.65 10 -m for aluminum1.68 10 -m for copper

where is the resistivity

c R

r

r

r

r

-

-

´=

= ´ W

= ´ W


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8

16

2.65 10 -m for aluminum

5 10 -m for quartz (insulator!)

160 -m for top soil

900 -m for sand/gravel

20 -m for salt marsh

What is resistance of a mile long, one inch diameter,

circular wir

r

r

r

r

r

-= ´ W

= ´ W

= W

= W

= W

8

2

e made out of aluminum ?

2.65 10 1609 R= 0.083

mile0.0128

-´ ´ W=´


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6

2

What is resistance of a mile long, one inch diameter,

circular wire made out of topsoil?

160 1609 R= 500 10

mile0.0128

In order to achieve 0.08 with our dirt wiremile

we would need a cross sectio

p

´ W= ´

´W

6 2

nal area of

160 1609 3.2 10 (i.e., a radius of about 1000 m)0.08

But what the ground lacks in , it makes up for in A!

m

r

´ = ´


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Calculation of grounding resistance

• Because of its large cross sectional area the earth is

actually a pretty good conductor.

• Devices are physically grounded by having aconductor in physical contact with the ground;

having a fairly large area of contact is important.

• Most of the resistance associated with establishing

an earth ground comes within a short distance of

the grounding point.


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Sequence Networks of Balanced- load

AD-connected

balanced load and its

equivalent Y-connected

load ( =∆

).

® I0 = 0

+

-

V0∆

3

Z0 = ¥

® I++

-V+

Z+ = ∆

® I-+

-V-

Z- = ∆

Sequence networks:

∆

3∆

3


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Sequence Networks of Generators

Key point: generators only

produce positive sequence

voltages; therefore only the

positive sequence has avoltage source.

Sequence networks:

¬ I0

+

-

V0

¬I

+

+

-

V+

¬ I-

+

-V-

Zg0

3Zn Zg+

Eg+

Zg-

Zero-sequence network Positive-sequence network Negative-sequence network


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• During a fault Zg+ » Zg

-» Xd

².

• The zero sequence impedance is usually

substantially smaller.

• The value of Zn depends on whether the generator

is grounded.

Sequence Networks of Generators, cont’d


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Sequence Networks of Motors

• The sequence networks of synchronous motors are the

same as the synchronous generators (with a source Em+ in

the positive-sequence network), except the currents go

into the networks.

• The sequence networks of the induction motors are the

same as the synchronous motors, except there is no

voltage source (Em+ = 0).


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Sequence Networks of Transformers

Ideal Y-Y transformer:

Sequence networks:® I

H

+ or IH

-

+

-

EH+ or EH

-

+

-

Positive and Negative-sequence network (p.u.)Zero-sequence network (p.u.)

® IX+ or IX

-

EX+ or EX

-


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• When IA0 = IB

0 = IC0 = I0 is applied to an ideal

transformer, I N = 3I0 flows through Z N and also

through Zn.

• Note that if either one of the neutrals is ungrounded,

then no zero sequence can flow in either the high- or

low-voltage windings. For example, if the high-

voltage winding has an open neutral, then I N = 3I0 = 0,

which in turns forces I0

= 0 on the low-voltage side.

Sequence Networks of Transformers, cont’d


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Sequence

Networks of

Practical

Transformers

(p.u.)


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Sequence Networks of Transformers, cont’d

Notes on sequence networks of practical transformers:

• In (a) double-grounded Y-Y: positive- and negative

sequence networks are the same as p.u. transformer model,

and 3Z N and 3Zn are added only in zero-sequence network.

• In (b) grounded-Y/D: positive- and negative sequence

networks are the same as transformer model. Zero-

sequence currents can enter Y side if it is grounded and

flow through the winding and also the Dwinding (since itwill be inducted) but cannot leave the Dside.

• In (c) D- D: zero-sequence currents cannot enter or leave D

windings, but can circulate within them.


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Power in Sequence Networks

Total complex power delivered to a 3-phase load, in

terms of phase voltages and currents is:

= ∗ +

∗ + ∗

= .∗ = ∗ = ( ∗)∗

∗ =

3 0 0

0 3 0

0 0 3

→ = 3

∗

→ = 3 ∗ + ∗ + ∗ = 3

Ss = total complex power delivered to sequence networks.

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