Electrical Power Systems

180 Electrical Power Systems
Iteration-1
P 3 = |V 3||V 1||Y 31| cos(q 31 – d 3 + d 1) + |V 3||V 2| |Y 32|
cos (q 32 – d 3 + d 2) + |V 3| 2 |Y 33| cos q 33
Q 2 = –|V 2 ||V 1 ||Y 21 | sin (q 21 – d 2 + d 1 ) – |V 2 | 2 |Y 22 |
sin q 22 – |V 2||V 3||Y 23| sin(q 23 – d 2 + d 3)
P 2(sch) = P g2 – PL 2 = 0.50 – 0.0 = 0.50 pu
Q 2(sch) = Q g2 – QL 2 = 1.0 – 0.0 = 1.0 pu
P 3(sch) = P g3 – PL 3 = 0.0 – 1.50 = –1.50 pu
p = 0
|V 1| = 1.0, d 1 = 0.0
|V2 | (0) ( 0)
= 1.0, d = 0.0
2
( 0)
|V3| = 1.0, d = 0.0
3
(0)
P2(cal) (0)
P3(cal) (0)
Q2(cal) = 1 × 1 × 1 × 11.18 cos (116.6°) + (1) 2 × 29.065 cos (–63.4°)
+ 1 × 1 × 17.885 × cos (116.6°) » 0.0
= 1 × 1 × 15.81 cos (108.4°) + 17.885 cos(116.6°)
+ (1) 2 × 33.615 cos (–67.2°) = 0
= –11.18 sin (116.6°) – 29.065 sin (–63.4°) – 17.885
sin (116.6°) = 0.0
( 0)
0
\ DP2 = P2(sch) – P2(cal) (0)
DP = –1.5
3
= 0.50
(0) (0)
DQ = Q2(sch) – 2
Q2(cal) = 1.0
d
P 2
2
d
P 2
d
3
P 3
d
2
P 3
26 -16
0
\ J1 =
3
= 11.18 sin(116.6°) + 17.885 sin(116.6°) = 25.988 » 26
= –17.885 sin(116.6°) = –16
= –17.885 sin(116.6°) = –16
= 15.81 sin(108.4°) + 17.885 sin(116.6°) = 31
L
NM
-16
31
O
QP