Document

Stability
l
The ability of the power system to remain in synchronism
and maintain the state of equilibrium following a
disturbing force
u
Steady-state stability: analysis of small and slow disturbances
n
u
Transient stability: analysis of large and sudden disturbances
n
1
gradual power changes
faults, outage of a line, sudden application or removal of load
Generator Dynamic Model
l
Under normal conditions, the relative position of the rotor
axis and the stator magnetic field axis is fixed
u
u
u
the angle between the two is the power angle or torque angle, d
during a disturbance, the rotor will accelerate or decelerate w.r.t.
the rotating stator field
acceleration or deceleration causes a change in the power angle
Pe
Pe
Te =
=
w e 2p (60 Hz )
Pm
= Tm
w rotor
Taccelation = DT = Tm - Te
d 2q m
J
= DT = Tm - Te q m = w ms t + d m
2
dt
2
w rotor poles
=
2
w ms
Generator Dynamic Model
dq m
dd m
d 2q m d 2d m
wm =
am =
= w ms +
=
2
dt
dt
dt
dt 2
d 2d m
d 2q m
J
=J
= Tm - Te
2
2
dt
dt
d 2d m
Jw m
= w mTm - w mTe = Pm - Pe
2
dt
2 WKE
2
1
1
WKE = 2 Jw m = 2 Mw m
M=
= Jw m
wm
w m » w ms
3
2 WKE
®M »
= Jw ms
w ms
Generator Dynamic Model
d 2d m
M
= Pm - Pe
2
dt
poles
d = de =
dm ®
2
p d 2d
p 2 WKE
M 2 =
dt
2
2 w ms
2 WKE d 2d
= Pm - Pe
2
w s dt
4
2
p d d
M 2 = Pm - Pe
dt
2
d 2d
2 WKE d 2d
=
2
dt
w s dt 2
2 WKE d 2d Pm Pe
®
=
2
SB SB
w s S B dt
Generator Dynamic Model
2 WKE d 2d
= Pm ( pu ) - Pe ( pu )
2
w s S B dt
WKE kinetic energy in MJ at rated speed
=
=H
SB
machine power rating in MVA
2 H d 2d
= Pm ( pu ) - Pe ( pu )
2
w s dt
H d 2d
®
= Pm ( pu ) - Pe ( pu )
2
p f dt
H d 2d
®
= Pm ( pu ) - Pe ( pu )
2
180 f dt
5
(radians )
(degrees)
Synchronous Machine Model
E
Pmax
Xd¢
Pe
VT
Pm
Round E ¢ = E ¢ Ðd
Rotor
Machine VG = VG Ð0°
Model
B= 1
X d¢
0
Pe0
d0
p/2
power angle curve
p
d
E ¢ VG
Pe = E ¢ VG B cos(d - 90°) =
sin d = Pmax sin d
X d¢
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The Swing Equation
H d 2d
= Pm - Pe
2
p f 0 dt
Dynamic Generator Model
Pe = Pmax sin d
Synchronous Machine Model
H d 2d
= Pm - Pmax sin d
2
p f 0 dt
Pm
H
p f0
7
Forming the Swing Equation
Pe
E
ws
Xd¢
VT
Transient Stability
The ability of the power system to remain in synchronism
when subject to large disturbances
Lyapunov energy functions
simplified energy method: the Equal Area Criterion
Time-domain methods
8
Large power and voltage angle oscillations do not permit
linearization of the generator swing equations
numerical integration of the swing equations
Runga-Kutta numerical integration techniques
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:
Method provides physical insight to the dynamic behavior
of machines
9
one machine connected to an infinite bus
two machines connected together
relates the power angle with the acceleration power
Equal Area Criterion
For a synchronous machine connected to an infinite bus
H d 2δ
= Pm − Pe = Paccel
2
π f 0 dt
π f0
d 2δ π f 0
(
)
=
P
−
P
=
×Paccel
m
e
2
dt
H
H
The energy form of the swing equation is obtained by
multiplying both sides by the system frequency (shaft
rotational speed)
 dδ
2
 dt
10
2
 d δ
 2
 dt
 π f0
 dδ 
 =
(Pm − Pe ) 2 
H
 dt 

Equal Area Criterion
 d 2δ
2 2
 dt
 dδ

 dt
 π f0
 dδ 
(Pm − Pe ) 2 
=
H

 dt 
The left hand side can be reworked as the derivative of
the square of the system frequency (shaft speed)
2

d  dδ   2π f 0
dδ
(
)
=
−
P
P
 

m
e
dt  dt  
H
dt
 dδ  2  2π f 0
(Pm − Pe )dδ
d 
 =
H
 dt  
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Equal Area Criterion
Integrating both sides with respect to time,
2
2π f 0
 dδ 

 =
H
 dt 
2π f 0
dδ
=
dt
H
δ
∫δ (P
0
δ
∫δ (P
0
m
− Pe )dδ
− 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
12
m
δ0
Equal Area Criterion
Consider a machine operating at equilibrium
the power angle, δ = δ0
the electrical load, Pe0 = 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
=ω =
dt
H
13
δ
∫δ (P
0
m
− Pe )dδ > 0
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
δ0
δ1
U Potential = ∫ (Pm1 − Pe )dδ − ∫
14
(Pm1 − Pe )dδ
=0
Equal Area Criterion
Pe
d
Pm1
Pm0
0
15
c A b
1
A2 e
a
δ
δ0 δ1 δmax
π
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
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this is known as the equal area criterion
the rotor angle will oscillate back and forth between δ and δmax at
its natural frequency
Equal Area Criterion - ∆P mechanical
Pe
Pm1
c
b
A2
d
A1
a
Pm2
0
17
δ
δ0
δ1
δmax
π
• The result is that the rotor swings to a maximum angle
– at which p
point the acceleration energy
gy area and the deceleration
energy area are equal
– this is known as the equal area criterion
– the rotor angle
g will oscillate back and forth between δ and δmax at
its natural frequency
Equal Area Criterion - ∆P mechanical
δ1
δ max
δ0
δ1
Pm1 (δ 1 − δ 0 ) − ∫ Pmax sin δ dδ = ∫
Pmax sin δ dδ − Pm1 (δ max − δ 1 )
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
18
Example
p
A 60 Hz synchronous generator having inertia constant H = 9.94
MJ/MVA and a transient reactance X’d = 0.3 pu is connected to an
infinite bus through the following network. The generator is
delivering 0.6 pu real power at 0.8 power factor lagging to the
infinite bus at a voltage of 1 pu.
pu
(a) The maximum power input that can be applied without loss of
synchronism.
( ) Repeat (a)
(b)
( ) with zero initial power input. Assume the generator
internal voltage remains constant at the value computed in (a).
3-Phase Fault
1
G
19
2
inf
Equal Area Criterion - 3 phase fault
Pe
f
e
A2
Pm
a
g
d
A1
0
20
b
δ
c
δ0 δc
δmax
π
Equal Area Criterion - 3 phase fault
δc
δ max
(
Pmax sin δ − Pm ) dδ
δ
Pm (δ c − δ 0 ) = Pmax (cos δ c − cos δ max ) − Pm (δ max − δ c )
∫δ
0
Pm dδ = ∫
c
Pm
(δ max − δ c ) + cos δ max
cos δ c =
Pmax
21
Critical Clearing Time
Pm
(δ max − δ c ) + cos δ max
cos δ c =
Pmax
Pe
e
Pm
A2
a
d
cos δ max =
f
Pm
Pmax
A1
0
22
b
δ0
δ
c
δc
δmax
π
Critical Clearing Time
H d 2δ
= Pm − Pe = Pm ¬ Pe = 0
2
π f 0 dt
23
d 2δ π f 0
Pm
=
2
dt
H
t
π f0
dδ π f 0
Pm ∫ dt =
Pm t
=
0
dt
H
H
π f0
δ=
Pm t 2 + δ 0
2H
2 H (δ c − δ 0 )
tc =
π f 0 Pm
3-Phase Fault
1
G
24
2
inf
Example
p
A 60-Hz synchronous generator having inertia constant
H = 5 MJ/MVA and a direct axis transient reactance X’d = 0.3 pu
i connected
is
t d to
t an infinite
i fi it bus
b through
th
h a purely
l reactive
ti circuit
i it as
shown. Reactances are marked on the diagram on a common
system base. The generator is delivering real power P = 0.8 pu
and
d Q = 0.074
0 0 4 pu to the
h infinite
i fi i bus
b at a voltage
l
off V = 1 pu
A temporary three-phase fault occurs at the sending end of the
point F. When the fault is cleared, both lines are intact.
line at p
Determine the critical clearing angle and the critical fault clearing
time.
F
Equal Area Criterion
Pe
f
e
Pm
a
d
A2
Pe Pre-fault
Pe Post-fault
Pe during fault
g
A1
c
b
0
25
δ
δ0
δc
δmax
π
Critical Clearing Time
Pe
Pe Pre-fault
Pe Post-fault
Pe during fault
e
Pm
a
d
A1
A2
f
c
b
0
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δ
δ0
δc
δmax
π
Critical Clearing Time
δc
δ max
δ0
δc
Pm (δ c − δ 0 ) − ∫ P2 max sin δ dδ = ∫
P3 max sin δ dδ − Pm (δ max − δ c )
Pm (δ max − δ c ) + P3 max cos δ max − P2 max cos δ 0
cos δ c =
P3 max − P2 max
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Steady State Stability
l
l
l
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The ability of the power system to remain in synchronism
when subject to small disturbances
Stability is assured if the system returns to its original
operating state (voltage magnitude and angle profile)
The behavior can be determined with a linear system
model
Assumption:
u
u
u
28
the automatic controls are not active
the power shift is not large
the voltage angles changes are small
Example
Delta, degree
30
25
20
15
10
0
0.5
1
1.5
t, s ec
2
2.5
3
0
0.5
1
1.5
t, s ec
2
2.5
3
60.1
f, Hz
60.05
60
59.95
59.9
59.85
29
Steady State Stability
l
Simplification of the swing equation
H d 2d 0
H d 2 Dd
+
= Pm - Pmax [sin d 0 cos Dd + cos d 0 sin Dd ]
2
2
p f 0 dt
p f 0 dt
Substitute the following approximations
Dd << d
cos Dd » 1
sin Dd » Dd
H d 2d 0
H d 2 Dd
+
= Pm - Pmax sin d 0 - Pmax cos d 0 × Dd
2
2
p f 0 dt
p f 0 dt
Group steady state and transient terms
H d 2d 0
H d 2 Dd
- Pm + Pmax sin d 0 = - Pmax cos d 0 × Dd
2
2
p f 0 dt
p f 0 dt
30
Steady State Stability
l
Simplification of the swing equation
H d 2 Dd
H d 2d 0
- Pm + Pmax sin d 0 = - Pmax cos d 0 × Dd
2
2
p f 0 dt
p f 0 dt
H d 2 Dd
0=
+ Pmax cos d 0 × Dd
2
p f 0 dt
Steady state term is equal to zero
dPe
dd
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d0
d
=
Pmax sin d
dd
= Pmax cos d 0 = Ps
d0
H d 2 Dd
+ Ps × Dd = 0 Second order equation.
2
p f 0 dt
The solution depends on the roots of the
characteristic equation
Stability
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Stability Assessment
u
u
When Ps is negative, one root is in the right-half s-plane, and the
response is exponentially increasing and stability is lost
When Ps is positive, both roots are on the jw axis, and the motion
is oscillatory and undamped, the natural frequency is:
p f0
s =PS
H
2
p f0
wn =
PS
H
32
a
Root locus
jw
S-plane
Damping Torque
dd
Damping force is due to air-gap interaction
PD = D
dt
H d 2 Dd
dDd
+D
+ PS Dd = 0
2
p f 0 dt
dt
d 2 Dd p f 0 dDd p f 0
+
D
+
PS Dd = 0
2
dt
H
dt
H
d 2 Dd
dDd
2
z
w
w
+
2
+
n
n Dd = 0
2
dt
dt
D p f0
z=
2 H PS
33
Characteristic Equation
2
2
n
s + 2zw n s + w = 0
D
z =
2
p f0
<1
H PS
for normal operation conditions
s1 , s2 = -zw n ± j w n 1 - z
wd = wn 1 - z 2
34
2
complex roots
the damped frequency of oscillation
Laplace Transform Analysis
dDd
x1 = Dd , x2 =
dt
1 ù é x1 ù
é x&1 ù é 0
ê x& ú = ê- w 2 - 2zw ú ê x ú = x& = Ax
nû ë 2û
ë 2û ë n
L {x& = Ax} ® sX( s) - x(0) = AX( s )
X( s ) = (sI - A ) x(0)
-1
és
( sI - A ) = ê 2
ëw n
35
-1 ù
s + 2zw n úû
é s + 2zw n 1ù
ê -w 2
ú
s
n
û x ( 0)
X( s ) = ë2
s + 2zw n s + w n2
Laplace Transform Analysis
Dd ( s ) =
(s + 2zw n )Dd 0
s 2 + 2zw n s + w n2
w n2 Dd 0
Dw ( s ) = 2
s + 2zw n s + w n2
Dd (t ) =
Dd 0
1-z 2
Dw (t ) = 36
e -zw nt sin (w d t + q ), q = cos -1 z
w n Dd 0
1-z 2
e -zw nt sin (w d t )
d (t ) = d 0 + Dd (t ), w (t ) = w 0 + Dw (t )
Example
l
37
A 60 Hz synchronous generator having inertia constant H =
9.94 MJ/MVA and a transient reactance X¢d = 0.3 pu is
connected to an infinite bus through the following network.
The generator is delivering 0.6 pu real power at 0.8 power
factor lagging to the infinite bus at a voltage of 1 pu.
Assume the damping power coefficient is
D = 0.138 pu. Consider a small disturbance of 10° or 0.1745
radians. Obtain equations of rotor angle and generator
frequency motion.
Example
X12 = 0.3
X'd = 0.3
G
inf
Xt = 0.2
38
X12 = 0.3
V = 1.0
Example
Delta, degree
30
25
20
15
10
0
0.5
1
1.5
t, s ec
2
2.5
3
0
0.5
1
1.5
t, s ec
2
2.5
3
60.1
f, Hz
60.05
60
59.95
59.9
59.85
39