Lecture 24 : Real and Reactive Power Scheduling
Objectives
In this lecture you will learn the following
Cost minimization considering network constraints and losses.
An illustrative example
Optimal Power Flow
Until now, we have considered fairly simple examples pertaining to the economic dispatch problem. In an actual
system, there are many variables (other than generated power) that need to be scheduled. These may be set
under the direction of a system operator. If the quantity which is to be scheduled belongs to an independent
entity, a system operator may "buy" the quantity based on the price at which it is sold.
The formulation of the minimum cost problem, by including the network, reactive power demand, voltage and
current constraints is known as an "optimal power flow". Note that optimal power flow takes into account of losses
in the network.
Let us consider an optimal power flow problem in which :
a) Control variables (which can be scheduled independently) are the generated real and reactive power
b) Load demand is fixed
c) Cost is a function of only the real power (this assumes that the service of providing reactive power for
maintaining voltages is free!)
d) Real powers (control variables) are constrained by maximum limits.
e) Voltage magnitudes (auxiliary variable) is constrained by maximum and minimum limits.
In actual practice, several additional constraints and control variables are present, but for the sake of simplicity
let us consider only the ones mentioned above.
Optimal Power Flow - An example
Consider a 2 generator system which supplies 2 loads at buses
1 and 2. Compute the minimum cost real and reactive power
schedule of the generators if the cost of real power (power is
in per-unit) is:
C1 = 50 *P1 + 300 * P1²
C2 = 50 *P2 + 250 * P2²
Reactive power is assumed to be free. The following limits are
applicable: Voltages at both buses should be between 0.975
pu and 1.025 pu. Generated real power for each generator
should be less than 3 pu. Reactive power capability is
assumed to be unlimited.
R+j X = 0.02 +j 0.1, PL1 + j QL1 = 4.0 + j*2.0, PL2+ jQL2 =
1.5 +j*0.75 pu