Electrical Power Systems

\ (4) 2 2
P = 0.00005 × (188.95) + 0.00008 × (185.58) L
\ (4)
P = 4.54 MW.
L
Error
\ DP g (4) = 0.01
DP g (4) = 370 + 4.54 – (188.95 + 185.58)
Optimal System Operation 431
This value is quite small and solution has converged. Optimum dispatch for l = 109.196
Rs/MWhr
P g1 = 188.95 MW
P g2 = 185.58 MW
Real power loss P L = 4.54 MW.
16.9 GENERAL METHOD OR INDING PENALTY ACTORS
Power loss expression can be written as:
where
P L =
n = total number of bus bars
n m
å Pi = å Pgi –
n
Pdi i=1 i=1 i=1
m = total number of generator bus bars
P gi = real power generation at bus i,
i, = 1, 2, ..., m
P di = real power load at bus i,
i, = 1, 2, ..., n
P i = net real power injected at bus i,
i, = 1, 2, ..., n.
rom eqn. (16.70), we can write
P L = P 1 + P 2 + ... + P m + P m+1 + ... + P n
\
d
P L
K
P l
=
d
K
P 2
+
d
K
å ...(16.70)
Pm K
+ ... +
d
Pm+l K
+
d
Pn K
+ ... +
d
K = 2, 3, ..., n ...(16.71)
since q l = constant (= 0º), we don’t include K = 1,
Now for given, d2 , d3 , ..., dn , we can compute Pi dK
explicitly from the load flow eqns. (7.15).
,