Economic Operation of Power Systems

Economic Operation of Power Systems

Overview
A good business practice is the one in which the production cost is minimized without sacrificing the quality. This is not any different in the power sector as well. The main aim here is to reduce the production cost while maintaining the voltage magnitudes at each bus. In this chapter we shall discuss the economic operation strategy along with the turbine-governor control that are required to maintain the power dispatch economically.
A power plant has to cater to load conditions all throughout the day, come summer or winter. It is therefore illogical to assume that the same level of power must be generated at all time. The power generation must vary according to the load pattern, which may in turn vary with season. Therefore the economic operation must take into account the load condition at all times. Moreover once the economic generation condition has been calculated, the turbine-governor must be controlled in such a way that this generation condition is maintained. In this chapter we shall discuss these two aspects.

Section I: Economic Operation Of Power System
In an early attempt at economic operation it was decided to supply power from the most efficient plant at light load conditions. As the load increased, the power was supplied by this most efficient plant till the point of maximum efficiency of this plant was reached. With further increase in load, the next most efficient plant would supply power till its maximum efficiency is reached. In this way the power would be supplied by the most efficient to the least efficient plant to reach the peak demand. Unfortunately however, this method failed to minimize the total cost of electricity generation. We must therefore search for alternative method which takes into account the total cost generation of all the units of a plant that is supplying a load.
Economic Distribution of Loads between the Units of a Plant
To determine the economic distribution of a load amongst the different units of a plant, the variable operating costs of each unit must be expressed in terms of its power output. The fuel cost is the main cost in a thermal or nuclear unit. Then the fuel cost must be expressed in terms of the power output. Other costs, such as the operation and maintenance costs, can also be expressed in terms of the power output. Fixed costs, such as the capital cost, depreciation etc., are not included in the fuel cost.
The fuel requirement of each generator is given in terms of the Rupees/hour. Let us define the input cost of an unit- i , fi in Rs./h and the power output of the unit as Pi . Then the input cost can be expressed in terms of the power output as
Rs./h
(5.1)
The operating cost given by the above quadratic equation is obtained by approximating the power in MW versus the cost in Rupees curve. The incremental operating cost of each unit is then computed as
Rs./MWh
(5.2)
Let us now assume that only two units having different incremental costs supply a load. There will be a reduction in cost if some amount of load is transferred from the unit with higher incremental cost to the unit with lower incremental cost. In this fashion, the load is transferred from the less efficient unit to the more efficient unit thereby reducing the total operation cost. The load transfer will continue till the incremental costs of both the units are same. This will be optimum point of operation for both the units.
Do an example
The above principle can be extended to plants with a total of N number of units. The total fuel cost will then be the summation of the individual fuel cost fi , i = 1, ... , N of each unit, i.e.,
(5.3)
Let us denote that the total power that the plant is required to supply by PT , such that
(5.4)
where P1 , ... , PN are the power supplied by the N different units.
The objective is minimize fT for a given PT . This can be achieved when the total difference dfT becomes zero, i.e.,
(5.5)
Now since the power supplied is assumed to be constant we have
(5.6)
Multiplying (5.6) by λ and subtracting from (5.5) we get
(5.7)
The equality in (5.7) is satisfied when each individual term given in brackets is zero. This gives us
(5.8)
Also the partial derivative becomes a full derivative since only the term fi of fT varies with Pi, i = 1, ..., N . We then have
(5.9)
Generating Limits
It is not always necessary that all the units of a plant are available to share a load. Some of the units may be taken off due to scheduled maintenance. Also it is not necessary that the less efficient units are switched off during off peak hours. There is a certain amount of shut down and start up costs associated with shutting down a unit during the off peak hours and servicing it back on-line during the peak hours. To complicate the problem further, it may take about eight hours or more to restore the boiler of a unit and synchronizing the unit with the bus. To meet the sudden change in the power demand, it may therefore be necessary to keep more units than it necessary to meet the load demand during that time. This safety margin in generation is called spinning reserve . The optimal load dispatch problem must then incorporate this startup and shut down cost for without endangering the system security.
The power generation limit of each unit is then given by the inequality constraints
(5.10)
The maximum limit Pmax is the upper limit of power generation capacity of each unit. On the other hand, the lower limit Pmin pertains to the thermal consideration of operating a boiler in a thermal or nuclear generating station. An operational unit must produce a minimum amount of power such that the boiler thermal components are stabilized at the minimum design operating temperature.
Example 5.2
let us consider a generating station that contains a total number of three generating units. The fuel costs of these units are given by
Rs./h
Rs./h
Rs./h

The generation limits of the units are

The total load that these units supply varies between 90 MW and 1250 MW. Assuming that all the three units are operational all the time, we have to compute the economic operating settings as the load changes.
The incremental costs of these units are
Rs./MWh
Rs./MWh
Rs./MWh
At the minimum load the incremental cost of the units are
Rs./MWh
Rs./MWh
Rs./MWh
Since units 1 and 3 have higher incremental cost, they must therefore operate at 30 MW each. The incremental cost during this time will be due to unit-2 and will be equal to 26 Rs./MWh. With the generation of units 1 and 3 remaining constant, the generation of unit-2 is increased till its incremental cost is equal to that of unit-1, i.e., 34 Rs./MWh. This is achieved when P2 is equal to 41.4286 MW, at a total power of 101.4286 MW.
An increase in the total load beyond 101.4286 MW is shared between units 1 and 2, till their incremental costs are equal to that of unit-3, i.e., 43.5 Rs./MWh. This point is reached when P1 = 41.875 MW and P2 = 55 MW. The total load that can be supplied at that point is equal to 126.875. From this point onwards the load is shared between the three units in such a way that the incremental costs of all the units are same. For example for a total load of 200 MW, from (5.4) and (5.9) we have

Solving the above three equations we get P1 = 66.37 MW, P2 = 80 MW and P3 = 50.63 MW and an incremental cost ( λ ) of 63.1 Rs./MWh. In a similar way the economic dispatch for various other load settings are computed. The load distribution and the incremental costs are listed in Table 5.1 for various total power conditions.
Table 5.1 Load distribution and incremental cost for the units of Example 5.1
PT (MW)
P1 (MW)
P2 (MW)
P3 (MW)
λ (Rs./MWh)
90
30
30
30
26
101.4286
30
41.4286
30
34
120
38.67
51.33
30
40.93
126.875
41.875
55
30
43.5
150
49.62
63.85
36.53
49.7
200
66.37
83
50.63
63.1
300
99.87
121.28
78.85
89.9
400
133.38
159.57
107.05
116.7
500
166.88
197.86
135.26
143.5
600
200.38
236.15
163.47
170.3
700
233.88
274.43
191.69
197.1
800
267.38
312.72
219.9
223.9
906.6964
303.125
353.5714
250
252.5
1000
346.67
403.33
250
287.33
1100
393.33
456.67
250
324.67
1181.25
431.25
500
250
355
1200
450
500
250
370
1250
500
500
250
410
At a total load of 906.6964, unit-3 reaches its maximum load of 250 MW. From this point onwards then, the generation of this unit is kept fixed and the economic dispatch problem involves the other two units. For example for a total load of 1000 MW, we get the following two equations from (5.4) and (5.9)

Solving which we get P1 = 346.67 MW and P2 = 403.33 MW and an incremental cost of 287.33 Rs./MWh. Furthermore, unit-2 reaches its peak output at a total load of 1181.25. Therefore any further increase in the total load must be supplied by unit-1 and the incremental cost will only be borne by this unit. The power distribution curve is shown in Fig. 5.1.
Fig.5.1 Power distribution between the units of Example 5.2.
Example 5.3
Consider two generating plant with same fuel cost and generation limits. These are given by


For a particular time of a year, the total load in a day varies as shown in Fig. 5.2. Also an additional cost of Rs. 5,000 is incurred by switching of a unit during the off peak hours and switching it back on during the during the peak hours. We have to determine whether it is economical to have both units operational all the time.
Fig. 5.2 Hourly distribution of load for the units of Example 5.2.
Since both the units have identical fuel costs, we can switch of any one of the two units during the off peak hour. Therefore the cost of running one unit from midnight to 9 in the morning while delivering 200 MW is
Rs.
Adding the cost of Rs. 5,000 for decommissioning and commissioning the other unit after nine hours, the total cost becomes Rs. 167,225.
On the other hand, if both the units operate all through the off peak hours sharing power equally, then we get a total cost of
Rs.
which is significantly less that the cost of running one unit alone.
Economic Sharing of Loads between Different Plants
So far we have considered the economic operation of a single plant in which we have discussed how a particular amount of load is shared between the different units of a plant. In this problem we did not have to consider the transmission line losses and assumed that the losses were a part of the load supplied. However if now consider how a load is distributed between the different plants that are joined by transmission lines, then the line losses have to be explicitly included in the economic dispatch problem. In this section we shall discuss this problem.
When the transmission losses are included in the economic dispatch problem, we can modify (5.4) as
(5.11)
where PLOSS is the total line loss. Since PT is assumed to be constant, we have
(5.12)
In the above equation dPLOSS includes the power loss due to every generator, i.e.,
(5.13)
Also minimum generation cost implies dfT = 0 as given in (5.5). Multiplying both (5.12) and (5.13) by λ and combining we get
(5.14)
Adding (5.14) with (5.5) we obtain
(5.15)
The above equation satisfies when
(5.16)
Again since

from (5.16) we get
(5.17)


where Li is called the penalty factor of load- i and is given by
(5.18)
Consider an area with N number of units. The power generated are defined by the vector

Then the transmission losses are expressed in general as
(5.19)
where B is a symmetric matrix given by

The elements Bij of the matrix B are called the loss coefficients . These coefficients are not constant but vary with plant loading. However for the simplified calculation of the penalty factor Li these coefficients are often assumed to be constant.
When the incremental cost equations are linear, we can use analytical equations to find out the economic settings. However in practice, the incremental costs are given by nonlinear equations that may even contain nonlinearities. In that case iterative solutions are required to find the optimal generator settings.
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