School of Electrical Engineering & telecommunications

School of Electrical Engineering and Telecommunications

Power System Stability

Dr Jayashri Ravishankar

Stability definition• Power system stability is the ability of an electric

power system, for a given initial operatingcondition, to regain a state of operatingequilibrium after being subjected to a physicaldisturbance, with most of the system variablesbounded so that practically the entire systemremains intact.

• The disturbances mentioned in the definitioncould be faults, load changes, generatoroutages, line outages, voltage collapse or somecombination of these.

ClassificationPower system

stability

Rotor anglestability

Frequencystability

Voltage stability

Small signalstability

Transientstability

Long term

Short term

Large disturb--ance stability

Small disturb--ance stability

Steady statestability

Rotor angle stability• Steady state stability - restricted to small and gradual changes in the

system operating conditions.• The steady-state stability is usually performed with a power-flow

computer program, to ensure:1- phase angle across transmission lines not too large2- bus voltage close to nominal values3- Generators, transmission line, transformers & others not

overloaded• Small signal stability (dynamic stability) - ability of a power system to

maintain stability under continuous small disturbances – involves longertime period (several minutes)

• These small disturbances occur due to random fluctuations in loads andgeneration levels.

• In an interconnected power system, these random variations can leadcatastrophic failure as this may force the rotor angle to increase steadily.

Transient Stability• Involves major disturbances, such as:

loss of generation, line switching, faults,sudden load changes

• Objective : determine if machines returnto synchronous frequency with newsteady-state power angles (changes inpower flows, and bus voltages also ofconcern)

• An interesting analogy (by Elgerd):Mechanical system of a number ofmasses representing synchronousmachines are interconnected by anetwork of elastic strings representingtransmission lines

• Assume system initially at rest in steady-state with net forceon each string below its break point

• Let one of the strings be cut representing the loss oftransmission line

• Mechanical analogy continued: The masses undergo transientoscillations and the forces on the strings fluctuate

• System either settles down to a new steady-state or additional strings break

• This results in an even weaker network and eventual systemcollapse

• For a large power system, with many synchronous machinesinterconnected by complicated transmission networks, transientstability studies best performed with a digital computer program

• In many cases it is determined during first swing of machineangles following a disturbance

• During first swing (last 1 sec), mechanical output & internalvoltage of gen. unit assumed constant, however in multi-swingslasting several seconds, model of turbine-governors &excitation beside detailed machine model are employed toobtain accurate transient stability results over longer timeperiod.

Transient Stability

Assumptions• Balanced 3 phase systems & balanced

disturbances are considered (i.e. pos. seq.networks employed)

• Deviations of machine frequencies fromsynchronous frequency (50 Hz) are small, & dcoffset current and harmonics neglected

• Therefore network of transmission lines,transformers, and impedance loads isessentially in steady-state and voltages, currentsand powers computed from algebraic power-flowequations

Swing equationLet us consider a three-phase synchronous alternator thatis driven by a prime mover. The equation of motion of themachine rotor is given by

(1)whereJ - total moment of inertia of the rotor mass in kgm2

Tm - mechanical torque supplied by the prime mover in N-mTe - electrical torque output of the alternator in N-mTa - accelerating torque output of the alternator in N-m - angular position of the rotor in rad (mechanical angle

wrt to stator reference)

aem TTTdtdJ 2

2

The angular position is measured with a stationaryreference frame. To represent it with respect to thesynchronously rotating frame, we define

(2)

where, - angular position in rad with respect to thesynchronously rotating reference frame, rotating at msyn inrad/s. Taking the time derivative of the above equation weget,

(3)Here, are expressed in mechanical radians.Converting them to electrical radians,

tmsyn

dtd

dtd

msyn

msynsynePP 22

,

Defining the angular speed of the rotor as,we can write (3) as,

(4)

Thus, rotor angular speed is equal to the synchronous speed onlywhen ddt is equal to zero.Taking the derivative of (3) and then substituting in (1) gives,

(5)

Multiplying both sides by r

(6)Pm - mech torque supplied by the prime mover – mech losses MW Pe - electrical torque output of the alternator+electrical losses MW

dtd

msynm

dtd

m

aem TTTdtdJ

dtdJ 2

2

2

2

aemm PPPdtdJ 2

2

Define a normalized inertia constant:

Substituting this in (6),(7)

Dividing throughout by Srated and defining

Swing equation

rated

msyn

S

JH

2

21

ratingMVAGeneratorjoules-megainspeedssynchronouatenergykineticStored

aemmmsyn

rated PPPdtdSH 2

2

22

puin22

2

aempumsyn

PPPdtdH

msyn

mpu

• Assume that as the variation of the speed, evenduring transients, from synchronous speed is quite less.

• This assumption does not mean that the speed of the rotor has reached the synchronous speed but instead the kinetic energy during the transient does not change appreciably.

• Thus and hence, the swing equation becomes,

• In terms of electrical angle,

msynm

1pu

emmsyn

PPdtdH

2

22

)sin(

)(2

max2

2

2

2

PPHf

dtd

PPHdt

d

m

emsyn

• This can be further broken down into 2 first-order differentialequations as,

• When Pm = Pe there will be no speed change and there will be noangle change.

• When Pm > Pe there is more input mechanical power than theelectrical power output. In this case, as the energy has to beconserved, difference between the input and output powers willlead to increase in the kinetic energy of the rotor and speedincreases.

• When Pm < Pe then, the input power is less than the requiredelectrical power output. Again the balance power, to meet the loadrequirement, is drawn from the kinetic energy stored in the rotordue to which the rotor speed decreases .

)sin(

)(

max

PPHf

dtddtd

mm

sme

mme

P2

Swing equation of coherent machines• In studies involving large scale power systems with many

generating units, computation time is reduced by combining swingequations of individual machines.

• Coherent Machines are usually connected to same bus or areelectrically close. They are usually remote from networkdisturbances under study.

21

212

2

212

212

);(

11)()(

HHHHHPP

Hf

dtd

HHPPf

dtd

eqemeq

em

)(

)(

;

22

22

12

12

2121

em

em

eeemmm

PPH

fdt

d

PPHf

dtd

PPPPPP

Swing equation of Non-coherent machines• Non-coherent machines do not swing together. This is often the

case when one machine is a motor and other the generator.

eqeq em

eq

eemm

eemm

emem

PPH

fdt

d

HHHPHP

HHHPHPf

dtd

HHHH

HHHPHP

HHHPHPf

dtd

HPP

HPPf

dtd

221

2

21

1221

21

12212

212

21

21

21

1221

21

12212

212

2

22

1

112

212

)(

)(

)(

)(

)(

)(

;

222

22

2

111

21

22121

em

em

eemm

PPH

fdt

d

PPHf

dtd

PPPP

21

1221

21

1221

21

21 ;;HH

HPHPPHH

HPHPPHH

HHH eem

mmmeq eqeq

Example 1A 50 Hz, 4-pole turbogenerator is rated 500 MVA,22 kV and has an inertia constant (H) of 7.5MJ/MVA. Assume that the generator issynchronized with a large power system and has azero accelerating power while delivering a powerof 450 MW. Suddenly its input power is changed to475 MW. Find the speed of the generator in rpm atthe end of a period of 10 cycles. The rotationallosses are assumed to be zero.

Solution

The machines accelerates for 10 cycles, i.e., 2010 = 200 ms= 0.2 s, starting with a synchronous speed of 1500 rpm.Therefore at the end of 10 cyclesSpeed = 1500 + 43.63320.2 = 1508.7266 rpm

rpm/s 6332.4325693.460

rad/s mechanical 5693.42

1385.9

rad/s electrical 1385.92515

1002

2

22

2

em

s PPHdt

d

Given: f = 50 Hz, P = 4, H = 7.5 MJ/MVA, Pm = 475 MW, Pe = 450 MW

Equal area criterion (EAC)• Consider the situation in which the synchronous machine is

operating in steady state delivering a power Pe equal to Pmwhen a fault occurs in the system.

• Opening up of the circuit breakers in the faulted sectionsubsequently clears the fault. The circuit breakers takeabout 5/6 cycles to open and the subsequent post-faulttransient last for another few cycles.

• The input power, on the other hand, is supplied by a primemover that is usually driven by a steam turbine. The timeconstant of the turbine mass system is of the order of fewseconds, while the electrical system time constant is inmilliseconds.

• Therefore, for all practical purpose, the mechanical powerremains constant during this period when the electrical transientsoccur.

• The transient stability study therefore concentrates on the abilityof the power system to recover from the fault and deliver theconstant power Pm with a possible new load angle .

• Equal-area criterion states that A1=A2 for stability.

21

2

2

2

2

2

22

2

2

)()(

0)()(0)(

)(

)(2

2&2

0

00

00

AAdPPdPP

dPPdPPordPP

dPPdtdH

dtdPP

dtd

dtdH

dtd

dtdH

dtd

dtd

dtd

dtdPP

dtdH

c m

c

c m

c

meem

ememem

emsyn

emsynsyn

emsyn

Critical clearing angle• To achieve stability, we have to clear the fault below a

critical clearing angle, such that area of acceleration isequal to area of deceleration.

• During fault the electrical output power is zero.

0max

1

0maxmax

max0

max

max

coscos

coscos)cos(cos

cos

)sin(

0

0

mm

mc

mmmc

cmmcmcm

mm

mm

PP

PPPPPP

PPP

dPPdP

m

c

c

c m

c

Critical clearing time• During fault,

• The critical clearing time is,

02

2

2

2

tPHftP

Hf

dtdP

Hf

dtd

mmm

sec2)(2

0

02

m

cc

cmc

fPHt

tPHf

EAC when Pe 0• Due to fault the maximum

power that can be transferred to infinite bus reduces to say Pmax2

• After the fault is cleared the faulted line is disconnected from the system.

• The maximum power that can be transferred to the infinite bus post fault also reduced, due to increased reactance, say to Pmax3.

• Hence, we will have three P – curves : pre-fault, during-fault and post-fault.

Example 2• Transient stability during a temporary 3 ph. Fault (Pload

1.0 p.u. with p.f. 0.95 lag)• Find the critical clearing angle for the fault F that is three

phase solidly grounded fault.

SolutionPre-fault condition :Xeq=0.3+0.1+0.2||(0.1+0.2)

=0.52 p.u.I = 1.0/[1.0 x 0.95] -18.195E = Vbus+jXeq I

=1.2812 23.946Pe= [1.2812x1.0/0.52] sinδ

= 2.4638 sinδδ1=180 - δ0= 156.05

= 2.7236 rad

cyclesst

tP

Ht

rad

dAdA

cr

msyn

cr

crcrcr

cr

cr

4.10208.0)4179.05489.1()0.1)(502(

12

))((4

74.885489.1

)7236.2()]7236.2cos([cos4638.24179.0

)0.1sin4638.2(0.1

0

7236.2

24179.0

1

Numerical integration techniques• The equal area criterion is applicable to one machine &

an infinite bus or two machines.• For multi-machine stability problems, numerical

integration techniques can be used to solve swingequation for each machine.

• Methods available: Euler’s method, Modified Euler’smethod, Runge-Kutta method, Trapezoidal integration

• Each machine is modeled by a simplified model usingconstant terminal voltage and its direct axis transientreactance X’d

• All loads are modeled as constant admittances.

Computational Procedure1. Run a prefault power-flow to determine initial bus

voltages, machine currents and power outputs.2. Compute internal machine voltages.3. Compute bus admittance matrix including loads and

generator admittances.4. Set time t=05. If there is a switching operation, change in load, modify

admittance matrix and if a short circuit, set the faultedbus voltage to zero

6. Using internal machine voltages, compute machineelectrical power at time t.

7. Using Pe of step 6, compute preliminary estimates ofpower angles & machine speeds

8. Using internal machine voltages, compute preliminaryestimates of machine electrical powers Pe at (t+∆t) – thisis where the methods differ.

9. Using Pe (in step 8) and power angles and speed (in step7) compute final estimates of power angles & machinespeeds at (t+∆t)

10.Set time t= t+∆t stop if t ≥ T, else return to step 5.

Step 8 – Euler’s method Step 9 – Euler’s method

Methods of improving transient stability• Increasing the system voltage - Increases power transfer

- increases the difference between initial load angle andcritical clearance angle. This allows the machine toallows to rotate through large angle before reachingcritical clearance angle.

• Increase in the X/R ratio in the power system increases the power limit of the line.

• High speed circuit breakers improves clearing time • By Turbine fast valving helps in reducing the mechanical

input power when the generator is under acceleration during the fault

• Use of Auto Re-closing - quick fault recovery• Adding reactive power compensation – improves

voltage, hence power transfer capability and stabilitymargin. Figure shows the effect of shunt compensation.

Example 31. Consider an uncompensated SMIB power

system is operating in steady state with amechanical power input Pm equal to 0.5 perunit. Calculate the critical clearing angle.

2. Calculate the new maximum angle after addingan ideal shunt compensator at the midpoint.

3. Comment on your results.

Uncompensated systemSin δ0 = 0.5δ0 = 30° = 0.5236 rad δmax = 150° = 2.6180 rad

δcr = 79.56°

Compensated system2 sin(δ/2) = 0.5δ1 = 28.96° = 0.5054 rad=0.5 −= (2sin(2) − 0.5)Solving for the same ,δ2 = 104.34°The max angle has moved from 150° to 104°. The stability margin has increased significantly in the compensated system.