Module 5 : Real and Reactive Power Scheduling
Lecture 22 : Real Power Scheduling
Merit Order Dispatch
The simplest real power scheduling problem can be formulated by assuming the following:
The generators which are "committed" are known (the issue of whether to engage a generation unit and
a) for what duration, based on cost and prime mover/energy supply constraints is not considered here).
b) The cost of generation (per kW-hr) for various generators is a constant.
There are no inequality constraints considered except for the maximum power output of various
c) generators.
Power losses are neglected and the only equality constraint to be satisfied is that the total load is equal to
d) the generation.
Now, schedule the generation so that the cost per kW-hr is minimized.
Let us illustrate the problem and its solution by a simple example:
Suppose there are three generators, say, G1, G2 and G3 with maximum power output of 200 MW, 250 MW and 300
MW respectively.
The cost of energy for the three generators are 2000, 3000 and 2500 Rs/ MW-hr respectively.
The total load of 550 MW is to be shared by the three generators. Find the least cost schedule.
The solution for this problem is easy. We just order the generators as per their cost and utilize the cheapest
generator fully. Since the cheapest generator when load fully cannot accomodate all the load demand, the next
cheapest generator is utilized to the extent that the load demand is met or upto its maximum output.
Therefore the obvious schedule is:
G1 : 200 MW, G3 : 300 MW and G2 : 50 MW.
The average cost of electricity is C = (200*2000+300*2500+50*3000)/550 = 2364 Rs/MW-hr
Economic Dispatch
In the previous problem we assumed a constant cost of energy in Rs/MW-hr for the generators irrespective of the
power output of generators. In actual practice, there is a fixed and variable cost component. As a result, cost is a
function of power generated.
A cost function could have the form:
C = 40*P1 + 5 * P1² Rs/hr,
where P1 is the actual power generated.
Consider two generators with the following cost functions:
C1 = 50 *P1 + 3 * P1²
C2 = 50 *P2 + 2.5 * P2²
Lecture 22 : Real Power Scheduling
Merit Order Dispatch
The simplest real power scheduling problem can be formulated by assuming the following:
The generators which are "committed" are known (the issue of whether to engage a generation unit and
a) for what duration, based on cost and prime mover/energy supply constraints is not considered here).
b) The cost of generation (per kW-hr) for various generators is a constant.
There are no inequality constraints considered except for the maximum power output of various
c) generators.
Power losses are neglected and the only equality constraint to be satisfied is that the total load is equal to
d) the generation.
Now, schedule the generation so that the cost per kW-hr is minimized.
Let us illustrate the problem and its solution by a simple example:
Suppose there are three generators, say, G1, G2 and G3 with maximum power output of 200 MW, 250 MW and 300
MW respectively.
The cost of energy for the three generators are 2000, 3000 and 2500 Rs/ MW-hr respectively.
The total load of 550 MW is to be shared by the three generators. Find the least cost schedule.
The solution for this problem is easy. We just order the generators as per their cost and utilize the cheapest
generator fully. Since the cheapest generator when load fully cannot accomodate all the load demand, the next
cheapest generator is utilized to the extent that the load demand is met or upto its maximum output.
Therefore the obvious schedule is:
G1 : 200 MW, G3 : 300 MW and G2 : 50 MW.
The average cost of electricity is C = (200*2000+300*2500+50*3000)/550 = 2364 Rs/MW-hr
Economic Dispatch
In the previous problem we assumed a constant cost of energy in Rs/MW-hr for the generators irrespective of the
power output of generators. In actual practice, there is a fixed and variable cost component. As a result, cost is a
function of power generated.
A cost function could have the form:
C = 40*P1 + 5 * P1² Rs/hr,
where P1 is the actual power generated.
Consider two generators with the following cost functions:
C1 = 50 *P1 + 3 * P1²
C2 = 50 *P2 + 2.5 * P2²
