Electrical Power Systems

304 Electrical Power Systems
P max =
12 . ´ 10 .
= 3.0 pu
04 .
and K 2 = 3/4 = 0.75 for use in eqn. (11.55)
Now d m ¢ = p – d m (ig. 11.13)
sin ¢
dm = 1
30 .
\ d m ¢ = 0.34 radian.
\ d m = p – d m ¢ = p – 0.34 = 2.80 radian
substituting K1 , K2 , d0 and dm in eqn. (11.55), yields
cosdcr = 1
075 . b2. 8 - 0. 2527g´ 0. 25 - 0 + 0. 75cos b2. 8g
\ cosd cr = – 0.093
\ d cr = 95.34º.
rom ig. 11.17, when d = d cr = 95.34º, t cr = 0.245 sec. Hence the fault must be cleared within
50 ´ 0.245 = 12.25 cycles.
We can conclude that the transient stability of a synchronous generator, during and
subsequent to fault conditions, depends upon the rotor swing and the critical clearing time.
These are governed by the machine inertia and direct axis transient reactance. The overall
system stability can be improved by appropriate control schemes such as turbine valve control,
fast fault clearing time, appropriate excitation systems and ACTS devices.
EXERCISE
11.1. A 20 MVA, 50 Hz, 4 pole synchronous machine has inertia constant H = 9 MJ/MVA. ind the
kinetic energy at synchronous speed and the accelerating torque if the shaft input is 26800 hp and
the electrical power developed is 16 MW.
Ans: 180 MJ, 23628 NW–M.
11.2. In P 11.1, assume that the acceleration is constant for 15 cycles. Determine (a) the change in
torque angle in that period (b) speed at the end of 15 cycles.
Ans: (a) 8.36 elect-degree (b) 1504.6 rpm.
11.3. The steady state limit of a power system is 150 MW. A generator with constant excitation is
supplying 60 MW to the system. Compute the maximum permissible sudden increase in generator
output without causing instability.
Ans: 67.0 MW.
11.4. A synchronous machine connected to an infinite bus through a double circuit line as shown in ig.
11.19. ind the maximum initial power with both the lines are operating, so that the system can
continue to be stable when one of the lines is suddenly tripped. Ans: 1.14 pu