Electrical Power Systems

Complete algorithm is given below:
Step-1 : Choose an initial value of l, i.e., IC = (IC) 0.
Step-2 : Solve for P gi (i = 1, 2, ..., m) using eqn. (16.14).
m
Optimal System Operation 411
Step-3 : If å Pgi - PD<
Î(a small specified value), the optimal solution is reached. Otherwise
go to next step.
i=1
HG
Step-4 : If P - P
or
HG
If P - P
I
KJ
m
å gi D < 0, IC = (IC) 0 + DIC i=1
I
KJ
m
å gi D > 0, IC = (IC) 0 – DIC and go to step-2.
i=1
This is possible because P gi is monotonically increasing function of (IC).
Let us now consider the generator limits given by eqn. (16.9). During the iterative process,
min max
if a particular generator loading Pg, k reaches the limit Pg,k or Pg,k , its loading is held at this
fixed value and the balanced load is shared among the remaining generators on the basis of
equal incremental cost. We have already seen that l is the common value of incremental cost,
relates increased fuel cost rate (Rs/hr) to increased demand (MW). Suppose for a given demand
0 0
PD , optimal generations are Pgi and corresponding cost is 0
CT . Now assume that load increases
to P D = P D 0 + DPD and we have to obtain the new cost C T . We may use two-term Taylor series:
C T = C T 0 + DCT
\ C T = i=1
m
å
L
M
N
M
e j
dC P
0 i gi
i gi
dP
0
+
gi
C P
Therefore relating increments, we get,
we know,
m
å
DC T = i=1
l =
dC P
dP
0
i gi
dC P
dP
e j
i gi
0
\ DCT = l DPgi i=1
But DPD = DPgi i=1
m
m
gi
gi
e j
O
P
Q
P
D Pgi
... (16.15)
e j DPgi ... (16.16)
å ... (16.17)
å ,