Introduction to Power System Stability

Introduction Power System Stabilityy y__
• The tendency of a power system to develop restoring forces equal to
or greater than the disturbing forces to maintain the state ofg g
equilibrium is know as stability.
• If the forces tending to hold machines in synchronism with one
another are sufficient to overcome the disturbing forces, the systemg , y
is said to remain stable.
• Stability Studies:
– TransientTransient
– Dynamic
– Steady‐state
• The main purpose of transient stability studies is to determineThe main purpose of transient stability studies is to determine
whether a system will remain in synchronism following major
disturbances such as transmission system faults, sudden load
changes, loss of generating units, or line switching.
1

• Transient Stability Problems:
– First‐swing; short study period after disturbance, based on a reasonableFirst swing; short study period after disturbance, based on a reasonable
simple generator model, without control system.
– Multi‐swing; longer study period of time after disturbance, thus consider the
effect of generator control systems.
• In all stability studies, the objective is to determine whether or not the
rotors of the machines being perturbed return to constant speed
operation.
• To simplify calculation, the following assumptions must be made:
– Only synchronous frequency currents and voltages are considered in the
stator windings and the power system. Consequently dc offset currents and
harmonic components are neglectedharmonic components are neglected.
– Symmetrical components are used in the representation of unbalanced
faults.
– Generated voltage is considered unaffected by machine speed variation.Generated voltage is considered unaffected by machine speed variation.
2

Mechanical analog of power system transient stability
3

Rotor dynamics and the swing equationy g q
• Accelerating torque is the product of the moment of inertia of
the rotor times its angular accelerationthe rotor times its angular acceleration.
ema
m
TTT
d
J −==2
2
θ (14.1)
ema
dt2
J the total moment of inertia of the rotor masses in kg m2
J the total moment of inertia of the rotor masses, in kg-m
mθ the angular displacement of the rotor with respect to stationary axis, in mechanical radians
t time, in seconds
mT the mechanical or shaft torque supplied by the prime mover less retarding torque due to rotationalm
losses, N-m
eT the net electrical or electromagnetic torque, in N-m
aT the net accelerating torque, in N-m
4

• The mechanical torque Tm and the electrical torque Te are considered
positive for the synchronous generator. Whereas for motor is another
way round.
• Tm is the resultant shaft torque that tends to accelerate the rotor in
the positive direction of rotation as shown in the figure.
• Under steady‐state operation of generator Tm and Te are equal and
l ti t T i
mθ
accelerating torque Ta is zero.
• In this case there is no acceleration or deceleration of the rotor
masses and the resultant constant speed is the synchronous speed.
5

• is measured with respect to a stationary reference axis on the stator, it is
an absolute measure of rotor angle.
• Rotor angular position with respect to a reference axis which rotates at
mθ
g p p
synchronous speed is given by:
msmm t δωθ += (14.2)
• Where is the synchronous speed of the machine in mechanical radians
per second and is the angular displacement of the rotor, in mechanical
radians, from the synchronously rotating reference axis.
smω
mδ
• The derivatives of (14.2) with respect to time are:
dd mm δ
ω
θ
+= (14.3)
dtdt
smω +=
2
2
2
2
d
d
d
d mm δθ
=
(14.3)
(14.4)
22
dtdt
6

• Equation (14.3) shows that the rotor angular velocity         is
constant and equals the synchronous speed only when is
dt
d mθ
dδconstant and equals the synchronous speed only when          is
equal zero.
• Thus, represents the deviation of rotor speed from
dt
d mδ
dt
d mδ
synchronism and the units of measure are mechanical radians
per second.
• Eq (14 4) represents the rotor acceleration measured in
dt
• Eq. (14.4) represents the rotor acceleration measured in
mechanical radians per second‐squared
• Substitute (14.4) into  (14.1):( ) ( )
mN2
2
−−== ema
m
TTT
dt
d
J
δ (14.5)
7

• Eq. (14.5) can be converted into power by multiplying with
angular velocity in Eq (14 6) (**Power = Torque x Angularωangular velocity in Eq. (14.6) (**Power = Torque x Angular
velocity).
dt
d m
m
θ
ω =
(14.6)
mω
W2
2
ema
m
m PPP
dt
d
J −==
δ
ω (14.7)
Where:
P the shaft power input to the machine less rotational lossesPm the shaft power input to the machine less rotational losses
Pe the electrical power crossing its air gap
Pa the accelerating power which account for any unbalance between Pm and Pe
8

• Eq. (14.7) can also be written as in (14.8), whereby :mJM ω=
• M is inertia constant (joule‐seconds per mechanical radian)
W2
2
ema
m
PPP
dt
d
M −==
δ (14.8)
• M is inertia constant (joule‐seconds per mechanical radian)
• In machine data supplied for stability studies, another constant
related to inertia is called H constant:
MVAinratingmachine
speedssynchronouatmegajoulesinenergykineticstored
H =
(14.9)
MVAMJ
S
M
S
J
H
mach
sm
mach
sm
/2
1
2
1 2
ωω
==
MVA.inmachinetheofratingphasethreethe−machS
9

• Solving for M in Eq. (14.9);
radmechMJS
H
M mach
sm
/
2
ω
= (14.10)
• Substitute the above equation in (14.8), we find;
PPPdH −2
2 δ
mechanical radians
mechanical radians per seconds
• Eq. (14.11) can also be written as;
mach
em
mach
am
sm S
PP
S
P
dt
dH
==2
2 δ
ω
(14.11)
perunitPPP
dt
dH
ema
s
−==2
2
2 δ
ω
(14.12) Swing Equation
s
10

• For a system with an electrical frequency of f hertz, Eq. (14.12)
becomes ( in electrical radians)δbecomes (     in electrical radians)
(14.13)perunitPPP
dt
d
f
H
ema −==2
2
δ
π
δ
• If      in electrical degree
(14.14)perunitPPP
dH
==
2
δ
δ
• Eq. (14.12)                                                    can be written as the
t fi t d diff ti l ti
(14.14)perunitPPP
dtf
ema −==2
180
perunitPPP
dt
dH
ema
s
−==2
2
2 δ
ω
two first‐order differential equations:
perunitPP
d
dH
em −=
ω2
s
dt
d
ωω
δ
−=(14.15) (14.16)p
dt
em
sω dt
11

Further Considerations of the swing equationg q
• In a stability study of a power system with many synchronous
machines only one MVA base common to all parts of themachines, only one MVA base common to all parts of the
system can be chosen.
• Thus, H constant for each machine must be converted into per
unit base on common MVA base;
(14.17)mach
ht
S
HH =
• The constant moment inertia M is rarely used in practice and H
(14.17)
system
machsystem
S
HH
• The constant moment inertia M is rarely used in practice and H
is often used in stability study.
12

• In a stability study for a large system with many machines
geographically dispersed over a wide area it is desirable togeographically dispersed over a wide area, it is desirable to
minimize the number of swing equations to be solved.
• This can be done if the transmission line fault, or other
disturbance on the system, affects the machines within the
plant so that their rotors swing together.
• Thus the machine within the plant can be combined into a• Thus, the machine within the plant can be combined into a
single equivalent machine just as if their rotors were
mechanically coupled and only one swing equation need to be
written for them.
13

• Consider a power plant with two generators connected to the same
bus which is electrically remote from network disturbances the swingbus which is electrically remote from network disturbances, the swing
equations on the common system base are:
perunitPP
dt
dH
em 112
1
2
12
−=
δ
ω
(14.18)
dtsω
perunitPP
dt
dH
em
s
222
2
2
22
−=
δ
ω
(14.19)
• Adding the equation together, and denoting and by since the
rotor angle swing together;
1δ 2δ δ
dH 2
2 δ
where
peruniPP
dt
dH
em
s
−=2
2
2 δ
ω
(14.20)
21 HHH +=
PPPwhere 21
21 mmm PPP += 21 eee PPP +=
14

The Power‐Angle Equationg q __________________
• In the swing equation, the input mechanical power from the
prime mover P is assumed constantprime mover Pm is assumed constant.
• Thus, the Pe will determine whether the rotor accelerates,Thus, the Pe will determine whether the rotor accelerates,
decelerates, or remains at synchronous speed.
• Changes in Pe are determined by conditions on the transmission
and distribution networks and the loads on the system to which
the generator supply power.the generator supply power.
15

• Each synchronous machine is represented for transient stability
studies by its transient internal voltage E’ in series with the transienty g
reactance X’d as shown in the Figure below.
• Armature resistance is negligible so that the phasor diagram is as
shown in the figure.g
• Since each machine must be considered relative to the system, the
phasor angles of the machine quantities are measured with respect
to the common system reference.y
+
jXd
'
I
E'
jIXd
'
Vt
_
E'
I
δ
α Vt
j d
Reference
(a) (b)
16

• Consider a generator supplying power through a transmission
system to a receiving end system at bus 2system to a receiving‐end system at bus 2.
1I 2I E’1 is transient internal voltage of
generator at bus 1
'
1E '
2E
E’2 is transient internal voltage of
generator at bus 2
• The elements of the bus admittance matrix for the network
reduced to a two nodes in addition to the reference node is:
⎥
⎦
⎤
⎢
⎣
⎡
=
2221
1211
YY
YY
Ybas
(14.28)
17

• Power equation at a bus k is given by:
• Let k =1 and N=2 and substituting E’ for V
nkn
N
n
kkk
VYVjQP
1=
∗
∑=−

(14.29)
• Let k =1 and N=2, and substituting E 2 for V,
( ) ( )∗∗
+=+ '
212
'
1
'
111
'
111 EYEEYEjQP (14.30)
where
1I 2I
1
'
1
'
1 δ∠= EE 2
'
2
'
2 δ∠= EE
111111 jBGY += 121212 θ∠= YY
'
1E '
2E
18

• We obtain:
2
)(cos 122112
'
2
'
111
2'
11 θδδ −−+= YEEGEP
)(sin 122112
'
2
'
111
2'
11 θδδ −−+−= YEEBEQ
(14.31)
(14.32)
• If we let and , we obtain from (14.31) and
(14.32)
21 δδδ −=
2
12
π
θγ −=
)-(sin12
'
2
'
111
2'
11 γδYEEGEP +=
2
(14.33)
)-(cos12
'
2
'
111
2'
11 γδYEEBEQ −−= (14.34)
19

• Eq. (14.33) can be written more simply as
where
(14.35))sin(max γδ −+= PPP ce
• When the network is considered without resistance, all the
elements of Y are susceptances so both G and becomes
(14.36)11
2'
1 GEPc = 12
'
2
'
1max YEEP =
γelements of Ybus are susceptances, so both G11 and becomes
zero and Eq. (14.35) becomes;
δsinmaxPPe = (14.37) Power Angle Equation
γ
where , with X is the transfer reactance between
E’1 and E’2
maxe
XEEP '
2
'
1max =
1 2
20

Example 1: Power‐angle equation before faultp g q
The single‐line diagram shows a generator connected through parallel
transmission lines to a large metropolitan system considered as an infinite
bus. The machine is delivering 1.0 pu power and both the terminal voltage
and the infinite‐bus voltage are 1.0 pu. The reactance of the line is shown
based on a common system base. The transient reactance of the generator is
0 20 pu as indicated Determine the power‐angle equation for the system0.20 pu as indicated. Determine the power‐angle equation for the system
applicable to the operating conditions.
21

The reactance diagram for the system is shown:
The series reactance between the terminal voltage (Vt) and the infinite bus is:
unitper3.0
2
4.0
10.0 =+=X
The series reactance between the terminal voltage (Vt) and the infinite bus is:
The 1.0 per unit power output of the generator is determined by the power‐angle
1.0sin
3.0
(1.0)(1.0)
sin == αα
X
VVt
The 1.0 per unit power output of the generator is determined by the power angle
equation.
αV is the voltage of the infinite bus, and is the angle of the terminal voltage
relative to the infinite bus
22

Solve α
01 01
458.173.0sin == −
α
Terminal voltage, Vt : unitper300.0954.0458.170.1 0
jVt +=∠=
The output current from the generator is:
30
00.1458.170.1 00
j
I
∠−∠
= unitper729.8012.11535.00.1 0
∠=+= j
3.0j
pj
The transient internal voltage is
XIVE +=' XIVE t +=
)1535.00.1)(2.0()30.0954.0(' jjjE +++=
unitper44.28050.15.0923.0 0
∠=+= j
23

The power‐angle equation relating the transient internal voltage E’ and the
infinite bus voltage V is determined by the total series reactanceinfinite bus voltage V is determined by the total series reactance
unitper5.0
2
4.0
1.02.0 =++=X
Hence, the power‐angle equation is:
upPe .sin1.2sin
50
)0.1)(05.1(
δδ ==
5.0
δWhere is the machine rotor angle with respect to infinite bus
The swing equation for the machine isThe swing equation for the machine is
unitpersin10.20.1
180 2
2
δ
δ
−=
dt
d
f
H
H is in megajoules per megavoltampere, f is the electrical frequency of the system
and is in electrical degree δ
24

Example 1: Power‐angle equationp g q
The power‐angle equation is plotted:
Before fault
After fault
During fault
25

Example 2: Power‐angle equation During Faultp g q g
The same network in example 1 is used. Three phase fault
occurs at point P as shown in the Figure Determine the poweroccurs at point P as shown in the Figure. Determine the power‐
angle equation for the system with the fault and the
corresponding swing equation. Take H = 5 MJ/MVA
26

Approach 1:
Th di f h d i f l i h b lThe reactance diagram of the system during fault is shown below:
The value is admittance
per unitp
27

As been calculated in example 1, internal transient voltage remains
as (based on the assumption that flux linkage is°∠= 4428051'E
The Y bus is:
as (based on the assumption that flux linkage is
constant in the machine)
∠ 44.2805.1E
⎥
⎥
⎤
⎢
⎢
⎡
−
−
= 5025070
333.30333.3
jYb
⎥
⎥
⎦⎢
⎢
⎣ − 833.1050.2333.3
50.250.70jYbas
Since bus 3 has no external source connection and it may be removed by the node
li i ti d th Y b t i i d d telimination procedure, the Y bus matrix is reduced to:
[ ]5.2333.3
83310
1
52
333.3
570
0333.3
⎥
⎦
⎤
⎢
⎣
⎡
−⎥
⎦
⎤
⎢
⎣
⎡−
=busY
833.105.25.70 −⎥
⎦
⎢
⎣
⎥
⎦
⎢
⎣ −
⎥
⎤
⎢
⎡−
=⎥
⎤
⎢
⎡ 769.0308.21211
j
YY
⎥
⎦
⎢
⎣ −⎥
⎦
⎢
⎣ 923.6769.02221
j
YY
28

The magnitude of the transfer admittance is 0.769 and therefore,
The power‐angle equation with the fault on the system is therefore,
' ' '
max 1 2 12 (1.05)(1.0)(0.769) 0.808P E E Y per unit= = =
unitpersin808.0 δ=eP
The corresponding swing equation isThe corresponding swing equation is
5
180
10 0808
2
2
f
d
dt
δ
δ= −. . sin per unit
Because of the inertia, the rotor cannot change position instantly upon
occurrence of the fault. Therefore, the rotor angle is initially 28.440 and
the electrical power output is
δ
38504428i8080 °P 385.044.28sin808.0 =°=eP
29

The initial accelerating power is: Pa = − =10 0385 0615. . . per unit
and the initial acceleration is positive with the value given by
d
dt
f
f
2
2
180
5
0615 2214
δ
= =( . ) . elec deg / s2
30

Approach 2:
C h d li ( hi h i Y f ) i d l dCovert the read line (which in Y form) into delta to remove node
3 from the network:
j1.3
1
1
3
1
1 3
3
2
2
31
)4.0)(2.0()4.0)(3.0()2.0)(3.0( jjjjjj
R
++
j0.65 j0.8667
3.1
2.0
)4.0)(2.0()4.0)(3.0()2.0)(3.0(
j
j
jjjjjj
RAC =
++
=
65.0
4.0
)4.0)(2.0()4.0)(3.0()2.0)(3.0(
j
j
jjjjjj
RAB =
++
=
31
8667.0
3.0
)4.0)(2.0()4.0)(3.0()2.0)(3.0(
j
j
jjjjjj
RBC =
++
=

Approach 2:
Covert the read line which in Y form into delta to remove nodeCovert the read line, which in Y form into delta to remove node
3 from the network:
j0.1625
769.0)3.1/1(12 =−= jY12
'
2
'
1 YEEP = )(12 j
max (1.05)(1.0)(0.769) 0.808
0.808sine
P
P δ
= =
=
1221max YEEP
32
e

Example 3: Power‐angle equation After Fault Clearedp g q
The fault on the system cleared by simultaneous opening of
the circuit breakers at each end of the affected linethe circuit breakers at each end of the affected line.
Determine the power‐angle equation and the swing
equation for the post‐fault period
CB open CB open
33

Example 3: Power‐angle equation After Fault Clearedp g q
Upon removal of the faulted line, the net transfer admittance
across the system isacross the system is
y
j
j12
1
0 2 01 0 4
1429=
+ +
= −
( . . . )
. per unit 429.112 jY =or
The post‐fault power‐angle equation is
δδ sin5.1sin)429.1()0.1()05.1( ==eP )()()(e
and the swing equation is
5
180
10 1500
2
2
f
d
dt
δ
δ= −. . sin
34

Synchronizing Power Coefficients___________
• From the power‐angle
curve, two values of angle
satisfied the mechanical
Before fault
satisfied the mechanical
power i.e at 28.440 and
151.560.
H l th 28 440 After fault
• However, only the 28.440
is acceptable operating
point.
A t bl ti• Acceptable operating
point is that the
generator shall not lose
synchronism when small
During fault
synchronism when small
temporary changes occur
in the electrical power
output from the machineoutput from the machine.
35

Consider small incremental changes in the operating point parameters, that is:
Δ+= δδδ 0Δ+= eee PPP 0
(14.40)
Substituting above equation into Eq 14 37 (Power angle equation) δsinPP =Substituting above equation into Eq. 14.37 (Power‐angle equation)
)sincoscos(sin
)sin(
00max
0max0
ΔΔ
ΔΔ
+=
+=+
δδδδ
δδ
P
PPP ee
δsinmaxPPe =
)scoscos(s 00max ΔΔ δδδδ
(14.41)
Since        is a small incremental displacement from  Δδ
0δ
ΔΔ ≅ δδsin 1cos ≅Δδ
Thus, the previous equation becomes:
+=+ δδδ )cos(sin PPPP
(14.42)
ΔΔ +=+ δδδ )cos(sin 0max0max0 PPPP ee
36

At the initial operating point :0δ
sinδPPP == (14 43)0max0 sinδPPP em == (14.43)
Equation (14.42) becomes:
(14 44)
ΔΔ −=+− δδ )cos()( 0max0 PPPP eem
(14.44)
Substitute Eq. (14.40) into swing equation;
)(
)(2
02
0
2
Δ
Δ
+−=
+
eem
s
PPP
dt
dH δδ
ω
(14.45)
Replacing the right‐hand side of this equation by (14.44);
0)cos(
2
0max2
2
=+ Δ
Δ
δδ
δ
P
d
dH
(14.46))( 0max2 Δ
ω dts
37

Since        is a constant value. Noting that                    is the slope of the power‐
angle curve at the angle    , we denote this slope as Sp and define it as:
0δ 0max cosδP
0δ
(14.47)0max cos
0
δ
δ δδ
P
d
dP
S e
p ==
=
Where Sp is called the synchronizing power coefficient. Replacing Eq. (14.47) into (14.46);
0
2
Δ
δ
ωδ Sd ps (14 48)0
22
=+ Δ
Δ
δ
δ
Hdt
d ps (14.48)
The above equation is a linear, second‐order differential equation.
If Sp positive – the solution             corresponds to that of simple harmonic motion.
If Sp negative – the solution             increases exponentially without limit.
)(tΔδ
)(tΔδp g p y)(tΔδ
38

The angular frequency of the un‐damped oscillations is given by:
sradelec
H
Sps
n /
2
ω
ω =
(14.49)
which corresponds to a frequency of oscillation given by:
H
S
f
ps1 ω
(14.50)
Hz
H
f
ps
n
22
1
π
=
39

Example:
The machine in previous example is operating at when it is subjected
to a slight temporary electrical‐system disturbance. Determine the frequency
°= 44.28δ
to a slight temporary electrical system disturbance. Determine the frequency
and period of oscillation of the machine rotor if the disturbance is removed
before the prime mover responds. H = 5 MJ/MVA.
8466.144.28cos10.2 =°=pSThe synchronizing power coefficient is
The angular frequency of oscillation is therefore;
sradelec
H
S ps
n /343.8
52
8466.1377
2
=
×
×
==
ω
ω
3438
The corresponding frequency of oscillation is Hzfn 33.1
2
343.8
==
π
and the period of oscillation is s
f
T
n
753.0
1
==
fn
40

Equal‐Area Criterion of Stability______________
The swing equation is non‐linear in nature and thus, formal solution
cannot be explicitly found.cannot be explicitly found.
perunitPPP
dt
dH
ema −==2
2
2 δ
ω
To examine the stability of a two‐machine system without solving the
dtsω
swing equation, a direct approach is possible to be used i.e using equal‐
area criterion.
41

Consider the following system:
At point P (close to the bus), a three‐phase fault occurs and cleared by circuit
breaker A after a short period of time.
Thus, the effective transmission system is unaltered except while the fault is on.
The short‐circuit caused by the fault is effectively at the bus and so theThe short circuit caused by the fault is effectively at the bus and so the
electrical power output from the generator becomes zero until fault is clear.
42
three‐phase fault

To understand the physical condition before, during and after the fault,
power‐angle curve need to be analyzed.
Initially, generator operates at synchronous speed with rotor angle of
and the input mechanical power equals the output electrical power Pe.
0δ
Before fault, Pm = Pe
43

At t = 0, Pe = 0, Pm = 1.0 pu
Acceleration constant
The difference must be accounted for by a
rate of change of stored kinetic energy in
the rotor masses.
Speed increase due to the drop of Pe
constant acceleration from t = 0 to t = tc. 1.0 pu
For t<tc, the acceleration is constant given
by:
perunitP
dH
0
2
−=
ω
At t = 0, three phase fault occursd
Ps
2
2
δ ω
= (14.51)
perunitP
dt
m
s
0=
ω
dt H
Pm2
2
(14.51)
44

Acceleration constantWhile the fault is on, velocity increase
above synchronous speed and can beabove synchronous speed and can be
found by integrating this equation:
d tδ ω ω
∫
d
dt H
P dt
H
P ts
m
s
m
tδ ω ω
= =∫ 2 20
For rotor angular position,;
(14.52)
δ
ω
δ+s mP
t2 (14.53)
At t = 0, three phase fault occurs
δ δ= +s m
H
t
4
2
0
(14.53)
(14.52) 45

Acceleration constantEq. (14.52) & (14.53) show that the
velocity of the rotor increase linearlyvelocity of the rotor increase linearly
with time with angle move from to0δ cδ
At the instant of fault clearing t = tc,g ,
the increase in rotor speed is
d Ps mδ ω
dt H
tt t
s m
cc= =
2
angle separation between the generator
and the infinite bus is
(14.54)
At t = 0, three phase fault occurs
and the infinite bus is
( )δ
ω
δt
P
H
tt t
s m
cc= = +
4
2
0
(14.55)
(14.52) 46

When fault is cleared at , Pe
increase abruptly to point d
cδ
At d, Pe > Pm , thus Pa is
negative
Rotor slow down as P goesRotor slow down as Pe goes
from d to e
At t = tc, fault is cleared
47

1. At e, the rotor speed is again
synchronous although rotor angle has
advance to xδ
2. The angle is determined by the
fact that A1 = A2
xδ
x
3. The acceleration power at e is still
negative (retarding), so the rotor
cannot remain at synchronous speed
but continue to slow down.
4. The relative velocity is negative and
the rotor angle moves back from pointthe rotor angle moves back from point
e to point a, which the rotor speed is
less than synchronous.
6 In the absence of damping rotor would5. From a to f, the Pm exceeds the Pe
and the rotor increase speed again until
reaches synchronous speed at f
6. In the absence of damping, rotor would
continue to oscillate in the sequence f‐a‐e, e‐a‐
f, etc 48

In a system where one machine is swinging with respect to infinite bus,
equal‐area criterion can be used to determine the stability of the system
d i di i b l i i iunder transient condition by solving swing equation.
Equal‐area criterion not applicable for multi‐machines.
The swing equation for the machine connected to the infinite bus is
dH 2
2 δ
(14.56)em
s
PP
dt
dH
−=2
2
2 δ
ω
Define the angular velocity of the rotor relative to synchronous speed by
ω
δ
ω ωr s
d
dt
= = − (14.57)s
dt
49

Differentiate (14.57) with respect to t and substitute in (14.56) ;
(14.58)2H d
dt
P P
s
r
m e
ω
ω
= −
When rotor speed is synchronous, equals and is zero. ω sω rω
Multiplying both side of Eq. (14.58) by ; dtdr /δω =
H d dω δ
( )H d
dt
P P
d
dts
r
r
m e
ω
ω
ω δ
2 = − (14.59)
The left‐hand side of the Eq. can be rewritten to give
(14.60)
dt
d
PP
dt
dH
em
r
s
δω
ω
)(
)(
2
−=
50

Multiplying by dt and integrating, we obtain;
δ
(14.61)
∫ −=−
2
1
)()( 2
1
2
2
δ
δ
δωω
ω
dPP
H
emrr
s
Since the rotor speed is synchronous at and then ;δ δ 0== ωωSince the rotor speed is synchronous at and , then ;1δ 2δ 021 == rr ωω
Under this condition, (14.61) becomes
(14.59)( )P P dm e− =∫ δδ
δ
0
1
2
and are any points on the power angle diagram provided that there are
points at which the rotor speed is synchronous.
1δ 2δ
51

In the figure, point a and e correspond to and 1δ 2δ
If perform integration in two steps;If perform integration, in two steps;
0)()(
0
=−+− ∫∫ δδ
δ
δ
δ
δ
dPPdPP
x
c
c
emem
(14.63)
δδ
δ
δ
δ
δ
dPPdPP
x
c
c
meem ∫∫ −=− )()(
0
(14.64)
Fault period
Area A1
Post‐fault period
Area A2
The area under A1 and A4 are directly proportional toThe area under A1 and A4 are directly proportional to
the increase in kinetic energy of the rotor while it is
accelerating.
The area under A and A are directly proportional toThe area under A2 and A3 are directly proportional to
the decrease in kinetic energy of the rotor while it is
decelerating. 52

Equal‐area criterion states that whatever kinetic energy is added to the
rotor following a fault must be removed after the fault to restore the rotor
to synchronous speedto synchronous speed.
The shaded area A1 is dependent upon the time taken to
clear the fault.
If the clearing has a delay, the angle increase.
As a result, the area A2 will also increase. If the increase
cδ
As a result, the area A2 will also increase. If the increase
cause the rotor angle swing beyond , then the rotor
speed at that point on the power angle curve is above
synchronous speed when positive accelerating power is
i t d
maxδ
again encountered.
Under influence of this positive accelerating power the
angle will increase without limit and instability results.g y
53

There is a critical angle for clearing the fault in order to satisfy the
requirements of the equal‐area criterion for stability.
This angle is called the critical clearing angle
The corresponding critical time for removing the fault is called critical
crδ
The corresponding critical time for removing the fault is called critical
clearing time tcr
Power‐angle curve showing the
critical‐clearing angle . Area A1
and A2 are equal
crδ
54

The critical clearing angle and critical clearing time tcr can be calculated
by calculating the area of A1 and A2.
crδ
( )A P d Pm m cr
cr
1 00
= = −∫ δ δ δδ
δ
( )∫
δ
(14.65)
(14 66)( )A P P dmcr
2 = −∫ max sinmax
δ δδ
δ
)()cos(cos maxmaxmax crmcr PP δδδδ −−−=
(14.66)
)()( maxmaxmax crmcr
Equating the expressions for A1 and A2, and transposing terms, yields
( )( )/δ δ δ δP P + (14.67)( )( )cos / cosmax max maxδ δ δ δcr mP P= − +0
55

From sinusoidal power‐angle curve, we see that
(14 68)δ π δ elec rad (14.68)δ π δmax = − 0 elec rad
P Pm = max sinδ0
(14.69)
Substitute and in Eq. (14.67),
simplifying the result and solving for
maxδ maxP
crδ
( )[ ]δ π δ δ δ= −
cos sin cos1
2 (14 70)( )[ ]δ π δ δ δcr = − −cos sin cos0 0 02 (14.70)
In order to get tcr, substitute critical angle equation into (14.55) and then solve to
obtain tcr;
δ
ω
δcr
s m
cr
P
H
t= +
4
2
0
( )t
H
Pcr
cr
s m
=
−4 0δ δ
ω
(14.71) (14.72)
56

Equal‐Area Criterion of Stability ‐Example_____
Calculate the critical clearing angle and critical clearing time for the system shown
below. When a three phase fault occurs at point P. The initial conditions are the
same as in Example 1 and H = 5MJ/MVAp /
P Pe = =max sin . sinδ δ210
Solution
The power angle equation is
The initial rotor angle is
δ0
0
28 44 0 496= =. . elec rad
d h h l h f h l land the mechanical input power Pm is 1.0 pu. Therefore, the critical angle is
calculated using Eq. (14.70)
( )[ ]δ πcr = − × −−
cos . sin . cos .1 0 0
2 0 496 28 44 28 44 = =81697 14260
. . elec rad
( )[ ]δ π δ δ δcr = − −−
cos sin cos1
0 0 02
and the critical clearing time is
( )× −4 5 1426 0496. . = 0 222s( )
tcr =
×377 1
= 0.222s
57

Further Application of the Equal‐Area Criterion
• Equal‐Area Criterion can only be applied for
the case of two machines or one machine
and infinite busand infinite bus.
• When a generator is supplying power to an
infinite bus over two parallel lines, opening
one of the lines may cause the generator to
lose synchronismlose synchronism.
• If a three phase fault occurs on the bus on
which two parallel lines are connected, no
power can be transmitted over either the
lineline.
• If the fault is at the end of one of the lines, CB will operate and power can
flow through another line.
• In this condition, there is some impedance between the parallel buses and
the fault. Thus, some power is transmitted during the fault.
58

Further Application of the Equal‐Area Criterion
Considering the transmitted of power during fault, a general Equal‐area
criterion is applied;criterion is applied;
By evaluating the area A1 and A2 as in the previous approach, we can find that;
1max2maxmax coscos))(/(
cos
rrPP oom
cr
−+−
=
δδδδ
δ (14.73)
12 rr
cr
−
( )
59

Further Application of the Equal‐Area Criterion ‐ Example
Determine the critical clearing angle for the three phase fault described in the
previous example.
The power‐angle equations obtained in the previous examples areThe power‐angle equations obtained in the previous examples are
δδ sin1.2sinmax =P δδ sin808.0sinmax1 =Pr
δδ sin5.1sinmax2 =Pr
Before fault: During fault:
After fault:
Hence
385.0
1.2
808.0
1 ==r 714.0
1.2
5.1
2 ==r
rad412.219.138
5.1
0.1
sin180 1
max =°=−°= −
δ
)44.28cos(385.0)19.138cos(714.0)496.0412.2)(1.2/0.1(
cos
°−°+−
δ
127.0
385.0714.0
)()())((
cos
=
−
=crδ
°= 726.82crδ
To determine the critical clearing time, we must obtain the swing curve of
versus t for this example.
δ
60

STEP‐BY‐STEP SOLUTION OF SWING CURVE____
For large systems we depend on the digital computer to determine δ versus t
for all the machines in which we are interested; and δ can be plotted versus tfor all the machines in which we are interested; and δ can be plotted versus t
for a machines to obtain the swing curve of that machine.
The angle δ is calculated as a function of time over a period long enough tog p g g
determine whether δ will increase without limit or reach a maximum and
start to decrease.
Although the latter result usually indicates stability, on an actual system
where a number of variable are taken into account it may be necessary to plot
δ versus t over a long enough interval to be sure δ will not increase again
ith t t i t l lwithout returning to a low value.
61

By determining swing curves for various clearing times the length of time
permitted before clearing a fault can be determined.
Standard interrupting times for circuit breakers and their associated relays
are commonly 8, 5, 3 or 2 cycles after a fault occurs, and thus breaker
speeds may be specified.
Calculations should be made for a fault in the position which will allow the
least transfer of power from the machine and for the most severe type of
fault for which protection against loss of stability is justified.
A number of different methods are available for the numerical evaluation
of second order differential equations in step by step computations forof second‐order differential equations in step‐by‐step computations for
small increments of the independent variable.
The more elaborate methods are practical only when the computations are
62
The more elaborate methods are practical only when the computations are
performed on a digital computer. The step‐by‐step method used for hand
calculation is necessarily simpler than some of the methods recommended
for digital computers.

In the method for hand calculation the change in the angular position of the
rotor during a short interval of time is computed by making the following
assumptions:assumptions:
•The accelerating power Pa computed at the beginning of an interval is
constant from the middle of the preceding interval to the middle of theconstant from the middle of the preceding interval to the middle of the
interval considered.
•The angular velocity is constant throughout any interval at the valueg y g y
computed for the idle of the interval.
**neither of the assumptions is true, since δ is changing continuously and both Pa
and ω are functions of δ.
63
As the time interval is decreased, the computed swing curve approaches the true
curve.

and
Figure 14.4 will help in visualizing the assumptions.
The accelerating power is computed for the pointsThe accelerating power is computed for the points
enclosed in circles at the ends of the n-2, n-1, and n
intervals, which are the beginnings of the n-1, n and
n+1 interval.
The step curve of Pa in Fig. 14.4 results from the
assumption that Pa is constant between midpoints of
the intervals.
Similarly, ωr, the excess of the angular velocity ω
over the synchronous angular velocity ωs, is shown
as a step curve that is constant throughout the
interval at the value computed for the midpoint.e v e v ue co pu ed o e dpo .
Between the ordinates (n-3/2) and (n-1/2) there
64
Between the ordinates (n 3/2) and (n 1/2) there
is a change of speed caused by the constant
accelerating power.

The change in speed in the product of the
acceleration and the time interval, thus:
ω ω
δ
r n r n a n
d
dt
t
f
H
P t, / , / ,1 2 3 2
2
2 1
180
− = = −Δ Δ (1)
The change in δ over any interval is the
product of ω, for the interval and the time of
the interval. Thus, the change in δ during the
1 i l in‐1 interval is:
Δ Δδ δ δ ωn n n r nt− − − −= − =1 1 2 3 2, /
(2)
and during the nth interval
65
Δ Δδ δ δ ωn n n r nt= − =− −1 1 2, /
(3)
Fig. 14.14

Substracting Eq (2) from Eq. (3) and
substituting Eq. (1) in the resulting
equation to eliminate all values of ω,
(4)
yields
Δ Δδ δn n a nkP= +− −1 1,
where
( )k
f
t=
180 2
Δ
(5)
( )
H
66
Fig. 14.14

Equation (4) is the important for the step-by-step solution of the swing equation with the necessary
assumption enumerated, for it shows how to calculate the change in δ for the previous interval and
the accelerating power for the interval in equation are known.
Equation (4) shows that (subject to the stated assumptions), the change in torque angle during a
given interval is equal to the change in torque angle during the preceding interval plus the
accelerating power at the beginning of the interval times k.
The accelerating power is calculated at the beginning of each new interval. The solution progresses
through enough intervals to obtain points for plotting the swing curve.
Greater accuracy is obtained when the duration of the interval is small. An interval of 0.05s isy
usually satisfactory.
The occurrence of a fault causes a discontinuity in the accelerating power Pa which is zero before
the fault and a definite amount immediately following the fault.y g
The discontinuity occurs at the beginning of the interval, when t=0. Reference to Fig. 14.14 shows
that our method of calculation assumes that the accelerating power computed at the beginning of an
interval considered.
67
When the fault occurs, we have two values of Pa at the beginning of the interval, and we must take
the average of these two values as our constant accelerating power.

EXAMPLE SOLUTION OF SWING CURVE____
Given a two‐bus system of 50 Hz system, with machine 1 delivering 0.9 p.u power
to infinite bus and the machine 1 has H = 4.5 MJ/MVA. Three phase fault
occurred at the middle of the line is cleared in 0.2s. The power angle equationsp g q
are given as below
Pre-fault: δsin0.2=eP
During fault: δsin091.1=eP
After fault cleared: δsin714.1=eP
Compute the swing curve using step-by-step method.
68

Solution:
Remember! Mechanical power doesn’t change during transient stability studies.
At the beginning machine 1 delivering 0 9 pu power At this steady stateAt the beginning, machine 1 delivering 0.9 pu power. At this steady state
condition, Pe = Pm = 0.9 pu.
Wh th f lt t 0 th t l i t th i iti l lWhen the fault occurs at t=0s the rotor angle is at the initial value:
deg7426
9.0
sin
9.0sin0.2
1
elec°=⎟
⎞
⎜
⎛
=
=
−
δ
δ
deg74.26
0.2
sin0 elec=⎟
⎠
⎜
⎝
=δ
5)05.0(
54
)50(180
)(
180 22
==Δ= t
H
f
k
69
5.4H

At the beginning of the first interval there is a discontinuity in the acceleration
power. Just before fault occurs, that is t=0-s Pa=0 since the machine is operating inp a p g
synchronism and the rotor angle is the initial steady-state rotor position 0δ
At t=0+s immediately after the fault has occurred;
409.0491.09.0
491.0)74.26sin(091.1
0
0
=−=−=
=°=
+
+
ema
e
PPP
P
The average value of Pa at t = 0 is 2045.02/409.0 ==avg
aP
Similarly, for representing any change in switching condition during stability
analysis, the accelerating power to be used is the average of the accelerating power
just before and just after the change in switching condition.
70

°=×= 02251204505kP
Then we can find
=×= 0225.12045.05akP
°=°+=Δ 0225.10225.101δ
tΔ
The interval
tΔis the change in rotor angle as time advances over the first interval from 0 to
At the end of the first interval, °=°+°=Δ+= 7625.270225.174.26101 δδδ
st 05.0=Δ °=°−=−= 9590.1)7625.27sin(091.19.0(5)(51 PePmkPAt st 05.0Δ 9590.1)7625.27sin(091.19.0(5)(51, PePmkPa
°=°+°=+Δ=Δ 9815.29590.10225.11,12 akPδδ
δδδ
At
and it follows that the increase in rotor angle over the second time interval is
°=°+°=Δ+= 744.309815.27625.27212 δδδHence, at the end of the second interval
The subsequent steps are shown in the following Table.
71

T (Sec) Pmax sin Po (pu) Pa (pu)
kPa (Elec
Degree (Elec Degree) (Elec Degree)
0‐ 2.0 0.450 0.9 0.0 26.74
0+ 1.091 0.450 0.491 0.409 26.74
0 Avg 1 5455 0 2045 1 0225
δδΔ
δ
0 Avg 1.5455 0.2045 1.0225
1.0225
0.05 1.091 0.465863 0.508257 0.391743 1.958716 27.7625
2.981216
0.1 1.091 0.511259 0.557783 0.342217 1.711084 30.74372
4.692301
0.15 1.091 0.579859 0.632626 0.267374 1.33687 35.43602
6.029171
0.2‐ 1.091 0.662 0.722 0.178 41.46
0.2+ 1.714 0.662 1.135 ‐0.235 41.46
0.2Avg 0.815 ‐0.0285 ‐0.1425
5 8866715.886671
0.25 1.714 0.735539 1.260714 ‐0.36071 ‐1.80357 47.34667
4.083099
0.3 1.714 0.781917 1.340206 ‐0.44021 ‐2.20103 51.42977
1.882069
0.35 1.714 0.801971 1.374579 ‐0.47458 ‐2.37289 53.31184
‐0.49082
0.4 1.714 0.796824 1.365756 ‐0.46576 ‐2.32878 52.82101
‐2.8196
0.45 1.714 0.766133 1.313152 ‐0.41315 ‐2.06576 50.00141
‐4.88536
0.5 1.714 0.70861 1.214557 ‐0.31456 ‐1.57278 45.11605
72
‐6.45815

0.5 1.714 0.70861 1.214557 ‐0.31456 ‐1.57278 45.11605
‐6.45815
0.55 1.714 0.624737 1.0708 ‐0.1708 ‐0.854 38.6579
‐7.31215
0.6 1.714 0.520262 0.891729 0.008271 0.041356 31.34575
‐7.27079
0.65 1.714 0.407981 0.69928 0.20072 1.003601 24.07496
‐6.26719
0.7 1.714 0.305863 0.524249 0.375751 1.878756 17.80777
‐4.38843
0.75 1.714 0.232106 0.397829 0.502171 2.510854 13.41934
‐1.87758
0 8 1 714 0 200108 0 342984 0 557016 2 785078 11 541760.8 1.714 0.200108 0.342984 0.557016 2.785078 11.54176
0.907497
0.85 1.714 0.215602 0.369542 0.530458 2.652288 12.44925
3.559785
0.9 1.714 0.275824 0.472762 0.427238 2.13619 16.00904
5 6959755.695975
0.95 1.714 0.369874 0.633964 0.266036 1.330182 21.70501
7.026157
1 1.714 0.480758 0.824019 0.075981 0.379907 28.73117
7.406064
73
1.05 1.714 0.589787 1.010896 ‐0.1109 ‐0.55448 36.13724

At t=0.2s, fault is cleared. Hence, the average of accelerating power just before and after should be
0 2
1 091 i (41 46 ) 0 223°
considered.
At t=0.2- s (just before fault cleared);
0.2
0.2 0.2
1.091sin(41.46 ) 0.7223
0.9 0.7223 0.1777
e
a m e
P
P P P
−
− −
= ° =
= − = − =
At t=0.2+ s (just after fault cleared);
2348.01348.19.0
1348.1)46.41sin(714.1
2.02.0
2.0
−=−=−=
=°=
++
+
ema
e
PPP
P
23480177702020 +
PP
(j );
02855.0
2
2348.01777.0
2
)(
2.02.0
−=
−
=
+
=
+−
aa
a
PP
averagePHence, average Pa;
The subsequent steps are shown in the following Table.
74

Introduction to Power System Stability

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