ECEN 667 Power System Stability
Transcript of ECEN 667 Power System Stability
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ECEN 667
Power System Stability
1
Lecture 22: Power System Stabilizers
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
Texas A&M University, [email protected]
Special Guest Lecturer: TA Iyke Idehen
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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
• We’ll be doing power system stabilizers today, and will
pickup with modal analysis next time
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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
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Block Diagram of System with a PSS
4Image Source: Kundur, Power System Stability and Control
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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
0
0
2sins
Md ep
E VH dT D
dt X X
d
orgE E K
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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
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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
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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
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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
Δx = AΔx BΔy EΔu
0 = CΔx+DΔy
-1
sysΔx = A BD C Δx EΔu A Δx EΔu
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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
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SMIB System (Flux Decay Model)
• SMIB with a flux decay machine model and a fast
exciter
11
~
)2/()( d j
qd 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
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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
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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 j
qd 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
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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 j
qd ejII
0V
ejXeR
jteV
Machine equations
14
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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
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Linearization (contd)
• Final Steps involve
1. 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Heffron–Phillips 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 BΔu
ΔuSMIB Model
26
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Block Diagram
• K1 to K6 are affected by system loading
27
On diagram
v is used to
indicate what
we've been
calling pu
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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
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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
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SMIB Results
• Doing the SMIB gives a matrix that closely matches the
book’s 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
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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
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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
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State-Space Analysis
• With no exciter there
are three loops
– Top complex pair of
eigenvalues loop
– Bottom loop through
Eq’ contributes positive
damping
• Adding the exciter gives
33
5 6
1fd fd A A q A ref
A
E E K K K K E K VT
d
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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 K6
can be
verified for
previous case
using the
PowerWorld
SMIB matrix
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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
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Root Locus for Varying KA
• This can be considered a feedback system, with the
general root locus as shown
36
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Frequency Domain Analysis
• Alternatively we could use frequency domain analysis
• Ignoring TA gives
• If K5 is positive, then the
response is similar to the
case without an exciter
• If K5 is negative, then with
a 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
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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
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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
sKj
H
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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
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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
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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