2. Introduction
Economic operation is very important for a power system to return a profit on the
capital invested.
Two things put pressure on power companies to achieve maximum possible efficiency.
(a) Rates fixed by regulatory bodies and
(b) the importance of conservation of fuel place
Maximum efficiency minimizes the cost of electrical energy consumed by the
consumers. Also it reduces the rising prices for fuel, labor, supplies and maintenance.
Operational economics involving power generation and delivery of the power.
Delivery can be subdivided into two parts.
(i) One dealing with minimum cost of power production called Economic dispatch.
(ii) Other dealing with minimum loss of the generated power delivery to the loads.
For any specified load condition, economic dispatch (i) determines the power output
of each plant. (ii)Minimizes the overall cost of fuel needed to serve the system load.
3. Introduction
The economic dispatch problem can be solved by means of the optimal power flow
(OPF) program.
We first study the most economic distribution of power within the plant.
The method that we develop applies to economic scheduling of plant output for a
given loading of the system without considering of transmission losses.
Next we express transmission loss as a function of the outputs of the various plants.
Then we determine how the output of each of the plants of a system is scheduled to
achieve the minimum cost of power delivered to the load.
Because the total load of the power system varies throughout the day, coordinated
control of the power plant outputs is necessary to ensure generation to load balance
so that the system frequency will remain as close as possible to the nominal
operating value, usually 50 or 60 Hz.
Also because of the daily load variation, the utility has to decide on the basis of
economics which generator to start up, which generators to shut down, and in what
order. The computational procedure for making such decisions, called unit
commitment.
4. Distribution of load between units within a plant
Power plant consisting of several generating units which are constructed by
investing huge amount of money. Fuel cost, staff salary, interest and
depreciation charges and maintenance cost are some of the components of
operating cost.
Fuel cost is the major portion of operating cost and it can be controlled.
Therefore we shall consider the fuel cost alone for further considerations.
To get different output power, we need to vary the fuel input. Fuel input can
be measured in tones/hr or millions of BTU(British Thermal Unit)/hr.
Knowing the cost of the fuel, in terms of Rs/tone or Rs/Millions of BTU, input
to the generating unit can be expressed as Rs/hr.
Let Ci Rs/h be the input cost to generate a power of Pi MW in unit i.
Fig.1 shows a typical input output curve of a generating unit. For each
generating unit there shall be a minimum and maximum power generated as
Pi,min and Pi,max.
5. If the input-output curve of Unit ‘I’ is quadratic , we can write
Where, Ci = Input cost, Pi – Output power in MW, α, β and ϒ are cost coefficient.
6. Incremental cost curve
As we shall see the criterion for distribution of the load between any two
units is based on whether increasing the generation of one unit, and
decreasing the generation of other unit by the same amount results in an
increase or decrease in total cost. This can be obtained if we can calculate the
change in input cost ΔCi for a small change in power ΔPi.
7. Thus while deciding the optimal scheduling, we are concerned with dCi/dPi,
Incremental cost (IC) which is determined by the slopes of the input-output
curves. Thus the incremental cost curve is the plot of dCi/dPi versus Pi. The
dimension of dCi/dPi is Rs/MWh.
Incremental cost curve
8. The fig.2 shows that the incremental cost is quite linear with
respect to power output over an appreciable range.
In analytical work, the curve is usually approximated by one or
two straight lines.
The dashed line in the fig-2 is a good representation of the
curve.
Incremental cost curve
9. Economic loading neglecting transmission losses
Consider a system having two generating units having costs C1 and C2
10. The above concept can be extended to a system with any number of units.
Thus optimum economy is achieved if every unit operates at the same cost.
11.
12.
13.
14.
15.
16. Distribution of load between different plants
Since plants are generally long distance apart in an integrated system, it becomes essential to
consider transmission losses in deciding the load allocation to different plants.
This leads to the dispatch of power in an economic way so as to make the overall cost to be the
minimum.
Let there be ‘m’ no of plants in a system integrated by transmission line.
PG1, PG2..PGm - Generation output of the plants in MW
PD -Total load in MW
PL - Losses in transmission lines
Where, Cn – Fuel cost of nth plant in Rs/hr
PGn – output of nth plant in MW.
17. Our objective is to obtain a minimum cost for a fixed system load PD subject to the power
balance constraint of equation (13).
We now present the procedure for solving such minimization problems called the
method of Lagrange multipliers.
The new cost function C is formed by combining the total fuel cost and the equality
constraint of equation(13) in the following manner.
The cost function C is often called the lagrangian, and we shall see that the parameter λ
which we now call Lagrange multiplier is the effective incremental fuel cost of the system
when transmission line losses are taken into account. C and λ are expressed in Rs/hr .
18. For minimum cost we require the derivative of C with respect to each PGn equal to zero
Since PD is fixed and the fuel cost of any one unit varies only if the power output of that
unit is varied. Therefore equation (15) becomes
Because Cm depends on only PGn, the partial derivative of Cm can be replaced by the full
derivative and equation (16) then gives
19. where Ln is called the penalty factor for plant n and is given by
Equation (17) is called the exact coordinate equation.
From equation (17),
So, minimum fuel cost is obtained when the incremental fuel cost of each
plant multiplied by its penalty factor is the same for all the plants.
20. Representation of transmission losses
The transmission losses depend on line currents and line resistances. It is possible to
represent these losses as a function of plant loadings.
Let ra, rb, rc be the resistances of line a, b and c respectively. The transmission loss is,
Since, I1 and I2 are in same phase,
Let P1 and P2 be the power outputs and V1 and V2 be the bus voltage and cos Φ1 and
cos Φ2 be the power factors of sources 1 and 2 respectively. Then
21. Substituting these values in transmission line equation we get,
Where,
Representation of transmission losses
22. The transmission loss equation:
To include the effect of transmission losses in deciding the load allocation, it is
necessary to represent the loss as a function of plant loading. The general form of loss
equation is
Where, PL – transmission losses in pu
P-plant loading in pu
B-Loss coefficient
For two generating source system
23. For a three generating source system’
The matrix representation of above loss equation is
Where for a total of k sources
Where, 𝑃𝑇
is the transposition of P. The B coefficients depend on the system network
parameters, configuration, plant power factor and voltages etc.
The transmission loss equation:
24. Automatic load dispatching
Economic load dispatching is the distribution of the load among the
generating units in such a manner so as to minimize the cost of supplying
the minute to-minute requirements of the system.
In a large interconnected system it is humanly impossible to calculate and
adjust such generations.
Hence, the help of digital computer system along with analogue devices is
used.
The whole process is carried out automatically; hence this process is
known as automatic load dispatching.
25. Unit Commitment
The total load in the power system varies throughout the day and reaches
different peak value from one day to another.
Different combination of generators are to be connected in the system to
meet the varying load. When the load increases, the utility has to decide in
advance the sequence in which the generator units are to be brought in.
Similarly when the load decreases, the operating engineer need to know in
advance the sequence in which the generating units are to be shut down.
The problem of finding the order in which the units are to be shut down over
a period of time (say one day), so the total operating cost involved on that day
is minimum, is known as Unit Commitment(UC).
Thus UC problem is economic dispatch over a day. The period considered
may be a week, a month or a year.
26. Constraints on UC problem
Spinning reserve: There may be sudden increase in load, more than what
was predicted. Further there may be a situation that one generating unit may
have to be shut down because of fault in generator or any of its auxiliaries.
Some system capacity has to be kept as spinning reserve
i) to meet an unexpected increase in demand and
ii) to ensure power supply in the event of any generating unit suffering a
forced outage
Minimum up time: When a thermal unit is brought in, it cannot be turned off
immediately. Once it is committed, it has to be in the system for a specified
minimum up time.
Minimum down time: When a thermal unit is decommitted, it cannot be
turned on immediately. It has to remain decommitted for a specified
minimum down time.
27. Crew constraint: A plant always has two or more generating units. It may not be
possible to turn on more than one generating unit at the same time due to non-
availability of operating personnel.
Transition cost: Whenever the status of one unit is changed some transition cost is
involved and this has to be taken into account.
Hydro constraints: Most of the systems have hydroelectric units also. The operation
of hydro units, depend on the availability of water. Moreover, hydro projects are
multipurpose projects. Irrigation requirements also determine the operation of hydro
plants.
Nuclear constraint: If a nuclear plant is part of the system, another constraint is
added. A nuclear plant has to be operated as a base load plant only.
Must run unit: Sometime it is a must to run one or two units from the consideration
of voltage support and system stability.
Fuel supply constraint: Some plants cannot be operated due to deficient fuel supply.
Transmission line limitation: Reserve must be spread around the power system to
avoid transmission system limitation, often called bottling of reserves.
Constraints on UC problem
28. Solving the unit commitment problem
The objective of unit commitment is not economical to run all the units available all
the time.
To determine the units of plants that should operate for a particular load is the
problem of unit commitment.
This problem is important for thermal power plant because the operating cost and
start up time are high and hence their on-off status is important.
A simple approach to the problem is to impose priority ordering, wherein the most
efficient unit is loaded first, and then followed by the less efficient units in order as
the load increases.
Finding the most economical combination of units that can supply this load demand
is to try all possible combination of units that can supply this load; to divide the load
optimally among the units of each combination by use of the co-ordination equation,
so as to find most economical operating cost of combination then to determine the
combination which has the least operating cost among all these.
These combinations can be solved by dynamic programming method.
29. Solution of unit commitment problem by Forward
Dynamic Programming (FDP)
In FDP, we start with the first duration obtain the unit commitment schedule
for this duration, go to second duration till we reach the last duration.
In FDP, the initial conditions are easily specified, calculation can be carried
out for any desired length of time and the previous history of each unit can be
calculated at every stage.
The operating cost for any stage needs method of economic dispatch. This is
due to the fact that for any given combination of units, the operating cost is
minimum if all the units in this combination are operating at equal
incremental cost.
Two other variables enter the strategy for UC by Forward dynamic
programming.
This is because a no of possible combinations exist for every state. Let this be
denoted by K.
Another variable is the no. of paths or strategies to save at every step. Let this
variable be denoted as L.
30. Procedure
The stage in load cycle is specified as i. We start with i=1 i.e. first stage.
The operating cost for this stage is computed. This has to be repeated for all
possible combination K.
We now go to (i+1)th stage.
The no. of feasible paths in duration (i-1) is found and stored.
The minimum total cost is calculated using the formula
This has to be repeated for all states in the ith interval.
The lowest cost paths (number L) are saved in computer memory.
Check whether the program has reached the last stage in load cycle. If yes
optimal schedule is printed.
Otherwise i is uploaded to (i+1) and program is repeated.
