power sys dynamics and stability

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



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




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.




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.




Outlines: System Modeling

Synchronous machine.

Transformer.

Transmission line.

Cable.

Loads.




Outlines: System Analysis

Basic stability concepts:

Nonlinear systems.

Equilibrium points.

Stability regions.

Power Flow:

System model.

Equations and solution techniques.

Contingency analysis.




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.




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.




Outlines: Frequency Stability

Definitions.

Basic concepts.

Protections and controls.

Real case example: October 2003 Italian blackout.

Real case example: November 2006 European blackout.




Outlines: Software Tools

Outlines.

UWPFLOW.

Matlab.

PSAT.




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.




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.




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.




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.




Evaluation

Alternatively, the students can develop a “user defined model” in PSAT.

Interested students are invited to contact Dr. Federico Milano.




Contents

Introduction

Generator Modeling

Transmission System Modeling

Load Modeling

Power Flow Outlines

Stability Concepts

Voltage Stability

Angle Stability

Frequency Stability

Software Tools

Projects




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



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




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




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




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




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




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




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




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 )




Synchronous Machine

Mechanical equations:

2

pJ

d

dtωr +

2

pDωr = Tm − Te

d

dtθr = ωr

Te =p

2(iqsλds − idsλqs)




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.




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




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.




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




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




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




Subtransient Model

Steady-state equations:

ea − vqs = rsiqs − xdids

−vds = rsids + xqiqs


d

dt∆ωr =

1

M[Pm − vasias cos(θvas

− θias) − D∆ωr]

d

dtδ = ∆ωr = ωr − ωs




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




Subtransient Model

Some definitions:

xF = xLF + xMd

xD = xLD + xMd

xQ1 = xLQ1 + xMq

xQ2 = xLQ2 + xMq




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

)




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




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




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.




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




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




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




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




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




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

]




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.]




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]




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.




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




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




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




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.




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




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




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




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.




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.




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




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]




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)]




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





Transformers:

Single phase:

Detailed model

Phasor model

Three phase:

Phase shifts

Models

June 26, 2008 Transmission System Modeling - 1




Transmission Lines:

Single phase:

Distributed parameter model

Phasor lumped model

Three phase:

Distributed parameter model

Reduced models

Underground cables




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.





++

v1

i1

v2

i2

N1 N2

λm

λℓ1 λℓ2

leakage

core

magnetizing





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





Equivalent circuit:

++ ++

− −−−

N1 : N2

a : 1

aii

e e/av1 v2

r1 r2

im

i1 i2

Gm

Lm

Lℓ1 Lℓ2





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





Phasor equivalent circuit:

+++

−−−

a : 1

V1 V2aV2

I1 I2I2/a

Im

Ym

Zℓ1 a2Zℓ2





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)





Phasor equivalent circuit with the approximation Zm ≫ Zℓ1:

+ ++

− −−

a : 1

V1 V2aV2

I1 I2I2/a

Im

Ym

Zℓ





Or Π form:

Z =Zℓ

a

Y1 = (1 − a)1

Zℓ

Y2 = (a2ZℓYm + a2 + a)1

Zℓ

≈ (a2 − a)1

Zℓfor Ym ≈ 0





Π equivalent circuit:

++

−−

V1 V2

I1 I2

Y1 Y2

Z





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





Equivalent circuit neglecting Zm:

+ ++

−−

a : 1

V1 V2aV2

I1 I2 = −I1/aZℓ





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.





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.





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.





Phasor equivalent circuit with complex tap ratio a:

+ ++

− −−

a : 1

V1 V2aV2

I1 I2I2/a∗

Im

Ym

Zℓ




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





Per unit parameters:

Zℓ = j0.1

Zm = 10Zℓ = j1

Ym =1

Zm= −j1

a =8/8 kV

80/80 kV= 1






A = 1(1 + j0.1(−j1)) = 1.1

B =j0.1

1= j0.1

C = 1(−j1) = −j1

D =1

1= 1






Z =j0.1

1= j0.1

Y1 = (1 − 1)1

j0.1= 0

Y2 = (12(j0.1)(−j1) + 12 − 1)1

j0.1= −j1




Three-phase Transformers

+

+

+

+

+

+

− −

−

−−

−

a : 1

a : 1

a : 1

T1

T2

T3

a1 a2

b1 b2

c1 c2

aVa2

aVb2

aVc2

Va2

Vb2

Vc2





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





∆ − Y : a =√

3a∠30

apu =√

3apu∠30

Y − ∆ : a =√

3a∠ − 30

apu =√

3apu∠ − 30

Y − Y : a = a

∆ − ∆ : a = a





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.





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




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.




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




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]





l [H/m]

c [F/m]

x dx

i

vv1 v2

i1 i2

didv+ +++

− −

−

−





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)





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)++

−−





Example:

E

t = 0

v1 v2

i1 i2 = 0

Trans.

Line

+ + +

−−−

v1(t) = E

i2(t) = 0





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





Example:

2E

E/Zc

−E/Zc

t

t

v2

i1

τ

τ

2τ

2τ

3τ

3τ

4τ

4τ

5τ

5τ

6τ

6τ





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





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





This can be converted into the Π equivalent circuit:

++

−−

V1 V2

I1 I2

Y′

1/2 Y′

2/2

Z′





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




Three-phase Transmission Line

nn

a b c

dα

2R

dβ

dγ

d × d

dδ





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





−∂

∂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





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





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





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]


8 cos 8φ − b9a9 cos 9φ + . . .

a = 4π√

510−4D√

f/ρ





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, . . .





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]


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





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





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).





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





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





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.





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





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 .





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





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 .





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





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




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




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




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




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




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




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




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




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




Example 2

nn

a b c

50′

1′

45′

110′

40′ × 40′

28′8′′




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




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




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




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




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




Cables

Steel pipe

Skid wires

Metallic tapes

Paper/oil insulation

Screen

Conductor

(filled with insulating oil)

(stranded copper)




Cables

Single-phase and three-phase cables:

conductor

sheath

SF6 gas




Cables

Model for a single-phase cable:

VC VS VA

I1

I2

I3




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




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




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




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




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)




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.




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




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




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



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).




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.




RLC Loads, Resistor

Ideal, linear resistors (R), inductors (L) and capacitors (C).

Resistor:

+ −

i

v

R

Time domain:

v = Ri

Phasor domain:

V = RI




RLC Loads, Inductor

Inductor:

+ −

i

v

L

Time domain:

v = Ldi

dt

Phasor domain:

V = jωLI

= ZLI




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




RLC Loads, Capacitor

This yields the equivalent resistive circuit:

+ −ik+1

vk+1

hk

2L∆t





Capacitor:

+ −

i

v

C

Time domain:

i = Cdv

dt

Phasor domain:

V = −j1

ωCI

= ZCI





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





This yields the equivalent resistive circuit:

+ −ik+1

vk+1

hk

∆t2C




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




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]




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




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)




Induction Motor

Rotor transformation equations:

v0r

vdr

vqr

= Kr

var

vbr

vcr

i0r

idr

iqr

= Kr

iar

ibr

icr




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




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




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




Induction Motor


2

pJ

d

dtωr +

2

pDωr = Tm − Te

d

dtθr = ωr

d

dtθ = ω

Te =3

2

p

2(iqsλds − idsλqs)




Induction Motor

Equivalent circuit representation of these equations:

+ +

+

− −

−

iqs iqr

vqs vqrM

rs r′rLls L′

lr

ωλds (ω − ωr)λ′

dr

Ns : Nr




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




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

]




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)




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




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




Induction Motor

Or equivalently:

−

+

Ias I′ar

Vasjxm

rs jxls jx′

lr

r′r1

σ

Plus mechanical equations.




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




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.




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.




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




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)




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)




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.




Example

The Z load model is then:

PL =21 MW

100 MVA= 0.21

= PZ + PIM

PZ = V 2LG = 0.06

⇒ PIM = 0.15




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




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




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.




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.




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




Power Models

This results in the following time response:

P

PL0

V

VL0

t

t

tf

tf




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




Example

Measurements vs. model:




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




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




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




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)




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

]




Power Flow Outlines

Power Flow:

System model.

Equations.

Solution techniques:

Newton-Raphson.

Fast decoupled.

June 26, 2008 Power Flow Outlines - 1



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




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




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




Power Flow Model

Hence, if Q reaches a limit:

P = unknown

Q = constant

V = unknownδ = 0




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




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




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.




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




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




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





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





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





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





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





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.





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.





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.




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.




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




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




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.




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)




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)




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




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′.




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.




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.




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.




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




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)




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)




Example


∆P2 = P2 −3∑

k=1


0 = P1/2 −3∑

k=1

B2k sin(δ2 − δk)

= 0.5P1 + 10 sin(δ2) + 10 sin(δ2 − δ3)




Example


∆Q2 = Q2 −3∑

k=1


0 = Q2 +

3∑

k=1

B2k cos(δ2 − δk)

= Q1 + 10 cos(δ2) − 19.8 + 10 cos(δ2 − δ3)




Example


∆P3 = P3 −3∑

k=1


0 = −0.9 −3∑

k=1

B3k sin(δ3 − δk)

= −0.9 + 10 sin(δ3) + 10 sin(δ3 − δ2)




Example


∆Q3 = Q3 −3∑

k=1


0 = Q −√

1 − 0.92 +

3∑

k=1

B3k cos(δ3 − δk)

= Q − 0.436 + 10 cos(δ3) + 10 cos(δ3 − δ2) − 19.8




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




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;




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’))



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


z0 =

-0.2501 -1.2613 12.0000 7.0658 4.8031 20.1667




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’))



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




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



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)




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.




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




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




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.




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)




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




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




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




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.




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.




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.




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




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.




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.




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.




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




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




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




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




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




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




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




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.




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




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);




Equilibria

This generates the equilibrium point for λ = PL = 1:

>> lambda = 1;

>> z0 = fsolve(@dae_eqs,[0 0 1 0 1],optimset(’Display’,’iter’))



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


z0 =

0.7539 0 1.0959 0.5236 0.2679




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])




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]




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]




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.




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




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)




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.




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




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




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




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




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




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




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.




Voltage Stability Outlines

Definitions.

Basic concepts.

Continuation Power Flow.

Direct Methods.

Indices.

Controls and protections.


Examples.

June 26, 2008 Voltage Stability - 1



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





“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.”





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).




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





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)





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)





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)





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





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

)





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.





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





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





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





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





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





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):





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





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.





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.





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





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]





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





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.





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





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


Voltage collapse

V2

V1

δω

t [s]





Collapse problems can be avoided using shunt or series compensation:

V1∠δ1 V2∠δ2

jxL

PG + jQG PL + jQL

−jxC





In this case, the system equations are:


:

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





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]


10BL + V10V2BL cos δ





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





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





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]





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]




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.




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.




CPF Predictor

Tangent vector method:

DzF |1dz

dλ︸︷︷︸

t1

− ∂F

∂λ

∣∣∣∣1

⇒ ∆z1 =k

‖t1‖︸︷︷︸

∆λ1

t1

(∆z1, ∆λ1)

(z1, λ1)

(z2, λ2)




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.




CPF Predictor

Secant method:

∆z1 = k(z1a − z1b)

∆λ1 = k(λ1a − λ1b)

(∆z1, ∆λ1)

(z1a, λ1a)

(z1b, λ1b)

(z2, λ2)




CPF Predictor

This method is faster but can have convergence problems with “sharp

corners” (e.g. limits):

(z1a, λ1a)

(z1b, λ1b)

(z2, λ2)




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.




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(·).




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




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




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




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.




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.




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).




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




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




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.




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.




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.




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.




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:




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.




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





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.





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.





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.





For example, for WECC, systems should be operated a minimum 5%

“distance” away from the maximum loadibility point when contigencies are

considered:




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.




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:





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.




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.




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




Example


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:




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




Example


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




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




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




Example

Batch file for Windows (run3area.bat):

rem -1- Run base case power flow


rem -2- Obatin PV curves and maximum loading


rem - with bus voltage limits enforced


rem - with current limits enforced





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.




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




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




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.





3 Interconnections, 10 Regional Reliability Councils:





Regions and control areas of the North American Electric Reliability

Council (NERC):





Reliability coordinators (some are also system and market operators,

such as the IMO and PJM):





Reliability coordinator and system/market operators in the Ohio area,

where the system collapse started:





Four causes:

Inadequate system understanding: FirstEnergy, ECAR

Inadequate situational awareness: FirstEnergy

Inadequate tree-trimming: FirstEnergy

Inadequate diagnostic support: MISO, PJM





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.





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.





Lost of the Eastlake 5 unit and beginning of reactive power problems.





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.





Line outages:





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.





The tipping point in Ohio:





Effect of a line trip: increasing loading on other lines.





Effect of a line trip: decreasing voltages leading to voltage collapse.





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.





The system goes “haywire”:





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).





Sequence of events:





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.





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.




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




Angle Stability Outlines

Definitions.

Small disturbance:

Hopf Bifurcations.

Control and mitigation.

Practical example.

Transient Stability

Time Domain.

Direct Methods.

Equal Area Criterion.

Energy Functions.


June 26, 2008 Angle Stability - 1



Angle Stability Definitions




231, 2003:


StabilityStability

Stability

Stability

Stability

Voltage

Transient


Power System



Long Term

Long Term

Short Term Short Term

Short Term




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.




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).




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




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′




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




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.




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




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




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




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




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]




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.




Hopf Bifurcation II

Single-machine dynamic-load system with SVC:

E∠δ V ∠0

V

Vref

jX

jBC

PG + jQG PL + jQT




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




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 )




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.




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




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




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)




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




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.




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.

)




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.




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





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)





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.





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





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)





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.





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





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)





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





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.




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.




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




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





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.





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





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





As a results of the undamped oscillations, the system split into four large

islands.





PSS solution:

San Onofre

(addition)Palo Verde

(tune existing)





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





7.5 million customers experienced outages from a few minutes to nine

hours.

Total load loss: 35,500 MW.




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.




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.




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.




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.




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





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





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.





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]





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]




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”.




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).




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.




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.




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.




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.




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




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.




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




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).




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




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




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).




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.





Pre-contingency or initial conditions:

PGpre= PL =

E′V

Xpresin δspre

QL = − V 2

Xpre+

E′V

Xprecos δspre





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





⇒ 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





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





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





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





⇒ 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





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





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)





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]





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]





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]





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





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 .





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





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]





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.




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.




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.





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





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.




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.





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





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





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.





The line trip and generator limits yield a voltage collapse associated with

a limit-induced bifurcation problem:





PV curves:





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.





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.




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



Frequency Stability Definitions




231, 2003:


Stability

Stability

StabilityStability

Stability

Voltage

Transient


Power System



Long Term

Long Term

Short TermShort Term

Short Term




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.




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





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)





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

)





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)





Define:

δ = δG − δ2

δ′ = δ1 − δ2

ω = ωG

⇒ δ = ω − ω2





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


Xsin δ

0 = V 22

(1

XL− 1

X

)

+E′V2

Xcos δ − V1V2

XLcos δ′





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


Xsin δ

0 = V 22

(1

XL− 1

X

)

+E′V2

Xcos δ − V1V2

XLcos δ′





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.





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′





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





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]





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.





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.





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).





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.





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.





Imports from France:





Frequency in Italy:





Frequency and voltages in Europe:




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





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





Area 1 : The frequency drops to 49 Hz, which causes an automatic load

schedding.

Area 2 : Real power surplus of 6000 MW.





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




Contents

Overview

UWPFLOW

Matlab

PSAT

June 26, 2008 Software Tools - 1



Overview

Software packages for power system analysis can be basically divided

into two classes of tools:

Commercial softwares.

Educational/research-aimed softwares.




Overview

Commercial softwares:

PSS/E

EuroStag

Simpow

CYME

PowerWorld

Neplan




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.




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.




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.




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)




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




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).




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.




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.




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.




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.




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.




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




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




UWPFLOW Example


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:




UWPFLOW Example


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




UWPFLOW Example


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




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




UWPFLOW Example

Batch file for UNIX (run3area):

echo -1- Run base case power flow


echo -2- Obatin PV curves and maximum loading


echo - with bus voltage limits enforced


echo - with current limits enforced





UWPFLOW Example

Batch file for Windows (run3area.bat):

rem -1- Run base case power flow


rem -2- Obatin PV curves and maximum loading


rem - with bus voltage limits enforced


rem - with current limits enforced





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




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




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




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




Matlab Example


:

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




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

)


1

XL+

V1V2

XLcos δ

with

x = [ω, δ, V2]T y = V1

p = QGmin,max λ = Pd




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.




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;




Matlab Example

Initialization and main routine:

clear all


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);




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])




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


V2

V1

δω

t [s]




Matlab Example

Differential equations without limits:

function dx = example(t,x)


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;




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


Voltage collapse

V2

V1

δω

t [s]




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.




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.




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




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.




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.




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




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.




Getting Started

Main PSAT GUI:




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.




Simulink Library

PSAT-Simulink Library:




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.




Data Format Conversion

GUI for data format conversion:




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.




User Defined Models

GUI for user defined models:




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.




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.




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.




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.




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.




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.





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 -





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.




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/




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




PSAT Example

Power flow report:




PSAT Example

Continuation power flow analysis (GUI):




PSAT Example

Continuation power flow analysis (plots):




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




PSAT Example

Optimal power flow analysis (GUI):




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.




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.




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]




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




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



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.




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.




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.




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.




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/




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




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.




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.


power sys dynamics and stability

Documents