Power System Stability Introduction

Institute of Energy Technology 1
INTRODUCTION TO STABILITY
 What is stability
 the tendency of power system to restore the state of
equilibrium after the disturbance
 mostly concerned with the behavior of synchronous
machine after a disturbance
 in short, if synchronous machines can remain
synchronism after disturbances, we say that system is
stable
 Stability issue
 steady-state stability – the ability of power system to
regain synchronism after small and slow disturbances
such as gradual power change
 transient stability – the ability of power system to
regain synchronism after large and sudden
disturbances such as a fault

2
The Synchronous Generator System

POWER ANGLE
 Power angle
 relative angle r
between rotor mmf and
air-gap mmf (angle
between Fr and Fsr),
both rotating in
synchronous speed
 also the angle r
between no-load
generated emf E and
stator voltage Esr
 also the angle 
between emf E and
terminal voltage V, if
neglecting armature
resistance and leakage
flux

5
Phasor Diagram
Phasor diagram of a cylindrical-rotor synchronous generator,
for the case of lagging power factor
Lagging PF: |Vt|<|Ef| for overexcited condition
Leading PF: |Vt|>|Ef| for underexcited condition

DEVELOPING SWING EQUATION
 Synchronous machine operation
 consider a synchronous generator with
electromagnetic torque Te running at synchronous
speed ωsm.
 during the normal operation, the mechanical torque
Tm = Te
 a disturbance occur will result in
accelerating/decelerating torque Ta=Tm-Te (Ta>0 if
accelerating, Ta<0 if decelerating)
 introduce the combined moment of inertia of prime
mover and generator J
 by the law of rotation --
 m is the angular displacement of rotor w.r.t.
stationery reference frame on the stator
ema
m
TTT
dt
d
J 2
2


 Derivation of swing equation
 m = ωsmt+m, ωsm is the constant angular velocity
 take the derivative of m, we obtain –
 take the second derivative of m, we obtain –
 substitute into the law of rotation
 multiplying ωm to obtain power equation
dt
d
dt
d m
sm
m 



2
2
2
2
dt
d
dt
d mm 

ema
m
TTT
dt
d
J 2
2

ememmm
mm
m PPTT
dt
d
M
dt
d
J  

 2
2
2
2

 Derivation of swing equation
 swing equation in terms of inertial constant M
 relations between electrical power angle  and
mechanical power angle m and electrical speed and
mechanical speed
 swing equation in terms of electrical power angle 
 converting the swing equation into per unit system
em
m
PP
dt
d
M 2
2

numberpoleiswhere
2
,
2
p
pp
mm  
em PP
dt
d
M
p
2
2
2 
s
puepum
s
H
MPP
dt
dH



2
where,
2
)()(2
2


10
EXAMPLE I
 A 3-phase, 60-Hz, 500-MVA, 15-kV, 32-pole hydrolelctric
generating unit has an H constant of 2.0 p.u.-s.
 (a) Determine ωsyn and ωmsyn.
 (b) Write the swing equation of the unit.
 (c) The unit is initially operating at pm.pu = pe.pu = 1.0, ω =
ωsyn , and δ = 10 degrees when a 3-phase-to-ground
bolted short circuit at the generator terminals causes pe.pu
to drop to zero. Determine the power angle 3 cycles after
the shoty circuit commences. Assume pe.pu remains
constant at 1.0 p.u. Also assume ωp.u(t) = 1.0 in the swing
equation.

11
EXAMPLE III
 A 3-phase, 60-Hz, 400-kV,13.8-kV, 4-pole steam turbine
generating unit has an H constant of 5.0 p.u.-s. The generating
unit is initially operating at pm.pu = pe.pu = 0.7 p.u., ω = ωsyn,
δ = 12 degrees when a fault reduces the generation of electrical
power by 70%.
 Determine the power angle δ five cycles after the fault
commences. Assume the the accelerating power remains
constant during the fault. Also assume ωp.u.(t) = 0 1.0 in the
swing equation.
 Repeat (b) for a bolted 3-phase fault at the generator terminals
that reduces the electrical power output to zero. Compare the
power angle with that determined in (a).

12
Example II
 A Power plant has two 3-phase, 60-Hz generating
units with the following ratings:
 Unit 1: 500-MVA, 15-kV, 0.85 power factor, 32-
pole, H1 = 2.0 p.u.-s
 Unit 2: 300-MVA, 15-kV, 0.90 power factor, 32-
pole, H1 = 2.5 p.u.-s
 Determine the per unit swing equation of each
unit on a 100-MVA system base.
 If the units are assumed to “swing together” that
is, δ1(t) = δ2(t), combine the two swing equations
into one equivalent swing equations.

SYNCHRONOUS MACHINE MODELS FOR
STABILITY STUDY
 Simplified synchronous machine model
 the simplified machine model is decided by the proper
reactances, X’’d, X’d, or Xd
 for very short time of transient analysis, use X’’d
 for short time of transient analysis, use X’d
 for steady-state analysis, use Xd
 substation bus voltage and frequency remain constant is referred
as infinite bus
 generator is represented by a constant voltage E’ behind direct
axis transient reactance X’d
Zs
ZLjX’d
E’
Vg V

STABILITY STUDY
 Converting the network into  equivalent circuit
 for the conversion, please see Eq.11.23
 use  equivalent line model for currents
 real power at node 1
y10
y12
E’ V
y20
1 2
I1 I2




















V
E
y
y
I
I '
yy
yy
122012
121210
2
1
 12121111
2
cos'cos'   YVEYEPe

STABILITY STUDY
 Real power flow equation
 let –y12 = 1 / X12
 simplified real power equation:
 Power angle curve
 gradual increase of generator power output is
possible until Pmax (max power transferred) is reached
 max power is referred as steady-state stability limit at
=90o
sin
'
12X
VE
Pe 
0
Pe
Pm
π/2 π
Pmax
0

Pe
12
max
'
X
VE
P 

STABILITY STUDY
 Transient stability analysis
 condition: generator is suddenly short-circuited
 current during the transient is limited by X’d
 voltage behind reactance E’=Vg+jX’dIa
 Vg is the generator terminal voltage, Ia is prefault
steady state generator current
 phenomena: field flux linkage will tend to remain
constant during the initial disturbance, thus E’ is
assumed constant
 transient power angle curve has the same form as
steady-state curve but with higher peak value,
probably with smaller X’d

SYNCHRONOUS MACHINE MODELS
INCLUDING SALIENCY
 Phasor diagram of salient-pole machine
 condition: under steady state with armature
resistance neglected
 power angle equation in per unit
 voltage equation in per unit
 E is no-load generated emf in pu, V is generator
terminal voltage in pu
E
V XdId
jXqIq
Id
Iq
Ia

a
 2sin
2
sin
2
qd
qd
d XX
XX
V
X
VE
P


   sincoscos addd IXVIXVE

SYNCHRONOUS MACHINE MODELS
INCLUDING SALIENCY
 Calculation of voltage E
 starting with a given (known) terminal voltage V and
armature current Ia, we need to calculate  first by
using phasor diagram and then result in voltage E
 once E is obtained, P could be calculated
 Transient power equation
 for salient machine
 this equation represents the behavior of SM in early
part of transient period
 calculate  first, then calculate |E’q|:
 see example 11.1
   sincos ad IXVE








 



sin
cos
tan 1
aq
aq
IXV
IX
 2sin
2
sin '
'
2
'
'
qd
qd
d
q
e
XX
XX
V
X
VE
P


   sincos ''
adq IXVE

STEADY-STATE STABILITY – SMALL
DISTURBANCE
 Steady-state stability
 the ability of power system to remain its synchronism
and returns to its original state when subjected to
small disturbances
 such stability is not affected by any control efforts
such as voltage regulators or governor
 Analysis of steady-state stability by swing
equation
 starting from swing equation
 introduce a small disturbance Δ
 derivation is from Eq.11.37 (see pg. 472)
 simplify the nonlinear function of power angle 



sinmax)()(2
2
0
PPPP
dt
d
f
H
mpuepum 

DISTURBANCE
 Analysis of steady-state stability by swing equation
 swing equation in terms of Δ
 PS=Pmax cos0: the slope of the power-angle curve at 0,
PS is positive when 0 <  < 90o (See figure 11.3)
 the second order differential equation
 Characteristic equation:
 rule 1: if PS is negative, one root is in RHP and system
is unstable
 rule 2: if PS is positive, two roots in the jω axis and
motion is oscillatory and undamped, system is
marginally stable
0cos 02
2
0




 mP
dt
d
f
H
0max cos0

  P
d
dP
PS 
02
2
0




 SP
dt
d
f
H
SP
H
f
s 02 


DISTURBANCE
 rule 2 (continued): the oscillatory frequency of the
undamped system
 Damping torque
 phenomena: when there is a difference angular velocity
between rotor and air gap field, an induction torque will
be set up on rotor tending to minimize the difference of
velocities
 introduce a damping power by damping torque
 introduce the damping power into swing equation
Sn P
H
f0
 
dt
d
DPd


02
2
0






 SP
dt
d
D
dt
d
f
H

STABILITY ANALYSIS ON SWING EQUATION

 Analysis of characteristic equation

 for damping coefficient
 roots of characteristic equation
 damped frequency of oscillation
 positive damping (1>>0): s1,s2 have negative real part
if PS is positive, this implies the response is bounded
and system is stable
02 2
2
2








nn
dt
d
dt
d
02 22
 nnss 
1
2
0

SHP
fD 

2
21 1-s,s   nn j
2
1   nd

STABILITY ANALYSIS ON SWING EQUATION
 Solution of the swing equation

 roots of swing equation
 rotor angular frequency
 response time constant
 settling time:
 relations between settling time and inertia constant H:
increase H will result in longer tS, decrease ωn and 
02 2
2
2








nn
dt
d
dt
d
   





 






 
tete d
t
d
t nn
sin
1
,sin
1 2
0
02
0
   tete d
tn
d
tn nn






 
sin
1
,sin
1 2
0
02
0 






Df
H
n 0
21

 
4St

SOLVING THE SWING EQUATION USING STATE
SPACE MATRIX
 State space approach
 state space approach can solve multi-machine system
 let x1=Δ, x2=Δω=Δ


 taking the Laplace transform, from Eq.11.52
 solution of the X(s)
)()(
2
10
2
1
2
2
1
tAxtx
x
x
x
x
nn























)()(
10
01
2
1
2
1
tCxty
x
x
y
y


















    








n
2
n
1
2s
1-
),0()(

s
AsIxAsIsX
22
2
)0(
12
)(
nn
n
n
s
x
s
s
sX














SOLVING THE SWING EQUATION USING STATE
SPACE MATRIX
 State space approach

 taking the inverse Laplace transform with initial state
x1(0)=Δ0, x2(0)=Δω0=0
 state solution: x1(t)=Δ(t), x2(t)=ω(t)

 
 222
221
2
)()(
2
)()(
nn
nn
s
u
ssx
ss
u
ssx










   





 






 
tete d
t
d
t nn
sin
1
,sin
1 2
0
02
0
   tete d
tn
d
tn nn






 
sin
1
,sin
1 2
0
02
0 







STEADY STATE STABILITY EXAMPLE
 Example 11.3
 using the state space matrix to solve  and ω
 the original state 0=16.79o, new state after ΔP is
imposed =22.5o
 the linearized equation is valid only for very small
power impact and deviation from the operating state
 a large sudden impact may result in unstable state
even if the impact is less than the steady state power
limit
 the characteristic equation of determinant (sI-A) or
eigenvalue of A can tell the stability of system
 system is asymptotically stable iff eigenvalues of A are
in LHP
 in this case, eigenvalues of A are -1.3  6.0i

TRANSIENT STABILITY
 Transient stability
 to determine whether or not synchronism is maintained
after machine has been subject to severe disturbance
 Severe disturbance
 sudden application of loads (steel mill)
 loss of generation (unit trip)
 loss of large load (line trip)
 a fault on the system (lightning)
 System response after large disturbance
 oscillations of rotor angle result in large magnitude that
linearlization is not feasible
 must use nonlinear swing equation to solve the
problem

EQUAL AREA CRITERION
 Equal area criterion
 can be used to quickly predict system stability after
disturbance
 only applicable to a one-machine system connected to
an infinite bus or a two-machine system
 Derivation of rotor relative speed from swing
equation
 starting from the swing equation with damping
neglected
 for detailed derivation, please see pp.486
 the swing equation end up with

poweronacceleratiPPPP
dt
d
f
H
aaem
o
,2
2



  




o
dPP
H
f
dt
d
em
o2

 Synchronous machine relative speed equation

 the equation gives relative speed of machine with
respect to the synchronous revolving reference frame
 if stability of system needs to be maintained, the speed
equation must be zero sometimes after the disturbance
 Stability analysis
 stability criterion
 consider machine operating at the equilibrium point o,
corresponding to power input Pm0 = Pe0
 a sudden step increase of Pm1 is applied results in
accelerating power to increase power angle  to 1
  




o
dPP
H
f
dt
d
em
o2
  0



o
dPP em

 Stability analysis
 the excess energy stored in rotor
 when =1, the electrical power matches new input
power Pm1, rotor acceleration is zero but relative speed
is still positive (rotor speed is above synchronous
speed),  still increases
 as long as  increases, Pe increases, at this time the
new Pe >Pm1 and makes rotor to decelerate

 rotor swing back to b and the angle max makes
 |area A1|=|area A2|
  1
1
AareaabcareadPP
o
em 



  21 de
max
1
AareabareadPP em 




 Equal area criterion (stable condition)
A2 A2max
A1
0
1 max
Pm1
Pm0
Equal Criteria: A1 = A2
A1 < A2max Stable
A1 = A2max Critically Stable
A1 > A2max Unstable
t0
Pm1
0
t
max
1
a
bc
d
e

APPLICATION TO SUDDEN INCREASE OF
POWER INPUT
 Stability analysis of equal area criterion
 stability is maintained only if area A2 at least equal to
A1
 if A2 < A1, accelerating momentum can never be
overcome
 Limit of stability
 when max is at intersection of line Pm and power-angle
curve is 90o <  < 180o
 the max can be derived as (see pp.489, figure 11.12)
 max can be calculated by iterative method
 Pmax is obtained by Pm=Pmaxsin1, where 1 = -max
  0maxmaxmax coscossin   o

SOLUTION TO STABILITY ON SUDDEN
INCREASE OF POWER INPUT
 Calculation of max

 max can be calculated by iterative Newton Raphson
method
 assume the above equation is f(max) = c
 starting with initial estimate of /2 < max
(k) < , Δ gives
 where

 the updated max
(k+1)
max
(k+1) = max
(k) + Δ max
(k)
  0maxmaxmax coscossin   o
 
)(
max
max
)(
max)(
max
kd
df
fc k
k






  )(
max0
)(
max
max
cos
)(
max
kk
kd
df

 


APPLICATION TO THREE PHASE FAULT
 Three phase bolt fault case
 a temporary three phase bolt fault occurs at sending end of line at
bus 1
 fault occurs at 0, Pe = 0
 power angle curve corresponds
to horizontal axis
 machine accelerate,
increase  until fault cleared at c
 fault cleared at c shifts operation
to original power angle curve at e
 net power is decelerating, stored
 energy reduced to zero at f
 A1(abcd) = A2(defg)
Pe
A1
0
c max
Pm
a
b c
f
e
d g
A2

F
1

- NEAR SENDING END
 when rotor angle reach f, Pe>Pm
rotor decelerates and retraces
along power angle curve passing
through e and a
 rotor angle would swing back and
forth around 0 at ωn
 with inherent damping, operating
point returns to 0
 Critical clearing angle
 critical clearing angle is reached when further increase in c cause
A2 < A1
 we obtain c
Pe
A1
0
c max
Pm
a
b c
f
e
d g
A2

  




dPPdP
c
c
mm  
max
0
sinmax
  max0max
max
coscos  
P
Pm
c

- NEAR SENDING END
 Critical clearing time
 from swing equation
 integrating both sides from t = 0 to tc
 we obtain the critical clearing time
mP
dt
d
f
H
2
2
0


 
m
c
c
Pf
H
t
0
02

 


- AWAY FROM SENDING END
 a temporary three phase fault occurs away from sending end of bus
1
 fault occurs at 0, Pe is reduced
 power angle curve corresponds
to curve B
 machine accelerate, increase 
from 0 (b) until fault cleared at c
(c)
 fault cleared at c shifts operation
to curve C at e
 net power is decelerating, stored
energy reduced to zero at f
 A1(abcd) = A2(defg)
F
1
Pe
A1
0
c max
Pm
a
b
c
f
e
d g
A2

A
C
B

 when rotor angle reach f, Pe>Pm
rotor decelerates and rotor angle
would swing back and forth
around e at ωn
 with inherent damping, operating
point returns to the point that Pm
line intercept with curve C
 Critical clearing angle
 critical clearing angle is reached when further increase
in c cause A2 < A1
 we obtain c
    
max
0
maxmax3max20 sinsin





c
c
cmcm PdPdPP
 
max2max3
0max2maxmax30max coscos
cos
PP
PPPm
c





Pe
A1
0
c max
Pm
a
b
c
f
e
d g
A2

A
C
B

 The difference between curve b and curve c
is due to the different line reactance
 curve b: the second line is shorted in the middle
point (Fig. 11.23)
 curve c: after fault is cleared, the second line is
isolated
 See example 11.5
 use power curve equation to solve max and then
c

NUMERICAL SOLUTION OF NONLINEAR
EQUATION
 Euler method
 tangent evaluation:
 updated solution:
 drawback: accuracy
 Modified Euler method
 tangent evaluation:
 updated solution:
 feature: better accuracy, but time step Δt should
be properly selected
0x
dt
dx
t
dt
dx
xxxx x  0001
2
10
p
xx
dt
dx
dt
dx

2/10001 t
dt
dx
dt
dx
xxxx xx 







NUMERICAL SOLUTION OF NONLINEAR
EQUATION
 Higher order equation
 use state space method to decompose higher
order equation
 use modified Euler method to solve state space
matrix
 for swing equation of second order, use 22
state space matrix to solve
 see pp. 504
)()(
2
10
2
1
2
2
1
tAxtx
x
x
x
x
nn
























NUMERICAL SOLUTION OF SWING
EQUATION
 Swing equation in state variable form

 use modified Euler method
 the updated values (see Ex. 11.6)
aP
H
f
dt
d
dt
d
0





t
dt
d
P
H
f
dt
d
t
dt
d
dt
d
i
p
i
p
i
i
p
i
i
p
ia
i
p
i
p
i

















1
0
11
where
where,
11
1
tdt
d
dt
d
tdt
d
dt
d
p
ii
p
ii
i
c
ii
c
i 









 














 


2
,
2
11






MULTIMACHINE SYSTEMS
 Multi-machine system can be written similar to
one-machine system by the following assumption
 each synchronous machine is represented by a
constant voltage E behind Xd (neglect saliency and flux
change)
 input power remain constant
 using prefault bus voltages, all loads are in equivalent
admittances to ground
 damping and asynchronous effects are ignored
 mech = 
 machines belong to the same station swing together
and are said to be coherent, coherent machines can
equivalent to one machine

METHOD TO SOLVE MULTIMACHINE
SYSTEMS
 Solution to multi-machine system
 solve initial power flow and determine initial bus voltage
magnitude and phase angle
 calculating load equivalent admittance
 nodal equations of the system
 electrical and mechanical power output of machine at
steady state prior to disturbances
idii
i
ii
i
i
i IjXVE
V
jQP
V
S
I ''
**
*
, 


20
i
ii
i
V
jQP
y




















'
0
m
n
mm
t
nm
nmnn
m E
V
YY
YY
I
   

m
j
jiijijjiiimiei YEEIEPP
1
''*
cosRe 

MULTIMACHINE SYSTEMS TRANSIENT
STABILITY
 Classical transient stability study is based on the
application of the three-phase fault
 Swing equation of multi-machine system

 Yij are the elements of the faulted reduced bus
admittance matrix
 state variable model of swing equation
 see example 11.7
  eimi
m
j
jiijijjimi
ii
PPYEEP
dt
d
f
H
 1
''
2
2
0
cos 


 eimi
i
i
i
i
PP
H
f
dt
d
ni
dt
d



0
,,1,





Power System Stability Introduction

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