Power System Transient Stability

Power System Transient Stability

Submitted By:Santosh Kumar Gupta Assistant ProfessorEE Department SIT Sitamrhi

Introduction Power SystemStability

• The tendency of a power system to develop restoring forces equal toor greater than the disturbing forces to maintain the state ofequilibrium is know asstability.

• If the forces tending to hold machines in synchronism with oneanother are sufficient to overcome the disturbing forces, the systemis said to remainstable.

• Stability Studies:

– Transient

– Dynamic

– Steady‐state

• The main purpose of transient stability studies is to determine

disturbances such as transmission system faults, suddenwhether a system will remain in synchronism following major

loadchanges, loss of generating units, or line switching.

2

The Power‐AngleEquation

• In the swing equation, the input mechanical power from the

prime mover Pm is assumedconstant.

• Thus, the Pe will determine whether the rotor accelerates,

decelerates, or remains at synchronousspeed.

• Changes in Peare determined by conditions on the transmission

and distribution networks and the loads on the system to which

the generator supplypower.

3


+

djX '

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

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

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

I

E'

djIX'

Vt

_E'

I

Vt

Reference

4(a) (b)


• Consider a generator supplying power through a transmission

system to a receiving‐end system at bus 2.

I1I2 E’1 is transient internal voltage of

generator at bus 1

1E '

2E '

E’2 is transient internal voltage of

generator at bus 2

5

• The elements of the bus admittance matrix for the network

reduced to a two nodes in addition to the reference node is:

21 22 YY =Y11 Y12

Ybas

(14.28)


• Power equation at a bus k is given by:

kn n

N

Yn=1

k k kVjQ VP − = (14.29)

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

' '

1 12 2

' '

1 1 1 11 1 E (Y E )P + jQ = E (Y E ) + (14.30)

whereI1

I2

'

1

'

1 1EE = 222 = E ' E '

11+ jBY11 =G11Y12 = Y12 12

E '

6

1 2E '


• We obtain:

122 12Y cos (1 − 2 − )E '

2

P = E ' G + E ' 1 1 11 1

12

2'

1 1Y sin (1 − 2 − )E 'B + E '

11 1 2 12Q = − E

(14.31)

(14.32)

21• If we let =

(14.32)

−2

− , we obtain from (14.31)and12

and =

121 211

2E '

G + E ''

11P = E (14.33)

1 2 12

7

11

2

Y sin ( - )

Y cos ( - )E 'B − E ''

11Q = − E (14.34)


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

where(14.35)Pe = Pc + Pmax sin( − )

• When the network is considered without resistance, all the elements of Ybus are susceptances,so both G11 and becomes zero and Eq. (14.35)becomes;

(14.36)' 2

G11cP = E1

P = E ' E ' Ymax 1 2 12

Pe = Pmax sin (14.37) Power AngleEquation

where , with X is the transfer reactancebetweenXE '

2max 1P = E '

E’1 andE’2

8

Example 1: Power‐angle equation beforefault

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 systemapplicable to the operatingconditions.

9


The reactance diagram for the system is shown:

X = 0.10 +0.4

= 0.3 per unit

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

2

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

0.3

10

(1.0)(1.0)sin =1.0sin =

X

Vt Vequation.

is the angle of the terminalvoltageV is the voltage of the infinite bus, and

relative to the infinitebus


Solve

= sin−1 0.3 = 17.4580

Terminal voltage, Vt : Vt = 1.017.458 = 0.954 + j0.300 perunit0

The output current from the generatoris:

1.017.4580 − 1.000

I = = 1.0 + j0.1535 = 1.0128.7290 perunitj0.3

The transient internal voltage is

E'= Vt + XI

E' = (0.954 + j0.30) + j(0.2)(1.0 + j0.1535)

= 0.923 + j0.5 = 1.05028.440 per unit

11


The power‐angle equation relating the transient internal voltage E’and the

infinite bus voltage V is determined by the total series reactance

2X = 0.2 + 0.1 +

0.4 = 0.5 per unit

Hence, the power‐angle equation is:

eP =(1.05)(1.0)

sin = 2.1sin p.u0.5

Where is the machine rotor angle with respect to infinite bus

The swing equation for the machineisd 2

= − 1.0 2.10 sin per unit

12

180 f dt2

H

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

and is in electricaldegree

Example 1: Power‐angle equation

The power‐angle equation isplotted:

Before fault

After fault

During fault

13

Example 2: Power‐angle equation DuringFault

The same network in example 1 is used. Three phase faultoccurs at point P as shown in the Figure. Determine the power‐

angle equation for the system with the fault and the

corresponding swing equation.TakeH = 5 MJ/MVA

14


Approach 1:

The reactance diagram of the system during fault is shown below:

The value is admittance

per unit

15


The Y bus is:

As been calculated in example 1, internal transient voltage remainsas E'= 1.0528.44 (based on the assumption that flux linkage is

constant in themachine)

3.333 Y = j 0

3.333

2.50

−3.333 0

− 7.50

2.50 − 10.833

bas

Since bus 3 has no external source connection and it may be removed by the node elimination procedure, the Y bus matrix isreduced to:

−

3.333 1 3.333 2.5=

−3.333 0bus

Y 2.5 −10.833 0 − 7.5

Y12 = j

− 2.308 0.769 Y11

21

16

Y22 0.769 − 6.923

Y


The magnitude of the transfer admittance is 0.769 and therefore,

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

Pe = 0.808 sin per unit

The corresponding swing equationis

P = E ' E ' Y '

max 1 2 12= (1.05)(1.0)(0.769) = 0.808 per unit

180 f dt2

17

5 d 2= 1.0 − 0.808 sin per unit

Because of the inertia, the rotor cannot change position instantly upon

occurrence of the fault. Therefore, the rotor angle is initially 28.440 and

the electrical power output is

Pe = 0.808 sin 28.44= 0.385


The initial accelerating power is:Pa = 1.0 − 0.385= 0.615 per unit

and the initial acceleration is positive with the value given by

dt 2 5

18

d 2 180 f(0.615) = 22.14 f= elec deg / s2


Approach 2:

Covert the read line (which in Y form) into delta to remove node

3 from thenetwork:j11.3

1 3

32

2

j0.65 j0.8667

j0.2R

AC=

( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)= j1.3

j0.4R

AB=

( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)= j0.65

=( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)

= j0.8667

19j0.3

RBC


Approach 2:

Covert the read line, which in Y form into delta to remove node

3 from thenetwork:

j0.1625

Y12 = −(1/ j1.3) = 0.769P = E ' E ' Ymax 1 2 12

Pmax = (1.05)(1.0)(0.769) =0.808

Pe = 0.808sin

20

Example 3: Power‐angle equation After FaultCleared

The fault on the system cleared by simultaneous opening of

the circuit breakers at each end of the affected line.

Determine the power‐angle equation and the swing

equation for the post‐faultperiod

CBopen

21

CBopen

Example 3: Power‐angle equation After FaultCleared

Upon removal of the faulted line, the net transfer admittance

across the systemis

j(0.2 + 0.1 +0.4)= − j1.429 per unit

12y =

1 Y12 = j1.429or

The post‐fault power‐angle equation is

Pe= (1.05)(1.0)(1.429)sin =1.5sin

and the swing equationis

5 d 2

22

180 f dt2= 1.0−1.500 sin

Synchronizing PowerCoefficients

• From the power‐anglecurve, two values ofangle

Before faultsatisfied the mechanical

After fault

power i.e at 28.440 and151.560.

• However, only the 28.440

is acceptable operatingpoint.

• Acceptable operatingpoint is that the

During fault

generator shall not losesynchronism when smalltemporary changes occurin the electrical poweroutput from themachine.

23


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

= 0 +Pe = Pe0 + Pe

(14.40)

Pe0

= Pmax sinSubstituting above equation into Eq. 14.37 (Power‐angle equation) Pe

(14.41)

+ Pe= Pmax sin(0 + )

= Pmax(sin0 cos + cos0 sin)

Since is a small incremental displacement from 0

sin cos 1

Thus, the previous equationbecomes:

(14.42)0

24

Pe0 + Pe = Pmax sin 0 + (Pmax cos )


At the initial operating point 0 :

0Pm = Pe0 = Pmax sin (14.43)

Equation (14.42)becomes:

Pm− (Pe0 + Pe) = −(Pmax cos 0 )(14.44)

Substitute Eq. (14.40) into swingequation;

e0 e− (P + P )m

s

0 = Pdt 2

2H d 2 ( + )

(14.45)

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

+ (Pmax cos 0 ) = 02H d 2 (14.46)

dt 2

25

s


Since 0 is a constant value. Noting that Pmax cos0 is the slope of thepower‐

angle curve at the angle0, we denote this slope as Sp and define it as:

(14.47)0max= P cos

d=

dPe

=0

S p

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

= 0+ s Sp

d 2

dt 2 2H(14.48)

The above equation is a linear, second‐orderdifferential equation.

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

If Spnegative – the solution (t) increases exponentially without limit.

26


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

elec rad / s2H

s Sp

n =

(14.49)

which corresponds to a frequency of oscillation given by:

s p1 S(14.50)

Hz

27

n2 2H

f =

Synchronizing PowerCoefficientsExample:

The machine in previous example is operating at = 28.44when it is subjected

to a slight temporary electrical‐system disturbance. Determine the frequency

and period of oscillation of the machine rotor if the disturbance is removed

before the prime mover responds. H = 5 MJ/MVA.

S p = 2.10 cos 28.44=1.8466The synchronizing power coefficient is

The angular frequency of oscillation istherefore;

elec rad / s2H 25

n

377 1.8466=8.343=

s S p =

The corresponding frequency of oscillationis nf =8.343

= 1.33 Hz2

and the period of oscillationis T =1

= 0.753 s f n

28

Equal‐Area Criterion ofStability

The swing equation is non‐linear in nature and thus, formal solution

cannot be explicitlyfound.

Pa = Pm − Pe perunitdt 2

2H d 2=

s

29

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

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

area criterion.

Equal‐Area Criterion ofStabilityConsider the following system:

At point P (close to the bus), a three‐phase fault occurs and cleared by circuit

breaker 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 the

electrical power output from the generator becomes zero until fault is clear.

three‐phase fault

30


To understand the physical condition before, during and after the fault,

power‐angle curve need to beanalyzed.

Initially, generator operates at synchronous speed with rotor angleof 0

and the input mechanical power equals the output electrical powerPe.

Before fault, Pm = Pe

31


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

The difference mustbe accounted for by a

rate of change of stored kinetic energy in

the rotor masses.

Speed increase due to the drop of Pe

constant acceleration from t=0 to t= tc. 1.0pu

For t<tc, the acceleration is constantgiven

by:

− 0 perunit2H d

= P

At t = 0, three phase faultoccursd 2 = s P (14.51)

m

s dt

m

32

dt 2 2H


Acceleration constantWhile the fault is on, velocity increase

above synchronous speed and can be

found by integrating thisequation:

dP ts

m

s

mP dt == t

dt 0 2H 2H

For rotor angularposition,;

(14.52)

(14.53)

At t = 0, three phase faultoccurs

+

(14.52)33

=s Pm t2

4H 0

Equal‐Area Criterion of Stability

Acceleration constantEq. (14.52) & (14.53) show that the

velocity of the rotor increase linearly

with time with angle move from0 to c

At the instant of fault clearing t =tc,

the increase in rotor speedis

ct=tc

d=

s Pmt dt 2H (14.54)

At t = 0, three phase faultoccurs

angle separation between thegenerator

and the infinite bus is

(14.52)34

Ps m

ct=tc(t) = t +

4H

2

0(14.55)


When fault is cleared at c ,Pe

increase abruptly to pointd

At d, Pe > Pm , thus Pa is

negative

Rotor slow down as Pe goes

from d toe

At t = tc, fault iscleared

35


1. At e, the rotor speed is again

synchronous although rotor angle has

advance to x

2. The angle is determined by thex

fact that A1 = A2

3.The acceleration power at e is still

negative (retarding), so the rotor

cannot remain at synchronous speed

but continue to slowdown.

4.The relative velocity is negative and

the rotor angle moves back from point

e to point a, which the rotor speed is

less than synchronous.

5. From a to f, the Pm exceeds the Pe

and the rotor increase speed again until

reaches synchronous speed atf

6. In the absence of damping, rotor would

continue 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

undertransient condition by solving swing equation.

Equal‐area criterion not applicable formulti‐machines.

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

(14.56)s

= Pm −Pedt 2

2H d 2

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

r s =d

= − (14.57)dt

37


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

(14.58)2H dr

= Pm − Pe

s dt

When rotor speed issynchronous, equals s andr is zero.

Multiplying both side of Eq. (14.58) by r = d / dt ;

d

dt dts

r

r

m e= (P − P )d

H2 (14.59)

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

(14.60)dt

38

de− P )m

s dt

H d (2 )

r = (P


Multiplying by dt and integrating, weobtain;

(14.61)

2

22 (

− ) = (P − P )dH

m er1r2

s 1

Since the rotor speed is synchronous at

39

= 0 ;1 and 2 ,then r1 = r2

Under this condition, (14.61)becomes

(14.59)Pm e

( − P )d = 01

2

1 and 2 are any points on the power angle diagram provided that there are

points at which the rotor speed is synchronous.


In the figure, point a and e correspond to1 and 2

If perform integration, in twosteps;

c x

(Pm − Pe ) d + (Pm − Pe ) d = 0

0 c

(14.63)

c x

(Pm − Pe ) d = (Pe − Pm )d0 c

(14.64)

Faultperiod

Area A1

Post‐faultperiod

Area A2

The area under A1 and A4 are directly proportional to

the increase in kinetic energy of the rotor while it is

accelerating.

The area under A2 and A3 aredirectly proportional tothe decrease in kinetic energy of the rotor while it is

decelerating. 52


Equal‐area criterion states that whatever kinetic energy is added to the

rotor following a fault must be removed after the fault to restore the rotor

to synchronous speed.

The shaded area A1 is dependent upon the time takento

clear the fault.

If the clearing has a delay, the angle c increase.

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

cause the rotor angle swing beyond max , then the rotor

speed at that point on the power angle curve is above

synchronous speed when positive accelerating power is

again encountered.

Under influence of this positive accelerating power the

angle will increase without limit and instabilityresults.

41


There is a critical angle for clearing the fault in order to satisfy the

requirements of the equal‐area criterion for stability.

This angle is called the critical clearing anglecr

The corresponding critical time for removing the fault is called critical

clearing timetcr

Power‐angle curve showing the

critical‐clearing angle cr . Area A1

and A2 are equal

42

Equal‐Area Criterion of Stability

The critical clearing angle cr and critical clearing time tcr can be calculated

by calculating the area of A1 andA2.

m m cr

cr

1A =

0P d = P ( − )0

(14.65)

( m2A = crmaxP sin − P )dmax (14.66)

= Pmax (cos cr − cos max ) − Pm (max − cr )

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

43

(14.67)cos = − + coscr Pm / Pmax max 0 max


From sinusoidal power‐angle curve, we seethat

(14.68)max 0= − elec rad

Pm = Pmax sin0(14.69)

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

cr= cos−1 ( − 2 )sin cos

0 0 0 (14.70)

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

obtain tcr;

cr = s m P

crt +4H

2

0 cr

44

t =s m

4H(cr − 0 ) P

(14.71) (14.72)

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/MVA

Solution

The power angle equation is Pe = Pmax sin = 2.10 sin

The initial rotor angle is0 = 28.44 = 0.496 elec rad

0

and the mechanical input power Pm is 1.0 pu.Therefore, the critical angle is

calculated using Eq.(14.70) cr ( )sin − cos −1

0 0 0 = cos − 2

cr = cos (− 2 0.496) sin 28.44 − cos 28.44 = 81.697 = 1.426 elec rad−1 0 0 0

and the critical clearing timeis

4 5(1.426 − 0.496)cr

45

t =377 1

= 0.222s

Further Application of the Equal‐AreaCriterion

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

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

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

• If the fault is at the end of one of the lines, CB will operate and power can

flow through anotherline.

• In this condition, there is some impedance between the parallel buses and

the fault.Thus, some power is transmitted during the fault.

46

Further Application of the Equal‐AreaCriterion

Considering the transmitted of power during fault, a general Equal‐area

criterion is applied;

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

cr=

(Pm / Pmax )(max − o ) + r2 cos max − r1cos ocos (14.73)

2 1r − r

47

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 arer1Pmax sin = 0.808sin Pmax sin = 2.1sin

r2Pmax sin = 1.5sin

During fault:Before fault:

After fault:

Hence

1r =0.808

= 0.3852.1 2.1

2r =1.5

=0.714

1.5max

= 180 − sin−1 1.0= 138.19 = 2.412 rad

=(1.0/ 2.1)(2.412 − 0.496) + 0.714cos(138.19) − 0.385cos(28.44)

0.714 − 0.385cos cr

=0.127

cr = 82.726

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

48

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Power System Transient Stability

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