# Economic operation of power system

3. May 2018
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### Economic operation of power system

• 1. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 1 Chapter-04 Economic Operation of Power System 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. 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.
• 2. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 2 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. 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. A power plant may have several generating units. If the input-output characteristic of different generator is identical, then the generating units can be equally loaded. But generating units will generally have different input-output characteristic. This means that for a particular input cost, the generator power Pi will be different for different generating units in a plant. 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
• 3. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 3 can be obtained if we can calculate the change in input cost ΔCi for a small change in power ΔPi. 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. Plot of Ic versus power o/p is shown in fig.2. 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. Economical division of plant load between generating units in a plant Let total load in a plant is supplied by two units and that the division of load between these units is such that the incremental cost of one unit is higher than that of the other unit. Now suppose some of the load is transferred from the unit with higher incremental cost to the unit with lower incremental cost. Reducing the load on the unit with higher incremental cost will result in greater reduction of cost than the increase in cost for adding the same amount of load to the unit with lower incremental cost.
• 4. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 4 The transfer of load from one to other can be continued with a reduction of total cost until the incremental costs of the two units are equal. The same reason can be extended to a plant with more than two generating units also. In this case, if any two units have different incremental costs, then in order to decrease the total cost of generation, decrease the output power in units having higher incremental cost and increase the output power in units having lower incremental cost. When this process is continued, a stage will reach where incremental costs of all the units will be equal. Now the total cost of the generation will be minimum. Thus the economical division of load between units within a plant is that all units must operate at the same incremental cost. Economy loading neglecting transmission losses Consider a system having two generating units having costs C1 and C2. C1 and C2 are the fuel costs of the two units in Rs/hr. Total output PT is equal to active power demand and is constant. It is desired to find P1 and P2 so that CT is minimum. From equation (5) For minimum total cost CT, Combining equation (6), (7) and (8) we have So it is concluded that the loads should be so allocated that the two units operate at equal incremental costs. The above concept can be extended to a system with any number of units.
• 5. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 5 Thus optimum economy is achieved if every unit operates at the same cost.
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• 9. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 9 Distribution of load between different plants Since plants are generally long distance a parts 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.
• 10. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 10 Our objective is to obtain a minimum cost for a fixed system load PD subject to the power balance constraint of equation (3). 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(3) 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 fule cost of the system when transmission line losses are taken into account. C and λ are expressed in Rs/hr For minimum cost we require the derivative of C with respect to each PGm to equal 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 (5) becomes Because Cm depends on only PGn, the partial derivative of Cm can be replaced by the full derivative and equation (6) then gives where Ln is called the penalty factor for plant n and is given by Equation (7) is called the exact coordinate equation.
• 11. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 11 From equation (7) So minimum fuel cost is obtained when the incremental fuel cost od each plant multiplied by its penalty factor is the same for all the plants. 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. Fig. showa a simple system having two sources. Derive an expression for the transmission loss and express it as a function of plant loadings. Assume the currents I1 I2 are in phase. 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 Substituting these values in transmission line equation we get Where,
• 12. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 12 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 a two generating source system 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. Example:1 The incremental cost characteristic of the two units in a plant are IC1 = 0.1 P1+8.0 Rs/MWh
• 13. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 13 IC2 = 0.15 P2 + 3.0 Rs/MWh When the total load is 100 MW, what is the optimum sharing of load? Solution: But P1 + P2 = 100 MW Optimum sharing of load is Example:2 A power system consisting of two generators of capacity 210MW each supplies a total load of 310 MW at a certain time. The respective incremental fuel cost of Generator-1 and Generator- 2 are: Where, powers PG in MW and costs C in Rs/hr. Determine (i) the most economical division of load between the generators and (ii) the saving in Rs/day thereby obtained compared to equal load sharing between the machines. Solution: Case-1: the most economical division of load between the generators
• 14. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 14 Case-2: the saving in Rs/day thereby obtained compared to equal load sharing between the machines. Th The negative sign indicates a decrease in cost as output is decreasing. Therefore the net increase in cost = 858.619-788.972=69.647 per hr. 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. Constraints on UC problem a. 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
• 15. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 15 ii) to ensure power supply in the event of any generating unit suffering a forced outage. b. 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. c. 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. d. 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. e. Transition cost: Whenever the status of one unit is changed some transition cost is involved and this has to be taken into account. f. 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. g. 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. h. Must run unit: Sometime it is a must to run one or two units from the consideration of voltage support and system stability. i. Fuel supply constraint: Some plants cannot be operated due to deficient fuel supply. j. Transmission line limitation: Reserve must be spread around the power system to avoid transmission system limitation, often called bottling of reserves. 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. Solution of unit commitment problem by Forward Dynamic Programming (FDP)
• 16. Prepared by Balaram Das, EE Dept., GIET, Gunupur Page 16 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. 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 • Tcost (i,n)=min[least total cost to reach state(i,n)+ operating cost for state(i,n)+transition cost from state (i-1,m) to state(i,n)] • 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. Automatic Generation Control (AGC) Almost all generating companies have tie line interconnections to neighboring utilities. Tie lines allow the sharing of generation resources in emergencies and economies of power production under normal conditions of operation. For purpose of control the entire interconnected system is subdivided into control areas which usually confirm to the boundaries of one or more companies. The net