Power System Dynamics [1 ed.] 9781781830451, 9781906574840

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POWER SYSTEM DYNAMICS

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POWER SYSTEM DYNAMICS RAU V GURUPRASADA Ex-Professor Department of Electrical Engineering Indian Institute of Technology, Kharagpur India

New Academic Science Limited NEW ACADEMIC SCIENCE

The Control Centre, 11 A Little Mount Sion Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk e-mail: [email protected]

Copyright © 2013 by New Academic Science Limited The Control Centre, 11 A Little Mount Sion, Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk • e-mail: [email protected]

ISBN : 978 1 781830 45 1 All rights reserved. No part of this book may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the copyright owner. British Library Cataloguing in Publication Data A Catalogue record for this book is available from the British Library Every effort has been made to make the book error free. However, the author and publisher have no warranty of any kind, expressed or implied, with regard to the documentation contained in this book.

Dedicated to My Dear Wife Mrs. V. Lakshmi Rau

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Preface The importance of the subject area “Power System Dynamics” cannot be over emphasised for an electrical engineering student. It was nine decades ago that electrical engineers first faced the problem of power system stability. Since then, power system dynamical performance under various disturbance conditions has been an active area of research and numerous papers and conference reports on the subject are published each year. The number of books dealing with the difficult subject area of power system dynamics are rather few especially for classroom teaching. The present book on ‘Power System Dynamics’ has primarily grown out of the courses offered by me to M.Tech. and B.Tech. (Hons.) students for several years at IIT, Kharagpur. The book aims at providing comprehensive and indepth information in the subject area of power system dynamics. The student is expected to be familiar with the basics of control systems, power systems and electrical machines especially synchronous machines. The layout of the lessons is intended to give the reader a capability to understand various important aspects of the power system dynamics. The lessons cover power system modelling, application of frequency domain and time domain techniques and Lyapunov direct method in detail for the analysis of dynamical behaviour of power system subjected to small and large disturbances. The mathematical treatment has been kept as simple as possible. To aid the student to comprehend the difficult subject area, several illustrative examples are presented wherever necessary. In addition, a number of exercise problems with a few typical solved examples, have been compiled and provided for the practice of students. While providing technical information, the author also confesses to an attempt to stimulate fresh thinking on the subject in the minds of the students and sincerely hopes he has succeeded in this pleasant task. RAU V GURUPRASADA

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Acknowledgement While the task of making my acknowledgement is a pleasant one, it is also somewhat daunting as it is difficult to do justice to all the individuals who have helped and encouraged me in one way or another, directly or indirectly. My debt to my alma mater. The Indian Institute of Technology, Kharagpur is deep. I am thankful to the authors of research papers, books and monographs, which I consulted for preparing my lectures, out of which the present book has evolved. I am thankful to numerous students who have taken my courses over the years and provided constructive and unbiased feed back. I must particularly thank my former students Dr. Suresh Jangamshetti and Mr. Debajyoti Roy who have helped in the computerisation of the manuscript of this book in the early stages. I must thank my dear son, Mr. Anandaswarup Vadapalli, a busy software engineer who has thoroughly proof read the manuscript, assisted in compiling the exercise problems and word processed the manuscript to its final form. I am grateful to friends and close relatives particularly my father-in-law and elder sister and brothers who encouraged and provided inspiration during the writing of the book. I must specially thank my wife Mrs V. Lakshmi Rau who extended help at various stages of book preparation. RAU V GURUPRASADA

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Contents Preface Acknowledgement 1.

FUNDAMENTAL CONCEPTS OF ELECTRIC POWER SYSTEMS

1.1 1.2 1.3 1.4 1.5 1.6

2.

POWER SYSTEMS UNDER SMALL DISTURBANCE

2.1 2.2 2.3 2.4 2.5 3.

Introduction Power System Configuration Components, Power Angle Relationships Synchronous Generator Power Relations Loads System Model Representation References Transient Processes Definitions of Stability, Stability Limit Steady State/Dynamic Stability, Transient Stability—Definitions Power System Dynamics under Small Disturbance, Simple Stability Criteria — Direct and Indirect Indirect Stability Criteria for Simple System References

DYNAMICAL ANALYSIS OF SIMPLE POWER SYSTEMS

3.1 3.2

Swing Equation of Single Machine — Infinite Bus System with No Control Equipment Dynamical Analysis of a Simple Power System subjected to Small Disturbances References

vii ix 1–7

1 2 2 3 5 6 7 9–12

9 9 10 10 11 12 13–20

13 14 20

Contents

xii 4.

DYNAMIC STABILITY ANALYSIS OF UNCONTROLLED POWER SYSTEM INCLUDING FIELD TRANSIENT

4.1 4.2 4.3

5.

SELF-EXCITATION PHENOMENA

5.1 5.2

6.

Linearisation and Derivation of Characteristic Equation Hurwitz’s Criterion: Testing for Instabilities like Aperiodic Instability, Self-excitation and Self-oscillations References

BLOCK DIAGRAM REPRESENTATION AND EFFECT OF EXCITATION CONTROL ON ELECTRIC POWER

8.1 8.2

9.

Basic Structure of a Regulator Voltage Regulator Speed Governing Systems References

DYNAMIC STABILITY ANALYSIS OF A POWER SYSTEM WITH CONTROLLED EXCITATION

7.1 7.2

8.

Parameters of Self-excitation Asynchronous and Synchronous Self-excitation References

VOLTAGE AND SPEED REGULATION SYSTEMS

6.1 6.2 6.3 7.

Linearised Dynamical Equations for Simple Power System with Field Flux Decay Operating Matrix Equation and Characteristic Equation of the Power System Routh’s Criterion for Stability Analysis References

Block Diagram Representation of a Linearised Power System Model Effect of Excitation Regulation on Electrical Power Output and Stability References

DYNAMIC STABILITY BOUNDARIES FOR A POWER SYSTEM USING D-PARTITION METHOD

9.1 9.2

Dynamic Stability Boundaries in Parameter Plane Basis of D-partition Method, Neimark’s Hatching Rule and its Application to Power System References

21–28

22 25 25 28 29–33

29 31 33 35–39

35 36 37 39 41–51

41 47 51 53–58

53 55 58 59–67

59 60 67

Contents 10. TIME DOMAIN FORMULATION OF DYNAMIC STABILITY PROBLEM FOR A SIMPLE POWER SYSTEM

xiii

.

10.1 Uncontrolled System 10.2 System with Regulating Equipment References 11.

STATE-SPACE FORMULATION OF DYNAMIC STABILITY PROBLEM FOR A MULTI-MACHINE POWER SYSTEM

11.1 Multi-machine Power Systems References 12. TRANSIENT STABILITY AND CONCEPTS

12.1 Transient Stability 12.2 Dynamic Modelling of a Synchronous Generator under Transient Conditions References 13. TRANSIENT STABILITY STUDY

13.1 A Typical Illustrative Example References 14. MODELLING AND METHODS OF TRANSIENT STABILITY ANALYSIS

14.1 Transient Stability Analysis Methods—Direct and Indirect 14.2 Modelling of the Power System for Analysis References 15. DETAILED INDIRECT TRANSIENT STABILITY METHODS

15.1 Analogue Simulation of Power System 15.2 Digital Simulation of Power System References 16. DIRECT METHOD FOR TRANSIENT STABILITY ASSESSMENT

16.1 Direct Method for Transient Stability Assessment using Equal Area Criterion 16.2 Illustrative Examples for Various Types of Faults References 17. APPLICATION OF EQUAL AREA CRITERION TO MULTI-MACHINE POWER SYSTEM TRANSIENT STABILITY ASSESSMENT

17.1 Multi-machine Stability Problems Reference

69–78

69 70 78 79–83

79 83 85–90

85 86 90 91–94

91 94 95–98

95 96 98 99–103

99 100 103 105–111

105 106 111 113–115

113 115

Contents

xiv 18. LYAPUNOV DIRECT METHOD

18.1 Concept of Lyapunov Stability, Asymptotic Stability and Instability 18.2 Lyapunov’s Symptotic Stability Theorem References 19. METHODS OF CONSTRUCTION OF LYAPUNOV FUNCTIONS

19.1 Sylvester’s Criteria for Definiteness and Semi-Definiteness 19.2 Construction of Lyapunov Functions for a Power System References 20. GENERATION OF LYAPUNOV FUNCTIONS

20.1 20.2 20.3 20.4

Energy Metric Algorithm for Generation of Lyapunov Functions Application to Power System Evaluation of Region of Asymptotic Stability Application of Lyapunov Method to Linear System References

21. EXERCISE PROBLEMS

21.1 Problems with Solutions 21.2 Unsolved Problems References

117–124

117 120 124 125–134

125 127 134 135–141

135 136 137 139 141 143–159

143 151 159

Fundamental Concepts of Electric Power Systems

1

Chapter 1

Fundamental Concepts of Electric Power Systems z z z z z z

1.1

Introduction Power system configuration Components, power angle relationships Synchronous generator power relations Loads System model representation

INTRODUCTION

Basic objective of a Power System is to supply electrical energy to various loads throughout a given service area at constant frequency, voltage, high reliability meeting real and reactive power variations. Electrical Power System Consists of : I. Power elements which generate, transform, distribute and consume the electrical energy. These are: z Generators z Transformers, Rectifiers, Inverters z Power Transmission Lines, Distribution Networks z Loads II. Control elements which change the operating condition of the power system. These are: z Excitation Regulators z Speed Governors z Relays z Circuit Breakers z Supplementary Controllers For steady state and dynamical performance analysis of a power system, two types of power system configuration are chosen: (a) Simple and (b) Complex.

Power System Dynamics

2

1.2

POWER SYSTEM CONFIGURATION Simple This simple configuration is also called single machine infinite bus system. z

Complex In a complex configuration as shown in figure below several machines are connected to an infinite bus. z

1.3

COMPONENTS, POWER ANGLE RELATIONSHIPS

Taking the simple configuration we will examine some of the fundamental features like power transmission capacity of an electrical power system. Static Transmission Capacity We consider a transmission link i–j and examine its load capacity I=

Vi – Vj

Bus i

Z

Bus j I

Line Powers

Vi

Z = R + jX Sij = Pij + jQij

Vj

Sij = Pij + jQij = Vi I * Sij = Vi

Vi* – V j* Z*

2

=

Vi – Vi V j ε jδ R – jX

S ji = Pji + jQ ji = V j ( – I ) *

Vi

2

V j* – Vi* V j – Vi V j ε jδ S ji = V j = R – jX Z*

Vj

Fundamental Concepts of Electric Power Systems

3

Rationalising and Separation Yields

Pij = Qij = Pji =

Qji =

1 2

2

2

2

2

2

2

2

R +X 1 R +X 1 R +X 1 R +X

(R V (X V

2

2

i

(R V

(

2 j

X Vj

) sin δ )

– R Vi V j cos δ + X Vi V j sin δ

i

2

– X Vi V j cos δ – R Vi V j

– R Vi V j cos δ – X Vi V j sin δ

)

– X Vi V j cos δ + R Vi V j sin δ

)

For R = 0, Pij = – Pji =

Vi V j X

sin δ

If Vi and Vj are held constant,

Pij = Pm sin δ; Pm = dPij dδ

1.4

Vi V j X

(Static transmission capacity or static stability limit)

= Pm cos δ (Electrical stiffness or synchronizing coefficient)

SYNCHRONOUS GENERATOR POWER RELATIONS

Let us now look at generator portion of the power system. Consider the m/c phasor diagram:

SG = PG + jQG = V I cos δ + j V I sin δ

E − I d X d = V cos δ and I q X q = V sin δ

I q = I sin ψ and I d = I cos ψ

Power System Dynamics

4

I cos φ = I q cos δ + I d sin δ PG = VI q cos δ + VI d sin δ PG =

V2 EV V2 sin 2 δ + sin δ − sin 2 δ Xd 2Xq 2Xd

EV V2  1 1  + sin δ – PG =   sin 2δ Xd 2  Xq Xd  If Xd = Xq then PG =

EV sin δ Xd

If E and V are constants, then PG = Pm sin δ The power angle diagram of the machine is shown

Synchronizing Power Coefficient:

dPG = Pm cos δ dδ

Similarly for reactive power: QG

EV V2 cos – δ = Xd Xd

QG > 0 for E cos δ > V (Machine is overexcited) QG < 0 for E cos δ < V (Machine is underexcited)

Fundamental Concepts of Electric Power Systems

5

A synchronous generator can be represented as

1.5

LOADS

Usually in a power system loads constitute equipment like motors, furnaces and lighting etc. Loads can be classified into: 1. Industrial Loads: They have a predictable duty cycle and load constancy. 2. Domestic Loads: These are random in character whose average pattern can be recognized by statistics. z In steady state, although loads are time variant, variations are slow. z A typical load consumes reactive power (IM). z A typical load is symmetric. Consider a typical R-L Load: For this load:

P + jQ = V 2 Y* Y* is the complex conjugate of load admittance

P =

Q =

RV

2

R 2 + ( 2π fL )

2

2π fL V

2

R 2 + ( 2π fL )

2

P = P( f , V

) and

Q = Q( f , V

)

P decreases with increase in f and, Q increases with increase in f.

>0

>0

Power System Dynamics

6

In a power system in operation: z Frequency constitutes a sensitive indicator of real power balance. z Unchanged voltage profile indicates that balance is kept between produced and consumed reactive power. To understand this quantitatively, we consider the system shown

Assumptions: z Vi is kept constant and is the reference. z Z = jX

V j = Vi − IZ = Vi − V j = Vi −

P − jQ jX Vi

X X Q− j P Vi Vi Vi X

Vj

1.6

Vj

–j

X Vi

Vi

Q

P

SYSTEM MODEL REPRESENTATION Looking at the Power System as a whole, in symmetrical steady state: z A synchronous generator can be modeled as a source behind an impedance or reactance. z A transmission line can be modeled as a lumped reactance. z A load can be modeled as a reactance.

Fundamental Concepts of Electric Power Systems Thus a simple model of a power system can be viewed as:

or

For this system P=

Q = where

EV sin δ X EV V2 cos δ − X X

X = X d + Xt + Xl

Qg = Q + I 2 X Qg = Q + Qg =

( E cos δ − V )2 + ( E sin δ )2 X

2

E EV cos δ − X X

REFERENCES z z z z z z

“Electrical Transmission and Distribution” Westinghouse Reference Book, Westinghouse, 1964. S.B. Crary, “Power System Stability”, Vols. I and II, Wiley, New York, 1945, 1947. E.W. Kimbark, “Power System Stability”, Vols. I, II and III, Wiley, New York, 1948, 1950, 1956. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. R.T. Byerly and E.W. Kimbark, “Stability of Large Electric Power Systems”, IEEE Press Book, IEEE, New York, 1974.

7

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Chapter 2

Power Systems Under Small Disturbance z z z z z

2.1

Transient processes Definitions of stability, stability limit Steady state/dynamic stability, transient stability—definitions Power system dynamics under small disturbance, simple stability criteria— direct and indirect Indirect stability criteria for simple system

TRANSIENT PROCESSES

A Power System is always subjected to normal Transient Processes. That is there is always a small variation in the parameters of the system. Considering a simple model, we define two fundamental concepts for a power system.

2.2

DEFINITIONS OF STABILITY, STABILITY LIMIT

Stability It is that attribute of the system or a part of the system which enables it to develop restoring forces between the elements there of equal to or greater than the disturbing forces so as to restore a state of equilibrium.

Power Systems Dynamics

10 Stability Limit

It is the maximum power flow possible through a particular point in the system, when the entire system or part of the system to which stability refers is operating with stability. Depending upon the type of disturbance there are three kinds of stability associated with power system dynamics.

2.3

STEADY STATE/DYNAMIC STABILITY, TRANSIENT STABILITY—DEFINITIONS 1. Steady State or Dynamic Stability: It is the ability of the system to restore to its initial condition after a small disturbance. 2. Transient Stability: It is the ability of the system to restore its initial condition after a large disturbance or go to a condition close to the initial one. 3. Resultant Stability: If after a large disturbance the synchronous operation of the system is violated and restored after an allowable interval of time the system is said to have resultant stability.

2.4

POWER SYSTEM DYNAMICS UNDER SMALL DISTURBANCE, SIMPLE STABILITY CRITERIA—DIRECT AND INDIRECT

Consider Power System Dynamics under small disturbance. We now derive simple criteria for stability. Simple Criteria A small random disturbance causes an imbalance or deviation in generator power (change in E, V, X, Pt ). The deviation of generator power ∆P associated with random disturbance can be defined as:

∂P ∆δ = K1∆δ , ∂δ K1 is the synchronising power coefficient. ∆P =

If K1 = 0, a critical condition. An infinitely small change ∆P results in extremely large values of ∆δ . When K1 > 0, ∆δ is finite, positive. Therefore the System is Stable. When K1 < 0, ∆δ is negative (there is increase in power demand which is not met). Therefore the System is Unstable. From physical considerations, it is clear that the system will return after a small disturbance provided K1 > 0.

Power Systems under Small Disturbance ∂P > 0. ∂δ

∴ The stability criteria is

2.5

11

INDIRECT STABILITY CRITERIA FOR SIMPLE SYSTEM δ

Ecr

E

Operating condition is varied due to change in E with δ

EV sin δ = Const. X

P = Pt = Differentiating P w.r.t. E, we get

dP V EV dδ sin δ + cos δ = dE X X dE

dδ – tan δ = dE E

P being constant,

δ = 90°

It can be seen that at

dδ = −∞ dE dQ g dE

As δ → 90°

=

1 X

V    2E –  cos δ  

dQg dδ

⇒ –∞

∴ Instability in the form of drift will take place.

100%

Power Systems Dynamics

12

REFERENCES z z z z z

S.B. Crary, “Power System Stability”, Vols. I and II, Wiley, New York, 1945, 1947. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002.

Chapter 3

Dynamical Analysis of Simple Power Systems z z z z

3.1

Swing equation of single machine — infinite bus system with no control equipment Dynamical analysis of a simple power system subjected to small disturbances Concept of characteristic equation and criteria for system stability with and without damping Discussion on aperiodic instability, self-oscillations and general principles for dynamic stability analysis

SWING EQUATION OF SINGLE MACHINE—INFINITE BUS SYSTEM WITH NO CONTROL EQUIPMENT

Considering a simple model of the power system (single m/c infinite bus system).

Any difference between ‘Shaft Torque’ and “Electromagnetic Torque” developed must cause acceleration or deceleration of the machine. Ta = Tt − Te Pa = Pt − Pe Pa = Ta ω = I αω = M α

Pa is in mega watts if M is in mega joule-second per electrical degree. Angular acceleration α is electrical degree per second2. In literature H and M are referred to as inertia constants. M=

H πf

H is in units of seconds and M is in sec/electrical radian.

Power System Dynamics

14 If θ is the angular position of rotor

α =

d 2θ

dt 2 since θ is continuously changing with time, it is more convenient to measure angular position with respect to a reference axis that is rotating at synchronous speed. If δ is the angular displacement in electrical degrees from synchronous rotating reference axis, ωs is the synchronous speed in electrical degrees per second. θ = ω st + δ

dθ dδ = ωs + dt dt dδ (Relative velocity of the rotor w.r.t. synchronous reference frame). ω − ωs = dt d 2θ dt 2 ∴

M

d 2δ dt 2

=

d 2δ dt 2

= Pa = Pt − Pe (Swing Equation)

The angle δ for a machine connected to ∞ bus is torque or power angle. It is the difference between the internal angle of the machine and the angle of the synchronously rotating reference frame which in this case is ∞ bus. For a 2-machine system two swing equations are necessary. The angular momentum M may be treated as constant since the speed of the machine does not differ much from synchronous speed unless stability limit is exceeded. If damping is not neglected and the damping power is considered proportional to equation of motion is modified as

d 2δ

dδ , the dt

dδ = Pa = Pt − Pe dt dt where D is the damping constant. M

3.2

2

+D

DYNAMICAL ANALYSIS OF A SIMPLE POWER SYSTEM SUBJECTED TO SMALL DISTURBANCES

No Damping Considering The Power Angle Diagram:

Dynamical Analysis of Simple Power Systems

15

Initial steady state is specified as Pt = Pe0 = Pm sin δ 0

∴

d 2δ0

= Pm sin δ 0 − Pm sin δ 0 dt 2 For a small disturbance which increases

M

...(a)

δ 0 ⇒ δ 0 + ∆δ , (assuming Pt constant)

M

d 2 ( δ 0 + ∆δ ) dt 2

= Pm sin δ 0 − Pm sin (δ 0 + ∆δ )

Expanding sin(δ 0 + ∆δ ) in Taylor series around δ 0

M

d 2δ0 dt

2

+M

d 2 ∆δ dt

2

Pm sin δ 0 − Pm sin δ 0 −

=

dPe ∆δ dδ

Comparing equations (a) and (b), we get ∴

M

d 2 ∆δ dt 2

Now,

= −

K1 =

dPe ∆δ dδ

dPe EV = cos δ dδ X δ =δ

0

The value of K1 depends on initial condition K1 = 0 at δ = 90°

The equation of motion of the generator is

M or

d 2 ∆δ dt 2

( Ms

2

+ K1∆δ = 0

)

+ K1 ∆δ = 0, where s is Laplace’s operator.

The solution of the equation is given by

∆δ = A1ε s1t + A2 εs 2t where s1, s2 are the roots of the characteristic equation Ms2 + K1 = 0 A1 and A2 are found from initial conditions at t = 0, ∆δ = ∆δ 0 and

d ∆δ =0 dt

...(b)

Power System Dynamics

16 This gives,

A1 + A2 = ∆δ 0

A1s1ε s1t + A2 s2 ε s2t = 0

and

At t=0 A1s1+A2 s2 = 0 The roots of the characteristic equation are

s1, 2 = ± j

K1 M ∴ A1 = A2 =

s1 = − s2

i.e.,

(

∆δ 0 2

)

1 s t st  ∆δ = ∆δ 0  ε 1 + ε 2  2 

∴

The character of the motion depends on the sign of K1. If K1 > 0, the roots are imaginary. Any disturbance appearing in the system will result

K1 M

in continuous oscillations with a frequency

∆δ = ∆δ 0 cos

Dd0

K1 t M

t

If K1 < 0, both roots are real and one of them is always positive. Any small disturbance results in aperiodic rise of angle. Dd

Dd

Dd0

Dd0 K1 > 0 t

K1 < 0 t

The angle δ = 90° corresponding to K1= 0 determines the Steady State Stability Limit. That is the boundary between two kinds of motion. z Oscillatory (δ0 < 90°) z A periodic with rising amplitude (δ0 > 90°) At δ0 > 90° an aperiodic violation of steady state stability is observed (sometimes referred as drift).

Dynamical Analysis of Simple Power Systems Damping Taken into Account The equation governing the transient process is Ms 2 ∆δ + Ds∆δ + K1∆δ = 0

or

D K s∆δ + 1 ∆δ = 0 M M Writing it in terms of a standard second order differential equation s 2 ∆δ +

d 2 ∆δ 2

+

D d ∆δ K1 + ∆δ = 0 M dt M

dt In the control engineering terminology this equation can be written as d 2 ∆δ dt

πf K1 = H

where ω n = ξ=

and

2

D 2

πf HK1

+ 2ξω n K1 M

d ∆δ + ω n2 ∆δ = 0 dt

is the natural frequency of oscillation

is the dimensionless damping ratio.

The characteristic equation of the power system is Ms 2 + Ds + K1 = 0

The roots of the equation are

s1, 2 =

− D ± D 2 − 4 MK1 2M

 K  D 2  −D ± − 1 −   2M  M  2M   = α ± jγ

s1, 2 = s1, 2 where

α=−

D is the damping factor 2M

K  γ =  1 − α2  is the frequency of oscillations of synchronous generator. M 

17

Power System Dynamics

18 For K1 > 0, the system is always stable. If

K1 < α 2 , s1, 2 are real and negative. M

The character of the dynamics is If

Dd

Dd0 t

K1 > α 2 , s1, 2 are complex with negative real parts. M

Dd

The Transient process assumes the form of decaying oscillations. If K1 < 0, the relation between

K1 and α has no effect on M

Dd0

the nature of the process.

t

One root is always positive while the other is negative. The transient will be as shown at K1 = 0 one root is zero and the other is equal to –

D . M

The presence of zero root implies critical case.

Dd

Dd0

Self-oscillations

t

It was assumed that the damping torque produced tends to decelerate the rotor. Damped oscillations are characterized by a damping factor α. Now consider a general situation where the damping coefficient is defined as: D = D 1 + D2 where, D1

(

)

= f r, E 2 ≅ 0

if r = 0 (r is circuit resistance)

= –ve if r ≠ 0 D2 =

V 2 ( xd − xd′ ) xd xd′

Td′ sin 2 δ 0 (due to synchronising forces)

D1 is dependent on the presence of large resistance in stator circuit. For low resistance D = D2. In general case: If |D1| > |D2| then the equivalent damping term gives additional acceleration to the rotor.

Dynamical Analysis of Simple Power Systems

19

The roots of the characteristic equation are

K  s1, 2 = α ± –  1 − α 2  M 

α =

Deq M

, Deq = ( – D1e + D2e )

The roots will have positive real parts. At K1 > 0 and

K1 > α 2, the roots are complex and the angle δ rises in an oscillating manner. M

At K1 < 0 and

K1 < α 2, the oscillation changes to aperiodic rise in angle because both roots are M

real and positive.

When K1 < 0, roots are real and one root is positive. Hence aperiodic rise of angle takes place though slowly. The increase of oscillations is known as phenomenon of self-oscillations. Self-oscillations A similar effect leading to self-oscillations can be caused by speed governors or excitation regulators if parameter settings are changed due to improper adjustment resulting in Torque applied to the shaft with a rise of speed or reduced power supplied by the generator to the network.

Power System Dynamics

20

Summarising, self oscillations appear when an additional torque accelerating the rotor develops in the system. This may happen because of: z High resistance of the circuit. z Improper choice of parameters of automatic speed governors or excitation regulators directly or indirectly responding to change in speed. z Additional energy that may appear due to electromagnetic energy stored in capacitor or inductor and released during transient process. From the analysis presented so far the following conclusions are drawn: The negative real part of all the roots of the characteristic equation is a necessary and sufficient condition of steady state stability or dynamic stability of the electrical power system. The dynamic stability analysis of a power system must include the following main steps: z Describe the transient process in the form of non-linear differential equations. z Linearise the equations using first-order approximation. z Calculate the derivatives for the operating condition considered. z Obtain the characteristic equation. z Estimate the dynamical behaviour using methods of determining sign of real roots and real parts of complex roots of characteristic equation.

REFERENCES z z z z z z z

W.D. Stevenson, “Elements of Power System Analysis”, McGraw-Hill, New York, 1962. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. I.J. Nagarath and D.P. Kothari, “Modern Power System Analysis”, Tata McGraw-Hill, 1980. M.L. Soni, P.V. Gupta, U.S. Bhatnagar and A. Chakrabarti, “A Text Book on Power System Engineering”, Dhanpatrai & Co. (P) Ltd., New Delhi, 1998.

Chapter 4

Dynamic Stability Analysis of Uncontrolled Power System Including Field Transient z z z z

Linearised dynamical equations for simple power system with field flux decay Operating matrix equation and characteristic equation of the power system Routh's criterion for stability analysis Conditions for aperiodic instability, self-oscillations and self-excitation E

Vt Xe

Pt

V

Until now only the electromechanical equation of motion was considered.

M s 2δ + D sδ = Pt − Pe Here Pt is constant and Pe is dependent only on δ, all other system variables remain constant. Now consider the analysis of this uncontrolled power system taking into account the transient process in the field winding. The field winding is fed with a constant voltage Vf from an exciter. The governing equation is: V f = rf i f +

d ψ ffd dt

This equation can be transformed as = if +

1 d ψ ffd rf dt

i fe = i f +

1 d ψ ffd rf dt

Vf rf ∴

i fe is the steady state field current.

Power System Dynamics

22 Rewriting the equation as

(

xafd i fe –i f

)

=

xafd d ψ ffd rf

dt

We have

Eqe –E

=

Eqe –E =

Defining,

∴

xafd x ffd d ψ ffd rf x ffd

dt

L ffd ω xafd d ψ ffd rf

x ffd xafd

E′q = ωψ ffd

Eqe –E = Td 0

dt

x ffd

d Eq′

where Td 0 =

dt

L ffd rf

This expresses the field voltage Vf in terms of pu field voltage Eqe required to produce a pu flux linking armature and field. The basic differential equation of the transient process in the rotor winding d Eq′

Eqe = E + Td0

dt

= E + Td0 s Eq′

We now have the governing equations for the transient process as

Ms 2δ + Dsδ = Pt − Pe Eqe = E + Td0 Pe =

4.1

d Eq′ dt

= E + Td0 s Eq′

EV sin δ X dε

LINEARISED DYNAMICAL EQUATIONS FOR SIMPLE POWER SYSTEM WITH FIELD FLUX DECAY Linearising about an operating point characterised by E0 and δ0

Pe0 =

E0V sin δ 0 X dε

Dynamic Stability Analysis of Uncontrolled Power System Including Field Transient Pe0 + ∆P = Ms 2 δ 0 + Dsδ 0 =

E0V ∂P ∂P sin δ 0 + e ∆δ + e ∆E X dε ∂δ ∂E Pt − Pe0

Ms 2 ( δ 0 + ∆δ ) + Ds (δ 0 + ∆δ ) = Pt − Pe0 − ∆P ∴

∂P  ∂Pe  ∆δ + e ∆E  Ms 2 ∆δ + Ds ∆δ = −  ∂E  ∂δ  Eqe = E0 + Td0

dEq′ dt

Eqe = E0 + ∆E + Td0 ∴

0 =

(

d Eq′ + ∆Eq′

)

dt

∆E + Td0 s ∆Eq′

The phasor diagram for the above equations is:

E′q = E – I d ( X d − X d′ ) Id =

E − V cos δ X dε = X d + Xe X dε

23

Power System Dynamics

24

E′q = E –

E − V cos δ X dε

X d′ ε V cos δ + X dε X dε

E′q = E

( X d − X d′ ) ( X d − X d′ )

∴ E′q is a non-linear function of E and δ ∂Eq′

∴

∆Eq′ =

∴

∆E + Td0 s

∂δ

∆δ +

∂Eq′ ∂E

∂Eq′ ∂E

∆E

∆E + Td0 s

∂Eq′ ∂δ

∆δ = 0

System linearised differential equations are:

∂Pe  ∂Pe  2 ∆E = 0  Ms + Ds +  ∆δ + ∂δ ∂E Td0 s

∂Eq′

∂Eq′   ∆δ +  1 + Td0 s ∆E = 0 ∂δ ∂E  

E0V ∂Pe cos δ 0 = K1 = X dε ∂δ

Using

∂Pe V sin δ 0 = b1 = ∂E X dε Now

∂Eq′ ∂E

=

∂Eq′ X′ X d′ ε = Td0 d ε = Td′ ; denoting Td0 X dε ∂δ X dε

the operating equations become,

( Ms Td′

2

)

+ Ds + K1 ∆δ + b1∆E = 0

X d ε ∂Eq′ s ∆δ + (1 + Td′ s ) ∆E = 0 X d′ ε ∂δ

Dynamic Stability Analysis of Uncontrolled Power System Including Field Transient

4.2

25

OPERATING MATRIX EQUATION AND CHARACTERISTIC EQUATION OF THE POWER SYSTEM

In matrix form the operating equations are:

 Ms 2 + Ds + K1  ∂Eq′  X dε Td′ X ′ s ∂δ dε 

b1   ∆δ      0   1 + Td′ s    =       ∆E 

The principal determinant of the matrix gives the characteristic equation of the system as

Ms 2 + Ds + K1 ∂Eq′ X Td′ d ε s ∂δ X d′ ε

b1 1 + Td′ s

= a0 s 3 + a1s 2 + a2 s + a3 where

a1 = M + DTd′

a0 = MTd′

∂Eq′ X d ε   a2 = Td′  K1 − b1 ∂δ X ′  + D = D + Td′ C2  dε  where

C2 =

K1 − b1

a3 =

K1 E

Pt

∂Eq′ X d ε ∂δ X d′ ε

Xe V

Having obtained the characteristic equation as a0 s 3 + a1s 2 + a 2 s + a3 = 0, we now examine the stability of the power system. The coefficients are dependent upon partial derivatives which are functions of the initial condition.

4.3

ROUTH’S CRITERION FOR STABILITY ANALYSIS

Now let us apply the Routh’s Stability Criterion to the system. For stability it is required that all the coefficients of the characteristic equation should be positive and in the first column of the Routh’s array, all the elements should be positive.

Power System Dynamics

26 Routh’s Array

s3 s2

a0

a2

⋅

a1

a3

⋅

a1a2 − a0 a3 ⋅ a1

s1

s0 a 3

⋅ ⋅

⋅

This gives as conditions of stability a2> 0

a3 > 0

a3 =

a1a2 − a0 a3 =

a1

a3

> 0. a0 a2 Let us examine when these conditions are violated so that the positive real roots or complex roots with positive real parts may appear.

a0 > 0 a1 > 0

K1 =

∂Pe EV = cos δ ∂δ X d ε

a3 is positive only for δ < 90°. Consequently as in earlier analysis, system is stable only for δ < 90°. Aperiodic Instability, Self-oscillations and Self-excitation At K1 < 0 there appears Aperiodic Instability. a1, a2 may become negative if Td′ is negative and its absolute value is greater than

Td′ >

D M . and C2 D

M D ; D C2

a0 = M T′d is negative if Td′ is negative. Thus regardless whether a1 and a2 are +ve or –ve, the system stability is determined by the sign of a0. The time constant Td′ (when r = 0) Td′ =

X d′ + X e Td Xd + Xe 0

It is clear that Td′ is negative if the external reactance Xe is negative (capacitive) so that the condition X d′ < X e < X d is satisfied. The instability in this case is called self-excitation. Consider a situation when ∆δ = 0 (i.e, at synchronous speed when no fluctuation in angle can occur). The system is governed by

Dynamic Stability Analysis of Uncontrolled Power System Including Field Transient

∆E + Td0 s

∂Eq′ ∂E

27

∆E = 0

∆E + Td′ s∆E = 0, whose solution is

or

t

∆E = ∆E0ε – Td′ If Td′ is negative then

∆E = ∆E0 ε

t |Td′ |

t |Td′ |

∆ Id = ∆ Id 0 ε

Voltage and current increase exponentially. Finally a1a2 − a0 a3 > 0

(M +

DTd′ )( D + Td′ C2 ) − MTd′ K1 ≥ 0

DTd′   D   C2  1 + − K1 ≥ 0 1+    M   C2Td′  C2 =

∵

K1 − b1

Eq′ = E ∂Eq′ ∂δ C2

= −

∂Eq′ X d ε ∂δ X d′ ε

X d′ ε V ( X d − X d′ ) + cos δ X dε X dε V ( X d − X d′ )

= K1 +

X dε

sin δ

V 2 ( X d − X d′ ) X d ε ⋅ X d′ ε

sin 2 δ

For small values of D the condition approximates to C2 – K1 ≥ 0 For D negative the condition becomes violated and an accelerating torque appears on the shaft of the generator and self-oscillations will set in. This instability sets in when r is not small or negligible and at δ ≅ 0 i.e, under lightly loaded conditions.

Power System Dynamics

28

Self-oscillation is stimulated by a decrease in rotor angle (light load), by an increase of resistance (r) and also increase in excitation because D1 ∝ E 2 r

REFERENCES z z z z z

P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976.

Chapter 5

Self-excitation Phenomena z z z

5.1

Parameters of self-excitation Asynchronous and Synchronous self-excitation Summary of instabilities and various operating conditions

PARAMETERS OF SELF-EXCITATION

In previous lessons — it has been shown that, if the time constant Td′ (when r = 0) becomes –ve, Instability (Self-excitation) sets in. Td′ =

X d′ + X e Td Xd + Xe 0

It is clear that Td′ is –ve if the external reactance Xe is –ve

X d′ < X e < X d

and

Resistance is inevitably present in stator circuit. If r is not small and negligible and Xe is –ve, the expression for Td′ gets modified as: Td′ =

( )T d + ( X d − X e )( X q − X e )

r 2 + ( X d′ − X e ) X q − X e r

2

0

Consider the numerator, r2 + (X′d – Xe)(Xq – Xe). For this to be –ve’ Xe should be such that (X′d – Xe) (Xq – Xe) should be –ve and |(X′d – Xe)(Xq – Xe)| > r2. The critical value of Xe is that for which r2 + (X'd – Xe)(Xq – Xe) = 0 There can be several values of r and Xe which satisfy this condition thereby defining a boundary. The equation can be written as:

(

)

r 2 + X e2 − X e X q + X d′ + X q X d′ = 0

Power System Dynamics

30 2

2

or

X q + X d′    X q − X d′  r +  Xe − −  2 2    

or

X q + X d′    X q − X d′  r2 +  Xe − =    2 2   

2

2

= 0 2

Xe

This is the equation of circle. Outside the boundary the numerator will not be –ve.

2 Xq + Xd'

Similarly considering the denominator of Td′ equation r2 +

( X d − X e )( X q − X e )

Xd'

2

r

=0

Xe

2

or

Xq

Xq – Xd'

Xd + Xq    Xd − Xq  r2 +  Xe − =    2 2   

2

Xd Xd – Xq 2

This equation also defines a circular boundary inside which the denominator becomes –ve. Combining both diagrams, we get

Xd + Xq 2

Xq r

Xe Xd – Xq

Xd

2 II Xq – Xd'

Xd + Xq 2

Xq

2 Xq + Xd' 2

I

Xd' r

The figure indicates the regions I and II because of variation of parameters where self-excitation occurs.

Self-excitation Phenomena

5.2

31

ASYNCHRONOUS AND SYNCHRONOUS SELF-EXCITATION

The region-I is called region of asynchronous self-excitation. In this region the current variations are in the form of beats. X d′ < X e < X q i

t

In region-II self-excitation is called synchronous self-excitation. It is accompanied by an exponential rise of current. i

Without saturation

Xq < Xe < Xd

t With saturation

It is of interest to examine whether self excitation will occur at the synchronous speed of the generator. i.e., when ∆δ = 0, in the governing equations.

( Ms

Td′

2

)

+ Ds + K1 ∆δ + b1∆Eq = 0

X d ε ∂Eq′ ∆δ + (1 + Td′ s ) ∆Eq = 0 s X d′ ε ∂δ

We have only one differential equation to analyse, i.e, ∆Eq + Td′ s∆Eq = 0 The solution of this equation is

If Td′ is negative, and

∆Eq = ∆Eq 0 ε

∆Eq = ∆Eq 0 ε ∆ I d = ∆I d 0 ε

−

t Td′

t Td′

t Td′

Power System Dynamics

32 Summarising

There are three basic types of instabilities in a simple uncontrolled system subjected to a small disturbance. I. Aperiodic Disturbance of Dynamic Stability (Drift) P = Pm bus

Hazardous initial conditions

Heavy generator loads, system is close to maximum allowable values of power and angle. Pm P

t

90°

II. Self-excitation (i) Series condenser of large value is used to compensate stator reactance (for any load). (ii) Unloaded generators connected to transmission line.

Compensation is increased. Usually no δ change occurs. (b)

Self-excitation Phenomena

33

III. Self-oscillation r V P

t

Low load operation of generator connected to an high resistance network. Self-oscillations are usually more intensive in salient pole machines and at high excitations.

REFERENCES z z

V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983.

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Chapter 6

Voltage and Speed Regulation Systems z z z

Regulating equipment, basic structure Automatic voltage regulator, simple and complex configurations, block diagrams and transfer functions Speed governing systems for thermal and hydel systems, block diagrams and transfer functions

So far the dynamic stability characteristics of a simple uncontrolled power system has been studied. In practice the system exists with regulating equipment. Hence it is necessary to study the effect of regulator parameters on the dynamic stability of the system. Therefore a study of the basic features of regulators is taken up.

6.1

BASIC STRUCTURE OF A REGULATOR

The regulator compares some machine quantity with a reference quantity say output voltage with reference voltage and the difference of the two is used to adjust field voltage and flux. z Regulator minimises any steady state and transient variation in the controlled quantity. z This involves negative feedback. z It means dynamic stability of the system can also be improved.

There are two regulating equipment associated with power systems: 1. Excitation System-Voltage Regulator. 2. Speed Governing System.

Power System Dynamics

36

6.2

VOLTAGE REGULATOR

Following the basic structure as envisaged, the simplest system would be:

The equation governing the performance of the system is: Eqe Vr − Vt

=

KA (1 + TA s ) (1 + Te s )

Normally amplifiers are very fast and hence time delay in amplification can be neglected. ∴

Eqe Vr − Vt

=

KA (1 + Te s )

Linearising the equations

∆Eqe =

K A ( −∆Vt )

(1 + Te s )

The regulator complexity can be increased by introducing stabilisation loops, like amplifier stabilisation loop, exciter stabilisation loop and also the number of amplifier stages.

One stage amplification and exciter stabilisation

Governing equation for the one stage amplification and exciter stabilisation are:

Eqe =

K A (Vr − Vt − Vs )

(1 + Te s )

Voltage and Speed Regulation Systems

Vs =

37

K s s Eqe

(1 + Te s )

The configuration of a complex Automatic Voltage Regulator (A.V.R.) is

6.3

SPEED GOVERNING SYSTEMS

A Speed Governing System (SGS) includes: I. Speed Governor (SG) II. Speed Control Mechanism (SCM) III. Governor Controlled Valves (GCV) IV. Speed Changer (SC) Interchange power

Frequency

Automatic generation control UG SC

SG

SCM

GCV

Turbine

M/C

I. The Speed Governor includes only those elements that are directly responsive to the speed which position or influence the action of other elements of the speed governing system. II. The Speed Control Mechanism includes all equipment such as relays, servo motors, pressure or power amplifying devices, levers and linkages between speed governor and governor controlled valves. III. Governor Controlled Valves include those valves that control the energy input to the turbine and that are normally actuated by speed governor through the medium of speed control mechanism.

Power System Dynamics

38

IV. Speed Changer is a device by means of which speed governing system may be adjusted to change the speed or power output of the turbine operation. The inputs for the speed governing system are the speed signal and speed changer position. The signal UG from governor speed changer is determined by automatic generation control system. It represents a composite load and speed reference and is assumed constant over the interval of study. The signals call for a change in governor controlled valve position which will change input to the turbine. The change in steam flow or input to the turbine causes a change in turbine torque which can in conjunction with characteristics of the machine determine the change in system speed. The schemes of speed governing system are of two types: z Speed governing system for steam turbines. z Speed governing system for hydraulic turbines. Steam Turbines The schematic diagram shows steam turbine speed governing system (SGS) Speed changer signal

UG

+

1 1+ Tg1s

1 1+ Tg2 s

PV

Pt

M/C

– Valve position Kg

The dynamic response of a non-reheat turbine may be approximated to a single time lag Tg 2 1 ∆Pt = 1+T s ∆Pv g1

where ∆ Pv = Change in valve position. ∆ Pt = Change in turbine power.

The speed governor response is given by:

–Kg ∆ω = ∆ Pv 1+Tg1 s The overall transfer function of the steam governing system is:

(

–Kg

)(

∆Pt = 1 + T s 1 + T s g1 g 2

)

∆ω

Voltage and Speed Regulation Systems

39

Hydraulic Turbine In these systems large inertia of water is used as a source of energy and causes a considerably greater time lag in response of the prime mover torque to a change in gate position compared to steam turbines. The transfer function of a hydro turbine may be approximated by: ∆Pt =

– Tω s ∆Pv 0.5 Tω s + 1

where Τω is the nominal starting time of water in penstock in secs. The block diagram of a hydraulic governor in terms of small deviations from steady state values is: Gate servo

Gate actuator w

1 Ta s

s

af

a

1 Tg s

g

–Tw ws 1+ 0.5 Tw ws

h

1 = Governor open loop gain Tg

s, dt = Governor permanent droop and Governor temporary droop d dtTr s 1+ Tr s

Tr = Dash pot time constant

Tg = Hydraulic turbine gate constant a f = Dash pot feedback signal h = Hydraulic head

a = Gate actuator signal g = Gate movement

REFERENCES z z z z

V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. R.T. Byerly and E.W. Kimbark, “Stability of Large Electric Power Systems”, IEEE Press Book, IEEE, New York, 1974.

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Chapter 7

Dynamic Stability Analysis of a Power System with Controlled Excitation z z z

7.1

Linearisation and derivation of characteristic equation Hurwitz’s criterion: Testing for instabilities like aperiodic instability, self-excitation and self-oscillations Limits on the voltage regulator, gains for system dynamic stability, stability boundaries

LINEARISATION AND DERIVATION OF CHARACTERISTIC EQUATION

Basic equations that govern the dynamics are:

d 2δ dδ M 2 + D = Pt − Pe dt dt Eqe = E + Td

dEq′ 0

dt

Eqe = G ( s ) (Vr − Vt ) G(s) is the transfer function of the AVR. Neglecting damping and linearising these equations, we get

Ms 2 ∆δ + ∆Pe = 0 Td0 s∆Eq′ + ∆E = ∆E qe

∆E qe = – G ( s ) ∆Vt

Power System Dynamics

42

Expressing the increments in the system variables in terms of ∆δ and ∆E, we have

∆Pe = ∆Eq′ =

∆Vt =

∂Pe ∂P ∆δ + e ∆E ∂δ ∂E

∂Eq′ ∂δ

∆δ +

∂Eq′ ∂E

∆E

∂Vt ∂V ∆δ + t ∆E ∂δ ∂E

∂V  ∂V  ∆Eqe = −G ( s )  t ∆δ + t ∆E  ∂E  ∂δ  Substituting these in linearised system equations, we get

( Ms

2

)

+ K1 ∆δ + b1∆E = 0 ∂Eq′ ∂Eq′   ∂Vt  ∂Vt  Td0 s ∂δ + G ( s ) ∂δ  ∆δ + Td0 s ∂E + 1 + G (s ) ∂E  ∆E = 0    

It should be noted that

Td0

∂Eq′ ∂E

= Td0

X d′ε = Td′ = K3Td0 X dε

The characteristic determinant of the system is

Ms 2 + K1 D( s ) =

Td′

b1

X d ε ∂Eq′ ∂V s + G ( s) t X d′ε ∂δ ∂δ

Td′s + 1 + G ( s )

∂Vt ∂E

Considering a simple voltage regulator with no stabilisation loops

Ms 2 + K1 D( s ) =

b1

∂V KA t X d ε ∂Eq′ ∂δ Td′ s + X d′ε ∂δ (1 + TA s ) (1 + Te s )

∂Vt ∂E Td′s + 1 + (1 + TA s ) (1 + Te s ) KA

D( s ) = D0 ( s ) + D1 ( s ) D( s ) =

a e1 + e2

b d1 + d 2

=

a

b

e1

d1

+

a

b

e2

d2

Dynamic Stability Analysis of a Power System with Controlled Excitation Simplifying,

Ms 2 + K1 D0 ( s ) =

D1 ( s) =

b1

(1 + TA s ) (1 + Te s )Td′ Ms 2 + K1

b1

∂Vt ∂δ

KA

KA

X d ε ∂Eq′ s X d′ε ∂δ

(1 + TA s ) (1 + Te s )(1 + Td′s )

∂Vt ∂E

5 4 3 2 D0 ( s ) = a0 s + a1s + a2 s + a3 s + a4 s + a5

where

a0 = M Td′ Te TA a1 = M (Td′ Te + Td′ TA + Te TA ) a2 = M (Td′ + Te + TA ) + Td′ TATeC2 ∂E ′ X

b ∂E ′

q q dε 1 C2 = K1 − b1 ∂δ X ′ = K1 − K ∂δ 3 dε

a3 = M + (Td′ Te + Td′ TA + Te TA ) C2 a4 = (Te + TA ) K1 + Td′ C2 a5 = K1

And D1 ( s ) = K A M

∂Vt 2 ∂V   ∂V s + K A  K1 t − b1 t   ∂E ∂E ∂δ 

usually TA is small and can be taken as ≅ 0. Hence

D( s) = A0 s 4 + A1s 3 + ( A2 + ∆A2 ) s 2 + A3 s + A4 + ∆A4 where

A0 = M Td′ Te A1 = M (Td′ + Te )

A2 + ∆ A2 = M + Td′ Te C2 + K A M

∂Vt ∂E

43

Power System Dynamics

44

A3 = Td′C2 + Te K1 ∂V   ∂V A4 + ∆ A4 = K1 + K A  K1 t − b1 t  ∂δ   ∂E At this stage, it should be remembered that,

K1 =

V ∂PE EV ∂P cos δ ; b1 = E = sin δ = X dε ∂δ ∂E X d ε

E′ = E

∂Eq′ ∂δ

= −

X d′ε V ( X d − X d′ ) cos δ + X dε X dε V ( X d − X d′ )

C2 = K1 +

X dε

sin δ

V 2 ( X d − X d′ ) X d ε X d′ε

sin 2 δ

Expressing E in terms of E′q as



V ( X d − X d′ )



X dε

E =  Eq′ − in PE =

X cos δ  d ε  X d′ε

EV sin δ and C2, we have X dε PEq′ =

∴

∂PEq′ ∂δ

=

Eq′V X d′ ε Eq′V X d′ ε

sin δ −

cos δ −

V 2 ( X d − X d′ ) 2 X d ε X d′ ε

V 2 ( X d − X d′ ) cos 2δ = C 2 X d ε X d′ ε

Let us now define

∂PEq′ ∂Eq′

= b2 =

sin 2δ

V sin δ X d′ε

Dynamic Stability Analysis of a Power System with Controlled Excitation Remembering the phasor diagram E' q

Vq

Iq

E'

IX

V

d

IX

e

IX '

I

de

de

Vt Id

E

We have

E = Vt cos δ t + I d X d

Id =

and

∴

Vt cos δ t − V cos δ Xe

E = Vt cos δ t

X dε X − V cos δ d Xe Xe

∴ The electrical power output in terms of Vt is PVt =

∴

∂PVt ∂Vt

V Vt cos δ t V 2 sin 2δ sin δ − Xd Xe 2X dε X e

= b3 =

V cos δ t sin δ Xe

Then we have

Xe Xe V sin δ b1 = = X d ε V cos δ t sin δ X d ε cos δ t b3 Differentiating Vt with respect to E

1=

X ∂Vt cos δ t d ε ∂E Xe

q

45

Power System Dynamics

46

Xe 1 b1 ∂Vt = X cos δ = b 3 dε t ∂E

∴

Therefore in the characteristic equation

∆A2 = K A M 

∂Vt b = K AM 1 ∂E b3 ∂V

∂V 

∆A4 = K A  K1 t − b1 t  ∂δ   ∂E

To evaluate

The following steps are followed

K1

∂Vt EVX e cos δ = ∂E X d2ε cos δt =

K1

∴

 X dε X  − V cos δ d  Vt cos δ t Xe Xe  

∂Vt VVt V 2 X d cos 2 δ cos δ − = ∂E Xdε X d2ε cos δt

Vt cos δ t =

Now

VX e cos δ X d2ε cos δ t

EX e X + V cos δ d X dε X dε

∂ ( cos δ t ) ∂Vt X cos δ t + Vt = 0 − Vsin δ d ∂δ ∂δ X dε

∂Vt 1 = ∂δ cos δt

∴

b1

then

∴

K1

 ∂ ( cos δt ) X  − Vsin δ d   −Vt ∂δ Xdε  

∂Vt VVt sin δ ∂ ( cos δ t ) V 2 X d sin 2 δ − = − ∂δ ∂δ X d ε cos δt X d2ε cos δt

∂Vt ∂V VVt V 2 X d cos 2δ VVt sin δ ∂ ( cos δ t ) − b1 t = + cos δ − ∂E ∂δ ∂δ X dε X d2ε cos δ t X d ε cos δ t

Dynamic Stability Analysis of a Power System with Controlled Excitation

47

Now we have

∂PVt It can be seen that

∂δ

= C3 =

VVt Xe

∂ (cos δ t )  V 2 X d  δ δ + δ − cos cos sin cos 2δ t  ∂δ  X d ε X e

∂V ∂V b1 b1 ∂PVt C3 = K1 t − b1 t = b3 ∂E ∂δ b3 ∂δ ∴

∆A4 = K A

Summarising

b1 C3 b3

D( s) = A0 s 4 + A1s 3 + ( A2 + ∆A2 ) s 2 + A3 s + A4 + ∆A4 = 0

A0 = MTd′Te , A1 = M (Td′ + Te ) A2 + ∆A2 = M + Td′TeC2 + K A M

b1 b3

A3 = Td′C2 + Te K1

A4 + ∆A4 = K1 + K A

b1 C3 b3

∂PE ∂PE ; b1 = ; ∂δ ∂E ∂PEq′ ∂PEq′ ; b2 = ; C2 = ∂Eq′ ∂δ

K1 =

C3 = 7.2

∂PVt ∂δ

;

b3 =

∂PVt ∂Vt

.

HURWITZ’S CRITERION: TESTING FOR INSTABILITIES LIKE APERIODIC INSTABILITY, SELF-EXCITATION AND SELFOSCILLATIONS

If a system characteristic equation is given by:

a0 s n + a1s n – 1 + … + an = 0 For the system to be stable it is both necessary and sufficient that all coefficients are positive and that the n determinants be positive. Where the determinants are taken as principal minors of the arrangement.

Power System Dynamics

48

a1

a0

0

0

a3

a2

a1

a0 0

a5

a4

a3

a7

a6

•

•

In other words: ∆1 = a1 > 0 ; ∆ 2 =

0

0

0

•

•

0

0

•

•

a2 a1 a0

0

0

•

•

a5

a4 a3 a2

a1 a0

•

•

•

•

•

•

a1

a0

a3

a2

0

•

•

•

•

>0

a1

a0

0

∆3 = a3

a2

a1 > 0…

a4

a3

a5

0

∆n > 0

For a 3rd order system this is characterised by: a1

a0

0

a3

a2

a1 > 0 ⇒

0

0

a3

a1

a0

a3

a2

> 0; a3 > 0

For a 4th order system

a1

a0

0

0

a3 a2

a1

a0

0 a4

a3

a2

0

0

a4

0

>0

Using the above, for the power system example, the Hurwitz’s determinant is written as:

∆ Hur =

A1

A0

0

0

A3

A2 + ∆A2

A1

A0

0

A4 + ∆A4

A3

A2 + ∆A2

0

0

0

A4 + ∆A4

≥0

Expanding

∆ Hur = A1 {( A4 + ∆A4 )  A3 ( A2 + ∆A2 ) − A1 ( A4 + ∆A4 )} − A3  A0 A3 ( A4 + ∆A4 ) ≥ 0

Dynamic Stability Analysis of a Power System with Controlled Excitation

49

2 2 ∆ Hur = A1 A3 ( A4 + ∆A4 )( A2 + ∆A2 ) − A12 ( A4 + ∆A4 ) − A0 A ( A4 + ∆A4 ) ≥ 0 3

Let us now examine various conditions of stability: 1. A0 and A1 are positive if Td′ > 0. 2. A3 is positive if C2 is positive and for all values of K1

∂PEq′ ∂PE Te Te > or Td′ ∂δ ∂δ Td′

C2 > – K1

Greater the value of Te greater should be the value of C2 and be +ve. If C2> 0 is not satisfied, system stability is violated. 3.

A2 + ∆A2 = M + Td′TeC2 + K A M

4.

A4 + ∆A4 = K1 + K AC3 where

( K A )min

b1 > 0 is always met if C2 > 0 and KA > 0 b3

b1 > 0 is always obeyed if for all values of K1 if KA > (KA) min b3 − K1 b3 = C3 b1

If KA < (KA)min. Aperiodic instability will set in (KA)min sets the limit for the minimum allowable voltage regulator gain. 5. Now setting ∆Hur ≥ 0, we have 2

∆ Hur

  b1  b1  =  KA  P +  KA  Q + R ≥ 0 b  b    3

3

where 2 P = M C3 (Te + Td′ ) ( K1 − C3 )Te + (C2 − C3 )Td′ 

∴

where

P = C3 F 2 F = M (Te + Td′ ) ( K1 − C3 ) Te + ( C2 − C3 ) Td′ 

{

}

Q = K1 M 2 (Te + Td′ ) ( K1 − C3 ) Te + (C2 − C3 ) Td′  +

∴

where

{

}

C3 M (C2 − K1 ) Td′  M (Te + Td′ ) + Te2 ( K1Te + C2Td′ ) 

Q = K1 F + C3G 2 G = M (C2 − K1 ) Td′  M (Te + Td′ ) + Te ( K1Te + C2Td′ )

Power System Dynamics

50

{

}

2 R = K1 M (C 2 − K1 ) Td′  M (Te + Td′ ) + Te ( K1Te + C2Td′ ) 

R = K1G

∴

Setting ∆Hur= 0 for finding critical values of KA, we have

KA

−Q ± Q 2 − 4 PR b1 = 2P b3

∴

− ( K1 F + C3G ) ± b KA 1 = b3

∴

KA

− ( K1 F + C3G ) ± ( K1 F − C3G ) b1 = 2C3 F b3

KA

b1 G = − b3 F

KA

M (C2 − K1 ) Td′  M (Te + Td′ ) + Te2 ( K1Te + C2Td′ )  G b1 − − = = F b3 M 2 (Te + Td′ ) ( K1 − C3 ) Te + (C2 − C3 ) Td′ 

∴

or −

b1 = b3

(C2 − K1 ) 1 + 



( K A )max

=

Te2 ( K1Te + C2Td′ )   M (Te + Td′ ) 

(C3 − K1 )   d (C3 − C2 ) 

T

(C3 − C2 ) 1 + Te′ 

∴

2C3 F

K1 C3



KA

( K1F + C3G )2 − 4 K1GC3 F

(C2 − K1 ) 1 + Te2 ( K1Te + C2Td′ )    M (Te + Td′ )  (C3 − C2 )  T (C − K1 ) 1+ e 3 Td′ (C3 − C2 )

⋅

b3 b1

− K1 b ⋅ 3 C3 b1 z An increase of KA > (KA)max results in violation of ∆Hur > 0, causing periodic oscillations: Self-oscillations in the system. z A decrease of KA < (KA)min results in Aperiodic Instability. We can plot a variation of (KA)min and (KA)max with respect to δ and prepare a chart for stable operation of the system.

( K A )min

=

Dynamic Stability Analysis of a Power System with Controlled Excitation

KA

(KA)max

51

t

Unstable Stable (KA)min Stable Unstable

t

REFERENCES z z z z z z z

P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983. A.E. Fitzgerald, C. Kingsley and S. Umans, “Electric Machinery”, McGraw-Hill, New York, 1982.

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Chapter 8

Block Diagram Representation and Effect of Excitation Control on Electric Power z z

8.1

Block diagram representation of a linearised power system model Effect of excitation regulation on electrical power output and stability of the system by frequency response analysis

BLOCK DIAGRAM REPRESENTATION OF A LINEARISED POWER SYSTEM MODEL

When studying the transient processes and stability of power system, it is necessary to have mathematical models of synchronous generators, excitation control systems and turbine speed governors. In studying dynamic stability of a power system, we use linearised system equations. It is advantageous to use linear mathematical models in the form of block diagrams and transfer functions. In the block diagram the element to be controlled is localised which in our case is synchronous machine with excitation regulator or turbine with speed governor. The forward paths over which the main energy flow is transmitted and feedback paths controlling an additional energy to the system output are clearly indicated in the block diagram, thereby giving a clear idea of the transient process. Let us set up the block diagram of a simple controlled power system. The basic linearised equations are:

( Ms

2

)

+ Ds ∆δ = ∆Pt − ∆Pe ∆Pe =

∂Pe ∂P ∆δ + e ∆Eq′ = C ∆δ + b ∆E ′ 2 2 q ∂δ ∂Eq′

∆Eqe = ∆E + sTd0 ∆Eq′ E = ∴

∆E = ∆E =

V ( X d − X d′ ) 1 X′ cos δ where K3 = d ε Eq′ − K3 X d′ ε X dε V ( X d − X d′ ) 1 sin δ∆δ ∆Eq′ + K3 X d′ ε

1 ∆Eq′ + K 4 ∆δ K3

Power System Dynamics

54

 1  + sTd0  ∆Eq′ + K 4 ∆δ ∆Eqe =   K3  ∴

∆Eq′ =

K3

(1 + K3Td s )

∆Eqe −

0

∆Eqe =

∆Vt =

∆Pt =

K3 K 4

(1 + K3Td s)

∆δ

0

− K A ∆Vt (1 + TA s ) (1 + Te s ) ∂Vt ∂V ∆δ + t ∆Eq′ = K5 ∆δ + K 6 ∆Eq′ ∂δ ∂Eq′ − K g ∆ω

(1 + Tg s )(1 + Tg s) 1

2

Using these equations we have the block diagram as:

1

1

Ms

s

K3 1+ K 3T S d0

Block Diagram Representation and Effect of Excitation Control on Electric Power

8.2

55

EFFECT OF EXCITATION REGULATION ON ELECTRICAL POWER OUTPUT AND STABILITY

The basic equations from the block diagram are ∆Pe = C2 ∆δ + b2 ∆Eq′

∆Eq′ =

K3 K3 K 4 ∆ E qe − ∆δ 1 + K 3Td 0 s 1 + K 3Td 0 s

∆Vt =

K5 ∆δ + K6 ∆Eq′

 bK K  b2 K3 ∆Eqe ∆Pe =  C2 − 2 3 4  ∆δ + 1 + K3Td0 s  1 + K3Td0 s 

∴

For unregulated machine: ∆Eqe = 0 For regulated machine: ∆Eqe =

I.

(

− K A K5 ∆δ + K6 ∆Eq′

(1 + Te s )

)

Considering the unregulated machine b2 K 3 K 4 ∆Pe ( s ) = C2 − 1 + K T s ∆δ 3 d0

To study the frequency response, we have b2 K 3 K 4 ∆Pe ( jω) = C2 − 1 + jK 3Td 0 ω ∆δ 2  b2 K3 K 4  ωb2 K3 K 4Td0 ∆Pe =  C2 − 1 + ω 2 K 2T 2  + 1 + ω 2 K 2T 2 ∆δ  3 d0  3 d0

Real Part

C2 −

b2 K3 K 4

1 + ω 2 K32Td20 gives the synchronising component which is reduced by the demagnetising action

of the armature reaction.

Power System Dynamics

56 Imaginary Part: ωb2 K32 K 4Td0 1 + ω 2 K32Td20

gives the damping component.

It can be seen that positive damping is introduced by the armature reaction. The equation governing dynamics is Ms 2 ∆δ + ∆Pe = 0

[for D = 0]

 bK K  Ms 2 ∆δ +  C2 − 2 3 4  ∆δ = 0 1 + sK 3Td0  

or simplifying, we have

MK3Td0 s3 + Ms 2 + C2 K3Td0 s + ( C2 − b2 K 3K 4 ) = 0 From Routh’s Criterion for system to be stable (ii) b2K3K4 > 0 (i) C2 – b2 K3 K4 > 0; (i) Synchronising power coefficient C2 must be greater than the demagnetising component of electrical power. (ii) This condition is satisfied when K3 > 0, K4 > 0 b2 > 0, II. Considering the system with a voltage regulator

∆Eq′ =

(

− K3 K A K5 ∆δ + K 6 ∆Eq′

(1 + K3Td s )(1 + Te s )

)−

0

  K A K3 K 6  = ∆Eq′  1 +  1 + K3Td0 s (1 + Te s ) 

(

∴

∴

∆Eq′ = −

)

K3 K 4

(1 + K3Td s )

∆δ

0

− K A K3 K5 − K3 K 4 (1 + Te s )

(1 + K T s )(1 + T s ) 3 d0

K3 K 4 (1 + Te s ) + K A K3 K5

(

)

K3Td0 Te s 2 + s K3Td0 + Te + 1 + K A K3 K6

e

∆δ

b2 K 4 (1 + Te s ) + b2 K A K5 ∆Pe ( s ) = C2 − ∆δ  1  Te  2  K + K A K 6 + Td0 Te s  + s  Td0 + K  3 3

∆δ

Block Diagram Representation and Effect of Excitation Control on Electric Power

57

The effect of b2 K4 (1+Tes) is small compared to b2 KA K5. Thus b2 K A K5 ∆Pe ( s ) ≅ C2 −  1  ∆δ Te  2  K + K A K 6 + Td 0 Te s  + s  Td0 + K  3 3

Putting s = jω

∆Pe ( jω) b2 K A K 5 = C2 − ∆δ  1  Te  2  K + K A K 6 − Td0 Te ω  + j ω  Td 0 + K  3 3 Let

where

∆Pe ( jω) = PS + jPd ∆δ

PS = C2 −

Pd =

 1  + K A K6 − Td0 Te ω 2  b2 K A K5   K3  2

 1 Te  2 2  K + K A K6 − Td0 Te ω  + ω  Td0 + K  3 3

2

 T  ωb2 K A K5  Td0 + e  K3   2

 1 Te  2 2  K + K A K6 − Td0 Te ω  + ω  Td0 + K  3 3

2

Note that Pd has the same sign as K5. The latter quantity may be negative at some operating conditions. In such cases the regulator reduces the inherent system damping. At very low frequencies and large KA PS ≅ C2 −

b2 K5 K6

Which is higher than the value obtained for an unregulated machine: PS = C2 − b2 K 3 K 4 . Therefore, where as the regulator improves the synchronising forces in the machine at low frequencies of oscillations, it reduces the inherent system damping when K5 is negative, a common condition for synchronous generators operated near rated load. The characteristic equation of the system is: Ms2 ∆δ + ∆ Pe = 0

Power System Dynamics

58     b2 K A K5  2   Ms + C2 −  ∆δ = 0  1  Te   2   K + K A K 6 + Td0 Te s  + s  Td0 + K    3 3 

 M  T  MTd0 Te s 4 + M  Td0 + e  s3 +  + MK A K 6 + C2Td0 Te  s 2 K3    K3  C   CT  b K  +  C2Td0 + 2 e  s +  2 + K A K 6  C2 − 2 5   = 0 K3  K6      K3

REFERENCES z z z z

P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983.

Chapter 9

Dynamic Stability Boundaries for a Power System Using D-Partition Method z z

9.1

Dynamic stability boundaries in parameter plane Basis of D-partition method, Neimark's hatching rule and its application to power system

DYNAMIC STABILITY BOUNDARIES IN PARAMETER PLANE

To investigate the sign of real roots or real parts of complex roots of characteristic equation, several criteria of stability have been evolved and broadly classified as: 1. Algebraic (Routh, Hurwitz) 2. Frequency domain criteria (D-partition, Nyquist etc.) We have been investigating the stability problem associated with a single machine-infinite bus system with controlled excitation.

D( s ) = A0 s 4 + A1s 3 + ( A2 + ∆A2 ) s 2 + A3 s + A4 + ∆A4 = 0 A0 = M Td′ Te ;

A1 = M (Td′ + Te ) ;

A2 + ∆A2 = M + Td′ TeC2 + K A M

b1 b3

A3 = Td′C2 + Te K1

b1 b3 For any operating point specified by δ there are two values of the AVR gain (KA)max and (KA)min, above and below these values respectively the system will be unstable. KA – δ diagram depicting these limiting values gives the stability boundaries. A4 + ∆A4 = K1 + K AC3

Power System Dynamics

60

Now we will consider a situation when two parameters i.e., KA, Te of the AVR are variable. One would like to determine the adjustable parameters of the AVR (KA, Te) ensuring the stability of the system in a specified operating condition. For this method of Domain Separation (D-Partition) in the plane of two parameters is used.

9.2

BASIS OF D-PARTITION METHOD, NEIMARK’S HATCHING RULE AND ITS APPLICATION TO POWER SYSTEM

Characteristic equation of any system can be written as: s n + a1s n – 1 + a2 s n – 2 + " + an = 0

Location of roots of this equation depends on the numerical values of the coefficients ai. A change in of even one of the coefficients changes the distribution of roots. For the sake of simplicity and concreteness, consider s 3 + a1s 2 + a2 s + a3 = 0

Now consider a coordinate system, whose axes measure the coefficients a1, a2, a3. a1 M" M' a '1

a "1

a2 a'3 a '2 a3

a "2

a "3

Dynamic Stability Boundaries for a Power System Using D-Partition Method

61

In the 3-dimensional polynomial space the point M ′ ( a1′ , a2′ , a3′ ) corresponds to the polynomial s 3 + a1′s 2 + a2′ s + a3′ . Similarly M ′′(a1′′, a2′′, a3′′) ⇒ s 3 + a1′′s 2 + a2′′s + a3′′. If the space is n dimen-

sional, each point of such space is characterised by a polynomial of nth degree. A 3rd order equation has 3 roots. When ‘s’ assumes the value of a root, the polynomial becomes zero. Thus roots are also called zeroes of the polynomial. In the 3-dimensional polynomial space, each point corresponds to 3 points in the plane of roots or complex plane.

The point M ′ ( a1′ , a2′ , a3′ ) in polynomial space corresponds to point (M1, M2, M3) in root plane. Similarly M ′′ ( a1′′, a2′′, a3′′) ⇒ ( N1 , N 2 , N 3 ) The point M ′ corresponds to stable system. The point M ′′ corresponds to unstable system. We introduce now the following notation. If a polynomial of nth degree has ‘r’ roots to the right and n – r roots to the left of the imaginary axis in the complex plane, the whole region of the polynomial space containing polynomials with such root distribution is denoted by D(n – r, r). Hence for stable system the symbolic notation is D(n,0). Thus M ′ is in a domain D(3, 0), M ′′ is in a domain D(1, 2). Now consider our system characteristic equation D(s), which can be written as: Te D1 ( s ) + K A D2 ( s ) + D0 ( s ) = 0

where

4 3 2 Te D1( s ) = Te  MTd′ s + Ms + Td′C2 s + K1s 

K A D2 ( s ) = K A

b1  2 Ms + C3  b3 

3 2 D0 ( s ) = MTd′ s + Ms + Td′C2 s + K1

Power System Dynamics

62

Te, KA are parameters whose effect on stability is being studied. Let us assume that for a certain value of Te and KA there are r roots situated to the right and n – r to the left of the imaginary axis — in other words we have D(n – r, r). If Te and KA are continuously varied, the roots of the equation correspondingly vary their position on complex root plane. At a certain value of Te and KA at least one root will fall on the imaginary axis. A further variation of Te and KA in the same direction will take this root across imaginary axis to the left side, giving rise to D(n – r, r) ⇒ D(n – r – 1, r + 1) Thus the values of KA and Te for which one root falls on the imaginary axis are limiting values in the sense that, at this point at least one root is situated on the imaginary axis. If the root on the imaginary axis is denoted by s = jω , the limiting values of Te and KA can be determined from Te D1( jω) + KAD2( jω) + D0( jω) = 0 ...(1) If D1 ( jω ) = P1 (ω ) + jP2 (ω ) D2 ( jω ) = Q1 (ω ) + jQ2 (ω ) D0 ( jω ) = R1 (ω ) + j R2 (ω )

we have Te P1 (ω ) + K AQ1 (ω ) + R1 (ω ) = 0 Te P2 (ω) + K AQ2 (ω) + R2 (ω) = 0

...(2)

Solving these equations

– R1 (ω ) Q1 (ω ) – R2 (ω ) Q2 (ω ) Te = P1 (ω ) Q1 (ω ) P2 (ω ) Q2 (ω )

P1 (ω ) – R1 (ω ) P (ω) – R2 (ω) KA = 2 P1 (ω ) Q1 (ω ) P2 (ω) Q2 (ω)

All the limiting values of Te and KA can be found by varying ω from – ∞ to + ∞. The locus of points in the Te – KA plane when ω varies from – ∞ to + ∞ divides the whole plane into regions where all polynomials have the same number of zeroes in L.H.S. of imaginary axis. This curve is called the boundary of D-partition by the two real parameters Te and KA. Usually equations (2) cease to be linearly independent for ω = 0 and ω = ∞. Hence the D-partition boundary previously obtained is supplemented by special lines whose equations are obtained by substitution of ω = 0 and ω = ∞ in (1). If the system can be made stable at certain values of Te and KA then with the help of D-partition boundary all such values of Te and KA can be determined. Evidently the region of stability is that region where the number of zeroes to the left of imaginary axis will be equal to the order of the characteristic equation.

Dynamic Stability Boundaries for a Power System Using D-Partition Method

63

Let us now consider the power system example under discussion, where

( ) P (ω) = −ω ( M ω − K ) Q (ω ) = − ( M ω − C ) b b

2 2 P1 (ω ) = Td′ M ω − C2 ω 2

2

1

2

1

1

3

3

Q2 (ω ) = 0

(

− R2 (ω) = ωTd′ M ω 2 − C2

− R1 (ω ) = M ω 2 − K1 ;

Now where

∆τ ∆ KA = k ; ∆ ∆ P1 Q1 b = −ω M ω 2 − K1 M ω 2 − C3 1 ∆ = b3 P2 Q2 – R1 Q1 b = ωTd′ M ω 2 − C2 M ω 2 − C3 1 ∆τ = b3 – R2 Q2 P1 – R1 2 = ω 3Td′ 2 M ω 2 − C2 + ω M ω 2 − K1 ∆k = P2 – R2 Te =

(

)(

(

Te =

(

−Td′ M ω 2 − C2

)

)

(

2

)

2

)

(M ω − K ) (M ω − K ) + ω T ′ (M ω − C ) = − ( M ω − K )( M ω − C ) 2

KA

)

)(

(

∴

)

1

2

2

1

2

1

d

2

2

2

2

b3 b1

2

3

Te and KA here are even functions of ω. Hence the gain locus drawn for ω = – ∞ to 0 coincides with locus drawn for ω = 0 to ∞. Assuming the following data:

M = 0.054;

Td′ = 4.1;

K1 = 0;

C2 = 0.101;

C3 = 0.212;

b1 = 0.647 b3

0.101   Te = − 4.1 1 −  0.054ω 2  KA = −

Varying ω from 0 to ± ∞ .

(

)

0.0834 ω 2 + 482 0.054 ω 2 − 0.101 0.054 ω 2 − 0.212

2

Power System Dynamics

64 1.

ω = 0; Te ⇒ ∞ ; K A = 23.2

2.

ω12 =

0.101 = 1.87; ω1 = 1.37; Te = 0; K A = 1.4047 0.054

3.

ω 22 =

0.212 = 3.94; ω 2 = 1.98; Te = −2.15; K A = ± ∞ 0.054

4.

ω ⇒ ∞; Te = − 4.1; K A = − ∞

5. 6.

dK A (ω ) d ω2

= 0; ω 32 = 1.77; ω3 = 1.33; Te = 0.226; K A = 1.4

ω 24 = 6.08; ω 4 = 2.46; Te = −2.84; K A = −218

A plot of these results may have the appearance shown below KA

ω=0 Te

ω=∞

Now the following questions become important: 1. In which region the largest number of zeroes exist on the left of the imaginary axis? 2. Is this region a stable region? To answer these questions we use Neimark’s Hatching rule: 1. Pursuing the D-partition boundary curve from ω = – ∞ to 0 and then thence to ω = ∞, the left hand side of the curve is hatched if ∆ > 0. For all values of ω for which ∆ < 0 the right hand side is hatched. 2. When traversing the curve from the hatched side to the unhatched side one root to the left of the imaginary axis is lost.

Dynamic Stability Boundaries for a Power System Using D-Partition Method

65

3. Conversely, when traversing the boundary curve from the unhatched side to the hatched side one root to right of the imaginary axis is lost. Performing hatching in accordance with the above principles, the region where the polynomials have the largest number of roots in left half of the plane can be directly found. We now discuss the hatching of special lines. At ω ≡ 0 and ω ≡ ∞. We obtain not points but straight lines. For ω ≡ 0 the equations of the straight line is obtained by equating the free term of the characteristic equation to zero and for ω ≡ ∞ by equating to zero the coefficient associated with highest power of ω. In case of our specific example the special lines are coordinate axis themselves (i.e., KA = 0; Te = 0) The hatching of the lines is done as follows: Near the intersection of the special straight line and the D-partition curve they are so hatched that the hatched side of the straight line faces the hatched side of the D-partition curve further hatching of the straight line remains unchanged. Following these rules the parameter plane plot and there from the dynamic stability region for the power system is obtained. These are shown in the subsequent figures.

KA KA

– 4.1

–2.84 –2.15

.. 0.226 4 –218

Te

66

Power System Dynamics

Dynamic Stability Boundaries for a Power System Using D-Partition Method

(

∆ = – 0.0347 ω 3 0.054 ω 2 − 0.212 0 < ω < ω2

For

ω2 < ω < ∞

67

)

∆ > 0 shade left side ∆ < 0 shade right side

REFERENCES z z z

M.V. Meerov, “Introduction to Dynamics of Automatic Regulating of Electrical Machines”, Butterworth, 1961. V.A. Venikov,“Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983.

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Chapter 10

Time Domain Formulation of Dynamic Stability Problem for a Simple Power System z z

A time domain formulation of power system dynamic stability problem with and without regulating equipment Rigorous formulation of state-space equations for a power system for a dynamic stability study

First step in time domain problem formulation is writing system performance equations in state space. System physical variables are chosen as state variables and system performance equations are expressed as first order differential equations.

10.1

UNCONTROLLED SYSTEM

The basic equations are

M

d 2 ∆δ dt

2

+D

d ∆δ = ∆Pt − ∆Pe dt ∆Pe = C2 ∆δ + b2 ∆Eq′ ∆Vt = K5 ∆δ + K6 ∆Eq′

K 3Td 0

d ∆Eq′ dt

= − K3 K 4 ∆δ − ∆Eq′ + K3∆Eqe

Designating ∆δ, ∆ω, ∆Eq′ as state variables and ∆Pt and ∆Eqe as input variables, we have

d ∆δ = ∆ω dt d ∆ω M = −C2 ∆δ − b2 ∆Eq′ − D ∆ω + ∆Pt dt or

d ∆ω C2 D b 1 ∆δ − ∆ω − 2 ∆Eq′ + ∆Pt = − dt M M M M

Power System Dynamics

70 d ∆Eq′ dt

= −

K4 1 1 ∆δ − ∆Eq′ + ∆Eqe Td0 K 3Td0 Td0

Writing in matrix form:

 ∆δ      0     ∆ω  =  −C2    M    −K    4 ∆ ′ E  q   Td0 ∆Vt = [ K5

1 −D M 0

0

  ∆δ    0   0    −b2   1    ∆ω  +  M  M   −1     0 K3Td0   ∆Eq′  

  0   ∆Pt    0     ∆E  1   qe  Td0 

 ∆δ    K 6 ]  ∆ω   ∆Eq′   

These equations are in the standard form:

X = AX + BU

10.2

Y = CT X

SYSTEM WITH REGULATING EQUIPMENT

Assuming a first-order voltage regulator, the governing equation is given by: Te

or

or

d ∆Eqe dt d ∆Eqe

= −∆Eqe − K A ∆Vt + K A ∆Vr = −

∆Eqe K A K A K5 K K ∆δ − A 6 ∆Eq′ − + ∆Vr Te Te Te Te

dt Assuming a steam turbine speed governing system, the governing equation is given by:

Tg1

d ∆Pv = − K g ∆ω − ∆Pv + ∆U G dt

Tg 2

d ∆Pt = ∆Pv − ∆Pt dt − K g ∆ω 1 d ∆Pv 1 − ∆Pv + ∆U G = dt Tg1 Tg1 Tg1

d ∆Pt 1 1 ∆Pv − ∆Pt = dt Tg 2 Tg 2

Time Domain Formulation of Dynamic Stability Problem for a Simple Power System

71

Choosing a state variable set X T =  ∆δ ∆ω ∆Eq′ ∆Eqe ∆Pv ∆Pt 

and input vector U T = [ ∆Vr ∆U G ] and output as ∆Vt , we have

1 0  0  −C −D −b2 2  M M  M  ∆δ   − K − 1   4 0   ∆ω  K3Td0  Td 0  ∆E ′    q  = − K A K5 − K A K6  0  ∆E   Te Te  qe    ∆Pv  −Kg    0 0  Tg1  ∆Pt    0 0  0 

∆Vt =

[ K5

0

0

0

0

1 Td0

0

−1 Te

0

0

0

1 Tg1

0

1 Tg2

K6

   0     ∆δ   0 0   ∆ω   0      ∆Eq′   K A   0  ∆E +  Te   qe     ∆P    v   0 0   ∆P   t     0 −1   Tg2  0 1 M

0

0

0  0  0   ∆Vr     0      1   ∆U G  Tg1   0 

  0]  X   

Rigorous Formulation of State Space Equations ib q-

q-

ax

ψb

ax

is

is

ax

d-

xis

is

Vf

i f ψ f

d-a

ψa ia

ψc

ic

(a) The layout of the windings of the synchronous machine without amortissuer windings

(b) Synchronous machine representation in d- and q-axes

Power System Dynamics

72

Equations of Park’s transformation are: Kd =

2  K a cosθ + Kb cos (θ – 120) + Kc cos ( θ + 120) 3

Kq =

2  K a sin θ + Kb sin ( θ – 120) + K c sin (θ + 120) 3

1 ( K a + Kb + K c ) 3 The symbol K is replaced by i, ψ or V, to give the transformation for the current, flux linkages or voltage respectively. The transformation resolved the stator quantities into components along the direct-and quadrature-axis. Park’s equations for an ideal synchronous machine connected to an ∞ bus via a transmission system are as follows: 1. Generator K0 =

d-axis

Vd = sψ d − ψ q ω − rd id ψ d = xafd i f − xd id

q-axis

Vq = sψ q + ψ d ω − rq iq ψ q = − xq iq

Field

V f = r f i f + sψ ffd ψ ffd = x ffd i f − xafd id

The equations expressing relationship between machine fluxes and currents can now be expressed in matrix form as:

Time Domain Formulation of Dynamic Stability Problem for a Simple Power System  ψ ffd   x ffd     ψ d  =  xafd  ψ  0  q  

− xafd − xd 0

73

0  i f    0  id   − xq  iq 

Following assumptions are made: z The speed change during a disturbance is a negligible fraction of the fundamental speed; i.e., ω is assumed to be constant and in p.u. system is equal to unity. z The voltages induced in the armature viz., sψd and sψq, are negligible compared to the voltages ωψd and ωψq, generated by fluxes ψd and ψq, rotating at synchronous speed. It is usually more convenient to express the field voltage Vf in terms of p.u. field voltage Eqe required to produce a p.u. flux linking armature and field.

Eqe =

xafd rf

V f = xafd i fe

xafd sψ ffd Eqe = xafd i f + rf

or

Eqe = E + Td0 sEq′ By substitution, we get

ψ ffd

2    x ffd  xafd ψd +  xd – =  id  xafd  x ffd     

ψ ffd = x ffd i f – where

xafd

( xd –xd′ ) id

2   xafd xd′ =  xd –  x ffd  

ψ fd =

xafd x ffd

ψ ffd = xafd i f – ( xd –xd′ ) id

E′q = ωψ fd ∴

x ffd

or

Eq′ = ψ fd

[∵

ω = 1]

Eqe = E + Td0 s  xafd i f − ( xd − xd′ ) id  Eqe = xafd i f + Td0 s xafd i f − Td0 s ( xd − xd′ ) id

(

)

Eqe = 1 + Td0 s xafd i f − Td0 s ( xd − xd′ ) id

Power System Dynamics

74

(

)

Eqe = 1 + Td 0 s E − Td0 s ( xd − xd′ ) id Now Vd = – rd id − ψ q = − rd id + xq iq

Vq = ψ d − rq iq = xafd i f − xd id − rq iq Therefore the voltage equations can be written as:

(

)

 Eqe   1 + Td0 s xafd    0 Vd  =  V   xafd  q  

− Td0 s ( xd − xd′ ) − rd − xd

0  i f    xq  id   − rq  iq  

Transmission System In a manner similar to that for the generator, when the transmission system components are resolved into the generator d- and q-axes, we have Vdb = V sin δ

and Vqb = V cos δ

Vd = Vdb + re id − sψ dt + ψ qt ω

Vq = Vqb + re iq − sψ qt − ψ dt ω ψ dt = − X e id

where

ψ qt = − X eiq

Vd = V sin δ + reid − X e iq

∴

Vq = V cos δ + reiq + X e id Other Equations The electromagnetic torque Te = ψ d iq − ψ q id Pe = ω (Te ) = Te in p.u.

or

Pt = ω (Tt ) = Tt in p.u.

Similarly

The equation of motion is

( Ms

2

)

+ Ds δ = Tt – (ψ d iq – ψ q id )

Generator terminal voltage and current are Vt2 = Vd2 + Vq2

and

I 2 = id2 + iq2

All these equations define the state of the machine. All of them are non-linear and linearisation about operating point is necessary for dynamic stability studies.

 Ms 2 + Ds 0 0 0 0  ∆Tt    ∆E   0 0 0 0 0  qe      –Vd    0 0 1 0 Vt  0         0 0 0 1 0  0      ..... ..... ..... ..... ..... .....  =  0   –V cos δ 0 0 0 1    0 0 0 0  0   V sin δ  0   0 0 0 0 0    0 0 0 0 1  0    0   0 0 0 0 0     0   0 0 0 0 0     0   0 1 0 0 0 

(

)

–xe xd

–xafd

0 0 –xafd –xafd

0 0 0 0

( xd − xd′ )

xd

0

rd

– re

.....

I

..... 0 0

..... 0 1 1

– id

0

0

–Td0 s ( xd − xd′ )

0

iq

–id   ∆δ 

  0   ∆Eq′     0 0 0   ∆V   t   – iq  0 0  ∆I   I  ..... ..... .....      ∆ V d 0 0  xe  0 0   ∆Vq  –re   0 0   ∆i f  rq  0   ∆id  –xq 0    0 1   ∆iq  xq   0 1 0   ∆ψ d    0 0 0   ∆ψ q 

ψd

–ψ q

0

0

1 + Td0 s xafd

0

0

Vt

–Vq

0

0

Such a linearisation yields the following operating equations in matrix form

Time Domain Formulation of Dynamic Stability Problem for a Simple Power System 75

Power System Dynamics

76

With the governing equations of the system established, it should be possible to choose a set of state variables from among the physical variables of the machine. It is inevitable that a number of algebraic equations arise containing excess variables of the machine. These are eliminated before a definite set of differential equations can be obtained. This can be done by matrix elimination procedure. The operating matrix equations can be written as

 ∆Tt    ∆E    qe    0       0    =            0           

M1

M2

M3

M4

  ∆δ    ∆E ′   q    ∆Vt      ∆I        ∆Vd             ∆ψ q 

By matrix reduction, we get  ∆Tt   ∆δ     ∆E      ∆Eq′  –1  qe  M − M 2M 4 M 3   0  =  1      ∆Vt       ∆I   0   

This equation can be written as  Ms 2 + Ds − A1  ∆Tt    ∆E   − A3  qe   0  =  −A 5     − A7  0 

− A2

0

− A4 (1 + Tdz′ s )

0

− A6

1

− A8

0

0   ∆δ    0   ∆Eq′   0   ∆Vt    1   ∆I 

where A coefficients are obtained as

  V  A1 =  − R R − x x ′   ψ q + iq xd′ Rq cos δ − xq sin δ  d q  q d

(

)(

(

+ ψ d + id xq

)

) ( Rd sin δ + xd′ cos δ)

Time Domain Formulation of Dynamic Stability Problem for a Simple Power System

77

  1 A2 =    ψ d + id xq Rd + Rd Rq + xq xe iq − ψ q xq  − R R − x x ′  d q q d

(

)

(

)

  1 A3 =   V ( xd − xd′ ) Rq cos δ − xq sin δ   Rd Rq + xq xd′  

(

A4 = −

A5 =

)

Rd Rq + xd xq Rd Rq + xq xd′

1 V  Vd Rd Rq + xq xd′ Vt 

{( x x′ + r R ) cos δ + (r x

+Vq

q d

d

q

e q

)

− rd xe sin δ

{(r x′ − r x ) cos δ − ( x′ x e d

q e

d q

)

} }

+ rq Rd sin δ  

A6 =

Vq Vd  1 re xq − rd xe + re Rd + xe xq   Rd Rq + xq xd′  Vt Vt 

A7 =

iq i  1 V  d − Rq cos δ + xq sin δ + ( xd′ cos δ + Rd sin δ )  Rd Rq + xq xd′  I I 

A8 =

iq   id 1 xq + Rd   Rd Rq + xq xd′  I I 

(

)

(

(

)

)

 Rd Rq + xq xd′  Tdz′ =  R R + x x  Td 0  d q q d where

Rd = rd + re ; Rq = rq + re ; xd = xd + xe ; xq = xq + xe ; xd′ = xd′ + xe .

From the reduced matrix equation, choosing a state variable set X, input vector U, and output vector Y, as given below, X T =  ∆δ ∆ω

∆Eq′ 

U T =  ∆Eqe ∆Tt  Y T = [ ∆δ ∆ω ∆Vt ]

Power System Dynamics

78 we can write   ∆δ  0    A1     ∆ω  M   =   − A3      A T′  4 dz  ∆Eq′  1  y1  0 y   2 =   A5  y3 

1 –D M 0

0 1 0

0   ∆δ   0    A2     0    M  ∆ω +    −1     −1 Tdz′   ∆E ′   A4Tdz′   q  

 0  1   ∆Eqe    M     0   ∆Tt  

0   x1  0   x2  A6   x3 

The power system equations are in the form:

X = AX + BU Y = CX The solution for this matrix differential equation is obtained as: t

At A(t–τ ) BU (τ ) d τ X (t) = ε X (0) + ∫ ε 0

t

X (t) = φ(t ) X (0) + ∫ φ(t − τ ) BU (τ ) d τ

or

0

where φ(t) = ε is called the state transition matrix. Stability of the system is described by the eigen values of the system matrix A. For absolute stability all the eigen values of A should have negative real parts. At

REFERENCES z z z z z

P. M. Anderson and A. A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002.

Chapter 11

State-Space Formulation of Dynamic Stability Problem for a Multi-Machine Power System z z z z

11.1

State space formulation for dynamic stability study of a multi-machine power system Equation formulation, linearisation Synchronising power coefficients between two nodes of a multi-machine power system Formulation of state-space matrix equations and stability evaluation

MULTI-MACHINE POWER SYSTEMS

We now consider a simplified state space formulation for multi-machine power system using lower order models for each individual machine. Consider a 2-machine system 1

2 I2

I1 y12 E1

E2

y10

 I1  Y11  I  = Y  2  21

Y12   E1  Y22   E2 

Y11 = Y11 ∠θ11 = y12 + y10 Y12 = Y12 ∠θ12 = − y12 * 2 P1 = Re E1I1 = E1 Y11cos θ11 + E1E2Y12 cos (θ12 − δ12 )

Therefore, electrical power output of ith machine in an n-machine system is n

(

2 Pei = Ei Gii +∑ Ei E j Yij cos θij – δij j=1 j ≠i

)

Power System Dynamics

80 n

(

2 = Ei Gii +∑ Ei E j Bij sin δ ij +Gij cos δ ij j=1 j ≠i

where

δij Ei Yii Yij

)

δi – δj Constant voltage behind transient reactance of machine i Gii + jBii diagonal element of the network admittance matrix Y Gij + jBij off-diagonal element of Y

= = = =

The initial operating condition is perturbed, then

δ ij = δij 0 + ∆δij sin δij = sin δ ij 0 cos ∆δ ij + cos δ ij 0 sin ∆δ ij ≅ sin δij 0 + ∆δij cosδij 0

cos δij ≅ cos δij 0 − ∆δij sin δ ij 0 n

∆Pei =

∴

∑ Ei E j ( Bij cos δij 0 − Gij sin δ ij 0 ) ∆δ ij j =1 j ≠i

For a given initial condition the term in the parenthesis is a constant n

∆Pei =

thus

Ps ij =

where

∑ Psij j =1 j ≠i

∆δ ij

∂Pij ∂δ ij

(

= Ei E j Bij cos δ ij 0 − Gij sin δ ij 0 δ ij 0

)

is the change in electrical power of machine i due to a change in the angle between machines i and j with all other angles held constant. It is the synchronising power coefficient between nodes i and j. The differential equations governing the motion of the multi-machine system are

Mi Mi

d 2 ∆δ i dt 2

d 2 ∆δi

+ ∆Pei = 0

i = 1, 2,..., n

+ ∑ PSij ∆δ ij = 0

i = 1, 2,..., n

dt 2 n

j =1 j ≠i

State-Space Formulation of Dynamic Stability Problem for a Multi-Machine Power System 81 These equations are not a set of n-independent equations since

∑ δij = 0. They are actually a

set of (n –1) independent equations. Choosing one of the machines as reference (say nth machine), subtract the nth equation from the ith equation. Then we have, n

1 d 2 ∆δ i d 2 ∆δ n − + 2 2 M dt dt i

∑ PSij ∆δij − j =1 j ≠i n

1 d 2 ∆δ in + 2 Mi dt

or

∑ PSij ∆δij − j =1 j ≠i

1 Mn

n −1

1 Mn

n −1

∑ PSnj ∆δ nj

= 0

j =1

∑ PSnj ∆δ nj

= 0

j =1

i = 1,2,..., n − 1

For the sake of concreteness, consider a 3-machine system:

M1

d 2 ∆δ1

d 2 ∆δ 2

M2

dt 2

M3

+ PS12 ∆δ12 + PS13 ∆δ13 = 0

dt 2

d 2 ∆δ3 dt 2

d 2 ∆δ1 dt

2

d 2 ∆δ 2 dt

2

d 2 ∆δ3 dt 2

+ PS21 ∆δ 21 + PS23 ∆δ 23 = 0 + PS31 ∆δ31 + PS32 ∆δ 32 = 0

+ + +

PS12 M1 PS21 M2 PS31 M3

∆δ12 + ∆δ 21 + ∆δ31 +

PS13 M1 PS23 M2 PS32 M3

∆δ13 = 0

...(1)

∆δ 23 = 0

...(2)

∆δ 32 = 0

...(3)

Subtracting eqn. (3) from eqn. (1) and (2), we get

d 2 ∆δ13 dt

2

d 2 ∆δ 23 dt 2 We know

+

−

PS  PS  PS ∆δ12 +  13 + 31  ∆δ13 + 32 ∆δ 23 = 0 M1 M3  M1 M 3 

PS12

PS21 M2

∆δ12 +

PS   PS ∆δ13 +  23 + 32  ∆δ 23 = 0 M3  M 2 M3 

PS31

∆δ12 + ∆δ 23 + ∆δ 31 = 0

Power System Dynamics

82

∆δ12 = − ( ∆δ 23 + ∆δ 31 ) = ∆δ13 − ∆δ 23

∴

Using these we can write

d 2 ∆δ13 dt

2

d 2 ∆δ 23 dt 2 or

PS PS  PS   PS  PS +  12 + 13 + 31  ∆δ13 +  32 − 12  ∆δ 23 = 0  M1 M1 M 3   M 3 M1  PS PS  PS   PS  PS +  31 − 21  ∆δ13 +  21 + 23 + 32  ∆δ 23 = 0  M3 M2   M1 M 2 M 3  d 2 ∆δ13 dt 2 d 2 ∆δ 23 dt 2

+ α11 ∆δ13 + α12 ∆δ 23 = 0

+ α 21∆δ13 + α 22 ∆δ 23 = 0

α’s are dependent on machine inertias, synchronising power coefficients, etc. Choosing ∆δ13, ∆δ23, ∆ω13 and ∆ω23 as state variables, we have  ∆δ 13  0  0    0  0  ∆δ 23   ""  "" """   =   ∆ω 13   – α11 – α12    – α 21 – α 22  ∆ω 23 

0   ∆δ13    0 1   ∆δ 23  # # """"" ""    0 0   ∆ω13  # 0 0   ∆ω 23  # #

1

The general form of above matrix is given below,

 X 1   0 # I   X1       "  =  " # " "   X   A # 0   X 2   2 i.e., 2(n – 1) state variables and system matrix of the order (2n – 2) by (2n – 2) in case of a n-machine system. The dynamic response of the system depends on the eigen values of the characteristic matrix (M).

State-Space Formulation of Dynamic Stability Problem for a Multi-Machine Power System 83 I   −λI # """#"""  M =   =0  A # − λI 

i.e.,

M =

λ2I – A = 0

This has 2(n – 1) roots and if they occur as complex, they will be n – 1 complex conjugate pairs. For a 3-machine example:  λ 2 0  α11 α12   +  =0 M =  2 α α 22    0 λ   21 

 λ 2 + α11  M =  α 21

(λ

2

)(

 =0  λ 2 + α 22  α12

)

+ α11 λ 2 + α 22 − α12α 21 = 0

λ 4 + λ 2 (α11 + α 22 ) + α11α 22 − α12α 21 = 0 ∴

λ2 =

− (α11 + α 22 ) ±

(α11 + α 22 )2 − 4 (α11α 22 − α12α 21 ) 2

Examining the above, we realise that values of λ2 are negative real or λ = ± jβ or ± jγ. Then the free response will be of the form C1 cos (βt + φ1 ) + C2 cos ( γt + φ2 )

where C1, C2, φ1 and φ2 are constants.

REFERENCES z z z z z

P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002.

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Chapter 12

Transient Stability and Concepts z z z

12.1

Problem of transient stability and concepts Procedure for transient stability analysis Dynamic modelling of a synchronous generator under transient conditions

TRANSIENT STABILITY

The problem of transient stability in power system dynamics is concerned with the capacity of the system to recover after a large and sudden disturbance. The disturbances are usually different faults or changes in the system loading, system generation, network interconnections, etc. Each one of these may give completely different transient stability results. Therefore transient stability should be investigated for all a priori selected disturbances and all adopted steady state conditions. To get a good feel of the nature of the problem consider the following simple analogue of a power system.

A number of masses (Generators) are suspended from a “network” consisting of elastic strings (Electric Transmission Lines). The system is in static steady state with each string loaded below its break point (Static Stability Limit). At this point one of the strings is suddenly cut (sudden loss of electric transmission line). As a result the masses will experience transient coupled motions and the forces in the strings will fluctuate. The sudden disturbance may cause one of the two end effects: z The system will settle down to a new equilibrium state characterised by a new set of string forces (Line Powers). z Due to the transient forces one additional string will break, causing a still weaker network, resulting in an ensuing chain reaction of broken strings and eventual system collapse.

Power System Dynamics

86

If the system has inherent strength to survive the disturbance and settle in a new steady state, we refer to it as “Transient Stable” for the fault in question. It should be noted that the system may be transient stable following the loss of one particular link but unstable following another or others. Depending on the nature and duration of fault the ensuing rotor transient may be over in a second or they may continue and grow in severity over the next few seconds or even minutes ending eventually in total system collapse or recovery. Hence, it is fruitful to divide the entire transient period into the following three distinct time intervals: z Initial Interval: This extends approximately throughout the first second following the fault. The rotor dynamics in this interval are completely uncontrolled in the sense that behaviour of the generators is essentially beyond the influence of P-f (Load-frequency) and Q-V (Reactive Power Voltage) controllers. The only “control” available is switching operations including removal and reclosure of faulty line sections, insertion of capacitors, disconnection of faulty generators. z Intermediate Interval: Following the initial interval and lasting approximately 5 seconds — In this interval the effect of P-f and Q-V controllers is felt. z Final Interval: Lasts perhaps for several minutes after the onset of fault. In this period, long term effects including thermal time constants of steam and nuclear core systems and permanent loss of generating equipment, load shedding, etc. are experienced. The events of the first two intervals are of importance in that order as the system will survive the initial “shock” i.e., whether system integrity is preserved or if it is going to pieces. If the system survives the “shock” the danger is not over. Due to permanent loss of the equipment the system may be “Loosing frequency” at a faster rate or slow rate. This requires resorting to secondary control aiming at the recovery of frequency. One major control force used is “Load shedding”. By disconnecting low priority load, the downward frequency trend is reversed. For investigation of Power System Transient Stability the following steps are required: z Determine the initial pre-fault state. z Initiate fault. z Compute the post-fault transient motions of the generators and resulting power flows in the line. z If these power flows do not exceed the stability limit of the lines, the system would be judged stable for the fault in question.

12.2

DYNAMIC MODELLING OF A SYNCHRONOUS GENERATOR UNDER TRANSIENT CONDITIONS

Basic Assumptions 1. Even though Power System consists of large number of generators performing coherently, the generators are now treated individually.

Transient Stability and Concepts

87

2. As a consequence of this individuality of generator dynamics, some generators perform fast swings and others swing slowly. It is more practical to use δ i ’s which convey better, the stability states of various m/c’s than do δ i ’s . For example, voltages of two terminal buses of a particular line may be momentarily separated by 90° electrical but characterised by equal velocities. Based on velocity information we would be ignorant of the extremely dangerous situation of the line. 3. Due to large inertias, velocity deviations are small. Therefore the static portion (lines and transformers) of electrical network is considered to be 50 Hz steady state. Hence individual generators will be described by swing equations which are mutually coupled via algebraic equations describing the lines and transformers. 4. In case of balanced disturbances, algebraic equations referred to are equations relating currents or voltages at various network buses, i.e., SLFE (Static Load Flow Equations). 5. In case of unbalanced disturbances, we use equations that relate positive sequence current or voltage at various buses. This follows from the fact that only positive sequence components result in synchronising forces with in the machine. Equation of Motion Consider the ith generator in an n-unit system. Generator receives via the shaft the mechanical input or turbine power Pti. It delivers electrical power output Pei via the bus bar to network. If these are equal, generator runs at synchronous speed. But on the contrary if a difference exists, the system gets accelerated or decelerated. The equation relating this phenomenon is

d 2δ i

d δi = Pti − Pei dt dt This equation is valid for steady state and transient operation. Mi

2

+ Di

Transient Turbine Power The rotor dynamics of ith generator depends entirely on the difference Pti – Pei. When it is +ve rotor accelerates and when it is –ve, rotor will decelerate. The changes in turbine power Pti are entirely dependent on the control actions initiated by the load-frequency (P-f ) controllers. In the initial interval (1sec) the turbine power is essentially constant. For this initial interval or transient period the postfault power is equal to the pre-fault value Pti = Ptio . In subsequent periods, the controller dynamics play a part. Transient Generator Power The typical situation is as follows — A generator is operating in nominal steady state into a large network. A short circuit occurs somewhere in this network, as a result, the voltages at all its buses will experience sudden changes. The bus voltage at a particular generator will change from pre-fault value Vt° to post-fault value Vtf. The following assumptions are made: 1. Vt° and Vtf are both sinusoidal and possess three phase symmetry. The post-fault voltage Vtf may not be in phase with Vt°.

Power System Dynamics

88

2. The change in voltage is instantaneous, compared to the slow rotor transient. As a result of sudden changes in voltage, there will be corresponding changes in armature currents, thus in real and reactive power outputs.

The phasor diagram shows the voltage and current situations before and immediately after the fault. z

Pre-fault current picture: The pre-fault voltage components are Vdo , Vqo . The steady state generator currents are I o , I do , I qo .

I qo

=

Vdo Xq

;

I do

=

E o − Vqo Xd

Real parts of (Vto I o*) gives the pre-fault generator power. z Post-fault current picture: As a result of fault the d and q components of voltage change by ∆Vd and ∆Vq. These changes tend to bring about corresponding changes in d and q components of stator current. z d-Direction current change: There is a magnetic coupling between d-current component and the rotor current. The effect of this magnetic coupling is reduction of effective reactance from steady state value Xd to the transient X'd X d′ = X d −

2 X afd

X ffd

.

Transient Stability and Concepts

89

A very gradual change in voltage difference |Eo| – |Vq| would bring about a current change determined by reactance Xd. A very sudden change | ∆Vq| will on the contrary give cause to much larger current change which will be computed from

∆I d =

z

X d′ q-Direction current change: Things are simpler in q-direction. The armature flux associated with the Iq current component is not linked with rotor field winding and thus free to change. Hence the change in the Iq is computed from ∆I q =

z

∆Vq

∆Vd Xq

Transient power formula: In the post-fault condition, assume that suddenly the armature current was changed to zero by disconnecting the generator from network. Since d and q components of current would suddenly change by Ido and Iqo, the voltage changes are X'd Ido and XqIqo. The generator terminal voltage would thus change instantaneously to E q′.

E′q = E o − ( X d − X d′ ) I do I do

=

E o − Vqo Xd

Therefore E q′ in terms of pre-fault variables is E q′ =

X d′ E o + ( X d − X d′ )Vqo Xd

The phasor diagram clearly indicates that the transient post-fault phasors Eq′ , Vtf, jX 'd Idf etc. are related to each other in a similar manner as the static pre-fault voltage phasors Eo, Vto, jXd Ido etc. In fact the static and transient phasor diagrams become of identical form if we substitute Eo ⇒ Eq′ and Xd ⇒ X d′ Using this observation we can derive generator power under transient conditions as Pet

Neglecting saliency

=

2

Vt f Eq′

Vf  1 1  − sin δ + t   sin 2δ X d′ 2  X q X d′ 

X q = X d′ Pet

=

Vt f Eq′ X d′

sin δ

Power System Dynamics

90 The post-fault phasor diagram will be

REFERENCES 1. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. 2. S.B. Crary, “Power System Stability”, Vols. I and II, Wiley, New York, 1945; 1947. 3. E.W. Kimbark, “Power System Stability”, Vols. I, II, and III, Wiley, New York, 1948; 1950, 1956. 4. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. 5. V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. 6. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. 7. G.W. Stagg and A.H.El-Abiad, “Computer Methods in Power System Analysis”, McGraw-Hill, New York, 1968.

Chapter 13

Transient Stability Study z z

13.1

Typical illustrative example for transient stability study Idea of critical clearing time

A TYPICAL ILLUSTRATIVE EXAMPLE

Stability analysis depends on solution of swing equation. Transient generator power is a non-linear function of δ. Since no analytical solutions exist for non-linear differential equation, we therefore resort to numerical computational techniques. Before we go into detailed analysis of the solution, let us consider an example. Simple Power System ∞ bus V = 1.0 p.u. xe

M = 0.0106, f = 60 Hz, Xd= Xq = 0.9 pu, X'd= 0.3 p.u. X tr+Line = Xe= 0.1 p.u., V = 1.0 p.u. Initial nominal point is characterised by: o o o E = 1.5 p.u. Pe = Pt = 0.75 p.u. using the static power formula Peo = ∴ ∴

E oV 1.5*1.0 sin δ 0 = sin δ 0 Xd + Xe 0.9 + 0.1

0.75 = 1.5 sin δ 0 or δ 0 = 30o

Vqo = V cos30° = 0.866 p.u.

Power System Dynamics

92 ∴

E′q =

( X d′ + X e ) E o + ( X d − X d′ )Vqo (Xd + Xe )

E′q =

0.4 *1.5 + 0.6 * 0.866 = 1.12 p.u. 1.0 VEq′

sin δ +

 V2  1 1 − sin 2δ  2  ( X d + X e ) ( X d′ + X e ) 

Pet

=

Pet

1.0 *1.12 1.02  1 1  sin δ + − =   sin 2δ 0.4 2 1.0 0.4 

X d′ + X e

Pet = 2.8 sin δ − 0.75 sin 2δ

Exercise: Plot static and transient power vs δ (10°¸ 20°¸ 30°¸ 40°¸ 50°¸ 60°¸ …) Before proceeding further, consider a small disturbance like dropping a local load corresponding to 1° change in static power angle i.e., rotor will assume new angle 29°. For small angular deviation ∆δ o  ∂Pe  Pet = Pe +   ∆δ ∂δ  δ 0

∂Pet = 2.8 cos δ δ = δ − 1.5 cos 2δ δ = δ 0 0 ∂δ = 1.67 p.u. Pet = 0.75 + 1.67 ∆δ

∴

If D = 0, the differential equation governing the transient during small disturbance is

0.0106 or

d 2 ∆δ dt 2 d 2 ∆δ dt 2

+ 1.67 ∆δ = 0 + 157 ∆δ = 0 f =

1 157 = 2.1 Hz 2π

Now suppose the line between generator and network is suddenly subject to a solid 3-φ short circuit. The protective relaying will command disconnection of the line. A fraction of a second later the line will be “re-closed” and assuming the fault is now removed, we will be back to normal operation.

Transient Stability Study

93

However, for the period of “fault-on” the generator output Pe will be zero and the turbine power which is unchanged will accelerate the rotor of the generator. During the acceleration the swing equation is of the form

d 2δ

M or

dt 2

= Pto = Pacc

d 2δ

= 70.754 rad/sec2 (In our example) dt 2 Integrating and converting to electrical degrees δ(t ) = 30° + 20271 . t2

The rotor position increases as square of t, i.e., we experience a runaway situation. Note that during acceleration Pacc is independent of δ. At the moment of reclosure the generator power jumps to the value given by Pet = 2.8 sin δ − 0.75 sin 2δ The swing equation at the moment of reclosure is

0.0106

d 2δ dt 2

= 0.75 − ( 2.8 sin δ − 0.75 sin 2δ )

Instantaneously positive accelerating power changes to a negative decelerating power which is dependent on δ. The solution of the swing equation after the reclosure (or clearing of the fault) gives variation of δ vs t which allows us to conclude whether the system is transient stable or not. The methods for solution of swing equation are discussed in detail in the subsequent chapters.  crit II

120°

90°

Rotor swing following reclosure after 220 msec (13 cycles)

Rotor swing following reclosure after 100 msec (6 cycles)

Parabolic runaway

60°

I max

30°

0

0.1

0.2

0.3

0.4

t (sec)

Power System Dynamics

94

3.0 max 2.0 Nominal operating point 1.0 Pt°

30

60

90

120

150

180 crit

The immediate effect of torque reversal is speed reduction. The dotted portions of swing curves indicate how δ begins to decrease at the moment of reclosure. If the reclosure does not occur too late the speed of the run away rotor is halted and then reversed. This is shown in I. The rotor will settle down close to old steady state following a series of swings, i.e., System is Transient Stable. If the reclosure takes place too late the rotor will have attained too high a speed and torque reversal will only slow down the rotor although not enough to stop it before it reaches the critical angle δcrit. If the rotor is permitted to swing beyond this the power difference Pt – Pe will again become +ve and deceleration will change to acceleration and now we will loose the rotor for good. This is exemplified by case II.

REFERENCES z z z z z z z

O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. W.D. Stevenson, “Elements of Power System Analysis”, McGraw-Hill, New York, 1962. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. I.J. Nagarath and D.P. Kothari, “Modern Power System Analysis”, Tata McGraw-Hill, 1980. M.L. Soni, P.V. Gupta, U.S. Bhatnagar and A. Chakrabarti, “A Textbook on Power System Engineering”, Dhanpatrai & Co. (P) Ltd., New Delhi, 1998.

Chapter 14

Modelling and Methods of Transient Stability Analysis z z

14.1

Transient stability analysis methods — direct, indirect Detailed modelling of the power system for analysis — Voltage regulator equations; Speed Governing equations

TRANSIENT STABILITY ANALYSIS METHODS—DIRECT AND INDIRECT

In the numerical example presented, there is obviously a critical time limit beyond which the line must not remain open if stability is to be preserved. How to find this limit? In practice two methods are used: 1. Indirect solution methods, and 2. Direct solution methods. 1. Indirect Solution Methods: This is by far the most useful. The swing equations are solved during and after the fault either by analogue simulation or by some numerical digital computer oriented step by step integration methods. Stability or lack of it is determined from the looks of the resulting swing curves. The mathematical modelling of a power system is essential for transient stability analysis. Mathematical modelling involves the formulation of performance equations in the form of differential and algebraic equations for synchronous m/c’s, excitation systems, speed governing systems, loads and transmission networks. Accurate modelling of the disturbances is of importance for correct transient stability assessment in power systems. State Space models derived from above mathematical equations are used in transient stability analysis. Owing to the large size of interconnected power systems, solution to problem of transient stability analysis becomes computationally difficult and it is also uneconomical to represent the entire systems in detail. Under such conditions, study through simplified dynamic equivalent is inevitable due to the complexity involved. There are two main approaches to the problem of constructing dynamic equivalents: (a) Coherency based approach, and (b) Modal approach. Most commonly used approach is coherency based approach. However, construction of dynamic equivalents is beyond the scope of this course of lectures and is not taken up here.* *

Interested scholars may refer to the author’s book “Coherent Generators” by V.Guruprasada Rau and Md. Yeakub Hussain, Allied Publishers, New Delhi, 1998.

Power System Dynamics

96

2. Direct Analysis Methods: In these analysis methods, the question of stability can be settled without actually solving the system differential equations. The direct analysis methods fall under two categories. (i) Equal Area Method: This method is of great historical interest and is one of the first methods to explain transient stability phenomena. It is true that it can be applied with success only to a simple model of power system, i.e., a single generator connected to an infinite bus through a transmission line. However, it can be extended to predict stability in case of multi-machine power systems also. (ii) Lyapunov Direct Method: Particularly popular in other fields like control engineering, it has recently been applied to study stability of power systems. This method requires power system equations in state space form. For the sake of completeness, we will recapitulate the mathematical equations of power systems. The electromechanical equation of motion that describes the dynamics of the ith machine of an n m/c power system is:

Mi

d 2δ i 2

= Pti − Pei − Di ω i , ωi =

dt The electrical power output Pei is given by:

d δi dt

n

Pei =

∑ Ei E jYij cos (θij − δij ), δij = δi − δ j j =1

Yij ∠θij is the transfer admittance between nodes i and j.

14.2

MODELLING OF THE POWER SYSTEM FOR ANALYSIS

The mechanical input Pti can be treated as a constant or variable quantity depending upon the speed of response of prime mover governor system. If the time constants of the governor system are considered to be large compared to period of transient disturbance under study, Pti will be constant and equal to pre-fault value during the period of interest i.e., n

Pti =

∑ Ei E jYij cos (θij − δijo ) j =1

δijo = δio − δ oj at pre-fault operating point. Using these equations the basic second order model of the power system can be written as

d 2δ i dt 2

Di 1 ωi + = − Mi Mi

n

∑

j =1

(

Ei E j Gij cos δ ijo

− cosδ ij

)

+

1 Mi

n

∑ Ei E j Bij (sin δijo − sin δij ) j =1

Modelling and Methods of Transient Stability Analysis

97

If transfer conductances are neglected the basic second order model becomes

d 2δ i dt

2

=

1 Mi

n

∑ Ei E j Bij (sin δijo − sin δij ) − Mi ωi D

i

j =1

where Bij = Yij sin θij . A more realistic synchronous machine model for transient conditions can be represented by the inclusion of the effect of flux decay in the model. The equation governing the variation of the internal voltages of the ith generator is dEqi′ dt dEqi′ dt

(

)

=

1 Eqei − Ei Td0i

=

1 Eqei − Eqi′ − ( X di − X di′ ) I di Td0i

(

)

In case of a single m/c (m/c i) connected to an infinite bus I di =

∴

dEqi′ dt

=

Eqi′ − V cos δi X di′ + X ei

V is ∞ bus voltage

Eqei −1( X di + X ei ) X di − X di′ Eqi′ + V cos δi + Td0i ( X di′ + X ei ) Td0 ( X ei + X di′ ) Td0i

dEqi′

= −η1Eqi′ + η2 cos δ + η3 Eqe dt In case of multi-machine systems the flux decay equation can be written in terms of internal voltage E i′ as

Eqei dEi′ E ′ X − X di′ − i + di = dt Td0i cos φi Td0i Td0i

n

∑ Bij E j cos δij j =1

−

n tan φi X di − X di′ ) ∑ Bij E j sin δij ( Td0i j =1

φi is the angle between Ei′ and q-axis and is usually small. Thus,

Eqei dEi′ E ′ X − X di′ − i + di = dt Td0i cos φi Td0i Td0i

n

∑ Bij E j cos δij j =1

Power System Dynamics

98 n

−η1i Ei′ + η2i ∑ Bij E j cos δ ij + η3i Eqei dEi′ = j =1 dt j ≠i

or

where

η1i =

1 − ( X di − X di′ ) Bii  , η = X di − X di′ 2i Td0i Td0i

1 η3i = T cos φ d0i i

If the action of the voltage regulator is not considered Eqe will be constant and the above equation can be written as n dEi′ o o E E −η − − η ′ ′ = i i 1i 2i ∑ Bij E j cos δ ij − cos δ ij dt j =1

(

)

(

)

j ≠i

If the transient stability is determined by the events within the ‘initial’ transient period, the equations derived so far are adequate for the stability study. Should our study extend to “intermediate” transient interval, then the effects of the voltage regulator and speed governor will be felt and they must be taken into account. Voltage Regulator Equations dEqei dt

=

K (V − Vri ) −1 Eqei − Ai ti Tei Tei

Speed Governing System Equations U Gi 1 1 dPvi = − T K gi ωi − T Pvi + T g1i g1i g1i dt

dPti P P = vi − ti dt Tg 2i Tg 2i

REFERENCES z z z

O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. G.W. Stagg and A.H.El-Abiad, “Computer Methods in Power System Analysis”, McGrawHill, New York, 1968.

Chapter 15

Detailed Indirect Transient Stability Methods z z

Analogue simulation of power systems Digital simulation of power system equations

The solution of swing equation and other equations can be obtained by integration using either analogue or digital simulation.

15.1

ANALOGUE SIMULATION OF POWER SYSTEM

Analogue Simulation For illustration we consider second order model of the power system, i.e.,

d 2δ

dδ = Pt – Pe dt dt we write this equation into simpler coupled first-order differential equations by introducing two dynamic state variables x1 and x2 as x = δ ; x = δ = ω M

2

+D

1

2

(i.e., rotor angular position and rotor angular velocity) These variables constitute a state vector

 x1  δ  X =   =   x2  δ  Using these variables the second order swing equation can be written as: x1 = x2

1 ( Pt − Pe − Dx2 ) M The generator power Pe is a function of x1. Therefore, we can write these equations in more general form as: x1 = f1 ( x1 , x2 ) x2 =

x 2 = f 2 ( x1 , x2 )

Power System Dynamics

100

In this specific case f1 is a linear function of x2 and independent of x1, whereas f 2 is linear in x2 and non-linear in x1. In vector form the equations can be expressed as:

X = F ( X ) In the present case analogue simulation requires two integrators and two function generators as shown in the simulation diagram. Initial x (o) position 1 –x1

x1

1

Recorder

–f1(x1, x2)

–x2

1

Initial x (o) 2 velocity x2

–f2(x1, x2) Function generator

Note that we need two integrators per generator. The “non-linear function generator” must generate with sufficient accuracy non-linear functions of the type given by Pet = 2.8 sin δ − 0.75 sin 2δ (refer to numerical example)

The beauty of analogue simulation is that once programming is done and debugged the solution of the swing equation is obtained by all integrators and function generators working simultaneously in ‘parallel’. As a result the solution time is same for a single generator case as for a multi generator case. No Reclosure

Reclose at 0.215 sec

Reclose at 0.21 sec Reclose at 0.1 sec

30° 0.1 0.21

15.2

DIGITAL SIMULATION OF POWER SYSTEM

The parallel operation feature of the analogue computer stands in sharp contrast with integration by means of digital computer.

Detailed Indirect Transient Stability Methods

101

The general procedure of digital simulation is as follows: The independent variable t is discretised into time elements t(0), t(1), t(2), …, t(r) … not necessarily equidistant. Starting with a known initial state x(0) we can compute by means of some appropriate algorithm the new states x(1), x(2), …, x(r) … As always in numerical analysis, the accuracy of the method depends on the quality of the algorithm used. We discuss here the simplest of them. Euler Numerical Integration Method Before discussing the vector case, let us consider the scalar case: x = f ( x) (r ) t = t

For

x ( r ) ≈

∆x ( r ) = f (x(r ) ) ∆t

Using this formula we can perform integration of x = f ( x) as given below: t = t (0) and x = x(0) compute

Step 1: For

∆ x(0) = f ( x (0) ) ∆t (0) t (1) = t + ∆t we therefore have the new state

Step 2: For

x(1) = x (0) + f ( x (0) ) ∆t (1) t (2) = t + ∆t , we have

Step 3: For

(1) (1) x(2) = x + f ( x ) ∆t

Clearly the computational algorithm is

x( r +1) = x ( r ) + f ( x ( r ) ) ∆t

r = 0, 1, 2 ,...

This can be readily extended to vector case. For the ith component of the x vector, we have ( r +1)

xi

(

)

= xi( r ) + fi x1( r ) , x2( r ) ,..., xn( r ) ∆t

or in vector form, (r ) (r ) X ( r +1) = X + F ( X ) ∆t

i = 1, 2, ..., n

Power System Dynamics

102 Graphical Display of Euler Method Exact solution

Approximate solution (1)

∆x (0)

∆x

x

(2)

(1)

x

(0)

x

(1)

(0)

(2)

t

t

∆t

t ∆t

Integration starts

Flow Chart for Euler’s Method Euler’s method is simple but not particularly accurate. This is because state variables at the end of an interval are computed on the basis of the derivative in the beginning of the interval, an error will be introduced which is more pronounced the faster the derivative is changing within the interval ∆t. Read Initial State X

(0)

Set time count r = 0 (r)

r=r+1

Compute X (r +1)

Compute X

Print X No

(r)

= F (X )

=X

(r)

(r)

+X

∆t

(r + 1)

Is r ≥ rend

Yes

Stop

Modified Euler Method The accuracy of the Euler method can be improved remarkably by the obvious modification of using an average value for the derivative throughout the interval. It is possible to repeat the upgrading sequence of X(r +1) N times or until no further improvement in X(r +1) can be detected. Another important point to be remembered is the appropriate choice of step size ∆t. If not chosen properly, it could lead to errors in computation of swing curves.

Detailed Indirect Transient Stability Methods

103

There are a large variety of integration algorithms in addition to what has been described so far. But none of them possess the inherent simplicity of the Euler methods and most of them require vast amount of computation per time interval.

(r )

X ave = [ X

X

( r +1)

(r )

=X

+ X

(r )

( r +1)

(r )

] /2

+ X ave

t

REFERENCES z z z z z z z

O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. W.D. Stevenson, “Elements of Power System Analysis”, McGraw-Hill, New York, 1962. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. I.J. Nagarath and D.P. Kothari, “Modern Power System Analysis”, Tata McGraw-Hill, 1980. M.L. Soni, P.V. Gupta, U.S. Bhatnagar and A. Chakrabarti, “A Textbook on Power System Engineering”, Dhanpatrai & Co. (P) Ltd., New Delhi, 1998. G.W. Stagg and A.H.El-Abiad, “Computer Methods in Power System Analysis”, McGraw-Hill, New York, 1968.

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Chapter 16

Direct Method for Transient Stability Assessment z z z

16.1

Direct method for transient stability assessment, equal area criterion for single machine-infinite bus power system Illustrative examples for various types of faults General case, derivation of expression for critical clearing angle

DIRECT METHOD FOR TRANSIENT STABILITY ASSESSMENT USING EQUAL AREA CRITERION

Equal Area Criterion: Of great historical interest is the fact that a direct stability analysis method, the so called Equal Area Method proved very useful in the first attempts that were made to explain Transient Stability phenomenon in Power Systems. In this method it is not necessary to plot and inspect the swing curves to determine whether the torque angle of the machine increases indefinitely or oscillates around an equilibrium position. It is true that equal area method has been applied with real success only to a single machine system, however it can be extended to multi machine systems also in a limited way. Formulation for Single Machine System: The swing equation for the system is

M

d 2δ dt 2

Multiplying both sides by

M

= Pt − Pe = Pa

dδ we have dt

dδ d 2δ d δ ( Pt − Pe ) dt 2 dt = dt 2

or

 dδ  d   dt  1 M dt 2

2

or

 dδ  d   dt 

= ( Pt − Pe ) =

dδ dt

2 ( Pt − Pe ) d δ M

( D = 0)

Power System Dynamics

106 Integrating δ

2

2  dδ    = M dt

dδ = dt

or

δ

∫

∫ ( Pt − Pe ) d δ

δ0

δ0

2 ( Pt − Pe ) d δ M

δ0 is the torque angle when the machine is operated synchronously before the disturbance occurs at dδ = 0. The angle δ will cease to change and machine will again be operating at which time dt dδ synchronous speed after a disturbance when =0 dt δ

or when

∫

δ0

2 ( Pt − Pe ) M

dδ = 0

dδ dt becomes zero, but the fact that δ has momentarily stopped changing may be taken to indicate stability, which corresponds to the interpretation that the swing curve indicates stability when the angle δ reaches a maximum δm and starts to decrease.

However, the machine will not remain at rest with respect to the infinite bus the first time

16.2

ILLUSTRATIVE EXAMPLES FOR VARIOUS TYPES OF FAULTS

Single line diagram of a power system is shown below T1

tr–1

G

T2 ∞ bus

tr–2

The positive sequence impedance diagram of this is

G

∞

Direct Method for Transient Stability Assessment

107

I. Reduction In Output Power Due To Line Opening Opening of one of the lines due to fault will allow the load to be supplied over the remaining line. With one line opened the positive sequence diagram of the system becomes:

∞

G

The opening of the line increases the reactance between generator and the bus. The generator is accelerated because of the reduced power output resulting from increased transfer reactance. Acceleration increases the torque angle. Power output curves against δ are shown below: A2

Pt

Pe (2 lines) Pe (1 line)

A1

δ0

δe

δm

δ

Pt is the mechanical power input to the generator, δ0 is the torque angle corresponding to initial condition. When one line is opened the power output drops from the point determined by δ0 and the upper curve to δ0 and lower curve. The excess power input over power output causes acceleration which increases δ. From δ0 → δe the generator accelerates and from δe → δm the generator is subjected to deceleration. From δm the torque angle decreases and thus oscillates between δ0 and δm. The maximum swing δm is found by graphic interpretation of the equation: δ

∫

δ0

2 ( Pt − Pe ) M

dδ = 0

When this equation is satisfied, given by

dδ = 0 and maximum value of δ is reached. The shaded area A1 is dt δe

A1 =

∫ ( Pt − Pe ) d δ

δ0

Power System Dynamics

108 Similarly A2 is given by δm

A2 =

∫ ( Pe − Pt ) d δ

δe δe

A1 − A2 =

∫ ( Pt − Pe ) d δ

δ0

−

δm

∫ ( Pe − Pt ) d δ

δe

δm

A1 − A2 =

∫ ( Pt − Pe ) d δ

δ0

The equation (describing equal area criterion) is satisfied when A1= A2 and hence the name. II. Short Circuit Faults Short circuit faults often cause loss of stability even though they may be removed by isolating the fault from the rest of the system. A 3-φ fault at one end of the double circuit line is represented in the following figure.

P

∞

When the fault is cleared by opening the circuit breakers at both ends of faulted circuit power is again transmitted from generator to ∞ bus over the remaining line. To this case equal area criterion is applied as shown.

Direct Method for Transient Stability Assessment

109

The maximum angle of swing is fixed by the condition A1 = A2. If the clearing of the faults takes place later than δc, it may be impossible to make area A2 above Pt equal to A1 and system will lose stability. Thus for any Pt there is a critical clearing angle and unless the fault is cleared before δ equals critical clearing angle, the m/c will lose synchronism. It is evident that larger values of Pt require quicker clearing of fault. III. General Case When a 3-φ fault occurs at some point on a double circuit line other than on the paralleling buses or at the extreme ends of the line, there is some impedance between paralleling buses and the fault. Therefore, some power is transmitted while the fault is still on the system. Regardless of their location, short circuit faults not involving 3-phases allow the transmission of some power, because they are represented by connecting some impedance between the fault point and the reference bus in the +ve sequence impedance diagram. The larger the impedance shunted across the +ve sequence network to represent the fault, the larger the power transmitted during the fault. The power transmitted during the fault may be calculated after reducing the network that represents the fault condition to a ∆-connected circuit between the internal voltage of the generator and the ∞ bus. Two examples of such reduction are shown in the following figures.

E

V

E

3-φ fault at the middle of one of two parallel lines

V

Fault other than 3-φ fault at the end of one line Xb

E

Xa

Xc

Equivalent circuit diagram

V

Power System Dynamics

110

If power is transmitted during the fault, the equal area criterion is applied as shown in figure below. Pmsin δ

Pe

r2Pmsin δ Pt r1Pmsin δ

δ0

δc

δm

δ

Pmsin δ = Power transmitted before fault r1Pmsin δ = Power transmitted during fault r2 Pmsin δ = Power transmitted after the fault is cleared by switching at δ = δc For the case presented δc is evidently the critical clearing angle since A2= A1 for the torque angle δm where Pt intersects r2Pmsin δ. The power transmitted during fault helps to reduce the value of A1 for any given clearing angle. Thus smaller is r1 greater is the disturbance to the system. In order of increasing severity (decreasing r1Pm) the various faults are: 1. Single line-to-ground fault, 2. Line-to-line fault, 3. Double line-to-ground fault, and 4. Three phase fault. The single line-to-ground fault occurs most frequently and 3-phase fault is least frequent. For complete reliability system should be designed for transient stability for 3-phase faults at worst locations. Usually, for a practical reasons, designing is done for double line-to-ground faults. An expression for critical clearing angle is derived from the figure as follows: A1 = Pt ( δ c − δ 0 ) −

δc

∫ r1Pm sin δ d δ

δ0

A1 = Pt (δ c − δ 0 ) + r1Pm (cos δ c − cos δ 0 )

and δm

A2 =

∫ r2 Pm sin δ d δ − Pt (δ m − δc )

δc

Direct Method for Transient Stability Assessment

111

A2 = r2 Pm ( cos δ c − cos δ m ) − Pt (δ m − δ c )

For stability

A1 = A2 or

Pt δ c − Pt δ 0 + r1 Pm cos δ c − r1 Pm cos δ 0

= r2 Pm cos δ c − r2 Pm cos δ m − Pt δ m + Pt δ c ∴

∴

(r1 − r2 ) Pm cos δ c

= Pt (δ 0 − δ m ) + r1Pm cos δ 0 − r2 Pm cos δ m

−1 δ c = cos

Pt (δ m − δ0 ) + r2 cos δ m − r1 cos δ 0 Pm r2 − r1

To evaluate δc it should be noted that Pt = Pm sin δ 0 –1 δ 0 = sin

Pt Pm

Pt = r2 Pm sin δ m

δ 0 < 90o ;

δ m = sin –1

Pt r2 Pm

δ m > 90o.

REFERENCES z z z z z

W.D. Stevenson, “Elements of Power System Analysis”, McGraw-Hill, New York, 1962. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. I.J. Nagarath and D.P. Kothari, “Modern Power System Analysis”, Tata McGraw-Hill, 1980. P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977.

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Chapter 17

Application of Equal Area Criterion to MultiMachine Power System Transient Stability Assessment z

17.1

Application of equal area criterion to multi-machine power system transient stability

MULTI-MACHINE STABILITY PROBLEMS

The essence of the method is considering one machine at a time and treating rest of the system as an ∞ bus. For this we assume that: (i) We will consider one generator node at a time and test it for transient stability for the case when the most heavily loaded line connected to it is lost. (ii) The line and generator resistances are neglected. (iii) Transient period is short i.e., Pt , Vi are unchanged at various nodes. i For generator at node i , the dynamic behaviour after tripping of the faulty line is given by

Mi or

Mi

d 2δ i dt

2

d 2δ i dt 2

n

ViV j

j =1

X ij

= Pti − ∑ n

sin δij

(

δij = δ i − δ j

= Pti − ∑ γ ij sin δi − δ j j =1

)

γ ij = Capacity of the transmission line between nodes i and j Consider the following diagram showing the relative positions of voltage phasors corresponding to different nodes with respect to an arbitrary reference. δi – δ1 γ i 1 sin(δi – δ1) γ i 2 sin(δi – δ 2 ) γ i 3 sin(δi – δ 3 )

δi – δ 2

Power System Dynamics

114

Draw vectors A1= γi1, A1= γi2,... along the voltage vectors V1,V2,… etc. Then the expression

∑ γ ij sin (δi − δ j ) j

is simply the sum of the projection of the vectors A1, A2,… on the axis POP' ⊥ lar to Vi. This in turn will be a projection of the vector Yi which is vectorial sum of A1, A2, …

Yi =

∑ γ ij2 + ∑ 2γ ij j

j ,k j ,k ≠ i

(

γ ik cos δ j − δ k

)

Note that Yij is a function of γij and relative angular difference between other nodes not involving i. If we consider Yi direction as reference

Mi

d 2δ i dt

2

= Pti − Yi sin δ i

The equation is similar to that of a single machine connected to ∞ bus with the difference that equivalent transmission capacity Yi linking node i to remainder of the system is not strictly constant but will vary with angular differences δj – δk that do not remain constant. However, it can be seen that value of Yi consists of two parts, one depending on γij which remains constant at its post-fault value, and other depending on cosines of relative angles between nodes other than i. The primary effect of change in capacity Yi will be on angles of branches connected to i whereas δj – δk j, k ≠ i will not be affected as severely. Thus change from pre-fault to post-fault values in the relative angles between nodes not involving i is often so small that amplitude Yi sin δi represents the familiar sinusoidal P – δ curve of a single m/c connected to ∞ bus and the transient stability of node i can be analysed by equal area criterion. Thus if we use post-fault values of γij and post-fault δj – δk and denote this value of Yi as Yi*, then Yi*sin δi will approximately represent post-fault power angle curve at node i. Similarly if we can use pre-fault values of γij and pre-fault steady state values of δj – δk to calculate Yi and denote it as Yi°, then Yi° sin δi represents approximately pre-fault power angle curve at node i.

Application of Equal Area Criterion to Multi-Machine Power System ....

115

For stability A1 ≤ A2, Defining S = (A2–A1)/A2 For critical stability S = 0, i.e., net accelerating torque is equal to net restoring torque and the transient stability limit has been reached. For S > 0, greater the value of S, more stable the system is * ° ° and it operates below transient stability limit. The value Z = (Yi – Yi ) / Yi represents relative reduction in the equivalent admittance. Usually Z% and Pti / Yi° % are plotted.

REFERENCE z

M.W. Siddiqee and John Peschon, “Application of Equal Area Criterion to MultiMachine Power System Stability Problems”, Presented at IEEE, PES Winter Meeting, 1969.

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Chapter 18

Lyapunov Direct Method z z

18.1

Lyapunov direct method, concepts of Lyapunov stability, asymptotic stability and instability Asymptotic stability theorem, proof

CONCEPT OF LYAPUNOV STABILITY, ASYMPTOTIC STABILITY AND INSTABILITY

Mathematical Basis A system is defined by

X = F ( X, t )

or

X = AX

Equilibrium state is that state of the system for which X =0 i.e.,

x1 = f1 ( x1 , x2 , … , xn ) = 0

x2 = f 2 ( x1 , x2 , … , xn ) = 0 xn = f n ( x1 , x2 , …, xn ) = 0

Number of equilibrium states are independent of n. For a linear system, the equilibrium condition is unique if |A| ≠ 0. If we wish to examine behaviour of the system in the vicinity of an equilibrium point (x1e, x2e, …,xne), we can use the coordinate transformation. y1 = x1 − x1e y2 = x2 − x2e yn = xn − xne Then the state equilibrium equations become

Power System Dynamics

118 y1 = g1 ( y1 , y2 ,…, yn ) y2 = g 2 ( y1 , y2 ,…, yn ) yn = g n ( y1 , y2 ,…, yn )

i.e., the system has an equilibrium point at the origin. Lyapunov’s Stability study is based on this kind of formulation of equations. Stability: The most significant interpretation attached to equilibrium condition is stability. Stability is commonly understood as a situation where the system is in equilibrium and when perturbed returns in time to equilibrium or close to it. Now consider a spherical region of radius k about an equilibrium state of Xe as ||X – Xe|| ≤ k, where ||X – Xe|| is called the Euclidean Norm. 1

X – Xe

2 2 2 = ( x1 − x1e ) + ( x2 − x2 e ) + … + ( xn − xne )  2  

In a 2-dimensional case, consider a circle of radius of δ at the origin. x1 x0 ε

ε δ t

xe S(δ)

x2

S(ε)

S(δ) consists of all points such that ||X – Xe|| ≤ δ (X0 is such a point). S(ε) consists of all points such that ||X – Xe|| ≤ ε. The equilibrium state Xe is said to be Stable in the sense of Lyapunov, if corresponding to each S(ε) there is an S(δ) such that trajectories starting in S(δ) do not leave S(ε) as t increases indefinitely. The real numbers δ depends on ε and in general on t = t0. If δ does not depend on t0, then the equilibrium state is said to be uniformly stable. Asymptotic Stability: An Xe of the system is said to be asymptotically stable if it is stable in the sense of Lyapunov and every solution starting in S(δ) converges without leaving S(ε) to Xe as t increases indefinitely.

Lyapunov Direct Method

119

Lim X – X e = 0 t →∞

In practice asymptotic stability is more important than mere stability. Since asymptotic stability is a local concept (i.e., valid for particular Xe) its simple establishment does not ensure that system will operate properly. Some knowledge of the size of the largest region of asymptotic stability is usually necessary. This region is called the domain of attraction. All trajectories originating in the domain of attraction are asymptotically stable. Xe is said to be asymptotically stable at large if it is stable and every solution of system equations converges to Xe as t ⇒ ∞. Obviously such situation entails only one equilibrium state for the system. Instability An equilibrium state Xe is said to be unstable if for some ε > 0 and any real δ > 0 no matter how small, there is always a state X0 in S(δ) such that the trajectory starting at this state leaves S(ε). Phase Plane Interpretation x2

x2

S(ε)

x2

S(ε) S(δ)

S(δ) x0

x1

δ

x0

x1

x1

δ

1. Stable: Trajectories tend towards an orbit as t ⇒ ∞ 2. Asymptotic Stability: Trajectories tend towards the origin as t ⇒ ∞ 3. Unstable: Trajectories tend towards ∞ as t ⇒ ∞ One of the problems of stability analysis is to find necessary and sufficient conditions on the system parameters so that convergence of solution takes place. One approach is to obtain the explicit solution of the related differential equations:

X = F ( X,U ) ; (By numerical methods) X = AX ;

X = AX + BU

Considering a linear system For

X = AX + BU

if X = PZ

Z = P –1 APZ + P –1 BU

P –1 AP = D where D is the diagonal matrix of eigenvalues of A

Power System Dynamics

120 t

Dt Dt − Dτ P B U dτ Z (t) = ε Z (0) + ε ∫ ε 0

t

X = Pε Dt Z (0) + ∫ Pε D (t – τ ) P B U d τ 0

It is very clear that for Asymptotic Stability all the eigen values of A matrix should have negative real values. Lyapunov Direct Method answers the questions of stability without explicit solution of related differential equations. The underlying idea of Lyapunov method is to find a positive definite scalar function V(X) with time rate of change V(X) for every possible state X of the system –ve, except for an equilibrium state where it attains its minimum value. Then the function V(X) will decrease along the solution X(t) of the system until it assumes its minimum value and system reaches the equilibrium. Lyapunov’s method is formulated on the concept of system energy. It is well known that during a transient total energy of a non-conservative physical system would continuously decrease until it finally assumes it’s minimum value at an equilibrium point. In other words, the rate of change of system energy for an isolated physical system is negative for every possible system state except for equilibrium condition. In view of this, a dissipative system perturbed from its equilibrium condition to a neighbouring state returns to the original state.

18.2

LYAPUNOV’S ASYMPTOTIC STABILITY THEOREM

In neighbourhood of origin if there exists a differentiable, real, variable function V that satisfies the following conditions: 1. V(x1, x2, …, xn) > 0 V(x1, x2, …, xn) = 0 for xi = 0, i = 1, 2,…, n i.e., V(X) is positive definite. 2.

n dV ∂V dxi = V =∑ < 0 for all xi ≠ 0 and t > 0 dt i =1 ∂xi dt

i.e., V is negative definite and not identically zero on the trajectory of the given equation or system. 3. V ⇒ ∞ for all ||X|| ⇒ ∞, then the point X = 0 is asymptotically stable in the sense of Lyapunov and V(X) is called a Lyapunov Function for the system. Proof: Consider x1 = a11x1 + a12 x2 x2 = a 21x1 + a22 x2

Lyapunov Direct Method

121

A candidate Lyapunov function V(X) is chosen as V(X) = x12 + x22 which is +ve definite for all values of states x1 and x2 except the origin. Thus V(X) is a +ve definite function of X. Geometric interpretation of such a function is the concave cup shown in figure below (left). V(X) V increases

x2 x(0)

x2 xe

x1

x1

If the state or phase plane (x1 – x2) is considered, in this plane V(X) = constant represents a family of concentric circles which surround the origin as shown above (right). Now, or

V (X ) =

∂V ∂V x1 + x2 = 2 x1 x1 + 2 x2 x2 ∂x1 ∂x2

V ( X ) = 2 x1 ( a11 x1 + a12 x2 ) + 2 x2 ( a21 x1 + a22 x2 ) V ( X ) = 2a11 x12 + 2a12 x1 x2 + 2a21 x1 x2 + 2a22 x 22  a11 a12   x1   a11 a12   x1  + [ x1 x2 ]  = [ x1 x2 ]        a12 a22   x2   a21 a22   x2 

for V ( X ) to be negative definite a11 < 0 2 a11a22 − a12 >0 It is obvious that a22 < 0 and

a11 < 0 2 a11a22 − a21 >0

a11 > a12

a11 > a21

a22 > a12

a22 > a21

Power System Dynamics

122 If the original system equations are: x1 = f1 ( x1 , x2 )

Conditions for V ( X ) are to be established

x2 = f 2 ( x1 , x2 )

depending on f1 and f2.

V ( X ) = 2 x1 f1 ( x1 , x2 ) + 2 x2 f 2 ( x1 , x2 ) If V ( X ) is negative at all points of phase space, except at the origin, then a point in state space will move towards decreasing values of V(X). A geometric interpretation of Lyapunov function V(X) is a measure of the distance of a state X from the origin of the state space. If the distance between instantaneous state and the origin is continuously decreasing as t increases i.e., V ( X ) < 0 and hence X(t) tends to the origin. Since Lyapunov function is regarded as defining distance from the origin in state space, its derivative is a quantitative estimate of the speed with which the origin is approached. We would look at it in another way. Consider Lienard’s equation

x + f ( x) x + g ( x) = 0 This represents a dissipative system. If we convert this into a conservative or dissipationless system then f ( x) x = 0 then the total energy or Hamiltonian of the system is constant. In phase plane the system equations then become, x1 = x2 x2 = − g ( x1 ) we assume g(0) = 0 and g(x1) ≠ 0 for x1 ≠ 0. In this case equilibrium point is the origin. The system trajectories enclose and are defined by the following differential equation of the integral curves.

Cross multiplying

dx2 − g ( x1 ) = dx1 x2

x2 dx2 + g ( x1 ) dx1 = 0

Integration yields x22 + 2

x1

∫ g ( τ) d τ

= H (constant)

0

The trajectories are concentric family of circles depending upon g(x). The total energy for system, also the Hamiltonian is composed of two scalar energy functions x2 K.E. = 2 2

P.E =

x1

∫ g (τ) d τ 0

Lyapunov Direct Method

123

dH = H = x2 x2 + g ( x1 ) x1 = − x2 g ( x1 ) + x2 g ( x1 ) = 0 dt For the non-conservative case f ( x) x > 0. Therefore system state equations are x1 = x2 x2 = − f ( x1 ) x2 − g ( x1 )

The equation of the trajectories of the system is

dx2 f ( x1 ) x2 + g ( x1 ) = − dx1 x2 or

x2 dx2 + g ( x1 ) dx1 + x2 f ( x1 ) dx1 = 0

Now the total energy H for the system is x2 H = 2 + 2

x1

∫ g ( x1 ) dx1 0

2 H = x2 x2 + g ( x1 ) x2 = − f ( x1 ) x2

or

If f(x1) > 0 then H < 0 The total energy is always decreasing along any solution except x2 = 0. The trajectory of motion is from higher energy contour to that of lower energy contour. Hence by integrating the trajectory equation function it is possible to construct a Lyapunov function. It has been seen that the Lyapunov function V(X) and its derivative V ( X ) are of quadratic form. A quadratic function of 2 variables is 2 2 Q = a11 x1 + a22 x2 + 2a12 x1 x2

This can be written as  a11 x2 ]   a12 In 3 variables, general quadratic form is

[ x1

a12   x1  a22   x2 

Q = a11 x12 + a22 x22 + a33 x32 + 2a12 x1 x2 + 2a23 x2 x3 + 2a31 x3 x1

[ x1

x2 x2 ]

 a11   a12  a13

a12 a22 a23

a13   a23  a33 

 x1  x   2  x3 

Power System Dynamics

124

Q = X T AX In general where A is symmetric Suppose A is diagonal then Q = X T AX becomes, for example Q = [ x1 x2 x2 ]

0  a11 0    0 a22 0   0 0 a33 

 x1    2 2 2  x2  = a11 x1 + a22 x2 + a33 x3  x3 

REFERENCES z z z z

J. Lasalle and S. Lefschetz, “Stability by Lyapunov’s Direct Method with Applications”, Academic Press, New York, 1961. V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. M.A. Pai, “Power System Stability”, North Holland, Amesterdam, 1981.

Chapter 19

Methods of Construction of Lyapunov Functions z z z

19.1

Sylvester’s criterion for definiteness and semi-definiteness Construction of Lyapunov functions Variable gradient method — basis, step wise procedure, application to power system examples

SYLVESTER’S CRITERIA FOR DEFINITENESS AND SEMIDEFINITENESS

Positive Definiteness Necessary and sufficient conditions for Q = X TAX to be positive definite, where A is a real symmetric matrix of order n × n. (i) |A| is positive, and (ii) Successive principal minors of |A| be positive. i.e., a11 > 0;

a11 a12 a12 a22

a11 a12 a13 > 0;

|A| > 0

a12 a22 a23 > 0 a13 a23 a33

Negative Definiteness Q = X TAX is negative definite where A is a n × n real symmetric matrix, if and only if, 1. Successive principal minors of even order are +ve, 2. Successive principal minors of odd order are –ve, and 3. |A| > 0 if n = Even, |A| < 0 if n = odd Positive Semi-Definiteness Q = X TAX is positive semi-definite, if 1. All principal minors are non-negative

Power System Dynamics

126

a11 ≥ 0;

a11 a12 a12 a22

a11 a12 a13 ≥ 0;

a12 a22 a23 ≥ 0 a13 a23 a33

2. |A| = 0. Negative Semi-Definiteness Q = X TAX is negative semi-definite if 1. All principal minors of even order are non –ve a11 a12 a12 a22

≥0

2. All principal minors of odd order are non +ve a11 a12 a13 a11 ≤ 0;

a12 a 22 a23 ≤ 0 a13 a23 a33

3. |A| = 0. Examples: 1.

X T AX = 10 x12 + x22 + 4 x32 + 4 x1 x2 + 6 x1 x3 + 2 x2 x3

10 2 3   A =  2 1 1  3 1 4  ∴ X T AX is a positive definite 2.

1. A > 0 2. 10 > 0;

10 2 2 1

>0

x12 − 3 x22 + x32 + 4 x1 x2 + 8 x1 x3

1 2 A =  2 – 3  4 0

4 0  1

1. A > 0 2. 1 > 0

1

2

2 –3

0

Choice of α11 =

α 21 =

2

(1 − x1x2 )2 x12

(1 − x1x2 )2

, α12 =

− x12

(1 − x1x2 )2

, α 22 = 2

Give

V = −2 x12 − 2 x22 , V =

For stability

x12 + x22 1 − x1x2

(1 − x1x2 ) > 0 REFERENCES

z z

V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976.

Chapter 20

Generation of Lyapunov Functions z z z z

20.1

Energy metric algorithm for generation of Lyapunov functions, basis and stepwise procedure Application to power system examples Evaluation of region of asymptotic stability, illustrative numerical examples Application of Lyapunov function approach to test the asymptotic stability of a linearised system

ENERGY METRIC ALGORITHM FOR GENERATION OF LYAPUNOV FUNCTIONS

The Energy Metric Algorithm is based upon a generalisation of total energy of the system and Lyapunov function is obtained from the line integral of a composite differential one form derived directly from the system equations. The procedure is as follows: Step 1: Describe the system as a set of first-order differential equations. xi = fi ( X )

i = 1, 2,...,n

Step 2: Form a set of differential equations of integral curves by taking quotients of the first-order equations there by eliminating dt in each case

fi ( X ) dxi = , j >i f j (X ) dx j There will be

n(n – 1) such equations 2

Step 3: The differential equations of integral curves are converted to the same number of differential one forms:

f j ( X ) dxi − fi ( X ) dx j = 0 Step 4: By addition and substitution these differential one forms are reduced to a single one form. W = w1 ( X ) dx1 + w2 ( X ) dx2 + " + wn ( X ) dxn Step 5: A line integration of the resulting one form is taken along the elbow path (0 → x1 → x2 → … → xn). The result of the line integration is the V function given as:

Power System Dynamics

136 V = ∫ W dX

V =

x1

x2

0

0

∫ W1 (τ1,0,",0) d τ1 + ∫ W2 ( x1, τ 2 ,0,",0) d τ 2 + ... +

xn

∫ Wn ( x1 , x2 ," xn–1 , τ n ) d τ n 0

Step 6: The generated V function and its derivative V are tested for the sign definiteness and conditions for the same are established. The Energy Metric Algorithm has been applied to both single machine and multi machine power systems.

20.2

APPLICATION TO POWER SYSTEM

Single machine connected to ∞ bus (neglecting transfer conductance). The equations are,

M

d 2δ 2

+D

dδ = Pt − Pm sin δ , dt

dt Defining the state variables as

Pm =

EV X

x1 = δ – δ o dδ =ω x2 = x1 = dt Step 1: The state equations are x1 = x2

or

x2 = −

D 1 x2 −  Pm sin ( x1 + δ 0 ) – Pt  M M

x2 = −

P D x2 − m sin ( x1 + δ 0 ) – sin δ 0  M M

Step 2:

dx1 dx2 = Steps 3 and 4:

x2 Pm D sin ( x1 + δ 0 ) – sin δ 0  − x2 − M M 

Pm D  sin ( x1 + δ 0 ) – sin δ 0   dx1 W ( X ) = x2 dx2 +  x2 + M M 

Generation of Lyapunov Functions

137

W ( X ) = Mx2 dx2 +  Dx2 + Pm (sin ( x1 + δ 0 ) – sin δ 0 ) dx1

or Step 5:

V = ∫ W ( X )dX =

x1

x2

0

0

∫ Pm (sin ( τ + δ 0 ) – sin δ 0 ) d τ + ∫ M τ d τ

Mx22 + Pm cos δ 0 − Pm cos ( x1 + δ 0 ) − x1Pm sin δ 0 2 ∂V ∂V x1 + x2 V = ∂x1 ∂x2 V =

∴

(

)

= Pm sin ( x1 + δ 0 ) – sin δ 0 x2

P  D  + Mx2  − x2 − m sin ( x1 + δ 0 ) – sin δ 0  M  M 

(

∴ Also

2 V = − Dx2

)

which is negative definite for D > 0

M ω2 + ( Pm cos δ 0 − Pm cos δ − Pt x1 ) 2 V > 0 for X ≠ 0 if ( Pm cos δ + Pt x1 ) < 0 V =

(cos ( x1 + δ0 ) + x1 sin δ0 ) < 0

or

In this method which is flexible, particular substitution and mathematical operation chosen in Step 4, determine the terms that appear in Lyapunov function, especially in case of higher order models of power systems.

20.3

EVALUATION OF REGION OF ASYMPTOTIC STABILITY

Taking for illustration single machine – ∞ bus system, we have V =

1 Mx22 + Pm cos δ 0 − cos ( x1 + δ 0 ) − x1 sin δ 0 2

(

)

2 V = − Dx2

The unstable equilibrium state nearest to stable equilibrium state δ0 is given by x1 = 0;

i.e., when

x2 = 0, and

sin ( x1 + δ 0 ) = sin δ 0

x2 = 0

Power System Dynamics

138

x1 = Π – 2δ 0 ∴ This point is also called as Saddle Point (x1s). Then V(x1s, 0) gives the critical value Vcr. The state plane trajectory corresponding to Vcr value will define the regions of Asymptotic Stability or Region of Attraction.

Now where

Vcr =

1 Mx22 + Pm (cos δ 0 − cos ( x1 + δ 0 ) − x1 sin δ 0 ) 2

Vcr = Pm  2cos δ 0 − ( Π − 2δ 0 ) sin δ 0 

Vcr curve is a closed curve and gives the largest stability boundary. The phase plane diagram shown below illustrates these points more clearly (Illustrative numerical example at the beginning of transient stability lectures is considered here). These parabolic trajectories are traced during fault for different initial operating points before reclosure. x2 = δ

2 1

0

Saddle point

x1 =

δ

These trajectories are traced after reclosure

Separatrix

For a typical operating point ‘O’. Following onset of fault generator traces the bold face trajectory. Early reclosure at 1 will give stable trajectory and a late reclosure at 2 will give unstable trajectory.

Generation of Lyapunov Functions

139

Separatrix defining the region of attraction. Trajectories inside the separatrix are stable, those outside are unstable. Vcr corresponds to separatrix.

20.4

APPLICATION OF LYAPUNOV METHOD TO LINEAR SYSTEM

The linearised power system equations can be directly related with the V function by using a theorem as shown below. To relate linearised system equations directly with the V function, we use the following theorem. Given the system equations as: X = AX . The linear system given above is asymptotically stable if and only if given any symmetric positive definite matrix Q there exists a symmetric positive definite matrix P which is the unique solution of AT P + PA = –Q Proof: To prove the sufficiency of the above result let us assume that a symmetric positive definite matrix P exists which is the unique solution of the above equation. Consider the scalar function V(X) = X TPX V(X) > 0 for X ≠ 0

V(0) = 0 The time derivative of V(X) is

V ( X ) = X T PX + X T PX = X T AT PX + X T PAX

(

)

T T = X A P + PA X T = – X QX

Since Q is positive definite V ( X ) is negative definite. Hence the system is asymptotically stable. To show that the result is necessary, suppose the system is asymptotically stable and P is negative definite. Consider the scalar function

V ( X ) = – X T PX T T V ( X ) = –  X PX + X PX 

= X T QX >0

Power System Dynamics

140

This is a contradiction. Hence the conditions of positive definiteness of P are necessary and sufficient for asymptotic stability of the system. It can be shown that solution of ∞

T

A P + PA = –Q

is

P=

∫ε

AT t

Qε At dt

0

This integral exists if Re λ i < 0 ∞ ∞ T  T AT ∫ ε A t Qε At dt +  ∫ ε A t Qε At dt  A 0  0

∞

=

∫ 0

(

∞

=

T

T

)

AT ε A t Qε At + ε A t Qε At A dt

d

∫ dt ε

AT t

0

Qε At  dt  ∞

AT t At =  ε Qε  = – Q  0

Illustrative example: Determine the stability of equilibrium state of the system. x1 = – x1 – 2 x 2

x2 = x1 – 4 x2  –1 – 2  A =    1 – 4

The equilibrium state is x1 = 0, x2 = 0 or X = 0. Let us solve the following equation: – AT P + PA = – I  –1 1   P11  – 2 – 4  P    12

P12   P11 + P22   P12

1 0  Q = 0 1    P12   –1 2   –1 0  =      P22   1 – 4  0 –1

If P matrix turns out to be positive definite then XTPX is a Lyapunov function and the system is asymptotically stable.

Generation of Lyapunov Functions  – P11 + P12  –2 P + 4 P 11 12 

141

– P12 + P22   – P11 + P12 + –2 P12 – 4 P22   – P12 + P22

–2 P11 – 4 P12   –1 0  = –2 P12 – 4 P22   0 –1

–2 P11 + 2 P12 = –1 –4 P12 – 8 P22 = –1 –2 P11 – 5 P12 + P22 = 0

By solving the equations, we get P11 =

23 –7 11 ; P12 = ; P22 = 60 60 60

 23  P =  60  –7  60

–7  60   11  60 

By Sylvester’s criterion this P matrix is positive definite. Hence we conclude that the system is asymptotically stable.  23  V ( x) = X T PX = [ x1 x2 ]  60  –7  60

=

–7  60   x1   11   x2  60 

1  23x12 –14 x1 x2 + 11x22   60

2 2 V ( x) = – x1 – x2

REFERENCES z z z z z

V. Guruprasada Rau and N. Krishnamurthy, “Stability Analysis of Power Systems”, Omega Scientific Publishers, New Delhi, 1994. K. Ogata, “Modern Control Engineering”, Prentice Hall of India (P) Ltd., New Delhi, 1976. E. Handschin, “Real Time Control of Electric Power Systems”, Elsevier, New York, 1972. O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. Yao-nan Yu, “Electric Power System Dynamics”, Academic Press, New York, 1983.

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Chapter 21

Exercise Problems 21.1

PROBLEMS WITH SOLUTIONS

Problem 1 Consider a synchronous machine characterised by the following parameters: X d = 1.0, X q = 0.7, X d′ = 0.4 per unit and negligible armature resistance. The machine is connected directly to an infinite bus of voltage 1.0 per unit. The generator is delivering a real power of 0.6 per unit at 0.8 power factor lagging. Determine the voltage behind transient reactance and the transient power angle equation for the following cases (a) Neglecting the saliency effect. (b) Including the effect of saliency. Solution −1 θ = cos 0.8 = 36.87°

S =

0.6 ∠36.87° = 0.75∠36.87°p.u. 0.8

The pre-fault steady state current is Ia =

S* V*

= 0.75∠ − 36.87° p.u.

(a) With saliency neglected, the voltage behind transient reactance is E ′ = V + jX d′ I a = 1.0 + ( j 0.4)(0.75∠ − 36.87°) = 1.204∠11.49° p.u.

The transient power angle curve is given by Pe =

E′ V X d′

sin δ =

(1.204)(1) sin δ = 3.01sin δ . 0.4

Power System Dynamics

144

(b) When the saliency effect is considered, the initial steady state power angle is 1 δ = tan −

X q I a cos θ V + X q I a sin θ

= tan −1

(0.7)(0.75)(0.8) = 17.71° 1.0 + (0.7)(0.75)(0.6)

The steady state excitation voltage E is given by

E = V cos δ + X d I a sin(δ + θ) = 1.564p.u. The transient voltage is E′q =

X d′ E + ( X d − X d′ ) V cos δ = 1.197 p.u. Xd

And so the transient power angle equation is Pe = 2.99sin δ − 0.5357 sin 2δ . The transient power angle curves for the two cases are:

Problem 2 A 50 Hz synchronous generator having inertia constant H = 9.94 MJ/MVA and a transient reactance of 0.3 p.u. is connected to an infinite bus through a purely reactive circuit as shown in the figure. Reactances are marked on the diagram on a common system base. The generator is delivering real power of 0.6 p.u., 0.8 pf lagging to the infinite bus at a voltage of V = 1 p.u. Assume that the per unit damping coefficient is D = 0.138. Consider a small disturbance of 0.1745 radians. Obtain the equations describing the rotor angle and the generator frequency.

Exercise Problems

145 1 X = 0.2

X = 0.3

2

V = 1.0 8

E X = 0.3

X = 0.3

Solution The transfer reactance between the generated voltage and the infinite bus is

X = 0.3 + 0.2 +

0.3 = 0.65 2

The per unit apparent power is S =

0.6 ∠ cos −1 0.8 = 0.75∠36.86° 0.8

The current is

I =

S* V

*

=

0.75∠ − 36.87 = 0.75∠ − 36.87 1.0∠0

The excitation voltage is E ′ = V + jIX = 1.35∠16.79

Thus the initial operating angle is 0.2931 radians. The synchronising power coefficient is Ps = Pmax cos δ0 = 1.9884

The undamped angular frequency of oscillation and the damping ratio are ωn =

ξ =

Πf 0 Ps = 5.61rad / sec H D Πf 0 = 0.1945 2 HPs

The linearised force free equation which determines the mode of oscillation is

d 2 ∆δ dt

2

+ 2.18

d ∆δ + 31.42∆δ = 0 dt

The motion of the rotor relative to the synchronously revolving field in electrical degrees and the frequency excursion in Hz are shown below.

Power System Dynamics

146 30

Delta, degree

25 20 15 10

0

0.5

1

1.5 t, sec

2

2.5

3

0

0.5

1

1.5

2

2.5

3

50.1 50.05

f, Hz

50 49.95 49.9 49.85

t, sec

Problem 3 The generator of problem 2 is operating in the steady at δ0 = 16.79° when the input power is increased by a small amount ∆P = 0.2 p.u. The generator excitation and the infinite bus bar voltage are the same as in problem 2. Obtain the step response for the rotor angle and the generator frequency. Solution Substituting for H, δ0, ξ, ωn evaluated in problem 2 and expressing the power angle in degree, we get, solving the differential equation with ∆P = 0.2 p.u. δ

 180 )(50 )(0.2 )  ( = 16.79° + 1− (9.94 )(5.61)2 

1 1 − (0.1945 )

2

e

−1.09t

 sin (5.5t + 77.6966° )  

–1.09t sin (5.5t + 77.6966°) δ = 16.79° + 5.753 1 – 1.019 e

Exercise Problems

147

The step response of the rotor angle and the frequency are plotted over a range of 0 to 3 seconds and the result is 26 24

Delta, degree

22 20 18 16

0

0.5

1

1.5 t, sec

2

2.5

3

0

0.5

1

1.5

2

2.5

3

Frequency, Hz

50.1

50.05

50 49.95

t, sec

Problem 4 A 50 Hz synchronous generator having inertia constant H = 5 MJ/MVA and a direct axis transient reactance X d′ = 0.3 per unit is connected to an infinite bus through a purely reactive circuit as shown in figure. Reactances are marked on the diagram on a common system base. The generator is delivering real power Pe = 0.8 per unit and Q = 0.074 per unit to the infinite bus at a voltage of V = 1 per unit. A temporary three phase fault occurs at the sending end of the line at point F. When the fault is cleared, both lines are intact. Determine the critical clearing angle and the critical fault clearing time. 1

X = 0.3

X = 0.2

X = 0.3

V = 1.0 8

E′

2

X

F

X = 0.3

Power System Dynamics

148 Solution The current flowing into the infinite bus is

I =

S* V

*

=

0.8 − j 0.074 = 0.8 − j 0.074 p.u. 1.0∠0°

The transfer reactance between the internal voltage and the infinite bus before fault is X 1 = 0.3 + 0.2 + 0.15 = 0.65 The transient internal voltage is E ′ = V + jX 1 I = 1.0 + ( j 0.65)(0.8 − j 0.074) = 1.17 ∠ 26.387° p.u.

Since both the lines are intact when the fault is cleared, the power-angle equation before and after the fault is Pmax sin δ =

(1.17)(1.0) sin δ = 1.8 sin δ 0.65

The initial operating angle is given by 1.8 sin δ0 = 0.8

⇒ Also

δ 0 = 26.388° = 0.46055 rad δ max = 180° – δ 0 = 153.612° = 2.681 rad

Since the fault is at the beginning of the transmission line, the power transfer during the fault is zero, and the critical clearing angle is cos δ c =

0.8 ( 2.681 – 0.46955) + cos 153.61° = 0.09106 1.8

Thus the critical clearing angle is –1 δc = cos ( 0.09106) = 84.775° = 1.48 rad

And the critical clearing time is tc =

2 H ( δc – δ0 ) πf 0 Pm

= 0.28 sec

Exercise Problems

149

Pm

Problem 5 In the system of problem 4, a three phase fault at the middle of one line is cleared by isolating the faulted circuit simultaneously at both ends. The fault is cleared in 0.3 sec. Obtain the numerical solution of the swing equation for 1.0 sec using the modified Euler method with a step size of 0.01 sec. From the swing curve determine the system stability. Solution For the purpose of understanding the procedure, the computations are performed for one step. From problem 4, the power angle curve before the occurrence of the fault is given by P1max = 1.8sin δ

and the generator is operating at the initial power angle δ0 = 26.388° = 0.46055 rad ∆ω0 = 0

The fault occurs at point F at the middle of one line, resulting in the circuit shown in the figure.

Power System Dynamics

150

Using the delta-star (∆-Y) conversion, reactance between the generator and infinite bus during fault and post fault can be computed as 1.8 p.u. and 0.8 p.u. respectively. The accelerating power equation during fault and post fault will be Pa = 0.8 − 0.65sin δ (faulted); Pa = 0.8 – 1.4625 sin δ (post fault)

Applying the modified Euler’s method, the derivatives at the beginning of the step are

dδ =0 dt ∆ω0 d ∆ω π(50) (0.8 − 0.65sin 26.388°) = 16.056 rad / sec2 = dt δ0 5 At the end of the first step the predicted values are (step size = 0.01 sec)

δ1p = 0.46055 + (0)(0.01) = 0.46055 rad = 26.388°

∆ω1p = 0 + (16.056)(0.01) = 0.1605 rad / sec Using the predicted value of δ1p , ∆ω1p the derivatives of the end of the interval are determined by dδ p = ∆ω1 = 0.1605 rad / sec 1 dt ∆ω p d ∆ω π(50) (0.8 − 0.65 sin 26.388°) = 16.056 rad / sec2 = 1 dt δ p 5

Exercise Problems

151

Then, the average value of the two derivatives is used to find the corrected value δ1c = 0.46055 + ∆ω1c = 0.0 +

0 + 0.1605 (0.01) = 0.4615 rad 2

16.056 + 16.056 (0.01) = 0.1605 rad / sec 2

The process is continued for the successive steps, until at t = 0.3 sec the fault is cleared. We know the post fault accelerating power equation. The process is continued with the new accelerating power equation until the specified final time of 1.0 sec. The swing curve of the system is One-machine system swing curve. Fault cleared at 0.3 s 100

Delta, degree

80

60

40

20

0

–20

0

0.2

0.4

0.6

0.8

1

1.2

1.4

t, sec

21.2

UNSOLVED PROBLEMS

Problem 1 The machine of Problem 2 is delivering real power of 0.6 per unit, at 0.8 power factor lagging to the infinite bus bar. The infinite bus bar voltage is 1.0 per unit. Determine (a) The maximum power input that can be applied without loss of synchronism.

Power System Dynamics

152

(b) Repeat (a) with zero initial power input. Assume the generator internal voltage remains constant at the value computed in (a). [Ans. 1.684 p.u., 1.505 p.u.] Problem 2 A four pole 60 Hz synchronous generator has a rating of 200 MVA, 0.8 power factor lagging. The moment of inertia of the rotor is 45,100 kg/m2 . Determine M and H. [Ans. M = 8.5 MJ.rad/sec, H = 4.0 MJ/MVA] Problem 3 A two pole, 60 Hz synchronous generator has a rating of 250 MVA, 0.8 power factor lagging. The kinetic energy of the machine at synchronous speed is 1080 MJ. The machine is running steadily at synchronous speed and delivering 60 MW to a load at power angle of 8 electrical degrees. The load is suddenly removed. Determine the acceleration of the rotor. If the acceleration computed for the generator is constant for a period of 12 cycles, determine the value of the power angle and the rpm at the end of this time. [Ans. 100 rpm/sec, 20°, 3620 rpm] Problem 4 Determine the kinetic energy stored by a 250 MVA, 60 Hz, two pole synchronous generator with an inertia constant H of 5.4 MJ/MVA. Assume the machine is running steadily at synchronous speed with a shaft input of 331,100 hp. The electrical power developed suddenly changes from its normal value to a value of 200 MW. Determine the acceleration or deceleration of the rotor. If the acceleration computed for the generator is constant for a period of 9 cycles, determine the change in power angle in that period and the rpm at the end of 9 cycles. [Ans. 62.667 rpm/sec, 28.2 degrees, 3609.4 rpm] Problem 5 The swing equations for two interconnected synchronous machines are written as

H1 d 2 δ1 = Pm1 − Pe1 πf 0 dt 2 H 2 d 2 δ2 = Pm 2 − Pe 2 πf 0 dt 2 Denote the relative power angle between the two machines by δ = δ1 − δ2 . Obtain a swing equation equivalent to a single machine in terms of δ. Problem 6 A sending end bus transfers power P0 through a transmission line of pu impedance 0.1. If the stability margin is 30%, find the operating power angle as well as the magnitude of P0. Assume constant 1.00 p.u. bus voltage at both the ends. [Ans. 7.00 p.u.]

Exercise Problems

153

Problem 7 A synchronous generator supplies power to a large grid through a parallel transmission line network. Suddenly a fault occurs at one of the lines and the power output drops considerably. Breakers at both the sides clear the fault and the power transferred is restored to a new value. Find the expression for critical clearing angle, initial angle and maximum power angle. Problem 8 A 60 Hz, 6 pole generator with H = 4.0 p.u. supplying 1 p.u. electrical power at an internal voltage of 1.2 p.u. is connected to an infinite bus having voltage 1.0 p.u. through a line of 0.3 p.u. reactance. Find the power angle. A three phase short circuit occurs on the line. Using step by step algorithm, calculate the swing curve data up to the fifth interval when step size is 0.05 sec. [Ans. 14.48 degrees, 17.86 degrees, 28.00 degrees, 44.89 degrees, 68.53 degrees and 98.93 degrees] Problem 9 A synchronous motor of negligible resistance is receiving 25% of power that it is capable of receiving from an infinite bus. If the load on the motor is suddenly doubled, calculate the maximum value of power angle during the swinging of the motor around its new equilibrium position. [Ans. 46.5 degrees or 0.81 elect. rad.] Problem 10 A 50 Hz, three phase synchronous generator delivers 1.00 p.u. power to an infinite busbar through a network in which resistance is negligible. A fault occurs which reduces the maximum power transferable to 0.40 p.u., whereas before the fault this power was 1.80 p.u., and after the clearance of the fault 1.30 p.u. By the use of the equal area criterion, determine the critical angle. [Ans. 55.20 degrees] Problem 11 A synchronous generator is operating at an infinite bus and supplying 45% of its peak power capacity. As soon as a fault occurs, the reactance between the generator and the line becomes four times its value before the fault. The peak power that can be delivered after the fault is cleared is 70% of the original maximum value. Determine the critical clearing angle. [Ans. 73 degrees] Problem 12 A 60 Hz, H = 3.0 alternator is supplying power of 1.5 p.u., the steady state stability limit being 3.0 p.u. During a three phase fault the maximum power transfer is 1.2 p.u. and after the fault clearing the maximum power limit is 2.0 p.u. Determine the rotor angle and the angular frequency of the machine at the end of 0.02 sec. Use Modified Euler’s method. [Ans. 0.53 rad, 1.131 rad/sec] Problem 13 An alternator (3 ph, 50 Hz) delivers 1.00 p.u. of active power to an infinite bus through a transmission network when a fault occurs. The maximum power which can be transferred during pre fault, during fault and post fault conditions are 1.75 p.u., 0.4 p.u. and 1.25 p.u. Find the critical clearing angle. [Ans. 51.6 degrees]

154

Power System Dynamics

Problem 14 A synchronous generator has an internal voltage of 1.2 p.u. and is connected to an infinite bus operating at a voltage of 1.0 p.u. through a 0.3 p.u. reactance. A 3 phase short circuit occurs on the line. Subsequently, circuit breakers operate and the reactance between the generator and the bus becomes 0.4 p.u. Calculate the critical clearing angle. [Ans. 95.4 degrees] Problem 15 Using the step by step algorithm plot the swing curve of the machine in problem 14. Find the critical time in cycles for an approximately set circuit breaker. [Ans. 14.7 cycles] Problem 16 A synchronous motor connected to an infinite bus is driving a load corresponding to its rated capacity. If the load is suddenly increased by 1.4 times the rated load, determine whether or not the drive is stable. Calculate the maximum additional load that can be thrown suddenly on the shaft of the motor without affecting the stability of the drive. [Ans. 0.74 times rated load] Problem 17 Consider a transmission line characterised by R = 0. The line is operated at its maximum capacity and with both the terminal voltages equal to 1.0 p.u. Define Pmax as base power equal to 1.0 p.u. (a) Prove the reactive power losses in the line amount to 2 p.u. (b) Prove that voltage measured midline is 0.71 p.u. Problem 18 A 50 Hz generator is delivering 50 % of the power that it is capable of delivering through a transmission line to an infinite bus. A fault occurs that increases the reactance of the bus to 500% of its value before the fault. When the fault is isolated, the maximum power that can be delivered is 75% of the original maximum value. Determine the critical clearing angle for the condition described. [Ans. 67.44 degrees] Problem 19 Station A transmits 50 MW of power to station B through a tie line. The maximum steady state capacity of the line is 100 MW. Determine the allowable sudden load that can be switched on without loss of stability. [Ans. 36.6 MW] Problem 20 A synchronous machine is connected to a large system (an infinite bus) through a long transmission line. The direct axis transient reactance is 0.20 p.u. The infinite bus voltage is 1.0 p.u. The transmission line impedance is Z = 0.20 + j0.60 p.u. The synchronous machine is to be represented by a constant

Exercise Problems

155

voltage behind transient reactance with E = 1.10 p.u. Calculate the minimum and maximum steady state load delivered at the infinite bus (for stability). Problem 21 Consider the two machine system shown above X1 = X2 = 1.0 p.u. E1 and E2 are subject to perfect regulator action such that V is held at 1.0 p.u. and power factor is unity. Determine the effect of excitation control under these conditions on Pe supplied by machine – 1 to machine – 2. [Ans. P = tan δ/2, δ = δ1+δ2] I

E1Ðd1

X1

X2 VÐ0

E2Жd2

Problem 22 Consider problem 21 with E2 and V being held constant at 1.0 p.u. and power factor variable. Find the effect of excitation control on Pe. [Ans. P = sin δ2] Problem 23 Compute the transmission capacity of two different lines both 100 miles long. One line is rated at 150 kV and consists of one conductor/phase with line reactance 0.8 Ω/mile. The second line is rated at 765 kV and consists of 4 conductors bundle/phase with line reactance 0.55 Ω/mile. Find the number of lines required, rated at 140 kV, to handle power that is handled by a single 765 kV line. [Ans. 281.25 MW (3 ph), 106.40 MW (3 ph), 38 lines rated at 150 kV] Problem 24 The static transmission capacity of a line (equal to Vi Vj /X) was derived on assumption of zero real line losses. Repeat the analysis without the assumption. Find the maximum for receiving end power assuming Vi, Vj have constant magnitude. Problem 25 Consider a Turbo Generator running initially over excited with E = 1.5 and corresponding PG = 0.25 p.u. The synchronous reactance is 1 p.u. and the infinite bus voltage is 1 p.u. (a) If there is 100% increase in torque, the real power output increases to 0.5 p.u. Find the reactive power QG. [Ans. 0.528 p.u.] (b) Assuming an increase of 20% in the field current with initial conditions remaining same, calculate the reactive power. [Ans. 0.944]

Power System Dynamics

156 Problem 26

The equivalent circuit for the above system is

The parameters and initial conditions are as follows: X1 = 0.735; X2 = 0.0606; S1 = 1 + j0.485; S2 = 4.5+j2.93; E1 = 1.54∠28.5°; E2 = 1.21∠13.1°; St = 5.5 + j3.415; cos φt = 0.85; sin φt = 0.527. (a) Find the minimum value of E, V at which the generator with P = 1 still operates with stability. [Ans. 0.735, 0.477] (b) Calculate maximum power Pm. [Ans. 2.1] (c) Determine the percentage stability margins. For the purpose of calculations, assume a large network (V = 1.0 p.u.) is connected at I. [Ans.

P –P V1 – V2 = 52.3%, m = 110%] P V1

Problem 27

Consider a synchronous generator connected to an infinite bus through a tie line as shown in the figure. The system data is: M = 0.0106; D = 0 ; f = 50 Hz; Xd = Xq = 0.9 p.u.; X d′ = 0.3 p.u.; Xe = 0.1 p.u.; V∞ = 1.0 p.u. The initial nominal operating point is characterised by E = 1.5 p.u.; PE = Pt = 0.75 p.u.

Exercise Problems

157

(a) Plot the static and transient power vs. δ. (b) Derive the dynamical equation for a small perturbation in the operating point resulting in a change of 1° in rotor angle. What is the frequency of oscillations due to the transient? [Ans.

d 2 ∆δ dt 2

+ 157 ∆δ = 0, f = 2.1Hz ]

Problem 28 For the system described in problem 27, assume D = 0.03 and Td0 = 3 seconds. Determine the characteristic equation of the system. Determine the series compensation that can be included without violating system stability. Problem 29 Assume the power system described in problem 27 to be provided with a voltage regulator having T.F. = KA/(1 + s). The terminal voltage of the generator is given by 1.06∠25° at the specified operating conditions. Determine the upper and lower limits of KA for dynamic stability. Apply Routh’s or Hurwitz’s criterion. Problem 30 A single machine-infinite bus power system (as in Prob. 27) has the following data: Xd = 0.9, Xq = 0.7, X d′ = 0.3 p.u. Plot zones of asynchronous and synchronous self-excitation in the r-Xe plane. Problem 31 Derive formula for critical clearing angle with zero power transmitted during fault. Problem 32

Consider the power system shown above. Reactances are marked on the diagram. Determine the critical clearing angle for the generator for a 3-phase fault at point P when the generator is delivering 1.0 p.u. power. Voltage behind transient reactance is 1.25 p.u. for generator and V∞ = 1.0 p.u. [Ans. 51.6 degrees]

Power System Dynamics

158 Problem 33 j0.4 j0.1

 Bus V = 1.00 f = 50 Hz.

G

j0.4

The machine is delivering 0.8 p.u. power with terminal voltage 1.05 p.u. The inertia constant of the machine is H = 5 MJ/MVA (M = 2 H/ω). A fault is applied at the sending end of transmission line. Consider 2 types of faults. (a) A 3-ph fault that presents a balanced impedance of j0.1 to neutral, and (b) A 3-ph fault with zero impedance. Calculate: (i) δ as a function of time (swing curve) assuming that fault is cleared in 9 cycles and examine transient stability. (ii) Apply equal area criterion and find out transient stability. Calculate critical clearing angles in both cases. [Ans. 31.6°, 74.43°] Problem 34 For the illustrative numerical example of chapter 20 draw the state plane trajectories for the following conditions: (a) During fault (b) Post-fault Also determine the critical clearing angle. Problem 35 For the system in problem 34 assume D = 0.2 p.u. Taking into account damping explain how the damping will change the trajectories. Problem 36 Consider a system characterised by M = 1 p.u., D = 0.13 p.u., Pm (post-fault) = 1.0 p.u., Pm (fault) = 0.2 p.u. and Pt = 0.4 p.u. The stable equilibrium point is δ0 = 0.412 rad. Determine the stability boundary of the above system.

Exercise Problems

159

REFERENCES z z z z z z z z

O.I. Elgerd, “Electric Energy System Theory”, McGraw-Hill, New York, 1971. W.D. Stevenson, “Elements of Power System Analysis”, McGraw-Hill, New York, 1962. V.A. Venikov, “Transient Phenomena in Electric Power System”, Mir Publishers, Moscow, 1977. C.L. Wadhwa, “Electrical Power Systems”, New Age International (P) Ltd., 2001. Hadi Saadat, “Power System Analysis”, Tata McGraw-Hill, New Delhi, 2002. I.J. Nagarath and D.P. Kothari, “Modern Power System Analysis”, Tata McGraw-Hill, 1980. M.L. Soni, P.V. Gupta, U.S. Bhatnagar and A. Chakrabarti, “A Textbook on Power System Engineering”, Dhanpatrai & Co (P) Ltd., New Delhi, 1998. P.M. Anderson and A.A. Fouad, “Power System Control and Stability”, Iowa State University Press, Ames, Iowa, 1977.