power systems stablity

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EXPERT SYSTEMS AND SOLUTIONS

Email: [email protected]

[email protected]

Cell: 9952749533www.researchprojects.info

PAIYANOOR, OMR, CHENNAI

Call For Research Projects Final

year students of B.E in EEE, ECE, EI,

M.E (Power Systems), M.E (Applied

Electronics), M.E (Power Electronics)

Ph.D Electrical and Electronics.

Students can assemble their hardware in our

Research labs. Experts will be guiding the projects.



Energy Conversion Lab

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 synchronous

machine after a disturbance in short, if synchronous machines can remain

synchronism after disturbances, we say that system isstable

Stability issue

steady-state stability ± the ability of power system toregain synchronism after small and slow disturbancessuch as gradual power change

transient stability ± the ability of power system toregain synchronism after large and suddendisturbances such as a fault




POWER ANGLE

Power angle relative angle Hr

between rotor mmf andair-gap mmf (anglebetween Fr and Fsr),

both rotating iinsynchronous speed also the angle Hr

between no-loadgenerated emf E andstator voltage Esr

also the angle Hbetween emf E andterminal voltage V, if neglecting armatureresistance and leakage

flux




DEVELOPING SWING EQUATION

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 inaccelerating/decelerating torque Ta=Tm-Te (Ta>0 if accelerating, Ta<0 if decelerating)

introduce the combined moment of inertia of prime

mover and generator J by the law of rotation --

Um is the angular displacement of rotor w.r.t.stationery reference frame on the stator

ema

mT T T

dt

d J !!

2

2U





Derivation of swing equation Um = smt+Hm, sm is the constant angular velocity

take the derivative of Um, we obtain ±

take the second derivative of Um, we obtain ±

substitute into the law of rotation

multiplying m to obtain power equation

d t

d

d t

d m sm

m H [ !

2

2

2

2

d t

d

d t

d mm H !

emam T T T

d t

d J !!

2

2H

ememmmmm

m P P T T d t

d M

d t

d J !!! [ [

H H [

2

2

2

2





Derivation of swing equation swing equation in terms of inertial constant M

relations between electrical power angle H and

mechanical power angle Hm and electrical speed andmechanical speed

swing equation in terms of electrical power angle H

converting the swing equation into per unit system

emm P P

d t

d M !

2

2H

number poleishere2

,2

p p p

mm [ [ H H !!

em P P d t d M

p!

2

2

2 H

s

pue pum

s

H M P P

d t

d H

[

H

[

2 here,

2)()(2

2

!!




SYNCHRONOUS MACHINE MODELS FORSTABILITY STUDY

Simplified synchronous machine model the simplified machine model is decided by the proper

reactances, X¶¶d, X¶d, or Xd

for very short time of transient analysis, use X¶¶d for short time of transient analysis, use X¶d for steady-state analysis, use Xd

substation bus voltage and frequency remain constant is referredas infinite bus

generator is represented by a constant voltage E¶ behind direct

axis transient reactance X¶d

Zs

ZL jX ¶d

E¶

Vg V





Real power flow equation let ±y12 = 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 atH=90o

Hsin'

12 X

V E P

e!

H0

PePm

/2

Pmax

0H

Pe

12

max

'

X

V E P !





Transient stability analysis condition: generator is suddenly short-circuited

current during the transient is limited by X¶d

voltage behind reactance E¶=Vg+jX¶dIa

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 X¶d




SYNCHRONOUS MACHINE MODELSINCLUDING SALIENCY

Calculation of voltage E starting with a given (known) terminal voltage V andarmature current Ia, we need to calculate H first byusing phasor diagram and then result in voltage E

once E is obtained, P could be calculated

Transient power equation for salient machine

this equation represents the behavior of SM in earlypart of transient period

calculate H first, then calculate |E¶q|:

see example 11.1

UH H ! sincos ad I X V E ¹¹

º

¸

©©

ª

¨

!

U

UH

sin

costan 1

aq

aq

I X V

I X

H H 2sin2sin '

'2

'

'

qd

qd

d

q

e X X

X X V

X

V E P

!

UH H ! sincos ''

ad q I X V E




STEADY-STATE STABILITY ± SMALLDISTURBANCE

Steady-state stability the ability of power system to remain its synchronism

and returns to its original state when subjected tosmall disturbances

such stability is not affected by any control efforts

such as voltage regulators or governor Analysis of steady-state stability by swing

equation starting from swing equation

introduce a small disturbance H

derivation is from Eq.11.37 (see pg. 472)

simplify the nonlinear function of power angle H

HHT

sinmax)()(2

2

0

P P P P d t d

f H m pue pum !!




STEADY-STATE STABILITY ± SMALLDISTURBANCE

Analysis of steady-state stability by swing equation swing equation in terms of H

PS=Pmax cosH0: the slope of the power-angle curve at H0,

PS is positive when 0 < H < 90o

(S ee figure 11.3) the second order differential equation

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 andmotion is oscillatory and undamped, system ismarginally stable

0cos 02

2

0

!((

H H H

T m P

dt

d

f 0max cos

0H

H H P

d

dP P S !!

02

2

0

!((

H H

T S P

d t

d

f

H

S P

H

f s 02 T

!




STABILITY ANALYSIS ON SWING EQUATION

Characteristic equation:

Analysis of characteristic equation

for damping coefficient

roots of characteristic equation

damped frequency of oscillation

positive damping (1> ^>0): s1,s2 have negative real part

if PS is positive, this implies the response is bounded

and system is stable

02 2

2

2

!((

(

H [ H

:[ H

nnd t

d

d t

d

02 22 ! nn s s [: [

12

0 !S HP

f D T^

2

211-s,s ^ [ ^ [ s!

nn j

21 ^ [ [ ! nd




STABILITY ANALYSIS ON SWING EQUATION

Solution of the swing equation

roots of swing equation

rotor angular frequency

response time constant

settling time:

relations between settling time and inertia constant H :

increase H will result in longer t S , decrease n and ^

02 2

2

2

!((

(

H [ H

:[ H

nnd t

d

d t

d

U[

^

H H H U[

^

H H

^[^[

(!

(!(

t et e

d

t

d

t nn sin1

,sin1 2

0

02

0

t et e d

t nd

t n nn [ ^

H [ [ [ [

^

H [ [

^ [ ^ [ sin

1 ,sin

1 2

002

0

(!

(!(

D f

H

n 0

21

T ^ [ X !!

X 4$S t




SOLVING THE SWING EQUATION USING STATESPACE MATRIX

State space approach state space approach can solve multi-machine system

let x1=H, x2==Hd

taking the Laplace transform, from Eq.11.52

solution of the X(s)

)()( 2

1 0

2

1

22

1t Axt x

x

x

x

x

nn

!!"¼½

»

¬-

«

¼½

»

¬-

«

!

¼½

»

¬-

«

^ [ [

)()( 1 0

0 1

2

1

2

1t Cxt y

x

x

y

y!!"¼

½

»¬-

«¼½

»¬-

«!¼

½

»¬-

«

¼½

»¬-

«

!!

n

2

n

1

2s

1- ),0()(

^ [ [

s A s I x A s I s X

22

2

)0(

1 2

)(

nn

n

n

s

x s

s

s X

[ ^ [

[

^ [

¼½

»¬-

«

!




SOLVING THE SWING EQUATION USING STATESPACE MATRIX

State space approach

taking the inverse Laplace transform with initial statex1(0)=H0, x2(0)=0=0

state solution: x1(t)=H(t), x2(t)=(t)

222

221

2

)()(

2)()(

nn

nn

s

u s s x

s s

u s s x

[ ^ [

[

[ ^ [ H

(!(!

(!(!

U[ ^

H H H U[ ^

H H ^ [ ^ [

(!

(!( t et e d t

d t nn sin

1 ,sin

1 2

00

2

0

t et e d

t nd

t n nn [ ^

H [ [ [ [

^

H [ [

^ [ ^ [ sin

1 ,sin

1 2

002

0

(!

(!(




STEADY STATE STABILITY EXAMPLE

Example 11.3 using the state space matrix to solve H and

the original state H0=16.79o, new state after P isimposed H=22.5o

the linearized equation is valid only for very smallpower impact and deviation from the operating state

a large sudden impact may result in unstable stateeven if the impact is less than the steady state power limit

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

system is asymptotically stable iff eigenvalues of A arein LHP

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




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)

System response after large disturbance oscillations of rotor angle result in large magnitude that

linearlization is not feasible

must use nonlinear swing equation to solve theproblem




EQUAL AREA CRITERION

Equal area criterion can be used to quickly predict system stability after

disturbance

only applicable to a one-machine system connected toan infinite bus or a two-machine system

Derivation of rotor relative speed from swingequation

starting from the swing equation with dampingneglected

for detailed derivation, please see pp.486

the swing equation end up with

power onacceler ati P P P P d t

d

f

H

aae

mo

,2

2

!"!!H

T

´ !H

H H

T H

o

d P P H

f

dt

d em

o2





Synchronous machine relative speed equation

the equation gives relative speed of machine withrespect to the synchronous revolving reference frame

if stability of system needs to be maintained, the speedequation must be zero sometimes after the disturbance

Stability analysis stability criterion

consider machine operating at the equilibrium point Ho,corresponding to power input Pm0 = Pe0

a sudden step increase of Pm1 is applied results inaccelerating power to increase power angle H to H

1

´ !H

HH

TH

o

d P P H

f

dt

d em

o2

0!´H

H H o

d P P em





Stability analysis the excess energy stored in rotor

when H=H1, the electrical power matches new inputpower Pm1, rotor acceleration is zero but relative speedis still positive (rotor speed is above synchronousspeed), H still increases

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

new Pe >Pm1 and makes rotor to decelerate

rotor swing back to b and the angle Hmax makes

|area A1|=|area A2|

1 1

Aareaabcaread P P o

em!!´

H

H H

21 de max

1

Aareabaread P P em !!´

H

H H





Equal area criterion (stable condition) A2 A2max

A1

H0H1 Hmax

Pm1

Pm0

Equal Criteria: A1 = A2

A1 < A2max Stable A1 = A2max Critically Stable A1 > A2max Unstable

t 0

Pm1

H0

t

Hmax

H1a

bc

d

e




APPLICATION TO SUDDEN INCREASE OFPOWER 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 beovercome

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

curve is 90o < H < 180o

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

Hmax can be calculated by iterative method

Pmax is obtained by Pm=PmaxsinH1, where H1 = T-Hmax

0maxmaxmax coscossin H H H H H !o




SOLUTION TO STABILITY ON SUDDENINCREASE OF POWER INPUT

Calculation of Hmax

Hmax can be calculated by iterative Newton Raphson

method

assume the above equation is f(Hmax) = c starting with initial estimate of T/2 < Hmax

(k) < T, H gives

where

the updated Hmax(k+1)

Hmax(k+1) = Hmax

(k) + Hmax(k)

0maxmaxmax coscossin H H H H H !o

)(max

max

)(

max)(

max

k d

d f

f c k k

H H

H H

!(

)(

max0

)(

max

max

cos)(

max

k k

k d

d f H H H

H H

!




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 H0, Pe = 0 power angle curve corresponds

to horizontal axis

machine accelerate,

increase H until fault cleared at Hc

fault cleared at Hc shifts operationto original power angle curve at e

net power is decelerating, stored

energy reduced to zero at f

A1(abcd) = A2(defg)

FPe

A1

H0Hc Hmax

Pma

b c

f

e

d g

A2

H

1




APPLICATION TO THREE PHASE FAULT- NEAR SENDING END

Three phase bolt fault case when rotor angle reach f, Pe>Pm

rotor decelerates and retraces

along power angle curve passing

through e and a

rotor angle would swing back andforth around H0 at n

with inherent damping, operating

point returns to H0

Critical clearing angle critical clearing angle is reached when further increase in Hc cause A2 < A1

we obtain Hc

Pe

A1

H0Hc Hmax

Pm a

b c

f

e

d g

A2

H

HHHH

H

H

Hd P P d P

c

c

mm ´´ !max

0

sinmax

max0max

max

coscos H H H H ! P

P mc




APPLICATION TO THREE PHASE FAULT- NEAR SENDING END

Critical clearing time from swing equation

integrating both sides from t = 0 to tc we obtain the critical clearing time

m P d t

d

f

H !

2

2

0

H

T

m

cc

P f

H t

0

02

T

H H !




APPLICATION TO THREE PHASE FAULT- AWAY FROM SENDING END

Three phase bolt fault case a temporary three phase fault occurs away from sending end of bus

1

fault occurs at H0, Pe is reduced power angle curve corresponds

to curve B

machine accelerate, increase H

from H0 (b) until fault cleared at Hc

(c) fault cleared at Hc shifts operation

to curve C at e

net power is decelerating, stored

energy reduced to zero at f

A1(abcd) = A2(defg)

F

1

Pe

A1

H0Hc Hmax

Pm

a

bc

f

e

d g A2

H

A

C

B





Three phase bolt fault case when rotor angle reach f, Pe>Pm

rotor decelerates and rotor angle

would swing back and forth

around e at n

with inherent damping, operatingpoint returns to the point that Pm

line intercept with curve C

Critical clearing angle

critical clearing angle is reached when further increasein Hc cause A2 < A1

we obtain Hc

´´ !max

0maxmax3max20 sinsin

H

H

H

H H H H H H H H H

c

c

cmcmP d P d P P

max2max3

0max2maxmax30max coscoscos

P P

P P P mc

!

H H H H H

Pe

A1

H 0 H c H max

Pm

a

bc

f

e

d g A2

H

A

C

B





The difference between curve b and curve cis 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 Hmax and then

Hc




NUMERICAL SOLUTION OF NONLINEAREQUATION

Euler method tangent evaluation:

updated solution:

drawback: accuracy Modified Euler method

tangent evaluation:

updated solution:

feature: better accuracy, but time step t should

be properly selected

0 xd t

d x

t d t

d x x x x x

x(!(!

0001

2

10p

x xd t

d x

d t

d x

2/10001 t

dt

dx

dt

dx x x x x

x x(¹

º

¸©ª

¨!(!




NUMERICAL SOLUTION OF NONLINEAREQUATION

Higher order equationuse state space method to decompose higher

order equation

use modified Euler method to solve state spacematrix

for swing equation of second order, use 2v2

state space matrix to solve

see pp. 504

)()( 2

1 0

2

1

22

1t A xt x

x

x

x

x

nn

!!"¼½»¬

-«¼

½»¬

-«

!¼

½»¬

-«

^[[




NUMERICAL SOLUTION OF SWINGEQUATION

Swing equation in state variable form

use modified Euler method

the updated values (see Ex. 11.6)

a P H

f

d t

d

d t

d

0T [

[ H

!(

(!

t dt

d P

H

f

dt

d

t dt

d

dt

d

i p

i

p

i

i p

i

i

p

ia

i

p

i

p

i

(!!(

((

(!((!

(

(

[ H H

H [

H H H

T [

[ [ [ [

H

10

11

where

where,

11

1

t d t

d

d t

d

t d t

d

d t

d p

ii

p

ii

i

c

ii

c

i(

¹¹¹¹

º

¸

©©©©

ª

¨ (

(

(!((

¹¹¹¹

º

¸

©©©©

ª

¨

!

((

2

,2

11

H H [ [

[ [

[ [

H H

H H




MULTIMACHINE SYSTEMS

Multi-machine system can be written similar toone-machine system by the following assumption

each synchronous machine is represented by a

constant voltage E behind Xd (neglect saliency and flux

change) input power remain constant

using prefault bus voltages, all loads are in equivalent

admittances to ground

damping and asynchronous effects are ignored Hmech = H

machines belong to the same station swing together

and are said to be coherent, coherent machines can

equivalent to one machine




MULTIMACHINE SYSTEMS TRANSIENTSTABILITY

Classical transient stability study is based on theapplication of the three-phase fault

Swing equation of multi-machine system

Yij are the elements of the faulted reduced bus

admittance matrix

state variable model of swing equation

see example 11.7

eimi

m

j jii ji j jimi

ii P P Y E E P

dt

d

f

H

!! §!1

''

2

2

0 cos H H U

H

T

eimi

i

i

i

i

P P H

f

dt

d

nidt

d

!(

!(!

0

,,1 ,

T[

[H -

power systems stablity

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