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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 1 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 <I12 will increase (counter-clockwise). Therefore, the phase angle of the current with respect to E2 will decrease. Choice (3) is the answer. Fig. 2.17 The power system of solution of problem 2.26 2.27. Based on the information given in the problem, we have the following speciﬁcations: Load 1 : P1 ¼ 2 p:u:, PF ¼ 0:8 Lagging ð1Þ Load 2 : P2 ¼ 2 p:u:, PF ¼ 0:8 Leading ð2Þ Load 3 : P3 ¼ 2 p:u:, PF ¼ 1 ð3Þ ð4Þ 2 Solutions of Problems: Fundamental Concepts in Power System Analysis 31 The total active power of the loads can be calculated as follows: PTotal ¼ P1 þ P2 þ P3 ¼ 2 þ 2 þ 2 ¼ 6 p:u: ð6Þ The relation below holds between the active and reactive power of a load: Q ¼ P tan ðθÞ ð7Þ The total reactive power of the loads can be calculated as follows: QTotal ¼ Q1 þ Q2 þ Q3 ¼ P1 tan ðθ1 Þ þ P2 tan ðθ2 Þ þ P3 tan ðθ3 Þ ¼ P1 tan cos 1 ðPF 1 Þ þ P2 tan cos 1 ðPF 2 Þ þ P3 tan cos 1 ðPF 3 Þ ¼ 2 tan cos 1 ð0:8Þ þ 2 tan cos 1 ð0:8Þ þ 2 tan cos 1 ð1Þ ¼ 2 tan cos 1 ð0:8Þ 2 tan cos 1 ð0:8Þ þ 2 tan ð0Þ ¼ 0 p:u: ð8Þ The total complex power is: STotal ¼ PTotal þ QTotal ¼ 6 þ j0 ¼ 6 p:u: ð9Þ The total current is: ð10Þ The equivalent admittance of the loads can be determined as follows: Choice (1) is the answer. Fig. 2.18 The power system of solution of problem 2.27 2.28. Based on the information given in the problem, the connection of the three capacitors is delta. Moreover, we have: V L ¼ 400 V, f ¼ 50 Hz, Q ¼ 600 kVAr ð1Þ As we know, the reactive power of a single-phase capacitor can be determined as follows: QC,ph ¼ V rms,ph 2 ¼ ωCV rms,ph 2 ¼ 2πfCV rms,ph 2 Xc ð2Þ 32 2 Solutions of Problems: Fundamental Concepts in Power System Analysis Therefore, the reactive power of three-phase capacitor with the delta connection is: QC,3ph ¼ 3QC,ph ¼ )C¼ 3V rms,ph 2 ¼ 3ωCV rms,ph 2 ¼ 6πfCV rms,ph 2 Xc ð3Þ QC,3ph 600 kVAr ¼ ) C ¼ 4000 μF 6πf V rms,ph 2 6 3:14 50 4002 Choice (2) is the answer. 2.29. Based on the information given in the problem, we have the following speciﬁcations: G : 22 kV, 90 MVA, X G ¼ 18% ð1Þ T1 : 22=220 kV, 50 MVA, X T1 ¼ 10% ð2Þ T2 : 220=11 kV, 40 MW, X T2 ¼ 6% ð3Þ T3 : 22=110 kV, 40 MW, X T3 ¼ 6:4% ð4Þ T4 : 110=11 kV, 40 MW, X T4 ¼ 8% ð5Þ M : 10:45 kV, 66:5 MVA, X M ¼ 18:5% ð6Þ TL1 : 220 kV, 48:4 Ω ð7Þ TL2 : 110 kV, 65:5 Ω ð8Þ SB ¼ 100 MVA, V B1 ¼ 22 kV ð9Þ As we know, base MVA is applied for the whole power system; however, base voltage might be different in each zone. Figure 2.19.2 shows the zones with the related base voltages that can be determined as follows: V B2 ¼ 22 kV 220 ¼ 220 kV 22 ð10Þ 11 ¼ 11 kV 220 ð11Þ 110 ¼ 110 kV 22 ð12Þ V B3 ¼ 220 kV V B4 ¼ 22 kV Now, we need to update the per unit (p.u.) value of the impedances based on the new base MVA and voltages as follows: 2 SB,new V B,old ð13Þ X new,p:u: ¼ X old,p:u: SB,old V B,new X G,new,p:u: ¼ 0:18 2 100 22 ¼ 0:2 p:u: 90 22 ð14Þ X T1,new,p:u: ¼ 0:10 2 22 100 ¼ 0:2 p:u: 22 50 ð15Þ 2 Solutions of Problems: Fundamental Concepts in Power System Analysis 100 220 2 ¼ 0:15 p:u: 40 220 ð16Þ 2 100 11 ¼ 0:16 p:u: 40 11 ð17Þ 100 110 2 ¼ 0:2 p:u: 40 110 ð18Þ 2 100 10:45 ¼ 0:25 p:u: 66:5 11 ð19Þ X T2,new,p:u: ¼ 0:06 X T3,new,p:u: ¼ 0:064 X T4,new,p:u: ¼ 0:08 X M,new,p:u: ¼ 0:185 33 The impedance of the lines has been given in Ohms. Therefore, to present them in per unit, we need to determine the base impedance for their zones as follows: ðV B Þ2 ðV Þ2 ) ZB ¼ B ZB SB ð20Þ Z B2 ¼ ð220 kV Þ2 ¼ 484 Ω 100 MVA ð21Þ Z B4 ¼ ð110 kV Þ2 ¼ 121 Ω 100 MVA ð22Þ SB ¼ Therefore: X TL1,new,p:u: ¼ 48:4 ¼ 0:1 p:u: 484 ð23Þ X TL2,new,p:u: ¼ 65:5 ¼ 0:5 p:u: 121 ð24Þ Now, by using (14)–(19) and ((23)) –((24)), we can draw the impedance diagram of the power system which is illustrated in Fig. 2.19.3. Since the equivalent impedance is requested, the power supply of the generator and the EMF of the motor are turned off (short-circuited) in the diagram. The impedance seen from the ﬁrst bus can be calculated as follows: Zeq,p:u: ¼ ð j0:2Þ ¼ ð j0:2Þ ð j0:2 þ j0:1 þ j0:15Þ ð j0:16 þ j0:5 þ j0:2Þ þ j0:25 ð j0:45Þð j0:86Þ ð j0:45Þ ð j0:86Þ þ j0:25 ¼ ð j0:2Þ þ j0:25 ð j0:45Þ þ ð j0:86Þ ð j0:2Þ ð j0:55Þ ¼ ð j0:2Þð j0:55Þ ð j0:2Þ þ ð j0:55Þ Zeq,p:u: j0:14 p:u: Choice (1) is the answer. 34 2 Solutions of Problems: Fundamental Concepts in Power System Analysis Fig. 2.19 The power system of solution of problem 2.29 2.30. Based on the information given in the problem, we have the following speciﬁcations: G : 100 V ð1Þ T1 : 200=400 V, 1 kVA, X T1 ¼ 0:1 p:u: ð2Þ Line : ZLine ¼ j8 Ω ð3Þ T2 : 200=200 V, 2 kVA, X T2 ¼ 0:1 p:u: ð4Þ 2 Solutions of Problems: Fundamental Concepts in Power System Analysis 35 Load : ZLoad ¼ j6 Ω ð5Þ V B1 ¼ 100 V, SB ¼ 1 kVA ð6Þ The base voltage in the second and third zones can be calculated as follows: 400 ¼ 200 V 200 ð7Þ 400 200 ¼ 200 V 200 200 ð8Þ V B2 ¼ 100 V V B3 ¼ 100 V Now, we need to update the per unit (p.u.) value of the impedances based on the new base MVA and voltages as follows: 2 S V B,old ð9Þ X new,p:u: ¼ X old,p:u: B,new SB,old V B,new 1 200 2 X T1,new,p:u: ¼ 0:1 ¼ 0:4 p:u: 1 100 ð10Þ 1 200 2 ¼ 0:05 p:u: X T2,new,p:u: ¼ 0:1 2 200 ð11Þ To present the impedance of the line in per unit (p.u.) value, we need to determine the base impedance in the second zone, as follows: SB ¼ ðV B2 Þ2 ðV Þ2 ð200 V Þ2 ) Z B2 ¼ B2 ¼ ¼ 40 Ω Z B2 SB 1 kVA ZLine,p:u: ¼ ZLine j8 ¼ j0:2 p:u: ) ZLine,p:u: ¼ Z B2 40 ð12Þ ð13Þ Likewise, to present the impedance of the load in per unit value, we need to determine the base impedance in the third zone, as follows: SB ¼ ðV LB Þ2 ðV Þ2 ð200 V Þ2 ) Z B3 ¼ LB3 ¼ ¼ 40 Ω ZB SB 1 kVA ZLoad,p:u: ¼ ZLoad j6 ¼ j0:15 p:u: ) ZLoad,p:u: ¼ 40 Z B3 ð14Þ ð15Þ The voltage of the generator in per unit can be determined, as follows: VG,p:u: ¼ VG 100 ¼ 1 p:u: ) VG,p:u: ¼ 100 V B1 ð16Þ Figure 2.20.3 shows the impedance diagram of the power system based on (10), (11), (13), (15), and (16). The current can be calculated as follows: Ip:u: ¼ VG,p:u: 1 ) Ip:u: ¼ j1:25 p:u: ) Ip:u: ¼ 1:25 p:u: ¼ j0:4 þ j0:2 þ j0:05 þ j0:15 ZTotal,p:u: Choice (4) is the answer. 36 2 Solutions of Problems: Fundamental Concepts in Power System Analysis Fig. 2.20 The power system of solution of problem 2.30 3 Problems: Transmission Line Parameters Abstract In this chapter, the problems concerning with transmission line parameters are presented. The subjects include the Geometrical Mean Distance (GMD) and Geometrical Mean Radius (GMR) of conductors and the inductance and capacitance of single-phase and three-phase transmission lines bundled with a variety of arrangements. 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. 3.1. What is the main purpose of conductors bundling in transmission lines? Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) Decreasing inductive reactance of transmission line 2) Decreasing resistance of transmission line 3) Decreasing Corona power loss by reducing effective electric ﬁled around conductors 4) Decreasing Corona power loss by reducing effective magnetic ﬁeld around conductors 3.2. Determine the Geometrical Mean Radius (GMR) of the conductors with the arrangements shown in Fig. 3.1. The Geometrical Mean Radius (GMR) of each conductor is r0. Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large pﬃﬃﬃﬃ 1) r 0 D pﬃﬃﬃﬃ 2) D r 0 pﬃﬃﬃﬃﬃﬃﬃ 3) r 0 D qﬃﬃﬃ 0 4) Dr Fig. 3.1 The power system of problem 3.2 3.3. Which one of the following choices is correct about the effect of bundling of conductors of a transmission line on its inductance, capacitance, and characteristic impedance? Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ 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_3 37 38 3 1) 2) 3) 4) Problems: Transmission Line Parameters Decrease, decrease, no change Increase, decrease, increase Decrease, increase, decrease Increase, increase, no change 3.4. Determine the Geometrical Mean Radius (GMR) of the conductors with the arrangements shown in Fig. 3.2. The Geometrical Mean Radius (GMR) of each conductor is r0. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 8 1) 2r 0 6 D6 p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 2) 23 r 0 3 D ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p 4 3) 22 r 0 D p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 8 4) 2r 0 2 D6 Fig. 3.2 The power system of problem 3.4 3.5. Determine the Geometrical Mean Radius (GMR) of the conductors with the arrangements shown in Fig. 3.3. The radius of each conductor is r. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 1.722r 2) 1.834r 3) 1.725r 4) 1.532r Fig. 3.3 The power system of problem 3.5 3.6. Figure 3.4 shows a single-phase transmission line including two conductors (“1” and “3”) for sending power and one conductor (“2”) for receiving power. The Geometrical Mean Radius (GMR) of each conductor is r0. Calculate the inductance of the line in H/m. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 2 107 ln Dr0 qﬃﬃﬃﬃﬃ r0 2) 2 107 ln 2D qﬃﬃﬃﬃ D 3) 2 107 ln 2r0 4) 107 3 ln Dr0 ln ð2Þ 3 Problems: Transmission Line Parameters 39 Fig. 3.4 The power system of problem 3.6 3.7. Figure 3.5 shows a single-phase transmission line. Herein, conductor “1” is for sending power, and conductors “2” and “3” are for receiving power. The Geometrical Mean Radius (GMR) of each conductor is r0. Calculate the inductance of the line in H/m. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) 6 107 ln Dr0 2) 2 107 ln Dr0 3) 3 107 ln Dr0 4) 4 107 ln Dr0 Fig. 3.5 The power system of problem 3.7 3.8. What difference can we see in the capacitance of a transmission line if we change the conductor arrangements from the two-bundling to the three-bundling, as can be seen in Fig. 3.6? The Geometrical Mean Radius (GMR) of each conductor is r0 and D > 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<D ð4Þ For case 1: L1 ¼ 2 10 7 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃﬃﬃ p p 3 D D 2D D32 7 ln ¼ 2 10 ln r0 r0 C1 ¼ 2πε0 p3 ﬃﬃ ln Dr0 2 ð5Þ ð6Þ For case 2: L2 ¼ 2 10 7 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃﬃﬃ p p 3 D D 2D D32 7 ln ¼ 2 10 ln r0 r0 ð7Þ 50 4 Solutions of Problems: Transmission Line Parameters C2 ¼ 2πε0 p3 ﬃﬃ ln Dr0 2 ð8Þ For case 3: L3 ¼ 2 10 7 p ﬃﬃﬃﬃﬃﬃﬃﬃ! ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p 3 3 Ddd Dd 2 7 ln ¼ 2 10 ln r0 r0 2πε0 pﬃﬃﬃﬃﬃﬃ 3 C3 ¼ ln Dd 2 r0 ð9Þ ð10Þ For case 4: L4 ¼ 2 107 ln ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p 3 DDD D ¼ 2 107 ln 0 0 r r C4 ¼ 2πε0 ln Dr0 ð11Þ ð12Þ From (5) to (12), we can conclude that: 2 10 7 p ﬃﬃﬃﬃﬃﬃﬃﬃ! ﬃﬃﬃ p 3 D D32 Dd 2 7 7 ln ln ln < 2 10 < 2 10 ) L3 < L4 < L2 ¼ L1 r0 r0 r0 2πε0 2πε 2πε0 p3 ﬃﬃ < D0 < pﬃﬃﬃﬃﬃﬃ ) C1 ¼ C2 < C4 < C3 3 D 2 ln 0 Dd 2 ln r0 r ln r0 ð13Þ ð14Þ As can be seen in (13) and (14), arrangement 3 has the least inductance and the most capacitance. Choice (3) is the answer. Fig. 4.10 The power system of solution of problem 4.12 4.13. Based on the information given in the problem, we know that the Geometrical Mean Radius (GMR) of each conductor is r0 and: D ¼ 4r 0 ð1Þ 4 Solutions of Problems: Transmission Line Parameters 51 As we know, the inductance of a transmission line can be determined as follows: GMD GMR ð2Þ GMD2b ln GMR2b ð3Þ GMD3b GMR3b ð4Þ L ¼ 2 107 ln Therefore: L2b ¼ 2 10 7 L3b ¼ 2 107 ln Where: GMR2b GMR3b ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃ Using ð1Þ pﬃﬃﬃﬃﬃﬃﬃﬃ 4 ¼ ðr 0 D Þ2 ¼ r 0 D 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 ¼ ¼ ) GMR2b ¼ 4r 0 2 ¼ 2r 0 qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ pﬃﬃﬃﬃﬃﬃﬃﬃﬃ Using ð1Þ p ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 3 3 9 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 ¼ ) GMR3b ¼ 16r 0 3 ¼ 23 r 0 ðr 0 D DÞ3 ¼ r 0 D2 ¼ ð5Þ ð6Þ The Geometrical Mean Distance (GMD) will not change, since only the bundling is changed. Therefore: GMD2b ¼ GMD3b ð7Þ Therefore: L3b L2b ¼ 2 107 ln GMD3b GMR3b 2 107 ln GMD2b GMD3b GMR2b ¼ 2 107 ln GMR2b GMR3b GMD2b ð8Þ Solving (7)–(8): L3b L2b ¼ 2 107 ln GMR2b GMR3b Solving (5), (6), and (9): 0 1 2r L3b L2b ¼ 2 107 ln 4 ¼ 2 107 ln 23 23 r 0 2 L3b L2b ¼ 107 ln ð2Þ 3 Choice (1) is the answer. Fig. 4.11 The power system of solution of problem 4.13 ð9Þ 5 Problems: Transmission Line Model and Performance Abstract In this chapter, the problems concerning with the transmission line model and performance are presented. The subjects include transmission line models, transmission line voltage regulation, transmission line compensation, and features of transmission matrix. 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. 5.1. Which one of the parameters below can be ignored for a short transmission line? Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) Resistance 2) Inductance 3) Reactance 4) Capacitance 5.2. Based on Ferranti effect, which one of the following terms is correct? Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) The voltage in the receiving end increases when the transmission line is operated in no-load or low-load conditions. 2) The voltage in the receiving end increases when the transmission line is operated in full-load condition. 3) The voltage in the receiving end increases when the transmission line is short-circuited. 4) The voltage in the receiving end decreases when the transmission line is operated in full-load condition. 5.3. Which one of the matrices below belongs to a transmission matrix of a real transmission line? Difﬁculty level ● Easy ○ Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large j 1 1) 0 j 1 j 2) 2 1 1 2 3) 3 1 1 j 4) 0 1 # 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_5 53 54 5 Problems: Transmission Line Model and Performance 5.4. Two power systems have the transmission matrices below. If these systems are cascaded, determine their equivalent transmission matrix: 1 j2 1 0 ½T 1 ¼ , ½T 2 ¼ 0 1 j2 1 Difﬁculty level Calculation amount 3 j2 1) j2 1 5 j2 2) j2 1 2 j2 3) j2 2 4) None of them ● Easy ● Small ○ Normal ○ Normal ○ Hard ○ Large 5.5. Calculate the characteristic impedance of a long lossless transmission line that has the inductance and capacitance of about 1 mH/meter and 10 μF/meter, respectively. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) 20 Ω 2) 5 Ω 3) 10 Ω 4) 40 Ω 5.6. At the end of a transmission line with the characteristic impedance of ZC = (1 j) Ω, a load with the impedance of ZL = (1 + j) Ω has been connected. Which one of the following components needs to be installed in parallel to the load to remove the reﬂected waves of the voltage and current? Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ● Small ○ Normal ○ Large 1) A capacitor with the reactance of 0.5 2) A capacitor with the reactance of 1 3) An inductor with the reactance of 0.5 4) An inductor with the reactance of 1 5.7. As is shown in Fig. 5.1, a medium transmission line has been presented by its T model. Calculate the charging current of the line (ICharging). Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) Only ZVR(1 + 0.5YZ)1 2) Only YVS(1 + 0.5YZ)1 3) YVR or YVS(1 + 0.5YZ)1 4) Only YVR(1 + 0.5YZ)1 Fig. 5.1 The power system of problem 5.7 5 Problems: Transmission Line Model and Performance 55 5.8. Figure 5.2 shows the single-line diagram of a short transmission line. Determine its transmission matrix. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1 þ YZ 1 1) Z Y 1 þ YZ Z 2) Y 1 1 þ YZ 1 3) Y Z 1 þ YZ Y 4) Z 1 Fig. 5.2 The power system of problem 5.8 5.9. Determine the characteristic impedance of a transmission line that the relation below is true for its parameters: R G ¼ L C Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ● Normal ○ Large 1) RL. 2) 1. 3) 0. 4) It is equal to the characteristic impedance of a lossless transmission line. 5.10. Figure 5.3 shows the single-line diagram of a short transmission line that a resistor with the resistance of R has been installed in its middle point. Determine its transmission matrix. Difﬁculty level ○ Easy ● Normal ○ Hard Calculation amount ○ Small ○ Normal ● Large 1) 2) 3) 4) R þ Z ZðR þ ZÞ Z RþZ 1 þ Z ZðR þ ZÞ Z RþZ 2 3 Z Z 1þ Z Rþ 6 2R 4R 7 4 5 1 Z 1þ 2 R 2R 3 Z Z Z Rþ 1þ 6 2R 2R 7 4 5 1 Z 1þ R 4R 56 5 Problems: Transmission Line Model and Performance Fig. 5.3 The power system of problem 5.10 5.11. Calculate the characteristic impedance of a long transmission line that its transmission matrix is as follows: 2 1 62 ½T ¼ 4 3 j 4 Difﬁculty level Calculation amount pﬃﬃ 1) 33 Ω pﬃﬃ 2) 2 3 3 Ω 3) 34 Ω 4) 12 Ω ○ Easy ○ Small ○ Normal ● Normal 3 j 1 2 7 5 ● Hard ○ Large 5.12. Calculate the charging current (ICharging) of a long transmission line. Difﬁculty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large ðγlÞ 1) VS tanh Zc 2) ZcVS tanh (γl) ðγlÞ 3) VS coth Zc 4) ZcVS coth (γl) 5.13. In a long transmission line, consider the deﬁnitions below, and choose the correct relation between ZC, ZS.C., and ZO.C.. 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 Difﬁculty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) ZC = ZS.C. 2 ZO.C. 2) ZC = ZO.C. 2 ZS.C. pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3) ZC = ZS:C: ZO:C: 4) ZC = 12 ðZS:C: þ ZO:C: Þ 5.14. In a long transmission line, the impedance measured from the beginning of the line, when its end is open circuit, is the reciprocal of the impedance measured from the beginning of the line, when its end is short circuit. Which one of the following relations is correct among the parameters of the transmission matrix of this line? A ½T ¼ C Difﬁculty level Calculation amount ○ Easy ○ Small ○ Normal ● Normal ● Hard ○ Large B D 5 Problems: Transmission Line Model and Performance 57 1 1) A þ B ¼ AB 1 2) A þ B ¼ BA pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3) A ¼ 1 B2 pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4) A ¼ B2 1 5.15. In a no-load and lossless transmission line, which one of the following relations is correct? Herein, VR, VS, β, γ, and l are the voltage of receiving end, voltage of sending end, phase constant, propagation constant, and length of line, respectively. Difﬁculty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ● Normal ○ Large 1) VR = sinVðSβlÞ 2) VR = 3) VR = 4) VR = VS sin ðγlÞ VS cos ðβlÞ VS cos ðγlÞ 5.16. A factory is supplied by an ideal transformer through a short transmission line. At the bus of the factory, a shunt capacitor has been installed to correct its power factor. Which one of the transmission matrices below is correct for this power system? Difﬁculty level ○ Easy ○ Normal ● Hard Calculation amount ○ Small ○ Normal ● Large " # 1 0 1 þ ZY Z 1) a Y 1 0 a " # a 0 1 þ ZY Y 2) 1 0 Z 1 a# " a 0 1 þ ZY Z 3) 1 0 Y 1 a# " 1 0 1 þ ZY Y 4) a Z 1 0 a Fig. 5.4 The power system of problem 5.16 6 Solutions of Problems: Transmission Line Model and Performance Abstract In this chapter, the problems of the ﬁfth chapter are fully solved, in detail, step by step, and with different methods. 6.1. In a short transmission line, the capacitance of the line can be ignored. Choice (4) is the answer. 6.2. Based on Ferranti effect, the voltage in the receiving end increases when the transmission line is operated in no-load or low-load conditions. Choice (1) is the answer. 6.3. The two-port of transmission line is symmetric and bidirectional. Therefore, the transmission matrix of a real transmission line has the following characteristics: ( A B A¼D ð 1Þ ½T ¼ ) C D detð½T Þ ¼ 1 ) AD BC ¼ 1 ð 2Þ Now, we need to check these characteristics for each choice, as follows: Choice 1: Condition (1) is not true, as j 6¼ (j). Choice 2: Condition (2) is not true, as AD BC ¼ 1 j2 6¼ 1. Choice 3: Condition (2) is not true, as AD BC ¼ 1 6 6¼ 1. Choice 4: Both conditions are true, as 1 ¼ 1 and AD BC ¼ 1 0 ¼ 1. Therefore, the transmission matrix of choice 4 belongs to a real transmission line. Choice (4) is the answer. 6.4. Based on the information given in the problem, we have: ½T 1 ¼ 1 0 1 , ½T 2 ¼ 1 j2 j2 0 ð1Þ 1 As we know, for the cascaded transmission systems, the relation below holds about their transmission matrices: ½T Total ¼ ½T 1 ½T 2 ð2Þ Solving (1) and (2): ½T Total ¼ 1 0 j2 1 1 j2 0 3 ) ½T Total ¼ 1 j2 j2 1 Choice (1) 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_6 59 60 6 Solutions of Problems: Transmission Line Model and Performance 6.5. Based on the information given in the problem, we know that the line is lossless. Therefore: 8 R ¼ 0, X ¼ Lω < Z ¼ R þ jX 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 ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼) Z = jLω G ¼ 0, B ¼ Cω : 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 ¼ ¼ ¼) Y = jCω Y ¼ G þ jB ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ð 1Þ ð 2Þ Moreover: L ¼ 1 mH=meter, C ¼ 10 μF=meter ð3Þ As we know, the characteristic impedance of a transmission line can be calculated as follows: rﬃﬃﬃﬃ Z ZC ¼ Y ð4Þ Solving (1)–(4): rﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃ sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Z jLω L 103 1 = ZC = = = = Y jCω C 102 10 106 ZC ¼ 10 Ω Choice (3) is the answer. 6.6. Based on the information given in the problem, we have: ZC = ð1 jÞ Ω ð1Þ ZL = ð1 þ jÞ Ω ð2Þ If a transmission line is loaded by an impedance, which is equal to its characteristic impedance, the reﬂected waves of the voltage and current will be eliminated. Therefore, we need an impedance (Z) to install it parallel to the load impedance (ZL) to achieve the goal (ZC): ZL kZ = ZC ) ð1 þ jÞkZ = 1 j ) ð1 þ jÞ Z =1 j ð1 þ jÞ þ Z ) Z þ jZ = ð1 þ jÞ þ Z 2 j þ 1 jZ ) 2jZ = 2 )Z¼ 1 Ω ) XC ¼ 1 Ω j Therefore, the component is a capacitor with the reactance of 1 Ω. Choice (2) is the answer. 6.7. As we know, the transmission matrix of the T model of a medium transmission line can be presented as follows: VS IS 2 ZY 1þ 6 2 ¼4 Y 8 3 ZY ZY ZY > < 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 # 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_7 69 70 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 # 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_8 75 76 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. # 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_9 85 86 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: ð0Þ ð0Þ ð0Þ ΔQ2 ¼ QSch 2 Q2 ¼ Q2,G Q2,L Q2 ¼ ð0 0:5Þ 0 ¼ 0:5 p:u: ð15Þ 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