Unit IV ECONOMIC OPERATION OF POWER SYSTEMSText BookR4. Allen J. Wood, Bruce F. Wollenberg, Power Generation, Operation and Control, John Wiley and Sons,

4.1. 1Economic considerations, Load curve and load-duration curve

4.2. 3Load factor, diversity factor

4.3. 4Unit commitment (UC) problem, and its Constraints

Priority list methodR4:139

Dynamic programming (qualitative treatment only)R4:240, 141

4.4. Economic dispatch problemR4: 29

4.5. Incremental cost curve Coordination equations without loss and with loss(No derivation of loss coefficients)

4.6. Solution by direct method and -iteration methodR4: 39

4.7. Base point and participation factors.R4: 5

4.8. Solution by direct method and -iteration method Tutorial

4.9. Tutorial Load curve and load-duration curve, cost curve

4.0 ECONOMIC OPERATION OF POWER SYSTEMS

The art of determining the per unit (i.e., one kWh) cost of production of electrical energy is known as economics of power generation.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. Operation strategy along with the turbine-governor control is 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.

4.1 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.

4.1.1 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

(4.1)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

(4.2)

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.

Example 4.1: Consider two units of a plant that have fuel costs of

Rs./h and Rs./h

Then the incremental costs will be

Rs./MWh and Rs./MWh

If these two units together supply a total of 220 MW, then P1 = 100 MW and P2 = 120 MW will result in an incremental cost of

Rs./MWh and Rs./MWh

This implies that the incremental costs of both the units will be same, i.e., the cost of one extra MW of generation will be Rs. 90/MWh. Then we have

Rs./h and Rs./h

and total cost of generation is

Rs./h

Now assume that we operate instead with P1 = 90 MW and P2 = 130 MW. Then the individual cost of each unit will be

Rs./h and Rs./h

and total cost of generation is

Rs./h

This implies that an additional cost of Rs. 75 is incurred for each hour of operation with this non-optimal setting. Similarly it can be shown that the load is shared equally by the two units, i.e. P1 = P2 = 110 MW, then the total cost is again 10,880 Rs./h.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.,

(4.3)Let us denote that the total power that the plant is required to supply by PT, such that

(4.4)whereP1, (, PN are the power supplied by the N different units.

The objective is minimizefT for a given PT. This can be achieved when the total difference dfT becomes zero, i.e.,

(4.5)Now since the power supplied is assumed to be constant we have

(5.6)Multiplying (4.6) by ( and subtracting from (4.5) we get

(4.7)The equality in (4.7) is satisfied when each individual term given in brackets is zero. This gives us

(4.8)Also the partial derivative becomes a full derivative since only the term fi of fT varies with Pi, i = 1, (, N. We then have

(4.9)

4.1.2Generating LimitsIt 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

(4.10)

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 4.2: 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 (4.4) and (4.9) 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 (() 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.

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 (4.4) and (4.9)

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. 4.1.

Table 4.1 Load distribution and incremental cost for the units of Example 5.1

PT (MW)P1 (MW)P2 (MW)P3 (MW)( (Rs./MWh)

9030303026

101.42863041.42863034

12038.6751.333040.93

126.87541.875553043.5

15049.6263.8536.5349.7

20066.378350.6363.1

30099.87121.2878.8589.9

400133.38159.57107.05116.7

500166.88197.86135.26143.5

600200.38236.15163.47170.3

700233.88274.43191.69197.1

800267.38312.72219.9223.9

906.6964303.125353.5714250252.5

1000346.67403.33250287.33

1100393.33456.67250324.67

1181.25431.25500250355

1200450500250370

1250500500250410

Fig.41 Power distribution between the units of Example 4.2.

Example 4.3: 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. 4.2 Hourly distribution of load for the units of Example 4.2.

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.

4.1.3Economic Sharing of Loads between Different PlantsSo 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 (4.4) as

(4.11)wherePLOSS is the total line loss. Since PT is assumed to be constant, we have

(4.12)In the above equation dPLOSS includes the power loss due to every generator, i.e.,

(4.13)Also minimum generation cost implies dfT = 0 as given in (4.5). Multiplying both (4.12) and (4.13) by ( and combing we get

(4.14)Adding (4.14) with (4.5) we obtain

(4.15)The above equation satisfies when

(4.16)Again since

from (4.16) we get

(4.17)

whereLi is called the penalty factor of load-i and is given by

(4.18)

Example 4.4: Let us consider the two generating units of Example 4.1. It is assumed that the transmission loss is defined in terms of the two units as

MW

Therefore

and

Let us assume that the incremental cost is ( = 150 Rs./MWh. We then have

Rearranging the above two equations we get

The solution of the above equation produces P1 = 168.78 MW and P2 = 198.41 MW. The total power loss is then

Therefore the total power supplied to the load is

MW

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

(4.19)

whereB 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.

4.2 Important Terms and Factors

(i) Connected load. It is the sum of continuous ratings of all the equipment connected to supply system. Unit : Watt (W).

(ii) Maximum demand. It is the greatest demand of load on the power station during a given period. The knowledge of maximum demand is very important as it helps in determining the installed capacity of the station. Unit: Watt (W).

(iii) Demand factor. It is the ratio of maximum demand on the power station to its connected load. The knowledge of demand factor is vital in determining the capacity of the plant equipment.

(iv) Average load. The average loads occurring on the power station in a given period (day or month or year) is known as average load or average demand.

(v) Load factor. The ratio of average load to the maximum demand during a given period.

Example 4.1

The maximum demand on a power station is 100 MW. If the annual load factor is 40%, calculate the total energy consumed in a year.

Solution:

Energy consumed in a year = Max. demand x hours in a year x L.F.

Example 4.2

A generating station has a connected load of 43 MW and a maximum demand of 20 MW and the unit generated being 61.5 x 106 kWh per annum. Calculate:

(i) the demand factor

(ii) load factor

Solution:

(i)

(ii)

Example 4.3

A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours and is shut down for the rest of each day. It is also shut down for maintenance for 45 days each year. Calculate the annual load factor.

Solution:

The generated unit on working hour:

Operation days in a year = 365 45 = 320 days in a year

The unit generated in a year:

= 500 MWh x 320 = 160000 MWh

Annual load factor

Examples 4.4

A diesel station supplies the following load to the various consumers:

Industrial consumer

: 1500 kW

Commercial establishment: 750 kW

Domestic power

: 100 kW

Domestic light

: 450 kW

If the maximum demand of the station is 2500 kW and the number of kWh being supplied per year is 45 x 105, determine:

(i) Average load in a year

(ii) Annual load factor

Solution:

(i) Annual average load

(ii) Annual load factor

Example 4.5

A power station has the following daily load cycle:

Time in hours6-88 - 1212 - 1616 - 2020 - 2424 - 6

Load in MW204060205020

(i) Plot the load curves

(ii) Plot the load duration curves

(iii) Calculate the total energy being supplied in a day

(iv) Calculate the daily load factor

Solution:

1. The load curves

2. The load duration curves

3. Total energy supplied

4. Daily load factor

4.3 UNIT COMMITMENT

Unit refers to generating unit and to commit a generating unit is to turn it on. That means, start and bring the unit up to speed, synchronize it to the power system to make it deliver power to the network. To commit enough units and leave them online, whatever be the load demand is uneconomical. It is expensive to run more generating units than what is required and decommitting or turning them off, when they are not needed, can save considerable money.

For example, assume that there are three units and each units minimum and maximum MW limits and fuel costs are known. Let unit 1 be the most economic and unit 3 the most expensive. Then the optimum commitment will be to first supply from unit 1 and the load can be supplied by operating closer to its best efficiency and then when the load increases the second unit will be committed with both units run at their best efficiency possible so that the cost is minimum. The third unit is committed when the load increases sufficiently so that it is economical to commit unit 3 than to supply the load from two units.

Figure shows a unit commitment schedule in the case of a simple peak - valley pattern of load variation.

Unit 1: Min = 150 MW, Max = 600 MW Most economical

Unit 2 Min = 100 MW, Max = 400 MW

Unit 3:Min = 50 MW, Max = 200 MW least economical.

But the unit commitment problem is not just meeting the load and is much more than this as there are other constraints to be considered.

4.3.1 CONSTRAINTS IN UNIT COMMITMENT:

Unit commitment problem can have many constraints depending on different rules imposed on scheduling by individual power system, power pool, reliability council etc.

4.3.1.1 Spinning reserve is the total amount of generation available from all units synchronized (i.e., spinning) on the system, minus the present load and losses. When a generator is lost, there must be ample reserve on other units to compensate up for the loss in a specified time period. Reserve is calculated as a percentage of forecasted peak demand, or it must be capable of making up the loss of the most heavily loaded unit in a given period of time. It is also sometimes calculated as a function of the probability of not having sufficient generation to meet the load. Reserve capacity must be allocated based on unit being fast-responding or slow-responding.

Unit commitment problem include scheduled reserves or off-line reserves that can be brought on-line, synchronized, and brought up to full capacity quickly.

Reserve must be spread around the power system so that they can be made use of even in the case of islanding when system gets disintegrated due to some problem.

4.3.1.2Thermal units require a team of personnel to operate them, especially when turned on and turned off. Since temperature changes can only be gradual some hours are required to bring the unit on-line. Due to these restrictions in the operation of a thermal plant, various constraints arise, such as

Minimum up time: once the unit is running, it should not be turned off immediately

Minimum down time: once the unit is decommitted, there is a minimum time before it can be recommitted. Crew constraints: In a plant with more than one unit there may not be enough personnel to attend both the units if both are turned on or off at the same time and hence at the same time both can not be turned on or off. A certain amount of energy is expended to bring the unit on-line. This is not generated and is included in the unit commitment problem as a start-up cost.4.3.1.3Hydro-Constraints

Unit commitment if separated from the scheduling of hydro-units as a separate hydrothermal scheduling or coordination problem may not result in an optimal solution.4.3.1.4Must Run

Some units must-run during certain times of the year for voltage support on the transmission network or for supply of steam for uses outside the steam plant itself.

4.3.1.5 Fuel Constraints

Some units may have limited fuel, or else have constraints that require them to burn a specified amount of fuel in a given time, presenting a challenging unit commitment problem.4.4 UNIT COMMITMENT SOLUTION METHODS:

The most talked about techniques for the solution of unit-commitment problem are :

Priority-list schemes

Dynamic Programming (DP)

Lagrange relation(LR)

4.4.1 PRIORITY LIST METHOD:

Priority-List Methods consist of a simple shut-down rule obtained by an exhaustive enumeration of all unit combinations at each load level or obtained by noting the full-load average production cost of each unit . The full-load average production cost is the net heat rate at full load multiplied by the fuel cost .

Typical shut-down rules

at each hour when load is dropping, determine whether dropping the next unit on the list leaves sufficient generation to supply the load plus the spinning-reserve requirements if the supply is not sufficient, keep the unit committed determine the number of hours before the unit is needed again

if the time is less than the minimum shut-down time for the unit, keep it committed perform a cost comparison

the sum of the hourly production costs for the next number of hours with the next unit to be dropped being committed

and the sum of the restart costs for the next unit based on the minimum cost of cooling the unit or banking the unit

Advantages:

are flexible and allow for the consideration of practical operating constraints

feasible solutions if there are any are usually obtained

computational requirements in terms of memory and running time are modest.

Disadvantages :

This method cannot guarantee the optimal solutions or even furnish an estimate of the magnitude of their sub-optimality.

4.4.2 DYNAMIC PROGRAMMING METHODS :

The following assumptions are made in this implementation of the DP approach a state consists of an array of units with specified units operating and the rest decommitted (off-line). A feasible state is one in which the committed units can supply the required load and meets the minimum capacity for each period

start-up costs are independent of the off-line or down-time i.e., it is a fixed amount w.r.t. time

no unit shutting-down costs a strict priority order will be used within each interval

a specified minimum amount of capacity must be operating within each intervalThere are two methods of DP. They are

1. Forward dynamic programming

2. Reverse dynamic programming

The forward DP approach :

The forward DP approach runs forward in time from the initial hour to the final hourRecursive algorithm to compute the minimum cost in Kth hour with Ith Combination is,

F cost (J,K) = min[ P cost (J,K) + Scost (J-1,L; J,K) + Fcost(J-1, L) ]

Where,

F cost (J,K) = Least total cost to arrive at state (J, K) ;

P cost (J,K) = Production cost for state (J, K) ;

Scost (J-1,L; J,K)= Transition cost from state (J-1,L) to State (J, K) ;

Fcost = Kth Combination in Jth hour

Disadvantages :

The dynamic -programming method of solution of the unit commitment problem has many disadvantages for large power systems with many generating units . This is because of the necessity of forcing the dynamic-programming solution to search over a small number of commitment states to reduce the number of combinations that must be tested in each time period .ECONOMIC DISPATCH PROBLEM3.1 THE ECONOMICDISPATCH PROBLEMINCREMENTAL COST CURVE COORDINATION EQUATIONS WITHOUT LOSS AND WITH LOSS(NO DERIVATION OF LOSS COEFFICIENTS)SOLUTION BY DIRECT METHOD AND -ITERATION METHODBASE POINT AND PARTICIPATION FACTORS.SOLUTION BY DIRECT METHOD AND -ITERATION METHOD TUTORIALTUTORIAL LOAD CURVE AND LOAD-DURATION CURVE, COST CURVEtime (hrs)

2 4 6 8 10 12 14 16 18 20 22 24

70

60

50

40

30

20

10

Load in MW

2 4 6 8 10 12 14 16 18 20 22 24

60

50

40

30

20

10

Hours duration

Load in MW

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