ECEN 667

Power System Stability

1

Lecture 22: Power System Stabilizers

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

Texas A&M University, overbye@tamu.edu

Special Guest Lecturer: TA Iyke Idehen

mailto:overbye@tamu.edu

Announcements

Read Chapter 8 Homework 7 is posted; due on Tuesday Nov 28 Final is as per TAMU schedule. That is, Friday Dec 8

from 3 to 5pm

Well be doing power system stabilizers today, and will pickup with modal analysis next time

2

Overview of a Power System

Stabilizer (PSS)

A PSS adds a signal to the excitation system to improve the rotor damping

A common signal is proportional to speed deviation; other inputs, such as like power, voltage or acceleration, can be used

Signal is usually generated locally (e.g. from the shaft)

Both local mode and inter-area mode can be damped. When oscillation is observed on a system or a planning

study reveals poorly damped oscillations, use of

participation factors helps in identifying the machine(s)

where PSS has to be located

Tuning of PSS regularly is important

3

Block Diagram of System with a PSS

4Image Source: Kundur, Power System Stability and Control

Power System Stabilizer Basics

Stabilizers can be motivated by considering a classical model supplying an infinite bus

Assume internal voltage has an additional component

This will add additional damping if sin(d) is positive In a real system there is delay, which requires

compensation 5

dts

dd

00

2sinsM

d ep

E VH dT D

dt X X

d

orgE E K

Example PSS

An example single input stabilizer is shown below (IEEEST)

The input is usually the generator shaft speed deviation, but it could also be the bus frequency deviation, generator electric

power or voltage magnitude

6

VST is an

input into

the exciter

The model can be

simplified by setting

parameters to zero

Example PSS

Below is an example of a dual input PSS (PSS2A) Combining shaft speed deviation with generator electric

power is common

Both inputs have washout filters to remove low frequency components of the input signals

7

IEEE Std 421.5 describes the common stabilizers

Washout Parameter Variation

The PSS2A is the most common stabilizer in both the 2015 EI and WECC cases. Plots show the variation in

TW1 for EI (left) and WECC cases (right); for both the

x-axis is the number of PSS2A stabilizers sorted by

TW1, and the y-axis is TW1 in seconds

8

PSS Tuning: Basic Approach

The PSS parameters need to be selected to achieve the desired damping through a process known as tuning

The next several slides present a basic method using a single machine, infinite bus (SMIB) representation

Start with the linearized differential, algebraic model with controls u added to the states

If D is invertible then

9

x = Ax By Eu

0 = Cx+Dy

-1 sysx = A BD C x Eu A x Eu

PSS Tuning: Basic Approach

Low frequency oscillations are considered the following approach

A SMIB system is setup to analyze the local mode of oscillation (in the 1 to 3 Hz range)

A flux decay model is used with Efd as the input

Then, a fast-acting exciter is added between the input voltage and Efd

Certain constants, K1 to K6, are identified and used to tune a power system stabilizer

10

SMIB System (Flux Decay Model)

SMIB with a flux decay machine model and a fast exciter

11

~

)2/()( d jqd ejII 0V

ejXeR

jteV

' '

' ' '

'

[ ( ( ) )]

( ( ) )

pu s

pu M q q d d d q pu

q q d d d fd

do

1T E I X X I I D

2H

1E E X X I E

T

d

Stator Equations

Assume Rs=0, then the stator algebraic equations are:

' '

( /2)

( /2)

sin( ) 0 (1)

cos( ) 0 (2)

( ) (3)

(4)

Expand the right hand side of (4)

sin( ) cos( ) (5)

sin( ) cos( )

(1) (2)

q q t

q t d d

j j

d q t

j j

d q t

d q t t

d t q t

X I V

E V X I

V jV e V e

V jV V e e

V jV V jV

V V and V V

and beco

d

d

d

d

d d

d d

' '

0 (6)

0 (7)

q q d

q q d d

me X I V

E V X I

12

Network Equations

The network equation is (assuming zero angle at the infinite bus and no local load)

( / )

( / )

( / )

( )( )

( )( )

j 2

d qj 2

d q

e e

j 2

d q

d q

e e

V jV e V 0I jI e

R jX

V jV V eI jI

R jX

d

d

d

Simplifying,

~

)2/()( d jqd ejII 0V

eXeR

jteV

sin

cos

e d e q d

e d e q q

R I X I V V

X I R I V V

d

d

13

Complete SMIB Model

' ' '

'

' '

' '

( ( ) )

note in the book

[ ( ( ) )]

sin

cos

q q d d d fd

do

pu s pu

s

pu M q q q d d q pu

q q d

q q d d

e d e q d

e d e q q

1E E X X I E

T

1T E I X X I I D

2H

X I V 0

E V X I 0

R I X I V V

X I R I V V

d

d

d

Stator equations

Network equations

~

)2/()( d jqd ejII

0V

ejXeR

jteV

Machine equations

14

Linearization of SMIB Model

''

Step 1: Linearize the stator equations

Step 2: Linearize the network equations

cos

sin

Step 3: Equate the

d dq

q q qd

d de e

q qe e

V I 00 X

V I EX 0

V IR X V

V IX R V

dd

d

''

righthand sides of the above equations

( ) cos

( ) sin

de e q

q qe d e

I 0R X X V

I EX X R V

dd

d

Notice this is equivalent to a generator at the infinite

bus with modified resistance and reactance values15

Linearization (contd)

Final Steps involve1. Linearizing Machine Equations

2. Substitute (1) in the linearized equations of (2).

'

'

'

Solve for ,

cos sin ( )( )

sin cos ( )

The determinant is ( )( )

d q

d e q e q e q

q e e d e

2

e e q e d

I I

I X X R V V X X E11

I R R V V X X

R X X X X

d d

d d d

( , , , , ) (2)d q fd MI I E Tx f x

16

Linearization of Machine

Equations

Substitute (3) in (2) to get the linearized model.

Symbolically we have

'

' '

' '

'

' ' '

( )

( )

( ) ( )

do

oq s

do do

1

Tq q

s

I Dpu pu

2H 2H

1 1d dT T

d fd

q Mo o o1 1 1 1q d q d q q q2H 2H 2H 2H

0 0E E

0 0 1

0

X X 0 0I E

0 0 0 0I T

I X X X X I E 0

d d

( )

( )

d q

d q

I 2

3

x A x B u C

I D x

17

Linearized SMIB Model

Excitation system is not yet included. K1 K4 constants are defined on next slide

' '

' ' '

'

4q q fd

3 do do do

s pu

s2 1pu q pu M

K1 1E E E

K T T T

DK K 1E T

2H 2H 2H 2H

d

d

d

18

K1 K4 Constants

K1 K4 only involve machine and not the exciter.

}]sincos)}{(){(

}cossin)){(([1

)())((1

cossin)()(

))((1

1

'''

'

1

'''

2

'

4

'

3

dd

dd

dd

eed

o

q

o

dqd

eeqqd

o

q

o

qe

o

dqdeeqqd

o

q

o

q

eeqdd

eqdd

RXXEIXXV

RXXXXVIK

ERIXXRXXXXIIK

RXXXXV

K

XXXX

K

19

Manual Control

Without an exciter, the machine is on manual control. The previous matrix will tend to have two complex

eigenvalues, corresponding to the electromechanical

mode, and one real eigenvalue corresponding to the flux

decay

Changes in the operating point can push the real eigenvalue into the right-hand plane

This is demonstrated in the following example

20

Numerical Example

Consider an SMIB system with Zeq = j0.5, Vinf = 1.05, in which in the power flow the generator has

S = 54.34 j2.85 MVA with a Vt of 115

Machine is modeled with a flux decay model with (pu, 100 MVA) H=3.2, T'do=9.6, Xd=2.5, Xq=2.1, X'd=0.39, Rs=0, D=0

21

Saved as case B2_PSS_Flux

Initial Conditions

The initial conditions are

22

( / ) .( ) ..

( ) ( ) .

( . )( . )

. .

j j 2

G d q

j j

t s q G

1 15 1 05 0I e I jI e 0 5443 18

j0 5

0 angle of E where E V e R jX I e

E 1 15 j2 1 0 5443 18

1 4788 65 5

d

d

( / )

( / )

. . ,

. , . .

.

Hence, . , .

j j 2

d q G

d q

j j 2

d q

d q

I jI I e e 0 5443 42 48

I 0 4014 and I 0 3676

V jV Ve e 1 39 48

V 0 7718 V 0 6358

d

d

SMIB Eigenvalues

With the initial condition of Q = -2.85 Mvar the SMIB eigenvalues are -0.223 j7.374, -0.0764

Changing the Q to 20 Mvar gives eigenvalues of-0.181 j7.432, -0.171

Change the Q to 20 Mvar gives eigenvalues of-0.256 j6.981, +0.018

23

This condition

is unstable

in Eqp

Including Terminal Voltage

The change in the terminal voltage magnitude also needs to be include since it is an input into the exciter

q

t

o

q

d

t

o

dt

q

o

qd

o

dtt

qdt

qdt

VV

VV

V

VV

VVVVVV

VVV

VVV

222

222

22

refVrefV

tV

tV

Differentiating

24

Computing

While linearizing the stator algebraic equations, we had

Substitute this in expression for Vt to get

'

''

d q q

q qd

V 00 X E

V EX 0 d

'

'

'

'

[ sin cos ( )]

[ ( cos ( )sin )]

( )

t 5 6 q

o

d5 q e d e

t

o

q

d e q e

t

o ooq qd

6 q e d q e

t t t

V K K E

V1K X R V V X X

V

VX R V V X X

V

V VV1K X R X X X

V V V

d

d d

d d

25

HeffronPhillips Model

(from 1952 and 1969 )

Add a fast exciter with a single differential equation

Linearize

This is then combined with the previous three differential equations to give

( )A fd fd A ref tT E E K V V

'

( )

( )

A fd fd A ref t

A fd fd A ref A 5 6 q

T E E K V V

T E E K V K K K Ed

[ ]

sys

T

M refT V

x A x Bu

uSMIB Model

26

Block Diagram

K1 to K6 are affected by system loading

27

On diagram

v is used to

indicate what

we've been

calling pu

Add an EXST1 Exciter Model

Set the parameters to KA = 400, TA=0.2, all others zero with no limits and no compensation

Hence this simplified exciter is represented by a single differential equation

28

( )A fd fd A REF t sT E E K V V V

Vs is the input

from the stabilizer,

with an initial

value of zero

Initial Conditions (contd)

From the stator algebraic equation,' '

. ( . )( . ) .

q q d dE V X I

0 63581 0 39 0 4014 0 7924

' '

' '

( )

0.7924 (2.5 0.39)0.4014 1.6394

1.63941 1.0041

400

377, ( )

(0.7924)(0.3676) (2.1 0.39)(0.4014)(0.3676)

0.5436 ( sin ) /

fd q d d d

fd

REF t

A

s M q q q d d q

t e

E E X X I

EV V

K

T E I X X I I

checks with VV X

29

SMIB Results

Doing the SMIB gives a matrix that closely matches the books matrix from Example 8.7

The variable order is different; the entries in the w column are different because of the speed dependence in the swing

equation and the Norton equivalent current injection

30

The values are

different from the

book because of

speed dependence

included in the

stator voltage

and swing

equations

Computation of K1 K6 Constants

Calculating the values with the formulas gives'

'

'

' ' '

1 5 6

( )( ) .

( )( ).

.

( )[( )sin cos ] .

[ ( )( ) ( ) ] .

Similarly K ,K ,and K are calculat

2

e e q e d

d d q e

3

3

d d4 q e e

o o o o

2 q q d q q e e d q d e q

R X X X X 2 314

X X X X11 3 3707

K

K 0 296667

V X XK X X R 2 26555

1K I I X X X X R X X I R E 1 0739

d d

ed as

. , . , .1 5 6K 0 9224 K 0 005 K 0 3572

31

Damping of Electromechanical Modes

Damping can be considered using either state-space analysis or frequency analysis

With state-space analysis the equations can be written as

The change in Efd comes from the exciter

32

' '

4

3 do do doq q

s fd

pu s pu2 1

K1 10

K T T TE E

0 0 0 E

D 0K K

2H 2H 2H

d d

State-Space Analysis

With no exciter thereare three loops

Top complex pair ofeigenvalues loop

Bottom loop throughEq contributes positive

damping

Adding the exciter gives

33

5 61

fd fd A A q A ref

A

E E K K K K E K VT

d

B2_PSS_Flux Example

The SMIB can be used to plot how the eigenvalues change as the parameters (like KA) are varied

34

''

4

3 do do doqq

s

refs2 1pupu

A

fdfdA 6 A 5 A

A A A

K1 10

0K T T TEE

00 0 0

V0DK K0

K2H 2H 2HEE

K K K K 1 T0

T T T

dd

Values for K1 to K6can be

verified for

previous case

using the

PowerWorld

SMIB matrix

Verified K Values

The below equations verify the results provided on the previous slide SMIB matrix with the earlier values

35

11

22

3

3

44

55

65

0.1441 0.2882 3.2 0.9222

0.1677 0.3354 3.2 1.0732

1 10.3511 0.297

0.3511 9.6

0.236 0.236 9.6 2.266

10.069 0.210.069 0.005

400

714.5 0.714.5

do

do

A

A

A

A

KK

H

KK

H

KK T

KK

T

K KK

T

K KK

T

20.357

400

Root Locus for Varying KA

This can be considered a feedback system, with the general root locus as shown

36

Frequency Domain Analysis

Alternatively we could use frequency domain analysis Ignoring TA gives

If K5 is positive, then theresponse is similar to the

case without an exciter

If K5 is negative, then witha sufficiently large KA the

electromechanical modes

can become unstable

37

'

( )

( )

q 3 4 A 5

A 3 6 3 do

E s K K K K

s 1 K K K sK Td

Frequency Domain Analysis

Thus a fast acting exciter can be bad for damping, but has over benefits, such as minimizing voltage

deviations

Including TA

Including the torque angle gives

38

'

( )

( )

3 4 A A 5q

A 3 6 3 do A

K K 1 sT K KE s

s K K K 1 sK T 1 sTd

( )( )( )

( ) ( )

qe2

E sT sK H s

s sd d

Torque-Angle Loop

The torque-angle loop is as given in Figure 8.11;with damping neglected it has two imaginary

eigenvalues

39

11,2

2

sKjH

Torque-Angle Loop with

Other Dynamics

The other dynamics can be included as in Figure 8.12

40

2

1

2( ) 0

s

Hs K H sd d d

H(s) contributes

both synchronizing

torque and

damping torque

Torque-Angle Loop with

Other Dynamics

Previously we had

If we neglect K4 (which has little effect at 1 to 3 Hz) and divide by K3 we get

41

'

( )( )

( )

2 3 4 A A 5e

A 3 6 3 do A

K K K 1 sT K KT sH s

s K K K 1 sK T 1 sTd

( )( )

( )

e 2 A 5

2AA 6 do do A

3 3

T s K K KH s

s T1K K s T s T T

K K

d

Synchronizing Torque

The synchronizing torque is determined by looking at the response at low frequencies (0)

With high KA, this is approximately

42

(j ) 2 A 5

A 6

3

K K KH 0

1K K

K

(j ) 2 5

6

K KH 0

K

2

1

2( ) 0

s

Hs K H sd d d

The synchronizing

torque is dominated

by K1; it is enhanced

if K5 is negative