Power System Analysis

Mehdi Rahmani-Andebili

Power System Analysis Practice Problems, Methods, and Solutions

Power System Analysis

Mehdi Rahmani-Andebili

Power System Analysis Practice Problems, Methods, and Solutions

Mehdi Rahmani-Andebili State University of New York Buffalo, NY, USA

ISBN 978-3-030-84766-1 ISBN 978-3-030-84767-8 https://doi.org/10.1007/978-3-030-84767-8

(eBook)

# The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

Electric Power System Analysis is one of the fundamental courses of Electric Power Engineering major which is taught for junior students. The subjects include fundamental concepts in power system analysis, transmission line parameters, transmission line model and performance, modeling of power system components, and determination of network impedance and admittance matrices, load flow, and economic load dispatch. Like the previously published textbooks, this textbook includes very detailed and multiple methods of problem solutions. It can be used as a practicing textbook by students and as a supplementary teaching source by instructors. To help students study the textbook in the most efficient way, the exercises have been categorized in nine different levels. In this regard, for each problem of the textbook a difficulty level (easy, normal, or hard) and a calculation amount (small, normal, or large) have been assigned. Moreover, in each chapter, problems have been ordered from the easiest problem with the smallest calculations to the most difficult problems with the largest calculations. Therefore, students are suggested to start studying the textbook from the easiest problems and continue practicing until they reach the normal and then the hardest ones. On the other hand, this classification can help instructors choose their desirable problems to conduct a quiz or a test. Moreover, the classification of computation amount can help students manage their time during future exams and instructors give the appropriate problems based on the exam duration. Since the problems have very detailed solutions and some of them include multiple methods of solution, the textbook can be useful for the underprepared students. In addition, the textbook is beneficial for knowledgeable students because it includes advanced exercises. In the preparation of problem solutions, it has been tried to use typical methods of electrical circuit analysis to present the textbook as an instructor-recommended one. In other words, the heuristic methods of problem solution have never been used as the first method of problem solution. By considering this key point, the textbook will be in the direction of instructors’ lectures, and the instructors will not see any untaught problem solutions in their students’ answer sheets. The Iranian University Entrance Exams for the Master’s and PhD degrees of Electrical Engineering major is the main reference of the textbook; however, all the problem solutions have been provided by me. The Iranian University Entrance Exam is one of the most competitive university entrance exams in the world that allows only 10% of the applicants to get into prestigious and tuition-free Iranian universities. Butte, MT, USA

Mehdi Rahmani-Andebili

v

Contents

1

Problems: Fundamental Concepts in Power System Analysis . . . . . . . . . . . . .

1

2

Solutions of Problems: Fundamental Concepts in Power System Analysis . . .

13

3

Problems: Transmission Line Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37

4

Solutions of Problems: Transmission Line Parameters . . . . . . . . . . . . . . . . . .

43

5

Problems: Transmission Line Model and Performance . . . . . . . . . . . . . . . . . .

53

6

Solutions of Problems: Transmission Line Model and Performance . . . . . . . .

59

7

Problems: Network Impedance and Admittance Matrices . . . . . . . . . . . . . . . .

69

8

Solutions of Problems: Network Impedance and Admittance Matrices . . . . . .

75

9

Problems: Load Flow and Economic Load Dispatch . . . . . . . . . . . . . . . . . . . .

85

10

Solutions of Problems: Load Flow and Economic Load Dispatch . . . . . . . . . .

91

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

vii

About the Author

Mehdi Rahmani-Andebili is an Assistant Professor in the Electrical Engineering Department at Montana Technological University, MT, USA. Before that, he was also an Assistant Professor in the Engineering Technology Department at State University of New York, Buffalo State, NY, USA, during 2019–2021. He received his first M.Sc. and Ph.D. degrees in Electrical Engineering (Power System) from Tarbiat Modares University and Clemson University in 2011 and 2016, respectively, and his second M.Sc. degree in Physics and Astronomy from the University of Alabama in Huntsville in 2019. Moreover, he was a Postdoctoral Fellow at Sharif University of Technology during 2016–2017. As a professor, he has taught many courses such as Essentials of Electrical Engineering Technology, Electrical Circuits Analysis I, Electrical Circuits Analysis II, Electrical Circuits and Devices, Industrial Electronics, Renewable Distributed Generation and Storage, and Feedback Controls. Dr. Rahmani-Andebili has more than 100 single-author publications including textbooks, books, book chapters, journal papers, and conference papers. His research areas include Smart Grid, Power System Operation and Planning, Integration of Renewables and Energy Storages into Power System, Energy Scheduling and Demand-Side Management, Plug-in Electric Vehicles, Distributed Generation, and Advanced Optimization Techniques in Power System Studies.

ix

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Problems: Fundamental Concepts in Power System Analysis

Abstract

In this chapter, the problems concerned with the fundamental concepts of power system analysis are presented. The subjects include phasor representation of signals, voltage and current in power system, impedance and admittance, singlephase and three-phase power systems, complex power and its components, power generation and consumption concepts, per unit (p.u.) system, and power factor correction. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 1.1. What is the phasor representation of the voltage signal of Difficulty level Calculation amount 1) 1 V 2) 3) 0 V 4) 

● Easy ● Small

1.2. Represent the current signal of Difficulty level Calculation amount 1) 1 A 2) 3) 0 A 4) 

○ Normal ○ Normal

pffiffiffi 2 cos ðt Þ?

○ Hard ○ Large

pffiffiffi 2 sin ðt Þ in phasor domain.

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large



1.3. Define the signal of cos(2t + 30 ) in phasor domain. Difficulty level Calculation amount

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large





1.4. Represent the signal of 10 sin (t  60 ) in phasor form. Difficulty level Calculation amount

● Easy ● Small

○ Normal ○ Normal

○ Hard ○ Large

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Power System Analysis, https://doi.org/10.1007/978-3-030-84767-8_1

1

2

1 Problems: Fundamental Concepts in Power System Analysis

 



1.5. In the single-phase power system of Fig. 1.1, the voltage and current are as follows:   vðt Þ ¼ 110 cos ωt þ 30 V   iðt Þ ¼ 0:5 cos ωt  30 A Determine the impedance, resistance, and reactance of the system seen from the beginning of the line. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

 

Fig. 1.1 The power system of problem 1.5

1.6. In the single-phase power system of Fig. 1.1, the voltage and current are given as follows: pffiffiffi vðt Þ ¼ 100 2 cos ðt Þ V iðt Þ ¼

pffiffiffi   2 cos t  30 A

Determine the admittance, conductance, and susceptance of the system seen from the beginning of the line. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large

  1.7. The impedance of a generator, with the rated specifications of 20 kV and 200 MVA, is Z ¼ j0.2 p. u. Determine its reactance in percent if 21 kV and 100 MVA are chosen as the base voltage and power. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 11% 2) 10.5% 3) 11.7% 4) 9.07%

1

Problems: Fundamental Concepts in Power System Analysis

3

1.8. The reactance of a generator, with the nominal specifications of 14 kV and 500 MVA, is 1.1 p. u. Determine its impedance in percent if 20 kV and 100 MVA are chosen as the base voltage and power. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 30.8% 2) 10.78% 3) 60.8% 4) 57.8% 1.9. In the power bus of Fig. 1.2, determine the i3(t) if we know that i1(t) = 10 cos (10t) A, i2(t) = 10 sin (10t) A, and pffiffiffi   i4 ðt Þ = 10 2 cos 10t þ 45 A. Difficulty level Calculation amount pffiffiffi 1) 10 2 A  2) 3) 4) 0 A

● Easy ○ Small

○ Normal ● Normal

○ Hard ○ Large

Fig. 1.2 The power system of problem 1.9

1.10. In the single-phase power bus of Fig. 1.3, Vrms = 200 V and the equivalent impedance of the loads are Z1 = (8  j6) Ω and Z2 = (3 + j4) Ω. Calculate the total active power consumed in the bus. Difficulty level Calculation amount 1) 8 kW 2) 15 kW 3) 7.5 kW 4) 9 kW

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

Fig. 1.3 The power system of problem 1.10

1.11. Calculate the instantaneous power of a single-phase power system that its voltage and current are vðt Þ ¼ pffiffiffi pffiffiffi   110 2 cos ð120πt Þ V and iðt Þ ¼ 2 2 cos 120πt  60 A. Difficulty level ○ Easy ● Normal Calculation amount ○ Small ● Normal 1) 110 W  2) 220 cos (240πt  60 )W  3) 55 + 110 cos (240πt  60 ) W  4) 110 + 220 cos (240πt  60 ) W

○ Hard ○ Large

4

1 Problems: Fundamental Concepts in Power System Analysis

1.12. In the single-phase power system of Fig. 1.4, calculate the active and reactive powers transferred from bus 1 to bus 2. Consider the following data:

Difficulty level Calculation amount pffiffiffi 1) 10 W, 10 3 VAr pffiffiffi 2) 5 W, 4 3 VAr 3) 5 W, 4 VAr pffiffiffi 4) 5 3 W, 5 VAr

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

Fig. 1.4 The power system of problem 1.12

. Which one of the following choices 1.13. In the power system of Fig. 1.5,  is true? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) The first electric machine is generating reactive power, and the second electric machine is consuming reactive power. Moreover, the first and the second electric machines are working as a generator and a motor, respectively. 2) The first electric machine is consuming reactive power, and the second electric machine is generating reactive power. Moreover, the first and the second electric machines are working as a motor and a generator, respectively. 3) Both electric machines are generating equal reactive power which is consumed by the reactance of the line. Moreover, the first and the second electric machines are working as a generator and a motor, respectively. 4) Both electric machines are generating equal reactive power which is consumed by the reactance of the line. Moreover, the first and the second electric machines are working as a motor and a generator, respectively.

Fig. 1.5 The power system of problem 1.13

1.14. In the power bus of Fig. 1.6, the base voltage and power are 20 kV and 100 MVA, respectively. If a reactor is connected to this bus, determine its reactance in per unit (p.u.). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0.25 2) 0.5 3) 0.75 4) 2

1

Problems: Fundamental Concepts in Power System Analysis

5

Fig. 1.6 The power system of problem 1.14

1.15. Figure 1.7 shows the single-line diagram of a power system with the following specifications. Calculate the resistance of the load in per unit (p.u.) if the nominal quantities of the generator are chosen as the base quantities: G : 20 kV, 300 MVA T1 : 20=200 kV, 375 MVA T2 : 180=9 kV, 300 MW Load : 9 kV, 180 MW Difficulty level Calculation amount 1) 1.25 p. u. 2) 1.35 p. u. 3) 1.45 p. u. 4) 1.55 p. u.

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

Fig. 1.7 The power system of problem 1.15

1.16. Figure 1.8 illustrates the single-line diagram of a power system with the given information. Calculate P and Q in per unit   (p.u.). In this problem, assume that sin(15 )  0.25 and cos(15 )  0.96. Difficulty level ○ Easy ● Normal Calculation amount ○ Small ● Normal 1) P ¼ 0.5 p. u. , Q ¼ 0.08 p. u. 2) P ¼ 0.8 p. u. , Q ¼ 0.5 p. u. 3) P ¼ 0.8 p. u. , Q ¼  0.5 p. u. 4) P ¼ 0.5 p. u. , Q ¼  0.08 p. u.

○ Hard ○ Large

Fig. 1.8 The power system of problem 1.16

6

1 Problems: Fundamental Concepts in Power System Analysis

1.17. Calculate the complex power delivered to a factory that includes two loads with the following specifications: Inductive Load : P1 ¼ 60 kW, Q1 ¼ 660 kVAr Capacitive Load : P2 ¼ 240 kW, PF ¼ 0:8 Difficulty level ○ Easy Calculation amount ○ Small 1) (180 + j840) kVA 2) (300 + j480) kVA 3) (300 + j840) kVA 4) (180 + j480) kVA

● Normal ● Normal

○ Hard ○ Large

1.18. Figure 1.9 shows the single-line diagram of a balanced three-phase power system, in which a synchronous generator has been connected to a no-load transmission line through a transformer. Calculate the Thevenin reactance seen from the end of the transmission line. In this problem, the rated quantities of the generator are considered as the base values: G : 20 kV, 300 MVA, X G ¼ 20% T1 : 20=230 kV, 150 MVA, X T ¼ 0:1 p:u: Line : 176:33 km, X Line ¼ 1 Ω=km Difficulty level Calculation amount 1) 0.9 p. u. 2) 1.2 p. u. 3) 1.3 p. u. 4) 1.4 p. u.

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

Fig. 1.9 The power system of problem 1.18

1.19. For the three-phase power system of Fig. 1.10, the following specifications have been given. Determine the voltage drop of the line in percent: Line : Z ¼ ð10 þ j40Þ Ω=phase Load : V ¼ 100 kV, S ¼ 50 MVA, PF ¼ 0:8 Lagging Difficulty level Calculation amount 1) 8% 2) 16% 3) 19% 4) 24%

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

1

Problems: Fundamental Concepts in Power System Analysis

7

Fig. 1.10 The power system of problem 1.19

1.20. In the power system of Fig. 1.11, calculate the impedance of the load in per unit (p.u.) for the following specifications. In this problem, 20 kV (in the generator side) and 3 MVA are chosen as the base voltage and power: G : 20 kV, 3 MVA, 3% T1 : 20=230 kV, 3 MVA, 5% T2 : 230=11 kV, 3 MVA, 5% Load : 11 kV, 0:2 MVA, 0:8 Lagging M : 11 kV, 1 MVA, 5% C : 0:5 MVA Difficulty level ○ Easy Calculation amount ○ Small 1) (12 + j9) p. u. 2) (18 + j15) p. u. 3) (15 + j12) p. u. 4) (12.75 + j7.9) p. u.

● Normal ● Normal

○ Hard ○ Large

Fig. 1.11 The power system of problem 1.20

1.21. In the single-phase power bus of Fig. 1.12, the characteristics of the loads are as follows. Determine the total power factor of the bus: Load 1 : P1 ¼ 25 kW, Q1 ¼ 25 kVAr Load 2 : S2 ¼ 15 kVA, cos ðθ2 Þ ¼ 0:8 Leading Load 3 : P3 ¼ 11 kW, cos ðθ3 Þ ¼ 1 Difficulty level Calculation amount 1) 0.94 Lagging 2) 0.94 Leading 3) 0.6 Lagging 4) 0.6 Leading

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

8

1 Problems: Fundamental Concepts in Power System Analysis

Fig. 1.12 The power system of problem 1.21

1.22. In the single-phase power bus of Fig. 1.13, determine the capacitance of the shunt capacitor that needs to be connected to the bus to adjust its power factor at one for the following data: Load : S ¼ 20 kVA, cos ðθÞ ¼ 0:8 Lagging V rms = 200 V, f ¼ 50 Hz, π ffi 3 Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1 μF. 2) 1 mF. 3) 0.5 mF. 4) It is impossible to adjust the power factor of the bus at one.

Fig. 1.13 The power system of problem 1.22

1.23. In the single-phase power system of Fig. 1.14, three loads have been connected to the power bus in parallel. Determine the capacitance of the shunt capacitor that needs to be connected to the bus to adjust its power factor at one for the following specifications. Moreover, calculate the current of the line after connecting the shunt capacitor to the bus: Load 1 : ð8  j16Þ Ω Load 2 : ð0:8 þ j5:6Þ Ω Load 3 : S ¼ 5 kVA, cos ðθÞ ¼ 0:8 Lagging V rms = 200 V, f ¼ 60 Hz

Difficulty level Calculation amount 1) 100 μF, 20 A 2) 55 μF, 25 A 3) 800 μF, 25 A 4) 530 μF, 30 A

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

1

Problems: Fundamental Concepts in Power System Analysis

9

Fig. 1.14 The power system of problem 1.23

1.24. In the power system of Fig. 1.15, determine the reactive power of the shunt capacitor to keep the voltage of its bus at 1 p.u. In this problem, assume that cos(sin1(0.1))  0.995. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) 1.05 p. u. 2) 1.15 p. u. 3) 1.5 p. u. 4) 2.2 p. u.

Fig. 1.15 The power system of problem 1.4

1.25. In the three-phase power system of Fig. 1.16, two balanced three-phase loads with the star and delta connections have been connected to a three-phase power supply. Calculate the line voltage of the loads for the following specifications: E rms = 4 V, Z1 = j2 Ω, Z2 = ð2 þ j2Þ Ω, Z3 = j3 Ω, Z4 =  j6 Ω Difficulty level Calculation amount pffiffiffi 1) 2 3 V 2) pffiffiffi 3) 6 3 V 4)

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

Fig. 1.16 The power system of problem 1.25

10

1 Problems: Fundamental Concepts in Power System Analysis 

1.26. In the power system of Fig. 1.17, δ ¼ 15 . If the value of δ increases and E1 and E2 are kept constant, which one of the following choices is correct? In this problem, assume that I12 always lags E2 and Z ¼ jX, E1 = E1 < δ, E2 = E2 < 0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) |I12| will increase and its phase angle with respect to E2 will increase. 2) |I12| will decrease and its phase angle with respect to E2 will decrease. 3) |I12| will increase and its phase angle with respect to E2 will decrease. 4) |I12| will decrease and its phase angle with respect to E2 will increase.

Fig. 1.17 The power system of problem 1.26

1.27. Three loads with the following specifications, resulted from the load flow simulation, have been connected to the power bus shown in Fig. 1.18. If all the loads are modeled by an admittance, determine it in per unit (p.u.): Load 1 : P1 ¼ 2 p:u:, PF ¼ 0:8 Lagging Load 2 : P2 ¼ 2 p:u:, PF ¼ 0:8 Leading Load 3 : P3 ¼ 2 p:u:, PF ¼ 1

 Difficulty level Calculation amount 1) 6 p. u. 2) (2  j) p. u. 3) (2 + j) p. u. 4) (2  j2) p. u.

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

Fig. 1.18 The power system of problem 1.27

1.28. At the end of a three-phase power system, 400 V, 50 Hz, three capacitor banks (with triangle configuration) have been connected to the system. Determine the capacitance of each bank if they deliver 600 kVAr to the system. Difficulty level Calculation amount

○ Easy ○ Small

○ Normal ● Normal

● Hard ○ Large

1

Problems: Fundamental Concepts in Power System Analysis

1) 2) 3) 4)

11

5000 μF 4000 μF 0.004 μF 0.005 μF

1.29. The single-line diagram of a balanced three-phase power system is shown in Fig. 1.19. In this problem SB = 100 MVA and VB = 22 kV in the first bus. Calculate the impedance seen from the first bus if the following specifications are given: G : 22 kV, 90 MVA, X G ¼ 18% T1 : 22=220 kV, 50 MVA, X T1 ¼ 10% T2 : 220=11 kV, 40 MW, X T2 ¼ 6% T3 : 22=110 kV, 40 MW, X T3 ¼ 6:4% T4 : 110=11 kV, 40 MW, X T4 ¼ 8% M : 10:45 kV, 66:5 MVA, X M ¼ 18:5% TL1 : 220 kV, 48:4 Ω TL2 : 110 kV, 65:5 Ω Difficulty level Calculation amount 1) j0.14 2) j0.2 3) j0.22 4) j0.4

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

Fig. 1.19 The power system of problem 1.29

1.30. In the power system of Fig. 1.20, calculate the current of the load in per unit (p.u.) for the following specifications. In this problem, 100 V (in the generator side) and 1 kVA are chosen as the base voltage and power: G : 100 V T1 : 200=400 V, 1 kVA, X T1 ¼ 0:1 p:u:

12

1 Problems: Fundamental Concepts in Power System Analysis

Line : ZLine ¼ j8 Ω T2 : 200=200 V, 2 kVA, X T2 ¼ 0:1 p:u: Load : ZLoad ¼ j6 Ω Difficulty level Calculation amount 1) 0.25 p. u. 2) 1.5 p. u. 3) 0.5 p. u. 4) 1.25 p. u.

○ Easy ○ Small

○ Normal ○ Normal

● Hard ● Large

Fig. 1.20 The power system of problem 1.30

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

Abstract

In this chapter, the problems of the first chapter are fully solved, in detail, step by step, and with different methods.

2.1. As we know, cos(t) is usually chosen as the reference phasor. Hence, its phase angle is zero. Moreover, the amplitude of a phasor is normally shown in root-mean-square (rms) value. Therefore, the phasor representation of the signal of pffiffiffi 2 cos ðt Þ can be calculated as follows. Herein, “ ” is the symbol of phase angle.

Choice (1) is the answer. 2.2. The relation below holds between the signals of sin(t) and cos(t).   sin ðt Þ ¼ cos t  90 The signal of cos(t) is usually chosen as the reference phasor. In addition, pffiffiffi the amplitude of a phasor is normally shown in root-mean-square (rms) value. Therefore, the phasor of the signal of 2 sin ðt Þ can be represented as follows.







Choice (4) is the answer. 

2.3. The phasor of cos(2t + 30 ) can be defined as follows.

Herein, the signal of cos(t) is chosen as the reference phasor, and the amplitude of the phasor is presented in root-meansquare (rms) value. Choice (4) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Power System Analysis, https://doi.org/10.1007/978-3-030-84767-8_2

13

14

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

2.4. As we know, the relation below exists between the signals of sin(t) and cos(t).       sin ðt Þ ¼ cos t  90 ) sin t  60 ¼ cos t  150 

Therefore, the phasor of 10 sin (t  60 ) can be represented as follows.





Herein, the signal of cos(t) is chosen as the reference phasor, and the amplitude of the phasor is presented in root-meansquare (rms) value. Choice (3) is the answer. 2.5. Based on the information given in the problem, we have the following specifications:   vðt Þ ¼ 110 cos ωt þ 30 V

ð1Þ

  iðt Þ ¼ 0:5 cos ωt  30 A

ð2Þ

Transferring to phasor domain: ð3Þ



ð4Þ

The impedance is defined as follows: Z¼

V I

Solving (3)–(5):

 The real and imaginary parts of impedance are called resistance and reactance, respectively. Thus: R ¼ RealfZg ¼ 110 Ω pffiffiffi X ¼ ImagfZg ¼ 110 3 Ω Choice (2) is the answer.

Fig. 2.1 The power system of solution of problem 2.5

ð5Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

15

2.6. Based on the information given in the problem, we have the following specifications: pffiffiffi vðt Þ ¼ 100 2 cos ðt Þ V iðt Þ ¼

ð1Þ

pffiffiffi   2 cos t  30 A

ð2Þ

Transferring to phasor domain: ð3Þ ð4Þ

 The admittance is defined as follows: Y¼

I V

ð5Þ

Solving (3)–(5):







The real and imaginary parts of admittance are called conductance and susceptance, respectively. pffiffiffi G ¼ RealfYg ¼ 0:005 3 mho B ¼ ImagfYg ¼ 0:005 mho Choice (3) is the answer. 2.7. Based on the information given in the problem, we have the following specifications: V ¼ 20 kV, S ¼ 200 MVA, Z ¼ j0:2 p:u:

ð1Þ

V B ¼ 21 kV, SB ¼ 100 MVA

ð2Þ

The impedance of the generator has been presented in per unit (p.u.) value based on its rated quantities. Now, we need to update its per unit value based on the new base MVA and voltage as follows: Znew,p:u: ¼ Zold,p:u: 

 2  2 SB,new V B,old 100 MVA 20 kV   ¼ j0:2  ¼ j0:0907 200 MVA 21 kV SB,old V B,new

ð3Þ

) Znew,percent ¼ Znew,p:u:  100 ) Znew,percent ¼ j9:07% Choice (4) is the answer. 2.8. Based on the information given in the problem, we have the following specifications: V ¼ 14 kV, S ¼ 500 MVA, X ¼ 1:1 p:u:

ð1Þ

V B ¼ 20 kV, SB ¼ 100 MVA

ð2Þ

16

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

The impedance of the generator has been presented in per unit (p.u.) based on its rated quantities. We need to update its per unit value based on the new base MVA and voltage as follows: X new,p:u:

 2  2 SB,new V B,old 100 MVA 14 kV  ¼ X old,p:u:   ¼ 1:1  ¼ 0:1078 500 MVA 20 kV SB,old V B,new

ð3Þ

) X new,percent ¼ X new,p:u: 3 100 ) X new,percent ¼ 10:78% Choice (2) is the answer. 2.9. For this problem, we only need to apply KCL in the bus as follows. KCL can be applied in this bus, since all the currents have the same angular frequency (ω ¼ 10 rad/sec): 2 i1 ðt Þ þ i2 ðt Þ þ i3 ðt Þ þ i4 ðt Þ ¼ 0

ð1Þ

Based on the information given in the problem, we know that: i1 ðt Þ ¼ 10 cos ð10t Þ A

ð2Þ

i2 ðt Þ ¼ 10 sin ð10t Þ A

ð3Þ

pffiffiffi   i4 ðt Þ ¼ 10 2 cos 10t þ 45 A

ð4Þ

It is better to represent the currents in phasor domain, as can be seen in the following. Herein, the signal of cos(t) is ” is chosen as the reference phasor, the amplitude of the phasor is presented in root-mean-square (rms) value, and “ the symbol of phase angle: I1 þ I2 þ I3 þ I4 ¼ 0

ð5Þ ð6Þ ð7Þ



ð8Þ Solving (5)–(8):



 





By transferring back to time domain, we can write: i3 ðt Þ ¼ 0 A Choice (4) is the answer.

Fig. 2.2 The power system of solution of problem 2.9

ð8Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

17

2.10. As we know, active power is consumed by resistance of load and can be calculated for a single-phase system as follows: P ¼ RðI rms Þ2

ð1Þ

Based on the information given in the problem, we have the following specifications: V rms ¼ 200 V

ð2Þ

Z1 ¼ ð8  j6Þ Ω

ð3Þ

Z2 ¼ ð3 þ j4Þ Ω

ð4Þ

Using Ohm’s law for the first load: I rms,1 ¼

V rms 200 ¼ 20 A ¼ jZ1 j j8  j6j

ð5Þ

V rms 200 ¼ ¼ 40 A jZ2 j j3 þ j4j

ð6Þ

Applying Ohm’s law for the second load: I rms,2 ¼ Solving (1), (3), and (5): P1 ¼ R1 ðI rms,1 Þ2 ¼ 8  202 ¼ 3200 W

ð7Þ

P2 ¼ R2 ðI rms,2 Þ2 ¼ 3  402 ¼ 4800 W

ð8Þ

Solving (1), (4), and (6):

Therefore: PTotal ¼ P1 þ P2 ¼ 3200 þ 4800 ¼ 8000 W ) PTotal ¼ 8 kW Choice (1) is the answer.

Fig. 2.3 The power system of solution of problem 2.10

2.11. Instantaneous power of a single-phase power system can be calculated as follows: pðt Þ ¼ vðt Þiðt Þ

ð1Þ

Based on the information given in the problem, we have the following specifications: pffiffiffi vðt Þ ¼ 110 2 cos ð120πt Þ V

ð2Þ

18

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

pffiffiffi   iðt Þ ¼ 2 2 cos 120πt  60 A

ð3Þ

pffiffiffi pffiffiffi     pðt Þ ¼ 110 2 cos ð120πt Þ  2 2 cos 120πt  60 ¼ 440 cos ð120πt Þ cos 120πt  60

ð4Þ

Solving (1)–(3):

From trigonometry, we know that: 1 cos ðaÞ cos ðbÞ ¼ ð cos ða þ bÞ þ cos ða  bÞÞ 2

ð5Þ

Solving (4) and (5):         ) pðt Þ ¼ 110 þ 220 cos 240πt  60 W pðt Þ ¼ 220 cos 240πt  60 þ cos 60 Choice (4) is the answer. 2.12. Based on the information given in the problem, we have the following specifications: ð1Þ The current flowing from bus 1 to bus 2 can be calculated as follows: 



ð2Þ

The complex power transferred from bus 1 to bus 2 can be calculated as follows:

Choice (4) is the answer.

Fig. 2.4 The power system of solution of problem 2.12

2.13. Based on the information given in the problem, we have the following specifications: 

ð1Þ

The active and reactive powers flowing in the transmission line from bus 1 to bus 2 can be calculated as follows: P12 ¼

jV 1 jjV 2 j 200  200 sin ð30  0Þ ¼ 4000 < 0 W sin ðθ1  θ2 Þ ¼ 5 X

ð2Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

jV 1 jjV 2 j 200  200 sin ð0  ð30ÞÞ ¼ 4000 > 0 W sin ðθ2  θ1 Þ ¼ 5 X

ð3Þ

jV 1 j 200 ðjV 1 j  jV 2 j cos ðθ1  θ2 ÞÞ ¼ ð200  200 cos ð30  0ÞÞ  1071 VAr > 0 5 X

ð4Þ

jV 2 j 200 ð200  200 cos ð0  ð30ÞÞÞ  1071 VAr > 0 ðjV 2 j  jV 1 j cos ðθ2  θ1 ÞÞ ¼ 5 X

ð5Þ

P21 ¼ Q12 ¼ Q21 ¼

19

As can be noticed from (2) and (3), P12 < 0 and P21 > 0. Therefore, the active power flows from bus 2 to bus 1. In other words, the first and the second electric machines are working as a motor and a generator, respectively. However, as can be noticed from (4) and (5), Q12 ¼ Q21 > 0. Thus, the reactive power is generated by both machines and ultimately consumed in the transmission line. Choice (4) is the answer.

Fig. 2.5 The power system of solution of problem 2.13

2.14. Based on the information given in the problem, we have the following specifications: V B ¼ 20 kV, SB ¼ 100 MVA

ð1Þ

V R ¼ 20 kV, QR ¼ 200 MVAr

ð2Þ

The reactance of the reactor can be calculated as follows: QR ¼

ðV R Þ2 ðV Þ2 ð20 kV Þ2 ¼2Ω ) XR ¼ R ¼ XR QR 200 MVAr

ð3Þ

The base impedance in the bus can be calculated as follows: SB ¼

ðV B Þ2 ðV Þ2 ð20 kV Þ2 ) ZB ¼ B ¼ ¼4Ω ZB SB 100 MVA

The reactance of the reactor in per unit (p.u.) can be determined by using (3) and (4): X R,p:u: ¼

XR 2 ) X R,p:u: ¼ ¼ 0:5 Ω 4 ZB

Choice (2) is the answer.

Fig. 2.6 The power system of solution of problem 2.14

ð4Þ

20

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

2.15. Figure 2.7 shows the single-line diagram of the power system with the indicated zones. Based on the information given in the problem, we have the following specifications: G : 20 kV, 300 MVA

ð1Þ

T1 : 20=200 kV, 375 MVA

ð2Þ

T2 : 180=9 kV, 300 MW

ð3Þ

Load : 9 kV, 180 MW

ð4Þ

V B1 ¼ 20 kV, SB ¼ 300 MVA

ð5Þ

The resistance of the purely resistive load can be calculated as follows: P¼

ðV Þ2 ðV Þ2 ð9 kV Þ2 )R¼ ¼ ¼ 0:45 Ω R P 180 MW

ð6Þ

The base voltage in the third zone can be calculated as follows: V B3 ¼ 20 kV 

200 9  ¼ 10 kV 20 180

ð7Þ

The base impedance in the third zone can be calculated as follows: SB ¼

ðV B3 Þ2 ðV Þ2 ð10 kV Þ2 1 ) Z B3 ¼ B3 ¼ ¼ Ω ZB SB 300 MVA 3

ð8Þ

The resistance of the load in per unit (p.u.) can be determined by using (6) and (8): Rp:u: ¼

R 0:45 ) Rp:u: ¼ 1 ) Rp:u: ¼ 1:35 p:u: ZB 3

Choice (2) is the answer.

Fig. 2.7 The power system of solution of problem 2.15

2.16. Based on the information given in the problem, we have the following specifications:     sin 15  0:25, cos 15  0:96

ð1Þ ð2Þ



ð3Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

21

Z ¼ j0:5 p:u:

ð4Þ

The active power flowing through the transmission line can be calculated as follows: P12 ¼

      jV 1 jjV 2 j 11 sin 0  15 sin ðθ1  θ2 Þ ) P12 ¼ ¼ 2 sin 15 0:5 X

ð5Þ

Solving (1) and (5): P12 ¼ 2  0:25 ) P12 ¼ 0:5 p:u

ð6Þ

The reactive power flowing through the transmission line can be calculated as follows: Q21 ¼

       jV 2 j 1  1  cos 15  0 ¼ 2 1  cos 15 ðjV 2 j  jV 1 j cos ðθ2  θ1 ÞÞ ¼ 0:5 X

ð7Þ

Solving (1) and (7): Q21 ¼ 2ð1  0:96Þ ¼ 0:08 p:u

ð8Þ

In bus 2, we can write: Q21 þ Q ¼ 0 ) Q ¼ Q21 ) Q ¼ 0:08 p:u: Choice (4) is the answer.

Fig. 2.8 The power system of solution of problem 2.16

2.17. Based on the information given in the problem, we have the following specifications: Inductive Load : P1 ¼ 60 kW, Q1 ¼ 660 kVAr

ð1Þ

Capacitive Load : P2 ¼ 240 kW, PF 2 ¼ 0:8

ð2Þ

S1 ¼ P1 þ jQ1 ¼ ð60 þ j660Þ kVA

ð3Þ

From (1), we can write:

In addition, from (2), we can write:

 

ð4Þ

22

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

In (4), a negative sign was applied in phase angle for the complex power, as the power factor of the load is capacitive: STotal ¼ S1 þ S2 ¼ ð60 þ j660Þ kVA þ ð240  j180Þ kVA ) STotal ¼ ð300  j480Þ kVA Choice (2) is the answer. 2.18. Based on the information given in the problem, we have the following specifications: G : 20 kV, 300 MVA, X G ¼ 20%

ð1Þ

T1 : 20=230 kV, 150 MVA, X T ¼ 0:1 p:u:

ð2Þ

Line : 176:33 km, X Line ¼ 1 Ω=km

ð3Þ

V B1 ¼ 20 kV, SB ¼ 300 MVA

ð4Þ

The impedance of the generator will not change as its rated values have been chosen as the base quantities. Hence: X G,p:u: ¼ 0:2 ð5Þ

ð5Þ

Now, we need to update the per unit (p.u.) value of the transformer’s impedance based on the new base MVA and voltage as follows: X new,p:u:

 2 SB,new V B,old ¼ X old,p:u:   SB,old V B,new

X T,new,p:u: ¼ 0:1 

 2 300 20  ¼ 0:2 p:u: 150 20

ð6Þ

ð7Þ

To present the impedance of the line in per unit (p.u.) value, we need to determine the base impedance in the second zone (see Fig. 2.9.2), as follows: V B2 ¼ 20 kV  SB ¼

230 ¼ 230 kV 20

ðV B2 Þ2 ðV Þ2 ð230 kV Þ2 ) Z B2 ¼ B2 ¼ ¼ 176:33 Ω ZB SB 300 MVA

ZLine,p:u: ¼

176:33 km  1 Ω=km ZLine ) ZLine,p:u: ¼ ¼ 1 p:u: Z B2 176:33

ð8Þ ð9Þ ð10Þ

Figure 2.9.3 shows the impedance diagram of the power system by using (5), (7), and (10). The Thevenin reactance, seen from the end of the transmission line, can be calculated as follows: X Thevenin,p:u: ¼ X G,p:u: þ X T,new,p:u: þ X Line,p:u: ¼ 1 þ 0:2 þ 0:2 ) X Thevenin,p:u: ¼ 1:4 p:u: Choice (4) is the answer.

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

23

Fig. 2.9 The power system of solution of problem 2.18

2.19. Based on the information given in the problem, we have the following specifications: Line : Z ¼ ð10 þ j40Þ Ω=phase

ð1Þ

Load : V ¼ 100 kV, S ¼ 50 MVA, PF ¼ 0:8 Lagging

ð2Þ

First, we should solve the problem for the single-phase system. The voltage of the load is chosen as the reference. Hence: V2,ph

  100 ¼ pffiffiffi < 0 kV 3

ð3Þ

The current of the load can be calculated as follows: ð3Þ Applying KVL:

ð4Þ

jV1,L j ¼

pffiffiffi 3 V1,ph ¼ 116:725 kV

ð5Þ

24

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

The voltage drop of the line in percent can be calculated as follows: ΔV% ¼

116:725 kV  ð100 < 0Þ kV jV1,L j 2 jV2,L j  100 ¼  100 ) ΔV% ¼ 16% jV2,L j ð100 < 0Þ kV

Choice (2) is the answer.

Fig. 2.10 The power system of solution of problem 2.19

2.20. Based on the information given in the problem, we have the following specifications: V B1 ¼ 20 kV, SB ¼ 3 MVA

ð1Þ

G : 20 kV, 3 MVA, 3%

ð2Þ

T1 : 20=230 kV, 3 MVA, 5%

ð3Þ

T2 : 230=11 kV, 3 MVA, 5%

ð4Þ

Load : 11 kV, 0:2 MVA, 0:8 Lagging

ð5Þ

M : 11 kV, 1 MVA, 5%

ð6Þ

C : 0:5 MVA

ð7Þ

The base voltage in the third zone can be calculated as follows: V B3 ¼ 20 kV 

230 11  ¼ 11 kV 20 230

ð8Þ

The impedance of the load can be calculated as follows: ð9Þ The base impedance in the third zone can be calculated as follows: SB ¼

ðV B3 Þ2 ðV Þ2 ð11 kV Þ2 121 ) Z B3 ¼ B3 ¼ Ω ¼ 3 Z B3 SB 3 MVA

The impedance of the load in per unit (p.u.) can be determined by using (9) and (10):

Choice (1) is the answer.

ð10Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

25

Fig. 2.11 The power system of solution of problem 2.20

2.21. The power factor of the bus can be calculated as follows: PF Total ¼

P Pi PTotal ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P P STotal ð Pi Þ2 þ ð Qi Þ2

ð1Þ

Based on the information given in the problem, we have the following specifications: Load 1 : P1 ¼ 25 kW, Q1 ¼ 25 kVAr

ð2Þ

Load 2 : S2 ¼ 15 kVA, cos ðθ2 Þ ¼ 0:8 Leading

ð3Þ

Load 3 : P3 ¼ 11 kW, cos ðθ3 Þ ¼ 1

ð4Þ

P2 ¼ S2 cos ðθ2 Þ ) P2 ¼ 15  0:8 ¼ 12 kW

ð5Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Q2 ¼ S2 sin ðθ2 Þ ) Q2 ¼ 15  1  ð0:8Þ2 ¼ 9 kVAr

ð6Þ

From (3), we can write:

In (6), a negative sign was applied in the formula, as the power factor of the load is leading. From (4), we can conclude the following term, since the power factor is unit: Q3 ¼ 0 Solving (1)–(7): 25 þ 12 þ 11 48 PF Total ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ) PF Total ¼ 0:94 ð25 þ 12 þ 11Þ2 þ ð25 þ ð9Þ þ 0Þ2 ð48Þ2 þ ð16Þ2 Since ∑Qi ¼ 16 kVAr > 0, the total power factor is lagging. Choice (1) is the answer.

Fig. 2.12 The power system of solution of problem 2.21

ð7Þ

26

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

2.22. Based on the information given in the problem, we have the following specifications: Load : S ¼ 20 kVA, cos ðθÞ ¼ 0:8 Lagging

ð1Þ

V rms ¼ 200 V, f ¼ 50 Hz, π ffi 3

ð2Þ

Since the final power factor of the bus must be unit, the whole reactive power of the load must be supplied by the shunt capacitor. In other words, the net reactive power of the bus must be zero: QNet ¼ QLoad þ ðQC Þ ¼ 0 ) QC ¼ QLoad

ð3Þ

From (1), we can write: QLoad ¼ SLoad sin ðθLoad Þ ¼ 20 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  ð0:8Þ2 ¼ 12 kVAr

ð4Þ

Solving (3) and (4): QC ¼ 12 kVAr

ð5Þ

As we know, the reactive power of a single-phase capacitor can be determined as follows: QC ¼

V rms 2 ¼ ωCV rms 2 ¼ 2πfCV rms 2 Xc

ð6Þ

Solving (2), (5), and (6): 12000 ¼ 2  3  50  C  2002 ) C ¼

12000 ) C ¼ 1 mF 12  106

Choice (2) is the answer.

Fig. 2.13 The power system of solution of problem 2.22

2.23. Based on the information given in the problem, we have the following specifications: Load 1 : ð8  j16Þ Ω

ð1Þ

Load 2 : ð0:8 þ j5:6Þ Ω

ð2Þ

Load 3 : S ¼ 5 kVA, cos ðθÞ ¼ 0:8 Lagging

ð3Þ

V rms ¼ 200 V, f ¼ 60 Hz

ð4Þ

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

27

The total reactive power of the loads must be supplied by the shunt capacitor, since the final power factor of the bus is adjusted at one. In other words, the net reactive power of the bus must be zero: QNet ¼ Q1 þ Q2 þ Q3 þ ðQC Þ ¼ 0 ) QC ¼ Q1 þ Q2 þ Q3

ð5Þ

From (1) and (4), we can write: ðV rms Þ2 2002 ¼ ¼ ð1 þ j7Þ kVA  Z1 ð8  j16Þ

ð6Þ

ðV rms Þ2 2002 ¼ ¼ ð1  j2Þ kVA  Z2 ð0:8 þ j5:6Þ

ð7Þ

S1 ¼ From (2) and (4), we have: S2 ¼ From (3), we can write:

ð8Þ Solving (5)–(8): QC ¼ 7 þ ð2Þ þ 3 ¼ 8 kVAr

ð9Þ

As we know, reactive power of a single-phase capacitor can be determined as follows: QC ¼

V rms 2 ¼ ωCV rms 2 ¼ 2πfCV rms 2 Xc

ð10Þ

Solving (4), (9), and (10): 8000 ¼ 2  3:14  60  C  2002 ) C ¼

8000 ) C ¼ 530 μF 15:072  106

As it was mentioned earlier, after connecting the shunt capacitor to the power bus, the net reactive power of the bus is zero because its power factor is unit. The current of the line can be calculated as follows:

) I ¼ 30 A Choice (4) is the answer.

Fig. 2.14 The power system of solution of problem 2.23

28

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

2.24. Based on the information given in the problem, we have the following specifications: jV 1 j ¼ jV 2 j ¼ 1 p:u:

ð1Þ

θ1 ¼ 0

ð2Þ

X ¼ 0:1 p:u:

ð3Þ

P12 ¼ 1 p:u:

ð4Þ

  cos sin 1 ð0:1Þ  0:995

ð5Þ

The active power flowing through the transmission line can be calculated as follows: P12 ¼

jV 1 jjV 2 j 11 sin ð0  θ2 Þ ) sin ðθ2 Þ ¼ 0:1 ) θ2 ¼ sin 1 ð0:1Þ ð6Þ sin ðθ1  θ2 Þ ) 1 ¼ 0:1 X

ð6Þ

The reactive power flowing through the transmission line can be calculated as follows: Q21 ¼

jV 2 j 1 ðjV 2 j  jV 1 j cos ðθ2  θ1 ÞÞ ¼ ð1  cos ðθ2  0ÞÞ ¼ 10ð1  cos ðθ2 ÞÞ X 0:1

ð7Þ

Solving (5)–(7):          Q21 ¼ 10 1  cos sin 1 ð0:1Þ ¼ 10 1  cos  sin 1 ð0:1Þ ¼ 10 1  cos sin 1 ð0:1Þ ¼ 10ð1  0:995Þ ¼ 0:05 p:u:

ð8Þ

To keep the voltage of the bus at 1 p.u., the whole reactive power of the bus must be compensated by the shunt capacitor. In other words, the net reactive power of the bus must be zero: QNet ¼ Q21 þ ðQC Þ þ QLoad ¼ 0 ) QC ¼ Q21 þ QLoad ¼ 0:05 þ 1 ) QC ¼ 1:05 p:u: Choice (1) is the answer.

Fig. 2.15 The power system of solution of problem 2.24

2.25. Based on the information given in the problem, we have the following specifications: E rms ¼ 4 V, Z1 ¼ j2 Ω, Z2 ¼ ð2 þ j2Þ Ω, Z3 ¼ j3 Ω, Z4 ¼ j6 Ω

ð1Þ

To solve this problem, we should convert the triangle (delta) connection to the star (wye) connection (see Fig. 2.16.2) and analyze the single-phase system shown in Fig. 2.16.3. We can connect the neutral node of the loads to each other, as the system is a balanced system.

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

29

As we know, the relation below exists between the impedance of a balanced triangle (delta) connection and its equivalent balanced star (wye) connection: 1 1 ZY ¼ ZΔ ) Z04 ¼ Z4 ¼ j2 Ω 3 3

ð2Þ

Since the power system is a balanced system, no current flows through the neutral line and Z2. Hence, no voltage drop occurs across Z2. Therefore, we can ignore this impedance in the diagram of the single-phase system, as is illustrated in Fig. 2.16.3. Applying voltage division rule:



 



 

ð3Þ

As we know, the relation below holds between the phase and line voltage:

ð4Þ

Choice (4) is the answer.

Fig. 2.16 The power system of solution of problem 2.25

30

2

Solutions of Problems: Fundamental Concepts in Power System Analysis

2.26. Based on the information given in the problem, I12 always lags E2. Moreover, we have: δ ¼ 15



ð1Þ

E 1 ¼ Const:, E2 ¼ Const:

ð2Þ

Z ¼ jX

ð3Þ

E1 ¼ E 1 < δ

ð4Þ

E2 ¼ E 2 < 0

ð5Þ

The current in the transmission line can be calculated as follows: I12 ¼

E 1 < δ  E2 < 0 E1 cos ðδÞ þ jE 1 sin ðδÞ  E 2 ðE1 cos ðδÞ  E 2 Þ þ jE 1 sin ðδÞ ¼ ¼ jX jX jX

ð6Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðE 1 cos ðδÞ  E2 Þ2 þ ðE1 sin ðδÞÞ2 jI12 j ¼ X

¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E 1 2 cos 2 ðδÞ  2E 1 E 2 cos ðδÞ þ E2 2 þ E 1 2 sin 2 ðδÞ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E1 2 þ E 2 2  2E 1 E 2 cos ðδÞ ¼ X X   E 1 sin ðδÞ π < I12 ¼ tan 1  2 E 1 cos ðδÞ  E2

ð7Þ ð8Þ

As can be noticed from (7), by increasing δ and keeping E1 and E2 constant, cos(δ) will decrease, and consequently |I12| will increase. Moreover, as can be noticed from (8), by increasing δ and keeping E1 and E2 constant, sin(δ) and cos(δ) will increase and decrease, respectively, and consequently r0. Herein, the distance between the phases is not changed. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) It will decrease. 2) It will not change. 3) It will increase. 4) It can decrease or increase.

Fig. 3.6 The power system of problem 3.8

3.9. Figure 3.7 illustrates two single-phase transmission lines. The Geometrical Mean Radius (GMR) of each conductor is r0. In Fig. 3.7 (b), conductors “2” and “3” are for sending power, and conductor “1” is for receiving power. What relation should be held between D and r0 so that the inductances of the transmission lines become equal? Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) D ¼ 94 r 0 0 2) D ¼ 37 16 r 3) D ¼ 52 r 0 0 4) D ¼ 21 8 r

40

3

Problems: Transmission Line Parameters

Fig. 3.7 The power system of problem 3.9

3.10. Figure 3.8 shows a single-phase line including two conductors (“2” and “3”) for sending and one conductor (“1”) for receiving power. The Geometrical Mean Radius (GMR) of each conductor is r0. Calculate the capacitance of the line in F/m. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 4πε0 1) 2 ln ðrD0 Þ 0 2) 3 ln2πε2D ð r0 Þ 3) 3 ln4πε0D ð2r0 Þ 4πε0 4) 3 ln ðrD0 Þ

Fig. 3.8 The power system of problem 3.10

3.11. Figure 3.9 illustrates two three-phase transmission lines. The Geometrical Mean Radius (GMR) of each conductor is r0 and r0 < d. What relation should be held between d and r0 so that the inductance of the transmission lines become equal? Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 0 1) d ¼ pr ffiffi2. 2) d ¼ 2r0. pffiffiffi 3) d ¼ 2r 0 . 4) No possible relation can be found.

Fig. 3.9 The power system of problem 3.11

3.12. Which one of the arrangements of a three-phase transmission line, shown in Fig. 3.10, has the least inductance and the most capacitance? The Geometrical Mean Radius (GMR) of each conductor is r0. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large

3

Problems: Transmission Line Parameters

41

Fig. 3.10 The power system of problem 3.12

3.13. What difference can we see in the inductance of a transmission line if we change the conductor arrangements from the two-bundling to three-bundling, as can be seen in Fig. 3.11? The Geometrical Mean Radius (GMR) of each conductor is r0 and D ¼ 4r0. Herein, the distance between the phases is kept constant. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) A decrease about 23  107 ln ð2Þ 2) No change 3) An increase about 23  107 ln ð2Þ 4) An increase about 32  107 ln ð2Þ

Fig. 3.11 The power system of problem 3.13

4

Solutions of Problems: Transmission Line Parameters

Abstract

In this chapter, the problems of the third chapter are fully solved, in detail, step by step, and with different methods.

4.1. Decreasing Corona power loss is the main purpose of conductors bundling in transmission lines which is caused by reducing effective electric filed around conductors. Choice (3) is the answer. 4.2. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0. Therefore: GMR ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi ðr 0  DÞ  ðr 0  DÞ ) GMR ¼ r 0 D

22

Choice (3) is the answer.

Fig. 4.1 The power system of solution of problem 4.2

4.3. As we know, the inductance and capacitance of a transmission line can be determined as follows: L # ¼ 2  107 ln C"¼



GMD GMR"



2πε  0  ln

GMD GMR "

Bundling of conductors of a transmission line can increase its Geometrical Mean Radius (GMR). Therefore, the inductance and the capacitance of the transmission line will decrease and increase, respectively. Moreover, based on the relation below, the characteristic impedance will decrease: rffiffiffiffiffiffiffi L# ZC # ¼ C" Choice (3) is the answer.

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Power System Analysis, https://doi.org/10.1007/978-3-030-84767-8_4

43

44

4 Solutions of Problems: Transmission Line Parameters

4.4. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0. The Geometrical Mean Radius (GMR) of the bundled conductors can be determined as follows: GMR ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðD11 D12 D13 D14 Þ  ðD22 D21 D23 D24 Þ  ðD33 D31 D32 D34 Þ  ðD44 D41 D42 D43 Þ

44

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 4 4 4 4 ðD11 D12 D13 D14 Þ ¼ D11 D12 D13 D14 ¼ r 0  D  D  D 2 ¼ r 0 D3 2

44

GMR ¼

p ffiffiffiffiffiffiffiffiffiffiffiffiffi 8 2r 0 2 D6

Choice (4) is the answer.

Fig. 4.2 The power system of solution of problem 4.4

4.5. Based on the information given in the problem, we know that the radius of each conductor is r. Therefore, the Geometrical Mean Radius (GMR) of each conductor is: r 0 ¼ re4 1

Therefore, the Geometrical Mean Radius (GMR) of the bundled conductors is: GMR ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðD11 D12 D13 D14 Þ  ðD22 D21 D23 D24 Þ  ðD33 D31 D32 D34 Þ  ðD44 D41 D42 D43 Þ

44

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 4 4 1 7 1 7 4 4 ðD11 D12 D13 D14 Þ ¼ D11 D12 D13 D14 ¼ r 0  2r  2r  2r 2 ¼ e4 22 r 4 ¼ e16 28 r

44

GMR ¼ 1:722r Choice (2) is the answer.

Fig. 4.3 The power system of solution of problem 4.5

4.6. Based on the information given in the problem, we know that conductors “1” and “3” are for sending power and conductor “2” is for receiving power. Moreover, the radius of each conductor is r0. To calculate the inductance of a single-phase transmission line, we need to calculate the sum of the inductances of power sending and power receiving lines, as they are connected in series. Therefore: LTotal ¼ L13 þ L2 ¼ 2  107 ln

    GMD13 GMD2 þ 2  107 ln GMR13 GMR2

ð1Þ

4

Solutions of Problems: Transmission Line Parameters

45

GMD13 ¼ GMD2 ¼ GMR13 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi DD¼D

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi r 0  2D  r 0  2D ¼ 2Dr 0

22

GMR2 ¼ r 0

ð2Þ ð3Þ ð4Þ

Solving (1)–(4): LTotal ¼ 2  10

7

      D D D D 7 7 ln pffiffiffiffiffiffiffiffiffi þ 2  10 ln 0 ¼ 2  10 ln pffiffiffiffiffiffiffiffiffi  0 r 2Dr 0 2Dr0 r

¼ 2  10

7



3

D2 ln 1 3 22 r 0 2

 ¼ 2  10

7



   32 D 1 ln 0 þ ln 1 r 22

    D LTotal ¼ 107 3 ln 0  ln ð2Þ r Choice (4) is the answer.

Fig. 4.4 The power system of solution of problem 4.6

4.7. Based on the information given in the problem, we know that conductor “1” is for sending power and conductors “2” and “3” are for receiving power. Moreover, the Geometrical Mean Radius (GMR) of each conductor is r0. To calculate the inductance of a single-phase transmission line, we need to individually calculate the inductances of power sending line and power receiving line and then add them up, as they are in series. Therefore: LTotal ¼ L1 þ L23 ¼ 2  10

7

    GMD1 GMD23 7 ln þ 2  10 ln GMR1 GMR23

GMD1 ¼ GMD23 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi DD¼D

GMR1 ¼ r 0 GMR23 ¼

LTotal ¼ 2  10

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 4 0 r  D  r0  D ¼ r0 D

       32 D D D D D 7 7 ln 0 þ 2  10 ln pffiffiffiffiffiffiffi ¼ 2  10 ln pffiffiffiffiffiffiffi  0 ¼ 2  107 ln 0 0 0 r r r rD rD LTotal ¼ 3  107 ln

Choice (3) is the answer.

ð2Þ ð3Þ

Solving (1)–(4): 7

ð1Þ

  D r0

ð4Þ

46

4 Solutions of Problems: Transmission Line Parameters

Fig. 4.5 The power system of solution of problem 4.7

4.8. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0. As we know, the capacitance of a transmission line can be determined as follows: C¼

2πε0   ln GMD GMR

ð1Þ

Therefore: 2πε  0 

ð2Þ

2πε  0 

ð3Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 4 ðr 0  DÞ2 ¼ r 0 D

ð4Þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 3 9 ðr 0  D  DÞ3 ¼ r 0 D2

ð5Þ

C2b ¼

C3b ¼

ln

ln

GMD2b GMR2b

GMD3b GMR3b

Where: GMR2b ¼ GMR3b ¼

The Geometrical Mean Distance (GMD) will not change, since only the bundling is changed. Therefore: GMD2b ¼ GMD3b ¼ GMD

ð6Þ

As can be noticed from (4) and (5): GMR2b

Using ð6Þ < GMR3b ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) ln



GMD GMR2b





GMD > ln GMR3b

 )

ln

2πε  0 <

Therefore: C2b < C3b Choice (3) is the answer.

Fig. 4.6 The power system of solution of problem 4.8

GMD2b GMR2b

2πε  0  ln

GMD3b GMR3b

4

Solutions of Problems: Transmission Line Parameters

47

4.9. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0. Moreover, in Fig. 4.7 (b), conductors “2” and “3” are for sending power, and conductor “1” is for receiving power. In addition: La ¼ Lb

ð1Þ

To calculate the inductance of a single-phase transmission line, we need to calculate the sum of the inductances of power sending line and power receiving line, as they are connected in series. Therefore:     2 2D 2D ¼ 2  107 ln 0 La ¼ 2  2  107 ln 0 r r Lb ¼ L12 þ L3 ¼ 2  10

7

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3D  2D 3D  2D 6D2 7 7 p ffiffiffiffiffiffiffi ln ¼ 2  10 ln þ 2  10 ln 3 0 0 r rD r0 2

ð2Þ ð3Þ

Solving (1)–(3): 2  10

7

 3 3 1  2  2 2D 6D2 2D 6D2 2D2 3 9 7 ln 0 ¼ 2  10 ln ¼ ) ¼ ) D ¼ r0 ) 3 3 1 0 0 0 0 r r 1 4 r2 r2 r2

Choice (1) is the answer.

Fig. 4.7 The power system of solution of problem 4.9

4.10. Based on the information given in the problem, we know that conductors “2” and “3” are for sending power and conductor “1” is for receiving power. Moreover, the Geometrical Mean Radius (GMR) of each conductor is r0. To calculate the capacitance of a single-phase transmission line, we need to determine the equivalent capacitance of the capacitance of power sending line and the capacitance of the power receiving line, since they are connected in series. Thus: C Total ¼

C1 C23 C1 þ C23

ð1Þ

C1 ¼

2πε  0 

ð2Þ

2πε  0 

ð3Þ

C23 ¼

ln

ln

GMD1 GMR1

GMD23 GMR23

Where: GMD1 ¼ GMD23 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi DD¼D

ð4Þ

48

4 Solutions of Problems: Transmission Line Parameters

GMR1 ¼ r 0 GMR23 ¼

ð5Þ

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 4 0 r  D  r0  D ¼ r0 D

ð6Þ

Solving (1)–(6): 2πε0 ln ðrD0 Þ

CTotal ¼

ln

2πε0  pDffiffiffiffi 0

2πε0

r D

2πε0 ln ðrD0 Þ

þ 2πε0  ln pDffiffiffiffi 0 r D

¼

ln ðrD0 Þ ln

ln ðrD0 Þþ ln ln ðrD0 Þ ln



pDffiffiffiffi





r0 D

pDffiffiffiffi

¼

 pDffiffiffiffi 0 r0 D

ln

D  r0

2πε0 þ ln



pD ffiffiffiffiffi r0 D

¼

2πε0  3 ln Dr0 2

r D

C Total ¼

4πε0   3 ln Dr0

Choice (4) is the answer.

Fig. 4.8 The power system of solution of problem 4.10

4.11. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0 and: r0 < d

ð1Þ

L1 ¼ L2

ð2Þ

As we know, the inductance of a three-phase transmission line can be determined as follows:   GMD GMR

ð3Þ

  GMD1 GMR1

ð4Þ

  GMD2 ln GMR2

ð5Þ

L ¼ 2  107 ln Therefore: L1 ¼ 2  107 ln

L2 ¼ 2  10

7

Where: GMD1 ¼

p p ffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 D  D  2D ¼ D 2

ð6Þ

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 DDD¼D

ð7Þ

GMD2 ¼

GMR1 ¼ r 0

ð8Þ

4

Solutions of Problems: Transmission Line Parameters

49

GMR2 ¼

ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 3 0 r  d  d ¼ r0 d2

ð9Þ

Solving (2)–(9): 2  10

7

pffiffiffi ffiffiffi  p   pffiffiffi 2 2 1 D32 D D32 D 7 03 3 23 ) r 0 ¼ d p ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ln ¼ ¼ d 2 ¼ 2  10 ln p ) ) r 0 0 3 3 r r r0 d2 r0 d2

ð10Þ

Equations (1) and (10) do not have any solution. Choice (4) is the answer.

Fig. 4.9 The power system of solution of problem 4.11

4.12. As we know, the inductance and the capacitance of a three-phase transmission line can be determined as follows: L ¼ 2  107 ln C¼

  GMD GMR

2πε0   ln GMD GMR

ð1Þ ð2Þ

Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0. The transmission lines are not bundled. Hence: GMR1 ¼ GMR2 ¼ GMR3 ¼ GMR4 ¼ r 0

ð3Þ

Moreover, as can be noticed from Fig. 4.10.3, we have: d  < VS ¼ 1 þ V I þ Z 1 þ Z 1þ R V 2 4 R 4 7 R ) 5   ZY ZY > IR : I IS ¼ YVR þ 1 þ 1þ 2 R 2

ð1Þ

6

Solutions of Problems: Transmission Line Model and Performance

61

The charging current (ICharging) of a transmission line is achieved when the line is in the no-load condition. In other words: IR ¼ 0, IS ¼ ICharging

ð2Þ

  ZY VR VS ¼ 1 þ 2

ð3Þ

ICharging ¼ YVR

ð4Þ

Solving (1) and (2):

Solving (3) and (4):   ZY 1 ICharging ¼ YVS 1 þ 2 Choice (3) is the answer.

Fig. 6.1 The power system of solution of problem 6.7

6.8. As we know, transmission matrix is presented in the following form:


VS IS






A ¼ C

B D




VR IR

 ð1Þ

Applying KVL: VS ¼ ZIS þ VR

ð2Þ

IS þ VR Y þ IR ¼ 0 ) IS ¼ VR Y þ IR

ð3Þ

VS ¼ ZðVR Y þ IR Þ þ VR ¼ VR ðZY þ 1Þ þ ZIR

ð4Þ

Applying KCL in the receiving end:

Solving (2) and (3):

Solving (1), (3), and (4):


VS IS

Choice (2) is the answer.




¼

1 þ ZY

Z

Y

1




VR IR



62

6 Solutions of Problems: Transmission Line Model and Performance

Fig. 6.2 The power system of solution of problem 6.8

6.9. Based on the information given in the problem, we have: R G ¼ L C

ð1Þ

As we know, the characteristic impedance of a transmission line can be calculated as follows: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi rffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u R u þ jω Z R þ jωL t L L = ZC = = Y G þ jωC C GC þ jω

ð2Þ

Solving (1) and (2): rffiffiffiffi L ZC ¼ C

ð3Þ

The characteristic impedance of a lossless transmission line (R ¼ 0, G ¼ 0) can be determined as follows: rffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffi rffiffiffiffi Z 0 þ jωL jωL L ZC = = = = Y 0 þ jωC jωC C

ð4Þ

By comparing (3) and (4), it is concluded that choice (4) is the answer. 6.10. As we know, the transmission matrix is presented as follows:


VS IS






A ¼ C

B D




VR IR

 ð1Þ

This power system should be considered as the three cascaded sub-systems, as is illustrated in Fig. 6.3.2–4. Then, the relation below holds about their transmission matrices: ½T Total  ¼ ½T 1   ½T 2   ½T 3 

ð2Þ

Note that since the resistor has been installed in the middle point, the impedance of the line (Z) is equally divided. The transmission matrix of the first or the third sub-system (see Fig. 6.3.2 and Fig. 6.3.4) can be determined as follows: Applying KVL: 1 VS ¼ ZIR1 þ VR1 2

ð3Þ

IS ¼ IR1

ð4Þ

Applying KCL:

6

Solutions of Problems: Transmission Line Model and Performance

63

Solving (1), (3), and (4):


VS

"

 ¼

IS

#
"  1 Z VR1 1 ) ½T 1  ¼ ½T 3  ¼ 2 I R1 1 0

1 0

1 Z 2 1

# ð5Þ

The transmission matrix of the second sub-system (see Fig. 6.3.3) can be determined as follows: Applying KVL: VS2 ¼ VR2

ð6Þ

Applying KCL: IS2 þ

VR2 1 þ IR2 ¼ 0 ) IS2 ¼ VR2 þ IR2 R R

ð7Þ

Solving (1), (6), and (7):


VS2 IS2

"

 ¼

1 1 R

# " 1 0
VR2  ) ½T 2  ¼ 1 1 IR2 R

1

# " 1 1 Z  1 2 1 R

0

# ð8Þ

1

Solving (2), (5), and (8): " ½T Total  ¼

0

2 6 ) ½T Total  ¼ 4

1þ 1 R

Z 2R

0 1

#

" 

1 0

1 Z 2 1

#

 3 Z Z Rþ 4R 7 5 Z 1þ 2R

Choice (3) is the answer.

Fig. 6.3 The power system of solution of problem 6.10

64

6 Solutions of Problems: Transmission Line Model and Performance

6.11. Based on the information given in the problem, we have: 2

3

1 62 ½T  ¼ 4 3 j 4

j 1 2

7 5

ð1Þ

As we know, the transmission matrix of a long transmission line is as follows:


VS IS




VS IS






¼

2

A

B

C

D

cosh ðγlÞ ¼4 1 sinh ðγlÞ ZC




VR

 ð2Þ

IR

3 ZC sinh ðγlÞ
V  5 R cosh ðγlÞ IR

ð3Þ

By considering (2) and (3), we can write: rffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi cosh ðγlÞ  ZC sinh ðγlÞ AB ZC = ZC = = 1 1 CD Z sinh ðγlÞ  cosh ðγlÞ Z C

C

Moreover, as we know, A ¼ D in the transmission matrix of a transmission line. Therefore, the characteristic impedance of a transmission line can be determined as follows: rffiffiffiffi B ZC ¼ C

ð4Þ

Solving (1) and (4): sffiffiffiffi pffiffiffi j 2 3 ZC = 3 ) ZC = Ω 3 4j Choice (2) is the answer. 6.12. As we know, the transmission matrix of a transmission line is as follows:


VS IS



2

cosh ðγlÞ ¼4 1 sinh ðγlÞ ZC

8 3 ZC sinh ðγlÞ
V  < VS = cosh ðγlÞVR þ ZC sinh ðγlÞIR 5 R ) 1 : IS = cosh ðγlÞ sinh ðγlÞVR þ cosh ðγlÞIR IR ZC

ð1Þ

The charging current (ICharging) of a transmission line is achieved when the line is in the no-load condition. In other words: IR ¼ 0, IS ¼ ICharging

ð2Þ

VS = cosh ðγlÞVR

ð3Þ

Solving (1) and (2):

ICharging ¼

1 sinh ðγlÞVR ZC

ð4Þ

6

Solutions of Problems: Transmission Line Model and Performance

65

Solving (3) and (4): cosh ðγlÞ VS 1 ¼ tanh ðγlÞVS ) ICharging ¼ ZC ICharging Z1 sinh ðγlÞ C

Choice (1) is the answer. 6.13. Based on the information given in the problem, we have: ZC: Characteristic impedance ZS.C.: The impedance seen from the beginning of the transmission line if its end is short circuit ZO.C.: The impedance seen from the beginning of the transmission line if its end is open circuit As we know, the transmission matrix of a transmission line is as follows:


VS IS



2

cosh ðγlÞ ¼4 1 sinh ðγlÞ ZC

8 3 ZC sinh ðγlÞ
V  < VS = cosh ðγlÞVR þ ZC sinh ðγlÞIR R 5 ) 1 : IS = cosh ðγlÞ sinh ðγlÞVR þ cosh ðγlÞIR IR ZC

ð1Þ

Therefore: ZS:C: ¼

 Z sinh ðγlÞIR VS  ¼ C ¼ ZC tanh ðγlÞ IS VR ¼0 cosh ðγlÞIR

ð2Þ

 cosh ðγlÞVR VS  ¼ ¼ ZC cothðγlÞ IS IR ¼0 Z1 sinh ðγlÞV R

ð3Þ

ZO:C: ¼

C

By considering (2) and (3), we can write: ZS:C: ZO:C: ¼ ZC tanh ðγlÞ  ZC cothðγlÞ

ð4Þ

tanh ðγlÞcothðγlÞ ¼ 1

ð5Þ

From trigonometry, we know that:

Solving (4) and (5): ZS:C: ZO:C: ¼ ðZC Þ2 ) ZC ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ZS:C: ZO:C:

Choice (3) is the answer. 6.14. Based on the information given in the problem, we have: ZO:C: ¼

1 ZS:C:

ð1Þ

As we know, the transmission matrix is presented as follows:


VS IS




¼

A

B

C

D




VR IR

 ð2Þ

66

6 Solutions of Problems: Transmission Line Model and Performance

Therefore: ZS:C:

 VS  BI B ¼  ¼ R¼ IS VR ¼0 DIR D

ð3Þ

ZO:C:

 VS  AVR A ¼  ¼ ¼ IS IR ¼0 CVR C

ð4Þ

A 1 A D = ) = C DB C B

ð5Þ

Solving (1), (3), and (4):

Since the two-port of transmission line is symmetric and bidirectional, its transmission matrix has the following features: A¼D

ð6Þ

AD  BC ¼ 1

ð7Þ

Solving (5)–(7): A2  B2 ¼ 1 ) ðA þ BÞðA  BÞ ¼ 1 ) A þ B ¼

1 AB

Choice (1) is the answer. 6.15. As we know, the transmission matrix of a transmission line is as follows:


VS IS



8 3 cosh ðγlÞ ZC sinh ðγlÞ
V  < VS = cosh ðγlÞVR þ ZC sinh ðγlÞIR 5 R ) ¼4 1 1 : IS = sinh ðγlÞ cosh ðγlÞ sinh ðγlÞVR þ cosh ðγlÞIR IR ZC ZC 2

ð 1Þ ð 2Þ

In a lossless transmission line, the attenuation coefficient is zero (α ¼ 0). Therefore: γ ¼ α þ jβ ¼ jβ

ð3Þ

Solving (1)–(3): 8 < VS = cosh ðjβlÞVR þ ZC sinh ðjβlÞIR ) 1 : IS = sinh ðjβlÞVR þ cosh ðjβlÞIR ZC

ð 4Þ ð 5Þ

From trigonometry, we know that: cosh ðjβlÞ ¼ cos ðβlÞ

ð6Þ

sinh ðjβlÞ ¼ j sin ðβlÞ

ð7Þ

Solving (4)–(7): 8 < VS ¼ cos ðβlÞVR þ jZC sin ðβlÞIR ) 1 : IS ¼ j sin ðβlÞVR þ cos ðβlÞIR ZC

ð 8Þ ð9Þ

6

Solutions of Problems: Transmission Line Model and Performance

67

Moreover, in a no-load transmission line, we have: IR ¼ 0

ð10Þ

Solving (8)–(10): 8 < VS ¼ cos ðβlÞVR ) 1 : IS ¼ j sin ðβlÞVR ZC

ð11Þ ð12Þ

From (11), we can write: VR ¼

VS cos ðβlÞ

Choice (3) is the answer. 6.16. As we know, the transmission matrix is presented as follows:


VS




¼

IS

A

B

C

D




VR

 ð1Þ

IR

This power system should be considered as the three cascaded sub-systems, as is shown in Fig. 6.4.2–4. Then, the total transmission matrix can be determined as follows: ½T Total  ¼ ½T 1   ½T 2   ½T 3 

ð2Þ

As we know, for the ideal transformer, shown in Fig. 6.4.2, the relations below can be written: VS ¼ aVR1

ð3Þ

1 IS ¼ IR1 a

ð4Þ

Thus, the transmission matrix of the first sub-system is as follows:


VS

"

 ¼

IS

a 0

# " # a 0 0
VR1  ) ½T 1  ¼ 1 1 0 IR1 a a

ð5Þ

The transmission matrix of the second sub-system can be determined as follows: Applying KVL: VS2 ¼ ZIR2 þ VR2

ð6Þ

IS2 ¼ IR2

ð7Þ

Applying KCL:

Solving (1), (6), and (7):


VS2 IS2




¼

1

Z

0

1




VR2 IR2




 ) ½T 2  ¼

1

Z

0

1

 ð8Þ

68

6 Solutions of Problems: Transmission Line Model and Performance

The transmission matrix of the third sub-system can be determined as follows: Applying KVL: VS3 ¼ VR

ð9Þ

IS3 þ YVR þ IR ¼ 0 ) IS3 ¼ YVR þ IR

ð10Þ

Applying KCL:

Solving (1), (9), and (10):


VS3 IS3




¼

1

0

Y

1




VR IR




 ) ½T 3  ¼

1

0

Y

1



Solving (2), (5), (8), and (11): " ½T Total  ¼

a 0

#
0 1 1  0 a "

½T Total  ¼

Z 1






#
1 0 1 þ ZY a Z 0 a

1

0

Y

1

Y





1

Choice (3) is the answer.

Fig. 6.4 The power system of solution of problem 6.16

ð11Þ

7

Problems: Network Impedance and Admittance Matrices

Abstract

In this chapter, the problems of network impedance and admittance matrices are presented. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations. 7.1. For the power system illustrated in Fig. 7.1, determine Z22 of the network impedance matrix ([ZBus]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) j0.6 Ω 2) j0.06 Ω 3) j0.4 Ω 4) j0.15 Ω

Fig. 7.1 The power system of problem 7.1

7.2. The network impedance matrix ([ZBus]) and the result of load flow simulation problem are presented in the following. If a capacitor with the reactance of 3.4 p. u. is connected to the fourth bus, determine its updated voltage: 2

0:20

6 0:15 6 ½ZBus  ¼ j6 4 0:25 0:24

0:15

0:25

0:30

0:13

0:13 0:14

0:15 0:25

0:24

3

0:14 7 7 7 p:u: 0:25 5 0:40

Difficulty level Calculation amount

○ Easy ● Small

● Normal ○ Normal

○ Hard ○ Large

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7

1) 2) 3) 4)

Problems: Network Impedance and Admittance Matrices

0.95 p. u. 0.98 p. u. 1.02 p. u. 1.20 p. u.

7.3. In a three-bus power system, the voltage of the second bus is about , and the network impedance matrix is as follows. If an inductor with the reactance of 2.7 p. u. is connected to the second bus, determine the voltage variation of the third bus: 2

0:2

6 ½ZBus  ¼ j4 0:15 0:1 Difficulty level Calculation amount 1) 0.075 p. u. 2) 0.06 p. u. 3) 0.12 p. u. 4) 0.15 p. u.

○ Easy ● Small

● Normal ○ Normal

0:15 0:3

0:1

3

7 0:15 5 p:u:

0:15 0:25

○ Hard ○ Large

7.4. For the power system shown in Fig. 7.2, determine the network admittance matrix ([YBus]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2 3 20 15 15 6 7 1) j4 15 25 10 5 p:u: 2

15 25

6 2) j4 10 2

5 20

6 3) j4 5 10 2 15 6 4) j4 30 40

10 10 35 5 5 30 15 30 20 20

30 3 5 7 5 5 p:u: 15 3 10 7 15 5 p:u: 35 3 40 7 20 5 p:u: 35

Fig. 7.2 The power system of problem 7.4

7

Problems: Network Impedance and Admittance Matrices

71

7.5. For the power system shown in Fig. 7.3, determine the network impedance matrix ([ZBus]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 2 3 2 1 6 7 1) j4 30 30 5 p:u: 1 2 2 30 30 3 1 1  6 15 7 p:u: 2) j4 15 5 1 1  2 15 315 2 2 6 15 15 7 3) j4 5 p:u: 2 2 2 15 15 3 2 1   6 30 7 p:u: 4) j4 30 5 1 2   30 30

Fig. 7.3 The power system of problem 7.5

7.6. For the power system shown in Fig. 7.4, determine the detriment of the network impedance matrix ([ZBus]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0.5 2) 0.5 3) 0.2 4) 0.2

Fig. 7.4 The power system of problem 7.6

7.7. For the power system shown in Fig. 7.5, determine the value of ZZ1222 , belonging to [ZBus], if the base voltage in the transmission line and the base MVA are 50 kV and 100 MVA, respectively. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0.5 2) 0.75 3) 1 4) 2

72

7

Problems: Network Impedance and Admittance Matrices

Fig. 7.5 The power system of problem 7.7

7.8. In a three-bus power system shown in Fig. 7.6, determine the sum of the diagonal components of the network admittance matrix ([YBus]). Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) j60 p. u. 2) j20 p. u. 3) j30 p. u. 4) j10 p. u.

Fig. 7.6 The power system of problem 7.8

7.9. The impedance diagram of a three-phase four-bus power system is shown in Fig. 7.7. If the lines of 2–4 and 1–3 are removed from the system, the network admittance matrix can be presented in the form of [YBus, New] ¼ [YBus] + [ΔYBus]. Determine [ΔYBus]. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 2

10 0 10 6 0 10 0 6 1) j6 4 10 0 10 2

0 10 0 10 10 0 10

6 0 6 2) j6 4 10

0 10 6 0 6 3) j6 4 10 2

2

3 0 10 7 7 7 p:u: 0 5

0 10

6 0 6 4) j6 4 10 0

10 0

0 10

10 0 10

0 10 0

0 10 0

10 0 10

10 0

0 10

10

0

0

3

10 7 7 7 p:u: 0 5 10 3 0 10 7 7 7 p:u: 0 5 10 3 0 10 7 7 7 p:u: 0 5 10

7

Problems: Network Impedance and Admittance Matrices

73

Fig. 7.7 The power system of problem 7.9

7.10. The network admittance matrix of a four-bus power system is presented in the following. Determine the updated network admittance matrix if the second and the third buses are short-circuited: 2

5

6 6 4 6 ½YBus  ¼ j6 6 3 4 2 Difficulty level ○ Easy Calculation amount ● Small 2 3 5 7 2 6 7 1) j4 7 16 5 5 p:u: 2 5 20 2 3 10 7 2 6 7 2) j4 7 10 5 5 p:u:

○ Normal ○ Normal

4

3

10

2

2

10

1

4

2

3

7 1 7 7 7 p:u: 4 7 5 20

● Hard ○ Large

10 3 6 6 7 3) j4 5 10 5 5 p:u: 6 5 20 2 3 0 10 20 6 7 4) j4 10 10 6 5 p:u: 20 6 20 2

2 16

5 5

7.11. The network admittance matrix of a power system is presented in the following. There are two parallel similar lines between the buses. If one of them is disconnected from bus 1 and then grounded, determine the updated network admittance matrix:  ½YBus  ¼

j10

j10

j10

j10

 p:u:

74

7

Difficulty level ○ Easy Calculation amount ○ Small   5 5 1) j p:u: 5 10   20 20 2) j p:u: 20 20   20 5 3) j p:u: 5 10   5 5 4) j p:u: 5 5

○ Normal ● Normal

● Hard ○ Large

Problems: Network Impedance and Admittance Matrices

8

Solutions of Problems: Network Impedance and Admittance Matrices

Abstract

In this chapter, the problems of the seventh chapter are fully solved, in detail, step by step, and with different methods.

8.1. As we know, Znn is the Thevenin impedance seen from the n’th bus. To find the Thevenin impedance, we need to turn off the generator, as is shown in Fig. 8.2. Now, we can write: Z22 ¼ ð j0:3Þkðð j0:2Þkð j0:2Þ þ j0:2Þ ¼ ð j0:3Þkð j0:3Þ Z22 ¼ j0:15 p:u: Choice (4) is the answer.

Fig. 8.1 The power system of solution of problem 8.4

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8 Solutions of Problems: Network Impedance and Admittance Matrices

8.2. Based on the information given in the problem, we have: 2

0:20

0:24

3

6 0:15 6 ½ZBus  ¼ j6 4 0:25

0:15

0:25

0:30 0:13

0:13 0:15

0:14 7 7 7 p:u: 0:25 5

0:24

0:14

0:25

0:40

ZC ¼ j3:4 p:u: If an inductor or a capacitor with the impedance of Z is connected to the ith bus, the updated voltage in the jth bus can be calculated as follows:   2 Vi,Old Vj,New = Vj,Old þ Zji Zii þ Z In this problem, the capacitor is connected to the fourth bus, and the updated voltage of the fourth bus is also requested. Thus, using the network impedance matrix and the result of load flow simulation problem, we can write:

-

-

-

V4,New ¼ 1:02 p:u: Choice (3) is the answer. 8.3. Based on the information given in the problem, we have:

2

0:2 6 ½ZBus  ¼ j4 0:15 0:1

3 0:15 0:1 7 0:3 0:15 5 p:u: 0:15 0:25

X L ¼ 2:7 p:u: If an inductor or a capacitor with the impedance of Z is connected to the ith bus, the updated voltage in the jth bus can be calculated as follows:   2 Vi,Old Vj,New = Vj,Old þ Zji Zii þ Z Herein, the inductor is connected to the second bus, and the voltage variation of the third bus is requested. Thus, we can write:

Choice (2) is the answer.

-

-

8

Solutions of Problems: Network Impedance and Admittance Matrices

77

8.4. Figure 8.2 shows the power system. The components of the network admittance matrix ([YBus]) can be determined as follows: y11 ¼

1 1 1 þ þ ¼ j10  j5  j10 ¼ j25 p:u: j0:1 j0:2 j0:1

y22 ¼

1 1 1 þ þ ¼ j20  j10  j5 ¼ j35 p:u: j0:05 j0:1 j0:2

y33 ¼

1 1 1 þ þ ¼ j5  j5  j5 ¼ j15 p:u: j0:2 j0:2 j0:2 

y12 ¼ y21

1 ¼ j0:1





y13 ¼ y31

1 ¼ j0:2 

y23 ¼ y32 ¼ 

1 j0:2

¼ j10 p:u:  ¼ j5 p:u:  ¼ j5 p:u:

Therefore: 2

25 6 ½YBus  ¼ j4 10 5

10 35 5

3 5 7 5 5 p:u: 15

Choice (2) is the answer.

Fig. 8.2 The power system of solution of problem 8.1

8.5. Building network impedance matrix ([ZBus]) is time-consuming. Hence, the best way is to determine the network admittance matrix ([YBus]), and then [ZBus] ¼ [YBus]1: y11 ¼

1 1 1 þ þ ¼ j10  j5  j5 ¼ j20 p:u: j0:1 j0:2 j0:2

y22 ¼

1 1 1 þ þ ¼ j10  j5  j5 ¼ j20 p:u: j0:1 j0:2 j0:2

78

8 Solutions of Problems: Network Impedance and Admittance Matrices



y12 ¼ y21

1 1 þ ¼ j0:2 j0:2

 ¼ ðj5  j5Þ ¼ j10 p:u:

Therefore:  ½YBus  ¼ j

½ZBus  ¼ ½YBus 1



20

10

10

20

  20 ¼ j 10

10 20

1

p:u: 2

3 2 1 6 7 ) ½ZBus  ¼ j4 30 30 5 1 2 30 30

Choice (1) is the answer.

Fig. 8.3 The power system of solution of problem 8.5

8.6. Building network impedance matrix ([ZBus]) is time-consuming. Therefore, the best way is to determine the network admittance matrix ([YBus]), and then [ZBus] ¼ [YBus]1. As is illustrated in Fig. 8.4.2, we need to turn off the generators: 1 1 y11 ¼ þ ¼ j2 p:u: j j 1 1 y22 ¼ þ ¼ j3 p:u: j j0:5 y12 ¼ y21

  1 ¼ ¼ j p:u: j

Therefore:  2 ½YBus  ¼ j 1

½ZBus  ¼ ½YBus 1

  2 ¼ j 1 

) detð½ZBus Þ ¼ Choice (3) is the answer.



1 3

1



3 1

p:u: 2

3 65 ) ½ZBus  ¼ j4 1 5

  6 1   ¼ 0:2 25 25

3 1 57 5 2 5

8

Solutions of Problems: Network Impedance and Admittance Matrices

79

Fig. 8.4 The power system of solution of problem 8.6

8.7. Based on the information given in the problem, we have: V Line,B ¼ 50 kV, SB ¼ 100 MVA X Line ¼ 12:5 Ω The base voltage in the zone of the line can be calculated as follows: Z Line,B ¼

ðV Line,B Þ2 ð50 kV Þ2 ¼ ¼ 25 Ω SB 100 MVA

Thus, the per unit (p.u.) value of the reactance of the line is: X Line,p:u: ¼

X Line 12:5 ¼ 0:5 Ω ) Z Line,p:u: ¼ j0:5 p:u: ¼ 25 Z Line,B

Now, the impedance diagram of the system is known and illustrated in Fig. 8.5.2. The network admittance matrix of the system can be determined as follows: y11 ¼ y22 ¼

1 1 þ ¼ j4 p:u: j0:5 j0:5 

y12 ¼ y21

1 ¼ j0:5

 4 ½YBus  ¼ j 2

 ¼ j2 p:u: 2



4

p:u:

2

3 1 6 7 p:u: 5 1 3

Then, the network impedance matrix is:

½ZBus  ¼ ½YBus 1

1 63 ¼ j4 1 6

80

8 Solutions of Problems: Network Impedance and Admittance Matrices

Therefore: j1 Z12 Z ¼ 16 ) 12 ¼ 0:5 Z22 Z22 j3 Choice (1) is the answer.

Fig. 8.5 The power system of solution of problem 8.7

8.8. The impedance diagram of the power system is shown in Fig. 8.6. The network admittance matrix of the system can be determined as follows: y11 ¼ y22 ¼ y33 ¼

1 1 þ ¼ j10  j10 ¼ j20 p:u: j0:1 j0:1

  1 ¼ j10 p:u: y12 ¼ y13 ¼ y21 ¼ y23 ¼ y31 ¼ y32 ¼  j0:1 2

20 6 ½YBus  ¼ j4 10 10

10 20 10

3 10 7 10 5 p:u: 20

Therefore, the sum of the diagonal components of the network admittance matrix is: Sum of the diagonal components ¼ j20  j20  j20 ¼ j60 p:u: Choice (1) is the answer.

Fig. 8.6 The power system of solution of problem 8.8

8

Solutions of Problems: Network Impedance and Admittance Matrices

81

8.9. Based on the information given in the problem, [YBus] belongs to the power system shown in Fig. 8.7.1. Moreover, [YBus, New] is related to the system that the lines of 2–4 and 1–3 have been removed from it. The impedance diagram of the primary system is shown in Fig. 8.7.2. The network admittance matrix of this system can be determined as follows: y11 ¼ y22 ¼ y33 ¼ y44 ¼

1 1 1 þ þ ¼ j10  j10  j10 ¼ j30 p:u: j0:1 j0:1 j0:1 

y12 ¼ y13 ¼ y14 ¼ y21 ¼ y23 ¼ y24 ¼ y31 ¼ y32 ¼ y34 ¼ y41 ¼ y42 ¼ y43 2

30

6 10 6 ½YBus  ¼ j6 4 10 10

10

10

30

10

10 10

30 10

1 ¼ j0:1

 ¼ j10 p:u:

3

10

10 7 7 7 p:u: 10 5 30

Figure 8.7.3 illustrates the impedance diagram of the updated system. The network admittance matrix of this system can be determined as follows: 1 1 þ ¼ j10  j10 ¼ j20 p:u: j0:1 j0:1

y11 ¼ y22 ¼ y33 ¼ y44 ¼



1 ¼ j0:1

y12 ¼ y14 ¼ y21 ¼ y23 ¼ y32 ¼ y34 ¼ y41 ¼ y43

 ¼ j10 p:u:

y13 ¼ y31 ¼ y24 ¼ y42 ¼ 0 p:u: 2

20 6 10 6 ½YBus,New  ¼ j6 4 0

10 20

0 10

10 0

20 10

10

3 10 0 7 7 7 p:u: 10 5 20

Therefore: 2

20

6 10 6 ½ΔYBus  ¼ j6 4 0 10

10

0

10

20 10

10 20

0 10

0

10

20

2

10

6 0 6 ½ΔYBus  ¼ j6 4 10 0 Choice (2) is the answer.

3

2

30

7 6 10 7 6 7  j6 5 4 10 10

0

10

10 0

0 10

10

0

0

10

10

10

30 10

10 30

10 10

10

10

30

3

10 7 7 7 p:u: 0 5 10

3 7 7 7¼ 5

82

8 Solutions of Problems: Network Impedance and Admittance Matrices

Fig. 8.7 The power system of solution of problem 8.9

8.10. By short-circuiting two buses of a power system, their corresponding components in the network admittance matrix ([YBus]) are added up. Therefore, for the second and the third buses, we have:

8

Solutions of Problems: Network Impedance and Admittance Matrices

2

5 6 ½YBus,New  ¼ j4 7

83

7 16

2

5

3 2 7 5 5 p:u: 20

Choice (1) is the answer. 8.11. Based on the information given in the problem, the network admittance matrix is as follows:  ½YBus  ¼

j10

j10

j10

j10

 ð1Þ

p:u:

From this [YBus], we can figure out that the power system has only two buses. Moreover, we know that there are two parallel similar lines between the buses. Now, it is better to draw the single-line diagram of the system which is shown in Fig. 8.8.1. The network admittance matrix of the primary system (see Fig. 8.8.1) can be formed as follows:  ½YBus  ¼

yþy ðy þ yÞ

  ðy þ yÞ 2y ¼ yþy 2y

 2y p:u: 2y

ð2Þ

By solving (1) and (2), we can write: 2y ¼ j10 ) y ¼ j5 p:u:

ð3Þ

Figure 8.8.2 shows the admittance diagram of the power system. Note that each quantity presents the admittance of the line. Based on the information given in the problem, one of them is disconnected from bus 1 and then grounded. Figure 8.8.3 illustrates the updated system. Now, the network admittance matrix of the updated system is as follows:  ½YBus  ¼

j5

ðj5Þ

ðj5Þ j5 þ ðj5Þ



 ) ½YBus  ¼

j5

j5

j5

j10

Choice (1) is the answer.

Fig. 8.8 The power system of solution of problem 8.11

 p:u:

9

Problems: Load Flow and Economic Load Dispatch

Abstract

In this chapter, the problems concerned with the load flow and economic load dispatch are presented. The subjects include Gauss-Seidel load flow, DC load flow (DCLF), Decoupled Load flow (DLF), Newton-Raphson load flow (NRLF), Jacobian matrix determination, and economic load dispatch. In this chapter, the problems are categorized in different levels based on their difficulty levels (easy, normal, and hard) and calculation amounts (small, normal, and large). Additionally, the problems are ordered from the easiest problem with the smallest computations to the most difficult problems with the largest calculations.

9.1. In a load flow problem, which type of the bus has a known active power? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) Load bus 2) Voltage-controlled bus 3) All buses except slack bus 4) None of them 9.2. To speed up the algorithm of Gauss-Seidel load flow, an accelerating factor (α) is usually used. Which one of the following relations presents that? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large ðkþ1Þ

ð kÞ

1) Vi,Acc = Vi

ðkþ1Þ

þ αΔVi

ðkþ1Þ

ðkÞ

ðkþ1Þ

ðkþ1Þ

ðkþ1Þ

2) Vi,Acc = αVi þ ΔVi
 ðkþ1Þ ðkþ1Þ ðkÞ 3) Vi,Acc = α Vi 2 Vi 4) Vi,Acc = Vi

ð kÞ

þ αΔVi

9.3. Which one of the following choices is correct about the DC load flow (DCLF), Decoupled Load flow (DLF), and Newton-Raphson load flow (NRLF)? Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) DLF is faster than DCLF, and DCLF is faster than NRLF. 2) DCLF is not appropriate for the AC power systems, and DCLF has more convergence probability compared to NRLF. 3) DLF and NRLF can achieve the same results but with different iterations. DCLF is faster than DLF and DLF is faster than NRLF. 4) DCLF is appropriate for the systems with the high value of XR. NRLF always converges.

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9

Problems: Load Flow and Economic Load Dispatch

9.4. Use DC load flow to determine the active power flowing through the line. Herein, SB ¼ 100 MVA. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 32.2 MW 2) 85.6 MW 3) 41.7 MW 4) 65.4 MW

Fig. 9.1 The power system of problem 9.4

9.5. In the power system, shown in Fig. 9.2, determine δ. Do not use DC load flow approximation. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large  1) 60  2) 30  3) 90  4) 0

Fig. 9.2 The power system of problem 9.5

9.6. Calculate P12 by using DC load flow. Herein, assume π  3. Difficulty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1.5 p. u. 2) 2 p. u. 3) 3 p. u. 4) 3.5 p. u.

Fig. 9.3 The power system of problem 9.6

9.7. Use DC load flow to determine PG2. Herein, assume π  3. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 0.2 p. u. 2) 0.25 p. u. 3) 0.6 p. u. 4) 0.75 p. u.

9

Problems: Load Flow and Economic Load Dispatch

87

Fig. 9.4 The power system of problem 9.7

9.8. Determine the inverse matrix of Jacobian matrix considering the following terms: 8 < P2 ¼ δ2 þ 3jV2 j : Q2 ¼ 0:1δ2 þ 1 jV1 j þ jV2 j 5 Difficulty level Calculation amount   3 1 1) 1 0:1   1 3 2) 0:1 1   0:1 1 3) 1 3   1 3 4) 0:1 1

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

Fig. 9.5 The power system of problem 9.8

9.9. Use DC load flow to determine the phase angle of bus 4. Herein, assume π  3. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large  1) 45  2) 36  3) 30  4) 15

Fig. 9.6 The power system of problem 9.9

88

9

Problems: Load Flow and Economic Load Dispatch

9.10. In a power plant, the power loss coefficients for the two power generation units are L1 ¼ $1.5/MW, L2 ¼ $1.8/MW. Calculate the total generation of the units if Lagrange Multiplier (λ) is about $300/MWh, and the generation cost functions of the units are as follows: (

C 1 ¼ 0:2P2G1 þ 100PG1 þ 5500 C 2 ¼ 0:1P2G2 þ 100PG2 þ 4000

Difficulty level Calculation amount 1) 250 MW 2) 583.3 MW 3) 425.5 MW 4) 720 MW

○ Easy ○ Small

● Normal ● Normal

○ Hard ○ Large

9.11. In a power plant, the generation cost functions of the units are as follows: (

C 1 ¼ 0:0075P2G1 þ 50PG1 þ 1000 C 2 ¼ 0:005P2G2 þ 45PG2 þ 3000

Solve the economic load dispatch problem for the load demand of 1000 MW. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) PG1 ¼ 900 MW, PG2 ¼ 100 MW 2) PG1 ¼ 750 MW, PG2 ¼ 250 MW 3) PG1 ¼ 600 MW, PG2 ¼ 400 MW 4) PG1 ¼ 200 MW, PG2 ¼ 800 MW 9.12. In a power plant, the generation cost functions of the units are as follows: (

C 1 ¼ 0:05P2G1 þ 50PG1 þ 1500 C 2 ¼ 0:075P2G2 þ 40PG2 þ 2000

Solve the economic load dispatch problem for the total load of 1400 MW. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) PG1 ¼ 400 MW, PG2 ¼ 1000 MW 2) PG1 ¼ 500 MW, PG2 ¼ 900 MW 3) PG1 ¼ 800 MW, PG2 ¼ 600 MW 4) PG1 ¼ 700 MW, PG2 ¼ 700 MW 9.13. In a power plant, the generation cost functions of the units are as follows: (

C 1 ¼ 135P2G1 þ 100000PG1 C 2 ¼ 115P2G2 þ 85000PG2

Solve the economic load dispatch problem for the total load of 1000 MW. Difficulty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large

9

Problems: Load Flow and Economic Load Dispatch

1) 2) 3) 4)

89

PG1 ¼ 430 MW, PG2 ¼ 570 MW PG1 ¼ 570 MW, PG2 ¼ 430 MW PG1 ¼ 500 MW, PG2 ¼ 500 MW PG1 ¼ 536 MW, PG2 ¼ 464 MW

9.14. The single-line diagram of a power system is shown in Fig. 9.7. The voltage of bus 1 is about ð 0Þ

SB ¼ 100 MVA. Calculate V2 using Gauss-Seidel load flow after one iteration if V2 ¼ Difficulty level ○ Easy Calculation amount ○ Small 1) (0.936  j0.08) p. u. 2) (0.940  j0.08) p. u. 3) (0.8  j0.91) p. u. 4) (0.836  j0.2) p. u.

○ Normal ○ Normal

● Hard ● Large

Fig. 9.7 The power system of problem 9.14

9.15. Use Newton-Raphson load flow (NRLF) to determine the voltage of load bus after one iteration. Difficulty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large 1) 0.95 <  0.12 rad 2) 0.98 <  0.1 rad 3) 0.93 <  0.12 rad 4) 0.9 <  0.1 rad

Fig. 9.8 The power system of problem 9.15

and ð0Þ

and V3 ¼

Solutions of Problems: Load Flow and Economic Load Dispatch

10

Abstract

In this chapter, the problems of the ninth chapter are fully solved, in detail, step by step, and with different methods.

10.1. The buses are categorized in three types: • Load bus (P-Q bus): In this bus, the active and reactive powers are known. • Voltage-controlled bus (P-V bus): In this bus, the active power and the magnitude of voltage are known. • Slack bus (reference bus): In this unique bus, only the primary value of magnitude and phase angle of voltage are known. Therefore, active power is known in all buses except in slack bus. Choice (3) is the answer. 10.2. To speed up the algorithm of Gauss-Seidel load flow, an accelerating factor (α) is usually applied, as follows: ðkþ1Þ

ð kÞ

Vi,Acc ¼ Vi

ðkþ1Þ

þ αΔVi

Choice (1) is the answer. 10.3. DLF and NRLF can achieve the same results but with different iterations. Moreover, DC load flow (DCLF) is faster than Decoupled Load flow (DLF), and DLF is faster than Newton-Raphson load flow (NRLF). Choice (3) is the answer. 10.4. In DC load flow, the relation below is applied, in which X and δ are in per unit (p.u.) and radian, respectively: P12 ¼

1 ðδ  δ 2 Þ X 12 1

ð1Þ

Based on the information given in the problem, we have: X 12 ¼ 0:4 p:u: 

δ1 ¼ 25 , δ2 ¼ 10

ð2Þ 

SB ¼ 100 MVA

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Power System Analysis, https://doi.org/10.1007/978-3-030-84767-8_10

ð3Þ ð4Þ

91

92

10 Solutions of Problems: Load Flow and Economic Load Dispatch

Solving (1)–(3): P12 ¼

1 1 π ð25  10Þ  ¼ 0:654 p:u: ðδ  δ 2 Þ ¼ X 12 1 0:4 180

ð5Þ

PMW ¼ Pp:u:  SB ) P12 ¼ 0:654  100 ¼ 65:4 MW Choice (4) is the answer.

Fig. 10.1 The power system of solution of problem 10.4

10.5. Based on the information given in the problem, we have: P12 ¼ PL ¼ 10 p:u:

ð1Þ

X 12 ¼ 0:05 p:u:

ð2Þ

δ2 ¼ 0



ð3Þ

Herein, we are not allowed to use DC load flow. The active power flowing through the transmission line can be calculated as follows: P12 ¼

jV 1 jjV 2 j sin ðδ1  δ2 Þ X

ð1Þ

Therefore: 10 ¼

 11 sin ðδ  0Þ ) sin ðδÞ ¼ 0:5 ) δ ¼ sin 1 ð0:5Þ ¼ 30 0:05

Choice (2) is the answer.

Fig. 10.2 The power system of solution of problem 10.5

10.6. In DC load flow, the relation below is applied, in which X and δ are in per unit (p.u.) and radian, respectively: P12 ¼

1 ðδ  δ 2 Þ X 12 1

ð1Þ

Based on the information given in the problem, we have: π3

ð2Þ

10

Solutions of Problems: Load Flow and Economic Load Dispatch 

δ1 ¼ 30 ¼ 30 

93

π ¼ 0:5 rad 180



δ2 ¼ 30 ¼ 30 

π ¼ 0:5 rad 180

X 12 ¼ 0:5 p:u:

ð3Þ ð4Þ ð5Þ

Solving (1) and (3)–(5): P12 ¼

1 1 ð0:5  ð0:5ÞÞ ðδ  δ2 Þ ¼ X 12 1 0:5 P12 ¼ 2 p:u:

Choice (2) is the answer.

Fig. 10.3 The power system of solution of problem 10.6

10.7. Based on the information given in the problem, we have: π3

ð1Þ



δ1 ¼ 0 ¼ 0 rad 

δ2 ¼ 12 ¼ ð12Þ 

π ¼ 0:2 rad 180

X 12 ¼ 0:5 p:u:

ð2Þ ð3Þ ð4Þ

Since there is no power loss in the lines, the total power generation will be equal to the total power demand. Hence: PG1 þ PG2 ¼ PL ) PG1 þ PG2 ¼ 1 p:u:

ð5Þ

As we know, in DC load flow, the relation below is applied, in which X and δ are in per unit (p.u.) and radian, respectively: P12 ¼

1 1 ðδ  δ2 Þ ) PG1 ¼ P12 ¼ ð0  ð0:2ÞÞ ¼ 0:4 p:u: X 12 1 0:5

Solving (5) and (6): 0:4 þ PG2 ¼ 1 ) PG2 ¼ 0:6 p:u: Choice (3) is the answer.

ð6Þ

94

10 Solutions of Problems: Load Flow and Economic Load Dispatch

Fig. 10.4 The power system of solution of problem 10.7

10.8. Based on the information given in the problem, we have: 8 <

P 2 ¼ δ 2 þ 3j V 2 j : Q2 ¼ 0:1δ2 þ 1 jV1 j þ jV2 j 5

ð1Þ

Jacobian matrix is defined as follows: 2

∂P2 6 ∂δ2 ½J  ¼ 6 4 ∂Q 2 ∂δ2

3 ∂P2 ∂jV2 j 7 7 ∂Q2 5 ∂jV2 j

ð2Þ

Herein, bus 1 is considered as the slack bus. Solving (1) and (2):  ½J  ¼

1

3

0:1

1



Choice (2) is the answer.

Fig. 10.5 The power system of solution of problem 10.8

10.9. In DC load flow, the relation below is applied, in which X and δ are in per unit (p.u.) and radian, respectively: PSR ¼

1 ðδ  δR Þ X SR S

ð1Þ

Based on the information given in the problem, we have: π3

ð2Þ

δ1 ¼ 0 rad

ð3Þ

X 13 ¼ 0:1 p:u:

ð4Þ

P13 ¼ PL1 þ PL2 þ PL3 ¼ 1 þ 1 þ 2 ¼ 4 p:u:

ð5Þ

P34 ¼ 2 p:u:

ð6Þ

10

Solutions of Problems: Load Flow and Economic Load Dispatch

95

Solving (1)–(5) for line 1–3: 1 1 ðδ  δ3 Þ ) 4 ¼ ð0  δ3 Þ ) δ3 ¼ 0:4 rad X 13 1 0:1

ð7Þ

1 1 ðδ  δ4 Þ ) 2 ¼ ð0:4  δ4 Þ ) δ4 ¼ 0:6 rad X 34 3 0:1

ð8Þ

P13 ¼ Likewise for line 3–4: P34 ¼

δ4 ¼ 0:6 

 180 ¼ 36 π

Choice (2) is the answer.

Fig. 10.6 The power system of solution of problem 10.9

10.10. Based on the information given in the problem, we have: L1 ¼ $1:5=MW, L2 ¼ $1:8=MW

ð1Þ

λ ¼ $300=MWh

ð2Þ

(

C 1 ¼ 0:2P2G1 þ 100PG1 þ 5500 C 2 ¼ 0:1P2G2 þ 100PG2 þ 4000

ð 3Þ ð 4Þ

If power loss exists in a power generation system, the conditions to have an economic load dispatch are as follows: λ ¼ L1

∂C 1 ∂C2 ¼ L2 ∂PG1 ∂PG2

ð5Þ

Solving (1)–(5): 300 ¼ 1:5ð0:4PG1 þ 100Þ ¼ 1:8ð0:2PG2 þ 100Þ  )

300 ¼ 1:5ð0:4PG1 þ 100Þ ) 0:4PG1 þ 100 ¼ 200 ) PG1 ¼ 250 MW

ð 6Þ

300 ¼ 1:8ð0:2PG2 þ 100Þ ) 0:2PG1 þ 100 ¼ 166:66 ) PG2 ¼ 333:3 MW

ð 7Þ

PG,Total ¼ PG1 þ PG2 ¼ 250 þ 333:3 ¼ 583:3 MW Choice (2) is the answer.

96

10 Solutions of Problems: Load Flow and Economic Load Dispatch

10.11. Based on the information given in the problem, we have: PDemand ¼ 1000 MW (

ð1Þ

C1 ¼ 0:0075P2G1 þ 50PG1 þ 1000

ð2Þ

C2 ¼

ð3Þ

0:005P2G2

þ 45PG2 þ 3000

If the power generation system is lossless, the conditions to have an economic load dispatch are as follows: λ¼

∂C 1 ∂C2 ¼ ∂PG1 ∂PG2

ð4Þ

Solving (2)–(4): 0:015PG1 þ 50 ¼ 0:01PG2 þ 45 ) 0:015PG1  0:01PG2 ¼ 5

ð5Þ

Using (1) and considering the fact that the total power generation must be equal to the total load demand, we can write: PG1 þ PG2 ¼ 1000

ð6Þ

Solving (5) and (6): PG1 ¼ 200 MW, PG2 ¼ 800 MW Choice (4) is the answer. 10.12. Based on the information given in the problem, we have: PDemand ¼ 1400 MW (

ð1Þ

C1 ¼ 0:05P2G1 þ 50PG1 þ 1500

ð 2Þ

C2 ¼ 0:075P2G2 þ 40PG2 þ 2000

ð 3Þ

If the power generation system is lossless, the conditions to have an economic load dispatch are as follows: λ¼

∂C 1 ∂C2 ¼ ∂PG1 ∂PG2

ð4Þ

Solving (2)–(4): 0:1PG1 þ 50 ¼ 0:15PG2 þ 40 ) 0:1PG1  0:15PG2 ¼ 10

ð5Þ

Using (1) and considering the fact that the total power generation must be equal to the total load demand, we can write: PG1 þ PG2 ¼ 1400 Solving (5) and (6): PG1 ¼ 800 MW, PG2 ¼ 600 MW Choice (3) is the answer.

ð6Þ

10

Solutions of Problems: Load Flow and Economic Load Dispatch

97

10.13. Based on the information given in the problem, we have:

(

PDemand ¼ 1000 MW

ð1Þ

C 1 ¼ 135P2G1 þ 100000PG1

ð 2Þ

C2 ¼

ð 3Þ

115P2G2

þ 85000PG2

If the power generation system is lossless, the conditions to have an economic load dispatch are as follows: λ¼

∂C 1 ∂C2 ¼ ∂PG1 ∂PG2

ð4Þ

Solving (2)–(4): 270PG1 þ 100000 ¼ 230PG2 þ 85000 ) 270PG1  230PG2 ¼ 15000

ð5Þ

Using (1) and considering the fact that the total power generation must be equal to the total load demand, we can write: PG1 þ PG2 ¼ 1000

ð6Þ

Solving (5) and (6): PG1 ¼ 430 MW, PG2 ¼ 570 MW Choice (1) is the answer. 10.14. Based on the information given in the problem, we have: ð1Þ SB ¼ 100 MVA

ð2Þ ð3Þ

First, we need to build the network admittance matrix, as follows: 1 1 þ 1 ¼ j80  j30 ¼ j110 p:u: j0:0125 j 30

ð4Þ

1 1 þ 1 ¼ j20  j30 ¼ j50 p:u: j0:05 j 30

ð5Þ

1 1 þ ¼ j80  j20 ¼ j100 p:u: j0:0125 j0:05

ð6Þ

y11 ¼

y22 ¼ y33 ¼

y12 ¼ y21

! 1 ¼ ¼ j30 p:u: 1 j 30

ð7Þ

98

10 Solutions of Problems: Load Flow and Economic Load Dispatch



y13 ¼ y31

1 ¼ j0:0125 

y23 ¼ y32

1 ¼ j0:05

 ¼ j80 p:u:

ð8Þ

 ¼ j20 p:u:

ð9Þ

Therefore: 2

110

6 ½YBus  ¼ j4 30 80

30

80

3

7 50 20 5 p:u: 20 100

ð10Þ

Now, we need to define all the quantities in per unit (p.u.) value: SL2,p:u: ¼

SL2 400 þ j320 ¼ ¼ ð4 þ j3:2Þ p:u: 100 SB

ð11Þ

SL3,p:u: ¼

SL3 300 þ j270 ¼ ð3 þ j2:7Þ p:u: ¼ 100 SB

ð12Þ

Based on Gauss-Seidel load flow, we have: 0 ðkþ1Þ

Vi

=

PSch i

B 1B B yii @

þ jQSch i ðkÞ Vi

1

! 

n X

ð kÞ C yij Vj C C A j¼1 j 6¼ i

ð13Þ

Where: PSch ¼ Pi,G  Pi,L i

ð14Þ

QSch ¼ Qi,G  Qi,L i

ð15Þ

In (14) and (15), positive and negative signs are considered for the generation power and load demand, respectively. Now, for the second bus, we can write: 0 ð1Þ V2

B 1 B = B y22 @

PSch 2

þ jQSch 2 ð0Þ V2

1

! 

3 X

  ð0Þ C  y2j Vj C 1 4  j3:2  4  j46:8  ð j30  1 þ j20  1Þ ¼ C¼ j50 1 j50 A j¼1 j 6¼ 2 ð1Þ

V2 ¼ ð0:936  j0:08Þ p:u: Choice (1) is the answer.

10

Solutions of Problems: Load Flow and Economic Load Dispatch

99

Fig. 10.7 The power system of solution of problem 10.14

10.15. Based on the information given in the problem, we have: ð0Þ V1 ¼ 1 p:u: ð0Þ

δ1 = 0

ð1Þ



ð2Þ

X 12 ¼ 0:1 p:u:, S2,L ¼ ð1 þ j0:5Þ p:u:

ð3Þ

First, we need to determine the network admittance matrix, as follows: y11 ¼ y22 ¼

1 ¼ j10 p:u: j0:1 

y12 ¼ y21 ¼ 

1 j0:1

ð4Þ

 ¼ j10 p:u:

ð5Þ

Therefore:  10 ½YBus  ¼ j 10

10 10

 p:u:

ð6Þ

The primary estimation for the magnitude of voltage and phase angle of the load bus are as follows: ð0Þ V2 ¼ 1 p:u: ð0Þ

δ2 = 0



ð7Þ ð8Þ

Based on Newton-Raphson load flow (NRLF), the relations below are held for a load bus: ðk Þ

Pi =

n X

   ðkÞ ðkÞ ðk Þ ðk Þ jVi jjVj jjyij j cos θij  δi  δ j

ð9Þ

j¼1

ðk Þ

Qi = 2

n X

j¼1

   ðkÞ ðkÞ ðk Þ ðk Þ jVi jjVj jjyij j sin θij  δi  δ j

ð10Þ

100

10 Solutions of Problems: Load Flow and Economic Load Dispatch

For bus 2, we can write:    ð0Þ ð0Þ ð0Þ ð0Þ ð 0Þ ð0Þ 2 P2 = jV2 jjV1 jjy21 j cos θ21  δ2  δ1 þ jV2 j jy22 j cos ðθ22 Þ

ð11Þ

   ð0Þ ð0Þ ð0Þ ð0Þ ð0Þ ð0Þ 2 Q2 = 2 jV2 jjV1 jjy21 j sin θ21  δ2  δ1  jV2 j jy22 j sin ðθ22 Þ

ð12Þ

By applying the primarily estimated quantities in (11), we have: ð0Þ

P2 ¼ 1  1  10 cos ð90  ð0  0ÞÞ þ 12  10 cos ð90Þ ¼ 0

ð13Þ

Likewise for the reactive power: ð0Þ

Q2 ¼ 1  1  10 sin ð90  ð0  0ÞÞ  12  10 sin ð90Þ =  10 þ 10 ¼ 0

ð14Þ

Then: ð 0Þ

ð0Þ

ð0Þ

ΔP2 ¼ PSch 2  P2 ¼ ðP2,G  P2,L Þ  P2 ¼ ð0  1Þ  0 ¼ 1 p:u:

ð15Þ

 ð0Þ ð0Þ ð0Þ ΔQ2 ¼ QSch 2  Q2 ¼ Q2,G  Q2,L  Q2 ¼ ð0  0:5Þ  0 ¼ 0:5 p:u:

ð16Þ

In (15) and (16), positive and negative signs are considered for the generation power and load demand, respectively. By considering bus 1 as the slack bus, the Jacobian matrix is as follows:

½J ð0Þ ¼



J1 J3

J2 J4

ð0Þ

2

∂P2 6 ∂δ2 ¼6 4 ∂Q 2 ∂δ2

3ð0Þ ∂P2 ∂jV2 j 7 7 ∂Q2 5 ∂jV2 j

ð17Þ

Solving (11), (12), and (17):    ð0Þ ð0Þ ð0Þ ð 0Þ ð 0Þ J 1 ¼ jV2 jjV1 jjy21 j sin θ21  δ2  δ1

ð18Þ

   ð0Þ ð0Þ ð0Þ ð0Þ ð0Þ J 2 ¼ jV1 jjy21 j cos θ21  δ2  δ1 þ 2jV2 jjy22 j cos ðθ22 Þ

ð19Þ

   ð0Þ ð0Þ ð0Þ ð0Þ ð0Þ J 3 ¼ jV2 jjV1 jjy21 j cos θ21  δ2  δ1

ð20Þ

   ð0Þ ð0Þ ð0Þ ð0Þ ð0Þ J 4 ¼ 2 jV1 jjy21 j sin θ21  δ2  δ1  2jV2 jjy22 j sin ðθ22 Þ

ð21Þ

By applying the primarily estimated quantities in (18)–(21), we have: ð0Þ

J 1 ¼ 1  1  10 sin ð90  ð0  0ÞÞ ¼ 10

ð22Þ

J 2 ¼ 1  10 cos ð90  ð0  0ÞÞ þ 2  1  10  cos ð90Þ = 0

ð23Þ

ð0Þ

ð0Þ

J 3 ¼ 1  1  10 cos ð90  ð0  0ÞÞ ¼ 0

ð24Þ

10

Solutions of Problems: Load Flow and Economic Load Dispatch

101

ð0Þ

J 4 ¼ 1  10 sin ð90  ð0  0ÞÞ 2 2  1  10  sin ð90Þ = 10

ð25Þ

Therefore: ½J ð0Þ ¼ "

ð0Þ

ΔP2

ð0Þ

ΔQ2

#

" ¼ ½J 

ð0Þ

ð0Þ

Δδ2

#

ð0Þ

ΔjV2 j

" )

ð0Þ

Δδ2

ð0Þ

ΔjV2 j

#

 ¼



10

0

0

10

10 0

1 "

0 10

 ð26Þ ð0Þ

ΔP2

ð0Þ

ΔQ2

#

     1 0:1 1 10 0 ¼ ¼ 100 0 10 0:5 0:05

Finally, we can write: ð1Þ

ð0Þ

ð0Þ

δ2 ¼ δ2 þ Δδ2 ¼ 0 þ ð0:1Þ ¼ 0:1 rad ð1Þ

ð0Þ

ð0Þ

jV2 j ¼ jV2 j þ ΔjV2 j ¼ 1 þ ð0:05Þ ¼ 0:95 p:u: ð1Þ

) V2 = 0:95 < 0:12 rad Choice (1) is the answer.

Fig. 10.8 The power system of solution of problem 10.15

ð27Þ

Index

A Accelerating factor (α), 85, 91 Active power, 4, 17, 18, 21, 91 Admittance, 2, 10, 15

B Balanced three-phase power system impedance, 11, 32–34 single-line diagram, 6, 22, 23 Base impedance, 20, 22 Base quantities, 5 Base voltage, 20, 24, 35, 79 Bundling, 37, 39, 41, 43, 46, 51

C Capacitance, 39, 40, 43, 46, 47, 49, 50, 59 Capacitor, 76 Complex power, 6, 18, 21, 31 Conductance, 2, 15 Conductors, 37–41, 43, 44, 47–49 Conductors bundling, 37, 43 Consuming power, 4 Corona power loss, 37, 43 Current, 1, 2, 8, 11

G Gauss-Seidel load flow, 85, 89, 91, 98 Generating reactive power, 4 Generation cost functions, 88, 97 Generator, 2–5, 15, 16 Geometrical Mean Distance (GMD), 46, 51 Geometrical Mean Radius (GMR) bundled conductors, 44 conductors, 37–41, 43–50

I Impedance, 2, 7, 11, 14, 15, 24, 32–35 Inductance, 38–41, 43–45, 47, 51 Inductor, 76 Instantaneous power, 3, 17

J Jacobian matrix, 87, 94, 100

K KCL, 16

D DC load flow (DCLF), 85, 91 active power, 86, 91, 92 determine δ, 86, 92 P12, 86, 92, 93 PG2, 86, 87, 93, 94 phase angle, 87, 94, 95

L Lagrange multiplier (λ), 88 Load bus (P-Q bus), 85, 89, 91, 99 Long transmission line model characteristic impedance, 56, 64, 65 charging current, 56, 64 open circuit, 56, 65 short circuit, 56, 65 transmission matrix, 64 Lossless transmission line, 54, 55, 57, 62, 66 Low-load transmission line, 53

E Economic load dispatch, 95, 96 load demand, 88, 96 power generation system, 96, 97 total load demand, 88, 96, 97 Electric filed, 37, 43 Electric machine, 4, 19 Equivalent admittance of loads, 31 Equivalent impedance of load, 3, 17

M Magnetic field, 37 Medium transmission line model charging current, 54, 60, 61 Motor, 4

F Ferranti effect, 53, 59 Full-load transmission line, 53

N Network admittance matrix, 97, 99 diagonal components, 72, 80 four-bus power system, 73, 82 power system, 70, 73, 75, 77, 83

# The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 M. Rahmani-Andebili, Power System Analysis, https://doi.org/10.1007/978-3-030-84767-8

103

104 Network admittance matrix (cont.) primary system, 81, 83 three-phase four-bus power system, 72, 73, 81, 82 updated system, 83 Network impedance matrix, 79 load flow simulation problem, 69, 76 power system, 69, 71, 75, 77–79 three-bus power system, 70, 76 Newton-Raphson load flow (NRLF), 85, 89, 91, 99–101 No-load transmission line, 6, 53, 57, 67 Nominal specifications, 3

O Ohm’s law, 17

P Phase angle, 10, 13, 16, 29, 30, 87, 94, 95, 99 Phase constant, 57 Phasor domain, 1, 13–16 Phasor representation, 1 Power factor, 7, 8, 25, 27 Power generation system, 95, 96 Power generation units, 88 Power loss coefficients, 88, 95 Power plant, 88 Power system base voltage, 4, 5, 19 base voltage, transmission line, 71, 72, 79, 80 complex power, 6, 21 current signal, phasor domain, 1, 13 electric machine, 4, 19 equivalent impedance of load, 3, 17 impedance, 2 load current, per unit (p.u.), 11, 12, 34–36 load impedance, per unit (p.u.), 7, 24, 25 loads, 10, 30, 31 phase angle, 10, 29, 30 phasor representation, voltage signal, 1, 13 power bus, 3, 16 reactance, 2 reactive power, shunt capacitor, 9, 28 resistance, 2 signal, phasor domain, 1, 13, 14 single-line diagram, 5, 20, 21, 89, 97, 99 Propagation constant, 57

R Rated quantities, 6, 15, 16 Reactance, 2–4, 6, 14 Reactive power, 4, 9, 18, 19, 28, 100 Resistance, 2, 5, 14, 37 Root-mean-square (rms) value, 13, 14, 16

S Short transmission line model capacitance, 59 ideal transformer, 57, 67, 68

Index parameters, 53 single-line diagram, 55, 56, 61–63 Shunt capacitor, 8, 9, 26–28, 57 Single-phase capacitor reactive power, 26, 27, 31 Single-phase power system active and reactive power, 4, 18 current, 2, 14, 15 instantaneous power, 3, 17 load characteristics, 7, 8, 25 shunt capacitor, 8, 9, 26, 27 voltage, 2, 14, 15 Single-phase system, 29 Single-phase transmission lines, 39, 40, 47 capacitance, 47 conductors, 38–40, 44–48 inductance, 44, 45, 47 Slack bus, 85, 91, 94, 100 Synchronous generator, 6, 22, 23

T Thevenin impedance, 75 Thevenin reactance, 6, 22 Three-bus power system, 70, 72, 76 Three-phase capacitor, 31 Three-phase power system balanced three-phase loads with star and delta connections, 9, 28, 29 capacitor banks, 10, 31 voltage, 6, 7, 23, 24 Three-phase transmission lines, 40, 41, 49 capacitance, 40, 49, 50 inductance, 48 Time domain, 16 Transformer, 6 Transmission line model base voltage, 71, 79, 80 characteristic impedance, 54, 55, 60, 62 charging current, 61 Ferranti effect, 59 reflected waves, 54, 60 transmission matrix, 59, 60 Transmission line parameters capacitance, 37, 39, 43, 46 characteristic impedance, 37 conductors bundling, 37, 43 GMR conductors, 37, 38, 43, 44 inductance, 37–41, 43, 51 Transmission line parameters capacitance, 39 Transmission matrix, 53–57, 60–63, 65–67 cascaded transmission systems, 59 long transmission line, 64 medium transmission line, 60 sub-system, 67, 68 transmission line, 59, 64–66 Triangle configuration, 10

V Voltage, 2–4, 6, 7, 9, 11, 15, 16, 19, 23, 24, 29, 35, 53, 99 Voltage-controlled bus (P-V bus), 85, 91