economic dispatch and unit commitment explained, Study notes of Engineering

economic dispatch and unit commitment in power systems explained

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----------------------- Page 1-----------------------
Chapter 4
ECONOMIC DISPATCH AND UNIT COMMITMENT
----------------------- Page 2-----------------------
1 INTRODUCTION
A power system has several power plants. Each power plant has several
generating units. At any point of time, the total load in the system is met by
the generating units in different power plants. Economic dispatch control
determines the power output of each power plant, and power output of
each generating unit within a power plant , which will minimize the overall
cost of fuel needed to serve the system load.
F 0
B 7
We study first the most economical distribution of the output of a
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----------------------- Page 1-----------------------

Chapter 4

ECONOMIC DISPATCH AND UNIT COMMITMENT

----------------------- Page 2-----------------------1 INTRODUCTION

A power system has several power plants. Each power plant has several generating units. At any point of time, the total load in the system is met by the generating units in different power plants. Economic dispatch control determines the power output of each power plant, and power output of each generating unit within a power plant , which will minimize the overall cost of fuel needed to serve the system load. F 0B 7 We study first the most economical distribution of the output of a

power plant between the generating units in that plant. The method we develop also applies to economic scheduling of plant outputs for a given system load without considering the transmission loss. F 0B 7 Next, we express the transmission loss as a function of output of the various plants. F 0B 7 Then, we determine how the output of each of the plants of a system is scheduled to achieve the total cost of generation minimum, simultaneously meeting the system load plus transmission loss. ----------------------- Page 3-----------------------2 INPUT – OUTPUT CURVE OF GENERATING UNIT

Power plants consisting of several generating units are constructed investing huge amount of money. Fuel cost, staff salary, interest and depreciation charges and maintenance cost are some of the components

P in MW i Pi min P i max Output

Fig.1 Input-Output curve of a generating unit ----------------------- Page 5----------------------- If the input-output curve of unit i is quadratic, we can write F 0 Ci (^) 3 Dαi i P2 i iF 02 B β P (^) iF 02 Bγ Rs / h (1)

A power plant may have several generator units. If the input-output characteristic of different generator units are identical, then the generating

units can be equally loaded. But generating units will generally have different input-output characteristic. This means that, for particular input cost, the generator power P will be different for different generating units i in a plant. 3 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 the 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 ΔC for a small change in power ΔP. i i dC ΔC dC Since i = i we can write Δi C = (^) i i ΔP

The unit that has the input – output relation as F 0 C (^) i 3 Dα (^) iP2 i (^) i F 02 Biβ P (^) iF 02 Bγ Rs / h (1)

has incremental cost (IC) as

F 0dC IC (^) i 3 D i F 03 D (^2) i αi P F 02 Biβ (2) dPi Here α , β and γ are constants. i i i ----------------------- Page 7----------------------- A typical plot of IC versus power output is shown in Fig.2.

IC in Rs / MWh i

Linear approximation

Actual incremental cost

P in MWi

Fig.2 Incremental cost curve

----------------------- Page 8----------------------- This figure shows that 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

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. 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.

This is illustrated through the characteristics shown in Fig. 3 ----------------------- Page 10-----------------------

IC 2

IC IC

P

P 2 P 1 i

Fig. 3 Two units case Initially, IC > IC. Decrease the output power in unit 2 by ΔP and 2 1 increase output power in unit 1 by ΔP. Now IC ΔP > IC ΔP. Thus there 2 1 will be more decrease in cost and less increase in cost bringing the total cost lesser. This change can be continued until IC = IC at which the 1 2

Thus the economical division of load between units within a plant is that all units must operate at the same incremental cost. Now we shall get the same result mathematically.

----------------------- Page 12----------------------- Consider a plant having N number of generating units. Input-output curve of the units are denoted as 1 C (P ),C (P )......... ..., C 1 2 2 N (^) N (P ). Our problem is,

for a given load demand P , find the set ofD P (^) is which minimizes the cost

function C = C (P ) F 02 BC (P ) F 02 B.......... .. F 02 BC (P ). (3) T 1 1 2 2 N N

subject to the constraints

P F 02 D(P F 02 BP F 02 B.......... .. F 02 BP ) F 03 D 0 (4)

D 1 2 N

and P F 0A 3P F 0A 3P i = 1,2,…….., N (5) i min i i max

Omitting the inequality constraints for the time being, the problem to be solved becomes N Minimize C F 03 DF 0E 5C (P ) (6) T (^) F 0 i i i (^) 3 D 1 N subject to P F 02 DF 0E 5P F 03 D 0 (7) D (^) F 0i i (^) 3 D 1 This optimizing problem with equality constraint can be solved by the method of Lagrangian multipliers. In this method, the Lagrangian function

L ( P , P , P , λ ) = C (P ) F 02 BC (P ) F 02 BC (P ) F 02 Bλ( P F 02 DP F 02 DP F 02 DP ) 1 2 3 1 1 2 2 3 3 D 1 2 3 Necessary conditions for a minimum are F 0B 6L F 0B 6C F 03 D + λ ( - 1 ) = 0 F 0B 6P F 0B 6P 1 1 F 0B 6L F 0B 6C F 03 D + λ ( - 1 ) = 0 F 0B 6P F 0B 6P 2 2 F 0B 6L F 0B 6C F 03 D + λ ( - 1 ) = 0 F 0B 6P F 0B 6P 3 3 F 0B 6L F 03 DP F 02 DP F 02 DP F 02 DP F 03 D 0 F 0 D^1 2 B 6λ ----------------------- Page 14----------------------- Generalizing the above, the necessary conditions are

F 0B 6C

F 0i^ F 03 Dλ^ i = 1,2,…….,N^ (9) B 6Pi N and P F 02 DF 0E 5P F 03 D 0 (10) D (^) F 0 i i (^) 3 D 1 F 0B 6C Here (^) F 0 i is the change in production cost in unit i for a small change in B 6Pi

generation in unit i. Since change in generation in unit i will affect the production cost of this unit ALONE, we can write F 0B 6C dC i (^) F 0 i F 0 3 D^ (11) B 6P^ dP

N

and P F 02 DF 0E 5P F 03 D 0 (14) D (^) F 0i i (^) 3 D 1 The above two conditions give N+1 number of equations which are to be solved for the N+1 number of variables λ, P , P ,......... ..,P 1 2 N. Equation (13)

simply says that at the minimum cost operating point, the incremental cost for all the generating units must be equal. This condition is commonly known as EQUAL INCREMENTAL COST RULE. Equation (14) is known as POWER BALANCE EQUATION. ----------------------- Page 16----------------------- It is to be remembered that we have not yet considered the inequality constraints given by P F 0A 3P F 0A 3P i = 1,2,…….., N i min i i max

Fortunately, if the solution obtained without considering the inequality constraints satisfies the inequality constraints also, then the obtained solution will be optimum. If for one or more generator units, the inequality constraints are not satisfied, the optimum strategy is obtained by keeping these generator units in their nearest limits and making the other generator units to supply the remaining power as per equal incremental cost rule. ----------------------- Page 17----------------------- EXAMPLE 1The cost characteristic of two units in a plant are: C = 0.4 P 2 + 160 P + K Rs./h 1 1 1 1 C = 0.45 P 2 + 120 P + K 2 2 2 2 Rs. / h

where P and P are power output in MW. 1 2 Find the optimum load allocation

between the two units, when the total load is 162.5 MW. What will be the