This PPT shows how to use equal area criterion to analyze trasient stability for various cases like sudden rise in mechanical input to the turbine, three-phase fault at generator bus and at the mid of the line etc.

1. Equal Area Criterion for Transient Stability
Assessment
Prof. V. B. Pandya
Asst. Prof. (EE)

2. Roadmap
 Concept of equal area criterion
 Transient stability assessment using
equal area criterion for sudden increase
in mechanical input to turbine
 First swing stability and instability for
sudden increase in mechanical input
 Fault at generator terminals – cleared
quickly
 Sudden short circuit of one of parallel
lines

3. Concept of Equal Area Criterion
• Single machine connected to infinite bus (SMIB) system
can be used to assess transient stability by means of
simple criterion known as Equal Area Criterion without
going for numerical solution of swing equation
• For a single machine connected to infinite bus system
2
2
1
( )
sin
a
m e a
s
e
L
d
V
P
d
P P where P Accelerating Power
dt M M
d
dt
E
P
X X



 
   
 

 
jXL
+
E’/δ
Pe
jXd’
V/0
Fixed (Infinite Bus)
Pm

4. Concept of Equal Area Criterion
• If system is unstable,  continues to increase indefinitely
with time and machine loses synchronism.
• On the other hand,  performs oscillations whose amplitude
decreases in actual practice because of damping. These
two situations are shown in below fig.
• It can be visualized that for a stable system, indication of
stability will be given by observation of the first swing,
where  will go to maximum and will start decreasing.
This fact can be stated as a stability criterion.

5. Concept of Equal Area Criterion
• Thus criterion for stability mathematically
0 .
0 .
d
i e system is stable
dt
d
i e system is unstable
dt




• This leads to concept of equal area criterion as follows:
0
0 0
2
2
2
2
2
1
( )
1
2 2 ( )
2
( )
2 2
( )
a
m e
m e
m e
m e a
P
d
P P
dt M M
d d d
P P
dt dt dt M
Integrating above
d
P P d
dt M
d
P P d P d
dt M M


 
 

  



 
  
    
 
 
 
 
   

 

6. Concept of Equal Area Criterion
• Here δ0 is initial rotor angle before it begins to swing due to perturbation
0 0
0
2 2
( ) 0
0
m e a
a
For stable system
d
P P d P d
dt M M
P d
 
 



 

   
 
 

• The condition of stability can be stated as:
 “The system is stable if the area under Pa (accelerating power)-
curve reduces to zero at some value of power angle”.
 In other words, the positive (accelerating) area under Pa- curve
must equal the negative (decelerating) area and hence the name
“equal area” criterion of stability.

7. Concept of Equal Area Criterion
• Ans:- Synchronous machine relative speed
– The equation gives relative speed of
machine with respect to the synchronous
revolving reference frame
– If stability of system needs to be maintained,
the relative speed equation must be zero
sometimes after the disturbance
 
2
o
o
m e s
f
d
P P d
dt H




  
   

• What does dδ/dt signify?

8. Sudden Change in Mechanical Input
1 2
0 1
1 1 2 1
1 2
( ) , ( )
m e e m
A P P d A P P d
A A
 
 
 
   

 
Pe
0 0
e m
P P

A
δ
δ0
Π
• Consider an SMIB system with machine operating at A (δ0,Pe0)
• Increase Pm0 suddenly to Pm1.
1
m
P
δ2
B
δ1
• Pm1> Pe0 =>Pa increases i.e dω/dt = (Pm-Pe)/M =>ω increases from ωs and
since ω> ωs and dδ/dt = ω- ωs => δ0 increases to δ1 and hence
Pe = E’ V sin δ /(Xd’+XL)=>Pe increases to point B from A.
• At B Pm1 = Pe now so Pa=0 and rotor starts decelerating such that ω stops
increasing but still ω> ωs so δ1 further increases till δ2 at C where ω = ωs i.e.
dδ/dt = 0.
C
A2
•As the generator continues to slow, dδ/dt
becomes negative due to ω < ωs. Since ω<
ωs, δ (δ2) decreases towards B being first swing
stable! Here Area A1=A2 i.e.
State point will overshoot point
B (where Pa=0) also due to
rotor inertia towards A &
oscillates around B
2
0
0
a
P d


 


9. First swing stability- Qualitative Behavior
Pm0=Pe0
Pe
δ
δ0
A
B C
Time
δ
First swing
Stable at δ2
δ1
δ0
0
0
δ1
Stable Equilibrium
δ2
s
d
dt

 
 

10. First swing stability- Qualitative Behavior
Unstable Equilibrium
max 1
  
 
•Any further increase in Pm means the area available for
A2 (decelerating) is less than A1 (accelerating) so that the
excess kinetic energy causes rotor angel δ to increase
beyond point C and decelerating power becomes
accelerating power resulting into instability of system.
Pe
0 0
e m
P P

A
δ0
Π
1
m
P
δ2 = δmax
B
δ1
C
ω-ωsyn
δ
Time
A2

11. Fault at generator terminals – cleared quickly
A, AB : Fault occurs Pm>Pe ω ↑ δ↑ (δ0 → δ1)
B, CD : Fault Cleared Pm < Pe, ω > ωsyn, dδ/dt ↓ but δ↑(δ1 → δ2)
D Pm<Pe ω =0 ↓ δ mom. constant
DE : Pm<Pe ω < ωsyn ↓ δ ↓ Swings back!
Pm=constant
δ
P
δ0
D
Pe during fault, from A to B
A B
C
E
Pe
∞
Pm
δ1 δ2
Pe - pre and post fault
Pmax
A1
A2

12. Fault at generator terminals – Critical clearing angle(δcr)
and critical clearing time(tcr)
If clearing is delayed such that δ2 becomes δmax as shown
where A1=A2. Any further delay in clearing fault results into A1>A2
making system unstable.
Pm=constant
Pe - pre and post fault
δ
P
δ0
Pmax D
Pe during fault, from A to B
A B
C
E
δcr δmax
A1
A2

13. Derivation of δcr and tcr when Pe=0 during fault
   
 
   
 
 
0
max
max 0
max 0
1 0
2 max
2 max max max
1 2
max 0 max
max
max 0 max 0
0
sin
0
sin
cos cos
cos cos (1)
sin (1)
cos 2 s
cr
cr
e
m m cr
m
cr m cr
m
cr
m
cr
P P
A P d P
A P P d
A P P
for stability A A
P
P
put and P P in




  

  
 
   
   
   
  
 

   
 
   

       
  
  


0 0
in cos
 

 
 

14. Derivation of δcr and tcr when Pe=0 during fault
 
2
2
2
0
2
0
0
, .
1
0
int .
1
2
1
2
2
m e
m
cr m cr
cr
cr
m
During fault swing equ
d
P as P
dt M
egrating twice above equ
P t
M
P t
M
M
t
P

 
 
 
 
 
  

 

15. Sudden short circuit of one of parallel lines
Case -1 : Fault at generator end of line

16. 1
1 2
1
1
'
( ) Pr sin
||
( ) 0
'
( ) sin
eI
d
eII
eIII
d
E V
i efault P
X X X
ii During fault P
E V
iii Postfault P
X X







A1
A2

17. Case -2 : Fault away from generator end on line (middle)

18. max
0
1
1 2
1
max
1
1 max
1 max 2 max
' '
( )Pr , sin ( ) , sin
||
'
( ) , sin ... sin
( sin ) , ( sin )
cr
cr
eI eII
d II
m
eIII
d III
m II III m
E V E V
i efault P ii During fault P
X X X X
P
E V
iii Postfault P and
X X P
A P P d A P P d
 
 
 
  
   

 

  

   
 
δmax

19. Derivation of δcr
max
0
max max
0 max 0
max max max
cos cos 0
( ) (cos cos )
0
(cos cos ) ( )
cr
cr
m II III m
m cr II cr
III cr m cr
P P P P
OR
P P
P P
 
 
   
   
   
   
  
 

 
   
 
   
max max 0 max max
max max
( ) cos cos
180
cos
m cr III III
cr
III II
P P P
P P
    

  



20. Case -3 : Reclosure

21. If CB of line 2 are successfully reclosed,
power transfer once again becomes
Case -3 : Reclosure
1
1 2
max
'
sin
||
sin
eIV
d
I
E V
P
X X X
P





Since reclosure restores power, changes of
stable operation improves.

22. Case -3 : Reclosure
1
1 max
max
sin m
I
P
P
   
  
For critical clearing angle,
0
max
max
max max
( sin )
( sin ) ( sin )
,
&
cr
rc
cr rc
m II
III m I m
rc cr
P P d
P P d P P d
where
t t
Time between clearing reclosure


 
 
 
   


 
  
 


 

23. Example
• A generator operating at 50 Hz delivers 1
pu power to an infinite bus through a
double circuit transmission line in which
resistance is neglected. A fault takes
place reducing the maximum power
transfer to 0.5 pu whereas before fault,
this power was 2.0 pu. and after fault
clearance, it is 1.5 pu. By equal area
criterion determine the critical clearing
angle.

24. Example
0
max max max
1 ,
2, 0.5, 1.5
m e
I II III
P P initial loading
P P P
 
  
1
0
max
sin 0.532
m
I
P
rad
P
 
 
1
max
max
sin
2.41
m
III
P
P
rad
  
 


25. Example
   
max max 0 max max
max max
( ) cos cos
180
cos
1.0(2.41 0.523) 0.5cos0.523 1.5cos2.41
1.5 0.5
0.337
70.3
m cr III III
cr
III II
P P P
P P
rad

    

  


  



