ECE 422 Power System Operations & Planning
7 ‐ Transient Stability
Spring 2015
Instructor: Kai Sun
1
References
•Saadat’s Chapter 11.5 ~11.10
•EPRI Tutorial’s Chapter 7
•Kundur’s Chapter 13
2
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 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
3
Single‐Machine Infinite Bus (SMIB) System
Pe=
=Pmaxsin
Pmax= E’EB/XT
• Swing Equation:
2 H d 2d
= Pm - Pe = Pm - Pmax sin d
2
w0 dt
Pa is called Accelerating Power
Pm= Pe
• The rotor will accelerate if Pm increases, or Pe decreases
4
Power Angle Relationship
Pe=
=Pmaxsin
Question: What if both
circuits are out of service?
5
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 equilibrium point 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
=0, 
b 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 the new equilibrium point b
with a constant amplitude.
6
Equal‐Area Criterion (EAC)
D w r2 =
2 H w0
D w r2 =
2
2w0
If max exists where d/dt=0:
ò
ò
( Pm - Pe )d d
2
1 æç Dwr ö÷
÷
Jç
2 çè w0 ÷÷ø
Moment of inertia
in p.u.
(Note: all losses are neglected)
w0
H
=ò
dmax
d0
( Pm - Pe )d d
( Pm - Pe )d d
At max, r=0 and
the integral=0
d =dmax
d1
= ò ( Pm - Pe )d d + ò
d0
= |area A1 |
dmax
d1
( Pm - Pe )d d
-|area A2 |= 0
7
• 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), so it will lose stability. • For the case with a step change in Pm, the new Pm does matter for transient stability. 8
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?
9
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
(k )
max
=d
(k )
max
+ Dd
(k )
max
where Dd
=
(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
10
Response to a three‐phase fault
Pe=
=Pmaxsin
• 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
11
Stable
Unstable
12
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 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 - dc ) + cos dmax
Pmax
Pm
(p - d0 - dc ) - cos d0
Pmax
= -0
13
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
14
• 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 - dc ) + P3max cos dmax - P2 max cos d0
P3max - P2 max
•
|A1| = Vke(c), 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
15
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
16
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
17
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, denoted by Vcr, to judge stability
– EAC is a direct method for a SMIB or two‐machine system
18
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
19
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
dx
 f ( x, t )
dt
x(t1)
x1
x0
dx
dt
t
 f ( x0 , t0 )
x0
t0
x 
dx
dt
t
x1  x0  x  x0 
x0
dx
dt
t1
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
20
Modified Euler (ME) Method
dx
 f ( x, t )
dt
• Modified Euler method consists of two steps:
(a) Predictor step:
x1p  x0 
dx
dt
x (t1)
x1c
t
x1p
x0
x0
Slope at the beginning of t
t
The derivative at the end of the t is estimated using x1p
dx
dt
xp
t0
t1
 f ( x1p , t1 )
1
Estimated slope at the end of t
(b) Corrector step:
x1  x0 
c
dx
dt
x0
dx

dt
xp
1
t
x i 1  xi 
c
dx
dt

xi
2
dx
dt
xp
i 1
t
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.
21
Runge‐Kutta (R‐K) Methods
dx
 f ( x, t )
dt
• General formula of the 2nd order R-K method:
(error is on the order of t3)
k1  f ( x0 , t0 ) t
x(t1)
x0 + k1
k2  f ( x0   k1 , t0  t ) t
x0
t
x1  x0  a1k1  a2k2
At Step i+1:
t0
k 1  f ( xi , ti ) t
t1
The ME method is a special case
with a1=a2=1/2, ==1
k 2  f ( xi   k1 , ti  t ) t
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 
22
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 |max/min| 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
23
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.
x(t1)=x(t0)+|A|+|B|
f
B
f(x1,t1)
f(x0,t0)
Compared to ME method:
A
x n 1  xn 
t
t0
Δt
 f  x0 ,t0  + f  x1 ,t1 
2
Δt
xn+1 = xn +  f  xn ,tn  + f  xn+1 ,tn+1 
2
x1 = x0 +
t1
t
 f  xn ,tn  + f  xnp1 ,tn+1 

2 
24
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)
DE
I(x, V) = YN V
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
25
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
26
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
27
• 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
28
• 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
29
• To simplify the analysis, all nodes other than the generator internal nodes are
eliminated as follows
I1
I2
where
Im
where ij is the angle of Yij
needs to be updated whenever the
network is changed.
2
0
30