Post on 19-Mar-2022
Power System Transient Stability
Submitted By:Santosh Kumar Gupta Assistant ProfessorEE Department SIT Sitamrhi
Introduction Power SystemStability
• The tendency of a power system to develop restoring forces equal toor greater than the disturbing forces to maintain the state ofequilibrium is know asstability.
• If the forces tending to hold machines in synchronism with oneanother are sufficient to overcome the disturbing forces, the systemis said to remainstable.
• Stability Studies:
– Transient
– Dynamic
– Steady‐state
• The main purpose of transient stability studies is to determine
disturbances such as transmission system faults, suddenwhether a system will remain in synchronism following major
loadchanges, loss of generating units, or line switching.
2
The Power‐AngleEquation
• In the swing equation, the input mechanical power from the
prime mover Pm is assumedconstant.
• Thus, the Pe will determine whether the rotor accelerates,
decelerates, or remains at synchronousspeed.
• Changes in Peare determined by conditions on the transmission
and distribution networks and the loads on the system to which
the generator supplypower.
3
The Power‐AngleEquation
+
djX '
• Each synchronous machine is represented for transient stabilitystudies by its transient internal voltage E’ in series with the transientreactance X’das shown in the Figure below.
• Armature resistance is negligible so that the phasor diagram is asshown in thefigure.
• Since each machine must be considered relative to the system, thephasor angles of the machine quantities are measured with respectto the common systemreference.
I
E'
djIX'
Vt
_E'
I
Vt
Reference
4(a) (b)
The Power‐AngleEquation
• Consider a generator supplying power through a transmission
system to a receiving‐end system at bus 2.
I1I2 E’1 is transient internal voltage of
generator at bus 1
1E '
2E '
E’2 is transient internal voltage of
generator at bus 2
5
• The elements of the bus admittance matrix for the network
reduced to a two nodes in addition to the reference node is:
21 22 YY =Y11 Y12
Ybas
(14.28)
The Power‐AngleEquation
• Power equation at a bus k is given by:
kn n
N
Yn=1
k k kVjQ VP − = (14.29)
• Let k =1 and N=2, and substituting E 2 for V,
' '
1 12 2
' '
1 1 1 11 1 E (Y E )P + jQ = E (Y E ) + (14.30)
whereI1
I2
'
1
'
1 1EE = 222 = E ' E '
11+ jBY11 =G11Y12 = Y12 12
E '
6
1 2E '
The Power‐AngleEquation
• We obtain:
122 12Y cos (1 − 2 − )E '
2
P = E ' G + E ' 1 1 11 1
12
2'
1 1Y sin (1 − 2 − )E 'B + E '
11 1 2 12Q = − E
(14.31)
(14.32)
21• If we let =
(14.32)
−2
− , we obtain from (14.31)and12
and =
121 211
2E '
G + E ''
11P = E (14.33)
1 2 12
7
11
2
Y sin ( - )
Y cos ( - )E 'B − E ''
11Q = − E (14.34)
The Power‐AngleEquation
• Eq. (14.33) can be written more simply as
where(14.35)Pe = Pc + Pmax sin( − )
• When the network is considered without resistance, all the elements of Ybus are susceptances,so both G11 and becomes zero and Eq. (14.35)becomes;
(14.36)' 2
G11cP = E1
P = E ' E ' Ymax 1 2 12
Pe = Pmax sin (14.37) Power AngleEquation
where , with X is the transfer reactancebetweenXE '
2max 1P = E '
E’1 andE’2
8
Example 1: Power‐angle equation beforefault
The single‐line diagram shows a generator connected through paralleltransmission lines to a large metropolitan system considered as an infinitebus. The machine is delivering 1.0 pu power and both the terminal voltageand the infinite‐bus voltage are 1.0 pu. The reactance of the line is shownbased on a common system base. The transient reactance of the generator is0.20 pu as indicated. Determine the power‐angle equation for the systemapplicable to the operatingconditions.
9
Example 1: Power‐angle equation beforefault
The reactance diagram for the system is shown:
X = 0.10 +0.4
= 0.3 per unit
The series reactance between the terminal voltage (Vt) and the infinite bus is:
2
The 1.0 per unit power output of the generator is determined by the power‐angle
0.3
10
(1.0)(1.0)sin =1.0sin =
X
Vt Vequation.
is the angle of the terminalvoltageV is the voltage of the infinite bus, and
relative to the infinitebus
Example 1: Power‐angle equation beforefault
Solve
= sin−1 0.3 = 17.4580
Terminal voltage, Vt : Vt = 1.017.458 = 0.954 + j0.300 perunit0
The output current from the generatoris:
1.017.4580 − 1.000
I = = 1.0 + j0.1535 = 1.0128.7290 perunitj0.3
The transient internal voltage is
E'= Vt + XI
E' = (0.954 + j0.30) + j(0.2)(1.0 + j0.1535)
= 0.923 + j0.5 = 1.05028.440 per unit
11
Example 1: Power‐angle equation beforefault
The power‐angle equation relating the transient internal voltage E’and the
infinite bus voltage V is determined by the total series reactance
2X = 0.2 + 0.1 +
0.4 = 0.5 per unit
Hence, the power‐angle equation is:
eP =(1.05)(1.0)
sin = 2.1sin p.u0.5
Where is the machine rotor angle with respect to infinite bus
The swing equation for the machineisd 2
= − 1.0 2.10 sin per unit
12
180 f dt2
H
H is in megajoules per megavoltampere, f is the electrical frequency of the system
and is in electricaldegree
Example 1: Power‐angle equation
The power‐angle equation isplotted:
Before fault
After fault
During fault
13
Example 2: Power‐angle equation DuringFault
The same network in example 1 is used. Three phase faultoccurs at point P as shown in the Figure. Determine the power‐
angle equation for the system with the fault and the
corresponding swing equation.TakeH = 5 MJ/MVA
14
Example 2: Power‐angle equation DuringFault
Approach 1:
The reactance diagram of the system during fault is shown below:
The value is admittance
per unit
15
Example 2: Power‐angle equation DuringFault
The Y bus is:
As been calculated in example 1, internal transient voltage remainsas E'= 1.0528.44 (based on the assumption that flux linkage is
constant in themachine)
3.333 Y = j 0
3.333
2.50
−3.333 0
− 7.50
2.50 − 10.833
bas
Since bus 3 has no external source connection and it may be removed by the node elimination procedure, the Y bus matrix isreduced to:
−
3.333 1 3.333 2.5=
−3.333 0bus
Y 2.5 −10.833 0 − 7.5
Y12 = j
− 2.308 0.769 Y11
21
16
Y22 0.769 − 6.923
Y
Example 2: Power‐angle equation DuringFault
The magnitude of the transfer admittance is 0.769 and therefore,
The power‐angle equation with the fault on the system is therefore,
Pe = 0.808 sin per unit
The corresponding swing equationis
P = E ' E ' Y '
max 1 2 12= (1.05)(1.0)(0.769) = 0.808 per unit
180 f dt2
17
5 d 2= 1.0 − 0.808 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
Pe = 0.808 sin 28.44= 0.385
Example 2: Power‐angle equation DuringFault
The initial accelerating power is:Pa = 1.0 − 0.385= 0.615 per unit
and the initial acceleration is positive with the value given by
dt 2 5
18
d 2 180 f(0.615) = 22.14 f= elec deg / s2
Example 2: Power‐angle equation DuringFault
Approach 2:
Covert the read line (which in Y form) into delta to remove node
3 from thenetwork:j11.3
1 3
32
2
j0.65 j0.8667
j0.2R
AC=
( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)= j1.3
j0.4R
AB=
( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)= j0.65
=( j0.3)( j0.2) + ( j0.3)( j0.4) + ( j0.2)( j0.4)
= j0.8667
19j0.3
RBC
Example 2: Power‐angle equation DuringFault
Approach 2:
Covert the read line, which in Y form into delta to remove node
3 from thenetwork:
j0.1625
Y12 = −(1/ j1.3) = 0.769P = E ' E ' Ymax 1 2 12
Pmax = (1.05)(1.0)(0.769) =0.808
Pe = 0.808sin
20
Example 3: Power‐angle equation After FaultCleared
The fault on the system cleared by simultaneous opening of
the circuit breakers at each end of the affected line.
Determine the power‐angle equation and the swing
equation for the post‐faultperiod
CBopen
21
CBopen
Example 3: Power‐angle equation After FaultCleared
Upon removal of the faulted line, the net transfer admittance
across the systemis
j(0.2 + 0.1 +0.4)= − j1.429 per unit
12y =
1 Y12 = j1.429or
The post‐fault power‐angle equation is
Pe= (1.05)(1.0)(1.429)sin =1.5sin
and the swing equationis
5 d 2
22
180 f dt2= 1.0−1.500 sin
Synchronizing PowerCoefficients
• From the power‐anglecurve, two values ofangle
Before faultsatisfied the mechanical
After fault
power i.e at 28.440 and151.560.
• However, only the 28.440
is acceptable operatingpoint.
• Acceptable operatingpoint is that the
During fault
generator shall not losesynchronism when smalltemporary changes occurin the electrical poweroutput from themachine.
23
Synchronizing PowerCoefficients
Consider small incremental changes in the operating point parameters, that is:
= 0 +Pe = Pe0 + Pe
(14.40)
Pe0
= Pmax sinSubstituting above equation into Eq. 14.37 (Power‐angle equation) Pe
(14.41)
+ Pe= Pmax sin(0 + )
= Pmax(sin0 cos + cos0 sin)
Since is a small incremental displacement from 0
sin cos 1
Thus, the previous equationbecomes:
(14.42)0
24
Pe0 + Pe = Pmax sin 0 + (Pmax cos )
Synchronizing PowerCoefficients
At the initial operating point 0 :
0Pm = Pe0 = Pmax sin (14.43)
Equation (14.42)becomes:
Pm− (Pe0 + Pe) = −(Pmax cos 0 )(14.44)
Substitute Eq. (14.40) into swingequation;
e0 e− (P + P )m
s
0 = Pdt 2
2H d 2 ( + )
(14.45)
Replacing the right‐hand side of this equation by (14.44);
+ (Pmax cos 0 ) = 02H d 2 (14.46)
dt 2
25
s
Synchronizing PowerCoefficients
Since 0 is a constant value. Noting that Pmax cos0 is the slope of thepower‐
angle curve at the angle0, we denote this slope as Sp and define it as:
(14.47)0max= P cos
d=
dPe
=0
S p
Where Sp is called the synchronizing power coefficient. Replacing Eq. (14.47) into (14.46);
= 0+ s Sp
d 2
dt 2 2H(14.48)
The above equation is a linear, second‐orderdifferential equation.
If Sp positive – the solution (t)corresponds to that of simple harmonic motion.
If Spnegative – the solution (t) increases exponentially without limit.
26
Synchronizing PowerCoefficients
The angular frequency of the un‐damped oscillations is given by:
elec rad / s2H
s Sp
n =
(14.49)
which corresponds to a frequency of oscillation given by:
s p1 S(14.50)
Hz
27
n2 2H
f =
Synchronizing PowerCoefficientsExample:
The machine in previous example is operating at = 28.44when it is subjected
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.
S p = 2.10 cos 28.44=1.8466The synchronizing power coefficient is
The angular frequency of oscillation istherefore;
elec rad / s2H 25
n
377 1.8466=8.343=
s S p =
The corresponding frequency of oscillationis nf =8.343
= 1.33 Hz2
and the period of oscillationis T =1
= 0.753 s f n
28
Equal‐Area Criterion ofStability
The swing equation is non‐linear in nature and thus, formal solution
cannot be explicitlyfound.
Pa = Pm − Pe perunitdt 2
2H d 2=
s
29
To examine the stability of a two‐machine system without solving the
swing equation, a direct approach is possible to be used i.e using equal‐
area criterion.
Equal‐Area Criterion ofStabilityConsider 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 the
electrical power output from the generator becomes zero until fault is clear.
three‐phase fault
30
Equal‐Area Criterion ofStability
To understand the physical condition before, during and after the fault,
power‐angle curve need to beanalyzed.
Initially, generator operates at synchronous speed with rotor angleof 0
and the input mechanical power equals the output electrical powerPe.
Before fault, Pm = Pe
31
Equal‐Area Criterion ofStability
At t = 0, Pe = 0, Pm = 1.0puAcceleration constant
The difference mustbe 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. 1.0pu
For t<tc, the acceleration is constantgiven
by:
− 0 perunit2H d
= P
At t = 0, three phase faultoccursd 2 = s P (14.51)
m
s dt
m
32
dt 2 2H
Equal‐Area Criterion ofStability
Acceleration constantWhile the fault is on, velocity increase
above synchronous speed and can be
found by integrating thisequation:
dP ts
m
s
mP dt == t
dt 0 2H 2H
For rotor angularposition,;
(14.52)
(14.53)
At t = 0, three phase faultoccurs
+
(14.52)33
=s Pm t2
4H 0
Equal‐Area Criterion of Stability
Acceleration constantEq. (14.52) & (14.53) show that the
velocity of the rotor increase linearly
with time with angle move from0 to c
At the instant of fault clearing t =tc,
the increase in rotor speedis
ct=tc
d=
s Pmt dt 2H (14.54)
At t = 0, three phase faultoccurs
angle separation between thegenerator
and the infinite bus is
(14.52)34
Ps m
ct=tc(t) = t +
4H
2
0(14.55)
Equal‐Area Criterion ofStability
When fault is cleared at c ,Pe
increase abruptly to pointd
At d, Pe > Pm , thus Pa is
negative
Rotor slow down as Pe goes
from d toe
At t = tc, fault iscleared
35
Equal‐Area Criterion ofStability
1. At e, the rotor speed is again
synchronous although rotor angle has
advance to x
2. The angle is determined by thex
fact that A1 = A2
3.The acceleration power at e is still
negative (retarding), so the rotor
cannot remain at synchronous speed
but continue to slowdown.
4.The relative velocity is negative and
the rotor angle moves back from point
e to point a, which the rotor speed is
less than synchronous.
5. From a to f, the Pm exceeds the Pe
and the rotor increase speed again until
reaches synchronous speed atf
6. In the absence of damping, rotor would
continue to oscillate in the sequence f‐a‐e,e‐a‐
f,etc 48
Equal‐Area Criterion ofStability
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
undertransient condition by solving swing equation.
Equal‐area criterion not applicable formulti‐machines.
The swing equation for the machine connected to the infinite bus is
(14.56)s
= Pm −Pedt 2
2H d 2
Define the angular velocity of the rotor relative to synchronous speed by
r s =d
= − (14.57)dt
37
Equal‐Area Criterion ofStability
Differentiate (14.57) with respect to t and substitute in(14.56) ;
(14.58)2H dr
= Pm − Pe
s dt
When rotor speed issynchronous, equals s andr is zero.
Multiplying both side of Eq. (14.58) by r = d / dt ;
d
dt dts
r
r
m e= (P − P )d
H2 (14.59)
The left‐hand side of the Eq. can be rewritten to give
(14.60)dt
38
de− P )m
s dt
H d (2 )
r = (P
Equal‐Area Criterion ofStability
Multiplying by dt and integrating, weobtain;
(14.61)
2
22 (
− ) = (P − P )dH
m er1r2
s 1
Since the rotor speed is synchronous at
39
= 0 ;1 and 2 ,then r1 = r2
Under this condition, (14.61)becomes
(14.59)Pm e
( − P )d = 01
2
1 and 2 are any points on the power angle diagram provided that there are
points at which the rotor speed is synchronous.
Equal‐Area Criterion ofStability
In the figure, point a and e correspond to1 and 2
If perform integration, in twosteps;
c x
(Pm − Pe ) d + (Pm − Pe ) d = 0
0 c
(14.63)
c x
(Pm − Pe ) d = (Pe − Pm )d0 c
(14.64)
Faultperiod
Area A1
Post‐faultperiod
Area A2
The area under A1 and A4 are directly proportional to
the increase in kinetic energy of the rotor while it is
accelerating.
The area under A2 and A3 aredirectly proportional tothe decrease in kinetic energy of the rotor while it is
decelerating. 52
Equal‐Area Criterion ofStability
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 speed.
The shaded area A1 is dependent upon the time takento
clear the fault.
If the clearing has a delay, the angle c increase.
As a result, the area A2 will also increase. If the increase
cause the rotor angle swing beyond max , then the rotor
speed at that point on the power angle curve is above
synchronous speed when positive accelerating power is
again encountered.
Under influence of this positive accelerating power the
angle will increase without limit and instabilityresults.
41
Equal‐Area Criterion ofStability
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 anglecr
The corresponding critical time for removing the fault is called critical
clearing timetcr
Power‐angle curve showing the
critical‐clearing angle cr . Area A1
and A2 are equal
42
Equal‐Area Criterion of Stability
The critical clearing angle cr and critical clearing time tcr can be calculated
by calculating the area of A1 andA2.
m m cr
cr
1A =
0P d = P ( − )0
(14.65)
( m2A = crmaxP sin − P )dmax (14.66)
= Pmax (cos cr − cos max ) − Pm (max − cr )
Equating the expressions for A1 and A2, and transposing terms, yields
43
(14.67)cos = − + coscr Pm / Pmax max 0 max
Equal‐Area Criterion ofStability
From sinusoidal power‐angle curve, we seethat
(14.68)max 0= − elec rad
Pm = Pmax sin0(14.69)
Substitute max and Pmax in Eq. (14.67), simplifying the result and solving for cr
cr= cos−1 ( − 2 )sin cos
0 0 0 (14.70)
In order to get tcr, substitute critical angle equation into (14.55) and then solve to
obtain tcr;
cr = s m P
crt +4H
2
0 cr
44
t =s m
4H(cr − 0 ) P
(14.71) (14.72)
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/MVA
Solution
The power angle equation is Pe = Pmax sin = 2.10 sin
The initial rotor angle is0 = 28.44 = 0.496 elec rad
0
and the mechanical input power Pm is 1.0 pu.Therefore, the critical angle is
calculated using Eq.(14.70) cr ( )sin − cos −1
0 0 0 = cos − 2
cr = cos (− 2 0.496) sin 28.44 − cos 28.44 = 81.697 = 1.426 elec rad−1 0 0 0
and the critical clearing timeis
4 5(1.426 − 0.496)cr
45
t =377 1
= 0.222s
Further Application of the Equal‐AreaCriterion
• Equal‐Area Criterion can only be applied forthe case of two machines or one machineand infinitebus.
• When a generator is supplying power to aninfinite bus over two parallel lines, openingone of the lines may cause the generator tolose synchronism.
• If a three phase fault occurs on the bus onwhich two parallel lines are connected, nopower can be transmitted over either theline.
• If the fault is at the end of one of the lines, CB will operate and power can
flow through anotherline.
• In this condition, there is some impedance between the parallel buses and
the fault.Thus, some power is transmitted during the fault.
46
Further Application of the Equal‐AreaCriterion
Considering the transmitted of power during fault, a general Equal‐area
criterion is applied;
By evaluating the area A1 and A2 as in the previous approach, we can find that;
cr=
(Pm / Pmax )(max − o ) + r2 cos max − r1cos ocos (14.73)
2 1r − r
47
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 arer1Pmax sin = 0.808sin Pmax sin = 2.1sin
r2Pmax sin = 1.5sin
During fault:Before fault:
After fault:
Hence
1r =0.808
= 0.3852.1 2.1
2r =1.5
=0.714
1.5max
= 180 − sin−1 1.0= 138.19 = 2.412 rad
=(1.0/ 2.1)(2.412 − 0.496) + 0.714cos(138.19) − 0.385cos(28.44)
0.714 − 0.385cos cr
=0.127
cr = 82.726
To determine the critical clearing time, we must obtain the swing curve of versus t for thisexample.
48