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
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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