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CHAPTER 9Transient Stability9.1 TRANSIENT STABILITYEQUAL AREA CRITERIONEqual area criterion (EAC) is the graphical interpretation of transient stability of the system. This method is only applicable to a single machine connected to an infinite bus system or a two-machine system. The concept of EAC is derived from the fact that the stored kinetic energy in the rotating mass tries to substantiate the imbalance between the machine output and input. A single machine connected to a large system such as an infinite bus is considered to explain the EAC concept.9.1.1 Mathematical Approach to EACConsider a synchronous machine connected to an infinite bus. Neglecting the damping effect, the swing equation is rewritten below.

wherePais the accelerating power, which is due to imbalance between mechanical input and electrical power output. From the above equation,

Multiplying both sides of the above equation by 2.we get

The above equation can be written in the modified form as:

or

Integrating both sides,

or

Under stable conditions of the machine, the variations in the rotor speed (or) must be equal to zero. This is possible only when the right hand side terms ofEquation (9.3)is zero.i.e

The Swing CurveIt is a graph indicating the variations inwith timet.

Fig 9.1Swing Curve for Stable and Unstable Power SystemsThe swing equation indicates that the rotor shall either accelerate or decelerate when there exists an imbalance between mechanical input and electrical power output. Due to this, the rotor may run above or below the synchronous speed, indicating system instability as shown inFigure 9.1.Stability/Instability Conditions

Graphical Interpretation of EACConsider the power-angle curve of an alternator as shown inFigure 9.2Initially the machine is operating at the equilibrium pointA, wherePe0=Pm0and=0as shown inFigure 9.2. Now, the mechanical input to the machine is suddenly increased fromPm0toPm1. AsPm1>Pc0, the value of accelerating powerPais positive and hence the rotor starts accelerating. Due to this acceleration of the rotor,increases as shown inFig 9.3.

Fig 9.2Power Angle Curve of an Alternator

Fig 9.3Rotor Movement from PointAtoBWith an increase in, Peincreases. Whenreaches1(at pointB), the accelerating powerPabecomes zero asPe=Pm1. At pointB, the rotor speed is above synchronous speed and due to inertia it continues to move forward beyond pointB. For>1,Pe>Pm1, andPAis negative. Hence the rotor starts decelerating. Beyond pointB, the rotor speed is decreases due to deceleration.The rotor speed reachesNsat pointD. At this point, asPe>Pm1, the rotor continues to decelerate andstarts decreasing as shown inFig 9.4.

Fig 9.4Rotor Movement from PointDtoBWhen the rotor reaches pointBfromD,Peagain equals toPm1, hencePabecomes zero. However, due to inertia the rotor continues to move beyond pointBtowardsAand acceleration begins.The result is that the rotor swings fromAtoDviaB, and fromDback toA.Considering losses (which give damping effect to the rotor), the rotor finally settles down at a new equilibrium pointBafter performing the swing as explained.It can be observed that the rotor accelerates when it moves fromAtoBand decelerates when it is moves fromBtoD. From the laws of mechanics the machine swings stably when the excess energy stored in the rotating mass while it accelerates is equal to the energy it gives up during deceleration.Mathematically,|areaA1| = |areaA2|

This is known as the equal area criterion.From EAC, the condition for stability can be stated as follows:The accelerating area (positive) under thePcurve must be equal to decelerating (negative) area.9.1.2 Application of Equal Area CriterionNow we consider the various types of disturbances on a single machine connected to an infinite bus (SMIB) system and analyse its stability by using the EAC criterion.Case-1: Mechanical input to the rotor is suddenly increasedConsider the SMIB system shown inFigure 9.5

Fig 9.5An SMIB SystemElectrical power transmitted to the infinite bus is given by

Initially the system is operating stably, at which

This initial steady point is marked as pointAon theP-curve shown inFigure 9.6.

Fig 9.6Power Angle Curve of an SMIB System for a Sudden Increase in Mechanical Input to the GeneratorNow, the mechanical input to the generator is suddenly raised topm1frompm0by opening the steam valve. Consequent to this change, the following sequence of changes takes place.Stage-1:At point AThe system is under steady-state with perfect balance betweenPm0andPe0andPa=Pm0Pc0= 0. The speed of the rotor equals to the synchronous speedNS.Now, the mechanical input to is raised suddenly toPm1fromPm0.Stage-2:Movement from point A to BThough the mechanical input to the rotor is increased, the generator cannot generate extra electrical power, sincecannot change instantly due to rotor inertia. Hence, the extra mechanical inputPm1Pm0is stored as kinetic energy within the rotor and therefore the rotor starts accelerating. The speed of the rotor is now aboveNSand hencePenow starts increasing. But the accelerating power begins to reduce as the imbalance betweenPmandPestarts reducing due to an increase in thePevalue.Stage-3:Reaching point BOn reaching pointB,value equals to1and the electrical power outputPeis raised and made equal toPm1. The rotor now stops accelerating asPavalue is zero. At this point it should be remembered the speed of the rotor is aboveNS.ThoughPavalue is zero,continues to increase beyond1as the forward movement of rotor continues due to inertia of the rotor.Stage-4:From point B to COn leaving pointB,increases beyond1. Due to this, the machine now generates electrical power output more thanPe1. ThePavalue is now negative asPe>Pm1. Due to this the rotor now starts decelerating. It should be noted that though the rotor is decelerating, it still moves in the forward direction (increasing) as the speed is more thanNS.Stage-5:Reaching Point COn reaching pointC,value reaches2. Now the speed of the rotor is equal toNS.As2>1, the electrical power output is more than the mechanical input and hence thePavalue is negative. The rotor now decelerates in the reverse direction (decreasing) and the negative value ofPabegins to reduce.Stage-6:From point C to BOn leaving pointC,Pevalue begins to reduce asvalue is decreasing. This is because the rotor is moving in the backward direction. The decelerating powerPabegins to reduce. However the rotor speed is now lesser thanNS, as the rotor is moving in backward direction.Stage-7:Reaching point BOn reaching pointB, thePevalue becomes equal toPm1and hencePavalue is zero. The rotor speed is still less thanNsand the rotor continues to move in the backward direction (decreasing) due to inertia.Stage-8:From point B to AOn leaving pointB,value becomes less than1, and consequently,Pevalue is less thanPm1. Now thePavalue is positive and the rotor begins to accelerate with the rotor speed again starting to increase.Stage-9:Reaching point AWhen the rotor reaches pointA,value is equal to0and the rotor speed rises toNS.It should be understood that while the rotor is moving from pointBtoA, though its movement is backward it is actually accelerating. Once it reaches the pointA, the movement in the backward direction stops and it begins moving in the forward (increasing) direction.This completes one cycle of operation and Stages 1 to 9 are repeated again. If the system damping is considered, these oscillations gradually die out and the machine will settle to the new equilibrium pointBwherePm1=Pe=Pmaxsin1The steps related to the swinging of the rotor are depicted inTable 9.1.The areasA1andA2are given by

For the system to be stableA1=A2Table 9.1Steps related to the swinging of the rotor

Condition of maximum swingFrom the steady-state stability point of view, the system reaches the critical stable point when= 90. However, the system can maintain transient stability beyond= 90, as long as the condition of EAC is satisfied.Figure 9.7illustrates this.Limitation to increase inPmvalueConsider the swing curve shown inFigure 9.7.For the given value of0, an increased value ofPmis such that, ifgoes beyondmthe system loses transient stability. At pointB,Pe1=Pm1=Pmaxsin1. At pointC,Pc1=Pm1=Pmaxsinmax. It can be easily verified that the above equations are valid only whenmax= 1. Considering thePcurve shown inFigure 9.7, ifPmvalue is increased beyondPm1as shown, then the rotor swings beyondm. However, ifbecomes greater thanmax, then the power transfer will be lesser thanPmand the rotor will experience further acceleration. This will further increaseand the generator will be out of synchronism (SeeFigure 9.8).

Fig 9.7The System has Transient Stability when the Rotor Swings beyond= 90

Fig 9.8Limiting Case for Increase inPmTherefore, the limiting value ofPm1for a givenPm0can be computed by the condition.

where

Case-1: Mathematical Equations for EAC

Accelerating area

Decelerating area

The system is stable whenA1=A2i.e.,pmax(cos0- cos2) =pm1(2-0). Substitutingpm1=pe1=pmaxsin1in the above equation,

For the limiting case,

Therefore,

For the given value of0, the value of1cannot be found directly.Equation (9.13)can only solved by the iteration process.Case-2: Three-Phase Fault on FeederConsiderFigure 9.9(a)where a generator is connected to an infinite bus bar through a radial feeder.Thep-curve is shown inFigure 9.9(b).Let the system shown inFigure 9.9(a)be operating at pointA, where the rotor speed is equal toNSand=0.Now, a 3-phase fault occurs at pointFon one of the outgoing feeders. Since the fault is closer to the generator and as resistance is being neglected,Peinstantly becomes zero. In other words, no electrical power is transferred to the infinite bus.

Fig 9.9(a)A Three-Phase Fault at PointFon an SMIB System

Fig 9.9(b)p-CurveNow, the entirePmbecomes equal toPa, and the rotor starts accelerating. SincePe= 0, the operating point shifts immediately to pointBand due to rotor acceleration,starts increasing. In timetcr(critical clearing time) when the rotor angle reaches tocr(critical clearing angle), the circuit-breaker installed at pointFclears the fault by isolating the faulted line. Upon removal of faulted line, the generator once again starts transmitting power to the infinite bus. Therefore, the operating point shifts fromCtoDinstantly. At pointD, sincePe>Pm, the rotor now begins to decelerate and the decelerating areaA2begins as the operating point moves alongDE.System is stable as long asA1A29.1.3 Determination of Critical Clearing AngleFor a given initial load there is a maximum value of clearing angle, known as critical clearing angle (cr) for stability to be maintained. If the actual clearing angle1is smaller thancr, the system is stable. If1>crthe system is unstable. It should be noted that when1=cr, the rotorswings up tomax, a permitted threshold value. Beyondmax, the stability of the system is lost. The circuit-breaker fault clearing time corresponding tocris known as the critical clearing time (tcr).If the actual fault clearing time is less thantcr, the system is stable, otherwise it is unstable.The formulas forcrandtcrcan be easily derived only for the case whenPe= 0. For other cases it is hard to establish the formulas.

Fig 9.10A Simple Case to Determine Critical Clearing AnglecrThe following equation can be easily verified in thep-curve shown inFigure 9.10

From the above,

Now, accelerating areaSincePeis zero,

Decelerating area

For the system to be stable,A2=A1, which yields the following relation

wherecr= critical clearing angle.By substitutingmax=0andPe=Pm=Pmaxsin0inEquation (9.16),We get,

Note: The value ofcrdepend upon initial loading condition of machine. In other words depends on the initial value of0.9.1.4 Determination of Critical Clearing Time [tcr]During the fault on the system,

Therefore,

Substituting

Integrating the above equation twice,

From the above,

Case-3: Loss of Faulted Parallel LineConsider the power system shown inFigure 9.11. The generator is connected to the infinite bus system through two parallel lines.

Fig 9.11Three-Phase Fault occurs at some Point on One Parallel Line of the SMIB SystemNow, consider a fault at pointFsome distance away from both ends on one parallel line as shown inFigure 9.11.9.1.5 Determination of Transfer Reactance Before, During and After Fault Conditions

Before fault

During fault

Let the transfer reactance during fault beXdu. Its value can be obtained by converting star reactancesX'd,X1andX21into delta. This is explained in the following figure.

Post fault:After some duration, the faulty parallel line is removed by circuit-breakers connected at both ends. The transfer reactance for this conditionXpois:

The steady-state power limits for these conditions are given below.0. Before fault1. During fault2. Post fault

Fig 9.12pCurve for Case-3 Study of EACNow consider thep-curve shown inFigure 9.12.Assume that the input powerPmis constant and that the machine is operating steadily, delivering power to the infinite bus at=0.Thep-curve for the pre-fault condition is marked asain the figure. During fault condition, equivalent transfer reactance between bus bars is increased, lowering the steady-state power limit. For this condition thep-curve is represented by the curveB. Finally curveCrepresents the post-faultp-curve.The following sequence of operations takes place.1. The system is steadily operating at pointa

2. Now a fault occurs at pointFon one of the parallel line as shown inFigure 9.11. The operating point shifts down to pointbon curveB. Due to excess mechanical input, the rotor starts accelerating towards pointC.3. By the time whenreachescr, the faulted line is removed from both ends.4. Now the operating point shifts up fromctoeon curveC. Now the rotor begins to decelerate.System is stable whenA1(accelerating) =A2(decelerating)i.e., Areaa b c d= Aread e fApplying EAC, for this caseecan be obtained as follows,

where

IntegratingEquation 9.19,

or

Case-4: Re-Closure Operation of Circuit BreakerConsidering the fault as temporary, circuit breakers in the faulted line are re-closed after some time. As the fault is cleared, the line is restored into service again.For this case,p-curves for before the fault and after re-closure are the same.i.e.,

Thep-curve is drawn for the case inFigure 9.13.The stability of the system improves, as the re-closure unit of the circuit breaker restores the second line back into service for transfer of electrical power.

Fig 9.13Case-4 of EAC: Application of the Re-Closure UnitThe following is the sequence of operations for this case study:1. Initially the system is operating stably at Pointa.

2. A fault occurs at Point F on the radial feeder and the operating point shifts down to Pointb. Due to higher value ofPm, the rotor begins to accelerate andincreases.3. After some time, the circuit breaker isolates the faulted line. The rotor angle reaches toc.The operating point shifts fromctoe, on the post-fault curve C. SincePe>Pm, the rotor now begins to decelerate.4. On reaching Pointf, when=r, the re-closure unit restores the second line back to service. Due to this the operating point shifts fromftogon the pre-faultp-curve.The accelerating area (A1) and the decelerating area are (A2) are marked on thep-curve. The maximum angle to which the rotor swings is2and is less thanm(maximum permissible rotor angle). Hence, the system is stable for the condition shown on thep-curve.Example 9.1A generator is generating 20% ofthe maximum power it is capable of generating. If the mechanical input to the generator is increased by 250% of the previous value, calculate the maximum value ofduring the swing of rotor around the new equilibrium point.Solution:Thep-curve is shown inFigure 9.14

Fig 9.14P-CurveAt Pointa,

At Pointb,

Now, we are required to find2.For the system to be stable,Accelerating areaA1= Decelerating areaA2

Equating(1)and(2),

The value of2may be found using the numerical method or the trial and error process. Here2determined by the trial and error process.From the condition2> 0.5235, the starting value of2is guessed as 0.552X= 0.52+ cos2X- 1.08 = error

0.551.12750.04752

0.571.12690.0469

0.61.125330.04533

0.71.14840.0348

0.751.106680.02668

0.771.10290.0229

0.781.104830.0248

0.791.100910.0209

0.81.09670.016

0.851.08490.0049

0.861.082430.00243

0.871.07890.0001734

From the tabulated results the approximate value of

Example 9.2Consider the power system shown inFigure 9.15

Fig 9.15An SMIB Power SystemThe p.u reactances are marked in the figure. A balanced three-phase fault occurs at the middle point of Line-2. The generator is delivering 1.0 p. u. power at the instant preceding to fault. By the use of equal area criterion, determine the critical clearing angle.Solution:Before fault condition:

During fault condition:For a fault at the middle of Line-2 the equivalent circuit is as shown inFigure 9.16below.Converting star reactances of 0.2, 0.4 and 0.2 p.u into delta, the transfer reactance can be obtained from this equivalent circuit.

Fig 9.16Determination of Transfer Reactance

Steady-state power limit for the condition is:

Post-fault condition:The fault in Line-2 is removed by opening the circuit breakers at both ends. The transfer reactance for this case is:

The three power angle curves are shown inFigure 9.17(a).

Fig 9.17(a)Power Angle Curves for a Fault at the Middle of Line-2Under steady-state conditions prior to fault,

Now, we are required to findcrwith 5 reference toFig 9.17(b)

Fig 9.17(b)Determination SCRThe maximum angle up to which the rotor can swing for a given value of0is:

Accelerating areaDecelerating areaFor the system to be stable,

By integration,

Solving the above equation,

Example 9.3In the above example, find the critical clearing angle if a three-phase fault occurs on Line-2, close to the generator.Solution:In this

Example 9.4Consider the system shown inFigure 9.18.

Fig 9.18An SMIB SystemThe per unit values of different quantities are:

The system is operating stably with a mechanical input ofPm=1.4 p.u. Now, one of the lines is suddenly switched off.0. Comment on the stability of the system1. If the system is stable, find the maximum value ofduring the swinging of the rotor.Solution:0. Thepcurve for this case is as shown inFigure 9.19.

Fig 9.19PCurveBefore Line-2 is switched off:Transfer reactance = 0.1 + (0.6||0.6) =0.4 p.u

After Line-2 is switched off:Transfer reactance = 0.1 + 0.6 = 0.7 p.u.

As the mechanical input is constant before and after Line-2 is switched off, the generator generates the same amount of electrical power output. The initial equilibrium point isaand the final equilibrium point iscas shown in the figure.i.e.,

From the above,1can be determined

Or,1= 54.77If stability is of interest, the rotor can swing to a maximum ofmax(up to pointf).At pointcandf, thePegenerated is same.

Solving,max=1= 2.186 rad or 125.3The accelerating areaA1is:

System stability depends on whether or not there is sufficient decelerating areaA2available. In other words, for the system to stable the condition isA1 A2Maximum decelerating area available,A2,max, can be found as:

SinceA2max>A1the system is stable.1. The actual value of rotor swing i.e.,2can be found by the condition

The above equation cannot be solved directly. Iterative methods are required to obtain2. An approximate value2is obtained by the trial and error process. The value of2is such that1