Al-Balqa Applied

University

Transient Stability Dr audih 0

 Part III
Stability of
power systems
Dr Audih 1
Power system stability refers to the ability of synchronous
machines to move from one steady-state operating point following
a disturbance to another steady-state operating , without losing
synchronism. There are three types of power stability:
1-Steady-state: involves slow or gradual changes in operating due to
variation on loads.
2-Transient: Involves major disturbances such as( loss of
generation due to faults and sudden load changes).This case
deviations from synchronous frequency (50 Hz) ,and changing
the machine power angles( our subject) .In many cases,
transient stability is determined during the first swing of
machine power angles following a disturbance( typically 1s).
3-Dynamic: After a fault, an individual synchronous machine may
remain in synchronism during the first swing . After this, strong
electromechanical oscillations occur may cause the system to have
natural oscillations. The system is said to be dynamically
Dr Audih
stable . 2
 Classification of stability
Classification is based on the following considerations:
physical nature of the resulting instability
size of the disturbance considered
processes, and the time span involved
Power
system
stability

3
 Usefully Formula
o o
2 P 
1o mechanical    electrical 1 electical    mechancal
o

p 2
P  No. of poles P
One revolution   360o elect.deg.
(mechanical )  2 n ' (mech.rad / sec) 2
o
2
n  speed in revolution / sec.(rps ) or 1 electical     360o (revolution )
' o

P 
2 .n
  2 
o
1
60 1 o elec . / sec     (rev / sec)
o
n  speed in revolution / min .(rpm )  P  360
o
(electrical)  Angular frequancy 2  1
1 elec . / min     
1 

o
o
(electrical)  2 f (elect .rad / sec)   
P 360 60 
o
(electrical)  360 f (elect .deg / sec )
o
2 60
  
 P  360
Dr Audih 4
 Rotation Mechanics-vs-Linear Mechanics
Rotational mechanical is very important in stability studies. The following table show us
comparison between the linear and rotational mechanical formula.
R otational m echanics L in ear m ec h an ic s

  A ngular deplacem ent ( rad .)

S = D ep alcem en t(d istan ce)(m )
d
  A ngular v elocity ( rad / sec)  ds
dt V  V elo city ( m / sec) 
dt
  A ngular accelaration ( rad / sec 2 )
2

a accelaratio n ( m / sec)
d  d 2
  dv d 2
dt dt 2 mv  
J  M om ent of inertia ( k g .m 2 ) dt dt
m  M ass ( k g )
T = T orque (N .m ) = J.
M  A ngular m om entum F  Fo rce  m .a ( N )
( k g .m 2 .rad / sec)  J  M  M o m e n tu m ( k g .m / sec) = m .v
1 1
K .E  J  2 (Joule) K .E .  .m .v 2
( Jo u le )
2 2
P  pow er (w att )  T  d
P = ( F .s )  F .v ( N .m / s )
d dt
W  w ork   Pe d    T T 
dt W o rk = F.s
Dr Audih 5
Transient stability limit: depends on the type of disturbance,
location and magnitude of disturbance, clearing time and the
method of clearing.
After every disturbance, the machines must adjust the relative
angles of their rotors to meet the condition of the power transfer
involved. The problem is mechanical as well as electrical.
The kinetic energy(KE) of an electric machine is given by:
1 M
KE = J . 2 ,  J  , w here J 
2 
1
KE  M  m ech
2
w h e re :
J -Is t h e M o m e n t o f In e r t i a i n ( k g .m 2 )
M -Is t h e a n g u la r M o m e n t u m i n J , a n d d e p e n d s o n t h e
s iz e o f th e m a c h in e a n d its ty p e .
 r a d m ech 
 -Is t h e r o t o r a n g u la r v e lo c i t y i n  
 sec . 
 P   r a d elect 
 e le c t     m e c h a n ic a l r o to r s p e e d in   o r ( 2 f ), H z
 2   sec . 
Dr Audih 6
The Inertia constant H is defined as the Mega Joules of stored energy
of the machine at synchronous speed per MVA of the machine.
K E (energy stored in rotor in M J )
H 
S ( M achine pow er rating in M V A )
The relation between the Angular Momentum M and the Inertia
constant H can be derived as follows.
1 1
K E  S .H  M  e l e c t ri c a l  M 2 f  M . f
2 2
K E  M . f  M . . f , J 
or
KE S .H Jo u le
M   , ( )
  f   f ra d e l e c t s e c
If th e p o w e r is e x p re s s e d in p e r u n it, th e n S = 1 .0 p u . a n d
H
M =
180o  f

Dr Audih 7
 SWING EQUATION : The differential
equation that relates the angular momentum
(M) the acceleration power (Pa) and the rotor
angle (δ) is known as swing equation.

Consider the generator shown in Fig.(a). It receives mechanical

power Pm at the shaft torque Tm and the angular speed ω via.
shaft from the prime-mover. It delivers electrical power Pe to the
power system network.
The generator develops electromechanical torque Te in
opposition to the shaft torque Ts. At steady state, Tm = Te
Dr Audih 8
 accelerating torque acting on the rotor is given by
Ta = Tm – Te
Multiplying by ω on both sides, we get power acceleration (Pa) :
Pa = Pm – Pe
In case of motor Ta = Te – Tm and Pa = Pe – Pm
Pa  T a .  J . .  M .
d  d 2
  = 2  angular acceleration
dt dt
d d 2
Pa  M .  M . 2 ,(   angular displacem ent of rotor ( rad .)
dt dt
d 2
 Pa  M . 2 , (   it ' s m ore convenient to m easure the angular position
dt
of rotor w ith respect to synchrou ously rotation fram e
t
(      t ) of reference  t  0  (   ) th e n ;
d 2 d 2 d 2
  Pa  M . 2
dt 2 dt 2 dt

Dr Audih 9
 d 2 S .H d 2 
Pa  Pm  Pe  M . 2  . K now n as sw ing equation or in per un it
dt 180 o .f dt 2
d 2 H d 2
Pa  M . 2  . or The swing equation is the differential equation that relates the
dt 18 0 o .f dt 2
angular momentum (M) the acceleration power (Pa) and the rotor
H d 2 2 H d 2 angle (δ).
Pa  .  .
   dt 2  dt 2
 
 2 
In case damping power (D) is to be P  P  P  M . d 2   D d 
a m e
included, then equation is dt 2 dt
Swing curve, plot of power angle δ vs time t, can be obtained by
solving the swing equation. Fig. (a) stable system and (b) unstable
system.

Dr Audih 10
Swing curves are used to determine the stability of the system. If
the rotor angle δ reaches a maximum and then decreases, then it
shows that the system has transient stability (figure-a). On the
other hand if the rotor angle δ increases indefinitely, then it shows
that the system is unstable(figure-b).
We are going to study the stability of a generator connected to
infinite bus and a synchronous motor drawing power from
infinite bus.
We know that the complex power is given by:
S=P + j Q = V. I* or P – j Q = V*.I
Thus real power P = Re {V*I}
Case 1-: Consider a generator connected to infinite bus.
Note A system having a constant voltage and constant frequency regardless of the load is called
an Infinite Busbar system. Thus, an infinite bus has a large power system and does not affect its
voltage and frequency. They both remain constant.
Dr Audih 11
Where:
V - is the voltage at infinite bus. V = V  0 o
E - is internal voltage of generator. E  E  o
X - is the total reactance
The internal voltage E leads V by angle δ, Thus:

V + j X I = E (for generator)

E  V E   V 0
o o
1
I
jX

jX

jX
 E .cos  j E .sin    V 
Dr Audih 12
 E .V
Electric output power Pe  ReV .I   .sin   Pmax .sin 
X

Case 2:- Consider a synchronous motor drawing power from

infinite bus

V – j X I = E (for motor)

Internal Voltage E lags the voltage V by angle δ. Thus E  E  

o

V  E V 0  E   
o o
1
I
jX

X 90o

jX

V   E .cos   j E .sin   
Dr Audih 13
As result of swing equation for generator is
d 2
Pa  Pm  Pe  Pm  Pm ax .sin   M . 2
dt
d 2
M . 2  Pm  Pmax .sin  swing equation for generator
dt
an d
d 2
M . 2  Pmax .sin   Pm for m o t o r
dt

Notice that the swing equation is second order nonlinear

differential equation since the power acceleration have sine term.

Dr Audih 14
The graphical plot of power angle equation in Pe  Pmax .sin 
for generator and motor is shown Fig.

Pe Pe  Pmax .sin 

Dr Audih 15
EXAMPLE 1: Generator per-unit swing equation and power
angle during a short circuit.
A three-phase, 60-Hz, 500-MVA, 15-kV, 32-pole hydro-electric
generating with H constant of 2.0 p.u. and D=0.

(a) Determine ωe and ωm in steady state conditions.

(b) Give the per-unit swing equation for this unit.
(c) The unit is initially operating at pm(pu)=pe(pu)=1 :ω=ωsyn,
and δ=10 when a three-phase-to-ground bolted short circuit at
the generator terminals causes pe(pu) to drop to zero for
t ≥ 0.Determine the power angle 3 cycles after the short circuit
commences. Assume pm(pu) remains constant at 1.0 per unit.
Also assume ωpu(t)=1 in the swing equation.

Dr Audih 16
S o lu ti o n :
( a )  e  2  f  2   6 0  3 7 7 rad / s
 P 
sin c e e    . m t h e n ;
 2 
 2   2 
m    . e     3 7 7  2 3 .5 6 rad / s
 P   32 
( b ) T h e p e ru n it s w i n g e q u ati o n f o r H = 2 p .u .
H d 2 2 d 2
Pm ( pu )  Pe ( pu )  . 2  . 2
o
180 . f dt 180  60 dt
o

(c ) Intial power angle is  (0)  10 o  0.1745 rad .

d  (0)
for t  0  0
dt
Dr Audih 17
Using Pm(pu) = 1 , Pe(pu) = 0 and (pu) (t )  1
H d 2 2 d 2 (t )
then the swing equation ( pu ) is .  .  Pm  Pe  1 , t  0
180o.f dt 2 180o  60 dt 2
d 2 (t ) 1
 Integrating twice and using the above intial conditions,
dt 2
 2 
 o 
 180  60 
d  (t )  180o  60 
we get from first integral for t :  t  0 
dt  2 
 180o  60  t 2 
And from second integral  (t )   .    (0)
 2  2
3 cycles
at t  3 cycle   0.05 second
60cycle / second
 180o  60 
 (0.05)     0.05  0.1745  0.2923rad  16.75
2 o

 4 
Dr Audih 18
Example 2:

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has

an inertia constant of 8.0 MJ/MVA.
Find:
(a) The stored energy in the rotor at synchronous speed.
(b) If the mechanical input is suddenly raised to 80 MW for an
electrical load of 50 MW, find rotor acceleration, neglecting
mechanical and electrical losses.
(c) If the acceleration calculated in part (b) is maintained for 10
cycles, find the change in torque angle and rotor speed in
revolutions per minute(rpm) at the end of this period.
Solution:
(a) Stored energy = S.H = 100 × 8 = 800 MJ

Dr Audih 19
 d 2
b  Pa  80  50  30 M W  M .
dt 2
S .H 100  8 4 M J .s
M    ,( )
1 8 0 o .f 180  50 45 e l e c .d e g re e
d 2 4 d 2
 Pa  M .  .  30 
dt 2 45 dt 2
d 2 30  45 e le c t d e g .
 2
  3 3 7 .5 , ( )
dt 4 sec .
d 2 e le c t d e g .
A n d s in c e   2
   3 3 7 .5 , ( )
dt sec .

cycles 10
( c ) tsec.    0.2 sec.
freq. 50
And changing in   is :
1
  . .( t ) 2 where   is tourqe angule and  is acceleration
2
1
     337.5  (0.2) 2  6.75 elect .deg .
2
Dr Audih 20
 rad 1
  in rpm angular velocity is  N m  m ( )  rpm / s
sec 2 (rad/ rev.)
2 1
for electrical in min .  e  (rad / sec)  o
 60  rpm / min
P 360 (rad / rev .)
2 1 60
for 4 pols  2  f    60  314.16   26.18rpm / min
4 360 o
2  360 o

120f
At the end of 10 cycle rotor speed is  + ns .t s 
P
120  50
  26.18  0.2  1505.23 rpm
4
**Note: We can also identify the mechanical speed of rotation in
rpm according to

Dr Audih 21
3

Dr Audih 22
 P
I   1.25
V .cos

b) Since we need Pe for

swing equation

Dr Audih 23
, Where Pi=Pm

Dr Audih 24
Dr Audih 25
Dr Audih 26
Dr Audih 27
 Further Considerations of the Swing Equation
The mega-voltampere (MVA) base Smach ,which is introduced
by the definition of H, with many synchronous machines only
one MVA base common to all parts of the system can be
chosen. In this case:
 S m ach 
H system  H m ach .  
S
 system 
For a large system with many machines over a wide area it
is desirable to minimize the number of swing equations to be
solved. In such cases the machines within the plant can be
combined into a single equivalent machine , and only one
swing equation must be written for them.
Consider a power plant with two generators connected to the
same bus which is electrically remote from the network
disturbances. Dr Audih 28
The swing equations on the common system base are
H 1 d 21
.  Pm1  Pe1 ( pu )
180o.f dt 2
H 2 d 22
o
. 2  Pm 2  Pe 2 ( pu )
180 .f dt
1   2   (since the rotors swing together ),then;
H equ . d 2 H equ . d 2 H equ . d 2 2H equ . d 2
. 2  . 2  . 2  . 2  Pm  Pe ( pu )
o
180 .f dt  .f dt     dt  dt
 
 2 
where : H equ . = H 1 + H 2 , Pm  Pm1  Pm 2 and Pe  Pe1  Pe 2
Machines swing together called coherent machines

Dr Audih 29
Example:
Tow synchronous machine the first one is 400MVA and H1=4.6MJ/MVA and
the second is1200MVA with H2=3MJ/MVA. The two machine operate in
parallel in power plant. Find Heq relative to a 100MVA base

K E  S .H

S 
H sys .   mach .  .H mach .
 S 
 sys . 

Dr Audih 30
Equal area criterion( alternative method for stability)
The accelerating power in swing equation will have sine term.
Pe  Pmax .sin 
Therefore the swing equation is non-linear differential equation .
For two machine system and one machine connected to infinite
bus bar, it is possible to say whether a system has transient
stability or not, without solving the swing equation. Using of
equal area in power angle diagram from sine wave which is
known as EQUAL AREA CRITERION.
A1  A2
P electrical

P mechanical

Dr Audih 31
Let us to find the condition which dδ/dt =0. The swing equation
for the generator connected to the infinite bus is: P  P  P  M . d 
2

a m e
dt 2
Multiplying both sides by 2dδ/dt , we get
d d 2 d 
2  Pm  Pe  .  2M . 2 .  m athm atically
dt dt dt
d
d 2  P m  Pe  .
d  
2 2
d d t  d . d  
2  P m  Pe  .  M . .    
dt  d t  d t   M dt  dt 
dt
m u ltip ly in g b o th s id e s b y w e get ;
d
2  P m  Pe  2 2
d  d   dt d d 
 .  .  .  or
M dt  dt  d  d   dt 
2  P m  Pe  d 
2
d 
   . or
 dt  M d

 2  P m  Pe  
2
d 
  
 dt 

o

 M
 .d 

Dr Audih 32
 2
 d  
d  
2 P m  Pe  
 2 P m  Pe   .d 
2
 dt   d  
M

d 
 
 dt 
 


o

 M 


Before the disturbance occurs,δ0 was the torque angle (initial

power angle). At that time dδ/dt= 0.
When disturbance occurs, dδ/dt ≠zero and δ starts changing.
Torque angle δ will change and the machine will again be
operating at synchronous speed after a disturbance, when back to
dδ/dt = 0 or

d 2 (P m - Pe )
dt
 
o M
.d   0

Note: The area above Pm is area 2 and down is area 1

Dr Audih 33
But for the area of accelerating
power Pa versus δ must be zero Pm
for some value of δ, the positive
(or accelerating) area A1 under
the graph must be equal to the Note:
negative area A2 ( decelerating). A1 is less then Pm.
A2 is greater than Pm

(a) corresponding to the initial steady-state operating point. At this

point, Pm0 = Pe0 and δ =δo ( prefault)
(b)When a sudden fault accrue Pa becomes positive the rotor
accelerates and the power angle δ1 begins to increase
At this point Pm = Pe and δ =δ1 (on fault)
(c)After the fault is cleared Pa is negative and δ reaches a
maximum value δ2, or point (c) and then swing back towards (b)
At this point Pm = Pe and δ =δ2 (pos tfault)
Dr Audih 34
which is ultimate steady-state stable operating point .In
accordance with eqn.
  

 P .d    P
a m  Pe .d    Pm  Pmax.sin .d  0
o
o o

Note: We have assumed that the machine is connected to a large power system so that
| Vt | and xd does not change field current maintains | Eg | constant.
The equal-area criterion requires that, for stability from figure
before , A1=A2 or
1 2

  P
m  Pmax .sin  .d     Pmax .sin   Pm .d  
1
o

 Pm (1  o )  Pmax (cos1  coso )  Pm (1  2 )  Pmax (cos1  cos2 )

But during th fault Pe = 0  Pm = Pmax sin1 then;
Dr Audih 35
 Pmax .sin1 ( 1 o )  Pmax (cos1  coso )  Pmax .sin1 ( 1 2 )  Pmax (cos1  cos2 ) 
sin1 (o )  coso  sin1 (2 )  cos2 ) or
This is stability equation
( 2  o )sin1  cos( 2 )  cos( 0 )  0 for equals area
Example: A synchronous generator supply 500MW per phase
with power angle =8o.By how much can the input shaft power
(Pm) be increased suddenly without loss of stability. Assume
Pmax remain constant
Solution :
Initially at  0  8o ,the power
Pe (0)  Pmax sin  0  500sin 8o  69.6MW
Let δm is power angle to which the rotor can swing before losing
synchronism. If this angle is exceeded Pm will again become
greater than Pe, and the rotor will once again be accelerated and
synchronism will be lost as shown in Fig.
Note: The area
above Pm is
area 2 and
down is area 1

Pm Pe

therefore, the equal area criterion requires that equation to be

satisfied with δmax replacing δ2.thus:
Dr Audih 37
From stability equation for equals area (   )sin  cos(   )  0 2 o 1 2 0

Where      (this the change in  for sability problem, replacing we get ;

2 1

(     )sin  cos(     )  0 Substituation   8  0.139 rad 

o

1 o 1 1 0 o

(  0.139   )sin  cos(  0.139   )  0

1 1 1   3.14

(3   )sin  cos(3   )  0
1 1 1

Solving eqn. iteratively , taking into acount

 cos(90   )  sin  
Area of
stability
1 1 problem

we get ;   50 .Now P  P

1 efect a

P  P .sin   δ1 δ2
π
efect max 1

 500sin(50 )  383.02MW
o

Dr Audih 38
 The machine was 69.6 MW. Hence, without loss of stability, the
system can accommodate a sudden increase of
Pef  Pe (0)  383.02  69.6  313.42 M W / phase or
 3  313.42  940.3M W / T hree phase of input shaft power

A1=A2 A1≈A2 A1≠A2

CRITICALLY
STABLE STABLE
UNSTABLE

Fig. shows three different cases: case (a) is STABLE. Case (b)
indicates CRITICALLY STABLE while case (c) falls under
UNSTABLE. Note that the areas A1 and A2 are obtained by
finding the difference between INPUT and OUTPUT.(Pm=Pe)
Dr Audih 39
* If a fault occurs in a system, δ begins to increase under the
influence of positive accelerating power, and the system will
become unstable.
* if δ becomes very large. There is a critical angle within which
the fault must be cleared. This angle is known as the clearing
angle.
Consider a system in figure operating with mechanical input Pm at
steady angle δ0 (Pm=Pe) at point 'a' on the power angle diagram
shown in the diagram in the next slide this is steady state.

Pm
If a three phase short circuit decrease power
occurs at the point F of the due to fault

outgoing radial line, the

terminal voltage goes to zero
and electrical power output
of the generator reduces to increase power
due to fault
zero(Pe = 0) and the state
point drops to 'b'.
The acceleration area A1 starts to increase while the state point
moves along bc. At time tc corresponding clearing angle δC, the
fault is cleared by the circuit breaker. tc is called clearing time
and δC is called clearing angle. After the fault is cleared, the
system becomes healthy and transmits power Pe = Pmax sinδ , the
state point shifts to "d" on the power angle curve. The rotor now
decelerates and the decelerating area A2 begins to increase while
the state point moves along de ( to maximum). Dr Audih 41
 For stability, the clearing angle, δC, must be such that area A1 =A2.
c
A 1  A 2 W h e re ; A1    P m  Pe .d 
0

A n d sin ce th e fau lt is clea re d at a rea A 1 .

T h en  Pe  0   A 1  P m (  c   o ) an d
1
A2     P m  Pe .d  (n e g ativ e s in c e p o w er d eaccelerat ed )
c

A 1  A 2 th u s ; 
1 

P m (  c   o )    Pe  P m .d   
c

 Pe  P m ax .sin  

1
P m ( c   o )    P m ax .s in   P m .d 
1

c Dr Audih 42
 1 1 1
   Pmax .sin   Pm .d     Pmax .sin  .d    Pm .d  
c c c

Back to equation we get ;

1 1
Pm (c  o )    Pmax .sin  .d    Pm .d  or
c c

Pm c  Pm o  Pmax   cos 1  cos c   Pm (1  c ) 

Pm c  Pm o  Pmax   cos 1  cos c   Pm 1  Pm c 
Pmax  cos c  cos 1   Pm 1  o 
At the intial Pinput = Poutput  Pm  Pe then 
 
Pm = Pmax .sino 
Dr Audih 43
Pm ax  cos  c  cos  1   Pm ax sin  o .  1   o  
 cos  c  cos  1     1   o  sin  o or
cos  c  cos  1   1   o  sin  o 
 c  c os  1  cos  1    1   o  sin  o 
W here :
 c  is clearing angle
 o  is intial pow er angle
 1  is pow er angle to w hich the rotor
ov ers hoots bey ond  c .
Dr Audih 44
In order to determine the clearing time, we rewrite the swing
equation in pu as:
H d 2
Pa  Pm  Pe  o
. 2 ( Pe  0, since we have a three phase fault )
180 . f dt
H d 2 d 2 H  . f .Pm
 Pm  . or  P 
 . f dt 2 dt 2
m
.f H
d 2  . f .Pm d   . f .Pm
  integrating   .t
dt 2 H dt H
 . f .Pm t 2
to find  the second integral    .  o
H 2
If t  tc and    c 

 . f .Pm 2 H  c   o 
c  .t   o
2
and tc 
2H
c
 . f .Pm

Dr Audih 45
The maximum allowable
value for the system to
remain stable are known as
critical clearing angle and
critical clearing time, from
the figure
 m    o and note
 m  1 in equation above x

and  x  o then from

 c  c o s  1  c o s  1   1   o  s in  o  w e g e t :

Dr Audih 46
 c r  c o s  1  c o s    o     o  o s in  o  
 c o s  1  c o s    o    s i n  o  
 2 o
 c o s  1  c o s   c o s  o    2  o  s i n  o  
 c o s  1  0  c o s  o    2  o  s i n  o  or

 cr  c o s 1    2 o s in  o  cos  o 
an d :

t cr 
4H  c r  o 
 .f . P m
W h ere :
 c r  i s c ri t i c a l c l e a r i n g a n g l e a n d
t cr  c ri t i c a l c le a ri n g t i m e .
Dr Audih 47
Example:-
50Hz, synchronous generator capable to supply 400MW
connected to a large power system and delivering 80MW when
three phase fault occurs at its terminals, determine:
(a) The time in which the fault must be cleared ,if the maximum
power angle is to be 85o
(b) The critical clearing angle if inertia constant is 7 pu

S o lu tio n :
8 5o  
W h en th e f au lt o ccu rs    1  8 5  o
 1 .4 8 3 5 rad .
1 8 0o
w e n eed to f in d th e in tial  o , th u s ;
Pe  P m ax .s in  o w h ere P m a x  4 0 0 M W
an d Pe  8 0 M W  P m  s tead y s tate  
Dr Audih 48
 Pe 80
T hen sin  o    0.2 
Pmax 400
11.54  
 o  sin (0.2)  11.54 
1 o
o
 0.201 rad . 
180
 o  11.54 o  0.20 1 radian
N ow to find  c , from e quation
 c  cos  1  cos  1   1   o  sin  o  
 cos  1  cos 1.4835   1.4835  0.201  sin  0.201   
 cos  1  0.0906 7   1 .48 3 5  0.201   0 .19867   

  c  cos  1 (0.345)  69.8o  1.21 8 rad .

Dr Audih 49
Sine Pe=80MW and before the fault Pe=Pm also in fault moment
pe=0 the Pm=80 ,if we take base of 100MW then Pm=0.8pu

The cleaning time t c is :

2 H  c   o  2  7 1.218  0.201
tc    0.336 sec .
 . f .Pm   50  0.8
(b ) For critical clearing angle  cr the equation is :
 cr  cos 1    2 o  sin o  cos  o  
 cos  1    2  0.2  sin (0.2)  cos(0.2)  
 cos  1 (  0.43)  115.46 o  2.01 rad .

Dr Audih 50
Dr Audih 51