small signal Disturbances

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chapter 3System Response to Small Disturbances

3.1 IntroductionThis chapter reviews the behavior of an electric power system when subjected tosmall disturbances. It is assumed the system under study has been perturbed froma steady-state condition that prevailed prior to the application of the disturbance.This small disturbance may be temporary or permanent. I f the system is stable, wewould expect that for a temporary disturbance the system would return to its initial

state, while a permanent disturbance would cause the system to acquire a new operatingstate after a transient period. I n either case synchronism should not be lost. Undernormal operating conditions a power system is subjected to small disturbances at ran-dom. I t is important that synchronism not be lost under these conditions. Thus systembehavior is a measureof dynamic stability as the system adjusts to small perturbations.We now define what is meant by a small disturbance. The criterion issimply thatthe perturbed system can be linearized about a quiescent operating state. A n exampleof this linearization procedure was given in Section 2.5. While the power-angle rela-tionship fora synchronous machine connected to an infinite bus obeys a sine law (2.33),it was shown that for small perturbations the changein power is approximately propor-tional to the changein angle(2.35). Typical examples of small disturbances are a smallchange in the scheduled generation of one machine, which results in a small change inits rotor angle 6, or a small load added to the network (say 1/100 of system capacityor less).I n general, the response of a power system to impacts is oscillatory. I f the oscil-lations are damped, so that after sufficient time has elapsed the deviation or the changein the state of the system due to the small impact is small (or less than some prescribedfinite amount), the system isstable. I f on the other hand the oscillations grow in magni-tudeor are sustained indefinitely, the system is unstable.For a linear system, modern linear systems theory provides a means of evaluationof its dynamic response once a good mathematical model is developed. The mathe-matical models for the various components of a power network will be developed ingreater detail in later chapters. Here a brief account is given of the various phenomenaexperienced n a power system subjected to small impacts, with emphasis on the qualita-tive description of the system behavior.

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54 Chapter 33.2 Types o f Problems Stud ied

The method of small changes, sometimes called the perturbation method [1.2.31,is very useful in studying two types of problems: system response to small impacts andthe distribution of impacts.3.2.1I f the power system is perturbed, it will acquire a new operating state.

System response to small impactsIf theperturbation is small, the new operating statewill not be appreciably different from theinitial one. I n other words, the state variables or the system parameters will usuallynot change appreciably. Thus the operation is in the neighborhood of a certainquiescent statex o . I n this limited range of operation a nonlinear system can be de-scribed mathematically by linearized equations. This is advantageous, since linear sys-tems are more convenient to work with. This procedure is particularly useful if the

system contains control elements.The method of analysis used to linearize the differential equations describing thesystem behavior is to assume small changes in system quantities such as b,, u,PA(change in angle, voltage, and power respectively). Equations for these variablesare found by making a Taylor series expansion about x o and neglecting higher orderterms [4 ,5 ,6 ] . The behavior or the motion of these changes is then examined. In ex-amining the dynamic performance of the system, it is important to ascertain not onlythat growing oscillations do not result during normal operations but also that the oscil-latory response to small impacts iswell damped.I f the stability of the system is being investigated, it is often convenient to assumethat the disturbances causing the changes disappear. The motion of the system is thenfree. Stability is then assured if the system returns to its original state. Such behaviorcan be determined in a linear system by examining the characteristic equation of thesystem. I f the mathematical description of the system is in state-space form, i.e., i fthe system is described by a set of first-order differential equations,

(3.1)= A X +BUthe free response of the system can be determined from the eigenvalues of theA matrix.

3.2.2 Distribution of power impactsWhen a power impact occurs at some bus in the network, an unbalance betweenthe power input to the system and the power output takes place, resulting in a transient.When this transient subsides and a steady-state condition is reached, the power impactis shared by the various synchronous machines according to their steady-state char-acteristics, which are determined by the steady-state droop characteristics of the variousgovernors [5 ,7 ] . During the transient period, however, the power impact is shared bythe machines according to different criteria. I f these criteria differ appreciably amonggroups of machines, each impact is followed by oscillatory power swings among groupsof machines to reflect the transition from the initial sharing of the impact to the finaladjustment reached at steady state.Under normal operating conditions a power system is subjected to numerous ran-dom power impacts from sudden application or removal of loads. As explained above,each impactwill be followed by power swings among groups of machines that respondto the impact differently at different times. These power swings appear as power oscil-


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System Respons e to Small Disturbances 55lations on the tie lines connecting these groups of machines. This gives rise to the termtie-line oscillations.In large interconnected power systems tie-line oscillations can become objectionableif their magnitude reaches a significant fraction of the tie-line loading, since they aresuperimposed upon the normal flow of power in the line. Furthermore, conditions mayexist in which these oscillations grow in amplitude, causing instability. This problemis similar to that discussed in Section 3.2.1. It can be analyzed if an adequate math-ematical model of the various components of the system is developed and the dynamicresponse of this model is examined. I f we are interested in seeking an approximateanswer for the magnitude of the tie-line oscillations, however, such an answer can bereached by a qualitative discussion of the distribution of power impacts. Such a discus-sion is offered here.3.3 The Unregulated Synchronous Machine

We start with the simplest model possible, i.e., the const ant - vol t age- behi nd- t r an-sient-reactance model. The equation of motion of a synchronous machine connectedto an infinite bus and the electrical power output are given by (2.18) and (2.41) re-spectively or

P, = Pc +PMsin(6- y) (3.2)(3.3)Letting 6

= 60 +6 ~ ,e = P,o+ P A, P,,, = Pm0and using the relationshipsin(6 - y ) = sin(& - y +6,) sin(& - y) + cos(6o -the linearized version of (3.2) becomes

where(3.4)

The system described by (3.4) is marginally stable (Le., oscillatory) for P, >0.Its response is oscillatory with the frequency of oscillation obtained from the roots ofthe characteristic equation (2H/wR)s2+P, = 0,which has the rootss =& j d P , w R / 2 H (3.6)

I f the electrical torque is assumed to have a component proportional to the speedchange, a damping term is added to (3.4) and the new characteristic equation becomes(2H/wR)s2 + (D/wR)s + P, = 0 (3.7)whereD isthe damping power coefficientin pu.The roots of (3.7) are given by

(3.8)


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56 Chapter 3Usually (D/wR)2 0 and for D >0. I f eitherone of these quantities is negative, the system is unstable.

Venikov [4] reports that a situation may occur where the machine described by(3.4) can be unstable under light load conditions if the network is such that t J o < y.This would be the case where there isappreciable series resistance (see[4],Sec. 3.2).From Chapter 2 we know that the synchronizing power coefficient P, is negativeif the spontaneous change in the angle 6 is .negative. A negative value of P, leads tounstable operation.3.3.1The model of constant main field-winding flux linkage neglects some importanteffects, among them the demagnetizing influenceof a change in the rotor angle 6. Toaccount for this effect, another model of the synchronous machine is used. It is notour concern in this introductory discussion to develop the model or even discuss it indetail, as this will be accomplished in Chapter 6. Rather, we will state the assump-tions made in such a model and give some of the pertinent results applicable tothis discussion. These results are found in de Mello and Concordia [8] and are basedon a model previously used by Heffron and Phillips (91. To account for the field con-ditions, equations for the direct and quadrature axis quantities are derived (see Chap-ter 4) . Major simplifications are then made by neglecting saturation, stator resistance,and the damper windings. The transformer voltage terms in the stator voltage equa-

tions are considered negligible compared to the speed voltage terms. Linearized rela-tions are then obtained between small changes in the electrical power Pea, the rotorangle a the field-winding voltage uFArand the voltage proportional to the mainfield-windingflux EA.For a machine connected to an infinite bus through a transmission network, thefollowings domain relations are obtained,

Demagnetizing effect of armature reaction

Pea = K16A + &EA (3.9)(3.10)

whereK, s the change in electrical power for a change in rotor angle with constantflux linkage in the direct axis, K 2 is the change in electrical power for a change inthe direct axis flux linkages with constant rotor angle, ri0 is the direct axis open cir-cuit time constant of the machine, K 3 is an impedance factor, and K4 is the demag-netizing effectof a changein therotor angle (at steady state). Mathematically, we writeK t = PeA/6AlEb=0 K 2 = peA/E;16~-0

K 3 = final valueof unit stepu, response = lim Ek(t)]6A-oI - -

v~1- 01 A - U ( I )1K 4 = - - im EA(r)K3 1-m (3.11)The constants K I ,K 2,and K 4 depend on the parameters of the machine, the exter-nal network, and the initial conditions. Note that K, s similar to the synchronizingpower coefficient P, used in the simpler machine model of constant voltage behind


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System Response to Small Disturbances 57

Fig. 3. Pr imitive linearizedblock diagram representationof a generator model.

transient reactance. Equations (3.9) and (3.10), with the initial equation (3.2), may berepresented by the incremental block diagramof Figure3.1.

For the case whereV = 0,(3.12)

(3.13)where we can clearly identify both the synchronizing and the demagnetizing compo-nents.Substituting in the linearized swing equation (3.4), we obtain the new characteristicequation,(with D = 0)

[ Z s ' +(K, -or we have the third-order system

s2 +!Q!K.$ +- IR (K, - K2K3K4)= 01 2H 2H K3Td0s3 +-36 0 (3.14)Note that all the constants(3.1 1) are usually positive. Thus from Routh's criterion [ IO ]this system is stable if K, - K2K3K4> 0 and K2K3 K4> 0 .The first of the above criteria states that the synchronizing power coefficient K,must be greater than the demagnetizing component of electrical power. The secondcriterion is satisfied if the constants K2, K3, and K4 are positive. Venikov [4] pointsout that i f the transmission network has an appreciable series capacitive reactance, it ispossible that instability may occur. This would happen because the impedance factorproducing the constant K, would become negative.

3.3.2 Effect o f small changes of speedI n the linearized version of (3.2) we are interested in terms involving changes ofpower due to changes of the angle6 and its derivative. The change in power due to


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58 Chapter 36, was discussed above and was found to include a synchronizing power componentand a demagnetizing component due the change in EL with 6,. The change in speed,W, = dsA/dl,causes a change in both electrical and mechanical power. In this casethe new differential equation becomes

(3.15)As in (3.7) the changein electrical power due to small changes in speed is in the formof

PL = (D/WR)WA (3.16)From Section2.3the change in mechanical power due to small changes in speed is alsolinear

PmA = apm/awlw ~W A (3.17)wherei3Pm/dw],, can be obtained from a relation such as the one given in Figure 2.3.I f a transient droopor regulation R is assumed, we may write inpu to the machine base

P mA = - ( ~/ W W A / W R ) PU (3.18)which is the equationof an ideal speed droop governor. The system block diagram withspeed regulation added is shown in Figure3.2.

I-,L.Fig. 3.2 Block diagram representationof the linearized model with speed regulation added.

The characteristic equationof the system now becomes

or(3.19)

+ - D +- +K iK37 ; o s +(K i - KZK3K4) = 0 (3.20)[R ( 1


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System Response to Small Disturbances 59Again Rouths criterion may be applied to determine the conditions for stability. Thisis left asanexercise (see Problem3.2).3.4 Modes of Oscillation of an Unregu lated Mul timachine System

The electrical power output of machine in ann-machine system is

where 6 =Ei =y..=y . . =

--

nPei = E:Gii + Ei Ej Yijcos(eij - 6,j)j - I

j +in

= E:Gii + EiE,(Bijsin6, +Gijcosaii) (3.2 )j - Ij + i

Si - 4constant voltage behind transient reactance for machineiG,i +jBii is a diagonal element of the network short circuit admittancematrix YGu +jBu is an off-diagonal element of the network short circuit admit-tancematrix YUsing the incremental model so that 6, = 6 +bijA, we compute

sin6, = sin Sij0 cosSijA +cosS,jo sin SijA Y sin ti i jo+6ijAcosSij0cos6, cosSij0 - a sin6

Finally, for PciA,n

P,, =C E,E,(B, COS a,, - Gijsin 6i jo)6 i jAFor a given initial condition sin Sijo and cosbij0are known, and the term in parenthesesin (3.22) isaconstant. Thus we write

(3.22)j - Ij 4 i

npeiA =Cj - I

j z iwhere

Psij s] = Ei E j(Bijcos6 - Gi jsin6ijo)8% d i jo

(3.23)

(3.24)is the change in the electrical power of machine i due to a change in the angle betweenmachines i and j , with all other angles held constant. Its units are W/rad or pupower/rad. It isa synchronizing power coefficient between nodes i and and is identicalto the coefficient discussed in Section2.5.2for one machine connected to an infinite bus.Wealso note that since (3.21) applies to any number of nodes where the voltagesare known, the linearized equations(3.22)and (3.23)can be derived for a given machinein terms of the voltages at those nodes and their angles. Thus the concept of the syn-chronizing power coefficients can be extended to mean the change in the electricalpower of a given machine due to the change in the angle between its internal EMF and


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60 Chapter 3any bus, with all other bus angles held constant. (An implied assumption is that thevoltage at the remote bus is also held constant.) This expanded definition of the syn-chronizing power coefficientwill be used in Section 3.6.Using the inertial model of the synchronous machines, we get the set of linearizeddifferential equations,--Hi d26iA +2 iEj(B,~os6,i,- G,sin6,0)6,A = 0 i = 1,2,. . n (3.25)WR dt2 j - 1j t ior

(3.26)j t i

The set (3.26) is not a set of n-independent second-order equations, sinceZb, = 0.From (3.26) for machine i ,Thus (3.26) comprises a set of ( n - I)-independent equations.

nPS,6,, = 0 i = 1,2,..., (3.27)

j t i

Subtracting thenth equation from the i th equation, we computen - I

2Hn j - 1 (3.28)-- j t iEquation (3.28) can be put in the form

Since6..J A = 6nA - (3.30)

(3.29) can be further modified asn - I

dt2 j - I+ c ,ajnA= o i = 1,2,..., - I (3.31)where the coefficientsaiidepend on the machine inertias and synchronizing power co-efficients.Equation (3.31) represents a set of n - 1 linear second-order differential equationsor a set of 2(n - 1) first-order differential equations. We will use the latter formula-tion to examine the free response of this system.Let xl , x2,.. .,xn- be the angles a InA ,&,,A,. . . c$,-,),,~ respectively, and letx,, . . x~,, -~e the time derivatives of these angles. The system equations are of theform


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System Response io Small Disturbances

X Ix2

xn-%+I...::2n- 61

or

0_ _ _ _ _ _ _ - - -...12

A 224 1 . 2

.... . . . . .

I 1 0 ... 0-1 0 1 ... 0I1. . . . . . . . . . . .IIIIIIIIII

I O 0 * a 1+-- - - - - - - - - - - - - -

0.I!

(3.32)*

(3.33)where U = the identity matrixX I = then - 1 vector of the angle changes6,,,X 2 = then - 1 vector of the speed changesdb,,,,/dt

To obtain the free response of the system, we examine the eigenvalues of the charac-teristic matrix [ l l , 121. This isobtained from the characteristic equation derived fromequating the determinant of the matrix to zero, as follows:- XU I Udet _ _ _ _ _ _ _ _ (3.34)[ A ; - Xu ]=detM =O

whereX is the eigenvalue. Since the matrix -XU is nonsingular, we compute the de-terminant of M asIM I = I -XU I I (-XU) - A(-XU)-U I

= ( - l ) - X - I -XU - (-1/X)-IA I = IX2U - A I (3.35)(See Lefschetz [121,p. 133.) The system described by IM I = 0. or I X2U - A I = 0,has2(n - 1) imaginary roots, which occur in n - 1 complex conjugate pairs. Thus thesystem hasn - 1 frequencies of oscillations.Example 3.1lated and classical model representationisused.Solution

Find the modes of oscillation of a three-machine system. The machines are unregu-

For an unregulated three-machine system, the system equations are given by

*Seetheaddendumon page650.


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62 Chapter 3Multiplying the above three equations by w,/2H, and subtracting the third equationfrom the first two, we get (noting that 6 = - a j i )

To obtain the eigenvaluesof this system, the characteristic equation isgiven by

det

Now by using(3.39,

-all - a12-a21 -a22

h2 +alldet[ a2 1

= o

I f we eliminate by noting that + + =0, the following two equationsare obtained:

or

T he state-space representation of the above system is


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System Respons e to Small Disturbances 63Examining the coefficientsa i i ,we can see that both values of Xz are negative realThe free responsewill be in the form 6, = C , cos (B r +&) +Czcos ( y t +c$~),quantities. Let these given values beX = i ja,whereC,,C,,

X = f y .and& are constants.

Example 3.2Consider the three-machine, nine-bus system of Example2.6, operating initially inthe steady statewith system conditions given by Figure 2.18 (load flow) and the com-puted initial values given in Example 2.6 for Ei/66, i = I , 2, 3. A small IO-MW load(about 3% of the total system load of 315 M W ) is suddenly added at bus 8 by adding athree-phase fault to the bus through a 10.0pu impedance. The system base is 100M V A .Assume that the system load after t = 0 is constant and consists of the original loadplus the I Opu shunt resistance at bus8.Compute the frequencies of oscillation that will result from this small disturbance.Then compare these computed frequencies against those actually observed in a digitalcomputer solution. Assume there are no governors active on any of the three turbines.Observe the system response for about two seconds.

SolutionFirst we compute the frequencies of oscillation. From (3.24)Psij = V, %(Si/cos6 - G sin a V,5B i , cos6,,

From Example2.6 we find the data needed to compute Psij with the results shown inTable3.1.

Table3.1. Synchro nizing Pow er Coeff ic ients of t he Ne t wor k of Example2.6Ij v vi Bij 4jti psi,12 I .0566 I .OS02 1.513 - 7.4598 1.601523 I .0502 1.0170 1.088 6.5563 1.154431 1.0170 I .0566 1.226 10.9035 I .2936

Note that the 6, are the values of the relative rotor angles at I = 0-. Since theseare rotor angles, theywill not change at the timeof impact, so these are also the correctvalues for t = O+. This is also true of angles at load buses to which appreciable inertiais connected. For loads that are essentially constant impedance, however, the voltageznglewill exhibit a step change.Also from Example 2.6 we know H i = 23.64, 6.40, and 3.01 for i = I , 2, 3 respec-tively. Thus we can compute the values of a i j from Example3.1 as follows:= (WR/2)(Ps12/HI +Ps13/H, +Ps31/H3) = 104.096

a 1 2 E (% /2)(Ps3z/H3 - Psiz/Hi)= 59.524a21= (0R/2)(Ps3& - Psz1/H2)= 33.841a22= (oR/2)(Ps2,/H2+Pa3/H2 +Ps32/H3)= 153.460

Then= -(1/2)[+1, +a 2 z ) d(a11+azz ) - 4(alI a22 - at~l ) ]= -(1/2)[-257.556 f (66336 - 55841)*] = -77.555, -180


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64 Chapter 3Now we can compute the frequencies and periods shown inTable 3.2.

Table32 Frequenciesof Oscillationofa Nine-Bus SystemQuantity Eigenvalue I Eigenvalue 2x 2 8. 807 kj 13. 416o rad/s 8.807 13.416f Hz 1. 402 2. 135Ts 0. 713 0. 468

Thus two frequencies, about 1.4 Hz and 2.1 Hz, should be observed in the inter-machine oscillations of the system. This can be approximately verified by an actual so-lution of the system by digital computer. The results of such a solution are shown inFigure 3.3, where absolute angles are given in Figure 3.3(a) and angle differences rela-tive to 6, are given in Figure 3.3(b). As might be expected, neither of the computedfrequencies is clearly observed since the response is a combination of the two frequen-cies. A rough measurement of the peak-to-peak periods in Figure 3.3(b) gives periods inthe neighborhood of 0.7 s.

Methods have been devised [3,1 ] by which a system such as the one in Example 3.2can be transformed to a new frame of reference called the Jordan canonical form. I nJ ordan form the different frequencies of oscillation are clearly separated. I n the form ofequations normally used, the variables 6,, and a,, (or other angle differences) contain

""CI4.0 I

-97.0 I 1 I I I0.0 0.500 1.000 1.500 2.000 2.500Time, I(a)

8.01 I I I 1 I 10.0 0.500 1.000 1.500 2.000 2.500Time, s

(b)Fig. 3.3 Unregulated response of the nine-bus system to a sudden load application at bus 8: (a) absoluteangles, (b) angles relative to 6 1 .


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System Response to Smal l Dis turb ances 65harmonic terms generally involving all fundamental frequenciesof oscillation. Hencewe have difficulty observing these frequencies in measured physical variables.Example3.3that in this form the system frequencies of oscillation are clearly distinguishable.Solut ion

Transform the system of Example 3.2 into the Jordan canonical form and show

The system equations for the three-machine problem are given by

or i = A x , wherex is defined by

. -

0

and thea coefficients are computed in Example3.2.vectorsE ,, E,, E ,, and E4. We then use these eigenvectors to definea matrix E .We now compute the eigenvectors of A , using any method [ I , 3, 111 and call these

-j 0.06266 i 0.14523 ; -0.145231 0.83069 ,1.00000 1. OOOOO1-j 0.07543 i -0.13831 0.13831I1.OoooO j

where the numerical values are found by a suitable computer library routine.j , = E - A E y = D y whereD = diag(X,,X,,X,,X,).

1.oooOO ! -0.95234 I -0.95234E = [E , E2 E E41 =

We now define the transformation x = E y to compute2 = E i = A x = A E yPerforming the indicated numerical work, we compute

or

-j3.5245 -j3.7008 0.2659 0.2792j3.5245 j3.7008 0.2659 0.2792

-j I .9221 j 1S967 0.2792 -0.23 19j1.9221 -j1.5967 0.2792 -0.2312

0.0 0.0 1I- = r-j13.2571 0.00.0 j13.2571 0.0 0.0D = E - I A E = 0.0 0.0 -j6.8854 0.0L 0.0 0o 0.0 j6.8854


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66 Chapter 3Substituting into =Dy,we can compute the uncoupled solution

yi = Ciexi'whereCi depends on the initial conditions.

This method of computing the distinct frequencies of oscillation is quite general andmay be applied to systems of any size. Forvery large systems this may not be practical,however, since the eigenvector computation may be too costly.Finally, we note that the simple model used here assumes that no damping exists.I n physical systems damping is usually present; therefore, the oscillatory response givenabove is usually damped. The magnitude of the damping, however, issuch that the fre-quenciesof oscillation given by the above equations are not appreciably affected.

i = 1,2,3,4

3.5 Regulated Synchronous MachineI n this section we examine the effect of voltage and speed control equipment on thedynamic performanceof the synchronous machine. Again we are interested in the freeresponse of the system. We will consider two simple cases of regulation: a simplevoltage regulator with one time lag and a simple governor with one time lag.3.5.1Referring to Figure 2.24, we note that a change in the field voltage uF, is pro-duced by changes in either VREFor y . I f we assume that V,,,, = 0and the transducerhas no time lags, uFA depends only upon K modified by the transfer function of the

excitation system. Analysis of such a system is discussed in Chapter 7. To simplify theanalysis, a rather simple model of the voltage regulator and excitation system is as-sumed. This gives the followings domain relation between the change in the excitervoltageu,, and the change in the synchronous machine terminal voltage y,:'FA = - K c/(l +7ts)1 y A (3.36)

Voltage regulator wi th one time lag

where K , = regulatorgain7, = regulator time constant

To examine the effect of the voltage regulator on the system response, we return tothe model discussed in Section 3.3 for a machine connected to an infinite bus through atransmission network. These relations are given in (3.9) and (3.10).To use (3.36), a relation between v, , 6,, and E: is needed. Such a relation is de-veloped in reference[8] and is in the form

(3.37)A = KS6, + &E :where K5 = y,/6,1EA = change in terminal voltage with change in rotor angle for

K6 = VIA/E6la, = change in terminal voltage with changein E' forconstant 6constantE'

The system block diagramwith voltage regulation added is shown in Figure3.4.UFA = -[K t/(l + 7es)l(KS6A +

From (3.36)and (3.37)(3.38)

Substituting in (3.10),we compute


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System Response to Small Disturbances 67

'mb REF

Fig. 3.4 System block diagram with voltage regulation.

rearranging,

From (3.39) and (3.9)

peA =

(3.39)

Substituting in thes domain swing equation and rearranging, we obtain the follow-ing characteristic equation:

Equation (3.41) isof the forms4 +03S3 +O*S2 +q s +cr , =0 (3.42)

Analysis of this fourth-order system for stability is left as an exercise (see Problem 3.7).


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68 Chapter 33.5.2Referring to Figure 2.24, we note that a change in the speed w or in the load orspeed reference [governor speed changer (GSC)] produces a change in the mechanicaltorque T,,,. The amount of change in T,,,depends upon the speed droop and upon the

transfer functions of the governor and the energy source.For the model under consideration it is assumed that GSCA =0and that the com-bined effect of the turbine and speed governor systems are such that the change in themechanical power in per unit is in theform

Governor with one time lag

(3.43)where Kg = gain constant = I /R

r g = governor time constantThe system block diagram with governor regulation is shown in Figure3.5.

Then the linearized swing equation in the s domain is in the form(with(3.44)wR in rad/s) SSA(S)( ~ W W R ) ~ ' ~ A S )-[&/(I +7gs)I - Ped($

The order of this equation will depend upon the expression used for PeA(s). I f we as-sume the simplest model possible, PeA(s)= PSGA(s),he characteristic equation of thesystem is given by(3.45)2H/wR)s2 +[ Kg/ ( I+T ~s ) ] S+ Ps = O

orS3(2H T g/W ~)+ s2(2H/WR) +(Kg+P s T g ) S +P, = O (3.46)The system is now of third order. Applying Routh's criterion, the system is stableif Kg >0and P, >0.Ifanother model is used for PeA(s),such as the model given by (3.9) and (3.10),the system becomes of fourth order, as shown in Figure3.5. Its dynamic response willchange. Information on stability can be obtained from the roots of the characteristicequationor from examining the eigenvalues of its characteristic matrix.

* .d

Fig. 3.5 Bl ock diagram of a system with governor speed regulation.


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S ys t em Response to S m a l l D i s t u r b an c e s 69

GSCA1

Fig. 3.6 Block diagram of a system wi th a governor and vol tage regulator.

I f both speed governor and voltage regulation are added simultaneously, as isusually the case, the system becomes fifth order, as shown in Figure3.6.3.6 Distributionof Power Impacts

I n this section we consider the effect of the sudden application of a small loadPLAat some point in the network. (See also [7,5].) To simplify the analysis, we also as-sume that the load has a negligible reactive component. Since the sudden change inload PLAcreates an unbalance between generation and load, an oscillatory transientresults before the system settles to a new steady-state condition. This kind of impactis continuously occurring during normal operation of power systems. The oscillatorytransient is in fact a spectrum of oscillations resulting from the random change inloads. These oscillations are reflected in power flow in the tie lines. Thus the scheduledtie-line flows will have random power oscillations superimposed upon them. Ourconcern hereisto make an estimate of the magnitude of these power oscillations. Notethat the estimates made by the methods outlined below are only approximate, yet theyare quite instructive.We formulate the problem mathematically using the network configuration of Fig-ure3.7 and the equations of Sections2.9 and 3.4. Referring to the (n + I)-port net-work in Figure3.7, the power into node i is obtained from(3.21) by adding nodek.

0- n. *pi = E ~ G ~ ~E,E,(B, sin6, +~,coss~,) v ~( B ~~inai, +cikossik)j - lj t i . k

For the caseof nearly zero conductancen

pi C E,B~~ insii + V,B~~inailj -j t i . k

(3.47)


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70 Chapter 31 (n + I)-port networknL-

Fig. 3.7 Network with power impact a t nodek.and the power into nodek (the load bus) is

(3.48)Here we assume.that the power network has a very high X/ R ratio such that theconductances are negligible. The machines are represented by the classical model ofconstant voltage behind transient reactance. We also assume that the network has beenreduced to the internal machine nodes (nodes I , 2, . . n of Figure 2.17) and the node k,where the impact P L A is applied.The mmedia te effect (assuming the network response to be fast) of the applicationof PL A is that the angle of bus k is changed while the magnitude of its voltage vkis unchanged, or V, &becomes v k /6ko +&A . Note also that the internal angles ofthe machine nodes d l , 2 , . . 6, do not change instantly becauseof the rotor inertia.3.6.1 LinearizationThe equations for injected power (3.47) and (3.48) are nonlinear because of thetranscendental functions. Since we are concerned only with a small impact P L A , welinearize these equations to find

Pi = Pi0 + Pia P k = P k O + P k Aand determine only thechange variables Piaand P k A .The transcendental functions are linearized by the relationssinbkj = sin(6kjo+6kjA) sin6kj0+(cos6kjO)6kjAcos6kj = cos(6kjO + 6kjA) cos6kjO - (sin6kjO)6kjA (3.49)

for any k , . Note that the order k j must be carefully observed since&j = - 6 j k . Sub-stituting (3.49) into (3.47) and (3.48) and eliminating the initial values, we computethe linear equations

j - lj 6 i . k

I - 1 j - lThese equations are valid for any time t following the application of the impact.

(3. 50)


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System Response to Small Disturbances 71362The instant immediately following the impact is of interest. I n particular, we wouldlike to determine exactly how much of the impact PLA is supplied by each generatorP iA , i = 1,2 ..., .At the instant r = O+ we know that a = 0 for all generators because of rotorinertias. Thus we can compute (with both i andj ndicating generator subscripts)

A spec ia l case: r =0

6J A = 0 6 i&A = 6 i A - 6 & ~ -6&.(o+) 6 & j A = 6& A - 6 j A = 6 & A ( o + )Thus (3.50)becomes

npiA(o+) = -psik6&A(o+) p& A(o+)= Ps&j6&A(O+) (3.51)/ - I

Comparing the above two equations at r = 0+,we note that at nodek(3.52)

This is to be expected since we are assuming a nearly reactive network. We also notethat at node i Pia depends upon Bikcos6p,. In other words, the higher the transfersusceptance Bik and the lower the initial angle 6iko. he greater the share of the im-pact picked up by machine i . Note also that PkA= -PLA, so the foregoing equa-tions can be written in terms of the load impact as

From(3.52) and (3.53)we conclude that(3.54)

It is interesting that at the instant of the load impact (i.e., at r = O?, the source ofenergy supplied by the generators is the energy stored in their magnetic fields and isdistributed according to the synchronizing power coefficients between i and k. Notethat the generator rotor angles cannot move instantly; hence the energy supplied by thegenerators cannot come instantly from the energy stored in the rotating masses. Thisisalso evident from the first equation of (3.51); Pia depends upon Psi&or Bik,whichdepends upon the reactance between generator i and nodek. Later on when the rotorangles change, the stored energy in the rotating masses becomes important, as shownbelow.Equations (3.52) and (3.55) indicate that the load impact PLA at a network bus kis immediately shared by the synchronous generators according to their synchronizingpower coefficients with respect to the busk. Thus the machines electrically close to thepointof impact will pick up the greater shareof the load regardlessof their size.Letusconsider next the deceleration of machine i due to the sudden increase in itsoutput power Pia. The incremental differential equation governing the motion ofmachinei is given by


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72 Chapter 32Hi dWiA-- PiA(t) = 0 i = 1,2, . . , t ~WR dt

and using (3.55)

Then if PLA isconstant for all t , we compute the acceleration in pu to be

(3.56)

(3.57)Obviously, the shaft decelerates for a positive load P L A . The pu deceleration of ma-chinei , given by (3.57), is dependent on the synchronizing power coefficient Psk andinertia H i . This deceleration will be constant until the governor action begins. Notethat after the initial impact the various synchronous machineswill be retarded at differ-ent rates, each according to its sizeH i and its electrical location given by P,ik.

3.6.3We now estimate the system behavior during the period 0< t < t , , where t , isthe time at which governor action begins. To designate this period simply, we referto time as t l , although there is no specific instant under consideration but a brieftime period of no more than a few seconds. Looking at the system as a whole, therewill be an overall deceleration of the machines during this period. To obtain themean deceleration, let us define an inertial center that has angle 8 and angularvelocitya,where by definition,

s ( l / C H , ) C G , H , i j A ( 1 / C H i ) C W i H 1 (3.58)

Average behavior prior to governor action ( t = 1, )

Summing the set (3.57) for all values of i , we compute(3.59)

(3.60)Equation (3.60) gives themean acceleration of all the machines in the system, which isdefined here as the acceleration of a fictitious inertial center.We now investigate the way in which the impact PLawill be shared by the variousmachines. Note that while the system as a whole is retarding at the rate given by(3.60), the individual machines are retarding at different rates. Each machine followsan oscillatory motion governed by its swing equation. Synchronizing forces tend topull them toward the mean system retardation, and after the initial transient decaysthey will acquire the same retardation as given by (3.60). I n other words, when thetransient decays, dwiA/dtwill be the same as dGA/dt as given by (3.60). Substitutingthis value of dwiA/dt n (3.56), at t = t, > t o ,

(3.61)Thus at the endof a brief transient the various machines will share the increase inload as a function only of their inertia constants. The time t , is chosen large enough


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System Response to Small Disturbances 73so that all the machineswill have acquired the mean system retardation. At the sametime t , is not so large as to allow other effects such as governor action to take place.Equation (3.61) implies that theH constants for all the machines are given to a commonbase. If they are given for each machine on its own base, the correct powers are ob-tained if H is replaced by HSB3/SsB,hereSBss the machine rating and S,, is thechosen system base.Examining(3.56)and (3.61), we note that immediately after the impact PLA(i.e.,att = 0+)the machines share the impact according to their electrical proximity to thepoint of the impact as expressed by the synchronizing power coefficients. After a brieftransient period the same machines share the same impact according to entirely differ-ent criteria, namely, according to their inertias.Example 3.4

Consider the nine-bus, three-machine system of Example2.6 with a small I O- MWresistive load added to bus8 as in Example3.2. Solve the system differential equationsand plot PtA nd wid as functions of time. Compare computed results against the-oretical values of Section3.6.

-2 1Fig. 3.8 PrAversus t following application of a 10MW resistive load at bus 8.

SolutionA nominal I O- MW0.1pu) load is added to bus 8 by applying a three-phase faultthrough a 10 pu resistance, using a library transient stability program. The resultingpower oscillations P,A, = 1, 2, 3, are shown in Figure 3.8 for the system operatingwithout governor action.The prefault conditions at the generators are given in Table3.1 and in Example 2.6.From the prefault load flow of Figure 2.19 we determine that V = 1.016 and a,,,, =0.7". A matrix reduction of the nine-bus system, retaining only nodes 1, 2, 3, and 8,gives the system data shown on Table3.3.


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74 Chapter 3Table3.3 Transfer Admittances andInitial Anglesof a Nine-Bus System

i j G i i B i i b o1-8 0.01826 2.51242 1.57172-8 -0.035 30 3.55697 19.03 153-8 -0.00965 2.61601 12.4752

From(3.24)we compute the synchronizing power coefficientsps ik = 6 v k ( B , k cos6ikO - Gk sin 6 i kO )

These values are tabulated in Table 3.4. Note that the error in neglecting the Giktermissmall.Table3.4. SynchronizingPowerCoefficients

psik psik(neglectingGjk ) (with Gik term)k18 2.6961 2.695528 3.5878 3.600138 2.6392 2.6414cpsik 8.923 I 8.9370

The values of p i A ( o + ) are computed from(3.55) as

where PLA(O+)= 10.0MW nominally.actual values determined from the stability study are shown in Table3.5.The results of these calculations and the

Table3.5. Initial Power ChangeatGeneratorsDue to I O- MWLoad Addedto Bus 8

I 3.02 1 3.016 2.8 2.749 2.7452 4.02 1 4.028 3.6 3.659 3.6653 2.958 2.956 2.7 2.692 2.6901O.OOO 10.000 9.1 9.100 9.100- - -

Note that the actual load pickup is only 9.1 MW instead of the desired I O MWThis is due in part to the assumption of constant voltage v k at bus 8 (actually, thevoltage drops slightly) and to the assumed linearity of the system. I f the computedP I A re scaled down by 0.91, the results agree quite well with values measured from thecomputer study. These values are also shown on the plot of Figure 3.8 at time t = O+and are due only to the synchronizing power coefficients of the generators with respectto bus8.The plots of P i a versus time in Figure 3.8 show the oscillatory nature of the powerexchange between generators following the impact. These oscillations have frequenciesthat are combinations of the eigenvalues computed in Example 3.2. The total, labeledZ P i A , averages about9.5MW.


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System Respons e to Small Disturbances 75

lime, I0 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1 .1 1.2 1.3 1.4 1,s 1.6 1:7 1,8 1:9 2:O

.-0"'t '%-0.04

r. 0.08.--0.10 a--0 .12 ,-

-0.14 ..-0.16-0.18 '-

tFig. 3.9 Speed deviation following application of a 10MW resistive load at bus 8.

Another point of interest in Figure 3.8 is the computed values of PiA(t1) thatdepend entirely on the machine inertia. These calculations are made fromPiA(tl) = (Hi/CHi)PLA = IOHi/(23.64 -I- .40 -I-3.01) = lOHi/33.05

= 7.15 M W i = 1= 1.94M W i = 2= 0.91 M W i = 3

and the results are plotted in Figure 3.8 as dashed lines. It is fairly obvious that thePiA(t) oscillate about these values of P,A(tl). It is also apparent that the system haslittle damping and the oscillations are likely to persist for some time. This is partly dueto the inherent nature of this particular system, but the same phenomenon would bepresent to some extent on any system.The second plot of interest is the speed deviation or slip as a function of time,shown in Figure 3.9. The computer program provides speed deviation data in Hz andthese units are used in Figure3.9. Note the steady deceleration with all units oscillatingabout the meanor inertial center. This iscomputed as

&A PL A 0. Odr 2C Hi 2(23.64 +6.40 +3.01)-I -= -

= -1.513 x pu/s = -0.570 rad/s2 = -0.0908 Hz/sThe individual machine speed deviationswiA are plotted in Figure 3.9 and show graphi-cally the intermachine oscillations that occur as the system slowly retards in frequency.The mean deceleration of about 0.09 Hz/s is plotted in Figure3.9as a straight line.I f the governors were active, the speed deviation would level off after a few secondsto a constant value and the oscillations would eventually decay. Since the governorshave a drooping characteristic, the speed would then continue at the reduced value as


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76 Chapter 3

z2 76.7

long as the additional load was present. I f the speed deviation is great, signifyinga substantial load increaseon the generators, the governors would need to be readjustedto the new load level so that additional prime-mover torque could be provided.

80. 0- - - -

- --

Example 3. 5Let us examine the effectof the above on the power flow in tie lines. Consider apower network composed of two areas connected with a tie line, as shown in Fig-ure 3.10. The two areas are of comparable size, say 1000 MW each. They are con-nected with a tie line having a capacity of 100 MW. The tie line is carrying a steadypower flow of 80MW from area I to area 2 as shown in Figure 3.10. Now let a loadimpact PLA = I O MW ( 1 % of the capacity of one area) take place at some point inarea I , and determine the distribution of this added load immediately after its applica-tion ( I = 0+)and a short time later ( t = t , ) after the initial transients have subsided.Becauseof the proximity of the groups of machines in area 1 to the point of impact,their synchronizing power coefficients are larger than those of the groups of machinesin area 2. I f we defineCPSi k J ar eaIPSI , Ps i kl ar ca2Psz,hen let us assume that P, , =2ps2. 9-Q0MW -

PM =10 M WFig. 3.10 T wo areas connected with a tie line.

SolutionSincePSI= 2Ps2, t the instantof the impact 2/ 3 of the IO-MW load will be sup-plied by the groups of machines in area 1, while 1/3 or 3.3 MW will be supplied bythe groups of machines in area 2. Thus 3.3 MW will appear as a reduction in tie-lineflow. In other words, at that instant the tie-line flow becomes76.7MW toward area2.At the endof the initial transient the load power impact PLA will be shared by themachines according to their inertias. Let us assume that the machines of area 1 are

t

0 t = OTime, I tl

Fig. 3.1 I T ie-line power.oscillations dueto the load impact in area I .


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System Response to Small Disturbances 77predominantly hydro units (with relatively small H), while the units of area 2 are oflarger inertia constants such that CHI J area22CHi],,e,I where all H 's are on a com-mon base. The sharing of the load among the groups of machines will now become6.7 M W contributed from area 2 and 3.3 M W from area I . The tie-line flow will nowbecome73.3 M W (toward area2).From the above we can see that in the situation discussed in this example a suddenapplication of a IO-MW load caused the tie-line flow to drop almost instantly by3.3 M W , and after a brief transient by 6.7 M W . The transition from 76.7-M W flowto 73.3-M W flow is oscillatory, and power swings of as much as twice the differencebetween these two values may be encountered. This situation is illustrated in Fig-ure3.1 I .The time t , mentioned above is smaller than the time needed by the various con-trollers to adjust the system generation to match the load and the tie-line flow to meetthe scheduled flow.Example 3.6Wenow consider a slightly more complex and more realistic case wherein the areaequivalents in Figure 3.10 are represented by their Thevenin equivalents and the tie-line impedance is given. The system data are given in Figure3.12 in pu on a 1000-M VAbase. The capacity of area I is20,000 M W and that of area2 is14,000 M W . The inertiaconstants of the machinesin the two areas are about equal.(a) Find the equations of power for P I and Pz .(b) Find the operating condition when P I = 100 M W . This would correspond ap-proximately to a 100-M W tie-line flow from area 1 to area2.(c) Find the synchronizing power coefficients.(d) Consider a sudden load addition to area 2, represented by the resistive load P4, ,at bus4. I f this load is 200 M W (1.43% of the capacity of area 2), find the distri-butionof this load at f = O+ and t = f l .

Ar a 1 eguivalent Tie litm Area 2 equivalentFig. 3.12 T wo areas connected by a tie line.

Sol uti onputeConsider the system as a two-port network between nodes 1 and 2. Then we com-

Z12 = 0.450+j1.820 = 1.875 /76.112" puplz = I / f 1 2 = 0.533/-76.112" = 0.128 - 0.518 puYlz = -712 = 0.533/103.888"GI1 = 0.128-

gl o =gzo = 0


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Chapter 3GI2 = -0.128PI = V:glo + V, V 2 ( G 12~~~612B12sin612) V:GI2

= 0+ I .O(-O.l28co~6~+0.518~in6~) 0.128= 0.128 +0.533sir1(6~- 13.796")P2 = V:gzo+ VI V 2( G12~~~621BI2sin

=0+ I.O(-O.I28c0s6~- 0.518sin6,) +0.128= 0.128 - 0.533sin(6, +13.796")

612 = 61 - 62 = 6,B12 = 0.518

- V:GzI

(b) Given that P I = 0.1 pu0.100 = 0.128 +0.533sin(dl - 13.796") 6 1 = 10.784"

(4 Pr12= K W B I Zos6 1 2 0 - G l2sind120)PS21 = K VZ(B2I cos6210 - G2I sin6 2 1 0 )

= l.O(O.518cos 10.784" +0.128sin 10.784") = 0.533= 1.0[0.518cos(- 10.784")+0.128sin(- 10.784")] = 0.485

(d) Now add the 200-MW load at bus4; P 4A = 200/1000 = 0.2pu.puteTo complete the problem, we must know the voltagep4at t = 0-. Thus we com-f12(0-)= (K - K) /Z , z = (1.0/10.784" - l.0~)/1.875/76.112"= 0.100/19.280"K(O-) = +(0.100+ j0.012)&2= 1.009+j0.004 = 1.009/0.252"

640 = 0.252' 6140 = 610 - 640 = 10.532" 6240 = 620 - 640 = -0.252"From the admittance matrix elements-Y 14 = -y14 = - /114 = -0.103 + j0.533

Y 24 = -y24 = -l/Fz4 = -9.858 +j1.183we compute the synchronizing power coefficients

Ps14= VI Y 4(B14~~~6140G14sin6140)= (1.009)(0.533 cos 10.532' + 0.103 sin 10.532') = 0.548= 1 .009 (1 .183~0~( -0 .? .52 " ) 9.858sin(-0.252")] = 1.150

Ps24 == VzV4(B24COS 6240 - G24sin 6240)Then the initial distributionof P 4A is

PIA( +) PS14(o.2)/(Ps14+PSz4) = (0.323)(0.2) = 0.0646PUPz~(o+) Ps24(0.2)/(Psi4 +Ps24) = (0.677)(0.2) = 0.1354 pu

The power distribution according to inertias is computed asPlA(fl) = 0.2[20,000ff/(20,000ff + 14,000H)J = 0.11765 puPzA(fl) = 0.2[14,000H/(20,000H + 14,000H)I = 0.08235 pu

In this example the synchronizing power coefficientsPSI.,s smaller thanPSz4,while theinertiaof area 1 is greater than that of area 2. Thus, while initially area1picks up only about


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System Response to Small Disturbances 79one thirdof the loadP,, at a later time t = t , it picks up about59% of the load andarea2 picks up the remaining 41%.I ngeneral, the initial distribution of a load impact depends on the point of impact.Problem3.10gives another example where the point of impact is in area I (bus3).

In the above discussion many factors have been neglected, e.g., the effect of thenetwork transfer conductances, the effect of the reactive component of the load impact,the fast primary controllers such as some of the modern exciters, the load frequency andvoltage characteristics, and others. Thus the conclusions reached above should beconsidered qualitative and as rough approximations. Yet these conclusions are basicallysound and give a good "feel" for what happens to the machines and to the tie-lineflows under the influence of small routine load changes.I f the system is made up of groups of machines separated by tie lines, they share

the impacts differently under different conditions. Hence they will oscillatewith respectto each other during the transient period following the impact. The power flow in theconnecting tieswill reflect these oscillations.The analysis given above could be extended to include governor actions. Followingan impact the synchronous machines will share the change first according to theirsynchronizing power coefficients, then after a brief period according to their inertias.The speed changewill be sensed by the prime-mover governors, which will act to makethe load sharing according to an entirely different criterion, namely, the speed governordroop characteristic. The transition from the second to the final stage is oscillatory(see Rudenberg [7), Ch. 23). The angular frequency of these oscillations can be esti-mated as follows. From Section 3.5.2, neglecting P I A , the change in the mechanicalpower PmAsof the form

(3.62)whereR isthe regulation and 7 , is the servomotor time constant. The swing equationfor machine becomes,h thes domain,

The characteristic equation of the system is given bys2 + (1/7si)s + 1/2HiRi~,i= 0 (3.63)

from which the natural frequencyof oscillation can be estimated.I t is interesting to note the order of magnitude of the frequency of oscillation in thetwo different transients discussed in this section. For a given machine (or a group ofmachines) the frequency of oscillation in the first transient is the natural frequency withrespect to the point of impact. These frequencies are determined by finding the eigen-values X of the A matrix by solving det (A - XU) =0, where U is the unit matrixandA is defined by (3. ).For the second transient, which occurs during the transition from sharing accordingto inertia to sharing according to governor characteristic, the frequency of oscillationi s given by Y : ~% 1/2HiRf7,,. Usually these two frequencies are appreciably different.


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80 Chapter 3Problems

3.

3.23.3

3.4

3.53.63.13.83.9

3.103.1 I

A synchronous machine is connected to a large system (an infinite bus) through a longtransmission line. The direct axis transient reactance x i = 0.20 pu. The infinitebus voltage is 1.0pu. The transmission line impedance is Zlinc= 0.20 + j0.60 pu. Thesynchronous machine is to be represented by constant voltage behind transient reactancewithE = 1.10pu. Calculate the mnimum and maximum steady-state load delivered atthe infinite bus (for stability). Repeat when there is a local loadof unity power factorhavingItload= 8.0pu.Use Rouths criterion to determine the conditions of stability for the system where thecharacteristic equation is given by(3.14).Compute the characteristic equation for the systemof Figure 3. I , including the dampingterm, and determine the conditions for stability using Rouths criterion. Compare theresults with thoseof Section3.3.1.Using 1 3 ~s the output variable in Figure 3.2, use block diagram algebra to reducethe system block diagram to forward and feedback transfer functions. Then determine thesystem stability and possible system behavior patterns by sketching an approximate root-locus diagram.Use block diagram algebra to reduce the system described by (3.45). Then determine thesystem behavior by sketching the root loci for variations in K ,.Give the conditions for stabilityof the system described by (3.20).A system described by (3.41) has the following data:H = 4, ria = 5.0, T , = 0.10, K I =4.8,K z = 2.6,K 3 = 0.26, K, = 3.30, K S = 0.1, and K b = 0.5. Find the maximum andminimum valuesof K, for stability. Repeat forK 5 = -0.20.Write the system described by(3.46) in state-space form. Apply Rouths criterion to (3.46).The equivalentpref uul r network is given in Table2.6 for the three-machine system dis-cussed in Section2.10and for the given operating conditions. The internal voltages andangles of the generators are given in Example2.6.(a) Obtain the synchronizing power coefficientsPSlz, S I ) , Sz3,and the correspondingcoefficientsa i j[see(3.3I)] or small perturbations about the given operating point.(b) Obtain the natural frequencies of oscillation for the angles 6 1 2 ~nd 6 1 3 ~ .Comparewith the periods of the nonlinear oscillations of Example2.7.Repeat Example3.6with the impact point shifted to area I and let P L ~ 100 MW asbefore.Repeat Problem3.10for an initial condition of PL A = 300MW.

ReferencesI . K orn, G . A ., and K orn, T . M . Mathematical Handbook for Scientists and Engineers.2. Hayashi, C . Nonlinear Oscil lations in Physical Systems. M cGraw-Hi ll, New York, 1964.3. Takahashi, Y ., Rabins. M . J ., and Auslander, D. M . Control and Dynamic Systems. Addison-Wesley,4. Venikov, V. A. Transient Phenomena in E lectric Power Systems. Trans. by B. Adkins and D. Ruten-5. Hore, R. A . Advanced Studies in Electrical Power System Design. Chapman and Hall, L ondon, 1966.6. Crary, S. B. Power System Stabili ty. Vols. 1 . 2. Wiley, New York, 1945, 1947.7. Rudenberg. R . Transient Performance of Electric Power Systems: Phenomena in Lumped Networks.M cGraw-H ill , New Y ork. 1950. (M I T Press, Cambridge, Mass., 1967.)8. de M ello. F. P.. and Concordia, C . Concepts of synchronous machine stabili ty as affected by excita-tion control. IEEE Trans. PAS-88:316-29, 1969.9. Heffron. W. G.. and Phillips, R. A . Effect of a modern amplidyne voltage regulator on underexcitedoperation of large turbine generators. AlE E Trans. 71 (Pt. 3):692-97, 1952.

McGraw-Hill,New Y ork, 1968.

Reading, M ass., 1970.berg. Pergamon Press, New Y ork, 1964.

10. Routh, E. J . Dynamics of a Systemof Rigid Bodies. M acmil lan, L ondon, 1877. (A dams Prize Essay.)1 1 . Ogata. K. State-Space Analysis of Control Systems. Prentice-Hall. Englewood Cliffs, NJ., 1967.12. L efschetz, S. Stability of Nonlinear Control Systems. Academic Press, New Y ork, L ondon, 1965.

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