ece 522 power systems analysis ii 3.2 –small‐signal...

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Spring 2018Instructor: Kai Sun

ECE 522 Power Systems Analysis II

3.2 – Small‐Signal Stability

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Content

•3.2.1 Small‐signal stability overview and analysis methods•3.2.2 Small‐signal stability enhancement

References:– EPRI Dynamic Tutorial– Chapters 12 and 17 of Kundur’s “Power System Stability and Control”– Chapter 3 of Anderson’s “Power System Control and Stability” – Joe H. Chow, “Power System Coherency and Model Reduction,”  Springer, 2013

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3.2.1 Small‐Signal Stability Overview and Analysis 

Methods

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Power Oscillations

• The power system naturally enters periods of oscillation as it continually adjusts to new operating conditions or experiences other disturbances. 

• Typically, oscillations have a small amplitude and do not last long. •When the oscillation amplitude becomes large or the oscillations are sustained, a response is required: – A system operator may have the opportunity to respond and eliminate harmful oscillations or, 

– less desirably, protective relays may activate to trip system elements.

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Small Signal Stability

• In this context, a disturbance is considered to be small if the equations that describe the resulting response of the system may be linearized for the purpose of analysis.

• It is convenient to assume that the disturbances causing the changes already disappear and details on the disturbance are unimportant

• The system is stable only if it returns to its original state, i.e. a stable equilibrium point (SEP). Thus, only the behaviors in a small neighborhood of the SEP are concerned and can be analyzed using the linear control theory.

Small signal stability (also referred to as small‐disturbance stability) is the ability of a power system to maintain synchronism when subjected to small disturbances

0

P0

P()

k∆+P0

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Classic‐Model SMIB System

Pe+jQe

Pt+jQt

PB+jQB

(Pe=Te)

With all resistances neglected:

eSeT KT

d dd

¶D » D = D

¶

Linearize swing equations at =0 :

Synchronizing torque coefficient:

max 0 0cos cosBS

T

E EK PX

d d¢

= =

0

2

r

rm e D r

m e D r

ddt

dH T T Kdt

T T K

dw w

ww

w

D= D

D= - - D

=D -D - D

max

max

sine e t B

B

T

T P P P PE EP

X

d= = = =¢

=

(Tm)

(Te)

KS

Sm D rT K Kd w»D - D - D

Note:

H in s

r in p.u.

in rad.

KD in p.u

KS in p.u/rad

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• Characteristic equation:

0 0

2 2 2s mD K TK

H H H w wd d d

DD + D + D =

• Apply Laplace Transform:

2 0

0

2

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2sD

m

Ks

THKs

H H

wd

wD

D = ´+ +

State‐space representation

2 0 02 2

sD KKs sH H

w+ + =

00 0

1 22 2

mS D

r r

TdK Kdt HH H

wd dw w

é ùé ù é ù é ùD Dê ú Dê ú ê ú ê úê ú= +ê ú ê ú ê úê úD D- - ë ûë û ë ûê úë û

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• It has two conjugate complex roots and its zero-input response is a damped sinusoidal oscillation:

• :2

1 2, 1n ns s j js w zw w z= =- -

A harmonic oscillator

2 22 0n ns szw w+ + = – Damping ratio

n – Natural frequency

2

( ) sin( )

sin( 1 )n

t

tn

x t Ae t

Ae t

s

zw

w j

w z j-

= +

= - +

11/n

The time of decaying to 1/e=36.8%:

s1

s2

0 0

(if 0 )2m

Ds TKH K d d d

w w- D D- D =D =

2 0 02 2

sD KKs sH H

w+ + =2 0D Ks s

M M+ + =

F M x K x D x = ⋅ =- ⋅ - ⋅

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Oscillation Frequency and Damping of an SMIB System

• How do n and change with the following?– if H (lower inertia)– if XT (stronger transmission)– if (lower loading)

2 22 0n ns szw w+ + =

2 2 0 0 0cos2 2

Bn S

T

E EKH HXw w d

w s w¢

= + = =

2 20 00

12 8 cos2

D TD

BS

K XKHE EK H

sz

w dws w

-= = =

¢+ 4D

nKH

s zw=- =-

22 0

212 16

Dn S

KKH Hw

w w z= - = -

21 2, 1n ns s j js w zw w z= =- -

0 max 0( cos cos )BS

T

E EK PX

d d¢

= =

Note the units:

r is in p.u.

is in rad.

KD is in p.u

KS is in p.u/rad

2 0 02 2

sD KKs sH H

w+ + =

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System Response after a Small Disturbance

2 0/rx w d w=D =D1x d=D

1 102

02 2

0 0

2 1n n

x xu

x x

ww w zw

é ù é ùé ù é ùê ú ê úê ú ê ú= + Dê ú ê úê ú ê ú- - ë ûë ûë û ë û

( ) ( ) ( )t t u t x Ax B

1

2

1 0( ) ( )

0 1x

t tx

y Cxé ùé ùê úê ú= = ê úê úë û ë û

( ) (0) ( ) ( )s s s U s X x AX B

2mTu

HD

D =

( ) uU ss

DD =

[ ]

02

02 2

2

( ) (0) ( )2

n

n

n n

ss

s U ss s

X x B＋

zw ww w

zw w

é ù+ê úê ú-ë û= D+ +

02

02 2

20( ) (0)

( )( ) (0)2

n

n

r rn n

ss s

us s ss

zw wd dw ww wzw w

é ù+ê ú é ùê úé ù é ùD D- ê úë ûê ú ê ú ê ú= + Dê ú ê ú ê úD D+ +ë û ë û ê úë û

Zero-input Zero-state

1( ) ( ) ( ) (0) ( )s s s U s Y X I A x B＋

Zero-input Zero-state

00 0

1 22 2

mS D

r r

TdK Kdt HH H

wd dw w

é ùé ù é ù é ùD Dê ú Dê ú ê ú ê úê ú= +ê ú ê ú ê úê úD D- - ë ûë û ë ûê úë û

(Assuming the angle and speed to be directly measured)

Apply Laplace transform:

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Zero‐input response

2

(0) sin( )1

in rad nte tzwdw q

zd -D

D= +

-

2

(0 in rad/s ) sin1

nr

tn e tzww

zw

dw-D

=--

D

1cos

• E.g. when the rotor is suddenly perturbed by a small angle (0)0 and assume r (0)=0

2 2

( 2 ) (0)( )2

n

n n

sss s

zw dd

zw w+ D

D =+ +

02

02 2

20( ) (0)/

( )( ) (0)2

n

n

r rn n

ss s

us s ss

zw wd dw ww wzw w

é ù+ê ú é ùê úé ù é ùD D- ê úë ûê ú ê ú ê ú= + Dê ú ê ú ê úD D+ +ë û ë û ê úë û

2 2( )2r

n n

uss s

Zero‐state response• E.g. when there is a small increase in

mechanical torque Tm (= Pm in pu)

02 2

11 sinin2 1

rad ntm

n

T e tH

02

sin in ra s2

d1

/ ntmr

n

T e tH

Inverse Laplace transform

pu2

mTuH

DD =

0 0 0/ ( ) / in pu r rw d w w w wD =D = -

20

2 2

(0) /( )2

nr

n n

ss sw d w

wzw w

DD =-

+ +

02 2( )

( 2 )n n

uss s s

Damping angle:

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Example• Exp. 11.2 & 11.3 in Saadat’s book• H=9.94s, KD=0.138pu, Tm=0.6 pu with PF=0.8.

Find the responses of the rotor angle and frequency under these disturbances

(1) (0)=10o=0.1745 rad(2) Pe=0.2pu

(0)=16.79+10=26.79o ()=16.79+5.76=22.55o

Zero‐input response: (0)=10o Zero‐state response: Pe=0.2pu

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Small‐Signal Stability of a Multi‐machine SystemInter‐area or intra‐area modes (0.1‐0.7Hz): machines in one part of the system swing against machines in other parts

Inter‐area model (0.1‐0.3Hz): involving all the generators in the system; the system is essentially split into two parts, with generators in one part swinging against machines in the other parts.

Intra‐area mode (0.4‐0.7Hz): involving subgroups of generators swinging against each other.

Local modes (0.7‐2Hz): oscillations involve a small part of the system

Local plant modes: associated with rotor angle oscillations of a single generator or a single plant against the rest of the system; similar to the single‐machine‐infinite bus system

Inter‐machine or interplant modes: associated with oscillations between the rotors of a few generators close to each other

Control or torsional modes (2Hz – )

Due to inadequate tuning of the control systems, e.g. generator excitation systems,  HVDC converters and SVCs, or torsional interaction (sub‐synchronous resonance) with power system control

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High & Low Frequency Oscillations•Whenever power flows, I2R losses occur. These energy losses help to reduce the amplitude of the oscillation. 

• High frequency (>1.0 HZ) oscillations are damped more rapidly than low frequency (<1.0 HZ) oscillations. The higher the frequency of the oscillation, the faster it is damped. 

• Power system operators do not want any oscillations. However, when oscillations occur, it is better to have high frequency oscillations than low frequency oscillations. 

• The power system can naturally dampen high frequency oscillations. Low frequency oscillations are more damaging to the power system, which may exist for a long time, become sustained (undamped) oscillations, and even trigger protective relays to trip elements

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11,600 MW loss

15,820MW loss

2,100 MW loss

970 MW lossBlackout on August 10, 19961. Initial event (15:42:03):

Short circuit due to tree contact Outages of 6 transformers and lines

2. Vulnerable conditions (minutes)Low‐damped inter‐area oscillationsOutages of generators and tie‐lines

3. Blackouts (seconds)Unintentional separation Loss of 24% load

200 300 400 500 600 700 8001100

1200

1300

1400

1500

Malin-Round Mountain #1 MW

Time in Seconds

0.264 Hz oscillationdamping ratio =3.46%

0.252 Hz oscillation

damping ratio 1%

0.276 Hzoscillation

damping ratio >7%

15:42:03 15:48:5115:47:36

Transient instability (blackouts)

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Oscillation Modes of a Multi‐machine System in the Classic Model

2 2 2

1 1 1

cos sin cos

, sin , cos

n n n

ei i ii ij i ii i j ij ij ij i ii i j ij ij ij ijj j jj i j i j i

ij i j ij ij ij ij ij ij

P E G P E G E E Y E G E E B G

B Y G Y

0

0 0

0 0

sin sin cos

cos cos sin

ij ij ij

ij ij ij ij

ij ij ij ij

Linearization at ij0:

2

20

2 1, 2, ,i imi ei

H d P P i ndt

2

210

2

210

0 0

2 0 1, 2, ,

2 0 co sin s

sij

i j ij ij ij

ni i

ijjj i

ni i

i ijjj i

j

H d i ndt

H

K

E E B Gddt

Synchronizing power coefficient

(Ignoring damping)

0compared to cosBS

T

E EKX

d¢

= 0

0 0 = cos sinij

ijsij i j ij ij ij ij

ij

PK E E B G

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2

210

2 0 1, 2, ,n

i isij ij

jj i

H d K i ndt

Note: There are only (n−1) independent equations because ij=0, so we need to formulate the (n−1) independent relative rotor angle equations with one reference machine, e.g., the n-th machine.

2 10 0 0 0

21 1

0, 1, , 12 2 2 2

n nin

sij sni in snj sij jnj ji n n ij i j i

d K K K K i ndt H H H H

2 1

21

0 0 0 0

1

0 1, 2, , 1

2 2 2 2

nin

ij jnj

n

ii sij sni ij snj sijji n n ij i

d i ndt

K K K KH H H H

2 1

2

20

1 12 0, 1, ,

2 21

ni

sij ijjij i

nn R

snj njjn

d Kdt

d K nd H H

it

Consider each in=i-n

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State‐space representation

1 2 1 1 2 1

1 2 2 1 2 1

Let , , , , , , and

, , , , , ,

n n n n n

n n n n n n n

x x x

x x x

1 1

2 2

1 1

11 12 1 1

21 22 2 11 1

2 2 2 21 1 1 2 1 1

1 0 00 1 0

0 0 1n n

nn n

nn n

n nn n n n

x xx x

x xx x

x x

x x

0

0

1

2

0 IXA 0X

• Its characteristic equation |2I-A|=0 has 2(n-1) imaginary roots, which occur in (n-1) complex conjugate pairs

• An n-machine system has (n-1) modes

Read Anderson’s Examples 3.2 and 3.3 about the linearization and eigen-analysis on the IEEE 9-bus system

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Formulation of General Multi‐machine State Equations• The linearized model of each dynamic

device:

xi – Perturbed values of state variablesii – Current injection into network from the devicev – Vector of network bus voltages

Bi and Yi have non-zero element corresponding only to the terminal voltage of the device and any remote bus voltages used to control the device

ii and v both have real and imaginary components

i i i i

i i i i

x A x B v

i C x Y v

• Such state equations for all the dynamic devices in the system may be combined into the form:

x is the vector of state variables of the complete systemAD and CD are block diagonal matrices composed of Ai and Ci associated with the individual devices

• Node equation of the transmission network:i = YNv

• The overall system state equation:

• Read Kundur’s sec. 12.7 for other related information, e.g. load model linearization and selection of a reference rotor angle

D D

D D

x A x B v

i C x Y v

1D D N D D

1D D N D D

( ) =

( )

x A x B Y Y C x Ax

A A B Y Y C

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Modal analysis (eigen‐analysis) on an n‐dimensional nonlinear system

Linearization at the equilibrium x0: consider a perturbation at x0 and u0

Equilibrium x0 (with u0):

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Eigen‐vectors

1 2

Similarly the row vector

Left eigenvector associated with

, , ,

j

j j j j

TT T Tn

λ

ψ

A

Ψ

The left and right eigenvectors corresponding to different eigenvalues are orthogonal(row vectors of -1 are left eigenvectors of A, so we may let = C -1

where C is a diagonal matrix or simply equal to I if normalized)

(usually true for a real-world system)

For any , the column vector satisfying is called a right eigenvector of associated with

i i i i i

i

λ λλ

AA

1 2

1 21

, , , ( , ,

Modal , )

if has distinct eigenvalues

matrix n

ndiag λ λ λ

ΦAΦ ΦΛ Λ

Φ AΦ Λ A

0 if , or 1j i i ii j ΨΦ I

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Free (zero‐input) response and stability Linearized system without external forcing x A x

• To eliminate the cross-coupling between the state variables, consider a new state vector z

iti i i i i

1

z z z t z 0 e

=

z Λz Φ AΦz Φz AΦz

i

1

n2 t

1 2 n i ii 1

n

z tz t

(t) (t) z 0 e

z t

x Φz

Free response is a linear combination of n modes

1

i i i

t t t

z 0 0 c

z Φ x Ψ x

x

i i

n nt t

i i i ii 1 i 1

t 0 e c ex x

i 1 n

nt t t

k ki i k1 1 kn ni 1

x ( t ) c e c e ... c e

(ci is the magnitude of the excitation of the ith mode)

• Each eigenvalue = j– A real eigenvalue (=0) corresponds to a non-oscillatory mode.

• a decaying mode has <0; a mode with >0 has aperiodic instability.– Complex eigenvalues (0) occur in conjugate pairs; each pair corresponds

to one oscillatory mode• Frequency of oscillation in Hz: f= /2• Damping ratio (rate of decay) of the oscillation amplitude

2 2

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Mode Shape and Mode Composition

• The variables x1, x2, …, xn are original state variables chosen to represent the dynamic performance of the system.

• Variables z1, z2, …, zn are transformed state variables; each is associated with only one mode. In other words, they are directly related to the modes.

• The right eigenvector i gives the mode shape of the ith mode, i.e. the relative activity of the original state variables when the ith mode is excited: The kth

element of i , i.e. ki , measures the activity of state variable xk in the ith mode

• The left eigenvector i gives the mode composition of the ith mode, i.e. what weighted composition of original state variables is needed to construct the mode: The kth element of i , i.e. ik, weights the contribution of xk’s activity to the ith mode

T1 2 n 1 2 nt t z (t), z (t),..., z (t) x Φz

T TT T T1 2 n 1 2 nt t , , , x (t), x (t),..., x (t) z Ψ x

n

k k i ii 1

x ( t ) z t

n

i ik kk 1

z ( t ) x ( t )

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Participation factor

n n

ik ki ikiik 1 k 1

1 p

ΨΦ I ΨΦ

• ki measures the activity of xk in the ith mode

• ik weights the contribution of this activity to the mode

• Participation factor pki= ikki measures the participation of the kth state variable xk in the ith mode.

• pki is dimensionless and hence invariant under changes of scale on the variables

• Question: considering =-1, why do we have to define the mode shape, mode composition, and participation factors, separately?

Learn Kundur’s Example 12.2 on a SMIB system