Eindhoven University of Technology

MASTER

A transient stability analysis by computer simulation of the industrial power system of DowChemical Terneuzen

Marteijn, P.A.M.

Award date:1986

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AFDELING DER ELEKTROTECHNIEK

Vakgroep Elektrische Energiesystemen

A trans.lent stability analysis bycomputer simulation of the industrialpower system of Dow Chemical Terneuzen.

P.A.M. Marteijn

EO.86.A.44

De Afdeling der Elektrotechniek van deTechnische Hogeschool Eindhoven aanvaardtgeen verantwoordelijkheid voor de inhoudvan stage- en afstudeerverslagen.

Afstudeerwerk verrichto.l.v.:

Ir. W.F.J. Kersten

Ir. R. Bergmann (Dow Chemical, Rotterdam)

Verantwoordelijk docent:

Prof.Ir. M. Antal

mei 1986

T E C H N I S C H E HOG ESC H 0 0 L E I N D H 0 V E N

anreq,uestedunit

- 2 -

PRE F ACE=============

In 1962 the Dow Chemical company expecting expandingeuropean markets, selected Terneuzen for itsproduction facilities.Infrastructure although not optium, was available.The nearby Schelde river, allowed import and exportby sea going vessels and was a source for cooling­water.The two first plants, producing polystyrene andpalyglycols included a power demand of 2 * 4 MW.This power was purchased of the PZEM, the companysupplying power in Zeeuws Vlaanderen.In 1964 the site activities were expanded, to includestyrene and ethylene oxide production facilities.Inhe~ently the purchased ~ower demand increased withanother 2 * 4 MW.The developed site powerg~id included 4 transformersdirectly connected to the PZEM grid to distribute the16 MW into the different ~lant area's.Up to now these t~ansfQrmers are still in function(See appendix 14i sub 1,4,8a and 8b·)A boilerplant was build, at the same time, for thesupply of the utility steam, as processheat sourceIn 1967 the site area increased by 200 I..It allowed a third phase expansion of the facilities,to include Light Hydro Carbon cracking, necessaryfor the production of ethylene, aromatics, butadieenand propylene as raw materials foT'" the "plastics",which now included also polyethylene facilities.With this expansion, the site power demand increasedto about 45 MW.The total power supply to Zeeuws Vlaanderen, includingDow Chemical, was transfered through one cable linkbeneath the Schelde river.An interruption in this single link, would cause ashut down oT the entire production with a lot of damages.Dow as a pioneer had also recognised the economicalbenefits of power steam generation on the basis ofgastu~bine boiler combined cycles.Safety and economic aspects were both important factorsfor the decision to build its own power plant.The power plant layout consisted of 3 gasturbinegenerators and three boilers of 35 bar.The power generator level was selected at 11 kV.Each of the generators had a nominal capacity of 15 MW.Increase of the demand in the polyethylene, withextra 14 MW and other process plant expansions,an extension of the generating facilities with aof 24 MW.

- 3 -

For the steam supply integration with the LHC plants,the boiler was designed to produce 95 bar steam, toenable an easier start up o~ the naphta cracking.In between LHC start up, the take of~ o~ this boileru.las minimum.To maintain the power production, the steam had to bereduced to 35 bar.Through pressure reduction, the unit could be kept onthe line in a standby ~unction, generating the powerneeded.In a ~ollowing phase o~ expansions, 1976, there was anincrease in the need ~or power only.This requirement could be fitted nicely with Dowsenergy saving e~~orts.

A 10 MW, 11kV gasturbine generator was added to 2existing identical boilers in the boilerplant build in1967.With the last major production ~acilities expansionsthe steampower production were extended with a 95 barboiler-steam turbine generator unit.The back pressure turbine generator o~ Stalhaval designhas a capacity o~ 25 MW at 6kV lev~l.

The installation o~ this unit permitted elimination o~

steam pressure reduction at the other 95 bar boilertrans~ering the energyloss into electrical power.

Through the years the power generation and distribution hasbecome very complex.Occuring ~aults can grow to severe iterruptions, causingproduction loss.The reliability o~ the system is a ~unction of the steamproduction, through the combined cycle operation, thepower generation, the electrical distribution system andthe protection system.Therefore the transient stability o~ the electrical powersystem o~ Dow Terneuzen can have an exponential in~luence

on the reliability o~ the whole system.

- 4 -

SUM MAR Y=============

Transient stability of an industrial Power system withgenerating machines depends on the behaviour of the rotorangles of these machines.In case the system is connected to an infinite bus, therotor angles with respect to the angle of this bus, controlthe stability limit.If there's no connection, then the angle displacement ofall synchronous machines with respect to each other areimportant for the power characteristics.For the DOW-Chemical site in Terneuzen, it is important toknow how the relatively large high voltage motors affectt~e stability of the whole system during a severe fault.E~uivalent models are derived to represent the dynamicbehaviour of the load and a suitable revision of theexisting program has been made.A model to include the excitation system in the step bystep calculation is also made, but not implemented in theprogram yet.Different network configurations have been investigated.The influence of the connection with the infin(te bus hasbeen analysed.

- 5 -

CON TEN T S------------------------------

PAGE

PREFACE

SUMMERY

CONTENTS

LIST OF SYMBOLS

2

4

5

8

Chapter 1: I N T ROD U C T ION 10

Chapter 2: THE 0 R Y

2. 1 STABILITY PRINCIPLES

2. 1. 1 Introduction

2.1.2 Power flaw single machine s'Jstem

2.1.3 Steady state stability

2.1.4 Transient stabilit'J

2.1.5 Multimachine s'Jstem

2.2 CALCULATION METHODS OF THE EXISTING PROGRAM

12

12

15

19

20

2.2.1 Assumptions made in the stabilit'J studies 25

2.2.2 Transferimpedances 26

2.2.3 Derivation of the Step by Step method 27

2.3 DERIVATION OF MODELS FOR A SUITABLE REVISION

2.3.1 Introduction

2.3.2 Saliency

2.3.3 Excitation

2.3.4 Load simulation

2.3.5 Composite load

31

32

35

37

38

- 6 -

2.3.6 High voltage induction motors

PAGE

39

Chap ter 3: PRO G RAM REV I S ION

3. 1 LOAD

3. 1. 1 Load composition

3. 1. 2 Spreatsheet program

45

45

3.2 SALIENCY AND EXCITATION

3.2.1 Crout algorithm 47

3.3 FRIENDLY USE OF THE PROGRAM

3.3.1 Plotting subroutine 48

3.3.2 The TSA command~ile 49

3.3.3 The program Transient Stability Analysis 49

3.4 INPUT AND OUTPUT

3.4.1 Input o~ the program 50

3.4.2 Output Or the program 51

Chap ter 4: RES U L T S

4. 1 LOAD COMPOSITION

4.1.1 Introduction

4.1.2 The three phase ~ault

4.1.3 Load close to the ~ault

4. 1. 4 Di~~erent load representations

53

53

54

55

4.2

- 7 -

4.1.5 Synchronous motor load

VARIATION OF CONFIGURATIONS

4.2.1 Connection with inTinte bus

4.2.2 The transient reactance

4.2.3 Position variation of a generator

PAGE

58

60

60

61

REFERENCES

APPENDIX:

1>

2)

3}

4)

5}

6)

7}

8)

9}

10)

11>

12)

13}

14)

15}

16)

Chap tel' 5: CON C L U S ION

Inertia constant

Per unit value's

Algorithm oT Crout

Flow diagram of the program TRASTA

Active power characteristic on IBM PC

Rea~tive power characteristic on IBM PC

Simulation induction motor on IBM PC

Printout of spreadsheet program on IBM PC

List of the program list Tiles on the VAX

System input for TRASTA

Generator and load program input data

Program output

Interaction user and CRT

One line DOW CHEMICAL Terneuzen

Impedance diagram correponding to the one line

Angular displacement plots ( 28 plots)

63

66

74

75

76

77

78

79

80

82

83

84

85

86

87

88

89

90

Xd =Xq =Xd ' =Xq' =Tdo =

- 8

LIS T a F S Y M B 0 L S===========================

E = complex voltageI = complex currentZ = impedanceP = real powerQ = reactive powerR = resistanceX = reactancet = time~ = moment of inertiaH = inertia constantG = machine ratingW = kinetic energyS := total MVAV = voltageM = torques = slip

Direct-axis synchronous reactanceQuadrature-axis synchronous reactanceDirect-axis transient reactanceQuadrature-axis transient reactanceDirect-axis transient open-circuit fieldtime constant. (in seconds>

s = angl e of current with res pect to reference8 = impedance anglea = 90 -8 = complement of the impedance angle<5 = angular displacement or voltage vector from ref. axis

f = frequencyW = 2wf = angular velocityWo = synchronous angular velocitycr = w - Wo = d ev i a t ion 0 fan gu 1a r vel 0 citY

k = acceleration constantn = efficiency

PROGRAM VARIABLES

ED = voltage behind the transient reactanceEB = bus voltageEi = field excitation voltageEQ = voltage behind XqEq' = quadrature component of voltage corresponding the

field flux linkageEx = exciter voltageZii = transfer impedanceAi = angle

PM = active powerGM = reactive power

SUBSCRIPT

m = mechanicale = electricalre = real partim = imaginary partd = direct axis'l = 'luadrature axisr = rateda = starting

- 9 -

- 10 -

1 N T ROD U C T ION=======================

This report contains the study of the transientstability of the electric~l distribution system ofDOW CHEMICAL TERNEUZEN, by program simulation on aVAX 750 computer.The second chapter is devided into three parts.This theoretical chapter starts with the discusion of thestability phenomena by description of a single machineconnected to an infinite bus,After that, the calculation method for a multimachinesystem based on this theory is analysed.Before derivation of the models for the revision of theexisting program, the calculation methods of this existingprogram are discussed.E~uivalent models for dynamic representation of the load,saliency effect and excitation of the synchronous machinesare derive-d. The- high voltage motor load is representedseperately by a steady state model and the rest of theload is simul~ted by a polynominal e~uation.

This steady state model represents the dynamical behaviourof the asynchronous machines, caused by the mechanical slipvariation oT these motors, due to the occuring disturbances.In chapter 3 the existing program is revised and the loadmodels and the saliency effect are implemented.The implemente-d models need extra system data and machinedata and therefore subroutines have had to be written.Together with these subroutines for extra data, provisionshave been made to link these model calculations to theexisting program.The individual steady state model of the high voltage motorhas been analysed by a spreadsheet program on an IBMpersonal computer. (appendix 5,6,7,8 )The program revision has not only been an extension bynew models, but the program has become more friendly forthe user.Angular displacement of the rotors OT the synchronousmachines can be shown graphicaly on the Printronixlineprinter.In the case that the distribution system is connected tothe infinite bus, the angular displacement of themachines relative to the infinite bus is drawn.Without connection a reference angle is needed.Using a command file procedure, control datacan be changed by interactive use of the program on theCRT terminal. The inputfile has been made easier to handle.Two separated directories have been created.In th e d i rec tory 'TRASTA', th e program can be used, wh enconstant MVA representation is re~uired.

Dynamic simulation of the load can be done by the programversion stored in subdirectory 'WORK'.

- 11 -

In each directory, specific system configurations can bestored in the system inputs.A list of drawn plots will be made in each directory.In chapter 4, the results of the program runs with differentload representations are discussed.The asynchronous high voltage motors and the stabilitylimit of the large synchronous motor of the LDPE factoryare analysed,The affect of this machine on the other machines will beshown.The influence of the connection with the infinite bus ofPZEM is investigated, together with different systemconfigurations.The conclusion of the results is discussed in chapter 5 andthe report will close with additional information in theappendices. The ,plots mentioned in chapter 4, can be foundat the end of the report.The report does not contain program listings of the programand the new subroutines, because this is DOW confidentialinformation. If a third person is interested, it is alwayspossible to contact the DOW CHEMICAL company.

I wish to finish this introduction, by thanking the DOWCHEMICAL company for the opportunity given me, to make anindependant study during the months spent in Terneuzen.I also wish to thank Ronald Bergman of EE~CS Rotterdam andWim Kersten of the university of technology in Eindhovenfor their support to obtain my degree at the Eindhovenuniversity.Finally I wan~ to thank all DOW employees, who have takenthe time to give me information, assistance etc.,when I have asked for it.The last year has been very instructive for me andhas given me a lot of practical experiences.

- 12 -

2 THE a R Y===========

2. 1 STABILITY PRINCIPLES

1. 1 Introduction

The stability of a system of interconnected dynamiccomponents is its ability to return to normal or stablecondition, after having been subjected to some form ofdisturbance (1,3,6)The phenomena of stability of a power system concernsdifferent conditions.Synchronous stability may be devided into two regimes,steady state and transient stabilityThe former is concerned with the ability to retainsynchronism when subject to small disturbances.Transient stability deals with large and sudden networkdisturbances such as brought about by faults.The transient stability limit of the system is always belowthe steady state stability limit.Stability performance involves the system power flowcharacteristics, and power angle curves.Steady state stability calculation uses the e~uivalent

synchronous reactance of the generators.The transient reactance is used for severe changes duringtransient stability studi~s.

Transient stability is concentrated among the mechanicalbehaviour of the machines during the first second of th~

dis turbanc e.The behaviour of the rotor angles of the generators iscalculated using the accelerating power in combination withthe inertia of the machine.The criterion for stability is known as the e~ual-area

criterion.

1.2 Power rlow of a single machine system

A single machine system can be represented by two supplyvoltages E1 and E2, connected to each other by an impedanceZ, like considered in figure 2.1The aspects that affect the transfer of active and reactivepower will be examined.For a symmetrical 3-phase system, a single phase model canbe use-d'.

I-I"[ =., ejo

1

l)

-z o

o

figure 2. 1

- 13 -

The complex representation of the vector diagram is shownin figure 2.2. (1,3,6,15)

x

R

figure 2.2

The current I is equal to the voltage between El and E2divided by the impedance Z

I. Z - E2 = 0 ==)-I = 2. 1

Figure 2.2 5how~ that the proJection of the current onEl is equal to

E1-. cos El andZ

E..1.. cos(012 + e )Z

2. 2

The active power output of E1 is the product of E1 and thisproJection (the in phase component)The delivered power output is:

PI [ E1 E2(0 12 + e~= E1 T cos o - r cos

E12 E1E2cos (0 12=r cos 0 Z + 0)

E 2 E1E2PI

1 .sin (0 12 - a)=r S1n a + -z--

. , wh ere 0 = 90 - a

2. 3

- 14 -

When alpha is constant, the power output varies with thesine o-F the d isp lacement o-F ang Ie 012 between the two sources.The power input P2 o-F E2 is e~ual to:

P2 = E2 [:~ cos (e - .12) - ~2 cos :JE1E2 E 2

P2 = -r- sin (012 + a) - ;. sin a

The power output o-F E1 is at a maximum when sin( 012 - a ) =1,i. e. 0 1 2 = 90 0 + a oThe received P2 at E2 is at a maximum when 0 12 = 90 - aIn case there's no reactance, then the power can onlychange, when the lengths o-F E1 and E2 are not e~ual.

I-F IElI/IE21 = 1 , there's no power trans-Fer.The di-Fference Pl - P2 is the power dissipated in theconnecting impedance Z.

2.4

E1

2 + E 2 .E 1E2. ~os (012+ e) f. cos (e - 012)JPI - P2

2 cos e= Z - --r-

cos e ~ 2 2 - 2 E1E2 cos 012J= Z E1 + E2

cos e 2= Z (IZ)

= IlZ cos e

= IZR2.5

The reactive power output o-F E1 is the product o-F thevoltage E1 and the ~uadrature component op the current.The reactive powe-r is:

Only the active power is important to analyse the e~uation

o-F motion, to determine the stability limit.The reactive power flow is important -For the calculationo-F the bus voltages o-F the network system.

2.6

- 15 -

L.3 Steady state stability

A balanced electrical power output with respect to themechanical turbine power needs to recover when there'sa disturbance of this equilibrium.A disturbance means a net torque, which producesaccelaration or decelerationFigure 2.3 shows the power angle curve corresponding tothe example of figure 2.1.Source E2 is considered as an infinite bus.

p

I

~.Q

---------

II

I ,

I C1 I~

190

figure 2.3

The angle ~ shows the equilibrium point between mechanicaland electrlcal power.For the angle below 90+ a. a sudden increase of mechanicalpower is not responded by a sudden increase or deliveredelectrical power. So there is a surplus Or mechanical powerThe rotor will accelerate and the angle increases, so theelectrical power increa;es and can lead to a newequi I ibrium.Ir the angle reaches 90+ a • then an increase Or the anglecauses an decrease Or electrical power.The dirrerence between the mechanical and electrical powerdoesn't decrease any more. This means an unstable situation.The angle 90+u is the THEORETICAL STABILITY LIMIT.If the synchronous machine has a rotation velocity of 3000rpm, corresponding the frequency of 50 Hz., then theelectrical an9ular displacement is equal to the mechanicalrotor ang Ie.

- 16 -

The kinetic .energy of the machine is eq,ual to ! .Jw 2

where .J is the moment of inertia of the combinatioJ ofturbine. generator and gearbox.w

Qis the angular velocity.

During a small disturbance. the rotating eq,uipment accele­rates 01' dec.elerates and the rotor velocity changes.If the machine accelerates. the new angular velocity iseq,ual to

W = wQ + ~dt 2. 7

The kinetic energy is increased. due to higher rotationve~ocity:

- do~ .J. ( WQ +-

dt) 2

= W + liW

2.8

Ii W - WI - W = J ~Q~dt

Ii P = doW = .J wQd20 = Pm - Pedt dt

where Pm is mechanical power and Pe is electrical power.Using some other notations. the eq,uation of motion of themachine rotor becomes

2.9

2.10

2. 11

H = Inertia constant; stored energy at synchronousspeed per volt-ampere of the rating of themachine MVA.(See appendix 1>

o = Rotor angle of the machineG - Rating (MVA)f = freq,uency

In eq,uation 2.. I I a negative change in pOUler output resultsin an increase in c .

- 17 -

The stored energ~ GH of the turbine-generator combinationis equal to ~ J Wo •

The differential equation of motion shows that the smallchange in mechanical power causes an oscillation to a newequilibrium.This is shown in figure 2.4

deSCJ= dt

figure 2.4

If all damping is neglected, then the oscillationsho~n in figure 2. 4 ~ill not end in the ne~ equilibriumpoint b2 .The oscillation will sustain between the angles andDepending on the amount of damping the motion ofoscillation will end in the equilibrium pointThe effect of extra damper ~indings is sho~n in figur~ 2.6.

The STEADY STATE STABILITY LIMIT is based on the equal­are-a criterion.In figure 2.5 this limit is sho~n. together with an examplein the case the stability limit is passed.The power system forms a group of interconnectedelectromechanical elements the motion of which may berepresented by the appropriate differential equationsWith large disturbances in the system the equations are NONLINEAIR. but with small changes the equations may belinearized with little loss of accuracy.If any of the roots of the characteristic equation of thesystem have positive real terms then the quantity(rotor angle) increases continuously with time and theoriginal steady condition is not re-astablished.The presence of oscillations are indicated by the imaginaryparts.

- 18 -

pt

stable

0-' ::

unstable

figure 2.5

This complex behaviour is a combination of three energysource~. The first one is the turbine axis,responsible forthe received mechanical power.The electrical power of the generator is the second energysource.The third source is the Inertia of the rotating equipment.Excitation systems. damper windings and the magneticcurrent of transformers improve this interaction.(figure 2.6 and figure 2. j )

Transformer

-(,

Damper

figure 2.6 figure 2.7

- 19 -

1.4 Transient Stability.

Transient stability is concerned with the effect of largedisturbances. These are usually due to faults the mostsevere of which is the three-phase short circuit.When a fault occurs at the terminals of a synchronousgenerator the power output of the machine is greatlyreduced a& it is supplying a mainly inductive circuit.However, the input power to the generator from the turbinedoesn't change during this sho~t period, this excess energyis stored in the rotating parts of the machine.The gain of speed will increase the rotor angle continiouslyand if the fault persists long enough, synchronism will belast.The criterion for stability is that the area between theP - 0 curve and line representing the power must be zero.This e~ual-area criterion is based on the assumption thatthe rotor must be able to transfer to the system all theenergy gained from the turbine during the accelarationperiod.Figure 2.8 shows the critical clearing angle for Po duringa three phase fault on one of two parallel lines whichconnect a generator to an infinite busbar.The single generator is represented by an e~uivalent supplyvoltage and the network is assumed to have only reactivecomponents.For this case all resistances are neglected.The mechanical power Po is assumed constant, and the faultwill be removed by disconnecting the faulted line.The fault is cleared in a time coresponding to 01 and theshaded area between 00 and °1 indicates the stored energyThe moment 01 is reached, the rotor speed is the maximumduring the whole swing.When the fault is removed. the rotor angle will stillincrease, because the angle speed is higher than the&ynchronous speed. when 6 = 01 .In this critical case the speed will be e~ual to thesynchronous &peed. when 0=02. If not, the system isunstab Ie.

IIII

0 80 81 &z 180·

S-

figure 2.8

- 20 -

Beyond 8 energy ~ill again be absorbed by therotor fro~ the turbine and the rotor speed ~ill increasec ont i nuous 1y.The critical clearing time for a simple case canbe determined as follows.Applying the equal-area criterion to figure 2.8

Fl f02(Po P . sin 8 )d 8+ (Po P .sin 8)d8=O 2.121 2

8 2 8 1

OJj81 82

[ Po 8 + P . cos + [P08+ P2 .cos8J = 01

8 0 8 1

as 8 = 1800 - arcsin(Po/P2 ) 2.132

cos 8 =1

Po ( 80

- 8t + Pl

• cos 80

- P2 . cos 82

PI - P2

2.14

Hence the critical clearing angle 8 1 is determined.In most practical situations the net~ork will be morecomplex than shown in figure 2.8 .Numerical methods are used to determine the criticalclearing time. The angle swing curve is the result of thistime angle calculation.The step by step method (1,4) simulates the rotorbehaviour during and after the fault.

1.5 Multimachine system.

If a generator is connected to an infinite bus, onlythe parameters of this single machine change, so a formalsolution is possible.In most practical situations, more generators are involvedand it isn't possible to find a formal solution, becauseof the complexity of the system e~uations.

In a multimachine system, the output and hence the accele­rating power o·f each machine depends upon the angularpositions of all the machines of the system. Thus for an-machine system there are 'n' simultaneous differentiale~uatiDns like equation 2.11

- 21 -

2.15

a multimachine system changes to:n

LE.. E.

__L -L_. sin ( <5. - <5 . - a. )lZCi,J)l 1 J 1J

j =1

proportional to the inertia

Pm.1

=

,where H~ is the variableconstant (appendix 1).El ... En are the driving point voltages of all machine nodes.The transfer impedances are determined in the polairnotation, where lZ(i, J): is the length of the vector andthe corresponding impedance angle.At each node 'i' a generator is connected, with acorresponding voltage Ei and angleTherefore only one generator is allowed at each generatorbus, for calculation purposes.The driving point voltages are calculated from the initialbus voltages and are constant during the whole program, ifsaliency and excitation are negligible.The derivation of this assumption will be explained inchap ter 2.More detailed information of this driving point voltage,when saliency and excitation are taken into account isdiscussed in chapter 3.If the electrical power generated by a machine is determined,then the change in rotor angle is calculated, in combina­tion with the inertia constant of that machine.This change is used for the correction of the rotor anglefor the next time interval.

A formal solution of such a set of e~uations is not feasible,because the calculation of the active power transfer of eachmachine is a complex function of all the other machines.The most feasible and widely used way of solving thesecomplex swing e~uations is the step by step solution.If a large disturbance occurs, then the machines react andcause a change of the power distribution in the system.The reaction is based on the e~uation of motion for onegenerator connected to an infinite bus.In the multimachine system, the po~er output of themachines is not only a function of the network situation,but depends also on the angular displacements of all othersynchronous machines.All rotor angles and therefore the rotor behaviour of allmachines are inter-related.Using the step by step method, it's possible to calculatethe new situation of a machine, corresponding to the anglesituation of all the other machines for that moment.The machine calculation is based on superposition.The total ~ower transfer of a machine is the summation ofall power transfers between this machine and all otherindividual machinp.s.The equation of motion ~Qr

- 22 -

2.16Z L =

Due to the changes Or all the angles. the voltages Or the busseshave new values and affect the load situation.The load is represented as being an impedance to groundparallel to the generator.Ir the load is equal to P + JG. then the correspondingadmittance is equal to

where V is the bus voltage.The transrer impedances are determined rrom the networkreactances and these load impedances.The nodes of the calculations are derived from the placeswhere a generator is connected. The transfer impedances arecalculated by a Gauss Seidel iteration.The transfer impedance Z(i. J) gives information of the linkbetween an excitation on node Iii and the response on thenode IJ/. caused by this excitation.The equation for the Gauss Seidel iteration is derived fromKirchoff/s law of currents.This law states that the sum of all the currents flowinginto a Junction (node) is zero.In the case of the multimachine system, all generato~s areshortcircuited and a voltage supply of 1 per unit isimplemented on the node Iii if the transfer impedancesZ(i, 1> .... Z(i, n) need to be determined.An example is shown in figure 2.9, where three machinesare interconnected.Only the voltage source of the machine is shortcircuited.The impedance between the neutral bus and the bus node isequal to the transient reactance of the generator, withsome additional reactance of the connection of the machineto the bus.

- 23 -

If a load is connected to the busbar, then this impedanceis e~ual to the parallel value of the transient reactanceand the corresponding load impedance.If the transfer impedance Z(i, x) needs to be determined,then the current Ix, caused by Ei has to be calculated.The impedance Z(i, x) is e~ual to Ei/lx, where Ei = 1 p. u.So Z(i, x) = 1/Ix = Zx/ExThe impedance Zx is easy to determine, if machine data andload data are known.The voltage Ex can be derived from the current equationsfor all nodes by a Gauss Seidel iteration.For the voltage E1 in figure 2.9, the e~uation is e~ual to

L I = 0

111( Ek - E1 ). - + ( E; - E1 )· + ( Ex - E1 ). - = 0

Z2 Zl +Z; Z3

2.17

1 1 1= Ek · - + E;. ( ) + Ex'-

Z2 Zl+2; Z3

iterE 1 =

-1) .

1E.. (---) +

, Z +Z1 ;

1E .-J

x Z3

2. 18

If El is calculated, then the new value of El,El(new), is .qual to

El(new) = E1(iter) + AFR. CE1(iter) - El(old)] 2.19

Where El(iter) is the voltage El calculated from equation2.18 E1<old) is the the value of El in the previousiteration step. AFR is the iteration acceleration factor.For each iteration step all voltages are calculated.The iteration procedure will proceed, as long as one ormore of the the voltage~ are not within the accuracy limits.If the voltage distribution, due to the 1 per unit supply isknown, the transfer impedances Z(i, 1> ... Z(i, n) can be derived.This iteration procedure is repeated for all machine nodes.After this a matrix of all transfer impedances is known.The matrix is in polair notation of the correspondingadmi ttances.This matrix in combination with equation 2.16 can be usedto calculate the new powerflow condition ~f the network,and the new value of power transfer of all machines canbe determined, using the machine angles of the formercalculation step.When the new power transfer condition of all synchronousmachines are solved, the new rotor angles can be calculated,related on the power transfer and the inertia constant ofthe machine. The mechanical power of this calculation,equation 2. 16, is constant.

- 24 -

The new angles are stored for the program output and thetime is increased by the time increment.Before a new step is started, the bus voltage of allgenerator busses are calculated from the voltage behind thetransient reactance of th~ generators in combination withthe power transfer of the machine and the impedance of thelink between this voltage and the busbar.If no important change of voltage has occured, then thesame matrix Or transfer impedances can be used, and onlythe new power transfer conditions and new rotor angles needto be calculated.If there's a variation of voltage on one or more busses.the new load situation is calculated and the program restart­at the beginning for a new step and determines the newmatrix of transfer imp~dances.

In figure 2.10 a flow diagram is drawn and shows the orderof calculation for a multimachine system in a step by stepmethod.

FLOW DIAGRAM

no

Calculate vo~tage on

all nodes due to 1 pu

supply on node

STOP+----<T>----..---~

Calculate transfer

impedances Zl" •• Zln

reactive power

figure 2. 10

- 25 -

2.2 CALCULATING METHODS OF THE EXISTING PROGRAM

2.1 Assumptions made in stability studies

The swing e~uation governing the motion of each machine ofa system (e~. 2.11) is the base e~uation for the graph ofthe solution. known as a swing curve.In a multimachine system with more than two machines. thisswing curve is derived by calculating the power of eachmachine and it's corresponding rotor angle displacement foreach time interval. (eq,. 2.16)For a transient stability studie. machines are included bythe eq,uivalent circuits.Depending on the accuracy of the machine performance. theeq,uivalent circuit of the machines are more sophisticatedand more differential e~uations need to be solved.Synch~onous machines are classified into two principaltypes. round-rotor machines and salient-pole machines.The word saliency is used as a short expression for thefact that th& rotor of a synchronous machine has differentelectric and magnetic properties on two axes 90 electricaldegrees apart. the direct axis and the q,uadrature axis.This difference between the two axes is present not onlyin salient pole machines but also. to lesser extent. inround-rotor machinesThe effect of saliency is discussed in chapter 3In a transient stCJbility study. the generator. or a largesynchronous machine. is commonly represented by their direct­axis transient reactance Xd' in series with a constant­voltage powersource. (figure 2.11>'The effect of the excitation system response is neglected.the flux linkages are held constant corresponding to thevoltage behind the transient reactance (Xd').The effect of the excitation system on this flux linkageis also determined in chapter 3The driving point voltage ED is calculated from the initialbus voltage EB.The reactance between the voltage source ED and the bus­voltage is 'I'.I is not only the impedance of the connection between thebus and the generator. but the impedance 'I' is inclusivethe tra·nsient reactance of the generator.

I

Xd ' 1ED EB

1figure 2.11

- 26 -

The program calculates the real and imaginary parts ofvoltage, current, power and impedances in separate equations.The values of real and imaginary parts are calculatedseperately.If the busvoltage, impedance, active and reactive powertransfer are known, the ED can be calculated:

z = R + jXI = Ire + jIimED= Ere + jEimS = PM + jGM

Ire = PM/EB

I im = -GM/EB

ED = EB + I.Z

Ere = EB + Ire.R - I im. X

jEim = j ( Ire. X + Iim.R)

IEOI (Ere2 Eim4 ) i= +

total impedancecurrentvoltage sourcetotal MVA

2.20

2.21

2.22

2.23

2.24

2.25

If 6~ is the angle of the corresponding generatorbus, then

OED = 0EB + arctan (Eim/Ere)2.26

This voltage source ED is constant in magnitude, but onlychange in angular position.This source is equal to the driving point voltage in thepower tranfer equation 2. 16.The driving point voltages, together with the angle displa­cements and tne transrer admittance matrix are used todetermine the different power flows between all synchronousmach ines.

Damping has been simulated by calculating the rotorloss&sfor O. 16 sec ond s, of th e generator c onnec ted to th e b us onwhich the three phase fault is submitted.The electrical power transfer of this machine will be small,and hence accelerating power is strongly increased.

Saliency is neglected, the same as governor performancesThe mechanical shaft torques are held constant during thewhole program.

2.3 Trans~erimpedances

The power output of the synchronous machines depends onthe configuration of the admittance matrix.The transfer impedance matrix is analysed using the Gauss­Seidel iteration method. (11,15,17)

- 27 -

The body o~ the p~og~am ~uns with the conve~ted admittanceso~ the t~ans~~~ impedances.In the existinQ. p~og~am, loads a~e ~ep~esented by shuntimp edanc es, co~~esp ond i ng to a cons tant MVA load.In case the busvoltage is deminished to 0.5 p. u., this loadimpedance ~emains unchanged and co~~espondes to the loadat t = a seconds.The initial load impedances a~e used in the t~ans~e~

impedance mat~ix calculation.The pe~ unit admittances ~ep~esenting the active and~eactive loads, a~e equal to the pe~ unit input values o~

the active and ~eactive loads.These inputs a~e in pe~ unit MW and pe~ unit MVa~ (100 MVAbase). When at t=O seconds, the bus voltages a~e 1 p. u. ,this assumption is pe~mitted. (appendix 2 )Due to voltage changes, at each time inte~val thisadmittance is co~~ected, co~~esponding a constant MVA load.

!. 3 De~ivation of the step by step method.

The swing eq,uation (eq,. 2. 16) gove~ns the motion of eachmachine of a system.The solution of this eq,uation gives the angle as a functionof time.' The o~de~ of calculation is shown in the flowdiag~am in figu~e 2.10.A g~aph of the solution is known as a swing cu~ve.

Inspection of this cu~ves of all the machines of a s~stem

will show whethe~ the machines will ~emain in syn~hronism

afte~ a distu~bance.

Successive calculation Or the ~oto~ angle and the poweroutput of the gene~ato~ for each time inte~val is the baseof the step by step method.These step by step solutions attain good accu~acy and thecomputations a~e simple.The accu~acy of the calcultations also depends on the methodthat is used. (1,4)Most of the er~o~s caused by assuming the accele~ation to beconstant du~ing a time inte~val can be eliminated by usingthe value of accele~ation at the middle instead of at thebeginning of the interval. If the accele~ation at thebeginning is used,diffe~ent solutions will be found withva~ious values of At. (figu~e 2.12)

Figu~e 2.13 shows the successive time inte~vals o~ a step­by-step calculation,n-2, n-3/2,n-l,n-1/2,n, etc.

- 28 -

{J .~~. ·'til

/ V '. \'~i/~ \'

(f \), \\.I \ \'.

I~V 1\ \,

~.'

\ \ /;f .. \~\

7'~,)\. . , l 6:

.\~~.' V 3

../ " /!-

3 • • l2 ~. 0.05 sec.4 <>----._---6 .11- 0.10 lee.50-----o4'-0.150ec:.6--- .11- 020 oec.7_~a_ i1t-O.25sec.

I 1 I I I

150

middle

oo 0.1 0.2 0.3 0.4 05 0.6 0.7 0.8 0.9 1.0 l.l

Time I lsecond,'

30

135

15

,'1/. "/~,/ 31'\ \(/~t\ i'\

f!# I"~ \, \'if 1'\Ii

I.V \\r; I \.

II ~, ;

~ ~ ,

f-- ,--,-- r\ ~I-- l'---,w-OOI61 ...

\J~~·OO~MII:.

'Ao----'IIllio.w -0,10 MIl:. II--f- l~--a~.O.l~-.

~\\

I

\ \~I

»

IS

'IIII.

,..

-I

-do u u u ~ 0) " ~ ~ ~ IA

nnw , (seconds.

beginning figure 2. 12

8

Time

IIIII 8n

JIII

figure 2. 13

At each time interval, the variation of the rotor angleis calculated and a correction of the angle is made forthe next interval.

2.27

The variation of the rotor angle during the time increment.is:

- 29 -

From thebecomes:

eq,uation 2.11 , the eq,uation o-F motionH d 20-- - P - P,...jdt 2 - m e

2. 28

2.29

2.30

~here Pm = mechanical po~er and Pe = electrical power output.All variables are in per unit no~ (base 100 MVA).'H' is still the inertia constant, but now in per unit ona base o-F 100 MVA.From eq,uation 2.30 the Tollowing expression Tor theinterval n-1 is obtained:

2.31

(d0) ~trrfat + -H- (Pm(II-1l - P e(n-l»)

II-H 2.32

(d 0) 1r'~1511_H

\cIE 1I-H = 180~t

~On_Ll _ 180 ~t(dO) + 180f(~t)2 (P Pn - - dt H m(lI-1) - e(II-I) )

r II-H

This eq,uation is valid Tor any time interval, except the-First. The eq,uation states that the incremental increasein angular displacement is eq,ual to the previous increm~n­

tal increase ~8n-~~ plus a constant times the acceleratingpo~er oT the previJus interval. This constant, ~hich ismultiplied by the accelerating po~er, ~ill be called theacceleration constant,

2.33

2.34

2.35

k = 2.36

- 30 -

For the rirst interval, the rollowing relation can bewritten:

(d0) 1 (do)dt n-~ = "2 6. dt n-I

. (dO) (dO)6.(ft =2-II-I dt II-j.i

Substituting e~. 2.38 in 2.31 I we have

(dO) "j6. tdt ,= 2H (P mCn- O - P~Cn-I) )

"-~1

Susbstituing ~~. 2.33 in 2.39 I

2.37

2.38

2.39

180j(Llt)22H (P rnCn-1)

2.40

E~uation 2.35 and 2.40 can be rewritten as Tollows:

where

P , for any intervala(n-l)

except the first

for the first interval

2.41

2.42

and

k _ 180j(6t)2

H

PQ (Il-I) = (Pm(II-1) - P c(Il-I»)

These e~uations make it possible to calculate the increasein angular displacement, when the preceding acceleratingpDwer is known.

All variable~ are given in per unit on 100 MVA base.Appendix 4 shows a rlow diagram or the complete existingprogram.

2.43

- 31 -

2.3 DERIVATION OF MODELS FOR A SUITABLE REVISION

3. 1 Introd uc t i on

With the transient stability program EE7004 of Dow Chemical,it is possible to analyse different system configurations.Due to the fact, that the research is done by simulation,it's important that the program in combination with itsinput and output is easy to handle.By the insertion of various related values, on can findthe critical clearing time, corresponding a specificsystem configuration.Transient stability is analysed by the swing curves, theangular displacement of the synchronous machines versustime.The old program only had a numeric output, that gave theinformation needed, but not a graphical view.By addition of a subroutine PLOTTSA, it's possible to makea graphical output of the generators selected.Then it's easy to determine, whether the system is stable orunstable.There's a graphical difference between the situation of anindustrial plant connected with an infinit grid andwithout connection.

The program Transient stability analysis represents allgenerators by constant voltage source behind the directtransient reactance. Saliency and excitation has beenneglected. It is interesting to know the influence thesaliency effect and the excitation system of the difrerentgenerators, have on the transient stability or the totalsy stem.Therefore a model is developed to simulate these aspectswhich has to be suitable ror the calculation method ofthe existing program. (2)

On an industrial plant with generating units of 10 to 20 MW. ,the individual motor loads are relatively large.The dynamic behaviour of these induction motors mightaffect the motion of the synchronous machines during faultsituations.In the program transient stability the load is representedas con5tant MVA load.The induction motor behaviour can only be simulated if theprogram is revised and recalculates the load situation for.each interval.Composite load is simulated as runction of the busvoltage.The high voltage induction motors are simulated with asteady-state model.

- 32 -

2 Saliency

In the case o~ turbo-altenators the assumption, that thedirect axis reactance is equal to the quadrature-axis valueis reasonably accurate.For the salien~ pole machine the reactance Xq is not equalto Xd and Xq' is equal to Xq, but Xq' is di~~erent ~rom Xd'.The vector diagram o~ the salient-pole synchronous machineis shown in ~igure 2.14.Since a salient-pole machine has no quadrature-axis ~ield

circuit, the excitation voltage always lies on thequadrature axis in the transient state as well as in thestead y state.The period o~ mechanical oscillation o~ a machine during adisturbance is o~ the order o~ 1 second, and the behaviouro~ the machine during the ~irst quarter-period o~ten su~~ices

to show whether the system is stable or unstable.The subtransient time constant o~ a machine is only about0.03 to 0.04 seconds, which is short compared with theperiod o~ mechanical oscillation. There~ore, in stabilitystudies the subtransient phenomena are disregarded.The only electrical transient which needs be considered ina stability study is the 'transient component'.Its time constant is about 0.5 to 10 seconds, usually beinglonger than the period o~ mechanical oscillation.There~ore the transient component cannot be ignored althoughits decrement is ~requently ignored.The transient component can be taken into account in anumber o~ ways wich di~~er as to the accuracy o~ theassumptions and the amount o~ calculation required.One o~ the assumptions re9arding a salient pole machine is:

, , ,,

Id(Xd-Xd') I~ )

,I

II ld.!<.

Id\

\

\ nf.

~igure 2. 14

- 33 -

Ei ~ per unit field excitation voltage corresponding tothe field excitation of the generator

EG = Voltage behind the reactance X~

E~'~ Voltage corresponding the field flux linkages

The flux linkage of the field winding may be assumedconstant (constant E~') or if more accuracy is desired:Field decrement and voltage-regulator action may be calcu­lated (varying E~')

The saliency of the salient pole machines increases thecomplexity of the calculations. but can be important. ifa greater ac~uracy is desired.

A salient pole machine can be represented by a constantreactance equal to the quadrature-axis synchronousreactance Xq. in series with a power source whose voltageis EG.The vector sum EG always lies on the quadrature axis andtherefore represents correctly the angular position of thefi e 1d struc ture.

The magnitude of EG is not constant. even if Eq' isconstant. The relation between EG and E~' is:

EG = E~' + j ( X~- Xd ' >. I d

If Eq' is constant. EG changes with any change in Id.caused by ~winging of the machines or by circuit changes.In a general network the e~uations for the current flowfrom any point in a linea~ network in terms of thedriving point and transfer impedances of the network andthe applied voltages may be written as follows:

2.44

Z(l.l)

EG2

Z(l, 2)

EG3

Z(1.3) Z(l, n)2.45

Zij = Transrerimpedance withmachine reactance Xq

Equation 2.45 is an expression involving vector~uantities and may be written in the ~ollowing form.

EG L_--","n,,- (eli - 8 )Z(l,n) J1 In

=EG L1 (0-8) -

Z(l.l) 1 11

EG2 L(O-8)

Z(1,2) 2 12 .2.46

- 34 -

EGI,EG~,EG1" .EG" are scalar ~uantities,

o '0 '0 .... 0, are the angular positions of the vector123 n - - -voltages EG"

EG 1 .... EG~.

81{81t813" .8 m are the impedance angles of Z(I,I),Z(1, 2>. .... Z(1, n>.

1 ~ l3, is th e ang le between th e current vec tor and th ere~erence angle, then the current on the direct axisis:

Togeth.r with equation 2.46, we have for generator 1

2.47

EG_ ......1_. S in8 -Z(I,1) 11

EG___2_. sin (0 + 8 )+Z(1,2) 12 12

EGo----. 5 i n ( 0 + 8 )Z(i,n) 10 10

and substitution of e~. 2.48 in eq. 2.44 results in

2.48

EIt'1

j(Xd'-X~ ) EG:z.= [1 + 1 . sin 8 J. EG - J ( Xd' - X0_). [ • sin (0 + e )

Z(I,I) 11 1 1 '1 Z(I,2) 12 12

where

EG..... + n. sin(010+ 810)

Z(1,n)

2.49

The power output o¥ generator 1 is:

Pe = EG . lit 2.50

EG21= -----. cose 11

Z(I,1)

EG EG- __L __z... cos (r + 8 )

u12 -12Z(1,2)

EG lEG",------. cos (0 + 8 )Z ( 1, n ) In 1"

- 35 -

=EQ 2

1---. sin (l11 +Z(l.1>

E9, EQ2

.---.;--....:.. Sln( °12 - (l12)Z(1.2)

.... +EQ 1EQ n .-~-. sln( 0ln- (lin)Z(l.n)

o(l = 90 - 0

2.51

Equation 2.49 shows the relation between Eq' and EQ.Similar equations exist ~or all 'n' generators.These 'n' equations with 'n' unknown variables can besolved. by determining the inverse matrix.The calculation method ror the step by step procedureonly change in the ~act. that not only the voltage behindthe direct transient reGctance. but the voltage correspon­ding the ~ield ~lux linkage (Eq') is held constant.The voltage EQ is the drivingpoint voltage now.In case Xq = Xd' • the calculation is the same. becausethen EQ = Eq'When excitation is neglected. Eq' is constant and at eachinterval, the voltage EQ o~ each machine needs to be cal­culated, by solving the equations 2. 49 ~or all machines.I~ EQ is solved, then the power output o~ all generatorscan be determined according equation 2.51This part is added to the main program by the subroutineSALIEN.

i. 3 Excitation

In the existing program. the voltage corresponding to the~ield ~lux linkages (Eq'), is assumed to be constant.The a~~ect o~ the excitation system response on thetransient stability can only be determined. if this responseis included in the step by step calculations.This change in flux linkages at each time interval isdetermined by the difference of the exciter voltage Exand the field excitation, corresponding to the fieldcurrent, Ei • at the middle of the time interval.This voltage Ei can be derived from the driving pointvoltage EQ and the ~ield ~lux linkage.voltage Eq'.Refering to figure 2.14 the relation between Ei andEQ is

Ei = EQ + I d. J ( Xd - Xq ) 2.52

The two relations 2.44 and 2.52 yield

(Xd

(Xq 2.53

Xq)------. Eq,'

Xd ' )

(Xd

(Xq

Xd')-----.EQ

Xd')Ei =

- 36 -

When EG and Eq,' are known, it is possible to obtain Ei,for the step by step calculation of the multimachine system,directly from eq,uation 2.53.If Ei is determined, the change of Ell' at that timeinterval can be calculated from the difference of Ex andEi , together with Tdo'. (2)

=Ex n~Y2 Ei n+o

Tdo'2.54

where Ex is the per unit exciter voltage, and Tdo' is thedirect axis open circuit field time constant in seconds.When the change of field excitation is known, the correctionof Eq,' from the interval 'n' to 'n+1' can be made.

Ell'n+1

= Eq,'n

2.55

If the exciter voltage is eq,ual to the field excitation,then the Eq,' is constant and again constant field fluxlinkage is simulated.Determination of the effect of excitation can be madeby assuming a constant exciter voltage or with exciterresponse.

The definition of nominal per unit exciter response isshown in figure 2. 15.The exciter response is the rate of increase or decreaseof th~ main exciter voltage when resistance is suddenlyremoved, or inserted in the main exciter field circuit.The nominal exciter response is deTined as the numericalvalue obtained when the slope of the straight line in voltsper second devided by the nominal collector ring voltage,the voltage across the collector rings to generate ratedkVA in the main machine.This part of the program revision isn't implemented yet,but with some additional steps, it is easy to complete,together with the saliency effect

Actual Build-UpCurve

figure 2. 15

Time--_

- 37 -

Instead o~ constant excitation response, the voltageEx can be derived by an excitation system model.With these additional e~uations, the program can berevised, without important variation of the body ofthe existing mainprogram.

3.4 Load Simulation.

The load representation of the program is a constantMVA load, simulated by a voltage dependent impedance.Due to large di~turbances, the load doesn't behave exactlyas a constant MVA load and therefore can influence thestab i 1 i ty.

One of the most important limitations, in making arevision of an existing program, is that the basic prin­ciples of the calculation methods must be maintained.If not, it would be better to rewrite the program completely.Another purpose, when you are making a program, has to bethe practical use. A program can be very soph isticated,but unef~icient for the user.If detailed data are neccessary for the program, it could bevery difficult or even impossible to obtain this data.Then sophisicated revisions are useless and the program willnever be used.The last remark is very important, when you want to simulatethe dynamic behaviour of the load, it is well known fromvarious research papers, that it is very dif~icult toobtain the needed load data.An effici~nt choice of modeling is very important.

The dynamic load presentation is devided in two compositionsof load. (figure 2.16)The first group is the composi te load. (15, ... ,35)The composite load is a voltage depended load simulation.A separated part of this group is the collection of lowvoltage inductionmotors.

inductionmotors

extern

system

Staticload

figure 2. 16

- 38 -

The second group, is the group of High volage induction­motors, using a voltage and mechanical slip dependantstead y-state mod e 1. (30, .. , 55)This model is useful for transient stability studies,because the electrical transient behaviour of the induction­motors can be neglected.The electrical time constants of these motors are muchsmaller than the important time constants of the consideredgen-erators.The loadbehaviour is simulated each time interval by thesubroutine DYNLOAD in combination with INDUMOT.

I. 5 Comp osi te Load

The composite load input is represented as an polynominale~uation of active and reactive power depending on thekind of load. (18,22,23)Because an industrial plant has a relativelq large groupof induction motors, the low voltage motors are representedbq a polqnominal relation, where the active and reactivepower could fluctuate, due to sudden voltage changes.The polynominal e~uations represent the active andreactive power consumed by low voltage motors in case thevoltage is ~igher than 0.6 p. u.If the voltage drops below the 0.6 p. u level, the-'low voltage motors will trip within 30 msec. (vacuum­cantac.tors)E~uation 2.56 shows the parameters of the polynominalrelation. (18)

P'ind = roo 1*V + 0.7 + (0. 2/V) J*Po ind

Gind 2.56

( 0.7 < V < 1.0 )

The polynominal voltage dependent e~uation of therest of the load, is a weighted mean of the individualtypes of loads involved.This this polynominal combination assumes a distribution of:

50 ;. heating25 ;. airconditioning25 ;. lighting

The power-voltage relationship corresponding to thisassumption is written in e~uation 2.57

Pcomp

Gcomp

= r V2- 0.93~V + 1.3 - 0.37/VJ*P~comp

:l= r 12.5*V - 21. 5*V + 10J*G~omp

2.57

(0.7< V<1. 0)

- 39 -

Poind and Pocomp are the low voltage motor loadand the rest of the composite load at t = 0 seconds.If the voltage V is lower then 0.7, then the active andreactive load are the same as if the voltage is equal to0.7.For the low inductionmotor load the last assumption is notimportant, because when the voltage is below the value Or0.65 p. u.• then the low voltage induction motors are discon­nected. For the rest of the load. this assumption isn'timportant. because its only a few pprcent of the total load.The load is irrelevant in the case of the bus voltagebeing zero.Wh en th e vo 1tag e is zero, th e imp edanc e, wi th resp ec t tothe neutral bus dominates the fault impedance, becaus~ withthe occurence of a three phase fault, this impedance isequal to zero.When the fault is removed, then the bus voltages recover andthe load might affect the stability phenomena.

3.6 High voltage induction motors.

The H.V. motors are simulated with the steady state modelof an inductian motor.Because of the restriction of the load data collection, it'simportant to use typical induction motor data.In figure 2.17 the steady state model of the induction motoris shown.This passive impedance network represents the dynamic behaviourofthe ma chi n e, duri n 9 v 0 I tag evaria t ion.Not everyone is familiar with the numerical values of themodel impedances, but with 'more practical data of the machine.The input for the transient stability program doesn't needthe numerical valu~sl but some practical inTormation of theinduction machin~

The program calculates on the basis of these load date themodel impedances. (37)

EB x~

figure 2.17

- 40 -

Th& neccesary information is:

Ia/IrMa/MkMk/Mrcos</> rcos</>aSrn

the relative starting currentthe relative initial starting tor~ue

the relative breakdown tor~ue

nominal loadfactorJ starting loadfactor

nominal slipefficiency

This information is sufficient to calculate the numericalvalues of the steady state model impedance~ in per unitvalues.

TJr' COS q:Jr

1 - 5r

} [p.u.]2.58

2.60

2.59[p.u.]XI'

sm~, - cos~, l~: -V(:J 1IRi (5= 5,)~ X~(COSq:Jr- R')'5r

{(COSQJ, - RJ? + (sinlp, - Xl _ Xy} [p.U.]

2.61

The impedance Xl is half the input impedance when theslip i-s e~ual to one, run-up. (18)This results .in Xl being e~ual to ~ <Ir/Ia) p. u.If- an additional impedance is used, then then resistanceRl and reactance Xl can be adjusted.Th is is sh own by e~. 2. 62 and 2. 63, incase th e reac tanc eis a transformer.

2.62

X y X P r;0 = 1\10 + T 'N- lp·u.]

rT2.63

Nrt is the nominal load of transformer and Pr is the nominalmotorload.

- 41 -

The steady-state e~uivalent circuit contains two variables.the busvoltage and the induction motor slip.This repr~sentation permits the calculation of the electricalpower at each point in a step by step swing-curve calculation.taking into account mechanical transients. but assuming thatall electrical time constants are negligible.During a transient disturbance. the resistance Rv offigure 2. lB is varied in order to maintain the power e~ui­

librium. The reactance Xv remains constant.

EB

Rv

xv

figure 2. 18

The equivalent impedance of figure 2. 18. can be derivedtogether with e~uation 2.58 up to and including eq. 2.63The equivalent circuit gives the equation:

I

jX~ jX20 + R2/SZv = R1 + jX10 + 2.64

jX 1 + R'/S + jXlJ20 2

'X I + R" + jXlJ - jXlJJ 20 2= R1 + jX 10 + jXlJ 2.65

jX20 R" jXlJ2

R" - R'/S 2.662 - 2

X2lJ(R2 - j X20 + XlJ )

= R1 + j(X1o + XlJ) + 2.67

(R22 + (X 20 + XlJ)2 )

- 42 -

RII X2 X2 (X I + XlJ)2 ~ ~ 2cr= R1 + + j X1cr + XlJ - 2.68

RII 2 + (X 2cr + X~)2 RII 2 + (XI + X~) 22 2 2cr

= R + jXv v

R'/S X22 ~

Rv = R1 +

R~/S + (X 2a + X\l )2

X2 (XI + X\l)\l 2crX = Xlcr + X\l ­

V

2.69

2.70

2.71

When th~ equivalent resistance and reactance are kno~n,

it is easy to determine the consumed electrical active andreactive po~er,

v V (R - jXv )vI =-- =

Z (Rv + jXv) (R - ·X )v J v

S = V. IJ:

2.72

2.73

V2 (R + jXv v 2.74=R2 + X2v v

V2 RvP =

R2 + X2 2.75ev v

V2 XvQ =e

R2 + X2 2.76

v v

- 43 -

When all electrical time constants of an induction motorare neglecte'd, the steady state circuit can be used torepresent the motor.When the system is operating at a base frequency, theequivalent resistance only depends on the mechanical slipof the motor.Due to frequency changes, the resistance Rv can be efrectedby the electrical slip of the busvoltage.The electrical slip is determined, by using the frequenciesof the last succesive time intervals.

w n+1 - W n

2. ??=-------1/2((0 1 + w )n+ n

The actual slip for the resistance c~lculation is equal tothe difference of the mechanical slip and the electricalslip:

= Sm Se

2.78

The slip change of the rotor is caused by the difference ofeffective electrical power and the mechanical power of thedriven part of the motor.The first is equal to :

2.79

(n is efficiency)The latter depends on the kind of the driven machine.The characteristics of a driven blower is quit differentthan the load behaviour of a compressor:

p = P .(1_s)expm m 2.80

where 'exp' is part of -the input data, corresponding to thedriven machine.Together with the inertia Or the motor, the slip changecan be determined, followed by the calculation of slip andequivalent resistance for the next time interval.

/). t

2H(Pm - Pg)

!'It I2.81

+2.82

- 44 -

The subroutine INDUMOT takes care o~ the calculationso~ all H.V. motors.When all load types are determined, ~or each generator busthe total active and reactive load is derived by summationoT all individual loads.In combination with the busvoltage, the correspondingequivalent admittance is calculated and stored ~or thenext time interval o~ the mainprogram.

One Or the additional load data for the high voltage motoris: TTRIPThe high voltage motors have relays equipment for under­voltage protection. This protection can be simulated and thesetting of the relais is presented by the variable TTRIP.If the busvoltage decreases under 0.65 p. u.• the motor willbe disconnected when the voltage remains lower than 0.65per unit for longer than the time corresponding TTRIP.

Another simulated protection. is a protection in case themotor will stall.When the mo~or slip exceed a certain level and stillincreases. then thp. motor will be disconnected.

- 45 -

3 PRO G RAM REV I SID N==============================

L 1 LOAD

.. 1 Load composition

All load compositions are simulated with the subroutineDYNLoAD.The existing program input doesn't contain enough load datato simulate all load compositions.The additional load inTormation, is thereTore added to therest oT the input data by the subroutine LOREAD.A call to this routine at the beginning oT the main programmakes the load input complete and print them in the programoutput.The routine DYNLOAD uses the input Tor the diTTerent poly­nominal eq,ua·tions, to simulate the composite load.BeTore the algorithm to calculate the eq,uivalent admittances,a call to INDUMOT within the routine DYNLOAD achieves theresults OT the dynamic H.V.-motor load Tor that interval.

One OT the variables Tor the dynamic load simulation, is thevoltage OT the diTTerend busses.In the case loads connected to a bus without a generator,the voltage oT the bus needs to be determined.This means extra registration oT the high voltage busses,because not only the voltage oT the program nodes areimportant, but all load busses.These extra high voltage busses are also req,uired by theroutine LOREAD and registrated corresponding to the numbersOT the existin~ generator nodes.At each time interval the voltage oT the high voltagebusses are determined using the voltage oT the generatornode, to which the bus is connected, as a reTerenceThe use oT all subroutines is outlined in the Tlow diagramin appendix 4.

L.2 Spreadsheet program

To expres the individual behaviour OT the steady stateeq,uivalent circuit oT the induction machine, thecorresponding eq,uations are analysed and characteristics aredrawn, using a spreadsheet program on a IBM persenalcomputer. (appendix 5 and 6)

(PMOT)(GMOT)(PCOMP)(GCOMP)

- 46 -

A p~intout of the sp~eadsheet p~og~am is shown in appendix 8.Seve~al cases. such as voltage dips a~e simulated. andthese ~esults a~e plotted. (appendix 7)

The analysis which can be de~ived fo~m this sp~eadsheet

p~og~am. is that the time inte~val of 0.04 seconds.~easonable fo~ the t~an~ient stability p~og~am. is to large.Therefore the subroutine INDUMOT determines the mechanicalsituation step by step. dividing each time interval in 5smaller intervals. by which the solution is more graduated.The spreadsheet program made it possible to analysedifferent kinds of aspects. assuming other parameters to beheld constant.This spreadsheet program has been very useful for writingthe subroutine INDUMOT.

For the input of the program. the additional load data isseparated in three catagories:

1) EXTRA H. V. BUS:Number or busName of busCorresponding generato~ busResistance connection to generator busReactance connection to generator bus

2) COMPOSITE LOADName corresponding H. V. busCor~esponding H. V. busActive L. V. motor loadReactive L. V. Motor loadActive Composite loadReactive Composite load

3) H. V. INDUCTIONMOTORSName motorInitial active power (PINMO)Value Ia/IrValue Ma/MrValue Mk/MrValue. cos rValue cos aValue Slip (in %)Value ETTiciencyValue Inertia (Kgm)Value rotor speed (rpm)Value Mechanical machine (exponent speed)Value Protection (TTRIP)

All this additional load data needs to be terminated by'END LOAD' . (See appendix 11 )

- 47 -

3.2 SALIENCY AND EXCITATION

2.1 Crout Algorithm

The existing program is based on the voltag~ behindthe transient reactance or the generatorsFor representation of saliency effect or excitation,more generator data is needed. This additional informationis obtained by calling the subroutine GEREADThe driving point voltage is changed from ED to EG.For each time interval this voltage EG needs to be adJusted.Equation 2.44 shows the relation Eq' = rAJ.EG andis calculated for all generators.The inverse matrix Ainv corresponding the relation

EG = CAinvJ.Eq'

is derived using the Crout algorithm (15 ), based onthe LU-decomposition. (Appendix 3)The subroutine to simulate the excitation by variationor Eq' has not been implemented yet.If the input value of Xd' is equal to the value of Xq,then Eq' is equal to EG and the reaction is the sameas the behaviour of the original program. ( ED )The new revision of the program asks for thefoliowing data CI2):

EB(*)ATC·.)

HK(*)PM(*)QM(*)PL(*)XD(*)XQ(*)XDD(*):XQQ(*) :TDO(*) :

Initial bus voltage in per unit on 100 MVA baseInitial angular displacement of the correspon­ding busvoltage. (in degrees)Inertia in per unit on 100 MVA base.Initial generated active power.Initial generated reactive power.Machine rating for rotorlossesDirect-axis synchronous reactanceQuadrature-axis synchronous reactanceDirect-axis transient reactanceQuadrature-axis transient reactanceDirect-axis transient open-circuit field timeconstant. (in seconds)

See appendix 11

equal to zero. or the99.0 per unit, then

with no connection to the

- 48 -

3.3 FRIENDLY USE OF THE PROGRAM.

,3.1 Plotting subroutine

To change the numerical output of the program to a suitableplot. the angles of all generators and the correspondingtime intervals have to be stored.Not only the absence of a plotter in the system. but alsothe large number of plots. has made it necessary to finda more convenient solution.The application program PLOTTSA. together with special plotsubroutines. has given the opportunity to make a graphicaldisplay of the output on the Printronix lineprinter.In case the system is connected with an infinite bus. thenthe absolute angular displacement of all machines are drawn.If there's no connection. a reference angle need to bespecified.The reference angle is a weighted mean of all machine anglesThe relative angular displacement with respect to thisreference angle is drawn.Tog.ther with the plot of this machine behaviour. someadditional information is written into the plot. by means ofregistration and recognition.Date and time are responsible for the registration.The corresponding outputs are chronologicaly stored and caneasy be found if additional information is required.Depending on the system conditions. the absolute Or therelative angular displacement is plotted.In the plot the situation with respect to the infinite busis wri~ten at the top of the plotThere are two critera for the system sim~lation. withoutconnection to the infinite bus.If the inertia of a generator istransient reactance is equal tothe program simulates the systeminfinite bus.The last possibility gives the opportunity to disconnectthe infinite bus. but to keep this generatorbus for otherpurposes. like the load distribution.At the end of th. program, a call to this subroutinecauses some interactive questions and system informationon the screen.There's an opportunity to give 40 characters of comment toexplain the typical system situation for that specificprogram run.The plot is completed with the name of the person who hasma de the stu dy and the s i mu I ate d fa u I t bus, tog e ther withthe chosen system input.

Not all generators need tQ be drawn, only the machinesthat are selected. When the plotfile is closed, it can besubmitted to the printqueue and printed without formfeed onthe printronix system lineprinter.

- 49 -

3.2 The T.S.A. command-file

For the interactive use of the program. a syst~m commandprocedure is written. the menu composition gives theopportunity to organise the creation of the input. runningof the program and handling of the output.During the analysis. it has been possible to use two programrevisions. The version in directory [ELE. 15.TRASTAJ is basedon a constant MVA load.In the directory tELE. 15. WORKJ the last revision is used.including the load representation mentioned before.Although the version in 'TRASTA' is based on constant MVA.the same input file TRASTA.INP can be used in 'TRASTA' and'WORK' as we II.The name of the command-file is TSA.COM in the directory[ELE. 15. COMJ.

3.3 Th. program Transient Stability Analysis.

To find the critical clearing time for a specific systemconfiguration. diff~rent combinations of number and siz~

of the time intervals has to be taken.Only these two variables have to be .changed. so thepossibility to incr.ase or decrease. without changing theinput file has been implemented.

All global variables are stored in a common bloc.Only this common bloc needs to changed if some additionalvariables are necessery.This common bloc is stored in the separated file TRASTA.CMNEach program or subroutine includes this file.Appendix 9 show a list of listfiles with correspondingdirectory.

- 50 -

3.4 INPUT AND OUTPUT

1J. 1 Input of the program

The original program reads all the input data from aninput data file.The first line contains common information for the headingof the program.The second line is filled with control data to run thesimulation program; Number of interval, fault bus, length oftime interval etc.The rest of the input is divided into three blocs.The first and last bloc contain information of the systembranches.The resistance and reactance of the branches are given andin the case of a load the active and reactive power.All values are given in pel' unit on 100 MVA base.The composition of the input file for the program includingthe dynamic representation of the high voltage inductionmotors is shown in figure 3.1.The system input files are almost the same a's the inputcomposition of figure 3.Instead of bloc 'B' in figure 3.1, in the system input filethis line only contains the word 'interaktief'.The progTam TRASTABI makes a conversion of this systeminput file to the input file TRASTA. INP for the program.This conversion program ask ro~ the control data on theCRT terminal and puts them in the right place in the finalinput file for the program. .When starting the Transient stability program, you firsthave to chose the directory.After that, the program asks for a system input.This system input has a typical name and will not bedestroyed by creation of other input files.From this input file the final input file for the TSAprogram will be made and is always named TRASTA. INP.When a new final input is created, the previous TRASTA. INPcan be destroyed (after the fifth input), because the lastfoul' versions of a file remain in the memory.This is of little importance as a new input file can be madeq,uickly. using the original system input.This procedure is shown in figure 3.2The interaction of the user with the CRT terminal is shownin appendix 13

- 51 -

common in~ormation ~or

A registration

control data: fault bus. numberB of intervals. length of interval

and additional switch

data o~ sytem branches duringC fault. load data for constant

MVA load representation

generator data (3.2. 1)o and intial condition

infurmation o~ extra highE voltage busses for load

representation

composite load OTF all high voltage busses

data high voltage inductinG motors

new control data ~or newH situation (Tault cleared)

data sllstem branches aTterI clearing the Tault

figure 3.1

4.2 The output of the program

At the end of a program run. the choice can be made, whethera plotfile has to be made or not. See figure 3.2

- 52 -

If this choice is confirmed. a call to the subroutinePLOTTSA is made and the output is stored in the plotfilePLOTFILE.. PRT.This file can be printed on the Printronix lineprinter.(list without 'form feed')The produced plot gives the graphical presentation of theangular displacement of the machines, selected during therun of the subroutine PLOTTSA.Some additional information for registration and recognitonis also written on the plot.Together ~ith this plot output a numerical output TRASTA.OUTis stored and can be listed an the printer.The numerical output contains the same information as theplot.The registration can be da.ne by date and time. (3.3.1)All input data will be registered in the numerical output.The output results of each time interval are stored inTRASTA.OUT and most of the additional information of therun, that has appeared on the CRT terminal.

START

helpexit (

5 print trasta.out on screen6 print trasta.out on printer7 print plotfile.prt

select inputfile~'----'t""'-l editor

createnew TRASTA.INP

figure 3.2

- 53 -

4 RES U L T S--------------------------

. 1 LOAD CONTRIBUTION

. 1 Introduction

To analyse the behaviour of the load distribution onthe transient stability, various load situations aredetermined and related differences have been madeThe new program revision, has made it possible todetermine different load representations.The same load situations are simulated with thesedifferent combinations of representation.The effect of a high voltage induction motoron the stability, with respect to the constant MVAload representation has been determined.The reaction of th~ dynamic load on the rotor motionof the machines close to a three fault place and machinesconnected at a larger distance are analysed.Due to undervoltage or stalling of the high voltageinduction m~tors, some load may be disconnected andaffect the stability limit.The synchronDus motor load is treated, as being a generator.Not only the stability limit of the synchronous motoritself, but also the effect of the pole slipping of thesynchronous machine on the rest of the generators issimulated .

. 2 The three phase fault

In all the studies a three phase ~ault is simulated, becaus~

this is the most severe cas~.

In practical situations, the faults not always is a threephase fault, or even if it's a three phase fault, thefault is often combined with an arc on the fault place.The fault impedance is not e~ual to zero (ideal three phasefault), but in case of an arc. the fault impedance isresistive.In case faults other than a three phase fault have to besimulated, then an extra load impedance can be connected.This fault impedance depends on the fault place and the typeof fault. This fault reactance 'Xf' can be derived by themethod of symmetrical components.

- 54 -

type of fault Xf============ ==

3 0 Three phase 0

L-L Line to line X2

L-G line t.O ground X2 + Xo

L-L-G Double line to X2. X~ground (X2+Xo)

where X2 = reactance o~ the negative phase se~uence

Xo = reactance oT the zero phase se~uence

In this study only the three phase ~ault is analysed,because the critical clearing time o~ all other ~ault

situations is higher than in the case o~ an ideal threephase ~ault. The most severe possibility has been analysed .

. 3 Load close to the fault·

In the case of a fault to earth in a electrical powersystem, then the load close to the fault will notaffect the g~nerators during the fault

figure 4. 1

This is easely shown in the single line diagram OTfigure 4.1, where two parallel impedances represent thefault impedance (Xf) and ~he corresponding load impedance(Xl) respectively.

- 55 -

These impedances connect the generator bus to the neutralbus.In the case of a severe fault, the fault impedance is muchsmaller than the load impedance. If there's a three phasefault, as simulated by the program, the fault impedance ise~ual to zera and the e~fect of the load impedance istotaly negligible.During the fault, the load behaviour close to the fault,or the load ~uantity isn't important, because it doesn'taffect the machine at all.This is demonstrated in plot 1 and plot 2 (Appendix 16 )plot 1 demonstrates the same situation as plot 2, exceptthat in the study of plot 2 the load on the fault bus ise~ual to ze1'o.In spite of this difference in the input, the plots 1 and 2are completely the same, confirming previous statements.

1.4 The'different load representations

The influence of the different load situations on thetransient stability, is determined by calculating thecritical clearing time for the various load representationsand by analysing the dra~n plots o~ the angular displacementof the generating machines.

The system net~ork, that has been simulated is sho~n

in appendix 15.The corresponding one line diagram is sho~n in appendix 14During the investigation of the effect of the load distri­bution on the transient stability, this one line diagram~ill be used, ~ith some small variations.This system configuration is registered in the inputfile SYSW0006. INP.Plot 3,4 and 5 sho~s the difference of identical situations,simulated ~ith different load representations.In all three cases a fault on bus 06 has a criticalclearing time e~ual to 0.40 seconds.Plot 3 simulates the constant MVA load (TRASTA) and plo~ 4the dynamic representation (WORK).By using different values of the variable responsible forthe undervoltage protection (TTRIP), it's possible to createdifferent load situatians.If the high valtage load is an important part of the totalload on its bus and the setting TTRIP is e~ual to 0.3seconds, then in the studies sho~n in plot 3,4 and 5, thehigh volta~e motors on the fault bus ~ill be disconnectedbefore the three phase fault is removed.

- 56 -

In the c.ase variable TTRIP is eq,ual to 1. 0 second, then thehigh voltage motor load will remain connected to it's busduring the total test period o~ 1.6 seconds.The initial amount o~ load is in all three cases the same.Plot 5 illustrates the di~~erent angle curves o~ thegenerator connected to the fault bus G6 for the threesimulated possibilities, that are mentioned above.The maximum angle is in all three cases the same andthe influence aT the high voltage induction motoron the stability limit is very small.The rotor motion of the other machines is much smooth~r, ifthe load is represented by the induction motor model.This difference is shown in plot 3 and 4; in plot 3 the loadis simulated being a constant MVA load.

When a fault is created on the llkV system, then there's alarger effect on the behaviour of the machines, comparedwith a fault on the 6kV system (bus G6).When a three phase fault is simulated on the 11 kV system,then the critical machine is generator 1051 (gen 80 in theplots), connl!ctl!d to bus 80.The relative low inertia causes an oscillation around theother machines on the llkV system.If TTRIP is 0.2 seconds, then the induction motors close tothe fault bus are disconnected, before the fault is removed.This causes a faster return of the generators, afterremoving the fault. Then there's no affect on the stabilityI imi t.In case the induction motors are still connected. when thefault is removed, then the angles of generator 2,3 and 4don't return that fast and the maximum angular displacementis higher. This is shown in plot 6 and 7.In plot 7, variable TTRIP is eq,ual to 0.4 seconds.The fault is removed within 0.25 seconds and the busvoltage will return before the time of the undervoltageprotection has run out, and the protection will reset.The motors are still connected. but the slip has increasedand the voltage has recovered.In plDt 7 the oscillation freq,uency of the machine 1051 hasdecreased. with respect of the freq,uency shown in plot 6.In plot 6 all induction motors close to those generators aredisc onnec ted.Also plot 6 shows a more smooth behaviour of the machine 6,the machine on the 6kV level.Due to the extra load of the high voltage motors, thecritical clearing time has increased.Figure 4.2 shows the relation between the critical clearingtime and the setting TTRIP of the undervoltage protection.In th~ case this time being lower than the time the faulthas taken tD be removed, then the critical clearing time is0.285 s·econds.When it's hi~her, the induc~ion motors can affect thegenerator behaviour, when the voltage recovers.

- 57 -

The stability limit is reached. when there is a fault forlonger than 0.25 seconds.Constant MVA load representation and composite loadrepresentation also cause a critical clearing time of0.29 ~econds. The angular displacement is shown inplot 8, composite load and plot 9. constant MVA.Plot 9 and plot 10 both represent a simulation of constantMVA l~ad. The input corresponding to plot 10 contains aload shedding of all loads, when the fault is removed.A combination of composite load and induction motorload corresponding to the practical situation is simulatedin plot 11.

T critical(msec)

284

240

,1

,.2

,.3

,.4

,.5

,.6

,.7

,.8

TTRIP (seconds)

figure 4.2

Dynamic representation of load has a much smaller affect.if there's no connection with the infinite bus.The original program in the directory tELE.15.TRASTAJcalculates a critical clearing time of 0.36 seconds.The revised program in the directory tELE.15.WORKJloses synchronism in case the clearing time is higherthan 0.37 seconds.

- 58 -

Due to the,dynamic representation, when there's noconnection with the infinite bus, the stabilitylimit is increased a little.When there's no connnection with the infinite bus, thefirst generator losing synchronism when a fault on the11 kV system is simulated, is the generator 1051 (gen ad>.In chapter 3 it is stated, that without connection to theinfinite bus, it's of little value to draw the absoluteangular displacsment.The stability limit is determined by the behaviour of themachines with resp~ct to each other.Now the relative angular displacement is most important.In plot 12, the illustration shows what happens if theabsolute angles are drawn.All machines show an increase of angular displacement.The augmentation of the frequency doesn't m~an that therelative angular displacements of the generators isaffec ted that muc h, that th e sync hron ism is lost.Therefore a weighted mean of all angles is calculated andby relating all angles to this mean value, it is possibleto illustrate the relative behaviour of the machines andto analyse the stability limit.Plot 13 is a simulation of the system without connectionto the infinite bus. /Due to the fact that all angles are related to the weightedmean valu·e of all machines, it's possible, that theillustrati~n of a generator changes, while the machin~ angledoesn't change. This is caused by the fluctuation of there~erence angle. In plot 13 the heavy fluctuation of genera­tor 1051 (gen ad) is respon~ible for changes in the weightedmean; the reference angle.More important is the difference of the angles of theindividual machines, because the absolute difference inangular displacement of two machin~s, is equal to thedifference of the relative values with respect to thereferenc e ang I e.

(A1 - A2) = (A1-Aref) - (A2-Aref)] 4.1

These differences form the stability criterion, in casethere's no connection to an infinite bus .

. 1.5 Synchronous motor load

The LOPE-factory contains induction motor load, and alsosome large s~nchronous motors.The representation of the synch~onQus motor is the sameas the generator representation.When there's a fault, the synchronous motor tends todecelerate.

- 59 -

The contribution of the motor opposes the mechanical powerof the machine. This causes a relative large deceleratingpower. The inertia of a machine is a combination of themoment of inertia and the speed of the machine (Appendix 1).and the inertia is proportional to the speed of the motorsq,uared.A speed of 200 rpm means that the per unit effective inertiaof the synchronous machine is much smaller than the inertiaof the generators.This relative high decelerating power in combination withthe low inertia means that the angular displacementincreases morv rapidly.In plot 14 and 15. the large synchronous machine of theLOPE is simulated by generator SM.A fault in the 11 kV system needs to be removed within O. 1sec onds. if not. th e sync hronous motor wi 11 go out of step.Severe faults in the 6 kV system will not cause aninstability of the synchronous machine of the LOPE. (plot 16)Due to pole slipping of the synchronous machine pulsatingpower is produce~ by this motor.In plot 15 and plot 17 is shown. the effect of th~se

pulsations on the rest of the synchronous machines.The behaviour of the relative large generators has not beenaffected by the pole slipping of the synchronous machine.The angular displacement or generator 1051 is influenced alittle. especially when generator SM is connected to bus 80.(plot 17)The critical clearing time for a three phase fault on bus SAfor the synchronous motor HP401 is about 0.085 seconds.When for a identical sy~tem configuration the transientr~actance of the synchronous machine (gen SM) is changedfrom 1.0 per unit to 2.0 per unit. but ~ti11 connected tothe bu~ SA. then the critical clearing time increases to0.115 seconds.Another possibility to increase the reactance between thesynchronous motor and the fault bus SA. is a connectionof th~ synchronous motor to bus 80. instead of theconnection to bus SA.When the same system configuration is repeated. and onlythis connection is changed. then the critical clearing timeis increased to O. 125 sec ond s.Th ese three sj"tuat ion s are sh own in plot 18. 19 and 20.The situation of a connection of the synchronous machineto bus 80 will not increase above the critical clearingtime of 0.85 seconds. if the fault is on bus 80.The stability limit of the synchronous motor will increase.when the inertia of the machine is increased.In plot 21 the inertia of the machine 0.12 instead of 0.06per unit. and the critical clearing time is changed to0.120 seconds (gen 8M connected to bus SA)If the fault is simulated on bus aD and gen. SM isconnect~d to bus 80. then the critical clearing timei seq,ua 1 toO. 1 sec 0 n d s. " (p lot 22)

- 60 -

4.2 VARIATION OF CONFIGURATIONS

2.1 The cannecti.on with the- infinite bus

In 4.1.3 it's mentioned, that the absolute angulardisplacement i. plotted, when there's a connectionto the inrinite bus 01' the relative angle, if noconnection present.The angular displacement of the infinite bus is assumedto be constant, due to the fact that the inertiaconstant of this "generator" is assumed to be infinite.For any disturbance in the distribution system of DOWTerneuzen, the change in the angular displacement of thePZEM grid is negligable.Therefore the angular displacements with respect to this'generator' give not only the information of the machinewith respect to this infinite bus, but also the behaviourof all machines with repect to each other, be~ause allmachines try to follow the angular displacement of theinfinite bus.If no connection is present, one has to find a way torepresent the angular displacement of all machine withrespect to each other.A presentation relative tD the weighted mean of all theangles can be sufficient for examination if the machines'stay together',or 'walk away' from each other.

If there's no connection with the infinite bus, thecritical ~learing time is much higher then in the caseof a connec~ion with the public grid.For example, if the critical clearing time for a certainsystem configuration is eq,ual to 0.24 seconds, thecritical clearing time of the same system configuration,but without the connection to the infinite bus, is eq,ual to0.37 seconds. (plot 13)

2.2 The transient reactance

In the case that the transient reactance of the synchronousmachine is increased, the stability limit increases.What will happen if the reactances of the generatorsare changed?In plot 23, the two generators G2 and G3 are in the samecondition, except for a difference of transient reactance.Under nQrmal condition, there's no difference in transientreactance. The twa machines are bath connected to bus SA.

- 61 -

The intial generation oT active and reactiv~ power is inboth cases the same.The inertia constant of the two machines are both 1.64 perunit.If a Tault is simulated on bus SA, then the machine G2 andG3 will react exactly the same.When the transient reactance of one of the two machinesis decreased from 2.5 per unit to 1. 0 per unit, thenthere's a difference in the behaviour of the angulardisplacements of both machines (plot 23)The maximum o~ the angular displacement of Generator G2is lower than the maximum ~f generator G3.

If the transient reactance of generator 1051 (gen 8D) is2.21 per unit (normal condition), the angular displacementof this machine varies acutely due to the three phasefault. (plot 24)If the reactance is decreased to a value of 1.21 per unit,the variation caused by the fault, shown in plot 25, isless.A decrease of reactance is only possible if some additionalreactance is installed in the line.Increase of reactance is always possible. Increase oT thetransient reactance i~ the same as an addition of a seriesreac tance.

2.3 Position vaTiation of a generator.

In the case of a fault on the 11 kV system the criticalmachine is generator 1051.The combination of inertia constant and generated powerof this machine differs from the other generators connectedto th e same bus.The mechanical movement of the rotor of this machine, whenthe electrical power collapses, is about the same asgenerater G6 on the 6 kV system.This is shown by connecting machine 1051 on the 6 kVsystem and simulating a three phase fault on the 11 kV system,plot 26, or the 6 kV system, plot 27.When machine 105.1 is connected to bus 50 kV, then if thesystem is not connected to the infinite bus of PZEM, thena fault on the 11 kV system will not cause instabilityaT the generators.Even if the fault is cleared after 0.5 seconds the an~ular

displacements of all machines stay together.This is shown in plot 28.As generator 1051 is the critical generator in the 11 kVsystem, the critical clearing time is investigated forindentical ~ystem configurations, that only differ from th~

'place of connection of machine 1051.

- 62 -

The system is connected to the infinite bus and generator1051 is connected to bus SA, bus 50kV and bus G6 respectivelyIn all the three cases, a fault is simulated on bus SA,bus 50kV a-nd bus G6 repec.tively.The result are shown in figure 4.3

============================GENERATOR 1051 CONNECTED ON:BUS SA, BUS G6 OR BUS 50KV============================

Fault bus\

\

\_------\

Connected \1

BUS 50KV BUS G6 BUS SA

=================== =:=========== ============ ============BUS 50KV 265 msec 400 msec 300 msec

BUS G6 330 msec 260 msec 300 msec

BUS SA 370 msec 380 msec 250 msec

figure 4.3

- 63 -

5 CON C L U S ION=========:=========

In chapter four is shown, that the critical clearingtime is independent of the load situation on the faultbus during the fault.The behaviour of the synchronous generator can be affectedby the load situation, when the fault is cleared and thevoltage reCQvers.If the .system is connected to the infinite bus, then thecritical clearing time will decrease, caused by the dynamicalbehaviour of the asynchronous machines on that bus.When no connection is provided, the critical clearing timewill increase a little,The dynamic behaviour of the induction motors have astabilising effect on the busses I when the voltage dipis less sever~ then on the fault bus.The total affect of the load on the transient stability isrestricted to a change of 1 or 2 cycles in critical clearingtime (critical clearing time is about ten cycles for a faulton the 11 kV system ).A fault on bus G6 shows a very little difference in angulardisplacement for different load representations.

The mechanical motion of the synchronous generators duringthe fault has become slower, when the dynamical representa­tion of the load is used.This means, that the time the critical angle is reachedhas become later and when the mechanical movement of themachine is relative slow, it might be possible that theeffect of control units can affect the motion beforethe critical angle is reached. (0.5 to 1.0 seconds)In that case the decelerating angle increase, caused bythe dynamical load, can have a positive influence on thestability, even if the critical clearing time is decreased.Important for the critical clearing time of a first swinginstability, is the impedance corresponding to the loadsituation on the faulted bus, the moment the- fault iscleared.

The critical clearing time of the synchronous motor of 6 MWis less than 0.10 seconds. In the case of the severe faultin the 11 kV system, during the first SWing, synchronism willbe lost.It's possible to increase the stability limit of this machineby an increase of the reactance between the machine and thefault bus SA.

When a fault is simulated on the llkV system, the angulardisplacement Qscillation of generato~ 1051 is differentfrom th'e other generators.The behaviour of the generators with respect to each otheris clearlv shawn, when a fault is simulated, withoutconnection to the infinite bus.

- 64 -

The behaviour of machine G6 and machine 1051 are identicaland therefore, from the stability kind of view, the mutualbehaviour during a fault would be better if machine 1051is placed near machine G6.The critical clearing time of a fault direct on the bus towh ich generator 1051 is connected, doesn' change.Figure 4.3 shows that this time will be about 0.26 secondsin case a fault is on the generator bus of machine 1051.If ma chi n e 1051 i.s conn e c ted tot he 50 kV bus, the~ynchronism of this machine will sustain, when a fault issimulated on bus G6. The critical machine is G6~ andthis machine will lose syncrhonism if the fault is clearedafter 0.4 seconds.In the case a three phase fault is simulated on the 11kVsystem, the synchronism of machine 1051 will also sustainand one of the machines on the 11 kV system will losesunchronism. ( T(cr. )=0.300 seconds)When generator 1051 is connected to the 50 kV system.then the power pulses of the pole slipping of the largesynchronous motors don't affect this machine that muchany more.

The critical ~learing time for a ideal three phase faulton the 11kV system is about 0.25 seconds.If there's a fault in the 11 kV system and no differentialprotection is able to detect this fault, then the faultwill be cleared in about O. 5 se~onds.

Depending on the type of fault this might be to long andsynchronisme of one or more generators can be lost.If any severe fault in the llkV system or in the 6kV systemcan be cleared by any protection within 0.25 seconds,synchronism of all six generator will not be lost and thethe first swing transient stability limit will not beexceeded,

- 66 -

REF ERE N C E S=========~=========

(1) Kimbar k, E. W.POWER SYSTEM STABILITY,Volume 1John Wiley & Sons, Inc. ,New York, 1948

(2) Kimbar k, E. W.POWER SYSTEM STABILITY,Volume 2John Wile-y & Sons, Inc. ,New York, 1950

(3) Crary, S. B.POWER SYSTEM STABILITY,Volume 1John Wiley & Sons, Inc. ,New York, 1945

(4) Crary, S. B.POWER SYSTEM STABILITY,Volume 2John Wiley & Sons, Inc. ,New York, 1947

(5) Anderson P.M. and Fouad, A.A.POWER SYSTEM CONTROL AND STABILITYThe Iowa State university press, 1977

(6) Weedy, B. M.ELECTRIC POWER SYSTEMSJohn Wiley & Sons, Inc., New York, 1967

(7) Venikov, V. A.TRANSIENT PHENOMENA IN ELECTRICAL POWER SYSTEMSPergamon press Ltd., 1965

(8) Edited by: Byerly, R. T., Kimbark, E.W.STABILITY OF LARGE ELECTRIC POWER SYSTEMSIEEE-~ress, 1974

(9) Stagg and El-AbiadCOMPUTER METHODS IN POWER SYSTEM ANALYSISMcGrawHill, 1968

(10) Anderson, P. M.ANALYSIS OF FAULTED POWER SYSTEMS

(11) MarteiJn, P.A.M.DYNAMISCHE STABILITEITS ANALYSEReport of project on DOW CHEMICALTerneuzen, 1984.

(12) Kimbark, E.W.POWER SYSTEM STABILITY, Volume III.

(13) Rentzsch H.HANDBUCH FUR ELEKTROMOTORENBrown, Boveri & Cie, 1968

- 67 -

(14) Walker .J.H.LARGE SYNCHRONOUS MACHINESClaredon press. OxPord. 1981

(15) Antal. M.ElektriscITe opwekking en transmissie (part 1)University o~ Technology Eindhoven1982

(15i Veltkamp. G.W. and Geurts. A. .J.NUMERIEKE METHODEN.University oP Technology Eindhoven.1980

(16) Klamt • .J.Berechnung und bemes;ung elektrischer maschinen

(17) Pai 1'1.1.Computer techniques in power system analysisMcgraw-Hill. 1979

(18) Adler. R. B.STEADY-STATE VOLTAGE POWeR CHARACTERISTICS FOR POWERSYSTEM LOADSIEEE paper 70 cp 706-pwr

(19) Power system working group.SYSTEM LOAD DYNAMICS - SIMULATION EFFECTS ANDDETERMINATION OF LOADCONSTANTS.Ieee committee report. ieee pas 92 1973. p600

(20) Shacksha~t G. et al.GENERAL-PURPOSE MODEL OF POWER-SYSTEM LOADS.PROC. lEE. VOL 124. 1977. p715

(21) Sa b irS. A. Y. eta l.DYNAMIC LOAD MODELS DERIVED FROM DATA ACQUIRED DURINGSYSTEM TRANSIENTS.IEEE PAs-i01. 1982. p3365.

(21) Concordia. C.LOAD REPRESENTATION IN POWER SYSTEM STABILITY STUDIES.IEEE PAS-I01. 1982, p969.

(22) Ket, M. H. et al.DYNAMIC MODELING OF LOADS IN STABILITY STUDIES.IEEE trans.. PAS-8S p756. 1969;

( 23 ) I 1 ice to, F. eta l.BEHAVIOR OF LOADS DURfNG VOLTAGE DIPS ENCOUNTERED INSTABILITY STUDIES. ~IELD AND LABORATORY TESTS.IEEE trans.. vo191 p2470

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(24) Maur i c i 0, W. et a l.EFFECT OF LOAD CHARACTERISTICS ON THE DYNAMIC STABILITYOF POWER SYSTEMSIEEE Winter meeting, New York, 1972

(25) Nelles, D.BEDEUTUNG DER SPANNUNGS- UND FREGUENZABHANGIGKEITEN VONLASTEN IN NETZPLANUNG UND NETZBETRIEB.etzArchiv Bd. 7 H. 1, 1985

(26) Gentile, T et al.DETERMINING LOAD CHARACTERISTICS FOR TRANSIENTPERFORMANCE.EPRI Report EL-8S0, Project 849-1, Palo Alto,Cal i of! 0 r n ie, 1981 .

(27) Chen, M. S:DETERMINING LOAD CHARACTERISTICS FOR TRANSIENTPERFORMANCE.EPRI Report EL-849, project rp849-3, Palo Alto,Calirornie, 1979.

(28 ) Rob e r t, J.LOAD BEHAVIOR AND TRANSIENT STABILITYIEEE ~inter power meeting, New York, 1970 (70 cp)

(29) Hi 11, D. J.STABILITY ANALYSIS OF MULTIMACHINE POWER NETWORKS WITHLINEAR FREQUENCY DEPENDENT LOADS.IEEE CAS-29, 1982 p840i

(30) Meyer, F.J. and Lee, K. Y.DYNAMIC POWER SYSTEM LOAD MODELING BY PARAMETERESTIMATION.Elektric power systems research,7, 1984 p231i

(31) Rogers, G. J. et al.AN AGGREGATE INDUCTION MOTOR MODEL FOR INDUSTRIALPLANTS.IEEE PAS-103, April 1984. p683;

(32) Lahoud, M. A. et al.DYNAMIC BEHAVIOUR OF A MULTI-INDUCTION MOTOR SYSTEM.TRANS. S.A. INST. OF ELECTR. ENG., 1982, pl06.

(33) Lahoud, M. A. and Harley R.G.TRANSMISSION LINE RESONANCES AS INFLUENCED BY INDUCTIONMOTOR LOADMODELLING IN A MULTIGENERATOR-MULTIMOTORPOWERSYSTEM.Elektric power systems res~arch,7, 1984 p6S;

(34) Bauman, H. A. et al.SYSTEM LOAD SWINGSAlEE Transactions. Vol 60, 1941, p541

- 69 -

(35) Ramamurthi, V. and Venkata S.S.LOAD REPRESENTATION THROUGH DYNAMIC CHARACTERISTICS INTRANSIENT STABILITY STUDIES.IEEE PES Winter meeting, 1979.

(36) Ueda, R. and Takata S.EFFECTS OF INDUCTION MACHINE LOAD ON POWER SYSTEMIEEE, PAS-l00, No 5 , May 1981

(37 ) Mil de, F.RECHENZEITSPARENDE NETZVEREINFACHUNG FUR TRANSIENTEUNTERSUCHUNGEN IN NETZEN MET UBERWIEGENDER MOTORENLAST.Brown-boveri technik 3-85 p130;

(38) Lahoud, M. A. et al.INDUCTION MOTOR MODELLING FOR SHORT THERM TRANSIENTSTUDIES IN A MULTIGENERATOR-MULTIMOTOR POWER SYSTEM.Elektric power systems research, 7, 1984 p53i

(39) Khalil, M. O. and El-Motti Zaher, M. A.DETERMINATION OF THE EGUIVALENT CIRCUIT PARAMETERS OFDEEP-BAR CAGE INDUCTION MOTOR FROM EXPERIMENTAL ORGIVEN CATALOGUES DATAetz ARCHIV SD 6, 1984 p33;

(40) Williams, A. ..J. et al.EVALUATING THE EFFECTS OF MOTOR STARTING ON INDUSTRIALAND COMMERCIAL POWER SYSTEMS.lAS '77 ANNUAL;

(41) Berg G. .J.NEW INDUCTION MOTOR LOAD MODEL FOR SYSTEM TRANSIENTSTABILITY STUDIES.IEEE PES ~JINTER POWER MEETING 3-8- '80 ; A 80-009-1.

(42) Brereton D.S. et al.REPRESENTATION OF INDUCTION-MOTOR LOAD DURING POWER­SYTEM STABILITY STUDIES.AlEE TRANS. 76, 1957, p451.

(43) Stanley H.C.AN ANALYSSIS OF THE INDUCTION MACHINE.AlEE VOL 57, 1938, p751.

(44) Cavin, R.K.TRANSIENT LOAD MODEL OF AN INDUCTION MOTOR.IEEE PES WINTER MEETING, NEW YORK, 1973. p1399.

(45) Akhtar, M. Y. et al.FREGUENCY-DEPENDENT DYNAMIC REPRESENTATION OFINDUCTION-MOTER LOADS.PROC. lEE, VOL 115, 1968, p802.

- 70 -

(46) Kalsi, S. S. et al.TRANSIENT STABILITY OF POWER SYSTEMS CONTAINING BOTHSYNCHRONOUS AND INDUCTION MACHINES.PROC. lEE, VOL 118, 1971, P 1467.

(47) Shankle, D. F. et a1.TRANSIENT-STABILITY STUDIES I, SYNCHRONOUS ANDINDUCTION MACHINES.AlEE, VOL 73 pt. III, 1954, p1563.

(48) I 1 i ceto F. et a 1.DYNAMIC EGUIVALENTS OF ASYNCHRONOUS MOTOR LOADS INSYSTEM STABILITY STUDIES.IEEE PES Ulinter meeting New York, 1974, p1650.

( 49 > Ha kim, M. M. A. eta 1.DYNAMIC SINGLE-UNIT REPRESENTATION OF INDUCTION MOTOR GROUPS.IEEE PAS-95, 1976, p155.

(50) Berg, G. J.INDUCTION MOTOR LOAD REPRESENTATIONIEEE PES summermeeting, 1979.

(51 >. Arnold, C. P. et alMODELLING INDUCTION MOTOR START-UP IN A MULTI-MACHINETRANSIENT STABILITY PROGRAMME.IEEE PES Summer meeting, 1979.

( 52 ) Ma ginn iss, F. J. eta 1.TRANSIENT PERFORMANCE OF INDUCTION MOTORS.AlEE trans., vol63, p641.

(53) Wiegand, H.A. and Eddy, L.B.CHARACTERISTICS OF CENTRIFUGAL PUMPS AND COMPRESSORSWHICH AFFECT THE MOTOR DRIVER UNDER TRANSIENT CONDITIONS

(54) Kelly, A. R.INDUCTION-MOTOR MODEL FOR INDUSTRIAL POWER SYSTEMCOMPUTATIONSAlEE Winter general meeting, NeUl York, 1962.

(55) Brickell, M. R.SIMULATION OF STAGED TEST ON THE ONTARIO HYDRONORTHWESTERN REGION.IEEE PaUl. Industry computer applications conference,1979, p357i

( 56) La h0 ud, M. A. eta 1.ANALYSIS TECHNIGUES FOR THE DYNAMIC BEHAVIOUR OF AMULTIGENERATOR SYSTEM.TRANS. S.A. INST. OF ELECTR. ENG., 1983, p83.

- 71 -

<57) ~oshi, S.S. and Tamaskar, D.G.DYNAMIC BRAKING AS AID FOR AUGMENTATION OF TRANSIENTSTABILITY LIMITS OF A POWER SYSTEM.~ournal of the inst. of engineers. VOL64. ELL 1983, p56

(58) Nitayanandan, B. N. a·nd Nagappan, K.AGGREGATION OF COHERENT MACHINES FOR TRANSIENTSTABILITY OF LARGE SCAL POWER SYSTEMS.~ournal of the inst. af engineers. VOL64.EL1,1983,p63

<59) Fouad, A.A. and Stanton, S.E.TRANSIENT STABILITY OF A MULTI-MACHINE POWER SYSTEMpart I: investigation of system trajectoriespart II: critical transient energie.IEEE PAS-100, 1981 p3409i

(60) Dommel, H.W. and Sato, M.FAST TRANSIENT STABILITY SOLUTIONS.IEEE PAS-91, 1972, p1643i

(61) Hatcher, W.L. et al.A FEASIBILITY STUDY FOR THE SOLUTION FOR TRANSIENTSTABILITYP~oblems by multiprocesso~ structu~es.

IEEE PAS-96, 1977,. p1789i

(62)SIMULATION OF STAGED TEST ON THE ONTARIO HYDRONORTHWESTERN REGION.IEEE POW. Industry Computer Applications Conference, 1979,P357i

(63) Brasch, F. M. et al.THE USE OF A MULTIPROCESSOR NETWORK FOR THE TRANSIENTSTABILITY PROBLEM.IEEE PICA conference , 1979

(64) Hatziargyriou, N. and Efthymiadis, A.E.AN ADVANCED ALGORITHM FOR FAULT LEVEL CALCULATIONS ININDUSTRIAL POWER SYSTEMS.MELECON 83, ATHENS

(65) Hamdan, A. M. A.TORGUE-ANGLE LOOPS IN MULTIMACHINE POWER SYSTEMS.Inter. ~. Control, 1984, VOL39,p1261i

(66) Ya-e, H. and Spalding B. D.TRANSIENT STABILITY ANALYSIS OF MULTIMACHINE POWERSYSTEMS BY THE METHOD OF HYPERPLANES.IEEE PAS-96 p276

(67) Brown, H. E. et al.TRANSIENT STABILITY SOLUTION BY IMPEDANCE MATRIX METHOD.IEEE PAS-84,1965, p1204

- 72 -

(68) Day, J. E. et al.GENERALISED COMPUTER PROGRAM FOR POWER-SYSTEM ANALYSIS.PROC. lEE, VOLl12, 1965, p2261.

(69) Ness, van J. E. et al.USE OF THE EXPONENTIAL OF THE SYSTEM MATRIX TO SOLVETHE TRANSIENT STABILITY PROBLEM.IEEE, VOL PAS-89, 1970, p83.

(70) Stanton, K.N. and Talukdar, S.N.NEW INTEGRATION ALGORITHMS FOR TRANSIENT STABILITYSTUDIES.IEEE Summer Power Meeting. Dallas, 1969.

(71) Humpage W. D. et al.PREDICTOR-CORRECTOR METHODS OF NUMERICAL INTEGRATIONIN DIGITAL COMPUTER ANALYSES OF POWER-SYSTEMTRANSIENT STABILITY.PROC. lEE, VOL 112, 1965, p1557.

(72 ) 0 I i ve , D. W.NEW TECHNIGUES FOR THE CALCULATION OF DYNAMIC STABILITY.Westingh~use electric corp, East Pittsburgh, PA.

(73) Bauer, ,D.L et al.SIMULATION OF LOW FREGUENCY UNDAMPED OSCILLATIONS INLARGE POWER SYSTEMS.IEEE PAS-94, 1975, ,207.

(74) IEEE Committee reportCOMPUTER REPRESENTATION OF EXCITATION SYSTEMSIEEE Trans. Power Appl SlJst., June 1968

(75) Ramarao, N.IMPROVED STABILITY WITH LOW TIME CONSTANT ROTATINGEXCITER

(76) Tanaka,. H.EFFECT OF AVR CHARACTERISTICS ON DYNAMIC STABILITY.Electr. eng. in .Japan oct '63.

(77) Lokay, H. E.EFFECT OF TURBINE-GENERATOR REPRESENTATION IN SYSTEMSTABILITY STUDIESIEEE PAS-84, 1965, p933.

(78) BalJne, J. P. et al.STATIC EXCITER CONTROL TO IMPROVE TRANSIENT STABILITY.IEEE PAS-94,1975, pl141.

(79) Bemello, F.P.CONCEPTS OF SYNCHRONOUS MACHINE STABILITY AS AFFECTEDBY EXCITATION CONTROLIEEE PAS-S8, 1969, p316

- 73 -

(80) Peneder. F. and Bertschi. R.STATISCHE ERREGUNGSSYSTEME MIT UNO OHNE KOMPOUNOZUSATZBROWN-BOVERI TECHNIK 7-85. p343

( 81) Ab da 11 a 0: H. eta1.COORDINATED STABILIZATION OF A MULTIMACHINE POWER SYSTEMPAS-103, 1984 p483; eigenwaarden en model excitatie.

(82) Ph ad ke G. A.DIGITAL SIMULATION OF ELECTRICAL TRANSIENT PHENOMENAIEEE Service center. 81 EH0173-5-PWR, 1980

- 74 -

APPENDIX 1

INERTIA CONSTANT

Equation of motion: WR2 . a = P pound-feet"3'2":"2"

rpm d20= • d't"260f

2160.frpm

= number of poles

per unit uni t (P) = (base) kVA-4

1.421rpm 1 10

1 42 10-4. 1 rpm 1

(base) kVA

(unit kW = unit kVA)

P=---P(unit)

= P p.u.

WR2 = moment of inertia in pound.(feet)2

GD2= moment of inertia in Kg.m2

H = 0.231 1 WR2 1 rpm2 1 10-6

kVA (base) kW.sec/kVA

PER UN IT VALUES

- 75 -

APPENDIX 2

definition: The actual. value (in any unit)The base or reference value in the same unit

=

=

E2 ETransformer:' Xohm = SN ° 100

2= £. E Sb =

100 S;-0 E2S

J..... ° _b_100 SN

Load: , where G = conductivity

- 76 -

APPENDIX 3

THE CROUT ALGORITHM

When ~n~ equatrions of ~n~ vari ab 1es need to be solved, then A~ = ~

is equi va1ent with L~ = Q. and Ux = ~, where Lis a not singular matrix.

Ltt 0 UII UI2 U1n

L21 L22U •

U22u2n

L • 0. . . .

Lnt Ln2 LnnUnn

for k := I step I until n do

begin for j :~ k step I until n do

begin s "~A.. "" kj ,

for t :~ I ~ I until k - I do s :~

Ukj

: .. s

s - Lkt l( U l.j ;

end

for i :s k + 1~

begin s :~ Aik

;

for t :s 1 step

Lik

:s s/Ukk

until n do

until k -.1 do s :s s - L )( U "it 10k'

end

:~ I step I until n do

begin s := bk

;

tor ~ :s I~ I until k - I do s :~ s - L )(kt c~;

Xn • C IUn nn

Xn_ 1 ~ (c - U x )/Un-I n-I,n n n-t,n-I

----------------------------------n

• (ck - r Uk]" xJ")/Ukk •j-k+1

[n pseudo-ALGOL:

for k :- n step -I until I do

begin S :- ck ;ex. ;

I

- 77 -

Flow diagram of theProgram Transient

stability analysis

sort noues ana put them ir<3scenuing orUl.lr.

rna"'" talltds

STOP

APPENDIX 4

new cir LJ new intervalCT=T+OTJpiJrame::ers

IT; NHHltlJTJ

calculate admittancesJfrom circuit pdrdlnet"rs

calculate transfor irnp,,­

dances lly iteration

read ~enoratDrd~td

rBuLl LOi'ld ddtil (LmlEAO)'read uxtra gen. uota (GEREADIcalculdtu drivin oint volt.

calculdtu active powercalculate reactive power

cdlculate

Print output in TRASTA.DUT

Calculate new angles

calculate new voltages ofextra high voltage busses

calculate new Llrivin~

voltdges ISALIEN)

111l.;,l'Uc.HifoJ r fur Ilt::xl

calculaU! corres­ponLlin[l ddrnittdn

calculate new luadsituation IOYNLOAOJ

calculate new slipinc1uction rnntoz:,; I INDlJMfI rJ

Power/ Sl1p for motor CC6B (par. EBJ

co~o:Irn.~ ~-o»r-..-..-zrncm:Dr-»zc---m<:•

*' OSplJ

Ul

10080

-b 0.6 pu

60

Sl1p96e:t 0.7 pu

40·

INDUCTION MOTOR MODEL

o O.8pu

20

OJ

0.09

0.08

.... 0.07~:>~

0 0.060,...c:

0.050:JS-L. 0.04Q)

~0n.. 0.03

0.02

0.01

0

0

-r-- EE: 1ptJ )( 0.9pu

INDUC·TION MOTOR .MODELReactIve/Slip for motor CC6B (par. EBJ

OJ7

OJ6

OJ5

OJ4

OJ3

.... OJ2~:>~ OJ]::> OJ::>....L.: 0.090:J O.OB~3) 0.07>..~

0.06Jub

0.05c0.04

0.03

0.02

0.01

0

0

f- EB -1 pu

c0~0J:m~ -....J- 1.0

0».....-..-zmcm:D

:t:o ...""0""0 »rrI:zc Z......x CO'l ........

m•20 40 60 80 100 <•

O.8puSlip 96

Q s 0.7 pu *' O.5pu

INDUCTION MOTOR MODEL

•:34Slip sImulatIon durIng fault CCC6B)

:32

30

2800

26 ~..., 24 0c:cu J:.., 22I" mc:Q 20 ~l)I - 00

c: 18 0a

0Q) 16 »e1.- r-.:t 14 -~ Z.., 12 m9-..

10 0f) m

8 :D):> r6 "" »rt"I:z

4 Cl Z.....x

02 ~

......ED0 •

0 0.2 0.4 0.6 .sec. <•Tlmel £B - 0.0/1.0f:f

-I Inertia H e InertIa 2 H

o

-

-

-

-

-

-

-L

•-

~\ ......

I I J 1 J I

Power simulation during fault CCCBB)

c0~0J:m~ OJ- I-'

0»r---...zmcm::D

::t:o r--0-0 »IT1:z

ZCl.....X C-............

m•<•O.B0.40.2

INDUCTION MOTOR MODEL

o

0.01

0.07

0.08

.:J O.oBQ..q,~ 0.D5ciIenkI..

0.04en~~

0IUCD

0.0.30:,en~~

0 0.02'(

+ Active powerTlmel Pmech-Constant~ Reactive power Cpu)

TRANSIENT STABILIT¥ ANALYSIS

~V-INDUCTION MOTOR MODEL

- 82 -

APPENDIX 8

HY: P. Mf·.. J<TEIJI'-!

X2R, X,0--- - - - --\,.-- -- .. --l-r -....·.,1 -- --- --- I -_._.- - -- -_.- I -- _. - - ..i .---- --- -' I - - - - .- - - .- j

_1_ _1_EB I I I I

1XMU I I I I

I I 1 I'--- 'I I

o-_·_-~-----------------'------------------·

F<2/S

ref 1.2: RECHENZEITSPARENDE NETZVEREINfACHUNG FUR TRANSIENTE UNIERSUCHUNIN NEIZEN MET UBERWIEGENDER MOIOR£NLAST

MOTOR: P(nom) . 1 .?O f<Pt1: 1500 f(:~.)m2: 330.

:U'VIR Mi~nlR M~(/Mr< COSF IF:: C()~3f IA Sr % 1~ENIIEt1Et··;

6.80 1 .40 2.:~() 0.06 () • ] 0 0 .. 60 O. 96

In p.u. (machine vermogen)

Rl Xl XMU R2' X2'0.01897 0.073529 3.29443 0.00617 0.20078

r~ 1,1 .:lnd XV .~He in p.u. on IOU ,'_jl)(1

EEl: 1 0.9 O.f3 O.'? o'. 6 (I • ',5

E f: = 1 pu .:) .. \:} puSr- % R2'/S 1;: I,) XV P 0 P Q

0.100 6.17198 68.316 180.349 0.OOlB4 0.00485 0.001:;,9 0.003930.20G 3.0<3599 1 1;3. ~502 :llO.359 0.00453 0.00440 0.0036'7 0.003570.300 2.0~'j73:J 102.264 6a .. 385 0.00676 0.00452 0.00547 O.OOJ660.400 1.5429S' 137.585 45.789 0.00897 0.0046";) 0.00726 0.003800.500 1.23440 '75.092 33.049 0.01116 0.00491 0.00904 O.0039B0.600 1.02\366 65.189 25.:36:'3 0.01332 0.OO~j18 0.01079 O.OO-UO0./'00 0.88171 57.379 20. -432 0.01547 0.00551 0.0J.253 0.01)·4-160.800 0.7'7150 ~51.14? 17.103 0.O17~)9 0.0051303 O.()l424 0.OO4'.7(i0.900 0.68~j78 46.0':15 14.758 0.01968 0.00630 0.01594 0.005101.000 0.61720 41. 9 :3~::i 13.0~':iO 0.02174 O.O067? 0.01'761 0.OO54B1.100 0.56109 38. 4~:;8 11.768 0.0237:3 0.00:?28 0.01926 O.OOS/_F)1 .. 200 0.51433 35.51~':i 10.7133 0.02~j?B 0.00783 0.020138 0.0063':11.]00 0.474/''J 32.994 10.010 ().0277~) O.()()1342 0.02~!:!El O.OOCiB21 .400 0.44086 30.Bl:3 9 .39:3 0.02969 0.00905 0.02-105 0.007331.500 0.41147 28. ~'I07 8.893 0.03160 0.00972 0.02560 0.0078'?1.600 0.38575 :~7. :~:3() 8.481 0.03348 0.01043 0.02712 0.008451.700 0.36306 25.'742 8 . 140 O. 03~j32 0.01117 O.02B61 O. OO"()~.:j

1 .800 0.34289 24. L1l4 ·7.B~)2 0.0:3712 O,()l194 0.03007 0.0096:·'1.900 (I. :::::~!484 23.221 :7 .608 O. «~88':) 0.01274 0.03150 0.010322.000 0.:30860 22.144 7.400 0.0406J O.013~)7 0.0:3290 0.01100

- 83 -

APPENDIX 9

LISTINGS FOR THE PROGRAM TRANSIENT STABILITY ANALYSIS.

TRASTA.LIS MAIN [ELE. 15. TRASTAJ

GEREAD.LIS SUB CELE. 15. TRASTAJ

LOREAD.LIS SUB CELE. 15. TRASTAJ

SALIEN.LIS SUB CELE.15.TRASTAJ

PLOTTSA.LIS SUB [ELE.1S.PLOTJ

TRASTAWO.LIS MAIN [ELE.15.WORK]

HVTENTSA.LIS SUB [ELE.1S.WORKJ

DYNLOAD.LIS SUB CELE.15.WORK]

INDUMOT.LIS SUB [ELE.1S.WORKJ

TRASTABI.LIS MAIN CELE. 15.WORKJ/CELE. 15.TRASTA]

TIME. LIS MAIN [ELE. lSJ

TSAHEAD MAIN [ELE. 15J

ANGLECALC.LIS MAIN [ELE. lS.WORK]/[ELE. lS.TRASTA]

TSA.COM COM CELE.1S.COM]

TRASTA.CMN SUB [ELE.1S.COMJ

CLPLK.LIS MAIN [ELE. lS]

MAIN Mainprogram. fileSUB Subroutin~ fileCOM Commandfile

- 84 -

SYSW000650. 0

TERNEUZEN APPENDIX 10

1. 70 6. 80 1. 40 2. 20 O. 86 O. 301.70 6. 80 1. 40 2. 20 O. 136 0.301. 70 6. 80 .1. 40 2. 20 O. 86 O. 301. 70 6. 80 1. 40 2.20 O. 86 o 30O. 48 6. 00 1. 10 2. 60 O. 82 O. 35O. 48 6. 00 1. 10 ~. 60 O. 82 O. 351. 70 6. 60 1. 40 2. 20 O. 86 O. 301. 70 6. 80 1.40 2. 20 O. 86 O. 301. 70 6 60 1. 40 2. 20 O. 86 O. 301. 70 6 80 1.40 2. ~o O. 86 O. 30O. 48 6. 00 1. 10 2. 60 O. 8~ O. 35

M.'1500 ;;> 1.01500 2 1 01500 2 1.01500 2 1.01500 ;;> 1.01500 2 1 01500 2 1.01500 2 1.01500 2 1.01500 2 1.01500 2 1.0

7. 6.0650. 1100.0800.0000.0550BUS NO78910111213141516TOTAL

.06506. 86.87.010. 0

TOTAL 47O. 60 O. 96 330O. 60 O. 96 330o 60 O. 96 330O. 60 O. 96 3301. 00 0 95 381 00 O. 95 38O. 60 O. 96 330o 60 0 96 330o 60 0 96 330O. 60 O. 96 3301. 00 0.95 38

SM 7 3S~I 4.9SM 4. 9

. 16002. 52. 51.08.7385.28052. 21, 1600.2400

1200.0000.0800

1. 642. 52. 51 087385

.28052. 211. 642. 04O. :; 1o 00O. 21

3 O. 0207 O. 04073 O. 0994 O. 92963 O. 0994 O. 92963 0.2219 1 20473 0.28;'10 1 ·15293 O. 2428 1. 40435 O. 1836 1 79405 o 1810 1.79085 0.21:i6 1. 39405 0.2156 1. 3940

O. 070 O. 000 0 020 O. 000O. 020 O. 016 0 010 O. 010O. 020 O. 010 0 005 0 005O. 0:58 O. 030 0 070 0 000o. 000 o. 000 o. 000 0 000O. 000 o 000 0 000 0 oeoo 000 O. 000 o. 000 o. 000o. 000 o 000 O. 000 0 000o. 000 O. oeo o. 000 o. 000O. 003 O. 006 O. 002 0 004O. 003 0.006 o 002 O. 004O. 003 O. 006 O. 002 O. 004O. 003 O. 006 O. 002 O. 004

INTERAKTIEr-G2 BUS XG2 .0205 2. eoo

DUS SA .0205 O. 10003 BUS XG3 0;:)05 2. ~OO

DUS SA .0205 O. 100G4 DUS SA .0170 1.009

LOAD .42 . 15G6 BUS 06 .0173 .7385

LOAD 12 .00PZ DUS 50KV .0010 .2085

BUS SA .0392 .5975LOAD 12 .093DUS 06 .0050 .5000

80 BUS 80 .0057 2.214LO,'D .08 .OS5BUS 50KV .0912 . 1548GENERATOR G2 1. 09 4. 9GENr:RATOR G2 2. 5 2. 5GENERATOR G3 2. 5 2. 5GENERATOR G4 1. 08 1. 08GENeRATOR G6 .7385 .7305GENERATOR PZ .2805 · ;>805GENERATOR 80 2.21 2.21GENERATOR G3 1. 09 4. 9GENF:RATOR G4 1 08 4. 2GENERATOR G6 1.00 O. 00GENERATOR PZ 1.00 O. 00GENERATOR 80 1. 08 4. 20

EXTRA IHISSENBUS 36ABUS 2111'1BUS 214nBUS 18Anus 34BUS 18nBUS IIIBUS 41\BUS 8/,nus 813CLWiP L.llt,Dnus 51\ 3BUS G6 4BUS 80 6DUS 36':!, 3nus 214A 3BUS 214B 3BUS 18,\ 3BUS 34 3BUS 180 3BUS 11\ 5BUS 4A 5BUS 8A 5BUS 8B 5I NOU. 110,..1MBUS5; 1 42~ll3tJS5; 2 43i1DUS5; 3 44MBUS5i4 438/4Ai 3 1439/4A;4 1440G101A 1541G301A 1542/813; 1 1643/8Bi2 1644/8B,3 16END LOAD

GENERATORBUS XG~

GENERATORBUS XG3GENERATORBUS SAGENERATORDUS G6OENE.RATCJRBUS 50lWAUS :iOIWBUS 501WGENERATORBUS 80BUS 80

INTERAKTIEFGENERATOR G2 BUS XG2 .0205 2. 500BliS XG~ BUS SA · 0~05 0 100GENERATOR G3 BUS XG3 · 0~05 2 500nus XG3 DUS SA .0205 O. 100GENERATOR G4 BUS SA .0170 1.069BUS SA LOAD .22 · 16GENERATOR G6 nus G6 .0173 . 73ASBUS 06 l.OAD · 14 · 10GENE.RATOR PI BUS 501W .0010 .2805BUS 50lW BUS SA · 039;! .5975BUS 501W LOAD .00 .00BUS 501W DUS 06 .0050 .5000GENERATOR 80 BUS 80 .0057 2. ~14BUS 80 LOAD .01 .01BUS 80 DUS 50KV .0912 1548

- 85 -APPENDIX 11

TRANSIENT STABILITY

DATE 10!-DEC-85 TIME 10:01:30

GENERATOR DATA

GENERATOR Xd Xd' Xll XII ' T'doGENERATOR QO! 2. 5000 2. 5000 O!. 5000 2. 5000 o. 8000GENERATOR G3 2. 5000 2. 5000 2. 9000 2.5000 0.8000GENERATOR G4 1.0800 1.0800 1. 0800 1.0800 7.0000GENERATOR Go O. 7389 O. 7389 0.7385 O. 7385 10. 0000GENERATOR 80 1.0000 1.0000 1. 0000 1.0000 7. 0000GENERATOR SM 1. 1000 1. 1000 1. 1000 1. 1000 8.0000

GENERATOR VOLTS ANGLE INERTIA MW MVAR RATINGGENERATOR G2 . 1. 1000 4.2400 1.0400 0.1500 0.0000 0.0000GENERATOR G3 1.0900 4. 1000 1.0400 0.1500 0.0000 O. 0000GENERATOR G4 1.0900 4. 1000 2. 0400 O. 1500 0.0038 O. 0000GENERATOR Go 1.0000 0.0000 O. 5100 O. 1030 0.0880 O. 0000GENERATOR 80 1.0900 4. 1000 O. 2100 O. 0800 O. 0550 0.0000GENERATOR 8M 1.0900 3. 8000 0.0000 -0.0000 0.0010 O. 0000

TRANSIENT STABILITY

DATE 10!-DEC-85

LOAD DATA

EXTRA H. V. BUS

TIME 10:01:30

PAGE 1

BUSNO BUSNAME GEN. NO R X7 BUS 30A 4 O. 0207 0.04078 BUS 214A 4 O. 0994 O. 92909 BUS 214B 4 O. 0994 0.9290

10 BUS 18A 4 0.2219 1. 204711 BUS 34 4 O. 2820 1. 492912 BUS 18B 4 O. 2428 1.404313 BUS lA. 0 O. 1890 1.794814 BUS 4A 0 O. 1810 1.790815 BUS 8A 0 0.2150 1.394010 BUS 6B 0 0.2150 1.3940

COMPOSITE LOAD

BUS GEN HV-BUS PMOT GMOT PCOMP GCOMP4 4 BUS SA O. 0000 0.0000 o. 0000 O. 0000

10 4 BUS 18A O. 0000 O. 0000 0.0000 O. 000011 4 BUS 34 O. 0000 O. 0000 O. 0000 O. 0000

8 4 BUS 214A O. 0000 0.0000 O. 0000 O. 00009 4 BUS 214B O. 0000 0.0000 O. 0000 O. 0000

12 4 BUS 18B O. 0000 O. 0000 O. 0000 0.00004 4 BUS 30 0.0000 O. 0000 0.0000 O. 00009 5 BUS 0 O. 0000 O. 0000 O. 0000 O. 00000 0 BUS 50KV O. 0000 0.0000 O. 0000 O. 0000

13 0 BUS lA 0.0000 0.0000 0.0000 O. 000014 0 BUS 4A O. 0000 0.0000 O. 0000 0.000015 0 BUS 8A O. 0000 O. 0000 O. 0000 O. 000010 0 BUS 8B O. 0000 0.0000 O. 0000 O. 0000

H. V. INDUKTIEMOTOREN

BUS GEN MOTOR MW Ia/Ir Ma/Mr Mk/Mr cosGr cosGa Sr N.tha H11 4 MPI01 1.090 O. 000 0.000 O. 000 O. 000 O. 000 O. 000 O. 000 O. 00011 4 MPI02 1. 090 0.000 0.000 0.000 O. 000 O. 000 O. 000 O. (JOO O. 000

9 4 M<;331 0.810 0.000 0.000 0.000 O. 000 0.000 O. 000 O. 000 0.0009 4 MP371B 0.390 0.000 0.000 0.000 0.000 O. 000 0.000 O. 000 O. 0009 4 MC131 0.390 0.000 0.000 0.000 0.000 0.000 O. 000 O. 000 O. 0009 4 MX113A O. 520 0.000 0.000 0.000 0.000 0.000 o. 000 O. 000 0.0009 4 MX113B O. 520 O. 000 O. 000 o. 000 0.000 o. 000 O. 000 o. 000 0.0009 4 MCI03 ? 1. 000 0.000 0.000 0.000 0.000 O. 000 O. 000 O. 000 0.000

10 4 MX112A O. 520 0.000 0.000 0.000 o. 000 0.000 0.000 O. 000 0.00010 4 MX112B O. 520 0.000 0.000 0.000 0.000 O. 000 0.000 O. 000 0.00010 4 MP371A O. 390 0.000 0.000 0.000 0.000 O. 000 0.000 0.000 O. 00010 4 MC231 O. 400 0.000 0.000 0.000 0.000 0.000 O. 000 0.000 O. 00013 o MPI03 1.090 0.000 0.000 0.000 0.000 0.000 O. 000 0.000 0.00013 o MPI04 1.090 O. 000 0.000 0.000 o. 000 O. 000 0.000 O. 000 O. 000

8 4 MC431 0.920 0.000 0.000 0.000 0.000 O. 000 0.000 O. 000 O. 0008 4 MP471 O. 360 0.000 0.000 0.000 O. 000 0.000 O. 000 O. 000 0.000

12 4 GBM890 O. 500 0.000 O. 000 0.000 O. 000 0.000 O. 000 O. 000 0.00015 o 7ME-18SA O. 550 0.000 0.000 0.000 0.000 o. 000 o. 000 O. 000 O. 000

- 86 -

, APPENDIX 12

PAGE 1~OB NUMBER

841055DIVISION

TERNEUZENDATE BY CHEC~ED BY

12-12-85 P.MARTEI~N P.M.FAULT BUS

SA

ROTORPOWER LOSS0.01010.01000.01950.07090.0198O. 1172

REFERENCE ANGLE

ANGLETIME GENERATOR DEGREES0.000 GENERATOR G2 19.570.000 GENERATOR G3 19.730.000 GENERATOR G4 11.380.000 GENERATOR G6 4.000.000 GENERATOR 80 7.770.000 GENERATOR SM 0.60

FREQUENCY 50.0000 CPS

0.018 GENERATOR G2 19.700.018 GENERATOR G3 19.860.018 GENERATOR G4 11.470.018 GENERATOR G6 4.090.018 GENERATOR 80 8.180.018 GENERATOR SM -3.71

FREQUENCY 49.9703 CPS

0.036 GENERATOR G2 20.070.036 GENERATOR G3 20.230.036 GENERATOR G4 11.750.036 GENERATOR G6 4.080.036 GENERATOR 80 9.410.036 GENERATOR SM -16.62

FREQUENCY 49.9026 CPS

0.054 GENERATOR G2 20.690.054 GENERATOR G3 20.850.054 GENERATOR G4 12.220.054 GENERATOR G6 3.870.054 GENERATOR 80 11.430.054 GENERATOR SM -38.16

FREQUENCY 49.8321 CPS

0.072 GENERATOR G2 21.570.072 GENERATOR G3 21.730.072 GENERATOR G4 12.870.072 GENERATOR G6 3.480.072 GENERATOR 80 14.260.072 GENERATOR SM -68.30

FREQUENCY 49. 7626 CPS

DELTAANGLE

0.00000.00000.00000.00000.00000.0000

O. 1243O. 12440.09330.09190.4183

-4.3071

0.37340.37370.2800

-0.01121.2221

-12.9164

O. 62240.62290.4667

-0.21102.0244

-21.5327

0.87140.8722O. 6534

-0.38372.8276

-30. 1413

O. 00990.0098O. 0194O. 12100.0221O. 1171

REFERENCE

0.00990.00980.0194O. 13790.02220.1173

REFERENCE

0.0100O. 00980.0194O. 13320.0222O. 1171

REFERENCE

0.01000.00980.0194O. 13320.0222O. 1171

REFERENCE

BUSREACTIVE VOLTS

0.6345 0.05110.6271 0.05081.2456 0.00000.6761 0.60250.4751 0.72771. 0171 0.0473

12.24742 DEGREES

0.6348 0.05060.6273 0.05031.2458 0.00000.7389 0.5617O. 4788 O. 72451. 0174 O. 0470

ANGLE 12.05478 DEGREES

0.6351 0.04990.6276 0.04961.2461 0.0000O. 7650 O. 54520.4792 0.72411. 0176 0.0468

ANGLE 11.42339 DEGREES

0.6347 0.05060.627:3 0.05031.2458 0.0000O. 7576 O. 54980.4788 0.72451. 0174 O. 0470

ANGLE 10.33573 DEGREES

0.6347 0.05060.6273 0.05031. 2458 O. 00000.7576 0.54990.4788 0.72451. 0174 0.0469

ANGLE 8.79721 DEGREES

FAULT REMOVED FROM GENERATOR G4 AT 0.0810

- 87 -

11111111 IIII 1111111111III 11111111 I ; 111111111 P1111 1111 II 11111111 II II

'1111111111~111111111111111:::::::::III::::II: II: '1111!:' 11111'111'1111111111111111111111111111111111 ' ,I III

'111111 1111 II III II II III iii iiI d 11111 iii iii 1111'1111111111111111111 C HE,. I r

"1111111111111: II1I i I~ ( I 1 ~ ~ [ I ! ; I

II:: 1111;1111111111111' APPENDIX 13\I 111111/ 111111111111111111'1111"

'. '1/'.11' '1IIIIIIIIIIIIIIIIIillllllllll lllll \

II' .. :( \: :./~: 1//1: III: I: I: I:: Illll: II: llllllil DIIIIIIIIIIII11111 L II 11/111 11111111111111111111111IT.rllllfi II 111111/

111111 II II II11 II I ! III II II II 1111111 III II III II II II II. III II II111111 II Iii II II II II III IIII II II II II II III i III II II II IIllllll 11111 II II II III; 11111111 III IIIII I; 1:.11 II II II t.~ II II111111 II !I II II II II IIIII II III II JIll 1111 I 1111 II IIlll

11III II1IIII II

11111 IIIIIIIII

IIlllP lIIIIIII

III IIII

III11111

~EAVE (HE

LEAVE THE PROGRA~ ('()?

OI;;:ECTORY ?

*****~~**,w*'**********".*"r~**"t, TRANSIENT STABILITY SIHULATION ,**'~'****'***'****.,.,*~'***.* ••*'**['(,T':: l~-MA (-86

you CAN SELECT ONE OF THE NE!T OPTIONS:

1. CALCULATE NEI~ VOL T,;GE:ANGLE~. SELECT SYSTEH INPUT !SC~EENlj. rF:,~S TA. I Ne' ON THE 5CF:EE~'~. RUN T~E ~ROGRAM rRAS1~.J. TR1~STA.OUT Orl THE SCREEN6. TRAST4.QUT ON THE SYSTEHPPINTER7. PLOTFILE.PRT ON THE SYSTEMPRINTERO. LIST OF PLOTFILES ON THE SCREEN9. FILE HANDLING

In. COMPILE/LINk/RUN/SUBMITH. HELPC. DIRECT LINE COMMAN[IR. RESTARTEo EX IT

TYPE 1. ~. 3... 9. R. H OR E I : 4

'******'***'*"**'*"**'*'**"*****'* TRANSIENT STABILITY SIMULATION *'*"***"************'******'*'*'***[lATE 12-MA'(-B6 TIME 15:25:50

DO YOU WANT TO CHANGE SYSTEH INPUT 'Yl? :

DO YOU WANT TO CRE~TE A NEW INPUT IY'? :

SYSBD002. HIP; 3SYSSH002.INP;4TRASTA.INP;271

DIRECTORY DATAO:CELE.15.TRASTAJ

SYS~.INP;~6 SYSBD001.INP;3SYSSHOOO.INP;4 SYSSH001.INP;6SYSTEM.INP;60 TRASTA.INP;272TRASTA.INP;~69

TOTAL OF ·13 FILES.

WHAT'S THE FILE YOU WANT TO USE? SYSSM001.INP

SYSBD'003. I NP; 2SYSTEM. INf';61TRASTA.INP;270

SYSSHOOl TERNEUZEN50.0

I NTERAKT I EFGENERATOR G2 BUS XG2 .0205 2.~0()[IllS XG2 [IUS SA .0205 0, J 00GENERATOR G3 BUS XG3 .0205 ~;. ~OO[IllS XG3 [lUS SA .0205 ).100BUS Sf' LOAO .4000 .1BOBGH1EI\'ATOR G4 BUS SA .0170 1.0B9GENERATOR G6 BUS G6 .0173 .73B5llUS G6 Ultl[l .1030 .OBBOGENERATOR f'Z BUS 50KV .0010 .2085[lliS 50kl) [IUS SA .0392 .5975IIUS 50~:1) LOAD .1400 .1400[IUS 50KV BUS G6 .0050 .5000GENERATOR 8D BUS B[' .0057 2.214BIIS 8[1 LOAD .OBOO .0550BUS 80 BUS SA .0912 .554BGENERATOR SH BUS 36 • 1111 1.000IIUS 36 BUS SA .0207 .1007DUS 36 LOMI .0300 .0300

GENERATOR G2 1.05 3.6 1. 64 1600 .0800GENERATOR G2 2.5 2.5 .) .. co 6.8.:.. ~ • .JGENERATQ.R G3 2.5 ., .. ., .. .. 6.8~'.J .... .,J • .JGOIEF:ATOR G4 1.0B 1. OB LOB .OB 7.0GENERATOR G6 .73B5 .73B5 .73B5 .73B5 10.0GENERATOR f'Z .:!BOS .2805 .2B05 .~B05GENERATOR BII 1.00 1.00 1.00 1.00 7.6GENERATOR SH 1.('10 1.00 1.00 1.00 B.OGENERATOR G3 1.05 3.6 1. 64 .1600 .OBI)()GFtlERAlOR 04 1. O~ :! ,El6 ~.04 .20(1) .1038llENFR,HIlR G6 I. ('10 0.00 0.51 .1030 .OB8()l1EtiERATOR F'Z 1 .00 0.00 0.00 .0:;00 • OB6~;I,EHERATlJR Btl 1.04 2.B6 0.21 .OBOO .0550«nlERATOR SH 1.04 2.Bo 0.06 -.060 .0010EXTRA II1ISS~ I~ flUS NllIllIS 3b,' " 0.0:'07 0.040 7 R....... " .....

- 88 -

APPENDIX 13

'_\JV..J

1.00 7.61.00 8.0.1600 .08t)().200() .1038.1030 .0880.0500 .0865.0800 .0550-.060 .0010

BUS NO89101 112131 41~1617

i:oo~1. 001. 642.040.510.000.210.06

1. 001.003.62.860.000.002.862.86

1. 001.001.051 .041.001.001. 04t .04

0.04070.92960.92961.20171.45291 • '10,1.\1.79461.79081.39401.314.)

62

.020~ 2.5Q()

.0205 0.100

.0:05 ~.50()• ('205 o.t 00.1000 .OROO.0170 1.089.0173.738501030 .0880.0010 .2085.0392 .':.975.1400 .1400.0050 .5000.0057 2.214.0350 .0260.0912 .5541'101111 1.000.0'207 .1007.0100 .0100

U tH Ut'·l'HfUGEGEVENS IN *u** ••• n***uF'. MAInE UNF'M:I!:I\ .M.nus SAYESB(l.012

EXTRA BUSSEN!HIS 36AIlIIS 211AIIIIS :'14[1HUS 1 BA"US 3·l[IUS 181l[IUS 1A[IllS 4A[IUS SABUS SEtEND LOAn

Ul:.I'Il:.KA 1UI-: 6[1GENEF:ATOR SMGF:NERATl1R G3GHIERATOR G4GENERA fOR G6GENERATOR PZI;EI~ERATOR 8[1('ENERATOR SM

4 0.02074 0.09'144 0.09<;'4" 0.2:1 Q

'l 0.28:'04 0.24266 O.18~66 0.1810b 0.21~66 0.2156

HITERAKTIEF1;[NEr.:,HOR G::' fillS Xli:~

!IlIS :<G2 11115 SAGENERAlllR 133 [IllS XG3~US XG3 BUS SA[IllS !::A LOAfII;U!EY{I1(If.: 04 fIllS SAOENERAfUR 136 flUS G6\lUS lib LlJA[1GENERATOR PZ BUS 50KVBliS 50~:IJ BUS SA«US 50t.'1) LilAD!:U~:: 50M' [ellS G6GENERATOR 8D BUS 8DBUS 8D LOADnUS an BUS SAGENER~TOR SM BUS 36[<1/536 flUS SA['lIS .56 LOAD

,**•••*****•••**.t LEES DE VOLGEHD**.****.**.******STUDY (IY

CHEO.rfl BYTIlE FAULT [(\)S ISADD.SWITCHING (YES/NO)NlI~l[tER OF UlTER'lflLSIHE nME HICRUIEtlT'liE FAUL f [IUS 15NUM~EI( or ItlTERVALSF.tHl OF ltlPUffWtl

[10 \Oll WANT TO RUN THE PROGRAM TRASTA (YIN) ?

CHANGE NINTIDT ? (YIN): YNEW NINT ? OLD= 8 : Q

NEW DT ? OLD=0.012 : 6.010

NEW NINT = 9 NEW DT 0.0100

FAULT BUS SA

CONNECTION WITH INFINITE BUS I

FAULT REMOVED FROM GENERATOR 134 AT 0.0850 SECONDSNEW NINT ? OLD=62: .

NEW NINT " 62

NO FAULT

DO YOU PLOT? (YIN)COMMENT (MAX 40 CHAR) •PLOT GENERATOR G2? (N):

NEW DT = 0.0100

YTRASTA; EXTRA INFORMATION

ANGLE GENERATOR G2 IS PLOTTEDPLOT GENERATOR G3? (N) : NPLOT GENERATOR G4? on: NF'LOT GENERATOR G6? nil: NF'LOT GENERATOR F'Z? (N) : NPLOT GENERATOR 8[I? (to:

ANGLE GENERATOR 8D IS PLOTTED

ANGLE GENERATOR SM IS PLOTTEDPLOT GENERATOR SM? (N):

ARCHIVE PLOTFILE ?(N) :Y

PROGRAM TRANSIENT STABILITY ANALYSIS COMPLETED

WAKE UP; THE PROGRAM IS ENDEfI 'I

ONE- LINE

Ok\!

L

58

US SA GS

(Xl

D.1.0

G

1 cb L511 5.. L

L: L

I ­I-

leA

L.. Load

G= generator

AM=osyn. motorSlvl .. syn. motor

BUS 50 kV

\0o

......Ul

I. V D w

o HWOI1nr

BUS 6

HWHval'

0.04 .......0:0' I ::JOO0:o, ,. 2000:o, • 200o:a' I 3000:-01 I :tOO0:0' I. :JOO

· :J2Q 1770· 0170 1.0••· DO" ::. ='1".2060 &100.0'91' . "".0.,2 .72.'0"0 .0:1'00

· DOlO . ~D8'·O~.' .,,71.1400 .1'00· 00'0 . '000.1111 I. lOS.02DJ 0"0'0300 0:1000

'''-Io-e,,. H.

I·,

LOAD l.O

• ".oss n.""'U.UN'00lUI SA 10CUI( .... 'o- 01 IUS leilUll Ie I lUI SACiN£JI"'DR C2 lUi 1'2lUI ICiI DUI IA;EN£A.. IDR C3 IUS .;2lUI Ie) I"" SAIUS ..... LD4DClNi."'DR C4 DUI IAUIIIIt"'OR .0 lUI eDlUI." LD4.IUS .0 1\11 IAC(""''''ElIlI.,GI'I •• OUS "6lUI., LD4D'.NI•• IQIII "' lUI 'DKVlUI 'QlIl.V I"" SAlui ,QlIl.V LD40tUS 30M" IUS Cl6C.....llotDR lit IUS ~.

lUI,). Dill SAlUll '). LbAD

14 HW

14 Hur

""I 9.0 HWS.l Hur

t"VGBO

12.21.'4 .1440 .90Iii. 3 •.•4 .....o.oe.o.=.:r '064 .1"60.00404. 10 • e' .I '''0 O,.~

• Iii 0 ill 0.00 'l""000900 .!'"0.00 0 00 '.~.". D. O' -. &.

i

11.6 I1W ~S.l Hur

14 I1W14 H.I'I"

CE:N(."'QIII 0'1 1.31C(N[.,llo1Dl11 ca 1.21CEH£."A'DIt 82 1.;00;[H[." 11M 0-. I 0'C[N(."IO* .0 I 0'CINE."" 11M C6 I. 00CIN[_AIGI' rl 1.00".NEIIA11M ... 1.0"

LOAD

lU HWU Hvar

G2

14.4 HW

8:9 Hnr

rv

G1

BUS XG1B_U_S-+-X_G_2 BUS ~G3 -!"B_U-.p~

LOAD!

j 8.l HW

1.1 Hvar

241 HW

BUS SA

BUS 36

147 Hnr

tHW

LOAD~ •••

LOAD

ANGLE240.

180.

120.

60.0

o .o

-120.

2.

TR AS T A: ALL LOA 0 t::: E EP 5' 0 N L HI ECONNECTION WITH INFINITE BUS

TIME.

x O. 1

-180.

FAULT CLEARED0. 3121121121 :5<__ 0. TRANSIENT STABILITY ANALYSIS

FAULT BUS SA :3 PHASE FAULTD~te~ 1-APR-S6 Time~11~49~41

BY P.MARTEIJN IN,SYSW0011

»-0-0IT1::zo.....x

ANGLE240.

180.

120.

60.0

60.0

120.

2.

TRASTA:NO LOAD ON SA DURING FAULTCONNECTION WITH INFINITE BUS

~~.. ~,4 • 6 • 8 • 1 Q1 • .;, l'~~.o' N. QI 1 6 • QI

". _sen/ -

,,~

G2

TIMEsec.

x: O. 1

""0ro-i

N

180.

FAULT CLEAREDQJ.3Q1Q1Q1 sac. TRANSIENT STABILITY ANALYSIS

FAULT BUS SA :3 PHASE FAULTD~te, 1-APR-86 Tim~,11,52,26

BY P.MARTEIJN IN,SYSW0011

»""0""0,.":zCJ.......X

AHGLE240.

SYSWOOOB/TRASTA/CONSTAHT MVA

-0ro-t

180. w

12 Cl.

60. (~

Cl

-60. (~

2 .. 8.

,/\. /" /'\ ....

...-/

)16. "

TIME

>< O. 1

\ __9~n G

)::a-0-0J'TI:zo.....x

TRANSIENT STABILITY ANALYSIS

FAULT BUS Gt :3 PHASE F.AULTD~t~,lS-MAR-86 Tim~,11,43,17

FA ULTC LEAR E[f------------------------,0. 3800 ~ec.

-1 8(~.

-1'20.

J:'. '0,' n M c.nT r Til.'

ANGLE240.

SYSWO~~6/WORK/AM; TTRIP=l.~ SEC.CONNECTION WITH INFINITE BUS

180.

120.

60.0

se.n G6G2~_ .. ge.en TIME0

0 2. 6. 10.0 12. 16. 0 lB'ec.

x D. 1

60.0

120.

):lo""0""0...,::zo......X

TRANSIENT STABILITY ANALYSISFAULT BUS G6 ;3 PHASE FAULTD~te;lg-MAR-86 Timer10 r 32r13BY P.MARTEIJN

F AUL TeL E ARE1..1-----------------------,QJ. 38 QJ QJ lB'ec.

180.

Ai'lGLE?4~1.

180.

SYSW0006/WORK/AMr TTRIP=1.0 SEC.

120.

(:1

6 (J. (:1

/I

;//

I'.l

.//

...~.'/

')c.. •

.~.,

,I

1 liJ .. 1:;:;1

..-..I

II

J,J

\...

T I ME-

120.

1 8 (:1.

F FI U L. TeL [ FI P E [I-.----_ ------------------,

QJ. 3 S f)1 QJ $' .....:;:0.. T P A '.J 9 I E H T S TAB I LIT Y A N A L Y SIS

FAULT BUS 66 :3 PHASE FAULT

Dot~,18-MAR-86 Tlm~r11r34,23

r;:. 'I.." r', r'1 ~ ,-., T r T , •.•

»-0-0rT1:z:o.....x

......m

ANGLE240.

DYN.LOAD. TTRIP~~.2 AM 2 T/M ~~ DISCONNECONNECTION WITH INFINITE BUS

180.

oee.TIME

>c: O. 116. 01'" • 012. 0

9li!n G•........9~n 029~n 8D

, gen G6

10. 08.6.'" .2.o

120.

60.0

60.0

120.

180.

FAULT CLEAREDlL 2259 sec. TRANSIENT STABILITY ANALYSIS

FAULT BUS SA ;3 PHASE FAULTD~t~,24-MAR-86 Tim~,0g,34,02

BY P.MARTEIJN IN,SYSW0006

):0\J\JfTl:zC......X

....0'1

ANGLE240.

DYN,LOAD. ~ITH TTRIP m ~,4 SECCONNECTION ~ITH INFINITE BUS

180.

sec.TIME

x O. 116. 01 -4 , 0

80

06

12. 0

se.... 02~_~.... n 04

10,Ql8,6,4 ,2,o

120.

60.0

60.0

120.

180.

FAULT CLEA~ED

0,2250 sec. TRANSIENT STABILITY ANALYSISFAULT BUS SA :3 PHASE FAULTD~te,24-MAR-86 Tlme,10,25,17BY P.MARTEI~N IN,SYSW0006

)::a-C-C/TI:zo.....x:

......O"l

ANGLE

240.

180.

SYS'~'0007 /OI'IL Y COrlPOSTTE LOAD

-0ra-l

120.

6(1. (21

o

60.0

1'20.

180.

FRULT CLERl=iEDB .. 225"5 sec., T1::: A IJ S I E I-J T S TAB I LIT Y A N A L Y SIS

FAULT BUS SA ~3 PHASE FAULT

DQte,14-nAR-86 Time,13,05,58

B Y P. n A I=!.T E I ~II'l

)::0-0-0IT1:zo.....x

......0'\

ANGLE240.

1- 80.

5Y5VO~~B/TPA5TA/DIF, VITH g.~~ (LOAD)

80

""0ro-I

120.

60.0

60.0

120.

2 . ... . 6. 8. 10.0

sen 80

TIME

)l( O. 1

180.

rAULT CLEAJ:lEDta. 2255 :Bec. TRANSIENT STABILITY ANALYSIS

rAULT BUS SA l3 PHASE rAULT

D~te~14-nAR-86 Tlme,09,10,32BY P.MARTEIJU

)::0""0""0JTl:z'='.....X

......Ol

240.

180.

120.

60.0

60.0

ANGLE

2.

SYSWOOOB~TRASTA/LOAD-~AFTER .226 SEC

.,-ge n 80

16. QJ

TIME

)C: O. 1

"'0ra-f

I-'aa

120.

180.

rRULT CLEAP,ED(3.2255 seo. TRAN$IENT STABILITY ANALYSIS

rAULT ~us SA :3 PHASE rAULT

Det~r14-nAR-86 Tlm~,0g,00,15

B Y P. 11 F1 R TEL J rJ

~"'0"'0rrl:zo.....x

I-'O'l

;~N GL E

240.

5'·.''3....,0005 ,J C On E: T ,.~ E (: ')ri POS I T E / Ttl Dun OT

"'t:lro--f

..........

2.

~ ~-----....... --~~===~-+-.:::::::._.....,.... ....-~--=:::::=--,-- --,-__~_.:::;::===--_....- -.--""--__--=-se n Tq~ E

6 • 8. 1 1~1. 1,/1 1 2 • 0 1 "' • 121 1 6 • 121 "",e c •

>0:0. 1

180.

120.0204

60. (21.....

80 0.....

60.0

120.

180.

FAULT CLEArlEDlL 2255 SE-:Q. TRANSIENT STABILITY ANALYSIS

FAULT PUS SA 13 PHASE FAULTDot.,14-MAR-86 Tlm.,13,1(21,26p y P. M Ft R TEl ~H'

::t:o"'t:l"'t:lrrI:zc......x

.....'"

FQOn PZEn <GEN P2)·ANGLE240.

180.

120.

60.0

60.0

120.

2, ... ,

SY ST En

6, 8, II,L 0 12.0 14. 0 16.0

TInE~ec.

>c: O. 1

t-'N

»""0""0..":zCJ......x

t-'oN

1 80.

FAULT CLEAQED., C.L2625 .&0,

GEN,' UN ST AS L~

TQANSIENT STABILITY ANALYSI9FAULT BU9 9A &3 PHASE FAULTD~t.,28-NOV-85 Tlm.,09,30r17BY P.MART t-l

ANGLE240.

SYS~O~~5)~ORK)ISLAND <1~~~.~388)

180.

-0ra-l

......w

12.0.

120.

......0'1

TRANSIENT STABILITY ANALYSISFAULT BUS SA ;3 PHASE FAULTD~tlii!>r 7-MAR-86 Timlii!>r13r28r28BY P.MARTEIJN

F AUL TeL EARE0.----------------------,0. 3686 eoac.

180.

240.

180 .

. 120.

ANGLE Gl DISCONNECTED.

TRANSIENT STABILITY ANALYSISFAULT BUS SA .3 PHASE FAULTD6t~~23-DEC-85 Time,11,47,17

BY P. nARTELJN

60.0

60.0

12(1.

1 8(1.

FftULT CLEAREDQJ. 082 5 :5Oe.c.

8. 1 QJ. '"

~90Ct 1'", sn

12 • QI 1 .... 0 16. 0

TIMEelite.

x O. 1

)::a""0""0IT1:zc.....><

......o~

I

240.

180.

120.

ANGLE ~UNNING IN ISLAND: G1 DISCONNECTED

TRANSIENT STABILITY ANALYSISFAULT BUS SA :3 PHASE FAULTD~t~,23-DEC-85 Tim.,13,31,05BY p.nARTEIJN

60.0

60.0

120.

180.

FAULT CLEARED0. 082 5 SOlCoC.

sen G2

10.121 12.0 14. 0 16.0TInE1518C.

)< O. 1

»""0""0rn::zo......x

ANGLE240.

180.

120.

60.0

oo

-60.0

2.

.35FDult,.

10.0

8"'" B

TInE••0.

)II O. 1

"ro-I

- 120.

-180.

TRANSIENT STABILITY ANALYSISStudy,Sf(V.

DDt.,30-0CT-85 TLm.,11,06,57By PDt~Lok nD~t.l ,.,

240.

180.

120.

ANGLE GEN 8M CONNECTED TO BUS 80

-0ro-i

......

.......

60.0

- 60.0

0.0 12.0 14. 0 16. (21

TInE••c.

)< O. 1

......o.......

- 1. 20.

-180.

rAULT CLEARED0.1050 sac. TRANSIENT STABILITY ANALYSIS

rAUlT BUS SA ;3.PHASE rAUlTDDt.,23-~AN-86 Tima,10,31,32

~-0-0JTl:zCI......x

BY p.nARTEI~U

tl/'I GL E

240.

180.

Tr:lASTA/SYNCHRON. nOTOR ... >:0' .. L QeOi'll" Eel I (11'1 ''r/ I 1 H T I'j F I I'l I 1 E 8 US

"0ra-I

......OJ

TRANSIENT STABILITY ANALYSIS

FAULT BUS SA ;3 PHASE FAULT

D~t~,10-nAY-86 Tim.r13,04r16

B Y P. MAR TEl .J IJ I H , S YSSM (113 1

/\120.

60.13

6'=.7.1.0

120.

180.

~(

" __ ~-:l.~n1_------ /"'-~-""'-'3 <=:- ..-,

-------,~-;;.~rr-f-- _/'. --/' 'I

'fl \ 2 .1 \ 4. I J'. 18 .

\ j \) \j\, J '--'

F N,I:J LTC LEA ~1 E D(1. (is ~5 ~.~C.

8D

•12.11) 14.·21 16. QJ

1 I rl E~ec ..

>0:0. 1

)::0"0"0/TI:zo......x

......aOJ

'240.ANGLE TRASTA/SYN.MOT.,XD' ~ 2.0

C Or'l NEe T I ON '~.n T H H~ FIN I T E 8 US

t 8fL

­o1.0

TIME

xO. 116 • (I)

I

14. 121I

12.121

E:'JD

8M

l;-----+--.--+---+-.-----+-.-------1f-~.------.--

1/) \ 2. II R\ \ -"

\\"~9e: n

t 2: 10.

12 (1.

60.0

18(1.

FAULT CLEAnED0. 1 1 ~i 5 3'~C. TRANSIENT STABILITY ANALYSIS

FAULT LUS SA ;3 PHASE FAULT

D~ t <::: r t (1- nAY - 86 T i. m~ r 1 3.' 1 5 r 52

BY P. MARTEl_IN IIlrSY9SM001

ANGLE

240.TRAS'TA/S''i'N. MOT .. PEACTAI'ICE 35 -6A:2. iJCONNECTION WITH INFINITE BUS

180.

"'t:Jro--i

No

1 2 f'J.

............o

:::'10>0.

lInE

x 0 .. 1113. 12112 .. @

G48DG6

1 (1, 0/6.} \) \

\ )1,•....-'

F'=..;-.-:r-=JI---+-r-------1P-'l""'--+---r+---~-----r-----.- -----,.----­1 ·L C1~.l \ 2,/

\ I\ I

I J\)

6(1.0

120.

60.\)

180.

FnUL T CLEAJ:!ED<:J. 1 (,"14 5 t:'fe.c. TI:(HIl9IENT 8TFtBILITY AIlALYSIS

FAULT BUS SA :3 PHASE FAULT

Oot~,10-MAY-86 Tlm~,13,09,11

BY P.MARTEIJN IN,SYSSMG01

;x:..."'t:J"'t:JfT1:zo-­x

......O'l

240.

180.

12(1.

ANGLE TRASTA/SYNCHR MOTOR ON BUS SA ,XO' -1.~CONNECTION WITH INFINITE BUS

In E"-.1 ,/\ ; 0.\ 'l.. PL-\""Cro-i

N......

16 • Q)

60.0

120.

so

k--+--+--r----;---+-r---'l;---+-..---+--r----r-,--------,,-----'.---------T----1,8. 1(1,\,) 12.0 14.12)Lgen 8M

TIMEscec ..

x O. 1

.....

......

......

T~iAI,J.sIEI_lT STABILITY AIJALYSIS

FAULT BUS SA ;3 PHASE FAULT

D~~.,10-MAY-86 Tim~,12,43,38

FFtUL T CLEAPED(1. filS2 =, ,-=, ...0.

BY P. MARTE I_Jl'1 I 1,1, BY SSM';:'01

):­""C""C/T1:zC.....x

......en

'..:.' .:J (I.

1 EO.

12(:1.

1=1 1'1 G L F -J [';, ,=, ·.··il ,:, .' c: [['j <?,':; ;J i I ::: L J:~,' :::0 H

(:IJr'li'IEt,111=:Ofl './11H l!j:rr'jITE

-0ro-I

NN

(,;. o. (1

1.:.0. ,':1

TInE

........

........N

1 :: C: .

FilUL I CL£FIRE.:DG. 1(145 s.",,~,;,. T1=,AfISIEf'IT STAEILITY AI-IALY5IS

tF1ULT £'-1980 :3 PHASE tAULT

:x:­-0-0fT1:zo.....x

P,( P. I1APTE I·_IN I II ~ BY SSt10CH........O'l

ANGLE240.

180.

SYSWIOG6/WORK/GEN 2, XO' -1.00

""0rC>--i

NW

120.

613.0

60.0

20.

............W

1 80.

FAULT CLEA~IED

Ql.231QJ ,-"eo. TRANSIENT STABILITY ANALYSIS

FAULT BUS SA :3 PHASE FAULT

[It:JL'''''' 14 - IIAR- 86 T i m&{ 15 .. 55 .. 20

FY P. MhRTE I Jt-j

):0""0""0l'TI:zCI......X

......0\

ANGLE240.

8Y8WO~~6/TRA8TA/D2 CON./8EE WO;12MA/11.3

\5I<ec.

TIME

)<; O. 116. QI

sen 80

o

t-dt"'"0

180. t-i

N.p-

120.

60.0

60.0

120.

180.

FAULT CLEAREDQJ. 2 3 1 QJ \5I<e c • TRANSIENT STABILITY ANALYSIS

FAULT BUS SA ~3 PHASE FAULTD=te,13-MAR-86 Tim~,08,50,22

BY P. MARTEI •.m

ANGLE

240.

180.

120.

60.0

60.0

20.

2.

SYS\.y'O~~fj.!GEN 80; Y.D' = 1. 21 PU

Iii 'Sj

4. 6. 8. 10.0 12.(2) 14.0 16.0

G2l.' ~

lIME

x O. 1

""0.­a--l

N(Jl

180.

FAULT CLEAJ:fEO0. 2 5 3 5 ece c • TRANSIENT STABILITY ANALYSIS

FAULT BUS SA ;3 PHASE FAULT

Dt:ltoa..·17-I1AR-86 Ti.m .... 1'0gr30r46

Bl' P.MARTEIJN

:x:­""0""0(T'l:zo.....x

240.

180.

At/GLE 'w'OQK / 11251 ON 512 KV /AMr 2-0.6CONNECTION 'w'ITH INFINITE BUS

""Cra-I

N0'1

120.

60.0

""-9"''''' G2~9 .... n l.i 4

TIrlE

..........0'1

60.0

120.

2. 4. 6. 8. 12.0 1 4 • QJ 16. QJ sec.

)<0. 1

180.

FAULT CLEARED0."2850 sec. TRANBIEHT STAB'ILITY ANALYSIS

FAULT BUS SA ;3 PHASE FAULTD~t ... , 2-APR-86 Tlm... ,16,26,18BY P.MARTEIJN IN,SYSW0011

:l;o""C""CITI:zo.....x

.....0'1

ANGLE240.

VO~K / AM;2-~.5 GEN 1051 ON 5QKVCONNECTION WITH INFINITE BUS

180.

\Jr­o--i

120.

60.0

121

60.0

2. 12. QJ 16. QJ

Jsen 06

\.

TIMEsec.

x O. 1

......

......""-J

120.

......0'1

»\J\J/Tl:zCI......X

TRANSIENT STABILITY ANALYSISFAULT BUS 06 ~3 PHASE FAULTD~te:, 2-APR-86 Time,16,13,00BY P.MARTEI..JU IN,SYSW0011

FA UL TeL E ARE t-L---------------------,QJ. 399121 sec.

180.

ANGLE24'21.

18'21.

12'21 .

NO PZ: GEN 1~51 ON 5~KV (SYSWO~~B)

NO CONNECTION WITH INFINITE BUS

""t:Iro--t

NCO

6'21.'21

6'21.'21

12'21.

18'21 .

FA UL T C L~::l..L.l..I:-L"",---- --,

'21.54'210 a TRANSIENT STABILITY ANALYSISFAULT BUS SA ;3 PHASE FAULTD~te,27-MAR-86 Time,11,eJ7,llBY P.MARTEIJN IN,SYSWOeJl1

TIMEl'ec.

>< O. 1

)::a""t:I""t:IfT1::zo......X

............CO,