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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 Power Systems I 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 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: s2 =  p f0 PS H p f0 wn = PS H Root locus Splane a Power Systems I jw 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 + + PS Dd = 0 D 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 s 2 + 2zw n s + w n = 0 D z = 2 p f0 <1 H PS for normal operation conditions complex roots s1 , s2 = zw n j w n 1  z 2 wd = wn 1  z
2 the damped frequency of oscillation Power Systems I Laplace Transform Analysis
dDd x1 = Dd , x2 = dt & 1 x1 x1 0 & x =  w 2  2zw x = x = Ax n 2 &2 n & L {x = Ax} sX( s )  x(0) = AX( s ) X( s ) = (sI  A ) x(0)
1 s ( sI  A ) = 2 w n 1 s + 2zw n Power Systems I s + 2zw n 1 w 2 s n x ( 0) X( s ) = 2 s 2 + 2zw n s + w n Laplace Transform Analysis
Dd ( s ) =
2 s 2 + 2zw n s + w n (s + 2zw n )Dd 0 2 w n Dd 0 Dw ( s ) = 2 2 s + 2zw n s + w n Dd (t ) = Dd 0 1z 2 e zw nt sin (w d t + q ), q = cos 1 z e zw nt sin (w d t ) Dw (t ) = Power Systems I w n Dd 0 1z 2 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 Xd = 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 X'd = 0.3 G Xt = 0.2 X12 = 0.3 inf X12 = 0.3 V = 1.0 Power Systems I Example
30 Delta, degree 25 20 15 10 0 0.5 1 1.5 t, s ec 2 2.5 3 60.1 60.05 f, Hz 60 59.95 59.9 59.85 0 0.5 1 1.5 t, s ec 2 2.5 3 Power Systems I ...
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 Spring '11
 THOMASBALDWIN
 Volt

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