Electrical Power Systems

0
Dd2
\
Dd3
0
( )
( ) NM
\ 0
Dd2 L
Q2
= –11.18 sin(116.6°) – 2 × 29.065 sin(–63.4°)
| V |
2
O
Q
P =
– 17.885 sin (116.6°) = 26
L
NM
O
Q
-1
L
N
26 -16
0.50
-16
31P
M-1.50
( ) = –0.015 rad = –0.86°
( 0)
Dd = –0.056 rad = –3.2°
3
D|V2 | 0 = DQ2
Q
V
0 ( )
2
| 2|
O
QP
1
= = 0.0384
26
\ () 1
d = 0 – 0.86°= –0.86°
2
Iteration-2
() 1
d = –3.2°
3
|V2| (1) = |V2| (0) + D|V2| (0) = 1.0 + 0.0384 = 1.0384
(1)
P2(cal) (1)
Similarly, P3(cal) p = 1
= 1.0384 × 11.18 cos(116.6° + 0.86° – 0°)
+ (1.0384) 2 × 29.065 cos(–63.4°)
+ 1.0384 × 17.885 cos (116.6° + 0.86° – 3.2°)
= 1.049
= –1.78
(1)
Q2(cal) = 0.79
\ () 1
DP = 0.5 – 1.049 = –0.549
2
() 1
DP3 = –1.5 – (–1.78) = 0.28
\
L
NM
() 1
DQ2 = 1 – 0.79 = 0.21
Dd
Dd
( 1)
2
( 1)
3
O
Q
1
\ Dd2 P =
L
NM
26 -16
O
QP
-1
L
NM
-0.549
-16
31 0. 28
() = –0.0228 rad = –1.3°
O
QP
() 1
Dd = –0.0027 rad = –0.15°
3
D|V2 | (1) = 021 .
26
= 0.008
( 2)
( 1)
( 1)
d2 = d2 2
+Dd = –0.86° – 1.3° = –2.16°
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