Power System Stablity

Introduction Power System Stabilityy y__• The tendency of a power system to develop restoring forces equal to

or greater than the disturbing forces to maintain the state ofg gequilibrium is know as stability.

• If the forces tending to hold machines in synchronism with oneanother are sufficient to overcome the disturbing forces, the systemg , yis said to remain stable.

• Stability Studies:– TransientTransient– Dynamic– Steady‐state

• The main purpose of transient stability studies is to determineThe main purpose of transient stability studies is to determinewhether a system will remain in synchronism following majordisturbances such as transmission system faults, sudden loadchanges, loss of generating units, or line switching.

1

Introduction Power System Stabilityy y__• Transient Stability Problems:

– First‐swing; short study period after disturbance, based on a reasonableFirst swing; short study period after disturbance, based on a reasonablesimple generator model, without control system.

– Multi‐swing; longer study period of time after disturbance, thus consider the effect of generator control systems.

• In all stability studies, the objective is to determine whether or not therotors of the machines being perturbed return to constant speedoperation.

• To simplify calculation, the following assumptions must be made:– Only synchronous frequency currents and voltages are considered in the

stator windings and the power system. Consequently dc offset currents andharmonic components are neglectedharmonic components are neglected.

– Symmetrical components are used in the representation of unbalancedfaults.

– Generated voltage is considered unaffected by machine speed variation.Generated voltage is considered unaffected by machine speed variation.

2

Introduction Power System Stabilityy y__

Mechanical analog of power system transient stability

3

Rotor dynamics and the swing equationy g q

• Accelerating torque is the product of the moment of inertia of the rotor times its angular accelerationthe rotor times its angular acceleration.

emam TTTdJ −==2

2θ (14.1)emadt 2

J the total moment of inertia of the rotor masses in kg m2 J the total moment of inertia of the rotor masses, in kg-mmθ the angular displacement of the rotor with respect to stationary axis, in mechanical radians

t time, in seconds

mT the mechanical or shaft torque supplied by the prime mover less retarding torque due to rotationalm

losses, N-m eT the net electrical or electromagnetic torque, in N-m

aT the net accelerating torque, in N-m

4


• The mechanical torque Tm and the electrical torque Te are consideredpositive for the synchronous generator. Whereas for motor is anotherway round.

• Tm is the resultant shaft torque that tends to accelerate the rotor inthe positive direction of rotation as shown in the figure.

• Under steady‐state operation of generator Tm and Te are equal andl ti t T i

mθ

accelerating torque Ta is zero.• In this case there is no acceleration or deceleration of the rotor

masses and the resultant constant speed is the synchronous speed.

5

Rotor dynamics and the swing equationy g q• is measured with respect to a stationary reference axis on the stator, it is

an absolute measure of rotor angle.• Rotor angular position with respect to a reference axis which rotates at

mθ

g p psynchronous speed is given by:

msmm t δωθ += (14.2)

• Where is the synchronous speed of the machine in mechanical radiansper second and is the angular displacement of the rotor, in mechanicalradians, from the synchronously rotating reference axis.

smωmδ

• The derivatives of (14.2) with respect to time are:

dd mm δωθ+= (14.3)

dtdt smω +=

2

2

2

2

dd

dd mm δθ

=

(14.3)

(14.4)22 dtdt

6


• Equation (14.3) shows that the rotor angular velocity is constant and equals the synchronous speed only when is

dtd mθ

dδconstant and equals the synchronous speed only when is equal zero.

• Thus, represents the deviation of rotor speed from

dtd mδ

dtd mδ

synchronism and the units of measure are mechanical radiansper second.

• Eq (14 4) represents the rotor acceleration measured in

dt

• Eq. (14.4) represents the rotor acceleration measured inmechanical radians per second‐squared

• Substitute (14.4) into (14.1):( ) ( )

mN2

2

−−== emam TTT

dtdJ δ (14.5)

7


• Eq. (14.5) can be converted into power by multiplying with angular velocity in Eq (14 6) (**Power = Torque x Angularωangular velocity in Eq. (14.6) (**Power = Torque x Angular velocity).

dtd m

mθω = (14.6)

mω

W2

2

emam

m PPPdt

dJ −==δω (14.7)

Where:

P the shaft power input to the machine less rotational lossesPm the shaft power input to the machine less rotational losses

Pe the electrical power crossing its air gap

Pa the accelerating power which account for any unbalance between Pm and Pe

8


• Eq. (14.7) can also be written as in (14.8), whereby :mJM ω=

• M is inertia constant (joule‐seconds per mechanical radian)

W2

2

emam PPP

dtdM −==δ (14.8)

• M is inertia constant (joule‐seconds per mechanical radian)

• In machine data supplied for stability studies, another constant related to inertia is called H constant:

MVAinratingmachinespeedssynchronouatmegajoulesinenergykineticstoredH =

(14.9)MVAMJS

M

S

JH

mach

sm

mach

sm/2

121 2 ωω

==

MVA.in machine the of rating phase three the−machS9


• Solving for M in Eq. (14.9);

radmechMJSHM machsm

/2ω

= (14.10)

• Substitute the above equation in (14.8), we find;

PPPdH −22 δmechanical radians

mechanical radians per seconds

• Eq. (14.11) can also be written as;mach

em

mach

am

sm SPP

SP

dtdH

==2

2 δω (14.11)

perunitPPPdtdH

emas

−==2

22 δω

(14.12) Swing Equations

10


• For a system with an electrical frequency of f hertz, Eq. (14.12) becomes ( in electrical radians)δbecomes ( in electrical radians)

(14.13)perunitPPPdtd

fH

ema −==2

2δπ

δ

• If in electrical degree

(14.14)perunitPPPdH==

2δ

δ

• Eq. (14.12) can be written as the t fi t d diff ti l ti

(14.14)perunitPPPdtf ema −==2180

perunitPPPdtdH

emas

−==2

22 δω

two first‐order differential equations:

perunitPPddH

em −=ω2

sdtd ωωδ

−=(14.15) (14.16)pdt em

sω dt

11

Further Considerations of the swing equationg q

• In a stability study of a power system with many synchronous machines only one MVA base common to all parts of themachines, only one MVA base common to all parts of the system can be chosen.

• Thus, H constant for each machine must be converted into per unit base on common MVA base;

(14.17)machht

SHH =

• The constant moment inertiaM is rarely used in practice and H

(14.17)system

machsystem SHH

• The constant moment inertia M is rarely used in practice and His often used in stability study.

12


• In a stability study for a large system with many machinesgeographically dispersed over a wide area it is desirable togeographically dispersed over a wide area, it is desirable tominimize the number of swing equations to be solved.

• This can be done if the transmission line fault, or otherdisturbance on the system, affects the machines within theplant so that their rotors swing together.

• Thus the machine within the plant can be combined into a• Thus, the machine within the plant can be combined into asingle equivalent machine just as if their rotors weremechanically coupled and only one swing equation need to bewritten for them.

13


• Consider a power plant with two generators connected to the samebus which is electrically remote from network disturbances the swingbus which is electrically remote from network disturbances, the swingequations on the common system base are:

perunitPPdt

dHem 112

12

12−=

δω

(14.18)dtsω

perunitPPdt

dHem

s222

22

22−=

δω

(14.19)

• Adding the equation together, and denoting and by since therotor angle swing together;

1δ 2δ δ

dH 22 δ

where

peruniPPdtdH

ems

−=2

22 δω

(14.20)

21 HHH +=PPPwhere 21

21 mmm PPP += 21 eee PPP +=14

The Power‐Angle Equationg q __________________

• In the swing equation, the input mechanical power from theprime mover P is assumed constantprime mover Pm is assumed constant.

• Thus, the Pe will determine whether the rotor accelerates,Thus, the Pe will determine whether the rotor accelerates,decelerates, or remains at synchronous speed.

• Changes in Pe are determined by conditions on the transmissionand distribution networks and the loads on the system to whichthe generator supply power.the generator supply power.

15


• Each synchronous machine is represented for transient stabilitystudies by its transient internal voltage E’ in series with the transienty greactance X’d as shown in the Figure below.

• Armature resistance is negligible so that the phasor diagram is asshown in the figure.g

• Since each machine must be considered relative to the system, thephasor angles of the machine quantities are measured with respectto the common system reference.y

+

jXd'

IE'

jIXd'

Vt

_E'

I

δα Vt

j d

Reference

(a) (b)16


• Consider a generator supplying power through a transmissionsystem to a receiving end system at bus 2system to a receiving‐end system at bus 2.

1I 2I E’1 is transient internal voltage of generator at bus 1

'1E '

2EE’2 is transient internal voltage of generator at bus 2

• The elements of the bus admittance matrix for the networkreduced to a two nodes in addition to the reference node is:

⎥⎦

⎤⎢⎣

⎡=

2221

1211

YYYY

Ybas(14.28)

17


• Power equation at a bus k is given by:

• Let k =1 and N=2 and substituting E’ for V

nkn

N

nkkk VYVjQP

1=

∗ ∑=−

(14.29)

• Let k =1 and N=2, and substituting E 2 for V,

( ) ( )∗∗+=+ '

212'1

'111

'111 EYEEYEjQP (14.30)

where1I 2I

1'1

'1 δ∠= EE 2

'2

'2 δ∠= EE

111111 jBGY += 121212 θ∠= YY

'1E '

2E

18


• We obtain:2

)(cos 122112'2

'111

2'11 θδδ −−+= YEEGEP

)(sin 122112'2

'111

2'11 θδδ −−+−= YEEBEQ

(14.31)

(14.32)

• If we let and , we obtain from (14.31) and (14.32)

21 δδδ −=212

πθγ −=

)-( sin 12'2

'111

2'11 γδYEEGEP +=

2

(14.33)

)-( cos 12'2

'111

2'11 γδYEEBEQ −−= (14.34)

19


• Eq. (14.33) can be written more simply as

where(14.35))sin(max γδ −+= PPP ce

• When the network is considered without resistance, all the elements of Y are susceptances so both G and becomes

(14.36)11

2'1 GEPc = 12

'2

'1max YEEP =

γelements of Ybus are susceptances, so both G11 and becomes zero and Eq. (14.35) becomes;

δsinmaxPPe = (14.37) Power Angle Equation

γ

where , with X is the transfer reactance betweenE’1 and E’2

maxe

XEEP '2

'1max =

1 2

20

Example 1: Power‐angle equation before faultp g q

The single‐line diagram shows a generator connected through paralleltransmission lines to a large metropolitan system considered as an infinitebus. The machine is delivering 1.0 pu power and both the terminal voltageand the infinite‐bus voltage are 1.0 pu. The reactance of the line is shownbased on a common system base. The transient reactance of the generator is0 20 pu as indicated Determine the power‐angle equation for the system0.20 pu as indicated. Determine the power‐angle equation for the systemapplicable to the operating conditions.

21


The reactance diagram for the system is shown:

The series reactance between the terminal voltage (Vt) and the infinite bus is:

unitper 3.024.010.0 =+=X

The series reactance between the terminal voltage (Vt) and the infinite bus is:

The 1.0 per unit power output of the generator is determined by the power‐angle

1.0 sin3.0

(1.0)(1.0) sin == ααXVVt

The 1.0 per unit power output of the generator is determined by the power angle equation.

αV is the voltage of the infinite bus, and is the angle of the terminal voltage relative to the infinite bus

22


Solve α01 01 458.173.0sin == −α

Terminal voltage, Vt : unitper 300.0954.0458.170.1 0 jVt +=∠=

The output current from the generator is:

3000.1458.170.1 00

jI ∠−∠= unitper 729.8012.11535.00.1 0∠=+= j

3.0jpj

The transient internal voltage is

XIVE +=' XIVE t +=

)1535.00.1)(2.0()30.0954.0(' jjjE +++=

unitper 44.28050.15.0923.0 0∠=+= j

23


The power‐angle equation relating the transient internal voltage E’ and the infinite bus voltage V is determined by the total series reactanceinfinite bus voltage V is determined by the total series reactance

unitper 5.024.01.02.0 =++=X

Hence, the power‐angle equation is:

upPe .sin1.2sin50

)0.1)(05.1( δδ ==5.0

δWhere is the machine rotor angle with respect to infinite bus

The swing equation for the machine isThe swing equation for the machine is

unitper sin 10.20.1180 2

2

δδ−=

dtd

fH

H is in megajoules per megavoltampere, f is the electrical frequency of the system and is in electrical degree δ

24

Example 1: Power‐angle equationp g q

The power‐angle equation is plotted:

Before fault

After fault

During fault

25

Example 2: Power‐angle equation During Faultp g q g

The same network in example 1 is used. Three phase faultoccurs at point P as shown in the Figure Determine the poweroccurs at point P as shown in the Figure. Determine the power‐angle equation for the system with the fault and thecorresponding swing equation. Take H = 5 MJ/MVA

26


Approach 1:

Th di f h d i f l i h b lThe reactance diagram of the system during fault is shown below:

The value is admittanceper unitp

27


As been calculated in example 1, internal transient voltage remainsas (based on the assumption that flux linkage is°∠= 4428051'E

The Y bus is:

as (based on the assumption that flux linkage isconstant in the machine)

∠ 44.2805.1E

⎥⎥⎤

⎢⎢⎡

−−

= 5025070333.30333.3

jYb

⎥⎥⎦⎢

⎢⎣ − 833.1050.2333.3

50.250.70jYbas

Since bus 3 has no external source connection and it may be removed by the nodeli i ti d th Y b t i i d d telimination procedure, the Y bus matrix is reduced to:

[ ]5.2333.383310

152

333.3570

0333.3⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−=busY

833.105.25.70 −⎥⎦⎢⎣

⎥⎦

⎢⎣ −

⎥⎤

⎢⎡−

=⎥⎤

⎢⎡ 769.0308.21211 j

YY⎥⎦

⎢⎣ −⎥

⎦⎢⎣ 923.6769.02221

jYY

28


The magnitude of the transfer admittance is 0.769 and therefore,

The power‐angle equation with the fault on the system is therefore,

' ' 'max 1 2 12 (1.05)(1.0)(0.769) 0.808P E E Y per unit= = =

unitper sin 808.0 δ=eP

The corresponding swing equation isThe corresponding swing equation is

5180

10 0 8082

2fddtδ

δ= −. . sin per unit

Because of the inertia, the rotor cannot change position instantly uponoccurrence of the fault. Therefore, the rotor angle is initially 28.440 andthe electrical power output is

δ

38504428i8080 °P 385.044.28sin808.0 =°=eP

29


The initial accelerating power is: Pa = − =10 0385 0 615. . . per unit

and the initial acceleration is positive with the value given by

ddt

ff

2

2180

50 615 2214

δ= =( . ) . elec deg / s2

30


Approach 2:

C h d li ( hi h i Y f ) i d l dCovert the read line (which in Y form) into delta to remove node3 from the network:

j1.3

113

11 3

322

31)4.0)(2.0()4.0)(3.0()2.0)(3.0( jjjjjjR ++

j0.65 j0.8667

3.12.0

)4.0)(2.0()4.0)(3.0()2.0)(3.0( jj

jjjjjjRAC =++

=

65.04.0

)4.0)(2.0()4.0)(3.0()2.0)(3.0( jj

jjjjjjRAB =++

=

31

8667.03.0

)4.0)(2.0()4.0)(3.0()2.0)(3.0( jj

jjjjjjRBC =++

=


Approach 2:

Covert the read line which in Y form into delta to remove nodeCovert the read line, which in Y form into delta to remove node3 from the network:

j0.1625

769.0)3.1/1(12 =−= jY12

'2

'1 YEEP = )(12 j

max (1.05)(1.0)(0.769) 0.8080.808sine

PP δ

= =

=

1221max YEEP

32

e

Example 3: Power‐angle equation After Fault Clearedp g q

The fault on the system cleared by simultaneous opening ofthe circuit breakers at each end of the affected linethe circuit breakers at each end of the affected line.Determine the power‐angle equation and the swingequation for the post‐fault period

CB open CB open

33

Example 3: Power‐angle equation After Fault Clearedp g q

Upon removal of the faulted line, the net transfer admittanceacross the system isacross the system is

yj

j121

0 2 01 0 41429=

+ += −

( . . . ). per unit 429.112 jY =or

The post‐fault power‐angle equation is

δδ sin5.1sin)429.1()0.1()05.1( ==eP )()()(e

and the swing equation is

5180

10 15002

2fddtδ

δ= −. . sin

34

Synchronizing Power Coefficients___________

• From the power‐anglecurve, two values of anglesatisfied the mechanical Before faultsatisfied the mechanicalpower i.e at 28.440 and151.560.

H l th 28 440 After fault• However, only the 28.440is acceptable operatingpoint.A t bl ti• Acceptable operatingpoint is that thegenerator shall not losesynchronism when small

During fault

synchronism when smalltemporary changes occurin the electrical poweroutput from the machineoutput from the machine.

35


Consider small incremental changes in the operating point parameters, that is:

Δ+= δδδ 0Δ+= eee PPP 0(14.40)

Substituting above equation into Eq 14 37 (Power angle equation) δsinPP =Substituting above equation into Eq. 14.37 (Power‐angle equation)

)sincoscos(sin)sin(

00max

0max0

ΔΔ

ΔΔ

+=+=+

δδδδδδ

PPPP ee

δsinmaxPPe =

)scoscos(s 00max ΔΔ δδδδ

(14.41)

Since is a small incremental displacement from Δδ 0δ

ΔΔ ≅ δδsin 1cos ≅Δδ

Thus, the previous equation becomes:

+=+ δδδ )cos(sin PPPP (14.42)ΔΔ +=+ δδδ )cos(sin 0max0max0 PPPP ee

36


At the initial operating point :0δ

sinδPPP == (14 43)0max0 sinδPPP em == (14.43)

Equation (14.42) becomes:

(14 44)ΔΔ −=+− δδ )cos()( 0max0 PPPP eem

(14.44)

Substitute Eq. (14.40) into swing equation;

)()(202

02

ΔΔ +−=

+eem

s

PPPdt

dH δδω

(14.45)

Replacing the right‐hand side of this equation by (14.44);

0)cos(20max2

2

=+ ΔΔ δδδ P

ddH (14.46))( 0max2 Δω dts

37


Since is a constant value. Noting that is the slope of the power‐angle curve at the angle , we denote this slope as Sp and define it as:

0δ 0max cosδP

0δ

(14.47)0max cos

0

δδ δδ

PddPS e

p ===

Where Sp is called the synchronizing power coefficient. Replacing Eq. (14.47) into (14.46);

02

Δ δωδ Sd ps (14 48)022 =+ Δ

Δ δδHdt

d ps (14.48)

The above equation is a linear, second‐order differential equation.

If Sp positive – the solution corresponds to that of simple harmonic motion.

If Spnegative – the solution increases exponentially without limit.

)(tΔδ

)(tΔδp g p y)(tΔδ

38


The angular frequency of the un‐damped oscillations is given by:

sradelecHS ps

n /2ω

ω =(14.49)

which corresponds to a frequency of oscillation given by:

HS

f ps1 ω (14.50)Hz

Hf ps

n 221π

=

39

Synchronizing Power Coefficients___________Example:

The machine in previous example is operating at when it is subjectedto a slight temporary electrical‐system disturbance. Determine the frequency

°= 44.28δto a slight temporary electrical system disturbance. Determine the frequencyand period of oscillation of the machine rotor if the disturbance is removedbefore the prime mover responds. H = 5 MJ/MVA.

8466.144.28cos10.2 =°=pSThe synchronizing power coefficient is

The angular frequency of oscillation is therefore;

sradelecHS ps

n /343.8528466.1377

2=

××

==ω

ω

3438The corresponding frequency of oscillation is Hzfn 33.1

2343.8

==π

and the period of oscillation is sf

Tn

753.01==

f n

40

Equal‐Area Criterion of Stability______________

The swing equation is non‐linear in nature and thus, formal solution cannot be explicitly found.cannot be explicitly found.

perunitPPPdtdH

ema −==2

22 δω

To examine the stability of a two‐machine system without solving the

dtsω

swing equation, a direct approach is possible to be used i.e using equal‐area criterion.

41

Equal‐Area Criterion of Stability______________Consider the following system:

At point P (close to the bus), a three‐phase fault occurs and cleared by circuitbreaker A after a short period of time.

Thus, the effective transmission system is unaltered except while the fault is on.The short‐circuit caused by the fault is effectively at the bus and so theThe short circuit caused by the fault is effectively at the bus and so theelectrical power output from the generator becomes zero until fault is clear.

42

three‐phase fault

Equal‐Area Criterion of Stability______________To understand the physical condition before, during and after the fault,power‐angle curve need to be analyzed.

Initially, generator operates at synchronous speed with rotor angle ofand the input mechanical power equals the output electrical power Pe.

0δ

Before fault, Pm = Pe

43


At t = 0, Pe = 0, Pm = 1.0 puAcceleration constant

The difference must be accounted for by a rate of change of stored kinetic energy in the rotor masses.

Speed increase due to the drop of Peconstant acceleration from t = 0 to t = tc. 1.0 pu

For t<tc, the acceleration is constant givenby:

perunitPdH 02−=

ω

At t = 0, three phase fault occursdPs

2

2δ ω= (14.51)

perunitPdt m

s0=

ω

dt HPm2 2

(14.51)

44


Acceleration constantWhile the fault is on, velocity increase above synchronous speed and can beabove synchronous speed and can be found by integrating this equation:

d tδ ω ω∫

ddt H

P dtH

P tsm

sm

tδ ω ω= =∫ 2 20

For rotor angular position,;

(14.52)

δω

δ+s mPt 2 (14.53)

At t = 0, three phase fault occurs

δ δ= +s m

Ht

42

0(14.53)

(14.52) 45


Acceleration constantEq. (14.52) & (14.53) show that thevelocity of the rotor increase linearlyvelocity of the rotor increase linearlywith time with angle move from to0δ cδ

At the instant of fault clearing t = tc,g ,

the increase in rotor speed is

d Ps mδ ωdt H

tt ts m

cc= =2

angle separation between the generator and the infinite bus is

(14.54)

At t = 0, three phase fault occurs

and the infinite bus is

( )δω

δtPH

tt ts m

cc= = +4

20 (14.55)

(14.52) 46


When fault is cleared at , Peincrease abruptly to point d

cδ

At d, Pe > Pm , thus Pa isnegative

Rotor slow down as P goesRotor slow down as Pe goesfrom d to e

At t = tc, fault is cleared

47

Equal‐Area Criterion of Stability______________1. At e, the rotor speed is againsynchronous although rotor angle hasadvance to xδ

2. The angle is determined by thefact that A1 = A2

xδ

x

3. The acceleration power at e is stillnegative (retarding), so the rotorcannot remain at synchronous speedbut continue to slow down.

4. The relative velocity is negative andthe rotor angle moves back from pointthe rotor angle moves back from pointe to point a, which the rotor speed isless than synchronous.

6 In the absence of damping rotor would5. From a to f, the Pm exceeds the Peand the rotor increase speed again untilreaches synchronous speed at f

6. In the absence of damping, rotor wouldcontinue to oscillate in the sequence f‐a‐e, e‐a‐f, etc 48


In a system where one machine is swinging with respect to infinite bus,equal‐area criterion can be used to determine the stability of the system

d i di i b l i i iunder transient condition by solving swing equation.

Equal‐area criterion not applicable for multi‐machines.

The swing equation for the machine connected to the infinite bus is

dH 22 δ(14.56)em

sPP

dtdH

−=2

22 δω

Define the angular velocity of the rotor relative to synchronous speed by

ωδ

ω ωr sddt

= = − (14.57)sdt

49


Differentiate (14.57) with respect to t and substitute in (14.56) ;

(14.58)2H ddt

P Ps

rm eω

ω= −

When rotor speed is synchronous, equals and is zero. ω sω rω

Multiplying both side of Eq. (14.58) by ; dtdr /δω =

H d dω δ( )H ddt

P Pddts

rr

m eωω

ω δ2 = − (14.59)

The left‐hand side of the Eq. can be rewritten to give

(14.60)dtdPP

dtdH

emr

s

δωω

)()( 2

−=

50


Multiplying by dt and integrating, we obtain;δ

(14.61)∫ −=−2

1

)()( 21

22

δ

δ

δωωω

dPPHemrr

s

Since the rotor speed is synchronous at and then ;δ δ 0== ωωSince the rotor speed is synchronous at and , then ;1δ 2δ 021 == rr ωω

Under this condition, (14.61) becomes

(14.59)( )P P dm e− =∫ δδ

δ 01

2

and are any points on the power angle diagram provided that there are points at which the rotor speed is synchronous.

1δ 2δ

51

Equal‐Area Criterion of Stability______________In the figure, point a and e correspond to and 1δ 2δ

If perform integration in two steps;If perform integration, in two steps;

0)()(0

=−+− ∫∫ δδδ

δ

δ

δ

dPPdPPx

c

c

emem(14.63)

δδδ

δ

δ

δ

dPPdPPx

c

c

meem ∫∫ −=− )()(0

(14.64)

Fault periodArea A1

Post‐fault periodArea A2

The area under A1 and A4 are directly proportional toThe area under A1 and A4 are directly proportional tothe increase in kinetic energy of the rotor while it isaccelerating.

The area under A and A are directly proportional toThe area under A2 and A3 are directly proportional tothe decrease in kinetic energy of the rotor while it isdecelerating. 52

Equal‐Area Criterion of Stability______________Equal‐area criterion states that whatever kinetic energy is added to therotor following a fault must be removed after the fault to restore the rotorto synchronous speedto synchronous speed.

The shaded area A1 is dependent upon the time taken to clear the fault.

If the clearing has a delay, the angle increase.

As a result, the area A2 will also increase. If the increase

cδ

As a result, the area A2 will also increase. If the increasecause the rotor angle swing beyond , then the rotorspeed at that point on the power angle curve is abovesynchronous speed when positive accelerating power is

i t d

maxδ

again encountered.

Under influence of this positive accelerating power theangle will increase without limit and instability results.g y

53

Equal‐Area Criterion of Stability______________There is a critical angle for clearing the fault in order to satisfy therequirements of the equal‐area criterion for stability.

This angle is called the critical clearing angle

The corresponding critical time for removing the fault is called critical

crδ

The corresponding critical time for removing the fault is called criticalclearing time tcr

Power‐angle curve showing thecritical‐clearing angle . Area A1and A2 are equal

crδ

54

Equal‐Area Criterion of Stability______________The critical clearing angle and critical clearing time tcr can be calculatedby calculating the area of A1 and A2.

crδ

( )A P d Pm m crcr

1 00= = −∫ δ δ δδ

δ

( )∫δ

(14.65)

(14 66)( )A P P dmcr2 = −∫ max sinmax δ δδ

δ

)()cos(cos maxmaxmax crmcr PP δδδδ −−−=

(14.66)

)()( maxmaxmax crmcr

Equating the expressions for A1 and A2, and transposing terms, yields

( )( )/δ δ δ δP P + (14.67)( )( )cos / cosmax max maxδ δ δ δcr mP P= − +0

55


From sinusoidal power‐angle curve, we see that

(14 68)δ π δ elec rad (14.68)δ π δmax = − 0 elec rad

P Pm = max sinδ0(14.69)

Substitute and in Eq. (14.67), simplifying the result and solving for

maxδ maxP

crδ

( )[ ]δ π δ δ δ= −cos sin cos1 2 (14 70)( )[ ]δ π δ δ δcr = − −cos sin cos0 0 02 (14.70)

In order to get tcr, substitute critical angle equation into (14.55) and then solve to obtain tcr;

δω

δcrs m

cr

PH

t= +4

20

( )t

HPcr

cr

s m=

−4 0δ δω

(14.71) (14.72)

56

Equal‐Area Criterion of Stability ‐Example_____Calculate the critical clearing angle and critical clearing time for the system shown below. When a three phase fault occurs at point P. The initial conditions are the same as in Example 1 and H = 5MJ/MVAp /

P Pe = =max sin . sinδ δ210

Solution

The power angle equation is

The initial rotor angle is

δ0028 44 0 496= =. . elec rad

d h h l h f h l land the mechanical input power Pm is 1.0 pu. Therefore, the critical angle is calculated using Eq. (14.70)

( )[ ]δ πcr = − × −−cos . sin . cos .1 0 02 0 496 28 44 28 44 = =81697 14260. . elec rad

( )[ ]δ π δ δ δcr = − −−cos sin cos10 0 02

and the critical clearing time is

( )× −4 5 1426 0 496. . = 0 222s( )tcr = ×377 1

= 0.222s

57

Further Application of the Equal‐Area Criterion

• Equal‐Area Criterion can only be applied forthe case of two machines or one machineand infinite busand infinite bus.

• When a generator is supplying power to aninfinite bus over two parallel lines, openingone of the lines may cause the generator tolose synchronismlose synchronism.

• If a three phase fault occurs on the bus onwhich two parallel lines are connected, nopower can be transmitted over either thelineline.

• If the fault is at the end of one of the lines, CB will operate and power canflow through another line.

• In this condition, there is some impedance between the parallel buses andthe fault. Thus, some power is transmitted during the fault.

58

Further Application of the Equal‐Area Criterion

Considering the transmitted of power during fault, a general Equal‐areacriterion is applied;criterion is applied;

By evaluating the area A1 and A2 as in the previous approach, we can find that;

1max2maxmax coscos))(/(cos rrPP oomcr

−+−=

δδδδδ (14.73)12 rrcr −

( )

59

Further Application of the Equal‐Area Criterion ‐ Example

Determine the critical clearing angle for the three phase fault described in the previous example.

The power‐angle equations obtained in the previous examples areThe power‐angle equations obtained in the previous examples are

δδ sin1.2sinmax =P δδ sin808.0sinmax1 =Pr

δδ sin5.1sinmax2 =PrBefore fault: During fault:

After fault:

Hence

385.01.2

808.01 ==r 714.0

1.25.1

2 ==r

rad412.219.1385.10.1sin180 1

max =°=−°= −δ

)44.28cos(385.0)19.138cos(714.0)496.0412.2)(1.2/0.1(cos °−°+−δ

127.0385.0714.0

)()())((cos

=−

=crδ

°= 726.82crδ

To determine the critical clearing time, we must obtain the swing curve of versus t for this example.

δ60

STEP‐BY‐STEP SOLUTION OF SWING CURVE____

For large systems we depend on the digital computer to determine δ versus tfor all the machines in which we are interested; and δ can be plotted versus tfor all the machines in which we are interested; and δ can be plotted versus tfor a machines to obtain the swing curve of that machine.

The angle δ is calculated as a function of time over a period long enough tog p g gdetermine whether δ will increase without limit or reach a maximum andstart to decrease.

Although the latter result usually indicates stability, on an actual systemwhere a number of variable are taken into account it may be necessary to plotδ versus t over a long enough interval to be sure δ will not increase again

ith t t i t l lwithout returning to a low value.

61

STEP‐BY‐STEP SOLUTION OF SWING CURVE____By determining swing curves for various clearing times the length of timepermitted before clearing a fault can be determined.

Standard interrupting times for circuit breakers and their associated relaysare commonly 8, 5, 3 or 2 cycles after a fault occurs, and thus breakerspeeds may be specified.

Calculations should be made for a fault in the position which will allow theleast transfer of power from the machine and for the most severe type offault for which protection against loss of stability is justified.

A number of different methods are available for the numerical evaluationof second order differential equations in step by step computations forof second‐order differential equations in step‐by‐step computations forsmall increments of the independent variable.

The more elaborate methods are practical only when the computations are

62

The more elaborate methods are practical only when the computations areperformed on a digital computer. The step‐by‐step method used for handcalculation is necessarily simpler than some of the methods recommendedfor digital computers.

STEP‐BY‐STEP SOLUTION OF SWING CURVE____In the method for hand calculation the change in the angular position of therotor during a short interval of time is computed by making the followingassumptions:assumptions:

•The accelerating power Pa computed at the beginning of an interval isconstant from the middle of the preceding interval to the middle of theconstant from the middle of the preceding interval to the middle of theinterval considered.

•The angular velocity is constant throughout any interval at the valueg y g ycomputed for the idle of the interval.

**neither of the assumptions is true, since δ is changing continuously and both Paand ω are functions of δ.

63

As the time interval is decreased, the computed swing curve approaches the true curve.

STEP‐BY‐STEP SOLUTION OF SWING CURVE____and

Figure 14.4 will help in visualizing the assumptions.

The accelerating power is computed for the pointsThe accelerating power is computed for the points enclosed in circles at the ends of the n-2, n-1, and nintervals, which are the beginnings of the n-1, n and n+1 interval.

The step curve of Pa in Fig. 14.4 results from the assumption that Pa is constant between midpoints of the intervals.

Similarly, ωr, the excess of the angular velocity ωover the synchronous angular velocity ωs, is shown as a step curve that is constant throughout the interval at the value computed for the midpoint. e v e v ue co pu ed o e dpo .

Between the ordinates (n-3/2) and (n-1/2) there

64

Between the ordinates (n 3/2) and (n 1/2) there is a change of speed caused by the constant accelerating power.

STEP‐BY‐STEP SOLUTION OF SWING CURVE____

The change in speed in the product of theacceleration and the time interval, thus:

ω ωδ

r n r n a nddt

tf

HP t, / , / ,1 2 3 2

2

2 1180

− = = −Δ Δ (1)

The change in δ over any interval is the product of ω, for the interval and the time of the interval. Thus, the change in δ during the 1 i l in‐1 interval is:

Δ Δδ δ δ ωn n n r nt− − − −= − =1 1 2 3 2, / (2)

and during the nth interval

65

Δ Δδ δ δ ωn n n r nt= − =− −1 1 2, / (3)

Fig. 14.14

STEP‐BY‐STEP SOLUTION OF SWING CURVE____Substracting Eq (2) from Eq. (3) andsubstituting Eq. (1) in the resultingequation to eliminate all values of ω,

(4)

yields

Δ Δδ δn n a nkP= +− −1 1,

where

( )kf

t=180 2Δ (5)( )

H

66Fig. 14.14

STEP‐BY‐STEP SOLUTION OF SWING CURVE____Equation (4) is the important for the step-by-step solution of the swing equation with the necessaryassumption enumerated, for it shows how to calculate the change in δ for the previous interval andthe accelerating power for the interval in equation are known.

Equation (4) shows that (subject to the stated assumptions), the change in torque angle during agiven interval is equal to the change in torque angle during the preceding interval plus theaccelerating power at the beginning of the interval times k.

The accelerating power is calculated at the beginning of each new interval. The solution progressesthrough enough intervals to obtain points for plotting the swing curve.

Greater accuracy is obtained when the duration of the interval is small. An interval of 0.05s isyusually satisfactory.

The occurrence of a fault causes a discontinuity in the accelerating power Pa which is zero beforethe fault and a definite amount immediately following the fault.y g

The discontinuity occurs at the beginning of the interval, when t=0. Reference to Fig. 14.14 showsthat our method of calculation assumes that the accelerating power computed at the beginning of aninterval considered.

67

When the fault occurs, we have two values of Pa at the beginning of the interval, and we must takethe average of these two values as our constant accelerating power.

EXAMPLE SOLUTION OF SWING CURVE____Given a two‐bus system of 50 Hz system, with machine 1 delivering 0.9 p.u powerto infinite bus and the machine 1 has H = 4.5 MJ/MVA. Three phase faultoccurred at the middle of the line is cleared in 0.2s. The power angle equationsp g qare given as below

Pre-fault: δsin0.2=eP

During fault:

δsin091.1=eP

After fault cleared: δsin714.1=eP

Compute the swing curve using step-by-step method.

68

EXAMPLE SOLUTION OF SWING CURVE____

Solution:

Remember! Mechanical power doesn’t change during transient stability studies.

At the beginning machine 1 delivering 0 9 pu power At this steady stateAt the beginning, machine 1 delivering 0.9 pu power. At this steady statecondition, Pe = Pm = 0.9 pu.

Wh th f lt t 0 th t l i t th i iti l lWhen the fault occurs at t=0s the rotor angle is at the initial value:

deg74269.0sin

9.0sin0.2

1 elec°=⎟⎞

⎜⎛=

=

−δ

δ

deg74.260.2

sin0 elec=⎟⎠

⎜⎝

=δ

5)05.0(54

)50(180)(180 22 ==Δ= tH

fk

69

5.4H


At the beginning of the first interval there is a discontinuity in the accelerationpower. Just before fault occurs, that is t=0-s Pa=0 since the machine is operating inp a p gsynchronism and the rotor angle is the initial steady-state rotor position 0δ

At t=0+s immediately after the fault has occurred;

409.0491.09.0

491.0)74.26sin(091.10

0

=−=−=

=°=+

+

ema

e

PPP

P

The average value of Pa at t = 0 is 2045.02/409.0 ==avgaP

Similarly, for representing any change in switching condition during stabilityanalysis, the accelerating power to be used is the average of the accelerating powerjust before and just after the change in switching condition.

70


°=×= 02251204505kP

Then we can find

=×= 0225.12045.05akP

°=°+=Δ 0225.10225.101δ

tΔThe interval

tΔis the change in rotor angle as time advances over the first interval from 0 toAt the end of the first interval, °=°+°=Δ+= 7625.270225.174.26101 δδδ

st 05.0=Δ °=°−=−= 9590.1)7625.27sin(091.19.0(5)(51 PePmkPAt st 05.0Δ 9590.1)7625.27sin(091.19.0(5)(51, PePmkPa

°=°+°=+Δ=Δ 9815.29590.10225.11,12 akPδδδδδ

At

and it follows that the increase in rotor angle over the second time interval is

°=°+°=Δ+= 744.309815.27625.27212 δδδHence, at the end of the second interval

The subsequent steps are shown in the following Table.

71


T (Sec) Pmax sin Po (pu) Pa (pu)kPa (ElecDegree (Elec Degree) (Elec Degree)

0‐ 2.0 0.450 0.9 0.0 26.740+ 1.091 0.450 0.491 0.409 26.74

0 Avg 1 5455 0 2045 1 0225

δδΔδ

0 Avg 1.5455 0.2045 1.02251.0225

0.05 1.091 0.465863 0.508257 0.391743 1.958716 27.76252.981216

0.1 1.091 0.511259 0.557783 0.342217 1.711084 30.743724.692301

0.15 1.091 0.579859 0.632626 0.267374 1.33687 35.436026.029171

0.2‐ 1.091 0.662 0.722 0.178 41.460.2+ 1.714 0.662 1.135 ‐0.235 41.460.2Avg 0.815 ‐0.0285 ‐0.1425

5 8866715.8866710.25 1.714 0.735539 1.260714 ‐0.36071 ‐1.80357 47.34667

4.0830990.3 1.714 0.781917 1.340206 ‐0.44021 ‐2.20103 51.42977

1.8820690.35 1.714 0.801971 1.374579 ‐0.47458 ‐2.37289 53.31184

‐0.490820.4 1.714 0.796824 1.365756 ‐0.46576 ‐2.32878 52.82101

‐2.81960.45 1.714 0.766133 1.313152 ‐0.41315 ‐2.06576 50.00141

‐4.885360.5 1.714 0.70861 1.214557 ‐0.31456 ‐1.57278 45.11605

72

‐6.45815


0.5 1.714 0.70861 1.214557 ‐0.31456 ‐1.57278 45.11605‐6.45815

0.55 1.714 0.624737 1.0708 ‐0.1708 ‐0.854 38.6579‐7.31215

0.6 1.714 0.520262 0.891729 0.008271 0.041356 31.34575‐7.27079

0.65 1.714 0.407981 0.69928 0.20072 1.003601 24.07496‐6.26719

0.7 1.714 0.305863 0.524249 0.375751 1.878756 17.80777‐4.38843

0.75 1.714 0.232106 0.397829 0.502171 2.510854 13.41934‐1.87758

0 8 1 714 0 200108 0 342984 0 557016 2 785078 11 541760.8 1.714 0.200108 0.342984 0.557016 2.785078 11.541760.907497

0.85 1.714 0.215602 0.369542 0.530458 2.652288 12.449253.559785

0.9 1.714 0.275824 0.472762 0.427238 2.13619 16.009045 6959755.695975

0.95 1.714 0.369874 0.633964 0.266036 1.330182 21.705017.026157

1 1.714 0.480758 0.824019 0.075981 0.379907 28.731177.406064

73

1.05 1.714 0.589787 1.010896 ‐0.1109 ‐0.55448 36.13724


At t=0.2s, fault is cleared. Hence, the average of accelerating power just before and after should be

0 2 1 091 i (41 46 ) 0 223°

considered.

At t=0.2- s (just before fault cleared);

0.2

0.2 0.2

1.091sin(41.46 ) 0.7223

0.9 0.7223 0.1777e

a m e

P

P P P

−

− −

= ° =

= − = − =

At t=0.2+ s (just after fault cleared);

2348.01348.19.0

1348.1)46.41sin(714.12.02.0

2.0

−=−=−=

=°=++

+

ema

e

PPP

P

23480177702020 +PP

(j );

02855.02

2348.01777.02

)(2.02.0

−=−

=+

=+−

aaa

PPaveragePHence, average Pa;

The subsequent steps are shown in the following Table.

74

Engineering

Power System Stablity