transient phenomena in electrical power systems. problems and illustrations

Transient Phenomena in Electrical Power Systems PROBLEMS AND ILLUSTRATIONS

N. D. ANISIMOVA, V. A. VENIKOV, V. V. EZHKOV, L. A. ZHUKOV, S. V. NADEZHDIN

M. N. ROZANOV, D. A. FEDOROV, A. N. TSOV'YANOV

Edited by

V. A. V E N I K O V

Translated from the Russian by R. KAPLAN

Translation edited by BERNARD ADKINS

PERGAMON PRESS OXFORD · LONDON · E D I N B U R G H · NEW YORK

PARIS · FRANKFURT

Pergamon Press Ltd., Headington Hill Hall, Oxford 4 & 5 Fitzroy Square, London W.l

Pergamon Press (Scotland) Ltd., 2 & I Teviot Place, Edinburgh 1

Pergamon Press Inc., 122 East 55th St., New York 22, N.Y.

Pergamon Press GmbH, Kaiserstrasse 75, Frankfurt-am-Main

Copyright © 1965 Pergamon Press Ltd.

First English edition 1965

Library of Congress Catalog Card No. 64-21902

This is a translation from nepexoflHbie npoueccu ajieKTpHMecKHx CHCTeM B npHMepax H HJiJiiocTpauHflx published by Energiz,

Moscow in 1962.

2017

PREFACE

THIS book, which was compiled for the attention of design engineers, operation engineers, apprentices and students, should assist them to master the technique of calculating the different transient processes of electrical systems. Concrete examples show the character of the processes and the order of magnitude is derived in some typical cases.

The work, which was carried out by the authors in the Department of Electrical Power Systems at the Moscow Power Institute under the guidance of the head of the department, Professor V. A. Venikov, reflects the experience of more than 25 years of teaching and scientific research work in the field of transient phenomena in electrical power systems, which was conducted in this department. This book reflects, therefore, the scientific path which was started by the academician S. A. Lebedev and Professor P. S. Zhdanov, at the Moscow Power Institute. During the last ten years the work has been continued by the editor of this book and his collaborators.

Many of the problems and examples given in this book were prepared for insertion in the book Electromechanical Transient Phenomena in Electrical Power Systems by V. A. VENIKOV, but were not included there. Some of them were specially completed for the present work, but to some small extent they are derived from Soviet and foreign literature. The origin of many problems in this book was exercise material for the course "Transient Phenomena in Electrical Power Systems" and, lastly, the research students of the faculty solved some of the problems as checking exercises.

Thus, besides the co-authors, a large number of the members and students of the "Electrical Power Systems" faculty also took a definite part in the selection of the material for the present collection of solutions and checking problems.

The authors earnestly request all readers to inform them of any errors they find, and to send their ideas and suggestions in order that they may be considered in the second edition, on which the department has already started to work.

n i

INTRODUCTION

THE present work was planned not as an ordinary book of problems but as a book of examples and illustrations, which show, by concrete examples, the development of diflFerent types óf transient electromechanical processes in electrical power systems.

This book is published as a supplement and continuation of the book Electromechanical Transient Phenomena in Electrical Power Systems by V. A. VENIKOV, where numerical examples were not given. However, the present work is also of independent value. Thus, one of its main problems is the creation of ideas clear enough for students or engineers engaged on problems of transient processes to have methods of calculation applicable to diflFerent sets of assumptions available.

It is necessary to point out that this dependence on assumptions, which are made as the basis of the calculations, often decisively influences the correspondence between the results of the calculations and the experimental data.

The reader should keep this fact in mind, as well as the fact that in many cases the degree of correspondence between the calculated characteristics of a process and the characteristics obtained experimentally, depends on the accuracy of the original data. Therefore, in order to access the probable degree of correspondence between the calculated results and the experimental data, it is desirable to carry out the calculation for various initial conditions in order that the result obtained should relate to a range of operating conditions.

The authors of this work have made their aim not merely to create a book of problems with solutions but, if possible, also to show by the numerical examples the expediency of application of the diflFerent formulae, the accuracy of the results obtained in the sense of conformity between the results of analytical examination of the phenomenon and its real physical development and, finally, to demonstrate to the reader the numerical values, which are characteristic, for diflFerent cases, in diflFerent stages of the transient process.

In conformity with this objective, the work should have contained a large quantity of illustrative material, obtained experimentally: oscillograms, curves and characteristics for the illustration of the most typical development of one or other process including those for which calculated

Transient Phenomena in Electrical Power Systems: Problems

examples are not given or are only given in a brief form. However, under the conditions of preparation of the manuscript for the press, it has not been possible to bring the experimental material into this edition, though this will be done in the second edition which the authors hope will not be long delayed.

The work consists of nine chapters, The first and the second chapters are devoted to the units of measurement, constructing the equations of the system and its elements, such as frequency regulators, turbine governors, transformers, lines and so on. In the second chapter are examples of the construction of equivalent circuits and the determination of the steady-state operation of a system and the original condition which precedes the transient process.

The third and fourth chapters deal with different characteristics of generators, synchronous condensers and loads of electrical systems.

The fifth chapter examines the general criteria of stability used in calculations of the conditions in electrical systems.

Problems of static stability and the effect of large oscillations on stability are given in the sixth, seventh and eighth chapters.

Special problems on the variation of operating conditions, frequency variation and the flow of power between systems are dealt with in the ninth chapter.

Every chapter has a short introduction, indicating the nature of the problems.

TRANSLATOR'S NOTE

THE terms "static stability" and "dynamic stability", which are literal translations of the Russian expressions, are used instead of "steady-state stability" and "transient stability" which are more common in English usage.

CHAPTER 1

SYSTEM OF UNITS AND THE DERIVATION OF THE EQUATIONS OF THE

ELEMENTS IN ELECTRICAL POWER SYSTEMS

IN THE present chapter, the construction of equations to characterize separate elements in electrical power systems is considered for transformers, electrical transmission lines, and regulating equipment (excitation regulators and turbine speed governors). Also given here are some characteristics of induction motors, which play a part in the general equivalent circuits of some systems.

As a rule, all calculations of transient processes and of the steady-state conditions preceding them are carried out in a system of relative units. Incorrect equations may result in serious errors, and make the subsequent calculations valueless. The greatest attention should, therefore, be paid to the unit system in which the calculations are made. A problem, covering all the different methods of writing the equations of transient processes, which can either be expressed wholly in relative units, or partly in relative and partly in actual units, is given in the present chapter. In a number of cases the latter methods are more convenient.

Problem 1.1

An electric circuit is given (Fig. 1.1). Required: to work out an equation and expression for a transfer function for this circuit.

vout

FIG. 1.1.

1

2 Transient Phenomena in Electrical Power Systems: Problems

Solution: In the case when the inductance does not depend on the current and is a constant, the differential equation takes the form:

In the operational form: vjip) =i(Ä1+Jla)+L/w·.

The output voltage is:

^out = = iR2 ·

The transfer function is:

w(p) = _ VoutJP) _ k vdP) Tp + V

where

R2 L k = ^ ^ : T = R± "f" J^2 **1 ~~ ^ 2

Problem 1.2t

A carbon excitation regulator circuit is given in Fig. 1.2.

Required: to form an equation and to find the transfer function for this circuit allowing for the mass of the moving system, the friction and the reaction of the carbon pile.

We assume that the characteristics of the electromagnet, the springs and the carbon pile are coordinated (the regulator is adjusted as an astatic one).

t The problem is copied from the book Symposium of Examples and Problems on the Linear Theory of Automatic Control by V. I. ANISIMOVA, A. A. VAVILOV and A .V. FATEEV (Gosenergoizdat, 1959).

FIG. 1.2.

System of Units and Derivation of Equations 3

Solution. The equation of the electromagnet's circuit under changing conditions will take the form:

Av(p) = r'eAie(p) + tePA0e(p)9 (1.1)

where re = re+rm; te = the number of turns in the winding of the electromagnet.

The magnetic flux Φβ is expressed by the equation

Φ =tA or when it varies:

ΔΦΛΡ) = f-AiJLp)-%*LARJj>). (1.2) t* , . , . tJa

For small deflections of the angle of rotation of the moving system, we have:

ARJp) = -cyA<z(p), (1.3)

where Aoc = the variation of the angle of rotation of the armature ; cx = the coefficient, determined by the empirical or the theore

tical characteristic Rm = / (a) at the point with coordinates ao> Rmo-

Solving jointly (1.1)—(1.3), we get:

Δν{ρ) = r'e(Tep +1) Aie(p) ^ φ ^ TePAoc(p), (1.4.)

v/here Te = —e—r = the time constant of the electromagnet's circuit. Rm<fe

The equation of motion of the armature is : (Jp*+hp)Aoc(p) = AMe(p)-AMsp(p)+AMc(p)9 (1.5)

where J = the moment of inertia of the moving system; h = the coefficient of damping;

AMe, AMsp, AMC = the deviations of the torques, created by the electromagnet, the spring and the reaction of the carbon pile.

The values AMe(p), AMsp(p), AMc(p)are determined by the expressions: AMe(p) = c2Aie(p)+czAoc(p);

AMsp(p) = cAAoc(p); AM (p) =c5A<z(p),

4

where


Co =

CA =

dMe

dMsp

«0

doc

c* =

Cu =

dMe

l«o

doc

dMc doc

<x**<x0

«0

The coefficients c2, c3, c4, c5, are determined by the characteristics Me = /(/„, a); Msp = /(a); Mc = /(a) at the operating point with the coordinates α0, ιΛ.

Substituting the values AMe(p), AMsp(p), AMc(p) in the equation of motion of the armature (1.5), we get:

(T*p*+T2p+c)Mp) =-^Aie{p\ (1.6)

where

^ 1 = cA + c5-cz

Cnl 2**0 car 2'eO

If the variation of the resistance in the carbon pile, because of the current flowing in it, is ignored, then at small deflections of the angle of rotation:

Are(p) = c.Aaip\ (1.7) where Arc = the variation in the resistance of the carbon pile;

c% = the coefficient, determined by the characteristic rc = /(a), at the operating point a0.

Solving jointly equations (1.4), (1.6) and (1.7), we get the equation for the regulator:

l{T9p + lUT*p* + Ttp+c)e'+T9p\ Arc(p) = kpAv(p\ where

c'=*=!L k - c « c

rjt eleO

The regulator transfer function is: Δτ€(ρ) W(p) = Av(p) (Tep + l)(T*p* + T2p + c)c' + Tep·

If the characteristics of the magnet, the springs and the carbon pile are coordinated in the working range of the angle of rotation of the armature,


then c = 0 and the equation of the regulator takes the form: [(Tep + l)(T*p*+T2p)c'+Tep]Arc(p) = kpAv(p).

The transfer function of the regulator in this case is determined by the expression:

W(p) = [(Tep + l)(T*p*+T2p)c'+Tep]

If the effect of the mass and the friction of the moving system are ignored, then with the astatic setting of the regulator, we shall have:

TepAre(p) =kpAv(p); kn W(p) = TeP '

Problem 1.3t

A direct-current generator with separate excitation is operating under zero load conditions (rL = «>) (Fig. 1.3).

FIG. 1.3.

The design parameters of the generator are: teg = the number of turns in the excitation winding (per pole);

reg = the resistance of the excitation winding; pg = the number of pairs of poles;

a = (1.12-1.25) = the leakage coefficient; N = the number of active conductors of the armature; a = the number of pairs of parallel branches of the armature wind

ing; n — the speed of the generator armature;

t The problem is copied from the book Symposium of Examples and Problems on the Linear Theory of Automatic ControlbyV. I. ANISIMOVA, A. A. VAVILOV and A. V. FATEEV (Gosenergoizdat, 1959).


0g =/(2ai e g) = the machine magnetization curve, given in analytical or graphic form, for the magneto-motive force (m.m.f.) of one pole.

Required: to form an equation and to find an expression for the transfer function of the generator.

When forming an equation for the generator, it is assumed that the speed of the generator is constant, the leakage flux varies according to the same law as the resultant flux 0g, and the hysteresis effects, the eddy currents and the armature reaction are ignored.

Solution. The flux of the generator 0g is a function of the generator magnetizing force, i.e.,

Φ, = / ( ! < , ) .

Expanding the function f{Eate^ by Taylor's series, we get, under changing conditions:

A0g = c'g(EAat^\ (1.8)

where c' = ———— = the coefficient determined from the * | b(Eat^) |ra,e>g0

characteristic 0g = /(Ztffeg)at the working point; EAateg = the increase in the resultant magnetizing force of the generator.

In the case under consideration:

A0g = cgEAat^g = cgt^Aicg9 or

The equation of the e.m.f. of the excitation circuit under changing conditions is:

àve(p) = r^Ai^(p)+2pgagt^pA0g(p\ (1.9)

where reg = the number of turns in the excitation winding per pole; pg = the number of pairs of poles.

The relation between the flux and the e.m.f. of the generator is determined by the expression:

Aeg(p) = cgA0g(p), (1.10) where

c =PgNXn * αχ60 '


N = the number of active conductors in the armature ; a = the number of pairs of parallel branches in the armature wind

ing; n = the speed of the armature in r.p.m. In solving jointly the equations (1.8)—(1.10), we finally get:

(Tgp + \)âeg{p)=kgAve{p\ (1.11) km W(p) = Τ,Ρ + Ι'

where T = —* * eg * = the time constant of the generator;

/ c c kg — e g * * = The conversion factor of the generator.

Problem 1.4

For a transmission system, a diagram of which is shown (Fig. 1.4), determine the zero load e.m.f. of the generator Ed. Calculations to be carried out in normal and relative units.

T-l T-2 50=l20-»-j80mva

Θ h®>+ ■ K® 1 150 MW 200mva 1 = 200 km I80mva Vc = ll8kV cosp=0-85 k,=IO-5/2A2 x0=0A ohm/km k2 = 220/l2l Unom=IO-5kV Vk ,r|3% V k 2 = l3%

xd»l·!

FIG. 1.4.

Solution. The electromotive force will be found from the expressiont

Èd = Ye+yJ*ï*ie9 (1.12) where xc = xd+xT^1+xl+xT_2.

We reduce the voltage Vc, the reactances and the current /0 to the same voltage level. We take the generator voltage as this voltage level.

220 10·^ tc = vckjcx = 1 1 8 | g x ^ = 9-31 kV.

t The small circles over the symbols indicate that the given values are referred to the same voltage level.


*™(ohm) = Τ ^ χ ^ Γ 2 = ϊοοχΊ80=35 ohm·

/10-5\2 xT-2 = xr_2 (ohm) k\ = 35 I ^r- j = 0-066 ohm.

j , = Xolk\ = θ · 4 χ 2 0 θ / ^ Υ = 0-151 ohm.

P * - I V P 2 . L _ 13 10-5» 100AS'r_1 100 A 200 ^Γ-Χ = τ ^ Χ τ τ ^ = ^ Χ Α ^ - = 0-072 ohm.

xe = 0-686+0-072+0-151+0-066 = 0-975 ohm.

70 = - p L - = ^ 0 - · 7 ' 8 0 =0-587-70-392 = 0-705^ -33-75° kA y/3Vc V3><118

i 171 242 *· = 7 · ξ ξ « 0 ' 7 0 5 " 33'75°22ÖXio^5 = 8 · 9 4 " 33-75° k A

We calculate the values of the voltage Vc, the reactances and the current IQ in relative units, assuming that 5basic = 150 MVA and Fbasic = 15 kV.

In this case:

/basic = -β*^ = - 7 ^ — = 5-78 kA. V3^basic V3X15

V2 1S2

Ύ — basic __ ^ _ __ i . c 0 u m ^basic — ^ ~" 1 ς π ~" O I i m ·

^basic 1 D U

K 9-31 ^ = T*- = 131 = °·62·

y basic 1 ; ?

Xr_2 0-066 Λ Λ „ „ *r -2 = y 1 ^ = -TTc- = 0-044.

^basic x D

s-^i_. i»JL.Moi . ^basic A J

Ϊ Γ _ , = | ^ = ^ = 0-048. ^basic A ^

xd __ 0-686 7 -'basic

v = 0-044+0-101+0-048+0-457 = 0-65.

xd = —*- = -J^ = 0-457.

^ A _ » 4 . < - » · » · = , . 5 4 5 , _ 3 3 , 7 5 . , 'basic ^ 7 0

8


Substituting numerical values into the expression for Ed9 we get:

in normal units

Èd = 9 ·31+ν3χ8·94^-33·75°χ0·975^90° = 9-31 + 15-1 ^56-25° = 9-31 + 15-1 (cos 56-25+y sin 56-25°) = 9-31+8-38+712-55 = 17-69+/12-55 = 21-7^35-35° kV;

in relative units $d = 0-62 + 1-545 ^ -33-75°x0-65 ^90° = 0-62 + 1-005 ^56-25°

= 0-62 + 1-005 (cos 56-25° +j sin 56-25°) = 1-448 ^35-35°;

or in normal units we get: Èd = $dVbuic = 1-448 ^35-35°x 15 = 21-7 ^35-35° kV.

Problem 1.5

The equation of motion of the generator is given in the form : d2ô

r / Z ? = P r - P - f (1.13)

where the mechanical constant Tj and the time / are expressed in radians, the angle ò in radians; the power of the prime mover PT and the electromag-netic power of the generator P d in relative units.

Required: to obtain some other forms of the given equation, in which the inertia constant would be expressed in kW/sec, the time in radians or seconds, the angle in radians or electrical degrees, and the power in relative units or kilowatts.

Solution· To obtain the indicated forms of the equation of motion, we make use of the following relations:

2*74GD2n2

(a) TAnd] = p X 1 0 - H ; (1.14) "* basic

0>) 'trad] = « V > W ( U 5 > (C) *[rad] = 3gQ ä[el.degree.i; 0 · 1 6 )

(d) ^...uni«,, = 0 ^ - , (1.17) ^basic [kW


where GD2 = the flywheel torque, T.m2; n = speed, r.p.m.;

Pbasic = basic power, kW; ω0 = 2nf0 = 314 = synchronous speed, 1/sec.

Substituting the above relations into the original equation of motion (1.13), any of the required forms of writing the equation of motion may be obtained (see table below).

In conclusion, it is noted that when calculating the dynamic stability, the form (6) in the table is generally used, where the angle is expressed in electrical degrees, the time and inertia constant in seconds, and the power in relative units; and when analysing static stability, taking into account automatic excitation control the form (7) is used, where the angle is expressed in radians, the inertia constant and the time in seconds, and the power in relative units.

Serial No.

1

2

3

4

5

6

7

8

9

Form of writing fundamental equation

_ d2<5 Jdt*~ T~ el

d2<5 7 > 0 d7 r = i>J,~/>eJ

d*(5 ^ωο^ί>ΜΐοΎρτ = ^r—^ei

ΊλΡ *i-p p ω0 ^ d / 1 * el

^ p ά*δ -p P 360/0 b M l cd/1 ~Γτ β1

360/0 di* - / * ei

Tj ά*δ

TJ°>O dfô P XH72" ~ FT~F*I

d*<5

Dimensions of the quantities

Angle

rad.

rad.

rad.

rad.

el. deg.

el. deg.

rad.

rad.

rad.

Time

rad.

rad.

rad.

sec

sec

sec

sec

rad.

sec

Constant of

inertia

rad.

sec

sec

sec

sec

sec

sec

kWsec

sec

Power

rei. units

rei. units

kW

kW

kW

rei. units

rei. units

rei. units

rei. units

CHAPTER 2

EQUIVALENT CIRCUITS AND THE DETERMINATION OF THE STEADY-STATE

QUANTITIES AND THE INITIAL CONDITIONS OF A TRANSIENT PROCESS

IN THIS chapter the construction of the equivalent circuits for stations and systems is considered, and examples are given of the determination of the flow of power, the voltage and the e.m.f. in different elements of a system.

Attention should be paid to the fact that, when solving problems associated with transient conditions, the equivalent circuits of the systems, as a rule, are constructed in such a way that the loads are represented either by constant impedances or by means of static characteristics. In this, such calculations differ from the calculations which are made for the preliminary design of electrical systems and from circuit calculations, whenever the loads are replaced by constant power, determined at the mean theoretical (nominal) voltage.

Problem 2.1

At a hydro-electric power station 15 hydrogenerators are installed. These are connected to the high-voltage station busbars across five groups of single-phase step-up transformers with split low-voltage windings (Fig. 2.1).

The parameters of the generators are shown in Table 2.1. Generators 1-6 each produce Ροα_β) = 100 MW at cos çtyi-e) = 0-85,

generators 7-12 produce 90 MW at cos <Po(7-i2) = 0*85, and generators 13-15 have a load of P(K13_15) = 100 MW at cos Po<i3-i5) = 0*8.

The voltage on the station busbars, in the condition under consideration is equal to 410 kV.

The nominal capacity of each single-phase transformer is 123-5 MVA; all the transformers are operating on the main busbars. The reactance of the transformers is xT = 14-5 per cent.

11


Required: (1) to find the parameters of the equivalent circuit of the generating station, in which all the hydro-generators and transformers are represented by one equivalent generator and transformer respectively;

(2) to calculate the conditions of the system.

GI G2 G3 GA G5 G6 G7 G8 G9 GIO Gil GI2 GI3 GIÀ GI5

FIG. 2.1.

TABLE 2.1

Number of generator

1-12

13-15

Nominal active power, MW

105

105

Nominal voltage,

kV

13-8

13-8

COS Vnom

0-85

0-85

Reactances, %

Xd

55

55

* t

30

30

Time constants, sec

Ti

14-7

100

Tdo

5*30

6-30

Solution. Calculations may be carried out in absolute as well as relative units ; all the parameters of the station elements and its conditions must be reduced to one level of transformation.

The magnitude of the reactances (in ohms) is determined according to the rating of the generators and transformers by the formula:

V2 x (ohm) = x (relat. units) - ^ , (2.1)

where Ynom = the nominal voltage of the circuit element reduced to the level of the transformation adopted for the calculation.

Equivalent Circuits and Determination 13

The reactances of the equivalent circuit for the transformers with split windings are determined in accordance with Fig. 2.2 and are such that:

XLX = = XL2 = XLZ = = 3*Γ· (2·2)

Fio. 2.2.

For the conditions of the problem when reducing the reactances to the generator voltage, the reactance of the three-phase transformer group is :

14-5 13-82 Λ „ , , .

^ = W X 33023^ = 0 - ° 7 4 6 o h m ;

therefore xL1 = % = % = 3x0-0746 = 0-224 ohm. The reactances of the generators are:

^ = x w x ° - 8 5 = 0'848ohm' x"=mx w x 0 " 8 5 = °'463 ohm·

The generators and transformers installed in the station are of the same type and the reactances of the generator, equivalent to all the 15 sets of the station, may be determined to be:

0-848 15

xde = —ττ~~ == 0-0565 ohm;

0-463 Λ Μ Λ Ο L * a e = —TE— = 0-0308 ohm; qe 15 '

0-224 Λ Λ 1 , Λ L Λ:Τβ = = 0-0149 ohm.

The inertia constant of the equivalent generator is given by the sum of the inertia constants, reduced to one basic power.

c c o 7* T nom (1) · 7* ünom (2) · _·_ rp °nom 0i) /* i\ JJe — 7 / l ~ ö h i / 2 - ö Γ - . . + ■«/,,-« . {Là)

°basic °basic °basic


Therefore, with the equal capacity of the generators of the station, the inertia constant of the equivalent generator, reduced to the nominal power of one generator (ShuSe = Snom), is:

TM1) = 14-7x12 + 10x3 = 206-5 sec, and reduced to the power of the equivalent generator (Sbasic = 155nom)is:

Tje = ^ - x 206-5 = 13-8 sec.

The time constant of the excitation winding of the equivalent generator may be calculated as the weighted value of the corresponding time constant of the generators of the station, i.e.

12x5-30 + 3x6-30 Tdoe = ^ = 5-50 sec.

The power delivered from the terminals of the equivalent generator is:

p0e = (6+3)x 100+6x90 = 1440 MW; Q0e = (6 x 100+6x90) tan v>i_12+-3x 100 tan^13_15

= 1140x0-62+300x0-75 = 931 MVAR.

The voltage at the terminals of the equivalent generator is determined by the equation in which the symbols are assumed in accordance with Fig. 2.3:

where J $t v = the reduced voltage of the high-voltage busbars of the station.

E, q« v

Vst.v ^ va I

© — * - ( © — *- '

P e 0 * ) Q i 0 FIG. 2.3.

Therefore, (Λ,^ 13-8V irr 931 x0-0149 V /1440χ0·0149 \ 2

(4">*42θ) -(»V i r — ) H y. ) ■ which reduces to

V* - 208-8 V*+653 = 0.


Solving this equation, we find Vg = 14-3 kV. The fictitious synchronous e.m.f. of the equivalent generator may be

calculated from the formula

which for the conditions of the problem will enable us to find

/ Γ Λ . . , 931x0-0308 Y /1440 x 0-0308 Y~| , . , . „ Eqe = ^[(14-3+ 14.3 j + ( 14.3 ) J = 16-6 kV.

The angle of displacement of the vector representing this e.m.f. in relation to the voltage vector Vg is :

1440x0-0308

aE,, = arctan ™xWaM = *»*>. 1 4 - 3 + Ϊ Φ 3 —

Problem 2.2

Figure 2.4 is a diagram of an electrical system in which two generating stations and a load, receiving supply from the busbars of one of them, are linked by a double-circuit transmission line.

Required: (1) to form an equivalent circuit of this system and to calculate its parameters ; (2) to ascertain the effect of the nominal voltage of the transmission Une upon the relative value of the reactance of the generators G-1 in the total impedance of the system, including the reactances of the generators G-1, of the lines and of the transformers T-1 and T-2. We solve the problem by representing the generators G-1 in the equivalent circuit by the reactances xd, xq9 and xd, taking the parameters of the electrical system to be those shown in the Table 2.2.

G-? - T-i

®fOQ{ T-2 , G-2

FIG. 2.4.


TABLE 2.2

The elements of the system

Hydrogenerators G-l

Transformers T-l

Transformers T-2

Line

Loads

Turbogenerators G-2

Parameters

j Pnom (MW) cos <pnom

Vnom (kV) xd(%) x9{%)

Snom (MVA)

V(%)

Snom (MVA) vk(e-c)(%) vk(e-L)(%) vk(c-L)(%)

V (%)

Length (km) Type of conductor.

P'LZ (MW) cos <p'L2

P'z't (MW) cos φ^ VLL (kV)

Pnom (MW) cos <pnom

*"nom (kV) χΛ (%)

TAe values of parameters, used in different cases of calculation

Line voltage

110 kV

50 0-8 6-3

120 60 30

63 10-5 3

63 10-5 17 6 3

100 A C - 1 2 0

400 0-8

50 0-85

35

600 0-85

10-5 180 28

Line voltage 220 kV

300 0-85

10-5 110 65 28

300 12 3

280 12-9 18-2 5-2 3

270 AC - 4 0 0

950 0-85

230 0-8

35

1200 0-8

10-5 170 30

Line voltage 400 kV

1050 0-85

13-8 55 29 19

1250 14 3

1200 12-9 18-2 5-2 3

1000 ACO-3x480

4200 0-85

1000 0-85

110

5000 0-85

10-5 180 30

Solution. The calculation for setting up the generalised equivalent circuit can generally be carried out by using a mean transformation-ratio.


We calculate in relative units, reduced to basic conditions. Here:

î f = x , ( i n . rei. units) ^ * ; (2.5)

"nom

_ Vk% s y ^basic __ XT% Ky ^basic . n ^

^ - T ö ö x s n o m - I ö ö x s n o m ' ( 2 · 6 )

(2.8)

(2.9)

(2.10)

x0 = 0-144 log 5=SH.+0-0157 ohm/km;

i 0 = — ~ p — x IO"8 1 /(ohm/km) (2.11) log-

formulae

* l =

*c =

»■/ =

(2.8) and

kxx0l

W kTr0l

(2.9)

C f "basic .

1/2 ' 7 basic

K2 r basic . °basic

V2 . r basic

»^basic

'mean

where -Dmean = the mean geometrical distance between the line phase conductors;

r = the radius of the conductors. Correction factors kr, kx, kb are introduced when the length of the

lines exceeds 300 km (ref. 3, page 21). The reactances of the equivalent circuit of the triple-wound transformer

are calculated from the rated values in the following manner:

vk(e-c)% +V/c(e-L)% " Vk(c-L)% XT(e)%

ΧΓ(ή% = »*<*--c)% + *>*(<

2

:-L)%~ 2

-»« · -

y

-L)% . 1

r _ Vk(e-L)% +VHc-L)% ~ ^fc(g-c)% çr(D% — 2 (2.12)

where the reduction of these reactances to the basic conditions is accomplished in accordance with (2.6).


The load ratings are calculated in relative units according to the formula;

(2.13) ■'basic '-'basic

The equivalent circuit of the system (Fig. 2.4) is shown (Fig. 2.5).

E| Xg| Χγ,

Θ-Ί^ <ΊΠ

♦pj^jcfc

FIG. 2.5.

For a nominal 110 kV, voltage line with a distance between the phase conductors of 4 m and with the conductors layed out horizontally:

1-26x400 x0 = 0-144 log - 0-765 +0-0157 = 0-422 ohm/km;

* o = 7-58

1-26x400 X10-« = 2-69X10-" l(ohm/km).

log 0-765

50 Assuming the basic power 5basic = — = 62-5 MVA, we have;

0*8

? / = 1 x 0 - 4 2 2 x 1 0 0 ^ =0-10;

t>, = 2x2-69 x 10-«xlOO-^g- =0-1135;

r, = 4 - x 0 ' 2 7 x 1 0 0 62-5 1152 = 00639.

The remaining parameters of the equivalent circuit are:

xdl = 1-2 g ^ = 1-20; 62-5 xgi = 0-60;

10-5 62-5

x'dl = 0-30;

^ = 1 Ö Ö X : 6 3 - = ( H 0 4 ;

^ = T 0 Ö X 6 l 5 = 0-0302:


10-5 + 17-6 62-5 ■X-z=- = 0-1065; •r2(" 2x100 " 6 3

10-5+6-17 62-5 ΛΛΛΛ>10 * ™ ° 2X100 X ^ =-0-00248;

17+6-10-5 62-5 Λ rt,„ y ^ > = 2x100 X - 6 T = 0 ' 0 6 2 ;

^ = 0-0302;

xd2 = 1 -8^0-85 =0-1595;

^ = 0 - 2 8 ^ 0 - 8 5 = 0 - 0 2 4 8 ;

S[2 =g^(400+y400x0-75) = 6-40+./4-80;

S " = 6^5 (50+V'50x 0-62) = 0-80+y0-496.

Representing the generators G-1 in the equivalent circuit by reactance xd, the total reactance of the transmission system is:

xdz = xdi+*Ti+xi+xT2(e) + xT2(L) = ! ' 2 0+0104+0' 10 +0-1065 + + 0-062 = 1-573,

where the reactance xd is :

^ - x 100 = j - ^ x 100 = 76-2 per cent. χάΣ 1-573

Representing the generators in the equivalent circuit by the reactances xql and x'dl, we find:

xqE =0·60+0·104+0·10+0·1065+0·062 =0-973; χ'„Σ = 0-30+0-104+0-10+0-1065 +0-062 = 0-673,

where

^ i X 100 = ; ° - ^ x 100 = 61-6 per cent,

^ x 100 = ^^ x 100 = 44-6 per cent. χάΣ 0·673

The results of the calculations of the parameters of the equivalent circuit of the system and the determined total reactances and values xg/xs


at the line voltages 220 kV and 400 kV are given in Table 2.3. As basic power in both these calculations we have taken the full nominal power of the generators G-l.

100 200 300 400 kV

FIG. 2.6.

Figure 2.6 shows the effect of the reactance of the generators on the total reactance of the transmission system when the nominal voltage of the line varies.

TABLE 2.3

Value in relat. units

Xdl x*x

t Xdi xTl x i

*l

ài ΧΤ2{β) ΧΤ2{β) XT2iL)

At line

r nom = 220 k V

110 0-65 0-28 0141 0-372 00704 0-221 0196 000378 00661

voltage

V r nom

\ = 400kV

0-55 0-29 j 019 0139 0-83 00466 1-197 0160 000309 00544

Value in relat. units

^ d 2

X'd2

s* S',2 Χ*Σ Χ*Σ

t χίΣ χάΐΙχάΣ xJxrt Xd\lXdΣ

At line voltage

Vnom = 220 kV

0-40 00705 2-69+yl-67 0-652+70-489 1-816 1-366 0-996 0-605 0-476 0-281

Vnom = 400 kV

0-377 00628 3-40+y2-ll 0-8104-7Ό-502 1-733 1-473 1-373 0-317 0197 0138

Equivalent Circuits and Determination

Problem 2.3

21

The power of the power station G-l is transmitted to the busbars of a power system along a 220 kV line (Fig. 2.7). The voltage on the busbars of the system in all operating conditions remains unchanged and equal to 232 kV.

U=const

YPLI+jaL, tp L 2 * ja L 2

FIG. 2.7.

Power to the same busbars is supplied by the power station G-2, part of the power of which is also transmitted to an intermediate load.

The parameters of the system elements, and the operating quantities are given below.

Station

G-l

G-2

Pm, MW

200

300

GENERATORS

cos <pnom

0-8

0-85

'noun ** *

10-5

10-5

*d> %

28

30

TRANSFORMERS

Station

1

2

Snom,MVA

240

360

vk, %

V*-4 =105

12

K j*

10-5/121/242

10-5/248

Capacity of windings, %

100/66, 7/100


LINES

Line

7-1

/-2

1-3

'nom ™ '

220

220

220

/, km

150

100

70

Type of conductor

AC-300

AC-400

AC-300

x09 ohm/km

0-422

0-414

0-422

b0, l(ohm.km)

2-69 x IO""6

2-73x10-«

2-69x10-«

Load

Load

1 2

LA L-2

P,AfPF

120 150 50 70

cos φ

0-75 0-85 0-85 0-9

*nom y K r

110 220

Required: to form an equivalent circuit of the system, to calculate its operating conditions and to form an equivalent transformation of the circuit by changing it from two generating stations to one. When solving the problem, the generating stations in the equivalent circuit are represented by transient reactances; the ohmic resistances are to be neglected.

Solution. The equivalent circuit under consideration, is shown in Fig. 2.8, where the values of some of the parameters of the elements of the system are shown. These are calculated in relative units, in accordance with the formulae (2.5) - (2.13) at Sbasic = 240 MVA and Fbasic = 220 KV. The given data shows that half of the capacitive susceptance of the lines /-l and 1-2 is very near to the value of the inductive susceptance of the magnetization of the transformers T-l and T-2 respectively. Their algebraic sum determines a reactance which is incomparably lower than the other reactances of the circuit. This makes it possible to omit these and the other susceptances in the calculation without introducing any noticeable error into the result.


VL2 =

The capacitive susceptances linked in Fig. 2.8 with the busbars 3, are connected directly to the busbars of fixed voltage and therefore they cannot have any effect on the value of the equivalent impedance of the circuit. These susceptances need not be taken into account in the reduction of the circuit. However, their influence should be allowed for when determining the flow of power through the inductive reactances xl± and xl2.

The charging powers Q'cl and Q'cS are: Q'cl = 1·0552χ 0-0407 = 0-0453; Q'cS = 1·0552χ 0-0379 = 0-0422,1

therefore S[ = 0-50+/0-441-./0-0453 = 0-50+/0-396; 5; = 0-625+7*0-383-;Ό·0422 = 0-625 +./Ό-346.

The charging powers Q^ and Q"2 can be allowed for by a corresponding change in the reactive power of the load L-2. For this, it is first necessary to determine the voltage on the busbars L-2 by using a formula similar to (2.4):

/Γ/ι ncc , 0-346 x 0-1465 \ 2 / 0-625x0-1465 \ 2 1 1 ΛΜ = V L ^ 0 5 5 * 1-055 ) + ( 1-055 ) J = Μ ° 3 ·

This voltage determines the charging powers Q'^ and β^ as :

Qä = l-1032x 00379 = 0-0461; β^ = 1-1032X 0-055 = 0-0670,

consequently, S'L2 = 0-292+y0-141-./Ό-670-./Ό-0461 = 0-292+./Ό-028.

Figure 2.9 shows a theoretical diagram of the system, in which the conductances are omitted or allowed for in accordance with the previous calculations. It also contains the values of the reactances of all the elements of the system.

The calculation is carried out taking into account the actual transformation ratios in accordance with the formula (Ref. 3, page 102).

xg = Xg^^x-^^- (2.14) ^nom " basic

The basic voltage, brought to the busbars of the generators G-l is:

be,ic(G-i) = 2 2 0 ^ = 9-55 kV;

^2

, Xil

X TI(L

)

i 0-0

75

U =

l-05

5 sco

nst

j 007

56

0-20

8* j

0129

S, =

0-5

0 +

j0-4

4l

s j 0

-314

^£U

j 0-04

07

sXS 2

= 0-

625*

j 0-3

88

L «α

^

_Lt»

iii =

-J00

248

* =

"Ψ

Ml

b-.

/2=i

003

79

jOI4

65\

^^

g j0-l

03

_ TJ

QÎ2

Ac

I -L

b c

2/2=j

0-05

5

~ φ

0-29

2*

k 2/2

=j

0-05

5 -L

b c

3/2=

j003

79^ +

j0. |A

, φ

FIG

. 2.

8.

jO 32

5 00

756

, J0

130

^ ^

' -

^ "'-

Ϊ&

«L

2 E<

j 010

15

j 0-2

58

r Ì a' 5"

FIG

. 2.

9.



that brought to the busbars of the generators G-2 is:

Consequently:

^».,*««> = 2 2 0 ^ = 9-33 kV.

χ· - 0 . 2 8 1 0 ' 5 2 * 2 4 ° -0-325-* Λ - 0 2 8 Ί 5 0 ~ Χ 9 ΐ 5 * 3 2 5 '

10-52 240 353 X9-332 x'd2 = 0-30 - ^ - χ ^ , = 0-258.

The load impedances are determined by the formula: V2

ZL=— (cos q>L +j sin <pL). (2.15)

0-292 As cos w" = zJZ± = 0-996; V(0-2922+0-0282)

Λ.ΛΟΟ

sin φ'ν = ^-^ = 0-0955, a V(0-2922+0-0282)

then in accordance with the formula (2.15): 1-103»

V(0-2922+0-0282) Z'L2 = , .„ (0-996+70-0955) = 4-15+./Ό-397

So

FIG. 2.10.

The impedance ZL1, for which the voltage on the load busbars was determined beforehand, is calculated similarly.

For the replacement of the two generating stations in the circuit of Fig. 2.9 by one equivalent to them, we transform the stars l-Ll-3 and


2-L2-3 into equivalent deltas (Fig. 2.10), by using the formulae:

ΖχΖ3. Z13 — Z1-f-Z3+-

-ΊΟ z1+zLL+ LI

^30(1) — ^Ll + Z s H-Zi "

\ (2.16)

We get: «..«, .Λ„„„ yO-401 xyO-444 ^ 0 . _ 0-401x0-444 Z18 =;O-401+yO-444+J - , . , ., , . =y0-845—

5-35+73-31

= -0-0241 +70-86 = 0-86^91-6;

6-28^31-8°

7 -n AM .< « j . -i ?i , y'0'401 (5-35+73-31) Z10 = ;0-401 +5-35 +/3-31+J j ^ ^

= 10-2+y6-71 = 12-2^33-4°; , , * . . , , , ·η.ΛΛ (5-35+y3-31)/0-444 2W) = 5-35+y3-31+70-444+^ j L _ ^

= 11-26+77-41 = 13-45^33-4°.

In a similar way the following impedances are determined:

Z-3 = -0-01625+70-611 = 0-611 ^91-5°; ΖΜ = 17-25+72-11 = 17-4^7-0°;

30(2) = 5-47+70-669 = 5-5 ^7-0°. The powers S j ^ and S30(2) in the circuit of Fig. 2.10 are determined

only by the voltage on busbar 3 and by the impedances Z j ^ , Z30(2), where

San — T~~ (2.17) ^30

consequently, c 1055' soa) ~i3-45^-33-4° = 0-0827 (0-8348+70-5505)

= 0-069+70-0455; 1-0552

S*W) = 5 . 5 ^ _ 7 . 0 o = 0-201 +70-02465.


In transforming the circuit in Fig. 2.9 the incoming power to busbar 3 from the generators G-l and G-2 cannot vary, therefore

S1Z = ^1+*>30(1)>·

»>2S = O a + «$30(2)·

Hence:

S1Z = 0-50 +/Ό-396+0-069 +./Ό-0455 = 0-569+./Ό-442;

,$23 = 0-625 +;Ό·346+0·201+;0·02465 = 0·826+;Ό·371.

The powers obtained make it possible to determine the e.m.f. of the generators G-l and G-2, whereupon:

E' = ^ 1 < I 5 5 + ^ ( - 0 I 0 ^ < H 4 2 X 0 - 8 6 J +

Γ0·569χ0·86-0·442(-0·0241)Ί2 , ,„ + [ ms^ J - ^

E' = ^•1^s+0-826(-0-01^+0-371x(HillJ +

-0-371 (-0-01625) I 2

1-055 J Γ0-826Χ 0-611-0-371 (-0-01625) I 2

έ3 = |·40<20·1· 0·358<9Ι·5°

&—t—ran—

7 85<22 ·6°^ 3-99<l4-6*Lf|

Fio. 2.11.

The angles of displacement of the vectors of these e.m.f.' s relative to the vector of the constant voltage Ϋζ are:

<513 = 18-7°; <523 = 21-2°.

By joining the points 1 and 2 of the circuit, Fig. 2.10, we get the required equivalent circuit of the system, shown in Fig. 2.11.

28

Here:


1 1 ^1S v 1 1

0-86 ^91-6° + 0-611 ^91-5° 1 1

-0-01625 +./Ό-611 -00241 +./Ό-86 0-526

2-8^

^183-1°

-91-5° - 0 3 5 8 -

1-471^91-6° 0-526^183-1°

91r5°;

Z10 = 7-35 ^22-6°; Z30 = 3-99 ^ 14-6°

The equivalent e.m.f. is determined by the formula:

^ 1 ^ 1 3 "^-^2-^23 E,= L13

i.e. 1-48^18-7° 1-345^21-2°

Ee = 0-86^91-6° ' 0-611^91-5° 2-8^-91-5°

= 1-40^20-1°.

(2.18)

Problem 2.4

Figure 2.12 shows a section of a diagram of an electrical system, including a generating station, which delivers power through an external inductive reactance xext. A portion of the station power is consumed by the loads L-l and L-2, which are represented in the circuit of Fig. 2.12 by a constant ohmic resistance RL1 and an impedance ZL2.

U n o m = , I O k Vv Substation A *cxt

Unom=35kV

FIG. 2.12.

The circuit is characterized by the parameters, the values of which are given below.


GENERATORS

Pnom = 100 MW; cos <pnom = 0-8; Vnom = 10-5 kV;

xd = 1#7.

TRANSFORMERS

Snom = 120 MVA; KMe_c) = 10-5 per cent; Vkie_L) = 17 per cent; Ykic-L) = 6 per cent; KT = 10-5/38-5/121.

LOAD

RL1 = 123 ohms; ZL2 = 200+/140 ohms.

System condition is characterized by a voltage VA = 117 kV maintained on the substation busbars, and by a transmission of power, P0=30 MW at cos φ0 = 0-9, from the station. The external inductive reactance xext = 40 ohms.

Required: to represent the section of the system by an equivalent circuit, not containing parallel circuits, and to determine its parameters.

Solution. The calculations are carried out in per-unit form, assuming the basic values Sbasic = 125 MVA; Fbasic = 117 kV.

The basic voltages on other stages of transformation are:

10-5 121 ^basic(io-5)= 117 T ^ - = 10-15 kV;

,38-5 121 Fbasic(35) = 1 1 7 ^ = 37-2 kV.

The parameters of the circuit and the given operating quantities with the assumed basic conditions are :

_ 10-52 125 _ y < , ~ 1 ' 7 W X Ï O : Ï 5 2 " 1 ' 8 2 ;

_ 10-5 + 17-6 1212 125 *™ - 2x100 Xl2ÖXTTß * °"12 ;

_ 10-5+6-17 121* 125 ' *™ 2x100 Χ Ί 2 0 " Χ Ϊ Ϊ 7 * Α ί 0 ;


17+6-10-5 1212 125 nn,n, * * » " 2X100 X-Ï20XÏT7* = ° ' 0 6 9 6 ;

125 1172 ?ex« = 4 0 T P « = 0-365;

125 37·22 ^ = 123 5 ^ 5 = 1 1 - 1 ;

125 ZL2 = (200+;140)XyT^ = 1-83+./1-28 = 2-23 ^35°;

S0 = j^5(30 +/30X0-483) = 0'24+y0-1163 = 0-267 ^25-8°.

The data obtained are marked on the equivalent circuit (Fig. 2.13).

EA XJ XT, S ' *I*H_ x , !» . ■ " TL—^-h — ^ β «xt —»- I .

Θ τ>—np—»Mr» » T*—K j 1-82 J00696 J0I2 j 0-365 I

I l ·Ι·«|| | SL2|| IZ^I-WjWe Pf i W Ύ

FIG. 2.13.

We find successively, the voltages at the junction points of the circuit and the power flowing in its different sections :

, 0-1163x0-365 .0-24x0-365 va = i + j +j 1

= 1-044+J0-0875 = 1-044^4-8°;

SO = So+JS^+Su = 0-24+yO-1163+y°^Ç-2x

i -0442 X0-365+" = 0-641 +0-./422 = 0-766 33-4°;

tb = 1 - 0 4 4 + ° ^ ^ ^ ° ^ ™ - l-093+y00736 - 1-093.3-9O;

$'0' =0-641 +yo-422+yY^gYxO-12 + 1093 11-1

= 0-749-fy'0-487 = 0-894 ^33-2°.

2


The e.m.f. of the generator is : 0-487 (1-82+0-0696) .0-749(1-82+00696) Eä = 1-093+ j ^ 3 +j ^

= 1-936+71-295 = 2-32.33-8°. The displacement of the e.m.f. vector with respect to the vector of the

voltage on the busbars of substation A, is determined by: òEd = 4-8+3-9+33-8 = 42-5°.

Ed V*d**ri b Ê „ *Tc β

0— v_ r 0—nn- w—

ÇJ FIG. 2.14.

To obtain the required equivalent circuit, not containing parallel circuits, we join the circuit of the generating station with the load circuit L-1 (Fig. 2.14, a). We get:

1 _1_ 7ΐ·89 + 11·1 YA = ΤΤΤ^ + ΓΠΤ = 009-y0-529 = 0535 . -80-3°;

ÉA « ÈJ± = 2-32 .42-5° 1 . 8 9 , 9 0 χ 01

5 3 5 ^ _ 8 0 . 3 ο - 2-30 .32-8°;

Z* = 0-535 J - S O · ^ 0 · 3 1 3 ^ 1 · 8 4 · Joining the e.m.f. circuit Eel with the load circuit L-2 (Fig. 2.14, b), we

can get; Y l i 1

Λ 0-313+;i-84+y0-12 2-23^35°

- i r â + 2 ¥ W =°·879 -59'40· E< = 2'3° - 3 2 ' 8 ° 1-98.80-9X0-879.-59-4° = 1>323 « "*'>

z ' = i + ^ 3 6 5 = o -4" 5 9 · 4 ^ 0 · 3 6 5

= 0-578 +j 1-344 = 1-465 .66-7°.


The equivalent circuit is shown in Fig. 2-15.

ΕνΙ·323<ΙΙ·3· Κ65<667° VA=KOe

Θ—nn—I — ►

éosO-24* 10-II63

FIG. 2.15.

Problem 2.5

Figure 2.16 shows a diagram of an electrical system in which two power stations, connected with each other and the loads by a 220—110 kV. network, operate in parallel with the constant voltage busbars. The parameters of the elements of this system, and also the values of the loads and some of the lines in the condition considered, are shown below.

GENERATORS

Station

HES^

HPP5

Type of generators

650 CB 32

132

TB-50-2

P ■*■ nom MW

45

50

r nom kV

10-5

6-3

cos <pnom

0-85

0-8

Reactances, %

Xd

113

183-9

xq

70-8

183-9

x'd

30-3

20

* l

46-4

16-4

Tj, sec

6

10

TRANSFORMERS

Station or substation

HESA HPPG

1

2 3

" n o m »

MVA

180 90

180

60 31-5

JYJ»

10-5/242+5% 6-3/121+5%

220/121/38-5-5%

110/11 110/11

vk9 %

12 10-5

> V - i = 1 5

**(·—D — 23 Vm(c-L) = 8

10-5 10-5


The power S2_c = 16-2-./47-0 MVA; SB_C = 58-8-./37-0 MVA, is transmitted along the lines leading to the substation of the receiving system. At the heating and power plant B only two generators are in operation, power, PB = 15 MW at cos φΒ = 0-95, being given out from the high-voltage busbars.

LOADS ON STATIONS AND SUBSTATIONS

Station or Substation

1 2 3

HPP£

r n o m , KY

35 10 10 6-3

Pnom,MrV

115 70 25 60

c o s <pnom

0-95 0-95 0-95 0-8

Required: (1) to form an equivalent curcuit of the system; (2) to calculate its parameters; (3) to calculate the operating quantities in relative units for the given conditions. When calculating, the actual transformation ratios are to be taken into account, neglecting the ohmic resistance and the conductances of the lines and transformers (see Problem 2.3) with V = 117-5 kV.

Solution. As basic values we take power S^^ = 135 MVA and voltage basic = 10-5 kV at the level of the generator voltage HES A. In other

transformation stages we have the following basic voltages:

Cb.,ic(22o) = 10-5 ^ - = 254-1 kV;

»ne 254-1 121 , , Λ , „ f i c a i o ) = 1 0 , 5 -IÖTX22Ö = 1 4° k V ;

Ut basic (35) , r t c 254-1 36-6 . , . , , .

= 10-5WX22Ö = 4 2 ' 4 k V ;

TT 1A . 254-1 121 tfbasic<e-3)= lO-5-πϊ^-χ

6-3 10-5 ~220 127-05 = 6-93 kV.


E

ft

f, SO

ri

2


The parameters of the generators HES A in relative units, reduced to the basic conditions, are:

135 *dA = 113-^x0-85 =2-88;

xqA = 0-708^x0-85 = 1-805;

x'dA = 0-303^x0-85 = 0-773;

X2A =0 -464^x0-85 = 1-18;

45 0-85x135 TJA = 6 „ „„ „ = 2-35 sec.

We find the parameters of the generators at the heating and power plant B to be:

^ = ** = 1-839 x 0-8 ( | ± J = 3-30;

xi.-0-2^x0.8 ( J ± j =0-359;

x» = 0-164^x0-8 ( | ± Y =0-294; so

0-8x135 TV, = 10 = 4-63 sec.

Each low-voltage winding of the transformers HES A has a reactance : - 1 2 135 _ _

*™«-3ÎOOXÏ80"U*2 7 ;

The reactance of each transformer ΗΡΡ,5 is equal to : 10-5

XTB - loo 135/127-05 V Λ , „

xw(-û<r) =°·13

The reactances of transformers, installed in substations 1, 2 and 3, are determined in the following manner:

15+23-8 135/ 220 \* ΛΛΟ„ ^ « = ^ a ô ô - X Î 8 o ( 2 5 4 T j = 0 0 8 4 ;

x-71(c) _ 15+8-23 135/121 y

2x100 X 180 ( 140 J '


23+8-15 135/36-6V ηηΛΛΟ *™ » ^ T Ô Ô - X Ï 8 o ( 4 2 4 j = 0 0 4 4 8 ;

νΓ2 10-5 135/110\2 . , , _ _

= lôôX-6o(ï4ôJ = ( H 4 5 5 ;

10-5 135 /100\2 Λ .__

= TOOX3T5(ï4ôj =°·277· Xj-Z —

For transmission lines of 220 kV, with conductor ACO-400 with 7 m spacing between the phase conductors, x0 = 0-421 ohm/km.

The inductive reactance of 1 km of 110 kV line with spacings of 4 m between the phase conductors of lines with single conductors, and of 4e3 m of lines of split conductors are:

with conductors ACO-500 x0 = 0*379 ohm/km;

with conductors 2 x ACO-400 x0 = 0-294 ohm/km (300 mm space between the conductors in phase);

with conductors ACO-300 x0 = 0-395 ohm/km;

with conductors AC-240 x0 = 0-40 ohm/km;

with conductors AC-185 x0 = 0-408 ohm/km.

Consequently, in the relative units for the lengths, shown in Fig. 2.16:

x ^ _ 1 = 0 - 4 2 1 x 2 5 0 J ^ F = 0-22;

*!_, = 0 - 4 0 x 3 0 ^ = 0-0826;

135 x3_B = 0 - 3 9 5 x 4 0 - ^ = 0-109;

135 xB-c = 0-294x60-^- = 0-1215;

x2-c = 0 - 4 0 8 x 6 5 - ^ = 0-1825;

*χ_2 = 0 - 3 7 9 x 6 5 ^ - = 0-170.


The equivalent circuit of the system and the reactance values determined above, are given (Fig. 2.17).

A0-5I8+J0-I70

j 0-773 jO-27

jO-773 jO-27

J0773 jO-27

J0 773 jO-27

*TA(c)' , jO-22 j0084 xT|(c)=0

0-852+J0-28

xTA(«)sUjO-22 j 0-084 r è

0I85+J006I

FIG. 2.17.

The load powers in the system at the substations, in relative units are:

Sx = y ^ (115+y 115x0-329) = 0-852 +7*0-28;

* · ~Î35 (70+y70x0-329) = 0-518+/0-170;

Sz = - ^ (25 +725x0-329) =0-185+70-061.

The load power at the busbars of the generating voltage of the heating and power plant B is:

SB = j ^ (60+7*60x0-75) = 0-445+7*0-333.


In the assumed system of relative units, the voltage of the busbars of the receiving system is:

The power, transmitted into the system by the lines leading to the substation, is given by:

■Si-c = Ï 5 5 (16-2-/47) = 0-12-./0-348;

Î*-c = Ï 5 J (58-8-7*37) = 0-435-/Ό-274.

The power given out by the high-voltage busbars of the heating and power plant B is :

SB = ^ (15 +7*15x0-329) = 0-111+./Ό-0365.

The final equivalent circuit of the system (Fig. 2.18) is simplified by combining the reactances of the low-voltage windings of the two transformers of the substations 1, and replacing each unit of the HES A by one equivalent generator, operating in a unit with one transformer.

| 0 - 5 l 8 + j0-170

k

<3 jO-0727

jO-1825

Vd =0-838

FIG. 2.18.


Problem 2.6

An electric system is given in the form of the equivalent circuit, shown in Fig. 2.18.

Required: (1) to transform the circuit, representing the load by constant impedances : (2) to determine the transient e.m.f. and the inertia constants of the stations in the following cases:

(a) The system includes three generating stations, the heating and power plant B and the receiving system being combined into one equivalent station;

(b) The system in question is represented by a circuit comprising two generating stations, one of which is a unit, three generators-transformer a t H E S ^ ;

(c) Station HES A and heating and power plant B are replaced by one equivalent generating station, operating on the busbars of the system.

The impedances of the circuit elements are shown in Fig. 2.18 (the numerator gives the number, and the denominator gives the magnitude); the load voltage on busbars d and the power flowing to the points d and e are known.

Solution: In calculating the electric conditions we take the voltage Vd, and the power arriving at the point d, as the original data. These quantities determine the voltage at the points c and e of the circuit and the losses of reactive power in the reactances 6 and 7. Here:

"■ - j{[™+i=i%^i!t!%^y\ - ™>

0·4352+0·2742

^ = 0-838» " " » M I S -0-0459.

The amounts of power, flowing across the reactances 6 and 7 at points c and e respectively, are:

$%, = 0·12-;Ό·348+y0-0353 = 0-12-^0-313; S%i = 0-435-yO-274+y'0-0459 = 0-435 -yO-229.


The displacement of the vectors of voltages Yc and Ve with respect to the vector of the voltage Vd are found as :

òcd = arc tan

0-12x0-1825 0-838

0-838- 0-348x0-1825

òed = arc tan

0-838

0-435x0-1215 0-838

0-838- 0-274x0-1215

= 1-5°;

= 4-5°.

0-838

We calculate the condition of circuit cB as :

'2 /η · ι ι ΐχ0·065\2 ' y^J[{—0-^m^Ji—, 80 = 0-803;

0·1112+0·03652

Sfa = 0-111 +70-0365 +0-444 +./Ό-333+ 0T80Y χ ° · ° 6 5

= 0-555 +/Ό-3695 +yO-00136 = 0-555+./0-371;

0-895;

òef = arc tan

0-111x0-065 0-80

0-0365x0-065 = 0-6°; 0-80 +

ôfg = arc tan -

0-80

0-555x0-179 0-803

0-803 + 0-371x0-179 0-803

= 7-9°;

àBd = 4-5+0-6+7-9 = 13-0°.

From the power balance at the point e, we find:

S'ge = 0-435-./0-229-0-111-/O-0365 = 0-324-/0-266; therefore,

« * . - * 5 £ g * £ x 0 . 1 0 . - 0-0299;


$*e = 0-324-./Ό-266+./Ό-0299 = 0-324-y0-236; /(ΤΛΟΛ (-0-266) χ0·109~|2 / 0-324 x 0-109 \21 V, = ^{[θ-80+i gig J +( 0,80 j }

= V[(0-80-0-0363)2+0-04412] = 0-765; 0-0441

^ = 4-5°+arctan 0 , 8 0 _ 0 . 0 3 6 3 = 7-8° .

We find the voltage at point h from the equation:

V, - 0-765 » (K, + » ™ f f U M ) ' + ( » I M f f 1 Î M Y ] . from which it follows that:

0·7652Κ2 = (Κ2+0·00845)2+0·02562, or

KJ-0-570KÎ+0-0007264 = 0. Hence, we have :

Vh = Vl>285 + V(° , 2 8 5 2 - ° · 0 0 0 7 2 6 ) ] = °·753. Consequently,

^ , = 0 - 8 50

!7

+5 ° r ' x O - 1 3 8 5 ^ 000928;

$*ag = 0-324 -70-236+0-185 +./Ό-061 +./Ό-009 = 0-509-/Ό-166.

Continuing the calculation, we get:

= V[(0-765-0-0179)2+0-05482] =0-75;

^ = 7 - 8 ° + a r c t a n 0 . 7 6 5 - 0 4 0 1 7 9 = 1 2 0 ° ;

. 0·5092+0·1662 nnMie ΛΛΛΜ Mii = ö^ jä X 0-0826 = 0-0403;

$aag = 0-509-;Ό· 166+y0-04 = 0-509 -j0-126.

We determine the voltage at the point b from the equation

/ Γ Λ , 0·28χ0·0224\2 / 0-852 X 0-0224 V v. = 0-75 = ^ y t + j j — j + ( — T T —

42" Transient Phenomena in Electrical Power Systems: Problems

hence Vb = 0·741 ; therefore 0-8522+0-282

0-741s " f t - LZn Χ0·0224 = 0-0328;

£» = 0-852+;Ό·28+70-0328 = 0-852+/Ό-313.

The power flow through the impedance 6, determines the voltage at point c:

,, . j |«l+fc^[+(ü^)} - 0,62; therefore, the voltage at point k can be found from the equation :

Vc = 0-762 = ^ [ ( Κ . + 0 · 1 7 0 ^ 0 7 2 7 ) 2

+ ( ° ' 5 1 8 Χ ^ 0 7 2 7 ) 2 ] .

Solving this equation, we find Vk = 0-744; therefore,

A,. - 0 · 1 2 ; ^ 4 8 ' X0..825 - 0-0353;

$'„ = 0-12-y0-348 +0-518 +/Ό-170+/Ό-0353 + +/Ό-039 = 0-638-j0-104;

. 0-6382+0-1042 rtI_ n t _

^ = 0-7622 X ° · 1 7 = ° ' 1 2 2 ;

$· . = 0-638-y0-104+/O-122 = 0-638 +/Ό-018. Assuming that the units of the hydro-electric station A are loaded uni

formly, we can determine from the power balance at the point a that :

^ ,« = S Ata = i (0-638 +/0-018 +0-509-

-./Ό-126+0-852+/Ό-313) = 1-0+/0-103. Therefore, the e.m.f. of the generators of HES A is :

= V[(0-75+0-0895)2+0-872] = 1-21; 0*87 òAd = 12-0+arctan0.75+( ) .0895 = 12-0+46-1 = 58-1».


The voltages found on the load busbars of the system enable us to calculate the impedances, representing the load in the equivalent circuit.

Using the formula (2.15):

ZL1 = Z13 = ? 3 ^ x 0-95 (0-95 +y0-3123) = 0-583 +./Ό-1915.

Similarly:

0-852

ZL2 = zu = 0-966+./Ό-318;

ΖχΛ = Ζ 1 5 = 2-77 +7'0·91;

Zu* = 2Γ1β = 0-933+7Ό-70.

212 I W/0*966*j 0-318

»k

' 5/j00727

Ej^l-21^58-1· l/jO-652 ©-♦ ΠΡ—

Ε^Ι·2!Ζ58· | · 2/J0-652 0-« η^-

13/0-583

Ε£=0·895Ζ!3·0β

Fio. 2.19.

The equivalent circuit, comprising the determined load impedances, is shown in Fig. 2.19. ·

We now proceed to the transformation of the circuit obtained.


(a) Combination of the heating and power plant B and the receiving system into one equivalent station

The star efBO is replaced in the equivalent circuit by an equivalent delta eOB:

Z17 = ZBe = yO-065 +70479 +J °Q1%*^™ = -0·008+7Ό·25.

Similarly:

18 = Zeo = 1-27+71-02;

Z19 = ZB0 = 3-50+y 2-81.

The stars ageO and acdO are transformed into deltas aeO and adO respectively, the impedances of the deltas being determined as :

Z M = Zae =J0.0S26+m09+ ^·0826χ,0·10?

Similarly:

j0· 1385+2-77+y'0-91

= -0-00285+y'0-1927.

Z2l=ZM= 4*7+J1-926;

Z22 = Ze0 = 6-42+72-54;

2» = Zad = -0-0276+y0-364;

Z2i = Λ ο = 1-866+70-925;

Z2S = Zd0 = 2-005+70-994.

The transformed circuit is given (Fig. 2.20). After adding the parallel circuits and transforming triangle ade into an equivalent star, the circuit takes the form shown in Fig. 2.21.

The impedances of the sides of the equivalent star adem are equal to : _ (-0-0276+70-364) (-0-00285+70-1927)

27 ~ am ~ -0-0276+70-364-0-00285+70-1927 +/0-1215 = -0-0045+70-1035;

Z28 = Zdm = -0-00182+70-0651; Z29 = Zem = 0-00109+70-0346.

Addition of the parallel circuits makes it possible to determine the im-


i l/jO-652 Z24 I866+J0-925 2-005+j 0-994 Z25

&+-Φ

2/J0-652

-0-0276 + JO-364

-QZl· - 000285 ♦! 0-1927

0-583* jO-2139 ^ 7/J0 1215

U-87*j 1-926

-ran—ii·-l-27*j 1-02

6-42*j 2 54 b J |zJ -0-008«-j0-25

Φ

3-50 *j 2-81

75"

FIG. 2.20.

l/j 0-652

^ A,

350+J2-8I

B i — Η 5 Ί li"·

FIG. 2.21.


pedances :

•^30 — AJ0£ — ' 1

1 1 1 0-583+/0-214 ' 4-87+71-926 l-866+y'0-925

= 0-41+/Ό-163; _ (1-27+;l-02) (6-42+y2-54) _

Z51 - Zrtr - 1 . 2 7 + 7 · 1 . 0 2 + 6 . 4 2 + 7 · 2 . 5 4 - 1 079+70 772.

Continuing the transformation, we substitute the star meBO in the circuit shown in Fig. 2.21 by an equivalent delta, the impedances of which are:

Z& = ZmB = 0-00109 +70-0346-0-008 +/Ό-25 + (000109 +70-0346) (-0-008 +/Ό-25)

+ : 1-079+70-772

Z33 = Zm = 1-235 +/Ό-904;

Zzi = Zgo = 8-46+77-08.

= -0-0122+70-288;

EUl-2k5e-le l/jO-652 0 - . ΠΓ-

E'A= I-2IÌ58-I· 2/j 0-652

6 H -rjpv

5 OAl+j 0-163 Z30 Z32 1-0-0122+] 0-288 / /

(a)

ε^ = »·21Ζ.5β·!β l/jO-652

Θ-: ^ —

B|—Γζ^ΓΗΐΐ' /

É'B = 0-895<I30°

-0-0045+j 0-1035 -0-00158*j0-0531 Ec(|)= 0*845 <L2'4e

-Œh É^s 1-21^58-1· 2/J0-652

θ- - Ί Γ -

-ŒZJ—?—θ

Π r i ζ37Π| Ι 3 + ί° ·6 β 6

ζ30ηοΛ1+]ο,β3ζ n l 2 3 ^ 0 9 0 ^ g

(b)

FIG. 2.22.


If after this we add the parallel impedances Z19 and Z34, the circuit will then take the form, shown in Fig. 2.22(a), where:

(3-50+72-81) (8-46+/7-08) 3-50+72-81+8-46+./7-08

7 - i-'-"'-rv^°1M°-™-r/'"°; _ 2 . 4 8 , / 2 . 0 •^35 - -j.cr> , n . o i , Q.A* , .-7.no — L *° +JZ U-

In the circuit obtained, the two voltage sources, the heating and power plant B and the receiving system, are connected to the common point of the circuit m without intermediate loads. This allows the points B and d to be connected and the required circuit to be obtained, in which HPP B and the receiving system are combined into one equivalent station (Fig. 2.22(b)).

Here:

_ (-0-00182 +70-0651) (-0-0122 +./0-288) ^ " " · ~ -0-00182 +70-0651 -0-0122+y0-288

= -0-00158+70-0531; Z„ = Za\\Zu = 1-13+70-686.

The electromotive force of the equivalent generating station is determined as:

0-895^13-0° 0-838^0° -0-00122+70-288 _-0-00182 +j0-00651

1 Éjo = vw"~^v"w . « w ^ - r ^ ^ w . = 0 . 8 4 5 ^ 2.4o

-0-00158+70-0531

The inertia constant of the equivalent station TJ(X) = <».

(b) Combination of three generators of HES A with an equivalent generating station

We replace the star amnO in the circuit, Fig. 2.22(b), by an equivalent delta with sides

Z38 = Zm = -0-0Ο45+7Ό-1035-0-ΟΟ158+/Ό-0531 + (-0-0045+70-1035) (-0-00158+70-0531)

1-235+/Ό-904

Z3» = Z a 0 = 3-61+72-80; ZÌO = Zna = 1-876+71-41.

= -0-00913+70-1585;

http://-7.no


The circuit obtained, after this transformation, is given (Fig. 2.23(a)). Joining points A± and n in this circuit, we arrive at the required circuit, which contains two generating stations as required by the conditions of the problem. This circuit is given Fig. (2.23(b)):

Ε^=1·2Ι£58·Ιβ 1/jO 652 ©-« <T^

Ε1=Ι·2!^58·Ιβ 2/j0-652 Θ—· ^

A,

0-009I3+J0-I585 Εβ(ι)=0·845Ζ2·4°

« * — r a n ■ nA e»

3 · 61+j 2-8 I

l-876+j I-4IJ H3+J0-686

Ε^Ι·2ΙΖ.58·Ι° l/jO-652 -0-006*jOI27 έβ(2)=0·823£Ι6·0° θ η^- -0Ε> -θ

·376*]0·16θΜζ42 Ζ43 M 0709*jO-A65

(b) ψ W

FIG. 2.23.

The parallel resistances connected at points a and n, are replaced by the geuivalent impedances, Z42 and Z43 respectively, whereupon

7 - 7 _ 70-652 (-0-00913+;Ό·1585) _ Z "

- Z^

> - ,Ό·652-0·00913+,Ό-1585 ~ - Ο ^ + ^ ^ ,

Z42 = 0-376+./0-160; Z a = 0-709+y0-465.

We determine the equivalent e.m.f. of the generating station:

1-21^58-1° 0-845^2-4° + ■ É„ = ° · 6 5 2 " 9 ° ° -0/00913+yO-1585 = 0 . 8 2 3 , ^

-0-006+./0-127

The inertia constant of this generating station is also infinitely large, in as much as one of the sources of power, combined in the equivalent generating station and receiving system is considered as infinitely large.


Εί = Ι·2ΙΖ.58·Ιβ -00772*] 0-455

z25 2005+J0-994

-000l82*j0-065l

-00I22+J0-288

Bt—nST—in-2-48* j 2-0

È'B = 0-895ZI30° (a)

Ε«(3)=0·958^32·8β -00I575+J0I77 -000l82*j 00651 Vd =0-838ZL0e

FIG. 2.24.

(c) Combining HES A and HPP B into one equivalent generating station

The corresponding transformations in the case under consideration should be carried out with respect to circuit Fig. 2.22(a). Combining points A± and A2 and replacing the star AamO by an equivalent delta, this circuit can be reduced to the form shown in Fig. 2.24(a), where :

Zu = ZAm = 7O-5x0-652-0-0045+/0-1035 +

. 70-326(-0-0045 +/Ό-1035) Λ „ „ ^ .Λ . . . -1- 0-41+70-163 = - ° - ° 7 7 2 + ^ 4 5 5 ;

Z« = Zil0 = 1-72+70-946; ZiG = Zm0 =0-533+70-323.

By joining points A and B, we get:

( - 0-0772 +7*0-455) ( -0-0122 +7*0-288) Zu — Z.(a\ — *(3) -0-0772 +./Ό-455 -0-0122 +./Ό-288

= -0-01575 +y'0-177; Z48 = 1-03+./Ό-655.


Adding, in parallel, the impedances which are connected at point m> we arrive at the required circuit, shown in Fig. 2.24(b), where

7 - 0-235+^904) (0-533+70-323) _ A M , ,A „ „ ^ 4 9 ~ 1-235+./Ό-904+0-533+/Ό-323 - " ' ^ + ^ ^ 1 ^ ·

The electromotive force of the equivalent generating station is:

1-21 ^58-1° 0-895^13-0° - + -Ä Ä = -0-0772+y0-455 ■ -0·0122+;0·288 = ^ ^

-0-01575+./Ό-177

and its inertia constant is:

r« -6 sS+ , 0ra - '♦'+MS - 2 3 · 3 5 "«■

Problem 2.7

An electrical system is shown in the form of an equivalent circuit in Fig. 2.25.

Required: to calculate the self and the mutual admittances of the circuit. The calculations are carried out: (a) by the method of transforming the circuit; (b) by the method of unit currents.

OZI—\

FIG. 2.25.

Solution. When solving the problem by the transformation method, we bring the circuit to a T-shape form. For this we replace the delta abO of the circuit, Fig. 2.25, by an equivalent star:

_ 10xy-2 720(15-/2) _ *ac Ì0+S+J2 152+22 " ° + ; '

Zbc = 0-0874 +./Ό-655; ΖΛ = 3-28-./Ό-437.


The impedances of the arms of the T-shape circuit, Fig. 2.26, are: Z, =yl +0-175+71-31 = 0-175+72-31 = 2-31 ^ 85-7°; Z„ =j2 +0-0874 +/Ό-655 = 0-0874+/2-655 = 2-655^88,1°;

Zra = 3·28-;Ό·437 = 3-305 ^ -7-6°. The required admittances are determined (Ref. 3, pp. 30, 31) as:

1 Yl2 =

Ζτ+Ζη + Z|Zn

'III

Yn = 1

Zi +

*9Ά —

Zumili Z\\ +■ Zm

1

Zn + Z\2m

Zi + Zi III

In the

FIG. 2.26.

; conditions of the problem: 1

0-175+/2-31 + 0-0874+y2-655 + —

1 ~ 5-18^107-9° '

Y - 1 11 4-25^66-9°

Y - l J e _ 4-35^72° '

^85-7° 3-305

x 2-655^ ^-7-6°

88-1°

ya = ^j8 = 0 1 9 3 ; "12 = 9 0 ~ 1 0 7 ' 7 = - 1 7 ' 7 ° ;


yn = _ = 0-236; a n = 90-66-9 = 231°;

y22 = — = 0-230; a22 = 90-72 = 18-0°.

In order to calculate the admittances Yn and Y12 by the unit current method, we connect point 2 with the return conductor of the circuit (Fig. 2.27) and we assume that in this condition the current I20 = 1-0 ^0. In this case:

K6 = /20Z2 = Ixj2=j2;

hb = 4>+Λο = l+yO-4; zlFe6 = fbZs = (l+y0-4)xy2 = -0-8 +J2;

Va = Vb+AVab =J2-0-S+j2 = -0-8+y4;

A = Λ* + Αο = 1+;Ό·4-0·08+7θ·4 = 0·92+>0·8 = 1-22 ^ 41-0°; /iFx = / A = (0·92+./Ό·8)χ./1 = -0·8+;Ό·92;

Ϋχ = Ϋα+ΔΫλ = -0-8+74-0-8+;0·92 = -1-6 + +J4-92 = 5-18^108°.

Γ|—QQ-J—OD-T—OD f

(α)

I'ab h a - * b

-f—QD—f-GD-f—OD h i ΙΠ ΠΙ ^Φ

(b)

Fia. 2.27.


Consequently, in accordance with the calculation:

1 1 r» =·

Yu =

V± 5-18^108° •MÌO

1 1 1 V± 5-18^108° 4-25 ^ 67-0° tx 1-22 ^41°

In order to determine the admittance y22, it is necessary to connect point 1 with the return lead (Fig. 2.27(b)) and introduce current /10 = 1 ^c0 in the impedance Zx. Carrying out the calculation with this condition for the circuit, Fig. 2.27(b), we get successively:

K0 = l x ; l =jl;

ha = 1+yO-l; ΔΫ* = (1+;Ό·1)χ;2 = -0-2+;2;

Γ ; = ; Ι - Ο · 2 + ; 2 = -o-i+fl-,

Πο = ~°Ί+β = -0·04+;Ό·6;

/2 = 1+;Ό·1-0·04+;Ό·6 = 0·96+;Ό·7 = 1-19 ^36-1°; AV3 = (0-96+y0-7)xy2 = -1·4+;'1·92;

V2 = - 0-2+/3-1-4+./1-92 = -V6+J4-92 = 5-18^108°; 1 1

*22 5-18 1-19

The mutual admittance we

^12 =

^108° ^36-1°

: obtain

1 V2 ~ 4>

4-35

as:

1 518 ^

^71-9°

108°

This calculation may serve as a reliable check of the correctness of all the previous calculations.


Problem 2.8

Figure 2.28 shows an equivalent circuit of an electrical system, which includes two generating stations and complex and reactive loads. In practical calculations the calculated circuits of systems with unsymmetrical short-circuits are reduced to circuits of this type.

Required: to determine the self and the mutual admittances of the generating stations and the loads, which are necessary for calculations of their operating conditions.

@4 * γ 0·2Ζ36·8β Ζ

-L?

jOI JOI

J2

+^

FIG. 2.28.

Solution. The calculation is carried out by the unit current method. Short-circuiting the generating station 2 and assuming that in this circuit the current /M = 10 ^ 0 (Fig. 2.29), we find:

1* = j205

Va = lOxyO-l =7l-0;

Ac = £ = 0-5;

tba = 10+0-5 = 10-5;

ζ^6β = 10·5χ;Ό·1 =/1·05;

Vb = yi-0+/ï-05 =y205;

= 10-25^53-2° = 10-25(0-6 +/Ό-8) = 6-15+y8-2; 0-2^36-8°

A = 10-5+6-15+/8-2 = 16-65+J8-2 = 18-6^26-2°;

ΔΫΧ = (16-65 +/ 8-2) xy2 = /3-33-16-4;

Ϋχ =y'2-05 +733-3 -16-4 =y35-35-16-4 = 38-9^114-9°.


TP f-

f U '20?

FIG. 2.29.

Figure 2.30 shows a vector diagram which illustrates the results obtained. The data found make it possible to calculate the self admittance Yn and the mutual admittances Yn, Γ13 and Yu, where :

Yn = 38-9^114-9° = 2-09^88-7° = ° ' 4 7 8 ^ ~ 8 8 ' 7 ° ;

18-6^26-2° Y12 = 0-257^-114-9°;

«ii = 90-88-7 = 1-3°; a12 = 90-114-9 = -24-9°; Yu = 0-263^-51-7°; Yu = 0-01285^-114-9°.

f IM

ε X o c* n

ε

•^

i

(Ì20xReZ,2) = OK \

V

-90?

\%^^

h

+

'20

FIG. 2.30.


The remaining admittances are determined by using the circuit shown in Fig. 2.31; assuming that in this circuit /10 = 1-0 ^0°, we get:

Vb = Ixj2=j2;

1L = Γο- = 10^53-2° = 6-0+./8-0; 0-2^36-8° tab = 1-0+6-0+78-0 = 7-0+./8-0;

AVab = (7-0+j8-0)xy0-l = -0-8+yO-7; V'a =J2-0-S+jO-7 = -0-8+./2-7;

0-8+y'2-7 /' -J2 - 1-41 ^16-5°

/2 = 7-0+./8-0+./Ό-4+1-35 = 11-85 ^45-2°; AV2 = (8·35+/8·4)χ./Ό·1 = y'0-835-0-84;

V2 = -0·8+;2·7-0·84+;0·835 = 3-89 ^114-9° In order to check the calculations, we determine that:

1 Y21 = 3-89^114-9° 0-257^-114-9° = F, 12-

Consequently, the calculations are correct and it is possible to proceed with confidence to calculation of the unknown admittances.

Î22 — 1

3-89^114-9 - = 3-05^-69-7°;

11-85^45-2° θ22 = 90-69-7 = 20-3°;

^23 —

Yu =

1 3-89^114-9°

10^53-2° 1

3-89^114-9°

= 2-57^-61-7°;

= 0-362 -98-4°.

FIG. 2.31.

(2.19)


The condit ions of the generat ing s tat ions are determined by the relationships (ref. 3, p . 29) :

Px = E\yn sin a u +£ ,1 £ ,

2 i y 1 2 sin (<51 2-a1 2); P2 = E\y22 sin oc22-EìE2y12 sin (<512+a12); öl = £?Λΐ C 0 S «11-^1^12 C 0 S (δ12-α12); Q2 = E\y22 cos a22 -E^E^ cos (δ12 + a1 2), J

i.e. />x = 0·478£2 sin 1·3°+0·257£,

1£2 s i n (*ΐ2+24·9°) :

Ρ 2 = 3 · 0 5 £ | sin 2 0 · 3 ° - 0 · 2 5 7 £ Α sin (<512-24·9°); ρ χ = 0-478£?cos 1 · 3 ° - 0 · 2 5 7 £ 1 £ 2 cos (<312+24·9°); β 2 = 3·05£· | cos 20 ·3° -0 ·257£ ,

1 £ 2 cos (<512-24·9°). The condit ions at the loads can be determined in the following manne r :

Sz = / } Z 8 and $4 = / J Z 4 , where (ref. 2, p . 46)

* 3 = El *13 +^2 *22 >

h = ^1^14+^2^24·

Problem 2.9

Figure 2.32 shows an equivalent circuit of a very simple electrical system.

V(=IZ.£ V2=Vor

l i ΠΠ Ì2 Po --0-8

FIG. 2.32.

Required: to determine the angle of displacement of the vectors of the voltage applied at points 1 and 2 of the circuit, with respect to the voltage value V2 when transmit t ing power P0 = 0-8.

Solution. In the circuit under considerat ion, containing reactance only, the expression

P = ^ ^ 2 sin <5, (2.20)

where δ is the required angle.

6262


Therefore, for the conditions of the problem,

. « Ρχ 0-8x1-2

The results of the calculation, represented in the graph (Fig. 2.33) show that a reduction of not more than 4 per cent is possible in the voltage V2 in relation to the voltage Vx with the assumed conditions.

dcgrets

90

70

50

30 0-90 0-96 10 II 1-2 13

FIG. 2.33.

With a larger reduction in the voltage at point 2, the required power P0 cannot be transmitted.

Problem 2.10

Figure 2.34 shows five equivalent circuits of separate sections of an electrical system.

Required: to construct power-angle characteristics assuming that the voltages applied at points 1 and 2 are equal to unity.

Solution. We calculate the self and mutual admittances of the first of the circuits under consideration. These admittances are (see


Problem 2.7):

Y - 1

J 2+jl

Y l

1 2 ~ . , ., yixyi

= 0-544^-77-5°;

= 0-485 ^ 104°

Hence we have also: a„ = 90-77-5 = 12-5°; a12 = 9 0 - 104 = - 14(

I 1/1-0 2/110 I

= 3/-j20

I 1/10 2/-JI 0

1/11-0 ' 2/jl-O -ΛΓ1—?—ΊΓ>-

! 3/-J2-0

:£: 3/-J20

(c) Jr

(«)

FIG. 2.34.

In view of the symmetry of the circuit Yn = Y229

Y22 = 0-544 ^ -77-5°; a12 = 12-5°. The power-angle characteristics are determined by the equations:

Λ = ^ÏJii sinall+E1E2yì2 sin (<312-a12); P2 = - ^ Ί ^ sin a22+£'i£,2>;i2sin(*i2+ai2); ôi = E\yxl cos an-jE^JE^M cos (<512-a12); Q2 = -£fj>22 C 0 S «22+^1^12 C 0 S (δΐ2+α12λ

Substituting the values of the admittances and angles a in these formulae, we get with Ελ = E2 = \:

Pl = 0-544 sin 12-5°+0-485 sin (<512 + 14°) = 0-118+0-485 sin (<312 + 14°);


P2 = -0-118+0-485 sin (<512-14°); ß t = 0-544 cos 12-5°-0-485 cos (<512 +14°)

= 0-53-0-485 cos (<512 + 14°); Q2 = -0-53+0-485 cos (<512-14°).

In accordance with the expressions obtained, we calculate the values of the active and reactive powers at different angles δ12 and form the functions shown in Fig. 2.35(a).

For the circuit as in Fig. 2.34(b)-(e), we get, by means of similar calculations, the parameters listed in Table 2.4.

TABLE 2.4

Circuit

b c d e

>Ίι

0-544 0-333 0-446 0-446

^22

0-544 0-333 10 10

yit

0-485 0-666 0-893 0-893

αιι> degrees

167-5 0

26-5 153-5

α22, degrees

167-5 0

531 126-9

α12> degrees

-166 0

26-5 153-5

In accordance with these data, the active and reactive power, which flows in the circuits, is determined by the following formulae: Circuit b

Λ =0-111 +0-485 sin (<512 + 166°); P2 = -0-118+0-485 sin (<512-166°); Ô! = -0-53-0-485 cos (<512 + 166°); Q2 = 0-53+0-485 cos (<512-166°).

Circuit c

Circuit d

p1 = p2 = 0-666 sin δ12; öi = 0-333-0-666 cos <512; Q2 = -0-333+0-666 cos ό12.

Px = 0-199+0-893 sin (<512--26-50); p2 = -0-799+0-893 sin (<512+26-5°); Qx = 0-399-0-893 cos (<512-26-5°); Q2 = -0-60+0-893 cos (δ12+2(5-5°).


Circuit e Px = 0-199+0-893 con (<312-153-5°); p2 = -0-799+0-893 sin (<512 +153-5°); Ö! = -0-399-0-893 cos (<512-153-5°); Q2 = 0-60+0-893 cos (<512 +153-5°).

The power-angle characteristics in these cases are shown in Fig. 2.35. Comparing the values of the angles a12 in the circuits and the power

diagrams considered, it follows: (1) With a series connection of the ohmic resistance (circuits d and

e) the displacement of the power characteristics at the beginning and the end of the circuit is greater the larger the angle δ12.

(2) With parallel connection of the ohmic resistance (circuits a and b), the displacement of the active power characteristics at the beginning and the end is greater at smaller values of the angle δ12, which is associated with the change in voltage at the point of connection of the ohmic resistance.

In the equations which determine the power-angle characteristics, the method of connection of the ohmic resistance into the circuit is associated with the sign of the angle a12 (with series connection of the ohmic resistance <z12 > 0, with parallel connection a12 < 0).

(3) With the transmission of power through a T-form circuit, where an inductive reactance is included in the series circuit, and an ohmic resistance in the parallel circuit, the reactive powers at the beginning and the end of the circuit with equal voltages at the end, have opposite directions at all angles.

Replacing the inductive reactances by capacitive ones (circuit b), the signs of the reactive powers are changed. In this case, the capacitive elements play the part of the reactive power generators. The active powers at the ends also change sign, and the increase of the angle <512 from 0 to 180° causes a flow of power from point 2 to point 1, whereas with the inductive reactance in the circuit, this power flows from point 1 to point 2 at the same angles.

In the equations for the angular characteristics, this difference is reflected in the values of the angles a12, which in the latter case lie in the second and fourth quadrants.

(4) The connexion of a capacitance into the parallel branch leads to the result that the reactive power at the beginning of the circuit, as well as the reactive power at the end, changes not only its value but also its direction according to the angle <512.


1-0

0-8

0-6

0-4

Al 0

-0-4

-0-6

-0-8

-1-0

RQ

/ /

'«12

/ /

5.

/ /

Ύ —

y sP|

— ^ 6 0 ~ 90 120 15018C

^

■ ^

—

S, αΝ V

\ \

a^ 2J "V2IO_2W_27q_300/âegd

/

S A

!·0

0-8

0-6

0-4

0-2

(b) o

-0-2

-0-4

-0-6

-0 -8

-1-0

IO

0-8

0-6

0-4

0-2

0

0-2

0-4

0-6

0-8

1-0

[RQ~ * *

k

^ ^

\ H \ \

N

$·^

V s Sy60 90 120 I50yfe0 210 240 270 300 degrees 1

X \ £±

A

(e)

[pua U t

3 0 /

F).Q

\ 90 I2Ò 150 '

\ ° 2

\X

12 V 210 240 270/Ndegrecsl

Fio. 2.35.

62


1 o*

<uT

CM

CM

5 * •o

o

>r

©

§ o

CN

8

o co

7*

Oy <M

a.

— — — O

o r · J- <■ c

-^f

3 o

a.

0 U

O

D

^ <SI

1 τ

x^ r * C

s CM

CM

o CM

o 00 o in

o —

o to

3 C-

CM a.

SI ?^T; 1 1

D 00 CM<

<M

-> r SI > r u 3 OC

1 Ί

Flo.

2.35

(co

nt.

).


At small angles, in the series circuit, small currents flow, and produce a negligible loss of reactive power, whereas the capacitive element in the circuit brings about a "generation" of power greater than the loss. With an increase of the angle <512, the loss of reactive power increases but the "generation" of power drops, since the voltage in the capacitive element diminishes.

Problem 2.11

An electrical system is represented by a quadripole, the generalized parameters of which have the following values :

À = 0-945 +y0-005;

È = 0-072+/Ό-925;

C=y0-115;

Ù = 0-942+/Ό-004.

Required: (1) to determine the self and mutual admittances of the circuit; (2) to construct power-angle characteristics for the beginning and the end of the quadripole, from which we find the steady-state quantities with voltages at the beginning and the end, to respectively equal :

Vx = 2-4^25°; V2= 1^0°.

Solution. The generalized parameters of the quadripole are obtained from the self and mutual admittances by the relations:

Λ 0 · 9 4 2 + ; 0 · Μ 4 = 1 . 0 2 ^ [ 11 È 0-072 +/0-925

hence :

Ya = Y21 = -S; = = 1-08 ^ -85-5°; È 0-072 +./0-925

Y„ = A= 0-945+y0-05 = 1 , 0 2 1 , _ 8 5 . 2 o . È 0-072 +/0-925

«u = 90-85-3 = 4-7°; a22 =90-85-2 =4-8°; a12 = 90-85-5 = 4-5°.


The equations of the power-angle characteristics may be written in the form:

Pt = 2·42χ 1-02 sin 4·7°+2·4χ 1 X 1-08 sin (<512-4-5°) = 0-480+2-59 sin (δ12-4·5°);

P2 = -1-021 sin 4-8°+2-4x lxl-08 sin (<512 +4-5°) = -0-0855+2-59 sin (<312 +4-5°);

Ô! = 2-42x 1-02 cos 4-7°-2-59 cos (δ12-4·5°) = 5-85-2-59 cos (<512-4-5°);

Ô2 = -1-021 cos 4-8°+2-59 cos («512 +4-5°) = -1-02+2-59 cos (<312+4-5°).

FIG. 2.36.

From the results of the calculations we plot the curves (Fig. 2.36) and read off the quantities at the given angle ò12 = 25°. Here we get:

P1(0) = 1-4; P2(0) = 117; Qm = 3-45; <22(0) = 1-25.


Problem 2.12

Figure 2.37 shows the equivalent circuit of a system which contains a generating station operating with constant-voltage busbars. A portion of the station capacity is transmitted to an intermediate load. The parameters of the system and other quantities are given in the circuit drawing.

xd=l-8 l/jO-6 2/J0-5 v=I=const @ np—±—or—\

-T * H V L « 5,L I L 0-55*j0-25

ZL Ι·6*]Ί·2=2·0*36·'β·

Fio. 2.37.

Required: (1) to construct the active power-angle characteristic of the generating station, representing the load by a constant impedance; (2) ditto, but combining the generating station and the loads in the circuit into one equivalent station.

Solution. (1) We determine the quantities in the system at the given transmitted power. The voltage at the point of connection of the load is:

+J 0-55x0-5 = Vi2s+j0.2js = 1·157^13·7°.

The loss of power in reactance 1 is :

ΔΡγ = 0;

^ 1 = ^ ö i , 1 = ^ ! ^ i x 0 - 5 = 0 , 8 2 .

The power consumed by the load at voltage V2 = 1-157 is:

Pw> = ^RL = ^ o ? x 1-6 = 0-535;

V2 1-1S72


From the power balance at point L, we determine the power flowing to the load from the side of the generating station;

Su. = Po+Puoi+ÄQo+AQi+Quo)) = 0·55+0·535+/(0·25+0·182+0·401) = 1-085+/0-833.

The power S^ enables us to calculate the generating station e.m.f. in the condition under consideration:

Zldio) - yL-i y rj y

0-833 (0-6 + 1-8) . 1 085 (0-6+ 1-8) - 1-157+ ^ +j ^

= 2-88+/2-25 = 3-66^38°;

àEy = i2(o) = ^vlv+àEyL = 13-7+38 = 51-7°.

The self and mutual admittances at the generating station are :

Yn = jO-Sjhó+jW = 2-82^88-5° = ° ' 3 5 5 ^ " 8 8 ' 5 ° ;

3 ■ l-6+jh2+j0-5

hence we have: a n = 90-88-5 = 1-5°; a12 = 90-98-4 = -8-4°.

The equation of the power characteristic is : pi = ^(ο)Λι s i n o u + ^ ^ u sin (<312-a12)

= 3-662x 0-355 sin 1-5° +3-66 x 1x0-304 sin (<312+8-4°) = 0-12454-1-11 sin (<512+8-4°).

As a result of the calculations we get the curve 1, shown in Fig. 2.38. (2) Combining the generating station and the load, we determine the

impedance and the e.m.f. of the equivalent station to be:

p _ 3-66 ^51-7°x 1-22 ^60-8° _ E' 2-4^90° * 8 6 ^225 ■


Figure 2.39 shows a transformed equivalent circuit. In this circuit, which includes series-connected impedances only, the self and mutual admittances are equal to each other.

ΙΆ

1-6

K

1-2

10

0-8

0-6

0·Α

0-2

ri

/ '

/ 2

\

\ *u\

30 60 90 120 150 degrees

F I G . 2.38.

0'595*j 1-065 2/J0-5 V=|

Θ QZJ-^-^-4 έ„=Ι·86Ζ22·5°

FIG. 2.39

We determine their numerical value:

^11 = ^22 = ^12 = 1 1

0-595 +j 1-065 +./Ό-5 1-67 ^69-2° = 0-599^-69-2°;

therefore, «i i = «22 = «12 = 9 0 - 6 9 - 2 = 20-8°.

The active power of the equivalent generating station here is determined by the expression:

P± = 1·862χ0·599 sin 20·8° + 1·86χ 1x0-599 sin (<512-20-8°) = 0-735 + 1-112 sin (<512-20-8°).

The characteristic 2 in Fig. 2.38 is constructed from the calculated results.


Comparison of the power characteristics 1 and 2 shows that the combination of the generating station with the load results in a considerable error. This error is connected with the different effects of the ohmic resistance in the circuits, to which the characteristics 1 and 2 refer. In the original T-form circuit, with an increase of the angle <512, the voltage at the point of connection of the load diminishes, and consequently the active power delivered to the load also dimishes. However, in the transformed circuit, in which the ohmic resistance is connected in the series circuit, an increase of the angle <512 is accompanied by an increase in the loss of active power in this resistance due to the increase of the current flowing in it. The difference in the effect of the ohmic resistance under consideration, is reflected in the theoretical formulae by the difference in the magnitudes of the angles an and a12 and the difference in the sign of the angle a12.

With a purely reactive load, these angles are equal to zero and, therefore, in this case, we can expect the active power characteristics to be the same for the original and the transformed circuits,

We shall check the correctness of this conclusion with numerical values for this problem; taking Z2 =71-2. For the original T-form equivalent circuit:

1-1572 , ΛΛη ß L ( 0 ) = - j ^ - = 1-117;

SÌL = ^)+y(ôo+422+ÔL(o)) = 0·55+/(0·25+0·182 + Μ17) = 0-55+/1-549;

* ι .en 1-549x2-4 , .0-55x2-4 Α r„ ΛΛ rQ Eld{0) = 1-157+ 1 1 5 7 +j M 5 7 = 4-53 ^14-6°;

àEv = «ΐ2(ο) = àVLv+ÒEv£ = 13-7 + 14-6 = 28-3°;

Y» = '/Mx/O-S = ° ·2 5 6 ^ " 9 0 ° ;

-Pi = Eld(0)Vyu sin ò12 = 1-16 sin ò12. When combining the generating station with the reactive load, the

impedance and the e.m.f. of the equivalent station are : _J2-4xjl-2 _

4·53.28·3°χ;0·8 = 1 . 5 ^ 2 8 . 3 Ο y2-4


The mutual admittance and the active power have the values:

r - = 7ô^W = 7F3 = ° · 7 6 9 ^ - 9 0 0 ' therefore

Pi = 1-51 x0-769 sin <512 = 1-16 sin b12. A complete agreement has been obtained in the expression for active

power.

Problem 2.13

An electrical system, comprising a station, in which two generators are installed with different parameters, a network, and busbars of constant voltage, has the equivalent circuit shown in Fig. 2.40. The parameters of the elements of the system and the steady-state quantities are given on the circuit.

G-l l/jO-2

® ' f o - ^ Xdlr1"6 0-25+j 0-116

G-2 2/J0-I5 © «TP ^

xd2=l-2 0-5«.j(K

FIG. 2.40.

Required: to construct the power-angle characteristics of the system in the cases:

(1) The electric power station is represented in the calculated circuit by two generators;

(2) Two generators of the electric power station are replaced by an equivalent one.

Solution. The active power characteristics of the generating stations in the system are determined by the general equations:

Λ = Ε&η sinan+^Î^Via sin (ô12-a1^+ElEzy1^ sin (<513-a13)+ . . .; P2 = E\y22 sin a22 + £ 2 ^ 2 i s i n (δ2ΐ-a2i) +^2^23 s i n (^23 - « 2 3 ) + · · · ;

0-75*j0-3

Pm = E%ymm *™ *mm + ^ ^ 1 ^ 1 Sin ( 3 m l - a m l ) +

-Γ-ΕΊιΑΡϋΛ S Ì n ( Ô /M3- a m3)+ · · · -


In the system under consideration there are three generating stations, one of which is a receiving system, and there is no ohmic resistance in the equivalent circuit. Therefore, it may be assumed that a u = a22 = 033 = «12 = «13 = 0.

In these conditions, the required characteristics are determined by the equations:

Pi = EyEJb^ sin Än+JEiKeixe sin <$lc:

P2 = -ExE2bxl sin d12+E2Vcb2c sin ó2c;

Pc = ν€Εφΐ€ sin <Slc + VE2b2c sin d2c.

(1) We determine the voltage at the junction point of the circuit, on the high-voltage busbars of the station and the e.m.f. of the generators:

0-3x0-4 .0-75x0-4 bus = M + 'L! +J—TT* =1-24^12-8°;

. Λ _ 0-116x1-8 .0-25x1-8 Λ AC ΛΑεο Edl = 1-24 + — j ^ — + j 1 2 4 = 1-45^14-5°;

* 1 <™ . 0-4x1-35 , . 0-5x1-35 , _, l o 0 Ed2 = 1-24+ V24 +j V24 = 1-76^18°:

*ic = àbusc + ôlc = 12-8 + 14-5 = 27-3°;

<52c = 12-8 + 18 = 30-8°;

<5l2 = 14-5-18 = -3-5°.

We calculate the mutual admittances, assuming:

*i = Xdi+Xi = 1-6+0-2 = 1-8;

*n = *d2+*2 = 1-2+0-15 = 1-35.

Then

*w = = *« o 1 »« = ° 1 0 8 2 · * + * + » 1.8 + 1-35 + i ^ ^ " xc 0-4

Similarly, we get: ble = 0-366; b2e = 0-488.


Substituting the obtained data into the equations for the power, we get:

i>! = 1-45 x 1-76x0-1082 sin δ12 + 1·45χ 1-1x0-366 sin ôlc

= 0-276 sin <312+0-583 sin <5lc; P2 = -1-45 x 1-76 x 0-1082 sin <312 +1-76 x 1-1x0-488 sin ô2c

= -0-276 sin «5x2+0-945 sin <32c; Pc = 1· 1 x 1-45 x0-366 sin ôlc + l- lx 1-76x0-488 sin ò2c

= 0-583 sin òle+0-945 sin b2c. To check the correctness of the expressions obtained, we calculate the

powers Pl9 P2 and Pc at the angles which were determined earlier for the normally steady-state condition. We have:

Px = 0-276 sin (-3-5°)+0-583 sin 27-3° = 0-2499; P2 = -0-276 sin (-3-5°)+0-945 sin 30-8° = 0-5006; Pc = 0-583 sin 27-3°+0-945 sin 30-8° = 0-7505.

The results calculated in accordance with the formulae obtained, agree with the initial data.

(2) We combine the circuits of the generators in the equivalent circuit and determine the reactances and the e.m.f. of the equivalent generator:

1-8x1-35 A „ t

^ = Τ8Τ ΐ^ = ° · 7 7 1 ;

1-45^14-5° 1-76^18°

Êde= l-^— ρ ^ = 1-56+/0-467 = 1-625 ^16-7°;

Γ8 + Γ35 bex = 16-7 + 12-8 =29-5°.

Ede *e ^ s ' c * 0 4

Θ ^—+-^Γ" \cc

FIG. 2.41.

Figure 2.41 shows the equivalent circuit of the system after transformation.

Problem 2.14

Figure 2.42 shows the basic circuit of an electrical system, in which two generating stations with commeasurable power are operating in parallel with a common load.


In the condition under consideration, power P2 = 400 MW is transmitted from station G-2 to the load L-2, where a voltage of 118 kV is maintained on the busbars of this load, and a voltage of 10*5 kV is maintained on the load busbars L- l .

G-i , T-i

L-l

. T-2 , T-3 G-2

FIG. 2.42.

The parameters of the generators, transformers, lines and loads are given below. In calculating the parameters in relative units, the basic values of the power are taken as Sbaaic = 270 MVA and the voltage Kbasic= 220 kV.

Required: to calculate the system conditions and to determine the self and the mutual admittances of its equivalent circuit.

Solution. (1) The equivalent circuit of the system is given in Fig. 2.43, where are also given the required operating quantities, which are express-

Va = H28 1 \ 2

j Ö ^ \ j 0-165

Vb=0-975

j 0-125 6

jO-0493 j 0-421

Α-βδ^ΟΤρτΙΐ

0-044*jOI76 . b 5 ab

OI85*jOU5 0-298

ib 1 1-48 CTI I

!*ί0·224ΓΊγ2·04*]Ί·53

FIG. 2.43.

ed in relative units. From the active power balance at the point b of this circuit, the active power, which is transmitted to the load L-2 by the generating station G-l, can be determined:

PL· = 2035 1-48 = 0-555.


GENERATORS

Generating station

G-l

G-2

Steady-state power, MVA

275

700

VuomkV

10-5

10-5

cos (pnom

0-8

0-85

Synchronous reactance xd

%

0-79

1-20

relat. units

0-985

0-421

TRANSFORMERS

Transformer substation

T-l T-2 T-3

Power SwtmfMVA

270 270 700

Transformation ratio

248/10-5 220/121

115-5/10-5

Reactances

%

13 12-5 14

relat. units

0165 0125 00493

LINE

Type of conductor

AC-300

Length, km

150

ohm/km

0105

ohm/km

0-42

Double circuit active resistance

ohm

7-88

relat. units.

0044

Reactance of double circuit line

ohm

31-5

relat. units

0176

LOAD

Load

L-l

L-2

Power

MW

50

550

relat. units

0185

2035

« w ç w

0-85

0-8

Resistance, relat. units

4-95+#-07

0-298 +/0-224


The reactive power g^ is related to the voltage values at points a and b and the power values by the equations:

Qìb = - vlybb cos <zbb+ Va Vbyab cos (aab+ocab); Pbab= - vlybb sin α ω + Va Vbyab sin (dab+ <zab)9

in which the unknown variable quantities are the reactive power Q^, and the angle of displacement between the vectors Va and Vb.

In so far as the circuit between the points a and b consists of series connected impedances only, the admittances Ybb and Y^ are equal to each other and are determined by the total impedance between the points a and b:

Y -Y - 1 l Jbb — Jab — yx2+Zs+/*4 ./Ό-165+0-044+./Ό-176+./Ό-125

= 2-14^-84-6°, whence

0Lbb =<zab = 90-84-6 = 5-4°.

By substituting the given voltage values into the expression for the active power, we get, with the calculated admittance:

0-555 = -0·9752χ2·14 sin 5·4°+0·975χ 1-128x2-14 sin (ôab+5-4°), or

0-555+0-1915 = 2-345 sin (δβο+5·40), hence we have

0-555+0-1915 2-345 òab = arc sin ^^ 5-4 = 13-1°

From the expression for the reactive power at the above angle δ^, we get:

(& = -0-9752x2-14cos5-4o+2-345cos(13-r + 5-4o) = -2-02+2-345x0-9483 =0-20.

From the reactive power balance at the point b, we can now determine:

Qbv> = 1-53-0-20 = 1-33. Therefore, S^, = 0-555 +JV33.

The electromotive force of the generating station G-2, which determines the flow of this power through the reactance is :

x9 = Χ5 + Λ:6 = 0-0493+0-421 = 0-47.


Hence

0-975 0-975

To determine the e.m.f. of the generating station G-l, it is necessary to determine first the power Sl9 generated by the station's generators. The loss of power in the impedance is:

Z10 = jx2 + Zs-hyx4 = 0-044+70-466 amounts to:

^ , n = ° - 5 5 r±° : 2 0 2 X0044 = 0-01605; 10

4Qio =

0-9752

0-5552+0-202 X 0-466 = 0170;

0-975*

Si = Sit+ASu+Su = 0-555+/Ό-20+0-016+./0-17 + + 0-185+/0-115 = 0-756+/Ό-485.

Therefore,

* , ,ΛΟ ,0-485x0-985 .0-756x0-985 , . . , . . „ , Eäx = 1128+ H 2 8 +j v m = 1-553+70661

= 1-685^23-1°; *i = Kb + àia = 13-1+23-1 = 36-2°;

δΐ2(β) = 36-2-23-8 = 12-4°.

FIG. 2.44.

Figure 2.44 shows a vector diagram, illustrating the calculated results. (2) In order to determine the self-and mutual admittances we apply the

unit current method. For this we short-circuit the generator G-2 and assu-


me that the current in it is equal to Ig = 1-0^0° (Fig. 2.45(a)). As a result of successive calculation of the conditions in such a circuit, we find:

/8 = yO-47

Vb =70-47;

r = 1-26^53-2° = 0-755+/101; 0-373^36-8

/10 = 1-10+0-755+71 01 = 1-755+/1-01; AV10 = (1 -755 +/1 01) (0044 +/Ό-466) = -0-393 +/Ό-863 ;

Ϋα = -0-393 +/Ό-863+/Ό-47 = -0-393+71-333 = 1-39^106-4°;

A = l5

3^Jn'^ = ° ' 2 3 9 ^ 7 4 ' 6 ° = ° · 0 6 3 5 +^° · 2 3 0 ; Λ = 1-755+71-01+0-0635+7Ό-230 = 1-819+71-24 = 2-2^34-3°;

AYr = (1-819+71-24) Χ7Ό-985 = 7*1-79-1-22; ΡΊ = -0-393+71-333-1-22+71-79 = -1-613 +/3-12=3-51 ^117-4°.

l i 4-95*j307fcj| I ' = 5·82£3Ι·8β n f

Therefore,

Y - A -

/2 Γ,.=

2-2^34-3° 3-51^117-4°

1 ΡΊ 3-51^117-3°

« u = 90-83-1 = 6-9°;

= 0-626^-83-1°;

= 0-285^-117-4°

'12 = 90-117-4 = -27-4°.


Short-circuiting the generating station G-1 (Fig. 2.45(b)) and assuming tx = 1Ό ^.0°, we find by similar calculations:

Yd = 70-985;

U = 5 . 3 2 ^ . 3 0 = 0169 ^58-2° = 00891 +/Ό1435;

/10 = 10+0-089 +7O· 1435 = 1-089 +/Ό- 1435; ΔΫ^ = (1 089+y0-1435X0044+70-466) = -0-019+/0-514;

Vb =yO-985-0019+;0-514 = -0-019+/1-499 = 1-5^90-7°; 1-5 ^90-7°

'« = 0-373^36-8° = 4 0 2 - 5 3 - 9 ° = 2-365+A25;

/ , = l-089+y0-1435+2-365+y3-25 = 3·45+/3·39 = 4-83 ^44-5°; AV9 = (3-45+y3-39)xy0-47 =71-62-1-595;

Ϋ2 = -0·019+/1·499-1·595+7ΐ·62 = -1-614+73-119 = 3-51 ^117-4°. From these data we find:

^ = 3 - 5 l J l l 7 ^ = 0 > 2 8 5 " - 1 1 7 ' 4 O > 022 = 90-72-9 = 17-1°; a21 =90-117-4 = -27-4°.

Since, as a result, it is found that Y12 = Y2i, we can assume that the calculation is correct.

In conclusion we make another check, which will allow us not only to check correctness of the calculated admittances of the circuit, but also the absence of errors in the calculation of the normal conditions.

For this it is necessary to determine the output of the generating stations under the given conditions according to the formulae for their active power:

Λ(ο) = ^ΐι^ιι sin"li+EdiZd&u sin (<512(o)-a12) = 1·6852χ0·626 sin 6-9°-f 1-685X 1-765x0-285 sin (12·4°+27·4°)

= 0-757; ^2(0) = ^3^22 sina22-£^2>>12 sin (<512(0)+a12)

= l-7652x 1-375 sin 17-1°-l-765x 1-685x0-285 sin (12·4°-27·4°) = 1-488.


The calculated results agree well with the power values obtained earlier from the condition of balance at the junction points.

Problem 2.15

A generating station G-l is connected to constant-voltage busbars through an external impedence Zext = 0·125+/Ό·3 (Fig. 2.46). Power S0 = 0-8 +/0-4 is transmitted into the system from the generating station, and a voltage V = 1 is maintained on the busbars of the receiving substation. The reactances of the generating station, expressed in relative units with the same basic conditions, under which the impedance Z ^ and power S0 were calculated, are: xd = 1-5; x'd = 0-3.

G-t 0-l25*j0-3 V s |

Θ EST)- \ s0=o-e*jo-4

Fio. 2.46.

Required: (1) to determine the limits of the transmitted power and the static stability margin with the generating station operating without an automatic excitation control system and with an inertialess proportional-type of regulator; (2) to find the effect of the power factor on the static stability margin.

Ed JI-5 0-125+J0-3 Yel

Θ — ^ ËE]—f ► ■

(a) S0*0-8+j(K

^ jO-3 <H25*jO-3 V=l

θ—^—-{SEI—} (b) S 0 = 0-8*j<K

F I G . 2 .47.

Solution. (1) With the generating station operating without an automatic excitation control system, this station is represented in the equivalent circuit of this system by its synchronous reactance. The equivalent circuit appears as shown in Fig. 2.47(a), where:

Zc =jxd+Zext = 7Ί·5+0·125+./Ό·3 = 0-125+/1-8.


The self and mutual admittances of this circuit, which is composed of elements connected in series, are equal to each other:

therefore, a u =a1 2 = 90-86 =4°

The synchronous e.m.f. of the generating station is found to be:

Λ ,0-8x0-125 +0·4χ1·8 .0·8χ 1-8-0-125x0-4 Ed = 1 + - +7 j

= 2-29^37-3°. The power limit of the system under consideration is determined as:

^max(i) = Eton sin a n + 3 , ^ 2 = 2-292x 0-555 sin 4° + + 2-29x1x0-555 = 1-47.

As the loss of active power in the circuit amounts to : 0-82+0-42

l2 ΔΡ = -—-2- xO-125 = 0-1,

the power of the generating station in the given conditions is : ^(o) = Ρ*+ΔΡ = 0-8+0-1 = 0-9,

and the static stability margin is: u _ ^max(i)-^g(o) _ 1-47-0-9

When the excitation of the generators is controlled with the aid of inertialess regulators of a proportional type, for approximate calculations, the generating stations may be represented in the equivalent circuits by their transient reactances. In this case (Fig. 2.47(b))

Zc = jx'd+Zext =yO-3+0-125+yO-3 = 0-125+yO-6. Therefore,

Γ " = ^ = 0 - Π 2 5 ^ 6 = 1 ' 6 3 " - 7 8 · 3 ° ;

a u =Xl2 =90-78-3 = 11-7°;

_, , 0-8x0-125+0-4x0-6 .0-8x0-6-0-4x0-125 E = 1 + ί +j j

= 1-41^17-8°;


Λη.χ<2> = 1·412χ1·63 sin 11·7° + 1·41χ lxl-63 =2-96; _ 2-96-0-9 _

ί(2) ~ (Η)

(2) When the power factor varies, the static stability margin depends on the value of the generating station e.m.f. Thus, when transmitting reactive power Q0 = 0-6 into the system, for which:

COSÇ) = 0-8

JW+o-e2) = 0-8,

then

, 0·8χ0·125+0·6χ1·8 , .0·8χ1·8-0·6χ0·125 Ed = 1 + -. +j :

= 2-57^32°;

Pm„ = 2·572χ0·555 sin 4°+2-57x 1x0-555 = 1-681.

^Ath quad

K

1-2

1-0

0-8

0-6

0-2 rant

*«

(Is qua dran )cos *J 0-8 0-9 10 09 0-8

FIG. 2.48.

07 0-6

In this case:

ΔΡ = 0·82+0·62

l2 x 0-125 =0-125;

Pg(0) = 0-8+0-125 =0-925;

1-681-0-925 n o l _ ke = 7ΓΚ^ = 0-817. e 0-925

The results of similar calculations allow us to plot the characteristics, Fig. 2.48.


Problem 2.16

An electrical system, of which Fig. 2.49 is the equivalent circuit and the values of parameters are given on the circuit, has the loads shown.

Pe«30MW

V0«l20kV 0 < · |

Ph = 60MW

FIG. 2.49.

Required: to determine the distribution of voltages and currents.

Solution. Measurements on a design model enabled us to determine the coefficients of distribution, Cij9 of the load currents.

VALUES OF THE COEFFICIENTS Cik

1 ^ \ ^ ^

a

b

I

0-55

0-33

2

- 0 - 4 5

0-33

3

0-45

0-67

Figure 2.49 shows the assumed positive directions of the currents in the branches of the equivalent circuit.

As a first approximation it is assumed that the voltages at all the points of the electrical circuit under consideration are equal :

Va=Vb=VQ = 120 kV.

The calculation is carried out in accordance with the following formula:

Vj = Vo-t^(t Zkcikcjk


The first approximation gives: 30 V'a = 1 2 0 - ^ (40 xO-552+20 xO-452+30x0-45»)-

an _^(4Οχ0·55χ0·33-20χ0·45χ0·33+30χΟ·45χ0·67)

= 1 2 0 - ^ x 1 2 - 2 3 - ^ x 1 3 - 3 4 = 120-12-2 = 107-8 kV;

30 120' V'b = 1 2 0 — ^ (40x0-55 X 0-33-20x0-45 X 0-33+ 30x0-45 X 0-67)-

—^L(40x0-332+20x0-332+30x0-672) = 120-

_ : ä X l 3 : 3 4 ~ S X 2 0 ' 0 0 = 120_13'3 = 106'7kV-After the second approximation we get:

V; = 1 2 0 — j | ^ - x l 2 - 2 3 - ^ y X 13-34 = 120-13-7 = 106-3 kV;

V'h' = 120—il^-x 1 3 - 3 4 — - ^ x 2000 = 120-150 = 1050 kV.

After the third approximation, which was carried out similarly, we have:

V'M" = 106-1 kV; Vi" = 104-8 kV.

The second approximation already gives a correct result. The current in the line L-l is determined in accordance with the fol

lowing formula, which gives the current in any branch k of the given circuit:

1 n §

where i is the number of the point with the given load : i = l , . . ., n;

cik is the coefficient of the load distribution current, applied at the point i, for the branch k of the circuit (which is assumed to be passive);

k = 1, . . ., τη·


The current in any other line is determined in a similar way. Substituting the numerical values we get:

/ = - L ( J L X 0-55 + — x 0-33 W = 199a. V3 \1061 104-8 )

The power at the transmitting end of the line L-1 is: p = V3/ÎK0 = 41-4 MW.

The power at any other point of the electrical circuit under consideration is determined in a similar way.

C H A P T E R 3

THE CHARACTERISTICS OF GENERATORS AND SYNCHRONOUS

CONDENSERS

THE characteristics of generators and synchronous condensers play a big part in calculating the transient phenomena in electrical power systems.

The problems encountered in calculating such characteristics include the calculation of the e.m.f. for the given parameters of the machine and a given flow of active and reactive power.

When solving various problems, it is often necessary to determine the flow of active and reactive power by replacing the generator by an e.m.f. (synchronous, transient or terminal) before a reactance. Examples in determining the corresponding operating quantities and the choice of characteristics are given in the present chapter.

The majority of the examples are given on the assumption that the iron saturation of the generators need not be taken into account. In more accurate calculations of the conditions, however, this assumption is found to be too basic and therefore some examples of how the saturation effects on the operation of a hydro or turbo-generator can be taken into account are given in the closing part of this chapter.

Problem 3.1

A hydrostation is connected to an infinite power system by a transmission line of 110 kV. The voltage at the substations óf the receiving system is kept constant (K=const)

Ed Xg Ï 9 X«t E q

d Xg I * *«*t I

d o n

FIG. 3.1.

in all the conditions which are to be used. Figure 3.1 shows the equivalent circuit of the system.

85


The parameters of the system elements and its original condition in relative units are:

F = l ; = 0-65; ^ = 1 - 8 ; *- = ! · ! ; *; = 0·55; />0=0·53; Q0 = 0-4.

The basic values are taken to be Sbasic = 300 MVA; Kbasic = 115 kV.

Required: to form the functions P = f(S) and Q = ψ(δ) in order to determine the permissible loads on the generators of the system and to obtain the characteristic of the system in the following cases:

(a) with a constant excitation current. (b) with a constant resultant flux linkage (E'd=const);

FIG. 3.2.

(c) with constant voltage at the generator terminals. The calculation of the characteristics does not allow for the saturation

of the generator.

Generators and Synchronous Condensers 87

Solution. In order to form the power characteristics, we calculate the e.m.f.s of the machine; Eq, Ed, E'd, and the voltage on the busbars of the generator Vr In addition it is expedient to make use of the relations obtained from the vector diagram of the salient pole machine (Fig. 3.2).

Neglecting the active resistance in the system, the e.m.f. Eq is determined by the expression

£ « = Ί{^-Ψ-)'^-Ψ-)\ <"> where xqE = xq+xext = 1-1+0-65 = 1-75;

From (3.1) we determine the angle Ò:

δ = arc tan /$<* ; (3.2) u +%l0xqS

A , 0-53x1-75 _ . 0 0 = a r C t a n r T Ö ^ T 7 5 = 2 8 · 6 ·

We determine the e.m.f. E' and δ' from (3.1) and (3.2), using x^r instead

x'ds = x'd+Xtxt = 0-55+0-65 = 1-2;

r -j[(F + a^) , +W1i

E' = V[(l +0-4X l-2)2+(0-53x 1-2)2] = 1-62; δ' = 23-3°. From the vector diagram, Fig. 3.2, we determine:

E'd = E'co%{ò-òr) = 1-62 cos (28-6°-23-3°) = 1-62x0-996 = 1-61.

Using x^ in the expressions (3.1) and (3.2) instead of xq£, we determine Vg and bc;

Vt = 1-3^15-3°; Vgd = F,cos(o-óc) = 1-3 cos (28-6°-15-3°) = 1-265.

We determine the e.m.f. Ed at zero load.

Ed = Eq?*-ii. -Ε'ά^>-Ξ± ; (3.3) Xq~Xd Xq~~Xd

_. , 0 . 1-8-0-55 1-8 — 11 E" - 1 9 3 1-1-0-55 - 1 ' 6 1 TT^sT = 2 3 3 ·

We now find the power characteristics of the hydrogenerators.


(a) With constant excitation current (Ie = const), which corresponds to the constant open circuit e.m.f. Ed, we have:

PEi = ^ s i n ó + ^ X ^ ^ s i n 2 ò ; (3.4)

*dL = *</+*ext = 1-8+0-65 = 2-45; 2-33x1 . k l2 1-8 — 1-1 . „.

^ = - T 4 r S m 3 + T X 2 - 4 5 x l - 7 5 S m 2 3 ;

pEi = 0-95 sin ò+0-082 sin 2<5. (3.4a)

In the expression for the power characteristic of the salient pole machine, there are two components.

The second component, which is a function of a double angle, is sometimes called the reluctance power. With a large external reactance, the reluctance power of the machine may prove to be considerably smaller than the main (the first component) power. Its calculation in such cases may be unnecessary. Hence, in approximate calculations of static stability the salient pole machines are replaced by non-salient pole machines with synchronous reactance xd and e.m.f. Ed.

The error from such an approach does not exceed several per cent, as can be seen from Fig. 3.3.

Defining more accurately, by allowing for the saliant pole effect, it is possible to present the hydro-generators in the form of non-salient pole machines, but having an imaginary e.m.f. (behind the reactance xq)9 which changes from one condition to another.

The characteristics of the active and reactive powers of the equivalent generators will then apply to the hydrogenerator. In this case, we find:

E V PE = d i s i n o ; 9 X „

XqL with δ = 28-6° and Eq = 1-93,

PEq = P0 = 0-53.

We determine the reactive power of the generator with Ed = const from:

QEi=^cosô-Çx^±^E + Çx^^coS2ô; (3.5) χάΣ *■ XdEXqE Δ XdZXqZ

2-33x1 $ l2 2-45 + 1-75 l2 1-8-1-1 . . QE<=-Ï4TCOSÔ-2X 2-45X1-75 + Τ Χ 2 · 4 5 Χ 1-75 COS 2 Ó ;

QEd = 0-95 cos <5-0-49+0-082 cos 20. (3.5a)


Relation QEi = Q(S) is given in Fig. 3.4. The power characteristics contain three components : a constant, which

does not depend on the angle à (1), the fundamental component (2) and the double frequency component (3).

1-0 r 08 l· 0-6

0-4

0-2

0

-0-2

-0-4

-0-6

-0-8

- 1 0

-1-2

-1-4

Γ (J

h

6

k

l·

h

Ed

vA \9oe

o Y — V —

°ΕΛ

,°· 3 ^

180e

2

V I

FIG. 3.4.

(b) With the constant resultant flux linkage (fi'd = const), we find the active power:

PE- = =£- sin ò~x?*-=*- sin 2Ó;

Ps> = 1-61x1

1-2 . χ l2 1-1-0-55 . -x

sino-—-————- sin 2(5; 2 1-75 X 1-2

(3-6)

/V = !*34 s i n Λ—0-131 sin 2<5.

The power characteristic is given in Fig. 3.5. This also has two components: the fundamental and the double frequency component. But in this case, the maximum point of the power characteristic P . is at an angle beyond 90°.

The value of the maximum power is greater here than it was with the excitation at constant current.

90 Transient Phenomena in Power Systems: Problems

For an approximate calculations it may be assumed, that E'V

P„> ^^4-sino; Ei κάΣ

i.e. the second harmonic is ignored.

K r

1-2

10

0 *

0-6 po

0-4

0-2

n

P

-

- i

^ v ~

/ / ύ

F I G .

*

ΊΗΪ·

3.5.

^ P Q

" ^ 4 6 ιβο·

1-4

1-2

10

0-8

0-6

0-4

0-2

0

-0-2

-0-4

-0-6

-0·β

- 1 0

- ! · 2

- |·4

- Ι · 6

- Ι · β

-2-0

Fio. 3.6.

The reactive power of the generators:

xqLxdE * XdL 2 X x.<x'„ 2 X " "' C 0 S 2 0 ' 1-61x1 . 1« QF- - —m— cos δ-— X

XqLXdS

1-75 + 1-2 l2 1-1-0-55 . l -75xl-2~ 2"X 1 - 7 5 x 1 - 2 C O S Z '

(3.7)

1-2 """ " 2 Q„. = 1-34 cos d-0-7-0-131 cos 20.

The reactive power characteristic is given in Fig. 3.6.


(c) The characteristics of the machine with constant voltage at the generator terminal is determined by the expressions (3.6) and (3.7) changing from xdr to x^ and Ed to Vgd.

-08 «-

- 2 0

-2-8

FIG. 3.7. FIG. 3.8.

After substitution we get: PVgd = 1-95 sin Ó-0-483 sin 2δ;

Qvgd = 1-95 cos <5-1-05-0-483 cos 2<5.


The power characteristics are given in Fig. 3.7 and 3.8. As would be expected, the power characteristic Pv has a maximum

at a larger angle than the characteristics Ed = const or E'd = const. The maximum power value in this case is also increased.

The voltage on the generators in comparison with the previous cases is held constant at all values of angle d. Naturally, this brings about a change in the flow of the reactive power, which is evident by comparing the corresponding curves which are shown in Figs. 3.4, 3.6 and 3.8.

Problem 3.2

A synchronous machine is connected to busbars of fixed voltage V through xc.

FIG. 3.9.

Required: to find currents Id, Iqy the active and reactive powers at different points of the equivalent circuit and the angles which correspond to the maximum internal power, which are to be expressed by Edy Ed, Exd> Vgd.

Non-salient pole salient pole FIG. 3.10.


Solution. In accordance with the equivalent circuit, which is shown in Fig. 3.9, and the vector diagram (Fig. 3.10),t we find the expression for the separate operating characteristics in a general form, summarizing the results in Table 3.1, and after that we make the numerical calculations.

We calculate the actual case in accordance with Table 3.1, using the following parameters:

xa=l-78; * ,= 1·06; x'd=Q-455; J x = 0-2; xc = 0-8; Κ=1·0; Ρ=0·9; Ôz, = 0-3;

xrfr = 2-58; χ,£=1·86; χ'„Σ = 1-255; Jx £ =1.0 .

NON-SALIENT POLE

/ = o-9-yo-3 1·0+(0·9_/0·3)χ./2·58 1-773 +/2-32 =2-92^ 52-65°; 1-0+(0-9-/0-3)xy 1-225 l-376+yi-13 = 1-78^39-4°; 1-780 cos (52-65-39-4) 1-780 cos 13-25° = 1-733; i-o+(o-9-yo-3)xyi-o

= 1-3+/0-9 = 1-582^34-7°; E* = 1-582 cos (52-65-34-7)

= 1-582 cos 17-95° = 1-505; Vt = 1 -0+(0-9 -/Ό-3) xyO 8

= l-24+yO-72 = l-433^30-15°; = 1-433 cos (52-65-30-15) = 1-433 cos 22-5° = 1-325;

Ed =

E' =

E'ä

Er =

gd

h = Ed-Vcosò

xdL

2-92-10 cos 52-65° 2-58 = 0-896;

. E'd—cos ò *d — —,

xdZ

1-733-0-607 1-255 = 0-896.

SALIENT POLE

/ = 0·9-/Ό·3 Eq = 1 0+(0-9 -j0-3)j 1-86

= l-557+7l-673=2-29^47-l° £" = 1 -0+(0-9 -y'0-3)xy 1-225

= 1-376 +j 1-13 = 1-78^39-4° E'd= 1-780 cos (47-1-39-4)

= 1-780 cos 7-7° = 1-762; Ex= l-0+(0-9-yO-3)xyi-0

= 1-3+./Ό-9 = 1-582^34-7°; Exd= 1-582 cos (47-1 -34-7)

= 1-582 cos 12-4° = 1-545; V, = 1·0+(0·9-/Ό·3)χ./Ό·8

= 1·24+./Ό·72 = 1·433^30·15° Vgd = 1-433 cos (47-1-30-15)

= 1-433 cos 16-95° = 1-37; Ea — V cos à

XqS

2-29-1-0 cos 47-Γ 1-86 = 0-865;

Ed = Eq+Id{xd-xq) = 2-29+0-865x0-72 = 2-913;

Ed-Vcosò h =

xdZ

2-913-10 cos 47-lc

2-58 = 0-865.

t Here the generator current is multiplied by / 3 .

TABL

E 3.

1. T

HE

OPE

RATI

NG Q

UANT

ITIE

S O

F A

SYN

CHRO

NOUS

MAC

HINE

IN

A V

ERY

SIM

PLE

SYST

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Generators and Synchronous Condensers 9S


NON-SALIENT POLE SALIENT POLE

. Exd-VcosÒ E'd-Vcosô d ~ ~Ai d ~ 7

''■»r χάΣ 1-505-0-607 non, 1-762-0-68

10 = 0-896; = 1-255 = ° · 8 6 5 '

T _ Vgd— F cosò _ Exd— Kcos δ Id 7C

Id - 1 χ ~ Σ — = 1^25^607 = 0 . 8 9 6 . = 1-545-0-68 = 0 . 8 6 5 .

K ! i n A = ^ = 0 . 3 0 8 Κ , , -Fcos* 9 χάΣ 2-58 " xc

_ 1-37-0-68 _ = ^— = 0-865, K s m ô = 0 ^ 3 2 =

« *,* 186

P DERIVED FROM Ed

2-92 2-913 p = τά x °'795 = °'90· p = ^ i8~ x ° ' 7 3 2 +

+ 2 Χ 2 0 5 8 7 2 Χ 1 · 8 6 Χ 0 · 9 9 7 = 0 · 9 0 ·

i> DERIVED FROM E'd

i > = ^ X 0 - 7 9 5 - 1-762X0-732 _ r 1-255 X U / y D ^ 1-255

1-325x0-965 = Q9Q _ 0-605x0-997 = Q9Q 2x2-58x1-255 2x1-86x1-255

P DERIVED FROM V& 1-325x0-795 1-37x0-732

0-8 0-8 1 - 7 8 X ° · 9 6 5 =0-90. - . 1 0 f X r 9 7 =0-903. 2x2-58x0-8 ' 2x1-86x0-8

P DERIVED FROM E xd 1-505x0-795 „ 1-545x0-732

p = vo p= FÖ _ 1-58X0-965 = _ 0-86x0-997 2x2-57x1-0 2x1-86x1-0


Qg DERIVED FROM Ed


_2·922χ0·8 2·9132χ0·8 Qg - 2-582 üg 2·582 +

-^W6(1-ré) ( 1 + 0 ' 0 7 3 ) = = 1020.

Ô , DERIVED FROM E'd

_ 1-733» x 0-8 n _ 1·7622χ0·8 ^* ~ 1-2552 * ~ 1-2552 +

- H l i ' - i r o ) « 1 - « ^ - - ^ ( ' - r a î ) · ' - 0 · 0 7 3 ) -

516^ 2-58^ (l+°-264> - m ^ i 1 - ^ ) ^ ^ · 0 7 3 ) = 1-022. = 1-02.

Qg DERIVED FROM Vgd

1-3252 1-325x0-607 Λ 1-372 1-37x0-68 Ö* — n.s n.8 δ* ~~ 0-8 0-8 ** 0-8 0-8

5-4 (' ~ïS)(l + 0 · 2 6 4 ) - 2XT86 ( l - m)(1 +°-°7 3 )

= 1-021. = 1-019. ANGLE 5max

(The angle corresponding to the maximum point of the power characteristic.)

DERIVED FROM Ed

3=90°. 0=81-9°.

DERIVED FROM E'd

0 = 104-65°. 6=99-6°.



DERIVED FROM E Δχ

0 = 108-7°

> = 113·3°. DERIVED FROM Vt gd

ó=io5-r

0 = 109°.

Problem 3.3

A synchronous condenser is connected to busbars of fixed voltage, V, through a pure reactance.

Required: to determine expressions for the current and the reactive power of the condenser derived from Ed, Kc k , Ed and Exd.

Solution. We assume that the circuit of the condenser does not contain an active resistance. In addition we shall have a direct-axis current component only, which is independent of whether we consider a salient pole or a non-salient pole machine; the expressions for Id and Qak will be equal.

By using the expressions obtained previously (Table 3.1) and the vector diagrams (Fig. 3.2 and 3.10), we easily get the expressions for 7rft and Qck9 given in Table 3.2, from which expressions for current and power may be determined, by substituting correspondingly in terms of Ed, Ed

and Vgd.

TABLE 3.2. OPERATING QUANTITIES OF A SYNCHRONOUS CONDENSER, EXPRESSED IN TERMS OF THE E.M.F. E^.

■ c * x

<&-\—Φ 1

Component current Id

Reactive power Qc k

ΔχΣ

ΔΧΣ ΔΧΣ ΔΧΣ O-'Ä)· ΔχΣ ' ΔχΣ)

t Here as in Problem 3.2 the current is multiplied by γ3.


Problem 3.4

A non-salient pole generator operates as a synchronous condenser, supplying reactive power through an impedance Z = r+jxc to busbars of fixed voltage F (Fig. 3.11).

CK Xc (g) ny\—

V

-P*Ì°C.K(U

F I G . 3.11

1

f I V

I

Required: to determine the angle of displacement of the condenser e.m.f. Ed in relation to the voltage V, the output from the condenser to the busbars, and the critical angle <5cr.

Solution. The active power of the condenser is zero. Therefore,

Pc k = ^ s i n a + ^ s i n ( 5 - a ) = 0 , zdE zdE

from which we find:

sin (Ó—a) = Edr or δ = arc sin

Ô c . k ( L ) = ^ : - c o s ( o + a ) - l - ^ .

(-fch V*x.

ύάΣ

The critical angle is found from the condition

dp dò Ed=const ZdE

cos (<5CT-a) = 0,

from which

âcr = y + « ·


Problem 3.5

A generator is connected through an impedance Z = r+jxc to busbars of fixed voltage (Fig. 3.12).

FIG. 3.12.

Required: to obtain expressions for the active and reactive power, given out to the busbars V, and also the direct and quadrature-axis components of current (Id, I^\ in the two cases:

(a) the generator is of a non-salient pole type and, (b) the generator has salient poles.

Solution. The solution is given in a general form, using the vector diagram of the salient pole generator (Fig. 3.13). The results are given in Table 3.3.

Problem 3.6

In the conditions of problem 3.5, the system has the following parameters:

xd =7*1-78; xq = ;1·06; x'd = 70-455;

Ax =;0·2; xc =y0-8; xdE =72-58;

xqL = 7*1-86; xdE =7*1-255; ΔχΣ =7*10;

r = 0-2.

The operating quantities are the same as in Problem 3.2.

Required: to find the values of the e.m.f.s Ed, Ed, Exd> and the voltages Vg and Vgd, and also the components of current Id, IqJ

t Here, as in Problem 3.2, the current is multiplied by γ3.


Solution. We find the e.m.f. for non-salient pole and salient pole machines. For clarity we give the solutions side by side.

' d * «

. « IMAXT

IH ( Χ,Γ *β>

Inr^ v94/

v \ IqXql

FIG. 3.13.

Ix'

ix,r

dr ιάχ,

1*.

NON-SALIENT POLE

/ = 0-9-J0-3 = 0-948 ^ -18-4°; Èd = V+t{r+jxds) = 1-0 +

+(0-9-y0-3)(0-2+j2-58) = 1-954 + +/2-262 =2-99^49-1°;

= 1-89^34-55°; Ed = 1-89 cos (49-1 -34-55) = 1-83;

Ex = l-0+(0-9-y0-3) (0-2+/Ί-0) = 1-48+/0-84 = 1-705^29-55°;

£^=1-705008(49-1-29-55) = 1-61; Vg = 1·0+(0·9-/0·3)(0·2+/Ό·8)

= 1-42+./Ό-66 = 1-57 ^24-9°; Vgd = 1-57 cos 24-2° = 1-43.

SALIENT POLE

/ = 0·9-;Ό·3 = 0-948 ^ -18-4°; 4 = V+î(r+jxdZ) = 1-0 +

+(0-9-y0-3)(0-2+yi-86) = 1-74 + -h/1-61 = 2-38^42-7°;

È = V+î{r+jxdL) = 1-89^34-55°;

Ed = l-89cos(42-7-34-55) = l-871; Ex = 1-705^29-55°;

£:^=l-705cos(42-7-29-55) = l-662; Vg = 1-57^24-9°;

Vgd = 1 -57 cos (42-7 - 24-9) = 1 -492 ; Id = 0-948 sin 61-1° = 0-831;

Ed = 2-38+0-831 x 0-72 = 2-98.


TABLE 3.3. CONDITION OF THE SYNCHRONOUS MACHINE, OPERATING ON BUSBARS OF FIXED VOLTAGE THROUGH AN IMPEDANCE Z = R+JXE WHEN EXPRESSED IN TERMS OF

THE E.M.E.£Ltf

Vfl X V

βΗ—Φ—CZ3—I

Current Id

Current / ,

Active power P

Reactive power

Conventional

symbolism

*d;*q

Non-salient pole

(Exd — V cos ά)χάΣ — V sin or

ΔζΣ

V sin δ ΔχΣ+(Exd - V cos ô)r

ΔζΣ

^ z ^ s M Ó - H O - r * - ^ -ΔζΣ ΔζΣ

V* χά-Δχ . _ χ d . sin 2δ

2 ΔζΣ

-^τ-ζάΣ cos ( ö + a j χ ΔζΣ 2

ΔζΣ 2 ΔζΣ

ζ*Σ = Υ(^+Χ*Σ); tan ad = - ί -

P;Q

Salient pole

(Exd — V co só )*^ — F sin òr

ΔζΣ

Ksin óz1jcr + ( £ X ( i - Kcos ô)r

Δζ\

ΖΙΖΓ zlzl V2 χ9-Δχ . ΛΛ x - i — — sin 2ó 2 zlzi

-*L_zy 2 :cos((5+ag) - l i — — s -ζϊζ^ 2 zlzl

-*——- cos 2(5 2 Jz£

*gr = ^ (r 2 +^îr) i tan aq = - ^

J z r = / ( r 2 + z l * r * , r )

We now proceed to determine the separate components of the current :


Id DERIVED FROM Ed

ζ4Σ = 0-22+2-582 = 6-71; ζάΣ = 0-22+2-58 X 1-86 = 4-84; cos ô = 0-655; sin ò = 0-756; cos δ = 0-735; sin δ = 0-678;

h h (2-99 - 0-655) X 2-58-0-756x0-2 (2·98-0·735)χ 1-86-0-678x0-2

6-71 = 0-878.

4-84 = 0-835.



Id DERIVED FROM E'd

z'd\ = 0-22 +1-255x2-58 = 3-28 z'd% - 0-22 +1-255 X 1-86 = 2-37 h h

(1-83-0-655) X 2-58-0-756 X 0-2 _ (1-871-0-735) X 1-86-0-678x0-2 3-28 ~ 2-37

= 0-878. = 0-834. Id DERIVED FROM Vgd

ζ\Σ = 0-22 +0-8x2-58 = 2-104 ζ\Σ = 0·22+0·8χ 1-86 = 1-53 h h

(1-43-0-655) X 2-58-0-756x0-2 (1-492-0-735) X 1-86-0-678x0-2 2-104

= 0-88. /,, DERIVED FROM Exd

1-53 = 0-833.

Azi = 0-2 + 1-0x2-58 = 2-62 Δζ\ = 0·22 + 1·0χ 1-86 = 1-90 h h

(1-61-0-655)X2-58-0-756x0-2 _ (1-66-0-735)X 1-86-0-678x0-2 2-62 ~ 1-90

= 0-883. = 0-883. Iq DERIVED FROM Ed

_ 0-756x2-58 _ 0-678x2-58 • ~ 6-71 + "~ 4-84 +

, (2-99-0-655) x 0-2 , (2-98-0-735) X 0-2 . . . . + ^ = 0-361. + — = 0-455.

Iq DERIVED FROM Ed

_ 0-756x1-255 _ 0-678x1-255 " 3-28 + " ~ 2-37 +

+ (1·83-0·655)χ0·2 = ^ + (1-871 -0-735) X 0-2 = 0 . 4 5 4 3-28 2-37

/ , DERIVED FROM Vgd

_ 0-756x0-8 _ 0-678x0-8 * 2-104 + "~ 1-53 +

(1-43-0-655) x 0-2 _ (1-492-0-735) X 0-2 + 2ÖÖ4 ° 3 6 1 · + Ï13 = °'456·


L DERIVED FROM E. xd

T 0-756x1-0 /„ = — ^ — +

+

q 2-62 (1-61 -0-655) X 0-2

2-62

_ 0-678x1-0

= 0-360. +

1-90 (1-66-0-735) X 0-2

1-90 = 0-454.

Problem 3.7

The system considered is the same as in Problems 3.5 and 3.6.

Required: to find expressions for the loss of the active and reactive power in impedances xc and r, through which the generator is connected to the busbars of fixed voltage. The expressions are to be obtained in terms of the e.m.f.s Ed,Edi Exd and voltage Vgd.

Solution· In order to obtain the solution, we make use of the results obtained earlier. In Table 3.4. we summarize the expressions for the loss in the active power.

TABLE 3.4. POWER LOSSES IN THE ACTIVE RESISTANCE r

Derived from Exd

Non-salient pole Salient pole

rzdEExd ΔζΣ

+ r{r2+âx\)V2

ΔζΣ

r(xìE -ΔχΣ)ν2 cos2 δ _ 1_

ΔζΣ

^(χάΣ-ΔχΣ)ν2ύη2δ ΔζΣ

2rzlLExdVcos δ ΔζΣ

2r2(xdS — ΔχΣ)Εχάνsin δ ΔζΣ

rz^Exd ΔζΣ

, Kr'+zixDr», H 7 l·-

ΔζΣ r(x$E-AxE)V2cos2o

1 ΔζΣ

r2(xqi:-AxE)V2sm2ô ΔζΣ

2rzrfExd V cos δ ΔζΣ

2r2(xqE — ΔχΣ)Εχά V sin ι ΔζΣ

Generators and Synchronous Condensers

Conventional symbols

105

ΑζΣ = Υ&2+ΑΧΣΧ*Σ)

ζ9Σ = Y(r2+x2rt)

ΑζΣ = Ϋ^*+ΑΧΣΧ9Σ)

We summarize in Table 3.5 the expressions for the loss of reactive power.

TABLE 3.5 Loss OF POWER AQ IN THE REACTIVE RESISTANCE

Derived from Exd

Non-salient pole

Χ,ΖΙΣΕΙ*

Azi , Xc(r* + Axl)V* |

Azi

, χΛΧαΣ~ΑχΣ)ν2 sin 2(5

2*czJr£'z<i Kcos<3

ΊχΑΧάΣ — ΔχΣ)ΕχΛνύΆ δ Azi

Salient pole

Χ^ΣΕχά Azi

ì xe(r* + Axi)V2 l

| xix^-Axj^cos^a Azi

χΛχςΣ — ΑχΣ)ν2 sin là Azi

Ix^E^V cos δ Azi

_ 2*/(χ,Γ — r )g z d Ks in 3

Conventional symbols

zdZ = V(r*+xîE)

ΑζΣ = Ϋγ* + ΑχΣχαΣ)

Ζ9Σ = Υ('2 + Χ2<Σ)

ΑζΣ = / (r 2 + zf*r*,r)


Problem 3.8

The system considered is the same as in Problems 3.5, 3.6 and 3.7.

Required: to determine the numerical values of the active power losses from the expressions derived from Ed and Éd, for non-salient pole and salient pole machines.

Solution. In accordance with the general expressions obtained in Problem 3.7, we find the losses, ΔΡ, for non-salient and salient pole machines.

For clarity we carry out the calculations side by side.

FOR NON-SALIENT POLE MACHINE FOR SALIENT POLE MACHINE

ΔΡ DERIVED FROMiSj

A» 0-2 ^ΛΛ, 0-2 0-4 0·2χ3·5χ2·982 Δρ = 7 ^ Τ Χ 2 · 9 9 2 + Γ ^ + 7 ^ Χ T^TÔ = 0-265; 6-71 ' 6-71 * 6-71 4·842

χ 2-99x0-655 =0-182. 0-2x6-71 = 0-057;

4·842

0-2(1-862-2-582)χ0·7352

4·842

= -0-015; 0-04x0-72x0-997 ΠΛΛ1

Φ842" ~ -0-001;

0-4 χ 3-5x2-98x0-735 — 4·842 —

0-08 X 0-72 X 2-98x0-678

ΔΡ DERIVED FROM E'd

4·842

ΔΡ = 0-181.

-0-130;

= 0-005

0-2x6-71 Χΐ-832 . . . . 0·2χ3·5χ1·872

= 0-417; ™ ^ = 0-436; 3·282 ' 2·372

0-2x1-62 . . „ , 0-2x1-62 —ψ2&~ = ° - 0 3 0 1 ;

2.372 = ° · ° 5 7 8 ; 0-2 χ 5-07 χΟ-6552 . Α ,Α, 0-2 χ 1-89 χΟ-7352

Λ Α „ . , 3 ^ = ° · 0 4 0 4 ; 2Ö7* = ° · ° 3 6 4 ;


0-04x 1-33x0-99 = _ 0-04 x 0-605 x 0-997 =

3-282 ' 2-372

0-4x6-71 X 1-83x0-655. 0-4 X 3-5 x 1-87x0-735 3-282 ' 2 372 = -0 -342;

= -0-298 0-08 X 0-605x1-87x0-678 0-08 x 1-33x1-83x0-756 ψγβ—

= -0-0109; 3-282

= -0-0136; zip = 0-182. ΔΡ = 0-4924-0-3116 = 0-181.

Problem 3.9

The same system is considered as in Problems 3.5, 3.6 and 3.7.

Required: to determine numerical values of the reactive power loss in reactance xe from the expressions derived from Ed, for salient pole and non-salient pole machines.

Solution. In accordance with the general expressions, which were obtained in Problem 3.7, we find the losses AQ.

FOR NON-SALIENT POLE MACHINE FOR SALIENT POLE MACHINE

°£x2-99 2 = 1-066; ° ^ x2-982 = 1-06;

°'8 -0-119- 0-8X6-71 _ Q 2 2 o . 6 T - 0 1 1 9 ' —£w 0 2 2 8 ' 2x0-8x2-99x0-655 . .._ 0-8X3-21 xO-7352

nncn frTi = - ° ' 4 6 7 Φ8Ρ = - ° · 0 5 9 ' AQ = 0-718. 0-8 X 0-2x0-72x0-997

4-842

= -0-0049; 1 -6 X 3-5x2-98x0-735

4-842

= -0-523; 1 -6 x 0-2 X 0-72 x 2-98 x 0-678

4-842

= 0-0198; AQ = 0-721.


Problem 3.10t

The calculations are to be carried out for a turbo-generator and hydro-generator both neglecting and allowing for saturation.

Neglecting saturation. (a) Turbogenerator

Nominal values of voltage and of the direct-axis synchronous and transient reactances, as well as the values of the active and apparent power, are given. It is required to determine Ed,Ed, <5, Ε', δ'.

V = 10-5 kV; xd = 150 per cent; xd = 18 per cent;

i>nom = 24 MW; S n o m = 3 0 M V A .

Solution. 10·52Χ150 __, u

Xd= 30X100 = 5 ' 5 1 o h m s >

, 10·52χ18 η _ .

^ = ^ ü ö r = ° - 6 6 o h m ;

Ônom=V302-24* = 18MVA;

Similarly, replacing xd by xd9 we find:

^[('0·-^Η^)Ί=η·73:

t anó ' =0-130 i.e. δ' = 7°23';

E'd =E'cos(Ò-ò'); E'd = 11-73 cos (32°10'-7°23') = 11-73 cos 24°47' = 11-73x0-91 = 10-67.

t Taken from ref. 6.


(b) Hydrogenerator

It is required to determine Ed,E'd, δ,Ε', δ'; given: V = 13-8 kV; xd = 80 per cent; xq = 50 per cent, x'd = 30 per cent; Pnom = 54,000 kW; Snom = 60,000 kVA; Qnom = 26150 kVA;

, 13·82χ30 η ο . . *'=-60x100" = 0 9 5 o h m '

13·82Χ80 . . . . ^ = -6Ö3üöo- = 2-5 4 o h m s ;

13-8X50 ' χ ·"ίΟχΤΟΟ-1·5 9 ο 1 , , ω·

As in the calculations for the turbogenerator, replacing xrfby xg, we find:

tan ό=0·370; ò = 26°15'; / Γ / „ 0 26-15x0-95 \ 2 / 54x0-95 \2~1

* = 7[(13 '8+ 13-8 j + ( - l 3¥ - J J = 16'°3;

·, we:

Ed

Ed = find:

= *«

tan Ò '=0-238 -- 16-03 cos 6°55'

Xd-x'd xq-x'd

=

~Ed—_ Λ9

; δ' = 13°20'; = 16-03x0-993 =

—2- = 1 7 - 9 2 — -■x'd 0-64 44-52-23-62 = 20-90.

= 15-9.

•15-91 0-95 0-64

Allowing for saturation.

(a) Turbogenerator Nominal values for the turbogenerator are:

V = 10-5 kV; xs =15 per cent; 5nom = 30 MVA; P n o m = 2 4 M W ; ßn o m = 18MVA;

10·52χ 15 Xs = 3QX100 = 0-55 ohm; 7nom = 1-651 A, or 1-651 kA

The internal e.n.f. is determined by the expression :

Further,


Simplifying the problem, assuming r=0, we find:

/ΓΛ,τ* , 18χ0·55\2 /24χΟ·55\Π , , . , . , , E<=vL(lcr'5+_iöT-j +{-w5-) j = n ' 5 1 kv;

tan/3 24x0-55 10·52 + 18χ0·55

= 0-110 i.e. £ = 6°15'.

According to the open-circuit characteristic, we find 7rot 0 corresponding toEt = 11-51 kV. Let:

/rot0 = 120A.

Further, we determine Ev (i.e. Et on the short-circuit test) :

Ev = V3/nomx, = V3x 1-651 xO-55 = 1-57 kV, and find l'xoi from the open-circuit characteristic, corresponding to Ev = 1-57 kV (/r'ot = 30 A) and ITOtk or the short-circuit characteristic for:

L· = 4cm = 1651 A. Let/sx = 150 A. The factor referring the rotor current to the stator equivalent current,

expressed in kAx^/3, is determined in accordance with Fig. 3.14, where:

a = nSr* 0Γ a = ± ^# = 13.76xl0-3xV3=23-8xl0-3.

The rotor current is determined by the formulae:

/ r o t =VCC7^0 °°s /,"*"^τ "*" c^«0 **" "^^y i * Q P where Ir = —— ; /_ = —— and a = the factor referring the rotor cur-V 3 F V 3 K

rent to the stator equivalent current (see above).


In deriving this formula, the diagram in Fig. 3.15 is used. After the current 7rot0 has been determined, its vector is set at an angle of 90° to the

vector E{. To it, in parallel with the vector 7, is added the vector —, a

which takes into account the demagnetizing action of the stator reaction (in the longitudinal and transverse axes). Then the closing vector Jrot determines the rotor current. Projecting the vector components of 7rot0

and — on the abscissa and ordinate axes, we get, in order to determine

Ir/*

90e \

Y

H1/

\$7\ I I

xrot hoc/fcc

FIG. 3.15.

/rot, the formula given above. The angular displacement of the rotor Ò is determined as follows:

t a n ä =g ^ s i n ^ ° .

a^rotO COS ß+Ip

Substituting the values given, we find:

18xl03 \ 2

-J[( 120cos6°15' + 10-5x23-80

= 220 A;

tan Ò = 0-5706 i.e. δ = 29°45'

1),*("»^Α)'

(b) Hydrogenerator

Given nominal values: V = 13-8 kV; Pnom = 54 MW; Qnom = 26-15 MVAp; 5nom = 60 MVA; 7nom = 2514 A, or 2-514 kA; x = 1-59 ohms;


13-82X15 60X100

are as determined above: Eq= 17-92, δ=20°15';

xs = 15 per cent; xs = ^ <rk/x = 0-48 ohm; Eq and Ò for this case

( ^ - K c o s 3 ) x g - Vsin or _ 17-92-13-8 cos 20°15/

r2+jc2 1-59

= 3-132 (kAxV3). Further, we find:

Eid = 13-8 cos 20°15'+3-132x0-48 = 14-44 kV. In accordance with the open-circuit characteristic, the rotor current

corresponding to Eid = 14-44 kV is: Jrot0 = 180 A.

Further, we determine Ev: Ev = y/3Inomxs = V3x 2-514x0-48 = 2-087 kV,

and the corresponding current of the rotor /r'otJk = 40 A. From the short-circuit characteristic at a nominal current we determine ITOtk = 400 A. Hence

α = 2 ' 5

Α ^ Χ ^ 3 = 6-98 x IO"3 V3 = 12-08 x 10~3. 400—40 v

The full rotor-current is determined as follows :

CHAPTER 4

LOAD CHARACTERISTICS AND THEIR STABILITY

THE load of an electrical power system, containing synchronous or induction motors, may be the origin of the electrical power system's instability. In the examples, which are given in this chapter, the stability of induction motors and the stability of a complex load, containing induction motors, are considered.

It should be noticed that the load stability is not always determined by the voltage on its terminals. If the total capacity of the load motors is comparable with the total capacity of the generators of the combined system, then the voltage on the load busbars does not appear to be independently variable and for a simplified determination of the load stability criterion, it is necessary to use the generator's e.m.f. for the system, the value of which in the present system may be considered as independently variable.

Problem 4.1

For practical calculations of steady conditions, and of transient phenomena in electrical power systems, using equivalent circuits, the load is often represented simply as a constant impedance. In the system of Fig. 4.1(a), the busbars of voltage V have a load SL. The active part, PL, of this load remains constant, but the reactive part varies.

Required: (1) to represent the load SL in the form of a constant impedance ZL; (2) to find and to represent diagrammatically the variations of ZL in relation to the variation of the loading power factor cos tpL.

Calculations are to be carried out in relative units.

Solution. The load impedance is determined as: V2

ZL = — (cos <pL ±j sin (pL), (4.1) p

where SL = — — . cos <pL

113


Therefore, „ V2 cos wL , . . ZL = -—— (cos cpL ±j sin ψι).

08 06 0-4 0-2 0+cos<jpL

FIG. 4.1.

Assuming that

Sbasic = PL and Fbasic = V,

Vha< then 7 — basic · men ^basic — ~ ^basic

Z V2 S ZL. = ^ - = -p- X - τ ^ - cos <pL(cos <pL ±y sin <pj.

^basic *L Y basic Allowing for the accepted basic conditions :

ZL« = cos 9x,(cos yL±_jsin <pL) = cos q>Le±J9L,

i.e.

| ZL· | = cos tpL.; RL* = cos2 yL\ xL. = ±cos <pL sin yL

= ± | — sin 2^^

Load Characteristics and their Stability 115

The calculated results are given in Table 4.1 and are shown in Fig. 4.1(b).

TABLE 4.1

COS^j 2L* RT* XL*

1 1 1 0

±0-9 0-9 0-81

±0-392

±0-8 0-8 0-64

±0-48

±0-7 0-7 0-49

±0-50

±0-6 0-6 0-36

±0-48

±0-5 0-5 0-25 0-433

±0-4 0-4 0-16

±0-366

±0-3 0-3 0-09

±0-287

±0-2 0-2 0-04

±0-195

±0-1 0-1 0-01

±0-099

Problem 4.2

An induction motor of 875 kW is connected (Fig. 4.2(a)) to the busbars of a substation with an output of ~ 8000 kW with a voltage 6 kV. The parameters of the induction motor, as referred to the nominal voltage and output, are given in the equivalent circuit (Fig. 4.2(b)).

V=6kV

V

^ JL

R2=003 (b)

Fio. 4.2.

Required: (1) to construct a characteristic of reactive power for the motor with a change in voltage at the substation busbars, Q = /(K), and to determine the critical voltage at which the pull out of the motor takes place, assuming that the mechanical output of the motor remains constant; (2) to derive the dependence of the active power on the slip function P = f(s), with different values of the voltage on the busbars (V = 1-1; 1; 0-9; 0-7; KCT); (3) to determine the critical slip sa and the maximum active power Pmax at V = 1.

On the busbars of the works substation, an increase in voltage may take place (if, for example, with a small load, the condenser batteries were not switched off), as well as considerable decrease of voltage in emergency conditions.


In practice it is important to know how the induction motor behaves in these conditions, at what bxisbar voltages its pull out takes place, and how the reactive power from the system varies.

<M 02 03 W M M 07 0-6 0*9 1-0

FIG. 4.3.

Solution. The voltage on the substation busbars, which are of large capacity, may be taken as independent of the motor conditions. Therefore, in order to determine the active power taken by the motor from the circuit, the expression may be used:

n „ R2 V2R2s p = / 2 _ 2 = 2 (sxs)2+m

(4.2)

For the condition that P = PmeCh = 1» i* *s easy» fr°m the expression (4.2), to determine the current Is and the corresponding voltage V for a given slip s.

(a) In accordance with the values V and I* obtained, we find the reactive power used by the motor:

β=βμ + & =£+£*, · (4.3)


The calculated results are shown in Fig. 4.3. (b) The critical slip sa and the critical voltage VCT are determined in

accordance with the known (ref. 3, p. 83) relations:

R2 __ 0-03 xv "" 0-12 = 0-25;

V^ = V(2 /W = V(2x 1 X 0-12) = 0-489.

(4.4)

(4.5)

(c) The relations P =f(s) at different voltages V, are determined in the following way: considering the slip s in (4.2), we determine the values of the active power Pv

01 0-2 0-3 0-4 05 0-6 0-7 0-8 0-9 !-0

FIG. 4.4.

The values of power Pv with other voltages on the busbars and similar slips, are easily obtained by the simple calculation:

Ρν = Ρλν*,

where Pv = the value of the active power at any voltage V. Px = the value of the active power at V = 1.

The relations P = f(s\ obtained are plotted in Fig. 4.4. (d) The maximum value of active power at nominal voltage V = 1 is :

1 P = ^ = max 2x< 2x0-12 = 4-17.


Problem 4.3

An induction motor of260kW output, at a voltage of6kV, cos φ =0-85, is supplied from the busbars of a powerful substation.

V

x s sO-25

(a)

FIG. 4.5.

Required: to construct the characteristic showing the reactive power of the motor in dependence on the busbars' voltage.

<H 0-2 0-3 0-4 0-5 0-6 0-7 08 0-9 10

FIG. 4.6.

The equivalent circuit of the motor and its parameters are given in Fig. 4.5(a). The dependence of the mechanical power of the motor on the slip s is shown in Fig. 4.5(b); it is given, approximately, by the relation:

mech = 1-as, = 1-1-4*·. which is also given in Fig. 4.5(b).

(4.6)


In this expression, sm is taken in relation to the synchronous axis, which is rotating at constant speed, i.e.

do d0 Λ s* = -j- = -=- = ΩΑ * dt dt * (4.7)

Solution. Considering diflFerent values of mechanical power of the motor Pmtch in accordance with (4.6), we find the slip s = 0-714 (1-Pmech)·

In accordance with the expression (4.2), assuming that there are no losses in the motor, i.e. Pmech = P9 we determine the values of voltage V and current Is corresponding to the assumed power Pmoch and slip s.

07 r Q

FIG. 4.7.

Figure 4.6 shows the characteristics of the mechanical and electrical power of the motor in relation to the slip for two voltage values at the motor busbars: V = 1 and V = 0·525.

We calculate the reactive power Q used by the motor, from the expression (4.3) and display it in the diagram (Fig. 4.7.)


It is easy to see that the motor at any voltage value will be stable, since

— ^ ί > 0 always. This is linked with the usual characteristic as as

form Pmcch = <p(j).

Problem 4.4

Figure 4.8 shows the equivalent circuit and the parameters of an induction motor, which is connected to busbars, the voltage of which is slowly decreasing.

T V

i __2* _/^

] R^OOol

FIG. 4.8.

Required: to derive a relation giving the reactive power in terms of the voltage when the motor is operating and when it is at a standstill ; Pmech = const = 1.

Solution. We find the slip of the motor at variable voltage, by solving the equation (4.2) for s.

The calculated results are given in Table 4.2.

TABLE 4.2

V

s

LO

00525

0-98

0-9

0069

0-766

0-8

0091

0-571

0-7

0142

0-364

The critical value of the voltage is :

VCI =72x0-22x1 =0-662.

Load Characteristics and their Stability

The critical value of the slip is:

121

In order to obtain the values of the slip s at the corresponding values of the voltage V, which agree with the expression (4.3), we determine the reactive power taken by the motor.

For P = 1, the results of calculations are given in Table 4.3 and in Fig. 4.9.

5-2 rQ

4-8 h

4-4 h

4-0 h

3-6 h z'2.r 2 '8Γ 2-4 h

2-Ol·

l-6h

l-2h

0-8 h

0-4 h

Motor at standstill

L \t

Pull out ]/ process

Operation on stable f / \ operation on stab

e=0227N^Ï22Î* r i S t i C

FIG. 4.9.

TABLE 4.3

V

s

Q,

Q.

Q = β . + β μ

vo

0-0525 0-98

0-555

0-231

0-786

4-40

4-95

0-9

0069 0-766

0-450

0-310

0-760

3-30

3-75

Ö-S

0091 0-571

0-356

0-403

0-759

2-50

2-86

0-7

0-142 0-364

0-272

0-625

0-897

1-59

1-86

0-662

0-227

0-243

0-995

1-24


The reactive power taken at a standstill is determined for s = 1. The calculated results are given in Table 4.4 and displayed in Fig. 4.9.

V Qß

Q.(s = 1) β = β . + β μ

10 0-555 4-32 4-88

0-9 0-450 3-50 3-95

TABLE 4.4

0-8 0-356 2-76 312

0-7 0-272 211 2-38

0-6 0-200 1-560 1-76

0-5 0139 108 1-22

0-4 0089 0-69 0-779

Problem 4.5

An electrical power system is represented by an equivalent circuit, which contains one equivalent station and a complex load, comprising induction motors (60 per cent) and lighting load (40 per cent) (Fig. 4.10). With the nominal voltage on the load busbars, the active load power />L(0) = 0·9; the reactive power 0L(0) = 0·7. The reactance of the system is JC„ = 0-8.

Ee xc«o-e

* ^ ητ^ QC=QL«>4Q * P L

FIG. 4.10.

0-7 0-8 0-9 10 II

FIG. 4.11.

Required: to check the stability of the load in the indicated system by forming the functions Ee =f(V) and Qe =f(Ee) and using the criteria for complex load stability:

dEe dV

dße dE/

The static characteristics of the complex load PL, QL =f(V) are given in Table 4.5 and in Fig. 4.11.

v.

10

0-9


TABLE 4.5

P*

10

0-941

Ö*

10

0-885

v.

0-8

| 0-7

* ·

0-893

0-855

Ö*

0-844

0-88

123

Solution. For the equivalent circuit (Fig. 4.10) we write expressions for Ee a n d ß , :

or

*-7("^)·*(¥)' (4.8)

The reactive power output from the equivalent station is,

Qe = QL+4Q = QL+'^^XC, (4.9)

(4.10) where

PL=P.x 0-9 and QL = Qmx 0-7, corresponding to the active and reactive load powers at different busbar voltages.

By using Table 4.5 and expressions (4.8), (4.9) and (4.10), we can find Ee and Qe in relation to V.


TABLE 4.6

y Pl=P,xO-9 β* = β.Χ0·7

ew V+QBxJV Ptxjy

E. PÌ+QÌ

PÌ+QÌ X,

V Q. = QL+àQ

10 0-9 0-7 0-56 1-56 0-72 1-72 1-3

104

1-74

0-95 0-874 0-653 0-55 1-50 0-736 1-67 1188

105

1-703

0-90 0-848 0-62 0-551 1-451 0-755 1-634 1104

109

1-71

0-80 0-804 0-591 0-591 1-391 0-804 1-61 0-997

1-25

1-841

0-75 0-785 0-596 0-635 1-385 0-838 1-62 0-972

1-382

1-98

0-70 0-77 0-616 0-704 1-404 0-88 1-66 0-97

1-588

2-204


In Figs. 4.12 and 4.13 the relations^ =f(V) and Qe =/(£,,) are formed and the critical value EQ = 1-61 is found.

Problem 4.6

The load in a system is represented by an equivalent induction motor. Figure 4.14 gives the parameters and the equivalent circuit of the system, including the motor.

x =0-5 V Xs = 0 2 5

PQ I Q; 7?. P.

i ÌQ^ Rf00,M' ?P0 0I2A T

FIG. 4.14.

1·611-6

~=odU

1·5 V0·7 0·8 o·g 1·0

FIG. 4.12.

"6

1·70 EcO="72

FIG. 4.13.


Required: to examine the effect on the system, load stability of compensating the reactive load power with the aid of static condensers, by considering these three versions:

(a) compensation is absent (cos φ = 0-89); (b) compensation corresponds to cos <p = 0-95; (c) reactive load power is fully compensated (cos φ = 1).

It is assumed that an active power P = 1 is used by the equivalent motor with voltage V = 1 on its busbars. When solving the problem make use of the criteria dEJd V and dQJdEe.

Solution. We find the amount of slip s for different values of the voltage, V, on the motor busbars at constant power P = 1, by using the expression (4.2). Then we determine the reactive power Qs in accordance with the corresponding values of V and s and expression (4.3).


TABLE 4.7

V s,% Q.

10 1-34 0-268

0-9 1-71 0-344

0-8 2-40' 0-485

0-75 303 0-611

0-707 4-96 10

In Table 4.7, the voltage V = 0-707 and the slip s = 4-96 per cent, correspond to the critical values:

Vc = j2P0xs = V2X 1XO-25 = 0-707;

- = § = ^ — · The power is ζ)μ = ν*\Χμ. The full reactive power is Q = ο 5 + ο μ ·

We find the electromotive force of the equivalent generator, Ee in accordance to (4.8), and the power Qe in accordance with (4.9).

The calculated results for case (a) (without compensation) are given in Table 4.8.

We determine the power of the condenser bank, which is necessary to reduce the power factor, by compensation (case (b)) to 0-95. In this case,


TABLE 4.8

V βμ Q.

Q = Q^Q. àQ

Q. = Q+àQ E.

1 0 0-244 0-268 0-512 0-631 114 1-35

0-90 0197 0-344 0-541 0-799 1-34 1-32

0-80 0156 0-485 0-641 110 1-74 1-35

0-75 0137 0-611 0-748 1-39 2 1 4 1-41

0-707 0122 1 0 112 2-26 3-38 1-66

under normal condition (V = 1), the reactive loading power Q is:

Q = P tan arc cos 0-95 = 1 x 0-329 = 0-329.

Therefore, the power of the condenser battery (Fig. 4.15(a)) is:

Qc = Ο μ + δ , - Ô = 0-244+0-268-0-329 = 0-183,

and the reactance of the battery xc is :

V2 1 x = — = —— = 5-47 c Qc 0-183

We find the impedance of the equivalent shunt (Fig. 4.15(b)) to be:

J e JXp-JXc y4-l-y5-47 J

After the impedance of the equivalent shunt xe has been determined, we find Qe and Ee again.


TABLE 4.9

V Q„ = νηχ,

Q. Q = β χ , + Q .

àQ Qe = Q + AQ

Ee

1 0 0061 0-268 0-329 0-554 0-883 1-27

0-9 00495 0-344 0-393 0-712 1105 1-25

0-8 0039 0-485 0-524 0-994 1-52 1-29

0-75 00343 0-611 0-645 1-26 1-90 1-35

0-707 00305 1 0 103 2 0 6 3 09 1 61


With cos φ = 1 (the full compensation of reactive load capacity), (case (c)), the reactance of an equivalent shunt xe is (Fig. 4.15(b) and (c)):

X = - — 3-74 e 0-268 '

where xe corresponds to the capacitance.

v=i v=i v=i

3 (a)

Ö?0-329 1 Οβ=

(b)

Fio. 4.15.

0?68 "oîo

If= QS*(H6Ô

(c)

The calculated results for case (c) are given in Table 4.10.

TABLE 4.10

V Q~ Q.

Q = β , - ß * . àQ

Q. = Q + AQ E.

10 0-268 0-268 0 0-5 0-5 112

0-9 0-216 0-344 0128 0-627 0-755 112

0-8 0171 0-485 0-314 0-856 117 117

0-75 0150 0-611 0-461 108 1-54 1-25

0-707 0134 100 0-866 1-57 2-62 1-50

The required functions Ee =f(V) and Qe =f(Ee) are represented for all cases in Figs. 4.16 and 4.17.

Figure 4.16 also gives the function V =f{Ee). From Fig. 4.16, it can be seen that the critical value of the voltage is VCT = 0-9 instead of VCT = 0-707, which is obtained if the equivalent motor is connected to the busbars of a powerful system (x = 0).

However, in the conditions under consideration (comparable power of the generators and the load), the change in the voltage on the motor terminals, with its corresponding change of Ee does not characterize the stability and cannot produce the effect of static condensers on the stability. The effect of connecting the condensers for an improvement of the power factor of the induction motors on their stability, may be found by determ-

128

Y IO

08

06

04

0·2

0

Transient Phenomena

- 3

-2

-

- 1

cos<p = i

vc r=o* T

-

dQ c — = 00

d E * /v

M2fcosfsl 1 1 1

in Electrical Power Systems: Problems

cos<f= 0*5 coscf=089

J& Y

\ | cos<f .= 0-89

| \ 1 | cos <f:; 0-95 1 1

1 M·«, |Ι·32 , t , ^

1-1

I-7rE,

1-2 1-3 1-4

F I G . 4.16.

1-5 1-6 17

It,

Ecr=I·32c

dE~:odV

COS'f'= I

0·1 0'8

FIG 4.17.0·9 1·0


ining the load stability margins for the initial and critical values of the e.m.f. of the equivalent generator:

km*rg=Ee0ZEecTx 100 Per cent.

For the cases considered, we get :

1-35—1-32 (a) cos φ = 0-89; fcmarg = — — — x 100 = 2-22 per cent;

(b) cos φ = 0-95; Â:marg = ^ ^ ) ' 2 5 X 100 = 1-57 per cent;

1-12—1-12 (c) cos φ = 1-0; fcmarg = — — — X 100 = 0 per cent.

Thus, the critical Voltage on the motor busbars FCT = 0-9 may be obtained at different unfavourable conditions of the system; for example, by decreasing the excitation current of the generators.

As follows from the calculations given, we decrease the safety coefficient of the load stability by about 1*5 times, by improving cos <p to 0-95; the improvement of cos φ to 1 in the given system will, in practice, cause the motors to pull out.

Conclusion. The compensation of the load reactive power with batteries of static condensers may sometimes lead to a considerable lowering of the induction motors' stability margin and with a high compensation level may lead to a "collapse of voltage".

Problem 4.7

A remote hydrostation G is supplying consumers on the substation on a double-circuit transmission of 110 kV. The load on the substation L is represented in the form of one equivalent induction motor, which is connected to the busbars of the substation's secondary voltage (Fig. 4.18). The busbars of substation L at 110 kV are connected with the busbars of the receiving system by a short line /2. The line impedance, because of its insignificant length, may be ignored. In the presence of such a connection, VL = Vc = const and does not depend on the transmission operating conditions or on the motor.


With the normal original condition along the line /2 (through the switch Bx)9 the flow of the active and reactive power is zero and this linkage serves as a reserve.

GKE>-£ £=50 MW cosf=ae xTI=OI xd = H x'rO-3 d

!,=95km XQsO'Alohm/km in relative units xl=0-2 for 1 circuit

FIG. 4.18.

The parameters of the circuit under consideration are given in Fig. 4.18, where the reactances of the generator, transformers and the motor are given in relative units at 5baaic = 62-5 MVA and Kbasic = 100 kV:

S**n — " T I — S' >T2 Smot = 62-5 MVA.

Required: (1) to find the critical voltage VCT, i.e. the voltage on the busbars of the substation N at which a pull out of the equivalent motor will take place (it is also to be assumed that the mechanical load of the motor does not depend on the slip and is equal to the electrical power of the standard condition); (2) to determine the normal slip, s09 of the motor at VL = 1 and the slip at the critical voltage Fcr, i.e. critical slip ^cr; (3) to find the maximum pull out moment of the motor Pmax at the normal voltage; (4) to determine the margin of stability in terms of the power and slip. It can be assumed in the calculations that the mechanical load of the motor does not depend on the slip.

Solution. The power used by the motor is determined approximately as: V2 R»

P =

xl+ ?)' X- (4.11)

where χσ = the total reactance of the motor xs plus the external reactance up to the point where the voltage is considered as constant.


In our example (see Fig. 4.19) *; = x5 = xT2 = 0-2+0-1 = 0-3.

But this equation gives all the values m which we are interested. The critical voltage is:

KCT = V(2/W = J(2^x0-3\ = VO-48 = 0-693 * 0-7.

xT2

h

FIG. 4.19.

In solving the equation (4.2) with respect to the slip s, we find the normal slip of the motor s0 at V = 1 :

Pa = V2Rzs

or,

Poxy-V*R2s+PoR* = 0. Hence,

„ V2R, RI n

·*0Λσ Λ<τ

1 2 X 0·03 ' + ί^τττί = 0 ; *2 -0-04165+001 = 0 , ■ ( $ ) ■ - * 0-8X0-32

512 = 0-208± V(00436-0-010) = 0-208±0-1822; sx = 0-0258 % 2-6 per cent; Ì2 = 0-3902 % 39 per cent,

i.e. the normal slip of motor s0 = 2-6 per cent. The second root corresponds to the unstable portion of the charac

teristic. The critical motor slip is:

R2 _ 0-03 ~ 0-3 ' ~ "cr sa = -* = —r- = 0-1, or 5cr= 10 per cent.


The maximum pull out torque at V = 1 is:

- — - h°2 - er raax ~~ 2χσ~ 2x0-3 "

The margin of torque is :

km = ^ ± ο =1 · 6 7 - ° · 8 ^ 1 · 0 9 .

0-8 The margin of slip is:

k s^-s^ = 10-*6 = 2 . 8 5

* o 2*6

Figure 4.20 shows the curve, giving the variation of the torque (power) of the motor in relation to the slip at V = 1 and Fcr = 0-693.

At 5 = 1,

2-0

1-6

Ί-2

Ci'fK

0-4 [

- P

PmeK= '-67

J S =01

P m.ch = c o n s t

^ ν ^ ^ ν = ι·ο

^^νΐ^7^ 1

0-5 s , 10

l2

(0-3)a + (0-03)s

FIG. 4.20.

χΟ-03 = Λ-L = 0-38. 0-0909

Problem 4.8

In the system, the circuit and the parameters of which are given in Fig. 4.18, as a result of a breakdown, a section of the system was disconnected (by switching off BJ and the load of the substation N was found to


be connected to the remote HES G only. Since no power flowed through the switch Swx in the original condition the balance of power in the remaining section of the system was retained.

Required: (1) to determine whether the motors will operate stably after the system has been divided; (2) to check the stability for the case when one transmission circuit is switched off simultaneously with the division of the system.

Two cases are to be considered for each condition. (a) generators HES G have no automatic excitation regulators and,

therefore, the stability is checked for the condition Ed = const; (b) Automatic excitation regulators of proportional type, without a

dead zone, are used with the generators HES G corresponding, approximately, to E = const.

At the instant of switching off, the voltage on the busbars does not change, but depends on the rating of the motor, which becomes a function of its slip s.

In the case of switching off, simultaneously, one transmission circuit, the voltage VL will change.

V, = const

scr2 scrl

FIG. 4.21.

Solution. In order to secure the stability, it is necessary that:

which Fig. 4.21 shows. The requirement Pmax > P0 may be written in another form:

E2 E2

P = — ^ P — 3SL max 2x ° ~ 2x '

hence E> Ea, where ECT = ^(2Ρ0χ).


Here E = the actual e.m.f., obtained in any working condition: having the value Ed, E' or Vg9 depending on the control system.

The equivalent circuit of the transmission takes the form given in Fig. 4.22.

In this circuit, the resultant impedance of the system and the actual e.m.f. of the condition preceding the switching off are calculated:

Condition

Standard

One transmission circuit is switched off

x9 up to point G

Ed = const

l - l + 0 - l + ^ - + 0 - l +

+ 0-2 = 1-6 11 + 0 1 +0-2+0-1 +

+0-2 = 1-7

E' — const

0 · 3 + 0 · 1 + ^ - + 0·1 +

+ 0-2 = 0-8 0-3+0-1+0-2+0-1 +

+ 0-2 = 0-9

In order to determine the e.m.f., we find the reactive power, which is consumed by the equivalent motor, taken from the busbars of the substation N at the voltage VL = 1.

On the basis of the equivalent circuit (Fig. 4.22):

0 = V2

■».+*+($)' ■ (Xj2 + Xs),

where x-n+x. = 0-1+0-2 = 0-3;

R9 003 0026

= 1-16;

0 = Is

0·32 + 1·162 Χ0·3 = 0-695 X 0-3 =0-208.

FIG. 4.22.


As a check, we find:

l2

135

P° 0-32 + M62 Xl-16 = 0-695x1-16 =0-8.

Then:

Ed = V[(l +0-208 Χΐ·3)2+(0·8χ1·3)2] = 164;

E' = V[(l+0-208 xO-5)2+(0-8 xO-5)2] = 1-17. The calculated results are given in Table 4.11. For conditions a-1 and a-2 we find the critical load voltage, which

corresponds to the critical values of the e.m.f., determined above (EdCT andO

At Ed = Edct » Λ) — ôo

Actually,

0 "~ E%„

JL· χ2 + Ι - 2

F2

X -ZÉSL - p , 2 — V-2J.V-2 ->« ~ ° '

V ' . since Ä2/5cr = x.

Y ΛΓ2+;

TABLE 4.11

2x

R*

Generators have no autom. excit.

control

Generators have autom. excit. control

Condition

a-1 normal

b-1 one circuì t

switched oflf

a-2 normal

Kept constant

£ , = 1-64

£ , = 1-64

£ ' = 1-17

Xc

1-6

1-7

0-8

So

0-026

0026

0-026

Ε„ = γ2Ρ9χ„ where

P0 = 0-8

1-60

1-65

113

s - * ! Xe

where R2 = 0-03

0019

0018

0037

Calculated results

unstable ^crÔ»

although EÊd

unstable sa^s0; EÊd

stable Ser^s0; EÊ'


In this case:

where x = the resistance from Ed to VL. For condition a-1 :

/Γ / ι r 0 · 8 χ 1 · 3 \ 2 / 0 · 8 χ 1 · 3 \ 2 " 1 , , c

i.e.

VL.« = M S > VL = 1.

For condition a-2:

/ΓΛ , , 0 · 8 χ 0 · 5 \ 2 / 0 · 8 χ Ο · 5 \ 2 Ί noc

Vu* = 7L(M3—TÏT-) +(-TÏ3-J J » °·85' i.e.

^.cr = 0-85 < VL = 1.

Calculations show that the reason for instability of the load — the pull out of the motors — may be because of the direct lowering of the voltage, which seldom occurs if the supply is from busbars of constant voltage, or because of changes in the circuit, the effect of which is that the point of application of the constant e.m.f. (Ed, E') is drawn out behind a larger reactance. The latter case leads to an increase of the critical voltage and, simultaneously, to the lowering of the critical slip of the motor.

With VL = const, the stability of the motor was very good, and the critical voltage was low. But this was only the case when VL did not depend on the motor load. When the supply to the motors is taken from generators, the power of which is commensurable with the power of the motor, VL = f(P, Q), the dependence of the busbar voltage on the rating of the motor is brought about. Thus, when VCT = 85 per cent the pull out of the loading motors can occur when automatic excitation control (£" = const) is used.

If there is no automatic excitation control (at Ed = const) the critical voltage is larger than the normal voltage of the motor, VLcr = 115 per cent, which (as in the relation sCT s0) indicates the impossibility of the motor working stably.


Problem 4.9

The circuit (Fig. 4.23(a)) of a system is similar to the circuit in Fig. 4.18. The original condition and the parameters of the system elements are given in Fig. 4.23(a) and repeat those given in Fig. 4.18. The only exception is the load of substation N, which is represented in the form of a complex load applied to the 110 kV busbars. The power used by the complex load at VL = 1, is given in Fig. 4.23(b).

Required: (1) to check whether the complex load, which has typical characteristics, in the system (Fig. 4.23(a)), will be stable after pulling out the switch B; (2) to determine the load stability margin in terms of the voltage.

The calculations are to be carried out for two conditions of transmission:

(a) both transmission circuits are in operation; (b) one transmission circuit is disconnected by the switch B simul

taneously, under the condition, that there is no automatic excitation control on the generators G.

X Vc=const

Pa 50 MW tOS<fa08

XfO-2

I=95 km xo=0AI ohm/km

V,= l

GD-+ P0 = 50MW

(a) <«<P*°1

f ΠΡ 1

'. = 0-3 I χ' = 0·3 Sbasic--62*5 M V A

P0:0-8 Q0=0-6

Vbasic5"0 k V (b)

FIG. 4.23.

For the condition of power transmission by both circuits, the calculation is to be repeated with the condition that automatic excitation regulators which are of the proportional type without a dead zone are arranged on the generators G.

The static characteristics of the complex load are given in Table 4.12.


TABLE 4.12

y.

10

0-9

p.

1000

0-941

G* v.

1000

0-885

0-8

0-7

* ·

0-893

0-855

Ö*

0-844

0-880

In the previous problems (4.7 and 4.8) it was assumed that the load consisted of an equivalent induction motor only, where the effect of the circuit magnetization was ignored.

The load is usually a combination of loads for different customers. Motors, which constitute 60-70 per cent of the power, are not alike in their characteristics. As was seen in the previous problems, the essential effect is produced by the magnetizing current. Consequently, when calculating, it is necessary to make use of the static characteristics PL, QL =f(VL) for the standard complex load (Appendix 2).

In order to determine the complex load stability, the criterion d £ / d K > 0 is applied.

Solution. We calculate the relationships Ed=f(V) and E' =f(V) in accordance with (4.8), setting different voltages V on the busbars of the complex load and determining the corresponding power values P and Q :

P=PoP. and Q = Q0Qm9

where P0 and Q0 = the composite load power at VL = 1 ; P^ and Qm = the relative changes in power, taken from Table

4.12. In the circuit for determining Ed, where the generators G are used

without an automatic excitation control (xd) the total reactance of the system :

(a) with two transmission circuits is:

xc = 1 - 1 + 0 - 1 + ^ = 1-3;

(b) with one transmission circuit is:

xc = 1-1+0-1+0-2 = 1-4.


In the circuit for determining E', where the generators G are used with an automatic excitation control (x'd) the total reactance of the system with two transmission circuits is:

xc =0-3+0-1 4 - ^ = 0-5.

The calculated results are given in Table 4.13 and are shown in Fig. 4.24.

TABLE 4.13

V

10 0-9 0-8 0-7

Normal

E4

206 1-995 1-998 212

condition

E'

1-36 1-227 1-21 118

One circuit disconnected

E4

2-50 209 210

~~

The values Ed = 2-06 and E' = 1-36 at V = 1 for the normal condition are the original values, which remain practically constant because there is no flow of power across the switch, as was previously assumed. As follows from Fig. 4.24, the load stability, with transmission at normal working conditions (with both circuits switched on), is maintained in the absence of as well as in the presence of the automatic excitation control.

The load stability margins in terms of voltage are: (1) with no automatic excitation control:

*marg = ^ ^ x 100 = - ^ ^ - X 100 = 15 percent;

(2) With automatic excitation control:

1-0-0-7 ίΛΛ ΛΛ *marg = — p ö — x 1 0° = 3 0 Ver c e n t ·

Thus, the automatic excitation control increases the load stability margin. When one· circuit is disconnected, the system without the automatic excitation control is unstable.


Problem 4.10

The complex load of a system consists of induction motors, using 70 per cent of the active power, and a lighting load, using the remaining 30 per cent. With normal voltage on the busbars (VL = 1), the total

| f » = l ; 0^0-525

'fight50·3 Pasyn=°·7

(a)

Q=0-525 xs=0-2

(b)

FIG. 4.25.

active power of the load is (in relative units) PLm = 1, and the total reactive power is QLm = 0-525.

The circuit and the load parameters are given in Fig. 4.25.

IVcr =O'S5 V

1·3

1·2

0·7 0-8 o.g 1·0

FIG. 4.24.


Required: (1) to determine the changes of the active and reactive power of the load for frequency changes in the system with limits ±10 per cent, assuming that the voltage on the load busbars is constant (VL = 1);

1-0

0-9

08

07

1 2 3 4 5 6 7 FIG. 4.26.

(2) to determine the changes of the active and reactive load power with a changes of frequency in the system and of the voltage on the load busbars simultaneously.

The changes of the frequency and the voltage in the system are shown in Fig. 4.26 and given in Table 4.14.

TABLE 4.14

Moment of time 0

10

10

1

0-95

0-968

2

0-90

0-935

3

0-85

0-905

4

0-80

0-872

5

0-75

0-842

6

0-70

0-81

Solution· (1) From the expression (4.2) we find the slip sv with which the equivalent induction motor is operating, when a power P± = 0-7 is taken from the system at the busbar voltage VL = Vx = 1;

%) -ΤΓχΓ-'ω + ^-) - 0 ,

2 12χ0·02 /0-02Υ . J ( 1 ) " 0 - 7 X 0 - 2 * J < » + ^ Ö T J - 0 ;

sfo-O-lUsM+O-Ol = 0; i(1) = 0-357-V(0-127-0-01) = 0-357-0-3426 = 00144.


(2) We find, by using expression (4.3), the original data, the calculated slip s{l), and the reactance of the magnetizing branch χμ of the induction motor-(Fig. 4.25(b)) that:

12χ0·2 l2 0-525 = — +

^ 00144 ) ^

hence χμ = 2*36. (3) We calculate the changes of the total active power P of the com

plex load at the constant voltage on busbars (VL = 1) and the changes of frequency in the system within the given limits.

(a) The active power, which is used by the lighting load, does not depend on the frequency changes in the system, i.e. Plight = 0·3 = const.

(b) The active power taken by the equivalent induction motor during the frequency changes in the system may be calculated by the relation:

/ p = p x asyn/ * asyn/x f Jl (4.12)

where Ρωγη = the active power used by the motor during the frequency changes.

^asyn/£ = ^ e a c t i v e power used by the motor at the normal frequency (A = i).

The calculated results are given in Table 4.15 and in Fig. 4.27.

ML

P

1-2

0-84

11

0-77

TABLE 4.15

1-0

0-70

0-95

0-665

0-90

0-63

0-85

0-595

0-80

0-56

(4) We now go over to the calculations of the changes in the reactive load power at the constant voltage and the changes in frequency.

(a) We assume that the reactive power taken by the lighting load (incandescent lamps) is zero.

(b) The total reactive power used by the equivalent induction motor at any frequency/, may be determined by the expression:

Qr = v2 ■ + JL m-m (4.13)


where sf = the slip of the motor, which is using active power PMyn f at the corresponding frequency / and voltage VL = 1.

PL 10

0-7

1

-0-60

0-55

-

0*50

p asyr

> ^

- ^0000^00^0^

' _ I . f

**îiehr^^ ^

Q|fQo«yn

Î ! _ _ . . _ . ! ^ 1

M 09 10 M

Fio. 4.27.

1-2 Ι·3

The slip of the motor sf is calculated from expression (4.2), which applies for the frequency / which is different from the normal:

s}-P y

V2R2 R\ 0.

Assuming f/fL = 1-2; from Table 4.15 we take Patyn{12) determine the required values.

(c) The slip sf - J(1.2):

(4.14)

0-84 and

12χ0·02 0-022

ί(1'2) 0-84xÖ-22xl-22 0-22xl-22

s0L-2)-0414-ya2)+°·00694 = °; *<ι·2) = 0-0175 (1-75 per cent). (d) The full reactive power is:

l 2

0(1 2) 2-36x1-2 +

+ l2x0-2xl-2

GÄ 02 y 0175 )

= 0-353+0176 =0-529. +0-22xl-22

The calculations are carried out similarly for other frequencies. The calculated results are given in Table 4.16 and are shown in Fig. 4.27.

(5) We calculate the changes of the total active load power with given simultaneous changes of voltage and frequency in the system.


TABLE 4.16

/ · *, % ß/. Q. Q

1-20 1-75 0-353 0176 0-529

115 1-67 0-369 0155 0-524

110 1-60 0-385 0137 0-522

10 1-44 0-423 0102 0-525

0-95 1-33 0-447 0083 0-530

0-90 1-29 0-472 0074 0-546

0-85 1-20 0-499 0061 0-560

0-80 1-13 0-53 0051 0-581

(a) The change of the lighting load power is determined by the change in the voltage. This may be calculated according to the expression

L-6

(4.15) Might — Might (1) 2LY-<

where / ^ t u ) = 0-3 = the active power of the lighting load at V=VL = 1.

(b) The active power taken by the equivalent induction motor at constant torque on the shaft does not depend on the change in the voltage. Therefore the active power of the motor only changes because of the change in the frequency, according to expression (4.12).

The calculated results are given in Table 4.17 and in Fig. 4.28. The voltage and the frequency at a given moment of time are taken from the known characteristic of the dynamic process.

TABLE 4.17

Instant

v, f,

flight p

*light ~ "·» Myn

0

10 10 0-3 0-7 10

/

0-95 0-968 0-280 0-688 0-968

2

0-90 0-935 0-256 0-654 0-910

3

0-85 0-905 0-231 0-633 0-864

4

0-80 0-872 0-213 0-610 0-823

5

0-75 0-842 0189 0-589 0-778

6

0-70 0-81 0169 0-567 0-736

(6) The change in the reactive load power with simultaneous changes of voltage and frequency in the system may be determined from the expressions (4.13) and (4.14), which apply at different moments of time with corresponding values of V^ fm and Pasyn.

As an example, for the first instant: K„ =0-95; / . =0-968; = 0-688;


the slip of the motor is 0-951X0-02J1 O022

Γι) 0-688 (0-2 x 0-968)2 (0-2 x 0-968)2 = 0;

^-0-712^+001034 = 0; sa) = 0-0149(1-49 per cent). The full reactive power of the motor is :

0-952

Q -Qp+Qs - 2-36 x 0-968+

0-952χ0-2χ0-968 = 0 . 3 9 5 + 0 . 0 9 5 = ^ / 0-02 V (^00149 )

+(0-2x0-968)2

i-o

04

08

0-7

0-6

05

0-A

0-31

0-21

01

0

PjQ P L s f î i g h t + P -

V»(fJ

0* 07 0-8 FIG. 4.28.

0-9 10

At other instants, the calculations are carried out similarly. The calculated results are given in Table 4.18.

TABLE 4.18

Instant

s,% Ωμ Q. Q

0

1-40 0-423 0102 0-525

1

1-49 0-395 0095 0-490

2

1-63 0-367 0098 0-465

3

1-80 0-338 0103 0-441

4

213 0-311 0122 0-433

5

2-32 0-283 0123 0-406

6

2-39 0-256 0109 0-365


Problem 4.11

The load, which consists of induction and synchronous motors, is supplied from constant-voltage busbars.

Required: to compare the degrees of stability of synchronous motors with and without automatic excitation control at different loads and values of cos<p and, of induction motors at different loads and leakage reactances.

Solution. The degree of stability may be estimated by the voltage Fcr, at which the stability is disturbed, i.e. the pull out of induction motors and falling out of synchronization of the synchronous motors.

For induction motors, the critical voltage is :

VCT = yJ(2mPLXl)9 (4.16) where m = the load of the motor in nominal units ;

PL = the nominal power of the motor; jc, = the total leakage reactance of the stator and the rotor.

0-8 rE,

0-6

(K

0-2

06 0 7 08 0-9 1-0

FIG. 4.29.

0-6 07 0-8 0-9 10

FIG. 4.30.

For the synchronous motor, the critical voltage can be determined from the power formula:

D E v · Λ P = sin o. x

(4.17)


Taking the motor power equal to the load on its shaft mPL at sin à = 1, we get:

For the e.m.f. E and the reactance x, when there is no automatic excitation control, the synchronous e.m.f. Ed and the synchronous reactance xd must be taken, and with automatic excitation control, (an example of current compounding,) the transient e.m.f. E' and the transient reactance xd are taken.

The electromotive force of a synchronous motor can be determined from the formula:

where Λ) = mPL\ Q0 = P0 tan <p.

The calculated values of the critical voltage (given that V0 = 1) according to the above formulae are given in Fig. 4.29 for synchronous motors and in Fig. 4.30 for induction motors.

Comparing the results, it can be seen that the degree of stability for synchronous motors, as a rule, is higher than that for induction motors. Therefore, the replacing of a section of induction motors by synchronous ones, may serve as a means of increasing the stability at points of industrial load. Besides, an installation of synchronous motors, which operate at the loading cos φ, makes it possible to reduce or even to give up completely the use of static condensers, which also increases the degree of load stability.

Problem 4.12

The load, which is represented by the equivalent unregulated synchronous and induction motors, is supplied by a system which is represented by an inductive reactance with a constant e.m.f. behind it (Fig. 4.31).

Required: to determine the voltage at the motor terminals at which the load stability will be disturbed.

The parameters of the motors are as follows :

induction motor: PL = 0-5;

xl = 0-4; m = 1;


synchronous motor: case I -cos φ =

- Ρ = 0·5; xd = 2\ 0-8 (lead); m = 1;

case II— PL = 0-5;

xd = 3;

external reactance:

EC

cos φ = 1 ; m = 1 ;

xext = 0-218.

'ΤΡ» J xext

(a)

ΓΓί VVSynM

rT^i, ^ ) 1 M

iiV-J< ^ ^ ·

x«xt P;Q

(b)

FIG. 4.31.

Solution. The analysis of the given systems stability is carried out with the aid of the criterion determined by the method of small oscillations. According to the criterion, the system is stable, if the following condition is fulfilled:

K . 2VcosômotOT-Ed XdCosômotor

IVxp2 2Kcosó-£c x2 — R2 xextcos<5 ,0 . (4.20)

From the intersection of the characteristics K =f(V) with the abscissa axis, the value of the critical voltage may be found.

In order to form a function K =f(V), it is necessary, taking a range of different values of voltage in each case, to find :

(1) ômot and Q1 for the synchronous motor with the conditions : Ed = const and Px = const;

(2) s and Q2 for the induction motor with the condition P2 = const; (3) P and Q for the balance of power at the junction point ; (4) Ec and à from P and Q determined above.


The following known relations are also used; for the synchronous motor:

xd

. V2 EdV and ô i = - — cosó„ xd xd

FIG. 4.32.

for the induction motor:

P* = V2Rs

2 x2s2+R2

for the branch of the system:

and Ô2 = V2XjS2

#

xfs2+R2''

E V P = f £ l sin δ and Q = +—— cosò. x~

The calculated results are given as curves in Fig. 4-32 from which, using the parameters of the synchronous motor, for case I, VCT = 0-69, and with the parameters for case II, VCT = 0-85. It also follows from Fig. 4.32 that K = 0 when dEJdV = 0, i.e. in the present case, the stability may also be estimated by using the practical criterion dEJdV> 0.

CHAPTER 5

THE CRITERIA FOR STABILITY OF SYSTEM OPERATION (THE SOLUTION OF THE

CHARACTERISTIC EQUATIONS AND THE DETERMINATION OF THE TYPE OF ROOT

WITHOUT SOLUTION)

IN THIS chapter the general problem of analysing the stability of electrical power systems is considered. The solution of the differential equations for the relative motion of any of the stations of a system, is reduced to solving certain characteristic equations. The character of the roots of the equation shows whether steady operation will take place or whether instability may be expected. From the character of the roots, the kind of instability (oscillating or aperiodic) may also be determined. The direct solution of the characteristic equation consists in solving an algebraic equation and in finding the roots of the equation as numerical values. After finding the roots, a curve which shows the changes of any variable in relation to time, may be plotted, and the stability or instability of the system is then clearly shown. Such a method of solution, however, is too laborious, and generally, in practice, the method of analysing the roots of the characteristic equation, without solving the equation, is used. Examples of applying the different methods for analysing the roots of a characteristic equation are given in this chapter.

Problem 5.1

A station is connected to a receiving system of large power through a transmission line (Fig. 5.1).

GK£) d>\ FIG. 5.1.

The parameters of the station and the transmission line (in relative units) are:

xdE = 1-5; £ , = 1-07; V = I; 7> = 15 sec; Λ, = 60. 150

Criteria for Stability of System Operation 151

Required: (1) to check the stability of the system, to find the frequency and the period of natural oscillation in different conditions both allowing for and ignoring the damping torque; (2) to plot the change of the angle with time when the rotor is deflected by 1° from the position of the steady condition at 0°, 60°, 90° and 100°.

Solution. To ascertain the nature of the transient process, to determine whether the system is stable or unstable for a small disturbance, and to construct curves for the transient process, by the small oscillation method.

The equation of free (disturbed) motion for small disturbances (deflection ΔΟ of the angle) may be written in the following form:

TjP2+PdP + SEd = 0, (5.1)

where c ft?Ed EdV -SE"=-dT = l^cosô· ( 5 · 2 )

By solving this equation for the general case, we get:

ΔΟ = Q e ^ ' + Qe'*'. (5.3) Here px and p2 (the roots of the equation (5.1)) are:

4Tj Tj ) 2Tj *

or, putting Pd = o. ^Ed 2Tj H' Tj '

Pi.2 = -/*±V(02-a) = -ß±Jv- (5·4) The constants Cx and C2 are determined from the initial conditions; i.e.

from

and (J«) ;_o=0.

The first condition gives: Q + C2 =Δδ0.

From the second condition we get: P\Cx+p2C2 = 0.

Pl,2 w


By solving them together, we get :

C\ = — Λ<50, and C2=-^— Αδ0. Pi -Pi Pi —P2

Substituting Cx and C2 in (5.3), we get:

Δά = ^\{ρι*ρ«-Ρ^)· (5.5)

The stability of the system may be determined from the sign of the real parts of the roots p± and/?2 or by the stability condition, which follows from the Hurwitz criterion, i. e. the sign of the free term of equation (5.1). Both of these conditions lead to the criterion of stability : S^ ^ 0, hence, allowing for (5.2), it follows that the stability of the system is disturbed at β0=90°.

By using equation (5.5), we can plot a curve of angle variation against time.

We now proceed to determine the frequency and the period of natural oscillation of the system.

Without taking into account the damper torque

Ignoring the damping torque when Pd = 0, the roots (5.4) are:

Pi,2 = -ß±Jr= ±.\zffi- <5·6) Also:

Cx — C2 — —2~ .

Then equation (5.3) becomes:

Λί = ^ ( e * ' + e*') . (5.7)

It is obvious that with SEd> 0, i.e. with δ < 90°

Λ.2 = ±JV-Then

ΔΟ = ^ (e^'+e-* 1 ) = AÒQ cos yt, W

i.e. the system is stable.

With à > 90; SEd<0 and Δο = Δδ0 (-—- + - ^


i.e. the system is unstable; <5=90° is a critical case. It is the boundary between the stable and unstable conditions.

We consider the precise conditions, with angles δ=0, 60, 90 and 100°. (a) 0 = 0 The synchronizing power is :

SEd = ^ c o s a0 = \ ^ 1 cos 0° = 0-713.

The angular frequency of oscillation is:

y = Ιξη = I0'11* = 1 2 3 x 10"2 1/rad or 0-614 — . r yj Tj yj 15x314 ' sec

The dependence of the real and imaginary parts of the roots y and β is shown in Fig. 5.2.

The oscillation period is : ^ 2π 2x3-14 __ _, T = — = -r^—77Γ-7Γ = 510 rad = 1-63 sec. y 1-23 χ 1 0 ~ 2

The equation of motion of the generator rotor is: ΔΟ = Zl<50 cos 1-23 xlO-2*,

where / is expressed in radians. The curve ΔΟ =f(t) is shown in Fig. 5.3 (curve 1).

(b) δ0=60° SEd=0-356; y=0-87x 10"2; Γ=2·3 sec;

/=0-435 1/sec; Αδ=Αδ0 cos 2-72/,

where t is expressed in seconds. The curve Aô=f(t) is also shown in Fig. 5.3 (curve 2). (e) ó0=90°

There is no practical point in considering this curve closely as it is a limiting condition and the slightest disturbance leads to the system with angles ό>90° or δ < 90°. We consider the condition that <5=91°. In this case the synchronizing power is near to zero and therefore,

y ^ 0 ; plt2^0; Γ^οο.

Substituting p1 2 ^ 0 in the equation of motion, we get:


Thus, with δ0%90°, the angle ΔΟ varies so slowly that at first it can be considered as maintaining its original value (Fig. 5.3, curve 3).

Comparing curves zio at different initial angles <50, we can see that with an increase of ό0, i.e. of transmitted power, the period of natural oscillation increases,

*io~:

1-2

10

0-8

0-6

0 4

0-2

0

-0 -2

- 0-4

- 0 - 6

- 0 - 8

- 1 - 0

- 1-2

FIG. 5.2.

Fio. 5.3.

and reaches an infinite value at <50=90°. (d) ô0 = 100°

The synchronizing power is :

SEd = ^ ? c o s 100° = -0-124.

120°

Criteria for Stability of System Operation

The roots of the equation are :

155

Ed M - ± / - f = ± . -0124

15x314 = ±0-514xl0-2.

The variation of the angle Jo in this case is given by the equation

2 where t is expressed in radians, or:

Ah — ° (e°'514xl<>"" I e-fr»nxio~*'\

where t is expressed in seconds. The curve Δδ =/(/) is shown in Fig. 5.3. The angle increases indefinitely,

as is basically determined by the first term of the equation of motion. The second term decays with a time constant

1 r = 1-61 = 0-62 sec,

which has a noticeable effect only at the beginning of the process.

Taking into account the damper torque

With PJT^O, the roots of equation (5.1) are determined by expression (5.4). The nature of the variations of the real and imaginary parts of the

degrees

FIG. 5.4.


roots for our actual case is shown in Fig. 5.4. The imaginary part in the condition becomes zero when the expression in (5.4) inside the root changes its sign, i.e. when:

$Ed = Pd

or when :

a„ = a r c c o s | - x ^ .

In the case under consideration, the imaginary part becomes zero when <$0=74-4°. With angles <50 ■< 74-4°, the damping characteristic zio is periodic, since y 5*0, and with <50> 74-4° it is aperiodic, since y=0.

For a periodic process (5.5), it is possible to obtain a more convenient form for making calculations, namely:

Δδ = ^ e~* Va sin (yt +ψ0)9 (5.9) y

where,

y;0 = a r c t a n j :

« = ^ [see (5.4)].

The curves showing the transient changes, calculated for the conditions when ό=0, 60, 80, 90, and 100°, are given in Fig. 5.5. When <5=0 and 60°, the character of the process is periodic. The period of oscillation is, however, larger than the value obtained when the damper torque is ignored.

FIG. 5.5.


With <50=80°, the process is aperiodic. With <30=90°, the angle also remains, practically, unchanged for the duration of time given, as it is when the damper torque is ignored.

With <50 = 100°, the angle increases indefinitely, the presence of the damper torque, however, leads to a slowing down of the increase of the angle.

Problem 5.2

In the system of Fig. 5.6, station 1, which has a local load, supplies the system with a part of its output, which is commensurable with the power of the station.

G H -^Γ* iK Θ FIG. 5.6.

The parameters lows:t

of the system and of the original condition are as fol-

z n =0-771 z12 = 1-065 z22 =0-185 £Ί = 1-449

/>10 = 1-333 7^ = 12 sec;

an =4-4°; a12 = —18-6°; α22 = 36Ό £2 = 1-093

7*20 = 3-496

Γ / 2 = 30 sec.

Required: to determine: (1) the power limit; (2) the stability limit.

Solution« The limiting power of the transmitting station is equal to:

' - = ? « i " « u + ^ . (5.10) z l l z12

This power value occurs at an angle <512 = 90+a12 =90— 18-6° = 71-4°. n 1-4492 . A AQ 1-449x1093 Λ Me P = sin 4-4° 4- τ-τ^. = 1 695.

0-771 1-065

t Generator reactances are included in the self and mutual reactances, such that the e.m.f.s behind them, may be assumed to be constant.


We construct the power characterictics of stations 1 and 2 (Fig. 5.7). The limit of stability, which does not coincide with the limit of power, is

easily determined by equating the relative acceleration to zero (ref. 3, p. 198) :

1 άΡλ 1 dP2 Λ

< / i d<5 12 TJ2"àà\2 (5.11)

*&*

50°

FIG. 5.7.

where

d/>j _ EXE2

do 12 612

dP2 EXE2

cos(<512-a12);

cos(<512+a12). dô12 z12

Substituting the synchronizing power values in (5.11) we get:

E E — ^ - [cos (ôl2-a12)+m cos (<312+a12)] = 0, Z12J Jl


where

From the last relation we can get the value of the relative angle, ô12, at which the stability limit occurs :

2EXE2 ^ 1 +m *12

<$ia = arc tan =3=5- X f-!— . (5.12) 1 —m

For the system under consideration, the limiting angle, which is in accordance with the condition of stability, is :

2x1-449x1-093 1+0-4 1-065 X 1-0-4 «H = 7 ^ x 4 - ^ - 7 - = 82· ic

Thus, the limit of stability occurs at a larger angle than the angle corresponding to the power limit. This case has a definite practical importance; it indicates that in the system of two stations there is a possibility of operating on limiting power without loss of the stability due to small sudden changes.

Depending on the ratio of the inertia constants of the station, the stabi-di>! dP2 lity limit may vary over the range, —— = 0 to —— = 0 (Fig. 5.7). An

do12 dd12 increase of m leads to an increase of the critical angle.

Problem 5,3

In a station, which is connected to a system of infinite power through a transmission line, excitation controls, which can be actuated by variations of the direct voltage component, are installed.

We note, that the direct-axis voltage component is equal, approximately, to the full voltage of the generator and, therefore, the results obtained in this problem may be extended to stations which are provided with electronic excitation controls which are actuated by the voltage variation. Replacing the full voltage by its direct-axis component, simplifies the solution of the problem.

The parameters of the system and the original condition are as follows:

xdE = 1-486; x'dE =0-848; xc =0-504; Γ, = 7·5 sec;

r;=2-85sec; Te=2 sec; kv = 30; P0 = l; £ ^ = 1-972;

Kc = l; <30=49,0°.


Required: to find the limit of the static stability (PIim and <5lim).

Solution. By using the small variation method for the given system, we get a characteristic equation of the fourth order. Analysing the latter with the aid of the Hurwitz criterion (Ref. 3, p. 235), the maximum permissible coefficient of amplification when actuated by variation of the longitudinal voltage, may be determined :

v v ' 1 + T (T*-LT\ ($E'dT'd + SEdTe) r _ ΧάΣ~ΧαΣ., ij\id^1e) (1\Χ\

χάΣ xc j , 1 e \χάΣ xc)xdE T'd(Xd-Xc)xd

Equating the maximum permissible coefficient to the one given, and also working out the expressions for synchronizing power and substituting in them :

Ed = Ed0+kv(Vgd- Vgdo)

xr (5.14)

l+—ky χάΣ

we get a quadratic equation for cos δ:

[ Je Vc(Xte — Xc)(Td + Te) j j,2 xdE~~xdE 7 - Π € 0 5 2 ( 3 _ (χάΣ+Χ<^ν)ΧάΣ χάΣΧάΣ J ^

K (Edo +kyVgdo) (T'd + Te) λ 2 xdE —xdE , v +xk 0\im~yc f 1d^r *>άΣ ^*Ό*>ν ΛάΣΛάΣ

IC Γΐ I Γ^β^ΧαΣ~Χ^Χ'άΣΛ ΧάΣ~Χο Λ Tj(Td+Te) _ Q V I Td (ΧάΣ - Χα)χάΣ\ ΧάΣ - Χ'θΣ J ^ Ο +

By solving this equation, we find ôVltn. The direct-axis voltage component of the generator in the original

condition is:

ΧάΣ ΧάΣ

0*504 , „ , , 1-486-0-504 t Λ , c r = T^zxl ,972+ r^z x 1x0-656 = 1102.

1-486 1-486


The characteristic P =/(<$) takes the form:

EdVc . £ Ed . £ i> = -^—csin δ = — — sin δ; xdL 1-486

161

and using (5.14), we get:

Ed =

t ^ . ™Λ 1 M 1-486-0-504 Λ > 1-972 + 30 1-102 ^-τ^ x l x cosò \ 1-486 ; ϊ 0-504 „Λ 1 + Ϊ ^ Χ 3 °

3503-19-8 cose 11-2

We summarize the calculated results in Table 5.1.

TABLE 5.1

(5°

COS Ô sin ô

Ed P

60

0-500 0-866 2-24 1-305

75

0-259 0-966 2-67 1-734

90

0 1 3-12 210

105

0-259 0-966 3-58 2-33

120

0-500 0-866 401 2-34

50 60 70 80 90 100 110 degrees

FIG. 5.8.


The characteristic P=f(S) is shown in Fig. 5*8. We find the critical angle by solving equation (5.15). Substituting the parameters of the system and the initial conditions in it, we get :

Γ (l-486-0-504)xl2(2-85+2) 1-486-0-848 , / Ι . fl |_J" (1·486+0·504χ30)χ 1-486 + 1 X 1-486x0-848 XZ 8 D J C 0 S , ün~

1(1·972+30χ1·102)(2·85+2) , , . 1-486-0-848 __ _ LÌ ί. COS O 1 Ζ V V (1-486+0-504x30) Iün 1-486x0-848

Λ ο < , ί , η Γ , 2 1-486-0-504 0-8481 χ 2 · 8 5 + {3 0[1 +2Τ5> <0·848-0·504 Χ Γ 4 8 6 ] Χ

0-848-0-504 ,1 7-5(2-85+2) η Λ 1-486-0-848 ] Λ 22χ314

or

hence

cos2 <5Um- 1-41 cos ólim-0-0176 = 0,

cos ólim = -0-046 and ôIim=92-60.

From the characteristic P=f(S) we determine, at angle δ=92·6°, the limiting power for static stability:

Plim=2-14.

Problem 5.4

The station is connected to a system of infinite power through a transmission line.

Required: (1) to form a domain separation curve D\ (2) to determine the stability area with excitation control of the generators depending on the variation and the first derivative of the voltage.

Solution. The operating condition of the controlled system under consideration, is described by the following characteristic equation :

<*oP7 +αιΡ6 +<*2Ρ5 +azP* +aiP3+asP2 + aeP +<*? + + {Δ2ρ*+Διρ*+Λη){Κιρ + Κ») = 0. (5.16)


Assuming that all the coefficients of this characteristic equation are known and are:

a0 = 11x10-«; ax = 2-91 x 10-*; a2 = 32-7xl0"4; a9 = 5-13 Xl0-2; a4 =22-5xl0~2; a5 = - 0 0 4 ;

To construct the domain separation curve D, expressed in coordinates K0 and Ku we reduce the characteristic equation to the form (Ref. 5, p. 271):

a, = - 1 1 0 ; a7 = - 1 0 7 ; 4 , = 0-645x10-*; ζί5 = 1-16X10-2; Δη = 0-021.

KoP(p)+K1Q(p)+S(p)=0. (5.17) Here

?ι(ω)=0; 1 ?2(ω)=0, j

(5.18)

Ρ(ρ)=Δ3ρ*+Δ5ρ*+ΔΊ; Q(p) = (Δ3ρ*+Δ5ρ2+Δ,)ρ;

S(p) = a0p7+a1pi+a2p5+a3pi+aip3+asp2+atp+aT Substituting^ (orp and separating the real and imaginary parts, we get

the two equations : ν>

1(ω)+^1β1(ω)+51(ω) = Α0/,2(ω)+ΑΓ102(ω)+5'2ι

where Ρχ(ω) = Δ3ωΑ — Δ3ω%+ΔΊ; Ρ2(ω) = 0; δι(ω) = 0; ρ2(ω) = (Δ3ω*-Δ&ω2+ΔΊ)ω; Si(°>) = —αιω*+α3ω4—α5ω2+α7; S2(o>) = — α0ω7+α2ω5—α4ω3+α6ω.

From (5.18) we find the equation of the domain separation curve D in the parametric form:

* o =

-Sx(fl>) βι(ω) - 5 2 ( ω ) β2(ω)

K,=

Ρ^ω) -5χ(ω) />2(ω) -52(ω)

Δ > " ΐ ^

The determinant of the equation system (5.18) is:

Δ = Λ(») ôi(«) Ρ2(ω) β2(ω)

(5.19)

(5.20)


Expanding the polynomials P{co), Q(co) and S(co) in (5.19), we get an equation of the domain separation curve D:

_ - αχω* + αζωΑ — α5ω2 + αΊ \

' " ' **-**+*> '■ ,5.2!) κ __ - a0co« + 02

ω* — α4°>2 + αι I 1 ~~ 4ω4-4>ω2+^7 ' )

The domain separation curve D includes a separate straight line, since with ω 0 =0 the numerators and the denominator in (5.19) are reduced to zero. We get the equation for the separate straight line from (5.19):

__ SiK) & K ) y Λ ° - Λ(ω0) Λ Κ ) *'

or with ω0 = 0

K0 = -%- (5.22)

Expanding the equation (5.20), we find the determinant:

Δ = (Ζΐ3ω4-ζ15ω2+^7)ω.

In varying co form — o© to 0, the determinant is negative, and therefore, the domain separation curve D is shaded on the right.

With ω =0, the main determinant changes its sign, therefore the separate straight line must be shaded. The straight line is shaded in such a manner that the shading of the straight line and the same domain separation curve Z>, near the singular point (=0) , are directed to the side.

Figure 5.9 gives the domain separation curve D and the separate straight line, formed in accordance with equations (5.21) and (5.22). These are shaded according to the instructions given above. Since the curve is traced twice when co goes from - œ to -f °° one side is cross-hatched. The domain separation curve D has two breaks, at values of ω, for which the denominators are reduced to zero (5.21).

In order to determine which of the areas obtained is the range stability, we make use of Routh's criterion.

The calculation of Routh's criterion for the first set of coordinates, i.e. with Ko=0 and ^ = 0 , is shown in Table 5.2.

Since, for the coefficients in the column a0-cl8 the number of sign changes is three, the point with co-ordinates K 0 = 0 and K ^ O is found in the area in which the characteristic equation has three roots with positive real parts.


Because the number of roots with positive real part is smaller by one on the shaded side, than on the other side of the boundary line, and also because the domain separation curve D is hatched twice, it follows

700 800 900 1000 ^ ^ 1 1 0 0

FIG. 5.9.

that the number of roots with positive real parts in each of the areas (Fig. 5.9) can be determined. The area, where the number of roots with positive real parts is zero, is the region of stability.

Problem 5.5

A station is connected to a system of infinite power through a transmission line.

Required: to determine the permissible values of the amplification factors in the regulating system of the generators.

Solution. We assume that the operating condition under the consideration (0=80°) is described by the characteristic equation, all the coefficients


TABLE 5.2

= 3-78xlO-1

= 0-222

" Cu

= 117

= -0-488

= -0-795 x xlO-*

a0 = l l x l O - 6

ax = 2-91 x 10-4

^13 = α2 — λ0α3

= 13-lxlO-4

CU ~ a 3 — ^ 1 ^ 2 3 - 11-2X10-4

^15 = ^ 2 3 — ^ 2 4 = -0-23 x IO"2

= 0-289

C17 — C25"~^4 C 2e = 0184

C18 = C2S = - 1 0 7

a2 = 32-7 x 10-4

a3 = 5-13xlO-2

C23 = ai — ha5

= 22-6xl0-2

2« = <**-hcsz - 0195

C25 = C33 ^2C34 = 0192

C26 = C34 = - 1 0 7

0

aK = 22-5 x 10-2

as = - 0 0 4

^33 = ο β — λ 0 ο 7

= - 1 0 6

^34 = 0 ? = - 1 0 7

0

aG = -1-10

a 7 = - 1 0 7

0

of which are known:

(a0P*+<ΊΡ*+<hPz +a3p2+aiP+as) (Tvp+1) +

+ (A3Ip2+à5l)(Typ + l)KOI+03vP2+4sv)Kov = 0,

where a0 = 0-135 x 10"2;

a! = 3-86 xlO-2;

02 = 0-240;

a3 = 0190;

a4 = 0-920;

a5 = 0080;

Δ31 = -0-0235;

J w = 0-285;

A3V = 00355;

zl5f, = 0-250.

Thè time constant of the voltage compensator is Tv—2 sec. By applying the method of domain separation which was used to solve


the previous problem, we get:

0·0054ωβ - 0·958ω4 + 3·44ω2 +0-920

167

Κον —

^ο/ —

-0·071ω2 +0-500

-0·0786ω4+0·620ω2-1·08 0·047ω2+0-570

The equation of the separate straight line is :

0080 +0-285*0, +0-250#OK = 0.

The domain separation curve D, constructed in accordance with the calculated results, is shown in Fig. 5.10. Taking into account that in the

FIG. 5.10.

absence of control excitation (Kov=Kor=0)9 the system must be stable (as=SEd>0)9 it may be concluded that the closed area, in which the points Kov=0 and Koi=0 are located, is the area of stability.


Problem 5.6

A station is connected to a system of infinite power through a transmission line.

Required: to choose the setting of a forced excitation control which is to react to variations of the first and second derivative of the line current.

Solution. Assuming that the characteristic equation of the system and its coefficients are known:

a0P* + «iP4+a«Pz + a*P2+Q\P+#5 +

+(4/>2+4) (K2lP2+KUP+*0/) = o.

The values of the coefficients are given in Table 5.3, where <5Z is the angle between the voltage vectors at the beginning of the line and the receiving system.

TABLE 5.3

Factors

*o «1 û ,

«3

0^=60°

0179 509

320 33-4

<5,=90°

0179 509

29-3 -27-4

Factors

ai

*5

4 A

<5,=<50°

160 8-4

- 4 0 2 97

(5,=90Ο

-207 -62-6 -7-26 134

The amplification factor for the current variation is chosen so that the free term of the characteristic equation is positive.

From

we have:

Assuming a margin:

Äo7=0-5.


In accordance with (5.19) we find the domain separation curve D in the form:

Ku = -

K2i — —

α0ω4 —a2ct)z+a4 -Αζω*+Α5

;

^ω 4 — αζω2+α5 (~ζ13ω2+ζ15)ω2""ω2

*r 0 / (5.23)

The calculated domain separation curves j9,in accordance with (5.23) are shown in Fig. 5.11. The check by Routh's criterion showed that the area which is bounded by the curve D> is the area of stability. The control must be set to operate inside the area common to all the conditions (in Fig. 5.11 it is shaded).

- i

- 2 L

pK.i

t

0-4

i Vv6°· 1 δρθθ·^

, 1 K2I 0* 1 1-2

FIG. 5.11.

Problem 5.7

The system of Fig. 5.12 consists of a synchronous machine operating on capacitive load. The parameters of the system are:

*Λ*=1'00; *ίΑί=0·79; ^ = 0 - 5 8 ; *ext=<H0; ;cc=0-80; R=002; 7 ^ = 1570 rad.

Ί Fio. 5.12.


Required: to ascertain, whether self-excitation of the synchronous machine arises with the given parameters, or whether the operating condition of the system is stable. The check is to be carried out by using the stability criteria of Routh, Mikhaylov and Hurwitz.

Solution. The characteristic equation of the system of equations, which describe the electromagnetic transient process in the given system, may be represented in the form:

a0P^+alpé+a2p^+anp1+aAp+a^ = 0, (5.24) where

<*2 = Tdç&XqX'a+Xcx'd+XqXc) + Td0 R2 + R(xQ+xd) ; I ( 5 2 5

3 = xcXd+xqXc+2xqXd+R2+TdoK(2xc+xq+x'd); I ^4 = T^(xe^xqXxe^x^ + T ^ + R(2xe+xq+x^); <*5 = (xc - Xd)(xc - xq) + * 2 · I

Here: xd = xdM + *ext> Xd — XdM +Xcxt>

Xq — XqM+Xexf

Substituting the parameters of the system in (5.25), we get:

a0=950; ^ = 50-3; α2 = 3870; a3 = 103; α 4 = - 1 6 · 3 ; α5 =0-0274.

(1) To check the stability by the Mikhaylov criterion, it is necessary to construct a curve (Mikhaylov hodograph). We get the equation of this hodograph by substituting jco for p in the characteristic equation (5.24) :

D(a>) = V{a>)+jU(a>), (5.26) where

V(co) = α4ω4-Λ3ω2+α5; U(co) = α0ω5 — α3ω3+α4ω.

Substituting the above values of the coefficients of the characteristic equation in (5.26) and using different values of ω from 0 to ~ we can design a hodograph (Fig. 5.13, where (a) is the beginning of the hodograph and (b) is the continuation).

According to the Mikhaylov criterion, the system is stable if the Mikhaylov hodograph, starting with co=0at the point on the positive real semi-axis, passes through n quadrants in succession, where n is the order of the


(a) (b)

F I G . 5.13.

TABLE 5.4

a, _ 950 ^ a, 50-3

= 18-9

- g * -0-0261

3 C13 C U

!824 = Ϊ 0 3 ^ 1 8 · 7

λβ - £ ϋ = -5-99

en = *o = 950

c12 = ai = 50-3

^i3 = <*ζ — λζα9

= 3870-18-9103 = 1824

<α* = α% — heiz = 103+0-261 X x 16-78 = 103-5

C15 = C23 — KCU = -16-78-18-7X x 00274 = -17-3

eie = c2i = 00274

c21 = a, = 3870

c« = Û3 = 103

C 2 3 = 0 4 — ^ 5

= -16-26-18-9 X x 0 0 2 7 4 = -16-78

c24 = e§ = 00278

0

-

c81 = aê

= -16-26

^3J = a% = 00274

0

-

-

-


characteristic equation. In this case, this condition is not fulfilled and, therefore, the system is not stable.

(2) In order to check the stability of the system by Routh's criterion, we calculate a table of coefficients (Table 5.4). It is seen from the table, that not all the coefficients in column cn-c i e are positive and, therefore, the system is not stable. There are two roots with positive real parts since, in the column c u =c i e , the sign of the coefficients changes twice.

(3) The Hurwitz criterion requires positive values of all the terms of the characteristic equation. In this case this condition is not satisfied, since a4 < 0 and, therefore the system is unstable. The synchronous machine will be self-exciting, i.e. the current and the voltage of the machine, when connected to the capacitance, will fluctuate spontaneously.

Problem 5.8

A system consists of a generator, connected through a line, with a longitudinal compensating capacity, to a system of infinite power (Fig. 5.14).

CH3DI « ? Fro. 5.14.

Required: to construct the boundary line of the stability area showing the dependence of the angle δ0 on the relative e.m.f. of the generator Eq0/Vc

(Κβ = 1·33). For this purpose, the characteristic equation of the system of equations,

which describes the electromechanical and the electromagnetic processes, is required. This takes the form:

<ΌΡ7 +a1pe+a2p5+a2pi+a4/>3 +asp2+a6p+a7 = 0, (5.27) where

α0 = 2·35χ107; ax = 1136x10'; α2 = 5·46χ107; α8 = 1·47χ107; α4 = 1·13χ107;

as = {ΐ2-1·04£2,+[2·78 cos (δ0-16·5°) + + 0-76 sin (ò0-16-5°)]^ +1-78 sin2 (δ0-16·5°)}χ IO3;

at = 2-5 X 103[0-856£·^, cos (δ0- 16·5°)+0·34Ε^ sin (δ0-16-5°) + +0-795 sin2 (Ò0-16-5°)];

α7 = 2·28£,ο cos (δ0-16-5°)-0-1216^ sin (<50-16-5°)--0-284 sin2 (ò0-16-5°).


Solution. The region of stability is determined by using Routh's criterion. The domain separation method is not applicable in this case, as the parameters E# and ò0 in the characteristic equation do not appear linearly.

8 - 4

80

40

- 4 0

-eo

-120

'CI2

-

4 1

8 l

12 1

16 2 o / 24 V

28

90

80

70

60

50

40

30

20

10

0

-10

-20

-30

-40

-50

-60

Λ

-

-

-

-^*

-

--

--L

FIG.

1

Fio.

5.15.

0

1

5.16.

Eq/Vc

2


Using Routh's criterion, it is possible to determine the limiting values of the angles ô0, for the various values of the excitation of the generator, at which the the system is stable. The Routh table for the characteristic equation of the seventh order, is given in Problem 5.4 (Table 5.2).

(1)2^=0. With δ0 = 16·5°, the coefficients αβ and a7 of the characteristic equation are zero. The coefficients cu-cl e are positive, and c17 and c18 are also zero.

It follows from this, that at this point of the diagram (δ0 ; E^9 the system is operating on the stability boundary line.

With <50^16·5°, αΊ is always less than zero, i.e. the system is unstable. (2) E^=0-5 Vc. The calculated results of the Routh table (column cu-c18)

for the given E^ and various values of angles <50 are given in Table 5.5.

TABLE 5.5

*0

*11 C12

Cl3 C14 Clo CIB cn Cm

26-5

2-35 X l 0 7

M 3 6 X 1 0 7

2-42 X 107

0-941 X 107

1124X10 7

1 1 3 9 x l 0 4

94 1-475

21-5

2-35 x lO 7

1136X10 7

2-42 x lO 7

0-941 x 107

1· 124 XlO7

1143 XlO4

- 6 0 1-501

16-5

2-35 x lO 7

1136X10 7

2-42 x lO 7

0-941 x 107

1124 x lO 7

1 1 4 6 x l 0 4

- 7 1 1-517

6-5

2-35 x 107

1136X10 7

2-42 x l O 7

0-941 x 107

1124 x lO 7

1-147 XlO4

- 1 0 4 1-487

1-5

2-35 x lO 7

1136X10 7

2-42 x lO 7

0-941 X 107

1· 124 XlO7

1-146 x lO 4

- 8 0 1-464

-8-5

2-35 XlO7

1136X10 7

2-42 x lO 7

0-941 x 107

1-124 XlO7

1-141 XlO4

58 1-355

It can be seen from Table 5.5, that the disturbance of stability is associated with the change in sign of the coefficient c17. By making c17 =/(<50), we can determine the limiting values of the angle <50. From Fig. 5.15 it follows that the system is stable with ό0>21·8°, and with ό0<—3°. In the range of angles between —3° and 21·8° the system is unstable.

Performing similar calculations for Εφ = 1-0; 1-35 and 1-88, we can construct the stability boundary line in coordinates δ0 and Εφ\νο (Fig. 5.16). The areas I are the stability areas; self-oscillation occurs in area II.

CHAPTER 6

STATIC STABILITY. POWER LIMITS AND LIMITS OF STABILITY

IN SIMPLE systems the limit of the power and the limit of static stability generally coincide. However, in more complicated systems, these two limits have to be considered separately.

In the present chapter, different methods of determining the transmitting power limit and the limit according to the static stability of the system with different assumptions, are discussed. In such problems, either the load calculation is made with constanti mpedance, which is much simpler and allows the use of the superimposition method, or the loads are calculated according to their static characteristics, which complicates the solution of the problem, since it requires a trial and error method. Besides that, the problems of determining static stability are classified into calculations, allowing for the excitation controls, where the controls themselves are represented by definite characteristic equations; and by calculations, in which the excitation regulators are ignored or allowed for approximately (by assuming a constant e.m.f., applied in front of one or other of the reactances which replace the generator).

Problem 6.1

In the system, of Fig. 6.1, the station G, which is provided with an automatic excitation regulator of a proportional type, is transmitting power over a transmission line of 220 kV, of 220 km length, into a system of considerably larger power than that of the station. The parameters of the transmission line and the initial operating condition are as follows :

JC;=0-461; JC^ =0-197; ^=0-7; ^^=0-142; Ρο=0·583;

cos φ =0-85; Kc = l.

Required: to examine the effect of the following factors on the margin of static stability :

175


(a) the power factor at the busbars of the receiving system is cos<p0; (b) the transient reactance of the transmitting station's generator is xd\ (c) the length of the line is /; (d) the division of the conductors is n (n is the number of conductors

per phase).

0-qp-{ [-CUM FIG. 6.1.

The excitation control of a proportional type is to be represented by a constant e.m.f. behind the transient reactance.

Solution. In the system under consideration, the limit of the transmitting power, which coincides, in this case, with the stability limit, may be determined from the formula

F'V ^ m = ^ , (6.1) KdE

where

* - My<+8tH^)'} Q0 = iYtan<p0 ;

t t xi ΧάΣ = xd~^XTL~^'~^"^'xT2 ·

Accordingly, the margin of static stability is:

k, = P]im~P° . (6.2)

Substituting the parameters of the system and the original condition into the formulae (6.1) and (6.2), we get

Q0 =0-583x0-62 =0-362; x'a = 0-461+0-197+0-350+0-142 = 115;

E' =V[(l+0-362xl-15)2+(0-583xl-15)2] = 1-57; 1-57 1-15 P*n=—< = 1-365;

1-365-0-583 , , . *' = 0-583 = 1 ' 3 4 ·

Static Stability 177

^^0^0^ Lead 1 t 1

0-7 08 · 0 9

20

16

1-2

0-8,

0-4

0

λ -

1-0 0-9

lag t

0-8 , C 0 * P ,

vo-7 ae

Fio. 6.2.

1-8 r

1-2 J -

1-0 h

0-81— _L 0-2 0-4

FIG. 6.3.

0-6 400 600

Fio. 6.4.

1-6 r k S î * n / * l

0-8 χ η/*Ι

0AL » » n

Fio. 6.5.

Results of similar calculations, made for different values of cos <p0, / and n, are given in Figs. 6.2-6.5.


Analysing the given curves, the following conclusions may be made: (1) An increase in reactive power as a result of the increase of the e.m.f.

of the generator, leads to an increase in the stability. (2) The lowering of the transient reactance increases the stability margin.

However, an increase in stability because of the lowering of xd is not feasible, as this leads to a considerable increase in the cost of the generator : it is possible to eliminate the effect of xd on the static stability by using automatic excitation control having a fast action.

(3) An increase in the length of the line leads to a decrease of the stability margin.

(4) A splitting of the line conductors of the transmission line leads to an increase in the stability margin, due to the decrease in the reactance of the line: in this case, the increase of the stability margin is insignificant, since the length of the lines, and, therefore, also the proportion of its reactance in the total reactance, is small.

Problem 6.2

A station is connected through a transmission line to a system of infinite power. The circuit and the parameters of the transmission line and the original operating condition are given in Problem 6.1. In addition it is given that xd=VS.

Required: to determine the static stability margin of the system in the following cases:

(a) without an automatic excitation control; (b) with an automatic excitation control of a proportional type; (c) with an automatic excitation control with a fast action.

Solution, (a) without the automatic excitation control, the limit of the transmitting power is determined from the condition for the constancy of the e.m.f. behind the synchronous reactance £'</=const.

The total reactance of the transmission line is :

The synchronous e.m.f. is:


The limit of the transmitting power is :

P t a - ^ = 0-96. χάΣ

The static stability margin is:

K = Plim~Fo = ° · 6 5 · (b) When an automatic excitation control of a proportional type is

used with the generators, the transmitting power and the stability may be determined approximately from the constancy of the e.m.f. behind the transient reactance E'= const.

The static stability margin is determined as in the previous problem, and is now &, = 1·34.

(c) Automatic excitation controls with fast action, depending on their setting, provide constancy of voltage either on the terminals of the generators, or at the beginning of the line. We determine the stability, assuming Vg=const.

The total reactance of the system is :

*c = *ri + -y + *Γ2 = 0#689.

The voltage at the terminals of the generators is:

The limit of the transmitting power is:

lim Pun, = - * - * - = 1-90.

The static stability margin is: P. — P

ks = ^Im r° = 2-26.

Comparing the calculated results, we can see that the automatic excitation control of a proportional type, compared with the case when the control is absent, increases the static stability margin by 0-69, and with automatic excitation control with fast action by another 0-92. The increase of the limit of the transmitting power is conditioned by the fact that the automatic excitation control completely (automatic excitation control with fast action) or partly (automatic excitation control of propor-


tional type) eliminates the effect of the internal reactance of the generators on the transmitting power limit according to the given assumptions relating to the static stability.

Problem 6.3

A station without automatic excitation control is connected to a receiving system by a transmission line (Fig. 6.6). At the beginning of the line there is a reactor for parallel compensation, which has a reactance χΓ=20·2.

The parameters of the transmission line and the original quantities are given in Problem 6.1. The synchronous e.m.f. of the generators, determined from a calculation of the electrical operating condition of the transmission line, is 2sd=2-49.

Required: to determine how the margin of static stability changes when the reactor is switched on.

Solution. The power limit of the transmission line according to the conditions of static stability after switching on the reactor is given by :

i t a = ^ c (6.3) where (Fig. 6.6)

1 x (χά+χτι) I ~2~+*r2 — = χ4+χτι + -τ-+χτ2 + 7 . (6.4) Λ12 L xr

Substituting the parameters of the system and other values in (6.3) and (6.4), we get:

J_ yi2

Hence the static stability margin is: A:, =0-68.

Thus, switching on the shunt reactor at the beginning of the line leads, in the given system, to an increase in the static stability margin by 3 per cent (see Problem 6.2). The increase in the margin is caused by the increase of the e.m.f., which is a result of the increased reactive load of the generator.

— =2-54 and ^ = 0 - 9 8 .


In the presence of an automatic excitation control of a proportional type on the generators, the positive effect of the increase of reactive load, due to the reactor, would have been smaller, and with an automatic excitation control with fast action, which maintains the constancy of the voltage at the beginning of the line, switching on the reactor would not have any effect on the limit of the transmitting power.

T2

i HH

FIG. 6.6.

Problem 6.4

A station without an automatic excitation control is connected to a system of infinite power by a transmission line.

The equivalent circuit of the system is given in Fig. 6.7. The parameters of the transmission line and the original condition are given in Problem 6.1. In addition, we are given: χβ1=χβ2

=^'^^ *β = 12·5.

EjJ&Jfr

FIG. 6.7.

The synchronous e.m.f. of the generators, which is determined from the electrical operating condition of the transmission line, is Ed=2· 17.

The mutual admittance is v12 = . /VL 2-42

Required: to determine the change of the static stability margin, taking into account the charging capacity of the line and the magnetizing current of the transformers.


Solution. The limit of the transmitting power can be determined by the formula (6.3)

2-17 pUm = —— = 0-897. 2-42 lim

The stability margin is:

ks=0-54.

Thus, the calculation of the load capacity of the line shows a reduction in the stability margin (see Problem 6.2). This can be explained in the following way. The total impedance of the branches magnetizing the transformers and the capacity of the line, is mainly capacitative (xc < χμ). The surplus of the generated capacitative reactive power (ßc—βμ) relieves the generators of reactive power; this leads to a decrease in the e.m.f. of the generator, and, therefore, to a reduction of the stability margin. The presence of automatic excitation controls on the generators would have reduced the effect of the line load capacity and the magnetizing power of the transformers on the limit of the transmitting power.

Problem 6.5

A station is connected with a receiving system of commensurable power by à transmission line (Fig. 6.8(a)).

(a) '

Ε,—■ -τψ— T _/ - ö v_ E , h 1"«· χ, "2

(b)

Fio. 6.8.

The parameters of the system and the original condition of the equivalent circuit (Fig. 6.8(b)) are:

^=2-49; x2 =0-506; Edl=2-39; Ed2=2-52; UL0 = l;

5,LO=3-66+y2-27; S10=0-583 +./Ό-362; 5,

20 = 3-08+yi-91.


Required: to determine the limit of the transmitting power when supplying: (a) a constant impedance load: (b) a load with static characteristics

PL and QL=f(VL).

Solution, (a) Replacing the load by a constant impedance we get: V2

ZL = -^ cos (φ^ +j sin φ^)

= J_(0-85-h/O-527) = 0-197 +y 0-122.

The effective power limit, in this case, may be determined by formula (5.10) (see Problem 5.2).

The self and mutual admittances are :

1 ~.JXI+4X*ZL

;Ό·506(0· 197+70-122) _ = j 2 4 9 V506+0-197+70-122 " ° - 1 1 6 + ^ 6 2 '

or — = 2-62; a u = 2-5°; yu — =jxl+jx2+J-%-*-112 ^ L

or — = 7-46; a12 = -38-2°.

Substituting the expression obtained for the impedance in (5.10), we determine the actual power limit as:

FUim = Λ * Ε ? S i n «11+^12^1^2

= | f siD2.5.+?^±2 = 0.900. 2-62 7-46

(b) To determine the limit of transmitting power, allowing for the load by its static characteristics, we form a characteristic P1 =f(VL) according to the method given in (Ref. 3, p 284).

The first point of this characteristic corresponds to the original operating condition of the system.


Calculation of the second point

We assume Ρ2<Αο> namely: P2 = 0·95Ρ20 = 0-95x3-08 = 2-92.

(1) We assume the internal reactive power of the second station:

e; = Q» = Q»+*Q» = ο,ο+^tì^2^ VLO

3·082 + 1·912 = 1 9 1 + ^ xO-506 = 8-56.

Then the voltage on the load busbars is:

According to the typical static load characteristics (appendix 2), we find with VL» = 0-97, that PL, = 0-99 and QL„ = 0-96. Whence we find the power consumed by the load is:

SL = 0-99 X 3-66+./Ό-96Χ 2-27 = 3-62+/2-18.

The power of the first station is

Pj. = PL-P2 = 3-62-2-92 = 0-70;

ßi = QL-(U'X-AQ'J

= 2 · 1 8 - ( 8 · 5 6 - 2 · 9 2 ; + 28

25 6 2 XO-506) - 0-14.

The electromotive force of the first station is

Ι Γ / η η , 0 ·14χ2·49\ 2 /0 ·70χ2·49\ 2 Ί „ „„

(2) We assume that:

Q'2 = 0 · 9 5 χ β ^ 0 = 8-13

Making similar calculations, we get

VL = 1-04; P1= 0-85; £ Λ = 2-92.


(3) We assume that: Q2 = 1-05 xQ'20 = 9-00.

Then VL = 0-90; Px = 0-53; Edl = 1*86.

Plotting the curves Edl =f(VL) and Λ =f(VL)(Fig. 6.9), for a given value Edl = 2*39, we find

VL =0-985; P± =0-73.

32

2-8

2-4

2-0

"08 0-9 10 1-1

FIG. 6.9.

The calculated results for successive points of the characteristic Px =f(VL), are given in Table 6.1.

Using the data given in Table 6.1, we plot a curve Pi =f(VL) (Fig. 6.10), whence we get Plim = 0-74.

TABLE 6.1

Number of the point

1 2 3 4 5

Λ

0-583 0-73 0-74 0-72 0-35

vL

10 0-985 0-95 0-90 0-79


Comparing the power limits, which were determined by different methods of calculating the load, we can see that representing the load by its static characteristics, leads to a lowering of the transmitting power limit in comparison with the case of the load being replaced by a constant

0-3 <K 0-5 0-6

FIG. 6.10.

0-7 P. i m OB

impedance. This is explained by the fact that, for a load, represented by

constant impedance, the controlling effe

larger than they are for the actual load.

àP dß constant impedance, the controlling effects — and — are considerably

Problem 6.6

A long transmission line with an intermediate substation, which contains synchronous condensers (Fig. 6.11), has the following parameters for the equivalent circuit (Fig. 6.12) and for the original condition:

Xi = 0-552; xc = 4; y12 = y22 = 1-81 ; y22 = y22 = 1-56; xT = 0-4, with Ax = xd = 7-93, yl3 = 1-01; yu = 0-067; with Ax = x'd = 1-20, y13 = 0-880; yu = 0-304; with Ax = 0, ylz = 585; yu = 0-805.

FIG. 6.11.

Static Stability

The original quantities are:

i>0 = 1-1; Vx = V2= V3= 1;

187

= 37-4°.

Required: to determine the limits of the transmitting power in accordance with the static stability with various methods of controlling the excitation of the synchronous condenser, namely:

(a) without any control; (b) with an automatic excitation control of a proportional type; (c) with an automatic excitation control with fast action, which main

tains constant voltage at the synchronous condenser terminals;

YI2 i Y22 Y23 » Y22

(d) as in (c), but taking into account the operations of a voltage compensating device, which maintains the voltage at a constant level at the point of connection of the synchronous condenser to the line.

The voltage at the beginning and at the end of the line is to be taken as a constant.

Solution. The limit of transmitting power, when controlling the excitation, in accordance with sections (a), (b) or (c) as specified in the problem, considering the condition of static stability, may be determined as the maximum of the power characteristic:

P = Jia^i v3 s i n 2ô+yuVxEx sin <5. (6.5)

Without excitation control Ex = Ed and for the calculation of the mutual admittances y13 and yu, the synchronous condenser is replaced by its synchronous reactance, i.e. Ax = xd.

Flo. 6.12.


The value Ed is determined from the condition that the reactive power balances at the junction point:

Osynxon. = 2Qi = 2(jfa V\ -y12 V, V2 cos <50) ; (6.6)

Ed= ^ + ( ^ + ^ ^ 2 5 1 . (6.7)

20

1-6

1-2

0-8

0-4

0

- 0 *

FIG. 6.13.

After calculating the e.m.f. Ed from the formulae (6.6) and (6.7), and substituting it in (6.5), we plot the power characteristic (curve 1, Fig. 6.13).

The limit of the transmitting power, without an automatic excitation control on the synchronous condenser is 1*16.

Making similar calculations with an automatic excitation control of proportional type {Ex = E' ; Ax = x'd) and with an automatic excitation control with fast action, without a voltage compensating device (Ex = Vs

yntCon ; Ax = 0), we determine that the power limits are, respectively, 1-20 and 1-28 (curves 2 and 3, Fig. 6.13).

With an automatic excitation control with fast action and a voltage compensating device, which maintains the voltage at a constant level at the point connection of the synchronous condenser to the line, the power characteristic may be constructed from the formula

P; dP/dÔ

P = A s f i s s i n o .


The maximum of this characteristic (curve 4, Fig. 6.13) occurs at δ = 90°. We determine the limit of the stability from the criterion of stability

dP — > 0. Differentiating (6.5) with respect to δ for constant Ex, (in this do case, equal to Ksynxon), we get:

^ - = 2γ13ν, V3 cos 2ô+yuVx ^yn.con. cos <5, (6.8)

where

V = K.+x.. syn,cop· 'syn.con. γ1 ^ΛΤ y

From the characteristic of the synchronous power (Fig. 6.13), we determine the limiting angle, in accordance with the condition of the static

àP stability, which corresponds to —- = 0, and the corresponding power,

do from curve 4. This power, which is equal to 1*56, is the limit of the transmitting power for static stability. When the synchronous condensers have an automatic excitation control with fast action the limit of the static stability is determined by the constancy of the voltage on the synchronous condenser terminals, and by the voltage compensating device which maintains a constant voltage level on the line at the point of connection to the synchronous condenser.

Comparing the limits of the transmitting power with various automatic excitation controls, we can see that the most effective method of controlling the excitation of the intermediate synchronous condensers is the automatic excitation control with fast action, with a compensating device.

The increase of the transmitting power limit compared with the limit without control is 35 per cent.

CHAPTER 7

LARGE OSCILLATIONS AND DYNAMIC STABILITY CURRENTS UNDER

OSCILLATING CONDITIONS

IN THE present chapter, transient processes with large operational disturbances are considered for conditions where the variation of the speed is small. The most general method of calculating the relative motion of the generator rotors which take place during these disturbances in an electrical power system, is the numerical integration of the differential equations. With the aid of the method of successive intervals, the non-linear equations of the transient processes may be solved in several ways. This method of calculation, which is used manually as well as when using analogue or digital calculating machines, allows us, with sufficient practical accuracy, to indicate the effect of various factors in the calculations (reactance of the armature, the dynamic load characteristics, damper torque and so forth). The calculations obtained, however, are very laborious and even very careful calculations do not guarantee against accumulating error from one interval to the next.

Besides, there are several physical factors in the calculations, for which the representation is insufficiently accurate, which leads to a certain inaccuracy in the results, regardless of the mathematical exactitude in performing the calculations. Of the number of such factors one may mention, for example, incorrectly calculated original conditions (angle <5, e.m.f. and others) and parameters of the system (not accounting for saturation, the absence of necessary data concerning the load, ignoring the impedance at the point of short-circuiting and similar matters).

The effect of all these factors increases substantially in cases where the operating condition of the system is close to the limit of stability and may even change the final result: the condition, estimated according to the calculation to be unstable, may be found, in reality, to be stable, and vice versa.

An exact analytical method of determining the effect of all the specified factors has not yet been found. In many cases, therefore, it is expedient to make use of some other simplified method of calculation, but, in addition,

190

Large Oscillations and Dynamic Stability Currents 191

to check the effect of some factors during the course of the transient process by changing, within small limits, these or other parameters.

In carrying out the calculations for checking dynamic stability during operation existing systems, if the initial data for calculation are known with the greatest possible degree in accuracy, the necessity for a preliminary stage of calculation with the aid of simplified methods, may be avoided. Even in this case, however, the simplified methods of calculation may find an application.

The simplified methods of calculation are especially effective when investigating the effect of particular physical factors on the course of the transient process. Also, the calculations are not usually carried out for any particular electrical system, but for general circuits with standard parameters for their components.

Problem 7.1

In the transmission system of Fig. 7.1, one of its parallel lines is suddenly switched off. The quantities in the original condition and the parameters of the transmission line with 5basic = 220 MVA and with basic voltage at the 220 kV level, Fbasic = 209 kV, are as follows:

PQ = U ôo = 0-2; Vc = l; ^ = 0-295; xT_± = 0-138; xT_2 = 0-122; xx = 0-244 (for the two circuits).

(Ξ>-Κ&£ ^ΣΗ ' 10-5/242 kV I = 230km 220/121 kV

ρο··°ο FIG. 7.1.

The inertia constant of the transmitting station, reduced to the basic conditions, is Tj = 8-18 sec.

Required: to calculate the dynamic transient and to plot a curve of δ' against time.

The calculation is to be carried out without allowing for the ohmic resistances and the line loading power. For the calculation, it is to be assumed that the transient e.m.f. remains constant.


Solution. We form an equivalent circuit for the normal condition and determine the e.m.f. of the generator behind the transient reactance.

The total impedance of the system is : xcl = Xj+xr^+xt+xj.^ = 0-295+0-138+0-244+0-122 = 0-799.

We determine the value and the phase of the transient e.m.f., behind the transient reactance, by formula (7.1):

W[( E' — F c + % £ l Y + /P0*dX 2"

Substituting the numerical values, we get:

/ΓΛ 0-2X0-799Y /1X0-799Y

0·799 tanôi = ^ ^ - = 0 - 6 8 8 ; 1-16

(7.1)

= V(M62x0-7992) = l-41;

b'0 = 34-5°.

The equivalent circuit of the transmission line is given (Fig. 7.2).

K=i H"0;2

FIG. 7.2.

The amplitude of the power characteristic for the normal condition Pml is determined from expression (7.2) with sin a' = 1 :

Λ F'TI ^ sino'. (7.2)

Substituting the numerical values, we get: 1-41 x 1

After switching off one circuit of the transmission line, the reactance xI is doubled and the total reactance of the transmission system will be:

JCCII = 0-295+0-138 + 2X0-244+0-122 = 1-043. The amplitude of the power characteristic with one circuit switched off,

is: 1-41x1 p -E*V<

»i l l M l 1-043

= 1-35.


The power characteristics for both cases are given in Fig. 7.3. After one circuit has been switched off, the electrical power output of

the generator (point a) is smaller than the power generated by the turbine, PT = P0.

vel.

Ι·β

1-6

1-4

1-2

10

0-8

0-6

0-4

0-2

θ0 6ίηοχ

FIG. 7.3.

Under the effect of the surplus power torque of the turbine, the rotor of the generator starts to accelerate and the angle <5' increases. At point b, the electrical and mechanical powers are equal, but the rotor of the generator, due to its inertia, passes this point and reaches the point c, where the kinetic energy, stored by the rotor from the accelerated motion from a to by is completely used up in the section be.

The relative movement of the rotor of the generator is determined by a differential equation of the second order:

T,^=P0-P, (7.3)

where P = the electromagnetic power supplied to the system by the generator.

We solve this equation by the most general method; the method of successive intervals.

We take the duration of the interval used for calculation At as 0-05 sec. The electrical power output of the generator at the first instant after

switching off one circuit is reduced to the value:

Λο) = Pmii s i n δΌ = ! · 3 5 s i n 34-5° = 0-764.

(-

L

/

/ e d/bJE w 'a

r^Fbrake \

Faccel. \ \ °

\ s


The surplus of power at the start of the first interval is :

AP(0) = P 0 -P ( 0 ) = 1-0-764 = 0-236.

The increase in the angle during the first interval of time is:

Jä;,,=Wf!iix^,Ä>, (7.4) u ( i ) Tj

where 360χ50χ0·052

cr k= 8Ί8 = 5 ' 5 ;

^ ( ' 0 = 5.5 i ^ = 0-65°.

The value of the angle at the end of the first or the beginning of the second interval is

δ'ω = δ'0+Δδ'ω = 34-5+0-65 = 35-15°.

The electrical power output of the generator at the beginning of the second interval is :

Λι) = Λ«π s i n δω = l'35 s i n 35'15° = °·778·

The surplus of power at the beginning of the second interval is:

AP(1) = Ρ0-Ρω = 1-0-778 = 0-222.

The increase of angle during the second interval is:

Δο{2) = Aò'w+k ΑΡω = 0-65 + 5-5x0-222 = 1-87°.

The value of angle at the end of the second interval is

<5('2) = δ'ω+Aò'w = 35-15 + 1-87 = 37-02°.

We then proceed with the calculation of the third and subsequent inte r vais. The results of the calculations are given in Table 7.1.

The variation of the angle with time is shown in Fig. 7.4. During the oscillation process, the maximum value that the angle reaches is 62-4°; the dynamic stability of the transmission line is not disturbed.


After a few oscillations with gradual attenuation of the amplitude, due to the presence frictional losses (ignored in the calculation), a new condition will be established, which corresponds to point b in Fig. 7.3.

TABLE 7.1

/, sec

0 005 010 015 0-20 0-25 0-30 0-35 0-40 0-45 0-50 0-55 0-60 0-65

<5', degrees

34-50 3515 3702 39-92 43-56 47-58 51-62 55-34 58-46 60-76 6209 62-36 61-56 59-74

P, rei. units.

0-764 0-778 0-813 0-865 0-930 0-997 1058 1109 1150 1176 1-192 1195 1185

ΔΡ, rei. units

0-236 0-222 0-187 0135 0070 0003

- 0 0 5 8 -0-109 -0-150 -0-176 -0-192 -0-195 -0-185

—

Αδ', degrees

0-65 1-87 2-90 3-64 402 4-04 3-72 3-12 2-30 1-33 0-27

-0-80 -1-82

—

degrees 60

50

40

30

20

10

0-1 0-2 0·3 W 05 W 0-7 sec

FIG. 7.4.

Problem 7.2

In the transmission system of Fig. 7.5, a two-phase short-circuit to earth takes place on one of the circuits. A power of Pc = 225 MW, cos φ0 = 0-9, is transmitted to a system through the transmission line. The voltage on the system busbars is Vc = 118 kV and is kept constant. The neutrals of the step-up and the step-down transformers are soundly earthed.

[V

J 1 ! I ! 1 I L


The parameters of the electric transmission elements are:

FIG. 7.5.

GENERATOR

Snom = 400 MVA; Vnom = 10-5 kV; cos cpQom = 0-85; xd = 0-235; x2 = 0-164; 7> = 7 sec.

TRANSFORMER T-l

5n o m = 360 MVA; ek = 12 per cent; kx = ^ .

TRANSFORMER T-2

Snom = 340 MVA; e* = 12 per cent; k2 = 220 Î2Ï"

TRANSMISSION LINE (TWO CIRCUITS)

/ = 225 km; xx = 0-4 ohm/km; x0 = 3χτ — 1-2 ohm/km.

LOAD

PL = 50 MW; cos<p = 0-85.

Required: to determine the time limit for switching off the short-circuit, by making approximate calculations, ignoring the aperiodic torque components of the ohmic resistances and the line load power. It is assumed that the transient e.m.f. remains constant in the disturbed condition.

Solution. The calculation is carried out in relative units. For the basic power and voltage we assume 5basic = 255 MVA and Fbasic = 215 kV on a 220 kV level. Then the voltage on the system busbars, in relative units will be:

Vc , 118 .220 basic

F * C " T ± : / : ! ! ~ 2 1 5 X 1 2 1 - 1 ·


The power, transmitted to the system is : P 225

P.c = -p- = χ ^ = 0-882; Q*c = P.e tan <pc = 0-882x0-483 = 0-427, òbasic Z D D

where cpc = arc cos 0-9 = 25-8°. The power consumed by the load is :

P 50 P#L=p-= =0-196; Q.L = PmL tan <pL =0-196x0-62 =0-122,

^basic ZDD

where yL = arc cos 0-85 = 31-8°. We reduce the parameters of all the elements to the basic values :

, , _ x. VÌA* (l Y _ Q-235 „ "-S2 X 255 / 248 \» . *"" ^ n o ^ a s i c U J ~ 400X215^10^5J - 0 1 9 9 '

. . . . 10·52χ255 /248Υ Λ , „ ^ 0 ' 1 6 4 * 400x215* ( l o i ) = 0 ' 1 3 9 ;

similarly,

100 Λ * Λ V*^ 100 Λ 360 Λ 2152 * * r - i — i n n X c X τ /2 ' — 1 Λ Λ Χ ^ Λ Χ 0 ι ς 2 —0-113; basic

similarly, 12 255 2202

Λ Λ Λ , **r-2 = τ τ^Χτ^Χ^τ τ -9 = 0-094: *Γ"2 ~~100Α340Α2152

1 S! 1 955 ^ / = γ ^ ι / ^ = γ Χ 0 · 4 χ 2 5 5 χ | ^ = 0-281;

ΓΦ/ = 7>=ss5L = 7 χ ^ = 10-98 sec. òbasic Z D : )

I Z 3 4 p' jO-199 w , J0II3 jO-281 . i0-094 v<=

P0 Lj 0-882* j 0-427

6-07+J3-78

FIG. 7.6.

In further calculations the parameters, reduced to the basic conditions, are used; the asterisks by the symbols are omitted for brevity.

We form an equivalent circuit for the normal condition and determine the e.m.f. of the generator behind the transient reactance (Fig. 7.6).

xc = jt2+x3+jc4 = 0-113+0-281+0-094 = 0-488.


The voltage on the busbars of the generator is given by formula (7.1): 0-427 x 0-488 \ 2 / 0-882 x 0-488 \2~

GO -M i+ V + 1 \ 1 = V(l-2092+0-4312) = 1-284;

tan ô c = - ^ i = 0-356; ôc = 19-6°.

We determine the load impedance from:

F 2 1-2842

ZL =-+{PL+jQÙ = ο·ΐ962+0·1222 (°·196+·/°·122) = 6-07+y3-78.

The reactive power loss in the circuit up to the busbars of the generator voltage is:

P2 + 0 2 0-8822 4-0-4272

AQC = ÌJL±*L xc = +, x0-488 = 0-469.

The output of the generating station is : $0 = SL + $c+jAQc = 0-196+70-122+0-882+y0-427+yO-469

= 1-078 +j 1-018.

The electromotive force behind the transient reactance of the generator according to formula (7.1) is:

ΙΓί*«ΒΛ 1·018χ0·199\2 / 1-078x0-199 VT K = 7L( 1 2 8 4 + Γ284 ) + ( 1-284 ) J

= V(l-4422+0-1672) = 1-452; 0-167 T442

The angle between e.m.f. E'Q and the voltage Vc is : δ0 = 19-6+6-6 =26-2°.

We determine the self and mutual admittances of the equivalent circuit for the normal condition (Fig. 7.6), by the method of transposition:

ZL - z 1 + # % =7o-i99+4°-488(6-07+y3-78)

tan (ô'0-ôc) = ^ 1 = 0-116; δ^-δ0 = 6-6°

Zc+Z5 J ' ./Ό-488+6-07+./3-78

= 0-027+70-669 = 0-670 ^87-7°;

^ - z T ^ o - ^ T T ^ 1 · 4 9 ^ - 8 7 · 7 0 ·


The complementary angle is α^ = 2-3°.

Z12 - Z1 + Z c + - ^ - _y0 199+70 488+ 6 . 0 7 + y . 3 . 7 8

= -0-012+/0-694 = 0-694^91°;

1 7*2 0-694^91° = 1-44 ^ - 9 1 ° .

The complementary angle is a 2 = — 1°. An equivalent circuit of the transmission line for a two-phase short-

circuit to earth, differs from the equivalent circuit of the normal condition, by the insertion of a fault impedance at the short-circuit point, (ref. 2, p. 221) the impedance of which is equal to the resultant of the parallel impedances (relative to the short-circuit point) of the negative and zero sequence impedances of the circuits.

The impedances of the elements of the transmission line for currents of negative sequence are given in Fig. 7.7; the load impedance is assumed equal to 0-35 of the positive sequence impedance (ref. 2, p. 200).

FIG. 7.7.

The resultant impedances of the circuit for current of negative sequence, relative to the short-circuit point are :

^2Σ '(*♦*&)] (Z3+Z4) ;

Z.+ zxz, * yo-m+i0·139^12^'1·32* 0-006+/Ό-248; '^Ζχ + Ζ% jyJ "*'~Γ./0·139+2·12+/1·32

Ζ3+Ζ* =yO-281+yO-094 =j 0-375;

Z., = (0-006 + ^ > * ^ = 0-003+70.149. ^2Σ 0-006 +/Ό-248 +y0-375


The resultant impedances of the circuit for currents of zero sequence, relative to the short-circuit point (Fig. 7.8) are:

Ζ0Σ = Z2//(Z3+Z4); Zz+Zt = 70-843 +J0O94 =yO-937;

y(H13x;0-937 Ζ0Σ — 70-113+7*0-937 = 70-101.

jO-113 /I jO-843 jO-094

FIG. 7.8.

The resultant impedance of the fault impedance at the short-circuit point is :

Zlr — Za — ZtyyZéi (0-003+70-149) Χ7Ό-ΙΟΙ

Ζ2Σ+Ζ0Σ~ 0-003+7*0-149+7*0-101 2Γ^0Γ = 7*0-06.

The equivalent circuit of the transmission line, for the two-phase short-circuit to earth, is given in Fig. 7.9. For this, we determine the self and mutual admittances by the unit current method.

^ ^ J ^ iO-281

6-07* j378 y jO-06

J0-09A /r^-|-^4vc

FIG. 7.9.

Let the current in impedance Z4 be equal to unity then : Λ = 4 = i +J0.

The voltage at point b is : Vb = (1 +/0)yO-375 =yo-375.


The current flowing across the fault impedance is :

Vb = 70-375 Ze 7ΌΌ6 / e = ^ = ^ ^ = 6-25.

The current in branch 2 is : /2 = / 4 + / 6 = 1+6-25 = 7-25.

The drop in voltage in the impedance of section 2 is : AV2 = /2Z2 = 7-25 XyO-113 =./0-819.

The voltage at point a is: Va = Vb +ΔΫ2 = jO-375 +/Ό-819 =71194.

The current in section 5 is :

The current in section 1 is: lx = / 2 + / 5 = 7-25 +0-088 +70-141 = 7-388+y0-141.

The drop in the voltage in the impedance of section 1 is: ΔΫΧ = Λ^ι = (7-388+70-141) Χ7Ό-199 = -0-028+71-460.

The electromotive force at the point of connection of the transmitting station is:

È = Ϋα+ΔΫχ =7ΐ·194+(-0·028+7ΐ·460) = -0-028 +7*2-654.

The self admittance of the short-circuited system in relation to the transmitting station is :

È = -0-028+7-2-654 = ( Μ χ ) 3 = 0 . 3 6 2 ^8 9 . 5o Λ 7-338 +/Ό· 141

ym — _ L — _ _ _ : = 2-763 <a —89-5° 11 Zl$ 0-362 89-5° Z °. W 3 '

The complementary angle is a™ = 0*5°. The mutual admittance of the short-circuited system is :

= È β -0-028+7-2-654 = _ 0 . 0 2 8 ·2 .6 5 4 = 2.654^90-6°; A i+yo

ym _ : 0-377 : — 90-6° *12 " 2-654^90-6° - U J / / ^ y u o -

202 Transient Phenomena in Power Systems: Problems

The complementary angle a™ = —0-6°. In the équivalant circuit, showing the condition after the fault has been

disconnected, the line impedance is doubled (Fig. 7.10). The self and mutual admittances for it are determined in the same manner as those for the normal condition of the circuit:

jO-679 jO-113 jO-562 . jO 094 .

U6<)7*j3 7e

Fio. 7.I0.

y» = 1083 ^ -86-1°; aft = 3-9°; Fß = 1021 ^ -91-1°; a'12 = -1-1°.

We determine the electrical power characteristics of the generators of the transmitting station for various conditions from the expression :

P = E?yxl sin a u +EVcyli sin (Ó' -a12). (7.5) Normal condition:

Pl = 1·4522χ 1-49xsin 2-3° + 1-452x1x1-44 sin (δ' + Γ) = 0· 126+2-09 sin (<5' + l°);

Plm = 0-126+209 = 2-216.

Fault condition (two-phase short-circuit to earth) : Pm = 1·4522Χ2·763 xsin 0-5° + l-452x 1 χ0·337 sin (ό' +0-6°)

= 0-051 +0-547 sin (δ'+0·6°); i>m = 0-051 +0-547 = 0-598.

Post-fault condition:

Pu = 1·4522χ 1-083xsin 3-9° +1-452x Ix 1-021 sin (ô' + M°) = 0-155 + 1-483 sin (<5' + 1·1°); Ρ£ = 0-155 + 1-483 = 1-638.

We determine the critical angle for switching off the short-circuit, from the condition of equality of the acceleration and braking areas (Fig. 7.11), by the formula:

cosò' - f o ( ^ - g ó ) + ^ c o s ^ - f f f cos 0'0 C O S °sw.off pll—plll * \''°)


In this case the angles are in radians. The critical angle (Fig. 7.11) is determined by the formula:

P 1 ·078 ó;r = 180-arc sin j £ = 180-arc s i n - j ^ - = 138-8°,

cos δ' ΛΓΓ = sw.off

1·078(138·8-26·2)τ^τ+1·638 cos 138·8°-0·598 cos 26-2° loi)

1-638-0-598

= 0-341;

^w.ofr = 70°.

180 degrees

In order to determine the time limit for switching off the short-circuit, it is necessary to form a function δ' =/(/)· The latter may be determined from the equation of motion of the rotor (7.3). We solve this equation by the method of successive intervals (numerical integration). The duration of the intervals At we take as equal to 0-05 sec. The coefficient k in the equation (7.4) is:

360/zl/2 360χ50χ0·052

k = ■=, = ττττ^ = 4·1. Tj 10-98

30 60 90

FIG. 7.11.

t20 150


First interval (0-0-05 sec)

The output of the generators at the first instant after the short-circuit occurs, using formula (7.5), is:

P(0) = 1·4522χ2-763 sin 0-5° + 1-452x1x0-377 sin (26-2° +0-6°) = 0-051+0-547x0-451 =0-297.

The surplus power at the start of the interval is :

APm = P0-Pi0) = 1Ό78-0-297 = 0-781.

The increase of angle during the interval is : _ kAP(0) 4-1x0-781 _

ΔΟω - — 2 — " = 2 ~ '

The angle at the end of the interval is : δ'ω = ο'0+Δο'ω = 26-2 + 1-6 = 27-8°.

Second interval (0Ό5-0Ί sec) Ρω = 0-051+0-547 sin (27-8°+0-6°) = 0-311;

ΔΡω = 1-078-0-311 =0-767; Δδ{2) = 1-6+4-1x0-767 = 1-6 + 3-14 = 4-74°;

<5('2) =27-8+4-74 = 32-54°. We proceed with the calculation for the third and subsequent intervals.

The results of the calculations are given in Table 7.2.

TABLE 7.2

/, sec

0 005 010 015 0-20 0-25 0-30

<$', degrees

26-20 27-80 32-54 40-26 50-73 63-66 78-78

P, rei. units

0-297 0-311 0-350 0-409 0-478 0-543 —

zlP, rei. units

0-781 0-767 0-728 0-669 0-600 0-535 —

Δδ', degrees

1-60 4-74 7-72 10-47 12-93 1512 —

Using the calculated results, we plot a curve ò' =f(t) (Fig. 7.12). Knowing that <5s'woff

0-27 sec. = 70°, from the curve, we determine: /s,


The switching off time limit may also be determined from the nomogram (see Appendix 4). In plotting the nomogram, the transformed equation of the rotor motion is used (7.3):

dr2

where,

-a +sin <5' = T,

-•'M**} T — °

In our example „ 1-078

0-598 = 1-8; ò'0 = 26-2°; sin δ'0 = 0-44;

degrees Γδ'

δ. sw.ofF = 70°.

01 0-2 0-3 sec

FIG. 7.12.

From the nomogram No. 5 of Appendix 4, with T= 1-8 and <5swoff=70°, we determine: r = 1-15.

Substituting the numerical value for r, we find the switching off time limit is:

1-15 = /, sw.off ^ - x „ , , 8 ) , whence tSVf off = 0-275 sec.

Problem 7.3

In the circuit, shown in Fig. 7.13, a two-phase short-circuit at the beginning of one of the lines takes place at the instant t0. During an interval At, at time tl9 the two-phase short-circuit changes to three-phase, and then at the instant t2 the faulty line is disconnected. The system during this time remains stable.


Required: to determine the maximum angle to which the rotor will deflect during oscillations, and to find the margin of the system stability.

Solution. We construct a diagram showing the acceleration and braking areas. We also determine that, with a two-phase short-circuit, the output of the generator falls to a value corresponding to point 2 on the characteristic III (Fig. 7.14). Due to the effect of the surplus torque AM0 ^ AP0, the rotor of the generator accelerates.

FIG. 7.13.

FIG. 7.14.

At the instant rx (corresponding to angle δ[) with a three-phase short-circuit, the output of the generator drops to zero. Due to the effect of the full excess torque, which is equal to the torque of the turbine, the rotor continues to accelerate.

At the instant t2 (corresponding to angle ό£), after the damaged line is cut off, the output of the generator increases to a value which is determined by point 7 on the characteristic after the fault condition II. Here the


power output of the generator is greater than the power developed by the turbine, the generator brakes, but the angle <5' continues to increase corresponding to the energy stored by the generator/The increase of rotor angle continues up to point 8 (angle ô'max), where the kinetic energy, stored by the rotor in the process of acceleration, is fully used for braking. This corresponds to the equality of the acceleration and braking areas (F2LCCcL = FbTake). Then the angle ό' starts to decrease and after several oscillation cycles of the rotor, a new condition will be established, which is determined by point 10 on the characteristic after the fault condition II.

The ratio of the possible braking area 6-7-8-9-6 to the acceleration area 1-2-3-4-5-6-1 gives the margin of stability.

Problem 7.4

In the circuit (Fig. 7.15) a three-phase short-circuit at the end of one of the lines takes place at the instant tQ. During an interval At, at time tl9 a load resistance RL is connected to the busbars of the generator, and then, at the instant t2, the defective line and the load resistance are cut off simultaneously; the system, during this time, remains stable.

Required: to determine the maximum value of angle <5, in the process of oscillation.

FIG. 7.15.

Solution. We solve the problem in a general manner, by forming the areas of acceleration and breaking (Fig. 7.16). With the three-phase short circuit, the output of the generator drops to zero. Due to the surplus torque AM^P0 = P r , the rotor of the generator starts to accelerate. At the instant tx (corresponding to δ[\ a load resistance RL is connected to the voltage of the generator, which increases the electrical power output of the generator (characteristic III' in Fig. 7.16). After the defective circuit and the loading resistance are cut off simultaneously, the output of the generator changes to the values determined by the point a on the charac-


teristic II (Fig. 7.16). At the same time the rotor of the generator starts to brake. The increase of the relative angle of the rotor of the generator is continued because of the inertia. The maximum relative angle ôf

max of the rotor of the generator is determined by the condition of the equality of the acceleration and braking areas.

FIG. 7.16.

Problem 7.5

In a circuit (Fig. 7.17) a three-phase short-circuit on one of the lines takes place at instant t0. During a time interval At, at the instant tl9 a relay suddenly halves the inlet of steam into the turbine, as a result of which the power developed by the turbine is halved. At the instant t2, the defective line is cut off, and then, at the instant f3, the initial power of the turbine is rapidly restored.

Required: to determine the maximum angle during oscillation and the safety margin.

Solution. We work out the solution approximately, by constructing the acceleration and braking areas for the three-phase short-circuit on one of the lines. In this particular case, the output of the generator is not reduced to zero, since an electrical link is maintained between the generator and the system.


From the equivalent circuit of the system (Fig. 7.18(a) and (b)), it is clear that after transforming the reactances AXl9 Xx and Xl—AXl from delta to star, the impedance in the branch of the star Xk is not equal to zero.

The electrical output of the generator at the first instant after the short-circuit, is determined by point 2 on the fault condition characteristic (Fig. 7.19,111).

Starting with the instant i0, an accelerating force acts on the rotor of the generator, and the angle δ' increases.

At the instant tx (δ[ in Fig.7.19), the power, developed by the turbine is halved, which in turn reduces the accelerating torque acting on the rotor of the generator. At the instant t2 (à'2 in Fig. 7.19), the defective line is cut off, the output of the generator increases and the surplus torque starts to brake the rotor of the machine.

emessa Po

FIG. 7.17.

FIG. 7.18. FIG. 7.19.

The maximum displacement of the angle of the rotor of the generator max ls determined by the condition for equality of the acceleration

(1-2-3-4-5-6-7-1) and braking (5-8-9-10-11-12-5) areas; The safety margin we find as a ratio of the area 5-8-9-10-11-12-5 to the area 1-2-3-4-5-6-7-1.


Problem 7.6

In a circuit (Fig. 7.20) both lines, which link the generating station with busbars of constant voltage, are cut off suddenly. The parameters of the circuit and the operating quantities are given in Fig. 7.20.

(S-kiD-fc H 1

Tj=7s«c Po=°·5"

FIG. 7.20.

Required: to determine the maximum permissible time of interruption of the supply of energy for which the dynamic stability of the system will not be disturbed.

Solution. We investigate the problem by using the method of areas. The maximum power, which can be transmitted in the normal condition, is:

_KVC_ 1-55x1 m " X12 " 1-15 ~ l

With the lines switched off, it is :

/>»< = 0.

The critical angle, which is permissible during the oscillations of the generator after the lines are switched on (Fig. 7.21), is:

P 0-584 δ' = 180 -a rc s in^ = 180-arc sin-^—- = 154-4°. cr PI 1-35 m

By using the method of areas, we determine the limiting switching angle for the system:

C O S ^sw.oo = pn r m

0·584(154·4-25·6)-^ + 1·35 cos 154-4° - j ! » 0-0665,

« : . „ = 8«·2·.


In order to determine the time limit for switching on, we integrate the equation of motion of the rotor. In the result we get:

W-K)TA t =

^ 3 6 0 / (7.7)

AP

Substituting the numerical values, we get tswon: •2-25-6)7 /Γ (86-2-25-6)7 Ί _

^|_0·584χ180χ5θ] ~ 0-284 sec.

Problem 7.7

In the system, the circuit of which is given (Fig. 7.22(a)), a reactance xr is inserted in order to improve the dynamic stability with a short-circuit in the neutral of a transformer.

Required: to investigate the effect of xr on the value of the limiting angle for removing the short-circuit.

Solution. The reactance in the neutral of the transformers changes the resultant zero sequence impedance, affecting the dynamic stability of the transmission line only for unsymmetrical short-circuits to earth. With

FIG. 7.21.


the increase in the reactance, inserted in the neutral of the transformers, the resulting zero sequence reactance increases (Fig. 7.22(b):

χ = (*Γ-Ι+3ΧΓ) (*/Ο+*Γ-2)

The increase of χ0Σ leads to an increase of the short-circuit impedance, and with it, an increase in the amplitude of the power characteristic for the fault condition.

In Fig. 7.23, the power characteristics and the corresponding areas of acceleration and breaking are shown for the cases, where the neutrals of the transformers are soundly earthed (characteristic III, Fig. 7.23) and when the neutrals of the transformers are connected through a reactance xr (characteristic ΙΙΓ, Fig. 7.23).

XT! / XIÔ XT2

^Γ J ΊΓ ΟΓ η

1 (α) "~" (b)

FIG. 7.22.

Ό °s^offl°sw.oft2 °cr

FIG. 7.23.

From Fig. 7.23 it is clear that inserting a reactance in the neutrals of the transformers increases the limiting angle for removing the short-circuit from <5s'w.offl to <Cofî2 .

€>-<TO Q T P *


Problem 7.8

Figure 7.22 (a) shows the circuit of a transmission system.

Required: to examine the effect on the stability of inserting an ohmic resistance into the neutral of the transformers during a single-phase short-circuit.

Solution. The short-circuit to earth leads to losses of power in the ohmic resistances, which increase the load of the generators, and to a greater or lesser degree compensates the loss of power, transmitted to the receiving system. If the ohmic resistances are small, the power loss, which loads the generator, increases nearly proportionally to the value of the resistance and the dynamic stability increases rapidly (ref. 3, p. 436).

The increased loss, however, proves to be favourable only up to a certain limit. If the losses are too large, they cause a worsening of the dynamic stability.

'0 C9*dH °max ömin ös*off Ö0 (a) (b)

FIG. 7.24.

We study the problem in a general form, using the method of areas and constructing (Fig. 7.24) power characteristics for a single-phase short-circuit close to the transmitting station.

The characteristics, (Fig. 7.24(a)), refer to the case, when the ohmic resistance in the neutral of the transformer improves the dynamic stability.


In Fig. 7.24(b) we give an example, where the ohmic resistance lowers the dynamic stability. In the present case, when the original condition is disturbed, the rotor of the generator, due to excessive losses, starts to brake. After switching off the short-circuit (òswoff in Fig. 7.24(b)), the angle δ', due to the inertia of the rotor, continues to fall, till it reaches the value δ^. The angle ô'min is determined from the condition for equality in the areas of braking Fbrakel and acceleration F&cccll. In the second half-period of the oscillation, because the acceleration area - accei. 2 is larger than the possible braking area Fbràktt9 the parallel operation of the generator and the system will be disturbed.

Problem 7.9

A generator is switched on in parallel to a large electrical system by a precise synchronization method.

Required: to determine the permissible deflection in the angular velocity of rotation of the generator, when switched on according to the condition of synchronization during the first period of oscillation. The power loss at no load are taken to be zero.

Solution. We use the method of areas and examine the problem in a general case. We recall, that in switching on a synchronous machine in parallel, the additional supply of kinetic energy to the rotor, due to the relative motion, may be determined, approximately, from the expression (ref. 3, p. 108):

where ζ1ω=ω—ω0 is the difference between the angular velocity of rotation of the machine to be connected and the angular velocity of the electrical system.

Since the synchronous machine is excited, at the first instant after switching on, it starts either to supply (generating condition) or to consume (motoring condition) active power. The power characteristic of the synchronous machine is shown in Fig. 7.25. To obtain the relation between the angular velocity of the generator and its switching-on angle, we examine the four possible cases:

(a) Δω > 0; 0 < b[mM < 180° (Fig. 7.25(a)).


With switching-on angles 0-180°, the synchronous machine at the instant of switching on starts to operate as a generator.

For the switching on to be successful, it is necessary that the braking energy, proportional to the area Fbrake, is larger than the kinetic energy stored by the rotor due to the relative motion, i.e.

A - TjA(°2 ■ Ά brake -~ 2 '

Λ180 ^rake = Λη Sin *' d o ' = P m{\ +COS «' V

or

Ìleo

7>ζ1ω2

/»m(l+cos5;won)

from which we determine:

to=l("mil+™*^\ (b) Δω > 0; -180° < Con < 0° (Fig. 7.25(b)).

With switching-on angles from —180° to 0°, the synchronous machine at the instant of switching on, starts to operate as a motor with input:

P = Pm sin Ò[WM.

The input of the synchronous machine is reduced as the rotor accelerates. Thus, for successful synchronization, the braking energy, determined by the braking area Fbrake, must be larger than the sum of the kinetic energy of the rotor of the synchronous machine before switching-on and the energy, determined by the acceleration area Facccl.

The relation between the switching-on angle and the limiting angular velocity of the rotor, may be determined from the expression:

Δω2

2 Tj —^ l· ^accel. ^ ^brake ·

(c) Δω - 0 ; 0° < Con < 180° (Rg- 7.25(c)).

With switching-on angles 0—180°, the synchronous machine at the instant of switching on starts to operate as a generator with an output of active power P=Pm Sin Con-


PP

~F br

akc

~W>0

~w>

0

180)&~

won)()

0>8'

>-I8

QO

sw.o

n

-180

&~on

180

(a)

~(A)

~o~W

(O

180>

8'>0

O>S'

)-18

00sw

.on

swon &~w

.on&'

-180

-18

018

0

F ace

tl.

(e)

(d)

Flo.

7.2S

.


The relation between the switching-on angle and the limiting angular velocity of the rotor, may be determined from the expression:

Αω* *accel. ·

(d) Δω < 0; -180° -= ó;w.on ^ 0° (Fig. 7.25(d)). With switching-on angles from —180° to 0°, the synchronous machine

at the instant of switching on starts to operate as a motor and consumes active power P=Pm sin 3a'w#on.

The relation between the switching-on angle and the limiting angular velocity of the rotor, may be determined from the expression :

_ Δω*

7. 2-2

2-0

1-8

1-6

!·4

1-2

10

0-8

06

04

0-2

"S

" area of successful '3V . synchronization ^ \

I I 1 t ■ I i J 1 Λ t 20 «0 60 80 100 «0 KO 160 180 degrees

FIG. 7.26.

As an example, we examine the case of connecting the synchronous machine with Δω > 0. The switching-on angle varies from 0 to 180° (case a).

The parameters of the transmission line, in relative units are:

JCC=2-49; Ve = l; E' = l; r , = 10sec. The maximum of the power-angle characteristic of the synchronous

machine is :

J> = E'VC _ l x l

xc " 2-49 = 0-402.


Witt C = 0' we get:

The permissible slip of the synchronous machine 5=4-2-3 %. For other switching-on angles we have

Kw.om 45° 90°

135° 180°

s, percent 2-1 1-6 10 0

In Fig. 7.26 is given a curve of the permissible slip as a function of the switching-on angle.

Problem 7.10

In the transmission system (Fig. 7.27), a three-phase short-circuit, occurs at the middle of line at the instant t0. The operation of the switches takes place in the following order:

*! — switch Bx on; tt — switch B2 off; tz — defective line switched off; f4 — switch B2 on; /5 — switch B4 off.

Bt

FIG. 7.27.

The system remains stable and does not reach the critical angle.

Required: to determine the maximum angle of deflection of the rotor during the oscillations.


Solution. The short-circuit leads to a considerable reduction in the electromagnetic power given out by the station to the system. The equivalent circuit of the transmission system at the instant of short-circuit is given (Fig. 7.28(a)).

By switching on Bx (f=/i), the capacitance is shunted. This leads to an increase of the electromagnetic power given out to the system, since the mutual admittance between the transmitting station and the system increases. The equivalent circuit of the transmission system is given (Fig. 7.28(b)).

(o)

1 (b)

(0

L i h - ^ u ^

(d)

t«t

(«) (f)

FIG. 7.28.

At the instant t2, the switch B2 is opened and an ohmic resistance is thereby connected to the voltage of the generator in series with the shunting reactor. In addition the power loss in the ohmic resistance loads


the generator. The equivalent circuit in this case is given (Fig. 7.28(c)). At the instant /3, the defective line together with the short-circuit is switched. off. The equivalent circuit with the line cut off, is given (Fig. 7.28(d)).

At the instant f4, the shunting reactor is short-circuited by switch 2?3. Consequently , the loss in the ohmic resistance increases. The equivalent circuit in this case, also, is given (Fig. 7.28(e)).

And finally, at the instant i5, the ohmic resistance is cut off by the switch. 2?4 (the equivalent circuit is·given in Fig. 7.28(f)).

In Fig. 7.29, the power-angle characteristics and the acceleration and braking areas are given for the instants considered.

FIG. 7.29.

Problem 7.11

In an electrical system (Fig. 7.30) a sustained three-phase short-circuit develops on the load busbars.

Required: to determine in what conditions the system remains stable in the presence of the short-circuit.

Large Oscillations and Dynamic Stability Cnrrents 221

Solution· For any generator the equation of motion of the rotor may be written as :

d23' di2 Tj-^ = P0-Pmsmd'.

With a three-phase short-circuit on the load busbars, the electrical output of each of the generators decreases to zero. The equation of motion, in this case, takes the form:

τ d2'd'

Integrating this equation twice, we get:

dt - Tjt+Cl

Prior to the short-circuit (f =0) we have:

(SL- 0 or ò'=d'0.

3-2

f FIG. 7.30.

Substituting the velocity and angle at the instant, f0, in the previous equations, we find the constants of integration Cx and C2:

Cx=0; C2=à'0. The variation of angle with time is given by the equation :

If the angle is expressed in degrees, and the time in seconds, the last expression may be written as:

a ' = i ^ ß L , 2 + <5;. (7.8)


Similar equations may be obtained for each of the three stations in the system considered.

For the system under consideration to be stable with a sustained three-phase short-circuit on the load busbars, it is necessary that the relative angles between the e.m.f.s of the stations remain constant, i.e. δ'12 = const and ^3= const.

We write down the expressions for δ'12 and δ'13:

ò'lt = δ[-δ'2 = i S O / ^ ^ - ^ W i o - o ; , ) ;

δ'η = δ[-δ3 = ΐ 8 0 / ί * ^ - ^ + ( ο ; ο - 3 · 0 ) .

From these expressions it follows that the angles δ'12 and δ[ζ are constant if:

* 10 = * 20 __ *30

^/l ^72 ^/3

Here, the powers and the inertia constants are referred to the basic values. If the powers and the inertia constants are to be referred to the nominal powers of the generators, the condition for maintaining stability can be written in the form:

* 10 * 20 ^30

^Vl nom 1 TJ2Snom 2 Λ/3^ηοπι 3

Problem 7.12

Θ4<2>£ » 20

FIG. 7.31.

In a circuit (Fig. 7.31), a three-phase short-circuit occurs at the load busbars. The inertia constants of the generators G-1 and G-2, reduced to the basic power, are equal to 3 and 16 sec respectively.

The values of the e.m.f. and the power, supplied by the generators in the original condition, are given on the diagram.


Required: to determine the variation with time of the absolute and the relative angles of deviation of the e.m.f.

Solution. The variation of the absolute angles with time, are found from formula (7.8):

^ = 1 8 0 x 53 ° X ( H W 4 7 - 5 ° = 1β0*»+47·5·.

= 180x50x3 ί 1 + 3 7 . 5 ο = 1 6 9 0 ί2+37·5ο. 2 16

The variation of the relative angle with time is :

b'12 = 1800/2+47·5-(1690ί2+37·5) = 110/2 + 10°.

The curve of the variation of the angle b'u = f{t) is given in Fig. 7.32.

degrees -*/* 100 J-

1 80 L

60

40

20

0 L_

0-2

FIG.

*s.c. _J i 1 0-4 0 6 08 sec

7.32.

Problem 7.13

In a transmission system (Fig 7.33), one of two parallel lines is switched off suddenly. The parameters of the transmission elements and the operating quantities prior to switching off the line, are given on the diagram.

Required: to determine (approximately) the manner of variation the angle with time.

Solution. We carry out the calculation in a simplified manner, by replacing the upward branch of the power characteristic II of the condition after the fault, given by equation (7.3), by a straight line passing through


the origin and the point of intersection of the characteristic II with the straight line PT = P0 (Fig. 7.34).

The differential equation of motion of the motor can be written as: d*ó' di2

where

Tj^=P.-aà\

a = tan a.

GH<g>-£ %m<< TjzlOscc

(a) P0<Q0

χ^0·295 χΤ|=0·138 χ,=0·244 x^O-122

o — - ^ ^ - j φ } φ 1 φ }vc=l (b)

FIG. 7.33.

UjO-2

6'1 $'u

FIG. 7.34.

The roots of the characteristic equation are:

The solution of the complete equation takes the form:

Ò' = ^ . + C1e/,li + C2eP2f. a


The constants of integration Cx and C2 are determined from the initial conditions.

With 1=0, ό' = ό'ΐ and ^ = 0 . at

The constants of integration Cx and C2 for these conditions have the values:

Ça — C2 — ~~"2~'

rad

10

0-9

0-8

0-7

0-6

0

*-*·

-

1 50

1 1 1 100 150 200

Fro. 7.35.

\ 8'n

1 . t , 250 300 Ä C

where

Therefore, ΔΟ' = Ô'11 - 6'1.

P Ah1 P a 2 a

or

( * · )■

δ' =^—AÒ'cos a

In order to determine the unknowns Δδ' and a, which appear in this expression, we first determine the amplitudes of the power characteristics I and II:

PL = 1-41 x 1 E'Ve χΣ 0-295+0-138+0-244+0-122

1-41x1

= 1-765;

Pm 0-295+0-138+0-488+0-122 = 1-35;

sin Ò'1 = 4 r = PL 1-765 = 0-567; Ò'1 = 34-53° = 0-602 rad;


sin ό'" = j-L = 0-741 ; ό'" = 47-85° = 0-835 rad;

Δο' = ο'π_δ'ΐ = 0-835-0-602 = 0-233 rad;

α = tan« = ^ = ^ = 1 - 2 .

The variation of the angle with time is :

à' = τ^-0-233 cos l(—l^—\ t = 0-835-0-233 cos 0-0195*. 1-2 ^1314x10 I The curve of the variation of the angle with time is given in Fig. 7.35.

Problem 7.14

The transmission system, the circuit of which is shown in Fig. 7.36, supplies power Sc = 0-882 +/Ό-427 to the system. The voltage on the busbars of the system is Vc = 1. The neutrals of the transformers are soundly earthed.

TPL.QL r pc-0c

FIG. 7.36.

The parameters of the elements of the system, reduced to the basic conditions (Sbaaic = 255 MVA; Kbasic = 215 kV) are:

GENERATOR

xd = 1-188; xq = 0-679; x'd = 0-199; x2 = 0-139; Tj = 10-98 sec; Td0 = 5 sec.

TRANSFORMERS

jcr_! = 0-113; xr__2 = 0-094.

LINE

xx = 0-281; x10 = 3xx = 0-843.

LOAD

PL = 0-196; QL = 0-122; ZL = 6-07+y3-78; ZL2 = 0-35ZL = 2-12+yl-32.


Required: to determine the limiting switching time for a two-phase short-circuit to earth at the beginning of one of the transmission lines.

The calculations are carried out by allowing for the reaction of the armature and the action of the excitation regulators (forcing excitation). The generators are of the salient pole type. The variation of the e.m.f. of the generators, taking into account the action of the excitation regulators, is to be taken to be exponential. The ceiling of the excitation voltage is 2-5, and the excitation time constant of the exciter is Te=0-35 sec.

The ohmic resistances of the elements of the transmission line and the charging capacity of the line, are to be ignored.

Solution. We set up an equivalent circuit for the normal condition and determine the e.m.f. of the generator.

For a machine with salient poles, because of the inequality of the direct and the quadrature synchronous resistances, it is not possible to set up an equivalent circuit. To determine the currents and the power we replace the actual machine by an ideal one with equal reactances (xq) in the direct and transverse axes, and a fictitious e.m.f. Eq, which coincides in direction with Ed (ref. 3, p. 139). The equivalent circuit of the transmission system is given in Fig. 7.37.

jO-679 jO-113 . jO-281 ι jO-094 j

O — ^ \ - ^ _ ^ J \ - ^ - ^ - ^ ^ 1

\ 6-07+J3-78

FIG. 7.37.

The total reactance between the busbars of the generator and the busbars of infinite capacity is :

xc = *2+x3+;c4 = 0-113+0-281+0-094 = 0-488.

The voltage on the busbars of the generator according to formula (7.1) i s :

/Γ/ 0-427 x 0-488 \ 2 /0-882 +0-488 \ 2 1 y"'A{+—«—)+\—i—)J

= V(l-2092+0-4312) = ! · 2 8 4 :

tan òt = ^ ^ = 0-356; ôc= 19-6 .


The reactive power losses in the circuit up to the generator terminals

Jfc = ? ψ χ , . 0 · 882'+0.427»χ Μ 8 8 _ 0 4 6 9

The output of the generating station is: $0 = $L+$e+jAQc = 0-196+/Ό-122+0-882+/Ό-427+/Ό-469

= 1-078+71-018. The fictitious e.m.f. behind the transverse reactance of the generator

(Fig. 7.38) is: /I7, ΛΟΧ , 1-018x0-679 V , /l-078X0-679 V I E«= A\ 1284 ) ( 1284 J J

= V(l-8222+0-572) = 1-909;

tan (ô0-ôc) = - ° ^ = 0-313; ô0-èc = 17-4°.

The angle between the e.m.f. Eq{Ed) and the voltage of the system Vs is: 60 = 19-6 + 17-4 = 37°.

We determine the transient e.m.f. of the generator E'0 to be: /ΓΛ ΛΟΛ 1-018x0-199 Y /1-078x0-199\2Ί

*° = A\ 1284 ) ( 1-284 ) J = ν(1·4422+0·1672) = 1-452;

t a n ( ó ; - ó c ) = - ° ^ - =0116; ^ —ôc = 6-6°; δ'0 = 6-6 + 19-6 = 26-2°.

The projection of the transient e.m.f. on the direct axis is: E'M = E^cos(Ô0-Ô'0) = 1-452 cos (37°-26-2°) = 1-426.

The e.m.f. behind the synchronous reactance along the direct axis, is determined by

£d=Eq^^-Ed^L. (7.9)

Substituting the numerical values :

EA - ! 909 1188-0199 1-188-0-679 _ hda - l W 9 0-679-0-199 ! 42t> 0-679 - 0-199 λ *12"

We determine, by the unit current method, the self and mutual admittances for the equivalent circuit of the fault condition, for which we find


the total negative and zero sequence impedances relative to the short-circuit point.

The impedance of the elements of the transmission system for currents of negative sequence, are given in Fig. 7.39. The resultant impedance of the circuit relative to the short-circuit point, we find from the expression:

Z*r — "2Σ -^m< z*+ ZlZ* =y(H13 +

(zs+z*); y0-139(212+yi-32)

Z1+Z5 7O-139+212+/1-32 Z j + Z 4 =y0-281 +70-094 = 70-375; (0-006+yO-248)xy0-375

= 0006+70-248;

Z^ — 0-006+7Ό-248+7Ό-375 = 0-003+70· 149.

j 0-139 2_

joua • , JO-281

U 2l2*j|-32

ja094

Fio. 7.39.

.0-113

r - ^ I iO-843 f Ì0094

Fro. 7.38. Fro. 7.40.

The resultant impedance for currents of zero sequence relative to the point of short-circuit (Fig. 7.40) is:

Ζ0Σ = Z 3 / / ( Z 3 + Z 4 ) ; Z»+Zt = 70-843 +7ΌΟ94 =70-937;

ΖοΣ — jO- 113Χ7Ό-937 7O· 113+70-937 = 7*0-101.


The resultant impedance of the fault at the point of short-circuit is : y = z = Ζ2ΣΖ0Σ _ (0-003 +y0-149) xyO-101 ·

* * Ζ2Σ + Ζ0Σ 0003 +y0-149 +70-101 J

The equivalent circuit of the electrical system for a two-phase short-circuit to earth is given in Fig. 7.41. We determine for it, by the unit current method, the self and mutual admittances. Let the current in the impedance Z4 be equal to unity:

The voltage at point b is Vb = (l+y0)xyO-375 = y0-375.

The current flowing in the fault is:

/ e~z;-7öö6 - 6 2 5 · The current in branch 2 is:

Λ =Λ+Λ = 1+6-25 = 7-25.

FIG. 7.41.

The drop in voltage in the impedance of branch 2 is: AV2 = /2Z2 = 7-25 x.y0· 113 =;Ό·819.

The voltage at point a is: Va = Vb+AV2 =yO-375-hyO-819 =yl-194.

The current in branch 5 is:

The current in branch 1 is: Λ = 4 + / 5 = 7-25 +0088 +y(M41 = 7-338 +j0-141.

The drop in voltage in the impedance of branch 1 is: ΔΫΧ = txZx = (7-338 +/Ό· 141) χ;Ό·679 = -0-096 +/4·983.


Yg = ^ = 7 = 1-187 -89-8°.

yni = i l = -ΖΖ^ = 01617 -90-9°

The electromotive force at the point of connection of the transmitting station is:

È = Ϋα+ΔΫχ = yï-194 + (-0-096+y4-983) = -0-096+./6· 177. The self admittance of the short-circuited branch relative to the trans

mitting station is: 1±_ 7-338 +/Ό-141 È " -0096+jfrlll

The complementary angle is a"/ = 0-2°. The mutual admittance of the short-circuited branch is

A _ l+yo - 0096 +j6- 111

The complementary angle is a™ =— 0-9°. In the equivalent circuit of the network, for the post-fault condition,

when the fault is disconnected, the impedance of the transmission line is doubled (Fig. 7.42). The self and mutual admittances are:

YH = 0-712 ^ -87-5°; aft = 2-5°; Y{\ = 0-671 -92-4°; a1^ = -2-4°.

To determine the limiting time for switching off the short-circuit, taking into account the armature reaction, it is necessary to take various switching off times and in each case to find the function δ = /(ί). The manner of variation of the angle with time shows whether the system is stable or unstable.

_J 2 3 u E/ J0I99 J0II3 jO-562 jO-094

607*j378

FIG. 7.42.

We assume a switching off time zlfsw off = 0-2 sec. The e.m.f. at each instant, allowing for the forcing of the excitation, may be determined from:

t_

Ede == Edeoo — (Edeoo ~ EdeQ) Q * ,

where Edeoo = the e.m.f. corresponding to the ceiling excitation; Ede0 = the e.m.f. in the original condition, Ed0.


Assuming an interval At = 0-1 sec,t we determine the mean values of the e.m.f. Ede in each interval.

With / = 0-1 sec: 0 1

0-35 Ede = 2-5x2-422-(2-5x2-422-2-422)e °'35 = 3-338. The mean value Ede for the first interval is:

2-422 + 3-338 -de = 2-88.

The mean values, Ede, in the intervals are:

Number of

inter^ val

Ed.

1

2-88

2

3-669

3

3-959

4

4-705

5

504

6

5-292

7

5-481

Knowing Ej0 and δ0 for the normal condition we determine the values of the fictitious e.m.f. Eq(fS> for the instant when the condition of operation of the transmission system is disturbed, by the expression :

£««» - l-(*9-*>Îïcos< (7.10)

-£4(0) = 1-426-1 χ0·1617(0·679-0·199) cos (37°+0·9°) = 3-174. 1 - (0-679 - 0-199) x 1 · 187 cos 0-2°

The zero load e.m.f. at the instant when the condition is disturbed, is determined from expression (7.9):

, ,„,, 1188-0199 , . . , 1-188-0-679 * ' « = 3 · 1 7 4 Χ0·679-0·199-1 ·4 2 6 Χ0·679-0·199 = 5 ° 2 6 ·

The variation of the direct-axis component of the transient e.m.f. is:

(7.11) ΔΕ·Λ(Λ=Ε* J?*™ At; *do

Λ £ ; ω = 2 · 8 8 ~ 5 · ° 2 6 χ 0 · 1 = -0-043.

t To shorten the calculations, the interval is taken as equal to 0.1 sec. Generally, the intervals or a calculation are smaller than 0-1 sec.


The value of the direct-axis component of the transient e.m.f. at the end of the first or the beginning of the second interval is:

*;<!)= ^ o + ^ ; ( 1 ) ; (7.12) Ε'αω = 1-426-0-043 = 1-383.

The active power supplied to the circuit by the generators of the transmitting station at the beginning of the first interval, is determined from expression (7.5): P(0) = 3·1742χ 1-187 sin 0-2°+3-174x 1 xO-1617 sin (37°+0-9°) =0-364.

The surplus active power, which is acting during the first interval of time is:

JP(0) = 1-078-0-364 = 0-714.

The variation of the angle during the first interval of time is :

360χ50χ0 ·1 2 0-714 , , . 0-714 = 10-98 X~T- = 1 6 · 4 χ — = 5-85°.

The value of the angle at the end of the first or the beginning of the second interval of time is :

δω = δ0+Δδω = 37+5-85 = 42-85°.

We carry out the calculation for the second interval: _ 1-383-1 χ0·1617(0·679-0·199) cos (42-85°+0-9°) _

«(1) ~ 1 -(0-679-0-199) X 1-187 cos 0-2° -3 -086;

■,«„* 1 1 8 8 - 0 1 9 9 , „0„ 1188-0-679 Λ on, E^ = 3 0 8 6 X 0-679-0-199 - 1 ' 3 8 3 Χ 0-679-0-199 = * 8 9 1 ;

3-669-4-891 5

ΔΕά($>= c X0' 1 = -0-024;

££<2) = 1-383-0-024 = 1-359; P(1) = 3-0862x 1-187 sin 0·2° + 3·086χ 1x0-1617 sin (42-85°+0-9°) = 0-392;

ΑΡω = 1-078-0-392 = 0-686; j a ö ) = 5-85 + 16-4x0-686 = 17-Γ;

ί(2) = 42-85 + 17-1 = 59-95°.

We proceed to calculate the third interval (the interval after the short-circuit has been switched off).


The fictitious e.m.f. prior to removing the short-circuit is: _ 1-359-1 X01617(0-679-0-199) cos (59-95° +0-9°)

£ί<ο·2<) - 1-(0·679-0·199)χ 1-187 cos 0-2°

The electromagnetic power of the generator prior to switching off is : Λο-r) = 3-0722x 1-187 sin 0-2°+3-072x 1x0-1617 sin (59-95°+0-9°)

= 0-479. The surplus power is:

P(0.r) = 1-078-0-479 = 0-599. The fictitious e.m.f. after the short-circuit has been removed is :

1-359-1 x 0-671(0-679-0-199) cos (59-95°+2-4°) Eq(2) - 1 - (0-679 - 0-199) x 1 · 187 cos 2-4° = 1-837;

. «-,-, 1188—0199 , ,_n 1-188-0-679 „ „,„ **» - 1-837X 0-679-0-199 - 1 3 5 9 X 0-679-0-199 = 2 ' 3 4 3 ;

(3-959-2-343) X 0-1 dEj(3y — 0-032;

E'd(s) = 1-359+0-032 = 1-391. The electromagnetic power after the short-circuit has been removed is :

Ρ(2) = 1·8372χ0-712 sin 2-5° + l-837x 1x0-671 sin (59-95°+ 2-4°)= 1-199; ΑΡω = 1078-1199 = -0-121;

,-, , ^ A 0-599-0-121 Jô(3) = 17-1 + 16-4X = 2Γ;

δω = 59-95+21 = 80-95°. Further calculations are given in Table 7.3.

TABLE 7.3

Number of /Λ-terval

1 2

3 4 5 6

/, sec

0 0-0-1

0-1-0-2 0-2 0-2-0-3 0-3-0-4 0-4-0-5 0-5-0-6

Eq

1-909 3174 3086 3 072 1-837 2056 2194 2-34

E*

2-422 5026 4-891

— 2-343 2-76 3 0 3-25

E'd

1-426 1-383 1-359 1-359 1-391 1-43 1-47 1-511

ΔΡ

0-714 0-686 0-599

- 0 1 2 1 -0-422 -0-541 -0-624

zl<5, degrees

5-85 17-10

— 210 1408 5-23

- 5 0

<5, degrees

37 42-85 59-95 59-95 80-95 9503

100-26 95-26


It follows from the calculation, that with isw.off = 0*2 sec, the dynamic stability is not disturbed (Fig. 7.43).

We make a similar calculation, assuming isw.off = 0-3 sec. The results of the calculation are given in Table 7.4.

TABLE 7.4

Number of interval

1 2 3

4 5 6 7

/, sec

0 0-0-1

0-1-0-2 0-2-0-3 0-3 0-3-0-4 0-4-0-5 0-5-0-6 0-6-0-7

E,

1-909 3-174 3 086 3072 3123 2040 2-329 2-516 2-636

E*

2-422 5026 4-891 4-887

— 2-775 3-330 3-730

—

E'd

1-426 1-383 1-359 1-346 1-346 1-385 1-419 1-450

—

ΔΡ

0-714 0-686 0-599 0-525

-0-423 -0-484 -0-275 +Ò-034

Αδ, degrees

5-85 1710 26-90

— 27-75 19-80 15-29 15-85

degrees

37 42-85 59-95 86-85 86-85

114-60 134-40 149-69 165-54

It follows from the calculation, that with tswo{{ = 0-3 sec, the dynamic stability is disturbed (Fig. 7.43).

Thus, the time limit for removing the short-circuit is found to be between the limits 0-2-0-3 sec. For a more exact calculation of the time limit of switching off, it is necessary to carry out similar calculations for 0-2 < /swoff ·< 0-3 sec. The curves showing the variation of the e.m.f. 2?Λ, Ed, Ed for a two-phase short-circuit, switching off in 0-2 sec, are given in Fig. 7.44.

Problem 7.15

In the transmission system, the circuit of which is given in Fig. 7.45, a three-phase short-circuit occurs at the beginning of the unloaded line, lx, i.e. practically at the high-voltage busbars of the step-up transformer T-l. In 0-12 sec. the short-circuit is switched off by the switch B-l.

The generators of the transmitting station are of salient pole type. The original quantities in relative units are: Pl0 = 0-434; Ed0 = 1-122; E^ = 1-08; E'd0= 1-046; Vc = 1-01. The angle between the e.m.f. Ed0

and the system voltage is ôQ = 57-2° (see the vector diagram, Fig. 7.46).


The parameters of the transmission system, in relative units,t are as follows:

xd = 0-537; xq = 0-346; x'd = 0-19; Td0 = 5-2 sec, Tj = 13-8 sec.

dtgrccs

150

120

90

60

30

S

/

_ -!.. I t f 1 I t ! 1

rcl. units

01 0-2 0-3 (K 05 0-6 0 7 s«c

FIG. 7.43.

Î

0-1

Ed

1 t f

0-2 0-3 04

F IG . 7.44.

, « , 0-5 sac

G T-l V "? T-2 Ve

}<3ΣΗΗ

FIG. 7.45.

The reactances of the transmission line in the original and post-fault conditions are:

*n = *i2 = 2-193; a u = a12 = 0.

t When determining the self and mutual admittances, the salient pole generator is represented by its quadrature-axis reactance.

Large Oscillations and Dynamic Stability Currents

The reactance of the transmission line with the short-circuit are:

*lls .c . = O * 4 4 8 ; a l l s . c . = Gì *12s.c. = °°-

237

Required: to calculate the dynamic transient and to construct a curve showing the variation of angle Ò with time.

FIG. 7.46.

Solution. We carry out the calculation by the method of successive intervals, taking into account the armature reaction, but ignoring the torque due to the aperiodic components and ignoring the ohmic resistances. The duration of the interval, for the calculation at the start of the transient process, we take as 0-04 sec, then we change to a new interval of Ol sec.

First interval (0-0-04 sec)

The fictitious e.m.f. at the beginning of the first interval (from formula (7.10)) is:

1-046 1-046 ^α» -

1-(0·346-0·19)χ- 1 0-652 = 1-6.

^0-448

From formula (7.9) we determine the zero load e.m.f. : Λ , 0-537-0-19 Λ Λ ^ 0-537-0-346

E«* = 1 ' 6 Χ 0-346-0-19^ 1 ' ° 4 6 Χ 0-346-0-19 = 1-6x2-22-1-046x1-22 = 2-27.

The change of the transient e.m.f. E'd during the first (interval formula (7-11)) is:

004 AE'dil) = (1-22-2-27) x - ^ = (1·122-2·27)χ0·0077 = -0-009.

Αδω = t " o X^-w- = 2-08 x ^ f - = 0-452°;


The transient e.m.f. Ed at the end of the first interval is: 3i<i> = ^ o + ^ d ) = 1-046-0009 = 1-037.

The active power output of the generator to the system during the short-circuit is P = 0 .

The excess power is: AP(0)=P10-P = 0-434.

The change of the angle during the first interval and the angle at the beginning of the second interval (formula (7.4)) are:

360 X 50 x 004 2 0-434 „ ΛΟ 0-434 T2Ü X — - 2 - O B x - j -

δω = 57-2+0-452 = 57-652°. The calculations for the second and third intervals are carried out si

milarly. Second interval (0-04-0-08 sec)

Eqil} = 1-59; Ed(1) = 2-265; AE'd(2) = - 0 0 0 9 ; Ed(2) = 1-028; Αδ(2) = 1-355°; δ(2) = 59-007°.

Third interval (0-08-0-12 sec)

Eq(2) = 1-58; Ed(2) = 2-246; AEd(3) = -0-008; Fio) = 1-02; zl<5(3) = 2-258°; <5(3) = 61-265°.

Fourth interval (0-12-0-16 sec) with short-circuit removed

The excess power before removing the short-circuit is:

Λ / W ) = 0-434. The fictitious e.m.f. after the short-circuit is removed is:

1-02-1-01 χ — ί — (0-346-0-19) cos 61-265° £«<3) j

1-(0-346-0-19) X-2-193

102 -0-0774 cos 61-265° = 1063; 0-929

Ed(3) = 1-063x2-22 - 1 - 0 2 x 1 -22 = 117; AEdU) = (1-122-1·117)χ0·0077 « 0; Ed(i) = 1-02.

t In this case, the change of the angle and the angle at the instant of switching off the short-circuit, may be determined together for the whole time of the short-circuit, according to formula (7.4).


The active power output from the generator after the short-circuit is removed is:

P(3) = l e^ioj 0 1 s i n 6 1 , 2 6 5 ° = 1-063x0-461 sin 61-265° = 0-43;

ΑΡω = 0-434-0-43 = 0004.

We determine the variation of the angle in the fourth interval from the average value of the excess power before and after the short-circuit has been removed:

^ U ) = 2 - 2 5 8 + 2 - 0 8 x ° - 4 3 4 + 0 · 0 0 4 =2-714°;

6U) = 61-265+2-714 = 63-979°.

Fifth interval (0· 16-0-2 sec)

£,(4) = ^ - - 0 0 7 7 4 cos 63-979° = 1-066;

EdU) = 1-066x2-22-1-02x1-22 = 1-122;

AEd(i) = 0; Ed(i) = 1-02;

Pw = 1-066x0-461 sin 63-979° = 0-441;

APU) = 0-434-0-441 = -0-007;

Δδ{Ά) = 2·714-(2·08χ 0-007) = 2-7°;

ό(5) = 63-979+2-7 % 66-68°

Sixth interval (0·2-0·3 sec) (transfer to a larger interval for calculation At = 0.1 sec)

As before:

Eq(i) = 1-069; EäU) = 1-13;

AEdW = ( 1 - 1 2 2 - l - 1 3 ) x ^ = (1-122-1-I3)x0-0192 % 0;

EdM = 1-02; i>(5) = 0-452; ΔΡω = -0-018.

With the change of the length of the interval, within the limit of one condition, the change of the angle is calculated from the formula :

ΔΚ = M ^ + k „ à P ^ l - k ^ P - - * l + 3 k J , (7.13)


where

In our example:

k - 4 s -*A - Att '

_18000

.. . _ 01 18000

X oΛ% - 0 · 0 0 7 ( ΐ - ^ ) - ( Η ) 1 8 ( ΐ + 3 χ ^ ) =

4 χ 01 004

6-56°;

<5(β) = 66-68+6-56 = 73-24°. Further calculation of the variation of the angle with time has been

carried out with an interval At =0-1 sec; the calculated results are given in Table 7.5.

degrees

90

80

70

60

50

An

δ

-y

I I 1 1 t t 1

0 2 0 4 0 6 0 8 10 sec

FIG. 7.47.

TABLE 7.5

Number of interval

7 8 9

10 11 12 13 14

t, sec

0-3-0-4 0-4-0-5 0-5-0-6 0-6-0-7 0-7-0-8 0-8-0-9 0-9-1-0 1 0 - 1 1

Eq

1078 1083 1090 1094 1097 1100 1099 1097

Ed

1150 1160 1180 1190 1193 1196 1-204

—

E'd

1019 1018 1017 1016 1015 1013 1011

—

ΔΡ

- 0 0 4 2 - 0 0 5 6 - 0 0 6 6 - 0 0 7 0 -0071 -0071 - 0 0 6 9 - 0 0 6 8

Aô, degrees

601 5-28 4-42 3-51 2-68 1-75 0-85

- 0 0 4

à, degrees

79-25 84-53 88-95 92-46 9514 96-89 97-74 97-70


From the calculated results, the curve given in Fig. 7.47 is plotted showing the variation of the angle with time. The angle of displacement of the rotor in the first period of oscillation, ignoring the torque due to the aperiodic components and the ohmic resistances, reaches 98°.

Problem 7.16

In the transmission system, considered in the previous problem (Fig. 7.45), a three-phase short-circuit occurs at the high-voltage busbars of the step-up transformer and is switched off in 0-12 sec.

Required: to calculate the dynamic transient and to construct a curve of the angle à as a function of time, allowing, in the calculation, for the reaction of the stator current and the ohmic resistances of the transmission system, but not for the effect of the aperiodic components on the torque.

The parameters of the transmission system in relative unitst are: xd = 0-537; xq = 0-346; xd = 0-19; Td0 = 5-2 sec; Tj = 13-8 sec.

The impedances of the transmission system in the original and post-fault conditions are: Z n = Z12 = 2-193; α η =α12 = 3-45°. The impedances of the transmission system with the short-circuit are: Z l l s c# = 0-448; aiis.c. = 1*25°; Z12sx> = °o.

Solution. We calculate by the method of successive intervals. For the calculation, the duration of the interval, for the beginning of the transient process, is taken as 0-04 sec, then we proceed to a new interval of 0-1 sec.

First interval (0-0*04 sec)

We find the fictitious e.m.f. at the beginning of the first interval (formula (7.10)) is:

1-046 1046 Λ , «(0) = Ϊ = 0-6S2 =

1_(0.346-0·19)Χ7Γ4—cos 1-25° 0-448

t The self and the mutual admittances were determined by taking into account the ohmic resistances of the generator stator, transformers, transmission lines and the network.


Using formula (7.9), we determine the zero load e.m.f.: , , 0-537-0-19 , . . . 0-537-0-346

E™ = 1 ' 6 Χ 0 1 4 6 = 0 1 9 - 1 · 0 4 6 Χ 0-346-0-19

= 1-6x2-22 -1-046x1-22 = 2-27.

From formula (7.11), we determine the e.m.f. Ed during the first interval as:

AEd(1) = (1-122-2-27) x ^ = (1 · 122-2-27) x 0-0077 = -0-009.

We determine the transient e.m.f. E'd at the end of the first interval to be : Ed(1) = 1-046-0-009 = 1-037.

We determine the active power output from the generator at the beginning of the first interval (formula (7.5)) to be:

Λο) = 7rxLr s m 1-25° = 1·62χ0·0487 = 0-125. " 0-448

The excess power to be: APm = 0-434-0-125 = 0-309.

We determine the change of the angle during the first interval and the angle at the beginning of the second interval (formula (7.4)) as:

360x50xO-042 0-309 „ ΛΟ 0-309 IT* X — - 2 - 0 8 X - 2 -

δω = 57-2+0-322 = 57-522°.

Calculations for the second and third intervals are carried out similarly.

Second interval (0-04-0-08 sec)

£ · , ω = 1·59; Edil) = 2-265;

AEd(2) = -0-009; E'dW = 1-028; J>(1) = 0123; ΑΡω = 0-311;

Αδω = 0-969°; δω = 58-491°.


^«(2) = l-58; Ed(2) = 2-264;

AEd(3) = -0-008; Ed(3) = 1-02; P(2) = 0-121 ; ΑΡω = 0-313;

Αδ(3) = 1-62°; òw = 60-111°.

Αδω = — χ —^— = 2-08 x —^— = 0-322° ;


Fourth interval (0-12-0-16 sec) with short-circuit removed

We determine the excess power up to the moment of removing the short-circuit, by first finding the e.m.f. Eq prior to the removal of the short-circuit:

/r l'02 1 w

Λο-ia-) = 1·5642χ 0-0487 =0-119;

^Λο-ιΐ') = 0-434-0-119 = 0-315.

The fictitious e.m.f. after the short-circuit has been removed is:

1 0 2 - 1-01 x^-fôT (°'346-0-19) cos (60-111°-3-45°) A Ï <3) j

1 -(0-346-0-19) x —— cos 3-45°

1 0 2 -0-0774 cos (60-11 Γ-3·45°) = 1-057; 0-929

£rf(3) = 1-057x2-22-1-02x1-22 = 1-11;

AE'dii) = (1-122-1·11)χ0·0077 ^ 0; E'dU) = 102.

The active power output from the generator after the short-circuit has been removed is:

1 0 5 7 2 · , „ « , 1-057x1-01 . , „ 1 1 1 0 ..... (3) = 2193 s m 3 ' 4 5 + — 2 Ί 9 3 — s , n (60· 111°-3.45°)

= l-0572x 0-0274 + 1-057x0-461 sin (60-111°-3-45°) = 0-4381;

AP{Z) = 0-434-0-4381 = -0-0039.

We determine the change of the angle in the fourth interval from the mean value of the excess power before and after the short-circuit has been removed :

0-315-0-0039 2 Zl<5(4) = 1-62+2-08 x ^ — = 1-944°;

δΜ = 60-111 + 1-944 = 62055°


Fifth interval (0-16-0-2 sec)

1-02 E"w 0-929 -0-0774 cos (62-055°-3-45°) = 106;

EdW = 1-06x2-22-1-02x1-22 = 1-11; AEd(s) = (1-122-1-1 l)x0-0077 * 0; Ed(s) = 1-02;

P(4) = l-062x 0-0274+ 1-06x0-461 sin (62-055°-3-45°) = 0-448. APit) = 0-434-0-448 = -0014; J<5(5) = l-944-(2-08x 0-014) = 1-915°;

<3(5) = 62-055 + 1-915 = 63-97°.

Sixth interval (0·2-0·3 sec) (transfer to a larger interval for the calculation

At = 0-1 sec) As before:

£et t , = 1-062; Edls) = 1-12;

AE'dis) = (1·122-1·12)χ5ΐ1 = (1-122- 1·12)χ0·0192 % 0;

E'äW = 1-02; P(5) = 0-4569; AP(i) = -0-022. We determine the change of the angle with the change in the length

of the interval, within the limit of one operating condition, from formula (7.13). In our case:

A6m = 4-56° <5(e) = 63-97+4-56 = 68-53°.

Further calculation of the variation of the angle with time, is carried out with the interval At = 0-1 sec: the calculated results are given in Table 7.6.

TABLE 7.6

Number of interval

7 8 9

10 11 12

/, sec

0-3-0-4 0-4-0-5 0-5-0-6 0-6-0-7 0-7-0-8 0-8-0-9

Eq

1069 1072 1077 1079 1081 1081

Ed

1120 1136 1146 1156 1156

E'd

102 102 102 102 102 —

ΔΡ

- 0 0 4 4 0 - 0 0 6 0 4 - 0 0 7 0 0 - 0 0 7 7 9 - 0 0 7 9 9 - 0 0 8 0 9

degrees

3-99 3-20 2-29 1-27 0-23

- 0 - 6 0

à, degrees

72-52 75-72 7801 79-28 79-51 78-91


From the calculated results a curve showing the variation of the angle δ with time is plotted (Fig. 7.48 (curve 2)). The largest rotor displacement angle in the first period of oscillation, is 79-5°. For comparison, curve 1

degrees 90

80

70

60

50

0-2 04 0 6 0-8 10 sec

FIG. 7.48.

is plotted on the same diagram, showing the variation of the angle δ with time, obtained from a calculation when the ohmic resistances of the transmission system had not been taken into account (see Problem 7.15).

Problem 7.17

Figure 7.45 (p. 236) is a transmission circuit. In this transmission system, a three-phase short-circuit occurs at the beginning of the line. The duration of the short-circuit is: Atsc = 0-12 sec. In the original condition a hydro-generator was transmitting power (S20 = 825—y*415 MVA) to the receiving system through a step-up transformer and a two-circuit transmission line of length 1000 km. The voltage at the end of the transmission line was V20 = 415 kV.

The parameters of the transmission system, in relative units,t determined with basic power Shasic = 2010 MVA and voltage Vhasic = 412-5 kV are: xd = 0-537; ^ = 0-346; ^ = 0-19; x2 = 0-123; Td0 = 5-2 sec;

7> = 13·8 sec; rrot meant = 0-0075. The impedances of the transmission system for the original and post-

fault conditions are: Z n = Z12 = 2-193; a u = a12 = 3-45°.

t By rrot.mean is understood the equivalent resistance of the rotor circuits (excitation winding and damping circuits). In determining the self and mutual admittances, the generator is represented by its direct-axis impedance.

j i i i i Li


The impedances of the transmission system with the short-circuit are :

Zlls.c. =0-448; alls.c. = 1-25°; Z12s.c. = ~.

The equivalent impedance of the system (including the step-down transformer) is :

Zc = 0-0627 ^72-8°.

The total resistance and reactances up to the short-circuit point are:

xdL = 0-639; xqE = 0-448; χ'άΣ = 0-292; x £ = 0-223;

rst = 0-00982.

The time constants with external άΣ resistance and reactance up to the point of short-circuit taken into account are :

T'd = 2-38 sec; T'd' = 0-104 sec; Ta = 0-088 sec.

Required: to calculate the dynamic transient and to construct the curves of the variation of the angle δ with time. The calculation is to be carried out with additional braking torques, which depend on the effect of the

FIG. 7.49.

aperiodic components of the short-circuit current. The capacitance of the transmission line and the transformer magnetizing currents are to be ignored.


Solution. (1) Calculation of the normal condition:

v"> - Ms =1005; *- = ^ r 1 = °·41 --^2 0 6· ^ = 0 · 4 1 + ; 0 · 2 0 6

0 F20 1-005

È„o = P'vt+UZu-Zc) = 1 -005 + (0-456 ^ 26-7°) (2-193 86-55° - 0-0627 ^ 72-8°)

= 1-08^55-6°; Vc = V^-ioZc = 1-005-(0-456 ^26-7°) (0-0627 ^72-8°) = 101 ^ -1-6°.

The angle between the e.m.f. Ed0 and the voltage Vc is: K = *È«pn + <cYnYe = 55-6 + 1-6 = 57-2°.

To determine the e.m.f.s EM, EdQ and Ed0, we resolve the current/„ into components along the axes d and q (see the vector diagram, Fig. 7.49):

Id0 = I0 sin [57-2° - (26-7° +1-6°)] = 0-456 sin 28-9° = 0-22; 1^ = 0-456 cos 28-9° = 0-399;

Ed0 = E^+I^x^-x^) = 1 08+0-22(0-639-0-448) = 1122; Ed0 = E^-I^x^-x^) = l-08-0-22(0-448-0-292) = 1046; E'd'o = E^-I^x^-x'Jz) = l-08-0-22(0-448-0-232) = 1-032.

The voltage at the point of the short-circuit, in the original condition is:

^ko = £ « o - / Z U l x . = 1-08-0-456 ^-(55-6°-26-7°)x X0-448 ^88-75° % 1 ^ - 9 ° .

The active power output by the generator is :

Λο = ^ο+Ζ? Re ( Z n - Z c ) = 0-41+0-4562Re(2-193 ^86-55°-00627 ^72-8°) = 0-434.

(2) Calculation of the transient electromechanical process taking into account the additional braking torques :

With the short-circuit, besides the loss in the ohmic resistance of the stator circuit, due to the current of normal frequency, additional losses appear in the machine, in the stator circuit as well as in the rotor windings. Besides this, a normal-frequency pulsating torque appears. The effect of the pulsating torque is to produce additional braking of the generator.


The total loss in the ohmic resistances of the generator, is determined by the expression :

χάΣ \xdL χάΣ/

\ΧάΣ ΧάΣ/ J S a \ΧάΣ/ +1 The pulsating torque is (at ω % ω0 = 1 ; M = P):

t

l ΧάΣ \ΧάΣ ΧαΣ/ \ΧάΣ ΧάΣ/ )

t

XVkQ T<t sin (cot+ ôk).

The calculation is carried out by the method of successive intervals. The interval used for the calculation is At = 0·04 sec.

At each interval we calculate the mean values APR and AP„ and determine the resultant power from the equation :

Δ?Σ = P10-(4PR+AP„),

where P10 = the output power of the generator in the original condition. We determine APR at each interval, as a mean of the values at the beginn

ing and end of the interval. We calculate the value AP„ for the interval from the expression

1 T'* ΔΡ~ν> =ji\ Δρ~άί>

where tx and t2 are the instants corresponding to the beginning and the end of the nth interval.

The expression under the integral for calculating AP„(n)i is a sum of

functions of the form (e r s in (œt + ôj. W i t h / = 50 c/s and Γ > 0-03 sec:

i '2 _L i JA e T sin (œt + ôk)dt % e T cos (ω/2 -f àk) -f

Ί ω

1 -îî-H — e T cos (ωΛ +ôk). ω


With an interval At, which is a multiple of the period of the original condition, cos (œt + ôk) = cos ôk; whence:

J>" -4 · / * ^ c o s δk( -ψ -ψ) ezTsm(cot + ôk)dt % -\e r - e Γ / .

ω We calculate the values APR in each interval :

Ä<0) ~ L0-639 V 0 - 2 9 2 ° 6 3 9 \0-232 0-292/ J

000982+1 0^32) e OO88x0-0075 =0-334; x

For t=004 sec.

AP - I " 1 ' 1 2 2 i f 1 ' 0 4 6 1 - 1 2 2 > \ e - ^ + ^*<oo4) - |_ 0 . 6 3 9 + ^ 0 . 2 9 2 0 . 6 3 9 j e +

x 0-0075 =0-2246; Similarly,

^Vo8> = 0-1734; ΛΡΛ(0.12, = 0-1484.

The mean values of APR for the intervals are :

0-334+0-2246 2 ^ J K I ) = ^ = ° · 2 7 9 3 ! APR(2) = 0-199 ;

JPÄ(3) = 0-1609.

The mean values of ΔΡ^ for the intervals are:

lxcos9° Γ 1-122 ( — -—Ί ΔΡ~™ =314^04 L W (e ° ' 0 8 8 - e 00MJ +

f ±046 _ H 2 \ ( e - ( £ + i & ) _ e - ( 0 ^ + 0 ^ 8 ) ) V 0-292 0-639 j v y +

+ V 0-232 0-292 ) K / j - u 14«,

Similarly, dP„w = 0088; /J-PM3) = 0-0485.


The resultant powers for the intervals are : First interval (0-0-04 sec)

ΑΡΣω = P10-(APR+AP„) = 0·434-(0·279+0-144) = 0-011.

Second interval (0Ό4-0Ό8 sec) ΑΡΣ{2)= 0-153.


ΔΡΣ{Ζ) = 0-225. Knowing ΑΡΣ for each interval, we determine the increase of the angle

δ during the short-circuit (formula (7.4)) :

Mm _ 360X50X0.04' χ O g l _ 2 . „ 8 χ Ο^Π _ ^ , . .

Αδω = άδω+ΙεΔΡΣ(Λ = 0-011+2-08x0-153 = 0-329°;

zl<5(3) = 0-329+2-08x0-225 = 0-798°. The angle up to the instant of the switching off of the short-circuit is :

<5(3) = δ0+Αδω+Αδ{2) + Αδω = 57-2+0-011+0-329+0-798 = 58-338°.

We determine the transient e.m.f. up to instant of the switching off the short-circuit as:

p p E'd&) = Ed0 +^^(0-0-1» = Ed0^ ^ ^ ^ ' s . c -

*d0

In this equation the e .m.f. Ede = Ed0 = 1-122, provided that the generator is considered without an excitation control.

Ed(0) = the e.m.f. along the direct-axis at the first instant of the disturbed condition.

We find its value from the equation:

rr — rr xd~~~xd r?t xd~~xg ^d(0) — ^q(0) — TT — dO — ~ T *

where Ε^0)9 the calculated e.m.f. of the fictitious non-salient machine, is determined by formula (7.10):

I 046 * - - JS-3«-0^)C O S, . 2 5 . - 1 6 ·

0-448


Knowing the e.m.f. £ 0)» w e fiQd the e.m.f. Edl0): , . 0-537-0-19 , . . . 0-537-0-346

E™ = 1 · 6 Χ 0 3 4 6 - 0 1 9 - 1 0 4 6 Χ 0-346-0-19 = 1-6x2-22-1-046x1-22 = 2-27.

From the value Ed(m we find the e.m.f. E^ :

£,«3) = 1·046+(1·122-2·27)χ-5:^ = 1-02. 0-12 5-2

We now calculate the variation of the angle with time after the short-circuitt has been cleared.

Fourth interval (0-12-0Ί6 sec)

The calculated e.m.f. E^ at the beginning of the interval (after the short-circuit has been switched off) is :

1-02-1-01 x—J—(0-346-0-19) cos (58-34°-3-45°) E^ = ^ = = 1-052.

1 - (0-346-0-19) x y i — cos 3-45°

The electromotive force Ed(z }at the beginning of the interval is: Edi3) = 1052x2-22-1-02x1-22 = 1-092.

The change of the e.m.f. E'd during the interval is: 004 5-2 AE'dU) = (1-122-1-092)x^y = 00002 * 0.

The transient e.m.f. E'd at the end of the interval is : EdU) = 1-02.

The electromagnetic power produced by the generator, is determined from the e.m.f. Eq at the beginning of the interval (after the short-circuit has been switched off).

1-0522 1·0^2χ1 ·01 Λ» = y ^ j sin 3-45° + 1 U y X

31 U 1 sin (58-34° -3-45°) = 0-4273.

t When switching off the short-circuit, the new transient process which arises in connection with the new aperiodic component of the stator current, is ignored. To a first approximation, it is considered, that the new aperiodic current component is compensated by the continuing action of the first one, which appeared at the instant of the short-circuit.


We find the difference between the mechanical and the electromagnetic power at the beginning of the fourth interval is :

^ (3 ) = Λ0 -Λ3 ) = 0-434-0-4273 = 0-0067.

We determine the change in the angle ò during the interval. Because the beginning of the interval coincides with the sudden change in the condition (short-circuit switched off), the acceleration at the beginning of this interval is determined as a mean value of acceleration, obtained for the given instant depending on the preceding and the new conditions :

^ ( 4 ) = 0·798+2·08χ°·2 2 5 +2

0 · 0 0 6 7 =1-039°.

The angle at the end of the interval is : au) = 58-338 + 1-039 = 59-377°.

Fifth interval (0-16-0-2 sec)

Similarly, Eq(t) = 1-054; EdU) = 1-097; AEd(5) = 0; Ed(ò) = 1-02;

P{i) = 0-4335; APU> = 0-0005. The change of the angle during the interval is

zl<5(5) = 1-039+2-08x0-0005 = 1-04°; <5(5) = 59-377 + 1-04 = 60-417°.

Sixth interval (0-2-0-3 sec)

(transfer to the larger interval At = 0-1 sec for the calculation). Similarly,

Eq(5) = 1-055; Ed{i) = 1098;

AEdM = (1-122-1-098) x ^ = 0-00046 % 0;

EdW = 1-02; P(i) = 0-4385; APW = -0-0045. With the change of the interval within the range of one condition of

operation, the change ofthe angle is calculated according to Formula (7.13). In our case,

AÔ(6) = 2-549°; «5(6) = 62-966°.


We carry out further calculations of the change of the angle Ò with time with an interval At = 0-1 sec; the calculated results are given in Table 7.7.

TABLE 7.7

Number of interval

1 8 9

10 11 12

/, sec

0-3-0-4 0-4-0-5 0-5-0-6 0-6-0-7 0-7-0-8 0-8-0-9

E*

1-058 1060 1063 1064 1066 1066

Ed

1-106 1106 1116 1120

• 1122 —

E'd

102 102 102 102 102

—

ΔΡ

-00177 -00267 -00360 -00441 -00461 -00481

Δδ, degrees

2-318 1-970 1-500 0-925 0-324

-0-302

<5, degrees

65-284 67-254 68-754 69-679 70003 69-700

A curve is plotted from the calculated results of the variation of angle Ò with time (Fig. 7.50(a) curve 3). For comparison, on the same diagram are given a curve showing the variation of the angle δ with time, obtained by calculation, when the ohmic resistances of the transmission system are ignored (curve 1) (see Problem (7.15)), and a curve of the variation of the angle with time, obtained by calculation, allowing for the ohmic resist-

dcgrees [£

90 k

80 h

70 U

60 \-A

so y

0-2 0-4 0-6 0-8 l-0sec 0-2 0 4 0-6 0 8 10 sec

(a) (b)

FIG. 7.50.

ances, but ignoring the additional braking torques (curve 2) (see Problem 7.16). On the same diagram a dotted curve is plotted, showing the variation of the angle ò with time, obtained experimentally on the electrodynamic model at the Moscow Power Institute (curve 4), where a system with the same parameters was set up.

degrees 90

80

70

to

50

6

1

/ ^ ^

^τ^ 3 ond U

1 f ! ! w 1


The comparison of the theoretical and experimental curves permits us to conclude that calculations ignoring the ohmic resistances and ignoring the additional braking torques when the short-circuit is near to the generator, may lead to considerable errors.

With the increase in the distance to the point of the short-circuit, the effect of the additional braking torque diminishes.

For example, in Fig. 7.50(b), similar curves for a three-phase short-circuit at a distance of 300 km from the beginning of the line, are given.

The comparison of curves 2 and 3 shows that the maximum displacement of the angles in this case, are practically equal whether the additional braking torques are allowed for or not, and moreover, the calculated results agree, with the results of the experiment on a dynamic model (curve 4).

Problem 7.18

In a system, of which Fig. 7.51 shows the main circuit, and Fig. 7.52 shows the equivalent circuit, a single-phase short-circuit occurs at the beginning of the line. The operating quantities and the parameters of the

F I G . 7 .51 .

jO-179 , J0096 , 0-0A2UJ0-2Î9 ι jOUA jO-0143 . jO-0286

o—nr*—Ι—'ΤΓ 1 —4—γ—t—i—9—I—^T5^—t—^—j"-^^-^ • ■ I I " I _ ' Ε2 = 1·037/ΐΙ·0°

00907* j 0-068

Ei 81-215/ 9-2» R0=I197 φ - ί 5 7 9 dp

FIG. 7.52.

system elements are given in the equivalent circuit of the normal condition (Fig. 7.52). The inertia constants of the generators, referred to the basic conditions are: Tn =8,25 sec; TJ2 = 112 sec.

Required: to calculate the dynamic transient and to determine the limiting permissible time of switching off the short-circuit. The calculation is to be carried out in a simplified manner, with the e.m.f. behind the transient


reactance of the generator assumed to be constant. The ohmic resistances of the generators and transformers are to be ignored.

Solution. The theoretical equivalent circuit, which corresponds to the single-phase short-circuit, is derived from the equivalent circuit of the normal condition after an equivalent reactance has been connected, equal to the sum of the resulting (referred to the position of the short-circuit) negative and zero sequence reactances (Fig. 7.53). The self and mutual admittances of the theoretical circuit (Fig. 7.53), may be determined either by the conversion method, or by the unit current method.

In our case:

Yn =2-29 ^ - 8 8 - 2 ° Yoo = 7 -62^-58-7°

[ 12 0-8075 ^ - 9 9 - 3 °

1 1 = 1-8° «22 = 31-3° ;

«12 = - 9 - 3 ° .

We determine the characteristic powers of both stations with the short-circuits to be :

Λ = E[2yn s i n a u + £ i £ ^ i a sin (<5;2-α12) = 1·2152χ2·29 sin 1-8° + + 1-215x1-037x0-8075 sin (<5;2+9-3°) = 1-083 +

+1-017 sin (<5;2+9-3°); P2 = E2

2y22 sin α^ -Ε[Ε^12 sin (<5[2+a12) = 10372x X7-62 sin 31-3°-1-215x1-037x0-8075 sin ( Ó ; 2 - 9 - 3 ° )

= 4-265-1-017 sin ( Ó ; 2 - 9 - 3 ° ) ,

where δ'12 = the angle between vectors E[ and E2.

ÌO-179 J0O96 0-042UJ0-2ÎS j 0-114

0-0123+j 0-273 ==-j5-79 =b-j5-79

, - _ JO 0IA3 j00286

009074·] 0-068

FIG. 7.53.

The excess power with the short-circuit is :

AP1 =P10-P1 = l-197-[0-1083-f 1017 sin (Ó;2 + 9-3°) ]

= 1-089-1-017 sin (<5;2 + 9-3); ΔΡ2 = P20-P2 = 3·875-[4·265-1-017 sin ( Ó ; 2 - 9 - 3 ° ) ]

= -0-39 + 1-017 sin ( Ó ; 2 - 9 - 3 ° ) .

256 Transient Phenomena in Electrical Power Systems. Problems

We determine the absolute acceleration of both stations to be:

« 1 = 360/ΔΡι __ 360x50

- 8-25 [ / i [1-089-1-017 sin (δ;2 + 9·3°)]

«2

= [2-375-2-22 sin ( Ô ; 2 + 9 - 3 ° ) ] X IO3; 360/zJi>2 360x50

[ / 2 112 [-0-39 + 1017 sin ( Ó ; 2 - 9 - 3 0 ) ]

= [-0-0626+0-1635 sin (<5;2-9-3°)] x 103. The difference of the absolute accelerations gives the relative accelera

tion a12 = <*!— a2. The results of the calculated accelerations for different angles ò'12 are given in Table 7.8.

The curve of the relative acceleration against ò[2 obtained from these calculations, is given in Fig. 7.54 (curve 1).

The equivalent circuit of the transmission system for the post-fault condition, is given Fig. 7.55. The self and mutual admittances for this circuit are given by:

^22

^12

12

10

08

06

04

0-2

0

-0-2

-0-4

-0-6

1-112^-83-4° 7-34^-55-5°;

= 0-94^-97-8°;

« - 1 1 = 6-6° a22 = 34-5°; a12 = —7-8 .

C'2"

-

L

IO"3

V.

V F occe l .

itV1

LV ; \ JL r ί i

S^gsw.of! ^

I ! I

T Fbrake S'a

I 1 1 30 40 50 60 70 80 90 100 ilO degrees

F IG . 7.54.

We calculate the power characteristic in relation to the relative angle <5j2as:

Px = E[2yn sin a u +E[E2yl2 sin (<5J2-a12) = 0-189 + 1-182 sin (<5;2 + 7-8°);

P2 = E22y22 sin oc22-E[E2y12 sin (ô[2+oc12)

= 4-48-1-182 sin ( Ô ; 2 - 7 - 8 ° ) .

Large Oscillations and Dynamic Stability Currents

TABLE 7.8

257

^12, degrees

^ x l O - 3 a 2 x l 0 - 3 a12XlO-3

20

1-290 -00322

1-322

30

0-970 -00048

0-975

40

0-693 0-0208 0-672

50

0-467 0-0439 0-423

60

0-300 0-0638 0-236

70

0195 00799 0115

80

0155 00916 00634

90

0185 00987 00863

100

0-280 0-1009 0179

110

0-440 00981 0-342

120

0-658 00903 0-568

j0-179 J 0-096 0-0421+j 0-219 j 0-114 γ j0-0143 j 0-0286

ZZ -15·79 ZZ -157Ô y 0·0907+ΐ<Η>6β

FIG. 7.55.

TABLE 7.9

^ίί» degrees j 20 I

αχΧΐΟ-3 Ö J X I O - 3 αι2ΧΐΟ-3

0-991 -00571

1048

30

0-615 -00253

0-640

40

0-284 0-0040 0-280

50

0-012 0-0305

-0-0185

60

-0-195 0-0530

—0-248

70

-0-325 00710

-0-396 .

80

-0-383 00838

-0-467

90

-0-360 00911

-0-451

100

-0-260 00928

-0-353

110

- 0 0 8 7 00886

- 0 1 7 6

120

+0158 00788

+0079


The excess power is:

ΔΡλ =Ρ10-Ρ1 = 1197 —[0189 + 1182 sin (<5;24-7-8°)] = 1-008-1-182 sin (<5;2 + 7-8°);

ΔΡ>ι =^20-^2 = 3·875-[4·48-1-182 sin ( Ó ; 2 - 7 - 8 ° ) ]

= -0-605 + 1-182 sin (<5;2-7-8°).

We calculate the absolute acceleration of both stations from the equations:

EqU) = 1-054; EdU) = 1-097; ΔΕ^ = 0; Ed(5) = 1-02; P(4) = 0-4335; APU) = 00005.

We determine the relative acceleration as the difference between the absolute accelerations. The calculated values of the relative acceleration are given in Table 7.9.

From the calculations, a curve of relative acceleration, with the short-circuit switched off, is plotted (Fig. 7.54, curve 2).

The critical switching off angle, determined by the condition of equality of acceleration and braking areas, is 61Ό°.

We determine the time limit for switching off the short-circuit, by the method of successive intervals.

We take the value of the interval At as 0-05 sec. We obtain the relative acceleration a12 for various intervals of time, from curve 1 (Fig. 7.54).

First interval

«12(0) % i ° 4 ° ; _cc12{0)At* 1040X0-052 _

^ ^ ( D 2 = 2 ~ *

*ia<i> = ά'12(0)+Δδ[2ω = 28-2 + 1-30 = 29-50°.

Second interval αΐ2(ι) ^ 998;

JÓ; 2 ( 2 ) = Aô'12(1)+<z12(1)At2 = 1-30+998 xO-052 = 3-79°;

*Ì2(2) = «Ì2<i)+^Ì2<a> = 29-50 + 3-79 = 33-29°.

The calculated results are given in Table 7.10. A curve ò'12 =f(t) is given in Fig. 7.56. Taking the switching off angle

<5i2sw.off = 61-0° as the ordinate, we obtain the limiting switching off time isw.off = 0-277 as the abscissa.


TABLE 7.10

t, sec

000 005 010 015 0-20 0-25 0-30

degrees

28-20 29-50 33-29 39-27 4700 55-95 65-66

'«I*

1040 998 875 700 488 305

« 1 2 ^

2-60 2-49 219 1-75 1-22 0-76 —

degrees

1-30 3-79 5-98 7-73 8-95 9-71 —

degrees 60

50

40

30

20

0 0-05 0-10 015 0-20 0-25 sec

FIG. 7.56.

Problem 7.19

In a complex electrical system, the connections of which are given in Fig. 7.57, a two-phase short-circuit to earth occurs on one of the lines, between the substation a and station B, near the high-tension busbars of station B.

The switching off time of the short-circuit is 0*25 sec. The parameters of the generators, transformers, transmission lines and

loads, are given in Tables 7.11-7.14. Basic power is taken as Sbasic = 50 MVA and basic voltage, Fbasic = 110 kV on the high-voltage sides of the transformers.

xsw.offv L J L_ I J_ 1 NJ l

Required: to check the dynamic stability of the system by plotting curves of the variation of the absolute and relative angles with time.

Stat

ion

D

114-9

5 12

1 6-

3 St

atio

n A

60 k

m A

C-1

20

Stat

ion

C

Subs

tatio

n a

45 M

W

cos<

p=0

85

K£>

K9

Stat

ion

B

Fio.

7.5

7.



TABLE 7.11

Station

A B C D

Capacity of station

MVA

30 60 31-3 45

Transient reactance x'dt reL

units

0-421 0181 0-349 0-255

Reactance of negative sequence

xi9 reL units

0-312 0163 0-262 0188

Inertia constant Tj, sec

4-66 12-95

4-69 6-99

TABLE 7.12

Station

A B C D

Transformation ratic

121/6-3 121/6-3

114-95/10-5 114-95/10-5

Power, MVA

31-5 40 31-5 40

Reactances reL units

0-201 0158 0-201 0158

TABLE 7.13

Line

St. A-St. B Subst. £ - S u b s t . a Subst. a - S t . D Subst. a—Subst. b Subst. b-St. D Subst. 6 - S t . C

Resistance rei. units

00252 +y0-041 00316-ryO-0525 00576+y0-802 00513 + /00710 00295-fyO-0410 0-0210+yO-0350

Zero Sequence Reactances

0-2050 0-2620 0-2810 0-2480 01435 01225

I

TABLE 7.14

Load

St. B Subst. a Subst. b St. D

Power PL

reL units

0-30 0-90 0-84 0 1 6

cos φΣ

0-85 0-85 0-88 0-90

Load resistance rei. units

3090 +yl -915 0-834-fyO-517 0-940-fyO-506 5-900+y2-860


Solution. For a calculation of the dynamic stability, we combine stations C and D, which are remote from the short-circuit point, and introduce them into the equivalent circuit as one equivalent station.

The original equivalent circuit with four generating stations, in which the generators are represented by their transient reactances, is given (Fig. 7.58).

The sequence of reducing this circuit to a three station circuit, is as follows:

We transform the delta abc into a star abcO:

Zfl0 15~z3+z4+z5 _ (0-0513 +/0-071)(0-0576 +/0·0802) " 00295-

similarly,

00295 +y0-041 +00513 +yO-071 +00576 +/Ό-0802 = 0-02135 +J0-02975;

ZM = Ζ1β = 0-0109+./Ό-0152; ΖΛ = Z17 = 0-01225 +./001705.

We add impedances Z12, Z13, Z14 and Z2, Ζχ7 in series :

Z18 = Z12+Zlz+Zu = (0-021+7*0-035)+70-201 +/Ό-349 = 0·021+/Ό·585;

Z19 = Z2+Z17 = yO-158+(0-01225+yO-01705) = 0-01225+/0-175.

A portion of the transformed circuit is given (Fig. 7.59). We divide the load Sz between the points C and O :

à* per - ^ 3 ~ -

= (0·84+;0.459) 0-0109-70-0152 (0-0109 -y<M>152)+(0-021 -yO- 585)

= 0-0143 +/Ό-0261; similarly,

^^=0-825+70-432.

We divide the load St between the points D and O :

•SUV = 0-0638+y0-0342; 55per = 0-0961+./Ό-04305.


E~=1

024~

5,:0

·J.jO

·186

______

_11_

jO'18

1

JL

jO'I5

878

90·

02S

2.jO

·041

jO'2

01jO

·421

~E~

:1'2

,110

'7°

0'4+

j00 2

48

V=lo I

3!r

d'

0·9.

jO·5

35

S2:0

-9+j

O'5

59

12O·

021.

j0-0

35

IjO

-255

II..

EO:I-

187~

~0'6

+jO

'417 S

,=0'1

6+j0

-077

4

E~:10

1SLz

:1o

FIG

.7.

58.


We add the impedances Ζ1β, Z18 and Zly Z19 in series:

20 = Ζ1β+Ζ1δ = 0-0319+/0-6;

Z21 = Zx+Z19 = 0-01225+/Ό-43.

We add the impedances Z20 and Z21 in parallel:

Z± = 3 ° ^ = 0-00547+./Ό-25. ^22 Z M + Z 20"τ"^-21

The loads at the points O and i7 are determined by summation of the corresponding transformed loads (Fig. 7.60):

$o = &> = ^ per+^ per = 0-9211+./Ό-475;

£F = £e = $ξ^ + £ V = 00781+J0-0603.

00109 «■ J0OI52

53=0-84*j 0-459

Π * 1J 0021+j 0-585

FIG. 7.59.

22 [5 00055 + j 0-25 0 0 2 l 3 5 * i 0 02975

F ' . 0 ' a i — ^ — a — 1 — f 0-92IUJ 0-499 y j 0024 yo-9 + jO-5

S6 =0-0781 ♦] 00603 S5=0-92IUj 0-475 S2=0-9*j0-559

r j V v ^ T v J + j 0 - 5 3 5 S5=0-92IUj0-475 S2=0

FIG. 7.60.


To determine the power and the e.m.f. of the equivalent generating station F, we calculate the power loss and the voltage in the impedances Z15 and Z22 ;

_ 5?5 _ 0024' 15 - y- "it - -ρο2Γ ΔΡ» = ψα R* = ^ Γ x 0-02135 « 0;

àQn = η&χ» = - ^ - Χ 0 · 0 2 9 7 5 * 0.

The power flowing to the point O from the left is,

££2 = 0-921 l+yO-499;

Λ1Ζ — -^15-^15+ QlSX15 ^ Λ · ZJK15 — — ~~ U,

δΚ15 = P » * n - g " * » « 0

and, therefore, F0 = Fe = 1-02^-4-7°;

P22Ä22 +ß22x22

0-9211 X 0-0055 +0-499 X 0-25 = 0-1275; 102 _P 2 2 x 2 2 - ß22P22 0-9211x0-25-0-499x0-0055 _

22 VQ = Π02 ° ' 2 2 2 ;

VF = EF = V[(*O+^K22)2 + ÓF|a]

= V[(l-02+0-1275)2+0-2222] = 1-17;

x ♦ ° ' 2 2 2 (5f0 = arctan-py5-7-= 11°.

The phase of the e.m.f. vector of the generating station Fis: à'P=K + ÒF0 = -4-7 + 11 =6-3°.

The values and phases of the e.m.f. behind the transient reactances of the generating station, in the transformed circuit are:

for station A 1-24 ^ 10-7°; for station B 1-24 ^9-2°; for station F 1-17 ^6-3°.


The reference axis for the angles is assumed to be the direction of the voltage vector on the high-voltage busbars of the generating station A.

The power output to the system from the equivalent generating station Fis:

SF = £22 + ZL$22 + £e = 0-9211+y0-499 + , 0-92112+0-4992

A A A g g , .0-92112+0-4992 w A ^ ,

+ Πο^ ™°55+J ÎW2 χ0*25 +

+0-0781 +/Ό-0603 = 1-005+/0-8233. 22 15 6 7 8 9

Μ7/6Α — I-005-J0-8233

0-0055+J0-25 0O2l35*j0^)2975 0-03I6+J0O525Ο0252*]ΟΟ4ΙiO-201 jO-421

[ I0-96+J845 [j0-886+j0*56[j(>834*ja517<^ 10

j 0-158

0-4* i 0-248

26 3 0 9 * j 1-915

0-84 *j 0-631 H ]^βϊ

FIG. 7.61.

We determine the impedances of the transformed loads, and also the loads Sx and S2:

1-172 V2 Z03 = — (cos φ + / sin w) = ——

Se V(0-07812+0-06032) = 10-96+78-45;

(0-7915+/Ό-61)

V2 Z24 = -^- (cos φ +j sin φ) = 0-886 +/Ό-456 ;

V2 Z2s =-=r-(cos <p+/sing?) = 0-834 +70-517;

V2 Z2e = -^-(cos<p+/sin<p) = 3-09 + 71-915.

The equivalent circuit of the system, reduced to three generating stations, is given (Fig. 7.61).

To calculate the fault impedance for a two-phase short-circuit, it is necessary to determine the impedances for the negative and zero sequence currents. These impedances must be determined relative to the short-circuit point.


The negative sequence circuit is given (Fig. 7.62). The negative sequence impedances of the transmission lines and transformers have the same values as for the positive sequence. The negative sequence impedances of the loads are assumed to be 35 per cent of the positive sequence impedances.

As a result of transforming the circuit, we get that the total negative sequence impedance of the system, relative to the short-circuit point, is

Ζ2Σ = 0-0326 +./0-0946.

FIG. 7.62.

The zero sequence circuit is given (Fig. 7.63). The zero sequence reactances of the transformers are equal to the positive sequence reactances. The generators and the loads do not appear in the zero sequence circuit. The zero sequence reactaces of the transmission lines are assumed to be 350 per cent for single circuits, and 500 per cent for a double-circuit of the corresponding positive sequence reactances.

After transforming the circuit we get that the total zero sequence impedance, related to the short-circuit point, is :

Ζ0Σ =0-0351+y0-0929.

Adding the impedances of the negative and zero sequences, we obtain the value of the fault impedance:

Zk = / 2 r f ° r = 0-0086 +y0-0475. Ζ,2Σ ~Γ Ζ0Σ


An equivalent circuit for a two-phase short-circuit to earth is given (Fig. 7.64).

j 0-158

«i i i—ny^ 0-0576«.j0-28l O03l6*j0-26^ 0-0252+J0-205 J0-201

- a ny—|p..

djO-1 158

0021 *j 01225

<3 j 0-201

FIG. 7.63.

0 00086+j 00475

0-055 +j 0-25 0-02135* j 0-02975 00316+j 00525 I o-

IO-96*j8A5 0-886 +j 0-465

00252 + J0-663 -o2

0-834+j 0-517 / \ 0-158

309*j 1-915

3 0-181

FIG. 7.64.

For the convenience of further indications on the diagram, station A has the number 2, station B—3, station i7—1.


For all the stations, the self and mutual admittances are determined by the unit current method. The calculated results are given below:

Yn = 2-88 ^ -78-2°; a n = 11-8°; Y12 = Γ η = 0-1415 ^ -92-5°; a12 = — 2·5°;

Y22 = 1-43 ^ -87-4°; a^ = 2-6°; Yn = Yn = 0-277 ^ -95-8°; a13 = —5-8°;

Γω = 2-77 ^ -87-9°; 033 = 2-1°; 7» = r32 = 0-156 ^ -99-8°; % = -9-8°.

The active powers of the stations with the short-circuit are:

Pi = E'iyii s inan+JEû sin ( ^ - a ^ H ^ i ^ i s sin (^3-al 3) = 0-815+0-206 sin (δ'12+2-5°) +0-4045 sin (<$ί3+5·8°);

P2 = Ε[Ε&21 sin (δ21 -oî) +E22y22 sin a22 +

+^£ÌF23 sin (δ^-α^) = 0-206 sin (3^+2-5°) + +0-0997+0,2395 sin (<323+9-8°);

P3 = EÊ[yzl sin ( « » - « « ) + £ & « sin (ä32-a32) + +^2^33 sina^ = 0-4045 sin (0^ + 5-8°) +

+0-2395 sin (<5£2+9-8°)+0-1555. The active power outputs from the stations into the circuit under the

normal condition (Fig. 7.61) are: P10 = 1-005; P20 = 0-4; P^ = 0-84.

The excess active power with the short-circuit is: APX = Ρ 1 0 - Λ = 0-19-0-206 sin (δ'12+2-5°)-0-4045 sin (<3ί3+528°); ΔΡ2 = ^20-^2 = 0-3003-0-206 sin (<521+2-5°)-0-2395 sin (023+9-8°); ΑΡΖ = Pgo-Ps = 0-6845-0-4045 sin (0^ + 5-8°)-0-2395 sin (<532+9·8°). We calculate the angule of displacement of the rotors of individual stati

ons, by the method of successive intervals, taking an interval At = 0-05 sec. For station 1 we find the change of the angle during the first interval,

from the equation

A&> _ ^Ι^Λ(θ) Δ0ιω 2 '

where , _ 360/Λ;2 360χ50χ0·052 . _ kl ~ Tn - 4-69+6-99 = 3 ' 8 6 ;


APm = 0-19-0-206 sin [(6-3°-10-7°)+2-5°]-

-0-4045 sin [(6-3°-9-2°)+5-8°] = 0-1764;

Αδ'ιω = 3-86X °'ΙΊ264 = 0-3402°.

The angle at the end of the first interval is : ai<i) = 6-3+0-3402 = 6-6402°.

Similarly for stations 2 and 3 :

Αδ'2ω = 1-105°; δ2(1) = 10-7 + 1-105 = 2-805°;

Αδ3(1) = 1-023°; δ3(1) = 9-2 + 1-023 = 10-223°.

The calculation of the increase of the angles in the subsequent intervals up to the moment of short-circuit, is carried out by the formulae:

Λ*ίω - Λ ^ , + Μ Λ θ . - ι ) = Αδ'1<η_1)+3-86ΑΡ1(η_1);

M*ò ^ ^ ( „ - D + M ^ W - D ^^(„- i )+3-48 z l / ^ , .

After clearing the short-circuit, the fault impedance is not included in the equivalent circuit (Fig. 7.64) and the impedance of the defective part of the line is doubled. The remaining portion of the circuit remains unchanged.

The self and mutual admittances for the circuit of the post-fault condition are:

Yu = 2-175 ^ -72-1°; an = 17-9°; Yn = Y21 = 0-451 ^ -92-2°; a12 = -2-2°;

y22 = 1-205 ^ -85-4°, a22 = 4-6°; Y13 = Yn = 0-8725 ^ -95-4°;

*i3 = - 5 , 4 ° ;

Y33 = 1-902 ^ -80-2°; a33 = 9-8°; r23 = r32 = 0-605 ^ -100-5°;

a23 = -10-5°.

The active powers of the generating stations are determined by the same formulae as for the fault condition :

Px = 0-922+0-658 sin (δ'η+2-2°) + 1-27 sin (<5J3+5-4°); P2 = 0-658 sin (<521+2-2°) +0-1489+0-924 sin (<523 +10-5°);

P3 = 1-270 sin (<531+5-4°)+0-924 sin (532 +10-5°)+0-496.


The powers of the prime movers in the post-fault condition remain the same. The excess power is:

ΔΡΧ = 0-083-0-6585 sin (δ'12+2-2°)-1-27 sin (^3 + 5-4°);

AP2 = 0-2511-0-6585 sin (δ'21+2-2°)-0-924 sin (< 3 + 10·5°);

ΔΡΖ = 0-344-1-27 sin (δ31 + 5·4°)-0·924 sin (<532 +10-5°).

The constants k in the post-fault condition, remain the same as with the short-circuit.

The angular displacements in the first interval, after switching off the short-circuit, are calculated by the formulae:

Ah' - Ay -ulr ^I<5)+zlfi(s) . Δ01(6) - ^ d l < 5 > + * l 2 '

dô~(6) = Ζΐ<5Ι,0+£: AP'(s)+AP'i

3(5) " 3(5)

where ΔΡ' and ΔΡ" = the excess powers, before switching off and at the first moment after switching off the short-circuit respectively.

01 0-2 0-3 0< 0-5 0-6 sec

Fio. 7.65.

The angular displacements in the seventh and the subsequent intervals, are calculated similarly to those during the short-circuit.

From the calculated results, curves of absolute and relative angle variation with time are plotted in Fig. 7.65 and 7.66. From the diagram it is clear that the stability of the system is maintained.


degrees

16

8

0

- 8

-16

-24

*mn

-

-

-

-

SCH

?K ~~ i , ^ ^ J y

0-2 0-3 OÎ^^jO / jAi L-^Se sec

Fio. 7.66.

Problem 7.20

In the system, the circuit of which is given (Fig. 7.67), a two-phase short-circuit to earth occurs at the beginning of the line and is switched off in 0-2 sec.

Δ φ

FIG. 7.67.

The elements of the system have the following parameters:

GENERATOR

Pg = 280 MW cos yg = 0-8 Vg = 10-5 kV x'd = 0-35 xd = 1-1 xq = 0-65 x2 = 0-4

Td0 = 5 sec Tj = 6 sec

TRANSFORMER TX

STl = 350 MVA kTl = 10-5/242 *r, = 14%

TRANSFORMER T2

STl = 330 MVA kTi = 220/121 xrt = 12%

F /

* i

*o

LINE

= 220 kV = 225 km = 0-42 ohm/km = 3-5*!


The original condition of the system is characterized by the following parameters:

P0 = 200 MW; cos <p0 = 0-97; V = 120 kV.

Required: to determine the value of the short-circuit current allowing for the change of rotor position (allowing for oscillations).

Solution. We assume the basic values Vy^^ = 120 kV and Sb„ic = 230 MVA.

The parameters of the system in relative units are: xn = 0-113; xh = 0-288 (for two circuits) xi0 = 3-5x0-228 =0-798; xTt = 0085;

x'd = 0-283; xq = 0-525; xd = 0-889; x2 = 0-323; Tj = 9-13 sec.

We calculate the original condition prior to the short-circuit to be: P0 = 0-957; Q0 = P0tan φ0 = 0-957x0-251 = 0-240;

χ'„Σ = 0-283+0-113+0-228+0-085 = 0-709; E'z =V[(1+0-240 xO-709)2+(0-957 xO-709)2] = 1-353;

t a n ó i = - ^ = 0-58; à'o = 30-1°;

xqL = 0-525+0-113+0-228+0085 = 0-951;

E# =V[(l+0-240x0-951)2+(0-957x0-951)2] = 1-53;

t a n ó 0 = ^ - = 0 - 7 4 1 ; ò0 = 36-5°;

Ed0 = 1-353 cos (36·5°-30·1°) = 1-343;

C e *à *<* E" *à * ? td — hq— ZT· hi xq~xd

0-889-0-283 «X 0-525-0-283 ■Sx

0-889-0-525 0-525-0-283 = 2-5\Eq-\-50E'd;

Ed0 = 2-51 X 1-53-1-50x1-343 = 1-820. We set up a complex equivalent circuit for the short-circuited condi

tion. The negative and zero sequence circuits are made-up as shown (Fig. 7.68 (a)).


The resultant negative sequence reactance is: _ (0-323+0-113) (0-228+0-085)

*2Σ~ 0-323+0-113+0-228+0-085

The resultant zero sequence reactance is:

0-113(0-798+0-085) Χ0Σ — 0-113+0-798+0-085 = 0-100.

0-323 0113 0-228 0-085

0-113 / 0-798 UUÖD

ΠΡ—f ^-n^-^ 0-798 0085

(a)

1 0638

3

2 0-313

0-0640

(b)

FIG. 7.68.

The fault reactance is : 0-182x0100

*k = 0-182+0-100 00540.

We determine the self and mutual admittances for the complex equivalent circuit (Fig. 7.68(b)), in which the generator is represented by the reactance xq:

n„0 0-313x0-0640 Λ ,Λ, * n = 0 - 6 3 8 + 0 . 3 1 3 + 0 . 0 6 4 0 = 0-691;

χ,, =0-638+0-313 + 0-638x0-313 0-0640

= 4-08;

0 - = Φ 0 8 = ° · 2 4 5 ;


χ1 3 =0·638+0·0640+° · 6 3 80 .Χ3^6 4 0 =0-833;

^ = 0.313+0-06404-°·3130

Χ637640 =0-408;

b»=öm = 2·45· We carry out the calculation by the method of successive intervals, using

an interval At = 0-05 sec. At the first instant after the occurrence of the short-circuit, we have:

P _ 5 w - Vyufx,-x'd) cos "Sp * « ( 0 ) - 1 - ( * , - * > u

£;0-0-24S (0-525-0-283) cos ό0 1-(0-525-0-283) X 1-45

= 1·54£·;0-0·0913 cos ò0 = 1-54x1-343-0-0913 cos 36-5° = 1-997;

Ed(0) = 2-51 x 1-997-1-50x 1-343 = 2 99;

Λο) = E,to)vyn s i n δο = 0·245£,(0) sin ô0

= 0-245 X 1-997 sin 36-5° = 0-291; àPl0) = 0-957-0-291 = 0-666.

The increase of angle during the first interval is :

. . 360x50x2-5xl0"3 0-666 „ . . 0-666 , _,„ A\i) = pria X ^ — = 2·93χ—2~ = 1>64 ·

The angle at the end of the first interval is: Οω =36-5° + 1-64° = 38-1°.

The change of the e.m.f. E'd during the first interval is :

AEd(1) = (1-82-2-99) x ° ^ = -0-012.

At the end of the first interval the e.m.f. E'd is:

Ed{1) = 1-343-0-012 = 1-331.


The positive sequence current in the short-circuited branch may be determined by using the general expression for the current in a complex system:

Λ = Λ1+Λ2 = ^ > Ί 3 *ά-90*+-^γ» ^ - 9 0 ° -

l*-qS*n&S£·

Fio. 7.69.

Resolving the current /3 into two components on the direct and transverse axes of the generator (Fig. 7.69), we can obtain the expressions for the currents Idz and IqZ:

Id* = -r-yiz+-j-y22 cos <5; V3 v3

IqZ = T-^23 S Ì n Ô>


or

Id =Eâ-X 1-20 +4-X2-45 cos Ò = 0·694£β+1·415 cos Ô;

/„. = —J-X2-45 sin ô = -1-415 sin δ. V3

01 0-2 0-3 sec

FIG. 7.70.

At the first instant after the short-circuit:

/3(o) = 0-694X 1-997 + 1-415 cos 36-5° = 2-52; Iq3(0) = -1-415 sin 36-5° = -0-841;

h(o) = V(4(0)+4(0)) = V(2'522+0-8412) = 2-66.

In order to determine the total short-circuit current, we calculate the coefficient m:

, If 0-182 X 0-100 \ _ - V 3

N/^1~(0-182+0-100)2J - 1-52·

The total current at the short-circuit point is Isx- = ml3. For the first instant after the short-circuit:

S.C. ( 0 ) = 1-52X2-66 = 4-05.


The calculated results for the subsequent intervals, starting with the second, are given in Table 7.15, and the function 7SC# = f(t) is plotted (Fig 7.70).

TABLE 7.15

Interval

Ε,- 1·54£;<1,_1)-0·0913 cos a^a Et = 2-51E,-l,5E^_u P = 0-245E, sin δ(Λ_1}

ΔΡ = 0-957 -P M = Μ^+Α-ΚΔΡ^»

<5 = à(n_u + AÔ Ι^ = 0·694£·,+ 1-415 cos δ(η_ν

/„ = -1-415 sin a,._„ h = Ufi,*®

I... = 1 52/, ^ Ε ί - α - β Σ - ^ , ^ χ Ι Ο - «

Ed — £*(„_!)-h £*

2

1-978 2-97 0-300 0-657 4-88

430 2-49

-0-873 2-64 401

- 0 0 1 1 1-320

3

1-968 2-97 0-329 0-628 7-98

510 2-40

-0-965 2-58 3-92

- 0 011 1-309

8

1-980 307 0-457 0-500

20-71 1300

0-899 -1-33

1-61 2-44

- 0 0 1 2 1-251

9

1-989 311 0-372 0-585

23-59 153-8

0-469 - 1 0 8

118 1-79

- 0 0 1 2 1-239

It should be noted that in so far as the short-circuit is switched off in 0-2 sees., the function, calculated for t > 0-2 sec. depends on this condition.

Problem 7.21

The system is given diagramatically (Fig. 7.71).

Required: to make an approximate analysis of the stability for the simplest (single-phase short-circuit at the beginning of the line). The equivalent circuit of the system is given (Fig. 7.72).

E, 225 MW

( ~ ) C3D—f If 220 kV 255 km 280 MVA Δ i ^

FIG. 7.71.

Solution. We carry out the calculation in a simple way, using a nomo-graphical method.

In this method, any system is referred to the station circuit where busbars are of constant voltage. After this, nomograms are used. By means

K00 MVA


of known methods of conversion, the system is reduced to the circuit, containing two terminal stations of known power, which are joined to each other by à circuit configuration independent of any loading quantities. In the same circuit, for the standard, fault and post-fault conditions,"!*" the self and mutual total admittances Yll9 Y12, Y22

and> t^e e.m.f.s. E[ and E2, behind the transient reactances are determined.

_ jO-179 jO-096 0-O42UJ0-219 . j 0-114 v jO-0143 j0-0286 r

li-^T—0Γ—t-^H=zi—*-nr> f-^—ns^ --J5-79 ZI-J579 "D 00907* j 0-068 r

FIG. 7.72.

The motion of the rotor of the equivalent generator of each station is determined by the equations:

Tn dF2~

T d2^

Κ — δί',

_ P (Ed2

z l l

z 2 2

12 ~ya

HiHn sinau——= sin

Z12 E'E'

sin a22 + -Λ-2- s i n Z12

(%

(«il

,—a12),

+«12).

where

The equation of the relative motion is determined as the difference of two equations given above:

d2a; d2(32 _ Px P2 (E[)2 sin q n (E2)2 sin a22

d/2 d/2 " Γ Λ Γ/ 2 ζ'ηΤΛ + ζ22Γ/2

^ sin (δ'12-α12) J E ^ sin (Ó;2 +a12) Z12"*Vl Z 12 ' /2

ΓηΓ, J1ÂJ2

- [^'--'■-^'»+'-]-[7>2i>m sin (<5ί2-α12)+Γ/1Ρ„ sin (^ 2 +a 1 2 ) ] ,

t The fault condition is marked by one dash, the post-fault by two dashes.


where p _ W . p _ ( ^ ) 2 s i n a n . p _(^) 2 s ina 2 2 -*/* — ~Γϊ > * 1 1 ~~ 17 » "*22 — Zi ·

z 1 2 z l l z 2 2

(T T ) Multiplying the two parts of the last equation by ^-^— , multi-

plying and dividing, simultaneously, the first term of the equation by TJ2 and noting

%£μ = TJ0 and £? = m, -* T I ■* TO. * 1

we get

[ Jlx J2 x Jl

Wl2_TJ0l TJ0~fir = f~ [mPi— P2 — mP-u+PzÄ —

- = % - \Tn sin (δ'12-<χ12)+Τη sin (δ12+α12)]. ■L ri-I p lJl*J2

Expanding sin (<5ί2—α12) and sin (<5ί2+α12), and expressing them as the sum and the difference of the sines of δ12 and a12, we substitute for the variables, symbolizing

-^—-^- tan a12 ;= tan a = — ;

transposing F'F' F'F' x' 1 F'F'

ώ12 ^12 ^12 /I 1 _ L / *M2 c;2 HS M2

We symbolize mPi-Pi = A;

mPn = B?-±, xn

B = m{E[)\ where


The functions R/x are determined from nomogram No. 1 in Appendix 3. Also Pn = (Ε^)2Κη and P22 = (E2)2K22.

K12 — K 1

or

K12 — [arc tan — ^ xn J ( 1 +m2+2m cos 2 arc tan

(1 +m)z M%n T n X\2

are de-The values μ, sin μ and cos μ in relation to m = and

termined from nomogram No. 2 in Appendix 3. The coefficient K12 is determined from Nomogram No. 3 in Appendix 3. Inserting the symbols assumed above, we get:

CU i / 2 [ X\\ X\2 J -X-12

Symbolizing

ÂJ2\_ *11 *22 J

we get „ d2e nt E:E2 V . Λ

^0^2 = P« ilf 12 ' or

■*■ Tt\X-\ JQ-*12 X d20 = I'fM -sing.

Symbolizing

and

E[E2KX2 at2 Ε[Ε2ΚΛ2

J" — *0*12

- J ( E[E2Kl2

2>UKl2E[E2


we finally have for the theoretical circuit (Fig. 7.73):

To determine the value r, corresponding to the angle <5', we can use nomograms No. 1-10 of Appendix 4, where T is taken as equal to Γ.

FIG. 7.73.

The conversion to time t is effected according to the expression r

t //314 yj\Tn

P' where

P — ^1^2^12

To determine the time limit for switching off the short-circuit, it is necessary to know the critical angle of switching off the short-circuit 0lim, which can be determined conveniently from the rule of equal areas in the power characteristic.

The power characteristics P =f(ò'12) now take the form shown (Fig. 7.74).

We compare the acceleration area F&ccci and the area of braking - brake in accordance with the equation (for brevity the index 12 with the angle <5' is dropped everywhere, e.g. <5iim = ô[2iim):

•k i2 Je i2

whence ô[im = arc cos x

X V[rf+r22-2r1r2cos(^-/u0]

r2 sin μ" — rx sin u' -arc tan 2 ^ 1 r

r2 cos μ —rx cos μ


where

and

_ x12P0K[2 __ Ε[Ε2ύη(δ'0 + μ) . *12 °0 -*M2 *12°0

r2 — - " p " * " ~" v " P " Λ12» •*12r0 Λ 12 x"P'

^ = 1 8 0 - / ' - a r c sin — sin(«; + /i) ;

ό^ = the maximum angle during the oscillations.

Post-fault condition

Fault condition

FIG. 7.74.

For analytical calculations it is convenient to present the formula as follows :

π cos(<5;im + y ) = — X

X [*;-j|*;+(|^ where

V[r?+r|-2r1r2 cos (μ"-μ')]

Γ2$ίημ" — /^sin/z' y = arc tan r2cos/*"—A^COS μ'

For a three-phase short-circuit: γ = μ"; r1 =0.

An approximate solution may be obtained from the nomogram in Appendix 5, which gives an approximate value of the switching off angle, and from

f π — arc sin— — δ'0 lsin δ'0 — rxcos ö'0-\-r2cosi π —arc sin ° j ôum^arccos

r*-ri


When determining rx and r2, we assume μ^Ο. The sequence for determining the limiting values of the switching off

angle and the time of switching off the short-circuit, is as follows: (1) The equivalent circuit of the system under consideration, is reduced

to a circuit containing two terminal stations joined together by a circuit of indefinite configuration, with any number of loads.

(2) For the normal condition, the angle ò'12 0 , the power of station P10 and P20 and the e.m.f.s, behind the transient reactances E[ and E2 are determined.

(3) Calculations are made for the self and mutual admittances Yn, Y12 and Y22, for the normal fault and post-fault conditions (fault conditions are marked by one dash, post-fault conditions by two dashes). The loads are replaced by constant impedances; calculations of the total impedances for the system are made analytically or by model circuits.

T (4) Calculations are made of the value m = —^- and the ratios

-fln ^12 ^22

x n xl2 x22

(5) We determine Kll9 K^ and K12 with the aid of nomograms. The results of the calculations, specified in paragraphs (3), (4) and (5), are conveniently summarized in a table.

(6) We determine the values of

TJ0, A9 B, P0, P'09 P'0', Γ', rl9 r2 at μ=0.

Further, we determine (/Ί—r2), y. We now consider a numerical example. Assuming that the e.m.f.s are:

E[ = 1-215 ^39-2°; E'2 = 1-037^11°. The relative angle between the e.m.f.s of the stations is :

0'0 = δ'120 = 39-2-11 =28-2°.

The active powers of the generators of both stations for the normal condition are:

P10 = 1-197 and i>20=3-875.

A power of 200 MVA is taken as a unit of power. The inertia constants of the generators of both stations, reduced to

the basic power are: ΓΛ = 8-25 sec; r/2 = 112sec.


In the equivalent circuit for the fault condition (single-phase short-circuit) at point a (Fig. 7.72) an equivalent impedance equal to the sum of the negative and zero sequence impedances is inserted:

Zk = Z2+Z0 = 0-0121 +/Ό-184+0-0004 +/Ό-089 = 0-0125+7*0-273.

Table 7.16 gives the values of the impedances, the ratio R'/X', the coefficients Kll9 K22, K12 and, μ. Κη and K22 were determined from nomogram No. 1 in Appendix 3. The values K12 and μ were determined analytically.

For example, with:

m = — = 112

ΚΛ» —

TJX 8-25

l+m2+2mcos2

= 13-6,

fare tan ^ - 1

Ήϊ)Ί (1 +m)2

Γ l + 13-62+2x 13-6 cos 19°24'1 / 0-1312

(1 + 1

8-25x112 J0~ 8-25 + 112

1312\ 7672)

= 7-7;

Λ = mP1-P2 = 13-6 X 1-197-3-875 = 12-425; B = m(E[)2 = 13-6 X 1-215* = 20-2;

TABLE 7.16

Condition

Normal Fault Post-fault

'11

00605 00137 01032

*11

0-68 0-4354 0-8927

*»

0072 0-0687 00773

*22

0-112 0-1117 0-1126

rit

-0131 -0199 -0-1452

*12

0-767 1-22 1055

*11

0089 00316 0116

0-64 0-615 0-685

xl2

- 0 1 7 1 - 0 1 6 3 - 0 1 3 7

«ii

0088 0032 0114

K2t

0-454 0-446 0-466

Kl2

0-975 0-980 0-982


Ρ9=Ζρ\Α-Β^+(βί)*ψ\ 1J2\_ XU x22 J

7·7 Γ „ , „ „Λ„ 0-088 , Λ„„, 0-4541 = Π 2 [ 1 2 · 4 2 5 - 2 0 ' 2 Χ ^ 6 8 - + 1 ·°3 7 * <ΜΪ2J = ° · 9 8 ;

Ρ· _ Ζ ΐ Γ ΐ 2 . 4 2 5 - 2 0 · 2 χ - ^ 2 - + 1 · 0 3 ^ χ - ^ ^ 1 - 1-05-^ 0 _ 1 1 2 [ 0-4354 + U J / Χ 0-1117 J ~ 1 υ 5 '

/»·· -7'7|~12-d25 20-2X ° · 1 1 4 Ι1·08^< °'4 6 6 1-0-975-F° _ 1 1 2 [ 1 2 4 2 5 U 2 X 0-8927 + 1 U8 X 0-1126 J - ° y 7 5 '

r = ^ » = 1-05x1-22 ^ â ^ i z 1-215 X 1-037x0-98 '

_ l e s i n a ; , _ 1-215x1-037 sin 28-2° _ 1 ~ « ' " " 1-22x0-975 ΧΟ 98 - 0 487,

_ i n s i t i ò'0 _ 1-215x1-037 sin 28-2° 2 " ^ Ϊ Ρ Γ " " " 1-055X0-975 Χ ° 9 8 2 = ° ' 5 6 7 '

Γ 2 -Γ! = 0-567-0-487 = 0-08.

From the nomogram, using the known values of r2, (r2—rx) and δ'0, we determine ô'um ss 58°.

From the nomogram, using the values sin δ'0=0-472, 7" = 1-03 and ölim=58°, we determine the limiting relative time for switching off:

Tum = 1-45,

and

P'm _ E[E'zK[2 1.215 X 1-037x0-98 TJO x[*TJ0 1-22X7-7 = 0-132.

Further, we can determine tììm = 0-24 sec. We now determine the value of the limiting angle analytically. The values, determined above, remain unchanged, with the exception

of rx and r2, which, in this case, must be determined after allowing for μ:


similarly we determine for the values μ'=8° and μ"=6·75°: = Ε1Εί^β±Ε_ = 1-215X 1-037 sin (28 2o + 8 . 4 0 )

xi2Po 1-22x0-975 _Ε[Ε^ίη(δ,

0+μ) ν„ r2 — v" P" 1 2

_ 1-215x1-037 sin (28-2°+8-4°) _ - 1-055x0-975 X0 9 8 5 - 0 715,

r, sin u" - r, sin u' 0-715 sin 6-75° - 0-617 sin 8° v = arc tan — - ■— = arc tan — 7 r2 cos μ''-rx cos μ' 0-715 cos 6-75-0-617 cos 8°

= arc tan 0-015 = 1°;

ò'm = 180-/i"-arcsin|—sin(ói + /i)l

= 180°-6-75o-arcsinrp^jsin(28-2o+8-4°)1 = 116-75°.

Further, we determine: cosier)

V[0-6172+0-7152-2x0-617x0-715 cos (6-75°-8°)] -0-617 cos (28-2° + 8-4°)+0-715 cos (116-75°+6-75°) V[0-6172+0-7152-2x0-617x0-715 cos (6-75°-8°)]

= 0-059+0-0065ó,'im. ôlim is determined by method of interpolation, for which the equations

obtained are rewritten as follows: N = cos (ô[im +1») - 0-0065«^ ; N0 = 0-059.

We take different values for ôlim and determine the corresponding values N:

(a) ό ^ = 65°; Ny = 0-423-0-0065x65° = 0; (b) ô'um = 62°; N2 = 0-454-0-0065x62° = 0-051; (c) ô'iim = 61-5°; N3 = 0-477-0-0065x61-5° = 0-077;

hence, we determine: «i« = 61-7°;

θ'0 = 28-2+8-4 = 36-6°; sin 36-6° =0-596;

0,'im= 61-7+8-0 = 69-7.


From the nomogram No. 4 we determine for sin 0Ο%Ο·6, 0^=70° and 7=1-03:

τ=1·7. P

From the nomogram No. 5, we determine for -^- = 0·132 and τ = 1·7:

^ = 0 - 2 7 sec.

Thus, we obtain the following results: (a) calculation by nomograms 3 ^ = 58°; tlim =0-24 sec; (b) calculation by the formula derived above and by the nomograms:

*L=61-7e; 'iim=0-25 sec; (c) calculation by the method of successive intervals:

«L=61°; iim =0-277 sec.

Thus, the accuracy of the calculations is quite adequate.

Problem 7.22

In the system, the circuit of which is similar to the one given in Fig. 7.71, a three-phase short-circuit occurs at the station busbars.

The parameters of the system are: 7^=16 sec; Td0=5.65 sec: Te=0; Ede max = 5 " 4 .

*Σ

Xd X1

1

x* xt

Normal condition

1*32 119 105 0-79

During s.c.

0-52 0-39 0-25 —

After s.c. is switched off

1-41 1-28 114 0-88

Required: to determine the limiting transmitting power with the short-circuit removed in ^=0-12 sec.

We solve this problem by using the standard characteristics for determining the limit of the dynamic stability, allowing for the effect of the excitation controls as given by V.l. Gorushkin. These standard curves are given in Appendix 6; they enable us to solve a number of complex problems, in particular to determine at once the limiting power values for dynamic stability.


The characteristics give an approximate solution of the differential equation relative to the motion of the generator rotor, when the generator is connected through an external impedance to the busbars of constant voltage at the limiting load:

d2<5 Tjjp+PiEÏ, δ)=Ρ09

where P(Ed, δ) = the electromagnetic power of the generator, expressed in terms of the e.m.f. E'd, the angle δ and parameters of the circuit.

To analyse the rotor movement during the transient process, after the short-circuit has been removed, the initial conditions of the motion, ôk, sk and Édk (angle, the slip and e.m.f.) at the instant of the removal of the short-circuit, must be determined.

During the short-circuit, the duration of which does not exceed 0-2 sec, the change of the electromagnetic torque does not affect the rotor acceleration substantially and the acceleration may be assumed to be unchanged and equal to its value at the beginning of the short-circuit process. Besides, the slip sk is proportional to the duration of the short-circuit t\, and the increase of angle is proportional to t\.

The electromotive force Edk at the moment of switching off, may be determined from

E'äk = J r (tk(Ede-Ed) dt+E^0, ^ o j o

where Ed0 is the original value Ed. When integrating, it may be assumed that the e.m.f. Ed is equal to

its value at the beginning of the short-circuit process. The voltage of the exciter Ε^ varies according to a rule, which depends on the system of automatic excitation control.

The slip at the end of the short-circuit, does not, generally, exceed 0-5-1 per cent and the angle changes by a few degrees only. The electromotive force Ed increases by a small percentage when a large force is applied.

For an approximate solution, the differential equation, given above, is rearranged.

The electromagnetic torque of the generatori is represented in the form: P = φ(Ε^9 Ò) sin δ ^ ϊψ(Ε^9 ôk) +

+(ä).<'-«+(£).«-*)],h* g' y χ χ*

t The stator losses are ignored, φ(Ε19 ô) = —γ-—V2 —-—τ— cos ô. xd χχ'


where, in the expressions for the partial derivatives, the variablesEd=Edk and ô = ôk are substituted.

The differential equation of the rotor motion now takes the form:

Assuming, that the e.m.f. Ed varies in proportion to the angle i.e., that Ε·ά=Ε·άί<+ο{δ-δά,

we get an equation:

r , ç + { K £ i , «+[(S),+*(&)J<4-«}*· -'·■ Dividing this equation by (p(E'dfc, ôk) and changing the unit of time,

we get the equation of rotor motion in the form :

d2<5 .

where

and

m - P° <P(Edk> àk)

The slip at the instant of removing the short-circuit, referred to the new unit of time r is :

?(Edk> ôk)

On the basis of these relations, standard curves, which are given in Appendix 6, are constructed.

The standard curves give the limiting values rnT=mmax, which give dynamic stability at a given angle òk, parameter Sk and coefficient κ. If m r o n m a x , then the system is stable; if mT^mmaxy a disturbance of the dynamic stability takes place.

Determining, with the aid of the standard curves, the value of the coefficient κ, at which the dynamic stability is ensured (i.e. the equality mT=mM


is maintained), it is not difficult, from the relation κ=ψ (E'dk, ôk, b) to determine the minimum value bmin, required of the coefficient b consistent with the dynamic stability.

Knowing bnun, it is not difficult to determine the required amplitude of forcing AEde max and the voltage ceiling of the exciter Ede max :

^4max % bTdQSm*x> 1

^ e m a x ^ ^deO ~^^^de max * J

The calculations based on the curves, result in an approximate solution. The error in determining the limit of the dynamic stability, however, does not, generally, exceed 1-2 per cent, which is quite sufficient for practical purposes. The necessary forcing amplitude is determined with an accuracy within 10-20 per cent, which is sufficient for a provisional estimated

An analysis of the standard curves, permits us to determine some very simple relations, which are useful for the general evaluation of the effect of excitation control on the dynamic stability.

At m0 we mark the value mmax for κ=0 (i.e. with a sinusoidal electromagnetic torque-angle characteristic).

The limit of the dynamic stability mmax may be given, with adequate approximation, in the form:

»w = i*+n- - »* +^7ÔJ x [(Ü)k+b (H)J ' (A)

where η depends on the angle òk only, and is :

η =0·73-0·53(^-0·7) (òk is in radians).

The values m0 in relation to the angle òk, at different original slips Sk are given in Appendix 7.

The second term of wmax in equation (A), i.e.,

V fd<p\ <p{Edk,òk)\dò)k>

is practically independent of the parameters of the excitation control system.

t When greater accuracy is required the method of successive approximation may be used; the result obtained by assuming the linear dependence of Έ'ά on ô, may be used as a first approximation.


The increase in the limit of the dynamic stability resulting from forcing and excitation control, is determined by the value of the third term in (A):

or as a fraction of the nominal transmitted power:

If the losses in the transmission line are ignored, then

BE'd - x'd-Accordingly,

Xd

Thus, estimating numerically the effect of regulating and forcing the excitation on the dynamic stability, it is sufficient to know the value of the angle òk (in order to determing the coefficient η) and the coefficient b. The angle òk depends on the angle in the pre-fault condition, and also on the value and duration of the sudden reduction of power during the short-circuit. Usually for such calculations, the angle òk for long-distance transmission, does not exceed 70°-75°; the coefficient η, at the same time remains within the limits 0-47-0-41. When òk decreases, the coefficient η increases, which indicates, in this case, an increase in the effectiveness of the excitation forcing. The angle ôk, practically, does not depend on the parameters of the control system and the value of the excitation forcing.

The only parameter, by which the forcing effect is taken into account, is the coefficient b. The value of this coefficient, with given parameters of the system under consideration, depends on the amplitude of forcing. With three or four-fold forcings and with the usual parameters of the generators 6=0-2—0·4, and the coefficient ^=0-4 — 0-5. The limit of the transmitting power is increased, as a result of the excitation forcing, by (0-08-0-20) V/x'd9 (in the power units of the nominal transmitting condition).

For example, in the transmission Kuybyshev-Moscow, for which the time constant of the exciter was 0-12 sec and the forcing was retarded


by 0-06 sec, forcing, with a ratio of four to one increased the dynamic stability limit by 18 per cent of the limit when there was no forcing.

The increase of the dynamic stability limit by an amount of the order of (0·1-0·2) Vlxdy increases the transmitting capacity by 10-20 per cent in comparison with the case, when there is no automatic control, or by 7-14 per cent in comparison with the case when an exciter with a minimum ceiling voltage is used.

We now continue with a numerical calculation: assuming P = l at Vg = Vc = l. For the normal condition we have:

JSÏo =1-16; δ0=70°; ^ = 1-35.

At the moment of removing the short-circuit;

bk = 78° = 1-35 rad;

sk = 0-0075 rad/sec;

Edk = 1-20.

Further, we find:

b = **+-* = ^05 Td0 5-65 X 314x0-0075 '

*-A^[(S)M&)J"rè<^+MMxM, ,-M4i

s<W(^)=°W(^H2· From the characteristics, given in Appendix 6 (two lower figures on

page 408), we find: m =0-94. Then we determine the limiting power tobe: Pm = mmq>{E'dk, «*) = 0-94X 1-05 =0-98.

Problem 7.23

A remote hydroelectric station is connected through a double-circuit transmitting line to busbars with a high power system load (Fig. 7.75). The parameters of the elements and of the original condition, in relative units, are given in the equivalent circuit (Fig. 7.76). Regulators, which are mounted on the hydrogenerators of the hydroelectric stations, maintain Vrd equal to its value in the normal condition. The ceiling voltage is fourfold. The time constant of the exciter winding is Te=0-4 sec. The time constant of the excitation winding is ^ = 4 - 5 sec.


Required: to check the system stability with a two-phase short-circuit to earth and to determine the possibility of applying an automatic re-closing. The following cycle is the time of the t0 short-circuit at point K-\ ; ^=0-15 sec is the time of the switch-off of the defective line, At =0-25 sec is the zero current interval, ί2=0·4 sec is the time of reclosing the line from both sides.

G VQ T-l f L I \ T_2 Vc=const eiob-j: K-l

Fio. 7.75.

0-2+J006 V =const

Jay^p 1 1 —- , r^-^p 1 xg x^O-MS L—ny> 1 _ x^O-14 χ|2=0·26 I

111 x;=0-246 XQlS3xj xT3=005

xL=0037

|RL = 0-88

FIG. 7.76.

To solve this problem, two cases must be considered: (1) the short-circuit is eliminated after the automatic reclosing (suc

cessful automatic reclosing); (2) the short-circuit is not eliminated after automatic reclosing (unsuc

cessful automatic reclosing) and a re-opening occurs in 0-15 sec.

Solution. Assume that on the busbars in the original condition voltage VL = 1^0°. We find the e.m.f. of the station generators and the voltage on the busbars of the system substation in the original condition:

Èd0 = 1+005Χ1·198 + .0·92Χ1·198 = hQ6+jhm = , . ^ ^ .

E, = 1+0·05Χ0·529 + .0·92Χ0·529 = 1<ββ+β^ = M 3 6 ^ 2 0 > .

E'd0 = 1-136 cos (46°10'-25°20') = 1-136x0-935 = 1062.


The voltage on the busbars of the system substation is :

Yc = χ + 0-06X0-4+y0-2x0-4 = 1 . 0 2 4 + ; 0 . 0 8 _ h027 ^yy.

We consider this voltage to be constant and not dependent on the transmitting conditions.

The angle in the original condition is: àEd-vc = 46°10/-4°30' = 41°40'.

We determine the self and mutual admittances for different conditions.

(1) Normal (I)

The equivalent circuit for the normal condition is given in Fig. 7.77·

yi _ ! _ . 12 " A, »

Z[2 =yl-198+yO-4+;-OQ , ^ , , = -0-539 +/1-652 = 1-74^107°; 0-88+70-087 Y{2= 0-575; a } 2 = - 1 7 ° ;

yi -±. 11 Zli '

Ζ\χ =]1·™+Ζ?;::':Γή, = 0-138+71-488 = 1-50^84*40'; yO-4(0-88+;Ό·087) 7Ό-4+0-88+/Ό-087

Ylx = 0-667; α}χ = 5°20°.

f2J Two-phase short-circuit to earth (III)

The equivalent circuits of the negative sequence, zero sequence and positive sequence in a fault condition are given in Fig. 7.78(a), (b) and (c).

The equivalent impedance of negative sequence circuit is :

Ζ2Γ = 0-072+70-202 = 0-214 ^70°20'. The equivalent impedance of the zero sequence circuit is :

ZOE =jx0i: =70-109. The total shunt impedance of the short-circuit is :

Z - Z Il Z _ ./Ό·109(0·072+7Ό·202) _ ZA -

Ζ2Σ II Ζ0Γ - 0 . 0 7 2 + y o . 1 0 9 + y 0 . 2 0 2 - « 0085+70 073

= 0-0735 ^'83°20'.


The mutual admittance in a two-phase short-circuit to earth condition (Fig. 7.78(c)) is:

1 1 Y® = ζϊϋ = 9 72 = ° ' 1 0 6 ; a i" = - 1 4 ° --12

χ,=1·198 χπ=0·Α

► x ^ O - 0 8 7

Rm=0-88

Fio. 7.77.

j 0-356 jO-143 jO-14

Φ' ► j 0 0 5

>0Axj0037 = J0-0I5

0-4x0-88 = 0-352

(a) jOK jO-4 Vc E j0-915 jO-143 , _ . . ,_ .

► j 0-073 J O j 0-087

0 0 8 5 Π 0-88

(c)

FIG. 7.78.

jO-143

rr^~ jO-42

Q>

(b)

jO-IA

► jO-05

FIG. 7.79.

The self-admittance in this condition is:

yni — 11 ~ zff

1 1-122

= 0-892; a"1 = 0°15'.


NOTE. The self and mutual admittances in this condition were determined by the unit current method.

(3) Post-fault condition—the defective circuit of the transmission line is switched off (II)

The equivalent circuit for condition II is shown in Fig. 7.79. The self admittance is :

The mutual admittance is:

J_ 1 711 "12

y - = ZH = T 8 9 5 = ° - 5 2 8 ; «ί'2 = -18°30'

In order to calculate the transient process with successful automatic reclosing, it is also necessary to know the relation between the e.m.f. behind the synchronous reactance Ed and the voltage K^and the direct-axis e.m.f. component behind the transient reactance Ed, which remained unchanged with transition from one condition to another.

Similarly to the equation (2.41) in ref. 3, by substituting the generator voltage Vgd9 we obtain the following equation:

E Vgd-Vcyi2XdCos(ò-oc12) d l-\yn*rfCOsau

whence Vgd = Ed(\ -ynxd cos a n ) + Vcy12xd cos (<5-a12).

We now write the equation for Ed for various conditions: (1) The normal condition (I)

Ei = K ^Ji2fa-^)cos(g-qj2) d i -y\\{Xd-Xd)cos αΐι i -yliixd-x'd)cos αίι

E'd 1·03χ0·575(0·915-0·246) 1-0-667(0-915-0-246) cos 5°20' 1-0·667(0·915-0·246) cos 5°20

xcos(â + 17°) = l-80£^-0-713cos(ô + 17o);

Vgd = ■E'iil -y\i*d cos axn) + Vcy\2xd cos ( δ - α ^

= £ i d ( l -0 -667x0-915 cos 5°20') + 1·03χ0·575χ0·915 cos (0 + 17°)

= 0-393£j+0-542 cos (5 + 17°).


Check. Ed0 = l-80xl-062-0-713cos(41°40' + 17o) = 1-91-0-371 = 1-539;

1-539 — 1-529 AEd = ^ 2 « = 00065(AEdper cent = 0-65 per cent).

(2) The fault condition (III)

Em=__lk 1-03X0-106X0-669 * 1-0-892X0-669 cos 15' 1-0-892x0-669 cos 15' M + ;

= 2 -47^-0-176 cos (0+14°) ;

V™ = £™0 -0-892X0-915 cos 15') +1-03 X 0-106x0-915 cos (0+14°)

= 0·184£ΠΙ+0·097 cos (δ + 14°).

(3) The post-fault condition (II) „„ E'd 1-03x0-528x0-669

d 1-0-612x0-669 cos 4°50' 1 -0-612 x 0-669 cos 4°50' A

Xcos(3 + 18°30') = 1-69^-0-615 cos (ô + 18°30');

K5 = £y(l-0-612x0-915cos4°50') + l-03x0-528x0-915x

xcos (<S + 18°30') = 0·44^+0·497 cos (<5 + 18°30').

For all conditions in each time interval:

where

ΔΕ'Λ =^Ç^At = % Ζ ^ χ 0 · 0 5 = 0 - 0 1 1 ( ^ - ^ ) . ■i do 4°

Here Efc = the forced synchronous e.m.f., which varies with the functioning of the forcing excitation, in accordance with the rule:

Ede = Ed0(4-3e") . The calculation is now carried out by the method of successive intervals. The time interval for the calculation is At =0-05 sec. _. , . 360/zJ/2 360X50X0-052 . . . The constant k = —— = — = 3-75.

1 j 12

It is assumed that the regulator has a time delay Treg = 0-05 sec. As an illustration we give the calculations for several time intervals.


(I) The initial interval (from the instant of short-circuit until t=0-05 sec)

The synchronous e.m.f. Ed at the instant of the short-circuit is :

Em = 2-47x1-062-0-176 cos (41°40' + 14°) =2-521.

The electrical output into the system is :

Pm = 2-522 x 0-004+2-52 x 0-106 sin (41°40' +14°) = 0-246.

The increase of the angle during the original interval is :

Αδω = 0-5kAP{0) = 0-5x3-75(0-92-0-246) = 1-260° = 1°16'.

The angle at the end of the original interval is:

<5i = άω+Αδω = 4Γ40/ + 1°16/ = 42°56\

The voltage Vf}0 at the incoming side of the regulator is :

K$0) = 0-184x2-52+0-097 cos (41°40' + 14°) = 0-518.

Thus, Kj2J0)= 0-518 -< Vgd0 = 0-82 and the excitation forcing affects the operation.

The effect of the forcing will not be immediately apparent due to the time delay in the circuit of the regulator (TTeg = 0-05 sec).

We further find the change of the e.m.f. E'd\

AE'd{Q) = 0-01U(Ede-Ed\Q) = 0-0111(1-529-2-521) = -0-011.

(2) The second interval (0-05-0-10 sec)

The transient e.m.f. is:

E'd = 1-062-0-011 = 1-051.

The synchronous e.m.f. is:

Ed = 2-47x1-051 -0-176 cos 56°24' = 2-504.

The electrical output is :

Pe = 2-5042x 0-004+2-504x0-106 sin 56°24' = 0-248.

The increase of the angle is :

Αδ(2) = 1-26+ 3-75(0-92-0-248) = 3-78° = 3°47'.

300 Transient Phenomena in Electrical Power Systems : Problems

The value of the angle at the end of the second interval is:

<5(2) = <5(1)+zl<5(2) = 42°56'+3°47' = 46°43'.

The change of the transient e.m.f. is:

ΔΕ'ά = 0·0111(1·529-2·504) = -0-0108.

The voltage at the input side of the regulator is :

Vgd = 0·184χ2·504+0·097 cos 56°24' = 0-514.

Further steps in the calculation are carried out similarly. The calculated results are summarized in the Tables 7.17, 7.18, and 7.19

and displayed in Fig. 7.80.

TABLE 7.17

/

[0] 0

0-05 010 [0-15] 0-15 0-20 0-25 0-30 0-35 [0-40] 0-40 0-45 0-50 0-55 [0-40] 0-40 0-45 0-50

[0-55] 0-55 0-60 0-65 0-70 0-75

Λ

0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92 0-92

P.

0-92 0-246 0-248 0-255 0-267 0-904 0-998 108 116 1-22 1-26 1-69 1-75 1-78

1-26 0-324 0-326 0-323 0-308 1-29 1-39 1-28 1-17 106

ΔΡ

0 0-674 0-672 0-665 0-653 \ 0016 /

—0078 - 0 1 6 -0-24 -0-30 -0-34 \ -0-77 / -0-83 -0-86

-0-34 \ +0-596 / +0-594 +0-597 +0-612 \ -0-370 / -0-470 -0-360 -0-250 - 0 1 4 0

kAP

0 1-20 2-52 2-49

1-25 -0-292 -0-615 -0-881 -1-12

- 2 0 9 -3-10 -3-22

+0-475 +2-23 +2-24

+0-45 -1-75 -1-36 -0-945 -0-532

Δ6

0 1-20 3-78 6-27

7-52 7-23 6-62 5-74 4-62

2-53 -0-57 -3-79

+ 509 7-32 9-56

1001 8-26 6-89 5-95 5-42

Ò

4Γ40' 41°40/

42°56' 46°43' 52°59' 52°59' 60°30' 67°44' 74°21' 80°05' 84°42' 84°42' 87°14' 86°40' 82°53' 84°42' 84°42' 89°48' 97°07'

106°4Γ 106°4r 11604r 124°58/

131052' 137°49'

Variable

Fault condition WVSLJVAA 11VS11

Post-fault condition

Sl lP f f^ f l l l KJ UWVW31 U l

aiitnmatif CL U LWind tiW

reclosing

Unsuccesful automatic reclosing


TABLE 7.18

/

0 005 010 015 0-20 0-25 0-30 0-35 0-40 0-45 0-50 0-55 0-60 0-65 0-70 0-75

-0Ό5

_ 0 005 010 015 0-20 0-25 0-30 0-35 0-40 0-45 0-50 0-55 0-60 0-65 0-70

Χ = ΊΓ4

_ —

0125 0-250 0-375 0-500 0-625 0-750 0-875 100 1125 1-250 1-375 1-500 1-625 1-750

e~*

_ —

0-882 0-779 0-686 0-606 0-534 0-472 0-417 0-368 0-324 0-286 0-253 0-223 0196 0174

3e~x

_ —

2-65 2-34 206 1-82 1-60 1-415 1-250 1105 0-971 0-858 0-758 0-669 0-588 0-522

4-3Q-*

_ —

1-35 1-66 1-94 218 2-40 2-585 2-750 2-895 3029 3-142 3-242 3-331 3-412 3-478

Ede — EdeQ X

x(4-3e~x)

1-529 1-529 2062 2-54 2-96 3-33 3-67 3-95 4-21 4-43 4-63 4-81 4-96 509 5-22 5-32

2 0 L I 1 I 1 1 I I I [ I l i i i i ID10 0 0 5 0-10 015 O20 0 2 5 O30 0-35 OAO 0 4 5 0 5 0 0 5 5 O60 0 6 5 s e c

FIG. 7.80.

302 Transient Phenomena in Electrical Power Systems: Problems Va

riab

lesi

n~

Ico

s~

IIa

=tl

Ed

-V'-I

-(X

IZlJE

d-

V o

TA

BL

E1.

19

V,d

0'01

11x

(Ed

,-E

d)IE

~IE

d.IE

dIE~

-EdI

[0]

1'06

21'

531·

53-

-0'

881

41°4

0'58

°40'

0·85

40-

520

1·06

21·

532·

52-0

'99

-0'0

110'

518

41°4

0'55

°40'

0'82

60'

564

Faul

t0'

051·

051

1'53

2·50

--0,

976

-0'0

110-

514

42°5

6'56

°56'

0'83

80'

547

cond

ition

0'10

1'04

02·

062·

48-0

,32

-0·0

030·

504

46°4

3'60

°43'

0·87

20'

489

[0'1

5]1·

031

2·54

2·49

--

0·48

952

°59'

66°5

9'0'

920·

391

--

0·15

1·03

72'

54I-

55+

0'99

+0'0

110·

841

52°5

9'71

°29'

0-94

80-

318

0'20

1'04

82·

961·

65+1

-31

+0·0

140·

822

60°3

0'79

°0·

982

0-19

10·

251·

062

3·33

1·75

+1'

58+0

·018

0·80

567

°44'

86°1

4'0-

998

0·06

7Po

st-f

ault

0'30

108

03·

671·

86+

1'81

+0·0

200·

790

74°2

1'92

°51

'0·

999

-0·0

50co

nditi

on0'

351'

100

3·95

1·95

+2'

00+

0'02

20·

785

80°0

5'98

°35'

0·98

9-0

,149

[0'4

0]1·

122

4·21

2·04

--

0'78

484

°42'

103°

12'

0·97

4-0

-228

--

0'40

1'12

24'

212·

162·

050'

023

0·74

484

°42'

101°

42'

0-97

9-0

-203

Succ

essf

ul0'

451·

145

4'43

2-24

2'19

0-02

40·

745

81°1

4'10

4°14

'0·

970

-0-2

46au

tom

atic

0'50

1·16

94·

632·

272'

370·

034

0·76

386

°40'

103°

40'

0'97

2-0

·236

recl

osin

g0'

551'

195

4·81

2'28

--

0·80

182

°53'

99°5

3'0-

985

-0·1

72--

[0'4

0]1'

122

4'21

2'04

--

0'78

484

°42'

103°

12'

0·97

4-0

·228

0'40

1-12

24'

212·

801'

410·

016

0·49

984

°42'

98°4

2'0·

989

-0-1

510'

451'

138

4'43

2·85

1'58

0·01

80·

501

89°4

8'10

3°48

'0·

971

-0-2

380'

501-

156

4·63

2-92

1·71

0·01

90-

503

97°0

7'II

I°07

'0·

933

-0,3

60U

nsuc

cesf

ul[0

'55]

1·17

54'

812·

99-

-0'

501

106°

41'

120°

41'

0·86

0-0

'510

auto

mat

ic0'

551·

175

4·81

2·34

2·47

0'02

70·

744

106°

41'

125°

11'

0·81

7-0

,576

recl

osin

gQ

-60

1·20

24'

962·

472'

490·

028

0-73

311

6°42

'13

5°12

'0-

705

-0,7

100-

651·

230

5·09

2·57

2-52

0·02

80·

732

124°

58'

143°

28'

0·59

5-0

·804

0'70

1·25

85'

222·

662-

560·

028

0·73

813

1°52

'15

0°22

'0'

494

-0'8

69


The analysis shows, that high-speed automatic reclosing on the double-circuit transmission line, for the example considered, should not be used. If the switching off of the defective line does not lead to a disturbance of stability, an unsuccessful automatic reclosing would lead to a falling out of step.

CHAPTER 8

ASYNCHRONOUS OPERATION, RESYNCHRONIZATION AND SELF-

SYNCHRONIZATION

IN THE present chapter, the behaviour of electrical systems when asynchronous operation occurs, is considered.

During the last ten years, in the electrical systems of the Soviet Union (and following it, in a number of systems in other countries), non-synchronous running of machines which fall out of step and the subsequent re-synchronization, without interference from the serving personnel or with some action on their part by the governor control of the dropped out unit (resulting in overall stability), is beginning to be widely used.

A fairly accurate and simple method of calculating the asynchronous condition and the subsequent re-synchronization, has not yet been worked out.

The simplest method, giving a rough approach to the evaluation of the slip occurring during asynchronous operation, is confined to the determination of the point of intersection of the turbine power curve and the asynchronous torque (Problem 8.5); the condition for bringing about resynchronization, considered in a simple, very approximate way, is assumed to be that the resynchronization takes place, if the momentary slip passes through zero (Problem 8.4).

More exact calculations, allowing for the speed governor of the turbine, are given in Problem 8.1, where, however, simple assumptions relating to the electrical power (torque) of the synchronous machine during asynchronous operation are made.

The results nearest to reality, are obtained by the method used in the example of Problem 8.2, where the maximum number of controlling factors is taken into account.

In order to help the reader to examine this complicated question, the formulae in this chapter are given in the most complete form.

304

Asynchronous Operation 305

Problem 8.1

A hydrostation is connected with a system of infinite power by transmission line (Fig. 8.1). The parameters of the transmission system are as follows: the torque of the turbine in the original condition ΜΓΟ = 0·75; MTnom= 1; 7^ = 12-5 sec. The speed governor has a rigid feed back An« =0-75; <5L<r. =0-0665; 7 > 7 sec; 0 = 1; kn=0; an=a =0-4. The asynchronous torque (mean value) is taken to be constant and equal to -"^asyn (mean) "*^·

E'd0-<5>[ QQ_[v=consl

FIG. 8.1

Required: to determine conditions, under which a station which has fallen out of synchronization may be re-synchronized.

Solution. We solve the problem approximately, allowing for the action of the speed governor, but at a constant asynchronous torque.

As simplified assumptions, we take: (1) The transient processes (electromagnetic) in the machine, which

is fitted with an excitation control, are ignored in the calculation, but are indicated by the condition Éd—const.

(2) The periodic components of the electromagnetic torque, which have a double frequency, are ignored.

The equation of motion of the rotor of the generator may then be written in the form:

A 2β\ TJ = ~^2 = Mr-^asyn(mean)-M12 SHI <5, (8.1)

or d.s

TjS di = Mr~Masyn(mean)-^12 S h l Ó ' ^8'2)

The values entered on the right side of equation (8.2), are complex functions of time and slip, hence the solution of the equation, without simplifying assumptions, can be carried out only by the method of successive intervals. The solution of this problem may be facilitated considerably if we assume the following :


(a) The synchronous torque produces an instantaneous deflection of the rotor which varies periodically from the mean value. The amplitude of this deflection rapidly diminishes with the increase in speed, since the rotor possesses a considerable amount of mechanical energy.

(b) The speed regulator responds only slightly to the periodic variation of the instantaneous slip because of its inertia, and therefore it is possible, approximately, to assume that the synchronous torque does not have a substantial effect on the change of the turbine torque.

In accordance with these assumptions, the effect of the synchronous torque on the motion of the rotor, can be determined, independently of the other torques'acting on the rotor shaft.

In this case, integrating (8.2), we get:

Γ /Τ = f (Mr-Masyn.<mean)) d<5+M12 COS Ô + C, (8.3)

where the left side allows, approximately, for the kinetic energy of the rotor, and the right side for the operating forces applied to the rotor, where the non-periodic components give the mean operational value, which corresponds to the mean value of the kinetic energy of the rotor,

s2 equal to Tj - ^ ^ . Therefore (8.3) may be represented in the form

TJ~^TJ ^ψ + M12 cos Ò, (8.4)

where

T/ΐψ = f (Mr-Masyn(mean))dô + C. (8.5)

At the instant when the kinetic energy of the rotor, in the process of braking, is zero, the instantaneous slip takes the zero value, i.e. the necessary condition for resynchronizing is fulfilled. This condition may be expressed mathematically by the inequality

τ/^ψ+Μ12 cos 0^0. (8.6)

In the course of one period of asynchronous operation with the generator approaching synchronizm, it is possible to assume, that the mean slip is constant. The necessary condition for synchronization may then be written:


The method of calculating the mean slip is based on the solution of the rotor's equation of motion, allowing for the speed of the regulator. For this, the known equations are applied.

(a) with the valve ports not fully opened:

T. ™1+βΜτ+ψ^ Saem = ßMT0; (8.8) u * umean.rot

^ . r o t = «I.ro^f.v. · ( 8 . 9 )

(b) With the valve ports fully opened:

MT = Mm±MTmBOm-—^. (8.10)

In the equation of the rotor motion, the characteristic of the asynchronous torque may be presented in the form of a piecewise linear function; then

7>%^+(an+Mmean) = MT9 (8.11)

where (an-\-krismean) = the function of the linearly piecewise section of the asynchronous torque characteristic.

By solving together (8.8), (8.11) and (8.10), we get:

TjTs i^aa+(ßTJ+krtsmcan) ^ +

+ (^I^+ßk')s = VW™-*»), (8.12) and

r y ^ + ^ m e a i l = M n o ] - a r t ± M r . n o m - ^ . (8.13)

In the process of calculation, when passing over from one linear section of the characteristic of the asynchronous torque to the next, the initial conditions for the subsequent section will be the final conditions of the preceding one. Knowing the curve of the mean slip variation, it is possible to determine the changes of the turbine torque with time, using the equation (8.11).

In the asynchronous condition, the turbine distributor, during a certain time interval, may be completely closed ; then the mean slip, in this period, is determined from the equation

Tj^n+knSmeaa = -(MTjr+aJ. (8.14)


The manner of variation of the valve opening with time, forgiven functions Smcan = fifji) and MT = f2(t), may be determined by applying the equation for the speed regulator valve:

σ = η-βμ.

Thus, the rotor slip calculation is carried out by solving three linear equations: (8.12), (8.13) and (8.14). Each of these equations is used in relation to the condition of the set (the opening of the valve and the distributor of the turbine).

From the calculation of the normal condition we determine the limit of the transmitting power:

pe = Me = -4— sin Ò = M} 12 sin

where

TUT E'äV

The possibility of resynchronizing the generator is determined from (8.7), by the condition

Tj

If, during asynchronous operation, this inequality holds, the momentary slip will take the zero value and, therefore, the generator may resynchron-ize. It may be noted, however, that this is only a necessary condition for synchronization, as after several synchronous oscillations or immediately after j i n s t = 0, a "collapse" of synchronization may take place. Determination of the sufficient conditions for resynchronization is not considered here.

(1) We consider the first case. We assume, that, with the falling out of the generator from synchronism, the relative motion of the rotor starts with the valve ports not fully opened and is determined by (8.12):

= (fTm + ^ " W ° = β(Μτο-°η). (8.15)


The equation (8.15), allowing for the assumed variables and the original data, will take the form:

/Mr. —Y V r.vAot /

J < rp Q ^mean \_τ mean r / "-T.nom 1 p \* η

J s dì2 J at ' z—*— I mean ~~ Μτο~α' or

β ^mcan , * w ^mean , ^*Γ. nom — ίΛ/ . π)

at2 + Γ , Χ di + 7 > r ^ , A . r o t m e a n ~ l Γ0 }TjTt

A particular integral of this equation is : __ ^f-vA-rot

MT.nom (Mn-a). (8.17)

We note, that Smtan st. corresponds to the point of intersection of the static characteristics of the speed regulator and the asynchronous torque. The roots of the characteristic equation (8.16) are.

1 /Γ / 1 \ 2 \/ΐ "Λ (8.18) /Ί.2 — 2T^^ll\2Ts

Substituting the numerical values,

^mean st.

1

0-75x0-0665 1

\\( 1 V

/ T/^/if,

we get:

(0-75-0-4) :

A.rotJ'

= 0-0175;

1 η,ζ 2x7^yJ[\2x7j 12-5x7x0-75x0-0665 J

= -0-071 ±y0-47= -a±yft ; a =0071; β1 = 0-47.

Therefore, the complete solution: imean = e-«(* sinß1t + Ccosß1t)+*mean „.. (8.19)

The integration constants B and C are determined by the initial conditions.

Let 'S'mean=Smean0 at i=0. Then from the equation (8.19) we get: C = Jmean 0 ~ ^mean st. '> ( 8 · 2 0 )

= MT0-(kjm^0+an). (8.21) m,r Differentiating the right side of equation (8.19) and equating it to the

right side of (8.21), we determine B:

B = MT0 " fa, + M m e a n o) + TjCcC ^ ^

Tjßi


For the given problem: MT0-a+TjacC

B- Tjß, ' Thus, we have, putting S ^ o =0:

C = 0-00175 = -00175; _ 0·75-0·4+12·5χ0·071(-0·0175) . A „ β = 1- -—- ._ = U-U57. 12-5x0-47

Therefore, rnean = e"0'0 7 1 C 0 * 0 5 7 s i n ftf-0-0175 COS ßxt) +0-0175,

or •Wan = 0-06e-°071< sin (0-47/-17°)+0-0175.

The turbine torque is determined by the equation:

MT = 7> ^^+(an+A:^m e a n). (8.23)

Differentiating the equation (8.19) and substituting into (8.23), we get: MT = TjQ^{C2QOsß1t-B2unß1t)Hdn+knsmtJy

where C2 = ßxB-*C\ B2 = ocB+ß^; (8.24)

B2 = ocB+ßxC = 0-071 χ0·057+0·47(-0·0175) = -0-0042 % 0; C2 = ß±B-ocC = 0·47χ0·057-0·071(-0·0175) = 0-0282.

Therefore, Μτ = 12-5e-0'071' 0-0282 cos&/+0-4 = 0-35e"0071' cos 0-47/+0-4.

(8.25) Knowing the turbine torque MT = /(/), it is possible to determine the

opening of the distributor:

M(0 = - ( Λ / Γ 0 - Μ Γ ) ^ .

We can now find the opening of the valve ports:

a = -JssHL-flti. (8.26) dL. rot

As, in the given example ß=l, then

°L. rot


The calculated results are given in Table 8.1. The calculation shows that the valve ports do not fully open (Table 8.1).

TABLE 8.1

/, sec

0 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

W

0 00271 00495 00605 0062 00545 00395 00226 00065

- 0 0 1 0 - 0 0 1 1 - 0 0 1 0 -00045

00075 00175 00295 0033 0035 0034

^metn ^mean

δΖ^ ~ 0 0665

0 0-41 0-75 0-91 0-93 0-82 0-59 0-34 01

- 0 1 5 - 0 1 6 5 - 0 1 5 - 0 0 6

011 0-26 0-44 0-5 0-27 0-51

MT

0-75 0-705 0-61 0-482 0-355 0-255 0195 0185 0-23 0-295 0-375 0-45 0-51 0-53 0-525 — 0-45

— 0-37

μ

0 0034 0105 0-2 0-3 0-37 0-416 0-423 0-39 0-34 0-28 0-225 018 0165 017

— 0-225

— 0-285

σ

0 -0-376 —0-645 -0-71 -0-63 —0-45 - 0 1 7 4

0083 0-29 0-49 0-445 0-375 0186 0055

- 0 1 1 — 0-475 —

-0-225

(2) We now calculate the mean slip in the case when A/asyn = 0; for this the initial conditions are the same:

Put = -0-071 ±70-47 = -cc±jß1; a =0071; ß1 = 0-47;

_ i"f.r.òlim (Μτο-α) = 0-75x0-0665 st. M \ — i» - / .

mT. nom l xO-75 = 00375.

Therefore, the complete solution takes the form :

*mean = β — (B sin ßtt + C COS A O + W n »t., (8.27)

where 0-75 +12-5 x0-071(-0-0375)

B = 12-5x0-47 = °"122;

C = imean0-*s.. = 0 - 0 - 0 3 7 5 = - 0 - 0 3 7 5 .


Substituting the values of B and C in (8.27), we get:

*mean = e"0'071' ((M22 sin ftf-0-0375 cos ß±t) +0-0375, or

*mean = 0*125 e"0071' sin (0-47/- 17°)+0-0375;

MT = Tj e-"' (C2 cos ßit-B2 sin ßxt) [see (8.23)],

where C2 = ßxB-<zC = 0·47χ0·122-0·071(-0·0375) == 0-06; B2 = χΒ+β^ = 0-071 χ0·122+0·47(-0·0375) % 0.

Therefore, ΜΓ = 12-5 e-°0 7 1 ix0-06 cos ßtt, Μτ = 0-75e-°071i cos 0·47ί;

or

μ = - ( Μ Γ 0 - Μ Γ ) - ^ - ; σ = - ^ - μ . mT.nom uL.TOt

The calculated results are given in Table. 8.2.

TABLE 8.2

/, sec

0 1 1-4 2

*?mean

0 0057 0073 01010

*?mean ^mean

<5j.rot 0-0665

0 0-850 112 1-52

MT

0-75 0-65 0-59 0-45

μ

0 0075 0 1 2 0-225

0 -0 -775 - 1 0 - 1 - 3

It can be seen from Table 8.2, that the valve ports are fully open at / = l-4 sec. For this,,

MT = Mno) = 0-59; Smean = 5m e a n 0 = 0073,

where the symbol 0 refers to the instant of full opening of the valve ports. The later calculation is carried out on the assumption that the valve

ports are fully open. In this case, the mean slip is determined from (8.13),

Tj ^ f 2 + k j a m = MT0-an - MT. nora — ^ - . (8.28)


But since, in the given example Afasyn = 0, (8.28) will be written in the form:

as AJ fa — IVITQ mT.

t >r,Ts

Integrating, we get:

^mean 0 + : M. TO t-Tj 27>7>f.r.

Substituting the numerical values in (8.29), we get:

(8.29)

^.mean= 0 - 0 7 3 + ^ - r - 2 x 1 2 - 5 x 7 x 0 - 7 5 '

or *mean = 0073+0*048*-0Ό076* 2 .

The turbine torque at cr= — 1 is determined from:

ΜΓ = Λ/Γο"τά:=0'59-7^75;

μ = -(Μτο-Μτ)μίτ; σ °mean δ X. rot

μ.

(8.30)

(8.31)

(8.32)

We carry out the calculation from formulae (8.30), (8.31) and (8.32) and obtain the values of μ and a. The calculated results are given in Table 8.3.

TABLE 8.3

/, sec

0 1 2 3 4 4-9 5

*?mean

0073 0113 0138 0148 0144 0123 0122

•?mean ^mean

<5z.rot 0-0665

112 1-7 208 2-2 217 1-86

MT

0-50 0-38 019 0027

- 0 1 5 7 -0-33

μ

012 0-28 0-42 0-54 0-68 0-81 0-81

<T

- 1 0 -1-42 -1-66 -1-66 -1-49 - 1 0 5

From Table 8.3 it can be seen that, after the lapse of f=4-9 sec, after the instant the ports of the valve are fully opened, the turbine distributor has been found to be fully closed. This is shown by the fact that the turbine


torque MT is equal to MTTm = 0-33 (at μ/Γ# = 0-75). The opening of the turbine distributor at this instant is 1 :

μ =0-81 + μ0 =0-81+0-19 = 1;

μ corresponds to MT0 = 0-75.

The turbine distributor will be fully closed up to the time when σ = — 1. When a is equal to zero, the speed regulator will start to act on the opening of the distributor. The opening will continue until a is within the limits 0<<r*s£l.

With the distributor fully closed, the motion of the rotor will be determined by equation (8.14):

r /% S B + *-»— = -WT.T+an).

In the given example AfMyn = 0; hence this equation becomes: as

Tj as " " M ^ ·

Integrating, we get: 0-33

"12-5 t =0·123-0·0264ί,

where SmmΔ = the slip at the instant when the distributor is fully closed.

TABLE 8.4

s δ -μ

t, sec

0 1 1-6 2 3

■ rnean

0123 0068 0054 0042 0015

/ •Tut«*

00665

1-86 102 0-81 0-63 0-225

,. rot

MT

-0-33 -0-33 -0-33 -0-33 -0-33

μ

0-81 0-81 0-81 0-81 0-81

(T

- 1 0 5 -0-21

0 018 0-59

It is seen from the table, that after a lapse of t = 1-6 sec, from the instant that the distributor is fully closed, the turbine torque starts to rise. This rise starts with the valve ports not fully closed.


The further motion of the rotor is determined, as before, by equation (8.12). The original conditions will be as follows:

"ΊιΐθβηΟ = 0-054; MT01 = -0-33 (Table 8.4);

*mean = e - 0 ' 0 " ' ( * ™ϊ ß,t + C COS ßtt) + J œ e a n s t . ,

-0-33 +12-5 x 0071x0-0165 (8.33)

B =

where ^ — Smean0 ^mean st,

,-. * n *n = -00535; 12-5x0-47

= 0054-00375 = 00165.

or

Substituting the numerical values in (8.33), we get:

■fme«, = e-0'071' (-0-0535 sin ßxt+00165 cos ßj) +00375,

*mea„ = -00560e"0071'sin (0-47* -17°)+0-0375; ΜΓ = Tj e-" (C2 cos ßxt-B2 sin ßj),

where

or

C2 = ßxB-<*C = 0·47(-0·Ο535)-Ο·Ο71χΟ·0165 = -00263; Ä, = aß = ßxC = 0071(-00535)-0-47xO-0165 = 0-004;

MT = 12-5 e"0071'(-0-0263 cos i-0-004 sin ßxt),

MT= -0-33e"0071' cos (0-47/-8°);

μ = -(MT0-MT) Ht M. ο=-ψ^-μ.

T. nom "L· rot


TABLE 8.5

/, sec

0 1 2 3 4 5 6

•?me»n

0054 00285 00085 00025 00045 00025 00065

•?mean

àL.rot

0-81 0-43 0128

- 0 0 3 8 - 0 0 6 8

0038 01

MT

-0-33 -0-29 -0-2 - 0 0 8

0043 014 0195

f*

0-81 0-78 0-71 0-62 0-535 0-46 0-42

σ

0 0-35 0-58 0-66 0-603 0-42 0-32


When constructing the curves ^mean = /(/) and MT = f(t), the time values in every section should be added.

(3) We now examine the case when Masyn#mean= 0 - 4 2 - S ^ . With the valve ports fully opened, the motion of the rotor is determined by the equation (8.12):

+(irrS!i-+ßk·)5— = PWTO-"»)-\ nr°L. rot /

In the given case, this equation is written as follows:

TjTs^^HTj-Ts)^ + i^^--kymeaa = MT0-a.

A particular integral is

MTQ-a -^Γ. nom___£

(8.34)

(8.35)

("f.Arot Substituting the numerical values in (8.35) we get:

0-75-0-42 1 - 1

= 0-0175. (8.36)

0-0665x0-75 The roots of the characteristic equation (8.34) are :

= - l 2 · 5 - 7 ± i\( l 2 · 5 - 7 v l—( ι Ai

2x12-5x7 xy L ^ 2 x l 2 - 5 x 7 y 12-5x7^0-75x00665 ^J

= -0-03315 ±y0-465 = -x±jß1; a = 00315; βγ = 0-465.

The general solution of (8.34) takes the form : *mean = e " " iß *& ft/ + C COS ßLt) + î m e a n s t . ,

where

mean B = Tjßy

^ "" ^mean 0 ^mean st. '

Asynchronous Operation

Substituting the numerical values, we get:

_ 0·75-0·42 + 12·5χ0·0315(0·0175) B = 12-5x0465 = ° ' 0 5 7 ;

C = 0-00175 = -00175. Therefore,

*mean = e"0'0315'(0057 sin &Ì-0-0175 cos &0+0-0175, or

mean = 006 e"00315'sin (0-465*-17°)+0-0175; MT = Tj e - ' (C2 cos ß1t-B2 sin ß1t)+MMsyUm mean ,

where B2 =<zB+ßC = 0·0315χ0·057+0·465(-0·0175) = -0-006; C2 = ßB-<zC = 0·465χ0·057-0·0315(-0·175) = 0-027,

whence Μτ = 12-5e00315'(0-027 cosßxt+0-006 sin ß±t) +3/asyn.

= 0-34e-00315 cos (0-465*-14°)+0-42-.ymcail . The solution is summarized in a table, similar to Table 8.5. The course of the resynchronization process is shown in Fig. 8.2.

317

Examples

FIG. 8.2.


Problem 8.2

A hydroelectric station js linked with a system of infinite power (constant voltage V= 1) by a transmission line. The reactance of the transmission line, which is represented in the equivalent circuit by series elements (Fig. 8.3 (a)), and the direct-axis synchronous reactance of the generator, are totally represented by a reactance Arc = l-0. The synchronous e.m.f., in the set condition of operation is Ed0 = l-2, and the generators give out the nominal active power, which is taken as the basic unit (Pnom =l).The sets are characterised by the following constants: 7^=16 sec; 7^=4 sec.

Required: to determine the approximate duration of the asynchronous operation of the station generators, which occurs after a loss of stability from the normal condition, and to determine the period of the cycle of operation of the generators without excitation, if it is assumed that the automatic switches of the generators are opened when the stability is disturbed, and closed again at the instant when the excitation is restored and successful ^synchronization takes place.

The time constants of the turbine control system are characterized by the values: the time constant of the servomotor is Ts=6 sec; the operating time of the distributor valves from the full-load position to the free running position is 7^=4-8 sec.

The braking torque at the closing of the distributor is Afbr r=—0-4; the relative opening of the distributor, which relates to the free running of the set, is /Jf.r.=0-8 (μ=0 when the distributor is fully opened; with the distributor fully closed μ = 1).

The initial slip of the asynchronous operation is provisionally taken to be zero; the isodromic linking is ignored; due to the great length of the transmission line, the asynchronous torque is considered as independent of the slip and is taken as Masyn = ±0-1.

Solution. We consider several stages in the transient process during each of which some group of factors has the predominant effect. The characteristics of the transient process at each stage are, moreover, calculated by allowing for the essential factors while the others are ignored for the first approximation.

(1) Time period 0^t*^T's

The operation of the sets in this period of time, is determined, principally, by the variable torque of the turbine, the variation of which occurs


o o —

(c)

FlO

.8.

3.


because of the effect of the speed control system. It is assumed that the asynchronous braking torque of the generator is constant (with appreciable ohmic resistances present in the transmitting system, the braking torque is determined also by the "inherent" power of the generator).

If, for practical reasons, in the evaluation of the transient process duration rate, the presence of the isodromic linking is ignored as is suggested by the conditions of the problem, then the calculation of the transient process period under consideration, as well as the subsequent periods, may be determined.

Then for the first stage:*

[(1-01)'-2ί6^8'!] = 5·62'-0·65Λ 100 16

The slip at the end of the period, i.e. at t=T's, is determined, from the equation obtained, as :

-s ,% = 5-62X4-8-0-65X4-82 = 12-0.

The maximum slip value takes place at M ^ M ^ at the instant

whence

' , « , = T.Pt*. MT.L = 6X0-8X—p— = 4-32 sec,

- * « , % = 5 0 ( Μ ™ - Μ ^ η ) 2 ^ · Γ MT.LTJ

100., . . . . 6x0-8 -2- ( 1-°-1 ) X-f^Î6- = 1 2 - 2 s e C ·

(2) Time period T^t^t,

The factors, which determine the principal operation of the set in the second period, are the synchronous torque of the generator and the turbine torque, which, when the opening of the distributor is larger than the opening for free running, produces a braking effect.


The slip is then determined by the expression:

S/°- 7>|_ 2 X Τ,{\-μ(.τ)

,, , ~,Λ 100Γ °·

Χ2(1(-"^χ6-°·^-4·8)] + 12

4 2 - X

= -0·625(/-4·8)-1·04(ί-4·8)2 + 12.

From this we determine the slip at the end of the period, at t = Ts :

-ST.% = -0-625(6-4-8)-1-04(6-4·8)2 +12 =9-75.

(3) Time period Γ, «£ ί «e tAGB

The braking of the set continues with a fully closed distributor. Prior to the commencement of the functioning of the distributor, for the usual conditions, the rotor is slowed down steadily.

The slip at this instant is determined by the expression :

1001 Tr -sAGB% = — I -(Mrr+Masyn)(rJ-r;)+

100 16

?[-, MT.T (T,-Tlf 1

2 X ΤΛΙ-μ^)]

| - _ ( _ 0 . 4 - 0 . 1 ) X ( 6 - 4 . 8 ) - 0 . 4 x 2 ^ | y = 2-25.

The duration of the third period of the transient process, may be calculated according to the equation:

- * A G B % = (MT.T + M,syn)(tAGB-Ts)—- sTt. 100 Tj

Substituting the known values here, we get :

100 2-25 = ( -0 -4 -0 .1 ) ( f A G B -6 )x^+9 .75 ,

hence, we determine the instant at which the distributor should again be opened and automatic breaker of the generator closed :

'AGB = 8*4 s e c ·


Therefore, the approximate duration of the operation of the generators, without excitation, is 8*4 sec. The duration of the asynchronous condition, cannot be-determined yet, since, after the automatic generator breaker has been closed, the operation of the sets is determined, not only by the increasing turbine torque and the constant asynchronous moment, but also by the slowly alternating, gradually increasing, synchronous torque. This case compels us to calculate the fourth period of the transient process, regardless of the fact that the calculated results of the third period were obtained on the assumption that the slip has a zero value with /J=/Jfreerun.

The analysis of the fourth period is also necessary in order to ascertain if the attainment of zero slip by the set is not only a necessary, but also a sufficient condition for a successful resynchronization.

(4) Time period t > tAGB

In the final period of the transient process, the motion of the generators is determined by the following equations :

(a) the equation of the relative motion of the rotor:

7 > ^ | = MT-Mci = MT-(M^+M,y,

(b) the power equation of the turbine under control when :

'AGB^'^>AGB+o;- :r ; )

takes the form

"'—»[■-lÄ]: for

Μτ = MTL — = ; H'f.x1 s

and for / > tAGB + Ts,

MT = MT±m ; (c) the equation of the synchronous power of the generator is:

Edv · x Ms = , a

ft/x sinä; ■k-n


(d) the equation for the synchronous e.m.f. of the generator is:

Ed = Jl-e;0. The solution of this set of equations is carried out by the method of

successive intervals. We take the time interval for the calculation to be At = 0-1 sec; the

initial angle we assume, provisionally, to be, ό0=0.

First interval

The excess torque at the beginning of the interval is :

AM0 = M r -M a s y n = -0 .4-0-1 = -0-5.

The change of angle during the first interval is :

where , 360f0At2 360χ50χ0·12 „ „ , Λ k = — ^ = — = 11-25 el. degrees; Tj 16

Abx = ^ χ 3 6 0 χ 5 0 χ 0 · 1 - 1 1 · 2 5 ? ^ = 40-5-2-8 = 37-7°.

We determine the increase of the e.m.f. for the first interval as well as for the subsequent one by assuming a strong forcing eflfect for which

ΔΕα1 = *deÇ, *«> At =*-^χ0Λ = 0 1 . Ed(tf-Edn A. 4 - 0 rd

The angle at the end of the first interval is : δ1 = δ0 + 01 = 2,1-T.

Second interval

The slip at the beginning of the interval is : Aôt 100 37-7x100

S P e r CCnt = Ή7Χ170Λ = 0-1X360X50 = 2Λ

The value of the e.m.f. at the beginning of the interval is: Edi = Edl = 0-1.


The synchronous torque at the beginning of the interval is :

ΜΛ = / ^ Γ ο / χ sin \ = 1 ( 10 1 1

0Χ

01

2 1 ) sin 37-7° = 0-0626.

{ ιοο) The turbine torque at the beginning of the interval is :

Μ" = -0 ·4[1-(ΐ^6]= =-0 ·3 6 7 · The excess torque is :

ΔΜ1 = M n - M a s y n - M , = -0-367-0-1 -0-0626 = -0-530.

The increase of the e.m.f. during the second interval is : 4—0-1 AEd2 = — ^ - i = 0-0975.

The increase of angle during the second interval is : Αδ2 =Aô1+kAM1 = 37-7-11-25x0-530 = 37-7 = 5-95 = 31-8°.

The angle at the end of the second interval is :

<52 = 37-7+31-8 = 69-5°.

The characteristics, obtained from the equations and the calculated results of the first three stages of the transient process, are shown (Fig. 8.3 (c)).

It is not difficult to see that, in the case considered, a successful ^synchronization takes place, and that the whole transient process lasts about 15 sec. The slip reaches the first zero value approximately 9 sec after the start of the asynchronous operation.

Problem 8.3

An electrical station, in which three turbo-generators are installed, is operating in parallel with a power system circuit (as shown in Fig. 8.4 (a)). The busbar voltage is kept constant and equal, in relative units to F = = 1 · 11 at a basic voltage Fbasic = 115 kV.

The parameters of the transformed circuit, Fig. 8.4(a), in relative units with Kbasic = 115 kV and basic power Sbasic=225 MVA, are given in Fig. 8.4(b). The parameters of the generating station, in this circuit, have the following values: xd=6-l; .χ^=0·72: ^ ' = 0 - 4 5 ; Λ^' = 0 · 6 3 ; TJ = 7 sec; Te=0-57 sec; Td0 = 3060 rad ; Td0= 75-4 rad; Τ^ = = 760 rad. The generators of the electrical station are provided with a relay


forcing excitation, which guarantees the maximum value of the forced e.m.f., £Λ/=2·5 Ed *do-

Required: to check the possibility of applying automatic reclosing to the line 7=1, without checking the synchronization with two turbo-generators operating in the station, assuming that:

(1) the automatic reclosing is accomplished after the three-phase short-circuit at the point shown in Fig. 8.4, has been cleared.

(2) the short-circuit is cleared, by the switches B± and B2, 0-2 sec after its appearance.

(3) the reclosing of the switches takes place after a no-current pause of 0·4 sec.

(4) The forcing of excitation starts after a delay of 0-05 sec from the instant the short-circuit appears until the reclosing of the switches Bx and Bo.

Solution. The decision whether automatic reclosing without synchronization control may be used, depends on: (1) to what degree the impulse currents and their corresponding mechanical torques, which appear at the moment of switching, will affect the generators of the electrical system; (2) the time required to re-establish synchronization.

TG-40500/II0

-Κ£Η 60 km TG-3I500/II0

80 km / ^ * - N ,V*con«t K3D-T

^

FIG. 8.4.


Before finding the magnitude of the impulse currents, which depend on the angle of deviation of the synchronous machine rotors, it is expedient to evaluate the effect of the currents, corresponding to the most unfavourable condition (5 = 180°), on the generators. For this, the periodic component of the switching current is calculated as follows :

_„ _ 2 E y ^ 2 - l xcL XcE

where, in the given problem,

xcL = 0-45 +0-3 4-0-545 +0-75 = 2045.

Therefore,

'- = we = 1025· In actual units, this current at the generator voltage level, calculated

from the mean transformation ratio, is:

l'ä = l'a* ΛΜ* = 1025 , 2 2 5 i £ = 12-7 kA. >/3>W V3X 115 10-5

At the same time, a periodic current flows in the circuit of each of the generators

ii(D = 'dit) = 0-5 X 12-7 = 6-35 kA.

The nominal current is :

IL = -, — = 3-44 kA, V3x 10-5x0-8

therefore,

= 1-85, Id 6-35 IL 3-44

which is permissible, since the limiting value is 3-5. In order to determine the possibility of using automatic reclosing with

out synchronization control, the calculation is carried out by the method of successive intervals. The self and mutual admittances of the system circuit for the normal, fault and post-fault conditions of operation and the values of the constant coefficients in the formulae from which calculations are carried out, have the values indicated in Table 8.6.

Asynchronous Operation

TABLE 8.6

327

Condition

II. Fault III. Post-fault/switches Bx and B, are off

>Ίι

0138 0147 0125

>12

010 0 0

a n

2-9° 0-7°

14-5°

a12

-16°

With the normal condition: pm = Ε&η ( i ) s i n «ii (i) +EdVyi2 (i) s i n (ò-«12 (i))

= £20-138 sin 2-9° +£^1-11x0-1 sin (Ò +16°) = 0-00699£3+0-l l l sin (0 + 16°).

With the fault condition: p u = E$y11(ll) sina1 1 ( I 1 ) = E% 0-147 sin 0-7° = 0-00179£2.

With the post-fault condition :

Λ π = ^ π α ι ΐ ) sina1 1 ( I I I ) = £J0-125 sin 14-5° = 0 0 3 1 3 ^ . (8.39)

The change of the e.m.f. (ref. 3) is : E'd cos(ó-a12)

Ed = l - ( ^ - ^ ) ^ n C O s a n -vyvt- 1

-Je

Xd"xd

dE'd Ed + Td0-^ ·

T -yu cos a n

(8.37)

(8.38)

(8.40)

(8.41)

For an exciter with independent excitation, with a damping resistance inserted in the circuit of the exciter's excitation, and with forcing, we get:

Edef = ^demax-AEdee Γ · .

With the forcing removed: (8.42)

/ -oe £ * rf = (** ,-0-β -Eäeo) « T> +Ede0 . (8.43)

For a strict approach to the solution of these equations, various time constants Te should be considered (with inserted and shunted damping resistance).

In the conditions of the problem :

Ede = £ , + 3 0 6 0 aEd d ' (rad)

£•,+9-75 άΕ', at, (sec)


hence, for the finite difference we have:

^Eld=Ef^At. (8.44)

Equation (8.44) takes the form : E'

E" = l - (6 - l -0 -72)Î0-138cos2-9 o _ 1 ' 1 1 X 0 ' 1 X

χ cos (0 + 16°) = 3 . 8 5 £ ( , _ 2 . 3 1 c o s ( δ + 16°); (8.45)

^ T ^ 7 2 - ° - 1 3 8 c 0 s 2 · 9 0

in the fault condition : E'

Ed = 1-(6·1 -0-72) X0-147 cos 0-7° = 4 ' 7 6 ^ ; ( 8 ' 4 6 )

in the post-fault condition: E'

E" = 1 -(6-1 -0-72)0*125 cos 14-5° = 2 ' 8 6 ^ · ( 8 , 4 7 )

The equations (8.41) and (8.42), in the conditions of the problem, have the following numerical coefficients :

t_ t__

Edei = 2-5x2-44-(2-5x2-44-2-44) e ° " = 6-1-3-66e °·57 ;

( 0·β \ f-0-6

6-l-3-66e °'57-2-44je 0'57+2-44

/ -0-6 = 2-39 e °'57 +2-44.

These characteristics are given in Fig. 8.5(a). With reclosing, the generators also develop asynchronous power.

Therefore, for further calculation, it is necessary to calculate the constant coefficients in the formulae of the asynchronous power, which depend on the currents in the damping circuit. It is unnecessary to determine the asynchronous power which corresponds to the current in the excitation winding, since in calculating the synchronous e.m.f. in accordance with equation (8.41), the expression (8.37) includes the synchronous, as well as the asynchronous components of the power produced by the current in the excitation winding.


The asynchronous power of the turbo-generator damper winding may be calculated from the formula (refs. 3 and 8):

P 4-P "* asyn [d\*A asyn [q\

= -v2 x 2[1 -(Xd-Xd)yn cosau] [l-(xj-x'ä^n c o s <*u]

X i+fs^y Γ1 ± V[l + W ) 2 ] sin (23-2a1 2-arctan ^ Y l -

yìziXg-x'g)

Here

-V2- sTL

X

2[1 - (x , -x;*)y u cos a u ] 1 +(JT;')

Γΐ T V[l + ( < ' ) 2 ] sin ( 2 0 - 2 ^ - a r c tan ^ Ί .

7If i .,· l-ixd-x'dlyiicosau d0 1 - (xd - x ^ u cos a u '

7? =7£[1-(*,-*;')*! cos «„].

E-d.

t 01 0-2 0-3 0-4 0-5 0-6 0-7 0 8 0-9 10 I I sec

(a)

s, per

cent 08

0-6

0-4

0-2

0

0-2

0-4

nc

60

50

- 4 0

- 3 0

- 20

- 10

ίδ~

<N ) l 0 2 0 3 0 4 0 5 0(

sV

j/o 7 0· 8 0 9 1· 0 !· t

2 sec

(b)

Fio. 8.5.


Substituting the given parameters, we get for s «= 0 : l-(6-l-0-45)x(H38cos2-9° _

ld * / 3 ' 4 1 - ( 6 - 1 - 0 - 7 2 ) X 0 1 3 8 C O S 2 - 9 ° '

rq' = 760[l-(6-l-0-63)x0138cos2-9°] = 188; i'asynW = - l ' H * X

0·12(0·72-0·45) %

2[1 -(6-1 -0·72)χ0·138cos 2-9°] [1 -(6-1 -0-45)x0-138cos 2-9°]'

xn^Ss?[ ' -^ 1 + < i x 6 4 · 8 » Xsinf 2ό+2x16-arctan—ττ~Λ \~ y sx64-8J\

0-l2(6-l-0-63)2 sxm 1·112χ - - — x v

2[1 -(6-1-0-63)2X0-138 cos 2-9°] 1 +(*x 188)2

xri+V[l+(5Xl88)2]xsin('2Ó+2xl6-arctan—Töö-)] =

Tl-V[l+(iX64-8)2]x 1·92ί 1+(ίΧ64·8)

X s i n ( 2 Ô + 3 2 ° - a r C t a n ^) ] -- n | x w [ l W [ 1 + ( i X l 8 8 ) 2 ^ xsinf 2ό + 32°-arctan——— ) . ^ i x l 8 8 y j

In calculating the transient process, we assume that at the instant of the short-circuit, the following remain constant : the turbine power ΡΓ=0·222, d0=25-8°, and the e.m.f. E'd, whence,

Eäo =3^5 [ £ Λ + 2 · 3 1 COS (25-8° + 16°)]

= ^ ( 2 - 4 4 + 2-31x0-7455) = 1-08.

For the purpose of calculation we take a time interval of At =0-1 sec. In this case

360fAt2 18000X0-12

k - —fj— - ^ - 25-7.


First interval

We determine the e.m.f. at the instant of the short-circuit from (8.46):

Em = 4·76^0 = 4-76 X 1-08 = 5-15. The output of the generators at this instant, corresponding to (8.38), is:

P0 = 0·00179£3[0] = 0-00179 X5-152 = 0-0475,

and excess power is : AP0 = PT-P0 = 0-222-0-0475 = 0-174.

The increase in angle during the first interval is :

, ^ o *>^ 0-174 _ . , 0 Δδχ = k-^ = 25-7 x — — = 2-24°.

We calculate the change of the e.m.f. E'd, and for this we find, from the curve, (Fig. 8.5(a)), the mean e.m.f. Ede after the interval, for which we assume that the forcing action starts with a delay of 0-05 sec. In these conditions Edel=2- 51 and

ΔΕ'Λ = £ %7 7 f ' [ 0 l x (H = 2 , 5 g~ 55 ' 1 5 x0- l = -0-0271.

Therefore at the end of the first interval, òl = δ0+Δδ1 = 25-8+2-24 = 28-04°;

E'A = ΕΜ+ΔΚΙ = 1Ό8-0-027 = 1-053; zJ<5xxl00 2-24x100

* = —Ϊ80003Γ = "18 000x0-1 » -0-124 per cent.

Second interval Edl = 4·76£;1 = 4-76 X 1-053 = 5-01;

/»! = 0·Ο0Π9Ε^ = 0-00179 X 5-012 = 0-045; AP1 = PT-PX = 0-222-0-045 = 0-177

(we consider the turbine power to be unchanged because of the inertia of the control system);

Αδ2 = Δοχ+kAP! = 2-24+25-7x0-177 = 6-79°. From the characteristic (Fig. 8.5), we determine: £A8=3-0; therefore


The system condition at the end of the second interval is characterized by the parameters:

62 = δ^Δδζ = 28-04+6-79 = 34-83°,

E'd2 = 1-053-0-0206 = 1-032;

zio. x 100 6-79x100 * " ~ 1 8 Ö Ö W » - 18000X0-1 = - ° · 3 7 7 P e r C e n t ·

We relate them to the values: e.m.f. Ed2 = 4-76x1-032 = 4-92 and the power, specified by the currents in the excitation winding,

P2 = 0-00179 X4-922 = 0-0434; the excess power,

ΔΡ'2 = 0-222-0-0434 = 0-179.

Third interval

This time interval covers the beginning of the period, following the instant of the clearing of the short-circuit :

Edi2] = 2-86£,;2 = 2-86 x 1-032 = 2-96;

P2 = 0-0313£|2 = 0-0313 X2-962 = 0-274;

ΔΡ'2' = 0-222-0-274 = -0-052;

Αδ3 = AÔ2+kAP>tAP> = 6-79+25-7 ° · 1 7 9 - 0 · 0 5 2 = 8-4°;

Ε„ζ-ΕΛΆ 3-52-2-96 9-75 9-75 ΔΕ·Λ =d'3a-7l2ìAt = t ~ g Χθ-1 = 0-00575;

δ3 = δ2+Αδ3 = 43-23°;

Ε·αζ = Ε'Λ + ΔΕΛ =1-038; s3 = - ^QOOXO-I = 0-466per cent.

The calculation of the remaining intervals of the condition is summarized in Table 8.7.

In calculating the whole of the post-fault condition, the power of the turbines in the electrical stations was assumed to be constant, since the very insignificant departure from the synchronous speed, may be considered to be in the insensitive range of the speed governors.


TABLE 8.7

Serial No

1

2

3

4

5

6

7

8

9

10

11

12

Calculation value

K Ed

ÉÌ

Λπ

AP = PT-Pm

kAP

AÔ

E*.

Ea.-Ed

AE'd

Ô

s, per cent

Working out order

l ( / i - l ) + 10/i

2 - 8 6 x l ( / i - l )

(2/02

00313 x3/i

0-222-4/1

25-7 x 5/1

7 ( / I - 1 ) + 6/I

According to Fig 8.5

8/1-2/1

9/ΐ0·1

9-75

l l ( / i - l ) + 7 / i

7/1 x 100

1800

Number of interval and time, sec

3 (OS)

1038

8-4

43-43

4 (0-4)

1048

2-965

8-8

0-275

- 0 0 5 3

- 1 - 3 6

7 0 4

3-94

0-97

000995

50-27

-0-391

5 (Ö-5)

1061

3 0

9 0

0-282

0 0 6

- 1 - 5 4

5-5

4-28

1-28

00128

55-77

-0 -306

6 (0-6)

1076

3 04

9-24

0-289

- 0 0 6 7

- 1 - 7 2

3-78

4-58

1-54

00158

59-55

- 0 - 2 1

Seventh interval

The reclosing, which restores the standard circuit condition, takes place at the beginning of the seventh interval.

Prior to reclosing:

Edt = 2-86 EM = 2-86 X 1-076 = 3-07;

Pe = 0-0313 X3-072 =0-295;

ΔΡ'% = 0-222-0-295 = -0-073.


After the line is switched on, (8.45), we get:

Ed[Q] = 3·85£;β-2·31 cos (όβ + 16°) = 3-85x1-076-2-31 cos (59-6 + 16°) = 3-565.

From equation (8.37), we also have: P6 = 0-00699£J[6]+0-111^ sin (<5e + 16°)

= 0-00699x3-5652-0-ll 1x3-565 sin (59-6 + 16°) =0-472. We calculate the asynchronous power which depends on the currents in

the damper windings. With s=.y6, it is:

'—=-τ+|^τΙτ^Ι3^«[ι-^ι+<-2· ΐχ ιο"·χ64·8>"3χ

xsin( 2 x 59-6+32-arc tan --2-1 x l 0 - 3 x 64-8

■ +2(5-2(Γχ,'ο-»Τΐ8ν['+Λ» +0-1 * io-x mnx

x s i n ( 2 x 5 9 - 6 + 3 2 - a r c t a n _ 2 1 x l | ) _ 3 x l 8 8 ) | = 3-96xl0~3x 0] x [1 -1-065 sin (151-2+82-3)] +46-8 X 10~3[1 +1-18 X

xsin (151-2+68-3)] = 7-35x 10"3 + 11·7χ 10~3 = 0-019. Then

AP'6' = 0-222-(0-472+0-019) = -0-269;

Aô7 = Aôa+kAP<+2

AP<' = 3 -78+25-7X- 0 · 0 7 3 - 0 · 2 6 9 = -0-52°;

EM-EM _ 4-75-3-57 T75~X U 1 ~~ΪΪΤ J ^ ^ ^ Z ^ X O - I = 1 ί £ = £ 2 ί χ θ . 1 =0-0121;

δ7 = δΛ+Δδ7 = 59-55-0-52 = 59-03°; K-7 =Ε'Μ+ΔΕ'Λ = 1-076+0-012 = 1-088;

0-53x100 - „ . - . Sl = 1800 = ρβΓ Cen

The calculated results of the subsequent intervals are given in Table 8.8 and Fig. 8.5(b). These show the possibility of applying automatic reclos-ing without a synchronization control. The reasons for such a conclusion are:


(1) the small oscillation amplitude of the angle Ò during the transient process :

(2) the insignificant slip values; (3) a quick pull into synchronism and intensive oscillation damping.

Problem 8.4

À hydro-electric station is transmitting energy into a powerful system. The busbar voltage in the receiving substation may be assumed to be constant. The principal circuit and its parameters are given in Fig. 8.6. The reactances of the equivalent generator of the hydro-electric station with the same basic units (SbaSiC=600 MVA; Fbasic = 115 kV) are: ^=3-72; ^=2-42; JC^=1-13; x'J =0-778; ^=0-906. The time constants of the hydro-generator circuits are: 7 ^ = 5 sec; 7^=0-08 sec; T^ = = 0· 13 sec. The speed governors of the hydro-turbines have identical characteristics; the coefficient of irregularity of all the turbine governors is σ = 1·2 per cent.

V=const

Xg xcxt =1-594

FIG. 8.6.

Required: to calculate the condition in the transmission system with asynchronous operation of the hydro-generators, the excitation of which, in this condition, is removed in order to lessen the power oscillations.

Solution. The powers and the voltages in the system with asynchronous operation of one of the generating stations, are determined by the slip at which the generators of the station operate.

The mean slip of the steady asynchronous condition depends on the relation between the characteristic of the mean asynchronous power of the generator and the static power characteristic of the turbine under control.


The mean asynchronous power values of the hydro-generators with a purely reactive connection with busbars of constant voltage, may be calculated from (ref. 3, page 163):

P - V2 S Xd~X* V Td I X'd-X'd y asynmean 2 xdx'd X 1 + T O 2 + jc>y X

7)2J· For the problem conditions, using this formula we get:

T» _ Γ » *«+*«* _ 0 , , 0-906 + 1-594 _ Γ* - Γ β 0 ^ + 3 ί - - ° 1 3 x 2-42 + 1-594 - 0 0 8 1 s e c ·

The reactances in the asynchronous power formula include the external reactance xcxt . Expressing the time constants in relative units (radians), we get:

_ _l ·^ Γ 3-72-1-13 2-56x314 ·* asvn mean ^ " I is τ \ . i cr\A\ / i i -> . i Γ Λ > Ι \ S^· ϊ M asynmean , | (3.72 +1-594) (1-13 + 1-594) ~ 1 + ( 2 - 5 6 X 3 1 4 ) V

1-13-0-778 0-0695x314 ' (l-l3 + l-594)(0-778 + l-594)~l+(0-00695x3l4)2.y

2-42-0-906 0-081x314

2

(2-42 + l -594) (0-906 + l -594) ~ l+(0-08lx3l4)V 71-fo 0-62* 1-915

1 +64-5 x 10V 1 +0-0475 X 10V 1 +0-0645 x 10V ' An asynchronous power characteristic, calculated from the equation

obtained, is shown in Fig. 8.7. Also plotted in it is a static power characteristic of the controlled turbine.

The coefficient of irregularity of the turbine is σ = 1·2 per cent. The point of intersection of the characteristics of the turbine and the generator determines the mean value of the slip of the established asynchronous operation, to be ^ = —0-0115. From the diagram, with this slip, it follows that:

asyn mean =0036 ; Pasyn ( / ) mean =0-0095 ; Pasyn w mcan = 0-0065 ;

^asyn(g) mean = 0 - 0 2 0 .


The value of the active asynchronous power, which contains both a mean and a variable component, may be calculated from(ref. 3, page 163):

4 1 1 - — i — " " — — — - . 2 3

FIG. 8.7.

-2 5-10

P = —i P * asyn ^ ** '■

asyn (/*) mean

Here

Ti+V[i+OT]x

X sin (ΐδ- arc tan sTl, I +jPasyn (d) mean

+ V[l +(sTÏ)2] sin (20-arc t a n ^ Y U

+P>syn (,) »can [ l - VD + « ' ) 2 ] sin f 2Ä-arc tan-±Λ~\

ίΓ; = -1·15χ2·56χ314χ10-2 = -9-25; sT'd' = -1·15χ6·95χ314χ10-4 = -0-251; sT'q' = -1-15x8-1 x314xl0"4 -0-292;

arc tan -

C\TC tan -

arc tan -

3CT25 = 1 ? 3 · 8 '

_ 104-1°· -0-251 " im l '

À . . = 106-3°. -0-292


TABLE

No

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15 16

17

18

19

20

21

22

23

24

25

26

27

Calculation value

E'

Ô

<5 + 16°

cos(<5+16°)

sin(<5 + 16°)

3-855J 2-31 cos ((5 + 16°)

Ed

El 000699 £ |

0111 £d sin (5+16°)

Λ 5x64-8

(5χ64·8)2

1+(5Χ64·8)2

/ [ 1+(5χ64 '8 ) 2 ]

1

5X64-8

1 m sss ari* tan "f/j. — a l W l a l l — — ^ —

* JX64-8

20

^ = 2 0 + 3 2 - ^ ,

siny^

/ [1+(5Χ64 ·8 ) 2 ]χ x sin ψά

1 - / [ 1 + ( 5 Χ 6 4 · 8 ) 2 ] χ x sin ψά

1-925

1-925

" 1+(5Χ64·8)2

Ρ

5X188

Working out order

1(/ ι -1)+46/ι

2 ( / ι -1 )+43/ ι

2 ( / ι - 1 ) + 16°

cos 3/1

sin 3/ι

3 · 8 5 Χ 1 ( / Ι - 1 )

2-31x4/1

6/1 —7/1

(8/02

0-00699x9/1

0111x8/1X5/1

10/1 + 11/1 4 7 ( / ι - 1 ) χ 6 4 · 8

(13/1)2

1 + 14/1

/15/I

1

Τϊη'

arc tan 17/i

2 χ 2 ( / ι - 1 )

19/1+32° —18/1

sin 20/1

16/1x21/1

1-22/1 !

1-92x47/1

24/1

Ϊ 5 Λ

25/1 χ 23/1

47/1x188

Number of interval n

6 1 (0-6)

1088

5902

-------------—

-

-

— --—

—

-

-

-

7 1 (0'7)

1097

52-38

7 5 0

0-2588

0-9659

4 1 9

0-598

3-59

12-9

0 0 9

0-385

0-475

190-5 x IO"4

0000364

10004

1 0

52-5

88-9

118

611

0-8755

0-8755

01245

5 -64x10-*

- 5 - 6 5 x 1 0 - *

-0 -702 x IO""4

5 5 3 x 1 0 - *

1 8 1 (0'8)

1105

44-18

68-4

0-3681

0-9298

4-22

0-851

3-37

11-35

00792

0-347

0-426

2 4 5 · 5 χ 1 0 ~ 3

00602 106

103

4 0 5

76· 1

104-8 60-7

0-8720

0-925

0075

7 -28x10-3

— 6-87x IO"3

- 0 - 5 1 5 x 1 0 - 3

0-713


and time ty sec

9 (0'9)

1112 36-24 60-2 0-4970 0-8678 4-25 115 310 9-60 0067 0-298 0-365

295X10-3

0087 1087 1042

3-39

73-6

88-4 46-8

0-7290

0-760

0-24

8-75X10-3

- 8 0 5 x 1 0 - 3

- 1 - 9 3 x 1 0 - 3 0-855

10 U'O)

112 29-60 52-2 0-6129 0-7902 4-29 1-42 2-87 8-24 00575 0-252 0-310

2 8 5 x l 0 - 3

00815 1082 104

3-51

74-1

72-4 30-3 0-5045

0-524

0-476

8-45x10-3

- 7 - 8 x 1 0 - 3

-3-71x10-3 0-826

11 (I'D

1128 25-22 45-7 0-6984 0-7157 4-31 1-615 2-69 7-24 00505 0-214 0-265

236x10-3 00556 1056 1027

4-24

76-7

59-4 14-7 0-2623

0-27

0-73

6-99x10-3

-6 -62x10-3

-4 -84x10-3 0-685

12 (1'2)

1135 22-23 41-2 0-7524 0-6587 4-34 1-74 2-60 6-78 0048 0191 0-239

I6IXIO-3

00259 1026 1011

6-2

80-9

50-4 1-5 00262

00265

0-973

4-75x10-3

-4 -64x10-3

-4 -50x10-3 0-466

13 \ d'3)

1141 20-37 38-3 0-7848 0-6198 4-37 1-81 2-56 6-55 00457 1 0176 0-222

104-2x10-3 00109 1011 1005

9-59

84

44-6 -7-4 -01888

-01892

1189

309x10-3

- 3 0 5 x 1 0 - 3

-3-62x10-3 0-303

14 d'4)

— 36-4 0-8049 0-5934 4-40 1 86 2-54 6-45 00451 0167 0-212

— — — -

-

-

— ---

-

-

-

— —

8_8


Continuation

No

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

Calculation value

(5X188)*

1+(5X188)*

/ [ l + ( 5 x 1 8 8 ) * ]

1 5x188

1 »n — UTC tsin ——.——-* a r C t a n * x l 8 8

φ9 = 2δ+32*-Ψ§Λ

sin y>q

/ [ l+( jx l88)«]s inv> f

i + ^ [ l + ( 5 X l 8 8 ) 2 ] x x sin ψ9

25-75^

25-755 " 1+(5χ188) 2

P

p

ΔΡ = PT-Ptt

kAP

Δδ

E" Ε*-Ε,

AK

Δδ ,

"~Έ%

Working out order

(27/1)2

1+28/1

γ29η

1 27Ü

arc tan 31Λ

19/ι+32°-32/ι

sin 33/1

30/fx34/i

1 + 35/1

25-75x47/1

37/1 29^

38/1x36/1

26/1 + 39/i

0 - 2 2 2 - 12/I-40/I

25-7x41/1

4 3 ( / i - l ) + 4 2 / i

Fig. 8-5

44/I-8/I

45/1 xO-1

9-75

43/f 1800

6 (0'6)

—

-

-

-

-

-

—

—

-

-

-

-

-

-

-

~"

Number of interval n

7 (0'7)

000306

1003

1:0

181

86-8

63-2

0-8926

0-8926

1-89

75*6x10-*

- 7 5 - 4 x 1 0 - *

- 1 4 2 - 5 x 1 0 - *

- 0 0 1 4 8

-0 -238

-6 -1 1

- 6 - 6 4

4-43

0-84

00086

0-379

8 (O'S)

0-51

1-51

1-23

1-4

54-5

82-3

0-9910

1-220

2-22

9 7 - 5 X 1 0 - 3

—64-5 x 10~ 3

- 1 4 3 X 1 0 - 3

- 0 -1 4 4

- 0 0 6 0

- 1 - 5 4

- 8 - 2 0

4 1 0

0-73

00075

0-455


of Table 8.8

and time t, sec

9 (0'9)

0-731

1-731

1-315

117

49-5

70-9

0-9449

1-24

2-24

117x10-3

-67-5x10-3

- 1 5 1 x 1 0 - 3

- 0 1 5 3

001

0-257

- 7 94

3-84

0-74

00076

0-44

10 d'O)

0-683

1-683

1-3

1-21

50-5

53-9

0-8080

105

205

113x10-3

- 6 7 x 1 0 - 3

-137-5x10-3

- 0 1 4 2

0054

1-39

-6-55

3-62

0-75

00077

0-364

(1Ί)

0-47

1-47

1-21

1-46

55-7

35-7

0-5835

0-705

1-705

93-7x10-3

-63-8x10-3

- 1 0 9 x 1 0 - 3

- 0 1 2 4

0081

208

-4-47

3-43

0-74

00076

0-248

12 (1'2) ■

0-215

1-215

11

215

651

17-3

0-2974

0-327

1-33

63-9x10-3

-52-5x10-3

- 7 0 0 x 1 0 - 3

-00745

0058

1-48

-2-99

3-27

0-65

000666

0161

13 (1'3)

00902

109

1043

3-3

73-2

3-4

00593

00619

1062

41-5x10-3

- 3 8 X 1 0 - 3

-40-4x10-3

- 0 0 4 4

0044

113

-1-86

314

0-58

000595

01035

14 (1'4)

—

-

-

-

-

—

-

-

-

—

-

—

-

-

-

-

-

-

-

-


Therefore,

^yn = -0-0095 [l+V(l+9-252)xsin (20-173-8°)]-

-0-0065 [l+V(l+0-2512) sin (2δ-104-1°)]-

-0-020 [1 - Vu +0-2922) sin (26-106-3°)]

= 0-034-0-0882 sin (20-173-8°)-0-0067 sin (20-104-1°) +

+ 0-0208 sin (20-106-3°).

In the asynchronous condition, the hydro-generators give out active power because of their normal action. This power is determined by

2 *dXq

= 7X(3-72 + l3594)"(22-422+l-594)SÌP 2δ = ° 0 3 0 4 SÌ° 2 *

The reactive power in the system with asynchronous operation of the hydro-generator is derived from the asynchronous power (Q^^), and the magnetizing power Qli, thus (ref. 3, page 166):

μ 2 \_xd xq \xq xdJ J

Q =.YlUL-i\\Wf + 2 \\xd xd)\\Hsrdy

+ ——— cosf 2<5-arc tan ) + V[l+W)«] V sT'd)\

( i i \ r (sry)2 *ry /„, t 1 \n ( ] v d h— - cosi 2ô-arctan ]

- — - i S cos Ì2Ò-&TC t a n — i l l .

+


In these formulae, the reactances also include the reactances of the external circuit. Therefore,

* · ■ · ■ !

2 [3-72 +1-594 2-42 +1-594 ^2-42 + 1-594

3-72+1 *, «ηΛ«»2*1 = -0-2188+0-0304 cos 2δ; + 1-594/ I

U«y> 2(^1-13 + 1-594 3-72 + 1-594 / [_ 1 +(-9-25)2 r

+ , (-9'25> cos(2ô-173-8°)l + /' -V[l+(-9·25)2] J V0-778 + 1-594

-104·1°)Ί i \ \ (-0-2: 1-13 + 1-594/ | _ l + ( - 0

cos

-0-251)2 (-0-251) -cos(2ô 251)2 VE1 +(-0-251)2]

-0-292)2

292)2

(-0-292)

x ^0-906+ 1-594 2-42 + 1-594^ [ ΐ + ( - 0 ·

V[l+(-0-292)2] cos (23-106-3°

or Ô»yn = -00961 -0-089 cos (20-173-8°)-

0-00667 cos (2Ó-104-l°)+0-0211 cos (20-106-3°). The power characteristics, obtained from these formulae, are shown in

Fig. 8.8. We calculate the voltage at the generating station busbars, by using the equation:

08

06

04

0-2

hO-3

0-2

01

0

k H

Κ)·2

Κ)·3

-0-4

U-5

*— — -?^90 ^* ·—"* · 180 ^ — - ~ ^ ^ 2 7 0 ^ v degrees

Γ ^ < 0 a « y n ^*

FIG. 8.8.


With 0=0, Vg = 1 + (-0·2008)χ1·594+.0·03χ1·594 = ^ ^

1 1 The values of the voltage Vg, calculated for other values of the angle <5,

are shown in the curves given in Fig. 8.8

Problem 8.5

Required: to check the possibility of connecting the electrical station, operating in the system shown in Fig. 8.9, into the circuit, by the self-synchronization method.

G-l

<SKO{

I3 8W

FIG. 8.9.

Installed in the station are turbo-generators TB — 2-100 — 2, which have the following parameters : Pnom = 100 MW; cos<pnotn=0-9; Fnom = 13-8 kV; ^ = ^ = 1-8;

χ2=0·168; ^=0-203; x'J =0-138; *;'=0-1725; 7^ = 13 sec; r ; = l-46 sec; 77=0-18 sec; ^'=0-292 sec; Γα=0·38 sec.

Eg X g 00IÖ7+J0I2I

V=I03

FIG. 8.10.

Simplified equivalent circuits are characterized by the parameters shown (Fig. 8.10(a) and (b)). These circuits are obtained by determining the parameters in relative units, taking the nominal power and voltage of one generator as the basic values. The circuit, Fig. 8.10 (a), characterizes the system with simultaneous switching of two machines: the circuit, Fig 8.10 (b), refers to the case of switching one turbo-generator into the circuit when the second one is in operation.


Solution· The possibility of inserting the machines into the circuit by the self-synchronizing method is limited by the stator current values and the electromagnetic torque produced in the machines at the instant of switching.

With a purely reactive connection between the generator and the constant voltage busbars, and connecting the machine when .s=0, the stator current components may be calculated from (ref. 9).

I L ^ r \XdS XdEJ t_ t

+ ( 4 r - 7 - V r^]cosó0--Le r"*cos(ô0+o}; \XdE ΧάΣ/ J χάΣ )

t

1 e r e r s in(ô0 + o } . XqZ )

The reactances, appearing in these equations, are calculated by allowing for the external reactance associated with the generator.

It is permissible to connect the machines by the self-synchronization method if the current, determined by the given equations, is smaller or equal to the stator current with a three-phase short-circuit on its terminals. The three-phase short-circuit current components of the unloaded generator, may be determined from :

\ xd xdJ xd

Xq

The phase currents (for the phase a) are related to the current components id and iq by the expression

ia = id c o s (θο + 0 + iq s i n (θο + 0-


We calculate the time constants appearing in these equations. In the case of simultaneous switching of two generators we get:

^ x'dr+x*xt 10 0-5X0-203 + 0-121 0 0„ ΤάΣ = Td°£+€< = l 3 X 0-5x1-8+0-121 = 2 · 8 3 sec'

r . _ r i4xfÄ±5a _ 0 l f i 0·203(0·5ΧΟ·138+0·121) _ Τ„Σ - TdχΙ,ΧX,ds+Xni - 0 l8X<H38(0-5x0-203+0-l2l) ~ 0 2 2 6 s e c >

Xa . , ΧςΣ «" Xc. XqE+Xa

rptt __ rptt *q s . ΛςΣ « -*ext 1 qL — l q —i * :

^,v> 1·8(0·5 χ(Μ725+0·121) Λ^1Λ = ° < 2 9 2 Χ 0 , 1 7 2 5 ( 0 - 5 Χ 1 > 8 + 0 - 1 2 1 ) = °'6 1 9 SeC·

To calculate the time constant ΤαΣ, we find, approximately, the reactance of the stator winding of the generator:

' • - ïSfc- 51^38 -•«1«»·'»· Therefore,

r _ *«+* .« _ 0-5x0-168+0-121 _ βΓ ~ 314(rst+reJ ~ 314(0-5x0001405 + 0-0187) ~ u u : > : , ' s e c ·

When one generator is switched on, the same time-constants are :

_, „0-203+0115 T<* = 13 1-8+0115 = 2 1 6 s e c »

0-203(0-138+0115) 0-138(0-203+0115) ÌJU; = O-lö A ! , o m <,rt, , rt 1 t e N = 0-21 sec;

1-8(0»1725+0115) n 4 V - . Γ«* = ° 2 9 2 01725(1-8+0-115) = ° ' 4 5 7 SCC'

0168+0115 314(0-001405+000057) ΤαΣ = „,,ηηη.ΑΖ, . „ ^ ^ =0-127 sec.

Allowing for the resistance values and the time constants for the conditions of the problem, we get the following theoretical formulae to determine the stator current of one generator:


(a) When two generators are switched on in accordance with the self-synchronization method :

, < i = " 2 X l { | _ 0-5x1-8+0-121 + ( 0-5 X 0-203 +0-121 ~

I )<Γ™ + ( I 0-5x1-8+0-121/ τ\0·5χ0·138+0·121

! | -"ölw L s _ 0-5x0-203+0-121/ J °

- 0 · 5 χ 0 · 1 3 ' 8 + 0 · 1 2 ΐ ' " , " ^ ' ° ' ^ + "}

= (θ-489 + l-755 e"*"" +0-385 e"5™) cos δ0-t

-2-63e O0337cos(o0+0;

Ιβ = Τ Χ ΐ { [ θ · 5 χ 1 · 8 + 0·121+(θ·5χΟ·1725 + 0·121~

- csx-Lo-ni ) '"^] 5 "^-

= (θ·489 + 1·925βΠΓ*ϊ5) sin δ0-2-42 e"5'533' sin (δ0 + 0· (b) When one generator is switched on and the second is already operat

ing:

id= 1 - 0 3 j 1 . 8 + 0 . 1 1 5 + ^0.203+0.115"l-8+0-115/e +

+ V0-138+0-115 "0-203+0-115 J e ° 2 1 c o s < 5 o -

-0-138+0-115 e " ^ C ° S ^ + 4 = (o-538+2-7e~?ïî+0-835e~°:iï) cos a0-4-07e_ÏÏTÎ"cos (δ(

'· = 1*03{l-8+0-115 + ( 0-1725+0-115-1·8 + 0·115/>

X e " ^ S i n δ°~ 0-1725+0-115 e " ^ S Ì n (*° + 4 = (θ·538+3-04 e" 5 ^) sin ô0-3-58e_FÏ"sin (δ0+ί).

X


(c) With a three-phase short-circuit on the terminals of an unloaded generator (the voltage on open-circuit is taken to be V^ = Ed0 = 1-05):

= -0-584-4-58 e r 4 e-2-42e 0*184-7-6e ^ c o s i ;

1-05 -dnï lq 0-1725

0-38 sin t = 6-09 e °'38 sin t.

2V l l |

o |

I

z\ 3

A

5

6

7

8

9

10

II

12 j

13 1

14 I

15'

rU

r J^^^^\. / ^ ^ ^ 0 0 0 5 Λ. \ Ö0I 0 0 Ι5 /γ

r l \ \ u y- t 1 v >■ /

1 \ > - ^ î / h l \ //"' j. \ \ y **■ 1 ^"Ί 1 '

L 1 ' Γ 1 ' \ \ ' 1 f Γ 1 / i ( 3 >

1 » *c .

l· » ' L l ' 1 \ ' [ \ ' ί v

/ \ i ( 2 ) // vfc y « . \ \ /o 02 \ \ \

' 003

\ /

1 ^ a ( 0Ό35/ /

1/ i

S* 1 ^ 11 il

1/

fj\ /sec

FIG. 8.11.

In Fig. 8.11 are given the characteristics, which are plotted from these expressions using the equation connecting the currents id and iq with the current of phase a.


When calculating, it is assumed that <50=45°, since the largest electromagnetic torque is obtained with this switching on angle.

The electromagnetic torque, which appears with the machine switched on in accordance with the self-synchronization method, may be calculated from:

m j = 0 V2 i -^ ) e T'"-(j i_\ τ',·Σ / j ι_\ τ;·Σ s in

\Χ'άΣ X'ds) \Χ<ιΣ xqZJ J

ΤιΣ + Τ*Σ

i +

X

xsin d0xcos(d0 + t) +

T'dr + Tt

■ \χάΣ χάΣ/

ΛΣΖΙ'Σ

Τ*Σ+Τ.Σ

+(Ar-A-V *«*" '-(4-—)c~M \χαΣ xdLJ \ΧςΣ xdz) J

X

1 / 1 1 \ "~r~~ xcos δ0 sin (i0 + / ) _ _ —-_ —-le βΓ sin 2(ί0 + ί) ζ \ x j r xqEj

The maximum value of the torque after a three-phase short-circuit, may be determined by the expression

m, -=£ 2[^s i n i4(^-T)s i n 2 i] The previously calculated currents with the machines switched on in

accordance with the self-synchronization method showed that the most difficult theoretical case is the switching on of one generator while the other is operating.

In accordance with the formula given above, we have, for this case,


within the conditions of the problem considered:

m,-o = 1 0 3 2 {γ [ - ( 0 ·203+0 ·115~ 1-8+0-115je 21'~

J 1 I V ^ + VO-138+0-115 0-203+0-115/

+ (θ·1725+0-115 - i r a ï ï s ) ^ ] SÌn ^ +

+ \(—l X—\ e~ra? -LV°-138+0-115 1-8+0-115/

/ 1 1 \ 0-457+0127 Π _ { i L _ _ | e 0-457x0127 γ

\ 0-1725+0-115 1-8+0-115/ J

K i 1 \ - »le+o-iw ,

(MW+o-in-ra+ÖÖB)' * " * " " +

/ 1 1 \ 0-21+0127 i f t [ 1 - 0-21x0-127 _

\0·138+0·115 0-203+0-115/

_ (___! ! "j e-ô^l x VO-1725+0-115 1-8+0-115/ J

6osina + 0 - i ( 0 , 1 3 8 ^ 0 , 1 1 5 - o . 1 7 2 51

+ 0 . 1 1 5 ) e - ^ X

Xsin2(50+o| = l-034(-l<31e~^-0-403e~°7" + l-48e~°:ï")x

X sin 2ό0 + (3·43 β~δΊ"-2·966_δ:δ»55) sin δ0 cos (ô0+t) +

+ (2-62 e-5:"+0-805 e"°":o^-2-96 e"»71") χ

xsm

xcos

xcos Ò0 sin (<50+ί)-0·235 e 00e35sin 2(«5, ,+ol The theoretical formula to determine the electromagnetic torque with a

three-phase short-circuit on the terminals of the generator, for the para-


meters of this machine, takes the form:

/w. 1-052 * ,„0 sin t——( Λ , - n - Λ , „ - ìsin 2/ |_ 0-138 2 \^0·138 01725 y J

= 7-98sinf-0-798sin2/. In Fig. 8.12 are given the characteristics of the electromagnetic torque

obtained with ό0=45°. These characteristics, as well as the characteristics, Fig. 8.11, indicate the possibility of switching into the circuit by the self-synchronization method one or two machines, simultaneously.

Im

# i

1 1 1 1 f f 1 1 / /

If f li 1

P 0-005

\ \ \

\ " Μ

\ο ·ο ι Ι^^δι t ms.s.

\ \ \ \ \ \

315 J

1 1 / / / / / /

/ / /

/

**> / / /

5

oor^y

\ \ \ \ \ \ \ \ \ ■ 1 1

!

• 5^^J°2^

l \ \ \ | \ \ \ \ \

V \

t

Ό35 ^ j 1 1 / 1 1 / f r

/ / /

/ '**

FIG. 8.12.

In both cases, the stator currents and the electromagnetic torques are considerably smaller than for the three-phase short-circuit, calculated for the synchronous generators.

It should be stressed, that for practical evaluation of the possibility of realizing the self-synchronization of the generators, it is unnecessary to determine the instantaneous current values and the electromagnetic torque.


In solving the given problem, these values have been calculated for the purpose of illustrating the phenomenon clearly. In practical cases, the decision on the possibility of self-synchronization may be based only on the determination of the periodic component of the subtransient current and comparing it with the permissible value:

(a) for the self-synchronization of turbo-generators: I" <= 5/

(b) for the self-synchronization of hydro-generators:

CHAPTER 9

VARIATION OF THE QUANTITIES IN THE SYSTEM (FREQUENCY, FLUX, EXCHANGE OF

POWER, VOLTAGE, ETC.)

WITH different variations in the system configuration, the connection and disconnection of loads, the line sections, transformers, generators and other elements of the system, variation of the operating conditions takes place. If the system, with the combinations described above, remains stable, statically as well as dynamically, then a problem arises in determining the new parameters for the condition established after the transient process. The present chapter is dedicated to various calculations of this kind.

Problem 9.1

The active power used by the equivalent load of the system (Fig. 9.1), with nominal voltage KL = 1, is equal, in relative units, to 0-8. The load power factor is 0-9. The system parameters are: xd=hl; χΓ=0·18; xc=0-6.

Θ-<2>-^—L E« ' k

FIG. 9.1.

Required: to determine the voltage variation on the load busbars VL with a change in the coupling reactance xc, assuming that the active load power PL remains constant; and that its reactive power is QL:

(a) does not depend on the voltage variation; (b) varies in proportion to the square of the voltage; (c) varies in proportion to the cube of the voltage.

353


The calculations are carried out, assuming that the equivalent e.m.f. of the station is constant (Ed=const), and determined from the initial conditions PL=0-S; cos ç>=0-9; VL = l.

Solution. We determine the e.m.f. of the equivalent station. Also:

QL = 0-8 tan (arc cos 0-9) = 0-8 x 0-484 = 0-387.

The total reactance of the system is:

*dE = **+*r+*c = 1-1+0-18+0-6 = 1-88; Ed = Λ/[(1+0·387χ1·88)1+(0·8χ1·88)1] = 2-29;

(a)gL=0-387=const. We determine VL from the reactance xc(xdE):

or

Vi-2 (β-Ωχ.χ,Λ Vl+XafQl+FS* = 0. (9.1)

toi l\fê-QLXd^-xh{Pl+QÏ>\

/fW^-ß**«)-^]. (9.2) For xc=0-3; χ</Γ = 1·58, we get:

VI =?^?!_ο·387χ1·58± /Γ2·292/ '^^-0·387χ1·58Ν)-0·82χ1·582Ί

= 201 ±1-43;

^ = 3 - 4 4 ; F22=0-58; FL1 = l-855; Κ£2=0·762.

The calculated results for other values of χάΣ are given in Table 9.1. The critical value XEctm, may be determined by equating the term under

the root sign in (9.2) to zero :

Hence

VL~ 2

= Y-QLxdr±

^■(j$-Qi*er\-n*k:=0,

Variation of the Quantities in the System

TABLE 9.1

355

x*

Χ4Σ y* vLt

0-5

1-58 1-855 0-762

0-45

1-73 1-775 0-865

0-55

1-83 1-71 0-95

0-6

1-88 1-67 100

0-65

1-93 1-62 106

0-75

203 1-48 1-22

**2;c = 2-055 V^ = 1-35

—

or xdL^ p2 XdE 4 p 2 "" U#

For xc =0-3, ^ + 3 - 1 7 ^ - 1 0 - 7 3 = 0 .

Hence, taking the positive root only, we get: ΧάΣοτ =2-055, which corresponds to (9.1);

(b)ÔL = 0-387 V\ With such a condition, from (9.1) we get the following equation :

£? w2 , 0-64*3* Vt-L l+0-774x,r+0-15xî VÌ + κάΣ 1+0-774x^+0-15^ = 0.

Solving for various xdz, we determine the voltage values VL, which are given in Table 9.2.

TABLE 9.2

Xc

χάΣ

VLI y*

0-3

1 58 1-28 0-61

0-45

1-73 118 0-704

0-6

1-88 100 0-87

^ror = l-9 VLcr = 0-935

~~

(c) QL =0-387 V\ In this case for VL we get an equation of the sixth order

xdL XdE xdE η+^ n+^F n-il· vi+4-27 = o,


the solution of which is possible by one of the approximate methods, for example as indicated in ref. 13.

The results of the solution of this equation are given in Table 9.3.

TABLE 9.3

Xo

ΧΛΣ Vu VL*

0-3

1-58 117 0-59

0-45

1-73 111 0-67

0-6

1-88 100 0-77

Figure 9.2 gives the functions VL=f(xdE) for all three cases (1) QL = =0-387; (2) QL=0-387 V\; (3) QL=0-387 Vf).

1-9

1-8

17

1-6

1-5

1-4

1-3

1-2

!·! 10

0-9

0-8

07

0-6

0-5 I 1·

k ^ --

---

1 2 1-3

\ X ^

> \ \

\ I

> 2 /

^ > \ / y v

/ ) \

' / ' y^^Z

xdE 1 1 1 1 1 1 1 1

M 1-5 1-6 1-7 1-8 1-9 2Ό 21

FIG. 9.2.

Problem 9.2

The loading of the electrical system, Fig. 9.1, follows the cycle, shown in the diagram (Fig. 9.3).

The parameters of the system are: ^=0-6 ; Λ: Γ =0 ·1 ; A:C=0-3.

Variation of the Quantities in the System 357

Required: to determine the voltage on the load busbars VL for all the stages of the given diagram, taking into account, that the reactive load power:

(a) does not depend on the change of voltage; (b) varies in proportional to the square of the voltage; (c) varies in proportional to the cube of the voltage. The value of the e.m.f. Ed remains constant and is determined from the

initial condition: FL = 1; PL=0-527; ßL=0-527; cos<pL=0-85.

0'85

0-6

0·2

I I

■ 1 — ! I 1

- t r

FIG. 9.3.

The power factor for all the stages in the diagram with nominal voltage KL = 1 is 0-85 (tan <pL=0-62).

Solution. The total reactance of the system is A ^ =0-6-f(M 4-0-3 = 1'0. The electromotive force of the equivalent generator in the original

condition is :

Ed = V[(l+0-527)2+0-852] = 1-75.

Unlike the previous problem, we determine the voltage VL by a grapho-analytical method.

(a) QL=const. 0 ) ^ = 1; GL=0-62 . Taking different values of the voltage VL, we determine the correspond

ing values of the e.m.f. Ed (Table 9.4). By plotting the values obtained for Ed on a diagram, Ed=f(VL), (Fig.

9.4, curve Ax), we find that the curve obtained does not intersect with the straight line £^=1-75. Therefore, the condition considered cannot exist.


TABLE 9.4

VL

1-3 1-2 11 10 0-9 0-8

("«^r 316 2-94 2-74 2-63 2-53 2-48

m' 0-591 0-695 0-83 1-00 1-24 1-56

Ed

1-935 1-90 1-895 1-91 1-94 201

When determining VL from the equations, i.e. using the method of the previous problem (9.1), we should presumably have had a root expression in (9.2) with a negative sign.

(2) PL=0-6; QL=0-372.

The calculations for this case are given in Table 9.5.

TABLE 9.5

vL

11 1-3 1-4 1-5

('.^j' 206 2-52 2-77 305

m 0-3 0-214 0184 0160

Ed

1-54 1-65 1-72 1-792

From Fig. 9.4. (curve A2) we determine, that the condition exists with KL = l-44.

(3) PL=0-2; QL=0-124.

By making similar calculations, we determine that VL = \-61 (Fig. 9.4 curve A3).

(b) QL=nQDom

Here QDOm= the reactive load power with KL = l(cosyL=0'85).

0 ) Λ . = 1; ônom=0-62.



TABLE 9.6

VL

1 0-95 0-9 0-85 0-8 0-7 0-65

VÎ(l+Qnomxdi:)*

2-63 2-37 212 1-9 1-68 1-29 111

(ΨΥ 100 110 1-24 1-38 1-56 204 2-37

Ed

1-91 1-86 1-83 1-81 1-80 1-825 1-865

The condition cannot be satisfied in this case (Fig. 9.4, curve B^. (2)PL = 0-6; ß n o m = 0-372.

From Fig. 9.4, (curve 52) we find that VL = 1-22.

(3) PL = 0-2; Qnom = 1-24.

VL = 1-56 (Fig. 9.4, curve B3).

(c) QL = VlQoom. (l)PL = l; Qaom=0-62 (Table 9.7).

TABLE 9.7

VL

10 0-9 0-8 0-75

{VL+VlQnomxdEf

2-63 1-96 1-44 1-21

m 100 1-24 1-56 1-78

Ed

1-91 1-79 1-73 1-73

VL = 0-85 (Fig. 9.4, curve Cx).

In a similar manner, for the remaining two cases, we get: PL = 0-6; VL = 1-17 (Fig. 9.4, curve C2); PL = 0-2; VL = 1-48 (Fig. 9.4, curve C3).


The values of voltages on the loading busbars VL, which were obtained in the above calculations, are given in Table 9.8, from which it can be seen that the smallest increase in voltage, when the active power consumed by the load is reduced, is obtained in the case when the reactive power is proportional to the cube of the voltage.

Ed = l-75

TABLE 9.8

PL

QL = const Ql = ^zßiiom QL — VLQXMTO.

1

0-85

0-85

1 1 1

0-6

1-44 1-22 1-17

0-2

1-67 1-56 1-48

Problem 9.3

In the system, Fig. 9.5(a), one circuit of the transmission line was suddenly disconnected, which caused the machines to oscillate. The corresponding change of the mutual angle <512 is shown in Fig. 9.6.

0-5

FIG. 9.4.


The equivalent circuit is given in Fig. 9.5(b). The parameters of the system are:

E[ = 1-50; x'd = 0-28; xT2 = 0-09; ZL = (Μ3+./Ό-0802; E'2 = 1-34;

xT1 = 0-12; xc = 0-11; ΓΛ = 12 sec; P10 = 1-3;

jt; = 0-44; Vm = 0-94; Γ/2 = 9 sec.

(ΞΚ2>Ρ (a)

f—/'ϋν- -^Γ-Eî

Required: (1) to determine the value of angle δ'12, obtained after the oscillation has been damped out; (2) to determine the change of voltage on the load VL during the transient process.

FIG. 9.5.

FIG. 9.6.


Solution. (1) We calculate the original condition prior to the disconnection of the circuit:

_ .-Q 7 1 i m Κ0·13 +j 00802) _ Z u - / 0 7 1 V l l+0-13+;0 -0802 " ° ' 7 7 7 ^ 8 7 8 '

^11 = i~ = 0^77 = 1 - 2 8 5 ; *u = 9 ° - 8 7 · 8 = 2 - 2 ° ;

Zt t-y0.71+y0.11 + 0 ^ i ^ l . l 7 3 ^111-7·;

Λ2 = — = γ-ί=τ = 0-852; α12 = 90°-111-7° =-21-7° ; z12 i · i / j

y0-71(0-13+;O0802) yO-71+0-13+yO0802 Z22 =y0-ll + ^ „ Vn.n ■ ■■».»,«■, = 0-224 ^62-8°;

22 = ^ = 0^4 = 4 ' 4 6 ; "** = 9 0 " 6 2 · 8 = 2 7 · 2 ° ·

From the power equation we determine the angle δ^: 1-3 = 1 ·52 X 1 -285 sin 2-2° +1 ·5 X 1-34x0-852 sin (δ120 +21 -7°).

Hence, δ'120 = 22-3°; Ρ20 = 1·342χ 4-46 sin 27-2°-1-5x1-34x0-852 sin (22-3°-21-7°) = 3-64.

(2) We determine the value of the angle δ'120 at which the oscillation ceases for the condition when the circuit is disconnected. This occurs when <x12=0;

_ /0-ll(0-13+y0-0802) _ Zn -•/°93+/<Μ1+0·13+7Ό·0802 - 1-0 ^88-3 ,

y'n = {^ = 1-0; «ύ = 90-88-3 = 1-7°;

z;2 = y 0 - 9 3 + y o - n + 0 ^ 00 8 ^ = 1.50^112.3·;

y'n =^z = 0-667; x'12 = 90-112-3 = -22-3°;

_ f ( H 1 , yO-93(0-13+;00802) _ Z*2 - y 0 1 1 V93+0-13+yO-0802 ~ ° 2 2 6 ^ 6 1 3 '

J>« = ô^26 = 4 ' 4 2 ; *22 = 9 0 _ 6 1 ' 3 = 2 8 - 7 ° ;


P[ = 1-52 sin 1-7° +1-5 x 1-34x0-667 sin (<5;°+22-3°) = 0-067+ 1-34 sin (ói»+22-30);

P'2 = 1·342χ4·42 sin 28-7°-1-34 X 1-5x0-667 sin (<5;g-22-3°) = 3-82-1-34 sin {ò'x%-22-3°).

The excess power is: ΔΡΧ = 1 -3-[0-067 + 1-34 sin (δ£+22-30)]; ΔΡ2 = 3-64-[3-82-1-34 sin (δ^-22-30)].

The absolute accelerations of each station are :

AP, 1-3Γ0-067 + 1-34 sin (<3J°+22-3°)] el.degrees * = m^TZ = — 1 2 — Ì " 1 8 0 0 0 - l e ^ -

363

[ / i

AP <*2 = 3 6 0 / ^ = 3·64-[3·82-1·34 sin («;§-22·3°)] 18 000 el.degrees

[ / 2 sec*

0-5 f

-0-5 L.

FIG. 9.7.

In Table 9.9 and in Fig. 9.7 are given the absolute accelerations ocl9 oc2 and the relative acceleration a12 as functions of the angle δ'£.

The value of the angle δ'£0 = 34° corresponds to the relative acceleration a12 being equal to zero. The absolute accelerations αα and a2 are then

364 Transient Phenomena on Electrical Power Systems: Problems

equal to each other, but diflfer from zero. This indicates that the absolute angles δχ and <52 will increase continuously and the frequency in the system will rise continuously. The latter is the result of the assumption, when calculating, that the outputs of the turbines are constant and that there is

0-93 h

0-92

κ 12 22 24 26 2β 30 32 34 36 38 40 42 degrees

FIG. 9.8.

094

0-93

0-92

0-91

an unbalance between the load power (which is changed because of the change in voltage) and the output of the turbine.

(3) We determine the change of the load voltage VL with time.

,,0

FIG. 9.9.

,,5 2-0 sec


The method of calculation is as follows. Taking various values of the angle δ'£, we determine the active and

reactive powers given out by one of the stations for example, the first) and then, by allowing for the drop in the voltage in the corresponding reactances, we determine VL. After this, by using the function VL =f(à'^ and the curve β2=/(0> (FiS- 9-6)> w e determine VL=f(t). For example, for «2=22-3°:

P, = 1-52 sin 1·7° + 1·5χ 1-34x0-667 sin (22·3°+22·3°) = 1-007;

Qx = l-52cos l-7°-l-5x 1-34x0-667 cos (22-3°+22-3°) = 1-297;

VL _ ^ 1 . 5 _1-297X0-93J + ^007X0-93J-| = 0 .9 3 5

Further calculations are given in Table 9.10. The calculated results, in the form of a curve VL =f(ô'£\ are given in Fig. 9.8. From Figs. 9.6 and 9.8, we determine the required function, VL=f(t), which is given in Fig. 9.9.

Problem 9.4

A three-phase short-circuit occurs (Fig. 9.10) at the beginning of the line, connecting a hydro-electric station with a receiving system. During

HES 105 T 220 &m>-l·^ =Ξ4=Θ Y pot°s% Ί Pc

E'c

110

FIG. 9.10.

Pc

f L 2

he short-circuit, the power, supplied from the generators of the hydroelectric station to the load Ll9 changes when the line is subsequently s witched off. In the hydro-electric station there are a relay forcing excitation system and speed governors. The short-circuit is cleared in 0-2 sec dy the switches B— 1 and B—2.

The parameters of the system are as follows;


ON ON W

CÛ

<

«n vo O

o © 10 1

«o

S •o ò ^ '

^ O O Ó T j-

vo Ό Ò m

O Ô m

*n m

vo © <N

"*" co Tf fS Ò <N

n

1 ^ H

X

V Ö

• o ON i o r r ON o 6

1

oo co r* oo VO VO © ό

1

ON ON *Ti OO r f co © ô

1

Tf T f m r*» <N © ό ό 1

m co Q vo © CN Ô Ô 1

co m oo co ^ «o Ò Ô 1

VO VO «O ON m r-ô © 1

? Γ ° 2 x x

« «

m <

v ^

^

*« <N VO

6 II

*\τ*

eh c3

on ? I

Λ ^ ^ ^ II

~ *n ^ 1 ^

X + Il S •H ^

01 ? 1

, + + ejj î^ "O ^ V© w o

M5 * is ' 35 + 1 ** ·

*i 7 +

"" ° *o <N <N +

1 o·* <T

** ^ o

1 /"■* 1 o *o

<N <N + o** «o* w

•S £

°2Î «o

«o m ON

©

r j -< N VO

Ò

Γ-« ON VO

6

f » ON «S ψ*4

t^·

s

c* r* ©

^ o r ©

m es <N

co co ON

Ö

<N T t

vo Ò

TJ· r-vo ©

0 0 co co

^

TJ-m O

^ H oo VO Ò

m <N r» 6

T f <N

CJ co ON

Ö

1-H

vo vo ©

VO «r> vo ©

ON «O r o

^

r-VO O

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1 ^ T f r-©

o ^

r* ON

©

VO 1 ^ r·* ©

© ON T f

©

OO ( N vp yL*

co «o es

vo VO TJ-

©

m 0 0 oo ©

<s T f

VO

ON

©

ON 0 0 r* ©

»o VO T f ©

© r vo ^

CM t ^ es

· *■ co T f ©

^ H © ON ©

<N T t

VO

^ ON

© '

T f ON r-©

»r> «o

^ ©

f^ 0 0 VO vL*

,-* oo f S

^ H (N T t ©

r* © ON " ©

0 0 f S T T


HYDRO-ELECTRIC STATION

PG = 4 x 3 5 MW; Tj = 6-5 sec;

cos cpQ = 0-85; Td0 = 4-5 sec;

VG = 10-5 kV; Te = 0-3 sec;

xd = 1-15; Edeao = 2-5;

Λ^ = 0-7; a = 002 ;

^ = 0-37; Ts = 5 sec;

/? = 1.

TRANSFORMER

S r = 1 6 0 M V A ; Vk 220/10 = 13%;

A:r = 248/121/10-5 kV; Ffte(M10 = 20%;

^A: 110/10 = ' /o·

LINE LOAD LX LOAD L2

y4C = 300; PL1 = 50 MW; PL2 = 490 MW; / = 200 km; cos <pL1 = 0-8. cos cpL2 = 0-85.

*0 = 0-414 ohm/km.

RECEIVING SYSTEM

Pc=700 MW; cos <pc=0-8; xc=0-35 (referred to the power of the system with the generators replaced by their transient reactances).

ORIGINAL CONDITION

P0 = 83-5 MW; cos<p0 = 0-93; F0 = 221 kV.

Required: to determine the change of the frequency on the load busbars Lx.

Solution. (1) We determine the parameters of the equivalent circuit in the original condition in relative units, assuming:

5basic = 165 MVA; Fbasic = 10-5 kV.


The reactances of the equivalent generator of the hydro-electric station are:

- 1C 10·52χ0·85 165 *' = 1 ' 1 5 Χ 4X35 X I Ö ^ = 1 ' 1 5 ;

xq = Q-l\ JC = 0-37.

TRANSFORMER

13+20-7 ^220 = 2 = 13 per cent.

Vkll0 = 13 — 13 = 0 per cent;

Vk io = 20-13 = 7 per cent;

Λ 1„ 10-52 165 Λ„Λ XT220 = ° , 1 3 x - Î 6 Ô " x ï o ^ = °*134;

165 160 xT10 = 0-07 x — = 0-0722.

LINE

x, = 0414X200X^1 ( ™ Y = 0-222.

RECEIVING SYSTEM

Λ , _ 2202Χ0·8 165 /10-5V Λ Λ „ ^ - 0 . 3 5 χ — T O O — X ï ô T 2 ( w ) = ° · ° 5 2 ·

The power, transmitted along the line is :

83*5 P0 = -J2Y = 0-506; tan <p0 = 0-396; 165

β0 =0-506x0-396 =0-2.

LOAD Ι^

P n = i ° - = 0-303 ; βχ,ι = 0-303 X 0-75 = 0-227.

Variation of the Quantités in the System

LOAD L2

369

490 Put = ^ = 2-97; tan yL7k = 0-62; sin q>L2 = 0-527;

β ^ = 2-97x0-62 = 1-84.

To determine the impedance of load 2^, we find the voltage at point a of the equivalent circuit (Fig. 9.11) :

221 v ! 2 ! - 0-891 ■ K ° -10-5 X 248 - 0 8 y i '

ΛΥΛ O«, , 0-2x 0-356V , /0-506Χ0-356 \2Ί Λ ηΜ Va = ^ 0 8 9 1 + - ö ^ - j +( 0.891 ) J =0-992;

Ä t 0^02 ÔK.-K» = arc tan ö ^ - = 11-7 ;

O'QQl2 ν 0 · 8 ^ ι = Λ , Λ , (0·8+;Ό-6) = 2-08+71-56;

^ £ 2 —

0-303 0-8912x0-85

2-97 (0-85+7*0-527) = 0-193+/Ό-120.

V1"03

E« i£Z ! i°2Z22 I I 2_

a8094.jO-56^/|J 0-303*j 0-227

3 A jW3A K j 0-222

- ^ — ΐ — α τ ^ 0506*j02 2 97*j 1*84

208*j h56

J0y4|

j 0-052

7 0i93tjWZ

Fio. 9.11.

ft

The power loss in reactances 3 and 4 (Fig. 9.11) is:

J Q = 0 ^ 5 1 x 0 . 3 5 6 , 0-33. The output of the hydro-electric station is :

Sg = 0-506+/Ό-2+./Ό· 133 +0-303 +/Ό-227 = 0-809+/Ό-560. The voltage on the busbars of the generator is :

so - > / [ · n M 0-560 Λ Λ„„Λ2

9 9 2 + ¥ 9 9 2 X ° · 0 7 2 2 ) + / 0-809 ^ 0-992 X 00722 Υ Ί = 103.


The electromotive force behind the quadrature-axis synchronous reactance is:

^[(ο-*ο·770Μ^χΗ>ΐ56; άε,-ν. = arc tan ^ - = 23-7°; δΜ = 11-7+23-7 = 35-4°.

The electromotive force behind the transient reactance is : % - M°-

* * 0 - 3 5 9 , £ , o

^ - r e = a r c t a nT^4r = 1 6 · 1 ;

= 1-29;

E'M = 1-29 cos (23-7°-16-1°) = 1-28.

The electromotive force behind the direct-axis synchronous reactance is :

F — P Xd Xj C-' X(l X1 Xq~Xd

, „ 1-15-0-37 , „ 1-15-0-7 = 1 · 5 6 χ - ο ^ ^ - 1 · 2 8 χ Μ ^ 3 Τ = 1 · 9 3 ·

The output to the receiving system is : Se = 2·97+/1·84-0·506-/Ό·2 = 246+71-64.

The transient e.m.f. of the receiving system is : E< = J[{°-m + M X 0 ' 0 5 2 J + ( M X W 2 J ] = °-"7;

Λ . 0-1435 ôw = arc tan - ^ ^ - = 8-3 ;

δ120 = 35-4-8-3 = 27-Γ. (2) We determine the self and mutual impedances. With a short-circuit at point K we have:

Variation of the Quantités in the System 371

After removing the connection with the system: Z[[ =j0-7+;Ό·0722+2-08+7*1-56 = 3-313 ^48-3°;

^ ί = 3Ί3 = 0 ' 3 1 9 ; α π = 4 1 β 7 β · z'A = °°; y'\2 = °-

(3) We now proceed to calculate the transient process by the method of successive intervals (step-by-step method).

First interval We take the length of the interval to be At =0-1 sec. At the first instant after the disturbance :

p _ ^do 1 > 2 8 _ 9 .n 9 . 1-- ^ i i ( ^ - ^ ) c o s a i i

£J(0).= 2-36x2-02-1-1-11(0-7-

-1-365x1-28 P o) =2-022x 1-11 sin 0-3° =

Λ<7

Ξ* F , _ ν ' ^9(0) γ

-0-37) cos = 3-01; 0-0236; x'é

q~Xd

0-3°

Since

and

then

= l-28x2-212-2-02xl-12 = 0-46.

t a n avs = -ΤΓ-" gd

p

tan «5 - Λ<ο)*« - Ο · 0 2 3 6 Χ ° · 7 - ο . ο 1 7 8 . t a n °v.,., - "e—Î7 2-02~"Λ* ~ '

= 1°; '«" ^(O^MO) 2-02x0-46

ra(o>

0-46 cos Γ ^ ( o ) = — — = 0-46 < 1-03.

The exciter voltage will increase according to the rule : t

Ede = Edeoo — {Edeoo — Ed0) Q T· .

For the first interval, if the delay is ignored, E^ will increase to

Ε^α) = 2·5χ1·93-(2·5χ1·93-1·93)β °'3 = 2-75.


The mean voltage of the exciter for the first interval is :

1-93 4-2-75

The change of Éd for the first interval is :

ΔΕ>α) ^ ^ A J J ^ X O · ! = -0.015.

E'd at the end of the first interval is:

^ ω = ^ o + ^ ( i ) = 1-28-0-015 = 1-265.

The excess power of the hydro-electric station is:

AP(1) = PT0-Pm = 0-809-0-0236 = 0-785.

The change of angle for the first interval is:

where , _ 360/Λ*2

i t 165 c Tj = 6-5 x r-^ = 6-5 sec.

, 360χ50χ0·12 „__ * = *5 = 2 7 ' 7 ;

ζ1ό(1) = 2Ί·Ίχ^-ψ-= 10-8°.

The angle at the end of the first interval is

δω = δ0+Δδα) = 35-4 + 10-8 = 46-2°.

The change of the relative speed or frequency for the first interval is :

The relative speed or frequency at the end of the first interval is : ωα) =/(i) = 1+Λω(1) = 1-006.


We determine the change of turbine power due to the action of the speed governor, from (ref. 3, p. 147)

ΔΡηη) = kT1Aò0ipiHykT2APIXl^1y υ(η)

fcn —

Κτ9. —

-*r.nom

σ (Γ '+Τ^)3 6 0 / ο ' ßAt

Γ2 ~ At „ Τ1 + γ β

We assume the nominal power of the turbine to be equal to the nominal power of the generator:

-Pr.nom = 0*85, then

0*85 kn = —^-p- = 0-468 x 10-3 1/degrees;

0-02 ( 5 + ^ ) x 360x50

0J_ 0-1 kn = ^, = 0-0198.

5 + - 2 The change of the turbine power for the first interval is:

APni) = 0-468 x 10-3x 10-8 x 1-006 = 0-00509. The power of the turbine at the end of the first interval is:

pr(1) = PTQ-APTiX) = 0-809-0-00509 = 0-804.

Second interval

1-265 1-265 _ *2) 1 -1-11(0-7-0-37) cos 0-3° ~ 0-634 '

Ed(X) = 2-36x1-995-1-365x1-265 = 2-97; Pg(2) = l-9952xl-ll sin 0-3° = 0-0230;

APW = 0-804-0-0230 = 0-781; 0*781

Δδ(2) = 10-8+27-7 xJW6 = 32-3°;


32-3 1800

Δω2 = TöT^C = °'0 1 8; ω2 = 1 +0-018 = 1-018.

ΔΡΤ(2) = 0-468X IO"3X32-3x1-018-0·0198χΟ·00509χ4^| = 0-0153; 1-UOo PTm = 0-804-0-0153 = 0-789;

Vgdm= 1·265χ2·12-1·995χ1·12 =0-44; , 0-0230x0-7 n m f i , _ . ,„ t&nÒy

> = 1-995X0-44 = ° · 0 1 8 3 5 ' , = 1;

0-44 cos Γ ^ 2 ) = ZZZÏÔ = 0-44;

_0-2 Ede(2) =4-83-2-9 e °'3 = 3-35;

2-75 + 3-35 xsi/e(2) — 2 " J U : ) '

3-05 — 2-97 ^ ( 2 ) = Φ5 X0*1 = 0-00178;

E«t) = 1-265+0-00178 = 1-267.

Third interval

At the beginning of the third interval, i.e. in 0-2 sec, the connection with the system is broken.

Prior to the disconnection we have :

E _ 1-267 _ 2 . 0 .

Edl3) = 2x2-36-1-267x1-365 = 2-99; Ρ^3) = 22x 1-11 sin 0-3° = 0-0231 ; ΔΡ{%) = 0-789-0-0231 = 0-766.

After the disconnection :

E = L 2 ^ - 1>267 - ,.37. ** 1-0-319(0-7-0-37) cos 41-7° 0-921 '

EdiS) = 1-37x2-36-1-267x1-365 = 1-51; P(3) = 1·372χ0·319 sin 41-7° = 0-398;

ΔΡ{^ = 0-789 -0-398 = 0-391;


^ ( 3 ) = 32·3+27·7χ0·Ι66+η0;3

ο91=48·0°;

2x1-018 Λ 4 8

Δ(°™ = T8ÖÖ = 0-027;

ω(3) = 1+0-027 = 1-027; 1-027 APns) = 0-468xl0-3x48xl-027-0-0198x0-0153x^öTg- = 0-0227;

pr(3) = 0-789-0-0227 = 0-766; Vgd(3) = 1·267χ2·12-1·37χ1·12 = 1-15;

n.108 v M òr = 10°; t χ 0-398x0-7

tan Oym = , , , ,„ , , c — 0-177; 1-37x1-15

V - M S - 1 1 7 K*(3> * cos 10° - * 17·

When the voltage on the busbars of the generator Vg exceeds the normal (Vg0= 1-03), the forcing of the excitation ceases, and the voltage of the exciter starts to drop according to the law:

Ede = Ed0 + [Edem- Ed0] e TÄ , where the time scale commences at the beginning of the third interval.

1'5

1·0

0·5

o 05 1·0

FIG_ 9.12_

1·5 s~c

TABL

E 9.

Inte

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9 0-

921

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2·

36£, ,-

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F ^•dt

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If it is assumed that Tel = Te = 0*3 sec, then the value of the exciter voltage, at the end of the first interval, is determined as :

~0·3 Edei3) = 1-93+(3-35-1-93)e °'3 = 2-95; 3-35+2-95 ETnean — £<fe(3) - = 3-15;

ΔΕ'ά(3) = 3 '154.5

1 '51Χ0·1 = 0-0364;

E'm) = 1-267+0-036 = 1-303.

i-o 1-0

0-8

0-6

<K

0-2

PT

-

- ■ .

pg

.

1 1 1 L , , 0-5 10

Fio. 9.13.

1-5

With the aid of Table 9.11, the calculation for the subsequent intervals is carried out similarly.

The calculated results are given in Fig. 9.12 and 9.13.

Problem 9.5

In the system, the equivalent circuit of which is shown in Fig. 9.14, a sudden increase in reactive load of amount Q'L occurs as a result of a remote short-circuit.

- ^ TL0

<f

x 2 E;

r 2 0

ÏP,,Q L'"L

FIG. 9.14.


The parameters of the system and of the original condition, in relative units are:

Xx = 1-13; x2 = 0-139; E[ = 1-806 .28-6°;

£j = 1-207 .13-7°; VL0 = 1 . 0 ° ; P10 = 0-765;

/»a, = 2-07; PL = 2-83; QL = 1-76; Q'L = 1-38;

Tn = 10 sec; TJ2 = 32 sec.

Required: to determine the amplitude of the oscillation of the relative angle δ12, and also the voltage on the load busbars, VL, during the oscillations.

Solution. We determine the self and mutual admittances of the equivalent circuit, taking into account the increased load :

SL = 2-83 +j 1 -76 +j 1-38 = 4-23 . 4 8 ° ;

sin ç>L = 0-743 ; cos tpL = 0-669 ;

ZL = - ^ (0-669+/Ό-743) = 0-158+/0-1755;

_ / M 3 1 7Ό·139(0·158+;Ό·1755) _ Z " -Jl 13+0·158+7Ό·1755+,Ό·139 ~ 1U ^ 8 8 8 '

yn = Vn = ° ' 8 2 ; "" = h2°;

Z - jQ-13? l yi-13(P-lS8+70-175S) ,. ZM -yO 139+ 0.i58+y0-1755+yi-13 " ° 3 2 ? 6 9 ° '

722 = Ö327 = 3 ' ° 6 ; ^ 2 = 2 1*°°;

Z12 = y i . l 3 + y 0 - 1 3 9 + 0 4 ^ i ^ = 1-816 .104-2°;

Λι = 3 ^ = °-550; «u = -14-2°.


The power characteristics of the stations are:

P1 = 1·8062χ0·82 sin 1-2° +1-806X 1-207x0-55X

xsin (ό°2 +14-2°) = 0-054 + 1-20 sin (δ°12 +14-2°);

P2 = l-2072x 3-06 sin 21-0°-1-806 X 1-207x0-55 X

xsin (<5°2-14-2°) = 1-595-1-20 sin (δ°12-14-2°);

Qx = l-8062x 0-82 cos 1-2°-1-806 X 1-207 x 0-55 X

X cos (δ°12 +14-2°) = 2-67-1-20 cos (δ°12 + 14-2°).

The excess power is :

ΔΡΧ = 0-765-[0-054+ 1-20 sin (ÔÎ2 + 14-2°)]

= 0-711-1-20 sin (0?2 + 14-2°);

ΔΡ2 = 2-07-[1-595-1-20 sin (δ°12-14·2°)]

= 0-475 + 1-20 sin (<5Î2-14-2°).

The accelerations of the stations are:

379

«i = 360/0 ΔΡ1 360x50 [0-711 -1-20 sin (δ°12 +14-2°)] Tn 10

= [1-28-2-16 sin (<5?2 + 14-2°)]x 10»;

360/0 ΔΡ*

[ / 2

360x50 32 [0-475 + 1-20 sin (<5J2-14-2°)]

= [0-267+0-675 sin (a;a-14-2°)]x 103;

α12 = «2-x2 = [l-013-2-16sin(^2-14-2°)-

-0-675 sin (ôj°2- 14-2°)] X IO3.

Assuming ój2, we determine the relative acceleration a12x 10-3 (Table 9.12).

TABLE 9.12

(5°

α12ΧΐΟ"3

12

0086

13

0040

14

- 0 0 0 5

15

- 0 0 4 9

16

- 0 0 9 6


Examining Fig. 9.15, in which the function 0L12=f{b°12) is plotted, we determine from the equality of the braking areas FT and Fy that the varia-

0Ό8

0-06

004

002

0

-002

-0-04

-0Ό6

-0Ό8

FIG. 9.15.

tion of the relative angle ò°12 during the oscillations, lies between the limits :

^2max = <5Î20 = 28.6o-13-7o = 14.9°; aï2min = 12-90.

An angle δί20 = 13·9° corresponds to the steady condition after the damping of the oscillations.

It is obvious that the maximum value of the angle, <512 = 14-9°, corresponds to the minimum value of the voltage VL, and vice versa.

With<5°2 = 14-9°:

Px = 0-054 + 1-20 sin (14-9°+ 14-2°) = 0-638; Q1 = 2-67-1-20 cos (14-9°+ 14-2°) = 1-62;

With Ô°12 = 12-9°: J»x = 0-599; Qx = 1-60;

VL*~ = 0-890.


It can be seen that for the transient process considered, the difference between the maximum and minimum values of the voltage on the load VL, amounts to less than 1 per cent.

Problem 9.6

In the system, the equivalent circuit of which is shown in Fig. 9.16, the switches B-1 and B-2 are opened simultaneously, thus reducing the load on stations 1 and 2.

The original values, in relative units are: Pu = 0-583; P20 = 3-08; Tn = 3-02 sec; TJ2 = 16 sec; σ = 0-01 ; Ts = 4 sec.

I B - I Ù Ù B - 2 2

T r20 L

FIG. 9.16.

Required: to determine the frequency in the divided sections of the system: (a) ignoring the operation of the speed governor: (b) allowing for the operation of the turbines' speed governors by an approximate method.

Solution, (a) ignoring the turbine speed governors. We determine the variation of the absolute angles of both stations :

δχ =lS%P^t2 + K = 180X50X0-583 fl+^ = 1 . 7 4 0 / 1 + 4 7 . 5 „ . 1 j \ J*U2

a, = Ιψ*> f + J , = 180χ50χ3·08,2 + 3 7 . 5 , = „w+yi-F. 7/2 16

During the short-circuit, the relative angle remains constant: Ô12 = Ô1-Ô2 = 10°.

Obviously this occurs if,

ζμ = JüL = const.

Actually, 0-583 3-08 302 - 16 = 0193.


The excess torque on the turbine shaft, determines the increase of the frequency of the systems, which can be determined from the expression for either of the absolute angles.

Hence, differentiating the expression for the angle dl9 we get the variation of the angular velocity and frequency in relative units:

1 άδχ P10 0-583 Λ 1Λ„ /Λ „χ Λο , 0-583 Τη 3-02

4/; = Δω+.

= 0-193/;

ΔΡΤ = Γ·^ Αω+.

The corresponding function Af^ =f(t) is given in Fig. 9.17. (b) Allowing for the turbine speed control. In this case we use the

method of successive intervals, taking At=0-1 sec. Assuming that the nominal power of the turbine is equal to the power given out by the generator in the original condition, the change in power APT, under the action of the governor can be determined as follows :

Pj^At

or ._ 0-583x0-1 . . n . , « . , -

ΔΡτ= 0-01Χ4 Λ / . - Ο - Ι θ Λ / . .

The variation of frequency Af, may be determined from (9.3), allowing for the variation of the turbine output:

Af, =3^2"^rX0-l = 0-O331Pr.

First interval Afm = 0-0331x0-583 = 0-0193;

APT(1) = 0-162x0-0193 = 0-00313; Pni) = 0-583-0-003 = 0-580;

/(1) = 1+0-0193 = 1-019.

Second interval Afa) = 0-0331x0-580+0-0193 = 0-0385;

APT(2) = 0-162x0-0385 = 0-00624; Pr(2) = 0-580-0-006 = 0-574; /(2) = 1+0-0385 = 1-038.

Variation of the Quantités in the System 383

All the calculations are summarized in Table 9.13. The frequency variation, allowing for the turbine control, is shown in Fig. 9.17 (curve 2).

14

12

10

8

6

U

2

01 0-2 0-3 0-4 0-5 0-6 0-7 sec

FIG. 9.17.

TABLE 9.13

Inter' val

1 2 3 4 5 6 7 8

/, sec

01 0-2 0-3 0-4 0-5 0-6 0-7 0-8

+0-0ÖJ57i>T(Ä_1)

00193 00385 00576 00763 0-0946 01124 01296 01461

ΔΡΤ = 0-162Af

000313 000624 000932 00124 00153 00182 00210 00236

-ΔΡΤ

0-580 0-574 0-565 0-553 0-538 0-520 0-499 0-475

/=1 + +4/*

1019 1038 1058 1076 1095 1112 1130 1146

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390 Appendixes

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392 Appendixes Ap

pend

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Appendix 1 393 Ap

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6643

1·67

091·

6775

1·68

421·

6909

1·69

771·

7045

1·71

131·

7182

1·72

5130

°60

°1·

7321

1·73

911'

7461

1·75

321·

7603

1·76

751·

7747

1·78

201·

7893

1·79

6629

611·

8040

1·81

151·

8190

1·82

651·

8341

1·84

181·

8495

1·85

721·

8650

1·87

2828

621·

8807

1·88

871·

8967

1·90

471·

9128

1·92

101·

9292

1·93

751·

9458

1·95

4227

631·

9626

1·97

111·

9797

1·98

831·

9970

2·00

572·

0145

2·02

332·

0323

2·04

1326

642·

0503

2·05

942-

0686

2·07

782·

0872

2·09

652·

1060

2·11

552·

1251

2·13

4825

°65

°2·

1445

2·15

432·

1642

2·17

422·

1842

2·19

432·

2045

2·21

482·

2251

2·23

5524

662·

2460

2·25

662·

2673

2·27

812·

2889

2·29

982·

3109

2·32

202·

3332

2·34

4523

672·

3559

2·36

732·

3789

2·39

062·

4023

2·41

422·

4262

2·43

832·

4504

2·46

2722

Deg

rees

1·0

0·9

0·8

0·7

0·6

0·5

0·4

0·3

0·2

0·1

cot

394 Appendixes Ap

pend

ix10

(con

td.)

tan

0·0

0·1

0·2

0.3

·10·

40·

50·

60·

70·

80·

9D

egre

es

682·

4751

2·48

762·

5002

2·51

292·

5257

2·53

862·

5517

2·56

492·

5782

2·59

1621

692·

6051

2·61

872·

6325

2·64

642·

6605

2·67

462·

6889

2·70

342·

7179

2·73

2620

°70

°2·

7475

2·76

252·

7776

2·79

292·

8083

2·82

392·

8397

2·85

562·

8716

2·88

7819

712·

9042

2·92

082·

9375

2·95

442·

9714

2·98

873·

0061

3·02

373·

0415

3·05

9518

723·

0777

3·09

613·

1146

3·13

343·

1524

3·17

163·

1910

3'21

063·

2305

3·25

0617

733·

2709

3·29

143·

3122

2·33

323·

3544

3·37

593·

3977

3·41

973·

4420

3·46

4616

743·

4874

3·51

053·

5339

3·55

763·

5816

3·60

593·

6305

3·65

543·

6806

3·70

6215

°75

°3·

7321

3·75

833·

7848

3·81

183·

8391

3·86

673·

8947

3'92

323·

9520

3·98

1214

764·

0108

4·04

084·

0713

4·10

224·

1335

4·16

534·

1976

4·23

034·

2635

4·29

7213

774·

3315

4·36

624·

4015

4·43

744·

4737

4·51

074·

5483

4·58

644·

6252

4·66

4612

784·

7046

4·74

534·

7867

4·82

884·

8716

4·91

524·

9594

4·00

455·

0504

5·09

7011

795·

1446

5·19

295·

2422

5·29

245·

3435

5·39

555·

4486

5·50

265·

5578

5·61

4010

°

80°

5·67

135·

7297

5·78

945·

8502

5·91

245·

9758

6·04

056·

1066

6·17

426·

2432

981

6·31

386·

3859

6·45

966·

5350

6·61

226·

6912

6·77

206·

8548

6·93

957·

0264

882

7·11

547·

2066

7·30

027·

3962

7·49

477·

5958

7·69

967·

8062

7·91

588·

0285

783

8·14

438·

2636

8·38

638·

5126

8·64

27.

8·77

698·

9152

9·05

799·

2052

9·35

726

849·

5144

9·67

79·

845

10·0

210

·20

10·3

910

·58

10·7

810

·99

11·2

05°

85°

11·4

311

·66

11·9

112

·16

12·4

312

·71

13·0

013

·30

13·6

213

·95

486

14·3

014

·67

15·0

615

·46

15·8

916

·35

16·8

317

·34

17·8

918

·46

387

19·0

819

·74

20·4

521

·20

22·0

222

·90

23·8

624

·90

26·0

327

·27

288

28·6

430

·14

31·8

233

·69

35·8

038

·19

40·9

244

·07

47·7

452

·08

189

57·2

963

·66

71·6

281

·85

95·4

911

4·6

143·

219

1·0

286·

557

3·0

0°

Deg

rees

1·0

0·9

0·8

0·7

0·6

0·5

0·4

0·3

0·2

0·1

cot

APPENDIX 2

TYPICAL STATIC CHARACTERISTICS OF THE LOADS OF ELECTRICAL SYSTEMS

116

112

ΙΌβ

K)4

10

096

092

VvQ* ^Q*(6kV)

Q«(II0 k V ^

Q*(IIOkVH

|Q*C6I<V)1 :P (6 ondllO kV) 1

1 1 1, L 45 A6 47 46 49 50 51 52 Hz

0-95

0-90

0-85

080

0-75 075

v*

1050 1025 1000 0-975 0-950 0-925 0-900

P. 6,110 kV

1033 1017 1000 0-984 0-969 0-954 0-941

6 kV

1130 1063 1000 0-945 0-900 0-863 0-833

Q* 110 kV

1090 1045 1000 0-962 0-930 0-905 0-885

V*

0-875 0-850 0-825 0-800 0-775 0-750

p* 6,110kV

0-923 0-916 0-904 0-983 0-882 0-872

Q* 6 kV

0-807 0-785 0-767 0-751 0-743 0-735

110 kV

0-870 0-858 0-849 0-844 0-844 0-848

Regulation at nominal voltage:

for a load at 6 k V ^ = 0-6; ^ = 2-3;

for a load at 110 k V ^ = 0-6; ^ = 1-6.

395

APPENDIX 3

NOMOGRAMS FOR DETERMINING THE SELF AND MUTUAL ADMITTANCES

AND AUXILIARY QUANTITIES

X 0 02

005-

o-H

4-89"

02 H

0-3-1

85M

1 - 80°

H70"

0-4-4

0-5 H

3-60° 06-4-

65

07-

0-8 -4 _ ï -50°

I 0-9-f-R 3 - a

I-

-55w

45"

0O2 - p 1

0Ό5-[_3ο

- 4 °

- 5 ° ο·Η

0-2-f

•75" 4-15"

0 3 4 ]

4-20°

0-4 4 ]

H io°

4-25° 0-5 H

0 6 H -

0 7-

08 ^

09-3-

R "3-I

^30°

35"

4 - 4 0 °

45°

JL cos* X sin a

0Ό2 -T- 002

0O5-4-

0 2 ^ a :

-01

-02

07-4 :

0-8-4:

09-4

03—I-[ -03

0-4-+

=H04

05 - i

3-05 0-6 -±

•06

sin Ψ cos oc 069

_R. sin Ψ X cosa

0O2-

005

01·

-0999

j-0-998 0997 0996 0995

H 099

0 2 - t -

0-3 H

0-44

0-5 H

0-6-4

07-4 -

0-8-4-

0^9-3 JL 4-R

1 - ^

-098

-097

-096

-095

4 - 0 9

4 - 0 8 5

•08

c o s * sin a

3- 071

002 - p 002

+ 004 005-

01-

j - 0 0 6

1-O08

01

0-2-4 4 - 0 2

0 3 H

h 03

0-4-4

05-

06-4

0-7-4-

0-8-3

0-9-1

I-

-04

045 046 047

048

4 - 0 4 9

-05

- I l ; ct = 90-¥

NOMOGRAM NO. 1.

396

Appendix 3 397

NOMOGRAM NO. 2 .

398 Appendixes

16--59

55- Formula *<* " % Λ,

F-' ' j i

A-5-

K,,= ■•m->z cos 2 CC

r 2 'JI "12

*50

^0-6

0-9-4-

3-5-

2-5-

1-5-

I ·

0-7

NOMOGRAM NO. 3.

APPENDIX 4

NOMOGRAMS FOR DETERMINING THE ANGLE y AS A FUNCTION OF TIME %

180°

160·

140°

120°

100·

80°

60°

40·

20°

0 ·

180·

160·

K0°

120·

100°

80°

60°

40°

20°

ölim

2-50

i n f i l i I / ' / 1/ p i

*~l Lz/Mß/

/ /// / s Τ:2·0 / / / / / /

Ι·75 f f f 1 f*■»

F*Y

0-30

010

1—

/<'/

J>$ 2sS

«^

N^

V

h s>>

sin (5g : 0

4>

sd 3 4 5 6

NOMOGRAM NO. 1.

püm

T*20

y t _

175

/ J

i

f Ψ

ff /i // //

/i?

ZMRL / / ////Λ IIS Uli

P /i fl ^ Q.;

/ 7

1

y y^ Vrft

r ^

î

1

^Λ ©*

?3

Γ^

η·7·>ι

sin fyO-l

!

1

!

!

τ 1 3 4 5 6

NOMOGRAM Ν θ . 2.

399

8 9

400 Appendixes

180e

160°

140e

120·

100°

80°

60°

Α0·

20e

Plim

2-50 Tr20

Ι·75 Ι·50

o l l / κΐ|

7[/// ri ΊΙ

lì i

un ml fifth

Τ=0·40 045

1 k

_ p 0-25

7X7 1 J£/À> /À

"?3

0-35

/C>

Λι >r oj*J 0-70

Φ };e>

sS^

0-30

sin 5J=0-2

|τ 2 3 A 5 6

NOMOGRAM NO. 3 .

180°

160°

K0°

120°

100°

80°

60°

40°

20°

0°

δίιιτ)!

τ=:

π / L£^

T=30 -4 1 il ί 2-50 >0-1-75 1-50

ÌIIWM wem ■

~*+ \Ά % &

* ·*

f -*

0-50

V f/*r / / /A / / cv/ Ύ<

^ y

^

yf<&

045

Γ ο

^ '

75

(V7n

<*0

^

0-A0

6^N

^^

0-30

sin 5^=0-3

rj 3 A 5

NOMOGRAM Ν θ .

Lsg-o

Pi

\- V S

l* 7*

è 0 c u >

2 i- 1

»

Ó II

c

ΓΝ|

hl fM

C

- o

> y

0 i c>y V «s /#l ί tai é

B

o o

Us 1 le 1 1 ■ 1 li-

ΝΛΝΧΙ

H > r

> 1

ί ) I 1 <N

ΐ] h

E |«o* o O

co îfe

e^ Γ

τ

o S

in ô «O

c

o o

& o o CM

σ § il fSl

o o co

ώ| ώ/ώΐ3

è 1

fe Tj &

Appendix 4

NOMOGRAM No. S.

401

NOMOGRAM No.6.

402 Appendixes

180·

160°

140·

120·

100°

80·

60·

40e

20·

0·

κ* 2-50 T»2 Ι·75 1-50 Ι·30

**>

—1

!

W ^Ίτ

J& V

^^*

0·β5

tôS

^ ^

^^ 08s)

k Ε£ sai

ι&

sin θ0'=0·6

= - ]

Τ 3 4 5 6 7

NOMOGRAM Ν θ . 7.

180e

160°

140°

120°

100°

80e

60°

40°

20°

0e

Him

T=2_

175-

" 1 lUM o

Ι*Ί\ι /κ y / / / / / ' /

l

I I

07

!

! i

/s i

<2>

/4 Ψ 090

f

2g5 ggo 22

! I

!

sin$£=0 7

! j

; I

j ! v j !

, ^ 1

4 5 6 7

NOMOGRAM NO. 8.

Appendix 4 403

$!im

T=2-

175

m

st

wff-fy *ΊΠΠ

|

ft

]$'

4' V

Ψ

1

/ψ

Jrt&—

09C

08b

sin <S' =0-8 0

T A 5 6 7 8

NOMOGRAM NO. 9.

10

[«Vim feit / j f /

Yfff8ffî&

' I : I

i i

! j

/y

ra*v / /

ί ^ , Λ ^ >

sir

1 / 1

I I V j /

' I ^ά' 0-9

I

! 1 i

I j ;

; I

1

! 1

1 I

r

NOMOGRAM NO. 10.

APPENDIX 5

NOMOGRAMS FOR DETERMINING THE CRITICAL SHORT-CIRCUIT

SWITCHING ANGLE

404

APPENDIX 6

NOMOGRAMS FOR CALCULATING DYNAMIC STABILITY

405

406 Appendixes

1 "

,

/

Γ-Γ u

!

r\i ri I ϊ lì ì f l i i ì i

o o

M ri hn \ ni

o

S i " /j -se

i l ' ih

Al

/ / / /

1

o ~ cr

7 evi -ô

i 7 /

/ f J

Jjj

j e j n i

- 0 / -se/

f *p 1 ,j<

«O

/

ε 1 1

— c

"*/

/ i y /

M / /

? —

ò — H

/ h 1/ / / 7 / /

J

ò

/ . 7 / / h J

(ΓΟ

ò ~i JE /

f

-CM

ό

JE

~o "1 oc

S ° 2 I t* 1 ° i J*i

I *° J (

lui

Je

/ / f /

I

_il

/ c

' / 7 / f 1 / /

J

0

e.

1 i /

/ ' ; / /

M

/ / ' / /

/ /

co "~ ô"

JE

/ \

/ / /

LL

Ó L. ·· _

r ,

cb~î

/

in

l-l ε 1

' Γ ! . 1

1

l i 1 1

VI je

/ / 4/ Y /Λ Λ/ /

J

— i —

to

^ ΐ

/ / ,

H 1

ô

1 1, / / ,

Λ 1

I ^

ό

ό

—

1 | i i

Appendix 6 407

co

Γ " '

u / L n 1

- /co

\ΊΦΙ I

» -E u

1 Π 1 1

'/ li 1 1

i^L h t

ti

■> I ó l ó /

lì

1 l 4 1 1

in \

/A A /

o

0 0

i

1 L

1

'/ h f I

Info kni 1 ai If 77 f

1 E i | V 0

fi

4uzh>L rr\

«fi

%

ó 1 1

Mi: i il

/ M/ / / I

1 E"H 11

» 1 ò Ir

///ϊ [ / / / / /

—irci)

II oc

1 1 I l i 1

CM

ò—

/ /

L

"1 ó

- OC —

<7»

Γ » "

i

y.

co ò

I h 1 1

T

1 1 ir h lì

in > *?t l

"" 1 L r 1

1

1

/ à f / 7 1

/ /

Il _

f ò II

—JC -n / , U

t\ ! /

Ó

0 "S 1

/ri

g/ _ X.

/> 0 ΛΛ

χ·

p CTI

JMILtì lì m Vi / x"

li O co

1

7?

O

//

// X" _ p co

X" Il J

o K>

o **

o co

o OD

._

/ X

O fs

X"

ί o 'xV

X"

I i V 1

ΓΓΓΓ // //

/j //

χ- _ p CO

/

x-

^1

/

/i //

' /

/ //

' o •CD

3 1 -1

o ro

O 1^

o en

o CD

__

X" II O à

x-y tii

.X" . O en

9

Ψ II J f o CO

/ /

x* It J

o l\>

o .Γ»

o CO

o CD

- o

'χ- i ii I

■^irli p "ö en

/x-o «

π J

i / /

1 II TT o

cö

l·3

/

x-

rô

Sdxipudddy 80t>

Appendix 6 409

Ι·3

II

0-9

07 0 0-2 0-4 0-6 0-8 1-0

APPENDIX 7

MAXIMUM LOAD CURVES

ι·ι

1-0

0-9

0-8

0-7

mT

1

k=0· 8

|ôk-l-5

>k=0-7'

k^Pfc&r _5KJ

maximum load

APPENDIX

THE PRINCIPAL PARAMETERS OF

I. Turbo-generators at 3000 r.p.m.

Type of turbo

generator

T2-6-2

T2-12-2

TVC-30

TV-60-2 TVF-60-2 TVF-60-2 TV-2-100-2 TVF-100-2 TV2-150-2 TVV-150-2 TVF-200-2 TGV-200 TVV-200-2 TVV-300-2 1GV-300

5, MVA

7-5

15

37-5

75 75 75 117-5 117-5 166-5 176-5 235 235 235 353 353

cos φ

0-8

0-8

0-8

0-8 0-8 0-8 0-85 0-85 0-9 0-85 0-85 0-85 0-85 0-85

kV,

315 6-3 6-3

1 10-5 6-3 10-5 10-5 10-5 6-3 13-8 10-5 18 18 H 15-75 15-75 20

0-85 1 20

I ILL Λ

1375 688 1375 825

3440 2065 4125 4125 6900 4925 6475 5350 5670 12350/2 8630 8630 10200 10200

hot.L

113 102 94 156 146 255 617 688 268 640 321 810| 860 700 1150 1205 1070

£-2 256 250 247 456 455 717 1500 1700 650 1650 670|

2200 2240 1750 2680 2920 2890

135

190

180

239 200 230 319 270 427 . 390 400 450 320 445 440

•Si

1 96 96-4 97-2 96-6

98-3

98-5 98-5 98-5 98-78 98-7 98-9 98-7 98-8 98-87 98-6 98-7 98-79

Ci

1-3

2-4

5-3

13-5 8-85 8-85 21 13 33-5 17-8 1 26-4 26-5 i 22-4 30 31-1

D.C. resistance of the stator phase winding at 15°C/ohm

000313 00124 000478 00141

000537

000224 00065 00070 000145 000103 000141 000236 0000414 000108 000121 000102 00013

* Symbols for the time constants Td0 = the time constant of the excitation winding with the stator winding disconnected ; rj8 = the time constant of the excitation winding with a three-phase short-circuit of

the stator winding; rj2 = the time constant of the excitation winding with a two-phase short-circuit of

the stator winding; T'dl = the time constant of the excitation winding with a single-phase short-circuit of

the stator winding;

410

8

SYNCHRONOUS MACHINES

(calculated values)

D.C. resistance

of the rotorphase winding at 15°C/ohm

0-326 0-341 0-559 0-559

0-331

0-226 00996 00996 j 0-335 0122 0-436 0118 0124 0194 008745 0131 0116

Reactances, %

Xd

12-0

11-5 13-2

15-2

13-2 17-6 1 22-45 13-8 18-3 12-2 21 16-5 19 19-2 17-3 19-6

x'd

17-2

17-5 20-2

25-7

24 271 33 0 20-3J 28-3 18 31 25 27 28 26 28

Xd

165

187 208

252-7

220 218 196 180 179-5 148-7 171 195 185 190 170 192

* 2

14-7

14-3 161

18-6

191 21-5 27-4 16-8 22-3 14-2 25-5 20 23-2 23-4

210 24

*o

6-7

5-5 7-3

7-2

6-7 7-7 5-5 8-2 9,5 7 10-5 100 8,3 10-3 8-7 9

Time constants, sec*

Tdo

71

8-20 7-9

10-3

11-7 6-5 4-7 13 6-2

11-9 51 712 6-5 60 6 0 6-65

7 Ì

0-8

0-75 0-76

102

1-3 0-8 0-79 1-46 0-97 1-44 0-9 0-91 0-96 0-9 0-91 0-97

Tit

1-26

1-25 1-27

1*63

213 1-32 1-27 2-44

T56 2-4 1-44 -

1-59 1-46 1-47 1-53

T'dl

1-44

1-44 1-48

1-84

2-39 1-48 1-36 2-88 1-76 2-7 1-63 -

1-78 1-67 1-66 1-72

7ÏÏ

01

0094 0095

0-127

0161 01015 0098 0182 0121 018 0113 0114 0121 0111 0114 0121

Tdz

0128

0159 0172

0-21

0-258 0-3 0-338 0-386 0-417 0-42 0-406 0-513 0-467 0-42 0-357 0-42

Tal

0104

0128 0145

0167

0-202 0-231 0-237 0-32 0-337 0-35 0-327 0-427 0-366 0-342 0-295 0-332

of the turbo-generators. T'd'3 — the time constant of the alternating component of the transient current ; Td3 = the time constant of the alternating component with a three or a two-phase

short-circuit; Tal = the time constant of the alternating component with a single-phase short-circuit.

411

II. H

ydro

-gen

erat

ors

Type

«-ÌS

-C

B^

.40

»S-

"m"

CB

-.

4

CB

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5,

26*3

27

30

40

50

52

68-7

5

68-7

5

66

71-5

kV

10-5

10-5

10-5

105

10*5

10-5

105

13-8

10-5

13-8

cos <

p L

0-8

0-8

0-8

0-8

08

0-8

08 0-8

0-85

0-8

Effic

ienc

y %

- - - - - - 97-4

96-7

5

Nom

inal

sp

eed,

r.p

.m.

125

150

100

100

136-

4

88-2

428-

6

62-5

375 62

5

tm2

3-0

3-2

7 8 7-8

15 0-75

50 1-2

47

Reac

tanc

e %

*d

70-6

105 95-3

81-5

1-82

73-8

113-

6

73

94-2

63-3

Xq

X(t

- - -

56-2

46-4

31

27*5

30-3

31

32

27-5

28

27-4

21-3

28-4

x'd

_

- - - -

_

- 20-7

T dr. sec - - - - - - - - 7-55

4-3

412 Appendixes

Tab

le I

L

(con

td.)

Type

«w-»

»s

-»3

?-«

-^

ο·

<«

CO

CB

|?.2

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353

590

590

590

590

590

kV

13-8

13-8

13-8

15-7

5

15-7

5

15-7

5

15-7

5

15-7

5

15-5

7

15-7

5

cos

(pL

0-8

0-85

0-85

0-85

0-85

0-85

0-85

0-85

0-85

0-85

Eff

icie

ncy

/o

97-3

6

97-6

3

97-4

2

98-2

5

98-3

5

98-1

98-2

98-2

98*2

98-2

Nom

inal

sp

eed,

r.p

.m.

83-3

125 68

2

125

214 83

-3

90-8

90-8

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185

160

170

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%

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27

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27

30

7λ

, se

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6-75

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50

Appendix 8 413

Tabl

e IL

(co

ntd.

)

Type

BC

O»H

*L«

BC

C!^

.«

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590

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824

824

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kV

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5

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5

15-7

5

15-7

5

200

200

200

200

15-7

5

15-7

5

cos

<pL

0-85

0-85

0-85

0*85

085

0-85

0-85

0-85

0-85

0-85

Effic

ienc

y %

98-2

98-2

98-5

98*3

98-4

98-4

98-4

98*4

98-3

98-2

Nom

inal

sp

eed,

r.p

.m.

93-8

93-8

93-8

83-3

125 75

77

115

100

100

tm2

200

+ 21

0

150+

160

230

+ 24

0

220

120

370

350

130

160

145

Reac

tanc

e %

Xd

123

133

111

176

158

176

185

163

165

200

x 9 - - - - - - - -

Xd 43

47

34

44

48

46

44

49

37

48

x'd 30

33

23

30

29

30

31

30

26-5

29

Té.. sec

7-6

80

8-33

5-1

5-0

51

50

4-9

51

5-5

414 Appendixes

HI.

Sync

hron

ous

cond

ense

rs

Type

KC

300

00-1

1

KC

150

00-1

1

KC

150

00-6

KC

750

0-6

KC

500

0-6

KC

B-7

5000

-11

KC

B-3

7500

-11

V,kV

105

105

66

66 63

105

10-5

n,

r.p.m

.

600

750

750

1000

1000

750

750

Loss

, kW

600

340

340

200

150

1026

600

Reac

tanc

e %

*σ

16-3

121

9-7

90

100

12-4

14*3

*o

10 8 8 6 6 10

11

Xba 39

25

20

18

17

x 2 27

19

15

16

21

194

24

x'd 48

34

28

25

25

36

45

x'd 27

18

15

15

16

19-4

24

Xd

177

194

158

168

157

238

211

x<

107

117 94

82

77

128

128

15°C

I oh

m

0013

0-02

8

0-01

2

0029

0042

0-00

435

0Ό97

fb

15°C

f oh

m

0-23

9

0-12

3

0123

0175

009

0-01

03

~0-

2

Td„

sec 8-8

7-3

7-5

4-7

44

10-2

8-7

Roto

r cu

rren

t

h.i>

a

580

565

560

401

485

1220

750

h.h*

o

185

186

206

145

182

350

225

GZ)

2 , tm

z

105

24-8

- 86 3-7

230

58-3

AppendiX 8 415

REFERENCES

1. GLAZUNOV, A. A., Electrical Networks and Systems, Gosenergoizdat, 1960. 2. UL'YANOV, S. A., Short-circuits in Electrical Systems, Gosenergoizdat, 1952. 3. VENIKOV, V. A. Electromechanical Transient Phenomena in Electrical

Power Systems, Gosenergoizdat, 1958. 4. SHDANOV, P. S., Stability of Electrical Power Systems, Gosenergoizdat,

1958. 5. VORONOV, A. A., Elements of Automatic Control Theory, Voyenizdat, 1954. 6. MARKOVICH, I. M., Power Systems and their Operation, Gosenergoizdat,

1957. 7. Electrical Reference Book, Gosenergoizdat, 1953. 8. VENIKOV, V. A. and ZHUKOV, L. A., Transient Phenomena in Electrical Power

Systems, Gosenergoizdat, 1953. 9. MAMIKONYANTS, L. G., Currents and Torques in Synchronous Machines,Thc

work of TSNIEL, edition IV, Gosenergoizdat, 1956. 10a. CHESNOV, M. P., Investigation of Asynchronous and Resynchronizing

Conditions of Generators, Elektrichestvo, 1960, No. 6. 10b. CHESNOV, M. P., The Effect of Turbine Speed Governors on the Resyn-

chronization of Generators Dissertation Moscow Power Institute, 1960. 11a. VENIKOV, V. A. and GORUSHKIN, V. I., Lectures for the course "Stability

of electrical power systems" — simplified methods of calculating the dynamic stability of electrical power systems. Moscow, edition VZEI, 1959.

l ib » GORUSHKIN, V. I., The effect of excitation control and forcing on the dynamic distant transmission stability, Manifold "Problemy energetici" Academy of Science USSR, 1959.

12. VENIKOV, V.A. and SOLDATKINA, L.A. Guidance to special designing course in transmission system, Moscow Power Institute, 1950,

13. VENIKOV, V.A., Analysis of transient phenomena in electrical power systems, with the aid of Gorev - Park equations, Lectures, edition VZEI, 1955.

416

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