Electrical Power Systems

Load low Analysis 179
Use deoupled load flow method to solve the problem. Perform three iterations. Write a
computer program and check whether you are converging or not.
Solution: If voltage controlled buses are contained in the power system, the Newton-
Raphson iterative solution process is actually simplified because the order of the Jacobian is
reduced by one for each voltage controlled bus. Therefore complete equations of this system can
be written as:
L
NM
DP
2
DP
3
DQ
2
O
QP
(p)
=
L
NM
P2 P2 P2
d2
d3
| V2|
P3 P3 P3
d2
d3
| V2|
Q2 Q2 Q2
d d | V |
or decoupled load flow case we can write
L
NM
DP
2
DP
3
O
QP
(p)
=
L
NM
HG
(p)
and DQ2 =
d
P 2
2
d
d
P 2
3
P 3
2
P 3
d
3
2
P2 P
d2
d
P3 P
d d
2
Q2
| V |
2
I
3
(p)
× KJ
O
QP
(p)
2
3
3
3
L
NM
2
Dd
Dd
D|
V |
O
QP
2
3
(p)
2
O
QP
(p)
(p)
L
NM
Dd
2
Dd3
D|
V |
= |V 2||V 1||Y 21|sin(q 21 – d 2 + d 1) + |V 2| |V 3| |Y 23|
sin (q 23 – d 2 + d 3)
= –|V 2| |V 3| |Y 23| sin (q 23 – d 2 + d 3)
2
O
QP
(p)
= –|V 3||V 2||Y 32| sin(q 32 – d 3 + d 2)
= |V 3 ||V 1 ||Y 31 | sin(q 31 – d 3 + d 1 ) + |V 3 ||V 2 ||Y 32 |
sin(q 32 – d 3 + d 2 )
Q2
| V | = –|V1||Y21| sin(q21 – d2 + d1) – 2|V2||Y22| 2
sin q 22 – |V 3||Y 23| sin(q 23 – d 2 + d 3)
P 2 = |V 2 ||V 1 ||Y 21 | cos(q 21 – d 2 + d 1 ) + |V 2 | 2 |Y 22 |
cos q 22 + |V 2 ||V 3 ||Y 23 | cos(q 23 – d 2 + d 3 )