ece 522 power systems analysis ii 3.2 –small‐signal...
TRANSCRIPT
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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
12
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