module 6 stability

7/24/2019 Module 6 Stability

1/59

Ontoseno Penangsang1)

1) Electrical Department, Sepuluh Nopember of Institute Technology

Surabaya, 60111, Indonesia


2/59

Introduction

Synchronous machine model for

stability study

Developing Swing equation

Steady-state stability small

disturbance

Stability analysis on Swing equation

Transient stability

Equal Area Criterion


3/59

INTRODUCTION TO STABILITY

What is stability

the tendency of power system to restore the state of

equilibrium after the disturbance

mostly concerned with the behavior of synchronousmachine after a disturbance

in short, if synchronous machines can remainsynchronism after disturbances, we say that system isstable


4/59

Stability issue

steady-state stability the ability of power system to

regain synchronism after small and slow disturbancessuch as gradual power change

transient stability the ability of power system toregain synchronism after large and sudden

disturbances such as a fault


5/59

POWER ANGLE

Power angle

relative angle rbetweenrotor mmf and air-gap

mmf (angle between Frand Fsr), both rotating insynchronous speed

also the angle rbetweenno-load generated emf E

and stator voltage Esr also the angle between

emf E and terminalvoltage V, if neglectingarmature resistance and

leakage flux


6/59

SYNCHRONOUS MACHINE MODELS

FOR STABILITY STUDY

Simplified synchronous machine model

the simplified machine model is decided by

the proper reactances,Xd, Xd, or Xd for very short time of transient analysis,

use Xd

for short time of transient analysis, use Xd

for steady-state analysis, use Xd


7/59


FOR STABILITY STUDY

Simplified synchronous machine model

substation bus voltage and frequency

remain constant is referred as infinite bus generator is represented by a constant

voltage E behind direct axis transientreactance Xd

Zs

ZLjXd

E

Vg V


8/59

Converting the network into equivalent circuit for the conversion, please see Eq.11.23

use equivalent line model for currents

real power at node 1

V

E

y

y

I

I '

yy

yy

122012

121210

2

1

121211112

cos'cos' YVEYEPe

y10

y12

E V

y20

1 2

I1 I2


FOR STABILITY STUDY


9/59

Real power flow equation

lety12 = 1 / X12 simplified real power equation:

Power angle curve

gradual increase of generator power output ispossible until Pmax (max power transferred) is reached

max power is referred as steady-state stability limit at

=90o

sin'

12X

VEPe

0

PePm

/2

Pmax

0

Pe

12

max

'

X

VEP


FOR STABILITY STUDY


10/59

Transient stability analysis condition: generator is suddenly short-circuited

current during the transient is limited by Xd

voltage behind reactance E=Vg+jXdIa

Vg is the generator terminal voltage, Ia is prefault

steady state generator current

phenomena: field flux linkage will tend to remain

constant during the initial disturbance, thus E is

assumed constant

transient power angle curve has the same form as

steady-state curve but with higher peak value,

probably with smaller Xd


FOR STABILITY STUDY


11/59

Phasor diagram of salient-pole machine

condition: under steady state with armature

resistance neglected


FOR STABILITY STUDY

qdXjI

qjIX

ddXjI

VaI

dI

qdd XXjI

E

qqXI

qI


12/59

Phasor diagram of salient-pole machine

power angle equation in per unit

voltage equation in per unit

E is no-load generated emf in pu, V is generatorterminal voltage in pu

2sin

2

sin2

qd

qd

d XX

XXV

X

VEP

sincoscos addd IXVIXVE


FOR STABILITY STUDY


13/59

Calculation of voltage E

starting with a given (known) terminal voltage V and

armature current Ia

, we need to calculate first by

using phasor diagram and then result in voltage E

once E is obtained, P could be calculated

sincos ad IXVE

sin

costan 1

aq

aq

IXV

IX


FOR STABILITY STUDY


14/59

Transient power equation

for salient machine

this equation represents the behavior of SM in early

part of transient period calculate first, then calculate |Eq|:

see example 11.1

2sin2

sin'

'2

'

'

qd

qd

d

q

eXX

XXV

X

VEP

sincos '' adq IXVE


FOR STABILITY STUDY


15/59

Synchronous machine operation

consider a synchronous generator with

electromagnetic torque Te running at synchronousspeed sm.

during the normal operation, the mechanical torque

Tm = Te

a disturbance occur will result in

accelerating/decelerating torque Ta=Tm-Te (Ta>0 if

accelerating, Ta


16/59

Synchronous machine operation

introduce the combined moment of inertia of prime

mover and generator J

by the law of rotation --

m is the angular displacement of rotor w.r.t. stationeryreference frame on the stator

emam TTT

dt

dJ

2

2

DEVELOPING SWING EQUATION


17/59

Derivation of swing equation

m = smt+m, sm is the constant angular velocity

take the derivative of m, we obtain

take the second derivative of m, we obtain

dt

d

dt

d msm

m

2

2

2

2

dt

d

dt

d mm



18/59


substitute into the law of rotation

multiplying m to obtain power equation

emam TTT

dt

dJ

2

2

ememmmmm

m PPTTdt

dM

dt

dJ

2

2

2

2



19/59


swing equation in terms of inertial constant M

relations between electrical power angle and

mechanical power angle m and electrical speed and

mechanical speed

emm PP

dtdM 2

2

numberpoleis where2

,2

ppp

mm



20/59


swing equation in terms of electrical power angle

converting the swing equation into per unit system

em PPdtdM

p

2

22

s

puepum

s

HMPPdtdH

2where,2 )()(2

2



21/59


swing equation in terms of inertial constant M

relations between electrical power angle and

mechanical power angle m

and electrical speed and

mechanical speed

emm PP

dt

dM

2

2

numberpoleis where2

,2

ppp

mm



22/59


swing equation in terms of electrical power angle

converting the swing equation into per unit system

em PPdtdM

p

2

22

s

puepum

s

HMPP

dt

dH

2where,

2)()(2

2



23/59


swing equation plot, electrical power angle as a

function of time


0 0.5 1 1.5 2 2.5 310

15

20

25

30

t, sec


24/59

STEADY-STATE STABILITY

SMALL DISTURBANCE

Steady-state stability

the ability of power system to remain its

synchronism and returns to its original statewhen subjected to small disturbances

such stability is not affected by any control

efforts such as voltage regulators or governor


25/59


SMALL DISTURBANCE

Analysis of steady-state stability by swingequation

starting from swing equation

introduce a small disturbance derivation is from Eq.11.37 (see pg. 472)

simplify the nonlinear function of power angle

sinmax)()(2

2

0

PPPPdt

d

f

Hmpuepum


26/59

Analysis of steady-state stability by swingequation

swing equation in terms of

PS=Pmax cos0: the slope of the power-angle curve at 0,

PS is positive when 0 < < 90o (See figure 11.3)

the second order differential equation

0cos 02

2

0

mP

dt

d

f

H0max cos0

Pd

dPPS

02

2

0

SP

dt

d

f

H


SMALL DISTURBANCE

STEADY STATE STABILITY


27/59

Characteristic equation:

rule 1: if PS is negative, one root is in RHP and

system is unstable

rule 2: if PS is positive, two roots in the j axis and

motion is oscillatory and undamped, system is

marginally stable

rule 2 : the oscillatory frequency of the undamped

system

SPHfs 02


SMALL DISTURBANCE

Sn PH

f0


28/59

Damping torque

phenomena: when there is a difference angular

velocity between rotor and air gap field, an inductiontorque will be set up on rotor tending to minimize the

difference of velocities

introduce a damping power by damping torque

introduce the damping power into swing equation

dt

dDPd

02

2

0

SP

dt

dD

dt

d

f

H


SMALL DISTURBANCE


29/59

STEADY STATE STABILITY EXAMPLE

Example 11.3

using the state space matrix to solve and

the original state 0=16.79o, new state afterP isimposed =22.5o

the linearized equation is valid only for very small

power impact and deviation from the operating state

a large sudden impact may result in unstable state

even if the impact is less than the steady state power

limit


30/59

STEADY STATE STABILITY EXAMPLE

Example 11.3

the characteristic equation of determinant (sI-A) oreigenvalue of A can tell the stability of system

system is asymptotically stable iff eigenvalues of A are

in LHP

in this case, eigenvalues of A are -1.3 6.0i

STABILITY ANALYSIS


31/59

STABILITY ANALYSIS

ON SWING EQUATION

Characteristic equation:

Analysis of characteristic equation

for damping coefficient

02 22

2

nn

dt

d

dt

d

0222

nnss

12

0 SHP

fD

STABILITY ANALYSIS


32/59

STABILITY ANALYSIS

ON SWING EQUATION

Analysis of characteristic equation

roots of characteristic equation

damped frequency of oscillation

positive damping (1>>0): s1,s2 have negative realpart if PS is positive, this implies the response is

bounded and system is stable

2

21 1-s,s nn j

21 nd

STABILITY ANALYSIS


33/59

Solution of the swing equation

roots of swing equation

rotor angular frequency

02 22

2

nndt

d

dt

d

tete d

t

d

t nn sin1

,sin1 2

002

0

tete dtn

d

tn nn

sin

1,sin

1 20

02

0

STABILITY ANALYSIS

ON SWING EQUATION

STABILITY ANALYSIS


34/59

Solution of the swing equation

response time constant

settling time:

relations between settling time and inertia constant H:

increase Hwill result in longertS, decrease n and

Df

H

n 0

21

4St

STABILITY ANALYSIS

ON SWING EQUATION

TRANSIENT STABILITY


35/59

TRANSIENT STABILITY

Transient stability

to determine whether or not synchronism is maintained

after machine has been subject to severe disturbance

Severe disturbance

sudden application of loads (steel mill)

loss of generation (unit trip)

loss of large load (line trip)

a fault on the system (lightning)

TRANSIENT STABILITY


36/59

TRANSIENT STABILITY

System response after large disturbance

oscillations of rotor angle result in large magnitude thatlinearlization is not feasible

must use nonlinear swing equation to solve the

problem

TRANSIENT STABILITY


37/59

TRANSIENT STABILITY

Time

Rotor

Angle

DisturbanceStable

Unstable

EQUAL AREA CRITERION


38/59


Equal area criterion

can be used to quickly predict systemstability after disturbance

only applicable to a one-machine system

connected to an infinite bus or a two-

machine system



39/59


Derivation of rotor relative speed from swingequation

starting from the swing equation with damping

neglected

for detailed derivation, please see pp.486

the swing equation end up with

poweronacceleratiPPPPdt

d

f

Haaem

o

,2

2

o

dPPH

f

dt

dem

o2



40/59

Synchronous machine relative speedequation

the equation gives relative speed of machine with

respect to the synchronous revolving reference

frame

if stability of system needs to be maintained, the

speed equation must be zero sometimes after the

disturbance

o dPPH

f

dt

dem

o2




41/59

Stability analysis

stability criterion

consider machine operating at the equilibrium point o,

corresponding to power input Pm0

= Pe0

a sudden step increase of Pm1 is applied results in

accelerating power to increase power angle to 1

0

o

dPP em




42/59

Stability analysis

the excess energy stored in rotor

when =1, the electrical power matches new input

power Pm1, rotor acceleration is zero but relative speedis still positive (rotor speed is above synchronous

speed), still increases

11

AareaabcareadPPo

em




43/59

Stability analysis

as long as increases, Pe increases, at this time the

new Pe >Pm1 and makes rotor to decelerate

rotor swing back to b and the angle max makes |area A1|=|area A2|

21 demax

1

AareabareadPP em




44/59

A2 A2max

A1

0 1 max

Pm1

Pm0

Equal Criteria:A1 = A2

A1 < A2max Stable

A1 = A2max Critically Stable

A1 > A2max Unstable

t0

Pm1

0

t

max

1a

bc

d

e

Equal area criterion (stable condition)


APPLICATION TO SUDDEN INCREASE


45/59

APPLICATION TO SUDDEN INCREASE

OF POWER INPUT

Stability analysis of equal area criterion stability is maintained only if area A2 at least equal to A1 if A2 < A1, accelerating momentum can never be

overcome

Limit of stability when max is at intersection of line Pm and power-angle

curve is 90o < < 180o

the max can be derived as (see pp.489, figure 11.12)

max can be calculated by iterative method

Pmax is obtained by Pm=Pmaxsin1, where 1 = -max

0maxmaxmax coscossin o

SOLUTION TO STABILITY ON


46/59


SUDDEN INCREASE OF POWER INPUT

Calculation of max

max can be calculated by iterative Newton Raphson

method

assume the above equation is f(max) = c

0maxmaxmax coscossin o



47/59



Calculation of max

starting with initial estimate of /2 < max(k) < , gives

where

the updated max(k+1)

max(k+1) = max

(k) + max(k)

)(max

max

)(max)(

max

kd

df

fc kk

)(max0)(maxmax

cos)(max

kk

kd

df



48/59



0 20 40 60 80 100 120 140 160 1800

0.5

1

1.5

2

Equal-area criterion applied to the sudden change in power

Power angle, degree

Power,perunit



49/59



0 20 40 60 80 100 120 140 160 1800

0.5

1

1.5

2

Equal-area criterion applied to the sudden change in power

Power angle, degree

Power,perunit

APPLICATION TO


50/59

APPLICATION TO

THREE PHASE FAULT

Three phase bolt fault case a temporary three phase bolt fault occurs at sending end of line at

bus 1

fault occurs at 0, Pe = 0

power angle curve corresponds

to horizontal axis

machine accelerate,

increase until fault cleared at c fault cleared at c shifts operation

to original power angle curve at e

net power is decelerating, stored

energy reduced to zero at f

A1(abcd) = A2(defg)

FPe

A1

0 c max

Pma

b c

f

e

d g

A

2

1

APPLICATION TO THREE PHASE FAULT


51/59


- NEAR SENDING END

Three phase bolt fault case

when rotor angle reach f, Pe>Pm

rotor decelerates and retraces

along power angle curve passingthrough e and a

rotor angle would swing back and

forth around 0 at n

with inherent damping, operating

point returns to0

Pe

A1

0 c max

Pma

b c

f

e

d gA2



52/59

Critical clearing time critical clearing angle is reached when further increase

in c causeA2 < A1

we obtainc

from swing equation

integrating both sides from t = 0 to tc

we obtain the critical clearing time

mP

dt

d

f

H2

2

0

m

cc

Pf

Ht

0

02


- NEAR SENDING END

dPPdP c

c

mm

max

0 sinmax

max0maxmax

coscos P

Pmc



53/59


- NEAR SENDING END

0 20 40 60 80 100 120 140 160 1800

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Application of equal area criterion to a critically cleared system

Power angle, degree

Power,p

erunit

Pm

Critical clearing angle = 84.7745

CB trip CB reclose



54/59


- AWAY FROM SENDING END


a temporary three phase fault occurs away from sending end of bus 1

fault occurs at 0, Pe is reduced

power angle curve corresponds

to curve B

machine accelerate, increase

from 0 (b) until fault cleared at c (c) fault cleared at c shifts operation

to curve C at e

net power is decelerating, stored energy reduced to zero at f

A1(abcd) = A2(defg)

F

1Pe

A1

0 c max

Pma

b

c

f

e

d gA

2

A

C

B



55/59


when rotor angle reach f,

Pe>P

mrotor decelerates and

rotor angle would swing back

and forth around e at n

with inherent damping, operating

point returns to the point that Pmline intercept with curve C

Pe

A1

0 c max

Pm a

bc

f

e

d g

A

2

A

C

B





56/59

Critical clearing angle

critical clearing angle is reached when further increasein c causeA2 < A1

we obtainc

max

0maxmax3max20 sinsin

c

c

cmcm PdPdPP

max2max3

0max2maxmax30maxcoscos

cosPP

PPPmc





57/59

The difference between curve b and curve c is due

to the different line reactance

curve b: the second line is shorted in the middle point

(Fig. 11.23)

curve c: after fault is cleared, the second line is isolated

See example 11.5

use power curve equation to solve max and then c





58/59

0 20 40 60 80 100 120 140 160 1800

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Application of equal area criterion to a critically cleared system

Power angle, degree

Power,per

unit

Pm

Critical clearing angle = 98.8335

P = 1.80 sin

P = 1.46 sin

r2 = 0.8

P = 0.65 sin

r1=0.36

1= 26.390 m= 146.58

0c= 98.80

0




59/59

Most common transient stability

analysis is numerical simulation

Apply a fault and integrate the differential

equations of all the generators while the fault is on.

Remove the fault after some time (representing

the relay setting time plus breaker clearing time).

Continue integrating and observe the response

and monitor bus voltage levels. Stop when stability

condition has been determined.

module 6 stability

Documents