Module 6

Power system stability

6.1 Introduction

In general terms, power system stability refers to that property of the power system which enablesthe system to maintain an equilibrium operating point under normal conditions and to attain astate of equilibrium after being subjected to a disturbance. As primarily synchronous generatorsare used for generating power in grid, power system stability is generally implied by the abilityof the synchronous generators to remain in ’synchronism’ or ’in step’. On the other hand, if thesynchronous generators loose synchronism after a disturbance, then the system is called unstable.The basic concept of ’synchronism’ can be explained as follows.

In the normal equilibrium condition, all the synchronous generators run at a constant speed andthe difference between the rotor angles of any two generators is constant. Under any disturbance, thespeed of the machines will deviate from the steady state values due to mismatch between mechanicaland electrical powers (torque) and therefore, the difference of the rotor angles would also change.If these rotor angle differences (between any pair of generators) attain steady state values (notnecessarily the same as in the pre-disturbance condition) after some finite time, then the synchronousgenerators are said to be in ’synchronism’. On the other hand, if the rotor angle differences keep onincreasing indefinitely, then the machines are considered to have lost ’synchronism’. Under this ’outof step’ condition, the output power, voltage etc. of the generator continuously drift away from thecorresponding pre-disturbance values until the protection system trips the machine.

The above phenomenon of instability is essentially related with the instability of the rotor anglesand hence, this form of instability is termed as ’rotor angle instability’. Now, as discussed above,this instability is triggered by the occurrence of a disturbance. Depending on the severity of thedisturbance, the rotor angle instability can be classified into two categories:

• Small signal instability: In this case, the disturbance occurring in the system is small. Suchkind of small disturbances always take place in the system due to random variations of the loadsand the generation. It will be shown later in this chapter that under small perturbation (ordisturbance), the change in the electrical torque of a synchronous generator can be resolved intotwo components, namely, a) synchronizing torque (Ts) - which is proportional to the change inthe rotor angle and b) damping torque (Td), which is proportional to the change in the speed

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of the machine. As a result, depending on the amounts of synchronizing and damping torques,small signal instability can manifest itself in two forms. When there is insufficient amount ofsynchronizing torque, the rotor angle increases steadily. On the other hand, for inadequateamount of damping torque, the rotor angle undergoes oscillations with increasing amplitude.These two phenomena are illustrated in Fig. 6.1. In Fig. 6.1(a), both the synchronizingand damping torques are positive and sufficient and hence, the rotor angle comes back to asteady state value after undergoing oscillations with decreasing magnitude. In Fig. 6.1(b), thesynchronizing torque is negative while the damping toque is positive and thus, the rotor angleenvelope is increasing monotonically. Fig. 6.1(c) depicts the classic oscillatory instability inwhich Ts is positive while Td is negative.

Figure 6.1: Influence of synchronous and damping torque

In an integrated power system, there can be different types of manifestation of the small signalinstability. These are:

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a. Local mode: In this type, the units within a generating station oscillate with respect tothe rest of the system. The term ’local’ is used because the oscillations are localized in aparticular generating station.

b. Inter-area mode: In this case, the generators in one part of the system oscillate withrespect to the machines in another part of the system.

c. Control mode: This type of instability is excited due to poorly damped control systemssuch as exciter, speed governor, static var compensators, HVDC converters etc.

d. Torsional mode: This type is associated with the rotating turbine-governor shaft. Thistype is more prominent in a series compensated transmission system in which the me-chanical system resonates with the electrical system.

• Transient instability: In this case, the disturbance on the system is quite severe and suddenand the machine is unable to maintain synchronism under the impact of this disturbance. Inthis case, there is a large excursion of the rotor angle (even if the generator is transiently stable).Fig. 6.2 shows various cases of stable and unstable behavior of the generator. In case 1, underthe influence of the fault, the generator rotor angle increases to a maximum, subsequentlydecreases and settles to a steady state value following oscillations with decreasing magnitude.In case 2, the rotor angle decreases after attaining a maximum value. However, subsequently,it undergoes oscillations with increasing amplitude. This type of instability is not caused bythe lack of synchronizing torque; rather it occurs due to lack of sufficient damping torque in thepost fault system condition. In case 3, the rotor angle monotonically keeps on increasing dueto insufficient synchronizing torque till the protective relay trips it. This type of instability, inwhich the rotor angle never decreases, is termed as ’first swing instability’.

Figure 6.2: Illustration of various stability phenomenon

Apart from rotor angle instability, instability can also occur even when the synchronous generatorsare maintaining synchronism. For example, when a synchronous generator is supplying power to aninduction motor load over a transmission line, the voltage at the load terminal can progressively

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reduce under some conditions of real and reactive power drawn by the load. In this case, loss ofsynchronism is not an issue but the challenge is to maintain a stable voltage. This type of instabilityis termed as voltage instability or voltage collapse. We will discuss about the voltage instability issuelater in this course.

Now, for analysing rotor angle stability, we have to first understand the basic equation of motionof a synchronous machine, which is our next topic.

6.2 Equation of motion of a synchronous machine

The equation of motion of a synchronous generator is based on the fact that the accelerating torqueis the product of inertia and its angular acceleration. In the MKS system, this equation can bewritten as,

Jd2θmdt2

= Ta = Tm − Te (6.1)

In equation (6.1),J → The total moment of inertia of the rotor masses in Kg −m2

θm → The angular displacement of the rotor with respect to a stationary axis, in mechanicalradians

t→ Time in secondsTa → The net accelerating torque, in N-mTm → The mechanical or shaft torque supplied by the prime mover less retarding torque due

to rotational losses, in N-mTe → The net electrical or electromagnetic torque in N-m

Under steady state operation of the generator, Tm and Te are equal and therefore, Ta is zero.In this case, there is no acceleration or deceleration of the rotor masses and the generator runsat constant synchronous speed. The electrical torque Te corresponds to the air gap power of thegenerator and is equal to the output power plus the real power loss of the armature winding.

Now, the angle θm is measured with respect to a stationary reference axis on the stator andhence, it is an absolute measure of the rotor angle. Thus, it continuously increases with timeeven with constant synchronous speed. However, in stability studies, the rotor speed relative tothe synchronous speed is of interest and hence, it is more convenient to measure the rotor angularposition with respect to a reference axis which also rotates at synchronous speed. Hence, let usdefine,

θm = ωsmt + δm (6.2)

In equation (6.2), ωsm is the synchronous speed of the machine in mechanical radian/sec. andδm (in mechanical radian) is the angular displacement of the rotor from the synchronously rotatingreference axis. From equation (6.2),

dθmdt

= ωsm +dδmdt

or,dδmdt

= dθmdt

− ωsm (6.3)

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d2θmdt2

= d2δmdt2

(6.4)

Equation (6.3) shows that the quantitydδmdt

represents the deviation of the actual rotor speedfrom the synchronous speed in mechanical radian per second. Substituting equation (6.4) intoequation (6.1) one gets,

Jd2δmdt2

= Ta = Tm − Te (6.5)

Now. let us define the angular velocity of the rotor to be ωm = dθmdt

. From equation (6.5) we get,

Jωmd2δmdt2

= ωmTa = ωmTm − ωmTe

Or,

Jωmd2δmdt2

= Pa = Pm − Pe (6.6)

In equation (6.6), Pa, Pe and Pm denote the accelerating power, electrical output power and theinput mechanical power (less than the rotational power loss) respectively.

The quantity Jωm is the angular momentum of the rotor and at synchronous speed, it is knownas the inertia constant and is denoted by M . Strictly, the quantity Jωm is not constant at alloperating conditions since ωm keeps on varying. However, when the machine is stable, ωm does notdiffer significantly from ωsm and hence, Jωm can be taken approximately equal to M . Hence, fromequation (6.6) we obtain,

Md2δmdt2

= Pa = Pm − Pe (6.7)

Now, in machine data, another constant related to inertia, namely H-constant is often encoun-tered. This is defined as;

H = stored kinetic energy in megajoules at synchronous speedmachine rating in MVA

Or,

H = 12Jωsm2

Smc= 1

2MωsmSmc

MJ/MVA = 12MωsmSmc

sec. (6.8)

In equation (6.8), the quantity Smc is the three phase MVA rating of the synchronous machine.Now, from equation (6.8),

M = 2HSmcωsm

MJ/mech. rad (6.9)

Substituting for M in equation (6.7), we get,

2Hωsm

d2δmdt2

= PaSmc

= Pm − PeSmc

(6.10)

In equation (6.10), both δm and ωsm are in mechanical units. Now, the corresponding quantities

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in electrical units are given as,ωs =

P

2 ωsm; δ = P2 δm; (6.11)

In equation (6.11), P is the number of pole in the generator, ωs is the synchronous speed of themachine in electrical radian/sec. and δ (in electrical radian) is the angular displacement of the rotorfrom the synchronously rotating reference axis. Substituting equation (6.11) in equation (6.10) weget,

2Hωs

d2δ

dt2= Pa = Pm − Pe per unit (6.12)

Equation (6.12) is known as the swing equation of the synchronous machine. As this is asecond order differential equation, it can be written as a set of two first order differential equationsas below.

2Hωs

dω

dt= Pm − Pe per unit (6.13)

dδ

dt= ω − ωs (6.14)

In equations (6.13) - (6.14), the quantity ω is the speed of the synchronous machine and isexpressed in electrical radian per second. Now, in the above two equations, no damping of themachine is considered. If damping is considered (which opposes the motion of the machine), a termproportional to the deviation of the speed (from the synchronous speed) is introduced in equation(6.13). Therefore, the modified equation becomes;

2Hωs

dω

dt= Pm − Pe − d(ω − ωs) per unit (6.15)

In equation (6.15), d is called the damping co-efficient. However, in the presence of damping, equation(6.14) does not change. Therefore, in the presence of damping, this pair of equations ((6.14) and(6.15)) describe the motion of the synchronous machine.

With this introduction of motion of synchronous machine, we are now ready to address the variousstability issues. From the next lecture we will start with transient stability analysis.

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6.3 Transient stability analysis

Before solving the differential equations to determine the transient stability or instability of thesystem, it is necessary to compute the initial conditions.

6.3.1 Computation of initial conditions

Let us consider an ‘n’ bus power system with ‘m’ generators (m < n). Without loss of generality, itis assumed that the ‘m’ generators are located at first ‘m’ buses of the system. Towards computationof initial conditions, initially the load flow solution of the system is computed. From the load flowsolution, following quantities are availbale:

a. Vi∠θi for i = 1,2,⋯⋯ n

b. PLi, QLi for i =m + 1,m + 2,⋯⋯ n

c. PLi, QLi, PGi, QGi for i = 1,2,⋯⋯m

In the above, PLi andQLi denote the real and reactive power load at bus ‘i’ respectively. Similarly,PGi and QGi denote the real and reactive power generation at bus ‘i’ respectively. Further, thequantities Vi and θi denote the voltage magnitude and angle of ith bus respectively. With theseinformation, following calculations are carried out:

(i) At any bus ‘i’, the loads are converted to equivalent admittance as;

yLi =PLi − jQLi

Vi2 for i = 1,2,⋯⋯ n (6.16)

(ii) Augment the YBUS matrix of the system as;

YBUS(i, i) = YoldBUS(i, i) + yLi for i = 1,2,⋯⋯ n (6.17)

where, YoldBUS is the original YBUS matrix of the system used in load flow calculation.

(iii) At the generator buses, the generators are represented as equivalent voltage sources behindthe direct axis transient reactances as,

Ei = Vi + jx∕diIi = jx∕diPGi − jQGi

V ∗i

= Ei∠δi for i = 1,2,⋯⋯m (6.18)

In equation (6.18), the quantity x∕di denotes the direct axis transient reactance of the ith machine.While performing the transient stability analysis, the magnitude ∣Ei∣ is held constant. The equivalentdiagram of the ‘n’ bus power system is shown in Fig. 6.3.

With the initial conditions computed as above, we are now ready to solve the transient stabilityproblem. Basically, the transient stability problem is solved by two techniques: i) partition explicit(PE) method and ii) simultaneous implicit (SI) method. In the PE method, the network algebraicequations and the generator differential equations are solved separately. In the SI method, these

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Figure 6.3: Equivalent representation of generators for transient stability analysis

algebraic and differential equations are solved together. In this course, however, we are going todiscuss PE method only. Before, discussing the PE method, it is necessary to describe the networkalgebraic equations.

6.3.2 Network algebraic equations

The network algebraic equations are represented as;

IBUS = YBUSVBUS (6.19)

In equation (6.19), YBUS is the bus impedance matrix of the system and

IBUS = [I1 I2 ⋯⋯ Im Im+1 ⋯⋯ In]T

VBUS = [V1 V2 ⋯⋯ Vm Vm+1 ⋯⋯ Vn]T

(6.20)

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The voltage current relationship of the generator reactance is given by,

vk = jx∕diik ⇒ ik = ykvk (6.21)

In equation (6.21), ik and vk are the current through and voltage across the generator reactancerespectively and yk = 1

jx∕di= −j/x∕di. Now, at the generator terminals, Ik = ik = ykvk; or, Ik =

yk(Ek − Vk) as (vk = Ek − Vk). Or,Ik + ykVk = ykEk (6.22)

From equations (6.22) and (6.23) we get,

(Y11 + y1) V1 + Y12V2 + ⋯ + Y1mVm + ⋯ + Y1nVn = y1E1

Y21V1 + (Y22 + y2) V2 + ⋯ + Y2mVm + ⋯ + Y2nVn = y2E2

⋮ + ⋮ + ⋯ + ⋮ + ⋯ + ⋮ = ⋮Ym1V1 + Ym2V2 + ⋯ + (Ymm + ym) Vm + ⋯ + YmnVn = ymEm

Y(m+1),1V1 + Y(m+1),2V2 + ⋯ + ⋯⋯⋯ + ⋯ + Y(m+1),nVn = 0⋮ + ⋮ + ⋯ + ⋮ + ⋯ + ⋮ = ⋮

Yn1V1 + Yn2V2 + ⋯ + ⋯⋯⋯ + ⋯ + YnnVn = 0

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭(6.23)

From equation (6.23), the voltages V1, V2 ⋯⋯ Vn can be solved, for known values of E1,E2 ⋯⋯ En. With these known terminal voltages, the electrical power output of each generatorcan be calculated as;

Pei = Re (EI I∗i ) for i = 1, 2, ⋯m (6.24)

Where,Ii = yi (Ei − Vi) for i = 1, 2, ⋯m (6.25)

With the above equations, we are now in a position to discuss the PE method.

6.3.3 Partition explicit solution scheme

In the PE method, the numerical integration of the generator differential equations are carriedout separately from the solution of the network algebraic equations. For numerical integration ofdifferential equations, the total simulation time (tT ) is divided into N intervals, each interval beingof duration ∆t seconds. Thus, ∆t = tT

N. Now, the major steps for solving the transient stability

problem, with PE method, are as follows.1. Obtain the load flow solution of the given system. Thereafter, compute the internal

voltages of all the generators (Ei, for i = 1, 2, ⋯⋯ m) using equations (6.16) - (6.18). Please notethat the magnitudes Ei would be kept constant at these calculated values throughout the simulation.

2. With the values of Ei obtained above, solve the equation set (6.23) to obtain the terminalvoltages at all the buses (Vi, for i = 1, 2, ⋯⋯ n). Subsequently, the electrical power output of allthe generators (Pei, for i = 1, 2, ⋯⋯m) are computed from equations (6.24) - (6.25).

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3. Under steady state condition, mechanical power input to each generator is equal to itsgenerator electrical output power (neglecting losses). Therefore, Pmi = Pei, for i = 1, 2, ⋯⋯ m.Also, under steady state, all the generators are assumed to operate at synchronous speed (ωs).

4. Thus, after the above three steps, at t = 0, the variables pertaining to generators (Ei, δi,ωi, Pmi for i = 1, 2, ⋯⋯m) and the network bus voltages (Vi, for i = 1, 2, ⋯⋯ n) are all known.

5. The simulation process advances to t = ∆t. At this instant, first the network equationsgiven in the equation set (6.23) are solved to compute the bus voltages and subsequently, the outputelectrical power of each generator is calculated by using equations (6.24) - (6.25). Now, if thesteady state condition is still maintained, i.e. if there is no change in the network (as comparedto the network condition at t = 0), then the calculated values of Pei would be again equal to thecorresponding value of Pmi.

6. With the solution of the network equations at hand, we should now solve the solve thegenerator differential equations for calculating the values of δi and ωi at t = ∆t. Towards that end,let us first re-write the swing equations of ith machine for convenience in equations (6.26) - (6.27)below.

2Hi

ωs

dωidt

= Pmi − Pei (6.26)

dδidt

= ωi − ωs (6.27)

Now, from equation (6.26), as Pmi = Pei,dωidt

= 0. In other words, there is no change in the

speed of the generator and hence ωi = ωs. As a result, from equation (6.27),dδidt

= 0 and hence, theangle δi would also be maintained at the value calculated at t = 0. Therefore, under steady statecondition, at t = ∆t, both δi and ωi would be maintained at the values calculated at t = 0.

7. Assume that the steady state condition continues from t = 0 to t = (k − 1)∆t, where, k isa positive integer. Following the arguments described at steps 5 and 6, it can be easily seen that atthe end of t = (k − 1)∆t sec., both δi and ωi would still be maintained at the values calculated att = 0.

8. Now, let us assume that a three phase to ground short circuit fault occurs in the systemat t = to = k∆t at the `th bus. To accommodate this fault condition in the network equations,the element Y`` is increased manyfold to reflect very high admittance from bus ‘`’ to ground. Withthis imposed condition, the network bus voltages are calculated from equation set (6.23). Subse-quently, the output electrical power of each generator is calculated by using equations (6.24) - (6.25)corresponding to time t = to.

9. With the values of Pei calculated in step 8, the swing equations (6.26) - (6.27) are integratedto obtain the values of δi and ωi at t = to. Now, for integrating the swing equations, initial values ofδi and ωi are required. These initial values are taken to be equal to the values of δi and ωi obtainedat the end of t = (k − 1)∆t. For brevity, the value of δi (i = 1, 2, ⋯⋯ m) calculated at t = to isdenoted as δi(to). With this value of δi(to), the voltage behind the transient reactance is updatedas Ei = Ei∠δi(to), for i = 1, 2, ⋯⋯m, where the magnitude Ei has been taken to be equal to theconstant value calculated at step 1.

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10. The simulation advances to t = to + ∆t. If the fault still persists, the values of Pei arecalculated as described in step 8. However, for this purpose, the values of Ei in equation set (6.23)are taken to be equal to the latest values calculated at the end of t = to sec.

11. With the values of Pei calculated at t = to+∆t, step 9 is repeated to update the variablesωi, δi and Ei at the end of t = to + ∆t. Again, please note that for integrating the differentialequations, the latest values of ωi and δi (calculated at t = to) have been taken as the initial values.

12. Steps 10 and 11 are repeated to update the variables at t = to + 2∆t, to + 3∆t, to +4∆t, ⋯⋯ till the fault clears.

13. At t = tcl (tcl = p∆t, p being a positive integer and p > k), the fault clears. This conditionis imposed in the network equations by restoring Y`` to its original pre-fault value and subsequently,steps 8 - 12 are repeated to obtain the variations of δi and ωi. By observing the variation of δi, thestability of the system is assessed.

In step 9, the generator differential equations are numerically integrated. In the next lecture, wewill look into two such numerical integration techniques, namely, i) modified Euler’s method and ii)4th order Runga-Kutta technique.

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6.3.4 Modified Euler’s method of integration

Before discussing the application of Euler’s method for solving the swing equations, let us first reviewthe basic Euler’s method of numerical integration. Let the general from of a differential equation isgiven by;

dy

dx= f(x, y); y(xo) = yo; (6.28)

In equation (6.28), x and y are independent and dependent quantities respectively and xo and yoare initial values of x and y respectively. For the purpose of numerical integration, the independentaxis (x-axis) is divided into intervals of length ‘h’ such that discrete points on the independent axisare xo, xo + h, xo + 2h, ⋯⋯ etc. As indicated in equation (6.28), the value of y at x = xo is yo.The task is to calculate the values y1, y2, ⋯⋯ corresponding to the x co-ordinates xo + h, xo + 2h,⋯⋯ respectively. Once these values are obtained, the smooth curve representing the solution of thedifferential equation given in equation (6.28) can be plotted.

In the modified Euler’s method, the values y1, y2, ⋯⋯ are calculated in two steps:

PredictorIn this step, the approximate value of y1 (denoted as y(1)1 ) as;

y(1)1 = yo + hdy

dx∣x=xo

= yo + hf(xo, yo) (6.29)

Correctora) With the calculated value of y(1)1 , calculate the approximate value of

dy

dxat x = xo + h as;

dy

dx∣x=xo+h

= f(xo + h, y(1)1 ) (6.30)

b) With this updated value ofdy

dxat x = x1 = xo + h, the final value of y1 is calculated as;

y1 = yo +h

2 [ dydx

∣x=xo

+ dy

dx∣x=xo+h

]

or, y1 = yo +h

2[f(xo, yo) + f(xo + h, y(1)1 )] (6.31)

With this final value of y1 obtained at x = xo + h = x1, the above two steps are repeated tocalculate y2 at x = xo + 2h = x2 and subsequently, this process is repeated to obtain the completesolution of the differential equation.

Now, for our application, let us note that the independent axis (x-axis) denotes time. Therefore,as already discussed, for solving the differential equations, the time axis is divided into intervals of du-ration ∆t sec. (i.e. h = ∆t). Further, let us also assume that the values of δi and ωi (i = 1,2,⋯⋯m)have already been obtained at t = to and these values are denoted as δio and ωio respectively. More-over, the initial values of Pei (denoted as P (o)ei ) are also assumed to be calculated utilising the values

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of δio. With thse known values, the values of δi and ωi at t = to +∆t are calculated as follows.

Predictor step

In this step, the approximate values of δi and ωi are calculated as (denoted as δ(1)i and ω(1)irespectively);

δ(1)i = δio +∆t dδidt

∣t=to

= δio +∆t(ωio − ωs) (6.32)

ω(1)i = ωio +∆t dωidt

∣t=to

= ωio +∆t ωs2Hi

(Pmi − P (o)ei ) (6.33)

Corrector step

With the new values of δ(1)i (i = 1,2,⋯⋯ m) obtained in the predictor step, the values of Pei(i = 1,2,⋯⋯ m) are updated using equations (6.23)-(6.25) (after appropriately incorporating thenetwork conditions in the equation set (6.23)). Let these updates values of Pei be denoted as P (1)ei .Thereafter, the derivatives at the end of the present time step are calculated as follows:

dδidt

∣t=to+∆t

= ω(1)i − ωs (6.34)

dωidt

∣t=to+∆t

= ωs2Hi

(Pmi − P (1)ei ) (6.35)

With the above new derivative values obtained, the final values of δi and ωi at t = to+∆t (denotedas δi1 and ωi1 respectively) are calculated as;

δi1 = δio +∆t2 [ dδi

dt∣t=to

+ dδidt

∣t=to+∆t

] (6.36)

ωi1 = ωio +∆t2 [ dωi

dt∣t=to

+ dωidt

∣t=to+∆t

] (6.37)

Proceeding further, for calculating δi and ωi at t = to+2∆t, the quantities δio and ωio are replacedby δi1 and ωi1 respectively and equations (6.32)-(6.37) are followed again.

6.3.5 Runga Kutta 4th order method of integration

Let us again consider the same general form of a differential equation as in equation (6.28):

dy

dx= f(x, y); y(xo) = yo; (6.38)

Again, the meanings of different notations used in equation (6.38) are same as those in equation

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(6.28). In RK 4th order method, the value y1 (corresponding to x = xo + h) is calculated as;

y1 = yo +h

6(k1 + 2k2 + 2k3 + k4) (6.39)

In equation (6.39),k1 = hf(xo, yo) (6.40)

k2 = hf (xo +h

2 , yo +k1

2 ) (6.41)

k3 = hf (xo +h

2 , yo +k2

2 ) (6.42)

k4 = hf (xo + h, yo + k3) (6.43)

Now, for solving the transient stability problem with RK 4th order method, let us again assumethat the value of δi and ωi (i = 1,2,⋯⋯ m) are known at t = to (denoted as δio and ωio respec-tively). Moreover, the values of Pei are also assumed to be known (calculated utilising the valuesof δio). From these initial values, the procedure of calculation of δi1 and ωi1 (values of δi and ωi att = to +∆t) is as follows.

Calculation of first estimate of the derivatives

In this step, the first estimates of the derivatives for the ith machine (i = 1,2,⋯⋯ m) arecalculated as:

dδidt

∣(1)

= ωio − ωs (6.44)

dωidt

∣(1)

= ωs2Hi

[Pmi − P (o)ei ] (6.45)

With these first estimates of the derivatives, the values of δi and ωi (i = 1,2,⋯⋯m) are updatedas:

δ(1)i = δio +12∆t dδi

dt∣(1)

(6.46)

ω(1)i = ωio +12∆t dωi

dt∣(1)

(6.47)

With these values of δ(1)i (i = 1,2,⋯⋯ m), the values of Pei (i = 1,2,⋯⋯ m) are updatedfrom equations (6.23) - (6.25). Let these newly calculated values of Pei be denoted as P (1)ei (i =1,2,⋯⋯m). We now proceed to the next step.

Calculation of second estimate of the derivatives

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The second estimates of the derivatives are calculated as:

dδidt

∣(2)

= ω(1)i − ωs (6.48)

dωidt

∣(2)

= ωs2Hi

[Pmi − P (1)ei ] (6.49)

With these second estimates of the derivatives, the values of δi and ωi (i = 1,2,⋯⋯ m) areupdated as:

δ(2)i = δio +12∆t dδi

dt∣(2)

(6.50)

ω(2)i = ωio +12∆t dωi

dt∣(2)

(6.51)

With these values of δ(2)i , the values of Pei are again updated from equations (6.23) - (6.25). Letthese newly calculated values of Pei be denoted as P (2)ei (i = 1,2,⋯⋯ m). We now proceed to thenext step.

Calculation of third estimate of the derivatives

The third estimates of the derivatives are calculated as;

dδidt

∣(3)

= ω(2)i − ωs (6.52)

dωidt

∣(3)

= ωs2Hi

[Pmi − P (2)ei ] (6.53)

With these third estimates of the derivatives, the values of δi and ωi (i = 1,2,⋯⋯m) are updatedas;

δ(3)i = δio +∆t dδidt

∣(3)

(6.54)

ω(3)i = ωio +∆t dωidt

∣(3)

(6.55)

With these values of δ(3)i , the values of Pei are again updated from equations (6.23) - (6.25). Letthese newly calculated values of Pei be denoted as P (3)ei (i = 1,2,⋯⋯ m). We now proceed to thenext step.

Calculation of fourth estimate of the derivatives

264

The fourth estimates of the derivatives are calculated as:

dδidt

∣(4)

= ω(3)i − ωs (6.56)

dωidt

∣(4)

= ωs2Hi

[Pmi − P (3)ei ] (6.57)

After the fourth estimates are obtained, we are now in a position to calculate δi1 and ωi1(i = 1,2,⋯⋯m).

Calculation of final valuesThe final, updated values are calculated as:

δi1 = δio +∆t6 [ dδi

dt∣(1)+ 2 dδi

dt∣(2)+ 2 dδi

dt∣(3)+ dδidt

∣(4)

] (6.58)

ωi1 = ωio +∆t6 [ dωi

dt∣(1)+ 2 dωi

dt∣(2)+ 2 dωi

dt∣(3)+ dωidt

∣(4)

] (6.59)

Proceeding further, for calculating δi and ωi at t = to+2∆t, the quantities δio and ωio are replacedby δi1 and ωi1 respectively and equations (6.44)-(6.59) are followed again.

In the next lecture, we will illustrate the applicaion of Modified Euler’s method for transientstability calculation.

265

6.3.6 Example with Modified Euler’s method

As an illustration of the Modified Euler’s method, let us consider a three machine, 9 bus system.The schematic diagram of this system is shown in Fig. 6.4. The bus data of this system is givenin Table A.7 while the line data are given in Table A.8. Further, the values of x∕di, di (dampingconstant) and Hi (inertia constant) for all the three generators are given in Table 6.1.

With the data given in Tables A.7 and Table A.8, the load flow solution of this system has beencarried out and the load flow results are given in Table 6.2. With the help of load flow results andvalues of x∕di, the magnitudes and angles (∣E∣ and δ) of the internal voltages of all the generatorshave been calculated by utilising equations (6.16)-(6.18) and are also shown in Table 6.2. It is tobe noted that, throughout the transient stability simulation, the values of ∣Ei∣ (i = 1, 2, 3) are tobe kept constant at the values given in Table 6.2. Also, at steady state, the speed of all generatorsare assumed to be equal to ωs (i.e. ωio = ωs; i = 1, 2, 3). Now, following the arguments given instep 7 of sub-section 6.3.3, under steady state, the values of δi (i = 1, 2, 3) will remain constantat those values given in Table 6.2. Similarly, under steady state, the values of ωi (i = 1, 2, 3) willall be equal to ωs (= 376.9911184307752 rad./sec for a 60 Hz. system). Moreover, for transientstability simulation, a time step of 0.001 sec. (∆t = 0.001) has been taken. Further, to start with,the damping of the generators have been neglected.

Figure 6.4: 3 machine, 9 bus system

Now, let us assume that a three phase to ground fault takes place at bus 7 at t = 0.5 sec. Tosimulate this fault, the element Y77 is increased 1000 times to represent very high admittance toground. With this modification in Y77, the equation set (6.23) are solved to calculate the faulted

266

Table 6.1: Machine data for 9 bus system

Gen. x∕di di Hino. (p.u)1 0.0608 0.0254 23.642 0.1198 0.0066 6.403 0.1813 0.0026 3.01

Table 6.2: Load flow result for 9 bus system

Gen. PG QG ∣V ∣ θ ∣E∣ δno. (MW) (MVAR) (p.u) (deg.) (p.u) (deg.)1 71.64102147 27.0459235334 1.04 0.0 1.0566418430 2.27164584042 163.0 6.6536603184 1.025 9.2800054816 1.0502010147 19.73158576933 85.0 -10.8597090709 1.025 4.6647513331 1.0169664112 13.1664110346

values of Vi (i = 1, 2, 3). These values are shown in second column of Table 6.3. With thesecalculated values of the generator terminal voltages, the values of P (o)ei (i = 1, 2, 3) have beencalculated using equations (6.24)-(6.25) and are shown in third column of Table 6.3. Now, from Fig.6.4, bus 7 is the terminal bus of generator 2 (just after the transformer) and therefore, for any shortcircuit fault at bus 7, the real power output of generator 2 is expected to fall drastically. Indeed,from Table 6.3, after the fault, P (o)e2 is indeed very low (0.0003691143 p.u.). Also, note that thevalues of δio and ωio (i = 1, 2, 3), which are to be used for calculation at this time instant, are allequal to the corresponding steady state values as there was no fault prior to this time instant.

Table 6.3: Calculations with Euler’s method for predictor stage at t = 0.5 sec. (damping = 0)

Gen.V (p.u) P (o)e

no. (p.u)1 0.8515307492 - 0.0053320553i 0.67917489392 0.3391142785 + 0.1215867116i 0.00036911433 0.6169489129 + 0.0743553183i 0.3821503052

With these values of P (o)ei and ωio, the initial estimates ofdδidt

anddωidt

(i = 1, 2, 3) arecalculated from equations (6.27) and (6.26) respectively and are shown in columns 2 and 3 of Table6.4 respectively. Finally, the values of δ(1)i and ω(1)i (i = 1, 2, 3) are calculated by using equations(6.32) - (6.33) and are shown in columns 4 and 5 of Table 6.4 respectively. From Tables 6.1 - 6.4 itis observed that the steady state values of δi (for i = 1, 2, 3; shown in the last column of Table 6.1)are equal to the corresponding values of δ(1)i (for i = 1, 2, 3; shown in the fourth column of Table

6.4). This is due to the fact that at this predictor stage,dδidt

(i = 1, 2, 3) are all equal to zero.

However, due to fall in Pei (i = 1, 2, 3), from equation (6.33), the values of ω(1)i (i = 1, 2, 3) are all

267

greater than ωs.

Table 6.4: Calculations with Euler’s method for predictor stage at t = 0.5 sec. (damping = 0)

Gen. dδ

dt

dω

dt

δ(1) ω(1)

no. (deg.) (rad/sec.)1 0 0.2968990107 2.2716458404 376.99141532972 0 47.9965914231 19.7315857693 377.03911502213 0 29.2982026019 13.1664110346 377.0204166333

Once the calculations pertaining to the predictor stage at t = 0.5 sec. are over, we move onto the calculations pertaining to the corrector stage. Towards this goal, initially the values of Pei(i = 1, 2, 3) are updated using the values of δ(1)i (i = 1, 2, 3) in equations (6.23) - (6.25). Asdiscussed earlier, the updated value of Pei is denoted as P (1)ei . Now, as the values of δ(1)i (i = 1, 2, 3)are equal to the corresponding steady state values, the values of P (1)ei (i = 1, 2, 3) are also equalto the values given in Table 6.3. Subsequently, with the values of P (1)ei and ω(1)i (i = 1, 2, 3), thevalues of

dδidt

anddωidt

(i = 1, 2, 3) at the end of the present time step are calculated from equations(6.34) - (6.35) and are shown in columns 2 and 3 of Table 6.5 respectively. Lastly, the final valuesof δi and ωi (i = 1, 2, 3) at t = 0.5 sec. are calculated by using equations (6.36) - (6.37), which areshown in columns 4 and 5 of Table 6.5 respectively.

Table 6.5: Calculations with Euler’s method for corrector stage at t = 0.5 sec. (damping = 0)

Gen. dδ

dt

dω

dt

δ ωno. (deg.) (rad/sec.)1 0.0002968990 0.2968990107 2.2716543459 376.99141532972 0.0479965914 47.9965914231 19.7329607703 377.03911502213 0.0292982026 29.2982026019 13.1672503663 377.0204166333

With these final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.5 sec. at hand, we nowincrement the time by ∆t (= 0.001 sec.) and repeat the calculations for predictor and correctorstages for t = 0.501 sec. The detailed calculations are shown in Tables 6.6 - 6.9. Initially, with thevalues of δi and ωi (i = 1, 2, 3) just obtained, the generator terminal voltages and the generatoroutput electrical powers are calculated and are shown in Table 6.6. With these newly calculated

values of Pei (i = 1, 2, 3), the values ofdδidt

,dωidt

, δ(1)i and ω(1)i (i = 1, 2, 3) corresponding to the

predictor stage are calculated and are shown in Table 6.7. With these updated values of δ(1)i and ω(1)i(i = 1, 2, 3), the values of the generator terminal voltages and Pei (i = 1, 2, 3) are re-calculatedcorresponding to the corrector stage and are shown in Table 6.8. Lastly, using these updated values

of Pei (i = 1, 2, 3), the values of dδidt

anddωidt

corresponding to the corrector stage are calculated andare shown in Table 6.9. Finally, the values of δi and ωi (i = 1, 2, 3) at t = 0.501 sec. are calculatedby using equations (6.36)-(6.37), which are shown in columns 4 and 5 of Table 6.9 respectively.

268

Table 6.6: Caculations with Euler’s method for predictor stage at t = 0.501 sec. (damping = 0)

Gen.V (p.u) Pe

no. (p.u)1 0.8515306823 - 0.0053313633i 0.67916502412 0.3391113605 + 0.1215948479i 0.00036912883 0.6169476981 + 0.0743625981i 0.3821597463

Table 6.7: Calculations with Euler’s method for predictor stage at t = 0.501 sec. (damping = 0)

Gen. dδ

dt

dω

dt

δ(1) ω(1)

no. (deg.) (rad/sec.)1 0.0002968990 0.2969777087 2.2716713570 376.99171230742 0.0479965914 47.9965909948 19.7357107724 377.08711161313 0.0292982026 29.2976113769 13.1689290296 377.0497142447

Table 6.8: Caculations with Euler’s method for corrector stage at t = 0.501 sec. (damping = 0)

Gen.V (p.u) Pe

no. (p.u)1 0.8515305484 - 0.0053299791i 0.67914528442 0.3391055238 + 0.1216111204i 0.00036915793 0.6169452683 + 0.0743771578i 0.3821786281

Table 6.9: Calculations with Euler’s method for corrector stage at t = 0.501 sec. (damping = 0)

Gen. dδ

dt

dω

dt

δ ωno. (deg.) (rad/sec.)1 0.0005938767 0.2971351045 2.2716798648 376.99171238612 0.0959931824 47.9965901381 19.7370857735 377.08711161273 0.0585958139 29.2964289348 13.1697683443 377.0497136535

For subsequent time instants, the calculations proceed exactly in the same way as describedabove. The fault is assumed to be cleared at t = 0.6 sec. To simulate this event (clearing of fault),the value of Y77 is restored to its pre-fault value and subsequently, the values of Pei (i = 1, 2, 3)are calculated from equations (6.23)-(6.25). Please note that, while doing so, the latest values of δi(i = 1, 2, 3) obatained at t = 0.599 sec. are used in equation set (6.23). For subsequent instances(beyond t = 0.6 sec.), the calculations proceed in the identical manner and finally, the simulationstudy is stopped at t = 5.0 sec. The variations of δi (i = 1, 2, 3) with respect to the center of inertia(COI) are shown in Fig. 6.5. For calculating the value of δi with respect to the COI (denoted as

269

δCOIi ) at each time step, the following expression has been used:

δCOIi = δi − δCOI ; for i = 1, 2, 3; (6.60)

where,

δCOI =

m

∑i=1Hiδi

m

∑j=1Hj

(6.61)

In equation (6.61), ‘m’ denotes the number of generators in the system (in our present case, m= 3). At each time step, with the final calculated values of δi (i = 1, 2, 3) calculated at the endof corrector step, the value of δCOI is computed and thereafter, each value of δCOIi (i = 1, 2, 3)is computed with the help of equation (6.60). With the values of δCOIi thus obtained for all timesteps, the plots shown in Fig. 6.5 are obtained. These plots show that the generators experiencesustained oscillations. This is due to the fact that in this study, the damping of the generators havebeen neglected.

Figure 6.5: Variation of δCOI1 (with no damping) obtained with Euler’s method

Let us now consider the damping of the generators. As discussed earlier (in the context ofequations (6.14) and (6.15)), when damping is considered, an extra term (representing damping)is introduced in the differential equation corresponding to rate of change of speed. However, therewould be no change in the differential equation representing the rate of change of generator angle.Therefore, the set of differential equations for ith generator is given by;

dδidt

= ωi − ωs (6.62)

2Hi

ωs

dωidt

= Pmi − Pei − di(ωi − ωs) (6.63)

In equation (6.63), the extra term di(ωi − ωs) represnts the damping of the generator. Therewould be, of course, no change in the algebraic equations and the set of algebraic equations would

270

still be represented by equation set (6.23). With these sets of differential and algebraic equations,the calculations proceed in identically the same way as described above and the variations of δCOIi

(i = 1, 2, 3) for the same fault considered above are shown in Fig. 6.6. Comparison of thesethree plots with those shown in Fig. 6.5 shows that when damping of the generators are taken intoconsideration, the oscillations in all the three generators reduce gradually with time, which, indeedshould be the case. As the generator oscillations are decreasing with time, the generators will remainin synchronism and therefore, the system is stable.

Figure 6.6: Variation of δCOI1 (with damping) obtained with Euler’s method

We will now discuss the application of Runga Kutta (RK) 4th order method of integration forsolving the transient stability problem in the next lecture.

271

6.3.7 Example with Runga Kutta 4th order method

Again, as an example, 3 machine, 9 bus system shown in Fig. 6.4 is again considered. Initially, thedamping of the generators are neglected (i.e. di = 0 for i = 1,2, 3). The load flow results and theinitial values of the magnitudes and angles (∣E∣ and δ) of the internal voltages of all the generatorsare same as those already given in Table 6.2. As before, it is again assumed that at t = 0.5 sec.,a three phase to ground short circuit fault takes place at bus 7. The faulted generator terminalvoltages and generator output powers are same as those shown in Table 6.3. With these values of

P (o)ei and ωio, the estimatesdδidt

∣(1)

anddωidt

∣(1)

(i = 1, 2, 3) are calculated from equations (6.44)and (6.45) respectively and are shown in columns 2 and 3 of Table 6.10 respectively. With these firstestimates of the derivatives, the values of δ(1)i and ω(1)i (i = 1,2, 3) are calculated from equations(6.46) - (6.47) and are shown in columns 4 and 5 of Table 6.10 respectively.

Table 6.10: Calculations with RK method for first estimate at t = 0.5 sec. (damping = 0)

Gen. no. dδidt

∣(1) dωi

dt∣(1) δ(1) ω(1)

(deg.) (rad/sec.)1 0 0.2968990107 2.2716458404 376.991266882 0 47.9965914231 19.7315857693 377.015116723 0 29.2982026019 13.1664110346 377.00576753

With the values of δ(1)i (i = 1,2, 3) obtained above, the generator terminal voltages and outputpowers are updated from equations (6.23) - (6.25) and are shown in Table 6.11. Please note thatfollowing the notations used earlier, the output powers calculated at this stage are denoted as P (1)ei

(i = 1,2, 3).

Table 6.11: Caculations with RK method for second estimate at t = 0.5 sec. (damping = 0)

Gen.V (p.u) P (1)e

no. (p.u)1 0.8515307492 - 0.0053320553i 0.67917489392 0.3391142785 + 0.1215867116i 0.00036911433 0.6169489129 + 0.0743553183i 0.3821503052

With the values of ω(1)i and P (1)ei (i = 1,2, 3) calculated above, the quantitiesdδidt

∣(2)

anddωidt

∣(2)

(i = 1, 2, 3) are calculated from equations (6.48) - (6.49) and are shown in columns 2 and 3 of Table6.12 respectively. Subsequently, the values of δ(2)i and ω(2)i (i = 1,2, 3) are calculated from equations(6.50) - (6.51) and are shown in columns 4 and 5 of Table 6.12 respectively.

Again, with the values of δ(2)i (i = 1,2, 3) obtained above, the generator terminal voltages andoutput powers are updated from equations (6.23) - (6.25) and are shown in Table 6.13. Please note

272

Table 6.12: Calculations with RK method for second estimate at t = 0.5 sec. (damping = 0)

Gen. no. dδidt

∣(2) dωi

dt∣(2) δ(2) ω(2)

(deg.) (rad/sec.)1 0.0001484495 0.2968990107 2.2716500931 376.99126688022 0.0239982957 47.9965914231 19.7322732698 377.01511672643 0.0146491013 29.2982026019 13.1668307004 377.0057675320

that following the notations used earlier, the output powers calculated at this stage are denoted asP (2)ei (i = 1,2, 3).

Table 6.13: Caculations with RK method for third estimate at t = 0.5 sec. (damping = 0)

Gen.V (p.u) P (2)e

no. (p.u)1 0.8515307158 - 0.0053317093i 0.67916995902 0.3391128195 + 0.1215907798i 0.00036912163 0.6169483055 + 0.0743589582i 0.3821550258

Proceeding further, with the values of ω(2)i and P (2)ei (i = 1,2, 3) calculated above, the quantitiesdδidt

∣(3)

anddωidt

∣(3)

(i = 1, 2, 3) are calculated from equations (6.52) - (6.53) and are shown in

columns 2 and 3 of Table 6.14 respectively. Subsequently, the values of δ(3)i and ω(3)i (i = 1,2, 3) arecalculated from equations (6.54) - (6.55) and are shown in columns 4 and 5 of Table 6.14 respectively.

Table 6.14: Calculations with RK method for third estimate at t = 0.5 sec. (damping = 0)

Gen. no. dδidt

∣(3) dωi

dt∣(3) δ(3) ω(3)

(deg.) (rad/sec.)1 0.0001484495 0.2969383597 2.2716543459 376.99141536912 0.0239982957 47.9965912089 19.7329607703 377.03911502193 0.0146491013 29.2979069891 13.1672503663 377.0204163377

Again, with the values of δ(3)i (i = 1,2, 3) obtained above, the generator terminal voltages andoutput powers are updated from equations (6.23) - (6.25) and are shown in Table 6.15. Please notethat following the notations used earlier, the output powers calculated at this stage are denoted asP (3)ei (i = 1,2, 3).

Lastly, with the values of ω(3)i and P (3)ei (i = 1,2, 3) calculated above, the quantitiesdδidt

∣(4)

anddωidt

∣(4)

(i = 1, 2, 3) are calculated from equations (6.56) - (6.57) and are shown in columns 2

273

Table 6.15: Calculations with RK method for final estimate at t = 0.5 sec. (damping = 0)

Gen.V (p.u) P (3)e

no. (p.u)1 0.8515306823 - 0.0053313633i 0.67916502412 0.3391113605 + 0.1215948479i 0.00036912883 0.6169476981 + 0.0743625981i 0.3821597463

and 3 of Table 6.16 respectively. Finally, the values of δi and ωi (i = 1,2, 3) at the end of t = 0.5sec. are calculated from equations (6.58) - (6.59) and are shown in columns 4 and 5 of Table 6.16respectively.

Table 6.16: Calculations with RK method for final estimate at t = 0.5 sec. (damping = 0)

Gen. no. dδidt

∣(4) dωi

dt∣(4) δi ωi

(deg.) (rad/sec.)1 0.0002969383 0.2969777087 2.2716543463 376.99141535602 0.0479965912 47.9965909948 19.7329607703 377.03911502203 0.0292979069 29.2976113769 13.1672503634 377.0204164363

After the final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.5 sec. are obtained, weincrement the time by ∆t (= 0.001 sec.) and repeat the calculations for t = 0.501 sec. Towardsthis goal, the quantities δi and ωi (i = 1, 2, 3) shown in Table 6.16 are substituted for δio and ωio(i = 1, 2, 3) in equations (6.44) - (6.59). With these values of δio (i = 1, 2, 3), equations (6.23) -(6.25) are solved to calculate the initial values of Pei (i = 1, 2, 3) at t = 0.501 sec. The results areshown in Table 6.17.

Table 6.17: Calculations with RK method for initial estimate at t = 0.501 sec. (damping = 0)

Gen.V (p.u) P (o)e

no. (p.u)1 0.8515306823 - 0.0053313633i 0.67916502412 0.3391113605 + 0.1215948479i 0.00036912883 0.6169476981 + 0.0743625981i 0.3821597462

With the values of P (o)ei (i = 1, 2, 3) thus obtained, calculations pertaining to the first estimateare carried out by using equations (6.44) - (6.47) and the results are shown in Table 6.18. Sub-sequently, the values of P (1)ei (i = 1,2, 3) are calculated and the results are shown in Table 6.19.

With the values of P (1)ei , δ(1)i and ω(1)i (i = 1,2, 3) obtained as above, the calculations pertainingto second estimate are performed to obtain P (2)ei , δ(2)i and ω(2)i (i = 1,2, 3). The results are shown

274

Table 6.18: Calculations with RK method for first estimate at t = 0.501 sec. (damping = 0)

Gen. no. dδidt

∣(1) dωi

dt∣(1) δ(1) ω(1)

(deg.) (rad/sec.)1 0.0002969252 0.2969777084 2.2716628526 376.99156384482 0.0479965912 47.9965909948 19.7343357714 377.06311331753 0.0292980055 29.2976113792 13.1680896895 377.0350652419

Table 6.19: Calculations with RK method for first estimate at t = 0.501 sec. (damping = 0)

Gen.V (p.u) P (1)e

no. (p.u)1 0.8515306154 - 0.0053306712i 0.67915515442 0.3391084422 + 0.1216029842i 0.00036914343 0.6169464833 + 0.0743698779i 0.3821691871

in Tables 6.20 and 6.21.

Table 6.20: Calculations with RK method for second estimate at t = 0.501 sec. (damping = 0)

Gen. no. dδidt

∣(2) dωi

dt∣(2) δ(2) ω(2)

(deg.) (rad/sec.)1 0.0004454140 0.2970564057 2.2716671065 376.99156388422 0.0719948867 47.9965905664 19.7350232719 377.06311331733 0.0439468112 29.2970201614 13.1685093468 377.0350649463

Table 6.21: Calculations with RK method for second estimate at t = 0.501 sec. (damping = 0)

Gen.V (p.u) P (2)e

no. (p.u)1 0.8515305819 - 0.0053303251i 0.67915021962 0.3391069830 + 0.1216070523i 0.00036915063 0.6169458758 + 0.0743735177i 0.3821739075

Proceeding further, using the values of P (2)ei , δ(2)i and ω(2)i (i = 1,2, 3), the calculations pertainingto third estimate are performed to obtain P (3)ei , δ(3)i and ω(3)i (i = 1,2, 3). The results are shown inTables 6.22 and 6.23.

Using the values of P (3)ei , δ(3)i and ω(3)i (i = 1,2, 3), the fourth estimates of the derivatives arecomputed and subsequently, the final values of δi and ωi (i = 1, 2, 3) corresponding to t = 0.501sec. are obtained. The calculations are shown in Table 6.24.

275

Table 6.22: Calculations with RK method for third estimate at t = 0.501 sec. (damping = 0)

Gen. no. dδidt

∣(3) dωi

dt∣(3) δ(3) ω(3)

(deg.) (rad/sec.)1 0.0004454534 0.2970957537 2.2716798689 376.99171245172 0.0719948865 47.9965903523 19.7370857735 377.08711161243 0.0439465156 29.2967245580 13.1697683133 377.0497131608

Table 6.23: Calculations with RK method for third estimate at t = 0.501 sec. (damping = 0)

Gen.V (p.u) P (3)e

no. (p.u)1 0.8515304815 - 0.0053292869i 0.67913541532 0.3391026052 + 0.1216192565i 0.00036917253 0.6169440533 + 0.0743844372i 0.3821880684

Table 6.24: Calculations with RK method for final estimate at t = 0.501 sec. (damping = 0)

Gen. no. dδidt

∣(4) dωi

dt∣(4) δi ωi

(deg.) (rad/sec.)1 0.0005940209 0.2972137971 2.2716798685 376.99171243862 0.0959931816 47.9965897098 19.7370857735 377.08711161243 0.0585947300 29.2958377565 13.1697683161 377.0497132593

For subsequent time instants, the calculations proceed exactly in the same way as describedabove. As in the case with Euler’s method, in this case also, the fault is assumed to be cleared at t =0.6 sec. and finally, the simulation study is stopped at t = 5.0 sec. The variations of δi (i = 1, 2, 3)with respect to the center of inertia (COI) are shown in Fig. 6.7 below. Please note that in thisfigure, no damping of the generators has been considered.

The simulation studies have also been carried out by considering the damping of the generators.The variations of δi (i = 1, 2, 3) with respect to the center of inertia (COI) for this case are shownin Fig. 6.8 below. Comaprison of Figs. 6.7 - 6.8 with Figs. 6.5 - 6.6 reveals that the responsesobtained with these two methods are almost identical to each other.

With this example, we are now at the end of discussion of transient stability analysis. From thenext lecture, we will start the discussion of small signal stability analysis.

276

Figure 6.7: Variation of δCOI1 (with no damping) obtained with Runga-Kutta method

Figure 6.8: Variation of δCOI1 (with damping) obtained with Runga-Kutta method

277

6.4 Small signal analysis

For small signal analysis of a multimachine power system, we need to linearise the differential equa-tions of the machines and form the system A matrix. Thereafter, by computing the eigenvalues ofthe A matrix, the system stability can be assessed. Now, let us consider an ‘n’ bus power systemhaving ‘m’ generators. Each generator is represented by its classical model. Further, without anyloss of generality, it is assumed that the generators are connected at the first ‘m’ buses of the system.Now, for linearising the differential equations, let us recall the swing equations of each generatorhere for ready reference.

dδidt

= ωi − ωs for i = 1,2,⋯⋯m (6.64)

2Hi

ωs

dωidt

= Pmi − Pei for i = 1,2,⋯⋯m (6.65)

Linearising equation (6.64) for the ith generator, one can get,

d∆δidt

= ∆ωi (6.66)

Now, let us define,∆δ = [∆δ1, ∆δ2, ⋯⋯ ∆δm]T (6.67)

∆ω = [∆ω1, ∆ω2, ⋯⋯ ∆ωm]T (6.68)

In equations (6.67) and (6.68), the vectors ∆δ and ∆ω denote the vectors of perturbed valuesof rotor angle and machine speed respectively. Please note that the size of each of these two vectorsis (m × 1).

Now, collecting equation (6.66) for all the ‘m’ generators and re-writing them in matrix notationwe can obtain,

d

dt∆δ = F1∆ω (6.69)

In equation (6.69), F1 is a (m×m) identity matrix. Now, linearising equation (6.65) for the ith

generator, one can get,2Hi

ωs

d∆ωidt

= −∆Pei (6.70)

For performing linearisation of Pei, its expression its required. This expression can be derived asfollows. From equation (6.25), one can write,

Ii =Ei∠δi − Vi∠θi

jx∕di= Ei

x∕diej(δi−π/2) − Vi

x∕diej(θi−π/2)

278

Or,EiI

∗i = Eie

jδi I∗i =Ei

x∕diejπ/2 − EiVi

x∕diej(π/2+δi−θi)

Therefore,Pei = Re (EI I

∗i ) =

EiVix∕di

sin(δi − θi) (6.71)

Linearisation of equation (6.71) yields,

∆Pei = k1i∆δi + k2i∆θi + k3i∆Vi (6.72)

Where,

k1i =EiVix∕di

cos(δi − θi); k2i = −EiVix∕di

cos(δi − θi); k3i =Ei

x∕disin(δi − θi); (6.73)

In equation (6.73), the constants k1i, k2i and k3i are evaluated using the values of Ei, Vi, δi andθi at the current operating point for i = 1,2,⋯⋯m.

Now, again let us define,∆θg = [∆θ1, ∆θ2, ⋯⋯ ∆θm]T (6.74)

∆Vg = [∆V1, ∆V2, ⋯⋯ ∆Vm]T (6.75)

∆Pe = [∆Pe1, ∆Pe2, ⋯⋯ ∆Pem]T (6.76)

In equations (6.74) and (6.75), the vectors ∆θg and ∆Vg denote the vectors of perturbed valuesof generator terminal voltage angles and magnitudes respectively. Similarly, the vector ∆Pe denotesthe perturbed values of the generator real powers. Please note that the size of each of these threevectors is also (m × 1).

Now, collecting equation (6.72) for all the ‘m’ generators and re-writing them in matrix notationwe can obtain,

∆Pe =K1∆δ +K2∆θg +K3∆Vg (6.77)

In equation (6.77), the size of each of the matrices K1, K2 and K3 is (m ×m). Moreover, allthese three matrices are diagonal matrices and are given by;

K1 = diag (k11, k12, ⋯⋯ k1m)K2 = diag (k21, k22, ⋯⋯ k2m)K3 = diag (k31, k32, ⋯⋯ k3m)

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭(6.78)

279

6.4.1 Linearisation of network equations (at the generator buses)

To illustrate the procedure for linearisation of the network equations, let us first consider the firstequation of the equation set (6.23) below.

(Y11 + y1) V1 + Y12V2 +⋯ + Y1mVm +⋯ + Y1nVn = y1E1 (6.79)

Now, let,Yij = Yij exp (jαij); Vi = Vi exp (jθi); for i, j = 1,2,⋯⋯ n; (6.80)

andyk = yk exp (jβk); Ek = Ek exp (jδk); for k = 1,2,⋯⋯m; (6.81)

Utilising equations (6.80) and (6.81) in equation (6.79), we get,

Y11V1 exp (j (α11 + θ1)) + y1V1 exp (j (β1 + θ1)) + Y12V2 exp (j (α12 + θ2)) +⋯+Y1mVm exp (j (α1m + θm)) +⋯ + Y1nVn exp (j (α1n + θn)) = y1E1 exp (j (δ1 + β1))

(6.82)

Taking the real part of both the sides of equation (6.82), we get,

Y11V1 cos (α11 + θ1) + y1V1 cos (β1 + θ1) + Y12V2 cos (α12 + θ2) +⋯+Y1mVm cos (α1m + θm) +⋯ + Y1nVn cos (α1n + θn) = y1E1 cos (δ1 + β1)

(6.83)

Linearising equation (6.83), we get,

Y11 cos (α11 + θ1)∆V1 − Y11V1 sin (α11 + θ1)∆θ1 + y1 cos (β1 + θ1)∆V1−y1V1 sin (β1 + θ1)∆θ1 + Y12 cos (α12 + θ2)∆V2 − Y12V2 sin (α12 + θ2)∆θ2

+ ⋯ + Y1m cos (α1m + θm)∆Vm − Y1mVm sin (α1m + θm)∆θm +⋯+Y1n cos (α1n + θn)∆Vn − Y1nVn sin (α1n + θn)∆θn = −y1E1 sin (δ1 + β1)∆δ1

(6.84)

Equation (6.84) can be re-written as,

a11∆V1 + a12∆V2 +⋯ + a1m∆Vm +⋯ + a1n∆Vn+b11∆θ1 + b12∆θ2 +⋯ + b1m∆θm +⋯ + b1n∆θn = g1∆δ1

(6.85)

In equation (6.85),

a11 = Y11 cos (α11 + θ1) + y1 cos (β1 + θ1); a12 = Y12 cos (α12 + θ2);a1m = Y1m cos (α1m + θm); a1n = Y1n cos (α1n + θn);

b11 = −Y11V1 sin (α11 + θ1) − y1V1 sin (β1 + θ1); b12 = −Y12V2 sin (α12 + θ2);b1m = −Y1mVm sin (α1m + θm); b1n = −Y1nVn sin (α1n + θn);

g1 = −y1E1 sin (δ1 + β1);

(6.86)

Again, taking the real part of both sides of the second equation of the equation set (6.23) (after

280

substituting equations (6.80) and (6.81) into it), we get,

Y21V1 cos (α21 + θ1) + y2V2 cos (β2 + θ2) + Y22V2 cos (α22 + θ2) +⋯+Y2mVm cos (α2m + θm) +⋯ + Y2nVn cos (α2n + θn) = y2E2 cos (δ2 + β2)

(6.87)

Linearising equation (6.87), one can get,

a21∆V1 + a22∆V2 +⋯ + a2m∆Vm +⋯ + a2n∆Vn+b21∆θ1 + b22∆θ2 +⋯ + b2m∆θm +⋯ + b2n∆θn = g2∆δ2

(6.88)

In equation (6.88),

a21 = Y21 cos (α21 + θ1); a22 = Y22 cos (α22 + θ2) + y2 cos (β2 + θ2);a2m = Y2m cos (α2m + θm); a2n = Y2n cos (α2n + θn);

b21 = −Y21V1 sin (α21 + θ1); b22 = −Y22V2 sin (α22 + θ2) − y2V2 sin (β2 + θ2);b2m = −Y2mVm sin (α2m + θm); b2n = −Y2nVn sin (α2n + θn);

g2 = −y2E2 sin (δ2 + β2);

(6.89)

Similarly, continuing with linearisation of the real parts of first ‘m’ equations (corresponding tothe generator buses) of the equation set (6.23), we get,

[A1 B1]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [G] [∆δ] (6.90)

In equation (6.90),

A1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

a11 a12 ⋯ a1m a1,(m+1) ⋯⋯ a1n

a21 a22 ⋯ a2m a2,(m+1) ⋯⋯ a2n

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮am1 am2 ⋯ amm am,(m+1) ⋯⋯ amn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m × n) matrix (6.91)

B1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

b11 b12 ⋯ b1m b1,(m+1) ⋯⋯ b1n

b21 b22 ⋯ b2m b2,(m+1) ⋯⋯ b2n

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮bm1 bm2 ⋯ bmm bm,(m+1) ⋯⋯ bmn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m × n) matrix (6.92)

281

G =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

g1 0 ⋯ 0 ⋯⋯ 00 g2 ⋯ 0 ⋯⋯ 0⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮0 0 ⋯ 0 ⋯⋯ gm

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m ×m) diagonal matrix (6.93)

The elements of the matrices A1, B1 and G are given as;

aii = Yii cos (αii + θi) + yi cos (βi + θi); i = 1,2,⋯⋯m;

aij = Yij cos (αij + θj); i = 1,2,⋯⋯m; j = 1,2,⋯⋯ n; i ≠ j;

bii = −YiiVi sin (αii + θi) − yiVi sin (βi + θi); i = 1,2,⋯⋯m;

bij = −YijVj sin (αij + θj); i = 1,2,⋯⋯m; j = 1,2,⋯⋯ n; i ≠ j;

gi = −yiEi sin (δi + βi); i = 1,2,⋯⋯m;

(6.94)

Further, the vectors ∆θL and ∆VL are defined as;

∆θL = [∆θm+1, ∆θm+2, ⋯⋯ ∆θn]T (6.95)

∆VL = [∆Vm+1, ∆Vm+2, ⋯⋯ ∆Vn]T (6.96)

Please note that the size of each of the above two vectors is ((n −m) × 1). So far, we haveconsidered only the algebraic equations at the generator buses. However, for completing the smallsignal model, the algebraic equations at the load buses all need to be linearised. We will discuss thisissue in the next lecture.

282

6.4.2 Linearisation of network equations (at the load buses)

Now, let us consider the algebraic equations at the load buses. For this, please recollect that theloads have been assumed to be connected at the last (n−m) buses. The real part of the pth equation(p = (m + 1), (m + 2),⋯⋯ n) of the equations set (6.23) is given by,

Yp1V1 cos (αp1 + θ1) + Yp2V2 cos (αp2 + θ2) +⋯ + YpnVn cos (αpn + θn) = 0 (6.97)

Linearising the above equation we get,

Yp1 cos (αp1 + θ1)∆V1 − Yp1V1 sin (αp1 + θ1)∆θ1 + Yp2 cos (αp2 + θ2)∆V2

−Yp2V2 sin (αp2 + θ2)∆θ2 +⋯ + Ypn cos (αpn + θn)∆Vn − YpnVn sin (αpn + θn)∆θn = 0(6.98)

Or,ap1∆V1 + ap2∆V2 +⋯ + apn∆Vn + bp1∆θ1 + bp2∆θ2 +⋯ + bpn∆θn = 0 (6.99)

In equation (6.99),

apj = Ypj cos (αpj + θj); j = 1,2,⋯⋯ n;bpj = −YpjVj sin (αpj + θj); j = 1,2,⋯⋯ n;

(6.100)

Please note that equation (6.99) can be written for all the ‘n−m’ load buses (by varying ‘p’ from‘m+ 1’ to ‘n’). In that case, in equation (6.100) also, ‘p’ will vary from ‘m+ 1’ to ‘n’. Collecting allthese ‘n −m’ equations and putting them in a matrix form, we get,

[R1 S1]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [Φ1] [∆δ] (6.101)

In equation (6.101), the matrix Φ1 is a (n −m) ×m null matrix and the matrices R1 and S1

are given by,

R1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

r11 r12 ⋯⋯ r1n

r21 r22 ⋯⋯ r2n

⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮

r(n−m),1 r(n−m),2 ⋯⋯ r(n−m),n

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a ((n −m) × n) matrix (6.102)

283

S1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

s11 s12 ⋯⋯ s1n

s21 s22 ⋯⋯ s2n

⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮

s(n−m),1 s(n−m),2 ⋯⋯ s(n−m),n

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a ((n −m) × n) matrix (6.103)

The elements of the matrices R1 and S1 are given by,

rij = a(i+m),j; i = 1,2,⋯⋯ (n −m); j = 1,2,⋯⋯ n;sij = b(i+m),j; i = 1,2,⋯⋯ (n −m); j = 1,2,⋯⋯ n;

(6.104)

In equation (6.104), the expressions of the co-efficients a(i+m),j and b(i+m),j are given by equation(6.100). Now, taking the imaginary part of both the sides of equation (6.82), we get,

Y11V1 sin (α11 + θ1) + y1V1 sin (β1 + θ1) + Y12V2 sin (α12 + θ2) +⋯+Y1mVm sin (α1m + θm) +⋯ + Y1nVn sin (α1n + θn) = y1E1 sin (δ1 + β1)

(6.105)

Linearising equation (6.105), we get,

Y11 sin (α11 + θ1)∆V1 + Y11V1 cos (α11 + θ1)∆θ1 + y1 sin (β1 + θ1)∆V1+y1V1 cos (β1 + θ1)∆θ1 + Y12 sin (α12 + θ2)∆V2 + Y12V2 cos (α12 + θ2)∆θ2

+ ⋯ + Y1m sin (α1m + θm)∆Vm + Y1mVm cos (α1m + θm)∆θm +⋯+Y1n sin (α1n + θn)∆Vn + Y1nVn cos (α1n + θn)∆θn = y1E1 cos (δ1 + β1)∆δ1

(6.106)

Equation (6.106) can be re-written as,

c11∆V1 + c12∆V2 +⋯ + c1m∆Vm +⋯ + c1n∆Vn+d11∆θ1 + d12∆θ2 +⋯ + d1m∆θm +⋯ + d1n∆θn = h1∆δ1

(6.107)

In equation (6.107),

c11 = Y11 sin (α11 + θ1) + y1 sin (β1 + θ1); c12 = Y12 sin (α12 + θ2);c1m = Y1m sin (α1m + θm); c1n = Y1n sin (α1n + θn);

d11 = Y11V1 cos (α11 + θ1) + y1V1 cos (β1 + θ1); d12 = Y12V2 cos (α12 + θ2);d1m = Y1mVm cos (α1m + θm); d1n = Y1nVn cos (α1n + θn);

h1 = y1E1 cos (δ1 + β1);

(6.108)

Again, taking the imaginary part of both sides of the second equation of the equation set (6.23)(after substituting equations (6.80) and (6.81) into it), we get,

Y21V1 sin (α21 + θ1) + y2V2 sin (β2 + θ2) + Y22V2 sin (α22 + θ2) +⋯+Y2mVm sin (α2m + θm) +⋯ + Y2nVn sin (α2n + θn) = y2E2 sin (δ2 + β2)

(6.109)

284

Linearising equation (6.109), one can get,

c21∆V1 + c22∆V2 +⋯ + c2m∆Vm +⋯ + c2n∆Vn+d21∆θ1 + d22∆θ2 +⋯ + d2m∆θm +⋯ + d2n∆θn = h2∆δ2

(6.110)

In equation (6.110),

c21 = Y21 sin (α21 + θ1); c22 = Y22 sin (α22 + θ2) + y2 sin (β2 + θ2);c2m = Y2m sin (α2m + θm); c2n = Y2n sin (α2n + θn);

d21 = Y21V1 cos (α21 + θ1); d22 = Y22V2 cos (α22 + θ2) + y2V2 cos (β2 + θ2);d2m = Y2mVm cos (α2m + θm); d2n = Y2nVn cos (α2n + θn);

h2 = y2E2 sin (δ2 + β2);

(6.111)

Continuing with linearisation of the imaginary parts of first ‘m’ equations (corresponding to thegenerator buses) of the equation set (6.23), we get,

[C1 D1]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [H] [∆δ] (6.112)

In equation (6.112),

C1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 ⋯ c1m c1,(m+1) ⋯⋯ c1n

c21 c22 ⋯ c2m c2,(m+1) ⋯⋯ c2n

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮cm1 cm2 ⋯ cmm cm,(m+1) ⋯⋯ cmn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m × n) matrix (6.113)

D1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

d11 d12 ⋯ d1m d1,(m+1) ⋯⋯ d1n

d21 d22 ⋯ d2m d2,(m+1) ⋯⋯ d2n

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮dm1 dm2 ⋯ dmm dm,(m+1) ⋯⋯ dmn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m × n) matrix (6.114)

H =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

h1 0 ⋯ 0 ⋯⋯ 00 h2 ⋯ 0 ⋯⋯ 0⋮ ⋮ ⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮ ⋮ ⋮0 0 ⋯ 0 ⋯⋯ hm

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a (m ×m) diagonal matrix (6.115)

285

The elements of the matrices C1, D1 and H are given as;

cii = Yii sin (αii + θi) + yi sin (βi + θi); i = 1,2,⋯⋯m;

cij = Yij sin (αij + θj); i = 1,2,⋯⋯m; j = 1,2,⋯⋯ n; i ≠ j;

dii = YiiVi cos (αii + θi) + yiVi cos (βi + θi); i = 1,2,⋯⋯m;

dij = YijVj cos (αij + θj); i = 1,2,⋯⋯m; j = 1,2,⋯⋯ n; i ≠ j;

hi = yiEi cos (δi + βi); i = 1,2,⋯⋯m;

(6.116)

Now, let us consider the imaginary parts of the algebraic equations at the load buses. The realpart of the pth equation (p = (m + 1), (m + 2),⋯⋯ n) of the equations set (6.23) is given by,

Yp1V1 sin (αp1 + θ1) + Yp2V2 sin (αp2 + θ2) +⋯ + YpnVn sin (αpn + θn) = 0 (6.117)

Linearising the above equation we get,

Yp1 sin (αp1 + θ1)∆V1 + Yp1V1 cos (αp1 + θ1)∆θ1 + Yp2 sin (αp2 + θ2)∆V2

+Yp2V2 cos (αp2 + θ2)∆θ2 +⋯ + Ypn sin (αpn + θn)∆Vn + YpnVn cos (αpn + θn)∆θn = 0(6.118)

Or,cp1∆V1 + cp2∆V2 +⋯ + cpn∆Vn + dp1∆θ1 + dp2∆θ2 +⋯ + dpn∆θn = 0 (6.119)

In equation (6.119),

cpj = Ypj sin (αpj + θj); j = 1,2,⋯⋯ n;dpj = YpjVj cos (αpj + θj); j = 1,2,⋯⋯ n;

(6.120)

Again, equation (6.119) can be written for all the ‘n−m’ load buses (by varying ‘p’ from ‘m+1’to ‘n’). In that case, in equation (6.120) also, ‘p’ will vary from ‘m + 1’ to ‘n’. Collecting all these‘n −m’ equations and putting them in a matrix form, we get,

[U1 V1]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [Φ2] [∆δ] (6.121)

In equation (6.121), the matrix Φ2 is a (n −m) ×m null matrix and the matrices U1 and V1

286

are given by,

U1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u11 u12 ⋯⋯ u1n

u21 u22 ⋯⋯ u2n

⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮

u(n−m),1 u(n−m),2 ⋯⋯ u(n−m),n

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a ((n −m) × n) matrix (6.122)

V1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

v11 v12 ⋯⋯ v1n

v21 v22 ⋯⋯ v2n

⋮ ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋮

v(n−m),1 v(n−m),2 ⋯⋯ v(n−m),n

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⇒ is a ((n −m) × n) matrix (6.123)

The elements of the matrices U1 and V1 are given by,

uij = c(i+m),j; i = 1,2,⋯⋯ (n −m); j = 1,2,⋯⋯ n;vij = d(i+m),j; i = 1,2,⋯⋯ (n −m); j = 1,2,⋯⋯ n;

(6.124)

In equation (6.124), the expressions of the co-efficients c(i+m),j and d(i+m),j are given by equation(6.120).

Now, combining equations (6.90), (6.101), (6.112) and (6.121), one can write,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A1 B1

R1 S1

C1 D1

U1 V1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

GΦ1

HΦ2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

[∆δ] (6.125)

Or,

[J1]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [J2] [∆δ] (6.126)

Where,

[J1] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A1 B1

R1 S1

C1 D1

U1 V1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

; [J2] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

GΦ1

HΦ2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.127)

Now, let us recollect that the dimensions of various sub-matrices are as follows: A1 → (m × n),B1 → (m × n), G → (m ×m), R1 → ((n −m) × n), S1 → ((n −m) × n), Φ1 → ((n −m) ×m),C1 → (m × n), D1 → (m × n), H → (m ×m), U1 → ((n −m) × n), V1 → ((n −m) × n),

287

Φ2 → ((n−m)×m). Therefore, the size of the matrix J1 is (2n×2n) and that of the matrix J2 is(2n ×m). Hence, matrix J1 is a square matrix and hence invertible. Thus, from equation (6.126),

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆Vg∆VL∆θg∆θL

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [J1]−1 [J2] [∆δ] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A2

B2

C2

D2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

[∆δ] (6.128)

Hence, from equation (6.128),

[∆Vg] = [A2] [∆δ] ; [∆θg] = [C2] [∆δ] ; (6.129)

Please note that the dimension of both the matrices A2 and C2 is (m×m). We are now ready toform the system state matrix which we will discuss in the next lecture. Further, in the next lecture,we will also look at an example of small signal stability analysis.

288

6.4.3 Formation of system state matrix

From equation (6.70), let us recall the following linearised equation for ith generator,

2Hi

ωs

d∆ωidt

= −∆Pei (6.130)

Collecting equation (6.130) for all the ‘m’ generators, the following matrix equation can bewritten;

2ωs

HMd

dt∆ω = −∆Pe (6.131)

In equation (6.131), the matrix HM is a (m ×m) diagonal matrix and is given by,

HM = diag (H1, H2, ⋯⋯Hm) (6.132)

Substituting equation (6.77) into equation (6.131) we can get,

2ωs

HMd

dt∆ω = − (K1∆δ +K2∆θg +K3∆Vg) (6.133)

Further, substituting equation (6.129) into equation (6.133) one can write,

2ωs

HMd

dt∆ω = − (K1 +K2C2 +K3A2)∆δ (6.134)

Or,d

dt∆ω =K4∆δ (6.135)

Where,K4 = −

ωs2 H−1

M (K1 +K2C2 +K3A2) (6.136)

Again, recalling equation (6.69), we have,

d

dt∆δ = F1∆ω (6.137)

Combining equations (6.135) and (6.137), we can write,

d

dt[∆δ∆ω

] = [ O F1

K4 O] [∆δ

∆ω] (6.138)

Equation (6.138) is the required state space equation of the system and the matrix A = [ O F1

K4 O]

is the corresponding state matrix. The eigenvalues of the matrix A gives the necessary informationabout the small signal stability of the system under study.

Now, let us consider the damping of the generators. Towards this goal, let us now linearize

289

equation (6.63) to obtain,2Hi

ωs

d∆ωidt

= −∆Pei − di∆ωi (6.139)

Collecting equation (6.139) for all the ‘m’ generators, the following matrix equation can bewritten;

2ωs

HMd

dt∆ω = −∆Pe −DM∆ω (6.140)

In equation (6.140), the matrix DM is a (m ×m) diagonal matrix and is given by,

DM = diag (d1, d2, ⋯⋯ dm) (6.141)

Following the same procedure as in equations (6.133) and (6.134), we can get,

2ωs

HMd

dt∆ω = − (K1 +K2C2 +K3A2)∆δ −DM∆ω (6.142)

Or,d

dt∆ω =K4∆δ +K5∆ω (6.143)

Where,K5 = −

ωs2 H−1

MDM (6.144)

Combining equations (6.137) and (6.144), we can write,

d

dt[∆δ∆ω

] = [ O F1

K4 K5] [∆δ

∆ω] (6.145)

Equation (6.145) is the required state space equation of the system when damping of the gener-

ators are considered and the matrix A = [ O F1

K4 K5] is the corresponding state matrix.

6.5 Example of small signal stability

As an illustration of small signal stability, let us consider the three machine, 9 bus system (m = 3and n = 9) shown in Fig. Fig. 6.4. The complete data of this system are given in Tables A.7, A.8and 6.1. Further, the different quantities pertaining to the initial condition are given in Table 6.2.The load flow solution of the system is shown in Table 6.25.

Initially, the damping of the machines are neglected (di = 0; i = 1, 2, 3). With these initialvalues, the different matrices are calculated as follows:

K1 =⎡⎢⎢⎢⎢⎢⎢⎣

18.0599 0 00 8.8364 00 0 5.6864

⎤⎥⎥⎥⎥⎥⎥⎦(6.146)

290

Table 6.25: Load flow solution of the 3 machine system

Bus No. Voltage (p.u.)1 1.04002 1.0116 + 0.1653i3 1.0216 + 0.0834i4 1.0250 - 0.0397i5 0.9932 - 0.0693i6 1.0106 - 0.0651i7 1.0236 + 0.0665i8 1.0158 + 0.0129i9 1.0317 + 0.0354i

K2 =⎡⎢⎢⎢⎢⎢⎢⎣

−18.0599 0 00 −8.8364 00 0 −5.6864

⎤⎥⎥⎥⎥⎥⎥⎦(6.147)

K3 =⎡⎢⎢⎢⎢⎢⎢⎣

0.6889 0 00 1.5902 00 0 0.8293

⎤⎥⎥⎥⎥⎥⎥⎦(6.148)

A1 =⎡⎢⎢⎢⎢⎢⎢⎣

0.0000 0 0 0.6715 0 0 0 0 00 3.9262 0 0 0 0 −1.0380 0 00 0 1.8364 0 0 0 0 0 −0.5856

⎤⎥⎥⎥⎥⎥⎥⎦(6.149)

B1 =⎡⎢⎢⎢⎢⎢⎢⎣

35.1608 0 0 −17.7955 0 0 0 0 00 24.6293 0 0 0 0 −16.3777 0 00 0 23.0684 0 0 0 0 0 −17.6066

⎤⎥⎥⎥⎥⎥⎥⎦(6.150)

G =⎡⎢⎢⎢⎢⎢⎢⎣

17.3653 0 00 8.2516 00 0 5.4618

⎤⎥⎥⎥⎥⎥⎥⎦(6.151)

R1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0.0000 0 0 1.7844 −0.5547 −1.2622 0 0 00 0 0 −0.9153 2.5634 0 −1.5727 0 00 0 0 −1.5342 0 3.0558 0 0 −1.47300 −2.5802 0 0 −0.7691 0 5.0984 −1.7909 00 0 0 0 0 0 −2.5024 4.0411 −1.49020 0 −1.3878 0 0 −0.9200 0 −1.2792 3.5391

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦(6.152)

The matrix S1 is given in parts in equations (6.153) and (6.154) below, as the width of the pageis not sufficient enough to accomodate the complete matrix S1.

291

The matrix S1(∶,1 ∶ 5) (comprising of the elements in columns 1 to 5 for all rows) is given by;

S1(∶,1 ∶ 5) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−18.0556 0 0 40.4237 −11.62000 0 0 −11.9486 17.98580 0 0 −10.8507 00 −16.1854 0 0 −6.01690 0 0 0 00 0 −17.4335 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.153)

The matrix S1(∶,6 ∶ 9) (comprising of the elements in columns 6 to 9 for all rows) is given by;

S1(∶,6 ∶ 9) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−10.7481 0 0 00 −6.0372 0 0

16.5710 0 0 −5.72020 36.0958 −13.8936 00 −13.9138 23.9677 −10.0539

−5.7307 0 −9.9240 33.0882

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.154)

C1 =⎡⎢⎢⎢⎢⎢⎢⎣

−33.8085 0 0 17.3481 0 0 0 0 00 −24.0286 0 0 0 0 15.9663 0 00 0 −22.5058 0 0 0 0 0 17.0548

⎤⎥⎥⎥⎥⎥⎥⎦(6.155)

D1 =⎡⎢⎢⎢⎢⎢⎢⎣

0.0000 0 0 0.6889 0 0 0 0 00 4.0244 0 0 0 0 −1.0648 0 00 0 1.8823 0 0 0 0 0 −0.6046

⎤⎥⎥⎥⎥⎥⎥⎦(6.156)

H =⎡⎢⎢⎢⎢⎢⎢⎣

0.6889 0 00 2.9596 00 0 1.2777

⎤⎥⎥⎥⎥⎥⎥⎦(6.157)

The matrix U1 is given in parts in equations (6.158) and (6.159) below, as the width of the pageis not sufficient enough to accomodate the complete matrix U1.

The matrix U1(∶,1 ∶ 5) (comprising of the elements in columns 1 to 5 for all rows) is given by;

U1(∶,1 ∶ 5) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

17.3611 0 0 −39.4074 11.67100 0 0 11.6482 −18.06470 0 0 10.5779 00 15.7906 0 0 6.04330 0 0 0 00 0 17.0083 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.158)

292

The matrix U1(∶,6 ∶ 9) (comprising of the elements in columns 6 to 9 for all rows) is given by;

U1(∶,6 ∶ 9) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

10.6138 0 0 00 5.8855 0 0

−16.3639 0 0 5.54100 −35.1890 13.6763 00 13.5642 −23.5930 9.7389

5.6591 0 9.7688 −32.0513

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.159)

V1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0.0000 0 0 1.8304 −0.5523 −1.2782 0 0 00 0 0 −0.9389 2.5522 0 −1.6133 0 00 0 0 −1.5737 0 3.0944 0 0 −1.52070 −2.6447 0 0 −0.7657 0 5.2297 −1.8194 00 0 0 0 0 0 −2.5669 4.1053 −1.53840 0 −1.4225 0 0 −0.9316 0 −1.2995 3.6536

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦(6.160)

The matrix J1 is given in parts in equations (6.161) - (6.163) below, as the width of the page isnot sufficient enough to accomodate the complete matrix J1.

The matrix J1(∶,1 ∶ 6) (comprising of the elements in columns 1 to 6 for all rows) is given by;

J1(∶,1 ∶ 6) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0.0000 0 0 0.6715 0 00 3.9262 0 0 0 00 0 1.8364 0 0 0

0.0000 0 0 1.7844 −0.5547 −1.26220 0 0 −0.9153 2.5634 00 0 0 −1.5342 0 3.05580 −2.5802 0 0 −0.7691 00 0 0 0 0 00 0 −1.3878 0 0 −0.9200

−33.8085 0 0 17.3481 0 00 −24.0286 0 0 0 00 0 −22.5058 0 0 0

17.3611 0 0 −39.4074 11.6710 10.61380 0 0 11.6482 −18.0647 00 0 0 10.5779 0 −16.36390 15.7906 0 0 6.0433 00 0 0 0 0 00 0 17.0083 0 0 5.6591

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.161)

The matrices J1(∶,7 ∶ 12) (comprising of the elements in columns 7 to 12 for all rows) and

293

J1(∶,13 ∶ 18) (comprising of the elements in columns 13 to 18 for all rows) are given by;

J1(∶,7 ∶ 12) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 35.1608 0 0−1.0380 0 0 0 24.6293 0

0 0 −0.5856 0 0 23.06840 0 0 −18.0556 0 0

−1.5727 0 0 0 0 00 0 −1.4730 0 0 0

5.0984 −1.7909 0 0 −16.1854 0−2.5024 4.0411 −1.4902 0 0 0

0 −1.2792 3.5391 0 0 −17.43350 0 0 0.0000 0 0

15.9663 0 0 0 4.0244 00 0 17.0548 0 0 1.88230 0 0 0.0000 0 0

5.8855 0 0 0 0 00 0 5.5410 0 0 0

−35.1890 13.6763 0 0 −2.6447 013.5642 −23.5930 9.7389 0 0 0

0 9.7688 −32.0513 0 0 −1.4225

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.162)

J1(∶,13 ∶ 18) =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−17.7955 0 0 0 0 00 0 0 −16.3777 0 00 0 0 0 0 −17.6066

40.4237 −11.6200 −10.7481 0 0 0−11.9486 17.9858 0 −6.0372 0 0−10.8507 0 16.5710 0 0 −5.7202

0 −6.0169 0 36.0958 −13.8936 00 0 0 −13.9138 23.9677 −10.05390 0 −5.7307 0 −9.9240 33.0882

0.6889 0 0 0 0 00 0 0 −1.0648 0 00 0 0 0 0 −0.6046

1.8304 −0.5523 −1.2782 0 0 0−0.9389 2.5522 0 −1.6133 0 0−1.5737 0 3.0944 0 0 −1.5207

0 −0.7657 0 5.2297 −1.8194 00 0 0 −2.5669 4.1053 −1.53840 0 −0.9316 0 −1.2995 3.6536

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.163)

294

J2 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

17.3653 0 00 8.2516 00 0 5.46180 0 00 0 00 0 00 0 00 0 00 0 0

0.6889 0 00 2.9596 00 0 1.27770 0 00 0 00 0 00 0 00 0 00 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.164)

A2 =⎡⎢⎢⎢⎢⎢⎢⎣

0.0200 −0.0154 −0.00460.0594 −0.0764 0.01690.0499 −0.0146 −0.0353

⎤⎥⎥⎥⎥⎥⎥⎦(6.165)

C2 =⎡⎢⎢⎢⎢⎢⎢⎣

0.8328 0.0934 0.07380.1811 0.6881 0.13070.2272 0.2054 0.5673

⎤⎥⎥⎥⎥⎥⎥⎦(6.166)

HM =⎡⎢⎢⎢⎢⎢⎢⎣

23.6400 0 00 6.4000 00 0 3.0100

⎤⎥⎥⎥⎥⎥⎥⎦(6.167)

K4 =⎡⎢⎢⎢⎢⎢⎢⎣

−24.1837 13.5322 10.651444.3537 −77.5840 33.230278.3229 73.9174 −152.2403

⎤⎥⎥⎥⎥⎥⎥⎦(6.168)

F1 =⎡⎢⎢⎢⎢⎢⎢⎣

1 0 00 1 00 0 1

⎤⎥⎥⎥⎥⎥⎥⎦(6.169)

295

With all the above calculated matrices, the system state matrix is finally computed as;

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 1.0000 0 00 0 0 0 1.0000 00 0 0 0 0 1.0000

−24.1837 13.5322 10.6514 0 0 044.3537 −77.5840 33.2302 0 0 078.3229 73.9174 −152.2403 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.170)

The eigenvalues of the system state matrix are lastly computed and are shown in Table 6.26.

Table 6.26: Eigenvalues of the 3 machine system with damping neglected

No. Eigenvalue1 0.0000 +13.3602i2 0.0000 -13.3602i3 0.0000 + 8.6898i4 0.0000 - 8.6898i5 -0.0000 + 0.0000i6 -0.0000 - 0.0000i

From Table 6.26, following points can be noted:

• There are total 6 eigenvalues. This indeed should be the case as we have three machines in thesystem and each machine has two state variables.

• There are two zero eigenvalues. One zero eigenvalue is due to the absence of damping in thesystem and the other eigenvalue is corresponding to δ1 (which is taken to be zero to providethe reference for load flow calculation).

• All the non-zero eigenvalues appear in pairs.

• The real parts of all the non-zero eigenvalues are zero. In other words, upon a disturbance,the oscillations in the system would be persisting (i.e. the magnitudes of the oscillations wouldnot reduce with time). Indeed, from the plots shown in Figs. 6.5 and 6.7 it is observed thatupon a fault, the oscillations are indeed sustaining with a constant amplitude.

Now let us consider the damping of the generators. For this case, the matrices given above wouldremain the same. However, one extra matrix, K5 would appear (as shown in equation (6.145)) andthis matrix is computed as;

K5 =⎡⎢⎢⎢⎢⎢⎢⎣

−0.2025 0 00 −0.1944 00 0 −0.1628

⎤⎥⎥⎥⎥⎥⎥⎦(6.171)

296

Subsequently, the system state matrix is finally computed as;

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 1.0000 0 00 0 0 0 1.0000 00 0 0 0 0 1.0000

−24.1837 13.5322 10.6514 −0.2025 0 044.3537 −77.5840 33.2302 0 −0.1944 078.3229 73.9174 −152.2403 0 0 −0.1628

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(6.172)

The eigenvalues of this state matrix are shown in Table 6.27.

Table 6.27: Eigenvalues of the 3 machine system with damping neglected

No. Eigenvalue1 -0.0844 +13.3599i2 -0.0844 -13.3599i3 -0.0970 + 8.6893i4 -0.0970 - 8.6893i5 -0.00006 -0.1970

Comparison of Tables 6.26 and 6.27 reveals that:

• When damping is considered, there is only one zero eigenvalue corresponding to δ1.

• The second zero eigenvalue in Table 6.26 is replaced with a real, negative eigenvalue.

• The complex eigenvalues still appear in pairs.

• The real parts of the complex eigenvalues are negative implying that upon a disturbance, theoscillations in the system will be damped (i.e. the amplitudes of the oscillations will decreasewith time). In fact, Figs. 6.6 and 6.8 show that this is indeed the case.

With this example, we are now at the end of discussion of small signal stability analysis. Fromthe next lecture, we will start the discussion of voltage stability analysis.

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6.6 Voltage stability

Voltage collapses usually occur on power system which are heavily loaded or faulted or have shortageof reactive power. Voltage collapse is a system instability involving many power system components.In fact, a voltage collapse may involve an entire power system.

Voltage collapse is typically associated with reactive power demand of load not being met due toshortage in reactive power production and transmission.

Voltage collapse is a manifestation of voltage instability in the system. The definition of voltagestability as proposed by IEEE/CIGRE task force is as follows:

Voltage stability refer to the ability of power system to maintain steady voltages at all buses inthe system after being subjected to a disturbance from a given initial operating point. The systemstate enters the voltage instability region when a disturbance or an increase in load demand oralteration in system state results in an uncontrollable and continuous drop in system voltage.

A system is said to be in voltage stable state if at a given operating condition, for every bus inthe system, the bus voltage magnitude increases as the reactive power injection at the same bus isincreased.

A system is voltage unstable if for at least one bus in the system, the bus voltage magnitudedecreases as the reactive power injection at the same bus is increased. It implies that if, V-Qsensitivity is positive for every bus the system is voltage stable and if V-Q sensitivity is negativefor at least one bus, the system is voltage unstable. The term voltage collapse is also often used forvoltage instability conditions. It is the process, by which, the sequence of events following voltageinstability leads to abnormally low voltages or even a black out in a large part of the system.

The driving force for voltage instability is usually the loads and load characteristics, hence, voltagestability is sometimes also called load stability. In response to a disturbance, the power consumedby the loads tends to be restored by load dynamics. This in turn increases the stress on the highvoltage network by increasing the reactive power consumption and further reducing the voltage.

A major factor contributing to voltage instability is the voltage drop in the line impedances whenactive and reactive powers flow through it. As a result, the capability of the transmission networkfor power transfer and voltage support reduces. Voltage stability of a system is endangered whena disturbance increases the reactive power demand beyond the sustainable capacity of the availablereactive power resources.

The voltage stability has been further classified into four categories: Large disturbance voltagestability, small disturbance voltage stability, short term voltage satiability and long term voltagestability. A summary of these classifications is as follows:

• Large disturbance voltage stability: It refers to the system’s ability to maintain steady voltagefollowing large disturbances such as, system faults, loss of generation or circuit contingencies.This ability is determined by the system load characteristics and interaction of both continuousand discrete controls and protections. The study period of interest may be from few secondsto tens of minutes. This requires long term dynamic simulation study of the system to capturethe interactions of under-load tap changer and generator field current limiter.

298

If following a large disturbance and subsequent system control actions, voltages at all the busesin the system settle down at acceptable levels, the system is said to be large disturbance voltagestable.

• Small-disturbance voltage stability: This stability is concerned with the ability of the systemto maintain acceptable level of steady voltages, when subjected to small perturbations such asincremental changes in system load. This form of stability is also influenced by the character-istics of loads, continuous controls, and discrete controls at a given instant of time. The basicprocesses contributing to small disturbance stability are essentially of a steady state nature.Therefore, static analysis can be effectively used to estimate stability margins.

• Short term voltage satiability: It involves dynamics of fast acting load components such asinduction motors, electronically controlled loads and HVDC converters. The study period ofinterest is in the order of several seconds and the analysis requires solution of appropriatesystem differential equations.

• Long term voltage stability: The study of long term voltage stability involves the dynamics ofslower acting equipment such as tap changing transformers, thermostatically controlled loadsand generator current limiters. The study period of interest may extend to several or manyminutes, and requires long term dynamics system simulation.

Voltage instability may arise due many reasons, but some significant contributors are:

○ Increase in loading

○ Generators, synchronous condensers, or SVC reaching reactive power limits

○ Action of tap changing transformers

○ Load recovery dynamics

○ Line tripping or generator outages

Most of these changes have a significant impact on the reactive power production, consumptionand transmission in the system.

Some counter measures to prevent voltage collapse are:

● Switching of shunt capacitors

● Blocking of tap-changing transformers

● Redispatch of generation

● Load shedding

● Temporary reactive power overloading of generators

299

Figure 6.9: Simple radial power system

Voltage stability may occur in different ways. A simple case of voltage stability can be explainedby considering the two terminal network of Fig. 6.9.

In this system, the network is represented by an equivalent generator that can be modeled in thesteady state by an equivalent voltage source E behind the equivalent impedance Zg.In general, thegenerator, transformer and line impedances are combined together and represented as ZL. The loadimpedance is ZD and V is the receiving end or load voltage. The current I in the circuit is givenby:

I = E

ZL + ZD

= E

ZL∠θ +ZD∠φ

= E

(ZL cos θ +ZD cos φ) + j(ZL sin θ +ZL sin φ)(6.173)

The magnitude of current is

I = E√(ZL cos θ +ZD cos φ)2 + (ZL sin θ +ZL sin φ)2

which may be written as:

I = E

ZL√FL

(6.174)

300

where,

FL = 1 + [ZDZL

]2

+ 2 [ZDZL

] cos (θ − φ)

WhereThe magnitude of the receiving and voltage is given by:

V = ZDI

= E√FL

[ZDZL

] (6.175)

The power supplied to the load is

PL = V Icos φ

PL = (ZDFL

) [ EZL

]2

cos φ (6.176)

The plots of I,V and PL are shown in Fig. 6.10 as a function ofZLZD

ratio for a specific value of

θ and φ.

Figure 6.10: Receiving end voltage, Current and Power as a function of Load

An explanation of the chracteristics of Fig. 6.10 is as follows:

• As the load demand is increased by reducing ZD, the load power PL increases rapidly at firstand then slowly, before reaching a maximum value and then starts decreasing. There is thus, amaximum value of active power that can be transmitted through an impedance from a constantvoltage source.

301

• The transmitted power reaches a maximum when the voltage drop in the line is equal to theload voltage V. This occurs, when the load impedance ZD is equal to line impedance ZL. AsZD is gradually reduced, current in the line I increases and load voltage V decreases. Initially,for high values of ZD, the enhancement in I is more than the reduction in V, and hence loadpower PL increases rapidly with reduction in ZD. As ZD approaches ZL, the effect of theenhancement in I is only slightly greater than that of the reduction in V, hence increase in PLis slow. Finally, when ZD is quite less than ZL the reduction in V dominates over increase inI and hence, PL decreases.

• The critical operating condition, corresponding to maximum power, is the limiting point ofsatisfactory operation. For higher load demand, control of power by varying load would beunstable, as a reduction in load impedance will reduce power. The load characteristics decideswhether the system voltage decreases progressively and the system will become unstable. Witha constant impedance static load characteristic, the system stabilizes at power and voltage levelslower than the desired values. For a constant power load characteristic, the system becomesunstable through collapse of load bus voltage. If the load is supplied by transformers withautomatic under load-tap- changing (ULTC), the tap changer will try to raise the effectiveload impedance ZD as seen from the system. This will lower the load bus voltage still furtherand lead to a progressive reduction of voltage. This is the ease of simple voltage instability.

From the study of voltage stability the relationship between PL and V is important and this willbe discussed in the next lecture.

302

6.7 Relation between PL, QL and V

Neglecting the resistance of generator transformer and transmission line, the equivalent circuit ofthe system and its phasor diagram are shown in Fig. 6.11 (a) and (b).

Figure 6.11: Equivalent circuit of the system(a) and the phasor diagram (b)

From the phasor diagram of Fig. 6.11 (b), it can be observed that IX cos φ = E sin δ andIX sin φ = E cos δ − V

PL(V ) = V I cos φ = V IX cos φ

X= EVX

sin δ

(6.177)

QL(V ) = V I sin φ = V IX sin φ

X= EVX

cos δ − V2

X

The angle δ between E and V phasor can be eliminated using the identity sin2 δ + cos2 δ = 1and thus one can obtain:

303

[EVX

]2

= [PL(V )]2 + [QL(V ) + V2

X]

2

(6.178)

This static power-voltage equation determines all the possible network solutions when the voltagecharacteristics PL(V ) and QL(V ) are taken into account.

For an ideally stiff load, the power demand of the load is independent of voltage and is constantPL(V ) = PL and QL(V ) = QL, where, PL and QL are the real and reactive power demand of theload at the rated voltage V.

For stiff load the equation (6.178) can now be written as:

[EVX

]2

= [PL]2 + [QL +V 2

X]

2

(6.179)

Substituting QL = PL tan φ in equation (6.179) one can obtain,

P 2L + P 2

L tan2 φ + 2 PL tan φ

V 2

X= [EV

X]

2

− [V2

X]

2

(6.180)

Substituting tan φ = sin φcos φ

and noting that sin2φ + cos2φ = 1 equation (6.180) can be furthersimplified as

P 2L + 2 PL

V 2

Xsin φ cosφ = V

2

X2 (E2 − V 2) cos2 φ (6.181)

Adding and subtracting (V2

Xsin φ cos φ)

2

to the right hand side of equation (6.181), gives

(PL +V 2

Xsin φ cos φ)

2

− (V2

X)

2

sin2 φ cos2 φ = (V2

X2) (E2 − V 2) cos2 φ

After simplifications the equation can be expressed as:

PL +V 2

Xsin φ cos φ = V

Xcos φ

√E2 − V 2 cos2 φ (6.182)

The voltage at the load bus can be expressed in per unit as V/E. Equation (6.182) can beexpressed as:

PL = −(E2

X)(VE

)2

sin φ cos φ + (E2

X)(VE

) cos φ

√1 − (V

E)

2

cos2 φ

Orp = −v2 sin φ cos φ + v cos φ

√1 − v2 cos2 φ (6.183)

Where p = PL(E2/X) , and v =

V

EEquation (6.183) describes a family of curves with φ as a parameter. One such P-V curve is

shown, for a particular value of power factor cos φ in Fig. 6.12.The Power-Voltage curve (PV-curve) presents load voltage as a function of load real power. For

304

Figure 6.12: PV curve for lagging power factor load

static load (PL =constant) as shown in the figure, two operating points (A) and (B) are possible.Point (A) represents low current high voltage solution and is the desirable operating point, whilepoint (B) represents high current low voltage solution. Operation at point B is possible, although,perhaps non-viable due to low voltage and high current condition. Further, with system initiallyat ‘point A’, if the load is increased then from the curve it can be seen that the voltage will drop.Increase in load will result in an increase in the current flowing in the transmission line, hence thevoltage will reduce and this is a perfectly normal response of the system. Hence, point A and theupper portion of PV curve represent stable system operation region. At point B, however, as theload is increased the system voltage increases which is not possible at all. Hence, point B and thelower portion of PV curve represent unstable operating region. Power systems are operated in theupper part of the PV-curve. As the load increases point (A) and (B) come closer and coincide atthe tip of P-V curve. This point is called the maximum loading point or critical point. Furtherincrease in the load demand results in no intersection between the load-characteristic and PV curveand hence, represents voltage instability this is shown in Fig. 6.13.

The impact of large a disturbance on voltage stability can also be explained with the help of PVcurves. Suppose, a large disturbance causes the loss of a transmission line resulting in increases inreactance X or loss of generator resulting in reduction in E. The post-disturbance and pre-disturbancePV characteristics along with load characteristic are shown in Fig. 6.14. The large disturbance causesthe network characteristic to shrink drastically, so that the post disturbance PV curve and the loadcharacteristic do not intersect at all. This causes voltage instability leading to a voltage collapse.

Assuming a smooth increase in load, the point where the load characteristic becomes tangent tothe network PV characteristic defines the loadability limit of the system. Any increase in load beyondthe loadability limit results in loss of voltage stability, and system can no longer function. In Fig.6.13, the point where the load characteristic is tangent to network PV curve, coincides with the max-imum deliverable power for a constant power load. However, a loadability limit need not necessarilycoincide with the maximum deliverable power, as it is dependent on the load characteristic.

305

Figure 6.13: Changes in the operating point wih increasing load

Figure 6.14: Loss of voltage stability due to a large disturbance

Figure 6.15: Maximum deliverable power and loadabilty limit for polynomial load

This is shown in Fig. 6.15 for a polynomial load P = P0 (V

V0)α

, where, P0 represents the basevalue of load active power at rated voltage V0 = 1 p.u. and α represents the voltage exponent. α = 0represents a constant power load.

Equation (6.183) when plotted for different values of φ gives a family of PV curves. Because of

306

Figure 6.16: PV curve drawn for different values of power factor

their characteristic shape, these curves are referred to as nose curves. The following observationscan be made regarding the curves shown in Fig. 6.16:

• For a given load below the maximum, there are two possible solutions- one with higer voltageand lower current and the other with lower voltage and higher current. The former correspondsto ’normal’ operating conditions, with the load voltage V closer to the generator voltage.

• As the load is more and more compensated (corresponding to smaller tan φ ), the maximumdeliverable power increases, and the voltage at which this maximum occurs also increases.

• For over-compensated loads (tan φ < 0 , leading power factor), there is a part of the upperPV curve along which the voltage increases, as the load power increases. This can be explainedas follows: when tan φ is negative then, with more active power consumption more reactivepower is produced by the load. At low values of load, the voltage drop due to increased activepower is offset by the increase in voltage due to increased reactive power. The more negativetan φ is, the larger is the portion of PV curve where this voltage rise occurs.

The usefulness of the nose curve is high in practice as the difference between a particular loadand maximum load, determined by the peak of the characteristic, is equal to stability margin for agiven power factor.

From equation (6.183), when QL = 0 ( and hence φ = 0), p = v√

1 − v2. To find the value of PLat the peak of the nose curve set

dp

dv= 0 and the solution of the resulting equation gives the values

of v = ±12 and p = 1

2 . Hence, PL = E2

2X . Note thatE2

Xis the short circuit power at the load bus,

as it is the product of no load voltage E and the short circuit current (E/X). The maximum powerlimit for a lossless line, with unity power power, thus corresponds to half the short circuit power.

PV-curves (nose curves), illustrate the dependency of the voltage on real power of a compositeload assuming that the power factor is a parameter. The QV curves discussed next are derivedassuming that the voltage is a parameter.

307

For given value of V, equation (6.179) describes a circle in the (PL −QL) plane as shown in Fig.6.17 (a). The centre of the circle lies on the QL-axis and is shifted vertically down from the originby (V 2/X) , the radius of the curve is (EV /X) . Increasing the voltage V produces a family ofcircles of increasing radius and increasing downward shift, bounded by an envelope as shown in Fig.6.17 (b).

Figure 6.17: QP curves for stiff load (a) one circle for a given V (b) family of curves for differentvoltages and their envelope

For each point inside envelope, for example, point A, there are two possible solutions to equation(6.179), at voltages V1 and V2, as it lies on both the circles. For any point on the envelope, say pointB, there is only one value of V for which the equation (6.179) is satisfied. By determining the valuesof PL and VL for which only one solution to equation (6.179) exists, the equation of the envelopecan be developed. Rearranging equation (6.179), one gets:

(V2

X)

2

+ (2QL −E2

X)(V

2

X) + (P 2

L +Q2L) = 0 (6.184)

This quadratic equation in (V 2/X) has only one solution when

D = (2QL −E2

X)

2

− 4 (P 2L +Q2

L) = 0 (6.185)

Solving for QL one gets

QL = (E2

4X ) − [ P 2L

(E2/X)] (6.186)

This represents the equation of an inverted parabola that crosses PL axis at E2/2X and has itsmaximum at

308

PLmax = 0; QLmax =E2

4X (6.187)

Thus the maximum reactive power supplied to a load with PL = 0 i.e. a purely reactive loadis equal to one fourth of the short-circuit power. Also a point with coordinates PL = E2/2X andQL = 0 corresponds to the peak of the nose curve for φ = 0 and QL = 0.

The parabola as described equation (6.186) defines the shape of envelope in Fig. 6.17 (b) thatencloses all the possible solutions to the network equation (6.179). Every point (PL,QL) inside theparabola satisfies two possible network solutions corresponding to two distinct values of load voltageV, and each point on the parabola satisfies only one network solution corresponding to only onepossible value of voltage. Further, there are no network solutions outside the parabola implying thatit is not possible to deliver power for a (PL,QL) point lying outside the parabola.

Various stability criteria to assess the system voltage stability are discussed in the next lecture.

309

6.8 Criteria for assessing voltage stability

With reference to Fig. 6.17 (b), for each point inside the envelope of network solution, there aretwo voltage solutions, one with a higher value of voltage and the other with a lower value. It thenbecomes necessary, to find out which of these two solutions represents a stable operating point. Thiscan be done by employing voltage stability criteria which are discussed next.(a) The d∆Q/dV criterion

The classical voltage stability criterion is based on the capability of the system to supply reactivepower for a given amount of load real power demand. For the explanation of this criterion, it isconvenient to notionally separate the real and reactive power demands of the load as represented inFig. 6.18.

Figure 6.18: Equivalent circuit for determining the reactive power characteristics of the system

Let PL(V ) and QL(V ) be the load real and reactive power demands respectively.Also, let PS(V ) and QS(V ) be the real and reactive powers supplied by the source to the load.As the real power demand is always connected to the transmission link, PL(V ) = PS(V ) also

during normal operation QL(V ) = QS(V ) , but for the purpose of stability analysis, this linkbetween QL(V ) and QS(V ) is theoretically separated, hence, QS(V ) is treated as the reactivepower supplied to the load and is assumed to be independent of the load reactive power demandQL(V ) .

The real and reactive load powers are given as

PL(V ) = PS(V ) = EVX

sin δ (6.188)

and

QS(V ) = EVX

cos δ − V2

X(6.189)

Squaring and adding the above two equations and using the identity sin2 δ + cos2 δ = 1, andthen solving for QS(V ) one can obtain,

310

QS(V ) =⎡⎢⎢⎢⎢⎣

√{EVX

}2

− {PL(V )}2⎤⎥⎥⎥⎥⎦− (V

2

X) (6.190)

This equation determines the reactive power-voltage characteristic of a system, and shows howmuch reactive power will be supplied by the source if the system is loaded only with the real powerPL(V ) and the load voltage is treated as a variable. For a constant power load PL(V ) = PL =constant equation (6.186) takes the form of an inverted parabola as shown in Fig. 6.18. The firstterm of the equation (6.186) depends on the equivalent system reactance X and the load real powerPL and has the effect of shifting the parabola downwards and towards the right.

Figure 6.19: QS − V Characteristics for PL = 0 and PL > 0

• For PL = 0, QS(V ) = (EVX

)− V2

Xand the parabola crosses the horizontal axis at V = E and

V = 0. For finding the maximum value Qmax, set the derivativedQS(V )dV

= 0. On solving the

resulting equation, the value of V at which Qmax occurs isE

2 and Qmax is equal toE2

4X .

• Similarly, for PL > 0,the maximum values ofQS(V ) occurs at voltage V =√

(E2 )2

+ (PL(V )XE

)2

which is greater thanE

2Next, the QS(V ) and QL(V ) characteristics can be drawn on the same diagram as shown in

Fig. 6.20 (a). At equilibrium the supply must equal the demand i.e., QL(V )=QS(V ) and the twopossible equilibrium points VS and VU are obtained. This situation is similar to the one shown inFig. 6.17 (b).

The stability of the two equilibrium points can be evaluated using small perturbation method andthe fact that an excess of reactive power produces an increase in voltage while a deficit of reactivepower decreases the voltage.

Now, consider the equilibrium points of Fig. 6.20(a), and assume a small reduction in voltage∆V . At point ‘S’, this reduction will result in the supplied reactive power QS(V ) being greaterthan the reactive power demand QL(V ). This excess reactive power will try to increase the voltage

311

Figure 6.20: QS(V ) and QL(V ) Characteristics for (a) two equilibrium points ’S’ and ’U’ (b) theclassical stability representation

and therefore, force the voltage to return to point ‘S’. If there is a small increase in ∆V , then atpoint ‘S’ the supplied reactive power QS(V ) becomes smaller than the load reactive power QL(V ).This deficit in reactive power brings the voltage back to point ‘S’ and thus, it can be concluded thatthe equilibrium point ‘S’ is a stable operating point.

At point ‘U’, the other equilibrium point, a small reduction in voltage results in QL(V ) becominggreater than QS(V ). This deficit of reactive power, further reduces the voltage, and thus, the systemfails to return to the equilibrium point ‘U’. Hence, the system is unstable at point ‘U’. Similarly,an increase in the voltage in the vicinity of ‘U’ results in QS(V ) becoming greater than QL(V ).This excess of reactive power increases the voltage further. Again the system fails to come back toequilibrium point, and hence, the equilibrium point ‘U’ is unstable.

From Fig. 6.20(b), it can be observed that at the two equilibrium points ‘S’ and ‘U’ the derivative

of the surplus reactive powerd (QS −QL)

dVis of opposite sign. It is negative at the stable point ‘S’

and positive at the unstable point ‘U’. Hence, it can be concluded that for stability:

d∆QdV

= d

dV(QS −QL) < 0

ordQS

dV< dQL

dV(6.191)

From Fig. 6.20(b), it can be clear that at point ‘U’,dQS

dV> dQL

dV, and hence, point ‘U’ is

not stable as per the criterion established in equation (6.191). While at point ‘S’ the condition ofequation (6.191) is satisfied and hence, it is a stable point.

For the simple system of Fig. 6.18, the supplied real and reactive powers are functions of twovariables V and δ

312

PS(V ) = PL(V ) = fP(V, δ)(6.192)

QS(V ) = fQ(V, δ)

Hence, the incremental values of ∆PS and ∆QS can be expressed as :

∆PL = ∆PS =∂PS∂V

∆V + ∂PS∂δ

∆δ(6.193)

∆QS =∂QS

∂V∆V + ∂QS

∂δ∆δ

∆δ can be written in terms of ∆PL from equation (6.193) as

∆δ = [∂PS∂δ

]−1

[∆PL −∂PS∂V

∆V ] (6.194)

Substituting ∆δ in the expression of ∆QS of equation (6.193) and dividing both sides of theresulting expression by ∆V one gets

∆QS

∆V = ∂QS

∂V+ ∂QS

∂δ[∂PS∂δ

]−1

[∆PL∆V − ∂PS

∂V] (6.195)

For small values of ∆V one can write

dQS

dV≈ ∂QS

∂V+ ∂QS

∂δ[∂PS∂δ

]−1

[dPLdV

− ∂PL∂V

] (6.196)

The partial derivatives are calculated from equation (6.188) and equation (6.189)

∂PS∂δ

= EV

Xcos δ

∂PS∂V

= E

Xsin δ

∂QS

∂δ= −EV

Xsin δ

∂QS

∂V= E

Xcos δ − 2V

X(6.197)

Substituting these partial derivatives into equation (6.196) gives

dQS

dV≈ EXcos δ − 2V

X− EVX

sin δX

EV cos δ[dPLdV

− EXsin δ]

313

Figure 6.21: QS(V ) and QL(V ) Characteristics indicating stable and unstable operating points

∂QS

∂V≈ E

Xcos δ− [2V

X+ dPLdV

tan δ] (6.198)

Hence, the stability criterion can be written as:

dQL

dV> [ E

Xcos δ− {2V

X+ dPLdV

tan δ}] (6.199)

The derivative of load active and reactive power w.r.t voltagedPLdV

anddQL

dVare calculated from

the load characteristics expressed in terms of V.For a load reactive power characteristic QL(V ) , different QS(V ) ,the supplied reactive power

can be plotted for different values of source voltage E. The resulting curves are shown in Fig. 6.21.The points of interaction of these curves correspond to the two possible equilibrium points ‘S’ and‘U’. The curve QS4 is tangential to the QL(V ) characteristic and the point ‘S′

4 represents the criticaloperating point. For any QS curve below QS4, the system operation is not possible. In the figureVcr represents the critical voltage and for normal stable operation it is necessary that V > Vcr.

(b) The dE/dV criterion

From equation (6.178) E can be expressed in terms of V as,

E(V ) =

¿ÁÁÀ(V + QL(V )X

V)

2

+ (PL(V )XV

)2

(6.200)

314

In this equation,QL(V )X

Vrepresents the voltage drop component in phase with V and

PL(V )XV

represents the voltage drop component in quadrature with V.A typical E(V) characteristic is shown in Fig. 6.22 with the normal operating point of the load

as ‘S’. V is large at this point and is much greater than both in phase and quadrature components ofthe voltage drop. In this case then, a drop in voltage V will result in a drop in the emf E(V). As Vis reduced further, the in phase and quadrature components of the voltage drop become prominentand below a certain value of V, they will force E(V) to rise. As a result each value of E(V) maycorrespond to two possible network solutions of voltage V. The stability of these two solutions canbe examined using small perturbation method.

Figure 6.22: The illustration of stability criteriondE

dV

Let us examine the system behavior at point ‘S’ as shown in Fig. 6.22 with the assumptionthat source emf is maintained at a constant value. A reduction in the load voltage by ∆V willcause a reduction in the required value of the emf E(V). As the available E is constant and greaterthan the required value of E(V), it will force the voltage to return to its initial value V. Thus, thesystem returns to the initial equilibrium point after the disturbance. Similarly, when the voltage isincreased by ∆V , the required emf E(V) to maintain the enhanced voltage (V +∆V ) is larger thanthe available source emf E. Hence, the voltage is again forced to return to its initial value V by theconstant emf E. Thus, it can be safety established that the point ‘S’ is a stable equilibrium point.

Next, consider the other equilibrium point ‘U’. As the voltage is reduced, a higher value of emfE(V) is required to maintain it. But as E is constant and less than the required value of emf E(V),this will result in further reduction in voltage. As a result, the voltage further reduces and movesaway from the equilibrium point. Similarly, if the voltage is increased by ∆V , then, the required emfE(V) is smaller than the available source emf E. This larger available emf E will cause the voltageto increase further and move further away from the initial equilibrium point ‘U’. Hence, it can beconcluded that point ‘U’ is an unstable equilibrium point.

From the above arguments, it is apparent that the system is stable if the equilibrium point lieson the right hand side of the characteristics, that is when :

dE

dV> 0

315

Figure 6.23: ThedE

dVcriterion of volatge stability

Vcr is the minimum system voltage at which the system can be operated stably. The usualsystem operation voltage is greater than Vcr.

(c) The dQS/dQL criterion

Let QS(V ) be the reactive power generation by the source and QL(V ) be the load demand.Then, QS(V ) = QL(V ) +line reactive power loss. From Fig. 6.11(b) one can write:

QS(V ) = EI sin (φ + δ) = EI sin φ cos δ + EI cos φ sin δ

= E

XIX sin φ cos δ + E

XIX cos φ sin δ (6.201)

Substituting IXsin φ = E cos δ − V and IX cos φ = E sin δ in equation (6.201), one gets:

QS(V ) = E

X(E cos δ − V ) cos δ + E

X(E sin δ) sin δ

QS(V ) = E2

X− EVX

cos δ (6.202)

Substituting the expression for QL(V ) from equation (6.177) in the above equation, we get:

QS(V ) = E2

X− V

2

X−QL(V )

Or

V 2

X= E

2

X−QL(V ) −QS(V ) (6.203)

Substituting this expression into equation (6.178) and rearranging the terms gives,

QL(V ) = − Q2S(V )

(E2/X) +QS(V ) − P 2L(V )

(E2/X) (6.204)

316

Figure 6.24: Generation and load characteristics

If the load is a constant real power load with PL(V ) = PL = constant, then equation (6.204)describes a horizontal parabola in the (QS,QL) plane and is shown in Fig. 6.24(a). The vertex ofthe parabola is at a constant QS value equal to E2/2X while the minimum value of QL dependson PL and for PL = 0, the minimum is at E2/4X . An increase in PL shifts the parabola to the leftalong the QL axis with no corresponding shift with respect to the QS axis.

Again a small perturbation in QL can be used to analyze the stability of the two equilibriumpoints. For a constant QL, less than the minimum value, the two equilibrium points are labeled as‘S’ and ‘U’ as shown in Fig. 6.24.

At the point ‘S’ a small increase ∆QL in the load reactive power is accompanied by an increase ingenerated reactive power QS and a small reduction in load reactive power causes a reduction in thegenerated reactive power. Thus, the balance between the load reactive demand and the generatedreactive power is always maintained. Hence, the equilibrium point‘S’ is always stable.

At the upper equilibrium point ‘U’, an increase in QL produce a reduction in QS , while areduction in QL causes an increase in QS. Thus, the changes in reactive generation are now inopposite direction to the changes in demand and hence the equilibrium point ‘U’ is unstable.

The stability criterion then can be stated as follows:A system is stable, if a small change in reactive load demand produces a change in the generation

which has the same sign. In other words the derivative dQS/dQL is positive i.e. :

dQS

dQL

> 0 (6.205)

Further, at the maximum loading point at the nose of QS −QL characteristic of Fig. 6.24, thederivative dQG/dQL tends to infinity.

It is worth noting that the QG −QL characteristic is a parabola only for ideally stiff real powerload PL(V ) = P=constant. For voltage dependant loads an explicit expression for QL(QG) withPL(V ) cannot be obtained, and an iteractive procedure is required.

317

Methods of improving Voltage Stability

Voltage stability can be improved by adopting the following means:

• Enhancing the load reactive power support using shunt compensators.

• Line length compensation using series compensation.

• Load shedding during contingencies.

• Constructing additional transmission lines.

• Using FACTS controllers.

318

Appendix A

The system data

Table A.1: Bus data for 5 bus system

Bus Type ∣V ∣ θ PG QG PL QL QMIN QMAX Gsh Bsh

no. (p.u) (deg) (MW) (MVAR) (MW) (MVAR) (MVAR) (MVAR) (p.u) (p.u)1 1 1 0 0 0 0 0 0 0 0 02 2 1 0 50 0 0 0 -500 500 0 03 2 1 0 100 0 0 0 -500 500 0 04 3 1 0 0 0 115 60 -500 500 0 05 3 1 0 0 0 85 40 -500 500 0 0

Base MVA = 100

Table A.2: Line data for 5 bus system

Branch no. From bus To bus R (p.u) X (p.u) B/2 (p.u) Tx. Tap1 1 2 0.042 0.168 0.041 02 1 5 0.031 0.126 0.031 03 2 3 0.031 0.126 0.031 04 3 4 0.031 0.126 0.031 05 3 5 0.053 0.210 0.051 06 4 5 0.063 0.252 0.061 0

319

Table A.3: Bus data for 14 bus system

Bus Type ∣V ∣ θ PG QG PL QL QMIN QMAX Gsh Bsh

no. (p.u) (deg) (MW) (MVAR) (MW) (MVAR) (MVAR) (MVAR) (p.u) (p.u)1 1 1.06 0 0 0 0 0 0 0 0 02 2 1.045 0 18.3 5.857 0 0 -500 500 0 03 3 1 0 0 0 119 8.762 0 0 0.0002 0.5024 3 1 0 0 0 47.79 3.9 0 500 0 05 3 1 0 0 0 7.599 1.599 0 0 0 06 2 1.07 0 11.2 44.2 0 0 -500 500 0 07 3 1 0 0 0 0 0 0 0 0 08 3 1 0 0 0 0 12.9 0 0 0.0023 0.13259 3 1 0 0 0 29.499 16.599 0 0 0 0.063310 3 1 0 0 0 9 5.799 0 0 0 011 3 1 0 0 0 3.501 1.8 0 500 0 012 3 1 0 0 0 6.099 1.599 0 0 0 013 3 1 0 0 0 13.5 5.799 0 0 0 014 3 1 0 0 0 14.901 5.001 0 0 0 0

Base MVA = 100

Table A.4: Line data for 14 bus system

Branch no. From bus To bus R (p.u) X (p.u) B/2 (p.u) Tx. Tap1 1 2 0.0194 0.0592 0.0528 02 1 5 0.054 0.223 0.0492 03 2 3 0.047 0.1979 0.0438 04 2 4 0.0581 0.1763 0.0374 05 2 5 0.0569 0.1738 0.0339 06 3 4 0.067 0.171 0.0346 07 4 5 0.0134 0.0421 0.0128 08 4 7 0 0.209 0 19 4 9 0 0.5562 0 110 5 6 0 0.2522 0 111 6 11 0.095 0.1989 0 012 6 12 0.1229 0.2557 0 013 6 13 0.0661 0.1302 0 014 7 8 0 0.1762 0 115 7 9 0 0.011 0 116 9 10 0.0318 0.0845 0 017 9 14 0.127 0.2703 0 018 10 11 0.082 0.192 0 019 12 13 0.2209 0.1999 0 020 13 14 0.1709 0.3479 0 0

320

Table A.5: Bus data for 30 bus system

Bus Type ∣V ∣ θ PG QG PL QL QMIN QMAX Gsh Bsh

no. (p.u) (deg) (MW) (MVAR) (MW) (MVAR) (MVAR) (MVAR) (p.u) (p.u)1 1 1.05 0 0 0 0 0 0 0 0 02 2 1.0338 0 57.56 2.47 21.7 12.7 -500 500 0 03 3 1 0 0 0 2.4 1.2 0 0 0 04 3 1 0 0 0 7.6 1.6 0 0 0 05 2 1.0058 0 24.56 22.57 94.2 19 -500 500 0 06 3 1 0 0 0 0 0 0 0 0 07 3 1 0 0 0 62.8 10.9 0 0 0 08 2 1.023 0 35 34.84 80 30 -500 500 0 09 3 1 0 0 0 0 0 0 0 0 010 3 1 0 0 0 5.8 2 0 0 0 0.1911 2 1.0913 0 17.93 30.78 0 0 -500 500 0 012 3 1 0 0 0 11.2 7.5 0 0 0 013 2 1.0883 0 16.91 37.83 0 0 -500 500 0 014 3 1 0 0 0 6.2 1.6 0 0 0 015 3 1 0 0 0 8.2 2.5 0 0 0 016 3 1 0 0 0 3.5 1.8 0 0 0 017 3 1 0 0 0 9 5.8 0 0 0 018 3 1 0 0 0 3.2 0.9 0 0 0 019 3 1 0 0 0 9.5 3.4 0 0 0 020 3 1 0 0 0 2.2 0.7 0 0 0 021 3 1 0 0 0 17.5 11.2 0 0 0 022 3 1 0 0 0 0 0 0 0 0 023 3 1 0 0 0 3.2 1.6 0 0 0 024 3 1 0 0 0 8.7 6.7 0 0 0 0.0425 3 1 0 0 0 0 0 0 0 0 026 3 1 0 0 0 3.5 2.3 0 0 0 027 3 1 0 0 0 0 0 0 0 0 028 3 1 0 0 0 0 0 0 0 0 029 3 1 0 0 0 2.4 0.9 0 0 0 030 3 1 0 0 0 10.6 1.9 0 0 0 0

Base MVA = 100

321

Table A.6: Line data for 30 bus system

Branch no. From bus To bus R (p.u) X (p.u) B/2 (p.u) Tx. Tap1 1 2 0.0192 0.0575 0.0528 02 1 3 0.0452 0.1852 0.0408 03 2 4 0.057 0.1737 0.0368 04 3 4 0.0132 0.0379 0.0084 05 2 5 0.0472 0.1983 0.0418 06 2 6 0.0581 0.1763 0.0374 07 4 6 0.0119 0.0414 0.009 08 5 7 0.046 0.116 0.0204 09 6 7 0.0267 0.082 0.017 010 6 8 0.012 0.042 0.009 011 6 9 0 0.208 0 112 6 10 0 0.556 0 113 9 11 0 0.208 0 014 9 10 0 0.11 0 015 4 12 0 0.256 0 116 12 13 0 0.14 0 017 12 14 0.1231 0.2559 0 018 12 15 0.0662 0.1304 0 019 12 16 0.0945 0.1987 0 020 14 15 0.221 0.1997 0 021 16 17 0.0824 0.1932 0 022 15 18 0.107 0.2185 0 023 18 19 0.0639 0.1292 0 024 19 20 0.034 0.068 0 025 10 20 0.0936 0.209 0 026 10 17 0.0324 0.0845 0 027 10 21 0.0348 0.0749 0 028 10 22 0.0727 0.1499 0 029 21 22 0.0116 0.0236 0 030 15 23 0.1 0.202 0 031 22 24 0.115 0.179 0 032 23 24 0.132 0.27 0 033 24 25 0.1885 0.3292 0 034 25 26 0.2544 0.38 0 035 25 27 0.1093 0.2087 0 036 27 28 0 0.396 0 137 27 29 0.2198 0.4153 0 038 27 30 0.3202 0.6027 0 039 29 30 0.2399 0.4533 0 040 8 28 0.0636 0.2 0.0418 041 6 28 0.0169 0.0599 0.013 0

322

Table A.7: Bus data for 9 bus system

Bus Type ∣V ∣ θ PG QG PL QL QMIN QMAX Gsh Bsh

no. (p.u) (deg) (MW) (MVAR) (MW) (MVAR) (MVAR) (MVAR) (p.u) (p.u)1 1 1.04 0 0 0 0 0 0 0 0 02 2 1.025 0 163 0 0 0 -500 500 0 03 2 1.025 0 85 0 0 0 -500 500 0 04 3 1 0 0 0 0 0 -500 500 0 05 3 1 0 0 0 125 50 -500 500 0 06 3 1 0 0 0 90 30 -500 500 0 07 3 1 0 0 0 0 0 -500 500 0 08 3 1 0 0 0 100 35 -500 500 0 09 3 1 0 0 0 0 0 -500 500 0 0

Base MVA = 100

Table A.8: Line data for 9 bus system

Branch no. From bus To bus R (p.u) X (p.u) B/2 (p.u) Tx. Tap1 2 7 0.0 0.0625 0.0 12 1 4 0.0 0.0576 0.0 13 3 9 0.0 0.0586 0.0 14 4 6 0.017 0.092 0.079 05 4 5 0.01 0.085 0.088 06 5 7 0.032 0.161 0.153 07 6 9 0.039 0.17 0.179 08 9 8 0.0119 0.1008 0.1045 09 8 7 0.0085 0.072 0.0745 0

323