Electrical Power Systems

176 Electrical Power Systems
L
NM
L
NM
\ DP
DP
\ Dd
Dd
Q 2(sch) = –1.102
Q 3(sch) = –0.452
( 0)
( 0)
DP2 = P2(sch) – P2( cal )
( 0)
( 0)
DP3 = P3(sch) – P3( cal )
( 0)
( 0)
DQ = Q2(sch) – Q
2
2(
cal )
( 0)
( 0)
DQ = Q3(sch) – Q
3
3(
)
( 0)
2
( 0)
3
( 0)
2
( 0)
3
O
QP
=
O
QP
=
L
NM
L
NM
52. 97 -3198
.
-3198
. 63. 48
52. 97 -3198
.
-3198
. 63. 48
= –2.556 – (–0.5) = –2.056
= –1.386 – (–0.44) = –0.946
= –1.102 – (–1) = –0.102
cal = –0.452 – (–1.503) = 1.051
O
QP
L
NM
-1
O L
QP
NM
Dd
Dd
( 0)
2
( 0)
3
O
QP
-2.
056
-0946
.
( 0)
\ Dd = –0.0687 radian = –3.936°
2
( 0)
\ Dd = –0.0495 radian = –2.837°
3
Similarly
L
NM
L
NM
D
\ DQ
DQ
( 0)
2
( 0)
3
\ D|
V2|
| V |
O
QP
=
O
QP
=
( 0)
( )
3 0
( )
V2 L
NM
L
NM
50. 97 -3198
.
-3198
. 60. 47
0 \ D| | = 0.01332
( 0 )
D| V3| = 0.0244
50. 97 -3198
.
-3198
. 60. 47
(1) ( 0)
( 0)
\ d = 2
d2 2
2nd Iteration
(1) ( 0)
( 0)
d3 = d3 3
O
QP
L
NM
- O
QP
2 0 ( )
D|
V |
3 0 ( )
D|
V |
1
L
NM
O
QP
O
QP
O
QP
-0102
.
1051 .
+Dd = –0.0687 radian = –3.936°
+Dd = –0.0495 radian = –2.837°
|V 2 | (1) = |V 2 | (0) + D|V 2 | (0) = 1.0 + 0.01332 = 1.01332
|V 3 | (1) = |V 3 | (0) + D|V 3 | (0) = 1.0 + 0.0244 = 1.0244
In load flow studies computation of elements of J 1 and J 4 matrices in each iteration is not
required because they do not change much. In this problem J 1 and J 4 as computed above,
assumed constant throughout the iterative process.