ECE 692 Advanced Topics on Power System Stability
4 – Transient Stability
Spring 2016
Instructor: Kai Sun
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Content
•Overview
•Transient stability analysis on an SMIB system
•Direct methods
•Time‐domain simulation
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Transient Stability
• The ability of the power system to maintain synchronism when subjected to a severe disturbance such as a fault on transmission facilities, loss of generation or loss of a large load.
– The system response to such disturbances involves large excursions of generator rotor angles, power flows, bus voltages, and other system variables.
– Stability is influenced by the nonlinear characteristics of the system
– If the resulting angular separation between the machines in the system remains within certain bounds, the system maintains synchronism.
– If loss of synchronism due to transient instability occurs, it will usually be evident within 2‐3 seconds of the initial disturbances
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Methods for Transient Stability Analysis
Analyzing a dynamic system’s transient stability following a given
disturbance
• Time-domain simulation solves the nonlinear differentialalgebraic equations (DAEs) with known initial values x=x0 and
t=t0 by using step-by-step numerical integration techniques (either
explicit or implicit).
x = f(x,u)
0 = g(x,u)
DE
AE
• Direct methods are based on Lyapunov’s second method and they
determine stability without explicitly solving the DAEs:
1. Define a Transient Energy Function (TEF) as a possible
Lyapunov function
2. Compare the TEF to a critical energy, denoted by Vcr, to judge
stability
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Direct Methods for Transient Stability Analysis
Lyapunov’s second method:
x = f(x, u)
(1)
Its equilibrium xs, which satisfies f(xs, u)=0, is stable if there
exists a positive definite function V(x), called a Lyapunov
function, such that its total derivative dV(x)/dt with respect to (1)
is not positive, i.e.
• V(x)0, with equality if and only if x= xs
• dV(x)/dt0
Its equilibrium xs is asymptotically stable if dV(x)/dt is negative
definite, i.e.
• dV(x)/dt0, with equality if and only if x= xs
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Transient Energy Function (TEF) Approach
• The TEF based methods are a special case of the more general Lyapunov’s 2nd
method. The TEF is one possible Lyapunov function.
• Consider a rolling ball analogy: two quantities are required to determine
whether the “ball” (state) will go outside the “bowl” (stability region)
– The initial kinetic energy of the ball
– The height of the rim at the crossing point (depending on the direction of
the initial motion)
Rim: Potential energy
boundary surface (PEBS)
(stable equilibrium point)
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Application to Power Systems
• Initially the system is operating at an SEP xs.
• If a fault occurs, the system gains kinetic energy VKE and potential energy VPE
during the fault-on period and hence moves away from xs .
• After fault clearing, VKE is converted into VPE.
• To avoid instability, the system must be capable of absorbing VKE at a time when
the system moves toward the old SEP xs or a new SEP.
• For a given post-disturbance network configuration, there is a maximum or critical
amount of transient energy Vcr that the system can absorb without loss of
synchronism
• Therefore, assessment of transient stability for the state of the system at fault
clearing (xcl) requires:
1. Estimate TEF V(xcl) that adequately describes the transient energy
responsible for separating one or more generators from the rest of the system
2. Estimate the critical energy Vcr.
3. Calculate the transient energy margin Vcr V(xcl)
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4.2 Transient Stability Analysis on an SMIB System
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SMIB System
D
2 H d 2d
= Pm - Pe = Pa
2
w0 dt
Pa is called Accelerating Power
E ¢EB
Pe = Pmax sin d =
sin d
XT
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Response to a step change in Pm
2 H d 2d
= Pa = Pm - Pe = Pm - Pmax sin d
w0 dt 2
Consider a sudden increase in Pm: Pm0 Pm1.
New SEP b satisfying Pe(1)=Pm1
• a b: Due to the rotor’s inertia,  cannot jump
from 0 to 1, so Pa=Pm1-Pe(0)>0 and r
increases from 0. When b is reached, Pa=0
but r >0, so  continues to increase.
• b c: >1 and Pa<0, so r decreases until c.
At c, r=0 and  reaches the peak value max.
Pa r
a >0
b 0
=0, 
=r, max
c <0 =0 , 



=max
•  c: At c, the rotor starts to decelerate (since
Pa<0) with r<0 and  decreases.
• With all resistances (damping) neglected,  and
r oscillate around new SEP b with a constant
amplitude.
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Equal‐Area Criterion (EAC)
• Neglect all losses:
r=d/dt=0 at max if it exists
D w r2 =
w0
H
0=
ò
( Pm - Pe )d d
æ Dw
1
(2 H w 0 )⋅ ççç r
2
è w0
ö÷
÷÷ =
÷ø
ò
d max
d0
( Pm - Pe )d d
d1
2
ò
( Pm - Pe )d d
= ò ( Pm - Pe )d d + ò
d0
dmax
d1
( Pm - Pe )d d
=| A1 | - | A2 |= 0
Moment of inertia in p.u.
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• Equal-Area Criterion (EAC): The stability is maintained only if a
decelerating area |A2|  the accelerating area |A1| can be located
above Pm (from b to d, i.e. the Unstable Equilibrium Point or UEP).
SEP
UEP
• If |A2|<|A1|,  will continue increasing at UEP (since r>0) and lose
stability.
• For the case with a step change in Pm, the new Pm does matter for
transient stability.
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Transient stability limit for a step change of Pm
Following a step change Pm0Pm, solve the transient stability limit of Pm:
• Assume | A1 |=| A2 | in order to solve the limit of Pm
d1
Pm (d1 - d0 ) - ò Pmax sin dd d = ò
d0
dmax
d1
Pmax sin dd d -Pm (dmax - d1 )
-Pmd0 + Pmax (cos d1 - cos d0 ) = -Pmax (cos dmax - cos d1 ) - Pmdmax
(dmax - d0 ) Pm = Pmax (cos d0 - cos dmax )
• At the UEP,
Pm = Pmax sin dmax
(dmax - d0 ) sin dmax + cos dmax = cos d0
SEP
UEP
• Solve max to calculate the transient
stability limit for a step change of Pm:
Pm = Pmax sin dmax
d1 = p - dmax
Questions: Can we increase Pm beyond that
transient limit? If yes, how much further can
we increase Pm?
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Solve max by the Newton‐Raphson method
(dmax - d0 )sin dmax + cos dmax = cos d0
• The nonlinear function form:
f (dmax ) = cos d0 = c
(k )
p / 2 < dmax
<p
• Select an initial estimate:
• Calculate iterative solutions by the N-R algorithm:
( k +1)
max
d
=d
(k )
max
+ Dd
(k )
max
where Dd
(k )
max
=
(k )
c - f (dmax
)
df
d dmax
(k )
c - f (dmax
)
= (k )
(k )
(dmax - d0 ) cos dmax
(k )
dmax
• Give a solution when a specific accuracy  is reached, i.e.
( k +1)
(k )
dmax
- dmax
£e
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Response to a three‐phase fault
E ¢EB
sin d
Pe = Pmax sin d =
XT
• Pe,during fault << Pe, post-fault
• Pe,post-fault <Pe,pre-fault for a permanent fault (cleared by tripping
the fault circuit)
• Pe, post-fault =Pe,pre-fault for a temporary fault
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Stable
Unstable
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Critical Clearing Angle (CCA)
• Consider a simple case
– A three-phase fault at the sending end
– Pe, during fault=0 if all resistances are
neglected
– Critical Clearing Angle c
ò
dc
d0
Pm d d = ò
dmax
dc
( Pmax sin d - Pm )d d
|A1|
|A2|
Integrating both sides:
Pm (dc - d0 ) = Pmax (cos dc - cos dmax ) - Pm (dmax - dc )
cos dc =
Pm
(dmax - d0 ) + cos dmax
Pmax
=
Pm
(p - 2d0 ) - cos d0
Pmax
= -0
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Critical Clearing Time (CCT)
• Solve the CCT from the CCA:
Since Pe, during fault=0 for this case, during the fault:
2 H d 2d
= Pm
2
w0 dt
t
dd
w0
w
=
Pm ò dt = 0 Pmt
0
dt
2H
2H
d=
tc =
w0
Pmt 2 + d0
4H
4 H (dc - d0 )
w0 Pm
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• For a more general case: Pe (during fault)>0
(pre-fault)
P3max
(post-fault) P3max
A3
A2
(fault-on)  P2max
A1
P2max
s
dc
(u)
Pm (dc - d0 ) - ò P2 max sin dd d = ò
d0
cos dc =
dmax
dc
P3max sin d d d - Pm (dmax - dc )
Pm (dmax - d0 ) + P3max cos dmax - P2max cos d0
P3max - P2max
•
|A1| = VKE(c), i.e. the kinetic energy at c
•
|A1|+|A3| = V(c)=VKE(c)+VPE(c), total energy at c
•
|A2|+|A3| = VPE(u)=Vcr, i.e. the largest potential energy
•
If and only if V(c)Vcr (i.e. |A1||A2|), the generator is stable
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Factors influencing transient stability
• How heavily the generator is loaded. • The generator output during the fault. This depends on the fault location and type
• The fault‐clearing time
• The post‐fault transmission system reactance
• The generator reactance. A lower reactance increases peak power and reduces initial rotor angle.
• The generator inertia. The higher the inertia, the slower the rate of change in angle. This reduces the kinetic energy gained during fault; i.e. the accelerating area A1 is reduced.
• The generator internal voltage magnitude (E’). This depends on the field excitation
• The infinite bus voltage magnitude EB
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EAC for a Two‐Machine System
• Two interconnected machines respectively with H1 and H2
– The system can be reduced to an equivalent SMIB system w0
w0
d 2d1
=
(
P
P
)
=
Pa1
m1
e1
dt 2
2 H1
2 H1
d 2d12 d 2d1 d 2d2 w0 Pa1 Pa1
= 2 - 2 = ( - )
2
2 H1 H2
dt
dt
dt
d 2d 2
w0
w0
=
(
P
P
)
=
Pa 2
m2
e2
dt 2
2H 2
2H 2
H 2 Pa1 - H 1 Pa 2
H 2 Pm1 - H 1 Pm 2 H 2 Pe1 - H 1 Pe 2
2 H 1 H 2 d 2 d12
=
=
2
w 0 H 1 + H 2 dt
H1 + H 2
H1 + H 2
H1 + H 2
Pe12
2 H12 d 2d12
= Pm ,12 - Pe,12
2
w0 dt
ò
d12,max
d12 ,0
w0
( Pm,12 - Pe,12 )d d = 0
H12
Pm12
12,0
12,max
12
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An Example with Multiple Faults
Pe
Pe,0
Pe,1
Pe,3=Pe,2
Pm

d1 d2 c1 c2 d3
c3
max
(0)
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