UNIVERSITY OF NAIROBI SCHOOL OF ENGINEERING DEPARTMENT OF ELECTRICAL AND INFORMATION ENGINEERING

Transcription

1 UNIVERSITY OF NAIROBI SCHOOL OF ENGINEERING DEPARTMENT OF ELECTRICAL AND INFORMATION ENGINEERING DETERMINATION OF A COEFFICIENTS FOR ECONOMIC DISPATCH FUNCTIONS PROJECT INDEX: 131 SUBMITTED BY OUMA KENNEDY NGALA F17/39351/2011 SUPERVISOR: MR. P. M. MUSAU EXAMINER: DR. C. WEKESA PROJECT REPORT SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENT FOR THE AWARD OF THE DEGREE OF BACHELOR OF SCIENCE IN ELECTRICAL AND ELECTRONIC ENGINEERING OF THE UNIVERSITY OF NAIROBI 2015 DATE SUBMITTED: 13 TH MAY 2016

2 DECLARATION OF ORIGINALITY NAME OF STUDENT: OUMA KENNEDY NGALA REGISTRATION NUMBER: F17/39351/2011 COLLEGE: Architecture and Engineering FACULTY/SCHOOL/INSTITUTE: Engineering DEPARTMENT: Electrical and Information Engineering COURSE NAME: Bachelor of Science in Electrical and Electronic Engineering TITLE OF WORK: DETERMINATION OF A COEFFICIENTS FOR ECONOMIC DISPATCH FUNCTIONS 1) I understand what plagiarism is and I am aware of the university policy in this regard. 2) I declare that this final year project report is my original work and has not been submitted elsewhere for examination, award of a degree or publication. Where other people s work or my own work has been used, this has properly been acknowledged and referenced in accordance with the University of Nairobi s requirements. 3) I have not sought or used the services of any professional agencies to produce this work. 4) I have not allowed, and shall not allow anyone to copy my work with the intention of passing it off as his/her own work. 5) I understand that any false claim in respect of this work shall result in disciplinary action, in accordance with University anti-plagiarism policy. Signature:.. Date:..

3 CERTIFICATION The report has been submitted to the Department of Electrical and Information Engineering, University of Nairobi with my approval as supervisor: Mr. Peter Moses Musau. Date:..

4 DEDICATION I would like to dedicate this project to my mum and dad for being patient and always believing in me.

5 ACKNOWLEDGEMENT I wish to appreciate the Almighty God for His guidance and amazing grace throughout my life. I extend my gratitude and thanks to my supervisor Mr. P.M Musau, for his constant support throughout the course of my project. I am very thankful to him for always being there to help me shape the problem and provide insights towards the solution. I also appreciate my classmates Waswa Lewis, Keter Wycliffe and Bundi Kotonya for taking their time to brainstorm and problem solve with me in regards to this project. And to the teaching staff and non-teaching staff at the University of Nairobi, Department of Electrical and Information Engineering, I appreciate their selfless effort that enabled me to achieve my academic goals during the entire course of my studies. Finally I thank my family for their patience and understanding.

6 DETERMINATION OF A-COEFFICIENTS FOR ECONOMIC DISPATCH FUNCTIONS DECLARATION OF ORIGINALITY... ii CERTIFICATION...iii DEDICATION... iv ACKNOWLEDGEMENT... v LIST OF FIGURES...viii LIST OF TABLES... ix ABSTRACT... x CHAPTER 1: Introduction Summary Determination of A-coefficient for Economic Dispatch functions Problem Statement Objective... 2 CHAPTER 2: LITERARTURE REVIEW Input-Output Characteristics of Thermal Units Classical Economic Dispatch Neglecting Losses Classical Economic Dispatch Considering Line Losses Power Flow Analysis... 7 CHAPTER 3: LOSS COEFFICIENTS Transmission Loss Coefficients Determining A-coefficients Steps of Determining A-coefficients Flow chart CHAPTER 4: RESULTS AND ANALYSIS Results Results for the 14 Bus Test System Results for the 30 Bus Test System Results for the 57 Bus Test System Analysis CHAPTER 5: CONCLUSION AND RECOMMENDATION FOR FURTHER WORK Conclusion Recommendation for further work... 33

7 REFERENCES PROGRAM LISTING Function for Calculating power factor angle Function for Calculating Function for Calculating Reactance Function for Calculating A Coefficients... 37

8 LIST OF FIGURES Figure 2.1 Input Output Characteristics of a Generating Unit... 3 Figure 3.1 Power System Network Figure 3.2 :Power Triangle Figure 4.1 :One line diagram of IEEE 14 bus system Figure 4.2 :One line diagram of IEEE 30 bus system Figure 4.3 :One line diagram of IEEE 57 bus system... 26

9 LIST OF TABLES 4.1 The transmission losses A-coefficients for the 14 bus test system The transmission losses A-coefficients for the 30 bus test system The transmission losses A-coefficients for the 57 bus test system... 27

10 ABSTRACT The complexity and size of electrical power systems is rapidly increasing due to the higher demand. The areas being served by the power system require to be served in a controlled and coordinated manner at the minimum cost possible. In an interconnected power system the objective is to minimize the total operating cost of a power system while meeting the total real load plus transmission losses within generator limits of real power as well as total reactive load plus reactive transmission losses within generator limits of reactive power.[4][8] The objective of economic dispatch is to determine the power output of a power system under the constraints of load demand and transmission losses that will give minimal operational cost.for accurate representation of the response of a power system (power output), transmission losses have to be calculated [3]. To determine the transmission losses we need the transmission losses A coefficients and transmission losses B coefficients. The A coefficients are used to calculate the transmission reactive power losses while the B coefficients are used to calculate the transmission real power losses. In this project we will determine the transmission losses A coefficients using MATLAB as the simulation software. The A coefficients will be determined for the IEEE 14, 30 and 57 bus systems.

11 CHAPTER 1: Introduction Investigating the formulation and determination of transmission losses A-coefficients using available optimization methods. Key words: Determination, A coefficients, Economic dispatch, functions, optimization. Determination: this is the process of establishing something exactly, typically by calculation or research. Economic dispatch: this is the operation of generation facilities to produce energy at the lowest cost possible while serving the required load and taking into consideration the transmission and operational constraints. The primary object of economic dispatch problem is to reduce the operational cost of the system fulfilling the load demand with limit of constraints. Function: this is the relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. A-coefficients: these are the loss coefficients in the economic dispatch function with transmission losses taken into account. These coefficients are used in the calculation of reactive power loss of a power system. Optimization: this is the selection of the best element from an available set. Optimization entails choosing elements from a given set to maximize or minimize a given function thus finding the best value that suits the required function. The primary object of economic dispatch problem is to reduce the operational cost of the system fulfilling the load demand with limit of constraints. Thus optimization techniques or methods are required. For accurate representation of the response of a power system, transmission losses have to be calculated [3]. To determine the transmission losses we need the transmission losses A coefficients and transmission losses B coefficients. The A coefficients are used to calculate the transmission reactive power losses while the B coefficients are used to calculate the transmission real power losses. In this project we will determine the transmission losses A coefficients using MATLAB as the simulation software.

12 1.1 Summary Determination of A-coefficient for Economic Dispatch functions. This can be defined as the process of establishing one of the inputs known as the A-coefficient in the economic dispatch function which is the relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. This will be done through the research of available optimization techniques Problem Statement The main purpose of this project is to investigate the formulation and determination of transmission losses A-coefficients. MATLAB is to be used for the creation of a program for the investigation Objective To determine A-coefficients for economic dispatch functions. 2

13 CHAPTER 2: LITERARTURE REVIEW Power systems have become very large and energy planning is now very complex. The primary objective of any economic dispatch problem is to reduce the operating cost of the power system while fulfilling the load demand within the limit of constraints, this optimization methods/techniques are required. From the view point of optimization, various techniques including traditional and modern techniques have been developed. The factors influencing the cost of generation are generator efficiency, fuel cost and transmission losses. The most efficient generator may not give minimum cost since it may be located in a place where fuel cost is high. Transmission costs may be high if the plant is located far from the load centers. The economic dispatch problem basically determines the generation of different plants to minimize total operating cost. The economic dispatch problems is defined as that which minimizes the total operating cost of a power system while meeting the load plus transmission losses within the generator limits. 2.1 Input-Output Characteristics of Thermal Units This is the fundamental curve for a thermal plant. It is the plot of the input energy rate in British Thermal units per hour (Btu/h) or cost of fuel used per hour (Kshs/h) against the power output of plant. It is also known as the fuel cost curve. Figure 2.1 Input Output Characteristics of a Generating Unit 3

14 It can be observed from the Input-Output characteristics of a generating unit that the power output is limited by the maximum and minimum capacities of the generating unit. That is Generally, the input-out characteristic is non-linear. It is usually given by the quadratic function. (2.1) where a, b, and c are the coefficients of the input-outputs characteristics. 2.2 Classical Economic Dispatch Neglecting Losses We assume that the generators needed to meet a particular land demand are known. The total fuel cost is given by ( ) (2.2) where is the number of generators is the fuel cost of running generator i is the power generated by generator i The load demand is given by (2.3) This is a constrained optimization problem and it can be solved by the Lagrange multiplier method. First of all the Lagrange function should be formed by adding the constraint function to the objective function after the constraint function has been multiplied by an undetermined multiplier. ( ) (2.4) 4

15 Where is the Lagrangian Multiplier The necessary conditions for the extreme value of the Lagrange function are to set the first derivative of the Lagrange function with respect to each of the independent variable equal to 0. Therefore (2.5) Or Where =1, 2 (2.6) Since the fuel consumption function of each generating unit is only related to its own power output, Equation (2.6) can be written as (2.7) Where =1, 2 From Equation (2.7) we have (2.8) Equation (2.7) is the principle of equal incremental rate of economic power operation for multiple generating units. 2.3 Classical Economic Dispatch Considering Line Losses The total fuel cost is given by ( ) (2.9) 5

16 Considering losses, we have that (2.10) Where is the transmission power loss is the total demand is the power generated by generator i The Lagrange function is now written as (2.11) The necessary conditions for the extreme value of the Lagrange function are to set the first derivate of the Lagrange function with respect to each of the independent variable equal to 0. ( ) (2.12) Since the fuel consumption of each generating unit is only related to its own power output, equation (2.12) can be written as: ( ) (2.13) Therefore where =1, 2.m ( ) (2.14) where =1, 2.m 6

17 Let ( ) (2.15) where =1, 2.m is known as the penalty factor of the generating unit. It is also known as the correction coefficient for network losses. Replacing in equation (2.14), we now have (2.16) or where =1, 2.m (2.17) Equation (2.16) is known as the coordination equation of economic power operator. 2.4 Power Flow Analysis Power flow algorithms include the Newton Raphson method, Gauss Seidel method, DC power flow method and decoupled power flow methods. Newton method and decoupled power flow methods are used due to their good convergence and fast calculation speed. Power flow is also known as load flow. This is the name given to a network solution that shows currents voltages real and reactive power flows at every bus in the system. [6] In the power flow problem, the relationship between voltage and current at each bus is nonlinear also the relationship between real and reactive power consumption at a bus or the generated real power and scheduled voltage magnitude at a generator bus. Thus the power flow calculation involves the solution of nonlinear equations. 7

18 Generally for a network with n independent buses, we can write the following n equations (2.18) The matrix form is [ ] [ ] [ ] (2.19) OR [ ][ ] [ ] (2.20) where is the bus current injection vector is the bus voltage vector is the admittance matrix For the Y admittance matrix, its diagonal element is the self admittance of bus which equals the sum of all branch admittances connecting to bus The off diagonal element of the bus admittance matrix admittance between buses and. is the negative of branch Equation (2.20) can be written as Where is the bus impedance matrix 8

19 [ ] The bus current can be represented by the bus voltage and power as (2.21) Where is the complex power injection vector is the real power output of the generator connecting to bus i is the reactive power output of the generator connecting to bus i is the real power load connected to bus i is reactive power load connected to bus i. In the power flow problem, load demands are known variable, we define the power injections as (2.22) Substituting (2.21) into (2.18) (2.23) where =1, 2 (2.24) Substituting (2.22) and (2.23) into (2.24) where =1, 2 9

20 Therefore where =1, 2 (2.25) If we divide (2.25) into real and imaginary parts we get 2 equations for each bus with four variables i. Bus Real Power P ii. Reactive Power Q iii. Voltage V iv. Voltage Angle From the variables, we can have 3 types of buses: 1) PV Bus This bus is also known as a Generator bus or voltage controlled bus. Real power P and voltage V are known. The bus reactive power Q and angle of voltage are unknown. The bus connected to a generator is a PV Bus. 2) PQ bus In this bus real power P and reactive power Q are known. The magnitude and angle of voltage (V, ) are unknown.the bus connected to a load is a PQ bus. 3) Slack bus: This bus is also known as a swing bus or reference bus. The magnitude and angel of voltage (V, ) are known. Bus real power P and reactive power Q are unknown. Since the power loss of the network is unknown during power flow calculation, it is necessary to have one bus at which complex power is unspecified/unknown so that it can supply the difference or balance the system power. The slack bus is usually the largest generator. 10

21 CHAPTER 3: LOSS COEFICIENTS 3.1 Transmission Loss Coefficients The total system loss is the sum of the bus power [4] i.e. (3.1) An exact transmission loss formula was derived by Dopazo et al (1967) where NB is the number of buses Real power loss of the system Reactive power loss of the power system is the voltage of the th bus is the injected complex power at the ith bus. Consider the power system network below Figure 3.1 Power System Network 11

22 For the network (3.2) Where Substituting Equation (3.2) into Equation (3.1) (3.3) But ( ) Therefore Equation (3.3) becomes ( ) (3.4) where is the voltage angle at the th bus is the power factor angle and is given by 12

23 Equation (3.4) is expanded to give ( ) ( ) (3.5) Since Therefore is a symmetric matrix we have [ ] [ ] (3.6) Taking into account equation (3.6), equation (3.5) becomes ( ) (3.7) From trigonometric identities ( ) Equation (3.7) becomes ( ) (3.8) But 13

24 Therefore Equation (3.8) becomes ( ) ( ) (3.9) Separating the real and imaginary parts ( ) (3.10) ( ) (3.11) For the determination of transmission loss A-coefficients, we take into consideration the reactive power loss of the power system,. Consider the power triangle below Figure 3.2 : Power Triangle 14

25 Reactive power injected at the th bus is given by [4] Therefore (3.12) But Therefore (3.13) Equation (3.12) into (3.13) (3.14) Also (3.15) Equation (3.14) and (3.15) into Equation (3.11) ( ) (3.16) 15

26 This can be written as (3.17) where And ( ) and are the reactive power outputs of generator connected bus and bus respectively and are the reactive power loads connected bus and respectively and are the reactive powers injected at bus and bus j respectively It can be seen that A coefficients are used to calculate the reactive power loss of the power system. 3.2 Determining A-coefficients We now have for the A-coefficients ( ) From the data sheets for IEEE Bus Test Systems we are provided with: 1) The bus voltages and. 2) The line data which is used to formulate the matrix. The reactance is extracted from the matrix. 16

27 3) The complex power at each bus( ) which is used to calculate the power factor angles and.from the power triangle in Fig.3,we have that And * ( ) ( ) + * ( ) ( ) + 4) The voltage angle which is used to calculate as shown below Also Using the data provided and calculated we can now find the A coefficients for the required bus system. 17

28 3.2.1 Steps of Determining A-coefficients. 1. Choose the bus system which you would like the A coefficients to be calculated using matrixa(num) where num is the number of buses (14, 30 or 57 bus systems). 2. The line data and bus data is obtained for the chosen bus system. 3. Formulate the matrix using the line data and extract for the bus system chosen. is the imaginary part of the matrix. 4. Calculate the power factor angles and using the bus data of the bus system chosen. And * ( ) ( ) + * ( ) ( ) + 5. Extract the voltage angles and from the bus data of the bus system chosen then calculate and. Also 6. Extract the bus voltages and from the bus data of the bus system chosen. 7. Formulate the A coefficient matrix for the chosen bus system using the extracted and calculated variables. ( ) 18

29 3.3 Flow chart START Input number of buses matrixa( num ) Is num=14, 30 or 57? Obtain line data and bus data for (num) Using line data formulate Z BUS and extract X ij Calculate φ i and φ j using bus data Obtain δ i and δ j from bus data Obtain θ i using θ i δ i φ i Obtain θ j using θ j δ j φ j Obtain V i and V j from bus data Calculate A coefficients using calculated and obtained variables STOP 19

30 CHAPTER 4: RESULTS AND ANALYSIS 4.1 Results The transmission losses A- coefficients were determined for the 14 Bus Test System, 30 Bus Test System and the 57 Bus Test System. The determination was done using MATLAB as the simulation software Results for the 14 Bus Test System Figure 4.1 : One line diagram of IEEE 14 bus system 20

31 The transmission losses A-coefficients for the 14 bus test system was as shown below 4.1 Columns 1 through 7 -Inf Inf -Inf Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf Inf Columns 8 to 14 Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf

32 4.1.2 Results for the 30 Bus Test System Figure 4.2 : One line diagram of IEEE 30 bus system 22

33 The transmission losses A-coefficients for the 30 bus test system was as shown below 4.2 Column 1 to 11 -Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf -Inf -Inf -Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf

34 Column 12 to 22 -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf Inf 24

35 Column 23 to 30 -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf

36 4.1.3 Results for the 57 Bus Test System Figure 4.3 : One line diagram of IEEE 57 bus system 26

37 The transmission losses A-coefficients for the 30 bus test system was as shown below. Columns 1 through Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf Inf Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf Inf Inf 27

38 Columns 12 through 22 Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf 28

39 Columns 23 to 33 Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf

40 Columns 34 to 44 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf Inf Inf -Inf Inf -Inf

41 Columns 45 to 55 -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf Inf Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf Inf Inf Inf -Inf Inf -Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf -Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf Inf -Inf Inf

42 Columns 56 to 57 -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf Inf -Inf Inf -Inf Inf -Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf -Inf -Inf -Inf Inf -Inf

43 4.2 Analysis The transmission losses A coefficients were determined for the 14 bus test system, 30 bus test system and 57 bus test system as shown in Table 4.1, Table 4.2 and Table 4.3 respectively using MATLAB as the simulation software. The order of the A coefficient matrix was determined by the number of buses in the system. As seen in the results, the 14 bus test system has an A coefficient matrix of order 14x14, the 30 bus test system has an A coefficient matrix of order 30x30 and the 57 bus test system has an A coefficient matrix of order 57x57. The infinity values were due to the calculation of the power factor angle.in some cases, resulted in a zero making the denominator of the equation used to calculate the A coefficients a zero. That is ( ) When is a zero, the denominator is a zero resulting in A becoming infinity. It can also be seen that the number of infinity values increase as the number of buses increase. CHAPTER 5: CONCLUSION AND RECOMMENDATION FOR FURTHER WORK 5.1 Conclusion The project investigated and determined the transmission loss A coefficients. The A coefficients are defined as seen in the project and are used to calculate the reactive transmission power losses as shown in equation (3.17). 5.2 Recommendation for further work 1. The A coefficients can be determined for larger bus systems such as the IEEE 116 and IEEE 300 bus systems. 2. Equation 3.17 can be expressed in terms of Kron s formula and the A coefficients formulated. 33

44 REFERENCES [1] DP. Kothari and I.J. Nagrath, Modern Power Systems Analysis Third Edition. [2] Modern Optimization Techniques with Applications in Electric Power Systems. [3] Benedict E., Collins T., Gotham D., Hoffman S., Karipides D., Pekarek S. and Ramabhadran R, Losses in Electric Power Systems (1992) ECE Technical Reports Paper 266. [4] D.P Kothari and J.S Dhillon, Power Systems Optimization. [5] Vara Prasad, Z-Bus Matrix building Algorithm for a balanced power system using inspection method. [6] W.D Stevenson Jr., Elements Of Power System Analysis New York 1982 [7] Jizhong Zhu, Optimization of Power System Operation Second Edition. [8] FERC Economic Dispatch: Concepts, Practices and Issues-Presentation to Joint Board for The Study of ED(200511) [9] Josef Kallrath, Panos, M. Pardalos, Steffen Rebennack and Max Scheldt, Optimization in the Energy Industry. [10] Electricalnotes.wordpress.com-Importance of Reactive Power for Systems. [11] Debapriya Das, Electrical Power Systems. [12] A.J Wood, Power Generation Operation and Control. [13] Ratandeep Gupta, Rashmi Chandra, Vikas Chaudhary and Nitin Saxena, Optimal Load Dispatch Using B-Coefficient. 34

45 PROGRAM LISTING Function for Calculating power factor angle %This function calculates the power factor angle,phi function [ phi ] = phifxn( num ) %To get the reactive power generated at a bus i(qi), %we subtract the reactive power load at the bus(qli)from %the reactive power output of the generator connected to the bus(qgi) thus Qi = QGi - QLi %To get the real power injected at a bus(pi), %we subtract the real power load at the bus i(pli)from %the real power output of the generator connected to the bus(qgi) thus Pi = PGi - PLi %QGi is given in column 6 of busdatas %QLi is given in column 8 of busdatas %PGi is given in column 5 of busdatas %PLi is given in column 7 of busdatas busdat = busdatas(num); %to find phi,phi=arctan(qi/pi) phi= zeros(num,1); for n = 1:num phi(n) = atan2d((busdat(n,6)-busdat(n,8)),(busdat(n,5)-busdat(n,7))); end end Function for Calculating function [ theta ] = theta( num ) %This function calculates theta %Theta=delta-phi.Delta is given in column 4 of bus data while phi is %extracted from the phi function busdat = busdatas(num); phi = phifxn(num); delta = zeros(num,1); for n = 1:num delta(n) = busdat(n,4); end % calculating theta theta = delta - phi; end 35