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Power System Dynamics and Stability 1
Universidad de Castilla-La Mancha
Power System Dynamics and Stability
Dr. Federico Milano
E-mail: [email protected]
Tel.: +34 926 295 219
Departamento de Ingenierıa Electrica, Electronica, Automatica y
Comunicaciones
Escuela Tecnica Superior de Ingenieros Industriales
June 26, 2008 Introduction - 1
Power System Dynamics and Stability 2
Universidad de Castilla-La Mancha
Programa de Doctorado
Technical and Economical Management of
Generation, Transmission and Distribution Electric
Energy Systems
Area de Ingenierıa Electrica de la E.T.S. de Ingenieros Industriales de la
Universidad de Castilla - La Mancha
June 26, 2008 Introduction - 2
Power System Dynamics and Stability 3
Universidad de Castilla-La Mancha
Note
This course is partly based on the course ECE664 hold by Prof. Dr. C.
Canizares at the University of Waterloo, Ontario, Canada.
I wish to sincerely thank Prof. Dr. C. Canizares for his courtesy in sharing
this material.
June 26, 2008 Introduction - 3
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Universidad de Castilla-La Mancha
Objectives
Understand the modeling and simulation of power systems from phasor
analysis to electromagnetic transients.
Discuss the basic definitions, concepts and tools for stability studies of
power systems.
Familiarize with basic concepts of computer modelling of electrical power
systems.
June 26, 2008 Introduction - 4
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Universidad de Castilla-La Mancha
Outlines: System Modeling
Synchronous machine.
Transformer.
Transmission line.
Cable.
Loads.
June 26, 2008 Introduction - 5
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Outlines: System Analysis
Basic stability concepts:
Nonlinear systems.
Equilibrium points.
Stability regions.
Power Flow:
System model.
Equations and solution techniques.
Contingency analysis.
June 26, 2008 Introduction - 6
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Outlines: Voltage Stability
Definitions.
Basic concepts:
Saddle-node bifurcation.
Limit-induced bifurcation.
Continuation Power Flow (CPF).
Direct methods.
Indices.
Protections and controls.
Real case example: August 2003 North American blackout.
June 26, 2008 Introduction - 7
Power System Dynamics and Stability 8
Universidad de Castilla-La Mancha
Outlines: Angle Stability
Definitions.
Small-disturbance:
Hopf Bifurcations.
Control and mitigation.
Practical applications.
Transient Stability (large-disturbance):
Time domain.
Direct Methods:
Equal Area Criterion.
Energy Functions.
Real case example: May 1997 Chilean blackout.
June 26, 2008 Introduction - 8
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Universidad de Castilla-La Mancha
Outlines: Frequency Stability
Definitions.
Basic concepts.
Protections and controls.
Real case example: October 2003 Italian blackout.
Real case example: November 2006 European blackout.
June 26, 2008 Introduction - 9
Power System Dynamics and Stability 10
Universidad de Castilla-La Mancha
Outlines: Software Tools
Outlines.
UWPFLOW.
Matlab.
PSAT.
June 26, 2008 Introduction - 10
Power System Dynamics and Stability 11
Universidad de Castilla-La Mancha
References
P. Kundur, Power system stability and control, Mc Graw Hill, 1994.
P. Sauer and M. Pai, Power system dynamics and stability, Prentice Hall,
1998.
A. R. Bergen and V. Vittal, Power systems analysis, Second Edition,
Prentice-Hall, 2000.
C. A. Caizares, Editor, Voltage stability assessment: concepts, practices
and tools, IEEE-PES Power System Stability Subcommittee Special
Publication, SP101PSS, May 2003.
June 26, 2008 Introduction - 11
Power System Dynamics and Stability 12
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References
P. M. Anderson and A. A. Fouad, Power system control and stability, IEEE
Press, 1994.
J. Arrillaga and C. P. Arnold, Computer analysis of power systems, John
Wiley, 1990.
I. S. Duff, A. M. Erisman and J. K. Reid, Direct Methods for Sparse
Matrices, Oxford Science Publications, 1986.
J. Stoer and R. Bulirsch, Introduction to Numerical Analysis, Second
Edition, Springer-Verlag, 1993.
June 26, 2008 Introduction - 12
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Universidad de Castilla-La Mancha
References
M. Ilic and J. Zaborszky, Dynamics and Control of Large Electric Power
Systems, Wiley, New York, 2000.
C. A. Canizares, UWPFLOW, available at www.power.uwaterloo.ca
F. Milano, PSAT, Power System Analysis Toolbox, available at
www.power.uwaterloo.ca
Journal papers and technical reports.
Course notes available on line.
June 26, 2008 Introduction - 13
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Evaluation
Two projects are required.
The projects concentrates in the various topics discussed in class.
These will require the use of MATLAB, PSAT and UWPFLOW (the last
two are free software for stability studies co-developed at the University of
Waterloo, Canada).
Reproducing examples presented in the slides using UWPFLOW and
MATLAB.
Stability analysis of the IEEE 14-bus test system using PSAT.
June 26, 2008 Introduction - 14
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Evaluation
Alternatively, the students can develop a “user defined model” in PSAT.
Interested students are invited to contact Dr. Federico Milano.
June 26, 2008 Introduction - 15
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Contents
Introduction
Generator Modeling
Transmission System Modeling
Load Modeling
Power Flow Outlines
Stability Concepts
Voltage Stability
Angle Stability
Frequency Stability
Software Tools
Projects
June 26, 2008 Introduction - 16
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Generator Modeling
Generator overview.
Synchronous machine.
Dynamic models of generators for stability analysis:
Subtransient model
Transient model
Basic control models
Steady-state model.
June 26, 2008 Generator Modeling - 1
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Generator Overview
Generator components:
+ ++−− −
Excitation
system
Fuel
Power set-point Ref ω Ref V
Boiler
Firing
control
Turbine
Governor
Generator
Steam at pressure, P
Enthalpy, h
Torque
at speed, ω
Power at voltage, V
Current, I
Generator:
Synchronous machine: AC stator and DC rotor.
Excitation system: DC generator or static converter plus voltage
regulator and stabilizer
June 26, 2008 Generator Modeling - 2
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Generator Overview
Generator with DC exciter and associated controls:
Regulator powerM
Limitersensing
Stabilizer
Other
amplifierTransistor
Magneticamplifier
amplifierMagnetic
MG set 90−52
Compensator
PT’s
CT
sliprings
Regulatortransfer
M
Amplidynefield
dcExc
acGen
Exciter field rheostat(manual control)
PMG
sensing
Referenceand voltage
sensing
Otherinputs
Stationauxiliarypower
Field breakerCommutator
June 26, 2008 Generator Modeling - 3
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Generator Overview
Generator with static exciter and associated controls:
Synmachine
Stabilizer
Rectifiercurrentlimit
amplifierTrinistat power
Excitationbreaker
Excitationpower
Power rectifier
Linearreactor
slip rings
FDRtransformercurrent
Excitation power
Excitationpowerpotentialtransformer
Gatecircuitry adjuster
Base
Regulatortransfer
controlmanual
PT’s
Compensator
amplifiermixingSignal
Othersensing
sensingLimiter
Reference& voltagesensing
adjusterVoltage
Otherinputs
Voltagebuildupelement
power inputAuxiliary
start up CT
June 26, 2008 Generator Modeling - 4
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Synchronous Machine
θr
ωr
qr
dr
ar
F
F ′
D
D′
a
a′
b
b′
c
c′
Q1Q2
Q′1
Q′2
DC field
Damperwindings
Effects of induced currentsin the rotor core
June 26, 2008 Generator Modeling - 5
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Synchronous Machine
Electrical (inductor) equations:
v = ri +dλ
dt
λ = L(θr)i +dλ
dt
Te =1
2iT
dL(θr)
dθri
Mechanical (Newton’s) equations:
Jdωr
dt+ Dωr = Tm − Te
dθr
dt= ωr
June 26, 2008 Generator Modeling - 6
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Synchronous Machine
Stator equations:
v0s
vds
vqs
= −
rs
rs
rs
i0s
ids
iqs
−ωr
0
λqs
−λds
− d
dt
λ0s
λds
λqs
Rotor equations:
vF
0
0
0
=
rF
rD
rQ1
rQ2
iF
iD
iQ1
iQ2
+d
dt
λF
λD
λQ1
λQ2
June 26, 2008 Generator Modeling - 7
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Synchronous Machine
Magnetic flux equations:
λ0s
λds
λqs
λF
λD
λQ1
λQ2
=
L0
Ld Md Md
Lq Mq Mq
Md LF Md
Md Md LD
Mq LQ1 Mq
Mq Mq LQ2
i0s
ids
iqs
iF
iD
iQ1
iQ2
June 26, 2008 Generator Modeling - 8
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Synchronous Machine
Transformation equations:
vas
vbs
vcs
= PT
v0s
vds
vqs
ias
ibs
ics
= PT
i0s
ids
iqs
P =
√
2
3
1/√
2 1/√
2 1/√
2
cos θr cos(θr − 2π3 ) cos(θr + 2π
3 )
sin θr sin(θr − 2π3 ) sin(θr + 2π
3 )
June 26, 2008 Generator Modeling - 9
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Synchronous Machine
Mechanical equations:
2
pJ
d
dtωr +
2
pDωr = Tm − Te
d
dtθr = ωr
Te =p
2(iqsλds − idsλqs)
June 26, 2008 Generator Modeling - 10
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Synchronous Machine
3-phase short circuit at generator terminals:
t
t
ia(t) √2|Ea(0)|
x′′d
√2|Ea(0)|
x′d
√2|Ea(0)|
xd
Steadystate
Subtransient
Transient
component
component
Ea(0) is the open-circuit RMS phase voltage.
June 26, 2008 Generator Modeling - 11
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Synchronous Machine
Assuming balanced operation (null zero sequence), the detailed machine
model can be reduced to phasor models useful for stability and
steady-state analysis.
Phasor models are based on the following assumptions:
The rotor does not deviate “much” from the synchronous speed, i.e.
ωr ≈ ωs = (2/p)2πf0.
The rate of change in rotor speed is ”small”, i.e. |dωr/dt| ≈ 0
June 26, 2008 Generator Modeling - 12
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Synchronous Machine
Subtransient models capture the full machine electrical dynamics,
including the “few” initial cycles (ms) associated with the damper
windings.
Transient models capture the machine electrical dynamics starting with
the field and induced rotor core current transient response.
Damper windings transients are neglected. Steady-state models capture
the machine electrical response when all transients have disappeared
after “a few” seconds.
June 26, 2008 Generator Modeling - 13
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Subtransient Model
External phase voltages and currents:
v2as = v2
qs + v2ds
θvas= tan−1
(vds
vqs
)
+ δ
i2as = i2qs + i2ds
θias= tan−1
(ids
iqs
)
+ δ
June 26, 2008 Generator Modeling - 14
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Subtransient Model
Subtransient “internal” voltages associated with the damper windings (D
and Q1):
d
dte′′q =
1
T ′′d0
[e′q + (x′d − x′′
d)ids − e′′q ]
d
dte′′d =
1
T ′′q0
[e′d − (x′q − x′′
q )iqs − e′′d ]
e′′q − vqs = rsiqs − x′′dids
e′′d − vds = rsids + x′′q iqs
June 26, 2008 Generator Modeling - 15
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Subtransient Model
Transient “internal” voltages associated with the field (F ) and rotor-core
induced current windings (Q2):
d
dte′q =
1
T ′d0
[ef + (xd − x′d)ids − e′q]
d
dte′d =
1
T ′q0
[−(x′q − x′′
q )iqs − e′d]
e′q − vqs = rsiqs − x′dids
e′d − vds = rsids + x′qiqs
June 26, 2008 Generator Modeling - 16
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Subtransient Model
Steady-state equations:
ea − vqs = rsiqs − xdids
−vds = rsids + xqiqs
Mechanical equations:
d
dt∆ωr =
1
M[Pm − vasias cos(θvas
− θias) − D∆ωr]
d
dtδ = ∆ωr = ωr − ωs
June 26, 2008 Generator Modeling - 17
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Subtransient Model
The subtransient reactances (x′′q , x′′
d ) and open circuit time constants
(T ′′q0, T ′′
d0), as well as the transient reactances (x′q , x′
d) and open circuit
time constants (T ′q0, T ′
d0) are directly associated with the machine
resistances and inductances:
xd = ω0Ld = xℓ + xMd
xq = ω0Lq = xℓ + xMq
x′d = xℓ +
xMdxLF
xLF + xMd
x′q = xℓ +
xMqxLQ2
xLQ2 + xMq
x′′d = xℓ +
xMdxLF xLD
xMdxLF + xMdxLD + xLF xLD
x′′q = xℓ +
xMqxLQ1xLQ2
xMqxLQ1 + xMqxLQ2 + xLQ1xLQ2
June 26, 2008 Generator Modeling - 18
Power System Dynamics and Stability 35
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Subtransient Model
Some definitions:
xF = xLF + xMd
xD = xLD + xMd
xQ1 = xLQ1 + xMq
xQ2 = xLQ2 + xMq
June 26, 2008 Generator Modeling - 19
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Subtransient Model
Time constants:
T ′d0 =
xF
ω0rF
T ′q0 =
xQ2
ω0rQ2
T ′′d0 =
1
ω0rD
(
xLD +xMdxLD
xD
)
T ′′q0 =
1
ω0rQ1
(
xLQ1 +xMqxLQ1
xQ1
)
June 26, 2008 Generator Modeling - 20
Power System Dynamics and Stability 37
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Subtransient Model
Typical machine parameters:
2-pole 4-pole
Conventional Conductor Conventional Conductor
Cooled Cooled Cooled Cooled
xd 1.7-1.82 1.7-2.17 1.21-1.55 1.6-2.13
x′d .18-.23 .264-.387 .25-.27 .35-.467
x′′d .11-.16 .23-.323 .184-.197 .269-.32
xq 1.63-1.69 1.71-2.16 1.17-1.52 1.56-2.07
x′q .245-1.12 .245-1.12 .47-1.27 .47-1.27
x′′q .116-.332 .116-.332 .12-.308 .12-.308
T ′d0 7.1-9.6 4.8-5.36 5.4-8.43 4.81-7.73
T ′′d0 .032-.059 .032-.059 .031-.055 .031-.055
T ′q0 .3-1.5 .3-1.5 .38-1.5 .36-1.5
T ′′q0 .042-.218 .042-.218 .055-.152 .055-.152
xℓ .118-.21 .27-.42 .16-.27 .29-41
rs .00081-.00119 .00145-.00229 .00146-00147 .00167-00235
M 5-7 5-7 6-8 6-8
June 26, 2008 Generator Modeling - 21
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Subtransient Model
Typical machine parameters:
Salient-pole Combustion Synchronous
Dampers No dampers Turbines Compensator
xd .6-1.5 .6-1.5 1.64-1.85 1.08-2.48
x′d .25-.5 .25-.5 .159-.225 .244-.385
x′′d .13-.32 .2-.5 .102-.155 .141-.257
xq .4-.8 .4-.8 1.58-1.74 .72-1.18
x′q = x′
q = x′q .306 .57-1.18
x′′q .135-.402 .135-.402 .1 .17-.261
T ′d0 4-10 8-10 4.61-7.5 6-16
T ′′d0 .029-.051 .029-.051 .054 .039-.058
T ′q0 − − 1.5 .15
T ′′q0 .033-.08 .033-.08 .107 .188-.235
xℓ .17-.4 .17-.4 .113 .0987-.146
rs .003-.015 .003-.015 .034 .0017-006
M 6-14 6-14 18-24 2-4
June 26, 2008 Generator Modeling - 22
Power System Dynamics and Stability 39
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Subtransient Model
In practice most of these constants are determined from short-circuit
tests.
All the e voltages are “internal” machine voltages directly assocated with
the “internal” phase angle δ.
The internal field voltage ef is directly proportional to the actual field dc
voltage vF , and is typically controlled by the voltage regulator.
The mechanical power Pm is controlled through the governor.
June 26, 2008 Generator Modeling - 23
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Subtransient Model
A simple voltage regulator model (IEEE type 1):
+ +
+−
−
−
vref
V
vm
Ka
Kf s
vfvr
vr max
vr min1
1
Trs + 1
Tas + 1
Tf s + 1
Tes + 1
Se
June 26, 2008 Generator Modeling - 24
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Subtransient Model
A static voltage regulator model:
+
+
+
+
+
+
−
−
−
−
vref
VTVC
VS
IT
If
If ref
Ef
Vuxl
V ∗efl
VA max
VA min
Vmax
Vmin
HVGate
0
Vc = |Vt| + (Rc + jXc)It
1+sTC1+sTB
1+sTC11+sTB1
11+sTR
KA1+sTA
KIR
sKF1+sTF
|VT |VR min
|VT |VR max − KC If
June 26, 2008 Generator Modeling - 25
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Subtransient Model
A simple governor model (hydraulic valve plus turbine):
Governor Servo Reheat
+
+
+
−
ωref
ω
1/R1
Torder
T∗in Tin
Tmin
Tmax
Tmech
Tss + 1
T3s + 1
Tcs + 1
T4s + 1
T5s + 1
June 26, 2008 Generator Modeling - 26
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Subtransient Model
A governor-steam turbine model:
K1
− +
+
−
+
−
+
++
+
+
1
1 + st_LP1+sT_IP
1
F_HP
F_LP
F_IP P_MAX
0.0
1.0
1
sT_C3
11
1 + sT_C2
1 + sT_HP
1
w_ref P_ref
P_mech
11 − K_RH
sT_RH
K_HP
1
1+ sT_R T_C1
1 + sT_1
1 + sT_2
A
P_GV
June 26, 2008 Generator Modeling - 27
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Example
For a 200 MVA, 13.8 kV, 60 Hz generator with the following p.u. data:
rs = 0.001096
xℓ = 0.15
xd = 1.7 ⇒ Ld = 0.00451 Md = 0.00411
x′d = 0.238324 ⇒ LF = 0.00438
x′′d = 0.184690 ⇒ LD = 0.00426
xq = 1.64 ⇒ Lq = 0.00435 Mq = 0.00395
x′q = xq ⇒ LQ2 = 0
x′′q = 0.185151 ⇒ LQ1 = 0.00405
T′d0 = 6.194876 ⇒ rF = 0.000742
T′′d0 = 0.028716 ⇒ rD = 0.0131
T′q0 = 0 ⇒ rQ2 = 0
T′′q0 = 0.074960 ⇒ rQ1 = 0.0540
p = 2 M = 10 D = 0
June 26, 2008 Generator Modeling - 28
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Example
A three-phase fault from open circuit conditions, i.e. before the fault
vas = 13.8/√
3 kV and ias = 0, and after the fault vas = 0, is
simulated using the detailed machine equations:
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2−2
−1
0
1
2
t [s]
va
s,v
bs
,vcs
[kV
]
June 26, 2008 Generator Modeling - 29
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Example
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11
2
3
4
5
6
7
8
9
10
t [s]
i F[p
.u.]
June 26, 2008 Generator Modeling - 30
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Example
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−20
0
20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−20
0
20
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−20
0
20
i a[p
.u.]
i b[p
.u.]
i c[p
.u.]
t [s]
t [s]
t [s]
June 26, 2008 Generator Modeling - 31
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Transient Model
Obtained by eliminating electromagnetic differential equations and the
damper winding dynamic equations.
For this reason, the damping D in the mechanical equations, which is
typically a small value, is assumed to be large to indirectly model the
significant damping effect of these windings on ωr .
It is the typical model used in stability studies.
June 26, 2008 Generator Modeling - 32
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Transient Model
Neglecting the induced currents in the rotor (winding Q2):
x′q = xq T ′
q0 = 0 ⇒ e′d = 0
This leads to the transient equations:
d
dte′a =
1
T ′d0
[ef + (xd − x′d)ids − e′a]
=1
T ′d0
[ef − ea]
e′a∠δ = vas∠θvas+ rsias∠θias
+jx′dids∠(δ + π/2) + jxqiqs∠δ
ea∠δ = vas∠θvas+ rsias∠θias
+jxdids∠(δ + π/2) + jxqiqs∠δ
June 26, 2008 Generator Modeling - 33
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Transient Model
The phasor diagram in this case is:
e′a = vas + rsias + jx′dids + jxqiqs
ea = vas + rsias + jxdids + jxqiqs
iqs
ids
vas
ias
δ
ℑ
ℜ
e′a
ea
rsiasjxdids
jxqiqs
j(xd − x′d)ids
June 26, 2008 Generator Modeling - 34
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Transient Model
For faults near the generator terminals, the q axis has little effect on the
system response, i.e. iqs ≈ 0.
This results in the classical voltage source and transient reactance
generator model used in simple stability studies:
e′a∠δ
ias∠θias
vas∠θvas
rs jx′d
+
−
Pmωr
June 26, 2008 Generator Modeling - 35
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Transient Model
A further approximation in some cases is used by neglecting the field
dynamics, i.e. T ′d0 = 0.
In this case, e′a is a “fixed” variable controlled directly through the voltage
regulator via ef .
The limits in the voltage regulator are used to represent limits in the field
and armature currents.
These limits can be “soft”, i.e. allowed to temporarily exceed the hard
steady-state limits, to represent under- and over-excitation.
June 26, 2008 Generator Modeling - 36
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Steady-state model
When all transient are neglected, the generator model becomes:
ea∠δ = vas∠θvas+ rsias∠θias
+jxdids∠(δ + π/2) + jxqiqs∠δ
ea = ef
ias∠θias= ids∠δ + jiqs∠δ
June 26, 2008 Generator Modeling - 37
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Steady-state model
For a round rotor machine (xd = x′d), the steady-state model leads to
the classical short circuit generator model:
ea∠δ
ias∠θias
vas∠θvas
rs jxd
+
−
Pmωr
June 26, 2008 Generator Modeling - 38
Power System Dynamics and Stability 55
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Steady-state model
Based on this simple model, the field and armature current limits can be
used to define the generator capability curves (for a given terminal
voltage vas = vt):
ef limit
ia limitvtea
x
− v2t
x
vti∗a
P
PmaxPmin
Q
Qmax
Qmin
June 26, 2008 Generator Modeling - 39
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Steady-state model
Considering the voltage regulator effect, the generator can be modeled
as a constant terminal voltage within the generator reactive power
capability, delivering constant power (Pm).
This yields the PV generator model for power flow studies:
P
QV
where P = Pm = constant, and V = vt = constant for
Qmin ≤ Q ≤ Qmax; otherwise, Q = Qmax,min and V is allowed to
change.
June 26, 2008 Generator Modeling - 40
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Example
The generator data are:
PGj0.1
v∞ = 1∠0va
θr(t) = ω0t + π/2 + δ
xd = xq = 0.9, x′d = 0.2
T ′d0 = 2 s
M = large ⇒ δ = constant
The generator is operating in steady-state delivering PG = 0.5 at
ea = 1.5. At t = 0 there is a fault and the line is disconnected. Find
va(t) for t > 0.
June 26, 2008 Generator Modeling - 41
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Example
Steady-state conditions:
ea∠δ = vas∠θvas + (rs + jxd)ias∠θias
= v∞ + j(xd + xL)ias
⇒ ias =1.5∠δ − 1
j(0.9 + 0.1)
= 1.5 sin δ − j(1.5 cos δ − 1)
PG = ℜv∞i∗as = ℜi∗as= 1.5 sin δ
⇒ δ = sin−1(PG/1.5) = 19.47
ias = 0.65∠ − 39.64
June 26, 2008 Generator Modeling - 42
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Example
iqs = ias cos(θias− δ)∠δ = 0.334∠19.47
ids = ias sin(θias− δ)∠(δ + π/2) = 0.558∠ − 70.53
⇒ ea∠δ = vas + rsias + jx′dids + jxqiqs
= v∞ + j(xd + xL)ias + jx′dids + jxqiqs
= 1 + j0.1(0.65∠ − 39.64) + j0.2(0.558∠ − 70.53)
j0.9(0.334∠19.47)
= 1.110∠19.47
d
dte′a =
1
T ′d0
[ea + (xd − x′d)ids − e′a]
June 26, 2008 Generator Modeling - 43
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Example
Transient:
d
dte′a(t) =
1
T ′d0
[ef + (xd − x′d)ids − e′a(t)]
ef = ea → steady-state
= 1.5
ids = 0 → open line
e′a(0) = 1.110
⇒ d
dte′a(t) =
1
2[1.5 − e′a(t)]
June 26, 2008 Generator Modeling - 44
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Example
Solution:
d
dte′a(t) = −0.5e′a(t) + 0.75
⇒ e′a(t) = Ae−0.5t + B
e′a(0) = A + B = 1.110
e′a(∞) = B = 0.75/0.5 = 1.5
⇒ e′a(t) = −0.390e−0.5t + 1.5
va(t) = e′a(t) → since ias = ids = iqs = 0
t
e′a
1.5
1.11
0
June 26, 2008 Generator Modeling - 45
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Transmission System Modeling
Transformers:
Single phase:
Detailed model
Phasor model
Three phase:
Phase shifts
Models
June 26, 2008 Transmission System Modeling - 1
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Transmission System Modeling
Transmission Lines:
Single phase:
Distributed parameter model
Phasor lumped model
Three phase:
Distributed parameter model
Reduced models
Underground cables
June 26, 2008 Transmission System Modeling - 2
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Single-phase Transformers
The basic characteristics of this device are:
Flux leakage around the transformer windings is represented by a
leakage inductance Lℓ.
The core is made of magnetic material and is represented by a
magnetization inductance (Lm ≫ Lℓ), but saturates.
Losses in the windings (Cu wires) and core (hysteresis and induced
currents) are represented with lumped resistances (r and Gm).
Steps up or down the voltage/current depending on the turn ratio
a = N1/N2 = V1/V2.
June 26, 2008 Transmission System Modeling - 3
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Single-phase Transformers
++
v1
i1
v2
i2
N1 N2
λm
λℓ1 λℓ2
leakage
core
magnetizing
June 26, 2008 Transmission System Modeling - 4
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Single-phase Transformers
The equivalent circuit is:
v = ri +dλ
dt= ri + L
di
dt
⇒
v1
v2
=
r1
r2
i1
i2
+
Lℓ1 + Lm Lm/a
Lm/a Lℓ2 + Lm/a2
d
dt
i1
i2
June 26, 2008 Transmission System Modeling - 5
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Single-phase Transformers
Equivalent circuit:
++ ++
− −−−
N1 : N2
a : 1
aii
e e/av1 v2
r1 r2
im
i1 i2
Gm
Lm
Lℓ1 Lℓ2
June 26, 2008 Transmission System Modeling - 6
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Single-phase Transformers
The phasor equivalent circuit is:
V1 = (Zℓ1 + Zm)I1 + Zm1
aI2
aV2 = ZmI1 + (a2Zℓ1 + Zm)1
aI2
⇒
V1
V2
=
Zℓ1 + Zm Zm/a
Zm/a Zℓ2 + Zm/a2
I1
I2
June 26, 2008 Transmission System Modeling - 7
Power System Dynamics and Stability 69
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Single-phase Transformers
Phasor equivalent circuit:
+++
−−−
a : 1
V1 V2aV2
I1 I2I2/a
Im
Ym
Zℓ1 a2Zℓ2
June 26, 2008 Transmission System Modeling - 8
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Single-phase Transformers
This can also be readily transformed into a ABCD input-output form based
on the following approximation, since Zm ≫ Zℓ1 (Zℓ1 ≈ a2Zℓ2):
V1
I1
=
a(1 + ZℓYm) Zℓ/a
aYm 1/a
V2
−I2
=
A B
C C
V2
−I2
Zℓ = r1 + jXℓ1 + a2(r2 + jXℓ2)
June 26, 2008 Transmission System Modeling - 9
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Single-phase Transformers
Phasor equivalent circuit with the approximation Zm ≫ Zℓ1:
+ ++
− −−
a : 1
V1 V2aV2
I1 I2I2/a
Im
Ym
Zℓ
June 26, 2008 Transmission System Modeling - 10
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Single-phase Transformers
Or Π form:
Z =Zℓ
a
Y1 = (1 − a)1
Zℓ
Y2 = (a2ZℓYm + a2 + a)1
Zℓ
≈ (a2 − a)1
Zℓfor Ym ≈ 0
June 26, 2008 Transmission System Modeling - 11
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Single-phase Transformers
Π equivalent circuit:
++
−−
V1 V2
I1 I2
Y1 Y2
Z
June 26, 2008 Transmission System Modeling - 12
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Single-phase Transformers
Neglecting Zm (Ym is small given the core magnetic properties):
1. Time domain:
v1 = ri1 + Ldi1dt
+ av2
i2 = −ai1
2. Phasor domain:
V1 = ZℓI1 + aV2
I2 = −aI1
June 26, 2008 Transmission System Modeling - 13
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Single-phase Transformers
Equivalent circuit neglecting Zm:
+ ++
−−
a : 1
V1 V2aV2
I1 I2 = −I1/aZℓ
June 26, 2008 Transmission System Modeling - 14
Power System Dynamics and Stability 76
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Single-phase Transformers
Certain transformers have built-in Under-Load Tap Changers (ULTC).
This is either operated manually (locally or remote controlled) or
automatically with a voltage regulator; the voltage control range is limited
(≈ 10%) and on discrete steps (≈ 1%).
The time response is in the order of minutes, with 1-2 min. delays, due to
ULTCs being implemented using electromechanical systems.
June 26, 2008 Transmission System Modeling - 15
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Single-phase Transformers
These are typically used to control the load voltage side, and hence are
used at subtransmission substations.
Nowadays, power electronic switches are used, leading to Thyristor
Controller Voltage Regulators (TCVR), which are faster voltage controllers
and are considered Flexible AC Transmission systems (FACTS).
These types of transformers are modelled using the same transformer
models, but a may be assumed to be a discrete controlled variable
through a voltage regulator with a dead-band.
June 26, 2008 Transmission System Modeling - 16
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Single-phase Transformers
Transformers with special connections and under-load tap changers can
also be used for phase shift control and are known as Phase Shifters.
These control the phase shift difference between the two terminal
voltages within approximately ±30, thus increasing the power capacity
of a transmission line (e.g. interconnection between Ontario and
Michigan).
Phase shifters are modeled using a similar model but the tap ratio is a
phasor as opposed to a scalar:
a = a∠α
A Π equivalent cannot be used in this case.
June 26, 2008 Transmission System Modeling - 17
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Single-phase Transformers
Phasor equivalent circuit with complex tap ratio a:
+ ++
− −−
a : 1
V1 V2aV2
I1 I2I2/a∗
Im
Ym
Zℓ
June 26, 2008 Transmission System Modeling - 18
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Single-phase Transformers (Example)
Single phase 8/80 kV, 30 MVA transformer with Xℓ = 10% and
Xm ≈ 10Xℓ.
Detailed model parameters:
a = 8/80 = 0.1
Lℓ1 =Xℓ1
ω0≈ Xℓ
2ω0
=0.1
2 · 377
(8 kV)2
30 MVA= 0.283 mH
Lℓ2 =Lℓ1
a2= 28.3 mH
Lm = 10Xℓ
ω0= 10
0.1
377
(8 kV)2
30 MVA= 5.66 mH
June 26, 2008 Transmission System Modeling - 19
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Single-phase Transformers (Example)
Per unit parameters:
Zℓ = j0.1
Zm = 10Zℓ = j1
Ym =1
Zm= −j1
a =8/8 kV
80/80 kV= 1
June 26, 2008 Transmission System Modeling - 20
Power System Dynamics and Stability 82
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Single-phase Transformers (Example)
Per unit parameters:
A = 1(1 + j0.1(−j1)) = 1.1
B =j0.1
1= j0.1
C = 1(−j1) = −j1
D =1
1= 1
June 26, 2008 Transmission System Modeling - 21
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Single-phase Transformers (Example)
Per unit parameters:
Z =j0.1
1= j0.1
Y1 = (1 − 1)1
j0.1= 0
Y2 = (12(j0.1)(−j1) + 12 − 1)1
j0.1= −j1
June 26, 2008 Transmission System Modeling - 22
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Three-phase Transformers
+
+
+
+
+
+
− −
−
−−
−
a : 1
a : 1
a : 1
T1
T2
T3
a1 a2
b1 b2
c1 c2
aVa2
aVb2
aVc2
Va2
Vb2
Vc2
June 26, 2008 Transmission System Modeling - 23
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Three-phase Transformers
The 3 single-phase transformers form a 3-phase bank that induces a
phase shift, depending on the connection:
Vab1 = aVa2
Vab2 =√
3∠30Va2
⇒ Vab1 =√
3a∠30Vab2
a =√
3a∠30
apu =√
3apu∠30
June 26, 2008 Transmission System Modeling - 24
Power System Dynamics and Stability 86
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Three-phase Transformers
∆ − Y : a =√
3a∠30
apu =√
3apu∠30
Y − ∆ : a =√
3a∠ − 30
apu =√
3apu∠ − 30
Y − Y : a = a
∆ − ∆ : a = a
June 26, 2008 Transmission System Modeling - 25
Power System Dynamics and Stability 87
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Three-phase Transformers
In balanced, “normal” systems, the net phase shift between the
generation and load sides is zero, and hence is neglected during system
analyses:
Generator
sideside
Load
∆∆ Y Y
In these systems, the p.u. per-phase models of the transformers are
identical to the equivalent single-phase transformer models.
June 26, 2008 Transmission System Modeling - 26
Power System Dynamics and Stability 88
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Three-phase Transformers
For “smaller” transformers (e.g. load transformers), integral designs are
preferred to transformers banks:
core
3 φ windows
a b c
vabc1
vabc2
=
r13×3
r23×3
iabc1
iabc2
+
L113×3
L123×3
L213×3L223×3
d
dt
iabc1
iabc2
June 26, 2008 Transmission System Modeling - 27
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Saturation
The magnetization inductance Lm changes with the magnetization
current due to saturation of the magnetic core.
Saturation occurs due to a reduction on the number of “free” magnetic
dipoles in the enriched core.
This results in the core behaving more like air than a magnet, i.e.
magnetic “conductivity” decreases.
It is typically represented using a piece-wise linear model.
June 26, 2008 Transmission System Modeling - 28
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Saturation
The magnetization inductance Lm changes with the magnetization
current due to saturation of the magnetic core:
λm
Lm1
Lm2
imims
Lm(im) =
Lm1 for im ≤ ims
Lm2 for im > ims
June 26, 2008 Transmission System Modeling - 29
Power System Dynamics and Stability 91
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Single-phase Transmission Line
Alossless line can be represented using a series of lumped elements:
l =µ0
2πln
D
R′
c =2πǫ0
ln(D/R)
D → distance between wires
R′ → wire GMR
R → wire radius
µ0 = 4π × 10−7 [H/m]
ǫ0 = 8.854 × 10−12 [F/m]
June 26, 2008 Transmission System Modeling - 30
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Single-phase Transmission Line
l [H/m]
c [F/m]
x dx
i
vv1 v2
i1 i2
didv+ +++
− −
−
−
June 26, 2008 Transmission System Modeling - 31
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Single-phase Transmission Line
The equations for this line are:
dv = −ldxdi
dt⇒ ∂v
∂x= −l
∂i
∂t
di = −cdxdv
dt⇒ ∂i
∂x= −c
∂v
∂t
These are D’Alambert equations with solution:
i1(t) =1
Zcv1(t) −
[
i2(t − τ) +1
Zcv2(t − τ)
]
︸ ︷︷ ︸
I2(t−τ)
i2(t) =1
Zcv2(t) −
[
i1(t − τ) +1
Zcv1(t − τ)
]
︸ ︷︷ ︸
I1(t−τ)
June 26, 2008 Transmission System Modeling - 32
Power System Dynamics and Stability 94
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Single-phase Transmission Line
where:
Zc =√
l/c → chracteristic impedance
s =1√lc
→ wave speed
τ =d
s→ travelling time for line length
Distributed parameter equivalent circuit:
r/2r/2
Zc Zcv1 v2
i1 i2
I1(t − τ)I2(t − τ)++
−−
June 26, 2008 Transmission System Modeling - 33
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Single-phase Transmission Line
Example:
E
t = 0
v1 v2
i1 i2 = 0
Trans.
Line
+ + +
−−−
v1(t) = E
i2(t) = 0
June 26, 2008 Transmission System Modeling - 34
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Single-phase Transmission Line
Example:
t I1(t − τ) I2(t − τ) i1 i2
0 0 0 E/Zc 0
τ 2E/Zc 0 E/Zc 2E
2τ 2E/Zc 2E/Zc −E/Zc 2E
3τ 0 2E/Zc −E/Zc 0
4τ 0 0 E/Zc 0
5τ 2E/Zc 0 E/Zc 2E
June 26, 2008 Transmission System Modeling - 35
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Single-phase Transmission Line
Example:
2E
E/Zc
−E/Zc
t
t
v2
i1
τ
τ
2τ
2τ
3τ
3τ
4τ
4τ
5τ
5τ
6τ
6τ
June 26, 2008 Transmission System Modeling - 36
Power System Dynamics and Stability 98
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Single-phase Transmission Line
Phasor model from the distributed line model:
d
dxV = −(r + jωl)I = −zI
d
dxI = −(jωc)V = −yI
⇒ d
dx
V
I
=
0 −z
−y 0
V
I
June 26, 2008 Transmission System Modeling - 37
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Single-phase Transmission Line
The solution to this set of linear dynamical equations is:
d
dx
V1
I1
=
cosh γd Zc sinh γd
1/Zc sinh γd cosh γd
V2
−I2
where
cosh γd =eγd + e−γd
2
sinh γd =eγd − e−γd
2γ =
√zy → propagation constant
Zc =√
z/y → characteristic impedance
June 26, 2008 Transmission System Modeling - 38
Power System Dynamics and Stability 100
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Single-phase Transmission Line
This can be converted into the Π equivalent circuit:
++
−−
V1 V2
I1 I2
Y′
1/2 Y′
2/2
Z′
June 26, 2008 Transmission System Modeling - 39
Power System Dynamics and Stability 101
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Single-phase Transmission Line
Π equivalent circuit:
Z′ = zd︸︷︷︸
Z
sinh γd
γd
Y′ = yd︸︷︷︸
Y
tanh γd
γd
for d < 250 km ⇒ Z′ ≈ Z Y′ ≈ Y
for d < 100 km ⇒ Z′ ≈ Z Y′ ≈ 0
June 26, 2008 Transmission System Modeling - 40
Power System Dynamics and Stability 102
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Three-phase Transmission Line
nn
a b c
dα
2R
dβ
dγ
d × d
dδ
June 26, 2008 Transmission System Modeling - 41
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Three-phase Transmission Line
Typically the phase wires are bundled (e.g. 4 wires/phase) and the guard
wires are grounded at every tower, i.e. correspond to the neutral.
For a total of N wires, the per unit length equations are:
∂
∂x
2
6
6
6
6
6
6
4
v1
v2
.
.
.
vN
3
7
7
7
7
7
7
5
=
2
6
6
6
6
6
6
4
r11 r12 . . . r1N
r21 r22 . . . r2N
.
.
....
. . ....
rN1 rN2 . . . rNN
3
7
7
7
7
7
7
5
2
6
6
6
6
6
6
4
i1
i2...
iN
3
7
7
7
7
7
7
5
+
2
6
6
6
6
6
6
4
l11 l12 . . . l1N
l21 l22 . . . l2N
.
.
....
. . ....
lN1 lN2 . . . lNN
3
7
7
7
7
7
7
5
∂
∂t
2
6
6
6
6
6
6
4
i1
i2...
iN
3
7
7
7
7
7
7
5
June 26, 2008 Transmission System Modeling - 42
Power System Dynamics and Stability 104
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Three-phase Transmission Line
−∂
∂x
2
6
6
6
6
6
6
4
i1
i2...
iN
3
7
7
7
7
7
7
5
=
2
6
6
6
6
6
6
4
g11 g12 . . . g1N
g21 g22 . . . g2N
.
.
....
. . ....
gN1 gN2 . . . gNN
3
7
7
7
7
7
7
5
2
6
6
6
6
6
6
4
v1
v2
.
.
.
vN
3
7
7
7
7
7
7
5
+
2
6
6
6
6
6
6
4
p11 p12 . . . p1N
p21 p22 . . . p2N
.
.
....
. . ....
pN1 pN2 . . . pNN
3
7
7
7
7
7
7
5
−1
∂
∂t
2
6
6
6
6
6
6
4
v1
v2
.
.
.
vN
3
7
7
7
7
7
7
5
cN×N = p−1N×N
June 26, 2008 Transmission System Modeling - 43
Power System Dynamics and Stability 105
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Three-phase Transmission Line
In phasor form, these equations are:
d
dx
2
6
6
6
6
6
6
4
V1
V2
.
.
.
VN
3
7
7
7
7
7
7
5
= −
2
6
6
6
6
6
6
4
z11 z12 . . . z1N
z21 z22 . . . z2N
.
.
....
. . ....
zN1 zN2 . . . zNN
3
7
7
7
7
7
7
5
2
6
6
6
6
6
6
4
I1
I2
.
.
.
IN
3
7
7
7
7
7
7
5
d
dx
2
6
6
6
6
6
6
4
I1
I2
.
.
.
IN
3
7
7
7
7
7
7
5
= −
2
6
6
6
6
6
6
4
y11 y12 . . . y1N
y21 y22 . . . y2N
.
.
....
. . ....
yN1 yN2 . . . yNN
3
7
7
7
7
7
7
5
2
6
6
6
6
6
6
4
V1
V2
.
.
.
vN
3
7
7
7
7
7
7
5
⇒ d
dxV = −[z]I
d
dxI = −[y]V
June 26, 2008 Transmission System Modeling - 44
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Three-phase Transmission Line
The image method is useful for computing line parameters:
i
i
j
j
2Ri
2Rj
dij
Dij
hi
hi
hj
hj
φij
Earth
ρ [Ωm]
image
image
June 26, 2008 Transmission System Modeling - 45
Power System Dynamics and Stability 107
Universidad de Castilla-La Mancha
Three-phase Transmission Line
The line parameters zij and yij can be computed using Carson’s
formulas:
rii = riint + ∆rii riint → from tables
rij = ∆rij
∆rij = 4ω10−4π/8 − b1a cos φ
+b2[(c2 − ln a)a2 cos 2φ + φa2 sin 2φ]
+b3a3 cos 3φ − d4a
4 cos 4φ − b5a5 cos 5φ
+b6[(c6 − ln a)a6 cos 6φ + φa6 sin 6φ]
+b7a7 cos 7φ − d8a
8 cos 8φ − b9a9 cos 9φ + . . .
a = 4π√
510−4D√
f/ρ
June 26, 2008 Transmission System Modeling - 46
Power System Dynamics and Stability 108
Universidad de Castilla-La Mancha
Three-phase Transmission Line
D =
2hi for i = j
Dij for i 6= jφ =
0 for i = j
φij for i 6= j
b1 =
√2
6b2 =
1
16
bk = bk−2s
k(k + 2)dk =
π
4bk
c2 = 1.3659315 ck = ck−2 +1
k+
1
k + 2
s =
+1 k = 1, 2, 3, 4, 5, . . .
−1 k = 5, 6, 7, 8, 13, . . .
June 26, 2008 Transmission System Modeling - 47
Power System Dynamics and Stability 109
Universidad de Castilla-La Mancha
Three-phase Transmission Line
xii = ωµ0
2πln
2hi
Ri+ xiint + ∆xii xiint → from tables(≈ 0)
xij = ωµ0
2πln
Dij
dij+ ∆xij
∆xij = 4ω10−41/2(0.6159315 − ln a) − b1a cos φ − d2a2 cos 2φ
+b3a3 cos 3φ − b4[(c4 − ln a)a4 cos 4φ + φa4 sin 4φ]
+b5a5 cos 5φ − d6a
6 cos 6φ + b7a7 cos 7φ
−b8[(c8 − ln a)a8 cos 6φ + φa8 sin 8φ] + . . .gij = gij ≈ 0
pii =1
2πǫ0ln
2hi
Ripij =
1
2πǫ0ln
2Dij
dij
June 26, 2008 Transmission System Modeling - 48
Power System Dynamics and Stability 110
Universidad de Castilla-La Mancha
Three-phase Transmission Line
f (Hz) R′ac/R′
dc L′ac/L′
dc f (Hz) R′ac/R′
dc L′ac/L′
dc
2 1.0002 0.99992 4000 7.1876 0.15008
4 1.0007 0.99970 6000 8.7471 0.12258
6 1.0015 0.99932 8000 10.0622 0.10617
8 1.0026 0.99879 10000 11.2209 0.09497
10 1.0041 0.99812 20000 15.7678 0.06717
20 1.0164 0.99254 40000 22.1988 0.04750
40 1.0632 0.97125 60000 27.1337 0.03879
60 1.1347 0.93898 80000 31.2942 0.03359
80 1.2233 0.89946 100000 34.9597 0.03004
100 1.3213 0.85639 200000 49.3413 0.02124
200 1.7983 0.66232 400000 69.6802 0.01502
400 2.4554 0.47004 600000 85.2870 0.01227
600 2.9421 0.38503 800000 98.4441 0.01062
800 3.3559 0.33418 1000000 110.0357 0.00950
1000 3.7213 0.29924 2000000 155.5154 0.00672
2000 5.1561 0.21204 4000000 219.8336 0.00475
June 26, 2008 Transmission System Modeling - 49
Power System Dynamics and Stability 111
Universidad de Castilla-La Mancha
Three-phase Transmission Line
The [z] parameters depend on the line frequqnecy ω, i.e. this is a
frequency dependent model.
These number of conductors, and hence equations, can be reduced
based on the following observations:
The voltage of all Nb conductors in a phase bundled are at the same
voltage (e.g. v1 = v2 = · · · = vNb = va).
The current in each phase is shared approximately equally by each
conductor in the bundle (e.g. i1 = i2 = · · · = iNb = ia/Nb).
The voltage in the guard wires is zero (e.g. vg = 0).
June 26, 2008 Transmission System Modeling - 50
Power System Dynamics and Stability 112
Universidad de Castilla-La Mancha
Three-phase Transmission Line
This reduces the matrices to:
[z]N×N →
zaa zab zac
zab zbb zbc
zac zbc zcc
[y]N×N →
yaa yab yac
yab ybb ybc
yac ybc ycc
⇒ d
dtVabc = −[zabc]Iabc
d
dtIabc = −[yabc]Vabc
June 26, 2008 Transmission System Modeling - 51
Power System Dynamics and Stability 113
Universidad de Castilla-La Mancha
Three-phase Transmission Line
A line is transposed to balance the phases.
The length of the barrel B must be much less than the wavelength
(s/f ≈ 5000 km @ 60 Hz). B ≈ 50 km.
a
b
c
1
1
1
2
2
2 3
3
3
B
B/3B/3B/3
[[zabc] =
zaa zab zac
zab zbb zbc
zac zbc zcc
[yabc] =
yaa yab yac
yab ybb ybc
yac ybc ycc
June 26, 2008 Transmission System Modeling - 52
Power System Dynamics and Stability 114
Universidad de Castilla-La Mancha
Three-phase Transmission Line
The phasor equations can be diagonalized using eigenvalue (modal)
analysis techniques:
[zm] = TTI [z]N×NTI (diagonal matrix)
[ym] = TTV [z]N×NTV (diagonal matrix)
Vm = T−1V V = TT
I V
Im = T−1I I = TT
V I
There are N modes of propogation, one for each eigenvalue.
June 26, 2008 Transmission System Modeling - 53
Power System Dynamics and Stability 115
Universidad de Castilla-La Mancha
Three-phase Transmission Line
Diagonalization of a 3-phase transposed line through sequence
transformation:
V0pn = T−1S Vabc
I0pn = T−1S Iabc
TS =1√3
1 1 1
1 a2 a
1 a a2
⇒ T−1S =
1√3
1 1 1
1 a a2
1 a2 a
a = 1∠120
June 26, 2008 Transmission System Modeling - 54
Power System Dynamics and Stability 116
Universidad de Castilla-La Mancha
Three-phase Transmission Line
Transformation of zabc into z0pn:
[z0pn] = T−1S [zabc]TS
=
zs + 2zm 0 0
0 zs − zm 0
0 0 zs − zm
=
z0 0 0
0 zp 0
0 0 zn
zp = zn z0 ≈ 3zp
Similar for [y0pn] = T−1S [yabc]TS .
June 26, 2008 Transmission System Modeling - 55
Power System Dynamics and Stability 117
Universidad de Castilla-La Mancha
Three-phase Transmission Line
Diagonalization of a 3-phase transposed line through 0αβ
transformation:
V0αβ = T−1Vabc
I0αβ = T−1Iabc
T =1√3
1√
2 0
1 −1/√
2√
3/2
1 −1/√
2 −√
3/2
⇒ T−1 = TT
June 26, 2008 Transmission System Modeling - 56
Power System Dynamics and Stability 118
Universidad de Castilla-La Mancha
Three-phase Transmission Line
Transformation of zabc into z0αβ :
[z0αβ ] = TT [zabc]T
=
zs + 2zm 0 0
0 zs − zm 0
0 0 zs − zm
=
z0 0 0
0 zα 0
0 0 zβ
zα = zβ = zp z0 ≈ 3zα
Similar for [y0αβ ] = TT [yabc]T .
June 26, 2008 Transmission System Modeling - 57
Power System Dynamics and Stability 119
Universidad de Castilla-La Mancha
Three-phase Transmission Line
These simplifications lead to the following per-phase (positive sequence),
per-unit length formulas:
r =rtables
Nb
l =µ0
2πln
Dm
R′b
c =2πǫ0
ln Dm
Rb
June 26, 2008 Transmission System Modeling - 58
Power System Dynamics and Stability 120
Universidad de Castilla-La Mancha
Three-phase Transmission Line
where:
Dm is the GMD of the 3 phases:
Dm = 3√
dabdacdbc
R′b is the GMR of the bundled and wires:
R′b = Nb
√
R′d12d13 · · · d1Nb
Rb is the GMR of the bundled:
Rb = Nb
√
Rd12d13 · · · d1Nb
June 26, 2008 Transmission System Modeling - 59
Power System Dynamics and Stability 121
Universidad de Castilla-La Mancha
Example 1
24.14 km transposed distributed line:
0.741 Ω
251.2 Ω
70.68 mH
70.16 mH
219.1 mH325/
√3 kV
Bus 1 Bus 2 Bus 315 miles
1 ms
R′
0 = 0.3167 Ω/km R′
1 = 0.0243 Ω/km
L′
0 = 3.222 mH/km L′
1 = 0.9238 mH/km
C′
0 = 0.00787 µF/km C′
1 = 0.0126 µF/km
June 26, 2008 Transmission System Modeling - 60
Power System Dynamics and Stability 122
Universidad de Castilla-La Mancha
Example 1
2
664
187.79 cos(377t)
187.79 cos(377t − 2π/3)
187.79 cos(377t + 2π/3)
3
775
=
2
664
vBUS1a
vBUS1b
vBUS1c
3
775
+
2
664
0.714 0 0
0 0.714 0
0 0 0.714
3
775
2
664
i1a
i1b
i1c
3
775
+d
dt
2
664
0.07068 0 0
0 0.07068 0
0 0 0.07068
3
775
2
664
i1a
i1b
i1c
3
775
2
664
vBUS10
vBUS1α
vBUS1β
3
775
=1√
3
2
664
1 1 1√
2 −1/√
2 −1/√
2
0p
3/2 −p
3/2
3
775
| z
T−1
2
664
vBUS1a
vBUS1b
vBUS1c
3
775
2
664
i10
i1α
i1β
3
775
= T−1
2
664
i1a
i1b
i1c
3
775
June 26, 2008 Transmission System Modeling - 61
Power System Dynamics and Stability 123
Universidad de Castilla-La Mancha
Example 1
i10(t) =1
Zc0
[
vBUS10 −r0
2i10(t)
]
− I20(t − τ0)
i20(t) =
1
Zc0
[
vBUS20− r0
2i20
(t)]
− I10(t − τ0)
r0 = 0.3167 × 24.14 = 7.6451 Ω
ZC0=
√
3.222 × 10−3
0.00787 × 10−6= 639.85 Ω
τ0 = 24.14√
3.222 × 10−3 0.00787 × 10−6 = 0.12156 ms
June 26, 2008 Transmission System Modeling - 62
Power System Dynamics and Stability 124
Universidad de Castilla-La Mancha
Example 1
i1α(t) =1
Zcα
[
vBUS1α − rα
2i1α(t)
]
− I2α(t − τα)
i2α(t) =
1
Zcα
[
vBUS2α− rα
2i2α
(t)]
− I1α(t − τα)
rα = 0.0243 × 24.14 = 0.5866 Ω
ZCα=
√
0.9238 × 10−3
0.0126 × 10−6= 270.77 Ω
τα = 24.14√
0.9238 × 10−3 0.0126 × 10−6 = 0.08236 ms
June 26, 2008 Transmission System Modeling - 63
Power System Dynamics and Stability 125
Universidad de Castilla-La Mancha
Example 1
i1β(t) =
1
Zcβ
[
vBUS1β− rβ
2i1β
(t)]
− I2β(t − τβ)
i2β(t) =
1
Zcβ
[
vBUS2β− rβ
2i2β
(t)]
− I1β(t − τβ)
rβ = rα
ZCβ= ZCα
τβ = τα
June 26, 2008 Transmission System Modeling - 64
Power System Dynamics and Stability 126
Universidad de Castilla-La Mancha
Example 1
2
664
vBUS2a
vBUS2b
vBUS2c
3
775
=1√
3
2
664
1√
2 0
1 −1/√
2p
3/2
1 −1/√
2 −p
3/2
3
775
| z
T
2
664
vBUS20
vBUS2α
vBUS2β
3
775
2
664
i1a
i1b
i1c
3
775
= T
2
664
i20
i2α
i2β
3
775
2
664
vBUS2a
vBUS2b
vBUS2c
3
775
= −
2
664
251.2 0 0
0 251.2 0
0 0 251.2
3
775
2
664
i2a
i2b
i2c
3
775
−d
dt
2
664
0.28926 0 0
0 0.28926 0
0 0 0.28926
3
775
2
664
i2a
i2b
i2c
3
775
June 26, 2008 Transmission System Modeling - 65
Power System Dynamics and Stability 127
Universidad de Castilla-La Mancha
Example 1
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02−300
−200
−100
0
100
200
300
t [s]
[kV
]vaBUS2
vasource
June 26, 2008 Transmission System Modeling - 66
Power System Dynamics and Stability 128
Universidad de Castilla-La Mancha
Example 1
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02−400
−300
−200
−100
0
100
200
300
t [s]
[kV
]
vaBUS2
vbBUS2
vcBUS2
June 26, 2008 Transmission System Modeling - 67
Power System Dynamics and Stability 129
Universidad de Castilla-La Mancha
Example 2
nn
a b c
50′
1′
45′
110′
40′ × 40′
28′8′′
June 26, 2008 Transmission System Modeling - 68
Power System Dynamics and Stability 130
Universidad de Castilla-La Mancha
Example 2
Line data:
4 Drake wires per phase (Nb = 4)
927 kcmil = 469.8 mm2
ACSR, 24 Al/13 steel, 3 layers
R = 0.554 in = 1.407 cm
R′ = 0.425 in = 1.080 cm
rdc = 0.1032 Ω/mile = 0.0645 Ω/km
rac@25C = 0.1061 Ω/mile = 0.0663 Ω/km
rac@100C = 0.1361 Ω/mile = 0.0851 Ω/km
d = 200 km
f = 60 Hz → ω = 377 rad/s
June 26, 2008 Transmission System Modeling - 69
Power System Dynamics and Stability 131
Universidad de Castilla-La Mancha
Example 2
r =rac@100C
Nb= 0.02134 Ω/km
Dm =3√
45′45′90′ = 17.27 m
R′b =
4√
1.080 30.48 30.48 43.11 cm = 0.1442 m
Rb =4√
1.407 30.48 30.48 43.11 cm = 0.1541 m
l = 2 × 10−7 ln17.27
0.1442= 0.9573 × 10−3 H/m
c =2π8.854 × 10−12
ln 17.270.1541
= 0.01179 µH/m
June 26, 2008 Transmission System Modeling - 70
Power System Dynamics and Stability 132
Universidad de Castilla-La Mancha
Example 2
z = r + jωl
= 0.3615∠86.62 Ω/km
y = jωc
= 4.444 × 10−6∠90 S/km
γ =√
zy
= 0.00217∠88.31 km−1
Zc =
√z
y
= 285.21∠ − 1.69 Ω
June 26, 2008 Transmission System Modeling - 71
Power System Dynamics and Stability 133
Universidad de Castilla-La Mancha
Example 2
A = D = cosh γd
=eγd + e−γd
2= 0.9686∠0.11
B = Zc sinh γd
= Zceγd − e−γd
2= 71.46∠86.65 Ω
C =1
Zcsinh γd
= 8.787 × 10−4∠90.03 S
June 26, 2008 Transmission System Modeling - 72
Power System Dynamics and Stability 134
Universidad de Castilla-La Mancha
Example 2
Z = zd
= 72.30∠86.62 Ω
Y = yd
= 8.89 × 10−4∠90 S
Z′ = Zsinh γd
γd
= 71.33∠86.65 Ω ≈ Z
Y′ = Ytanh γd/2
γd/2
= 9.032 × 10−4∠89.95 S ≈ Y
June 26, 2008 Transmission System Modeling - 73
Power System Dynamics and Stability 135
Universidad de Castilla-La Mancha
Cables
Steel pipe
Skid wires
Metallic tapes
Paper/oil insulation
Screen
Conductor
(filled with insulating oil)
(stranded copper)
June 26, 2008 Transmission System Modeling - 74
Power System Dynamics and Stability 136
Universidad de Castilla-La Mancha
Cables
Single-phase and three-phase cables:
conductor
sheath
SF6 gas
June 26, 2008 Transmission System Modeling - 75
Power System Dynamics and Stability 137
Universidad de Castilla-La Mancha
Cables
Model for a single-phase cable:
VC VS VA
I1
I2
I3
June 26, 2008 Transmission System Modeling - 76
Power System Dynamics and Stability 138
Universidad de Castilla-La Mancha
Cables
Model for a single-phase cable:
V1 = Vcore − Vsheath = VC − VS
V2 = Vsheath − Varmour = VS − VA
V3 = VA
IC = I1
IS = I1 − I2
IA = I2 − I3
June 26, 2008 Transmission System Modeling - 77
Power System Dynamics and Stability 139
Universidad de Castilla-La Mancha
Cables
Loop equations:
d
dx
V1
V2
V3
= −
z11 z12 0
z12 z22 z23
0 z23 z33
I1
I2
I3
d
dx
I1
I2
I3
=
y1 0 0
0 y2 0
0 0 y3
V1
V2
V3
June 26, 2008 Transmission System Modeling - 78
Power System Dynamics and Stability 140
Universidad de Castilla-La Mancha
Cables
The elements of the impedance matrix z arecomputed as follows:
z11 = zCext + zCS + zSint
z12 = −zSmut
z22 = −zSext + zSA + zAint
z23 = −zAmut
z33 = −zAext + zAE + zE
June 26, 2008 Transmission System Modeling - 79
Power System Dynamics and Stability 141
Universidad de Castilla-La Mancha
Cables
These impedances are calculated using the formulas:
q
r
zext =ρm
2πrD[I0(mr)K1(mq) + K0(mr)I1(mq)]
zint =ρm
2πqD[I0(mq)K1(mr) + K0(mq)I1(mr)]
zmut =ρ
2πqrD
zins = y−1 = jωµ0
2π
r
q
June 26, 2008 Transmission System Modeling - 80
Power System Dynamics and Stability 142
Universidad de Castilla-La Mancha
Cables
D = I1(mr)K1(mq) + K1(mr)I1(mq)
m =
√
jωµ
ρ
Where In and Kn are modified Bessel functions, as follows:
In(x) =
∞∑
k=0
1
k!Γ(n + k + 1)
(x
2
)n+2k
Kn(x) =π
2
I−n(x) − In(x)
sin(nπ)
June 26, 2008 Transmission System Modeling - 81
Power System Dynamics and Stability 143
Universidad de Castilla-La Mancha
Cables
The Gamma function Γ(n) is defined as follows:
Γ(n) =
∫ ∞
0
e−xxn−1dx
Γ(n + 1) = nΓ(n)
Γ(n + 1) = n! for n = 1, 2, 3, . . .
For asymmetric cables a finite element method is needed to compute
these impedances.
June 26, 2008 Transmission System Modeling - 82
Power System Dynamics and Stability 144
Universidad de Castilla-La Mancha
Cables
From the relations between loop voltages/currents and node
voltage/currents, the node equations are:
d
dx
VC
VS
VA
= −
zCC zCS zCA
zCS zSS zSA
zCA zSA zAA
IC
IS
IA
e.g. zCC = z11 + 2z12 + z22 + 2z23 + z33
d
dx
IC
IS
IA
=
y1 −y1 0
−y1 y1 + y2 −y2
0 −y2 y2 + y3
VC
VS
VA
June 26, 2008 Transmission System Modeling - 83
Power System Dynamics and Stability 145
Universidad de Castilla-La Mancha
Cables
For a three-phase cable made of 3 single-phase cables:
d
dx[V] = −[z]9×9[I]
d
dx[I] = −[y]9×9[V]
[z] =
[zaa]3×3 [zab]3×3 [zac]3×3
[zab]3×3 [zbb]3×3 [zbc]3×3
[zac]3×3 [zbc]3×3 [zcc]3×3
[y] =
[yaa]3×3 0 0
0 [ybb]3×3 0
0 0 [ycc]3×3
June 26, 2008 Transmission System Modeling - 84
Power System Dynamics and Stability 146
Universidad de Castilla-La Mancha
Load Modeling
RLC models.
Induction motors.
Detailed models.
Phasor models.
Aggregated models:
Impedance models.
Power models.
Induction motor power models.
June 26, 2008 Load Modeling - 1
Power System Dynamics and Stability 147
Universidad de Castilla-La Mancha
Load Classification
By demand level:
Residential: lighting and heating (RL + controls); AC (motor +
controls); appliances (small motors + controls).
Commercial: similar types of devices as residential.
Industrial: motor drives (induction and dc motor-based mostly); arc
furnaces; lighting; heating; others (e.g. special motor drives).
By type:
RLC + controls.
Drives: ac/dc motors + electronic controls.
Special (e.g. arc furnace).
June 26, 2008 Load Modeling - 2
Power System Dynamics and Stability 148
Universidad de Castilla-La Mancha
Remarks on Load Classification
Most controls are implemented using a variety of power electronic
converters.
Aggregate load models are necessary at the transmission system
modeling level.
Only large loads can be represented with their actual models.
June 26, 2008 Load Modeling - 3
Power System Dynamics and Stability 149
Universidad de Castilla-La Mancha
RLC Loads, Resistor
Ideal, linear resistors (R), inductors (L) and capacitors (C).
Resistor:
+ −
i
v
R
Time domain:
v = Ri
Phasor domain:
V = RI
June 26, 2008 Load Modeling - 4
Power System Dynamics and Stability 150
Universidad de Castilla-La Mancha
RLC Loads, Inductor
Inductor:
+ −
i
v
L
Time domain:
v = Ldi
dt
Phasor domain:
V = jωLI
= ZLI
June 26, 2008 Load Modeling - 5
Power System Dynamics and Stability 151
Universidad de Castilla-La Mancha
RLC Loads, Inductor
The inductor time domain model can be discretized using the trapezoidal
integration method as follows:
i =1
L
∫
vdt
⇒ ik+1 = ik +∆t
2L(vk+1 + vk)
=∆t
2Lvk+1 + (ik +
∆t
2Lvk)
︸ ︷︷ ︸
hk
June 26, 2008 Load Modeling - 6
Power System Dynamics and Stability 152
Universidad de Castilla-La Mancha
RLC Loads, Capacitor
This yields the equivalent resistive circuit:
+ −ik+1
vk+1
hk
2L∆t
June 26, 2008 Load Modeling - 7
Power System Dynamics and Stability 153
Universidad de Castilla-La Mancha
RLC Loads, Capacitor
Capacitor:
+ −
i
v
C
Time domain:
i = Cdv
dt
Phasor domain:
V = −j1
ωCI
= ZCI
June 26, 2008 Load Modeling - 8
Power System Dynamics and Stability 154
Universidad de Castilla-La Mancha
RLC Loads, Capacitor
The capacitor time domain model can be discretized using the trapezoidal
integration method as follows:
v =1
C
∫
idt
⇒ vk+1 = vk +∆t
2C(ik+1 + ik)
ik+1 =2C
∆tvk+1 − (ik +
2C
∆tvk)
︸ ︷︷ ︸
hk
June 26, 2008 Load Modeling - 9
Power System Dynamics and Stability 155
Universidad de Castilla-La Mancha
RLC Loads, Capacitor
This yields the equivalent resistive circuit:
+ −ik+1
vk+1
hk
∆t2C
June 26, 2008 Load Modeling - 10
Power System Dynamics and Stability 156
Universidad de Castilla-La Mancha
Induction Motor
Arbitrary 0dq reference:
r
s
ω
θ
qd
d
θr
ωr
as
bs
cs
ar
br
cr
a′
s
b′s
c′s
0
June 26, 2008 Load Modeling - 11
Power System Dynamics and Stability 157
Universidad de Castilla-La Mancha
Induction Motor
Electrical (inductor) equations:
[vabcs] = [rabcs
][iabcs] +
d
dt[λabcs
]
[vabcr] = [rabcr
][iabcr] +
d
dt[λabcr
]
λabcs
λabcr
=
Labcs
Lsr(θr)
Lsr(θr) Labcr
︸ ︷︷ ︸
L(θr)
iabcs
iabcr
︸ ︷︷ ︸
[i]
Te =1
2[i]T
d[L(θr)]
dθr[i]
June 26, 2008 Load Modeling - 12
Power System Dynamics and Stability 158
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Induction Motor
Mechanical (Newton’s) equations:
Jd
dtωr + dωr = Tm − Te
d
dtθr = ωr
Stator transformation equations:
v0s
vds
vqs
= Ks
vas
vbs
vcs
i0s
ids
iqs
= Ks
ias
ibs
ics
June 26, 2008 Load Modeling - 13
Power System Dynamics and Stability 159
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Induction Motor
Where the transformation matrix Ks is as follows:
Ks =2
3
1/2 1/2 1/2
sin θ sin(θ − 2π/3) sin(θ + 2π/3)
cos θ cos(θ − 2π/3) cos(θ + 2π/3)
K−1s =
1 sin θ cos θ
1 sin(θ − 2π/3) cos(θ − 2π/3)
1 sin(θ + 2π/3) cos(θ + 2π/3)
June 26, 2008 Load Modeling - 14
Power System Dynamics and Stability 160
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Induction Motor
Rotor transformation equations:
v0r
vdr
vqr
= Kr
var
vbr
vcr
i0r
idr
iqr
= Kr
iar
ibr
icr
June 26, 2008 Load Modeling - 15
Power System Dynamics and Stability 161
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Induction Motor
Where the transformation matrix Kr is as follows:
Kr =2
3
1/2 1/2 1/2
sinβ sin(β − 2π/3) sin(β + 2π/3)
cos β cos(β − 2π/3) cos(β + 2π/3)
K−1r =
1 sinβ cos β
1 sin(β − 2π/3) cos(β − 2π/3)
1 sin(β + 2π/3) cos(β + 2π/3)
β = θ − θr
June 26, 2008 Load Modeling - 16
Power System Dynamics and Stability 162
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Induction Motor
Stator equations:
v0s
vds
vqs
=
rs 0 0
0 rs 0
0 0 rs
i0s
ids
iqs
+ ω
0
−λqs
λds
+d
dt
λ0s
λds
λqs
Rotor equations referred to the stator:
0
0
0
=
a2rr︸︷︷︸
r′r
0 0
0 r′r 0
0 0 r′r
i′0r
i′dr
i′qr
+ (ω − ωr)
0
−λ′qr
λ′dr
+d
dt
λ′0r
λ′dr
λ′qr
a =Ns
Nr
June 26, 2008 Load Modeling - 17
Power System Dynamics and Stability 163
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Induction Motor
Magnetic flux equations:
λ0s
λds
λqs
λ′0r
λ′dr
λ′qr
=
Lls 0 0 0 0 0
0 Lls + M︸ ︷︷ ︸
Ls
0 0 M 0
0 0 Ls 0 0 M
0 0 0 L′lr 0 0
0 M 0 0 L′lr + M︸ ︷︷ ︸
L′r
0
0 0 M 0 0 L′r
i0s
ids
iqs
i′0r
i′dr
i′qr
June 26, 2008 Load Modeling - 18
Power System Dynamics and Stability 164
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Induction Motor
Mechanical equations:
2
pJ
d
dtωr +
2
pDωr = Tm − Te
d
dtθr = ωr
d
dtθ = ω
Te =3
2
p
2(iqsλds − idsλqs)
June 26, 2008 Load Modeling - 19
Power System Dynamics and Stability 165
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Induction Motor
Equivalent circuit representation of these equations:
+ +
+
− −
−
iqs iqr
vqs vqrM
rs r′rLls L′
lr
ωλds (ω − ωr)λ′
dr
Ns : Nr
June 26, 2008 Load Modeling - 20
Power System Dynamics and Stability 166
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Induction Motor
−
− −
+
+ +ids idr
vds vdrM
rs r′rLls L′
lr
ωλqs (ω − ωr)λ′
qr
Ns : Nr
−
+
i0s i0r
v0s v0r
rs r′rLls L′
lr Ns : Nr
June 26, 2008 Load Modeling - 21
Power System Dynamics and Stability 167
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Induction Motor
Assuming a balanced, fundamental frequency (ω = ω0) system, the
model can be reduced to a p.u. “transient” model (3rd order model):
Vas = VasR+ jVasI
Ias = IasR+ jIasI
d
dtE′
R = ω0σE′I −
1
T ′0
[E′R + (xS − x′)IasI
]
d
dtE′
I = −ω0σE′R − 1
T ′0
[E′I + (xS − x′)IasR
]
June 26, 2008 Load Modeling - 22
Power System Dynamics and Stability 168
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Induction Motor
VasR− E′
R = rsIasR− x′IasI
VasI− E′
I = rsIasI+ x′IasR
d
dtσ =
1
H(TL − Te + Dω0 − Dσ)
Te =1
ω0(E′
RIasR+ E′
IIasR)
TL = f(σ)
June 26, 2008 Load Modeling - 23
Power System Dynamics and Stability 169
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Induction Motor
where:
σ =ω0 − ωr
ω0→ slip
xs = xls + xm → stator reactance
x′ = xls +x′
lrxm
x′lr + xm
→ transient reactance
T ′0 =
x′lr + xm
ω0r′r→ open circuit transient time constant
June 26, 2008 Load Modeling - 24
Power System Dynamics and Stability 170
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Induction Motor
If the electromagnetic transients are neglected, the system can be
reduced to the following quasi-steady state equivalent circuit model:
−
+
Ias I′ar
Vasjxm
rs r′rjxls jx′
lr
r′r1−σ
σ
June 26, 2008 Load Modeling - 25
Power System Dynamics and Stability 171
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Induction Motor
Or equivalently:
−
+
Ias I′ar
Vasjxm
rs jxls jx′
lr
r′r1
σ
Plus mechanical equations.
June 26, 2008 Load Modeling - 26
Power System Dynamics and Stability 172
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Induction Motor
This simplification can be justified from the point of view of the
torque-speed characteristic:
10ωrωc
= 1 − σ
Te
Te max
Steady state
Transient
characteristic
start up
June 26, 2008 Load Modeling - 27
Power System Dynamics and Stability 173
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Induction Motor
If the mechanical dynamics are ignored, the slip σ becomes a fixed value,
and hence the equivalent circuit can be reduced to a simple equivalent
reactive impedance, i.e. a Z load.
For loads with multiple IMs, a equivalent motor model can be used to
represent these motors.
Neglecting the motor dynamic equations in large or equivalent aggregate
motors can lead to significant modeling errors, as the transient model and
mechanical time constants can be on the same range as the generator
time constants.
June 26, 2008 Load Modeling - 28
Power System Dynamics and Stability 174
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Induction Motor
Double-cage IMs can be modeled by introducing an additional inductance
on the rotor side, which leads to a “subtransient” model.
Since rotor cores are laminated, eddy currents do not play a significant
role on the system dynamics.
Controls can also be modeled in detail by representing the converters
with ideal electronic switches plus their control systems.
For balanced, fundamental frequency systems, approximate equivalent
models of the IM and its controls can be readily implemented.
June 26, 2008 Load Modeling - 29
Power System Dynamics and Stability 175
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Induction Motor
For example, an IM with voltage-fed field oriented control can be
represented by:
dωr
dt=
P
2J(Te − TL)
diqs
dt= −
(
L′2rrs + M2r′r
L′rσ
)
iqs − ωids +Mr′rL′
rσλ′
qr
−M
σωrλ
′dr +
L′r
σvqs
dids
dt= −
(
L′2rrs + M2r′r
L′rσ
)
ids + ωiqs +Mr′rL′
rσλ′
dr
+M
σωrλ
′qr +
L′r
σvds
June 26, 2008 Load Modeling - 30
Power System Dynamics and Stability 176
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Induction Motor
dλ′qr
dt= −(ω − ωr)λ
′dr −
r′rL′
r
λ′qr +
r′rM
L′r
iqs
dλ′dr
dt= (ω − ωr)λ
′qr −
r′rL′
r
λ′dr +
r′rM
L′r
ids
Te =3pM
4L′r
(iqsλ′dr − idsλ
′qr)
σ = LsL′r − M2
dTref
dt= kq3
(pTL
2J− 3p2M
8JL′r
iqsλ′dr
)
+ kq4(ωref − ωr)
June 26, 2008 Load Modeling - 31
Power System Dynamics and Stability 177
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Induction Motor
duqs
dt=
3pMkq1
4L′r
[r′rL′
r
iqsλ′dr −
r′rM
L′r
idsiqs +M
σλ′2
drωr
+
(
L′2rrs + M2r′r
L′rσ
)
iqsλ′dr −
L′r
σλ′
druqs]
+kq1kq3
(pTL
2J− 3p2M
8JL′r
iqsλ′dr
)
+kq2
(
Tref− 2pMaL′
r
iqsλ′dr
)
+kq1kq4(ωref − ωr)
duds
dt= kd1
(r′rL′
r
λ′dr −
Mr′rL′
r
ids
)
+ kd2(λref − λ′dr)
June 26, 2008 Load Modeling - 32
Power System Dynamics and Stability 178
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Example
A 21 MW load at 4 kV and 60 Hz is made of:
An inductive impedance load with G = 0.06047, B = −0.03530.
An aggregated induction motor model with rs = 0.07825,
xls = 0.8320, r′r = 0.1055, x′lr = 0.8320, xm = 16.48.
This data is all in p.u. on a 100 MVA, 4 kV base.
June 26, 2008 Load Modeling - 33
Power System Dynamics and Stability 179
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Example
The Z load model is then:
PL =21 MW
100 MVA= 0.21
= PZ + PIM
PZ = V 2LG = 0.06
⇒ PIM = 0.15
June 26, 2008 Load Modeling - 34
Power System Dynamics and Stability 180
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Example
ZIM = rs + jxls +jxM (r′r/σ + jx′
lr)
r′r/σ + j(xM + x′lr)
= 0.07825 + j0.832 +−13.7114 + j1.7386/σ
0.1055/σ + j17.312
=
(
0.07825 +28.652/σ
0.01113σ2 + 299.71
)
+j
(
0.832 +0.18342/σ2 + 237.37
0.01113/σ2 + 299.71
)
=(0.00087/σ2 + 28.652/σ + 23.452) + j(0.19628/σ2 + 486.73)
0.01113/σ2 + 299.71
June 26, 2008 Load Modeling - 35
Power System Dynamics and Stability 181
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Example
PIM = V 2LGIM = GIM
0.15 =(0.01113/σ2 + 299.71)(0.00087/σ2 + 28.652/σ + 23.452)
(0.00087/σ2 + 28.652/σ + 23.452)2 + (0.19628/σ2 + 486.73)2
⇒ σ = 0.0191 (by trial-and-error)
⇒ ZIM = 4.6221 + j3.0742
YL = (G + jB) +1
ZIM
ZL =1
YL= 3.3654 + j2.1597
June 26, 2008 Load Modeling - 36
Power System Dynamics and Stability 182
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Impedance Models
Ignoring “fast” and “slow” transients, certain loads can be represented
using an equivalent impedance.
A Z load model is typically used for a variety of dynamic analysis of
power systems.
Using these load models, an equivalent impedance can be readily
obtained for all loads connected at a particular bus at the transmission
system level.
ULTCs are used to connect distribution systems (subtransmission and LV
systems and the loads connetced to these) to the transmission system to
control the steady state voltage on the load side.
June 26, 2008 Load Modeling - 37
Power System Dynamics and Stability 183
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Power Models
Hence, for “slow” dynamic analysis of balanced, fundamental frequency
system models:
PL = V 2LGL = PL0
(VL
VL0
)2
QL = V 2LBL = QL0
(VL
VL0
)2
VL ≈ VL0 ⇒
PL ≈ PL0
QL ≈ QL0
This is typically referred as a constant PQ model.
June 26, 2008 Load Modeling - 38
Power System Dynamics and Stability 184
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Power Models
Load recovery of certain loads (e.g. thermostatic) with respect to voltage
changes can be modeled as:
dx(t)
dt= −x(t)
Tp+ PL0
[VL(t)
VL0
]Nps
− PL0
[VL(t)
VL0
]Npt
PL(t) =x(t)
Tp+ PL0
[VL(t)
VL0
]Npt
dy(t)
dt= −y(t)
Tq+ QL0
[VL(t)
VL0
]Nqs
− QL0
[VL(t)
VL0
]Nqt
QL(t) =y(t)
Tq+ QL0
[VL(t)
VL0
]Nqt
June 26, 2008 Load Modeling - 39
Power System Dynamics and Stability 185
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Power Models
This results in the following time response:
P
PL0
V
VL0
t
t
tf
tf
June 26, 2008 Load Modeling - 40
Power System Dynamics and Stability 186
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Example
Identification of LD1-LD4 loads at paper mill in Sweden:
LD1 LD2
LD3 LD4
G1 G2
Net 894 MVA, 30 kV
1
2
3
4
5
6
June 26, 2008 Load Modeling - 41
Power System Dynamics and Stability 187
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Example
Measurements vs. model:
June 26, 2008 Load Modeling - 42
Power System Dynamics and Stability 188
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Power Models
Certain loads have been shown to behave in steady state as follows:
PL = KP V αP
L fβP
L ≈ PL0
(VL
VL0
)αP
QL = KQVαQ
L fβQ
L ≈ PL0
(VL
VL0
)αQ
If f ≈ f0, then the following approximation holds:
PL ≈ PL0
(VL
VL0
)αP
QL ≈ PL0
(VL
VL0
)αQ
June 26, 2008 Load Modeling - 43
Power System Dynamics and Stability 189
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Power Models
Load αP αQ βP βQ
Filament lamp 1.6 0 0 0
Fluorescent lamp 1.2 3.0 -0.1 2.8
Heater 2.0 0 0 0
Induction motor (half load) 0.2 1.6 1.5 -0.3
Induction motor (full load) 0.1 0.6 2.8 1.8
Reduction furnace 1.9 2.1 -0.5 0
Aluminum plant 1.8 2.2 -0.3 0.6
June 26, 2008 Load Modeling - 44
Power System Dynamics and Stability 190
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Power Models
ZIP model:
PL = PLZ
(VL
VL0
)2
+ PLI
(VL
VL0
)
+ PLP
QL = QLZ
(VL
VL0
)2
+ QLI
(VL
VL0
)
+ QLP
Jimma’s model:
PL = PLZ
(VL
VL0
)2
+ PLI
(VL
VL0
)
+ PLP
QL = QLZ
(VL
VL0
)2
+ QLI
(VL
VL0
)
+ QLP+ KV
dVL
dt
June 26, 2008 Load Modeling - 45
Power System Dynamics and Stability 191
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Induction Motor Power Models
Walve’s model:
PL = Kpff + Kpv
[
VL + TdVL
dt
]
≈ PL0 + Kpv
[
(VL − VL0) + TdVL
dt
]
QL = Kqff + KqvVL
≈ QL0 + Kqv(VL − VL0)
June 26, 2008 Load Modeling - 46
Power System Dynamics and Stability 192
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Induction Motor Power Models
Mixed model:
PL = Kpff + Kpv
[
V αL + Tpv
dVL
dt
]
≈ PL0 + Kpv
[
(VL − VL0)α + Tpv
dVL
dt
]
QL = Kqff + Kqv
[
V βL + Tqv
dVL
dt
]
≈ QL0 + Kqv
[
(VL − VL0)β + Tqv
dVL
dt
]
June 26, 2008 Load Modeling - 47
Power System Dynamics and Stability 193
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Power Flow Outlines
Power Flow:
System model.
Equations.
Solution techniques:
Newton-Raphson.
Fast decoupled.
June 26, 2008 Power Flow Outlines - 1
Power System Dynamics and Stability 194
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Power Flow Model
The steady-state operating point of a power system is obtained by solving
the “power flow” equations.
Power flow system model corresponds to the steady state model.
Generator:
Generates and injects power P in the system while keeping the
output voltage V constant within active and reactive power limits
(capability curve):
Pmin ≤ P ≤ Pmax
Qmin ≤ Q ≤ Qmax
June 26, 2008 Power Flow Outlines - 2
Power System Dynamics and Stability 195
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Power Flow Model
Thus, it is modeled as a PV bus:
P = constant
Q = unknown
V = constantδ = unknown
When Q reaches a limit it becomes a PQ bus:
P = constant
Q = constant
V = unknownδ = unknown
June 26, 2008 Power Flow Outlines - 3
Power System Dynamics and Stability 196
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Power Flow Model
Slack bus:
The phasor model needs a reference bus.
A “large” generator is typically chosen as the the reference bus, as it
should be able to take the power “slack”:
P = unknown
Q = unknown
V = constantδ = 0
Pslack =∑
L
PL + Plosses −∑
G
PG
June 26, 2008 Power Flow Outlines - 4
Power System Dynamics and Stability 197
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Power Flow Model
Hence, if Q reaches a limit:
P = unknown
Q = constant
V = unknownδ = 0
June 26, 2008 Power Flow Outlines - 5
Power System Dynamics and Stability 198
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Power Flow Model
Load:
Loads are typically connected to the transmission system through
ULTC transformers.
Thus, most loads in steady state represent a constant power demand
in the system, and hence are modeled as a PQ bus:
P = constant
Q = constant
V = unknown
δ = unknown
June 26, 2008 Power Flow Outlines - 6
Power System Dynamics and Stability 199
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Power Flow Model
Transmission system:
AC transmission lines and transformers in steady state are basically
modeled using the following model:
Vi Vk
a∠α : 1
a∠αVk
SiSk
Ii Ik
Ik/a∠αZ
Y2Y1
June 26, 2008 Power Flow Outlines - 7
Power System Dynamics and Stability 200
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Power Flow Model
Hence:
Si = ViI∗i
= Vi
(Vi − a∠αVk)
(1
Z
)
︸ ︷︷ ︸
Y
+ViY1
∗
= Vi
(Y + Yi)
∗︸ ︷︷ ︸
Y∗ii
V∗i + (−a∠αY)∗
︸ ︷︷ ︸
Y∗ik
V∗k
Similarly for Sk.
June 26, 2008 Power Flow Outlines - 8
Power System Dynamics and Stability 201
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Power Flow Model
Thus, for an N bus system interconnected through an ac transmission
system, an N × N bus admittance matrix can be defined:
I1
I2
...
Ii
...
IN
=
Y11 Y12 · · · Y1i · · · Y1N
Y21 Y22 · · · Y2i · · · Y2N
......
. . ....
. . ....
Yi1 Yi2 · · · Yii · · · YiN
......
. . ....
. . ....
YN1 YN2 · · · YNi · · · YNN
V1
V2
...
Vi
...
VN
I︸︷︷︸
node injections
= Ybus V︸︷︷︸
node voltages
June 26, 2008 Power Flow Outlines - 9
Power System Dynamics and Stability 202
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Power Flow Model
where:
Ybus =
Yii =∑N
k=11
Zik+∑
Yi
= sum of all the Y’s connected to node i
Yij = − 1Zij
= negative of the Y between nodes i and j
June 26, 2008 Power Flow Outlines - 10
Power System Dynamics and Stability 203
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Power Flow Equations
In steady state, a system with n generators G and m loads L can be
modeled as:
SG1
SGn
SL1
SLm
VG1
VGn
VL1
VLm
Ybus (N × N)
N = n + m
Transmission
System
June 26, 2008 Power Flow Outlines - 11
Power System Dynamics and Stability 204
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Power Flow Equations
Hence the power injections at each node are defined by:
Si = Pi + jQi
= ViI∗i
= Vi
N∑
k=1
Y∗ikV
∗k
= Vi∠δi
N∑
k=1
(Gij − jBij)Vk∠ − δk
Si =
SGifor generator buses
−SLifor load buses
June 26, 2008 Power Flow Outlines - 12
Power System Dynamics and Stability 205
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Power Flow Equations
This yields two equations per node or bus:
∆Pi(δ, V, Pi) =
Pi −N∑
k=1
ViVk[Gik cos(δi − δk) + Bik sin(δi − δk)] = 0
∆Qi(δ, V, Qi) =
Qi −N∑
k=1
ViVk[Gik sin(δi − δk) − Bik cos(δi − δk)] = 0
And two variables per node:
PQ buses → Vi and δi
PV buses → δi and Qi
Slack buses → Pi and Qi
June 26, 2008 Power Flow Outlines - 13
Power System Dynamics and Stability 206
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Power Flow Equations
These equations are referred to as the power mismatch equations.
The equations are typically subjected to inequality constraints
representing control limits:
0.95 ≤ Vi ≤ 1.05 for all buses
Qmini≤ Qi ≤ Qmaxi
for generator buses
June 26, 2008 Power Flow Outlines - 14
Power System Dynamics and Stability 207
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Power Flow Equations
In summary, typical power flow data are as follows:
Bus Parameters Variables
PQ P , Q V , θ
PV P , V Q, θ
slack V , θ P , Q
June 26, 2008 Power Flow Outlines - 15
Power System Dynamics and Stability 208
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Power Flow Equations
In “constrained” power flow analysis, standard buses can degenerate if
some limits is reached:
Bus Parameters Variables
PQ P , Vmax,min Q, θ
PV P , Qmax,min V , θ
slack Qmax,min, θ P , V
Continuation power flow techniques are by far more accurate and robust
than standard power flow analysis if limits are taken into account.
June 26, 2008 Power Flow Outlines - 16
Power System Dynamics and Stability 209
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Power Flow Equations
The slack bus can be single or distributed.
This refers to losses.
For single slack bus model, all system losses are “cleared” by the
slack bus.
For distributed slack bus model losses are shared (equally or
proportionally) among all or part of the generators:
Continuation power flow techniques are by far more accurate and robust
than standard power flow analysis if limits are taken into account.
June 26, 2008 Power Flow Outlines - 17
Power System Dynamics and Stability 210
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Power Flow Equations
The distributed slack bus model is based on a generalized power center
concept.
This is practically obtained by including in power flow equations a variable
kG and rewriting the system active power balance as follows:
nG∑
i
(1 + kGγi)PGi−
nP∑
i
PLi− Plosses = 0
The parameters γi allow tuning the weight of the participation of each
generator to the losses.
For single slack bus model, γi = 0 for all generators but one.
June 26, 2008 Power Flow Outlines - 18
Power System Dynamics and Stability 211
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Power Flow Solution
The power flow equations can be represented as
F (z) = 0
There are 2 equations per bus, with 2 known variables and 2 unknown
variables per bus; the problem is of dimension 2N .
Since these equations are highly nonlinear due to the sine and cosine
terms, Newton-Raphson (NR) based numerical techniques are used to
solve them.
June 26, 2008 Power Flow Outlines - 19
Power System Dynamics and Stability 212
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Power Flow Solution
A “robust” NR technique may be used to solve these equations:
1. Start with an initial guess, typically V 0i = 1, δ0
i = 0, Q0Gi = 0,
P 0slack = 0 (flat start).
2. At each iteration k(k = 0, 1, 2, . . .), compute the “sparse” Jacobian
matrix:
∂F
∂z
∣∣∣zk
= Jk =
∂F1
∂z1|zk . . . ∂F1
∂zN|zk
.... . .
...
∂FN
∂z1|zk . . . ∂FN
∂zN|zk
2N×2N
June 26, 2008 Power Flow Outlines - 20
Power System Dynamics and Stability 213
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Power Flow Solution
3. Find ∆zk by solving the following linear set of equations (the sparse
matrix Jk is factorized and not inverted to speed up the solution process):
Jk∆zk = −F (zk)
4. Computes the new guess for the next iteration, where α is a step control
constant to guarantee convergence (0 < α < 1):
zk+1 = zk + α∆zk
June 26, 2008 Power Flow Outlines - 21
Power System Dynamics and Stability 214
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Power Flow Solution
5. Stop when:
‖F (zk+1)‖ = max |Fi(zk+1)| ≤ ǫ
This is basically the technique used in MATLAB’s fsolve() routine,
based on either numerical or actual Jacobians.
June 26, 2008 Power Flow Outlines - 22
Power System Dynamics and Stability 215
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Power Flow Solution
If only the unknown bus voltage angles and magnitudes are calculated
using NR’s method (the generator reactive powers and active slack power
are evaluated later):
Jk∆zk = −F (zk)
⇒
H M
N L
∆δk
∆V k/V k
= −
∆P (δk, V k)
∆Q(δk, V k)
June 26, 2008 Power Flow Outlines - 23
Power System Dynamics and Stability 216
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Power Flow Solution
Where:
H =∂∆P
∂δ
∣∣∣(δk,V k)
M = V∂∆P
∂V
∣∣∣(δk,V k)
N =∂∆Q
∂δ
∣∣∣(δk,V k)
L = V∂∆Q
∂V
∣∣∣(δk,V k)
June 26, 2008 Power Flow Outlines - 24
Power System Dynamics and Stability 217
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Power Flow Solution
Assuming:
(δi − δk) < 10, then
cos(δi − δk) ≈ 1
sin(δi − δk) ≈ δi − δk
The resistance in the transmission system are small, i.e. R ≪ X ,
then Gij ≪ Bik.
The M and N matrices may be neglected, and:
H ≈ V kB′V k
M ≈ V kB′′V k
June 26, 2008 Power Flow Outlines - 25
Power System Dynamics and Stability 218
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Power Flow Solution
Thus the linear step equations may be decoupled and reduced to:
B′∆δk = −∆P (δk, V k)/V k
B′′∆V k = −∆Q(δk, V k)/V k
where B′′ is the imaginary part of the Ybus matrix, and B′ is the
imaginary part of the admittance matrix obtained by ignoring the system
resistances, i.e. B′′ 6= B′.
June 26, 2008 Power Flow Outlines - 26
Power System Dynamics and Stability 219
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Power Flow Solution
The fast Decoupled iterative method is then defined as follows:
1. Start with an initial guess, typically δ0i = 0, V 0
i = 1.
2. Solve for ∆δk → B′∆δk = −∆P (δk, V k)/V k.
3. Update → δk+1 = δk + ∆δk.
4. Solve for ∆V k → B′′∆V k = −∆Q(δk, V k)/V k.
5. Update → V k+1 = V k + ∆V k.
6. Compute unknown generator powers and check for limits.
7. Repeat process for k = 1, 2, . . ., until convergence.
June 26, 2008 Power Flow Outlines - 27
Power System Dynamics and Stability 220
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Power Flow Solution
This technique requires a relatively large number of iterations as
compared to the robust NR method.
It is significantly faster, as there is no need to re-compute and re-factorize
the Jacobian matrix every iteration.
It is sensitive to initial guesses and there is no guarantee of convergence,
particularly for systems with large transmission system resistances.
June 26, 2008 Power Flow Outlines - 28
Power System Dynamics and Stability 221
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Example
For the following system:
P1P2
Q
1 2
3
All lines:
200 MVA
200 MVA
138 kV
X = 0.1 p.u.
B = 0.2 p.u.
0.9 p.f. lagging
V1 = 1, δ1 = 0, V2 = 1, V3 = 1, and P2 = P1/2.
Determine the voltage phasor angles δ2 and δ3 and the shunt Q by
solving the PF equations.
June 26, 2008 Power Flow Outlines - 29
Power System Dynamics and Stability 222
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Example
The Ybus matrix is:
Y11 = Y22 = Y33 =2
jX+ 2
(
jB
2
)
= −j19.8
Y12 = Y13 = Y23 = − 1
jX= j10
⇒ Ybus = j
−19.8 10 10
10 −19.8 10
10 10 −19.8
= j
B11 B12 B13
B21 B22 B23
B31 B32 B33
June 26, 2008 Power Flow Outlines - 30
Power System Dynamics and Stability 223
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Example
Mismatch equation ∆P1:
∆P1 = P1 −3∑
k=1
V1Vk[G1k cos(δ1 − δk) + B1k sin(δ1 − δk)]
0 = P1 −3∑
k=1
B1k sin(−δk)
= P1 + B12 sin(δ2) + B13 sin(δ3)
= P1 + 10 sin(δ2) + 10 sin(δ3)
June 26, 2008 Power Flow Outlines - 31
Power System Dynamics and Stability 224
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Example
Mismatch equation ∆Q1:
∆Q1 = Q1 −3∑
k=1
V1Vk[G1k sin(δ1 − δk) − B1k cos(δ1 − δk)]
0 = Q1 +
3∑
k=1
B1k cos(−δk)
= Q1 − 19.8 + 10 cos(δ2) + 10 cos(δ3)
June 26, 2008 Power Flow Outlines - 32
Power System Dynamics and Stability 225
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Example
Mismatch equation ∆P2:
∆P2 = P2 −3∑
k=1
V2Vk[G2k cos(δ2 − δk) + B2k sin(δ2 − δk)]
0 = P1/2 −3∑
k=1
B2k sin(δ2 − δk)
= 0.5P1 + 10 sin(δ2) + 10 sin(δ2 − δ3)
June 26, 2008 Power Flow Outlines - 33
Power System Dynamics and Stability 226
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Example
Mismatch equation ∆Q2:
∆Q2 = Q2 −3∑
k=1
V2Vk[G2k sin(δ2 − δk) − B2k cos(δ2 − δk)]
0 = Q2 +
3∑
k=1
B2k cos(δ2 − δk)
= Q1 + 10 cos(δ2) − 19.8 + 10 cos(δ2 − δ3)
June 26, 2008 Power Flow Outlines - 34
Power System Dynamics and Stability 227
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Example
Mismatch equation ∆P3:
∆P3 = P3 −3∑
k=1
V3Vk[G3k cos(δ3 − δk) + B3k sin(δ3 − δk)]
0 = −0.9 −3∑
k=1
B3k sin(δ3 − δk)
= −0.9 + 10 sin(δ3) + 10 sin(δ3 − δ2)
June 26, 2008 Power Flow Outlines - 35
Power System Dynamics and Stability 228
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Example
Mismatch equation ∆Q3:
∆Q3 = Q3 −3∑
k=1
V3Vk[G3k sin(δ3 − δk) − B3k cos(δ3 − δk)]
0 = Q −√
1 − 0.92 +
3∑
k=1
B3k cos(δ3 − δk)
= Q − 0.436 + 10 cos(δ3) + 10 cos(δ3 − δ2) − 19.8
June 26, 2008 Power Flow Outlines - 36
Power System Dynamics and Stability 229
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Example
Thus 6 equations and 6 unknowns, i.e. δ2, δ3, P1, Q1, Q2, and Q, can
be solved using MATLAB’s fsolve() routine:
>> global lambda
>> lambda = 1;
>> z0 = fsolve(@pf_eqs,[0 0 0 0 0 0], optimset(’Display’,’iter’))
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 7 0.945696 18 1
1 14 0.000661419 0.705901 0.0205 1
2 21 9.98637e-18 0.0257331 2.52e-09 1.76
Optimization terminated: first-order optimality is less than options.TolFun.
z0 =
-0.0100 -0.0500 0.6000 -0.1870 -0.1915 0.2565
June 26, 2008 Power Flow Outlines - 37
Power System Dynamics and Stability 230
Universidad de Castilla-La Mancha
Example
Where lambda is used to simulate constant power factor load changes
and pf eqs.m is:
function F = pf_eqs(z)
global lambda
d2 = z(1);
d3 = z(2);
P1 = z(3);
Q1 = z(4);
Q2 = z(5);
Q = z(6);
F(1,1) = P1 + 10*sin(d2) + 10*sin(d3);
F(2,1) = Q1 - 19.8 + 10*cos(d2) + 10*cos(d3);
F(3,1) = 0.5*P1 - 10*sin(d2) - 10*sin(d2-d3);
F(4,1) = Q2 + 10*cos(d2) - 19.8 + 10*cos(d2-d3);
F(5,1) = -0.9*lambda - 10*sin(d3) - 10*sin(d3-d2);
F(6,1) = Q - 0.436*lambda + 10*cos(d3) + 10*cos(d3-d2) - 19.8;
June 26, 2008 Power Flow Outlines - 38
Power System Dynamics and Stability 231
Universidad de Castilla-La Mancha
Example
Observe that as the load (lambda) is increased, convergence problems
show up:
lambda = 20;
z0 = fsolve(@pf_eqs,[0 0 0 0 0 0], optimset(’Display’,’iter’))
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 7 396.67 360 1
1 14 246.008 1 143 1
.
.
.
9 70 2.68094e-07 0.179729 0.000311 25.9
10 77 7.43614e-16 0.00127159 1.64e-08 25.9
Optimization terminated: first-order optimality is less than options.TolFun.
z0 =
-0.2501 -1.2613 12.0000 7.0658 4.8031 20.1667
June 26, 2008 Power Flow Outlines - 39
Power System Dynamics and Stability 232
Universidad de Castilla-La Mancha
Example
Convergence problems develop until the equations fail to converge:
lambda = 22;
z0 = fsolve(@pf_eqs,[0 0 0 0 0 0], optimset(’Display’,’iter’))
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 7 480.33 396 1
1 14 310.93 1 168 1
.
.
.
90 595 0.00622095 0.0211144 0.0322 0.0211
91 602 0.00621755 0.0211144 0.0155 0.0211
Maximum number of function evaluations reached: increase options.MaxFunEvals.
z0 =
-0.3322 -1.7225 13.1600 11.8633 8.5479 29.1072
June 26, 2008 Power Flow Outlines - 40
Power System Dynamics and Stability 233
Universidad de Castilla-La Mancha
Stability Concepts Outlines
Basic stability concepts
Nonlinear systems:
Ordinary differential equations (ODE)
Differential algebraic equations (DAE)
Equilibrium points:
Definitions.
Stability:
Linearization.
Eigenvalue analysis.
Stability regions.
June 26, 2008 Stability Concepts Outlines - 1
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ODE Systems
Nonlinear systems are represented by a nonlinear set of differential
equations:
x = s(x, p, λ)
where
x → n state variables (e.g. generator angles)
p → k controllable parameters (e.g. compensation)
λ → ℓ uncontrollable parameters (e.g. loads)
s(·) → n nonlinear functions (e.g. generator equations)
June 26, 2008 Stability Concepts Outlines - 2
Power System Dynamics and Stability 235
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ODE Systems
For example, for a simple generator-system model:
Generator
Infinite bus (M = ∞)
AVR
System
E′∠δ V ∠0V1∠δ1 V2∠δ2
jxL jxthjx′G
PG + jQG PL + jQL
The generator is modeled as a simple d axis transient voltage behind
transient reactance.
June 26, 2008 Stability Concepts Outlines - 3
Power System Dynamics and Stability 236
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ODE Systems
The generator has only the mechanical dynamics:
δ = ω = ωr − ω0
ω =1
M(PL − PG − Dω)
where
PG =E′V
x′G + xL + xth
sin δ
=V1V
xL + xthsin δ1
June 26, 2008 Stability Concepts Outlines - 4
Power System Dynamics and Stability 237
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ODE Systems
If the AVR is modeled, V1 may be assumed to be kept constant by
varying E′, with the generator’s reactive power within limits:
QG =V 2
1
xL + xth− V1V
xL + xthcos δ1
⇒ QG min ≤ QG ≤ QG max
June 26, 2008 Stability Concepts Outlines - 5
Power System Dynamics and Stability 238
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ODE Systems
If the AVR is not modeled:
x = [δ, ω]T → state variables
p = [E′, V ]T → controlled parameters
λ = PL → uncontrolled parameters
Hence, assuming p = [1.5, 1]T , M = D = 0.1, and x = 0.75:
s(x, p, λ) =
x1 = x2
x2 = 10λ = 20 sinx1 − x2
These are basically the pendulum equations.
June 26, 2008 Stability Concepts Outlines - 6
Power System Dynamics and Stability 239
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DAE Systems
Differential Algebraic Equation (DAE) models are defined as:
x = f(x, y, p, λ)
0 = g(x, y, p, λ)
where:
y → m algebraic variables (e.g. load voltages)
f(·) → n nonlinear differential equations (e.g. generator equations)
g(·) → m nonlinear algebraic equations (e.g. reactive power
equations)
June 26, 2008 Stability Concepts Outlines - 7
Power System Dynamics and Stability 240
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DAE Systems
For example, for the generator-infinite bus example with AVR, for
QG min ≤ QG ≤ QG max:
δ = ω
ω =1
M
(
PL − E′V
xsin δ − Dω
)
0 =E′V
xsin δ − V1V
xL + xthsin δ1
0 = −QG − V 21
x′G
+V1E
′
x′G
cos(δ1 − δ)
0 = QG − V 21
xL + xth+
V1V
xL + xthcos δ1
June 26, 2008 Stability Concepts Outlines - 8
Power System Dynamics and Stability 241
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DAE Systems
Thus, for:
x = [δ, ω]T
y = [E′, δ1, QG]T
p = [V1, V ]T = [1, 1]T
λ = PL
M = D = 0.1, x = 0.75, x′G = 0.25, xL + xth = 0.5
f(x, y, p, λ) =
x1 = x2
x2 = 10λ − 13.33y1 sinx1 − x2
g(x, y, p, λ) =
0 = 1.333y1 sinx1 − 2 sin y2
0 = −y3 − 4 + 4y1 cos(y2 − x1)
0 = y3 − 2 + 2 cos y2
June 26, 2008 Stability Concepts Outlines - 9
Power System Dynamics and Stability 242
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DAE Systems
If QG = QG max or QG = QG min:
x = [δ, ω]T
y = [E′, δ1, V1]T
p = [QG, V ]T = [±0.5, 1]T
λ = PL
M = D = 0.1, x = 0.75, x′G = 0.25, xL + xth = 0.5
f(x, y, p, λ) =
x1 = x2
x2 = 10λ − 13.33y1y3 sinx1 − x2
g(x, y, p, λ) =
0 = 1.333y1 sinx1 − 2y3 sin y2
0 = ∓0.5 − 4y23 + 4y1y3 cos(y2 − x1)
0 = ±0.5 − 2y23 + 2y3 cos y2
June 26, 2008 Stability Concepts Outlines - 10
Power System Dynamics and Stability 243
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DAE Systems
If the Jacobian matrix Dyg = [∂gi/∂yi]m×m is nonsingular, i.e.
invertible, along the trajectory solutions, the system can be transformed
into an ODE system (Implicit Function Theorem):
y = h(x, p, λ)
⇒ x = f(x, h(x, p, λ), p, λ)
= s(x, p, λ)
In practice, this is a purely “theoretical” exercise that is not carry out due
to its complexity.
The system model should be revised when Dyg is singular.
June 26, 2008 Stability Concepts Outlines - 11
Power System Dynamics and Stability 244
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Equilibria
For the ODE system, equilibria are defined as the solution x0 for given
parameter values p0 and λ0 of the set of equations
s(x0, p0, λ0) = 0
There are multiple solutions to this problem, i.e. multiple equilibrium
points.
June 26, 2008 Stability Concepts Outlines - 12
Power System Dynamics and Stability 245
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Equilibria
The “stability” of these equilibria is defined by linearizing the nonlinear
system around x0, i.e.
∆x =
[∂si
∂xj(x0, p0, λ0)
]
n×n
x − x0︸ ︷︷ ︸
∆x
= Dxs|0∆x
where Dxs|0 = Dxs(x0, p0, λ0) = ∂s/∂x|0 is the system Jacobian
matrix evaluated at the equilibrium point.
June 26, 2008 Stability Concepts Outlines - 13
Power System Dynamics and Stability 246
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Equilibria
From linear system theory, the linear system stability is defined by the
eigenvalues µi of the Jacobian matrix Dxs|0, which are defined as the
solutions of the equation:
Dxs|0v = µv → right e-vector
Dxs|T0 w = µw → left e-vector
⇒ det(Dxs|0 − µIn) = 0
anµn + an−1µn−1 + . . . + a1µ + a0 = 0
June 26, 2008 Stability Concepts Outlines - 14
Power System Dynamics and Stability 247
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Equilibria
There are n complex eigenvalues, left and right eigenvectors associated
with the system Jacobian Dxs|0.
In practice, these eigenvalues are not computed using characteristic
polynomial but other more efficient numerical techniques, as the costs
associated with computing these values is rather large in realistic power
systems.
June 26, 2008 Stability Concepts Outlines - 15
Power System Dynamics and Stability 248
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Equilibria
These eigenvalues define the small perturbation stability of the ODE
system, i.e. the “local” stability of the nonlinear system near the
equilibrium points:
Stable equilibrium point (s.e.p.): The system is locally stable about x0
if all the eigenvalues µi(Dxs|0) are on the left-half (LH) of the
complex plane.
Unstable equilibrium point (u.e.p.): The system is locally unstable
about x0 if at least one eigenvalue µi(Dxs|0) is on the right-half
(RH) of the complex plane.
June 26, 2008 Stability Concepts Outlines - 16
Power System Dynamics and Stability 249
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Equilibria
The equilibrium point x0 is a bifurcation point if at least one eigenvalue
µi(Dxs|0) is on the imaginary axis of the complex plane.
Some systems have equilibria with eigenvalues on the imaginary axis
without these being bifurcation points; for example, a lossless
generator-infinite bus system with no damping.
June 26, 2008 Stability Concepts Outlines - 17
Power System Dynamics and Stability 250
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Equilibria
For example, for the simple generator-infinite bus example with no AVR:
x1 = x2
x2 = 10λ − 20 sinx1 − x2
the equilibrium points can be found from the steady-state (power flow)
equations:
0 = x20
0 = 10λ − 20 sinx10 − x20
June 26, 2008 Stability Concepts Outlines - 18
Power System Dynamics and Stability 251
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Equilibria
which leads to the solutions:
x10
x20
=
sin−1(λ/2)
0
⇒
δ0
ω0
=
sin−1(PL/2)
0
This yields basically three equilibrium points (other solutions are just
“repetitions” of these three):
s.e.p → −π/2 < x1s < π/2
u.e.p.1 → x1u1 = x1s + π
u.e.p.2 → x1u2 = x1s − π
June 26, 2008 Stability Concepts Outlines - 19
Power System Dynamics and Stability 252
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Equilibria
This corresponds to the intersections of PG = E′V/x sin δ with PL:
Stable
UnstableUnstable
BifurcationEV/x PG
PL
δ, (x1)−π π
π/2δµ2
(xµ2) (xµ1)(xs)
δµ1δs
June 26, 2008 Stability Concepts Outlines - 20
Power System Dynamics and Stability 253
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Equilibria
The stability of these equilibria is determined using the system Jacobian:
Dxs|0 =
∂s1/∂x1|0 ∂s1/∂x2|0∂s2/∂x1|0 ∂s2/∂x2|0
=
0 1
−20 cos x10 −1
June 26, 2008 Stability Concepts Outlines - 21
Power System Dynamics and Stability 254
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Equilibria
⇒ det(Dxs|0 − µI2) = det
−µ 1
−20 cos x10 −1 − µ
= µ2 + µ + 20 cos x10
= 0
⇒ µ1,2 = −1
2± 1
2
√1 − 80 cos x10
PL = 1 ⇒ µ1,2(Dxs|xs) = −0.5 ± j4.132
µ1,2(Dxs|xu1/u2) =
3.192
−5.192
June 26, 2008 Stability Concepts Outlines - 22
Power System Dynamics and Stability 255
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Equilibria
For this system, the equilibria are:
stable if∂PG
∂δ> 0
unstable if∂PG
∂δ< 0
bifurcation point for
∂PG
∂δ= 0 ⇒ δ = π/2
June 26, 2008 Stability Concepts Outlines - 23
Power System Dynamics and Stability 256
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Equilibria
For DAE systems, the equilibria z0 = (x0, y0) for parameter values p0
and λ0 are defined as the solution to the nonlinear, steady state problem:
F (x0, y0, p0, λ0) = 0
f(x0, y0, p0, λ0) = 0
g(x0, y0, p0, λ0) = 0
In this case, the linearization about (x0, y0) yields:
∆x = Dxf |0∆x + Dyf |0∆y
∆0 = Dxg|0∆x + Dyg|0∆y
June 26, 2008 Stability Concepts Outlines - 24
Power System Dynamics and Stability 257
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Equilibria
Hence, by eliminating ∆y from these equations, one obtains:
Dxs|0 = Dxf |0 − Dyf |0Dyg|−10 Dxg|0
Observe that, as mentioned before, the nonsingularity of the Jacobian
Dyg|0 in this case is required.
The same local stability conditions apply in this case based on the
eigenvalues of Dxs|0.
June 26, 2008 Stability Concepts Outlines - 25
Power System Dynamics and Stability 258
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Equilibria
For the generator-infinite bus example with AVR example (within QG
limits), the steady-state solutions are obtained from solving the
steady-state or power flow equations:
0 = x20
0 = 10λ − 13.33y10 sinx10 − x20
0 = 1.333y10 sinx10 − 2 sin y20
0 = −y30 − 4 + 4y10 cos(y20 − x10)
0 = y30 − 2 + 2 cos y20
June 26, 2008 Stability Concepts Outlines - 26
Power System Dynamics and Stability 259
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Equilibria
Which in MATLAB format are:
function f = dae_eqs(z)
global lambda
x10 = z(1);
x20 = z(2);
y10 = z(3);
y20 = z(4);
y30 = z(5);
f(1,1) = x20;
f(1,2) = 10*lambda - 13.33 * y10 * sin(x10) - x20;
f(1,3) = 1.333 * y10 * sin (x10) - 2 * sin(y20);
f(1,4) = -y30 - 4 + 4 * y10 * cos(y20 - x10);
f(1,5) = y30 - 2 + 2 * cos(y20);
June 26, 2008 Stability Concepts Outlines - 27
Power System Dynamics and Stability 260
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Equilibria
This generates the equilibrium point for λ = PL = 1:
>> lambda = 1;
>> z0 = fsolve(@dae_eqs,[0 0 1 0 1],optimset(’Display’,’iter’))
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 6 102 133 1
1 12 1.52385 1 10.2 1
2 18 0.0050436 0.298687 0.219 2.5
3 24 2.72816e-05 0.0607967 0.0534 2.5
4 30 2.0931e-13 0.000413599 4.89e-06 2.5
5 36 4.98474e-28 5.13313e-08 2.23e-13 2.5
Optimization terminated: first-order optimality is less than options.TolFun.
z0 =
0.7539 0 1.0959 0.5236 0.2679
June 26, 2008 Stability Concepts Outlines - 28
Power System Dynamics and Stability 261
Universidad de Castilla-La Mancha
Equilibria
And the symbolic Jacobian matrices:
syms x1 x2 y1 y2 y3 real
z = [x1 x2 y1 y2 y3];
F = dae_eqs(z);
f = F(1:2);
g = F(3:5);
Dxf = jacobian(f,[x1,x2])
Dyf = jacobian(f,[y1,y2,y3])
Dxg = jacobian(g,[x1,x2])
Dyg = jacobian(g,[y1,y2,y3])
June 26, 2008 Stability Concepts Outlines - 29
Power System Dynamics and Stability 262
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Equilibria
And the symbolic Jacobian matrices:
Dxf =
[ 0, 1]
[ -1333/100*y1*cos(x1), -1]
Dyf =
[ 0, 0, 0]
[ -1333/100*sin(x1), 0, 0]
Dxg =
[ 1333/1000*y1*cos(x1), 0]
[ -4*y1*sin(-y2+x1), 0]
[ 0, 0]
Dyg =
[ 1333/1000*sin(x1), -2*cos(y2), 0]
[ 4*cos(-y2+x1), 4*y1*sin(-y2+x1), -1]
[ 0, -2*sin(y2), 1]
June 26, 2008 Stability Concepts Outlines - 30
Power System Dynamics and Stability 263
Universidad de Castilla-La Mancha
Equilibria
These generate the following eigenvalues at the equilibrium point:
Dxf =
[ 0, 1]
[ -1333/100*y1*cos(x1), -1]
Dyf =
[ 0, 0, 0]
[ -1333/100*sin(x1), 0, 0]
Dxg =
[ 1333/1000*y1*cos(x1), 0]
[ -4*y1*sin(-y2+x1), 0]
[ 0, 0]
Dyg =
[ 1333/1000*sin(x1), -2*cos(y2), 0]
[ 4*cos(-y2+x1), 4*y1*sin(-y2+x1), -1]
[ 0, -2*sin(y2), 1]
June 26, 2008 Stability Concepts Outlines - 31
Power System Dynamics and Stability 264
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Equilibria
These generate the following eigenvalues at the equilibrium point:
x1 = z0(1); x2 = z0(2); y1 = z0(3); y2 = z0(4); y3 = z0(5);
A = vpa(subs(Dxf),5);
B = vpa(subs(Dyf),5);
C = vpa(subs(Dxg),5);
D = vpa(subs(Dyg),5);
Dxs = A - B * inv(D) * C;
ev = vpa(eig(Dxs),5)
ev =
-.50000+3.5698*i
-.50000-3.5698*i
Hence, this is a s.e.p.
June 26, 2008 Stability Concepts Outlines - 32
Power System Dynamics and Stability 265
Universidad de Castilla-La Mancha
Equilibria A u.e.p. can be computed as well from these equations (neglecting QG
limits):>> z0 = fsolve(@dae_eqs,[3 0 0.1 3 3],optimset(’Display’,’iter’));
z0 =
2.7465 0 1.9491 2.6180 3.7321
>> x1 = z0(1); x2 = z0(2); y1 = z0(3); y2 = z0(4); y3 = z0(5);
>> A = vpa(subs(Dxf),5);
>> B = vpa(subs(Dyf),5);
>> C = vpa(subs(Dxg),5);
>> D = vpa(subs(Dyg),5);
>> Dxs = A - B * inv(D) * C;
>> ev = vpa(eig(Dxs),5)
ev =
4.2892
-5.2892
June 26, 2008 Stability Concepts Outlines - 33
Power System Dynamics and Stability 266
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Equilibria
Associated with every s.e.p. xs there is a stability region A(xs), which
basically corresponds to the region of system variables that are all
attracted to xs, i.e. x(t → ∞) → xs:
stable
unstable(stability region
boundary)
x1(0)
x2(0)
x1(t)
x2(t)
xs
A(xs)
∂A(xs)
June 26, 2008 Stability Concepts Outlines - 34
Power System Dynamics and Stability 267
Universidad de Castilla-La Mancha
Equilibria
if A(xs) is known, the stability of a system for large perturbations can be
readily evaluated.
However, determining this region is a rather difficult task.
This can realistically be accomplished only for 2- or 3-dimendional
systems using “sophisticated” nonlinear system analysis techniques.
June 26, 2008 Stability Concepts Outlines - 35
Power System Dynamics and Stability 268
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 0.2
x1
x 2
∂A(xs)
xs
xu1
xu2
June 26, 2008 Stability Concepts Outlines - 36
Power System Dynamics and Stability 269
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 0.5
x1
x 2
∂A(xs)
xs x
u1x
u2
June 26, 2008 Stability Concepts Outlines - 37
Power System Dynamics and Stability 270
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 1.0
x1
x 2
∂A(xs)
xs
xu1
xu2
June 26, 2008 Stability Concepts Outlines - 38
Power System Dynamics and Stability 271
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 1.5
x1
x 2
∂A(xs)
xs
xu1
June 26, 2008 Stability Concepts Outlines - 39
Power System Dynamics and Stability 272
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 1.8
x1
x 2
∂A(xs)
xs
xu1
June 26, 2008 Stability Concepts Outlines - 40
Power System Dynamics and Stability 273
Universidad de Castilla-La Mancha
Equilibria
−10 −8 −6 −4 −2 0 2 4 6 8 10−15
−10
−5
0
5
10
λ = 1.9
x1
x 2
∂A(xs)
xs
xu1
June 26, 2008 Stability Concepts Outlines - 41
Power System Dynamics and Stability 274
Universidad de Castilla-La Mancha
Equilibria
In real systems, trial-and-error techniques are usually used:
A contingency yields a given initial condition x(0).
For the post-contingency system, the time trajectories x(t) can be
computed by numerical integration.
If x(t) converges to the post-contingency equilibrium point xs, the
system is stable, i.e. x(0) ∈ A(xs).
If it diverges, the system is unstable.
June 26, 2008 Stability Concepts Outlines - 42
Power System Dynamics and Stability 275
Universidad de Castilla-La Mancha
Voltage Stability Outlines
Definitions.
Basic concepts.
Continuation Power Flow.
Direct Methods.
Indices.
Controls and protections.
Practical applications.
Examples.
June 26, 2008 Voltage Stability - 1
Power System Dynamics and Stability 276
Universidad de Castilla-La Mancha
Voltage Stability Definitions
IEEE-CIGRE classification (IEEE/CIGRE Joint Task Force on Stability)
Terms and Definitions, “Definitions and Classification of Power System
Stability”, IEEE Trans. Power Systems and CIGRE Technical Brochure
231, 2003:
Voltage StabilityVoltage StabilityAngle Stability
Stability Stability
Stability
Stability
Stability
Voltage
Transient
FrequencyRotor Angle
Power System
Large SmallSmall Disturbance
Disturbance Disturbance
Long Term
Long Term
Short Term
Short Term
Short Term
June 26, 2008 Voltage Stability - 2
Power System Dynamics and Stability 277
Universidad de Castilla-La Mancha
Voltage Stability Definitions
“Power system stability is the ability of an electric power system, for a
given initial operating condition, to regain a state of operating equilibrium
after being subjected to a physical disturbance, with most system
variables bounded so that practically the entire system remains intact.”
“Voltage stability refers to the ability of a power system to maintain steady
voltages at all buses in the system after being subjected to a disturbance
from a given initial operating condition.”
June 26, 2008 Voltage Stability - 3
Power System Dynamics and Stability 278
Universidad de Castilla-La Mancha
Voltage Stability Definitions
The inability of the trnasmission system to supply the demand leads to a
“voltage collapse” problem.
This problem is associated with the “disappearance” of a stable
equilibrium point due to a saddle-node or limit-induced bifurcation point,
typically due to contingencies in the system (e.g. August 14, 2003
Northeast Blackout).
June 26, 2008 Voltage Stability - 4
Power System Dynamics and Stability 279
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The concept of saddle-node and limit induced bifurcations can be readily
explained using the generator-load example:
V1∠δ1 V2∠δ2
jxL
PG + jQG PL + jQL
June 26, 2008 Voltage Stability - 5
Power System Dynamics and Stability 280
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Neglecting for simplicity losses, electromagnetic dynamics, and the
transient impedance in the d-axis transient model, the generator can be
simulated with:
δ1 = ω1 = ωr − ω0
ω1 =1
M(Pm − PG − DGω1)
June 26, 2008 Voltage Stability - 6
Power System Dynamics and Stability 281
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The load can be simulated using the mixed models.
For P , neglecting voltage dynamics (Tpv = 0) and voltage dependence
(α = 0):
PL = Kpff2 + Kpv[V α2 + TpvV2]
⇒ PL = Pd + DLω2
δ2 = ω2 =1
DL(PL − Pd)
June 26, 2008 Voltage Stability - 7
Power System Dynamics and Stability 282
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For Q, neglecting frequency dependence (Kqf = 0) and voltage
dependence (β = 0)
QL = Kqff2 + Kqv[V β2 + TqvV2]
⇒ QL = Qd + τ V2
V2 =1
τ(QL − Qd)
June 26, 2008 Voltage Stability - 8
Power System Dynamics and Stability 283
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The transmission system yields the power flow equations:
PL = −V1V2
XLsin(δ2 − δ1)
PG =V1V2
XLsin(δ1 − δ2)
QL = − V 22
XL+
V1V2
XLcos(δ2 − δ1)
QG =V 2
1
XL− V1V2
XLcos(δ1 − δ2)
Finally, since the system is lossless:
Pm = Pd
June 26, 2008 Voltage Stability - 9
Power System Dynamics and Stability 284
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Hence, defining:
δ = δ1 − δ2
ω = ω1
⇒ δ = ω − ω2
the system equations are:
ω =1
M
(
Pd − V1V2
XLsin δ − DGω
)
δ = ω − 1
DL
(V1V2
XLsin δ − Pd
)
V2 =1
τ
(
− V 22
XL+
V1V2
XLcos δ − Qd
)
June 26, 2008 Voltage Stability - 10
Power System Dynamics and Stability 285
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Observe that these equations also represent a generator-dynamic load
system with no AVR and with XL including X ′G, where V1 would stand
for E′G.
The steady-state load demand may be assumed to have a constant
power factor, i.e.
Qd = kPd
If generator reactive power limits are considered, and neglecting X ′G,
one has a DAE system.
June 26, 2008 Voltage Stability - 11
Power System Dynamics and Stability 286
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For QGmin≤ QG ≤ QGmax
:
ω =1
M
(
Pd − V10V2
XLsin δ − DGω
)
δ = ω − 1
DL
(V10V2
XLsin δ − Pd
)
V2 =1
τ
(
− V 22
XL+
V10V2
XLcos δ − Qd
)
0 = QG −−V 210
XL+
V10V2
XLcos δ
with
x = [ω, δ, V2]T y = QG
p = V10 λ = Pd
June 26, 2008 Voltage Stability - 12
Power System Dynamics and Stability 287
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For QG = QGmin,max:
ω =1
M
(
Pd − V1V2
XLsin δ − DGω
)
δ = ω − 1
DL
(V1V2
XLsin δ − Pd
)
V2 =1
τ
(
− V 22
XL+
V1V2
XLcos δ − Qd
)
0 = QGmin,max− V 2
1
XL+
V1V2
XLcos δ
with
x = [ω, δ, V2]T y = V1
p = QGmin,max λ = Pd
June 26, 2008 Voltage Stability - 13
Power System Dynamics and Stability 288
Universidad de Castilla-La Mancha
Voltage Stability Concepts
All the equilibrium point for the system with and without limits can be
obtained solving the power flow equations:
0 = Pd − V10V20
XLsin δ0
0 = kPd +V 2
20
XL− V10V20
XLcos δ0
0 = QG0 −V 2
10
XL+
V10V20
XLcos δ0
June 26, 2008 Voltage Stability - 14
Power System Dynamics and Stability 289
Universidad de Castilla-La Mancha
Voltage Stability Concepts
And the stability of these equilibrium points come from the state matrix:
Without limits, or for QGmin≤ QG ≤ QGmax
:
Dxs|0 =
−DG
M −V10V20
MXLcos δ0 − V10
MXLsin δ0
1 −V10V20
DLXLcos δ0 − V10
DLXLsin δ0
0 −V10V20
τXLsin δ0 −2 V20
τXL+ V10
τXLcos δ0
June 26, 2008 Voltage Stability - 15
Power System Dynamics and Stability 290
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For QG = QGmin,max:
Dxs|0 =
−DG
M −V10V20
MXLcos δ0 − V10
MXLsin δ0
1 −V10V20
DLXLcos δ0 − V10
DLXLsin δ0
0 −V10V20
τXLsin δ0 −2 V20
τXL+ V10
τXLcos δ0
−
− V10
MXLsin δ0
− V10
DLXLsin δ0
V20
τXLcos δ0
[
−2V10
τXL+
V20
τXLcos δ0
]−1
0
−V10V20
XLsin δ0
V10
XLcos δ0
T
June 26, 2008 Voltage Stability - 16
Power System Dynamics and Stability 291
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Assume XL = 0.5, M = 1, DG = 0.01, DL = 0.1, τ = 0.01,
k = 0.25.
With the help of MATLAB and the continuation power flow routine of PSAT,
for the system without limits and V1 = V10 = 1, the power flow solutions
yield a “PV” or “nose” curve (bifurcation diagram):
June 26, 2008 Voltage Stability - 17
Power System Dynamics and Stability 292
Universidad de Castilla-La Mancha
Voltage Stability Concepts
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
λ = Pd
x3
=V
2s.e.p.
u.e.p.
SNB
June 26, 2008 Voltage Stability - 18
Power System Dynamics and Stability 293
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The saddle-node bifurcation point SNB corresponds to a point where the
state matrix Dxs|0 is singular (one zero eigenvalue).
This is typically associated with a power flow solution with a singular PF
Jacobian DzF |0.
This is not always the case, as for more complex dynamic models, the
singularity of the state matrix does not necessarily correspond to a
singularity o the PF Jacobian, and vice versa.
June 26, 2008 Voltage Stability - 19
Power System Dynamics and Stability 294
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Observe that the SNB point corresponds to a maximum value
λmax = Ps max ≈ 0.78, which is why is also referred to as the
maximum loading or loadability point.
For a load greater than Pd max, there are no PF solutions.
This point is also referred to as the voltage collapse point.
June 26, 2008 Voltage Stability - 20
Power System Dynamics and Stability 295
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Modeling a “contingency” at λ0 = Pd0 = 0.7 by increasing
XL = 0.5 → 0.6:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
λ = Pd
x3
=V
2
operatingpoint
contingency
XL = 0.5
XL = 0.6
June 26, 2008 Voltage Stability - 21
Power System Dynamics and Stability 296
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The dynamic solution yields a voltage collapse:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
0
1
2
3
4
5
6
Voltage collapse
Operating point Contingency
V2
V1
δω
t [s]
June 26, 2008 Voltage Stability - 22
Power System Dynamics and Stability 297
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For the system with limits and QGmax,min= ±0.5, the “PV” or “nose”
curve is:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.70
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
λ = Pd
x3
=V
2
s.e.p.
u.e.p.
LIB
June 26, 2008 Voltage Stability - 23
Power System Dynamics and Stability 298
Universidad de Castilla-La Mancha
Voltage Stability Concepts
In this case, the maximum loading or loadability point
λmax = Pd max ≈ 0.65 corresponds to the point where the generator
reaches its maximum reactive power limit QG = QG max = 0.5, and
hence losses control of V1.
This is referred to a limit-induced bifurcation or LIB point.
“Beyond” the LIB point, there are no more power flow solutions, due to
the limit recovery mechanism of the AVR.
In this case the LIB point is also a voltage collapse point.
June 26, 2008 Voltage Stability - 24
Power System Dynamics and Stability 299
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Modeling a “contingency” at λ0 = Pd0 = 0.6 by increasing
XL = 0.5 → 0.6:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
λ = Pd
x3
=V
2
operatingpoint
contingency
XL = 0.5
XL = 0.6
June 26, 2008 Voltage Stability - 25
Power System Dynamics and Stability 300
Universidad de Castilla-La Mancha
Voltage Stability Concepts
The dynamic solution also yields a voltage collapse:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
0
1
2
3
4
5
6
Operating point Contingency
Voltage collapse
V2
V1
δω
t [s]
June 26, 2008 Voltage Stability - 26
Power System Dynamics and Stability 301
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Collapse problems can be avoided using shunt or series compensation:
V1∠δ1 V2∠δ2
jxL
PG + jQG PL + jQL
−jxC
June 26, 2008 Voltage Stability - 27
Power System Dynamics and Stability 302
Universidad de Castilla-La Mancha
Voltage Stability Concepts
In this case, the system equations are:
For QGmin≤ QG ≤ QGmax
:
ω =1
M(Pd − V10V2BL sin δ − DGω)
δ = ω − 1
DL(V10V2BL sin δ − Pd)
V2 =1
τ[−V 2
2 (BL − BC) + V10V2BL cos δ − kPd]
0 = QG − V 210BL + V10V2BL cos δ
where
BL =1
XLBC =
1
XC
June 26, 2008 Voltage Stability - 28
Power System Dynamics and Stability 303
Universidad de Castilla-La Mancha
Voltage Stability Concepts
For QG = QGmin,max:
ω =1
M(Pd − V10V2BL sin δ − DGω)
δ = ω − 1
DL(V10V2BL sin δ − Pd)
V2 =1
τ[−V 2
2 (BL − BC) + V10V2BL cos δ − kPd]
0 = QGmin,max− V 2
10BL + V10V2BL cos δ
June 26, 2008 Voltage Stability - 29
Power System Dynamics and Stability 304
Universidad de Castilla-La Mancha
Voltage Stability Concepts
These equations generate the PV curves:
Without limits:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
λ = Pd
V2
Bc = 0
Bc = 0
Bc = 0.5
XL = 0.5
XL = 0.6
XL = 0.6
June 26, 2008 Voltage Stability - 30
Power System Dynamics and Stability 305
Universidad de Castilla-La Mancha
Voltage Stability Concepts
With limits:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
λ = Pd
V2
Bc = 0
Bc = 0
Bc = 0.5
XL = 0.5
XL = 0.6
XL = 0.6
June 26, 2008 Voltage Stability - 31
Power System Dynamics and Stability 306
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Applying compensation at t = 1.3 s without limits:
0 1 2 3 4 5 6 7 8 9 10−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Apply compensation
Contingency
Operating point
V2
V1
δω
t [s]
June 26, 2008 Voltage Stability - 32
Power System Dynamics and Stability 307
Universidad de Castilla-La Mancha
Voltage Stability Concepts
Applying compensation at t = 1.3 s with limits:
0 1 2 3 4 5 6 7 8 9 10−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Apply compensation
Contingency
Operating point
V2
V1
δω
t [s]
June 26, 2008 Voltage Stability - 33
Power System Dynamics and Stability 308
Universidad de Castilla-La Mancha
Continuation Power Flow
These PV or nose curves are obtained using a continuation power flow.
This technique “traces” the solutions of the power equations
F (z, p, λ) = 0
as λ changes.
June 26, 2008 Voltage Stability - 34
Power System Dynamics and Stability 309
Universidad de Castilla-La Mancha
Continuation Power Flow
The algorithm steps are:
λ
λ
λ1
λ2
z1
z2
1. Predictor3. Corrector
2. Parametrization
(∆z1, ∆λ1)
(z0, λ0)
These methods “guarantee” convergence.
June 26, 2008 Voltage Stability - 35
Power System Dynamics and Stability 310
Universidad de Castilla-La Mancha
CPF Predictor
Tangent vector method:
DzF |1dz
dλ︸︷︷︸
t1
− ∂F
∂λ
∣∣∣∣1
⇒ ∆z1 =k
‖t1‖︸︷︷︸
∆λ1
t1
(∆z1, ∆λ1)
(z1, λ1)
(z2, λ2)
June 26, 2008 Voltage Stability - 36
Power System Dynamics and Stability 311
Universidad de Castilla-La Mancha
CPF Predictor
Good method to follow “closely” the PV curves, but relatively slow.
The tangent vector t defines “sensitivities” at any power flow solution
point.
This vector can also be used as an index to predict proximity to a
saddle-node bifurcation, as opposed to using the smallest eigenvalue,
which changes in highly non linear fashion.
June 26, 2008 Voltage Stability - 37
Power System Dynamics and Stability 312
Universidad de Castilla-La Mancha
CPF Predictor
Secant method:
∆z1 = k(z1a − z1b)
∆λ1 = k(λ1a − λ1b)
(∆z1, ∆λ1)
(z1a, λ1a)
(z1b, λ1b)
(z2, λ2)
June 26, 2008 Voltage Stability - 38
Power System Dynamics and Stability 313
Universidad de Castilla-La Mancha
CPF Predictor
This method is faster but can have convergence problems with “sharp
corners” (e.g. limits):
(z1a, λ1a)
(z1b, λ1b)
(z2, λ2)
June 26, 2008 Voltage Stability - 39
Power System Dynamics and Stability 314
Universidad de Castilla-La Mancha
CPF Parametrization
Used to avoid singularities during the predictor step.
Methods:
Local : interchange a zi ∈ z with λ, i.e. “rotate” the PV curve.
Arc length (s): assume z1(s) and λ1(s); thus solve for ∆z1 and
∆λ1:
DzF |1∆z1 +∂F
∂λ
∣∣∣∣1
+ ∆λ1 = 0
∆zT1 ∆z1 + ∆λ2
1 = k
No real need for parametrization in practice if step cutting is needed.
June 26, 2008 Voltage Stability - 40
Power System Dynamics and Stability 315
Universidad de Castilla-La Mancha
CPF Corrector
The idea is to add an equation to the equilibrium equations, i.e. solve for
(z, λ) at a given point p:
F (z, p, λ) = 0
ρ(z, λ) = 0
These equations are nonsingular for the appropriate choice of ρ(·).
June 26, 2008 Voltage Stability - 41
Power System Dynamics and Stability 316
Universidad de Castilla-La Mancha
CPF Corrector
Perpendicular intersection technique:
λ
z
(∆z1, ∆λ1)
(z1, λ1)
(z2, λ2)
F (z, p, λ) = 0
(z1 + ∆z1 − z)T ∆z1 + (λ1 + ∆λ1 − λ)∆λ1 = 0
June 26, 2008 Voltage Stability - 42
Power System Dynamics and Stability 317
Universidad de Castilla-La Mancha
CPF Corrector
Local parametrization technique:
λ
z
(∆z1, ∆λ1)
(z1, λ1)
(z2, λ2)
F (z, p, λ) = 0
λ = λ1 + ∆λ1 or zi = zi1 + ∆zi1
June 26, 2008 Voltage Stability - 43
Power System Dynamics and Stability 318
Universidad de Castilla-La Mancha
Direct Methods
Allow to directly find the maximum loading point (saddle-node or limit
induced bifurcation).
This problem can be set up as an optimization problem:
Max. λ
s.t. F (z, p, λ) = 0
zmin ≤ z ≤ zmax
pmin ≤ p ≤ pmax
June 26, 2008 Voltage Stability - 44
Power System Dynamics and Stability 319
Universidad de Castilla-La Mancha
Direct Methods
If limits are ignored, the solution for given values of the control paramters
(p = p0) to the associated optimization problem, based on the
Lagrangian and KKT conditions, is given by:
F (z, p0, λ) 0 → steady stae solution
DTz F (z, p0, λ)w = 0 → singularity condition
DTλ F (z, p0, λ)w = −1 → w 6= 0 condition
This yields a saddle-node bifurcation point, and corresponds to the “left
eigenvector” (w) saddle-node equations.
June 26, 2008 Voltage Stability - 45
Power System Dynamics and Stability 320
Universidad de Castilla-La Mancha
Direct Methods
The solution to these nonlinear equations do not converge if the
maximum loading point is a limit-induced bifurcation.
In this case, the otpimization problem
Max. λ
s.t. F (z, p0, λ) = 0
zmin ≤ z ≤ zmax
must be solved using “standard” optimization techniques (e.g. Interior
Point Method).
The solution to this optimization problem may be a saddle-node or
limit-induced bifurcation.
June 26, 2008 Voltage Stability - 46
Power System Dynamics and Stability 321
Universidad de Castilla-La Mancha
Direct Methods
Observe that if the control parameters p are allowed to change, the
problem is transformed into a maximization of the maximum loading
margin.
Other optimization problems can be set up to not only
maximize/guarantee loading margins but at the same time minimize costs
(e.g. maximize social welfare).
June 26, 2008 Voltage Stability - 47
Power System Dynamics and Stability 322
Universidad de Castilla-La Mancha
Direct Methods
For example, for an OPF-based auction in electricity markets, the
following multi-objective optimization problem can be posed:
Min. G = − ω1(CTDPD − CT
S PS) − ω2λc
s.t. f(δ, V, QG, PS , PD) = 0 → PF equations
f(δc, Vc, QGc, λc, PS , PD) = 0 → Max load PF eqs.
λcmin≤ λc ≤ λcmax
→ loading margin
0 ≤ PS ≤ PSmax→ Sup. bid blocks
0 ≤ PD ≤ PDmax→ Dem. bid blocks
June 26, 2008 Voltage Stability - 48
Power System Dynamics and Stability 323
Universidad de Castilla-La Mancha
Direct Methods
With the phyiscal and security limits:
Iij(δ, V ) ≤ Iijmax→ Thermal limits
Iji(δ, V ) ≤ Ijimax
Iij(δc, Vc) ≤ Iijmax
Iji(δc, Vc) ≤ Ijimax
QGmin≤ QG ≤ QGmax
→ Gen. Q limits
QGmin≤ QGc
≤ QGmax
Vmin ≤ V ≤ Vmax → V “security” lim.
Vmin ≤ Vc ≤ Vmax
June 26, 2008 Voltage Stability - 49
Power System Dynamics and Stability 324
Universidad de Castilla-La Mancha
Direct Methods
More information about this probem can be found in:
F. Milano, C. A. Canizares and M. Invernizzi, “Multi-objective
Optimization for PRicing System Security in Electricity Marktes”, IEEE
Trans. on Power Systems, Vol. 18, No. 2, May 2003, pp. 596-604.
June 26, 2008 Voltage Stability - 50
Power System Dynamics and Stability 325
Universidad de Castilla-La Mancha
Indices
Indices have been developed with the aim of monitoring proximity to
voltage collapse for on-line applications.
For example, the minimum real eigenvalue of the system Jacobian can be
used to “measure” proximity to a saddle-node bifurcation, since this
matrix becomes singular at that point.
Many indices have been proposed, but the most “popular/useful” are:
Singular value.
Reactive power reserves.
June 26, 2008 Voltage Stability - 51
Power System Dynamics and Stability 326
Universidad de Castilla-La Mancha
Indices
The singular value index consists of simply monitoring the singular value
of the Jacobian of the power flow equations as λ changes, e.g.
June 26, 2008 Voltage Stability - 52
Power System Dynamics and Stability 327
Universidad de Castilla-La Mancha
Indices
Observations:
Computationally inexpensive.
Highly nonlinear behavior, especially when control limits are reached.
Cannot really be used to detect proximity to limit-induced bifurcation.
Useful in some OPF-applications to help represent voltage stability as
a constraints.
June 26, 2008 Voltage Stability - 53
Power System Dynamics and Stability 328
Universidad de Castilla-La Mancha
Indices
The reactive-power-reserve index consists of monitoring the “difference”
between the reactive power generator output and its maximum limit, e.g.
for a generator bus system:
June 26, 2008 Voltage Stability - 54
Power System Dynamics and Stability 329
Universidad de Castilla-La Mancha
Indices
Observations:
Computationally inexpensive.
Highly nonlinear behavior.
Only works for the “right” generators, i.e. the generators associated
with the limit-induced bifurcation.
Cannot really be used to detect proximity to a saddle-node bifurcation.
June 26, 2008 Voltage Stability - 55
Power System Dynamics and Stability 330
Universidad de Castilla-La Mancha
Voltage Stability Applications
These concepts and associated techniques are applied to real power
systems through the computation of PV curves.
These curves are obtained either through a simple series of “continuous”
power flows or using actual CPFs.
In both cases, these “nose” curves are computed with respect to load
changes, which are defined as follows:
PL = PL0 + λ∆PL
QL = QL0 + λ∆QL
June 26, 2008 Voltage Stability - 56
Power System Dynamics and Stability 331
Universidad de Castilla-La Mancha
Voltage Stability Applications
Generator power changes are then defined, with the exception of the
slack bus, as:
PG = PG0 + λ∆PG
For a “distributed” slack bus model:
PG = PG0 + (λ + kG)∆PG
for all generators, where kG is a variable in the power flow equations
replacing the variable power in the slack bus.
June 26, 2008 Voltage Stability - 57
Power System Dynamics and Stability 332
Universidad de Castilla-La Mancha
Voltage Stability Applications
The Available Transfer Capability (ATC) of the tranmsission system is
typically obtained from (NERC definition):
ATC = TTC + ETC + TRM
The ATC can be associated to the maximum loading margin λmax of the
system if N-1 contingency criteria are taken into account.
June 26, 2008 Voltage Stability - 58
Power System Dynamics and Stability 333
Universidad de Castilla-La Mancha
Voltage Stability Applications
TTC or Total Transfer Capability is the maximum loading level of the
system considering N-1 contingency criteria, i.e. the λmax for the worst
realistic single contingency.
ETC or Existent Transmission Commitments represents the “current”
loading level plus any reserved transmission commitments.
TRM or Transmission Reliability Margin which is an additional margin
defined to represent other contingencies.
June 26, 2008 Voltage Stability - 59
Power System Dynamics and Stability 334
Universidad de Castilla-La Mancha
Voltage Stability Applications
For example, for WECC, systems should be operated a minimum 5%
“distance” away from the maximum loadibility point when contigencies are
considered:
June 26, 2008 Voltage Stability - 60
Power System Dynamics and Stability 335
Universidad de Castilla-La Mancha
Control and Protections
To improve system loadability, i.e. avoid voltage stability problems, the
most common control and protections techniques are:
Increase reactive power output from generators, especially in the
“critical” area (the area most “sensitive” to voltage problems).
Introduce shunt compesation through the use of MSC, SVC or
STATCOM (see slides 304 and 305).
Use undervoltage relays.
June 26, 2008 Voltage Stability - 61
Power System Dynamics and Stability 336
Universidad de Castilla-La Mancha
Secondary Voltage Regulation
One way of improving reactive power support is to coordinate the reactive
power outputs of generators.
The French and Italians typically referred to as “secondary voltage
regulation” or control.
The basic structure and controls are:
June 26, 2008 Voltage Stability - 62
Power System Dynamics and Stability 337
Universidad de Castilla-La Mancha
Secondary Voltage Regulation
June 26, 2008 Voltage Stability - 63
Power System Dynamics and Stability 338
Universidad de Castilla-La Mancha
Secondary Voltage Regulation
June 26, 2008 Voltage Stability - 64
Power System Dynamics and Stability 339
Universidad de Castilla-La Mancha
Secondary Voltage Regulation
Observations:
Control areas and associated pilot buses and controlled generators
must be “properly” identified.
Control “hierarchy” is important, i.e. PQR is “slower” than AVR, and
RVR is “slower” than PQR.
It is relatively inexpensive compared to shunt compensation solutions.
June 26, 2008 Voltage Stability - 65
Power System Dynamics and Stability 340
Universidad de Castilla-La Mancha
Undervoltage Relays
These relays are installed on sub-transmission substations (loads) to
shed during “long duration” voltage dips.
The idea is to reduce load demand on the system to increase the
loadability margin (see slide 334).
Operation is somewhat similarly to taps in a LTC:
Discrete load shedding steps (e.g. 1-2% of total load).
Activated with a time delay (e.g. 1-2 mins.) after the voltage dips
below certain values (e.g. 0.8-0.9 p.u.)
The lower the voltage, the faster and larger the load shed.
June 26, 2008 Voltage Stability - 66
Power System Dynamics and Stability 341
Universidad de Castilla-La Mancha
Example
For a 3-area sample system:
R = 0.01 p.u.
X = 0.15 p.u.
100 MW
v3<d3
150 MW
100 MW
150 MW
50 MVAr
Bus 2
Bus 3
50 MW
40 MVAr
150 MW
56 MVAr
V2∠δ2
1.02∠0
Area 1
50 MVAr
60 MVAr
June 26, 2008 Voltage Stability - 67
Power System Dynamics and Stability 342
Universidad de Castilla-La Mancha
Example
For a 3-area sample system:
Bus ∆PG ∆PL ∆QL
Name (p.u.) (p.u.) (p.u.)
Area 1 1.5 0 0
Area 2 0 1.5 0.56
Area 3 0.5 0.5 0.40
Using UWPFLOW to obtain the system PV curves, for a distributed slack
bus model:
June 26, 2008 Voltage Stability - 68
Power System Dynamics and Stability 343
Universidad de Castilla-La Mancha
Example
Data file in EPRI format (3area.wsc):
HDG
UWPFLOW data file, WSCC format
3-area example
April 2000
BAS
C
C AC BUSES
C
C | SHUNT |
C |Ow|Name |kV |Z|PL |QL |MW |Mva|PM |PG |QM |Qm |Vpu
BE 1 Area 1 138 1 150 60 0 0 0 150 0 0 1.02
B 1 Area 2 138 2 150 56 0 50 0 100 0 0 1.00
B 1 Area 3 138 3 50 40 0 50 0 100 0 0 1.00
June 26, 2008 Voltage Stability - 69
Power System Dynamics and Stability 344
Universidad de Castilla-La Mancha
Example
Data file in EPRI format (3area.wsc):
C
C AC LINES
C
C M CS N
C |Ow|Name_1 |kV1||Name_2 |kV2|||In || R | X | G/2 | B/2 |Mil|
L 1 Area 1 138 Area 2 1381 15001 .01 .15
L 1 Area 1 138 Area 3 1381 15001 .01 .15
L 1 Area 2 138 Area 3 1381 15001 .01 .15
C
C SOLUTION CONTROL CARD
C
C |Max| |SLACK BUS |
C |Itr| |Name |kV| |Angle |
SOL 50 Area 1 138 0.
END
June 26, 2008 Voltage Stability - 70
Power System Dynamics and Stability 345
Universidad de Castilla-La Mancha
Example
Generator and load change file (3area.k):
C
C UWPFLOW load and generation "direction" file
C for 3-area example
C
C BusNumber BusName DPg Pnl Qnl PgMax [ Smax Vmax Vmin ]
1 0 1.5 0.0 0.0 0 0 1.05 0.95
2 0 0.0 1.5 0.56 0 0 1.05 0.95
3 0 0.5 0.5 0.40 0 0 1.05 0.95
June 26, 2008 Voltage Stability - 71
Power System Dynamics and Stability 346
Universidad de Castilla-La Mancha
Example
Batch file for UNIX (run3area):
echo -1- Run base case power flow
uwpflow 3area.wsc -K3area.k
echo -2- Obatin PV curves and maximum loading
uwpflow 3area.wsc -K3area.k -cthreearea.m -m -ltmp.l -s
echo - with bus voltage limits enforced
uwpflow 3area.wsc -K3area.k -c -7 -k0.1
echo - with current limits enforced
uwpflow 3area.wsc -K3area.k -c -ltmp.l -8 -k0.1
June 26, 2008 Voltage Stability - 72
Power System Dynamics and Stability 347
Universidad de Castilla-La Mancha
Example
Batch file for Windows (run3area.bat):
rem -1- Run base case power flow
uwpflow 3area.wsc -K3area.k
rem -2- Obatin PV curves and maximum loading
uwpflow 3area.wsc -K3area.k -cthreearea.m -m -ltmp.l -s
rem - with bus voltage limits enforced
uwpflow 3area.wsc -K3area.k -c -7 -k0.1
rem - with current limits enforced
uwpflow 3area.wsc -K3area.k -c -ltmp.l -8 -k0.1
June 26, 2008 Voltage Stability - 73
Power System Dynamics and Stability 348
Universidad de Castilla-La Mancha
Example
PV curves (threearea.m):
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
50
100
150
L.F. [p.u.]
Profiles
kVArea 3 138
kVArea 2 138
kVArea 1 138
⇒ λmax ≈ 1.6 p.u.
June 26, 2008 Voltage Stability - 74
Power System Dynamics and Stability 349
Universidad de Castilla-La Mancha
Example
Hence, if contingencies are not considered:
TTC =∑
PL0 + λmax
∑
∆PL
= 670 MW
ETC =∑
PL0
= 350 MW
TRM = 0.05TTC
= 33.5 MW
⇒ ATC = 286.5 MW
June 26, 2008 Voltage Stability - 75
Power System Dynamics and Stability 350
Universidad de Castilla-La Mancha
Example
The singular value index obtained with UWPFLOW is as follows:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
0.1
0.2
0.3
0.4
0.5
0.6
0.7
L.F. [p.u.]
Ful
l mat
rix s
ing.
val
ue in
dex
June 26, 2008 Voltage Stability - 76
Power System Dynamics and Stability 351
Universidad de Castilla-La Mancha
August 14 2003 Blackout
This information is extracted from the final report of the US-Canada Joint
Task Force.
The full details of the final report can be found on the Internet at:
https://reports.energy.gov/B-F-Web-Part1.pdf
The report is titled: U.S.-Canada Power System Outage Task Force.
“Final Report on the August 14, 2003 Blackout in the United States and
Canada: Causes and Recommendations.” Washington DC: USGPO,
April 2004.
June 26, 2008 Voltage Stability - 77
Power System Dynamics and Stability 352
Universidad de Castilla-La Mancha
August 14 2003 Blackout
3 Interconnections, 10 Regional Reliability Councils:
June 26, 2008 Voltage Stability - 78
Power System Dynamics and Stability 353
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Regions and control areas of the North American Electric Reliability
Council (NERC):
June 26, 2008 Voltage Stability - 79
Power System Dynamics and Stability 354
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Reliability coordinators (some are also system and market operators,
such as the IMO and PJM):
June 26, 2008 Voltage Stability - 80
Power System Dynamics and Stability 355
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Reliability coordinator and system/market operators in the Ohio area,
where the system collapse started:
June 26, 2008 Voltage Stability - 81
Power System Dynamics and Stability 356
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Four causes:
Inadequate system understanding: FirstEnergy, ECAR
Inadequate situational awareness: FirstEnergy
Inadequate tree-trimming: FirstEnergy
Inadequate diagnostic support: MISO, PJM
June 26, 2008 Voltage Stability - 82
Power System Dynamics and Stability 357
Universidad de Castilla-La Mancha
August 14 2003 Blackout
At 12:15 EDT, MISO began having problems with its state estimator; it did
not return to full functionality until 16:04.
Sometime after 14:14, FirstEnergy began losing its energy management
system (EMS) alarms but did not know it.
At 14:20, parts of FE’s EMS began to fail first remote sites, then core
servers but FE system operators did not know this and FE IT support staff
did not tell them.
Without a functioning EMS, FE operators did not know their system was
losing lines and voltage until about 15:45.
June 26, 2008 Voltage Stability - 83
Power System Dynamics and Stability 358
Universidad de Castilla-La Mancha
August 14 2003 Blackout
FirstEnergy did not do sufficient system planning to know the
Cleveland-Akron area was seriously deficient in reactive power supply
needed for voltage support.
At 13:31 EDT, FE lost its Eastlake 5 unit, a critical source of real and
reactive power for the Cleveland-Akron area.
FE did not perform contingency analysis after this event.
June 26, 2008 Voltage Stability - 84
Power System Dynamics and Stability 359
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Lost of the Eastlake 5 unit and beginning of reactive power problems.
June 26, 2008 Voltage Stability - 85
Power System Dynamics and Stability 360
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Between 15:05 and 15:41 EDT, FE lost 3 345 kV lines in the
Cleveland-Akron area under normal loading due to contact with too-tall
trees but did not know it due to EMS problems.
Line loadings and reactive power demands increased with each line loss.
Between 15:39 and 16:08 EDT, FE lost 16 138kV lines in the
Cleveland-Akron area due to overloads and ground faults.
June 26, 2008 Voltage Stability - 86
Power System Dynamics and Stability 361
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Line outages:
June 26, 2008 Voltage Stability - 87
Power System Dynamics and Stability 362
Universidad de Castilla-La Mancha
August 14 2003 Blackout
At 16:05.57 EDT, FE lost its Sammis-Star 345 kV line due to overload.
This closed a major path for power imports into the Cleveland-Akron area
and initiated the cascade phase of the blackout.
June 26, 2008 Voltage Stability - 88
Power System Dynamics and Stability 363
Universidad de Castilla-La Mancha
August 14 2003 Blackout
The tipping point in Ohio:
June 26, 2008 Voltage Stability - 89
Power System Dynamics and Stability 364
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Effect of a line trip: increasing loading on other lines.
June 26, 2008 Voltage Stability - 90
Power System Dynamics and Stability 365
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Effect of a line trip: decreasing voltages leading to voltage collapse.
June 26, 2008 Voltage Stability - 91
Power System Dynamics and Stability 366
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Cascade:
Definition: A cascade on an electric system is a dynamic, unplanned
sequence of events that, once started, cannot be stopped by human
intervention.
Power swings, voltage fluctuations and frequency fluctuations cause
sequential tripping of transmission lines, generators, and automatic
load-shedding in a widening geographic area.
The fluctuations diminish in amplitude as the cascade spreads.
Eventually equilibrium is restored and the cascade stops.
June 26, 2008 Voltage Stability - 92
Power System Dynamics and Stability 367
Universidad de Castilla-La Mancha
August 14 2003 Blackout
The system goes “haywire”:
June 26, 2008 Voltage Stability - 93
Power System Dynamics and Stability 368
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Higher voltage lines are better able to absorb large voltage and current
swings, buffering some areas against the cascade (AEP, Pennsylvania).
Areas with high voltage profiles and ample reactive power were not
swamped by the sudden voltage and power drain (PJM and New
England).
After islanding began, some areas were able to balance generation with
load and reach equilibrium without collapsing (upstate New York and
southern Ontario).
June 26, 2008 Voltage Stability - 94
Power System Dynamics and Stability 369
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Sequence of events:
June 26, 2008 Voltage Stability - 95
Power System Dynamics and Stability 370
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Sequence of events:
June 26, 2008 Voltage Stability - 96
Power System Dynamics and Stability 371
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Sequence of events:
June 26, 2008 Voltage Stability - 97
Power System Dynamics and Stability 372
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Sequence of events:
June 26, 2008 Voltage Stability - 98
Power System Dynamics and Stability 373
Universidad de Castilla-La Mancha
August 14 2003 Blackout
When the cascade was over at 4:13 pm, as many as 50 million people in
the northeast U.S. and the province of Ontario had no power.
This is a “good” example of a voltage instability problem triggered by a
series of contingencies.
June 26, 2008 Voltage Stability - 99
Power System Dynamics and Stability 374
Universidad de Castilla-La Mancha
August 14 2003 Blackout
June 26, 2008 Voltage Stability - 100
Power System Dynamics and Stability 375
Universidad de Castilla-La Mancha
August 14 2003 Blackout
Had the system properly monitored and NERC recommended operating
guidelines followed, the system might have been saved.
An operating rule regarding max. system loadability margins similar to
WECC’s might have given the operators a better picture of the situation,
but without proper monitoring, these would have probably not worked
either.
In the absence of these, under-voltage relays with ULTC blocking might
have saved the system, as these would have automatically shed load
when the voltages started to collapse in First Energy’s (FE) region in
Ohio.
June 26, 2008 Voltage Stability - 101
Power System Dynamics and Stability 376
Universidad de Castilla-La Mancha
Voltage Stability Report
Much more information regarding the issue voltage stability can be found
in:
C. A. Canizares, editor, “Voltage Stability Assessment: Concepts,
Practices and Tools,” IEEE-PES Power System Stability Subcommittee
Special Publication, SP101PSS, August 2002. IEEE-PES WG Report
Award 2005.
This is a 283-page report, published after 5 years in preparation, and
coauthored by several voltage stability from around the world.
More details about the report can be found at:
http://thunderbox.uwaterloo.ca/˜claudio/claudio.html/#VSWG
June 26, 2008 Voltage Stability - 102
Power System Dynamics and Stability 377
Universidad de Castilla-La Mancha
Angle Stability Outlines
Definitions.
Small disturbance:
Hopf Bifurcations.
Control and mitigation.
Practical example.
Transient Stability
Time Domain.
Direct Methods.
Equal Area Criterion.
Energy Functions.
Practical applications.
June 26, 2008 Angle Stability - 1
Power System Dynamics and Stability 378
Universidad de Castilla-La Mancha
Angle Stability Definitions
IEEE-CIGRE classification (IEEE/CIGRE Joint Task Force on Stability)
Terms and Definitions, “Definitions and Classification of Power System
Stability”, IEEE Trans. Power Systems and CIGRE Technical Brochure
231, 2003:
Voltage StabilityVoltage StabilityAngle Stability
StabilityStability
Stability
Stability
Stability
Voltage
Transient
FrequencyRotor Angle
Power System
Large SmallSmall Disturbance
Disturbance Disturbance
Long Term
Long Term
Short Term Short Term
Short Term
June 26, 2008 Angle Stability - 2
Power System Dynamics and Stability 379
Universidad de Castilla-La Mancha
Angle Stability Definitions
“Rotor angle stability refers to the ability of synchronous machines of an
interconnected power system to remain in synchronism after being
subjected to a disturbance. It depends on the ability to maintain/restore
equilibrium between electromagnetic torque and mechanical torque of
each synchronous machine in the system.”
In this case, the problem becomes apparent through angular/frequency
swings in some generators which may lead to their loss of synchronism
with other generators.
June 26, 2008 Angle Stability - 3
Power System Dynamics and Stability 380
Universidad de Castilla-La Mancha
Small Disturbance
“Small disturbance (or small signal) rotor angle stability is concerned with
the ability of the power system to maintain synchronism under small
disturbances. The disturbances are considered to be sufficiently small
that linearization of system equations is permissible for purposes of
analysis.”
This problem is usually associated with the appearance of undamped
oscillations in the system due to a lack of sufficient damping torque.
Theoretically, this phenomenon may be associated with a s.e.p.
becoming unstable through a Hopf bifurcation point, typically due to
contingencies in the system (e.g. August 1996 West Coast Blackout).
June 26, 2008 Angle Stability - 4
Power System Dynamics and Stability 381
Universidad de Castilla-La Mancha
Hopf Bifurcation I
For the generator-load example, with AVR but no QG limits:
Generator
E′∠δ V1∠δ1
V10
V2∠δ2
jxLjx′G
PG + jQG PL + jQL
+
−Kvs
June 26, 2008 Angle Stability - 5
Power System Dynamics and Stability 382
Universidad de Castilla-La Mancha
Hopf Bifurcation I
The DAE model is:
ω =1
M
(
PmE′V2
Xsin δ − DGω
)
δ = ω − 1
DL
(E′V2
Xsin δ − Pd
)
E′ = Kv(V10 − V1)
V2 =1
τ
(
−V 22
X+
E′V2
Xcos δ − kPd
)
0 =V1V2
XLsin δ′ − E′V2
Xsin δ
0 = V 22
(1
XL− 1
X
)
+E′V2
Xcos δ − V1V2
XLcos δ′
June 26, 2008 Angle Stability - 6
Power System Dynamics and Stability 383
Universidad de Castilla-La Mancha
Hopf Bifurcation I
Observe that the algebraic constraint can be eliminated, since:
V1r = V1 cos δ′
V1i = V1 sin δ′
Thus:
0 =V1iV2
XL− E′V2
Xsin δ ⇒ V1i =
E′XL
Xsin δ
and
0 = V 22
(1
XL− 1
X
)
+E′V2
Xcos δ − V1rV2
XL
⇒ V1r = V2
(
1 − 1
XL
)
+E′XL
Xcos δ
June 26, 2008 Angle Stability - 7
Power System Dynamics and Stability 384
Universidad de Castilla-La Mancha
Hopf Bifurcation I
This yields the following equations, which are better for numerical time
domain simulations:
ω =1
M
(
PmE′V2
Xsin δ − DGω
)
δ = ω − 1
DL
(E′V2
Xsin δ − Pd
)
E′ = Kv
(
V10 −√
V 21r + V 2
1i
)
V2 =1
τ
(
−V 22
X+
E′V2
Xcos δ − kPd
)
Observe that in this case, Pm = Pd, i.e. generation and load are
assumed to be balanced.
June 26, 2008 Angle Stability - 8
Power System Dynamics and Stability 385
Universidad de Castilla-La Mancha
Hopf Bifurcation I
The PV curves for M = 0.1, DG = 0.01, DL = 0.1, τ = 0.01,
Kv = 10, X ′G = 0.5, V10 = 1, k = 0.25 are:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
xL = 0.6
xL = 0.5
HB
HB
OPV2
Pd
June 26, 2008 Angle Stability - 9
Power System Dynamics and Stability 386
Universidad de Castilla-La Mancha
Hopf Bifurcation I
The eigenvalues for the system with respect to changes in Pd for
xL = 0.5:
−5 −4 −3 −2 −1 0 1 2 3−3
−2
−1
0
1
2
3
Real
Imag
June 26, 2008 Angle Stability - 10
Power System Dynamics and Stability 387
Universidad de Castilla-La Mancha
Hopf Bifurcation I
There is a Hopf bifurcation for Pd = 0.65, xL = 0.5:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8−10
−8
−6
−4
−2
0
2
4
6
HB
ℜ(µ)
Pd
June 26, 2008 Angle Stability - 11
Power System Dynamics and Stability 388
Universidad de Castilla-La Mancha
Hopf Bifurcation I
A Hopf bifurcation with eigenvalues µ = ±jβ yields a periodic
oscillation of period:
T =2π
β
Hence, for the example:
µ ≈ ±j3
⇒ T ≈ 2 s
June 26, 2008 Angle Stability - 12
Power System Dynamics and Stability 389
Universidad de Castilla-La Mancha
Hopf Bifurcation I
The contingency xL = 0.5 → 0.6 yields:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6−0.5
0
0.5
1
1.5
2
2.5
ωδE′
V2
V1
t [s]
June 26, 2008 Angle Stability - 13
Power System Dynamics and Stability 390
Universidad de Castilla-La Mancha
Hopf Bifurcation I
Notice that in this case no oscillations are observed, which is a
“trademark” of Hopf bifurcations and small-disturbance angle instabilities.
The reason for this is that the oscillation period is 2 s (typical in practice
where these kinds of oscillations are in the 0.1-1 Hz range), but the bus
voltage collapses well before the oscillations appear, which is atypical
and is probably due to the chosen impedances and time constants.
This example stresses the point that angle instabilities do lead to voltage
collapse, and vice versa, voltage instabilities lead to angle/frequency
oscillations, even though the reason behind each stability problem are
fairly different.
June 26, 2008 Angle Stability - 14
Power System Dynamics and Stability 391
Universidad de Castilla-La Mancha
Hopf Bifurcation II
Single-machine dynamic-load system with SVC:
E∠δ V ∠0
V
Vref
jX
jBC
PG + jQG PL + jQT
June 26, 2008 Angle Stability - 15
Power System Dynamics and Stability 392
Universidad de Castilla-La Mancha
Hopf Bifurcation II
The total reactive power absorbed by the load and the SVC is as follows:
QT (V, δ) = −V 2
X+
EV
Xcos(δ) + V 2BC
The SVC controller is modeled as a first order pure integrator.
V
+
−
Vref
1/sTBC
June 26, 2008 Angle Stability - 16
Power System Dynamics and Stability 393
Universidad de Castilla-La Mancha
Hopf Bifurcation II
The resulting differential equations of the SMDL system with SVC are as
follows:
δ = ω
ω =1
M[Pd − EV
Xsin(δ) − Dω]
V =1
τ[−kPd + V 2(BC − 1
X) +
EV
Xcos(δ)]
BC =1
T(Vref − V )
June 26, 2008 Angle Stability - 17
Power System Dynamics and Stability 394
Universidad de Castilla-La Mancha
Hopf Bifurcation II
BC is the equivalent susceptance of the SVC; T and Vref are the SVC
time constant and reference voltage, respectively.
In the following, it is assumed that T = 0.01 s and Vref = 1.0 p.u.
Observe that also in this case it is possible to deduce the set of ODE, i.e.
the algebraic variables can be explicitly expressed as a function of the
state variables and the parameters.
June 26, 2008 Angle Stability - 18
Power System Dynamics and Stability 395
Universidad de Castilla-La Mancha
Hopf Bifurcation II
The state matrix of the system is as follows:
A =
0 | 1 | 0 | 0
− EVMX cos(δ) | − D
M | − EMX sin(δ) | 0
−EVτX sin(δ) | 0 | 1
τX [E cos(δ) − 2V + 2V BCX] | V 2
τ
0 | 0 | − 1T | 0
June 26, 2008 Angle Stability - 19
Power System Dynamics and Stability 396
Universidad de Castilla-La Mancha
Hopf Bifurcation II
Eigenvalue loci:
−0.4 −0.2 0 0.2 0.4 0.6 0.8−400
−300
−200
−100
0
100
200
300
400
Real
Imag
inar
y
Hopf BifurcationP
d=1.4143
June 26, 2008 Angle Stability - 20
Power System Dynamics and Stability 397
Universidad de Castilla-La Mancha
Hopf Bifurcation II
A complex conjugate pair of eigenvalues crosses the imaginary axis for
Pd = 1.4143, thus leading to a Hopf bifurcation.
The HB point is:
(δ0, ω0, V0, BC0, Pd0) = (0.7855, 0, 1, 1.2930, 1.4143)
June 26, 2008 Angle Stability - 21
Power System Dynamics and Stability 398
Universidad de Castilla-La Mancha
Hopf Bifurcation II
Bifurcation diagram Pd-δ:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
Pd (p.u.)
δ (r
ad)
Pdmax
Hopf Bifurcation
June 26, 2008 Angle Stability - 22
Power System Dynamics and Stability 399
Universidad de Castilla-La Mancha
Hopf Bifurcation II
We simulate a step change in Pd from 1.41 p.u. to 1.42 p.u. for t = 2 s.
For t > 2 s the system does not present a stable equilibrium point and
shows undamped oscillations (likely an unstable limit cycle), as expected
from the P -δ curve.
For t = 2.57 s, the load voltage collapses.
Note that, in this case, the generator angle shows an unstable trajectory
only after the occurrence of the voltage collapse at the load bus.
June 26, 2008 Angle Stability - 23
Power System Dynamics and Stability 400
Universidad de Castilla-La Mancha
Hopf Bifurcation II
Time domain simulation results:
1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.60.75
0.8
0.85
t (s)
δ (r
ad)
1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.60
0.5
1
1.5
t (s)
V (
p.u.
)
June 26, 2008 Angle Stability - 24
Power System Dynamics and Stability 401
Universidad de Castilla-La Mancha
Hopf Bifurcation II
The use of the SVC device gives a birth to a new bifurcation, namely a
Hopf bifurcation.
This Hopf bifurcation cannot be removed by simply adjusting system
parameters.
However SVC and load dynamics can be coordinated so that the
loadability of the system can be increased.
June 26, 2008 Angle Stability - 25
Power System Dynamics and Stability 402
Universidad de Castilla-La Mancha
Control and Mitigation
For the IEEE 14-bus test system:
Bus 14
Bus 13
Bus 12
Bus 11
Bus 10
Bus 09
Bus 08
Bus 07
Bus 06
Bus 05
Bus 04
Bus 03
Bus 02
Bus 01
Breaker
June 26, 2008 Angle Stability - 26
Power System Dynamics and Stability 403
Universidad de Castilla-La Mancha
Control and Mitigation
Generator speeds for the line 2-4 outage and 40% overloading:
0 5 10 15 20 25 300.998
0.9985
0.999
0.9995
1
1.0005
1.001
time (s)
June 26, 2008 Angle Stability - 27
Power System Dynamics and Stability 404
Universidad de Castilla-La Mancha
Control and Mitigation
This has been typically solved by adding Power System Stabilizers (PSS)
to the voltage controllers in “certain” generators, so that equilibriun point
is made stable, i.e. the Hopf is removed.
Kw
vSI vs
vs max
vs min
Tws
Tws + 1
T1s + 1
T2s + 1
T3s + 1
T4s + 1
1
Tǫs + 1
FACTS can also be used to address this problem.
June 26, 2008 Angle Stability - 28
Power System Dynamics and Stability 405
Universidad de Castilla-La Mancha
Control and Mitigation
For the IEEE 14-bus test system with PSS at bus 1:
Bus 14
Bus 13
Bus 12
Bus 11
Bus 10
Bus 09
Bus 08
Bus 07
Bus 06
Bus 05
Bus 04
Bus 03
Bus 02
Bus 01
Breaker
June 26, 2008 Angle Stability - 29
Power System Dynamics and Stability 406
Universidad de Castilla-La Mancha
Control and Mitigation
Generator speeds for the line 2-4 outage and 40% overloading:
0 5 10 15 20 25 300.998
0.9985
0.999
0.9995
1
1.0005
1.001
time (s)
June 26, 2008 Angle Stability - 30
Power System Dynamics and Stability 407
Universidad de Castilla-La Mancha
Control and Mitigation
Data regarding this system are available at:
http://thunderbox.uwaterloo.ca/∼claudio
/papers/IEEEBenchmarkTFreport.pdf
More details regarding this example can be found in:
F. Milano, “An Open Source Power System Analysis Toolbox”,
accpeted for publication on IEEE Trans. On Power Systems, March
2004.
June 26, 2008 Angle Stability - 31
Power System Dynamics and Stability 408
Universidad de Castilla-La Mancha
Control and Mitigation
For the IEEE 145-bus, 50-machine test system:
1 93
33
5
4
3
34
2 114
113
6
7
9
12
104
8
11
66
111
72
13
64 65
6897
69
124
17 22
14 25
58
59
21 20 19 18
31 26
83 78 30 23
27
77
29 28
67
50
121120131
130
142
9460
76
138
136
134
143144145
89
79
92 90
137
139
36
125
99
61 86
62
63
122
AREA 2 AREA 1
35
38
49
135
141
51
52
54
140
129
132
115116
117
118
101
102
15 16
112
71
70
10 32
39
127
40
44
45
4143
84
87 88
73
81
80
24
108 109
91
96
98
100 103
107
7574
106
105
82
110
119126
12395
133
48
555356
57 128
46
85
49
47
37
42
June 26, 2008 Angle Stability - 32
Power System Dynamics and Stability 409
Universidad de Castilla-La Mancha
Control and Mitigation
For an impedance load model, the PV curves yield:
0 0.002 0.004 0.006 0.008 0.01 0.0120.82
0.84
0.86
0.88
0.9
0.92
0.94
L.F. (p.u.)
Vol
tage
(p.
u.)
Base CaseLine 79−90 Outage
2 2.5 3 3.5 4 4.5 5 5.5 6
x 10−3
0.9
0.91
0.92
0.93
0.94
λ
Vol
tage
(p.
u.)
(b)
Base CaseLine 79−90 Outage
HB
HB Operating point
Constant impedance load line
Operating point
HB HB
Constant impedance load line
(a)
June 26, 2008 Angle Stability - 33
Power System Dynamics and Stability 410
Universidad de Castilla-La Mancha
Control and Mitigation
Indices based on the singular values have been proposed to predict Hopf
bifucations:
0 0.002 0.004 0.006 0.008 0.01 0.0120
0.005
0.01
0.015
λ
Hop
f Ind
ices
HBI2 Base Case
HBI1 Base Case
HBI2 Line 79−90 Outage
HBI1 Line 79−90 Outage
HB HB
June 26, 2008 Angle Stability - 34
Power System Dynamics and Stability 411
Universidad de Castilla-La Mancha
Control and Mitigation
More details regarding this example can be found in:
C. A. Canizares, N. Mithulananthan, F. Milano, and J. Reeve, “Linear
Performance Indices to Predict Oscillatory Stability Problems in Power
Systems”, IEEE Trans. On Power Systems, Vol. 19, No. 2, May 2004,
pp. 1104-1114.
June 26, 2008 Angle Stability - 35
Power System Dynamics and Stability 412
Universidad de Castilla-La Mancha
Small Disturbance Applications
In practice, some contingencies trigger plant or inter-area frequency
oscillations in a “heavily” loaded system, which may be directly
associated with Hopf bifurcations.
This is a “classical” problem in power systems and there are many
examples of this phenomenon in practice, such as the August 10, 1996
blackout of the WSCC (now WECC) system.
June 26, 2008 Angle Stability - 36
Power System Dynamics and Stability 413
Universidad de Castilla-La Mancha
Small Disturbance Applications
Observe that the maximum loadability of the system is reduced by the
presence of the Hopf bifurcation.
This leads to the definition of a “dynamic” ATC value.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ETC ATC TRM
TTC
WorstContingency
HBHB
OP
V2
Pd
June 26, 2008 Angle Stability - 37
Power System Dynamics and Stability 414
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
This information was extracted from a presentation by Dr. Prabha Kundur,
President and CEO of PowerTech Labs Inc.
The material is availbale at:
toronto.ieee.ca/events/oct0303/prabha.ppt
June 26, 2008 Angle Stability - 38
Power System Dynamics and Stability 415
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
System conditions:
High ambient temperatures in Northwest, and hence high power
transfers from Canada to California.
Prior to main outage, three 500 kV line sections from lower Columbia
River to loads in Oregon were out of service due to tree faults.
California-Oregon interties loaded to 4330 MW north to south.
Pacific DC intertie loaded at 2680 MW north to south.
2300 MW flow from Britsh Columbia.
June 26, 2008 Angle Stability - 39
Power System Dynamics and Stability 416
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
Growing 0.23 Hz oscillations caused tripping of lines:
3000
2900
2800
2700
2600
2500
2400
23007471686562595653504743403128252219161260 3
1
2
3
4
4 5
Time in seconds
June 26, 2008 Angle Stability - 40
Power System Dynamics and Stability 417
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
Event 1: 14:06:39
→ Big Eddy-Ostrander 500 kV LG fault - flashed to tree
Event 2: 14:52:37
→ John Day-Marion 500 kV LG -flashed to tree
Event 3: 15:42:03
→ Keeler-Alliston 500 KV - LG - flashed to tree
Event 4: 15:47:36
→ Ross-Lexington 500 kV - flashed to tree
Event 5: 15:47:36-15:48:09
→ 8 McMary Units trip
June 26, 2008 Angle Stability - 41
Power System Dynamics and Stability 418
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
As a results of the undamped oscillations, the system split into four large
islands.
June 26, 2008 Angle Stability - 42
Power System Dynamics and Stability 419
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
PSS solution:
San Onofre
(addition)Palo Verde
(tune existing)
June 26, 2008 Angle Stability - 43
Power System Dynamics and Stability 420
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
PSS solution:
3000
3000
2800
2800
2600
2600
2400
2400
2200
2200
0
0
18 32 50 68 75
15 30 45 61 72
With existing controls
Eigenvalue = 0.0597 + j1.771
Frequency = 0.2818 Hz
Damping = −0.0337
With PSS modifications
Eigenvalue = −0.0717 + j1.673
Frequency = 0.2664 Hz
Damping = −0.0429
Time in seconds
Time in seconds
June 26, 2008 Angle Stability - 44
Power System Dynamics and Stability 421
Universidad de Castilla-La Mancha
August 10, 1996 WSCC Blackout
7.5 million customers experienced outages from a few minutes to nine
hours.
Total load loss: 35,500 MW.
June 26, 2008 Angle Stability - 45
Power System Dynamics and Stability 422
Universidad de Castilla-La Mancha
Transient Stability
“Large disturbance rotor angle stability or transient stability, as it is
commonly referred to, is concerned with the ability of the power system to
maintain synchronism when subjected to a severe disturbance, such as a
short circuit on a transmission line. The resulting system response
involves large excursions of generator rotor angles and is influenced by
the nonlinear power-angle relationship”.
The system nonlinearities determine the system response; hence,
linearization does not work in this case.
June 26, 2008 Angle Stability - 46
Power System Dynamics and Stability 423
Universidad de Castilla-La Mancha
Transient Stability
For small disturbances, the problem is to determine if the resulting steady
state condition is stable or unstable (eigenvalue analysis) or a bifurcation
point (e.g. Hopf bifurcation).
For large disturbances, the steady state condition after the disturbance
can exist and be stable, but it is possible that the system cannot reach
that steady state condition.
June 26, 2008 Angle Stability - 47
Power System Dynamics and Stability 424
Universidad de Castilla-La Mancha
Transient Stability
The basic idea and analysis procedures are:
Pre-contingency (initial conditions): the system is operating in
“normal” conditions associated with a s.e.p.
Contingency (fault trajectory): a large disturbance, such as a short
circuit or a line trip forces the system to move away from its initial
operating point.
Post contingency (fault clearance): the contingency usually forces
system protections to try to “clear” the fault; the issue is then to
determine whether the resulting system is stable, i.e. whether the
system remains relatively intact and the associated time trajectories
converge to a “reasonable” operating point.
June 26, 2008 Angle Stability - 48
Power System Dynamics and Stability 425
Universidad de Castilla-La Mancha
Transient Stability
Based on non linear theory, this analysis can be basically viewed as
determining wheter the fault trajectory at the “clearance” point is outside
or inside of the stability region of the post-contingency s.e.p.
June 26, 2008 Angle Stability - 49
Power System Dynamics and Stability 426
Universidad de Castilla-La Mancha
Time domain analysis
Given the complexity of power system models, the most reliable analysis
tool for these types of studies is full time domain simulations.
For example, for the generator-load example:
Generator
E′∠δ V1∠δ1 V2∠δ2
jxLjx′G
PG + jQG PL + jQL
−jxC
June 26, 2008 Angle Stability - 50
Power System Dynamics and Stability 427
Universidad de Castilla-La Mancha
Time domain analysis
The ODE for the simplest generator d-axis transient model and
neglecting AVR and generator limits is:
ω =1
M(Pd − E′V2B sin δ − DGω)
δ = ω − 1
DL(E′V2B sin δ − Pd)
V2 =1
τ[−V 2
2 (B − BC) + E′V2B cos δ − kPd]
where
B =1
X=
1
X ′G + XL
June 26, 2008 Angle Stability - 51
Power System Dynamics and Stability 428
Universidad de Castilla-La Mancha
Time domain analysis
The objective is to determine how much time an operator would have to
connect the capacitor bank BC after a severe contingency, simulated
here as a sudden increase in the value of the reactance X , so that the
system recovers.
In this case, and as previously discussed in the voltage stability section,
the contingency is severe, as the s.e.p. disappears.
Full time domain simulations are carried out to study this problem for the
parameter values M = 0.1, DG = 0.01, DL = 0.1, τ = 0.01,
E′ = 1, Pd = 0.7, k = 0.25, BC = 0.5.
June 26, 2008 Angle Stability - 52
Power System Dynamics and Stability 429
Universidad de Castilla-La Mancha
Time domain analysis
A contingency X = 0.5 → 0.6 at tf = 1 s, with BC connection at
tc = 1.4 s yields a stable system:
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
V2
ωδ
E′
tc
tf
t [s]
June 26, 2008 Angle Stability - 53
Power System Dynamics and Stability 430
Universidad de Castilla-La Mancha
Time domain analysis
If BC is connected at tc = 1.5 s, the system is unstable:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
1
2
3
4
5
6
V2
ωδ
E′
tc
tf
t [s]
June 26, 2008 Angle Stability - 54
Power System Dynamics and Stability 431
Universidad de Castilla-La Mancha
Direct Methods
Time domain analysis is expensive, so direct stability analysis technique
have been proposed based on Lyapounov’s stability theory.
The idea is to define an “energy” or Lyapounov function ϑ(x, xs) with
certain characteristics to obtain a direct “measure” of the stability region
A(xs) associated with the post-contingency s.e.p. xs.
A system’s energy is usually a good Lyapounov function, as it yields a
stability “measure”.
June 26, 2008 Angle Stability - 55
Power System Dynamics and Stability 432
Universidad de Castilla-La Mancha
Direct Methods
The rolling ball example can used to explain the basic behind these
techniques:
~v
h
m
u.e.p.1
u.e.p.2
s.e.p.
There are 3 equilibrium points: one stable (“valley” bottom), two unstable
(“hill” tops).
June 26, 2008 Angle Stability - 56
Power System Dynamics and Stability 433
Universidad de Castilla-La Mancha
Direct Methods
The energy of the ball is a good Lyapounov or Transient Energy Function
(TEF):
W = Wkinetic + Wpotential
= WK + WP
=1
2mv2 + mgh
= ϑ([v, h]T , 0)
The potential energy at the s.e.p. is zero, and presents local maxima at
the u.e.p.s (WP1 and WP2).
The “closest” u.e.p. is u.e.p.1 since WP1 < WP2.
June 26, 2008 Angle Stability - 57
Power System Dynamics and Stability 434
Universidad de Castilla-La Mancha
Direct Methods
The stability of this system can then be evaluated using this energy:
if W < WP1, the ball remains in the “valley”, i.e. the system is
stable, and will converge to the s.e.p. as t → ∞.
If W > WP1, the ball might or might not converge to the s.e.p.,
depending on friction (inconclusive test).
When the ball’s potential energy WP (t) reaches a maximum with
respect to time t, the system leaves the “valley”, i.e. unstable
condition.
June 26, 2008 Angle Stability - 58
Power System Dynamics and Stability 435
Universidad de Castilla-La Mancha
Direct Methods
The “valley” would correspond to the stability region when friction is
“large”.
In this case, the stability boundary ∂A(xs) corresponds to the “ridge”
where the u.e.p.s are located and WP has a local max. value.
The smaller the friction in the system, the larger the difference between
the ridge and ∂A(xs).
For zero friction, ∂A(xs) is defined by WP1.
June 26, 2008 Angle Stability - 59
Power System Dynamics and Stability 436
Universidad de Castilla-La Mancha
Direct Methods
The direct stability test is only a necessary but not sufficient condition:
ϑ(x, xs) < c ⇒ x ∈ A(xs)
ϑ(x, xs) > c ⇒ Inconclusive!
where the value of c is usually associated with a local maximum of a
“potential energy” function.
June 26, 2008 Angle Stability - 60
Power System Dynamics and Stability 437
Universidad de Castilla-La Mancha
Direct Methods
For the simple generator-infinite bus example, neglecting limits and AVR:
Generator System
E′∠δ V ∠0V1∠δ1 V2∠δ2
jxL jxthjx′G
PG + jQG PL + jQL
δ = ω = ωr − ω0
ω =1
M
(
PL − E′V
Xsin δ − Dω
)
X = X ′G + XL + Xth
June 26, 2008 Angle Stability - 61
Power System Dynamics and Stability 438
Universidad de Castilla-La Mancha
Direct Methods
The kinetic energy in this system is defined as:
WK =1
2Mω2
And the potential energy is:
WP =
∫
(Tc − Tm)dδ
≈∫
(Pc − Pm)dδ → in p.u. for ωr ≈ ω0
≈∫ δ
δs
(PG − PL)dδ ≈∫ δ
δs
(E′V
X− PL)dδ
≈ −E′V B(cos δ − cos δs) − PL(δ − δs)
where δs is the s.e.p. for this system.
June 26, 2008 Angle Stability - 62
Power System Dynamics and Stability 439
Universidad de Castilla-La Mancha
Direct Methods
With WP presenting a very similar profile as the rolling ball example:
E′VX
PG
WF
WF1
WF2
min
max
max
stable
unstableunstable
δs
δs δu1
δu1
δu2
δu2
June 26, 2008 Angle Stability - 63
Power System Dynamics and Stability 440
Universidad de Castilla-La Mancha
Direct Methods
Hence, the system Lyapounov function of TEF is:
TEF = ϑ(x, xs)
= ϑ([δ, ω]T , [δs, 0]T )
=1
2Mω2 − E′V B(cos δ − cos δs)
−PL(δ − δs)
Thus, using similar criteria as in the case of the rolling ball:
If TEF < WP1 ⇒ system is stable.
If TEF > WP1 ⇒ inconclusive for D > 0 (“friction”).
If TEF > WP1 ⇒ unstable for D = 0 (unrealistic).
June 26, 2008 Angle Stability - 64
Power System Dynamics and Stability 441
Universidad de Castilla-La Mancha
Direct Methods
This is equivalent to compare “areas” in the PG vs. δ graph (Equal Area
Criterion or EAC):
PG
PL
δ(0) = δspreδspost
δ(tc)δu1post
pre-contingency
post-contingency
contingency (fault)
δ
fault clearing time
June 26, 2008 Angle Stability - 65
Power System Dynamics and Stability 442
Universidad de Castilla-La Mancha
Direct Methods
Thus, comparing the “acceleration” area:
Aa =
∫ δ(tc)
δspre
(PL − PGfault)dδ
=
∫ δ(tc)
δspre
(
PL − E′V
Xfault
)
dδ
versus the “deceleration” area:
Ad =
∫ δspost
δ(tc)
(PGpost− PL)dδ
=
∫ δspost
δ(tc)
(E′V
Xpost− PL
)
dδ
June 26, 2008 Angle Stability - 66
Power System Dynamics and Stability 443
Universidad de Castilla-La Mancha
Direct Methods
In conclusion:
If Aa < Ad ⇒ system is stable at tc.
If Aa > Ad ⇒ inconclusive for D > 0.
If Aa > Ad ⇒ unstable for D = 0 (unrealistic).
June 26, 2008 Angle Stability - 67
Power System Dynamics and Stability 444
Universidad de Castilla-La Mancha
Direct Methods: Example 1
A 60 Hz generator with a 15% transient reactance is connected to an
infinite bus of 1 p.u. voltage through two identical parallel transmission
lines of 20% reactance and negligible resistance. The generator is
delivering 300 MW at a 0.9 leading power factor when a 3-phase solid
fault occurs in the middle of one of the lines; the fault is then cleared by
opening the breakers of the faulted line.
Assuming a 100 MVA base, determine the critical clearing time for this
generator if the damping is neglected and its inertia is assumed to be
H = 5 s.
Assuming D = 0.1 s, determine the actual critical clearing time.
June 26, 2008 Angle Stability - 68
Power System Dynamics and Stability 445
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Pre-contingency or initial conditions:
PGpre= PL =
E′V
Xpresin δspre
QL = − V 2
Xpre+
E′V
Xprecos δspre
June 26, 2008 Angle Stability - 69
Power System Dynamics and Stability 446
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Where:
Xpre = 0.15 +0.2
2= 0.25
PL =300 MW
100 MVA
3 =E′
0.25sin δspre
QL = 3 tan(cos−1 0.9)
1.4530 = − 1
0.25+
E′
0.25cos δspre
June 26, 2008 Angle Stability - 70
Power System Dynamics and Stability 447
Universidad de Castilla-La Mancha
Direct Methods: Example 1
⇒ E′ipre
= E′ sin δspre
= 0.75
E′rpre
= E′ cos δspre
= 1.3633
E′ =√
E′2rpre
+ E′2ipre
= 1.5559
δspre= tan−1
(
E′ipre
E′rpre
)
= 28.82 = 0.5030 rad
June 26, 2008 Angle Stability - 71
Power System Dynamics and Stability 448
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Fault conditions:
PGfault=
E′V
Xfaultsin δ
=1.5559
Xfaultsin δ
where, using a Y-∆ circuit transformation due to the fault being in the
middle of one of the parallel lines:
jXfault
E′∠δ V ∠0
j0.1 j0.1
j0.2
j0.15
June 26, 2008 Angle Stability - 72
Power System Dynamics and Stability 449
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Xfault =0.15 × 0.2 + 0.1 × 0.2 + 0.15 × 0.1
0.1⇒ PGfault
= 2.394 sin δ
Aa =
∫ δ(tcc)
δspre
(PL − PGfault)dδ
=
∫ δ(tcc)
0.503
(3 − 2.394 sin δ)dδ
= 3(δ(tcc) − 0.503) + 2.394(cos δ(tcc) − cos(0.503))
= 3δ(tcc) + 2.394 cos δ(tcc) − 3.6065
June 26, 2008 Angle Stability - 73
Power System Dynamics and Stability 450
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Post contingency conditions:
Xpost = 0.15 + 0.2 = 0.35
⇒ PGpost=
E′V
Xpostsin δ
= 4.446 sin δ
⇒ 3 = 4.446 sin δspost
δspost= 42.44
= 0.7407 rad
June 26, 2008 Angle Stability - 74
Power System Dynamics and Stability 451
Universidad de Castilla-La Mancha
Direct Methods: Example 1
⇒ Ad =
∫ π−δspost
δ(tcc)
(PGpost− PL)dδ
=
∫ 2.4
δ(tcc)
(4.446 sin δ − 3)dδ
= −4.446(cos 2.4 − cos δ(tcc)) − 3(2.4 − δ(tcc))
= 3δ(tcc) + 4.446 cos δ(tcc) − 3.9215
June 26, 2008 Angle Stability - 75
Power System Dynamics and Stability 452
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Aa = Ad
= 3δ(tcc) + 2.394 cos δ(tcc) − 3.6065
= 3δ(tcc) + 4.446 cos δ(tcc) − 3.9215
⇒ δ(tcc) = 81.17
= 1.4167 rad
June 26, 2008 Angle Stability - 76
Power System Dynamics and Stability 453
Universidad de Castilla-La Mancha
Direct Methods: Example 1
During the fault:
δ = ω
ω =1
M
(
PL − E′V
Xfaultsin δ
)
M =H
πf
=5 s
π60 Hz
= 0.0265 s2
⇒ δ = ω
ω = 37.70(3 − 2.394 sin δ)
June 26, 2008 Angle Stability - 77
Power System Dynamics and Stability 454
Universidad de Castilla-La Mancha
Direct Methods: Example 1
Integrating these equations numerically for δ(0) = δspre= 28.82:
0 0.05 0.1 0.15 0.2 0.25 0.3 0.3520
40
60
80
100
120
140
160
180
200
220
δ[d
eg]
t [s]
June 26, 2008 Angle Stability - 78
Power System Dynamics and Stability 455
Universidad de Castilla-La Mancha
Direct Methods: Example 1
For D = 0.1 and a clearing time of tc = 0.27 s, the system is stable:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
50
100
150
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−10
−5
0
5
10
δ[d
eg]
ω[d
eg]
t [s]
t [s]
June 26, 2008 Angle Stability - 79
Power System Dynamics and Stability 456
Universidad de Castilla-La Mancha
Direct Methods: Example 1
For a clearing time of tc = 0.28 s, the system is unstable; hence
tcc ≈ 0.275 s:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
500
1000
1500
2000
2500
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
10
20
30
40
δ[d
eg]
ω[d
eg]
t [s]
t [s]
June 26, 2008 Angle Stability - 80
Power System Dynamics and Stability 457
Universidad de Castilla-La Mancha
Direct Methods: Example 2
Generator-motor, i.e. system-system, cases may also be studied using
the EAC method based on an equivalent inertia
M = M1M2/(M1M2), and damping
D = MD1/M1 = MD2/M2.
For the generator-load example neglecting the internal generator
impedance and assuming an “instantaneous” AVR:
V1∠δ1 V2∠δ2
jxL
PG + jQG PL + jQL
June 26, 2008 Angle Stability - 81
Power System Dynamics and Stability 458
Universidad de Castilla-La Mancha
Direct Methods: Example 2
The “energy” functions, with or without generator limits, can be shown to
be:
WK =1
2Mω2
WP = −B(V1V2 cos δ − V10V20 cos δ0)
+1
2B(V 2
2 − V 220) +
1
2B(V 2
1 − V 210)
−Pd(δ − δ0) + Qd ln
(V2
V20
)
− QG ln
(V1
V10
)
The stability of this system can then be studied using the same “energy”
evaluation previously explained for TEF = ϑ(x, x0) = WK + WP .
June 26, 2008 Angle Stability - 82
Power System Dynamics and Stability 459
Universidad de Castilla-La Mancha
Direct Methods: Example 2
Thus for V1 = 1, XL = 0.5, Pd = 0.1, and Qd = 0.25Pd, the
potential energy WP (δ, V2) that defines the stability region withr espect
to the s.e.p. is:
0
0.5
1
1.5
2
−400
−200
0
200
4000
1
2
3
4
5
6
7
8
V2δ
WP
s.e.p.
u.e.p.saddle
node
June 26, 2008 Angle Stability - 83
Power System Dynamics and Stability 460
Universidad de Castilla-La Mancha
Direct Methods: Example 2
Simulating the critical contingency XL = 0.5 → 0.6 for Pd = 0.7 and
neglecting limits, the “energy” profiles are:
0.9 1 1.1 1.2 1.3 1.4 1.5 1.6−0.4
−0.3
−0.2
−0.1
0
0.1
0.2 WpWk+Wp
TE
F
t [s]
June 26, 2008 Angle Stability - 84
Power System Dynamics and Stability 461
Universidad de Castilla-La Mancha
Direct Methods: Example 2
The “exit” point on ∂A(xs) is approximately at the maximum potential
energy point.
Thus, the critical clearing time is:
tcc ≈ 1.42 s
A similar value can be obtained through trial-and-error.
June 26, 2008 Angle Stability - 85
Power System Dynamics and Stability 462
Universidad de Castilla-La Mancha
Direct Methods: Conclusions
The advantages of using Lyapounov functions are:
Allows reduced stability analysis.
Can be used as an stability index.
The problems are:
Lyapounov functions are model dependent; in practice, only
approximate “energy” functions can be found.
Inconclusive if test fails.
The post-perturbation system state must be known ahead of time, as
the energy function is defined with respect to the corresponding s.e.p.
Can only be used as an “approximate” stability analysis tool.
June 26, 2008 Angle Stability - 86
Power System Dynamics and Stability 463
Universidad de Castilla-La Mancha
Transient Stability Applications
In practice, transient stability studies are carried out using time-domain
trial-and-error techniques.
These types of studies can now be done on-line even for large systems.
The idea is to determine whether a set of “realistic” contingencies make
the system unstable or not (contingency ranking), and thus determine
maximum transfer limits or ATC in certain transmission corridors for given
operating conditions.
June 26, 2008 Angle Stability - 87
Power System Dynamics and Stability 464
Universidad de Castilla-La Mancha
Transient Stability Applications
Thus, the maximum loadability of the system may be affected by the
“size” of the stability region, leading to the definition of a “true” ATC value.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ETC ATC TRM
TTC
WorstContingency
HBHB
OP
V2
Pd
June 26, 2008 Angle Stability - 88
Power System Dynamics and Stability 465
Universidad de Castilla-La Mancha
Transient Stability Applications
Critical clearing times are not really an issue with current fast acting
protections.
Simplified direct methods such as the “Extended Equal Area Criterion” (Y.
Xue et al., “Extended Equal Area Criterion Revisited”, IEEE Transaction
on Power Systems, Vol. 7, No. 3, Aug. 1992, pp. 1012-1022) have been
proposed and tested for on-line contingency pre-ranking, and are being
implemented for practical applications through an E.U. project.
June 26, 2008 Angle Stability - 89
Power System Dynamics and Stability 466
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
An example of an application of transient analysis techniques can be
found in L.S. Vargas and C. A. Canizares, “Time Dependence of Controls
to Avoid Voltage Collapse”, IEEE Transaction son Power Systems, Vol.
15, No. 4, November 2000, pp. 1367-1375.
This paper discusses the May 1997 voltage collapse event of the main
power system in Chile.
June 26, 2008 Angle Stability - 90
Power System Dynamics and Stability 467
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
Most of Chile lost power in a major blackout on Friday evening, snarling
rush hour traffic in the capital. The blackout began at about 7:20 pm, and
power was back in about a third of the affected areas at 9:00 pm.
At 9:00 pm lights were gradually coming on in parts of the capital, home
to 5 million people or one third of the country’s population. Television
reports said power went out as far as Puerto Montt, a city some 600 miles
south of Santiago, and in areas the same distance to the north.
Source Reuters
June 26, 2008 Angle Stability - 91
Power System Dynamics and Stability 468
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
Main system characteristics:
Extension: 756626 km2.
Inhabitants: 14.5 mil.
National consumptions: 33531 GWh.
National peak load: 5800 MW.
Installed capacity: 8000 MW.
Frequency: 50 Hz.
Trans. Level: 66/110/154/220/500 kV.
Four interconnected systems: SING, SIC, AISEN, MAGALLANES
June 26, 2008 Angle Stability - 92
Power System Dynamics and Stability 469
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
June 26, 2008 Angle Stability - 93
Power System Dynamics and Stability 470
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
Initial state of SIC system:
2500 MW load.
Power flow south-north near 1000 MW (900 MW through 500 kV lines
and 100 MW through 154 kV lines).
Events:
Line 154 kV trips.
Major generator in the south hits reactive power limits and losses
voltage control.
Operator tries to recover falling voltages by connecting a capacitor
bank near Santiago.
June 26, 2008 Angle Stability - 94
Power System Dynamics and Stability 471
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
June 26, 2008 Angle Stability - 95
Power System Dynamics and Stability 472
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
The line trip and generator limits yield a voltage collapse associated with
a limit-induced bifurcation problem:
June 26, 2008 Angle Stability - 96
Power System Dynamics and Stability 473
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
PV curves:
June 26, 2008 Angle Stability - 97
Power System Dynamics and Stability 474
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
The connection of the capacitor bank after the generator limits are
reached did not save the system, as the “faulted” system trajectories had
“left” the stability region of the post contingency operating point.
June 26, 2008 Angle Stability - 98
Power System Dynamics and Stability 475
Universidad de Castilla-La Mancha
Chilean Blackout (11/07/2003)
If the capacitor bank is connected before the generator limits are reached,
the system would have been saved, as the “faulted” system trajectories
were still within the stability of the post-contingency operating point.
June 26, 2008 Angle Stability - 99
Power System Dynamics and Stability 476
Universidad de Castilla-La Mancha
Frequency Stability Outlines
Definitions.
Basic Concepts.
Practical applications, controls and protections.
Italian Blackout (28/10/2003)
European Blackout (4/11/2006)
June 26, 2008 Frequency Stability - 1
Power System Dynamics and Stability 477
Universidad de Castilla-La Mancha
Frequency Stability Definitions
IEEE-CIGRE classification (IEEE/CIGRE Joint Task Force on Stability)
Terms and Definitions, “Definitions and Classification of Power System
Stability”, IEEE Trans. Power Systems and CIGRE Technical Brochure
231, 2003:
Voltage StabilityVoltage StabilityAngle Stability
Stability
Stability
StabilityStability
Stability
Voltage
Transient
FrequencyRotor Angle
Power System
Large SmallSmall Disturbance
Disturbance Disturbance
Long Term
Long Term
Short TermShort Term
Short Term
June 26, 2008 Frequency Stability - 2
Power System Dynamics and Stability 478
Universidad de Castilla-La Mancha
Frequency Stability Definitions
“Frequency Stability refers to the ability of a power system to maintain
steady frequency following a severe system upset resulting in a
significant imbalance between generation and load.”
Thus, frequency stability analysis concentrates on studying the overall
system stability for sudden changes in the generation-load balance.
June 26, 2008 Frequency Stability - 3
Power System Dynamics and Stability 479
Universidad de Castilla-La Mancha
Frequency Stability Concepts
For the generator-load example, with AVR but no QG limits:
Generator
E′∠δ V1∠δ1
V10
V2∠δ2
jxLjx′G
PG + jQG PL + jQL
+
−Kvs
June 26, 2008 Frequency Stability - 4
Power System Dynamics and Stability 480
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Neglecting losses and electromagnetic dynamics, the generator with a
very simple AVR and no limits can be modeled using a d-axis transient
model:
δG = ωG = ωr − ω0
ωG =1
M(Pm − PG − DGωG)
E′ = Kv(V10 − V1)
June 26, 2008 Frequency Stability - 5
Power System Dynamics and Stability 481
Universidad de Castilla-La Mancha
Frequency Stability Concepts
The load can be simulated using “simplified” mixed models and constant
power factor:
δ2 = ω2 =1
DL(PL − Pd)
V2 =1
τ(QL − kPD
︸︷︷︸
Qd
)
June 26, 2008 Frequency Stability - 6
Power System Dynamics and Stability 482
Universidad de Castilla-La Mancha
Frequency Stability Concepts
The transmission system yields the power flow equations
(X = XL = X ′G):
PG = PL =E′V2
Xsin(δG − δ2)
=V1V2
XLsin(δ1 − δ2)
QL = −V 22
X+
E′v2
Xcos(δ2 − δG)
= − V 22
XL+
V1V2
XLcos(δ2 − δ1)
QG =V 2
1
XL− V1V2
XLcos(δ1 − δ2)
June 26, 2008 Frequency Stability - 7
Power System Dynamics and Stability 483
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Define:
δ = δG − δ2
δ′ = δ1 − δ2
ω = ωG
⇒ δ = ω − ω2
June 26, 2008 Frequency Stability - 8
Power System Dynamics and Stability 484
Universidad de Castilla-La Mancha
Frequency Stability Concepts
This yields the DAE model:
ω =1
M
(
PmE′V2
Xsin δ − DGω
)
δ = ω − 1
DL
(E′V2
Xsin δ − Pd
)
E′ = Kv(V10 − V1)
V2 =1
τ
(
−V 22
X+
E′V2
Xcos δ − kPd
)
0 =V1V2
XLsin δ′ − E′V2
Xsin δ
0 = V 22
(1
XL− 1
X
)
+E′V2
Xcos δ − V1V2
XLcos δ′
June 26, 2008 Frequency Stability - 9
Power System Dynamics and Stability 485
Universidad de Castilla-La Mancha
Frequency Stability Concepts
And the equilibrium equations:
0 = PmE′V2
Xsin δ − DGω
0 = DLω − E′V2
Xsin δ − Pd
0 = V10 − V1
0 = −V 22
X+
E′V2
Xcos δ − kPd
0 =V1V2
XLsin δ′ − E′V2
Xsin δ
0 = V 22
(1
XL− 1
X
)
+E′V2
Xcos δ − V1V2
XLcos δ′
June 26, 2008 Frequency Stability - 10
Power System Dynamics and Stability 486
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Hence:
x = [ω, δ, E′, v2]T
y = [V1, δ′]T
p = [Pm, V10]
λ = Pd
Observe that in this analysis, Pm 6= Pd to study the effect of
generator-load imbalances in the system.
June 26, 2008 Frequency Stability - 11
Power System Dynamics and Stability 487
Universidad de Castilla-La Mancha
Frequency Stability Concepts
In normal operating conditions Pm = Pd ⇒ ω = 0, from the first 2
equilibrium equations.
Hence, these equations can be replaced by the following 4 power flow
equations, with 4 unknowns (E′, V2, δ, δ′):
0 = Pd − E′V2
Xsin δ
0 = −kPd − V 22
X+
E′V2
Xcos δ
0 = Pd − V10V2
XLsin δ′
0 = kPd +V 2
2
XL− V10V2
XLcos δ′
June 26, 2008 Frequency Stability - 12
Power System Dynamics and Stability 488
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Simulating a 50% generation and load reduction, respectively, for
M = 0.1, DG = 0.01, DL = 0.1, τ = 0.01, Kv = 10, XL = 0.5,
X ′G = 0.5, V10 = 1, Pd0 = 0.7, k = 0.25.
Initial solution:
ω = 0
δ = 0.7266
E′ = 1.3463
V2 = 0.7826
V1 = 1.0000
δ′ = 0.4636
June 26, 2008 Frequency Stability - 13
Power System Dynamics and Stability 489
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Time domain simulation:
0 1 2 3 4 5 6 7 8 9 10−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
50% generator drop
50% load drop ω
δ
E′V2V1
t [s]
June 26, 2008 Frequency Stability - 14
Power System Dynamics and Stability 490
Universidad de Castilla-La Mancha
Frequency Stability Concepts
The generation reduction yields a ≈0.5 Hz frequency drop and a load
voltage increase.
The load reduction yields a ≈0.5 Hz frequency increase, and an even
larger load voltage increase, as the reactive power demand drops.
The system reaches new s.e.p.s in both cases, as expected.
Observe that the AVR keeps the generator terminal fairly stable and close
to its set value V10.
June 26, 2008 Frequency Stability - 15
Power System Dynamics and Stability 491
Universidad de Castilla-La Mancha
Frequency Stability Concepts
These frequency excursions due to generation-load imbalances are
typical.
It might lead to unstable conditions due to device protections such as
frequency relays in generators and loads.
Frequency problems may be solved manually by operators or
automatically through controls and/or protections.
Generator governors automatically regulate local frequency excursions.
June 26, 2008 Frequency Stability - 16
Power System Dynamics and Stability 492
Universidad de Castilla-La Mancha
Frequency Stability Concepts
Centralized frequency regulators, such as automatic Area Control Error
(ACE) regulators, may be used to regulate power exchanges among
control areas by controlling the frequency deviations on the interties.
Examples of frequency instabilities are:
The Italian blackout of Tuesday, October 28, 2003 (Material courtesy
of Prof. Alberto Berizzi, Politecnico di Milano, Italy).
The European blackout of Saturday, November 4, 2006 (Material
courtesy of Prof. Edmund Handschin, University of Technology,
Dortmund, Germany).
June 26, 2008 Frequency Stability - 17
Power System Dynamics and Stability 493
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 18
Power System Dynamics and Stability 494
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 19
Power System Dynamics and Stability 495
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
3:10.47 ETrans (operator of interties between Switzerland and rest of
Europe) lets the GRTN (Italian operator) know of the Mettlen-Lavorgo line
trip (1320 MW) and the overloading of the Sils-Soazza line (1650 MW),
and requests a 300 MW demand reduction to relief the overload.
3:18.40 ETrans contacts EGL (Switzerland operator) requesting the tripping
of a transformer in Soazza.
3:21.00 GRTN reduces the power imports by 300 MW.
June 26, 2008 Frequency Stability - 20
Power System Dynamics and Stability 496
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
3:22.03 ATEL (Swiss energy company) changes the connection of the
transformer at Lavorgo.
3:25.22 Protections trip the Sils-Soazza line (1783 MW). This is basically the
beginning of the cascading events that follow, severing Italy from the rest
of Europe and leading to the collapse of the Italian system.
June 26, 2008 Frequency Stability - 21
Power System Dynamics and Stability 497
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 22
Power System Dynamics and Stability 498
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
Imports from France:
June 26, 2008 Frequency Stability - 23
Power System Dynamics and Stability 499
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
Frequency in Italy:
June 26, 2008 Frequency Stability - 24
Power System Dynamics and Stability 500
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
Frequency and voltages in Europe:
June 26, 2008 Frequency Stability - 25
Power System Dynamics and Stability 501
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 26
Power System Dynamics and Stability 502
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 27
Power System Dynamics and Stability 503
Universidad de Castilla-La Mancha
Italian Blackout (28/10/2003)
June 26, 2008 Frequency Stability - 28
Power System Dynamics and Stability 504
Universidad de Castilla-La Mancha
European Interconnected Systems
June 26, 2008 Frequency Stability - 29
Power System Dynamics and Stability 505
Universidad de Castilla-La Mancha
European Interconnected Systems
System Capacity Peak Load Energy Population
GW GW TWh Mio.
Nordel 94 66 405 24
UPS/IPS 337 215 1285 280
UKTSOA 85 66 400 65
Nordel 600 390 2530 450
June 26, 2008 Frequency Stability - 30
Power System Dynamics and Stability 506
Universidad de Castilla-La Mancha
European Interconnected Systems
450 million people served
2530 TWh used
600 GW installed capacity at 500 e/kW = 300 Ge
230.000 km HV network at 400 000 e/km = 90 Ge
Approx. 5.000.000 km MV+LV network
1500 e investment per EU citizen
Largest man-made system
June 26, 2008 Frequency Stability - 31
Power System Dynamics and Stability 507
Universidad de Castilla-La Mancha
European Blackout (4/11/2006)
June 26, 2008 Frequency Stability - 32
Power System Dynamics and Stability 508
Universidad de Castilla-La Mancha
European Blackout (4/11/2006)
June 26, 2008 Frequency Stability - 33
Power System Dynamics and Stability 509
Universidad de Castilla-La Mancha
European Blackout (4/11/2006)
Area 1 : The frequency drops to 49 Hz, which causes an automatic load
schedding.
Area 2 : Real power surplus of 6000 MW.
June 26, 2008 Frequency Stability - 34
Power System Dynamics and Stability 510
Universidad de Castilla-La Mancha
European Blackout (4/11/2006)
22:11:00
frequency50.0
49.3
49.2
49.4
49.5
49.6
49.7
49.8
49.9
50.1
49.1
49.0
48.9
f [Hz]
22:10:00 22:10:10 22:10:20 22:10:5022:10:30 22:10:40
June 26, 2008 Frequency Stability - 35
Power System Dynamics and Stability 511
Universidad de Castilla-La Mancha
European Blackout (4/11/2006)
June 26, 2008 Frequency Stability - 36
Power System Dynamics and Stability 512
Universidad de Castilla-La Mancha
Contents
Overview
UWPFLOW
Matlab
PSAT
June 26, 2008 Software Tools - 1
Power System Dynamics and Stability 513
Universidad de Castilla-La Mancha
Overview
Software packages for power system analysis can be basically divided
into two classes of tools:
Commercial softwares.
Educational/research-aimed softwares.
June 26, 2008 Software Tools - 2
Power System Dynamics and Stability 514
Universidad de Castilla-La Mancha
Overview
Commercial softwares:
PSS/E
EuroStag
Simpow
CYME
PowerWorld
Neplan
June 26, 2008 Software Tools - 3
Power System Dynamics and Stability 515
Universidad de Castilla-La Mancha
Overview
Commercial software packages follows an “all-in-one” philosophy and are
typically well-tested and computationally efficient.
Despite their completeness, these softwares can result cumbersome for
educational and research purposes.
commercial softwares are “closed”, i.e. do not allow changing the source
code or adding new algorithms.
June 26, 2008 Software Tools - 4
Power System Dynamics and Stability 516
Universidad de Castilla-La Mancha
Overview
For research purposes, the flexibility and the ability of easy prototyping
are often more crucial aspects than computational efficiency.
At this aim, there is a variety of open source research tools, which are
typically aimed to a specific aspect of power system analysis.
An example is UWPFLOW which provides an extremely robust algorithm
for continuation power flow analysis.
June 26, 2008 Software Tools - 5
Power System Dynamics and Stability 517
Universidad de Castilla-La Mancha
Overview
C anf FORTRAN are very fast but requires keen programming skills and
are not suitable for fast prototyping.
Several high level scientific languages, such as Matlab, Mathematica and
Modelica, have become more and more popular for both research and
educational purposes.
At this aim, there is a variety of open source research tools, which are
typically aimed to a specific aspect of power system analysis.
Matlab proved to be the best user choice.
June 26, 2008 Software Tools - 6
Power System Dynamics and Stability 518
Universidad de Castilla-La Mancha
Overview
Matlab-based power system analysis tools:
Power System Toolbox (PST)
MatPower
Voltage Stability Toolbox (VST)
Power Analysis Toolbox (PAT)
Educational Simulation Tool (EST)
Power system Analysis Toolbox (PSAT)
June 26, 2008 Software Tools - 7
Power System Dynamics and Stability 519
Universidad de Castilla-La Mancha
Overview
Comparison of Matlab-based power system analysis softwares:
Package PF CPF OPF SSA TD EMT GUI GNE
EST X X X X
MatEMTP X X X X
MatPower X X
PAT X X X X
PSAT X X X X X X X
PST X X X X
SPS X X X X X X
VST X X X X X
June 26, 2008 Software Tools - 8
Power System Dynamics and Stability 520
Universidad de Castilla-La Mancha
Overview
The features illustrated in the table are:
power flow (PF)
continuation power flow and/or voltage stability analysis (CPF-VS)
optimal power flow (OPF)
small signal stability analysis (SSA)
time domain simulation (TD)
graphical user interface (GUI)
graphical network editor (GNE).
June 26, 2008 Software Tools - 9
Power System Dynamics and Stability 521
Universidad de Castilla-La Mancha
Overview
An important but often missed issue is that the Matlab environment is a
commercial and “closed” product, thus Matlab kernel and libraries cannot
be modified nor freely distributed.
To allow exchanging ideas and effectively improving scientific research,
both the toolbox and the platform on which the toolbox runs should be
free (Richard Stallman).
An alternative to Matlab is the free GNU/Octave project.
June 26, 2008 Software Tools - 10
Power System Dynamics and Stability 522
Universidad de Castilla-La Mancha
UWPFLOW
UWPFLOW is a research tool that has been designed to calculate local
bifurcations related to system limits or singularities in the system
Jacobian.
The program also generates a series of output files that allow further
analyses, such as tangent vectors, left and right eigenvectors at a
singular bifurcation point, Jacobians, power flow solutions at different
loading levels, voltage stability indices, etc.
June 26, 2008 Software Tools - 11
Power System Dynamics and Stability 523
Universidad de Castilla-La Mancha
UWPFLOW Features
Adequate handling of generators limits, with generators being able to
recover from a variety of limits, including S limits.
Steady state models of generators and their control limits (AVR and
Primemover limits) are included.
Voltage dependent load models for voltage stability analysis are also
included.
June 26, 2008 Software Tools - 12
Power System Dynamics and Stability 524
Universidad de Castilla-La Mancha
UWPFLOW Features
Either BPA/WSCC ac-dc (HVDC systems) input data formats (and
variations) or IEEE common format may be used.
Detailed and reliable steady state models of SVC, TCSC and STATCOM
models, and their controls with the corresponding limits are included.
Secondary voltage control, as defined by ENEL (elecetricity company of
Italy), can be modeled and simulated in the program.
June 26, 2008 Software Tools - 13
Power System Dynamics and Stability 525
Universidad de Castilla-La Mancha
UWPFLOW Features
The program is able to compute the minimum real eigenvalue and the
related right and left eigenvectors and several voltage stability indices.
The program generates a wide variety of output ASCII and MATLAB (.m)
files as well as IEEE common format data files.
The program has being designed to automatically run script files.
June 26, 2008 Software Tools - 14
Power System Dynamics and Stability 526
Universidad de Castilla-La Mancha
UWPFLOW Usage
Like any other UNIX program, i.e., command-line options (-option) with
redirection of output (>) from screen into files:
uwpflow [-options] input_file [[>]output_file]
For example, to generate the program help:
uwpflow -h
June 26, 2008 Software Tools - 15
Power System Dynamics and Stability 527
Universidad de Castilla-La Mancha
UWPFLOW Example
3-area sample system:
R = 0.01 p.u.
X = 0.15 p.u.
100 MW
v3<d3
150 MW
100 MW
150 MW
50 MVAr
Bus 2
Bus 3
50 MW
40 MVAr
150 MW
56 MVAr
V2∠δ2
1.02∠0
Area 1
50 MVAr
60 MVAr
June 26, 2008 Software Tools - 16
Power System Dynamics and Stability 528
Universidad de Castilla-La Mancha
UWPFLOW Example
For a 3-area sample system:
Bus ∆PG ∆PL ∆QL
Name (p.u.) (p.u.) (p.u.)
Area 1 1.5 0 0
Area 2 0 1.5 0.56
Area 3 0.5 0.5 0.40
Using UWPFLOW to obtain the system PV curves, for a distributed slack
bus model:
June 26, 2008 Software Tools - 17
Power System Dynamics and Stability 529
Universidad de Castilla-La Mancha
UWPFLOW Example
Data file in EPRI format (3area.wsc):
HDG
UWPFLOW data file, WSCC format
3-area example
April 2000
BAS
C
C AC BUSES
C
C | SHUNT |
C |Ow|Name |kV |Z|PL |QL |MW |Mva|PM |PG |QM |Qm |Vpu
BE 1 Area 1 138 1 150 60 0 0 0 150 0 0 1.02
B 1 Area 2 138 2 150 56 0 50 0 100 0 0 1.00
B 1 Area 3 138 3 50 40 0 50 0 100 0 0 1.00
June 26, 2008 Software Tools - 18
Power System Dynamics and Stability 530
Universidad de Castilla-La Mancha
UWPFLOW Example
Data file in EPRI format (3area.wsc):
C
C AC LINES
C
C M CS N
C |Ow|Name_1 |kV1||Name_2 |kV2|||In || R | X | G/2 | B/2 |Mil|
L 1 Area 1 138 Area 2 1381 15001 .01 .15
L 1 Area 1 138 Area 3 1381 15001 .01 .15
L 1 Area 2 138 Area 3 1381 15001 .01 .15
C
C SOLUTION CONTROL CARD
C
C |Max| |SLACK BUS |
C |Itr| |Name |kV| |Angle |
SOL 50 Area 1 138 0.
END
June 26, 2008 Software Tools - 19
Power System Dynamics and Stability 531
Universidad de Castilla-La Mancha
UWPFLOW Example
Generator and load change file (3area.k):
C
C UWPFLOW load and generation "direction" file
C for 3-area example
C
C BusNumber BusName DPg Pnl Qnl PgMax [ Smax Vmax Vmin ]
1 0 1.5 0.0 0.0 0 0 1.05 0.95
2 0 0.0 1.5 0.56 0 0 1.05 0.95
3 0 0.5 0.5 0.40 0 0 1.05 0.95
June 26, 2008 Software Tools - 20
Power System Dynamics and Stability 532
Universidad de Castilla-La Mancha
UWPFLOW Example
Batch file for UNIX (run3area):
echo -1- Run base case power flow
uwpflow 3area.wsc -K3area.k
echo -2- Obatin PV curves and maximum loading
uwpflow 3area.wsc -K3area.k -cthreearea.m -m -ltmp.l -s
echo - with bus voltage limits enforced
uwpflow 3area.wsc -K3area.k -c -7 -k0.1
echo - with current limits enforced
uwpflow 3area.wsc -K3area.k -c -ltmp.l -8 -k0.1
June 26, 2008 Software Tools - 21
Power System Dynamics and Stability 533
Universidad de Castilla-La Mancha
UWPFLOW Example
Batch file for Windows (run3area.bat):
rem -1- Run base case power flow
uwpflow 3area.wsc -K3area.k
rem -2- Obatin PV curves and maximum loading
uwpflow 3area.wsc -K3area.k -cthreearea.m -m -ltmp.l -s
rem - with bus voltage limits enforced
uwpflow 3area.wsc -K3area.k -c -7 -k0.1
rem - with current limits enforced
uwpflow 3area.wsc -K3area.k -c -ltmp.l -8 -k0.1
June 26, 2008 Software Tools - 22
Power System Dynamics and Stability 534
Universidad de Castilla-La Mancha
UWPFLOW Example
PV curves (threearea.m):
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
50
100
150
L.F. [p.u.]
Profiles
kVArea 3 138
kVArea 2 138
kVArea 1 138
June 26, 2008 Software Tools - 23
Power System Dynamics and Stability 535
Universidad de Castilla-La Mancha
UWPFLOW Example
The singular value index obtained with UWPFLOW is as follows (-0
option):
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
0.1
0.2
0.3
0.4
0.5
0.6
0.7
L.F. [p.u.]
Ful
l mat
rix s
ing.
val
ue in
dex
June 26, 2008 Software Tools - 24
Power System Dynamics and Stability 536
Universidad de Castilla-La Mancha
Matlab Overview
Matlab is a general purpose environment for mathematical and
engineering analysis.
Is vector/matrix based. A variable is by default a matrix.
Is an “interpreted” language, thus can be slow for heavy applications.
Is not open source. The GNU-Octave project provides a good alternative:
http://www.octave.org
June 26, 2008 Software Tools - 25
Power System Dynamics and Stability 537
Universidad de Castilla-La Mancha
Matlab Example
Generator-load example (see the introductory example in the Voltage
Stability section starting from slide 279): :
V1∠δ1 V2∠δ2
jxL
PG + jQG PL + jQL
June 26, 2008 Software Tools - 26
Power System Dynamics and Stability 538
Universidad de Castilla-La Mancha
Matlab Example
For QGmin≤ QG ≤ QGmax
:
ω =1
M
(
Pd − V10V2
XLsin δ − DGω
)
δ = ω − 1
DL
(V10V2
XLsin δ − Pd
)
V2 =1
τ
(
− V 22
XL+
V10V2
XLcos δ − Qd
)
0 = QG −−V 210
XL+
V10V2
XLcos δ
with
x = [ω, δ, V2]T y = QG
p = V10 λ = Pd
June 26, 2008 Software Tools - 27
Power System Dynamics and Stability 539
Universidad de Castilla-La Mancha
Matlab Example
For QG = QGmin,max :
ω =1
M
(
Pd − V1V2
XLsin δ − DGω
)
δ = ω − 1
DL
(V1V2
XLsin δ − Pd
)
V2 =1
τ
(
− V 22
XL+
V1V2
XLcos δ − Qd
)
0 = QGmin,max− V 2
1
XL+
V1V2
XLcos δ
with
x = [ω, δ, V2]T y = V1
p = QGmin,max λ = Pd
June 26, 2008 Software Tools - 28
Power System Dynamics and Stability 540
Universidad de Castilla-La Mancha
Matlab Example
Assume XL = 0.5, M = 1, DG = 0.01, DL = 0.1, τ = 0.01,
k = 0.25.
The time domain integration can be solved with the help of MATLAB.
June 26, 2008 Software Tools - 29
Power System Dynamics and Stability 541
Universidad de Castilla-La Mancha
Matlab Example
Differential equations without limits:
function dx = example(t,x)
global M DL DG tau k Pd V10
if t <= 1
XL = 0.5;
else
XL = 0.6;
end
delta = x(1);
omega = x(2);
V2 = max(x(3),0);
dx(1,1) = omega - (V10*V2*sin(delta)/XL-Pd)/DL;
dx(2,1) = (Pd-V10*V2*sin(delta)/XL-DG*omega)/M;
dx(3,1) = (-V2*V2/XL+V10*V2*cos(delta)/XL-k*Pd)/tau;
June 26, 2008 Software Tools - 30
Power System Dynamics and Stability 542
Universidad de Castilla-La Mancha
Matlab Example
Initialization and main routine:
clear all
global M DL DG tau k Pd V10
XL = 0.5;
M = 0.1;
DL = 0.1;
DG = 0.01;
tau = 0.01;
k = 0.25;
Pd = 0.7;
V10 = 1;
x0 = [0.4636; 0.0000; 0.7826];
tmax = 2;
[t,x] = ode23(’example’,[0 tmax],x0);
June 26, 2008 Software Tools - 31
Power System Dynamics and Stability 543
Universidad de Castilla-La Mancha
Matlab Example
Graphical commands:
figure
plot(t,x(:,1),’b-’)
hold on
plot(t,x(:,2),’g--’)
plot(t,max(x(:,3),0),’c-.’)
plot([0 tmax],[V10 V10],’r:’)
legend(’delta’,’omega’,’V2’,’V1’)
xlabel(’t [s]’)
ylim([-1 6])
June 26, 2008 Software Tools - 32
Power System Dynamics and Stability 544
Universidad de Castilla-La Mancha
Matlab Example
The dynamic solution without limits:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
0
1
2
3
4
5
6
Voltage collapse
Operating point Contingency
V2
V1
δω
t [s]
June 26, 2008 Software Tools - 33
Power System Dynamics and Stability 545
Universidad de Castilla-La Mancha
Matlab Example
Differential equations without limits:
function dx = example(t,x)
global M DL DG tau k Pd V10
if t <= 1, XL = 0.5; else, XL = 0.6; end
delta = x(1);
omega = x(2);
V2 = max(x(3),0);
V10 = 1;
Q = V10*V10/XL - V10*V2*cos(delta)/XL;
if Q > 0.5
a = 1/XL;
b = -V2*cos(delta)/XL;
c = -0.5;
V10 = (-b + sqrt(b*b - 4*a*c))/2/a;
end
dx(1,1) = omega - (V10*V2*sin(delta)/XL-Pd)/DL;
dx(2,1) = (Pd-V10*V2*sin(delta)/XL-DG*omega)/M;
dx(3,1) = (-V2*V2/XL+V10*V2*cos(delta)/XL-k*Pd)/tau;
June 26, 2008 Software Tools - 34
Power System Dynamics and Stability 546
Universidad de Castilla-La Mancha
Matlab Example
The dynamic solution with limits:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
0
1
2
3
4
5
6
Operating point Contingency
Voltage collapse
V2
V1
δω
t [s]
June 26, 2008 Software Tools - 35
Power System Dynamics and Stability 547
Universidad de Castilla-La Mancha
PSAT Features
PSAT has been thought to be portable and open source.
PSAT runs on the commonest operating systems
PSAT can perform several power system analysis:
1. Continuation Power Flow (CPF);
2. Optimal Power Flow (OPF);
3. Small signal stability analysis;
4. Time domain simulations.
June 26, 2008 Software Tools - 36
Power System Dynamics and Stability 548
Universidad de Castilla-La Mancha
PSAT Features
PSAT deeply exploits Matlab vectorized computations and sparse matrix
functions in order to optimize performances.
PSAT also contains interfaces to UWPFLOW and GAMS which highly
extend PSAT ability to solve CPF and OPF problems, respectively.
June 26, 2008 Software Tools - 37
Power System Dynamics and Stability 549
Universidad de Castilla-La Mancha
Synoptic Scheme
CommandHistory
OutputText
ResultsSave Graphic
Output
SavedResults
SimulinkModels
Other DataFormat
PlottingUtilities
AnalysisStatic
DataFiles
LibrarySimulink
Settings
SimulinkModel
Conversion Power Flow &
InitializationState Variable
ConversionUtilities
Time DomainSimulation
Small SignalStability
DynamicAnalysisGAMS
UWpflow
Interfaces
User DefinedModels
Power FlowOptimal
Power FlowContinuation
Output
PSAT
Input
June 26, 2008 Software Tools - 38
Power System Dynamics and Stability 550
Universidad de Castilla-La Mancha
PSAT Features
In order to perform accurate and complete power system analyses, PSAT
supports a variety of static and dynamic models.
Dynamic models include non conventional loads, synchronous machines
and controls, regulating transformers, FACTS, wind turbines, and fuel
cells.
June 26, 2008 Software Tools - 39
Power System Dynamics and Stability 551
Universidad de Castilla-La Mancha
PSAT Features
Besides mathematical algorithms and models, PSAT includes a variety of
additional tools, as follows:
1. User-friendly graphical user interfaces;
2. Simulink library for one-line network diagrams;
3. Data file conversion to and from other formats;
4. User defined model editor and installer;
5. Command line usage.
June 26, 2008 Software Tools - 40
Power System Dynamics and Stability 552
Universidad de Castilla-La Mancha
PSAT Features
Not all features are available on GNU-Octave:
Function Matlab GNU/Octave
Continuation power flow yes yes
Optimal power flow yes yes
Small signal stability analysis yes yes
Time domain simulation yes yes
GUIs and Simulink library yes no
Data format conversion yes yes
User defined models yes no
Command line usage yes yes
June 26, 2008 Software Tools - 41
Power System Dynamics and Stability 553
Universidad de Castilla-La Mancha
Getting Started
PSAT is launched by typing at the Matlab prompt:
>> psat
which will create all structures required by the toolbox and open the main
GUI.
All procedures implemented in PSAT can be launched from this window
by means of menus, buttons and/or short cuts.
June 26, 2008 Software Tools - 42
Power System Dynamics and Stability 554
Universidad de Castilla-La Mancha
Getting Started
Main PSAT GUI:
June 26, 2008 Software Tools - 43
Power System Dynamics and Stability 555
Universidad de Castilla-La Mancha
Simulink Library
PSAT allows drawing electrical schemes by means of pictorial blocks.
The PSAT computational engine is purely Matlab-based and the Simulink
environment is used only as graphical tool.
A byproduct of this approach is that PSAT can run on GNU/Octave, which
is currently not providing a Simulink clone.
June 26, 2008 Software Tools - 44
Power System Dynamics and Stability 556
Universidad de Castilla-La Mancha
Simulink Library
PSAT-Simulink Library:
June 26, 2008 Software Tools - 45
Power System Dynamics and Stability 557
Universidad de Castilla-La Mancha
Other Features
To ensure portability and promote contributions, PSAT is provided with a
variety of tools, such as a set of Data Format Conversion (DFC) functions
and the capability of defining User Defined Models (UDMs).
The set of DFC functions allows converting data files to and from formats
commonly in use in power system analysis. These include: IEEE, EPRI,
PTI, PSAP, PSS/E, CYME, MatPower and PST formats. On Matlab
platforms, an easy-to-use GUI handles the DFC.
June 26, 2008 Software Tools - 46
Power System Dynamics and Stability 558
Universidad de Castilla-La Mancha
Data Format Conversion
GUI for data format conversion:
June 26, 2008 Software Tools - 47
Power System Dynamics and Stability 559
Universidad de Castilla-La Mancha
User Defined Models
The UDM tools allow extending the capabilities of PSAT and help
end-users to quickly set up their own models.
Once the user has introduced the variables and defined the DAE of the
new model in the UDM GUI, PSAT automatically compiles equations,
computes symbolic expression of Jacobians matrices and writes a Matlab
function of the new component.
Then the user can save the model definition and/or install the model in
PSAT.
If the component is not needed any longer it can be uninstalled using the
UDM installer as well.
June 26, 2008 Software Tools - 48
Power System Dynamics and Stability 560
Universidad de Castilla-La Mancha
User Defined Models
GUI for user defined models:
June 26, 2008 Software Tools - 49
Power System Dynamics and Stability 561
Universidad de Castilla-La Mancha
Command Line Usage
PSAT is provided with a command line version. This feature allows using
PSAT in the following conditions:
1) If it is not possible or very slow to visualize the graphical environment
(e.g. Matlab is running on a remote server).
2) If one wants to write scripting of computations or include calls to PSAT
functions within user defined programs.
3) If PSAT runs on the GNU/Octave platform, which currently neither
provides GUI tools nor a Simulink-like environment.
June 26, 2008 Software Tools - 50
Power System Dynamics and Stability 562
Universidad de Castilla-La Mancha
Power System Model
The standard power system model is basically a set of nonlinear
differential algebraic equations, as follows:
x = f(x, y, p)
0 = g(x, y, p)
where x are the state variables x ∈ Rn; y are the algebraic variables
y ∈ Rm; p are the independent variables p ∈ R
ℓ; f are the differential
equations f : Rn × R
m × Rℓ 7→ R
n; and g are the algebraic
equations g : Rm × R
m × Rℓ 7→ R
m.
June 26, 2008 Software Tools - 51
Power System Dynamics and Stability 563
Universidad de Castilla-La Mancha
Power System Model
PSAT uses these equations in all algorithms, namely power flow, CPF,
OPF, small signal stability analysis and time domain simulation.
The algebraic equations g are obtained as the sum of all active and
reactive power injections at buses:
g(x, y, p) =
gp
gq
=
gpm
gqm
−∑
c∈Cm
gpc
gqc
∀m ∈ M
where gpm and gqm are the power flows in transmission lines, M is the
set of network buses, Cm and [gTpc, g
Tqc]
T are the set and the power
injections of components connected at bus m, respectively.
June 26, 2008 Software Tools - 52
Power System Dynamics and Stability 564
Universidad de Castilla-La Mancha
Component Models
PSAT is component-oriented, i.e. any component is defined
independently of the rest of the program as a set of nonlinear
differential-algebraic equations, as follows:
xc = fc(xc, yc, pc)
Pc = gpc(xc, yc, pc)
Qc = gqc(xc, yc, pc)
where xc are the component state variables, yc the algebraic variables
(i.e. V and θ at the buses to which the component is connected) and pc
are independent variables. Then differential equations f are built
concatenating fc of all components.
June 26, 2008 Software Tools - 53
Power System Dynamics and Stability 565
Universidad de Castilla-La Mancha
Component Models
These equations along with Jacobians matrices are defined in a function
which is used for both static and dynamic analyses.
In addition to this function, a component is defined by means of a
structure, which contains data, parameters and the interconnection to the
grid.
June 26, 2008 Software Tools - 54
Power System Dynamics and Stability 566
Universidad de Castilla-La Mancha
Component Models: Example
Let’s consider the exponential recovery load (ERL).
The set of differential-algebraic equations are as follows:
xc1= −xc1
/TP + P0(V/V0)αs − P0(V/V0)
αt
xc2= −xc2
/TQ + Q0(V/V0)βs − Q0(V/V0)
βt
Pc = xc1/TP + P0(V/V0)
αt
Qc = xc2/TQ + Q0(V/V0)
βt
where and P0, Q0 and V0 are initial powers and voltages, respectively,
as given by the power flow solution.
Observe that a constant PQ load must be connected at the same bus as
the ERL to determine the values of P0, Q0 and V0.
June 26, 2008 Software Tools - 55
Power System Dynamics and Stability 567
Universidad de Castilla-La Mancha
Component Models: Example
Component Data:
Column Variable Description Unit
1 - Bus number int
2 Sn Power rating MVA
3 Vn Active power voltage coefficient kV
4 fn Active power frequency coefficient Hz
5 TP Real power time constant s
6 TQ Reactive power time constant s
7 αs Static real power exponent -
8 αt Dynamic real power exponent -
9 βs Static reactive power exponent -
10 βt Dynamic reactive power exponent -
June 26, 2008 Software Tools - 56
Power System Dynamics and Stability 568
Universidad de Castilla-La Mancha
Component Models: Example
Exponential recovery loads are defined in the structure Erload, whose
fields are as follows:
1. con: exponential recovery load data.
2. bus: Indexes of buses to which the ERLs are connected.
3. dat: Initial powers and voltages (P0, Q0 and V0).
4. n: Total number of ERLs.
5. xp: Indexes of the state variable xc1.
6. xq: Indexes of the state variable xc2.
June 26, 2008 Software Tools - 57
Power System Dynamics and Stability 569
Universidad de Castilla-La Mancha
PSAT Example
This section illustrates some PSAT features for static and dynamic
stability analysis by means of the IEEE 14-bus test system.
All data can be retrieved from the PSAT web site:
http://www.power.uwaterloo.ca/∼fmilano/
June 26, 2008 Software Tools - 58
Power System Dynamics and Stability 570
Universidad de Castilla-La Mancha
PSAT Example
IEEE 14-bus test system:
Bus 14|V| = 1.0207 p.u.<V = −0.2801 rad
Bus 13|V| = 1.047 p.u.<V = −0.2671 rad
Bus 12|V| = 1.0534 p.u.<V = −0.2664 rad
Bus 11|V| = 1.0471 p.u.<V = −0.2589 rad
Bus 10|V| = 1.0318 p.u.<V = −0.2622 rad
Bus 09|V| = 1.0328 p.u.<V = −0.2585 rad
Bus 08|V| = 1.09 p.u.
<V = −0.2309 rad
Bus 07|V| = 1.0493 p.u.<V = −0.2309 rad
Bus 06|V| = 1.07 p.u.<V = −0.2516 rad
Bus 05|V| = 1.016 p.u.<V = −0.1527 rad
Bus 04|V| = 1.012 p.u.<V = −0.1785 rad
Bus 03|V| = 1.01 p.u.<V = −0.2226 rad
Bus 02|V| = 1.045 p.u.<V = −0.0871 rad
Bus 01|V| = 1.06 p.u.<V = 0 rad
Breaker
Breaker
June 26, 2008 Software Tools - 59
Power System Dynamics and Stability 571
Universidad de Castilla-La Mancha
PSAT Example
Power flow report:
June 26, 2008 Software Tools - 60
Power System Dynamics and Stability 572
Universidad de Castilla-La Mancha
PSAT Example
Continuation power flow analysis (GUI):
June 26, 2008 Software Tools - 61
Power System Dynamics and Stability 573
Universidad de Castilla-La Mancha
PSAT Example
Continuation power flow analysis (plots):
June 26, 2008 Software Tools - 62
Power System Dynamics and Stability 574
Universidad de Castilla-La Mancha
PSAT Example
Nose curves at bus 14 for different contingencies for the IEEE 14-bus test
system:
1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.80
0.2
0.4
0.6
0.8
1
λc
Vol
tage
[p.u
.]
Base Case
Line 2-4 Outage
Line 2-3 Outage
June 26, 2008 Software Tools - 63
Power System Dynamics and Stability 575
Universidad de Castilla-La Mancha
PSAT Example
Optimal power flow analysis (GUI):
June 26, 2008 Software Tools - 64
Power System Dynamics and Stability 576
Universidad de Castilla-La Mancha
PSAT Example
Comparison between OPF and CPF analysis:
Contingency BCP λ∗ MLC ALC
[MW] [p.u.] [MW] [MW]
None 259 0.7211 445.8 186.8
Line 2-4 Outage 259 0.5427 399.5 148.6
Line 2-3 Outage 259 0.2852 332.8 73.85
Because of the definitions of generator and load powers PG and PL, one
has λc = λ∗ + 1.
June 26, 2008 Software Tools - 65
Power System Dynamics and Stability 577
Universidad de Castilla-La Mancha
PSAT Example
Time domain simulation:
It has been used a 40% load increase with respect to the base case
loading, and no PSS at bus 1. A Hopf bifurcation occurs for the line
2-4 outage resulting in undamped oscillations of generator angles.
A similar analysis can be carried on the same system with a 40% load
increase but considering the PSS of the generator connected at bus 1.
In this case the system is stable.
June 26, 2008 Software Tools - 66
Power System Dynamics and Stability 578
Universidad de Castilla-La Mancha
PSAT Example
Time domain simulation (without PSS):
0 5 10 15 20 25 300.998
0.9985
0.999
0.9995
1
1.0005
1.001
1.0015
1.002ω1 - Bus 1
ω2 - Bus 2
ω3 - Bus 3
ω4 - Bus 6
ω5 - Bus 8
Gen
erat
orS
peed
s[p
.u.]
Time [s]
June 26, 2008 Software Tools - 67
Power System Dynamics and Stability 579
Universidad de Castilla-La Mancha
PSAT Example
Eigenvalue analysis (with PSS):
−1 −0.8 −0.6 −0.4 −0.2 0 0.2−10
−8
−6
−4
−2
0
2
4
6
8
10
Real
Imag
June 26, 2008 Software Tools - 68
Power System Dynamics and Stability 580
Universidad de Castilla-La Mancha
Project 1
Reproduce the examples illustrated in the following slides:
Voltage Stability: Slides 341-350 (using UWPFLOW)
Angle Stability: Slides 444-456 (using Matlab)
Frequency Stability: 479-489 (using Matlab)
The software UWPFLOW is freely available at:
http://thunderbox.uwaterloo.ca/∼claudio/software/pflow.html
June 26, 2008 Projects - 1
Power System Dynamics and Stability 581
Universidad de Castilla-La Mancha
Project 1
Voltage stability (slides 341-350):
Write the 3area.wsc data file in the EPRI data format (the format
is fully described in the UWPFLOW documetation).
Write the generator load change file 3area.k using the format
described in the UWPFLOW documentation.
Write a batch file for running the simulations as described in the
slides.
June 26, 2008 Projects - 2
Power System Dynamics and Stability 582
Universidad de Castilla-La Mancha
Project 1
Voltage stability (slides 341-350):
Run a base case power flow, a continuation power flow with and
without enforcing voltage and current limits. Compute also the singular
value index.
Report power flow results as given by UWPFLOW and plots PV
curves and the singular value index using Matlab.
June 26, 2008 Projects - 3
Power System Dynamics and Stability 583
Universidad de Castilla-La Mancha
Project 1
Angle stability (slides 444-456):
Write a Matlab function with the system differential equations.
Use a Matlab script file to initialize and solve the time domain
simulation (function ode23).
Using a trial-and-error technique find the clearing time tc of the
system for D = 0.1, D = 0.05 and D = 0.2.
For each value of the damping D, provide plots of the rotor angle δ
and the rotor speed ω.
June 26, 2008 Projects - 4
Power System Dynamics and Stability 584
Universidad de Castilla-La Mancha
Project 1
Frequency stability (slides 479-489):
Write a Matlab function with the system differential equations.
Use a Matlab script file to initialize and solve the time domain
simulation (function ode23).
Run the time domain integration for a 25%, 50% and 60% generation
drop at t = 1 followed by a 25%, 50% and 60% load drop at t = 5,
respectively.
For each value of the generation and load drop, provide plots of ω, δ,
E′, V2 and V1.
June 26, 2008 Projects - 5
Power System Dynamics and Stability 585
Universidad de Castilla-La Mancha
Project 2
Reproduce the results for the IEEE 14-bus tests system illustrated in the
paper:
F. Milano, “An Open Source Power System Analysis Toolbox”,
accepted for publication on the IEEE Transactions on Power Systems,
March 2005, 8 pages.
The full paper as well as the software PSAT is available at:
http://www.power.uwaterloo.ca/∼fmilano/
June 26, 2008 Projects - 6
Power System Dynamics and Stability 586
Universidad de Castilla-La Mancha
Project 2
The IEEE 14-bus test system is provided wintin the PSAT main
distribution (folder tests).
For the base case power flow, the continuation power flow and the
optimal power flow routines, use the file:
d 014 dyn l10.mdl
For the time domain simulations without PSS, use the file:
d 014 dyn l14.mdl
For the time domain simulations with PSS, use the file:
d 014 pss l14.mdl
June 26, 2008 Projects - 7
Power System Dynamics and Stability 587
Universidad de Castilla-La Mancha
Project 2
Hints:
For static analyses (PF, CPF, OPF), disable loading dynamic
components by checking the box “Discard dynamic components” in
the GUI Settings (within the menu “Edit” in the main window).
To simulate a line outage in static analyses, uncheck the box “initially
close” box (within the Simulink block mask) of the breaker of the line
that one wants to keep out.
June 26, 2008 Projects - 8
Power System Dynamics and Stability 588
Universidad de Castilla-La Mancha
Project 2
Hints:
For OPF analysis, disable the “base case” powers in the GUI for OPF
Settings and set the weighting factor to 1 (maximization of the ditance
to voltage collapse).
For time domain simulations, remember to uncheck “Discard dynamic
components” box and do not allow the conversion to PQ buses when
the program asks for.
June 26, 2008 Projects - 9