Transient Stability

The ability of the power system to remain in synchronism

when subject to large disturbances
 Large power and voltage angle oscillations do not permit
linearization of the generator swing equations

Lyapunov energy functions
 simplified energy method: the Equal Area Criterion

Time-domain methods
 numerical integration of the swing equations
 Runga-Kutta numerical integration techniques

Power Systems I
Equal Area Criterion

Quickly predicts the stability after a major disturbance
 graphical interpretation of the energy stored in the rotating
masses
 method only applicable to a few special cases:
 one machine connected to an infinite bus
 two machines connected together

Method provides physical insight to the dynamic behavior
of machines
 relates the power angle with the acceleration power

Power Systems I
Equal Area Criterion

For a synchronous machine connected to an infinite bus
H d 2δ
= Pm − Pe = Paccel
π f 0 dt 2

d 2δ π f 0 π f0
2
= (Pm − Pe ) = ⋅ Paccel
dt H H

The energy form of the swing equation is obtained by
multiplying both sides by the system frequency (shaft
rotational speed)
 dδ   δ  π f0  dδ 
2
2
d
 2  = (Pm − Pe ) 2 
 dt  dt  H  dt 

Power Systems I
Equal Area Criterion

 d 2δ  dδ  π f 0  dδ 
2 2  = (Pm − Pe ) 2 
 dt  dt  H  dt 

The left hand side can be reworked as the derivative of
the square of the system frequency (shaft speed)

d  dδ   2π f 0
 dδ
2

  = (Pm − Pe )
dt  dt   H dt
 dδ  2  2π f 0
d   = (Pm − Pe )dδ
 dt   H

Power Systems I
Equal Area Criterion

Integrating both sides with respect to time,
 dδ  2π f 0
2
δ
  =
 dt  H ∫δ (P
0
m − Pe )dδ

dδ 2π f 0 δ

dt
=
H ∫δ (P
0
m − Pe )dδ

The equation gives the relative speed of the machine.
For stability, the speed must go to zero over time
dδ
=0
dt t →∞
δ
0 = ∫ (Pm − Pe )dδ
δ0
Power Systems I
Equal Area Criterion

Consider a machine operating at equilibrium
 the power angle, δ = δ0

 the electrical load, P

e0 = Pm0

Consider a sudden increase in the mechanical power
input
 Pm1 > Pe0 ; the acceleration power is positive

 excess energy is stored in the rotor and the power frequency

increases, driving the relative power angle larger over time
δ1
U Potential = ∫ (Pm1 − Pe )dδ > 0
δ0

dδ 2π f 0 δ

dt
=ω =
H ∫δ (P
0
m − Pe )dδ > 0

Power Systems I
Equal Area Criterion
 with increase in the power angle, δ, the electrical power
increases
Pe = Pmax sin δ
 when δ = δ1, the electrical power equals the mechanical power,
Pm1
 acceleration power is zero, but the rotor is running above
synchronous speed, hence the power angle, δ, continues to
increase
 now Pm1 < Pe; the acceleration power is negative (deceleration),
causing the rotor to decelerate to synchronous speed at δ = δmax
 an equal amount of energy must be given up by the rotating
masses
δ1 δ max
U Potential = ∫ (Pm1 − Pe )dδ − ∫ (Pm1 − Pe )dδ =0
δ0 δ1
Power Systems I
Equal Area Criterion

Pe

A2 e
Pm1 c A b
1

Pm0 a

δ
0 π
δ0 δ1 δmax
Power Systems I
Equal Area Criterion

The result is that the rotor swings to a maximum angle
 at which point the acceleration energy area and the deceleration
energy area are equal
δ1
∫δ (P
0
m1 − Pe )dδ = area abc = area A1
δ max
∫δ (P
1
m1 − Pe )dδ = area bde = area A2

area A1 = area A2
 this is known as the equal area criterion
 the rotor angle will oscillate back and forth between δ and δmax at
its natural frequency

Power Systems I
Equal Area Criterion - ∆P mechanical

Pe

A2
Pm1 c b d

A1

a
Pm2
δ
0 π
δ0 δ1 δmax
Power Systems I
 Equal Area Criterion - ∆P mechanical

δ1 δ max
Pm1 (δ 1 − δ 0 ) − ∫ Pmax sin δ dδ = ∫ Pmax sin δ dδ − Pm1 (δ max − δ 1 )
δ0 δ1

Pm1 (δ max − δ 0 ) = Pmax (cos δ 0 − cos δ max )

Pm1 = Pmax sin δ max
(δ max − δ 0 )sin δ max = cos δ 0 − cos δ max
→ Pm1 = Pmax sin δ 1
Function is nonlinear in δmax
Solve using Newton-Raphson

Power Systems I
3-Phase Fault

1 2

G inf

Power Systems I
Equal Area Criterion - 3 phase fault

Pe
f
e
A2
a d g
Pm

A1

b c δ
0 π
δ0 δc δmax
Power Systems I
Equal Area Criterion - 3 phase fault

δc δ max
∫δ Pm dδ = ∫
δ
( Pmax sin δ − Pm ) dδ
0 c

Pm (δ c − δ 0 ) = Pmax (cos δ c − cos δ max ) − Pm (δ max − δ c )

cos δ c =
Pm
(δ max − δ c ) + cos δ max
Pmax

Power Systems I
Critical Clearing Time

Pe cos δ c =
Pm
(δ max − δ c ) + cos δ max
Pmax
e
Pm
A2 cos δ max =
Pmax
a d f
Pm

A1

b c δ
0 π
δ0 δc δmax
Power Systems I
Critical Clearing Time

H d 2δ
= Pm − Pe = Pm ← Pe = 0
π f 0 dt 2

d 2δ π f 0
2
= Pm
dt H
dδ π f 0 t π f0
= Pm ∫ dt = Pm t
dt H 0 H
π f0
δ= Pm t 2 + δ 0
2H
2 H (δ c − δ 0 )
tc =
π f 0 Pm
Power Systems I
3-Phase Fault

1 2

G inf

Power Systems I
Equal Area Criterion

Pe Pe Pre-fault
f
Pe Post-fault
e
Pe during fault
A2
a d g
Pm
A1
c
b

δ
0 π
δ0 δc δmax
Power Systems I
Critical Clearing Time

Pe Pe Pre-fault
Pe Post-fault
e
Pe during fault
A2
a d
Pm f
A1
c
b

δ
0 π
δ0 δc δmax
Power Systems I
 Critical Clearing Time

δc δ max
Pm (δ c − δ 0 ) − ∫ P2 max sin δ dδ = ∫ P3 max sin δ dδ − Pm (δ max − δ c )
δ0 δc

Pm (δ max − δ c ) + P3 max cos δ max − P2 max cos δ 0

cos δ c =
P3 max − P2 max

Power Systems I