EE2351 POWER SYSTEM ANALYSIS

EE2351POWER SYSTEM ANALYSIS

EE2351 A.Ahamed RiazudeenAP/EEE

1

A.Ahamed RiazudeenEEE DEPARTMENT

UNIT I

INTRODUCTION


2

Power system network


3

SINGLE LINE DIAGRAM

It is a diagrammatic representation of a power system inwhich the components are represented by their symbols.


4

It is a diagrammatic representation of a power system inwhich the components are represented by their symbols.

COMPONENTS OF A POWER SYSTEM

1.Alternator2.Power transformer

3.Transmission lines

4.Substation transformer

5.Distribution transformer

6.Loads


5

1.Alternator2.Power transformer

3.Transmission lines

4.Substation transformer

5.Distribution transformer

6.Loads

MODELLING OF GENERATOR ANDSYNCHRONOUS MOTOR


6

1Φ equivalent circuit of generator 1Φ equivalent circuit of synchronous motor

MODELLING OF TRANSFORMER


7

2 2 1

1 1 2

' 201 1 2 1 2

' 201 1 2 1 2

E N IK

E N I

RR R R R

KX

X X X XK

=Equivalent resistance referred to 1o

=Equivalent reactance referred to 1o

MODELLING OF TRANSMISSION LINE


8

Π type T type

MODELLING OF INDUCTION MOTOR


9

'

'

'

1( 1 )r

S r

S r

Rs

R R R

X X X

=Resistance representing load

=Equivalent resistance referred to stator

=Equivalent reactance referred to stator

per unit=actual value/base value

Let KVAb=Base KVA

kVb=Base voltage

Zb=Base impedance in Ω


10

2 2

1000

b bb

bb

kV kVZ

KVAMVA

Changing the base of per unit quantities

Let z = actual impedance(Ω)= base impedance (Ω)bZ

. 2 2

* bp u

b b b

b

Z MVAZ ZZ

Z kV kV

MVA


11

. 2 2

* bp u

b b b

b

Z MVAZ ZZ

Z kV kV

MVA

Let , ,

, ,

&

&b old b old

b new b new

kV MVB

kV MVB

represent old base values

represent new base values

,. , 2

,

. , ,

2

,

,. , 2

,

2

, ,. , . , 2

,,

*(1)

*(2)

*(3)

* *

b oldp u old

b old

p u old b old

b old

b newp u new

b new

b old b newp u new p u old

b oldb new

Z MVAZ

kV

Z MVAZ

kV

Z MVAZ

kV

kV MVAZ Z

MVAkV


12

,. , 2

,

. , ,

2

,

,. , 2

,

2

, ,. , . , 2

,,

*(1)

*(2)

*(3)

* *

b oldp u old

b old

p u old b old

b old

b newp u new

b new

b old b newp u new p u old

b oldb new

Z MVAZ

kV

Z MVAZ

kV

Z MVAZ

kV

kV MVAZ Z

MVAkV

ADVANTAGES OF PER UNIT CALCULATIONS

The p.u impedance referred to either side of a 1Φtransformer is same

The manufacturers provide the impedance value in p.u The p.u impedance referred to either side of a 3Φ

transformer is same regardless of the 3Φ connections Y-Y,Δ-Y

p.u value always less than unity.


13

The p.u impedance referred to either side of a 1Φtransformer is same

The manufacturers provide the impedance value in p.u The p.u impedance referred to either side of a 3Φ

transformer is same regardless of the 3Φ connections Y-Y,Δ-Y

p.u value always less than unity.

IMPEDANCE DIAGRAM

• This diagram obtained by replacing each component bytheir 1Φ equivalent circuit.

Following approximations are made to draw impedancediagram

1. The impedance b/w neutral and ground omitted.2. Shunt branches of the transformer equivalent circuit

neglected.


14

• This diagram obtained by replacing each component bytheir 1Φ equivalent circuit.

Following approximations are made to draw impedancediagram

1. The impedance b/w neutral and ground omitted.2. Shunt branches of the transformer equivalent circuit

neglected.

REACTANCE DIAGRAM

It is the equivalent circuit of the power system in whichthe various components are represented by theirrespective equivalent circuit.

Reactance diagram can be obtained after omitting allresistances & capacitances of the transmission line fromimpedance diagram


15

It is the equivalent circuit of the power system in whichthe various components are represented by theirrespective equivalent circuit.

Reactance diagram can be obtained after omitting allresistances & capacitances of the transmission line fromimpedance diagram

REACTANCE DIAGRAM FOR THE GIVEN POWER SYSTEMNETWORK


16

PROCEDURE TO FORM REACTANCEDIAGRAM FROM SINGLE DIAGRAM

1.Select a base power kVAb or MVAb

2.Select a base voltage kVb

3. The voltage conversion is achieved by means of transformer kVb on LTsection= kVb on HT section x LT voltage rating/HT voltage rating

4. When specified reactance of a component is in ohmsp.u reactance=actual reactance/base reactancespecified reactance of a component is in p.u


17

1.Select a base power kVAb or MVAb

2.Select a base voltage kVb

3. The voltage conversion is achieved by means of transformer kVb on LTsection= kVb on HT section x LT voltage rating/HT voltage rating

4. When specified reactance of a component is in ohmsp.u reactance=actual reactance/base reactancespecified reactance of a component is in p.u

2

, ,. , . , 2

,,

* *b old b newp u new p u old

b oldb new

kV MVAX X

MVAkV

p.u. calculation of 3 winding transformer


18

Zp=Impedance of primary windingZs’=Impedance of secondary windingZt’=Impedance of tertiary windingShort circuit test conducted to find out the above 3impedances

'

' '

' '

1

21

21

2

p ps pt st

s ps st pt

t ps pt st

Z Z Z Z

Z Z Z Z

Z Z Z Z


19

psZ

ptZ

= Leakage impedance measured in 1o with 2o short circuitedand tertiary open.

= Leakage impedance measured in 1o with tertiary shortcircuited and 2o open.

'stZ = Leakage impedance measured in 2o with tertiary short

circuited and 1o open and referred to primary

PRIMITIVE NETWORKIt is a set of unconnected elements which provides informationregarding the characteristics of individual elements. it can berepresented both in impedance & admittance form


20

BUS ADMITTANCE(Y BUS) MATRIX

Y BUS can be formed by 2 methods

1.Inspection method

2.Singular transformation

11 12 1

21 22 2

1 2

n

n

n n nn

Y Y Y

Y Y Y

Y Y Y


21

Y BUS =11 12 1

21 22 2

1 2

n

n

n n nn

Y Y Y

Y Y Y

Y Y Y

INSPECTION METHOD

For n bus systemDiagonal element of Y BUS

Off Diagonal element of Y BUS1

n

ii ijj

Y y


22

For n bus systemDiagonal element of Y BUS

Off Diagonal element of Y BUS

ij ijY y

1

n

ii ijj

Y y

SINGULAR TRANSFORMATION METHOD

Y BUS =

Where [y]=primitive admittance

A=bus incidence matrix

TA y A


23

ALGORITHM FOR FORMATION OF THEBUS IMPEDANCE MATRIX

• Modification of Zbus matrix involves any one of the following 4 cases

Case 1:adding a branch impedance zb from a new bus p tothe reference bus

Addition of new bus will increase the order the Zbus matrix by 1

(n+1)th column and row elements are zero except the diagonaldiagonal element is zb


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• Modification of Zbus matrix involves any one of the following 4 cases

Case 1:adding a branch impedance zb from a new bus p tothe reference bus

Addition of new bus will increase the order the Zbus matrix by 1

(n+1)th column and row elements are zero except the diagonaldiagonal element is zb

,

0

0original

bus newb

zZ

z

Case 2: adding a branch impedance zb from a new bus pto the existing bus qAddition of new bus will increase the order the Zbus matrix by 1The elements of (n+1)th column and row are the elements ofqth column and row and the diagonal element is Zqq+Zb

Case 3:adding a branch impedance zb from an existing bus p tothe reference bus

The elements of (n+1)th column and row are the elements ofqth column and row and the diagonal element is Zqq+Zb and(n+1)th row and column should be eliminated using the followingformula


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Case 2: adding a branch impedance zb from a new bus pto the existing bus qAddition of new bus will increase the order the Zbus matrix by 1The elements of (n+1)th column and row are the elements ofqth column and row and the diagonal element is Zqq+Zb

Case 3:adding a branch impedance zb from an existing bus p tothe reference bus

The elements of (n+1)th column and row are the elements ofqth column and row and the diagonal element is Zqq+Zb and(n+1)th row and column should be eliminated using the followingformula

( 1) ( 1),

( 1)( 1)

1,2... ; 1,2..j n n kjk act jk

n n

Z ZZ Z j n k n

Z

Case 4:adding a branch impedance zb between existing buses h and qelements of (n+1)th column are elements of bus h column –bus q column and elements of (n+1)th row are elements ofbus h row – bus q row the diagonal element=and (n+1)th row and column should be eliminated using the followingformula

2b hh qq hqZ Z Z Z


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Case 4:adding a branch impedance zb between existing buses h and qelements of (n+1)th column are elements of bus h column –bus q column and elements of (n+1)th row are elements ofbus h row – bus q row the diagonal element=and (n+1)th row and column should be eliminated using the followingformula

( 1) ( 1),

( 1)( 1)

1,2... ; 1,2..j n n kjk act jk

n n

Z ZZ Z j n k n

Z

UNIT II

POWER FLOW ANALYSIS


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BUS CLASSIFICATION

1.Slack bus or Reference bus or Swing bus:|V| and δ are specified. P and Q are un specified, and to becalculated.

2.Generator bus or PV bus or Voltage controlled bus:P and |V| are specified. Q and δ are un specified, and to becalculated

3.Load bus or PQ bus:P and Q are specified. |V| and δ are un specified, and to becalculated


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1.Slack bus or Reference bus or Swing bus:|V| and δ are specified. P and Q are un specified, and to becalculated.

2.Generator bus or PV bus or Voltage controlled bus:P and |V| are specified. Q and δ are un specified, and to becalculated

3.Load bus or PQ bus:P and Q are specified. |V| and δ are un specified, and to becalculated

ITERATIVE METHOD

The above Load flow equations are non linear andcan be solved by following iterative methods.

1.Gauss seidal method2.Newton Raphson method3.Fast Decoupled method

1

*

*1

n

p pq qq

p p p P p

np p

pq qqP

I Y V

S P jQ V I

P jQY V

V


29

The above Load flow equations are non linear andcan be solved by following iterative methods.

1.Gauss seidal method2.Newton Raphson method3.Fast Decoupled method

1

*

*1

n

p pq qq

p p p P p

np p

pq qqP

I Y V

S P jQ V I

P jQY V

V

GAUSS SEIDAL METHOD

For load bus calculate |V| and δ from Vpk+1 equation

11 1

*1 1

1

( )

p np pk k k

p pq q pq qkq q ppp P

P jQV Y V Y V

Y V


30

For generator bus calculate Q from QPK+1 equation

11 * 1

1

1*Im ( )p n

k k k kp P pq q pq q

q q p

Q V Y V Y V

11 1

*1 1

1

( )

p np pk k k

p pq q pq qkq q ppp P

P jQV Y V Y V

Y V

• Check Qp,calk+1 with the limits of Qp

• If Qp,calk+1 lies within the limits bus p remains as PV bus

otherwise it will change to load bus• Calculate δ for PV bus from Vp

k+1 equation• Acceleration factor α can be used for faster convergence

• Calculate change in bus-p voltage

• If |ΔVmax |< ε, find slack bus power otherwise increasethe iteration count (k)

• Slack bus power=


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• Check Qp,calk+1 with the limits of Qp

• If Qp,calk+1 lies within the limits bus p remains as PV bus

otherwise it will change to load bus• Calculate δ for PV bus from Vp

k+1 equation• Acceleration factor α can be used for faster convergence

• Calculate change in bus-p voltage

• If |ΔVmax |< ε, find slack bus power otherwise increasethe iteration count (k)

• Slack bus power=

1 1k k kp p pV V V

G LS S

NEWTON RAPHSON METHOD

1

1

1

1 2

3 4

co s( )

s in ( )

n

i i i j ij ij i jj

n

i i j ij ij i jj

n

i i j ij ij i jj

k sch ki i i

k sch ki i i

P Q V V Y

P V V Y

Q V V Y

J JP

J J VQ

P P P

Q Q Q


32

1

1

1

1 2

3 4

co s( )

s in ( )

n

i i i j ij ij i jj

n

i i j ij ij i jj

n

i i j ij ij i jj

k sch ki i i

k sch ki i i

P Q V V Y

P V V Y

Q V V Y

J JP

J J VQ

P P P

Q Q Q

• Calculate |V| and δ from the following equation

• If

• stop the iteration otherwise increase the iteration count (k)

1

1

k k ki i

k k ki i iV V V

ki

ki

P

Q


33

• Calculate |V| and δ from the following equation

• If

• stop the iteration otherwise increase the iteration count (k)

ki

ki

P

Q

FAST DECOUPLED METHOD J2 & J3 of Jacobian matrix are zero

1

4

1

4

'

''

1'

1''

0

0

i

i

JP

VJQ

PP J

QQ J V V

V

PB

V

QB V

V

PB

V

QV B

V


34

1

4

1

4

'

''

1'

1''

0

0

i

i

JP

VJQ

PP J

QQ J V V

V

PB

V

QB V

V

PB

V

QV B

V

1

1

k k ki i

k k ki i iV V V

This method requires more iterations than NR

method but less time per iteration

It is useful for in contingency analysis


35

This method requires more iterations than NR

method but less time per iteration

It is useful for in contingency analysis

COMPARISION BETWEEN ITERATIVEMETHODS

Gauss – Seidal Method1. Computer memory requirement is less.2. Computation time per iteration is less.3. It requires less number of arithmetic operations to

complete an iteration and ease in programming.4. No. of iterations are more for convergence and rate of

convergence is slow (linear convergencecharacteristic.

5. No. of iterations increases with the increase of no. ofbuses.


36

Gauss – Seidal Method1. Computer memory requirement is less.2. Computation time per iteration is less.3. It requires less number of arithmetic operations to

complete an iteration and ease in programming.4. No. of iterations are more for convergence and rate of

convergence is slow (linear convergencecharacteristic.

5. No. of iterations increases with the increase of no. ofbuses.

NEWTON – RAPHSON METHOD

Superior convergence because of quadratic convergence. It has an 1:8 iteration ratio compared to GS method. More accurate. Smaller no. of iterations and used for large size systems. It is faster and no. of iterations is independent of the no. of buses. Technique is difficult and calculations involved in each iteration are

more and thus computation time per iteration is large. Computer memory requirement is large, as the elements of jacobian

matrix are to be computed in each iteration. Programming logic is more complex.


37

Superior convergence because of quadratic convergence. It has an 1:8 iteration ratio compared to GS method. More accurate. Smaller no. of iterations and used for large size systems. It is faster and no. of iterations is independent of the no. of buses. Technique is difficult and calculations involved in each iteration are

more and thus computation time per iteration is large. Computer memory requirement is large, as the elements of jacobian

matrix are to be computed in each iteration. Programming logic is more complex.

FAST DECOUPLED METHOD

It is simple and computationally efficient. Storage of jacobian matrix elements are60% of NR

method computation time per iteration is less. Convergence is geometric,2 to 5 iterations required for

accurate solutions Speed for iterations is 5 times that of NR method and 2-3

times of GS method


38

It is simple and computationally efficient. Storage of jacobian matrix elements are60% of NR

method computation time per iteration is less. Convergence is geometric,2 to 5 iterations required for

accurate solutions Speed for iterations is 5 times that of NR method and 2-3

times of GS method

UNIT III

FAULT ANALYSIS-BALANCED FAULT


39

Need for fault analysis To determine the magnitude of fault current throughout

the power system after fault occurs. To select the ratings for fuses, breakers and switchgear. To check the MVA ratings of the existing circuit breakers

when new generators are added into a system.


40

Need for fault analysis To determine the magnitude of fault current throughout

the power system after fault occurs. To select the ratings for fuses, breakers and switchgear. To check the MVA ratings of the existing circuit breakers

when new generators are added into a system.

BALANCED THREE PHASE FAULT

All the three phases are short circuited to each other and to earth. Voltages and currents of the system balanced after the symmetrical fault

occurred. It is enough to consider any one phase for analysis.

SHORT CIRCUIT CAPACITY

It is the product of magnitudes of the prefault voltageand the post fault current.

It is used to determine the dimension of a bus bar and theinterrupting capacity of a circuit breaker.


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BALANCED THREE PHASE FAULT

All the three phases are short circuited to each other and to earth. Voltages and currents of the system balanced after the symmetrical fault

occurred. It is enough to consider any one phase for analysis.

SHORT CIRCUIT CAPACITY

It is the product of magnitudes of the prefault voltageand the post fault current.

It is used to determine the dimension of a bus bar and theinterrupting capacity of a circuit breaker.

Short Circuit Capacity (SCC)0

2

,1

1.

,3

3.

6

3

6,

/

*10

3 * *10

F

TF

T

bT

T T p u

b

T p u

f

L b

SCC V I

VI

Z

SVSCC MVA

Z Z

SSCC MVA

Z

SCCI

V


42

0

2

,1

1.

,3

3.

6

3

6,

/

*10

3 * *10

F

TF

T

bT

T T p u

b

T p u

f

L b

SCC V I

VI

Z

SVSCC MVA

Z Z

SSCC MVA

Z

SCCI

V

Procedure for calculating short circuit capacityand fault current

Draw a single line diagram and select common base SbMVA and kV

Draw the reactance diagram and calculate the total p.uimpedance from the fault point to source (Theveninimpedance ZT)

Determine SCC and If


43

Draw a single line diagram and select common base SbMVA and kV

Draw the reactance diagram and calculate the total p.uimpedance from the fault point to source (Theveninimpedance ZT)

Determine SCC and If

ALGORITHM FOR SHORT CIRCUIT ANALYSISUSING BUS IMPEDANCE MATRIX

• Consider a n bus network. Assume that three phase faultis applied at bus k through a fault impedance zf

• Prefault voltages at all the buses are

• Draw the Thevenin equivalent circuit i.e Zeroing all voltage sourcesand add voltage source at faulted bus k and draw the reactancediagram

1

2

(0)

(0)

.(0)

(0)

.

(0)

busk

n

V

V

VV

V


44

• Consider a n bus network. Assume that three phase faultis applied at bus k through a fault impedance zf

• Prefault voltages at all the buses are

• Draw the Thevenin equivalent circuit i.e Zeroing all voltage sourcesand add voltage source at faulted bus k and draw the reactancediagram

1

2

(0)

(0)

.(0)

(0)

.

(0)

busk

n

V

V

VV

V

(0)kV

• The change in bus voltage due to fault is

• The bus voltages during the fault is

• The current entering into all the buses is zero.the current enteringinto faulted bus k is –ve of the current leaving the bus k

1

.

.

.

busk

n

V

VV

V

( ) (0)bus bus busV F V V


45

• The change in bus voltage due to fault is

• The bus voltages during the fault is

• The current entering into all the buses is zero.the current enteringinto faulted bus k is –ve of the current leaving the bus k

( ) (0)bus bus busV F V V

11 1 1

1

1

. . 0

. . . . . .

. . ( )

. . . . . .

. . 0

( ) (0) ( )

( ) ( )

(0)( )

( ) (0) ( )

bus bus bus

k n

bus k kk kn k

n nk nn

k k kk k

k f k

kk

kk f

i i ik k

V Z I

Z Z Z

V Z Z Z I F

Z Z Z

V F V Z I F

V F Z I F

VI F

Z Z

V F V Z I F


46

11 1 1

1

1

. . 0

. . . . . .

. . ( )

. . . . . .

. . 0

( ) (0) ( )

( ) ( )

(0)( )

( ) (0) ( )

bus bus bus

k n

bus k kk kn k

n nk nn

k k kk k

k f k

kk

kk f

i i ik k

V Z I

Z Z Z

V Z Z Z I F

Z Z Z

V F V Z I F

V F Z I F

VI F

Z Z

V F V Z I F

UNIT IV

FAULT ANALYSIS – UNBALANCEDFAULTS


47

FAULT ANALYSIS – UNBALANCEDFAULTS

INTRODUCTION

UNSYMMETRICAL FAULTSo One or two phases are involvedo Voltages and currents become unbalanced and each phase is to be

treated individuallyo The various types of faults are

Shunt type faults1.Line to Ground fault (LG)2. Line to Line fault (LL)3. Line to Line to Ground fault (LLG)

Series type faultsOpen conductor fault (one or two conductor open fault)


48

UNSYMMETRICAL FAULTSo One or two phases are involvedo Voltages and currents become unbalanced and each phase is to be

treated individuallyo The various types of faults are

Shunt type faults1.Line to Ground fault (LG)2. Line to Line fault (LL)3. Line to Line to Ground fault (LLG)

Series type faultsOpen conductor fault (one or two conductor open fault)

FUNDAMENTALS OF SYMMETRICALCOMPONENTS

Symmetrical components can be used to transformthree phase unbalanced voltages and currents tobalanced voltages and currents

Three phase unbalanced phasors can be resolved intofollowing three sequences1.Positive sequence components2. Negative sequence components3. Zero sequence components


49

Symmetrical components can be used to transformthree phase unbalanced voltages and currents tobalanced voltages and currents

Three phase unbalanced phasors can be resolved intofollowing three sequences1.Positive sequence components2. Negative sequence components3. Zero sequence components

Positive sequence componentsThree phasors with equal magnitudes, equally displaced from oneanother by 120o and phase sequence is same as that of originalphasors.

Negative sequence componentsThree phasors with equal magnitudes, equally displaced from oneanother by 120o and phase sequence is opposite to that of originalphasors.

Zero sequence componentsThree phasors with equal magnitudes and displaced from one anotherby 0o

1 1 1, ,a b cV V V

2 2 2, ,a b cV V V


50

Positive sequence componentsThree phasors with equal magnitudes, equally displaced from oneanother by 120o and phase sequence is same as that of originalphasors.

Negative sequence componentsThree phasors with equal magnitudes, equally displaced from oneanother by 120o and phase sequence is opposite to that of originalphasors.

Zero sequence componentsThree phasors with equal magnitudes and displaced from one anotherby 0o

2 2 2, ,a b cV V V

0 0 0, ,a b cV V V

RELATIONSHIP BETWEEN UNBALANCED VECTORSAND SYMMETRICAL COMPONENTS

0 1 2

0 1 2

0 1 2

0

21

22

2

2

1 1 1

1

1

1 1 1

1

1

a a a a

b b b b

c c c c

a a

b a

c a

V V V V

V V V V

V V V V

V V

V a a V

a aV V

A a a

a a


51

0 1 2

0 1 2

0 1 2

0

21

22

2

2

1 1 1

1

1

1 1 1

1

1

a a a a

b b b b

c c c c

a a

b a

c a

V V V V

V V V V

V V V V

V V

V a a V

a aV V

A a a

a a

Similarly we can obtain for currents also

SEQUENCE IMPEDANCE

Impedances offered by power system components to positive,negative and zero sequence currents.

Positive sequence impedanceThe impedance of a component when positive sequencecurrents alone are flowing.

Negative sequence impedanceThe impedance of a component when negative sequencecurrents alone are flowing.

Zero sequence impedanceThe impedance of a component when zero sequence currentsalone are flowing.


52

Impedances offered by power system components to positive,negative and zero sequence currents.

Positive sequence impedanceThe impedance of a component when positive sequencecurrents alone are flowing.

Negative sequence impedanceThe impedance of a component when negative sequencecurrents alone are flowing.

Zero sequence impedanceThe impedance of a component when zero sequence currentsalone are flowing.

SEQUENCE NETWORK

SEQUENCE NETWORK FOR GENERATOR


53

positive sequence network negative sequence network Zero sequence network

SEQUENCE NETWORK FORTRANSMISSION LINE


54


SEQUENCE NETWORK FORTRANSFORMER


55


SEQUENCE NETWORK FOR LOAD


56


SINGLE LINE TO GROUND FAULT

b

c

fa a

a1 a2 a0 a

a11 2 0

I 0

I 0

V Z I

I I I I /3

I3

af

E

Z Z Z Z


57

b

c

fa a

a1 a2 a0 a

a11 2 0

I 0

I 0

V Z I

I I I I /3

I3

af

E

Z Z Z Z

Consider a fault between phase a and

ground through an impedance zf

LINE TO LINE (LL) FAULT

a2 a1

a0

fa1 a2 a1

a11 2

1 2

I 0

I I

V V I

I I

I 0

V V Z I

I3

I I3

a

c b

fb c b

af

ab c f

Z

E

Z Z Z

jE

Z Z Z


58

a2 a1

a0

fa1 a2 a1

a11 2

1 2

I 0

I I

V V I

I I

I 0

V V Z I

I3

I I3

a

c b

fb c b

af

ab c f

Z

E

Z Z Z

jE

Z Z Z

Consider a fault between phase b and cthrough an impedance zf

DOUBLE LINE TO GROUND (LLG) FAULT

a0

a1 a2 a0

fa0

fa0 a1 a0

a11 2 0 2 0

I 0

I I I 0

V=V (I I ) 3Z I

V V V 3Z I

I( 3 )/( 3 )

fb c b c

b

af f

Z

E

Z Z Z Z Z Z Z


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a0

a1 a2 a0

fa0

fa0 a1 a0

a11 2 0 2 0

I 0

I I I 0

V=V (I I ) 3Z I

V V V 3Z I

I( 3 )/( 3 )

fb c b c

b

af f

Z

E

Z Z Z Z Z Z Z

Consider a fault between phase b and cthrough an impedance zf to ground

UNBALANCED FAULT ANALYSIS USINGBUS IMPEDANCE MATRIX

SINGLE LINE TO GROUND FAULT USING Zbus

Consider a fault between phase a and ground throughan impedance zf at bus k

For a fault at bus k the symmetricalcomponents of fault current


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For a fault at bus k the symmetricalcomponents of fault current

0 1 21 2 0

V (0)I I I

3k

k k k fkk kk kkZ Z Z Z

LINE TO LINE (LL) FAULT

Consider a fault between phase b and c through an impedance zf

0

1 21 2

I 0

V (0)I I

k

kk k f

kk kkZ Z Z


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0

1 21 2

I 0

V (0)I I

k

kk k f

kk kkZ Z Z

DOUBLE LINE TO GROUND (LLG) FAULT

Consider a fault between phase b and c through an impedance zf to ground

12 0

12 0

1 12

2

1 10

0

V (0)I

( 3 )3

V (0) II

V (0) II

3

I ( ) I I

kk f

kk kkkk f

kk kk

k kk kk

kk

k kk kk f

kk

b ck k k

Z Z ZZ

Z Z Z

Z

Z

Z

Z Z

F


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12 0

12 0

1 12

2

1 10

0

V (0)I

( 3 )3

V (0) II

V (0) II

3

I ( ) I I

kk f

kk kkkk f

kk kk

k kk kk

kk

k kk kk f

kk

b ck k k

Z Z ZZ

Z Z Z

Z

Z

Z

Z Z

F

BUS VOLTAGES AND LINE CURRENTSDURING FAULT

0 0 0

1 0 1 1

2 2 2

0 00

0

1 11

1

2 22

2

( ) 0 I

( ) (0) I

( ) 0 I

( ) ( )I

( ) ( )I

( ) ( )I

i ik k

i i ik k

i ik k

i jij

ij

i jij

ij

i jij

ij

V F Z

V F V Z

V F Z

V F V F

Z

V F V F

Z

V F V F

Z


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0 0 0

1 0 1 1

2 2 2

0 00

0

1 11

1

2 22

2

( ) 0 I

( ) (0) I

( ) 0 I

( ) ( )I

( ) ( )I

( ) ( )I

i ik k

i i ik k

i ik k

i jij

ij

i jij

ij

i jij

ij

V F Z

V F V Z

V F Z

V F V F

Z

V F V F

Z

V F V F

Z

UNIT V

STABILITY ANALYSIS


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UNIT V

STABILITY ANALYSIS

STABILITY The tendency of a power system to develop restoring

forces equal to or greater than the disturbing forces tomaintain the state of equilibrium.

Ability to keep the machines in synchronism with anothermachine


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STABILITY The tendency of a power system to develop restoring

forces equal to or greater than the disturbing forces tomaintain the state of equilibrium.

Ability to keep the machines in synchronism with anothermachine

CLASSIFICATION OF STABILITY

Steady state stabilityAbility of the power system to regain synchronism after small and

slow disturbances (like gradual power changes)

Dynamic stabilityAbility of the power system to regain synchronism after smalldisturbances occurring for a long time (like changes in turbinespeed, change in load)

Transient stabilityThis concern with sudden and large changes in the networkconditions i.e. . sudden changes in application or removal of loads,line switching operating operations, line faults, or loss of excitation.


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Steady state stabilityAbility of the power system to regain synchronism after small and

slow disturbances (like gradual power changes)

Dynamic stabilityAbility of the power system to regain synchronism after smalldisturbances occurring for a long time (like changes in turbinespeed, change in load)

Transient stabilityThis concern with sudden and large changes in the networkconditions i.e. . sudden changes in application or removal of loads,line switching operating operations, line faults, or loss of excitation.

Steady state limit is the maximum power that can betransferred without the system become unstable whenthe load in increased gradually under steady stateconditions.

Transient limit is the maximum power that can betransferred without the system becoming unstable whena sudden or large disturbance occurs.


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Steady state limit is the maximum power that can betransferred without the system become unstable whenthe load in increased gradually under steady stateconditions.

Transient limit is the maximum power that can betransferred without the system becoming unstable whena sudden or large disturbance occurs.

Swing Equation for Single Machine InfiniteBus System

• The equation governing the motion of the rotor of asynchronous machine

whereJ=The total moment of inertia of the rotor(kg-m2)

=Singular displacement of the rotorTm=Mechanical torque (N-m)Te=Net electrical torque (N-m)Ta=Net accelerating torque (N-m)

2

2m

a m e

dJ T T T

dt


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• The equation governing the motion of the rotor of asynchronous machine

whereJ=The total moment of inertia of the rotor(kg-m2)

=Singular displacement of the rotorTm=Mechanical torque (N-m)Te=Net electrical torque (N-m)Ta=Net accelerating torque (N-m)

m

• Where pm is the shaft power input to the machinepe is the electrical powerpa is the accelerating power

2 2

2 2

2

2

m sm m

m msm

m m

mm a m e

t

d d

dt dt

d d

dt dt

dJ p p p

dt


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• Where pm is the shaft power input to the machinepe is the electrical powerpa is the accelerating power

2 2

2 2

2

2

m sm m

m msm

m m

mm a m e

t

d d

dt dt

d d

dt dt

dJ p p p

dt

2

2

2

2

2

2

2

20

20 0

2 max2

20

2

2

2

2

2

sin

m

ma m e

machinesm

m a m e

sm machine machine

a m es

s

a m e

m a

a

J M

dM p p p

dtH

M S

d p p pH

dt S S

H dp p p

dt

f

H dp p p

f dt

f fdp p p

dt H Hd

dt

fd dp

dt H dt

H=machine inertia constant


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2

2

2

2

2

2

2

20

20 0

2 max2

20

2

2

2

2

2

sin

m

ma m e

machinesm

m a m e

sm machine machine

a m es

s

a m e

m a

a

J M

dM p p p

dtH

M S

d p p pH

dt S S

H dp p p

dt

f

H dp p p

f dt

f fdp p p

dt H Hd

dt

fd dp

dt H dt

δ and ωs are in electricalradian

p.u

p.u

Swing Equation for Multimachine System

2

2

systema m e

machinesystem machine

system

H dp p p

f dt

SH H

S

p.u

m a ch in eS =machine rating(base)

systemS =system base


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2

2

systema m e

machinesystem machine

system

H dp p p

f dt

SH H

S

p.u

Rotor Angle Stability

• It is the ability of interconnected synchronous machinesof a power system to maintain in synchronism. Thestability problem involves the study of the electromechanical oscillations inherent in power system.

• Types of Rotor Angle Stability1. Small Signal Stability (or) Steady State Stability2. Transient stability


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• It is the ability of interconnected synchronous machinesof a power system to maintain in synchronism. Thestability problem involves the study of the electromechanical oscillations inherent in power system.

• Types of Rotor Angle Stability1. Small Signal Stability (or) Steady State Stability2. Transient stability

Voltage Stability

It is the ability of a power system to maintain steadyacceptable voltages at all buses in the system undernormal operating conditions and after being subjected toa disturbance.

The major factor for instability is the inability of thepower system to meet the demand for reactive power.


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It is the ability of a power system to maintain steadyacceptable voltages at all buses in the system undernormal operating conditions and after being subjected toa disturbance.

The major factor for instability is the inability of thepower system to meet the demand for reactive power.

• Mid Term StabilityIt represents transition between short term and longterm responses.Typical ranges of time periods.1. Short term : 0 to 10s2. Mid Term : 10 to few minutes

3. Long Term : a few minutes to 10’s of minutes• Long Term Stability

Usually these problem be associated with1. Inadequacies in equipment responses.2. Poor co-ordination of control and protection equipment.3. Insufficient active/reactive power reserves.


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• Mid Term StabilityIt represents transition between short term and longterm responses.Typical ranges of time periods.1. Short term : 0 to 10s2. Mid Term : 10 to few minutes

3. Long Term : a few minutes to 10’s of minutes• Long Term Stability

Usually these problem be associated with1. Inadequacies in equipment responses.2. Poor co-ordination of control and protection equipment.3. Insufficient active/reactive power reserves.

Equal Area Criterion

• This is a simple graphical method to predict the transientstability of two machine system or a single machineagainst infinite bus. This criterion does not require swingequation or solution of swing equation to determine thestability condition.

• The stability conditions are determined by equating theareas of segments on power angle diagram.


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• This is a simple graphical method to predict the transientstability of two machine system or a single machineagainst infinite bus. This criterion does not require swingequation or solution of swing equation to determine thestability condition.

• The stability conditions are determined by equating theareas of segments on power angle diagram.

Power-angle curve for equal area criterion


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Power-angle curve for equal area criterion

multiplying swing equation by on both sides

Multiplying both sides of the above equation by dt and then integrating between two arbitrary angles δ0 and δc

Once a fault occurs, the machine starts accelerating. Once the fault is cleared, themachine keeps on accelerating before it reaches its peak at δc ,

The area of accelerating A1


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The area of deceleration is given by A2

If the two areas are equal, i.e., A1 = A2, then the power system will be stable

Critical Clearing Angle (δcr) maximum allowable value of the clearing time and angle forthe system to remain stable are known as critical clearing time and angle.

δcr expression can be obtained by substituting δc = δcr in the equation A1 = A2

Substituting Pe = 0 in swing equation


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Substituting Pe = 0 in swing equation

Integrating the above equation

Replacing δ by δcr and t by tcr in the above equation, we getthe critical clearing time as


79

Factors Affecting Transient Stability

• Strength of the transmission network within the systemand of the tie lines to adjacent systems.

• The characteristics of generating units including inertia ofrotating parts and electrical properties such as transientreactance and magnetic saturation characteristics of thestator and rotor.

• Speed with which the faulted lines or equipments can bedisconnected.


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• Strength of the transmission network within the systemand of the tie lines to adjacent systems.

• The characteristics of generating units including inertia ofrotating parts and electrical properties such as transientreactance and magnetic saturation characteristics of thestator and rotor.

• Speed with which the faulted lines or equipments can bedisconnected.

Numerical Integration methods

Modified Euler’s method

Runge-Kutta method


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MODIFIED EULER’S METHOD

• Using first derivative of the initial point next point is obtained• the step

• Using this x1p dx/dt at x1

p=f(t1, x1p)

• Corrected value is

1 0p dX

X X tdt

1 0t t t


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0 1

1

1 0

1

2

2

p

pi i

X XP

X Xci i

d x d x

d t d tX X t

d x d x

d t d tX X t

Numerical Solution of the swing equation

• Input power pm=constant• At steady state pe=pm,

• At synchronous speed

10

1 m ax

'

1 m ax1

s in mp

p

E Vp

X


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• Input power pm=constant• At steady state pe=pm,

• At synchronous speed

10

1 m ax

'

1 m ax1

s in mp

p

E Vp

X

0

'

2 max2

0

E Vp

X

2

20

20 0

2max2

20

2

sin

a m e

m a

a

H dp p p

f dt

f fdp p p

dt H Hd

dt

fd dp

dt H dt

The swing equation


84

2

20

20 0

2max2

20

2

sin

a m e

m a

a

H dp p p

f dt

f fdp p p

dt H Hd

dt

fd dp

dt H dt

Applying Modified Eulers method to above equation

1 0t t t

1

1

i

i

pi i

pi i

dt

dt

dt

dt

• The derivatives at the end of interval

1

11

1

0

pi

ppi

i

pi

a

d

dt

fdp

dt H

The corrected value


85

1

1

1

1

2

2

pi i

pi i

ci i

ci i

d d

dt dtt

d d

dt dtt

Runge-Kutta Method• Obtain a load flow solution for pretransient conditions• Calculate the generator internal voltages behind transient reactance.• Assume the occurrence of a fault and calculate the reduced admittance matrix• Initialize time count K=0,J=0• Determine the eight constants

1 1

1 2

1 12 1

1 12 2

2 23 1

2 23 2

3 34 1

3 34 2

1 2 3 4

( , )

( , )

( , )2 2

( , )2 2

( , )2 2

( , )2 2

( , )2 2

( , )2 2

2 2

k k k

k k k

k kk k k

k kk k k

k kk k k

k kk k k

k kk k k

k kk k k

k k k

k

K f t

l f t

K lK f t

K ll f t

K lK f t

K ll f t

K lK f t

K ll f t

K K K K

1 2 3 4

6

2 2

6

k

k k k k

kl l l l


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1 1

1 2

1 12 1

1 12 2

2 23 1

2 23 2

3 34 1

3 34 2

1 2 3 4

( , )

( , )

( , )2 2

( , )2 2

( , )2 2

( , )2 2

( , )2 2

( , )2 2

2 2

k k k

k k k

k kk k k

k kk k k

k kk k k

k kk k k

k kk k k

k kk k k

k k k

k

K f t

l f t

K lK f t

K ll f t

K lK f t

K ll f t

K lK f t

K ll f t

K K K K

1 2 3 4

6

2 2

6

k

k k k k

kl l l l

• Compute the change in state vector

• Evaluate the new state vector

• Evaluate the internal voltage behind transient reactance using the relation

• Check if t<tc yes K=K+1• Check if j=0,yes modify the network data and obtain the new reduced

admittance matrix and set j=j+1• set K=K+1• Check if K<Kmax, yes start from finding 8 constants

1

1

k k k

k k k

1 2 3 4

1 2 3 4

2 2

6

2 2

6

k k k k

k

k k k k

k

K K K K

l l l l

1 1 1cos sink k k k kp p p p pE E j E


87

• Compute the change in state vector

• Evaluate the new state vector

• Evaluate the internal voltage behind transient reactance using the relation

• Check if t<tc yes K=K+1• Check if j=0,yes modify the network data and obtain the new reduced

admittance matrix and set j=j+1• set K=K+1• Check if K<Kmax, yes start from finding 8 constants

EE2351 POWER SYSTEM ANALYSIS

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