# A Traditional Approach to Solve Economic Load Dispatch Problem of Thermal Generating Unit Using MATLAB Programming

DOI : 10.17577/IJERTV2IS90218

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#### A Traditional Approach to Solve Economic Load Dispatch Problem of Thermal Generating Unit Using MATLAB Programming

Ashish Dhamanda1, Arunesh Dutt2, Surya Prakasp, A.K.Bhardwaj4 1Ph.d Student (EE), SSET, SHIATS, Allahabad, U.P, India

2Assistant Professor (EEE), SGIET, NGBU, Allahabad, U.P, India

3Assistant Professor (EE), SSET, SHIATS, Allahabad, U.P, India

4Associate Professor (EE), SSET, SHIATS, Allahabad, U.P, India

Abstract

Economic load dispatch (ELD) problem is one of the most important in power system operation and planning. So Many models by using different techniques have been used to solve these problems. The main objective of the ELD problems is to determine the optimal combination of power outputs of all generating units so as to meet the required demand at minimum cost while satisfying the constraints. This paper presents a traditional approach to solve the ELD problem using Lambda iteration method (LIM) in MATLAB environment for two generator units and four separate cases has to be considered with and without transmission losses and generator constraints. Result obtained for the proposed method is compared with the all cases and find out the optimum case.

#### Keywords: Economic Load Dispatch (ELD), Satisfying constraints, Lambda Iteration Method (LIM), MATrix LABoratory (MATLAB).

1. Introduction

The sizes of electric power system are increasing rapidly to meet the power demand so it becomes necessary to operate the plant units most economically and with large interconnection of the electric networks, the energy crisis in the world and continuous rise in the prices, it is very essential to reduce the running charges of the electric energy i.e. reduce the fuel consumption for meeting a particular load demand. The main factor controlling the most desirable load allocation between various generating units is the total running cost [1].

The operating cost of a thermal plant is mainly the cost of the fuel. Fuel supplies for thermal can be coal, natural gas, oil, or nuclear fuel. The other costs such as costs of labour, supplies, maintenance, etc. being difficult to be determined and approximate, are assumed to vary as a fixed percentage of the fuel cost. Therefore, these costs are included in the fuel cost. Thus, the operating cost of a thermal plant, which is mainly the fuel cost, is given as a function of generation. This function is defined as a nonlinear function of plant generation. The cost of generation depends upon the system constraint for a particular load demand it means the cost of generation is not fixed for a particular load demand but depends upon the operating constraint of the sources [1],[2]. The ELD problem has been solved via many traditional optimization methods, including: Gradient-based techniques, Newton methods, linear programming, and quadratic programming [5]. The economic operation of a thermal unit, input-output modeling characteristic is significant. For this function considers a single unit consisting of a boiler, a turbine, and a generator as shown in figure 1 [3],[4].

Fig. 1 Simple Model of Thermal Generation System

2. ELD Problem Formulation

ELD is an important function in modern power system to schedule the power generator outputs with respect to the load demands, and to operate the power system most economically, the main objective of economic load dispatch is to allocate the optimal power generation from different units at the lowest possible

Where Pgimin, and Pgimax are the minimum and maximum limit of power generation of a ith plant.

2.3 Transmission Loss

The transmission loss can be calculated by the following equations:

cost while satisfying the system constraints.ELD problem can be mathematically formulated as follows-

PL = NG

i = 1

NG Pgi Bij Pgj (6)

j = 1

2.1 Objective Function

Minimize F(Pgi) = NG

i = 1

Fi(Pgi) (1)

Where Pgi and Pgj are the real power generations at ith and jth buses respectively. Bij are the loss coefficients or B- coefficients.1

2.4 Incremental Fuel Cost

Subject to:

The energy balance equation

The incremental fuel cost can be obtained from the following equation:

NG i = 1

Pgi = Pd (2)

(IC)i = (2 ai Pgi + bi) /hr (7)

Fi(Pgi) = NG

i = 1

(ai Pgi2 + bi Pgi + ci) /hr (3)

Where IC is incremental fuel cost. a is actual incremental cost curve. b is approximated (linear)

Where ai, bi and ci are the cost coef icients of ith units.

1. Constraints Function

Constraints details are given below:

1. #### Equality Constraints

The sum of real power generation of all the various units must always be equal to the total real power demand on the system.

incremental cost curve. Pg is total power generation [4].

Incremental fuel cost curve are shown in figure 2 as follows-

Pd = NG

i = 1

Pgi (4)

Where Pgi is the total real power generation. Pd is the total real power demand.

2. #### Inequality Constraints

Inequality constraints for power generating units are as follows:

Pgimin <= Pgi <= Pgimax (i = 1, 2., NG) (5)

Fig 2. Incremental Cost Curve of Generator i

For dispatching purposes, the cost is usually approximated by one or more quadratic segments, so the fuel cost curve in the active power generation, takes up a quadratic form.

3. Lambda Iteration Method

Lambda iteration method is more conventional to deal with the minimization of cost of generating the power at any demand. For more number of units, the Lambda iteration method is more accurate and incremental cost curves of all units are stored in memory.

Algorithm for Lambda iteration method is given below:

1. Guess the initial value of 0 with the use of cost curve equations.

2. Calculate Pgi0

i

i

3. Calculate NG Pg 0.

i = 1

i

i

4. Check whether NG Pg 0 = Pd

i = 1

PL(pu) = 0.0346 P1(pu 2 + 0.00643 P2(pu)2 (10)

)

)

Let us consider =12

5. Simulation And Result

The Lambda iteration method is applied in four cases with two generating unit to find out the minimum cost for any demand. The optimal results with the conventional Lambda iteration method will get.

In the first case transmission losses and generator constraints are neglected, in second case generator constraints are consider without transmission losses, in third case transmission losses are consider without generator constraint and the fourth case with transmission losses and generator constraint. All these simulation are done on MATLAB environment. The tables for each case are as follows-

NG Pgi d

NG Pgi d

 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.6684 260 3152.4 2 10.1391 400 4538.9 3 10.7443 580 6418.4 4 11.1142 690 7620.6

3152.4

 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.6684 260 2 10.1391 400 4538.9 3 10.7443 580 6418.4 4 11.1142 690 7620.6

0 – P (atolerancevalue)

i = 1

1. If NG Pg 0 < Pd, set a new value for, i.e.,

i = 1 i

= + and repeat from step (2) till the tolerance value is satisfied.

2. If NG Pg 0 > Pd, set a new value for, i.e.,

i = 1 i

= and repeat from step (2) till the tolerance value is satisfied.

3. Stop [3].

4. Numerical Example

Two generating units considered are having different characteristic. Their cost function characteristics are given by following equations-

F1 = 0.004P12+ 9.2P1+ 420 /r (8)

F2= 0.0029P22+ 8.5P2 + 350 /r (9)

The unit operating ranges are- 100 MW P1 200 MW

150 MW P2 500 MW

The transmission line losses can be calculated by the given expression

Table. 1: ELD without transmission line losses and generator constraints (For Case 1)

 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.4280 260 3164.2 2 10.1391 400 4538.9 3 10.7443 580 6418.4 4 11.3420 690 7631.3

Table. 2: ELD without transmission line losses and with generator constraints (For Case 2)

 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.6236 260 3161.0 2 10.6508 400 4540.8 3 11.7426 580 6661.1 4 12.3427 690 7637.3
 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.6236 260 3161.0 2 10.6508 400 4540.8 3 11.7426 580 6661.1 4 12.3427 690 7637.3

Table. 3: ELD with transmission line losses, without generator constraints (For Case 3)

10000

9000

8000

F u e l C o s t ( R / h r )

F u e l C o s t ( R / h r )

7000

6000

5000

4000

 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.9927 260 3189.8 2 10.7291 400 4639.8 3 11.7426 580 6661.1 4 12.4023 690 7988.8
 S. No. Lambda Power Demand (MW) Fuel Cost (F) /hr 1 9.9927 260 3189.8 2 10.7291 400 4639.8 3 11.7426 580 6661.1 4 12.4023 690 7988.8

3000

200 250 300 350 400 450 500 550 600 650 700 750 800

Power Demand (Mw)

Table. 4: ELD with transmission line losses and generator constraints (For Case 4)

From the above tables the response of four separate cases can be obtained:

10000

9000

8000

F u e l C o s t ( R / h r )

F u e l C o s t ( R / h r )

7000

6000

5000

4000

3000

200 250 300 350 400 450 500 550 600 650 700 750 800

Power Demand (Mw)

Graph. 1: Between Power Demand (Mw) and Fuel Cost ( /hr) (For Case 1)

Graph. 2: Between Power Demand (Mw) and Fuel Cost ( /hr) (For Case 2)

10000

9000

8000

F u e l C o s t ( R / h r )

F u e l C o s t ( R / h r )

7000

6000

5000

4000

3000

200 250 300 350 400 450 500 550 600 650 700 750 800

Power Demand (Mw)

Graph. 3: Between Power Demand (Mw) and Fuel Cost ( /hr) (For Case 3)

10000

9000

8000

F u e l C o s t ( R / h r )

F u e l C o s t ( R / h r )

7000

6000

5000

4000

3000

200 250 300 350 400 450 500 550 600 650 700 750 800

Power Demand (Mw)

Graph. 4: Between Power Demand (Mw) and Fuel Cost ( /hr) (For Case 4)

Comparison of above stated cases and respective graphs the combined responses can be obtained between Power Demand (Mw) and Fuel Cost ( /hr) as follow:

10000

9000

Case 3

6. Conclusion

For solving economic load dispatch problem of thermal generating units, we considered two generating units and each generating unit have four different cases. The first case is economic load dispatch(ELD) without transmission loss and generator constraints, second case is ELD without transmission loss and with generator constraints, third case is ELD with transmission loss and without generator constraints, and fourth case is ELD with transmission loss and with generator constraints. For each case a separate table and corresponding response we have obtained and the combined response of all separate cases also obtained after comparison of the above cases we find that the first case (ELD without transmission line losses and generator constraints) give the optimal value in comparison to the other cases. Thus we can conclude that the Lambda iteration method gives the better result and useful to solve ELD problem.

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8000

F u e l C o s t ( R / h r )

F u e l C o s t ( R / h r )

7000

6000

5000

4000

3000

Case 1

Case 4

Case 2

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200 250 300 350 400 450 500 550 600 650 700 750 800

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Graph. 5: Between Power Demand (Mw) and Fuel Cost ( /hr)

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