Electrical Power Systems

188 Electrical Power Systems
rom ig. 8.2, voltage behind subtransient reactance (generator)
E g ² = V 0 + j (0.15 + 0.10 + 0.08 + 0.10) × I 0
E² g = 0.86880º + j0.43 × 0.7194 36. 87 º
\ E² g = 0.7266 19. 9º
pu
Similarly,
E ² m = 0.8688 0º – j 0.15 × 0.7194 36. 87 º
\ E ² m = 0.9374 -528 . º pu
rom ig. 8.3,
ault current
I g ² =
E 0. 7266 19. 9º
=
j (. 015+ 028 . ) 043 . 90º
\ I² g = 1. 689 - 70. 1º
pu
\ I g ² = (0.575 – j 1.588) pu
I ² m = Em
²
g ²
=
j015
.
09374 528
. - . º
015 . 90º
\ I ² m = 625 . -95. 28º
pu
\ I m ² = (–0.575 –j6.223) pu.
\ I f = –j7. 811 pu.
Base current (generator and motor)
I f = I g ² + I m ² = 0.575 – j1.588 – 0.575 – j 6.223
20 ´ 1000
IB = = 912.085 Amp.
3 ´ 1266 .
\ I g ² = 912.085 × 1.689 -70 1
.º= 1540. 5 - 70. 1º
Amp.
\ I ² m = 912.085 × 6.25 -95. 28 º = 5700.5 -95. 28 º Amp.
\ I f = 912.085 × (–j7·811) = 7124.3 -90 º Amp.
Example 8.2: Solve Ex-8.1 using Thevenin’s Theorem.
Solution: The detailed derivation for this is given in Chapter-4, Section-4.8.2.
igure 8.4 shows the Thevenin’s equivalent of the system feeding the fault impedance.
X ² = j (0.1 + 0.08 + 0.01) = j0.28
X dg ² = j0.15, X dm ² = j0.15