electricmotor3

EXPERT SYSTEMS AND SOLUTIONS

Email: [email protected]@yahoo.com

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 is stable

Stability issue steady-state stability – the ability of power system to

regain synchronism after small and slow disturbances such as gradual power change

transient stability – the ability of power system to regain synchronism after large and sudden disturbances such as a fault


POWER ANGLE

Power angle relative angle δr

between rotor mmf and air-gap mmf (angle between Fr and Fsr), both rotating iin synchronous speed

also the angle δr between no-load generated emf E and stator voltage Esr

also the angle δ between emf E and terminal voltage V, if neglecting armature resistance and leakage flux


DEVELOPING SWING EQUATION

Synchronous machine operation consider a synchronous generator with

electromagnetic torque Te running at synchronous speed ωsm.

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<0 if decelerating)

introduce the combined moment of inertia of prime mover and generator J

by the law of rotation --

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

emam TTT

dt

dJ −==

2

2θ


DEVELOPING SWING EQUATION

Derivation of swing equation θm = ωsmt+δm, ωsm is the constant angular velocity

take the derivative of θm, we obtain –

take the second derivative of θm, we obtain –

substitute into the law of rotation

multiplying ωm to obtain power equation

dt

d

dt

d msm

m δωθ +=

2

2

2

2

dt

d

dt

d mm δθ =

emam TTT

dt

dJ −==

2

2δ

ememmmmm

m PPTTdt

dM

dt

dJ −=−== ωωδδω

2

2

2

2


DEVELOPING SWING EQUATION Derivation of swing equation

swing equation in terms of inertial constant M

relations between electrical power angle δ and mechanical power angle δm and electrical speed and mechanical speed

swing equation in terms of electrical power angle δ

converting the swing equation into per unit system

emm PP

dt

dM −=

2

2δ

number pole is where2

,2

ppp

mm ωωδδ ==

em PPdt

dM

p−=

2

22 δ

spuepum

s

HMPP

dt

dH

ωδ

ω2

where,2

)()(2

2

=−=


SYNCHRONOUS MACHINE MODELS FOR STABILITY 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 referred as infinite bus

generator is represented by a constant voltage E’ behind direct axis transient reactance X’d

Zs

ZLjX’dE’

Vg V



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

use π equivalent line model for currents

real power at node 1

y10

y12

E’ V

y20

1 2I1 I2

+−

−+=

V

E

y

y

I

I '

yy

y y

122012

121210

2

1

( )12121111

2cos'cos' θδθ −+= YVEYEPe



Real power flow equation let –y12 = 1 / X12

simplified real power equation: Power angle curve

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

max power is referred as steady-state stability limit at δ=90o

δsin'

12X

VEPe =

δ0

PePm

π/2 π

Pmax

0δ

Pe

12max

'

X

VEP =



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 MODELS INCLUDING SALIENCY

Phasor diagram of salient-pole machine condition: under steady state with armature

resistance neglected

power angle equation in per unit

voltage equation in per unit

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

E

V XdId

jXqIq

Id

Iq

Ia

θ δa

δδ 2sin2

sin2

qd

qd

d XX

XXV

X

VEP

−+=

( )θδδδ ++=+= sincoscos addd IXVIXVE


SYNCHRONOUS MACHINE MODELS INCLUDING SALIENCY

Calculation of voltage E starting with a given (known) terminal voltage V and

armature current Ia, we need to calculate δ first by using 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 early part of transient period

calculate δ first, then calculate |E’q|: see example 11.1

( )θδδ ++= sincos ad IXVE

+

= −

θθ

δsin

costan 1

aq

aq

IXV

IX

δδ 2sin2

sin'

'2

'

'

qd

qd

d

q

e XX

XXV

X

VEP

−+=

( )θδδ ++= sincos ''adq IXVE


STEADY-STATE STABILITY – SMALL DISTURBANCE

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

and returns to its original state when subjected to small 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 Δδ derivation is from Eq.11.37 (see pg. 472) simplify the nonlinear function of power angle δ

δδπ

sinmax)()(2

2

0

PPPPdt

d

f

Hmpuepum −=−=



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

PS=Pmax cosδ0: the slope of the power-angle curve at δ0, PS is positive when 0 < δ < 90o (See 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 jω axis and

motion is oscillatory and undamped, system is marginally stable

0cos 02

2

0

=∆+∆ δδδπ mPdt

d

f

H0max cos

0δ

δ δ Pd

dPPS ==

02

2

0

=∆+∆ δδπ SPdt

d

f

H

SPH

fs 02 π−=



Characteristic equation: rule 2 (continued): the oscillatory frequency of the

undamped system

Damping torque phenomena: when there is a difference angular velocity

between rotor and air gap field, an induction torque 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

Sn PH

f0πω =

dt

dDPd

δ=

02

2

0

=∆+∆+∆ δδδπ SPdt

dD

dt

d

f

H


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>ζ>0): s1,s2 have negative real part if PS is positive, this implies the response is bounded and system is stable

02 22

2

=∆+∆+∆ δωδςωδnn dt

d

dt

d

02 22 =++ nnss ωςω1

20 <=SHP

fD πζ

221 1-s,s ζωζω −±= nn j

21 ζωω −= 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 tS, decrease ωn and ζ

02 22

2

=∆+∆+∆ δωδςωδnn dt

d

dt

d

( ) ( )θωζ

δδδθωζ

δδ ζωζω +−

∆+=+−

∆=∆ −− tete dt

dt nn sin

1 ,sin

1 2

002

0

( ) ( )tete dtn

dtn nn ω

ζδωωωω

ζδωω ζωζω sin

1 ,sin

1 2

002

0 −−

−∆+=

−∆−=∆

Df

H

n 0

21

πζωτ ==

τ4≅St


SOLVING THE SWING EQUATION USING STATE SPACE MATRIX

State space approach state space approach can solve multi-machine system let x1=Δδ, x2=Δω=Δδ′

taking the Laplace transform, from Eq.11.52

solution of the X(s)

)()( 2

1 0

2

1

22

1 tAxtxx

x

x

x

nn

==>

−−

=

ζωω

)()( 1 0

0 1

2

1

2

1 tCxtyx

x

y

y==>

=

( ) ( )

+

=−−= −

n2n

1

2s

1- ),0()(

ζωωs

AsIxAsIsX

22 2

)0(

1 2

)(nn

n

n

s

xs

s

sXωζω

ωζω

++

−

+

=


SOLVING THE SWING EQUATION USING STATE SPACE MATRIX

State space approach

taking the inverse Laplace transform with initial state x1(0)=Δδ0, x2(0)=Δω0=0

state solution: x1(t)=Δδ(t), x2(t)=ω(t)

( )

( )222

221

2)()(

2)()(

nn

nn

s

ussx

ss

ussx

ωζωω

ωζωδ

++∆=∆=

++∆=∆=

( ) ( )θωζ

δδδθωζ

δδ ζωζω +−

∆+=+−

∆=∆ −− tete dt

dt nn sin

1 ,sin

1 2

002

0

( ) ( )tete dtn

dtn nn ω

ζδωωωω

ζδωω ζωζω sin

1 ,sin

1 2

002

0 −−

−∆+=

−∆−=∆


STEADY STATE STABILITY EXAMPLE Example 11.3

using the state space matrix to solve δ and ω the original state δ0=16.79o, new state after ΔP is

imposed δ=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

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 are in LHP

in this case, eigenvalues of A are -1.3 ± 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 the

problem


EQUAL AREA CRITERION Equal area criterion

can be used to quickly predict system stability after disturbance

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

Derivation of rotor relative speed from swing equation

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

=>=−=δπ

( )∫ −=δ

δδπδ

o

dPPH

f

dt

dem

o2


EQUAL AREA CRITERION Synchronous machine relative speed equation

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

Stability analysis stability criterion

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

a sudden step increase of Pm1 is applied results in accelerating power to increase power angle δ to δ1

( )∫ −=δ

δδπδ

o

dPPH

f

dt

dem

o2

( ) 0=−∫δ

δδ

o

dPP em


EQUAL AREA CRITERION Stability analysis

the excess energy stored in rotor

when δ=δ1, the electrical power matches new input power Pm1, rotor acceleration is zero but relative speed is still positive (rotor speed is above synchronous speed), δ still increases

as long as δ increases, Pe increases, at this time the new Pe >Pm1 and makes rotor to decelerate

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

( ) 1 1

AareaabcareadPPo

em ==−∫δ

δδ

( ) 21 de max

1

AareabareadPP em ==−∫δ

δδ


EQUAL AREA CRITERION Equal area criterion (stable condition)

A2 A2max

A1

δ0δ1 δmax

Pm1

Pm0

Equal Criteria: A1 = A2

A1 < A2max StableA1 = A2max Critically StableA1 > A2max Unstable

t0

Pm1

δ0

t

δmax

δ1a

bc

d

e


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 δmax is at intersection of line Pm and power-angle

curve is 90o < δ < 180o the δmax can be derived as (see pp.489, figure 11.12)

δmax can be calculated by iterative method Pmax is obtained by Pm=Pmaxsinδ1, where δ1 = π-δmax

( ) 0maxmaxmax coscossin δδδδδ =+− o


SOLUTION TO STABILITY ON SUDDEN INCREASE OF POWER INPUT

Calculation of δmax

δmax can be calculated by iterative Newton Raphson

method assume the above equation is f(δmax) = c

starting with initial estimate of π/2 < δmax(k) < π, Δδ gives

where the updated δmax

(k+1)

δmax(k+1) = δmax

(k) + Δ δmax(k)

( ) 0maxmaxmax coscossin δδδδδ =+− o

( )

)(max

max

)(max)(

max

kddf

fc kk

δδ

δδ −=∆

( ) )(max0

)(max

max

cos)(

max

kk

kd

df δδδδ δ

−=


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 δ0, Pe = 0 power angle curve corresponds to horizontal axis machine accelerate, increase δ until fault cleared at δc

fault cleared at δc shifts operation

to original power angle curve at e net power is decelerating, stored energy reduced to zero at f A1(abcd) = A2(defg)

F Pe

A1

δ0δc δmax

Pma

b c

f

e

d gA2

δ

1



- 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 and forth around δ0 at ωn

with inherent damping, operating point returns to δ0

Critical clearing angle critical clearing angle is reached when further increase in δc cause

A2 < A1

we obtain δc

Pe

A1

δ0δc δmax

Pma

b c

f

e

d gA2

δ

( ) δδδδ

δ

δ

δdPPdP

c

c

mm ∫∫ −= max

0

sinmax

( ) max0maxmax

coscos δδδδ +−=P

Pmc



- NEAR SENDING END Critical clearing time

from swing equation

integrating both sides from t = 0 to tc

we obtain the critical clearing time

mPdt

d

f

H =2

2

0

δπ

( )m

cc Pf

Ht

0

02

πδδ −=



- 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 δ0, Pe is reduced power angle curve corresponds to curve B machine accelerate, increase δ from δ0 (b) until fault cleared at δc

(c) fault cleared at δc 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

δ0δc δmax

Pma

bc

f

e

d gA2

δ

A

C

B



- AWAY FROM SENDING END 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 ωn

with inherent damping, operating point returns to the point that Pm

line intercept with curve C Critical clearing angle

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

we obtain δc

( ) ( )∫∫ −−=−− max

0maxmax3max20 sin sin

δ

δ

δ

δδδδδδδδδ

c

c

cmcm PdPdPP

( )max2max3

0max2maxmax30max coscoscos

PP

PPPmc −

−+−= δδδδδ

Pe

A1

δ0δc δmax

Pma

bc

f

e

d gA2

δ

A

C

B



- AWAY FROM SENDING END The difference between curve b and curve c

is due to the different line reactancecurve 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 δmax and then

δc


NUMERICAL SOLUTION OF NONLINEAR EQUATION

Euler method tangent evaluation: updated solution:drawback: accuracy

Modified Euler method tangent evaluation: updated solution:

feature: better accuracy, but time step Δt should be properly selected

0xdt

dx

tdt

dxxxxx x ∆+=∆+=

0001

2

10pxx dt

dxdtdx +

2/10001 t

dt

dx

dt

dxxxxx xx ∆

++=∆+=


NUMERICAL SOLUTION OF NONLINEAR EQUATION

Higher order equationuse state space method to decompose higher order equation

use modified Euler method to solve state space matrix

for swing equation of second order, use 2×2 state space matrix to solve

see pp. 504

)()( 2

1 0

2

1

22

1 tAxtxx

x

x

x

nn

==>

−−

=

ζωω


NUMERICAL SOLUTION OF SWING EQUATION

Swing equation in state variable form

use modified Euler method

the updated values (see Ex. 11.6)

aPH

f

dt

ddt

d

0πω

ωδ

=∆

∆=

tdt

dP

H

f

dt

d

tdt

d

dt

d

ipi

pi

ipi

ipia

ipi

pi

∆+==∆

∆∆+∆=∆∆=

∆+

++∆

++

+

ωδδ

δω

δδδπω

ωωωωδ

10

11

where

where,

11

1

tdtd

dtd

tdtd

dtd

pii

pii

icii

ci ∆

∆+∆

+∆=∆∆

++= +

∆∆

+ 2 ,

2 11

δδωωωω

ωω

δδ

δδ


MULTIMACHINE SYSTEMS Multi-machine system can be written similar to

one-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 δmech = δ machines belong to the same station swing together

and are said to be coherent, coherent machines can equivalent to one machine


METHOD TO SOLVE MULTIMACHINE SYSTEMS

Solution to multi-machine system solve initial power flow and determine initial bus voltage

magnitude and phase angle

calculating load equivalent admittance

nodal equations of the system

electrical and mechanical power output of machine at steady state prior to disturbances

idiii

ii

i

ii IjXVE

V

jQP

V

SI ''

**

*

, +=−==

20

i

iii

V

jQPy

−=

=

'

0

m

n

mmtnm

nmnn

m E

V

YY

YY

I

{ } ( )∑=

+−===m

jjiijijjiiimiei YEEIEPP

1

''* cosRe δδθ


MULTIMACHINE SYSTEMS TRANSIENT STABILITY

Classical transient stability study is based on the application 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

jjiijijjimi

ii PPYEEPdt

d

f

H −=+−−= ∑=1

''2

2

0

cos δδθδπ

( )eimii

i

ii

PPH

f

dt

d

nidt

d

−=∆

=∆=

0

,,1 ,

πω

ωδ

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