Unformatted text preview: Steady State Stability
l l l l 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 the automatic controls are not active
the power shift is not large
the voltage angles changes are small Power Systems I Steady State Stability
l Swing Equation H i d 2d i
= Pmi  Pmax sin d
2
p f 0 dt
l Small disturbance modeling d = d 0 + Dd Consider a small deviation H d 2 (d 0 + Dd )
= Pm  Pmax sin(d 0 + Dd )
2
dt
p f0
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
Power Systems I 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 Power Systems I 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 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
Power Systems I
characteristic equation Stability
l Stability Assessment
u u When Ps is negative, one root is in the righthalf splane, 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
PS
s =H
2 p f0
wn =
PS
H Power Systems I Root locus jw a Splane Damping Torque
dd
Damping force is due to airgap 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
d Dd
2
+ 2z wn
+ w n Dd = 0
dt 2
dt
D p f0
z=
2 H PS
Power Systems I 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 Power Systems I 2 complex roots the damped frequency of oscillation Laplace Transform Analysis
x1 = Dd , x2 = dDd
dt
1 ù é x1 ù
&
ú ê x ú = x = Ax
 2zw n û ë 2 û &
é x1 ù é 0
ê x ú = ê w 2
ë &2 û ë n
&
L {x = Ax} ® sX( s )  x(0) = AX( s )
1 X( s ) = (sI  A ) x(0)
és
( sI  A ) = ê 2
ëw n Power Systems I 1 ù
s + 2zw n ú
û é s + 2zw n 1ù
ê w 2
sú
n
û x ( 0)
X( s ) = ë
2
s 2 + 2zw n s + w n Laplace Transform Analysis
Dd ( s ) = (s + 2zw n )Dd 0
2
s 2 + 2zw n s + w n 2
w n Dd 0
Dw ( s ) = 2
2
s + 2zw n s + w n Dd (t ) = Dd 0
1z 2 Dw (t ) = Power Systems I e zw nt sin (w d t + q ), q = cos 1 z w n Dd 0
1z 2 e zw nt sin (w d t ) d (t ) = d 0 + Dd (t ), w (t ) = w 0 + Dw (t ) Example
l 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. Power Systems I Example X12 = 0.3 X'd = 0.3
G inf
Xt = 0.2 Power Systems I 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 Power Systems I ...
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This note was uploaded on 02/01/2012 for the course EEL 4213 taught by Professor Thomasbaldwin during the Fall '11 term at FSU.
 Fall '11
 THOMASBALDWIN
 Volt

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