ecen 667 power system stability - thomas overbye
TRANSCRIPT
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ECEN 667 Power System Stability
Lecture 23: Small Signal Stability,
Oscillations
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
Texas A&M University
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1
Announcements
• Read Chapter 8
• Homework 5 is due Today
• Homework 6 is due on Tuesday December 3
• Final is at scheduled time here (December 9
from1pm to 3pm)
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Example: Bus 4 SMIB Dialog
• On the SMIB dialog, the General Information tab
shows information about the two bus equivalent
PowerWorld case B4_SMIB
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Example: Bus 4 SMIB Dialog
• On the SMIB dialog, the A Matrix tab shows the Asys
matrix for the SMIB generator
• In this example A21 is showing
, ,. cos .
. .
.
4 pu E 4
4 4 4
P1 1 11 2812 23 94
2H 6 0 3 0 22
0 3753
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Example: Bus 4 with GENROU Model
• The eigenvalues can be calculated for any set of
generator models
• The example can be extended by replacing the bus 4
generator classical machine with a GENROU model
– There are now six eigenvalues, with the dominate response
coming from the electro-mechanical mode with a frequency
of 1.84 Hz, and damping of 6.9%
PowerWorld case B4_SMIB_GENROU
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Relation to the Signal-Based Approaches
• The results from the SMIB analysis can be compared
with the signal-based approach by applying a short self-
clearing fault to disturb the system
• This can be done easily in PowerWorld by running the
transient stability simulation, looking at the Results
from RAM page, right-clicking on the desired signal(s)
and selecting Modal Analysis Selected Column.
• The next slide shows the results for the Bus 4 Generator
rotor angle (which would not be directly observable)
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Bus 4 Generator Rotor Angle
• Notice that the main mode, at 1.84 Hz with 7%
damping closely matches
Not all modes will be easily observed in all signals
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Example: Bus 4 with GENROU Model and Exciter
• Adding an relatively slow EXST1 exciter adds
additional states (with KA=200, TA=0.2)
– As the initial reactive power output of the generator is
decreased, the system becomes unstable
Case is saved as B4_GENROU_EXST1
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Example: Bus 4 with GENROU Model and Exciter
• With Q4 = 25 Mvar the eigenvalues are
• And with Q4=0 Mvar the eigenvalues are
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Example: Bus 4 with GENROU Model and Exciter
• Graph shows response following a short fault when Q4
is 0 Mvar
• This motivates trying to get additional insight into how
to increase system damping, which is the goal of modal
analysis
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Modal Analysis - Comments
• Modal analysis (analysis of small signal stability
through eigenvalue analysis) is at the core of SSA
software
• In Modal Analysis one looks at:
– Eigenvalues
– Eigenvectors (left or right)
– Participation factors
– Mode shape
• Power System Stabilizer (PSS) design in a multi-
machine context can be done using modal analysis
method (we’ll see another method later)
Goal is to
determine
how the various
parameters
affect the
response of
the system
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Eigenvalues, Right Eigenvectors
• For an n by n matrix A the eigenvalues of A are the
roots of the characteristic equation:
• Assume l1…ln as distinct (no repeated eigenvalues).
• For each eigenvalue li there exists an eigenvector
such that:
• vi is called a right eigenvector
• If li is complex, then vi has complex entries
det[ ] 0l l A I A I
i i ilAv v
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Left Eigenvectors
• For each eigenvalue li there exists a left
eigenvector wi such that:
• Equivalently, the left eigenvector is the right
eigenvector of AT; that is,
t t
i i ilw A w
t
i i ilA w w
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Eigenvector Properties
• The right and left eigenvectors are orthogonal i.e.
• We can normalize the eigenvectors so that:
, ( )t t
i i i j0 0 i j w v w v
, ( )t t
i i i j1 0 i j w v w v
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Eigenvector Example
2
2
1,2
11 11 21 11
1 1 1
21 11 21 21
2 2
1 4 1 4, 0 0
3 2 3 2
3 (3) 4(10) 3 493 10 0 5, 2
2 2
4 55
3 2 5
,
42
3
v v v v
v v v v
Similarly
ll
l
l l l
l
A A I
Av v v
v
21 11
1
choose 1 1
1
1
v v
v
Right Eigenvectors 51 l
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Eigenvector Example
• Left eigenvectors
51 l1 1 11 21 11 21
11 21 11
21 11
11 21 21
1 2 2
1 2 1 2
1 1 2 2 2 1 1 2
1 45 [ ] 5[ ]
3 2
3 54, 3
4 2 5
3 12
4 1
1 4 3 1
1 3 4 1
7 , 7 , 0 , 0
t t
t t t t
w w w w
w w wLet w then w
w w w
Verify
l
w A w
w w
v v w w
w v w v w v w v
We would like to make
This can be done in many ways.
1.t
i iw v
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Eigenvector Example
• It can be verified that WT=V-1 .
• The left and right eigenvectors are used in computing the participation factor matrix.
3 11
4 17
3 4 1 4 1 01
1 1 1 3 0 17
T
Let
Then
Verify
W
W V I
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Modal Matrices
• The deviation away from an equilibrium point can be
defined as
• If the initial deviation corresponds to a right
eigenvector, then the subsequent response is along this
eigenvector since
Δx AΔx
i i ilAv v
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Modal Matrices
• From this equation ( ) it is difficult to
determine how parameters in A affect a particular x
because of the variable coupling
• To decouple the problem first define the matrices
of the right and left eigenvectors (the modal
matrices)
1 2 1 2[ , ..... ] & [ , ,..... ]
when ( )
n n
iDiag l
V v v v W w w w
AV VΛ Λ
Δx AΔx
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Modal Matrices
• It follows that
• To decouple the variables define z so
• Then
• Since is diagonal, the equations are now uncoupled
with
• So
x Vz x Vz AΔx AVz
1 V AV Λ
1 z V AVz WAVz Λz
( ) ( )t t x Vz
i i iz zl
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Example
• Assume the previous system with
1 4
3 2
1 4
1 3
5 0
0 2
-1
A
V
V AV
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Modal Matrices
• Thus the response can be written in terms of the
individual eigenvalues and right eigenvectors as
• Furthermore with
• So z(t) can be written as using the left eigenvectors
as
( ) ( ) i
nt
i i
i 1
t z 0 el
x v
1 T Δx= VZ z V x W x
( )
( ) ( ) [ .... ]
( )
1
t t
1 2 n
n
x t
t t
x t
z W x w w w
Note, we are
requiring that
the eigenvalues
be distinct!
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Modal Matrices
• We can then write the response x(t) in terms of the
modes of the system
• So ci is a scalar that represents the magnitude of
excitation of the ith mode from the initial conditions
( ) ( )
( ) ( )
so ( )
Expanding ( ) ...
i
n1 2
t
i i
t
i i i
nt
i i
i 1
tt t
i i1 1 i2 2 in n
z t w x t
z 0 w x 0 c
t c e
x t v c e v c e v c e
l
ll l
x v
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Numerical example
, ( )
Eigenvalues are ,
Eigenvectors are ,
Modal matrix
. .Normalize so
. .
1 1
2 2
1 2
1 2
x x0 1 10
x x8 2 4
4 2
1 1
4 2
1 1
4 2
0 2425 0 4472
0 9701 0 8944
l l
x
v v
V
V
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Numerical example (contd)
Left eigenvector matrix is:
. .
. .
1
1 1
2 2
1 3745 0 6872
1 4908 0 3727
z z4 0
z z0 2
T
T
W V
z = W AVz
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Numerical example (contd)
, ( ) ( )
( ) .,
( )
.( ) ( ) ; ( ) ( ) , ( )
( ) ( )
( ) ( )
. .( )
. .
1
1 1
1
2 2
2
4t 2t T
1 1 2 2
1 1
2 2
1 1 2
z 4z 0 V 0
z 0 4 123z 2z
z 0 0
4 123z t z 0 e z t z 0 e 0
0
x t z t1 1
x t z t4 2
0 2425 0 4472c z t c
0 9701 0 8944
z x
C W x
x = Vz
( ) ( ) i
2t
2 i i i
i 1
z t c z 0 el
v
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Because of the initial
condition, the 2nd mode does
not get excited
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Mode Shape, Sensitivity and Participation Factors
• So we have
• x(t) are the original state variables, z(t) are the transformed variables so that each variable is associated with only one mode.
• From the first equation the right eigenvector gives the “mode shape” i.e. relative activity of state variables when a particular mode is excited.
• For example the degree of activity of state variable xk
in vi mode is given by the element Vki of the the right eigenvector matrix V
) ( ), ( ) ( )tt t t t x( Vz z W x
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Mode Shape, Sensitivity and Participation Factors
• The magnitude of elements of vi give the extent of
activities of n state variables in the ith mode and
angles of elements (if complex) give phase
displacements of the state variables with regard to
the mode.
• The left eigenvector wi identifies which
combination of original state variables display only
the ith mode.
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Eigenvalue Parameter Sensitivity
• To derive the sensitivity of the eigenvalues to the
parameters recall Avi = livi; take the partial derivative
with respect to Akj by using the chain rule
Multiply by
[ ]
i ii i i
kj kj kj kj
t
i
t t t ti ii i i i i i i
kj kj kj kj
t t ti ii i i i i i
kj kj kj
A A A
A A A
A A A
ll
ll
ll
i
i
vvAv A v
A
w
vvAw v w A w v w
A
vAw v w A I w v
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Eigenvalue Parameter Sensitivity
• This is simplified by noting that
by the definition of wi being a left eigenvector
• Therefore
• Since all elements of are zero, except the kth
row, jth column is 1
• Thus
( )t
i i 0l w A I
t ii i
kj kjA A
l
Aw v
kjA
A
iki ji
kj
W VA
l
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Sensitivity Example
• In the previous example we had
• Then the sensitivity of l1 and l2 to changes in A are
• For example with
1,2
1 4 1 4 3 11, 5, 2, ,
3 2 1 3 4 17l
A V W
1 23 3 4 31 1
,4 4 4 37 7
iki ji
kj
W VA
l l l
A A
1,2
1 4ˆ ˆ, 5.61, 1.61,3 3
l
A
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Eigenvalue Parameter Sensitivity
• This is simplified by noting that
by the definition of wi being a left eigenvector
• Therefore
• Since all elements of are zero, except the kth
row, jth column is 1
• Thus
( )t
i i 0l w A I
t ii i
kj kjA A
l
Aw v
kjA
A
iki ji
kj
W VA
l
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Participation Factors
• The participation factors, Pki, are used to determine how
much the kth state variable participates in the ith mode
• The sum of the participation factors for any mode or any
variable sum to 1
• The participation factors are quite useful in relating the
eigenvalues to portions of a model
ki ki kiP V W
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Participation Factors
• For the previous example with Pki = VkiWik and
• We get
3 41
4 37
P
1 4 1 4 3 11, ,
3 2 1 3 4 17
A V W
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PowerWorld SMIB Participation Factors
• The magnitudes of the participation factors are
shown on the PowerWorld SMIB dialog
• The below values are shown for the four bus
example with Q4 = 0
Case is saved as B4_GENROU_Sat_SMIB_QZero
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Oscillations
• An oscillation is just a repetitive motion that can be
either undamped, positively damped (decaying with
time) or negatively damped (growing with time)
• If the oscillation can be written as a sinusoid then
• And the damping ratio is defined as (see Kundur 12.46)
The percent damping is just the damping
ratio multiplied by 100; goal is sufficiently
positive damping
2 2
cos sin cos
where and tan
t t
2 2
e a t b t e C t
bC A B
a
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Power System Oscillations
• Power systems can experience a wide range of
oscillations, ranging from highly damped and high
frequency switching transients to sustained low
frequency (< 2 Hz) inter-area oscillations affecting an
entire interconnect
• Types of oscillations include
– Transients: Usually high frequency and highly damped
– Local plant: Usually from 1 to 5 Hz
– Inter-area oscillations: From 0.15 to 1 Hz
– Slower dynamics: Such as AGC, less than 0.15 Hz
– Subsynchronous resonance: 10 to 50 Hz (less than
synchronous)
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Example Oscillations
• The below graph shows an oscillation that was
observed during a 1996 WECC Blackout
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Example Oscillations
• The below graph shows oscillations on the
Michigan/Ontario Interface on 8/14/03
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Fictitious System Oscillation
Movie shows
an example
of sustained
oscillations in
an equivalent
system
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Forced Oscillations in WECC (from [1])
• Summer 2013 24 hour data: 0.37 Hz oscillations
observed for several hours. Confirmed to be forced
oscillations at a hydro plant from vortex effect.
• 2014 data: Another 0.5 Hz oscillation also
observed. Source points to hydro unit as well. And
0.7 Hz. And 1.12 Hz. And 2 Hz.
• Resonance is possible when a system mode is
poorly damped and close. Resonance can be
observed in model simulations
1. M. Venkatasubramanian, “Oscillation Monitoring System”, June 2015
http://www.energy.gov/sites/prod/files/2015/07/f24/3.%20Mani%20Oscillation%20Monitoring.pdf
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Inter-Area Modes in the WECC
• The dominant inter-area modes in the WECC have
been well studied
• A good reference paper is D. Trudnowski,
“Properties of the Dominant Inter-Area Modes in
the WECC Interconnect,” 2012
– Four well known modes are
NS Mode A (0.25 Hz),
NS Mode B (or Alberta Mode),
(0.4 Hz), BC Mode (0.6 Hz),
Montana Mode (0.8 Hz)
Below figure from
paper shows NS Mode A
On May 29, 2012