Electrical Power Systems

Similarly, the change in the power angle for the first interval is
Power System Stability 301
Dd 1 = w r (avg). Dt ...(11.62)
and so d 1 = d 0 + Dd 1 ...(11.63)
11.9 EVALUATION O P a AND w r(AVG)
When using the step-by-step technique, Pa is assumed to be the constant over the step interval
and equal to its value at the beginning of the interval. Thus
P0 = Pa(n – 1) +
...(11.64)
as shown in ig. 11.16. If a discontinuity occur during a step interval (such as might be
caused by the clearing of a fault), the standard approach is to simply redefine the intervals at
that point so that the discontinuity occurs at the end (beginning) of a step interval. Then
eqn. (11.64) may be used as previously indicated. Average speed over an interval is given as:
w r(n, n–1) = w r (avg) =
wr(n) + wr(n
-1)
2
11.10 ALGORITHM OR THE ITERATIONS
...(11.65)
Returning now to eqn.(11.62), we see that d1 gives us one point on the swing curve. The
algorithm for the iterative process is as follows:
Pa (n–1) = Pi – Pe (n–1)
...(11.66)
P e (n–1) =
a (n–1) =
||| E V|
x
sind (n–1)
Pa(n-1) f
H
180
...11.67)
b g ...(11.68)
Dw r(n) = a (n–1) Dt ...(11.69)
w r(n, n–1) = w r (avg) = w r (n–1) + Dw r(n)
2
...(11.70)
Dd (n) = wr (n, n–1) Dt ...(11.71)
d (n) = d (n–1) + Dd (n)
...(11.72)
Example 11.13: The kinetic energy stored in the rotor of a 50 Hz, 60 MVA synchronous
machine is 200 MJ. The generator has an internal voltage of 1.2 pu and is connected to an
infinite bus operating at a voltage of 1.0 pu through a 0.3 pu reactance. The generator is
supplying rated power when three-phase short circuit occurs on the line. Subsequently circuit
breakers operate and the reactance between the generator and the bus becomes 0.4 pu. Using the
step-by step algorithm, plot the swing curve for the machine for the time before the fault is
cleared.