EE543 Power System Stability

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EE543Power System Stability
EUMP Distance Education Services

• Satish J Ranade

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Transients and Dynamics

• Time scale Phenomenon Result
• µS Lightning Overvoltage
• mS Switching Insulation Failure
• mS Abnormal Transient Fault
• .1 S Breaker Operations
  Instability
• 1 S Mechanical Dynamics
• Many Seconds Load Dynamics Collapse

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Power System Stability

• Definition (Ch.13)
• The ability of a power system to reach a new
  steady state or equilibrium after a disturbance.
• Interconnected synchronous generators must settle
  to a common, constant speed
• Voltages and power flow must settle to reasonable
  values ( otherwise relays will trip breakers)

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Power System Stability
A disturbance, e.g., fault causes generator
speeds, system voltages and power flow to change
over time
Stability ? post-disturbance quantities become
constant
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Power System Stability

• Manifestations Angle Stability

Pm Pe
Large System Infinite Bus Voltage and frequency
constant
V
Fault Occurs Generator Terminal voltage V goes
to zero Generator electrical real power output Pe
goes to zero Turbine is still putting out
mechanical power Pm Generator speeds up builds
up kinetic energy Fault is cleared Can generator
get back to constant --synchronous speed? Time
scale of 1-10 Seconds
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Power System Stability

• Manifestations Angle Stability

P
KE Builds up
P
Excess KE Needs to be removed
Can generator get back to constant --synchronous
speed? Only if it can get rid of excess
KE Excess KE needs to go into the infinite bus
through the line? Will it? What happens if it
cant? Stability means returning to synchronous
speed In a multi-machine system it means
settling at a common speed
SPEED
Line Real Power P
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Power System Stability

• Manifestations Angle Stability

P
KE Builds up
P
Excess KE Needs to be removed
The process of power transfer across the line to
get rid of excess KE is inherently
oscillatory The infinite bus is trying to
synchronize the generator or bring it back
into step Stability requires synchronizing
torque In addition damping torque to make
oscillation decay this comes from the machine
as well as from control systems
SPEED
Line Real Power P
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Power System Stability

• Manifestations Voltage Stability

Line Opens Load Voltage Drops Many loads keep
power constant --Current goes up Voltage drops
further Reactive power loss goes
up . Generator hits limit Generator
voltage drops Voltage collapses Time scale
of 1 seconds to minutes to hours
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Power System Stability

• Terminology

First Swing (Transient Stability) Generator
speeds swing around to common speed (Better
definition later..)Little or no control action
from exciters.. 1 Second Transient
Stability Multiple Swings 1-5 Sec Field action
most important Mid-Term Past 1 second control
action is significantIssue is oscillations and
damping. This term is not used as much any
more Long Term Past 1 second and including all
control action Includes voltage stability
effects This has become the standard study
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Power System Stability

• Terminology

Rotor Angle Stability Refers to conditions in
which generator dynamics is significant and
voltages less important Steady state
Stability Slow incremental changes that
ultimately makes the system unstable (
associated with maximum power transfer) Small
signal stability Response to small changes that
can be analyzed using linear models.
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Power System Stability

• Terminology

Voltage Stability Inability to maintain voltage
because of reactive power deficit Voltage
Collapse Voltage instability leading to
low-voltage profile
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Power System Stability

• Comment on Terminology

Its all one big ball of wax! Distinctions are
made for a number of reasons Ease of analysis
or computation To emphasize/identify
components and controls that have major
impact A systems may be more prone to one
type of stability than the other Modern
long-term stability simulations capture most of
the effects
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Power System Stability

• Purpose of Stability Study

Planning Transmission Requirements Voltage
Support ( VAR Supply) Design Controls
Excitation, Power system Stabilizers, FACTS
devices Relay Settings Load
Shedding Operations Operating Margins
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• Angle Stability Revisited-
• Machine Connected to Infinite Bus

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8
P
KE Builds up
P
Excess KE Needs to be removed
Fault makes generator electric Power
zero Generator accelerates. Can generator get
back to constant --synchronous speed? Only if it
can get rid of excess KE Excess KE needs to go
into the infinite bus through the line? Will it?
What happens if it cant? Stability means
returning to synchronous speed
SPEED
Line Real Power P
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Power System Stability

• First swing stability-Background needed

P
KE Builds up
P
Excess KE Needs to be removed

• Power flow characteristics in
• Network
• What governs power flow in a line?
• Dynamics of Turbine generator
• How does a generator change speed?
• How does generator dynamics affect
• Power flow?

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Power System Stability

• Modeling

P
KE Builds up
P
Excess KE Needs to be removed
Mechanical Model --- Swing Equation Network
Model --- Power Angle Equation
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Power System Stability

• Modeling Assumptions
• Generator dynamics

P
KE Builds up
P
Excess KE Needs to be removed
Mechanical power is constant Rotor speed changes
SLOWLY as compared to 60 Hz Voltage and current
can be represented by slowly varying phasors
Real power can be calculated from phasor
models System remains balanced (Some Unbalanced
faults can be handled)
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Power System Stability
Summary Lecture 1
P
KE Builds up
P
Excess KE Needs to be removed
Many forms of instability First Swing Transient
Stability Ability of system to reach a steady
state condition after a disturbance with
machines running at common speed Transient
Stability After disturbance machine speeds turn
back to a common speed once Key Aspect Excess
KE from accelarated machines should transfer
through the network to decelerated machines to
restore speed
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Power System Stability

• Lecture 2 Mechanical Aspects and Swing Equation

Pm Pe
Large System Infinite Bus Voltage and frequency
constant
V
Fault Occurs Generator Terminal voltage V goes
to zero,Generator electrical real power output Pe
goes to zero Turbine is still putting out
mechanical power Pm Generator speeds up builds
up kinetic energy Fault is clearedCan generator
get back to constant --synchronous speed?
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Power System Stability

• Lecture 2 Mechanical Aspects and Swing Equation

Pm Pe
Large System Infinite Bus Voltage and frequency
constant
V

• Can generator get back to constant --synchronous
  speed?
• How does speed change in a disturbance
• Mechanical system model
• SWING EQUATIONDYNAMICS
• How do speed changes ( angle changes) affect
  power flow?
• Machine model
• Power angle equation
• Given these models can energy be transferred to
  bring speeds back?

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First swing stability-Dynamics

• In steady state synchronous Machines run at
  synchronous speed

Pm Pe
Tm ?m Te
?m Mechanical position of rotor rad ?m
mechanical speed in radians per secondd
?m/dt am mechanical accelaration rad/sec2 d
?m/dt d2?m/dt2
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First swing stability-Dynamics

• In steady state synchronous Machines run at
  synchronous speed

Pm Pe
P number of poles ?msyn (2/P) ?syn ?syn 2
p f Electrical frequency in rad/s f
electrical frequency in Hz Position and speed
can be measured in mechanical (actual) or
electrical units ? (P/2)?m electrical radians
per second d (P/2) dm electrical radians per
second
Tm ?m Te
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First swing stability-Dynamics

• In steady state synchronous Machines run at
  synchronous speed

Pm Pe
Tm ?m Te
?m
4 poles at 60 Hz ?syn 377 el. Rad/sec
?msyn 188.5 Rad/sec (1800 rpm)
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Power System Stability
First swing stability-Dynamics
Pm Pe

• Generator dynamics

Tm Te J am
Tm Te
Tm Te J d?m /dt
Tm Te J d2?m/dt2
If mechanical torque Tm gt Electrical torque Te
speed ?m goes up If mechanical torque Tm lt
Electrical torque Te speed ?m goes down
Normally write this in terms of Power, speed in
electrical rad/s and units of PU
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First swing stability-Dynamics

Pm Pe

• Mechanical position
• and speed

Rotor spins CCW at speed
Tm ?m Te
?m Mechanical position with respect to
stationary reference (rad.) ?m Mechanical
speed with respect to stationary
reference(rad/sec) ?m d ?m / dt ?m ? ?m
dt
?m
?m
Fixed Reference
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First swing stability-Dynamics

Pm Pe

• Mechanical position
• and speed

Tm ?m Te
dm Mechanical position with respect to
reference rotating at synchronous speed
?msyn ?m Mechanical speed with respect to
rotating reference (rad/sec) ?m ? ?msyn dt
dm ?msyn t dm d ?m / dt ?m - ?msyn d
dm / dt
?m
?m
?msyn
Reference rotates At Synchronous Speed
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First swing stability-Dynamics

• Mechanical position
• and speed

Pm Pe
?m - ?msyn d dm / dt
Tm ?m Te
?m - ?msyn 0 gt dm constant ?m - ?msyn 0 gt
dm increases ?m - ?msyn 0 gt dm
decreases Think about watching the hash mark on
the rotor under a strobe. If the rotor turns
faster than the strobe the hash advances at the
difference speed
?m
?m
?msyn
Reference rotates At Synchronous Speed
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First swing stability-Dynamics
Pm Pe

• Back to Generator dynamics

Tm Te
Tm Te J d2?m/dt2
J d2dm /dt2 Tm Te Ta Ta accelerating
torque dd / dt ?m- ?msyn Using PT? and
dividing by Base MVA SB (?m/SB) J d2dm /dt2 Pm
- Pe Pa per unit
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Tm Te
(2?m/?msyn2) ((1/2 J?msyn2)/SB) J d2d /dt2 Pm -
Pe
(2?pu/?syn) ((1/2J?msyn2 )/SB) J d2d /dt2 Pm -
Pe Pa
?m/?msyn ?pu
Electrical
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Tm Te
(2?pu/?syn) ((1/2J?msyn2 )/SB) J d2d /dt2 Pm -
Pe Pa
H(1/2) Jmsyn2/SB KE at synchronous speed/ Base
MVA H is called machine inertia in seconds
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Finally, since ?syn 2pf ?pud? /dt ( pf/H
)(Pm Pe) dd / dt ? ?syn Equivalently,
?pud2 d / dt2 ( pf/H )( Pm Pe) dd / dt ?
?syn These are called the SWING
EQUATIONS Note ? and d are NOT in pu f is
electrical frequency in HZ
Tm Te
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Example ( text 13.1)
Tm Te
3 phase 60 Hz 500 MVA 15kV 32 Pole H2 S
Initially PmPe1 pu d 10 deg
Find electrical and mechanical synchronous speed
swing equation A fault reduces Pe to
zero find ? and d at 3 cycles
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Example ( text 13.1)
Tm Te
3 phase 60 Hz 500 MVA 15kV 32 Pole H2 S
Initially PmPe1 pu d 10 deg
?syn 377 el. Rad/s always for f60 Hz ?msyn
(1/P) ?syn 23.56 rad/s
Swing equation ?pud2 d / dt2 94.25( Pm Pe)
dd / dt ? 377
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Example ( text 13.1)
Tm Te
A fault reduces Pe to zero find ? and d at 3
cycles(.05s)
Assume ?pu1. Since Pe0 and Pm1
d? / dt 94.25 rad/s2 ?(0)377 rad/s ?
94.25 t 377 ?(0.05) 379.356 dd / dt ?
377 94.25t d(0)10deg d(t) 47.125 t2
do d(.05) 16.75 deg
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First swing stability-Dynamics
Pm Pe

• Generator dynamics

Example ( text 13.1)
Tm Te
Summary Speed increases at a rate dictated by
power imbalance and inertia Angle
increases depending on difference between machine
speed and synchronous speed
?(0)377 rad/s ? 94.25 t 377 ?(0.05)
379.356 d(0)10deg d 47.125 t2 do d(.05)
16.75 deg
?
377
d
t
3 .05 S
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Power System Stability

• Lecture 2 Mechanical Aspects and Swing Equation

Pm Pe
Large System Infinite Bus Voltage and frequency
constant
V

• Can generator get back to constant --synchronous
  speed?
• How does speed change in a disturbance
• Mechanical system model
• SWING EQUATIONDYNAMICS
• How do speed changes ( angle changes) affect
  power flow?
• Machine model
• Power angle equation
• Given these models can energy be transferred to
  bring speeds back?

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First swing stability-Swing Equation
SUMMARY Lecture 1

• A generator connected to an infinite bus through
  a line
• Initially PmPe

Pm Pe
Pm

jXL
jXd
(Infinite Bus)
Pe
V/0
E/d
Speed change is governed by the Swing Equation
d2d/dt2 (pf/H) (Pm-Pe)
dd /dt ?-?syn
Now need to know how Pe changes during a
disturbance
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Power System Stability

• Manifestations Angle Stability

P
KE Builds up
P
Excess KE Needs to be removed
Can generator get back to constant --synchronous
speed? Only if it can get rid of excess
KE Excess KE needs to go into the infinite bus
through the line? Will it? What happens if it
cant? Stability means returning to synchronous
speed In a multi-machine system it means
settling at a common speed
SPEED
Line Real Power P
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First swing stability-Swing Equation
SUMMARY Lecture 1

• A generator connected to an infinite bus through
  a line
• Initially PmPe

Pm Pe
Pm

jXL
jXd
(Infinite Bus)
Pe
V/0
E/d
Transient Stability means that after a
disturbance the generator will return to Steady
state operation at synchronous speed
Speed change is governed by the Swing Equation
d2d/dt2 (pf/H) (Pm-Pe)
dd /dt ?-?syn
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First swing stability-Swing Equation
SUMMARY Lecture 1

• A generator connected to an infinite bus through
  a line
• Initially PmPe

Pm Pe
Pm

jXL
jXd
(Infinite Bus)
Pe
V/0
E/d
Speed change is governed by the Swing Equation
d2d/dt2 (pf/H) (Pm-Pe)
dd /dt ?-?syn
Now need to know how Pe changes during a
disturbance Need Generator and Network Model
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Power System Stability
First swing stability-Generator Electrical

• Classical model
• Per phase model
• The generator reactance X Xd Transient
  Reactance
• E is the phasor induced voltage
• Magnitude of E freezes at the value
• just prior to disturbance

I
Vt/0
E/d
E/d Vtpf/0 Ipf (RajXd)
Vtpf , Itpf are terminal voltage and current
before the fault
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• System is Modeled with SLOWLY VARYING PHASORS

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• SLOWLY VARYING PHASOR
• Original amplitude and phase modulated wave

Write as a cosine at ?o and a tome varying phase
?(t)
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• SLOWLY VARYING PHASOR

Slowly Varying Magnitude Slowly Varying Phase
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• SLOWLY VARYING PHASOR

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Power System Stability
First swing stability-Power Transmission How
does power change with speed and rotor angle?

• The N- node network can be represented by the
  admittance matrix equation
• I Y V
• The current injected into node m
• Im ?Nm1 Ymn Vn
• The Complex power into node m is
• SmVm (?Nm1 Ymn Vn)

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Power System Stability
First swing stability-Power Transmission

• Let VmVm/?m VnVn/?n YmnYmn/?mn
• Starting with
• SmVm (?Nm1 Ymn Vn)
• PmRe(Sm)Vm ?nm1 Ymn Vn cos (?m - ?n -
  ?mn )
• QmIm(Sm)Vm ?nm1 Ymn Vn sin (?m - ?n -
  ?mn )

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Power System Stability
First swing stability-Power Transmission

• Real Power PmRe(Sm)Vm ?Nm1 Ymn
  Vn cos (?m - ?n - ?mn )
• Real power flow is controlled by voltage phase
  angles ?m, ?n
• For a line ( or generator with resistance and
  capacitance neglected, i.e., pure inductive
  reactance
• Vm/?m Pmn Vn/?n

jX


Y
1/jX
-1/jX
1/jX
-1/jX
Pmn Vm Vn sin (?m - ?n ) /X
Real power goes from higher phase angle to lower
phase angle To make power go down a line need to
make sending end phase angle larger
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First swing stabilityPower Angle Equation

Pm Pe

• A generator connected to an infinite bus through
  a line
• Initially PmPe

Pm
jXL
jXd
(Infinite Bus)


Pe
E/d
V/0
If d changes Pe changes lt Pe E V sin
(d) /(XXL) If Pm or Pe changes speed
changes ltPm Pe (H/ pf) d? /dt
How are the two related?
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First swing stabilityPower Angle Equation
Pm Pe

• A generator connected to an infinite bus through
  a line
• Initially PmPe Suppose we increase Pm then
  speed increases

jXL
jXd
Pm

Fixed (Infinite Bus)
Pe
V/0
E/d
If the speed of the generator increases the phase
angle d of induced voltage increases
Inf. Bus 60
Gen 60.6
d
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First swing stability
Pm Pe

• A generator connected to an infinite bus through
  a line
• Initially PmPe Suppose we increase Pm then
  speed increases

If the speed of the generator increases the phase
angle d of induced voltage increases
Inf. Bus 60
Gen 60.6
d
d can be constant only if the frequencies are
identical -- Synchronous
d d/dt ?-?syn
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Power System Stability
First swing stability

• Lecture 3
• Qualitative Analysis
• Examples of developing Swing Equations and
  Power Angle Curves
• Equal Area Criterion of Stability

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• Angle Stability Revisited-
• Machine Connected to Infinite Bus

8
8
P
KE Builds up
P
Excess KE Needs to be removed
Fault makes generator electric Power
zero Generator accelerates. Can generator get
back to constant --synchronous speed? Only if it
can get rid of excess KE Excess KE needs to go
into the infinite bus through the line? Will it?
What happens if it cant? Stability means
returning to synchronous speed
SPEED
Line Real Power P
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Power System Stability

• First swing stability-Background needed

P
KE Builds up
P
Excess KE Needs to be removed

• Power flow characteristics in
• Network
• What governs power flow in a line?
• Dynamics of Turbine generator
• How does a generator change speed?
• How does generator dynamics affect
• Power flow?

57
Power System Stability

• Modeling Assumptions

P
KE Builds up
P
Excess KE Needs to be removed
Mechanical power is constant Rotor speed changes
SLOWLY as compared to 60 Hz Voltage and current
can be represented by slowly varying phasors
Real power can be calculated from phasor
models System remains balanced (Some Unbalanced
faults can be handled)
58
Power System Stability
First swing stability-Generator Electrical

• Classical model
• The generator reactance X Xd
• E is the phasor induced voltage
• It is assumed that the magnitude
• of E freezes at the value
• just prior to disturbance

I
Vt/0
E/d
E/d Vtpf/0 Ipf (RajXd)
Vtpf , Itpf are terminal voltage and current
before the fault
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First swing stability Lecture 2 Summary

• A generator connected to an infinite bus through
  a line
• Initially PmPe

Pm Pe
jXL
jXd
Stability is governed by the Swing Equation
d2d/dt2 (pf/H) (Pm-Pe)
Swing Equation Power Angle Equation
dd /dt ?-?syn
For our system Pe E V sin (d) /(XXL)
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First swing stability-Equal Area Criterion

Pm Pe

• A generator connected to an infinite bus through
  a line. Initially PmPe

Stability is governed by the Swing Equation
Swing Equation Power Angle Equation
d2d/dt2 (pf/H) (Pm-Pe)
dd /dt ?-?syn
Pe E V sin (d) /(XXL)
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
At D ? is decreasing but gt 0 d increases further
say to point E By now suppose ? is back to zero
and decreasing Thus ? becomes lt 0 as the
generator continues to slow Since ?lt0 d
decreases towards B First swing stable!
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
P
Pe
E
C
D
Pm1
B
A
Pm
d
do
d1
First swing Stable
?-?syn
0
d
d1
d0
Time
0
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First swing stability-Equal Area
Criterion Stable Equilibriumsmall increase in
mechanical power
P
Pe
E
C
D
Pm1
B
A
Pm
d
do
d1
?-?syn
0
d
d1
Equal Speeds at Points A and E
d0
Time
0
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First swing stability-Equal Area CriterionBasic
Principle

• Two points on the trajectory
• A speed?1 angle d1 time t1
• B speed?1 angle d2 time t2gtt1
• d/dt(dd/dt)2 (dd/dt)(d2d/dt2)

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First swing stability-Equal Area CriterionBasic
Principle
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First swing stability-Equal Area CriterionBasic
Principle
If ?(t1 ) ?(t2)
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First swing stability-Equal Area CriterionBasic
Principle
Relates to changes in KE If change in KE can be
recovered 1 machine-infinite bus system will be
stable
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First swing stability-Equal Area
CriterionBasic Principle
P
Pe
E
C
Pm1
A
Pm
d
do
dx
d2
dx
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First swing stability-Equal Area
CriterionApplication

• Establish initial conditions
• Define sequence of events and network for each
  event
• Develop Power angle curves
• Apply EAC

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First swing stability-Equal Area Criterion
Example 1
Stability under small change in mechanical power
Xd0.2 XL0.2 pu.
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First swing stability-Equal Area Criterion
Example 1
Stability under small change in mechanical power
A 10 MVA, 0.8 pf lagging, 4160 V, 60Hz,
three-phase generator supplies 50 rated power
at .8 pf lagging to a 4160 V infinite bus.
Determine if the generator is first-swing stable
if the prime mover power is increased by 10
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Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
1. Initial
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First swing stability-Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
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First swing stability-Equal Area Criterion
Example 1 Stability under small change in
mechanical power
Xd0.2 XL0.2 pu.
2. Power angle equation before and after change
in Pm Pe Pmax Sin(d) Pmax EV8 /(XdXL)
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First swing stability-Equal Area Criterion Small
change in mechanical power
P
Pe
E
C
Pm1
A
Pm
d
do
dx
d2
dx
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Equal Area Criterion-Small change in mechanical
power
EAC
Remember
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Equal Area CriterionExample 2-Fault
2
1
8
3
The infinite bus receives 1 pu real power at 0.95
power factor lagging
A fault at bus 3 is cleared by opening lines from
1-3 and 2-3 when the generator power angle
dReaches 40 deg. Is the system first swing
stable?
78
Equal Area CriterionExample 2-Fault
Example 2
2
1
8
3
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Equal Area CriterionExample 2-Fault
Example 2
Data (Resistances are zero)
Xd0.3 Ztj0.1 Z1Z3j0.2 Z2j0.1 V81 pu
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Equal Area CriterionExample 2-Fault
Example 2
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
1. Initial
S(1/0.95)/acos(0.95) I(S/V8)
1.05/-18.2o Xeq (XdXt)X1(X2X3)0.52 pu
E/d V8 jXeq I 1.28/23.95o
81
Equal Area CriterionExample 2-Fault

• 2. Events
• Prefault Steady State
• Fault three phase fault at bus 3
• Post fault Line 2-3 1-3 open

82
Equal Area CriterionExample 2-Fault
3a. Pre-fault Power Angle curve
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
Xeq (XdXt)X1(X2X3)0.52 pu E/d V8
jXeq I 1.28/23.95o V81/0 Pe E V8 sin d /Xeq
2.46 sin d Note d 23.95o Pe1
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Equal Area CriterionExample 2-Fault
3b. During-fault Power Angle curve
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
Z3
Z2
3
1
Zt
Zth
jXd
Vth0.33 V 8 Zth j 0.066
Vth -
E -
Vt -
I
84
Equal Area CriterionExample 2-Fault
3c. During-fault Power Angle curve
Xeq0.466
Pe E V8 sin d /Xeq 0.915 sin d
1
Zt
Zth
jXd
Vth0.33 V 8 Zth j 0.066 Xd 0.2 Zt0.2
Vth -
E -
Vt -
I
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Equal Area CriterionExample 2-Fault
3c. Post Fault Line 1-3 2-3 out
I
1
2
jXd
Z1
Zt
V8 -
S
E -
Vt -
I
3
Xeq .2.2.2
Pe E V8 sin d /Xeq 2.14 sin d
86
Equal Area CriterionExample 2-Fault
Deceleration
4. Trajectories and ares
P
Pe
Acceleration
Pm
d
dm
do
dcl
87
Equal Area CriterionExample 2-Fault
Apply EAC
P
Pe
Pm
d
dm
do
dcl
88
Equal Area CriterionExample 2-Fault
Solve by trial and error Want Error 0
89
Equal Area CriterionExample 2-Fault
Rotor swings to 55 degrees then swings back-
STABLE
P
Pe
Pm
d
do
dcl
dm
55 40 24
23.95 40 55
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Equal Area CriterionExample 2-Fault
Basic Question How Quickly can the fault be
cleared Relay Time Breaker Operation Arc
Extinction If breaker fails then Backup
operates gt10 Leads to notion of critical
clearing time more the better
3-5
91
Equal Area CriterionExample 2-Fault with delayed
clearing
Fault cleared when rotor angle is 120 deg no
solution
92
Equal Area CriterionExample 2-Fault
Rotor swings past 156 degrees UN STABLE
P
Pe
Pm
d
do
dcl
dm
156 120 24
23.95 40 55
93
Equal Area CriterionExample 2-Fault-Critical
clearing angle
For stability dm lt p- d1 solve for dcl
94
Equal Area CriterionExample 2-Fault- Critical
Clearing
Rotor swings past 156 degrees UN STABLE
P
Pe
Pm
d
d0 d1
dmp- d1
dcl112.9
23.95
156 112 24
95
Summary Equal Area Criterion

• For simple systems
• 1 machine infinite bus
• 2 machines
• Provides convenient way to determine first swing
  stability
• For 2 points on the P-d curve that have equal
  speed
• Accelerating are Decelerating area

96
Summary Equal Area Criterion

• 1 machine to infinite bus
• Machine must return to synchronous speed
• EAC indicates stable or unstable
• Introduced notion of critical clearing

97
Next

• Critical Clearing Angle v. Critical Clearing
  Time See example 13.8
• Numerical Solution
• Multimachine System