Electrical Power Systems

286 Electrical Power Systems
Solution:
Equivalent circuit of the system is shown in ig. 11.5.
Using eqns. (i) and (ii)
ig. 11.5: Equivalent circuit of Example 11.6.
|E g | d = V t + jx d .I ...(i)
I =
Vt-V jx
= 11 1 0
|Eg | d = 11 . q j04
.
. q – º
j1
11 . q – 1
j1
+ HG I KJ
\ |E g | d = 1.1 cos q + j 1.1 sin q + 0.4 ´ 1.1 q – 0.4
...(ii)
\ |E g | d = (1.54 cos q – 0.4) + j 1.54 sin q ...(iii)
Maximum steady-state power capabitity (limit) is reached when d = 90º, i.e., real part of
eqn. (iii) is zero. Thus
1.54 cos q – 0.4 = 0
\ q = 74.9º
\ |E g | = 1.54 sin (74.9º) = 1.486 pu.
\
\ P max =
Vt = 1.1 74. 9º
| Eg| V|
=
( x + x)
1486 10 . ´ .
04 . + 1
\ P max = 1.061 pu. Ans.
11.6 EQUAL-AREA CRITERION
d
In the preceding discussions we have indicated that a solution to the swing equation for d (t),
leads to the determination of the stability of a single machine operating as part of a large power