Electrical Power Systems

Load low Analysis 167
\ 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+1 )
Q2 = –|V2| (p) (p)
.|V1||Y21|sin (q21 – d2 + d1) – (|V2| (p) ) 2 |Y22| sin q 22 –|V 2| (p) |V 3| (p) |Y 23|sin (q 23 – d 2 (p) + d3 (p) )
|Y 21 | = 22.36, q 21 = 116.56°,
|Y 22| = 58.13, q 22 = –63.4°
|Y 23| = 35.77, q 23 = 116.6°
\ p = 0
(1)
Q2 = –|V2| (o) (o)
|V1||Y21|sin (q21 – d2 + d1) – (|V2| (o) ) 2 |Y22| sin q 22 –|V 2| (o) |V 3| (o) |Y 23|sin (q 23 – d 2 (o) + d3 (o) )
|V 1| = 1.05, d 1 = 0.0°,
|V 2 | (o) = 1.0, d 2 (o) = 0.0°,
|V 3| (o) = 1.0, d 3 (o) = 0.0°
(1) 2
Q2 = –1 × 1.05 × 22.36 sin (116.56°) – (1.0) × 58.13 sin (–63.4°)
–1 × 1 × 35.77 sin (116.6° )
(1)
\ Q2 = –21 + 51.97718 – 31.98389
(1)
\ Q2 = –1.0067 pu MVAr.
The value of Q 2 (1) is taken as net reactive power injected at bus 2, i.e. Q2 = Q 2 (1) = –1.0067
pu MVAr.
Now, compute
P - jQ
Y
(p+1)
V = C2
1
Y22 (1)
VC2 (1)
2 2
22
Y
Y
21
22
(1)
\ VC2 p = 0
= 1
Y 22
L
NM
L
NM
P2-jQ2 p eV 2 j
( ) *
P2-jQ ( o)
V
2
*
e j
( p+
1)
2 1 ( )
( p)
21 1 23 3
- Y V - Y V
( o)
21 1 23 3
- Y V - Y V
= - + 2556 . j10067
.
5813 . - 634 . °
= 0. 04725 2219 . °
Y23
= - 0. 3846; Y
= -0.
6153
L
NM
22
0. 04725 2219 .
= 0. 3846 0. 6153
( ) *
°
+ V + V
o
V
e 2 j
O
QP
O
QP
( o)
1 3
O
QP