Electrical Power Systems

\ V = 138
Using eqn. (4.18),
.
3
= 796 . kv
Synchronous Machine: Steady State and Transient Operations 85
Ia = S
*
3f
*
3V
50 - 3180 . ´ 10
=
3 ´ 796 . 0º
\
Using eqn. (4.16)
Ia = 2093. 8 - 318 . º Amp.
( j3)
* 2093. 8
E = 796 . + -318
. º
1000
3
Amp.
\ E = 7. 96 + 6. 28 58. 2º = 1127 . + j 5. 33
\ E =12. 46 25. 3º
kV.
Therefore the excitation voltage magnitude is 12.46 kv and power angle is 25.3º.
(b) When the generator is delivering 22 MW, from eqn. (4.23),
-1L22 ´ 3
-1
66
d = sin
sin
3 ´ 1246 . ´ 796 .
297. 54
I a =
NM O QP = HG I KJ
12. 46 12. 8º - 7. 96 0º
(. 419 + j276
. )
=
j3
j3
\ Ia = 1672. 4 -56. 6º
Amp.
The power factor is cos (56.6º) = 0.55
(c) P max = 3
I a =
| E|| V|
3 ´ 1246 . ´ 796 .
=
= 99. 18 MW
xs 3
12. 46 90º - 7. 96 0º
(. 796 1246 . )
=
3
3
- - j
j
j
Ia = 49285 . 32. 5ºAmp
Ans.
= 12.8º
Example 4.2: A 80 MVA, 69.3 kV, three-phase, synchronous generator has a synchronous
reactance of 10 W per phase and ra » 0. The generator is delivering rated power at 0.8 pf lagging
at the rated terminal voltage to an infinite bus bar. Determine the magnitude of the generated
emf per phase and the power angle d.
Solution:
cos q = 0.8 \q = 36.87º
The rated voltage per phase is V =
\ V = 40. 01 0º
k V
S3f = 80 36. 87º
= (64 MW + j 48 MVAr)
69. 3
k V = 40. 01k
V
3