Electrical Power Systems

Optimal System Operation 433
The terms in matrix of eqn. (16.76) relate the bus powers P i to the angles d K and the matrix is
just the transpose of J 1 found in Chapter-7, Section-11.0. Right hand side of eqn. (16.76) can
easily be obtained from eqn. (7.41). It is
d
P 1
K
=|V 1 ||V K |[G 1K sin(d 1 – d K ) – B 1K cos(d 1 – d K )],
K = 2, 3, ..., n ...(16.77)
Thus, it is very easy to find out the penalty factors using eqn.(16.76).
Example 16.8: igure 16.9 shows a sample power system network. Given that
IC 1 = 4.0 + 0.60 P g1
IC 2 = 4.0 + 0.60 P g2
Y BUS =
L
M
NM
1-j10 0
- 1+ j10
ind out optimal generation scheduling.
Solution:
rom eqn.(7.15),
0
05 . - j50
.
- 05 . + j50
.
- 1+ j10
- 05 . + j50
.
15 . - j15
ig. 16.9: Sample power system network of Ex-16.8.
n
å i K iK cos (qiK – di + dK )
K=1
P i = V V Y
\ P i = V V Y
\ P i = V V Y
n
å i K iK cos {qiK – (di – dK )}
K=1
n
å i K iK [cosqiK cos(di – dK ) + sinqiK sin(di – dK )]
K=1
O
P
QP