ECE 422/522
Power System Operations &
Planning/Power Systems Analysis II :
7 - Transient Stability
Spring 2014
Instructor: Kai Sun
1
Transient Stability
• The ability of the power system to maintain synchronism
when subjected to a severe transient disturbance such as
a fault on transmission facilities, loss 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 because of transient instability occurs, it will
usually be evident within 2-3 seconds of the initial disturbances
2
Single-Machine Infinite Bus (SMIB) System
Pe=
=Pmaxsinδ
Pmax= E’EB/XT
• Swing Equation:
2 H d 2
 Pm  Pe  Pm  Pmax sin 
2
0 dt
Pa
is called Accelerating Power
Pm= Pe
• The rotor will accelerate if Pm increases,
or Pe decreases due to, e.g., a
contingency on the transmission path
3
Power Angle Relationship
Pe=
=Pmaxsinδ
Question: What if both
circuits are out of service?
4
Response to a step change in Pm
2 H d 2
 Pa  Pm  Pe  Pm  Pmax sin 
0 dt 2
Consider a sudden increase in Pm: Pm0→ Pm1.
New equilibrium point b satisfying Pe(δ1)=Pm1
• a→ b: Due to the rotor’s inertia, δ cannot jump
from δ0 to δ1, so Pa=Pm1-Pe(δ0)>0 and ωr
increases from ω0. When b is reached, Pa=0 but
ωr >ω0, so δ continues to increase.
• b→ c: δ>δ1 and Pa<0, so ωr decreases until c.
At c, ωr=ω0 and δ reaches the peak value δmax.
Pa ωr
a >0
=ω0, ↑
b 0
=ωr, max,↑
c <0
=ω0 , ↓
δ
↑
↑
=δmax, ↓
• ← c: At c, the rotor starts to decelerate (since
Pa<0) with ωr<ω0 and δ decreases.
• With all resistances (damping) neglected, δ and
ωr oscillate around the new equilibrium point b
with a constant amplitude.
5
Equal-Area Criterion (EAC)
r2 
0
H

( Pm  Pe )d 
2 H 0
r2   ( Pm  Pe )d 
2
20
If δmax exists where dδ/dt=0:
1  r 

J
2  0 
2

max
0
( Pm  Pe )d 
At δmax, ∆ωr=0 and
the integral=0
 max
1
Moment of inertia in p.u.
(Kundur’s Page 131)
  ( Pm  Pe )d   
0
max
1
( Pe  Pm )d 
= |area A1 | |area A2 | 0
(Note: all losses are neglected)
6
• Equal-Area Criterion (EAC): The stability is maintained only if a decelerating area
|A2| ≥ the accelerating area |A1| can be located above Pm (before d, i.e. the
Unstable Equilibrium Point or UEP).
UEP
SEP
• If |A2|<|A1|, δ will continue increasing at UEP (since ωr>ω0), so it will lose stability.
• For the case with a step change in Pm, ∆Pm or the new Pm does matter for
transient stability.
7
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 |:
1
Pm (1  0 )   Pmax sin  d   
0
max
1
Pmax sin  d  Pm (max  1 )
Pm0  Pmax (cos 1  cos 0 )  Pmax (cos max  cos 1 )  Pmmax
(max  0 ) Pm  Pmax (cos 0  cos max )
• At the UEP, Pm  Pmax sin max
SEP
UEP
(max  0 )sin max  cos max  cos 0
• Solve δmax to calculate the transient
stability limit for a step change of Pm:
Pm  Pmax sin max
1    max
Question: Can we increase Pm beyond that transient limit (with Pm<Pmax)?
8
Solve δmax by the Newton-Raphson method
(max  0 )sin max  cos max  cos 0
• The nonlinear function form:
f (max )  cos 0  c
(k )
 / 2  max

• Select an initial estimate:
• Calculate iterative solutions by the N-R algorithm:
( k 1)
max


(k )
max
 
(k )
max
where 
(k )
max

(k )
c  f (max
)
df
d max
(k )
c  f (max
)
 (k )
(k )
(max  0 ) cos max
(k )
max
• Give a solution when a specific accuracy ε is reached, i.e.
( k 1)
(k )
max
 max

9
Response to a three-phase fault
Pe=
=Pmaxsinδ
• Pe,during fault << Pe, post-fault
• Pe,post-fault <Pe,pre-fault for a permanent fault (cleared by tripping the fault circuit)
or Pe, post-fault =Pe,pre-fault for a temporary fault
10
Unstable
Stable
11
Critical Clearing Angle (CCA)
• Saadat’s Sec.11.6 (Example 11.5)
• Consider a simple case
– A three-phase fault at the sending end
– Pe, during fault=0 if all resistances are
neglected
– Critical Clearing Angle δc

c
0
Pm d   
max
c
( Pmax sin   Pm )d 
|A1|
|A2|
Integrating both sides:
Pm c  0   Pmax (cos c  cos max )  Pm (max  c )
cos c 

Pm
(max  c )  cos max
Pmax
Pm
(   0  c )  cos 0
Pmax
= π-δ0
12
Critical Clearing Time (CCT)
• Solve the CCT from the CCA:
Since Pe, during fault=0 for this case, during the fault:
2 H d 2
 Pm
2
0 dt
t
d
0


Pm  dt  0 Pmt
0
dt
2H
2H

tc 
0
Pmt 2  0
4H
4 H (c  0 )
0 Pm
13
• For a more general case: Pe (during fault)>0
(pre-fault)
P3max
(post-fault) ≤P3max
A3
A2
(fault-on) ≤ P2max
A1
P2max
δs
c
(δu)
Pm c  0    P2 max sin d   
0
cos c 
max
c
P3max sin  d   Pm (max  c )
Pm (max  c )  P3max cos max  P2 max cos 0
P3max  P2 max
•
|A1| = Vke(δc), the kinetic energy at δc
•
|A1|+|A3| = V(δc)=Vke(δc)+Vpe(δc), total energy at δc
•
|A2|+|A3| = Vpe(δu)=Vcr, i.e. the largest potential energy
•
If and only if V(δc)≤Vcr (i.e. |A1|≤|A2|), the generator is stable
14
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
15
EAC for a Two-Machine System
• Two interconnected machines respectively with H1 and H2
– The system can be reduced to an equivalent SMIB system
d 21
0
0
(
)
P
P
Pa1



m1
e1
2 H1
2 H1
dt 2
d 212 d 21 d 22 0 Pa1 Pa1
 2  2  (  )
2
2 H1 H 2
dt
dt
dt
d 2 2
0
0
(
)

P

P

Pa 2
m2
e2
2H 2
2H 2
dt 2
H 2 Pa1  H1Pa 2 H 2 Pm1  H1Pm 2 H 2 Pe1  H1Pe 2
2 H1H 2 d 212


=
2
0 H1  H 2 dt
H1  H 2
H1  H 2
H1  H 2
Pe12
2 H12 d 212
 Pm ,12  Pe ,12
2
0 dt

12,max
12,0
0
( Pm,12  Pe ,12 )d   0
H12
Pm12
16
δ12,0
δ12,max
δ12
Single Machine Equivalent Method (EEAC or SIME) [1]-[3]
• Based on generator trajectories obtained from time-domain simulations, split all
generators into two groups (fixed or dynamic) and build a 2-machine equivalent
and consequently a single machine equivalent, such that EAC can be applied.
1. Y. Xue, et al, "A Simple Direct Method for Fast Transient Stability Assessment of Large Power Systems". IEEE Trans. on PWRS,
PWRS3: 400–412, 1988.
2. Y. Xue, et al, "Extended Equal-Area Criterion Revisited". IEEE Trans. on PWRS, PWRS7: 10101022,1992.
3. M. Pavella, et al, “Transient Stability of Power Systems: a Unified Approach to Assessment and Control”, Kluwer, 2000.
17
Some examples from [3]
18
Methods for Transient Stability Analysis
Analyzing a system’s transient stability following a given
contingency
• Time-domain simulation:
– At present, the most practical available method of transient stability
analysis is time-domain simulation in which the nonlinear differential
equations are solved by using step-by-step numerical integration
techniques.
• Direct methods:
– Those methods determine stability without explicitly solving the
system differential equations. The methods are based on Lyapunov’s
second method, define a Transient Energy Function (TEF) as a
possible Lyapunov function, and compare the TEF to a critical
energy Vcr to judge stability
– EAC is a direct method for a SMIB or two-machine system
19
Numerical Integration Methods
• The differential equations to be solved in power system stability analysis are
nonlinear ODEs (ordinary differential equations) with known initial values
x=x0 and t=t0
dx
= f ( x, t )
dt
where x is the state vector of n dependent variables and t is the independent
variable (time). Our objective is to solve x as a function of t
• Explicit Methods
– In these methods, the value of x at any value of t is computed from the
knowledge of the values of x from only the previous time steps, e.g. Euler
method and R-K methods
• Implicit Methods
– These methods use interpolation functions (involving future time steps)
for the expression under the integral, e.g. the Trapezoidal Rule
20
dx
= f ( x, t )
dt
Euler Method
• The Euler method is equivalent to using the first two
terms of the Taylor series expansion for x around the
point (x0, t0), referred to as a first-order method (error is
on the order of ∆t2)
– Approximate the curve at x=x0 and t=t0 by its tangent
x(t1)
x1
x0
dx
dt
∆t
= f ( x0 , t0 )
x0
t0
∆x ≈
dx
dt
∆t
x1= x0 + ∆x= x0 +
x0
dx
dt
∆t
x0
– At step i+1
xi +1 =
xi +
dx
∆t
dt xi
• The standard Euler method results in inaccuracies
because it uses the derivative only at the beginning of the
interval as though it applied throughout the interval
21
t1
dx
= f ( x, t )
dt
Modified Euler (ME) Method
• Modified Euler method consists of two steps:
(a) Predictor step:
dx
x1p =
x0 +
dt
x (t1)
x1c
x1p
∆t
x0
∆t
x0
Slope at the beginning of ∆t
t0
t1
The derivative at the end of the ∆t is estimated using x1p
dx
dt
xp
= f ( x1p , t1 )
1
Estimated slope at the end of ∆t
dx
dx
dx
dx
+
+
dt xi dt x p
dt
dt x p
x0
c
i +1
c
1
∆t
x i +1 =
xi +
x1 =
x0 +
∆t
2
2
• It is a second-order method (error is on the order of ∆t3)
• Step size ∆t must be small enough to obtain a reasonably accurate solution, but at the
same time, large enough to avoid the numerical instability with the computer, e.g.
increasing round-off errors.
(b) Corrector step:
22
dx
= f ( x, t )
dt
Runge-Kutta (R-K) Methods
• General formula of the 2nd order R-K method:
(error is on the order of ∆t3)
=
k1 f ( x0 , t0 ) ∆t
k2
x(t1)
x0+αk1
f ( x0 + α k1 , t0 + β∆t ) ∆t
x0
∆t
x1 =x0 + a1k1 + a2k2
At Step i+1:
t0
=
k 1 f ( xi , ti ) ∆t
The ME method is a special case
with a1=a2=1/2, α=β=1
k=
f ( xi + α k1 , ti + β∆t ) ∆t
2
xi +1 =xi + a1k1 + a2k2
• General formula of the 4th order R-K method:
(error is on the order of ∆t5)
1
xi +1 =xi + ( k1 + 2k2 + 2k3 + k4 )
6
=
k1 f ( xi , ti ) ∆t
k2 = f ( xi +
k1
∆t
, ti + ) ∆t
2
2
k2
∆t
, ti + ) ∆t
2
2
=
k4 f ( xi + k3 , ti + ∆t ) ∆t
k3 = f ( xi +
23
t1
Numerical Stability of Explicit Integration Methods
• Numerical stability is related to the stiffness of the set of differential equations
representing the system
• The stiffness is measured by the ratio of the largest to smallest time constant,
or more precisely by |λmax/λmin| of the linearized system.
• Stiffness in a transient stability simulation increases with modeling more
details (more small time constants are concerned).
• Explicit integration methods have weak stability numerically; with stiff
systems, the solution “blows up” unless a small step size is used. Even after the
fast modes die out, small time steps continue to be required to maintain
numerical stability
24
Implicit Methods
• Implicit methods use interpolation functions for the expression under the
integral. “Interpolation” implies the function must pass through the yet
unknown points at t1
• The simplest implicit integration method is the Trapezoidal Rule method.
It uses linear interpolation.
• The stiffness of the system being analyzed affects accuracy but not
numerical stability. With larger time steps, high frequency modes and fast
transients are filtered out, and the solutions for the slower modes is still
accurate. For example, for the Trapezoidal rule, only dynamic modes
faster than f(xn,tn) and f(xn+1,tn+1) are neglected.
Δt
 f ( x0 ,t0 ) + f ( x1 ,t1 )
2
Δt
xn+1 = xn +  f ( xn ,tn ) + f ( xn+1 ,tn+1 )
2
x1 = x0 +
x(t1)=x(t0)+|A|+|B|
f
B
f(x1,t1)
f(x0,t0)
Compared to ME method:
A
x n +=
xn +
1
∆t
t0
t1
25
∆t
 f ( xn ,tn ) + f ( xnp+1 ,tn+1 )

2 
Kundur’s Example 13.1
26
27
28
Overall System Equations
• The overall system equations are expressed in the general form
comprising a set of first-order differential equations (dynamic
devices) and a set of algebraic equations (devices and network)
where
x
V
I
YN
x = f(x, V)
I(x, V) = YN V
DE
AE
state vector of the system
bus voltage vector
current injection vector
node admittance matrix. It is constant except for changes introduced by
network-switching operations; symmetrical except for dissymmetry
introduced by phase-shifting transformers (Kundur’s Sec. 6.2.3)
29
Solution of the Equations
• Schemes for the solution of equations DE and AE are
characterized by the following factors
– The manner of interface between the DE and AE. Either a
partitioned approach or a simultaneous approach may be used
– The integration method, i.e. an implicit method or explicit method,
used to solve the DE.
– The method used to solve the AE (power flow analysis), e.g. the
Newton-Raphson method.
• Most commercialized power system simulation programs
provide the Modified Euler, 2nd order R-K, 4th order R-K
and Trapezoidal Rule methods
30
A Simplified Model for Multi-Machine Systems
• Consider these classic simplifying assumptions:
– Each synchronous machine is represented by a voltage source E’ with
constant magnitude |E’| behind X’d (neglecting armature resistances, the effect
of saliency and the changes in flux linkages)
– The mechanical rotor angle of each machine coincides with the angle of E’
– The governor’s actions are neglected and the input powers Pmi are assumed to
remain constant during the entire period of simulation
– Using the pre-fault bus voltages, all loads are converted to equivalent
admittances to ground. Those admittances are assumed to remain constant
(constant impedance load models)
– Damping or asynchronous powers are ignored.
– Machines belonging to the same station swing together and are said to be
coherent. A group of coherent machines is represented by one equivalent
machine
31
• Solve the initial power flow and determine the initial bus voltage phasors Vi.
• Terminal currents Ii of m generators prior to disturbance are calculated by their
terminal voltages Vi and power outputs Si, and then used to calculate E’i
• All loads are converted to equivalent admittances:
• To include voltages behind X’di, add m internal generator buses to the n-bus
power system network to form a n+m bus network (ground as the reference for
voltages):
X’
E’1
E’2
E’m
d1
X’d2
X’dm
reduce
Ybus nxn Ybus
m×m
32
• Node voltage equation with ground as reference
Ibus is the vector of the injected bus currents
Vbus is the vector of bus voltages measured from the reference node
Ybus is the bus admittance matrix :
Yii (diagonal element) is the sum of admittances connected to bus i
Yij (off-diagonal element) equals the negative of the admittance between
buses i and j
Compared to the Ybus for power flow analysis, additional m internal
generator nodes are added and Yii (i≤n) is modified to include the load
admittance at node i
33
• To simplify the analysis, all nodes other than the generator internal nodes are
eliminated as follows
I1
I2
where
Im
needs to be updated whenever the
network is changed.
where θij is the angle of Yij
2
ω0
34
Homework #6
• Saadat’s 11.14-11.17, due on 4/24 (Thur)
35